# Jeph Jacques's comics discussion forums

## Fun Stuff => CHATTER => Topic started by: Jace on 17 Oct 2013, 23:45

Title: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Jace on 17 Oct 2013, 23:45
I don't have any math to do just yet.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 18 Oct 2013, 06:13
Can the title be "Math is Delicious!"?

I know, it's from the comic, but...
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 18 Oct 2013, 14:18
I hereby abuse my powers to decree that this thread can also be used for complaints about math.

I do this because I have to understand a bunch of math by Monday, including Taylor series and some other stuff that I forget, so I am complaining.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: LeeC on 18 Oct 2013, 14:28
I've never done calculus. My education in college and highschool was pretty much algebra, trig, and statistics.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: jwhouk on 18 Oct 2013, 18:06
I didn't care much for Calculus. It was too derivative.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: pwhodges on 19 Oct 2013, 01:28
It's better if you take an integrated approach to it.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ackblom12 on 19 Oct 2013, 01:30
I feel like if I was meant to specialize in math, I'd be given some kind of sine.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: pwhodges on 19 Oct 2013, 01:33
But you weren't, so you went off on a different tangent?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Jace on 19 Oct 2013, 06:30
Do you fuckers think that is acute exchange? This is serious stuff
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: pwhodges on 19 Oct 2013, 06:42
Well, we didn't want to be obtuse about it, because that wouldn't have solved anything.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: jwhouk on 19 Oct 2013, 08:40
Algebra was actually easy for me. It was simple as a + b = c!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 19 Oct 2013, 10:53
Well, this doesn't say much. It doesn't even imply that a, b and c are elements of a group!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: jwhouk on 19 Oct 2013, 18:46
Oh, don't get all quadratic on me.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 19 Oct 2013, 19:09
Wait, that one didn't even make any sense.

As for the Taylor series, they're just (infinite) polynomials impersonating another function.  The basic idea is that they'll have the same value for the function and all its derivatives at a given point, so they'll look just like it.  It's a nice trick, and ... if you give me an idea where you are with them, I can put you where you need to be.

The only real issue is the interval of convergence...
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 19 Oct 2013, 23:54
I keep forgetting that having a math teacher on the forum has certain perks :D

I haven't had time to look at them yet. I will elaborate what we need them for later when I'm not on the run. And hey, maybe I will figure it out on my own by then.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 20 Oct 2013, 05:48
Yeah, it's pretty awesome when you take an infinite series, do some magic, and end up with a neat little polynomial. Maths is* sexy!

I'm not doing any maths just now, but next semester I'll be doing category theory and introduction to logic, so that'll be fun.

Maths is, right? I mean, maths is written in plural form, but it is still just a single thing, right? Or is it maths are sexy?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Gridgm on 20 Oct 2013, 13:33
a strange thing happened halfway through my math course where i stopped being able to solve maths sums unless they had variables in them, i checked with some of my other friends in the course and they agreed similar things occurred

for example
Code: [Select]
7 + 5 = x, solve for x is easy

however

Code: [Select]
7 + 5 = i would stare at for a couple of minutes then either insert the x myself and solve it or get out a calculator
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Jace on 20 Oct 2013, 15:02
I don't want to study related rates stuff right now, it isn't that I can't do this, it's just that I don't like word problems that much. After this I should go over the next section, local linear approximation; differentials.  But I'm just not able to focus right now.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: pwhodges on 20 Oct 2013, 15:55
Maths is, right? I mean, maths is written in plural form, but it is still just a single thing, right? Or is it maths are sexy?

Is.  "Maths" is not the plural of "math"; it's a different abbreviation of "mathematics".  Many words end in s without being plural.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Jace on 20 Oct 2013, 17:09
Okay I'm making a mistake somewhere here.

Related rates. Circle area increasing at rate of 6mi^2/hr. Find radius increase when area is 9mi^2
A = pi*r^2
sqrt(A/pi) = r

Then I find the derivative, but I'm fucking that part up.

I end up with 1/[2*sqrt(pi*da\dt)] = dr\dt
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 20 Oct 2013, 20:08
So close!

The chain rule gives d/dt[f(g(t))] = f'(g(t))* g'(t), so when you take the derivative of sqrt(A/pi), you should be getting

{1/[2sqrt(A)*pi]}*{dA/dt}.  The derivative ofthe inner function is multiplied onto the outside of the composition - you have it inside with the inner function A.

By the way, one of the nicer tricks with related rate problems is that you don't need to solve for the variable first... you could leave it as

A = pi*r2

and take the derivative as it stands, getting

dA/dt = 2*pi*r*dr/dt

then plug in the values and solve for dr/dt.  It gives the same result, and is a little easier to work with (no square roots and such).

Prost!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Jace on 20 Oct 2013, 21:48
Well, I first tried to use the original equation, but then I had dr/dt and r as unknown variables when trying to solve, and realized that I needed to have A in there as my known values were A and dA/dt.

It probably didn't help that I was on my 9th hour of work when trying to do that problem. At least I was messing up doing the derivative (which is a simple thing to figure out the error in)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 21 Oct 2013, 07:31
But you do have r.  The circle's area is 9 mi2, so the radius is sqrt(9/pi).

I know, the 9th hour is not a pretty one...
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: de_la_Nae on 21 Oct 2013, 08:43
I used to be decent at math, but I haven't had to think in these terms in quite a while. My eyes are getting crossed just looking at the equations. D;
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: de_la_Nae on 21 Oct 2013, 08:43
Also I hate/love all of you for those earlier puns and wordplay.  :psyduck:
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Jace on 21 Oct 2013, 14:28
But you do have r.  The circle's area is 9 mi2, so the radius is sqrt(9/pi).

I know, the 9th hour is not a pretty one...

Ah, of course. Although I think solving for r put it in a more manageable form when doing the derivative and then plugging in numbers.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 21 Oct 2013, 20:29
hate/love :psyduck:

Glad you have your priorities straight!   :-D
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 27 Oct 2013, 09:07
I wish to use this opportunity to declare my hatred towards absolute-value inequalities.
Which also contain fractions in the abs(). Ohgod why  :psyduck:

Edit: okay, so I had one of those inequalities resolve into (5>0) and (x>5/4). Why does the fact that I got a statement which is always true not mean that x can be an arbitrary number?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 27 Oct 2013, 10:19
If the inequalities resolve into two inequalities, then both has to be followed. If one does not have x in it, then you can disregard it, and x>5/4 is the only restriction.

Try to plug in 5/4, values under 5/4, and values over 5/4 into the original inequality. If your solution is correct, only the group over 5/4 should give correct statements.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 27 Oct 2013, 11:30
Damn. I definitely need to learn how to prove stuff. They assume that we learned it in school, but I didn't.

Also, how can I show that (http://latex.codecogs.com/png.download?%7B%5Ccolor%7Bwhite%7D0.5%5Cgeq%5Cfrac%7B%5Cleft%7Cx%5Cright%7C%7D%7B1+x%5E2%7D%7D)? I can see that it has a global maximum at 1 and -1, but I just don't know how to show that.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Pilchard123 on 27 Oct 2013, 11:40
From the top of my head...

Differentiate y = right-hand side.

The maxima/minima will be where dy/dx = 0.
To find out if those points are maxima or minima, calculate d2y/dx2 at those points. It it is negative, then the point is a maximum (that is the right way around). If it is positive, then d2y/dx2 will be positive.

(This gets funny when there is a point of inflection around)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 27 Oct 2013, 11:48
I don't think we're allowed to use differentiation. We're only allowed to use stuff we've proven. And I don't want to prove differentiations for this one exercise.

A classmate just wrote me she found a solution. She didn't tell me yet, but I'm interested.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 27 Oct 2013, 13:14

I wish to use this opportunity to declare my hatred towards absolute-value inequalities.
That said, multiply both sides by the denominator. This does not change the unequality sign, because 1+x² is >0 in real numbers. Then use this site:
http://www.purplemath.com/modules/absineq.htm
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ev4n on 28 Oct 2013, 09:21
Friday's xkcd was great.  Got to love Bayes theorem for messing with people.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 29 Oct 2013, 10:05
All this math stuff is a little bit too fast for me. Problem is that I'm from a state where they have a slightly different focus than in most other states of Germany.

On Thursday I'll need to hand in the first exercise sheet from advanced mathematics. And these damn exercises are damn hard. Here's an example:

I have to examine the following sets for supremum, infimum, minimum and maximum, and prove everything. I'm not allowed to use differentiation and such. I also have to prove that the results I have are correct.

I mean, I can see that there's 2 = min A, but I have no idea how to prove it.

I've never in my life proven something in a mathematical sense, so I don't even know how to start.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 29 Oct 2013, 11:01
If you can't differentiate, can't you just draw the sets? Like A, that's just the curve of the function f(x)=x+(1/x) in the range (0, 5). Then you can just circle the stuff you are looking for. "Here, it's here. You can see it on the drawing". Since you're working within a limited domain, that's a valid proof.

I'll give some hints - though I do not guarantee their correctness, and I use the definitions of infimum and supremum from wikipedia, as I've never seen those terms before.

So for A, the minimum is 2, and the infimum is also 2 - if you draw it as a curve, the function goes from infinity at 0 to 2 at 1, to 5.2 at 5. Thus all values in A are >= 2. The maximum is infinity and the supremum is also infinity.

B is the union of [1,3] and [-1,-3] (you might need to prove that? I dunno). Minimum is -3, of course. For any number n<-3, n^2>9, as x^2 is a decreasing function for x<0, so the infimum is -3. The reverse goes for maximum and supremum of 3.

C is [1, 16) - show that for 0<x<1 and x >= 16, there is no y in [1,4) such that y^2=x. (and mention that the same goes for x<=0, but that's trivial)

D is {-1, 1/2, -1/3, 1/4, -1/5 ...}. Proofs will have to rely on the fact that the absolute value of elements are decreasing, so there's no x in D such that x < -1 or x > 1/2

E is all positive numbers, at least, otherwise that's too much for me to get into after a day of uni work.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 29 Oct 2013, 11:15
Sadly having drawn it won't suffice. We have to give the professor a mathematical proof of our statements, either by mathematical induction, proof by exhaustion or reductio ad absurdum.

This is too damn difficult for the first exercise sheet.

By the way, everything I'm allowed to use is written in this German script up to page 19.5 (page 28.5 of the PDF):
Inofficial script (http://www.danielwinkler.de/hm/hmscript.pdf)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Mlle Germain on 29 Oct 2013, 13:39
Hi Ankhtahr,
I don't have much time, so I won't try to solve those exercises right now (I probably also would have difficulties doing it, since once you have proved all the basic stuff in maths, you kind of just use most of them without remembering how exactly the proofs work on an elementary level - just like you did before having seen the formal treatment). But I think we come from the same part of Germany, so our curriculum at school would have been similar. So I just wanted to encourage you by saying that in my experience, (almost) nobody who does maths at uni knows how to do proofs at the beginning and  that people from other parts of Germany did not seem to have much of an advantage over me, although they might have seen sequences and series and such things before. I was very confused and annoyed about not knowing how proofs work during my first weeks of maths lectures, but after having seen more proofs in the lectures, you get used to the methods and to the way of thinking. Really don't worry, even if some people seem to understand everything and find it really easy, either they're just pretending or it will level out quite quickly. Uni maths just comes as a bit of a shock, since it's so different from school.
I hope you're going to love your maths lectures after a while; logical thinking and formal proofs are awesome! (That said, I'm going back to my differential geometry sheet...)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: NotAwesomeAnymore on 31 Oct 2013, 02:28
Ankhtahr, it sounds like you're doing an introduction to Real Analysis. Real Analysis is great, but I wish in my course we'd focussed more on how to prove inequalities, because it's involved in pretty much everything. For your |x|/1+x^2 question, you need to find a another function that it's always less than, and that you know is always less than 0.5.

As for advice on proofs, don't think that proving stuff is the same as the incomprehensible proofs your teachers probably skimmed through when you were younger. It's really just explaining your logic, applying the definitions and theorems you were recently taught in the class, so don't overthink it!

Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 01 Nov 2013, 19:23
Wow, a lot's gone by this week.  Sorry I didn't get here sooner...

One of the simplest forms of proof is the reductio ad absurdum, or proof by contradiction.  For example, in

you can try assuming that it's less than .5.  But that means 1 + x2 < 2|x|, and so 1 - 2|x| + x2 < 0.  But that means that (1 - |x|)2 < 0, which is a contradiction to x being real (real squares aren't negative).

Sometimes you just need to play with it for a bit.

The proof that 2 is the min for set A is similar, multiply through by x (since it's positive) and you get x2 + 1 > 2x, subtract the 2x and you get a perfect square that's positive.  So if you assume x + 1/x can be less than 2, you get the absurdum that the square is negative.

Really, you can prove a lot of stuff by saying "suppose not", and then showing that there's a problem!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 02 Nov 2013, 01:49
Yay, you're giving me hope.

Tht's exactly the way I solved these problems. It took me literally the whole night (as you can see in the university thread), but I finished everything.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 06 Nov 2013, 08:52
I seriously hope that the exercise sheets will get easier, when I'm more used to doing stuff like this. This is one of our current exercises:
We have to prove these:
(http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Ctext%7B%28i%29%20Is%20%7Dq%5Cin%5Cmathbb%7BR%7D%5Csetminus%5Cleft%5C%7B1%5Cright%5C%7D%5Ctext%7Band%20%7Dn%5Cin%5Cmathbb%7BN%7D_0%5Ctext%7B%20then%20%7D%5Csum_%7Bk%3D0%7D%5Enq%5Ek%3D%5Cfrac%7B1-q%5E%7Bn&plus;1%7D%7D%7B1-q%7D%7D%5C%5C%20%7B%5Ccolor%7BWhite%7D%5Ctext%7B%28ii%29%20%7D%5Csum_%7Bk%3D0%7D%5E%7Bn%7D%5Cbinom%7Bn%7D%7Bk%7D%5Cleft%28-1%5Cright%29%5Ek%3D0%5Ctext%7B%20for%20all%20%7D%20n%5Cin%5Cmathbb%7BN%7D%5Ctext%7B.%7D%7D%5C%5C%20%7B%5Ccolor%7BWhite%7D%5Ctext%7B%28iii%29%20%7D%5Csum_%7Bk%3D0%7D%5En%5Cbinom%7Bn%7D%7Bk%7D%3D2%5En%5Ctext%7B%20for%20all%20%7D%20n%5Cin%5Cmathbb%7BN%7D%5Ctext%7B.%7D%7D%5C%5C%20%7B%5Ccolor%7BWhite%7D%5Ctext%7B%28iv%29%2023%20is%20for%20all%20%7Dn%5Cin%5Cmathbb%7BN%7D_0%5Ctext%7B%20a%20divisor%20of%20%7D5%5E%7B2n%7D-2%5En%5Ctext%7B.%7D%7D)

The first one was easy. And seeing that all these are best proven by induction is pretty obvious. I'm currently working on the second one. I'm absolutely sure the solution involves the binomial theorem, which we've learnt as (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%28a&plus;b%29%5En%20%3D%20%5Csum_%7Bk%3D0%7D%5En%20%5Cbinom%7Bn%7D%7Bk%7D%20a%5E%7Bn-k%7D%20b%5Ek%20%7D).

The toughest exercise though is supposed to be this one:
(http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Ctext%7BIs%20%7Dn%5Cin%5Cmathbb%7BN%7D%2Cn%5Cgeq2%5Ctext%7B%20and%20is%20%7Da%2Cb%5Cin%5Cmathbb%7BR%7D%2Ca&plus;b%3E0%2Ca%5Cneq%20b%5Ctext%7B%20then%20%7D2%5E%7Bn-1%7D%28a%5En&plus;b%5En%29%3E%28a&plus;b%29%5En%20%7D)

Nobody I know has found a solution for it. Somebody found a solution for it online though.

Edit: I'm currently kinda stuck at (ii) at (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Csum_%7Bk%3D0%7D%5E%7Bn%7D%28%5Cbinom%7Bn&plus;1%7D%7Bk%7D%28-1%29%5Ek%20%29&plus;%28-1%29%5E%7Bn&plus;1%7D%3D0%7D). I don't know how to get the +1 out of the binomial coefficient. I know that (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Cbinom%7Bn&plus;1%7D%7Bk%7D%3D%5Cbinom%7Bn%7D%7Bk-1%7D%7D), but I don't know if that helps me any further.
Wait. I got that wrong. It's (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Cbinom%7Bn&plus;1%7D%7Bk%7D%3D%5Cbinom%7Bn%7D%7Bk-1%7D&plus;%5Cbinom%7Bn%7D%7Bk%7D%7D), and now I can see that it might help me.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Barmymoo on 06 Nov 2013, 09:10
I find this thread completely fascinating and completely baffling.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 06 Nov 2013, 09:29
Damnit, I just noticed that (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Cbinom%7Bn&plus;1%7D%7Bk%7D%3D%5Cbinom%7Bn%7D%7Bk-1%7D&plus;%5Cbinom%7Bn%7D%7Bk%7D%7D) is of course only valid for (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%201%5Cleq%20k%5Cleq%20n%7D). So it does not help me, as in the sum there is k=0.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 06 Nov 2013, 10:39
Do you have to pay to use the site that makes the latex maths symbols?

Anyways, (iii):
I'll use (n k) for (http://upload.wikimedia.org/math/3/8/2/382c5908d125a08662b2fedc22f4d02c.png) ("n choose k")

(n k) is the number of distinct subsets of size k of a set of size n

The sum from k=(0 to n) of (n k) is thus the number of subsets of any size of a set of size n*

This sum is now pretty easy to figure out: for every element of the original (size n) set, you make a distinct subset by either selecting, or not selecting each element of the original set. So, for every element you have two choices, and there are n elements - thus you can create a subset in 2^n different ways. The sum is equal to 2^n.

* the same as the number of elements in the power set of a set of size n, which is generally known to be 2^n, but I doubt you can use that fact.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 06 Nov 2013, 10:46
Damnit, I just noticed that (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Cbinom%7Bn&plus;1%7D%7Bk%7D%3D%5Cbinom%7Bn%7D%7Bk-1%7D&plus;%5Cbinom%7Bn%7D%7Bk%7D%7D) is of course only valid for (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%201%5Cleq%20k%5Cleq%20n%7D). So it does not help me, as in the sum there is k=0.

That's not a problem! For k=0, the sum of the expression is 0: (n 0)*1 + -1 = 1*1-1=0. Just write that, and you can prove that the rest also sums to 0 using that relation.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 06 Nov 2013, 11:01
The LaTeX stuff is from http://www.codecogs.com/latex/eqneditor.php (http://www.codecogs.com/latex/eqneditor.php). Stuff I had to figure out: It works better when you have a "\\" at the beginning, as the first line will look weird otherwise, you should put everything in "{\color{White} <formula>}" to make it readable on the forums, you should choose PNG over GIF for transparency, and you should use "inline" and "compressed" to make it fit into your text. Then just copy the image address, and paste it into [img] tags.

I'm currently stuck at (ii), not at (iii). I'm trying to prove the whole thing by induction, so after showing that it's true for one n, which is 0, I now want to show that it's true for n+1, somehow using the whole prove I've made for n.

So now I have this:
(http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Csum_%7Bk%3D0%7D%5E%7Bn&plus;1%7D%5Cbinom%7Bn&plus;1%7D%7Bk%7D%28-1%29%5Ek%3D0%7D)
and want it to include this again:
(http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Csum_%7Bk%3D0%7D%5E%7Bn%7D%5Cbinom%7Bn%7D%7Bk%7D%28-1%29%5Ek%7D)

So somehow I have to get the +1 out of the binomial coefficient. Changing it all to (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Csum_%7Bk%3D0%7D%5E%7Bn%7D%5Cbinom%7Bn&plus;1%7D%7Bk%7D%28-1%29%5Ek%20&plus;%20%5Cbinom%7Bn&plus;1%7D%7Bn&plus;1%7D%28-1%29%5E%7Bn&plus;1%7D%7D) is easy, by pulling the n+1th summand out of the sum.

In this case it won't be possible to use (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Cbinom%7Bn&plus;1%7D%7Bk%7D%3D%5Cbinom%7Bn%7D%7Bk-1%7D&plus;%5Cbinom%7Bn%7D%7Bk%7D%7D), as it would lead to (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Cbinom%7Bn%7D%7B-1%7D%20%7D).
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 06 Nov 2013, 12:04
Ok, in the meantime I proved (iii). That was rather simple. While I still have no idea how to continue with (ii), I'm also stuck at (iv). I've shown that for n=0: (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%205%5E%7B2n%7D-2%5E%7Bn%7D%3D23m%20%7D).
Now I have to show the same thing for n+1, so I'm stuck at (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%205%5Ctimes%205%5E%7B2n%7D-2%5Ctimes%202%5E%7Bn%7D%3D23m%20%7D), and don't see how I can prove that with what I've shown in the first step.

Wait. I just realised that is wrong. Lemme figure that out again.

Typing all this stuff out makes me realise many errors I wouldn't see otherwise. Yay!

Okay, so (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%205%5E2%5Ctimes%205%5E%7B2n%7D-2%5Ctimes%202%5E%7Bn%7D%3D23m%20%7D) is looking much better, and now I can definitely see where this is going: (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%205%5E2-2%3D23%20%7D). But how do I legally change it into that? I'm out of training, that's for sure.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 06 Nov 2013, 12:12
I'm coming with an answer for (ii), but the formatting is KILLING ME.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 06 Nov 2013, 13:43
Okay, got it.

First of all, your base case is n=1, as the definition is given for the natural numbers*. That's easy to check - write it out and you get 1 - 1 = 0. Also, before this gets to crazy, the goal is to show that the n+1 case is in fact the n case twice, and is thus equal to 0+0

Now, assume that the statement (http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20%5Csum%20_%7Bk%3D0%7D%5E%7Bn%7D%20%5Cbinom%7Bn%7D%7Bk%7D*%28-1%29%5E%7Bk%7D%20%3D%200%7D) holds for 1 to n**. Then write out for n+1:

(http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20%5Csum%20_%7Bk%3D0%7D%5E%7Bn&plus;1%7D%20%5Cbinom%7Bn&plus;1%7D%7Bk%7D*%28-1%29%5E%7Bk%7D%20%3D%20%5Cbinom%7Bn&plus;1%7D%7B0%7D*%28-1%29%5E%7B0%7D%20&plus;%20%5Cbinom%7Bn&plus;1%7D%7B1%7D*%28-1%29%5E%7B1%7D%20&plus;%20...%20&plus;%20%5Cbinom%7Bn&plus;1%7D%7Bn%7D*%28-1%29%5E%7Bn%7D%20&plus;%20%5Cbinom%7Bn&plus;1%7D%7Bn&plus;1%7D*%28-1%29%5E%7Bn&plus;1%7D%7D)

Apply the identity   (http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20%5Cbinom%7Bn&plus;1%7D%7Bk&plus;1%7D%20%3D%20%5Cbinom%7Bn%7D%7Bk%7D%20&plus;%20%5Cbinom%7Bn%7D%7Bk1%7D%7D)   to all terms of the sum, except the first and the last:

(http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20%5Cbinom%7Bn&plus;1%7D%7B0%7D%20*%20%28-1%29%5E%7B0%7D%20&plus;%20%28%5Cbinom%7Bn%7D%7B0%7D&plus;%5Cbinom%7Bn%7D%7B1%7D%29*%28-1%29%5E%7B1%7D%20&plus;%20...%20&plus;%20%28%5Cbinom%7Bn%7D%7B0%7D&plus;%5Cbinom%7Bn%7D%7B1%7D%29%20*%20%28-1%29%5E%7Bn%7D%20&plus;%20%5Cbinom%7Bn&plus;1%7D%7Bn&plus;1%7D%20*%20%28-1%29%5E%7Bn&plus;1%7D%7D)

Reduce (http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20%5Cbinom%7Bn&plus;1%7D%7B0%7D%20*%20%28-1%29%5E%7B0%7D%20%7D)   to   (http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%201%7D)   and   (http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5Cbinom%7Bn&plus;1%7D%7Bn&plus;1%7D*%28-1%29%5E%7Bn&plus;1%7D%7D)   to   (http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%28-1%29%5E%7Bn&plus;1%7D%7D)  . Also multiply in the   (http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%28-1%29%5E%7Bk%7D%7D)   terms:

(http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%201%20&plus;%20%5Cbinom%7Bn%7D%7B0%7D%28-1%29%5E%7B1%7D%20&plus;%20%5Cbinom%7Bn%7D%7B1%7D%28-1%29%5E%7B1%7D%20&plus;%20...%20&plus;%20%5Cbinom%7Bn%7D%7Bn-1%7D%28-1%29%5E%7Bn%7D%20&plus;%20%5Cbinom%7Bn%7D%7Bn%7D%28-1%29%5E%7Bn%7D%20&plus;%20%28-1%29%5E%7Bn&plus;1%7D%7D)

Reorder: for all of the pairs on the form   (http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20%28-1%29%5E%7Bt%7D%5Cbinom%7Bn%7D%7Bt-1%7D%20&plus;%20%28-1%29%5E%7Bt%7D%5Cbinom%7Bn%7D%7Bt%7D%7D)  , group the left hand and right hand terms with the other left hand and right hand terms:

(http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%201%20&plus;%20%28%20%5Cbinom%7Bn%7D%7B0%7D%28-1%29%5E%7B1%7D%20&plus;%20...%20&plus;%20%5Cbinom%7Bn%7D%7Bn-1%7D%28-1%29%5E%7Bn%7D%20%29&plus;%20%28%20%5Cbinom%7Bn%7D%7B1%7D%28-1%29%5E%7B1%7D%20&plus;%20...%20&plus;%20%5Cbinom%7Bn%7D%7Bn%7D%28-1%29%5E%7Bn%7D%20%29%20&plus;%20%28-1%29%5E%7Bn&plus;1%7D%20%7D)

Extract the -1 from the left grouping, and turn the left and right grouping into sums:

(http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%201%20-%28%20%5Csum%20_%7Bk%20%3D%200%7D%5E%7Bn-1%7D%20%5Cbinom%7Bn%7D%7Bk%7D%20*%20%28-1%29%5E%7Bk%7D%29&plus;%20%28%20%5Csum%20_%7Bk%20%3D%201%7D%5E%7Bn%7D%20%5Cbinom%7Bn%7D%7Bk%7D%20*%20%28-1%29%5E%7Bk%7D%29%20&plus;%20%28-1%29%5E%7Bn&plus;1%7D%7D)

Now, the right hand sum is only missing the k=0 term to be equal to 0 (by the original assumption). But the k=0 term is, as we saw as a part of the base proof, 1, so we can add the right hand sum and the 1, and get 0. We remove those two, and are left with:

(http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20-%28%20%5Csum%20_%7Bk%20%3D%200%7D%5E%7Bn-1%7D%20%5Cbinom%7Bn%7D%7Bk%7D%20*%20%28-1%29%5E%7Bk%7D%29&plus;%20%28-1%29%5E%7Bn&plus;1%7D%7D)

The left hand sum can be simplified a lot, using the original assumption:

(http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20-%28%20%5Csum%20_%7Bk%20%3D%200%7D%5E%7Bn-1%7D%20%5Cbinom%7Bn%7D%7Bk%7D%20*%20%28-1%29%5E%7Bk%7D%29%20%3D%20-%28%20%5Csum%20_%7Bk%20%3D%200%7D%5E%7Bn%7D%20%5Cbinom%7Bn%7D%7Bk%7D%20*%20%28-1%29%5E%7Bk%7D%29%20&plus;%20%5Cbinom%7Bn%7D%7Bn%7D*%28-1%29%5E%7Bn%7D%20%3D%20-0%20&plus;%20%5Cbinom%7Bn%7D%7Bn%7D*%28-1%29%5E%7Bn%7D%20%7D)

Meaning that:

(http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20-%28%20%5Csum%20_%7Bk%20%3D%200%7D%5E%7Bn-1%7D%20%5Cbinom%7Bn%7D%7Bk%7D%20*%20%28-1%29%5E%7Bk%7D%29%20&plus;%20%28-1%29%5E%7Bn&plus;1%7D%20%3D%20%28-1%29%5E%7Bn%7D%20&plus;%20%28-1%29%5E%7Bn&plus;1%7D%20%3D%200%20%7D)

So, the assumption for the sum being equal to 0 at n lead to the sum being equal to 0 at n+1, and the proof is done. Wow that got really fucking long. It sounded a lot easier in my head.

* Some definitions of the natural numbers include 0, but since the notes you have make specific mention of it when they want 0 included (with the subscript 0), that's not the case here
** n should not be used here, but from your examples it seems like you've learned that for induction, so I'll not go into the idiocy of some text books and lecturers of insisting that calling the general case n, and assuming n as a part of your proof makes sense by having two different variables named n aaaaaahrg
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 06 Nov 2013, 13:43
Oh holy shit is humans supposed to be able to read that?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 06 Nov 2013, 13:47
holy crap. Thanks a lot! I don't think I would have managed that on my own. I'm sorry for having you type all that out.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 06 Nov 2013, 14:06
Sorry to say, but I'm not really sure what you did at this step:

Apply the identity   (http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20%5Cbinom%7Bn&plus;1%7D%7Bk&plus;1%7D%20%3D%20%5Cbinom%7Bn%7D%7Bk%7D%20&plus;%20%5Cbinom%7Bn%7D%7Bk1%7D%7D)   to all terms of the sum, except the first and the last:

(http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20%5Cbinom%7Bn&plus;1%7D%7B0%7D%20*%20%28-1%29%5E%7B0%7D%20&plus;%20%28%5Cbinom%7Bn%7D%7B0%7D&plus;%5Cbinom%7Bn%7D%7B1%7D%29*%28-1%29%5E%7B1%7D%20&plus;%20...%20&plus;%20%28%5Cbinom%7Bn%7D%7B0%7D&plus;%5Cbinom%7Bn%7D%7B1%7D%29%20*%20%28-1%29%5E%7Bn%7D%20&plus;%20%5Cbinom%7Bn&plus;1%7D%7Bn&plus;1%7D%20*%20%28-1%29%5E%7Bn&plus;1%7D%7D)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Barmymoo on 06 Nov 2013, 14:21
As far as I can tell, he put some numbers into a visually pleasing pattern! I have so much respect for people who understand this stuff. It's all I can do to understand the biology I have to study, and at least that's expressed in words.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 06 Nov 2013, 14:38
Oh shit, it's supposed to say k+1 in the second binomial there, not k1. Erp.

So, example, you would turn the term   (http://latex.codecogs.com/png.latex?%7B%5Ccolor%7BWhite%7D%20%5C%5C%20%5Cbinom%7Bn&plus;1%7D%7B5%7D*%28-1%29%5E%7B5%7D%20%7D)   and turn it into  (http://latex.codecogs.com/png.latex?%7B%5Ccolor%7BWhite%7D%20%5C%5C%20%28%5Cbinom%7Bn%7D%7B4%7D%20&plus;%20%5Cbinom%7Bn%7D%7B5%7D%29*%28-1%29%5E%7B5%7D%20%7D). Do that for every term of the sum except the first and the last (since they're not applicable to the identity)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 06 Nov 2013, 14:40
Is the images? not working for anyone else? If I open them in a new tab and replace "White" with "Black", it works. Huh.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 06 Nov 2013, 14:43
'cause they're gifs and not pngs. That's why I recommended png.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 06 Nov 2013, 14:52
Fix'd!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Schmee on 08 Nov 2013, 07:18
This was in my uni's Maths and Stats society newsletter:
Quote
Several years ago, A/Prof Barry Hughes set a ﬁrst-year exam question which asked for a proof that there is no largest prime number.The following was sent to him by one of the graduate students who helped with the marking:

Hi Barry, just for educational value, I compiled a list of the most incorrect answers to question B4(b). It is entitled "15 good reasons why Pure Mathematics is not taught to first year students."
• Proof by example: “Let x be the largest prime. Then x = 91 but 91 + 6 = 97, which is prime. Therefore, 91 cannot be the largest prime number. Therefore there is no largest prime number.”
• Proof by oddness: “If n is the largest prime number, then n is odd. Then (n + 1)=2 is even. Therefore, (n + 1)=2 + n is odd. But (n + 1)=2 + n is not divisible by any number except itself. As it is bigger than n, the assumption is wrong, by contradiction.”
• Proof by intuition: “Prime numbers are integers that can be divided by themselves only; prime numbers are odd with the exception of 2. By intuition as n -> infinity, there will always be an odd number that cannot be divided by any other number besides itself.”
• Proof by sqrt(2): “that p is divisible by q, i.e. p/q = 2r, where r is an even number. Then p = 2qr so p2 = 4q2r2. But r2 does not exist and q! = 1. Therefore, q must exist. Since q exists, p must be divisible. Therefore, by contrary, there is no largest prime number.”
• Proof by superinduction: “2 is a prime number. Now assume N is the largest prime. But then N + 1 exists and is also prime. Therefore, by induction there is no highest prime number.
• Proof by the previous question: “Suppose N is the largest prime. Then let N = (n2)/2. Therefore n = sqrt(2N). But from above (1 + 2 + 3 + ... +n) > N. Hence, there is a larger prime number than N.”
• Proof by tutorial question: “Let m, n be two integers with m > n + 1. If k is even, mk + nk cannot be expressed in terms of (m + n)(polynomial in m and n) and so is prime. Therefore, as m and n can be any numbers, there is obviously no largest prime number.”
• Proof by having no idea what a prime is: "Say the largest prime possible is x, then 2x is also a prime since the statement is true for all natural numbers."
• Proof by experimental data: "Suppose n is the highest prime. Then 2n-1 is also prime. But 2n-1>n so there is no highest prime (Check: 2*2-1=3, 2*3-1=5, 2*5-1=11, 2*11-1=23, so true)"
• Proof by subscript: “If there is a highest prime, we can number all the primes p1, p2, ... , pn. But as there is no highest natural number, there is always an n + 1 so there must be a pn+1. Therefore, there is no highest prime.”
• Proof by inﬁnity: “Let n be the highest prime number. But infinity is greater than all numbers so infinity > n. If n is the highest prime this would mean infinity has factors. Therefore, we have a contradiction.”
• Proof by reverse logic: “All prime numbers are odd. Suppose there were a highest prime. Then we have a highest odd number. But if 2k+1 is the highest odd number, then 2k+3 = 2(k+1)+1 is also an odd number. Therefore, we have a contradiction and therefore we have a contradiction."
• Proof by denial: “Assume there is a largest prime M. We can add 1 to M until we get another prime number N (M + 1 + 1 + 1 + ... + 1 = N). But then N > M. Therefore, M is not the largest prime number, so there is no largest prime number.”
• Proof by formula: “As prime numbers are derived via the formula, we can assume it works for n = k giving the highest prime number. But then it also works for n = k + 1, so there is no highest prime number.”
• Proof by continuity: “Let x be the largest prime number. Then x > all other primes. But then (x + n), the next prime number, does not exist. However, numbers are continuous and so (x + n) does exist. Therefore, there is no x.”
These are all verbatim answers from the 150-odd papers I marked, i.e. about one in ten. My favourite is deﬁnitely number 10; I think that is quite ingenious.
— George Doukas
... That took a while to copy over, but I think you'll agree it was worth it.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Barmymoo on 08 Nov 2013, 07:54
What would be a correct answer to that question?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 08 Nov 2013, 10:44
Euclid had a good one.  If there's a largest prime, then you list them all and multiply all of the primes together into one big number.

Then add 1 to that result.  It's bigger than the biggest prime, so it's not prime.  And none of the primes will divide it, they'll all have a remainder of 1.  That means that it must be prime, since nothing divides it but 1 and itself.

And that's absurd - it can't be both prime and not prime, so the assumption (that there's a largest prime) is false, there isn't one.

Ankhtahr, did you get the 52n - 2n yet?  It's (52)n - 2n, which will always factor as

(52 - 2)((52)n-1 + (52)n-22 + ... + 2n-1), and so there's always a factor of 23.

(sometimes, it just looks like it needs induction...)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 08 Nov 2013, 11:04
Schmee: The first one cracked me up already. Where the hell did that 91 come from?

"Proof by having no idea what a prime is"

Bwahaha!

Carl: Yes, I had to turn in the exercise sheet on Thursday already.

Oh, and I'm so very sorry to say, but Snalin: I found a much, much simpler proof.

Like I had said, I suspected the whole thing could be solved with the binomial theorem ((http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%28a&plus;b%29%5En%20%3D%20%5Csum_%7Bk%3D0%7D%5En%20%5Cbinom%7Bn%7D%7Bk%7D%20a%5E%7Bn-k%7D%20b%5Ek%20%7D)). I noticed that you can just choose a = 1, and apply it on the term.

Which makes it look like this:
(http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Csum_%7Bk%3D0%7D%5E%7Bn&plus;1%7D%5Cbinom%7Bn&plus;1%7D%7Bk%7D%28-1%29%5Ek%20%3D%200%7D%20%5C%5C%5C%5C%20%7B%5Ccolor%7BWhite%7D%5Csum_%7Bk%3D0%7D%5E%7Bn&plus;1%7D%5Cbinom%7Bn&plus;1%7D%7Bk%7D1%5E%7Bn-k%7D%28-1%29%5Ek%20%3D%200%7D%20%5C%5C%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%281-1%29%5E%7Bn&plus;1%7D%20%3D%200%7D%20%5C%5C%5C%5C%20%7B%5Ccolor%7BWhite%7D0%5E%7Bn&plus;1%7D%20%3D%200%7D)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 08 Nov 2013, 11:18
"Therefore, we have a contradiction, and so therefore, we have a contradiction"

Pure gold!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 10 Nov 2013, 07:42
Quote
Prove: Each convergent series over K \in {\mathds{R}, \mathds{C}} is bounded.

Is there any sensible answer except "well. duh, if it converges to x, it is bounded by x"?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 10 Nov 2013, 09:49
Quote
Prove: Each convergent series over K \in {\mathds{R}, \mathds{C}} is bounded.

Is there any sensible answer except "well. duh, if it converges to x, it is bounded by x"?

False!

Just in R, the sequence 1, -1, 1/2, -1/2, 1/3, -1/3, ... converges to 0, but is bounded above by 1 and below by -1.  The limit is rarely a bound unless the sequence is monotone.

Though maybe I don't understand the question - what exactly is {\mathds{R}, \mathds{C}}?  It looks like the two-dimensional space R X C, which doesn't really change the answer, but I don't think that's what it means?

(Topology is almost nothing but a collection of counterexamples... I got  million of 'em!)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: PthariensFlame on 10 Nov 2013, 16:07
(Topology is almost nothing but a collection of counterexamples... I got  million of 'em!)

On the contrary (heh), only concrete topology is a collection of counterexamples; there are almost no counterexamples in synthetic topology!

EDIT:  Also, Barmymoo said I should probably cross-post this (http://forums.questionablecontent.net/index.php/topic,29433.0.html) from the ENJOY subforum.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 11 Nov 2013, 00:10
Carl: yes, I realized something similar yesterday in bed.

Sorry about the Latex. The exercise means to say that K is an element of the set containing the real number set and the complex number set. That is, K can be R or C. The cartesian product (aka the two-dimensional space) would have been denoted by R x C.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 11 Nov 2013, 04:55
Thanks!  I'm an old TeX haXer, but I was stupid-sleepy and couldn't make it out.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 19 Nov 2013, 06:16
Hey, it's business as usual, I'm sitting at university with some others, working on an exercise sheet, due the day after tomorrow..

We're sitting here for about 3 hours already, and we're still at the first of 7 exercises. We have to find accumulation points of a series. The first five are similar to this one, the last two are different, and probably more difficult.

I had only three hours of sleep last night, as I had to do another exercise sheet.

I'm loving this.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 19 Nov 2013, 16:46
That's the delirium talking.

Keep it up!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 19 Nov 2013, 17:23
I have never had to spend more than 4ish hours a day at Uni work. I'm not sure if it's super easy here, if I'm more naturally talented than I thought, or if I'm just good at managing my laziness.

Although I suppose that if I had wanted a good grade in the introductory maths courses, I would have had to work way hard. I didn't do that, and got a D. Aaaand that hasn't mattered a single bit. Calculus is dumb anyways.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 19 Nov 2013, 17:39
:x
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Haseo on 19 Nov 2013, 19:08
So I'm in introduction to Derivatives. I'm scared to get into AP Calc AB next semester...
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 19 Nov 2013, 21:39
Ankh, could you PM me links to one of each of the sheets you are doing?

*looks at sentence* *rewords* Could you show me the current worksheets for all of the courses you are taking?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 20 Nov 2013, 02:47
:x

No, but, like, I don't need it, anyways. Who ever needed derivatives? :D
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 20 Nov 2013, 05:22
Maths is, right? I mean, maths is written in plural form, but it is still just a single thing, right? Or is it maths are sexy?

Is.  "Maths" is not the plural of "math"; it's a different abbreviation of "mathematics".  Many words end in s without being plural.

Depends. It's "Maths" in the UK, and "Math" on the other side of the pond (or should I say elsewhere?).

But looks like you guys started having fun while I was gone from this subforum. I don't have anything against doing it here, but I also might recommend that we take it to Math.Stackexchange. (http://math.stackexchange.com/) There is a real-time editable TeX-interpreter (MathJax) there, so we don't need to use graphics. I'm sure Paul might be able to hack us into the cloud running MathJax (or whatever it is), but we won't have hundreds or thousands of qualified people answering here.

Take it from someone who typed the plainTeX-souce to his dissertation 23 years ago. With the VI-editor that came with Unix.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 21 Nov 2013, 06:50
Currently working on this sheet (https://own.halfur.de/public.php?service=files&t=6cd7e9a72bff60eede44c7d3c502df0c) (sorry for using this shitty owncloud-website, I don't have SSH-access to my server, so I can only use that if I don't want to upload it at any strange service.

Don't really think anyone here can help me with it, if I don't give a translation. I don't have the time to give a translation right now.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 21 Nov 2013, 08:10
You are doing weird stuff.

Re 2b)
Quote
Let G be a group and H1, H2 subgroups of G.

Show that H1 union H2 is a subgroup or find a counterexample

Should be true, imho. The neutral element of G is in H1uH2. For every element x of H1, -x in in H1 (H1 is subgroup), and therefore in H1uH2. The same for H2.
Associativity should be, uh, trivial.

Do you know where to start on the others?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 21 Nov 2013, 09:41
You are doing weird stuff.

Re 2b)
Quote
Let G be a group and H1, H2 subgroups of G.

Show that H1 union H2 is a subgroup or find a counterexample

This is false. Let G be the group S_3 of symmetries of an equilateral triangle. Let H1 and H2 be groups generated by a single reflection (or single transposition, if you are doing it in terms of permutations rather than symmetries of a triangle). There union then contains two distinct reflections, but not their composition (which is rotation or a 3-cycle again depending which way you view this group).

In terms of the group axioms it is the zeroth axiom that fails. This set (the union of H1 and H2) does not have an operation.

It is a "standard" exercise to show that the union of two subgroups is a subgroup, if and only if one of the subgroups is contained in the other. The idea is the same. If x is an element of H1 but not an element of H2, and y is an element of H2 but not of H1, then xy cannot be an element of either H1 or H2. Therefore it is not in the union.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 21 Nov 2013, 09:44
Don't really think anyone here can help me with it, if I don't give a translation. I don't have the time to give a translation right now.
If the topic of the sheet is something like the first course in abstract algebra (judging from Loki's post), I am qualified. I don't fully trust my mathematical German, but I have read a number of German math books (though it was a while ago). The problem is that my browser is convinced that following up your link is a bad idea, risky even.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Barmymoo on 21 Nov 2013, 10:17
Maths is, right? I mean, maths is written in plural form, but it is still just a single thing, right? Or is it maths are sexy?

Is.  "Maths" is not the plural of "math"; it's a different abbreviation of "mathematics".  Many words end in s without being plural.

Depends. It's "Maths" in the UK, and "Math" on the other side of the pond (or should I say elsewhere?).

That's what Paul is saying - unless for some odd reason you're pluralising the American word "math". If you're using the UK word "maths", it is a singular noun.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 21 Nov 2013, 10:27
Sorry, May, Paul. I've just heard this maths vs. math discussion too many times, and jumped to a conclusion.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 21 Nov 2013, 10:38
The problem is that my browser is convinced that following up your link is a bad idea, risky even.

Lemme guess, you're getting a certificate error. I'm using a free certificate issued by CACert. Go to cacert.org, click on "Root Certificates" and install the Class 1 and Class 3 Certificate (PEM is usually the best option). Then the message should disappear.

Or you could just ignore the message.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Is it cold in here? on 21 Nov 2013, 14:30
http://www.amazon.com/How-Solve-Aspect-Mathematical-Method/dp/4871878309/ref=sr_1_2?s=books&ie=UTF8&qid=1385072927&sr=1-2&keywords=how+to+solve+it
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 21 Nov 2013, 17:15
How far are you along, Ankh? Because I seem to remember this being your starting year, and abstract algebra the first semestre seems a bit crazy. Are you taking pure maths?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 21 Nov 2013, 21:43
I'll be teaching freshman abstract algebra this coming Spring, and Ankh's sheet could well come from my course. Which questions are giving you a headache?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 21 Nov 2013, 23:31
This is the math I need to do for Computer Science here. This is linear algebra. I also have advanced mathematics/analysis.

Currenly I'm still struggling with some of the definitions, but 1 c) and 4 are probably the most difficult.

I'm working a little bit ahead right now, This sheet is only due on tuesday, but I'd like to maybe sleep next week.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 22 Nov 2013, 03:35
Quote from: Ankhtahr's Teacher
Assignment 1: Let G and H be groups with respective neutral elements e G and  e H
Let f be a homomorphism of groups from G to H. Show that:

c) Let p: G -> G/ ker( f ) be the canonical projection mapping the element g to its coset g ker(f) (sorry, I don't know
what "coset" is in German). Then there exists a unique homomorphism f from  G/ ker( f ) to H such that
f = f o p (using underscore instead of the overline bar, sorry). Furthermore, f is injective and its image
is equal to that of f.

Here we have no choice in constructing the homomorphism f. If g ker(f) is an arbitrary element of the quotient group  G/ ker( f ), we have no choice. Because f = f o p, and g ker(f)=p(g), we must define f(g ker(f))=f(g). Can you follow this?

This means that the uniqueness part will be clear, once we get existence. But there are several catches before we are there. First and foremost you should check that f  is well defined (=wohldefiniert?). This is because distinct elements of the group G may give rise to the same coset. So we need to clear the following obstacle. If g ker(f) = g' ker(f) for two different elements g and g' of the group G, then we must be sure that f(g ker(f))= f (g'ker(f)). In other words, we need to check that f(g)=f(g'). Can you do that? Hint: here it is absolutely crucial that we are talking about cosets of ker(f).

Ok, if that is clear, then the rest is relatively easy. Namely you need to check that f is a homomorphism of groups. This is not very difficult, because you only need to recall what the group operation of G/ ker(f) is, and then utilize the fact that f was known to be a homomorphism of groups.

This exercise is designed to make you once more think through the meaning of the relevant definitions.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 22 Nov 2013, 03:54
Quote from: Ankhtahr's teacher
Assignment 4: Let M be a set and  P(M) its power set. For all elements A, B of P(M) define the
symmetric difference (Can't draw the Delta symbol here. I'm using '+' instead, because this will become
the addition of the ring you are creating. Similarly 'n' denotes intersection when standing alone).
A + B := (A U B) \ (A n B).
Show that (P(M),+,n) is a ring. You are allowed the assume the associativity of the symmetric difference.

Ok. So the first task is to show that (P(M),+) is a commutative group. Before we start let me recap. The set A+B  consists of those elements of M that belong to either A or B but not to both. This will be another subset of M, i.e. an element of P(M), so '+' is an operation on P(M). We were allowed to assume associativity, so let's skip that. A neutral element? Can you think of a subset S of M with the property that no matter which set A is, then S+A=A. Hint: S should not contain any elements of A, because those would be excluded from S+A.

Proving the existence of inverse is dependant on knowing what the neutral element is, so I stop here, and let you sweat a bit. :evil:

Edit: I am trying to guess a helpful level of chatter and hints here. If it is Greek to you, say so, and I will make another attempt. Getting used to the level of abstractness here requires a fair amount of work, and takes a little while. For example, I would guess that may be top 10% of the students in my course would make any serious headway in your Exercise #4. A vast majority of students here are unprepared to fathom that they can do addition and multiplication on sets. German high school may be a bit better in that regard. The exercise is not very difficult, but you have to absorb the rules of the game of definitions and axioms to stand a chance.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Pilchard123 on 22 Nov 2013, 05:59
(click to show/hide)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 22 Nov 2013, 06:05
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 25 Nov 2013, 18:42
I want to take this opportunity to thank you, Skewbrow, for saving me a little bit of sleep. While it still took me a lot of time, I don't think I would have come to any solution without your help. It's 03:40 am now, but I'm done with the exercise sheet, which is due tomorrow/today. I'll have to get up at 06:45, so I'll leave now, but I wanted to express my gratitude.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 28 Nov 2013, 18:49
And once again I have another linear algebra exercise sheet for Tuesday.

Here it is (https://own.halfur.de/public.php?service=files&t=f64ee31b08eeba0335744e5560287fe1)

I'm doomed.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 29 Nov 2013, 05:35
I'm impressed at the pace your course is going forward (but wouldn't try this at home). Apparently Germans know something about teaching abstract algebra that we don't.

Since you did well last week with very minimal hints (more like gentle probing), let's just roll with it

Quote from: Ankhtahr's teacher
Assignment 1:
A) Show that the set Q(31/2) of numbers of the form a+b31/2, with a and b ranging over Q, is a field with respect to the usual addition and multiplication of real numbers.
B) Show that the mapping that sends a+b31/2 to a-b31/2 is a well-defined automorphism of fields.
Hint: You are allowed to assume without proof that 31/2 is irrational, i.e. not an element of Q.

In part A your are basically asked to show that the set Q(31/2) is closed under addition, subtraction, multiplication and division. If you have covered a result called "subfield criterion", then that is basically what it says. You can forget about checking the field axioms. They hold in the reals, and this a subset, so you get the axioms for free. The only worry is that the basic arithmetic operations won't take you outside the set. Checking that ( a+b31/2 ) +/- (a'+b'31/2 )
with a,a',b,b' rational can be written in the form  c+d31/2  with c and d rational should be easy. Doing the same for  (a+b31/2 )*(a'+b'31/2 ) takes a bit more work. Expand the product and use the fact that 31/2*31/2=3 is rational. Closure under division is the hard part. As the set is closed under multiplication it suffices to check that the inverses, i.e. the numbers 1/( a+b31/2) can be written in that form. Hint: show that
( a+b31/2)( a-b31/2) is a non-zero rational number. This is where the "free" assumption about irrationality of 31/2 will help you out. If you get stuck, describe exactly how far you made it.

In part B the proof of this mapping being an automorphism parallels very much the proof that complex conjugation is an automorphism of fields. In the complex case the key is that i2=-1=(-i)2. Here the equatiions (31/2)2=3=(-31/2)2 play a similar role. Checking the conditions for this being a homomorphism amounts to just using that equation and expanding everything. (Note to others: that's a lot of work, I'm not doing Ankhtahr's homework here - just giving pointers). Before you go there you seem to be required to check that the mapping is well-defiined. Unlike last week, there were no choices made. You do need to check that if  (a+b31/2 ) is equal to (a'+b'31/2 ), then we must have both a=a' and b=b'. This is another spot where you can take advantage of the irrationality of 31/2.

I'm not gonna bother translating Assignment #2. Too many formulas. Part A you can just brute force (my students would use WolframAlpha for this, but it is good for you to do a bit of pencilwork). Part B is fun. The solution set of that equation is a circle in the complex plane (unless I made a mistake) - just write z=x-iy and see what you get. That pair of inequalities: the first requires that z is closer to zero than to 2i. The second requires that the distance to 2i is less than 3. Draw a picture!

I don't feel like reproducing the last one here either. I doubt the sanity of any teacher who expects to cover all the details needed here within a 90 minute problem session. Is s/he a rookie or something? I tried things like this when the ink on my dissertation would still stick to the fingers, but ... Might 's well get on with it. Part A is just identifying the additive and multiplicative neutral elements and checking out all the axioms. If you are meticulous, it will run a few pages. Part B is more of the same. For the part asking about the field of two elements I just tell you to check what happens with the polynomial X2-X.

This should keep you occupied for a while  :evil:
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 01 Dec 2013, 15:57
So, uhm. Ever know that kind of tasks where you know the answer, but you just don't know how to prove it? I have one of those.

Quote
Prove: every polynomial function p: R -> R (with R as reals) of an odd degree has at least one x-intercept.

Below are the thoughts I have got so far:

I suspect that I have to use Bolzano's theorem, showing that there are values a and b in the reals which satisfy p(a)<0 and p(b)>0, and therefore there must exist an x-intercept between a and b, provided the function is continuous (which a polynomial is. I hope I don't have to show that).

It is fairly easy to show that the condition is true for a function p~(x) = c*x2n+1, with c as a real constant and n an arbitrary natural number, but how do I show that it's also true for
p(x) = c2n+1*x2n+1 + (c2n*x2n + c2n-1*x2n-1 + ... + c1*x1 + c0)?
Ie that the bracketed term doesn't "set off" the signum of p(x) by too much?

It would be easier if the polynomial was symmetrical to the origin, but if I had that as a given, I'd already know it has an x-intercept.

Maybe I am approaching this from the wrong angle. Obviously, a function with c0 = 0 has an x-intercept at the origin.
Let c0 be positive without loss of generality. Then we have our b=0 with f(b)=c0>0.
Then, according to highschool analysis, p*(x) := p(x - c0) should have an x-intercept.
p*(x) = c2n+1*(x-c0)2n+1 + (c2n*(x-c0)2n + c2n-1*(x-c0)2n-1 + ... + c1*(x-c0)1 + c0)

I suspect I can then  somehow show that all those "-c0" terms cancel the final "+c0" out, but I am stuck on how. Probably has something to do with the dreaded Pascal triangle.

Help?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 01 Dec 2013, 21:15
Odd degree polynomial over the reals? What can you say about its limits as x goes to +/- infinity?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 01 Dec 2013, 21:45
...
Thanks.  :psyduck:

I don't remember what properties of polynomials we did or did not prove, so I might have to prove that your hint is applicable in a fairly roundabout way but that has definitely set me on the right track.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 10 Dec 2013, 12:07
Even though I feel bad for bothering you guys with so many exercise sheets of mine, I guess I don't have much of a choice. I really need the points now. (University thread for more info)

So yeah. After missing two lectures because of being sick, I'll have to turn in advanced mathematics tomorrow. The sheet is this one here (http://www.math.kit.edu/iana3/lehre/hm1inf2013w/media/blatt_7.pdf). We only have to do the tasks with a (K), so 25 and 27.
Sorry Skewbrow, advanced mathematics, not linear algebra. But I have no idea how to even approach these tasks. A friend of mine told me that it would be simple calculating, but he's basically always a week ahead in exercise sheets and I don't know how. My other friends don't really know either. Tomorrow evening (it's evening now as well) we'll be sitting in the university again, trying to do these tasks.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: PthariensFlame on 10 Dec 2013, 22:03
Ok, so I might actually be able to help you here (thank you, Google Translate!).

25 is all about reducing expressions to simplest form; the expressions just happen to contain limits in them, but the limits are incidental to the overall goal.  Most of them require you to invoke l'Hôpital's rule, possibly multiple times:  if f and g are differentiable functions and there exists some k such that f(k) = g(k) = 0 or f(k) = g(k) = infinity, then limx->k f(x)/g(x) = limx->k f'(x)/g'(x).

27 just requires you to inspect the given functions f and g and determine, in each case, what subset of the function's domain it's continuous on.  For example, the absolute value function is continuous over its whole domain, but its derivative is not (it's missing 0).
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 11 Dec 2013, 02:59
As a semi-helpful hint for 27a, you should know by now that if two functions f(x) and g(x) are continuous in x0, then their combinations are also continuous in x0, ie
1) f(x0)+g(x0)
2) f(x0)-g(x0)
3) f(x0)*g(x0)
4) f(x0)/g(x0) (provided that g(x0) is not 0 of course)
are continuous.

27b: The rational numbers Q are dense in R (please don't beat me up as I wasn't able to find the correct translation for "liegen dicht in R"). Which means that for any number p which is in R\Q, there is a number q in Q which is infinitely close to it (and vice-versa). What values would the function take between, say, 1 and 1 plus some infinitely small ε? Does this look like a continuous function?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 11 Dec 2013, 04:18
Well, we didn't have L’Hôpital's rule yet, so we're not allowed to use it. Actually we didn't even have differetiation yet, so we're not allowed to use that either...
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: PthariensFlame on 11 Dec 2013, 12:22
Well, we didn't have L’Hôpital's rule yet, so we're not allowed to use it. Actually we didn't even have differetiation yet, so we're not allowed to use that either...

Really?  I can't think of how to solve, e.g., 25a or 25d without l'Hôpital's rule; then again, my favorite calculus is definitely not the limit calculus.  :)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 12 Dec 2013, 02:23
What PhtariensFlame said. I'd say screw it and use it anyway rather than not getting any points at all. (Or, if you still have time to do so, ask your tutor or a jury of your peers if you can use it.)
Maybe you missed the introduction of it when you were sick?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: pwhodges on 12 Dec 2013, 02:35
Maybe the exercise is deliberately intended to get you to find out about the rule yourself - it's one style of teaching...
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 12 Dec 2013, 03:56
Nah, it says:

"Beachten Sie bitte, dass Sie (...) nur den bisher behandelten Stoff verwenden dürfen!"

Which basically translates to "only use the material that's already been used in the course". Or something like that. I almost know some German!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: BeoPuppy on 12 Dec 2013, 06:02
Maybe the exercise is deliberately intended to get you to find out about the rule yourself - it's one style of teaching...
Sounds like being pushed of a cliff to see if the students evolve wings or not.

(Thanks, T.Pratchett!)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 12 Dec 2013, 08:39
Nope, we found solutions for both which were relatively simple without it.

I still remember our solution for the d) involving simply showing that lim cos(x) is 1, so lim cos(x) -1 is 0, so lim sin(cos(x) -1) is 0.

And as far as I know the rule will be introduced in a few weeks.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 14 Dec 2013, 13:35
Turns out it's hard to get back into Linear Algebra after being forced to skip a week.

Exercise Sheet (https://own.halfur.de/public.php?service=files&t=16a945525f171bb768635c28aa23a337)

I have to read up on a lot right now. I don't even really know what vector spaces are.

I have until Monday evening for this sheet.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: PthariensFlame on 14 Dec 2013, 14:10
I don't even really know what vector spaces are.

A vector space is a module over a field, with some appropriate additional laws.  :P
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 14 Dec 2013, 14:56
Some informal thoughts:

You don't need to know what vector spaces are for the first one, just what linear (in)dependency is. Nevertheless, a vector space is a set of numbers which fulfills some axioms (https://en.wikipedia.org/wiki/Vector_space#Definition). A subspace is a subset of that set of numbers which also fulfills the axioms (ie is a vector space in itself).

Side note: A vector space is spanned by some vectors. Which means if you add vectors together or scale them up, you can get the whole vector space. Remember how you were calculating points in R² in school? You needed a*(1,0) + b*(0,1) to get to any point (a,b). The set {(1,0), (0,1)} was a basis for R². Incidentally, the set {(1,0), (0,1), (13,37)} would also span R², but you'd never need to use (13,37) to reach your points. The vectors {(1,0), (0,1)} were a bare minimum you needed, that's why a basis is also called a minimum spanning system. There are many spanning systems for a vector space. For example, you could use {(4,0),(0,2)}, because you can cale them down to {(1,0), (0,1)} and proceed from there as usual.

If you only used {(1,0)}, then you could get all vectors which were on the x-axis, but no other vectors. All the vectors (x,0) are in a subspace of R². Incidentally, if you were to omit the 0 (because it's kinda redundant, right), you'd have the number line (that is, R). Congratulations! You discovered that R is a subspace of R²!

Now, to some tasks.

1a) There are multiple approaches I can think of.
Because c1*v1+c2*v2+c3*v3=0 (I have substituted lambda with c) has at least one non-trivial solution, the set {v1, v2, v3} is linearly dependent. That means that you can "take away" at least one vector from it and they will still span the same space. <v1, v2, v3> = <v1, v2> = <v2, v3> = <v1, v3>.

Another way would be to start with a trivial case <v1>=<v2>. Then <v1, v2> = <v1> = <v2>. It would then follow from the equation that <v1> = <v3> = <v1, v3>. You'd then have to show that you get to the same conclusion even if <v1>!=<v2> (this is more formal, but harder and probably pointless).

1b) Would start with showing both directions separately.
=>: let (v_1, ...v_n) be a linearly indep. system. It then follows that it spans some subspace of V\{0}, let's call this subspace U. (v_1...v_s) then spans some subspace of U, let's call it U1. Similarly, (v_(s+1)...v_n) spans some subspace U2. Because (v_1, ...v_n) is lin. indep., every non-empty subset of it is also lin. indep. (this might be a bit of pain in the ass to prove. When in doubt, use induction). That is, for every s, {v1...vs} is linearly independent and {v_(s+1)...v_n} is linearly independent. In particular, for every v in U1, it is not in U2, except if it is 0. (I am doing it a bit sloppily right now.)
Because of this, the only element in U1 intersect U2 is 0.
<=: let's have <v1,...vs> intersect <v_(s+1), ...vn> = 0 for all s. Then it particularly follows that <v1> intersect <v2...vn> = 0. v1 is thus not in <v2...vn>. {v1} union {v2...vn}={v1...vn} is then lin. indep.

3b) Let V have n spanning vectors in it's basis and U have k spanning vectors in it's basis, with k<=n (ie U is k-dimensional). Then there are (n-k) vectors which are linearly independent (since they are a subset of a the linearly independent set of the basis vectors of V). They span a subspace W. None of the elements of W are in U, except 0 (does this look familiar?), and vice-versa. Thus, V\U=W.
I will leave the special case U=V as an exercise to you, because I am not sure if a complement can be an empty set.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: pwhodges on 14 Dec 2013, 15:16
Reading that (for nostalgia's sake, though it's already more set theory than I ever did), I had the thought that my maths prowess was ultimately limited by the fact that I picture mathematical relationships geometrically - I cannot abstract the ideas to the point that I can handle them without that crutch, and so I am also limited to ideas that can fit into a three- (or maybe four-) dimensional space in my head.

That may not make sense to you, but I assure you it did to me.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 14 Dec 2013, 15:32
I, on the other hand, really struggle with geometry. I'm pretty good at maths, but going to linear algebra lectures were often pointless, because they modelled something that's basically just elements consisting of ordered sets of numbers as points in space, which became confusing.

The same thing goes for linear programming. It's basically the use of the very powerful simplex algorithm to solve a class of programming problems. The thing is, there's an algebraic way of looking at how it works and a geometric way of looking at how it works, and the geometric one is always used in explanations. I absolutely hate it. I can always see how the maths can model the geometry, but working with the geometry is just adding a layer of complication on something that's as simple as just numbers.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 16 Dec 2013, 15:44
So we're sitting here, working on the exercise sheet.

We have a solution for the 2a), but to me it seems overly complex, and formally wrong. I can see what the creator of the solution wanted to ge to, but I don't think you're allowed to write it that way.

(click to show/hide)

I don't know how it can be done easier though. (this is not my solution)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 17 Dec 2013, 03:49
So there's another linear algebra sheet due Monday. But as I don't have time to go to university on monday I need to be done with it on Friday. Damn.

Here (https://own.halfur.de/public.php?service=files&t=2353ff2d9c76c3a35e87be8edbfcfa58)'s the exercise sheet.

I'm not happy about it.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 18 Dec 2013, 04:09
I've been gone for a couple weeks - nurturing that freshman course to a conclusion. My admiration for the level of freshman courses at your university is reinforced.

For 1A: As a homomorphism of groups f respects addition. Your task is to show that f(qv)=q f(v) for all rational numbers q and all vectors v. I would do this in steps as follows:
1) first prove it by induction whenever q is a positive integer (an example like this has probably been done).
2) a group homomorphism takes the additive inverse (=negative) to the additive inverse, so we can conclude that the claim holds whenever q is any integer.
3) next we do it in the case q=1/n for some positive integer n. Here a trick is needed. Write v=(n/n)v=n*((1/n)v). By step 1 we have f(v)=f(n*((1/n)v))=n f((1/n)v). Because W is a vector space over Q, this implies (multiply both sides by 1/n) that (1/n)f(v)=f((1/n)v), which is what we want for this step.
4) The general case q=m/n then follows by combining steps 2 and 3. Leaving this to you.

1B) Hint: Consider the Q-linear mapping f defined by f(1)=1, f(sqrt(3))=-sqrt(3). Show that it is not linear over Q(sqrt(3)).

2A) This you can do simply by using the subspace criterion. If phi1 and phi2 are two linear mappings sending all the elements of the subspace U to 0, then you can easily deduce that the same thing happens with the linear combination a*phi1+b*phi2, a and b in k.
2B) Barring suitable examples/auxiliary results in the course notes this makes your hands a bit dirty. Start with a basis of U, extend that to a basis B of V. You have probably covered the bit that in the finite dimensional case you get a basis for V* by mapping one of the elements of B, say b, to 1, and the rest to 0. Call that mapping phib. You should be able to show that the elements phib, b ranging over B, such that b is not an element of U, form a basis of U0.
2C) Many ways to do this. One is to find two linearly independent elements, v1 and v2,of V that are orthogonal to (1,1,2). Both of those give rise to an element in U0 by the recipe phii: x -> the inner product of x and vi, i=1,2. Another way (perhaps the intended way?) is to redo 2B in this specific instance: find such a basis of V that (1,1,2) is one of the basis vectors. Then describe in detail the two mappings in V* that map (1,1,2) and one of those other basis vectors to zero, and the remaining one to 1.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 22 Dec 2013, 13:49
You won't believe how grateful we are for you Skewbrow. Without you we would have been doomed. With you, we've been a bit less doomed. It still took us a long time, but otherwise we wouldn't have figured out such nice solutions at all. My whole exercise-sheet-group has worked on this, and we wouldn't have made it without you.

Thank you.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 09 Jan 2014, 15:58
This is not under time pressure, as it was for an assignment which has already been turned in, but:

Let's say you integrate a function over the reals, for example f(x)=x*exp(-x). The way I learned it, if you integrate a product of functions, you can call one function u(x) and the other v'(x) and then do integration/derivation voodoo with them, which I believe is the inverse chain rule method (https://en.wikipedia.org/wiki/Inverse_chain_rule_method).

Now, my question: How do I decide which of the two functions should be u and which should be v'? Is there a rule of thumb?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: snalin on 09 Jan 2014, 16:11
Try one assignment, and if that's too hard, try the other. That was always my way of figuring it out.

As the ordering of the factors are irrelevant, which one you choose for u and v' is up to you - the result must always be the same.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 10 Jan 2014, 03:00
Actually, thereis a rule of thumb, and a mnemonic for it.

The rule is that you want u to be something that gets simpler with derivation, and v' should be something that doesn't get much more difficult under integration.

So, for example, if there's a logarithm, it's automatically u, because it can't be integrated without this trick anyway.

So the rule of thumb;

Choice of u should be

L - logarithmic terms (derivatives are negative powers of x, much simpler)
I - inverse trig terms (simpler derivatives, rational expressions, some with roots)
A - algebraic terms (polynomials, actually - the powers all get knocked down by one each time this is done)
T - trig terms (rarely simpler, but at least you're just getting other trig terms)
E - exponentials (no simpler, but no more complex when integrated, either, which makes it a good choice for v')

LIATE's not too hard to remember - it's almost a word.  If you have any two of these, use the first one that appears in the list should be used as u, the other one as v' and you'll usually get something simpler.  For your example,

u = x --> du = dx
v' = e^(-x) --> v = -e^(-x)

so int[x*e^(-x) dx] = (x)(-e^(-x)) - int[-e^(-x) dx] = -xe^(-x) - e^(-x) + C

BTW, this technique of reversing the terms in the product rule to get something easier is usually called "Integration by Parts", at least here in the US.

And, lastly, this;

(http://imgs.xkcd.com/comics/borders.png)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Schmee on 10 Jan 2014, 04:52
(http://imgs.xkcd.com/comics/borders.png)

Alt-text: "Eventually, a UN is set up. And then a lone rebel runs along the line of flags at the front, runs back to his base, and gets a kajillion points."
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 27 Apr 2014, 05:40
Okay, so last time I did linear algebra was a year ago, so my memory is rusty.

I have a set of some five matrices in Q^(2x2). I am supposed to check whether this set a spanning set of Q^(2x2).
What I would do is transform a matrix ((a,b),(c,d)) with a,b as the first row and c,d as the second row into a column vector (a,b,c,d)^t, then use those column vectors to create a 4x5 matrix, which I would Gauss to check whether it has a rank of 4.

Iff it has a rank of 4, then my set is a spanning set.

Is what I am doing correct?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Schmee on 27 Apr 2014, 05:55
...I'm pretty sure that's right. I just finished a semester on linalg, and your method seems sound. I haven't heard of Gaussing a matrix, though - we probably used a different term. You use row operations to reduce it to a row echelon form, right?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 27 Apr 2014, 05:57
Yes. I was using a colloquial used here for performing a Gaussian elimination (http://en.wikipedia.org/wiki/Gaussian_elimination) (aka reducing it to row echelon form).

Thanks for doublechecking!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Schmee on 17 May 2014, 19:28
Okay, so I've hit upon a fairly nasty inhomogenous second-order DE, and I'm stuck. Any help?

We have y'' - y' - y = cosh(x), where y(0) = 0 and y'(0) = 0.
I can get the homogenous part to be y = Ae-x + Be2x, but then I get stuck - I can't think of a substitution for y to solve the second part.
y = c cosh(x) + d sinh(x) doesn't work, so the answer can't be in terms of sinh and cosh.
I've tried separating cosh(x) into 1/2(ex + e-x), and treating it as two separate functions, but again I can't find a solution.

Ideas?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 19 May 2014, 07:07
Didn't check, but you should be able to handle that ex/2 part of hyperbolic cosine with the usual ansatz Aex. On the other hand the e-x/2 part requires an ansatz of the form Axe-x because e-x is a solution of the homogeneous DE.

But wait just a minute! Recheck the solution to the homogeneous part. The solutions of r2-r-1=0 are r1=(1+sqrt(5))/2 and r2=(1-sqrt(5))/2. Or perhaps you meant that your DE is y''-y'-2y=cosh(x) ?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Schmee on 19 May 2014, 16:56
Don't worry - I've figured it out  :-)
You were right, it was y'' - y' - 2y, I mistyped it.

On another note, I'm only just realising how different the names of maths terms are between countries - I've never even heard of an ansatz, but we use the same method.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 19 May 2014, 17:04
In this context this word just means "starting point" or rather "starting method" in German. It comes from the verb "ansetzen", which itself comes from the two words "an" / "daran" = "onto" and "setzen" = "to set"/"to place" (in the context of the word Ansatz), so it refers to the way you initially place a tool on a work piece.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 20 May 2014, 00:19
English math terms have borrowed in quite a few places from German. Consider, for example, eigenvectors and eigenvalues.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 20 May 2014, 03:27
Indeed, it has. Actually I only relatively recently learned that many (or at least some) English texts use the word 'Ansatz' like here. That is, guess the form of the solution with some unknown coefficients, and then try and see if it works for some choices of the coefficients.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: pwhodges on 20 May 2014, 13:14
Someone on another forum I visit is doing some maths...:

(https://forum.evageeks.org/profilepic.php?u=7083&full)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: cesium133 on 20 May 2014, 13:19
English math terms have borrowed in quite a few places from German. Consider, for example, eigenvectors and eigenvalues.
Physics terms as well: gerade/ungerade, brehmstrallung (probably spelled that one wrong...), etc.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: GarandMarine on 26 May 2014, 21:17
(https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-ash3/t1.0-9/555307_723209670113_750753573_n.jpg)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: bainidhe_dub on 10 Aug 2014, 17:34
Can someone explain how to handle exponents in simple algebra equations?

According to the answer key, 25/2-23/2=23/2

But why??
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: cesium133 on 10 Aug 2014, 18:11
You can rewrite 25/2 as 2*2*21/2 (each time you multiply by two that adds 1 to the exponent).

So, in this case, 25/2-23/2 = 2*23/2-23/2 = 23/2 * (2-1) = 23/2
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 11 Aug 2014, 16:32
In other words, it's not the exponents that needed handling.  It needed to be broken down into components in order to be solved.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 11 Aug 2014, 16:34
So, just wondering, anyone here got any familiarity with first-order-logic?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 11 Aug 2014, 16:43
I know I took two semesters of logic, up through Godel's theorems, but that was back in the late 80's.  As for what I actually remember... well, let's see the question before I commit!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: bainidhe_dub on 11 Aug 2014, 19:02
In other words, it's not the exponents that needed handling.  It needed to be broken down into components in order to be solved.

Got it, that makes sense. It could also be written as 2x - x = x(2 - 1). I remember fucking nothing of how to do math. How to multiply fractions?? Simplify quadratics?? What all the letters in SohCahToa stand for?? Ugh.

Robert is starting at community college this fall and we were trying to remember/learn how to do things for the placement test. He took the test last week and missed the "start out in a for-credit class instead of remedial" score by 2 points, so we spent a while on trig functions, and then there was only 1 trig question when he retook it today. *shrug* We'll see. There's no benefit in taking a class he's not prepared for, but spending a semester in a class that won't count in the end - because of two points - would suck.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 11 Aug 2014, 23:17
I know I took two semesters of logic, up through Godel's theorems, but that was back in the late 80's.  As for what I actually remember... well, let's see the question before I commit!
Oh, I have no question at the moment - I just constantly hear the claim that barely any university outside of mine (at least in Germany) does mathematical logic as mandatory part of the curriculum. Thought I'd check if that was true. In case you are interested, the contents of the course are, roughly in that order:

* intro to Boolean logic
* completeness theorem for BL
* Horn formulas
* sequence calculus in BL
* compactness theorem for BL
* intro to first-order logic (FO)
* (finite and non-finite) axiomatization of structures in FO
* model-checking games
* theories (a thing so obscure there seems to be no definition of it on Wikipedia)
* Ehrenfeucht-Fraisse games
* sequence calculus in FO
* completeness theorem in FO
* compactness theorem in FO
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 14 Aug 2014, 23:17
I know I took two semesters of logic, up through Godel's theorems, but that was back in the late 80's.  As for what I actually remember... well, let's see the question before I commit!
Oh, I have no question at the moment - I just constantly hear the claim that barely any university outside of mine (at least in Germany) does mathematical logic as mandatory part of the curriculum. Thought I'd check if that was true. In case you are interested, the contents of the course are, roughly in that order:

* intro to Boolean logic
* completeness theorem for BL
* Horn formulas
* sequence calculus in BL
* compactness theorem for BL
* intro to first-order logic (FO)
* (finite and non-finite) axiomatization of structures in FO
* model-checking games
* theories (a thing so obscure there seems to be no definition of it on Wikipedia)
* Ehrenfeucht-Fraisse games
* sequence calculus in FO
* completeness theorem in FO
* compactness theorem in FO

:laugh: :laugh: :laugh: :laugh: :laugh: not a goddamn thing there sounds familiar  :laugh: :laugh: :laugh: :laugh: :roll:

OK, one or two of the first few topics.

Seriously, Math in the states, even at the graduate level vs. math in Germany is like a little kid trying to arm wrestle with The Hulk.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 24 Aug 2014, 10:28
Logic? Lost me there. It was very popular at ND in the late 80s (and presumable before and after that), but I never caught the bug, so there's a Gödel size hole in my education. Peasant's logic is good enough to get a PhD, so...

Mind you, topics like that are popular here, too. More than half the research done at our math department is on problems that originated from computer science. We do not even seek to cover all of what is known as Core Math in US grad schools. I'm not at all sure that it is a good thing.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: GarandMarine on 05 Sep 2014, 12:01

Well I tell y'all what. I might just be a dumb jarhead who ain't had much use for learnin these many years, especially that there "arithmetic" but even I can tell you that that's pretty stupid.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Barmymoo on 05 Sep 2014, 14:15
She's explained it in a ridiculous way, but that's how I add. It's not stupid, it's true - it is easier to add 10 and 5 than 9 and 6.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 06 Sep 2014, 10:38
Barmy's absolutely right, the common core is all about the way we think about math to actually learn it.  Well, the way people who are fairly comfortable with math think about it - I liked it when the reporter mentioned that she had to "just memorize it".  Memorization works to some extent, but memorization backed by some understanding is easier, faster, and lasts longer.

As far as explaining it, I get the feeling the teacher was trying to explain it at a meta-level.  As part of a news report, I get the feeling this was aimed at adults who are getting confused (and frustrated) when they see all this Common Core stuff that they never had to do.

The other (and much bigger) problem with the transition to the Common Core is the same one that killed New Math in the 60's - the teachers.  They're doing a much better job educating the teachers on what they need to do, but a bad implementation (locally) of a good idea is just going to be a mess.

A quick example; subtraction in the common core is initially taught by "counting up" from the smaller to the larger number.  That's the way most of us actually learned it, and I know several adults who are embarrassed that they still use that trick.  But it works...

So I saw an example the other day in some course materials  I was reviewing that took the example of 57 - 22, and counted up.

By 5's.

Now yes, you can land exactly on the result that way, but dammit, it's easier to go by 10's, it always is, it's a base 10 system!  Then do the last bit, the extra 5.  This is exactly the sort of crap that really makes a mess of things - it's not the common core itself (the ideas), it's bad implementations by teachers who aren't thinking about the math they probably learned through memorization!  And to top it off, they're often not flexible enough to see it any other way.  The same materials insisted that visual representations products always be row x column, so that

......
......
......
......

is 4 x 6, but

....
....
....
....
....
....

has to be 6 x 4!

As someone who has always looked at things sideways, I pointed out that they damn well better accept both answers for both arrays, you never know what a kid's going to call a row!

It's become another slow motion train wreck.  Toss in the whackos claiming it's federal mind control at work, and you've got a system doomed from the outset.

Hell, why change anything, ever?  "It were good enuf fer mah semi-literate grampa, it's good enuf fer me (and mah kids)!"

[crawls back into his hermit-hole, muttering and cursing the whole way]
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: GarandMarine on 06 Sep 2014, 10:46
Because government education is a waste of tax dollars at it's finest, and why let the market deliver efficient systems when we can waste time and money?

She's explained it in a ridiculous way, but that's how I add. It's not stupid, it's true - it is easier to add 10 and 5 than 9 and 6.

Goodie for you? Meanwhile 6+9 is 15. Want to have a brief race on simple addition and subtraction? Why weaken children by teaching them poor methods that require more work?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: bainidhe_dub on 06 Sep 2014, 11:00
How does it take more work to apply concepts you've already learned, rather than approaching everything as brand new? 9+6=15, yes, but kids have already been introduced to base-10, so it make sense to have them "find" the 10 that they're familiar with, and solve from there. I still solve 9+x by thinking of it as (x-1)teen. I don't try to pull the fact from my head fully-formed. I know perfectly well that I am stupid slow at basic arithmetic sometimes, so I use all the shortcuts I can.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 06 Sep 2014, 11:21
Fine, then homeschool while you're also trying to work two - three jobs to feed and house them.

Look, the foundng fathers were pretty clear (especially Jefferson) - an educated citizenry is in a democracy's best interest.  They left it to the states initially (as with so many other things), but it's grown far from the community-based schoolhouses.  The industrial revolution saw to that - automata need to be trained, after all!

There's little evidence that a market based educational system will deliver at all.  You get what you pay for, and the poor will be left behind (again).  An equal opportunity to education is absolutely necessary, and a foundation of our country's entire system of government.

As for your second point, this actually strengthens their ability to learn facts faster, remember them better, and extrapolate to do more than their parent's ever could.  No, we're not going to get a pack of Art Benjamins (http://youtu.be/M4vqr3_ROIk), but at least they'll stop with the fucking "I hate math" crap, because they'll actually understand what they're being asked to do.

And Baindhe, you're far from stupid slow if you know and use the shortcuts!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: GarandMarine on 06 Sep 2014, 11:47

There's little evidence that a market based educational system will deliver at all.

Which is why private schools of any kind are consistently the bottom tier in all forms of testing me... oh wait. The free market can absolutely deliver better, cheaper and more efficiently then government. Hell any one could do better then the DOE that's been consistently failing American students for... is it three generations now?

I'm fine with some of my taxes (if we absolutely must be taxed) going to educating other people's larva. However I will be damned if I don't argue for that money to be used as efficiently and effectively as possible. Between that and my mother's career in education, I think it explains why I'm so anti-union.

Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Barmymoo on 06 Sep 2014, 14:08
Private fee-paying schools are vastly disproportionately populated by children of wealthy families, who are already at an advantage for achievement. Stick the same kids in a public school and they'd still do well. Stick them in a school where there aren't any poor kids whose lack of educational support at home, or distracting personal circumstances, or simply generations of poor academic achievement telling in their critical thinking skills are holding everyone's progress back, and they will do even better. That's not proof that the market works. It's proof that society is incredibly unequal.

I absolutely don't deny that the American public school system is failing. It's an abysmal system and only seems to be getting worse. Public schools in other countries don't seem to have the same issue though. I would like to invite one of our Scandinavian forumites to input their experience in this discussion. It seems to me that most of the issues of publicly-funded services in the USA stem from not getting enough taxation funding, not too much. Obviously someone who has a blanket opposition to the mere concept of taxation wouldn't agree with that though.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 06 Sep 2014, 14:13
Guys? We have moved into whole another kind of math entirely.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 07 Sep 2014, 15:33
And now for something more light hearted - a song for all the math enthusisasts:

Lyrics (https://en.wikipedia.org/wiki/User:Nobi/Finite_Simple_Group_of_Order_Two)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ev4n on 08 Sep 2014, 09:27
Because government education is a waste of tax dollars at it's finest, and why let the market deliver efficient systems when we can waste time and money?

Aside:  Was reading recently that the free market education system in Sweden has turned out to be a total disaster.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 08 Sep 2014, 19:57
As are most charter schools in the US.  The only ones that get better results are the ones with some form of selective admission, which is what private schools do indirectly.

I think we can agree, though, that the US system is badly broken to the point where re-assembly may be required.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Skewbrow on 20 Sep 2014, 11:29
ROTFL! Thank you for sharing, Loki.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 28 Feb 2015, 02:27
A tweet in my timeline reminded me of how our math teacher in highschool explained asymptotes (https://en.wikipedia.org/wiki/Asymptote) to us.
Quote
It's like teenagers at a catholic sleepover party: they can get as close to each other as they wish, as long as there is no touching.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 28 Feb 2015, 11:48
I guess only vertical asymptotes are observant Catholics, then.

'cause other types can touch, they just usually don't... maybe they're just picky?   :roll:
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 28 Feb 2015, 11:49
*blink* how exactly does an asymptote touch whatever it is asymptotic to?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 28 Feb 2015, 11:51
Take a look at y = {sin(x)}/x.  Asymptotic to y = 0 as x -> infinity, and crosses it at every multiple of pi.

There are rational examples, too, like y = x/(x^2 + 4), which is also asymptotic to y = 0 as x -> infinity, but crosses it at the origin.

And it's not just the x-axis; any horizontal, slant or curvilinear asymptote can be crossed by a function that it's asymptotic to.

Just not the vertical ones.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 28 Feb 2015, 12:29
Huh. Yeah, I see now - thanks!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Pilchard123 on 28 Feb 2015, 12:38
Partially related, but useful in other cases too. https://www.desmos.com/calculator

You have dialled an imaginary number. Please rotate your handset by 90 degrees and try again.

(click to show/hide)

(click to show/hide)

(click to show/hide)

:claireface:
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 28 Feb 2015, 12:42
I got the third before clicking on the solution! (And I had already heard the second before)
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 28 Feb 2015, 15:28
Max Zorn - yes, the Zorn of Zorn's lemma - was a professor emeritus at Indiana when I was in grad school.  In his 80's, he would come to every colloquium and seminar, and always had an insightful comment or question or two.  He lunched at Mother Bear's pizza with his wife over a pitcher of beer, then spent the afternoons in the seminars.  I taped up a couple of white index cards behind the coffee pots for him so he could see when the coffee was still brewing and not pull the pot out too early.

The story around the department was that at a party sometime in the late 50's, he got very drunk, got up on a table, and loudly pronounced "It's not a lemma, and it's not mine!"
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 28 Feb 2015, 15:30
Woha. I am completely :psyduck: right now

Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 23 Mar 2015, 20:29
It's my birthday today.  My age is the next to last prime sum of two squares before the usual retirement age (65 in the US).

:-D
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 23 Mar 2015, 23:58
Happy birthday!

How'd one go about solving this analytically?

(click to show/hide)

You devious bastard. You nerdsniped me for at least 15 minutes. Congratulations :D
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Carl-E on 24 Mar 2015, 08:02
Welp, my work here is done!

I figured if anyone would crank it out, it would have been you... and Fermat's sum of two squares theorem was a nice easter egg!

Have fun proving it...
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 24 Mar 2015, 17:17
Welp, my work here is done!

I figured if anyone would crank it out, it would have been you... and Fermat's sum of two squares theorem was a nice easter egg!
o.O I am far from being the most math adept person on the forum (not counting you, of course). We have physicists amd actual rocket scientists here.

Quote
Have fun proving it...
Welp. Uh. I can do one direction. That's already half of the work, right? :-D The rest is left as an exercise.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: LTK on 08 Apr 2015, 08:15
This article piques my interest even though I have virtually no understanding of the kind of mathematics it deals with.

Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 07 Jul 2016, 10:37
This isn't math per se, but I think it's close enough to put it here.
So I'm writing my exam on theoretical computer science tomorrow, and I'm still having trouble with one, or rather two things: the pumping lemma for regular and for context free languages.

Can somebody here explain to me what e.g. a general approach to proving that a language isn't regular using the pumping lemma would be? I think I can grasp the general idea of the lemma, but I'm struggling with how to apply it to tasks. Here's an example task from a test exam:

A language L is defined as follows:

Prove that L is not regular, using the pumping lemma. (Hint: Use words in the form (https://latex.codecogs.com/png.download?%5Cinline%20%7B%5Ccolor%7Bwhite%7D%200%5En10%5En%7D))

I guess it'll go in the direction that pumping the prefix or the suffix of the middle one will break it, and pumping the middle won't work either, but how do I really approach that task?
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: Loki on 07 Jul 2016, 15:16
(click to show/hide)
(click to show/hide)
Good luck!
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: ankhtahr on 08 Jul 2016, 01:40
Thank you, I'm pretty sure I passed. I didn't have much trouble with the pumping lemma task.

In fact I'm much more sure I passed this exam than the technical exam I wrote on Wednesday, on which I spent a lot more time.
Title: Re: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Post by: LTK on 25 Oct 2017, 16:56
I think the Venn diagram of irrational nonsense (http://crispian-jago.blogspot.nl/2013/03/the-venn-diagram-of-irrational-nonsense.html) has appeared on the forum before but now there's an expanded version. You shouldn't waste your time reading it, just enjoy the beauty of it.

(http://3.bp.blogspot.com/-3Lufulseck4/UWF6MsaAosI/AAAAAAAAEk8/T8FNhnLbUug/s1600/ZghIM4Sh.jpg)

That's five categories and thirty-one areas between them! Fun fact: the number of areas in a Venn diagram of x categories is (2^x)-1, but you can also calculate it by taking the number of areas of a Venn diagram that's one smaller, doubling it, and adding one. This makes sense because every new category has to subdivide each area in two, as well as adding one area of its own that overlaps with no other. Makes me wonder how you could draw one that uses six. It's certainly possible if you abandon all symmetry but it'd be nice to do it with bilateral symmetry at least.