Topic: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math! (Read 63452 times)

Jace · « **on:** 17 Oct 2013, 23:45 »

I don't have any math to do just yet.

Carl-E · « **Reply #1 on:** 18 Oct 2013, 06:13 »

Can the title be "Math is Delicious!"?

I know, it's from the comic, but...

Loki · « **Reply #2 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.

LeeC · « **Reply #3 on:** 18 Oct 2013, 14:28 »

I've never done calculus. My education in college and highschool was pretty much algebra, trig, and statistics.

jwhouk · « **Reply #4 on:** 18 Oct 2013, 18:06 »

I didn't care much for Calculus. It was too derivative.

pwhodges · « **Reply #5 on:** 19 Oct 2013, 01:28 »

It's better if you take an integrated approach to it.

ackblom12 · « **Reply #6 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.

pwhodges · « **Reply #7 on:** 19 Oct 2013, 01:33 »

But you weren't, so you went off on a different tangent?

Jace · « **Reply #8 on:** 19 Oct 2013, 06:30 »

Do you fuckers think that is acute exchange? This is serious stuff

pwhodges · « **Reply #9 on:** 19 Oct 2013, 06:42 »

Well, we didn't want to be obtuse about it, because that wouldn't have solved anything.

jwhouk · « **Reply #10 on:** 19 Oct 2013, 08:40 »

Algebra was actually easy for me. It was simple as a + b = c!

Loki · « **Reply #11 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!

jwhouk · « **Reply #12 on:** 19 Oct 2013, 18:46 »

Oh, don't get all quadratic on me.

Carl-E · « **Reply #13 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...

Loki · « **Reply #14 on:** 19 Oct 2013, 23:54 »

I keep forgetting that having a math teacher on the forum has certain perks

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.

snalin · « **Reply #15 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?

Gridgm · « **Reply #16 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

Jace · « **Reply #17 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.

pwhodges · « **Reply #18 on:** 20 Oct 2013, 15:55 »

Quote from: snalin on 20 Oct 2013, 05:48

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.

Jace · « **Reply #19 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

Carl-E · « **Reply #20 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!

Jace · « **Reply #21 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)

Carl-E · « **Reply #22 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...

de_la_Nae · « **Reply #23 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;

de_la_Nae · « **Reply #24 on:** 21 Oct 2013, 08:43 »

Also I hate/love all of you for those earlier puns and wordplay.

Jace · « **Reply #25 on:** 21 Oct 2013, 14:28 »

Quote from: 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...

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.

Carl-E · « **Reply #26 on:** 21 Oct 2013, 20:29 »

Quote from: de_la_Nae on 21 Oct 2013, 08:43

hate/love

Glad you have your priorities straight!

Loki · « **Reply #27 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

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?

snalin · « **Reply #28 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.

ankhtahr · « **Reply #29 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 ${\color{white}0.5\geq\frac{\left|x\right|}{1+x^2}}$ ? I can see that it has a global maximum at 1 and -1, but I just don't know how to show that.

Pilchard123 · « **Reply #30 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)

ankhtahr · « **Reply #31 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.

Loki · « **Reply #32 on:** 27 Oct 2013, 13:14 »

Quote from: ankhtahr on 27 Oct 2013, 11:30

${\color{white}0.5\geq\frac{\left|x\right|}{1+x^2}}$

Quote from: Loki on 27 Oct 2013, 09:07

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

ev4n · « **Reply #33 on:** 28 Oct 2013, 09:21 »

Friday's xkcd was great. Got to love Bayes theorem for messing with people.

ankhtahr · « **Reply #34 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.

$\\ {\color{White}A:=\left\{x+\frac{1}{x}\mid0<x<5\right\}}\\ {\color{White}B:=\left\{x\in\mathbb{R}: 1\leq x^2\leq9\right\}}\\ {\color{White}C:=\left\{x\in\mathbb{R}: \exists y\in\left[1,4\right):y^2=x\right\}}\\ {\color{White}D:=\left\{(-1)^{n^2}+\frac{1}{n}:n\in\mathbb{N}\right\}}\\ {\color{White}E:=\left\{\frac{\left|x\right|}{1+x^2}:x\in\mathbb{R}\right\}}$

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.

snalin · « **Reply #35 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.

ankhtahr · « **Reply #36 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

Mlle Germain · « **Reply #37 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...)

NotAwesomeAnymore · « **Reply #38 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!

Carl-E · « **Reply #39 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

${\color{white}0.5\geq\frac{\left|x\right|}{1+x^2}}$

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!

ankhtahr · « **Reply #40 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.

ankhtahr · « **Reply #41 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:
$\\ {\color{White} \text{(i) Is }q\in\mathbb{R}\setminus\left\{1\right\}\text{and }n\in\mathbb{N}_0\text{ then }\sum_{k=0}^nq^k=\frac{1-q^{n+1}}{1-q}}\\ {\color{White}\text{(ii) }\sum_{k=0}^{n}\binom{n}{k}\left(-1\right)^k=0\text{ for all } n\in\mathbb{N}\text{.}}\\ {\color{White}\text{(iii) }\sum_{k=0}^n\binom{n}{k}=2^n\text{ for all } n\in\mathbb{N}\text{.}}\\ {\color{White}\text{(iv) 23 is for all }n\in\mathbb{N}_0\text{ a divisor of }5^{2n}-2^n\text{.}}$

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 $\\ {\color{White} (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k }$ .

The toughest exercise though is supposed to be this one:
$\\ {\color{White} \text{Is }n\in\mathbb{N},n\geq2\text{ and is }a,b\in\mathbb{R},a+b>0,a\neq b\text{ then }2^{n-1}(a^n+b^n)>(a+b)^n }$

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 $\\ {\color{White} \sum_{k=0}^{n}(\binom{n+1}{k}(-1)^k )+(-1)^{n+1}=0}$ . I don't know how to get the +1 out of the binomial coefficient. I know that $\\ {\color{White} \binom{n+1}{k}=\binom{n}{k-1}}$ , but I don't know if that helps me any further.
Wait. I got that wrong. It's $\\ {\color{White} \binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}}$ , and now I can see that it might help me.

Barmymoo · « **Reply #42 on:** 06 Nov 2013, 09:10 »

I find this thread completely fascinating and completely baffling.

ankhtahr · « **Reply #43 on:** 06 Nov 2013, 09:29 »

Damnit, I just noticed that $\\ {\color{White} \binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}}$ is of course only valid for $\\ {\color{White} 1\leq k\leq n}$ . So it does not help me, as in the sum there is k=0.

snalin · « **Reply #44 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

("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.

snalin · « **Reply #45 on:** 06 Nov 2013, 10:46 »

Quote from: ankhtahr on 06 Nov 2013, 09:29

Damnit, I just noticed that $\\ {\color{White} \binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}}$ is of course only valid for $\\ {\color{White} 1\leq k\leq n}$ . 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.

ankhtahr · « **Reply #46 on:** 06 Nov 2013, 11:01 »

The LaTeX stuff is from 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:
$\\ {\color{White} \sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^k=0}$
and want it to include this again:
$\\ {\color{White} \sum_{k=0}^{n}\binom{n}{k}(-1)^k}$

So somehow I have to get the +1 out of the binomial coefficient. Changing it all to $\\ {\color{White} \sum_{k=0}^{n}\binom{n+1}{k}(-1)^k + \binom{n+1}{n+1}(-1)^{n+1}}$ is easy, by pulling the n+1th summand out of the sum.

In this case it won't be possible to use $\\ {\color{White} \binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}}$ , as it would lead to $\\ {\color{White} \binom{n}{-1} }$ .

ankhtahr · « **Reply #47 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: $\\ {\color{White} 5^{2n}-2^{n}=23m }$ .
Now I have to show the same thing for n+1, so I'm stuck at $\\ {\color{White} 5\times 5^{2n}-2\times 2^{n}=23m }$ , 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 $\\ {\color{White} 5^2\times 5^{2n}-2\times 2^{n}=23m }$ is looking much better, and now I can definitely see where this is going: $\\ {\color{White} 5^2-2=23 }$ . But how do I legally change it into that? I'm out of training, that's for sure.

snalin · « **Reply #48 on:** 06 Nov 2013, 12:12 »

I'm coming with an answer for (ii), but the formatting is KILLING ME.

snalin · « **Reply #49 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 ${\color{White} \\ \sum _{k=0}^{n} \binom{n}{k}*(-1)^{k} = 0}$ holds for 1 to n**. Then write out for n+1:

${\color{White} \\ \sum _{k=0}^{n+1} \binom{n+1}{k}*(-1)^{k} = \binom{n+1}{0}*(-1)^{0} + \binom{n+1}{1}*(-1)^{1} + ... + \binom{n+1}{n}*(-1)^{n} + \binom{n+1}{n+1}*(-1)^{n+1}}$

Apply the identity ${\color{White} \\ \binom{n+1}{k+1} = \binom{n}{k} + \binom{n}{k1}}$ to all terms of the sum, except the first and the last:

${\color{White} \\ \binom{n+1}{0} * (-1)^{0} + (\binom{n}{0}+\binom{n}{1})*(-1)^{1} + ... + (\binom{n}{0}+\binom{n}{1}) * (-1)^{n} + \binom{n+1}{n+1} * (-1)^{n+1}}$

Reduce ${\color{White} \\ \binom{n+1}{0} * (-1)^{0} }$ to ${\color{White} 1}$ and ${\color{White} \binom{n+1}{n+1}*(-1)^{n+1}}$ to ${\color{White} (-1)^{n+1}}$ . Also multiply in the ${\color{White} (-1)^{k}}$ terms:

${\color{White} \\ 1 + \binom{n}{0}(-1)^{1} + \binom{n}{1}(-1)^{1} + ... + \binom{n}{n-1}(-1)^{n} + \binom{n}{n}(-1)^{n} + (-1)^{n+1}}$

Reorder: for all of the pairs on the form ${\color{White} \\ (-1)^{t}\binom{n}{t-1} + (-1)^{t}\binom{n}{t}}$ , group the left hand and right hand terms with the other left hand and right hand terms:

${\color{White} \\ 1 + ( \binom{n}{0}(-1)^{1} + ... + \binom{n}{n-1}(-1)^{n} )+ ( \binom{n}{1}(-1)^{1} + ... + \binom{n}{n}(-1)^{n} ) + (-1)^{n+1} }$

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

${\color{White} \\ 1 -( \sum _{k = 0}^{n-1} \binom{n}{k} * (-1)^{k})+ ( \sum _{k = 1}^{n} \binom{n}{k} * (-1)^{k}) + (-1)^{n+1}}$

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:

${\color{White} \\ -( \sum _{k = 0}^{n-1} \binom{n}{k} * (-1)^{k})+ (-1)^{n+1}}$

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

${\color{White} \\ -( \sum _{k = 0}^{n-1} \binom{n}{k} * (-1)^{k}) = -( \sum _{k = 0}^{n} \binom{n}{k} * (-1)^{k}) + \binom{n}{n}*(-1)^{n} = -0 + \binom{n}{n}*(-1)^{n} }$

Meaning that:

${\color{White} \\ -( \sum _{k = 0}^{n-1} \binom{n}{k} * (-1)^{k}) + (-1)^{n+1} = (-1)^{n} + (-1)^{n+1} = 0 }$

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

THESE FORUMS NOW CLOSED (read only)

Author Topic: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math! (Read 63452 times)

ackblom12