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Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
snalin:
--- Quote from: ankhtahr on 06 Nov 2013, 09:29 ---Damnit, I just noticed that is of course only valid for . So it does not help me, as in the sum there is k=0.
--- End quote ---
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:
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:
and want it to include this again:
So somehow I have to get the +1 out of the binomial coefficient. Changing it all to is easy, by pulling the n+1th summand out of the sum.
In this case it won't be possible to use , as it would lead to .
ankhtahr:
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: .
Now I have to show the same thing for n+1, so I'm stuck at , 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 is looking much better, and now I can definitely see where this is going: . But how do I legally change it into that? I'm out of training, that's for sure.
snalin:
I'm coming with an answer for (ii), but the formatting is KILLING ME.
snalin:
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 holds for 1 to n**. Then write out for n+1:
Apply the identity to all terms of the sum, except the first and the last:
Reduce to and to . Also multiply in the terms:
Reorder: for all of the pairs on the form , group the left hand and right hand terms with the other left hand and right hand terms:
Extract the -1 from the left grouping, and turn the left and right grouping into sums:
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:
The left hand sum can be simplified a lot, using the original assumption:
Meaning that:
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
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