Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: squidgetx on June 17, 2010, 06:22:40 pm
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k so basically I was wondering if proof by induction is valid if, for the first step, you prove for x=0 instead of x=1.
Using domino logic, it's really a matter of whether you decide to start counting from 0 or 1 (which is where I got the idea to ask you guys....;)
kthxbai
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I'm not exactly sure what you are asking...could you please explain more?
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Yeah, it should be valid.
meishe: squidgetx is referring to proofs by induction, which typically work by proving that statement p is true for x=1, and p is true for x+1 if p is true for x. Therefore, p is true for all natural numbers. He's asking if it's valid to start at x=0, rather than x=1. I said yes
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Ah ok. Well i don't quite know still but I'll look it up. Thanks though.
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Thanks
Meishe; quick explanation of proof by induction using domino's....heh heh
Statement 1: The 1st domino falls over
Statement 2: If a domino falls over, the one after it falls over
Therefore, all the dominoes fall over
So in proofs by induction, you prove that a statement is true for 1 (or 0, as calc84se says), then assume it is true for some constant, k, and the prove it's true for k+1
Calc84se, if what you've said is true, then I just proved Euler's formula w/o any calculus :D (kinda random I know but ever since I saw that xkcd where e^(pi i)=-1 I've never been completely satisfied with the explanations (mainly b/c i'm not as advanced in calculus yet >.<)
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Since we're talking about math, do you mind showing me your proof? I'd like to see it personally, as I'm always interested to learn something new about math. :D
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The way proofs by induction work, is that you have to "pretend" that a function works and then if this assumption leads to a truth then you can conclude that you were right in assuming the original function. The dominoes example would be more like this:
1. The first domino has fallen over.
2. Suppose that Nth has fallen over
3. You can show that if that is the case, then the N+1th domino will fall over
4. Since if this is true for the Nth domino, then it is true for the N+1th domino, and it is true for the first domino, they you can say that all dominoes past the first position will fall over.
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......Shat.
I kind of made an assumption that's not quite valid
Here it is though, if you want to see it
Ok, so Euler's Formula states
e^(xi)=cos x + i sinx (or cis x)
Step 1: prove for x=0
e^0i= cos0 + i sin0
1=1+0i
1=1 :D
Step 2 Assume true for x=k
e^(ki) = cis k
Step 3 Prove for x=k+1
e^(k+1)i=cis(k+1)
e^(ki+i)= //distribute i
(e^ki)(e^i)= // exponent property
(cis k)(cis 1)= // here's where i mess up. In my original proof, I did induction the standard way, except it's really hard to prove e^i=cis 1, so I just kind of assumed that it was true and used it here to sub it in as e^i=cis 1
(cos k +i sin k)(cos 1 + i sin 1)= //expand
cos k cos 1 + i sin 1 cos k + i sin k cos 1 + -1 sin k sin 1 //multiply out
cos k cos 1 -sin k sin 1 + i (sin 1 cos k + sin k cos 1) = //re order terms
cos (k+1) +i sin (k+1) //recognize cos and sin sum formulas
cis (k+1) =cis (k+1) :/
Subbing in pi gives e^ (pi i) = cos pi + i sin pi
= -1 + 0i
= -1
I always do this, I get an idea, I get lazy, skip a step, and then realize that it threatens the whole integrity of the proof >.<
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Not valid, then. Too bad... :(
It would have been amazing (and I would've been very happy for you) if that had worked.
Oh well, it was a nice try. Better luck next time. At least the answer to your original question was yes.
Have fun with math, anyway. :P