Proving F(x) = ∫[0,x] exp(-t^2) is odd.


by peripatein
Tags: 𕖮, expt2, proving
peripatein
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#1
Jan24-13, 03:18 PM
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Hi,
1. The problem statement, all variables and given/known data
Any idea how I might demonstrate that F(x)=∫ [0,x] exp(-t^2) dt is odd?


2. Relevant equations



3. The attempt at a solution
Will it be by showing that ∫[-x,x] exp(-t^2) = 0?
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Jan24-13, 03:22 PM
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Quote Quote by peripatein View Post
Hi,
1. The problem statement, all variables and given/known data
Any idea how I might demonstrate that F(x)=∫ [0,x] exp(-t^2) dt is odd?


2. Relevant equations
Wouldn't the definition of "odd function" be relevant?
Quote Quote by peripatein View Post



3. The attempt at a solution
Will it be by showing that ∫[-x,x] exp(-t^2) = 0?
No, I don't think so.
Dick
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Jan24-13, 03:23 PM
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Quote Quote by peripatein View Post
Hi,
1. The problem statement, all variables and given/known data
Any idea how I might demonstrate that F(x)=∫ [0,x] exp(-t^2) dt is odd?


2. Relevant equations



3. The attempt at a solution
Will it be by showing that ∫[-x,x] exp(-t^2) = 0?
That's not true. You do it by showing F(-x)=(-F(x)).

peripatein
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#4
Jan24-13, 03:30 PM
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Proving F(x) = ∫[0,x] exp(-t^2) is odd.


I know that a function is odd if F(x)=-F(-x). But how may I show that that is indeed the case for the above function?
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Jan24-13, 03:33 PM
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Quote Quote by peripatein View Post
I know that a function is odd if F(x)=-F(-x). But how may I show that that is indeed the case for the above function?
Try substituting -x for x and see what you get.
HallsofIvy
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#6
Jan24-13, 03:38 PM
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It's fairly easy to show that a function is odd if and only if its derivative is even. And, by the "fundamental theorem of Calculus" [itex]F'= e^{-x^2}[/itex].
peripatein
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#7
Jan24-13, 04:08 PM
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Actually, I did initially manage to show that the derivative was even. Problem is, I had no idea that would mean the function itself was odd!
For you, HallsofIvy, it might be fairly easy. I would have to mull over it. May I demonstrate that by simple integration on both sides, or is it more intricate than that?
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Jan24-13, 04:33 PM
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Quote Quote by peripatein View Post
Actually, I did initially manage to show that the derivative was even. Problem is, I had no idea that would mean the function itself was odd!
For you, HallsofIvy, it might be fairly easy. I would have to mull over it. May I demonstrate that by simple integration on both sides, or is it more intricate than that?
The way you show it is basically the same as jbunniii's suggestion. Write the integral for F(-x) and do the substitution t->(-t) in the integral.
peripatein
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#9
Jan24-13, 04:40 PM
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I did that too, Dick! Leaves me with a proof that int[0,x]exp(-t^2)=-int[0,-x]exp(-t^2), which I am not sure how to approach.
In my previous reply I was wondering how to demonstrate even derivative implies odd function.
Dick
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Jan24-13, 04:45 PM
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Quote Quote by peripatein View Post
I did that too, Dick! Leaves me with a proof that int[0,x]exp(-t^2)=-int[0,-x]exp(-t^2), which I am not sure how to approach.
In my previous reply I was wondering how to demonstrate even derivative implies odd function.
Take int[0,-x]exp(-t^2)dt. Do the u-substitution u=(-t), du=(-dt). And don't forget to change the limits. If t goes from 0 to -x, what are the limits for u? Since F'(x)=exp(-t^2) you can substitute any even function for exp(-t^2) and the same proof works.
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#11
Jan24-13, 05:01 PM
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Wouldn't that leave me to prove that int[0,x] exp(-u^2)du + int[0,-x]exp(-u^2)du=0?
Dick
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Jan24-13, 05:08 PM
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Quote Quote by peripatein View Post
Wouldn't that leave me to prove that int[0,x] exp(-u^2)du + int[0,-x]exp(-u^2)du=0?
That would be true. Why are you expressing it as a sum?
peripatein
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#13
Jan24-13, 05:13 PM
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I am sorry, but I fail to see how that got me closer to proving the original statement. I mean, how is this integral different than the one with dt? What's eluding me?
Dick
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Jan24-13, 05:16 PM
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Quote Quote by peripatein View Post
I am sorry, but I fail to see how that got me closer to proving the original statement. I mean, how is this integral different than the one with dt? What's eluding me?
I don't know. Can you show me what you get if you follow the suggestions in post 10? You should just get one transformed integral with a u variable instead of t. What is it?
peripatein
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#15
Jan24-13, 05:39 PM
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Could you please explain to me why I could substitute any even function? I understand that the proof would still work, but would that be rigorous enough? What would the formal justification for such a maneuver be?
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Jan24-13, 05:45 PM
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Quote Quote by peripatein View Post
Could you please explain to me why I could substitute any even function? I understand that the proof would still work, but would that be rigorous enough? What would the formal justification for such a maneuver be?
Mmm. I just meant that once you prove it then if they give you f(t) and tell you f(t) is even instead of exp(-t^2) the same proof works. I wasn't trying to say anything fancy.
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#17
Jan24-13, 06:04 PM
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Quote Quote by HallsofIvy View Post
It's fairly easy to show that a function is odd if and only if its derivative is even.
I disagree with the "if" part. That f'(x) is an even function is a necessary but not sufficient condition for f(x) being an odd function. Example: f(x)=1 has an even derivative but obviously isn't an odd function.

Having f(0)=0 is a sufficient but not necessary condition for f(x) being an odd function. Example: f(x)=1/x is odd, but f(0) does not exist.
peripatein
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#18
Jan24-13, 06:04 PM
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Okay, then I am still left with the seeming difficulty in seeing how the du substitution facilitated things. I believe you have expressed your approval of the expression I obtained having performed that substitution (post 12).


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