Register to reply

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

by peripatein
Tags: 𕖮, expt2, proving
Share this thread:
peripatein
#1
Jan24-13, 03:18 PM
P: 818
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?
Phys.Org News Partner Science news on Phys.org
Physical constant is constant even in strong gravitational fields
Montreal VR headset team turns to crowdfunding for Totem
Researchers study vital 'on/off switches' that control when bacteria turn deadly
Mark44
#2
Jan24-13, 03:22 PM
Mentor
P: 21,408
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
#3
Jan24-13, 03:23 PM
Sci Advisor
HW Helper
Thanks
P: 25,235
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
#4
Jan24-13, 03:30 PM
P: 818
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?
jbunniii
#5
Jan24-13, 03:33 PM
Sci Advisor
HW Helper
PF Gold
jbunniii's Avatar
P: 3,289
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
#6
Jan24-13, 03:38 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,683
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
#7
Jan24-13, 04:08 PM
P: 818
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?
Dick
#8
Jan24-13, 04:33 PM
Sci Advisor
HW Helper
Thanks
P: 25,235
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
#9
Jan24-13, 04:40 PM
P: 818
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
#10
Jan24-13, 04:45 PM
Sci Advisor
HW Helper
Thanks
P: 25,235
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.
peripatein
#11
Jan24-13, 05:01 PM
P: 818
Wouldn't that leave me to prove that int[0,x] exp(-u^2)du + int[0,-x]exp(-u^2)du=0?
Dick
#12
Jan24-13, 05:08 PM
Sci Advisor
HW Helper
Thanks
P: 25,235
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
#13
Jan24-13, 05:13 PM
P: 818
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
#14
Jan24-13, 05:16 PM
Sci Advisor
HW Helper
Thanks
P: 25,235
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
#15
Jan24-13, 05:39 PM
P: 818
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?
Dick
#16
Jan24-13, 05:45 PM
Sci Advisor
HW Helper
Thanks
P: 25,235
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.
D H
#17
Jan24-13, 06:04 PM
Mentor
P: 15,204
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
#18
Jan24-13, 06:04 PM
P: 818
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).


Register to reply

Related Discussions
I am asked to find ∫xf(x^2)dx if ∫f(x)dx = 9 ..? Calculus & Beyond Homework 3
Required to prove that ∫f(x)dx[b, a] =∫f(x−c)dx [b+c, a+c] Calculus & Beyond Homework 4
Triangle integral ∫∫dxdyf(x*y) how to reduce to one dimension? Calculus 1
∫∫ x^2 dA ; bounded by ellipse Calculus & Beyond Homework 6
[math analysis] sup f< sup g==>∫f^n<∫g^n Calculus & Beyond Homework 3