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

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In summary, the conversation discusses ways to demonstrate that the function F(x)=∫ [0,x] exp(-t^2) dt is odd by showing that ∫[-x,x] exp(-t^2) = 0 or by showing that the derivative of F(x) is even. The conversation also explores the possibility of proving this using integration and substitution. Ultimately, it is suggested that substituting u=-t and taking into account the change in bounds will lead to a transformed integral with a u variable instead of t, which can then be evaluated to show that F(x)=-F(-x), thus proving that F(x) is indeed an odd function.
  • #1
peripatein
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Hi,

Homework Statement


Any idea how I might demonstrate that F(x)=∫ [0,x] exp(-t^2) dt is odd?


Homework Equations





The Attempt at a Solution


Will it be by showing that ∫[-x,x] exp(-t^2) = 0?
 
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  • #2
peripatein said:
Hi,

Homework Statement


Any idea how I might demonstrate that F(x)=∫ [0,x] exp(-t^2) dt is odd?


Homework Equations

Wouldn't the definition of "odd function" be relevant?
peripatein said:

The Attempt at a Solution


Will it be by showing that ∫[-x,x] exp(-t^2) = 0?
No, I don't think so.
 
  • #3
peripatein said:
Hi,

Homework Statement


Any idea how I might demonstrate that F(x)=∫ [0,x] exp(-t^2) dt is odd?


Homework Equations





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)).
 
  • #4
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?
 
  • #5
peripatein said:
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.
 
  • #6
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].
 
  • #7
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?
 
  • #8
peripatein said:
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.
 
  • #9
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.
 
  • #10
peripatein said:
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.
 
  • #11
Wouldn't that leave me to prove that int[0,x] exp(-u^2)du + int[0,-x]exp(-u^2)du=0?
 
  • #12
peripatein said:
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?
 
  • #13
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?
 
  • #14
peripatein said:
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?
 
  • #15
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?
 
  • #16
peripatein said:
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.
 
  • #17
HallsofIvy said:
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.
 
  • #18
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).
 
  • #19
peripatein said:
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).

It wasn't approval. It was a confession that the equality you wrote is true even though I have no idea how you got it. What I wanted to know is what you got for the du integral following the steps in post 10.
 
  • #20
peripatein said:
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).

##F(x) = \int_0^x e^{-t^2}dt##

Hence ##F(-x) = \int_0^{-x} e^{-t^2}dt##

What you need to do now is to evaluate ##\int_0^{-x} e^{-t^2}dt##

To do that, substitute u = -t. You haven't actually done that yet. You will end up with an integral with respect to u. You also need to take into account what happens to the bounds of definite integration when you make the substitution (remember, everything needs to be expressed in terms of u).

Do this and write what you get.
 
  • #21
What I wrote, to which you acquiesced, was precisely what I obtained via that substitution.
 
  • #22
peripatein said:
What I wrote, to which you acquiesced, was precisely what I obtained via that substitution.

If you did then you showed that F(x)+F(-x)=0. That would pretty much finish it.
 
  • #23
Let me write it again, for clarity's sake:
int [0,x] exp(-u^2) du int [0,-x] exp(-u^2) du.
 
  • #24
There should be a "+" there, between the two integrals
 
  • #25
And that is not a proof! That remains to be proven! Once proven, and only then, could it be affirmed that F is indeed odd.
 
  • #26
peripatein said:
And that is not a proof! That remains to be proven! Once proven, and only then, could it be affirmed that F is indeed odd.

##\int_0^{-x} e^{-t^2}dt##

You STILL haven't evaluated this integral by the suggested substitution! The substitution u = -t hasn't actually been performed (or, if it has, it has not been done correctly). If you did it correctly, you'll find that the upper bound becomes "x" (rather than "-x").
 
  • #27
What I keep getting, when I perform the suggested substitution, is F(x)=-int[0,-x]exp(^-u^2) du, and -F(-x)=int[0,x]exp(-u^2) du.
 
  • #28
peripatein said:
What I keep getting, when I perform the suggested substitution, is F(x)=-int[0,-x]exp(^-u^2) du, and -F(-x)=int[0,x]exp(-u^2) du.

You should learn how to write integrals in LaTeX... They're pretty easy. The following would go between two sets of $ $ characters (no space between each pair):
\int_0^{-x} e^{-u^2}du

The above produces this output:
$$ \int_0^{-x} e^{-u^2}du $$
 
  • #29
peripatein said:
What I keep getting, when I perform the suggested substitution, is F(x)=-int[0,-x]exp(^-u^2) du, and -F(-x)=int[0,x]exp(-u^2) du.

You don't have to do the substitution for F(x). Just leave it as ##\int_0^x e^{-t^2}dt##.

You only have to do the sub for F(-x). You should get the result that this is equal to ##-\int_0^x e^{-u^2}du##. I'm assuming you got this because your second expression is equivalent to this (you just put a minus sign on it).

Now you need to recognise that in a definite integral, the independent variable and the element of integration are simply dummy variables, and can be replaced by any other symbol (as long as it's done consistently). In other words, ##-\int_0^x e^{-u^2}du = -\int_0^x e^{-t^2}dt##.

You should now be quite easily able to see that F(x) = -F(-x).
 

What is the definition of an odd function?

An odd function is a mathematical function that satisfies the property f(-x) = -f(x) for all values of x. This means that the function is symmetric about the origin and has a graph that is reflected across the y-axis.

What does it mean for a function to be odd?

A function is considered odd if it satisfies the property f(-x) = -f(x). This means that the function is symmetric about the origin and has a graph that is reflected across the y-axis.

How can I prove that F(x) = ∫[0,x] exp(-t^2) is odd?

To prove that F(x) is odd, you can use the definition of an odd function and show that F(-x) = -F(x) for all values of x. This can be done by substituting -x into the integral and using properties of definite integrals.

Are there any special techniques or strategies for proving a function is odd?

Yes, there are a few techniques that can be helpful in proving that a function is odd. These include using the properties of definite integrals, substitution, and manipulating the function algebraically to show that it satisfies the definition of an odd function.

Why is it important to prove that F(x) = ∫[0,x] exp(-t^2) is odd?

Proving that F(x) is odd is important because it allows us to make conclusions about the symmetry and behavior of the function. It also allows us to simplify calculations and make predictions about the values of the function at different points.

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