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Proving F(x) = ∫[0,x] exp(-t^2) is odd. |
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| Jan24-13, 03:18 PM | #1 |
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Proving F(x) = ∫[0,x] exp(-t^2) is odd.
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? |
| Jan24-13, 03:22 PM | #2 |
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| Jan24-13, 03:23 PM | #3 |
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| Jan24-13, 03:30 PM | #4 |
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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 | #5 |
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| Jan24-13, 03:38 PM | #6 |
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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].
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| Jan24-13, 04:08 PM | #7 |
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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? |
| Jan24-13, 04:33 PM | #8 |
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| Jan24-13, 04:40 PM | #9 |
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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. |
| Jan24-13, 04:45 PM | #10 |
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| Jan24-13, 05:01 PM | #11 |
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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?
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| Jan24-13, 05:08 PM | #12 |
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| Jan24-13, 05:13 PM | #13 |
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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?
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| Jan24-13, 05:16 PM | #14 |
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| Jan24-13, 05:39 PM | #15 |
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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 | #16 |
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| Jan24-13, 06:04 PM | #17 |
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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. |
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