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January 19th, 2010, 11:19 AM   #1
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Periodic Integration Proof/Substitution

Let f:R->R be a continuous function. Let T>0 be such that f(x+t) = f(x) for all x.

Use an appropriate substitution to prove that for all real numbers a,

.



..I was thinking:


, but I can't think of a better way to explain it.

Also, intuitively, I understand that f(x+T) = f(x), so for a periodic function F(x+T) = F(x), but again can't really translate that into a legible proof.
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January 19th, 2010, 01:33 PM   #2
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Re: Periodic Integration Proof/Substitution

Quote:
Originally Posted by semper
Let f:R->R be a continuous function. Let T>0 be such that f(x+t) = f(x) for all x.

Use an appropriate substitution to prove that for all real numbers a,

.



..I was thinking:


, but I can't think of a better way to explain it.

Also, intuitively, I understand that f(x+T) = f(x), so for a periodic function F(x+T) = F(x), but again can't really translate that into a legible proof.
Two step process.
First: there is an integer n and some b between 0 and T so that b=a-nT.
Therefore by periodicity the integral from b to b+T is the same as the integral from a to a+T (the integrands are identical).
Second: The integrand from 0 to b is the same as the integrand from T to b+T, so the integral from 0 to T is the same as the integral from b to b+T.
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January 19th, 2010, 03:07 PM   #3
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Re: Periodic Integration Proof/Substitution

Very helpful. Thank you for your insight.
semper is offline  
January 19th, 2010, 03:50 PM   #4
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Each original T should be t.

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January 19th, 2010, 11:08 PM   #5
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Re: Periodic Integration Proof/Substitution

...Hey! You're in Dr. J's class aren't you?

Thank goodness someone posted this up, I was having major problems with this too.
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