What is the sum of the Fourier series for g(x) at x = pi/2 and x = 3pi/2?

[nicolayh]

Homework Statement

The function g(x) is defined as follows:

g(x) = \left\{ \begin{array}{rcl}<br /> {-\pi e^x} &amp; \mbox{for}<br /> &amp; -\pi &lt; x &lt; 0 \\<br /> {\pi e^{ -x}} &amp; \mbox{for} &amp; 0 &lt; x &lt; \pi<br /> \end{array}\right.

And the Fourier series for g(x) is as follows:

<br /> <br /> \sum_{n=0}^\infty \frac{2n}{n^2+1}(1 - (-1)^n e^{-\pi})\sin{nx}<br /> <br />

What is the sum of this series given x = \frac{\pi}{2} and x = \frac{3\pi}{2}?

The Attempt at a Solution

We've tried googeling, adressing the textbook on the subject (Kreyzig's Advanced Engineering Mathematics), but have yet to find a solution to this problem. Any help would be greatly appreciated! :)

 

[Dick]

Assuming the Fourier transform is correct, your function g(x) is continuous at x=pi/2. Can't you find the sum of the series by finding the value of g(pi/2)? Your other value is outside the range of definition of g(x) but your transform should converge to a periodic extension of g(x), right?

 

[nicolayh]

Thank you very much! :)

We managed to get the first one right (pi/2) by simply inserting the value into the function as suggested.

However, we are a bit more puzzeled about the second part of the task. How do we go about finding that periodic extension of g(x)? (and how do we show that the Fourier series converges to it?

 

[HallsofIvy]

Since the Fourier series for g involves only sines, g is an odd function: that is it is extended exactly as it is with period 2\pi.

For x between \pi and 2\pi, g(x)= -\pi e^x.

 

[nicolayh]

We get it now, thank you both very much! :D

 

[LCKurtz]

Since the Fourier series for g involves only sines, g is an odd function: that is it is extended exactly as it is with period 2\pi.

For x between \pi and 2\pi, g(x)= -\pi e^x.

Surely you don't mean that. It is an appropriate translation of -\pi e^x to the new interval.