sofia keith

Posted on Friday, 05 December, 2003 - 09:10 pm:

please can someone help....as this problem is too difficult...

Q. The fuction f(x) is periodic with period 4 and is defined by

f(x)=
-1 when -2< x < -1
x when -1< x < 1
1 when 1< x < 2

Find the fourier series of f.

Thanks...

Kerwin Hui

Posted on Friday, 05 December, 2003 - 10:17 pm:

Note that

f

is odd, so you need only calculate the sine part of the series.

Now assume

$f (x) = \sum_{m = 1}^{\infty} b_{m} \sin (\frac{m π x}{2})$

Integrate $f (x) \sin (n π x / 2)$ in two different ways to obtain the value of $b_{n}$ .

Kerwin

sofia keith

Posted on Friday, 05 December, 2003 - 10:21 pm:

i'm sorry..but could u show me the integration part...as that's the bit where i'm going wrong somewhere..i've kept on trying it...bit it's not coming out right...thanks for the help so far though...

Kerwin Hui

Posted on Friday, 05 December, 2003 - 10:58 pm:

Well, one way is
LaTeX Image

The other way:
LaTeX Image

Equating, we have
LaTeX Image

More explicitly:
LaTeX Image

Kerwin

sofia keith

Posted on Saturday, 06 December, 2003 - 02:28 pm:

I'm sorry...but how u did u get the answer on the 3rd line..i.e equating it?

Kerwin Hui

Posted on Saturday, 06 December, 2003 - 07:43 pm:

In the result from the second line, integrate by parts in the first integral and the second integral should not pose any problem. The limit zero will not contribute anything and the

\cos (n π / 2)

cancels.

Kerwin