Question

need help on these fourier series questions http://i.imgur.com/Q6KAm.jpg

Answer 1

simply apply fourier series formula...

Answer 2

Which one is giving you trouble.

Answer 3

the first one

Answer 4

i didn't understand how to solve it when absolute values are included

Answer 5

Well clearly the function is even. So the only terms in Fourier series will be the coefficients of the cosine functions. All the sin functions will be zeroed out.

Answer 6

Now, what's the formula for the coefficients of the cosine functions? Write that down with this function and we can figure how to get rid of the absolute value sign.

Answer 7

1/L\[\int\limits_{\pi/2}^{\pi} \sin(x)*\cos(n)*\pi*xdx\]

Answer 8

will it be like this

Answer 9

\[ a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) \ dx \]

Answer 10

Now in that integral, f(x) = 0 when x > pi/2 and when x < pi/2

Hence

\[ a_n = \frac{1}{\pi} \left( \int_{-\pi/2}^0 (-\sin x) \cos(nx) \ dx + \int^{\pi/2}_0 \sin x \cos(nx) \ dx \right) \]
\[ = \frac{1}{\pi} \left( \int^{\pi/2}_0 \sin x \cos(nx) \ dx + \int^{\pi/2}_0 \sin x \cos(nx) \ dx \right) \]
\[ = \frac{2}{\pi} \int^{\pi/2}_0 \sin x \cos(nx) \ dx \]

Now evaluate that.

Answer 11

thank you