Question

find the fourier series expansion for the following function
f(x)=x sin x for -pi 13 years ago

Answer 1

can you draw the function?

Answer 2

can you tell if it is odd even or neither?

Answer 3

what i know when i test f(-x) is x is odd then sin x is odd when odd n odd it become even right?

Answer 4

yeah it turns out to be even

Answer 5

so you only have to find \(a_0\) and , \(a_n\)

Answer 6

can you set you the \(a_0\) integral?

Answer 7

what do u mean? is it right i use formula \[2 \div \pi \int\limits_{\pi}^{- \pi} xsinx dx\]

Answer 8

\[a_0=\frac1l\int\limits_{a}^{a+2l} f(x)\,\text dx\]
\[a_0=\frac1\pi\int\limits_{-\pi}^{\pi} x\sin(x)\,\text dx\]

Answer 9

now we note that the function even, so
\[=\frac2\pi\int\limits_0^{\pi} x\sin(x)\,\text dx\]

Answer 10

(you were close ),

can you integrate that?

Answer 11

i get \[a _{0}=2\] is it correct? then how to find \[a _{n}?\]

Answer 12

yes! good work

Answer 13

\[a_n=\frac1l\int\limits_{a}^{a+2l} f(x)\cos\left(\frac{n\pi}lx\right)\,\text dx\]

Answer 14

its a bit trickier

Answer 15

(and use the fact its even first


then you might need to use some trig formula like
\[2\sin(A)\cos(B)=\sin(A+B)+\sin(A-B)\]

Answer 16

why we used full range equation? even though we know that it is even..