Question

can someone help me with the following Fourier
series question
F(x) = {-9 when - π 9 years ago

Answer 1

i have worked out the fourier series i just need to work out b for n = 1,2,3,4,5 if this makes any more sense

Answer 2

happy to pay for the correct answers

Answer 3

I'm afraid that doesn't make any more sense. Are you trying to work out the coefficients of the sine series?

Answer 4

Your function is odd, so I'm assuming that's the case, as both \(a_0\) and \(a_n\) are zero for all \(n\in\mathbb{N}\).

You have
\[b_n=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin(nx)\,\mathrm{d}x=\frac{9}{\pi}\left(\int_0^\pi\sin(nx)\,\mathrm{d}x-\int_{-\pi}^0\sin(nx)\,\mathrm{d}x\right)\]Are you having trouble computing these integrals?

Answer 5

Replacing \(x\mapsto -x\), you can consolidate the integral to get
\[b_n=\frac{18}{\pi}\int_0^\pi \sin(nx)\,\mathrm{d}x\]Surely you know the chain rule?

Answer 6

i have bn = 18/nπ [1-(-1)n]

just been asked to work out b for 1,2,3,4,5 and really cant remember how to do it

Answer 7

b for n = 1,2,3,4,5 if you could show me how to do them 5 it would be much appreciated

Answer 8

That's just a matter of replacing \(n\) with each of those values. For instance,
\[b_1=\frac{18}{1\times\pi}(1-(-1)^1)=\frac{18}{\pi}(1+1)=\frac{36}{\pi}\]

Answer 9

so what would b 2 and 3 be? just so im clear

Answer 10

For even values of \(n\), notice that \((-1)^n=1\), so both \(b_2\) and \(b_4\) would be \(0\), since
\[\frac{18}{n\pi}(1-1)=0\]
When \(n\) is odd, \((-1)^n=-1\). So for instance, when \(n=3\), you get
\[\frac{18}{3\pi}(1-(-1)^3)=\frac{6}{\pi}(1+1)=\frac{12}{\pi}\]

Answer 11

so for b2 its 18/2π (1-1) = 0

Answer 12

and b5 would be 18/5π (1-(-1)5) = 3.6/π (1+1) = 7.2/π ??? is this correct

Answer 13

????

Answer 14

That's correct, yes