Question

can someone help me with this im really stuck
determine the fourier series for the function
f (x) = {-9 when -pie < x < 0
{+9 when 0 < x < pie
periodic over 2 pie

Answer 1

Hi @jonnywktn,
At which step are you stuck?

Answer 2

can you show your steps, whatever you have tried till now?

Answer 3

im not sure how to do it at all struggling to understand it was looking for help from start to finish if possible

Answer 4

is this something you could help with please?

Answer 5

yes,
do you know the formulas?

if not, please check this:
http://www.sosmath.com/fourier/fourier1/fourier1.html

especially from the part that says
"This theorem helps associate a Fourier series to any $2\pi$-periodic function."

Answer 6

all I have is what ive posted I was just looking for someone to complete the question so I could work through it as im struggeling to understand or follow notes online

Answer 7

can you solve integrals of the type:

\(\Large \int \limits_{-\pi }^{\pi } f(x) \cos nx \: dx\)
?

Answer 8

I can provide you the entire solution, but you might not learn anything from it, you might just copy it...

Answer 9

the next part it to change the limits to 8 and 0 so I need to work through the first part and then do the other so im not just going to copy it I would appreciate the entire solution and answer for the one posted so I can learn from that to enable me to complete the rest
thanks

Answer 10

are you able to do this please
thank you

Answer 11

The fourier series of f(x) is represented by
\(\Large f(x) = a_0 + \sum \limits_{n=1}^\infty a_n \cos nx + b_n \sin nx\)

The constants, \(a_0, a_n, b_n\) for the periodic function with period = \(2\pi\) can be calculated as :
\(a_0 = \dfrac{1}{2\pi}\int \limits_{-\pi}^{\pi}f(x)dx\)
\(a_n = \dfrac{1}{\pi}\int \limits_{-\pi}^{\pi}f(x) \cos nx dx\)
\(b_n = \dfrac{1}{\pi}\int \limits_{-\pi}^{\pi}f(x) \sin nxdx\)

here, f(x) = -9 \((-\pi < x < 0)\)
+9 \((0 < x < \pi)\)


\(~\\ \Large a_0 = \dfrac{1}{2\pi}\int \limits_{-\pi}^{\pi}f(x)dx \\ \Large = \dfrac{1}{2\pi}[\int \limits_{-\pi}^{0}f(x)dx+\int \limits_{0}^{\pi}f(x)dx ]\\ \Large = \dfrac{1}{2\pi}[\int \limits_{-\pi}^{0}-9dx+\int \limits_{0}^{\pi}9dx ] \\ \Large = \dfrac{1}{2\pi} [[-9x]^0_{{-\pi }}+[9x]_0^\pi ] \\ \Large = \dfrac{1}{2\pi}[0-9\pi +9\pi +0] \\ \Large \bbox[yellow ]{a_0=0} \)

Answer 12

\(\Large a_n = \dfrac{1}{\pi}\int \limits_{-\pi}^{\pi}f(x) \cos nx dx \\ ~\\ \Large = \dfrac{1}{\pi}[\int \limits_{-\pi}^{0}f(x)\cos nx dx+\int \limits_{0}^{\pi}f(x)\cos nxdx ]\\ \Large = \dfrac{1}{\pi}[\int \limits_{-\pi}^{0}-9\cos nx dx+\int \limits_{0}^{\pi}9 \cos nx dx ] \\ \Large = \dfrac{1}{\pi} [[\dfrac{-9}{n}\sin nx]^0_{{-\pi }}+[\dfrac{9}{n}\sin nx]_0^\pi ] \\ \Large = \dfrac{1}{n\pi}[0-9\sin n\pi +9\sin n\pi +0] \\ \Large Since, \sin n\pi = 0\\ \Large \bbox[yellow ]{a_n=0} \)

Since, f(x) is an odd function, we got \(a_0 =0, a_n =0\).
[Sometimes, we can use this directly, for odd functions, a0, an are 0, for even functions, bn =0]

Answer 13

\(\Large b_n = \dfrac{1}{\pi}\int \limits_{-\pi}^{\pi}f(x) \sin nx dx \\ ~\\ \Large = \dfrac{1}{\pi}[\int \limits_{-\pi}^{0}f(x)\sin nx dx+\int \limits_{0}^{\pi}f(x)\sin nxdx ]\\ \Large = \dfrac{1}{\pi}[\int \limits_{-\pi}^{0}-9\sin nx dx+\int \limits_{0}^{\pi}9 \sin nx dx ] \\ \Large = \dfrac{1}{\pi} [[\dfrac{9}{n}\cos nx]^0_{{-\pi }}+[\dfrac{-9}{n}\cos nx]_0^\pi ] \\ \Large = \dfrac{9}{n\pi}[1-\cos (-n\pi) -\cos n\pi +1] \\ \Large = \dfrac{9}{n\pi}[2-2\cos (n\pi)]\\ \Large Since, \cos n\pi = (-1)^n\\ \Large \bbox[yellow ]{b_n=\dfrac{18}{n\pi}[1-(-1)^n]\: } \)

Answer 14

Hence, the fourier series of f(x) is
\(\Large f(x) = a_0 + \sum \limits_{n=1}^\infty a_n \cos nx + b_n \sin nx \\ \Huge f(x) = \sum \limits_{n=1}^\infty \dfrac{18}{n\pi}(1-(-1)^n) \sin nx \)

Thats your final answer :)

Answer 15

you can express it as
\(\large f(x) = \dfrac{36}{\pi} [\sin x + \dfrac{\sin 3x}{3}+\dfrac{\sin 5x}{5}+...]\)

Answer 16

Let me know if you have any doubts @jonnywktn :)

Answer 17

sorry for the late reply I think I follow that part but I have been asked to work out b when n = 1,2,3,4 & 5
would this be something you could help with
note I managed to complete this part for the other question so thanks for your help on that bit its much appreciated

Answer 18

sure,
\(\Large b_n=\dfrac{18}{n\pi}[1-(-1)^n]\)

n=1, (-1)^n = -1 ,
1-(-1) = 1+1= 2

so b1 = 36/1\(\pi\) = 36/\(\pi\)

got this?

Answer 19

I think so could you show me n=5 just to confirm I understand

Answer 20

going for n=2,

(-1)^2 = 1
1-1 = 0
so, b2 = 0

doubts??


knowing how to get b1 and b2,
can you please give a try to find others??

Answer 21

sure,
\(\Large b_n=\dfrac{18}{n\pi}[1-(-1)^n]\)

n=5, (-1)^5 = -1 ,
1-(-1) = 1+1= 2

so b5 = 36/5\(\pi\)

Answer 22

in all,
b2 = b4 = 0
b1 = 36/pi
b3 = 36/3pi = 12/pi
b5 = 36/5pi

Answer 23

so would you divide out from that so the answer is 36/5pi or would you leave it as that?

Answer 24

leave it as it is.