Question

fourier

Answer 1

show that

\(sgn(\sin \theta) sin^2 \theta = \frac{8}{\pi} \sum\limits_{k=1}^{\infty} \frac{sin ((2k-1)\theta)}{(2k-1)(3+4k-4k^2)}\)

Answer 2

\(sgn(\sin \theta) =\dfrac{|sin \theta|}{\sin\theta}\times \sin^2\theta\)

can that just be
\(sgn(\sin \theta) =|\sin\theta|.\sin\theta\)
that would be more like sin^2\theta \theta but we need to treat the sign lol idk i can see it more clear from here xD

lets roll fourier in next step!

Answer 3

\[sgn(\sin \theta) =\dfrac{|sin \theta|}{\sin\theta}\]

no need for the extra bit :-))

Answer 4

its even function anyway so we need to find \( b_n\) only xD right ?

Answer 5

then Fourier would be

\(f(x)=\sum b_k \sin (k\theta)\)

Answer 6

HOW could it be odd :O, i made a test and i got even xD

Answer 7

\(\Large b_k=\dfrac{1}{\pi}\int_{-\pi}^{\pi}|\sin \theta| \sin\theta ~d \theta=\dfrac{2}{\pi}\int_{0}^{\pi}|\sin \theta| \sin\theta ~d \theta \)

and since \(\pi \ge \theta \ge 0\) then \(\sin \theta \ge 0\)

so
\(\Large b_k=\dfrac{2}{\pi}\int_{0}^{\pi} \sin^2\theta ~d \theta \)

Answer 8

first try to know if ur function is odd\even by testing then do ur summation.

test
if f(x)=f(-x) even
if f(x)=-f(x) odd
if neither then Fourier failed , if both anything work.

Answer 9

i have forgotten a part ... pardon

Answer 10

\(\Large b_k=\dfrac{1}{\pi}\int_{-\pi}^{\pi}|\sin \theta| \sin\theta \color{red}{\sin k\theta }~d \theta=\dfrac{2}{\pi}\int_{0}^{\pi}|\sin \theta| \sin\theta \color{red}{\sin k\theta } ~d \theta \)

and since π≥θ≥0 then sinθ≥0 then

\(\Large b_k=\dfrac{2}{\pi}\int_{0}^{\pi} \sin^2\theta \color{red}{\sin k\theta } ~d \theta \)

Answer 11

cool so far ? or have a question ?

Answer 12

why only in 0< x < pi

surely a full period is needed, @ikram002p ??

Answer 13

its only as translating limits of integral to evaluate b_k and not Fourier itself
http://prntscr.com/9pdwk9

Answer 14

just away to make integral of \( b_k\) solvable without the sh!ty \( | sin \theta|\)

Answer 15

could we recall this on skype ? *i didn't get ur question*

Answer 16

you don't need to deal with integrals and differentials of sgn(x) and |x| its crap :P
just simplify to get an easy method or integral hahaha

Answer 17

Naa with fourier , you just need to recall even\odd functions properties and life will be easy :P

Answer 18

so here is full description for the question
\(sgn(\sin\theta)\) is odd, and \(\sin \theta\) is odd
\(f(\theta)=sgn(\sin\theta)\sin \theta \)
test
\(f(-\theta)=sgn(\sin-\theta)\sin- \theta =sgn(-\sin\theta)\times-\sin \theta=-sgn(\sin\theta)\times-\sin \theta \\=sgn(\sin\theta)\sin \theta =f(\theta)\)

so its even.

Answer 19

oh done with it already @IrishBoy123 ?

Answer 20

i was gonna repeat everything xD

Answer 21

damn :'( wadle wadle i made the wrong test

Answer 22

ok we need \(a_n\) xD

grapes... a little

sorry to confuse u :-\

Answer 23

@IrishBoy123 lets start over okay ?

Answer 24

\(f(\theta)=sgn(\sin\theta)\sin\theta=|\sin\theta|\sin\theta \) is an odd function.

\(a_0=0,a_k=0\), so we need \(b_k \)
\( b_k=\dfrac{1}{\pi}\int_{-\pi}^{\pi}|\sin \theta| \sin\theta \color{red}{\sin k\theta }~d \theta=\dfrac{2}{\pi}\int_{0}^{\pi}|\sin \theta| \sin\theta \color{red}{\sin k\theta } ~d \theta\\ =\dfrac{2}{\pi}\int_{0}^{\pi} \sin^2\theta \color{red}{\sin k\theta } ~d \theta \)

ermm

Answer 25

i liked the song @IrishBoy123 <3

Answer 26

\(b_k=\dfrac{2}{\pi} \dfrac{2\cos(\pi k)-2}{k^3-4k} =\frac{4}{\pi} \dfrac{\cos(\pi k)-1}{k^3-4k}\)

Answer 27

u wanna learn how to do the integral or how to move from here and evaluate the series ?

Answer 28

both. @ikram002p

Answer 29

\(b_k =\dfrac{2}{\pi}\int_{0}^{\pi} \sin^2\theta \sin k\theta ~d \theta \\ =\dfrac{2}{\pi}\int_{0}^{\pi} (1-\cos^2\theta) \sin k\theta ~d \theta =\dfrac{2}{\pi}[\int_{0}^{\pi} \sin k\theta ~d \theta-\int_{0}^{\pi} \cos^2\theta \sin k\theta ~d\theta ]\)

Answer 30

yep, but over a half period ?

Answer 31

\( \int_{0}^{\pi} \sin k\theta ~d \theta \) let \(u=k\theta ,d\theta=1/k ~du\) thus
\( \int_{0}^{\pi} \sin k\theta ~d \theta= \frac{1}{k} \int_{0}^{n\pi} \sin u ~d u= \frac{1}{k} [ 1-cos(k\theta)]\) \)


\(\int_{0}^{\pi} \cos^2\theta \sin k\theta ~d\theta \) integrate by parts ,

Answer 32

i thought u wanna see how to solve, listen b_k is just a constant we don't care about its period, what matter us is \(sin(k\theta)\) in the other part of series which is periodic on (-pi,pi) by default.. ..
this is a reference if u feel dull about the formula
http://mathworld.wolfram.com/FourierSeries.html