Question

\[\begin{align*} % S(x)
S(x)
&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((2r-1)x)}{1-(2r-1)^2}\right)\\
\\
S(\tfrac\pi2)&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((\pi r-\tfrac\pi2)}{1-(2r-1)^2}\right)\\
1&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1-(2r-1)^2}\right)\\
\frac\pi2-1&=2\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1-(2r-1)^2}\\
\frac\pi4-\frac12&=\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1-(2r-1)^2}\\
\end{align*}\]

Answer 1

um , i think i made a mistake somewhere

Answer 2

sin r pi = 0

Answer 3

well sine of πr is going to be zero, so

Answer 4

so the last two lines are not true

Answer 5

and the third line isnt right either

Answer 6

the second line is not true .. check again.

Answer 7

is the first line right?

Answer 8

sorry ... it seems i made mistake I didn't see 2r there.
I wouldn't now about the first line since it is your question. What is your function??

Answer 9

why did you put \( S(\pi/2) = 1 \)

Answer 10

the function is f(x)=|sin(x)|

Answer 11

I think the summation is mostly zeros, except for r=1 where it is 0/0
it's pi at r=1 ?

Answer 12

hmm i forgot it could do that

Answer 13

so have i made any mistake then?

Answer 14

as far as I can see, your very first line, evaluated at x=pi/2 gives
2/pi *(1 + 2*pi/4) = 2/pi +1

Answer 15

maybe you don't want that leading 1?

Answer 16

\[\begin{align*} % a_0
a_0&=\frac1\pi\int\limits_{-\pi}^\pi |\sin(x)| \,\text dx\\
&=\frac2\pi\int\limits_{0}^\pi \sin(x) \,\text dx\qquad\text{(even integrand)}\\
&=\frac{2}\pi\left(-\cos(x)\Big|_0^\pi\right)\\
&=\frac{2(-(-1-1))}\pi
\\
&=\frac{4}\pi
\end{align*}\]

\begin{align*} % a_n
a_n&=\frac1\pi\int\limits_{-\pi}^\pi|\sin(x)|\cos(nx)\,\text dx\\
&=\frac2\pi\int\limits_{0}^\pi\sin(x)\cos(nx)\,\text dx\qquad\text{(even integrand)}\\
&=\frac2\pi\int\limits_{0}^\pi\frac{\sin(x+nx)-\sin(x-nx)}2\,\text dx\\
&=\frac1\pi\int\limits_{0}^\pi{\sin\left((1+n)x\right)-\sin\left((1-n)x\right)}\,\text dx\\
&=\frac1\pi\left(\frac{-\cos((1+n)x)}{1+n}-\frac{\cos((1-n)x))}{1-n}\Big|_0^\pi\right)\\
&=\frac1\pi\left(\frac{1-\cos((1+n)\pi)}{1+n}+\frac{1-\cos((1-n)\pi))}{1-n}\right)\\
&=\frac1\pi\left(\frac{1-(-1)^{n+1}}{1+n}+\frac{1-(-1)^{n-1})}{1-n}\right)\\
\\&=\frac1\pi\left(\frac{1+(-1)^{n}}{1+n}+\frac{1+(-1)^{n})}{1-n}\right)\\
\end{align*}

\begin{align*} % b_n
b_n&=\frac1\pi\int_{-\pi}^\pi|\sin(x)|\sin(nx)\,\text dx\\
&=0\qquad\text{(odd integrand)}
\end{align*}

\begin{align*} % S(x)
S(x)&=\frac{2}\pi+\frac1\pi\sum\limits_{n=1}^\infty\left(\frac{1+(-1)^{n}}{1+n}+\frac{1+(-1)^{n})}{1-n}\right)\cos(nx)\\
&=\frac{2}\pi+\frac2\pi\sum\limits_{n=2,4,6\dots}^\infty\left(\frac{1}{1+n}+\frac{1}{1-n}\right)\cos(nx)\\
&=\frac{2}\pi\left(1+2\sum\limits_{n=2,4,6\dots}^\infty\frac{\cos(nx)}{1-n^2}\right)\\
&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((2r-1)x)}{1-(2r-1)^2}\right)\\
\\&=\frac4\pi\left(\frac12-\frac{\cos(2x)}{3}+\frac{\cos(4x)}{15}+\frac{\cos(6x)}{35}+\dots\right)
\end{align*}
\]

Answer 17

a_0 = 2/pi
http://www.wolframalpha.com/input/?i=integrate+ |sin%28x%29|%2Fpi+from+0+to+pi

Answer 18

http://www.wolframalpha.com/input/?i=%E2%88%AB+%7Csin%28x%29%7C%2F%CF%80+from+-%CF%80+to+%CF%80

Answer 19

it seems that you are taking 2pi as period, although it works, pi should also work fine.

Answer 20

Plot[2/Pi - Sum[2/Pi * 1/(4 n^2 - 1) Cos[n x], {n, 1, 20}], {x, 0,
4 Pi}]
however the period seems to be 2pi

Answer 21

sorry the result was correct .. i forgot it was for even case. On Mathematica
Plot[2/Pi - Sum[2/Pi * 1/(4 n^2 - 1) Cos[2 n x], {n, 1, 20}], {x, 0,
2 Pi}]

Answer 22

so the final series is
\[ |\sin x| = {2 \over \pi} +{2 \over \pi} \sum_{n=1}^\infty \frac 1{1 - 4 n^2} \cos (2nx) \]

Answer 23

In case you didn't find your mistake, it happens when you change the index from n=2,4,6... to r=1,2,3...
you should substitute n=2r (so that r= n/2 giving 1,2,3...)
you then get exper's result

Answer 24

ah yes

\[\begin{align*} % S(x)
S(x)&=\frac{2}\pi+\frac2\pi\sum\limits_{n=1}^\infty\left(\frac{1+(-1)^{n}}{1-n^2}\right)\cos(nx)\\
&=\frac{2}\pi\left(1+2\sum\limits_{n=2,4,6\dots}^\infty\frac{\cos(nx)}{1-n^2}\right)\\
&=\frac{4}\pi\left(\frac12+\sum\limits_{r=1}^\infty\frac{\cos(2rx)}{1-(2r)^2}\right)\\
\\&=\frac4\pi\left(\frac12-\frac{\cos(2x)}{3}-\frac{\cos(4x)}{15}-\frac{\cos(6x)}{35}-\dots\right)
\end{align*}\]

Answer 25

\[\begin{align*} % S(x)
S(x)
&=\frac{4}\pi\left(\frac12+\sum\limits_{r=1}^\infty\frac{\cos(2rx)}{1-(2r)^2}\right)\\
\\
S(\tfrac\pi2)&=\frac{4}\pi\left(\frac12+\sum\limits_{r=1}^\infty\frac{\cos(\pi r)}{1-4r^2}\right)\\
1&=\frac{4}\pi\left(\frac12+\sum\limits_{r=1}^\infty\frac{(-1)^{r}}{1-4r^2}\right)\\
\frac\pi4&=\frac12+\sum\limits_{r=1}^\infty\frac{(-1)^{r}}{1-4r^2}\\
\end{align*}\]

thats better

thanks to both of you