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 2 years ago
\[\begin{align*} % S(x)
S(x)
&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((2r1)x)}{1(2r1)^2}\right)\\
\\
S(\tfrac\pi2)&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((\pi r\tfrac\pi2)}{1(2r1)^2}\right)\\
1&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1(2r1)^2}\right)\\
\frac\pi21&=2\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1(2r1)^2}\\
\frac\pi4\frac12&=\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1(2r1)^2}\\
\end{align*}\]
 2 years ago
\[\begin{align*} % S(x) S(x) &=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((2r1)x)}{1(2r1)^2}\right)\\ \\ S(\tfrac\pi2)&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((\pi r\tfrac\pi2)}{1(2r1)^2}\right)\\ 1&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1(2r1)^2}\right)\\ \frac\pi21&=2\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1(2r1)^2}\\ \frac\pi4\frac12&=\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1(2r1)^2}\\ \end{align*}\]

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UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0um , i think i made a mistake somewhere

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0well sine of πr is going to be zero, so

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0so the last two lines are not true

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0and the third line isnt right either

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1the second line is not true .. check again.

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0is the first line right?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1sorry ... 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??

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1why did you put \( S(\pi/2) = 1 \)

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0the function is f(x)=sin(x)

phi
 2 years ago
Best ResponseYou've already chosen the best response.1I think the summation is mostly zeros, except for r=1 where it is 0/0 it's pi at r=1 ?

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0hmm i forgot it could do that

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0so have i made any mistake then?

phi
 2 years ago
Best ResponseYou've already chosen the best response.1as far as I can see, your very first line, evaluated at x=pi/2 gives 2/pi *(1 + 2*pi/4) = 2/pi +1

phi
 2 years ago
Best ResponseYou've already chosen the best response.1maybe you don't want that leading 1?

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0\[\begin{align*} % a_0 a_0&=\frac1\pi\int\limits_{\pi}^\pi \sin(x) \,\text dx\\ &=\frac2\pi\int\limits_{0}^\pi \sin(x) \,\text dx\qquad\text{(even integrand)}\\ &=\frac{2}\pi\left(\cos(x)\Big_0^\pi\right)\\ &=\frac{2((11))}\pi \\ &=\frac{4}\pi \end{align*}\] \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(xnx)}2\,\text dx\\ &=\frac1\pi\int\limits_{0}^\pi{\sin\left((1+n)x\right)\sin\left((1n)x\right)}\,\text dx\\ &=\frac1\pi\left(\frac{\cos((1+n)x)}{1+n}\frac{\cos((1n)x))}{1n}\Big_0^\pi\right)\\ &=\frac1\pi\left(\frac{1\cos((1+n)\pi)}{1+n}+\frac{1\cos((1n)\pi))}{1n}\right)\\ &=\frac1\pi\left(\frac{1(1)^{n+1}}{1+n}+\frac{1(1)^{n1})}{1n}\right)\\ \\&=\frac1\pi\left(\frac{1+(1)^{n}}{1+n}+\frac{1+(1)^{n})}{1n}\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})}{1n}\right)\cos(nx)\\ &=\frac{2}\pi+\frac2\pi\sum\limits_{n=2,4,6\dots}^\infty\left(\frac{1}{1+n}+\frac{1}{1n}\right)\cos(nx)\\ &=\frac{2}\pi\left(1+2\sum\limits_{n=2,4,6\dots}^\infty\frac{\cos(nx)}{1n^2}\right)\\ &=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((2r1)x)}{1(2r1)^2}\right)\\ \\&=\frac4\pi\left(\frac12\frac{\cos(2x)}{3}+\frac{\cos(4x)}{15}+\frac{\cos(6x)}{35}+\dots\right) \end{align*} \]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1a_0 = 2/pi http://www.wolframalpha.com/input/?i=integrate+sin%28x%29%2Fpi+from+0+to+pi

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=%E2%88%AB+%7Csin%28x%29%7C%2F%CF%80+from+%CF%80+to+%CF%80

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1it seems that you are taking 2pi as period, although it works, pi should also work fine.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1Plot[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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1sorry 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}]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1so the final series is \[ \sin x = {2 \over \pi} +{2 \over \pi} \sum_{n=1}^\infty \frac 1{1  4 n^2} \cos (2nx) \]

phi
 2 years ago
Best ResponseYou've already chosen the best response.1In 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

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0ah yes \[\begin{align*} % S(x) S(x)&=\frac{2}\pi+\frac2\pi\sum\limits_{n=1}^\infty\left(\frac{1+(1)^{n}}{1n^2}\right)\cos(nx)\\ &=\frac{2}\pi\left(1+2\sum\limits_{n=2,4,6\dots}^\infty\frac{\cos(nx)}{1n^2}\right)\\ &=\frac{4}\pi\left(\frac12+\sum\limits_{r=1}^\infty\frac{\cos(2rx)}{1(2r)^2}\right)\\ \\&=\frac4\pi\left(\frac12\frac{\cos(2x)}{3}\frac{\cos(4x)}{15}\frac{\cos(6x)}{35}\dots\right) \end{align*}\]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0\[\begin{align*} % S(x) S(x) &=\frac{4}\pi\left(\frac12+\sum\limits_{r=1}^\infty\frac{\cos(2rx)}{1(2r)^2}\right)\\ \\ S(\tfrac\pi2)&=\frac{4}\pi\left(\frac12+\sum\limits_{r=1}^\infty\frac{\cos(\pi r)}{14r^2}\right)\\ 1&=\frac{4}\pi\left(\frac12+\sum\limits_{r=1}^\infty\frac{(1)^{r}}{14r^2}\right)\\ \frac\pi4&=\frac12+\sum\limits_{r=1}^\infty\frac{(1)^{r}}{14r^2}\\ \end{align*}\] thats better thanks to both of you
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