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

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UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
um , i think i made a mistake somewhere
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
sin r pi = 0
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
well sine of πr is going to be zero, so
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
so the last two lines are not true
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
and the third line isnt right either
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
the second line is not true .. check again.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
is the first line right?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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??
 one year ago

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

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
the function is f(x)=sin(x)
 one year ago

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

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
hmm i forgot it could do that
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
so have i made any mistake then?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
maybe you don't want that leading 1?
 one year ago

UnkleRhaukus Group TitleBest 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*} \]
 one year ago

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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}]
 one year ago

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

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

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

UnkleRhaukus Group TitleBest 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
 one year ago
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