A community for students.
Here's the question you clicked on:
 0 viewing
 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
\[\begin{align*} % S(x) S(x) &=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((2r1)x)}{1(2r1)^2}\right)\\ \\ S(\tfrac\pi2)&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\cos((\pi r\tfrac\pi2)}{1(2r1)^2}\right)\\ 1&=\frac{2}\pi\left(1+2\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1(2r1)^2}\right)\\ \frac\pi21&=2\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1(2r1)^2}\\ \frac\pi4\frac12&=\sum\limits_{r=1}^\infty\frac{\sin(\pi r)}{1(2r1)^2}\\ \end{align*}\]

This Question is Closed

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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.