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Let $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}=f(x)$, where $f(u)=c\sin u$.Find $c$.

Trial:$\sum_{k=1}^{\infty}\frac{c \sin (x+k\pi)}{2^k}=c\sin x$.Then $c=0$ is a solution. Is there any other solution of $c$? Mainly I am interested in the sum $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}$. please help.

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  • $\begingroup$ Clearly if any $c\ne0$ works, then every $c$ does... Also $\sin(x+k\pi)=(-1)^k\sin(x)$, so you can cancel $\sin(x)$ and just have a geometric series left. $\endgroup$ – hmakholm left over Monica Jan 1 '13 at 11:25
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Using a little trigonometry and the sum of a convergent infinite geometric series:

$$\sin(x+k\pi)=\sin x\cos k\pi+\sin k\pi\cos x=\sin x\cos k\pi=(-1)^k\sin x\Longrightarrow$$

$$c\sin x=\sum_{k=1}^\infty\frac{c\sin(x+k\pi)}{2^k}=c\sin x\sum_{k=1}^\infty\left(-\frac{1}{2}\right)^k=c\sin x\frac{-\frac{1}{2}}{1+\frac{1}{2}}=$$

$$=-\frac{c\sin x}{3}\Longrightarrow c\sin x=-\frac{c\sin x}{3}\Longleftrightarrow c=0\,\,\,\vee\,\,\,\sin x=0\Longleftrightarrow$$

$$\Longleftrightarrow c=0\,\,\,\vee\,\,\,x=n\pi\,\,\,,n\in\Bbb Z$$

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  • $\begingroup$ Happy new year, Don. :-) $\endgroup$ – mrs Jan 1 '13 at 11:42
  • $\begingroup$ Ditto for you, @Babak. $\endgroup$ – DonAntonio Jan 1 '13 at 12:08
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Note that $\sin(x+k\pi)=(-1)^k \sin x$ so that $$\sum_{k=1}^\infty \frac{\sin(x+k\pi)}{2^k}= \sin x \cdot \sum_{k=1}^\infty \frac{(-1)^k}{2^k}=(-1/3)\sin x,$$ on using the sum of a geometric series.

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If $$T_k=\frac{f(x+k\pi)}{2^k}=\frac{c\sin(x+k\pi)}{2^k}$$

$$T_{k+1}=\frac{f\{x+(k+1)\pi\}}{2^k}=\frac{c\sin\{x+(k+1)\pi\}}{2^{k+1}}=-\frac{T_k}2$$

So, $T_2=-\frac{T_1}2,T_3=-\frac{T_2}2=\left(-\frac12\right)^2T_1$ and $T_1=\frac{c\sin(x+\pi)}2=-\frac{c\sin x}2$

$$\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}=T_1\sum_{k=0}^{\infty}\left(-\frac12\right)^k=\left(-\frac{c\sin x}2\right)\frac1{1-\left(-\frac12\right)}=-\frac c3\sin x$$

Now equate this with $f(x)=c\sin x$

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