Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
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. –  Henning Makholm Jan 1 '13 at 11:25

3 Answers 3

up vote 2 down vote accepted

Using a little trigonometry and the sum of a convergent infinite geometric series:

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

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

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

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

share|cite|improve this answer
Happy new year, Don. :-) –  Babak S. Jan 1 '13 at 11:42
Ditto for you, @Babak. –  DonAntonio Jan 1 '13 at 12:08

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.

share|cite|improve this answer

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


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 $f(x)=c\sin x$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.