I want to calculate the sum of $$\sum_{n\geq 0}\frac{\sin n}{n!}.$$

I think I am supposed to use the Taylor polynomial of $\ e^x$ but I don't know how to solve it.

Thanks for your help.

  • $\begingroup$ @graydad I see it's been edited. $\endgroup$ – Akiva Weinberger Jan 23 '15 at 0:05

Since $\sin x =\Im(e^{ix})$, we have: $$ \sum_{n\geq 0}\frac{\sin n}{n!} = \Im\left(e^{e^i}\right) = e^{\cos 1}\sin(\sin 1).$$

[$\Im(z)$ (also written "$\operatorname{Im}(z)$") means the imaginary part of $z$.]

  • 1
    $\begingroup$ Perhaps it's better to write \text{Im} , as that notation for "Imaginary" sucks...and many just don't understand it. $\endgroup$ – Timbuc Jan 22 '15 at 23:45
  • $\begingroup$ @JackD'Aurizio thanks for your answer. Could you show me the second equality using baby steps please? $\endgroup$ – user1118686 Jan 22 '15 at 23:55
  • $\begingroup$ @user1118686 You need to know this formula first: $e^{ix}=\cos(x)+i\sin(x)$. Do you? $\endgroup$ – Akiva Weinberger Jan 22 '15 at 23:55
  • $\begingroup$ @user1118686: we have that $\Im$ is a linear operator and for any $z$, $$\sum_{n\geq 0}\frac{z^n}{n!}=e^{z},$$ so we plug in $z=e^{i}$ and take the imaginary part. $\endgroup$ – Jack D'Aurizio Jan 22 '15 at 23:57
  • $\begingroup$ $\displaystyle\sum_{n\geq0}\frac{\sin n}{n!}=\sum_{n\geq0}\frac{\Im(e^{ix})}{n!}= \Im\sum_{n\geq0}\frac{e^{ix}}{n!}=\Im\left(e^{e^i}\right)=e^{\cos 1}\sin(\sin 1)$ $\endgroup$ – Akiva Weinberger Jan 22 '15 at 23:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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