# how can we convert sin function into continued fraction?

how can we convert sin function into continued fraction ?

for example

http://mathworld.wolfram.com/EulersContinuedFraction.html

how can we convert sin to simmilar continued fraction ??

and what about sinh and cosh ? arcsin ? arctan ? cos ? arccos ??

in general , how can convert any function to continued fraction ???

my friend asked me this question , so i hope that you help me to be enabled to help him

thanx for all of you

#### Solutions Collecting From Web of "how can we convert sin function into continued fraction?"

We will proceed as in this answer.

Define
$$P_n(x)=\sum_{k=0}^\infty\frac{4^{k+n}-\sum\limits_{j=1}^n\binom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k\tag{1}$$
Then
\begin{align} \frac{P_{n-1}(x)}{P_n(x)} &=\frac {\displaystyle\sum_{k=0}^\infty\frac{4^{k+n-1}-\sum\limits_{j=1}^{n-1}\binom{2k+2n-1}{2j-1}}{(2k+2n-1)!}(-x^2)^k} {\displaystyle\sum_{k=0}^\infty\frac{4^{k+n}-\sum\limits_{j=1}^n\binom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\\[12pt] &=\color{#C00000}{-x^2+}\frac {\displaystyle\sum_{k=0}^\infty\frac{\color{#C00000}{\binom{2k+2n-1}{2n-1}}}{(2k+2n-1)!}(-x^2)^k} {\displaystyle\sum_{k=0}^\infty\frac{4^{k+n}-\sum\limits_{j=1}^n\binom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\\[12pt] &=-x^2+\frac {\displaystyle\sum_{k=0}^\infty\color{#C00000}{\frac{2n(2n+1)\binom{2k+2n+1}{2n+1}}{(2k+2n+1)!}}(-x^2)^k} {\displaystyle\sum_{k=0}^\infty\frac{4^{k+n}-\sum\limits_{j=1}^n\binom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\\[12pt] &=\color{#C00000}{2n(2n+1)}-x^2\color{#C00000}{-}\frac {\displaystyle\sum_{k=0}^\infty\frac{2n(2n+1)\color{#C00000}{\left[4^{k+n}-\sum\limits_{j=1}^{n+1}\binom{2k+2n+1}{2j-1}\right]}}{(2k+2n+1)!}(-x^2)^k} {\displaystyle\sum_{k=0}^\infty\frac{4^{k+n}-\sum\limits_{j=1}^n\binom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\\[12pt] &=2n(2n+1)-x^2\color{#C00000}{+2n(2n+1)x^2}\frac {\displaystyle\sum_{k=0}^\infty\color{#C00000}{\frac{4^{k+n+1}-\sum\limits_{j=1}^{n+1}\binom{2k+2n+3}{2j-1}}{(2k+2n+3)!}}(-x^2)^k} {\displaystyle\sum_{k=0}^\infty\frac{4^{k+n}-\sum\limits_{j=1}^n\binom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\\[12pt] &=2n(2n+1)-x^2+2n(2n+1)x^2\color{#C00000}{\left/\frac{P_n(x)}{P_{n+1}(x)}\right.}\tag{2} \end{align}
As I suggested in chat, consider
\begin{align} \sin(x) &=\frac{\sin(2x)}{2\cos(x)}\\ &=\frac {\displaystyle x\sum_{k=0}^\infty\frac{4^k(-x^2)^k}{(2k+1)!}} {\displaystyle\sum_{k=0}^\infty\frac{(-x^2)^k}{(2k)!}}\\ &=x\left/\left(\frac {\displaystyle\sum_{k=0}^\infty\frac{(-x^2)^k}{(2k)!}} {\displaystyle \sum_{k=0}^\infty\frac{4^k(-x^2)^k}{(2k+1)!}} \right)\right.\\ &=x\left/\left(1+x^2\left/\frac{P_0(x)}{P_1(x)}\right.\right)\right.\tag{3} \end{align}
$(2)$ and $(3)$ lead us to the continued fraction
$$\sin(x)=\cfrac{x}{1+\cfrac{x^2}{2\cdot3-x^3+\cfrac{2\cdot3x^2}{\ddots\lower{6pt}{2n(2n+1)-x^2+\cfrac{2n(2n+1)x^2}{P_n(x)/P_{n+1}(x)}}}}}\tag{4}$$

I learned how to do this from this document (look for Theorem I)