Jtdylktuy

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

We will proceed as in this answer.

Define
$$
P_n(x)=sum_{k=0}^inftyfrac{4^{k+n}-sumlimits_{j=1}^nbinom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^ktag{1}
$$
Then
$$
begin{align}
frac{P_{n-1}(x)}{P_n(x)}
&=frac
{displaystylesum_{k=0}^inftyfrac{4^{k+n-1}-sumlimits_{j=1}^{n-1}binom{2k+2n-1}{2j-1}}{(2k+2n-1)!}(-x^2)^k}
{displaystylesum_{k=0}^inftyfrac{4^{k+n}-sumlimits_{j=1}^nbinom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\[12pt]
&=color{#C00000}{-x^2+}frac
{displaystylesum_{k=0}^inftyfrac{color{#C00000}{binom{2k+2n-1}{2n-1}}}{(2k+2n-1)!}(-x^2)^k}
{displaystylesum_{k=0}^inftyfrac{4^{k+n}-sumlimits_{j=1}^nbinom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\[12pt]
&=-x^2+frac
{displaystylesum_{k=0}^inftycolor{#C00000}{frac{2n(2n+1)binom{2k+2n+1}{2n+1}}{(2k+2n+1)!}}(-x^2)^k}
{displaystylesum_{k=0}^inftyfrac{4^{k+n}-sumlimits_{j=1}^nbinom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\[12pt]
&=color{#C00000}{2n(2n+1)}-x^2color{#C00000}{-}frac
{displaystylesum_{k=0}^inftyfrac{2n(2n+1)color{#C00000}{left[4^{k+n}-sumlimits_{j=1}^{n+1}binom{2k+2n+1}{2j-1}right]}}{(2k+2n+1)!}(-x^2)^k}
{displaystylesum_{k=0}^inftyfrac{4^{k+n}-sumlimits_{j=1}^nbinom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\[12pt]
&=2n(2n+1)-x^2color{#C00000}{+2n(2n+1)x^2}frac
{displaystylesum_{k=0}^inftycolor{#C00000}{frac{4^{k+n+1}-sumlimits_{j=1}^{n+1}binom{2k+2n+3}{2j-1}}{(2k+2n+3)!}}(-x^2)^k}
{displaystylesum_{k=0}^inftyfrac{4^{k+n}-sumlimits_{j=1}^nbinom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\[12pt]
&=2n(2n+1)-x^2+2n(2n+1)x^2color{#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)}{2cos(x)}\
&=frac
{displaystyle xsum_{k=0}^inftyfrac{4^k(-x^2)^k}{(2k+1)!}}
{displaystylesum_{k=0}^inftyfrac{(-x^2)^k}{(2k)!}}\
&=xleft/left(frac
{displaystylesum_{k=0}^inftyfrac{(-x^2)^k}{(2k)!}}
{displaystyle sum_{k=0}^inftyfrac{4^k(-x^2)^k}{(2k+1)!}}
right)right.\
&=xleft/left(1+x^2left/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}{2cdot3-x^2+cfrac{2cdot3x^2}{ddotslower{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)
https://people.math.osu.edu/sinnott.1/ReadingClassics/continuedfractions.pdf

Interestingly, this looks to be a translation of Euler's original work. I find him easy to understand, he avoids confusing his message in clumsy notation.