Expand in Taylor series $\frac{1}{1-\sin{x}}$
I have an idea that $\frac{1}{1-\sin{x}} = 1 + \sin {x} + \sin^2 {x} + \sin^3 {x} + \dots$
But I don't know what to do next. Every sine expands in infinity series...
Can anybody help?
Expand in Taylor series $\frac{1}{1-\sin{x}}$
I have an idea that $\frac{1}{1-\sin{x}} = 1 + \sin {x} + \sin^2 {x} + \sin^3 {x} + \dots$
But I don't know what to do next. Every sine expands in infinity series...
Can anybody help?
$$\sin^n(x)=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^n=\frac1{(2i)^n}\sum_{k=0}^n\binom nke^{i(2k-n)x}=\\ \frac1{(2i)^n}\sum_{k=0}^n\binom nk\sum_{j=0}^\infty\frac{i^j(2k-n)^j}{j!}x^j,$$
and
$$\frac1{1-\sin(x)}=\sum_{n=0}^\infty\frac1{(2i)^n}\sum_{k=0}^n\binom nk\sum_{j=0}^\infty\frac{i^j(2k-n)^j}{j!}x^j.$$
Not the simplest on Earth, but usable for the first few terms.
If you're looking for a closed form, see OEIS sequence A099612.
The coefficient of $x^n$ is
$$ \dfrac{(\cos(\pi n/2) - \sin(\pi n/2))\; (4^{n+2}-2^{n+2}) }{n!} \left( \dfrac{\zeta(-n-1,3/4) - \zeta(-n-1,1/4)}{2^{-n-1}-2} - \zeta(-n-1)\right)$$ where the two-argument $\zeta$ is the Hurwitz zeta function.
But if this is a homework exercise, it's likely just asking you to compute the first few terms.
You can use Maclaurin's expansion to expand in powers of x as follows:
where $f(x)=1/(1-sinx )$