4
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ en.wikipedia.org/wiki/Formal_power_series#Composition_of_series $\endgroup$
    – vadim123
    Oct 26, 2015 at 17:07
  • $\begingroup$ Have you tried doing it from first principles? As in actually evaluating the derivatives and seeing if there's a pattern. $\endgroup$
    – kbau
    Oct 26, 2015 at 17:08
  • $\begingroup$ @vadim123 It's also seems difficult to calculate explicit coefficients... $\endgroup$ Oct 26, 2015 at 17:16
  • $\begingroup$ @kbau I tried $ 1 + x + x^2 + \frac{5}{6}x^3 + \frac{2}{3}x^4 + \frac{61}{120}x^5 + \frac{17}{45}x^6 + \dots$ and I can't see a pattern $\endgroup$ Oct 26, 2015 at 17:19
  • 1
    $\begingroup$ Note that $\frac{1}{1-\sin x}=\sec^2x + \sec x \tan x$, by multiplying by the conjugate. Both $\sec x$ and $\tan x$ are standard, if not too "nice", series, see wiki, so you can multiply twice and then add. $\endgroup$
    – vadim123
    Oct 26, 2015 at 17:32

3 Answers 3

3
$\begingroup$

$$\sin^n(x)=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^n=\frac1{(2i)^n}\sum_{k=0}^n\binom nke^{i(2k-n)x}=\\ \frac1{(2i)^n}\sum_{k=0}^n\binom nk\sum_{j=0}^\infty\frac{i^j(2k-n)^j}{j!}x^j,$$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

You can use Maclaurin's expansion to expand in powers of x as follows:

  • $f(x)$=$f(0)$+$xf'(0)$+$(x^2$/$2!$)$f''(0)$+...

where $f(x)=1/(1-sinx )$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .