# Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$ [duplicate]

Determine the Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$.

$f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}} = \arcsin(x)\frac{1}{\sqrt{1-x^2}}$

It is known:

(1.) $\arcsin(x) = \sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$

(2.) $\frac{d}{dx} (\arcsin(x)) = \frac{1}{\sqrt{1-x^2}} = \frac{d}{dx}(\sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}) = \sum_{n=0}^\infty\frac{(2n+1)!!x^{2n+2}}{2^{n+1}(n+1)!}$

(3.) In third step I multiplied these two series, but I am not sure whether it is correct: $\arcsin(x)\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m+1)!!x^{2m+2}}{2^{m+1}(m+1)!})\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$

EDIT: How did you get this sum $\sum_{n=0}^{\infty}\frac{4^n (n!)^2}{(2n+1)!}x^{2n+1}$ ?

And, in case I have to determine the product of the (some other) series, which one of these should I write?

(a) $\sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m+1)!!x^{2m+2}}{2^{m+1}(m+1)!})\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$

or

(b) $= \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m-1)!!x^{2m+1}}{2^mm!(2m+1)})\frac{(2n+1)!!x^{2n+2}}{2^{n+1}(n+1)!}$

We have, by the extended binomial theorem: $$\frac{1}{\sqrt{1-x^2}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n} x^{2n},\tag{1}$$ hence by integrating termwise: $$\arcsin x=\sum_{n\geq 0}\frac{1}{(2n+1)4^n}\binom{2n}{n} x^{2n+1}=x\cdot\phantom{}_2 F_1\left(\frac{1}{2},\frac{1}{2};\frac{3}{2};z^2\right)\tag{2}$$

Now we may notice that: $$\sum_{a+b=N}\frac{1}{(2a+1)4^a}\binom{2a}{a}\frac{1}{(2b+1)4^b}\binom{2b}{b} = \frac{4^N}{(N+1)(2N+1)\binom{2N}{N}}\tag{3}$$ We may find the Taylor series of $\arcsin^2(z)$ by exploiting the Catalan's convolution formula, the hypergeometric identity: $$\phantom{}_2 F_1\left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};z\right)^2 = \phantom{}_3 F_2\left(a,b,\frac{a+b}{2};\frac{a+b+1}{2},a+b;z\right)\tag{4}$$ the Lagrange's inversion theorem or the rather simple technique shown in this answer. So we have: $$\arcsin^2(x) = \sum_{n\geq 0}\frac{4^n}{(n+1)(2n+1)\binom{2n}{n}}x^{2n+2}\tag{5}$$ and by differentiating $(5)$:

$$\frac{\arcsin x}{\sqrt{1-x^2}} = \sum_{n\geq 0}\frac{4^n}{(2n+1)\binom{2n}{n}}x^{2n+1}. \tag{6}$$

Do not multiply, tackle it directly by looking at the Taylor recipe...

$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}\frac{4^n (n!)^2}{(2n+1)!}x^{2n+1}$$
I would substitute $x = \sin(x)$ and simplify to $1/\cos(x)$ and proceed from there.
• So, how do you get the Taylor series of $\arcsin^{10}(x)$ in this way? Jun 24 '15 at 16:19
• $\arcsin^{10}(x) = \arcsin^{10}(sin(t)) = t^{10}$, where $x = \sin(t)$ or $t = \arcsin(x)$, then the taylor expansion of the expression can be found as the tenth monomial of the taylor expansion of $\arcsin(x)$. Jun 24 '15 at 16:29
• I do not get it. You have some function, $f(x)$, for which you have to compute the Taylor series. You compute the Taylor series of another function, $g(x)=f(\sin x)$. And so, how do you recover the Taylor series of $f(x)$? Please do that with $f(x)=\arcsin(x)^{10}$. What is the coefficient of $x^{20}$, just to say? Jun 24 '15 at 16:32
• The trick is to use substitution. Say that $\arcsin(x) = a_0+a_1x + a_2x^2 + \cdots$, then we can calculate $(a_0+a_1x+a_2x^2)^{10}$. If you need as high a polynomial as 20th degree, you probably are using the wrong approximation anyway and should be switching basis to something more suitable. Jun 24 '15 at 16:39
• That still doesn't answer the question. Given the Taylor series of $\arcsin(x)$, how do you find the coefficient of $x^n$ in the Taylor series of $\arcsin(x)^2$ ? By squaring, ok, I get it, but what is the final outcome? Jun 24 '15 at 16:47