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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)!}$

purgerica
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3 Answers3

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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 $(5)$:

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

Community
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Jack D'Aurizio
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Do not multiply, tackle it directly by looking at the Taylor recipe...

Just a short answer for your check:

$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}\frac{4^n (n!)^2}{(2n+1)!}x^{2n+1}$$

Math-fun
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  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – muaddib Jun 24 '15 at 17:43
  • I thought this is easier and somewhat funnier than multiplying all the terms. Many thanks anyhow. Of course I respect Jack and enjoyed his approach. – Math-fun Jun 25 '15 at 08:02
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I would substitute $x = \sin(x)$ and simplify to $1/\cos(x)$ and proceed from there.

mathreadler
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  • So, how do you get the Taylor series of $\arcsin^{10}(x)$ in this way? – Jack D'Aurizio 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 $\arcsin(x)$. – mathreadler 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? – Jack D'Aurizio 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. – mathreadler 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? – Jack D'Aurizio Jun 24 '15 at 16:47
  • You can evaluate the expression in your own favourite way. One kind of fast way would to split it into evaluating powers 2,4 and 8 by successive squaring and then multiplying the result from 8 and 2. But these are all itty bitty practical details. – mathreadler Jun 24 '15 at 16:55
  • So, after all that nasty manipulations, you get that the coefficient of $x^n$ in the Taylor series of $\arcsin(x)^2$ is... what? – Jack D'Aurizio Jun 24 '15 at 17:00
  • Ah you mean algebraically? I'm so used to doing this numerically that I did not even think of it. – mathreadler Jun 24 '15 at 17:02
  • I do not think that the OP was requesting a numerical approximation of the Taylor series of $\frac{\arcsin x}{\sqrt{1-x^2}}$, but the Taylor series in its closed form. Yes, the point is to prove that the coefficient of $x^n$ has a nice closed form. – Jack D'Aurizio Jun 24 '15 at 17:04