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I have been playing with Taylor and Maclaurin series lately and stumble on this beautiful identity. I don't know to expand the left hand side to yield the right hand side: How to prove: $\dfrac{1}{\sqrt{1-x^2}} =1+\dfrac{1}{2}x^2+\dfrac{1 \cdot3}{2\cdot4}x^4+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^6+\dfrac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}x^8...$ I can only expand this as followed: $$\frac{1}{\sqrt{1-x^2}} = 1+\frac{1}{2}x^2+\frac{3}{8}x^4+\frac{5}{16}x^6...,$$

How can you prove this by using Maclaurin series? I need two proofs, one in Maclaurin series and one in binomial theorem. Please don't use the sigma notation too much as I cannot see the pattern.

James Warthington
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  • Just a thought, you might be able to use the fact that the LHS of your equation is the derivative of arcsin(x)... – JG123 Nov 03 '19 at 00:00
  • @JG123 How do I get the beautiful coefficient this way? – James Warthington Nov 03 '19 at 00:02
  • You want people to give you two proofs? – Calvin Khor Nov 03 '19 at 00:06
  • @Calvin Khor, I just need to see how can I expand the first series by using binomial series and Maclaurin series. I am new to this so I don't know how to. – James Warthington Nov 03 '19 at 00:08
  • Well if you just want proof, just replace the LHS of the equation with d(arcsin(x))/dx and integrate both sides. See if the RHS of the resultant equation matches the Maclaurin series of arcsin(x). There is probably a more elegant way to prove your identity but this method came to me first. – JG123 Nov 03 '19 at 00:09
  • That gives me a different series, namely $\arcsin(x)= x+\frac{1}{6}x^3+\frac{3}{40}x^5+\frac{5}{122}x^7...$ – James Warthington Nov 03 '19 at 00:25
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    Can you simplify $\binom{-1/2}{n}$? Also, your comment about sigma notation strikes me as odd, since the whole point of sigma notation is to say what the pattern is (with formulas). – anon Nov 03 '19 at 00:48

3 Answers3

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Very simple: expand $$\frac1{\sqrt{1-u}}=(1-u)^{-\tfrac12}$$ with the binomial series, and substitute $u=x^2$.

Note the general term of the binomial series is $$(-1)^n\frac12\cdot\frac32\dotsm\frac{2n-1}2\,\frac{(-u)^n}{n!}= \frac{1\cdot 3\dotsm(2n-1)}{2^n n!}\,u^n=\frac{(2n-1)!!}{(2n)!!}\,u^{n}.$$ From the final formula, you can deduce instantly the Taylor series for $\arcsin x$.

Bernard
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  • That is the binomial coefficient $\binom{-1/2}{n}$ – GEdgar Nov 03 '19 at 00:11
  • If you wish, but I prefer an explicit form. – Bernard Nov 03 '19 at 00:13
  • @Bernard can you do this in more steps? I have been using the binomial series and I can only expand it as the second series instead of the first one. – James Warthington Nov 03 '19 at 00:15
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    I don't see very well what to add – except perhaps, that the factor in the general term comes from $$-\frac12\Bigl(-\frac12-1\Bigr)\Bigl(-\frac12-2\Bigr)$$ and so on. Also you have to dispatch the powers of $2$ in the denominator among the factors of $n!$ to obtain $(2n)!!$. – Bernard Nov 03 '19 at 00:18
  • @Bernard The general binomial series is $(1+x)^k=1+kx+\frac{k(k-1)}{2!}x^2+\frac{k(k-1)(k-2)}{3!}+\frac{k(k-1)(k-2)(k-3)}{4!}...$ – James Warthington Nov 03 '19 at 00:32
  • How can I use this to obtain the first series? – James Warthington Nov 03 '19 at 00:33
  • It's this general formula that I appled, taking into account the denominators were systematically $2$. To obtain the first series, just substitute $x^2$ to $u$. – Bernard Nov 03 '19 at 00:42
  • @Bernard, can you do this in more details, I used the formula and get the series 2, not 1. – James Warthington Nov 03 '19 at 00:56
  • $(1-u)^{-\frac{1}{2}}=1+\frac{1}{2}u+\frac{-\frac{1}{2}(-\frac{1}{2}-1)u^2}{2!}+\frac{-\frac{1}{2}(-\frac{1}{2}-1)(-\frac{1}{2}-2)}{3!}...=1+\frac{1}{2}u+\frac{-\frac{1}{2}(-\frac{1}{2}-1)}{2!}u^2+\frac{-\frac{1}{2}(-\frac{1}{2}-1)(-\frac{1}{2}-2)}{3!}u^3$ – James Warthington Nov 03 '19 at 00:58
  • I don't see how this leads up to series 1 – James Warthington Nov 03 '19 at 00:59
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I have finally been able to derive the series after some manipulations:

The general formula for binomial series is: $(1+x)^k=1+kx+\dfrac{k(k-1)}{2!}x^2+\dfrac{k(k-1)(k-2)}{3!}x^3+\dfrac{k(k-1)(k-2)(k-3)}{4!}x^4...$

$ \dfrac{1}{\sqrt{1-x^2}}=(1-x^2)^{-\frac{1}{2}}=1+(-\dfrac{1}{2})(-x^2)+\dfrac{(-\dfrac{1}{2})(-\dfrac{1}{2}-1)}{2!}(-x^2)^2+\dfrac{(-\dfrac{1}{2})(-\dfrac{1}{2}-1)(-\dfrac{1}{2}-2)}{3!}(-x^2)^3+\dfrac{(-\dfrac{1}{2})(-\dfrac{1}{2}-1)(-\dfrac{1}{2}-2)(-\dfrac{1}{2}-3)}{4!}(-x^2)^4...$

$=1+\dfrac{1}{2}x^2+\dfrac{(-\dfrac{1}{2})(-\dfrac{3}{2})}{2!}(x^4)+\dfrac{(-\dfrac{1}{2})(-\dfrac{3}{2})(-\dfrac{5}{2})}{3!}(-x^6)+\dfrac{(-\dfrac{1}{2})(-\dfrac{3}{2})(-\dfrac{5}{2})(-\dfrac{7}{2})}{4!}(x^8)...$

$=1+\dfrac{1}{2}x^2+\dfrac{1 \cdot3}{2^2\cdot2!}x^4+\dfrac{1\cdot 3\cdot 5}{2^3\cdot 3!}x^6+\dfrac{1\cdot 3\cdot 5\cdot 7}{2^4\cdot 4!}x^8...$

$=1+\dfrac{1}{2}x^2+\dfrac{1 \cdot3}{2\cdot4}x^4+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^6+\dfrac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}x^8...$

$$=\sum_{n=0}^{\infty} \dfrac{(2n-1!!)}{2^nn!}x^{2n}$$

James Warthington
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At the site https://math.stackexchange.com/a/4657751, it was established that \begin{equation*} \biggl[\frac{1}{\sqrt{1-x^2}\,}\biggr]^{(k)} =\frac1{\sqrt{1-x^2}\,}\frac{k!}{2^k} \frac1{x^k}\sum_{j=0}^{k}\frac{2^{j}(2j-1)!!}{j!}\binom{j}{k-j}\frac{x^{2j}}{(1-x^2)^{j}}, \quad k\ge0. \end{equation*} Therefore, we derive \begin{align*} \lim_{x\to0}\biggl[\frac{1}{\sqrt{1-x^2}\,}\biggr]^{(k)} &=\frac{k!}{2^k} \lim_{x\to0}\sum_{j=0}^{k}\frac{2^{j}(2j-1)!!}{j!}\binom{j}{k-j}\frac{x^{2j-k}}{(1-x^2)^{j}}\\ &=\begin{cases} 0, & k=2m+1\\ (2m)! \dfrac{(2m-1)!!}{(2m)!!}, & k=2m \end{cases} \end{align*} for $k,m\ge0$. Consequently, we acquire \begin{equation*} \frac{1}{\sqrt{1-x^2}\,} =\sum_{m=0}^\infty \dfrac{(2m-1)!!}{(2m)!!}x^{2m}, \quad |x|<1. \end{equation*}

qifeng618
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