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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.

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  • $\begingroup$ Just a thought, you might be able to use the fact that the LHS of your equation is the derivative of arcsin(x)... $\endgroup$ – JG123 Nov 3 '19 at 0:00
  • $\begingroup$ @JG123 How do I get the beautiful coefficient this way? $\endgroup$ – James Warthington Nov 3 '19 at 0:02
  • $\begingroup$ You want people to give you two proofs? $\endgroup$ – Calvin Khor Nov 3 '19 at 0:06
  • $\begingroup$ @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. $\endgroup$ – James Warthington Nov 3 '19 at 0:08
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    $\begingroup$ 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). $\endgroup$ – runway44 Nov 3 '19 at 0:48
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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 $\arcsin x$.

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  • $\begingroup$ That is the binomial coefficient $\binom{-1/2}{n}$ $\endgroup$ – GEdgar Nov 3 '19 at 0:11
  • $\begingroup$ If you wish, but I prefer an explicit form. $\endgroup$ – Bernard Nov 3 '19 at 0:13
  • $\begingroup$ @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. $\endgroup$ – James Warthington Nov 3 '19 at 0:15
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    $\begingroup$ 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 $(2n)!!$. $\endgroup$ – Bernard Nov 3 '19 at 0:18
  • $\begingroup$ @Bernard The general binomial series is $(1+x)^k=1+kx+\frac{k(k-1)}{2!}x^2+\frac{k(k-1)(k-2)}{3!}+\frac{k(k-1)(k-2)(k-3)}{4!}...$ $\endgroup$ – James Warthington Nov 3 '19 at 0:32
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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...$

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

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

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

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

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