# How to prove this beautiful series by using Taylor and Maclaurin series

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.

• 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 3 '19 at 0:00
• @JG123 How do I get the beautiful coefficient this way? – James Warthington Nov 3 '19 at 0:02
• You want people to give you two proofs? – Calvin Khor Nov 3 '19 at 0:06
• @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 3 '19 at 0:08
• 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). – runway44 Nov 3 '19 at 0:48

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 $$\arcsin x$$.

• That is the binomial coefficient $\binom{-1/2}{n}$ – GEdgar Nov 3 '19 at 0:11
• If you wish, but I prefer an explicit form. – Bernard Nov 3 '19 at 0:13
• @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 3 '19 at 0:15
• 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 3 '19 at 0:18
• @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 3 '19 at 0:32

I have finally been able to derive the series after some manipulations:

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

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

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

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

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

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