2 Zigmas Apart

By expanding $\arcsin x$ into Taylor Series , we will obtain:

$\arcsin x = x + \dfrac{1}{2}(\dfrac{x^3}{3}) + \dfrac{1\cdot 3}{2\cdot 4}(\dfrac{x^5}{5}) + \dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}(\dfrac{x^7}{7}) + \cdots$

Taking derivative of both sides, we will get:

$\dfrac{\mathrm{d}\arcsin x}{\mathrm{d}x} = \dfrac{1}{\sqrt{1-x^2}} = 1 + \dfrac{1}{2}(x^2) + \dfrac{1\cdot 3}{2\cdot 4}(x^4) + \dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}(x^6) + \cdots = \displaystyle \sum_{n=0}^{\infty} \dfrac{(2n-1)!!}{(2n)!!}x^{2n}$

If we substitute $y= x^2$ , we will get:

$\dfrac{1}{\sqrt{1-y}} = \displaystyle \sum_{n=0}^{\infty} \dfrac{(2n-1)!!}{(2n)!!}y^n$

And from the geometric series summation, we have a generating function of:

$\displaystyle \sum_{n=0}^{\infty} y^n = \dfrac{1}{1-y}$

If we let $S = \dfrac{1}{\sqrt{1-y}}$ , we can come up with a quadratic equation of:

$S^2 = 2 + S$

$S^2 - S - 2 = (S+1)(S-2) = 0$ .

Thus, $S = 2$ and $S^2 = 4$ .

Then solving for $S^2 = 4 = \dfrac{1}{1-y}$ , we will get $y=\dfrac{3}{4} = 0.75$ .

Thus, $\boxed{0.75}$ is the solution.

Using the Generalised Binomial Theorem. we have $(1 - x)^{-\frac12} \; = \; \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!} x^n \hspace{2cm} |x| < 1$ so we need to find $0 < x < 1$ such that $(1 - x)^{-1} \; = \; 2 + (1-x)^{-\frac12}$ which leads to $x=\boxed{\tfrac34}$ .

Since it is putting $x=1$ , the first line is a divergent series, and so is not really relevant to what follows.

Another way to derive the generating function

$Let\quad { a }_{ n }\quad =\quad \frac { (2n-1)!! }{ (2n)!! } \Rightarrow \quad { a }_{ n+1 }=\frac { 2n+1 }{ 2n+2 } { a }_{ n }\\ \therefore \quad \quad (2n+2){ a }_{ n+1 }{ x }^{ n }=(2n+1){ a }_{ n }{ x }^{ n }\Rightarrow \quad \quad \sum _{ n=0 }^{ \infty }{ (2n+2){ a }_{ n+1 }{ x }^{ n } } =\quad \sum _{ n=0 }^{ \infty }{ (2n+1){ a }_{ n }{ x }^{ n } } \\ Define\quad y=f\left( x \right) =\sum _{ n=0 }^{ \infty }{ { a }_{ n } } { x }^{ n }\\ \therefore \quad 2y'=\quad 2xy'+y\Rightarrow \quad \int _{ 0 }^{ y }{ \frac { dy }{ y } } =\quad \int _{ 0 }^{ x }{ \frac { dx }{ 2(1-x) } } \Rightarrow y=\frac { 1 }{ \sqrt { 1-x } } \\ \\$