Euler's work II

Corollary to Euler's expansion for $arctan(x)$

Today we show the proof of this result:
\[\frac{\arcsin(w)}{\sqrt{1-w^2} } = \sum_{n=0}^{\infty} \frac{2^{2n} n!^2}{(2n+1)!}w^{2n+1}\]
We have proved it earlier but this time we also prove the Euler's expansion for $\arctan(x)$ on which this proof relies on.

Proof

Recall the famous Euler's formula for $arctan(x)$ (Proof below) which states that:
\[\arctan (x) = \frac{x}{1+x^2} \sum_{n=0}^{\infty}\left[\frac{(2n)!!}{(2n+1)!!} \left(\frac{x^2}{1+x^2}\right)^{n}\right]\]
Now consider the following identity for inverse trigonometric functions:
\[\arcsin(w) = \arctan\left(\frac{w}{\sqrt{1-w^2}}\right)\] Then if $x= \displaystyle \frac{w}{\sqrt{1-w^2}}$ we have \begin{align*} \arcsin(w) =& \arctan\left(\frac{w}{\sqrt{1-w^2}}\right) = w\sqrt{1-w^2}\left[1+\frac{2}{3}w^2 +\frac{2}{3}\frac{4}{5}w^4 + \frac{2}{3}\frac{4}{5}\frac{6}{7}w^6...\right]\\ \Longrightarrow \frac{\arcsin(w)}{\sqrt{1-w^2} }=& w+\frac{2}{3}w^3+\frac{2}{3}\frac{4}{5}w^5+\frac{2}{3}\frac{4}{5}\frac{6}{7}w^7+...\\ =& \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} w ^{2n+1} \end{align*} Now from the definition of double factorial for even integers: \[ (2n)!! = 2^{n} n! \] And for odd integers: \[ (2n+1)!! = \frac{(2n+1)!}{2^n n!} \] Hence \[ \frac{\arcsin(w)}{\sqrt{1-w^2} } = \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} w ^{2n+1} = \frac{2^{2n} n!^2}{(2n+1)!}w^{2n+1}\] Therefore \[\boxed{\frac{\arcsin(w)}{\sqrt{1-w^2} } = \sum_{n=0}^{\infty}\frac{2^{2n} n!^2}{(2n+1)!}w^{2n+1}}\]

Appendix.

But now it is time to prove Euler's formula for $\arctan(x)$. We know that there are fancy proofs for this formula but we want to show a proof more in line with the spirit of Euler. We need this lemma:

Proposition If $\displaystyle (1+x^2)(1-y^2) =1$ then
\[\sum_{k=0}^{\infty} a_{2k+1} x^{2k+1} = \sqrt{1-y^2} \sum_{n=0}^{\infty} \Delta^n a_{0} y^{2n+1} \]
where \[\Delta^n a_{0} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} a_{2k+1} \]
Proof
If $\displaystyle (1+x^2)(1-y^2)=1$ then $\displaystyle x^2=\frac{y^2}{1-y^2}$
Hence
\begin{align*} \sum_{k=0}^{\infty} (-1)^k a_{2k+1} x^{2k+1} = &\sum_{n=0}^{\infty} (-1)^k a_{2k+1} x^{2k}\cdot x\\ =& \sum_{k=0}^{\infty} (-1)^k a_{2k+1} \left(\frac{y^2}{1-y^2}\right)^{k}\frac{y}{\sqrt{1-y^2}}\\ =& \sum_{k=0}^{\infty} (-1)^ka_{2k+1} \frac{y^{2k}}{(1-y^2)^{k+1}}\frac{y(1-y^2)}{\sqrt{1-y^2}}\\ =& \sum_{k=0}^{\infty} (-1)^ka_{2k+1} \frac{y^{2k}}{(1-y^2)^{k+1}} y\sqrt{1-y^2}\\ =& \sum_{k=0}^{\infty} (-1)^ka_{2k+1} \left[\sum_{n=0}^{\infty}\binom{n}{k} y^{2n}\right] y\sqrt{1-y^2}\\ =& \sum_{n=0}^{\infty}\left[\sum_{k=0}^{\infty} \binom{n}{k} (-1)^ka_{2k+1} \right] y^{2n+1} \sqrt{1-y^2}\\ =& \sum_{n=0}^{\infty}\Delta^n a_{0} y^{2n+1} \sqrt{1-y^2}\\ \end{align*} Hence \[\boxed{\sum_{k=0}^{\infty} a_{2k+1} x^{2k+1} = \sqrt{1-y^2} \sum_{n=0}^{\infty} \Delta^n a_{0} y^{2n+1} \quad \textrm{ if } (1+x^2)(1-y^2)=1 }\]

Now, we will apply the Lemma to $arctan(x)$

Recall the series expansion for $arctan(x) $
\[ \arctan(x) = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{2k+1} \] Therefore \[\displaystyle a_{2k+1} = \frac{1}{2k+1}, \quad \Delta^n a_{0} = \sum_{k=0}^{n} (-1)^k \binom{n}{k}\frac{1}{2k+1}\] Hence \begin{align*} \arctan(x) = & \sum_{j=0}^{\infty} \frac{(-1)^k x^{2k+1}}{2k+1}\\ =& \sqrt{1-y^2} \sum_{n=0}^{\infty} \left[\sum_{k=0}^{n} (-1)^k \binom{n}{k}\frac{1}{2k+1}\right] y^{2n+1} \\ =&\sqrt{1-y^2} \sum_{n=0}^{\infty} \left[\sum_{k=0}^{n} (-1)^k \binom{n}{k} \int_{0}^{1} t^{2k} dt \right] y^{2n+1}\\ =&\sqrt{1-y^2} \sum_{n=0}^{\infty} \left[ \int_{0}^{1}\sum_{k=0}^{n} (-1)^k \binom{n}{k} t^{2k} dt \right] y^{2n+1}\\ =&\sqrt{1-y^2} \sum_{n=0}^{\infty} \left[ \int_{0}^{1} (1-t^2)^n dt \right] y^{2n+1}\\ =&\sqrt{1-y^2} \sum_{n=0}^{\infty} \left[\frac{1}{2} \int_{0}^{1} (1-w)^n w^{-\frac{1}{2}} dt \right] y^{2n+1}\\ =&\sqrt{1-y^2} \sum_{n=0}^{\infty} \left[\frac{1}{2} B\left(n+1,\frac{1}{2}\right) \right] y^{2n+1}\\ =&\sqrt{1-y^2} \sum_{n=0}^{\infty} \left[\frac{1}{2} \frac{\sqrt{\pi}{n!}}{\Gamma \left(n+\frac{3}{2}\right)} \right] y^{2n+1}\\ \end{align*} From Legendre duplication formula we can write: \[\Gamma\left(n+\frac{3}{2}\right) = \frac{\Gamma(2n+2)\sqrt{\pi}}{2^{2n+1} \Gamma(n+1) } = \frac{(2n+1)! \sqrt{\pi}}{ 2^{2n+1} n!} \] Then \[ \arctan(x) = \sqrt{1-y^2} \sum_{n=0}^{\infty} \left[ \frac{2^{2n} n!^2}{(2n+1)!} \right] y^{2n+1} = \sqrt{1-y^2} \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} y^{2n+1}\] Given that $\displaystyle y=\frac{x}{\sqrt{1+x^2}}$ Therefore \[\boxed{\arctan (x) = \frac{x}{1+x^2} \left[\frac{(2n)!!}{(2n+1)!!} \left(\frac{x^2}{1+x^2}\right)^n\right]}\]

This type of proof is called Euler's transform