Pi and the AGM: Exercises 1.2

Exercise 1.

Proof of (1.2.5). The given functional equation yields the following power series equation: $$\sum_{k=0}^\infty d_{\lfloor k/2\rfloor}x^k = \sum_{k=0}^\infty d_k \bigg(\frac{4x}{(1+x)^2}\bigg)^k$$ We may expand $4x/(1+x)^2$ in each term of the RHS and apply the following power series formula: $$\frac{1}{(1-y)^m} = \sum_{k=0}^\infty \binom{m+k}{k} y^k$$ and reorder the summation to obtain the following formula: $$\sum_{k=0}^\infty d_{\lfloor k/2\rfloor}x^k = \sum_{n=0}^\infty x^n \sum_{k=0}^n d_k 4^{k} \binom{n+k}{n-k}(-1)^{n-k}$$ which yields the following recurrence, by equating coefficients: $$d_{\lfloor n/2\rfloor} = \sum_{k=0}^n d_k 4^{k} \binom{n+k}{n-k}(-1)^{n-k}$$ This can be used to solve for $d_n$ in terms of previous terms of $(d_n)$. But there's no way in hell I'm going to slog through solving this recurrence explicitly - not when the proof of the desired equality given by (1.2.10) is just as valid. Good on Gauss though, I guess.

Proof of (1.2.7). Recall the following Maclaurin expansion: $$\frac{1}{\sqrt{1-x}} = \sum_{n=0}^\infty \binom{2n}{n}\frac{x^n}{4^n}$$ which yields the following series for the integrand: $$\frac{1}{\sqrt{1-x^2\sin^2\vartheta}} = \sum_{n=0}^\infty \binom{2n}{n}\frac{x^{2n}}{4^n}\cdot \sin^{2n}\vartheta$$ Thus, exchanging sum and integral, we get the following: $$K(x) = \sum_{n=0}^\infty \binom{2n}{n}\frac{x^{2n}}{4^n}\cdot \int_0^{\pi/2}\sin^{2n}(x) ~ dx$$ Note that regardless of the value of $x\in [-1,1]$, the integrand of the integral in each term of the sum is strictly positive everywhere on the interval of integration. This fact allows us to justify the exchange of summation and integral without any problems. Finally, the integral in term $n$ can be integrated as follows, either using integration by parts to establish a recurrence, or by using the Beta function: $$\int_{0}^{\pi/2}\sin^{2n}(x) ~ dx = \frac{\pi}{2^{2n+1}}\binom{2n}{n}$$ Hence: $$K(x) = \frac{\pi}{2}\sum_{n=0}^\infty \binom{2n}{n}^2\frac{x^{2n}}{4^{2n}}$$ which is equivalent to the desired expansion.

Proof of (1.2.10). Let's perform the suggested substitution $u= \tfrac{1}{2}(t-ab/t)$ in the integral. Note that as $t$ ranges over $\mathbb R$, $u$ ranges over $\mathbb R$ twice, so for this integral, we have $$\int_\mathbb R f(u(t)) dt = 2\int_\mathbb R f(u) du$$ Now let's work on rearranging the integrand. Note that with this substitution, we have that $$du = \frac{1}{2}\Big(1 + \frac{ab}{t^2}\Big)dt$$ and hence $$\begin{align*}\tfrac{\pi}{2}T(a,b) &= \int_\mathbb R \frac{dt}{\sqrt{(a^2+t^2)(b^2+t^2)}} \\ &= \int_\mathbb R \frac{1}{\sqrt{t^2 + (a^2 + b^2) + \frac{a^2 b^2}{t^2}}}\frac{dt}{t} \\ &= \int_\mathbb R \frac{1}{\sqrt{t^2 + (a^2 + b^2) + \frac{a^2 b^2}{t^2}}} \cdot \frac{2}{t + \frac{ab}{t}} \cdot\frac{du}{dt} dt \\ &= \int_\mathbb R \frac{1}{\sqrt{t^2 + (a^2 + b^2) + \frac{a^2 b^2}{t^2}}} \cdot \frac{2}{\sqrt{t^2 + 2ab + \frac{a^2b^2}{t^2}}} \cdot\frac{du}{dt} dt \\ &= \int_\mathbb R \frac{2}{\sqrt{(4u^2 + (a+b)^2)(4u^2 + 4ab)}}\cdot \frac{du}{dt} dt \\ &= \frac{1}{2}\int_\mathbb R \frac{1}{\sqrt{(u^2 + (\tfrac{a+b}{2})^2)(u^2 + (\sqrt{ab})^2)}}\cdot \frac{du}{dt} dt \\ &= \int_\mathbb R \frac{du}{\sqrt{(u^2 + (\tfrac{a+b}{2})^2)(u^2 + (\sqrt{ab})^2)}} \\ &= \tfrac{\pi}{2}T\big(\tfrac{a+b}{2}, \sqrt{ab}\big)\end{align*}$$ which establishes the desired equality.

Exercise 2. Deriving the differential equation for $f(x) = M(1-x,1+x)^{-1}$ is straightforward given the formula for the coefficients $(d_n)$ established in the first exercise: $$d_n = \binom{2n-1}{n}^2 \frac{1}{4^{2n-1}}$$ Note that this implies the following recurrence: $$(2n+2)^2 d_{n+1} = (2n+1)^2 d_n$$ If we let $\mathcal D$ be the derivative operator and $X$ denote the operator consisting of multiplication by $x$, then note that $(X\mathcal D)^2$ transforms the coefficients in the series for $f$ into the LHS of the above, and $X^2 (DX)^2$ transforms them into the RHS, that is:

$$\begin{align*}(X\mathcal D)^2f(x) &= \sum_{n=0}^\infty (2n+2)^2 x^{2n+2}d_{n+1} \\ X^2 (DX)^2f(x) &= \sum_{n=0}^\infty (2n+1)^2 x^{2n+2}d_n\end{align*}$$

if we take the convention that $d_0 = 1$. This implies that the following operator annihilates $f$: $$X^2 (DX)^2 - (X\mathcal D)^2 = (X^4 - X^2)\mathcal D^2 + (3X^3-X)\mathcal D + X^2$$ Which implies the differential equation we were seeking: $$(x^3 -x) f'' + (3x^2-1)f + xf = 0$$

It remains to be proven that the same differential equation applies to $M(1,x)^{-1}$. We know from (1.2.6) that $$\frac{1}{M(1,x)} = \frac{1}{M(1+\sqrt{1-x^2}, 1-\sqrt{1-x^2})}$$ hence, if we can show that $f(\sqrt{1-x^2})$ is also a solution to the above differential equation when $f$ is, we will be done. Let's start by expressing the derivatives of $f(x) = g(\sqrt{1-x^2})$ in terms of the derivatives of $g$. We have: $$\begin{align*}f(x) &= g(\sqrt{1-x^2}) \\ f'(x) &= -\frac{x}{\sqrt{1-x^2}}\cdot g'(\sqrt{1-x^2}) \\ f''(x) &= \frac{1}{(1-x^2)^{3/2}}\cdot g'(\sqrt{1-x^2}) + \frac{x^2}{1-x^2}\cdot g''(\sqrt{1-x^2})\end{align*}$$ If we substitute these expressions for $f'',f',f$ in the differential equation for $f$, we get the following: $$-x^3 g''(\sqrt{1-x^2}) + \frac{2x-3x^2}{\sqrt{1-x^2}}g'(\sqrt{1-x^2}) + xg(\sqrt{1-x^2}) = 0$$ Substituting $x$ for $\sqrt{1-x^2}$ yields: $$-(1-x^2)^{3/2}g''(x) + \frac{(3x^2-1)\sqrt{1-x^2}}{x}g'(x) + \sqrt{1-x^2}g(x) = 0$$ and finally, multiplying this by $x/\sqrt{1-x^2}$ yields the desired differential equation: $$(x^3 - x)g'' + (3x^2 - 1)g' + xg = 0$$ establishing that $M(1,x)^{-1}$ is another solution to the same differential equation as $M(1-x,1+x)^{-1}$.