Calculus Infinite Series Homework Problem.

Question

I have the problem "Let $f(x) = x + \frac{2}{3} x^3 + \frac{2 \cdot 4}{3 \cdot 5} x^5 + \dots + \frac{2 \cdot 4 \dotsm 2n}{3 \cdot 5 \dotsm (2n + 1)} x^{2n + 1} + \dotsb$ on the interval $(-1,1)$ of convergence of the defining series. (a) Prove that $(1 - x^2) f'(x) = 1 + xf(x).$ (b) Prove that $f(x) = \frac{\arcsin x}{\sqrt{1 - x^2}}.$" I don't exactly know how to prove it without doing part b first, then part a. Would prefer if someone could prove it using the infinite series. If anyone can help me with proving part a, then part b, that'd be greatly appreciated.

Answer 1

If you are given a power series $f(x) = \sum_{n = 0}^\infty a_n (x - x_0)^n$ with radius of convergence $R$, then $f$ is differentiable on $(x_0 - R, x_0 + R)$ and the derivative is $f'(x) = \sum_{n = 1}^\infty n a_n (x - x_0)^{n - 1}$.
In other words, you can differentiate term-by-term.

I shall leave it to you to verify that your series does converge on $(-1, 1)$. (For example, use an appropriate version of the root test.)

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Recall the notion of a double factorial: We have $$(2n)!! = 2 \cdot 4 \cdot \cdots \cdot (2n)$$ and $$(2n + 1)!! = 1 \cdot 3 \cdot \cdots \cdot (2n + 1).$$

By convention, we have $(-1)!! = 0!! = 1$.

In particular, for $n \ge 1$, we have the identity $\frac{n!!}{n} = (n - 2)!!$.

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Now, in your case, we have

$$f(x) = \sum_{n = 0}^\infty \frac{(2n)!!}{(2n + 1)!!}x^{2n + 1}.$$

and thus, \begin{align} f'(x) &= \sum_{n = 0}^\infty \frac{(2n)!!}{(2n + 1)!!} \cdot (2n + 1) \cdot x^{2n} \\ &= 1 + \sum_{n = 1}^\infty \frac{(2n)!!}{(2n - 1)!!} x^{2n}. \end{align}

Thus, we get \begin{align} x^2f'(x) &= x^2 + \sum_{n = 1}^\infty \frac{(2n)!!}{(2n - 1)!!} x^{2n + 2} \\ &= x^2 + \sum_{n = 2}^{\infty} \frac{(2n - 2)!!}{(2n - 3)!!} x^{2n} \\ &=\sum_{n = 1}^{\infty} \frac{(2n - 2)!!}{(2n - 3)!!} x^{2n}. \end{align}

Subtracting from the earlier expression, we get \begin{align} (1 - x^2)f'(x) &= 1 + \sum_{n = 1}^{\infty}\left[\frac{(2n)!!}{(2n - 1)!!} - \frac{(2n - 2)!!}{(2n - 3)!!}\right] x^{2n}. \end{align}

On the other hand, we have the expression \begin{align} 1 + xf(x) &= 1 + \sum_{n = 0}^\infty \frac{(2n)!!}{(2n + 1)!!}x^{2n + 2} \\ &= 1 + \sum_{n = 1}^{\infty} \frac{(2n - 2)!!}{(2n - 1)!!} x^{2n}. \end{align}

I leave it to you to check that $$\frac{(2n)!!}{(2n - 1)!!} - \frac{(2n - 2)!!}{(2n - 3)!!} = \frac{(2n - 2)!!}{(2n - 1)!!}$$ for all $n \ge 1$. Thus, we have shown that $$(1 - x^2)f'(x) = 1 + xf(x).$$

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Now, we wish to conclude that $f(x) = \frac{\arcsin x}{\sqrt{1 - x^2}}$. For this, simply note that $g(x) := \frac{\arcsin x}{\sqrt{1 - x^2}}$ also satisfies the first order ODE

$$y' - \frac{x}{1 - x^2}y = \frac{1}{1 - x^2}$$

on $(-1, 1)$.

The above is a linear first-order ODE with continuous coefficients on $(-1, 1)$. Thus, it has a unique solution which satisfies $y(0) = 0$.
Since $f$ and $g$ both satisfy the ODE with same initial condition, we see that $f = g$. Thus,

$$\sum_{n = 0}^\infty \frac{(2n)!!}{(2n + 1)!!}x^{2n + 1} = \frac{\arcsin x}{\sqrt{1 - x^2}}.$$