Exercise 8.3.12

Question

Our work thus far shows that the Taylor series in (5) is valid for all $\left|x\right| < 1$, but note that $\arcsin(x)$ is continuous for all  $\left|x\right| \leq 1$. Carefully explain why the series in (5) converges uniformly to $\arcsin (x)$ on the closed interval $[-1,1]$.

Ulisse Mini · Accepted Answer

If we can show that the power series converges absolutely for $x = 1$, then by Theorem 6.5.2 the series uniformly converges over $[-1, 1]$, and is therefore continuous. Taking limits on both sides approaching $\pm 1$ would lead us to conclude that the equality is valid for $|x| \leq 1$. We now show that $\sum \frac{c_n}{2n+1}$ converges absolutely.

By the Cauchy Condensation Test (Theorem 2.4.6), this series converges if
$$\sum^\infty_{n=0} 2^n \frac{c_{2^n}}{2 \cdot 2^n+1} = \sum^\infty_{n=0}\frac{2^n}{2 \cdot 2^n + 1} c_{2^n} \leq \frac{1}{2} \sum^\infty_{n=0} c_{2^n} $$
converges. We now show that $\sum^\infty_{n=0} c_{2^n}$ converges by comparison against a geometric series.

Let $m = 2^n$; we have
$$\frac{c_{2m}}{c_m}= \frac{(4m)! (m!)^2}{\left((2m)!\right)^3 2^{2m}} = \left(\prod^{m}_{i=1} \frac{2m+i}{m+i} \right) \left( \prod^{m}_{i=1} \frac{3m+i}{m+i} \right) \left(\frac{1}{4}\right)^m \leq \left(\frac{3}{2} \cdot \frac{4}{2} \cdot \frac{1}{4}\right)^m = \left(\frac{3}{4} \right)^m$$
which is less than $7/8 < 1$ for all $n > N$. This can readily be used to show convergence by comparison against $K\sum^\infty_{n=0} (7/8)^n$ with $K$ some constant.

Exercise 8.3.12

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Exercise 8.3.12

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