Exercise 6.7.6

Question

\begin{enumerate}[label=(\alph*)] 
\item Let
$$
c_{n}=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots 2 n}
$$
for $n \geq 1$. Show $c_{n}<\frac{2}{\sqrt{2 n+1}}$.
\item Use (a) to show that $\sum_{n=0}^{\infty} a_{n}$ converges (absolutely, in fact) where $a_{n}$ is the sequence of Taylor coefficients generated in Exercise 6.7.4.
\item Carefully explain how this verifies that equation (1) holds for all $x \in$ $[-1,1]$
     \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item We can show this by induction, if a bit inelegantly. The base case is trivial matter of computation. For the inductive case, we want to show $c_{n+1} = c_n \frac{2n+1}{2n+2} \leq \frac{2}{\sqrt{2n+1}} \sqrt{\frac{2n+1}{2n+3}}$. If we work through the algebra of the claim $\frac{2n+1}{2n+1} < \sqrt{\frac{2n+1}{2n+3}}$, we find that this is equivalent to
    $$8n^3 + 20n^2 + 14n + 3 < 8n^3 + 20n^2 + 16n + 4$$
which is clearly true for $n \geq 1$, and the inductive step is done.
    \item $\left|a_n\right| = c_n / (2n-1) < \frac{2}{(2n-1)\sqrt{2n+1}} \leq 2 \cdot (2n-1)^{2/3}$ which implies absolute convergence by comparison against an appropriate geometric series.
    \item (b) implies that the Taylor series of $\sqrt{1-x}$ converges absolutely at 1. With Theorem 6.5.2, the Taylor series converges uniformly over $[-1,1]$ and is therefore continuous. We also have that the Taylor series converges to $\sqrt{1-x}$ for $x \in (-1, 1)$, which is also continuous. Therefore taking limits as both functions approach $\pm 1$ gets us that they are equal over $[-1,1]$.
 \end{enumerate}

Exercise 6.7.6

Answers

Comments

Exercise 6.7.6

Answers

Comments

Add answer