Exercise 6.6.10

Question

Consider $f(x)=1 / \sqrt{1-x}$.
  \begin{enumerate}[label=(\alph*)] 
  \item Generate the Taylor series for $f$ centered at zero, and use Lagrange's Remainder Theorem to show the series converges to $f$ on $[0,1 / 2]$. (The case $x<1 / 2$ is more straightforward while $x=1 / 2$ requires some extra care.) What happens when we attempt this with $x>1 / 2 ?$
  \item Use Canchy's Remainder Theorem proved in Exercise 6.6.9 to show the series representation for $f$ holds on $[0,1)$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item We have $$f^{(n)}(x) = \frac{\prod^n_{i=1} (2i-1)}{2^n}\left(\frac{1}{1-x}\right)^n \sqrt{\frac{1}{1-x}}$$ The Taylor series is $$\sum^N_{n=0} \frac{\prod^n_{i=1} (2i - 1)}{2^n n!} x^n = \sum^N_{n=0} \left(\frac{x^n}{2^n} \prod^n_{i=1} \frac{2i-1}{i}\right) \leq \sum^N_{n=0} \left(\frac{x^n}{2^n} \prod^n_{i=1}2\right) = \sum^N_{n=0}x^n$$ so we know that the Taylor series at least converges to something for $x \in [0, 1)$. Lagrange's Remainder Theorem gives us $$E_N(x) = \frac{\left(\prod^{N+1}_{i=1} (2i-1)\right)\left(\frac{1}{1-c}\right)^{N+1}\sqrt{\frac{1}{1-c}}x^{N+1}}{2^{N+1}(N+1)!}$$ for some $c \in (0, x)$. For $x = 1/2$ and $c < x$: $$ \begin{aligned} \left|E_N(x)\right| & \leq \left(\prod^{N+1}_{i=1} \frac{2i-1}{i}\right) \left(\frac{1}{2-2c}\right)^{N+1} \left(\sqrt{\frac{1}{1-c}}\right) \frac{1}{2^{N+1}} \ & \leq 2^{N+1} d^{N+1} \left(\sqrt{\frac{1}{1-c}}\right) \frac{1}{2^{N+1}} = d^{N+1}\sqrt{\frac{1}{1-c}} \end{aligned} $$ where $d = 1/(2-2c) < 1$; this shows $E_N$ converges to 0 over $[0, 1/2]$. Writing $E_N$ in product notation, $$E_N(x) = \prod^{N+1}_{i=1} \frac{(2i - 1)x}{2i (1-c)}$$ If $x > 1/2$, then it's possible for $\frac{x}{1-c} > 1$. Then for some $I$, for $i > I$ we have $$\frac{x}{1-c} > \frac{2i}{2i - 1} > 1$$ and beyond that point, the terms in the product begin increasing, with the product as a whole growing exponentially and diverging. \item Plugging in Cauchy's Remainder Theorem, $$ \begin{aligned} E_N(x) &= \left( \prod^N_{i=1}\frac{2i-1}{i} \right) \frac{(2N + 1) (x-c)^N x}{2^N (1-c)^{N+1}} \sqrt{\frac{1}{1-c}} \ &\leq (2N+1) \left(d\right)^N \frac{x}{1-c}\sqrt{\frac{1}{1-c}} \end{aligned} $$ where $d = \frac{x-c}{1-c} < 1$. The first term is linear in $N$, the second is exponentially decaying in $N$, and the last two terms are constant, so the behaviour is dominated by exponential decay and $E_N(x)$ converges to 0. \end{enumerate}

Exercise 6.6.10

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Exercise 6.6.10

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