Exercise 6.6.2

Question

Starting from one of the previously generated series in this section, use manipulations similar to those in Example 6.6.1 to find Taylor series representations for each of the following functions. For precisely what values of $x$ is each series representation valid?
  \begin{enumerate}[label=(\alph*)] 
  \item $x \cos \left(x^{2}\right)$
  \item $x /\left(1+4 x^{2}\right)^{2}$
  \item $\log \left(1+x^{2}\right)$
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item We know $\cos(x) = 1 - x^2/2 + x^4/4! - \dots$ on all of $\mathbf{R}$ so
    $$
    x\cos(x^2)
    = x - \frac{x^5}{2!} + \frac{x^9}{4!} - \frac{x^{13}}{6!} + \dots
    = \sum_{n=0}^\infty (-1)^n \frac{x^{4n+1}}{(2n)!}
    $$
  \item
    Since
    $$
    \frac{x}{(1+4x^2)^2} = \frac{\mathrm d}{\mathrm dx}\, \frac{- 1/8}{1 + 4x^2}
    $$
    We can use the geometric series then differentiate
    $$
    \frac{-1/8}{1 - (-4x^2)} = (-1/8)\sum_{n=0}^\infty (-4x^2)^n = \sum_{n=0}^\infty \frac{(-1)^{n+1} (4x^2)^n}{8}
    $$
    Is valid when $|4x^2|<1$ or $|x| < 1/2$, differentiating gives
    $$
    \frac{x}{(1 - 4x^2)^2}
    \stackrel{?}{=} \sum_{n=0}^\infty \frac{(-1)^{n+1} 8x\cdot n(4x^2)^{n-1}}{8}
    = \sum_{n=0}^\infty (-1)^{n+1} xn(4x^2)^{n-1}
    $$
    Every $|x|<1/2$ converges by the ratio test, meaning the right-hand series converges for all $x \in (-1/2,1/2)$.
    Checking endpoints $x=1/2$ gives $\sum (-1)^{n+1}n/2$ which clearly doesn't converge, similarly $x=-1/2$ doesn't converge.

Finally, we must show our differentiated series converges to the right thing. Let $f_m = \sum_{n=0}^m (-1)^{n+1}xn(4^2)^{n-1}$ and $f(x) = x/{(1+4x^2)}$. We have $(f_m') 	o f'$ uniformly (by construction via geometric series), and since $(f_m)$ and $f$ agree at $x=0$ ($f_m(0) 	o 0$ and $f(0) = 0$) we must have $(f_m) 	o f$ uniformly.
  \item
    We know the Taylor series for $\log(1 + x)$ is
    $$
    \log(1 + x) = x - \frac 12 x^2 + \frac 13 x^3 - \dots = \sum_{n=1}^\infty \frac {(-1)^n x^n}{n}
    $$
    Which converges on $(-1, 1]$. Substituting $x^2$ for $x$ gives
    $$
    \log(1 + x^2) = \sum_{n=1}^\infty \frac{(-1)^n x^{2n}}{n}
    $$
    Which converges on $[0, 1]$.
   \end{enumerate}

Exercise 6.6.2

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Exercise 6.6.2

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