(29.4) Let \(f\) and \(g\) be differentiable functions on an open interval \(I\). Suppose there are \(a,b \in I\) with \(a < b\) and \(f(a) = f(b) = 0\). Show that there exists \(x \in I\) so that
\[f'(x) + f(x)g'(x) = 0.\]Hint. Consider \(h(x) = f(x) e^{g(x)}\).
Proof. Let \(h(x) = f(x) e^{g(x)}\), so that \(h'(x) = f'(x) e^{g(x)} + f(x) e^{g(x)} g'(x) = e^{g(x)}(f'(x) - f(x) g'(x))\). By Rolle’s theorem, as \(h(a) = h(b) = 0\), there exists \(c\) between \(a, b\) so that \(h(c) = 0\). But as \(e^{g(x)} \ne 0\) always, this implies that \(f'(c) - f(c) g'(c) = 0\).
(31.2) Consider the hyperbolic sine and cosine functions, \(\sinh(x) = \frac{e^x - e^{-x}}{2}\) and \(\cosh(x) = \frac{e^x + e^{-x}}{2}\).
Find the Taylor series for \(\sinh(x)\) and show that it converges to \(\sinh(x)\).
Find the Taylor series for \(\cosh(x)\) and show that it converges to \(\cosh(x)\).
We have that \(\sinh'(x) = \frac{e^x + e^{-x}}2 = \cosh(x)\) and that \(\cosh'(x) = \sinh(x)\) by basic calculation. Let \(c = 0\) be where we center our Taylor series, and notice that \(\sinh(0) = 0\) and \(\cosh(0) = 1\). So using the formula for Taylor series, we get that the Taylor series for \(\sinh(x)\) is,
\[\sum_{n=1}^\infty {\frac{x^{2n+1}}{(2n+1)!}}\]Let \(M > 0\) and consider the interval \((-M, M)\). Notice that \(\sinh(x) = \frac 12 (\vert e^x - e^{-x} \vert) \le \frac 12 (e^x + e^x) \le e^M\) on the interval \((-M, M)\). By a similar argument, \(\cosh(x) \le e^M\) on \((-M, M)\). By the Corollary to Taylor’s theorem, we deduce that \(\sinh\) converges to its Taylor series on \((-M, M)\). \(M> 0\) is arbitrary, so we may in fact deduce that \(\sinh\) converges to its Taylor series on all of \(\mathbb R\).
The Taylor series for \(\cosh(x)\) is,
\[\sum_{n=1}^\infty {\frac{x^{2n}}{(2n)!}}\]and the argument for showing that it converges to \(\cosh\) on \(\mathbb R\) is identical to part (a).
(31.6) We’re going to work through an alternative proof of Taylor’s theorem, for Taylor series centered at \(c = 0\).
Let \(f\) be defined on \((a, b)\) with \(a < 0 < b\) and assume that, for given \(n \in \mathbb N\), \(f^{(n)}\) exists on \((a, b)\). Let \(x > 0\), and take \(M\) to be the unique number so that
\[f(x) = \sum_{k=0}^{n-1}{\frac{f^{(k)}(0)}{k!} x^k + \frac{Mx^n}{n!}}\]is valid. Define
\[F(t) = f(t) + \sum_{k=1}^{n-1}{\frac{(x-t)^k}{k!}f^{(k)}(t) + M\frac{(x-t)^n}{n!}}\]for \(t \in [0, x]\). An earlier version had a typo, saying that \(F\) was a function of \(x\), instead of \(t\). The version now is correct.
Show that \(F\) is differentiable on \([0, x]\) and that furthermore
\[F'(t) = \frac{(x-t)^{n-1}}{(n-1)!}[f^{(n)}(t) - M].\]Show \(F(0) = F(x)\).
Apply Rolle’s Theorem to obtain \(y \in (0, x)\) so that \(f^{(n)}(y) = M\).
This is a simple calculation. With care, you will notice that all interior terms in the sum cancel to yield the desired expression.
A calculation shows that \(F(0) = f(x) = F(x)\).
As \(F(0) = F(x)\) there exists \(0 < y < x\) for which \(F'(y) = 0\) by Rolle’s theorem. \(F'(y) = 0\) implies that \(f^{(n)}(y) - M = 0\), from which we conclude that \(f^{(n)}(y) = M\).
Let \(f\) be differentiable on \(\mathbb R\) with \(a = \sup\{ \vert f'(x) \vert : x \in \mathbb R \} < 1\).
Select \(s_0 \in \mathbb R\) and define \(s_n = f(s_{n-1})\) for \(n \ge 1\).
Prove that \((s_n)\) is a convergent sequence. Hint. To show that \((s_n)\) is Cauchy, first show that \(\vert s_{n+1} - s_n \vert \le a \vert s_n - s_{n-1}\vert\) for \(n \ge 1\).
Show that \(f\) has a fixed point. In other words, there is a number \(s\) so that \(f(s) = s\).
Proof. Consider the interval from \(s_{n-1}\) to \(s_n\). The Mean Value Theorem says that there exists a number \(c\) between \(s_{n-1}\) and \(s_n\) so that
\[\left\vert\frac{s_{n+1} - s_n}{s_n - s_{n-1}}\right\vert = \left\vert\frac{f(s_n) - f(s_{n-1})}{s_n - s_{n-1}}\right\vert = \vert f'(c)\vert \le a < 1.\]In particular, we have that \(\vert s_{n+1} - s_n \vert \le a\vert s_n - s_{n-1}\vert\). From this, iterated, we see that for all \(N \in \mathbb N\), \(\vert s_{N+1} - s_N \vert \le a^N \vert s_1 - s_0 \vert\). The series \(\sum_{k=1}^\infty a^k\) is convergent as \(a < 1\), and say its value is \(A\).
Let \(\epsilon < 0\) and consider a natural number \(N > \frac{\log (\epsilon/(A\vert s_1 - s_0 \vert))}{\log a}\). Then for \(n, m > N\) we have that
\[\begin{align*} \vert s_n - s_m\vert &= \left\vert \sum_{k=m+1}^{n}{s_k - s_{k-1}} \right\vert \\ &\le \sum_{k=m+1}^n {\vert s_k - s_{k-1} \vert} \\ &\le \sum_{k=m+1}^n a^k {\vert s_1 - s_{0} \vert} \\ & = a^N \vert s_1 - s_0 \vert\sum_{k=m+1}^n {a^{k-N}} \\ &\le a^N \vert s_1 - s_0 \vert A < \epsilon. \end{align*}\]So \((s_n)\) converges.
Let \(\lim s_n = s\). Then notice that as \(f\) is differentiable, hence continuous, \(\lim_{n\to \infty} f(s_n) = f(\lim_{n\to \infty} s_n) = f(s)\) but also that \(\lim_{n\to \infty} f(s_n) = \lim_{n\to \infty} s_{n+1} = s\). So \(f(s) = s\).