# Harrison Chapman

## Math 317: Homework 8

Due: Friday, April 12, 2019
1. (29.4) Let $f$ and $g$ be differentiable functions on an open interval $I$. Suppose there are $a,b \in I$ with $% $ and $f(a) = f(b) = 0$. Show that there exists $x \in I$ so that

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 $f'(c) - f(c) g'(c) = 0$.

2. (31.2) Consider the hyperbolic sine and cosine functions, $\sinh(x) = \frac{e^x - e^{-x}}{2}$ and $\cosh(x) = \frac{e^x + e^{-x}}{2}$.

1. Find the Taylor series for $\sinh(x)$ and show that it converges to $\sinh(x)$.

2. Find the Taylor series for $\cosh(x)$ and show that it converges to $\cosh(x)$.

1. 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,

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 $\mathbb R$.

2. The Taylor series for $\cosh(x)$ is,

and the argument for showing that it converges to $\cosh$ on $\mathbb R$ is identical to part (a).

3. (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 $% $ 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

is valid. Define

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.

1. Show that $F$ is differentiable on $[0, x]$ and that furthermore

2. Show $F(0) = F(x)$.

3. Apply Rolle’s Theorem to obtain $y \in (0, x)$ so that $f^{(n)}(y) = M$.

1. This is a simple calculation. With care, you will notice that all interior terms in the sum cancel to yield the desired expression.

2. A calculation shows that $F(0) = f(x) = F(x)$.

3. As $F(0) = F(x)$ there exists $% $ for which $F'(y) = 0$ by Rolle’s theorem. $F'(y) = 0$ implies that $f^{(n)}(y) - M = 0$, from which we conclude that $f^{(n)}(y) = M$.

4. Let $f$ be differentiable on $\mathbb R$ with $% $.

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 $n \ge 1$.

2. Show that $f$ has a fixed point. In other words, there is a number $s$ so that $f(s) = s$.

1. 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

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 $% $, and say its value is $A$.

Let $% $ 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

So $(s_n)$ converges.

2. 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 $f(s) = s$.