A.5 Chapter 5

Fix $a\in\mathbb{R}$ and $h\in\mathbb{R}\setminus\{0\}$ and consider the difference quotient

From this, we see that

and so $f$ is differentiable at $a$ with $f^{\prime}(a)=m$. Since $a\in\mathbb{R}$ was arbitrary, we conclude $f\colon\mathbb{R}\to\mathbb{R}$ is differentiable.

Considering $h>0$, we have

while for $h<0$ we have

Since the one-sided limits do not agree, the limit $\displaystyle\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$ does not exist (by the result of Exercise 4.72).

Hence, $f$ is not differentiable at $0$.

(i) Using the result of Exercise 4.46, observe that

Thus, by definition, $\cos^{\prime}(0)=0$.

(ii) Fix $a\in\mathbb{R}$ and $h\in\mathbb{R}\setminus\{0\}$ and consider the difference quotient

Taking the limit as $h\to 0$ and applying the limit laws and the definition of the derivative, we see $\sin$ is differentiable at $a$ with

Here we used the fact that $\sin^{\prime}(0)=1$, which was already proven in Example 5.13, and $\cos^{\prime}(0)=0$, which was proven in part (i).

The same argument can also be applied to $\cos$. Alternatively, we can also derive the formula for $\cos^{\prime}$ from that of $\sin^{\prime}$ using the fact that $\cos(\theta)=\sin(\theta+\pi/2)$ for all $\theta\in\mathbb{R}$. Indeed, fix $a\in\mathbb{R}$ and $h\in\mathbb{R}\setminus\{0\}$ and consider the difference quotient

Taking the limit as $h\to 0$, we see that $\cos$ is differentiable at $a$ with

as required.

We prove by a very simple induction that $\exp\colon\mathbb{R}\to\mathbb{R}$ is $n$-times differentiable for all $n\in\mathbb{N}$ with $\exp^{(n)}=\exp$.

Base case: We know from Lemma 5.12 that $\exp$ is differentiable with $\exp^{\prime}=\exp$, which provides the base case for the induction.

Inductive step: Let $n\in\mathbb{N}$ and suppose, as an induction hypothesis, that $\exp$ is $n$-times differentiable with $\exp^{(n)}=\exp$. Since, again by Lemma 5.12, $\exp$ is differentiable with $\exp^{\prime}=\exp$, it follows that $\exp^{(n)}$ is differentiable with $\exp^{(n+1)}=(\exp^{(n)})^{\prime}=\exp^{\prime}=\exp$, as required. This closes the induction and completes the proof.

From earlier exercises and examples we know that

in particular, $p_{2}^{(n)}$ exists for all $n\in\mathbb{N}$. Therefore, the function $p_{2}$ is infinitely differentiable, and consequently it is also twice differentiable and three times differentiable.

For $x\in I\setminus\{a\}$, the value $F(x)$ corresponds to the gradient of the secant line between the points $(a,f(a))$ and $(x,f(x))$ on the graph of $f$; see Figure A.4. On the other hand, $F(a)$ corresponds to the gradient of the tangent line.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be the sine function $f(x):=\sin x$ for all $x\in\mathbb{R}$ and $a=0$. Then $f(0)=0$ and $f^{\prime}(0)=1$ and, correspondingly, we have

That is, $F=\mathrm{sinc}$ is precisely the $\mathrm{sinc}$ function from Example 4.44, which we know is continuous.

Let $I\subseteq\mathbb{R}$ be an open interval, $f$, $g\colon I\to\mathbb{R}$ be differentiable at $a\in I$. Thus, by definition

Let $\lambda\in\mathbb{R}$ and consider the difference quotients

and

Hence, by the limit laws,

and

Thus, by definition, both $\lambda f$ and $f+g$ are differentiable at $a$ and the derivatives are given by the above formulæ.

Consider the function $p_{n}\colon(0,\infty)\to(0,\infty)$ given by $p_{n}(x):=x^{n}$ for all $x>0$. We know that $f=p_{n}^{-1}$ is the inverse of $p_{n}$. Furthermore, from Example 5.11, we know that $p_{n}$ is differentiable with $p_{n}^{\prime}(x)=nx^{n-1}>0$ for all $x\in\mathbb{R}$. Since the derivative of $p_{n}$ is nonvanishing on $(0,\infty)$, we may apply the differential inverse function theorem to conclude that $f$ is differentiable with

as required.

An illustration of the proof of the Theorem 5.34. Here $x_{\ell}$ is a local maximum of the differentiable function $f$ and $h>0$. The secant line between $(x_{\ell},f(x_{\ell}))$ and a point $(x_{\ell}+h,f(x_{\ell}+h))$ lying to the right has a positive gradient. The secant line between $(x_{\ell},f(x_{\ell}))$ and a point $(x_{\ell}-h,f(x_{\ell}-h))$ lying to the left has a negative gradient. This observation forces $f^{\prime}(x_{\ell})=0$.

(i) Consider the function $f\colon[0,1]\to\mathbb{R}$ given by

Then $f(0)=f(1)=0$ and $f$ is differentiable on $(0,1)$, since on this interval it agrees with the linear function $\ell\colon\mathbb{R}\to\mathbb{R}$ given by $\ell(x):=x$. Note, however, $f$ is not continuous on $[0,1]$. Furthermore, for all $c\in(0,1)$ we have $f^{\prime}(c)=\ell^{\prime}(c)=1\neq 0$, so the conclusion of the Rolle’s theorem fails for this function. This shows that the hypothesis that $f$ is continuous on the closed interval $[a,b]$ is necessary to ensure the conclusion of the theorem always holds.

(ii) Consider the function $f\colon[-1,1]\to\mathbb{R}$ given by $f(x):=|x|-1$ for all $x\in[-1,1]$. Then $f(-1)=f(1)=0$ and $f$ is continuous on $[-1,1]$ and differentiable on $(-1,1)\setminus\{0\}$ with

Note, however, that $f$ is not differentiable at $0$. Furthermore, from the above we see $f^{\prime}(c)\neq 0$ for all $c\in(-1,1)\setminus\{0\}$. This shows that the hypothesis that $f$ is differentiable everywhere on the open interval $(a,b)$ is necessary to ensure the conclusion of Rolle’s theorem always holds.

We know that any polynomial $p\colon\mathbb{R}\to\mathbb{R}$ function is differentiable (and therefore also continuous). Since $f\colon[-3,3]\to\mathbb{R}$ given by $f(x):=x^{3}-9x$ is the restriction of a polynomial, $f$ is continuous on $[-3,3]$ and differentiable on $(-3,3)$. Furthermore,

Thus, $f$ satisfies the hypotheses of Rolle’s theorem so we know there exists some point $c\in(-3,3)$ satisfying $f^{\prime}(c)=0$.

In fact, by a simple computation we can show $f$ has precisely two such stationary points. Indeed,

and so $f^{\prime}$ has precisely two roots: $\sqrt{3}$ and $-\sqrt{3}\in(-3,3)$. See Figure A.5.

A typical example is $f\colon[0,1]\to\mathbb{R}$ given by $f(x):=\sin(5\pi x)$, which has graph sketched in Figure A.6. Conceptually, the graph of the function should either $3$ peaks and $2$ troughs or $2$ peaks and $3$ troughs, so any sketch which incorporates these essential features works.

Suppose $f\colon[a,b]\to\mathbb{R}$ satisfies the hypotheses of Rolle’s theorem, so that $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$. Using the first two conditions, we may apply the mean value theorem to deduce that there exists some $\xi\in(a,b)$ such that

However, since $f(b)=f(a)$ and $b-a>0$, we see that $f^{\prime}(\xi)=0$. Hence, we have shown there exists some $\xi\in(a,b)$ such that $f^{\prime}(\xi)=0$, which is the conclusion of Rolle’s theorem.

Suppose $f\colon[a,b]\to\mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$. As in (5.19), we define $h\colon[a,b]\to\mathbb{R}$ by

It follows by the continuity and differentiability laws that $h$ is also continuous on $[a,b]$ and differentiable on $(a,b)$. Furthermore,

and

Hence, we may apply Rolle’s theorem to $h$ to deduce that there exists some $\xi\in(a,b)$ with $h^{\prime}(\xi)=0$. On the other hand, by computing the derivative, we see that

and rearranging gives the desired result.

Let $a<b$ and $f\colon[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Given $a\leq x<y\leq b$, by the mean value theorem, there exists some $\xi\in(x,y)\subseteq(a,b)$ such that

1) Suppose $f^{\prime}(z)>0$ for all $z\in(a,b)$. Applying this to (A.16) with $z=\xi$, we see that $f(y)-f(x)>0$ and so $f(x)<f(y)$. Since this holds for arbitrary $a\leq x<y\leq b$, it follows that $f$ is strictly increasing on $[a,b]$.

2) We can use a similar argument to that used in part 1), but it is also possible to reduce the result to that of part 1) using a reflection. Indeed, suppose $f^{\prime}(z)<0$ for all $z\in(a,b)$, so that $-f^{\prime}(z)>0$ for all $z\in(a,b)$. By part 1), we know that $-f$ is strictly increasing on $[a,b]$, and so $f$ is strictly decreasing on $[a,b]$.

3) Suppose $f^{\prime}(z)=0$ for all $z\in(a,b)$. Applying this to (A.16) with $z=\xi$, we see that $f(y)-f(x)=0$ and so $f(x)=f(y)$. Since this holds for arbitrary $a\leq x<y\leq b$, it follows that $f$ is constant on $[a,b]$.

(i) Recall that $a\in I$ is a stationary point of $f$, so that $f^{\prime}(a)=0$ and, by definition,

Since $f^{\prime\prime}(a)<0$, if we define $\varepsilon:=-f^{\prime\prime}(a)/2$, then $\varepsilon>0$. By the $\varepsilon$-$\delta$ definition of a limit, there exists some $\delta>0$ such that

In particular,

for all $0<|h|<\delta$.

(ii) We know that $f$ is continuous and

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  strictly decreasing on $(a,a+\delta)$;

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  strictly increasing on $(a-\delta,a)$.

The two bullet points imply that if $0<|u-a|<|x-a|<\delta$, then $f(u)>f(x)$.

By continuity, $f(a)=\lim_{u\to a}f(u)$. It remains to show $\lim_{u\to a}f(u)\geq f(x)$ for all $x\in J:=(a-\delta,a+\delta)$.

Arguing by contradiction, suppose there exists some $x\in J$ such that $f(x)>\lim_{u\to a}f(u)$. Let $\varepsilon:=(f(x)-f(a))/2>0$. By the $\varepsilon$-$\delta$ definition of continuity, there exists some $\eta>0$ satisfying $0<\eta<\delta$ such that

We must have $|x-a|\geq\eta$, since $|f(x)-f(a)|=2\varepsilon>\varepsilon$. However, if $|u-a|<\eta\leq|x-a|<\delta$, then it follows by our earlier observation that $f(u)>f(x)$ and so

This is a contradiction and, hence, no such $x$ can exist.

By the differentiation laws and Example 5.13, we know $f$ is infinitely (and therefore twice) differentiable with $f^{\prime}(x)=\cos x-\sin x$. Thus,

so $\pi/4$ and $5\pi/4$ are both stationary points. Furthermore, $f^{\prime\prime}(x)=-\sin x-\cos x$ and so

Hence, by the second derivative test, $f$ has a local maximum at $\pi/4$ and a local minimum at $5\pi/4$.

(i) Take $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x):=x^{3}$ for all $x\in\mathbb{R}$. Then $f$ is twice differentiable with $f^{\prime}(x)=3x^{2}$ and $f^{\prime\prime}(x)=6x$ for all $x\in\mathbb{R}$. In particular, $f^{\prime}(0)=f^{\prime\prime}(0)=0$. However, $0$ is neither a local maximum nor a local minimum for $f$. Indeed, given any $\delta>0$, if we choose $-\delta<x<0<y<\delta$, then $f(x)<0<f(y)$ and so there is no open interval around $0$ upon which $0$ is either a maximum or a minimum point.

(ii) Take $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x):=-x^{4}$ for all $x\in\mathbb{R}$. Then $f$ is twice differentiable with $f^{\prime}(x)=-4x^{3}$ and $f^{\prime\prime}(x)=-12x^{2}$ for all $x\in\mathbb{R}$. In particular, $f^{\prime}(0)=f^{\prime\prime}(0)=0$. Furthermore, $0$ is a maximum point since $f(x)=-(x^{2})^{2}\leq 0=f(0)$ for all $x\in\mathbb{R}$.

(iii) Take $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x):=x^{4}$ for all $x\in\mathbb{R}$, the reflection of the function used in (ii). It automatically follows from the analysis in (ii) that $f^{\prime}(0)=f^{\prime\prime}(0)=0$ and $0$ is a minimum point.

We know from Example 5.13 that $\cos\colon\mathbb{R}\to\mathbb{R}$ is differentiable with $|\cos^{\prime}(a)|=|\sin(a)|\leq 1$ for all $a\in\mathbb{R}$. In light of this,

by the mean value inequality. Since $\sum_{k=1}^{\infty}1/k^{2}$ converges, by the comparison test the series $\sum_{k=1}^{\infty}(1-\cos(1/k^{2}))$ converges.

It suffices to consider $x>0$, since the case $x=0$ is immediate.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be given by $f(x):=(1+x)^{n}$ for all $x\in\mathbb{R}$. We know from Example 5.11 and the composition law that $f$ is differentiable with $f^{\prime}(x)=n(1+x)^{n-1}$ for all $x\geq 0$. Thus, by the mean value theorem, given $x>0$, there exists some $\xi\in(0,x)$ such that

where the last step follows since $\xi\geq 0$. Rearranging, we obtain the Bernoulli inequality.

Recall that $\exp(0)=1>0$. Suppose there exists some $x\in\mathbb{R}$ satisfying $\exp(x)<0$. Since $\exp\colon\mathbb{R}\to\mathbb{R}$ is continuous, it follows that there exists some $\min\{0,x\}<y<\max\{0,x\}$ satisfying $\exp(y)=0$. However, this contradicts Lemma 5.57. Hence, we must have $\exp(x)>0$ for all $x\in\mathbb{R}$.

Fix $y\in\mathbb{R}$ and recall the definition

this is well defined by Lemma 5.57 (in particular, $\exp(y)\neq 0$ and so there is no risk of dividing by $0$). We know from Lemma 5.12 that $\exp\colon\mathbb{R}\to\mathbb{R}$ is differentiable, and so by the chain rule and linearity of the derivative, $F$ is differentiable with

Finally,

as required.

(i) In the proof of Lemma 5.57, we demonstrated that $\exp(x)\exp(-x)=1$ for all $x\in\mathbb{R}$. From this, the claim is immediate.

(ii) Fixing $x\in\mathbb{R}$, we prove $\exp(kx)=\exp(x)^{k}$ for all $k\in\mathbb{N}$ by induction on $k$.

Base case: For $k=1$, the statement is tautological.

Inductive step: Suppose the result holds for some $k\in\mathbb{N}$ and consider

where the second identity is due to the multiplicative identity from Theorem 5.59. Applying the induction hypothesis, $\exp((k+1)x)=\exp(x)^{k}\exp(x)=\exp(x)^{k+1}$, which closes the induction and completes the proof.

(iii) Let $x\in\mathbb{R}$ and $k\in\mathbb{N}$. Applying part (ii) with $x/k$, we have

Thus, taking the $k$th root, $\exp(x/k)=\exp(x)^{1/k}$.

(iv) Fix $x\in\mathbb{R}$ and $q\in\mathbb{Q}$ and write $q=a/b$ where $a\in\mathbb{Z}$ and $b\in\mathbb{N}$. By first applying the result of part (ii) and then the result of part (iii), we have

as required.

(v) This is simply a special case of part (iv), taking $x=1$.

Using Definition 5.64, we have $f(x)=\exp(p\log x)$. We know that $\log$ is differentiable on $(0,\infty)$ by Example 5.31, and $\exp$ is differentiable on $\mathbb{R}$ by Lemma 5.12. Thus by linearity and the chain rule, $f$ is differentiable on $(0,\infty)$.

Applying the chain rule, for $x>0$ we have

Applying the multiplicative identity for $\exp$ (Theorem 5.59), this becomes

(i) Fix $k\in\mathbb{N}$. We shall show by induction on $0\leq j\leq k$ that

Base case: The $j=0$ case (where the left-hand side is just interpreted as $(x-a)^{k}$) is immediate, and serves as the base for the induction.

Inductive step: Now suppose the result holds for some $0\leq j\leq k-1$. Using the induction hypothesis,

By Example 5.11 and the chain rule,

This closes the induction and completes the proof.

(ii) As a special case of the above,

is a constant, and therefore

(iii) Recall the definition

Let $0\leq j\leq n$. In light of the above and the linearity property of the derivative,

Evaluating at $x=a$, we deduce that

as required.

The result is immediate if $x=0$, so we may assume $x\neq 0$.

Fix $x\in\mathbb{R}\setminus\{0\}$ and observe that

and so $a_{n+1}/a_{n}\to 0<1$ as $n\to\infty$. Hence, by the ratio test, the series $\sum_{n=1}^{\infty}\frac{|x|^{n+1}}{(n+1)!}$ converges. Thus, by the sequence test, $\lim_{n\to\infty}\frac{|x|^{n+1}}{(n+1)!}=0$, as required.

We first compute the Taylor polynomials $P_{2\ell}^{\cos,0}(x)$. A simple application of Example 5.13 shows that $\cos^{(j)}(0)=0$ if $j\in\mathbb{N}$ is odd and