## Problem 6.2.4

Suppose that $f:[0,1]\rightarrow[0,1]$ is continuous.Prove that there exists a number $c$ in $[0,1]$ such that $f(c)=c$.

### Solution

If $f(0)=0$ or $f(1)=1$, there is nothing to prove. Consider the function $g(x)=f(x)-x$. It is easy to see that $g(x)$ is continuous on $[0,1]$. We also have $g(0)=f(0)>0$ and $g(1)=f(1)-1<0$ as $f(0) \neq 0$ and $f(1)\neq 1$. Intermediate-Value theorem guarantees the existence of a $c$ in $[0,1]$ for which $g(c)=0$. Therefore, there exists a number $c$ in $[0,1]$ such that $f(c)=c$.

## Problem 6.2.4

A rock climber starts to climb a mountain at $7:00 A.M.$ on Saturday and gets to the top at $5:00 P.M.$ He camps on top and climbs back down on Sunday, starting at $7:00 A.M.$ and getting back to his original starting point at $5:00 P.M.$ Show that at some time of day on Sunday he was at the same elevation as he was at that time on Saturday.

### Solution

We have two functions $h_a(t)$ and $h_d(t)$ giving the height of the climber as a function of time for the ascent and descent. Let $t_s$ and $t_e$ be $7:00 A.M.$ and $5:00 P.M.$ respectively. We can assume that $h_a$ and $h_d$ are continuous on $[t_s,t_e]$. Define a function $h(t)=h_a(t)-h_d(t)$. We have $h(t_s)< 0$ and $h(t_e) < 0$. Intermediate-Value theorem guarantees the existence of a $t$ in $[t_s,t_e]$ for which $h(t)=0$.

## Problem 6.5.5

a. Show that $5x^4-4x+1=0$ has a root between $0$ and $1$.

b. If $a_0,a_1,\dots,a_n$ are real numbers satisfying

\begin{align*} \frac{a_0}{1} + \frac{a_1}{2} + \cdots + \frac{a_n}{n+1} = 0 \end{align*}

show that the equation $a_0+a_1x+ \dots+a_nx^n=0$ has atleast one real root.

### Solution

We use Rolle’s theorem for the following:

a. Let $f(x)=x^5-2x^2+x$. $f$ is continuous on $[0,1]$ and differentiable on $(0,1)$. We also have $f(0)=0$ and $f(1)=0$.Therefore there is a number $c$ in $(0,1)$ such that $f'(c)=5c^4-4x+1=0$.

b. Let $f(x)$ be

\begin{align*} a_0x + \frac{a_1x^2}{2} + \dots + \frac{a_nx^{n+1}}{n+1} = 0 \end{align*}

$f$ is continuous on $[0,1]$ and differentiable on $(0,1)$. We also have $f(0)=0$ and $f(1)=0$.Therefore there is a number $c$ in $(0,1)$ such that $f'(c) = a_0+a_1c+ \dots+a_n c^n=0$.

## Problem 6.5.6

a. Suppose that $f:[0,1] \rightarrow \mathbb{R}$ is differentiable, $f(0)=0$, and $f(x)>0$ for $x$ in $(0,1)$. Prove that there is a number $c$ in $(0,1)$ such that

\begin{align*} \frac{2f'(c)}{f(c)} = \frac{f'(1-c)}{f(1-c)} \end{align*}

b. Is there a number $d$ in $(0,1)$ such that

\begin{align*} \frac{3f'(d)}{f(d)} = \frac{f'(1-d)}{f(1-d)} \end{align*}

### Solution

We use Rolle’s theorem for the following:

a. Consider $g(x) = f^2(x)f(1-x)$. $g$ is differentiable on $(0,1)$.

We also have $g(0)=0$ and $g(1)=0$ .

Therefore there is a number $c$ in $(0,1)$ such that $g'(c) = -f^2(c)f'(1-c)+ 2f(c)f'(c)f(1-c) = 0$. Therefore, as $f(x)>0$ for $x>0$,

\begin{align*} \frac{2f'(c)}{f(c)} = \frac{f'(1-c)}{f(1-c)} \end{align*}

b. Consider $g(x) = f^3(x)f(1-x)$. $g$ is differentiable on $(0,1)$.

We also have $g(0)=0$ and $g(1)=0$ .

Therefore there is a number $d$ in $(0,1)$ such that $g'(d) = -f^3(d)f'(1-d)+ 3f^2(d)f'(d)f(1-d) = 0$. Therefore, as $f(x)>0$ for $x>0$,

\begin{align*} \frac{3f'(d)}{f(d)} = \frac{f'(1-d)}{f(1-d)} \end{align*}

## Problem 6.5.7

a. Cauchy mean-value theorem If $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$, then there is a number $c$ in $(a,b)$ such that

\begin{align*} [f(b)-f(a)]g'(c) = [g(b)-g(a)]f'(c) \end{align*}

b. Mean-value theorem If $f:[a,b] \rightarrow \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is a number $c$ in $(a,b)$ such that

\begin{align*} \frac{f(b)-f(a)}{b-a} = f'(c) \end{align*}

### Solution

a. Define $F(x)$ as follows

\begin{align*} F(x) = f(x) - \frac{f(b)-f(a)}{g(b)-g(a)}g(x) \end{align*}

We see that $F(a)=F(b)$. Rolle’s theorem guarantees the existence of $c$ in $(a,b)$ such that $F'(c)= 0$. Therefore,

\begin{align*} f'(c) - \frac{f(b)-f(a)}{g(b)-g(a)}g'(c) &= 0 \ \implies [f(b)-f(a)]g'(c) &= [g(b)-g(a)]f'(c) \end{align*}

b. Setting $g(x)=x$ in Cauchy’s mean value theorem we have

\begin{align*} [f(b)-f(a)] &= [b-a]f'(c) \ \implies \frac{f(b)-f(a)}{b-a} &= f'(c) \end{align*}

## Problem 6.5.8

a. Show that $x^3-3x+b=0$ cannot have more than one zero in $[-1,1]$, regardless of the value of $b$.

b. Let $f(x)=(x^2-1)e^{cx}$. Show that $f'(x)=0$ for exactly one $x$ in the interval $(-1,1)$ and that this $x$ has the same sign as parameter $c$.

### Solution

A useful corollary to Rolle’s theorem is that if $f$ is a continuous and differentiable function, say on the interval $[a,b]$, and if $x_1$ and $x_2$ are zeros of $f$, $a<x_1<x_2<b$, then $f'$ has a zero between $x_1$ and $x_2$. More generally, if $f$ has $n$ distinct zeros in $[a,b]$, then $f'$ has at least $n-1$ zeros(these are interlaced with the zeros of $f$), $f''$ has at least $n-2$ zeros(assuming $f'$ is continuous and differentiable on $[a,b]$), and so forth.

a. If $f(x)=x^3-3x+b$ has more than one zero in $[-1,1]$, then $f'(x)$ should have at least one zero in $(-1,1)$. But $f'(x) = 3x^2-3$ has no zeros in $(-1,1)$, therefore $f(x)$ cannot have more than zero in $[-1,1]$ irrespective of the value of $b$.

b. We have $f(-1)=f(1)=0$. Therefor, $f'(x)$ has at least one zero in $(-1,1)$. As the zeros of $f'(x)$ have to be interlaced with $f(x)$, $f'(x)$ cannot have more than one zero in $(-1,1)$. If $d$ is a root of $f'(x)=0$. we have

\begin{align*} (d^2-1)ce^{cd}+e^{cd}2d&=0 \ \implies c &= \frac{2d}{1-d^2} \end{align*}

As $(1-d^2)$ is positive for $d$ in $(-1,1)$, $d$ the zero of $f'(x)$ has the same sign as the parameter $c$.

## Problem 6.5.9

How many zeros does the function $f(x)=2^x-1-x^2$ have on the real line?

### Solution

Clearly there are no roots for negative $x$, since for such $x$, $2^x < 1$, whereas $1 + x^2 > 1$. There are certainly roots at $x = 0$ and $1$. Also $2^4 < 4^2 + 1$, whereas $2^5 > 5^2 + 1$, so there is a root between $4$ and $5$. We have to show that there are no other roots. Put $f(x) = 2^x - x^2 - 1$. Then $f ‘'(x) = (ln 2)^2 2^x - 2$. This is strictly increasing with a single zero. $f’(0) > 0$, so $f'(x)$ starts positive, decreases through zero to a minimum, then increases through zero. So it has just two zeros. Hence $f(x)$ has at most three zeros, which we have already found.

## Problem 6.5.10

Let $f(x)=a_0+a_1x+\dots+a_nx^n$ be a polynomial with real coefficients such that $f$ has $n+1$ distinct real zeros. Use Rolle’s therorem to show that $a_k=0$, for $0\leq k \leq n$.

### Solution

If $f$ has $n+1$ distinct real zeros, the equation $f^n(x)=n!a_n=0$ where $f^n(x)$ is the $n^{th}$ derivative should have at least one zero. This gives us $a_n=0$.By extending the same argument to the other derivatives $f^k(x)$ where $1\leq k \leq n-1$, we can show that all the coefficients of $f(x)$ are zero.

## Problem 6.7.3

Evaluate

\begin{align*} \lim_{n \to \infty} 4^n\left(1-cos\frac{\theta}{2^n} \right) \end{align*}

### Solution

We have

\begin{align*} \lim_{n \to \infty} 4^n\left(1-cos\frac{\theta}{2^n} \right) &= \lim_{n \to \infty} \frac{2sin^2 \frac{\theta}{2^{n+1}}}{4\frac{\theta^2}{4^{n+1}}}\theta^2 = \frac{\theta^2}{2} \end{align*}

## Problem 6.7.4

Evaluate the following limits

a. $\lim\limits_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$

b. $\lim\limits_{n \to \infty} \left(\frac{n+1}{n+2}\right)^n$

c. $\lim\limits_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^n$

d. $\lim\limits_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n^2}$

e. $\lim\limits_{n \to \infty} \frac{2p_nP_n}{p_n+P_n}$, where $p_n = \left (1+\frac{1}{n}\right)^n$ and $P_n = \left(1+\frac{1}{n}\right)^{n+1}$

### Solution

We use L’H^ospital’s Rule for the following

a. Let $y = \left(1 + \frac{1}{n}\right)^n$. We have

\begin{align*} log(y) &= nlog\left(1+ \frac{1}{n} \right) \ \implies log \lim_{n \to \infty}y &= \lim_{n \to \infty} nlog\left(1+ \frac{1}{n} \right) \ \implies log \lim_{n \to \infty}y &= \lim_{n \to \infty} \frac{\frac{1}{1+\frac{1}{n}}\left(\frac{-1}{n^2}\right)}{\frac{-1}{n^2}} \ \implies log \lim_{n \to \infty}y &= 1 \ \implies \lim_{n \to \infty}y &= e \end{align*}

b. Let $y = \left(\frac{n+1}{n+2}\right)^n$. We have

\begin{align*} log(y) &= nlog\left(\frac{n+1}{n+2}\right) \ \implies log \lim_{n \to \infty}y &= \lim_{n \to \infty} n log\left(\frac{n+1}{n+2}\right) \ \implies log \lim_{n \to \infty}y &= \lim_{n \to \infty} \frac{\frac{1}{1-\frac{1}{n+2}}\frac{1}{(n+2)^2}}{\frac{-1}{n^2}} \ \implies log \lim_{n \to \infty}y &= -1 \ \implies \lim_{n \to \infty}y &= e^{-1} \end{align*}

c. Let $y = \left(1 + \frac{1}{n^2}\right)^{n}$. We have

\begin{align*} log(y) &= n log\left(1+ \frac{1}{n^2} \right) \ \implies log \lim_{n \to \infty}y &= \lim_{n \to \infty} n log\left(1+ \frac{1}{n^2} \right) \ \implies log \lim_{n \to \infty}y &= \lim_{n \to \infty} \frac{\frac{1}{1+\frac{1}{n^2}}\left(\frac{-2}{n^3}\right)}{\frac{-1}{n^2}} \ \implies log \lim_{n \to \infty}y &= 0 \ \implies \lim_{n \to \infty}y &= 1 \end{align*}

d. Let $y = \left(1 + \frac{1}{n}\right)^{n^2}$. We have

\begin{align*} log(y) &= n^2log\left(1+ \frac{1}{n} \right) \ \implies log \lim_{n \to \infty}y &= \lim_{n \to \infty} n^2 log\left(1+ \frac{1}{n} \right) \ \implies log \lim_{n \to \infty}y &= \lim_{n \to \infty} \frac{\frac{1}{1+\frac{1}{n}}\left(\frac{-1}{n^2}\right)}{\frac{-2}{n^3}} \ \implies log \lim_{n \to \infty}y &= \infty \ \implies \lim_{n \to \infty}y &= \infty \end{align*}

e. We have

\begin{align*} \lim_{n \to \infty} p_n &= e \ \lim_{n \to \infty} P_n &= e \ \end{align*}

Therefore,

\begin{align*} \lim_{n \to \infty} \frac{2p_nP_n}{p_n+P_n} = \frac{2e^2}{2e} = e \end{align*}

## Problem 6.7.5

Let $0<a<b$. Evaluate

\begin{align*} \lim_{t \to 0} \left[\int_0^1 [bx+a(1-x)]^tdx \right]^{1/t} \end{align*}

### Solution

We first evaluate the integral.Let $y = bx+a(1-x)$, we have $dy = (b-a)dx$.

\begin{align*} \int_0^1 [bx+a(1-x)]^t dx = \frac{1}{b-a}\int_a^b y^t dy = \frac{1}{b-a}\left(\frac{b^{t+1}}{t+1} - \frac{a^{t+1}}{t+1} \right) \end{align*}

\begin{align*} \text{Let } z = \left[\frac{1}{b-a}\left(\frac{b^{t+1}}{t+1} - \frac{a^{t+1}}{t+1}\right)\right]^{1/t} \ \implies log(z) = \frac{1}{t}log\left[\frac{1}{b-a}\left(\frac{b^{t+1}}{t+1} - \frac{a^{t+1}}{t+1}\right)\right] \ \implies log \lim_{t \to 0} z = \lim_{t \to 0} \frac{1}{\frac{1}{b-a}\left(\frac{b^{t+1}}{t+1} - \frac{a^{t+1}}{t+1}\right)}\ \frac{1}{b-a}\left( \frac{b^{t+1}((t+1)log(b)-1)}{(t+1)^2} - \frac{a^{t+1}((t+1)log(a)-1)}{(t+1)^2} \right) \ \implies log \lim_{t \to 0} z = \frac{b\cdot log(b)-a\cdot log(a)}{b-a}-1\ \implies \lim_{t \to 0} z = \frac{b^{\frac{b}{b-a}}}{e a^{\frac{a}{b-a}}} \end{align*}

## Problem 6.7.6

Calculate

\begin{align*} \lim_{x \to \infty} x\int_0^x e^{t^2-x^2}dt \end{align*}

### Solution

We have

\begin{align*} \lim_{x \to \infty} x\int_0^x e^{t^2-x^2}dt = \lim_{x \to \infty} \frac{x\int_0^x e^{t^2}dt}{e^{x^2}} &= \lim_{x \to \infty} \frac{xe^{x^2} + \int_0^x e^{t^2}dt}{e^{x^2}2x} \ &= \frac{1}{2} + \lim_{x \to \infty} \frac{\int_0^x e^{t^2}dt}{e^{x^2}2x} \ &= \frac{1}{2} + \lim_{x \to \infty} \frac{e^{x^2}}{2e^{x^2} + 4x^2e^{x^2}} \ &= \frac{1}{2} \end{align*}

## Problem 6.7.7

Prove that the function $y=(x^2)^x$, $y(0)=1$, is continuous at $x=0$.

### Solution

We need to prove that $\lim_{x \to 0}(x^2)^x= 1$.

\begin{align*} \text{Let }y &= (x^2)^x\ \implies log(y) &= 2x log(x) \ \implies log \lim_{x \to 0}y &= 2 \lim_{x \to 0} \frac{log(x)}{\frac{1}{x}}\ \implies log \lim_{x \to 0}y &= 2 \lim_{x \to 0}\frac{\frac{1}{x}}{\frac{-1}{x^2}} = 0 \ \implies \lim_{x \to 0}y = 1 \end{align*}

## Problem 6.9.7

What function is defined by the equation

\begin{align*} f(x) = \int_0^x f(t) dt + 1 \end{align*}

### Solution

Differentiating both sides we get

\begin{align*} f'(x) = f(x) \end{align*}

The solution for the above differential equation is

\begin{align*} f(x) = ke^{x} \end{align*}

We have $f(0)=1$, therefore $k=1$.

## Problem 6.9.8

Let $f:[0,1] \rightarrow (0,1)$ be continuous.Show that the equation

\begin{align*} 2x - \int_0^x f(t) dt = 1 \end{align*}

has one and only one solution in the interval $[0,1]$.

### Solution

Let $F(x) = 2x - \int_0^x f(t) dt - 1$.

$F(x)$ is continuous on $[0,1]$.

We have $F(0)=-1$ and $F(1) = 1 - \int_0^1 f(t) dt > 0$.

Intermediate-Value Theorem guarantees the existence of value $c$ in $[0,1]$ such that $F(c)=0$.

We have $F'(x)= 2 - f(x)>0$ for $x$ in $[0,1]$ which means $F(x)$ is strictly increasing on $[0,1]$. Therefore, $F(x)$ cannot intersect the $x-axis$ more than once.

## Problem 6.9.9

Suppose that $f$ is a continuous function for all $x$ which satisfies the equation

\begin{align*} \int_0^x f(t) dt = \int_x^1 t^2f(t)dt + \frac{x^{16}}{8} + \frac{x^{18}}{9} + C \end{align*}

where $C$ is a constant.Find an explicit form of $f(x)$ and find the value of the constant $C$.

### Solution

Differentiating wrt $x$ on both sides, we have

\begin{align*} f(x) &= -x^2f(x) + 2x^{15}(1+x^2) \ \implies f(x) &= 2x^{15} \end{align*}

We also have $C + \int_0^1 x^2 2x^{15} dx = 0$. Therefore $C = -\frac{1}{9}$.

## Problem 6.9.10

Let $C_1$ and $C_2$ be curves passing through the origin.A curve $C$ is said to bisect in area the region between $C_1$ and $C_2$ if for each point $P$ of $C$ the two shaded areas $A$ and $B$ shown in the figure have equal areas. Determine the upper curve $C_2$ given that the bisecting curve has the equation $y=x^2$ and the lower curve $C_1$ has the equation $y=\frac{x^2}{2}$.

### Solution

Let $x=f(y)$ be the equation of the upper curve. We have

\begin{align*} \int_0^u t^2 - \frac{t^2}{2} dt = \int_0^{u^2} \sqrt{t} - f(t)dt \end{align*}

Differentiating both sides wrt $u$ we get,

\begin{align*} \frac{u^2}{2} &= \left(u - f(u^2)\right)2u \ \implies f(u^2) &= \frac{3}{4}u \ \implies f(y) &= \frac{3}{4}\sqrt{y} \ \implies x &= \frac{3}{4}\sqrt{y}\ \implies x^2 &= \frac{9}{16}y \ \implies y &= \frac{16}{9}x^2 \text{ is the upper curve C_2} \end{align*}

## Problem 6.9.11

Sum the series $1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{13} - \dots$

### Solution

Consider the function defined by the infinite series

\begin{align*} f(x) = x + \frac{x^3}{3} - \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} + \frac{x^{11}}{11} - \frac{x^{13}}{13} - \dots \end{align*}

for $0 < x \leq 1$. The series is absolutely convergent for $\vert x \vert <1$, and therefore we can rearrange the terms:

\begin{align*} f(x) = \left(x + \frac{x^9}{9} + \dots \right) + \left(\frac{x^3}{3} + \frac{x^{11}}{11} + \dots \right) - \left(\frac{x^5}{5} + \frac{x^{13}}{13} + \dots \right) + \dots \end{align*}

We have for $0<x<1$,

\begin{align*} f'(x) &= (1 + x^8 + \dots) + (x^2 + x^{10} + \dots ) - (x^4 + x^{12} + \dots) - \dots) + \dots \ &= (1+x^2-x^4-x^6)(1 + x^8 + \dots) \ &= \frac{(1+x^2)(1-x^4)}{1-x^8}= \frac{1+x^2}{1+x^4} \end{align*}

Integrating and noting that $f(0)=0$, we get

\begin{align*} f(x) = \frac{arctan(\sqrt{2}x+1) - arctan(1-\sqrt{2}x)}{\sqrt{2}} \end{align*}

Therefore,

\begin{align*} 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{13} - \dots = f(1) = \frac{\pi}{2\sqrt{2}} \end{align*}