Chapter 4.5, Problem 73E

(a)

To determine

To show: $g(a)=g\text{'}(a)=0$ and $g\text{'}\text{'}(x)=f\text{'}\text{'}(x)$

Answer to Problem 73E

The function is conformal except $z=i\pi$ .

Explanation of Solution

Given information:

Calculation:

  $y=f(x)$has a continuous second-order derivative and is the difference between $\Delta y=f\left(x\right)-f\left(a\right)$ and $f\text{'}\left(a\right)\Delta x=f\text{'}\left(a\right)\left(x-a\right)$

  $g\left(x\right)=f\left(x\right)-f\left(a\right)-f\text{'}\left(a\right)\left(x-a\right)$ then there is some real number between $x$ and $a$ such that

  $g\left(x\right)=\frac{1}{2}f\text{'}\text{'}\left(c\right){\left(x-a\right)}^2$

In order to show that $g\left(a\right)=g\text{'}\left(a\right)=0$ ,first find $g\text{'}(x)$ .

By the definition of absolute change in $g$ :

  $\begin{array}{l}\Delta g=g\left(a+\Delta x\right)-g\left(a\right)\\ \Delta g=g\text{'}\left(a\right)\Delta x\end{array}$

Differentiate $g\left(x\right)=f\left(x\right)-f\left(a\right)-f\text{'}\left(a\right)\left(x-a\right)$ with respect to $x$ gives:

  $\begin{array}{l}g\text{'}\left(x\right)=f\text{'}\left(x\right)-\frac{d}{dx}f\left(a\right)-\frac{d}{dx}\left(f\text{'}\left(a\right)\left(x-a\right)\right)\\ g\text{'}\left(x\right)=f\text{'}\left(x\right)-0-\left(x-a\right)\frac{d}{dx}f\text{'}\left(a\right)-f\text{'}\left(a\right)\frac{d}{dx}\left(x-a\right)\end{array}$

Product rule of differentiation is used.

  $g\text{'}\left(x\right)=f\text{'}\left(x\right)-0-f\text{'}\left(a\right)$

Differentiation of $f\text{'}(a)=0$ .

  $g\text{'}(x)=f^{\text{'}}\left(x\right)-f\text{'}\left(a\right)$

Now substitute $x=a$ in $g\text{'}(x)=f^{\text{'}}\left(x\right)-f\text{'}\left(a\right)$ gives:

  $\begin{array}{l}g\text{'}(a)=f^{\text{'}}\left(a\right)-f\text{'}\left(a\right)\\ g\text{'}(a)=0\mathrm{....}(1)\end{array}$

Now substitute $x=a$ in $g\left(x\right)=f\left(x\right)-f\left(a\right)-f\text{'}\left(a\right)\left(x-a\right)$ gives:

  $\begin{array}{l}g\left(a\right)=f\left(a\right)-f\left(a\right)-f\text{'}\left(a\right)\left(a-a\right)\\ g\left(a\right)=0\mathrm{....}(2)\end{array}$

By (1) and (2) it is proved that $g\left(a\right)=g\text{'}\left(a\right)=0$ .

Now differentiate $g\text{'}(x)=f^{\text{'}}\left(x\right)-f\text{'}\left(a\right)$ again with respect to $x$ gives:

  $\begin{array}{l}g\text{'}\text{'}(x)=f\text{'}\text{'}\left(x\right)-0\\ g\text{'}\text{'}(x)=f\text{'}\text{'}\left(x\right)\end{array}$

Since $f\text{'}(a)$ is constant hence $\frac{d}{dx}f\text{'}(a)=0$ .

This proves the second result.

(b)

To determine

To find: Use intermediate value theorem to show that for any real number $r$ between A and B there is some value of c in $\left[a,x\right]$ for which $r=\frac{1}{2}g\text{'}\text{'}\left(c\right)$ .

Explanation of Solution

Given information:

Calculation:

  $y=f(x)$has a continuous second order derivative and $g$ to be the difference between $\Delta y=f\left(x\right)-f\left(a\right)$ and $f\text{'}\left(a\right)\Delta x=f\text{'}\left(a\right)\left(x-a\right)$

  $g\left(x\right)=f\left(x\right)-f\left(a\right)-f\text{'}\left(a\right)\left(x-a\right)$ then there is some real number between $x$ and $a$ such that

  $g\left(x\right)=\frac{1}{2}f\text{'}\text{'}\left(c\right){\left(x-a\right)}^2$

Intermediate theorem states that for any function $f$ which is continuous over the interval $\left[a,b\right]$ the function will take any value $c$ between $f\left(a\right)$ and $f(c)$ over the interval .It means that for any value $L$ between $f\left(a\right)$ and $f\left(b\right)$ there is value $c$ in $\left[a,b\right]$ for which $f\left(c\right)=L$ .

Now let $A$ be the minimum value $\frac{1}{2}g\text{'}\text{'}\left(t\right)$ and $B$ be the maximum value $\frac{1}{2}g\text{'}\text{'}\left(t\right)$ for $t$ in the interval $\left[a,x\right]$

  $g$ is continuous on $\left[a,x\right]$ then there exist a number $A,B\in \left[a,x\right]$ such that $g\left(A\right)\le g\left(x\right)\le g\left(B\right)$ for all $x\in \left[a,x\right]$ can be written as the supremum $M$ and infimum $m$ of the range $g\left(x\right):A\le r\le B$ exist and there exist number $c\in \left[a,x\right]$ such that $g\left(A\right)=r=m$ and $g\left(B\right)=M$ .

Given that $m=\frac{1}{2}g\text{'}\text{'}\left(t\right)$ hence by the above definition it can be written that

  $r=\frac{1}{2}g\text{'}\text{'}\left(c\right)$

Hence proved.

(c)

To determine

To Prove: $g\text{'}\text{'}(t)-2A\ge 0$ and $g\text{'}\text{'}(t)-2B\le 0$ in the interval $\left[a,x\right]$

Explanation of Solution

Given information:

By the proof in part (b) $g$ is continuous on $\left[a,x\right]$ then there exist a number $A,B\in \left[a,x\right]$ such that $g\left(A\right)\le g\left(x\right)\le g\left(B\right)$ for all $x\in \left[a,x\right]$ can be written as the supremum $M$ and infimum $m$ of the range $g\left(x\right):r\in \left[a,x\right]$ exist and there exist number $A,B\in \left[a,x\right]$ such that $g\left(A\right)=m$ and $g\left(d\right)=M$ .

Calculation:

Given that $m=\frac{1}{2}g\text{'}\text{'}\left(t\right)$ hence by the above definition it can be written that

  $r=\frac{1}{2}g\text{'}\text{'}\left(t\right)$

Since $A\le r\le B$ it can be written that $A\le \frac{1}{2}g\text{'}\text{'}\left(t\right)$ .

  $\begin{array}{l}2A\le g\text{'}\text{'}\left(t\right)\\ g\text{'}\text{'}\left(t\right)\ge 2A\\ g\text{'}\text{'}\left(t\right)-2A\ge 0\end{array}$

Similarly,

  $\begin{array}{l}B\ge \frac{1}{2}g\text{'}\text{'}\left(t\right)\\ 2B\ge g\text{'}\text{'}(t)\\ g\text{'}\text{'}(t)\le 2B\\ g\text{'}\text{'}(t)-2B\le 0\end{array}$

(d)

To determine

To prove: $g\text{'}(t)-2A\left(t-a\right)\ge 0$ and $g\text{'}(t)-2B\left(t-a\right)\le 0$ in the interval $\left[a,x\right]$

Explanation of Solution

Given information:

Given that $g\text{'}\left(a\right)-2A\left(a-a\right)=g\text{'}\left(a\right)-2B\left(a-a\right)=0$

Calculation:

Let $g$ be a continuous function on $\left[a,b\right]$ and differentiable on $\left(a,b\right)$ then if $g\text{'}>0$ at each point of $\left(a,b\right)$ then $g$ increases on $\left[a,b\right]$ . If $g\text{'}<0$ at each point of $\left(a,b\right)$ then $g$ decreases on $\left[a,b\right]$ .

Using the given fact that $g\text{'}\left(a\right)-2A\left(a-a\right)=g\text{'}\left(a\right)-2B\left(a-a\right)=0$ the function $g\left(t\right)$

Can be written as $g\text{'}\left(t\right)-2A\left(t-a\right)$ and $g\text{'}\left(t\right)-2B\left(t-a\right)$

Since $g\left(A\right)=m$ is the minimum value on the interval $\left[a,x\right]$ .

The function is increasing on $\left[a,b\right]$ if $g\text{'}>0$ therefore $g\text{'}\left(t\right)-2A\left(t-a\right)>0$ .

Since $A$ is the minimum value is given the value start increasing after this point so, by the definition for increasing function :

  $g\text{'}\left(t\right)-2A\left(t-a\right)\ge 0$

Similarly,

Since $B$ is the maximum value is given the value start decreasing after this point so ,by the definition for decreasing function :

  $g\text{'}\left(t\right)-2B\left(t-a\right)\le 0$

Hence proved.

(e)

To determine

To Prove: $g\text{'}(t)-2A{\left(t-a\right)}^2\ge 0$ and $g\text{'}(t)-2B{\left(t-a\right)}^2\le 0$ in the interval $\left[a,x\right]$

Explanation of Solution

Given information:

Given that $g\left(a\right)-2A{\left(a-a\right)}^2=g\left(a\right)-2B{\left(a-a\right)}^2=0$

Calculation:

Let $g$ be a continuous function on $\left[a,b\right]$ and differentiable on $\left(a,b\right)$ then if $g\text{'}>0$ at each point of $\left(a,b\right)$ then $g$ increases on $\left[a,b\right]$ . If $g\text{'}<0$ at each point of $\left(a,b\right)$ then $g$ decreases on $\left[a,b\right]$ .

Using the given fact that $g\left(a\right)-2A{\left(a-a\right)}^2=g\left(a\right)-2B{\left(a-a\right)}^2=0$ the function $g\left(t\right)$

Can be written as $g\left(t\right)-2A{\left(t-a\right)}^2$ and $g\left(t\right)-2B{\left(t-a\right)}^2$

Since $g\left(A\right)=m$ is the minimum value on the interval $\left[a,x\right]$ .

The function is increasing on $\left[a,b\right]$ if $g\text{'}>0$ therefore $g\text{'}\left(t\right)-2A{\left(t-a\right)}^2>0$ .

Since $A$ is the minimum value is given the value start increasing after this point so ,by the definition for increasing function :

  $g\left(t\right)-2A{\left(t-a\right)}^2\ge 0$

Similarly,

Since $B$ is the maximum value is given the value start decreasing after this point so ,by the definition for decreasing function :

  $g\left(t\right)-2B{\left(t-a\right)}^2\le 0$ for all $t$ in $\left[a,x\right]$

Hence proved.

(e)

To determine

To Prove: $A\le \frac{g\left(x\right)}{{\left(x-a\right)}^2}\le B$ and $\left|\Delta y-f\text{'}\left(a\right)\Delta x\right|=\frac{1}{2}\left|f\text{'}\text{'}\left(c\right)\right|{\left|\Delta x\right|}^2$

Explanation of Solution

Given information:

From part (e) $g\left(t\right)-2A{\left(t-a\right)}^2\ge 0$ and $g\left(t\right)-2B{\left(t-a\right)}^2\le 0$

Calculation:

It is given $g\left(x\right)=\frac{1}{2}f\text{'}\text{'}\left(c\right){\left(x-a\right)}^2$ which can be written as :

  $\frac{g\left(x\right)}{{\left(x-a\right)}^2}=\frac{1}{2}f\text{'}\text{'}\left(c\right)\mathrm{...}(1)$

From part (b)

Now let $A$ be the minimum value $\frac{1}{2}g\text{'}\text{'}\left(t\right)$ and $B$ be the maximum value $\frac{1}{2}g\text{'}\text{'}\left(t\right)$ for $t$ in the interval $\left[a,x\right]$ and $r=\frac{1}{2}g\text{'}\text{'}\left(c\right)$ because $A\le r\le B$ .

Using (1) $A\le \frac{g\left(x\right)}{{\left(x-a\right)}^2}\le B$ .

Therefore, from part (b) there is some value $c$ on $\left[a,x\right]$ such that

  $\frac{g\left(x\right)}{{\left(x-a\right)}^2}=\frac{1}{2}f\text{'}\text{'}\left(c\right)=\frac{1}{2}g\text{'}\text{'}\left(c\right)$

Substitute the value of $g\left(x\right)$ by part (a)gives:

  $f\left(x\right)-f\left(a\right)-f\text{'}\left(a\right)\left(x-a\right)=\frac{1}{2}f\text{'}\text{'}\left(c\right){\left(x-a\right)}^2\mathrm{.....}(1)$

Given,

  $\Delta y=f\left(x\right)-f\left(a\right)\mathrm{...}(2)$

  $f\text{'}\left(a\right)\Delta x=f\text{'}\left(a\right)\left(x-a\right)\mathrm{...}(3)$

Subtract equation (2) from (3) gives:

  $\Delta y-f\text{'}\left(a\right)\Delta x=f\left(x\right)-f\left(a\right)-f^{\text{'}}\left(a\right)\left(x-a\right)$

Substitute the value of $f\left(x\right)-f\left(a\right)-f^{\text{'}}\left(a\right)\left(x-a\right)$ from equation (1) gives:

  $\Delta y-f\text{'}\left(a\right)\Delta x=\frac{1}{2}\left|f\text{'}\text{'}\left(c\right)\right|\Delta x^2$

Hence proved.