Chapter 4.3, Problem 62E

(a)

To determine

To prove: The equation for $x\ge 0$ .

Answer to Problem 62E

The equation is proved.

Explanation of Solution

Given information:

The equation is $e^x\ge 1+x$ .

Calculation:

Let $f(0)=e^x-1-x$

Compute the value $f(0)$ as follows:

  $\begin{array}{c}f(0)=e^0-1-0\\ =1-1\\ =0\end{array}$

Derivative of the function $f(x)=e^x-1-x:$

  $\begin{array}{c}f(x)=\frac{d}{dx}(e^x-1-x)\\ =\frac{d}{dx}(e^x)-\frac{d}{dx}(1)-\frac{d}{dx}(x)\\ =e^x-0-1\\ =e^x-1\\ \\ \end{array}$

So, $f(x)=e^x-1$

For $x\ge 0$ , we have $f(x)=e^x-1\ge 0$ .

Thus, the function $f(x)=e^x-1-x$ is increasing on $(0,\infty )$ .

Since, $f(0)=0$ and $f$ is increasing on $(0,\infty )$ , $f(x)\ge 0$ for $x\ge 0$ .

It follows that

  $\begin{array}{l}{e}^x-1-x\ge 0\\ e^x\ge 1+x\\ \end{array}$

Therefore, the statement $e^x\ge 1+x$is proved for $x\ge 0$ .

(b)

To determine

To prove: The equation for $x\ge 0$ .

Answer to Problem 62E

The equation $e^x\ge 1+x+\frac{1}{2}{x}^2$ is proved for $x\ge 0$ .

Explanation of Solution

Given information:

The given equation is $e^x\ge 1+x+\frac{1}{2}{x}^2$ .

Calculation:

Let $f(x)=e^x-1-x-\frac{1}{2}{x}^2.$

Compute the value $\begin{array}{l}f(0)=e^0-1-x-\frac{1}{2}(0^2)\\ =1-1\\ =0\end{array}$

Calculate the derivative of the function $f(x)=e^x-1-x-\frac{1}{2}{x}^2.$

  $\begin{array}{l}f(x)=\frac{d}{dx}(e^x-1-x-\frac{1}{2}{x}^2)\\ =\frac{d}{dx}(e^x)-\frac{d}{dx}(1)-\frac{d}{dx}(x)-\frac{1}{2}\frac{d}{dx}(x^2)\\ =e^x-0-1-\frac{1}{2}(2x)\\ =e^x-1-x\end{array}$

Thus, $f^{\text{'}}(x)=e^x-1-x$ .

Since. $f^{\text{'}}(x)=e^x-1\ge 0$ for $x\ge 0$ , so the function $f^{\text{'}}(x)=e^x-1-x$ is increasing on $(0,\infty )$ . And since $f^{\text{'}}(0)=0,$ therefore, $f^{\text{'}}(x)=e^x-1-x$ is positive for $x\ge 0$ .

So, the function $f(x)=e^x-1-x-\frac{1}{2}{x}^2.$ is increasing on $(0,\infty )$ .

Since, $f^{\text{'}}(0)=0,$ and $f$ is increasing on $(0,\infty )$ , $f(x)\ge 0$ for $x\ge 0$ .

It follows that

  $e^x\ge 1+x+\frac{1}{2}{x}^2$

Therefore, the statement $e^x\ge 1+x+\frac{1}{2}{x}^2$ is proved for $x\ge 0$ .

(c)

To determine

To prove: The equation for $x\ge 0$ and positive integer n.

Answer to Problem 62E

The equation is proved shown below.

Explanation of Solution

Given information:

The given equation is $e^x\ge 1+x+\frac{x^2}{2!}+\mathrm{.....}+\frac{x^n}{n!}$

Calculation:

Let the function $f$ be $f(x)=e^x-1-x-\frac{x^2}{2!}-\mathrm{.....}-\frac{x^n}{n!}$

First, prove that the statement is true for $n=1$ .

Let $f(x)=e^x-1-x$.

compute the value $f(0)$ as follows:

  $\begin{array}{l}f(0)=e^0-1-0\\ =1-1\\ =0\end{array}$

Calculate the derivative of the function $f(x)=e^x-1-x$:

  $\begin{array}{l}f(x)=\frac{d}{dx}(e^x-1-x)\\ =\frac{d}{dx}(e^x)-\frac{d}{dx}(1)-\frac{d}{dx}(x)\\ =e^x-0-1\\ =e^x-1\end{array}$

So, $f^{\text{'}}(x)=e^x-1$

For $x\ge 0$ , we have $f^{\text{'}}(x)=e^x-1\ge 0$ .

Thus, the function $f(x)=e^x-1-x$ is increasing on $(0,\infty )$ .

Since, $f(0)=0$ and $f$ is increasing on $(0,\infty )$ , $f(x)\ge 0$ for $x\ge 0$ .

It follows that

  $\begin{array}{l}{e}^x-1-x\ge 0\\ e^x\ge 1+x\end{array}$

Thus, $e^x\ge 1+x$for $x\ge 0$ .

Therefore, the statement is proved for $n=1$ .

Now, let $f(x)=e^x-1-x-\frac{x^2}{2!}-\mathrm{.....}-\frac{x^k}{k!}$

Assume that the statement is true for $n=k$ .

That means, $e^x\ge 1+x+\frac{x^2}{2!}+\mathrm{.....}+\frac{x^k}{k!}$for $x\ge 0$ .

That implies $f(x)=e^x-1-x-\frac{x^2}{2!}-\mathrm{.....}-\frac{x^k}{k!}\ge 0$for $x\ge 0$ .

Now. Let $f(x)=e^x-1-x-\frac{x^2}{2!}-\mathrm{.....}-\frac{x^k}{k!}-\frac{x^{k+1}}{k+1!}$.

Then, $f(x)=e^x-1-x-\frac{x^2}{2!}-\mathrm{.....}-\frac{x^k}{k!}\ge 0$by assumption, so,

  $f(x)=e^x-1-x-\frac{x^2}{2!}-\mathrm{.....}-\frac{x^k}{k!}-\frac{x^{k+1}}{k+1!}$is increasing on $(0,\infty )$ .

Thus, $e^x\ge 1+x+\frac{x^2}{2!}+\mathrm{.....}+\frac{x^k}{k!}+\frac{x^k+1}{k+1!}$.

Therefore, for $x\ge 0$ , $e^x\ge 1+x+\frac{x^2}{2!}+\mathrm{.....}+\frac{x^n}{n!}$for every positive integer $n$ , by mathematical induction.

Therefore, the statement $e^x\ge 1+x+\frac{x^2}{2!}+\mathrm{.....}+\frac{x^n}{n!}$is proved for $x\ge 0$ and for any positive integer $n$ , by using mathematical induction.