Chapter 2.5, Problem 47E

a.

To determine

To find: The first six polynomials in the given series.

Answer to Problem 47E

The first six polynomials of the series are said to be,

  $\begin{array}{c}{p}_0\left(x\right)=1\\ p_1\left(x\right)=1+x\\ p_2\left(x\right)=1+x+\frac{x^2}{2}\\ p_3\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}\\ p_4\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\\ p_5\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\\ p_6\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}\end{array}$

Explanation of Solution

Given information:

The series of polynomials is given as follows.

  $\begin{array}{l}{p}_0\left(x\right)=1\\ p_1\left(x\right)=1+x\\ p_2\left(x\right)=1+x+\frac{x^2}{2}\\ p_3\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}\end{array}$

The $n\text{th}$ polynomial in the series is constructed by adding the last term of the previous polynomial multiplied by $\frac{x}{n}$ to itself.

Calculation:

As given in the question, the last term of the fourth polynomial can be calculated as follows.

The last term of the third polynomial is $\frac{x^3}{6}$ . This should be multiplied by $\frac{x}{4}$ .

  $\frac{x^3}{6}\times \frac{x}{4}=\frac{x^4}{24}$

Then this is added to the third polynomial.

  $\begin{array}{c}{p}_4\left(x\right)=p_3\left(x\right)+\frac{x^4}{24}\\ p_4\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\end{array}$

The next two polynomials can be obtained according to the above process.

  $\begin{array}{c}{p}_5\left(x\right)=p_4\left(x\right)+\frac{x^4}{24}\times \frac{x}{5}\\ p_5\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\end{array}$

And

  $\begin{array}{c}{p}_6\left(x\right)=p_5\left(x\right)+\frac{x^5}{120}\times \frac{x}{6}\\ p_6\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}\end{array}$

b.

To determine

To graph:The first six polynomials in the series given.

Explanation of Solution

Given information:

The series of polynomials has been calculated as follows.

  $\begin{array}{c}{p}_0\left(x\right)=1\\ p_1\left(x\right)=1+x\\ p_2\left(x\right)=1+x+\frac{x^2}{2}\\ p_3\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}\\ p_4\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\\ p_5\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\\ p_6\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}\end{array}$

Graph:

The following graph shows all the six polynomials in the series.

  

Each line in different colors represents a polynomial in the series.

c.

To determine

To find:The derivative of the polynomial $p_6\left(x\right)$ .

Answer to Problem 47E

The derivative of the sixth polynomials of the series isfound to be,

  $p_6\text{'}\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}$

And it is identical to the fifth polynomial.

Explanation of Solution

Given information:

The fifth and sixth polynomial of the seriesaresaid to be,

  $\begin{array}{c}{p}_5\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\\ p_6\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}\end{array}$

Calculation:

The derivative of the sixth polynomial can be determined as follows.

  $\begin{array}{c}\frac{d\left(p_6\left(x\right)\right)}{dx}=\frac{d}{dx}\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}\right)\\ =\frac{d\left(1\right)}{dx}+\frac{d\left(x\right)}{dx}+\frac{d}{dx}\left(\frac{x^2}{2}\right)+\frac{d}{dx}\left(\frac{x^3}{6}\right)+\frac{d}{dx}\left(\frac{x^4}{24}\right)+\frac{d}{dx}\left(\frac{x^5}{120}\right)+\frac{d}{dx}\left(\frac{x^6}{720}\right)\\ =0+1+\frac{2x}{2}+\frac{3x^2}{6}+\frac{4x^3}{24}+\frac{5x^4}{120}+\frac{6x^5}{720}\\ p_6\text{'}\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\end{array}$

As calculated above, the derivative of the sixth polynomial is identical to the fifth polynomial. This can be plotted in the same figure.

  

Comparing these two functions, a conclusion can be made as these two polynomials closely vary for larger $x$ values and take different directions for minor $x$ values.

d.

To determine

To find:The function this series of polynomials is approaching.

Answer to Problem 47E

These polynomials approach to the exponential function $e^x$ when $n$ goes to infinity where

  $p_n\left(x\right)={\displaystyle \sum _{i=0}^n\frac{x^n}{n!}}$

Explanation of Solution

The series of polynomial which discussed in this question can be shortened as follows.

  $p_n\left(x\right)={\displaystyle \sum _{i=0}^n\frac{x^n}{n!}}$

Here by factorial $n$

  $\left(n!\right)$ , the product of all the natural number until $n$ is represented.

  $\begin{array}{c}\underset{n\to \infty }{\lim }{p}_n\left(x\right)=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=0}^n\frac{x^n}{n!}}\\ =1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\mathrm{...}+\frac{x^n}{n!}+\mathrm{...}\end{array}$

In a long run, when $n$ goes to infinity, this polynomial approaches to a special function that called as the exponential function. This exponential function is represented by $e$ .

If a function is said to be an exponential function of $x$ , then,

  $f\left(x\right)=e^x$

The exponential function is expanded/defined as follows.

  $e^x=$ $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\mathrm{...}+\frac{x^n}{n!}+\mathrm{...}$

Hence, this series of polynomials approaches the exponential function of $x$ .