Chapter 8.4, Problem 86E

To determine

To simplify: The difference quotient $\frac{f\left(x+h\right)-f\left(x\right)}{h}$ , where $f\left(x\right)=x^8$ by using the Binomial Theorem.

Answer to Problem 86E

The simplified form is $8x^7+28x^{6}h+56x^5{h}^2+70x^4{h}^3+56x^3{h}^4+28x^2{h}^5+8x{h}^6+h^7$ .

Explanation of Solution

Given information: The functionis $f\left(x\right)=x^8$ .

Theorem used:The Binomial Theorem: ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\cdots {+}_n{C}_r{x}^{n-r}{y}^r+\cdots +nx{y}^{n-1}+y^n$ , the coefficient of $x^{n-r}{y}^r$ is ${}_n{C}_r=\frac{n!}{\left(n-r\right)!r!}$ .

Calculation:

Here, $n=8$ .

First write out the Pascal’s triangle such that the row that begins with 1, 6.

  $\begin{array}{ccccccccccccc}& & & & & & 1& & & & & & \\ & & & & & 1& & 1& & & & & \\ & & & & 1& & 2& & 1& & & & \\ & & & 1& & 3& & 3& & 1& & & \\ & & 1& & 4& & 6& & 4& & 1& & \\ & 1& & 5& & 10& & 10& & 5& & 1& \\ 1& & 6& & 15& & 20& & 15& & 6& & 1\end{array}$

From the Pascal’s triangle, it is observed that the binomial coefficients of the expression are 1, 6, 15, 20, 15, 6 and 1.

Simplify $\frac{f\left(x+h\right)-f\left(x\right)}{h}$ , where $f\left(x\right)=x^8$ by using formula in Binomial theorem.

  $\begin{array}{c}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{{\left(x+h\right)}^8-x^8}{h}\\ =\frac{x^8+8x^{7}h+28x^6{h}^2+56x^5{h}^3+70x^4{h}^4+56x^3{h}^5+28x^2{h}^6+8x^1{h}^7+h^8-x^8}{h}\\ =\frac{8x^{7}h+28x^6{h}^2+56x^5{h}^3+70x^4{h}^4+56x^3{h}^5+28x^2{h}^6+8x^1{h}^7+h^8}{h}\\ =8x^7+28x^{6}h+56x^5{h}^2+70x^4{h}^3+56x^3{h}^4+28x^2{h}^5+8x{h}^6+h^7\end{array}$

Thus, the simplified form is $8x^7+28x^{6}h+56x^5{h}^2+70x^4{h}^3+56x^3{h}^4+28x^2{h}^5+8x{h}^6+h^7$ .