Chapter 8, Problem 16P

(a)

To determine

To Show: The MacLaurin series of the function $f\left(x\right)=\frac{x}{1-x-x^2}$ is ${\displaystyle \sum _{n=1}^{\infty }{f}_n{x}^n}$ .

Explanation of Solution

Given:

The $f_n$ is the Fibonacci number that is $f_1=1$ , $f_2=1$ and $f_n=f_{n-1}+f_{n-2}$ for $n\ge 3$ .

Calculation:

The given function is,

  $f\left(x\right)=\frac{x}{1-x-x^2}$

Then write the function as,

  $\begin{array}{c}\frac{x}{1-x-x^2}=c_0+c_{1}x+c_2{x}^2+\mathrm{...}\\ ={\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^n}\end{array}$

Multiply both the sides of the equation by $\left(1-x-x^2\right)$ as,

  $\begin{array}{l}\left(1-x-x^2\right)\frac{x}{1-x-x^2}=\left(1-x-x^2\right){\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^n}\\ x={\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^n}-{\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^{n+1}}-{\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^{n+2}}\end{array}$

Then, expand the above equation and factorize as,

  $x=c_0+\left(c_1-c_0\right)x+\left(c_2-c_1-c_0\right)x^2+\left(c_3-c_2-c_1\right)x^3\mathrm{...}$

Then, compare the coefficient and the power of $x$ as,

  $\begin{array}{l}{c}_0=0\\ c_1-c_0=1\\ c_1=1\\ c_2-c_1-c_0=0\end{array}$

Solve further as,

  $\begin{array}{l}{c}_2=c_1+c_0+0\\ c_2=1\\ c_3-c_2-c_1=0\\ c_3=2\end{array}$

Thus, the required sequence is,

  $c_n=c_{n-1}+c_{n-2}$

Thus, for $n\ge 3$ then each $c_n$ is equal to the $n$ th Fibonacci number. Then,

  $\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^n}={\displaystyle \sum _{n=1}^{\infty }{c}_n{x}^n}\\ ={\displaystyle \sum _{n=1}^{\infty }{f}_n{x}^n}\end{array}$

Hence, proved.

(b)

To determine

To Find: The writing $f\left(x\right)$ as the sum of the partial fraction and thereby obtaining the MacLaurin series in the different way and then find the explicit for the nth Fibonacci number.

Answer to Problem 16P

The explicit formula for the nth Fibonacci number is $\frac{1}{\sqrt{5}}{\displaystyle \sum _{n=1}^{\infty }\left[\frac{{\left(1+\sqrt{5}\right)}^n-{\left(1-\sqrt{5}\right)}^n}{2^n}\right]x_n}$ .

Explanation of Solution

Calculation:

The given function is,

  $f\left(x\right)=\frac{x}{1-x-x^2}$

Factorize the denominator as,

  $\begin{array}{c}f\left(x\right)=\frac{x}{-\left(1-x-x^2\right)}\\ =-\frac{-x}{\left(x^2+x+\frac{1}{4}-\frac{1}{4}-1\right)}\\ =\frac{-x}{\left[x+\frac{1+\sqrt{5}}{2}\right]\left[x+\frac{1-\sqrt{5}}{2}\right]}\end{array}$

Then, by the partial decomposition,

  $\begin{array}{l}\frac{-x}{\left[x+\frac{1+\sqrt{5}}{2}\right]\left[x+\frac{1-\sqrt{5}}{2}\right]}=\frac{A}{x+\frac{1+\sqrt{5}}{2}}+\frac{B}{x+\frac{1-\sqrt{5}}{2}}\\ -x=A\left(x+\frac{1-\sqrt{5}}{2}\right)+B\left(x+\frac{1+\sqrt{5}}{2}\right)\\ -x=\left(A+B\right)x+A\left(\frac{1-\sqrt{5}}{2}\right)+B\left(\frac{1+\sqrt{5}}{2}\right)\\ 0=A\left(\frac{1-\sqrt{5}}{2}\right)+B\left(\frac{1+\sqrt{5}}{2}\right)\end{array}$

Solve further as,

  $\begin{array}{l}0=\left(A+B\right)+\sqrt{5}\left(-A+B\right)\\ 0=-1+\sqrt{5}\left(-A-1-A\right)\\ 0=-1+\sqrt{5}\left(-2A-1\right)\\ A=\frac{1+\sqrt{5}}{-2\sqrt{5}}\end{array}$

Solve for B as,

  $B=\frac{1-\sqrt{5}}{2\sqrt{5}}$

Thus, the required expression as,

  $\frac{-x}{\left[x+\frac{1+\sqrt{5}}{2}\right]\left[x+\frac{1-\sqrt{5}}{2}\right]}=\frac{\frac{1+\sqrt{5}}{-2\sqrt{5}}}{x+\frac{1+\sqrt{5}}{2}}+\frac{\frac{1-\sqrt{5}}{2\sqrt{5}}}{x+\frac{1-\sqrt{5}}{2}}$

Then, solve the equation as,

  $\begin{array}{c}\frac{-x}{\left[x+\frac{1+\sqrt{5}}{2}\right]\left[x+\frac{1-\sqrt{5}}{2}\right]}=-\frac{1}{\sqrt{5}}{\displaystyle \sum _{n=0}^{\infty }{\left(-\frac{2}{1+\sqrt{5}}x\right)}^n}+\frac{1}{\sqrt{5}}{\displaystyle \sum _{n=0}^{\infty }{\left(-\frac{2}{1-\sqrt{5}}x\right)}^n}\\ =\frac{1}{\sqrt{5}}{\displaystyle \sum _{n=0}^{\infty }{\left(-\frac{2}{1-\sqrt{5}}x\right)}^n-{\left(-\frac{2}{1+\sqrt{5}}x\right)}^n}\\ =\frac{1}{\sqrt{5}}{\displaystyle \sum _{n=1}^{\infty }\left[\frac{{\left(-2\right)}^n{\left(1+\sqrt{5}\right)}^n-{\left(-2\right)}^n{\left(1-\sqrt{5}\right)}^n}{{\left(1-\sqrt{5}\right)}^n{\left(1+\sqrt{5}\right)}^n}\right]x_n}\\ =\frac{1}{\sqrt{5}}{\displaystyle \sum _{n=1}^{\infty }\left[\frac{{\left(1+\sqrt{5}\right)}^n-{\left(1-\sqrt{5}\right)}^n}{2^n}\right]x_n}\end{array}$

Then, from part (a),

  ${\displaystyle \sum _{n=1}^{\infty }{f}_n{x}^n}=\frac{1}{\sqrt{5}}{\displaystyle \sum _{n=1}^{\infty }\left[\frac{{\left(1+\sqrt{5}\right)}^n-{\left(1-\sqrt{5}\right)}^n}{2^n}\right]x_n}$ =

Thus, the explicit formula for the nth Fibonacci number is $\frac{1}{\sqrt{5}}{\displaystyle \sum _{n=1}^{\infty }\left[\frac{{\left(1+\sqrt{5}\right)}^n-{\left(1-\sqrt{5}\right)}^n}{2^n}\right]x_n}$ .