Q. 16E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: To find the first few terms in the power series for the quotient q(x) in Problem 15, treat the power series in the numerator and denominator as "long polynomials" and carry out long division. That is, perform

16. $1+x+\frac{1}{2}+...$ 

Step-by-Step Solution

Answer

The series solution for q(x) is q(x) = 1-(x/2)+(x²/4)-(x³/24)+....

Step 1: solution by long division method

 

We will solve the previous problem using the method of long division.

 q(x)=$\left(\sum _{n=2}^{\infty }\frac{1}{2^n}{x}^n\right)/\left(\sum _{n=0}^{\infty }\frac{1}{n!}{x}^n\right)$

Numerator = $\sum _{n=0}^{\infty }\frac{x^n}{2^n}=1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8}+...$

 Denominator = $\sum _{b=0}^{\infty }\frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+...$

Now, performing the long division.

1+x+(x²/2)+(x³/6)+...$1-(x/2)+\left(x^2/4\right)-\left(x^3/24\right)+....1+(x/2)+\left(x^2/4\right)+\left(x^3/8\right)+...$

                 

                            $\frac{-1\pm x\pm (x^2/2)\pm (x^3/6)\pm ...}{-(x/2)-(x^2/4)-(x^3/24)-...}$

                            $\frac{\pm (x/2)\pm (x^2/2)\pm (x^3/24)\pm ...}{(x^2/4)+(5x^3/24)+...}$

                            $\frac{-(x^2/4)\pm (x^3/4)\pm ...}{-x^3/24-...}$

                             $\frac{\pm x^3/24\pm ...}{0}$

Step 2: Conclusion

The series solution for q(x) is q(x) = 1-(x/2)+(x²/4)-(x³/24)+....