Question

Express the function as the sum of a power series by first using partial fractions. f(x)=x+62x2−9x−5

Answer 1

Answer:

$\frac{x+6}{2x^2-9x+5}=-\sum_{n=0}^{\infty} [(-2)^{n}x^{n} + \frac{x^{n}}{5^{n+1}}]$

when:

$|x|$

Step-by-step explanation:

In order to solve this problem, we must begin by splitting the function into its partial fractions, so we must first factor the denominator.

$\frac{x+6}{2x^2-9x+5}=\frac{x+6}{(2x+1)(x-5)}$

Next, we can build our partial fractions, like this:

$\frac{x+6}{(2x+1)(x-5)}=\frac{A}{2x+1}+\frac{B}{x-5}$

we can then add the two fraction on the right to get:

$\frac{x+6}{(2x+1)(x-5)}=\frac{A(x-5)+B(2x+1)}{(2x+1)(x-5)}$

Since we need this equation to be equivalent, we can get rid of the denominators and set the numerators equal to each other, so we get:

$x+6=A(x-5)+B(2x+1)$

and expand:

$x+6=Ax-5A+2Bx+B$

we can now group the terms so we get:

$x+6=Ax+2Bx-5A+B$

$x+6=(Ax+2Bx)+(-5A+B)$

and factor:

$x+6=(A+2B)x+(-5A+B)$

so we can now build a system of equations:

A+2B=1

-5A+B=6

and solve simultaneously, this one can be solved by substitution, so we get>

A=1-2B

-5(1-2B)+B=6

-5+10B+B=6

11B=11

B=1

A=1-2(1)

A=-1

So we can use these values to build our partial fractions:

$\frac{x+6}{(2x+1)(x-5)}=\frac{A}{2x+1}+\frac{B}{x-5}$

$\frac{x+6}{(2x+1)(x-5)}=-\frac{1}{2x+1}+\frac{1}{x-5}$

and we can now use the partial fractions to build our series. Let's start with the first fraction:

$-\frac{1}{2x+1}$

We can rewrite this fraction as:

$-\frac{1}{1-(-2x)}$

We can now use the following rule to build our power fraction:

$\sum_{n=0}^{\infty} ar^{n} = \frac{a}{1-r}$

when |r|<1

in this case a=1 and r=-2x so:

$-\frac{1}{1-(-2x)}=-\sum_{n=0}^{\infty} (-2x)^n$

or

$-\frac{1}{1-(-2x)}=-\sum_{n=0}^{\infty} (-2)^{n} x^{n}$

for: |-2x|<1

or: $|x|$

Next, we can work with the second fraction:

$\frac{1}{x-5}$

On which we can factor a -5 out so we get:

$-\frac{1}{5(1-\frac{x}{5})}$

In this case: a=-1/5 and r=x/5

so our series will look like this:

$-\frac{1}{5(1-\frac{x}{5})}=-\frac{1}{5}\sum_{n=0}^{\infty} (\frac{x}{5})^n$

Which can be simplified to:

$-\frac{1}{5(1-\frac{x}{5})}=-\sum_{n=0}^{\infty} \frac{x^n}{5^(n+1)}$

when:

$|\frac{x}{5}|$

or

|x|<5

So we can now put all the series together to get:

$\frac{x+6}{2x^2-9x+5}=-\sum_{n=0}^{\infty} [(-2)^{n}x^{n} + \frac{x^{n}}{5^{n+1}}}$

when:

$|x|$

We use the smallest interval of convergence for x since that's the one the whole series will be defined for.