Answer:
[tex]\frac{x+6}{2x^2-9x+5}=-\sum_{n=0}^{\infty} [(-2)^{n}x^{n} + \frac{x^{n}}{5^{n+1}}][/tex]
when:
[tex]|x|<\frac{1}{2}[/tex]
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.
[tex]\frac{x+6}{2x^2-9x+5}=\frac{x+6}{(2x+1)(x-5)}[/tex]
Next, we can build our partial fractions, like this:
[tex]\frac{x+6}{(2x+1)(x-5)}=\frac{A}{2x+1}+\frac{B}{x-5}[/tex]
we can then add the two fraction on the right to get:
[tex]\frac{x+6}{(2x+1)(x-5)}=\frac{A(x-5)+B(2x+1)}{(2x+1)(x-5)}[/tex]
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:
[tex]x+6=A(x-5)+B(2x+1)[/tex]
and expand:
[tex]x+6=Ax-5A+2Bx+B[/tex]
we can now group the terms so we get:
[tex]x+6=Ax+2Bx-5A+B[/tex]
[tex]x+6=(Ax+2Bx)+(-5A+B)[/tex]
and factor:
[tex]x+6=(A+2B)x+(-5A+B)[/tex]
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:
[tex]\frac{x+6}{(2x+1)(x-5)}=\frac{A}{2x+1}+\frac{B}{x-5}[/tex]
[tex]\frac{x+6}{(2x+1)(x-5)}=-\frac{1}{2x+1}+\frac{1}{x-5}[/tex]
and we can now use the partial fractions to build our series. Let's start with the first fraction:
[tex]-\frac{1}{2x+1}[/tex]
We can rewrite this fraction as:
[tex]-\frac{1}{1-(-2x)}[/tex]
We can now use the following rule to build our power fraction:
[tex]\sum_{n=0}^{\infty} ar^{n} = \frac{a}{1-r}[/tex]
when |r|<1
in this case a=1 and r=-2x so:
[tex]-\frac{1}{1-(-2x)}=-\sum_{n=0}^{\infty} (-2x)^n[/tex]
or
[tex]-\frac{1}{1-(-2x)}=-\sum_{n=0}^{\infty} (-2)^{n} x^{n}[/tex]
for: |-2x|<1
or: [tex] |x|<\frac{1}{2} [/tex]
Next, we can work with the second fraction:
[tex]\frac{1}{x-5}[/tex]
On which we can factor a -5 out so we get:
[tex]-\frac{1}{5(1-\frac{x}{5})}[/tex]
In this case: a=-1/5 and r=x/5
so our series will look like this:
[tex]-\frac{1}{5(1-\frac{x}{5})}=-\frac{1}{5}\sum_{n=0}^{\infty} (\frac{x}{5})^n[/tex]
Which can be simplified to:
[tex]-\frac{1}{5(1-\frac{x}{5})}=-\sum_{n=0}^{\infty} \frac{x^n}{5^(n+1)}[/tex]
when:
[tex]|\frac{x}{5}|<1[/tex]
or
|x|<5
So we can now put all the series together to get:
[tex]\frac{x+6}{2x^2-9x+5}=-\sum_{n=0}^{\infty} [(-2)^{n}x^{n} + \frac{x^{n}}{5^{n+1}}}[/tex]
when:
[tex]|x|<\frac{1}{2}[/tex]
We use the smallest interval of convergence for x since that's the one the whole series will be defined for.
