Question

Express the function as the sum of a power series by first using partial fractions. f(x) = (x + 7) /(2x^2 − 11x − 6 )

f(x) = ∞ n = 0 Find the interval of convergence. (Enter your answer using interval notation.)

Answer 1

Factorize the denominator:

$\dfrac{x+7}{2x^2-11x-6}=\dfrac{x+7}{(x-6)(2x+1)}$

Then we look for coefficients $a,b$ such that

$\dfrac{x+7}{(x-6)(2x+1)}=\dfrac a{x-6}+\dfrac b{2x+1}$

$\implies x+7=a(2x+1)+b(x-6)$

• If $x=6$, then $13=13a\implies a=1$.
• If $x=-\frac12$, then $\frac{13}2=-\frac{13}2b\implies b=-1$.

So we have

$f(x)=\dfrac12\left(\dfrac1{x-6}-\dfrac1{2x+1}\right)$

Recall that for $|x|<1$, we have

$\displaystyle\frac1{1-x}=\sum_{n\ge1}x^n$

Now,

$\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}$

$\implies\displaystyle\frac1{x-6}=-\frac16\sum_{n\ge1}\left(\frac x6\right)^n$

for $\left|\frac x6\right|<1$, or $|x|<6$, and

$\dfrac1{2x+1}=\dfrac1{1-(-2x)}$

$\implies\displaystyle\frac1{2x+1}=\sum_{n\ge1}(-2x)^n$

for $|-2x|<1$, or $|x|<\frac12$.

The interval of convergence for $f(x)$ is the intersection of these two intervals,

$\boxed{|x|<\dfrac12}$