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OlgaM077 [116]
3 years ago
12

Express the function as the sum of a power series by first using partial fractions. f(x) = 8 x2 − 4x − 12 f(x) = ∞ n = 0 find th

e interval of convergence. (enter your answer using interval notation.)
Mathematics
1 answer:
alexandr1967 [171]3 years ago
6 0

I'm guessing the function is

f(x)=\dfrac8{x^2-4x-12}=\dfrac8{(x-6)(x+2)}

which, split into partial fractions, is equivalent to

\dfrac1{x-6}-\dfrac1{x+2}

Recall that for |x| we have

\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n

With some rearranging, we find

\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n

valid for \left|\dfrac x6\right|, or |x|, and

\dfrac1{x+2}=\dfrac12\dfrac1{1-\left(-\frac x2\right)}=\displaystyle\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n

valid for \left|-\dfrac x2\right|, or |x|.

So we have

f(x)=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n-\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n

f(x)=\displaystyle-\sum_{n=0}^\infty\left(\frac{x^n}{6^{n+1}}+\frac{(-x)^n}{2^{n+1}}\right)

f(x)=\displaystyle-\sum_{n=0}^\infty\frac{1+3(-3)^n}{6^{n+1}}x^n

Taken together, the power series for f(x) can only converge for |x|, or -2.

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