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Viefleur [7K]
2 years ago
10

Find a power series representation for the function. (Give your power series representation centered at x = 0.) f(x) = x/6x^2 +

1 f(x) = sigma^infinity_n = 0 (-1)^n x^2n+1 6^n Determine the interval of convergence. (Enter your answer using interval notation.)
Mathematics
1 answer:
timama [110]2 years ago
3 0

Looks like your function is

f(x)=\dfrac x{6x^2+1}

Rewrite it as

f(x)=\dfrac x{1-(-6x^2)}

Recall that for |x|, we have

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

If we replace x with -6x^2, we get

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

By the ratio test, the series converges if

\displaystyle\lim_{n\to\infty}\left|\frac{(-6)^{n+1} x^{2(n+1)+1}}{(-6)^n x^{2n+1}}\right|=6|x^2|\lim_{n\to\infty}1=6|x|^2

Solving for x gives the interval of convergence,

|x|^2

We can confirm that the interval is open by checking for convergence at the endpoints; we'd find that the resulting series diverge.

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