Question

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.)

Answer 1

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 $|x|<1$, 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

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

Solving for $x$ gives the interval of convergence,

$|x|^2<\dfrac16\implies|x|<\dfrac1{\sqrt6}\implies -\dfrac1{\sqrt 6}<x<\dfrac1{\sqrt 6}$

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

Answer 2

The interval of the convergence is (-1/√6 < x < 1/√6). We can confirm that the interval is open by checking for convergence at the endpoints.

What is a function?

The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.

The given function is

$\rm f(x) = \dfrac{x}{6x^2 + 1} \\\\or \\\\f(x) = \dfrac{x}{1 - (-6x^2)}$

For |x| < 1, we have

$\rm \dfrac{1}{1-x} = \Sigma_{n=0}^{\infty} \ x^n$

If x is replaced with -6x², then we have

$\rm f(x)= \Sigma_{n=0}^{\infty} (-6x^2 )^n = \Sigma_{n=0}^{\infty} (-6)^n x^{2n+1}$

Then by the ratio test, the series converges if

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

Solving for x, the interval of convergence will be

$|x^2| < \dfrac{1}{6} \\\\|x| < \dfrac{1}{\sqrt6} \\\\-\dfrac{1}{\sqrt6} < x < \dfrac{1}{\sqrt6}$

We can confirm that the interval is open by checking for convergence at the endpoints.

More about the function link is given below.

https://brainly.com/question/5245372