in the following, use that we know: 1 / (1 - x) = sum(n = 0 to infinity) x^n, for |x| < 1. consider the…

in the following, use that we know: 1 / (1 - x) = sum(n = 0 to infinity) x^n, for |x| < 1. consider the function f(x) = x / (1 - 5x). give the power series representation for f(x) centered at 0. f(x) = sum(n = 0 to infinity) . give the open interval of convergence for this power series. (note: give your answer in interval notation.) submit answer next item
Answer
Explanation:
Step1: Substitute $5x$ into the known series
We know that $\frac{1}{1 - t}=\sum_{n = 0}^{\infty}t^{n}$ for $|t|<1$. Let $t = 5x$, then $\frac{1}{1-5x}=\sum_{n = 0}^{\infty}(5x)^{n}=\sum_{n = 0}^{\infty}5^{n}x^{n}$ for $|5x|<1$.
Step2: Multiply by $x$
Since $f(x)=\frac{x}{1 - 5x}$, we multiply the series for $\frac{1}{1-5x}$ by $x$. So $f(x)=x\sum_{n = 0}^{\infty}5^{n}x^{n}=\sum_{n = 0}^{\infty}5^{n}x^{n + 1}$.
Step3: Find the interval of convergence
We had $|5x|<1$. Solving for $x$, we divide both sides of the inequality by 5. So $|x|<\frac{1}{5}$, which in interval - notation is $(-\frac{1}{5},\frac{1}{5})$.
Answer:
$f(x)=\sum_{n = 0}^{\infty}5^{n}x^{n+1}$; $(-\frac{1}{5},\frac{1}{5})$