MHB Power Series: Find First 5 Terms of x^2/(1-5x) - Help Needed

[dmdenney]

For this function

f(x)=x^2/(1-5x).

The interval of convergence is (-1/5) < x < (1/5).

I tried to differentiate, but got it wrong.

Could someone please help?

 

[Opalg]

For this function

f(x)=x^2/(1-5x).

The interval of convergence is (-1/5) < x < (1/5).

I tried to differentiate, but got it wrong.

Could someone please help?

Hi dmdenney, and welcome to MHB!

To find the power series representation for this function, I would use the binomial expansion $(1-t)^{-1} = 1+t+t^2 + t^3 + t^4 + \ldots$ (valid for $-1<t<1$), and substitute $t=5x$.

 

[Prove It]

For this function

f(x)=x^2/(1-5x).

The interval of convergence is (-1/5) < x < (1/5).

I tried to differentiate, but got it wrong.

Could someone please help?

$\displaystyle \begin{align*} f(x) &= \frac{x^2}{1 - 5x} \\ &= x^2 \left( \frac{1}{1 - 5x} \right) \end{align*}$

Do you recall the geometric series $\displaystyle \begin{align*} \sum_{n = 0}^{\infty} r^n = \frac{1}{1 - r} \end{align*}$ provided $\displaystyle \begin{align*} |r| < 1 \end{align*}$? Do you see how the stuff in the brackets looks like the closed form of the geometric series? What is r in this case?