Question

use power series operations to find the taylor series at ( x = 0 ) for the following function.
( \frac { x ^ { 2 } } { 1 - 14 x } )
the taylor series for ( \frac { 1 } { 1 - x } ) is a commonly known series. what is the taylor series at ( x = 0 ) for ( \frac { 1 } { 1 - x } )?
( sum _ { n = 0 } ^ { infty } square ) (type an exact answer.)

Explanation:

Step1: Recall the geometric series formula

The geometric series formula is \(\frac{1}{1 - t}=\sum_{n = 0}^{\infty}t^{n}\), for \(|t|<1\).

Step2: Substitute \(t = 14x\) into the formula

When \(t = 14x\), we have \(\frac{1}{1-14x}=\sum_{n = 0}^{\infty}(14x)^{n}=\sum_{n = 0}^{\infty}14^{n}x^{n}\), for \(|14x|<1\) (i.e., \(|x|<\frac{1}{14}\)).

Step3: Multiply by \(x^{2}\)

Multiply the series for \(\frac{1}{1 - 14x}\) by \(x^{2}\). Using the property of power - series multiplication \(x^{2}\sum_{n = 0}^{\infty}14^{n}x^{n}=\sum_{n = 0}^{\infty}14^{n}x^{n + 2}\). Let \(k=n + 2\), when \(n = 0\), \(k = 2\). Then \(n=k - 2\) and the series becomes \(\sum_{k=2}^{\infty}14^{k - 2}x^{k}\). We can also write it as \(\sum_{n=0}^{\infty}14^{n}x^{n+2}\) (by keeping the index \(n\) as it is).

For the Taylor series of \(\frac{1}{1 - x}\), using the geometric series formula \(\frac{1}{1 - x}=\sum_{n=0}^{\infty}x^{n}\), for \(|x|<1\).

Answer:

For \(\frac{x^{2}}{1 - 14x}\): \(\sum_{n = 0}^{\infty}14^{n}x^{n+2}\)
For \(\frac{1}{1 - x}\): \(\sum_{n=0}^{\infty}x^{n}\)