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Question: How do you find a power series representation for \[\ln \left( 1-{{x}^{2}} \right)\] and what is the...

How do you find a power series representation for ln⁡(1−x2)\ln \left( 1-{{x}^{2}} \right) and what is the radius of convergence?

Explanation

Solution

In mathematics, a power series (in one variable) is an infinite series. Power series are useful in mathematical analysis, where they arise as Taylor series of infinitely differentiable functions. To solve this question, we should know how to write the power series/ MacLaurin series of a function of form ln⁡(1+x)\ln \left( 1+x \right). The power series representation for ln⁡(1+x)\ln \left( 1+x \right) is ∑n=0∞(−1)nxn+1n+1\sum\limits_{n=0}^{\infty }{\dfrac{{{(-1)}^{n}}{{x}^{n+1}}}{n+1}}. Also to find the radius of convergence, use the ratio test which states that if lim⁡x→∞∣an+1an∣<1\displaystyle \lim_{x \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|<1, then ∑n=0∞an\sum\limits_{n=0}^{\infty }{{{a}_{n}}} converges.

Complete step by step solution:
We are asked to express the power series expansion of ln⁡(1−x2)\ln \left( 1-{{x}^{2}} \right), and to find the radius of convergence. We know that power series representation for ln⁡(1+x)\ln \left( 1+x \right) is ∑n=0∞(−1)nxn+1n+1\sum\limits_{n=0}^{\infty }{\dfrac{{{(-1)}^{n}}{{x}^{n+1}}}{n+1}}.
To find the representation for ln⁡(1−x2)\ln \left( 1-{{x}^{2}} \right), we substitute −x2-{{x}^{2}} at the place of x in the summation formula, by doing this we get
⇒∑n=0∞(−1)n(−x2)n+1n+1\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{\left( -{{x}^{2}} \right)}^{n+1}}}{n+1}}
Using the exponential property (ab)n=anbn{{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}, the above expression can also be written as
⇒∑n=0∞(−1)n(−1)n+1(x2)n+1n+1\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{\left( -1 \right)}^{n+1}}{{\left( {{x}^{2}} \right)}^{n+1}}}{n+1}}
Again using the property (ab)n=anbn{{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}} in the reverse direction this time, we can simplify the above expression as
⇒∑n=0∞(−1)2n+1(x)2n+2n+1\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{2n+1}}{{\left( x \right)}^{2n+2}}}{n+1}}
As 2n+1 is an odd number, (−1)2n+1{{\left( -1 \right)}^{2n+1}} will always be −1-1. Thus, we get
⇒∑n=0∞−(x)2n+2n+1\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{-{{\left( x \right)}^{2n+2}}}{n+1}}
To find the radius of convergence, use the ratio test which states that if lim⁡x→∞∣an+1an∣<1\displaystyle \lim_{x \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|<1, then ∑n=0∞an\sum\limits_{n=0}^{\infty }{{{a}_{n}}} converges
⇒lim⁡x→∞∣−(x)2(n+1)+2(n+1)+1−(x)2n+2n+1∣<1\Rightarrow \displaystyle \lim_{x \to \infty }\left| \dfrac{\dfrac{-{{\left( x \right)}^{2\left( n+1 \right)+2}}}{\left( n+1 \right)+1}}{\dfrac{-{{\left( x \right)}^{2n+2}}}{n+1}} \right|<1
Simplifying the above expression, we get
⇒lim⁡x→∞∣−(x)2n+4n+2×n+1−(x)2n+2∣<1\Rightarrow \displaystyle \lim_{x \to \infty }\left| \dfrac{-{{\left( x \right)}^{2n+4}}}{n+2}\times \dfrac{n+1}{-{{\left( x \right)}^{2n+2}}} \right|<1
⇒∣x2∣<1\Rightarrow \left| {{x}^{2}} \right|<1
Solving the above inequality, we get
⇒−1<x<1\Rightarrow -1< x <1
Thus, the convergence radius of the given expression is 1.

Note:
To solve these types of problems, one should know the expansions of different types of series, for example, the Taylor series. The Maclaurin series is a special case of the Taylor series. Calculation mistakes should be avoided.