Answer
396.9k+ views
Hint: 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 \[\ln \left( 1+x \right)\]. The power series representation for \[\ln \left( 1+x \right)\] is \[\sum\limits_{n=0}^{\infty }{\dfrac{{{(-1)}^{n}}{{x}^{n+1}}}{n+1}}\]. Also to find the radius of convergence, use the ratio test which states that if \[\displaystyle \lim_{x \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|<1\], then \[\sum\limits_{n=0}^{\infty }{{{a}_{n}}}\] converges.
Complete step by step solution:
We are asked to express the power series expansion of \[\ln \left( 1-{{x}^{2}} \right)\], and to find the radius of convergence. We know that power series representation for \[\ln \left( 1+x \right)\] is \[\sum\limits_{n=0}^{\infty }{\dfrac{{{(-1)}^{n}}{{x}^{n+1}}}{n+1}}\].
To find the representation for \[\ln \left( 1-{{x}^{2}} \right)\], we substitute \[-{{x}^{2}}\] at the place of x in the summation formula, by doing this we get
\[\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{\left( -{{x}^{2}} \right)}^{n+1}}}{n+1}}\]
Using the exponential property \[{{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}\], the above expression can also be written as
\[\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{\left( -1 \right)}^{n+1}}{{\left( {{x}^{2}} \right)}^{n+1}}}{n+1}}\]
Again using the property \[{{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}\] in the reverse direction this time, we can simplify the above expression as
\[\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{2n+1}}{{\left( x \right)}^{2n+2}}}{n+1}}\]
As 2n+1 is an odd number, \[{{\left( -1 \right)}^{2n+1}}\] will always be \[-1\]. Thus, we get
\[\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{-{{\left( x \right)}^{2n+2}}}{n+1}}\]
To find the radius of convergence, use the ratio test which states that if \[\displaystyle \lim_{x \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|<1\], then \[\sum\limits_{n=0}^{\infty }{{{a}_{n}}}\] converges
\[\Rightarrow \displaystyle \lim_{x \to \infty }\left| \dfrac{\dfrac{-{{\left( x \right)}^{2\left( n+1 \right)+2}}}{\left( n+1 \right)+1}}{\dfrac{-{{\left( x \right)}^{2n+2}}}{n+1}} \right|<1\]
Simplifying the above expression, we get
\[\Rightarrow \displaystyle \lim_{x \to \infty }\left| \dfrac{-{{\left( x \right)}^{2n+4}}}{n+2}\times \dfrac{n+1}{-{{\left( x \right)}^{2n+2}}} \right|<1\]
\[\Rightarrow \left| {{x}^{2}} \right|<1\]
Solving the above inequality, we get
\[\Rightarrow -1< x <1\]
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.
Complete step by step solution:
We are asked to express the power series expansion of \[\ln \left( 1-{{x}^{2}} \right)\], and to find the radius of convergence. We know that power series representation for \[\ln \left( 1+x \right)\] is \[\sum\limits_{n=0}^{\infty }{\dfrac{{{(-1)}^{n}}{{x}^{n+1}}}{n+1}}\].
To find the representation for \[\ln \left( 1-{{x}^{2}} \right)\], we substitute \[-{{x}^{2}}\] at the place of x in the summation formula, by doing this we get
\[\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{\left( -{{x}^{2}} \right)}^{n+1}}}{n+1}}\]
Using the exponential property \[{{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}\], the above expression can also be written as
\[\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{\left( -1 \right)}^{n+1}}{{\left( {{x}^{2}} \right)}^{n+1}}}{n+1}}\]
Again using the property \[{{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}\] in the reverse direction this time, we can simplify the above expression as
\[\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{2n+1}}{{\left( x \right)}^{2n+2}}}{n+1}}\]
As 2n+1 is an odd number, \[{{\left( -1 \right)}^{2n+1}}\] will always be \[-1\]. Thus, we get
\[\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{-{{\left( x \right)}^{2n+2}}}{n+1}}\]
To find the radius of convergence, use the ratio test which states that if \[\displaystyle \lim_{x \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|<1\], then \[\sum\limits_{n=0}^{\infty }{{{a}_{n}}}\] converges
\[\Rightarrow \displaystyle \lim_{x \to \infty }\left| \dfrac{\dfrac{-{{\left( x \right)}^{2\left( n+1 \right)+2}}}{\left( n+1 \right)+1}}{\dfrac{-{{\left( x \right)}^{2n+2}}}{n+1}} \right|<1\]
Simplifying the above expression, we get
\[\Rightarrow \displaystyle \lim_{x \to \infty }\left| \dfrac{-{{\left( x \right)}^{2n+4}}}{n+2}\times \dfrac{n+1}{-{{\left( x \right)}^{2n+2}}} \right|<1\]
\[\Rightarrow \left| {{x}^{2}} \right|<1\]
Solving the above inequality, we get
\[\Rightarrow -1< x <1\]
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.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)