Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Power series. Click on this to access more such competitive quizzes on Algebra.

Power series Quiz 2

mathematics

 Find the centre of the power series   $\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}}{\left( 2n+1 \right)!}{{\left( \dfrac{2x-6}{5} \right)}^{n}}}$.

 To find the center of the series we need to equate the given function by 0 . Since it does not depend on the value of n therefore $x$ is the center of the series.

  $f\left( x \right)=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}}{\left( 2n+1 \right)!}{{\left( \dfrac{2x-6}{5} \right)}^{n}}} \\ \text{Equate the equation to 0 to find the center} \\ \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}}{\left( 2n+1 \right)!}{{\left( \dfrac{2x-6}{5} \right)}^{n}}}=0 \\ \dfrac{2x-6}{5}=0 \\ x=3 \text{ (Center of series)}$

 What is the power series representation of ${{e}^{2{{x}^{2}}}}+3cos2x$ ?

 First right the expansion of both the terms individually and then express it in the form of series and then keep it in the equation to get required power series.

  $ P={{e}^{2{{x}^{2}}}}+3cos2x \\ \text{Expansion of }e^{x}=1+\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}+..... \\ {{e}^{x}}=\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{n}}}{n!}} \\ {{e}^{2{{x}^{2}}}}=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( 2{{x}^{2}} \right)}^{n}}}{n!}} \\ \text{Expansion of }cosx = 1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+... \\ \cos x=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{x}^{2n}}}{2n!}} \\ \cos 2x=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{\left( 2x \right)}^{2n}}}{2n!}} \\ \text{Overall: } \\ \text{P=}\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( 2{{x}^{2}} \right)}^{n}}}{n!}}+3\left( \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{\left( 2x \right)}^{2n}}}{2n!}} \right) $ 

 What is the Radius of Convergence of the series  $\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}}{{{2}^{n}}}{{\left( \dfrac{x}{3} \right)}^{n}}}$

 To find the radius of convergence use the formula for radius of convergence and the using the properties of limit evaluate it.

  $ \dfrac{1}{R}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{|{{a}_{n+1}}|}{|{{a}_{n}}|} \\ \dfrac{1}{R}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{|{{\left( -1 \right)}^{n}}\times {{6}^{n}}|}{|{{6}^{n+1}}|} \\ \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{6} \\ R=6 $ 

 Find the convergence interval of  $\sum\limits_{n=0}^{\infty }{{{\left( x+5 \right)}^{n}}}$.   

 First find the radius of convergence and then find the center and using the interval formula find the interval of series.

  $ \dfrac{1}{R}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{|{{a}_{n+1}}|}{|{{a}_{n}}|} \\ \dfrac{1}{R}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{n=0}^{\infty }{{{\left( x+5 \right)}^{n}}} \\ \dfrac{1}{R}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{1}^{n+1}}}{{{1}^{n}}} \\ R=1 \\ x+5=0 \\ x=-5 \text{ (Center)} \\ \text{Interval }=\left( C-R,C+R \right) \\ \Rightarrow \left( -6,-4 \right) $ 

 Find the interval of convergence of the series  $\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}}{{{2}^{n}}}{{\left( \dfrac{4x+5}{3} \right)}^{n}}}$.

 Find the radius of convergence and then find the center of the curve and after that use the interval formula to evaluate the interval.

  $ \dfrac{1}{R}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{|{{a}_{n+1}}|}{|{{a}_{n}}|} \\ \dfrac{1}{R}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{|{{\left( -1 \right)}^{n+1}}\times {{4}^{n+1}}\times {{6}^{n}}|}{|{{6}^{n+1}}{{\left( -1 \right)}^{n}}\times {{4}^{n}}|} \\ \Rightarrow \dfrac{2}{3} \\ R=\dfrac{3}{2} \\ 4x+5=0 \\ x=\dfrac{-5}{4}\text{ (Center)} \\ \text{Interval }=\left( C-R,C+R \right) \\ \Rightarrow \left( \dfrac{-11}{4},\dfrac{1}{4} \right) \\ \Rightarrow \dfrac{-11}{4} < x < \dfrac{1}{4} $ 