Power Series MCQ Quiz in मल्याळम - Objective Question with Answer for Power Series - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept -

Let {an : n ≥ 1} n be a sequence of real numbers such that $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ is convergent and $\sum _{n=1}^{\mathrm{\infty }}\left|a_n\right|$ is divergent.

Let R be the radius of convergence of the power series $\sum _{n=1}^{\mathrm{\infty }}{a}_n{x}^n$.

Then we can conclude that R = 1 only.

Explanation -

According to the given concept,

option (ii) is correct.

Concept -

If the power series is given by $\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^{n^2}$

 then Radius of convergence R of the power series is 

$\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}(a_n{)}^{\frac{1}{n^2}}$

Explanation - 

Here $a_n=1$

$⟹\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}(1{)}^{\frac{1}{n^2}}$

⇒ R = 1

Hence the option (iii) is correct.

Given -

Let P(x) be a polynomial of degree d ≥ 2. The radius of convergence of the power series $\sum _{n=0}^{\mathrm{\infty }}p(n)z^n$  is R.

Concept -

If the power series is given by $\sum _{n=0}^{\mathrm{\infty }}{a}_n{Z}^n$

 then Radius of convergence R of the power series is 

$\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}\frac{a_{n+1}}{a_n}$

Explanation - 

Let $a_n=p(n)$

and $p(x)=b_0+b_{1}x+.....+b_n{x}^n$

$\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}\frac{p(n+1)}{p(n)}=1$

⇒ R =1 

Hence the option (2) is correct.

Given -

Consider the following power series in the complex variable z:

$\mathrm{f}(\mathrm{z})=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}$ n log n zn, g(z)=$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{{\mathrm{n}}^2}}{\mathrm{n}}{\mathrm{z}}^{\mathrm{n}}$.

If r, R are the radii of convergence of f and g respectively.

Concept -

If the power series is given by $\sum _{n=0}^{\mathrm{\infty }}{a}_n{Z}^n$

 then Radius of convergence R of the power series is 

$\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}\frac{a_{n+1}}{a_n}=li{m}_{n\to \mathrm{\infty }}{a}_n^{\frac{1}{n}}$

Explanation - 

(a) let a_n =  n log n

$\frac{1}{r}=li{m}_{n\to \mathrm{\infty }}\frac{a_{n+1}}{a_n}$

⇒ $\frac{1}{r}=li{m}_{n\to \mathrm{\infty }}\frac{(n+1)log(n+1)}{nlog(n)}$

⇒ $\frac{1}{r}=li{m}_{n\to \mathrm{\infty }}\frac{(n+1)log(n+1)}{nlog(n)}=1$

⇒ r = 1

(a) let an =  $\frac{e^{n^2}}{n}$

$\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}{a}_n^{\frac{1}{n}}$

⇒ $\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}[\frac{e^{n^2}}{n}{]}^{\frac{1}{n}}=\mathrm{\infty }$

⇒ R = 0

Hence option (ii) is correct.

Concept:

Radius of convergence of ∑(1 + b₀) a_n(z - z₀)ⁿ is $\frac{R}{(1+b_0)}$ where $\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}(|a_n|{)}^{1/n}$.

Explanation:

Series = $\sum _{n=0}^{\mathrm{\infty }}(1+b_0)(1+\frac{1}{n}{)}^{n^2}{z}^n$

$\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}(|a_n|{)}^{1/n}$

$\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}[(1+\frac{1}{n}{)}^{n^2}{]}^{1/n}$

$\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}[(1+\frac{1}{n}{)}^n]$

$\frac{1}{R}=e$

R = 1/e

So radius of convergence of the series is

$\frac{1}{(1+b_0)e}$

Option (1) is true.

Concept:

The region of convergence of ∑ a_nzⁿ is |z| < R where

 $\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\frac{a_{n+1}}{a_n}$

Explanation:

f(z) = ${\sum }_{n=0}^{\mathrm{\infty }}\frac{z^n}{2^n+1}$

a_n = $\frac{1}{2^n+1}$

$\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\frac{a_{n+1}}{a_n}$

$\Rightarrow \frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\frac{2^n+1}{2^{n+1}+1}$

$\Rightarrow \frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\frac{2^n(1+\frac{1}{2^n)}}{2^n(2+\frac{1}{2^n})}$

$\Rightarrow \frac{1}{R}=\frac{1}{2}$

So, R = 2

Hence region of convergence is |z| < 2

Option (3) is true.

Concept:

(i) ROC: A non-negative number R is said to be radius of convergent of the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n$ if the power series converges for |z| < R and diverges for |z| > R. 

(ii) By D’Alembert ratio test $\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\frac{a_{n+1}}{a_n}$

(iii) If the radius of convergence of $\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n$ is R the  the radius of convergence of $\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^{nk}$ is R^1/k

Explanation:

Series is $\sum _{n=0}^{\mathrm{\infty }}{3}^{-n}{z}^{2n}$

here a_n = 3^-n

So, ROC of $\sum _{n=0}^{\mathrm{\infty }}{3}^{-n}{z}^n$ is given by

$\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\frac{a_{n+1}}{a_n}$ = $\underset{n\to \mathrm{\infty }}{lim}\frac{3^n}{3^{n+1}}$ = $\frac{1}{3}$

So, R = 3

Then ROC of $\sum _{n=0}^{\mathrm{\infty }}{3}^{-n}{z}^{2n}$ is √3

Hence, the series converges for

|z| < √3

Option (2) is true.

Explanation:

D. Given Series is $\sum {(1+\frac{1}{n})}^n{z}^n$

$a_n=(1+\frac{1}{n}{)}^n$  

$(a_n{)}^{\frac{1}{n}}=(1+\frac{1}{n})$ 
 

$li{m}_{n\to \infty }(a_n{)}^{\frac{1}{n}}=li{m}_{n\to \infty }(1+\frac{1}{n})=1$

ROC = 1

⇒ (D) → IV (which is in only option 3) 

Hence Option(3) is the correct answer.

Concept:

The radius of convergence R of a power series ∑a_nx^_n is given by

$\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_n}{a_{n+1}}\right|$

Explanation:

$f(x)=\sum _{n=2}^{\mathrm{\infty }}\log (n)x^n$

Here a_n = log(n)

So, 

$\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{\log n}{\log (n+1)}\right|$

⇒ $\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\frac{\log n}{\log (n+1)}$

Using L'hospital rule

$\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{n}}{\frac{1}{n+1}}$

⇒ $\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}\frac{n+1}{n}$

⇒ $\frac{1}{R}=\underset{n\to \mathrm{\infty }}{lim}(1+\frac{1}{n})$

⇒ $\frac{1}{R}=1$

⇒ R = 1

So, the radius of convergence of the series f(x) is  1

Option (2) is true