Eigenvalues and Eigenvectors

Example 1.

 

We want to find all numbers x for which the power series

(1)   \begin{equation*}\sum _{n =0}^{\infty }nx^{n} \end{equation*}

converges. Define a_{n} =nx^{n} for every n and use Ratio test to write

(2)   \begin{align*}\lim _{n \rightarrow \infty }\genfrac{\vert }{\vert }{}{}{a_{n +1}}{a_{n}} =\lim _{n \rightarrow \infty }\genfrac{\vert }{\vert }{}{}{\left (n +1\right )x^{n +1}}{nx^{n}} \\ =\lim _{n \rightarrow \infty }\genfrac{\vert }{\vert }{}{}{\left (n +1\right )x}{n} \\ =\left \vert x\right \vert \lim _{n \rightarrow \infty }\frac{n +1}{n} =\left \vert x\right \vert \end{align*}

We know if \left \vert x\right \vert <1 then the infinite sum is finite by Ratio test and when \left \vert x\right \vert >1 it is divergent. However, Ratio test fails when \left \vert x\right \vert =1. We check these here. When x =1 , we get

(3)   \begin{equation*}\sum _{n =0}^{\infty }n \end{equation*}

and the divergent test tells us

(4)   \begin{equation*}\lim _{n \rightarrow \infty }n =\infty \neq 0 \end{equation*}

Therefore x =1 is not acceptable. Next when x = -1 , we get

(5)   \begin{equation*}\sum _{n =0}^{\infty }n\left ( -1\right )^{n} =1 -2 +3 -4 +5 -6 +7 - . . . \end{equation*}

Again the divergence test tells us

(6)   \begin{equation*}\lim _{n \rightarrow \infty }n\left ( -1\right )^{n} \neq 0 \end{equation*}

Therefore only when \left \vert x\right \vert <1 , the given power series converges. Therefore the interval of convergence is \left ( -1 ,1\right ) with center 0 and radius of convergence equal to 1.

 

Example 2.

 

Suppose we are given the power series

(7)   \begin{equation*}\sum _{n =0}^{\infty }\frac{\left (x +1\right )^{n}}{3^{n}} \end{equation*}

We would like to determine for what values of x , this power series is convergent. Use Ratio test to write

(8)   \begin{align*}\lim _{n \rightarrow \infty }\genfrac{\vert }{\vert }{}{}{\frac{\left (x +1\right )^{n +1}}{3^{n +1}}}{\frac{\left (x +1\right )^{n}}{3^{n}}} =\lim _{n \rightarrow \infty }\genfrac{\vert }{\vert }{}{}{\frac{\left (x +1\right )^{n}\left (x +1\right )}{3^{n}\left (3\right )}}{\frac{\left (x +1\right )^{n}}{3^{n}}} \\ =\lim _{n \rightarrow \infty }\genfrac{\vert }{\vert }{}{}{x +1}{3} \\ =\frac{\left \vert x +1\right \vert }{3}\end{align*}

We know when

(9)   \begin{equation*}\frac{\left \vert x +1\right \vert }{3} <1 \end{equation*}

the given power series is convergent that is

(10)   \begin{align*}\left \vert x +1\right \vert <3 \\ -3 <x +1 <3 \\ -4 <x <2\end{align*}

We know when \left \vert x +1\right \vert >3, the power series is divergent but what about when \left \vert x +1\right \vert =3? The Ratio test fails here. We check the endpoints of the interval of convergence x =2 and x = -4. When x =2 , we get

(11)   \begin{equation*}\sum _{n =0}^{\infty }\frac{\left (x +1\right )^{n}}{3^{n}} =\sum _{n =0}^{\infty }1 \end{equation*}

Since \lim _{n \rightarrow \infty }1 \neq 0, the divergence test tells us the power series is divergent. Next, when x = -4 , we get

(12)   \begin{equation*}\sum _{n =0}^{\infty }\frac{\left (x +1\right )^{n}}{3^{n}} =\sum _{n =0}^{\infty }\left ( -1\right )^{n} \end{equation*}

We have shown in the past that the series is divergent. So, the interval of convergence is \left ( -4 ,2\right ) , center at x = -1 and radius of convergence equal to 3.

 

Example 3.

 

We want to use geometric series to write f\left (x\right ) =\ln \left (1 -2x\right ) as a power series. Differentiate f and write

(13)   \begin{equation*}f^{ \prime }\left (x\right ) =\frac{1}{1 -2x}\left ( -2\right ) \end{equation*}

We know, when \left \vert 2x\right \vert <1

(14)   \begin{align*}\left ( -2\right )\frac{1}{1 -2x} =\left ( -2\right )(1 +\left (2x\right ) +\left (2x\right )^{2} +\left (2x\right )^{3} +\left (2x\right )^{4} + . . .) \\ =\left ( -2\right )(1 +2x +4x^{2} +8x^{3} +16x^{4} + . . .)\end{align*}

Now integrate over the interval \left [0 ,x\right ]

(15)   \begin{equation*}\int _{0}^{x}\left ( -2\right )\left (1 +2t +4t^{2} +8t^{3} +16t^{4} + . . .\right )dt = -2\left (x +\frac{2}{2}x^{2} +\frac{4}{3}x^{3} +\frac{8}{4}x^{4} + . . .\right ) \end{equation*}

On the other hand we know

(16)   \begin{equation*}\int _{0}^{x}\frac{ -2}{1 -2t}dt =\ln \left (1 -2x\right ) \end{equation*}

Therefore

(17)   \begin{equation*}\ln \left (1 -2x\right ) = -2\left (x +\frac{2}{2}x^{2} +\frac{4}{3}x^{3} +\frac{8}{4}x^{4} + . . .\right ) \end{equation*}

 

Example 4.

 

Let us assume that f\left (x\right ) =\ln x has a Taylor expansion at c =1 and we would like to construct this. Write

(18)   \begin{align*}f^{ \prime }\left (x\right ) =\frac{1}{x} \\ f^{ \prime \prime }\left (x\right ) = -\frac{1}{x^{2}} \\ f^{ \prime \prime \prime }\left (x\right ) =\frac{2}{x^{3}} \\ f^{\left (4\right )}\left (x\right ) = -\frac{\left (3\right )\left (2\right )}{x^{4}} \\ f^{\left (5\right )}\left (x\right ) =\frac{\left (4\right )\left (3\right )\left (2\right )}{x^{5}} \\ f^{\left (n\right )}\left (x\right ) =\left ( -1\right )^{n +1}\frac{\left (n -1\right ) !}{x^{n}}\end{align*}

We know if f has a Taylor expansion at c then

(19)   \begin{equation*}f\left (x\right ) =f\left (c\right ) +f^{ \prime }\left (c\right )\left (x -c\right ) +\frac{f^{ \prime \prime }\left (c\right )}{2 !}\left (x -c\right )^{2} +\frac{f^{ \prime \prime \prime }\left (c\right )}{3 !}\left (x -c\right )^{3} + . . . \end{equation*}

For this example, c =1 and so

(20)   \begin{align*}\ln x =f\left (1\right ) +f^{ \prime }\left (1\right )\left (x -1\right ) +\frac{f^{ \prime \prime }\left (1\right )}{2 !}\left (x -1\right )^{2} +\frac{f^{ \prime \prime \prime }\left (1\right )}{3 !}\left (x -1\right )^{3} + . . . \\ =0 +\left (1\right )\left (x -1\right ) +\frac{ -1}{2 !}\left (x -1\right )^{2} +\frac{1}{3 !}\left (x -1\right )^{3} + . . . \\ \,\end{align*}

 

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