University of Florida/Egm4313/IEA-f13-team10/R5

Report 5

Problem 1: Taylor Series Expansion of the log Function

Problem Statement

$y^{″}+6y^{\prime }+9y=log(3+4x)$
$y(0)=1,y^{\prime }(0)=.5$
Use the point
$\hat{x}=-1/2$
$\log x={\log }_{e}x=\ln x$

Solution

$f(x)=\log (x)$
$f(\hat{x})=\log (\hat{x})$

$\sum _{k=0}^{\mathrm{\infty }}\frac{(f^k)(\hat{x})}{k!}(x-\hat{x}{)}^k$
$f^1(x)=\frac{1}{1+x}=f^1(\hat{x})=\frac{1}{1+(\hat{x})}$
Set
$w(x)=3+4x$
$w^{\prime }(x)=4$

$f^2(x)=\frac{-1*w^{\prime }*4}{w^2}=\frac{-1}{w^2}$

$f^3(x)=\frac{1*2*w*w^{\prime }*4^2}{w^4}=\frac{1*2}{w^3}$

$f^4(x)=\frac{-1*2*3*w^2*w^{\prime }*4^3}{w^6}=\frac{-1*2*3}{w^4}$

$f^k(x)=\frac{(-1{)}^{k-1}*(k-1)!*4^{k-1}}{w^k}$

$\sum _{k=0}^{\mathrm{\infty }}\frac{(f^k)(\hat{x})}{k!}(x-\hat{x}{)}^k=\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^{k-1}(k-1)!}{k!(1+(\hat{x}){)}^k}(x-\hat{x}{)}^k$

For $log(3+4x)$ the series expansion results in,
$\sum _{k=0}^{\mathrm{\infty }}(-1{)}^{k+1}*(\frac{(2+4x{)}^k}{k})$

Plots of taylor series expansion: Up to order 4
File:Taylor1.JPG

Up to order 7
File:Taylor2.JPG

Up to order 11
File:Taylor3.JPG

Up to order 16
File:Taylor4.JPG

The visually estimated domain of convergence is from .8 to .2.
Now use the transformation of variable
$x⟶t\text{ such that }3+4x=1+t$
$x=\frac{t-2}{4}$

If $log(1+t)$ has a domain of convergence from $[-1,+1]$ then $log(3+4x)$ converges from $[\frac{-3}{4},\frac{-1}{4}]$

Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 2: Plots of Truncated Series

Number 1

Plot at least 3 truncated series to show convergence

$\sum _{m=0}^{inf}\frac{x^{2m+1}}{(2m+1)!}$

m=0:$x$

m=1:$x+\frac{x^3}{3*2*1}$

m=2:$x+\frac{x^3}{3*2*1}+\frac{x^5}{5*4*3*2*1}$

File:P2G.png

Number 2

Plot at least 3 truncated series to show convergence

$\sum _{m=0}^{inf}(\frac{2}{3}{)}^m{x}^{2}m$

m=0:$x$

m=1:$x+\frac{2}{3}{x}^2$

m=2:$x+\frac{2}{3}{x}^2+\frac{4}{9}{x}^4$

File:P1G2.png

Number 3

Find the radius of convergence for the taylor series of sinx, x = 0

The Taylor series of sinx is:

$\sum _{m=0}^{inf}\frac{(-1{)}^m(x{)}^{2m+1}}{(2m+1)!}$

The radius of convergence can be found by:

$R_c=[lim\left|\frac{d_{k+1}}{d_k}\right|{]}^{-1}$

$R_c=[lim\frac{(-1{)}^{m+1}}{(2m+3)!}*\frac{(2m+1)!}{(-1{)}^m}{]}^{-1}$

$R_c=\mathrm{\infty }$

Number 4

Find the radius of convergence for the taylor series of log(1+x), x = 0

The Taylor series of log(x+1) is:

$-\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n(x{)}^n}{n}$

The radius of convergence can be found by:

$R_c=[lim\left|\frac{d_{k+1}}{d_k}\right|{]}^{-1}$

$R_c=[lim\frac{(-1{)}^{n+1}}{n+1}*\frac{n}{(-1{)}^n}{]}^{-1}$
$R_c=1$

Number 5

Find the radius of convergence for the taylor series of log(1+x), x = 1

The Taylor series of log(x+1) is:

$-\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n(x-1{)}^n}{n}$

The radius of convergence can be found by:

$R_c=[lim\left|\frac{d_{k+1}}{d_k}\right|{]}^{-1}$

$R_c=[lim\frac{(-1{)}^{n+1}}{n+1}*\frac{n}{(-1{)}^n}{]}^{-1}$

$R_c=1$

Number 6

derive the expression for the radius of convergence of log(1+x) about any focus point

The taylor series of log(1+x) is:

$-\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n(x-\hat{x}{)}^n}{n}$

$R_c=[\frac{(-1{)}^{n+1}(\hat{x}{)}^{n+1}}{n+1}*\frac{n}{(-1{)}^n(\hat{x}{)}^n}{]}^{-1}$

Number 7

Find the Taylor series representation of log(3+4x)

$\sum _{k=0}^{\mathrm{\infty }}\frac{(f{)}^k(\hat{x})}{k!}*(x-\hat{x}{)}^k$
Expanding out 4 terms results in,
[ $r(x)=0+\frac{4*(x-\hat{x})}{3+4\hat{x}}-\frac{(4^2)*((x-\hat{x}{)}^2}{((3+4\hat{x}{)}^2)*(2!)}+\frac{2(4^2)*((x-\hat{x}{)}^3}{((3+4\hat{x}{)}^3)*(3!)}$
The series representation is
$\sum _{k=0}^{\mathrm{\infty }}(-1{)}^{k+1}*(\frac{(2+4x{)}^k}{k})$

Number 8

Radius of convergence of log(3+4x) about the point $\hat{x}=-.5$

$\hat{x}=-.5$
$\sum _{n=0}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{4^n}{n}(x+.5{)}^n$
$R_c=li{m}_{n\to \mathrm{\infty }}[(-1{)}^{n+2}\frac{4^{n+1}}{n+1}(.25{)}^{n+1}*\frac{n}{4^n(-1{)}^{n+1}(.25{)}^n}{]}^{-1}$
Cancelling some terms out, you get
$R_c=li{m}_{n\to \mathrm{\infty }}\frac{-(n+1)}{n}$
Using L'Hopitals Rule, you get
$R_c=li{m}_{n\to \mathrm{\infty }}\frac{1}{1}=1$

Number 9

Radius of convergence of log(3+4x) about the point $\hat{x}=-.25$

$\hat{x}=-.25$

$\sum _{n=0}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{4^n}{n}(x+.25{)}^n$
$R_c=li{m}_{n\to \mathrm{\infty }}[(-1{)}^{n+2}\frac{4^{n+1}}{n+1}(.0625{)}^{n+1}*\frac{n}{4^n(-1{)}^{n+1}(.0625{)}^n}{]}^{-1}$
Cancelling some terms out, you get
$R_c=li{m}_{n\to \mathrm{\infty }}\frac{-4(n+1)}{n}$
Using L'Hopitals Rule, you get
$R_c=li{m}_{n\to \mathrm{\infty }}\frac{4}{1}=4$

Number 10

Radius of convergence of log(3+4x) about the point $\hat{x}=1$

$\hat{x}=1$
$\sum _{n=0}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{4^n}{n}(x-1{)}^n$
$R_c=li{m}_{n\to \mathrm{\infty }}[(-1{)}^{n+2}\frac{4^{n+1}}{n+1}(1{)}^{n+1}*\frac{n}{4^n(-1{)}^{n+1}(1{)}^n}{]}^{-1}$
Cancelling some terms out, you get
$R_c=li{m}_{n\to \mathrm{\infty }}\frac{-(n+1)}{4n}$
Using L'Hopitals Rule, you get
$R_c=li{m}_{n\to \mathrm{\infty }}\frac{1}{4}=1/4$

Number 11

Radius of convergence of log(3+4x) about any given point $\hat{x}$

$\sum _{n=0}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{4^n}{n}(x-\hat{x}{)}^n$
$R_c=li{m}_{n\to \mathrm{\infty }}[(-1{)}^{n+2}\frac{4^{n+1}}{n+1}({\hat{x}}^2{)}^{n+1}*\frac{n}{4^n(-1{)}^{n+1}(({\hat{x}}^2{)}^n)}{]}^{-1}$

Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 3:

Problem Statement

Use the Determinant of the Matrix of Components and the Gramian to verify the linear independence of the two vectors $b_1$ and $b_2$.

${𝐛}_1=5{𝐞}_1-3{𝐞}_2$

${𝐛}_2=-2{𝐞}_1+4{𝐞}_2$

Solution

Determinant of the Matrix of Components

The Matrix of components of the vectors $b_1$ and $b_2$ is

$\left|\begin{array}{cc}5& -3\\ -2& 4\end{array}\right|=(5)(4)-(-3)(-2)=20-6=14\ne 0$

So the vectors $b_1$ and $b_2$ are linearly independent.

Gramian

For vectors, the Gramian is defined as:

$\mathrm{\Gamma }(b_1,b_2)=\left[\begin{array}{cc}⟨b_1,b_1⟩& ⟨b_1,b_2⟩\\ ⟨b_2,b_1⟩& ⟨b_2,b_2⟩\end{array}\right]$

where:

$⟨b_i,b_j⟩=b_i\cdot b_j$

For the given vectors, the dot products are:

$⟨b_1,b_1⟩=(5e_1-3e_2)\cdot (5e_1-3e_2)=(5)(5)+(-3)(-3)=25+9=34$

$⟨b_1,b_2⟩=(5e_1-3e_2)\cdot (-2e_1+4e_2)=(5)(-2)+(-3)(4)=-10-12=-22$

$⟨b_2,b_1⟩=(-2e_1+4e_2)\cdot (5e_1-3e_2)=(-2)(5)+(4)(-3)=-10-12=-22$

$⟨b_2,b_2⟩=(-2e_1+4e_2)\cdot (-2e_1+4e_2)=(-2)(-2)+(4)(4)=4+16=20$

So the Gramian matrix becomes: $\mathrm{\Gamma }(b_1,b_2)=\left[\begin{array}{cc}34& -22\\ -22& 20\end{array}\right]$

Finding the determinant of the Gramian matrix gives the Gramian:

$\mathrm{\Gamma }=(34)(20)-(-22)(-22)=680-484=156\ne 0$

So the vectors $b_1$ and $b_2$ are linearly independent.

Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 4: Wronskian and Gramian

Problem Statement

Use both the Wronskian and the Gramain to find whether the following functions are linearly independent. Consider the domain of these functions to be [-1, +1] for the construction of the Gramian matrix.

$f(x)=x,g(x)=x^5$

$f(x)=cos2x,g(x)=sin4x$

Solution

Wronskian:

$W(f,g):=\left[\begin{array}{cc}f& g\\ f^{\prime }& g^{\prime }\end{array}\right]=f{g}^{\prime }-g{f}^{\prime }$

Function is linearly independent if $W(f,g)\ne 0$

1) $f(x)=x,g(x)=x^5$

$f(x)=x,f^{\prime }(x)=1$
$g(x)=x^5,g^{\prime }(x)=5x^4$
$W(f,g):=\left[\begin{array}{cc}x& x^5\\ 1& 5x^4\end{array}\right]=5x^5-x^5=4x^5\ne 0$ so function is linearly independent.

2)$f(x)=cos2x,g(x)=sin4x$

$f(x)=cos2x,f^{\prime }(x)=-2sin2x$
$g(x)=sin4x,g^{\prime }(x)=4cos4x$

$W(f,g):=\left[\begin{array}{cc}cos2x& sin4x\\ -2sin2x& 4cos4x\end{array}\right]=cos2x*4cos4x+sin4x*2sin2x\ne 0$ so function is linearly independent.

Gramian:
$\mathrm{\Gamma }(f,g):=\left[\begin{array}{cc}<f,f>& <f,g>\\ <g,f>& <g,g>\end{array}\right]$

Function is linearly independent if $\mathrm{\Gamma }(f,g)\ne 0$

1) $f(x)=x,g(x)=x^5$
$<f,f>=\underset{-1}{\overset{1}{\int }}{x}^{2}dx=\frac{2}{3}$
$<f,g>=\underset{-1}{\overset{1}{\int }}{x}^{6}dx=\frac{2}{7}$
$<g,f>=\underset{-1}{\overset{1}{\int }}{x}^{6}dx=\frac{2}{7}$
$<g,g>=\underset{-1}{\overset{1}{\int }}{x}^{1}0dx=\frac{2}{11}$

$\mathrm{\Gamma }(f,g):=\left[\begin{array}{cc}\frac{2}{3}& \frac{2}{7}\\ \frac{2}{7}& \frac{2}{11}\end{array}\right]\ne 0$ so function is linearly independent.

2) $f(x)=cos2x,g(x)=sin4x$
$<f,f>=\underset{-1}{\overset{1}{\int }}cos2x*cos2xdx=\frac{1}{4}(4+sin4)=.818$
$<f,g>=\underset{-1}{\overset{1}{\int }}cos2x*sin4xdx=[.5co{s}^{2}x-1/12cos6x{]}_{-}^{1}1=0$
$<g,f>=\underset{-1}{\overset{1}{\int }}cos2x*sin4xdx=0$
$<g,g>=\underset{-1}{\overset{1}{\int }}sin4x*sin4xdx=1+\frac{sin8}{8}=.876$

$\mathrm{\Gamma }(f,g):=\left[\begin{array}{cc}.818& 0\\ 0& .876\end{array}\right]\ne 0$ so function is linearly independent.

Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.