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

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Report 5

Problem 1: Taylor Series Expansion of the log Function

Problem Statement

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

Solution

${\displaystyle f(x)=\log(x)}$
${\displaystyle f({\widehat {x}})=\log({\widehat {x}})}$

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

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

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

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

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

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

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

Plots of taylor series expansion: Up to order 4

Up to order 7

Up to order 11

Up to order 16

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

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

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

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

m=0:${\displaystyle x}$

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

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

Number 2

Plot at least 3 truncated series to show convergence

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

m=0:${\displaystyle x}$

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

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

Number 3

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

The Taylor series of sinx is:

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

The radius of convergence can be found by:

${\displaystyle R_{c}=[lim\left|{\frac {d_{k+1}}{d_{k}}}\right|]^{-1}}$

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

${\displaystyle R_{c}=\infty }$

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:

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

The radius of convergence can be found by:

${\displaystyle R_{c}=[lim\left|{\frac {d_{k+1}}{d_{k}}}\right|]^{-1}}$

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

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:

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

The radius of convergence can be found by:

${\displaystyle R_{c}=[lim\left|{\frac {d_{k+1}}{d_{k}}}\right|]^{-1}}$

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

${\displaystyle R_{c}=1}$

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:

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

${\displaystyle R_{c}=[{\frac {(-1)^{n+1}({\widehat {x}})^{n+1}}{n+1}}*{\frac {n}{(-1)^{n}({\widehat {x}})^{n}}}]^{-1}}$

Number 7

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

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

Number 8

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

${\displaystyle {\widehat {x}}=-.5}$
${\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}{\frac {4^{n}}{n}}(x+.5)^{n}}$
${\displaystyle R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}(.25)^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(.25)^{n}}}]^{-1}}$
Cancelling some terms out, you get
${\displaystyle R_{c}=lim_{n\rightarrow \infty }{\frac {-(n+1)}{n}}}$
Using L'Hopitals Rule, you get
${\displaystyle R_{c}=lim_{n\rightarrow \infty }{\frac {1}{1}}=1}$

Number 9

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

${\displaystyle {\widehat {x}}=-.25}$

${\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}{\frac {4^{n}}{n}}(x+.25)^{n}}$
${\displaystyle R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}(.0625)^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(.0625)^{n}}}]^{-1}}$
Cancelling some terms out, you get
${\displaystyle R_{c}=lim_{n\rightarrow \infty }{\frac {-4(n+1)}{n}}}$
Using L'Hopitals Rule, you get
${\displaystyle R_{c}=lim_{n\rightarrow \infty }{\frac {4}{1}}=4}$

Number 10

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

${\displaystyle {\widehat {x}}=1}$
${\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}{\frac {4^{n}}{n}}(x-1)^{n}}$
${\displaystyle R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}(1)^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(1)^{n}}}]^{-1}}$
Cancelling some terms out, you get
${\displaystyle R_{c}=lim_{n\rightarrow \infty }{\frac {-(n+1)}{4n}}}$
Using L'Hopitals Rule, you get
${\displaystyle R_{c}=lim_{n\rightarrow \infty }{\frac {1}{4}}=1/4}$

Number 11

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

${\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}{\frac {4^{n}}{n}}(x-{\widehat {x}})^{n}}$
${\displaystyle R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}({\widehat {x}}^{2})^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(({\widehat {x}}^{2})^{n})}}]^{-1}}$

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 ${\displaystyle b_{1}}$ and ${\displaystyle b_{2}}$.

${\displaystyle \mathbf {b} _{1}=5\mathbf {e} _{1}-3\mathbf {e} _{2}}$

${\displaystyle \mathbf {b} _{2}=-2\mathbf {e} _{1}+4\mathbf {e} _{2}}$

Solution

Determinant of the Matrix of Components

The Matrix of components of the vectors ${\displaystyle b_{1}}$ and ${\displaystyle b_{2}}$ is

${\displaystyle {\begin{vmatrix}5&-3\\-2&4\end{vmatrix}}=(5)(4)-(-3)(-2)=20-6=14\neq 0}$

So the vectors ${\displaystyle b_{1}}$ and ${\displaystyle b_{2}}$ are linearly independent.

Gramian

For vectors, the Gramian is defined as:

${\displaystyle \ {\boldsymbol {\Gamma }}(b_{1},b_{2})={\begin{bmatrix}\langle b_{1},b_{1}\rangle &\langle b_{1},b_{2}\rangle \\\langle b_{2},b_{1}\rangle &\langle b_{2},b_{2}\rangle \end{bmatrix}}}$

where:

${\displaystyle \ \langle b_{i},b_{j}\rangle =b_{i}\cdot b_{j}}$

For the given vectors, the dot products are:

${\displaystyle \ \langle b_{1},b_{1}\rangle =(5e_{1}-3e_{2})\cdot (5e_{1}-3e_{2})=(5)(5)+(-3)(-3)=25+9=34}$

${\displaystyle \ \langle b_{1},b_{2}\rangle =(5e_{1}-3e_{2})\cdot (-2e_{1}+4e_{2})=(5)(-2)+(-3)(4)=-10-12=-22}$

${\displaystyle \ \langle b_{2},b_{1}\rangle =(-2e_{1}+4e_{2})\cdot (5e_{1}-3e_{2})=(-2)(5)+(4)(-3)=-10-12=-22}$

${\displaystyle \ \langle b_{2},b_{2}\rangle =(-2e_{1}+4e_{2})\cdot (-2e_{1}+4e_{2})=(-2)(-2)+(4)(4)=4+16=20}$

So the Gramian matrix becomes: ${\displaystyle \ {\boldsymbol {\Gamma }}(b_{1},b_{2})={\begin{bmatrix}34&-22\\-22&20\end{bmatrix}}}$

Finding the determinant of the Gramian matrix gives the Gramian:

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

So the vectors ${\displaystyle b_{1}}$ and ${\displaystyle b_{2}}$ 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.

${\displaystyle f(x)=x,g(x)=x^{5}}$

${\displaystyle f(x)=cos2x,g(x)=sin4x}$

Solution

Wronskian:

${\displaystyle W(f,g):={\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}=fg'-gf'}$

Function is linearly independent if ${\displaystyle W(f,g)\neq 0}$

1) ${\displaystyle f(x)=x,g(x)=x^{5}}$

${\displaystyle f(x)=x,f'(x)=1}$
${\displaystyle g(x)=x^{5},g'(x)=5x^{4}}$
${\displaystyle W(f,g):={\begin{bmatrix}x&x^{5}\\1&5x^{4}\end{bmatrix}}=5x^{5}-x^{5}=4x^{5}\neq 0}$ so function is linearly independent.

2)${\displaystyle f(x)=cos2x,g(x)=sin4x}$

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

${\displaystyle W(f,g):={\begin{bmatrix}cos2x&sin4x\\-2sin2x&4cos4x\end{bmatrix}}=cos2x*4cos4x+sin4x*2sin2x\neq 0}$ so function is linearly independent.

Gramian:
${\displaystyle \Gamma (f,g):={\begin{bmatrix}&\\&\end{bmatrix}}}$

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

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

${\displaystyle \Gamma (f,g):={\begin{bmatrix}{\frac {2}{3}}&{\frac {2}{7}}\\{\frac {2}{7}}&{\frac {2}{11}}\end{bmatrix}}\neq 0}$ so function is linearly independent.

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

${\displaystyle \Gamma (f,g):={\begin{bmatrix}.818&0\\0&.876\end{bmatrix}}\neq 0}$ 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.