# University of Florida/Egm4313/IEA-f13-team10/R5
## Report 5
### Problem 1: Taylor Series Expansion of the log Function
#### Problem Statement

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

#### Solution

$f(x)=\log(x)$ $f({\widehat {x}})=\log({\widehat {x}})$
$\sum _{k=0}^{\infty }{\frac {(f^{k})({\widehat {x}})}{k!}}(x-{\widehat {x}})^{k}$ $f^{1}(x)={\frac {1}{1+x}}=f^{1}({\widehat {x}})={\frac {1}{1+({\widehat {x}})}}$ Set $w(x)=3+4x$ $w'(x)=4$
$f^{2}(x)={\frac {-1*w'*4}{w^{2}}}={\frac {-1}{w^{2}}}$
$f^{3}(x)={\frac {1*2*w*w'*4^{2}}{w^{4}}}={\frac {1*2}{w^{3}}}$
$f^{4}(x)={\frac {-1*2*3*w^{2}*w'*4^{3}}{w^{6}}}={\frac {-1*2*3}{w^{4}}}$
$f^{k}(x)={\frac {(-1)^{k-1}*(k-1)!*4^{k-1}}{w^{k}}}$
$\sum _{k=0}^{\infty }{\frac {(f^{k})({\widehat {x}})}{k!}}(x-{\widehat {x}})^{k}=\sum _{k=0}^{\infty }{\frac {(-1)^{k-1}(k-1)!}{k!(1+({\widehat {x}}))^{k}}}(x-{\widehat {x}})^{k}$ For $log(3+4x)$ the series expansion results in,
$\sum _{k=0}^{\infty }(-1)^{k+1}*({\frac {(2+4x)^{k}}{k}})$
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
$x\longrightarrow t\ {\text{ such that }}\ 3+4x=1+t$ $x={\frac {t-2}{4}}$
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:$x+{\frac {x^{3}}{3*2*1}}+{\frac {x^{5}}{5*4*3*2*1}}$

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

#### 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:
$R_{c}=[lim\left|{\frac {d_{k+1}}{d_{k}}}\right|]^{-1}$
$R_{c}=[lim{\frac {(-1)^{m+1}}{(2m+3)!}}*{\frac {(2m+1)!}{(-1)^{m}}}]^{-1}$
$R_{c}=\infty$

#### 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:
$-\sum _{n=1}^{\infty }{\frac {(-1)^{n}(x)^{n}}{n}}$
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:
$-\sum _{n=1}^{\infty }{\frac {(-1)^{n}(x-1)^{n}}{n}}$
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:
$-\sum _{n=1}^{\infty }{\frac {(-1)^{n}(x-{\widehat {x}})^{n}}{n}}$
$R_{c}=[{\frac {(-1)^{n+1}({\widehat {x}})^{n+1}}{n+1}}*{\frac {n}{(-1)^{n}({\widehat {x}})^{n}}}]^{-1}$

#### Number 7

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

#### Number 8

Radius of convergence of log(3+4x) about the point ${\widehat {x}}=-.5$
${\widehat {x}}=-.5$ $\sum _{n=0}^{\infty }(-1)^{n+1}{\frac {4^{n}}{n}}(x+.5)^{n}$ $R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}(.25)^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(.25)^{n}}}]^{-1}$ Cancelling some terms out, you get
$R_{c}=lim_{n\rightarrow \infty }{\frac {-(n+1)}{n}}$ Using L'Hopitals Rule, you get
$R_{c}=lim_{n\rightarrow \infty }{\frac {1}{1}}=1$

#### Number 9

Radius of convergence of log(3+4x) about the point ${\widehat {x}}=-.25$
${\widehat {x}}=-.25$
$\sum _{n=0}^{\infty }(-1)^{n+1}{\frac {4^{n}}{n}}(x+.25)^{n}$ $R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}(.0625)^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(.0625)^{n}}}]^{-1}$ Cancelling some terms out, you get
$R_{c}=lim_{n\rightarrow \infty }{\frac {-4(n+1)}{n}}$ Using L'Hopitals Rule, you get
$R_{c}=lim_{n\rightarrow \infty }{\frac {4}{1}}=4$

#### Number 10

Radius of convergence of log(3+4x) about the point ${\widehat {x}}=1$
${\widehat {x}}=1$ $\sum _{n=0}^{\infty }(-1)^{n+1}{\frac {4^{n}}{n}}(x-1)^{n}$ $R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}(1)^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(1)^{n}}}]^{-1}$ Cancelling some terms out, you get
$R_{c}=lim_{n\rightarrow \infty }{\frac {-(n+1)}{4n}}$ Using L'Hopitals Rule, you get
$R_{c}=lim_{n\rightarrow \infty }{\frac {1}{4}}=1/4$

#### Number 11

Radius of convergence of log(3+4x) about any given point ${\widehat {x}}$
$\sum _{n=0}^{\infty }(-1)^{n+1}{\frac {4^{n}}{n}}(x-{\widehat {x}})^{n}$ $R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}({\widehat {x}}^{2})^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(({\widehat {x}}^{2})^{n})}}]^{-1}$

#### 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 $b_{1}$ and $b_{2}$.
$\mathbf {b} _{1}=5\mathbf {e} _{1}-3\mathbf {e} _{2}$
$\mathbf {b} _{2}=-2\mathbf {e} _{1}+4\mathbf {e} _{2}$

#### Solution
##### Determinant of the Matrix of Components

The Matrix of components of the vectors $b_{1}$ and $b_{2}$ is
${\begin{vmatrix}5&-3\\-2&4\end{vmatrix}}=(5)(4)-(-3)(-2)=20-6=14\neq 0$
So the vectors $b_{1}$ and $b_{2}$ are linearly independent.

##### Gramian

For vectors, the Gramian is defined as:
$\ {\boldsymbol {\Gamma }}(b_{1},b_{2})={\begin{bmatrix}\langle b_{1},b_{1}\rangle &\langle b_{1},b_{2}\rangle \\\langle b_{2},b_{1}\rangle &\langle b_{2},b_{2}\rangle \end{bmatrix}}$
where:
$\ \langle b_{i},b_{j}\rangle =b_{i}\cdot b_{j}$
For the given vectors, the dot products are:
$\ \langle b_{1},b_{1}\rangle =(5e_{1}-3e_{2})\cdot (5e_{1}-3e_{2})=(5)(5)+(-3)(-3)=25+9=34$
$\ \langle b_{1},b_{2}\rangle =(5e_{1}-3e_{2})\cdot (-2e_{1}+4e_{2})=(5)(-2)+(-3)(4)=-10-12=-22$
$\ \langle b_{2},b_{1}\rangle =(-2e_{1}+4e_{2})\cdot (5e_{1}-3e_{2})=(-2)(5)+(4)(-3)=-10-12=-22$
$\ \langle b_{2},b_{2}\rangle =(-2e_{1}+4e_{2})\cdot (-2e_{1}+4e_{2})=(-2)(-2)+(4)(4)=4+16=20$
So the Gramian matrix becomes: $\ {\boldsymbol {\Gamma }}(b_{1},b_{2})={\begin{bmatrix}34&-22\\-22&20\end{bmatrix}}$
Finding the determinant of the Gramian matrix gives the Gramian:
$\ \Gamma =(34)(20)-(-22)(-22)=680-484=156\neq 0$
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:
$W(f,g):={\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}=fg'-gf'$
Function is linearly independent if $W(f,g)\neq 0$
1) $f(x)=x,g(x)=x^{5}$
$f(x)=x,f'(x)=1$ $g(x)=x^{5},g'(x)=5x^{4}$ $W(f,g):={\begin{bmatrix}x&x^{5}\\1&5x^{4}\end{bmatrix}}=5x^{5}-x^{5}=4x^{5}\neq 0$ so function is linearly independent.
2)$f(x)=cos2x,g(x)=sin4x$
$f(x)=cos2x,f'(x)=-2sin2x$ $g(x)=sin4x,g'(x)=4cos4x$
$W(f,g):={\begin{bmatrix}cos2x&sin4x\\-2sin2x&4cos4x\end{bmatrix}}=cos2x*4cos4x+sin4x*2sin2x\neq 0$ so function is linearly independent.
Gramian: $\Gamma (f,g):={\begin{bmatrix}<f,f>&<f,g>\\<g,f>&<g,g>\end{bmatrix}}$
Function is linearly independent if $\Gamma (f,g)\neq 0$
1) $f(x)=x,g(x)=x^{5}$ $<f,f>=\int _{-1}^{1}x^{2}dx={\frac {2}{3}}$ $<f,g>=\int _{-1}^{1}x^{6}dx={\frac {2}{7}}$ $<g,f>=\int _{-1}^{1}x^{6}dx={\frac {2}{7}}$ $<g,g>=\int _{-1}^{1}x^{1}0dx={\frac {2}{11}}$
$\Gamma (f,g):={\begin{bmatrix}{\frac {2}{3}}&{\frac {2}{7}}\\{\frac {2}{7}}&{\frac {2}{11}}\end{bmatrix}}\neq 0$ so function is linearly independent.
2) $f(x)=cos2x,g(x)=sin4x$ $<f,f>=\int _{-1}^{1}cos2x*cos2xdx={\frac {1}{4}}(4+sin4)=.818$ $<f,g>=\int _{-1}^{1}cos2x*sin4xdx=[.5cos^{2}x-1/12cos6x]_{-}^{1}1=0$ $<g,f>=\int _{-1}^{1}cos2x*sin4xdx=0$ $<g,g>=\int _{-1}^{1}sin4x*sin4xdx=1+{\frac {sin8}{8}}=.876$
$\Gamma (f,g):={\begin{bmatrix}.818&0\\0&.876\end{bmatrix}}\neq 0$ 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.