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

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.

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

$$ \sum_{m=0}^{inf}(\frac{2}{3})^mx^2m $$

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 Statement
Use the Determinant of the Matrix of Components and the Gramian to verify the linear independence of the two vectors $$b_1$$ and $$b_2$$. $$\mathbf b_1 = 5 \mathbf e_1 - 3 \mathbf e_2$$ $$\mathbf b_2 = -2 \mathbf e_1 + 4 \mathbf e_2$$

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:

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

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 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: $$\Gamma(f,g) := \begin{bmatrix}  &  \\  &  \end{bmatrix}$$ Function is linearly independent if $$\Gamma(f,g) \neq 0$$

1) $$ f(x) = x, g(x) = x^5 $$ $$ = \int_{-1}^{1}x^2dx = \frac{2}{3}$$ $$ = \int_{-1}^{1}x^6dx = \frac{2}{7}$$ $$ = \int_{-1}^{1}x^6dx = \frac{2}{7}$$ $$ = \int_{-1}^{1}x^10dx = \frac{2}{11}$$

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

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

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