University of Florida/Egm4313/s12.teamboss/R4

File:Captur44.JPG

Problem R4.1 Expanding Particular Solutions and the Coefficient Matrix A

Statement

4.1 from lecture notes R4.1 Lect. 7c pgs. 19-22

Given the general form of polynomial excitation.

$y^{″}+a{y}^{\prime }+by=\sum _{j=0}^n{d}_j{x}^j$

(1.1)

The particular solution that satisfies:

$y_p(x)=\sum _{j=0}^n{c}_j{x}^j$

(1.2)

The first and second derivative of the particular solution that solves the original polynomial excitation equation.

$y_{p^{\prime }}(x)=\sum _{j=0}^{n-1}{c}_{j+1}(j+1)x^j$

(1.3)

$y_{p^{″}}(x)=\sum _{j=0}^{n-2}{c}_{j+2}(j+2)(j+1)x^j$

(1.4)

The particular solutions are put into the polynomial excitation equation to give the general summation form:

$\sum _{j=0}^{n-2}[c_{j+2}(j+2)(j+1)+a{c}_{j+1}(j+1)+b{c}_j]x^j+a{c}_{n}n{x}^{n-1}+b[c_{n-1}{x}^{n-1}+c_n{x}^n]=\sum _{j=0}^n{d}_j{x}^j$

(1.5)

Obtain the equations associated with $d_1$, coefficients of $x$; $d_2$, coefficients of $x^2$; $d_{n-2}$, coefficients of $x^{n-2}$; $d_{n-1}$, coefficients of $x^{n-1}$; $d_n$, coefficients of $x^n$. Five total equations for coefficients.

Also set up the matrix $𝐀$ that satisfies $𝐀𝐜=𝐝$.

Solution

The given equation associated with $d_0$ taking $j=0$

$2c_2+a{c}_1+b{c}_0=d_0$

(1.6)

Taking $j=1$

$[c_{1+2}(1+2)(1+1)+a{c}_{1+1}(1+1)+b{c}_1]x^1=d_1{x}^{{}^1}$

(1.7)

$[6c_3+2a{c}_2+b{c}_1]x=d_{1}x$

(1.8)

The equation associated with $d_1$ , coefficients of $x$

$6c_3+2a{c}_2+b{c}_1=d_1$

(1.9)

Taking $j=2$

$[c_{2+2}(2+2)(2+1)+a{c}_{2+1}(2+1)+b{c}_2]x^2=d_2{x}^{{}^2}$

(1.10)

$[12c_4+3a{c}_3+b{c}_2]x^2=d_2{x}^2$

(1.11)

The equation associated with $d_2$ , coefficients of $x^2$

$12c_4+3a{c}_3+b{c}_2=d_2$

(1.12)

Taking $j=n-2$

$[c_{(n-2)+2}(n-2+2)(n-2+1)+a{c}_{(n-2+1)}(n-2+1)+b{c}_{n-2}]x^{n-2}=d_{n-2}{x}^{n-2}$

(1.13)

$[c_{n}n(n-1)+a{c}_{(n-1)}(n-1)+b{c}_{n-2}]x^{n-2}=d_{n-2}{x}^{n-2}$

(1.14)

The equation associated with $d_{n-2}$ , coefficients of $x^{n-2}$

$c_{n}n(n-1)+a{c}_{(n-1)}(n-1)+b{c}_{n-2}=d_{n-2}$

(1.15)

For $n-1$ (the summation term only goes to $j=n-2$)

$a{c}_{n}n{x}^{n-1}+b{c}_{n-1}{x}^{n-1}=d_{n-1}{x}^{n-1}$

(1.16)

$a{c}_{n}n+b{c}_{(n-1)}]x^{n-1}=d_{n-1}{x}^{n-1}$

(1.17)

The equation associated with $d_{n-1}$ , coefficients of $x^{n-1}$

$a{c}_{n}n+b{c}_{(n-1)}=d_{n-1}$

(1.18)

For $n$.

$b{c}_n{x}^n=d_n{x}^{{}^n}$

(1.19)

The equation associated with $d_n$ , coefficients of $x^n$

$b{c}_n=d_n$

(1.20)

Now set up the equation:

$𝐀𝐜=𝐝$

(1.21)

$\left[\begin{array}{c}Equationfor{d}_0\\ Equationfor{d}_1\\ Equationfor{d}_2\\ ...\\ Equationfor{d}_{n-2}\\ Equationfor{d}_{n-1}\\ Equationfor{d}_n\end{array}\right]\left[\begin{array}{c}{c}_0\\ c_1\\ c_2\\ ...\\ c_{n-2}\\ c_{n-1}\\ c_n\end{array}\right]=\left[\begin{array}{c}{d}_0\\ d_1\\ d_2\\ ...\\ d_{n-2}\\ d_{n-1}\\ d_n\end{array}\right]$

(1.22)

Therefore the matrix $𝐀$ that satisfies the matrix equation is:

$𝐀=\left[\begin{array}{ccccccc}b& a& 2& & & & \\ & b& 2a& 6& & & \\ & & b& 3a& 12& & \\ ...& ...& ...& ...& ...& ...& ...\\ & & & & b& a(n-1)& n(n-1)\\ & & & & & b& an\\ & & & & & & b\end{array}\right]$

(1.23)

Author

Solved and Typed By - Chris Stewart Egm4313.s12.team1.stewart (talk) -- 21:12, 11 March 2012 (UTC)

Reviewed By - --Egm4313.s12.team1.durrance (talk) 23:14, 11 March 2012 (UTC)

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Problem R4.2 Taylor Series Approximation of ODE with Excitation sin(x)

Statement

Consider the L2-ODE-CC with $\sin x$ as excitation (see R4.2 Lect. 7c pgs. 26-27):

$y^{″}-3y^{\prime }+2y=r(x)$
$r(x)=\sin x$

and the initial conditions

$y(0)=1,y^{\prime }(0)=0$

1) Use the Taylor series for $\sin x$ to reproduce the figure on p.7-24.

2) Let $y_{p,n}(x)$ be the particular soln corresponding to the excitation $r_n(x)$:

$y^{\prime }{\text{'}}_{p,n}+a{y}_{p^{\prime },n}+b{y}_{p,n}=r_n(x)$

Let $r_n(x)$be the truncated Taylor series of $\sin x$:

$r_n(x):=\sum _{k=0}^n\frac{(-1{)}^k{t}^{2k+1}}{(2k+1)!}=t-\frac{t^3}{3!}+...+\frac{(-1{)}^n{t}^{2n+1}}{(2n+1)!}$

Let $y_n(x)$ be the overall soln for the L2-ODE-CC: $y^{\prime }{\text{'}}_n+ay{\text{'}}_n+b{y}_n=r_n(x)$
With the same initial conditions as stated above.

Find $y_n(x)$ for n=3,5,9; plot these solns for x in the interval [0,4π].

3)Find the exact overall soln $y(x)$, and plot it in the above figure to compare with $y_n(x)$ for n=3,5,9.

Solution

To approximate the value of the excitation, the Taylor expansion must be found:

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

(2.0)

For the sine function, the Taylor series approximated to 13 places is:

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+\frac{x^{13}}{13!}$

(2.1)

Any lower approximation would include all of the terms above without any terms with a higher order than the desired order:

Plotting every order approximation of the Taylor series up to n=13 with the actual sine function produces the following:

Figure 4.2-1

File:Taylor sine.png

In order to find the overall solution for the L2-ODE-CC corresponding to the Taylor series expansion of the sine function, both the homogenous and particular solutions must be found. The homogenous equation can be found through this method:

$y^{″}-3y^{\prime }+2y=0$

(2.2)

${\lambda }^2-3\lambda +2=0\to \lambda =1,2$

(2.3)

$y_{h,n}(x)=C_1{e}^x+C_2{e}^{2x}$

(2.4)

Next, the excitation must be expanded to the desired n. The following shows the excitation expanded to n=3, 5, and 9:

$r_3(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}$

(2.5)

$r_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}$

(2.6)

$r_9(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+\frac{x^{13}}{13!}-\frac{x^{15}}{15!}+\frac{x^{17}}{17!}-\frac{x^{19}}{19!}$

(2.7)

Next, the particular solution must be found. The particular solution will be of the form:

$y_{p,n}=\sum _{i=0}^n{k}_i{x}^i$

(2.8)

Using the derivation discussed in R4.1, a matrix equation in the form Ak = d can be found, where k is the matrix containing the coefficients of the particular solution and d is the matrix containing the coefficients of each power of x in the expansion of the excitation. The general formula for A is:

$\left[\begin{array}{ccccccc}b& a& 2& & & & \\ & b& 2a& 6& & & \\ & & b& 3a& 12& & \\ & & & ...& ...& ...& \\ & & & & b& a(n-1)& n(n-1)\\ & & & & & b& an\\ & & & & & & b\end{array}\right]$

(2.9)

n=3
The matrix A is found as:

$A=\left[\begin{array}{cccccccc}2& -3& 2& & & & & \\ & 2& -6& 6& & & & \\ & & 2& -9& 12& & & \\ & & & 2& -12& 20& & \\ & & & & 2& -15& 30& \\ & & & & & 2& -18& 42\\ & & & & & & 2& -21\\ & & & & & & & 2\end{array}\right]$

(2.10)

Therefore, the matrix equation is:

$\left[\begin{array}{cccccccc}2& -3& 2& & & & & \\ & 2& -6& 6& & & & \\ & & 2& -9& 12& & & \\ & & & 2& -12& 20& & \\ & & & & 2& -15& 30& \\ & & & & & 2& -18& 42\\ & & & & & & 2& -21\\ & & & & & & & 2\end{array}\right]\left[\begin{array}{c}{k}_0\\ k_1\\ k_2\\ k_3\\ k_4\\ k_5\\ k_6\\ k_7\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ 0\\ -\frac{1}{6}\\ 0\\ \frac{1}{120}\\ 0\\ -\frac{1}{5040}\end{array}\right]$

(2.11)

$\left[\begin{array}{c}{k}_0\\ k_1\\ k_2\\ k_3\\ k_4\\ k_5\\ k_6\\ k_7\end{array}\right]=\left[\begin{array}{c}-0.1992\\ -0.3984\\ -0.3984\\ -0.0990\\ -0.0078\\ -0.0031\\ -0.0010\\ -9.9206\times 10^{-5}\end{array}\right]$

(2.12)

Therefore the particular solution is:

$y_{p,3}(x)=-9.9206\times 10^{-5}{x}^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992$

(2.13)

And the overall solution for n=3 is:

$y_3(x)=C_1{e}^x+C_2{e}^{2x}-9.9206\times 10^{-5}{x}^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992$

(2.14)

Using the initial conditions:

$1=C_1+C_2-0.1992\mathrm{\&}0=C_1+2C_2-0.3984$

(2.15)

Solving yields $C_1=2\mathrm{\&}{C}_2=-0.8008$. Therefore the overall solution is:

$y_3(x)=2e^x-0.8008e^{2x}-9.9206\times 10^{-5}{x}^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992$

(2.16)

n=5
The matrix equation is:

File:R4217.png

Therefore the particular solution is:

$y_{p,5}(x)=-1.2526\times 10^{-8}{x}^{11}-2.0668\times 10^{-7}{x}^{10}-1.0334\times 10^{-6}{x}^9-4.6503\times 10^{-6}{x}^8-1.1781\times 10^{-4}{x}^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2000$

(2.18)

And the overall solution for n=5 is:

$y_5(x)=C_1{e}^x+C_2{e}^{2x}-1.2526\times 10^{-8}{x}^{11}-2.0668\times 10^{-7}{x}^{10}-1.0334\times 10^{-6}{x}^9-4.6503\times 10^{-6}{x}^8-1.1781\times 10^{-4}{x}^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2000$

(2.19)

Using the initial conditions:

$1=C_1+C_2-0.2\mathrm{\&}0=C_1+2C_2-0.3999$

(2.20)

Solving yields $C_1=2.0001\mathrm{\&}{C}_2=-0.8001$. Therefore the overall solution is:

$y_5(x)=2.0001e^x-0.8001e^{2x}-1.2526\times 10^{-8}{x}^{11}-2.0668\times 10^{-7}{x}^{10}-1.0334\times 10^{-6}{x}^9-4.6503\times 10^{-6}{x}^8-1.1781\times 10^{-4}{x}^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2$

(2.21)

n=9
The matrix equation is:

File:R4222.png

Therefore the particular solution is:

$y_{p,9}(x)=-4.1103\times 10^{-18}{x}^{19}-1.1714\times 10^{-16}{x}^{18}-1.0543\times 10^{-15}{x}^{17}-8.9615\times 10^{-15}{x}^{16}-4.5405\times 10^{-13}{x}^{15}-9.1408\times 10^{-12}{x}^{14}-6.3985\times 10^{-11}{x}^{13}-4.1590\times 10^{-10}{x}^{12}-1.5021\times 10^{-8}{x}^{11}-2.2040\times 10^{-7}{x}^{10}-1.1020\times 10^{-6}{x}^9-4.9591\times 10^{-6}{x}^8-1.1904\times 10^{-4}{x}^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000$

(2.23)

And the overall solution for n=9 is:

$y_9(x)=C_1{e}^x+C_2{e}^{2x}-4.1103\times 10^{-18}{x}^{19}-1.1714\times 10^{-16}{x}^{18}-1.0543\times 10^{-15}{x}^{17}-8.9615\times 10^{-15}{x}^{16}-4.5405\times 10^{-13}{x}^{15}-9.1408\times 10^{-12}{x}^{14}-6.3985\times 10^{-11}{x}^{13}-4.1590\times 10^{-10}{x}^{12}-1.5021\times 10^{-8}{x}^{11}-2.2040\times 10^{-7}{x}^{10}-1.1020\times 10^{-6}{x}^9-4.9591\times 10^{-6}{x}^8-1.1904\times 10^{-4}{x}^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000$

(2.24)

Using the initial conditions:

$1=C_1+C_2-0.2\mathrm{\&}0=C_1+2C_2-0.4$

(2.25)

Solving yields $C_1=2\mathrm{\&}{C}_2=-0.8$. Therefore the overall solution is:

$y_9(x)=2e^x-0.8e^{2x}-4.1103\times 10^{-18}{x}^{19}-1.1714\times 10^{-16}{x}^{18}-1.0543\times 10^{-15}{x}^{17}-8.9615\times 10^{-15}{x}^{16}-4.5405\times 10^{-13}{x}^{15}-9.1408\times 10^{-12}{x}^{14}-6.3985\times 10^{-11}{x}^{13}-4.1590\times 10^{-10}{x}^{12}-1.5021\times 10^{-8}{x}^{11}-2.2040\times 10^{-7}{x}^{10}-1.1020\times 10^{-6}{x}^9-4.9591\times 10^{-6}{x}^8-1.1904\times 10^{-4}{x}^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000$

(2.26)

The following plot shows the overall solution of the ODE at n=3,5,9 over the domain [0,4π]:

Figure 4.2-2

File:FinODE.png

The near perfect overlap between all three graphs shows that the approximations converge quickly for very low values of n.

Next, the exact ODE will be calculated to find the accuracy of the above approximations. The homogeneous solution is the same as the homogeneous solution above, only with different coefficients. The particular solution will be in the form:

$y_p(x)=K\cos x+M\sin x$

(2.27)

${y_p}^{\prime }(x)=-K\sin x+M\cos x$

(2.28)

${y_p}^{″}(x)=-K\cos x-M\sin x$

(2.29)

Plugging these values into the ODE:

$-K\cos x-M\sin x+3K\sin x-3M\cos x+2K\cos x+2M\sin x=\sin x$

(2.30)

Separating sine and cosine terms yield two linear equations which can be used to solve for the unknown coefficients:

$K-3M=0\mathrm{\&}3K+M=1$

(2.31)

Solving these equations yield K = 0.3 and M = 0.1.

$y_p(x)=0.3\cos x+0.1\sin x$

(2.32)

Therefore the exact overall solution is:

$y(x)=C_1{e}^x+C_2{e}^{2x}+0.3\cos x+0.1\sin x$

(2.33)

$y^{\prime }(x)=C_1{e}^x+2C_2{e}^{2x}-0.3\sin x+0.1\cos x$

(2.34)

Using the initial conditions yield the equations:

$1=C_1+C_2+0.3\mathrm{\&}0=C_1+2C_2+0.1$

(2.35)

Solving these two equations yield $C_1=1.5\mathrm{\&}{C}_2=-0.8$. Therefore the exact overall solution is:

$y(x)=1.5e^x-0.8e^{2x}+0.3\cos x+0.1\sin x$

(2.36)

The following figure shows the plot of the exact solution over the plot in Fig. 4.2-2:

Figure 4.2-3

File:ExacODE.png

The overlap of the plots shows that the Taylor series approximation approach to the ODE is actually accurate to a very large degree with respect to the exact solution of the ODE.

Author

Solved and Typed By - Egm4313.s12.team1.armanious (talk) 04:02, 14 March 2012 (UTC)

Reviewed By - Chris Stewart Egm4313.s12.team1.stewart (talk) 04:52, 14 March 2012 (UTC)

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Problem R4.3

Statement

Consider the L2-ODE-CC (5) p7b-7 with log(1+x) as excitation:

$y^{″}-3y^{\prime }+2y=r(x)$

(3.0)

$r(x)=\log (1+x)$

(3.1)

And the initial conditions

$y(-\frac{3}{4})=1,y^{\prime }(-\frac{3}{4})=0$

(3.2)

1. Develop log(1+x) in Taylor series, about $\hat{x}=0$ to reproduce the figure on p.7-25

2. Let $r_n(x)$ be the truncated Taylor series, with n terms--which is also the highest degree of the Taylor (power) series -- of lg(1+x).

Find $y_n(x)$ , for n=4,7,11, such that:

with the same initial conditions (2).

Plot $y_n(x)$ for n = 4,7,11 for x in $[-\frac{3}{4},3]$.

3. Use the matlab command ode45 to integrate numerically (5) p.7b-7 with (1)-(2) p 7-28 o obtain the numerical soln for y(x). Plot y(x) in the same figure with $y_n(x)$

Solution

1. Developing the Taylor series
First, we take the generic Taylor series formula:

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

(3.3)

Then, log(1+x) is expanded into a power series:

$log(1+x)=\frac{1}{1!}(x{)}^1+\frac{-1}{2!}(x{)}^2+\frac{1}{3!}(x{)}^3+\frac{-1}{4!}(x{)}^4...$

(3.4)

2. Yn(x) at different n values First we create the characteristic equation in standard form:

${\lambda }^2-3\lambda +2=0$

(3.5)

Then, by setting it equal to zero, we can find what $\lambda$ equals:

$(\lambda -2)(\lambda -1)=0$

(3.6)

$\lambda =2,\lambda =1$

(3.7)

Given two, distinct, real roots, the homogeneous solution looks like this:

$y_h(x)=C_1{e}^{2x}+C_2{e}^x$

(3.8)

The particular solution at n=4 is of the form:

$y_p(x)=A_4{x}^4+A_3{x}^3+A_2{x}^2+A_{1}x+A_0$

(3.9)

It's derivative would look like this:

${y_p}^{\prime }(x)=4A_4{x}^3+3A_3{x}^2+2A_{2}x+A_1$

(3.10)

And the second derivative to follow would then become:

${y_p}^{″}(x)=12A_4{x}^2+6A_{3}x+2A_2$

(3.11)

Based on the coefficients, the following system of equations exists:

$2A_4=\frac{-1}{24}$

(3.12)

$-12A_4+2A_3=\frac{1}{6}$

(3.13)

$12A_4-9A_3+2A_2=\frac{-1}{2}$

(3.14)

$6A_3-6A_2+2A_1=1$

(3.15)

$2A_2-3A_1+2A_0=0$

(3.16)

The results of this set of equations make the coefficients of A's:

$A_4=\frac{-1}{48}$

$A_3=\frac{-1}{24}$

$A_2=\frac{-3}{16}$

$A_1=\frac{1}{16}$

$A_0=\frac{9}{32}$

The resulting particular equation looks like this:

$y_p(x)=\frac{-1}{48}{x}^4+\frac{-1}{24}{x}^3+\frac{-3}{16}{x}^2+\frac{1}{16}x+\frac{9}{32}$

(3.17)

By adding the particular and homogeneous solutions, we get the complete solution:

$\frac{-1}{48}{x}^4+\frac{-1}{24}{x}^3+\frac{-3}{16}{x}^2+\frac{1}{16}x+\frac{9}{32}+C_1{e}^{2x}+C_2{e}^x=y_{n=4}(x)$

(3.18)

We consider the initial conditions by taking the first derivative of the complete solution:

$4\frac{-1}{48}{x}^3+3\frac{-1}{24}{x}^2+2\frac{-3}{16}x+\frac{1}{16}+2C_1{e}^{2x}+C_2{e}^x=y^{\prime }$

(3.19)

By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants $C_1,C_2$:

$1=\frac{-1}{48}(-.75{)}^4+\frac{-1}{24}(-.75{)}^3+\frac{-3}{16}(-.75{)}^2+\frac{1}{16}(-.75)+\frac{9}{32}+C_1{e}^{2(-.75)}+C_2{e}^{-.75}=y_{n=4}(-.75)$

(3.20)

$0=4\frac{-1}{48}(-.75{)}^3+3\frac{-1}{24}(-.75{)}^2+2\frac{-3}{16}(-.75)+\frac{1}{16}+2C_1{e}^{2(-.75)}+C_2{e}^{(-.75)}=y{\text{'}}_{n=4}(-.75)$

(3.21)

Solving the equations proves that $C_1=-4.8322,C_2=4.0745$:
The resulting complete solution at n=4 with consideration for initial conditions then becomes:

$\frac{-1}{48}{x}^4+\frac{-1}{24}{x}^3+\frac{-3}{16}{x}^2+\frac{1}{16}x+\frac{9}{32}+-4.8322e^{2x}+4.0745e^x=y_{n=4}(x)$

(3.22)

The particular solution at n=7 is found using this matrix equation:

$\left[\begin{array}{ccccccc}2& -3& 2& 0& 0& 0& 0\\ 0& 2& -6& 6& 0& 0& 0\\ 0& 0& 2& -9& 12& 0& 0\\ 0& 0& 0& 2& -12& 20& 0\\ 0& 0& 0& 0& 2& -15& 30\\ 0& 0& 0& 0& 0& 2& -18\\ 0& 0& 0& 0& 0& 0& 2\end{array}\right]\cdot \left(\begin{array}{c}{A}_0\\ A_1\\ A_2\\ A_3\\ A_4\\ A_5\\ A_6\end{array}\right)=\left\{\begin{array}{c}0\\ 1\\ -1/2\\ 1/6\\ -1/24\\ 1/120\\ -1/720\end{array}\right\}$

(3.23)

The values for A then look like this:

$A_0=\frac{-21}{128}$ $A_1=\frac{-21}{64}$ $A_2=\frac{-21}{64}$ $A_3=\frac{-5}{96}$ $A_4=\frac{-5}{192}$ $A_5=\frac{-1}{480}$ $A_6=-6.944E-4$

The resulting particular equation looks like this:

$y_p(x)=-6.944E-4x^7+\frac{-1}{480}{x}^6+\frac{-5}{192}{x}^5+\frac{-5}{96}{x}^4+\frac{-21}{64}{x}^3+\frac{-21}{64}{x}^2+\frac{-21}{128}x$

(3.24)

By adding the particular and homogeneous solutions, we get the complete solution:

$-6.944E-4x^7+\frac{-1}{480}{x}^6+\frac{-5}{192}{x}^5+\frac{-5}{96}{x}^4+\frac{-21}{64}{x}^3+\frac{-21}{64}{x}^2+\frac{-21}{128}x+C_1{e}^{2x}+C_2{e}^x=y_{n=7}(x)$

(3.25)

We consider the initial conditions by taking the first derivative of the complete solution:

$(7)-6.944E-4x^6+(6)\frac{-1}{480}{x}^5+(5)\frac{-5}{192}{x}^4+(4)\frac{-5}{96}{x}^3+(3)\frac{-21}{64}{x}^2+(2)\frac{-21}{64}x+\frac{-21}{128}+2C_1{e}^{2x}+C_2{e}^x=y{\text{'}}_{n=7}(x)$

(3.26)

By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants $C_1,C_2$:

$1=-6.944E-4(-.75{)}^7+\frac{-1}{480}(-.75{)}^6+\frac{-5}{192}(-.75{)}^5+\frac{-5}{96}(-.75{)}^4+\frac{-21}{64}(-.75{)}^3+\frac{-21}{64}(-.75{)}^2+\frac{-21}{128}(-.75)+C_1{e}^{2(-.75)}+C_2{e}^{(-.75)}=y_{n=7}(-.75)$

(3.27)

$0=(7)-6.944E-4(-.75{)}^6+(6)\frac{-1}{480}(-.75{)}^5+(5)\frac{-5}{192}(-.75{)}^4+(4)\frac{-5}{96}(-.75{)}^3+(3)\frac{-21}{64}(-.75{)}^2+(2)\frac{-21}{64}(-.75)+\frac{-21}{128}+2C_1{e}^{2(-.75)}+C_2{e}^{(-.75)}=y{\text{'}}_{n=7}((-.75))$

(3.28)

Solving the equations proves that $C_1=-3.3921,C_2=3.5789$:
The resulting complete solution at n=7 with consideration for initial conditions then becomes:

$-6.944E-4x^7+\frac{-1}{480}{x}^6+\frac{-5}{192}{x}^5+\frac{-5}{96}{x}^4+\frac{-21}{64}{x}^3+\frac{-21}{64}{x}^2+\frac{-21}{128}x+(-3.3921)e^{2x}+(3.5789)e^x=y_{n=7}(x)$

(3.29)

The particular solution at n=11 is found using this matrix equation:

File:4.3eqn.JPG

(3.30)

The values for A then look like this:

$A_0=\frac{2481}{2048}$ $A_1=\frac{36635}{3072}$ $A_2=\frac{5897}{1024}$ $A_3=\frac{8987}{4608}$ $A_4=\frac{1417}{3072}$ $A_5=\frac{415}{4608}$ $A_6=\frac{33}{2560}$ $A_7=.001434$ $A_8=5.27E-5$ $A_9=3.44E-6$ $A_{10}=1.37786E-7$

The resulting particular equation looks like this:

$y_p(x)=1.37786E-7x^{11}+3.44E-6x^{10}+5.27E-5x^9+.001434x^8+\frac{33}{2560}{x}^7+\frac{415}{4608}{x}^6+\frac{1417}{3072}{x}^5+\frac{8987}{4608}{x}^4+\frac{5897}{1024}{x}^3+\frac{36635}{3072}{x}^2+\frac{2481}{2048}x$

(3.31)

By adding the particular and homogeneous solutions, we get the complete solution:

$1.37786E-7x^{11}+3.44E-6x^{10}+5.27E-5x^9+.001434x^8+\frac{33}{2560}{x}^7+\frac{415}{4608}{x}^6+\frac{1417}{3072}{x}^5+\frac{8987}{4608}{x}^4+\frac{5897}{1024}{x}^3+\frac{36635}{3072}{x}^2+\frac{2481}{2048}x+C_1{e}^{2x}+C_2{e}^x=y_{n=11}(x)$

(3.32)

We consider the initial conditions by taking the first derivative of the complete solution:

$(11)1.37786E-7x^{10}+(10)3.44E-6x^9+(9)5.27E-5x^8+(8).001434x^7+(7)\frac{33}{2560}{x}^6+(6)\frac{415}{4608}{x}^5+(5)\frac{1417}{3072}{x}^4+(4)\frac{8987}{4608}{x}^3+(3)\frac{5897}{1024}{x}^2+(2)\frac{36635}{3072}x+\frac{2481}{2048}+C_1{e}^{2x}+C_2{e}^x=y{\text{'}}_{n=11}(x)$

(3.33)

By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants $C_1,C_2$:

$1=1.37786E-7(-.75{)}^{11}+3.44E-6(-.75{)}^{10}+5.27E-5(-.75{)}^9+.001434(-.75{)}^8+\frac{33}{2560}(-.75{)}^7+\frac{415}{4608}(-.75{)}^6+\frac{1417}{3072}(-.75{)}^5+\frac{8987}{4608}(-.75{)}^4+\frac{5897}{1024}(-.75{)}^3+\frac{36635}{3072}(-.75{)}^2+\frac{2481}{2048}(-.75)+C_1{e}^{2(-.75)}+C_2{e}^{(-.75)}=y_{n=11}(-.75)$

(3.34)

$0=(11)1.37786E-7(-.75{)}^{10}+(10)3.44E-6(-.75{)}^9+(9)5.27E-5(-.75{)}^8+(8).001434(-.75{)}^7+(7)\frac{33}{2560}(-.75{)}^6+(6)\frac{415}{4608}(-.75{)}^5+(5)\frac{1417}{3072}(-.75{)}^4+(4)\frac{8987}{4608}(-.75{)}^3+(3)\frac{5897}{1024}(-.75{)}^2+(2)\frac{36635}{3072}(-.75)+\frac{2481}{2048}+C_1{e}^{2(-.75)}+C_2{e}^{(-.75)}=y{\text{'}}_{n=11}(-.75)$

(3.35)

Solving the equations proves that $C_1=56.1374,C_2=-32.64$:
The resulting complete solution at n=11 with consideration for initial conditions then becomes:

$1.37786E-7x^{11}+3.44E-6x^{10}+5.27E-5x^9+.001434x^8+\frac{33}{2560}{x}^7+\frac{415}{4608}{x}^6+\frac{1417}{3072}{x}^5+\frac{8987}{4608}{x}^4+\frac{5897}{1024}{x}^3+\frac{36635}{3072}{x}^2+\frac{2481}{2048}x+56.1374e^{2x}+-32.64e^x=y_{n=11}(x)$

(3.36)

Plot of all n=4 (red), n=7 (blue), n=11 (green).
File:IEA1.jpg

3. Plotting the actual y(x) against approximations

Author

Solved and Typed By - Egm4313.s12.team1.silvestri (talk) 23:17, 12 March 2012 (UTC)
Reviewed By ---Egm4313.s12.team1.rosenberg (talk) 03:10, 13 March 2012 (UTC)

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Problem R4.4

4.4 from lecture notes R4.1 Lect. 7c pgs. 29-30

Extend the accuracy of the solution beyond $\hat{x}=1$.

Part 1 Statement

1. Back up away a little from the brink of non-convergence at $x=1$ for the Taylor series of $log(1+x)$ about $\hat{x}=0$, and consider the point $x_1=0.9$.

Find the value of $y_n(x_1),y{\text{'}}_n(x_1)$ that will serve as initial conditions for the next iteration to extend the domain of accuracy of the analytical solution. Find $n$ sufficiently high so that $y_n(x_1),y{\text{'}}_n(x_1)$ do not differ from the numerical solution by more than $10^{-5}$.

Part 2 Statement

2. Develop $log(1+x)$ in Taylor series about $\hat{x}=1$ for $n=4,7,11$, and plot these truncated series vs the exact function.

What is now the domain of convergence? (by observation of your results.)

Part 2 Solution

First, we have the general form of a Taylor series:

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

(4.2.0)

Next, we set $f(x)=r(x)=log(1+x)$ and take several derivatives of $f(x)$:

$f(x)=log(1+x)\to f^{\prime }(x)=\frac{1}{(x+1)},f^{″}(x)=\frac{-1}{(x+1{)}^2},f^{‴}(x)=\frac{2}{(x+1{)}^3}...$

Using these derivatives, we can now develop $log(1+x)$ in Taylor series about $\hat{x}=1$:

$r(x)=log(1+x)=f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(1)}{n!}(x-1{)}^n=\frac{1}{0!}(x-1{)}^0+\frac{1}{1!}(x-1{)}^1-\frac{1}{2!}(x-1{)}^2+\frac{2}{3!}(x-1{)}^3-\frac{6}{4!}(x-1{)}^4+...$

(4.2.1)

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

(4.2.2)

Now the Taylor series for $n=4,7,11$ can be developed:

For $n=4$:

$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^3{f}^{(4)}(x-1{)}^4}{4!}=(x-1)-\frac{1}{2}(x-1{)}^2+\frac{1}{6}(x-1{)}^3-\frac{1}{24}(x-1{)}^4$

(4.2.3)

For $n=7$:

$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^6{f}^{(7)}(x-1{)}^7}{7!}=(x-1)-\frac{1}{2}(x-1{)}^2+\frac{1}{6}(x-1{)}^3-\frac{1}{24}(x-1{)}^4+\frac{1}{120}(x-1{)}^5-\frac{1}{720}(x-1{)}^6+\frac{1}{5040}(x-1{)}^7$

(4.2.4)

For $n=11$:

$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{10}{f}^{(11)}(x-1{)}^{11}}{11!}=(x-1)-\frac{1}{2}(x-1{)}^2+\frac{1}{6}(x-1{)}^3-\frac{1}{24}(x-1{)}^4+\frac{1}{120}(x-1{)}^5-\frac{1}{720}(x-1{)}^6+\frac{1}{5040}(x-1{)}^7-\frac{1}{40320}(x-1{)}^8+\frac{1}{362880}(x-1{)}^9-\frac{1}{3628800}(x-1{)}^{10}+\frac{1}{39916800}(x-1{)}^{11}$

(4.2.5)

Part 3 Statement

3. Find $y_n(x)$, for $n=4,7,11$, such that:
$y^{\prime }{\text{'}}_n+ay{\text{'}}_n+b{y}_n=r_n(x)$
for $x$ in $[0.9,3]$ with the initial conditions found i.e., $y_n(x_1),y{\text{'}}_n(x_1)$.

Plot $y_n(x)$ for $n=4,7,11$ for $x$ in $[0.9,3]$.

Part 3 Solution

Using the same characteristic in problem 4.3 (3.5), we see that the homogenous solution is the same (3.8):

$y_h(x)=C_1{e}^{2x}+C_2{e}^x$

(4.3.0)

The particular solution at n=4 is of the form:

$y_p(x)=A_4(x-1{)}^4+A_3(x-1{)}^3+A_2(x-1{)}^2+A_1(x-1)+A_0$

(4.3.1)

It's derivative would look like this:

${y_p}^{\prime }(x)=4A_4(x-1{)}^3+3A_3(x-1{)}^2+2A_2(x-1)+A_1$

(4.3.2)

And the second derivative to follow would then become:

${y_p}^{″}(x)=12A_4(x-1{)}^2+6A_3(x-1)+2A_2$

(4.3.3)

Based on the coefficients, the following system of equations exists:

$2A_4=\frac{-1}{4}$

(4.3.4)

$-12A_4+2A_3=\frac{1}{3}$

(4.3.5)

$12A_4-9A_3+2A_2=\frac{-1}{2}$

(4.3.6)

$6A_3-6A_2+2A_1=1$

(4.3.7)

$2A_2-3A_1+2A_0=0$

(4.3.8)

The results of this set of equations make the coefficients of A's:

$A_4=\frac{-1}{8}$

$A_3=\frac{-7}{12}$

$A_2=\frac{-17}{8}$

$A_1=\frac{-33}{8}$

$A_0=\frac{-65}{16}$

The resulting particular equation looks like this:

$y_p(x)=\frac{-1}{8}(x-1{)}^4+\frac{-7}{12}(x-1{)}^3+\frac{-17}{8}(x-1{)}^2+\frac{-33}{8}(x-1)+\frac{-65}{16}$

(4.3.9)

By adding the particular and homogeneous solutions, we get the complete solution:

$y_{n=4}(x)=C_1{e}^{2x}+C_2{e}^x+\frac{-1}{8}(x-1{)}^4+\frac{-7}{12}(x-1{)}^3+\frac{-17}{8}(x-1{)}^2+\frac{-33}{8}(x-1)+\frac{-65}{16}$

(4.3.10)

The particular solution at n=7 is found using this matrix equation:

$\left[\begin{array}{ccccccc}2& -3& 2& 0& 0& 0& 0\\ 0& 2& -6& 6& 0& 0& 0\\ 0& 0& 2& -9& 12& 0& 0\\ 0& 0& 0& 2& -12& 20& 0\\ 0& 0& 0& 0& 2& -15& 30\\ 0& 0& 0& 0& 0& 2& -18\\ 0& 0& 0& 0& 0& 0& 2\end{array}\right]\cdot \left(\begin{array}{c}{A}_0\\ A_1\\ A_2\\ A_3\\ A_4\\ A_5\\ A_6\end{array}\right)=\left\{\begin{array}{c}0\\ 1\\ -1/2\\ 1/3\\ -1/4\\ 1/5\\ -1/6\end{array}\right\}$

(4.3.11)

The values for A then look like this:

$A_0=\frac{-99}{2}$ $A_1=-99$ $A_2=-49$ $A_3=\frac{-95}{6}$ $A_4=\frac{-15}{4}$ $A_5=\frac{-13}{20}$ $A_6=\frac{-1}{12}$

The resulting particular equation looks like this:

$y_p(x)=\frac{-1}{12}(x-1{)}^7+\frac{-13}{20}(x-1{)}^6+\frac{-15}{4}(x-1{)}^5+\frac{-95}{6}(x-1{)}^4+-49(x-1{)}^3+-99(x-1{)}^2+\frac{-199}{2}(x-1)$

(4.3.12)

By adding the particular and homogeneous solutions, we get the complete solution:

$y_{n=7}(x)=C_1{e}^{2x}+C_2{e}^x+\frac{-1}{12}(x-1{)}^7+\frac{-13}{20}(x-1{)}^6+\frac{-15}{4}(x-1{)}^5+\frac{-95}{6}(x-1{)}^4+-49(x-1{)}^3+-99(x-1{)}^2+\frac{-199}{2}(x-1)$

(4.3.13)

The particular solution at n=11 is found using this matrix equation:

File:MatrixA.jpg

Part 4 Statement

4. Use the matlab command ode45 to integrate numerically $y^{″}-3y^{\prime }+2y=r(x)$ with $r(x)=log(1+x)$ and the initial conditions $y_n(x_1),y{\text{'}}_n(x_1)$ to obtain the numerical solution for $y(x)$.

Plot $y(x)$ in the same figure with $y_n(x)$.

Author

Solved and Typed By - --Egm4313.s12.team1.wyattling (talk) 19:44, 14 March 2012 (UTC)

Reviewed By - Egm4313.s12.team1.essenwein (talk) 19:52, 14 March 2012 (UTC)

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Contributing Members

Team Contribution Table
Problem Number	Lecture	Assigned To	Solved By	Typed By	Proofread By
4.1	R4.1 Lect. 7c pgs. 19-22	Chris Stewart	Chris Stewart	Chris Stewart	Jesse Durrance
4.2	R4.2 Lect. 7c pgs. 26-27	George Armanious	George Armanious	George Armanious	Chris Stewart
4.3	[Lecture link]	Emotion Silvestri	Emotion Silvestri	Emotion Silvestri	Steven Rosenberg
4.4	R4.1 Lect. 7c pgs. 29-30	Wyatt Ling	Wyatt Ling	Wyatt Ling	Eric Essenwein