University of Florida/Egm4313/s12.team5.R6

Report 6

R 6.1

Question

Find the (smallest) period of $cosn\omega x$ and $sinn\omega x$.
Show that these functions also have a period p.
Show that the constant $a_{o}$ is also a periodic function with period p.

Solution

Given (1) p.9-2: $f(x+np)=f(x)$ for all $n(...,-2,-1,0,1,2...)$ and where p is the period.
The smallest period occurs when n=1 thus we can use the equation as: $f(x+p)=f(x)$

For $f(x)=cos(n\omega x)$ we have:
$f(x+p)=cosn\omega (x+p)=cos(n\omega x+n\omega p)=cos(n\omega x)cos(n\omega p)-sin(n\omega x)sin(n\omega p)$
Thus, $cos(n\omega x)cos(n\omega p)-sin(n\omega x)sin(n\omega p)=cosn\omega x$
$cos(n\omega x)cos(n\omega p)=cos(n\omega x)(1)$
$cos(n\omega p)=1$
$n\omega p=2\pi$

$p={\frac {2\pi }{n\omega }}$, smallest period occurs when n=1 so, $p={\frac {2\pi }{\omega }}$

For $f(x)=sin(n\omega x)$ we have:
$f(x+p)=sinn\omega (x+p)=sin(n\omega x+n\omega p)=sin(n\omega x)cos(n\omega p)-cos(n\omega x)sin(n\omega p)$< br /> Thus, $sin(n\omega x)cos(n\omega p)-cos(n\omega x)sin(n\omega p)=sin(n\omega x)$
$sin(n\omega x)cos(n\omega p)=sin(n\omega x)(1)$
$cos(n\omega p)=1$
$n\omega p=2\pi$

$p={\frac {2\pi }{n\omega }}$, smallest period occurs when n=1 so, $p={\frac {2\pi }{\omega }}$

From (1) p9-5, we know that $\omega ={\frac {2\pi }{p}}$
Then, $p={\frac {2\pi }{\omega }}={\frac {2\pi }{\frac {2\pi }{p}}}=p$
Thus, the period of both of these functions is also $p$.

From (1) p.9-7, we know $a_{o}={\frac {1}{2L}}\int _{0}^{2L}f(x)dx={\frac {1}{p}}<f,1>$
Also, $a_{o}={\frac {1}{2\pi }}\int _{0}^{2\pi }f(x)dx={\frac {1}{p}}<f,1>$

Thus, $p=2\pi$ showing  $a_{o}$ is a periodic function with period $p$. 

Author

This problem is solved and uploaded by Radina Dikova

R 6.2

Question

Is the given function even or odd or either even nor odd? Find its Fourier series. Show details of your work.

$11.f(x)=x^{2},\;(-1<x<1),\;p=2$
$12.f(x)=1-{\frac {x^{2}}{4}},\;(-2<x<2),\;p=4$

Author

Solved and uploaded by William Knapper

R 6.3

Question

K 2011 p.491 pbs 15,17
Find if the graph is for an even or odd problem and then find the fourier series. Graph the resulting equations.
Problem 15:

Problem 17:

Solution

For problem 15:
f(-x) = -f(x), therefore the graph shows an odd function. Similarly, L = $\pi$.
Therefore we can use the Euler formula for an odd function:
$f(x)=\sum _{n=1}^{\infty }b_{n}\sin(nx)$
$b_{n}={\frac {2}{\pi }}\int _{0}^{\pi }f(x)\sin(nx)dx$
During $[0,{\frac {\pi }{2}}]$, f(x) = x. During $[{\frac {\pi }{2}},\pi ]$, $f(x)=\pi -x$
Therefore, $b_{n}={\frac {2}{\pi }}(\int _{0}^{\frac {\pi }{2}}x\sin(nx)dx+\int _{\frac {\pi }{2}}^{\pi }(\pi -x)\sin(nx)dx)$
Using integration by parts shows $b_{n}$ to be:
$b_{n}={\frac {2}{\pi n^{2}}}(2\sin({\frac {\pi }{2}}n)-\sin(\pi n))$
when, for the first integral:
$u=x,du=dx,dv=\sin(nx)dx,andv=-{\frac {1}{n}}\cos(nx)$
and for the second integral (where u, du, v, and dv are not the same as the first integral):
$u=\pi -x,du=-dx,dv=-\sin(nx)dx,andv={\frac {1}{n}}\cos(nx)$
Therefore, when n is odd, $b_{n}=+or-{\frac {4}{\pi n^{2}}}$ and when n is even, $b_{n}=0$
Therefore, the fourier series' are as follows:
For n=2 -> $f(x)={\frac {4}{\pi }}\sin(x)$
For n=4 -> $f(x)={\frac {4}{\pi }}\sin(x)+{\frac {4}{\pi 9}}\sin(3x)$
For n = 8 -> $f(x)={\frac {4}{\pi }}\sin(x)+{\frac {4}{\pi 9}}\sin(3x)+{\frac {4}{\pi 25}}\sin(5x)+{\frac {4}{\pi 49}}\sin(7x)$
A graph of the fourier series' is shown below:


For Problem 17: $f(-x)=f(x)$ Therefore, it is an even function and L = 1. Similarly, because it's an even function, the Euler equations for an even function can be used:
$f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}\cos(nx)$
$a_{0}={\frac {1}{\pi }}\int _{0}^{\pi }f(x)dx$
$a_{n}={\frac {2}{\pi }}\int _{0}^{\pi }f(x)\cos(nx)dx$
During [0,1], $f(x)=1-x$
Therefore:
$a_{0}=\int _{0}^{1}(1-x)dx={\frac {1}{2}}$
However, the period is not 2$\pi$, so in $a_{n}$ $\cos(nx)$ is replaced with $cos(n\pi x)$ to shift the function to a period of 2$\pi$. Similarly, ${\frac {2}{\pi }}$ is replaced with ${\frac {2}{1}}$ since L = 1.
Therefore:
$a_{n}=2\int _{0}^{1}(f(x)\cos(n\pi x))dx=2\int _{0}^{1}((1-x)\cos(n\pi x))dx$
Letting $u=1-x,du=-dx,dv=\cos(n\pi x)dx,v={\frac {1}{n\pi }}\sin(n\pi x)$ where u, du, v, and dv are not the same as was used in number 15 above.
Integration by parts then yields:
$a_{n}=2({\frac {-1}{n^{2}\pi ^{2}}}(\cos(n\pi )+{\frac {1}{n^{2}\pi ^{2}}})$
$Therefore,whennisodd,<math>a_{n}={\frac {4}{n^{2}\pi ^{2}}}$ and when n is even, $a_{n}=0$
Therefore, the fourier series's are as follows:
For n = 2 -> $f(x)={\frac {1}{2}}+{\frac {4}{\pi ^{2}}}\cos(\pi x)$
For n = 4 -> $f(x)={\frac {1}{2}}+{\frac {4}{\pi ^{2}}}\cos(\pi x)+{\frac {4}{9\pi ^{2}}}\cos(3\pi x)$
For n = 8 -> $f(x)={\frac {1}{2}}+{\frac {4}{\pi ^{2}}}\cos(\pi x)+{\frac {4}{9\pi ^{2}}}\cos(3\pi x)+{\frac {4}{25\pi ^{2}}}\cos(5\pi x)+{\frac {4}{49\pi ^{2}}}\cos(7\pi x)$

Author

This problem was solved and uploaded by John North

R 6.4

Question

Consider the L2-ODE-CC with the window function f(x) from p9-8 as excitation:

$\ y''-3y'+2y=r(x)\$ where r(x) = f(x)

and the initial conditions

$\ y(0)=1,y'(0)=0\$

1. Find $\ y_{n}(x)\$ such that:

$\ y_{n}''+ay_{n}'+by_{n}=r_{n}(x)\$

with the same initial conditions as above.

Plot $\ y_{n}(x)\$ for n = 3, 6, 9 for x in [0, 10]

2. Use the matlab command ode45 to integrate the L2-ODE-CC and plot the numerical soln to compare with the analytical soln.

Level 1: n = 0,1

Solution

$\ \lambda ^{2}-3\lambda +2=0=>(\lambda -2)(\lambda -1)=0\$

$\ \lambda _{1,2}=2,1\$

The Fourier series of a periodic function

$\ f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}\$ cos(nωx) + $\ b_{n}\$ sin(nωx)]

For an odd function the Fourier series will be the following:

$\ f(x)=\sum _{n=1}^{\infty }b_{n}\$ sin(nωx) $\ =\sum _{n=1}^{\infty }b_{n}sin({\frac {n*pi}{L}}x)dx\$

The independent variable t will be used to shift x to the left

$\ t=x-{\frac {1}{4}}\$

The period of oscillation and frequency of oscillation will be as follows:

$\ p=2L=4\$

ω = π/2

where $\ b_{n}={\frac {2}{L}}\int _{0}^{L}f(x)sin({\frac {n*pi}{L}}x)dx={\frac {1}{2}}\int _{0}^{2}f(x)sin({\frac {n*pi}{2}}x)dx\$

which comes to be: $\ b_{n}={\frac {A}{n*pi}}\$ (1 - cosnπ)

$\ y''-3y'+2y=r(x)\$ = 1/2 + (2/nπ) sin(nπ/2}t

The homogeneous equation for Y will be as follows:

$\ Y_{h}=C_{1}e^{2x}+C_{2}e^{x}\$

$\ Y=Y_{h}+Y_{p}=C_{1}e^{2x}+C_{2}e^{x}+C_{3}\$

Author

This problem was solved and uploaded by Mike Wallace

R 6.5

Question

Redo R4.2 to redisplay the particular solution, the homogenous solution, and the exact solution for n = 3,5,9 over the interval [0,20π]

Redisplay the particular solution, the homogenous solution, and the exact solution. Superpose each solution with the exact solution.

Redo R4.3 with the TA code over the interval [0.10]. Zoom in about x = -0.5, 0, and 0.5 and comment on the accuracy of different approximations.

Redo R4.4 with the TA code over the interval [0.9,10] for n = 4, 7. Zoom in about x = 1, 1.5, 2, 2.5 and comment on the accuracy of different approximations.

Solution

R 4.2 redo

The re-displayed functions for the homogenous solutions are:

$\displaystyle y_{3h}(x)=2e^{x}-0.8008e^{2x}\$

$\displaystyle y_{5h}(x)=2.0001e^{x}-0.8001e^{2x}\$

$\displaystyle y_{9h}(x)=2e^{x}-0.8e^{2x}\$

The re-displayed functions for the particular solutions are:

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

$\displaystyle y_{5p}(x)=-1.2526*10^{-8}x^{11}-2.0668*10^{-7}x^{10}-1.0334*10^{-6}x^{9}-4.6503*10^{-6}x^{8}-1.1781*10^{-4}x^{7}\$

$\displaystyle -0.0011x^{6}-0.0033x^{5}-0.0083x^{4}-0.0999x^{3}-0.3999x^{2}-0.399x-0.2000\$

$\displaystyle y_{9p}(x)=-4.1103*10^{-18}x^{19}-1.1714*10^{-16}x^{18}-1.0543*10^{-15}x^{17}-8.9615*10^{-15}x^{16}\$

$\displaystyle -4.5405*10^{-13}x^{15}-9.1408*10^{-12}x^{14}-6.3985*10^{-11}x^{13}-4.1590*10^{-10}x^{12}-1.5021*10^{-8}x^{11}-2.2040*10^{-7}x^{10}$

$\displaystyle -1.1020*10^{-6}x^{9}-4.9591*10^{-6}x^{8}-1.1904*10^{-4}x^{7}-0.0011x^{6}-0.0033x^{5}-0.0083x^{4}-0.1000x^{3}-0.4000x^{2}-0.4000x-0.2000\$

The re-displayed functions for the general solutions are:

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

$\displaystyle y_{5}(x)=2.0001e^{x}-0.8001e^{2x}-1.2526*10^{-8}x^{11}-2.0668*10^{-7}x^{10}-1.0334*10^{-6}x^{9}-4.6503*10^{-6}x^{8}\$

$\displaystyle -1.1781*10^{-4}x^{7}-0.0011x^{6}-0.0033x^{5}-0.0083x^{4}-0.0999x^{3}-0.3999x^{2}-0.399x-0.2000\$

$\displaystyle y_{9}(x)=2e^{x}-0.8e^{2x}-4.1103*10^{-18}x^{19}-1.1714*10^{-16}x^{18}-1.0543*10^{-15}x^{17}-8.9615*10^{-15}x^{16}\$

$\displaystyle -4.5405*10^{-13}x^{15}-9.1408*10^{-12}x^{14}-6.3985*10^{-11}x^{13}-4.1590*10^{-10}x^{12}-1.5021*10^{-8}x^{11}\$

$\displaystyle -2.2040*10^{-7}x^{10}-1.1020*10^{-6}x^{9}-4.9591*10^{-6}x^{8}-1.1904*10^{-4}x^{7}-0.0011x^{6}\$

$\displaystyle -0.0033x^{5}-0.0083x^{4}-0.1000x^{3}-0.4000x^{2}-0.4000x-0.2000\$

Plot for $\displaystyle y_{3}(x)\$

Plot for $\displaystyle y_{5}(x)\$

Plot for $\displaystyle y_{9}(x)\$

R 4.3 Redo

Matlab code:

x = 0:0.01:10;

y = log(1+x);

EDU>> x1 = 0:0.01:10;

EDU>> y1 = zeros(1,1001);

EDU>> for i = 1:4

for j = 1:1001

y1(j) = y1(j) - ((-x1(j))^i)/i;

end

end

EDU>> y2 = zeros(1,1001);

EDU>> for i = 1:7

for j = 1:1001

y2(j) = y2(j) - ((-x1(j))^i)/i;

end

end

EDU>> y3 = zeros(1,1001);

EDU>> for i = 1:11

for j = 1:1001

y3(j) = y3(j) - ((-x1(j))^i)/i;

end

end

EDU>> y4 = zeros(1,1001);

EDU>> for i = 1:16

for j = 1:1001

y4(j) = y4(j) - ((-x1(j))^i)/i;

end

end

EDU>> h = plot(x,y);

orange = [1 0.5 0.2];

EDU>> set(h,'Color',orange);

EDU>> hold on;

EDU>> plot(x1,y1,'r');

EDU>> plot(x1,y2,'g');

EDU>> plot(x1,y3,'b');

EDU>> plot(x1,y4,'c');

legend('log(1+x)','T_4','T_7','T_1_1','T_1_6');

EDU>> grid on;

EDU>> axis([0 10 -10 10])

Zoom plot about 0:

This plot makes it appear that only the 16 term approximation is in the window at point 0. This is, therefore, the most accurate of the other n solution approximations.

Zoom plot about 0.5:

This plot makes it appear that most of the approximations are very accurate and close to the exact solution. The only approximation that appears to deviate a little after 0.5 is the n = 4 approximation.

R 4.4 redo

Matlab code:

syms x

EDU>> f = log(x+1);

EDU>> fT1 = taylor(f,5,1);

EDU>> fT2 = taylor(f,8,1);

X = 0.9:0.1:10;

Y(:,1) = subs(fT1,'x',X);

EDU>> Y(:,2) = subs(fT2,'x',X);

EDU>> Y(:,3) = log(1+X);

EDU>> figure

EDU>> plot(X,Y);

axis([0.9 10 -10 10])

Zoom in about 1:

This shows that neither of the approximations are very close about x = 1 to the exact solution.

Zoom in about 1.5:

This shows that the n = 7 approximation is close about x = 1.5 to the exact solution.

Zoom in about 2:

This shows that all the approximations are very close to the exact solution about x = 2.

Zoom in about 2.5:

This shows that the approximations are very close about x = 2.5, however, the n = 4 approximation is beginning to deviate and soon will not be a good approximation.

Author

This problem was solved and uploaded by David Herrick

R 6.6

Question

Given $\displaystyle y_{p}''+4y_{p}'+13y_{p}=2e^{-2x}cos(3x)$

(1) Simplify the first term $\displaystyle y_{p}''$ on the lhs

(2) Simplify the second term $\displaystyle 4y_{p}'$ and combine with the simplified first term

(3) Finally, add the third term $\displaystyle 13y_{p}$

Solution

From Lecture Notes Sec.10 p.10-3:

Particular solution is of the form $\displaystyle y_{p}(x)=xe^{-2x}[Mcos(3x)+Nsin(3x)]$

From Lecture Notes Sec.10 p.10-3:

$\displaystyle y_{p}^{'}(x)=e^{-2x}[sin(3x)(-3mx-2nx+n)+cos(3x)(-2mx+m+3nx)]$

$\displaystyle y_{p}^{''}(x)=e^{-2x}[sin(3x)[6m(2x-1)-n(5x+4)]-cos(3x)[m(5x+4)+6n(2x-1)]]$

We want to substitute these derivatives back into the original ODE and verify that Wolfram Alpha's solution of $\displaystyle 6e^{-2x}[ncos(3x)-msin(3x)]$ is correct.

(1)

$\displaystyle y_{p}^{''}=e^{-2x}[sin(3x)[6m(2x-1)-n(5x+4)]-cos(3x)[m(5x+4)+6n(2x-1)]]$

$\displaystyle y''_{p}=e^{-2x}[12mxsin(3x)-6msin(3x)-5nxsin(3x)-4nsin(3x)-5mxcos(3x)-4mcos(3x)-12nxcos(3x)+6ncos(3x)]$

(2)

$\displaystyle y_{p}^{'}=e^{-2x}[sin(3x)(-3mx-2nx+n)+cos(3x)(-2mx+m+3nx)]$

$\displaystyle 4y'p=e^{-2x}[-12mxsin(3x)-8nxsin(3x)+4nsin(3x)-8mxcos(3x)+4mcos(3x)+12nxcos(3x)]$

$\displaystyle y''_{p}+4y'_{p}=e^{-2x}[-6msin(3x)-13nxsin(3x)-13mxcos(3x)+6ncos(3x)]$

(3)

$\displaystyle 13y_{p}=e^{-2x}[13mxcos(3x)+13nxcos(3x)]$

$\displaystyle y''_{p}+4y'_{p}+13y_{p}=e^{-2x}[-6msin(3x)+6ncos(3x)]$

Comparing this to Wolfram Alpha's answer:

$\displaystyle 6e^{-2x}[ncos(3x)-msin(3x)]=6e^{-2x}[ncos(3x)-msin(3x)]$

Author

Solved and uploaded by Joshua House

R 6.7

Question

(1) Find the separated ODEs for the heat equation

$\displaystyle {\frac {\partial u}{\partial t}}=\kappa {\frac {\partial ^{2}u}{\partial x^{2}}}$

Solution

Assuming $\displaystyle u(x,t)=F(x)\cdot G(t)$

Then:

$\displaystyle {\frac {\partial u(x,t)}{\partial t}}=F(x)\cdot {\dot {G}}(t)$

$\displaystyle {\frac {\partial ^{2}u(x,t)}{\partial x^{2}}}=F''(x)\cdot {G}(t)$

Substituting partial derivatives back into original PDE:

$\displaystyle F(x)\cdot {\dot {G}}(t)=\kappa F''(x)\cdot G(t)$

$\displaystyle {\frac {{\dot {G}}(t)}{\kappa G(t)}}={\frac {F''(x)}{F(x)}}=c$

* Where "c" is a constant, because if both sides were variables then they would never be equal one another (each side would be a function of a different variable.) Kreysig 2011, pp.546

Multiplying by the denominators to get two separate ODEs:

$\displaystyle F''(x)-cF(x)=0$
$\displaystyle {\dot {G}}(t)-\kappa cG(t)=0$

Author

Solved and uploaded by Joshua House

Contribution Summary

Problem 1 was solved and uploaded by: Radina Dikova

Problem 2 was solved and uploaded by: William Knapper

Problem 3 was solved and uploaded by: John North

Problem 4 was solved and uploaded by: Michael Wallace

Problem 5 was solved and uploaded by: David Herrick

Problem 6 was solved and uploaded by: Joshua House

Problem 7 was solved and uploaded by: Joshua House