University of Florida/Egm4313/s12.team5.R6

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 $a_o=\frac{1}{2L}\underset{0}{\overset{2L}{\int }}f(x)dx=\frac{1}{p}<f,1>$
Also, $a_o=\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}f(x)dx=\frac{1}{p}<f,1>$

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:
$f(x)=\sum _{n=1}^{\mathrm{\infty }}{b}_n\sin (nx)$
$b_n=\frac{2}{\pi }\underset{0}{\overset{\pi }{\int }}f(x)\sin (nx)dx$
During $[0,\frac{\pi }{2}]$, f(x) = x. During $[\frac{\pi }{2},\pi ]$, $f(x)=\pi -x$
Therefore, $b_n=\frac{2}{\pi }(\underset{0}{\overset{\frac{\pi }{2}}{\int }}x\sin (nx)dx+\underset{\frac{\pi }{2}}{\overset{\pi }{\int }}(\pi -x)\sin (nx)dx)$
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:
$f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos (nx)$
$a_0=\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}f(x)dx$
$a_n=\frac{2}{\pi }\underset{0}{\overset{\pi }{\int }}f(x)\cos (nx)dx$
During [0,1], $f(x)=1-x$
Therefore:
$a_0=\underset{0}{\overset{1}{\int }}(1-x)dx=\frac{1}{2}$
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:
$a_n=2\underset{0}{\overset{1}{\int }}(f(x)\cos (n\pi x))dx=2\underset{0}{\overset{1}{\int }}((1-x)\cos (n\pi x))dx$
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:

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

and the initial conditions

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

1. Find $y_n(x)$ such that:

${y_n}^{″}+a{y_n}^{\prime }+b{y}_n=r_n(x)$

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

$f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}[a_n$ cos(nωx) + $b_n$ sin(nωx)]

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

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

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 $b_n=\frac{2}{L}\underset{0}{\overset{L}{\int }}f(x)sin(\frac{n*pi}{L}x)dx=\frac{1}{2}\underset{0}{\overset{2}{\int }}f(x)sin(\frac{n*pi}{2}x)dx$

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

$y^{″}-3y^{\prime }+2y=r(x)$ = 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:

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

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

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

The re-displayed functions for the particular solutions are:

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

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

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

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

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

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

The re-displayed functions for the general solutions are:

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

$y_5(x)=2.0001e^x-0.8001e^{2x}-1.2526*10^{-8}{x}^{11}-2.0668*10^{-7}{x}^{10}-1.0334*10^{-6}{x}^9-4.6503*10^{-6}{x}^8$

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

$y_9(x)=2e^x-0.8e^{2x}-4.1103*10^{-18}{x}^{19}-1.1714*10^{-16}{x}^{18}-1.0543*10^{-15}{x}^{17}-8.9615*10^{-15}{x}^{16}$

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

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

$-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000$

Plot for $y_3(x)$

Plot for $y_5(x)$

Plot for $y_9(x)$

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 $y_{p^{″}}+4y_{p^{\prime }}+13y_p=2e^{-2x}cos(3x)$

(1) Simplify the first term $y_{p^{″}}$ on the lhs

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

(3) Finally, add the third term $13y_p$

Solution

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

Particular solution is of the form $y_p(x)=x{e}^{-2x}[Mcos(3x)+Nsin(3x)]$

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

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

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

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

(1)

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

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

(2)

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

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

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

(3)

$13y_p=e^{-2x}[13mxcos(3x)+13nxcos(3x)]$

$y^{\prime }{\text{'}}_p+4y{\text{'}}_p+13y_p=e^{-2x}[-6msin(3x)+6ncos(3x)]$

Comparing this to Wolfram Alpha's answer:

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

Author

Solved and uploaded by Joshua House

R 6.7

Question

(1) Find the separated ODEs for the heat equation

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

Solution

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

Then:

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

$\frac{{\partial }^{2}u(x,t)}{\partial x^2}=F^{″}(x)\cdot G(t)$

Substituting partial derivatives back into original PDE:

$F(x)\cdot \dot{G}(t)=\kappa F^{″}(x)\cdot G(t)$

$\frac{\dot{G}(t)}{\kappa G(t)}=\frac{F^{″}(x)}{F(x)}=c$

* 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:

$F^{″}(x)-cF(x)=0$
$\dot{G}(t)-\kappa cG(t)=0$

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