# University of Florida/Egm6341/s10.Team2/HW7

### problem 1: Stable Growth for Verhulst Equation

#### Statement

Pg.38-4 Show the case of stable growth for Verhulst equation

#### Author

Egm6341.s10.team2.lee 20:56, 23 April 2010 (UTC)

### problem 2: Solution of the Logistic equation

#### Solution

We have Verhulst model or the Logistic equation as P.38-3

     ${\displaystyle {\dot {x}}={\frac {dx}{dt}}=rx(1-{\frac {x}{x_{max}}})}$


Separating variables and integrating we have

     ${\displaystyle \int _{x_{0}}^{x}{\frac {x_{max}}{x(x_{max}-x)}}dx=\int _{t_{0}=0}^{t}rdt}$


which can be written as

     ${\displaystyle \int _{x_{0}}^{x}{\frac {1}{x}}dx+\int _{x_{0}}^{x}{\frac {1}{(x_{max}-x)}}dx=\int _{t_{0}=0}^{t}rdt}$


We get

     ${\displaystyle log_{e}({\frac {x}{x_{0}}})-log_{e}({\frac {x_{max}-x}{x_{max}-x_{0}}})=r(t-0)}$

     ${\displaystyle {\frac {x(x_{max}-x_{0})}{x_{0}(x_{max}-x)}}=e^{rt}}$


Rearranging we have

     ${\displaystyle x={\frac {x_{max}x_{0}e^{rt}}{x_{max}+x_{0}(e^{rt}-1)}}}$


#### Author

--Egm6341.s10.team2.niki 12:31, 23 April 2010 (UTC)

### problem 3: Hermite-Simpson Algorithm to solve for the non-linear first order differential equation i.e. Verhulst population growth model on Pg.38-3

#### Statement

Use the Hermit-Simpson Algorithm to integrate the Verhulst population growth model on Pg.38-3:

${\displaystyle {\frac {\mathrm {d} x}{\mathrm {d} t}}=rx\left(1-{\frac {x}{x_{max}}}\right)}$

Consider the following: ${\displaystyle x_{max}=10\;\;r=1.2\;\;t\in \left[0,10\right]}$ and
Case 1: ${\displaystyle \;x_{0}=2}$
Case 2: ${\displaystyle \;x_{0}=7}$

And also plot the two cases using matlab.

#### Solution

Matlab code

%Matlab code to integrate the the Verhulst population growth model
clear;
clc;
xmax=10;
r=1.2;
t=0:0.1:10;
h=0.1; %time step size
syms x;
syms x11;
f=r*x*(1-(x/xmax));
%case 1: when x0=2
for i=1:1:100
x1(1)=2;
f1=subs(f,x,x1(i));
f11=subs(f,x,x11);
x1half=((1/2)*(x1+x11))+((h/8)*(f1-f11));
f1half=subs(f,x,x1half);
F=x1(i)-x11+(h/6)*(f1+f11+(4*f1half));
guessx11(i)=x1(i);
dummy=1;
while dummy==1
newx11(i)=guessx11(i)-((1/(subs(diff(F(i)),x11,guessx11(i))))*(subs(F(i),x11,guessx11(i))));
if double(abs(newx11(i)-guessx11(i)))<=10^(-6)
x1(i+1)=newx11(i);
dummy=0;
break;
else
guessx11(i)=newx11(i);
dummy=1;
end
end
end
plot(t,x1);


THE PLOT SHOWING THE POPULATION CHANGE W.R.T TIME, HERE VERHULST MODEL IS USED TO COMPUTE THE POPULATION WHERE XO=INITIAL POPULATION=2

THE PLOT SHOWING THE POPULATION CHANGE W.R.T TIME, HERE VERHULST MODEL IS USED TO COMPUTE THE POPULATION WHERE XO=INITIAL POPULATION=7

### Problem4: Creating graphs, showing the bifurcation diagram for the logistic map and its sensitivity to initial conditions

#### Statement

Produce the diagrams showing the bifurcation diagram for the logistic map and its sensitivity to initial conditions, as shown on figures 15.6 and 15.7, of the book "Differential Equations: Linear,Nonlinear,Ordinary,Partial" by King,Billingham and Otto (Pg.455-456).

#### Solution

##### Creating Figure 15.6:Bifurcation Diagram

Logistic Map:

${\displaystyle x_{n+1}=rx_{n}(1-x_{n})}$

The Bifurcation Diagram shows the period doubling process of the logistic map. As the constant r increases the period is doubled to 2 periods and then stability is lost which then leads to a doubling of period to 4 and so on.

The plot is as follows:

###### Matlab Code

Matlab code used to create the graph as outlined in Pg. 455 of King et. al

for r= 0:0.005:4
x=rand(1);
for j = 1:100
x=r*x*(1-x);
end
xout=[];
for j=1:400
x=r*x*(1-x); xout=[xout x];
end
plot (r*ones(size(xout)),xout,'.','MarkerSize',3)
axis([0 4 0 1]), hold on, pause(0.01)
end

##### Creating Figure 15.7: Sensitivity to Initial Conditions

The Following Plot explores the big effect of a small change in the initial conditions. One can note the changes as 'n' becomes larger than 50. The graph was produced for the following Conditions:

${\displaystyle r=4\qquad 100iterations\qquad IC1=0.1\qquad IC2=0.1+10^{-16}}$

The Following Equation was used to produce the iterations for the graph:

${\displaystyle x_{n+1}=rx_{n}(1-x_{n})}$

###### Matlab Code

Matlab Code used to produce the aforementioned plot.

function [xn1 xn2]=fig157

r=4;

format long

x01=.1;
x02=0.1+(10^-16);

for j=0:100
k=j+1;
x1=r*x01.*(1-x01);
x2=r*x02.*(1-x02);

xn1(k,1)=x1;
xn2(k,1)=x2;

x01=x1;
x02=x2;

end

n=0:100;
plot(n,xn1,'o')
hold on
plot(n,xn2,'x')


#### Author

--Egm6341.s10.Team2.GV 16:28, 23 April 2010 (UTC)

### problem 5: Variation of the plot with the increase in the time step size

#### Solution

Matlab code

clear;
clc;
xmax=10;
r=1.2;
t=0:0.1:10;
h=0.1; %time step size
syms x;
syms x11;
f=r*x*(1-(x/xmax));
%case 1: when x0=2
for k=1:1:5
h=2*h;
for i=1:1:100
x1(1)=2;
f1=subs(f,x,x1(i));
f11=subs(f,x,x11);
x1half=((1/2)*(x1+x11))+((h/8)*(f1-f11));
f1half=subs(f,x,x1half);
F=x1(i)-x11+(h/6)*(f1+f11+(4*f1half));
guessx11(i)=x1(i);
dummy=1;
while dummy==1
newx11(i)=guessx11(i)-((1/(subs(diff(F(i)),x11,guessx11(i))))*(subs(F(i),x11,guessx11(i))));
if double(abs(newx11(i)-guessx11(i)))<=10^(-6)
x1(i+1)=newx11(i);
dummy=0;
break;
else
guessx11(i)=newx11(i);
dummy=1;
end
end
end
plot(t,x1);
hold on
end


THE PLOT SHOWING THE POPULATION CHANGE W.R.T TIME, HERE VERHULST MODEL IS USED TO COMPUTE THE POPULATION WHERE XO=INITIAL POPULATION=2

 color scheme in the plot below Color h-value Red h=0.2 Blue h=0.4 Green h=0.8 Magenta h=1.6 Brown h=3.2

### Problem6: Solving Ordinary Differential Equations using Euler Implicit Methods

#### Statement

Solve the following Group of ODE's as they relate to the problem mentioned in the lecture pages mentioned above. The problem is of the motion of an airplane through different maneuvers.

for a vector with the following parameters:

${\displaystyle z=[x,y,V,\gamma ]}$
${\displaystyle {\dot {z}}=[{\dot {x}},{\dot {y}},{\dot {V}},{\dot {gamma}}]}$

${\displaystyle {\dot {x}}=Vcos(\gamma )}$

${\displaystyle {\dot {y}}=Vsin(\gamma )}$

${\displaystyle {\dot {v}}=({\frac {T-D}{m}}cos(\alpha )+{\frac {L}{m}}sin(\alpha )-gsin(\gamma )}$

${\displaystyle {\dot {\gamma }}=({\frac {T-D}{mV}}sin(\alpha )+{\frac {L}{mV}}cos(\alpha )-{\frac {g}{V}}cos(\gamma )}$

Initial Conditions are as follows:

${\displaystyle \gamma _{0}=0\quad degrees}$

${\displaystyle V_{0}=272\quad meters/sec}$

${\displaystyle x_{0}=0\quad meters}$

${\displaystyle y_{0}=0\quad meters}$

The Physical Modeling Parameters to be used are

${\displaystyle m=1005\quad Kg}$
${\displaystyle g=9.81\quad m/sec^{2}}$
${\displaystyle Sref=0.3376\quad m^{2}}$
${\displaystyle A1=-1.9431\qquad A2=-0.1499\qquad A3=0.2359}$
${\displaystyle B1=21.9\qquad B2=0}$
${\displaystyle C1=3.312x10^{-9}\quad Kgm^{-5}}$
${\displaystyle C2=-1.142x10^{-4}\quad Kgm^{-4}}$
${\displaystyle C3=1.1224\quad Kgm^{-3}}$

${\displaystyle \rho =C1y^{2}+C2y+C3}$
${\displaystyle L={\frac {1}{2}}C_{l}\rho V^{2}Sref}$
${\displaystyle C_{l}=B1\alpha +B2}$
${\displaystyle D={\frac {1}{2}}C_{d}\rho V^{2}Sref}$
${\displaystyle C_{d}=A1\alpha ^{2}+A2\alpha +A3}$

The Inputs are defined as follows:

Input T:

for ${\displaystyle 0\leq t<27seconds=6000N}$
for ${\displaystyle 27\leq t<33seconds=1000N}$
for ${\displaystyle 33\leq t\leq 40seconds=6000N}$

Input ${\displaystyle \alpha }$:

for ${\displaystyle 0\leq t<21seconds=0.03}$
for ${\displaystyle 21\leq t<27seconds=0.09to0.13}$ as it varies linearly
for ${\displaystyle 27\leq t<33seconds=-0.13to-0.2}$ as it varies linearly
for ${\displaystyle 33\leq t\leq 40seconds=-0.2to-0.13}$ as it varies linearly

#### Solution

##### Defining the Inputs

The first part of the solution is to create a function that will calculate the thrust and angle of a attack at a particular time. The following Matlab Codes were used:

Thrust:

% Function that calculates the thrust for any given time, t
function T = thrust(t)

%for t=0:40
%   k=t+1;
if (t >= 0) & (t < 27),
T = 6000;
elseif (t >= 27) & (t < 33),
T = 1000;
else
T = 6000;
end;
%end


Angle of Attack:

% Function calculates the angle of attack from 0-40secs
function alpha = angleatak(t)

%for t=0:40
%   k=t+1;
if (t >= 0) & (t < 21),
alpha = 0.03;
elseif (t >= 21) & (t < 27),
alpha = ((0.13-0.09)/(27-21))*(t-21) + 0.09;
elseif (t >= 27) & (t < 33),
alpha = ((-0.2 + 0.13)/(33-27))*(t-27) - 0.13;
else
alpha = ((-0.13 +0.2)/(40-33))*(t-33) - 0.2;
end;
%end

##### Technique used to solve the problem

The overall goal is to find a solution for each of the states being considered (x,y,V,gamma).

From P 36.3, of the lectures, the following was established using the Hermite-Simpson Algorithm:

    ${\displaystyle Z_{i+1}=z_{i}+{\frac {h/2}{3}}\left[f_{i}+4f(g(z_{i},z_{i+1}))+f(z_{i+1})\right]\qquad \qquad EQUATION1}$


The Following is recognized and derived as:

    ${\displaystyle g(z_{i},z_{i+1})={\frac {1}{2}}(z_{i}+z_{i+1})+{\frac {h}{8}}\left[(f(z_{i})-f(z_{i+1})\right]\qquad \qquad EQUATION2}$


It is now necessary to define the following using the previously mentioned Eq. 1:

    ${\displaystyle F(Z_{i+1})=0=Z_{i+1}-z_{i}-{\frac {h/2}{3}}\left[f_{i}+4f(g(z_{i},z_{i+1}))+f(z_{i+1})\right]\qquad \qquad EQUATION3}$


The next step is to apply the Newton Raphson Method to solve the function ${\displaystyle F(Z_{i+1})=0}$

A Matlab Algorithm is used to apply the Newton Raphson Method as follows:

##### Newton Raphson Method

Function for obtaining ${\displaystyle f(z)}$

function f = fcalc(t,z)

% z(1) = x : horizontal position
% z(2) = y : vertical position(height)
% z(3) = v : velocity
% z(4) = gamma : angle to new x-axis

% Set parameters
m = 1005; % kg
g = 9.81; % m/s^2
Sr = 0.3376; % m^2
A1 = -1.9431;
A2 = -0.1499;
A3 = 0.2359;
B1 = 21.9;
B2 = 0;
C1 = 3.312e-9; % kg/m^5
C2 = -1.142e-4; % kg/m^4
C3 = 1.224; % kg/m^3

alpha = angleatak(t); % Call alpha
T = thrust(t); % Call thrust

Cd = A1.*alpha.^2 + A2.*alpha + A3;
Cl = B1.*alpha + B2;
rho = C1*z(4)^2 +C2*z(4) + C3; % density

D = 0.5*Cd*rho*z(3)^2*Sr; % Calculate D
L = 0.5*Cl*rho*z(3)^2*Sr; % Calculate L

f = zeros(4,1);
f(1) = z(2)*cos(z(4));
f(2) = z(2)*sin(z(4));
f(3) = (T-D)*cos(alpha)/m - L*sin(alpha)/m -g*sin(z(4));
f(4) = (T-D)*sin(alpha)/(m*z(3)) + L*cos(alpha)/(m*z(3))-(g*cos(z(4)))/z(3);


Function for obtaining ${\displaystyle df(z)}$

function df = dfcalc(t,z)

% z(1) = x : horizontal position
% z(2) = y : vertical position(height)
% z(3) = v : velocity
% z(4) = gamma : angle to new x-axis

% Set parameters
m = 1005; % kg
g = 9.81; % m/s^2
Sr = 0.3376; % m^2
A1 = -1.9431;
A2 = -0.1499;
A3 = 0.2359;
B1 = 21.9;
B2 = 0;
C1 = 3.312e-9; % kg/m^5
C2 = -1.142e-4; % kg/m^4
C3 = 1.224; % kg/m^3

alpha = angleatak(t); % Call alpha
T = thrust(t); % Call thrust

Cd = A1.*alpha.^2 + A2.*alpha + A3;
Cl = B1.*alpha + B2;
rho = C1*z(4)^2 +C2*z(4) + C3; % density

D = 0.5*Cd*rho*z(3)^2*Sr; % Calculate D
L = 0.5*Cl*rho*z(3)^2*Sr; % Calculate L

%Caculates DF
df=zeros(4,4);
df(1,1) = 0; %
df(1,2) = 0; %
df(1,3) = cos(z(4));
df(1,4) = -z(3)*sin(z(4));

df(2,1) = 0;
df(2,2) = 0;
df(2,3) = sin(z(4));
df(2,4) = z(2)*cos(z(1));

df(3,1) = 0;
df(3,2) = -Cd*(2*C1*z(4)+C2)*z(2)^2*Sr*cos(alpha)/(2*m) - Cl*(2*C1*z(4)+C2)*z(2)^2*Sr*sin(alpha)/(2*m);
df(3,3) = -Cd*rho*z(3)*Sr*cos(alpha)/m - Cl*rho*z(3)*Sr*sin(alpha)/m;
df(3,4) = -g*cos(z(4));

df(4,1) = 0;
df(4,2) = -Cd*(2*C1*z(2)+C2)*z(3)*Sr*sin(alpha)/(2*m) + Cl*(2*C1*z(2)+C2)*z(3)*Sr*cos(alpha)/(2*m);
df(4,3) = -T*sin(alpha)/(m*z(3)^2) - Cd*rho*Sr*sin(alpha)/(2*m) + Cl*rho*Sr*cos(alpha)/(2*m) + g*cos(z(4))/z(3)^2;
df(4,4) = g*sin(z(4))/z(3);


Function for using the Newton Raphson Operations until an error threshold is reached

This Matlab code was developed using the published code of Team 3 as part of their HW 7 Report( HW 7: Team 3)

% This function returns the solution of EOMs given
% in Subchan and Zbikowski, Optim. Control Appl. Meth. 2007; 28:311-353.
% This function uses Hermite-Simpson (HS) algorithm and Newton-Raphson (NR) method.
% Note:
% z_out(time,1) = gamma : angle b/w aircraft body and x-axis for a given time
% z_out(time,2) = v : aircraft velocity for a given time
% z_out(time,3) = x : x-coordinate of aircraft for a given time
% z_out(time,4) = y : y-coordinate of aircraft for a given time
% t_out = times corresponding to nodes
function [t,z] = NRplane

i = 0;

% This "while" loop is for increasing # of nodes to achieve the given tolerance
while (true),
i = i + 1; % # of iterations
n = 2^(i-1); % # of subinterval

tinit = 0.0; % initial time (sec)
tfinal = 40.0; % final time (sec)

%initial conditions
z0 = zeros(4,1);
z0(1) = 0;
z0(2) = 30;
z0(3) = 272;
z0(4) = 0;
f0 = fcalc(tinit,z0); % Call f(z) for t = 0

h = (tfinal - tinit)/n; % time step

z_out = zeros(n+1,4); % output solution
t_out = zeros(n+1,1); % output time evolution

% assign the initial conditions to the solution for t = 0
z_out(1,1) = z0(1); z_out(1,2) = z0(2);
z_out(1,3) = z0(3); z_out(1,4) = z0(4);
t_out(1) = tinit;

% We need the values of previous time step in HS algorithm
zj_p = zeros(4,1); % a column vector
zj_p(1) = z0(1); zj_p(2) = z0(2);
zj_p(3) = z0(3); zj_p(4) = z0(4);
tj_p = tinit;

% This "for" loop is for time evolution.
% "j = 1" corresponds to t = tinit and "j = n+1" corresponds to t = tfinal
for j = 2:n+1,
tj = tinit + (j-1)*h; % current time
zj = zj_p; % initial guess for NR method
k = 0;
% This "while" loop is for NR method.
while (true),
k = k + 1;
fb_p = fcalc(tj_p,zj_p); % Call f(z) for t = t(j)
fb = fcalc(tj,zj); % Call f(z) for t = t(j+1)

tj_md = 0.5*(tj_p + tj); % t = t(i+1/2)
g = 0.5*(zj_p + zj) + (h/8)*(fb_p - fb); % Obtain z for t = t(j+1/2)
fb_md = fcalc(tj_md,g); % Call f(z) for t = t(j+1/2)

% Calculate F(z) for t = t(j+1)
ff = zj - zj_p - (h/6)*(fb_p + 4.0*fb_md + fb);

dfb = dfcalc(tj,zj); % Call df/dz for t = t(j+1)
dg = 0.5*eye(4,4) - (h/8)*dfb; % Calculate dg/dz for t = t(j+1)
dfbdg = dfcalc(tj_md,g); % Call df/dz for t = t(j+1/2)

% Calculate dF/dz for t = t(j+1)
dff = eye(4,4) - (h/6)*(4.0*dfbdg*dg + dfb);
inv_dff = inv(dff); % obtain inverse of dF/dz
zj_f = zj - inv_dff*ff; % Calculte new solution z for t = t(j+1)

% Check tolerance
if (norm(abs(zj_f-zj),inf) < 1.e-6),
break;
end;

% Redo interation if the error is still larger than tolerance
zj = zj_f;
end;

% Assign the satisfied solution to the output matrices
t_out(j) = tj;
z_out(j,1) = zj_f(1); z_out(j,2) = zj_f(2);
z_out(j,3) = zj_f(3); z_out(j,4) = zj_f(4);

% Current values become previous values for the next time step
tj_p = tj;
zj_p = zj_f;
end;

% check tolerance
if (i ~= 1) & (norm(abs(zj_f - zj_fp),inf) < 1.e-2),
break;
end;

zj_fp = zj_f;
end;

##### Results of Methods Used and comparison to ODE45

Plot of Horizontal Position Vs. Time

Plot of Vertical Position(Height) Vs. Time

Plot of Velocity Vs. Time

Plot of Gamma Vs. Time

From each of these plots it can be concluded that the combination of the Hermite Simpson Algorith and the Newton Raphson Method is an effective way to solve the set of ordinary differential equations without much error, as compared to ODE45 solver of Matlab. The downside to this method is that it takes a high amount of computational resources and time.

#### Author

--Egm6341.s10.Team2.GV 23:41, 27 April 2010 (UTC)

### problem7 :Integration of Logistic equation by

#### Statement

Integrate the Logistic equation using the inconsistent Trapezoidal-Simpson approximation ${\displaystyle x_{i+1/2}=x_{i}+x_{i+1}}$

#### Solution

The Matlab code is a slight alteration of the code used to integrate above.

  %Matlab code to integrate the the Verhulst population growth model
clear;
clc;
xmax=10;
r=1.2;
t=0:0.1:10;
h=0.1; %time step size
syms x;
syms x11;
f=r*x*(1-(x/xmax));
%case 1: when x0=2
for i=1:1:100
x1(1)=2;
f1=subs(f,x,x1(i));
f11=subs(f,x,x11);
x1half=((1/2)*(x1+x11))%Trapezoidal_simpson approxiamtion;
f1half=subs(f,x,x1half);
F=x1(i)-x11+(h/6)*(f1+f11+(4*f1half));
guessx11(i)=x1(i);
dummy=1;
while dummy==1
newx11(i)=guessx11(i)-((1/(subs(diff(F(i)),x11,guessx11(i))))*(subs(F(i),x11,guessx11(i))));
if double(abs(newx11(i)-guessx11(i)))<=10^(-6)
x1(i+1)=newx11(i);
dummy=0;
break;
else
guessx11(i)=newx11(i);
dummy=1;
end
end
end
plot(t,x1);


for ${\displaystyle x_{0}=2}$

for ${\displaystyle x_{0}=7}$

#### Author

--Egm6341.s10.team2.niki 20:33, 28 April 2010 (UTC)

### Problem9: Eliminating t from the Parameterization of an ellipse

#### Statement

Given the parameterization for an ellipse as follows for the x and y coordinates, Eliminate 't' as a parameter.

${\displaystyle x=a\cos(t)\qquad \qquad y=b\sin(t)}$

#### Solution

${\displaystyle x=a\cos(t)}$
${\displaystyle x^{2}=a^{2}\cos ^{2}(t)}$
${\displaystyle y=b\sin(t)}$
${\displaystyle y^{2}=b^{2}\sin ^{2}(t)}$

${\displaystyle {\frac {x^{2}}{a^{2}}}=\cos ^{2}(t)}$
${\displaystyle {\frac {y^{2}}{b^{2}}}=\sin ^{2}(t)}$

${\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=\cos ^{2}(t)+\sin ^{2}(t)=1}$
At this point t can be eliminated as a parameter.

#### Author

--Egm6341.s10.Team2.GV 17:51, 23 April 2010 (UTC)

### problem11: To show the integral for elliptical circumference

#### Statement

Pg.42-2 To show the integral for elliptical circumference

#### Author

Egm6341.s10.team2.lee 20:58, 23 April 2010 (UTC)

### problem 13: Change of Variable

#### Statement

To show that ${\displaystyle I=\int _{-1}^{+1}f(x)dx=\int _{0}^{\pi }f(cos\theta )sin\theta d\theta }$

#### Solution

We have the integral as

${\displaystyle I=\int _{-1}^{+1}f(x)dx}$.

Using the transformation ${\displaystyle \displaystyle {x=cos\theta }}$ we have

${\displaystyle \displaystyle {f(x)=f(cos\theta )}}$

${\displaystyle \displaystyle {dx=-sin\theta d\theta }}$

and using the inverse transformation the limits ${\displaystyle \displaystyle {[-1,1]}}$ can be written ${\displaystyle \displaystyle {[\pi ,0]}}$

thus we have

${\displaystyle I=\int _{-1}^{+1}f(x)dx=\int _{\pi }^{0}-f(cos\theta )sin\theta d\theta =\int _{0}^{\pi }f(cos\theta )sin\theta d\theta }$

#### Author

--Egm6341.s10.team2.niki 18:47, 23 April 2010 (UTC)

### problem 14: Constants of the Cosine Series

#### Statement

We have the cosine series expressed as ${\displaystyle f(cos\theta )={\frac {a_{0}}{2}}+\sum _{k=1}^{\infty }a_{k}cos(k\theta )}$, we need to express teh constants ${\displaystyle a_{k}}$ as

${\displaystyle a_{k}={\frac {2}{\pi }}\int _{0}^{\pi }f(cos\theta )cos(k\theta )d\theta }$

#### Solution

We have the given expression for the cosine series as

${\displaystyle f(cos\theta )={\frac {a_{0}}{2}}+\sum _{k=1}^{\infty }a_{k}cos(k\theta )}$

multiplying both sides by ${\displaystyle cos(m\theta )}$ where k is not the same as m and integrating we get,

${\displaystyle \int _{0}^{2\pi }f(cos\theta )(cos(m\theta ))=\int _{0}^{2\pi }{\frac {a_{0}cos(m\theta )}{2}}+\int _{0}^{2\pi }\sum _{k=1}^{\infty }a_{k}cos(k\theta )cos(m\theta )}$

Using the property of orthogonality we know that ${\displaystyle \int _{0}^{2\pi }a_{k}cos(k\theta )cos(m\theta )}$ exists only when k = m i.e

${\displaystyle \int _{0}^{2\pi }f(cos\theta )(cos(k\theta ))d\theta ={\frac {a_{0}}{2}}\int _{0}^{2\pi }cos(m\theta )d\theta +a_{k}\int _{0}^{2\pi }cos^{2}(k\theta )d\theta }$

wkt,

${\displaystyle \int _{0}^{2\pi }cos(m\theta )d\theta =0}$ and

${\displaystyle \int _{0}^{2\pi }cos^{2}(k\theta )d\theta ={\frac {2\pi -0}{2}}=\pi }$

Thus we have by substituting and rearranging terms,

${\displaystyle a_{k}={\frac {2}{\pi }}\int _{0}^{\pi }f(cos\theta )cos(k\theta )d\theta }$

#### Author

--Egm6341.s10.team2.niki 14:43, 23 April 2010 (UTC)

### problem15: Expression for Clenshaw-Curtis Quadrature Rule

#### Statement

Pg.42-3 Simplification of Clenshaw-Curtis Quadrature Rule

## Signatures

solutions of problems 2,7,13,14 --Egm6341.s10.team2.niki 18:48, 23 April 2010 (UTC)

solutions of problems 4,6, 9 --Egm6341.s10.Team2.GV 20:58, 23 April 2010 (UTC)

solutions of problems 1,11,15 --Egm6341.s10.team2.lee 22:47, 23 April 2010 (UTC)

## Solution Authors and Reviewers

 HW Assignment # 7 Problem # Solution by Reviewed by Problem 1 JP NN Problem 2 NN SM Problem 3 SM GV Problem 4 GV JP Problem 5 SM PO Problem 6 GV JP Problem 7 NN & SM GV Problem 8 PO NN Problem 9 GV PO Problem 10 PO SM Problem 11 JP PO Problem 12 PO GV Problem 13 NN NN Problem 14 NN JP Problem 15 JP SM