1. (10 points) Use the bisection method with tol = 10−12 to find ALL ZEROS in the interval [0, 1] for the following function: f(x) = x3 − (a1 − exp(a2x))x2 + a3x− a4 with a1 = 1.42, a2 = −7.89, a3 =...

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There are 8 questions and I need to follow the process in the problem(to fill all the blank that required)


1. (10 points) Use the bisection method with tol = 10−12 to find ALL ZEROS in the interval [0, 1] for the following function: f(x) = x3 − (a1 − exp(a2x))x2 + a3x− a4 with a1 = 1.42, a2 = −7.89, a3 = 0.52, a4 = 0.047. Present numerical results in a table as follows: Here, [a, b] is the starting interval used in the bisection method to compute the zero. Put each zero found and the related computational data in one row. Use more rows if more than one zero are found. 2. (10 points) Consider the following nonlinear function: f(x) = sin ( a1 + (a2 − exp(a3x))x2 + a4x3 ) . (a) Use the secant method to compute the zeros of f in the interval [0, 1] with the following parameters: a_1 = 0.1; a_2 = -3.2; a_3 = -5; a_4 = -1; Use the following inputs for the secant method: tol = 10^(-12); nmax = 1000; Present numerical results in a table as follows: 1 zero found a b Iterations Used zero found Residual NOI Add more rows for multiple zeros found. Present your script for generating these numerical results. “NOI” in the table means the number of iteration actually used by the method. (b) Use Newton’s method to compute the zeros of f in the interval [0, 1] with the following parameters: a1 = 0.2; a2 = 4.5; a3 = -5; a4 = -1; Use the following inputs for the Newton’s method: tol = 10^(-12); nmax = 1000; Present numerical results in a table as follows: zero found Residual NOI Add more rows for multiple zeros found. Present your script for generating these numerical results. “NOI” in the table means the number of iteration actually used by the method. 3. (30 points) Consider the following nonlinear system for p,Q1, Q2 and Q3: Q1 +Q2 +Q3 = 0, p γ + c1sign(Q1)Q 2 1 + z1 = 0, p γ + c2sign(Q2)Q 2 2 + z2 = 0, p γ + c3sign(Q3)Q 2 3 + z3 = 0. and sign(x) is the so called sign function whose values are 1 or 0 or −1 depending whether x is positive, zero, or negative. We can put the given nonlinear system in the vector form f(x) = 0 by setting x =  x1 x2 x3 x4  =  Q1 Q2 Q3 p  (a) (10 points) Implement f(x) in a Matlab function whose interface is as follows: function f = fun_Qp(x, gam, c, z) and implement the Jacobian of f(x) in a Matlab function whose interface is as follows: 2 function J = fun_QpJ(x, gam, c, z) where gam is for the value of γ, and c, z are vectors for c = [c1, c2, c3] T , z = [z1, z2, z3] T . (b) (10 points) Find approximations to Q1, Q2, Q3, p by using the Broyden’s method with the initial matrix B0 specified in the following table to solve this nonlinear system with parameters: γ = 9791, z1 = −19, z2 = −7, z3 = 2, c1 = 3.721, c2 = 64.55, c3 = 21.273. Use the following inputs for the Broyden’s method: x0 = [1; 1; 1; 1*10^5]; tol = 10^(-8); nmax = 200; Present numerical results in the following table: B0 Identity Matrix approxJ_fdh approxJ_compl residual number of iter. Q2 p Present the scripts for generating the numerical results in the table above. Among the three choices for the initial matrix B0, which one is the best? Why? Is the approximate Jacobian by the complex variable method a good approximation to the exact Jacobian at the specified x0? Why? Why not? (c) (10 points) Find approximations to Q1, Q2, Q3, p by using Newton’s method to solve this nonlinear system with the following parameters: γ = 9790, z1 = −19, z2 = −7, z3 = 2, c1 = 3.73, c2 = 64.6, c3 = 21.2734. Use the following inputs for the Newton’s method: x0 = [1; 1; 1; 1*10^5]; tol = 10^(-8); nmax = 200; Present numerical results in the following table: Q1 Q3 number of iterations Are these approximation acceptable? Why or Why not? 3 4. (30 points) In designing the shape of a gravity-flow discharge chute that will minimize transit time of discharged granular particles, we need to solve the following system of nonlinear equations: sin(θn+1) vn+1 − sin(θn) vn = 0, n = 1, 2, · · · , 19, ∆y ∑20 i=1 tan(θi)−X = 0, X, ∆y, vn, n = 1, 2, · · · , 20 are parameters to be specified. (a) (10 points) Put this nonlinear system in vector form f(θ) = 0, in which θ = (θ1, θ2, · · · , θ20)T . Then, implement Matlab functions for this nonlinear function and its Jacobian such that their interfaces are as follows: function f = fun_gravity_flow(theta, v, X, Del_y) function J = fun_gravity_flow_J(theta, v, X, Del_y) Then, assume the parameters in this nonlinear system are such that g = 32.16 ft/s2, X = 1.9, ∆y = 0.19, vn = √ 2gn∆y, n = 1, 2, · · · , 20. Use your Matlab functions to compute f(θ(0)) and Jf (θ (0)) where θ(0) = (1, 2, 3, · · · , 20)T . Present numerical results in the following table: f11(θ (0)) f20(θ (0)) J(13, 13) J(13, 14) J(13, 16) J(20, 16) (b) (10 points) Use Newton’s method to solve this nonlinear system with the param- eters specified in Problem 4a. Use the following inputs for Newton’s method: theta0 = ones(20, 1); tol = 10^(-12); nmax = 200; Present numerical results in a table as follows: 4 θ3 θ13 θ18 Residual Number of iterations Also, preset the related Matlab script. (c) (10 points) Use Broyden’s method to this nonlinear system with the following parameters: Assume g = 32.16 ft/s2, X = 1.98,∆y = 0.251, vn = √ 2g(n+ 1/n)∆y, n = 1, 2, · · · , 20. Use the following inputs for the Broyden’s method: theta0 = ones(20, 1); tol = 10^(-12); nmax = 200; Justify the choice for the initial matrix B0 to be used in the Broyden’s method. Present numerical results in a table as follows: θ1 θ11 θ17 Residual Number of iterations Also, preset the related Matlab script. 5. (40 points) Consider the following nonlinear equations for ui, i = 1, 2, · · · , N : −2u1 + u2 + h2u21 − h2g1 = 0, ui−1 − 2ui + ui+1 + h2u2i − h2gi = 0, i = 2, 3, · · · , N − 1, uN−1 − 2uN + h2u2N − h2gN = 0, where h = 1 N + 1 , x1 = h, xi = xi−1 + h, i = 2, 3, · · · , N, gi = e −2xi ( sin(xi) ( e2xi + (e− exi)2 sin(xi) ) − 2exi+1 cos(xi) ) (a) (10 points) Let u = (u1, u2, · · · , uN)t be the unknown vector for the nonlinear system above. Implement the nonlinear function f(u) = 0 such that its interface is as follows: function f = fu(u, g, h) 5 Then implement the Jacobian for this nonlinear system such that its interface is as follows: function J = fu_J(u, g, h) Then, for N = 5, carry out computations to fill in the following table: f1(u (0)) f3(u (0)) J(2, 2) J(2, 3) J(2, 4) J̃(3, 3) J̃(3, 2) J̃(3, 1) where u(0) = [1, 1, 1, 1, 1]T , J is Jacobian of f , and J̃ is the approximate Jacobian by the complex variable method with EPS = 10^(-6). (b) (10 points) Solve this nonlinear system for N = 512 by Newton’s method. Present your numerical results by filling the following table: u15 u305 Residual Iterations used Use tol = 10−12, u(0) =  1 1 ... 1  Also, present your Maltab script used to generate data in this table. Make sure that your script uses the Matlab’s “varargin” functionality. (c) (10 points) Solve this nonlinear system for N = 1024 by Broyden’s method. Present your numerical results by filling the following table: u512 u750 Residual Iterations used 6 Use tol = 10−12, u(0) =  1 1 ... 1  Also, present your Maltab script used to generate data in this table. Make sure that your script uses the Matlab’s “varargin” functionality. (d) (10 points) Note that the solution in part 5b above provide data for a function u(x). Use this set of data to find approximations to u(x) and present these approximations in the following table: x u(x) π/8 π/4 Justify your choice of the method for computing these approximations and present the related script. 6. (40 points) Consider the following data for the trajectory in the x-y plane of a robot (from Problem 3.4 and Example 3.10): t 0 1 2 3 5 x 0 1 4 3 0 y 0 2 4 1 0 Let T1(t) = [ x1(t) y1(t) ] , t ∈ [0, 2], T2(t) = [ x2(t) y2(t) ] , t ∈ [2, 5] be the cubic spline interpolations with the natural boundary condition of the trajectory data in corresponding time intervals. Then download data file curve_intersects_RobotTraj.mat from Canvas for the trajectory of another moving object, and let T3(s) = [ x3(s) y3(s) ] , s ∈ [0, 5] be the cubic spline interpolation with the natural boundary condition for this set of data. In these functions, both t and s variables represent the time. 7 (a) (10 points) Make a plot of the trajectories by the three parametric curves defined above with red, blue, and green color, respectively. (b) (10 points) Find the x-y coordinates where T3(s) curve intersects with the T1(t) curve. Present your results in the following table x y t∗ s∗ where T1(t ∗) = [ x y ] = T3(s ∗). 8 (c) (10 points) Find the x-y coordinates where T3(s) curve intersects with the T2(t) curve. Present your results in the following table x y t∗ s∗ where T2(t ∗) = [ x y ] = T3(s ∗). (d) (5 points) Make a plot of the trajectories by the three parametric curves defined above with red, blue, and green color, respectively, together with the intersection points found above. Mark the two intersection points with red *. (e) (5 points) Discuss the concern about whether the moving object can collide with the robot. Why? Why not?
Answered Same DayNov 02, 2021

Answer To: 1. (10 points) Use the bisection method with tol = 10−12 to find ALL ZEROS in the interval [0, 1]...

Kshitij answered on Nov 05 2021
151 Votes
issue of science/5d.mat
x:[512x1 double array]
x1:[512x1 double array]
issue of science/Code_1.m
%% problem No.1
% Clear Screen
clc
clear all
disp('Bisection Method');
% Define constants
a1 = 1.42;
a2 = -7.89;
a3 = 0.52;
a4 = 0.047;
% Define Function
f = @(x) x^3 - (a1-exp(a2*x))*x^2 + a3*x - a4;
% starting interval
low = 0;
high = 1;
% tol
tol = 1e-12;
% Evaluate both ends of the interval
y1 = feval(f, low);
y2 = feval(f, high);
i = 0;
% Display error and finish if signs are not different
if y1 * y2 > 0
disp('Have not found a change in sign. Will not continue...');
return
end
while (abs(high - low) >= tol)
i = i + 1;
% Find a new v
alue to be tested as a root
m = (high + low)/2;
y3 = feval(f, m);
if y3 == 0
return
end

% Update the limits
if y1 * y3 > 0
low = m;
y1 = y3;
else
high = m;
end
end
% Show the last approximation considering the tolerance
w = feval(f, m);
fprintf('\n x = %f produces f(x) = %g \n %i iterations\n', m, y3, i-1);
fprintf(' Approximation with tolerance = %g \n\n\n', tol);
fprintf(' Zero Found | a | b | Iteration Used\n')
fprintf(' %0.4f | %d | %d | %d\n\n',m,0,1,i-1)
issue of science/Code_2_a.m
%% Problem No. 2 a
% clear screen
clc
clear all
%% Secant Method initilization
% Define constants
a_1 = 0.1;
a_2 = -3.2;
a_3 = -5;
a_4 = -1;
syms x;
% Define function
f=sin(a_1+(a_2 - exp(a_3*x))*x^2 + a_4*x^3);
% tolerance
epsilon = 1e-12;
%initial guess
x0 = 0;
x1 = 1;
% secant method
for i=1:100
f0=vpa(subs(f,x,x0));
f1=vpa(subs(f,x,x1));
y=x1-((x1-x0)/(f1-f0))*f1;
err=abs(y-x1);
if errbreak
end
x0=x1;
x1=y;
end
y = y - rem(y,1e-12);
% Display roots
fprintf('The Root is : %f \n',y);
fprintf('No. of Iterations : %d\n\n\n',i);
fprintf('Zero Found | Resudial | NOI\n')
fprintf(' %0.4f | %0.4e | %d\n\n',y,double(err),i)
issue of science/Code_2_b.m
%% Problem No. 2 b
% clear screen
clc
clear all
%% Newton's Method to find roots
% Define constants
a_1 = 0.1;
a_2 = -3.2;
a_3 = -5;
a_4 = -1;
syms x;
f=sin(a_1+(a_2 - exp(a_3*x))*x^2 + a_4*x^3); %Enter the Function here
g=diff(f); %The Derivative of the Function
n=11;
epsilon = 10^-(n+1)
x0 = -0.1;
for i=1:100
f0=vpa(subs(f,x,x0)); %Calculating the value of function at x0
f0_der=vpa(subs(g,x,x0)); %Calculating the value of function derivative at x0
y=x0-f0/f0_der; % The Formula
err=abs(y-x0);
if errbreak
end
x0=y;
end
y = y - rem(y,10^-n); %Displaying upto required decimal places
fprintf('The Root is : %f \n',y);
fprintf('No. of Iterations : %d\n',i);
issue of science/Code_3_b.m
%% problem 3. b
% clear screen
clc
clear all
%% Broyden's Method
x0=[1;1;1;1*10^5];
tol = 1e-8;
nmax = 200;
gam = 9791;
z=[-19 -7 2];
c=[3.721 64.55 21.273];
J =@(x) fun_QpJ(x,gam,c,z);
invJ =@(x) inv(fun_QpJ(x,gam,c,z));
F =@(x) fun_Qp(x,gam,c,z);
x1 = x0;
J1 = J(x0);
n=0;
delF=1;
while tol x2 = x1 - inv(J1) * F(x1);
delx = x2-x1;
delF = F(x2)-F(x1);
J2 = J1 + ((delF - J1*delx)./(abs(x2).^2))*delx';
x1=x2;
J1=J2;
n=n+1;
end
fprintf(' Resudial | NOI | Q2 | P\n')
fprintf(' %0.4e | %d | %0.4f | %0.3e\n\n',abs(mean(delF)),n,x1(2),x1(4))
issue of science/Code_3_c.m
%% Problem No. 3 c
% clear screen
clc
clear all
%% Newtons's Method to solve non linear equation
% initial guess
x0=[1;1;1;1*10^5];
% tol
tol = 1e-8;
% maximum iteration
nmax = 200;
% remaining constant
gam = 9791;
z=[-19 -7 2];
c=[3.721 64.55 21.273];
% anomly. function
J =@(x) fun_QpJ(x,gam,c,z);
invJ =@(x) inv(fun_QpJ(x,gam,c,z));
F =@(x) fun_Qp(x,gam,c,z);
x1 = x0;
n=0;
delF=1;
while tol x2 = x1 - inv(J(x1)) * F(x1);
x1=x2;
n=n+1;
end
fprintf(' Q1 | Q3 | NOI\n')
fprintf(' %0.4f | %0.4f | %d\n\n',x1(1),x1(3),n)
issue of science/Code_4_a.m
%% Problem No. 4 a
% clear screen
clc
clear all
%% using fuction fun_gravity_flow and fun_gravity_flow_J cmopute table
theta=ones(20,1);
g=32.16;
Del_y=0.19;
n=1:20;
v=sqrt(2*g.*n*Del_y);
v=v';
X=1.9;
% compute function and jacobian
J =fun_gravity_flow_J(theta, v, X, Del_y);
F =fun_gravity_flow(theta, v, X, Del_y);
% print Table
fprintf(' f11 | f20 | J(13,13) | J(13,14) | J(13,16) | J(20,16)\n')
fprintf(' %0.4f | %0.4f | %0.4f | %0.4f | %0.4f | %0.4f \n\n',F(11),F(20),J(13,13),J(13,14),J(13,16),J(20,16))
issue of science/Code_4_b.m
%% Problem No. 4 b
% clear screen
clc
clear all
%% Newtons's Method to solve non-linear equations
theta0=ones(20,1);
g=32.16;
Del_y=0.19;
n=1:20;
v=sqrt(2*g.*n*Del_y);
v=v';
X=1.9;
tol = 1e-12;
nmax = 200;
J =@(theta) fun_gravity_flow_J(theta, v, X, Del_y);
invJ =@(theta) inv(fun_gravity_flow_J(theta, v, X, Del_y));
F =@(theta) fun_gravity_flow(theta, v, X, Del_y);
x1 = theta0;
n=0;
delF=1;
while tol x2 = x1 - inv(J(x1)) * F(x1);
x1=x2;
n=n+1;
end
% Print Results
fprintf(' theta3 | theta13 | theta18 | Resudial | NOI\n')
fprintf(' %0.4f | %0.4f | %0.4f | %0.3e | %d\n\n',x1(3),x1(13),x1(18),abs(mean(F(x1))),n)
issue of science/Code_4_c.m
%% Problem No. 4 c
clc
clear all
%% Broyden's Method
theta0=ones(20,1);
g=32.16;
Del_y=0.251;
n=1:20;
v=sqrt(2*g.*((n+1)./n).*Del_y);
v=v';
X=1.98;
tol = 1e-12;
nmax = 200;
J =@(theta) fun_gravity_flow_J(theta, v, X, Del_y);
invJ =@(theta) inv(fun_gravity_flow_J(theta, v, X, Del_y));
F =@(theta) fun_gravity_flow(theta, v, X, Del_y);
x1 = theta0;
J1 = J(theta0);
n=0;
delF=1;
while tol x2 = x1 - inv(J1) * F(x1);
delx = x2-x1;
delF = F(x2)-F(x1);
J2 = J1 + ((delF - J1*delx)./(abs(x2).^2))*delx';
x1=x2;
J1=J2;
n=n+1;
end
% print Results
fprintf(' theta1 | theta11 | theta17 | Resudial | NOI\n')
fprintf(' %0.4f | %0.4f | %0.4f | %0.3e | %d\n\n',x1(1),x1(11),x1(17),abs(mean(F(x1))),n)
issue of science/Code_5_a.m
%% problem 5. a
clc
clear all
N = 5;
h=1/(1+N);
x(1)=h;
for i=2:N
x(i)=x(i-1)+h;
end
for i=1:N
g(i) = exp(-2*x(i))*(sin(x(i))*(exp(2*x(i))+(exp(1)-exp(x(i)))^2*sin(x(i)))-2*exp(x(i)+1)*cos(x(i)));
end
u=[1 1 1 1 1]';
g=g';
f=fu(u,g,h);
J=fu_J(u,g,h);
% print result
fprintf(' f1 | f3 | J(2,2) | J(2,3) | J(2,4) | J(3,3) | J(3,2) | J(3,1)\n')
fprintf(' %0.4f | %0.4f | %0.4f | %0.4f | %0.4f | %0.4f | %0.4f | %0.4f\n\n',f(1),f(3),J(2,2),J(2,3),J(2,4),J(3,3),J(3,2),J(3,1))
issue of science/Code_5_b.m
%% Problem 5 b
clc
clear all
%% Newtons's Method
N = 512;
h=1/(1+N);
x(1)=h;
for i=2:N
x(i)=x(i-1)+h;
end
for i=1:N
g(i) =...
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