1. Write an M file spline.m such that spline(m(k),m(k+1),p(k), q(k),x0(k), x0(k+1) , x ) computesSk(x) for x in the interval [x0(k) ,x0(k+1)]2. Write a matlab program in an M filespline_nat.mthat...

math4446/dspline.m
function Skp = dspline(mk,mkp1,pk,qk,xk,xkp1,x)
hk = xkp1-xk;
Skp = -mk./(2.*hk).*(x-xkp1).^2 + ...
mkp1./(2.*hk).*(x-xk).^2 - pk + qk;
end
math4446/gesolve.m
function x = gesolve(a,b,c,f)
d(1) = a(1);
z(1) = f(1);
n = length(f);
for i=2:n
d(i) = a(i) - b(i-1)*c(i-1)/d(i-1);
z(i) = f(i) - b(i-1)*z(i-1)/d(i-1);
end
%
%solve Ux =z or Backward substitution step
%
x(n) = z(n)/d(n);
for i=n-1:-1:1
x(i) = (z(i) - c(i)*x(i+1))/d(i);
end
x = x';
end
math4446/headbot.m
headb=[
-1,0
-0.95,-0.31225
-0.9,-0.43589
-0.85,-0.52678
-0.8,-0.6
-0.75,-0.66144
-0.7,-0.71414
-0.65,-0.75993
-0.6,-0.8
-0.55,-0.83516
-0.5,-0.86603
-0.45,-0.89303
-0.4,-0.91652
-0.35,-0.93675
-0.3,-0.95394
-0.25,-0.96825
-0.2,-0.9798
-0.15,-0.98869
-0.1,-0.99499
-0.05,-0.99875
0,-1
0.05,-0.99875
0.1,-0.99499
0.15,-0.98869
0.2,-0.9798
0.25,-0.96825
0.3,-0.95394
0.35,-0.93675
0.4,-0.91652
0.45,-0.89303
0.5,-0.86603
0.55,-0.83516
0.6,-0.8
0.65,-0.75993
0.7,-0.71414
0.75,-0.66144
0.8,-0.6
0.85,-0.52678
0.9,-0.43589
0.95,-0.31225
1,0 ];
math4446/headtop.m
headt=[
-1,0
-0.95,0.31225
-0.9,0.43589
-0.85,0.52678
-0.8,0.6
-0.75,0.66144
-0.7,0.71414
-0.65,0.75993
-0.6,0.8
-0.55,0.83516
-0.5,0.86603
-0.45,0.89303
-0.4,0.91652
-0.35,0.93675
-0.3,0.95394
-0.25,0.96825
-0.2,0.9798
-0.15,0.98869
-0.1,0.99499
-0.05,0.99875
0,1
0.05,0.99875
0.1,0.99499
0.15,0.98869
0.2,0.9798
0.25,0.96825
0.3,0.95394
0.35,0.93675
0.4,0.91652
0.45,0.89303
0.5,0.86603
0.55,0.83516
0.6,0.8
0.65,0.75993
0.7,0.71414
0.75,0.66144
0.8,0.6
0.85,0.52678
0.9,0.43589
0.95,0.31225
1,0];
math4446/leyebot.m
leyeb=[
-0.7,0.5
-0.69,0.46878
-0.68,0.45641
-0.67,0.44732
-0.66,0.44
-0.65,0.43386
-0.64,0.42859
-0.63,0.42401
-0.62,0.42
-0.61,0.41648
-0.6,0.4134
-0.59,0.4107
-0.58,0.40835
-0.57,0.40633
-0.56,0.40461
-0.55,0.40318
-0.54,0.40202
-0.53,0.40113
-0.52,0.4005
-0.51,0.40013
-0.5,0.4
-0.49,0.40013
-0.48,0.4005
-0.47,0.40113
-0.46,0.40202
-0.45,0.40318
-0.44,0.40461
-0.43,0.40633
-0.42,0.40835
-0.41,0.4107
-0.4,0.4134
-0.39,0.41648
-0.38,0.42
-0.37,0.42401
-0.36,0.42859
-0.35,0.43386
-0.34,0.44
-0.33,0.44732
-0.32,0.45641
-0.31,0.46878
-0.3,0.5];
math4446/leyetop.m
leyet=[
-0.7,0.5
-0.69,0.53122
-0.68,0.54359
-0.67,0.55268
-0.66,0.56
-0.65,0.56614
-0.64,0.57141
-0.63,0.57599
-0.62,0.58
-0.61,0.58352
-0.6,0.5866
-0.59,0.5893
-0.58,0.59165
-0.57,0.59367
-0.56,0.59539
-0.55,0.59682
-0.54,0.59798
-0.53,0.59887
-0.52,0.5995
-0.51,0.59987
-0.5,0.6
-0.49,0.59987
-0.48,0.5995
-0.47,0.59887
-0.46,0.59798
-0.45,0.59682
-0.44,0.59539
-0.43,0.59367
-0.42,0.59165
-0.41,0.5893
-0.4,0.5866
-0.39,0.58352
-0.38,0.58
-0.37,0.57599
-0.36,0.57141
-0.35,0.56614
-0.34,0.56
-0.33,0.55268
-0.32,0.54359
-0.31,0.53122
-0.3,0.5 ];
math4446/mouth.m
mouthd=[
-0.5,-0.2
-0.475,-0.27806
-0.45,-0.30897
-0.425,-0.3317
-0.4,-0.35
-0.375,-0.36536
-0.35,-0.37854
-0.325,-0.38998
-0.3,-0.4
-0.275,-0.40879
-0.25,-0.41651
-0.225,-0.42326
-0.2,-0.42913
-0.175,-0.43419
-0.15,-0.43848
-0.125,-0.44206
-0.1,-0.44495
-0.075,-0.44717
-0.05,-0.44875
-0.025,-0.44969
0,-0.45
0.025,-0.44969
0.05,-0.44875
0.075,-0.44717
0.1,-0.44495
0.125,-0.44206
0.15,-0.43848
0.175,-0.43419
0.2,-0.42913
0.225,-0.42326
0.25,-0.41651
0.275,-0.40879
0.3,-0.4
0.325,-0.38998
0.35,-0.37854
0.375,-0.36536
0.4,-0.35
0.425,-0.3317
0.45,-0.30897
0.475,-0.27806
0.5,-0.2];
math4446/nose.m
nosed=[
-0.2,0.44
-0.19,0.3776
-0.18,0.3184
-0.17,0.2624
-0.16,0.2096
-0.15,0.16
-0.14,0.1136
-0.13,0.0704
-0.12,0.0304
-0.11,-0.0064
-0.1,-0.04
-0.09,-0.0704
-0.08,-0.0976
-0.07,-0.1216
-0.06,-0.1424
-0.05,-0.16
-0.04,-0.1744
-0.03,-0.1856
-0.02,-0.1936
-0.01,-0.1984
0,-0.2
0.01,-0.1984
0.02,-0.1936
0.03,-0.1856
0.04,-0.1744
0.05,-0.16
0.06,-0.1424
0.07,-0.1216
0.08,-0.0976
0.09,-0.0704
0.1,-0.04
0.11,-0.0064
0.12,0.0304
0.13,0.0704
0.14,0.1136
0.15,0.16
0.16,0.2096
0.17,0.2624
0.18,0.3184
0.19,0.3776
0.2,0.44];
math4446/project2.docx
Matt 4446 Project 2
Clearing the workspace
clear;
close;
clc;
Seleting the format for results
format long e
Problem I
Testing the program
% Defining the given parameters
A = [
5 2 0 0
4 13 -4 0
0 7 15 6
0 0 12 -16
];
f = [
1
2
3
4
];
Getting the solution using MATLAB command
x = A\f
Now getting the solution using the Gaussian elimination and forward algorithm defined under gesolve
n = length(f);
a = zeros(n, 1);
= zeros(n-1, 1);
c = zeros(n-1, 1);
for i=1:n
a(i) = A(i, i);
if i ~= n
b(i) = A(i+1, i);
c(i) = A(i, i+1);
end
end
x = gesolve(a, b, c, f)
From the results above, we can see that the results are exactly the same from the algorithm to the results from MATLAB command.
Problem II
Testing the program
% Defining the given variables
x = [ -2,-1,0,1,2];
y = [ 2,1,0,1,2];
m =[ 0 , -6/7 , 24/7, -6/7, 0];
p = [ 2,8/7, -4/7, 8/7];
q =[ 8/7, -4/7 , 8/7, 2 ];
figure;
spline_nat(x, y)
Running the MATLAB smile program to plot the natural spline approximation of the smiling face.
figure;
smile
Problem III
Testing the code for a cubic function
x0 = [-1 2 5 8];
n = length(x0);
func = @(x) x.^3;
funcprime = @(x) 3*(x.^2);
y0 = func(x0);
fp = funcprime(x0);
fp0 = fp(1);
fp1 = fp(n);
spline_clam(x0,y0,fp0,fp1,func,funcprime)
From the plot above, as expected we can see that the e
or is almost equal to zero.
Now testing the code for
i = 0:10;
x0 = i*0.2;
n = length(x0);
func = @(x) x.*exp(x);
funcprime = @(x) exp(x) + x.*exp(x);
y0 = func(x0);
fp = funcprime(x0);
fp0 = fp(1);
fp1 = fp(n);
spline_clam(x0,y0,fp0,fp1,func,funcprime)
Yes, the average e
or tends to be about zero, which is very small, but the round-off e
or is not exactly equal to 0. This is still quite a good approximation, although it is not as reliable as in the cubic function.
Appendix
function x = gesolve(a,b,c,f)
d(1) = a(1);
z(1) = f(1);
n = length(f);
for i=2:n
d(i) = a(i) - b(i-1)*c(i-1)/d(i-1);
z(i) = f(i) - b(i-1)*z(i-1)/d(i-1);
end
%
%solve Ux =z or Backward substitution step
%
x(n) = z(n)/d(n);
for i=n-1:-1:1
x(i) = (z(i) - c(i)*x(i+1))/d(i);
end
x = x';
end
function Sk = spline(mk,mkp1,pk,qk,xk,xkp1,x)
hk = xkp1 - xk;
Sk = - mk./(6.*hk).*(x - xkp1).^3 + ...
mkp1./(6.*hk).*(x - xk).^3 + ...
pk.*(xkp1 - x) + ...
qk.*(x - xk);
end
function spline_nat(x0,y0)
n = length(x0);
x0=reshape(x0,1,n);
y0=reshape(y0,1,n);
h = x0(2:n) - x0(1:(n-1));
d = (y0(2:n) - y0(1:(n-1)))./h;
a = 2*(h(1:(n-2)) + h(2:(n-1)));
b = h(2:(n-2));
u = 6*(d(2:n-1) - d(1:n-2));
m = gesolve(a,b,b,u);
m = [0 m' 0];
p = y0(1:(n-1))./h - m(1:(n-1)).*h/6;
q = y0(2:n)./h - m(2:n).*h/6;
hold on
plot(x0, y0, '*');
for i = 1:(n-1)
x=linspace(x0(i),x0(i+1),1000);
Sk = spline(m(i),m(i+1),p(i),q(i),x0(i),x0(i+1),x);
plot(x,Sk,'linewidth',1.5);
end
xlabel('x');
ylabel('f(x)');
title('Natural spline for f(x)');
end
function Skp = dspline(mk,mkp1,pk,qk,xk,xkp1,x)
hk = xkp1-xk;
Skp = -mk./(2.*hk).*(x-xkp1).^2 + ...
mkp1./(2.*hk).*(x-xk).^2 - pk + qk;
end
function spline_clam(x0,y0,fp0,fp1,func,funcprime)
n = length(x0);
x0=reshape(x0,1,n);
y0=reshape(y0,1,n);
h = x0(2:n) - x0(1:(n-1));
d = (y0(2:n) - y0(1:(n-1)))./h;
a = 2*(h(1:(n-2)) + h(2:(n-1)));
a(1) = 3*h(1)/2 + 2*h(2);
a(n-2) = 3/2*h(n-1) + 2*h(n-2);
b = h(2:(n-2));
f=6*(d(2:n-1) - d(1:n-2));
f(1) = f(1) - 3*(d(1) - fp0);
f(n-2) = f(n-2) - 3*(fp1 - d(end));
m = gesolve(a,b,b,f);
m0 = 3*(d(1) - fp0)/h(1) - m(1)/2;
mn = 3*(fp1 - d(end))/h(end) - m(end)/2;
m =[m0 m' mn];
p = y0(1:(n-1))./h - m(1:(n-1)).*h/6;
q = y0(2:n)./h - m(2:n).*h/6;
figure(1);
clf
hold on
for i = 1:(n-1)
x=linspace(x0(i),x0(i+1),100);
Sk = spline(m(i),m(i+1),p(i),q(i),x0(i),x0(i+1),x);
F = func(x);
plot(x,Sk,'linewidth',1.5)
plot(x,F,'linewidth',1.5)
end
xlabel('x')
ylabel('f(x)')
title('Clamped cubic spline and f(x)')
pause
figure(2);
clf
clear F
hold on
for i = 1:(n-1)
x=linspace(x0(i),x0(i+1),100);
Skp = dspline(m(i),m(i+1),p(i),q(i),x0(i),x0(i+1),x);
F = funcprime(x);
plot(x,Skp,'linewidth',1.5)
plot(x,F,'linewidth',1.5)
end
xlabel('x')
ylabel('g(x)')
title('Derivative of clamped cubic spline, g(x)')
pause;
figure(3);
clf;
clear F;
hold on;
for i = 1:(n-1)
x=linspace(x0(i),x0(i+1),100);
Sk = spline(m(i),m(i+1),p(i),q(i),x0(i),x0(i+1),x);
F = func(x);
e = Sk - F;
plot(x,e,'linewidth',1.5)
end
xlabel('x')
ylabel('f(x)')
title('E
or of the clamped cubic spline for f(x)')
end
math4446/project2.mlx
[Content_Types].xml

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matla
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matla
document.xml
Matt 4446 Project 2 Clearing the workspace clear;
close;
clc; Seleting the format for results format long e Problem I Testing the program % Defining the given parameters
A = [
5 2 0 0
4 13 -4 0
0 7 15 6
0 0 12 -16
];
f = [
1
2
3
4
]; Getting the solution using MATLAB command A\backslash f x = A\f Now getting the solution using the Gaussian elimination and forward algorithm defined under gesolve n = length(f);
a = zeros(n, 1);
= zeros(n-1, 1);
c = zeros(n-1, 1);
for i=1:n
a(i) = A(i, i);
if i ~= n
b(i) = A(i+1, i);
c(i) = A(i, i+1);
end
end
x = gesolve(a, b, c, f) From the results above, we can see that the results are exactly the same from the algorithm to the results from MATLAB command. Problem II Testing the program % Defining the given variables
x = [ -2,-1,0,1,2];
y = [ 2,1,0,1,2];
m =[ 0 , -6/7 , 24/7, -6/7, 0];
p = [ 2,8/7, -4/7, 8/7];
q =[ 8/7, -4/7 , 8/7, 2 ];
figure;
spline_nat(x, y) Running the MATLAB smile program to plot the natural spline approximation of the smiling face. figure;
smile Problem III Testing the code for a cubic function x0 = [-1 2 5 8];
n = length(x0);
func = @(x) x.^3;
funcprime = @(x) 3*(x.^2);
y0 = func(x0);
fp = funcprime(x0);
fp0 = fp(1);
fp1 = fp(n);
spline_clam(x0,y0,fp0,fp1,func,funcprime) From the plot above, as expected we can see that the e
or is almost equal to zero. Now testing the code for f\left(x\right)=x\times \mathrm{exp}\left(x\right) i = 0:10;
x0 = i*0.2;
n = length(x0);
func = @(x) x.*exp(x);
funcprime = @(x) exp(x) + x.*exp(x);
y0 = func(x0);
fp = funcprime(x0);
fp0 = fp(1);
fp1 = fp(n);
spline_clam(x0,y0,fp0,fp1,func,funcprime) Yes, the average e
or tends to be about zero, which is very small, but the round-off e
or is not exactly equal to 0. This is still quite a good approximation, although it is not as reliable as in the cubic function. Appendix function x = gesolve(a,b,c,f)
d(1) = a(1);
z(1) = f(1);
n = length(f);
for i=2:n
d(i) = a(i) - b(i-1)*c(i-1)/d(i-1);
z(i) = f(i) - b(i-1)*z(i-1)/d(i-1);
end
%
%solve Ux =z or Backward substitution step
%
x(n) = z(n)/d(n);
for i=n-1:-1:1
x(i) = (z(i) - c(i)*x(i+1))/d(i);
end
x = x';
end
function Sk = spline(mk,mkp1,pk,qk,xk,xkp1,x)
hk = xkp1 - xk;
Sk = - mk./(6.*hk).*(x - xkp1).^3 + ...
mkp1./(6.*hk).*(x - xk).^3 + ...
pk.*(xkp1 - x) + ...
qk.*(x - xk);
end
function spline_nat(x0,y0)
n = length(x0);
x0=reshape(x0,1,n);
y0=reshape(y0,1,n);
h = x0(2:n) - x0(1:(n-1));
d = (y0(2:n) - y0(1:(n-1)))./h;
a = 2*(h(1:(n-2)) + h(2:(n-1)));
b = h(2:(n-2));
u = 6*(d(2:n-1) - d(1:n-2));
m = gesolve(a,b,b,u);
m = [0 m' 0];
p = y0(1:(n-1))./h - m(1:(n-1)).*h/6;
q = y0(2:n)./h - m(2:n).*h/6;
hold on
plot(x0, y0, '*');
for i = 1:(n-1)
x=linspace(x0(i),x0(i+1),1000);
Sk = spline(m(i),m(i+1),p(i),q(i),x0(i),x0(i+1),x);
plot(x,Sk,'linewidth',1.5);
end
xlabel('x');
ylabel('f(x)');
title('Natural spline for f(x)');
end
function Skp = dspline(mk,mkp1,pk,qk,xk,xkp1,x)
hk = xkp1-xk;
Skp = -mk./(2.*hk).*(x-xkp1).^2 + ...
mkp1./(2.*hk).*(x-xk).^2 - pk + qk;
end
function spline_clam(x0,y0,fp0,fp1,func,funcprime)
n = length(x0);
x0=reshape(x0,1,n);
y0=reshape(y0,1,n);
h = x0(2:n) - x0(1:(n-1));
d = (y0(2:n) - y0(1:(n-1)))./h;
a = 2*(h(1:(n-2)) + h(2:(n-1)));
a(1) = 3*h(1)/2 + 2*h(2);
a(n-2) = 3/2*h(n-1) + 2*h(n-2);
b = h(2:(n-2));
f=6*(d(2:n-1) - d(1:n-2));
f(1) = f(1) - 3*(d(1) - fp0);
f(n-2) = f(n-2) - 3*(fp1 - d(end));
m = gesolve(a,b,b,f);
m0 = 3*(d(1) - fp0)/h(1) - m(1)/2;
mn = 3*(fp1 - d(end))/h(end) - m(end)/2;
m =[m0 m' mn];
p = y0(1:(n-1))./h - m(1:(n-1)).*h/6;
q = y0(2:n)./h - m(2:n).*h/6;
figure(1);
clf
hold on
for i = 1:(n-1)
x=linspace(x0(i),x0(i+1),100);
Sk = spline(m(i),m(i+1),p(i),q(i),x0(i),x0(i+1),x);
F = func(x);
plot(x,Sk,'linewidth',1.5)
plot(x,F,'linewidth',1.5)
end
xlabel('x')
ylabel('f(x)')
title('Clamped cubic spline and f(x)')
pause
figure(2);
clf
clear F
hold on
for i = 1:(n-1)
x=linspace(x0(i),x0(i+1),100);
Skp = dspline(m(i),m(i+1),p(i),q(i),x0(i),x0(i+1),x);
F = funcprime(x);
plot(x,Skp,'linewidth',1.5)
plot(x,F,'linewidth',1.5)
end
xlabel('x')
ylabel('g(x)')
title('Derivative of clamped cubic spline, g(x)')
pause;
figure(3);
clf;
clear F;
hold on;
for i = 1:(n-1)
x=linspace(x0(i),x0(i+1),100);
Sk = spline(m(i),m(i+1),p(i),q(i),x0(i),x0(i+1),x);
F = func(x);
e = Sk - F;
plot(x,e,'linewidth',1.5)
end
xlabel('x')
ylabel('f(x)')
title('E
or of the clamped cubic spline for f(x)')
end
matla
output.xml
manual document ready 0.4 matrix x 4×1 4 1 double 1.338393927287255e-01
1.654015181781862e-01
1.713943268078306e-01
-1.214542548941270e-01
double double [[{"value":"{\"value\":\"1.338393927287255e-01\"}"},{"value":"{\"value\":\"1.654015181781862e-01\"}"},{"value":"{\"value\":\"1.713943268078306e-01\"}"},{"value":"{\"value\":\"-1.214542548941270e-01\"}"}]] 19 matrix x 4×1 4 1 double 1.338393927287255e-01
1.654015181781862e-01
1.713943268078306e-01
-1.214542548941270e-01
double double [[{"value":"{\"value\":\"1.338393927287255e-01\"}"},{"value":"{\"value\":\"1.654015181781862e-01\"}"},{"value":"{\"value\":\"1.713943268078306e-01\"}"},{"value":"{\"value\":\"-1.214542548941270e-01\"}"}]] 32 figure cba6917a-d32d-4f77-852e-b33351d568dd data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAA0gAAAJ2CAYAAABl4KKyAAAgAElEQVR4AezBD1zVhaH
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