I want the function P to look like this:
-1 + 0.6366*(x+pi/2) + (-0.000)*(x + pi/2)*(x)
and right now it looks like this
(5734161139222659*x)/9007199254740992 + (5734161139222659*pi)/18014398509481984 - (8131029572207409*x*(x + pi/2))/324518553658426726783156020576256 - 1.
How to convert S array so that the values are not symbolic?
syms P x
f = sin(x);
f = matlabFunction(f);
X = [-pi/2, 0, pi/2];
Y = f(sym(X));
P = MetN(X,Y,x)
P = matlabFunction(P);
function [P] = MetN(X,Y,x)
n = length(X);
for i = 1:n
A(i,1) = 1;
end
for i = 2:n
for j = 2: n
if i >= j
produs = 1;
for k =1:j-1
produs = produs * (X(i) - X(k));
end
A(i,j) = produs;
end
end
end
S = SubsAsc(A, Y);
S = double(S);
disp(S);
sym produs
P = double(sym(S(1)));
for i = 2:n
produs = 1;
for j = 1:i-1
produs = produs * (x - sym(X(j)));
end
disp(produs);
P = P + double(S(i))*produs;
end
end
function [x] = SubsAsc(A,b)
n = length(b);
x(1) = (1/A(1,1))*b(1);
for k = 2:n
s = 0;
for j = 1:k-1
s = s + A(k,j)*x(j);
end
x(k) = (1/A(k,k))*(b(k)-s);
end
end
The output you currently have is because symbolic uses exact arithmetic, so it outputs it as a rational number (hence the ugly fraction).
To have it output P using decimals, use vpa(). For instance output P using decimals to 5 significant digits
>> vpa(P, 5)
ans =
0.63662*x - 2.5056e-17*x*(x + 1.5708)
This will, however, also round pi, so you can't really have the best of both worlds here.
Related
I want to determine the Steepest descent of the Rosenbruck function using Armijo steplength where x = [-1.2, 1]' (the initial column vector).
The problem is, that the code has been running for a long time. I think there will be an infinite loop created here. But I could not understand where the problem was.
Could anyone help me?
n=input('enter the number of variables n ');
% Armijo stepsize rule parameters
x = [-1.2 1]';
s = 10;
m = 0;
sigma = .1;
beta = .5;
obj=func(x);
g=grad(x);
k_max = 10^5;
k=0; % k = # iterations
nf=1; % nf = # function eval.
x_new = zeros([],1) ; % empty vector which can be filled if length is not known ;
[X,Y]=meshgrid(-2:0.5:2);
fx = 100*(X.^2 - Y).^2 + (X-1).^2;
contour(X, Y, fx, 20)
while (norm(g)>10^(-3)) && (k<k_max)
d = -g./abs(g); % steepest descent direction
s = 1;
newobj = func(x + beta.^m*s*d);
m = m+1;
if obj > newobj - (sigma*beta.^m*s*g'*d)
t = beta^m *s;
x = x + t*d;
m_new = m;
newobj = func(x + t*d);
nf = nf+1;
else
m = m+1;
end
obj=newobj;
g=grad(x);
k = k + 1;
x_new = [x_new, x];
end
% Output x and k
x_new, k, nf
fprintf('Optimal Solution x = [%f, %f]\n', x(1), x(2))
plot(x_new)
function y = func(x)
y = 100*(x(1)^2 - x(2))^2 + (x(1)-1)^2;
end
function y = grad(x)
y(1) = 100*(2*(x(1)^2-x(2))*2*x(1)) + 2*(x(1)-1);
end
I am trying to solve this problem. But I keep getting an error.
This is my First Code.
% Program 3.3
function [L, U, P] = lufact(A)
[N, N] = size(A);
X = zeros(N, 1);
Y = zeros(N, 1);
C = zeros(1, N);
R = 1:N;
for p = 1: N-1
[max1, j] = max(abs(A(p:N, p)));
C = A(p,:);
A(p,:) = A(j + p - 1,:);
A(j + p -1, :) = C;
d = R(p);
R(p) = R(j + p -1);
R(j + p - 1) = d;
if A(p,p) == 0
'A is Singular. No unique Solution'
break
end
for k = p + 1:N
mult = A(k,p)/A(p,p);
A(k,p) = mult;
A(k,p + 1:N) = A(k, p + 1:N) - mult *A(p, p + 1:N);
end
I=(1:N)'*ones(1,N,1); J=I';
L = (I>J).*A + eye(N);
U = (J>=I).*A;
P = zeros(N);
for k=1:N
P(k,R(k))=1;
end
end
X(N) = Y(N)/A(N,N);
for k = N-1: -1: 1
X(k) = (Y(k) - A(k, k+1:N)*X(k+1:N))/A(k,k);
end
And This is my 2nd Code which I'm using to solve this problem.
function B = Ques3(A)
% Computes the inverse of a matrix A
[L,U,P] = lufact(A);
N = max(size(A));
I = eye(N);
B = zeros(N);
for j = 1:N
Y = forsub(L,P*I(:,j));
B(:,j) = backsub(U,Y);
end
But I keep getting an error in MATLAB,
>> Ques3(A)
Unrecognized function or variable 'forsub'.
Error in Ques3 (line 12)
Y = forsub(L,P*I(:,j));
I am trying to evaluate two matrixes which I defined outside of the function MetNewtonSist using subs and I get the error Undefined function or variable 'x' whenever I try to run the code.
[edit] I added the code for the GaussPivTot function which determines the solution of a liniear system.
syms x y
f1 = x^2 + y^2 -4;
f2 = (x^2)/8 - y;
J = jacobian( [ f1, f2 ], [x, y]);
F = [f1; f2];
subs(J, {x,y}, {1, 1})
eps = 10^(-6);
[ x_aprox,y_aprox, N ] = MetNewtonSist( F, J, 1, 1, eps )
function [x_aprox, y_aprox, N] = MetNewtonSist(F, J, x0, y0, eps)
k = 1;
x_v(1) = x0;
y_v(1) = y0;
while true
k = k + 1;
z = GaussPivTot(subs(J, {x, y}, {x_v(k-1), y_v(k-1)}),-subs(F,{x, y}, {x_v(k-1), y_v(k-1)}));
x_v(k) = z(1) + x_v(k-1);
y_v(k) = z(1) + y_v(k-1);
if norm(z)/norm([x_v(k-1), y_v(k-1)]) < eps
return
end
end
N = k;
x_aprox = x_v(k);
y_aprox = y_v(k);
end
function [x] = GaussPivTot(A,b)
n = length(b);
A = [A,b];
index = 1:n;
for k = 1:n-1
max = 0;
for i = k:n
for j = k:n
if A(i,j) > max
max = A(i,j);
p = i;
m = j;
end
end
end
if A(p,m) == 0
disp('Sist. incomp. sau comp. nedet.')
return;
end
if p ~= k
aux_line = A(p,:);
A(p,:) = A(k, :);
A(k,:) = aux_line;
end
if m ~= k
aux_col = A(:,m);
A(:,m) = A(:,k);
A(:,k) = aux_col;
aux_index = index(m);
index(m) = index(k);
index(k) = aux_index;
end
for l = k+1:n
M(l,k) = A(l,k)/A(k,k);
aux_line = A(l,:);
A(l,:) = aux_line - M(l,k)*A(k,:);
end
end
if A(n,n) == 0
disp('Sist. incomp. sau comp. nedet.')
return;
end
y = SubsDesc(A, A(:,n+1));
for i = 1:n
x(index(i)) = y(i);
end
end
By default, eps is defined as 2.2204e-16 in MATLAB. So do not overwrite it with your variable and name it any word else.
epsilon = 1e-6;
Coming to your actual issue, pass x and y as input arguments to the MetNewtonSist function. i.e. define MetNewtonSist as:
function [x_aprox, y_aprox, N] = MetNewtonSist(F, J, x0, y0, epsilon, x, y)
%added x and y and renamed eps to epsilon
and then call it with:
[x_aprox, y_aprox, N] = MetNewtonSist(F, J, 1, 1, epsilon, x, y);
function p = newton_hw(p0,tol,Nmax)
%NEWTON'S METHOD: Enter f(x), f'(x), x0, tol, Nmax
f = #(x) x*cos(x)-((sin(x))^2);
fp= #(x) -x*sin(x)+ cos(x)-2*sin(x)*cos(x);
p = p0 - (f(p0)/fp(p0));
y1=f(p);
fprintf('y1=%f',y1)
i = 1;
while (abs(p - p0) >= tol)
p0 = p;
p = p0 - f(p0)/fp(p0);
i = i + 1;
if (i >= Nmax)
fprintf('Fail after %d iterations\n',Nmax);
break
end
y=f(p);
fprintf('a=%f,y=%f,\n',p,y);
end
end
This is my question:
How to iterate for each of p0 = 0,.1,.2,...,49,5.
Iterating using user's step can be done this way:
for i = 0:0.1:5
Also any indexing can be done the same way: x = [0:2:50].
If your function works correctly (I suppose so), we can go this way:
k = 1;
for i = 0:0.1:5
res(k) = newton_hw(i,0.001,1000);
k = k+1;
end
But also we can do it at one line - style:
res = arrayfun( #(x) newton_hw(x, 0.001, 1000), I)
I'm trying to vectorize the 2 inner nested for loops, but I can't come up with a way to do this. The FS1 and FS2 functions have been written to accept argument for N_theta and N_e, which is what the loops are iterating over
%% generate regions
for raw_r=1:visual_field_width
for raw_c=1:visual_field_width
r = raw_r - center_r;
c = raw_c - center_c;
% convert (r,c) to polar: (eccentricity, angle)
e = sqrt(r^2+c^2)*deg_per_pixel;
a = mod(atan2(r,c),2*pi);
for nt=1:N_theta
for ne=1:N_e
regions(raw_r, raw_c, nt, ne) = ...
FS_1(nt-1,a,N_theta) * ...
FS_2(ne-1,e,N_e,e0_in_deg, e_max);
end
end
end
end
Ideally, I could replace the two inner nested for loops by:
regions(raw_r,raw_c,:,:) = FS_1(:,a,N_theta) * FS_2(:,N_e,e0_in_deg,e_max);
But this isn't possible. Maybe I'm missing an easy fix or vectorization technique? e0_in_deg and e_max are parameters.
The FS_1 function is
function h = FS_1(n,theta,N,t)
if nargin==2
N = 9;
t=1/2;
elseif nargin==3
t=1/2;
end
w = (2*pi)/N;
theta = theta + w/4;
if n==0 && theta>(3/2)*pi
theta = theta - 2*pi;
end
h = FS_f((theta - (w*n + 0.5*w*(1-t)))/w);
the FS_2 function is
function g = FS_gne(n,e,N,e0, e_max)
if nargin==2
N = 10;
e0 = .5;
elseif nargin==3
e0 = .5;
end
w = (log(e_max) - log(e0))/N;
g = FS_f((log(e)-log(e0)-w*(n+1))/w);
and the FS_f function is
function f = FS_f(x, t)
if nargin<2
t = 0.5;
end
f = zeros(size(x));
% case 1
idx = x>-(1+t)/2 & x<=(t-1)/2;
f(idx) = (cos(0.5*pi*((x(idx)-(t-1)/2)/t))).^2;
% case 2
idx = x>(t-1)/2 & x<=(1-t)/2;
f(idx) = 1;
% case 3
idx = x>(1-t)/2 & x<=(1+t)/2;
f(idx) = -(cos(0.5*pi*((x(idx)-(1+t)/2)/t))).^2+1;
I had to assume values for the constants, and then used ndgrid to find the possible configurations and sub2ind to get the indices. Doing this I removed all loops. Let me know if this produced the correct values.
function RunningFunction
%% generate regions
visual_field_width = 10;
center_r = 2;
center_c = 3;
deg_per_pixel = 17;
N_theta = 2;
N_e = 5;
e0_in_deg = 35;
e_max = 17;
[raw_r, raw_c, nt, ne] = ndgrid(1:visual_field_width, 1:visual_field_width, 1:N_theta, 1:N_e);
ind = sub2ind(size(raw_r), raw_r, raw_c, nt, ne);
r = raw_r - center_r;
c = raw_c - center_c;
% convert (r,c) to polar: (eccentricity, angle)
e = sqrt(r.^2+c.^2)*deg_per_pixel;
a = mod(atan2(r,c),2*pi);
regions(ind) = ...
FS_1(nt-1,a,N_theta) .* ...
FS_2(ne-1,e,N_e,e0_in_deg, e_max);
regions = reshape(regions, size(raw_r));
end
function h = FS_1(n,theta,N,t)
if nargin==2
N = 9;
t=1/2;
elseif nargin==3
t=1/2;
end
w = (2*pi)./N;
theta = theta + w/4;
theta(n==0 & theta>(3/2)*pi) = theta(n==0 & theta>(3/2)*pi) - 2*pi;
h = FS_f((theta - (w*n + 0.5*w*(1-t)))/w);
end
function g = FS_2(n,e,N,e0, e_max)
if nargin==2
N = 10;
e0 = .5;
elseif nargin==3
e0 = .5;
end
w = (log(e_max) - log(e0))/N;
g = FS_f((log(e)-log(e0)-w*(n+1))/w);
end
function f = FS_f(x, t)
if nargin<2
t = 0.5;
end
f = zeros(size(x));
% case 1
idx = x>-(1+t)/2 & x<=(t-1)/2;
f(idx) = (cos(0.5*pi*((x(idx)-(t-1)/2)/t))).^2;
% case 2
idx = x>(t-1)/2 & x<=(1-t)/2;
f(idx) = 1;
% case 3
idx = x>(1-t)/2 & x<=(1+t)/2;
f(idx) = -(cos(0.5*pi*((x(idx)-(1+t)/2)/t))).^2+1;
end