I am trying to using fminunc to obtain the optimal theta in logistic regression, however I keep getting that:
fminunc stopped because it cannot decrease the objective function
along the current search direction.
Searching online, I found that this is usually the result of a gradient error which I am implementing in logistic_costFunction.m. I re-checked my work but I cannot spot the root cause.
I am not sure how to solve this issue, any help would be appreciated.
Here is my code:
clear all; close all; clc;
%% Plotting data
x1 = linspace(0,3,50);
mqtrue = 5;
cqtrue = 30;
dat1 = mqtrue*x1+5*randn(1,50);
x2 = linspace(7,10,50);
dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));
x = [x1 x2]'; % X
subplot(2,2,1);
dat = [dat1 dat2]'; % Y
scatter(x1, dat1); hold on;
scatter(x2, dat2, '*'); hold on;
classdata = (dat>40);
%% Compute Cost and Gradient
% Setup the data matrix appropriately, and add ones for the intercept term
[m, n] = size(x);
% Add intercept term to x and X_test
x = [ones(m, 1) x];
% Initialize fitting parameters
initial_theta = zeros(n + 1, 1);
% Compute and display initial cost and gradient
[cost, grad] = logistic_costFunction(initial_theta, x, dat);
fprintf('Cost at initial theta (zeros): %f\n', cost);
fprintf('Gradient at initial theta (zeros): \n');
fprintf(' %f \n', grad);
%% ============= Part 3: Optimizing using fminunc =============
% In this exercise, you will use a built-in function (fminunc) to find the
% optimal parameters theta.
% Set options for fminunc
options = optimset('GradObj', 'on', 'MaxIter', 400);
% Run fminunc to obtain the optimal theta
% This function will return theta and the cost
[theta, cost] = ...
fminunc(#(t)(logistic_costFunction(t, x, dat)), initial_theta, options);
logistic_costFunction.m
function [J, grad] = logistic_costFunction(theta, X, y)
% Initialize some useful values
m = length(y); % number of training examples
grad = zeros(size(theta));
H = sigmoid(X*theta);
T = y.*log(H) + (1 - y).*log(1 - H);
J = -1/m*sum(T);
% ====================Compute grad==================
for i = 1 : m
grad = grad + (H(i) - y(i)) * X(i,:)';
end
grad = 1/m*grad;
end
sigmoid.m
function g = sigmoid(z)
% Computes thes sigmoid of z
g = zeros(size(z));
g = 1 ./ (1 + (1 ./ exp(z)));
end
Related
I am trying to solve the 1D ADE
This is my code so far:
clc; clear; close all
%Input parameters
Ao = 1; %Initial value
L = 0.08; %Column length [m]
nx = 40; %spatial gridpoints
dx = L/nx; %Length step size [m]
T = 20/24; %End time [days]
nt = 100; %temporal gridpoints
dt = T/nt; %Time step size [days]
Vel = dx/dt; %Velocity in each cell [m/day]
alpha = 0.002; %Dispersivity [m]
De = alpha*Vel; % Dispersion coeff. [m2/day]
%Gridblocks
x = 0:dx:L;
t = 0:dt:T;
%Initial and boundary conditions
f = #(x) x; % initial cond.
% boundary conditions
g1 = #(t) Ao;
g2 = #(t) 0;
%Initialization
A = zeros(nx+1, nt+1);
A(:,1) = f(x);
A(1,:) = g1(t);
gamma = dt/(dx^2);
beta = dt/dx;
% Implementation of the explicit method
for j= 1:nt-1 % Time Loop
for i= 2:nx-1 % Space Loop
A(i,j+1) = (A(i-1,j))*(Vel*beta + De*gamma)...
+ A(i,j)*(1-2*De*gamma-Vel*beta) + A(i+1,j)*(De*gamma);
end
% Insert boundary conditions for i = 1 and i = N
A(2,j+1) = A(1,j)*(Vel*beta + De*gamma) + A(2,j)*(1-2*De*gamma-Vel*beta) + A(3,j)*(De*gamma);
A(nx,j+1) = A(nx-1,j)*(Vel*beta + 2*De*gamma) + A(nx,j)*(1-2*De*gamma-Vel*beta)+ (2*De*gamma*dx*g2(t));
end
figure
plot(t, A(end,:), 'r*', 'MarkerSize', 2)
title('A Vs time profile (Using FDM)')
xlabel('t'),ylabel('A')
Now, I have been able to solve the problem using MATLAB’s pdepe function (see plot), but I am trying to compare the result with the finite difference method (implemented in the code above). I am however getting negative values of the dependent variable, but I am not sure what exactly I could be doing wrong. I therefore will really appreciate if anyone can help me out here. Many thanks in anticipation.
PS: I can post the code I used for the pdepe if anyone would like to see it.
I updated the question to clarify it more. Here is a graph:
For the curve in the attached photo, I hope to draw the curve. I have its equation and it is after simplification will be like this one
% Eq-2
(b*Y* cos(v) + c - k*X*sin(v))^2 + ...
sqrt(k*X*(cos(v) + 1.0) + b*Y*sin(v))^2) - d = 0.0
Where:
v = atan((2.0*Y)/X) + c
and b, c, d and k are constants.
from the attached graph,
The curve is identified in two points:
p1 # (x=0)
p2 # (y=0)
I a new on coding so accept my apologize if my question is not clear.
Thanks
So, after your edit, it is a bit more clear what you want.
I insist that your equation needs work -- the original equation (before your edit) simplified to what I have below. The curve for that looks like your plot, except the X and Y intercepts are at different locations, and funky stuff happens near X = 0 because you have numerical problems with the tangent (you might want to reformulate the problem).
But, after checking your equation, the following code should be helpful:
function solve_for_F()
% graininess of alpha
N = 100;
% Find solutions for all alphae
X = zeros(1,N);
options = optimset('Display', 'off');
alpha = linspace(0, pi/2, N);
x0 = linspace(6, 0, N);
for ii = 1:numel(alpha)
X(ii) = fzero(#(x)F(x, alpha(ii)), x0(ii), options);
end
% Convert and make an X-Y plot
Y = X .* tan(alpha);
plot(X, Y,...
'linewidth', 2,...
'color', [1 0.65 0]);
end
function fval = F(X, alpha)
Y = X*tan(alpha);
% Please, SIMPLIFY in the future
A = 1247745517111813/562949953421312;
B = 4243112111277797/4503599627370496;
V = atan2(2*Y,X) + A;
eq2 = sqrt( (5/33*( Y*sin(V) + X/2*(cos(V) + 1) ))^2 + ...
(5/33*( Y*cos(V) - X/2* sin(V) ))^2 ) - B;
fval = eq2;
end
Results:
So, I was having fun with this (thanks for that)!
Different question, different answer.
The solution below first searches for the constants causing the X and Y intercepts you were looking for (p1 and p2). For those constants that best fit the problem, it makes a plot, taking into account numerical issues.
In fact, you don't need eq. 1, because that's true always for any curve -- it's just there to confuse you, and problematic to use.
So, here it is:
function C = solve_for_F()
% Points of interest
px = 6;
py = 4.2;
% Wrapper function; search for those constants
% causing the correct X,Y intercepts (at px, py)
G = #(C) abs(F( 0, px, C)) + ... % X intercept at px
abs(F(py, 0, C)); % Y intercept at py
% Initial estimate, based on your original equation
C0 = [5/33
1247745517111813/562949953421312
4243112111277797/4503599627370496
5/66];
% Minimize the error in G by optimizing those constants
C = fminsearch(G, C0);
% Plot the solutions
plot_XY(px, py, C);
end
function plot_XY(xmax,ymax, C)
% graininess of X
N = 100;
% Find solutions for all alphae
Y = zeros(1,N);
X = linspace(0, xmax, N);
y0 = linspace(ymax, 0, N);
options = optimset('Display', 'off',...,...
'TolX' , 1e-10);
% Solve the nonlinear equation for each X
for ii = 1:numel(X)
% Wrapper function for fzero()
fcn1 = #(y)F(y, X(ii), C);
% fzero() is probably the fastest and most intuitive
% solver for this problem
[Y(ii),~,flag] = fzero(fcn1, y0(ii), options);
% However, it uses an algorithm that easily diverges
% when the function slope is large. For those cases,
% solve with fminsearch()
if flag ~= 1
% In this case, the minimum of the absolute value
% is searched for (which should be zero)
fcn2 = #(y) abs(fcn1(y));
Y(ii) = fminsearch(fcn2, y0(ii), options);
end
end
% Now plot the X,Y solutions
plot(X, Y,...
'linewidth', 2,...
'color', [1 0.65 0]);
xlabel('X'), ylabel('Y')
axis([0 xmax+.1 0 ymax+.1])
end
function fval = F(Y, X, C)
% Unpack constants
b = C(1); d = C(3);
c = C(2); k = C(4);
% pre-work
V = atan2(2*Y, X) + c;
% Eq. 2
fval = sqrt( (b*Y*sin(V) + k*X*(cos(V) + 1))^2 + ...
(b*Y*cos(V) - k*X* sin(V) )^2 ) - d;
end
I'm new to Matlab and am really struggling even to get to grips with the basics.
I've got a function, myspring, that solves position and velocity of a mass/spring system with damping and a driving force. I can specify values for the spring stiffness (k), damping coefficient (c), and mass (m), in the command window prior to running ode45. What I am unable to do is to define a forcing function (even something simple like g = sin(t)) and use that as the forcing function, rather than having it written into the myspring function.
Can anyone help? Here's my function:
function pdot = myspring(t,p,c,k,m)
w = sqrt(k/m);
g = sin(t); % This is the forcing function
pdot = zeros(size(p));
pdot(1) = p(2);
pdot(2) = g - c*p(2) - (w^2)*p(1);
end
and how I'm using it in the command window:
>> k = 2; c = 2; m = 4;
>> tspan = linspace(0,10,100);
>> x0 = [1 0];
>> [t,x] = ode45(#(t,p)myspring(t,p,c,k,m),tspan,x0);
That works, but what I want should look something like this (I imagine):
function pdot = myspring(t,p,c,k,m,g)
w = sqrt(k/m);
pdot = zeros(size(p));
pdot(1) = p(2);
pdot(2) = g - c*p(2) - (w^2)*p(1);
end
Then
g = sin(t);
[t,x] = ode45(#(t,p)myspring(t,p,c,k,m,g),tspan,x0);
But what I get is this
In an assignment A(:) = B, the number of elements in A and B must be the same.
Error in myspring (line 7)
pdot(2) = g - c*p(2) - (w^2)*p(1);
Error in #(t,p)myspring(t,p,c,k,m,g)
Error in odearguments (line 87)
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ode45 (line 115)
odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);
Horchler, thank you for the reply. I can do as you suggested and it works. I am now faced with another problem that I hope you could advise me on.
I have a script that calculates the force on a structure due to wave interaction using Morison's equation. I gave it an arbitrary time span to begin with. I would like to use the force F that script calculates as the input driving force to myspring. Here is my Morison script:
H = 3.69; % Wave height (m)
A = H/2; % Wave amplitude (m)
Tw = 9.87; % Wave period (s)
omega = (2*pi)/Tw; % Angular frequency (rad/s)
lambda = 128.02; % Wavelength (m)
k = (2*pi)/lambda; % Wavenumber (1/m)
dw = 25; % Water depth (m)
Cm = 2; % Mass coefficient (-)
Cd = 0.7; % Drag coefficient (-)
rho = 1025; % Density of water (kg/m^3)
D = 3; % Diameter of structure (m)
x = 0; % Fix the position at x = 0 for simplicity
t = linspace(0,6*pi,n); % Define the vector t with n time steps
eta = A*cos(k*x - omega*t); % Create the surface elevation vector with size equal to t
F_M = zeros(1,n); % Initiate an inertia force vector with same dimension as t
F_D = zeros(1,n); % Initiate a drag force vector with same dimension as t
F = zeros(1,n); % Initiate a total force vector with same dimension as t
fun_inertia = #(z)cosh(k*(z+dw)); % Define the inertia function to be integrated
fun_drag = #(z)(cosh(k*(z+dw)).*abs(cosh(k*(z+dw)))); % Define the drag function to be integrated
for i = 1:n
F_D(i) = abs(((H*pi)/Tw) * (1/sinh(k*dw)) * cos(k*x - omega*t(i))) * ((H*pi)/Tw) * (1/sinh(k*dw)) * cos(k*x - omega*t(i)) * integral(fun_drag,-dw,eta(i));
F_M(i) = (Cm*rho*pi*(D^2)/4) * ((2*H*pi^2)/(Tw^2)) * (1/(sin(k*dw))) * sin(k*x - omega*t(i)) * integral(fun_inertia,-dw,eta(i));
F(i) = F_D(i) + F_M(i);
end
Any further advice would be much appreciated.
You can't pre-calculate your forcing function. It depends on time, which ode45 determines. You need to define g as a function and pass a handle to it into your integration function:
...
g = #(t)sin(t);
[t,x] = ode45(#(t,p)myspring(t,p,c,k,m,g),tspan,x0);
And then call it I n your integration function, passing in the current time:
...
pdot(2) = g(t) - c*p(2) - (w^2)*p(1);
...
I am trying to implement the finite difference method in matlab. I did some calculations and I got that y(i) is a function of y(i-1) and y(i+1), when I know y(1) and y(n+1). However, I don't know how I can implement this so the values of y are updated the right way. I tried using 2 fors, but it's not going to work that way.
EDIT
This is the script and the result isn't right
n = 10;
m = n+1;
h = 1/m;
x = 0:h:1;
y = zeros(m+1,1);
y(1) = 4;
y(m+1) = 6;
s = y;
for i=2:m
y(i) = y(i-1)*(-1+(-2)*h)+h*h*x(i)*exp(2*x(i));
end
for i=m:-1:2
y(i) = (y(i) + (y(i+1)*(2*h-1)))/(3*h*h-2);
end
The equation is:
y''(x) - 4y'(x) + 3y(x) = x * e ^ (2x),
y(0) = 4,
y(1) = 6
Thanks.
Consider the following code. The central differential quotient is discretized.
% Second order diff. equ.
% y'' - 4*y' + 3*y = x*exp(2*x)
% (y(i+1)-2*y(i)+y(i-1))/h^2-4*(y(i+1)-y(i-1))/(2*h) + 3*y(i) = x(i)*exp(2*x(i));
The solution region is specified.
x = (0:0.01:1)'; % Solution region
h = min(diff(x)); % distance
As said in my comment, using this method, all points have to be solved simultaneously. Therefore, above numerical approximation of the equation is transformed in a linear system of euqations.
% System of equations
% Matrix of coefficients
A = zeros(length(x));
A(1,1) = 1; % known solu for first point
A(end,end) = 1; % known solu for last point
% y(i) y'' y
A(2:end-1,2:end-1) = A(2:end-1,2:end-1)+diag(repmat(-2/h^2+3,[length(x)-2 1]));
% y(i-1) y'' -4*y'
A(1:end-1,1:end-1) = A(1:end-1,1:end-1)+diag(repmat(1/h^2+4/(2*h),[length(x)-2 1]),-1);
% y(i+1) y'' -4*y'
A(2:end,2:end) = A(2:end,2:end)+diag(repmat(1/h^2-4/(2*h),[length(x)-2 1]),+1);
With the rhs of the differential equation. Note that the known values are calculated by 1 in the matrix and the actual value in the solution vector.
Y = x.*exp(2*x);
Y(1) = 4; % known solu for first point
Y(end) = 6; % known solu for last point
y = A\Y;
Having an equation to approximate the first order derivative (see above) you can verify the solution. (note, ddx2 is an own function)
f1 = ddx2(x,y); % first derivative (own function)
f2 = ddx2(x,f1); % second derivative (own function)
figure;
plot(x,y);
saveas(gcf,'solu1','png');
figure;
plot(x,f2-4*f1+3*y,x,x.*exp(2*x),'ko');
ylim([0 10]);
legend('lhs','rhs','Location','nw');
saveas(gcf,'solu2','png');
I hope the solution shown below is correct.
How can I solve a 2nd order differential equation with boundary condition like z(inf)?
2(x+0.1)·z'' + 2.355·z' - 0.71·z = 0
z(0) = 1
z(inf) = 0
z'(0) = -4.805
I can't understand where the boundary value z(inf) is to be used in ode45() function.
I used in the following condition [z(0) z'(0) z(inf)], but this does not give accurate output.
function [T, Y]=test()
% some random x function
x = #(t) t;
t=[0 :.01 :7];
% integrate numerically
[T, Y] = ode45(#linearized, t, [1 -4.805 0]);
% plot the result
plot(T, Y(:,1))
% linearized ode
function dy = linearized(t,y)
dy = zeros(3,1);
dy(1) = y(2);
dy(2) = y(3);
dy(3) = (-2.355*y(2)+0.71*y(1))/((2*x(t))+0.2);
end
end
please help me to solve this differential equation.
You seem to have a fairly advanced problem on your hands, but very limited knowledge of MATLAB and/or ODE theory. I'm happy to explain more if you want, but that should be in chat (I'll invite you) or via personal e-mail (my last name AT the most popular mail service from Google DOT com)
Now that you've clarified a few things and explained the whole problem, things are a bit more clear and I was able to come up with a reasonable solution. I think the following is at least in the general direction of what you'd need to do:
function [tSpan, Y2, Y3] = test
%%# Parameters
%# Time parameters
tMax = 1e3;
tSpan = 0 : 0.01 : 7;
%# Initial values
y02 = [1 -4.805]; %# second-order ODE
y03 = [0 0 4.8403]; %# third-order ODE
%# Optimization options
opts = optimset(...
'display', 'off',...
'TolFun' , 1e-5,...
'TolX' , 1e-5);
%%# Main procedure
%# Find X so that z2(t,X) -> 0 for t -> inf
sol2 = fminsearch(#obj2, 0.9879680932400429, opts);
%# Plug this solution into the original
%# NOTE: we need dense output, which is done via deval()
Z = ode45(#(t,y) linearized2(t,y,sol2), [0 tMax], y02);
%# plot the result
Y2 = deval(Z,tSpan,1);
plot(tSpan, Y2, 'b');
%# Find X so that z3(t,X) -> 1 for t -> inf
sol3 = fminsearch(#obj3, 1.215435887288112, opts);
%# Plug this solution into the original
[~, Y3] = ode45(#(t,y) linearized3(t,y,sol3), tSpan, y03);
%# plot the result
hold on, plot(tSpan, Y3(:,1), 'r');
%# Finish plots
legend('Second order ODE', 'Third order ODE')
xlabel('T [s]')
ylabel('Function value [-]');
%%# Helper functions
%# Function to optimize X for the second-order ODE
function val = obj2(X)
[~, y] = ode45(#(t,y) linearized2(t,y,X), [0 tMax], y02);
val = abs(y(end,1));
end
%# linearized second-order ODE with parameter X
function dy = linearized2(t,y,X)
dy = [
y(2)
(-2.355*y(2) + 0.71*y(1))/2/(X*t + 0.1)
];
end
%# Function to optimize X for the third-order ODE
function val = obj3(X3)
[~, y] = ode45(#(t,y) linearized3(t,y,X3), [0 tMax], y03);
val = abs(y(end,2) - 1);
end
%# linearized third-order ODE with parameters X and Z
function dy = linearized3(t,y,X)
zt = deval(Z, t, 1);
dy = [
y(2)
y(3)
(-1 -0.1*zt + y(2) -2.5*y(3))/2/(X*t + 0.1)
];
end
end
As in my comment above, I think you're confusing a couple of things. I suspect this is what is requested:
function [T,Y] = test
tMax = 1e3;
function val = obj(X)
[~, y] = ode45(#(t,y) linearized(t,y,X), [0 tMax], [1 -4.805]);
val = abs(y(end,1));
end
% linearized ode with parameter X
function dy = linearized(t,y,X)
dy = [
y(2)
(-2.355*y(2) + 0.71*y(1))/2/(X*t + 0.1)
];
end
% Find X so that z(t,X) -> 0 for t -> inf
sol = fminsearch(#obj, 0.9879);
% Plug this ssolution into the original
[T, Y] = ode45(#(t,y) linearized(t,y,sol), [0 tMax], [1 -4.805]);
% plot the result
plot(T, Y(:,1));
end
but I can't get it to converge anymore for tMax beyond 1000 seconds. You may be running into the limits of ode45 capabilities w.r.t. accuracy (switch to ode113), or your initial value is not accurate enough, etc.