I have a very simple problem.
I have
x=[ 10 25 50];
y=[ 1.2 3 7.5];
I know my curve fitting function
f(x)=(a*x+1)/(bx+c);
How can I get coefficient(a,b,c) solve in matlab and also plot this curve?
Rearrange y = f(x) to make a, b, and c the unknowns:
y = (ax + 1) / (bx + c)
y(bx + c) = ax + 1
ax - bxy - cy = -1;
This describes a system of simultaneous linear equations in a, b, and c when you substitute your three paired values of x and y.
x = [10, 20, 100];
y = [1.2, 0.7, 0.4];
coeffs = [x', (-x.*y)', -y'];
knowns = [-1, -1, -1]';
v = coeffs \ knowns; % v is [a; b; c]
Now you have the coefficients a, b, and c so you can plot the function.
Addendum: plotting
To plot a function, first choose the x-values of the data points
xt = 1:100;
Then calculate the y-values (assuming you've already got a, b, c)
yt = (a*x + 1) ./ (b*x + c)
Then just plot them!
plot(xt, yt);
Read the Matlab help on the plot function for customizing the style of the plot.
Related
In MATLAB, I am trying to write a program that will take 3 coordinates on a graph, (x,y), use those values to solve a system of equations that will find the coefficients of a polynomial equation, y = ax^2 + bx + c, which I can then use to plot a parabola.
To test my code, I figured I could start with a polynomial, graph it, find the minimum location of the parabola, use its immediate neighbors for my other 2 locations, then run those 3 locations through my code which should spit out the coefficients of my original polynomial. But for some reason, my resulting parabola is right shifted and my values for b and c are incorrect.
Does anyone see where my issue is? I am out of ideas
clear all; close all;
x = -10:10;
%Original Polynomial
y = 6.*x.^2 + 11.*x -35;
% Find 3 Locations
[max_n, max_i] = min(y)
max_il = max_i - 1 % left neighbor of max_ni
max_nl = y(max_il) % value at max_il
max_ir = max_i + 1 % left neighbor of max_ni
max_nr = y(max_ir) % value at max_ir
% Solve for coefficients
syms a b c
equ = (a)*(max_i)^2 + (b)*(max_i) + (c) == (max_n);
equ_l = (a)*(max_il)^2 + (b)*(max_il) + (c) == (max_nl);
equ_r = (a)*(max_ir)^2 + (b)*(max_ir) + (c) == (max_nr);
sol = solve([equ, equ_l, equ_r],[a, b, c]);
Sola = sol.a
Solb = sol.b
Solc = sol.c
% New Polynomial
p = (sol.a).*(x).^2 + (sol.b).*(x) +(sol.c);
%Plot
plot(x,y); grid on; hold on;
plot(x, p);
axis([-10 10 -41 40])
[max_np, max_ip] = min(p)
legend('OG', 'New')
You are confusing the index into your array y, and the corresponding x coordinate.
x = -10:10;
y = 6.*x.^2 + 11.*x -35;
[max_n, max_i] = min(y)
Here. max_i is the index into the y array, the corresponding x coordinate would be x(max_i).
I suggest you find three data points to fit your curve to as follows:
[~, max_i] = min(y);
pts_x = x(max_i + (-1:1));
pts_y = y(max_i + (-1:1));
then use pts_x(i) and pts_y(i) as your x and y values:
syms a b c
equ = a * pts_x.^2 + b * pts_x + c == pts_y;
sol = solve(equ, [a, b, c]);
I'm trying to numerically find the solution to A*cos x +B*sin x = C where A and B are two known square matrices of the same size (for example 100x100), and C is a known vector (100x1).
Without the second term (i.e. with a single matrix), I will use Jacobi or Gauss-Seidel to solve this problem and get x but here, I don't see how to proceed to solve this problem in Matlab.
May be, it would be useful to solve the problem as : A*X + B*sqrt(1-X^2) = C.
I would greatly appreciate any help, ideas or advices
Thanks in advance
If I understood you correctly, you could use fsolve like this (c and X are vectors):
A = ones(2,2);
B = ones(2,2);
c = ones(2,1);
% initial point
x0 = ones(length(A), 1);
% call to fsolve
sol = fsolve(#(x) A * cos(x) + B*sin(x) - c, x0);
Here, we solve the nonlinear equation system F(x) = 0 with F: R^N -> R^N and F(x) = A * cos(x) + B*sin(x) - c.
Only for the sake of completeness, here's my previous answer, i.e. how one could do it in case C and X are matrices instead of vectors:
A = ones(2,2);
B = ones(2,2);
C = ones(2,2);
% initial point
x0 = ones(numel(A), 1);
% call to fsolve
fsolve(#(x) fun(x, A, B, C), x0)
function [y] = fun(x, A, B, C)
% Transform the input vector x into a matrix
X = reshape(x, size(A));
% Evaluate the matrix equation
Y = A * cos(X) + B*sin(X) - C;
% flatten the matrix Y to a row vector y
y = reshape(Y, [], 1);
end
Here, the idea is to transform the matrix equation system F: R^(N x N) -> R^(N x N) into a equivalent nonlinear system F: R^(N*N) -> R^(N*N).
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 want to fit a polynomial to noisy data so that the approximated polynomial is always >= the original data. For example:
x = linspace (-2, 6);
y = (x-2).^2 + 1 + 2 * randn (size (x));
function ret = delta (P, x, y)
yP = polyval (P, x);
d = yP - y;
d (d < 0) *= 1000;
ret = sumsq (d);
endfunction
P0 = polyfit (x, y, 2);
f = #(P) delta (P, x, y);
[P, FVAL] = sqp (P0, f)
xi = linspace (min(x), max(x), 100);
yi = polyval (P, xi);
plot(x, y, xi, yi);
grid on
Is there a better way/method which also works with higher order polynomials?
The easies way would be to just use polyfit and then calculate max(y-yi) and add this as offset but this wouldn't be optimal...
EDIT: I want to use GNU OCtave but added "matlab" as tag because the language and functions are similiar.
EDIT: based on tvo's answer and real data:
x = [10 20 30 40 50 60 80 100];
y = [0.2372, 0.1312, 0.0936, 0.0805, 0.0614, 0.0512, 0.0554, 0.1407];
function ret = delta (P, x, y)
ret = sumsq (polyval (P, x) - y);
endfunction
f = #(P) delta (P, x, y);
h = #(P) polyval(P, x) - y;
P0 = polyfit (x, y, 3);
[P] = sqp (P0, f, [], h)
xi = linspace (min(x), max(x));
yi = polyval (P0, xi);
yio = polyval (P, xi);
plot(x, y, xi, yi, ";initial guess;", xi, yio, ";optimized;");
grid on
but as you can see, the optimized and evaluated poly has points < the orginal point which isn't allowed.
your method seems fine, and I see no reason it can't be used for higher-order polynomials actually. Please explain why if you think it can't be used.
You are using Octave's 'sqp' solver. Documentation is here: http://www.gnu.org/software/octave/doc/v4.0.1/Nonlinear-Programming.html
You may want to avoid multiplying the error by an arbitrary number (1000 in your example) when it is negative. This may fail for different data sets, especially if they are larger, i.e. more data points.
You could try to use the non-linear inequality constraint options that Octave's 'sqp' offers, i.e. the h(x)>=0 (see doc).
As objective function phi you could use a square norm error, as in your example, and add constraints of the form h(x)>=0 for every data point. Note that 'x' would be the polynomial coefficients you want to fit and h(x) is the polynomial evaluated at a specific data point.
For example:
phi = #(P) delta_mod (P, x, y); % mod: Don't increase the importance of negative residuals!!
h = #(P) polyval(P, x1) - y1;
Psol = sqp(P0, phi, [], h);
Notice that the constraint function 'h' ensures that the polynomial will lie above (x1,y1), and the objective function 'phi' will try to keep it as close as possible. You can extend 'h' to contain one constraint for every data point in your set (see doc).
I have to numerically evaluate, in Matlab, the integral of the product of two functions A(x,y) and B(x,y). These two functions are in 2-dimensional array form. The integral should read as following
I understand that the function such as "trapz" are for numerical integration of arrays of data, but I do not understand ho to adapt to my case.
thank you
Giuseppe
To guide you I will build some (square) matrix functions
%// Problem size.
n = 3;
%// Auxiliary constant matrices.
D1 = rand(n, n);
D2 = rand(n, n);
D3 = rand(n, n);
D4 = rand(n, n);
%// Matrix functions and product.
A = #(x,y) x^2*D1.^2 + y*D2;
B = #(y,z) (log(y)+z)*D3.^3 - z*sin(D4);
P = #(x,y,z) A(x,y)*B(y,z);
In this way P is a function of three variables, now we will integrate it in y, when x and z are both 0.
a = 0;
b = 1;
integral(#(y) P(0,y,0), a, b, 'ArrayValued', true)
To generalize the solution to arbitrary domains we can build a new function in (x,z)
Int = #(x,z) integral(#(y) P(x,y,z), a, b, 'ArrayValued', true);
and define a mesh
x = linspace(-1, 1, 11);
z = linspace(-2, 2, 21);
[X, Z] = meshgrid(x, z);
Now we can evaluate the integral on the whole mesh.
C = arrayfun(Int, X(:), Z(:), 'UniformOutput', false);
In this way C will contain all the integrals, stored in a 1D array of cells.
As a check we can get the result at (x,z) = (0,0) by calling
C{sub2ind([11 21], 6, 11)}