I must write using Laguerre's method a piece of code to find the real and complex roots of poly:
P=X^5-5*X^4-6*X^3+6*X^2-3*X+1
I have little doubt. I did the algorithm in the matlab, but 3 out of 5 roots are the same and I don't think that is correct.
syms X %Declearing x as a variabl
P=X^5-5*X^4-6*X^3+6*X^2-3*X+1; %Equation we interest to solve
n=5; % The eq. order
Pd1 = diff(P,X,1); % first differitial of f
Pd2 = diff(P,X,2); %second differitial of f
err=0.00001; %Answear tollerance
N=100; %Max. # of Iterations
x(1)=1e-3; % Initial Value
for k=1:N
G=double(vpa(subs(Pd1,X,x(k))/subs(P,X,x(k))));
H=G^2 - double(subs(Pd2,X,x(k))) /subs(P,X,x(k));
D1= (G+sqrt((n-1)*(n*H-G^2)));
D2= (G-sqrt((n-1)*(n*H-G^2)));
D = max(D1,D2);
a=n/D;
x(k+1)=x(k)-a
Err(k) = abs(x(k+1)-x(k));
if Err(k) <=err
break
end
end
output (roots of polynomial):
x =
0.0010 + 0.0000i 0.1434 + 0.4661i 0.1474 + 0.4345i 0.1474 + 0.4345i 0.1474 + 0.4345i
What you actually see are all the values x(k) which arose in the loop. The last one, 0.1474 + 0.4345i is the end result of this loop - the approximation of the root which is in your given tolerance threshold. The code
syms X %Declaring x as a variable
P = X^5 - 5 * X^4 - 6 * X^3 + 6 * X^2 - 3 * X + 1; %Polynomial
n=5; %Degree of the polynomial
Pd1 = diff(P,X,1); %First derivative of P
Pd2 = diff(P,X,2); %Second derivative of P
err = 0.00001; %Answer tolerance
N = 100; %Maximal number of iterations
x(1) = 0; %Initial value
for k = 1:N
G = double(vpa(subs(Pd1,X,x(k)) / subs(P,X,x(k))));
H = G^2 - double(subs(Pd2,X,x(k))) / subs(P,X,x(k));
D1 = (G + sqrt((n-1) * (n * H-G^2)));
D2 = (G - sqrt((n-1) * (n * H-G^2)));
D = max(D1,D2);
a = n/D;
x(k+1) = x(k) - a;
Err(k) = abs(x(k+1)-x(k));
if Err(k) <=err
fprintf('Initial value %f, result %f%+fi', x(1), real(x(k)), imag(x(k)))
break
end
end
results in
Initial value -2.000000, result -1.649100+0.000000i
If you want to get other roots, you have to use other initial values. For example one can obtain
Initial value 10.000000, result 5.862900+0.000000i
Initial value -2.000000, result -1.649100+0.000000i
Initial value 3.000000, result 0.491300+0.000000i
Initial value 0.000000, result 0.147400+0.434500i
Initial value 1.000000, result 0.147400-0.434500i
These are all zeros of the polynomial.
A method for calculating the next root when you have found another one would be that you divide through the corresponding linear factor and use your loop for the resulting new polynomial. Note that this is in general not very easy to handle since rounding errors can have a big influence on the result.
Problems with the existing code
You do not implement the Laguerre method properly as a method in complex numbers. The denominator candidates D1,D2 are in general complex numbers, it is inadvisable to use the simple max which only has sensible results for real inputs. The aim is to have a=n/D be the smaller of both variants, so that one has to look for the D in [D1,D2] with the larger absolute value. If there were a conditional assignment as in C, this would look like
D = (abs(D_1)>abs(D2)) ? D1 : D2;
As that does not exist, one has to use commands with a similar result
D = D1; if (abs(D_1)<abs(D2)) D=D2; end
The resulting sequence of approximation points is
x(0) = 0.0010000
x(1) = 0.143349512707684+0.466072958423667i
x(2) = 0.164462212064089+0.461399841949893i
x(3) = 0.164466373475316+0.461405404094130i
There is a point where one can not expect the (residual) polynomial value at the root approximation to substantially decrease. The value close to zero is obtained by adding and subtracting rather large terms in the sum expression of the polynomial. The accuracy lost in these catastrophic cancellation events can not be recovered.
The threshold for polynomial values that are effectively zero can be estimated as the machine constant of the double type times the polynomial value where all coefficients and the evaluation point are replaced by their absolute values. This test serves in the code primarily to avoid divisions by zero or near-zero.
Finding all roots
One approach is to apply the method to a sufficiently large number of initial points along some circle containing all the roots, with some strict rules for early termination at too slow convergence. One would have to make the list of the roots found unique, but keep the multiplicity,...
The other standard method is to apply deflation, that is, divide out the linear factor of the root found. This works well in low degrees.
There is no need for the slower symbolic operations as there are functions that work directly on the coefficient array, such as polyval and polyder. Deflation by division with remainder can be achieved using the deconv function.
For real polynomials, we know that the complex conjugate of a root is also a root. Thus initialize the next iteration with the deflated polynomial with it.
Other points:
There is no point in the double conversions as at no point there is a conversion into the single type.
If you don't do anything with it, it makes no sense to create an array, especially not for Err.
Roots of the example
Implementing all this I get a log of
x(0) = 0.001000000000000+0.000000000000000i, |Pn(x(0))| = 0.99701
x(1) = 0.143349512707684+0.466072958423667i, |dx|= 0.48733
x(2) = 0.164462212064089+0.461399841949893i, |dx|=0.021624
x(3) = 0.164466373475316+0.461405404094130i, |dx|=6.9466e-06
root found x=0.164466373475316+0.461405404094130i with value P0(x)=-2.22045e-16+9.4369e-16i
Deflation
x(0) = 0.164466373475316-0.461405404094130i, |Pn(x(0))| = 2.1211e-15
root found x=0.164466373475316-0.461405404094130i with value P0(x)=-2.22045e-16-9.4369e-16i
Deflation
x(0) = 0.164466373475316+0.461405404094130i, |Pn(x(0))| = 4.7452
x(1) = 0.586360702193454+0.016571894375927i, |dx|= 0.61308
x(2) = 0.562204173408499+0.000003168181059i, |dx|=0.029293
x(3) = 0.562204925474889+0.000000000000000i, |dx|=3.2562e-06
root found x=0.562204925474889+0.000000000000000i with value P0(x)=2.22045e-16-1.33554e-17i
Deflation
x(0) = 0.562204925474889-0.000000000000000i, |Pn(x(0))| = 7.7204
x(1) = 3.332994579372812-0.000000000000000i, |dx|= 2.7708
root found x=3.332994579372812-0.000000000000000i with value P0(x)=6.39488e-14-3.52284e-15i
Deflation
x(0) = 3.332994579372812+0.000000000000000i, |Pn(x(0))| = 5.5571
x(1) = -2.224132251798332+0.000000000000000i, |dx|= 5.5571
root found x=-2.224132251798332+0.000000000000000i with value P0(x)=-3.33067e-14+1.6178e-15i
for the modified code
P = [1, -2, -6, 6, -3, 1];
P0 = P;
deg=length(P)-1; % The eq. degree
err=1e-05; %Answer tolerance
N=10; %Max. # of Iterations
x=1e-3; % Initial Value
for n=deg:-1:1
dP = polyder(P); % first derivative of P
d2P = polyder(dP); %second derivative of P
fprintf("x(0) = %.15f%+.15fi, |Pn(x(0))| = %8.5g\n", real(x),imag(x), abs(polyval(P,x)));
for k=1:N
Px = polyval(P,x);
dPx = polyval(dP,x);
d2Px = polyval(d2P,x);
if abs(Px) < 1e-14*polyval(abs(P),abs(x))
break % if value is zero in relative accuracy
end
G = dPx/Px;
H=G^2 - d2Px / Px;
D1= (G+sqrt((n-1)*(n*H-G^2)));
D2= (G-sqrt((n-1)*(n*H-G^2)));
D = D1;
if abs(D2)>abs(D1) D=D2; end % select the larger denominator
a=n/D;
x=x-a;
fprintf("x(%d) = %.15f%+.15fi, |dx|=%8.5g\n",k,real(x),imag(x), abs(a));
if abs(a) < err*(err+abs(x))
break
end
end
y = polyval(P0,x); % check polynomial value of the original polynomial
fprintf("root found x=%.15f%+.15fi with value P0(x)=%.6g%+.6gi\n", real(x),imag(x),real(y),imag(y));
disp("Deflation");
[ P,R ] = deconv(P,[1,-x]); % division with remainder
x = conj(x); % shortcut for conjugate pairs and clustered roots
end
Related
I am writing a Matlab script that will approximate sin(x) and cos(x) using their Maclaurin polynomials.
When I input
arg = (5*pi)/4 I expect to get the correct approximations for
sin((5*pi)/4) = -0.7071067811865474617
cos((5*pi)/4) = -0.7071067811865476838.
Instead I get the following when running the script:
Approximation of sin(3.92699) >> -0.7071067811865474617
Actual sin(3.92699) = -0.7071067811865474617
Error approximately = 0.0000000000000000000 (0)
----------------------------------------------------------
Approximation of cos(3.92699) >> 0.7071067811865474617
Actual cos(3.92699) = -0.7071067811865476838
Error approximately = 0.0000000000000001110 (1.1102e-16)
I am getting the correct answers for sin but incorrect for cosine when the argument (angle) is in quadrant 3 or 4. The problem is that I am getting the wrong sign on the cos(arg) value. Where have I messed up?
CalculatorForSineCosine.m
% Argument for sine/cosine in radians.
arg = (5*pi)/4;
% Move the argument x so it's within [0, pi/2].
newArg = moveArgumentV2(arg);
% Calculate what degree we need for our Taylorpolynomial.
TOL = 0; % If 0, assume we want Machine Epsilon.
r = findDegreeV2(TOL);
% Plot nth degree Taylorpolynomial around x = 0 for sine.
% and calculate approximation of sin(x).
[approximatedSin, errorSin] = sin_taylorV2(r, newArg);
eS = num2str(errorSin); % errorSin in string format
% Plot nth degree Taylorpolynomial around x = 0 for cosine.
% and calculate approximation of cos(x).
[approximatedCos, errorCos] = cos_taylorV2(r, newArg);
eC = num2str(errorCos); % errorCos in string format
% Print out the result.
fprintf('\nApproximation of sin(%.5f)\t >> %.19f\n', arg, approximatedSin);
fprintf('Actual sin(%.5f)\t\t\t\t = %.19f\n', arg, sin(arg));
fprintf('Error approximately\t\t\t\t = %.19f (%s)\n', errorSin, eS);
disp("----------------------------------------------------------")
fprintf('Approximation of cos(%.5f)\t >> %.19f\n', arg, approximatedCos);
fprintf('Actual cos(%.5f)\t\t\t\t = %.19f\n', arg, cos(arg));
fprintf('Error approximately\t\t\t\t = %.19f (%s)\n\n', errorCos, eC);
sin_taylorV2.m
function [approximatedSin, errorSin] = sin_taylorV2(r, x)
%% sss
% Q_2n+1(x) where 2n+1 = degree of polynomial.
n = (r - 1)/2;
% Approximate sin(x) using its Taylorpolynomial.
approximatedSin = 0;
for k = 0:n
approximatedSin = approximatedSin + (((-1).^k) .* (x.^(2.*k+1)))./(factorial(2.*k+1));
end
% Calculate the error.
errorSin = abs(sin(x) - approximatedSin);
end
cos_taylorV2.m
function [approximatedCos, errorCos] = cos_taylorV2(r, x)
%% sss
% Q_2n+1(x) where 2n+1 = degree of polynomial and n = # terms.
n = (r - 1)/2;
% Approximate cos(x) using its Taylorpolynomial.
approximatedCos = 0;
for k = 0:n
approximatedCos = approximatedCos + (((-1).^k) .* (x.^(2.*k)))./(factorial(2.*k));
end
% Calculate the error.
errorCos = abs(cos(x) - approximatedCos);
end
moveArgumentV2.m
function newArg = moveArgumentV2(arg)
%% Moves the argument x to the interval [0, pi/2].
% Make use of sines periodocity and choose n as ceil( (x-pi)/2pi) )
n = ceil((arg-pi)/(2*pi));
x1 = arg - 2*pi*n; % New angle will be in [-pi, pi]
x2 = abs(x1); % Angle will be in [0, pi]
if (x2 < pi/2) && (x2 > 0)
x3 = x2;
else
x3 = pi - x2;
end
newArg = x3*sign(x1); % Angle will be in [0, pi/2]
end
I would like to notice two things in your code.
First, you don't need the moveArgumentV2(arg) function, as, if you remember, the radius of convergence for the Maclaurin/Taylor series of the sin(x)/cos(x) is the set of all real numbers. That means the series should converge for any real x, disregarding the round-off errors inherently to every arithmetic operations done in a computer.
As a matter of fact, following your code, we can write a function that approximates the cos as:
function y = mycos(x,n)
y = 0;
for k=0:n
term = (-1)^k*x.^(2*k)/factorial(2*k);
y = y + term;
end
end
Notice this function works for values outside the range [-pi,pi]:
x = -10*pi:0.1:10*pi;
ye = cos(x) % exact value
ya = mycos(x,100) % approximated value
plot(x,ye,x,ya,'o')
The values returned by the mycos function are close to the exact value given by the cos built-in function. This happens because I calculated the approximation with the first 100 terms. The error, however, for higher values of x, is extremely large if we use just a few terms.
ya = mycos(x,10) % approximated value with 10 terms only
plot(x,ye-ya); title('error')
The problem now is that we can't just increase the number of terms without running in another problem.
If we increase the number of points, the mycos function crumbles due to round-off errors, because of the factorial function that overflows. A good idea is to try to change your code in order to avoid the use of the factorial function. Notice the recurrence between sucessive terms in the Maclaurin expansion of the cos function, and you can create another function without the use of the factorial:
function y = mycos2(x,n)
term = 1;
y = 1;
for k=1:n
term = -term.*x.^2/(2*k-1)/(2*k);
y = y + term;
end
end
Here, we calculate each term in the series expansion from the previous calculated term. We avoid the calculation of the factorial and make use of what we already have. This speeds the code and avoids overflow. As a matter of fact, if we now calculate the cos approximation with 500 terms, we get:
x = -10*pi:0.5:10*pi;
ye = cos(x); % exact value
ya = mycos(x,500); % approximated value
ya2 = mycos2(x,500); % approximated value
plot(x,ye,x,ya,'x',x,ya2,'s')
legend('ye','ya','ya2')
Notice in this figure the x marks are the calculations done with the mycos function, while the o marks are done without using the factorial function. The first function crumbles for values outside the range [-2,2], but the second one runs just fine. It works even when I use 1e5 terms. Increasing the number of terms reduces the errors, so you can estimate how much terms you will use on an approximation, given a desired tolerance. If this number is greater than 170, the first function will not work properly.
factorial(170) returns 7.2574e+306, but factorial(171) returns Inf, so any value that should be calculated with more than 170 terms will have problems in the first function. Avoid the calculation of factorial at all costs.
This is what I tried:
x = -3*pi:0.01:3*pi;
y = x;
for ii=1:numel(y)
y(ii) = moveArgumentV2(y(ii)); % not vectorized
end
plot(sin(x))
hold on
plot(sin(y))
Both sin(x) and sin(y) produce the same plot. But:
plot(cos(x))
hold on
plot(cos(y))
Now we see that cos(x) and cos(y) are not the same! This is because moveArgumentV2 changes the angle to be in the first and fourth quadrant (in the range [-pi/2, pi/2]), which is what you need for the sin function, but is not adequate for the cos function.
I would modify sin_taylorV2 and cos_taylorV2 to call moveArgumentV2, so you don't rely on the caller to know what the valid input range is. In cos_taylorV2 you would need to call it this way:
x = moveArgumentV2(x+pi/2) - pi/2;
and in sin_taylorV2 you'd call it the same way you do now.
Or, better, write cos_taylorV2 in terms of sin_taylorV2, which we know to be correct. This avoids code duplication.
I am trying to find a solution for a non-linear system in MATLAB using the fsolve function. I have an idea of the region the solution is, so my initial points are a random variation inside this area. The code follows:
%MATLAB parameters
digits = 32;
format shortEng
%Initialization of solving process
min_norm = realmax;
options = optimoptions('fsolve', 'Algorithm', 'trust-region-dogleg',...
'Display', 'off', 'FunValCheck', 'on', 'Jacobian', 'on',...
'DerivativeCheck', 'on');
%Solving loop
while 1
%Refreshes seed after 100000 iterations
rng('shuffle');
for n=1:100000
%Initial points are randomly placed in solving region
x_0 = zeros(1, 3);
x_0(1) = 20*(10^-3)+ abs(3*(10^-3)*randn);
x_0(2) = abs(10^(-90) + 10^-89*randn);
x_0(3) = abs(10*randn);
%Solving
[x, fval] = fsolve(#fbnd, x_0, options);
norm_fval = norm(fval);
%If norm is minimum, result is displayed
if all(x > 0.0) && (norm_fval < min_norm)
iter_solu = x;
display(norm_fval);
display(iter_solu);
min_norm = norm_fval;
end
end
end
The function to be optimized and its jacobian follow:
function [F, J] = fbnd(x)
F(1) = x(1) - x(2)*(exp(7.56/(1.5*0.025))-1);
F(2) = x(1) - x(2)*(exp((x(3)*0.02)/(1.5*0.025))-1)-0.02;
F(3) = x(1) - x(2)*(exp((6.06+x(3)*0.018)/(1.5*0.025))-1)-0.018;
J = [[1.0, -3.5790482371355382991651020082214*10^87, 0.0];...
[1.0, 1.0-exp((8/15)*x(3)),-(8/15)*x(2)*exp((8/15)*x(3))];...
[1.0, 1.0-exp(0.48*x(3)+161.6), -0.48*x(2)*exp(0.48*x(3)+161.6)]];
end
When I turn DerivativeCheck on, I get the following error:
Objective function derivatives:
Maximum relative difference between user-supplied
and finite-difference derivatives = 1.
User-supplied derivative element (1,1): 1
Finite-difference derivative element (1,1): 0
For some reason, it things dF(1)/dx(1) should be zero instead of 1, when it is obvious it is 1. So I went ahead and replaced the J(1,1) by 0, and now I get the same problem with J(3,1):
Objective function derivatives:
Maximum relative difference between user-supplied
and finite-difference derivatives = 1.
User-supplied derivative element (3,1): 1
Finite-difference derivative element (3,1): 0
So, I replaced J(3,1) by 0 and it catches only minor precision errors. Is there an explanation why it thinks J(1,1) and J(3,1) should be zero? It seems really strange to me.
I'm basicaly trying to find the product of an expression that goes like this:
(x-(N-1)/2).....(x+(N-1)/2) for even value of N
x is a value that I will set at the beginning that changes too but that is a different problem...
let's say for the sake of argument that for now x is a constant (ex x=1)
example for N=6
(x-5/2)(x-3/2)(x-1/2)(x+1/2)(x+3/2)*(x+5/2)
the idea was to create a row vector every element of which is each individual term (P(1)=x-5/2) (P(2)=x-3/2)...etc and then calculate its product
N=6;
x=1;
P=ones(1,N);
for k=(-N-1)/2:(N-1)/2
for n=1:N
P(n)=(x-k);
end
end
y=prod(P);
instead this creates a vector that takes only the first value of the epxression and then
repeats the same value at each cell.
there is obviously a fundamental problem with my loop but I just can't see it.
So if anyone can help with that OR suggest a better way to calculate the product I would be grateful.
Use vectorized commands
Why use a loop when you can use vectorized commands like prod?
y = prod(2 * x + [-N + 1 : 2 : N - 1]) / 2;
For convenience, you may want to define an anonymous function for it:
f = #(N,x) reshape(prod(bsxfun(#plus, 2 * x(:), -N + 1 : 2 : N - 1) / 2, 2), size(x));
Note that the function is compatible with a (row or column) vector input x.
Tests in MATLAB's Command Window
>> f(6, [2,2]')
ans =
-14.7656
4.9219
-3.5156
4.9219
-14.7656
>> f(6, [2,2])
ans =
-14.7656 4.9219 -3.5156 4.9219 -14.7656
Benchmark
Here is a comparison of rayreng's approach versus mine. The former emerges as the clear winner... :'( ...at least as N increases.
Varying N, fixed x
Fixed N (= 10), vector x of varying length
Fixed N (= 100), vector x of varying length
Benchmark code
function benchmark
% varying N, fixed x
clear all
n = logspace(2,4,20)';
x = rand(1000,1);
tr = zeros(size(n));
tj = tr;
for k = 1 : numel(n)
% rayreng's approach (poly/polyval)
fr = #() rayreng(n(k), x);
tr(k) = timeit(fr);
% Jubobs's approach (prod/reshape/bsxfun)
fj = #() jubobs(n(k), x);
tj(k) = timeit(fj);
end
figure
hold on
plot(n, tr, 'bo')
plot(n, tj, 'ro')
hold off
xlabel('N')
ylabel('time (s)')
legend('rayreng', 'jubobs')
end
function y = jubobs(N,x)
y = reshape(prod(bsxfun(#plus,...
2 * x(:),...
-N + 1 : 2 : N - 1) / 2,...
2),...
size(x));
end
function y = rayreng(N, x)
p = poly(linspace(-(N-1)/2, (N-1)/2, N));
y = polyval(p, x);
end
function benchmark2
% fixed N, varying x
clear all
n = 100;
nx = round(logspace(2,4,20));
tr = zeros(size(n));
tj = tr;
for k = 1 : numel(nx)
disp(k)
x = rand(nx(k), 1);
% rayreng's approach (poly/polyval)
fr = #() rayreng(n, x);
tr(k) = timeit(fr);
% Jubobs's approach (prod/reshape/bsxfun)
fj = #() jubobs(n, x);
tj(k) = timeit(fj);
end
figure
hold on
plot(nx, tr, 'bo')
plot(nx, tj, 'ro')
hold off
xlabel('number of elements in vector x')
ylabel('time (s)')
legend('rayreng', 'jubobs')
title(['n = ' num2str(n)])
end
function y = jubobs(N,x)
y = reshape(prod(bsxfun(#plus,...
2 * x(:),...
-N + 1 : 2 : N - 1) / 2,...
2),...
size(x));
end
function y = rayreng(N, x)
p = poly(linspace(-(N-1)/2, (N-1)/2, N));
y = polyval(p, x);
end
An alternative
Alternatively, because the terms in your product form an arithmetic progression (each term is greater than the previous one by 1/2), you can use the formula for the product of an arithmetic progression.
I agree with #Jubobs in that you should avoid using for loops for this kind of computation. There are cases where for loops perform fast, but for something as simple as this, avoid using loops if possible.
An alternative approach to what Jubobs has suggested is that you can consider that polynomial equation to be in factored form where each factor denotes a root located at that particular location. You can use poly to convert these factors into a polynomial equation, then use polyval to evaluate the expression at the point you want. First, generate your roots by linspace where the points vary from -(N-1)/2 to (N-1)/2 and there are N of them, then plug this into poly. Finally, for any values of x, put this into polyval with the output of poly. The advantage of this approach is that you can evaluate multiple points of x in a single sweep.
Going with what you have, you would simply do this:
p = poly(linspace(-(N-1)/2, (N-1)/2, N));
out = polyval(p, x);
With your example, supposing that N = 6, this would be the output of the first line:
p =
1.0000 0 -8.7500 0 16.1875 0 -3.5156
As such, this is saying that when we expand out (x-5/2)(x-3/2)(x-1/2)(x+1/2)(x+3/2)(x+5/2), we get:
x^6 - 8.75x^4 + 16.1875x^2 - 3.5156
If we take a look at the roots of this equation, this is what we get:
r = roots(p)
r =
-2.5000
2.5000
-1.5000
1.5000
-0.5000
0.5000
As you can see, each term corresponds to one factor in your polynomial equation, so we do have the right mindset here. Now, all you have to do is use p with your values of x into polyval to obtain your results. For example, if I wanted to evaluate that polynomial from -2 <= x <= 2 where x is an integer, this is the result I get:
polyval(p, -2:2)
ans =
-14.7656 4.9219 -3.5156 4.9219 -14.7656
Therefore, when x = -2, the result is -14.7656 and so on.
Though I would recommend the solution by #Jubobs, it is also good to check what the issue is with your loop.
The first indication that something is wrong, is that you have a nested loop over 2 variables, and only index with one of them to store the result. Probably you just need a single loop.
Here is a loop that you may be interested in that should do roughly what you need:
N=6;
x=1;
k=(-N-1)/2:(N-1)/2
P = ones(size(k));
for n=1:numel(k)
P(n)=(x-k(n));
end
y=prod(P);
I tried to keep the code close to the original, so hopefully it is easy to understand.
I'm trying to create a matrix such that if I define a random number between 0 and 1 and a random location in the matrix, I want all the values around that to "diffuse" out. Here's sort of an example:
0.214 0.432 0.531 0.631 0.593 0.642
0.389 0.467 0.587 0.723 0.654 0.689
0.421 0.523 0.743 0.812 0.765 0.754
0.543 0.612 0.732 0.843 0.889 0.743
0.322 0.543 0.661 0.732 0.643 0.694
0.221 0.321 0.492 0.643 0.521 0.598
if you notice, there's a peak at (4,5) = 0.889 and all the other numbers decrease as they move away from that peak.
I can't figure out a nice way to generate a code that does this. Any thoughts? I need to be able to generate this type of matrix with random peaks and a random rate of decrease...
Without knowing what other constraints you want to implement:
Come up with a function z = f(x,y) whose peak value is at (x0,y0) == (0,0) and whose values range between [0,1]. As an example, the PDF for the Normal distribution with mu = 0 and sigma = 1/sqrt(2*pi) has a peak at x == 0 of 1.0, and whose lower bound is zero. Similarly, a bivariate normal PDF with mu = {0,0} and determinate(sigma) == [1/(2*pi)]^2 will have similar characteristics.
Any mathematical function may have its domain shifted: f(x-x0, y-y0)
Your code will look something like this:
someFunction = #(x,y) theFunctionYouPicked(x,y);
[x0,y0,peak] = %{ you supply these values %};
myFunction = #(x,y) peak * someFunction(x - x0, y - y0);
[dimX,dimY] = %{ you supply these values %};
mymatrix = bsxfun( myFunction, 0:dimX, (0:dimY)' );
You can read more about bsxfun here; however, here's an example of how it works:
bsxfun( blah, [a b c], [d e f]' )
That should give the following matrix (or its transpose ... I don't have matlab in front of me):
[blah(a,d) blah(a,e) blah(a,f);
blah(b,d) blah(b,e) blah(b,f);
blah(c,d) blah(c,e) blah(c,f)]
Get a toy example working, then you can tinker with it to be more flexible. If the function dictating how it decreases is random (with the constraint that points closer to (x0,y0) are larger than more distant points), it won't be an issue to make a procedural function instead of using strictly mathematical ones.
In response to your answer:
Your equation could be thought of as a model for gravity where an object instantaneously induces a force on another mass, then stops exerting force. Following that logic, it could be modified to a naive vector formulation like this:
% v1 & v2 are vectors that point from the two peak points to the point [ii,jj]
theMatrix(ii,jj) = norm( (r1 / norm( v1 )) * v1 / norm( v1 ) ...
+ (r2 / norm( v2 )) * v2 / norm( v2 ) ...
);
The most extreme type of corner case you'll run into is one where v1 & v2 point in the same direction as in the following row:
[ . . A X1 X2 . . ]
... where you want a value for A w/respect to X1 & X2. Using the above expression it'll boil down to A = X1 / norm(v1) + X2 / norm(v2), which will definitely exceed the peak value at X1 because norm(v1) == 1. You could certainly do some dirty stuff to Band-Aid it, but personally I'd start looking for a different function.
Along those lines, if you used Newton's Law of Universal Gravitation with a few modifications:
You wouldn't need an analogue for G, so you could just assume G == 1
Treat each of the points in the matrix as having mass m2 == 1, so the equation reduces to: F_12 == -1 * (m1 / r^2) * RHAT_12
Sum the "force" vectors and calculate the norm to get each value
... you'll still run into the same problem. The corner case I laid out above would boil down to A = X1/norm(v1)^2 + X2/norm(v2)^2 == X1 + X2/4. Since it's inversely proportional to the square of the distances, it'd be easier to Band-Aid than the linear one, but I wouldn't recommend it.
Similarly, if you use polynomials it won't scale well; you can design one that won't ever exceed your chosen peaks, but there wouldn't be a lower bound.
You could use the logistic function to help with this:
1 / (1 + E^(-c*x))
Here's an example of using the logistic function on a degree 4 polynomial with peaks at points 2 & 4; you'll note I gave the polynomial a scaling factor to pull the polynomial down to relatively small values so calculated values aren't so close together.
I ended up creating a code that wraps the way I want based on a dimension, which I provide. Here's the code:
dims = 100;
A = zeros(dims);
b = floor(1+dims*rand(1));
c = floor(1+dims*rand(1));
d = rand(1);
x1 = c;
y1 = b;
A(x1,y1) = d;
for i = 1:dims
for j = i
k = 1-j;
while k <= j
if x1-j>0 && y1+k>0 && y1+k <= dims
if A(x1-j,y1+k) == 0
A(x1-j,y1+k) = eqn(d,x1-j,y1+k,x1,y1);
end
end
k = k+1;
end
end
for k = i
j = 1-k;
while j<=k
if x1+j>0 && y1+k>0 && y1+k <= dims && x1+j <= dims
if A(x1+j,y1+k)==0
A(x1+j, y1+k) = eqn(d,x1+j,y1+k,x1,y1);
end
end
j = j+1;
end
end
for j = i
k = 1-j;
while k<=j
if x1+j>0 && y1-k>0 && x1+j <= dims && y1-k<= dims
if A(x1+j,y1-k) == 0
A(x1+j,y1-k) = eqn(d,x1+j,y1-k,x1,y1);
end
end
k=k+1;
end
end
for k = i
j = 1-k;
while j<=k
if x1-j>0 && y1-k>0 && x1-j <= dims && y1-k<= dims
if A(x1-j,y1-k)==0
A(x1-j,y1-k) = eqn(d,x1-j,y1-k,x1,y1);
end
end
j = j+1;
end
end
end
colormap('hot');
imagesc(A);
colorbar;
If you notice, the code calls a function (I called it eqn), which provided the information for how to changes the values in each cell. The function that I settled on is d/distance (distance being computed using the standard distance formula).
It seems to work pretty well. I'm now just trying to develop a good way to have multiple peaks in the same square without one peak completely overwriting the other.
I use convolution and for loops (too much for loops) for calculating the interpolation using
Lagrange's method , here's the main code :
function[p] = lagrange_interpolation(X,Y)
L = zeros(n);
p = zeros(1,n);
% computing L matrice, so that each row i holds the polynom L_i
% Now we compute li(x) for i=0....n ,and we build the polynomial
for k=1:n
multiplier = 1;
outputConv = ones(1,1);
for index = 1:n
if(index ~= k && X(index) ~= X(k))
outputConv = conv(outputConv,[1,-X(index)]);
multiplier = multiplier * ((X(k) - X(index))^-1);
end
end
polynimialSize = length(outputConv);
for index = 1:polynimialSize
L(k,n - index + 1) = outputConv(polynimialSize - index + 1);
end
L(k,:) = multiplier .* L(k,:);
end
% continues
end
Those are too much for loops for computing the l_i(x) (this is done before the last calculation of P_n(x) = Sigma of y_i * l_i(x)) .
Any suggestions into making it more matlab formal ?
Thanks
Yeah, several suggestions (implemented in version 1 below): if loop can be combined with for above it (just make index skip k via something like jr(jr~=j) below); polynomialSize is always equal length(outputConv) which is always equal n (because you have n datapoints, (n-1)th polynomial with n coefficients), so the last for loop and next line can be also replaced with simple L(k,:) = multiplier * outputConv;
So I replicated the example on http://en.wikipedia.org/wiki/Lagrange_polynomial (and adopted their j-m notation, but for me j goes 1:n and m is 1:n and m~=j), hence my initialization looks like
clear; clc;
X=[-9 -4 -1 7]; %example taken from http://en.wikipedia.org/wiki/Lagrange_polynomial
Y=[ 5 2 -2 9];
n=length(X); %Lagrange basis polinomials are (n-1)th order, have n coefficients
lj = zeros(1,n); %storage for numerator of Lagrange basis polyns - each w/ n coeff
Lj = zeros(n); %matrix of Lagrange basis polyns coeffs (lj(x))
L = zeros(1,n); %the Lagrange polynomial coefficients (L(x))
then v 1.0 looks like
jr=1:n; %j-range: 1<=j<=n
for j=jr %my j is your k
multiplier = 1;
outputConv = 1; %numerator of lj(x)
mr=jr(jr~=j); %m-range: 1<=m<=n, m~=j
for m = mr %my m is your index
outputConv = conv(outputConv,[1 -X(m)]);
multiplier = multiplier * ((X(j) - X(m))^-1);
end
Lj(j,:) = multiplier * outputConv; %jth Lagrange basis polinomial lj(x)
end
L = Y*Lj; %coefficients of Lagrange polinomial L(x)
which can be further simplified if you realize that numerator of l_j(x) is just a polynomial with specific roots - for that there is a nice command in matlab - poly. Similarly the denominator is just that polyn evaluated at X(j) - for that there is polyval. Hence, v 1.9:
jr=1:n; %j-range: 1<=j<=n
for j=jr
mr=jr(jr~=j); %m-range: 1<=m<=n, m~=j
lj=poly(X(mr)); %numerator of lj(x)
mult=1/polyval(lj,X(j)); %denominator of lj(x)
Lj(j,:) = mult * lj; %jth Lagrange basis polinomial lj(x)
end
L = Y*Lj; %coefficients of Lagrange polinomial L(x)
Why version 1.9 and not 2.0? well, there is probably a way to get rid of this last for loop, and write it all in 1 line, but I can't think of it right now - it's a todo for v 2.0 :)
And, for dessert, if you want to get the same picture as wikipedia:
figure(1);clf
x=-10:.1:10;
hold on
plot(x,polyval(Y(1)*Lj(1,:),x),'r','linewidth',2)
plot(x,polyval(Y(2)*Lj(2,:),x),'b','linewidth',2)
plot(x,polyval(Y(3)*Lj(3,:),x),'g','linewidth',2)
plot(x,polyval(Y(4)*Lj(4,:),x),'y','linewidth',2)
plot(x,polyval(L,x),'k','linewidth',2)
plot(X,Y,'ro','linewidth',2,'markersize',10)
hold off
xlim([-10 10])
ylim([-10 10])
set(gca,'XTick',-10:10)
set(gca,'YTick',-10:10)
grid on
produces
enjoy and feel free to reuse/improve
Try:
X=0:1/20:1; Y=cos(X) and create L and apply polyval(L,1).
polyval(L,1)=0.917483227909543
cos(1)=0.540302305868140
Why there is huge difference?