Related
I am trying to understand now fminunc (fmincon) works, however I keep getting error.
When I use documentation example with two variables
fun = #(x)3*x(1)^2 + 2*x(1)*x(2) + x(2)^2 - 4*x(1) + 5*x(2);
x0 = [1,1];
[x,fval] = fminunc(fun,x0);
everything works fine.
Hovewer, when I am trying to fit a plane for 3 points,
the code does not work
n0 = [ 0 1 -2;
1 2 1;
-2 -4 -4]
fun = #(x) [x(1) x(2) x(3)] * n0 - [1 1 1]
The task for fminunc is just an example. I know I can solve it easily analytically.
The cost function returns a scalar. What you have written returns a [1x3] matrix. You could try something like this if you want to minimise the euclidean distance
fun = #(x) sum(([x(1) x(2) x(3)] * n0 - [1 1 1]).^2);
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.
Hey i have an issuse with plotting my own function in scilab.
I want to plot the following function
function f = test(n)
if n < 0 then
f(n) = 0;
elseif n <= 1 & n >= 0 then
f(n) = sin((%pi * n)/2);
else
f(n) = 1;
end
endfunction
followed by the the console command
x = [-2:0.1:2];
plot(x, test(x));
i loaded the function and get the following error
!--error 21
Invalid Index.
at line 7 of function lala called by :
plot(x, test(x))
Can you please tell me how i can fix this
So i now did it with a for loop. I don't think it is the best solution but i can't get the other ones running atm...
function f = test(n)
f = zeros(size(n));
t = length(n);
for i = 1:t
if n(i) < 0 then
f(i) = 0;
elseif n(i) <= 1 & n(i) >= 0
f(i) = sin((%pi * n(i)/2));
elseif n(i) > 1 then
f(i) = 1;
end
end
endfunction
I guess i need to find a source about this issue and get used with the features and perks matlab/scilab have to over :)
Thanks for the help tho
The original sin is
function f = test(n)
(...)
f(n) = (...)
(...)
endfunction
f is supposed to be the result of the function. Therefore, f(n) is not "the value that the function test takes on argument n", but "the n-th element of f". Scilab then handles this however it can; on your test case, it tries to access a non-integer index, which results in an error. Your loop solution solves the problem.
Replacing all three f(n) by f in your first formulation makes it into something that works... as long as the argument is a scalar (not an array).
If you want test to be able to accept vector arguments without making a loop, the problem is that n < 0 is a vector of the same size as n. My solution would use logical arrays for indexing each of the three conditions:
function f = test(n)
f = zeros(size(n));
negative = (n<0);//parentheses are optional, but I like them for readability
greater_than_1 = (n>1);
others = ~negative & ~greater_than_1;
f(isnegative)=0;
f(greater_than_1)=1;
f(others) = sin(%pi/2*n(others));
endfunction
I am looking for a one-line function f = #(x) {something} that produces NaN if x >= 1, and either 0 or 1 if x < 1.
Any suggestions?
Aha, I got it:
f = #(x) 0./(x<1)
yields 0 for x < 1 and NaN for x>=1.
Here's a modification of Jason's solution that works for arrays. Note that recent versions of MATLAB do not throw divide-by-zero warnings.
>> f = #(x) zeros(size(x)) ./ (x < 1)
f =
#(x)zeros(size(x))./(x<1)
>> f(0:.3:2)
ans =
0 0 0 0 NaN NaN NaN
Update: a coworker pointed out to me that Jason's original answer works just fine for arrays.
>> f = #(x) 0./(x<1)
f =
#(x)0./(x<1)
>> f(0:.3:2)
ans =
0 0 0 0 NaN NaN NaN
Here's a less obvious solution (vectorized nonetheless):
f = #(x) subsasgn(zeros(size(x)), struct('type','()','subs',{{x>=1}}), nan) + 0
Basically its equivalent to:
function v = f(x)
v = zeros(size(x));
v( x>=1 ) = nan;
The +0 at the end is to always force an output, even if f called with no output arguments (returned in ans). Example:
>> f(-2:2)
ans =
0 0 0 NaN NaN
Here's a solution that won't risk throwing any divide-by-zero warnings, since it doesn't involve any division (just the functions ONES and NAN):
f = #(x) [ones(x < 1) nan(x >= 1)];
EDIT: The above solution is made for scalar inputs. If a vectorized solution is needed (which isn't 100% clear from the question) then you could modify f like so:
f = #(x) arrayfun(#(y) [ones(y < 1) nan(y >= 1)],x);
Or apply ARRAYFUN when calling the first version of the function f:
y = arrayfun(f,x);
I have the following program
format compact; format short g; clear; clc;
L = 140; J = 77; Jm = 10540; G = 0.8*10^8; d = L/3;
for i=1:500000
omegan=1.+0.0001*i;
a(1,1) = ((omegan^2)*(Jm/(G*J))*d^2)-2; a(1,2) = 2; a(1,3) = 0; a(1,4) = 0;
a(2,1) = 1; a(2,2) = ((omegan^2)*(Jm/(G*J))*d^2)-2; a(2,3) = 1; a(2,4) = 0;
a(3,1) = 0; a(3,2) = 1; a(3,3) = ((omegan^2)*(Jm/(G*J))*d^2)-2; a(3,4) = 1;
a(4,1) = 0; a(4,2) = 0; a(4,3) = 2; a(4,4) = ((omegan^2)*(Jm/(G*J))*d^2)-2;
if(abs(det(a))<1E-10) sprintf('omegan= %8.3f det= %8.3f',omegan,det(a))
end
end
Analytical solution of the above system, and the same program written in fortran gives out values of omegan equal to 16.3818 and 32.7636 (fortran values; analytical differ a little, but they're there somewhere).
So, now I'm wondering ... where am I going wrong with this ? Why is matlab not giving the expected results ?
(this is probably something terribly simple, but it's giving me headaches)
You're looking for too small of determinant values because Matlab is using a different determinant function (or some other reason like something to do with the floating point accuracy involved in the two different methods). I'll show you that Matlab is essentially giving you the correct values and a better way to approach this problem in general.
First, let's take your code and change it slightly.
format compact; format short g; clear; clc;
L = 140; J = 77; Jm = 10540; G = 0.8*10^8; d = L/3;
vals = zeros(1,500000);
for i=1:500000
omegan=1.+0.0001*i;
a(1,1) = ((omegan^2)*(Jm/(G*J))*d^2)-2; a(1,2) = 2; a(1,3) = 0; a(1,4) = 0;
a(2,1) = 1; a(2,2) = ((omegan^2)*(Jm/(G*J))*d^2)-2; a(2,3) = 1; a(2,4) = 0;
a(3,1) = 0; a(3,2) = 1; a(3,3) = ((omegan^2)*(Jm/(G*J))*d^2)-2; a(3,4) = 1;
a(4,1) = 0; a(4,2) = 0; a(4,3) = 2; a(4,4) = ((omegan^2)*(Jm/(G*J))*d^2)-2;
vals(i) = abs(det(a));
if(vals(i)<1E-10)
sprintf('omegan= %8.3f det= %8.3f',omegan,det(a))
end
end
plot(1.+0.0001*(1:500000),log(vals))
All that I've done really is logged the values of the determinant for all values of omegan and plotted the log of those determinant values as a function of omegan. Here is the plot:
You notice three major dips in the graph. Two coincide with your results of 16.3818 and 32.7636, but there is also an additional one which you were missing (probably because your condition of the determinant being less than 1e-10 was too low even for your Fortran code to pick it up). Therefore, Matlab is also telling you that those are the values of omegan that you were looking for, but because of the determinant was determined in a different manner in Matlab, the values weren't the same - not surprising when dealing with badly conditioned matrices. Also, it probably has to do with Fortran using single precision floats as someone else said. I'm not going to look into why they aren't because I don't want to waste my time on that. Instead, let's look at what you are trying to do and try a different approach.
You, as I'm sure you are aware, are trying to find the eigenvalues of the matrix
a = [[-2 2 0 0]; [1 -2 1 0]; [0 1 -2 1]; [0 0 2 -2]];
, set them equal to
-omegan^2*(Jm/(G*J)*d^2)
and solve for omegan. This is how I went about it:
format compact; format short g; clear; clc;
L = 140; J = 77; Jm = 10540; G = 0.8*10^8; d = L/3;
C1 = (Jm/(G*J)*d^2);
a = [[-2 2 0 0]; [1 -2 1 0]; [0 1 -2 1]; [0,0,2,-2]];
myeigs = eig(a);
myeigs(abs(myeigs) < eps) = 0.0;
for i=1:4
sprintf('omegan= %8.3f', sqrt(-myeigs(i)/C1))
end
This gives you all four solutions - not just the two that you had found with your Fortran code (though one of them, zero, was outside of your testing range for omegan ). If you want to go about solving this by checking the determinant in Matlab, as you've been trying to do, then you'll have to play with the value that you're checking the absolute value of the determinant to be less than. I got it to work for a value of 1e-4 (it gave 3 solutions: 16.382, 28.374, and 32.764).
Sorry for such a long solution, but hopefully it helps.
Update:
In my first block of code above, I replaced
vals(i) = abs(det(a));
with
[L,U] = lu(a);
s = det(L);
vals(i) = abs(s*prod(diag(U)));
which is the algorithm that det is supposedly using according to the Matlab docs. Now, I am able to use 1E-10 as the condition and it works. So maybe Matlab isn't calculating the determinant exactly as the docs say? This is kind of disturbing.
New answer:
You can investigate this problem using symbolic equations, which gives me the correct answers:
>> clear all %# Clear all existing variables
>> format long %# Display more digits of precision
>> syms Jm d omegan G J %# Your symbolic variables
>> a = ((Jm*(d*omegan)^2)/(G*J)-2).*eye(4)+... %# Create the matrix a
diag([2 1 1],1)+...
diag([1 1 2],-1);
>> solns = solve(det(a),'omegan') %# Solve for where the determinant is 0
solns =
0
0
(G*J*Jm)^(1/2)/(Jm*d)
-(G*J*Jm)^(1/2)/(Jm*d)
-(2*(G*J*Jm)^(1/2))/(Jm*d)
(2*(G*J*Jm)^(1/2))/(Jm*d)
(3^(1/2)*(G*J*Jm)^(1/2))/(Jm*d)
-(3^(1/2)*(G*J*Jm)^(1/2))/(Jm*d)
>> solns = subs(solns,{G,J,Jm,d},{8e7,77,10540,140/3}) %# Substitute values
solns =
0
0
16.381862247021893
-16.381862247021893
-32.763724494043785
32.763724494043785
28.374217734436371
-28.374217734436371
I think you either just weren't choosing values in your loop close enough to the solutions for omegan or your threshold for how close the determinant is to zero is too strict. When I plug in the given values to a, along with omegan = 16.3819 (which is the closest value to one solution your loop produces), I get this:
>> det(subs(a,{omegan,G,J,Jm,d},{16.3819,8e7,77,10540,140/3}))
ans =
2.765476845475786e-005
Which is still larger in absolute amplitude than 1e-10.
I put this as an answer because I cannot paste this into a comment: Here's how Matlab calculates the determinant. I assume the rounding errors come from calculating the product of multiple diagonal elements in U.
Algorithm
The determinant is computed from the
triangular factors obtained by
Gaussian elimination
[L,U] = lu(A) s = det(L)
%# This is always +1 or -1
det(A) = s*prod(diag(U))