I need help to add a function, from 1 to 10, using MATLAB. The function is ((1/n)*sin(n*pi*x)) where n goes from 1 to 10 and x stays as a variable. Ultimately I want to have a summation of ten sines (i.e K1*sin(pi*x)+K2*sin(2*pi*x)+k3*sin(3*pi*x)+...etc) where k is a constant. I would really appreciate any assistance. Thanks
Edit: Thanks to everyone who helped with my problem However I should have been more specific when asking my question. After getting the summation, I want to plot the sine series. I tried doing this but I kept getting an error saying that "conversion from sym to double is not possible" Now I tried doing a for loop to get my graph. My code is as follows:
n = 0:10;
while i <= n
for i = 1:length(n);
T = (1/n(i))*sin(n(i)*pi*x);
end
i = 1+i;
max = sum(T);
end
plot(x,max,'black')
However this doesn't work. I don't think that this is the proper way to get the sum of a double. I would really appreciate it if someone could help me again. Thanks again
Learn to exploit MATLAB's vector nature.
For one-off shots:
>> f = #(n,x) sin((1:n)*pi*x) * (1./(1:n).');
>> f(200, 0.5)
ans =
7.828982258896381e-001
To be able to evaluate f(n,x) with vector/matrix input x:
>> f = #(n,x) reshape( sin( bsxfun(#times, (1:n)*pi, x(:)) ) * (1./(1:n).'), size(x) );
>> f(15,rand(2))
ans =
5.077194963950054e-001 2.660834723822258e-001
1.416130930552744e+000 1.012255979042172e-001
Replace (1./(1:n).') by [K1 K2 K3 ...].' when you want to use other constants than 1/n.
From my understanding you are trying to sum the multivariable expression. In your case, you have two variables n and x. You want to sum the expression from n = 1 to 10 keeping x as a variable. You can do this in MATLAB by using symsum function, whose syntax is
symsum(expr,var,a,b)
Where expression expr defines the terms of a series, with respect to the symbolic variable var. The value of the variable var changes from a to b. If you do not specify the variable, symsum uses the default variable determined by symvar. If expr is a constant, then the default variable is x.
So in your case
expr = (1/n*sin(n*pi*x)
var = n
a = 1
b = 10
The simple code would be
>>syms n x
>>F = symsum ((1/sym('n'))*sin(sym('n')*pi*x), n, 1, 10)
Answer to Edit: MATLAB cannot convert the sys variable to double. You can instead substitute the value of sym variable with variable from MATLAB workspace. For example you can plot the above function for the range 0 to 10 by using following command.
>> x = 0:0.1:10;
>> plot(x, subs(F))
Related
Suppose I have a column vector of formulae like this
N =
4*k2 + 5*k3 + k1*x
7*k2 + 8*k3 + k1*y
and a column vector of symbolic variables like this
k =
k1
k2
k3
The formulae are linear with respect to k. I'd like to find a matrix M such that M*k equals N.
I can do this with N/k. However, that gives
[ (4*k2 + 5*k3 + k1*x)/k1, 0, 0]
[ (7*k2 + 8*k3 + k1*y)/k1, 0, 0]
which is correct, but not what I want. What I want is the matrix
x 4 5
y 7 8
which seems to me the simplest answer in that it involves no variables from k.
How do I convince Matlab to factor out the specified variables from a formula or a vector of formulae?
You can use coeffs, specifically the form
C = coeffs(p,vars) returns coefficients of the multivariate polynomial p with respect to the variables vars.
Since the first input needs to be a polynomial, you need to pass each component of N:
coeffs(N(1), k)
coeffs(N(2), k)
Or use a loop and store all results in a symbolic array:
result = sym('result', [numel(N) numel(k)]); % create symbolic array
for m = 1:numel(N)
result(m,:) = coeffs(N(m), k);
end
In your example, this gives
result =
[ 5, 4, x]
[ 8, 7, y]
Based on #LuisMendo's answer, I used coeffs. But there are a couple of problems with coeffs. The first is that its result doesn't include any coefficients that are 0. The second is that it doesn't seem to guarantee that the coefficients are ordered the same way as the variables in its second argument. I came up with the following function to replace coeffs.
Luckily coeffs returns a second result that lists the variables associated with each item in the first result. (It's more complicated if the formula is not linear.)
function m = factorFormula(f, v )
% Pre: f is a 1x1 sym representing a
% linear function of the variables in v.
% Pre: v is a column vector of variables
% Post: m is a row vector such that m*v equals f
% and the formulas in m do not contain the
% variables in v
[cx,tx] = coeffs(f,v)
n = size(v,1)
m = sym(zeros(1,n))
for i = 1:n
j = find(tx==v(i))
if size(j,2) == 1
m(i) = cx(j)
end
end
end
This only works for one formula, but it can be extended to a vector using the loop in #LuisMendo's answer or this equivalent expression in #Sanchises comment there.
cell2sym(arrayfun( #(f)factorFormula(f,k),N,'UniformOutput',false ) )
I hope there is a better answer than this.
I am having trouble using fminsearch: getting the error that there were not enough input arguments for my function.
f = #(x1,x2,x3) x1.^2 + 3.*x2.^2 + 4.*x3.^2 - 2.*x1.*x2 + 5.*x1 + 3.*x2 + 2.*x3;
[x, val] = fminsearch(f,0)
Is there something wrong with my function? I keep getting errors anytime I want to use it as an input function with any other command.
I am having trouble using fminsearch [...]
Stop right there and think some more about the function you're trying to minimize.
Numerical optimization (which is what fminsearch does) is unnecessary, here. Your function is a quadratic function of vector x; in other words, its value at x can be expressed as
x^T A x + b^T x
where matrix A and vector b are defined as follows (using MATLAB notation):
A = [ 1 -1 0;
-1 3 0;
0 0 4]
and
b = [5 3 2].'
Because A is positive definite, your function has one and only one minimum, which can be computed in MATLAB with
x_sol = -0.5 * A \ b;
Now, if you're curious about the cause of the error you ran into, have a look at fuesika's answer; but do without fminsearch whenever you can.
It is exactly what Matlab is telling you: your function expects three arguments. You are passing only one.
Instead of
[x, val] = fminsearch(f,0)
you should call it like
[x, val] = fminsearch(f,[0,0,0])
since you define the function f to accept a three dimensional vector as input only.
You can read more about the specification of fminsearch in the online documentation at http://mathworks.com/help/matlab/ref/fminsearch.html:
x = fminsearch(fun,x0) starts at the point x0 and returns a value x
that is a local minimizer of the function described in fun. x0 can be
a scalar, vector, or matrix. fun is a function_handle.
I've been searching the net for a couple of mornings and found nothing, hope you can help.
I have an anonymous function like this
f = #(x,y) [sin(2*pi*x).*cos(2*pi*y), cos(2*pi*x).*sin(2*pi*y)];
that needs to be evaluated on an array of points, something like
x = 0:0.1:1;
y = 0:0.1:1;
w = f(x',y');
Now, in the above example everything works fine, the result w is a 11x2 matrix with in each row the correct value f(x(i), y(i)).
The problem comes when I change my function to have constant values:
f = #(x,y) [0, 1];
Now, even with array inputs like before, I only get out a 1x2 array like w = [0,1];
while of course I want to have the same structure as before, i.e. a 11x2 matrix.
I have no idea why Matlab is doing this...
EDIT 1
Sorry, I thought it was pretty clear from what I wrote in the original question, but I see some of you asking, so here is a clarification: what I want is to have again a 11x2 matrix, since I am feeding the function with arrays with 11 elements.
This means I expect to have an output exactly like in the first example, just with changed values in it: a matrix with 11 rows and 2 columns, with only values 0 in the first column and only values 1 in the second, since for all x(i) and y(i) the answer should be the vector [0,1].
It means I expect to have:
w = [0 1
0 1
0 1
...
0 1]
seems pretty natural to me...
You are defining a function f = #(x,y) [0, 1]; which has the input parameters x,y and the output [0,1]. What else do you expect to happen?
Update:
This should match your description:
g=#(x,y)[zeros(size(x)),ones(size(y))]
g(x',y')
Defining an anonymous function f as
f = #(x,y) [0,1];
naturally returns [0,1] for any inputs x and y regardless of the length of those vectors.
This behavior puzzled me also until I realized that I expected f(a,b) to loop over a and b as if I had written
for inc = 1:length(a)
f(a(inc), b(inc))
end
However, f(a,b) does not loop over the length of its inputs, so it merely returns [0,1] regardless of the length of a and b.
The desired behavior can be obtained by defining f as
g=#(x,y)[zeros(size(x)),ones(size(y))]
as Daniel stated in his answer.
Just a simple question today. If I have an m*n matrix and I want to cycle through every value in it and apply a probability based function.
Basically, if the probability is p, then each value in the matrix has p chance of having the function applied to it.
I have the loop and the function itself all worked out, but I haven't found how to actually apply the probability itself.
Any advice would be greatly appreciated! Thanks in advance.
Here's your data matrix:
>> X = reshape(1:9, 3, 3);
and you want to (possibly) apply the following function to every element (note how I've vectorized it, so that it can take a matrix as an argument)
>> f = #(x) x.^2;
You want to apply the function with probability p
>> p = 0.25;
So generate some random numbers between 0 and 1, and see which ones are less than p
>> idx = rand(3,3) < p;
And now apply the function to the relevant indexes
>> X(idx) = f(X(idx));
Here's your result:
>> X
X =
1 16 7
2 5 64
3 6 81
The trick is that you can generate the random numbers first, and then apply the other formulas.
For example:
R = rand(m,n) < p
Now each value of R(row,col) corresponds to the outcome that you need to process your original matrix(row,col).
So I suggest applying your function to every cell and then setting the values to a default value based on some probability. So lets assume M is the result of applying to function to everycell:
default = NaN % Or 0 or whatever
p = 0.8;
M(rand(size(M)) > p) = default;
I think you might have to reshape m after this... not sure
M = reshape(M, m, n);
I'm trying to use MatLab code as a way to learn math as a programmer.
So reading I'm this post about subspaces and trying to build some simple matlab functions that do it for me.
Here is how far I got:
function performSubspaceTest(subset, numArgs)
% Just a quick and dirty function to perform subspace test on a vector(subset)
%
% INPUT
% subset is the anonymous function that defines the vector
% numArgs is the the number of argument that subset takes
% Author: Lasse Nørfeldt (Norfeldt)
% Date: 2012-05-30
% License: http://creativecommons.org/licenses/by-sa/3.0/
if numArgs == 1
subspaceTest = #(subset) single(rref(subset(rand)+subset(rand))) ...
== single(rref(rand*subset(rand)));
elseif numArgs == 2
subspaceTest = #(subset) single(rref(subset(rand,rand)+subset(rand,rand))) ...
== single(rref(rand*subset(rand,rand)));
end
% rand just gives a random number. Converting to single avoids round off
% errors.
% Know that the code can crash if numArgs isn't given or bigger than 2.
outcome = subspaceTest(subset);
if outcome == true
display(['subset IS a subspace of R^' num2str(size(outcome,2))])
else
display(['subset is NOT a subspace of R^' num2str(size(outcome,2))])
end
And these are the subset that I'm testing
%% Checking for subspaces
V = #(x) [x, 3*x]
performSubspaceTest(V, 1)
A = #(x) [x, 3*x+1]
performSubspaceTest(A, 1)
B = #(x) [x, x^2, x^3]
performSubspaceTest(B, 1)
C = #(x1, x3) [x1, 0, x3, -5*x1]
performSubspaceTest(C, 2)
running the code gives me this
V =
#(x)[x,3*x]
subset IS a subspace of R^2
A =
#(x)[x,3*x+1]
subset is NOT a subspace of R^2
B =
#(x)[x,x^2,x^3]
subset is NOT a subspace of R^3
C =
#(x1,x3)[x1,0,x3,-5*x1]
subset is NOT a subspace of R^4
The C is not working (only works if it only accepts one arg).
I know that my solution for numArgs is not optimal - but it was what I could come up with at the current moment..
Are there any way to optimize this code so C will work properly and perhaps avoid the elseif statments for more than 2 args..?
PS: I couldn't seem to find a build-in matlab function that does the hole thing for me..
Here's one approach. It tests if a given function represents a linear subspace or not. Technically it is only a probabilistic test, but the chance of it failing is vanishingly small.
First, we define a nice abstraction. This higher order function takes a function as its first argument, and applies the function to every row of the matrix x. This allows us to test many arguments to func at the same time.
function y = apply(func,x)
for k = 1:size(x,1)
y(k,:) = func(x(k,:));
end
Now we write the core function. Here func is a function of one argument (presumed to be a vector in R^m) which returns a vector in R^n. We apply func to 100 randomly selected vectors in R^m to get an output matrix. If func represents a linear subspace, then the rank of the output will be less than or equal to m.
function result = isSubspace(func,m)
inputs = rand(100,m);
outputs = apply(func,inputs);
result = rank(outputs) <= m;
Here it is in action. Note that the functions take only a single argument - where you wrote c(x1,x2)=[x1,0,x2] I write c(x) = [x(1),0,x(2)], which is slightly more verbose, but has the advantage that we don't have to mess around with if statements to decide how many arguments our function has - and this works for functions that take input in R^m for any m, not just 1 or 2.
>> v = #(x) [x,3*x]
>> isSubspace(v,1)
ans =
1
>> a = #(x) [x(1),3*x(1)+1]
>> isSubspace(a,1)
ans =
0
>> c = #(x) [x(1),0,x(2),-5*x(1)]
>> isSubspace(c,2)
ans =
1
The solution of not being optimal barely scratches the surface of the problem.
I think you're doing too much at once: rref should not be used and is complicating everything. especially for numArgs greater then 1.
Think it through: [1 0 3 -5] and [3 0 3 -5] are both members of C, but their sum [4 0 6 -10] (which belongs in C) is not linear product of the multiplication of one of the previous vectors (e.g. [2 0 6 -10] ). So all the rref in the world can't fix your problem.
So what can you do instead?
you need to check if
(randn*subset(randn,randn)+randn*subset(randn,randn)))
is a member of C, which, unless I'm mistaken is a difficult problem: Conceptually you need to iterate through every element of the vector and make sure it matches the predetermined condition. Alternatively, you can try to find a set such that C(x1,x2) gives you the right answer. In this case, you can use fminsearch to solve this problem numerically and verify the returned value is within a defined tolerance:
[s,error] = fminsearch(#(x) norm(C(x(1),x(2)) - [2 0 6 -10]),[1 1])
s =
1.999996976386119 6.000035034493023
error =
3.827680714104862e-05
Edit: you need to make sure you can use negative numbers in your multiplication, so don't use rand, but use something else. I changed it to randn.