I use matlab syms to define a function
f(x,y,z) = ||y-zx||^2
where x and y are vectors of R^5, z is a scalar in R as follows
syms x y [5,1] matrix;
syms z;
f = (y-z*x).'*(y-z*x)
Now, I have two questions:
How to expand the expression of f using MATLAB?
How to define the function t-> d f(x+tw,y,z)/d t via an output of diff?
Thanks a lot for any comments and suggestions.
What I tried are summarized below:
How to expand the expression of f using MATLAB?
It seems that
expand(f)
does not work! and the other functions such as simplify, collect and factor work neither. So I want to know how to expand and simplify this expression in the correct way?
How to define the function t-> d f(x+tw,y,z)/d t via an output of diff?
I have tried to define the function phi(t) = f(x+tw,y,z) first as
syms x y w [5,1] matrix;
syms z t;
f = #(x,y,z) (y-z*x).'*(y-z*x);
phi = #(t) f(x+t*w,y,z);
Then I compute the derivative using diff:
diff(phi(t),t);
But I don't know how to make the resulting expression as a function of t (so that I can evaluate the expression of phi'(1))? For the moment, I just copy the expression of phi'(t) computed by diff and define it manually. Hoping to get a better way to do so. Thanks a lot for any comments and suggestions.
Related
I am new to Octave (and matlab for that matter). I have a function that looks like this
I would like to plot g(x,0.5,5) say.
Here it is what I tried in Octave
I defined an anonymous function
f=#(n,x,t) 1./n.*log(n.*pi.*t).*sin(n.*pi.*x);
then another anonymous function
g=#(m,x,t)x.^2+sum(f([1:m],x,t));
Finally defined
x=-1:0.1:1;
plot(x,g(5,x,0.5))
but I get an error. Is this the right way of plotting this function? I must be doing a simple beginner error?
When you call f(n,x,t), you are passing a 1-by-5 vector for n and a 1-by-21 vector for x. These have different numbers of elements, and therefore can't be multiplied element-by-element. However, you can rewrite f to accommodate vectors for each and perform the sum from g by using matrix multiplication:
f = #(n, x, t) (1./n.*log(n.*pi.*t))*sin(pi.*n(:)*x);
g = #(m, x, t) x.^2 + f(1:m, x, t);
And now your plot will work:
x = -1:0.1:1;
plot(x, g(5, x, 0.5));
I need to draw the curve of x^3+y^3-3*axy=0 and calculate its area. I'm new to matlab so I don't know much. Any help would be appreciated.
Here's what I came up with already`
function makeres(a)
syms x y;
k=y.^3+x.^3-3.*x.*a.*y==0;
ezplot(k)
fun=#(x)(k);
integral(integral(k,0,5),0,5)
But I get an error on integral.
Try this:
syms x y;
k=y.^3+x.^3-3.*x.*a.*y==0;
ezplot(k)
fun=#(x,y)y.^3+x.^3-3.*x.*a.*y;
g = integral2(fun,0,5,0,5);
Your erros where in the last 2 lines:
a - you were defining k (which is an equation) as the function to be integrated
b - then you integrated k, not fun
c - it is better use integral2 for double integral
I'm don't know what a should be, it works for me when a is a number, an the code as a single script (not a defined function).
syms x
syms y
f = 2*x^2 + y
z = diff(f,x)
x = 3
SolvedDiffEq = z
Obviously z = 4x. But how do I solve to get the numerical answer of 12? I've played with vpa, double, subs and cannot figure it out.
This is a simple example. My actual code has very complex equations with many variables. Furthermore, I am trying to iteratively solve these equations so it is not practical to manually substitute each variable.
You are looking for symbolic substitution command, which is subs. In your example, subs(z) returns 12.
I want to plot relations like y^2=x^2(x+3) in MATLAB without using ezplot or doing algebra to find each branch of the function.
Does anyone know how I can do this? I usually create a linspace and then create a function over the linspace. For example
x=linspace(-pi,pi,1001);
f=sin(x);
plot(x,f)
Can I do something similar for the relation I have provided?
What you could do is use solve and allow MATLAB's symbolic solver to symbolically solve for an expression of y in terms of x. Once you do this, you can use subs to substitute values of x into the expression found from solve and plot all of these together. Bear in mind that you will need to cast the result of subs with double because you want the numerical result of the substitution. Not doing this will still leave the answer in MATLAB's symbolic format, and it is incompatible for use when you want to plot the final points on your graph.
Also, what you'll need to do is that given equations like what you have posted above, you may have to loop over each solution, substitute your values of x into each, then add them to the plot.
Something like the following. Here, you also have control over the domain as you have desired:
syms x y;
eqn = solve('y^2 == x^2*(x+3)', 'y'); %// Solve for y, as an expression of x
xval = linspace(-1, 1, 1000);
%// Spawn a blank figure and remember stuff as we throw it in
figure;
hold on;
%// For as many solutions as we have...
for idx = 1 : numel(eqn)
%// Substitute our values of x into each solution
yval = double(subs(eqn(idx), xval));
%// Plot the points
plot(xval, yval);
end
%// Add a grid
grid;
Take special care of how I used solve. I specified y because I want to solve for y, which will give me an expression in terms of x. x is our independent variable, and so this is important. I then specify a grid of x points from -1 to 1 - exactly 1000 points actually. I spawn a blank figure, then for as many solutions to the equation that we have, we determine the output y values for each solution we have given the x values that I made earlier. I then plot these on a graph of these points. Note that I used hold on to add more points with each invocation to plot. If I didn't do this, the figure would refresh itself and only remember the most recent call to plot. You want to put all of the points on here generated from all of the solution. For some neatness, I threw a grid in.
This is what I get:
Ok I was about to write my answer and I just saw that #rayryeng proposed a similar idea (Good job Ray!) but here it goes. The idea is also to use solve to get an expression for y, then convert the symbolic function to an anonymous function and then plot it. The code is general for any number of solutions you get from solve:
clear
clc
close all
syms x y
FunXY = y^2 == x^2*(x+3);
%//Use solve to solve for y.
Y = solve(FunXY,y);
%// Create anonymous functions, stored in a cell array.
NumSol = numel(Y); %// Number of solutions.
G = cell(1,NumSol);
for k = 1:NumSol
G{k} = matlabFunction(Y(k))
end
%// Plot the functions...
figure
hold on
for PlotCounter = 1:NumSol
fplot(G{PlotCounter},[-pi,pi])
end
hold off
The result is the following:
n = 1000;
[x y] = meshgrid(linspace(-3,3,n),linspace(-3,3,n));
z = nan(n,n);
z = (y .^ 2 <= x .^2 .* (x + 3) + .1);
z = z & (y .^ 2 >= x .^2 .* (x + 3) - .1);
contour(x,y,z)
It's probably not what you want, but I it's pretty cool!
I would like to numerically integrate a vector which represents a function f(x) over the range of x specified by bounds x0 and x1 in Matlab. I would like to check that the output of the integration is correct and that it converges.
There are the quad and quadl functions that serve well in identifying the required error tolerance, but they need the input argument to be a function and not the resulting vector of the function. There is also the trapz function where we can enter the two vectors x and f(x), but then it computes the integral of f(x) with respect to x depending on the spacing used by vector x. However, there is no given way using trapz to adjust the tolerance as in quad and quadl and make sure the answer is converging.
The main problem why I can't use quad and quadl functions is that f(x) is the following equation:
f(x) = sum(exp(-1/2 *(x-y))), the summation is over y, where y is a vector of length n and x is an element that is given each time to the function f(x). Therefore, all elements in vector y are subtracted from element x and then the summation over y is calculated to give us the value f(x). This is done for m values of x, where m is not equal to n.
When I use quadl as explained in the Matlab manual, where f(x) is defined in a separate function .m file and then in the main calling file, I use Q = quadl(#f,x0,x1,tolerance,X,Y); here X is a vector of length m and Y is a vector of length L. Matlab gives an error: "??? Error using ==> minus
Matrix dimensions must agree." at the line where I define the function f(x) in the .m function file. f(x) = sum(exp(-1/2 *(x-y)))
I assume the problem is that Matlab treats x and y as vectors that should be of the same length when they are subtracted from each other, whereas what's needed is to subtract the vector Y each time from a single element from the vector X.
Would you please recommend a way to solve this problem and successfully numerically integrate f(x) versus x with a method to control the tolerance?
From the documentationon quad it says:
The function y = fun(x) should accept a vector argument x and return a vector result y, the integrand evaluated at each element of x.
So every time we call the function, we need to evaluate the integrand at each given x.
Also, to parameterize the function call with the constant vector Y, I recommend an anonymous function call. There's a reasonable demo here. Here's how I implemented your problem in Matlab:
function Q = test_num_int(x0,x1,Y)
Q = quad(#(x) myFun(x,Y),x0,x1);
end
function fx = myFun(x,Y)
fy = zeros(size(Y));
fx = zeros(size(x));
for jj=1:length(fx)
for ii=1:length(Y)
fy(ii) = exp(-1/2 *(x(jj)-Y(ii)));
end
fx(jj) = sum(fy);
end
end
Then I called the function and got the following output:
Y = 0:0.1:1;
x0 = 0;
x1 = 1;
Q = test_num_int(x0,x1,Y)
Q =
11.2544
The inputs for the lower and upper bound and the constant array are obviously just dummy values, but the integral converges very quickly, almost immediately. Hope this helps!
I believe the following would also work:
y = randn(10,1);
func = #(x) sum(exp(-1/2 *(x-y)));
integral(func,0,1,'ArrayValued',true)