How to solve the function f(x)=ln(x^2)-0.7=0 with a known Matlab command?
clc;clear all;close all;
f(x)=ln(x^2)-0.7=0
B=sqrt f(x)
You can use symbolic variables together with the solve function:
syms x;
eqn = log(x^2) - 0.7 == 0;
solve(eqn,x)
The above code will output:
ans =
exp(7/20)
-exp(7/20)
Since the equation is quadratic, the solver returns two distinct solutions (often people forget that quadratic equations may have two specular solutions, one positive and one negative).
If you want to retrieve the numerical values (for example, in order to calculate their sqrt value):
sol = solve(eqn,x);
num = double(sol)
num =
1.4191
-1.4191
Put the following code into a MATLAB script, name it "main.m".
function b=main
clc
x=solveF()
y=f(x)
b=sqrt(y)
end
function y=f(x)
y=log(x^2)-0.7
end
function x=solveF()
g = #(x) abs(f(x)-0)
x = fminsearch(g, 1.0)
end
Then run it as:
main
You will get the results:
x =
1.4190
y =
-3.4643e-05
b =
0.0000 + 0.0059i
ans =
0.0000 + 0.0059i
You can define equations in matlab as such:
f = #(x) log(x^2)-0.7;
B = #(x) sqrt(f(x));
If you want to find the value of x satisfying a constraint you can design a function that will be equal to zero when the constraint is respecte, then call fminsearch to find x:
f_constraint = #(x) abs(f(x)-0);
x_opt = fminsearch(f_constraint, 1.3); % function handle, initial estimate
In your example, B(x_opt) should be equal to zero. This is not exactly the case as fminsearch estimated a solution.
Related
I have such an question, And I will do it in matlab. But, I get some errors:
Find the value of x ∈ [0, 1] that minimizes the largest eigenvalue of the matrix A(x) = xM +(1−x)P, where M is a 5×5 magic square and P is a 5 × 5 Pascal matrix.
My matlab code:
%Define Matrices
M = magic(5);
P = pascal (5);
% Define the variable x
syms x
%Define the given matrix A
>> A = x*M + (1-x)*P;
%Define the eigenvalue lambda as y;
syms y
%Find determinant of |A - lambda * I|
D = det (A - y*eye(5))
%Define Objective function
objective = #(y) y
%And Define the constraint
constraint = #(x,y) (-1)*D
%initial value x0 = (0:0.001:1);
%Minimization problem solving
x = fmincon(objective, constraint, x0)
I get this error;
Error using fmincon (line 221)
FMINCON requires the following inputs to be of data type double: 'X0'.
Or If I use another function: fminsearch
x = fminsearch(objective, constraint, x0)
In this case I get the following error:
Error using fminsearch (line 96)
FMINSEARCH accepts inputs only of data type double.
How can I deal with these errors ? Where is my mistake? How can I correct them?
I guess what you are looking for might be fminbnd, which helps to
Find minimum of single-variable function on fixed interval
n = 5;
M = magic(n);
P = pascal(n);
x = fminbnd(#(x) max(eig(x*M + (1-x)*P)),0,1);
such that
>> x
x = 0.79603
I suspect you did not show us the proper code, as you have sums there. I suspect you mean syms.
fmincon only works for numeric data, not symbolic data.
I am trying to calculate the integral of a function in Matlab and Mathematica that the software cannot do symbolically.
Here is my MatLab code so far, but I understand it may not be very helpful as is.
f = #(t) asin(0.5*sin(t));
a = #(t) sin(t);
F = int(f,t) % Matlab can't do this
F =
int(asin(sin(t)/2), t)
A = int(a,t) % This works
A =
-cos(t)
dt = 1/(N-1); % some small number
for i=1:N
F(i) = integral(f,(i-1)*dt,i*dt);
A(i) = integral(a,(i-1)*dt,i*dt);
end
Both of the calculations in the for loop give a rough approximation of f or a not their integrals after multiplying by dt.
On the math stack-exchange I found a question that derives a finite difference like method for the integral at a point. However, when I did the calculation in Matlab it output a scaled down version of f which was evident after plotting (see above for what I mean by scaled down). I think that's because for smaller intervals the integral basically approximates the function to varying degrees of accuracy (again see above).
I am trying to get either a symbolic equation for the integral, or an approximation of the integral of the function at each location.
So my question is then if I have a function f that MatLab and Mathematica cannot easily take the integral of
can I approximate the integral directly with an integral calculator besides the default ones? (int,integral,trapz)
or
can I approximate the function with finite differences first and then evaluate the integral symbolically?
Your code is nearly fine it's just that
for i=1:N
F(i) = integral(f,0,i*dt);
end
You could also do
F(1)=integral(f,0,dt)
for i=2:N
F(i) = F(i-1)+integral(f,(i-1)*dt,i*dt);
end
Second option is surely more efficient
Because the primitive is really F(x)=int(f(x), 0, x) (0 defines a certain constant ) and for sufficiently small dx you have shown that f(x)=int(f(x), x,x+dx)/dx i. You have proven that MATLAB intégral function does its job.
For example let's take = the function above will compute if you wish to compute just replace 0 above by the constant a you like.
now and so you should get F containing a discretization of
The accepted answer in general is by far the best method I would say but if certain restrictions on your functions are allowable then there is a second method.
For two functions f and g see below
T = 1; % Period
NT = 1; % Number of periods
dt = 0.01; % time interval
time = 0:dt:NT*T; % time
syms t
x = K*sin(2*pi*t+B); % edit as appropriate
% f = A/tanh(K)*tanh(K*sin(2*pi*t+p))
% g = A/asin(K)*asin(K*sin(2*pi*t+p))
formulas found here
f = A1/tanh(K1)*(2^(2*1)-1)*2^(2*1)*bernoulli(2*1)/factorial(2*1)*x^(2*1-1);
% |K1|<pi/2
g = A2/asin(K2)*factorial(2*0)/(2^(2*0)*factorial(0)^2*(2*0+1))*x^(2*0+1);
% |K2|<1
there are no such limitations in the accepted answer
N = 60;
for k=2:N
a1 = (2^(2*k)-1)*2^(2*k)*bernoulli(2*k)/factorial(2*k);
f = f + A1/tanh(K1)*a1*x^(2*k-1);
a2 = factorial(2*k)/(2^(2*k)*factorial(k)^2*(2*k+1));
g = g + A2/asin(K2)*a*x^(2*k+1);
end
MATLAB can calculate sin^n(t) for n being an integer.
F = int(f,t);
phi = double(subs(F,t,time));
G = int(g,t);
psi = double(subs(G,t,time));
Consider the following MWE in MATLAB:
f = #(t) integral(#(x) x.^2,0,t);
integral(f,0,1);
This yields the error
Error using integral (line 85) A and B must be floating-point scalars.
(and a bit more). How do I fix this? Is this even possible? I think the problem is the variable upper bound.
If you want to use integral then set 'ArrayValued' to true otherwise t would be an invalid end point in integral(#(x) x.^2,0,t). So it would be:
f = #(t) integral(#(x) x.^2,0,t);
integral(f,0,1,'ArrayValued',true)
% ans =
% 0.0833
Alternately, since you're doing double integration, so use the function dedicated for this purpose i.e. integral2. For your example, it would be:
f = #(t,x) x.^2 ;
integral2(f,0,1,0, #(t) t)
% ans =
% 0.0833
If you have Symbolic Math Toolbox, you can also use int as int(expr,var,a,b) but it would be slower. For your case, it would be:
syms x t;
f = x.^2;
req = int(int(f,x,0,t),t,0,1); % It gives 1/12
req = double(req); % Convert to double if required
Here is the equation in WolframAlpha returning me the correct answer.
In MATLAB, I've written the following:
mu = 305; %as a temporary example since it's greater than c, which is 300
syms x
eqn = ((1 + (x/(mu + 300)))^((1/2) + (150/mu)))*((1 - (x/(mu - 300)))^((1/2) - (150/mu))) - 0.2 == 0 %matlab shows the answer is -605
solve(eqn,x)
It's the same equation, except MATLAB substitutes for mu for me. MATLAB is returning the following:
eqn = logical 0
ans = x
Am I typing the equation in wrong somehow? Is that why it's showing me a logical zero when I'm not suppressing the equation? How do I get it to result in the same values as WolframAlpha?
I also want to note that Maple seems to hang on this same equation as well.
Unless you have a specific reason for using symbolic expressions, you can solve the equation that you have using fsolve as follows:
%Define your value of mu
mu = 305;
% Define the equation as an anonymous function
fn = #(x) ((1 + (x/(mu + 300)))^((1/2) + (150/mu)))*((1 - (x/(mu - 300)))^((1/2) - (150/mu))) - 0.2;
% Define the initial value for x so that fsolve can find the root nearest to that
x0 = 1;
root_x = fsolve(fn, x0);
This leads to the output root_x = 5.0000 + 0.0000i
You can also change the initial value of x0
x0 = 400;
root_x = fsolve(fn, x0);
This will return the output root_x = -4.9005e+02 - 2.6326e-05i
This method can be used to solve any of the equations that you might have.
I have a Matlab function that returns a polynomial of the form:
poly = ax^2 + bx*y + cy^2
where a, b, and c are constants, and x and y are symbolic (class sym).
I want to get the coefficients of the polynomial in the form [a b c], but I'm running into the following problem. If the function returns poly = y^2, then coeffs(poly) = 1. I don't want this – I want it to return [0 0 1].
How can I create a function that will give me the coefficients of a symbolic polynomial in the form that I want?
You can use sym2poly if your polynomial is a function of a single variable like your example y^2:
syms y
p = 2*y^2+3*y+4;
c = sym2poly(p)
which returns
c =
2 3 4
Use fliplr(c) if you really want the coefficients in the other order. If you're going to be working with polynomials it would probably also be a good idea not to create a variable called poly, which is the name of a function you might want to use.
If you actually need to handle polynomials in multiple variables, you can use MuPAD functions from within Matlab. Here is how you can use MuPAD's coeff to get the coefficients in terms of the order of variable they precede (x or y):
syms x y
p = 2*x^2+3*x*y+4*y;
v = symvar(p);
c = eval(feval(symengine,'coeff',p,v))
If you want to extract all of the information from the polynomial, the poly2list function is quite helpful:
syms x y
p = 2*x^2+3*x*y+4*y;
v = symvar(p);
m = eval(feval(symengine,'poly2list',p,v));
c = m(:,1); % Coefficients
degs = m(:,2:end); % Degree of each variable in each term
The polynomial can then be reconstructed via:
sum(c.*prod(repmat(v,[size(m,1) 1]).^degs,2))
By the way, good choice on where you go to school. :-)
There is also an alternative to this problem. For a given degree, this function returns a polynomial of that degree and its coefficients altogether.
function [polynomial, coefficeint] = makePoly(degree)
syms x y
previous = 0 ;
for i=0:degree
current = expand((x+y)^i);
previous= current + previous ;
end
[~,poly] = coeffs(previous);
for j= 1:length(poly)
coefficeint(j) = sym(strcat('a', int2str(j)) );
end
polynomial = fliplr(coefficeint)* poly.' ;
end
Hope this helps.