It's time to retire and replace IMSHOW Speaking as someone with 31+ years of experience developing and using imshow, I want to advocate for retiring and replacing it. The function imshow has behaviors and defaults that were appropriate for the MATLAB and computer monitors of the 1990s, but ...
plot(x) will replace plot(x,y)! %% x2 = 1:10; y2 = sin(x2); plot(x2,y2); hold on % change mode, add plot ,not replace. default value is replace! plot(x2); % a line. plot(x) will be added! plot(x2+y2);
%I replace my theta with the values that I want y_reduce=(subs(y_reduce,theta,theta_Val)); %I recreate the function %Fonction = @(Var_Search,theta) y_reduce Fonction = matlabFunction(y_reduce) Fonction = function_handle with value: @(e)[0.0,e.*sin(pi./3.6e...
What in your symbolic expression is your subs call actually trying to replace? I would change the name of the variable containing the numeric values you want to substitute into your symbolic expression to something else, like "xx = 5". Then substitute the value of xx for the symbolic ...
syms x y f = x + y; Replace y in this expression with the value 1. Here f itself does not change: subs(f, y, 1); f f = x + y To replace the value of f with the new expression, assign the result returned be subs to f: ...
We can also solve algebraic equations with Matlab. For example, what if we wanted to know the value of x where f1(x) = 3? With the above constants this problem becomes: find x such that In Matlab, we can use the solve command: solve(x^2-2*x-2) or solve(F1 – 3). ...
Expressions such aspi = sym(pi),theta = sym(‘1/10’) etc. create symbolic numbers which enable the program to avoid floating-point approximations. The variable is created as a workspace variable to hold the symbolic value given within the sym() function. It temporarily replaces the name fo...
The problem is, what it found is not terribly useful in terms of a solution. Effectively, in order to find a symbolic solution, it needs to solve for the roots of a degree 9 polynomial. And sadly, this is mathematically impossible to do in an algebraic form. That ...
Suppose I asked you to work with the sequence formed as 2*n*F_n + 1, where F_n is the n'th Fibonacci number? Part of me would not be surprised to find there is nothing simple we could do. But, then it costs nothing to try, to see where MATLAB can take me in an explorative...
It might seem like the expanded version of the equation is correct and produces the same value on substituting k with 1 in both the forms. When we replace k with 2 the values don’t match confirming that they are indeed not equivalent. ...