My goal is to compute the derivative of a function and use this result as another function. Clearly, it means :
f = #(x) (x-1)*(x-2); %A simple function
derivative = jacobian(f,x) %MATLAB output : "2*x - 3"
df = #(x) derivative %= #(x) 2*x - 3
df(2) %= "2*x -3" instead of 2*2 - 3
How can I do such a thing ? I tried syms x but it doesn't help.
You want matlabFunction:
g = matlabFunction(f) converts the symbolic expression or function f to a MATLAB function with handle g.
In your example:
>> syms x
>> f = #(x) (x-1)*(x-2);
>> derivative = jacobian(f,x);
>> df = matlabFunction(derivative)
df =
function_handle with value:
#(x)x.*2.0-3.0
>> df(2)
ans =
1
Related
I am working through an example using fmincon().
I define my objective function in objFun.m
function f=objFun(x)
f = 100*(x(2) - (x(1))^2)^2 + (1 - x(1))^2;
end
and I define an initial point x0
x0=[1; -1]
And if I run the objective function with that point as a test I get
>> objFun(x0)
ans =
400
But when I try to use it in fmincon() I get
>> [x, fval] = fmincon(objFun, x0, [1;2],1,[],[],[0; -inf],[inf, 0]);
Not enough input arguments.
Error in objFun (line 2)
f = 100*(x(2) - (x(1))^2)^2 + (1 - x(1))^2;
I suspect I'm missing something very simple here, but what?
you need to pass a handle to the function #objFun not the function itself, and your A and x0 matrix need to be transposed, ie: a row with 2 columns, each row in A is another linear constraint.
x0=[1, -1];
A = [1,2];
b = 1;
[x, fval] = fmincon(#objFun, x0, A,b,[],[],[0; -inf],[inf; 0]);
function f=objFun(x)
f = 100*(x(2) - (x(1))^2)^2 + (1 - x(1))^2;
end
I want to integrate x^2 from 2 to 4 with the trapezoidal integration method. For this, I defined a function trap that takes 4 arguments:
function y = trap( fn, a, b, h )
n = (b-a)/h;
x = a + [1:n-1]*h;
y = h/2*(feval(fn, a) + feval(fn, b) + 2*sum(feval(fn,x)));
and a function f
function y= f(x)
y=x^2
end
Now, by executing trap(f,2,4,0.1), I get the following error:
Not enough input arguments.
Error in f (line 2)
y=x^2
What is the origin of that error?
You have to call trap using the function handle #f, not f.
trap(#f,2,4,0.1)
function y = trap( fn, a, b, h )
n = (b-a)/h;
x = a + [1:n-1]*h;
y = h/2*(fn(a) + fn(b) + 2*sum(fn(x)));
end
function y= f(x)
y = x.^2;
end
which gives, as expected,
ans =
18.67
Also you needed element-wise multiplication in f(x) to compute y = x.^2.
And feval is not necessary. You can directly call fn(a) to evaluate the function.
I am trying to compute the taylor series of ln(x) for any value of x.
What I have so far is:
clear
clc
n = input('Enter number of iiterations (n): ' );
x = input('enter value of x (x): ');
y = zeros(1,n);
for i = 0:n
y(i+1)=sum + (-1)^(n+1)*(x-1)^n/n;
end
But this code seems to be broken and I can't figure out why. Any suggestions on how to improve?
This is a one liner in addition to the for-loop answer provided by #farbiondriven
For 0<x<1 :
sumLn = #(x, n)(sum(((-1).^(0:n-1)).*((x-1).^(1:n))./(1:n)));
sumLn(0.5,10)
ans =
-0.6931
>> log(0.5)
ans =
-0.6931
For x>0.5 :
sumLn = #(x, n)(sum( ((x-1)/x).^(1:n) ./ (1:n) ));
sumLn(2,10)
ans =
0.6931
log(2) =
0.6931
Note: The variable x in this formula is bounded as mentioned in this link.
Try this:
clear
clc
n = input('Enter number of iterations (n): ' );
x = input('enter value of x with abs value < 1 (x): ');
y = zeros(1,n+1);
y(1)=0;
for i = 1:n
y(i+1)= y(i) + ((-1)^(i+1)*(x-1)^i/i);
end
txt = sprintf('The output is: %f', y(n+1))
I suggest using built-in function and hopefully there is one. taylor(f,var) approximates f with the Taylor series expansion of f up to the fifth order at the point var = 0.
Specify Expansion Point :
Find the Taylor series expansions at x = 1 for these functions. The default expansion point is 0. To specify a different expansion point, use 'ExpansionPoint':
syms x
taylor(log(x), x, 'ExpansionPoint', 1)
ans =
x - (x - 1)^2/2 + (x - 1)^3/3 - (x - 1)^4/4 + (x - 1)^5/5 - 1
Specify Truncation Order :
The default truncation order is 6.
syms x
f = log(x);
t6 = taylor(f, x);
Use 'Order' to control the truncation order. For example, approximate the same expression up to the orders 8.
syms x
taylor(log(x), x, 'ExpansionPoint', 1, 'Order', 8);
I am new to matlab and want to calculate something like f(x)/f'(x). I want the user to input the function f(x), the parameter x and a value of x (suppose 5,so that I can evaluate f(5)/f'(5)) .Please suggest what I should do.
One approach is to use symbolic variables
function [ val ] = func( fun, num )
symfun = sym(fun);
dsymfun = diff(symfun);
y = symfun/dsymfun;
val = subs(y, num);
end
and then call it
e.g.
value = func('x^2', 5)
value =
5/2
Otherwise, you can provide your input as a symbolic variable:
function [ val ] = func( fun, num )
dfun = diff(fun);
y = fun/dfun;
val = subs(y, num);
end
and then write
syms x;
func(x^2, 5)
You can do this using polyder and polyval as follows:
f = input('Enter f(x): '); %e.g; Enter [1 3 4] if f(x)= x^2 + 3*x + 4
df = polyder(f); %f'(x)
x= input('Enter x: '); %Enter the value of 'x' e.g 5
fx_dfx= polyval(f,x)/ polyval(df,x) %f(x)/f'(x)
If you have Symbolic Math Toolbox, you can also do this using:
syms x; %Creating a symbolic variable x
f = input('Enter f(x): '); %Enter f(x) e.g: x^2 + 3*x + 4
f(x)= f; %Converting sym to symfun
df(x) = diff(f) %f'(x)
x_val= input('Enter x: '); %Enter the value of 'x' e.g 5
fx_dfx = double(f(x_val)/df(x_val)) %f(x)/f'(x)
I'm trying build a matlab function that will evaluate a function and vector that are sent in as parameters. I'm having a hard time trying to figure out how to send in the function so that it can be evaluated in the matlab function. I figured out how to do it without the function but I'm a little lost trying to evaluate it within a matlab function. Below is my code...
This is what I'm trying to do...
x = [x1 x2]';
f = x(x1)^2 + 2 * (x2)^2
x = [5 10];
f = (5)^2 + 2 * (10)^2 % which I would like to return 225, not a column vector
This is what I have and what I have tried...
x = [5 10]';
% without using a matlab function
% k = 1
% f = x(k)^2 + 2 * x(k + 1)^2; % returns the correct answer of 225
f = x^2 + 2 * x^2 % complains about the scalar 2
f = x.^2 + 2 * x.^2 % returns a column vector [75; 300]
function [value] = evalFunction(f,x)
value = f(x);
I've tried...
f = #(x) x.^2 + 2 * (x+1).^2;
value = evalFunction(#f,x) %Error: "f" was previously used as a variable
So I tried...
f = #(x) x.^2 + 2 * (x+1).^2;
value = evalFunction(f,x) %value = [97;342]
I'm new to matlab so any help is appreciated. I've been doing some research and found some stuff here on stackoverflow but can't seem to get it to work. I've seen there are other ways to do this, but I will eventually be adding more code to the matlab evalFunction function so I'd like to do it this way. Thanks!
Anonymous functions and function handles plus array indexing. Taking x as a 2-element vector, define and use your function like:
f = #(x) x(1).^2 + 2 * x(2).^2;
value = evalFunction(f,x) % but you can just do f(x) if that is all you need
However, if evalFunction does nothing other than evaluate f at x, then you don't need it at all. Just do f(x).
Alternately,
f = #(x1,x2) x1.^2 + 2*x2.^2;
value = evalFunction(f,x1,x2); % here your function will call it by f(x1,x2)
You are probably coming at this from a C background - in Matlab, x+1 is the entire vector x with 1 added - not the element offset by 1.
The function you need is
f = #(x)x(1).^2 + 2 * (x(2)).^2;
or, to be a little more "matlab-like":
f = #(x) [1 2] * x(1:2)'.^2;
Which performs the element-wise square of the first two elements of x as a column vector, and then does the matrix multiplication with [1 2], resulting in
1 * x(1) .^2 + 2 * x(2) .^2;
Which seems to be what you were asking for.
caveat: did not have opportunity to test this...