I have 3 equations:
f = (exp(-x.^2)).*(log(x)).^2
g = exp(-x.^2)
h = (log(x)).^2
The interval is:
x = 0.05:10
I am able to correctly plot the equations but when I try to find an integral, it says that there is an error.
The code I used to find an integral is:
integral(f,0,Inf)
integral(g,0,inf)
integral(h,0,10)
The integrals for f and g are from 0 to infinity and the integral for h is from 0 to 10. None of my code to find integrals works.
You need to define f,g,h as functions like shown below. See documentation of integral(), it takes a function as its first argument. Matlab integral documentation
x = 0.05:10
f = #(x) (exp(-x.^2)).*(log(x)).^2
g = #(x) exp(-x.^2)
h = #(x) (log(x)).^2
integral(f,0,Inf) % 1.9475
integral(g,0,inf) % 0.8862
integral(h,0,10) % 26.9673
h = #(x) (log(x)).^2
This syntax is called anonymous functions, basically they are nameless functions. In above case it takes x as input and returns log(x) squared.
From now on h is a function and it can be used like this.
h(1) % will be equal 0
For more on anonymous functions refer to matlab anonymous functions guide:
Anonymous Functions
Related
I want to make a function from the output of Matlab Hermite function (for example, if we had an output from Hermite function [8 0 -12 0] it would be 8x^3 - 12x polynomial) and then integrate this function using the Simpson's 3/8 Rule.
I have already created a function in Matlab that integrate any function using this rule and also I have created function that returns coefficients of Hermite's polynomial (with the recursion relation) in the vector form.
My questions:
If it's possible, in Hermite function I want from this output [8 0 -12 0] make this output 8x^3 - 12x. This output I will able to integrate. How can I do this?
Can I combine these two functions and integrate Hermite's polynomial without convention the output of the first function?
Code of Hermite polynomial function, where n is the order of the polynomial:
function h = hermite_rec(n)
if( 0==n ), h = 1;
elseif( 1==n ), h = [2 0];
else
h1 = zeros(1,n+1);
h1(1:n) = 2*hermite_rec(n-1);
h2 = zeros(1,n+1);
h2(3:end) = 2*(n-1)*hermite_rec(n-2);
h = h1 - h2;
end
Code of Simpson function, that integrate function using the Simpson 3/8 Rule. a is a lower limit of integral, b is a upper limit of integral:
n = 3;
h = (b-a)/(3*n); %3h = (b-a)/n
IS2=0;
for i=1:n
IS2 = IS2+(f(a+(3*i-3)*h) + 3*f(a+(3*i-2)*h) + 3*f(a+(3*i-1)*h) + f(a+(3*i)*h))*3*h/8;
end
end
Thank you for any advice!
To create a polynomial function given its coefficients, you can use polyval (see also anonynmous functions):
p = [1 2]; % example. This represents the polynomial x+2
f = #(x) polyval(p, x); % anonymous function of x, assigned to function handle f
Now f is a function that you can integrate numerically.
If you want to include this directly as part of your Hermite function, just add something like this at the end:
h = #(x) polyval(p, x);
Then the Hermite function will return a function (handle) representing the Hermite polynomial.
Consider the following code:
f = #(x) x.^2;
Is it possible to get the derivative of the function handle f as another function handle, without defining a symbolic variable?
No, to obtain a derivative function you need to use the Symbolic toolbox.
But you can get an approximation (finite difference approximation) by creating a function as follows:
f = #(x) x.^2;
d = 1e-6;
df = #(x) (f(x+d)-f(x))/d;
d here determines precision of the approximation. If you make it too small, you'll end up in the floating-point rounding error domain, so be careful!
Testing:
x = -2:0.01:2;
max(abs(df(x) - 2*x)) % returns 1.0006e-06
I am trying to integrate a function F which is defined as:
function F
x = -3:0.1:3;
F = zeros(1, length(x));
for i = 1:length(x)
if (1.4<= x(i)) && (x(i) <= 1.6)
F(i) = x(i).^2;
else
F(i) = 2;
end
end
end
but the integral function gives me an error saying that there are too many arguments. I think the problem that the function is defined as a points?
The problem with your function, is that integral has no way to pass the arguments you supply to your function F. The function doesn't know that it can just pull certain elements from the vector you created. If you rewrite your function such that for an input (or x value), the output of F is returned, then integral will work as you need given two values to integrate between.
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
As far as I can tell, no one has asked this.
I've been asked to compute the double integral of a function, and also the same double integral but with the order of integration swapped (i.e: first integrate for dydx, then dxdy). Here is my code:
%Define function to be integrated
f = #(x,y) y^2*cos(x);
%First case. Integration order: dydx
ymin = #(x) cos(x);
I = integral2(f,ymin,1,0,2*pi)
%Second case. Integration order: dxdy
xmin = #(y) asin(y)+2*pi/2;
xmax = #(y) asin(y)-pi/2;
B = integral2(f,xmin,xmax,-1,1)
The error I'm getting is this:
Error using integral2 (line 71)
XMIN must be a floating point scalar.
Error in EngMathsA1Q1c (line 5)
I = integral2(f,ymin,1,0,2*pi)
I'm sure my mistake is something simple, but I've never used Integral2 before and I'm lost for answers. Thank you.
Per the integral2 documentation, the variable limits are given as the second pair of limits. So your first integral should be
% Define function to be integrated
f = #(x,y) y.^2.*cos(x);
% First case. Integration order: dydx
ymin = #(x) cos(x);
I = integral2(f,0,2*pi,ymin,1);
The set of constant limits always goes first, and Matlab assumes the first argument of f is associated with the first set of limits while the second argument of f is associated with the second set of limits, which may be a function of the first argument.
I point out that second part because if you wish to switch the order of integration, you also need to switch the order of the inputs of f accordingly. Consider the following example:
fun = #(x,y) 1./( sqrt(2*x + y) .* (1 + 2*x + y).^2 )
A nice little function that is not symmetric in its arguments (i.e., fun(x,y) ~= fun(y,x)). Let's integrate this over an elongated triangle in the first quadrant with vertices at (0,0), (2,0), and (0,1). Then integrating with dA == dy dx, we have
>> format('long');
>> ymax = #(x) 1 - x/2;
>> q = integral2(fun,0,2,0,ymax)
q =
0.220241017339352
Cool. Now let's integrate with dA == dx dy:
>> xmax = #(y) 2*(1-y);
>> q = integral2(fun,0,1,0,xmax)
q =
0.241956050772765
Oops, that's not equal to the first calculation! That's because fun is defined with x as the first argument and y as the second, but the previous call to integral2 is implying that y is the first argument to fun, and it has constant limits of 0 and 1. How do we fix this? Simply define a new function that flips the arguments:
>> fun2 = #(y,x) fun(x,y);
>> q = integral2(fun2,0,1,0,xmax)
q =
0.220241017706984
And all's right with the world. (Although you may notice small differences between the two correct answers due to the error tolerances of integral2, which can be adjusted via options per the documentation.)
The error states that you can't pass in a function for the limits of integration. You need to specify a scalar value for each limit of integration. Also, there are some errors in the dimensions/operations of the function. Try this:
%Define function to be integrated
f = #(x,y) y.^2.*cos(x);%changed to .^ and .*
%First case. Integration order: dydx
%ymin = #(x) cos(x);
I = integral2(f,-1,1,0,2*pi)%use scalar values for limits of integration
%Second case. Integration order: dxdy
%xmin = #(y) asin(y)+2*pi/2;
%xmax = #(y) asin(y)-pi/2;
B = integral2(f,0,2*pi,-1,1)% same issue, must use scalars