I am trying to integrate x^3/(exp(x)-1) in the limit 0 to infinity with respect to x,and it should answer pi^4/15 but it instead of this ocatve is printing original integral in symbolic form. How to resolve this issue? I tried same integral on MATLAB mobile and it is giving correct pi^4/15
First of all:
pi^4 / 15 = 6.4939
I tried the below code in octave online
fun = #(x) (x.^3)./(exp(x)-1);
q = integral(fun, 0, Inf)
The answer is:
q = 6.4939
Related
Is there any method to find the root of a polynomial, not in matrix form, in MATLAB?
I know, to find roots of a polynomial (say, p(x) = x^.2 - 4), I should do the following:
p = [1 0 -4];
r = roots(p)
What I wanted to know if there is some way to find the root of a function (say p(x) = x^.2 - 4) already present in polynomial form (not in matrix form) in my matlab code? Like anything similar to r = roots(p(x)) (this doesn't work, of course).
Root is good
First of all the solution using roots is probably the one that will give you the most accurate and fastest results if you are indeed working with polynomials. I will acknowledge that it might be an issue if your function is not a polynomial.
Finding the roots of a function
If you don't want to use roots that means you will probably have to represent your polynomial as an anonymous function. Then you can use any root-finding algorithm on that function. Wikipedia has a few of them listed. What is tricky is that in general they don't guarantee that they will find one root, let alone all of them. So you might need as much prior information on your function as you can.
In matlab you can use fzero. The issue with it is that it only finds one zero and that it will only find zeros where the function changes sign (it wouldn't work on p(x) = x² for example). This is how you would implement it:
p = #(x) x.^2 - 4; % Define your polynomial as an anonymous function
x0 = 12; % Initial guess for the zero
% Find a root
fzero(p, x0)
>>> ans = 2
% Now with a different initial guess for a different solution
x0 = -12;
fzero(p, x0)
>>> ans = -2
As you can see this works only if you want to find a root and don't care which one it is.
Problem
The issue is that you polynomials with integer or rational coefficients have a way of finding the roots by using square-free factorization. Yet you can only apply that if you have some way of storing and accessing those coefficients in matlab. The anonymous functions don't allow you to do that. That's why roots works with a matrix and not an anonymous function.
I searched the forum and found this thread, but it does not cover my question
Two ways around -inf
From a Machine Learning class, week 3, I am getting -inf when using log(0), which later turns into an NaN. The NaN results in no answer being given in a sum formula, so no scalar for J (a cost function which is the result of matrix math).
Here is a test of my function
>> sigmoid([-100;0;100])
ans =
3.7201e-44
5.0000e-01
1.0000e+00
This is as expected. but the hypothesis requires ans = 1-sigmoid
>> 1-ans
ans =
1.00000
0.50000
0.00000
and the Log(0) gives -Inf
>> log(ans)
ans =
0.00000
-0.69315
-Inf
-Inf rows do not add to the cost function, but the -Inf carries through to NaN, and I do not get a result. I cannot find any material on -Inf, but am thinking there is a problem with my sigmoid function.
Can you provide any direction?
The typical way to avoid infinity in these cases is to add eps to the operand:
log(ans + eps)
eps is a very, very small value, and won't affect the output for values of ans unless ans is zero:
>> z = [-100;0;100];
>> g = 1 ./ (1+exp(-z));
>> log(1-g + eps)
ans =
0.0000
-0.6931
-36.0437
Adding to the answers here, I really do hope you would provide some more context to your question (in particular, what are you actually trying to do.
I will go out on a limb and guess the context, just in case this is useful. You are probably doing machine learning, and trying to define a cost function based on the negative log likelihood of a model, and then trying to differentiate it to find the point where this cost is at its minimum.
In general for a reasonable model with a useful likelihood that adheres to Cromwell's rule, you shouldn't have these problems, but, in practice it happens. And presumably in the process of trying to calculate a negative log likelihood of a zero probability you get inf, and trying to calculate a differential between two points produces inf / inf = nan.
In this case, this is an 'edge case', and generally in computer science edge cases need to be spotted as exceptional circumstances and dealt with appropriately. The reality is that you can reasonably expect that inf isn't going to be your function's minimum! Therefore, whether you remove it from the calculations, or replace it by a very large number (whether arbitrarily or via machine precision) doesn't really make a difference.
So in practice you can do either of the two things suggested by others here, or even just detect such instances and skip them from the calculation. The practical result should be the same.
-inf means negative infinity. Which is the correct answer because log of (0) is minus infinity by definition.
The easiest thing to do is to check your intermediate results and if the number is below some threshold (like 1e-12) then just set it to that threshold. The answers won't be perfect but they will still be pretty close.
Using the following as the sigmoid function:
function g = sigmoid(z)
g = 1 ./ (1 + e.^-z);
end
Then the following code runs with no issues. Choose the threshold value in the 'max' statement to be less than the expected noise in your measurements and then you're good to go
>> a = sigmoid([-100, 0, 100])
a =
3.7201e-44 5.0000e-01 1.0000e+00
>> b = 1-a
b =
1.00000 0.50000 0.00000
>> c = max(b, 1e-12)
c =
1.0000e+00 5.0000e-01 1.0000e-12
>> d = log(c)
d =
0.00000 -0.69315 -27.63102
I am trying to run fsolve to resolve a dissociation equation but I keep getting this error. I tracked it down to the x(1)^(1/2) term (removing the square root yields no error) but I can't find a way to solve the proper equation I need. The code is below.
T = 2000
Kp = exp(-deltaG/(Ru*T))
function [f]=func(x)
f(1) = 2-x(1)*4 / (3*x(1) - 1)*(x(1))^(1/2) - Kp
endfunction
x0 = [1]
[x,f_x] = fsolve(x0,func)
EDIT: More requested info
The error is
!--error 98 variable returned by scilab argument function is
incorrect
Ru is the gas constant, 8.315.
DeltaG is -135643.
Kp is 3.489e-3.
This is a book example, x should yield 0.3334.
What kind of solved this problem was that I updated scilab to version 6.0.1 from 5.5. The problem is that depending on the initial guess x0 the values of x get really absurd and x0 has to be so close to the real answer that it defeats the purpose of the calculation.
Also I don't have access to Maple, my other alternative would be MATLAB
Using symbolic calculus software like xcas or Maple one can solve your equation symbolycally
There are 3 solutions:
s1=((1/4)*t1+(1/4)*(Kp-2)^2/t1-(1/4)*Kp+1/2)^2
s2=(-(1/8)*t1-(1/8)*(Kp-2)^2/t1-(1/4)*Kp+1/2+(1/2*%i)*sqrt(3)*((1/4)*t1-(1/4)*(Kp-2)^2/t1))^2
s3=(-(1/8)*t1-(1/8)*(Kp-2)^2/t1-(1/4)*Kp+1/2-(1/2*%i)*sqrt(3)*((1/4)*t1-(1/4)*(Kp-2)^2/t1))^2
where
t=sqrt(-Kp*(Kp-4)*(Kp-2)^2)
t1=(-(Kp-2)*(Kp-2*sqrt(2))*(Kp+2*sqrt(2))+4*t)^(1/3);
depending on Kp values some solutions can be complex.
Using Mathematica I need to evaluate the integral of a function. Since it is taking the program too much to compute it, would it be possible to use parallel computation to shorten the time needed? If so, how can I do it?
I uploaded a picture of the integrand function:
I need to integrate it with respect to (x3, y3, x, y) all of them ranging in a certain interval (x3 and y3 from 0 to 1) (x and y from 0 to 100). The parameters (a,b,c...,o) are preventing the NIntegrate function to work. Any suggestions?
If you evaluate this
expr=E^((-(x-y)^4-(x3-y3)^4)/10^4)*
(f x+e x^2+(m+n x)x3-f y-e y^2-(m+n y)y3)*
((378(x-y)^2(f x+e x^2+(m+n x)x3-f y-e y^2-(m+n y)y3))/
(Pi(1/40+Sqrt[((x-y)^2+(x3-y3)^2)^3]))+
(378(x-y)(x3-y3)(h x+g x^2+(o+p x)x3-h y-g y^2-(o+p y)y3))/
(Pi(1/40+Sqrt[((x-y)^2+(x3-y3)^2)^3])))+
(h x+g x^2+(o+p x)x3-h y-g y^2-(o +p y) y3)*
((378(x-y)(x3-y3)(f x+e x^2+(m+n x)x3-f y-e y^2-(m+n y)y3))/
(Pi(1/40+Sqrt[((x-y)^2+(x3-y3)^2)^3]))+
(378 (x3 - y3)^2 (h x + g x^2 + (o + p x)x3-h y-g y^2-(o+p y)y3))/
(Pi(1/40+Sqrt[((x-y)^2+(x3-y3)^2)^3])));
list=List ## Expand[expr]
then you will get a list of 484 expressions, each very similar in form to this
(378*f*h*x^3*x3)/(Pi*(1/40+Sqrt[(x^2+x3^2-2*x*y+y^2-2*x3*y3+y3^2)^3]))
Notice that you can then use NIntegrate in this way
f*h*NIntegrate[(378*x^3*x3)/(Pi*(1/40+Sqrt[(x^2+x3^2-2*x*y+y^2-2*x3*y3+y3^2)^3])),
{x,0,100},{y,0,100},{x3,0,1},{y3,0,1}]
but it gives warnings and errors about the convergence and accuracy, almost certainly due to your fractional powers in the denominator.
If you can find a way to pull out the scalar multipliers which are independent of x,y,x3,y3 and then perform that integration without warnings and errors and get an accurate result which isn't infinity then you could perhaps perform these integrals in parallel and total the results.
Some of the integrands are scalar multiples of others and if you combine similar integrands then you can reduce this down to 300 unique integrands.
I doubt this is going to lead to an acceptable solution for you.
Please check all this very carefully to make certain that no mistakes have been made.
EDIT
Since the variables that are independent of the integration appear to be easily separated from the dependent variables in the problem posed above, I think this will allow parallel NIntegrate
independentvars[z_] := (z/(z//.{e->1, f->1, g->1, h->1, m->1, n->1, o->1, p->1}))*
NIntegrate[(z//.{e->1, f->1, g->1, h->1, m->1, n->1, o->1, p->1}),
{x, 0, 100}, {y, 0, 100}, {x3, 0, 1}, {y3, 0, 1}]
Total[ParallelMap[independentvars, list]]
As I mentioned previously, the fractional powers in the denominator result in a flood of warnings and errors about convergence failing.
You can test this with the following much simpler example
expr = f x + f g x3 + o^2 x x3;
list = List ## Expand[expr];
Total[ParallelMap[independentvars, list]]
which instantly returns
500000. f + 5000. f g + 250000. o^2
This is a very primitive method of pulling independent symbolic variables outside an NIntegrate. This gives absolutely no warning if one of the integrands is not in a form where this primitive attempt at extraction is not appropriate or fails.
There may be a far better method that someone else has written out there somewhere. If someone could show a far better method of doing this then I would appreciate it.
It might be nice if Wolfram would consider incorporating something like this into NIntegrate itself.
Keep getting the error Arguments are not sufficiently instantiated for the multiplication by addition rule I wrote as shown below.
mult(_, 0, 0). %base case for multiplying by 0
mult(X, 1, X). % another base case
mult(X, Y, Z) :-
Y > 1,
Y1 is Y - 1,
mult(X, Y1, Z1),
Z is X + Z1.
I am new to Prolog and really struggling with even such simple problems.
Any recommendations for books or online tutorials would be great.
I am running it on SWI-Prolog on Ubuntu Linux.
In your definition of mult/3 the first two arguments have to be known. If one of them is still a variable, an instantiation error will occur. Eg. mult(2, X, 6) will yield an instantiation error, although X = 3 is a correct answer ; in fact, the only answer.
There are several options you have:
successor-arithmetics, constraints, or meta-logical predicates.
Here is a starting point with successor arithmetics:
add(0,Y,Y).
add(s(X),Y,s(Z)) :- add(X,Y,Z).
Another approach would be to use constraints over the integers. YAP and SWI have a library(clpfd) that can be used in a very flexible manner: Both for regular integer computations and the more general constraints. Of course, multiplication is already predefined:
?- A * B #= C.
A*B#=C.
?- A * B #= C, C = 6.
C = 6, A in -6.. -1\/1..6, A*B#=6, B in -6.. -1\/1..6.
?- A * B #= C, C = 6, A = 2.
A = 2, B = 3, C = 6.
Meta-logical predicates: I cannot recommend this option in which you would use var/1, nonvar/1, ground/1 to distinguish various cases and handle them differently. This is so error prone that I have rarely seen a correct program using them. In fact, even very well known textbooks contain serious errors!
I think you got the last two calls reversed. Don't you mean:
mult(X,Y,Z):- Y>1,Y1 is Y-1, Z1 is X+Z, mult(X,Y1,Z1).
Edit: nevermind that, looked at the code again and it doesn't make sense. I believe your original code is correct.
As for why that error is occuring, I need to know how you're calling the predicate. Can you give an example input?
The correct way of calling your predicate is mult(+X, +Y, ?Z):
?- mult(5,0,X).
X = 0
?- mult(5,1,X).
X = 5
?- mult(5,5,X).
X = 25
?- mult(4,4,16).
yes
?- mult(3,3,10).
no
etc. Calling it with a free variable in the first two arguments will produce that error, because one of them will be used in the right side of an is or in either side of the <, and those predicates expect ground terms to succeed.