I have the following code:
for n:1 thru 11 do for j:1 thru 21 do v[n,j]:1/sqrt(dp)*
(sum(eigenfunctionsort[n,j]*exp(%i*2*%pi*m*x/dp),m,-10,10));
Where eigenfunctionsort is defined earlier,x is a variable I will integrate over later and I am summing over m.
When I print say v[1,1], I get a big long nasty equation. How can i have Maxima boil this down in to something meanigful so I can check my results.
Best,
Ben
Try the 'trigsimp' function or maybe map(trigsimp, your_expression). Not sure it will help, but it's worth a try. Also look at 'demoivre'.
I don't know what your vector eigenfunctionsort or your "big long nasty equation" looks like, but I often get complex eigenvalues and eigenvectors from Maxima even when I know they should be simple and real.
For example,
(%i1) A : matrix([1, 4, 1], [4, 1, 9], [1, 9, 1]);
(%i2) eigenvalues(A);
makes a mess. It can be simplified by applying rectform to transform the output to Cartesian form, followed by trigreduce to reduce the imaginary part of the result. Finally you can convert the
result into floating point:
(%i3) rectform(%)$
(%i4) trigreduce(%)$
(%i5) float(%);
Related
I'm trying to obtain a numerical solution to the following integral:
1
The correct answer is -0.324 + 0.382i but as seen below I am not getting a numerical answer and would appreciate help with the Maxima syntax.
2
Perhaps related to why I am not getting a numerical output are two specific questions:
I read that e and i in Maxima need to be preceded by % in input but should these also appear as %e and %i as seen in the Maxima output?
Why is dy missing at the end of the integral in the Maxima output?
Thank you!
Looks to me like your input is okay, however, the function to compute approximations to integrals is named quad_qags. (There are actually several related functions. See ?? quad_ for more info.) Also, a wrinkle here is that the integrand is a complex-valued function (of a real variable), and quad_qags can only work on real-valued integrands, so we'll have to work around it. Here's how I would arrange it.
myintegrand: exp(%i*(1 + %i*y))/(1 + %i*y + 1/(1 + %i*y));
result_realpart: quad_qags (realpart (myintegrand), y, 0, 6);
result_imagpart: quad_qags (imagpart (myintegrand), y, 0, 6);
result: result_realpart[1] + %i*result_imagpart[1];
I get 0.3243496676292901*%i + 0.3820529930785175 as the final result. That's a little different from what you said; maybe a minus sign went missing? or there's a missing or extra factor of %i?
A quick approximation
0.1 * lsum (x, x, float (rectform (makelist (ev (myintegrand, y = k/10), k, 0, 60))));
seems to show the result from quad_qags is reasonable.
(I know a question like this exists, but I wanted help with a specific example)
If the linear filter has even dimensions, how is the "center" defined? i.e. in the following scenario:
filter = np.array([[a, b],
[c, d]])
and the image was:
image = np.array([[0, 1, 0],
[1, 1, 1],
[0, 1, 0]])
what would be the result of correlation of the image with the linear filter?
Which of the elements of the even-sized filter is considered the origin is an arbitrary choice. Each implementation will make a different choice. Though a and d are the two most likely choices for reasons of similarity of the two image dimensions.
For example, MATLAB's imfilter (which implements correlation, not convolution) does the following:
f = [1,2;4,8];
img = [0,1,0;1,1,1;0,1,0];
imfilter(img,f,'same')
ans =
14 13 4
11 7 1
2 1 0
meaning that a is the origin of the kernel in this case. Other implementations might make a different choice.
Maxima does not seem to come up with an analytic solution to this equation which includes the error function. The independent variable here is "p" and the dependent variable to be solved for is "x".
see an illustration of equation follow link
(%i3) solveexplicit:true$ ratprint:false$ fpprintprec:6$
(%i4) eqn: (sqrt(%pi)*(25*2^(3/2)*p-25*sqrt(2))*erf(1/(25*2^(3/2)*x))*x+1)/(25*p) = 0.04;
(%i5) solve (eqn, x);
(%o5) []
(%i6) eqn, [p=2,x=0.00532014],numer;
(%o6) 0.04=0.04
Any help or pointing in the right direction is appreciated.
As far as I know, Maxima can't solve equations containing erf. You can get a numerical result via find_root:
(%i5) find_root (eqn, x, 0.001, 0.999), p=2;
(%o5) 0.005320136894034347
As for symbolic solutions, I worked with the equation a little bit. One can get it into the form erf(something/x)*x = otherstuff, or equivalently erf(y) = somethingelse*y where y = something/x and somethingelse = otherstuff/something if I'm not mistaken. I don't know anything in particular about equations of that form, but perhaps you can find something.
Yes, solve can only do polynominals. I used the series expansion for small values of x and the accuracy is good enough.
(%i11) seriesE: 1$
termE: erf(x)$
for p: 1 unless p > 3 do
(termE: diff (termE, x)/p,
seriesE: seriesE + subst (x=0, termE)*x^p)$
seriesE;
(%o11) -(2*x^3)/(3*sqrt(%pi))+(2*x)/sqrt(%pi)+1
However, the "Expression longer than allowed by the configuration setting!"
I'm trying to solve system of equations with roots in maxima, for example:
solve([sqrt(x) = 1, y = 1], [x,y]);
But maxima says that this system has no solutions. On the other hand, maxima is able to solve this equation:
solve([sqrt(x) = 1], [x]);
Can I solve systems like the above in maxima?
The built-in solve has serious limitations. The add-on function to_poly_solve can solve equations which contain radicals; I don't know what its limitations are.
(%i2) load (to_poly_solve);
(%o2) /usr/local/share/maxima/5.40.0/share/to_poly_solve/to_poly_solve.mac
(%i3) to_poly_solve ([sqrt(x) = 1, y = 1], [x,y]);
(%o3) %union([x = 1, y = 1])
%union means a union of solutions. Since there is only one solution found, %union could be simplified away; its presence is perhaps a little inconvenient, but not incorrect.
Hello i am trying to solve a algebric equation in maxima, the equation has alpha, delta and psi as variable. I want the alpha in equation to be solved in terms of psi and delta. I tried using the solve command but i am getting alpha in terms of alpha.
Here is the equation to solve Equation to solve
And this is the output from maxima
.
This is the code i am trying -->
solve([(sqrt(-4*alpha*delta*psi-4*delta*psi+alpha^2*delta^2)/(delta^2+delta)-(alpha*delta/(delta^2+delta)))/2-sqrt(4*alpha^2*delta^2+6*alpha*delta^2+3*delta^2+2*alpha^2*delta+4*alpha*delta+2*delta)/(3*delta^2+2*delta)+alpha*delta/(3*delta^2+2*delta)=0],alpha);
Thank you
The problem with that algebraic equation is that involves square roots,or radicals and normal polynomial, and that that type of equation is not easy to solve, take a look at this equation:
(%i30) solve(x=sqrt(x+6),x);
(%o30) x = sqrt{x+6}
So Maxima doesn't return any value, but for example other software Mathematica does.
Let's square both sides of equation and try to solve it
(%i31) solve(x^2=x+6,x);
(%o31) x=3 , x=-2
we get two solutions, let's try with first equation:
3 = sqrt(3+6) => 3 = sqrt(9) => 3 = 3
-2 = sqrt(-2+6) => -2 = sqrt(4) => -2 = 2 ??????
So the second solution is not valid,
maxima solve program in Macsyma/Maxima generally avoids methods that
produce false solutions, like "square both sides". It may still
make errors based on dividing by expressions that appear to be
non-zero, but actually ARE zero, and maybe some other similar
situations.
from this mailing list
In your case I will factor the equation to get a simplified version but with those free variable this will be difficult so, try to assume some values for psi and delta:
(%i26) solve(factor((sqrt(-4*alpha*delta*psi-4*delta*psi+alpha^2*delta^2)/(delta^2+delta)-(alpha*delta/(delta^2+delta)))/2-sqrt(4*alpha^2*delta^2+6*alpha*delta^2+3*delta^2+2*alpha^2*delta+4*alpha*delta+2*delta)/(3*delta^2+2*delta)+alpha*delta/(3*delta^2+2*delta))=0,alpha);
(\%o26) \left[ \alpha=\ifrac{\left(3\,\delta+2\right)\,\isqrt{\left(-4\,\alpha-4\right)\,\delta\,\psi+\alpha^2\,\delta^2}+\left(-2\,\delta-2\right)\,\isqrt{\left(4\,\alpha^2+6\,\alpha+3\right)\,\delta^2+\left(2\,\alpha^2+4\,\alpha+2\right)\,\delta}}{\delta^2} \right]
Expand your Equation and try to remove square roots or some assumptions:
Equation : (sqrt(-4*alpha*delta*psi-4*delta*psi+alpha^2*delta^2)/(delta^2+delta)-(alpha*delta/(delta^2+delta)))/2-sqrt(4*alpha^2*delta^2+6*alpha*delta^2+3*delta^2+2*alpha^2*delta+4*alpha*delta+2*delta)/(3*delta^2+2*delta)+alpha*delta/(3*delta^2+2*delta);