I'm trying to solve a set of symbolic equations in maxima.
I'm using "algsys" comand, and it works, but the answer is depended on "%r1".
I dont understand what %r1 is, and how can I get a explicit expression?
maxima code:
(%i4) eq1:diff(diff(Lag,diff(u(t),t)),t)-diff(Lag,u(t)),ratsimp;
(%i5) eq2:diff(diff(Lag,diff(v(t),t)),t)-diff(Lag,v(t)),ratsimp;
(%i6) eq3:diff(diff(Lag,diff(theta(t),t)),t)-diff(Lag,theta(t))-Tm,ratsimp;
(%i7) A:algsys([eq1, eq2, eq3], [diff(u(t),t,2),diff(v(t),t,2),diff(theta(t),t,2)]);
PS: cant post picture cause it's my first question...
FWIW %r1 is simply names for roots introduced by solve and algsys.
See http://maxima.sourceforge.net/docs/manual/en/maxima_20.html
Related
Introduction
I'm doing research in computational contact mechanics, in which I try to solve a PDE using a finite difference method. Long story short, I need to solve a linear system like Ax = b.
The suspects
In the problem, the matrix A is sparse, and so I defined it accordingly. On the other hand, both x and b are dense arrays.
In fact, x is defined as x = A\b, the potential solution of the problem.
So, the least one might expect from this solution is to satisfy that Ax is close to b in some sense. Great is my surprise when I find that
julia> norm(A*x-b) # Frobenius or 2-norm
5018.901093242197
The vector x does not solve the system! I've tried a lot of tricks discover what is going on, but no clues as of now. My first candidate is that I've found a bug, however I need more evidence to make this assertion.
The hints
Here are some tests that I've done to try to pinpoint the error
If you convert A to dense, the solution changes completely, and in fact it returns the correct solution.
I have repeated the proccess above in matlab, and it seems to work well with both sparse and dense matrices (that is, the sparse version does not agree with that of Julia's)
Not all sparse matrices cause a problem. I have tried other initial conditions and the solver seems to work quite well. I am not able to predict what property of the matrix can be causing this discrepancy. However;
A has a condition number of 120848.06, which is quite high, although matlab doesn't seem to complain. Also, the absolute error of the solution to the real solution is huge.
How to reproduce this "bug"
Download the .csv files in the following link
Run the following code in the folder of the files (install the packages if necessary
using DelimitedFiles, LinearAlgebra, SparseArrays;
A = readdlm("A.csv", ',');
b = readdlm("b.csv", ',');
x = readdlm("x.csv", ',');
A_sparse = sparse(A);
println(norm(A_sparse\b - x)); # You should get something close to zero, x is the solution of the sparse implementation
println(norm(A_sparse*x - b)); # You should get something not close to zero, something is not working!
Final words
It might easily be the case that I'm missing something. Are there any other implementations apart from the usual A\b to test against?
To solve a sparse square system Julia chooses to do a sparse LU decomposition. For the specific matrix A in the question, this decomposition is numerically ill-conditioned. This is evidenced by the cond(lu(A_sparse).U) == 2.879548971708896e64. This causes the solve routine to make numerical errors in turn.
A quick solution is to use a QR decomposition instead, by running x = qr(A_sparse)\b.
The solve or LU routines might need to be fixed to handle this case, or at least maintainers need to know of this issue, so opening an issue on the github repo might be good.
(this is a rewrite of my comment on question)
I am trying to integrate a Heaviside theta function with two signs inside and Mathematica won't give me a solution. Is there any way of improving the approach before just acknowledging that Mathematica cannot integrate it?
Some things you do really worry me.
You use X as a variable and as the name of a function. I've changed those to X and Xf.
You use ω as a variable and as the name of a function. I've changed those to ω and ωf.
You use = and not := in your function definitions. I've changed that.
With those changes I have
Xf[s1_,s2_,α_]:=(1+s2(-2+α)-α+s1(-1+2s2+α))/((-1+2 s1)(-1+2s2));
ωf[s1_,s2_,α_]:=((1-s2+s1(-3+4s2))(1+s2(-2+α)-α+s1(-1+2s2+α)))/((-1+2s1)(-1+s1+s2)(-1+2s2));
λ1[s1_,s2_,α_,X_,ω_]:=1/2(s1-2s2-3s1 X+3s2 X-α+s1 α+s2 α+ω-s1 ω-s2 ω-
1/2Sqrt[(4(α-ω+s2(2-3X-α-ω)+s1(-1+3X-α+ω))^2+8(2-5X+4X^2-2α+2X α+
s1(-2+4s2(1-2X)^2+7X-8X^2+2α-2X α-ω)+ω-s2(4-11X+8X^2-2α+2X α+ω)))]);
λ2[s1_,s2_,α_,X_,ω_]:=1/2(s1-2s2-3s1 X+3s2 X-α+s1 α+s2 α+ω-s1 ω-s2 ω+
1/2Sqrt[(4(α-ω+s2(2-3X-α-ω)+s1(-1+3X-α+ω))^2+8(2-5X+4X^2-2α+2X α+
s1(-2+4s2(1-2X)^2+7X-8X^2+2α-2X α-ω)+ω-s2(4-11X+8X^2-2α+2X α+ω)))]);
λ1simp[s1_,s2_,α_]:=λ1[s1,s2,α,Xf[s1,s2,α],ωf[s1,s2,α]];
λ2simp[s1_,s2_,α_]:=λ2[s1,s2,α,Xf[s1,s2,α],ωf[s1,s2,α]];
fint[s1_,s2_]:=HeavisideTheta[Sign[-λ1simp[s1,s2,α]]*Sign[-λ2simp[s1,s2,α]]];
Please check that very carefully to see if I've made any mistakes.
Now I want to look at your integrand and see what Mathematica sees.
Simplify[fint[s1,s2],1/2<=s1<=1&&1/2<=s2<=s1]
And it responds with
HeavisideTheta[CompexInfinity] 2s1==1||2s2==1||s1+s2==1
HeavisideTheta[True Sign[...]*Sign[...]]
so it looks like your integrand is blowing up at the boundary.
I check that with
Simplify[fint[1/2,s2]]
or
Simplify[fint[s1,1/2]]
and it responds with 1/0 and Indeterminate and HeavisideTheta[Indeterminate]
When it isn't at the boundary, for example
Simplify[fint[3/4,3/4]]
it returns
HeavisideTheta[Sign[5-4*Sqrt[7-28*α+25*α^2]]*Sign[5+4*Sqrt[7-28*α+25*α^2]]]
and that probably says that α is a free variable and we aren't able to determine the value of the Sign without more information.
So I think this is a strong hint where I would begin looking for why your integration is not simply completing.
If you are curious what that integrand looks like then try
α=1/4;
Plot3D[fint[s1,s2],{s1,1/2,1},{s2,1/2,s1}]
Sage seems to want to evaluate derivatives as far as possible using the chain rule. A simple example is:
var('theta')
f = function('f')(theta)
g = function('g')(theta)
h = f*g
diff(h,theta)
which would display
g(theta)*diff(f(theta), theta) + f(theta)*diff(g(theta), theta)
My question is, is there a way to control just how far Sage will take derivatives? In the example above for instance, how would I get Sage to display instead:
diff(f(theta)*g(theta))
I'm working through some pretty intensive derivations in fluid mechanics, and being able to not evaluate derivatives all the way like discussed above would really help with this. Thanks in advance. Would appreciate any help on this.
This would be called "holding" the derivative.
Adding this possibility to Sage has already been considered.
Progress on this is tracked at:
Sage Trac ticket 24861
and the ticket even links to a branch with code implementing this.
Although progress on this is stalled, and the branch has not been merged,
you could use the code from the branch.
I have a major problem: I am an absolute beginner in R programming :) using libaray PMCMRplus I want to perform a dunnetTest. However, I cant change the methods to adjust the p value. Whatever command I've used I will allways get a single-stept adjustment. Can please anyone help me?? The commant I am using is:
dunnettTest(v~Gruppe, alternative="greater", p.adjust.methods ="BH")
Maybe a little bit late but three inputs to your question that may help:
1.) For v~Gruppe have you specified what content should be used? You can define this by using $ such as v~Gruppe$columnofyourdataframe
2.) Also for v~Gruppe: according to the PMCMRplus documentation you have to use dunnettTest(x, g). You use v instead of x and did not define g.
3.) For alternative="greater" have you tried to further specify by using alternative = c("two.sided", "greater", "less")? This is also explained in the PMCMRplus documentation.
I need to know how to write a function to solve a simple linear equation like 2x +1 = 5. How would one do this? If anyone can show some code or point me to a site, it would be much appreciated.
How is the equation entered? In text-form? As coefficients?
ax + b = c of course has the solution x = (c - b) / a.
For parsing, you could use regular expressions.
Parse the expression into a simple tree, then use basic math to solve it. I'd start by converting it to postfix notation and then evaluating that.