User Portlet User Portlet

Tim Laska
Discussions
Hi Roberta, The expression may have been rewritten, but you can verify that the are equivalent by using Simplify. Take the right hand side of eqn 1 and compare the two cases for example: Simplify[-0.001*a[t] + 2.5*^-8*a[t]*(1.2*^6 - a[t]...
You can complete the square to verify that you have a circle of radius $\sqrt{5}$ centered at (3,1). x^2 + y^2 - 6 x - 2 y + 5 == (x - 3)^2 + (y - 1)^2 - 5 // Simplify (* True *) There are two solutions for $k$. You can use the...
Your transformation matrix operates an a single vertex. You will want to transpose q so that it is a list of vertices. You will then want to use [Map (/@) ](https://reference.wolfram.com/language/ref/Map.html) to map your transformation across all...
Thank you Gianluca! It is always good to know about the alternatives. It appears that there is some touchiness when using Circular objects within GeometricScene. If you want to use hypotheses like area and perimeter relations, you need to use...
Setting up the derivative as an anonymous function might help. newD = D[#, {{t, z}}] &; newD@f[t u/L , z/L] (*{(u*Derivative[1,0][f][(t*u)/L,z/L])/L,Derivative[0,1][f][(t*u)/L,z/\ L]/L}*)
Perhaps this [Vectors and Matrices Tutorial][1] will help get you started. [1]: https://reference.wolfram.com/language/tutorial/VectorsAndMatrices.html
You should look at the documentation for [ParametricNDSolveValue](https://reference.wolfram.com/language/ref/ParametricNDSolveValue.html) under Examples/Applications/Parameter Fitting for an example work flow on fitting parameters. If you still have...
I think you will need to pass a list of equations to DSolve versus a vector equation. You can accomplish this by adding Thread@@ like so s = DSolve[ Thread @@ {u[r, \[CurlyPhi], z] == {-1/r, 0, 0}}, {f, g, h}, {r, \[CurlyPhi],...
Roberto, I was able to get FindFit to work by using LocalAdaptive stepping on the NIntegrate and by setting the WorkingPrecision to 8. data = Table[{2 i, xmpardata[[i]]}, {i, 1, 25}]; phi[n_, a_, a2_][x_] := n*(x (1 - x))^(a -...
You are welcome Claudio. I am glad that it worked out for you and is more general than I imagined.