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Questions on exact, symmetric reversal of a definition or functional mapping (i.e. the original form is returned when applied twice). Use this tag for issues on inversion of Mathematica expressions, or general inversion of math constructs.

How about this? This samples the function g 100000 times an approximates the inverse of g by Interpolation: H2 = x \[Function] -x Log[x] - (1 - x) Log[1 - x]; g = x \[Function] H2[(1 + x)/2]; xlist … = Subdivide[0., 1., 100000]; ylist = Join[g[Most[xlist]], {0.}]; ginv = Interpolation[Transpose[{ylist, xlist}]]; Here is a plot of the approximate inverse function: Show[ Plot[g[x], {x, xlist[[1 …
answered Apr 23 '18 by Henrik Schumacher
The Moore-Penrose pseudoinverse of A is a right inverse only if A is surjective. But your A is not surjective since Transpose[A] has a nontrivial kernel: NullSpace[Transpose[A]] {{1,-1,-1,1 … }} But as generalized inverse, you have of course A.PseudoInverse[A].A == A PseudoInverse[A].A.PseudoInverse[A] == PseudoInverse[A] True True Addendum Actually, this is true for each …
answered Jul 17 '18 by Henrik Schumacher
Hmmm. I am not 100 % sure because I am not familiar with Hankel transforms. But I read in the docs that HankelTransform implicitly assumes that the input function is supported in $]0,\infty[$. So, if …
answered Dec 10 '18 by Henrik Schumacher
I just stumbled upon it! At least when storing the factorization in a LinearSolveFunction object, we can use it for the transposed solve by supplying a further (not documented?) string variable to it: …
answered Dec 12 '17 by Henrik Schumacher
Using the CholeskyDecomposition explicitly not only seems to remove the problem, it is also faster: Moreover, this gets rid of one of the matrix-matrix multiplications and, probably more important, it …
answered Feb 13 by Henrik Schumacher
Your function g is equal to -2 Sinh[x/2]: g[x_] := E^(-x/2) - E^(x/2); FullSimplify[-2 Sinh[x/2] == g[x]] True So you are looking for ginv[y_] := -2 ArcSinh[y/2]
answered Jul 21 '18 by Henrik Schumacher
[Y]; Inverse[Df[X]] ]; StreamDensityPlot[g[{x, y}], {x, -3, 5}, {y, -3, 5}] Addendum FindRoot can be very sensitive to the initial guess. By applying f to the points a sufficiently large … and fine grid, one can employ Nearest to obtain a ``coarse inverse'' of f that can be refined with FindRoot: ClearAll[f, Df]; Block[{X,x,y}, f[X_] = {2 Indexed[X, 1] + Sin[Indexed[X, 1] + Indexed …
answered Sep 1 by Henrik Schumacher
Out of curiosity, I tried to write my own version of inverse CDF for the normal distribution. I employ a qualitative approximation of the inverse CDF as initial guess and apply Newton iterations with … secant method would have been more appropriate? Edit Using expansions of the inverse CDF at $0$, $1/2$ and $1$, I was able to come up with a way better initial guess function g. …
answered Sep 2 '18 by Henrik Schumacher