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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.

4
votes
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
5
votes
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
5
votes
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
19
votes
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
5
votes
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
6
votes
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
3
votes
[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
7
votes
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