The inverse of a multivariate function ?

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Hamza K
Hamza K am 16 Jun. 2017
Kommentiert: Steven Lord am 16 Jun. 2017
Hello everyone,
I have a function F which depends on 4 variables (x1,x2,x3,x4).
for example F have this form :
F=[x2-x1;x2+x3;x4;x1+x3].
I needed to find the inverse of such function and for that I used the "finverse" function like this : finverse(F,{x1,x2,x3,x4}) but the result is unreadable, very ambiguous and might be false.
Do you have another way to find such inverse, or a precise method?
Thanks in advance, Hamza K.

Antworten (1)

John D'Errico
John D'Errico am 16 Jun. 2017
There is NO method that exists to solve for ALL/ANY general such nonlinear problem. Period. Even for a linear problem such as you show, the solution need not be trivial, since not all such problems need have a solution.
I assume you asked about finverse, from the symbolic toolbox. finverse does not appear to be designed to work on multi-variable problems, so you cannot use that tool.
You can use solve from the symbolic toolbox. Or, for a numerical result, use vpasolve or fsolve. There is absolutely no assurance that a solution exists, or that a solution (IF one does exist) can be found, or that any solution is unique. In fact, for a general nonlinear problem, the solution will usually not be unique.
Sorry, but the subject of your question is sufficiently broad that it can require an entire book(s) to be written about the various methods one can employ, or the various reasons those methods can fail.
In general, use solve/vpasolve/fsolve as you wish. But expect that any arbitrary such problem may well fail to yield a simple solution. Understanding the various methods is crucial for success.
  1 Kommentar
Steven Lord
Steven Lord am 16 Jun. 2017
To add to what John said, my guess is that your end goal is not to invert this multivariate function. If you describe your real purpose, why you're trying to invert the function or for what you would use that result, we may be able to offer a solution that doesn't require the "entire book(s)" John mentioned.

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