How to convert an expression to a function with lossless precision?
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Chenguang Yan
am 27 Sep. 2020
Kommentiert: Walter Roberson
am 28 Sep. 2020
I use matlabFunction to convert an expression to an anonymous function, but the convertion of matlabFunction has a loss of precision. Is there a lossless conversion method? If not, is there any conversion method to improve accuracy?
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Ameer Hamza
am 27 Sep. 2020
As Walter mentioned in his answer, you will lose precision once you convert from symbolic maths to finite-precision. To get exact results, you must stick to variable precision or symbolic mathematics.
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Walter Roberson
am 27 Sep. 2020
Is there a lossless conversion method?
NO, there is no lossless conversion method.
A few years ago I encountered a paper that showed an explicit function that had the property that for any given finite numeric precision, there were inputs to the function for which the error due to rounding could be made arbitrarily large compared to the actual value of the function at the location. The paper showed that no matter how many guard digits you used for rounding purposes, the rounding error could be arbitrarily large relative to the correct value.
I do not recall now what the demonstration function used was, but it was not a complicated function. Something along the lines of x+(1/x) but a bit more developed.
I have myself encountered nonlinear equalities that when I evaluated to a million digits, I still could not decide if the equality was true. The symbolic manipulations said that the equalities held, but because the expressions were so different I was trying to verify whether the symbolic package had made a conversion mistake... and I couldn't decide based on numeric techniques. The rounding issues even at a million digits were substantial.
So, NO. There cannot be any finite numeric technique that can give enough precision to solve arbitrary nonlinear least squares.
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Walter Roberson
am 28 Sep. 2020
The truth of that statement would depend on which solver you were using. Most of the Mathworks ones do not permit you to turn off everything else and use exact 0 as the termination. Solvers would, in many cases, narrow down to the double precision numbers that are on either side of the exact solution, and if exact 0 residue was required and everything else was turned off for termination, the solver would then cycle infinitely between the two adjacent numbers.
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madhan ravi
am 27 Sep. 2020
vpa(f_func(sym(sol1), sym(sol2)))
3 Kommentare
madhan ravi
am 27 Sep. 2020
When you use matlabFunction() , it converts it into double precision, why not use subs()?
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