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Topic:
XY functions that cannot be solved
Replies:
2
Last Post:
Jun 14, 1996 9:41 AM




Re: XY functions that cannot be solved
Posted:
Jun 13, 1996 11:43 AM


In article <tkidd.834456294@hubcap>, <tkidd@hubcap.clemson.edu> writes:
> bosd@cs.utwente.nl (Rene Bos) writes: > >> >Please define "closed form formula". > >> A function that is finite in length and allowing only > >> predetermined constants,+,,*,/,exponents, and roots. > >To me, this definition seems arbitrary. Why do you include roots, but exclude > >logarithms, hyperbolic functions, etc.? > Aren't all definitions arbitrary? I suppose roots did not even need to be > included, since xth roots are just 1/xth powers. >
stuff deleted...
> >What is 'determining directly'? How do you 'determine directly' what the > >square root of 2 is? You can say 'it's sqrt(2)', but this is just as directly > >as saying 'the value of Lambert's W for 2 is W(2)'. If you mean that you can > >get a good approximation for sqrt(2): the same holds for the Wfunction. > But it has to be expressible in terms of the above. > > This may be arbitrary, but that's the way it is.
Why? because *you* say so? What makes sqrt(2) closed form, while W(2) is not? Is it simply because you understand the former, but not the latter?
> Otherwise, like I said, there's a closed form for the inverse of any closed > form invertible function.
> Travis
First of all you need to define what you mean by "invertible function". Do you refer to the set of monotonic algebraic functions only? Or is a function considered to be invertible by restricting the range of the inverse to be some subset of the domain of the original function?
May I suggest you actually go out and *learn* some mathematics? There is NOT a closed form (under your definition) for any closed form invertible function. (e.g. an arbitrary Quintic is not solvable in terms of radicals)



