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

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Posts: 77
Registered: 12/6/04
Re: XY functions that cannot be solved
Posted: Jun 13, 1996 11:43 AM
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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
> >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

> >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 W-function.

> 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)

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