Re: Why is it so difficult?

mike3 wrote:
On Aug 15, 3:54 pm, "I.N. Galidakis" <morph...@xxxxxxxxxxxx> wrote:
[snip]

If y=x^^{1/n} and x is given, after one computes y, then indeed
y^^n=x.

But how does that mean it makes sense to define ^(1/n) x to
be the solution of ^n y = x? Why couldn't we define
^(1/(n^2)) x as the solution???

Because y=x^^{1/n^2} contrasts badly (at an operator level) with its inverse
operation: The inverse operation is y^^n, not y^^{n^2}. In other words, you need
to tetrate y, n-times. NOT n^2-times to get back x. Usually a consistent
mathematical concensus requires "inverse" operators to be related in the
simplest way possible, and in this case the "natural choice" is taken (as an
analogue) from the corresponding relationship between the operators "n-th root"
and "n-th power".

What problems does one
run into when attempting to use a definition like that
that make it less "natural" and more "contrived"?

But from what I've examined the definition of
"tetraroot" simply only requires that a solution
y to the equation ^n y = x exists, NOT that y =
^(1/n) x.

I am sorry, I don't understand your question. It seems that the notation
x^^{1/n} bothers you for some obscure (to me) reason.

Yeah, I guess you could define y instead as:

x^^{Gobledygookery-biscuit+i^56135+gazeebo*superdooper/n!*1/Gamma(Ronald
McDonald))},

to be that number, which if you tetrate it n-times you get back x. The notation
for the 1/n is merely a symbol place holder similar to the symbol place holder
x^{1/n} for a regular n-th root, reminiscent of the operation for which this is
a solution. Such a number exists always when x,y > 1, is unique, and no matter
how you want to call it, it can be calculated to any degree of accuracy.

[snip]
--
I.N. Galidakis

.

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