[QUOTE=rogue;261118]Notice how he says "it results in the indeterminate form 0/0", but then in the same sentence says "which can easily be determined to be 14". What kind of twisted logic is that? How can one say that it is "indeterminate", but is then "determined" at the same time?
[/QUOTE]

He's most likely a quantum physicist. A quantum result can be both determined and indeterminate at the same time.

[QUOTE]
My understanding is that since 0/0 is undefined, then 1^(0/0) is also undefined.
[/QUOTE]

(0/0) is "indeterminate". It is "undefined" [B][I]only[/I][/B] in
the sense that it is not a [B][U]particular[/U][/B] number.

The word "indeterminate" is in fact [B][I]synonymous[/I][/B]
with the words "unknown" and "variable".

(0/0) literally [B][I]means[/I][/B] "any number n such that n*0 = 0",
which [B][I]means[/I][/B] that if we encounter the expression 1^(0/0)
then we can immediately replace it with 1^n = 1.

In other words, in the expression 1^(0/0),
the exponent (0/0) is absolutely [B][I][U]trivial[/U][/I][/B]
and may be disregarded.

Thus, 1^(0/0) = 1^n = 1.

It is R.D. Silverman, CRGreathouse and everyone else
who disagrees with me who is wrong.
That's all there is to it!

Don.

Don Blazys, you are my hero. :bow:

Aw shucks!

[QUOTE=Don Blazys;261396]Aw shucks![/QUOTE]

don care to try :

[url]http://apps.facebook.com/arithmeticchallenge/?ref=bookmarks&count=0[/url]

of course : [url]http://apps.facebook.com/mathbxjb-jhdfccjznho/?start=1&target=home[/url] may be what you want.

Although [URL="http://en.wikipedia.org/wiki/Indeterminate_form"]this link[/URL] doesn't cover [TEX]1^{\frac{0}{0}}[/TEX] specifically, it provides the enough information to allow one to make the logical conclusion that [TEX]1^{\frac{0}{0}}[/TEX] is indeterminate.

Given that [TEX]1^{\infty}[/TEX] is indeterminate and that [TEX]{\frac{0}{0}}[/TEX] could be [TEX]{\infty}[/TEX], one can conclude that [TEX]1^{\frac{0}{0}}[/TEX] is indeterminate.

[QUOTE=Don Blazys;261389]Thus, 1^(0/0) = 1^n = 1.[/QUOTE]

So, praytell, where did you learn this gem? Not a textbook, nor a professor of mathematics... did you intuit it on your own? Find it in a crossword? Discern it from the ramblings of a priest on Mount Parnassus?

[QUOTE=rogue;261418]Given that [TEX]1^{\infty}[/TEX] is indeterminate and that [TEX]{\frac{0}{0}}[/TEX] could be [TEX]{\infty}[/TEX], one can conclude that [TEX]1^{\frac{0}{0}}[/TEX] is indeterminate.[/QUOTE]

I'm actually pretty uncomfortable with your imprecision here. This is of course the sort of thing that mathematicians say to each other, but when rigor is important (as, presumably, here) I think that more care should be taken.

In that sense, "[TEX]1^{\infty}[/TEX]" is a [i]form[/i] taken by certain [i]limits[/i], in particular
[TEX]\lim_x f(x)^{g(x)}[/TEX] where [TEX]\lim_xf(x)=1,\lim_xg(x)=\infty[/TEX]
(of course the precise path of x doesn't matter; for concreteness you can assume it increases without bound). Then we say that this [i]form[/i] of limit is indeterminate: that is, without studying f and g themselves the limit cannot be determined. This is unlike the case
[TEX]\lim_x f(x)^{g(x)}[/TEX] where [TEX]\lim_xf(x)=1,\lim_xg(x)=7[/TEX]
where nothing further need be known about f and g -- the limit exists and is 1.

So when rogue writes "[TEX]{\frac{0}{0}}[/TEX] could be [TEX]{\infty}[/TEX]" what this means, formally, is that a limit of the form
[TEX]\lim_x f(x)/g(x)[/TEX] where [TEX]\lim_xf(x)=0,\lim_xg(x)=0[/TEX]
is indeterminate, and further there exist f, g, and parameterizations of x such that the limit is [TEX]\infty.[/TEX] An example would be f(x) = x^2, g(x) = x, and x increases without bound.

Similarly, when we have a limit of the form
[TEX]\lim_xf(x)^{g(x)/h(x)}[/TEX] where [TEX]\lim_xf(x)=1,\lim_xg(x)=\lim_xh(x)=0[/TEX]
it is possible to choose f, g, h, and the path of x such that the limit has any value you choose. Exercise: find f, g, and h such that the limit is L.

[QUOTE=CRGreathouse;261424]...So when rogue writes "[TEX]{\frac{0}{0}}[/TEX] could be [TEX]{\infty}[/TEX]" ...[/QUOTE]

This was taken from the linked to website. I'm not making a claim for its veracity.

The point is that although it appears that 1^n = n for any n, in actuality it isn't. It is only true if n is defined and determinate.

[QUOTE=rogue;261426]This was taken from the linked to website. I'm not making a claim for its veracity.

The point is that although it appears that 1^n = n for any n, in actuality it isn't. It is only true if n is defined and determinate.[/QUOTE]

I always though for all x 1^x=1 that's the rule I've ever been taught or known.same applies with n except you can't say n must be this value when I use it here because n can take on all values.

[QUOTE=science_man_88;261427]I always though for all x 1^x=1 that's the rule I've ever been taught or known.same applies with n except you can't say n must be this value when I use it here because n can take on all values.[/QUOTE]
Did you look at this example?
[QUOTE=Condor;261219]Because we know the continuous extension of [TEX]\({e^X}\)^{\({\frac{e^X-1}{X^2}}\)}[/TEX] is [TEX]e[/TEX] as [TEX]X\to 0[/TEX]:
[code]X e^X e^X-1 X^2 (e^x)^((e^X-1)/X^2)
1.0000 2.7183 1.7183 1.0000 5.5749
0.3333 1.3956 0.3956 0.1111 3.2767
0.1000 1.1052 0.1052 0.0100 2.8625
0.0333 1.0339 0.0339 0.0011 2.7645
0.0100 1.0101 0.0101 0.0001 2.7320
0.0033 1.0033 0.0033 0.0000 2.7228
0.0010 1.0010 0.0010 0.0000 2.7196
0.0003 1.0003 0.0003 0.0000 2.7187
0.0001 1.0001 0.0001 0.0000 2.7184
etc.[/code][/QUOTE]
[tex]1^\infty[/tex] is an example of an indeterminate form, so 1^(0/0) is too.
It can be any value, as a limit, since X/Y can go to [tex]\infty[/tex] as X and Y go to 0. This example goes to 2.71828..., and is an explicit example of Don being wrong.