[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]

This is where you need to understand how to apply l'Hôpital's rule. [tex]1^{\infty}[/tex] is indeterminate, i.e. not equal to 1. Read [URL="http://en.wikipedia.org/wiki/Indeterminate_form"]this[/URL] again. You can search elsewhere. That [tex]1^{\infty}[/tex] is indeterminate is a well established fact. Maybe some of these websites (especially the wiki) should include [tex]1^{\frac{0}{0}}[/tex] as well.

[QUOTE=rogue;261432]This is where you need to understand how to apply l'Hôpital's rule. [tex]1^{\infty}[/tex] is indeterminate, i.e. not equal to 1. Read [URL="http://en.wikipedia.org/wiki/Indeterminate_form"]this[/URL] again. You can search elsewhere. That [tex]1^{\infty}[/tex] is indeterminate is a well established fact. Maybe some of these websites (especially the wiki) should include [tex]1^{\frac{0}{0}}[/tex] as well.[/QUOTE]

I don't have the math required to know the discussion. as I never got to functions in the number theory prerequisite thread I set up.

[QUOTE=science_man_88;261433]I don't have the math required to know the discussion. as I never got to functions in the number theory prerequisite thread I set up.[/QUOTE]

Do you understand slope? The "slope" along a curve f(x) at a particular point x (there are ways of making this precise) is called the derivative f'(x). L'Hospital's rule says that if f(x)/g(x) is indeterminate, you can look at f'(x)/g'(x) instead. That's the general idea -- to understand it better you'd need to learn what precisely a derivative is and how to take them.

[QUOTE=CRGreathouse;261441]Do you understand slope? The "slope" along a curve f(x) at a particular point x (there are ways of making this precise) is called the derivative f'(x). L'Hospital's rule says that if f(x)/g(x) is indeterminate, you can look at f'(x)/g'(x) instead. That's the general idea -- to understand it better you'd need to learn what precisely a derivative is and how to take them.[/QUOTE]

[TEX]\text slope = \frac{\text rise}{\text run}[/TEX] I could see a way to limit this if not calculate it x^2+x goes up 2+2*(x-1) every time x goes up 1. so the [TEX] \text slope = \frac{2+2*(x-1)}{1}[/TEX] going to the point and the [TEX]\text slope = \frac{2+(2*x)}{1}[/TEX] going from it to x+1 if you do a rough line.

Right, and in fact the derivative in this case is right between those: 2x + 1. So if you had f(x) = x^2 + x and g(x) = x, f(x)/g(x) is undefined at 0. The derivative of f(x) is f'(x) = 2x+1 and the derivative of g(x) is 1... since it's a line this is the slope at any point.

You can use L'Hospitals's rule to find that the limit of f(x)/g(x) as x approaches 0 is f'(x)/g'(x) = (2x+1) / 1 evaluated at x = 0, or 1.

If you use f(x) = x^2 + 2x instead, f'(x) = 2x + 2 and the limit is 2. So you can see that limits in this one indeterminate form 0/0 can be 1 or 2. In fact they can take on any real value, or they can be infinite, or they can fail to exist.

[QUOTE=CRGreathouse;261449]Right, and in fact the derivative in this case is right between those: 2x + 1. So if you had f(x) = x^2 + x and g(x) = x, f(x)/g(x) is undefined at 0. The derivative of f(x) is f'(x) = 2x+1 and the derivative of g(x) is 1... since it's a line this is the slope at any point.

You can use L'Hospitals's rule to find that the limit of f(x)/g(x) as x approaches 0 is f'(x)/g'(x) = (2x+1) / 1 evaluated at x = 0, or 1.

If you use f(x) = x^2 + 2x instead, f'(x) = 2x + 2 and the limit is 2. So you can see that limits in this one indeterminate form 0/0 can be 1 or 2. In fact they can take on any real value, or they can be infinite, or they can fail to exist.[/QUOTE]

the tangent part makes it sound complicated ( and it is in one sense) but the fact that I was even able to get the slope going in and out is kinda weird.

[QUOTE=science_man_88;261471]the tangent part makes it sound complicated ( and it is in one sense) but the fact that I was even able to get the slope going in and out is kinda weird.[/QUOTE]

kinda makes sense that the slope of a tangent line would be in between because the slope should go up gradual and smooth on a smooth version of the line.

[QUOTE=Don Blazys;261389](0/0) is "indeterminate".[/quote]
Only correct thing you've said.
[QUOTE]It is "undefined" [B][I]only[/I][/B] in the sense that it is not a [B][U]particular[/U][/B] number.[/QUOTE]
When f(x1)=g(x1)=0, f(x)/g(x) is undefined at x=x1 because the division operation is undefined if the divisor is 0. The result in not just "not a particular number," it is not a number at all. If you try this in a computer program, the computer has to put something in the memory location. What it puts is a value called "NAN" which means "Not A Number."

But you can sometimes interpret the function at x1 by using limits. The limit as x goes to x1 could be any number, or it could be unbounded, so you can't make any statements about it without knowing what f(x) and g(x) are. For example, if f(x)=e^x-1 and g(x)= x^t, then as x goes to 0:[list][*]If t=1/2, f(x)/g(x) goes to 0, and (e^x)^(f(x)/g(x) goes to 1.[*]If t=1, f(x)/g(x) goes to, and (e^x)^(f(x)/g(x) goes to 1.[*]If t=2, f(x)/g(x) is unbounded (i.e., goes to [tex]\infty[/tex]), but (e^x)^(f(x)/g(x) goes to 2.71828....[/list]Proving that this is wrong:
[QUOTE](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.[/QUOTE]
... since I shown you an example where what you call 1^(0/0) has to be replaced with 2.71828....

Quoting "Condor"
[QUOTE]
But you can sometimes interpret the function at x1 by using limits.
[/QUOTE]And so "Condor" lays yet another egg!
This has absolutely nothing to do with limits.
We are dealing strictly with positive integers.

Regarding the fact that 1^(0/0) = 1^n = 1,
Quoting "CRGreathouse":
[QUOTE]
So, praytell, where did you learn this gem?
Not a textbook, nor a professor of mathematics...
did you intuit it on your own?
[/QUOTE]My work is highly original, and I am indeed the
first mathematician to unequivocally state that
1^(0/0) = 1^n = 1. :smile:

For you, this may be hard, but for me, it's simple,
obvious, straightforward and absolutely elementary.

The [B][U]fact[/U][/B] is, 0 cannot divide any number [B][COLOR=blue]exept itself [/COLOR][/B]and
when it [B][I]does[/I][/B] divide itself, its quotient is [B][COLOR=red]any number[/COLOR][/B] n.

(0/0) literally [B][I]means[/I][/B] "[COLOR=red][B]any number [/B][/COLOR][COLOR=black]n[/COLOR] such that n*0 = 0.
Therefore, 1^(0/0) literally [B][I]means[/I][/B] 1^([COLOR=black]n[/COLOR][COLOR=black]) = 1[/COLOR][COLOR=black].[/COLOR]

It's really [B][I]that[/I][/B] simple, so anyone who has difficulty understanding that
1^(0/0) = 1^n = 1, [B][I][U]must[/U][/I][/B] be a retard! (Sorry, but that's the truth!)

Besides, in my proof, which you can find here:

[COLOR=Navy][U]httр://unsolvedрroblems.org/S14b.рdf[/U][/COLOR]

[B]there are no indeterminate forms[/B] because
logarithms are not even involved when either z = 1 or z = 2.

Thus, what you all think is an "issue" is in fact, utterly trivial!

Don

Once again Don decides to cherry pick, i.e. choosing to avoid certain arguments against his proof.

I have these questions for Don, which I expect a simple yes or no answer to.

1) Is 1^(0/0) determinate?
2) Is 1^n = 1 when n is undefined?
3) Is 1^n = 1 when n is indeterminate?

I see no point in responding to his latest post until he answers these basic questions.

[QUOTE=Don Blazys;261680]My work is highly original, and I am indeed the
first mathematician to unequivocally state that
1^(0/0) = 1^n = 1. :smile:

For you, this may be hard, but for me, it's simple,
obvious, straightforward and absolutely elementary.[/QUOTE]

Ah yes, so the "identity" which no other person believes is "obvious", and yet you don't give a proof, just a lot of blather.

[QUOTE=Don Blazys;261680](0/0) literally [B][I]means[/I][/B] "[COLOR=red][B]any number [/B][/COLOR][COLOR=black]n[/COLOR] such that n*0 = 0.
Therefore, 1^(0/0) literally [B][I]means[/I][/B] 1^([COLOR=black]n[/COLOR][COLOR=black]) = 1[/COLOR][COLOR=black].[/COLOR][/QUOTE]

Actually it doesn't -- the operation you describe is the inverse image of multiplication, not division. Let's use / to denote division and "/" to denote the inverse image of (real) multiplication. The two are similar:

6 / 3 = 2
6 "/" 3 = {2}

21 / 7 = 3
21 "/" 7 = {3}

but they have differences:

5 / 0 is undefined, i.e., meaningless
5 "/" 0 = {}, the empty set

0 / 0 is undefined, i.e., meaningless
0 "/" 0 = R, the set of real numbers.

So now what happens when you attempt to take 1 ^ (0 "/" 0)? You're raising a number to the power of a set, not an allowed operation. (For those who know set theory: yes, this notation is often used for the cardinality of the function from the set into a canonical ordinal with cardinality equal to the number, but this is obviously not what is intended here.)

If you'd like to define your own function here as well, a function "^" that takes a real number on the left and a set on the right, feel free -- but don't kid yourself into thinking it has any relation to the TZ conjecture.

[QUOTE=Don Blazys;261680][B]there are no indeterminate forms[/B] because
logarithms are not even involved when either z = 1 or z = 2.[/QUOTE]

The specific indeterminate form has been pointed out to you by many people.