Talk:Defined and undefined

Misleading title?

I don't like this page. The title is very misleading. It looks more like a list of undefined expressions. Also I don't see why 1^infty or infty^0 should be undefined. What's wrong with 1 in both cases? --MarSch 13:46, 28 October 2005 (UTC)[reply]

You don't know why 1^\infty and \infty^0 are undefined? Well, I guess we can stop the conversation here. :) Oleg Alexandrov (talk) 14:40, 28 October 2005 (UTC)[reply]

What an enlightening insight. --MarSch 15:05, 28 October 2005 (UTC)[reply]

Come on MarSch, this is elementary calculus. One has

e^{x}=\lim _{h\to \infty }\left(1+{\frac {1}{h}}\right)^{hx}.

On the right hand-side you have 1^\infty, while on the left hand-side you have a well-defined number. For \infty^0, take the log and you will get the indeterminate 0× ∞ which is probably familiar to you. Oleg Alexandrov (talk) 15:23, 28 October 2005 (UTC)[reply]

Thanks for your explanation. Okay so for x in ]0, 1[, x^∞ = 0 and for x > 1, x^∞ = ∞ So you loose some continuity. This is no reason not to define 1^∞ := 1. The second example also still works, if you also define 0 x ∞ := 0. --MarSch 15:46, 28 October 2005 (UTC)[reply]

And I agree with the merger, provided you do a good job at it. Oleg Alexandrov (talk) 15:23, 28 October 2005 (UTC)[reply]

What on earth are you talking about? The example I showed to you proves that

e^{x}=1^{\infty }

for any $x>0$ . Come on MarSch, I did not think I need to explain that to you. You cannot define $1^{\infty }$ any more than you can define

{\frac {0}{0}}.

Actually, the two indeterminates are equivalent. Watch this:

1^{\infty }=e^{\ln 1^{\infty }}=e^{\infty \cdot \ln 1}=e^{\frac {\ln 1}{1/\infty }}=e^{\frac {0}{0}}.

Oleg Alexandrov (talk) 16:52, 28 October 2005 (UTC)[reply]

article is problematic

I agree with one thing MarSch said in the above section—I don't like this article, and most especially I don't like the name. Adjectives make bad article titles; two adjectives with a conjunction are just out of bounds. Also the article doesn't specify well enough in what context the expressions are being considered (reals? naturals? complex numbers? extended complex numbers?) and is duplicative of indeterminate form.

I was very tempted to start an AfD, but I suppose there is some value in having this information available in elementary form. But the mathematical context needs to be specified, and the name just has to go. Alternative naming suggestions? --Trovatore 17:20, 13 February 2006 (UTC)[reply]

I think I don't like the name either. :) And the article is not well written either. But I don't have any ideas of what to do, neither of the defined kind, nor of the undefined one. :) Oleg Alexandrov (talk) 00:08, 14 February 2006 (UTC)[reply]

How about moving to implied domain of a function? We could include square roots of negative numbers and logs of nonpositive numbers. As it stands this is a bad article; one way or another we've got to get rid of it. --Trovatore 15:27, 4 March 2006 (UTC)[reply]

Or maybe "implicit domain" is better than "implied domain". --Trovatore 15:28, 4 March 2006 (UTC)[reply]

0⁰

In his edit summary, User:Oleg Alexandrov claims that

(A) One can find sequences x and y going to 0 such that x^y goes to anything you want.
(B) So 0^0 is indeed undefined

There is no doubt that (A) is true, but I don't see how (B) follows. All you can conclude from (A) is that there is no way to define 0^0 in such a way that the function x^y become continuous at 0^0. There is no mathematical proof that 0^0 is undefined (nor is there a proof that 0^0 is defined). Whether 0^0 is defined or not is a matter of convenience, and indeed different authors decide this question differently (though I do not have a reference at hand where 0^0 is declared as "undefined").

I therefore suggest to not include 0^0 in the list of "undefined expressions", and instead to write a short paragraph explaining that 0^0 is often defined as 1 when this is convenient (there is a famous quote by Donald Knuth, citing the importance of x^0 as summand in a polynomial, and the relative unimportance of the function 0^x) and left undefined when not needed (quote needed).

--Aleph4 23:50, 19 December 2006 (UTC)[reply]

That's a better explanation than the original removal. Thanks. Oleg Alexandrov (talk) 03:05, 20 December 2006 (UTC)[reply]

Ok, I added an explanation. Hopefully not too long. --Aleph4 13:06, 20 December 2006 (UTC)[reply]

The explanation here seems fine. I added a pointer to the main article on the subject. To avoid redundancy, please keep this section minimal and expand the main article if you have energy. And the main article may clarify Oleg's viewpoint, by the way. CMummert 14:13, 20 December 2006 (UTC)[reply]

Wow, I did not know that this was such a hot issue at Talk:Exponentiation...--Aleph4 12:06, 21 December 2006 (UTC)[reply]

I think the article there is pretty stable right now; there was some back-and-forth on the wording of the article, but everyone seems to accept the general principle that both conventions about 0^0 should be described, so it's just a matter of getting the correct tone. CMummert 13:21, 21 December 2006 (UTC)[reply]

The article is well written, but I think not simple enough. There are examples of indeterminat forms from Arithmatic, Calculcus and Set Theory, that all bastically say the same thing, but the Arithmatic idea is the simplest

The history about this is interesting also. Decades ago, I found I could crash a multi-user computer by evaluating 0^0. Then suddenly it worked. ( they upgraded the software ), later I could figure out what version they were running, by evaulating it. Then a patch came out for the older version, so when ever I came across the older version, I could tell them about the patch, and a few installations started improving their uptime. A few years later, I found out that my Computer Programming teacher had a degree in Math, and was actually quite bright in explainations. The classoom of course had two doors, one of which was never ever ever in use. He later put a sign on the door, n*0=0!

The Arithmatic idea is this:

Two rules. Any number times Zero is Zero, Any number times One is Any number, so what is 0 times One? Zero, because multiplication by zero is an operational axiom, and multiplication by one is an elemental axiom. Look at the axioms of Arithematic:

Any number multiplied by itself is a power of that number. (Operational axiom)

Any number muliplied by itself zero times is one, except zero. Why?

Because Any number times zero is zero is an operational axiom, and it is defined first. If we tried to nail down a definitionl, it would be either zero, from the zero rule, or one from the power rule ( distinct from calculus ). So we cannot arbitarly determine its value. Its Indeterminate, ( brilliant part now, this is what my teacher said ) If a class room has two doors, how, without knowing, did a petcualr student come in? There is no way arbitrarily to determine it.

Now, fast forward 5 years to AP Calculus. 0^0 is an indeterminate form. ok. What is the argument?

The Calculus form is the limits of the functions 1^x and x^0. x-> 0

The set theory comes from assuming that n^0 is complete over integers. Let it =1, then prove that n =0 is not part of the set of numbers that this operation holds )

There is also discussion in group theory. —Preceding unsigned comment added by 67.188.118.64 (talk) 18:21, 22 November 2007 (UTC)[reply]

A list of undefined expressions?

If we're making a list of undefined expressions, that's hardly all of them. If one plugs a value not within the domain of any given function, it is undefined, unless I'm wrong. 69.117.64.161 (talk) 01:53, 22 June 2008 (UTC)[reply]

All three of the expressions that it says are undefined in all conditions can be defined in some conditions as limits. For instance the limit as x goes to infinity of x-x is zero. The limit as the integer n goes to infinity of (-1)²ⁿ is 1. The limit of x/x is 1. The article definitely needs a good clean up. Dmcq (talk) 06:25, 3 June 2009 (UTC)[reply]

Is 0.999... defined or undefined?

There are many proofs for 1=0.999... but are both 1 an 0.999... defined? Tristan Tondino (talk) 19:27, 7 February 2010 (UTC)[reply]

See 0.999.... Dmcq (talk) 20:32, 7 February 2010 (UTC)[reply]

Yes, they are. Michael Hardy (talk) 02:04, 8 February 2010 (UTC)[reply]

Merged

I copied (actually, largely rewrote) the contents of this page to well-definition and merged. I hope I did it correctly. Tilmanbauer (talk) 12:41, 17 October 2010 (UTC)[reply]

I'm happy for the page to go in whatever fashion! It wasn't being looked after and was a bit crappy, being put in with that sounds a good idea to me. Dmcq (talk) 13:15, 17 October 2010 (UTC)[reply]