Talk:Indeterminate form

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[edit] comment by Taku

Isn't 0 over 0 is a determinate form because if you keep multiplying very small number very few times you just get 0. -- Taku 02:25, Nov 15, 2003 (UTC)

0/0 is indeterminate since if it had some definite value, say x, so 0x=0. But any number satisfies that equation for x, so 0/0 has no definite or determinable value. Dysprosia 02:28, 15 Nov 2003 (UTC)

Umm, elegant! Thanks a lot, Dysprosia. -- Taku
But outside the realm of limits, there is very little practical use in calling 0/0 indeterminate, versus just grouping it with a/0 undefined forms (a != 0)

In the sense of this article, 0/0 is indeterminate because if f(x) and g(x) both approach 0 as x approaches something, then f(x)/g(x) can approach a number that depends on which functions f and g are. (That does not detract from the elegance of Dysprosia's observation, however.) Michael Hardy 20:18, 15 Nov 2003 (UTC)

I've just moved some material from zero divided by zero to this page performing significant changes to it to make it apropriate for this page. I am planning to move more material in a few days time but not necessarily to this page and maybe to a subpage of this page. Barnaby dawson 19:56, 20 Sep 2004 (UTC)

After an edit clash, I've added to the second paragraph. Charles Matthews 11:05, 18 Mar 2005 (UTC)

I've now done further work on that discussion. Charles Matthews 11:16, 18 Mar 2005 (UTC)

[edit] Missing Infinity ^ Infinity

The form oo^oo is missing from the list.

How can ∞∞ be considered an indeterminate form? If f(x) and g(x) both approach ∞ as x approaches a, then f(x)g(x) can only approach one thing, ∞, regardless of which functions f and g are. So it's not an indeterminate form. Michael Hardy 23:34, 18 December 2005 (UTC)

It can be indeterminate in a weak sense, like 1/0 (but perhaps a bit stronger). For example, as x approaches 0 from the right, (1/x)1/x approaches infinity (from the left), while (1/x)-1/x approaches zero (from the right). But nothing of this form can converge to any non-zero finite value (because the logarithm must converge to infinity). I'm not sure if this should count as indeterminate or not; the precise definition that I've just written says that it is, but one could easily fix that so that it's not. --Toby Bartels 01:57, 23 May 2006 (UTC)

Remember that when one says ∞, one means "positive infinity" (i.e. increasing without bound). So your example is not the same because it raises "positive infinity" to "negative infinity". --Spoon! 11:33, 31 August 2006 (UTC)
What one means depends on context; compare the extended real line to the real projective line. See also this formal definition, which uses the Riemann sphere. —Toby Bartels 22:03, 20 September 2006 (UTC)

[edit] On merging with defined and undefined

User:Michael Hardy, at 15:37, 18 December 2005, wrote in this diff:

This merger proposal has no merit. Too many readers keep failing to see that "indeterminate form" does NOT mean something that is undefined. Maybe we should hit that point much harder.

(copied here for future reference by Oleg Alexandrov (talk) 00:16, 19 December 2005 (UTC))

To be honest I'm not exactly sure what MH's getting at there. An indeterminate form per se is not undefined, but its value is, and they're pretty much coextensive with the expressions treated at defined and undefined. The latter is just a bad article with a bad title, and I think the real point of the merger was to figure out some way to get rid of it, a project I'd support. --Trovatore 16:58, 13 February 2006 (UTC)
No -- they're NOT coextensive, since 1/0 is NOT an indeterminate form. 0/0 is an indeterminate form; 5/0 is not. That means that if f and g both approach 0 as x approaches something, then f/g could approach anything. That is not true if f approaches 5 and g approaches 0. Michael Hardy 22:08, 13 February 2006 (UTC)
OK, true, 1/0 is not an indeterminate form, but it's not enough by itself to justify a separate article from this one, particularly one with as bad a name as "defined and undefined". And all the indeterminate forms are denotationally undefined (unless you're going to take the 00=1 approach; I'd be against presenting that as the default approach, though it ought to be mentioned). --Trovatore 22:16, 13 February 2006 (UTC)

[edit] Determinate and inderminate discussion

The "Discussion" section had a problem paragraph (reading "By contrast "1/0" is not an indeterminate form because there is no range of different values that f/g could if f approaches 1 and g approaches 0."). There was at least a missing word (between could and if). I attempted to correct that and add more detail to clarify it, but then realized I still don't understand the distinction which is the whole point of the article.

Several times 1/0 is contrasted with 0/0, but the only characterization of this contrast is that there is "no range of different values" without elaboration. But f/g could have a limit of either +∞ or -∞ depending on whether the function g approaches 0 from the + or - side. So there seems to be more than one "value" possible among limits approaching 1/0. Just how is this determinate while 0/0 is not? Whatever the answer, I think the article needs improvement in explaining this. -R. S. Shaw 00:58, 13 March 2006 (UTC)

I think that the whole article is rather unclear in a few ways, first of all that this is fudnamentally about limits (although of course, that is not your problem, R. S. Shaw). With this in mind, I'm going to rewrite it a good deal right now, and I should be able to address your (Shaw's) problem at the same time. -- Toby Bartels 20:29, 22 May 2006 (UTC)

[edit] Lead sentence

The lead sentence says

In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression whose limit cannot be evaluated by substituting the limits of the subexpressions.

This is clearly wrong - an indeterminate form may be obtained by substituting limits of subexpressions, but the form itself is just a few symbols and doesn't have a "limit", it has a "value" (or more precisely, it does not have a value). That is, the question "What is the limit of 0/0?" is meaningless. CMummert 02:22, 29 December 2006 (UTC)

Yep. How about:
…an indeterminate form is an algebraic expression representing a limit that cannot be evaluated by substituting the limits of the subexpressions.
–EdC 17:02, 29 December 2006 (UTC)

[edit] Untitled comment

I think the situation is being looked into the wrong way. Say you had (x^2-9)/(x+3). Someone that didn't know that x+3 is a conjugate of x^2-9 would tell you that -3 gives an undefined answer. Yet someone who factored would get x-3, and say that all values are defined. I take this to mean that in a situation like x^2/x, 0/0 is 0. However, in the cases where it's 0(a)/0, where a is any number except 0, the result is a as 0/0 would negate eachother and equal one. An example is above. And don't gimme a "but my calculator says no." k, I want to know what you guys think. Also, another example is x/x, every number outside of zero gives a 1 answer, so it would be expected that 0/0 gives a one answer. This even works with x^2/x. The reason (0,0) is a point, and not (0,1), is because when you expand ((0)^2)/0 you get 0*0/0. 0/0 is one, and that leaves 0*1, which gives 0. —The preceding unsigned comment was added by 66.66.92.167 (talk • contribs) 07:11, 3 January 2007 (UTC)

Well, some of your terminology is nonstandard: you say "conjugate" when you mean "factor" and "negate" when you mean "cancel". But the "0"s in the expression 0/0 do not cancel each other. When you have, e.g., 3/3, the two "3"s cancel each other because you can divide both the numerator and the denominator by 3. But in 0/0 you cannot divide both the numerator and the denominator by 0. But I'm not sure what your question is otherwise. It is certainly true that x/x approaches 1 as x approaches 0, and x2/x approaches 0 as x approaches 0, and (x2 − 9)/(x + 3) approaches −6 as x approaches 3. But what is in the article that is being "looked into the wrong way"? Michael Hardy 00:45, 4 January 2007 (UTC)
Conjugate is being used the right way, just outside of complex numbers. I'm using the other factor to remain with a 2 variable-number combination (or however you want to say it). The article says that 0/0 is 0 when it's X^2/x (or just call it x if you want), but x/x is not 1 when x=0. I say it's being looked at the wrong way because if you have 0^2/0, it expands to (0/1 * 0/1) / 0/1. and since a top and the bottom 0/1 cancel out, you get (0/1 * 1) / 1 which is zero. However, when you have x/x, any real number / any real number is 1. It would also make sense that if you had (0/1) / (0/1) it would equal one. And if that doesn't make sense, a perfect example is limits. If you do the differential equation, you have to cancel out the h on the bottom (if my equation's the same as yours) with the hs' on top. This gives you the formula for any point. But in any situation, if you used the original formula with h=0, You get the numerator over 0, which be definition is indeterminate or at the minimum undefined, which refutes the ability to use limits on points. Therefore this means that division comes before multiplication when using the order of operation because otherwise, with the above x/x and differential equation, the equations wouldn't be true in any situation, so the only thing can be true is 0/0 is 1 —The preceding unsigned comment was added by 66.66.92.167 (talk) 23:12, 8 January 2007 (UTC).
No, the article says that x^2/x has the limit 0 as x tends to 0, and x/x has the limit 1. Neither formula has a value at x=0.
You can't cancel out zeroes from a fraction; you can only cancel out non-zero common factors.
I don't understand which differential equation you're referring to.
I don't think refute means what you think it means.
I don't understand your last point. In mathematics the order of operations is explicit. –EdC 00:32, 9 January 2007 (UTC)
You can cancel out zeroes, which is why the limits are what they are. The differential equation I'm referring to is the basic slope formula or whatever other areas call it. When you use f(x) instead of y, the equation is dy/dx = (f(x+h) - f(x)) / h. This is the equation used to find the slope of a line between two points. When this is done with any equation with a leading x power > 1, you result has xs and hs. To determine the "instantaneous veloctiy", "slope", whatever you want to call it, 0 is substituted for every h value because you want the slope at that point. Now this part is crucial. Slopes work, and they are proven to work, otherwise we wouldn't use them because they wouldn't work. To say that you can't divide by 0 makes the whole beginning equation moot. And yes, the definition of refute is "to prove to be false or erroneous, as an opinion or charge." You would be refuting the whole ability to find the slope on a given point, which is a significant part of calculas. I know you will rant about "Slopes work, and they are proven to work, otherwise we wouldn't use them because they wouldn't work." but there is no other way to word it. It's like telling me to prove adding works. It works because it does, it doesn't need a reason. —The preceding unsigned comment was added by 66.66.92.167 (talk) 02:17, 17 January 2007 (UTC).
No. As the article on differentiation describes, the slope of a line is found by taking the limit of the equation (f(x+h) - f(x)) / h as h tends to 0: \frac{dy}{dx} = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}. At no point is a division by 0 performed. Imagining that taking a derivative involves cancelling zeroes leads to the same mistakes that motivated the development of mathematical analysis on a rigorous footing. –EdC 03:11, 17 January 2007 (UTC)
Note that you can, if you believe in non-standard analysis, set h to be an infinitesimal. However, that's not the same as zero. –EdC 03:19, 17 January 2007 (UTC)

[edit] List given in introduction

I've changed "The indeterminate forms include [...]" to "The indeterminate forms are [...]" on the assumption that the list is exhaustive. Previous wording admitted interpretation of list as a subset.

If the given list is not exhaustive, I believe the best compromise is "The indeterminate forms (as described this article) are [...]". That is to say, whether or not the list is exhaustive, the current wording is needlessly ambiguous. 23:29, 7 March 2007 (UTC)

[edit] List given in introduction (second concern)

A separate critique is that it seems better to say "are expressed by/using/as" or "are referred to by/as" in place of "are". It may be confusing to many readers to say such-and-such is 0/0 as it might seem to imply (as we almost always intend when writing such expressions) that 0/0 represents a number rather than a form. I have not made this change; I don't know what wording best resolves this potential misreading. Please discuss here. 23:35, 7 March 2007 (UTC)

The real problem is that although numerous books use the term "indeterminate form", I have never seen published a definition of one. Can you think of a textbook that defines them? CMummert · talk 00:25, 8 March 2007 (UTC)
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