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November 16

definition and usage of "identity"

Mathworld says: An identity is a mathematical relationship equating one quantity to another (which may initially appear to be different).

Wikipedia says: An identity is an equality that remains true regardless of the values of any variables that appear within it

I guess this is distinguished from mere "equations" in that $\sin ^{2}x+\cos ^{2}x=1\qquad \qquad \!$ is an identity whereas ƒ(x) = sin(x) is an equation.

So my question is, both these definitions seem to apply to both Euler's formula ( $e^{ix}=\cos x+i\sin x\!$ ) and Euler's identity ( $e^{i\pi }=-1\,\!$ ), making them both identities. So is there a particular reason why Euler's identity is called such?

Thanks, --VectorField (talk) 07:03, 16 November 2008 (UTC)[reply]

I converted the mathematical symbols into LaTex for you (hopefully this makes it easier for other editors to read your message). Topology Expert (talk) 08:08, 16 November 2008 (UTC)[reply]

The "formula" is clearly an identity, it's a equation that holds for all values of x. The "identity" on the other hand just equates two constants, so I guess it is trivially an identity (it holds for all values of all the variables that appear, since there are no variables), but it's not what we usually mean when we talk about identities. I think the problem is that Euler just has far too many things named after him (interestingly, I think he was actually involved in the discovery of some of them, which may be unique in mathematics!) so you have to come up with different names to give them which results in stretching the definitions a bit to find something that hasn't already been used. --Tango (talk) 13:55, 16 November 2008 (UTC)[reply]

Yes, precisely... List of topics named after Leonhard Euler is a rather long article! Eric. 131.215.158.179 (talk) 00:40, 17 November 2008 (UTC)[reply]

1x1 matrix determinant

what is the determinant of a 1x1 matrix? ----~~The Successor of Physics~~ 14:46, 16 November 2008 (UTC) —Preceding unsigned comment added by Superwj5 (talk • contribs)

The number in the determinant. Now if you'd asked about a 0x0 determinant you might have got an argument - I'll stick my neck out and say 1. Dmcq (talk) 14:50, 16 November 2008 (UTC)[reply]

Just to be contrary, I'd say 0, a determinant is a sum over permutations, there are no permutations of no elements and an empty sum is 0. --Tango (talk) 15:25, 16 November 2008 (UTC)[reply]

There is exactly one permutation of a zero-element set (the empty function), and so the sum-over-permutations is the sum of one term, that term being the empty product, which is 1. More signally, det is by definition a homomorphism from the general linear group of the space to the multiplicative group of the underlying field, and in particular must map the identity map to 1. Algebraist 15:32, 16 November 2008 (UTC)[reply]

I'll confess, I'd never before heard of "the empty function"... --Tango (talk) 15:55, 16 November 2008 (UTC)[reply]

I've never seen a definition of function that doesn't include it. It isn't always emphasized, but it's always there. Algebraist 15:58, 16 November 2008 (UTC)[reply]

Sure, when you think about it, the empty set is a set so there must be a function from it to other sets, I'd just never thought about it. --Tango (talk) 16:20, 16 November 2008 (UTC)[reply]

The empty function is somewhat important in the category of sets, since it makes the empty set an initial object of that category. Aenar (talk) 22:19, 16 November 2008 (UTC)[reply]

It's more basic than that: if for some reason we disallowed the empty function, then the category of sets wouldn't even be a category, since the empty set would lack an identity morphism. Algebraist 23:26, 16 November 2008 (UTC)[reply]

On a different note, if the determinant of a 0x0 matrix was not 1, then we wouldn't have

\det(A\oplus B)=\det(A)\det(B)

. 76.126.116.54 (talk) 04:29, 17 November 2008 (UTC)[reply]

I guess you mean matrix multiply, and 0=0×0. By the way I just had a look at the determinant article and it actually does mention the 1x1 and 0x0 determinants. It even has a specific reference for saying the 0x0 determinant is 1! Now that's a thorough job. Dmcq (talk) 09:25, 17 November 2008 (UTC)[reply]

No, he used the symbol for direct sum, and I'm sure he meant direct sum. Algebraist 11:34, 17 November 2008 (UTC)[reply]

Thanks, yes it would follow then. Never thought of it actually meaning that. Dmcq (talk) 11:57, 17 November 2008 (UTC)[reply]

Yes, it should be very unconfortable to deny the status of mathematical object to the empty function. Notice that all "trivial" objects like the empty map, the number "0", the identical transformation,... had their big trubles to be accepted as the others, and in fact they came as the last, historically. Even "1" was looked on as a suspicious number initially (because, after all, it does not require the plural form!). As to det of the 0x0 matrix, all definitions and all formulas agree on 1 as its value. For instance, it's a a sum of exactly 0!=1 term(s) (indicized by the empty permutation), the term beeing an empty product (thus 1). --PMajer (talk) 12:01, 17 November 2008 (UTC)[reply]

The Derivative

A couple of us can't seem to settle this argument so we turn to wikipedia experts to resolve this issue. What is the derivative of f(x) with respect to x if
$f(x)=\int _{a(x)}^{b(x)}g(x,t)dt$ .
Thanks!--130.166.165.22 (talk) 22:15, 16 November 2008 (UTC)[reply]

You should look at Leibniz integral rule#Limits that are variable. Eric. 131.215.158.179 (talk) 22:33, 16 November 2008 (UTC)[reply]

So, $f'(x)=g(x,b(x))b'(x)-g(x,a(x))a'(x)+\int _{a(x)}^{b(x)}{\frac {\partial g(x,t)}{\partial x}}dt$ ? Wohoooooo! Thanks!--130.166.165.22 (talk) 22:43, 16 November 2008 (UTC)[reply]