Mathematics desk
< November 10	<< Oct \| November \| Dec >>	Current desk >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

November 11

mathematics

can we differentiate log log sin x? —Preceding unsigned comment added by Jituinfo007 (talk • contribs) 03:58, 11 November 2009 (UTC)[reply]

See chain rule. You will need to apply it 3 times. --Tango (talk) 04:07, 11 November 2009 (UTC)[reply]

No. Since |sin(x)|<=1, log(sinx)<=0, so we can't differentiate it because log of negative numbers are not real and log(0) is undefined. However we can differentiate it by treating it as a complex function. —Preceding unsigned comment added by Money is tight (talk • contribs) 04:11, 11 November 2009 (UTC)[reply]

Even in the reals, log log sin x denotes the function with empty domain by your argument, and we can certainly differentiate that: it equals its derivative. — Emil J. 12:36, 11 November 2009 (UTC)[reply]

Why does a function having an empty domain have a derivative equal to itself?--Shahab (talk) 15:20, 11 November 2009 (UTC)[reply]

The idea being that the derivative also has an empty domain and the same codomain. There's really only one such function, so they're equal. Rckrone (talk) 15:32, 11 November 2009 (UTC)[reply]

Why should the derivative have the same domain? For example the domain of log x and 1/x are different.-Shahab (talk) 15:37, 11 November 2009 (UTC)[reply]

The derivative of log x is not what you call 1/x, but the restriction of 1/x to (0,+∞). In order for a function to have a derivative in a point a, it has to be defined in a neighbourhood of a by the definition, so the derivative of f cannot have a larger domain than (the interior of) the domain of f. f being differentiable means that the derivative exists at each point of its domain, hence the domain of a differentiable function equals the domain of its derivative. — Emil J. 15:56, 11 November 2009 (UTC)[reply]

(Edit Conflict) If

f:D\to \mathbb {R}

where

D\subset \mathbb {R}

, for instance, the domain of the derivative of f is precisely the set of all points within the domain of f at which f is differentiable, and thus contained within the domain of f. Hence, if the domain of f is empty, the same must be true of the domain of the derivative of f. Your counterexample is false because you are presuming that there is some universal set which contains both the domain of the logarithm function and the domain of

f(x)={\frac {1}{x}}

- the domain of a function is simply an abstract set with no particular reference to any universal set (more precisely, one cannot infer a specific universal set from the domain alone). --PS T 15:58, 11 November 2009 (UTC)[reply]

Obviously a function and its derivative don't generally have the same domain: sometimes the domain of the derivative is smaller than that of the original function (e.g. look at the square-root function, defined at 0, and its derivative, undefined at 0). And Shahab's example is a silly mistake, as has already been explained above. But this one particular function—the empty function—has the same domain as its derivative. Michael Hardy (talk) 19:28, 11 November 2009 (UTC)[reply]

Personally, I don't see any silly mistake in Shahab's post, but rather a misunderstanding on the vocabolary of the other posts. Or maybe I'm myself misunderstanding the meaning of "silly mistake". In any case it is quite common, and not only in elementary calculus, to call "domain" of certain analytic functions what is elsewhere called "natural domain" or "maximal domain" to avoid confusion. --pma (talk) 20:56, 11 November 2009 (UTC)[reply]

I think "silly mistake" just means a mistake that is obviously a mistake once it is pointed out to you and that the person making could reasonably have not made. For former criterion is certainly satisfied, but the latter might not be. It is a very common mistake to forget that the definition of a function requires defining the domain and that two functions which are equal on the intersection of their domains are not necessarily the same function - it is a mistake worth avoiding. --Tango (talk) 22:45, 11 November 2009 (UTC)[reply]

I will probably be on the bad books of Michael Hardy with this post but I think that technicalities do not lie at the real heart of mathematics. It is certainly important that one appreciates subtle errors in proofs (which happen quite frequently), and that he/she understands the importance of precision, but one can be a mathematician without passing these criteria. Shahab has made a mistake which should be noted, but I do not see any advantages in the fact that an empty function equals its derivative, except for conventional conveniences. --PS T 03:34, 12 November 2009 (UTC)[reply]

but the empty function also solves

u''=-ku

and other equations as well... Will you acknowledge this at least..? ;-) --pma (talk) 04:40, 12 November 2009 (UTC)[reply]

You are right (as usual...I was hoping that I would not be caught :)). However, I would still think that classifying equations with an empty solution set is merely a conventional convenience - the empty function solves every equation on the empty domain (whether differential, algebraic or functional). Perhaps I shall acknowledge that the empty function is a trivial solution and nothing more. Luckily we are not living in finite group theory, in which case we would have a hard time in formulating a precise definition of what a "trivial solution/counterexample" is... :( --PS T 07:12, 12 November 2009 (UTC)[reply]

The OP doesn't explicitly say that the function's domain is the real numbers; although it is implied by the use of the variable x. Consider the function ƒ : C → ℛ, where ℛ is the Riemann sphere, given by ƒ(z) = log(log(sin(z))). Then the differential is given by

{\mbox{d}}f=\left({\frac {\cos z}{\sin z\cdot \log(\sin z)}}\right){\mbox{d}}z.

This has essential singularities for z ∈ {πn | n ∈ N}. Besides that I don't really see a problem. ~~ Dr Dec (Talk) ~~ 16:26, 12 November 2009 (UTC)[reply]

There are quite a few additional problems: this is a multivalued function (as is the original f), you forgot about poles in points π(n + 1/2) for integer n, and it does not have an essential singularity in the points πn you mention, it's much worse (there does not even exist a continuous branch defined on a punctured neighbourhood of any of these points). — Emil J. 17:02, 12 November 2009 (UTC)[reply]

You're right: I forgot about the poles you mention; thanks for pointing that out. Multivaluedness isn't a problem; if it were then complex analysis would be one big problem. See principal values, branch cuts, etc. Are the points z ∈ {πn | n ∈ N} branch points? Even if it is a mix of poles, essential singularities, and branch points; that has to be better than the anti-derivative being a function of empty this and empty that. ~~ Dr Dec (Talk) ~~ 17:29, 12 November 2009 (UTC)[reply]

(You mean Z rather than N. Yes, they are; the function behaves like 1/(z log z) for z → 0, and similarly for the other points, hence it inherits all branching problems of log z there.) You seem to be mixing things up. Money is tight wrote in his post that one can differentiate the function as a complex function, but not as a real function. I corrected him by pointing out that the real function, being empty, is also differentiable, and a rather off-topic discussion on what is the domain of a derivative ensued. Nobody ever claimed here that the complex version of the function or its derivative is empty, only the real version is. — Emil J. 18:07, 12 November 2009 (UTC)[reply]

You're right again: I did mean Z instead of N. I was thinking about the log function over the reals for too long and kind of got stuck on the log function working on the positive real numbers (hence positive integers). Thanks for pointing that out. I'm not getting mixed up though. I'm trying to answer the OP's question. The posts above seem to have shifted from the original question, and there seems to be an argumentative feel to it all. That's why I didn't indent my post: it's a fresh take. Let's just consider the function ƒ : C → C ∪ {∞} and there are less problems. Remember that our posts are not intended for ourselves; they ought to be directed at the OP. S/he was no doubt lost in this argumentative soup. ~~ Dr Dec (Talk) ~~ 19:33, 12 November 2009 (UTC)[reply]

Unrestricted Grammars

What kind of sequence would a(n) need to be so that there is no turing machine with symbols {0, 1} that accepts only strings 0^a(n) for all n? 66.202.66.78 (talk) 11:03, 11 November 2009 (UTC)[reply]

It depends on what you mean by "accepts only". If you mean that the machine always returns 0 when a string is in the sequence, and always returns 1 when the string is not in the sequence, then the sequences that have that property are the ones that enumerate decidable sets. If you mean that the machine always returns 0 for a string in the sequence and never halts for any string not in the sequence, the sequences with that property are the r.e. sequences. So the answer is either that a(n) does not enumerate a decidable set, or that a(n) is not an r.e. sequence, depending on what you meant. — Carl (CBM · talk) 12:35, 11 November 2009 (UTC)[reply]

Circle Theorem question

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

This is the theorem. Suppose the figure is like this, we construct CO and produce it to P. Then, angle OCA = angle OAC (OC = OA and converse of isosceles triangle theorem) and thus angle OCA = 1/2 angle AOP (exterior angle teorem). Similarly, angle OCB = 1/2 angle BOP. Therefore, angle ACB = 1/2 angle AOB. But if the figure is like this, how do we prove it? Srin ivas 14:27, 11 November 2009 (UTC)[reply]

We can do the same thing: extend CO to a point P on the other side of the circle. Using the same arguments angle OCA = 1/2 angle AOP, and angle OCB = 1/2 angle BOP. This time instead of adding these angles together, we subtract: angle OCA - angle OCB = angle ACB, etc. Rckrone (talk) 15:12, 11 November 2009 (UTC)[reply]

You can prove it using complex numbers. Start with a unix circle on the complex plane. The origin is the centre of the circle. The point C is at the location 1+0i and chose point A and B randomly on the unit circle. Then calculate the angles 203.41.81.1 (talk) 21:41, 12 November 2009 (UTC)[reply]

I am not that good to understand all that, 203.41.81.1, so what Rckrone is much easier for me. Thanks Rckrone. Srin ivas 06:18, 14 November 2009 (UTC)[reply]

You put forward a problem in Geometry. There are several methods of solving the problem. One method is to use Complex Number/Euclidean vector. It will solve your problem. But if what you want is a Geometry only method for solving your problem, then you need to state so in your description of your problem.122.107.207.98 (talk) 00:28, 15 November 2009 (UTC)[reply]