Wikipedia:Reference desk/Archives/Mathematics/2011 April 10

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April 10

Differentiation problem

I am differentiating

${\frac {1}{1+e^{-a(x+b)}}}$

I use the chain rule to get

${\frac {-1}{(1+e^{-a(x+b)})^{2}}}\times -ae^{-a(x+b)}={\frac {ae^{-a(x+b)}}{(1+e^{-a(x+b)})^{2}}}$

But the actual answer is ${\frac {ae^{a(x+b)}}{(1+e^{a(x+b)})^{2}}}$

What happened to the negative signs in the exponents? —Preceding unsigned comment added by 130.102.158.15 (talk) 03:55, 10 April 2011 (UTC)

I did this quickly mentally so may be wrong, but I think the two expressions are the same. To see this multiply both numerator and denominator by $(e^{a(x+b)})^{2}$. Zunaid 06:58, 10 April 2011 (UTC)

Correct. Trivial algebra shows that the two are the same. Michael Hardy (talk) 16:27, 10 April 2011 (UTC)

Set Theory

Is there a surjection bijection - in an arithmetical language - from the set of natural numbers on the set of prime numbers?

1. If there is - then: how would it look like (in an arithmetical language)?
2. If there isn't - then: wouldn't the continuum hypothesis be refuted by the set of prime numbers, on which there is no first-order surjection bijection from the set of natural numbers, and from which there is no first-order surjection on bijection from the set of real numbers? Note that when Paul Cohen proved that the continuum hypothesis does not derive from ZF, he proved that for first order mappings only.

HOOTmag (talk) 06:51, 10 April 2011 (UTC)

The primes are a subset of the natural numbers. So let $f:N\to P$ be the map defined by f(x)=x when x is prime, or f(x)=2 when x is not prime. I'm not sure if that will satisfy your requirement of "arithmetical language", but surely the condition that x be prime is expressible in an "arithmetical language"? Staecker (talk) 11:31, 10 April 2011 (UTC)

Yes, thank you, but I was wrong with the "surjection"; It should have been a "bijection". I've just fixed that (after reading your response), by striking out the "surjection" and by replacing it by "bijection". See above.

I'm still looking for an answer for my question in its new version, see above.

HOOTmag (talk) 12:43, 10 April 2011 (UTC)

The obvious bijection is the function that sends n to the nth smallest prime. I'm not sure what you mean by "in an arithmetical language", but this function is first-order definable in the natural numbers considered as a structure for the language with symbols for plus and times, for example. Algebraist 15:19, 10 April 2011 (UTC)

Is the term "nth smallest prime" first order definable? I'm really eager to know how you define the bijection. HOOTmag (talk) 19:32, 10 April 2011 (UTC)

Yes. Moreover, the function enumerating the primes is computable (in fact primitive recursive) so is simply (Delta 1 at most) definable in Peano arithmetic. To see it's primative recursive is fairly straightforward as bounded existentials and minimization is, and a fine computable bound for the next prime after n is 2n. As for what you are claiming, it's not quite true. Any finite set, for example, has a different cardinality than the naturals and reals, so there are no bijections. A counterexample to the continuum hypothesis would be a set A such that there is no surjection from the naturals to A and there is no injection from A to the reals. Under the assumption of AC, this is the same as saying the cardinality of A is strictly between the cardinality of the naturals and reals. Also, to not be misleading, what Trovatore said below is correct. This being first order definable doesn't really matter too much. There are continuum many functions from N to N, but only countably many definable ones in any countable language.Wgunther (talk) 20:20, 10 April 2011 (UTC)

Yes, and you can check that the original version of my question was really about no surjection from the set of natural numbers on A, and no surjection from A on the set of real numbers; I was wrong when I finally struck out the surjection and replaced it by bijection. Anyways, I think you made a mistake, in your statement about an injection from A to the reals, instead of a surjection from A to the reals (or an injection from the reals to A). HOOTmag (talk) 21:02, 10 April 2011 (UTC)

It was Euclid who first (as far as anyone knows, apparently) proved that there actually are infinitely many primes. See Euclid's theorem. That is essential to this problem. Michael Hardy (talk) 18:23, 10 April 2011 (UTC)

I don't see how this would bear upon the continuum hypothesis. The continuum hypothesis is false if there is a cardinality between that of the natural numbers and that of the reals. But the primes are a subset of the naturals, so their cardinality is not more than that of the naturals. Michael Hardy (talk) 18:25, 10 April 2011 (UTC)

"Between" - yes, and that's what I meant by bijection, which is a one-to-one correspondence between sets.

"The cardinality of the primes is not more that that of the reals": yes, but cardibality is defined by the existence of a bijection between all the sets belonging to the same cardinality. Now, notice that when Paul Cohen proved that the continuum hypothesis does not derive from ZF, he proved that for first order bijections only, so he meant that there may be an (unconstructible) set from which there is no first-order bijection to the set of natural numbers, and no first-order bijection to the set of real numbers. Note that Paul Cohen has never promised that there is no bijections at all, e.g bijections in higher orders! That's why I would like to know whether there is a first order bijection between the set of natural numbers and the set of prime numbers, because if there isn't - then the set of prime numbers refutes the continuum hypothesis (at least as far as first order languages are concerned), which is a stronger result than Cohen's, who didn't refute the continuum hypothesis but rather only proved (by referring to first order bijections) that the continuum hypothesis does not derive from ZF...

HOOTmag (talk) 19:32, 10 April 2011 (UTC)

No, sorry, I think you misunderstand the result. There is a "first-order" involved here, but it's not about the bijections themselves.

What Cohen proved is, there is no proof of CH from ZFC, in first-order logic. (The second-order version of ZFC, interpreted in full second-order logic, either implies or refutes CH, we just don't know which one.)

But the bijections involved are not restricted to first-order definable ones. (I assume that by "first-order bijection" you actually mean "first-order definable bijection"; your literal term "first-order bijections" simply doesn't mean anything in standard terminology.) --Trovatore (talk) 19:37, 10 April 2011 (UTC)

Are those mappings restricted to finitely definable ones? For my question to make sense, please check any random al real number, i.e. any infinite set S of random al natural numbers; Such a set is not constructible, i.e. it can't be obtained by using ZFC axioms only; However, it can be defined "separated" from the set of natural numbers (using the Axiom of Separation) - by an infinite first order formula: S = {x|(x=a_1) V (x=a_2) V ... V (x=a_n) V ...}, i.e. by an illegitimate formula in ZFC. Further, there is no (finitely definable) surjection from the set of natural numbers on S, and no surjection from S on the set of real numbers, so may S be an (unconstructible) counterexample to the continuum hypothesis (assuming that the very existence of S is consistent in ZFC)? HOOTmag (talk) 20:54, 10 April 2011 (UTC)

My response below, timestamped 21:10, 10 April 2011 (UTC), addresses this point. --Trovatore (talk) 21:17, 10 April 2011 (UTC)

My new question was about whether "those mappings [are] restricted to finitely definable ones", so I understand now that your answer is simply: "no", isn't it? HOOTmag (talk) 21:28, 10 April 2011 (UTC)

There certainly exists an injection from the set of primes to the set of natural numbers (the inclusion map). Therefore, by definition, the cardinality of the set of primes is no greater than the cardinality of the set of natural numbers.

I can give you many examples of sets which have neither a bijection to the set of natural numbers nor a bijection to the set of real numbers: the empty set, any finite set, the set of all functions from R to R, etc. In order for the continuum hypothesis to be false, there would need to exist a set S such that there exist injections from the set of natural numbers into S and from S into the set of real numbers, but there do not exist injections from the set of real numbers into S or from S into the set of natural numbers. Since the set of prime numbers plainly fails the last of these criteria, it cannot possibly be a counterexample to the continuum hypothesis. —Bkell (talk) 19:42, 10 April 2011 (UTC)

Yes, and you can check that the original version of my question was really about no surjection from the set of natural numbers on S, and no surjection from S on the set of real numbers. I was wrong when I finally struck out the surjection and replaced it by bijection.

Anyways, how about a random al real number, i.e an infinite set S of random al natural numbers? Such a set is not constructible, i.e. it can't be obtained by using ZFC axioms only; However, it can be defined "separated" from the set of natural numbers (using the Axiom of Separation) - by an infinite first order formula: S = {x|(x=a_1) V (x=a_2) V ... V (x=a_n) V ...}, i.e. by an illegitimate formula in ZFC. Further, there is no (finitely definable) injection from S to the set of natural numbers, and no injection from the set of real numbers to S, so may S be an (unconstructible) counterexample to the continuum hypothesis (assuming that the very existence of S is consistent in ZFC)?

HOOTmag (talk) 20:54, 10 April 2011 (UTC)

It's frustrating to discuss these things with you, because you keep saying things that sort of almost make sense, but not quite, and I have trouble pinpointing the exact thing that you're missing.

For the moment, let's just be clear on what I tried to point out before: The statement of CH has nothing to do with whether the various maps (bijections or surjections as you like) are definable or not. The result that says CH is not provable from ZFC, is talking about CH full stop, not about some version of CH restricted to definable maps. Where the "first-order" part comes in is not in regards to the meaning of CH, but only in regards to the methods of proof that are being considered.

So for example, one way of stating CH is there is a bijection between the real numbers and the countable ordinals. Now, it's tempting to think Cohen proved there could not be any definable such bijection, because if there were, you might think that you could just give the definition, and your proof of CH is complete.

But that's wrong. There "could" be a definable such bijection, for some value of "could". For example, if V=HOD and CH holds, then there is a definable bijection between the reals and the countable ordinals, and ZFC does not refute this situation. So you can give a definition that may define a bijection between the reals and the countable ordinals. What you can't prove, in ZFC alone, is that the definition works. --Trovatore (talk) 21:10, 10 April 2011 (UTC)

My new question was about whether "those mappings [are] restricted to finitely definable ones", so I understand now that your answer is simply: "no", isn't it? HOOTmag (talk) 21:26, 10 April 2011 (UTC)

Yes, the answer to that question is indeed "no". There could (in some sense of "could") be a non-definable counterexample to CH, but it cannot possibly be a set of natural numbers. --Trovatore (talk) 21:45, 10 April 2011 (UTC)

Now, let's assume the Axiom of Separation is replaced by a "stronger" one, which permits infinite formulas, e.g. formulas of the form: S = {x|(x=a_1) V (x=a_2) V ... V (x=a_n) V ...}. Would this change the independent status of CH? I assume this wouldn't, am I right? HOOTmag (talk) 08:51, 12 April 2011 (UTC)

Umm ... it's complicated.

First of all, as soon as you allow infinite formulas, you're outside the realm of first-order logic. FOL is rather special in having a computable complete deductive system of proof. If every model of a first-order sentence P is also a model of Q, then assuming P, you can actually prove Q, where a proof is a thing that can be checked by computer.

For infinitary logic this is no longer true, so you have to be careful about what you mean by "independent".

Let's say you mean in the sense of logical implication (model theory) warning: logical implication redirects to entailment, and the latter article is not relevant here, so what you're asking is, does every model of ZFC+your strengthened separation axiom, also satisfy CH, or also fail to satisfy CH?

In that case, it depends. Basically what your strengthened axiom guarantees is that your model can't leave out any of the "real" reals. However it still might leave out sets of reals. Also it could fail to be wellfounded, although I think it probably has to have a tall enough wellfounded part that this point is irrelevant to CH (I'd have to think about that some more though).

If I'm right about the wellfounded-part thing, then if CH is (Platonistically) false, then every model of your axiom knows that CH is false, because otherwise it would have a wellordering of its reals whose every initial segment is countable, but its reals are the genuine reals, and because it has all the genuine reals, countability is also sufficiently absolute.

However, if CH is actually true, then it might still be independent of ZFC+your new axiom, because your model could have all the reals, but leave out the wellordering.

So the bottom line is, it's complicated, and probably no one knows the exact answer to whether CH is independent of ZFC+your strengthened axiom. That's assuming that I've correctly interpreted what you mean by the strengthened axiom. --Trovatore (talk) 09:56, 12 April 2011 (UTC)

To sum up (provided that you're right about the wellfounded-part thing): CH is semi-decidable in the strengthened ZFC, i.e. if CH is false then it's refuted in the strengthened ZFC, but if CH is true then it's independent of strengthened ZFC. Did I interpret you correctly?

How about the strengthened ZF (i.e. without the Axiom of Choice)?

HOOTmag (talk) 11:20, 12 April 2011 (UTC)

The first part's right; if CH is false then it's refuted. For the other direction (true=>independent) I'm less sure; there's a step or two missing.

Without AC there's another definitional complication, which is, what do you mean by CH? CH is often stated in several equivalent forms, that aren't equivalent after you drop AC. My preferred take on it is to think of CH as meaning $2^{\aleph _{0}}=\aleph _{1}$, or equivalently "there is a wellordering of the reals whose every proper initial segment is countable", and for that sense I think you get the same answers as you get with AC. But if you mean "there is no set of cardinality strictly between $\aleph _{0}$ and $2^{\aleph _{0}}$", I'd have to think about it some more. --Trovatore (talk) 19:00, 12 April 2011 (UTC)

Thanks. HOOTmag (talk) 22:35, 12 April 2011 (UTC)

Is "randomal" a technical term? Randomal redirects to a city in Armenia, and Google is not helpful either. –Henning Makholm (talk) 22:39, 10 April 2011 (UTC)

AFAIK, not a technical term. My best guess in what was meant by was "randomal real" was a non-constructible real (in the L sense). Wgunther (talk) 22:57, 10 April 2011 (UTC)

Sorry, I meant random. A "random" real number, is a real number S which can't be obtained by using ZFC axioms only, yet can be "separated" from the set of natural numbers (using the Axiom of Separation) - by an infinite first order formula: S = {x|(x=a_1) V (x=a_2) V ... V (x=a_n) V ...}.

HOOTmag (talk) 08:37, 12 April 2011 (UTC)

Very easy maths question

Hi guys. This is probably the easiest question that has ever been asked on this forum but please bear in mind that I'm very stupid and I can't seem to find an answer anywhere.

I have an equation ½v² = g x h. I want to find the value of v but I'm really struggling to know what to do with the 1/2 and square, ie. to isolate the v - do they cancel each other out? If I times by v then I assume I get rid of the 1/2 but does that leave me with v³ or not?

I know this is embarrassingly easy but I have no intuitive maths skills at all.

Thanks

Pantscat (talk) —Preceding undated comment added 12:08, 10 April 2011 (UTC).

x^y is a common computer notation for x^y.

Your equation can be written ½ × v^2 = g × h. You can isolate v^2 by multiplying both sides by 2, since 2 × ½ = 1. Once you have isolated v^2, you can take the square root on both sides to isolate v. Consider whether v can be both positive and negative in your problem. For example, v^2 = 9 would have two integer solutions: v = 3, and v = -3. PrimeHunter (talk) 13:01, 10 April 2011 (UTC)

Thanks Prime Hunter. So does that leaves me with v = square root of 2gh? If so, does that mean I times the product of g x h by 2? The alternative is that I multiply 2g by 2h, which gives a totally different answer.

Yours dazed and confused Pantscat (talk) —Preceding undated comment added 13:35, 10 April 2011 (UTC).

Yes, your final answer is plus or minus the square root of 2gh. This can be interpreted in two ways, either 2(gh), which is the way that you describe, multiplying the product gh by 2, or (2g)h, which takes the product 2g and multiplies it by h (note: NOT by 2h). These should give the same answer. --COVIZAPIBETEFOKY (talk) 14:19, 10 April 2011 (UTC)

If you multiplied 2g by 2h, you would get 4gh. -- Meni Rosenfeld (talk) 15:56, 10 April 2011 (UTC)

If I understand the question correctly, the you would like to solve

${\frac {1}{2}}v^{2}=gh\,,$

with respect to v. Since both sides of the equation are equal, it follows that if we do the same thing to both sides of the equation then they will also be equal. So, first of all, multiply both sides by 2, this gives v² = 2gh. Next, we take the square root of both sides of the equation, not forgetting the ±, this gives

$v=\pm {\sqrt {2gh}}\,.$

The ± appears because, for example, 5² = 25 and (−5)² = 25. It follows that the square root of 25 is either +5 or −5.

So, whenever we use a square root, we need to add a ± to account for that fact. Fly by Night on Tour (talk) 18:30, 10 April 2011 (UTC)

Thanks, guys. All makes sense now. Pantscat (talk) —Preceding undated comment added 19:07, 10 April 2011 (UTC).

Map from z-plane to w-plane

If you consider the map from the z-plane to the w-plane of $C_{r}$, the circle centred at the origin, radius r in the z-plane, given by $w=f(C_{r})$ where $f(z)=(z-a)^{n}$ for some complex constant a and some integer n, how can you analytically determine the values of r at which the curve in the w-plane has a cusp (perhaps more explicitly, the values of r such that the curve has an extra loop for just greater than r than it has for just less than r). Perhaps the general case is hard to treat; if so, can an answer for n=3, 4 be given? Thanks. asyndeton talk 20:52, 10 April 2011 (UTC)

Intuitively, there'll be a cusp when your circle passes through an odd-degree root of the derivative of f. You'll probably want to consider roots of even degree too, though there will be no visible cusp in that case. –Henning Makholm (talk) 22:33, 10 April 2011 (UTC)

Thanks for that, it's very useful but apparently the use of the word 'cusp' in my question was misleading. Essentially, I want to know how to analytically determine at which values of r an extra loop will evolve. asyndeton talk 22:55, 10 April 2011 (UTC)

Then ignore whether the degree of the root is odd or even. (That's for the sense of "loop" I think would be most appropriate. You're welcome to suggest a more precise definition for what you're looking for). –Henning Makholm (talk) 23:14, 10 April 2011 (UTC)

I'm unsure how to describe it mathematically. Perhaps a second attempt would be: how do you analytically determine the values of r for which some part of the curve crosses over itself (inducing a loop, by my definition)? asyndeton talk 23:35, 10 April 2011 (UTC)

You know that "interesting" things happen to $f(C_{r})$ when $C_{r}$ passes through a root of $f(z)=0$. If $f(z)=(z-a)^{n}$ then the only root of $f(z)=0$ is at $z=a$, where there is a root of degree n. So pick values for a and n and try plotting $f(C_{r})$ for two values of r, one just less than |a|, and one just greater than |a|. Do this for a few different values of n to see how the behaviour depends on the order of the root. Gandalf61 (talk) 08:50, 11 April 2011 (UTC)

I know that I can get a feel for the behaviour by trying different values of 'a' and 'n', indeed this is what the project I am working on demands I do, but I'd like to get ahead of the game so-to-speak and determine the behaviour analytically, if possible. asyndeton talk 13:37, 11 April 2011 (UTC)

I think you are approaching this from the wrong direction. The aim of the project seems (to me) to be helping you to develop a visual intuition for how the curves $f(C_{r})$ behave. Without this visual inutuition, it will be very difficult to understand the verbal and mathematical explanations given in this thread (even though they are excellent explanations). Once you have developed a visual intuition then the analysis gives you a formal explanation of why the curves behave as they do. If you want to follow up on some of the concepts mentioned here, I recommend Tristan Needham's book Visual Complex Analysis. Gandalf61 (talk) 08:40, 13 April 2011 (UTC)

Interesting things happen elsewhere as well. Although the index of $f(C_{r})$ with respect to the origin only changes when $r=|a|$, the index of $f(C_{r})$ with respect to other points will change at other values of r. Specifically, the index with respect to p will increase by one for each new root of $f(z)=p$ enclosed by $C_{r}$. It should be possible to use this to answer the original question. Sławomir Biały (talk) 12:15, 11 April 2011 (UTC)

This looks interesting. Could you please define index for me though? It's a term I am unfamiliar with in this context. asyndeton talk 12:17, 11 April 2011 (UTC)

It's the winding number. Sławomir Biały (talk) 12:47, 11 April 2011 (UTC)

So, as $C_{r}$ passes through r=|a|, does the winding number of $f(C_{r})$ 'jump' from zero to n, where $f(z)=(z-a)^{n}$? asyndeton talk 00:03, 12 April 2011 (UTC)

The winding number of f(C_r) about the origin (!) jumps by n. For other points, the situation is more complicated. Sławomir Biały (talk) 00:16, 12 April 2011 (UTC)

It was noted above that "there'll be a cusp when your circle passes through an odd-degree root of the derivative of f". Clearly, when r is equal to a root of the derivative of f, f is at a minimum so I can see that, perhaps $f(C_{r})$ should locally be 'at a minimum', by which I mean that as r approaches the root, $f(C_{r})$ should travel along one side of the cusp then hit the point of the cusp when r is equal to the root and then travel along the other side of the cusp when r departs from the root. (That's not a brilliant explanation in any respect of what's going on but I'm sure you get the point.) I don't see why we need this to be a root 'of odd degree' of the derivative. Why won't even roots work? And could someone perhaps give me a better explanatino of what is actually going on? Thanks. asyndeton talk 19:21, 12 April 2011 (UTC)

If what you're interested in is the turning number of the curve, then this can only change when there is a singularity in the curve (such as, but not limited to, a cusp). These are the points of the parametric curve $\theta \mapsto f(re^{i\theta })$ where the derivative is zero (so only when r=|a|). If you are interested in where the number of loops with respect to given points (such as the origin, or some other point) changes, then the relevant notion is the winding number and singularities become less relevant. I think a basic question that you need to ask, and answer, first is "what is a loop?" Sławomir Biały (talk) 21:35, 12 April 2011 (UTC)

OK, this is becoming quite a challenge. I was under the, apparently rather naïve, impression that the winding number was the end of the story. Clearly this isn't the case. I only first encountered the winding number when asking a question on the Ref desk a few days ago, so I do not have a great comprehension of it. I'll have to get my head around both the winding number and turning number before I go too much further. Thanks for all your help. asyndeton talk 21:47, 12 April 2011 (UTC)

When $f'$ has a root of degree n at $a$, it implies that $f(z)$ near $a$ will look like $A(z-a)^{n+1}+C$ for some coefficients A and C. Of these C determines where in the w plane the image of z=a will be, and A determines the scale and orientation, but the essential shape of what happens is independent of both. For investigating the local behavior, the essential features will not be disturbed by setting A=1, C=0.

As long as we're only interested in the behavior near $a$, your $C_{r}$ can be approximated by a straight line through (or almost through) $a$. We can then move everything down to the origin by translating the z-plane by $-a$, and we're then left to study the behavior of $w=z^{n+1}$ when $z$ moves in a straight line that passes through (or close to) the origin. So z will approach 0 from some direction (say, at an angle $\theta$ from the positive axis), and then leave 0 towards the opposite direction $\theta +\pi$. In the w-plane these directions map to $(n+1)\theta$ and $(n+1)(\theta +\pi )$, respectively.

Now if $n$ (the degree of the root of $f'$, remember) is odd, then $n+1$ is even, so the difference between the two directions is $(n+1)\pi$ which is a multiple of $2\pi$ -- the directions are really the same direction. So w will approach the origin from some direction, come to a stop, and then leave it in the direction is just came from. That is a cusp! On the other hand, if $n$ is even and $n+1$ is odd, then the directions in the w plane differ by $\pi$ plus some multiple of $2\pi$. In this case, w approaches from some direction, comes to a stop (this is the singlularity Sławomir mentioned) but then accelerates away in the opposite direction it came from. The image of the line will look smooth (no cusp) even though there's really a singularity there.

For a deeper understanding of how this makes loop arise, consider a straight line in the z plane that passes close to the origin but not exactly through it. Its image in the w plane must wind around the origin $(n+1)/2$ times (so there are $n/2$ full turns of the curve here). Now compare this to a line that passes the origin on the other side. It now winds around the origin the same number of times, but in the opposite direction. The net difference in turning number is $n$. –Henning Makholm (talk) 01:35, 13 April 2011 (UTC)

integral

can someone tell me how to write on wikipedia the definite integral sign? i need both top and bottom limits, the middle function, and dx at the end thnx —Preceding unsigned comment added by 129.94.130.220 (talk) 22:51, 10 April 2011 (UTC)

Something like $\int _{-\pi }^{\pi -h}f(x)\,dx$. Or, if you prefer an upright d, $\int _{-\pi }^{\pi -h}f(x)\,\mathrm {d} x$. See also Help:Displaying a formula. –Henning Makholm (talk) 22:56, 10 April 2011 (UTC)

Or do what I do and see what somebody else has done by pressing edit but not saving the edit. Plagiarize! Let No One Else's Work Evade Your Eyes! Dmcq (talk) 20:03, 11 April 2011 (UTC)