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March 19

Why is the Integer zero-dimensional?

According to High-dimensional space, the Integers are zero-dimensional, Reals are one-dimensional, and Complex numbers are two-dimensional. Why are the integers zero-dimensional? They expand indefinitely in either direction along an imaginary 'line' (a 1D constuct) so I would have thought they were 1D themselves, much like the reals. --Iae (talk) 14:18, 19 March 2013 (UTC)

Maybe Zero-dimensional space would be helpful here... --CiaPan (talk) 16:29, 19 March 2013 (UTC)

Not really, for me. What part of that would relate to the integers? --Iae (talk) 18:30, 19 March 2013 (UTC)

Topologically, the integers are discrete — there is no way to approach an integer as a limit of other integers. Each integer just sits there all alone by itself. All discrete spaces are zero-dimensional, no matter how many points they have. --Trovatore (talk) 18:32, 19 March 2013 (UTC)

In integers, every integer number n (ie. every point of the space) creates a singleton {n}, which is a clopen set. Those singletons form a topological base for integers, and Zero-dimensional space says that a space with such base is zero-dimensional (with respect to the small inductive dimension). --CiaPan (talk) 06:37, 20 March 2013 (UTC)

Thanks. This runs counter to my intuition of dimensions, but I know little about topological spaces. Thanks for the links. --Iae (talk) 10:38, 20 March 2013 (UTC)

Fibonacci n

In Binet's formula

$F_{n}={\frac {\varphi ^{n}-\psi ^{n}}{\varphi -\psi }}={\frac {\varphi ^{n}-\psi ^{n}}{\sqrt {5}}}$

is it possible to isolate n? Pokajanje|Talk 15:45, 19 March 2013 (UTC)

I haven't worked it out in detail, but since $\psi =-{\frac {1}{\varphi }}$, you can mulitply through by $\varphi ^{n}$ and get a quadratic equation in $\varphi ^{n}$. Use the quadratic formula on that, being careful which root you use, and take logs. AndrewWTaylor (talk) 15:59, 19 March 2013 (UTC)

I can reduce things to $F_{n}{\sqrt {5}}=\varphi ^{n}+{\frac {1}{\varphi ^{n}}}$, but it seems like there's one variable too many. (And, sorry, the quadratic equation you describe isn't obvious to me.) Pokajanje|Talk 20:24, 19 March 2013 (UTC)

Multiply both sides of the following equation $F_{n}{\sqrt {5}}=\varphi ^{n}+{\frac {1}{\varphi ^{n}}}$ through $\varphi ^{n}$, and subtract the first side from both sides, then replace $\varphi ^{n}$ with t or x. You'll get a quadratic equation in t or x, whose roots are then computed through the consecrated formula. Extract then the logarithm base $\varphi$ from it/them in order to obtain n. — 79.113.221.97 (talk) 21:10, 19 March 2013 (UTC)

That would mean that $\varphi ^{n}={\frac {F_{n}{\sqrt {5}}-{\sqrt {{5F_{n}}^{2}-4}}}{2}}$. Therefore, $n=\log _{\varphi }({\frac {F_{n}{\sqrt {5}}-{\sqrt {{5F_{n}}^{2}-4}}}{2}})$ (I use that root because $F_{n}{\sqrt {5}}>{\sqrt {{5F_{n}}^{2}-4}}$ and $\varphi ^{n}$ cannot be negative.) Can that be reduced further? Pokajanje|Talk 22:14, 19 March 2013 (UTC)

Uhm... I thought this was about finding n when F_n is known... was I wrong ? :-\ — 79.113.221.97 (talk) 23:49, 19 March 2013 (UTC)

That's what I thought too. For all n, Fn is the closest integer to :${\frac {\varphi ^{n}}{\sqrt {5}}}\,.$

So, given a Fibonnacci number, multiply it by the square root of 5, then take log base phi, and round to the nearest integer, right? Bubba73 ^{You talkin' to me?} 03:09, 20 March 2013 (UTC)

I thought that formula I just found was the way to derive n from F_n. Pokajanje|Talk 03:45, 20 March 2013 (UTC)

To find $\varphi ^{n}$ you want to take the larger of the two roots of the quadratic (if n is odd both roots are positive, but the smaller root is $\varphi ^{-n}$). So

$\varphi ^{n}={\frac {F_{n}{\sqrt {5}}+{\sqrt {{5F_{n}}^{2}+(-1)^{n}4}}}{2}}\approx F_{n}{\sqrt {5}}$

$\Rightarrow n\approx \log _{\varphi }(F_{n}{\sqrt {5}})$

which is what Bubba73 said. Gandalf61 (talk) 14:32, 20 March 2013 (UTC)

I'm not sure it's appropriate to round it off in this case. Also, how did you get $+4(-1)^{n}$? The quadratic I get is $0=(\varphi ^{n})^{2}-\varphi ^{n}F_{n}{\sqrt {5}}+1$. Pokajanje|Talk 16:36, 20 March 2013 (UTC)

Your quadratic is not quite correct because $\psi =-{\frac {1}{\varphi }}$ so you have to take care with signs (or split into even and odd cases). Try a couple of numerical examples:

$\varphi ^{2}={\frac {2{\sqrt {5}}+6}{4}}={\frac {F_{2}{\sqrt {5}}+{\sqrt {{5F_{2}}^{2}+4}}}{2}}$

$\varphi ^{3}={\frac {8{\sqrt {5}}+16}{8}}={\frac {F_{3}{\sqrt {5}}+{\sqrt {{5F_{3}}^{2}-4}}}{2}}$

Gandalf61 (talk) 16:56, 20 March 2013 (UTC)

Rounding off as I described is appropriate because the procedure gets close to the proper integer very quickly. Bubba73 ^{You talkin' to me?} 18:21, 20 March 2013 (UTC)

Speed isn't that much of a factor here, because a computer's going to be doing all the calculations. Pokajanje|Talk 04:12, 21 March 2013 (UTC)

The method I gave is very fast. If you have calculated the square root of 5 and phi in advance, it takes one multiplication, one log, and one round. Bubba73 ^{You talkin' to me?} 04:15, 21 March 2013 (UTC)

Postscript: if you work it out, the difference between -4 and +4(-1^x) actually isn't that different. Plot the graphs of the functions and you'll see they're mostly identical. Pokajanje|Talk 03:21, 21 April 2013 (UTC)

Period of a repeating sequence

Hi,

I was wondering if anyone could help me prove something. Basically I’ve got a problem where I have a bunch of variables that vary periodically over time, forming a ‘sequence’, and I want to know when that sequence repeats. If the functions describing the variables are $f_{1}(t)\ldots f_{N}(t)$, then they can be put in a vector:

${\vec {x}}(t)={\begin{bmatrix}f_{1}(t)\\\cdots \\f_{N}(t)\end{bmatrix}}$

So now the idea is to find what the smallest value of ${\Delta }t$ such that ${\vec {x}}(t+{\Delta }t)={\vec {x}}(t)$.

One case I'm interested in is when the functions don't repeat within their periods (for example, they're saw-tooth waves). In this case working out when ${\vec {x}}(t)$ repeats is easy. For all the functions we can say:

$f_{n}(t+{\Delta }t)=f_{n}(t)\Leftrightarrow {\Delta }t=kT_{n},k\in \mathbb {Z}$

Where $T_{n}$ is the period of the nth function. Then for ${\vec {x}}(t)={\vec {x}}(t+{\Delta }t)$:

${\Delta }t=k_{1}T_{1}=\ldots k_{N}T_{N},k_{1},\ldots k_{N}\in \mathbb {Z}$

From which we can immediately conclude (by the definition of the greatest common divisor) that the smallest ${\Delta }t$ is:

${\Delta }t=\gcd(T_{1},\ldots T_{N})$

However, the other case I'm interested in is when the functions repeat once within their period, specifically when they are triangle waves. Intuitively I would believe that in this case (triangle waves) the sequence repeats when:

${\Delta }t={\gcd(T_{1},\ldots T_{N}) \over 2}$

But I can't figure out how to prove this. Can anybody think of a way?

203.167.139.208 (talk) 20:40, 19 March 2013 (UTC)

Your functional notation suggests that t is continuous, but your use of GCD suggests that it is discrete. If it's discrete, I'm not sure that there is a unique standard definition for a triangle wave. However, supposing that $f_{2}(t):=t\%2$ is one such (with the programmer's definition of % as the modulo operation) and $f_{4}(t):=\min(t\%4,4-t\%4)$ is another, then we have a counterexample because your $\Delta t={\frac {\gcd(2,4)}{2}}=1$ and obviously not all values are the same. (I unfortunately don't immediately have any idea for what it is in that case, so I'll first see if my interpretation is even correct.) --Tardis (talk) 05:53, 20 March 2013 (UTC)

Sorry, what I meant was least common multiple, not gcd, whoops! But I don't understand how either imply that t is discrete? It does imply that the periods T1...TN are integers, but even then, if the meaning of LCM() is extended I think it will still work with non-integers (i.e. if LCM(a,b) is taken to mean 'the smallest number that is an integer multiple of both a and b'). But yes, if t were discrete I wouldn't have expected my above intuition to be right because it's based on the fact that a (continuous) triangle wave is symmetrical. 118.93.174.101 (talk) 09:21, 20 March 2013 (UTC)

The implication arises because (with your definition, which is the only sensible one I know) neither $\gcd(1,\pi )$ nor ${\text{lcm}}(1,\pi )$ exist. Of course, you can restrict to rational periods (technically: commensurate periods), but you didn't say so, and that's relatively uncommon. Assuming that ${\vec {x}}(t+{\Delta }t)={\vec {x}}(t)$ is meant to apply at every t, your statement about triangle waves is still false, as the period cannot be smaller than the period of any component, and half the lcm can be (consider two periods, one a multiple of the other). You might also mean $\min\{\Delta t\in \mathbb {R} ^{+}:\exists t\;\forall k\in \mathbb {Z} \;\;{\vec {x}}(t+k\Delta t)={\vec {x}}(t)\}$, in which case I think it's rather more complicated with all the recrossings. --Tardis (talk) 13:04, 20 March 2013 (UTC)

Ok, fair enough, I think made too many unspoken assumptions in the question, and you comments has been helpful in pointing out the pitfalls in my argument above - but let me rephrase the question. If I have a set of triangle waves, each with an integer period, what is the 'period' of the whole set? By 'period' I mean what you said above ($\min\{\Delta t\in \mathbb {R} ^{+}:\exists t\;\forall k\in \mathbb {Z} \;\;{\vec {x}}(t+k\Delta t)={\vec {x}}(t)\}$). I do see what you mean by it being complicated by the recrossings - since there are points where x(t) takes on the same value it has before, but the sequence as a whole isn't 'repeating'; but that's why I'm asking for help :). 203.167.139.208 (talk) 19:56, 20 March 2013 (UTC)

(One more assumption: $T_{i}$ should be even, so define $H_{i}:=T_{i}/2$.) Well, now that we have a precise problem, what do we know? Each $f_{i}(t):=g_{i}((t+a_{i})\%T_{i})$ with $a_{i}\in [0,H_{i}),g_{i}(x):=\min\{x,T_{i}-x\}$ (the precise values of course do not matter, and only half the phases need be considered). To have $g_{i}(x)=g_{i}(y)$ we need $x=y$ or $x+y=T_{i}$. To have $f_{i}(t)=f_{i}(t+\Delta t)$ we then need $(t+a_{i})\%T_{i}=(t+\Delta t+a_{i})\%T_{i}$ (that is, $\Delta t=k_{i}T_{i}$) or $(t+a_{i})\%T_{i}+(t+\Delta t+a_{i})\%T_{i}=T_{i}$ (that is, $\Delta t=k_{i}T_{i}-2(t+a_{i})$). We can't neglect the simpler case by setting $t=-a_{i}$ in the complicated one because t must be shared for all i. In either case, $\Delta t$ must be even, so define $h:=\Delta t/2$.

Then we apply the Chinese remainder theorem: for any given t, we have a number of congruences of the form $h\equiv 0{\pmod {H_{i}}}$ or $h\equiv -t-a_{i}{\pmod {H_{i}}}$. If we take the latter case for all i, a solution exists iff $\forall i,j\;a_{i}\equiv a_{j}{\pmod {\gcd(H_{i},H_{j})}}$. We can choose the former case for any i by defining $a'_{i}=-t\%H_{i}$ (that is, we may conservatively shift it so that its minima are under consideration); any non-coprime pairs of (half-) periods will define a set of allowed restrictions (that is, sets of i for which we use $a'_{i}$) and values of t. In particular, if $H_{i}=H_{j}$ but $a_{i}\neq a_{j}$ we must substitute at least one one of them; the other must then have the value it would have were it substituted, so we may as well say that we must substitute both. (Then the two are equivalent and we can drop one of them.)

Beyond this, I don't see a particular approach beyond considering each $(H_{i},H_{j})$, finding the required substitutions (substituting both will always work, of course, and devolves to the sawtooth case) and constraints on t (which will eventually be subject to its own Chinese remainder problem from all the half-substituted pairs) via constraint propagation, and then finding h for each allowed system. As for actually finding the answer, it might be simpler to do it inductively by keeping a set of allowed $(t,h)$ pairs and extending/filtering it as you consider each function (extending to reach the new LCM trivial answer). --Tardis (talk) 03:02, 23 March 2013 (UTC)

See Almost periodic function. Bo Jacoby (talk) 14:27, 20 March 2013 (UTC).

March 20

Normal Distribution

If I have a sample of size n drawn from a normal distribution with mean μ and variance σ², what is the probability density function of the value of the k^th percentile P_k? — Preceding unsigned comment added by 123.136.64.14 (talk) 07:31, 20 March 2013 (UTC)

See Order statistic. Bo Jacoby (talk) 14:22, 20 March 2013 (UTC).

Integral of 1/x

I was taught that the integral of 1/x is ln|x|+C but does this actually mean that the integral of dx/x is ln|x| + C? Clover345 (talk) 16:01, 20 March 2013 (UTC)

Yes. Because 1 * dx = dx. Obviously. — 79.113.214.121 (talk) 16:30, 20 March 2013 (UTC)

No. It means

$\int {1 \over x}\,dx=\ln \left|x\right|+C.$

We talk about the integral of a function (meaning the indefinite integral in this case), not the integral of dx times a function. The dx goes with the integral and indicates that the integral is over x. RockMagnetist (talk) 20:42, 20 March 2013 (UTC)

$\int {1 \over x}dx=\int {dx \over x}$ — 79.113.214.121 (talk) 20:56, 20 March 2013 (UTC)

While the two statements are equivalent, I would say that the second is more accurate: dx/x is a differential form and so can be integrated. The bare statement "integral of 1/x" is meaningless without saying what measure the integral is taken with respect to. Sławomir Biały (talk) 21:54, 20 March 2013 (UTC)

This isn't really what you asked, but it's probably worth pointing out that the absolute-value sign in the expression $\ln |x|+C$ is kind of a cheat. If you're not careful, you might think it means that, say, you can write $\int _{-2}^{2}{\frac {dx}{x}}=\ln |x|{\bigg |}_{-2}^{2}=0$. But you can't — that definite integral is not defined. The version with the absolute value works for intervals completely contained within the negative reals or completely contained within the positive reals, but it doesn't work for intervals containing zero. --Trovatore (talk) 22:05, 20 March 2013 (UTC)

While the Lebesgue and improper Riemann integrals of ${\frac {1}{x}}$ are not defined over intervals containing zero, the formula still gives the correct Cauchy principal value for such integrals. In that sense, the formula $\int _{-2}^{2}{\frac {dx}{x}}=0$ is true (but you have changed the meaning of the $\int$). —Kusma (t·c) 10:43, 21 March 2013 (UTC)

Yeah, but Cauchy principal values are mostly a curiosity. They're not very robust under stuff you'd like to be able to do. In my view they're ordinarily best avoided unless you have a special problem with a clear rationale as to why that value is relevant. --Trovatore (talk) 14:58, 21 March 2013 (UTC)

Yes the principal value is a far more useful idea, but then you're dealing with complex numbers. Any idea who first called it that anyone? Cauchy didn't also do this did he? Dmcq (talk) 15:54, 21 March 2013 (UTC)

I really don't see any very direct connection between those two notions. The article you linked to seems to be just talking about making a branch cut so that you have a single-valued analytic function on a slightly reduced domain. Cauchy principal values, on the other hand, are ways of assigning a value to integrals that don't converge. They're similar to the various ways of assigning values to the sum of a divergent series, and have the same sorts of problems. --Trovatore (talk) 16:18, 21 March 2013 (UTC)

I'm just saying that for complex numbers you've got Cauchy's integral theorem and the constant is fixed within a branch, the integral is given by the difference in the principal values of the integral at the two points, in this case by the values of log z as given in the principal value article. It is true though that one can always go around 0 a few times and so not stay with a branch. Dmcq (talk) 17:53, 21 March 2013 (UTC)

Oh, true — I thought you were trying to find a commonality between the two senses of "principal value". If you take one of the two direct paths from −2 to 2, the integral will be either −iπ or iπ, depending on which one you take. --Trovatore (talk) 19:13, 21 March 2013 (UTC)

I’d take issue with the claim that Cauchy principal values are mostly a curiosity. They come up naturally in Fourier analysis, for one thing.—Emil J. 13:27, 26 March 2013 (UTC)

March 21

Number Theory Problems

• Are there any solutions to: x!y!=z! for any x,y and z.
• The number [100a+10b+c] is a prime. Prove that b^2-4ac cannot be a perfect square. Solomon7968 (talk) 11:06, 21 March 2013 (UTC)

1) Allowing some of x, y, z to be 0 or 1 gives lots of solutions. Otherwise I don't think it's possible. Staecker (talk) 11:36, 21 March 2013 (UTC)

1) There's also a family of solutions (n!-1)!n! = (n!)! e.g. 5!3! = 6!. Gandalf61 (talk) 12:40, 21 March 2013 (UTC)

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. See further our articles on factorial and quadratic formula. Pokajanje|Talk 15:51, 21 March 2013 (UTC)

Are x, y, and z supposed to be natural ? Either way, you might also want to check this out. Also, are a, b, and c supposed to be digits from 0 to 9 ? Frankly, I see no reason why b² - 4ac can not be a perfect square... all that is certain is that b² - 4ac + 4ap is supposed to be one, and the two conditions do not seem to be mutually exclusive... — 79.113.245.81 (talk) 21:52, 21 March 2013 (UTC)

There is a relatively easy non-trivial set of solutions to the factorials problem. To find it, you must remember that n! = n⋅(n-1)!.

There are, however, solutions outside this particular set (but which follow a similar logic). It burned my brain to find one, but I did: 6!7!=10! because 6!=720=8×9×10. Thank you and good night.

Oh, and if you use advice from the reference desk for HW at the university level, you must cite it on your paper (as with any person irl from whom you received help). Not doing so may be considered bad. 14:52, 22 March 2013 (UTC) (By User:SamuelRiv)

Another solution set can be found by setting y=z-2 and solving the quadratic eqn for z in terms of x (this method shown to me by a colleague). Then you get a condition for x alone (which may be related to Brocard's problem, suggesting finitely many solutions). The smallest solution I found via plug-and-chug was x=19, giving z=348776557. SamuelRiv (talk) 17:41, 23 March 2013 (UTC)

Re second question: a=0, b=2, c=3. [100a+10b+c]=23, a prime. b^2-4ac = 2^2-4*0*c = 4, a square number. Did I misunderstand the question? --NorwegianBlue ^talk 21:13, 22 March 2013 (UTC)

Given the way that the problem is written, I presumed that 'a' had to be greater than zero. Just because we've probably gone passed when the HW was due, I think the major points in the proof are something like

• if b^2+4ac is a perfect square then (-b+/-sqrt(b^2-4ac))/2a has two integer values, m & n.

• In that case the factorization of ax^2+bx+c is of the form x-m and x-n, which means that a*10^2+b*10+c can be factored into 10-m and 10-n.

• The piece that needs to be handled is making sure that for these values that m(or n) can't be 9 which would allow for a factorization, but would still allow b^2+4ac to be a square. I'm not quite sure how to do that.Naraht (talk) 21:41, 24 March 2013 (UTC)

March 22

0/0

why we are not able to divide zero by zero? in artical parity of zero,this tell that zero is even.if zero is even then why it is not divided by itself — Preceding unsigned comment added by Muhammad waqas arif (talk • contribs) 16:14, 22 March 2013 (UTC)

Please read Division by zero, especially the sections called In elementary arithmetic and Division as the inverse of multiplication. Zero is even, which means it can be divided by 2 with no remainder: 0/2=0. But no number, including zero, can be divided by zero. —Bkell (talk) 16:18, 22 March 2013 (UTC)

There are various rules for how to handle zero which conflict with each other in the case of 0/0:

• "Zero divided by anything equals zero" means the result should be zero.

• "Anything divided by itself equals one" means the result should be one.

• "Anything divided by zero equals either positive or negative infinity (the sign of the numerator)" means the result should be ±∞.

It can't be all of these things, so the operation is not permitted. StuRat (talk) 16:26, 22 March 2013 (UTC)

"Anything divided by zero equals either positive or negative infinity (the sign of the numerator)" is a rule in IEEE floating point, but it is not a rule in any mathematical context that I know of. It is true that for a nonzero constant c the limit of c/x as x approaches zero is ±∞, depending on the sign of c, but c/0 itself is not defined. —Bkell (talk) 16:41, 22 March 2013 (UTC)

See real projective line. --Trovatore (talk) 16:43, 22 March 2013 (UTC)

Sure, division by zero is allowed there, but there is no distinction between positive and negative infinity. —Bkell (talk) 16:45, 22 March 2013 (UTC)

I like to think of the tangent graph, which shows how the curve is asymptotic to both positive and negative infinity, at the same points: [1]. So, it's +∞ approaching from one direction and −∞ when approaching from the other. However, at the actual point, it is rather undefined as to whether it's +∞ or −∞, I agree. StuRat (talk) 01:07, 24 March 2013 (UTC)

Anyway, the contradiction among these "rules" is not the fundamental reason that 0/0 is not defined; 0/0 is undefined because there is not a unique solution to the equation 0x = 0. —Bkell (talk) 16:46, 22 March 2013 (UTC)

Which is very closely related - StuRat's examples are the solutions 0, 1 and +/- ∞. Rich Farmbrough, 00:53, 23 March 2013 (UTC).

An interesting subtlety is that, while 0/0 does not usually make sense, it does make sense to say that zero divides zero (because $a|b\iff \exists n(b=na)$). However, unaccountably, one convention is that zero does not divide zero.

That convention, to the extent a convention can be wrong, is just wrong. I can see no advantage whatsoever, and considerable disadvantage, to excluding 0|0. In particular, if you don't say 0|0, then you lose the lattice structure of divisibility on the naturals, with 0 as the greatest element. --Trovatore (talk) 16:48, 22 March 2013 (UTC)

${\frac {0}{0}}={\frac {\pm 0}{0^{2}}}=\lim _{x\to 0}{\frac {\pm x}{x^{2}}}=\lim _{x\to 0}{\frac {\pm 1}{x}}=\pm \infty$

${\frac {0}{0}}={\frac {a\cdot 0}{0}}=\lim _{x\to 0}{\frac {ax}{x}}=a$ , where a can be any real number whatsoever. — 79.113.218.110 (talk) 20:14, 22 March 2013 (UTC)

0 being even and 0/0 being undefined don't have any relation to each other. Double sharp (talk) 14:36, 23 March 2013 (UTC)

Correction: $a$ is not 'any' real number, it is 'every' real number, and positive and negative infinity. Plasmic Physics (talk) 01:16, 24 March 2013 (UTC)

a cannot be every real number, since that contradicts both logic and grammar. Nor is a x 0 defined when a = ±∞. — 79.113.210.71 (talk) 04:21, 24 March 2013 (UTC)

Care to explain the perceived contradiction? Plasmic Physics (talk) 04:45, 24 March 2013 (UTC)

My reasoning is that since the product of every element of the set of real numbers; and zero, is equal to zero; and zero divided by poitive and negative infinity, is equal to zero, it follows that zero divided by zero equals every element of the set of real numbers, and positive, and negative infinity. Plasmic Physics (talk) 04:54, 24 March 2013 (UTC)

As mentioned already, 0/0 is not defined. Nor is 0/∞. The example above shows how you can make a limit of the form 0/0 that actually converges to a for any a. You couldn't make it converge to all a simultaneously, because that would violate the definition of convergence.--80.109.106.49 (talk) 07:45, 24 March 2013 (UTC)

There must be something wrong with the convergence proof above, because my proof is definitely correct, unless the commutative principle is incorrect. Plasmic Physics (talk) 09:12, 24 March 2013 (UTC)

The limit of a function approaching a point is not necessarily equal to the value at the point. For the function which is 3 everywhere but 5 at 0 the limit at 0 is 3 but the value at 0 is 5. In the equation above 0/0 = limit x/x as x goes to 0 but that's not necessarily true. If there was a consistent result then that would be a good definition of 0/0 but there's nothing that's anywhere good enough. Dmcq (talk) 12:34, 24 March 2013 (UTC)

True, but division isn't exactly some random, custom-made, user-defined function. — 79.113.235.6 (talk) 18:57, 24 March 2013 (UTC)

I'm confused, is that comment for or against my comment? Plasmic Physics (talk) 12:33, 25 March 2013 (UTC)

Saying 0/0 = limit ax/x as x tends to 0 is wrong. That type equality is only true for functions which are continuous at the limit point. Continuity is a nice property and if there was a value that such limits consistently were equal to then that would be a good value to assign to 0/0. However as you have shown there isn't. It would be possible to assign some standard agreed value to 0/0 but that gives far more problems than it solves so we don't assign a value to 0/0. On the other hand if we had defined 5+x for all values except x=0 we would find that the limit as x tended to 0 for 5+x, 5+ax, 5+x² etc all were 5 and we would feel happy about assigning the value 5 to 5+0 so as to get a continuous function. For 0/0 if we knew we should be getting some definite value we'd assume we had made a small mistake could get rid of the zero's somehow, perhaps t is a limit we re trying to get? perhaps we kept a common factor of 0 in the numerator and denominator we should have cancelled out at an earlier stage? If we had a standard value then some maths software would just churn out that stupid value and we wouldn't have the foggiest idea we were totally wrong. Dmcq (talk) 13:25, 25 March 2013 (UTC)

Truth is self-coherent and self-contained, devoid of contradictions, and never at odds with itself. The di- in divergence means two. The con- in convergence means together. If one can show that the limit of a function in a point can have at least two different values, this means that the function is divergent. Because of the very meaning and definition of the word divergent. — 79.113.239.66 (talk) 18:47, 24 March 2013 (UTC)

The first element is actually dis, "apart": dis vergentia, "[to] bend/turn apart". Double sharp (talk) 11:26, 25 March 2013 (UTC)

dis- itself ultimately comes from the same PIE word for two, just like di-. — 79.113.235.6 (talk) 11:32, 25 March 2013 (UTC)

March 23

Nonconvex polyhedra

What's the smallest number of faces a concave polyhedron can have? A self-intersecting one? What if the faces themselves must be convex? Double sharp (talk) 14:37, 23 March 2013 (UTC)

I suppose that the empty set qualifies as a polyhedron: It is a subset of three dimensional space, all edges are straight and all faces are plane. So the answer is zero. Bo Jacoby (talk) 15:30, 23 March 2013 (UTC).

LOL. Excluding this, then? Double sharp (talk) 16:29, 23 March 2013 (UTC)

Possibly six: a tetrahedron minus a lower tetrahedron on the same base. And I can't see for now how that could be less than six. --CiaPan (talk) 23:11, 23 March 2013 (UTC)

You can do it with five faces if they don't have to be convex. Just think tertrahedron with one side chopped away a bit so it forms two faces. Dmcq (talk) 23:19, 23 March 2013 (UTC)

That would still be six faces, wouldn't it? The tetrahedron has four faces, and then you're forming two new faces from one of its edges. Maybe I'm not correctly visualizing what you're describing. —Bkell (talk) 23:35, 23 March 2013 (UTC)

No two faces from one face. Altogether you get six points, nine lines and five faces. Dmcq (talk) 01:33, 24 March 2013 (UTC)

Oh, I see it now. It's a pyramid with a base that is a concave quadrilateral. That has five faces, five vertices, and eight edges, right? —Bkell (talk) 05:30, 24 March 2013 (UTC)

Yes that shape is right. I had a line across the face but if it ends at one of the vertices yes you can cut it down to five vertices and eight edges, don't know why I didn't think of that, thanks. Dmcq (talk) 11:48, 24 March 2013 (UTC)

That's right, Dmcq, I must have limited myself somehow to use convex faces only. --CiaPan (talk) 19:03, 24 March 2013 (UTC)

mechanics

tell me the method to find the moment of inertia...write in detail — Preceding unsigned comment added by 204.14.79.224 (talk) 19:16, 23 March 2013 (UTC)

See moment of inertia and come back to us if you have a more specific question. Rojomoke (talk) 19:52, 23 March 2013 (UTC)

March 24

The symbol used when defining the alphabet of a language

Syntax (programming languages)#Syntax definition uses the ::= symbol to mean something like "is comprised of the following". However, I wished to know the name of that symbol, but could not find it. Technical restrictions prevent you from searching for it directly, and none of the articles I manually checked (e.g. Syntax (logic), Formal language and Alphabet (computer science)) make use of it. Could someone tell me (a) what it is called, and (b) whether it has its own WP article? Thanks! It Is Me Here ^{t / c} 18:56, 24 March 2013 (UTC)

See also Backus-Naur_form. I don't think that collection of 3 symbols has a name, but just means "is defined as". -- SGBailey (talk) 21:29, 24 March 2013 (UTC)

It is really quite difficult at present to find out what that symbol means. Might I suggest adding information on it to more articles, e.g. List of logic symbols? It Is Me Here ^{t / c} 12:47, 26 March 2013 (UTC)

March 25

Parent function

What is a parent function? Admittedly, it's been a while since I got my math degree, but I don't remember ever hearing the term. The only page I found on WP that mentions it is Isocline, but this fails to explain what it is. Mathworld (Wolfram.com) fails to have an entry for it, either. — Loadmaster (talk) 03:09, 25 March 2013 (UTC)

What is the context? —Bkell (talk) 03:11, 25 March 2013 (UTC)

My daughter's 7th grade homework (ha!), in which she has to determine the parent function of several functions, e.g., a set of parabolas. I just found this video, which explains that it's essentially the simplest of a family of functions, e.g., for n-degree polynomials it is y = xⁿ. WP ought to have an article for this. — Loadmaster (talk) 03:16, 25 March 2013 (UTC)

I just added a stub article for parent function. — Loadmaster (talk) 03:34, 25 March 2013 (UTC)

Hmm. I've never heard that terminology used before. I watched the video you linked, and I understand the concept in the examples that were given, but the "definition" given there is not very rigorous, and it's hard to extend it beyond very basic families of functions. I imagine the point of all of this is to lead into the graphing of functions via transformations, such as vertical and horizontal shifting and scaling (and I can't find a Wikipedia article about this, either). If that's the goal, then the idea is just that there are a handful of "fundamental" functions, like y = x, y = x², y = √x, and y = |x|, the graphs of which the student should have memorized; and then graphs of related functions can be found by transforming these "fundamental" graphs. So the term "parent function" seems to mean just "the function in this list of 'fundamental' functions that most closely resembles the given function." With that definition, though, for any list of "fundamental" functions there are many functions that have no parent function at all. For example, what should the parent function of f(x) = x³ + 2^x be?

I don't think we should have a Wikipedia article about this unless we can find a solid, rigorous definition of the term from a reliable source. —Bkell (talk) 03:53, 25 March 2013 (UTC)

Here's another question to ponder: According to the current definition in the parent function article ["a parent function is the simplest function of a family of functions that preserves the definition (or shape) of the entire family"], why should y = x⁴ be the parent function of y = x⁴ − 10x² + 9? The graphs of those functions have rather different shapes. —Bkell (talk) 03:58, 25 March 2013 (UTC)

I'd think it should be y = x⁴ - x², if the rule is to drop the coefficients and constants. StuRat (talk) 04:10, 25 March 2013 (UTC)

Why should that be the rule? That contradicts what the parent function article currently says: "For the family of n-degree polynomial functions for any given n, then, the parent function is simply xⁿ." This all seems very arbitrary to me. —Bkell (talk) 04:16, 25 March 2013 (UTC)

It also contradicts the video that Loadmaster posted, which is the only source for the article at the moment. That video clearly states, for example, that the parent function of y = −x² + 10x − 20 is y = x². —Bkell (talk) 04:23, 25 March 2013 (UTC)

Based on the [lack of] 'context' given I suppose the meaning of function x → x² being a 'parent' function for all ax² + bx + c might be that each function of the latter family can be obtained from the 'parent' function by a shift in its argument and value space or scaling. That however strongly limits an application area of the 'parent function' notion, because no such 'parent' function exists for polynomials of 3rd degree or higher. Possibly the notion is used only at 7th grade...? --CiaPan (talk) 06:43, 25 March 2013 (UTC)

Let's not take anything that the parent function article says as established fact, because I just wrote it. It's based on the video I found. There do not seem to be many authoritative sources out there in net-land to say anything definitive about the subject. (Further discussion about this should continue on the article talk page.) — Loadmaster (talk) 17:29, 25 March 2013 (UTC)

A Question of Convergence

I am unable to calculate $\lim _{n\to \infty }{\int _{0}^{\infty }{e^{-e^{x^{n}}}}}$. Even with all the aid of mathematical software, I can only understand its behaviour on the interval [0, 4], but whether afterwards it converges asymptotically towards a certain value, or whether -on the contrary- it continues to grow -slowly but steadily- towards +Infinity, is a mystery to me... Could anyone please help me ? — 79.113.235.6 (talk) 06:12, 25 March 2013 (UTC)

Dollars to donuts this thing converges to e⁻¹. I don't have a rigorous proof of that claim, but look at the graph of exp(−e^xn) for a few values of n, and you can see what's going on. —Bkell (talk) 06:27, 25 March 2013 (UTC)

Okay, here's an outline for how to deal with this thing. Split the interval of integration into two subintervals, [0, 1] and [1, ∞). For 0 ≤ x ≤ 1, show that $e^{-1}(1-x^{n})\leq e^{-e^{x^{n}}}\leq e^{-1}$, and for 1 ≤ x < ∞, show that $0\leq e^{-e^{x^{n}}}\leq e^{-e}x^{-n}$. Now use something like the squeeze theorem to show that the integral on [0, 1] approaches e⁻¹, and the integral on [1, ∞) approaches 0. —Bkell (talk) 06:55, 25 March 2013 (UTC)

Loss of information

What is it called when information about a number of function is lost throught some sort of manipulation, which makes the manipulation irreversable. For example, negative one squared equals one, the reverse: taking the square -root of one, does not equal negative. By performing the square function, you lose the negatie sign. Plasmic Physics (talk) 21:41, 25 March 2013 (UTC)

The function is not injective. —Bkell (talk) 22:06, 25 March 2013 (UTC)

Integer-only Mathematics

Is there a specific name for mathematics that only uses integers and all functions take integers and return integers? — Preceding unsigned comment added by 71.204.230.66 (talk) 23:44, 25 March 2013 (UTC)

See "Number theory".—Wavelength (talk) 00:11, 26 March 2013 (UTC)

Computability theory also deals with functions whose domain and range are the natural numbers, although its primary focus is the question of which of these functions can be computed by algorithms, rather than the kinds of questions that are usually asked about functions in, say, mathematical analysis. The calculus of finite differences is roughly an analogue of calculus in the domain of the integers. —Bkell (talk) 05:38, 26 March 2013 (UTC)

Modular arithmetic. StuRat (talk) 05:57, 26 March 2013 (UTC)

Related fields that deal with mathematical objects that can be indexed by the integers are discrete mathematics and combinatorics. Gandalf61 (talk) 08:45, 26 March 2013 (UTC)

The adjective Diophantine comes to mind, though I’m not sure it can be used for non-algebraic functions.—Emil J. 13:21, 26 March 2013 (UTC)

See "Category:Number theory".—Wavelength (talk) 15:28, 26 March 2013 (UTC)

Perspective correct interpolation of screen coordinates given desired texture coordinates on a plane

File:Iterpolation of screen coordinates based for given texture coordinates.png

Interpolation of screen cordinates for given texture coordinates (lower frame), initially given only four correspondences (upper frame).

I'm looking for a simple formula to allow perspective correct interpolated 2D screen cordinates to be calculated for given 2D texture coordinates, given that a 3D projection, distortionless camera is looking at a plane. Specifically, I don't know anything about the camera position, orientation or field of view (and I don't care about it directly) and I only have some reference points. It needs to be as fast as possible to solve.

The article on texture mapping shows the opposite: perspective correct interpolated 2D texture coordinates (U, V) given the 2D screen coordinates (X, Y).

Here is the problem described exactly:

Input: Four corresponding screen and texture coordinates (four sets of (X, Y) coordinates and four sets of corresponding (U, V) coordinates)

Output: A function that, given (U, V), it generates the appropriate (X, Y) coordinate, exact if the camera is distortionless.

I would also settle for the solution where the input (U, V) values are limited to (0, 0), (0, 1), (1, 0) and (1, 1). Eonzo (talk) 14:25, 26 March 2013 (UTC) Eon

If you don't supply the 4 corner points, and thus are doing extrapolation, then the error in the calculations will tend to magnify, the further outside the inputs you go. The problem is also simpler if 2 pairs of input coords vary only in U, and 2 pairs vary only in V.

I also note that your intermediate grid lines are not evenly spaced. This is correct in a perspective view, but does complicate the math. Would you settle for evenly spaced grid lines ? That is, where we can just linearly interpolate between the inputs ? (If we can't take this shortcut, then we'd probably do best to calculate the camera angles from the inputs, which could mean that many sets of inputs would fail to converge on a solution, and others would have multiple solutions. Very messy. Note that in the top pic, we can't tell which corner is closest, since that info isn't present, but we've added that info with the spacing of the grid lines in the 2nd pic, meaning we've added info not present in the input. If we spaced the grid lines further from the lower, right point, then that point would appear to be the closest.) StuRat (talk) 15:22, 26 March 2013 (UTC)

• The standard algorithm is explained at Texture mapping#Perspective correctness. Looie496 (talk) 16:34, 26 March 2013 (UTC)

• I'd think we could do better than the "affine" version displayed at that link, even without knowing depth info, like so:

      +-+-+
     /  |  \
    /---+---\
   /    |    \
  +-----+-----+

I see no reason why the grid lines can't all be straight, even if we do lack depth or camera placement info. We do, however, still have the problem of how the grid lines are spaced, as I mentioned above. StuRat (talk) 17:08, 26 March 2013 (UTC)

Why Are All Basic Irrationals So Close to Multiples of 1/7 ?

$\pi \approx {22 \over 7}=3{1 \over 7}\ \ \ \ \ \ \ \ \ \ e\approx {19 \over 7}=2{5 \over 7}\ \ \ \ \ \ \ \ \ \ \gamma \approx {4 \over 7}$

${\sqrt {2}}\approx {10 \over 7}=1{3 \over 7}\ \ \ \ \ \ \ \ \ \ {\sqrt {3}}\approx {12 \over 7}=1{5 \over 7}\ \ \ \ \ \ \ .$ 79.113.221.135 (talk) 15:16, 26 March 2013 (UTC)

Interesting observation, but there can't really be any reason for this. Mathematics just is what it is. There are "reasons why" when you get into theorems, but not general observations. IBE (talk) 15:55, 26 March 2013 (UTC)

Let's list the deviation of each:

$\pi \approx {22 \over 7}=3{1 \over 7}$ Deviation = 0.040%

$e\approx {19 \over 7}=2{5 \over 7}$ Deviation = 0.147%

$\gamma \approx {4 \over 7}$ Deviation = ?

${\sqrt {2}}\approx {10 \over 7}=1{3 \over 7}$ Deviation = 1.015%

$sqrt3\approx {12 \over 7}=1{5 \over 7}$ Deviation = 1.026%

The last two aren't particularly close. I'm not sure what you had in mind for the value of gamma. For the square root of 2, you'd do a lot better with 99/70, which has a deviation of only 0.005%. For the square root of 3, 97/56 also gives a deviation of only 0.005%. StuRat (talk) 16:13, 26 March 2013 (UTC)

I suppose that'd be the Euler-Mascheroni constant. With a deviation from $4 \over 7$ of around 1.00%. 129.234.186.45 (talk) 17:28, 26 March 2013 (UTC)

Independence number of a cycle

Is it true that the independence number of a cycle is $\lfloor n/2\rfloor$? It is easy to see that if the cycle is given by $v_{1}v_{2}\cdots v_{2m+1}$ then $\{v_{1},v_{3},\cdots ,v_{2m-1}\}$ form an independent set having $\lfloor n/2\rfloor$ elements in the case of an odd cycle and the alternate vertices form an independent set in the case of an even cycle. But where can I find a proof that this is the independent set of maximum cardinality? Thanks.--Shahab (talk) 17:10, 26 March 2013 (UTC)