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March 25

whaddya call this functional?

If you take a power series and each term c_n×xⁿ is replaced by {c_n÷n!} × xⁿ, then affine functions are fixed and each subspace <xⁿ> has eigenvalue (1/n!). It would send every series convergent on a disk of nonzero radius to a series convergent everywhere. I would think, though knowing zilch about functional analysis, that it would be called a uniformly continuous functional. Does this already have a name and can anyone tell me more or refer me? Thanks, Rich (talk) 05:05, 25 March 2008 (UTC)[reply]

Well, I don't know about you latter questions, in that post, but your series is the Maclaurin series for

e^{x}

at

a=0

, if the bounds for n are 0 and infinity. A math-wiki (talk) 10:34, 25 March 2008 (UTC)[reply]

Only if all the c_n's are 1... --Tango (talk) 11:59, 25 March 2008 (UTC)[reply]

Sum of diagonals

In

789
612
543

the sum of numbers on diagonals is 25. I made a program to brute-force compute the sum for 1001x1001 spiral, but now the problem is that the place from which this thing is is down so I can't check the answer I got (669,171,001). Also, is there some short way I could do this with just paper and pen? --212.149.216.233 (talk) 13:25, 25 March 2008 (UTC)[reply]

Yes and yes. The answer is indeed 669,171,001. To calculate it manually, first try to find the patterns in the numbers on the diagonal (for example, the sequence starting at the center and going to the bottom-left is given by the formula

4k^{2}+1\;\!

. Then you can express the sum as

1+\sum _{k=1}^{500}(16k^{2}+4k+4)

. To calculate this, the following formulae will be useful:

\sum _{k=1}^{n}k={\frac {n^{2}+n}{2}}

\sum _{k-1}^{n}k^{2}={\frac {2n^{3}+3n^{2}+n}{6}}

-- Meni Rosenfeld (talk) 13:54, 25 March 2008 (UTC)[reply]

I see the light now. Thank you! --212.149.216.233 (talk) 14:31, 25 March 2008 (UTC)[reply]

This spiral seems to be a very popular problem. I've first met it in this forum thread. Later, the sum of diagonals have been asked in a Project Euler problem, which I've learnt of in this mailing list thread. Somewhere in these, I give an explicit formula for computing any element of the spiral from its position (though I don't prove it). As that formula is a simple conditional with all branches being quadratic polynomials, you could easily write the sum of it as an explicit formula, though I didn't do that. Once you know that, you could just interpolate the polynomial formula for the sum from small values instead of calculating it from the first formula. Or, since the problem is now well-known, you could just look those small values up in the OEIS. – b_jonas 18:39, 25 March 2008 (UTC)[reply]

Exponential generating function

I've forgotten some of my power series skills. How does one efficiently figure out which function is given by, say, $f(x)=\sum _{n=0}^{\infty }{\frac {n^{2}x^{n}}{n!}}$ ? (In other words, how do we find the exponential generating function for the sequence {n²}?) I know which function it is, because I've worked it out a couple of different ways (including finding it in a list), but I feel that I'm missing the easy way to calculate it. Thanks in advance. -GTBacchus^(talk) 16:03, 25 March 2008 (UTC)[reply]

Either note that the operator x(d/dx) has the effect of multiplying each term in an exponential generating function by n, so your function is

x{\frac {d}{dx}}\left(x{\frac {d}{dx}}\left(e^{x}\right)\right)

or use the following identity:

{\frac {n^{2}}{n!}}={\frac {n}{(n-1)!}}=\left(1+{\frac {1}{n-1}}\right){\frac {1}{(n-2)!}}={\frac {1}{(n-2)!}}+{\frac {1}{(n-1)!}}

Gandalf61 (talk) 16:18, 25 March 2008 (UTC)[reply]

You should also check out this link. It is a link to the book "Generatingfunctionology", which is free online. It helped me quite a bit in my extremal combinatorics class, where we discussed generating functions. –King Bee ^{(τ • γ)} 18:51, 25 March 2008 (UTC)[reply]

I'll add that rule 2' on page 41 will give you an idea of how to find such an f that you seek. (The rule is basically what is described above by Gandalf.) –King Bee ^{(τ • γ)} 18:54, 25 March 2008 (UTC)[reply]

Cool; thank you both very much! -GTBacchus^(talk) 20:34, 25 March 2008 (UTC)[reply]

Sum of powers

It is possible to simplify $1^{3}+2^{3}+3^{3}+...+n^{3}$ to $(1+2+3+...+n)^{2}$ . (Proved by Mathematical Induction) It is possible to simplify $1^{2}+2^{2}+3^{2}+...+n^{2}$ similarly? How about $1^{m}+2^{m}+3^{m}+...+n^{m}$ ?

Yes. For any positive integer m, you can (rather easily) find a polynomial P of degree m such that

\sum _{k=1}^{n}k^{m}=nP(n)

In the case of

m=2

, you have

\sum _{k=1}^{n}k^{2}={\frac {2n^{3}+3n^{2}+n}{6}}

I don't know of a closed form formula for a general m. -- Meni Rosenfeld (talk) 18:39, 25 March 2008 (UTC)[reply]

A general approach might be to follow http://mathforum.org/library/drmath/view/56383.html ? 86.130.122.126 (talk) 19:37, 25 March 2008 (UTC)[reply]

There's also Faulhaber's formula and Bernoulli number, I guess. x42bn6 Talk Mess 19:39, 25 March 2008 (UTC)[reply]

Solving equation with inverse tangents

How would I begin to solve an equation of the following form for x?

$\tan ^{-1}(Ax+B)+C=\tan ^{-1}(Fx+G)+H\,$

--Bavi H (talk) 16:46, 25 March 2008 (UTC)[reply]

You could begin by applying tan to both sides and using a trigonometric identity. Algebraist 17:08, 25 March 2008 (UTC)[reply]

Or use the identity arctan(a)-arctan(b) = arctan((a-b)/(1+ab)). Collect the arctans on one side, apply the identity, and I believe that you'll get a quadratic in x involving tan(H-C). I assume that A, B, C, F, G and H are constants.—81.132.237.54 (talk) 19:43, 25 March 2008 (UTC)[reply]

Thanks to both of you for your help. I used Algebraist's advice and took the tan of both sides, then used the identity $\tan(a+b)={\frac {\tan a+\tan b}{1-\tan a\tan b}}$ . I was eventually able to see it would turn out to be a quadratic equation in terms of x, although I didn't complete all of the algebra myself.

In my question above, I used a general form with x representing the variable I'm solving for and A, B, C, F, G, H representing expressions not involving x. In fact, I was actually trying to solve the following equation for h

$\tan ^{-1}\left({\frac {L+h-E}{D}}\right)-\tan ^{-1}\left({\frac {{\tfrac {1}{2}}C}{D}}\right)=\tan ^{-1}\left({\frac {E-h}{D}}\right)-\tan ^{-1}\left({\frac {{\tfrac {1}{2}}E}{D}}\right)$

I created this equation while trying to solve a made-up problem about how high to hang a mirror on a wall so the angles above and below your reflection are equal. To begin solving this equation, I used another identity, -arctan(a) = arctan(-a), to change the differences into sums, then I was able to take the tan of both sides and use the tan(a+b) identity. Or I can see that the arctan a - arctan b identity that 81.132.237.54 mentioned does the same thing in one step.

I had originally tried to plug my equation into Quickmath to get some ideas about the solution. But the first time I entered my equation Quickmath couldn't find a solution. When I entered the general form I used in my question above, Quickmath found a complicated solution involving complex numbers. After the advice I got here, I was able to change my equation from a trig equation to a quadratic equation, but still got bogged down by all the algebra. Although I didn't completely solve it, I was more certain there was a simpler solution, so I tried Quickmath again. I re-entered my equation and Quickmath was able to solve it this time. (The first time I used .5C/D and .5Y/D -- Quickmath reserves E for the natural log base -- and Quickmath couldn't find the solution. This time I used C/2D and Y/2D and Quickmath was able to solve it.)

Thanks again to both of you for your help. Using your advice, I was more certain of the feasibility and method of the solution. I could have solved it myself if I had more determination to wade through the algebraic manipulation of all the variables of my problem, but gave up and went with an automated solution. --Bavi H (talk) 05:04, 26 March 2008 (UTC)[reply]

Parabolas and Flashlights - real world situation - moving the focus to change the light output

Ok... so most flashlights have a reflective curve in the shape of a parabola, to focus the light source. the light bulb itself is at the focus of the parabola.

my question: when you twist the top of a common flashlight and the beam narrows or widens, what is happening? clearly the metal reflective mirror is not physically changing, so i assume the light bulb (focus) is moving in and out... how does this in and out movement result in the beam changing?

Thanks...

The focus is a property of the reflector, so moves with it. If the reflector moves, the bulb changes position wrt the focus - if the reflector is unscrewed so that it moves out from the body of the torch, the bulb is then behind the focus and the beam will diverge. If it is possible to screw the reflector in past its normal position, i.e. with the bulb at the focus, the beam will converge to something approximating a point, and then obviously diverge thereafter.—81.132.237.54 (talk) 19:36, 25 March 2008 (UTC)[reply]

sphere question

can a 4" sphere fit through a 4" opening?

Depends—is its radius 4 inches? its diameter? its circumference? Strad (talk) 20:51, 25 March 2008 (UTC)[reply]

I assume that the opening is a circular hole cut out from a plane in Euclidean space. Let us define "fit through" as: there is a continuous path (curve) for the (centre of) the sphere such that at one point of the path the sphere is wholly on one side of the plane, at another point the sphere is wholly at the other side, and everywhere in between the sphere and the plane-with-a-hole are disjoint (they have no points in common).

Without loss of generality, let the plane be given as the set of points (x,y,z) such that z = 0, and let the radius of the circular hole be R. If we define the plane-with-a-hole as the closed set of points (x,y,z) in the plane such that x² + y² ≥ R², so that the circular edge of the hole has not been removed, then only (and precisely) spheres with a radius less than R will fit through. If also the edge of the disk cut out is removed, which means that we are left with the set of points in the plane for which x² + y² > R², then also the sphere with radius R will fit through, but still none with a larger radius.

The above equally applies if we give the plane a certain thickness, by taking the points such that abs z ≤ δ for some δ > 0.

In physical reality there are no closed sets, and any sphere will fit through any hole if you push hard enough – although one or both may not be quite the same as they were before the push. If there is no measurable difference between the radii within the limits of present technology, however, then a slight push should suffice, and there should be no appreciable change in quality of the sphere and the hole.

In a nice physics experiment, 's Gravesande's ring and ball,[1] a metal ball that would normally just fit through a ring is heated and placed on the ring, which is fixed at some height. Because the ball was heated, it has expanded in size and doesn't fit through. But as you wait, the ball will cool down and shrink, and suddenly it drops through. --Lambiam 09:41, 26 March 2008 (UTC)[reply]

Julian and Gregorian Calendars and Zeller's Congrunce

Hey everyone, recently I discovered Zeller's Congruence as well as Conway's Doomsday Rule. Zeller's Congruence is obviously powerful and useful (because I wanted to write a little program which would give the day of the week given a date and this was perfect). And FYI, my program takes care of identifying which calendar to use (Julian or Gregorian) based on the input date. Now my question is this, if I wanted to find out the day of the week on May 28, 585 B.C., would I plug in -585 for the year? The reason I ask is because as far as I know, there was no "year" numbered zero. We had 1 A.D. and then 1 B.C. but if we correspond years with integers, then what happens to the zero integer. Does Zeller's Congruence take this into account automatically or should I always add a year and plug in -584 instead (so 1 corresponds to 1 A.D., 0 corresponds to 1 B.C., -1 corresponds to 2 B.C., etc...)? Another question which was posed to me was that "A man was nearly 48 years old on celebrating his first birthday. Where, when, and what day of the week was it?" Any help from someone who knows more about Western calendars than me, would be appreciated! Thanks!
A Real Kaiser (talk) 21:41, 25 March 2008 (UTC)[reply]

The Julian calendar came into force in 45 BC, just in time to fix the date of Julius Caesar's murder as March 15, 44 BC. To find the day of the week for that date, you would have to treat 44 BC as −43 (making sure you don't compute remainders of negative numbers for modulo arithmetic). So y BC = −(y − 1).[2] I don't know whether in general earlier dates anachronistically but standardly use the Julian calendar, but on the NASA website maintained by Fred Espenak they do: [3]. For the Battle of Halys, see the blue swipe labelled "−0584 May 28" in this image. --Lambiam 10:56, 26 March 2008 (UTC)[reply]

How did a calander with Jesus' birth as its epoch appear before said date? *Max* (talk) 00:11, 1 April 2008 (UTC).[reply]

"Perfect" partition of the reals

This has been bothering me for years. I thought I would have figured it out by now, but I never did. There must be something completely obvious I am missing. So the question is: Can we find two sets A and B such that:

$A\cup B=\mathbb {R}$ ,
$A\cap B=\phi$ ,
$\mathrm {int} A=\mathrm {int} B\ (=\phi )\;\!$ and
$|A|=|B|\ (={\mathfrak {c}})$ ?

1 and 2 just means this is a partition, and 3 means the parts are dense in each other (between any two points of one part, there is a point of the other). So far this is satisfied by, say, the rationals and irrationals. But the asymmetry in their cardinalities is an eyesore, and while it seems intuitive that one could satisfy 4 as well, such a construction has so far eluded me. Thanks for any assistance resolving the issue. -- Meni Rosenfeld (talk) 22:45, 25 March 2008 (UTC)[reply]

How about the A as repeated copies of cantor set union the rational numbers, and B as it’s compliment? I don’t know if A would have any interior points or not though. GromXXVII (talk) 23:23, 25 March 2008 (UTC)[reply]

Hi Meni. I agree that this is tricky. You want a real analogy to the integer concepts of even and odd numbers. Bo Jacoby (talk) 23:32, 25 March 2008 (UTC).[reply]

Yes, that's a good way to look at it. -- Meni Rosenfeld (talk) 00:25, 26 March 2008 (UTC)[reply]

I think GromXXVII's construction works. All you have to do is add an uncountable number of points to the rationals, and even 1 copy of the cantor set would do it. The cantor set is totally disconnected, so it has no interior points, and throwing a few rationals in isn't enough to give it an interior, because they're measure 0. It's not a particularly symmetric construction, although it does satisfy the four conditions (I think). If you also require that both sets have the same measure, it gets a little harder, but you can still finesse it: of all the points in [0,1), put the rationals and the cantor set values in set A, and the rest in set B. On the interval [1,2), do the opposite. Continue alternating, and you've got two totally disconnected sets of unbounded measure that partition the reals. -GTBacchus^(talk) 00:08, 26 March 2008 (UTC)[reply]

So I am an idiot after all. I guess I never thought about this because subconsciously I was actually looking for something else, where there is even more symmetry. The last construction has a "global" symmetry, but I'd like to see something exhibiting more local symmetry - perhaps, the added requirement that even when restricted to any interval, the sets have the same measure. Any ideas? -- Meni Rosenfeld (talk) 00:25, 26 March 2008 (UTC)[reply]

Hmmm... another nice symmetry would be that any real translation of one of the two sets is either onto itself or onto its complement. The evens and odds have that property, if you look at integer translations. -GTBacchus^(talk) 00:43, 26 March 2008 (UTC)[reply]

What Meni wants can't be done with measurable sets. Suppose you had such a partition. Then the measure of A intersect (0,1) would have to be 0.5. So then cover A∩(0,1) with a countable collection of disjoint open intervals such that the sum of their lengths is less than 0.6. But then the measure of the intersection of A with each of those intervals is half the length of the interval, so the measure of A∩(0,1) as a whole is less than 0.3; this is a contradiction.

What you can get is a partition such that the intersection of A with any interval has inner measure zero and outer measure equal to the length of the interval (and similarly for the complement of A). I think the following works: Let U be an ultrafilter on the natural numbers, and then let x be in A just in case the set of positions in the binary expansion of x (past the binary point) where there's a 1, is an element of U. --Trovatore (talk) 00:58, 26 March 2008 (UTC)[reply]

You write "So then cover A∩(0,1) with a countable collection of disjoint open intervals such that the sum of their lengths is less than 0.6." I believe you can cover Q∩(0,1) with disjoint regions having this property, but doesn't countable additivity of Lebesgue measure imply that you can't cover the whole interval? (I know from experience that here be dragons, so I'm prepared to be wrong.) Tesseran (talk) 01:19, 26 March 2008 (UTC)[reply]

"The whole interval" is (0,1)? You're right; you can't cover that interval with a countable collection of disjoint open intervals whose lengths sum to less than 1. That's what keeps the whole concept of measure from trivializing. But by hypothesis A has measure 0.5, so by the definition of measure (or one of the definitions, anyway), for any ε>0, you can cover A with a collection whose lengths sum to less than 0.5+ε. --Trovatore (talk) 01:28, 26 March 2008 (UTC)[reply]

Note by the way that this is closely related to a question above, #Lebesgue_measure_of_a_subset_of_.5B0.2C_1.29_invariant_under_translation_by_Q, which once this discussion is archived should be reachable from the following link: Wikipedia:Reference desk/Archives/Mathematics/2008_March_24#Lebesgue_measure_of_a_subset_of_.5B0.2C_1.29_invariant_under_translation_by_Q. --Trovatore (talk) 03:05, 26 March 2008 (UTC)[reply]

How about letting A be the union of all positive rationals and all negative irrationals, and B the union of all positive irrationals and all negative rationals? And throw 0 into A. SmaleDuffin (talk) 13:46, 26 March 2008 (UTC)[reply]

That looks good to me. It's no more symmetric than the cantor set method, but it's a hell of a lot simpler. --Tango (talk) 14:26, 26 March 2008 (UTC)[reply]

Ok, thanks for all the replies! -- Meni Rosenfeld (talk) 20:19, 26 March 2008 (UTC)[reply]

Belated response to Bacchus

GTBacchus wrote above

Hmmm... another nice symmetry would be that any real translation of one of the two sets is either onto itself or onto its complement.

Turns out you can't do that, for quite trivial reasons (though it took me a long time to figure out that it was trivial :-). Suppose you have a partition of the reals into sets A and B with this property, and without loss of generality take 0 to be in A. Then for any real a in A, write a+A for the set $\{a+a'|a'\in A\}$ , and notice that a+A must equal A. That's because, by hypothesis, the only other possibility is that a+A equals B. But a itself is an element of a+A (because it equals a+0, and 0 is in A), and a is not an element of B, so a+A cannot equal B.

Similarly, for b taken from B, b+B must equal A. Otherwise b+B would have to equal B, which means b would be an element of b+B, which means b would equal $b+b'$ for some $b'$ in B. But the only possible such $b'$ is 0, and 0 is not an element of B.

Now notice that for any real number x, 2x must be an element of A, because either x is in A, in which case 2x is an element of x+A which in that case equals A, or x is in B, in which case 2x is an element of x+B which in that case again equals A.

But every real number is twice some real number. So every real number belongs to A, and B must be empty.

This is not a terribly robust argument, in that it depends strongly on the status of zero and on the translates of the sets being exactly equal to one of the two sets. I'm not sure what would happen if, say, you left zero out of the mix and demanded only that for any x, x+A (resp. x+B) be a subset of either A or B. --Trovatore (talk) 05:08, 28 March 2008 (UTC)[reply]

Ooh, thank you! I hadn't given the matter another thought, but this is cool. It's neat to apply your argument to A=the set of even integers and B=the set of odd integers. Everything makes sense, right up until "every x is twice some other number", which is not true in the integers, so it's clear why this is possible in the ring of the integers, and not in the field of the reals, nor in the rationals, because you didn't use completeness in your argument. It seems... that your argument would hold in any field, no? No, Z/2Z is a finite field with the partition A={0}, B={1}. I guess the desired partition is impossible in a field with more than two elements. -GTBacchus^(talk) 00:48, 29 March 2008 (UTC)[reply]

It probably fails for any field of characteristic 2: since 2 isn't a unit you can't half numbers, so not every number is twice another number. --Tango (talk) 01:19, 29 March 2008 (UTC)[reply]

Seems robust to me. The status of 0 is what it is, I see no problem with depending on it - how can you leave it out of the mix? It's a real number, so it has to go somewhere. And the sets being exactly equal was a requirement of the problem, so there's certainly no problem depending on that. --Tango (talk) 01:19, 29 March 2008 (UTC)[reply]

"Robust" in the sense of "easily generalizable to small variants of the original problem". --Trovatore (talk) 01:35, 29 March 2008 (UTC)[reply]

I don't think 0 itself matters. You can take any other point on the line as reference. Say you have a partition of the reals into sets A and B with this property, and without loss of generality take O to be in A. (O for origin.) Then for any real a in A, write a+A for the set

\{a+a'-O|a'\in A\}

, and notice that a+A must equal A. That's because, by hypothesis, the only other possibility is that a+A equals B. But a is an element of a+A (let a' = O), and a is not an element of B, so a+A cannot equal B. Similarly, for b taken from B, b+B must equal A. Otherwise b+B would have to equal B, which means b would be an element of b+B, which means b would equal b + b' - O for some b' in B. But the only possible such b' is O, and O is not an element of B. Now notice that for any real number x, 2x-O must be an element of A, because either x is in A, in which case 2x-O is an element of x+A which in that case equals A, or x is in B, in which case 2x-O is an element of x+B which in that case again equals A. But for every x, there is some y so that 2y-O=x. So every real number belongs to A, and B must be empty. Black Carrot (talk) 23:34, 29 March 2008 (UTC)[reply]

If it helps clarify that, the idea is to move the entire line over a bit, use the new 0 to do whatever you wanted, then move it back. For instance, a+A goes from a+a' to (a-O)+(a'-O)+O, which simplifies to a+a'-O. That's how you move the 0 point of any vector space around, and this is a vector space in one dimension. Black Carrot (talk) 23:40, 29 March 2008 (UTC)[reply]