I remember an old math teacher taught me a method, given numbers n1, n2, etc. , to determine a function f(x) for which f(1) = n1, f(2) = n2, etc., but I forget how. I recall that it had something to do with subracting the first two terms with each other, then subtracting this difference with the 3 number, and so on until you got 0, but I'm not sure about this. There were also a lot of factorials involved. Does anyone know what I'm talking about? —Preceding unsigned comment added by 74.15.138.134 (talk) 02:30, 16 June 2009 (UTC)[reply]

Note that, trivially, one can always define a function on the natural numbers to have those values (just definite it like that!). However, with the other information you have provided, it appears that you want a polynomial having those values restricted to the natural numbers. Let me give you the intuition behind this:

Suppose the difference between n₁ and n₂ is the same as the difference between n₂ and n₃ which is the same as the difference between n₃ and n₄ etc... Then the desired function is linear; that is, one may define f(x) = n₁ + d(x - 1), where d is the constant difference in question (d could be negative or positive). Observe that this function "works."

On the other hand, if the differences are increasing at a constant rate (and are not themselves constant), the desired polynomial will have a term of degree two. For instance, consider the following values of n_i:

n₁ = 1, n₂ = 4, n₃ = 9, n₄ = 16 etc...

Observe that the differences are: 3, 5, 7 etc... and that the differences are increasing at a rate of 2. Since the differences are not constant, the function cannot be linear. Therefore, it must be a quadratic since the differences are increasing at a constant rate (if this rate were not constant, we might have to jump to cubics, or even quintics). I think that you can see that the desired function is the "square function" - f(x) = x².

How would we generalize the above method if we did not know the values of the n_i but knew particular differences? Well, we use factorials. If the differences are constant, the function must be linear. If the differences are changing at a constant rate (like above), let d_i be the ith difference, and let (d') be the difference in the differences (i.e the rate of change of the differences). Then the co-efficient of the x² in the polynomial is given by (d')/2!. Notice that this agrees with the above where (d') = 2, and we found the co-efficient of the x² to be 1. See if you can compute the co-efficient of the x in the quadratic. In general, assuming you are familiar with differential calculus, one may observe that the co-efficients seem similar to the co-efficients of the taylor series expansion of an analytic function.

It might be interesting to see if you can generalize the above method to arbitrary n_i's where the difference of the difference of the.....differences between the n_i's is constant. Depending on how many differences it takes to achieve this constant, one may compute the degree of the desired polynomial. Hope this helps. --PST 04:52, 16 June 2009 (UTC)[reply]

Could it be the Lagrange form? (Igny (talk) 04:34, 16 June 2009 (UTC))[reply]

I think you are thinking of the Lagrange interpolation polynomial. You can compute it (the Newton form) using a divided difference table doing those subtractions like you describe. 67.122.209.126 (talk) 09:33, 16 June 2009 (UTC)[reply]

Rmk: According to the historian Florian Cajori, the mathematical term coefficient (from a composition of cum+ex+facio) has been introduced by François Viète. I've never seen it written this way above, with a dash... --pma (talk) 14:16, 16 June 2009 (UTC)[reply]

If a dog is tied to the corner of a shed 6m in width and 3m in length by a rope that is 5m, what is the area within the dog's reach? I know the answer is 62 m squared, but i can find no way to get it. —Preceding unsigned comment added by 174.6.144.211 (talk) 03:03, 16 June 2009 (UTC)[reply]

Seems like the answer should be 19.75π. There are two regions the dog can get to: one is three quarters of a circle with a radius of 5. This is the area the dog can get to with the rope as a straight line, and is 3/4*5²*π m² == 18.75π m². The other region is one around the corner of the shed, where the dog still has 2 m of rope (having spent 3 m on the shed side) -- this has an area of 1/4*2²*π == π m². —Preceding unsigned comment added by 24.106.180.134 (talk) 04:18, 16 June 2009 (UTC)[reply]

I just realized that 19.75π is about 62.04. So there we go. --24.106.180.134 (talk) 04:21, 16 June 2009 (UTC)[reply]

The answer puzzled me for a moment until I understood that the dog is outside the shed. —Tamfang (talk) 23:55, 19 June 2009 (UTC)[reply]

I'm having an argument with a friend about how many MLB games there are total in one season. The parameters of the argument are that there are 30 teams, which each play 162 games a season. We disregard both post-season games and any single game tie-breakers at the end of the season. I think there are 162*15 = 2430 games total -- there are 15 pairs of teams, and since each team has to play all of its 162 games with another team, we need to count only half of MLB to get the total. My friend argues that I'm double counting games that the 15 teams I choose play against each other, and that this is not necessarily equal to the number of games that the 15 teams I didn't pick play against each other. He thinks that the only way to know would be to count -- i.e. that simply knowing there are 30 teams and 162 games per team doesn't help you know the total unless you know a lot more about the exact distribution of opponents in the games that each team played. (For people who might not follow MLB: over the course of the season, each team will play the other teams in its division very often, the teams not in its division but in its league not as often, and teams in the league rarely if at all. There are two leagues of 3 divisions, which vary in size from 4 teams to 6. It is not the case that each team plays some set number of games per each opponent team, or that each team faces all of the other 29 teams as opponents.)

Are either of us correct? Is there some formula to quickly determine the total games play per season, or does one need to count?

--24.106.180.134 (talk) 04:39, 16 June 2009 (UTC)[reply]

As I see it, say the Yankees and the Red Sox played a game against each other today. Since their combined effort resulted in one game, we could say that each of them played half a game. Each team in that manner plays 162 half-games per season. So in the complete season, there are 162*30=4860 half-games or 2430 games. You don't have to worry about who played who. 208.70.31.186 (talk) 05:53, 16 June 2009 (UTC)[reply]

Yes, 2430 is correct, assuming you are counting scheduled games, there are often a few rainouts that aren't made up if they don't affect the standings. Sometimes all scheduled games are played, sometimes not. (And you disregard the All-Star game, and pre-season games).--SPhilbrickT 12:40, 16 June 2009 (UTC)[reply]

I have set up a voting page for my daughter's middle name (http://adalinewagner.com/nameme.html). If you don't want to look at it, it is a list of names sent to me by email. Each name has a rating from -2 to +2. Users rate each name in the list (or ignore the name and don't rate it at all). I was wondering what the most accurate algorithm for tallying up the votes may be. The catch is that not every person will rate each of the shown names. So, if 100 people vote, there may be a name with 100 ratings - which makes it obvious that you take the average. There may also be a name with only 1 rating. Taking the average will make that 1 vote be the entire vote. If I'm not explaining this well enough, please let me know. -- kainaw™ 13:26, 16 June 2009 (UTC)[reply]

I think it is a better idea to decide on the tallying method before presenting the choices. Once you have the choices it is often possible to present an "almost fair" tallying method that chooses any winner that is not universally hated. Since there are no fair tallying methods, you are forced to choose between almost fair ones. Probably your best bet given your current polling method is to treat this as a "primary"; any name whose sum or average is positive after the polls closed is moved on to the next round where you ask people to rate the names in order of preference. I would suggest using a Borda count for the final tallying, since it should produce a winner that is not despised by anyone and is moderately pleasing to most. JackSchmidt (talk) 14:05, 16 June 2009 (UTC)[reply]

The simplest way to resolve this is of course to treat each missing vote as a vote of 0 with a lower weight. A slight modification is to use the total average, or the user average, instead of 0. I don't think the problem is well-defined enough to speak in terms of "the most accurate algorithm" - for this you will need a model of the reasons people choose not to vote on a name. -- Meni Rosenfeld (talk) 15:06, 16 June 2009 (UTC)[reply]

Thanks. The main reason for not voting on a name is that the name was added after the person voted and the person has not returned to vote again. Previously, I used a zero for all non-votes. Now, after reading this, I am leaning towards multiplying the average of the ratings which exist by the percentage of people who voted, which will push names with very few votes to an average of zero. -- kainaw™ 15:15, 16 June 2009 (UTC)[reply]

Aren't these two methods identical? Anyway, I do recommend using a lower weight for a missing vote than for an actual 0 vote (which is equivalent to multiplying by some function of the percentage). -- Meni Rosenfeld (talk) 18:35, 16 June 2009 (UTC)[reply]

See this, which illustrates the main flaw with your approach. One way to address this flaw is to normalize each person's votes so that the average vote of each person is 0. So, if you look at all the names that a person voted on, and the average is 0.65, then you would subtract 0.65 from each of that person's votes so that their average is now 0. After doing that, proceed as before; the score of a name is the average of the scores that people who voted on it gave it. Eric. 128.12.170.55 (talk) 21:25, 16 June 2009 (UTC)[reply]

Is there any method of finding if a is an integer power of b (where a and b are integers) in faster than logarithmic time? Assuming that arithmetic operations are constant time. Angus Lepper^{(T, C)} 14:34, 16 June 2009 (UTC)[reply]

Logarithmic in what? If you start with calculating ${\displaystyle b^{2},b^{4},b^{8},b^{16}\ldots }$, you can do this using ${\displaystyle O(\log _{2}\log _{b}a)\,\!}$ operations. -- Meni Rosenfeld (talk) 14:53, 16 June 2009 (UTC)[reply]

Logarithmic in a (my a and b were the other way around from yours). e.g. if a isn't divisible by b, return false; if a = b, return true; otherwise, divide a by b and start over. But your suggestion's obviously faster. Thanks. Angus Lepper^{(T, C)} 15:15, 16 June 2009 (UTC)[reply]

Right, I've edited my post to match your notation. -- Meni Rosenfeld (talk) 16:00, 16 June 2009 (UTC)[reply]

(removing my comment on % operation, just now noticing the use of "power" not "multiple") -- kainaw™ 16:01, 16 June 2009 (UTC)[reply]

If a is a power of b, then the b^th root of a will be an integer. Taking a root is a O(log n) function in most computer implementations (any that are worth noting). Testing to see if the result is an integer is a O(1) operation. So, it is logarithmic. -- kainaw™ 16:05, 16 June 2009 (UTC)[reply]

I seriously doubt that the OP's assumption that arithmetic operations take constant time is supposed to apply also to powering. Anyway, "a is a power of b" does not mean that the b^th root of a is an integer, but that there exists an integer c such that the cth root of a is b. — Emil J. 16:16, 16 June 2009 (UTC)[reply]

DJB has a few papers on detecting perfect powers, that might or might not be what you want. See [1] and look for the word "powers". 67.122.209.126 (talk) 16:11, 16 June 2009 (UTC)[reply]

Perhaps Lookup table is helpful. 195.35.160.133 (talk) 14:38, 17 June 2009 (UTC) Martin.[reply]

Starting with any triangle it's possible to inscribe a circle, form a triangle from the points of contact, inscribe a circle, etc. Successive triangles will be smaller and it's intuitively obvious that they will tend to equilaterality. But can this be done with quadrilaterals? The largest one must be inscribable by a circle, so its opposite sides must sum to the same amount, and all subsequent ones must be both inscribable and circumscribable, i.e. as well as the condition on opposite sides, the opposite angles must sum to 180°. Is it possible to draw an initial non-square quadrilateral (which I suspect on grounds of symmetry would have to be circumscribable, so that the construction could continue outwards as well as inwards) from which an infinite number of nested circles/quadrilaterals can be drawn, with intuitively the latter tending to squareness as they get smaller?→86.148.185.203 (talk) 15:04, 16 June 2009 (UTC)[reply]

Try an isosceles trapezoid with bases 1 and 3 and sides 2. -- Meni Rosenfeld (talk) 15:50, 16 June 2009 (UTC)[reply]

The inscribed circles wouldn't be concentric in that case, would they? I think requiring concentric circes might force the inscribed polygons to be regular. -GTBacchus^(talk) 16:54, 16 June 2009 (UTC)[reply]

No, but I think the shape converges to a square anyway. -- Meni Rosenfeld (talk) 18:38, 16 June 2009 (UTC)[reply]

Do you mean that the circles are concentric in the trapezoid case, or just that concentricness doesn't force regularity? -GTBacchus^(talk) 23:25, 16 June 2009 (UTC)[reply]

I mean that the circles aren't concentric, and that concentricness isn't required for (edit: convergence to) regularity. -- Meni Rosenfeld (talk) 23:35, 16 June 2009 (UTC)[reply]

Ah, but if you insist that the circles be concentric, then is regularity forced? -GTBacchus^(talk) 08:28, 17 June 2009 (UTC)[reply]

Yes, if a polygon has a circumscribed and an inscribed circles which are concentric, then the polygon must be regular by symmetry. -- Meni Rosenfeld (talk) 10:46, 17 June 2009 (UTC)[reply]

Let's start from your construction with triangles, and let's try to explain the intuitively obvious fact you noticed, about the shape of the triangles, when the construction is iterated. It seems to me a nice geometrical example of some facts in finite dimensional spectral theory.

So, the best thing is looking just at the angles (α,β,γ) of the triangle, and look at how they transform in the construction, forgetting about the size, that is not relevant to the shape. There is an even more important reason to consider the angles (and not, e.g. the edges or the cartesian coordinates). The new triangle has angles ( (β+γ)/2 , (α+γ)/2 , (α+β)/2 ): it's a linear transformation. It is represented by a symmetric 3x3 matrix A, with 0 on the diagonal, and all remaining entries 1/2. The eigenvalues of A are 1 (simple) and -1/2 (double). Now, consider the initial triangle, with angles x:=(α,β,γ). Write this 3vector as x = u + v, where u:= (π/3, π/3, π/3) is in the eigenspace of 1, and v is in the eigenspace of -1/2. Then you have for the k-th iterate: A^kx= A^k(u + v)=u +(-1/2)^kv, that converges exponentially to u, that is, geometrically, the zoomed triangles approach exponentially the equilateral triangle. --pma (talk) 17:41, 16 June 2009 (UTC)[reply]

Then I tried to do the same with quadrilaterals... But the problem is that angles are not enough to describe the shape of a quadrilateral, so the linear iteration of angles only tell us half of the story, and the thing is a bit more complicated. What is true about angles is that: if under your construction a quadrilateral produces a sequence of circumscribable quadrilaterals, then the corresponding sequence of angles converge to (π/4,π/4,π/4,π/4). pma (talk) 20:31, 16 June 2009 (UTC)[reply]

A quadrilateral with angles all ${\displaystyle \pi /4}$ is a rectangle, and a rectangle has an inscribed circle only if it is a square. Surely this is enough to show that the sequence of shapes, if it exists, converges to a square. -- Meni Rosenfeld (talk) 22:41, 16 June 2009 (UTC)[reply]

Exact. And whether there is such a sequence remains unclear to me too. I'll think a bit about it in spare time.--pma (talk) 10:25, 17 June 2009 (UTC)[reply]

Of course, what we both meant was ${\displaystyle \pi /2}$... -- Meni Rosenfeld (talk) 15:48, 18 June 2009 (UTC)[reply]

I tried computing the result for my example. I'm suffering heavily from numerical accuracy issues, but it seems like the sequence indeed exists and converges to a square. -- Meni Rosenfeld (talk) 23:35, 16 June 2009 (UTC)[reply]

Okay, I take that back. After fixing said issues, it is clear that my example only goes as far as trapezoid, kite, trapezoid. The latter does not have an inscribed circle. So it seems that the hard part is finding an infinite sequence. -- Meni Rosenfeld (talk) 00:02, 17 June 2009 (UTC)[reply]

I agree, one more step forces the trapezoid to be a square. But then I'm inclined to think that any sequence must stop in 3 or 4 steps unless it's made by squares... --pma (talk) 22:30, 17 June 2009 (UTC)[reply]

(OP) Thanks for replies. This was prompted by a newspaper puzzle, the problem being to get the (all different) angles of the starting quadrilateral so that the eleventh one would have integral-degree angles. The answer is (26,58,154,122) on the basis of the angle progression, but I couldn't find a way of drawing this so that the next one could have a circle inscribed. So it looks as if the compiler isn't as smart as he thinks he is.→86.132.235.58 (talk) 11:44, 18 June 2009 (UTC)[reply]

Indeed, with these angles, the only way to have an inscribed circle is to have sides ${\displaystyle (1.,1.25579,0.416497,0.160706)}$, and for this quadrilateral the sequence doesn't continue. -- Meni Rosenfeld (talk) 15:48, 18 June 2009 (UTC)[reply]

Hi. I'm still studying for a complex qualifying exam, and I'm stuck on an integral problem. The task is to evaluate the improper real integral

${\displaystyle \int _{-\infty }^{\infty }{\frac {e^{ax}}{e^{x}+1}}}$, with ${\displaystyle 0<a<1}$.

The function is strictly positive, so we expect a positive, real answer. There's a hint suggesting the contour ${\displaystyle \gamma =\partial ([-R,R]\times [0,2\pi ])}$, and we'll let R go to infinity.

So, looking at the complex function ${\displaystyle f(z)={\frac {e^{az}}{e^{z}+1}}}$, it's got one simple pole in the interior of the contour, the pole is at ${\displaystyle z_{0}=\pi i}$, and we have ${\displaystyle \operatorname {Res} (f,z_{0})=e^{(a-1)\pi i}}$. Applying the residue theorem, we multiply by 2πi to obtain

${\displaystyle \int _{\gamma }f(z)\,dz=2\pi e^{(a-{\frac {1}{2}})\pi i}}$

So far, so good.

The next step is to break the contour into pieces that we can deal w/ separately. We take ${\displaystyle \gamma _{1}=[-R,R]}$, and the other three follow in the usual (anti-clockwise) order. It's easy to show that the short vertical sides of the rectangle, ${\displaystyle \gamma _{2}}$ and ${\displaystyle \gamma _{4}}$, become small as ${\displaystyle R\rightarrow \infty }$; that part is no problem. That leaves ${\displaystyle \gamma _{1}}$ and ${\displaystyle \gamma _{3}}$, and we can express the latter in terms of the former.

First, we parametrize ${\displaystyle \gamma _{1}(x)=x}$, with ${\displaystyle x\in [-R,R]}$, so ${\displaystyle \int _{\gamma _{1}}f(z)\,dz=\int _{-R}^{R}{\frac {e^{ax}}{e^{x}+1}}\,dx}$.

Then, ${\displaystyle \gamma _{3}(x)=2\pi i-x}$ with ${\displaystyle x\in [-R,R]}$ again. Now, we set up the integral:

${\displaystyle \int _{\gamma _{3}}f(z)\,dz=\int _{-R}^{R}f(\gamma _{3}(x)){\dot {\gamma _{3}}}(x)\,dx=-\int _{-R}^{R}f(2\pi i-x)dx=\int _{-R}^{R}f(2\pi i+x)\,dx}$

${\displaystyle =\int _{-R}^{R}{\frac {e^{a(2\pi i+x)}}{e^{2\pi i+x}+1}}\,dx=e^{a(2\pi i)}\int _{-R}^{R}{\frac {e^{ax}}{e^{x}+1}}\,dx=e^{a(2\pi i)}\int _{\gamma _{1}}f(z)\,dz}$

Putting all this together, we obtain

${\displaystyle 2\pi e^{(a-{\frac {1}{2}})\pi i}=\int _{\gamma }f(z)\,dz=\int _{\gamma _{1}}f(z)\,dz+\int _{\gamma _{2}}f(z)\,dz+\int _{\gamma _{3}}f(z)\,dz+\int _{\gamma _{4}}f(z)\,dz}$

${\displaystyle =(1+e^{a(2\pi i)})\int _{\gamma _{1}}f(z)\,dz+\int _{\gamma _{2}}f(z)\,dz+\int _{\gamma _{4}}f(z)\,dz}$

In the limit as R grows, this turns into

${\displaystyle (1+e^{a(2\pi i)})\int _{-\infty }^{\infty }f(x)\,dx=2\pi e^{(a-{\frac {1}{2}})\pi i}}$, which gives us

${\displaystyle \int _{-\infty }^{\infty }f(x)\,dx={\frac {2\pi e^{(a-{\frac {1}{2}})\pi i}}{1+e^{a(2\pi i)}}}}$,

...which is not, in general, a real number. In particular, for ${\displaystyle a={\frac {1}{2}}}$, we get zero in the denominator.

Can some kind and knowledgeable soul tell me what I'm doing wrong? I'm very grateful for any hints you can throw me. -GTBacchus^(talk) 16:07, 16 June 2009 (UTC)[reply]

If you call your original integral I_a then the integral along γ₃ is expressible in terms of I_1−a, not quite what you calculated. Following that through you get a relation between I_a and I_1−a. But there's also a trivial relationship between these two integrals which you can see if you replace x by −x in I_a. Putting the two relations together will give you a formula for I_a that looks like it should be a complex number, but is really real. Really. Chenxlee (talk) 18:22, 16 June 2009 (UTC)[reply]

Yeah, I got the γ₃ integral in terms of 1-a at one point, but that didn't seem helpful, so I went back and did a change of variable instead, replacing x with -x. Did I make an error in the change of variable? -GTBacchus^(talk) 18:57, 16 June 2009 (UTC)[reply]

It looks like when you made the substitution x → −x, you didn't change the limits of integration. On second thoughts you don't need to get the relation between I_a and I_1−a, once you fix your limits of integration you'll get γ₃ in terms of I_a and can finish your original method from there. Chenxlee (talk) 10:29, 17 June 2009 (UTC)[reply]

By gum, you're right. That just might do it... -GTBacchus^(talk) 13:29, 17 June 2009 (UTC)[reply]

I love it when a complex integral comes together, to quote Hannibal Smith. Chenxlee (talk) 22:52, 17 June 2009 (UTC)[reply]

Indeed. The final answer, I have finally been able to express as a real number that looks like a real number: ${\displaystyle \int _{-\infty }^{\infty }{\frac {e^{ax}}{e^{x}+1}}\,dx={\frac {2\pi \operatorname {sin} (\pi a)}{1-\operatorname {cos} (2\pi a)}}}$.

It only took the better part of a week to figure it out, too! Thanks very much for your help. :) -GTBacchus^(talk) 04:34, 18 June 2009 (UTC)[reply]

You can simplify the result to ${\displaystyle {\frac {\pi }{\sin(\pi a)}}}$. — Emil J. 11:43, 18 June 2009 (UTC)[reply]

Ah, so you can. That's interesting: When I was looking at the function as a color-plot the other day, it was smelling very much like some kind of imaginary cosecant.

I'm still getting used to the idea that the exponential function is essentially another trig function, only sideways. Maybe more accurately, the trig functions are sideways exponentials. In a manner of speaking. The numbers e and π sure seem to know a lot about each other, at least. -GTBacchus^(talk) 13:35, 18 June 2009 (UTC)[reply]

${\displaystyle e^{i\pi }+1=0}$ ( from De Moivre's formula) Dbfirs 16:53, 18 June 2009 (UTC)[reply]

Yes, that's a very good example of it. Euler, as usual, said it well. (I hope I wouldn't be studying for a Ph.D. qualifying exam in Complex Variables without knowing about that formula!) -GTBacchus^(talk) 17:18, 18 June 2009 (UTC)[reply]

Sorry, didn't mean to suggest you were unaware of it - I just love the equation and couldn't resist the opportunity to quote it! Dbfirs 00:31, 19 June 2009 (UTC)[reply]

Is the area of the minor sector bounded by the equations ${\displaystyle y=mx+c}$ and ${\displaystyle (x-a)^{2}+(y-b)^{2}=R^{2}}$ given by ${\displaystyle A=R^{2}(\arccos {\frac {c{\sqrt {4m^{4}+1}}-b}{R}})-(c{\sqrt {4m^{4}+1}}-b)({\sqrt {R^{2}-(c{\sqrt {4m^{4}+1}}-b)^{2}}})}$

If I'm wrong please tell me but don't tell me the answer. I have, in a sense, cheated because the two previous parts of the question asked for the area of the sectors defined by ${\displaystyle y=c}$ and ${\displaystyle (x-a)^{2}+(y-b)^{2}=R^{2}}$ and ${\displaystyle y=mx+c}$ and ${\displaystyle x^{2}+y^{2}=R^{2}}$ and wondered if I could possibly avoid the algebraic slog that would be needed and just find a formula that works in the two special cases m=0 and b=0. —Preceding unsigned comment added by 92.4.46.28 (talk) 17:44, 16 June 2009 (UTC)[reply]

a doesn't appear anywhere in your formula, and the answer clearly does depend on a, so it can't be right. You only need to solve the problem for ${\displaystyle a=b=0}$ and ${\displaystyle R=1}$, and then the general case can be reduced to it. -- Meni Rosenfeld (talk) 18:52, 16 June 2009 (UTC)[reply]

Actually you can also assume that ${\displaystyle m=0}$, to which other cases are reduced by rotation. -- Meni Rosenfeld (talk) 18:59, 16 June 2009 (UTC)[reply]

I may be missing something, but those equations don't seem to span a sector. Do you mean a segment? If so, I agree that it is best to reduce the problem to a simpler one. You can very easily work out how to generalise the solution for the special case that Meni describes. --Tango (talk) 19:24, 16 June 2009 (UTC)[reply]

How can I prove that the product of two contiguous sets is always itself contiguous? I can intuitively understand it, by visualising two contiguous subsets of ${\displaystyle \mathbb {R} }$ as contiguous horizontal and vertical lines, and their product (a subset of ${\displaystyle \mathbb {R} ^{2}}$ as a contiguous, filled rectangle. But I can't make a formal proof of it. The thing is, this used to be my homework. I couldn't do it, but then I saw someone else do it. I understood the solution fully, and then forgot it. It was an exam question less than one month later, and I couldn't remember how to do it. Can anyone at least give a hint? JIP | Talk 19:31, 16 June 2009 (UTC)[reply]

By 'continguous', do you mean connected, or perhaps path-connected? Algebraist 19:33, 16 June 2009 (UTC)[reply]

I mean connected, i.e. not able to be represented as the union of two disjoint non-empty subsets, of which either both are closed or both are open. JIP | Talk 19:41, 16 June 2009 (UTC)[reply]

Consider 'slices' of the product (i.e. sets of the form X×{y} or {x}×Y in the space X×Y). Note that if you could disconnect the product, then you could disconnect some slice, and that each slice is homeomorphic to the appropriate factor. Algebraist 19:46, 16 June 2009 (UTC)[reply]

What do you mean by "appropriate factor"? JIP | Talk 19:51, 16 June 2009 (UTC)[reply]

Sorry, I was being lazy. I mean that slices X×{y} (where y is any point in Y) are homeomorphic to X (via the obvious map), while slices {x}×Y are homeomorphic to Y. Algebraist 19:53, 16 June 2009 (UTC)[reply]

OK, now I understand the basic idea. I have to think about it some more to be able to make a formal proof. JIP | Talk 20:25, 16 June 2009 (UTC)[reply]

The following fact may be of use to you:

If U and V constitute a separation of a topological space, X, and if C is a connected non-empty subset of X, C must lie in either U or V (but not both).

Now note that each slice is connected and use the above fact.

After you have proved that, it may be an interesting exercise to show that the arbitrary product of connected sets is connected in the product topology (if you are familiar with it). If you need a hint, let us know. --PST 00:50, 17 June 2009 (UTC)[reply]

Just attempted to work out a summation and would appreciate someone checking it for me.

${\displaystyle \sum _{k=1}^{2n}(-1)^{[{\frac {k}{n}}]}2^{-k},}$

where ${\displaystyle {[{\frac {k}{N}}]}}$ denotes the integer part of k/N. I got

${\displaystyle 1-2({\frac {1}{2}})^{n-1}+({\frac {1}{2}})^{2n-1}+({\frac {1}{2}})^{2n}}$

Is this right? Thanks asyndeton talk 20:15, 16 June 2009 (UTC)[reply]

Yes, this is correct, but you can simplify it even further. -- Meni Rosenfeld (talk) 20:37, 16 June 2009 (UTC)[reply]

Heres how it goes: three men go to a hotel room and they split the cost of their room- the cost is 30 dollars so they each pay ten. Later the manager realizes that he charged them too much- the room cost only 25 dollars, so he gives a bellboy 5 dollars to give to the men. The bellboy steals 2 dollars out of the 5 so he gives the men 3 dollars- they each paid 9 dollars in effect. But 9 x 3 =27+2=29. Why doesn't it add to thirty. It makes sense if you look it from 30-3-2 is 25 but why doesn't it work it you flat add the amount of money each man paid (27) and the 2 dollars the bellboy stole? 24.171.145.63 (talk) 06:56, 17 June 2009 (UTC)[reply]

See Missing dollar riddle. AndrewWTaylor (talk) 07:52, 17 June 2009 (UTC)[reply]

The answer is that you're adding it up incorrectly. The $2 that the bellboy stole is part of the $9 apiece ($27 total) that the men paid. It's the $3 that the men got back that you want to add to get $30, or you can subtract the $2 that the bellboy stole to get the $25 that the hotel charged.

This is, of course, all included in the article that AndrewWTaylor linked to. --COVIZAPIBETEFOKY (talk) 14:17, 17 June 2009 (UTC)[reply]

--Another way of looking at it which may make more sense is as follows:

The manager originally charges the men $30, (10 x 3 = 30)

The manager returns 5 leaving him $25, therefore the men pay (25/3 = 8.33 per individual) When the bellboy returns 1 to each man, they would be paying (8.33 + 1 =9.33 per) or (9.33 x 3) or $28

Upon adding the $2 from the bellboy, it is clear to see how multiplication can and will work in this situation. --f5centvillian143--

What is the name of the math concept that says that all even integers are the sum of 2 primes? How could you possibly prove that since we are always coming up with larger primes, and one could select an arbitrarily large even number? 65.121.141.34 (talk) 14:50, 17 June 2009 (UTC)[reply]

Goldbach's conjecture. Proofs in mathematics rarely work by giving examples for every possibility, you come up with a logical argument that proves it for every possibility, even ones we haven't thought of yet. This could, for example, be done using mathematical induction, where we prove that if the conjecture holds for one possibility, then it must hold for the next so (if it holds for the first one) it must hold for all of them. --Tango (talk) 14:58, 17 June 2009 (UTC)[reply]

In case it wasn't clear, Goldbach's conjecture has not been proved. So, in fact, no one knows for sure whether it's true or not (although there is great statistical evidence in favor of its truth). That's not to say that it can't be proved, but that no one has ever constructed a proof. It is an open problem. --COVIZAPIBETEFOKY (talk) 15:17, 17 June 2009 (UTC)[reply]

No one knows "for sure" even those things that have been proved. --Trovatore (talk) 23:51, 23 June 2009 (UTC)[reply]

How to find the maximum capacity of a sack made by gluing together at edges two pieces of plastics or paper or something else. —Preceding unsigned comment added by 113.199.161.144 (talk) 18:40, 17 June 2009 (UTC)[reply]

Can we assume that the material of the sack does not stretch? 65.121.141.34 (talk) 18:53, 17 June 2009 (UTC)[reply]

That would be a good starting point. If the material stretches, that seems like something you would add later, as a complication, after working out the non-stretch case. That's kind of like how we describe projectile motion in a frictionless vacuum, and then learn to correct for air resistance later. On the other hand, the material had better be flexible, because gluing together two flat pieces of sheet metal along edges does not a sack make. -GTBacchus^(talk) 20:01, 17 June 2009 (UTC)[reply]

So, we're determining how much stuffing a pillow can hold? Black Carrot (talk) 05:28, 18 June 2009 (UTC)[reply]

Yes. This sounds messy to do rigorously. See isoperimetric inequality for a less constrained case. 208.70.31.186 (talk) 05:59, 18 June 2009 (UTC)[reply]

If there is not restriction on size of the pieces of plastic or paper, then there will be no maximum capacity of the sack. The volume of the sack is proportional to the cube of the linear size of the pieces, so the remaining problem depends on shape only. Choosing shapes that makes the sack approximately spherical makes the maximum volume for given surface area. Bo Jacoby (talk) 08:29, 18 June 2009 (UTC).[reply]

If the sack absolutely cannot stretch but can bend and it is glued all round the edge then it is a limiting form of a Flexible polyhedron. And by the Bellows conjecture (now proved) the volume will be zero. :) Dmcq (talk) 09:49, 18 June 2009 (UTC)[reply]

Actually that's only true if you glue the edges before you make the volume. You really need to fold the two sides first so the edges will join up and then glue them together. You might be interested in the problem of folding a shopping bag mentoned in rigid origami, there the top is open. Dmcq (talk) 09:58, 18 June 2009 (UTC)[reply]

Following on from Bo Jacoby, if the two pieces are circular and of diameter D then the maximum possible volume would be that of the sphere of diameter d where d = ${\displaystyle {\frac {2D}{\pi }}}$ i.e. ${\displaystyle {\frac {4D^{3}}{3\pi ^{2}}}}$, but it is not possible to stretch two circular pieces of most materials to meet on the circumference of the sphere, or to join them with zero overlap, so the maximum volume is "a bit less than this". How much less depends on the materials and how they are joined, but this idealised optimum is about 13.5% of ${\displaystyle D^{3}}$, so the best to expect in practice would be 10% of ${\displaystyle D^{3}}$, and most situations would give considerably less. Dbfirs 12:10, 18 June 2009 (UTC)[reply]

If you're going to use shaped pieces then how about just one long piece of ticker tape? Just go round and round putting in small pleats and you'll soon have a very good approximation to a sphere. You can do this with a balloon and wallpaper paste :) Dmcq (talk) 15:11, 18 June 2009 (UTC)[reply]

... or take that to the limit as the width of the ticker tape tends to zero, and start with a pre-formed sphere.

... but if you insist on starting with right-angles, start with two squares of side D, fold D/4 along each edge, cuting out the corners or using them for strengthening overlap. This produces a cube of side D/2, with volume ${\displaystyle D^{3}/8}$, giving a lower limit of 12.5% of ${\displaystyle D^{3}}$. This can be imporoved (perhaps to 15% of ${\displaystyle D^{3}}$ or more) by packing tightly and allowing the faces to bulge. The reason that this appears to be a better solution than the two circles is that you are starting with more material (though if you have to cut circles out of squares in the first place, then don't bother). I doubt whether many materials are manufactured in circular form, except possibly round tea bags? Dbfirs 16:42, 18 June 2009 (UTC)[reply]

One case is the paper bag problem. (I remembered reading some work on it but would never have found it if Dbfirs had not mentioned tea bags.) —Tamfang (talk) 21:36, 19 June 2009 (UTC)[reply]

Ah, thank you! So I under-estimated how much tea can be packed into a (very strong) square teabag. You might achieve 20% of ${\displaystyle D^{3}}$ with really tight and careful packing. Dbfirs 00:35, 21 June 2009 (UTC)[reply]

I solved this puzzle using brute force; my question is whether there is a more elegant solution that you could get to without a computer program. Essentially the problem is to find three three-digit numbers such that:

• between them, the three numbers use each of the digits 1-9 exactly once
• the product of the three numbers is a nine-digit number that uses each of the digits 1-9 exactly once
• the smallest of the three numbers is 7 times a prime
• the middle of the three numbers is 8 times a prime
• the largest of the three numbers is 9 times a prime

The answer is:

• 413 = 7 x 59
• 568 = 8 x 71
• 927 = 9 x 103
• 413 x 568 x 927 = 217,459,368

But how can you arrive at this solution, short of trying different permutations of three primes until you get it? At first I thought you might be able to narrow it down using divisibility rules, particularly for divisibility by 7, but I didn’t get anywhere with that. --Mathew5000 (talk) 23:11, 17 June 2009 (UTC)[reply]

This is a problem in constraint satisfaction. There are well known algorithms like unification for solving these problems, and logic programming languages like Prolog that have the algorithms built in. For an example of how to do it by hand, see the article SEND MORE MONEY. 208.70.31.186 (talk) 00:43, 18 June 2009 (UTC)[reply]

I wonder what's the less elegant solution. Checking all 9! permutations by prime factorization? --pma (talk) 10:44, 18 June 2009 (UTC)[reply]

It's not quite 9!, the order of the numbers doesn't matter, so you can divide by 6. --Tango (talk) 13:15, 18 June 2009 (UTC)[reply]

I meant all numbers with 9 different digits, form 123456789 to 987654321... ;) --131.114.73.61 (talk) 16:21, 18 June 2009 (UTC)[reply]

Surely the less elegant solution is to cycle through all 10 digits for each place giving 10⁹ possibilities? However one could do it on paper seeing there aren't all that many primes to check, 100<7p₁<800, 200<8p₂<900, 300<9p₃<1000, the numbers can't have repeats in them, each number is greater than the one before, the sum of the first two must be a multiple of 9. Dmcq (talk) 15:57, 18 June 2009 (UTC)[reply]

Let's see if we can reduce the size of the search space even further. There are 9x8x7=504 three digits numbers with (1) no repeated digits and (2) no 0 digit. But only approximately 1 in 9 of these are divisible by 9 - and testing for divisibility by 9 is simple. So we have about 56 choices for our first factor. For each choice of first factor we have approximately 6x5x4/8 = 15 possible second factors which are divisible by 8 - and, again, quite easy to eliminate non-qualifiers by inspection. That leaves 6 candidates for the final factor - let's say on average 1 of these is divisible by 7, but it takes a bit more work to filter out non-qualifiers here. So the size of our search space is now about 56x15 = 840 sets of candidate factors. Multiply each candidate set together and eliminate it if the product contains a repeated digit or a 0. But that's still a lot of work if you want to solve it "by hand" or just with a calculator. Gandalf61 (talk) 15:57, 18 June 2009 (UTC)[reply]

Well I just tried it and it took half an hour without a calculator just biro and paper. Now I must really go and do something slightly more useful. Dmcq (talk) 16:43, 18 June 2009 (UTC)[reply]

Very good; then maybe to make it more feasible, one thing is that the larger of the three factors X is larger then the cubic root of 123456789≈125*10⁶, so larger than 500-3, and certainly not larger than 987. Its quotient by 9 is a prime between 497/9≈55 and 987/9≈109. Removing from these well known primes the one that multiplied by 9 give a result with some repeated digits, [or 0], one is left with 61 67 71 73 89 97 103 107 109. These give 9 6 choices for X: 549,603,..,981. Then one lists similarly the possible choices for the medium factor Y and for the least factor Z, that are numbers with distinct digits respectively of the form Y=8p and Z=7p, with p a prime between 29 and 113. I guess some 15 possibilities for Y and around the same for Z. Then one goes on like in Gandalf's method, trying products X*Y*Z, but only with X>Y>Z and all digits different in X,Y,Z. So actually very few multiplications are to be effectively computed. I guess it's something one could do by hands, but since the OP already gave the answer, he spoiled the small satisfaction to try it... ;) It could be an exciting project in a class of kids that are learnig multiplication by hand, if they still learn it. --pma (talk) 17:50, 18 June 2009 (UTC)Oh I forgot that zero is not there, so it's even better[reply]

I did 46 multiplies in total of 7 and 8 with prime numbers and cut out the ones with repeats and zeroes. Then I got the digit sums of the multiples of 7 and 8 so I could match up pairs that gave a sum that was a multiple of 9 and had no repeat digits and with space for a multiple of 9 after them. This gave only four possibilities. I then did 12 multiples of 9 with primes in the possible range. This gave 5 possible triples. I then had to try all five of these possibilities which involved 10 big multiplies. So about 70 arithmetical operations in total, 10 of them big, plus lots of straightforward checking digits and comparisons. Two sheets of paper. Don't know why I took half an hour put like that. Dmcq (talk) 21:46, 18 June 2009 (UTC)[reply]

Thanks very much! --Mathew5000 (talk) 23:09, 19 June 2009 (UTC)[reply]

Is there a way to convert a floating point number to an integer (what int() would do) using only arithmetic functions (+ - / * % and ^)? And while I guess I could loop it subtracting fractions of an integer, I don't have any function to test whether the number is an integer (or floating point) until I've finished. This seems like the kind of problem that would have an answer, either that it's impossible or that it is possible. Anyone know which? Shadowjams (talk) 06:04, 18 June 2009 (UTC)[reply]

You cannot do it with any continuous combination of continuous operations, because there are no nonconstant continuous functions from the reals to the integers. — Carl (CBM · talk) 06:10, 18 June 2009 (UTC)[reply]

My pure math background is very outdated, so bear with me. When you say continuous operations, do you mean the arithmetic functions (and maybe others)? What kind of function do you need to actually do this? Shadowjams (talk) 06:29, 18 June 2009 (UTC)[reply]

All the arithmetical operations are continuous. In an actual computer, the floor operation can look directly at the bits of the floating point number to get the corresponding integer; it does not have to do any arithmetical manipulations. — Carl (CBM · talk) 06:36, 18 June 2009 (UTC)[reply]

Can you test whether your float is larger than a given integer? If you then a crude solution is to start on 1 and increment by 1 until the integer is larger than the float. Then subtract 1, and you have the answer. Taemyr (talk) 07:00, 18 June 2009 (UTC)[reply]

Hm Good point. Actually, I'm trying to do this in Yahoo Pipes, so no, I don't think that I can test any of the arithmetic conditions, although there might be a very convoluted way to use a regular expression to check what you're talking about, but I'm not sure I can do it an indefinite number of times, so there's no elegant solution apparently. Shadowjams (talk) 07:25, 18 June 2009 (UTC)[reply]

If the "%" you listed above is floating point mod, then floor(x) = x - (x % 1), for positive real numbers. Dragons flight (talk) 07:35, 18 June 2009 (UTC)[reply]

That would actually give the nearest integer using IEEE floating point standard remainder operation which gives the exact value of x-(round(x/y)*y) irrespective of rounding mode. The result of the mod can be positive or negative irrespective of whether the two operand are positive. Dmcq (talk) 10:11, 18 June 2009 (UTC)[reply]

Actually the whole business of the Modulo operation and Remainder is a bit of a mess and you need to check which one the compiler implements in any particular computer language even for integers. Dmcq (talk) 10:17, 18 June 2009 (UTC)[reply]

I see you want to do this in reality on a system that only provides those operations. What an idea, reality, must think about that. Oh okay then what you can do is multiply by the smallest possible Denormal number and then divide by it again. This will unfortunately round to nearest but for positive numbers you can compensate somewhat by subtracting just less than 0.5 to start with. Exactly 0.5 would turn 3 into 2. Dmcq (talk) 10:51, 18 June 2009 (UTC)[reply]

If you want a different rounding mode you can fix it up afterwards, e.g. if (result > n) --result; if you want to round to -∞. In C, nextafter(0,1) will return the smallest positive double (and there are similar functions for float and long double). Some implementations use a wider format for intermediate results (including many x86 compilers), and to get around that you would have to assign the intermediate result to a volatile variable or some such. Some implementations don't support denormals and on those this trick won't work. -- BenRG (talk) 11:24, 18 June 2009 (UTC)[reply]

The denormal article doesn't give the values for float of 1.5e-45 and double 5e-324. As BenRG says there can be problems and even if it works in C the performance can be dreadful. Try it out and you will feel the power of the dark side of computing. Dmcq (talk) 12:29, 18 June 2009 (UTC)[reply]

I think I'd try the floor(x) = x - (x % 1) of Dragons flight first. The system may not implement % for float but my guess is that they do. Dmcq (talk) 13:33, 18 June 2009 (UTC)[reply]

WRT Interior (topology), is it true that Set S comprises the interior points + the boundary points? Is the boundary of the complement of Set S the same as the boundary of Set S? If not, how can the exterior of S be the interior of the complement of S? -- SGBailey (talk) 10:19, 18 June 2009 (UTC)[reply]

No, a set does not have to include its boundary, unless it is closed. Yes, the boundaries of any set and its complement are the same. In any case, the exterior of a set is defined to be the interior of its complement, so it is meaningless to ask "how can be". — Emil J. 10:33, 18 June 2009 (UTC)[reply]

What do you mean by "Set S"? Do you mean the underlying set of the topological (sub)space S? If so, it doesn't have a boundary, that is a topological concept. --Tango (talk) 13:57, 18 June 2009 (UTC)[reply]

I don't know what I mean by Set S. See Interior (topology) where it is defined. -- SGBailey (talk) 15:40, 18 June 2009 (UTC)[reply]

Sometimes a set is just a set. For example, if our topological space is ${\displaystyle \mathbb {R} ^{2}}$ with its usual topology then S can be ${\displaystyle \{(x,y)|x^{2}+y^{2}<1\}\,\!}$. -- Meni Rosenfeld (talk) 16:01, 18 June 2009 (UTC)[reply]

I don't see "Set S" mentioned in the article, did you just mean "a set called S"? The way you wrote it made it look like "Set" was some operator acting on S. --Tango (talk) 01:56, 19 June 2009 (UTC)[reply]

Words 7 and 8 in the lead. Yes, it just means a set called S. -- Meni Rosenfeld (talk) 15:45, 19 June 2009 (UTC)[reply]

Helo maths lovers !

Please excuse my poor English, I'm French. I got very few informations for the following question in the French Wikipedia. It seems that your desk is more active.

When I was a student, I learnt methodes to calculate an approximative value for an integration (such the Simpson methode) because we were taught that many continuous functions have no known primitive.

The strange thing is that I can give only 2 or 3 examples of such functions :

The well knonw f(x)= exp(-x²) and others of the same type

g(x)=cos(x²)

and g(θ)= square root(a²cos²θ + b²sin²θ) with wich one can calculate the perimeter of an ellipse.

Who could give other examples ? thank you very much for your attentions. Joël DESHAIES , in Reims, France. --82.216.144.50 (talk) 13:56, 18 June 2009 (UTC)[reply]

"No known primitive" is an incorrect term. These are just functions whose primitive is not in a somewhat arbitrary family of functions which are called "elementary" and commonly featured in calculators. A primitive of ${\displaystyle \exp(-x^{2})\,\!}$ is ${\displaystyle {\tfrac {1}{2}}{\sqrt {\pi }}\mathrm {erf} (x)}$, where ${\displaystyle \mathrm {erf} \,\!}$ is the error function, a "non-elementary" function.

So the correct question should have been about functions whose primitives are not elementary. There is no shortage of examples - in fact, I think "most" functions have no elementary primitive. For instance, virtually any function of the form ${\displaystyle \exp(f(x))\,\!}$ where f is not linear or logarithmic (${\displaystyle \exp(x^{2}),\exp(x^{3}),\exp(\cos x),\exp(\exp(x)),\exp(x\log x)=x^{x},\ldots }$). Same goes if you use ${\displaystyle \cos }$ or ${\displaystyle \sin }$ instead of ${\displaystyle \exp }$. -- Meni Rosenfeld (talk) 14:49, 18 June 2009 (UTC)[reply]

(ec)Some of these functions, like ${\displaystyle x^{x}}$, do not even have a primitive in terms of special functions. But this is not because the primitive is somehow "unknown", but only because nobody deemed this primitive function important enough to bother with giving it a name. -- Meni Rosenfeld (talk) 15:01, 18 June 2009 (UTC)[reply]

What proportion of continuous functions are even integrable? --Tango (talk) 14:52, 18 June 2009 (UTC)[reply]

All of them. Algebraist 14:56, 18 June 2009 (UTC)[reply]

If they are defined over the whole real line, maybe, but what about functions with singularities? Not all improper integrals converge. --Tango (talk) 15:08, 18 June 2009 (UTC)[reply]

In the context of a primitive (i.e. ${\displaystyle \textstyle \int _{0}^{t}f(x)\,dx}$), it's enough for the function to be continuous on an open set, in which case one will get a primitive at every point of continuity. Characterizing which discontinuous functions are derivatives (even at their points of discontinuity) is a much harder question. — Carl (CBM · talk) 15:16, 18 June 2009 (UTC)[reply]

True, in the context of primitives it doesn't really matter. You wouldn't expect the function to have a primitive where it isn't defined. --Tango (talk) 15:35, 18 June 2009 (UTC)[reply]

It's still an interesting question (e.g. 1/x and ln(|x|) work). But Darboux's theorem (analysis) is the only result I know in the area. — Carl (CBM · talk) 16:04, 18 June 2009 (UTC)[reply]

See Nonelementary integral. A large proportion of named functions in applied mathematics were invented as the solution of an interesting integral. Dmcq (talk) 14:59, 18 June 2009 (UTC)[reply]

Another approach: elementary functions are smooth, so only smooth functions can have elementary primitives, while piecewise smooth functions might have piecewise elementary primitives (I think it's fair to say a piecewise elementary primitive is still a 'known' primitive). Thus any sufficiently horrible continuous function will not have a (piecewise) elementary primitive. Since almost all continuous functions are nowhere differentiable, this gives lots of examples. Algebraist 15:02, 18 June 2009 (UTC)[reply]

See the Risch algorithm article for some references about this. Finding elementary antiderivatives is sort of like finding roots of polynomials in terms of radicals, in that a lot of them can't be solved that way, and there's a theory sort of like Galois theory that tells you which ones are solvable and which ones aren't. 208.70.31.186 (talk) 17:40, 18 June 2009 (UTC)[reply]

Notice that there are functions that are not differentiable, or not even continuous, yet, whose integral function has a nice representation. To mention two classes of examples: functions defined by a Fourier series; or functions defined as solution of a functional equation: they may be very discontinuous, yet their integral function may be deduced from the Fourier series, resp. from the functional equation. --84.221.208.172 (talk) 22:57, 18 June 2009 (UTC)[reply]

In this thread it was explained that the torus double-covers the sphere with four "ramification points". In the article titled torus that is also pointed out.

I wondered if it's possible to do away with the ramification points and find a two-to-one continuous mapping from the torus to the sphere that's locally one-to-one everywhere, like a circle getting mapped twice around another circle.

So I thought: Jacobian elliptic functions are doubly periodic, so their domain can be considered a torus, and they map into the Riemann sphere and assume every value twice, so maybe that would do it, and the mapping would be conformal. But then I realized: these functions do have points where the derivative is zero. They assume every value twice if you count by multiplicities, and at those points the multiplicity is 2, and the mapping is neither locally one-to-one nor conformal at those points. And if I'm not mistaken, there are four of them.

So my question: is the thing I had in mind provably impossible?

If so, I think maybe that fact should be stated explicitly in the torus article, with a reference to the literature. Michael Hardy (talk) 18:23, 18 June 2009 (UTC)[reply]

This just follows from the Riemann-Hurwitz formula: if your map ${\displaystyle \mathbb {T} \to {\hat {\mathbb {C} }}}$ has degree ${\displaystyle d}$, then the number of branch points is

${\displaystyle b=d\chi ({\hat {\mathbb {C} }})-\chi (\mathbb {T} )=2\times 2-0=4}$

where ${\displaystyle \chi }$ is the Euler characteristic.

I guess another way of seeing this is that we know that any nonconstant elliptic function must have at least 2 poles (with multiplicities) so must have degree at least 2 as a map to the Riemann sphere. Then its derivative must have at least 3 (this is the case for the derivative of the Weierstrass ${\displaystyle \wp }$ function). It also has to be branched over infinity (though I'm not totally sure why yet), so that you get the required four branch points, and this is the minimum possible. --Xedi^Talk 21:06, 18 June 2009 (UTC)[reply]

The Riemann–Hurwitz article seems to cover it. Thank you.

Next task: what's the best way to incorporate this information into the torus article? Maybe I'll do that soon. Michael Hardy (talk) 02:28, 19 June 2009 (UTC)[reply]

It seems a local homeomorphism, a covering space map, is being described. A torus, or any other connected manifold, covering a sphere without ramification points is impossible because the sphere is simply connected - and so its own universal Covering space. The fundamental group of a nice space is the same as the automorphism group of its universal covering space. So no analyticity assumption is necessary.John Z (talk) 04:10, 20 June 2009 (UTC)[reply]

This last proposed reason why no such covering is possible seems simpler than the one that Xedi proposed.

If by "analyticity" you mean that it's conformal, I certainly never thought that was necessary. Michael Hardy (talk) 19:58, 20 June 2009 (UTC)[reply]

I cannot understand why Zeta[0] is -1/2. One would think that it is infinity but obviously this is not the case. 69.120.50.249 (talk) 01:45, 19 June 2009 (UTC)[reply]

One would only think it was infinity if one was using the definition

${\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}={\frac {1}{1^{s}}}+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+\cdots .\!}$

But that definition is valid only for s with real part greater than 1. Algebraist 01:51, 19 June 2009 (UTC)[reply]

What is the definition for other values? —Preceding unsigned comment added by 69.120.50.249 (talk) 02:01, 19 June 2009 (UTC)[reply]

As stated in our article, the zeta function is defined elsewhere by analytic continuation. Algebraist 02:03, 19 June 2009 (UTC)[reply]

While punching buttons on my calculator recently, I noticed that (using Mathematica syntax):

Integrate[x^x,{x,0,1}] ≈ 0.783430510909

and

-Sum[(-x)^(-x),{x,1,Infinity}] ≈ 0.783430510712.

I was wondering: is this near-miss a near miss because they are actually the same and the calculator is making rounding errors, or is it a near-miss because the integral and the sum actually give different results?

If it helps, the calculator is a TI-89 Titanium, OS version 3.0, hardware version 4.0. --72.197.202.36 (talk) 02:42, 19 June 2009 (UTC)[reply]

See Sophomore's dream. J Elliot (talk) 08:43, 19 June 2009 (UTC)[reply]

Congratulations. And another one if you can prove the result, think how to convert it into a series. You might also like Lambert W function. Dmcq (talk) 14:44, 19 June 2009 (UTC)[reply]

Spoiler alert: If you want to try proving the equality (which is correct, if it wasn't clear) yourself, don't follow the link to Sophomore's dream. -- Meni Rosenfeld (talk) 15:42, 19 June 2009 (UTC)[reply]

Two side remarks. The integral ${\displaystyle \scriptstyle \int _{0}^{1}(x\log \,x)^{n}\,dx}$, that appears in the quoted computation, may also be evaluated by a change of variable ${\displaystyle \scriptstyle x=e^{-t}}$: then another linear change of variables leads to the Euler integral ${\displaystyle \scriptstyle \int _{0}^{\infty }u^{n}e^{-u}du}$. Second remark: the same computation gives, more generally, the power series expansion ${\displaystyle \scriptstyle z\int _{0}^{1}x^{-zx}dx=\sum _{n=1}^{\infty }{\frac {z^{n}}{n^{n}}}}$ (the linked article reports ${\displaystyle \scriptstyle z=1}$ and ${\displaystyle \scriptstyle z=-1}$). --pma (talk) 19:22, 19 June 2009 (UTC)[reply]

Does anybody know a nice, closed-form way of generating Bernoulli numbers? I.e., B_n = what? --72.197.202.36 (talk) 03:59, 19 June 2009 (UTC)[reply]

Is this a serious question? I cannot believe it. Please read the article. There are several closed form formulas given there. — Preceding unsigned comment added by 92.227.144.52 (talk • contribs)

If no such method is available, are there any relatively simple algorithms to generate them? --72.197.202.36 (talk) 04:00, 19 June 2009 (UTC)[reply]

Bernoulli_number#Efficient_computation_of_Bernoulli_numbers contains an algoritm. Bo Jacoby (talk) 08:32, 19 June 2009 (UTC).[reply]

But this is not a 'relatively simple algorithm'. The Akiyama-Tanigawa algorithm is one and is given in the article in full length in pseudo code. Obviously people are talking here about articles without reading them. This is annoying and frustrating. — Preceding unsigned comment added by 92.227.144.52 (talk • contribs)

Why the disbelief of the OP being serious? WP:AGF and all that... I skimmed through the article, and didn't really see the nice, closed form formula, that I'd easily be able to implement (preferring simplicity above speed). Would anyone care to spell it out? --NorwegianBlue ^talk 21:12, 19 June 2009 (UTC)[reply]

Why the disbelief of the OP being serious? Are you serious? I copy verbatim from the article. If you want I also copy the formula. If you want I also read it loud. Here it is:

These relations lead to a simple algorithm to compute the Bernoulli number. The input is the first row, A_0,m = 1/(m + 1) and the output are the Bernoulli number in the first column A_n,0 = B_n . This transformation is shown in pseudo-code below.

Akiyama-Tanigawa algorithm for Bn

Enter integer n. For m from 0 by 1 to n do     A[m] ← 1/(m + 1)     For j from m by −1 to 1 do         A[j − 1] ← j × (A[j − 1] − A[j]) Return A[0]  (which is Bn).

— Preceding unsigned comment added by 92.230.245.194 (talk • contribs)

I wouldn't call that "closed form". Michael Hardy (talk) 22:05, 19 June 2009 (UTC)[reply]

It answers a question of a person who did not read the article: I quote: "If no such method is available, are there any relatively simple algorithms to generate them?" Would you also call it not a simple algorithm, Mr. Hardy? —Preceding unsigned comment added by 92.230.245.194 (talk) 23:45, 19 June 2009 (UTC)[reply]

Certainly simple, but not exactly what we mean by closed form. As far as I know, there are no simple closed forms for Bernoulli numbers. "Closed form" is a relative concept, of course, as relative concept as the one of elementary function &c. However, I met difficulty in following this thread, for one or more posters are not signing their contributions, and it's not clear where a post starts and ends, and who is replying to whom. --pma (talk) 08:51, 20 June 2009 (UTC)[reply]

I have added signatures to the unsigned entries, and also (per WP:IAR) taken the liberty of modifying the indentation, in order to improve the readability of this thread. --NorwegianBlue ^talk 11:08, 20 June 2009 (UTC)[reply]

The question may have been inspired by the news story Iraq-born teen cracks maths puzzle which begins

"STOCKHOLM (AFP) – A 16-year-old Iraqi immigrant living in Sweden has cracked a maths puzzle that has stumped experts for more than 300 years, Swedish media reported on Thursday. In just four months, Mohamed Altoumaimi has found a formula to explain and simplify the so-called Bernoulli numbers ..." 208.70.31.186 (talk) 11:04, 20 June 2009 (UTC)[reply]

Some more info. 208.70.31.186 (talk) 11:06, 20 June 2009 (UTC)[reply]

From the generating function formula ${\displaystyle {\frac {x}{e^{x}-1}}=\sum _{n=0}^{\infty }B_{n}{\frac {x^{n}}{n!}}}$ it follows that ${\displaystyle B_{n}=\left({\frac {d^{n}{\frac {x}{e^{x}-1}}}{dx^{n}}}\right)_{x=0}}$ which has the closed form the OP was asking for: B_n = what?. But perhaps the OP is not happy because the intermediate results are functions rather than numbers. Bo Jacoby (talk) 15:57, 20 June 2009 (UTC).[reply]

Closed, indeed, as it was Pandora's box... ;-) --pma (talk) 22:14, 20 June 2009 (UTC)[reply]

knowing that the coefficient of friction between the 25kg block and the incline is 0.25 determine the value of p for which motion of block up the inclined plane is impending —Preceding unsigned comment added by 81.199.50.67 (talk) 12:32, 19 June 2009 (UTC)[reply]

We are not here to do your homework. 65.121.141.34 (talk) 13:31, 19 June 2009 (UTC)[reply]

Especially when you don't post enough of the homework question to work out what you're talking about. Algebraist 13:37, 19 June 2009 (UTC)[reply]

What is p? What is the angle of incline? Describe the problem in full, make an attempt at it (a good start would be to describe the forces on the block), then ask for help.86.132.164.238 (talk) 13:41, 19 June 2009 (UTC)[reply]

For the motion of the block up the plane to be impending, the following equation must be satisfied (the force on the block up the plane must equate to the force on the block down the plane):

m*g*sin(Ə) = f*m*g*cos(Ə)

where m is the mass of the block, g is the acceleration due to gravity, f is the co-efficient of (static) friction and Ə is the angle of inclination. Since all variables except Ə are known, I am assuming Ə is p according to your question. Therefore, if we solve for Ə, we obtain:

tan(Ə) = f, since the m*g factor may be cancelled from both sides of the equation.

Therefore, tan(Ə) = 0.25 which, with the use of a calculator, gives Ə = 14 degrees approximately. Notice that since the mass of the block (m) cancelled on both sides of the equation, we did not need to know its value for the purposes of this computation. Hope this helps. --PST 22:17, 19 June 2009 (UTC)[reply]

For all these kinds of questions the method is the same: Resolve the forces parallel and perpendicular to the plane and equate them to the appropriate values using F=ma. Then solve for the value to be found. --Tango (talk) 22:34, 19 June 2009 (UTC)[reply]

Is it automatically unphysical to have a PCA reconstruction that has some stations negatively weighted? Note, that this is more than just a PCA, but PCA followed by some regressions. Would think that it could occur for both degeneracy and anticorrelation with the average (actual physical effects). Of course the summation must be positive, but is it automatically wrong if some of the stations have negative weights?

This is being debated on these blog threads. Unfortunatley, the debate has muddled particular examination of the Stieg Antarctic PCA-based recon with general absolute claims that negative weightings are bad, bad, bad.

Could you please adjuticate?

See here:

http://noconsensus.wordpress.com/2009/06/07/antarctic-warming-the-final-straw/#comment-6727

http://noconsensus.wordpress.com/2009/06/09/tired-and-wrong-again/#comment-6726

http://wattsupwiththat.com/2009/06/10/quote-of-the-week-9-negative-thermometers/#more-8362

http://www.climate-skeptic.com/2009/06/forgetting-about-physical-reality.html

Would appreciate some reference to a citation or an academician and/or a more in-depth answer than just "he or she is right" so that I can either get my opponent to understand where he is wrong, or understand it myself. Muchos gracias. 69.250.46.136 (talk) 18:00, 19 June 2009 (UTC)[reply]

Hi. I'm halfway through working out the answer to a question and am now stuck. Here's what I have

"Let ${\displaystyle f(x)=\sin(2nx)+\sin(4nx)-\sin(6nx)}$ where n is a positive integer and ${\displaystyle 0<x<{\frac {\pi }{2}}}$. Find an expression for the largest root of the equation f(x)=0, distinguishing between the cases where n is even and n is odd. You may assume that ${\displaystyle \sin(3\theta )=\sin \theta (4\cos ^{2}\theta -1)}$."

So after working through this I get solutions as follows:

${\displaystyle x={\frac {p\pi }{2n}}}$ provided ${\displaystyle 0<p<n}$

${\displaystyle x={\frac {p\pi }{n}}}$ provided ${\displaystyle 0<2p<n}$

${\displaystyle x={\frac {\pi }{n}}({\frac {1}{3}}+p)}$ provided ${\displaystyle 0<{\frac {2}{3}}+2p<n}$

${\displaystyle x={\frac {\pi }{n}}({\frac {2}{3}}+p)}$ provided ${\displaystyle 0<{\frac {2}{3}}+p<{\frac {n}{2}}}$

where p is some integer.

First of all, is what I have correct? Secondly, how do you distinguish between odd and even? I can't see what difference it will make to my answer. Thanks asyndeton talk 20:51, 19 June 2009 (UTC)[reply]

Have you tried graphing it for a few values of n? -GTBacchus^(talk) 20:48, 19 June 2009 (UTC)[reply]

You aren't allowed access to any graphing applications (or even a calculator) while sitting the paper this question comes from. Slightly hypocritical I know, since you don't have access to Wikipedia mathematicians either, but this way I can see how they intended you to approach it, which is more important to me than just reaching the right answer. asyndeton talk 20:51, 19 June 2009 (UTC)[reply]

If only there was away to draw a graph without a computer or a calculator ... Gandalf61 (talk) 20:56, 19 June 2009 (UTC)[reply]

I have graphed f(x) on my computer in the cases n=2,3. It hasn't helped. asyndeton talk 21:39, 19 June 2009 (UTC)[reply]

Let u = 2nx. Then the function is

${\displaystyle {\begin{aligned}&{}\quad \sin u+\sin(2u)-\sin(3u)\\&=(\sin u)(1+2\cos u-(4\cos ^{2}u-1))\\&=(2\sin u)(1+\cos u-2\cos ^{2}u)\end{aligned}}}$

and this is 0 iff either sin u = 0 (so u is an integral multiple of π) or cos u = either 1 or −1/2 (so that u is an integral multiple of 2π/3).

The condition that x is between 0 and π/2 says that u is between 0 and nπ.

If n is even then the largest multiple of 2π/3 that is less than nπ is π(n − 2/3). If n is odd, then put "1/3" there instead of "2/3". (Just draw the picture on the x-axis with n going from 0 to about 5 or so, and you'll see it. Never mind what the curve looks like; just graph the zeros as I described them above.) Michael Hardy (talk) 21:44, 19 June 2009 (UTC)[reply]

Hi. Just did a show that involving an approximation and I'm not convinced that what I've done is rigorous. Could someone please check it for me?

"Let ${\displaystyle {\frac {1}{y}}={\frac {2+3x^{2}}{3(1+x^{2})^{1.5}}}+{\frac {1}{3}}}$. Show that, for large positive x, ${\displaystyle y\approx 3-{\frac {9}{x}}}$."

My method goes like this. for large positive x ${\displaystyle 1+x^{2}\approx x^{2}}$ and so ${\displaystyle (1+x^{2})^{1.5}\approx x^{3}}$. Also for large positive x, ${\displaystyle 2+3x^{2}\approx 3x^{2}}$ so ${\displaystyle {\frac {2+3x^{2}}{3(1+x^{2})^{1.5}}}\approx {\frac {3x^{2}}{3x^{3}}}={\frac {1}{x}}}$ so ${\displaystyle {\frac {1}{y}}\approx {\frac {1}{x}}+{\frac {1}{3}}}$. Rearranging this gives ${\displaystyle y\approx {\frac {3x}{3+x}}={\frac {3x+9}{3+x}}-{\frac {9}{3+x}}=3-{\frac {9}{3+x}}}$. Finally, for large positive x, ${\displaystyle 3+x\approx x}$ so ${\displaystyle y\approx 3-{\frac {9}{x}}.}$

Is this a solid argument? Thanks 92.2.16.39 (talk) 10:09, 20 June 2009 (UTC)[reply]

Yes, I think that's spot on.. Rkr1991 (talk) 14:15, 20 June 2009 (UTC)[reply]

Yes, that looks fine to me. The statement "for large positive x, ${\displaystyle y\approx 3-{\frac {9}{x}}}$" isn't a rigorous statement, so there is no way your "proof" could be rigorous. What you've got is good enough, in my opinion. If the question had been to show that ${\displaystyle y=3-{\frac {9}{x}}+O({\frac {1}{x^{2}}})}$ or something (or whatever the appropriate notation is, I can never remember the differences between big-O, little-O, etc. without looking them up), then you could get a proper rigorous proof (probably along the same lines as the argument you gave). --Tango (talk) 14:26, 20 June 2009 (UTC)[reply]

You may prefer to simplify first and approximate afterwards.

${\displaystyle y={\frac {3(1+x^{2})^{1.5}}{2+3x^{2}+{(1+x^{2})^{1.5}}}}=3-9x^{-1}\cdot {\frac {2x^{-2}+3}{6x^{-3}+9x^{-1}+3{(1+x^{-2})^{1.5}}}}\approx 3-9x^{-1}}$.

Bo Jacoby (talk) 14:37, 20 June 2009 (UTC).[reply]

Please tell me about the algorithm of finding International Roughness Index(of a road surface)from a longitudinal profile data. —Preceding unsigned comment added by 113.199.158.140 (talk) 12:36, 20 June 2009 (UTC)[reply]

Googleing 'International Roughness Index' leads to the fortran program at http://www.umtri.umich.edu/content/IRIMain.f . Bo Jacoby (talk) 13:51, 20 June 2009 (UTC).[reply]

I wanted to have the algorithm and not the program. —Preceding unsigned comment added by 113.199.170.182 (talk) 02:47, 21 June 2009 (UTC)[reply]

What's the difference? A programming language is just a language for writing algorithms. --Tango (talk) 03:09, 21 June 2009 (UTC)[reply]

The program contains the following comment: 'For more information about the IRI and how this program works, see Sayers, M. W., "On the Calcluation of International Roughness Index from Longitudinal Road Profile." Transportation Research Record 1501, (1995) p. 1-12. See http://spyder.umtri.umich.edu/erd_soft/erd_file.html on the web for a description of ERD files'. So now it is time for the OP to do some homework. Bo Jacoby (talk) 10:20, 21 June 2009 (UTC).[reply]

I'm a little hazy on N&S conditions so wanted to check my thoughts with people who can tell me if I'm right or wrong. Let ${\displaystyle f(x)=x^{2}-2bx+c}$. I have to find N&S conditions on b and c a) for f(x) to have two distinct real roots and b) for f(x) to have two distinct positive real roots.

For a) I get that ${\displaystyle c-b^{2}<0}$ just from using the fact that if the turning point is below the x axis, you've got two distinct roots.

For b) I again get ${\displaystyle c-b^{2}<0}$, for the same reason as before. I then get c>0, which ensures that the roots are of the same sign and b>0, which says the minimum point occurs in the first quadrant (at least I think that's its name; I mean bottom right). So the y-intercept is positive, both the roots are of the same sign and the turning point is in the first quadrant, so between them, they say that the roots are distinct and positive.

Are they necessary and sufficient or not? Thanks 92.4.255.16 (talk) 17:52, 20 June 2009 (UTC)[reply]

"A is necessary for B" means "B implies A". "A is sufficient for B" means "A implies B". Therefore to prove that something is necessary and sufficient you need to prove it in both directions. Sometimes you can do both directions at once, but if you're not sure it is best to do them separately. You need to come up with a sequence of implications from A to B and then a sequence from B to A. While considering the graphs is a good idea in order to get an intuitive answer, your rigorous proof should probably be purely algebraic. --Tango (talk) 18:01, 20 June 2009 (UTC)[reply]

The quadratic formula gives you the roots of f as ${\displaystyle x={\frac {2b\pm {\sqrt {4b^{2}-4c}}}{2}}=b\pm {\sqrt {b^{2}-c}}}$. The answer to both questions follow naturally from there. For question a you are correct; the expression in the radical must be positive for the radical to be real. For question b, it is sufficient and necessary for the smaller root to be real and positive. I'll let you work the algebra to see where that gets you. --COVIZAPIBETEFOKY (talk) 18:24, 20 June 2009 (UTC)[reply]

Your solution to (b) is almost right. The x-coordinate of the vertex is ${\displaystyle {\frac {-b}{2a}}}$ (note the minus sign). So to be in the fourth quadrant, you need ${\displaystyle b<0}$. -- Meni Rosenfeld (talk) 19:37, 20 June 2009 (UTC)[reply]

Where has the 'a' come from? Are you possibly confusing the f(x) I gave above with the general form of a quadratic equation? 92.7.54.6 (talk) 19:41, 20 June 2009 (UTC)[reply]

Yes, COVIZAPIBETEFOKY was correct, and Meni is correct to say that you meant the fourth quadrant, and your conclusions were also correct, but see Tango's suggestion for a rigorous proof. Dbfirs 00:12, 21 June 2009 (UTC)[reply]

Of course, silly me. I was thinking about the general case, and didn't notice the differences in your case, most importantly that your b is already negated. So indeed it should be ${\displaystyle b>0}$. -- Meni Rosenfeld (talk) 08:50, 21 June 2009 (UTC)[reply]

Has anybody already seen an optimization problem like this one:

Find the maximum value of the 3-dimensional volume of the subset of the unit cube:

${\displaystyle {\hat {A}}:=\{(x,y,z)\in [0,1]\times [0,1]\times [0,1]\,:(x,y)\in A,(y,z)\notin A\}}$

among all measurable subsets ${\displaystyle \textstyle A}$ of the unit square ${\displaystyle [0,1]\times [0,1]}$.

Incidentally, I know the answer (it is hidden here in case somebody prefers not to see it), but I'd like to settle this and other similar but more difficult problems into a general theory/method, if there is any available. --pma (talk) 21:17, 20 June 2009 (UTC)[reply]

Cute problem. I don't recall having seen it before. Maybe I'll try to work out where it came from. I wonder if thinking of it in the language of probability could shed any light. Michael Hardy (talk) 17:48, 21 June 2009 (UTC)[reply]

I am told that I need to solve ${\displaystyle \int _{m}^{\infty }f(x)\,dx=0.5}$, and in the textbook example they get a polynomial with only m⁴, m² and m⁰ terms in it, which lets them solve a quadratic in m², take a square root and compare the results to the range which you are given at the start to find the right one, but I keep getting equations like ${\displaystyle \int _{m}^{3}{\frac {2}{9}}x\left(3-x\right)\,dx=0.5}$, where the p.d.f f(x) of the continuous random variable X was the above for ${\displaystyle 0\leq x\leq 3}$, which eventually gets me to ${\displaystyle {\frac {2}{27}}m^{3}-{\frac {1}{3}}m^{2}+0.5=0}$, but I don't know how to solve this. In another example, I had the p.d.f. f(y) of the continuous random variable Y being ${\displaystyle 12y^{2}\left(1-y\right)}$ for ${\displaystyle 0\leq y\leq 1}$, which got me to ${\displaystyle \int _{m}^{1}12y^{2}\left(1-y\right)\,dy=0.5\Rightarrow 3m^{4}-4m^{3}+0.5=0}$, but, again, I don't know what to do with this. It Is Me Here ^{t / c} 08:58, 21 June 2009 (UTC)[reply]

Well, ${\displaystyle \int _{0}^{3}{\frac {2}{9}}x\left(3-x\right)\,dx=1}$, and the symmetry of ${\displaystyle x\left(3-x\right)}$ suggests that

${\displaystyle \int _{0}^{3/2}{\frac {2}{9}}x\left(3-x\right)\,dx=\int _{3/2}^{3}{\frac {2}{9}}x\left(3-x\right)\,dx=0.5}$

so try m = 3/2 in your first problem. Your second problem is more difficult, as ${\displaystyle y^{2}\left(1-y\right)}$ doesn't have an obvious symmetry to exploit. Gandalf61 (talk) 12:21, 21 June 2009 (UTC)[reply]

What's worse, Mathematica seems unable to find a representation of the solution simpler than direct substitution in the quartic formula. Is there any chance they want you to find a numerical solution (which is 0.614272...)?

By the way, the first problem can also be solved by the rational root theorem. -- Meni Rosenfeld (talk) 13:21, 21 June 2009 (UTC)[reply]

See root-finding algorithm for methods to solve equations numerically. Or draw the conclusion that the median is not that interesting anyway and use the mean value instead! Your examples are beta distributions. By the way, 0.614272 is an approximate solution to ${\displaystyle 3m^{4}-4m^{3}+0.5=0}$. The J program p. 1 0 0 _8 6 does the trick. Bo Jacoby (talk) 13:42, 21 June 2009 (UTC).[reply]

OK, the actual question states: "The continuous random variable Y has p.d.f. f(y) defined by f(y) = 12y²(1 - y) for 0 ≤ y ≤ 1; f(y) = 0 otherwise. Show that, to 2 decimal places, the median value of Y is 0.61." It Is Me Here ^{t / c} 15:09, 21 June 2009 (UTC)[reply]

That's easy then. You know that ${\displaystyle F(y)}$ (the cumulative distribution function) is continuous and increasing, so you only need to show that ${\displaystyle F(0.605)<0.5}$ and ${\displaystyle F(0.615)>0.5}$. The intermediate value theorem will then say that there is a root ${\displaystyle 0.605<y<0.615}$, which is thus equal to 0.61 to two decimal places. -- Meni Rosenfeld (talk) 18:47, 21 June 2009 (UTC)[reply]

Not a massive fan of induction and so I would like to check my solution to a problem with you guys.

"Let ${\displaystyle S_{n}(x)=e^{x^{3}}{\frac {d^{n}}{dx^{n}}}(e^{-x^{3}})}$. Prove by induction on n that ${\displaystyle S_{n}(x)}$ is a polynomial."

Assume that ${\displaystyle S_{n}(x)}$ is a polynomial for n=k.

${\displaystyle e^{x^{3}}{\frac {d^{k}}{dx^{k}}}(e^{-x^{3}})=f(x)}$

So ${\displaystyle {\frac {d^{k}}{dx^{k}}}(e^{-x^{3}})=e^{-x^{3}}f(x)}$

Differentiate the above wrt x.

${\displaystyle {\frac {d^{k+1}}{dx^{k+1}}}(e^{-x^{3}})=e^{-x^{3}}f'(x)-3x^{2}e^{-x^{3}}f(x)}$

Factorise

${\displaystyle {\frac {d^{k+1}}{dx^{k+1}}}(e^{-x^{3}})=e^{-x^{3}}(f'(x)-3x^{2}f(x))}$

${\displaystyle f'(x)-3x^{2}f(x)}$ is still a polynomial for any f(x), so let us call it g(x).

So we have

${\displaystyle {\frac {d^{k+1}}{dx^{k+1}}}(e^{-x^{3}})=e^{-x^{3}}g(x)}$

So if the result is true for k, it is also true for k+1. Now let n=0 (the question doesn't actually state which set of numbers 'n' belongs to but I assume it's the non-negative integers), which gives us the case where ${\displaystyle S_{0}(x)=1}$. So by induction, ${\displaystyle S_{n}(x)}$ is a polynomial for all integers n, n≥0.

Is that airtight? Thanks. asyndeton talk 14:15, 21 June 2009 (UTC)[reply]

Yes. —JAO • T • C 14:42, 21 June 2009 (UTC)[reply]

It looks fine to me, although I would have started with n=1. While defining the zeroth derivative to be the identity makes sense, I would normally think of n be positive in the context of n-th derivatives. --Tango (talk) 14:56, 21 June 2009 (UTC)[reply]

well, I think yours is mainly a psycological limit. But if you work on it a bit, you too will be able to start with n=0  ;-) --pma (talk) 19:11, 21 June 2009 (UTC)[reply]

It's obvious that is does work with n=0, I just don't normally think of the identify being a derivative. It's a matter of definition, rather than any mathematical, of course. --Tango (talk) 00:05, 22 June 2009 (UTC)[reply]

Do the rules of double-deck cancellation hearts generalize well to hearts games with N decks and 4N-2 to 4N+2 players? NeonMerlin 23:46, 21 June 2009 (UTC)[reply]

Well my first thought is that there are two possible ways in which the game could be generalised, either we stick with the rule that if a second identical card is played then the two cards cancel, which leads to a game where if odd numbers of an identical card are played the last player to play that card (assuming it is the strongest one, otherwise it makes no difference) takes the trick whereas if even numbers are played the cards are ignored, this seems slightly artificial to me and as if it would lead to a confusing game.

Alternatively the other way in which the game could be generalized is to only apply the cancellation rule when all N of an identical card are played in one trick, otherwise if say K (<N) identical cards were played in one trick then the last player to play the identical card would count as having played it and the other players who played it would be ignored. This to some extent removes the excitement that double-deck cancelation hearts has, since especially when N is large the chance of a cancellation occuring are very slim.

Both possibilities have their pros and cons, when N is small (say 3 or 4) I think both variations would make interesting playing. —Preceding unsigned comment added by 86.129.82.211 (talk) 00:52, 22 June 2009 (UTC)[reply]

What is the exact definition of a linear topological space (the formal definition). I am unable to find it on the internet. Thanks. —Preceding unsigned comment added by 58.161.138.117 (talk) 04:42, 22 June 2009 (UTC)[reply]