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January 7

standard scrabble points

E=1, M=3, O=1, R=1, Y=4

so why E=1, M=2, O=1, R=1 Y=10? http://www.bbc.co.uk/news/health-16425522 — Preceding unsigned comment added by 81.147.58.70 (talk) 12:28, 7 January 2012 (UTC)

Scrabble has many editions for different languages, each with different letter distributions and points. The photo clearly isn't of an English-language edition. From the points shown, it could be a French edition. Qwfp (talk) 12:50, 7 January 2012 (UTC)

In fact of the scrabble editions mentioned, French is the only one that matches up. French and German both have 10 point 'Y's, but an 'M' in German is 3 points, not 2.

Two- or three-dimensional chaotic system or map with at least two parameters

The section header pretty much says it. I would prefer the system or map (I don't know if there is a difference between the two terms) to be continuous. I am currently taking calculus in high school so I don't understand how to apply terms like dyadic transformation and eigenvalues, so I would prefer the system to be in an explicit form. --Melab±1 ☎ 15:26, 7 January 2012 (UTC)

What have you attempted and where did you get stuck? Bo Jacoby (talk) 20:38, 7 January 2012 (UTC).

I am not stuck due to this is not being an assignment. I am interested in experimenting with initial conditions and I would rather use a continuous system with an exact and explicit solution as opposed to any of the discrete systems I could find that have exact and explicit solution. Connecting points generated discretely using a continuous equation just isn't satisfactory. --Melab±1 ☎ 20:50, 7 January 2012 (UTC)

Your question was posed and answered here[1], remember. Bo Jacoby (talk) 08:23, 8 January 2012 (UTC).

Perhaps we could help you better if you explained some of your goals. There is plenty of fun to be had playing with chaotic systems, but why the need for a three dimensional state variable and continuity? Why do you wish for exact solutions? Exact solutions for chaotic systems are usually found only in simple "toy" models, and even then are not so common. There is a reason most of this stuff was not studied well until modern computation for numeric integration became cheap. As an aside, I'd highly recommend this book, "Nonlinear dynamics and chaos" [2]. You should be able to handle at least the first few chapters, and it gives some very nice intuitive approaches. It deals with chaos, but also other important features of nonlinear dynamics, such as stable limit cycles. SemanticMantis (talk) 15:09, 8 January 2012 (UTC)

The classical analysis of the Geiger-Marsden experiment may be relevant. The orbit in 3-dimensional space depend critically on the impact parameter, so the actual orbit is unpredictable, even if the formula is well known. Bo Jacoby (talk) 12:42, 9 January 2012 (UTC).

I want it to be continuous because I want to be able to evaluate for any $t$ and not have to use only specific values and I want an exact solution simply because I do not like errors to have any chance of propagating. --Melab±1 ☎ 13:10, 9 January 2012 (UTC)

I just checked the book, some of the notation looks to be beyond me. Like on page 235, I don't understand what $h(x,{\dot {x}})$ solves. I also don't understand how $x(0=a)$ and ${\dot {x}}(0)=0$ fit into ${\ddot {x}}+x+\varepsilon h(x,{\dot {x}})$. --Melab±1 ☎ 22:06, 9 January 2012 (UTC)

The system doesn't need to model a physical process. --Melab±1 ☎ 22:16, 9 January 2012 (UTC)

Your enthusiasm is inspiring, but we must walk before we run. Strogatz starts from a (somewhat) elementary perspective, but you need to read (and fully understand) the first 234 pages before you tackle p. 235. Make sure you are confident with ch.1-3, and feel free to ask a new question if you need help with the book. As to the original question, I don't have the time to search out what you're looking for right now, nor am I totally sure that it exists. Perhaps after you've read a bit more (and made sure you ace your HS math class this semester), you can re-post the question :) SemanticMantis (talk) 22:59, 9 January 2012 (UTC)

Thank you, very much. :) I want to say that I think I may have come up with a discrete chaotic system a while ago, it just doesn't satisfy my requirements. --Melab±1 ☎ 23:16, 9 January 2012 (UTC)

January 8

Cardinality

I'm not sure if I've asked this question before. If I have, then forgive me. Given two differentiable manifolds X and Y, I want to know the cardinality of the space of smooth maps from X to Y, in terms of the cardinalities of X and Y. Forgive me if my question doesn't make perfect sense; I'm a novice in set theory. I have an idea that I'm trying to make sense of. I'm trying to express that the space of smooth maps from two-space to three-space is "much bigger" than the space of planes in three-space (which is diffeomorphic to the real projective plane RP²). I'm trying to make sense of "how many more" smooth surfaces there are than planes. Cardinality is an obvious way of doing it, but it seems very crude. Maybe there is another way of doing it? Hopefully you understand what I'm trying to get at. — Fly by Night (talk) 18:21, 8 January 2012 (UTC)

Cardinality is rarely a useful measure of size, except in discrete mathematics and naive set theory. In the case of smooth functions from one manifold to another, this always has the cardinality of the continuum, as do both X and Y, unless Y is zero dimensional (i.e., discrete points). Indeed, one can see that the cardinality of the set of smooth maps is not less than the continuum by considering just the constant maps into Y. For the opposite inequality, Y can be embedded into a ball in Rⁿ for sufficiently large n (by the Whitney embedding theorem). Smooth mappings from X to Y are thus contained in the separable Hilbert space $L^{2}(X)^{n}$, whose cardinality is the continuum.

There are many more useful ways to measure the "size" of a space than cardinality. Usually these involve introducing some topology on the space of interest, or taking some kind of quotient by an equivalence relation, or both. For instance, rather than counting mappings from X to Y, it might make more sense to count the homotopy classes of mappings. In the specific problem you are interested in, you can define a topology on the space of mappings in terms of which it becomes an infinite dimensional Frechet manifold. You probably also want to quotient by the group of automorphisms of X to get a meaningful comparison to the set of planes in R³, and that will complicate things. But, at any rate, one will be an infinite dimensional space, and the other a finite dimensional one. Sławomir Biały (talk) 11:52, 9 January 2012 (UTC)

Sławomir Biały has already given a nice answer, but let me expand a bit. I'd like to explain how you can see that the cardinality of smooth maps is at most continuum. The trick is that a manifold X is a separable space, which means that it has a dense subset of countable cardinality. Choose such a dense subset, say K. Now if you have even a map f from X to Y, then consider g = f|_K (the restriction of f to K) uniquely determines f. As this g has to be chosen as a function from a countable set (K) to a set of at most continuum cardinality (Y), the cardinality of such functions is also at most continuum. Note that all we are using here about the maps is that they're continuous, that is, we don't need to restrict to smooth maps. – b_jonas 15:37, 9 January 2012 (UTC)

Veering slightly off-topic here — whether a manifold has to be separable depends on the details of your definition of "manifold". An example of a non-separable manifold is the long line. Which by the way is one of my favorite topological spaces — I love the way that the construction itself can be pushed as far as you like through the ordinals, but if you go past ω₁, it's no longer a manifold --Trovatore (talk) 10:15, 10 January 2012 (UTC)

Thanks Sławomir and b_jonas. I've worked out what I need to do. I use the Whitney topologies on the space of map germs from X to Y. Assuming dim(X) < dim(Y), at a point p of X, I can write the image of X locally as the graph of a function. I get a map into a jet space, where any plane gets mapped to a point. — Fly by Night (talk) 17:39, 10 January 2012 (UTC)

Four color theorem

help me understand this theorem, does it mean that if you use less than four color on a map, two adjacent section will have the same color? If so, next question is does checkered pattern be accepted as a 'map'? becase you can do three colors on checkered pattern. MahAdik _usap 19:20, 8 January 2012 (UTC)

No it doesn't. It says that every separation of the plane can be coloured (i.e. assigned colour so that no two adjacent sections have the same colour) with at most four colours. The trivial separation, i.e. no boundaries, can be coloured with a single colour. If the split the plane into two or three pieces then you can colour it using two or three colours resp. The contrapositive of the theorem says that if you have less than four colours, say n, then there exists a separation which cannot be coloured with n colours. — Fly by Night (talk) 20:57, 8 January 2012 (UTC)

The contrapositive is that if a graph is not 4-colorable then it's not planar, or if a map requires more than 4 colors then its regions aren't contiguous. The statement you said, while true, is not equivalent to the four color theorem. Rckrone (talk) 01:45, 9 January 2012 (UTC)

Actually, that is the wrong contrapositive in this case, Fly By Night is saying that given a separation of the plane, then "If you have four, or more colours, you can colour it." The contrapositive would be, "If you cannot colour it, then you have less than four colours.", which is the same thing, basically, as his sentence (you would certainly be right if he were taking graphs as the basic object." Sorry for being contentious...Phoenixia1177 (talk) 16:44, 9 January 2012 (UTC)

I agree that the contrapositive depends on exactly how you state the theorem. However, what you said is also not the same as what Fly By Night said. You said that if a map is not n-colorable, then n < 4. Fly By Night's statement is a partial converse to that: if n < 4, then there exists a map that is not n-colorable. The 4 color theorem gives an upper bound on the number of colors needed, while his/her statement is that this bound is tight. Rckrone (talk) 02:09, 10 January 2012 (UTC)

I, suppose, that you are technically correct, but that the bound would need to be tight is rather obvious. More over, your version of the contrapositive is still off for the reasons I said. That said, I'm starting to feel really nit-picky and am going to stop:-) Phoenixia1177 (talk) 04:40, 10 January 2012 (UTC)

The issue isn't what statements are obvious, it's about what statements are logically equivalent. The statement "if you have less than four colours, say n, then there exists a separation which cannot be coloured with n colours," is not equivalent to the 4-color theorem. As a result it's not the contrapositive of any formulation of the 4-color theorem. Rckrone (talk) 05:29, 10 January 2012 (UTC)

...Wow, I'm feeling embarrassed, the entire time I've been reading that as something different. For some reason, my brain kept interpreting it as "If you have less than four colours, only then, and definitely then, do you have a separation that cannot be coloured.", I'm really not sure why (probably not paying attention closely...) At any rate, it seemed like you were complaining about the fact that this would not only claim you could always get by with four colours, but that sometimes they were required...which seemed unreasonable; obviously, you weren't being unreasonable, my apologies. Though, I still stand by my whining about whether you are talking about graphs or maps, etc. Sorry again:-) Phoenixia1177 (talk) 10:22, 10 January 2012 (UTC)

January 9

series acceleration

hello. What is the best form of series acceleration to apply to the (real-valued) Taylor series for e^x (which already converges pretty fast, but hey)? I had a glance at your series acceleration article but it doesn't seem to say which is best used when. This is preferably something that it is easy to write a computer program for. Thanks. 24.92.85.35 (talk) 22:56, 9 January 2012 (UTC)

I don't think there are many specific rules since it depends on factors independent of the specific series. For example if you're only doing a few values to a relatively low accuracy and the series converges reasonably quickly then it may be simpler just to go with the original series. Otherwise it's generally a tradeoff between accuracy, speed of convergence and how complex of an algorithm you're willing to program. The series for e^x already converges pretty rapidly unless x is large, and there are simple identities you can apply to reduce to the case where x is within given bounds. Instead of trying to accelerate the series it might be better to use a different method such as continued fractions to get more accuracy, but in many cases adding a few more terms to the series is just as effective.--RDBury (talk) 14:44, 10 January 2012 (UTC)

January 10

How soon would a dishwasher pay for itself in savings?

I was told that machine dishwashers are more efficient than washing by hand.

In this case, how much do we use on average to wash dishes by hand?

On the other... "hand," how efficient is a new dishwasher made this year, and by how much more is it efficient than the hand methods? How much does said dishwasher cost (and from what store?)

Therefore, assuming regular usage, how soon would the machine dishwasher pay for itself in savings? --70.179.174.101 (talk) 00:19, 11 January 2012 (UTC)

A comparison of the article and its comments here give some indication of the gap between the spin and the common experience. --Tagishsimon (talk) 00:58, 11 January 2012 (UTC)

Another factor is the amount of detergent used up. Hand dish washing detergent seems cheaper, to me, and you might use less, since I apply it directly to the dishes, and only as needed. StuRat (talk) 16:50, 12 January 2012 (UTC)

Polynomials

Hello. I am trying to prove succinctly but rigorously that if for some polynomial P(x+c)-P(x)=k, c and k constant, for all x, then P must be a polynomial of degree at most one. I already have a proof considering an indeterminate degree 'n' that involves sigma summation and binomial expansion but it is very ugly. Can anybody provide a hint? Thanks. 24.92.85.35 (talk) 01:06, 11 January 2012 (UTC)

Assume that $p$ has degree $n>1$, and say $p(x)=a_{n}x^{n}+\cdots +a_{1}x+a_{0}$. What is the coefficient of $x^{n-1}$ in $p(x+c)-p(x)$? --COVIZAPIBETEFOKY (talk) 01:43, 11 January 2012 (UTC)

Take a derivative of both sides of your equation and conclude that P' is a periodic function (and therefore it is bounded). The only bounded polynomials are constant, so P' must be constant. Sławomir Biały (talk) 10:52, 11 January 2012 (UTC)

Sławomir, that's a neat little proof. It is intuitively obvious that the only bounded polynomials are constant but how might one prove that? The sine function can be represented as an infinite power series. If you took the polynomial consisting of the first googol to the power of a googol terms in that power series then you'd have a polynomial. It's tempting to think that this polynomial might be periodic over some large interval. Presumably you'd need an argument involving the convergence of power series? What would you do next? — Fly by Night (talk) 19:53, 11 January 2012 (UTC)

You just consider x large enough and the polynomial will be dominated by the largest power. n times the largest coefficient (or 2 if smaller) will be quite big enough and more. Dmcq (talk) 20:11, 11 January 2012 (UTC)

I'm aware of the method, but I was asking Sławomir to give explicit details about what he would do. — Fly by Night (talk) 23:04, 11 January 2012 (UTC)

Actually, there's a simple proof that $P'$ is constant that doesn't rely on this fact. First note that for each integer j (by the argument I just gave) ${\frac {1}{c}}(P'((j+1)c)-P'(jc))=0$. By Rolle's theorem, this implies that $P''(x)$ has a zero in $[jc,(j+1)c]$. Thus $P''$ is a polynomial with infinitely many zeros, and so is therefore the zero polynomial.

To answer your question about how I would prove that the only bounded polynomials are constant: By the squeeze theorem, if $P'$ is bounded and $n>0$ then $\lim _{x\to \infty }P'(x)/x^{n}=0$. But if $P'$ had degree $n>0$, then the leading coefficient of $P'$ is $\lim _{x\to \infty }P'(x)/x^{n}\not =0$, a contradiction. Sławomir Biały (talk) 00:08, 12 January 2012 (UTC)

It's obvious when you put it like that :o) — Fly by Night (talk) 17:53, 12 January 2012 (UTC)

Thank you everybody, this was really helpful! Thank you especially Slawomir, for such a clever proof. Just to be sure I understand you: there is no real requirement that the j in your argument be an integer correct? It could in fact be any number? Thanks again! 24.92.85.35 (talk) 03:10, 12 January 2012 (UTC)

Right, j need not be an integer for the identity to hold, but it simplifies the argument a bit to assume this because it ensures that the intervals $[jc,(j+1)c]$ do not overlap. Sławomir Biały (talk) 10:06, 12 January 2012 (UTC)

Alternatively: prove that if $P(x)$ is a polynomial of degree $n$ then $P(x+c)-P(x)$ is a polynomial of degree $n-1$. This is straightforward for the polynomial $P(x):=x^{n}$ and generalizes immediately to any polynomial. So if $k$ is a non-zero constant, $P(x)$ has degree 1, and if $k=0$, then $P(x)$ is a constant polynomial, and that's it. --pma 13:55, 13 January 2012 (UTC)

January 11

Matrix multiplication on integers and overflow

While reading the description of YUV, this occurred to me. Normally, multiplication by a square, nonsingular matrix is one-to-one and onto, but because RGB and YUV values are constricted on an interval, the transform can overflow. This led me to the following question:

For the matrix equation A*x=b, what values of x, given (x1, x2, x3, ...) exist in an interval [0, 1], will produce values of b with (b1, b2, b3, ...) etc on the interval [0, 1]? — Preceding unsigned comment added by 68.40.57.1 (talk) 05:35, 11 January 2012 (UTC)

All possible values of x where Ax falls within a Cartesian product of intervals [0, 1], a hypercube, are given by inverting the matrix and applying that to the hypercube, which will give a distorted hypercube where bits may fall outside the source. You'd then need to find the intersection of that and the source hypercube which doesn't sound too nice. Dmcq (talk) 11:12, 11 January 2012 (UTC)

Rapid calculation of standard deviation in a time series

I came up with a rapid algorithm that calculates a running variance of the last N values of a noisy (chaotic) time series as each new data sample comes in, without relying on any loops. Basically it maintains two running sums:

• S = the sum of the last N values of x
• T = the sum of the last N values of x².

Each new data value x is added to S, and the square is added to T, while the value that was added n samples ago is subtracted from each. In this way the running sums require no loops. Based on this identity for variance:

${\frac {1}{N}}\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}={\frac {1}{N}}\left(\sum _{i=1}^{N}x_{i}^{2}\right)-{\overline {x}}^{2}$

...the standard deviation for my time series (N weighted, not N−1 weighted) is simply:

$\sigma ={\sqrt {\frac {NT-S^{2}}{N^{2}}}}$

So far this is working OK for me. However, I am concerned about errors that can occur when the two terms in the numerator are many orders of magnitude larger than their difference (or worse, the difference due to roundoff is negative). This hasn't happened yet in my application, but the possibility is there.

So I've spent several hours looking for rapid calculation techniques for time series, and found nothing. I do find single-pass methods for calculating variance (see Standard deviation#Rapid calculation methods for example, based on Welford's classic paper of 1962), but for a fixed-length variance of a time series, this would still require a loop every time the series gets a new data point.

Does anybody know of a loop-less rapid calculation of standard deviation of a time series that doesn't introduce the possibility of roundoff error? The only alternative I know of is exponential moving standard deviation:

$w={\frac {2}{N+1}}$

$\mu _{i}=wx_{i}+(1-w)\mu _{i-1}\$

$\sigma _{i}={\sqrt {wx_{i}^{2}+(1-w)\sigma _{i-1}^{2}}}$

...but this has some undesirable settling time properties for me, so I'd prefer not to use it.

Anyone know of any other alternatives? ~Amatulić (talk) 21:37, 11 January 2012 (UTC)

Would this Algorithms_for_calculating_variance#Compute_running_.28continuous.29_variance work? --NorwegianBlue ^talk 22:20, 11 January 2012 (UTC)

No, that's Welford's algorithm that I mentioned above. That section title is somewhat misleading. You could run that algorithm continuously on a time series but your result would be the variance of all the data contained within the time series, not the last N values as I'm trying to calculate.

My problem can be restated like this:

Calculate the variance or standard deviation inside a fixed-length window that slides over an infinitely long data set, using a rapid calculation algorithm that doesn't require loops AND doesn't introduce potentially catastrophic roundoff errors.

I have solved the first part but not that last part. ~Amatulić (talk) 23:23, 11 January 2012 (UTC)

Try searching for "recursive calculation of variance". Does it help? --HappyCamper 23:37, 11 January 2012 (UTC)

You can always remove roundoff errors by not introducing any when adding or subtracting by doing rounding before that stage. For instance in computing terms if you round the square to a float instead of a double before adding it to a double and you don't have too wide a range of values then double value will be the exact sum of the float values. Dmcq (talk) 12:54, 12 January 2012 (UTC)

HappyCamper: Thanks. My searches for recursive variance turned up mixed results, either a method that does what I'm already doing, or a description of exponentially weighted variance that I described above, or a method like Welford's algorithm, which isn't a sliding-window algorithm.

Dmcq: Intuitively, rounding off a double to a float seems like it would introduce even more error, which one might get if one started out with floats in the first place. After thinking about it, though, I see how errors in the low significant digits would get lopped off by rounding to floats, so that might work. What I'm doing now, just for safety, is to use as my numerator max(0, NT−S²). At least that will prevent the possibility of a negative argument in the square root. ~Amatulić (talk) 20:33, 12 January 2012 (UTC)

January 12

Power series reloaded

Hey again, sorry for so many questions! I will help someone else when I can, I promise! My question is, where does the power series for an exponential function a^x arise naturally? Alternately put, assuming we know nothing about Taylor series but assuming that we know infinite series "work", how can we discover it without constructing it from its derivatives? (I think the series for this particular function was known before Taylor series were developed, so someone apparently did it) What I'm thinking is something along the lines of a formula or discovery or something where it just "appears" (and possibly we do not know it is the series for a^x). Thanks, and sorry if I wasn't really clear. 24.92.85.35 (talk) 02:53, 12 January 2012 (UTC)

The exponential function y=e^x satisfis the linear differential equation y'=y and the initial condition y(0)=1. Substituting the power series y=Σ a_ixⁱ into the initial condition gives a₀=1. Substituting the power series into the differential equation gives Σia_ixⁱ⁻¹=Σa_ixⁱ=Σa_i−1xⁱ⁻¹. So ia_i=a_i−1. So a_i=a_i−1/i=a₀/i!=1/i!. So e^x=Σxⁱ/i! Bo Jacoby (talk) 08:22, 12 January 2012 (UTC).

Looks like the OP was asking about $a^{x}$ for a general a. But this is kind of moot since to expand this you usually first translate it to $\exp(x\ln a)$ and use the power series for exp. And the series for that can be found for example using exp'=exp or $e^{x}=\lim _{n\to \infty }\left(1+{\frac {x}{n}}\right)^{n}$. Both ways are fairly elegant, I don't think you'll find something more "natural". -- Meni Rosenfeld (talk) 10:38, 12 January 2012 (UTC)

How advantageous is it to replace a paper-towel dispenser with a wall dryer?

In a dorm restroom, how often does an average stack of paper towels get used up, how much is it to get another new stack, and therefore, how much will it cost per month and per year to keep the typical paper towel dispenser restocked?

On the other hand, if it gets replaced with a heated wall dryer, how much electricity would the typical new 2012-model wall dryer consume in each typical use, in a month, and a year? (Let us assume 9¢ a kilowatt-hour, as it was 8.44¢/KWh in Kansas last time I checked in May.)

(The initial purchase of a dryer will be made by donation, so the recipient won't have to worry about the procurement cost.)

Therefore, how far ahead will the dorm come out by switching from a paper towel dispenser to a heated wall-dryer? Thanks! --70.179.174.101 (talk) 13:23, 12 January 2012 (UTC)

How many paper towels are in an average stack? It depends. How often is an average stack used up? It depends. How often is the stack refilled? It depends. How much does a new stack cost? It depends. How much does the electricity for a heated dryer cost? It depends. That is the only thing that you gave values to work with. You need to make assumptions about everything else. Otherwise, any result is simply useless. -- kainaw™ 13:51, 12 January 2012 (UTC)

You could get an energy monitor to measure actual energy usage, or estimate it using the power rating of the device and multiply it by the time the drier is on. Finding out how much energy is used to make a paper towel is a harder, you could start at Carbon footprint to try and find out.--Salix (talk): 14:22, 12 January 2012 (UTC)

Another option I thought of - if this is real life - is to simply install the air dryer next to the paper towel dispenser. If the air dryer saves money and some people actually use it, then there will be an overall savings. Of course, there would be more savings if everyone simply stopped washing their hands! -- kainaw™ 14:27, 12 January 2012 (UTC)

Shouldn't questions like this be under Miscellaneous rather than maths? The work here is looking up things and knowing about the objects, why would anyone on this reference desk know anything like that? Dmcq (talk) 14:35, 12 January 2012 (UTC)

Well I did a bit of looking about and its tricky to find comprehensible data. The European Commission give full lifecycle inputs and output for some materials such as Corrugated board boxes, slightly more useful is the UK Enviornment Agency who have spreadsheet of CO2 equivilents for the construction industry[3]. Nearest equivalent to a paper towel I could find on that is particle board at 0.54 tonnes of CO2 per tonne of material. Grid electric is 0.00059368 tonnes of CO2 per kWh or 593.68g CO2 per kWh. I estimate 1 paper towel is 5g so 2.7g of CO2 equivalent (excluding transport costs). I estimate air drier runs at about 3kW (same as my kettle) and I spend 30 secs drying my hands using it, so thats 0.025 kWh or 15g of CO2. My estimates could easily be off by an order of magnitude so I can draw any conclusion.--Salix (talk): 15:48, 12 January 2012 (UTC)

The OP didn't raise any question about carbon, but about cost of the $ & £ sort. --Tagishsimon (talk) 15:54, 12 January 2012 (UTC)

Another issue is that paper towels can do things which hot air blowers can't, like allow you to open the bathroom door without having germs redeposited on your hand, or clean up water splashed all over the sink. So, if you remove the paper towels, you may find that toilet paper is now used for those purposes, and it's usage increase must then enter into the calcs. So, I second the idea of offering both the blower and towels. Paper towels can also both be recycled and made from recycled paper, so that's a plus. StuRat (talk) 16:42, 12 January 2012 (UTC)

My reply-all

Okay, I cannot calculate the paper towel usage myself, so try a typical 54-person dorm. There's a girl's floor, a guy's floor, and a "grab-bag" floor, each with a restroom. Can we establish an assumption with said dorm? If anyone has lived in or worked with a dorm, then they may be able to give a reasonable ballpark. I'm not looking for "carbon footprint" because the costs matter more to them, but I would have to assume the footprint is less with a wall dryer. (The outlet could turn in many ways and therefore even help dry a nearby sink, for example.)

About the cost of bundles of paper towels, let's choose a supplier easily available in the state of Kansas, preferably to the town of Manhattan. We can figure in their cost.

A dryer takes 30 seconds to dry, typically. Some decide to take a full minute. How many kWh are used in one minute? (Remember the figure of 9¢/kWh was cited.) Was anything else needed or are we good?

(Moreover, I hope my reply was well-formed enough because after having done plenty today, I am about ready to nap, so I'm not sure how my mental processes are right now. But I hope this was a good enough shot.) --70.179.174.101 (talk) 00:49, 13 January 2012 (UTC)

You need to estimate how often each person uses the restroom. This will depend on the nature of the dorm and the people within: if they're out all day and not back till late in the evening (or sleeping over with their significant others in deluxe off-campus accommodation, or visiting their parents for days at a time) then they may only use it twice a day, but if they're in the dorm all day studying hard and drinking lots of coffee then they may use it 8 or 10 times a day each. (This may of course change at different times of year.) You are far better placed than anybody else on this website to measure the usage of paper towels in your dorm. --Colapeninsula (talk) 11:17, 13 January 2012 (UTC)

January 13

Help with integral

How do you evaluate this integral? Could someone help?

${\frac {d}{dz}}\left(\int \!{\frac {{\it {f}}\left(z-r\right)}{{r}^{3}}}{dr}\right)$

deeptrivia (talk) 00:12, 13 January 2012 (UTC)

$\int \!{\frac {{\it {f'}}(z-r)dr}{r^{3}}}.$ Bo Jacoby (talk) 08:00, 13 January 2012 (UTC).

Is the integral in the question meant to be a definite integral with r ranging on a fixed interval (which does not depend on z)? – b_jonas 09:07, 13 January 2012 (UTC)

b_jonas and Bo Jacoby, thanks for your answer. This integral comes out as a part of a general solution to a PDE, where f can be any function. Since the solution you came up with is an integral of a derivative, could this be simplified further?

The PDE I'm trying to solve is

${\frac {\partial ^{2}}{\partial {r}^{2}}}v\left(r,z\right)+{\frac {{\frac {\partial }{\partial r}}v\left(r,z\right)}{r}}-{\frac {v\left(r,z\right)}{{r}^{2}}}+{\frac {\partial ^{2}}{\partial z\partial r}}v\left(r,z\right)-{\frac {{\frac {\partial }{\partial z}}v\left(r,z\right)}{r}}=0$

The solution Maple comes up with is:

$\left(\int \!{\frac {{\it {\_F3}}\left(-r+z\right)}{{r}^{3}}}{dr}+{\it {\_F4}}\left(z\right)\right)r$

Thanks a ton! deeptrivia (talk) 13:31, 13 January 2012 (UTC)