Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

September 5

Question on polynomials

Hey guys, how are ya all? ANyway, I'm stuck on a question and your help would be appreciated.

The question is this: Let k be a natural number and let r be a real number such that |r| < 1. Prove (by induction on k) that for any polynomial P of degree k there's a polynomial Q of degree k s.t. Q(n+1)r^(n+1) - Q(n)r^n = P(n)r^n.

The hint is this: Consider differences of succesive terms for n^k r^n, and use the inductive hypothesis.

OK. So that's the question and hint. I've been trying hard at this question since I've woken up but to no avail. So I've been trying for 5 hours. I really would appreciate an answer. Like for example I've tried it when P(x) = x^2, r = 1/2 and I've found that Q(x) = -2x^2 - 4x - 6 and it works. The reason I did this is 'cause it helps me to sum the series n^2 2^(-n). But I'm really stumped on this one. Help??? Thanks guys ... I've worked out about 7 pages of rough work so please don't say I'm lazy and I want my homework done for me. This isn't homework just independent study and it's for my own benefit but I'd really like a decent hint or an answer please. —Preceding unsigned comment added by 114.72.228.12 (talk) 05:02, 5 September 2010 (UTC)

Also you don't have to use the hint if you don't want to. Thanks guys ...

I've just run through the argument but only have a mess (though I think it's a correct mess). The point you might be missing (which is basically the hint restated) is that you don't want to evaluate what Q is if P = x^k; it suffices to show you can deal with x^k (in much the same way as the grounding case of the induction) and then the left overs are of less degree so can be absorbed into the rest of P and dealt with by Inductive Hypothesis. Also I don't see that you need |r|<1 (though you're in trouble if r=1). Hope that helps. 95.150.22.63 (talk) 14:17, 5 September 2010 (UTC)

Following from 95.150.22.63's bit about induction, I think it is pretty clear that we can rewrite as rQ(n+1)r^n - Q(n)r^n = x^k r^n. We ignore the r = 0 case, since it's rather meaningless, and so we divide through by r^n. Q is a polynomial of degree k, so let us express it as

$Q(n)=\sum _{i=0}^{k}a_{i}n^{i}$, so our equation is $r\sum _{i=0}^{k}a_{i}(n+1)^{i}-\sum _{i=0}^{k}a_{i}n^{i}=x^{k}$

Then expand the first term by the binomial theorem, and then equating coefficients of n^i for i = 0, ..., k, you get k+1 linear equations to solve for the k+1 coefficients a_0, ..., a_k. To calculate, you can start from a_k and work back, a_k = 1/(r-1) always, for example (a_i can be solved for directly once you know the values of a_{i+1}, ..., a_k). In any case, you have k+1 linear equations for k+1 unknowns, and now you want to be sure that for all |r|<1, these have a unique solution. Invrnc (talk) 14:31, 5 September 2010 (UTC)

Jensen's inequality?

Is this a case of Jensen's inequality, or some other inequality?

$(\mathbb {E} (|a-c|^{2}))^{1/2}\leq (\mathbb {E} (|a-b|^{2}))^{1/2}+(\mathbb {E} (|b-c|^{2}))^{1/2}$

—Preceding unsigned comment added by 130.102.158.15 (talk • contribs) 08:14, 5 September 2010 (UTC)

That's Minkowski's inequality, with $p=2$. —Bkell (talk) 15:43, 5 September 2010 (UTC)

Thanks! —Preceding unsigned comment added by 130.102.158.15 (talk) 22:32, 5 September 2010 (UTC)

Limit Question

I know it's possible to reason out that $\lim _{(x,y)\rightarrow (0,0)}\exp {\tfrac {-1}{x^{2}+y^{2}}}=0$, but is there any way to do it algebraically? Thanks. --Basho: banana tree (talk) 20:42, 5 September 2010 (UTC)

For every ε>0, you can choose an appropriate δ>0 (you will be able to express this value in terms of ε) such that for every point (x,y) within a distance of δ of (0,0), $|\exp {\tfrac {-1}{x^{2}+y^{2}}}|<\epsilon$. This is formally how you demonstrate a limit. Rckrone (talk) 22:34, 5 September 2010 (UTC)

(edit conflict) You are interested in how the function behaves as (x,y) tends towards (0,0). But there are many ways in which a point (x,y) can tend towards (0,0). Let's assume that (x,y) follows a line towards (0,0). In other words x = r⋅cos(θ) and y = r⋅sin(θ), where θ is a fixed parameter and r is the variable along the line. We consider:

$\lim _{(x,y)\to (0,0)}e^{\left({\frac {-1}{x^{2}+y^{2}}}\right)}=\lim _{r\to 0}e^{\left({\frac {-1}{(r\cos \theta )^{2}+(r\sin \theta )^{2}}}\right)}=\lim _{r\to 0}e^{\left({-1/r^{2}}\right)}=0.$

Notice that the penultimate expression is independent of θ and does not depend upon the sign of r since 1/(+r)² = 1/(−r)². — Fly by Night (talk) 23:04, 5 September 2010 (UTC)

Note that this is not a proof. There are functions which converge along every line towards the origin, and yet do not converge at the origin.--203.97.79.114 (talk) 09:53, 6 September 2010 (UTC)

That's the whole point of a limit. The function may not be defined at the origin, but you calculate the limit of the function as you tend towards the origin. In this example the function is indeterminate at the origin; but the limit exists and is well defined. — Fly by Night (talk) 12:59, 6 September 2010 (UTC)

I didn't say such functions were undefined; I said they did not converge. Consider the function ${\tfrac {xy}{x+y^{3}}}$. The limit as $(x,y)\rightarrow (0,0)$ along any line is $0$. The limit as $(x,y)\rightarrow (0,0)$ along the curve $x=y^{2}$ is $1$. The limit in this case does not converge, yet if you only considered approaching along a line, you would think it does. —Preceding unsigned comment added by 203.97.79.114 (talk) 13:18, 6 September 2010 (UTC)

I don't think that example works. Surely if we have $x=y^{2}$ then ${\tfrac {xy}{x+y^{3}}}={\tfrac {y}{1+y}}$ and then $\lim _{y\rightarrow 0}{\tfrac {y}{1+y}}=0$ ? (However, Meni's example below illustrates the point).Gandalf61 (talk) 13:36, 6 September 2010 (UTC)

Woops. ${\tfrac {xy^{2}}{x^{2}+y^{4}}}$ works if you want something elementary. Or Meni's, as you say. --203.97.79.114 (talk) 13:48, 6 September 2010 (UTC)

[ec] 203's point is that only looking at straight lines is insufficient; you need it to work for every curve. The canonical example is

$f(x,y)={\begin{cases}1&x>0,y=x^{2}\\0&{\textrm {otherwise}}\end{cases}}$

This function has no limit at the origin; but along any line, it converges to 0 at the origin.

Of course, in the OP's case, this objection can be considered a nitpick, since you have demonstrated that the function depends only on r, so finding the limit for a single curve suffices. -- Meni Rosenfeld (talk) 13:27, 6 September 2010 (UTC)

Strictly speaking, you cannot prove this statment (or any other statement involving limits) algebraically because it is a statement of analysis, not of algebra. The result depends on the topology that you use - with the usual topology on R² the statement is true, but with the discrete topology (for example) it is false. The choice of topology does not affect the algebraic properties of R² - therefore the result cannot be derived from algebraic properties alone. Gandalf61 (talk) 08:34, 6 September 2010 (UTC)

True, but we often teach an algebraic approach to limits that relies on analysis only for the fact that certain basic functions (addition, division where defined, etc) are continuous. Presumably that's what was being asked for.--203.97.79.114 (talk) 13:29, 6 September 2010 (UTC)

So what do we think the OP is looking for ? What would a proof of the OP's statement look like under this algebraic approach to limits ? Presumably it would not involve any δs, εs, open intervals or neighbourhoods ? Gandalf61 (talk) 13:51, 6 September 2010 (UTC)

I think it would be something like $\lim _{(x,y)\rightarrow (0,0)}\exp {\tfrac {-1}{x^{2}+y^{2}}}=\exp \lim _{(x,y)\rightarrow (0,0)}{\tfrac {-1}{x^{2}+y^{2}}}=\exp(-\infty )=0$. -- Meni Rosenfeld (talk) 14:55, 6 September 2010 (UTC)

dL

In physics you express uncertainty as dL or delta-L. WHat does this have to do with derivatives? 76.229.214.25 (talk)` —Preceding undated comment added 23:00, 5 September 2010 (UTC).

because if y=f(x), then dy=f '(x)dx is an approximate computation of the uncertainty of y based on the uncertainty of x. Bo Jacoby (talk) 08:13, 6 September 2010 (UTC).

September 6

Proving the sequence defining e converges

How would you go about proving that $\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$ converges? Or, say, that $2<\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}<3$, then that $2.70<\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}<2.72$ and so on, zooming in as far as you like? It Is Me Here ^{t / c} 17:54, 6 September 2010 (UTC)

You can use the binomial theorem, $\left(1+{\frac {1}{n}}\right)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}{\frac {1}{n^{k}}}$. It's easy to show that the kth term is less than $1/k!$ so the expression is less than $\sum _{k=0}^{\infty }{\frac {1}{k!}}$. In fact the kth term converges to $1/k!$, and it should be possible to bound the differences to make the entire expression converges.

There should be a result that states conditions under which $\lim _{n\to \infty }\sum _{k=0}^{\infty }a_{n,k}=\sum _{k=0}^{\infty }\lim _{n\to \infty }a_{n,k}$, but I can't remember or find the details. -- Meni Rosenfeld (talk) 18:17, 6 September 2010 (UTC)

The dominated convergence theorem gives convergence here. There's probably a separate name for dominated convergence applied to discrete measure spaces, but if so I don't know it. Algebraist 18:28, 6 September 2010 (UTC)

Also, you can show that the sequence $(1+1/n)^{n}$ is (strictly) increasing: use the inequality of arithmetic and geometric means with choice of $n+1$ numbers $x_{1}=x_{2}=\dots =x_{n}:=1+1/n$ and $\textstyle x_{n+1}:=1.$ To show that it is bounded, note that the argument above also holds more in general for the sequence $(1+c/n)^{n}$ for any real $c$, showing it is increasing as soon as $n+c>0.$ Since for n>|c|

$\left(1+{\frac {c}{n}}\right)^{n}\left(1-{\frac {c}{n}}\right)^{n}=\left(1-{\frac {c^{2}}{n^{2}}}\right)^{n}\leq 1$

and both sequences on the LHS are positive and increasing, they are bounded. So this actually gives the convergence of $(1+c/n)^{n}$ for all real $c$ together with the bounds: in particular for $c=1$ you should obtain, for all $n$

$\left(1+{\frac {1}{n}}\right)^{n}<e<\left(1+{\frac {1}{n}}\right)^{n+1}.$

Note however that the argument via the dominated convergence theorem for series shown by Algebraist also holds when $c$ is more generally a complex number, or a matrix, or even an element in a Banach algebra. Also, it can be shown that the convergence of the series is much faster. --pma 02:46, 7 September 2010 (UTC)

The Space of Oriented Geodesics

Consider the a 2-sphere is 3-space, centred at the origin. The geodesics are given by the great circles. For each great circle we get a unique plane, namely the plane containing the great circle (and passing through the origin). Given a plane passing through the origin we get a unique line passing through the origin, namely the line perpendicular to the given plane passing through the origin. So each great circle gives a line through the origin. The same procedure works in reverse too: a line through the origin gives a plane through the origin (the plane passing through the origin that is perpendicular to the line). A plane through the origin gives a great circle (the circle generated by intersecting the sphere with the plane). It follows that the space of unoriented geodesics is exactly RP², i.e. the real projective plane. Now consider the space of oriented geodesics of the sphere. This is exactly the sphere itself. Given a point p of the sphere, we get an unoriented geodesic by intersecting the sphere with the plane passing through the origin that is perpendicular to the chord joining the origin to p. Let us denote this intersection by C. Assume that g : S¹ → S² gives a smooth parametrisation of C, with parameter θ. We can give an orientation to C by insisting that

${\text{Volume}}\left\{\left[\,p,\,g(\theta ),\,g'(\theta )\,\right]\right\}>0\ {\text{ for all }}\ \theta .$

Clearly, if we replace p by −p then we must reverse the orientation of C, i.e. replace θ with –θ, in order to maintain the inequality above. This means that there is a one-to-one correspondence between the oriented geodesics of the 2-sphere and the 2-sphere itself. My question is this:

• What other manifolds M have the property that there is a one-to-one correspondence between the oriented geodesics of M and M itself?

— Fly by Night (talk) 18:14, 6 September 2010 (UTC)

If two sets have the same cardinality, then there is a one-to-one correspondence between them. The cardinality of the set of real numbers is c. The cardinality of the set of points in a manifold is also c. The cardinality of the set of oriented geodesics in a manifold is also c. So all manifolds M have the property that there is a one-to-one correspondence between the oriented geodesics of M and M itself. You did not require the correspondance to be continuous. Bo Jacoby (talk) 23:37, 6 September 2010 (UTC).

Could you please show how you prove that the space of oriented geodesics on M has the same cardinality as M? — Fly by Night (talk) 00:57, 7 September 2010 (UTC)

The set of all points in an infinite plane has the same cardinality as the set of all points in a finite line segment, namely, c.[1] What is the one-to-one correspondence between an infinite plane and a finite line segment? — Fly by Night (talk) 00:10, 7 September 2010 (UTC)

Well, finding a specific one is a little complicated. Not too bad, but not something you can draw on a napkin like the one between the rationals and the natural numbers.

The easiest approach to giving an explicit bijection is probably to find injections both ways, and then apply the method used to prove the Schröder–Bernstein theorem. Finding an injection from the line segment to the plane is trivial; you can get the reverse injection by something such as interleaving the decimal representations of the x- and y- coordinates of a point, after projecting the plane into a finite square. --Trovatore (talk) 00:39, 7 September 2010 (UTC)

Thanks Trovatore. Do you think this cardinality argument shows that there is a bijective correspondence between the space of oriented geodesics on M and M itself? I mean, how can Bo say that the space of oriented geodesics on M has the same cardinality as M? Geodesics are differential invariants, whereas cardinality is much weaker. Making small bumps on the sphere may stop is having any closed geodesics at all. — Fly by Night (talk) 00:46, 7 September 2010 (UTC)

N.B. I meant to say that there is a bijective correspondence between the 2-sphere and its space of oriented geodesics. And my question should have been: What other manifolds M have the property that there is a bijective correspondence between the oriented geodesics of M and M itself? — Fly by Night (talk) 00:10, 7 September 2010 (UTC)

"how can Bo say that the space of oriented geodesics on M has the same cardinality as M?" An oriented geodesic is defined by the two points (P, P+dP). The cardinality of pairs of points is c when the cardinality of points is c. Bo Jacoby (talk) 06:25, 7 September 2010 (UTC).

Why not try a stronger condition that just bijection, such as Homeomorphism, or Diffeomorphism both of which are exhibited by the sphere? This leads to a more general question of when we can give the space of all geodesics a manifold structure?--Salix (talk): 09:33, 7 September 2010 (UTC)

It seems like there is a bit of existing work on the space of geodesics [2] has a survey article. There are several results about the topological structure of the space, including conditions for it to be a manifold, but falls short of answering Fly by Night's question.--Salix (talk): 10:07, 7 September 2010 (UTC)

That's really interesting. Thanks Salix. This is more like what I had in mind, i.e. does the space of oriented geodesics have a manifold structure and is it the same as the original manifold. I know I didn't say as much but I think you hit the nail on the head. Thanks again! — Fly by Night (talk) 16:41, 7 September 2010 (UTC)

big-O?

I've got some calc homework and I need to solve a problem using the big-O. Unfortunately I forgot how (it's been a long labor day weekend ;) and your article is kind of written more for the person looking something up that they'd learnt years ago than me, just learning it now. I'm not going to give the actual problem because it would be kind of pointless if I didn't figure out how to do it myself, but can someone show me how in simple terms (using any problem)? Thanks 76.229.163.32 (talk) 19:59, 6 September 2010 (UTC)

If you won't give an explicit problem then you must be seeking a general solution. In that case; you'll find everything you need in the formal definition section. — Fly by Night (talk) 20:43, 6 September 2010 (UTC)

September 7

What does this expression count?

Hi. I'm working on a combinatorial problem, and I've encountered a lot of expressions with form similar to the following: ${\tbinom {9}{4}}-{\tbinom {8}{3}}+{\tbinom {7}{2}}-{\tbinom {6}{1}}+{\tbinom {5}{0}}$. Can anyone think of a situation where this counts something? If I can think about a real-world representation of it, I might be able to count that same quantity with a different technique, and thus establish some kind of identity.

Thanks in advance for any ideas. -GTBacchus^(talk) 02:06, 7 September 2010 (UTC)

You can often simplify that kind of expression mechanically with WZ theory, if that's of any interest. See the A=B book (online) linked in the references to that article. For your concrete example I'd put a few terms into OEIS and see what comes out. 67.122.211.178 (talk) 06:11, 7 September 2010 (UTC)

For example, you have this. -- Meni Rosenfeld (talk) 09:18, 7 September 2010 (UTC)

It looks like some form of the Inclusion-exclusion principle to me. 198.161.238.19 (talk) 18:02, 7 September 2010 (UTC)

What about the number of subsets S of {1,...,9} with card(S)=5 and such that max(S) has the same parity of 9?--pma 20:05, 7 September 2010 (UTC)

Good grief, not another homework question

If f is continuous on [0,2], and f(0) = f(2), prove that there is a real number x ∈ [1,2] such that f(x) = f(x-1).

Intuitively I know this should be true, but that hasn't helped me out much. The only approach I can think of is to show that you can find infinitely many pairs of numbers such that f(a) = f(b), and that a-b will range from 2 to 0, but this hasn't helped. Can anyone help me? 74.15.136.172 (talk) 23:02, 7 September 2010 (UTC)

Consider the function g on [1,2] with g(x)=f(x)-f(x-1). Algebraist 23:10, 7 September 2010 (UTC)

Got it, thanks! 74.15.136.172 (talk) 00:17, 8 September 2010 (UTC)

Dixon's Method of Factorization

Perhaps I'm being slow here, but under 'Method' on Dixon's factorization method, why is it the case that N = gcd(a − b, N) × gcd(a + b, N)? Surely N -divides- gcd(a − b, N) × gcd(a + b, N), since gcd(A,BC) divides gcd(A,B)gcd(A,C), but why must the equality hold here? Is it something to do with the method of computing a and b? Or perhaps I'm missing something?

Cheers, 92.40.243.116 (talk) 23:24, 7 September 2010 (UTC)

The equality is wrong, unless a + b and a − b are coprime (take N = lcm(a + b,a − b) for a counterexample).—Emil J. 10:03, 8 September 2010 (UTC)

September 8

Σ and ʃ

How are summation and integration are related? That is to say, if i can only perform a sum of a function from a to b how can I find the area, and if I can only integrate a function along [a, b] how can I find the sum of it (or it's sequence)? 24.92.78.167 (talk) 02:34, 8 September 2010 (UTC)

Two relations are the definition of the integral via Riemann integral#Riemann sums and the Integral test for convergence of series. -- 58.147.53.113 (talk) 05:21, 8 September 2010 (UTC)

There are also the Euler–Maclaurin formula which works both ways, and methods like Simpson's rule and Gaussian quadrature that can be used to approximate an integral using a sum. -- Meni Rosenfeld (talk) 08:06, 8 September 2010 (UTC)

They are related by Taylor's theorem. Write Δ for the difference operator (Δf (x) = f(x+1)-f(x) ) and write D for differentiation, so D = 1/ʃ and Δ = 1 / Σ. The classic game to play with Σ, ʃ, Δ and D is as follows: you can interpret Taylor as saying Δ = exp(D)-1. So Σ = 1/(exp(D)-1) = (1/D)(D / (exp(D)-1)) = ʃ (1 - (1/2) D + (1/12)D^2 - (1/720)D^4 ...) = ʃ + 1/2 + (1/12)D - (1/720)D^3 + ... This can actually be made rigourous, at the expense of some error terms and conditions on the functions f to which it is applied: you end up with the Euler-Maclaurin summation formula mentioned by Meni. The coefficients in the power series are closely related to the Bernoulli numbers. Tinfoilcat (talk) 17:59, 8 September 2010 (UTC)

September 9

Atmospheric carbon scenario difference projections again

I'm moving this back from the archives in hopes that someone else can help balance it against File:Extreme-weather-cost.gif: What is the optimal amount of money to spend on climate change mitigation which will minimize total financial losses?

There is a calc problem that I don't understand at Wikipedia:Reference_desk/Archives/Science/2010 August 27#What is the wind-water-solar climate change mitigation scenario atmospheric carbon projection? [copied below] Why Other (talk) 22:22, 27 August 2010 (UTC)

From what I can tell, since there seems to be a lot of jargon there that would be better understood by a atmospheric scientist than a mathematician, the model being used is that the atmosphere is a giant tank of some fluid where some impurity is being added at a given rate while at the same time it's being removed at a rate proportional to its concentration. The tank is assumed to be well mixed, meaning you don't have to worry about the concentration not being the same in different parts of the tank. This is a fairly standard problem in ODE's and the solution is a straightforward use of separation of variables. I hope that helps with the mathematical aspect of the model at least.--68.40.56.142 (talk) 15:04, 28 August 2010 (UTC)

Thanks! Why Other (talk) 03:24, 30 August 2010 (UTC)

In Jacobson, M.Z. (2009) "Review of solutions to global warming, air pollution, and energy security" Energy and Environmental Science 2:148-73 doi 10.1039/b809990c and Jacobson, M.Z. and Delucchi, M.A. (November 2009) "A Plan to Power 100 Percent of the Planet with Renewables" (originally published as "A Path to Sustainable Energy by 2030") Scientific American 301(5):58-65 what is the projected atmospheric carbon over time for their preferred wind-water-solar program?

What year do they start subtracting carbon and when do they reach 350 ppm?

I have asked also here, but I have been having better luck here at WP:RDS. Why Other (talk) 02:04, 27 August 2010 (UTC)

Per Dr. Jacobson, this is related to Eqn. 3 in http://www.stanford.edu/group/efmh/fossil/ClimRespUpdJGR%201.pdf

"[calculate] the time-dependent change in CO2 mixing ratio from a given anthropogenic emission rate, [and with that] the time-dependent difference in mixing ratio resulting from two different emission levels by subtracting results from the equation solved twice. Note that chi in the equation is the anthropogenic portion of the mixing ratio (this is explained in the text) and units of E need to be converted to mixing ratio. The conversion is given in the paper."

This almost might be ready for the math reference desk. Why Other (talk) 22:18, 27 August 2010 (UTC)

Maybe something like the scenario proposed by James Hansen's Alternative Scenario paper? ~AH1^(TCU) 18:38, 28 August 2010 (UTC)

Interesting, but the assumptions in that paper don't involve adjusting the CO2 emissions rate, which is what I am trying to do.

This is from a response to a pointer to this question I put on the Math Reference Desk:

"... This is a fairly standard problem in ODE's and the solution is a straightforward use of separation of variables....--68.40.56.142 (talk) 15:04, 28 August 2010 (UTC)"

I wish I had more experience with ODEs. Why Other (talk) 03:27, 30 August 2010 (UTC)

The separation of variables article describes it some, and a little more explanation is here. Basically the ODE is something like

${dy \over dx}=ay+b$

Separating variables gives you

${dy \over ay+b}=dx$

Integrating both sides,

${1 \over a}{\log(ay+b)}+C_{1}=x+C_{2}$

Exponentiating both sides, rearranging terms, and renaming some constants,

$y={ke^{ax}-b \over a}$

where $k=e^{a(C_{2}-C_{1})}$ or something like that. It's possible that I messed up that calculation somewhere (it's late and I haven't done this in a while) but it's a basic ODE technique.

That said, that linear model sounds oversimplistic for something as complicated as the earth/atmosphere system. See for example clathrate gun hypothesis. 67.119.3.248 (talk) 08:50, 30 August 2010 (UTC)

Can anyone take it further than that? I know compared to File:Extreme-weather-cost.gif this is "merely" an optimization problem in probability distributions, but I'm sure it has few enough degrees of freedom to have a singular optimum solution.

If you want to consider a less tractable, but still important problem: how much is knowing the answer worth? Why Other (talk) 05:40, 8 September 2010 (UTC)

Fitting a parabola and Euler's theorem

I have two questions. I know that to write an equation for a parabola given three distinct points A, B, and C (which do not include any special points like the vertex) you substitute the x and y of A, B, and C for x and y in $y=ax^{2}+bx<c$, then solve a system of equations. Is there a faster way to do this, using more sophisticated maths? I was think the mean value theorem and Rolle's Theorem could be used, if you find the slope of the lines AB, BC, and AC and set them all equal to $2xa+b$, but I got hung up because the 'x' is not the same in each so you have x₁ etc. And #2 Euler's method goes like: $y_{n+1}=y_{n}+\Delta x_{step}\cdot f(x_{n},y_{n}),\$ where f is the derivative of some function y(x). Why the $f(x_{n},y_{n})$? What are (xn, yn) there for, since f is presumably a function only in x and is ther any difference to just f(xn)? Thanks a lot. 24.92.78.167 (talk) 00:21, 9 September 2010 (UTC)

Lagrange polynomial may be useful here. Euler's method is a method for solving differential equations and, unless I misunderstood the question, is not applicable.--RDBury (talk) 06:51, 9 September 2010 (UTC)

Euler's method is a separate question entirely; as I understand it, it's why might f depend on y. Usually it's for solving dy/dx = f(x,y) and f really could depend on y (eg dy/dx = y). Of course, if you can solve for y analytically then it only really depends on x (and the initial conditions), but if you can solve analytically you're not going to use Euler's method. 95.150.23.63 (talk) 15:28, 9 September 2010 (UTC)

Underline notation for n!/(n-k)!

Binomial coefficient#Multiplicative formula states:

${\binom {n}{k}}={\frac {n^{\underline {k}}}{k!}}$

Clearly $n^{\underline {k}}={\frac {n!}{(n-k)!}}$, but I've never seen that notation before. Is it standard, and what is it called? -- 124.157.218.142 (talk) 02:42, 9 September 2010 (UTC)

Falling factorial.--203.97.79.114 (talk) 03:58, 9 September 2010 (UTC)

Thank you! And from Pochhammer symbol#Alternate notations, spoken "n to the k falling". -- 124.157.218.142 (talk) 04:07, 9 September 2010 (UTC)

SVD and Polysemy

I've read in a few papers on latent semantic indexing that singular value decomposition solves problems with polysemy in data. This claim is made, but not backed up. I understand clearly how synonymy and data sparsity are solved (the other two commonly noted data problems), but I do not clearly see how polysemy is solved. Any ideas? -- kainaw™ 04:07, 9 September 2010 (UTC)

Probably because of the occurrence of words of related meaning tend to be correlated with each other. Say w1 and w2 have meanings m1 and m2 (that are related, say both are in medicine), and w1 and x2 have meanings n1 and n2 (i.e. w1 is polysemous) where n1 and n2 are related to finance. You'll see {w1,w2} in medical documents and {w1,x2} in financial documents. 75.57.241.73 (talk) 07:53, 9 September 2010 (UTC)

Radom number generation

Hi there. If I wanted to generate lognormal random numbers (from the distribution LN(μ,σ²)) can I just generate normal numbers (from the distribution N(μ,σ²)) and use the expenential of the second set? Thanks. --Mudupie (talk) 14:48, 9 September 2010 (UTC)

The second sentence of lognormal says yes! Note of course that the mean of the result won't be $\mu$ or even $e^{\mu }$. --Tardis (talk) 16:24, 9 September 2010 (UTC)

Truncated lognormal distribution

Hello. Say I have a variable X that I know to be lognormally distributed. I would like to parametrise the distribution Y = X|X>k given some observations that I have. I only know the observations that are more than k but I know the number of observations that are smaller than k (although I don't know what those observations are). How do I parametrise Y? Thanks. --Mudupie (talk) 15:41, 9 September 2010 (UTC)

Consider log(X) which has a normal distribution. Find the number m such that the number of observations above m is equal to the known number of observations below k. Look up in a table of the normal distribution to find the number of standard deviations between log(m) and log(k), and you are done. Bo Jacoby (talk) 20:47, 9 September 2010 (UTC).

I corrected the above explanation several times, because it was unclear and incorrect, so let me try to be a little more precise. Let the total number of observations be N. Let the number of observations below k be K. The variable log(X) has normal distribution with mean value (log(m)+log(k))/2, and the K/N quantile is log(k). Bo Jacoby (talk) 21:03, 10 September 2010 (UTC).

Finding Remainder using Scientific Cal?

Is it possible for me to find out the remainder (long division?) using a scientific calculator?

My Scientific Calculator is Sharp EL-520W (EL-520WBBK).

The manual is here: http://ec1.images-amazon.com/media/i3d/01/A/man-migrate/MANUAL000031630.pdf

Say for example: 28,925.5 / 3,600 = 8.03486...

I have a fraction button which outputs either ${\tfrac {57,851}{7,200}}$ or $8{\tfrac {251}{7,200}}$.

The remainder is supposed to be 125.5.

How can I find that using my calculator?

--33rogers (talk) 18:23, 9 September 2010 (UTC)

Either repeatedly subtract the divisor from the quotient until the result is less than the quotient, or once you have the $8{\tfrac {251}{7,200}}$ solution, take the integer part (8 in this case), multiply that by the divisor, and subtract that from the quotient.

28,925.5 - (8 x 3,600) = 125.5

--Rojomoke (talk) 19:46, 9 September 2010 (UTC)

The manual doesn't (appear to) list all the keys, so it's hard to be sure: what you're looking for is the modulo operation. Be warned that some systems that provide such an operation define it only for integers, some systems define it in an odd way for negative numbers, and that the operators modulo and remainder are closely related but not identical. It's confusing; if you'd like to know more, just ask. --Tardis (talk) 15:19, 10 September 2010 (UTC)

e-values and e-vectors of a transition matrix

I am trying to find all the eigenvalues and eigenvectors of this matrix: $\left({\begin{array}{cccc}0&1&0&0\\p&0&(1-p)&0\\0&p&0&(1-p)\\0&0&1&0\end{array}}\right)$

I know that one of the eigenvalues is 1 with an eigenvector of all 1's...

but when I try to expand out the polynomial taken from $|P-\lambda I|=0$, it's not something I can work with. Are there actually four eigenvalues/eigenvectors, and is there a trick to getting the other three? —Preceding unsigned comment added by 130.102.158.15 (talk) 23:03, 9 September 2010 (UTC)

What is the polynomial you get? Rckrone (talk) 03:34, 10 September 2010 (UTC)

I got it to work out alright. Rckrone (talk) 03:50, 10 September 2010 (UTC)

my polynomial is $\lambda ^{4}+\lambda ^{2}(1-2p)+p(1-p)$ and when I try and solve this I get a very ugly thing...

$\pm {\sqrt {\frac {-(1-2p)\pm {\sqrt {(1-2p)^{2}-4p(1-p)}}}{2}}}$ —Preceding unsigned comment added by 130.102.158.15 (talk) 07:09, 10 September 2010 (UTC)

That is not the correct characteristic polynomial. One way to see this is by checking if $\lambda =1$ is a root. (And you know it's an eigenvalue, so it has to be a root of the characteristic polynomial.)

Once you have the correct polynomial, you can try to find the roots of that degree 4 polynomial. (We're actually lucky in this case, and can solve it using the quadratic equation-- often degree 4 is much harder.) But since you already know that 1 is an eigenvalue, you can factor out $\lambda -1$ and get just a degree 3 polynomial. There is one other root you should be able to guess, since the characteristic polynomial is actually a function of $\lambda ^{2}$; and if you factor that out too, you have just a degree two polynomial to solve to find the final two eigenvalues.140.114.81.55 (talk) 09:27, 10 September 2010 (UTC)

It should be noted that finding the characteristic polynomial is easier if you already know some of the eigenvectors. In this case you have the all 1's vector and the alternating ±1 vector are eigenvectors. If you use these to create new columns in $|P-\lambda I|=0$ you can factor out the $\lambda ^{2}-1$ and reduce the problem to a 2×2 matrics. The other factor comes out as $\lambda ^{2}-p(1-p)$.--RDBury (talk) 20:25, 10 September 2010 (UTC)

September 10

Markov chains

I am a bit confused about part of the article on Markov chains.

For the chain with transition matrix $P$, the limiting distribution $\pi$ is the solution to $\pi =\pi P$.

Is the stationary distribution simply an initial distribution that is equal to the limiting distribution? Is it possible for there to be a stationary distribution that is not also a limiting distribution? I'm not sure of the distinction. —Preceding unsigned comment added by 118.208.141.67 (talk) 02:55, 11 September 2010 (UTC)

basic probability / stats question

There is an big jar containing 1000's of very expensive diamonds and rubies, say $1000 each (i.e. it costs $1000 to draw one sample, so I don't want to draw too many). I want to know the approximate mix in the jar. I hand over $5000 to draw 5 stones, and I get 2 diamonds and 3 rubies. So the immediate guess that the jar has 40% diamonds. But how do I estimate the probability that the jar has more than 20% diamonds, based on this limited experiment? What if I go up to 10 samples? Thanks. 75.62.3.153 (talk) 03:59, 11 September 2010 (UTC)