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November 15

piecewise polynomial least squares

(I tried math.stackexchange and got no response for a week. Boo hoo.)

I have in mind a project involving a least-squares fit using piecewise polynomials; at a finite number of known arguments x_j, the k_jth derivative is discontinuous.

How many basis functions are needed? My guess is: xⁿ for 0≤n<min(k), and then, for each j,n such that k_j ≤ n ≤ the maximum degree, a pair of functions which are zero on one side and (x-x_j)ⁿ on the other. Is that right?

In general, I welcome any pointers that might reduce the number of wheels I'll reinvent. —Tamfang (talk) 07:00, 15 November 2015 (UTC)

Just trying to understand the problem here. You are trying to create a spline composed of multiple polynomial arcs, right ? The adjacent arc endpoints must have point continuity, of course, but how about tangent & curvature continuity, etc. ? Since you are using least squares method, I assume you don't need an exact fit. So, how many points would each arc run through ? (Just offhand, this method sounds like it would generate an extremely "lumpy" spline.) I assume you already know how the number of constraints relates to the degree of the polynomial ? StuRat (talk) 07:12, 15 November 2015 (UTC)

Sure, let's say I'm trying to create a spline composed of multiple polynomial arcs, and the degree of continuity is k_j-1. Maybe I like it runny lumpy; if it's lumpier than I like, I'll increase k_j. Rather than discrete points, my input is piecewise continuous, so the algo involves integrals rather than sums. Number of constraints, in the sense I think you mean, is not meaningful here. —Tamfang (talk) 08:51, 15 November 2015 (UTC)

If the function is on the domain $[x_{0},x_{m}]$, and the polynomials are of degree at most d, and for $1\leq j<m$ the derivatives at $x_{j}$ are expected to be continuous up to $k_{j}-1$ ($k_{j}$ constraints), then I'm pretty sure the number of degrees of freedom is $m(d+1)-\sum _{j=1}^{m-1}k_{j}$. -- Meni Rosenfeld (talk) 09:53, 15 November 2015 (UTC)

And I think the following basis functions will work (probably the same as what you wrote, but I think is clearer): Letting $k_{0}=0$, for each $0\leq j<m$ and $k_{j}\leq n\leq d$, the function which is 0 for $x\leq x_{j}$ and $(x-x_{j})^{n}$ for $x>x_{j}$. This also means their number can be rewritten as $\sum _{j=0}^{m-1}(d+1-k_{j})$. -- Meni Rosenfeld (talk) 10:06, 15 November 2015 (UTC)

Hm ... thanks, yes, I think that does work; the k₀=0 is a good gimmick (removing some special cases from the description). You've saved me some redundancy; I was thinking that for each discontinuity I'd need pairs of functions: zero on the left and (x-x_j)ⁿ on the right, (x-x_j)ⁿ on the left and zero on the right. —Tamfang (talk) 03:51, 24 November 2015 (UTC)

This is above my head (I don't know why I even look at this notice board!) but does Savitzky–Golay filter help? Thincat (talk) 09:00, 20 November 2015 (UTC)

That's interesting, but no. —Tamfang (talk) 03:03, 22 November 2015 (UTC)

November 19

Finite endomorphism ring

Is there an infinite abelian group with a finite endomorphism ring? GeoffreyT2000 (talk) 01:17, 19 November 2015 (UTC)

I don't think so. If an abelian group has a finite endomorphism ring, then it is necessarily a torsion abelian group. (Otherwise multiplication by an integer gives an obvious injection from the set of integers into the endomorphism ring.) In fact, for the same reason, the elements must have bounded order. By the first Prüfer theorem, a torsion abelian group of this kind is isomorphic to a direct sum of cyclic groups. Because the order of these cyclic groups is bounded, infinitely many direct factors are repeated by the pigeonhole principle, and so in that case the endomorphism ring is uncountably infinite. Sławomir
Biały 12:07, 19 November 2015 (UTC)

Probability distributions, frequency of events

Hi, I'd like to create an IID stochastic process $O_{n}\in \{0,1\}$, where 1 indicates that an event has occurred, and 0 that it has not. I had been using $O_{n}\thicksim B(p)$, where B(p) is the Bernoulli distribution with parameter p. This works ok, and I can vary the mean frequency of events, but the variance is p(1-p), and so the variance increases as I increase frequency - I am primarily interested in $p\in (0,0.5)$. What distribution should I use that would allow me to manipulate the frequency and variance independently? Or at least have a fixed variance for all frequencies? I need at the end to have a string of length N like 010...001, so counting events in an interval is not helpful. To be clear, I'd like to have the time between events (expected value of 1/p) have the same variance for all frequencies p. I can think of a few ways to generate strings with the necessary properties programmatically, but it would be far better if I could do it with a simple distribution. Any ideas? I feel like there must be something simple I'm forgetting about. Thanks, SemanticMantis (talk) 16:14, 19 November 2015 (UTC)

That's impossible. The only distribution with support $\{0,1\}$ is Bernoulli. To get what you described (which may or may not be what you want), you'll have to make your events dependent (contrary to the assumption of IID).

And the best way to do that is probably to start with the distribution you want for the time between successful events, with a given mean and variance (plenty of choice there - a good choice is the maximum entropy distribution), and simply running that and deriving the process (with 0's to fill in the gaps between successive 1's). But again, this will mean that there will be dependence between the events at given times (the dependence will be stronger for nearby events). -- Meni Rosenfeld (talk) 18:53, 19 November 2015 (UTC)

@Meni Rosenfeld: D'oh! Thanks, that makes sense, I forgot to leave out option "c) Is this impossible?" -- of course I see now that what I asked for is indeed impossible. To clarify, you're suggesting that I could instead create an IID process S_n for the spaces, then create O=1...1...1, where the number of zeros in the ... is S_n. Then O_n is not IID, but I could create S_n such that the mean and variance are independent. But can I do that with maximal entropy? E.g. I thought the geometric distribution had maximal entropy on {(0),1,2,...}, and that won't let me pick mean and variance independently. If I want to demand independent mean and variance, what are my options for support on \mathbb{N}? I think maybe I can re-parameterize the Beta-binomial distribution by mean and variance like you can do with the Beta distribution, but that's only quasi-independent, because once you pick the mean it bounds the allowable variances. I think I may well stick with my Bernoulli set up for now, but I'm interested in the time-dependent case as possible future refinement. SemanticMantis (talk) 19:54, 19 November 2015 (UTC)

I guess I could pick the mean \mu and then let S_n = DiscreteUniform(\mu-k, \mu+k). Variance could then at least be arbitrarily large or small. SemanticMantis (talk) 20:01, 19 November 2015 (UTC)

Yes, that is the process I proposed.

I meant, "maximum entropy for given mean and variance supported on positive integers" (geometric is max. ent. for given mean, without variance specification). This kind of distribution has the same form as the normal distribution, but the scale & shift parameters will not be exactly the mean and s.d. (since the restriction to positive integers changes the mean and variance for a given formula). You'd have to do a bit of work to find the correct parameters.

Alternatively, a combination of binomial distribution (for low variance) and negative binomial (for high variance) can work, and it's easier to find the parameters, but they don't cover the entire possibilities of mean & variance. -- Meni Rosenfeld (talk) 20:18, 19 November 2015 (UTC)

Note also that a uniform distribution will be fairly restrictive - you must keep it positive, so it limits how wide an interval you can take, and hence you can't have high variance. Negative binomial doesn't have this problem. -- Meni Rosenfeld (talk) 20:31, 19 November 2015 (UTC)

Thanks again Meni, very helpful. SemanticMantis (talk) 22:31, 19 November 2015 (UTC)

November 20

Calculus formula on back of jacket

I saw a guy on a train wearing a fairly plain black jacket but there was a mathematical formula embroided on the back.

${\delta V \over \delta t}+{1 \over 2}\sigma ^{2}S^{2}{\delta ^{2}V \over \delta S^{2}}+rS{\delta V \over \delta S}-rV=0$

Does anyone recognize what this is and why would someone have it on their jacket? I'm guessing it's some math "in joke". Vespine (talk) 03:36, 20 November 2015 (UTC)

It's the Black-Scholes equation. As for what the intended message of putting it on a jacket was, your guess is as good as mine.--Antendren (talk) 08:29, 20 November 2015 (UTC)

This was the equation that all the financial experts were using until it was realised recently that there was a flaw in the assumptions. Sorry I can't remember the subtle details. Dbfirs 23:10, 21 November 2015 (UTC)

That's as much of an answer as I was expecting, thank you!

Resolved

The standard Black-Scholes equation computes the theoretical price of a European call option, which is a simple type of financial derivative. The jacket formula you saw is the generalized form of the equation, which is a partial differential equation that estimates the theoretical price of any "derived" asset in terms of the change in its "strike" asset. It's not flawed per se, but it assumes the asset return is normally distributed, which is not always the case in practice. My guess is the person you saw is a quant. OldTimeNESter (talk) 21:31, 23 November 2015 (UTC)

November 21

Infinite Integral Sum

Hello! I'm a first year university student, and while practicing integration I made up a question $\int x^{2}e^{x^{2}}dx$ and tried to solve it using integration by parts. I got nowhere after a short while and, after looking the integral up, it turns out there is no elementary integral for the problem. However, integration by parts gives me this infinite series:

${\frac {1}{3}}x^{3}e^{x^{2}}-{\frac {2}{3}}\left({\frac {1}{5}}\right)x^{5}e^{x^{2}}+{\frac {2}{3}}\left({\frac {2}{5}}\right)\left({\frac {1}{7}}\right)x^{7}e^{x^{2}}-{\frac {2}{3}}\left({\frac {2}{5}}\right)\left({\frac {2}{7}}\right)\left({\frac {1}{9}}\right)x^{9}e^{x^{2}}+{\frac {2}{3}}\left({\frac {2}{5}}\right)\left({\frac {2}{7}}\right)\left({\frac {2}{9}}\right)\left({\frac {1}{11}}\right)x^{11}e^{x^{2}}\pm \cdots$

I was just wondering, is there a way to write this expression as an infinite sum? I can kind of see a pattern but I don't have enough experience with infinite sums to generate one (if it even exists!). Thanks for your help. 70.54.112.243 (talk) 04:15, 21 November 2015 (UTC)

You just wrote down an infinite sum, so it is not clear to me what you want. You may put the first term outside parentheses to get the expression

${\frac {1}{3}}x^{3}e^{x^{2}}\sum _{i=0}^{\infty }\prod _{k=1}^{i}{\frac {-2x^{2}}{2k+3}}$

Try also [[1]]

Bo Jacoby (talk) 05:59, 21 November 2015 (UTC).

This can also written without the product notation as $e^{x^{2}}\sum _{i=0}^{\infty }{\frac {(-1)^{i}4^{i+1}(i+2)!}{(2i+4)!}}x^{2i+3}$. -- Meni Rosenfeld (talk) 22:49, 21 November 2015 (UTC)

I'm sure you are familiar with Capital-sigma notation, but if you haven't seen Capital Pi notation before, you may wish to check out that article. (And while reading on the subject, you should familiarize yourself with Empty product.) Finaly, the erfi in the indefinite integral on the Wolfram Alpha page Bo Jacoby linked above is the imaginary error function. -- ToE 14:18, 21 November 2015 (UTC)

In J the function looks like this:

f=.13 : '3%~(y^3)*(^y^2)*+/*/\1,(-2*y^2)%3+2*1+i.10'

and a test run looks like this:

f&>0 0.5 1 1.5 2 2.5

0 0.0485128 0.627815 5.0843 46.7366 1398.4

Bo Jacoby (talk) 18:46, 22 November 2015 (UTC).

The Maclaurin series of this is straightforward to derive: $\int x^{2}e^{x^{2}}dx=\int x^{2}\sum _{i=0}^{\infty }{\frac {(x^{2})^{i}}{i!}}dx=\int \sum _{i=0}^{\infty }{\frac {x^{2i+2}}{i!}}dx=\sum _{i=0}^{\infty }{\frac {x^{2i+3}}{i!(2i+3)}}+C$. Using the doubling formulas for the gamma function, this can be easily shown to be equivalent to the series you derived. --Jasper Deng (talk) 23:33, 23 November 2015 (UTC)

November 22

A006355 - Number of binary vectors of length n containing no singletons - obvious proof?

I started looking at how many ways a season could go with every game being part of a 2+ game winning streak or a 2+ game losing streak. After calculating the examples for up to 6 games, I found OEIS: A006355, which has as one if its examples "a(6)=10 because we have: 000000, 000011, 000111, 001100, 001111, 110000, 110011, 111000, 111100, 111111. - Geoffrey Critzer, Jan 26 2014". So this is what I'm looking for. This sequence is a Fibonacci sequence starting with positive numbers: 0,2,2,4,6,10,16, etc. (So if you only count those where the team started out by winning, you get the standard Fibonacci series, offset)

So my question is, is there any obvious way to see in calculating that (for example) the number of 8 game seasons is equal to the sum of the number of 7 game seasons and the 6 game seasons?Naraht (talk) 08:49, 22 November 2015 (UTC)

Take all possible 7-digit sequences, repeat the last digit, and you get all possible 8-digit sequences ending in 000 or 111. Take all possible 6-digit sequences, append the opposite of the last digit twice, and you get all possible 8-digit sequences ending in 011 or 100. Combining these two cases, you get all possible 8-digit sequences. Egnau (talk) 13:17, 22 November 2015 (UTC)

A direct proof is this: Remove the initial and final digit from each sequence. Now for a given sequence, replace each 01 or 10 with a 2, and each digit between two of the same digit with a 1. For example 001110000110011111→0111000011001111→21211222111. This gives a 2 to 1 correspondence between the sequences of length n+2 and ways of writing n as a sum of 1's and 2's. If you count the latter you get the Fibonacci sequence. There are many similar correspondences between strings and sums satisfying certain criteria; some come from simple rewrite rules and some are more difficult. A surprising one is between strings of 0's and 1' that do not contain "101" (OEIS: A005251) and sums of 2's and 3's (OEIS: A000931) . --RDBury (talk) 15:23, 22 November 2015 (UTC)

@Egnau: That makes sense, thanx!Naraht (talk) 04:21, 24 November 2015 (UTC)

November 24

Path of steepest descent?

I'm wondering how, given a differentiable function of two spatial variables, how to compute the steepest paths of descent on its graph. All I know is that such a path must have the gradient as a tangent vector at every point, and thus should be itself a smooth curve. I also know that that means that the x and y components of the derivative of the function (of one parameter) describing the path should be parallel (as a vector) to the gradient. I'd much rather not have to use the calculus of variations.--Jasper Deng (talk) 00:16, 24 November 2015 (UTC)

I think what you're asking for are essentially orthogonal trajectories. --RDBury (talk) 00:51, 24 November 2015 (UTC)

Maths fields

Discrete mathematics is opposed to continuous mathematics, which is basically numerical analysis, which is opposed to symbolic manipulations. How is the relation between the discrete math, and symbolic manipulations? Is there something like an internal opposition in discrete math - symbolic discrete methods vs. numeric discrete methods? Or, is discrete math just symbolic manipulation (or a subset of this latter)?--3dcaddy (talk) 15:12, 24 November 2015 (UTC)

Er... why should continuous mathematics be identified with numerical analysis? Numerical analysis usually means approximating continuous problems by discrete ones. —Kusma (t·c) 15:23, 24 November 2015 (UTC)

I have changed the redirect to go to mathematical analysis instead. —Kusma (t·c) 15:30, 24 November 2015 (UTC)

Yes, that's right. Anyway, the first part of the question is still more confusing than explanatory. So, I am leaving the bottom part and striking it.--3dcaddy (talk) 16:13, 24 November 2015 (UTC)

I think you may be referring to something like the discrete element method which, despite the name, would fall under numerical analysis and therefore continuous mathematics and not discrete mathematics. Typically, the term discrete mathematics refers to finite (countable sometimes) structures while analysis or continuous mathematics refers to structures involving real numbers in some way. But there are areas that don't fall into either camp and our Areas of mathematics divisions cut across this line. Mathematics Subject Classification has main divisions for discrete math/algebra and analysis, but it also has divisions for foundations, geometry, and applied mathematics. But Lie groups, classified in with other areas of group theory under discrete math/algebra, might more properly be put under analysis. And while most of geometry would fall under continuous mathematics, there is a whole area of discrete geometry that wouldn't. The boundaries between different areas of mathematics are usually fuzzy and often based more on tradition than anything else, so it's difficult to give a precise meaning to the distinction. --RDBury (talk) 18:07, 24 November 2015 (UTC)

Why isn't complex numbers taught as vectors equivalent?

Teaching complex number to school kids should not be hard. Complex number addition is just equivalent to vector addition. Complex number multiplication is just treating one complex number as a vector and using the other complex number as a procedure to perform scaling and follow by an angle rotation. This can be done in the argand diagram. When intepreted in the geometrical format there is nothing special about I (imaginary number) which is nothing but a rotation of 90 degrees counter-clockwise in the argand diagram.

It is not just teachers in high school. Look at the wikipedia article on complex numbers, it does not mention that addition of two complex numbers has similar properties to vector addition nor does it mention that multiple of two complex number to be consider a form of scaling and rotation of a complex number (about the origin) in the complex plane. The smart student may figure this out all by themselves but there is no reason not to mention this to all readers of the wikipedia article.

This is nothing but a global elitist conspiracy to prevent the understanding of the true nature of complex numbers from the dumb masses. Instead of allowing them to understand complex numbers, the students are forced to memorize complex procedures (pun not intended) and endure deliberate obfuscation. Where as what the students need to know is what the complex numbers are and then the procedures for dealing with them which can be programmed into a smart phone similar to calculators which can perform addition , subtraction , multiplication and division. I can't even remember when I last did complex multiplication and complex division by hand. My point is that the conspiracy to prevent the understanding of complex numbers is prevents literally billions of people from using complex numbers. 175.45.116.59 (talk) 23:02, 24 November 2015 (UTC)

Note: The redirect Argand diagram -> Complex plane is tagged {{R with possibilities}}, and while "Argand diagrams" is bolded in the lede, they are not defined there and are not mentioned in the article's body. -- ToE 23:28, 24 November 2015 (UTC)