Wikipedia:Reference desk/Archives/Mathematics/2012 April 25

= April 25 =

Interpolation in 2D
So let us say that I have 9 points on a 2D mesh, making four units squares with x-y coordinates like this

(0,2),(1,2),(2,2)

(0,1),(1,1),(2,1)

(0,0),(1,0),(2,0)

and I also know the value of a function f(x,y) at these nine points as well. Now I need to do some interpolation (because I want the interpolated function to go through those nine known values). I need to know values at a bunch of points within those four squares. I am looking for something analogous to quadratic splines in 1D. At these nine points I only know the function values. I don't know any of the derivatives/partials. So the interpolating function could be a piecewise function made up of

f3(x,y), f4(x,y)

f1(x,y), f2(x,y)

for each of the four squares and at the boundaries the function values and the gradient matches so that the x,y partials at least are continuous. Another problem is that I will be interpolating at a LOT of points and I won't know any of them in advance so I need to compute the interpolating function before hand (preprocessing) and then just evaluate it at the needed points as they come up. So my question is how does this work in dimensions more than one? Should I assume f1,f2,f3,f4 to be a general quadratic in two dimensions like $$f1(x,y)=ax^2+bx^2+cxy+dx+ey+g$$

because I only need the first partials continuous? But then what should be the constraints? I am a little confused about the partials? How do I write the constraints so that the partials are equal? Would I just have a gigantic matrix at the end, solve the system, and have the interpolating coefficients? Something like Bilinear interpolation is too slow because it needs to be done everytime and my program spends a HUGE amount of time in it. I want to try doing all the work before hand and then just evaluate a polynomial when I need to. If it helps, I will eventually be doing this on a 4D mesh f(x,y,z,t) with maybe a 100 points in each direction so 100000000 points total. I can guess the matrix will grow horrendously large and just to solve THAT system will take some ingenuity. On that thought, is there ny way interpolation can be done independently on a square (so that the resulting linear systems stay small) but still have the gradient continuous for example? I know constant and linear interpolation do this but I just need the continuity of the gradient as well. I also saw the article on Bicubic interpolation but it seems a but more than what I need. So to keep the system as small as possible, I am trying to see something like a biquadratic interpolation. Any help/comments/suggestions will be very helpful. Thank you! 184.96.137.243 (talk) 05:14, 25 April 2012 (UTC)
 * A polynomial function of the form $$f_1(x,y)=Dxy+Cx+By+A$$ has 4 adjustable coefficients. The 4 equations f1(0,0)=A, f1(1,0)=C+A, f1(0,1)=B+A, f1(1,1)=D+C+B+A are easily solved. Filling each square with such a function provides a continuous - but not differentiable - interpolation. This is not what you ask for. The Discrete Fourier transform provides an interpolating trigonometric polynomial. It is differentiable everywhere. And the computation is feasible due the to FFT algorithm. This may be what you are looking for. Bo Jacoby (talk) 07:43, 25 April 2012 (UTC).
 * The common way of doing this is by using some sort of Spline. Spline interpolation may be the most useful reference. There is a vast literature on the subject and some efficient algorithms for solving your problem.--Salix (talk): 08:23, 25 April 2012 (UTC)


 * The simple trick if you want a polynomial is to start with the expression x(x-1)(x-2)y(y-1)(y-2). If you remove any x term and any y term, the resulting quadratic is zero at every point but one -- you can fit it to that point by multiplying by a coefficient.  That means you can create a quadratic polynomial that passes through all your points by adding up the nine quadratic terms you fit that way. Looie496 (talk) 20:18, 25 April 2012 (UTC)

Prime <--> Fibonacci
Is there any intelligable, meaningful result from the concurrence of any prime number/s and the matching number/s in the Fibonacci series? Benyoch...Don't panic! Don't panic!... (talk) 06:55, 25 April 2012 (UTC)
 * Take a look at our Fibonacci prime article. Gandalf61 (talk) 07:53, 25 April 2012 (UTC)


 * Thanks Gandalf. Was hoping for a graphic, but was good to see what you showed me. Benyoch...Don't panic! Don't panic!... (talk) 11:19, 28 April 2012 (UTC)

Best way to represent mathematically?
What is the best way to represent the concept of a multiple of 7 and any number that is within 2 units of it? (Or, in the general case, a number that is a multiple of n and any number that is within m units.) Thus, the set I am referring to would be for non-negative integers is: {0,1,2, 5,6,7,8,9, 12,13,14,15,16, ...}. This seems like it would be a useful/interesting thing to be able to reason about and must have been studied before, and I am hoping that someone can point me to the right algebraic (or other) concept. Nkot (talk) 14:36, 25 April 2012 (UTC)


 * There are several ways (all of which express the same idea), e.g.


 * Any number belonging to the following set: {7n-2, 7n-1, 7n, 7n+1, 7n+2 | n is an integer}.
 * n mod 7 is in {0,1,2,5,6}.


 * 77.125.249.41 (talk) 15:10, 25 April 2012 (UTC)


 * How about $$n^2 \ne 2 \mod 7$$ Gandalf61 (talk) 15:28, 25 April 2012 (UTC)


 * For the specific case, we have:

7a-2 ≤ x ≤ 7a+2


 * For the general case, we have:

na-m ≤ x ≤ na+m


 * (Note that some of the previous answers used "n" not as you had defined it, but as I used "a" in my examples.) StuRat (talk) 15:59, 25 April 2012 (UTC)


 * You could also say that x is an integer satisfying |7a - x| ≤ 2 for some integer a. Rckrone (talk) 17:57, 25 April 2012 (UTC)

Set Theory
Hi y'all.

Can you find a proposition (written in ZF language), having the following three properties:
 * 1) If ZF is consistent, then also - the combination of ZF with the proposition - is consistent.
 * 2) The proposition cannot be derived from ZF.
 * 3) The proposition is of the form: "There is a unique set such that...".

Thanx. 77.125.249.41 (talk) 14:40, 25 April 2012 (UTC)
 * Sure. "There is a unique set n such that n is a natural number, and n is the least Goedel number of a proof of 0=1 from ZF." --Trovatore (talk) 16:36, 25 April 2012 (UTC)
 * Remember that: one of my conditions is that the proposition can be expressed in ZF language. Are you sure your propostion satisfies that condition? (Btw, I'm from the restrahnt, rather than from the restrnt) 77.125.249.41 (talk) 17:26, 25 April 2012 (UTC)
 * Yes, I'm sure. --Trovatore (talk) 18:24, 25 April 2012 (UTC)
 * Btw, why did you choose 0=1, rather than the simpler one: 0=0? 77.125.249.41 (talk) 18:52, 25 April 2012 (UTC)
 * If you substitute 0=0, it's no longer independent of ZF. --Trovatore (talk) 18:55, 25 April 2012 (UTC)
 * Back to my original question: So, your suggested proposition is pretty long, isn't it? Could you find a proposition whose length is more "normal", i.e. not longer than a given "normal" limit? Let's start with: 1000 signs - as a "normal" limit (if you suspect a limit of 1000 signs may not be sufficient, then I'll agree to replace it by a larger one, but not by an extraordinarily large one, like the length of propositions dealing with Goedel numbers). 77.125.249.41 (talk) 20:23, 25 April 2012 (UTC)
 * I don't have a particularly good intuition for symbol counts in the minimalist language of set theory. Maybe what you really want is a more "natural", less "meta", question having that form?  How about "there exists an inaccessible cardinal", which can be rephrased into your form by simply writing it as "there exists a least inaccessible cardinal"? --Trovatore (talk) 21:31, 25 April 2012 (UTC)
 * Are you sure the class of inaccessible cardinals can have a bottom? Maybe it can't, so that your proposition is not consistent with ZF, i.e. does not satisfy the first of my three conditions? 77.125.249.41 (talk) 23:30, 25 April 2012 (UTC)
 * Not only can, has to (assuming there are inaccessibles). An inaccessible cardinal (in the usual setup) is literally an ordinal, and every nonempty class of ordinals has a least element. --Trovatore (talk) 00:04, 26 April 2012 (UTC)
 * Thankxs. You've helped me!
 * When you visit my restrnt (doh! restrahnt, of course), I'll give you a pizza - as a gift (unless you prefer ice cream).
 * 77.125.249.41 (talk) 11:30, 26 April 2012 (UTC)