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Do you know about operational equations?

  1. Oct 8, 2016 #1
    Many years ago, I encountered a problem involving four numerical data in a square array or a rectangular array. The standard method for interpolating that design is the bilinear equation.

    For example, let the array be ACIG as below left. If A=1, C=3, G=7, I=9, then the bilinear equation yields z = (1+2x+6y) in the (x, y) = (0 .. 1, 0 ..1) coordinate system. However, if the data are the squares of
    the cited data, the bilinear equation yields z = 35 at the center of the design. That is not a
    G I good estimate of the true value z = 25. I sought help at universities but was rebuffed by
    A C the remark that no other equation for the four-point rectangle is possible. A few years
    later I happened on the relation (e^x)F(x) = F(x+h) where is 'e' is now an operator and F(x) is any function of (x).

    This ultimately led to an interpolating equation that is exact on bilinear data and on the squares of bilinear data. That led to a nasty verbal interchange. It also resulted in two papers: "No Free Lunch: Comments on Silver" where the new equation was declared impossible. Read it and see: Quality Engineering 5(3) 369-373 (1993) by Norman R Draper and Dennis K J Lin. It also led to my rebuttal: Free Lunch, Bigger Menu, Better Food" by G. L. Silver in Quality Engineering 6(2) 307-310 (1994). Read it and see.

    Draper and Lin were never heard from again. After I arrived at Los Alamos National Laboratory (New Mexico) I added to the altercation: "Operational equations for data in rectangular array" by G. L. Silver, Computational Statistics and Data Analysis 28 (1998) 211-215. That citation gives the complete equation in the -1 .. 1 coordinate system. It is exact on (positive) bilinear numbers and on the squares of such numbers.

    For example, if A=1, C=3, G=7, I=9 then the equation is z = (5+x+3y) but if A=1, C=9, G=49, I=81 then z = (5+x+3y)^2. There are lots of operational equations. Most are "substitute and see" equations. They are probably pertinent to physics. You can join in the fun, too!
     
    Last edited by a moderator: Oct 13, 2016
  2. jcsd
  3. Oct 13, 2016 #2

    Stephen Tashi

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    If any of the papers you cite are available online, you should give links. Otherwise, it would be helpful if you stated the problem.

    You mention "design" but it isn't clear whether you are talking about data involving "the design of experiments" or simply some arbitrary data without any specific interpretation.

    You haven't made it clear why z = 25 is the "true value". What criteria is used to determine the true value at the center of the rectangle?
     
  4. Oct 13, 2016 #3
    Reply to Stephen Tashi about Operational Equations from G. L. Silver. The four-point rectangle ACIG has A=1 at lower left corner, C=3 at lower right corner, G=7 at upper left corner, I=9 at upper right corner. These are bilinear numbers. Let (x) be the x-coordinate and (y) be the y-coordinate in the -1..1 coordinate system for both (x) and (y). Let (z) be the altitude over the x,y plane. The bilinear equation for this design is z = (A+C+G+I)/4 + (I+C-A-G)(x/4) + (I+G-A-C)(y/4) + (I+A-C-G)(xy/4). (The bilinear equation is not new. It is Eq. (8) in the literature citation below.) The bilinear equation for this four-point design is z = (5+x+3y). At the center point of the rectangle we have (x,y) = (0,0) so z = 5. Now suppose the data are squared so that A=1, C=9, G=49, I=81. That is, we apply the squaring function to the four-point design. At the center point of the rectangle, the bilinear equation yields z = (35+10x+30y+6xy) so that at (x,y)=(0,0) and we have z=35. That is a poor approximation to the true value z=25 at (0,0). This deficiency attracted my attention with respect to work in another field. I wanted something more accurate. By chance, one day I saw the relationship exp(x)F(x) = F(x+h). That was years later but it reminded me of the rectangle problem so I used exp(x)F(x)=F(x+h) to find a new interpolating equation for the four-point rectangular design. The explicit equation is Eq. (7) in Computational Statistics and Data Analysis 28(1998)211-215. If you put A=1, C=9, G=49, I=81 into the cited Eq. (7) the result is an interpolating equation that factors to z = (5+x+3y)^2. (That is the square of the bilinear equation.) In summary: the bilinear equation is exact on four bilinear numbers in a rectangular array. The operational equation is exact on four POSITIVE bilinear numbers in a rectangular array AND on the squares of those numbers. That was supposed to be impossible! Make no mistake: emotions ran very high! In summary, the bilinear equation is to four data in a rectangular array what the straight line is to two data. The operational equation is to four data in a rectangular what the parabola is to three equidistant curvilinear data. That analogy is good enough. The derivation of the operational equation is suggested in the 1985 reference found on page 215 of the cited Comp. Stat. Data Analysis paper. (Do not bother with that citation. It isn't pertinent here.) In summary, there are three kinds of calculus: the differential calculus, the integral calculus, and the operational calculus. I applied the last one to geometry. (That is where the term "operational equation" comes from.) That is the story. I hope it helps.
     
  5. Oct 13, 2016 #4

    Stephen Tashi

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    What is a "bilinear number" ?


    I have no idea why we would expect to get the same answer from a formula if we replace the original inputs with their squares.

    I don't have access to the journals you mention, so unless you can describe the problem coherently, I must leave this thread to other forum numbers who have the appropriate subscriptions.
     
  6. Oct 13, 2016 #5
    2nd reply to Stephen Tashi. You write: "What is a "bilinear number"? So far as I am aware, there is no such thing as "a bilinear number." The phrase "bilinear numbers" refers to a set of numbers (in our case, a set of four numbers in a rectangular array) that change by a constant amount in the x-direction and by a constant amount in the y-direction. Consider our former case of the rectangle ACIG. Here, A=1 and C=3. The numbers 1 and 3, on the bottom of the rectangle, increase by a change of +2 in the positive x-direction. At the top of the rectangle the numbers G and I (7 and 9, respectively) also change by +2 also in the positive x-direction. On the left-hand side of the rectangle the number A=1 (lower left-hand corner) increases by +6 to render 7 at G (upper left-hand corner) in the positive y-direction. The number at C=3 (lower right hand corner) increases to I=9 in the upper right-hand corner, also in the positive y-direction. Again, the change is +6. The change in the x-direction is always +2 (left to right) and the change in the y-direction (bottom to top) is always +6. Thus, we have a set of four bilinear numbers. These numbers are linear in two dimensions: up from bottom to top and sideways from left to right. The set is "bilinear." (Incidentally, I did not originate this lingo.) You also said "I have no idea why we would expect to get the same answer from a formula if we replace the original inputs with their squares." I did not say you would get the same answer with different numbers. What I tried to say was this: "With a set of bilinear numbers in a rectangular array we get one result with the bilinear equation as the interpolating equation. If we square those numbers, then the operational equation renders the square (second power) of the original bilinear equation." That was thought to be impossible based only on four bilinear numbers in a rectangular array and nothing else. On another subject: I was asked to supply a literature citation so I am surprised by your complaint. I can do nothing about your inability to access the literature of mathematics. On that problem, I am mostly helpless. (Maybe you could ask a librarian, or go to a nearby university, or ask a Forum colleague for assistance.)

    As a courtesy, I will try to type the operational equation (for the four-point rectangle) into this letter. It is a tedious job, so if it fails the problem is likely a typographical mistake.

    z = ((C+G)(A-C+G-I)(A+C-G-I) - (C-G)^2(A-C-G+I)) / (2(A-C+G-I)(A+C-G-I)) + (I+C-A-G)(x/4) + (G+I-A-C)(y/4) + (A+I-C-G)(xy/4)
    + ((I+A-C-G)(I+C-A-G)x^2) / (8(G+I-A-C)) + ((I+A-C-G)(G+I-A-C)y^2) / (8(I+C-A-G))

    Remember, the coordinate system is -1 .. +1 in BOTH the x- and y-directions. If you still have trouble, ask a colleague (or someone on the PhysicsForum) to help you. Surely the management at PhysicsForum will have someone to assist the members. Do those things FIRST. This is a matter of public-school algebra so there should be a nearby colleague or member of the Forum to help with such problems. --- Schimmel

     
  7. Oct 14, 2016 #6

    DrClaude

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    Please use the reply feature.


    This is not how things work here. First, you are the one who brought up the topic for discussion. Second, I assure you that @Stephen Tashi has the mathematical knowledge to understand this work.

    Thread closed.
     
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