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Function Describing Caloric Target

  1. Feb 5, 2013 #1
    [Heavily edited for clarity.]

    So, I'm playing around with a function that describes the change of an individuals weight as a function of days passed given a number of constant variables: current weight, target weight, number of days until target weight will be reached, and daily calorie target.

    My function is derived from an equation that describes weight after a number of days given a constant caloric deficit, d:

    f(0) = a;
    f(x+1) = f(x) - (d/3500)
    f(n) = b;

    With this set of equations, we can solve for d (daily caloric deficit) given a (starting weight), b (ending weight), and n (number of days until ending weight is reached).

    That's fine, but I wanted to alter this set of equations to allow for variable values of d. So, I removed d as a constant variable and allowed it to equal g(x)-t where g(x) is a calculated caloric baseline. Now I want to solve for a constant value t (a constant daily caloric target) that would result in a decrease to weight b after n days given a starting weight a.

    f(0) = a; // where a is current weight
    g(x) = q + p*f(x); // this describes a linear relationship between caloric baseline and weight
    f(x+1) = (g(x)-t)/3500; // t is the constant caloric target, so g(x)-t is caloric deficit.
    f(n) = b; // n = number of days until target weight will be reached. b = target weight.

    Of course, f(x+1) and g(x) could be combined; I kept the separate to illustrate the model.

    Now, my question is... how would I solve for t if a, n, b, q, and p are all known? I intuitively known there is a way to solve this, but I can't figure it out, and it's bugging me. If anyone could help me figure this out, I'd appreciate it.

    Edit: by the way, if this isn't calculus... my bad.
     
    Last edited: Feb 5, 2013
  2. jcsd
  3. Feb 5, 2013 #2

    HallsofIvy

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    Not exactly "Calculus" but close to it. This is, technically, a "finite difference equation" but a simple one. If f(0)= a then f(1)= f(0+ 1)= a- d/3500, f(2)= f(1+ 1)= (a- d/3500)- d/3500= a- 2d/3500, f(3)= f(2+ 1)= (a- 2d/3500)- d/3500= a- 3d/3500, and, in general f(x)= a- 3x/3500.

    So g(x)= q+ pa- 3px/3500. But I don't understand what you mean by "line that approximates a given individuals caloric baseline as a function of weight".

    You have already said that f(x+1)= f(x)- d/3500 and that g(x)= q+ pf(x). So f(x+1)= (g(x)- t)/3500 is f(x)- d/3500= (q+ pf(x)- t)/3500= (q- t)/3500+ pf(x)/3500. Then f(x)- pf(x)/3500= (1- p/3500)f(x)= (d+ q- t)/3500 and f(x)= [(d+ q- t)/3500][3500/(3500- p)]
    f(x)= (d+ q- t)/(3500- p) which can't be correct because there is no "x" on the right side. Given that d, q, t, and p are constant, the right side stays the same no matter how x changes.
     
  4. Feb 5, 2013 #3
    Oh, I see. No, I guess I was confusing. The first function is actually describing a different model than the latter four, and for the sake of "solving the problem" doesn't exist. (It's describing a model with constant caloric deficit and a steadily decreasing daily target caloric intake, whereas the other four are describing steadily decreasing caloric deficit and constant daily caloric intake. I just provided the former equation as background info, to show where I was coming from before I found this problem.

    I'm just trying to find the equation that solves for the unknown value of t with respect to the other (known) constant values (d, q, t, p, and n) given only the latter four equations.

    If that makes sense. :\

    (I heavily edited my original post to clarify this.)

    EDIT: I just meant there is an approximately linear relationship between weight ("f(x)") and baseline caloric expenditure ("g(x)") when other factors (age, height, gender, etc...) are held constant. Incidentally, "g(x) = q + p*f(x)" is that line.
     
    Last edited: Feb 5, 2013
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