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Thread: Need to find the equation

  1. #1
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    Need to find the equation

    Hello. I'm not sure if this question goes to algebra, calculus, or other place, and if it is, please let me know, I am willing to fix.

    Here's my question. I have the following sets of numbers, and I need to find the formula to the question mark, so the formula is the same exact for all sets.

    For example if ? = /1.56/100

    Then that would wok for first set, but not the second, because 18/1.56/100 would equal .115 instead of .4

    25 ? =0.16
    18 ? =0.40
    15 ? =0.5
    10 ? =1
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  2. #2
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    Re: Need to find the equation

    25 * .0064 = .16

    Is that what you mean?
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  3. #3
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    Re: Need to find the equation

    But he said "so the formula is the same exact for all sets." 25 * .0064= 0.16 (rounded) but 18*.0064= 0.11 (rounded), not .40.

    It simply is not possible to "do the same thing", in this simple sense, to 25, 18, 15, and 10, and get .16, .40, 0.5, and 1, respectively.

    You could treat this as a function and seek a function, f(x), such that f(25)= 0.15, f(18)= 0.40, f(15)= 0.5. and f(10)= 1. There are an infinite number of functions that will give those values but, since there are 4 points there exist a unique third degree polynomial function that satisfies those.

    One way to find that function would be to write it as f(x)= ax^3+ bx^2+ cx+ d so that you have the four linear equations, a(25^3)+ b(25^2)+ c(25)+ d= 0.16, a(18^3)+ b(18^2)+ c(18)+ d= 0.40, a(15^3)+ b(15^2)+ c(15)+ d= 0.5, and a(10^3)+ b(10^2)+ c(10)+ d= 1 and solve for a, b, c, and d.

    Or you can use the "Lagrange interpolating formula":
    f(x)= 0.16(x- 18)(x- 15)(x- 10)/[(25- 18)(25- 15)(25- 10)]+ 0.40(x- 25)(x- 15)(x- 10)/[(18- 25)(18- 15)(18- 10)]+ 0.5(x- 25)(x- 18)(x- 10)/[(15- 25)(15- 18)(15- 10)]+ (x- 25)(x- 18)(x- 15)/[(10- 25)(10- 18)(10- 15)].

    That gives a cubic polynomial such that f(25)= 0.16, f(18)= 0.40, f(15)= 0.5, and f(10)= 1
    Last edited by HallsofIvy; Feb 14th 2018 at 04:06 AM.
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