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Thread: How to find the hidden formula behind a set of numbers

  1. #1
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    How to find the hidden formula behind a set of numbers

    Question: I'm looking for help in finding the original equation behind a set of given answers:

    when x = 100, the answer is 0.5 (1/2)
    when x = 200, the answer is 0.6666 (2/3rds)
    when x = 300, the answer is 0.75 (3/4ths)
    when x = 400, the answer is 0.8 (4/5ths)

    There is a pattern, and I tried approaching this problem as a Sequence, but could not get the differences to converge.

    I cheated, and found the hidden formula (shown below) which governs the above sequence, but I still want to know how to reverse-engineer this problem so I could have found the "Hidden Formula" without looking up the answer.

    Hidden Formula: x (x + 100)
    so when x = 100, the formula would be: 100 (100 + 100) = 100 200 = 0.5

    Thank you for any help you can give.
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  2. #2
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    Re: How to find the hidden formula behind a set of numbers

    Pick a big n > 1. For the first n elements of the sequence, write down a formula for drawing a polynomial through them. See if that formula matches the remaining elements a_{n+1}, a_{n+2}, .... Experiment with n. It would be easiest to do this programmatically.

    If the polynomials are no good, try regressing the sequence on functions of the form

    (x - mu)^k, k = ..., -2, -1, 0, 1, 2, ...

    k^x, log_k(x), k = 1, 2, ...

    etc.
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  3. #3
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    Re: How to find the hidden formula behind a set of numbers

    There will exist an infinite number of solutions because there exist an infinite number of graphs through any finite set of points. In particular, given n "data points" there always exists a unique polynomial of degree n- 1 that passes through those points. Here, you are given 4 points so there exist a unique cubic satisfying this.

    Any cubic can be written ax^3+ bx^2+ cx+ d. you must
    a(100)^3+ b(100)^2+ c(100)+ d= 1000000a+ 10000b+ 100c+ d= 1/2
    a(200)^3+ b(200)^2+ c(200)+ d= 8000000a+ 40000b+ 200c+ d= 2/3
    a(300)^3+ b(300)^2+ c(300)+ d=27000000a+ 90000b+ 300c+ d= 3/4
    a(400)^3+ b(400)^2+ d(400)+ d= 64000000a+ 160000b+ 400c+ d= 4/5

    four equations to solve for a, b, c, and d.
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