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Kinda weird Diff Eq problem

  1. Jul 20, 2011 #1
    The problem statement, all variables and given/known data

    Now that I'm working it out as I'm typing it, I think I may have solved this problem; could still be a mistake though...

    This was an extra credit problem on the final for my into to diff eq class. I never saw anything like it before and I didn't finish it because I got stuck on one part. It's copied pretty much exactly as it was on the paper.

    Define [x] by
    [2] = 5
    [3] = 8
    [5] = 14
    [-2] = -7

    [x]1 = [x]
    [x]2 = [[x]]
    [x]3 = [[[x]]]

    Solve dy/dx = [10]x

    The attempt at a solution

    I see that [x] is basically the function f(x) = 3x - 1. Then I can find a formula for [x]1, [x]2, ... by plugging f(x) into itself and finding a pattern.

    [x]1 = 3x - 1
    [x]2 = 9x - 4
    [x]3 = 27x - 13
    [x]4 = 81x - 40

    Then I got [x]n = 3nx - (3n - 1)/2
    Finding a closed form for the last term of the nth term of [x]n had me stuck and that was where I originally stopped on the problem. I used Wolframalpha to find a closed form, but is there a clever way of finding it just by looking at the numbers, without knowing the formula beforehand? I can't see it...

    Now if g(x) = [10]x, then g(x) = 3x(10) - (3x - 1)/2 and I can integrate that to solve dy/dx = [10]x.


    Something that kind of bothered me is that [10]x looks like it can only be defined for natural numbers, so does dy/dx = [10]x even make sense in the first place?
  2. jcsd
  3. Jul 20, 2011 #2


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    Let's look at this a bit further:

    First you were given [n] evaluated for a few integer values. Then you extended the pattern to all integers with [n] = 3n-1. Then you extended that to the reals with [x] =3x-1.

    You were given a definition for [x]n, presumably for n a natural number.

    You found a general form for the n-th term: [x]n = 3nx - (3n - 1)/2
    BTW: This can be written, [x]n = (3n(2x - 1) + 1)/2​

    Then you put some in constant, namely 10, for x and obtain [10]n = 3n(10) - (3n - 1)/2 = (3n(2(10) - 1) + 1)/2 = ((19)3n + 1)/2.

    So, does it make sense to extend this to real numbers? You can ask if it makes sense to extend it merely to rationals or for negative integers --- or for that matter, what does [10]0 mean?

    Well, you can plug 0 into your formula for [10]n to get [10]0 = 10.

    [10]-1 = 11/3 ...

    Taking [10]n, as being defined for the naturals and extending it to [10]x, defined on the reals, reminds me of developing exponential functions.

    In fact, since 3 = eln3, you can write [10]x as ((19)e(ln3)x + 1)/2

    So, yes, it seems to make sense to think of [10]x as a function on the reals.
  4. Jul 22, 2011 #3
    I thought it could be generalized to the reals, but all that was given was whole numbers as superscripts on [x]: [x]1 = [x], [x]2 = [[x]], [x]3 = [[[x]]], etc. so I thought maybe I should stick to whole numbers. But it's clear it has to work for all real numbers in order for it to be differentiable and the given problem to be solved, so I think I'm good with this problem now :smile:
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