- #1
Bohrok
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Homework Statement
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
...
and
[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.
Correct?
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?
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
...
and
[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.
Correct?
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?