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LordKinbote
Daedalian Member
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 Posted: Tue Oct 02, 2012 11:47 pm Post subject: 1 |
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This is a variation I wrote on a problem I've seen in math competitions.
What is the 100th positive integer a such that sqrt(a + sqrt(a + sqrt(a + sqrt(a + ...)))) is an integer?
(I guess I could make an image of that, but I think you guys get the general idea: infinite nested square roots.) |
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groza528
No Place Like Home
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| Posted: Wed Oct 03, 2012 4:43 am Post subject: 2 |
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| With some quick math... 10100? |
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Chuck
Daedalian Member
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| Posted: Wed Oct 03, 2012 4:59 am Post subject: 3 |
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| That's what I got with a computer search. |
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Thok
Oh, foe, the cursed teeth!
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| Posted: Wed Oct 03, 2012 6:16 am Post subject: 4 |
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| In general, the nth number is n^2+n. The proof is not so hard (well unless you include the step where you prove the infinite square root makes sense and converges at all. |
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DejMar
(Possibly a robot)
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| Posted: Wed Oct 03, 2012 10:36 am Post subject: 5 |
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The nth positive integer would be where a = (n^2 + n) with the expression sqrt(a + sqrt(a + sqrt(a + ...))) converging to the positive integer (n + 1).
The 100th positive integer would be where a = (100^2 + 100) = 10100, with the expression converging to (100 + 1) = 101. |
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L'lanmal
Daedalian Member
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| Posted: Wed Oct 03, 2012 9:39 pm Post subject: 6 |
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| Thok wrote: |
| In general, the nth number is n^2+n. The proof is not so hard (well unless you include the step where you prove the infinite square root makes sense and converges at all. |
Um, those would be the things which make it a proof. Although this could be one of those "application to a mathematician is theory to an engineer" things. Try the word "solution". You might have a solution like: Find the points fixed by the iteration by noting that they occur at x = sqrt(a+x), note they are attractors (stable fixed points and not unstable ones) when viewing this as a 1-dimensional difference equation, Solve for a and note that a is an increasing function of x, and plug in 100 for x.
But hold on, when x=1, then a=0 isn't exactly a positive integer, is it? And everyone else is getting x^2+x rather than x^2-x. What's wrong? |
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lostdummy
Daedalian Member
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| Posted: Thu Oct 04, 2012 8:20 am Post subject: 7 |
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| L'lanmal wrote: |
But hold on, when x=1, then a=0 isn't exactly a positive integer, is it? And everyone else is getting x^2+x rather than x^2-x. What's wrong? |
You stated reason in first part of that sentence ;p
Solution indeed is a=x(x-1), but when you want to index solution so that when you use n=1 you get first acceptable solution to problem (a>0), or when you use n=100 you get required solution, then you need to skip a=0, and thus you need to start at x=2.
So if we use substitution x=n+1, we get result where for n=1, a>0:
a=n(n+1)
If for example problem was 'find 100th number where a>10' then we would need to shift by 3, ie x=n+3, and thus a=(n+2)(n+3) |
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