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Topic: A small hint re: 2008 A4 (Read 8283 times) |
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Mickey1
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A small hint re: 2008 A4
« on: Mar 24th, 2011, 7:13am » |
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In the 2008 exam, problem A4
f(x) =
x if x =< e and
xf(ln x) if x > e.
How is the second equality, f(x)= xf(ln (x)) solved?
In the published solution on the archive it seems to be so self-evident that it doesn't require any explanation.
Grateful for a hint
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| « Last Edit: Mar 24th, 2011, 7:23am by Mickey1 » |
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towr
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Re: A small hint 2008 A4
« Reply #1 on: Mar 24th, 2011, 9:40am » |
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It's a definition, so it's not solved, but evaluated. You just reapply the definition until you have a value.
f(100) = 100 f( ln(100) )
.. f( ln(100) ) = ln(100) f( ln( ln(100) ) )
.. .. ln( ln(100) ) < e, so f( ln(ln(100)) ) = ln( ln(100) )
.. f( ln(100) ) = ln(100) ln( ln(100) )
f(100) = 100 ln(100) ln( ln(100) )
~= 703.292208
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| « Last Edit: Mar 24th, 2011, 9:42am by towr » |
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Mickey1
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Re: A small hint 2008 A4
« Reply #2 on: Mar 25th, 2011, 2:53pm » |
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Thank you.
Something similar actually came to me during the night, and I was thinking about
f (exp(x)) = exp(x)*f (ln(exp(x)))= exp(x)*x*f(x) =
exp(x)*x*f(ln(x)) = exp(x)*x * f(product(ln-k times- (x)) , k=1 to N), until ln-N times- (x) < e, and that is when you have the solution. I realize the exp(x) approach is not necessary but it helps to get you started. I meant to visit the riddle site about that but I had to sit all day in an emergency response bunker looking at dose rates from the Fukushima reactors, so I can't prove it.
Anyway, now the solution also makes sense.
PS These Putnam exams seem to be more difficult than the IMO problems. They seem to require more of the participants.
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ThudnBlunder
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Re: A small hint 2008 A4
« Reply #3 on: Mar 25th, 2011, 7:41pm » |
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on Mar 24th, 2011, 9:40am, towr wrote:It's a definition, so it's not solved, but evaluated. You just reapply the definition until you have a value.
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Then f(x) = f(x+1)/x is also a 'definition'; but f(x) can be expressed in terms of x, namely  (x).
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towr
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Re: A small hint 2008 A4
« Reply #4 on: Mar 26th, 2011, 1:39am » |
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True enough; you can often express the function differently, sometimes simpler. And admittedly it's not always as simple as recursively applying the definition to evaluate the function at a given point (because that depends on the kind of definition). But the point is it's not an equation to be solved, you're not looking for the conditions under which both sides are equal. Looking at the problem from the wrong perspective just makes it more difficult to understand.
Aside from that, shouldn't you include f(x) != 0 as part of the definition in your example if you want f(x) to be
Gamma(x) ?
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Mickey1
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Re: A small hint 2008 A4
« Reply #5 on: Mar 26th, 2011, 4:12am » |
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Moving on to closed forms, I found this on the internet
An expression is a closed-form expression if it can be expressed in terms of a bounded number of elementary functions. Informally is it in apposition to a recurrence relation, which defines a sequence of term from earlier terms in that sequence.
My question is
If N! is not a closed form then how can Gamma(N) or Gamma (x) deserve this adjective?
Would you agree that N! is not a closed form and what does this imply for Gamma(N)?
It seems that the integration concept hides something and allows the integral to pose as simple, masking a complicated perhaps non-closed procedure.
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ThudnBlunder
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Re: A small hint 2008 A4
« Reply #6 on: Mar 26th, 2011, 9:32am » |
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Mickey, you might find this post useful.
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| « Last Edit: Mar 26th, 2011, 2:01pm by ThudnBlunder » |
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Mickey1
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Re: A small hint 2008 A4
« Reply #7 on: Mar 27th, 2011, 12:57pm » |
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That was interesting. Definition-related research unfolding as a row between two professors.
I was surprised that, of all people, mathematicians did not seem to have stricter definitions for these things.
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