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Part 2: Let f(x) be the average number of straws it takes to break a camel's back if it can hold x units. Determine the asymptotics of f up to o(1), i.e., produce a simple function g so that the limit of f(x)-g(x) as x->infinity is 0.
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Solution to Part 2 in white:
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As blah_blah pointed out, f(x) satisfies an integeral equation, although I think it is better to write it differently: f(x) = 1 + Integral from x-1 to x of f(t) dt, with initial condition that f(x) = 0 on [-1,0]. This is even simpler if you consider g(x) = f(x)-2x, which satisfies
(1) g(x) = Integral from x-1 to x of g(t) dt,
with initial condition g(x) = -2x on [-1,0].
The integral equation satisfied by g looks like it smooths g out, and damps down the oscillations of g. Since g(x) is an average of values of g on the interval [x-1,x], g is bounded by the values on [-1,0], and then by the values on [0,1], etc. This can be used to show that g is asymptotic to a constant, although I won't give the details here.
The idea I had was to find a conserved quantity, some function h defined on [0,1], so that the integral from t=0 to t=1 of g(x+t) h(t) dt does not depend on x. Let's see what properties h might have to force
H(x) = integral from 0 to 1 of h(t) g(x+t) dt
to be constant. By Leibniz's rule (
link),
H'(x) = integral from 0 to 1 of g'(x+t) h(t) dt
Integrating by parts,
H'(x) = g(x+1)h(1) - g(x)h(0) - integral from 0 to 1 of g(x+t) h'(t) dt.
We'd like to choose h to force this to be 0, using what we know about g, equation (1). To get the RHS to involve the integral of g(t) on an interval of length 1, we can set h'=1. Setting h(0)=0 forces a term to drop out, so let h(t)=t.
H'(x) = g(x+1) - integral from t=0 to t=1 of g(x+t) dt
H'(x) = g(x+1)-g(x+1)
H'(x) = 0.
So, we have established that integral from t=0 to t=1 of t * g(x+t) dt is constant. If g is asymptotic to some constant c, then
Limit x->oo H(x)
= integral from 0 to 1 of t c dt
= c/2
H(-1)
= integral from 0 to 1 of t g(t-1) dt
= integral from 0 to 1 of t (-2 (t-1)) dt
= 1/3
Since H is constant, c=2/3.
So, f(x)=g(x)+2x is asymptotic to 2x+2/3. As a quick check, f(1)=e, which is not too much greater than 2 2/3 = 1/0! + 1/1! + 1/2! + 1/3!.
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