On 12 May 2001 22:07:07 GMT, [EMAIL PROTECTED] (Francis Dermot
Sweeney) wrote:
>I had got to a similar stage, and had tried showing the integral vanishes
>by complex analysis methods. I had thought that since ln r is harmonic,
>the value of the integral around the curve gives teh value at the center
>(*2pi), but the problem is that the origin is a point on the curve where
>the function fails to be harmonic.
That's exactly why that was not the argument I actually presented.
You can do it this way, but there's a detail or two. For 0<r<1 we have
that the integral of log(|1+re^{it}|) vanishes, and now some limiting
argument, like the Dominated Convergence Theorem if nothing more
elementary, shows that this remains true for r=1. (Quiz: What's the
value of the integral for r > 1?)
>Why, in your argument, does integral(log|1-exp(2it)| = I ? are n't you
>traversing the same circle twice, giving twice the
>integral. Maybe my complex analysis needs work.
Calculus, rather. If f(t) is a function defined on [0,2pi) and we
extend it to [0,4pi) by decreeing that f(t+2pi)=f(t) then the
integral of f(t) from 0 to 2pi is the same as the integral of
f(2t) from 0 to 2pi. (A simple change of variables shows that
the integral of f(2t) from 0 to pi = (int f(t), 0, 2pi)/2, and
similarly (int f(2t), pi, 2pi) = (int f(t), 0, 2pi)/2. Now apply the
fact that 1/2 + 1/2 = 1.)
>Francis
>
>[EMAIL PROTECTED] (David C. Ullrich) writes:
>
>>On 11 May 2001 19:17:40 GMT, [EMAIL PROTECTED] (Francis Dermot
>>Sweeney) wrote:
>
>>>Here is a problem that is quite tricky. Starting at a radius R_o, a hop
>>>is made of length from the current point to the origin (R_o), in a random,
>>>uniform direction, on a 2d plane. This take us to a new point, with
>>>distance to the
>>>origin R_1. The next hop is then of length R_1, in a random uniform
>>>direction, etc.
>
>>So (X_n) satisfies... oops, you use "X_n" for something else below.
>>So if H_n is the position after the n-th hop then
>
>>H_(n+1) = H_n + |H_n| * exp(i theta_n),
>
>>where the theta_n are i.i.d. uniformly distributed on [0,2 pi), right?
>>Or equivalently, making the notation nicer below,
>
>>H_(n+1) = H_n + H_n * exp(i theta_n)
>
>>>Show that X_n=log(R_n)-log(R_(n-1)) are i.i.d. random variables, with
>>>mean zero, and finite variance.
>
>>Well,
>
>>log(R_n) - log(R_(n-1)) = log(R_n/R_(n-1))
>
>> = log(|H_n|/|H_(n-1)|)
>
>> = log(|1+exp(i theta_(n-1))|),
>
>>which makes the iid and finite variance parts clear. To show it has
>>mean zero you need to show that the integral from 0 to
>>2 pi of log(|1+exp(it)|) equals 0. You could do that with a little
>>complex analysis or you could use a trick: Say this integral is I.
>>Then by symettry we also have I = integral from 0 to 2 pi of
>>log(|1-exp(it)|), and so again by symettry
>
>>I + I = integral(log(|1+exp(it)|) + log(|1-exp(it)|))
>> = integral(log(|1-exp(2it)|))
>> = I,
>
>>hence I = 0.
>
>
>>>Francis.
>>>
>>>
>>>--
>>>Francis Sweeney
>>>Dept. of Aero/Astro
>>>Stanford U.
>
>>David C. Ullrich
>>*********************
>>"Sometimes you can have access violations all the
>>time and the program still works." (Michael Caracena,
>>comp.lang.pascal.delphi.misc 5/1/01)
>
>--
>Francis Sweeney
>Dept. of Aero/Astro
>Stanford U.
David C. Ullrich
*********************
"Sometimes you can have access violations all the
time and the program still works." (Michael Caracena,
comp.lang.pascal.delphi.misc 5/1/01)
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