Re: Putnam 2005 -- some answers [SPOILER ALERT]
- From: rob@xxxxxxxxxxxxxx (Rob Johnson)
- Date: Mon, 05 Dec 2005 18:53:13 GMT
In article <dmuc6u$sqq$1@xxxxxxxxxxxxxxxxxx>,
rusin@xxxxxxxxxx (David Rusin) wrote:
>I posted the Putnam problems earlier this evening. I have only had time
>to work on some of the questions, so there is no answer here (yet) to the
>following problems: A-3, A-4, A-6, B-4, B-5.
>
>As always I will keep copies of solutions at
> http://www.math.niu.edu/~rusin/problems-math/
>
[...]
>A-5.
>
>Maple gives an answer with dilogarithms and things like that but the
>answer seems to be pi log(2) / 8 .
>
>Substitute the Taylor series ln(x+1) = x - x^2/2 + ... into the
>integral to get
>
> - sum_{n>=1} (-1)^n/n J_n where J_n = \int_0^1 x^n/(x^2+1) dx .
>
>(In order to justify term-by-term integration we could, for example,
>replace the Taylor series by the Taylor polynomial and note that the
>remainders go to zero. In fact, I will be working below with a partial
>sum of the series with the J's anyway, and then take the limit.)
>
>Clearly
> J_n + J_{n+2} = \int_0^1 (x^n + x^{n+2})/(x^2+1) dx = \int_0^1 x^n dx
>which is just 1/(n+1). We also dredge up ancient memories to compute
> J_0 = arctan(1) = pi/4
> J_1 = 1/2 log( 1^2 + 1) = log(2)/2.
>
>So we have formulas
> J_{2i} = (-1)^i [ pi/4 - sum (-1)^j/(2j+1)] (sum over 2j+1 < 2i )
> J_{2i+1} = (-1)^i [ln(2)/2 - sum (-1)^j/(2j) ] (sum over 2j < 2i+1 )
>
>So the Nth partial sum for our series for the integral expands to
>
> -pi/4 sum_{2i<=N} (-1)^i/(2i) + ln2/2 \sum_{2i+1<=N} (-1)^i/(2i+1)
>
> + sum_{2i<=N, 2j+1 <=N} (-1)^{i+j}/( (2i)(2j+1) )
>
>(In the last sum, the summands with the odd factor smaller than the even
>one come from a J_even and the others come from a J_odd. I wrote out
>a few more intermediate steps when doing this by hand but I'm a lazy typist.)
>
>That last double sum then factors into
> ( \sum_i (-1)^i/(2i) )( \sum_j (-1)^j/(2j+1) )
>and each of those factors duplicates another factor already shown,
>i.e. we have something of the form AB + CD + BD, which I write as
>(A+D)B + (C+B)D - AD :
>
> ( -pi/4 + ( \sum_j (-1)^i/(2i+1) ) ) sum_{2i<=N} (-1)^i/(2i)
> + ( ln2/2 + ( \sum_j (-1)^i/(2i) ) ) sum_{2i+1<=N} (-1)^i/(2i+1)
> - ( \sum_i (-1)^i/(2i) )( \sum_j (-1)^j/(2j+1) )
>
>Now as we let N -> oo we have "A+D" and "C+B" both tend to zero,
>so only the rest remains, which of course tends to (-A)(-C) = (ln2/2)(pi/4) .
I came up with a couple of methods that rely on cancelling parts that
are not indefinitely integrable (at least not that I know of).
Method 1
--------
Define
|\1 log(1+tx)
I(t) = | --------- dx
\| 0 1+x^2
Then I(0) is 0 and I(1) is the integral we seek. Also,
I'(t)
|\1 x dx
= | -------------
\| 0 (1+tx)(1+x^2)
1 |\1 x+t t
= ----- | ( ----- - ---- ) dx
1+t^2 \| 0 1+x^2 1+tx
1 1 pi
= ----- ( - log(2) + -- t - log(1+t) )
1+t^2 2 4
So
|\1 log(1+x)
| -------- dx
\| 0 1+x^2
= I(1)
|\1
= I(0) + | I'(t) dt
\| 0
|\1 1 1 pi
= 0 + | ----- ( - log(2) + -- t - log(1+t) ) dt
\| 0 1+t^2 2 4
pi pi |\1 log(1+t)
= -- log(2) + -- log(2) - | -------- dt
8 8 \| 0 1+t^2
pi
= -- log(2) - I(1)
4
Solving for I(1), we get
|\1 log(1+x) pi
| -------- dx = -- log(2)
\| 0 1+x^2 8
Method 2
--------
Let x = tan(u):
|\1 log(1+x)
| -------- dx
\| 0 1+x^2
|\pi/4
= | log(1+tan(u)) du
\| 0
|\pi/4
= | ( log(cos(u)+sin(u)) - log(cos(u)) ) du
\| 0
|\pi/4 1
= | ( log(cos(u-pi/4)) + - log(2) - log(cos(u)) ) du
\| 0 2
|\pi/4 1
= | - log(2) du
\| 0 2
pi
= -- log(2)
8
Rob Johnson <rob@xxxxxxxxxxxxxx>
take out the trash before replying
.
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