11
$\begingroup$

Resistance across opposite vertices of n-dimensional cube with each edge at one ohm resistance is $$R_n=\sum_{k=0}^{n-1}\frac1{(n-k){n\choose k}}=\frac1n\sum_{k=0}^{n-1}\frac1{{n-1\choose k}}.$$ The proof is simple: we can identify nodes with equal voltages, and there are ${n\choose k}$ nodes with a distance $k$ to a given point.

From another hand $$R_n=\frac1{2^n}\sum_{k=1}^{n}\frac{2^k}{k},$$ because both sums satisfy the same recurrence relation $$R_n=\frac1n+\frac12R_{n-1}.$$

We can consider the sum $\sum_{k=1}^{n}\frac{2^k}{k}$ as a partial sum of $p$-adic logarithm $$\log_p(1+x)=-\sum_{k=1}^{\infty}\frac{(-x)^k}{k}$$ at the point $x=2$ (it is well defined for $p=2$). Using two formulas for $R_n$ we can get a simple application. We can find the value of $2$-adic logarithm at the point $-2$: $$-\log_2(-2)=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{2^k}{k}= \lim_{n\to\infty}\frac{2^n}n\sum_{k=0}^{n-1}\frac1{{n-1\choose k}}=0.$$ It is not a surprise (see § 4.4.11 from Cohen (2007), Number theory, Volume I: Tools and Diophantine equations).

First question: why does $2$-adic logarithm arise in combinatorial problem?

Second question: do you know any more connections between combinatorial and $p$-adic objects?

$\endgroup$
  • 5
    $\begingroup$ That the series vanishes at $-2$ is unsurprising because, unlike the complex log, the $p$-adic log is a homomorphism, to a torsion-free group, so that it has to vanish at torsion points. You’ll also get zero when you plug in $x=i-1$. $\endgroup$ – Lubin Oct 29 '13 at 13:57
  • $\begingroup$ Yes, but does it have a combinatorial sencs? $\endgroup$ – Alexey Ustinov Oct 29 '13 at 14:16
  • 2
    $\begingroup$ Do you have access to MathSciNet? I typed in Anywhere: p-adic and Primary Classification: 05, and got 37 hits. $\endgroup$ – Gerry Myerson Oct 29 '13 at 22:45
  • $\begingroup$ Thank you, Gerry. I've found an example concerning p-adic properties of alternating sign matrices (arrays of 0, 1 and −1, such that the entries of each row and column add up to 1 and the non-zero entries of a given row/column alternate) $\endgroup$ – Alexey Ustinov Oct 30 '13 at 2:06
1
$\begingroup$

The following article https://arxiv.org/abs/0904.1757 (The Hypercube of Resistors, Asymptotic Expansions, and Preferential Arrangements, by Nicholas Pippenger. Published in Mathematics Magazine 83(N5) (2010), 331-346) might be relevant for this question (as well as for Asymptotic rate for $\sum\binom{n}k^{-1}$).

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.