# The ultimate pathogen vector

If civilization succumbs to a deadly pandemic, we will all know what the vector was. Every physician, nurse, dentist, hygienist, and health care worker is bound to check their smartphone sometime during the day before, during, or after seeing a patient and they are not sterilizing it afterwards.  The fully hands free smartphone could be the most important invention of the 21st century.

# Optimizing food delivery

This Econtalk podcast with Frito-Lay executive Brendan O’Donohoe from 2011 gives a great account of how optimized the production and marketing system for potato chips and other salty snacks has become. The industry has a lot of very smart people trying to figure out how to ensure that you maximize food consumption from how to peel potatoes to how to stack store shelves with bags of chips. This increased efficiency is our hypothesis (e.g. see here) for the obesity epidemic. However, unlike before where I attributed the increase in food production to changes in agricultural policy, I now believe it is mostly due to the vastly increased efficiency of food production. This podcast shows the extent of the optimization after the produce leaves the farm but the efficiency improvements on the farm are just as dramatic. For example, farmers now use GPS to optimally line up their crops.

# Analytic continuation continued

As I promised in my previous post, here is a derivation of the analytic continuation of the Riemann zeta function to negative integer values. There are several ways of doing this but a particularly simple way is given by Graham Everest, Christian Rottger, and Tom Ward at this link. It starts with the observation that you can write

$\int_1^\infty x^{-s} dx = \frac{1}{s-1}$

if the real part of $s>0$. You can then break the integral into pieces with

$\frac{1}{s-1}=\int_1^\infty x^{-s} dx =\sum_{n=1}^\infty\int_n^{n+1} x^{-s} dx$

$=\sum_{n=1}^\infty \int_0^1(n+x)^{-s} dx=\sum_{n=1}^\infty\int_0^1 \frac{1}{n^s}\left(1+\frac{x}{n}\right)^{-s} dx$      (1)

For $x\in [0,1]$, you can expand the integrand in a binomial expansion

$\left(1+\frac{x}{n}\right)^{-s} = 1 +\frac{sx}{n}+sO\left(\frac{1}{n^2}\right)$   (2)

Now substitute (2) into (1) to obtain

$\frac{1}{s-1}=\zeta(s) -\frac{s}{2}\zeta(s+1) - sR(s)$  (3)

or

$\zeta(s) =\frac{1}{s-1}+\frac{s}{2}\zeta(s+1) +sR(s)$   (3′)

where the remainder $R$ is an analytic function when $Re s > -1$ because the resulting series is absolutely convergent. Since the zeta function is analytic for $Re s >1$, the right hand side is a new definition of $\zeta$ that is analytic for $s >0$ aside from a simple pole at $s=1$. Now multiply (3) by $s-1$ and take the limit as $s\rightarrow 1$ to obtain

$\lim_{s\rightarrow 1} (s-1)\zeta(s)=1$

which implies that

$\lim_{s\rightarrow 0} s\zeta(s+1)=1$     (4)

Taking the limit of $s$ going to zero from the right of (3′) gives

$\zeta(0^+)=-1+\frac{1}{2}=-\frac{1}{2}$

Hence, the analytic continuation of the zeta function to zero is -1/2.

The analytic domain of $\zeta$ can be pushed further into the left hand plane by extending the binomial expansion in (2) to

$\left(1+\frac{x}{n}\right)^{-s} = \sum_{r=0}^{k+1} \left(\begin{array}{c} -s\\r\end{array}\right)\left(\frac{x}{n}\right)^r + (s+k)O\left(\frac{1}{n^{k+2}}\right)$

Inserting into (1) yields

$\frac{1}{s-1}=\zeta(s)+\sum_{r=1}^{k+1} \left(\begin{array}{c} -s\\r\end{array}\right)\frac{1}{r+1}\zeta(r+s) + (s+k)R_{k+1}(s)$

where $R_{k+1}(s)$ is analytic for $Re s>-(k+1)$.  Now let $s\rightarrow -k^+$ and extract out the last term of the sum with (4) to obtain

$\frac{1}{-k-1}=\zeta(-k)+\sum_{r=1}^{k} \left(\begin{array}{c} k\\r\end{array}\right)\frac{1}{r+1}\zeta(r-k) - \frac{1}{(k+1)(k+2)}$    (5)

Rearranging (5) gives

$\zeta(-k)=-\sum_{r=1}^{k} \left(\begin{array}{c} k\\r\end{array}\right)\frac{1}{r+1}\zeta(r-k) -\frac{1}{k+2}$     (6)

where I have used

$\left( \begin{array}{c} -s\\r\end{array}\right) = (-1)^r \left(\begin{array}{c} s+r -1\\r\end{array}\right)$

The righthand side of (6) is now defined for $Re s > -k$.  Rewrite (6) as

$\zeta(-k)=-\sum_{r=1}^{k} \frac{k!}{r!(k-r)!} \frac{\zeta(r-k)(k-r+1)}{(r+1)(k-r+1)}-\frac{1}{k+2}$

$=-\sum_{r=1}^{k} \left(\begin{array}{c} k+2\\ k-r+1\end{array}\right) \frac{\zeta(r-k)(k-r+1)}{(k+1)(k+2)}-\frac{1}{k+2}$

$=-\sum_{r=1}^{k-1} \left(\begin{array}{c} k+2\\ k-r+1\end{array}\right) \frac{\zeta(r-k)(k-r+1)}{(k+1)(k+2)}-\frac{1}{k+2} - \frac{\zeta(0)}{k+1}$

Collecting terms, substituting for $\zeta(0)$ and multiplying by $(k+1)(k+2)$  gives

$(k+1)(k+2)\zeta(-k)=-\sum_{r=1}^{k-1} \left(\begin{array}{c} k+2\\ k-r+1\end{array}\right) \zeta(r-k)(k-r+1) - \frac{k}{2}$

Reindexing gives

$(k+1)(k+2)\zeta(-k)=-\sum_{r'=2}^{k} \left(\begin{array}{c} k+2\\ r'\end{array}\right) \zeta(-r'+1)r'-\frac{k}{2}$

Now, note that the Bernoulli numbers satisfy the condition $\sum_{r=0}^{N-1} B_r = 0$.  Hence,  let $\zeta(-r'+1)=-\frac{B_r'}{r'}$

and obtain

$(k+1)(k+2)\zeta(-k)=\sum_{r'=0}^{k+1} \left(\begin{array}{c} k+2\\ r'\end{array}\right) B_{r'}-B_0-(k+2)B_1-(k+2)B_{k+1}-\frac{k}{2}$

which using $B_0=1$ and $B_1=-1/2$ gives the self-consistent condition

$\zeta(-k)=-\frac{B_{k+1}}{k+1}$,

which is the analytic continuation of the zeta function for integers $k\ge 1$.