Ramanujan summation

I just came across a way that the amazing mathematician Ramanujan developed of assigning a value to certain divergent series. I found it interesting, so I want to share a short summary of it here. It is based on the Euler–Maclaurin formula
\begin{align*} \sum_{k=\alpha}^\beta f(k) &= \int_\alpha^\beta f(t)\,dt + \frac{f(\alpha)+f(\beta)}{2} \\ &\quad+ \sum_{k=1}^n \frac{B_{2k}}{(2k)!}\Bigl(f^{(2k-1)}(\beta)-f^{(2k-1)}(\alpha)\Bigr) + R_n \end{align*}where $B_{2k}$ denotes the $(2k)$th Bernoulli number, $f$ has $2n+1$ continuous derivatives on $[\alpha,\beta]$ with $\alpha$, $\beta$, and $n\geq0$ being integers, and $R_n$ is the remainder term given by
\[ R_n = \int_\alpha^\beta \frac{B_{2n+1}(t-\lfloor t\rfloor)}{(2n+1)!}f^{(2n+1)}(t)\,dt . \] Continue reading

The importance of direction

Lately I’ve been reflecting on the impact that having a direction (or lack of it) has had in my life. For a long time now—since around the time I first set foot on a university campus—it’s been clear to me that I would be doing and writing about mathematics for the rest of my life. However, perhaps because I have never been particularly interested in thinking about money I never necessarily assumed that I would be paid to do mathematics. My expectation was that I might need to have a day job to support myself and my mathematical hobby.

Recently, I’ve been thinking that being paid to do mathematics would be great because it would allow me to devote more time to it. For example, I am quite proud of my latest paper—but it took a lot of time to write, and would have taken even longer if I didn’t already have a job as a postdoctoral researcher. One of my previous papers was written during evenings and weekends since at the time I was working full-time at an internship and it wasn’t ideal.

Thus, I’ve been doing the things necessary to apply to academic positions like putting together a research statement. During this process the importance of direction has been made abundantly clear to me and I can’t help but lament the amount of time I spent as an undergraduate and graduate student with unclear direction and ridiculous misconceptions.

For example, as an undergraduate student I studied the Ramanujan–Nagell equation and came up with a new solution for it that I hadn’t seen published anywhere. My thought was: this is so cool, I have to share this with the world. So I wrote a short report, gave a copy to one of my professors who I thought might be interested, and uploaded a copy to my website. But what to do beyond that? My thinking was: I don’t know how to get it published somewhere, but I do know how to make it accessible to anyone who wants to read it via the Internet. I told myself that 99% of the value comes from writing the report and I didn’t need the additional 1% “seal of approval” that comes from getting it published. Now I know this is totally backwards—at least when it comes to job applications almost all of a paper’s value is derived from being published.

Looking back, my advice to myself would absolutely be to try to get the report published. Maybe I could’ve gotten it published and maybe not, but either way it would have been a very valuable learning experience. Incidentally, the report does seem to be of value to some people: it’s been cited by two OEIS entries, two published papers, a PhD thesis, and Noam Elkies recently gave a talk referencing it (!!!):

We state several finiteness theorems, outline some of the connections among them, explain how a finiteness proof can be ineffective, and (time permitting) sketch Nagell’s proof and an even more elementary one discovered only 12 years ago by C. Bright.

Maybe I’ll come back to that result and get it formally published some day, but I already have more than enough papers on the table. At the very least it’s encouraging to know that I’m in no danger of running out of material to research and write about anytime soon. Even more importantly, I’ve learned how important direction is to achieving your dreams.

The value of the trigonometric harmonic series revisited

Shortly after my last post I realized there was a simpler way of determining the exact value of the series $\sum_{n=1}^\infty\cos n/n$. Instead of following the method I previously described which required the intricate analysis of some integrals, one can simply use the formula

\[ \sum_{n=1}^\infty\frac{a^n}{n} = -\ln(1-a) \]

which is valid for $a\in\mathbb{C}$ which satisfies $\lvert a\rvert\leq 1$ and $a\neq 1$. This comes from a simple rewriting of the so-called Mercator series (replace $x$ with $-x$ in the Taylor series of $\ln(1+x)$ and then take the negative).

Then we have

\sum_{n=1}^\infty\frac{\cos n}{n} &= \sum_{n=1}^\infty\frac{e^{in}+e^{-in}}{2n} \\
&= -\bigl(\ln(1-e^i)+\ln(1-e^{-i})\bigr)/2 \\
&= -\ln\bigl((1-e^i)(1-e^{-i})\bigr)/2 \\
&= -\ln(2-e^i-e^{-i})/2 \\
&= -\ln(2-2\cos1)/2 \\
&\approx 0.0420195

since $\lvert e^i\rvert=\lvert e^{-i}\rvert=1$, but $e^i\neq1$ and $e^{-i}\neq1$.

The value of the trigonometric harmonic series

I’ve previously discussed various aspects of the “trigonometric harmonic series” $\sum_{n=1}^\infty\cos n/n$, and in particular showed that the series is conditionally convergent. However, we haven’t found the actual value it converges to; our argument only shows that the value must be smaller than about $2.54$ in absolute value. In this post, I’ll give a closed-form expression for the exact value that the series converges to.

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The names in boxes puzzle

This is one of the best puzzles I’ve come across:

100 prisoners have their names placed in 100 boxes so that each box contains exactly one name. Each prisoner is permitted to look inside 50 boxes of their choice, but is not allowed any communication with the other prisoners. What strategy maximizes the probability that every prisoner finds their own name?

I heard about this puzzle years ago, spent several days thinking about it, and never quite solved it. Actually, I did think of a strategy in which they would succeed with probability over 30% (!), which was the unbelievably-high success rate quoted in the puzzle as I heard it posed. However, I ended up discarding the strategy, as I didn’t think it could possibly work (and probably wouldn’t have been able to prove it would work in any case).

Revisiting a lemma

We’ve discussed before the “trigonometric harmonic” series $\sum_{n=1}^\infty\cos n/n$. In particular, we showed that the series converges (conditionally). The argument involved the partial sums of the sequence $\{\cos n\}_{n=1}^\infty$, and we denoted these by $C(m)$. The closed-form expression we found for $C(m)$ involved the quantity $\cos m-\cos(m+1)$; in this post we show that this expression can also be written in the alternative form $2\sin(1/2)\sin(m+1/2)$.

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That harmonic series variant absolutely

Previously I discussed a variant on the harmonic series, $\sum_{n=1}^\infty\cos n/n$. Last time we showed that

\[ \sum_{n=1}^\infty\frac{\cos n}{n} = \sum_{n=1}^\infty\sum_{m=1}^n\cos m\Bigl(\frac{1}{n}-\frac{1}{n+1}\Bigr) , \]

and then showed that the series on the right converges absolutely, by comparison with the series $\sum_{n=1}^\infty3/n^2$.

Since the series on the right converges and the two series have the same value, the series on the left also converges. However, this does not imply that the series on the left also converges absolutely. As a trivial counterexample, if a conditionally convergent series sums to $c$ then $c\sum_{n=1}^\infty \href{http://en.wikipedia.org/wiki/Kronecker_delta}{\delta_{n,1}}$ is an absolutely convergent series which sums to the same value. 🙂

In this post, we answer the question of whether $\sum_{n=1}^\infty\cos n/n$ converges absolutely or not.

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The infinite hat problem

The infinite hat problem is a great puzzle. If you have a strong math background, you should try solving it before reading my solution below!

Here’s my strategy for the wizards: first, they agree on an ordering of themselves. Each wizard can be indexed by a natural number, since there are countably many of them. They then consider the set of all possible hat configurations $S$ with respect to that ordering. By the well-ordering theorem (which is equivalent to the axiom of choice) a well-ordering of $S$ exists; the wizards also agree on a specific well-ordering.

Note that this step is non-constructive because it relies on the axiom of choice. That is, such a well-ordering exists but there may not be a way to explicitly construct it. The point of the note about assuming the axiom of choice was a tip-off that the wizards need to make their decision based off of a set whose existence is only ensured by the axiom of choice.

Once the well-ordering has been chosen the wizards are ready to receive their hats. Once they are able to see everyone else’s hat, they each construct a subset $T$ of $S$ which contains the hat configurations which differ from the configuration they can see in only finitely many hats. The lack of knowledge about a wizard’s own hat is irrelevant to this construction, since that lack of knowledge only changes the configuration they see in finitely many hats. In particular, for every wizard their subset $T$ will consist of the true configuration along with all configurations which differ from the true configuration in finitely many hats, and therefore be the same for all wizards.

Now that all the wizards have constructed the same $T\subset S$, they use the well-ordering of $S$ to find the least element of $T$, and everyone guesses the hat colour which they have in the least element. Since every element of $T$ differs from the true configuration in finitely many hats, the configuration that the wizards guess will also differ in finitely many hats. Thus almost all wizards will choose correctly.

I heard about the problem on a list of good logic puzzles compiled by Philip Thomas. I purposely haven’t read his solution yet, since I didn’t want that to influence me while writing down my solution.