Pi is trancendental

First the rabbit. The introduction of this function is the part which I don’t know how to motivate. Let f be a polynomial and define

\displaystyle I(z,f)=\int_0^z e^{z-t}f(t)\,dt.

Integration by parts gives the recursion

\displaystyle I(z,f)=e^z f(0)-f(z)+I(z,f')

and therefore we have the formula

\displaystyle I(z,f)=e^z\sum_{j\geq 0} f^{(j)}(0) - \sum_{j\geq 0}f^{(j)}(z).

Now suppose (for want of a contradition) that \pi i \in \overline{\mathbb{Q}}. Let the set of Galois conjugates of \pi i be \{a_1,\ldots,a_k\}. Then we have \prod_{i=1}^k (1+e^{a_i})=0, expand this as \sum_{J\subset[k]}e^{\sum_{j\in J}a_j}=0 and rewrite as

\displaystyle (2^k-d)+\sum_{i=1}^d e^{\theta_i}=0

where the \theta_i are the nonzero exponents.

Now consider

\displaystyle \sum_{i=1}^d I(\theta_i,f)=(d-2^k)\sum_{j\geq 0}f^{(j)}(0)-\sum_{j\geq 0}\sum_{i=1}^df^{(j)}(\theta_i)

Let N be an integer such that N\pi i is an algebraic integer. Let p be a (large) prime and we choose to take

\displaystyle f(x)=N^{dp}x^{p-1}\prod_{i=1}^d(x-\theta_i)^p\in\mathbb{Z}[x].

There are absolute constants A and B (independent of p) such that

\displaystyle |\sum_{i=1}^d I(\theta_i,f)|\leq A B^p

(look at the integral definition of I(z,f) and apply the naive estimate).

Now consider \sum_{i=1}^d I(\theta_i,f). It is a Galois-invariant algebraic integer, hence an integer. We have

\displaystyle \sum_{i=1}^d I(\theta_i,f)=(d-2^k)f^{(p-1)}(0)+\text{higher order terms}.

Here higher order means at least p derivatives appearing. Each of these higher order terms is divisible by p!, hence by p. Since p is prime, for large enough p, \sum_{i=1}^d I(\theta_i,f) is not divisible by p, hence nonzero.

Now every term is divisible by (p-1)!, so we get the lower bound

\displaystyle |\sum_{i=1}^d I(\theta_i,f)|\geq (p-1)!

As there are infinitely many primes, we can send choose p large enough to get a contradiction, QED.

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I found the bike racks at Brisbane airport

There are about six of them at the domestic terminal, not under cover, on the ground floor, near the rental cars and the lift to the overpass to the terminals. Here is a map with their location, together with the cycle routes in and out.

As you can see from the map, openstreetmap also knows where these bike racks are.

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Single variable polynomial division in sage

As far as I can tell, this isn’t done purely within sage, and requires a call to maxima.

If you want to divide the polynomial a by b to produce a quotient and remainder, the code is.


The first element of the output is the quotient and the second is the remainder.

So for example

sage: R=QQ['x']
sage: a=x^210-1
sage: b=R.cyclotomic_polynomial(210)*(x-1)
sage: q,r=a.maxima_methods().divide(b)
sage: q
x^161 + 2*x^160 + 2*x^159 + x^158 - x^156 - x^155 - x^154 - x^153 - x^152 + x^150 + 2*x^149 + 2*x^148 + 2*x^147 + 2*x^146 + 2*x^145 + x^144 - x^142 - x^141 + x^139 + x^138 + x^137 + x^136 + x^135 + x^134 + x^133 + x^132 + x^131 + x^130 + x^129 + x^128 + x^127 + 2*x^126 + 3*x^125 + 3*x^124 + 2*x^123 + x^122 + x^116 + 2*x^115 + 3*x^114 + 3*x^113 + 3*x^112 + 3*x^111 + 3*x^110 + 2*x^109 + x^108 + x^105 + 2*x^104 + 2*x^103 + 2*x^102 + 2*x^101 + 2*x^100 + 2*x^99 + 2*x^98 + 2*x^97 + 2*x^96 + 2*x^95 + 2*x^94 + 2*x^93 + 2*x^92 + 2*x^91 + 2*x^90 + 2*x^89 + 2*x^88 + 2*x^87 + 2*x^86 + 2*x^85 + 2*x^84 + 2*x^83 + 2*x^82 + 2*x^81 + 2*x^80 + 2*x^79 + 2*x^78 + 2*x^77 + 2*x^76 + 2*x^75 + 2*x^74 + 2*x^73 + 2*x^72 + 2*x^71 + 2*x^70 + 2*x^69 + 2*x^68 + 2*x^67 + 2*x^66 + 2*x^65 + 2*x^64 + 2*x^63 + 2*x^62 + 2*x^61 + 2*x^60 + 2*x^59 + 2*x^58 + 2*x^57 + x^56 + x^53 + 2*x^52 + 3*x^51 + 3*x^50 + 3*x^49 + 3*x^48 + 3*x^47 + 2*x^46 + x^45 + x^39 + 2*x^38 + 3*x^37 + 3*x^36 + 2*x^35 + x^34 + x^33 + x^32 + x^31 + x^30 + x^29 + x^28 + x^27 + x^26 + x^25 + x^24 + x^23 + x^22 - x^20 - x^19 + x^17 + 2*x^16 + 2*x^15 + 2*x^14 + 2*x^13 + 2*x^12 + x^11 - x^9 - x^8 - x^7 - x^6 - x^5 + x^3 + 2*x^2 + 2*x + 1
sage: r

hat tip: this question

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A free semigroup

In the appendix here, we make the claim that the semigroup

\Gamma_A=\langle \begin{pmatrix}a & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix}b & 1 \\ 1 & 0 \end{pmatrix}\mid 1\leq a,b\leq A\rangle

is free on the given generators.

Here, I will give a proof. It is a simpler version of the ping-pong argument commonly used to prove that a group is free on a given set of generators.

First a simple observation. it suffices to prove that the set

\left\{ \begin{pmatrix}a & 1 \\ 1 & 0 \end{pmatrix}\mid 1\leq a\leq A\right \}

generates a free semigroup.

Now, consider the action of our semigroup on the interval (1,\infty) by fractional linear transformations:

\begin{pmatrix}a & b \\ c & d \end{pmatrix}\cdot x=\frac{ax+b}{cx+d}

In particular, note that

\begin{pmatrix}a & 1 \\ 1 & 0 \end{pmatrix}\cdot x=a+\frac{1}{x}.

Now suppose that

\begin{pmatrix}y_1 & 1 \\ 1 & 0 \end{pmatrix}\cdots\begin{pmatrix}y_m & 1 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix}z_1 & 1 \\ 1 & 0 \end{pmatrix}\cdots\begin{pmatrix}z_n & 1 \\ 1 & 0 \end{pmatrix}

Apply both sides to some x\in(1,\infty). The LHS lies in (y_1,y_1+1) and the RHS lies in (z_1,z_1+1). Therefore y_1=z_1 and the rest is an easy induction.

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Some easy singular Schubert varieties

I write w for both an element of the Weyl group and the corresponding point in G/B. Let X_w be a Schubert variety. Then T_eX_w\subset T_e(G/B) is an inclusion of \mathfrak{b} representations. The latter is generated by the -\theta-weight space, where \theta is the highest root.

Suppose w\geq s_\theta, where s_\theta is the reflection corresponding to \theta. Then the \mathbb{P}^1 connecting e and s_\theta lies in X_w (think of the corresponding SL2), so the -\theta-weight space lies in T_eX_w. Since this generates T_e(G/B), we have T_eX_w=T_e(G/B).

Therefore if w_0\neq w\geq s_\theta, then X_w is singular. This includes some of the first examples of singular Schubert varieties, for example the B2 singular Schubert variety and one of the A3 ones.

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BIRS (Banff, Canada)

I was fortunate to recently spend a week at the Banff International Research Station, for a conference on Whittaker Functions: Number Theory, Geometry and Physics. Rather than talk about the mathematics, I will concentrate on the facilities in this post. BIRS has been successfully running many maths workshops for many years now and are in high demand as they consistently do a good job overall.

Banff is a tourist town in the picturesque Canadian Rocky mountains. The standard way to get there is via Calgary airport, catching the Banff airporter bus (one-way 2 hours, CAD54 with BIRS discount applied). BIRS itself is part of the larger Banff Centre, which is mainly an arts venue situated on a hill with elevated views over the town.

Food and Lodging:
Participants are all accommodated onsite in clean and comfortable single rooms sharing a bathroom with one other participant. There were some participants complaining about the temperature but I found no problems. The meals are also onsite in a buffet arrangement, and generally of a good quality (and hence tempting to pig out on the dessert). Internet access exists and works. I didn’t subject this to any serious stress tests.

Academic Facilities:
Here is a picture of Manish Patnaik giving his talk.

Maniish Patnaik

You can see the tiered seating (good) and the lack of blackboard space (bad). There was space from an architectural point of view to put in blackboards that can be moved up and down, so for a dedicated conference centre this is especially disappointing. All talks were recorded by an automated system (it is possible that not all were remembered to be recorded, as this process required manually pressing a button at the start of each talk).

Unfortunately some people continue to turn down lights during slide talks, despite the slides being perfectly visible under normal lighting conditions. This is annoying because it strains the eyes to make notes, and makes it harder to work on other things when (some) speakers inevitably power through their slides at too rapid a pace.

There are also other small private discussion rooms equipped with whiteboards for participants’ use.

Stephen D. Miller:
Stephen D. Miller was originally scheduled to attend and give a talk, but got screwed over by an airline and ended up deciding not to come. He then recorded a talk which we watched (and you can too (mp4)) and skyped in at the end of the viewing to answer questions. This worked surprisingly well, which I suspect is more a function of the speaker being Stephen as opposed to the format.

I overheard some other participants remark that they were still disappointed that Stephen didn’t come as they were hoping to discuss informally with him. Experienced conference-goers will naturally and immediately recognise this sentiment, knowing that the informal discussions that take place at a conference can be even more important than the formal program.

It is a little known fact that it is possible to spend a couple of additional days at BIRS afterwards. A few of us availed ourselves of this welcome opportunity.

My talk:
I did not give a talk at this conference. I did however give a talk on my recent work on geometric extension algebras at the University of Alberta the following week.

The scenic route from Banff to Edmonton

The scenic route from Banff to Edmonton

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My 2016 Senate Election Vote

It is possible to check if your vote for the Senate was counted correctly in the 2016 Australian Federal election. To begin, go to this AEC website, download the Formal Preferences file for your relevant state or territory and unzip.

In my case, I download the Queensland file. The preferences I gave the first five candidates were 2, 1, 46, 47 and 20 so I type

grep 2,1,46,47,20 aec-senate-formalpreferences-20499-QLD.csv

and out comes

Griffith,South Brisbane,874,32,29,",,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,2,1,46,
109,48,49,32,33,34,60,61,114,115,116,117,118,119,18,4,120,121,58,59,24,25,26, 30,31,19,5,110,111,28,29,57,56,55,54,53,52,17,16,15,14,13,12,11,10,9,8,7,6,50,

The output begins with my electorate and polling place. I don’t know what the next three numbers mean. The numbers following the string of commas are my preferences, in order that the candidates appeared on the ballot paper.

Here is where it gets embarrassing. Eagle eyed readers will immediately have noticed that there is no 39 in this list, and there are two 82s. Nevertheless, I believe it is still formal.

Using the distribution of preferences from this other AEC site, it is possible to track my vote through the count. It passes through the hands of a few no-hopers before reaching Larissa Waters (Green) who is elected in 9th place.

Larissa Waters is elected at the 738th count with a surplus of only 181 votes. In theory my vote should then transfer at a paltry value of 181/209475 (about .0009) to Suzanne Grant (Xenophon team) but the vagaries of the inclusive Gregory system and the existence of loss by fraction makes it hard to say this with any certainty.

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