What are the chances that a bird will poop on my car?

This question was posed as a scenario of mathematics in everyday life. Some googling led to Randall “XKCD” Munroe’s post here. Borrowing similar ideas leads to my own computation below.
I park at an open air carpark for work. The easy assumptions are my car has a 4.3m x 1.75m footprint and I park for 8 hours a day. For data on birds, after some searching I found a 2016 article that estimates (provided I interpreted correctly) a total of 377 birds in 113 hectares of built-up area at another university in the country. So if we assume a similar bird population here and that birds poop uniformly everywhere once every hour,
[tex] \frac{377}{113 \times 10,000} \times 8 \times 4.3 \times 1.75 \approx 2\% [/tex]

In my opinion, the estimate is on the low side. In practical terms, it appears birds spend a large amount on time perched on trees. I also read that birds tend to clear their vowels to lighten their loads before they fly. So it is a catch-22 for me. I either park in the shade under trees and risk being bombarded or I park in the open and find myself a 40 degrees oven when I leave work on a typical sunny afternoon.

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Queen of Mathematics

Number theory has been called the Queen of Mathematics. Until some fifty years ago, it did not occur to anyone that number theory, especially the study of prime numbers, would have any immediate applications to business. More recently, the Queen has been relegated to be the object of a courtship, inspired by material gains, rather than awe. As a result, progress has been made in unexpected directions, which have required deeper investigations. — Papa Paulo

A.K.A. Paulo Ribenboim from his pseudo-novel-number theory text “Prime Numbers, Friedns Who Give Problems: A Trialogue with Papa Paulo” (p. 50).

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Triangular numbers modulo powers of 2 and its generalizations

Someone discussed with me an interesting problem that he was working on with his students. They found that the congruence
[tex] \frac{1}{2}X(X+1) \equiv a \pmod{n} [/tex]
has a solution for every [tex] 0 \le a < n [/tex] if and only if [tex] n =2^k[/tex].

My first instinct of course was to complete the squares for triangular numbers and reduce the problem to [tex] X^2 \equiv a \pmod{n} [/tex].
This turn out to work well for odd modulus and the solutions for triangular numbers and squares correspond. But when the modulus was a power of 2, completing the square would not work. A simple search found a few websites where the phenomenon was recorded and it seems a (perhaps original?) source is Knuth’s the Art of Computer Programming Volume 3, Section 6.4, Exercise 20. Knuth was talking about hashing but essentially the exercise is the above problem. I prefer to rephrase it as for every positive integer k, these two sets are identical:
[tex] \{ \frac{x(x-1)}{2} \pmod{2^k} \} = \{0, 1, \ldots, 2^k-1 \} [/tex]

Knuth gave a slick proof which can be easily adapted and generalized to the following:
For a prime [tex] p, 1 \le m < p, k \ge 1,[/tex]
[tex] \{ px^2+mx \pmod{p^k} \} = \{0, 1, \ldots, p^k-1 \} [/tex]

Proof: Suppose
[tex] px^2+mx \equiv py^2+my \pmod{p^k} [/tex]
[tex] (x-y) ( p(x+y)+m) \equiv 0 \pmod{p^k}[/tex]
Since [tex] p \nmid p(x+y)+m [/tex] we can conclude that
[tex] x \equiv y \pmod{p^k}[/tex].

So each of [tex]x = 0, 1, \ldots p^k-1 [/tex] gives rise to a different value of [tex] px^2+mx \pmod{p^k}[/tex].

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If there is some good inflammable stuff it will catch fire

Do not try to satisfy your vanity by teaching them great many things. Awake their curiosity. It is enough to open the minds, do not overload them. Put there just a spark. If there is some good inflammable stuff it will catch fire.

The quote appears at the end of chapter 14 of Polya’s Mathematical Discovery. Polya attributes the quote to Anatole France from Le jardin d’Epicure. Perhaps he translated the French into English. He further adds: There is a great temptation to paraphrase this passage: “Do not try to satisfy your vanity by teaching high school kids a lot of … just because you wish to make people believer that you understand it yourself …” Yet les us resist temptation.

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Catalan numbers

I really enjoyed reading Federico Ardila’s article in the Mathematical Intelligencer. Apparently there was a vote of 3030 members at an assembly of CUP (Not Cambridge University Press but the Candidatura d’Unitat Popular). The vote had to do with forming an alliance with another party and ultimately related to the independence of Catalonia. The amazing thing that happened was that the vote came out 1515 Yes and 1515 No.

The probability that a YES-NO vote of 2m persons ends up in a tie is [tex] \binom{2m}{m}/2^{2m} [/tex], closely related to the Catalan number [tex] \binom{2m}{m}/(m+1) [/tex]. I love the dual connections and of course Ardila did not fail to mention Stanley’s Enumerative Combinatorics. What I did not know was that Stanley even included a joke. Made my day.

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Truncatable Primes

A colleague asked about sequences of primes a(n) such that a(n+1) is obtained by appending a single digit (in base 10) to the right of a(n).
For example: 3, 31, 311 …

Some thinking lead to the conjecture that such sequences are of finite length and that it is possible to use an exhaustive search to find all of them. A natural question would be what is the longest possible sequence but I was unable to find any conclusive answer on the web. So I decided to write a simple (and not very efficient) recursion in maple to search for all such primes. Here’s my ugly code:

cat3prime:= proc(n)
local d, s, i; s:=n; d:=irem(n,10);
if isprime(s) then print(s); return(cat3prime(10*s+1));
else for i from d to 7 by 2 do
if isprime(s-d+i+2) then s:=s-d+i+2; print(s); return cat3prime(10*s+1); fi; od;
while (irem(s,10)=9) do s:=(s-9)/10; od;
if s=0 then return print(“search complete”);
else return cat3prime(s+2); fi; fi;
end proc;

The search yielded five sequences of length 8 and no other longer sequences:

Only one of the above sequences appears as is on OEIS but with a more careful search, I found a sequence called right-truncatable primes.


Which contains 83 primes, where successively truncating one digit from the right still results in a prime. I verified that my own maple search also yielded 83 primes and I guess the two lists must be identical. See the wikipedia entry on truncatable primes https://en.wikipedia.org/wiki/Truncatable_prime

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Twin corrections

Today is the 20th anniversary of the passing of Erdős and I would like to make two corrections. I had always thought the accent on Erdős’ name was ö , html code &#246 but it is actually Hungarian, html code &#337. The second is the coffee quote which I had attributed to him. I realised my mistake a number of years ago but never got a chance to correct it online. Both errors were perpetuated in this post from 2004. Here is a quote from Erdős’ paper “Child Prodigies”

In Hungary, many mathematicians drink strong coffee, in fact Rényi once said “a mathematician is a machine which turns coffee into theorems.”

Correction done but sadly I am still not quite sure how to pronounce his name correctly.

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Bollobas on solving problems

What you should be terrified of is a blank sheet in front of you after having thought about a problem for a little while. If after a session your wastepaper basket is full of notes of failed attempts, you may still be doing very well. Avoid pedestrian approaches, but always be happy to put in work. In particular, doing the simplest cases of a problem is unlikely to be a waste of time and may well turn out to be very useful.

Bela Bollobas, from Advice to Young Mathematicians

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Logical order

The most efficient logical order for a subject is usually different from the best psychological order in which to learn it

William Thurston from his book Three dimensional geometry and topology.

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One-seventh ellipse

Fun fact. It is well known that
[tex] \frac{1}{7} = 0.\overline{142857} [/tex].

It turns out that if the repeating digits are taken in sequence as (x,y) pairs in the following manner to form six points:
(1,4), (4,2), (2,8), (8,5), (5,7), (7,1),
then these six points actually lie on an ellipse defined by the following equation.
[tex] 19x^2+36xy+41y^2-333x-531y+1638=0 [/tex]

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Lakatos on Discovery

Discovery does not go up or down, but follows a zig-zag path: prodded by counterexamples, it moves from the naïve conjecture to the premises and the turns back again to delete the naïve conjecture and replace it by the theorem. Naïve conjecture and counterexamples do not appear in the fully fledged deductive structure: the zig-zag of discovery cannot be discerned in the end-product.

From his Proofs and Refutations.

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Notes from ICME13

Gila Hanna mentioned the carpet proof of the irrationality of [tex]\sqrt{2}[/tex]. A little digging reveals that it was due to Tennenbaum (1950s) and popularized by Conway (1990s). The original proof appeared in a book but the simple idea is described in this paper “Picturing Irrationality” by Steven Miller and David Montague in the Mathematics Magazine (2012). What is more interesting is the expanded version of their paper on arXiv called “Irrationality from the Book”. One can only infer the amount of editing and changes that took place from submission to publication.

This is a quote from Guershon Harel that I am trying to piece together from memory.

Some of us are constructing open neighbourhoods* but these neighbourhoods are in no way compact.

By neighbourhoods, he was referring to the impact of mathematics education on the curriculum.

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Logic joke

Three logicians walk into a bar. The bartender says:”Do you all want a drink?”
First logician says “I don’t know.”
Second says “I don’t know.”
Third says “Yes!”.

Meta: Why is this a joke?

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Hypergeometric functions

Hypergeometric functions are one of the paradises of nineteenth century mathematics that remain unknown to mathematicians of our day. Hypergeometric functions of several variables are an even better paradise: they will soon crop up in just about everything — Gian Carlo Rota

From p. 231 of Indiscrete Thoughts.

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Freeman Dyson

on how mathematics is permanent while physics is ephemeral.

it’s the beauty of mathematics, as opposed to physics, that it’s forever. I published my selected papers recently in one volume, and I found out that when you publish your selected papers most of the physics is ephemeral, that you don’t want to publish stuff that was written 10 or 20 years earlier, but the mathematics is permanent. So essentially everything I’ve ever published in mathematics is there, whereas only about a quarter of what I published on physics was worth preserving.

The quote is taken from a interview from 1998 via Web of Stories. Watch the video directly here. By the way, thumbs-up for providing transcripts. Great site!

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Halmos on how to study

It’s been said before and often, but it cannot be overemphasized : study actively. Don’t just read it; fight it! Ask your own question, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis?

Paul Halmos – I want to be a Mathematician: An Automathography (1985), p.69.

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What you have been obliged to discover by yourself leaves a path in your
mind which you can use again when the need arises. — G. C. Lichtenberg

I am not familiar with Lichtenberg but he was apparently a German physictists from the 1742 to 1799.

Added quote from Project Euler, the context is sharing one’s solution.

Real learning is an active process and seeing how it is done is a long way from experiencing that epiphany of discovery. Please do not deny others what you have so richly valued yourself.

Added quote attributed to Plato, not quite about the magic of discovery but somewhat related.

Do not train a child to learn by force or harshness; but direct them to it by what amuses their minds, so that you may be better able to discover with accuracy the peculiar bent of the genius of each.

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COMC problem of the week

A nice problem from the 2015 archive filed under week 9, dated 27th October 2015.

Define the function [tex]t(n)[/tex] on the nonnegative integers by [tex]t(0)=t(1)=0, t(2)=1,[/tex] and for [tex]n>2[/tex] let [tex]t(n)[/tex] be the smallest positive integer which does not divide [tex]n[/tex]. Let [tex]T(n)=t(t(t(n))).[/tex] Find the value of [tex]S[/tex] if [tex]S=T(1)+T(2)+T(3)+⋯+T(2006)[/tex].

My answer is 1171. Here’s my solution.
It is easy to see that [tex] t(n)=2 [/tex] for odd [tex] n \ge 3 [/tex]. Correspondingly, [tex] T(n)=0 [/tex]. Also [tex]T(2)=0[/tex] so we just need to sum over all even numbers from 4 to 2006.
Claim: [tex] t(n) =p^k [/tex], for some prime.
Pf of claim: Suppose [tex] t(n)= ab , \gcd(a,b)=1, a, b > 1[/tex]. Since [tex] a < ab \Rightarrow a \mid n [/tex]. Likewise we have [tex] b\mid n \Rightarrow ab \mid n [/tex] which is a contradiction.
Now if [tex] t(n) [/tex] is odd, [tex] t(t(n))=2 \Rightarrow T(n)=1 [/tex]. On the other hand if [tex] t(n) = 2^k, k > 1[/tex], [tex] t(t(n))=3 \Rightarrow T(n)=2 [/tex].

So each term in the sum contributes either 1 or 2. Now [tex] t(n)= 16 [/tex] is impossible as we would have [tex] 5, 7, 11, 13 \mid n \Rightarrow n \ge 5005 [/tex].
In the case where [tex] t(n)= 8 [/tex] we have [tex] 4, 3, 5, 7 \mid n \Rightarrow n = 420(2k+1) [/tex]. Within the range of 4 to 2006, the only possible candidates are 420 and 1260.
Next for [tex] t(n)= 4 [/tex] we have [tex] 2, 3 \mid n \Rightarrow n = 6(2k+1) [/tex] and there are 167 choices in the range.
Finally, there are 1002 even numbers with a positive contribution to the sum, so
[tex] S= 1002 + 167 + 2 = 1171 [/tex].

A verification using mathematica

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A Sunday Morning Geometry Problem

It’s 6.15 am on a Sunday morning, 18th October to be exact. The newspapers have not been delivered and the kids are still sleeping. Some peace and quiet for me to do some reading while sipping my morning coffee … hmmm … here’s a nice problem*.

In the diagram, the three sides of a triangle have been split into thirds, then the lines have been added. What is the area of the smaller shaded triangle as a fraction of the area of the large triangle?

This looks familiar. Now, where have I seen it before? I don’t recall an immediate solution or approach to this. Let’s try to understand the problem. I need to draw. A cursory search on the table and floor revealed a piece of paper and a purple coloured magic marker. From the way the question is asked and previous experience with problems of this sort, the proportion is an invariant. Can I draw a special triangle that helps to solve the problem immediately? An equilateral triangle? A right-angled triangle? I cannot seem to control the shape of the inner triangle. It looks like another approach is needed.

Clearly triangle BCD** has 1/3 of the area, as do the other pieces ACE and BFA. Ah … inclusion and exclusion. If I take away these three pieces I would get the shaded portion but I’m double counting. This would mean that the shaded area = sum of the areas of the 3 small triangles at the vertices. Is that right? Let me check that again. … Yes, that’s right. Now let’s label the areas.

Hmm… how to proceed? … Can’t seem to get anywhere? … I wonder what if I draw another three lines joining the vertices with the other dots G, H and I. Would I get something nice here? I should. It looks like I would get a hexagon. Ok… must focus on solving the simpler triangle problem first.

Uh-oh, I hear the door opening, the boy must be awake. I remember reading that Euler could write mathematics with kids on his lap and/or running circles around him. But he is Euler. He could even write mathematics half-blind. Then again, Euler didn’t have my kids.

Good Morning, Dad!

Ok, time to put this problem on hold.
It’s half an hour to noon. The kids are happy painting away in the living room and I steal some time to return a couple of emails. That ratio has been on my mind the whole morning even as I was playing and watching cartoons with the monkeys. What is that number? Since the computer is switched on, time to call in the heavy artillery — geogebra. I plotted a general triangle. Dividing the sides into three equal parts turns out to be not trivial. An easy way out is to use the distance function and move the points around until the ratio is 1:3. Geogebra automatically calculates the areas of the two triangles and I can compute the ratio directly in the input box to get the number 0.14. Hmm … looks weird. In rational terms that would be 7/50. Can this be right? How to split the triangles into 50 parts? Anyway I distorted the triangle and as expected the ratio remains the same 0.14. Hmm … . Oh time for lunch. I’ll leave the computer on and maybe return to it later.
It’s 3.30 in the afternoon. The girl is napping but the boy, as usual, cannot sleep and is playing with lego. Time to work on my geometry problem again. The purple marker and the sketch from the morning is still on the table. Let me make a neater sketch which may help me spot something that I missed. … Yes! I can draw an auxiliary line and the area will be 2X. Doing this for Y and Z, I now have 4 (X+Y+Z) with 3 more regions to compute. The 7/50 puzzles me, I cannot see how the fractions will sum to 7/50.

But it looks like I have enough the solve the problem. The unknown regions at the lower left and right have area A/3 -3Z-Y and A/3 – 3X-Z respectively. Now by focusing on the correct inner triangle (i.e. AKB in Diagram 1) we have
[tex] (X+Y+Z)+ (\frac{A}{3}-3Z-Y) +2Z = 2 ( \frac{A}{3} -3X-Z +Z) [/tex]
[tex] \Rightarrow 7X = \frac{A}{3}. [/tex]

By symmetry (in this case algebraic symmetry in the sense of the same computations), we must have X=Y=Z. So in fact
[tex] \frac{(X+Y+Z)}{A}=\frac{3X}{A} = \frac{1}{7} [/tex].
So I was deceived by geogebra. The ratio was not 0.14 but 0.142857 … which was rounded to two decimal place. There is a lesson here about using computers/software and not paying attention to the assumptions and possible pitfalls.

Ok problem solved. Very nice. There is the natural generalization of whether the problem has an invariant solution for other ratios besides 1:2 but that is for another day. The girl has woken up from her nap and it’s playtime again.
Monday, 19th Oct, 10.30 am. Lunch with colleagues. Between the few of us, we rummaged up a pen and some useless piece of paper and I explained the problem. D quickly recognized it as a problem he has previously solved, took over the paper and furiously calculated away. E quickly worked out the relation depicted in diagram 2. While D computed on, we mused about generalisations and whether adaptations to arbitrary quadrilaterals can be solved and by and by the conversation drifted to another topic.
12 noon. D came into my office and presented the following generalization. Given a triangle of size 1, with one side divided into the ratio 1:t and another into the ratio 1:s, where we only require t, s to be positive. By drawing the auxiliary line from F, the area A (and hence B) can always be computed.

We have the following pair of equations:
[tex] (t+1)A+B=\frac{1}{1+s}[/tex]
[tex] tA + (s+1)B = \frac{t}{1+t} [/tex]
The system can be solved if and only if
[tex] \frac{t+1}{t} \ne \frac{1}{1+s} [/tex].
The above is clear since the left side is strictly greater than 1 while the right is strictly less than 1. Let me check by substituting t=s=2, … Yes, A= 1/21.
20th Oct, Tuesday morning at the office. Perhaps it would be helpful to record down the thoughts and processes that went through my head over the past two days while thinking about the problem on my own and with colleagues. Writing is an integral part of discovery because it allows one to rethink the problem, consolidate the understanding, check the solutions and possibly further generalize — what Polya calls Look Back. … In both of the solutions, the drawing of the auxiliary line seems to be critical. As an experiment, let me try to label the three unknown regions in Diagram 2 as L, M, N. Assuming the total area is 1, I now have six variables and I can form 7 linear equations, for example of the form
[tex] X +L + Z =\frac{1}{3} [/tex]
[tex] M + Y+ (X+Y+Z)+ N =\frac{2}{3} [/tex]
[tex] X+Y+Z+L+M + N + (X+Y+Z) =1 [/tex].
Let’s see if this system can be solved. Oops, that last one there is just the sum of the previous two. Well I still have six equations with six variables. Instead of doing it by hand, I use matlab. … Ok, it does not work. The matrix has rank 3. … Back to writing things down. … Hey in D’s general solution, although the system is consistent would it be possible to end up with non-positive solutions? Ok, might as well try to get a closed form for A and B. … Uh-oh, time to go for a meeting. Have to put this aside again.
Tuesday 6.10 pm. At the classroom with the part-time Masters students. I just passed out quiz 4 and the students are working on it. Good. I have some time to continue with my calculations … The algebra is slightly messy but the end result is nice.
[tex] A = \frac{1}{(1+t)(1+s+st)}, B= \frac{st}{(1+s)(1+s+st)} [/tex]
Quite neat. Better check again for s=t=2. Yes, it agrees. How about s=t=1? A=B=1/6. That’s right, since 3 medians of a triangle are concurrent and results in six triangles of equal area. … 20 minutes have passed, time to put this aside again and get ready to collect the quiz and continue with the lesson.
Wednesday 21 Oct, 7.40 am at the office. No classes today, maybe I can finish up my writing before embarking on the other boring admin duties. I’m pretty sure the closed form I obtained was right, but let us substitute it into the equation to be sure. … Yes, the solutions check out. … Oh, I still have my hexagon generalization unresolved. Let’s try. … Hey, the general formula still works!

Using the ratios t=1, s=1/2, area of the three red triangles are all equal to A=2/15.
Similarly, using the ratios t=1/2, s=2, the area of the three green triangles are all equal to sB=1/6. Hence the hexagon has area = 1-1/2-2/5 = 1/10. Just to be sure, let me use geogebra to check on a general triangle.

I’m almost ready to move on to other things, but maybe it is a good time to do a search on google. The search terms “area triangle 1/7″ look promising. … As expected the problem is well known. The problem is sometimes called Feynman’s Triangle after the physicist but it certainly dates back further. In fact, there is a general theorem called Routh’s theorem. In diagram 4, assuming the third side is partitioned into the ratio 1:r, then the inclusion-exclusion argument would give the area of the middle triangle as
[tex] 1- \frac{1}{1+t} -\frac{1}{1+s} – \frac{1}{1+r} + \frac{1}{(1+t)(1+s+st)} + \frac{1}{(1+s)(1+r+rs)}+ \frac{1}{(1+r)(1+t+tr)} [/tex]
[tex] = \frac{(rst-1)^2}{(1+s+st)(1+r+rs)(1+t+tr)} [/tex].
In the above, I used Maple to simplify the algebra. Wikipedia has a very good write-up on the theorem which appears in Routh’s text from 1891, together with links to the corresponding pages of the text. The Feynman connection is also mentioned and many other references including several ingenious proofs of the 1/7 result can be found by following the various links.

At this point, 21st Oct 12:19pm, the narrative caught up with the writing in real time and I think I should stop here.

*I saw the problem in Issue 1 of chalkdustmagazine.
** There were no labellings in the original diagram and they were included for clarity.

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Byss and Abyss

Saw this quote attributed to Winston Churchill. A search on google books revealed that it came from chapter III of his autobiography, My Early Life: 1874-1904.

I had a feeling once about Mathematics, that I saw it all — Depth beyond depth was revealed to me — the Byss and the Abyss. I saw, as one might see the transit of Venus — or even the Lord Mayor’s Show, a quantity passing through infinity and changing its sign from plus to minus. I saw exactly how it happened and why the tergiversation was inevitable: and how the one step involved all the others. . . . But it was after dinner and I let it go!

A little digging into the context and the book seems to be in public domain. So here goes the entire bits leading to the above paragraph.



IT took me three tries to pass into Sandhurst. There were five subjects, of which Mathematics, Latin and English were obligatory, and I chose in addition French and Chemistry. In this hand I held only a pair of Kings — English and Chemistry. Nothing less than three would open the jackpot. I had to find another useful card. Latin I could not learn. I had a rooted prejudice which seemed to close my mind against it. Two thousand marks were given for Latin. I might perhaps get 400! French was interesting but rather tricky, and difficult to learn in England. So there remained only Mathematics. After the first Examination was over, when one surveyed the battlefield, it was evident that the war could not be won without another army being brought into the line. Mathematics was the only resource available. I turned to them — I turned on them — in desperation. All my life from time to time I have had to get up disagreeable subjects at short notice, but I consider my triumph, moral and technical, was in learning Mathematics in six months. At the first of these three ordeals I got no more than 500 marks out of 2,500 for Mathematics. At the second I got nearly 2,000. I owe this achievement not only to my own `back-to-the-wall’ resolution — for which no credit is too great; but to the very kindly interest taken in my case by a much respected Harrow master, Mr. C. H. P. Mayo. He convinced me that Mathematics was not a hopeless bog of nonsense, and that there were meanings and rhythms behind the comical hieroglyphics; and that I was not incapable of catching glimpses of some of these.

Of course what I call Mathematics is only what the Civil Service Commissioners expected you to know to pass a very rudimentary examination. I suppose that to those who enjoy this peculiar gift, Senior Wranglers and the like, the waters in which I swam must seem only a duck-puddle compared to the Atlantic Ocean. Nevertheless, when I plunged in, I was soon out of my depth. When I look back upon those care-laden months, their prominent features rise from the abyss of memory. Of course I had progressed far beyond Vulgar Fractions and the Decimal System. We were arrived in an `Alice-in-Wonderland’ world, at the portals of which stood `A Quadratic Equation.’ This with a strange grimace pointed the way to the Theory of Indices, which again handed on the intruder to the full rigours of the Binomial Theorem. Further dim chambers lighted by sullen, sulphurous fires were reputed to contain a dragon called the `Differential Calculus’. But this monster was beyond the bounds appointed by the Civil Service Commissioners who regulated this stage of Pilgrim’s heavy journey. We turned aside, not indeed to the uplands of the Delectable Mountains, but into a strange corridor of things like anagrams and acrostics called Sines, Cosines and Tangents. Apparently they were very important, especially when multiplied by each other, or by themselves! They had also this merit — you could learn many of their evolutions off by heart. There was a question in my third and last Examination about these Cosines and Tangents in a highly square-rooted condition which must have been decisive upon the whole of my after life. It was a problem. But luckily I had seen its ugly face only a few days before and recognised it at first sight.

I have never met any of these creatures since. With my third and successful examination they passed away like the phantasmagoria of a fevered dream. I am assured that they are most helpful in engineering, astronomy and things like that. It is very important to build bridges and canals and to comprehend all the stresses and potentialities of matter, to say nothing of counting all the stars and even universes and measuring how far off they are, and foretelling eclipses, the arrival of comets and such like. I am very glad there are quite a number of people born with a gift and a liking for all of this ; like great chess-players who play sixteen games at once blindfold and die quite soon of epilepsy. Serve them right! I hope the Mathematicians, however, are well rewarded. I promise never to blackleg their profession nor take the bread out of their mouths.

I had a feeling once about Mathematics, that I saw it all — Depth beyond depth was revealed to me — the Byss and the Abyss. I saw, as one might see the transit of Venus or even the Lord Mayor’s Show, a quantity passing through infinity and changing its sign from plus to minus. I saw exactly how it happened and why the tergiversation was inevitable: and how the one step involved all the others. It was like politics. But it was after dinner and. I let it go!

The practical point is that if this aged, weary-souled Civil Service Commissioner had not asked this particular question about these Cosines or Tangents in their squared or even cubed condition, which I happened to have learned scarcely a week before, not one of the subsequent chapters of this book would ever have been written. I might have gone into the Church and preached orthodox sermons in a spirit of audacious contradiction to the age. I might have gone into the City and made a fortune. I might have resorted to the Colonies, or ‘Dominions’ as they are now called, in the hopes of pleasing, or at least placating them; and thus had, a la Lindsay Gordon or Cecil Rhodes, a lurid career. I might even have gravitated to the Bar, and persons might have been hanged through my defence who now
nurse their guilty secrets with complacency. Anyhow the whole of my life would have been altered, and that I suppose would have altered a great many other lives, which in their turn, and so on. …

But here we seem to be getting back to mathematics, which I quitted for ever in the year 1894. …

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