Wednesday, November 17, 2010

Freshman Journal 1.1.R2

USP Chatterbox
The hell was I thinking when I said no? USP is like some heaven on earth!

Revision Round-up
Made understandable: L'Hopital's Rule, Fourier coefficients, Green/Stokes/Gauss Theorems
What still needs to be looked into: Taylor coefficients, Hessian determinant, Vector dot/cross products

Done: Flowcharts for Newtonian Mechanics + Thermodynamics
Made understandable: Adiabatic free expansion (!!!!!)
I don't need to take out notes anymore AHAHAHA

Memorise stuff.

Mathematical Thinking

Monday, November 15, 2010

Szymborska's Utopia

GEK1517 Tutorial 8 / Question 1: Discuss the metaphors and associations used in the following poem 'Utopia' by Ms. Wisława Szymborska (Nobel Laureate in Literature, 199[6]).

Ms. Wisława Symborska

I am sorry to find out, after a brief background search on Ms. Symborska, that her life displayed all the dweeb-patterns as seen in Jaromil, the Milan Kundera character I love to deride so much. This is not a very good poem; it makes an awkward reading, but the mathematical ideas expressed makes it relevant to the module and especially interesting.

This is reason enough for me to offer a free-rambling intepretation, so if you'll indulge me for one post--

Island where all becomes clear.
1. The island is Mathematics. Welcome to the world of Maths!

Solid ground beneath your feet.
2. "Solid ground" undefined terms, axioms of mathematics(?)
3. "Solid ground" everything else is groundless (?)

The only roads are those that offer access.
4. The only questions asked are constructive
aside: There are roads that "don't offer access"?!

Bushes bend beneath the weight of proofs.
5: Proofs open up new paths through ignorance (foliage)

The Tree of Valid Supposition grows here
with branches disentangled since time immemorial.

6: This vaguely refers to speculation, or intelligent guesses (?)

The Tree of Understanding, dazzlingly straight and simple,
sprouts by the spring called Now I Get It
7: This vaguely refers to the "eureka" moment as cherished by all bona fide nerds.

The thicker the woods, the vaster the vista:
the Valley of Obviously.

8: At first I thought this was stating the obvious, but then
9: Imagine looking across the valley, specifically at the trees on the other side. You can see all the features there very clearly, but you have not found a way to get to the other side! For that, you will need some "[bush bending] beneath the weight of proofs."

If any doubts arise, the wind dispels them instantly.
10: Possibility 1. The atmosphere discourages you from thinking critically.
11: Possibility 2. In this Utopia, nothing stays unknown or unresolved for long.

Echoes stir unsummoned
and eagerly explain all the secrets of the worlds.

12: Because you get all these dead mathematicians whispering stuff into your ears.

On the right a cave where Meaning lies.

On the left the Lake of Deep Conviction.
Truth breaks from the bottom and bobs to the surface.

13: Conclusions comes only sporadically. Once it is freed from the lake bottom, it bobs quite readily to the surface. Maybe this means that the first step is always the hardest.

Unshakable Confidence towers over the valley.
Its peak offers an excellent view on the Essence of Things.

14: Possibly uncharacteristic praise of dogmatism.
Corollary to 14: Possibly, after all this doubting and questioning and decosntruction of knowledge, being a good mathematician still takes a definite serving of balls and tough braincases.

For all its charms, the island is uninhabited,
and the faint footprints scattered on its beaches
turn without exception to the sea.

15: No one makes mathematics his/her whole reality.

As if all you can do here is to leave
and plunge, never to return, into the depths.

16: Corollary to (15).

Into unfathomable life.
17: Classical dweeb sign-off!
18: Real life is always where we end up in and what we are more used to, even though it is more complicated and much harder to understand than mathematics.


Please comment if you have some better way of deciphering the above poem!

Saturday, November 13, 2010

Turtle Crawl

GEK1517 Tutorial 6 / Question 2.
Consider the turtle's movement:
F means advancing 1 unit in the firection of the turtle's head:
+ / - means, respectively, turning the turtle's head to the right / left by d degrees - the position of the turtle remains unchanged.
Take d = 90˚and start with F + F + F + F (0[th] iteration). In the subsequent steps, let the rewriting rule be
F ↦ F - F + F
(a) Sketch the first two iterations.
(b) It is observed that in the first few iterations, the turtle always returns to the same position where it starts. Does this hold for all iterations? Present the key ideas critically and carefully to support your answers.

Key Ideas:
1: As given, Iteration 0 produces a closed loop.

2: (1) applies because in Iteration 0, F is reproduced 4 times, then joined together so that each F is rotated 90 degrees to the right w.r.t. the previous F. This forms a closed square loop.

3: In every subsequent iteration, the instances of F produce identical segments.
F + F + F + F
↦ F' + F' + F' + F'
↦ F" + F" + F" + F" ...

4. In every subsequent iteration, we find 4 identical segments joined end on end, each one rotated 90 degrees right w.r.t the previous segment. This also forms a closed loop.
Fn + Fn + Fn + Fn

4. If (4) is shown to be true, proposition holds for all iterations
Summary: (1) is given, (1) requires (2), (2)⇒(4) proven, ⇒ proposition holds


Iterations 3 and 4 for your enjoyment

Fibonacci's Rabbits

GEK1517 Tutorial 6 / Question 1.
Starting with the letter b, consider the rewriting rules
b → a; a → ab.
For example, the first few 'words' formed by the above rewriting rules are
b, a, ab, aba, abaab, abaababa, ...
Let C(n) be the number of letters in the n-th word (n is a positive integer). For instances [sic],
C(1) = 1, C(2) = 1, C(3) = 2, C(4) = 3, C(5) = 5, C(6) = 8, ...
Prove that
C(n+1) = C(n) + C(n-1)
for all integers n ≥ 2. (Caution: Math. induction may not work well here.)

Spoiler: Lecturer's solution is more elegant, so please bear with me here.

Let number of 'b's in n-th word: Bn.
Let number of 'a's in n-th word: An.

Because of the rewriting rules b → a; a → ab.
'b's in a word is generated only from 'a's in the previous word
∴ Bn = An-1
'a's in a word is generated from both 'a's and 'b's in the previous word.
∴ An = An-1 + Bn-1

Now total number of letters in n-th word is C(n)
Total number of letters in n-th word is sum of 'a's and 'b's.
∴ C(n) = An + Bn
C(n-1) = An-1 + Bn-1 = An
C(n+1) = An+1 + Bn+1

To show that C(n+1) = C(n) + C(n-1), express all terms in An and Bn.
C(n+1) = C(n) + C(n-1)
⇒ An+1 + Bn+1 (LHS)
= An + Bn + An = 2An + Bn (RHS)
∵ Bn = An-1,
Bn+1 = An.
∵ An = An-1 + Bn-1,
An+1 = An + Bn.
Sub back into LHS:
LHS = An + An + Bn
= 2An + Bn = RHS.


Friday, November 12, 2010


Self-respecting intellectuals listen up:
If someone says your ideas are "chim", do not take it as a compliment.
It is an INSULT.
Go back home and sleep on a nettle bed.
Hang a snake gall from the ceiling or something to remind yourself of the shame you had to bear.
Later on, make a comeback by presenting your ideas more clearly and intuitively.

Somewhat Random Quotables

It is only in the popular mind that knowledge equates with facts.
That is of course flattering, since facts are incontrovertible.
But it is also demeaning, since facts are meaningless;
they contain no narrative.

-- POLANYI John Charles

The idea that there is one people in possession to the truth,
one answer to the world's ills,
or one solution to humanity's needs,
has done untold harm throughout history, especially in the last century.

-- ANNAN Kofi (2001)

The certainty of mathematics is different from the certainty one finds in other fields; it is somehow purer and therefore more powerful. To be certain is to know. What an extraordinary feeling it is to know and to know that you know! What strength there is in that position! What confidence one has when one sees into the truth of some mathematical theorem! The angles of a triangle add up to two right angles. It is not a matter of "probably," "almost always," "in our experience," or "as far as we know." None of these usual caveats apply -- there is no quibbling. If you are not completely certain, if you have the slightest doubt, then you just don't get it. Mathematical truth has this certainty, this quality of inexorability. This is its essence.

-- BYERS William (2007), How Mathematicians Think
I just love it when mathematicians get emotional.

Friday, November 05, 2010

Theorems of Vector Calculus

Notes on the physical meanings of the theorems of Green, Stokes and Gauss covered in MA1505 Chapter 10

Those who are familiar with the Mathematics module (the one specially designed to traumatise engineering freshmen) would readily identify the following formulas to be the point where it all starts to fall apart (ignoring Fourier series).

At first glance, these formulae all seem to be coincidence, as what can there be to relate integration over a line with integration over a surface, or to relate integration over a surface with that over a volume? The notes give no helpful explanation either... but the underlying idea is quite simple to understand!

This following is a demonstration, not a proof. The goal is understanding, not certainty. If one is interested, there are pretty good proofs for Green's theorem as well as the theorems of Stokes and Gauss (TAYLOR Peter D., 1992).

Special case assumption: When you sum up small loops, they become one big loop.
As seen in this tile of four loops, the interior arrows cancel out and leave the outer boundary of the loop (which has nothing to cancel out with).

We apply this principle everywhere else rather thoughtlessly. Does anyone remember Carnot engines from Physics PC1431? How do you reduce a reversible process to a sum of Carnot cycles?

General assumption: The interior boundaries cancel out.
The important thing is that the R.H.S of the three theorems deal with summing up quantities for which the internal boundaries cancel out each other. This is why summations (integral) over a plane/volume can equate to summations over the plane/volume's outer boundary.
This general assumption is necessary to bring Gauss' Theorem into the discussion.

Stokes:A Loopy Proposition
Aside: In practise, it is easier to use Green's theorem to prove Stokes' Theorem rather than the other way round, for an idea how this is done, please refer to Thomas' Calculus (12th ed.), Chapter 16. However, for this section, only Stokes' theorem will be demonstrated (Green's theorem follows as its 2-D speical case, a fact which you may have learnt in class).

The basic idea of Stoke's theorem is to add up the curl (small loops) of the vector field to a closed-path integral (the big loop).
(1)It can be seen that in a 3-D space, many different surfaces can share the same boundary.

(2)As follows from the assumption, any closed loop can be split up into smaller loops, and then split up into even smaller loops, and so on ad infinitium. The sum of these loops, as expected, equals the big loop i.e. the boundary of the surface. This ensures that even for different surfaces, the integral can be the same if the surfaces share the same boundary!

(3)If the big loop can be thought of as the sum of infinitesimal loops, we can approximate each infinitesimal loop to the curl at a point (normal to the surface).

What is the meaning of curl?
Curl is what textbooks describe as "circulation density". Basically, it is how much vectors swirl in the immediate neighbourhood of a given point. The curl vector points upwards if the net circulation is counter-clockwise from above, and vice versa.

The big last step: Approximating infinitesimal loop to curl
After obtaining your infinitesimal surface dS, it is time to match the 2 quantities
(1) Line integral over the infinitesimal loop
(2) Curl over the point

Admittedly, I have not found a way to settle this step yet, but here are 2 important observations:

(1) The quantities a)line integral over infinitesimal closed loop and b)curl derive their values from the same 4 vectors in the neighbourhood of small surface area dS.

(2) Only the components of a) and b) aligned with the surface of dS are taken:

Gauss: Analogy using Jigsaw Pieces

The cutest formula in the bunch also looks the most devastating, not least for its prominent feature of a triple integral. However, the fact that features a divergence operator rather than a curl operator makes it infinity times friendlier to the brain.

Basically, what is says is:
The volume flow rate over a closed surface
is equal to
The net divergence in the volume enclosed by the surface.

What is the meaning of divergence?
Divergence at a point is a measure of the net tendency of nearby vectors to point inwards into the point or outwards away from the point.

The cube is a small volume dV in the space enclosed in the surface.
In the jigsaw analogy, we simplify each small volume dV into a jigsaw piece.
The tendency for outflow at one edge, contributing to positive divergence, is represented by a protruding knob.
The tendency of inflow at one edge, contributing to negative divergence, is represented by a hole.

The big step: Cancelling out the interior boundaries
The boundaries must cancel out in between volumes. This is because any vector which contributes to positive divergence in one small volume dV also contributes to the same amount of negative divergence in an adjacent dV, and vice versa.

This is the same as saying in the jigsaw analogy that 2 adjacent jigsaw pieces fit each other smugly.

The "divergence" is found by adding up the knobs (1 per knob) and holes (-1 per hole) along the edge. The "divergence" in the result sum can be thought of as the sum of the "divergences" in the individual jigsaw pieces. It can also be thought as the sum of knobs and holes along the edge of the result (i.e. by giving it a count).

A final observation: The key to understand why Gauss' Theorem works is the same key to understand why a completed puzzle always has straight edges (a particular case of zero divergence) even though all the pieces in it are jaggedy!


Freshman Journal 1.1.12

Maths tutorial this week has disappeared. I don't know if the venue had shifted without warning, but when I arrived there 15 minutes late, there wasn't a class in E1-06-03.
Lecturer went into an exam-question-crunch frenzy, which for all his virutosity is totally useless for revising concepts.
I am into Stokes' Theorem at the moment, partly to distract myself from the existential horror of never being able to get a girlfriend, and partly because it is awesome.
What secrets doest thou hide? O impenetrable formula!

Completed the Mastering Physics questions. Considering to finish the non-compulsory questions on the site along with those in the tutorials, but I hate doing stuff on the computer.
I realise that our Physics tutor (Ms. Qiu Leiju) is one of the best tutors I've had for the semester.
Syllabus sped in and out of Thermodynamics for the past weeks. The flame of physics was rekindled in me at the point when it went to the Second Law. Holding that disorderliness is ever-increasing and that all energy is eventually useless, this law is a sobering reminder of the mortality of humankind. Who knew that such a profound result could arise from the obscure fumblings of some 19th-century nerds?

The last lessons were concluded with 2 sessions of final presentations, a group photograph, and some wicked delicious cookies baked for us by our tutor, Mr Griner.

The whole syllabus of CM1417 is so random and counterintuitive that I find it quite unsettling that I will have to face the same sort of courses for all my subsequent semesters.

Monday, November 01, 2010

Cracking Tiles

GEK1517 Tutorial Three / Question Two: Suppose that there are m×n unit square tiles arranged into a rectangle (a square if m=n; here m and n are positive integers). A crack appears as a diagonal line of the rectangle (see the figure for m=3 and n=5. Speculate on the number of tiles that are ruined and need to be replaced. (Your answers should be in terms of m and n). Provide arguments to support your answers.

Step I: Panic

Step II: After calming down, try the shit out of as many lower values of m and n you could be bothered with.

Assume that the number of tiles follow a rule; nail the rule down to a symbol.
Number of tiles damaged = D(m,n)

Make the following observations which may or may not come useful later on.

m and n are interchangeable.

Extreme case I: Thin rectangle

Extreme case II: Square

After a bit of trying, arrive at the formula
D(m,n) = m + n - 1
which works for the largest number of the cases (including Extreme I cases)
Group cases according to how well they conform to D = m + n - 1 (mirror images are ruled out as trivial)

It should be very clear now what differentiates the two groups.
The left group ("conformist" rectangles) have cracked tiles of continguous shape. The cracks do not pass through a point in between tiles.
The right group ("nonconformist" rectangles) have cracks that pass through a point (or points).
Somehow, passing through a point affects the crack count!

For curiosity's sake, have a closer look at the nonconformist group.

The nonconforming rectangles do not follow a single rule! Evidently, something has to be found out about this group that dictates the D(m,n).

Step III: Observe that every crack starts and ends on a point.

Observe that, if the crack passes through a point in a middle of a rectangle, you can break a smaller rectangle from it with that middle point as a new (top-right or bottom-left) corner. We take bottom-left:

Observe that, if the new rectangle also has the crack passing through a point between tiles, the process can be repeated until there is no point.

A rectangle where the crack does not pass through a point is a conformist rectangle.
We have demonstrated that a nonconformist rectangle can be reduced to a conformist rectangle.

How do you get back the nonconformist rectangle?
Every conformist rectangle, a series of nonconformist rectangles can be constructed by replicating the conformist x times along the crack's diagonal:

What is the D for a rectangle of length xm and height xn?
It is quite clear that this quantity D for a nonconformist rectangle can be taken apart into the sum of cracks for x conformist rectangles.

We arrive at
D(xm,xn) = xD(m,n)
for nonconformists, where D(m,n) is of a conformist rectangle
Which, owing to D(m,n) = m + n - 1, becomes
D(xm,xn) = xD(m,n) = x(m + n - 1) = xm + xn - x

Step IV:
Observe that in nonconformist rectangles (D(xm,xn) = xm + xn - x) width and height have a common factor x
Observe that in conformist rectangles (D(m,n) = m + n - 1) the width and height have no common factors other than 1.
Observe that D(m,n) = m + n - 1 is equivalent to D(1m,1n) = 1m + 1n - 1, which is just a special case for the nonconformists' rule (x = 1).
Observe that x is the highest common factor (h.c.f) between the width xm and height xn; since m and n have no common factors, the common factors include x and possibly a number smaller than x which is also a factor of x (x is still the highest).

Obtain the unified rule for all rectangles of width m and height n:
D(m,n) = m + n - (h.c.f of m and n)
Verify that rule applies to all cases presented above.


Addendum: Ruohan's description of a proof
Set A {(a, an/m), where 0 ≤ a < m}.
Set B {(bm/n, b), where 0 ≤ b < n}.
Each point in A and B correspond to a cracked tile to the right of the point itself. The answer would then be A ∪ B = A + B - (A ∩ B).
Solving A ∩ B would then be solving the integer pairs for an = bm.

Lacuna: I asserted that a nonconformist rectangle be formed by duplicating the same conformist along the crack. I did not explain why a different conformist can not be used. To do so, I may slip into a circular argument.

Spoiler: Lecturer touched on how to count the cracked tiles, and did not arrive at the expression m + n - 1 by accident like I did.