Tuesday, June 11, 2013

I was asked a middle school physics homework problem by a "bear kid".
Jim was holding one end of a plank, and the other end of the plank was placed on a cylinder as shown in the figure (OK, let's not focus on why he had to do that). Jim pushed this plank forwards, and the cylinder began rolling and moving.  The length of the plank was L, the radius of the cylinder is r, and in the whole process the cylinder was clinging to both the floor and the plank so that there's no slip at all. Question: how far did Jim move when the position P reached the cylinder?
This problem suddenly reminds me another "rolling wheel" : two same round coins gear into each other, fix one, let the other roll around it for a complete revolution. Then the question is, for how many cycles has the rolling coin rotated?

Are you able to think out these two problems just by imagination?

I find it difficult. Many people find it difficult, too. If the answers easily come to your mind, perhaps you are very familiar to epitrochoids and hypotrochoids, or you have met these problems before, or, congratulations, you do have a powerful physics engine in your mind. Mm, this tweet won't discuss too much about math, but something that may be more interesting -- why do many people feel difficult?

Answers to the two problems

Before posting my guess, first putting the answers would be good.

The key point to the rolling wheel problems is "how much distance the wheel rolled, how far the center of the wheel has gone", i.e. when the wheel rotates for a $\theta$ angle (it can be larger than $2\pi$), the center of the wheel moves $r\theta$ far.
So, for the first problem, when the plank is $s$ distance ahead the cylinder, the cylinder must have rolled $s$ distance, and then at that moment the center will be $s$ distance away from the original point -- as a result, the plank will be moved $2s$ in total. Thus, the final answer should be $2L$.

Some of you may get it now ... but my imagination is poor, so I have to illustrate it for myself.
It's more clear now. And for the second problem,  to finish a revolution, the center of the rolling coin has to move around a larger circle with a doubled radius, so it will rotate for $$\frac{2\pi(2r)}{2 \pi r} = 2$$ cycles. Again, an illustration.

Interesting psychological effect

Something I find interesting is, if we don't roll a wheel in our mind, but a square, it will be much more easier. I guess the cause is we can easily find an anchor on a square, as a square is "angular".
We render this process, rolling a square, much more precisely and clearly in our mind. 

"Yes, the square moves forward s distance, and the plank, 2s distance. No doubt." You may state it firmly, full of self-confidence, even without seeing the illustration above.

This will happen again when you imagine a square rolling around another square with the same size.
You can easily (at least more easily than thinking a rolling circle) get it: the rolling square (the blue one) rotates for two cycles when it comes back to the original place.

How amazingly the anchor helps our imagination!

However, the anchoring, is considered as a cognitive bias in many psychological researches. Maybe you may argue they are not the same concept, but at least they look alike, and very probably have the same cause. Or alternatively, my examples are the cause of the anchoring effect -- the "anchors" help humans understand how some nature works, so it becomes a part in our intuitive physics engines. Anyway, this conjecture needs to be examined by well-designed experiments.

Rolling squares are the same as rolling wheels?

This section is for those who are interested in math. To some extent, yes, and in fact rolling all kinds of regular polygons are similar to rolling circles.
Let $Q$ be the center of a regular $n$-sided polygon, $MO$ and $OP$ be two adjacent sides, so $QMOP$ is a part of the whole polygon. When the polygon rotates forward for one move, $Q$ will reach $Q'$, and $P$ will reach $P'$. The length of one side of the polygon is $s$, and the circumradius is $r$. $\overline{QQ'}$ is the displacement of the center in one move, denoted by $x_2$; and $\overset{\frown}{QQ'}$ is the trajectory of the center in one move, denoted by $x_1$.

It's easy to prove that $\bigtriangleup QOQ'$, $\bigtriangleup Q'OP'$, $\bigtriangleup QMO$, and $\bigtriangleup QOP$  are congruent, and a useless but interesting fact, point $P$ just lies on $Q'P'$.

Thus, the trajectory and the displacement are respectively $$\begin{align}x_1 & = r \theta = \frac{2\pi}{n} r \\ x_2 & = s = 2r \sin \frac{\theta}{2} = 2r \sin \frac{\pi}{n} \text{.}\\ \end{align}$$
When the polygon rotates for one cycle, the total length of the trajectory of the center is $nx_1 = 2\pi r$ and the total displacement of the center is $n x_2 = 2rn \sin (\pi/n)$.

We can see that, for $n \rightarrow \infty$, i.e. the case of the polygon becoming a circle, the trajectory length keeps $2\pi r$, and the displacement is then $$2r \lim_{n \rightarrow \infty} n \sin \frac{\pi}{n} = 2 \pi r \text{.}$$ The limit can be calculated by Taylor series or by L'Hôpital's rule (after making $n \sin (\pi/n)$ to its continuous case $x \sin (\pi / x)$).

So, if we have to imagine a rolling wheel, we can first try to imagine a rolling square instead, as there are anchors on a square!

Sunday, June 9, 2013

So, why is 0.999... equal to 1? This wiki page has given an answer:  in fact, 0.999... and 1 are different representations of the same number. It seems not intuitive at all: 0.999... does look less than 1 to a very very tiny extent. Why can they be exactly equal?

A Primary School Level Proof

Few kids would doubt the correctness of this equation:
$$ \frac{1}{3}=0.333... $$
If we multiply 3 on both sides of the equal sign, we will get a surprising result:
$$\frac{1}{3} \times 3 = 0.333... \times 3 ~~\Rightarrow~~ 1=0.999...$$
Is there any flaw in this proof? Well, maybe inquisitive kids still have questions, but many of them would admit this proof. Every step is perfect, no cheating, no dividing zero. However, very possibly, they will still consider it to be weird, though they cannot tell why.
Where is the problem? Let's recall the reason for thinking 0.999...<1:
(Wrong Impression) There are countless 9s in the tail of 0.999..., when we write down one more 9, 0.999... will come a little closer to 1, but no matter how many 9s we have written down, 0.999... cannot exactly reach 1, so 0.999... should be throughout a little less than 1.
Wait! Is there any difference in the 0.333... case?! No doubt, 1/3 is the exact value that we equally divide 1 into three parts and take one of them. We could have the same logic as the 0.999... case:
(Wrong Impression) There are countless 3s in the tail of 0.333..., when we write down one more 3, 0.333... will come a little closer to 1/3, but no matter how many 3s we have written down, 0.333... cannot exactly reach 1/3, so 0.333... should be throughout a little less than 1/3.
Both of them are wrong impressions like Zeno's paradoxes. However, the previous one has a more serious impact in our minds. This is probably due to psychology problems:
  1. We can easily get 0.333... by doing a long division 1/3, but we cannot experience getting a 0.999... result by a long division.
  2. The image of 1 is more clear than 1/3 in our minds. We can directly compare 1 with 0.999... by considering 1 as its decimal expression 1.000....
In conclusion, proving 1=0.999... by 1/3=0.333... is circular reasoning. What we can obtain is nothing more than psychological self-satisfaction. What makes us confused in the 1=0.999... case actually persists in the 1/3=0.333... case, so it is illogical to treat 1/3=0.333... as a known condition.

You can also find some similar proofs. One of them is like a trapping game:
$$\begin{align} \frac{1}{9} & = 0.111... \\ \frac{2}{9} & = 0.222... \\ \frac{3}{9} & = 0.333... \\ & \vdots \\ \frac{8}{9} & = 0.888... \\ \frac{9}{9} & = \text{?} \end{align}$$
Mm, it seems, if we want to rigorously prove this equation, we have to deal with some "even more basic" concepts.

Another Primary School Level Proof

One magical trick we are taught in our primary schools, is converting a decimal number into its fractional expression (of course, only for rational numbers). For example,
$$\begin{align}x & = 0.121212... \\ 100x & = 12.121212... \\ 100x-x & = 12.121212... - 0.121212... \\ 99x & = 12 \\ \therefore x & = \frac{12}{99} \\ \end{align}$$
Have we missed something? Yes we have. Kids will ask, when they do multiplication to a finite decimal number
$$0.1212 \times 100 = 12.12 \text{,}$$
and then compare it with the original number
$$\begin{align}12 & . 1200 \\ 0 & . 1212\end{align} \text{,}$$
finally they will find that the decimal part cannot be canceled off. If intuitively (but wrongly) applying this phenomenon to infinite decimal numbers, wow, 12.121212...-0.121212... isn't precisely equal to 12!

"There will be a 00 at the infinite end," says a little boy.
$$\begin{align} 12 & .121212...121200 \\ 0 & .121212...121212\end{align}$$
The result seems a very very little less than 12. Is that true? Of course not. But we cannot discuss why here, with the limitation of primary school math knowledge. That's a college topic. So sorry kids, let's forget the rigor first, and see how 0.999... can be converted into a fraction.
$$\begin{align}x & = 0.999... \\ 10x & = 9.999... \\ 10x-x & = 9.999... - 0.999... \\ 9x & = 9 \\ \therefore x & = 1 \\ \end{align}$$
The result is just 1. This proof is perhaps the most rigorous proof at the primary school level -- it avoids circular reasoning -- but still, it lacks explanation for why 9.999...-0.999... is exactly 9.

Middle School Level Proof

So what is a decimal fraction? Wiki answers it again, and that's what we are taught in middle schools. A real number can be expressed as the sum of an infinite series
$$b_0 . b_1 b_2 b_3 ... := \sum\limits_{i=0}^{\infty} b_i \times 10^{-i} \text{.}$$
where $b_o \in \mathbb{Z}$, $b_i \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ for other $i$-s.

Unlike assuming some "known facts" in primary schools, more basic concepts are concerned at this level, e.g. the meaning of the decimal fraction expression. Once we know that, we are able to check the value of 0.999... just by the definition, i.e., an infinite series.

Yes, that's a geometric series.

Time for tricks again. The trick for calculating the sum of a geometric sequence is very similar to that "decimal to fraction" one.
$$\begin{align}S & = a_0 + \color{blue}{a_0 \times r^1 + a_0 \times r^2 + \cdots + a_0 \times r^{n-1}} \\ r \times S & = \color{blue}{a_0 \times r^1 + a_0 \times r^2 + \cdots + a_0 \times r^{n-1}} + a_0 \times r^n \\ r \times S - S & = a_0 \times r^n - a_0 \\ \therefore S & = \frac{a_0 \left( r^n - 1 \right)}{r-1} \\ \end{align}$$
When $|r|<1$, we can obtain the result of the geometric series by
$$\lim_{n \rightarrow \infty} \frac{a_0 \left( r^n - 1\right)}{r-1} = \frac{a_0}{1-r} \text{.}$$
Mm, it seems a perfect, elegant result that can explain why 0.999...=1
$$S = 0.999... = 0.9 + 0.9 \times 10^{-1} + 0.9 \times 10^{-2} + \cdots = \frac{0.9}{1-10^{-1}} = 1 \text{.}$$
Is that the whole story, and the world rests in harmony again? No, it's still far from the satisfaction for everyone. Compared with the primary school proofs, this version hasn't got much more superiority, but just left more deeper problems:
  • WTF is a real number?
  • What is a limit? Why can we apply it to calculate the sum of that infinitive sequence?
It's a long journey we come here, but let's move on, for the glory of rigor!

A College Level Proof

The world is cruel.

Although humans have known real numbers and used them in many ways for a long long time, the question "what is a real number" cannot be rigorously answered until the late 19th century. (What a shame!) At that time, two kinds of definitions of real numbers were developed. One is utilizing convergent rational sequences to define a real number, by Cantor, Méray, and Weierstrass; the other is utilizing a Dedekind cut to define a real number, by Dedekind (as the name has suggested). They are equivalent to each other, so we only discuss Cantor et al.'s definition below.

"What is a real number?" If we consider this question carefully, we may find that it's kind of philosophic: real numbers have already existed before human redefining them in the 19th century. So, are the "former real numbers" and the "current real numbers" the same thing?

In fact, the rational numbers are earlier defined in Euclid's Elements (but Euclid didn't call them rational numbers at that time). The modern statement for the definition of rational numbers is:
Any number that can be expressed as the quotient or fraction p/q of two integers, with the denominator q not equal to zero, is a real number.
And before the existence of the rigorous real number definition, the irrational numbers are simply considered as the not-rational numbers in the real numbers. Thus, neither the irrational number, nor the real number, is clear. In the bad old days, if someone had asserted 0.999... is not a real number, without an unambiguous definition, who could reasonably point out he/she was wrong?

So, the relationship between the "former real numbers" and the "current real numbers", is

  • the numbers considered as real numbers in the past, are real numbers according to the current definition, so theories based on the "former real numbers" won't collide with the new definition;
  • the numbers that we couldn't rigorously prove as real numbers in the past, can be examined by the current definition now.
At last, the real number field have had a specific, clear, and concrete "border". And this time, a simple RTFD (read the fxxking definition) would elegantly answer the original question, why 0.999...=1.

Oracles Cantor, Méray, and Weierstrass said, "Let there be a definition of real numbers", and there was a definition of real numbers.
Definition (real numbers): let $a_1$, $a_2$, ..., $a_n$, ..., be rational numbers. If for any positive rational number $\varepsilon$, there exists a natural number $N$, such that when $n,m \geq N$, the inequality $$\left| a_n - a_m \right| < \varepsilon$$ holds true. $\{a_n\}$ is then called a convergent sequence of rational numbers. If for an arbitrary positive rational number $\varepsilon$, there exists a natural number $N$, such that when $n \geq N$, two convergent sequences of rational numbers, $\{a_n\}$ and $\{b_n\}$, keep the relation $$\left| a_n - b_n\right| < \varepsilon$$ true, then we say $\{a_n\}$ is equal to $\{b_n\}$, and it is written as $\{a_n\}=\{b_n\}$. A convergent sequence of rational numbers is defined as a real number. And we provide that, equal convergent sequences of rational numbers are the same real number.
2 And Oracles saw that the definition of the convergent sequences of rational numbers was good. And Oracles proved the results of addition and multiplication of the convergent sequences of rational numbers are still convergent sequences of rational numbers.
Theorems for convergent sequences of rational numbers: 
  • If $\{a_n\}$ and $\{b_n\}$ are convergent sequences of rational numbers, then $\{a_n + b_n\}$ and $\{a_n b_n\}$ will be convergent sequences of rational numbers.
  • If $\{a_n\}$, $\{c_n\}$, $\{b_n\}$ and $\{d_n\}$ are convergent sequences of rational numbers, and $\{a_n\} = \{c_n\}$ and $\{b_n\} = \{d_n\}$, then $\{a_n + b_n\} = \{ c_n + d_n \}$ and $\{a_n b_n\} = \{c_n d_n\}$.
3 And Oracles said, "Let there be operations for real numbers", and it was so.
Definition (addition and multiplication of real numbers): let $a=\{a_n\}$ and $b=\{b_n\}$ are two real numbers, say the real number $\{a_n+b_n\}$ is the sum of $a$ and $b$, denoted by $a+b$; say the real number $\{a_n b_n\}$ is the product of $a$ and $b$, denoted by $ab$.
And Oracles saw it was good, and proved the rules of addition and multiplication operations. And Oracles made the zero element, the negative element, and the multiplicative inverse element. And there was subtraction and division.

4 And Oracles said, "real numbers should have orders." And Oracles made the definition of the order of real numbers.
Definition (the order of real numbers): let $a=\{a_n\}$ and $b=\{b_n\}$ be two real numbers. If there is a positive rational number $\delta$ and a natural number $N$, such that when $n \geq N$, $$a_n - b_n > \delta$$ holds true, then  say $b$ is less than $a$, denoted by $b<a$; or say $a$ is greater than $b$, denoted by $a>b$.
5 And Oracles said, "Let these definitions and theorems bring forth more theorems." And it was so.
Theorem (relation of real numbers): if $a$ and $b$ are two real numbers, among the relations $a=b$, $a<b$ and $a>b$, there must be one and only one holding true.
Theorem (real numbers are dense): if $a$ and $b$ are two real numbers, $a<b$, then there must be a rational number $c$ satisfying $a<c<b$.

Note that, a definition is not like a theorem. A definition is an axiom, is the root of the logic chain. Cantor et al.'s definition of real numbers, is based on the definition of the rational number (Oracle Euclid has already given a rigorous definition of rational numbers), and then, says that a convergent sequence of real numbers is a real number

"WTF?! One number can be defined by one sequence?" You may ask.

Yes. Although we can imagine a real number is the limit where the sequence converges today, however, that's taking the effect for the cause -- we haven't got the definition of the limit yet! Without the rigorous definition of real numbers, the limit of a real sequence cannot be rigorously defined.
Definition (the limit of a real sequence): let $\{a_n\}$ be a real sequence, if a real number $a$ satisfying: for an arbitrary positive real number $\varepsilon$, there exists a natural number $N$, such that when $n \geq N$, $$| a_n - a | < \varepsilon$$ holds true, then say the real sequence $\{a_n\}$ converges at $a$, denoted by $$\lim_{n \rightarrow \infty} a_n = a \text{.}$$
Now, the logic chain is clear:

  • First, Euclid defined rational numbers.
  • Second, Cantor et al. defined real numbers.
  • Third, the operations of real numbers are defined.
  • Fourth, the order of real numbers is defined.
  • In the end, the limit of a real sequence is defined.

Confusion is totally clear now.

Now, we can prove 0.999...=1 by some higher level theorems, e.g. Theorem (real numbers are dense):
You just cannot find a real number between 0.999... and 1, so, they are the same real number.
However, it is also easy to prove 0.999...=1 from the original definition of real numbers, i.e. to prove the two convergent sequences of rational numbers $$\begin{align} a & = \{ a_n = 0 . \underbrace{99...9}_{n ~~ 9s} \} \\ b & = \{ b_n = 1 . \underbrace{00...0}_{n ~~ 0s} \} \\ \end{align}$$ are [different representations of] the same real number. Once an arbitrary $\varepsilon$ is chosen, we can select the natural number $$N = \left\lceil \log \frac{1}{\varepsilon} \right\rceil + 1 \text{,}$$ and then when $n \geq N$, $$|a_n - b_n| = \frac{1}{10^n} < \varepsilon$$ holds true. So, undoubtedly, $\{a_n\}=\{b_n\}$, i.e. $a=b$. Without the rigorous basic definition, such a proof can hardly be rigorous.

We can even go further. Wiki mentions there is "non-uniqueness of decimal representation". Besides the 999-tailed style, we can also construct convergent sequences of rational numbers like these: $$\{ a_n = 0 . \underbrace{99...9}_{n ~~ 9s} r_1 r_2 ... r_K\}$$ $$\{ b_n = 1 . \underbrace{00...0}_{n ~~ 0s} r_1 r_2 ... r_K\}$$ where $r_1 r_2 ... r_K$ is an arbitrary finite length "tail number" with arbitrary value. Such $\{a_n\}$ and $\{b_n\}$ are the same as 1, too.

So, 0.999... is nothing special at all! It is only one water drop in the vast ocean of decimal representations that are equal to 1.