Download Mathematica 2014

MATHEMATICA
2014
1
DC MATHEMATICA 2014
EDITORS
Xiaofeng Xu
Leslie Leung
SUPERVISOR
Dr Purchase
COVER DESIGNED BY
Kenza Wilks
ARTICLE CONTRIBUTORS
Kaichun Liu
Charlie Sparkes
Mitchell Simmonds
Xiaofeng Xu
Leslie Leung
Ayman D’Souza
Faraz Taheri
Mr Ottewill
Harry Goodhew
Dr Purchase
Kit George
2
Table of contents
What is e?
By Kaichun Liu
A-Level Statistics in Investment Banking and Portfolio Theory
By Charlie Sparkes
Mathemagics
By Mitchell Simmonds
How to trap the prisoners?
By Xiaofeng Xu
An Introduction to Modular Arithmetic and its Applications
By Leslie Leung
Proof by Induction or Contradiction
By Ayman D’Souza
Cavalieri’s Principle
By Faraz Taheri
Where’s the policeman?
By Mr Ottewill
Hilbert's 6th Problem
By Harry Goodhew
BMO question – two equations, three unknowns
By Kit George
How calculators calculate
By Dr Purchase
New Mathematics for an Old Problem
By Mr Ottewill
3
What is e?
By Kaichun Liu
Before writing something really related to the topic, I would like to draw your
attention to π first.
What is π? If I ask my little sister who is 12 years old what π is she can definitely
tell me that π is a constant and has the value of 3.1415926…… I have no idea
how many decimal places she can count to but I am pretty sure it will be enough
for calculations.
However, what is the definition of π,or what does π actually represent? The
question is also easy and I will answer for my little sister – it is a mathematical
constant that is the ratio of a circle's circumference to its diameter.
All right, what if I replace π with e and ask you the same question? If you can
recall its value to the nineth decimal place immediately please text Dr. Purchase
and see if he is impressed. However, it is not enough, so far, you have only got 1
mark. WHAT IS THE DEFINATION? WHAT DOES e STAND FOR?!!!
To be honest, I myself actually had no idea. Thus I asked help from the mighty
Wikipedia.
Wikipedia generously provided me with a beautifully organized definition -- The
mathematical constant e is the base of the natural logarithm.
“ What is natural logarithm then?”
The natural logarithm, formerly known as the hyperbolic logarithm, is the logarithm to the base e,
where e is an irrational constant approximately equal to 2.718281828459.
Did you really tell me
anything?
So what is e?
Let us take a look at this example first.
Let’s assume that a cell will split into two in a day, so mathematically, after a day,
we will get two new cells. If we need to write a formulae of how many cells we
will get after x days, that will be 2
However, according to cell biology, each 12 hours, which the split is halfway
through, there are a number of new cells which equals the total number of the
original cells have been produced and have started splitting into next generation.
In fact, the growth speeds up as time goes—although the rate of splitting (the
ratio) remains at 100%, the base number is continuously growing, which makes
4
the 100% arithmetically bigger that the 100% a moment ago.
Thus a day can be divided into two parts, within each part the
total number of cells grows by 50%.
100% 2.25
1
2
In fact, if there are 100 cells at first, after 24 hours, we can get 225 cells.
If we can chage the rule of splitting and make it “producing 1/3 of the cells after
8 hours ”
The formulae would be
100% 1
2.37037 …
3
If we make the split carry on forever and continuesly, the formulae would be
100% ! 2.718281828
lim 1 “Coincidently”,the number we get here just equals the “e”
When the rate of increasing stays at 100%, within a unit of time the cells can only
increase to 271.8%
In fact e shows up whenever systems grow exponentially and continuously: population,
radioactive decay, interest calculations, and moreover it
represents the idea that all continually growing systems are scaled versions of a common
rate.
e shows up whenever systems grow exponentially and continuously: population,
radioactive decay, interest calculations, and more.
5
A-Level Statistics in Investment Banking
and Portfolio Theory
By Charlie Sparkes
Investment banking is often made out to be like an impossible, unpredictable
science that only those audacious enough to wear pin striped suits can ever
fathom. However, as you may know, a crucial part of an investment banker’s day
to day regime requires a lot of pondering over the magical relationship between
risk and return and how to optimize the performance of a portfolio based on past
and predicted data. Or, in S1 terms, this is the relationship between expected
return (E(R) - return), standard deviation (S - risk) and the correlation coefficients
(PMCC – just a coefficient in an equation).
So, as this is a Maths essay, here is the inevitable set of data around which my
explanations will revolve1.
% Return
Probability
Security A
Security B
0.25
20
45
0.50
10
25
0.25
0
5
Total = 1.00
Total = 30
Total = 75
E(RA) = 10%
E(RB) = 25%
Var(RA) = 50%
Var(RB) = 200%
SA = 7.07%
SB = 14.14%
Notation: R = % return, WA + WB = weight of portfolio invested in respective
securities = 1 (in total), P = % return on portfolio of A and B.
To calculate (a rather generalized) return on investments, using the data we can
derive that;
E(RP) = WAE(RA) + WBE(RB) – i.e the weighted average return on A and B.
So if the investor split his or her investment equally in A and B then, using the
data;
E(RP) = 0.5(10) + 0.5(25)
= 17.5% expected return on investment in the portfolio.
Although I may have made some sweeping assumptions here, portfolio theory
(and the law of large numbers) does suggest that for some overcompensating
discrepancies that have been made, there will be some other discrepancies to
counter these externalities, so the calculated expected returns value isn’t far off
1
An Introduction To The Stock Exchange – Janette Rutterford
6
the truth if you have a very large portfolio.
Risk on the other hand is very hard to quantify since it very much depends on
future events that even Nostradamus would struggle to predict, for example, very
few people categorically predicted the financial crash of 2007 which was why it hit
so many financial institutions, households, businesses (so basically everyone) hard.
However, the ability to assess the volatility of listed assets using past data to
perhaps try to model future predictions is an extremely important ability, for
example, a hedge fund manager called John Paulson made $3.7 billion during the
financial crisis2 as he saw the weakness in the subprime mortgage market first and
short sold every appropriate share/collateralized debt obligation (CDO) he could.
He’s now worth $11.5 billion3.
So, now I’ve established being very good at Maths can make you very rich, it’s
back to the calculations for risk or, in S1 lingo, the variance. The variance of any
security can be written as;
Vi = Si2 = ∑ x% P X x% ( E R = ∑%./
%.0 +R ,% ( E R , - P R ,% From this, we can sub in the equation WAA + WBB = P and then expand the
brackets to give;
VP = SP2 = WA2SA2 + WB2SB2 +
2WAWB ∑%./
%.0 +R 1% ( E R 1 -+R 2% ( E R 2 -p R % As you may have noticed, the final part of this long equation is actually the equation
for the covariance of returns for A and B (COVAB). The covariance is the absolute
value that measures how one variable changes with respect to another, where
positive values indicate that as one variable increases, so does the other variable, and
of course the opposite applies for a negative value. The product moment
correlation coefficient, as seen in S1, is simply a version of the covariance (so we call
the covariance the determinant of the PMCC) that is scaled by the product of the
standard deviations of A and B to give the PMCC a value that is always within the
parameter;
(1 4 PMCC 4 1
Thus, we can say that;
COV12 PMCC12 S1 S2 = ∑%./
%.0 +R 1% ( E R 1 -+R 2% ( E R 2 -p R % And subbing this into the equation for the variance of your portfolio gives;
VP = SP2 = WA2SA2 + WB2SB2 + 2WAWBSASBPMCCAB
This is known in the financial industry as the ‘Portfolio Risk Equation’.
The PMCC is now the main player in determining the risk of investing in these
two securities, therefore, by using the data from the table on page 1 we can deduce
that;
2
3
The Long Short – Michael Lewis
http://en.wikipedia.org/wiki/John_Paulson
7
PMCC =
∑ <: ∑ =: >
∑ <: @
∑ =: @
@
√ ∑ 1: ;
∑ 2@: ;
>
>
∑ 1: 2: ;
=
ABB
+√BB- √CBB
=1
(indicates a perfect positive correlation)
Therefore;
VP = SP2 = WA2SA2 + WB2SB2 + 2WAWBSASB
Simplified;
VP = SP2 = (WASA + WBSB)2
So;
SP = WASA + WBSB
Or to put in it simply, if the returns on A and B are perfectly positively correlated,
the risk of the portfolio return (in terms of standard deviation) is just the
weighted average of the risks of the constituent assets’ returns. At first
impressions then, this may seem to be a positive outcome as the risk is
quantifiable and linear and as is the expected returns. Thus, you can model your
risk (x-axis) against return (y-axis) as a uniform straight line and make informed
decisions on where to invest.
However, it is because the two securities have a PMCC of 1 that means investing
in these two assets simultaneously brings the same reward as investing all your
money in just A or B, hence, we can say that no risk is eliminated and that all this
maths has reaped no reward.
Or, we could look at the flip side, investment bankers may be interested in buying
two stocks that have a PMCC of -1, i.e. as one increases in value, the other
decreases in value by the same amount. If we plug PMCC = -1 into the ‘Portfolio
Risk Equation’, we get;
VP = SP2 = WA2SA2 + WB2SB2 - 2WAWBSASB
Simplified;
VP = SP2 = (WASA – WBSB)2
So;
SP = WASA – WBSB
This equation (that is slightly more complex for non-integer negative values of the
PMCC - but nevertheless the theory holds) is very interesting as it implies that risk
can be reduced or even completely nullified and the standard deviation would
therefore be 0.
In fact, for a given set of data (again, let’s use the table on page 1) there is a unique
combination of WA : WB that can be found because if the two securities (with a
PMCC of -1) ‘are combined in inverse proportion to the ratio of the standard
deviations of their respective returns, so the resultant portfolio will yield a constant
return and hence be riskless’1.
For example, using the data;
W1
S2
W2
S1
8
So;
W1
14.14
W2
7.07
And;
WB = 2WA
Since WA + WB =1, then;
WA =
&
WB =
0
And by plugging this into the equation for returns, we get;
0
E(RP) = (10) + (25) = 15%
So, if you have the exact securities A and B as above, and two thirds of your
portfolio consists of A and the rest B, then you will get 15% return with 0% risk.
Smashing. These equations will actually produce curves on a graph of return
against risk, so investors can choose shares with certain levels of risk and return,
with the variables being; S, E(RP), and WA:WB, that they feel will maximise profit
(whilst bearing in mind the relative risk associated with the investment).
To conclude though, I must mention that the chances that past data will perfectly
link to future market fluctuations and hence that the inductive reasoning in the
calculations will give you the undisputed values for risk and return is almost as
close to nil as you can get. Although a famous stock investor called John
Templeton did once say “The four most dangerous words in investment are; ‘this
time it’s different’”, suggesting market trends actually mean everything.
However, many theories that have been formulated over the years in many
different fields of academia have certain annoying parameters that are tricky to get
round. Nevertheless, I feel that the underlying principle here that is really
interesting is that if an investor can combine any number of securities and
combine them in a portfolio, they can expect to get the weighted average of the
returns with less than the weighted average of the risk (given the PMCC is less
than 1 which reduces the value of the portfolio risk equation and hence brings the
value of the standard deviation below that of the returns you get from the same
assets).
I also think it is worth mentioning that these ideas are not restricted to situations
with only two assets. The general equation for return (or expectation in S1) is
simple; you just add a summation sign (Σ) and set the limits between 1 and ‘n’ for
the % return and portion of portfolio taken by asset ‘i’. The equation for risk is as
follows;
%./
%./ G./
%.0
%.0 G./
V, F W% S% F F W% WG S% SG PMCC%G
I leave you with a quote from the late Phillip Fisher, an American stock investor,
9
that I think perfectly encapsulates the justification for my essay;
“The stock market is filled with people who know the price of everything but the
value of nothing”
YOU MAY BE INTERESTED IN:
Division by 11
Here are four ways for different types of numbers:
1. If the sum of every other digit, starting with the first, is equal to the sum
of every other digit starting with the second, then the number is evenly
divisible by 11. Try 13057. 1+0+7 = 3+5, therefore it should divide evenly
by 11. And indeed it does: 13057 / 11 = 1187.
2. If all the digits are the same and there's an even number of digits, then the
number is evenly divisible by 11. 333,333 - Yes. 3,333,333 - No.
3. If the number is a 3 digit number with different digits, add the two outside
digits. If the difference between the sum and the middle digit is 11 then 11
divides evenly into the 3 digit number. If the sum is the same as the middle
digit, then 11 will also divide evenly into the number. Try 484. 4 + 4 = 8
which equals the middle digit so 11 divides into 484 evenly. How about 913?
9 + 3 = 12 and 12 - 1 = 11 so 913 is evenly divisible by 11.
4. If the digits are different, count them from the right and then add the
numbers in the odd positions and the even positions. Subtract the smaller
number from the larger. If the difference is evenly divisible by 11, so is
your original number. Take the number 181,907. The numbers 8,9, and 7
are in the odd positions. They sum to 24. The numbers 1,1, and 0 are in
the even positions. They sum to 2. Subtract 2 from 24 to get 22. 22 divides
by 11 into 2, so 181,907 is evenly divisible by 11.
10
Mathemagics
By Mitchell Simmonds
After performing magic for many years, you learn that magic isn’t simply about
sleight of hand; it also about probability and maths. Whilst many will be familiar
with the standard: ‘Think of a number. Double it. Add 10. Halve it. Take away
your original number. Your answer is five,’ there is more to Mathemagics than that.
In fact, some of the best card tricks invented revolve around Mathematics.
The best way to explain how theory can be used in magic is to reveal a trick itself.
Again, many will be familiar with the 21 card trick in which three rows of seven
cards are laid out and a volunteer picks a card. The volunteer tells the magician
what row their card is in and this is repeated twice further. After this, the magician
can name the card. In my opinion this trick lacks imagination. It is dull, boring
and a waste of the audience’s time. Magic is all about amazing your spectators, and
giving them that ‘Wow!’ moment. The only way to do this is to make it appear that
maths is not part of the magic trick, even though it may be the very basis of the
trick.
Without further ado I will explain, Fitch Cheney’s Five-Card Twist, which I feel is a
perfect example of Mathemagics.
The magician exits the room leaving his assistant with the audience. She gives a full deck of
ordinary playing cards to an audience member and asks him/her to shuffle it and then pick out
any five cards.
The assistant then takes the cards, looks at them and places one card face down and the four
others face up.
The magician is then allowed back in; he takes a look at the cards and in an instant names the
card that is face down!
How is it done? The explanation is simple; the assistant uses a pre-arranged code
with the magician to tell him what the card is. As there are five cards and 4 suits, at
least two of the cards have to be of the same suit. The assistant therefore chooses
one of these cards to be the hidden one and places the others face up.
For example, if there are two hearts, the assistant hides one of these hearts and
the other heart is the first card on the left. Therefore, when the magician sees the
first card, he instantly knows the suit.
Now, the clever part - how does the magician use the other three cards to
determine the value of the card? There are 13 values in a deck of cards. The order
goes as follows: A,2,3,4,5,6,7,8,9,10,J,Q,K. You have to imagine these numbers as
a clock. So after King comes Ace, for example. It is a rule, that if you pick any two
cards, their values can be, at most, six positions apart. So, for example, if you pick
2 and 9, there is 9,10,J,Q,K,A,2- six positions apart.
The lowest card must be revealed, when the assistant turns over one card, so that
the card that is hidden is either 1,2,3,4,5 or 6 values above it. The magician must
then deduce what value it is from the other three cards and count that number
from the value of the first card to gain the value of the hidden card…simple!
11
Ace is low and King is high and the suits also go as follows:
Clubs
Hearts
Spades
Diamonds
Low---------------------------------------------High
So the Ace of Clubs is the lowest card and the King of Diamonds is the highest
card. The assistant must find the Lowest, Middle and Highest ranking cards- let’s
call these L, M and H. The order of these cards determines how many numbers to
add the lowest value card. So,
LMH=1 LHM=2 MLH=3 MHL=4 HLM=5 HML=6
The example therefore is as follows:
The spectator picks out 5 cards. The assistant puts the 3 Clubs on the far left. The other cards
are arranged as the 7 of Hearts, the 3 of Spades and the Ace of Diamonds. After carefully
thinking, the magician says that the card faced down is the 8 of Clubs! Everyone is amazed!
You may have to read it through a couple of times, to understand the trick, but I
hope that this article has changed your mind on Mathemagics and that you can see
quite how complex maths tricks can be.
YOU MAY BE INTERESTED IN:
Division by 6
The number has to be even. If it's not, forget it. Otherwise, add up the digits and
see if the sum is evenly divisible by 3. It it is, the number is evenly divisible by 6.
Try 108,273,288. The digits sum to 39 which divides evenly into 13 by 3, so the
number is evenly divisible by 6. If you want, you can keep adding numbers until
only one digit remains and do the same thing. So in this case, 3 + 9 = 12 and 1 +
2 = 3, and 3 is evenly divisible by 3!
12
How to trap the prisoners?
By Xiaofeng Xu
This is a story I have recently read from a book which draws my great interest.
It is said that once there was a cave an emperor or used to trap the prisoners. The
prisoners tried to find a way to escape but the guards were somehow always one
step ahead of them. The prisoners, paranoid as ever, tried to seek out the traitor
who had leaked out their escape plans, only to find out that it was due to the
shape of the cave.
How does this happen? Because the shape of the
cave is an ellipse, the prisoners are just
ust at one focus
of the ellipse and the guards are located at the other
focus of the ellipse. Therefore, the sound of the
prisoners will just be heard by the guards after
reflection.
Can this be proved by vigorous mathematics proof ? The answer is ‘yes’.
The rule I get from the story is that after reflection, the ray passing through one
focus of an ellipse will then pass through the other focus of the ellipse. This can
be shown by the diagram below.
Proof:
Firstly, after transferring or rotating, any ellipse
ellipse can be shown in the rectangular
coordinate system with its two foci on the x-axis.
x axis. Therefore, by proving that the
ellipse with the function
x 2 y2
+
= 1 ( A is greater than B ) obeys the previous
A2 B 2
rule I have mentioned, I can say that this is right for any single ellipse.
Here is the graph for the model. The line l is the tangent to the ellipse. The ray is
coming from the focus F1. At the point of contact T it reflects and passes through
the other focus F2. The normal line is drawn perpendicular to the tangent through
the point of contact. The incident angle is marked as α , and the reflective angle as
β.
(Picture drawn by Geometer’s Sketchpad)
13
My way to prove β is equal to α is to prove that the complementary angles of
β and α are equal. So two perpendiculars are drawn from the foci to the line l，
with the feet G and H respectively. The problem then obviously turns into
proving that the two triangles are similar.
Here come the details.
x 2 y2
From the equation 2 + 2 = 1 ( A > B ), I can derive the coordinates of the two
A B
foci. Let’s say C 2 = A2 − B 2 , so the coordinates of foci F1 and F2 are ( C , 0) and
( −C , 0) respectively.
Apparently, when y = 0 , the tangents(on the left and right) are parallel with the
y-axis. The ray will just reflect back in the same way as it comes. So it will pass
through both of the points.
When y ≠ 0 , we can get the derivative function of the ellipse by implicit
derivation(1).
dy
B2 x
=− 2
dx
A y
We assume the point of contact T have the coordinate (a, b) .
The gradient of the tangent: gradient = −
B2a
A2 b
So the line l, which is the tangent, has the equation B 2 ax + A2by − A2b2 − B 2 a 2 = 0
The distance between the right focus F1 and point T: F1T =
(a − C )2 + b 2
Similarly, the distance between the left focus F2 and point T: F2T =
(a + C )2 + b 2
Because we have the equation of the tangent and the coordinates of the two foci,
we can calculate the distance from the foci to the line.
This is the formula. The distance from a point (x1，y1) to a line A0 x + B0 y + C0 = 0 :
A0 x1 + B0 y1 + C0
A0 2 + B0 2
(2)
14
So the distance from the right focus to the line l: F1G=
The distance from the left focus to the line l: F2H=
=
B 2 aC − A2b 2 − B 2 a 2
( B 2 a ) 2 + ( A2 b ) 2
− B 2 aC − A2b 2 − B 2 a 2
( B 2 a ) 2 + ( A2 b ) 2
B 2 aC + A2b 2 + B 2 a 2
( B 2 a ) 2 + ( A2b ) 2
Because point T(a , b) is on the ellipse.
a2 b2
B2 2
2
2
2 2
2
2
+
=
1
⇒
B
a
+
A
b
=
A
B
⇒
b
=
B
−
a
A2 B 2
A2
B 2 aC − A2b 2 − B 2 a 2
B 2 aC − A2b 2 − B 2 a 2
( B 2 a ) 2 + ( A2b ) 2
F1G
=
= 2
F2 H
− B 2 aC − A2b 2 − B 2 a 2
B aC + A2b 2 + B 2 a 2
( B 2 a ) 2 + ( A2b ) 2
C
a− A
B aC − A B
A
= 2
=
C
B aC + A2 B 2
a+ A
A
2
2
2
What’s more, an interesting thing can be found that the ratio of F1T and F2T is
the same as the ratio above!
(a − C ) 2 + b 2
FT
1
=
F2T
(a + C ) 2 + b 2
=
a 2 − 2Ca + C 2 + b 2
a 2 + 2Ca + C 2 + b 2
Because point T(a , b) is on the ellipse.
a2 b2
B2 2
2
2
2 2
2
2
+
=
1
⇒
B
a
+
A
b
=
A
B
⇒
b
=
B
−
a
A2 B 2
A2
So,
B2 2
a − 2Ca + C + B − 2 a
A
B2
a 2 + 2Ca + C 2 + B 2 − 2 a 2
A
2
FT
1
=
F2T
2
2
15
=
A2 − B 2 2
a − 2Ca + C 2 + B 2
2
A
2
A − B2 2
a + 2Ca + C 2 + B 2
A2
=
C2 2
a − 2Ca + A2
2
A
C2 2
a + 2Ca + A2
2
A
C
a − A) 2
A
=
C
( a + A) 2
A
C
a− A
A
=
C
a+ A
A
(
FT
FG
1
= 1
and ∠F1GT = ∠F2 HT = 90o
F2T F2 H
So, Triangle F1GT is similar to Triangle F2HT. (3)
⇒ ∠FTG
= ∠F2TH
1
⇒ β =α
All considered, the proposition is now proved!
This proof shows a property of an ellipse, which is of great use in real life.
We can find it in modern medical equipments, where people use it to concentrate
high energy to break the stone in human bodies. It also has connection with sonar
system used to locate submarines. Exciting, isn’t it?
Finally, ellipses have a lot of properties. Other properties can all be proved
mathematically. This is, in my opinion, why mathematics is of great fun and
challenge. You may find the other properties through various ways and have a try
to prove it on your own!
Thank you!
(1) http://en.wikipedia.org/wiki/Implicit_function
(2) http://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line
(3) http://mathforum.org/library/drmath/view/57674.html
(prove triangles similar
16
An Introduction to Modular Arithmetic and
its Applications
By Leslie Leung
Modular arithmetic can be defined as a system of arithmetic for integers.4It is a
mathematical tool that is widely used in daily life as well as solving mathematical
problems.
Explanation of notation:
a I b mod n read as ′a is congruent to b modulo n′
It means that the difference between U and V is a multiple of
, or in other
words, when both U and V give the same remainder when divided by
.
The most common use of modular arithmetic is perhaps the 12-hour clock. A day
has 24 hours but a cycle of the clock lasts only 12 hours. Therefore, when the
time is 21:00, the clock shows 9 o'clock because 21 I 9 mod 12.
XYZ[Y\[ ]^\_Y`ab: deZY Y`f\ ga\_ Ye\ hi ea^[ jkajl _eam
hnn ea^[_ ZoY\[ baab?
Answer: 100 I 4 mod 12, so the clock shows 4 o'clock.
Before moving on to explore how modular arithmetic can be used to solve other
problems, a few 'laws of modular arithmetic' has to be explained.
1) Addition/Subtraction
If a I c mod n, b I d mod n, then a b I c d mod n
Also, a ( b I c ( d mod n
2) Multiplication
If a I c mod n, b I d mod n, then ab I cd mod n
Proof:
Let c a pn, d b qn, where p is an integer.
cd a pn b qn tu aqn bpn pqn
4
http://en.wikipedia.org/wiki/Modular_arithmetic
17
The only term in the expression that is not a multiple of n is UV. Therefore, using
the addition law, cd ab mod n
3) Power
If a I b mod n, then av I bv mod n
Proof:
Let b a pn
bv a pnv = tw +v0- av;0 pn0 ++v- av; pn +......+ pnv
is Ux .
From the binomial expansion, the only term not divisible by
y[azk\f h: de{ `_ Zb `bY\|\[ g`}`_`zk\ z{ ~ `o Zbg abk{ `o `Y_ g`|`Y _^f `_
g`}`_`zk\ z{ ~?
A quick check on a few examples can be shown to be consistent with this rule.
~€ 3x119
Digit sum check:
Digit sum check:
3 5 7 15 3x5
~i 10x3. . .2
3 2 5 3x1 2
This fact amazed me when I first encountered it, but I never understood why it
works until I read about modular arithmetic.
Proof:
Notice that each integer can be broken down into a sum of integers that are
multiples of powers of 10.
~€ 300 50 7
~i 30 2
We can express them individually as multiples of powers of 10.
300 3x10
50 5x10
7 7
Notice that 10 I 1 mod 3.
Using
the
power
law,
10v I 1v mod 3
,
thus
10v
I
18
1 mod 3 for all integers k.
Now, using the multiplication law, we have:
300 3x10 I 3x1 I 3 mod 3
50 5x10 I 5x1 I 5 mod 3
7 7x1 I 7x1 I 7 mod 3
Using the addition law, 357 I 0 mod 3 if and only if the digit sum is divisible
by 3.
Something for you to try:
Can you find any patterns within numbers divisible by 9 or 11 using modular
arithmetic?
y[azk\f i: Xeam YeZY, oa[ \}\[{ ƒa_`Y`}\ `bY\|\[ b,
hihb ( ib h„nnb ( ( b
`_ ZkmZ{_ g`}`_zk\ z{ innn. †[`Y`_e ‡ZYe\fZY`jZk ˆk{fƒ`Zg innn
This problem may look a bit daunting at first sight. However, modular arithmetic
can simplify the problem rather nicely. Taking modulo 2000 directly would lead
you to nowhere, but we can break down 2000 after which modular arithmetic may
be useful.
Solution:
We first express 2000 in its prime factors.
2000 2A x 5
If
the
expression
is
divisible
by
16 and 125 ,
it
is
divisible
by
2000 because 16 and 125 are coprime. This naturally leads us to take modulo
16 and modulo 125.
Using the power law,
121/ I 9/ mod 16
25/ I 9/ mod 16
1900/ I 12/ mod 16
(4/ I 12/ mod 16
The whole expression: 9/ ( 9/ 12/ ( 12/ I 0 mod 16
Similarly,
121/ I 121/ mod 125
25/ I 25/ mod 125
19
1900/ I 25/ mod 125
(4/ I 121/ mod 125
The whole expression: 121/ ( 25/ 25/ ( 121/ I 0 mod 125
We can now reach the conclusion required that the expression is divisible by 2000
for every positive integer n.
YOU MAY BE INTERESTED IN:
Division by 3
Add the number's digits. If the sum is evenly divisible by 3, then so is the number.
So, will 3 divide evenly into 2,169,252? Yes it will, because the sum of the digits is
27, and 27 is divisible by 3. If you want, you can keep adding numbers until one
digit remains. For example, keep going with 27. 2 + 7 = 9, which is also evenly
divisible by 3.
20
Proof by Induction or Contradiction
By Ayman D’Souza
Very often in Mathematics we find ourselves needing to prove something: a
formula, a fact, a theorem. But what do we do when we can no longer use a
simple proof, e.g. 9x must be divisible by 3 as 9x= 3*3*x? Very often, we can
use the method of Proof by Induction, or Proof by Contradiction. In this article I
will explain both methods by employing the use of examples.
Proof by Induction
We are given the arithmetic progression: 1 + 2 + 3 + ..... + n =
0
n (n +1)
If we already know that to find the value of this some we should take the average
and multiply it by the number of terms, the formula
0
n (n +1) makes perfect
sense. But how could prove it by induction?
Let us assume that the arithmetic progression is true for n = k, rewriting the
formula as:
{Formula A}: 1 + 2 + 3 + ..... + k =
0
k (k + 1)
If it is true for all values, we can fill the formula in with k + 1 as well as k, reading:
{Formula B}: 1 + 2 + 3 + ..... + k + (k + 1) =
Since we already know that 1 + 2 + 3 + ..... + k =
rewrite the {Formula B} above as:
0
k (k + 1) + (k + 1) =
0
(
k + 1) (k + 1) =
0
0
0
0
(k + 1) ((k + 1) +1)
k (k + 1) {Formula A}, we can
(k + 1) (k + 2)
(k + 1) (k + 2)
We can manipulate the right hand side of the formula to resemble the left hand side
if we halve (k + 2), and double
0
(k + 1) not affecting the overall value. Since we
have proved that the original statement to prove works for n + 1 (remember n
represents k), it logically works for any integer equal to or larger than one, and so we
have proved the formula to work for any value of n.
Proof by Contradiction
Often, is something is untrue, we can prove this by contradicting it. For example, let
us take the example √2 Š
‹
(assuming that
Š
‹
is cancelled down as far as possible)
which we can contradict to prove the square root of two is an irrational number,
similarly to pi.
21
√2 2=
U
V
Š@
‹@
2V = U
As U is double another number (a and b are integers), it must be even. Therefore,
let us call it 2n. We now have 2V = U , or :
2V = 2 2V = 4
V = 2
This tells us that V is an even number, as it is another number doubled, and even
squares are always the product of the same even number. As both a and b are even,
the fraction
Š
‹
can be further cancelled, but this is a contradiction with the original
statement. Thus, by disproving this statement, we can prove that √2 is irrational.
In conclusion, both methods are very powerful for complicated or simple proofs,
and the logic behind them cannot be questioned or doubted, making them
extremely useful for proving more difficult formulae and theorem.
YOU MAY BE INTERESTED IN:
Division by 4
If the number's last 2 digits are 00 or if they form a 2-digit number evenly
divisible by 4, then number itself is divisible by 4. How about 56,789,000,000?
Last 2 digits are 00, so it's divisible by 4. Try 786,565,544. Last 2 digits, 44, are
divisible by 4 so, yes, the whole number is divisible by 4.
22
Cavalieri’s Principle
By Faraz Taheri
Abstract
Cavalieri’s Principle, named after Bonaventura Cavalieri, is a geometrical method
to find the area or volume of an object on a plane or in a three-space respectively.
1 Introduction
Cavalieri’s Principle, aka method of indivisibles, presumes lines are made of an
infinite number of points, planes are made of an infinite number of lines and
volumes are made up of an infinite number of planes. And therefore it aims to
find the sum of infinite lengths or areas. It basically uses the concept of
integration without using the integral notation.
1.1 Two-dimensional space
Suppose the baselines of two regions on a plane are placed on a single line. If
every line drawn parallel to the baselines, makes 2 line segments of equal length,
the areas of those regions are equal.
23
1.2 Three-dimensional
dimensional space
Suppose the bases of two objects (with equal base area and height) in a space are
placed on a single plane. If every plane parallel to the bases, makes 2 regions of
equal area, the volumes of those objects are equal.
24
1.2.1 How to use the principle
The following
ollowing example illustrates how to use the principle:
Prove that any two pyramids with equal height and base area are of equal volume.
’“  
SP'1
h 2
'
QP ŽP1 ~ P1 Ž
= ! —
SP1
H •
œŠŠžŸ ¡Ÿ
Œ
2 ˜ ™ ′ ™ ′ ¢£££££¤
SP'2
¤ ¥0 ¥
h
š›
š@
QP ŽP'2 ~ P2 Ž
= ! –
S2
H •
SP1 = SP2
”
Let P and Q be parallel planes. The base of both pyramids are placed on Q. By
hypothesis, 0′  0 and ′   . Then we know that 0′ and 0 are similar;
as ′ and 
with the ratio of  ‘ . According to our assumption, SP1 = SP2 . Therefore,
h 2
SP'1 SP'2 . These steps could be repeated for every plane parallel to Q. The
H
25
Cavalieri’s Principle tells us the volumes of these pyramid equal one another.
2 Initial steps
To begin with, we need to define the volume of cube and cuboid.
2.1 Cube
We postulate the volume of a cube is U3, where U is the length of one side.
2.2 Cuboid
Suppose three values a, b, c which are the sides of a cuboid U ¦ V ¦ §. Then
we form a cube with the side length of U V §. According to our axiom, the
volume of this cube would be U V §3 :
U V §3 U3 V3 §3 3UV§ U V §
˜ U V §3 U3 V3 §3
U V3 U §3 V §3 6UV§
U V3 U §3 V §3 6UV§
Then we divide the big cube into 3 cubes which have a3, b3, c3 cubes intersecting.
Therefore, the remaining pieces are 6 cuboids; the volume of each would be
U ¦ V ¦ § . Accordingly, the volume of a cuboid would be : ¨©ª
«© ¦ ¬ª­ .
¦
3 Implementation
In this section we try to work out the volume of some common shapes using the
introduced principle.
3.1 Pyramid
Let’s define “perfect pyramid”. The base of a perfect pyramid is a rectangle and
the height is one of its sides.
26
As the figures above demonstrate, these three perfect pyramids could be put
together to form a cuboid :
The volumes of the pyramids are as follows :
1
¥0 ¦ U ¦ V ¦ §
3
1
¥ ¦ U ¦ V ¦ §
3
27
1
¦U¦V¦§
3
˜ ¥0 ¥ ¥
¥ ¥
Œ ® ¥0 ¥ ¥ ¯ ˜ ¥š°¡Š±Ÿ² 1 ¦ VCuboid 1 ¦ ³©U´Šµ ¦ ¨©ª
¥0 ¥ ¥
3
3
3.2 Cone
Place a cone next to a pyramid with the same base area on a plane.
’“  
S′1
h 2
QP
~ S1 Ž = ! —
S1
H •
œŠŠžŸ ¡Ÿ
Œ
2 ˜ S' S' ¢£££££¤
′
¤ ¥0 ¥
S2
h
1
2
QP ŽS'2 ~ S2 Ž = ! –
S2
H •
”
™0 ™
1
˜ ¥œ¶ ¦ ³©U´Šµ ¦ ¨©ª
3
ŽS'1
28
3.3 Sphere
3.3.1 Auxiliary Principle
The plane P intersects the center of a sphere with the radius of R. The plane Q is
parallel to P with the distance of h. Find the intersection area of Q in terms of R
and r.
3.3.2 Hemisphere
Œ · (
™ ¸ ¹ ˜ ™ ¸ ¦ · (
The image above illustrates a cylinder containing a cone and a hemisphere with
equal radii. The intersecting surface of each plane of the hemisphere equals the
excluding part of the cylinder.
For a random plane passing through the hemisphere, let ·š ·œ°žŸ²
°žŸ²
Then, the auxiliary principle is taken into account :
¡
(
29
™œ°žŸ²
Œ
™œ¶
¸ ¦ º ˜ ™»
1
¦ ¸ 3
¡
±Ÿµ¼½ ¡
2
¦ ¸ ¦ 3
3.3.3 Complete Sphere
Respecting the previous steps, the volume of a sphere would be :
™¾¼½
™¾¼½
¡
¡
™¾¼½
2 ¦ ™»
±Ÿµ¼½ ¡
2
2 ¦ ¦ ¸ ¦ 3
4
¦ ¸ ¦ ¡ 3
4 Further steps
What other volumes can be found by propagating Cavalieri’s Principle? Is it
possible to use this principle to work out the volume of a doughnut? Furthermore,
this technique might help you to solve the “The napkin ring problem” and find
the volumes of Cycloids.
5 Reference
Zill, D. (2009) Calculus: Early Transcendentals, Jones Bartlett Learning.
YOU MAY BE INTERESTED IN:
Division by 7
Multiply the last digit by 2. Subtract this answer from the remaining digits. Is this
number evenly divisible by 7? If it is, then your original number is evenly divisible
by 7. Try 364. 4, the last digit, multiplied by 2 = 8. 36, the remaining digits, minus
8 = 28. The last time I checked, 28 is evenly divisible by 7, and thus, so is 364!
30
Where’s the policeman?
By Mr Ottewill
The curve x 2 y 2 = x 2 + y 2 is said to represent a policeman on traffic duty.
The
graph looks like the following which illustrates the road junction.
y
4
2
-4
-2
0
2
4
x
-2
-4
The question is: where is the policeman?
See below for the answer.
YOU MAY BE INTERESTED IN:
Division by 8
If the number's last 3 digits are 000 or if they form a 3-digit number evenly
divisible by 8, then the number itself is divisible by 8. How about 56,789,000,000?
Last 3 digits are 000, so it's divisible by 8. Try 786,565,120. The last 3 digits, 120,
divide by 8 into 15, so yes, the whole number is divisible by 8.
Answer to ‘Where’s the policeman?’
x 2 y 2 = x 2 + y 2 is also true when x = 0 and y = 0, so the point (0, 0) is on the
graph as an isolated point – this point represents the policeman.
31
Hilbert’s 6th Problem
By Harry Goodhew
In 1900 David Hilbert presented 23 problems which he intended to be the focus
of mathematical thought for the coming century and beyond. Since then 17 of
the problems have been solved although some solutions have not been fully
accepted. The remaining 6 are either too vague to be solved or have resisted
solution so far, the latter include the Riemann hypothesis and the one which I will
be looking at, the 6th problem, the axiomatic treatment of physics.
This problem is somewhat different to the others that Hilbert proposed as it is
not a problem in itself but a general task, which he suggested, following his work
on the construction of geometric axioms. So far it has not been fully solved
despite exhaustive work from many mathematicians including Hilbert himself.
However, some fields within physics have succumbed to such axiomatic treatment,
including Mechanics and Statistics.
In order to advance the discussion of Hilbert’s 6th problem it is first necessary to
define an axiom, it is one of a set of statements, that are assumed to be true, from
which all other true statements within a system of logic can be proved. For
instance from the axioms of addition:
abba
§¿¿À U ªÁª [1]
UÄħªU ªÁª [2]
a bc abc
[3]
a0a
the existence of an identity
[4]
a (a0
the existence of an inverse
It can be proved that if abca, then bc
V 0 V [~] (U U V [ ] (U U V[i] (U § U
(U U § [h] (U U § [ ] 0 § [i] § [~]
■(the bracketed numbers indicate which axiom has been used)
Similarly all other true statements about field addition can be proved from these
axioms.
The usefulness of the existence of an equivalent set of axioms for physics is clear
as it would allow physical facts to be proved logically and provide a framework for
testing whether or not a proposition is true, as well as a way to ensure that the
system did not rely on any assumptions. Furthermore, the nature of such axioms
could reveal facts about the universe itself. Unfortunately, due to our lack of a
complete understanding of physics, epitomised by the inconsistency between
general relativity and the standard model, a solution to this problem remains
elusive.
However, the developments that have been made are worth discussing. In the
years following his presentation of the 23 problems, Hilbert devoted much of his
32
time to the mathematical treatment of physics, most notably general relativity
which has since been axiomatised5. Similar attempts have been made with the
standard model, though this is far from complete. Other physical theories, with a
somewhat longer history have been broken down to this, most basic level, below I
will discuss both mechanics and probability calculus as well as their applications in
more advanced physics.
Probability calculus is again worth defining, as the calculation of the probability
of an event. This study that lies on the border between mathematics and physics
can be expressed as just three, familiar, axioms:
0 4  È 4 1
 ™ 1
 È0 É È É … F  ȟ , ªÊ È0 , È … U© ¿À ÀUËË ©Ì§ËÀĪÁ©
∞
Ÿ.0
These may seem insufficient, to prove all that you have learnt about probability
calculus, however, it has been shown to be consistent and independent, and has
managed to prove all statements currently known to be true, although its
completeness cannot be proved in accordance with Gödel’s incompleteness
theorems. Furthermore, to demonstrate how powerful these seemingly inert
statements can be, I will show how they can prove the inclusion, exclusion
principle.
A/B is deÐined as A Ñ B′ visually:
³ ³/B É A Ñ B, which are mutually exclusive
Ò Ò/A É A Ñ B, which are also mutually exclusive
A/B
AÑB
 ³ + A/B É A Ñ B-  A/B  A Ñ B – 3rd axiom (if
B/A
you
include an infinite number of empty sets following A and B which all have
probability 0, which I have not shown for simplicity)
 Ò + B/A É A Ñ B-  B/A  A Ñ B-3rd axiom
 ³  Ò  A/B  A Ñ B  B/A  A Ñ B
 A/B É A Ñ B É B/A  ³ Ñ B – 3rd axiom
 ³ É Ò  ³ Ñ Ò – Which follows from the definition of A/B
From here, the inclusion, exclusion principle follows inductively.6
■
This system has clearly got a mathematical cleanness, but what about its relevance
5
An Axiom System for General Relativity Complete with respect to Lorentzian Manifolds
6
http://www.proofwiki.org/wiki/Inclusion-Exclusion_Principle
33
to the axiomatic treatment of physics? I will discuss here two of the countless
important extensionsto probability calculus, the kinetic theory of gases and, as
Hilbert called it, “the theory of compensation of errors”, which attempts to
calculate the errors associated with the use of measuring equipment in order to
reduce it. This study comes from any one of three axioms, the other two being
derivable as theorems from the one chosen to be an axiom, these are:
“If various values have been obtained from measuring a certain magnitude, the
most probableactual value of the magnitude is given by the arithmetical average
of the various measurements.
… The frequency of error in measuring a given magnitude is given by:
Õ
Ó © ;Ô
B
… The most probable value of the variables measured is obtained by minimising
the squares of the errors involved in each observation.”7
The fact that these axioms are interchangeable highlights something important
about the nature of axioms, that for a system to be logically coherent the choice
of axioms is arbitrary, which seems to undermine their importance. However, it
merely serves to demonstrate that the proposed axioms are logically equivalent.
Furthermore, in many physical theories those statements chosen to be axiomatic
are those that directly follow from experimental data. The theorems are
predictions that guide future research. This suggests that in many physical systems
this problem does not arise as only one set of axioms will follow directly from the
data. However, when there is a choice it may allow a problem to be simplified by
choosing an alternative equivalent system.
The calculus of probabilities is also relevant to the kinetic theory of gases which
again requires just a few additional assumptions; that gases are made up of point
particles that collide elastically, that their velocities are independent, and that they
are uniformly distributed throughout the volume of space that they occupy. These
three “axioms”8 can be used to derive the average potential energy, kinetic energy
and the mean free path of a molecule as well as the second law of
thermodynamics, and much more besides.
Another of the areas within physics that have been successfully axiomatised is
vector addition, vital for much of physics, as well as some areas that are more
commonly considered mathematics, these axioms are as follows:
³ÄħªU ªÁª Ê U­­ª ª : ^ } m ^ } m
Ö¿¿À U ªÁª Ê U­­ª ª : ^ } } ^
×­© ª ©Ë©¿© Ê U­­ª ª : Ø 0 Ù ¥| } 0 }’ } Ù ¥
× Á©Ä© ©Ë©¿© Ê U­­ª ª : ’ } Ù ¥ Ø (} | } (} n
Ö¿ÛU ªVªËª Ê Ä§UËU ¿ÀË ªÛ˪§U ª : U V} UV}
7
http://tau.ac.il/~corry/publications/articles/pdf/hilbert.pdf
8
I use the apostrophes here to indicate that these are not usually viewed as such but function in a
similar way to axioms
34
×­© ª ©Ë©¿© Ê Ä§UËU ¿ÀË ªÛ˪§U ª : 1} }
ÜªÄ ªVÀ ªÁª Ê Ä§UËU ¿ÀË ªÛ˪§U ª
ª ©ÄÛ©§
} U^ U}
ÜªÄ ªVÀ ªÁª Ê Ä§UËU ¿ÀË ªÛ˪§U ª
ª ©ÄÛ©§
V} U} V}9
Á©§ U­­ª ª : U ^
ʪ©Ë­ U­­ª ª : U
Again, from these axioms, it is possible to deduce a vast amount of physical
knowledge related solely to vector mathematics and, in conjunction with a few
others, can be extended to cover many related studies. For example, the dynamics
of point particles can be deduced from newton’s laws of motion which are almost
axioms:
³ VÝ©§ ªËË § ª À© U § Ä U Á©Ë§ª À Ë©ÄÄ U§ ©­ V© U ʧ©
­¿}
, ª§ Ê § Ä U ¿UÄÄ V©§¿©Ä Þ ¿Z
Þ
­
ÈÁ© U§ ª UÄ U ©ßÀUË U ­ ÛÛĪ © ©U§ ª
However, these are not quite axioms as they are not independent as the first one
can be derived from the second by setting F=0 you get a=0, which means that the
object will continue at constant velocity. Furthermore, the statement that every
action has an equal and opposite reaction comes from the law of conservation of
momentum, which, along with the other conservation laws is generally assumed to
be axiomatic for physics. Therefore, F=ma is the only axiom required, in fact,
along with the definitions of time and energy, as the duration of movement, and
the integral of force with respect to time, respectively, it is sufficient as the basis
of all of classical mechanics.10
I will now advance the discussion to newer theories, firstly, quantum field theory,
our principle theory of particle interactions. This has been described in an
axiomatic form by Arthur Wightman.These axioms are of principle concern to
both mathematicians and physicists, one of the millennium problems11 being the
realisation of these axioms in the case of Yang-Mills fields. However, they require
an understanding of complex mathematics and so stating them will offer very
little.
Similarly, general relativity requires a reasonable knowledge of vector mathematics,
however, it has been summarised by a set of axioms about the equivalence of
both inertial and accelerated observers. This seemingly innocuous statement
actually has a profound impact on physics at a grand scale as it is possible, with
these axioms to show that light has a fixed, finite speed which is the maximum
speed at which anything is capable of moving, a fairly unintuitive fact. Similarly,
these axioms can prove that mass bends space time in such a way that is seen as
gravity to an observer. Unfortunately the mathematics required for this proof was
9
http://en.wikipedia.org/wiki/Vector_space
10
http://www.marinsek.com/files/axiomatic_of_mechanics.pdf
11
Devlin, Keith J. (2003) [2002]. The Millennium Problems: The Seven Greatest Unsolved Mathematical
Puzzles of Our Time. New York: Basic Books
35
beyond even Einstein who required help from Hilbert himself.
These axioms are capable of describing the macroscopic movement of bodies at
the largest scale, and are consistent in themselves. However, the combination of
this and the axioms of quantum field theory does not represent the full
axiomatisation of physics as they are not consistent with each other, because
quantum field theory disrupts space time in such a way that general relativity
breaks down. One of the many proposed theories that attempts to overcome this
problem, is string theory, or more precisely supersymmeric string theory, often
shortened to superstring theory. In this model, all elementary particles are
replaced by one dimensional strings, as this is a continually developing theory it is
not fully reducible to axioms, however, you can find further reading here12about
such an attempt.
I believe that whilst many steps have been taken towards the completion of
Hilbert’s 6th problem, it remains to be seen whether a complete solution will ever
be found and looks likely to provide mathematicians and physicists with work for
many years to come. Indeed, physics, as a branch of mathematics is theoretically
inexhaustible, according to Gödel.
YOU MAY BE INTERESTED IN:
Division by 9
Sum the number's digits. If it divides by 9, you're in luck. As with the tests for 3
and 6, you can keep adding numbers until you're left with only one digit.
12
http://arxiv.org/pdf/hep-th/0011065v1.pdf
36
BMO Question: Two Equations,
Three Variables
By Kit George
xyz4018
x2y2z22yz2
Find all possible integer values of x, y and z.
SOLUTION
x2y2z22yz2
y2z2-2yz2-x2
y-z22-x2
y-z√ 2-x2
∴√ 2-x2 is a real number
∴ 2-x2 is positive
∴x242
∴-√24x4√2
x must be an integer
∴x-1, 0 or 1
∴y-z√1 or √2 1, -1 or √2
y and z are integers
∴y-z1
∴x-1 or 1
xyz4018
yz4018-x
∴yz4019, if x-1, or 4017, if x1
y-z-1
∴If x-1, y2010, z2009
or y2009, z2010
37
(In the original equations, y and z are completely interchangeable and so their
values can be swapped: I only took one factorisation of y2z2-2yz through the
extrapolation, although (z-y2 works in exactly the same way)
If x1, y2009, z2008
or y2008, z2009
YOU MAY BE INTERESTED IN:
Division by 15
If the number can be evenly divided by 3 and 5, the same can also be said for 15.
Use the methods for Division by 3 and Division by 5 above. If they both work,
your number is also evenly divisible by 15.
38
How calculators calculate
By Dr Purchase
Calculators do not compute values of certain functions (e.g. sin 47) by using a
series or polynomial expansion. Instead calculators use an algorithm known
as the CORDIC method (see Edwards et al. [1]). The circuits in calculators
can only perform the following elementary computations:
• additions;
• subtractions;
• comparisons (stating whether a number is positive or negative);
• storage of a finite number of constants;
• digit shifts.
The latter of these, digit shifts, deserves further clarification. The number
system we use in everyday life is known as base 10 and includes the numbers
0,1,2,3,4,5,6,7,8,9.
If we multiply a given number by 10 ± n (where n is a
positive integer) then this will shift a number n places to the left (positive
power) or n places to the right (negative power) keeping the decimal point
fixed.
For
example,
67.98426 × 10 −5 = 0.0006798426 .
3.56386 × 10 3 = 3563.86
or
Calculators (and computers) work in base 2
(binary) which include the numbers 0 and 1 only. To digit shift a binary
number we multiply it by 2 ± n . The above operations are the only ones a
calculator can perform due to hardware restrictions.
The CORDIC algorithm only uses the above operations to calculate values
such as
5
, sin 56,
7
2 , e 3.4 , etc.
The algorithm uses the following iterative
scheme:
x k +1 = x k − mδ k y k 2 − k
y k +1 = y k + δ k x k 2 − k
z k +1 = z k − δ k σ k
39
Where the constants m, δ k and σ k are specified in Table 1 according to
which functions are being computed.
Table 1
Function
m
Multiplication
0
σk
2−k
δk
Initial
constants
sgn(z k )
x0 , z 0
given,
Result
y n ≈ x0 y 0
y0 = 0
Division
0
2−k
− sgn( y k )
x0 , y 0
given,
zn ≈
y0
x0
z0 = 0
Sine
cosine
and
1
tan −1 2 − k
sgn(z k )
x0 = A, z0 = θ ,
x n ≈ cosθ
y n ≈ sin θ
y0 = 0
Inverse
tangent
1
tan −1 2 − k
− sgn( y k )
x0 , y 0
given,
y 
z n ≈ tan −1  0 
 x0 
z0 = 0
Hyperbolic
sine
and
cosine,
exponential
Hyperbolic
inverse
tangent,
square root
n
-1
tanh −1 2 − k
sgn(z k )
x 0 = B, z 0 = θ ,
x n ≈ cosh θ
y n ≈ sinh θ
y0 = 0
-1
tanh −1 2 − k
− sgn( y k )
x0 , y 0
eθ ≈ x n + y n
given, z ≈ tanh −1  y 0 
n
 
 x0 
z0 = 0
xn ≈
x02 − y 02
B
n
Where A = ∏ cos(σ j ) and B = ∏ cosh(σ j ) are pre-stored constants.
j =0
j =0
To illustrate how the CORDIC method works consider the following division
example. Suppose we wish to calculate 6.4 divided by 4. It follows from
Table 1 that
40
x k +1 = x0
y k +1 = y k + δ k x0 2 − k
z k +1 = z k − δ k 2 − k
where x0 = 4, y 0 = 6.4 and z n will approximate our quotient
6.4
(the larger
4
n, the number of iterations, then the better the approximation). The above
iteration scheme is encoded as a computer program (FORTAN 77) and the
iteration scheme is allowed to execute 50 times: the results are in displayed in
Graphs 1 and 2.
Graph 1 - plot of z sequence values against k
z sequence
2
1.5
1
0
5
10
15
20
25
k
30
35
40
45
50
41
Graph 2 - a plot of y sequence values against k
2.4
2
y sequence
1.6
1.2
0.8
0.4
0
-0.4
-0.8
0
5
10
15
20
25
30
35
40
45
50
k
It is not surprising that the y values have been driven to zero since
δ k = − sgn( y k ) .
Furthermore it can be shown that
n
y n +1 = y 0 + ∑ δ k x0 2 − k ,
k =0
n
z n +1 = −∑ δ k 2 − k .
k =0
Thus,
y 0 y n +1
=
+ z n +1 .
x0
x0
It follows that
z n +1 −
y
y0
y
= n +1 and hence z n +1 ≈ 0 , since the y n values
x0
x0
x0
are driven to zero for sufficiently large n.
The more interested reader should read the paper by Edwards and Underwood
[2] for a more detailed analysis on the convergence of the CORDIC method
for more complicated functions.
42
References
1. Edwards B.H., Jackson R.T., Underwood J.M., ‘How do calculators
calculate?’, Department of Mathematics, University of Florida
2. Edwards B.H., Underwood J.M., ‘How do calculators calculate
trigonometric functions?’, Clayton College and the University of Florida
YOU MAY BE INTERESTED IN:
Division by 33
If the number can be evenly divided by 3 and 11, the same can also be said for 33.
Use the methods for Division by 3 and Division by 11 above. If they both work,
your number is also evenly divisible by 33.
43
New Mathematics for an Old Problem
By Mr Ottewill
The Thwaites Conjecture
A few weeks back, Professor Sir Bryan Thwaites (OA) gave a lunchtime talk at the
College which covered, among other things, a conjecture which is sometimes
called the Thwaites conjecture, due to the fact that for many years now Sir Byan
has offered a £1000 prize for a proof or disproof of the conjecture.
(The
conjecture is also called the Collatz conjecture, or simply the 3n + 1 conjecture).
Stated as simply as possible, the conjecture says that if you start with a positive
integer and then repeatedly follow the formula ‘halve the number if it is even,
multiply by three and add one if it is odd’, reapplying the formula to each new
number formed, then you eventually end up with 1.
For example, starting with 7 gives: 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8,
4, 2, 1
The conjecture has been tested for billions of numbers and always seems to be
true, but to date no-one (including Sir Bryan) has proved whether or not it is true
for all positive integers.
On the day of the talk, Mr Kulatunge showed Sir Bryan a paper written by Craig
Alan Feinstein which claims to prove that the conjecture is unprovable. The
methods used in Feinstein’s paper are rather unusual. I hope here to indicate a
rough outline of the methods which Feinstein uses, along with some brief
comments on whether they are successful.
‘Standard’ work on the conjecture
It will be useful first of all to look briefly at some of the more standard
approaches to the conjecture.
The first thing to do is to note that, as multiplying by 3 and adding 1 always
produces an even number, we can combine this step with the step which always
follows, namely halving, giving rise to the rule that next number after an odd
number will be
3n +1
2
, not just 3n + 1 .
The next observation is often a rough, though fairly plausible, reason why the
conjecture is likely to be true. The argument goes that a randomly chosen
integer is equally likely to be odd or even. We can also argue that for large
44
3n +1
2
3n
2
, i.e. the + 1 in the numerator has little
effect on the value for large numbers.
Thus in the long run we are likely to be
numbers
multiplying by
is roughly equal to
3
2
roughly half the time, i.e. after an odd number, and by
rest of the time, i.e. for even numbers.
by
3
2
× 12 =
3
4
many times over.
As
3
4
1
2
the
The net result of this is to be multiplying
is less than 1, this means that over time
the sequence will get smaller and smaller, eventually resulting in 1.
Now, whilst a plausible argument, the previous paragraph is nowhere near the
rigor needed in Mathematics. There are too many assumptions, not least the
premise that there will be roughly the same number of odd and even numbers in a
sequence – this isn’t true for the sequence that starts with 64, for example.
A more rigorous approach (taken in ‘The 3x + 1 conjecture and its generalizations’
by Jeffrey C. Lagarais [1]) starts by labeling the next number in the sequences after
n as T (n) .
The sequence is then n , T (n) , T (T (n)) , T (T (T (n))) , ... where
T (n)
n
if n is even
2
3n + 1
=
if n is odd
2
=
We save time by writing T 2 (n) for T (T (n)) and T 3 (n) for T (T (T (n))) etc.
It is then noticed that any sequence produced by the conjecture can be
represented by the starting value plus a string of zeros and ones, e.g. (1, 0, 0, 1, 1,
0, 1, 1,...), where each zero or one is labeled xk (n) , i.e. in this case x0 (n) = 1,
x1 (n) = 0 , x2 (n) = 0 , x3 (n) = 1 etc., with xk (n) being 1 if T k (n) is odd and
xk (n) being 0 if T k (n) is even.
We then notice that a formula for T k (n) , i.e. the result of applying T k times
to n , is:
T k (n) = λ k (n) n + ρ k ( n)
where
λk (n) =
3 x0 ( n )+...+ xk −1 ( n )
2k
45
k −1
and ρk (n) = ∑ xi (n)
i =0
3xi +1 ( n)+...+ xk −1 ( n )
2 k −i
This looks pretty horrendous, but it is actually just an observation on the form of
the terms of the sequence. For example, suppose the sequence is a string of odd
numbers, then, starting with n we get:
T 0 ( n) = n
3n + 1
2
3n + 1
3×
+1
32 × n 3 1
2
2
T ( n) =
=
+ 2+
2
22
2
2
T 1 ( n) =
 32 × n 3 1 
3 × 
+ +  + 1
4
4 2
33 × n 32 3 1

T 2 ( n) =
= 3 + 3+ 2+
2
2
2 2
2
The formula with λ ’s and ρ ’s is just this with some terms taken out for the even
terms.
For example, we can see that the coefficient of n in the last line is just
the λk (n) as stated.
Many more results follow from having started out in the route above, all of which
can be found in the article by Lagarais (copies available in the Mathematics Office).
The article concludes by saying that no final proof or disproof of the conjecture
is known yet though.
Feinstein’s argument
The article shown by Mr Kulatunge to Sir Bryan, called ‘The Collatz 3n + 1
conjecture is unprovable’ by Craig Alan Feinstein [2], picks up from the above.
The essence of the argument is:
(i) First of all to notice that for any sequence of zeros and ones as mentioned, e.g.
(1, 0, 0, 1, 1, 0, 1, 1,...), there will be a value of n which produces this sequence
when applying T k times to n as described. (This is proved in the Lagarais
article).
46
(ii) Feinstein claims to show that, given the formula given above for T k (n) in
terms of λk (n) and ρ k (n) , to show that T k (n) = 1 for a given k and n we
would need to specify each of the zeros and ones in the sequence (1, 0, 0, 1, 1, 0,
1, 1, ...)
(iii) Feinstein then says that if there were a proof of the Thwaites conjecture, this
could be written in a document in a finite number of characters, e.g. typed up in
Word, needing a finite number of zeros and ones to store the file in binary on a
hard drive of a computer.
(iv) Lastly, Feinstein essentially argues from the idea of a random sequence of
zeros and ones, as defined in a moment, to the conclusion that the information
stored in the computer file would never be enough to include all of the
information in the Thwaites conjecture, as shown by the T k (n) formula in terms
of zeros and ones.
For anyone used to ‘standard’ mathematical proofs, the proof offered by Feinstein
is bizarre to say the least. The idea of using the length of a possible proof, and
how it might be stored on a computer, to show that no proof is possible is totally
unlike any other proof encountered in number theory. One may even be led to
suspect that Feinstein is a crank (this is hinted at in some internet discussion).
However, whether or not Feinstein’s argument is correct, the ideas that he uses are
actually rooted in a very rigorous and valid area of mathematics, albeit a fairly
recent one, as discussed next.
Algorithmic Information Theory
The area of mathematics / computer science called Algorithmic Information
Theory is largely the result Gregory Chaitin who has worked on the theory since
the 1960s. He has produced several serious textbooks on the theory (published
in the Springer-Verlag series, one of the most respected series of mathematics
textbooks) along with several more popular expositions of the theory. One of
the key ideas, used in Feinstein’s argument is that of ‘randomness’.
Chaitin states himself that the word ‘random’ is a poor choice, for there is nothing
truly random about the concept as used in the theory. The idea instead is that a
random number is one that cannot be stated more succinctly that simply writing it
out.
A good example of a non-random number is the number
1000000000000000000000000. This can be written using 25 digits as just shown,
or can be written much more succinctly as 1 × 1024. The reason why we can
47
write it in a very short form is clearly because it has an obvious pattern, i.e. all of
the zeros.
We can contrast this with a number such as
3476081933756308203956271 which, unless there is a fluke about the number, e.g.
it happens to be the 2053rd triangle number, can quite possibly not be specified in
any shorter form than by writing it in full.
This may sound all well and interesting (or not, depending on your tastes) but not
the basis of a proper mathematical theory. However, Chaitin has shown that a
rigorous branch of mathematics can be developed from this, which he has called
Algorithmic Information Theory.
As just one example of the results of the theory, Chaitin has proved that there are
indeed numbers which are random in exactly his sense of the word, i.e. for which
there is no quicker way to write the number than simply writing it out. The
culmination of this is his definition of a number which he labels Ω, i.e. a capital
Greek ‘omega’, presumably chosen due to the fact that this is the last letter of the
Greek alphabet, it being the ‘most random’ number and in some sense the last
thing which Mathematics can specify.
Chaitin’s proof that Ω is random is not simple, but the basis of it rests on Alan
Turing’s ideas of computability and the halting problem, first discussed in Turing’s
1936 paper, ‘On computable numbers, with an application to the
Entscheidungsproblem’ [3]. The ‘Entscheidungsproblem’ is the question of
whether a computer program could ever be written which would be able to
determine whether any computer program will come to a stop in a finite amount
of time.
To illustrate Turing’s argument, consider the computer program:
1. Let x = 10
2. If x = 10 then print x and stop, otherwise go back to line 1.
This clearly stops, or ‘halts’, after the second line.
Compare this to the program:
1. Let x = 10
2. If x = 15 then print x and stop, otherwise go back to line 1.
We can see that this second program will never stop.
48
Turing asked the apparently simple question: could a computer program ever be
written which would be able to look at any computer program and decide whether
or not the program would stop in a finite amount of time?
Turing’s beautiful proof that this wasn’t possible was to argue as follows.
(i) Suppose such a program did exist, i.e. one which would be able to print either
‘It stops’ or ‘It doesn’t stop’ if shown any program (and then itself stop,
presumably).
(ii) Take this program and modify it slightly so that instead of printing ‘It stops’
when it sees a program which will stop, it instead goes into an infinite loop (such
as in the first program above), otherwise printing ‘It doesn’t stop’ as before.
(iii) Now show the program from (ii) to itself. The problem is that the program
can’t determine that the program that it has just seen will stop – if it does this
then it goes into an infinite loop and so the program does not stop. On the
other hand, if the program thinks that the program it has seen won’t stop, then it
prints ‘It doesn’t stop’ and then stops, contradicting the fact that it doesn’t stop.
There is clearly a problem with (iii). The only place that the flaw seems to come
from is the first step, i.e. (i), i.e. in assuming that there could be a program which
could determine whether or not all programs stop. This is Turing’s famous result,
akin to the liar paradox, i.e. the statement ‘this statement is false’, and Gödel’s
incompleteness theorem showing that there will always be statements in
Mathematics which be neither proved nor disproved, essentially by encoding the
statement ‘this statement cannot be proved’.
Chaitin takes Turing’s idea of halting and applies it to a number, creating the next
digit in the number from the previous ones in such a way that one could never
specify the information in any shorter form. This is his number Ω.
Back to Feinstein’s argument
So, the question is where this all leaves Feinstein’s argument regarding the
Thwaites conjecture. The ideas behind the argument are based on Algorithmic
Information Theory, which uses the word ‘random’ in a very specific sense, and
which has produced some specific and apparently valid results.
The problem is partly that Feinstein’s article is so condensed that it is hard to see
some of the steps clearly. My own feeling is that it in step (ii) of Feinstein’s
argument that things go wrong.
He states that ‘in order to prove that T k (n) = 1 ,
it is necessary to specify the formula for T k (n) in the proof ’.
There is no
49
doubt some validity in this, i.e. proving the conjecture most likely involves getting
some grip on exactly when T k (n) = 1 .
However, it doesn’t seem to follow that
one will need to say exactly what the value of k is for any n , one only need to
show that there is a k . Similarly, it seems that one cannot argue that from the
fact that there is a formula for T k (n) involving information about all of the
intermediate numbers, e.g. the formula from Lagarais’s article, to it not being
possible for there to be a shorter formula. For example, the formula for the sum
of the first n terms of a geometric series can be written a + ar + ar 2 + ... + ar n−1 .
It doesn’t follow that there isn’t a simpler formula not involving the intermediate
numbers, in this case
a ( r n −1)
r −1
doing the trick.
Those are some initial thoughts on the proof. Feinstein would no doubt come
back with counter arguments – as Chaitin’s work shows, while basing arguments
on ideas such as ‘how much information is stored in a proof ’ is totally alien to any
standard mathematical techniques, there is a new branch of mathematics which
takes this seriously and is producing interesting results.
The final word should perhaps go to Chaitin. He makes the point in his book
‘Metamaths: the quest for Omega’ [4] that while he believes that the implications
for number theory of finding a fully random number (in his sense), Ω, is that
there will always be true mathematical facts which cannot be proved, he is not
certain that it is possible to show exactly which facts they are. He also tells the
story of how, for many years, when asked for a likely example of a true but
unprovable theorem, he would reply ‘Fermat’s Last Theorem’, adding how since
1995 and Andrew Wiles he has had to change his tune.
References
[1] Lagarias, J.C. (1985) The 3x + 1 problem and its generalizations, Amer. Math.
Monthly 92 (1985) 3-23, http://www.cecm.sfu.ca/organics/papers
[2] Feinstein, C.A. (2003), The Collatz 3n+1 Conjecture is Unprovable,
http://arxiv.org/abs/math/0312309
[3] Turing, A.M. (1936) 'On computable numbers, with an application to the
Entscheidungsproblem', http://www.turingarchive.org/browse.php/B/12
[4] Chaitin, G.J. (2006) Metamaths: The Quest for Omega, Vintage
MATHEMATIC
“A mathematical theory is not to be
considered complete until you have made it so
clear that you can explain it to the first man
whom you meet on the street.”
– David Hilbert –
– 1862-1943 –
-