XI

Topics Page No.

n Syllabus Guidelines 04

n Mathematics For Everyone 06

n Let do Some Number Magic 08

n Amazing Maths 10

n The Power of Partitions 12

n Find out Where You Can go Wrong 15

n Mathematical Horoscope 17

n Fruit Juice Testup 19

n Using Mathematical Equations to Fly a Broken Space Station 21

n Maths, Computers And Chess 27

CONTENTS

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Based on CBSE & ICSE Syllabus

UNIT-I: SETS AND FUNCTIONS 1. Sets:

Sets and their representations. Empty set. Finite & Infinite sets. Equal sets. Subsets. Subsets of the set of real numbers especially intervals (with notations). Power set. Universal set. Venn diagrams. Union and Intersection of sets. Difference of sets. Complement of a set.

2. Relations & Functions: Ordered pairs, Cartesian product of sets. Number of elements in the cartesian product of two finite sets. Cartesian product of the reals with itself (upto R R R ´´). Definition of relation, pictorial diagrams, domain, co-domain and range of a relation. Function as a special kind of relation from one set to another. Pictorial representation of a function, domain, co-domain & range of a function. Real valued function of the real variable, domain and range of these functions, constant, identity, polynomial, rational, modulus, signum and greatest integer functions with their graphs. Sum, difference, product and quotients of functions.

3. Trigonometric Functions: Positive and negative angles. Measuring angles in radians & in degrees and conversion from one measure to another. Definition of trigonometric functions with the help of unit circle. Truth of the identity sin 2 x + cos 2 x = 1, for all x. Signs of trigonometric functions and sketch of their graphs. Expressing sin (x+y) and cos (x+y) in terms of sinx, siny, cosx & cosy. Deducing the identities like following :

± ± ± = ± =

m m

tan tan cot cot 1 tan( ) ,cot( ) ,

1 tan tan cot cot

x y x y x y x Y

x y x y

+ − + − + = + = sin sin 2sin cos ,cos cos 2cos cos ,

2 2 2 2 x y x y x y x y

x y x y

sin sin 2 cos sin ,cos cos 2 sin sin . 2 2 2 2

x y x y x y x y x y x y

+ − + − − = − = −

Identities related to sin 2x, cos2x, tan 2x, sin3x, cos3x and tan3x. General solution of trigonometric equations of the type sinq = sina, cosq = cosa and tanq = tan a. Proofs and simple applications of sine and cosine formulae.

UNIT II: ALGEBRA 1. Principle of Mathematical Induction:

Processes of the proof by induction, motivating the application of the method by looking at natural numbers as the least inductive subset of real numbers. The principle of mathematical induction and simple applications.

2. Complex Numbers and Quadratic Equations: Need for complex numbers, especially 1 − , to be motivated by inability to solve every quadratic equation. Brief description of algebraic properties of complex numbers. Argand plane and polar representation of complex numbers. Statement of Fundamental Theorem of Algebra, solution of quadratic equations in the complex number system.

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3. Linear Inequalities: Linear inequalities. Algebraic solutions of linear inequalities in one variable and their representation on the number line. Graphical solution of linear inequalities in two variables. Solution of system of linear inequalities in two variables- graphically.

4. Permutations & Combinations: Fundamental principle of counting. Factorial n. Permutations and combinations, derivation of formulae and their connections, simple applications.

5. Binomial Theorem: History, statement and proof of the binomial theorem for positive integral indices. Pascal s triangle, general and middle term in binomial expansion, simple applications.

6. Sequence and Series: Sequence and Series. Arithmetic progression (A. P.), arithmetic mean (A.M.). Geometric progression (G.P.), general term of a G. P., sum of n terms of a G.P., geometric mean (G.M.), relation between A.M.

and G.M. Sum to n terms of the special series: 2 3 , n n and n ∑ ∑ ∑ . UNIT- III: COORDINATE GEOMETRY 1. Straight Lines:

Brief recall of 2D from earlier classes. Slope of a line and angle between two lines. Various forms of equations of a line: parallel to axes, point-slope form, slope-intercept form, two -point form, intercepts form and normal form. General equation of a line. Distance of a point from a line.

2. Conic Sections: Sections of a cone: circles, ellipse, parabola, hyperbola, a point, a straight line and pair of intersecting lines as a degenerated case of a conic section. Standard equations and simple properties of parabola, ellipse and hyperbola. Standard equation of a circle.

3. Introduction to Three -dimensional Geometry Coordinate axes and coordinate planes in three dimensions. Coordinates of a point in space. Distance between two points and section formula.

UNIT-IV: CALCULUS 1. Limits and Derivatives:

Derivative introduced as rate of change both as that of distance function and geometrically, intuitive idea of derivatives. Definition of derivative, limits, limits of trigonometric functions.

UNIT-V: MATHEMATICAL REASONING 1. Mathematical Reasoning:

Mathematically acceptable statements. Connecting words/ phrases - consolidating the understanding of if and only if (necessary and sufficient) condition , implies , and/or , implied by , and , or , there exists and their use through variety of examples related to real life and Mathematics. Validating the statements involving the connecting words-difference between contradiction, converse and contrapositive.

UNIT-VI: STATISTICS & PROBABILITY 1. Statistics:

Measure of dispersion; mean deviation, variance and standard deviation of ungrouped/grouped data. Analysis of frequency distributions with equal means but different variances.

2. Probability: Random experiments: outcomes, sample spaces (set representation). Events: occurrence of events, not , and & or events, exhaustive events, mutually exclusive events. Axiomatic (set theoretic) probability, connections with the theories of earlier classes. Probability of an event, probability of not , and & or events.

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• What is a number system? The number system consists of some important sets of numbers.

• Natural numbers : The numbers 1, 2, 3, ... are called the natural or counting numbers. The set of natural numbers is denoted by N. i.e. N = {1, 2, 3, ...}.

• Integers : The numbers 0, ±1, ±2, ... are called integers. The set of integers is denoted by Z. i.e. Z = {.... –3, –2, –1, 0, 1, 2, 3, ...}. The set of positive integers are known as Z + and the set of negative integers are known as Z –

The set of integers divisible by 2 is called even integers and the set of integers which are not divisible by 2 are called odd integers. 0 (zero) is an even integer. • Prime and composite integer : A positive integer which has no divisors other than 1 and itself is called a prime number and others are called composite integers. 1 is neither prime nor composite. There are infinitely many prime numbers. • Relatively prime number : The number p and q are called relatively prime number if h.c.f. of p and q is 1. e.g. 2 and 9 are relatively prime.

• Rational and irrational number : A number r is rational if it can be written as a fraction r = p/q, where p and q both are integers. In reality each number can be written in many different ways. But to be rational, a number ought to have at

least one fractional representation.

e.g. the number 2 2 

2 ) 1 5 ( 

2 1 5

−

+

+ may

not look like a rational but it simplifies to 3 = 3/1 and hence a rational number. The set of rational numbers is denoted by Q where Q = {p/q, p, q are integers, q ≠ 0}. The set Q of all rational numbers is equivalent to set N of all integers, where as the number  5 is called an irrational number. However the theory of irrational number belongs to calculus. Although with the help of arithmetic we may prove that  5 is not rational.

So, the numbers like  5 , 2 , 3 1/5 , π, e, log6 etc are irrational numbers.

Real and complex numbers : Collectively rational and irrational numbers are called real. Thus a set obtained by taking all rational and irrational numbers is called a set of real numbers and it is denoted by R. While it is true that there are infinitely many rational numbers and infinitely many irrational numbers, in a well defined sense, there are more irrational numbers than rational. Real numbers can also be treated as points on a line (also called the real line). Now between any numbers of one kind there always existed numbers of the same and of the other kind. However rational numbers form a countable set

Mathematics for Everyone

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where as irrational numbers form a set which is not countable. [A set A is called countable (or enumerable, or denumerable and even sometimes may be numerable) if A ~ N, i.e. if it is equivalent to set of all integers).

Complex numbers are pairs C = (x, y) of two real numbers. Since the situations of having to handle two numbers instead of one is quite complex, the terminology is well justified. The impetus for developing the theory of complex numbers came originally from unsolvability of a simple equation x 2 + 1 = 0 in the set of real numbers. In the set of complex numbers the equation has two solutions; ± i. The set of the complex number could also be defined by adding this new number. Thus the number of the form a + ib where i =  1 − and a, b are real numbers are known as complex number. Thus without breaking existing laws and by adding a single new number we actually get a set of numbers in which every polynomial equation with real coefficients has a solution. Complex numbers have unusual properties especially when it comes to differentiation. With real numbers differentiation is a starting point of calculus and real analysis. With complex numbers it leads to the analytic function theory.

Imaginary numbers : As we have defined a complex number C is a pair of (x, y) of two real numbers, x in the pair called the real part and y in the pair is the imaginary part. Now C = x + iy be a complex number;

here x → real part ; y → imaginary part.

Algebraic and transcendental numbers. Let us consider polynomial with whole (+ve integers and –ve integers) coefficients.

P n (x) = a n x n + a n – 1 x n – 1 + ... ax + a 0 = 0. Real roots of such equations are said to be algebraic. In other words, a number a is called algebraic if it satisfies an algebraic equation P n (a) = 0, for some polynomial P n (x) = 0 with integer coefficients. Rational numbers and integers are all algebraic. Indeed, for given rational number r = p/q, if we take n = 1 and P 1 (x) = qx – p. Obviously P 1 (r) = 0. 5 is also algebraic for it solves the

equation P 2 (x) = x 2 – 5 = 0. It can be proved that, for every n, the set of polynomials of order n is countable. The fundamental theorem of algebra claims that a polynomial of order n has exactly n (perhaps complex) roots. Then the set of all numbers that satisfy some equation of order n is countable. The subset of this set that consists of real number is as well countable. Thus the set of all algebraic numbers is countable.

Transcendantal numbers : Real numbers that are not algebraic are called transcendantal. Thus the set R of all real numbers is the union of two sets ‐ algebraic and transcendantal. The set R is not countable and therefore the set of all transcendantal numbers is not countable either. π and e are transcendantal numbers.

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11 Perfect numbers A perfect number is a number like 6 with the property that all its factors add up to the number itself. e.g. the factors of 6 are 1, 2 and 3 and 1 + 2 + 3 = 6. The next smallest perfect number is 28. The factors of 28 are 1, 2, 4, 7 and 14 and 1 + 2 + 4 + 7 + 14 = 28. i.e. any number n is perfect if it equals the sum of its divisors including 1 but excluding n. The next four perfect numbers are 496, 8128, 33550336, 8589869056. Only 37 perfect numbers have been found. All even perfect numbers must have the form 2 p – 1 (2 p – 1) where p and 2 p – 1 are prime numbers. e.g. for 6 and 28 we have p = 2 and p = 3. No one has yet discovered an odd perfect number but it is known that there is no odd perfect number less than 10 300 .

1 . What is special about this number 73939133.

Ans.: All the numbers 73939133 7393913 739391 73939 7393 739 737

are prime numbers. This is the biggest number with such property.

2 . What are twin primes? Ans.: The number of the form n – 1 and n + 1, both of which are primes are called twin primes. e.g. 17 and 19 or 101 and 103. The world’s largest known pair of twin primes is 190116 × 3003 × 10 5120 – 1 and 190116 × 3003 × 10 5120 + 1. This was discovered in 1995. It is believed that there are infinitely many prime numbers.

Lets do Some Number Magic Mathematics is not always high

and dry. You can play and have fun with math too. You may ask any person to think any 3 digited number and may guess its number. How?? Here it follows.

How to guess the numbers of 3 digits. Let the number be ABC, where A, B and C are the three digits. Find out the remainders from division by 11 of the numbers formed by the digits ABC, BCA, CAB in turn. Suppose the respective

remainders be a, b & c. Do the sum a + b, b + c & c + a. If any of these is odd, increase or decrease it by 11 in order to get a positive number less than 20. Divide each of these number by 2 and the obtained numbers are the digits A, B, C in order. Example : Let the number thought by your friend is 789. You are supposed to guess its number. A = 7, B = 8, C = 9

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ABC = 789, BCA = 897, CAB = 978 71 789 11

77

1911 8

1 897 8 11 88

1711 6

978 88 11 88

9888 10

a = 8 b = 6 c = 10 a + b = 14, b + c = 16, c + a = 18 Now your friend after performing the above operations tells you A = 14/2 = 7, B = 16/2 = 8, C = 18/2 = 9. So by the given procedure we obtain the 3 numbers 7, 8 & 9 in turn.

NUMBER FUN

1 . Find a number such that each digit except zero appears just once in the number and its square. There are two solutions (a)(567) 2 = 321,489 (b) (854) 2 = 729,316 2 . Funny number sequence: 1 + 4 + 5 + 5 + 6 + 9 = 2 + 3 + 3 + 7 +

7 + 8 1 2 + 4 2 + 5 2 + 5 2 + 6 2 + 9 2 = 2 2 + 3 2 + 3 2

+ 7 2 + 7 2 + 8 2

1 3 + 4 3 + 5 3 + 5 3 + 6 3 + 9 3 = 2 3 + 3 3 + 3 3 + 7 3 + 7 3 + 8 3

3. Funny square number: (69) 2 = 4761 It is interesting to see that the product 4761 has same digits of Kaprekar’s well known constant number 6174.

NUMBER PUZZLE

Find four numbers such that the product of any two of them is one less than a

perfect square. This problem belongs to famous mathematician Diophantus. Solution : Let a, b, c and d are the required numbers. There are six possible products of any two of these, namely ab, ac, ad, bc, bd, cd. Then if m and n are any arbitrary integers, the requirement can be stated. a = m; b = n(mn + 2); c = (n + 1)(mn + m+2) d = 4(mn + 1)(mn + m + 1)(mn 2 + mn+2n+1) ab + 1 = (mn + 1) 2 ; ac + 1 = (mn + n + 1) 2

ad + 1 = (2m 2 n 2 + 2m 2 n + 4mn + 2m + 1) 2

bc + 1 = (mn 2 + mn + 2n + 1) 2

bd + 1 = (2m 2 n 3 + 2m 2 n 2 + 6mn 2 +4mn+ 4n+1) 2

cd + 1 = (2m 2 n 3 + 4m 2 n 2 + 6mn 2 + 2m 2 n + 8mn + 4n + 2m + 3) 2

m and n are any integers. For example, m = 2 and n = 1 yield a= 2, b =4, c = 12, d =420, ab+1 = 3 2 , ac+1 =5 2 , ad + 1 = 29 2 , bc + 1 = 7 2 , bd + 1 = 41 2 , cd + 1 = 71 2 .

PERFECT SQUARES

Since n 10 ‐ 1 = 9 a , 10 n – 1 = 9a

b = 9a + 1 = 10 n . Let c = 8a + 1. Now, 4ab + c = 4a(9a + 1) + 8a + 1 =36a 2 + 12a + 1=(6a + 1) 2 , a =1, 11, 111, ... = (7 2 , 67 2 , .....), Put a = 1, 6a + 1 = 7, Put a = 11, 6a + 1 = 6×11 + 1 = 67

(a – 1)b + c = (a – 1)(9a + 1) + 8a + 1 = 9a 2 – 9a + a – 1 + 8a + 1 = 9a 2 = (3a) 2

16ab + c = 16a(9a + 1) + (8a + 1) = 144a 2 + 24a + 1 = (12a + 1) 2

(25a + 3)b + c = (25a + 3)(9a + 1) +8a+1 = 225a 2 + 52a + 3 + 8a + 1=225a 2 +60a +4 = (15a + 2) 2 .

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Amazing Maths How to find last digit of a number with high power & the remainder left

when such number is divided by some small number.

Since last few years, in many engineering and other entrance objective exams (like UPSEAT & MBA/MCA etc.), questions relating to “find the last/unit digit of a high powered number” is being asked. It has been observed that most of the students do these problems using the concept of Binomial Theorem, which becomes lengthy and time consuming in objective exams.

So, a much easier and time saving method of doing these problems is presented here.

First you should understand that Mod (or Modulo) is a mathematical function that gives remainder of the process of division and the statement that “x is the remainder when N is divided by b” is symbolically denoted as N≡x (Mod b). Now, to find the last digit of a number (N), we can simply find the remainder (x) obtained when N is divided by 10 (b in this case).

Now, one can practise the method by going through few solved examples:

Examples : • Find the last digit of 3 123 . 3 4 ≡ 1 (Mod 10) ⇒ (3 4 ) 30 ≡ 1 30 (Mod 10) or, 3 120 ≡ 1(Mod 10) .... (i)

Again, 3 3 ≡ 7 (Mod 10) ... (ii) Eqn. (i) × (ii) gives, 3 120 × 3 3 ≡ 1 × 7

(Mod 10) or 3 123 ≡ 7 (Mod 10)

∴ Last digit = 7. [* Note : When number (N) is raised to any power, the remainder (x) is also raised to the same power]. • Last digit of 11 132 = ? ; 11 ≡ 1 (Mod 10)

∴ 11 132 ≡ 1 132 (Mod 10) or, 11 132 ≡ 1 (Mod 10) ∴ last digit = 1.

• Last digit of 6 500 = ? 6 2 ≡ 6 (Mod 10)

∴ 6 4 ≡ 6 2 (Mod 10) ≡ 6 (Mod 10). [Note : Since 6 2 gives 6 as remainder on division by 10]

Similary, (6 2 ) 250 ≡ 6 (Mod 10). or, 6 500 ≡ 6 (Mod 10)

∴ Last digit = 6. • Last digit of 7 291 = ?

7 4 ≡ 1 (Mod 10) ∴ (7 4 ) 72 ≡ 1 72 (Mod 10)

or, 7 288 ≡ 1 (Mod 10) .... (i) Also, 7 2 ≡ 9 (Mod 10) .... (ii)

7 1 ≡ 7 (Mod 10) .... (iii) Eqns. (i) × (ii) × (iii) gives, 7 291 ≡ 1 × 7 × 9 (Mod 10)

≡ 63 (Mod 10) ≡ 3 (Mod 10) ∴ Last digit = 3.

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[Note : In general N ≡ N (Mod b) if N < b like in above example, 7 being less than 10, cannot be divided by 10. So, we say “7 on division by 10 leaves 7 (itself) as remainder” which is represented as “7 ≡ 7 (Mod 10)”]. Now the above idea can be extended to find the remainder when number with high power is divided by a smaller number.

Examples : • When 7 30 is divided by 5, what is the remainder?

7 4 ≡ 1 (Mod 5). So, (7 4 ) 7 ≡ 1 7 (Mod 5) or, 7 28 ≡ 1 (Mod 5) ... (i) Also, 7 2 ≡ 4 (Mod 5) ... (ii) Eqn. (i) × (ii) gives, 7 28 × 7 2 ≡ 1 × 4

(Mod 5) or, 7 30 4 (Mod 5) ∴ Remainder = 4.

Note : The above problem can be also solved by “Binomial Theorem” method as follows:

7 30 ≡ (7 2 ) 15 = (49) 15 = (50 – 1) 15

= 50 15 – 15 C 1 50 14 + 15 C 2 50 13 – 15 C 3

50 12 + ... + ( 15 C 15 (–1) 15 50 0 ) = 50 15 – 15 C 1 50 14 + 15 C 2 50 13 – 15 C 3 50 12

+ ... + (–1) = 50 15 – 15 C 1 50 14 + 15 C 2 50 13 – 15 C 3 50 12

+ ... + (–5 + 4) = (a multiple of 5) + 4.

∴ remainder = 4.

2 . When 5 20 is divided by 7, what is the remainder?

5 2 ≡ 4 (Mod 7) ∴ (5 2 ) 2 ≡ 4 2 (Mod 7)

or, 5 4 ≡16 (Mod 7) ≡ 2 (Mod 7). [Note : since 16, on division by 7 gives 2 as remainder] ∴ (5 4 ) 5 ≡ 2 5 (Mod 7)

or, 5 20 ≡ 32 (Mod 7) ≡ 4 (Mod 7) ∴ Remainder = 4. [Similarly, 32 when divided by 7 gives 4 as remainder] A more typical example 3. What is the remainder obtained when

64 × 65 × 66 is divided by 67? 64 ≡ –3 (Mod 67) ... (i) 65 ≡ –2 (Mod 67) ... (ii) 66 ≡ –1 (Mod 67) ... (iii) Eqns. (i) × (ii) × (iii) gives, 64 × 65 × 66 ≡ –6 (Mod 67) ≡

= − ( 6 1 6 7 6 ) 6 1

(Mod 67) *

∴ Remainder = 61. [ * Note: When (N) is divided by (b) such that b > N then “N ≡ –a (Mod b)” means “N is short of a to get divisible by b where a = b – N” Here, a is not the remainder. While as told earlier the “remainder of division of N by b where b > N will be N itself” i.e. “N ≡ N (Mod b)” (N = b – a).

You can find more of these articles in the magazine EduSys Exam Today Teachers, students can also contribute articles in the magazine.

For details call 011-26161014 or email : info@edusys.in

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Just a year before his death in 1920 at the age of 32, mathematician Srinivasa Ramanujan came upon a remarkable pattern in a special list of whole numbers.

The list represented counts of how many ways a given whole number can be expressed as a sum of positive integers. For example, 4 can be written as 3 + 1, 2 + 2, 2 + 1 + 1 and 1 + 1 + 1+1. Including 4 itself but excluding different arrangements of the same integers (2 + 1 + 1 is considered the same as 1 + 2 +1), there are five distinct possibilities, or so‐called partitions, of the number 4. Similarly, the integer 5 has seven partitions.

The list that Ramanujan perused gave for each of the first 200 integers, the number of their partitions, which range from 1 to 3,972,999,029,388.

Ramanujan noticed that, starting with 4, the number of partitions for every fifth integer is a multiple of 5. For instance, the number of partitions for 9 is 30 and for 14 is 135.

He discovered several more such patterns. Starting with 5, the number of partitions for every seventh integer is a multiple of 7, and starting with 6, the number of partitions for every 11 th

integer is a multiple of 11. Moreover, similar relationships occur where the interval between the chosen integers is a power of 5,7, or 11 or a product of these powers. Ramanujan went on to prove rigorously that these patterns hold not only for the 200 partition numbers in his table but also for all higher numbers.

It was a curious discovery. Nothing in the definition of partitions hinted that such relationships, called congruences, should exist or that the prime numbers 5,7, and 11 should play such a special role.

After many decades of nearly fruitless searching that yielded just one or two apparently isolated examples of large numbers that fit into the pattern,

Ramanujan consequences The idea that starting with 4, the number of partitions for every 5 th integer is a multiple of 5 can be expressed completely in the following mathematical form, when p(n) called partition function, is the function of partition of an integer n, indicates congruence and 0 (mod 5) means that the remainder after division by 5 must be 0. For n greater than or equal to 0

p(5N + 4) ≡ 0 (mod 5) Examples of new congruence discovered by Rhiannon L. Weaver.

p(11,864, 749 N + 56,062) ≡ 0 (mod 13) p(14375 N + 3474) ≡ 0 (mod 23).

The Power of Partitions Writing a whole number as the sum of smaller numbers

springs a mathematical surprise

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mathematicians came to believe that no other congruences exist. Those found by Ramanujan and the later mathematicians were thought to be little more than numerical flukes.

However, in a startling turnabout, mathematician Ken Ono of University of Wisconsin has now proved that infinitely many such relationships may occur. His work has been ranked as the most important work on partition congruences since the epic work of Ramanujan.

Moreover, Ono’s partition research has intriguing, unexpected links to complex mathematical ideas and methods that earlier led to proof of Fermat’s last theorem and the Taniyama‐Shimura conjecture. It has opened up new avenues for studying some of the most important, but difficult, questions in number theory.

Born into poverty, Ramanujan grew up in Southern India, and although he had little formal training in mathematics, he became hooked on mathematics. He spent the years between 1903 and 1913 filling notebooks with page after page of mathematical formulae and relationships that he had uncovered.

Ramanujan’s life as a professional mathematician began in 1914 when he accepted an invitation from the prominent British mathematician G.H. Hardy to come to Cambridge University. He spent 5 years in England, publishing many papers and achieving international recognition for his mathematical research.

Though his work was cut short by a mysterious illness that brought him back to India for the final year of his life, Ramanujan’s work has remained a subject of considerable interest. For the past 2 decades, Berndt has been going through Ramanujan’s scribbled notes, systematically deciphering, writing out, and proving each of the numerous theorems Ramanujan had formulated.

Two years ago, Berndt was examining an unpublished, handwritten Ramanujan manuscript on partitions. “It contained claims that had never been proved,” Berndt remarks. So he recruited Ono, an expert on partitions from a modern perspective, to help him fill in the details and provide the necessary proofs.

“I had never read any of Ramanujan’s notebooks, though I was familiar with a lot of what he had done through the writings of more modern mathematicians,”

According to Ono, the manuscript didn’t contain anything startling or new. However, Ramanujan had written out one mathematical expression, or identity, in a strange way. However, after translating the expression into modern mathematical language, it made perfect sense.

Yet the apparent awkwardness of Ramanujan’s original formulation bothered Ono. It got him to think deeply about the many different ways in which mathematicians can express identities. In the course of that rethinking, he obtained crucial insights that led him to tackle the question of partition congruences ‐ something that

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In a remarkable tour de force, Ono proved that partition congruences can be found not only for the prime numbers 5, 7, and 11 but also for all larger primes. To do so, he had to exploit a previously unsuspected link between partition numbers and complex mathematical objects called modular forms.

They are special types of mathematical relationships that involve so‐called complex numbers, which have a real and an imaginary component. In effect, the relationships represent particular types of repeating patterns, roughly analogous to the way a trigonometric function such as sine represents a repeating pattern for ordinary numbers.

Mathematicians have identified many different types of modular forms. The connection between certain modular forms and elliptic curves helped Andrew Wiles of Princeton University prove Fermat’s last theorem in 1993.

Applying modular‐form theory, as developed by other mathematicians, Ono uncovered a connection between partition numbers and specific modular forms. He used that relationship to establish, in effect, the existence of congruences involving prime‐number divisors among the partition numbers.

Unlike many previous advances in partition theory, Ono’s research involved no computation and relied heavily on theory.

Interestingly, although Ono proved the existence of these congruences, he provided only one explicit example. In this new congruence, the starting number is 111,247, with each successive step to the next integer going up by 59 4 × 13. The corresponding partitions are then multiples of the prime number 13.

Ono, however, didn’t have an effective recipe, or algorithm, for generating examples. But Rhiannon L. Weaver, an undergraduate at Penn State, found a simple criterion for identifying the start of a progression. She then developed an algorithm, working with primes from 13 to 31, to obtain more than 70,000 new congruences. The number of partitions for each integer from 1 to 45.

Integer Number o f Partitions

1 1 2 2 3 3 4 5 5 7 6 11 7 15 8 22 9 30 10 42 11 56 12 77 13 101 14 135 15 176 16 231 17 297 18 385 19 490 20 627 21 792 22 1,002

Integer Number o f Part it io ns

23 1,255 24 1,575 25 1,958 26 2,436 27 3,010 28 3,718 29 4,565 30 5,604 31 6,842 32 8,349 33 10,143 34 12,310 35 14,883 36 17,977 37 21,637 38 26,015 39 31,185 40 37,338 41 44,583 42 53,174 43 63,261 44 75,175 45 89,134

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Sign error are surely the most common errors of all. The great number of sign errors suggests that students are careless and unconcerned ‐‐ that students think sign errors do not matter. But sign errors certainly do matter, a great deal. Sign errors are just the symptom; there can be several different underlying causes. One cause is the “loss of invisible parentheses,” Another cause is the belief that a minus sign means a negative number. Is –x a negative number ? That depends on what x is. • Yes, if x is a positive number. • No, if x itself is a negative number. For instance, when x=– 6, then –x = 6 (or, for emphasis, – x = + 6). Misunderstanding this point causes some students to have difficulty in understanding the definition of the absolute value function.

Many students don’t know how to add fractions. They don’t know how to add (x/y) + (u/v), and some of them don’t even know how to add (2/3) + (7/9). It is hard to classify the different kinds of mistakes they make, but in many cases their mistakes are related to this one : Everything is additive. In advanced mathematics, a function or operation f is called additive if it satisfies

f (x + y) = f (x) + f (y) for all numbers x and y. This is true for certain familiar operations ‐ for instance. • the limit of a sum is the sum of the

limits. • the derivative of a sum is the sum of

the derivatives, • the integral of a sum is sum of the

integrals. But it is not true for certain other kinds of operations. Nevertheless, students often apply this addition rule indiscriminately. For instance, contrary to the belief of many students,

( ) + sin x y is NOT equal to ( ) ( ) + sin sin x y ( ) + 2 x y is NOT equal to ( ) ( ) + 2 2 x y

+ x y is NOT equal to + x y

+ 1

x y is NOT equal to + 1 1 x y

We do get equality holding for a few unusual and coincidental choices of x and y, but we have inequality for most choices of x and y. The “everything is additive” error is actually the most common occurrence of a more general class of errors. 3. Everything is commutative : In higher mathematics, we say that two operations commute if we can perform them in either order and get the same result. We’ve already looked at some

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11 examples with addition; here are some example with other operations. Contrary to some students’ beliefs,

log x is NOT equal to logx ,

sin 3x is NOT equal to3 sinx,

etc. Another common error is to assume that multiplication commutes with differentiation or integration. But actually, in general (uv)′ does not equal (u′ ) (v′ ) and ( ) ∫  uv does not equal ( )( ) ∫ ∫ u v . It is applicable only when the region of integration is a rectangle with sides parallel to the coordinate axes, and u (x) is a function that depends only on x (not on y), and v (y) is a function that depends only on y (not on x). Under those conditions,

∫ ∫ ∫ ∫ = [ ( ) ( )] ( ) ( )

b d b d

a c a c u x v y dydx u x dx v y dy

Undistributed cancel lation . In a sense, this is the reverse of the “loss of invisible parentheses”

( ) ( )( ) ( )

( )( ) + − + +

= = + +

2

3

3 7 2 9 1 3 7 6

x x x f x

x x

( ) ( ) ( ) + − + +

= + +

2

3

(3 7) 2 9 1 3 7 ( 6)

x x x x x

− + + = +

2

3 (2 9) ( 1)

( 6) x x

x (Wrong)

+ − + + =

+ +

2

3

(3 7) [(2 9) ( 1)] ( )

(3 7) ( 6) x x x

g x x x

− + + − + + = = + +

2 2

3 3 [(2 9) ( 1)] (2 9) ( 1)

( 6) ( 6) x x x x

x x (Correct)

Apparently some students think that f (x) and g (x) are the same thing ‐‐ or perhaps they simply don’t bother to look carefully enough at the top line of f (x), to discover that not everything in the top line of f (x) has a factor of (3x + 7). If you still don’t see what’s going on, here is a correct computation involving that first function f :

+ − + + =

+ +

+ − + + = +

2

3

2

3

(3 7)(2 9) ( 1) ( )

(3 7)( 6)

1 2 9 3 7 6

x x x f x

x x x x x

x Why would students make errors like these ? Perhaps it is partly because they don’t understand some of the basic concepts of fractions. Here are some things worth noting : • Multiplication is commutative – that

is, xy = yx. Consequently, most rules about multiplication are symmetric. For instance, multiplication distributes over addition both on the left and on the right : 

and 1 2 1 2

1 2 1 2

( ) ( ) ( ) ( ) ( ) ( )

x x y x y x y x y y xy xy

+ = +

+ = +

• Division is not commutative – in general, x/y is not equal to y/x. Consequently, rules about division are not symmetric. For instance.

but in general

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+ = + 1 2 1 2 ( ) / ( / ) ( / ) x x y x y x y but in general

+ ≠ + 1 2 1 2 /( ) ( / ) ( / ) x y y x y x y • Fractions represent division and

grouping (i.e., parentheses). For

instance, the fraction is the + +

a b c d same

thing as (a + b)/(c + d). If you omit either pair of parentheses from that last expression, you get something entirely different.

Dimens ional Error. Dimensional analysis doesn’t tell you the right answer, but it does enable you to instantly recognize the wrongness of some kinds of wrong answers. Just keep careful track of your dimensions, and then see whether your answer looks right. Here are some examples :

• If you’re asked to find a volume, and your answer is some number of square inches, then you know you’ve made an error somewhere in your calculations.

• If you’re asked to find an area or a volume, and your answer is a negative number, then you know you’ve made an error somewhere.

• Even if you don’t remember the formula for the volume of a sphere of radius r, you know that it has to have a factor of r 3 , not r 2 , so it couldn’t possibly be πr 2 .

• Even if you don’t remember the formula for the surface area of a sphere of radius r, you know it has to get small when r gets small. So it couldn’t possibly be something like 2 + 3r 2 .

Mathematical Horoscope AREAS (Mar 21‐Apr 19) Today a close friend who has let down her boundaries will have a dilemma. Though you have several axes to grind with her, you should not let this be a factor. If she can count on you, you will get closure. TORUS (Apr 20‐May 20) Today you will be in your prime. A clique of friends will give you a ring and hold an enjoyable function. You will feel connected to this circle of friends, and will be the center of attention.

M and I (May 21‐Jun 21) Today you will get a visit from an ex, though you won’t know why. The jerk is still square, a real number, quite a character. He will feed you a line about being a pair again, and this will cause an argument, making you tensor and tensor. ANSWER (Jun 22‐Jul 22) Don’t commute today. If you go by plane, the floor will drop out. If your mode is by car,

but in general

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11 you will hit a field of trees in the median. Even walking on your own digits will result in a funeral plot. E‐O (Jul 23‐Aug 22) Today life will throw you a curve. You and a loved one will diverge, and the distance will feel empty. Though you will see her less, her existence proves your perfect friendship is in a separate category. Don’t let it make a difference. ERGO (Aug 23‐Sep 22) On the surface, today will seem like an irrational day, continuously full of problems. The constant chaos will be far from ideal. But the day will be less negative and more normal than it first appears. Stay to your regular routine and minimize variation. E‐BAR (Sep 23‐Oct 22) Today you will intersect a radical from class, a tan gent of average height. He will be a cute guy, but obtuse. Help him by reading his abstract, so he can get his degree and reach his proper rank. SCORE‐PI‐O (Oct 23 ‐ Nov 21) You have reached a critical point in your life. You are on edge and have about reached your limit. This minor identity crisis has kept you in knots for some time. Seek a group of solid friends to set things right. SLANTED‐AREAS (Nov 22‐Dec 21) Today you should stay in bed and catch some Z’s. Stay horizontal and sleep like a log. This complete lack of activity will mean a minimum of stress, but it is also a product of the power you have. CHORD (Dec 22‐Jan 19) This period of your life will be similar to one year ago, when your life was simple and ordered. It is integral that you be careful, lest you do commit an improper error. Look for a sign, but stick to your roots. A‐SQUARY‐PLUS (Jan 20‐Feb 18) Keep an open mind today. Stop projecting and admit the magnitude of your problems for a second. If you are analytic, the origin of your exact difficulty will soon be clear, and your troubles will decrease by half. PI‐CEES (Feb 19‐Mar 20) Today will be an odd day. You will face many complex problems, all equally frustrating. But you will maximize your joy by finding value in all of them, and it will turn out to be a fine day.

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You have been invited to a fruit juice tasting party. There are several cans of juice, each of a different flavor. The cans are covered with paper so that their labels are hidden. You have a list of all the different flavors. You have to guess what flavor is in each can without looking at the labels. In other words, you have to match each flavour on the list with a can. You can’t guess a flavor more than once. This math puzzle is in two parts.

Part 1 To start with you have two cans of juice. What is the probability that you will guess correctly? (Since this is a math puzzle, let’s assume that you really can’t taste the difference so you are just making random guesses.) With three cans, what’s the probability of guessing all of them correctly? What if there are four cans of juice? How about 10? How about 100? Can you come up with a general formula for the probability of guessing all the flavors correctly with N cans of juice?

Part 2 Now let’s look at the probability of getting all the guesses wrong. If there are only two cans of juice then you either guess both correctly or both incorrectly. But with three cans, you could get one right and the other two wrong. What is the probability of getting all three wrong? With 10 cans of juice, what is the probability that you will get none of them right? How about with 100 cans? Can you come up with a general formula for the probability of getting no matches at all when trying to guess the flavors of N cans of juice? Finally, what is the probability of getting just one right?

Part 1 One way to think about this problem is to look at the number of possible ways there are to arrange the cans of juice in a row. If you have two cans, let’s say orange juice and apple juice, there are two possible arrangements: apple orange orange apple

One of these will be the correct arrangement. Since there are two possible guesses, the probability of guessing correctly is 1 out of 2 or 1/2.

Fruit Juice Testup

SOLUTION

Let’s add grape juice. There are now six possible arrangements: apple orange grape apple grape orange orange apple grape orange grape apple grape apple orange grape orange apple

Again, only one of these is the correct order so your chance of guessing correctly is 1 out of 6 or 1/6. If you add a fourth can you will find that the number

20 

11 of possible arrangements is 24 and, therefore the probability of a correct guess is 1/24. In general, the number of possible arrangements of N objects is

N × (N ‐1) × ... × 1 This is called the factorial of a number and is written N!

2! = 2 3! = 6 4! = 24

As N increases, N! increases very quickly. If you had 10 cans of juice, there would be 10! Possible arrangements. 10! = 3,628,800 so your chances of guessing correctly are rather small. With 100 cans of juice there are approximately 93, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000,0 00, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000 Possible arrangements In general, the probability of correctly guessing all the flavors when you have N cans of juice is 1/N! Part 2 If you have only two cans of juice, the probability of getting them both wrong is the same as the probability of getting them both right. This is because there are only two possible arrangements: apple orange orange apple One is all right. The other is all wrong. With 3 flavors there are 6 possible arrangements: apple orange grape apple grape orange orange apple grape orange grape apple grape apple orange grape orange apple Let’s say that the first one is correct. Then there are 2 arrangements that are all wrong: orange grape apple grape orange apple

The other arrangements have one can in the correct place and the other two wrong. So the probability of guessing all wrong is 2 / 6 or 1 / 3. In general, the question is: Given any sequence of n distinct objects, in how many ways can those objects be rearranged so that none of the objects is in its original position? This is known as the number of “derangements” of n, denoted by D(n). Another term for the number of derangements is the subfactorial denoted !n. D (1) = !1 = 0 and D(2) = !2 = 1 One way (not the most efficient) of determining the value of D(n) for any given n is by means of the inclusion/exclusion principle. For example, with n=5 we begin with 5! possible re‐arrangements, but of those we know that 4! leave the first object unmoved, and 4! leave the 2nd object unmoved, and so on. Thus there are 5 × 4! that leave one specific object unmoved, so we have to subtract this number. However, in doing so we have subtracted some of the arrangements twice, because some of them leave both the 1st AND the 2nd element unchanged (for example). So we need to add back the number of arrangements that leave any two of the objects unmoved, of which there are C(5, 2) × 3! (There’s an explanation of what C means at the bottom of this page.) But then we need to subtract the number of arrangements that leave THREE objects unmoved, and so on. The final result is D(5) = 1 × 5! ‐ 5 × 4! + 10 × 3! ‐ 10 × 2! + 5 × 1! ‐ 1 × 0! = 44 In general we have

n j / n \ D(n)= SUM (‐1) ( ) (n‐j)!

j=0 \ j / If we examine the effect of multiplying each term of this sum by n+1, we see that the value of D(n) can be computed from the value of D(n‐1) by the recurrence formula D(n) = n D(n ‐ 1) + (‐ 1)^n The first several values of D(n) for n =1, 2, 3, ... are 0, 1, 2, 9, 44, 265, 1854, 14833, ...

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On the 25th of May 1997 an unmanned cargo ship named Progress crashed into the Russian space station Mir, tearing a hole into Mir’s side and sending it hurtling through space. In his desperate attempt to save Mir and its crew, NASA astronaut Michael Foale had no choice but to get right down to the maths that governs space flight. The space station Mir, manned by NASA astronaut and scientist Michael Foale, Flight Engineer Alexander “Sasha” Lazutkin and Commander Vasily Tsibliev, had been put in orbit around Earth to test for the first time the feasibility of long‐term human habitation in space. On this particular day in May 1997, the unmanned cargo ship Progress, orbiting Earth a few kilometres away from the station, was due to dock with Mir, under remote‐control by Mir’s crew. But this wasn’t a routine docking manoeuvre. “What was so different and dangerous about that particular day,” says Michael Foale, “was that we had been given orders to disconnect the antennae systems on the Progress, so that the Progress did not know how fast it was coming into the space station or how far away it was. All we had was a view from a camera, mounted on the Progress, coming in towards Mir. This amounted to flying a cargo rocket weighing seven tons, with very, very weak thrusters, and a two‐second time delay between inputting commands at the controls and the Progress’s reaction. If you imagine guiding a seven ton truck into a narrow parking slot by remote control, only with the help of a small black‐and‐white monitor and subject to a time delay, you get some idea of the difficult task that was facing Vasily Tsibliev, who was to execute the docking manoeuvre. But there was another complication: viewing the world from the Mir was very different to viewing it from the Progress. To understand this, think of a plane starting from London and aiming straight at Perth in Australia. If the plane was to roll over the surface of the Earth along the straight line between London and Perth, it would eventually arrive. As soon as it takes to the air, however, it leaves the Earth’s frame of reference. While it is flying into the prescribed direction, the Earth underneath it rotates, causing the plane to miss its target. From the point of view of someone on Earth, the plane has flown along a curve, as if it had been diverted from its original path by some external force.

Using Mathematical Equations to Fly a Broken Space Station

22 

11 This fictitious force — fictitious because it is only a result of our viewpoint — is called the Coriolis force. The same difficulty applies in space: sitting within a space station you are in what is called a rotating frame of reference. Any space vehicle thrusting towards you is not part of this frame of reference and will feel the Coriolis force. You have to take account of this force when guiding the vehicle — or miss your target. “Vasily had not been trained on this. This was a poorly planned experiment, one which was going to affect our safety quite critically,” says Michael. The Progress was moving in at about six metres per second, meaning that Vasily could not really ascertain its speed until the very end. “It all happens in the last few seconds,” says Michael, “then suddenly the angles open up and the speed of the Progress becomes apparent. But when your thrusters are very small, as they are on the Progress, then you need to brake for a long time to take away your speed. Vasily’s recognition of the Progress’s speed came too late.” Sasha Lozutkin, who had seen through one of Mir’s windows that disaster was imminent, had time to shout out a warning — and then the Progress crunched straight into Mir’s side. “I didn’t feel the crunch because I was not touching anything, but I could see our equipment shake and hear a big bang,” says Michael. “That’s when I thought ‘this could be the moment we all die’. The de‐pressurisation alarm went off and this meant that there really was a hole in the station. I felt the pressure fall in my ears. I started heading for the Soyuz, Mir’s only life boat.” But meanwhile Sasha, who had seen the impact, had other plans. The Mir is made up of a central node to which five different modules attach by hatchways. The Progress had hit one of these modules, rather than the central node, and so Sasha had decided to simply close off the damaged module. As precious minutes and air were trickling away, the crew set about cutting eighteen thick cables which were preventing them from closing the door to the damaged module. Sparks flying, the last cable finally gave in and the hatchway’s door could be lifted into position. “I could immediately hear the door being sucked into place,” says Michael “and about a minute later I could feel the pressure stabilise. At this point the emergency was over.”

Silence The emergency was over, but Mir and its crew were far from safe. What they had just done was disconnect thirty per cent of the station’s power supply, which came from the solar arrays attached to the damaged module. “Within twenty to thirty minutes after the collision, every single piece of equipment on the station went dead,” says Michael. “It was a completely silent station. It’s quite eerie to be in a station in

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space where there is no sound except for your own breathing. And then it started to get cold. Really, really cold.”

There still were functional solar arrays attached to the undamaged modules, and with some exposure to the Sun, these arrays could provide sufficient power to keep operations going. But exposure to the Sun was precisely the problem. Before the impact, Mir had been orbiting Earth all the while maintaining the same attitude towards the Sun. Its solar panels had been facing straight into the Sun, slowly turning to keep track whenever Mir moved over the Earth’s sunny side. But Progess’s impact had knocked Mir out of its attitude to the Sun. Worse, it had also set it spinning about its own centre of gravity at a rate too fast for the solar arrays to keep up. There was only one way to save the station: the lifeboat called Soyuz, still attached to the station, had its own power and thrusters. Using these thrusters it might be possible to counteract Mir’s tumble and to manoeuvre it into a position that allowed its solar arrays to find the Sun. A risky plan, as Michael explains: “Remember, this is our life boat. It carries the fuel to take us back to Earth. We couldn’t afford to use our fuel up. But at the same time, we had no communications with the ground, because the station was dead, and we didn’t seem to be getting into a better situation. We had already gone a few orbits without power. So in the end the three of us agreed to give it a go.” Rule of thumb The initial attempts to halt Mir’s bad spin were a trial‐and‐error affair involving much shouting of commands and floating about the station. Without communications with the ground, the only way to ascertain how Mir was spinning was, quite literally, by rule of thumb: “I knew that the Moon was about half a degree in angular width. My thumb covered the Moon about three times, making one and a half degrees.” Stretching his thumb out against the stars and watching the stars drifting past it, Michael managed to get a rough idea of the rate and direction of Mir’s spin. Based on his thumb measurements, Michael shouted instructions to Vasily who was controlling the Soyuz’s thrusters. After four and a half hours and some nerve‐wrecking failed attempts, they finally saw a result. The Mir’s spin slowed and the station’s position became relatively stable with respect to the Sun. But this wasn’t the end of the problem. For a space station to keep facing the Sun in more or less the same way while it orbits Earth, powerful control mechanisms are needed. Once the power is down, these mechanisms stop functioning. The minutest tugs and pulls will cause the station to start drifting again and over the course of just a few hours it will lose its stable attitude toward the Sun — unless you put the right spin on it.

24 

11 The maths behind the spin As you can see in the image above, the space station Mir is quite a symmetrical object. It is set out along three axes, each perpendicular to the other two, with its centre of gravity located roughly at its central node. The crucial component of Mir’s movement was not its orbit around Earth, but the way it revolved about its own centre of gravity, as this was making it hard for the solar arrays to pick up energy. So for simplicity, imagine, for the moment, this centre of gravity being fixed in one place.

The Swiss mathematician showed in the 18th century that any movement of a rigid body that keeps one of its points fixed must be a rotation about an axis going through the fixed point. In other words, when Mir was set spinning, its initial movement was a rotation about an axis passing through its centre of gravity. However, in general such a rotation is not stable: as the object spins about one axis, it may start rotating about other axes too. In Mir’s case, this meant that a spin about an arbitrary axis afforded practically no chance of a stable attitude toward the Sun.

Euler’s equations of rotational motion describe exactly how the movement of a rotating body evolves over time, and capture how this movement changes if you apply certain external forces. His equations show that every body possesses three very special axes, and that the body’s movement is crucially determined by three numbers called the principal moments of inertia.

When you set an object spinning about an axis, you need to apply a certain amount of force to get it going in the first place. This amount depends on the way in which the object’s mass is distributed with respect to the axis: a wheel, for example, is harder to spin when most of its mass is located around its rim, than when it is concentrated close to the centre. Mathematicians use a number to measure the resistance of the object to being spun, and this number is called a moment of inertia. Each axis has its own moment of inertia.

Euler noticed that, no matter which object you are spinning, whether it’s a space station, a potato or an elephant, the same basic pattern always occurs. The axis that gives you the smallest moment of inertia (the least resistance to being spun) is perpendicular to the axis giving you the greatest moment of inertia. These two axes, together with the line that is perpendicular to both of them, are known as the object’s principal axes of inertia.

Euler s equations of motion Euler s equations describing the rotational motion of a rigid object in the absence of external forces are:

I 1

d  1 ω (t)/dt -  2 ω  2 ω (I 2

- I 3 ) = 0,

I 2

d  2 ω (t)/dt -  3 ω  1 ω (I 3

- I 1 ) = 0,

I 3

d  3 ω (t)/dt -  1 ω  2 ω (I 1

- I 2 ) = 0.

Here, I 1 , I

2 and I

3 are the principal moments of

inertia. The  1 ω ,  2 ω and  3 ω are functions of time and together describe the motion the body has performed after time t. There are similar equations describing the motion when external force, or torque, is applied.

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When an object has some symmetry, the principal axes often coincide with the symmetry axes, and this is the case for Mir: its principal axes of inertia are exactly the x, y and z‐axes shown in the image below. What Euler’s equations also show — and what Michael Foale knew — is that the two axes associated to the smallest and the largest moment of inertia give rise to stable rotations: spin Mir around one of these axes and it will keep on spinning nicely in the same direction. A spin about the third axis, however, is not stable. The right spin A spin around one of Mir’s stable axes could be Mir’s saviour, but it could also spell doom. If the axis was oriented so that the solar arrays face into the Sun whenever Mir passes on the Earth’s sunny side, then a stable spin would allow it to recharge its battery continually. If, however, the axis was oriented so that the solar arrays can’t ever pick up the Sun, then the station would be lost: a lot of energy would be required to brake out of the stable spin and energy was precisely what Mir didn’t have. Mir would be locked into the bad spin forever and, as Michael says, “that would be it, we would have to abandon the station.” So which of Mir’s principal axes were stable? Unfortunately, Michael and his colleagues did not know. Neither did ground control in Moscow, as it turned out later. Using a cracked and broken model of the Mir, and supplementing it with torches taped to its ends, Michael eventually ascertained that their best bet lay with the y‐axis. The Soyuz’s thrusters, because of their position, were unable to induce a rotation about the x‐axis, and rotation about the z‐axis would not have allowed the functioning solar arrays to pick up the Sun. “The best thing we could do was to rotate about the y‐axis and hope that it was stable.

Moments of inertia and stable spins

To demonstrate how a body spins about its three principal

axes, take a box-shaped object, for example an empty

cereal box. Its three principal axes are the ones shown in

the figure below.

If you throw the box up in the air giving it a spin about the

axis marked a in the figure, you will find that the box

keeps spinning in that direction until it falls to the floor.

The same happens if you give it a spin about the axis

marked c. These two axes have the smallest and largest moment of inertia and they are stable.

But if you give the box a spin about the axis marked b, then the box will do a wobble and a flip in

mid-air. This axis has the middle moment of inertia it is unstable.

c b

a

26 

11 “Eventually, we managed to get a rough spin about the y‐axis and to orient it in what we thought was the direction of the Sun. We could not know for sure because at the time we were passing the dark side of the Earth. But then we saw the rim of the Earth go blue, then red, and then the Sun popped out — and low and behold it was roughly in the direction of the y‐axis.”

Almost safe “At that point we managed to charge the station’s batteries continually while the Mir passed on the Earth’s sunny side, which takes about an hour. That was enough to bring the station back to life and to bring up communications with the ground. With Moscow’s help we could get accurate control of our orientation. After 30 hours we could finally relax, get some sleep and start dealing with other consequence of the collision.”

The y‐axis, as it turned out, was not stable. “I noticed that our rotation was stable for only about one and a half hours. Then the Mir would do a flip and the solar arrays were pointing exactly in the opposite direction. What we had discovered was that the y‐axis was the one with the middle moment of inertia.” Luckily though, with batteries charged and communications to the ground re‐established, this did not turn out to be a major problem.

But still, Michael’s adventure was far from over. After the accident, he remained on Mir for another three and a half months until a space shuttle came to get him in October. “After the initial emergency, we lost control again two or three times, because in rescuing Mir we had made its control systems very vulnerable. During this time my worst fear was that we would somehow get locked into a bad spin about the z‐axis, which wouldn’t allow us to recharge our batteries.” So, using a laptop with shaky batteries, Michael set about trying to solve Euler’s equations once and for all to get an accurate idea of the Mir’s rotation dynamics.

This whole incident holds many lessons for space craft design. But the clearest message is one that rarely gets mentioned: it’s the important role of maths. “Mathematics is everywhere in space flight,” says Michael. “When I think about a space craft in orbit, I always think in terms of the underlying maths and physics. I didn’t suddenly turn on my ability to think about maths when the accident happened; it was a continuation of my usual, mathematical way of thinking. You cannot be an effective mission designer, astronaut, flight controller or engineer if you don’t know maths.”

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The origins of chess Shaturanga, the ancestor of all chess‐like games, is thought to have been developed around 1,600 years ago by an Indian philosopher. Since then it has spread across the globe, and evolved into several distinct forms, with China, Japan, Korea, Thailand and Europe all playing similar but distinct games throughout the middle ages. Talking about a revolution International‐level players will spend hours thinking about a single position during a game, and spend their entire careers analysing and perfecting tactics for the “opening” first few moves. But the game is now undergoing another revolution. The processing power of computers has been increasing exponentially since the 1950s when the first chess‐playing programs appeared. By the mid‐80s computers were approaching Grandmaster level, and in 1997 Garry Kasparov, then World Champion, very famously lost an entire match to the IBM supercomputer Deep Blue (the computer won two games, lost one and drew the other three). It is only a matter of time before the machines become absolutely unbeatable, but even Kasparov himself is optimistic about the future for human players, who can use computers as excellent learning tools or in preparing for matches. In the lingo of game theory chess is a finite game with no random element, played between two players each with complete information. It is also zero‐sum, which means that one player’s gain or advantage is necessarily to the opponent’s detriment. As a consequence, it is theoretically possible to play a perfect game of chess ‐ that is, both players could always work out the exact sequence of optimum moves, right through to checkmate. When playing such a perfect game, a player would always know the best move to make regardless of how the opponent responded, which in turn could also always be predicted. The entire game would be transparent, with both players knowing exactly how the game would end 10, 30, or 100 moves in advance. It may be rational ‐ but is it fun? For example, at the start of a game of noughts‐and‐crosses there is a choice between nine permissible moves. Only 512 (2 9 ) different arrays of 0s and Xs can be made on a 3x3 grid, although the maximum number of games is actually less since rotational symmetry ensures some games are effectively identical. Also, of course, many of these arrangements are wins and so finish before all nine squares are filled. You can therefore know all possible games, and so be able to play the optimum response to any move from your opponent. In noughts‐and‐crosses, a draw is the only result possible between two rational players. Chess, however, is almost inconceivably more complex, and the pieces can be arranged on the 64 squares of the board in 10 44 distinct ways. One mathematician has calculated

Maths, Computers And Chess

28 

11 that there are about 10 (1050) different legal games, which is far more than the number of particles in the entire visible universe. This is effectively an infinite number of permutations, and so in all practical senses it is impossible to play chess perfectly. From most positions the player will have a choice of which piece to move, and often also which square to move it to. From the initial position White can choose between only 20 legal moves, but this balloons into many hundreds of possibilities, or “candidates”, in the middle game when more pieces have freedom of movement. The progression of a game can therefore be thought of as a flourishing bush, with numerous twigs coming off the same branching point to indicate each of the separate legal moves available from one position. At the next turn, each of these twigs itself branches off into many possibilities. As you look “deeper” into a position (that is, as you look a greater number of moves ahead), the number of possibilities that must be analysed increases exponentially. So the trick to playing chess efficiently is quickly to ignore lines of play that don’t seem to improve your position, and to focus your analysis on a few promising candidates in order to “prune the game‐tree”. The computation is especially difficult because you must find a move that not only serves you well, but that also limits the good options available to your opponent. The crux of the difference between humans and machines is the disparate ways that we prune this tree. Who is the better gardener? The brain of an experienced player seems to be able to prune the search‐tree almost intuitively, and Grandmasters typically study only a handful of candidates at a position. This extremely effective focusing means that those lines selected can be studied to a great depth. Skilled players have a good sense of “boardfeel”: without explicitly analysing a position they can spot which pieces are in commanding locations, which pawn structures are strong, where there are weaknesses in the opponent’s formation, and so on. Computers play like a human novice, albeit one that can check millions of moves a second. They apply a “brute force” technique to playing chess, and have no conception of this abstract boardfeel. A computer can only wander blindly along the branches of the search tree, until it stumbles across a sequence

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of moves that may prove beneficial. A good human player, however, will identify a threat or an objective and will think laterally and inventively about how to use the available pieces in a combined way to achieve the desired result. Although the programmed procedures, or algorithms, that computers use to reduce the search‐space are being developed all the time, this is still the weakest aspect of a chess‐playing computer. There are another two methods that programmers use to improve a machine’s game. The first is an “opening library” that contains established sequences of moves for the first few turns. This received wisdom is the result of centuries of analysis of the best way to advance your pieces in the earliest stage of a game. The second method is known as a “bottom‐up approach”. Many scenarios in the endgame are known to be an enforceable win for one of the players, such as a lone king versus a queen, or two rooks, or a rook and pawn against a bishop. A computer can then “unmove” pieces, or step backwards, to find all the positions from which these known wins can be reached. This approach has produced enormous endgame databases so that if a computer aims for one of the included positions it can then play an essentially perfect game. All endgames with less than six pieces have already been analysed exhaustively. On the whole these computer investigations have agreed with the analyses of chess masters through the centuries. Some surprising results have been found, however, such as for the endgame of rook and knight versus two knights. This is generally assumed to be a drawn game, but from a few special positions it is possible for White to capture one of the knights safely and ensure an eventual win. The position shown here requires the longest known sequence before capture: 243 moves. However, this exact position is very unlikely to arise, and a real game would be ended by the 50‐move rule, which states a game is a draw if 50 moves pass without a single capture or promotion. Tactics for beating a computer at chess Because a computer relies on a strictly mathematical, brute force approach in the middle game it can see only a limited number of moves in advance and is oblivious to patterns or general manoeuvres. This short‐sightedness is known as the horizon, and as long as a human player makes no direct threats that are within the computer’s horizon, it is unaware of the danger. The human can lure the computer into a devastating trap, or slowly prepare an overwhelming assault and only strike when it is too late for the computer to defend itself effectively.

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11 Another tactic that human players have successfully used takes advantage of the fact that a computer may assess positions in the game as if they have only just arisen. It will analyse millions of branches in the game tree all over again, even if it has seen an almost identical position very recently. A human will recognise that some moves hardly change the situation, and so will not bother reanalysing it. This difference can be exploited in clocked games where the players must make all of their moves within a specified time limit. The human can sometimes win simply by moving a piece back and forth between the same two squares, and the computer wastes all of its time re‐examining each position. Another common tactic in a timed game is to play quick moves that are mediocre but probably safe in that they don’t lose you a piece or leave you open to checkmate. An especially effective move is one that complicates the position on the board and looks as if it could be part of an elaborate attack. The computer cannot know that this is what you are trying, and so will devote a lot of its time carefully analysing the position. This is known as the enough rope principle. Trees and horizons The horizon effect is becoming less important as faster processors and more efficient tree‐pruning algorithms are developed. But the better programs may be susceptible to the inverse problem. This position comes from a game between a human playing as White against a computer. White has just placed the Black king in check with his queen on a8. The computer moved its rook to e8 and the audience laughed, thinking the program must have malfunctioned to play such a feeble move. Although the check has been blocked, the rook is completely undefended and the White queen can immediately capture this valuable piece. The more natural response might be to move the king out of check to g7 instead. It wasn’t until the program was tested later that an explanation was found. The machine had realised that if the king moved to g7 then it would be unable to prevent White forcing an elegant checkmate five moves later. But by sacrificing the rook checkmate is delayed, which the computer assessed as a gain. Any human player, however, would have moved the king anyway, because there is the chance that an imperfect opponent might not have seen this mating sequence. Sacrificing the rook is suicidal and effectively throws away the game. So, paradoxically, in certain situations a computer would perform better against humans if it played what it knew to be a weaker move! As we’ve seen, then, programming a computer to play chess is a natural extension to using maths to analyse positions. Chess is inconceivably mathematically complex, and yet both humans and computers are able to play, although by using very different methods. Despite a computer’s raw calculating power, human intelligence can still outwit the machines during the middle game. Our time is already up with draughts, however, and computers are close to solving the game.