Thanks Dr. Gowers’. These are invaluable insights into basics. Thanks for giving so much of your time.
Thanks a lot Prof. Gowers! Math should be sans ambiguities as far as possible…!
I hope my students and readers can appreciate the details in this blog article of Prof. Gowers.
GENERATING FUNCTIONS and RECURRENCE RELATIONS:
The concept of a generating function is one of the most useful and basic concepts in the theory of combinatorial enumeration. If we want to count a collection of objects that depend in some way on n objects and if the desired value is say, , then a series in powers of t such as is called a generating function for . The generating functions arise in two different ways. One is from the investigation of recurrence relations and another is more straightforward: the generating functions arise as counting devices, different terms being specifically included to account for specific situations which we wished to count or ignore. This is a very fundamental, though difficult, technique in combinatorics. It requires considerable ingenuity for its success. We will have a look at the bare basics of such stuff.
We start here with the common knowledge:
….(2i) where sum of the products of the ‘s taken r at a time. …(2ii)
Incidentally, the ‘s thus defined in (2ii) are called the elementary symmetric functions associated with the ‘s. We will re-visit these functions later.
Let us consider the algebraic identity (2i) from a combinatorial viewpoint. The explicit expansion in powers of t of the RHS of (2i) is symbolically a listing of the various combinations of the ‘s in the following sense:
represents all the 1-combinations of the ‘s
represents all the 2-combinations of the ‘s
and so on.
In other words, if we want the r-combinations of the ‘s, we have to look only at the coefficients of . Since the LHS of (2i) is an expression which is easily constructed and its expansion generates the combinations in the said manner,we say that the LHS of (2i) is a Generating Function (GF) for the combinations of the ‘s. It may happen that we are interested only in the number of combinations and not in a listing or inventory of them. Then, we need to look for only the number of terms in each coefficient above and this number will be easily obtained if we set each as 1. Thus, the GF for the number of combinations is n times;
and this is nothing but . We already know that the expansion of this gives as the coefficient of and this tallies with the fact that the number of r-combinations of the ‘s is . Abstracting these ideas, we make the following definition:
The Ordinary Generating Function (OGF) for a sequence of symbolic expressions is the series
If is a number which counts a certain type of combinations or permutations, the series is called the Ordinary Enumeration (OE) or counting series for for
The OGF for the combinations of five symbols a, b, c, d, e is
The OE for the same is . The coefficient of in the first expression is
(*) abcd+abce+ abde+acde+bcde.
The coefficient of in the second expression is , that is, 5 and this is the number of terms in (*).
The OGF for the elementary symmetric functions in the symbols is ….(2iv)
This is exactly the algebraic result with which we started this section.
The fact that the series on the HRS of (2iii) is an infinite series should not bother us with questions of convergence and the like. For, throughout (combinatorics) we shall be working only in the framework of “formal power series” which we now elaborate.
*THE ALGEBRA OF FORMAL POWER SERIES*
The vector space of infinite sequences of real numbers is well-known. If and are two sequences, their sum is the sequence , and a scalar multiple of the sequence is . We now identify the sequence with with the “formal” series
where only means the following:
In the same way, , where corresponds to the formal series:
we define: , and .
The set of all power series f now becomes a vector space isomorphic to the space of infinite sequences of real numbers. The zero element of this space is the series with every coefficient zero.
Now, let us define a product of two formal power series. Given f and g as above, we write where
, where .
The multiplication is associative, commutative, and also distributive wrt addition. (the students/readers can take up this as an appetizer exercise !!) In fact, the set of all formal power series becomes an algebra. It is called the algebra of formal power series over the real s. It is denoted by , where means the algebra of reals. We further postulate that in iff for all . As we do in polynomials, we shall agree that the terms not present indicate that the coefficients are understood to be zero. The elements of may be considered as elements of . In particular, the unity 1 of is also the unity of . Also, the element with belongs to , it being the formal power series with and all other ‘s zero. We now have the following important proposition which is the only tool necessary for working with formal power series as far as combinatorics is concerned:
Proposition : 2_4:
The element f of given by (2v) has an inverse in iff has an inverse in .
If is such that , the multiplication rule in tells us that so that is the inverse of . Hence, the “only if” part is proved.
To prove the “if” part, let have an inverse in . We will show that it is possible to find in such that . If such a g were to exist, then the following equations should hold in order that , that is,
So we have from the first equation. Substituting this value of in the second equation, we get in terms of the ‘s. And, so on, by the principle of mathematical induction, all the ‘s are uniquely determined. Thus, f is invertible in . QED.
Note that it is the above proposition which justifies the notation in , equalities such as
The above is true because the RHS has an inverse and
So, the unique inverse of is and vice versa. Hence, the expansion of as above. Similarly, we have
and many other such familiar expansions.
There is a differential operator in in , which behaves exactly like the differential operator of calculus.
Then, one can easily prove that is linear on , and further
from which we get the term “Taylor-MacLaurin” expansion
In the same manner, one can obtain, from , which in turn is equal to
the result which mimics the logarithmic differentiation of calculus, viz.,
The truth of this in is seen by multiplying the series on the RHS of (2vii) by the series for , and thus obtaining the series for .
Let us re-consider generating functions now. We saw that the GF for combinations of is .
Let us analyze this and find out why it works. After all, what is a combination of the symbols : ? It is the result of a decision process involving a sequence of independent decisions as we move down the list of the ‘s. The decisions are to be made on the following questions: Do we choose or not? Do we choose or not? Do we choose or not? And, if it is an r-combination that we want, we say “yes” to r of the questions above and say “no” to the remaining. The factor in the expression (2ii) is an algebraic indication of the combinatorial fact that there are only two mutually exclusive alternatives available for us as far as the symbol is concerned. Either we choose or not. Choosing “” corresponds to picking the term and choosing “not ” corresponds to picking the term 1. This correspondence is justified by the fact that in the formation of products in the expression of (2iv), each term in the expansion has only one contribution from and that is either or .
The product gives us terms corresponding to all possible choices of combinations of the symbols and — these are:
standing for the choice “not-” and “not-”
standing for the choice of and “not-”
standing for the choice of “not-” and .
standing for the choice of and .
This is, in some sense, the rationale for (2iv) being the OGF for the several r-combinations of .
We shall now complicate the situation a little bit. Let us ask for the combinations of the symbols with repetitions of each symbol allowed once more in the combinations.
To be discussed in the following article,
Combinatorics, Theory and Applications, V. Krishnamurthy, East-West Press.
Amazon India Link:
1) In how many ways can 5 men and 5 women be seated in a round table if no two women may be seated side by side?
2) Six generals propose locking a safe containing top secret with a number of different locks. Each general will be given keys to certain of these locks. How many locks are required and how many keys must each general have so that, unless at least four generals are present, the safe cannot be opened?
3) How many integers between 1000 and 9999 inclusive have distinct digits? Of these, how many are even numbers? How many consist entirely of odd digits?
4) In how many ways can 9 distinct objects be placed in 5 distinct boxes in such a way that 3 of these boxes would be occupied and 2 would be empty?
5) In how many permutations of the word AUROBIND do the vowels appear in the alphabetical order?
6) There is an unlimited supply of weights of integral numbers of grams. Using n or fewer weights, find the number of ways in which a weight of m grams can be obtained. Prove that there is a bijection of the set of all such ways on the set of increasing words of length or ordered letters.
7) How many distinct solutions are there of (a) in positive integers and (b) in non-negative integers?
8) A train with n passengers aboard makes m stops. In how many ways can the passengers distribute themselves among these m stops as alighting passengers? if we are concerned only with the number of alighting passengers at each stop, how would the answer be modified?
9) There are 16 books on a bookshelf. In how many ways can 6 of these books be selected if a selection must not include two neighbouring books?
10) Show that there are distinct throws of a throw with n non-distinct dice.
11) Given n indistinguishable objects and n additional distinct objects —- also distinct from the earlier n objects — in how many ways can we choose n out of the 2n objects?
12) Establish the following relations:
13) Prove the following identity for all real numbers x:
14) Express in terms of , , …by using the ‘s. Express in terms of , , …by using the ‘s.
15) A circular loop is divided into p parts, p prime. In how many ways can we paint the loop with n colours if we do not distinguish between patterns which differ only by a rotation of the loop? Deduce Fermat’s Little theorem: is divisible by p if p is prime.
16) In problem 15, prove that is also divisible by 2p if . Where is the hypothesis that p is prime used in Problem 15 or in this problem?
17) How many equivalence relations are possible on an n-set?
18) The complete homogeneous symmetric function of n variables , , , of degree r is defined as the summation being taken over all ordered partitions of r, where the parts are also allowed to be zero. How many terms are there in ?
Test yourself ! Improve your mettle in math !
1) Show that quadrilateral ABCD can be inscribed in a circle iff and are supplementary.
2) Prove that a parallelogram having perpendicular diagonals is a rhombus.
3) Prove that a parallelogram with equal diagonals is a rectangle.
4) Show that the diagonals of an isosceles trapezoid are equal.
5) A straight line cuts two concentric circles in points A, B, C and D in that order. AE and BF are parallel chords, one in each circle. If CG is perpendicular to BF and DH is perpendicular to AE, prove that .
6) Construct triangle ABC, given angle A, side AC and the radius r of the inscribed circle. Justify your construction.
7) Let a triangle ABC be right angled at C. The internal bisectors of angle A and angle B meet BC and CA at P and Q respectively. M and N are the feet of the perpendiculars from P and Q to AB. Find angle MCN.
8) Three circles with radii , with . They are placed such that lies to the right of and touches it externally; lies to the right of and touches it externally. Further, there exist two straight lines each of which is a direct common tangent simultaneously to all the three circles. Find in terms of and .
1) Assume that in a ABC and , we know that , , . Prove that without using the SSS congruence criterion.
2) Let be isosceles with base BC. Then, . Also, the median from vertex A, the bisector of , and the altitude from vertex A are all the same line. Prove this.
3) If two triangles have equal hypotenuses and an arm of one of the triangles equals an arm of the other, then the triangles are congruent. Prove.
4) An exterior angle of a triangle equals the sum of the two remote interior angles. Also, the sum of all three interior angles of a triangle is 180 degrees.
5) Find a formula for the interior angles of an n-gon.
6) Prove that the opposite sides of a parallelogram are equal.
7) In a quadrilateral ABCD, suppose that AB=CD and AD=BC. Then, prove that ABCD is a parallelogram.
8) In a quadrilateral ABCD, suppose that AB=CD and AB is parallel to CD. Then, prove that ABCD is a parallelogram.
9) Prove that a quadrilateral is a parallelogram iff its diagonals bisect each other.
10) Given a line segment BC, the locus of all points equidistant from B and C is the perpendicular bisector of the segment. Prove.
11) Corollary to problem 10 above: The diagonals of a rhombus are perpendicular. Prove.
12) Let AX be the bisector of in . Then, prove . In other words, X divides BC into pieces proportional to the lengths of the nearer sides of the triangle. Prove.
13) Suppose that in , the median from vertex A and the bisector of are the same line. Show that .
14) Prove that there is exactly one circle through any three given non collinear points.
15) An inscribed angle in a circle is equal in degrees to one half its subtended arc. Equivalently, the arc subtended by an inscribed angle is measured by twice the angle. Prove.
16) Corollary to above problem 15: Opposite angles of an inscribed quadrilateral are supplementary. Prove this.
17) Another corollary to above problem 15: The angle between two secants drawn to a circle from an exterior point is equal in degrees to half the difference of the two subtended arcs. Prove this.
18) A third corollary to above problem 15: The angle between two chords that intersect in the interior of a circle is equal in degrees to half the sum of the two subtended arcs. Prove this.
19) Theorem (Pythagoras): If a right triangle has arms of lengths a and b and its hypotenuse has length c, then . Prove this.
20) Corollary to above theorem: Given a triangle ABC, the angle at vertex C is a right angle iff side AB is a diameter of the circumcircle. Prove this.
21) Theorem: The angle between a chord and the tangent at one of its endpoints is equal in degrees to half the subtended arc. Prove.
22) Corollary to problem 21: The angle between a secant and a tangent meeting at a point outside a circle is equal in degrees to half the difference of the subtended arcs.
23) Fix an integer, . Given a circle, how should n points on this circle be chosen so as to maximize the area of the corresponding n-gon?
24) Theorem: Given and , suppose that and . Then, prove that and so . Prove this theorem.
25) Theorem: If , then the lengths of the corresponding sides of these two triangles are proportional. Prove.
26) The following lemma is important to prove the above theorem: Let U and V be points on sides AB and AC of . Then, UV is parallel to BC if and only if . You will have to prove this lemma as a part of the above proof.
27) Special case of above lemma: Let U and V be the midpoints of sides AB and AC, respectively in . Then, UV is parallel to BC and .
28) Suppose that the sides of are proportional to the corresponding sides of . Then, .
29) Given and , assume that and that . Then, .
30) Consider a non-trivial plane geometry question now: Let P be a point outside of parallelogram ABCD and . Prove that .
31) Given a circle and a point P not on the circle, choose an arbitrary line through P, meeting the circle at points X and Y. Then, the quantity depends only on the point P and is independent of the choice of the line through P.
32) You can given an alternative proof of Pythagoras’s theorem based on the following lemma: Suppose is a right triangle with hypotenuse AB and let CP be the altitude drawn to the hypotenuse. Then, . Prove both the lemma and based on it produce an alternative proof of Pythagorean theorem.
33) Prove the following: The three perpendicular bisectors of the sides of a triangle are concurrent at the circumcenter of the triangle.
34) Prove the law of sines.
35) Let R and K denote the circumradius and area of , respectively and let a, b and c denote the side lengths, as usual. Then, .
36) Theorem: The three medians of an arbitrary triangle are concurrent at a point that lies two thirds of the way along each median from the vertex of the triangle toward the midpoint of the opposite side.
37) Time to ponder: Prove: Suppose that in , medians BY and CZ have equal lengths. Prove that .
38) If the circumcenter and the centroid of a triangle coincide, then the triangle must be equilateral. Prove this fact.
39) Assume that is not equilateral and let G and O be its centroid and circumcentre respectively. Let H be the point on the Euler line GO that lies on the opposite side of G from O and such that . Then, prove that all the three altitudes of pass through H.
40) Prove the following basic fact about pedal triangles: The pedal triangles of each of the four triangles determined by an orthic quadruple are all the same.
41) Prove the following theorem: Given any triangle, all of the following points lie on a common circle: the three feet of the altitudes, the three midpoints of the sides, and the three Euler points. Furthermore, each of the line segments joining an Euler point to the midpoint of the opposite side is a diameter of this circle.
42) Prove the following theorem and its corollary: Let R be the circumradius of triangle ABC. Then, the distance from each Euler point of to the midpoint of the opposite side is R, and the radius of the nine-point circle of is . The corollary says: Suppose is not a right angled triangle and let H be its orthocentre. Then, , , , and have equal circumradii.
43) Prove the law of cosines.
44) Prove Heron’s formula.
45) Express the circumradius R of in terms of the lengths of the sides.
46) Prove that the three angle bisectors of a triangle are concurrent at a point I, equidistant from the sides of the triangle. If we denote the by r the distance from I to each of the sides, then the circle of radius r centered at I is the unique circle inscribed in the given triangle. Note that in order to prove this, the following elementary lemma is required to be proved: The bisector of angle ABC is the locus of points P in the interior of the angle that are equidistant from the sides of the triangle.
47) Given a triangle with area K, semiperimeter s, and inradius r, prove that . Use this to express r in terms of the lengths of the sides of the triangle.
Please be aware that the above set of questions is almost like almost like a necessary set of pre-requisites for RMO geometry. You have to master the basics first.