Category Archives: INMO

Graphs of trig raised to trig

Question: Consider the function

y=f(x)=x^{x}. Can you graph it? It is variable raised to variable. Send me your observations.

Now, consider the functions:

(\tan \theta)^{\tan \theta}, (\tan \theta)^{\cot \theta},

(\cot \theta)^{\tan {\theta}}, (\cot \theta)^{\cot \theta}.

Can you graph these? What is the difference between these and the earlier generalized case?

Now, consider the function:

Let 0 \deg < \theta < 45 \deg.

Arrange t_{1}=(\tan \theta)^{\tan \theta}, t_{2}=(\tan \theta)^{\cot \theta}

t_{3}=(\cot \theta)^{\tan \theta} and t_{4}=(\cot \theta)^{\cot \theta}

in decreasing order.

Kindly send your comments/observations.

More later,

Nalin Pithwa

Are complex numbers complex ?

You  might perhaps think that complex numbers are complex to handle. Quite contrary. They are easily applied to various kinds of engineering problems and are easily handled in pure math concepts compared to real numbers. Which brings me to another point. Mathematicians are perhaps short of rich vocabulary so they name some object as a “ring”, which is not a wedding or engagement ring at all; there is a “field”, which is not a field of maize at all; then there is a “group”, which is just an abstract object and certainly not a group of people!!

Well, here’s your cryptic complex problem to cudgel your brains!

Problem:

Prove the identity:

({n \choose 0}-{n \choose 2}+{n \choose 4}- \ldots)^{2}+({n \choose 1}-{n \choose 3}+{n \choose 5}-\ldots)^{2}=2^{n}

Solution:

Denote

x_{n}={n \choose 0}-{n \choose 2}+{n \choose 4}- \ldots and

y_{n}={n \choose 1}-{n \choose 3}+{n \choose 5}-\ldots

and observe that (1+i)^{n}=x_{n}+i y_{n}

Passing to the absolute value it follows that

|x_{n}+y_{n}i|=|(1+i)^{n}|=|1+i|^{n}=2^{n/2}.

This is equivalent to x_{n}^{2}+y_{n}^{2}=2^{n}.

More later,

Nalin Pithwa

Announcement: A Full Scholarship Program

We are Mathematics Hothouse, Bangalore, http://www.mathothouse.com We are pleased to announce that henceforth, every academic year, we will be admitting 5 students with full scholarship or 100% discount, from any part of India, who are talented, deserving or needy, to our program for RMO and INMO coaching. The coaching will be via on-line, live, video interactive Skype sessions mimicking traditional classroom or just classroom coaching or even correspondence course.

If you wish to apply, please write to mathhothouse01@gmail.com

Regards,

Nalin Pithwa

Quick Review of Trigonometric Optimization Methods

Let us review together the four general methods we can use for triangular optimization.

I) Trigonometric Method: 

The essence of this method is the observation that the cosine of an angle is at most one and that it equals 1 only when the angle is zero. This fact is applied to the difference between two of the angles A, B and C, holding the third angle fixed, to show that unless those two angles are equal, the objective functionn can be increased (or decreased as the case may be). Consequently, at an optimal solution, these two angles must be equal. If the objective function is symmetric (as is the case, in almost all problems of triangular optimization), then every two of A, B and C must be equal to each other and hence, the triangle ABC must be equilateral.

This method is elementary and easy to apply. Even when the objective function is only partially symmetric, that is, symmetric in two but not in all the three variables, it can be applied to those two variables, holding the third variable fixed. Suppose, for example, that we want to maximize f(A,B,C)=\cos{A}+2\cos{B}+\cos{C}. This is symmetric in A and C. So, by the same reasoning as for maximizing \cos{A}+\cos{B}+\cos{C}, which at an optimal solution we must have A=C. Then, B=\pi-2A, which makes f effectively a function of just one variable, viz., \cos{A}+2\cos(\pi-2A)+\cos{A}, which equals 2\cos{A}-4\cos^{2}{A}+2. This can be maximized as a quadratic in \cos{A} either by completing the square or using calculus. The maximum occurs when A=\cos^{-1}{1/4}. Thus, the maximum value of \cos{A}+2\cos{B}+\cos{C} for a triangle ABC is 1/4. The method, of course, fails if the function is not even partially symmetric. This is not surprising. Basically, in a triangular optimization problem, we are dealing with a function f(A,B,C) of three variables. Because of the constraint A+B+C=\pi, any one of the variables can be expressed in terms of the other two. This effectively makes f a function of two variables. Optimization of functions of several variables requires advanced methods. It is only when f satisfies some other conditions such as partial symmetry that we can hope to reduce the number of variables further so that elementary methods can be applied.

II) Algebraic Method: 

The essence of this method is to reduce the optimization problem to some inequality using suitable trigonometric formulae or identities. The inequality is then established using some standard inequality such as the AM-GM-HM inequality, or Jensen’s inequality, or sometimes, by doing some more basic work. The fundamental ideas are very simple, viz., (a) the square of any real number is non-negative and is zero only when that number is zero, and (b) the sum of two or more non-negative numbers is non-negative and vanishes if and only if each of the term is zero. When this method works, it works elegantly. But it is not always easy to come up with the right algebraic manipulations. Sometimes, certain simplifying substitutions have to be used. Still, it is an elementary method and deserves to be tried.

III) Jensen’s inequality:

This is a relatively advanced method. It is directly applicable when the objective function is, or can be recast, in a certain form, viz., h(A)+h(B)+h(C), where h is a function of one variable whose second derivative maintains the same sign over a suitable interval. But, even when h fails to do so, the method can sometimes be applied with a suitable conversion of the problem.

IV) Lagrange’s Multipliers:

This is a highly advanced method based on the calculus of functions of several variables. It is applicable to all types of objective functions, not just those that are symmetric or partially symmetric. When applied to triangular optimization problems with symmetric objective functions, the optimal solution is either degenerate or an equilateral triangle.

Naturally, for a particular given problem, some of these methods may work better than others. The method of Lagrange’s multipliers is the surest but the most mechanical of all the four. The algebraic method is artistic and sometimes gives the answer very fast. Jensen’s inequality also works fast once you are able to cast the objective function in a certain form. Such a recasting may involve some ingenuity sometimes. The trouble is that both these methods work only in the case of an  optimization problem where the objective function is symmetric. And, in such cases, the method of Lagrange’s multipliers makes a mincemeat of the problem. From an examination point of view, this is a boon if a question about triangular optimization is asked in a “fill in the blanks” or “multiple choice” form, where you don’t have to show any reasoning. if the objective function is symmetric, then the optimal solution is either degenerate or an equilateral triangle. But, degenerate triangles are often excluded from the very definition of a triangle because of the requirement that the three vertices of a triangle must be distinct and non-collinear and, in any case, such absurdities are unlikely to be asked in an examination! So, it is a safe bet to simply assume that the optimal solution is an equilateral triangle and proceed with further work (namely, calculating the value of the objective function for an equilateral triangle). This saves you a lot of time.

More later,

Nalin Pithwa

 

The Sieve — elementary combinatorial applications

One powerful tool in the theory of enumeration as well as in prime number theory is the inclusion-exclusion principle (sieve of Erathosthenes). This relates the cardinality of the union of certain sets to  the cardinalities of the intersections of some of them, these latter cardinalities often being easier to handle. However, the formula does have some handicaps, it contains terms alternating in sign, and in general it has too many of them!

A natural setting for the sieve is in the language of probability theory. Of course, this only means a division by the cardinality of the underlying set, but it has the advantage that independence of occurring events can be defined. Situations in which events are almost independent are extremely important in number theory and also arise in certain combinatorial applications. Number theorists have developed ingenious methods to estimate the formula when the events (usually divisibility by certain primes) are almost independent. We give here the combinatorial background of some of these methods. Their actual use, however, rests upon complicated number theoretic considerations which are here illustrated only by two problems.

It should be emphasized that the sieve formula has many applications in quite different situations.

A beautiful general theory of inclusion-exclusion, usually referred to as the theory of the Mobius function is due to L. Weisner, P. Hall and G. C. Rota.

Question 1: In a high school class of 30 pupils, 12 pupils like mathematics, 14 like physics and 18 chemistry, 5 pupils like both mathematics and physics, 7 both physics and chemistry, 4 pupils like mathematics and chemistry. There are 3 who like all three subjects. How many pupils do not like any of them?

Question 2: (a) The Sieve Formula: 

Let A_{1}, \ldots, A_{n} be arbitrary events of a probability space (\Omega, P). For each

I \subseteq \{ 1, \ldots , n\}, let

A_{I}= \prod_{i\in I}A_{i}, A_{\phi}=\Omega

and let \sigma_{k}=\sum_{|I|=k}P(A_{I}), \sigma_{0}=1

Then, P(A_{1}+\ldots + A_{n})=\sum_{j=1}^{n}(-1)^{j-1}\sigma_{j}

Question 2: (b) (Inclusion-Exclusion Formula)

Let A_{1}, \ldots, A_{n} \subseteq S, where S is a finite set, and let

A_{I}=\bigcap_{ j \in J}A_{j}, A_{\phi}=S. Then,

|S-(A_{1}\cup \ldots \cup A_{n})|=\sum_{J \subset \{ 1, \ldots n\}}(-1)^{|I|}|A_{I}|

Hints:

1) The number of pupils who like mathematics or physics is not 12+14. By how much is 26 too  large?

2) Determine the contribution of any atom of the Boolean Algebra generated by A_{1},, \ldots A_{n} on each side.

Solutions.

1) Let us subtract from 30 the number of pupils who like mathematics, physics, chemistry, respectively:

30-12-14-13.

This way, however, a student who likes both mathematics and physics is subtracted twice; so we have to add them back, and also for the other pairs of subjects:

30-12-14-13+5+7+4.

There is still trouble with those who like all three subjects. They were subtracted 3 times, but back 3 times, so we have to subtract them once more to get the result:

30-12-14-13+5+7+4-3=4/

2) (a) Let B=A_{1}A_{2}\ldots A_{k}\overline A_{k+1}\ldots \overline A_{n}

be any atom of the Boolean algebra generated by A_{1}, A_{2}, \ldots A_{n} (with an appropriate choice of indices, every atom has such a form.) Every event in the formula is the union of certain (disjoint) atoms; let us express each P(A_{I}) and P(A_{1}+A_{2}+\ldots +A_{n}) as the sum of the probabilities of the corresponding atoms. We show that the probability of any given atom cancels out.

The coefficient of P(B) on the left hand side is

1, if k \neq 0 and 0, if k=0

B occurs in A_{I} if I \subseteq \{ 1, \ldots k\}. so its coefficient on the right hand side is

\sum_{j=1}^{k}\left( \begin{array}{c}    k \\    j \end{array} \right) (-1)^{j}=1-(1-1)^{k}=1, k \neq 04, and latex 0 if k =0$.

Thus, P(B) has the same coefficient on both sides, which proves part a.

Solution (b):

Choose an element x of S by a uniform distribution. Then, A_{i} can be identified with the event that

x \in A_{I}, and we have

P(A_{i})=\frac{|A_{i}|}{|S|}

So, we have, by the above,

P(A_{1}+A_{2}+\ldots + A_{n})=\sum_{j=1}^{n}(-1)^{j-1}\sum_{|I|=j}\frac{|A_{I}|}{|S|}, where

\sum_{\phi \neq I \subseteq \{ 1, \ldots n\} }(-1)^{|I|-1}\frac{A_{I}}{|S|}, or equivalently

P(\overline{A_{1}}\ldots \overline{A_{n}})=1-P(A_{1}+\ldots + A_{n}), which in turn equals,

1- \sum_{\phi \neq I \subseteq \{ 1, \ldots , n\}}(-1)^{|I|-1}\frac{|A|}{|S|}

The assertion (b) follows on multiplying by |S|.

More later,

Nalin Pithwa

Jensen’s inequality and trigonometry

The problem of maximizing \cos{A}+\cos{B}+\cos{C} subject to  the constraints A \geq 0,

B \geq 0, C \geq 0 and A+B+C=\pi can be done if instead of the AM-GM inequality we use a stronger inequality, called Jensen’s inequality. It is stated as follows:

Theorem. 

Suppose h(x) is a twice differentiable, real-valued function on an interval [a,b] and that h^{''}(x)>0 for all a<x<b. Then, for every positive integer m and for all points x_{1}, x_{2}, \ldots x_{m} in [a,b], we have

h(\frac{x_{1}+x_{2}+\ldots+x_{m}}{m}) \leq \frac{h(x_{1})+h(x_{2})+h(x_{3})+\ldots+h(x_{m})}{m}

Moreover, equality holds if and only if x_{1}=x_{2}=\ldots=x_{m}. A similar result holds if

h^{''}(x)<0 for all a<x<b except that the inequality sign is reversed.

What this means is that the value of assumed by the function h at the arithmetic mean of a given set of points in the interval [a,b] cannot exceed the arithmetic mean of the values assumed by h at these points, More compactly, the value at a mean is at most the mean of values if h^{''} is positive in the open interval (a,b) and the value at a mean is at least the mean of values if h^{''} is negative on it. (Note that h^{''} is allowed to vanish at one or both the end-points of the interval [a,b].)

A special case of Jensen’s inequality is the AM-GM inequality.

Jensen’s inequality can also be used to give easier proofs of certain other trigonometric inequalities whose direct proofs are either difficult or clumsy. For example, applying Jensen’s inequality to the function h(x)=\sin{x} on the interval [0,\pi] one gets the following result. (IITJEE 1997)

If n is a positive integer and 0<A_{i}<\pi for i=1,2,\ldots, n, then

\sin{A_{1}}+\sin{A_{2}}+\ldots+\sin{A_{n}} \leq n \sin{(\frac{A_{1}+A_{2}+\ldots+A_{n}}{n})}.

More later,

Nalin Pithwa

Trigonometric Optimization continued

Prove that in any acute angled triangle ABC, \tan {A}+\tan{B}+\tan{C} \geq 3\sqrt{3} with equality holding if and only if the triangle is equilateral. (IITJEE 1998)

Proof:

Suggestion: Try this without reading further! It looks complicated, but need not be so!! Then, after you have attempted whole-heartedly, compare your solution with the one below.

The solution to the above problem is based on the well-known identity:

\tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}. For brevity, denote \tan{A}, \tan{B}, \tan{C}

by x, y and z respectively. As ABC is acute-angled, x, y, z are all positive and the AM-GM inequality which says

x+y+z \geq 3{(xyz)}^{1/3} can be applied. Taking cubes of both the sides and cancelling

x+y+z, (which is positive) this gives (x+y+z)^{2} \geq 27. Taking square root we get the desired inequality. If equality is to hold, then it must also hold in the AM-GM inequality, which can happen if and only if

x=y=z, that is, if and only if the triangle is equilateral.

Still, this approach requires some caution. Actually, there are so many trigonometric identities that there is no unanimity as to which ones among them are standard enough to be assumed without proof !! But, of course, the IITJEE Examinations, both Mains and Advanced are multiple choice only.

More later,

Nalin Pithwa