Category Archives: Cnennai Math Institute Entrance Exam

The Sandwich Theorem or Squeeze Play Theorem

It helps to think about the core concepts of Calculus from a young age, if you want to develop your expertise or talents further in math, pure or applied, engineering or mathematical sciences. At a tangible level, it helps you attack more or many questions of the IIT JEE Advanced Mathematics. Let us see if you like the following proof, or can absorb/digest it:

Reference: Calculus and Analytic Geometry by Thomas and Finney, 9th edition.

The Sandwich Theorem:

Suppose that g(x) \leq f(x) \leq h(x) for all x in some open interval containing c, except possibly at x=c itself. Suppose also that \lim_{x \rightarrow c}g(x)= \lim_{x \rightarrow c}h(x)=L. Then, \lim_{x \rightarrow c}f(x)=c.

Proof for Right Hand Limits:

Suppose \lim_{x \rightarrow c^{+}}g(x)=\lim_{x \rightarrow c^{+}}h(x)=L. Then, for any \in >0, there exists a \delta >0 such that for all x, the inequality c<x<c+\delta implies L-\in<g(x)<L+\in and L-\in<h(x)<L+\in ….call this (I)

These inequalities combine with the inequality g(x) \leq f(x) \leq h(x) to give

L-\in <g(x) \leq f(x) \leq h(x)<L+\in

L-\in <f(x)<L+\in

-\in <f(x)-L<\in….call this (II)

Therefore, for all x, the inequality c<x<c+\delta implies |f(x)-L|<\in. …call this (III)

Proof for LeftHand Limits:

Suppose \lim_{x \rightarrow c^{-}} g(x)=\lim_{x \rightarrow c^{-}}=L. Then, for \in >0 there exists a \delta >0 such that for all x, the inequality c-\delta <x<c implies L-\in<g(x)<L+\in and L-\in<h(x)<L+\in …call this (IV).

We conclude as before that for all x, c-\delta <x<c implies |f(x)-L|<\in.

Proof for Two sided Limits:

If \lim_{x \rightarrow c}g(x) = \lim_{x \rightarrow c}h(x)=L, then g(x) and h(x) both approach L as x \rightarrow c^{+} and as x \rightarrow c^{-} so \lim_{x \rightarrow c^{+}}f(x)=L and \lim_{x \rightarrow c^{-}}f(x)=L. Hence, \lim_{x \rightarrow c}f(x)=L. QED.

Let me know your feedback on such stuff,

Nalin Pithwa

Lagrange’s Mean Value Theorem and Cauchy’s Generalized Mean Value Theorem

Lagrange’s Mean Value Theorem:

If a function f(x) is continuous on the interval [a,b] and differentiable at all interior points of the interval, there will be, within [a,b], at least one point c, a<c<b, such that f(b)-f(a)=f^{'}(c)(b-a).

Cauchy’s Generalized Mean Value Theorem:

If f(x) and phi(x) are two functions continuous on an interval [a,b] and differentiable within it, and phi(x) does not vanish anywhere inside the interval, there will be, in [a,b], a point x=c, a<c<b, such that \frac{f(b)-f(a)}{phi(b)-phi(a)} = \frac{f^{'}(c)}{phi^{'}(c)}.

Some questions based on the above:

Problem 1:

Form Lagrange’s formula for the function y=\sin(x) on the interval [x_{1},x_{2}].

Problem 2:

Verify the truth of Lagrange’s formula for the function y=2x-x^{2} on the interval [0,1].

Problem 3:

Applying Lagrange’s theorem, prove the inequalities: (i) e^{x} \geq 1+x (ii) \ln (1+x) <x, for x>0. (iii) b^{n}-a^{n}<ab^{n-1}(b-a) for b>a. (iv) \arctan(x) <x.

Problem 4:

Write the Cauchy formula for the functions f(x)=x^{2}, phi(x)=x^{3} on the interval [1,2] and find c.

More churnings with calculus later!

Nalin Pithwa.

 

 

Could a one-sided limit not exist ?

Here is basic concept of limit :

Pappus’s theorem

Problem:

Given a point on the circumference of a cyclic quadrilateral, prove that the product of the distances from the point to any pair of opposite sides or to the diagonals are equal.

Proof:

Let a, b, c, d be the coordinates of the vertices A, B, C, D of the quadrilateral and consider the complex plane with origin at the circumcenter of ABCD. Without loss of generality, assume that the circumradius equals 1.

The equation of line AB is

\left | \begin{array}{ccc}    a & \overline{a} & 1 \\    b & \overline{b} & 1 \\    z & \overline{z} & 1 \end{array} \right | = 0.

This is equivalent to z(\overline{a}-\overline{b})-\overline{z}(a-b)=\overline{a}b-a\overline{b}, that is,

z+ab\overline{z}=a+b

Let point M_{1} be the foot of the perpendicular from a point M on the circumcircle to the line AB. If m is the coordinate of the point M, then

z_{M_{1}}=\frac{m-ab\overline{m}+a+b}{2}

and

d(M, AB)=|m-m_{1}|=|m-\frac{m-ab\overline{m}+a+b}{2}|=|\frac{(m-a)(m-b)}{2m}| since m \overline{m} = 1.

Likewise,

d(M, BC)=|\frac{(m-b)(m-c)}{2m}|, d(M, CD)=|\frac{(m-c)(m-d)}{2m}|

d(M, DA)=|\frac{(m-d)(m-a)}{2m}|, d(M, AC)=|\frac{(m-a)(m-c)}{2m}|

and d(M, BD)=|\frac{(m-b)(m-d)}{2m}|

Thus,

d(M, AB).d(M, CD)=d(M, BC).d(M, DA)=d(M, AC).d(M, BD) as claimed.

QED.

More later,

Nalin Pithwa

PS: The above example indicates how easy it is prove many fascinating theorems of pure plane geometry using the tools and techniques of complex numbers.

 

Maxima and Minima using calculus

Problem:

The vertices of an (n+1)-gon lie  on the sides of a regular n-gon and divide its perimeter into  parts of equal length. How should one construct the (n+1)- gon so that its area is :

(a) maximum

(b) minimum

Hint only:

[One of the golden rule of solving problems in math/physics is to draw diagrams, as had benn emphasized by the maverick American physics Nobel Laureate, Richard Feynman. He expounded this technique even in software development. So, in the present problem, first draw several diagrams.]

There exists a side B_{1}B_{2} of the (n+1) -gon that lies entirely on a side A_{1}A_{2} of the n-gon. Let b=B_{1}B_{2} and b=A_{1}A_{2}. Show that b=\frac{n}{n+1}a. Then, for x=A_{1}B_{1}, we have 0 \leq x \leq \frac{n}{n+1} and the area S of the (n+1) -gon is given by

S(x)=\frac{\sin{\phi}}{2}\Sigma_{i=1}^{n}(\frac{i-1}{n+1}a+x)(\frac{n-i+1}{n+1}a-x)

where \phi=\angle{A_{1}A_{2}A_{3}}. Thus, S(x) is a quadratic function of x. Show that S(x) is a minimal when x=0 or x=\frac{a}{n+1} and S(x) is maximal when x=\frac{a}{2(n+1)}.

Let me know if you have any trouble when you attempt it,

-Nalin Pithwa

 

Inclusion Exclusion Principle theorem and examples

Reference: Combinatorial Techniques by Sharad Sane, Hindustan Book Agency.

Theorem: 

The inclusion-exclusion principle: Let X be a finite set and let and let P_{i}: i = 1, 2, \ldots n  be a set of n properties satisfied by (s0me of) the elements of X. Let A_{i} denote the set of those elements of X that satisfy the property P_{i} . Then, the size of the set \overline{A_{1}} \bigcup \overline{A_{2}} \bigcup \ldots \bigcup \overline{A_{n}} of all those elements that do not satisfy any one of these properties is given by

\overline{A_{1}} \bigcup \overline{A_{2}} \bigcup \ldots \bigcup \overline{A_{n}} = |X| - \sum_{i=1}^{n}|A_{n}|+ \sum_{1 \leq i <j \leq n}|A_{i} \bigcup A_{j}|- \ldots + \{ (-1)^{k} \sum_{1 \leq i_{1} < i_{2}< \ldots < i_{k} \leq n}|A_{i_{1} \bigcup A_{i_{2}}} \ldots \bigcup A_{i_{k}}|\}+ \ldots+ (-1)^{n}|A_{1} \bigcup A_{2} \ldots \bigcup A_{n}|.

Proof:

The proof will show  that every object in the set X is counted the same number of times on both the sides. Suppose x \in X and assume that x is an element of the set on the left hand side of above equation. Then, x has none of the properties P_{i}. We need to show that in this case, x is counted only once on the right hand side. This is obvious since x is not in any of the A_{i} and x \in X. Thus, X is counted only once in the first summand and is not counted in any other summand since x \notin A_{i} for all i. Now let x have k properties say P_{i_{1}}, P_{i_{2}}, \ldots, P_{i_{k}} (and no  others). Then x is counted once in X. In the next sum, x occurs {k \choose 2} times and so on. Thus, on the right hand side, x is counted precisely,

{k \choose 0}-{k \choose 1}+{k \choose 2}+ \ldots + (-1)^{k}{k \choose k}

times. Using the binomial theorem, this sum is (1-1)^{k} which is 0 and hence, x is not counted on the right hand side. This completes the proof. QED.

More later,

Nalin Pithwa

 

 

 

 

 

 

 

 

 

The inclusion-exclusion principle for RMO and IITJEE Maths

Reference: Combinatorial Techniques by Sharad Sane:

The principle and its applications:

The inclusion-exclusion principle, is among the most basic techniques of combinatorics. Suppose we have a set X with subsets A and B. Then, the number of elements that are in A or B (or both), that is, the cardinality of A \cup B is given by |A|+|B|- |A \cap B|. The elements that are in both A and B were counted twice. To get rid of the over-counting, we must subtract. If \overline{A} and \overline{B} denote the complements of A and B respectively, then how many elements does the set \overline{A} \cup \overline{B} have? This number is |X|-|A|-|B|+|A \cap B|. The explanation is as before. From the set of all the elements we get rid of those that are in A or B. In doing so, we subtracted the elements of A \cap B twice. This has to be corrected by adding such elements once. Essentially, this way of over-counting (inclusion) correcting it using under-counting (exclusion) and again correcting (over-correcting) and so on is referred to as the inclusion-exclusion technique. As another example, consider the question of finding how many positive integers up to 100 are not divisible by 2, 3 or 5. We see that there are 50 integers that are multiple of 2, 33 that are multiples of 3 and 20 that are multiples of 5. This certainly amounts to over-counting as there are integers that are divisible by two of the given three numbers 2, 3 and 5. In fact, the number of integers divisible by both 2 and 3 is 16, the number of integers divisible by both 2 and 5, that is divisible by 10 is 10, the number of integers divisible by both 3 and 5 is the number of integers below 100 and divisible by 15 and that number is 6. Finally, the number of integers divisible by all three 2, 3, and 5 is just 3. Hence, the number of integers NOT divisible by any one of 2, 3 or 5 is

100-(50+33+20)+(16+10+6)-3=26

The technique used above is called the sieve method. This technique was known to the Greeks and is in fact, known as the Sieve of Eratosthenes. (Remember: in high school, you have used this method to find primes from 1 to 100).

Also, note that I am reminded of the technique of telescoping series when I think about inclusion-exclusion. There seems to be some similarity in spirit.

More later,

Nalin Pithwa

Solution to another S L Loney problem for IITJEE Advanced Mathematics

Exercise XXVII. Problem #32. The sides of a triangle are in AP and the greatest and least angles are \theta and \phi. Prove that

4(1-\cos{\theta})(1-\cos{\phi})=\cos{\theta}+\cos{\phi}

Proof:

Let a, b, c be in AP. Hence, a<b<c, which in turn implies c is greatest and a is the least.

Hence, \theta=\angle{C} and \angle{A}.

Want:

4(1-\cos{A})(1-\cos{C})=\cos{A}+\cos{C}

Given:

b-a=c-b

(b-a)^{2}=(b-c)^{2}. Hence,

b^{2}-2ab+a^{2}=b^{2}+c^{2}-2bc

b^{2}+a^{2}-c^{2}=b^{2}-2bc+2ab

2b=a+c

4b^{2}=a^{2}+c^{2}+2ac

a^{2}+c^{2}-b^{2}=3b^{2}-2ac

a^{2}+b^{2}-c^{2}=b^{2}-2bc+2ab

b^{2}+c^{2}-a^{2}=b^{2}-2ab+2bc

Now, we have 1-\cos{A}=1-\frac{b^{2}+c^{2}-a^{2}}{2bc} and this is equal to the following:

\frac{2bc-(b^{2}+c^{2}-a^{2})}{2bc}=\frac{2bc-(b^{2}-2ab+2bc)}{2bc}

= \frac{2bc-b^{2}+2ab-2bc}{2bc}=\frac{2ab-b^{2}}{2bc}.

Also, similarly, we have the following:

1-\cos{C}=1-\frac{a^{2}+b^{2}-c^{2}}{2ab}=1-\frac{b^{2}-2bc+2ab}{2ab}=\frac{2bc-b^{2}}{2ab}

\cos{A}+\cos{C}=\frac{b^{2}+c^{2}-a^{2}}{}+\frac{a^{2}+b^{2}-c^{2}}{2ab}

= \frac{b^{2}-2ab+2bc}{2bc}+\frac{b^{2}-2bc+2ab}{2ab}

=\frac{a(b^{2}-2ab+2bc)+c(b^{2}-2bc+2ab)}{2abc}

=\frac{ab^{2}-2a^{2}b+2abc+b^{2}c-2bc^{2}+2abc}{2abc}

=\frac{ab^{2}+b^{2}c-2a^{2}b-2c^{2}b+4abc}{2abc}

= \frac{ab+bc-2a^{2}-2c^{2}+4ac}{2ac}

=\frac{b(a+c)-2(a-c)^{2}}{2ac}

=\frac{2b^{2}-2(a-c)^{2}}{2ac}

=\frac{b^{2}-(a-c)^{2}}{2ac}

\frac{(b+a-c)}{(b-a+c)}{ac}

but it is given that b-a=c-b, hence, a+c=2b, a+c-b=b. So the above expression changes to

= \frac{(b+a-c)(c-b+c)}{ac}

= \frac{(2c-b)(a+b-c)}{ac}

= \frac{(2c-b)(a+b-\overline{2b-a})}{ac}

= \frac{(2c-b)(a-b+c)}{ac}

= \frac{(2c-b)(2a-b)}{ac}

= \frac{4ac-2bc-2ab+b^{2}}{ac}

= 4(1-\cos{A})(1-\cos{C}).

QED.

A solution to a S L Loney Part I trig problem for IITJEE Advanced Math

Exercise XXVII. Problem 30.

If a, b, c are in AP, prove that \cos{A}\cot{A/2}, \cos{B}\cot{B/2}, \cos{C}\cot{C/2} are in AP.

Proof:

Given that b-a=c-b

TPT: \cos{B}\cot{B/2}-\cos{A}\cot{A/2}=\cot{C/2}\cos{C}-\cos{B}\cot{B/2}. —— Equation 1

Let us try to utilize the following formulae:

\cos{2\theta}=2\cos^{2}{\theta}-1 which implies the following:

\cos{B}=2\cos^{2}(B/2)-1 and \cos{A}=2\cos^{2}(A/2)-1

Our strategy will be reduce LHS and RHS of Equation I to a common expression/value.

LHS=(\frac{2s(s-b)}{ac}-1)(\frac{\sqrt{\frac{(s)(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}})-\cos{A}\cot{(A/2)}

which is equal to

(\frac{2s(s-b)}{ac}-1))(\frac{\sqrt{\frac{(s)(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}})-(2\cos^{2}(A/2)-1)\frac{\cos{A/2}}{\sin{A/2}}

which is equal to

(\frac{2s(s-b)}{ac}-1))(\frac{\sqrt{\frac{(s)(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}})-(\frac{2s(s-a)}{bc}-1)\frac{\sqrt{\frac{s(s-a)}{bc}}}{\sqrt{\frac{(s-b)(s-c)}{bc}}}

which is equal to

(\frac{2s(s-b)}{ac}-1))\sqrt{s(s-b)}{(s-a)(s-c)}-(\frac{2s(s-a)}{bc}-1)\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}

which in turn equals

\sqrt{\frac{s}{(s-a)(s-b)(s-c)}}((\frac{2s(s-b)}{ac}-1)(s-b)-(\frac{2s(s-a)}{bc}-1)(s-a))

From the above, consider only the expression, given below. We will see what it simplifies to:

\frac{2s(s-b)^{2}}{ac}-(s-b)-\frac{2s(s-a)^{2}}{bc}+(s-a)

=\frac{2s(s-b)^{2}}{ac}-\frac{2s(s-a)^{2}}{bc}+b-a

=(\frac{2s}{c})(\frac{(s-b)^{2}}{a}-\frac{(s-a)^{2}}{b})+b-a

=\frac{2s(s-b)^{2}}{ca}-\frac{2s(s-a)^{2}}{bc}+c-b —- Equation II.

Now, consider RHS of Equation I. Let us see if it also boils down to the above expression after simplification.

RHS=\cot{(C/2)}\cos{C}-\cos{B}\cot{(B/2)}

=(2\cos^{2}{(C/2)}-1)\cot{(C/2)}-(2\cos^{2}({B/2})-1)\cot{(B/2)}

=(\frac{2s(s-c)}{ab}-1)\frac{\sqrt{\frac{s(s-c)}{ab}}}{\sqrt{\frac{(s-b)(s-a)}{ab}}}-(\frac{2s(s-b)}{ac})\frac{\sqrt{\frac{s(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}}

= \sqrt{\frac{s}{(s-a)(s-b)(s-c)}}((\frac{2s(s-c)}{ab}-1)(s-c)-(\frac{2s(s-b)}{ac}-1)(s-b))

From equation II and above, what we want is given below:

\frac{2s(s-c)^{2}}{ab}-(s-c)-\frac{2s(s-b)^{2}}{ac}+(s-b)=\frac{2s(s-b)^{2}}{ac}-\frac{2s(s-a)^{2}}{bc}+c-b

that is, want to prove that c(s-c)^{2}+a(s-a)^{2}=2b(s-b)^{2}

but, it is given that a+c=2b and hence, c=2b-a, which means a+c-b=b and b-a=c-b

that is, want to prove that

c(a+b-c)^{2}+a(b+c-a)^{2}=2b(a+c-b)^{2}=2b^{3}

i.e., want: c(a+b-c)^{2}+a(b+c-a)^{2}=2b^{3}

i.e., want: (2b-a)(a+b-2b+a)^{2}+a(b-a+2b-a)^{2}=2b^{3}

i.e., want: (2b-a)(2a-b)^{2}+a(3b-2a)^{2}=2b^{3}

Now, in the above, LHS=(2b-a)(4a^{2}+b^{2}-4ab)+a(9b^{2}+4a^{2}-12ab)

= 8a^{2}b+2b^{3}-8ab^{2}-4a^{3}-ab^{2}+4ab^{2}+9ab^{2}+4a^{3}-12a^{2}b

= 2b^{3}.

Hence, LHS+RHS.

QED.

Complex ain’t so complex ! Learning to think!

Problem:

If 1, \omega, \omega^{2}, \omega^{3}, \ldots, \omega^{n} are the nth roots of unity, then find the value of (2-\omega)(2-\omega^{2})(2-\omega^{3})\ldots (2-\omega^{n-1}).

Solution:

Learning to think:

Compare it with what we know from our higher algebra — suppose we have to multiply out:

(x+a)(x+b)(x+c)(x+d). We know it is equal to the following:

x^{4}+(a+b+c+d)x^{3}+(ab+ac+ad+bc+bd+cd)x^{2}+(abc+acd+bcd+abd)x+abcd

If we examine the way in which the partial products are formed, we see that

(1) the term x^{4} is formed by taking the letter x out of each of the factors.

(2) the terms involving x^{3} are formed by taking the letter x out of any three factors, in every possible way, and one of the letters a, b, c, d out of the remaining factor

(3) the terms involving x^{2} are formed by taking the letter x out of any two factors, in every possible way, and two of the letters a, b, c, d out of the remaining factors

(4) the terms involving x are formed by taking the letter x out of any one factor, and three out of the letters a, b, c, d out of the remaining factors.

(5) the term independent of x is the product of all the letters a, b, c, d.

Further hint:

relate the above to sum of binomial coefficients.

and, you are almost done.

More later,

Nalin Pithwa