Monthly Archives: January 2017

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.



Blaise Pascal (1623-1662)

Blaise Pascal exhibited his talents at an early age, although his father who made discoveries in analytic geometry, kept mathematics books away from him to encourage other interests. At 16 Pascal discovered an important result concerning conic sections. At 18 he designed a calculating machine, which he built and sold. Pascal, along with Fermat, laid the foundations for the modern theory of probability. In this work, he made new discoveries concerning what is now called Pascal’s triangle. In 1654, Pascal abandoned his mathematical pursuits to devote himself to theology. After this, he returned to mathematics only once. One night, distracted by a severe toothache, he sought comfort by studying the mathematical properties of the cycloid. Miraculously, his pain subsided, which he took as a sign of divine approval of the study of mathematics.

PS: The programming language, Pascal, was, of course, named after Blaise Pascal!

-Nalin Pithwa.

Who will be first to reach 100?

A calls 7, B 12, A 22, B 23, and so on. Each call is higher by any number from 1 through 10.

Whoever calls 100 wins. How does A win?

Cheers to the winner ! 🙂

Nalin Pithwa.

Alexandre-Theophile Vandermonde (1735-1796)

Because Alexandre-Theophile Vandermonde was a sickly child, his physician father directed him to a career in music. However, he later developed an interest in mathematics. His complete mathematical work consists of four papers published in 1771-1772. These papers include fundamental contributions on the roots of equations, on the theory of determinants, and on the knight’s tour problem. Vandermonde’s interest in mathematics lasted for only 2 years. Afterward, he published papers on harmony, experiments with cold, and the manufacture of steel. He also became interested in politics, joining the cause of the French revolution and holding several different positions in government.

More stories of mathematicians are everywhere, including the web, printed books, journals and my mind !! 🙂

Nalin Pithwa

PS: One of the best source of stories of mathematicians is, of course, the famous “Men of Mathematics” by E. T. Bell.


Some questions based on Rolle’s theorem

Problem 1:

Verify the truth of Rolle’s theorm for the following functions:

(a) y=x^{2}-3x+2 on the interval [1,2].

(b) y=x^{3}+5x^{2}-6x on the interval [0,1].

(c) y=(x-1)(x-2)(x-3) on the interval [1,3].

(d) y=\sin^{2}(x) on the interval [0,\pi].

Problem 2:

The function f(x)=4x^{3}+x^{2}-4x-1 has roots 1 and -1. Find the root of the derivative f^{'}(x) mentioned in Rolle’s theorem.

Problem 3:

Verify that between the roots of the function y=\sqrt[3]{x^{2}-5x+6} lies the root of its derivative.

Problem 4:

Verify the truth of Rolle’s theorem for the function y=\cos^{2}(x) on the interval [-\frac{\pi}{4},+\frac{\pi}{4}].

Problem 5:

The function y=1-\sqrt[5]{x^{4}} becomes zero at the end points of the interval [-1,1]. Make it clear that the derivative of the function does not vanish anywhere in the interval (-1,1). Explain why Rolle’s theorem is NOT applicable here.

Calculus is the fountainhead of many many ideas in mathematics and hence, technology. Expect more beautiful questions on Calculus !

-Nalin Pithwa

Frank P. Ramsey, Cambridge mathematician

Frank Plumpton Ramsey(1903-1930), son of the president of of Magdalene College, Cambridge, was educated in Winchester and Trinity College. After graduating in 1923, he was elected a fellow of King’s College, Cambridge, where he spent the remainder of his life. Ramsey made important contributions to mathematical logic. What we now call Ramsey theory began with his clever combinatorial arguments, published in the paper “On a Problem of Formal Logic.” Ramsey also made contributions to the mathematical theory of economics. He was noted as an excellent lecturer on the foundations of mathematics. His death at the age of 26 deprived the mathematical community and Cambridge University of a brilliant young scholar.

PS: The reader interested in Ramsey numbers should consult references like the following:

J. G. Michaels and K. H. Rosen, Applications of Discrete Mathematics, McGraw Hill.


Nalin Pithwa

Is there such a number?

Is there a number which when divided by 3 gives a remainder of 1; when divided by 4 gives a remainder of 2, when divided by 5, gives a remainder of 3, and when divided by 6, gives a remainder of 4?

Hope all of you have fun with math in the new year 2017 !!

Nalin Pithwa

A Three Digit Number

I am thinking of a three digit number — if you subtract 7 from the it, the result is divisible by 7; if 8, divisible by 8; and if 9, divisible by 9. What is the number?

🙂 🙂 🙂

Nalin Pithwa

System of Equations — IITJEE Foundation Maths

Some problems are presented below as warm-up exercises:

Question 1:

Solve : x+y = 15 and xy=36

Question 2:

Solve: x-y=12 and xy=85

Question 3:

Solve: x^{2}+y^{2}=74 and xy=35

Question 4:

Solve : x^{2}+y^{2}=185 and x+y=17

Some other standard set of systems of equations and techniques for solving them are presented below:

Any pair of equations of the form

x^{2} \pm pxy + y^{2}=a^{2}….Equation I

x \pm y = b…Equation II

where p is any numerical quantity, can be reduced to one of the cases in the warm-up exercises; for, by squaring II and combining with I, an equation to find xy is obtained, the solution can then be completed by the aid of equation II.

Example 1:

Solve x^{3}-y^{3}=999…Equation 1

and x-y=3…Equation 2.

By division, x^{2}+xy+y^{2}=333…Equation 3.

From Equation 2, x^{2} - 2xy + y^{2} = 9;

by subtraction, 3xy=324 and xy=108…Equation 4.

From Equation (2) and (4), x=12, y=9 or x=9, y = -12

Example 2:

Solve: x^{4}+x^{2}y^{2}+y^{4}=2613…Equation I

and x^{2}+xy+y^{2}=67…Equation 2

Dividing (1) by (2), x^{2}-xy+y^{2}=39…Equation 3.

From Equation (2) and (3), by addition, x^{2}+y^{2}=33; by subtraction xy=14, hence, we get the following: x = \pm 7, x = \pm 2, or y = \pm 2, y = \pm 7

Example 3:

Solve: \frac{1}{x} - \frac{1}{y} = \frac{1}{3}…Equation (1)

and \frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{5}{9}…Equation (2)

From (1), by squaring, \frac{1}{x^{2}} - \frac{2}{xy} + \frac{1}{y^{2}} = \frac{1}{9},

and by subtraction, \frac{2}{xy} = \frac{4}{9}

adding to (2), \frac{1}{x^{2}} + \frac{2}{xy} + \frac{1}{y^{2}} = 1

and hence, \frac{1}{x} + \frac{1}{y} = \pm 1.

Combining with (1), \frac{1}{x} = \frac{2}{3}, or -\frac{1}{3}; \frac{1}{y} = - \frac{1}{3}, or -\frac{2}{3}; so, the solutions are: x = \frac{3}{2}, -3 and y =3, -\frac{3}{2}.

The following method of solution may always be used when the equations are of the same degree and homogeneous.


Solve x^{2}+ xy + 2y^{2} = 74 …Equation 1

and 2x^{2} + 2xy + y^{2} = 73…Equation 2.

Put y = mx, and substitute in both equations. Thus,

x^{2}(1+m+2m^{2}) =74…Equation 3.

x^{2}(2 + 2m + m^{2}) =73…Equation 4.

By division, \frac{1+m+2m^{2}}{2+2m+m^{2}} = \frac{74}{73}

73 + 73m + 146m^{2} = 148 + 148m + 74m^{2}

Hence, 72m^{2} -75m  - 75=0; or 24m^{2} - 25m - 25 = 0; hence, (8m + 5)(3m - 5) = 0; so m = -\frac{5}{8} or \frac{5}{3}.

Case (i):

Take m= -\frac{5}{8} and substitute in either (3) or (4).

From (3): x^{2}(1- \frac{5}{8} + \frac{50}{64}) = 74

x^{2} = 64, so x = \pm 8; so y = mx = - \frac{5}{8}x = \mp 5.

Case (ii):

Take m = \frac{5}{3}, then from (3) x^{2}(1 + \frac{5}{3} + \frac{50}{9}) = 74, so we get x^{2} = 9, so x = \pm 3; and hence, y = mx = \frac{5}{3}x = \pm 5.

When one of the equations is of first degree and the other of a higher degree, we may from the simple equation, find the value of one of the unknowns in terms of the other, and substitute in the second equation.


Solve 3x - 4y = 5…Equation (1)

and 3x^{2} - xy - 3y^{2} = 21…Equation (2)

From (1), we have x = \frac{5+4y}{3}

and substituting in (2), 3\frac{(5+4y)^{2}}{9} - \frac{y(5+4y)}{3} - 3y^{2}= 21

75 + 120y +48y^{2} - 15y - 12y^{2} - 27y^{2} = 189

9y^{2} + 105y -114 = 0

3y^{2} + 35y - 38 = 0

(y-1)(3y+31) = 0

Hence, y = 1, -\frac{38}{3}, and x = 3, -\frac{137}{9}.

The examples we have presented above will be sufficient as a general explanation of the methods to be employed; but, in some cases, special artifices are necessary.


Solve: x^{2} + 4xy + 3x = 40 - 6y -4y^{2}…Equation (1)

and 2xy - x^{2} = 3…Equation (2)

From (1) and (2), we have x^{2} + 4xy + 4y^{2} + 3x + 6y = 40;

that is, (x+2y)^{2} + 3(x+2y) - 40 = 0;

or, ((x+2y)+8)((x+2y)-5) = 0, hence, x+2y = -8, 5


Combining x+2y =5 with (2), we obtain 2x^{2} - 5x + 3=0, so x = 1, \frac{3}{2}; and by substituting in x + 2y = 5, we get y = 2, -\frac{7}{4}.


Combining  x+2y = -8 with (2), we obtain 2x^{2} + 8x + 3 = 0; hence, x = \frac{-4 \pm \sqrt{10}}{2}, and y = \frac{-12 \mp \sqrt{10}}{4}


Solve: x^{2}y^{2} - 6x = 34 - 3y…Equation (1)

and 3xy + y = 2((9+x)…Equation (2)

From (1), we have x^{2}y^{2} -6x + 3y = 34;

From (2), we have 9xy - 6x + 3y = 54;

By subtraction, x^{2}y^{2} - 9xy + 20 = 0; that is, (xy - 5)(xy -4) = 0; so xy = 5, 4.


Substituting xy = 5 in (2), gives y -2x = 3. From these equations, we obtain x =1 or -\frac{5}{2}; y = 5, or -2.


Substituting xy =4 in (2), gives y - 2x = 6. From these equations, we obtain x = \frac{-3 \pm \sqrt{17}}{2}, and y = 3 \pm \sqrt{17}.

Hope you enjoyed it,

Nalin Pithwa.










Words of Wisdom from Anand Kumar, mathematician of Super 30

Reference: Today’s print edition, Mumbai, DNA Newspaper.

Help  technology reach masses: Anand Kumar to Kharagpur IITians.


Super 30 founder and mathematician, Anand Kumar on Saturday exhorted students at the Indian Institute of Technology, Kharagpur to use their knowledge and skills for the betterment of the society by contributing more to improving people’s living standard and increasing farm productivity.

“For the students graduating from top rung institutions like IIT, Kharagpur bringing about positive change in the society through their expertise should be the main motto. Merely earning good (salary) packages should not be the goal,” he said.

The Super 30 is a niche training centre for students from lower income groups, where Kumar and his associates train bright youngsters to crack the prestigious Joint Entrance Examinations for the IITs. These national entrance exams are generally considered one of the toughest in the country for students of science; the IIT’s have a notoriously low acceptance rate of less than 1 per cent, lower than some of the top universities anywhere in the world. About 14 lakh students are estimated to take the examination every year.

Kumar said the way technology is changing is also increasing inequity in society.

“(The) Time has come to make tecnnology accessible to the masses so that it can bring direct benefit to people’s life. And this is something techno-savvy youth can do the best,” he added.


Cheers to Anand Kumar !!

from Nalin Pithwa