Mathematics Hothouse

Applications of Derivatives: Tutorial Set 1: IITJEE Mains Maths

October 19, 2018 – 6:33 pm

“Easy” questions:

Question 1:

Find the slope of the tangent to the curve represented by the curve $x=t^{2}+3t-8$ and $y=2t^{2}-2t-5$ at the point $(2,-1)$ .

Question 2:

Find the co-ordinates of the point P on the curve $y^{2}=2x^{3}$ , the tangent at which is perpendicular to the line $4x-3y+2=0$ .

Question 3:

Find the co-ordinates of the point $P(x,y)$ lying in the first quadrant on the ellipse $x^{2}/8 + y^{2}/18=1$ so that the area of the triangle formed by the tangent at P and the co-ordinate axes is the smallest.

Question 4:

The function $f(x) = \frac{\log (\pi+x)}{\log (e+x)}$ , where $x \geq 0$ is

(a) increasing on $(-\infty, \infty)$

(b) decreasing on $[0, \infty)$

(d) decreasing on $[0, \pi/e)$ and increasing on $[\pi/e, \infty)$ .

Fill in the correct multiple choice. Only one of the choices is correct.

Question 5:

Find the length of a longest interval in which the function $3\sin(x) -4\sin^{3}(x)$ is increasing.

Question 6:

Let $f(x)=x e^{x(1-x)}$ , then $f(x)$ is

(a) increasing on $[-1/2, 1]$

(b) decreasing on $\Re$

(d) decreasing on $[-1/2, 1]$ .

Fill in the correct choice above. Only one choice holds true.

Question 7:

Consider the following statements S and R:

S: Both $\sin(x)$ and $\cos (x)$ are decreasing functions in the interval $(\pi/2, \pi)$ .

R: If a differentiable function decreases in the interval $(a,b)$ , then its derivative also decreases in $(a,b)$ .

Which of the following is true?

(i) Both S and R are wrong.

(ii) Both S and R are correct, but R is not the correct explanation for S.

(iii) S is correct and R is the correct explanation for S.

(iv) S is correct and R is wrong.

Indicate the correct choice. Only one choice is correct.

Question 8:

For which of the following functions on $[0,1]$ , the Lagrange’s Mean Value theorem is not applicable:

(i) $f(x) = 1/2 -x$ , when $x<1/2$ ; and $f(x) = (1/2-x)^{2}$ , when $x \geq 1/2$ .

(ii) $f(x) = \frac{\sin(x)}{x}$ , when $x \neq 0$ ; and $f(x)=1$ , when $x=0$ .

(iii) $f(x)=x |x|$

(iv) $f(x)=|x|$ .

Only one choice is correct. Which one?

Question 9:

How many real roots does the equation $e^{x-1}+x-2=0$ have?

Question 10:

What is the difference between the greatest and least values of the function $f(x) = \cos(x) + \frac{1}{2}\cos(2x) -\frac{1}{3}\cos(3x)$ ?

More later,

Nalin Pithwa.

By Nalin Pithwa | Posted in applications of maths, calculus, Inequalities, ISI Kolkatta Entrance Exam, KVPY | Comments (0)

Primality testing: recap

September 11, 2018 – 4:13 am

via Primality testing: recap

By Nalin Pithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Comments (0)

Dear IIT Students: Address to IIT Hyderabad by present President of India, Shri Ram Nath Kovind

August 6, 2018 – 6:55 am

Dear IIT students

By Nalin Pithwa | Posted in miscellaneous, motivational stuff | Comments (0)

Applications of Derivatives: A Quick Review

July 17, 2018 – 12:19 am

Section I:

The Derivative as a Rate of Change

In case of a linear function $y=mx+c$ , the graph is a straight line and the slope m measures the steepness of the line by giving the rate of climb of the line, the rate of change of y with respect to x.

As x changes from $x_{0}$ to $x_{1}$ , y changes m times as much:

$y_{1}-y_{0}=m(x_{1}-x_{0})$

Thus, the slope $m=(y_{1}-y_{0})(x_{1}-x_{0})$ gives the change in y per unit change in x.

In more general case of differentiable function $y=f(x)$ , the difference quotient

$\frac{f(x+h)-f(x)}{x+h-x} = \frac{f(x+h)-f(x)}{h}$ , where $h \neq 0$

give the average rate of change of y (or f) with respect to x. The limit as h approaches zero is the derivative $dy/dx = f^{'}(x)$ , which can be interpreted as the instantaneous rate of change of f with respect to x. Since, the graph is a curve, the rate of change of y can vary from point to point.

Velocity and Acceleration:

Suppose that an object is moving along a straight line and that, for each time t during a certain time interval, the object has location/position $x(t)$ . Then, at time $t+h$ the position of the object is $x(t+h)$ and $x(t+h)-x(t)$ is the change in position that the object experienced during the time period t to $t+h$ . The ratio

$\frac{x(t+h)-x(t)}{t+h-t} = \frac{x(t+h)-x(t)}{h}$

gives the average velocity of the object during this time period. If

$\lim_{h \rightarrow 0} \frac{x(t+h)-x(t)}{h}=x^{'}(t)$

exists, then $x^{'}(t)$ gives the instantaneous rate of change of position with respect to time. This rate of change of position is called the velocity of the object. If the velocity function is itself differentiable, then its rate of change with respect to time is called the acceleration; in symbols,

$a(t) = v^{'}(t) = x^{''}(t)$

The speed is by definition the absolute value of the velocity: speed at time t is $|v(t)|$

If the velocity and acceleration have the same sign, then the object is speeding up, but if the velocity and acceleration have opposite signs, then the object is slowing down.

A sudden change in acceleration is called a jerk. Jerk is the derivative of acceleration. If a body’s position at the time t is $x(t)$ , the body’s jerk at time t is

$j = \frac{da}{dt} = \frac{d^{3}x}{dt^{3}}$

Differentials

Let $y = f(x)$ be a differentiable function. Let $h \neq 0$ . The difference $f(x+h) - f(x)$ is called the increment of f from x to $x+h$ , and is denoted by $\Delta f$ .

$\Delta f = f(x+h) - f(x)$

The product $f^{'}(x)h$ is called the differential of f at x with increment h, and is denoted by $df$

$df = f^{'}(x)h$

The change in f from x to $x+h$ can be approximated by $f^{'}(x)h$ :

$f(x+h) - f(x) = f^{'}(x)h$

Tangent and Normal

Let $y = f(x)$ be the equation of a curve, and let $P(x_{0}, y_{0})$ be a point on it. Let PT be the tangent, PN the normal and PM the perpendicular to the x-axis.

The slope of the tangent to the curve $y = f(x)$ at P is given by $(\frac{dy}{dx})_{(x_{0}, y_{0})}$

Thus, the equation of the tangent to the curve $y = f(x)$ at $(x_{0}, y_{0})$ is $y - y_{0} = (\frac{dy}{dx})_{(x_{0}, y_{0})}(x-x_{0})$

Since PM is perpendicular to PT, it follows that if $(\frac{dy}{dx})_{(x_{0}, y_{0})} \neq 0$ , the slope of PN is

$- \frac{1}{(\frac{dy}{dx})_{(x_{0}, y_{0})}} = - (\frac{dx}{dy})_{(x_{0}, y_{0})}$

Hence, the equation of the normal to the curve $y = f(x)$ at $(x_{0}, y_{0})$ is

$y - y_{0} = - (\frac{dx}{dy})_{(x_{0}, y_{0})}(x-x_{0})$

The equation of the normal parallel to the x-axis is $y = y_{0}$ , that is, when $(\frac{dy}{dx})_{(x_{0}, y_{0})} = 0$ . The length of the tangent at $(x_{0}, y_{0})$ is PT, and it is equal to

$y_{0}\csc{\theta} = y_{0}\sqrt{1+\cot^{2}{\theta}} = y_{0}\sqrt{1+[(\frac{dx}{dy})_{(x_{0}, y_{0})}]^{2}}$

The length of the normal is PN and it is equal to $y_{0}\sec {\theta} = y_{0}\sqrt{1 + [(\frac{dy}{dx})_{(x_{0}, y_{0})}]^{2}}$

If the curve is represented by $x = f(t)$ and $y = g(t)$ , that is, parametric equations in t, then

$\frac{dy}{dx} = \frac{g^{'}(t)}{f^{'}(t)}$ where $g^{'}(t)= \frac{dy}{dt}$ and $f^{'}(t) = \frac{dx}{dt}$ . In this case, the equations of the tangent and the normal are given by

$y - g(t) = \frac{g^{'}(t)}{f^{'}(t)}[x - f(t)]$ and $[y-g(t)] g^{'}(t) + [x-f(t)]f^{'}(t) = 0$ respectively.

The Angle between Two Curves

The angle of intersection of two curves is defined as the angle between the two tangents at the point of intersection. Let $y = f(x)$ and $y=g(x)$ be two curves, and let $P(x_{0}, y_{0})$ be their point of intersection. Also, let $\psi$ and $\phi$ be the angles of inclination of the two tangents with the x-axis, and let $\theta$ be the angle between the two tangents. Then,

$\tan {\theta} = \frac{\tan{\phi}-\tan{\psi}}{1+\tan{\phi}\tan{\psi}} = \frac{g^{'}(x) - f^{'}(x)}{1+f^{'}(x)g^{'}(x)}$

Example 1:

Write down the equations of the tangent and the normal to the curve $y = x^{3} - 3x + 2$ at the point $(2,4)$ .

Solution 1:

$\frac{dy}{dx} = 3x^{2}-3 \Longrightarrow \frac{dy}{dx}_{(2,4)} = 3.4 - 3 = 9$ .

Hence, the equation of the tangent at $(2,4)$ is given by $y-4 = 9(x-2) \Longrightarrow 9x-y-14=0$ and the equation of the normal is $y - 4 = (-1/9)(x-2) \Longrightarrow x+9y -38=0$ .

Rolle’s Theorem and Lagrange’s Theorem:

Rolle’s Theorem:

Let $f(x)$ be a function defined on a closed interval $[a,b]$ such that (i) f(x) is continuous on $[a,b]$ , (ii) f(x) is derivable on $(a,b)$ , and (iii) f(a) = f(b). Then, there exists a $c \in (a,b)$ such that $f^{'}(x)=0$ .

For details, the very beautiful, lucid, accessible explanation in Wikipedia:

https://en.wikipedia.org/wiki/Rolle%27s_theorem

Lagrange’s theorem:

Let $f(x)$ be a function defined on a closed interval $[a,b]$ such that (i) $f(x)$ is continuous on $[a,b]$ , and (ii) $f(x)$ is derivable on $(a,b)$ . Then, there exists a $c \in [a,b]$ such that

$f^{'}(c) = \frac{f(b)-f(a)}{b-a}$

Example 2:

The function $f(x) = \log {\sin(x)}$ satisfies the conditions of Rolle’s theorem on the interval $[\frac{\pi}{6}, \frac{5\pi}{6}]$ , as the logarithmic function and $\sin (x)$ are continuous and differentiable functions and $\log {\sin (\frac{5\pi}{6})} = \log {\sin (\pi - \frac{\pi}{6})} = \log{\sin{(\frac{\pi}{6})}}$ .

The conclusion of Rolle’s theorem is given at $c=\frac{\pi}{2}$ , for which $f^{'}(c) = \cot (c) = \cot (\pi/2) =0$ .

Rolle’s theorem for polynomials:

If $\phi(x)$ is any polynomial, then between any pair of roots of $\phi(x)=0$ lies a root of $\phi^{'}(x)=0$ .

Monotonicity:

A function $f(x)$ defined on a set D is said to be non-decreasing, increasing, non-increasing and decreasing respectively, if for any $x_{1}, x_{2} \in D$ and $x_{1} < x_{2}$ , we have $f(x_{1}) \leq f(x_{2})$ , $f(x_{1}) < f(x_{2})$ , $f(x_{1}) \geq f(x_{2})$ and $f(x_{1}) > f(x_{2})$ respectively. The function $f(x)$ is said to be monotonic if it possesses any of these properties.

For example, $f(x) = e^{x}$ is an increasing function, and $f(x)=\frac{1}{x}$ is a decreasing function.

Testing monotonicity:

Let $f(x)$ be continuous on $[a,b]$ and differentiable on $(a,b)$ . Then,

(i) for $f(x)$ to be non-decreasing (non-increasing) on $[a,b]$ it is necessary and sufficient that $f^{'}(x) \geq 0$ ( $f^{'}(x) \leq 0$ ) for all $x \in (a,b)$ .

(ii) for $f(x)$ to be increasing (decreasing) on $[a,b]$ it is sufficient that $f^{'}(x)>0$ ( $f^{'}(x)<0$ ) for all $x \in (a,b)$ .

(iii) If $f^{'}(x)=0$ for all x in $(a,b)$ , then f is constant on $[a,b]$ .

Example 3:

Determine the intervals of increase and decrease for the function $f(x)=x^{3}+2x-5$ .

Solution 3:

We have $f^{'}(x) = 3x^{2}+2$ , and for any value of x, $3x^{2}+2>0$ . Hence, f is increasing on $(-\infty, -\infty)$ . QED.

The following is a simple criterion for determining the sign of $f^{'}(x)$ :

If $a,b \geq 0$ , then $(x-a)(x-b)>0$ iff $x > \max (a,b)$ or $x < \min(a,b)$ ;

$(x-a)(x-b)<0$ if and only if $\min(a,b) < x < \max(a,b)$

Maxima and Minima:

A function has a local maximum at the point $x_{0}$ if the value of the function $f(x)$ at that point is greater than its values at all points other than $x_{0}$ of a certain interval containing the point $x_{0}$ . In other words, a function $f(x)$ has a maximum at $x_{0}$ if it is possible to find an interval $(\alpha, \beta)$ containing $x_{0}$ , that is, with $\alpha < x_{0} < \beta$ , such that for all points different from $x_{0}$ in $(\alpha, \beta)$ , we have $f(x) < f(x_{0})$ .

A function $f(x)$ has a local minimum at $x_{0}$ if there exists an interval $(\alpha, \beta)$ containing $x_{0}$ such that $f(x) > f(x_{0})$ for $x \in (\alpha, \beta)$ and $x \neq x_{0}$ .

One should not confuse the local maximum and local minimum of a function with its largest and smallest values over a given interval. The local maximum of a function is the largest value only in comparison to the values it has at all points sufficiently close to the point of local maximum. Similarly, the local minimum is the smallest value only in comparison to the values of the function at all points sufficiently close to the local minimum point.

The general term for the maximum and minimum of a function is extremum, or the extreme values of the function. A necessary condition for the existence of an extremum at the point $x_{0}$ of the function $f(x)$ is that $f^{'}(x_{0})=0$ , or $f^{'}(x_{0})$ does not exist. The points at which $f^{'}(x)=0$ or $f^{'}(x)$ does not exist, are called critical points.

First Derivative Test:

(i) If $f^{'}(x)$ changes sign from positive to negative at $x_{0}$ , that is, $f^{'}(x)>0$ for $x < x_{0}$ and $f^{'}(x)<0$ for $x > x_{0}$ , then the function attains a local maximum at $x_{0}$ .

(ii) If $f^{'}(x)$ changes sign from negative to positive at $x_{0}$ , that is, $f^{'}(x)<0$ for $x<x_{0}$ , and $f^{'}(x)>0$ for $x > x_{0}$ , then the function attains a local minimum at $x_{0}$ .

(iii) If the derivative does not change sign in moving through the point $x_{0}$ , there is no extremum at that point.

Second Derivative Test:

Let f be twice differentiable, and let c be a root of the equation $f^{'}(x)=0$ . Then,

(i) c is a local maximum point if $f^{''}(c)<0$ .

(ii) c is a local minimum point if $f^{''}(c)>0$ .

However, if $f^{''}(c)=0$ , then the following result is applicable. Let $f^{'}(c) = f^{''}(c) = \ldots = f^{n-1}(c)=0$ (where f^{r} denotes the rth derivative), but $f^{(n)}(c) \neq 0$ .

(i) If n is even and $f^{(n)}(c)<0$ , there is a local maximum at c, while if $f^{(n)}(c)>0$ , there is a local minimum at c.

(ii) If n is odd, there is no extremum at the point c.

Greatest/Least Value (Absolute Maximum/Absolute Minimum):

Let f be a function with domain D. Then, f has a greatest value (or absolute maximum) at a point $c \in D$ if $f^(x) \leq f(c)$ for all x in D and a least value (or absolute minimum) at c, if $f(x) \geq f(c)$ for all x in D.

If f is continuous at every point of D, and $D=[a,b]$ , a closed interval, the f assumes both a greatest value M and a least value m, that is, there are $x_{1}, x_{2} \in [a,b]$ such that $f(x_{1})=M$ and $f(x_{2})=m$ , and $m \leq f(x) \leq M$ for every $x \in [a,b]$ .

Example 4:

a) $y=x^{2}$ , with domain $(-\infty, \infty)$ . This has no greatest value; least value at $x=0$

b) $y=x^{2}$ with domain $[0,2]$ . This has greatest value at $x=2$ and least value at $x=0$ .

c) $y=x^{2}$ with domain $(0,2]$ . This has greatest value at $x=2$ and no least value.

d) $y=x^{2}$ with domain $(0,2)$ . This has no greatest value and no least value.

Some other remarks:

The greatest (least) value of continuous function $f(x)$ on the interval $[a,b]$ is attained either at the critical points or at the end points of the interval. To find the greatest (least) value of the function, we have to compute its values at all the critical points on the interval $(a,b)$ , and the values $f(a), f(b)$ of the function at the end-points of the interval, and choose the greatest (least) out of the values so obtained.

We will continue with problems on applications of derivatives later,

Nalin Pithwa.

By Nalin Pithwa | Posted in applications of maths, calculus, IITJEE Mains | Comments (0)

Pre RMO Practice question: 2018: How long does it take for a news to go viral in a city? And, a cyclist vs horseman

June 7, 2018 – 1:23 am

Problem 1:

Some one arrives in a city with very interesting news and within 10 minutes tells it to two others. Each of these tells the news within 10 minutes to two others(who have not heard it yet), and so on. How long will it take before everyone in the city has heard the news if the city has three million inhabitants?

Problem 2:

A cyclist and a horseman have a race in a stadium. The course is five laps long. They spend the same time on the first lap. The cyclist travels each succeeding lap 1.1 times more slowly than he does the preceding one. On each lap the horseman spends d minutes more than he spent on the preceding lap. They each arrive at the finish line at the same time. Which of them spends the greater amount of time on the fifth lap and how much greater is this amount of time?

I hope you enjoy “mathematizing” every where you see…

Good luck for the Pre RMO in Aug 2018!

Nalin Pithwa.

By Nalin Pithwa | Posted in Pre-RMO | Comments (0)

A leaf out of Paul Erdos’ biography: My Brain is Open: by Bruce Schechter

May 20, 2018 – 5:29 am

Reference:

My Brain is Open: The Mathematical Journeys of Paul Erdos by Bruce Schechter, a TouchStone Book, Published by Simon and Schuster, New York.

Amazon India link:

https://www.amazon.in/My-Brain-Open-Mathematical-Journeys/dp/0684859807/ref=sr_1_1?ie=UTF8&qid=1526794050&sr=8-1&keywords=my+brain+is+open

Chapter One: Traveling.

The call might come at midnight or an hour before dawn — mathematicians are oddly unable to handle the arithmetic of time zones. Typically, a thickly accented voice on the other end of the line would abruptly begin:

“I am calling from Berlin. I want to speak to Erdos.”

“He’s not here, yet.”

“Where is he?”

“I don’t know.”

“Why don’t you know!” Click!

Neither are mathematicians always observant of social graces.

For more than sixty years mathematicians around the world have been roused from their abstract dreams by such calls, the first of the many disruptions that constituted a visit from Paul Erdos. The frequency of the calls would increase over the next several days and would culminate with a summons to the airport, where Erdos himself would appear, a short, frail man in a shapeless old suit, clutching two small suitcases that contained all of his worldly possessions. Stepping off the plane he would announce to the welcoming group of mathematicians, “My brain is open!”

Paul Erdos’ brain, when open, was one of the wonders of the world, an Ali Baba’s cave, glittering with mathematical treasures, gems of the most intricate cut and surpassing beauty. Unlike Ali Baba’s cave, which was hidden behind a huge stone in a remote desert, Erdos and his brain were in perpetual motion. He moved between mathematical meetings, universities, and corporate think tanks, logging hundreds of thousands of miles. “Another roof, another proof,” as he liked to say. “Want to meet Erdos?” mathematicians would ask. “Just stay here and wait. He’ll show up.” Along the way, in borrowed offices, guest bedrooms, and airplane cabins, Erdos wrote in excess of 1600 papers, books and articles, more than any other mathematician who ever lived. Among them are some of the great classics of the twentieth century, papers that opened up entire new fields and became the obsession and inspiration of generations of mathematicians.

The meaning of life, Erdos often said, was to prove and conjecture. Proof and conjecture are the tools with which mathematicians explore the Platonic universe of pure form, a universe that to many of them is as real as the universe in which they must reluctantly make their homes and livings, and far more beautiful. “If numbers aren’t beautiful, I don’t know what is,” Erdos frequently remarked. And although, like all mathematicians, he was forced to make his home in the temporal world, he rejected worldly encumbrances. He had no place on earth called home, nothing resembling a conventional year-round, nine-to-five job, and no family in the usual sense of the word. He arranged his life with only one purpose, to spend areas many hours a day as possible engaged in the essential, life-affirming business of proof and conjecture.

For Erdos, the mathematics that consumed most of his waking hours was not a solitary pursuit but a social activity, a movable feast. One of the greatest mathematical discoveries of the twentieth century was the simple equation that two heads are better than one. Ever since Archimedes traced his circles in sand, mathematicians, for the most part, have laboured alone — that is, until some forgotten soul realized that mathematics could be done anywhere. Only paper and pencil were needed, and those were not strictly essential. A table-cloth would do in a pinch, or the mathematician could carry his equations in his head, like a chessmaster playing blindfolded. Strong coffee, and in Erdos’ case even more powerful stimulants, helped too. Mathematicians began to frequent the coffeehouses of Budapest, Prague, and Paris, which led to the quip often attributed to Erdos:”A mathematician is a machine for turning coffee into theorems.” Increasingly, mathematical papers became the work of two, three, or more collaborators. That radical transformation of how mathematics is created is the result of many factors, not the least of which was the infectious example set by Erdos.

Erdos had more collaborators than most people have acquaintances. He wrote papers with more than 450 collaborators —- the exact number is still not known, since Erdos participated in the creation of new mathematics until the last day of his life, and his collaborators are expected to continue writing and publishing for years. The briefest encounter could lead to a publication — for scores of young mathematicians a publication that could become the cornerstone of their life’s work. He would work with anyone who could keep up with him, the famous or the unknown. Having been a child prodigy himself, he was particularly interested in meeting and helping to develop the talents of young mathematicians. Many of the world’s leading mathematicians owe their careers to an early meeting with Erdos.

Krishna Alladi, who is now a mathematician at the University of Florida, Gainesville, is one of the many young mathematicians whom Erdos helped. In 1974, when Alladi was an undergraduate in Madras, India, he began an independent investigation of a certain number theoretic function. His teachers could not help Alladi with his problem, nor could his father, who was a theoretical physicist and head of Madras Institute of Mathematics. Alladi’s father told some of his knowledgeable friends about his son’s difficulty, and they suggested that he write to Erdos.

Because Erdos was constantly on the move, Alladii sent a letter to the Hungarian Academy of Sciences. In an astonishingly short time, Alladi heard from Erdos, who said he would soon be lecturing in Calcutta. Could Alladi come there to meet him? Unfortunately, Alladi had examinations and could not attend, so he sent his father in his place to present the results of his research. After his father’s talk, Alladi recounts, “Erdos walked up to him and told him in very polite terms that he was not interested in the father but in the son.” Determined to meet with the promising young mathematician, Erdos, who was bound for Australia, rerouted his trip to stop briefly in Madras, which lies about 860 miles south of Calcutta.

Alladi was astonished that a great mathematician should change his plans to visit a student. He was nervous when he met Erdos at the airport, but that soon passed. “He talked to me as if he had known me since childhood,” Alladi recalls. The first thing Erdos asked was, “Do you know my poem about Madras?” And then he recited:

This the city of Madras

The home of the curry and the dhal,

Where Iyers speak only to Iyengars

And Iyengars speak only to God.

The Iyers and Iyengars are two Brahmin sects. The Iyers worship Shiva the Destroyer but will also worship in the temples of the Iyengars, who worship only Lord Vishnu, the Protector. Erdos explained that this was his variation on the poem about Boston and the pecking order among the Lowells, the Cabots, and God. Having put Alladi at ease, Erdos launched into a discussion of mathematics. Erdos was so impressed with Alladi, who was applying to graduate schools in the United States, that he wrote a letter on his behalf. Within a month, Alladi received the Chancellor’s Fellowship at the University of California, Los Angeles.

A celebrated magazine article about Erdos was called, “The Man Who Loved Only Numbers.” While it is true that Erdos loved numbers, he loved much more. He loved to talk about history, politics, and almost any other subject. He loved to take long walks and to climb towers, no matter how dismal the prospective view, he loved to play ping-pong, chess, and Go, he loved to perform silly tricks to amuse children and to make sly jokes and thumb his nose at authority. But, most of all, Erdos loved those who loved numbers, mathematicians. He showed that love by opening his pocket as well as his mind. Having no permanent job, Erdos also had little money, but whatever he had was at the service of others. If he heard of a graduate student who needed money to continue his studies, he would sent a cheque. Whenever he lectured in Madras, he would send his fee to the needy widow of the great Indian mathematician Srinivasa Ramanujan; he had never met Ramanujan or his wife, but the beauty of Ramanujan’s equations had inspired Erdos as a young mathematician. In 1984, he won the prestigious Wolf prize, which came with a cash reward of $at 50000, easily the most money Erdos had ever received at one time. He gave $30000 to endow a postdoctoral fellowship in the name of his parents at the Technion in Haifa, Israel, and used the remainder to help relatives, graduate students, and colleagues:”I kept only $720,” Erdos recalled.

In the years before the internet, there was Paul Erdos. He carried a shopping bag crammed with latest papers, and his brain was stuffed with the latest gossip as well as an amazing database of the world of mathematics. He knew everybody: what they were interested in; what they had conjectured, proved, or were in the midst of proving; their phone numbers; the names and ages of their wives, children, pets; and, much more. He could tell off the top of his head on which page in which obscure Russian journal a theorem similar to the one you were working on was proved in 1922. When he met a mathematician in Warsaw, say, he would immediately take up the conversation where they had left it two years earlier. During the iciest years of the Cold War Erdos’s fame allowed him freely to cross the Iron Curtain, so that he became vital link between the East and the West.

In 1938, with Europe on the brink of war, Erdos fled to the United States and embarked on his mathematical journeys. This book is the story of those adventures. Because they took Erdos everywhere mathematics is done, this is also the story of the world of mathematics, a world virtually unknown to outsiders.. Today perhaps the only mathematician most people can name is Theodore Kacznyski. The names of Karl Friedrich Gauss, Bernhard Riemann, Georg Cantor and Leonhard Euler, who are to Mathematics what Shakespeare is to literature and Mozart to music, are virtually unknown outside of the worlds of math and science.

For all frequent flier miles Erdos collected, his true voyages were journeys of the mind. Erdos carefully constructed his life to allow himself as much time as possible for those inward journeys, so a true biography of Erdos should spend almost as much time in the Platonic realm of mathematics as in the real world. For a layman this may seem to be a forbidding prospect. Fortunately, many of the ideas that fascinated Erdos can be easily grasped by anyone with a modest recollection of high school mathematics. The proofs and conjectures that made Erdos famous are, of course, far more difficult to follow, but that should not be of much concern to the reader. As Ralph Boas wrote, “Only professional mathematicians learn anything from proofs. Other people learn from explanations.” Just as it is not necessary to understand how Glenn Gould fingers a difficult passage to be dazzled by his performance of thee “Goldberg Variations,” one does not have to understand the details of Erdos’s elegant proofs to appreciate the beauty of mathematics. And, it is the nature of Erdos’s work that while his proofs are difficult, the questions he asks can be quite easy to understand. Erdos often offered money for the solution to problems he proposed. Some of those problems are enough for readers of this book to understand — and, perhaps, even solve. Those who decide to try should be warned that, as Erdos has pointed out, when the number of hours it takes to solve one of his problems is taken into account, the cash prizes rarely exceed minimum wage. The true prize is to share in the joy that Erdos knew so well, joy in understanding a page of the eternal book of mathematics.

— shared by Nalin Pithwa (to motivate his students and readers.)

By Nalin Pithwa | Posted in mathematicians, miscellaneous, motivational stuff, pure mathematics | Comments (0)

Diophantus of Alexandria: some trivia, some tidbits

May 13, 2018 – 4:03 am

The name/word Diophantine equation honours the mathematician Diophantus, who initiated the study of such equations. Practically, nothing is known of Diophantus as an individual, save that he lived in Alexandria sometime around 250 A.D. The only positive evidence as to the date of his activity is that the Bishop of Laodicea, who began his episcopate in 270, dedicated a book on Egyptian computation to his friend Diophantus. Although Diophantus’ works were written in Greek and he displayed the Greek genius for theoretical abstraction, he was most likely a Hellenized Babynolian. The only personal particulars we have of his career come from the wording of an epigram-problem (apparently dating from the 4th century). His boyhood lasted 1/6 of his life; his beard grew after 1/12 more; after 1/7 more he married; and his son was born 5 years later, the son lived to half his father’s age and the father died 4 years after his son. If x was the age at which Diophantus died, these data lead to the equation:

$\frac{1}{6}x + \frac{1}{12}x + \frac{1}{7}x + 5 + \frac{1}{2}x + 4=x$

with solution $x=84$ . Thus, he must have reached an age of 84, but in what year or even in what century is not certain.

The great work upon which the reputation of Diophantus rests is his Arithmetica, which may be described as the earliest treatise on algebra. Only six Books of the original thirteen have been preserved. It is in the Arithmetica that we find the first systematic use of mathematical notation, although the signs employed are of the nature of abbreviations for words rather than algebraic symbols in the sense with which we use them today. Special symbols are introduced to represent frequently occurring concepts, such as the unknown quantity in an equation and the different powers of the unknown up to the sixth power. Diophantus also had a symbol to express subtraction, and another for equality.

It is customary to apply the term Diophantine equation to any equation in one or more unknowns that is to be solved in the integers. The simplest type of Diophantine equation is the linear Diophantine equation in two variables:

$ax+by=c$

where a, b, c are given integers and a, b are not both zero. A solution of this equation is a pair of integers $x_{0}, y_{0}$ that, when substituted in to the equation, satisfy it; that is, we ask that $ax_{0}+by_{0}=c$ . Curiously enough, the linear equation does not appear in the extant works of Diophantus (the theory required for its solution is to be found in Euclid’s Elements), because he viewed it as trivial, most of his problems deal with finding squares or cubes with certain properties.

Reference:

Elementary Number Theory, David M. Burton, 6th edition, Tata McGraw Hill Edition.

More such interesting information about famous mathematical personalities is found in the classic, “Men of Mathematics by E. T. Bell”.

— Nalin Pithwa

By Nalin Pithwa | Posted in mathematicians, miscellaneous | Comments (0)

Richard Feyman: colourful life.

May 13, 2018 – 2:36 am

https://www.hindustantimes.com/more-lifestyle/feyn-balance-meet-one-of-the-world-s-most-influential-colourful-physicists/story-a6T3fQUHspE47zxYHGoHWO.html

Thanks to Hindustan Times and Rachel Lopez.

By Nalin Pithwa | Posted in motivational stuff, physicisrs | Comments (0)

Remembering Richard Feynman on his 100th birth centenary

May 13, 2018 – 2:32 am

https://www.hindustantimes.com/more-lifestyle/infectious-enthusiastic-relevant-a-physicist-s-take-on-richard-feynman/story-jD3jualkHCX5WuXcCp0ndL.html

Thanks to Hindustan Times and Prof. Arnab Bhattacharya, TIFR, Mumbai.

By Nalin Pithwa | Posted in motivational stuff, physicisrs | Comments (0)

Co-ordinate geometry practice for IITJEE Maths: Ellipses

May 6, 2018 – 11:19 pm

Problem 1:

Find the locus of the point of intersection of tangents to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}}=1$ , which are at right angles.

Solution I:

Any tangent to the ellipse is $y = mx + \sqrt{a^{2}m^{2}+b^{2}}$ ….call this equation I.

Equation of the tangent perpendicular to this tangent is $y=-\frac{-1}{m}x+\sqrt{\frac{a^{2}}{m^{2}}+b^{2}}$ …call this equation II.

The locus of the intersection of tangent lines (i) and (ii) is obtained by eliminating m between these equations. So, squaring and adding them, we get

$(y-mx)^{2}+(my+x)^{2}=a^{2}m^{2}+b^{2}+a^{2}+b^{2}m^{2}$

$\Longrightarrow (1+m^{2})(x^{2}+y^{2})=(1+m^{2})(a^{2}+b^{2})$

$\Longrightarrow x^{2}+y^{2}=a^{2}+b^{2}$

which is a circle with its centre at the centre of the ellipse and radius equal to the length of the line joining the ends of the major and minor axis. This circle is called the director circle of the ellipse.

Problem II:

A tangent to the ellipse $x^{2}+4y^{2}=4$ meets the ellipse $x^{2}+2y^{2}=6$ at P and Q. Prove that the tangents at P and Q of the ellipse $x^{2}+2y^{2}=6$ at right angles.

Solution II:

Let the tangent at $R(2\cos {\theta}, \sin{\theta})$ to the ellipse $x^{2}+4y^{2}=4$ meet the ellipse $x^{2}+2y^{2}=6$ at P and Q.

Let the tangents at P and Q to the second ellipse intersect at the point $S(\alpha,\beta)$ . Then, PQ is the chord of contact of the point $S(\alpha,\beta)$ with respect to ellipse two, and so its equation is

$\alpha x + 2\beta y=6$ ….call this “A”.

PQ is also the tangent at $R(2\cos {\theta}, \sin{\theta})$ to the first ellipse and so the equation can be written as $(2\cos{\theta})x+(4\sin{\theta})y=4$ ….call this “B”.

Comparing “A” and “B”, we get $\frac{2\cos{\theta}}{\alpha} = \frac{4\sin{\theta}}{2\beta} = \frac{4}{6}$

$\Longrightarrow \cos{\theta}=\frac{\alpha}{3}$ and $\sin{\theta}=\frac{\beta}{3}$

$\Longrightarrow \frac{\alpha^{2}}{9} + \frac{\beta^{2}}{9}=1 \Longrightarrow \alpha^{2}+\beta^{2}=9$

The locus of $S( \alpha, \beta)$ is $x^{2}+y^{2}=9$ , or $x^{2}+y^{2}=6+3$ , which is the director circle of the second ellipse.

Hence, the tangents at P and Q to the ellipse (ii) are at right angles (by the solution to the previous example).

Problem 3:

Let d be the perpendicular distance from the centre of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1$ to the tangent drawn at a point P on the ellipse. If $F_{1}$ and $F_{}{2}$ are the two foci of the ellipse, then show that $(PF_{1}-PF_{2})^{2}=4a^{2}(1-\frac{b^{2}}{d^{2}})$

Solution 3:

Equation of the tangent at the point $P(a\cos {\theta}, b\sin{\theta})$ on the given ellipse is $\frac{x\cos{\theta}}{a} + \frac{y\sin{\theta}}{b}=1$ . Thus,

$d= |\frac{-1}{\sqrt{\frac{\cos^{2}{\theta}}{a^{2}}+\frac{\sin^{2}{\theta}}{b^{2}}}}|$

$\Longrightarrow d^{2}=\frac{a^{2}b^{2}}{b^{2}\cos^{2}{\theta}+a^{2}\sin^{2}{\theta}}$

We know $PF_{1}+PF_{2}=2a$

$\Longrightarrow (PF_{1}-PF_{2})^{2}=(PF_{1}+PF_{2})^{2}-4PF_{1}PF_{2}$ …call this equation I.

Also, $(PF_{1}.PF_{2})^{2}=[ (a\cos{\theta}-ae)^{2}+(b\sin{\theta})^ {2}].[(a\cos{\theta}+ae)^{2} + (b\sin{\theta})^{2}]$ , which in turn equals,

$[a^{2}(\cos{\theta}-e)^{2}+a^{2}(1-e^{2})\sin^{2}{\theta} ]. [a^{2}(\cos{\theta}+e)^{2}+a^{2}(1-e^{2})\sin^{2}{\theta} ]$ , that is,

$a^{4}[ (\cos^{2}{\theta}+e^{2}) -2e\cos{\theta}+\sin^{2}{\theta} - e^{2} \sin^{2}{\theta} ]. [ (\cos^{2}{\theta}+e^{2}) + 2e \cos{\theta} + \sin^{2}{\theta} - e^{2}\sin^{2}{\theta}] = a^{4}[ 1-2e\cos{\theta}+e^{2}\cos^{2}{\theta} ] [1+ 2e \cos{\theta} + e^{2}\cos^{2}{\theta} ]$

that is,

$a^{4}[ (1+e^{2}\cos^{2}{\theta})^{2}-4e^{2}\cos^{2}{\theta}] = a^{4}(1-e^{2}\cos^{2}{\theta})^{2}$

$\Longrightarrow PF_{1}. PF_{2}=a^{2}(1-e^{2}\cos^{2}{\theta})$

Now, from I, we get $(PF_{1} - PF_{2})^{2} = 4a^{2}-4a^{2}(1-e^{2}\cos^{2}{\theta}) = 4a^{2}e^{2}\cos^{2}{\theta}$ ,

also, $1-\frac{b^{2}}{a^{2}} = 1 - \frac{b^{2}\cos^{2}{\theta} + a^{2}\sin^{2}{\theta}}{a^{2}} = \frac{(a^{2}-b^{2})\cos^{2}{\theta}}{a^{2}} = e^{2}\cos^{2}{\theta}$

Hence, $(PF_{1} - PF_{2})^{2} = 4a^{2}(1-\frac{b^{2}}{d^{2}})$

we will continue later,

Cheers,

Nalin Pithwa

By Nalin Pithwa | Posted in co-ordinate geometry, IITJEE Mains | Comments (0)

Mathematics Hothouse

Applications of Derivatives: Tutorial Set 1: IITJEE Mains Maths

Primality testing: recap

Dear IIT Students: Address to IIT Hyderabad by present President of India, Shri Ram Nath Kovind

Applications of Derivatives: A Quick Review

Pre RMO Practice question: 2018: How long does it take for a news to go viral in a city? And, a cyclist vs horseman

A leaf out of Paul Erdos’ biography: My Brain is Open: by Bruce Schechter

Diophantus of Alexandria: some trivia, some tidbits

Richard Feyman: colourful life.

Remembering Richard Feynman on his 100th birth centenary

Co-ordinate geometry practice for IITJEE Maths: Ellipses

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