# PROMYS India: Mehta Fellowships, Online July 4-Aug 14 2021 Challenging Math Training Program

PROMYS India

Program in Mathematics for Young Scientists

Dear Dr. Pithwa,

We are seeking mathematically talented students in Standards IX-XII as applicants to the Mehta Fellowships to PROMYS, a challenging summer program in mathematics. The Mehta Fellowships cover all costs of participation. PROMYS 2021 will run online from July 4 – August 14, 2021.

PROMYS is the parent program of PROMYS India whose launch has sadly been delayed by Covid-19. PROMYS has been running at Boston University in the U.S. since 1989. It will run as close to the in-person program as possible in 2021, with intense mathematical activity and social events to build community.

The 2021 Mehta Fellowship application is HERE on the PROMYS site (www.promys.org). The application deadline is March 30, 2021. Students should allow themselves plenty of time to work on the intriguing application problems.

Successful PROMYS applicants need to show unusual enthusiasm for thinking deeply about mathematics. They do not need to have taken any advanced courses or participated in competitions. Lectures, problem sets, and mathematical discussions at PROMYS are all in English.

PROMYS is an intensive six-week program which draws talented secondary school students from across the U.S. and around the world. Each year, PROMYS creates a collaborative and supportive mathematical community of 85 pre-university students, 25 undergraduate counselors, faculty, research mentors and guest mathematicians. The focus is on developing the mathematical habits of mind that support independence and creativity in facing unfamiliar mathematical challenges.

Approximately 50% of PROMYS alumni go on to earn doctorates. About half of these are PhDs in Mathematics.

Mathematical focus: First-year students focus primarily on a series of very challenging problem sets, a daily lecture, and exploration labs in Number Theory. Students receive daily feedback from their counselors on their Number Theory problem sets. There are also dozens of additional seminars, mini-courses, and guest lectures on a wide range of mathematical topics. Advanced seminars and mentored research are offered, and the program is individualized to ensure each student is fully challenged mathematically.

The Mehta Fellowship flyer is here. I hope you will share this information with mathematically talented students and with their teachers and schools.

Please let me know if you have any questions.

Yours sincerely,

Professor Ila Varma
Director of PROMYS India
“PROMYS is all about thinking hard and having fun. Thinking and exploring really helped me grow intellectually.” Vruddhi Shah, 2020 Mehta Fellow

“The path of discovery that one follows as a student of PROMYS is unique and will change your perception of mathematics,” Siddharth Sridhar, 2017 and 2018 Mehta Fellow

# More praise for the Queen of All Sciences !

This therefore is Mathematics: She reminds

you of the invisible forms of the soul; She

gives life to her own discoveries; She awakens

the mind and purifies the intellect; She

brings light to our intrinsic ideas.

— Proclus (412-485 CE)

# In praise of the Queen of all Sciences !

1. Computo ergo sum: D H Bailey
2. The only thing I know is that I don’t know anything. : Socrates.
3. Also, he made a molten sea of ten cubits from brim to brim, round in compass, and five cubits the height thereof, and a line of thirty cubits did compass it roundabout. : Old Testament I, Kings 7:23
4. We would be amiss, however, if we did not emphasize that the extended precision calculation of “pi” / $\pi$ has substantial application as a test of the “global integrity” of a supercomputer. : D.H, Bailey, J.M. Borwein, and P.B. Borwein
5. The purpose of computing is insight, not numbers. Richard Hamming.

# A cute bare basic proof of factorization of a^{3}+b^{3}+c^{3}-3abc by Gopal/Sravan Pithwa

This proof is not from Titu Andreescu. But original attempt by student Gopal/Sravan Pithwa. It had never occured to me a brute force proof would be so easy in this case :_-) Surely we learn a lot from children/students !!!

Prove that $a^{3}+b^{3}+c^{3}-3abic = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Proof:

LHS = $a^{3}+b^{3}+c^{3}-3abc = (a^{3}+b^{3})+ c^{3}-3abc$

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

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

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

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

$= (a+b)^{3}-3ab(a+b)+c^{3}-3abc$

$= (a+b)^{3}+c^{3}-3ab(a+b)-3abc$

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

$=(a+b+c)(a^{2} + b^{2}+c^{2}+2ab-ca-bc) - 3ab(a+b+c)$

$= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.

QED. Hence, $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

# Fibonacci Numbers: Recreational math

Reference: ICEEM Year 8 Math text book, Australian Mathematical Society.

A Fibonacci sequence is a sequence $F_{1}, F_{2}, F_{3}, \ldots$ of numbers in which each term from the third one onwards is the sum of the two terms that immediately precedes it. You have to have two numbers to start with, $F_{1}$ and $F_{2}$. These are called seeds. Then,

$F_{3}=F_{1}+F_{2}$

$F_{4}=F_{2}+F_{3}$

and so on. The classic Fibonacci sequence has 1 and 1 as its seeds. It first ten terms are:

1,1,2,3,5,8, 13, 21, 34 and 55.

Use a calculator when appropriate in the following:

Activity 1

Write out the classic Fibonacci sequence as far as its 25th term, $F_{25}$. Before you calculate $F_{11}$, make a rough guess of what $F_{25}$ will be. See how good your guess turns out to be.

Activity 2

Pick any two numbers as seeds and work out the first 20 terms for that Fibonacci sequence. Pick entirely different seed numbers from the person besides you (example, your friend, or teacher, or any one else imaginary :-)), and keep your list reasonably neat, as we will be coming back to it in a little while.

Activity 3

Swap your two seed numbers from Activity 2 around and figure out the first twenty terms in the new Fibonacci sequence. (If, your sequence in Activity 2 started 6, 11, 17, 28, 45, …) your new sequence will start 11, 6, 17, 23, 40, ….). Yes, you do get quite different numbers from Activity 2.

Activity 4:

It is now time to make a few observations about your Fibonacci sequences.

• The classic sequence in Activity 1 has two odd seeds. This gives a certain pattern of odd and even terms through out the sequence. What happens if you start with two even seeds, or an odd and even seed? Explain.
• Compare the 10th terms you generated in each of the sequences in Activity 2 and Activity 3. Which one is larger? Compare the 20th term as well. Can you explain what is happening?
• Use the calculator to divide the term $F_{10}$ by the term $F_{9}$ immediately before it. Write your answer down. Then, do the same with the second sequence. Now repeat the calculation with terms $F_{20}$ and $F_{19}$. Do you notice anything interesting? Did any of the “other people” who are doing this activity get the same ratio? They should all have got the same ratio. They should all get the same number, even though there may be v small diferences in the sixth decimal places.
• For the classic Fibonacci sequence, the first two terms larger than 1000000 are $F_{32}=1346269$ and $F_{33}=2178309$. Use these two values to see if what you noticed in the previous ratio calculations still holds for higher order terms in the classic Fibonacci sequence.

The number you obtained (to a good approximation) in the ratio calculations is famous and interesting enough to deserve its own Greek letter. It is called $\phi$ (pronounced to rhyme with spy). It is called the golden ratio or golden mean. It is a very interesting number with long history.

Search Google and you will discover some amazing facts about $\phi$. It appears in many different ways in geometry and architecture.

Now try calculating these values and see what you notice about them:

a. $\phi^{2}$

b. $\frac{1}{\phi}$

c. $(2\phi -1)^{2}$

Regards,

Nalin Pithwa.