The Spirit of Ramanujan
Go back to the Math Jam ArchiveAoPS instructor David Patrick will discuss the Spirit of Ramanujan Talent Initiative. We will discuss Ramanujan's life and some of the mathematics that he produced, and we will discuss the SoR program and how to apply. AoPS is a partner of SoR in 2020-21. We will be joined by Professor Ken Ono of the University of Virginia, who is the Director of the SoR program and was an associate producer and mathematical consultant for The Man Who Knew Infinity, a feature film about Ramanujan starring Dev Patel.
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Facilitator: Dave Patrick
Welcome to the 2020-21 Spirit of Ramanujan Math Jam!
woohoo
whoo hoo! so excited
I'm Dave Patrick, and I'll be leading our discussion tonight. Many of you know me from around AoPS: I've taught dozens of AoPS classes over the past 16 years, and I've written or co-written a few of our textbooks.
um hi
Wait a second...
OMG HI DAVE
DAVID PATRICK!!!!?
You're not Dave Patrick...
rrllly
Ur not Dave Patrick tho
I think they prepared this for someone else.
We are pleased to have a very special guest here with us tonight. Ken Ono (KenOno) is the Thomas Jefferson Professor of Mathematics at the University of Virginia and is also Vice President of the American Mathematical Society (AMS). The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society.
Say hi, Ken!
Hi everyone! Pleased to be here.
Hi Ken
Hi Ken!
Hello Ken!
Now, THAT's right! Hi!
Hello Ken!
hi, Ken!
hello ken!
hi
Hi!
Ken is also the Director of the "Spirit of Ramanujan" project, which AoPS is also a partner with, and which we'll be talking about tonight.
THANKS IN ADVANCE FOR ALL THE HARD WORK THE MODS ARE DOING
Speaking of which, today we have a teaching assistant: Victoria Li (mag1c).
HI!
Hi
Hi mag1c!
Hello!
hello!!
hello!
Hi Victoria
Hello mag1c!
Hi!
Hi Victoria!
Hello Victoria!
Hi Victoria!
Hi Victoria.
Hi!
Victoria studied mathemagics at HCSSiM and Colorado Math Circle, did research fellowships on AMO physics and genome modification off-target prediction at RSI and BioFrontiers Institute, studied mathematics at four universities in two countries, has an MS Applied Mathematics from the University of Colorado Boulder, and has been teaching math/programming/Chinese since 2010. As a learner and educator, she strives for:
1. Compassion (everyone can become a successful mathematician when taught by the right teacher)
2. Open-mindedness (there are many ways to solve a problem and many different learning styles)
3. Constant Improvement (learning never stops...)
In her free time she likes to ponder questions such as "What happened before the Big Bang (13.8 billion years ago)?", "Why do Fibonacci numbers occur in pinecones?", and "What are the implications of Gödel's Incompleteness Theorem..."
Victoria is here to try to answer any questions that you have, and Ken and I will also be answering questions at the end. We may have a lot of students here tonight, so please be patient! We may not be able to get to every question.
Wat is this?
what will this be about If not relevant I gotta go
Tonight, we're going to talk about the life and mathematics of Srinivasa Ramanujan.
Mathematics or Mathemagics?
so history?
All of the above.
But I'll let Dr. Patrick show you all that. I think his internet kicked back on.
I have a first day issue of that stamp.
Hopefully I shooed all the gremlins away that were gnawing at my internet cable...
As you may know, there was recently (2015) a motion picture about Ramanujan's life titled The Man Who Knew Infinity, starring Dev Patel and Jeremy Irons. The film is based on a biography of the same title written in 1991 by Robert Kanigel.
Did anybody here see the movie?
I did!
Meee!
Yes
That was a very entertaining movie! I saw it in theaters!
YES!
Me!
I saw it a week ago because of this math jam!
yes!
I loved the movie!
me!
Our guest Ken Ono was one of the mathematical consultants on the film.
Unfortunately as of November 2020 it does not appear to be available on any streaming service in the U.S.
But it is available for rent for $2.99 on a bunch of different sites.
Ramanujan was born in Erode, India in 1887. He grew up relatively poor and worked as a clerk. But from an early age he clearly had a love and aptitude for mathematics.
Srinivasa Ramanujan had his interest in mathematics unlocked by a book.
You're right! A particular book that furthered his study was A Synopsis of Elementary Results in Pure and Applied Mathematics by G.S. Carr, that Ramanujan obtained in 1903 at age 16. This book was written in 1886. It contained over 5000 theorems and was an attempt to summarize all the "basic" mathematics known at the time. (You can find it in Google Books!) Ramanujan read and studied it in great detail and it formed the foundation of his mathematical thought.
By age 11 he was the mathematical equal of university students who were boarding in his family home. He was able to continue his advanced study of math by reading as many books as he could get his hands on. He also, around age 15, discovered on his own how to solve a quartic equation.
(And don't worry if you don't know what "quartic equation" means -- I'll explain that in a little bit!)
Ramanujan spent so much time on mathematics that he generally did poorly in other subjects in school, and as such failed to graduate from university. He unfortunately also suffered from various illness (much of which were probably made worse due to poverty, and to the lack of "modern" medicine in India at the time), which also interfered with his formal studies.
Shoutout to G. H. Hardy for beginning Ramanujan's career! ;P
Indeed, we're coming to that!
Even though his mathematical talent was becoming acknowledged within India, he had trouble finding employment as a mathematician because of his lack of a formal university degree. Instead, he had to work as a clerk for low pay and rely on the additional financial support of others in the Indian mathematics community.
He sent multiple letters to Cambridge!
A major turning point in Ramanujan's life was his writing a letter to the prominent British mathematician G. H. Hardy in January 1913.
Was there a reply?
Did G. H. Hardy reply to him?
Good question, I think Dave left you with a cliff hanger.
Noo, I hate those!
Before telling you if he got a reply, here's how the letter began.
I have had no university education but I have undergone the ordinary school course. After leaving school I have been employing the spare time at my disposal to work at mathematics. I have not trodden through the conventional regular course which is followed in a university course, but I am striking out a new path for myself. I have made a special investigation of divergent series in general and the results I get are termed by the local mathematicians as 'startling'.
Enclosed with the letter was pages of Ramanujan's work -- he was hoping that Hardy would be able to mentor him and help him get his work improved and published to the wider world. He had written to other mathematicians in Britain, who had basically ignored his letters.
Did G.H Hardy listen?
(drumroll, please!)
Hardy took the letter seriously!
Yay!
yayayayay
Yay!!
YAAAY
Although a lot of the work was unproved and was poorly written, it was clear to Hardy that there was a lot of really good math in Ramanujan's letter! Hardy started a professional relationship with Ramanujan and was able to provide financial support. Indeed, on Hardy's recommendation, Ramanujan was finally able to get a scholarship to the University of Madras.
Finally, in 1914, Hardy arranged for Ramanujan to be able to travel and visit Trinity College at Cambridge University in England.
Do you know how long it took Ramanujan to travel from India to London? Remember, this was slightly over 100 years ago!
2 years
1 year?
4 years?
A year?
I said 100 years ago, not 500.
a month???
months?
About a month?
one month
one month?
27 days!
oh, wow
wow
wow?
Wow!
That's a lot faster than I expected
wow it would only take like 9 hours now
Ramanujan left India on March 17, 1914 by boat, and arrived in London on April 14. Then began a 5-year collaboration with Hardy that produced many important mathematical results. Also, Ramanujan was finally able to finish his formal education, and graduated from Cambridge with the equivalent of a Ph.D. in 1916.
With financial stability and a group of peers that respected his work, Ramanujan was able to professionally thrive in England. Many honors followed including perhaps the most prestigious: in 1918 he was elected a Fellow of the Royal Society, probably the highest honor that can be given for science in Britain. Ramanujan was only the second Indian so elected, and one of the youngest at age 31.
yay!
Unfortunately, for much of his time in England, Ramanujan was in very poor health. He spent much of his time in hospitals while in England, and nearly died in 1917. (The fact that World War I was going on, affecting supplies, probably didn't help.) There's a famous mathematical anecdote about Hardy visiting Ramanujan in the hospital that I'll share in a little bit.
c'mon devenware
Apparently Dave's Gremlins came to my house too.
lol
Atoms789 hi
Huh?
nuu
I'm now on a mobile hotspot, we got this.
lol I thought I lost connection
Nope, just me.
In early 1919, the war was over and Ramaujan's health had improved, so (now age 31) he decided to return to his native India, with his reputation firmly established and his financial situation stable. The hope was that he would continue his career as a full member of India's mathematical community.
But sadly, he died.
hope...
he got sick again
Sadly, upon returning to India, he almost immediately fell into poor health, and after about a year, he died on April 26, 1920, at an age of only 32.
oh.
If you're interested in reading a more thorough biography of Ramanujan, there is one on the "History of Mathematics" site maintained by the University of St. Andrews, at
http://www-history.mcs.st-and.ac.uk/Biographies/Ramanujan.html
The quotes that I've used above are from this page.
As an aside, this website is a great resource: it has biographies of nearly 3,000 mathematicians spanning over 35 centuries, from Ahmes (1680-1620 BC) to Mirzakhani (1977-2017), and includes dozens of living mathematicians as well.
is that the whole story of his lfe?
Well, we gave a brief summary! You can find much more detail on that page.
is this it? or is there more?
We still got math right?
Now let's talk about some the mathematics that Ramanujan studied.
Yay math!!!!
yessss
Yay Math
Ramanujan did a lot of mathematics in his relatively short life, and a great deal of it was quite advanced. But there are some objects of his study that we can discuss tonight.
MATH IS AS INTERESTING AS IT IS BECAUSE OF PEOPLE LIKE HIM!!
First, I mentioned earlier that one of Ramanujan's earliest advanced studies was, in his early teen years, his study of solutions to quartic and quintic equations.
To start, many of you already know how to solve quadratic equations, which is an equation involving a polynomial of degree 2.
In particular, how can we solve $ax^2 + bx + c = 0$?
we can use the quadratic formula
quardratic equation
factoring quadratic
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
quadratic formula
quadratic formula
We use the well-known quadratic formula: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
This is not too hard to prove using basic algebra. (We prove it in AoPS's Introduction to Algebra textbook and classes.)
complete the square
or complete the square
Yeah, that's how you do it!
The next step up is solving the cubic equation $ax^3 + bx^2 + cx + d = 0$. There is a formula for solving this as well, but it's a lot more complicated.
If we set$$\begin{array}{rcl}
p &=& -\frac{b}{3a} \\
q &=& p^3 + \frac{bc-3ad}{6a^2} \\
r &=& \frac{c}{3a}
\end{array}$$
cubic formula?
then
$$x = \sqrt[3]{q + \sqrt{q^2 + (r-p^2)^3}} + \sqrt[3]{q - \sqrt{q^2 + (r-p^2)^3}} + p.$$
wow
Wow that's big
wow long formula
Wow. That's complex
And we simplified it using our new variables...
holy cow that's complicated
cOMPLICATED
It's confusing!
Oh it gets even worse
Exactly!
But, you know who loves cubics?
me
me?
Then you and DPatrick have something in common!
The mention of cubics woke him up from his slumber.
I do love cubics.
And I hope the gremlins are gone for good this time.
But...I don't love cubics enough to have memorized that formula for solving them.
Most people don't bother to ever learn this formula nowadays: it's way too complicated to accurately memorize. (I certainly don't have it memorized -- I looked it up. I don't even guarantee that I typed it correctly above!)
can't there be up to 3 solutions?
Indeed, good observation!
That formula only give us one solution...
...or so it seems.
But there are three cube roots if you consider complex numbers!
Use polynomial division and the quadratic formula to find the other two.
we can then use quadratic formula
Or you can do that: once you have one root, you can divide it down to a quadratic.
You'll learn about all this stuff in algebra classes down the line if you don't know it already.
This formula, and other similar formulas for solving cubics, were discovered in the 1500s. Ramanujan would have learned this in his early teens.
is this mentioned because he worked with cubics?
Yes---and he took it one step further!
And he found the quartic?
quartics
What Ramanujan then undertook, and eventually successfully solved, was solving the quartic equation $ax^4 + bx^3 + cx^2 + dx + e = 0$.
You saw how complicated the cubic formula was...imagine how complicated the quartic formula is!
I'm not even going to write the formula down -- it would take way way too much space.
Can we see it?
$\boxed{\text{show me it}}$
I wish I could see it
Oh, OK...
yikes
aaaah
!!
that is horrifying
a wolfram alpha screenshot?
It's from http://pjk.scripts.mit.edu/pkj/wp-content/uploads/2018/11/quartic_formula.jpg
that is soooo long
He derived that himself??
Yes. It might now have been this equal formula, but it was something equivalent.
It's possible that Ramanujan thought that he had discovered something new. He almost certainly didn't know that the quartic had also been solved in the 1500s.
But it is an extremely complicated formula, and it's amazing that Ramanujan was able to discover it, essentially on his own, at age 15.
w==but you cant solve a plynomial to the 5thdegree right?
Good question! Ramanujan attempted to solve the quintic equation:
$$ ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0 $$
He did not have success. Does anyone know why?
Yes Quintics are impossible
IT'S IMPOSSIBLE
impossible?
Is it not possible?
Cuz it's impossible?
Quintics are impossible?
Abel-Ruffini I believe...
Abel–Ruffini theorem
In the early 1800s, the mathematicians Ruffini and Abel actually proved that a solution was impossible! That is, there is no possible formula (no matter how complicated) that can be used to plug in the coefficients of a quintic and output a solution. (Ruffini first came up with the result in 1799, but his proof was incomplete. Abel completed the proof in 1824.)
Wow
Cool
does that mean degree 5 and up are all impossible?
That's right. Degree 5 and up are all impossible.
A little later in the 1800s, the mathematician Galois developed a new branch of mathematics (that today is called Galois theory) that provided a more elegant proof that there is no formula to solve the quintic. It's part of a branch of mathematics called field theory, which you typically first learn about in an undergraduate abstract algebra course.
Ooh can we see the proof?
I used to teach this proof! But it came at the year of a year-long abstract algebra class taught at university.
So maybe we'll skip it today.
Again, Ramanujan probably didn't know any of this at the time. But more importantly, he didn't let his lack of success in this particular problem deter him from his continued pursuit of mathematics.
Also, sir what is ramanujan's biggest achievement
There are so many to choose from! But the next one I'll talk about might be the most famous.
One of the most famous stories regarding Ramanujan occurred when Hardy was visiting Ramanujan in the hospital during one of his many illnesses.
1729?
1729!!!
Because of Ramanujan, 1729 is now my favourite number
oh you mean 1729?
$1729$
Hardy recalled telling Ramanujan that he traveled to the hospital in taxicab number $1729$, and was disappointed that the cab number was so dull. Ramanujan immediately replied that on the contrary, $1729$ was a very interesting number.
Do you know why?
Sum of 2 cubes, 2 ways
smallest number that can be represented as the sum of 2 cubes in 2 ways
it is the least one that is the sum of two cubes in two different ways
$1729$ is the smallest positive integer than is the sum of $2$ positive pairs of $2$ positive cubes.
It is the smallest number that can be written as a sum of two positive cubes in two different ways! Can you find them?
Nice...
12^3 + 1^3 = 10^3 + 9^3 = 1729
12^3+1=10^3+9^3
12^3+1^3=10^3+9^3
${10}^3 + 9^3, 1^3 + {12}^3$
$12^3+1^3$ and $10^3+9^3$
$1^3+12^3=9^3+10^3=1729$
$1^3+12^3$ and $9^3+10^3$
1^3 + 12^3 and 9^3 + 10^3?
$12^{3}+1^{3}$ and $10^{3}+9^{3}$
$1729 = 1^3 + 12^3 = 1 + 1728$
$1729 = 9^3 + 10^3 = 729 + 1000$
For this reason, $1729$ is called the 2nd taxicab number. It's also sometimes called the Hardy-Ramanujan number.
is there a proof?
bashing is the proof
Right. The proof is essentially: you check that all of 1 through 1728 don't work.
why is it the second and not the first?
what is the first taxicab number?
Great question with a slightly silly answer.
The first is the smallest number that can be written as a sum of two positive cubes in at least one way.
Which is...?
2
wow so it is 2
2?
2
2
2
2
2?
2
$2$
2?
2
Right. The first taxicab number is $2 = 1^3 + 1^3$.
And continuing, the 3rd taxicab number is the smallest number that can be written as a sum of two positive cubes in three different ways. Any guess as to how big it is? (How many digits do you think it has?)
8?
8
It's the 8-digit number $87{,}539{,}319$, which is $167^3 + 436^3 = 228^3 + 423^3 = 255^3 + 414^3$.
It's likely that this was first discovered as recently as 1957.
More generally, the $n$th taxicab number is the smallest positive integer that can be written as a sum of two positive cubes in $n$ different ways.
what's the largest known taxicab number?
Have we discovered the 4th taxicab number yet?
Only the first 6 taxicab numbers are known:
$2$
$1729$
$87539319$
$6963472309248$
$48988659276962496$
$24153319581254312065344$
This is all pretty new! The last of these was discovered by computer search in 2008.
What is the practicality of the the taxicab number though? Does it translate to an applied purpose?
I am so glad you asked!
This is not just a fun game to find these numbers. Just five years ago in 2015, our guest Ken Ono and my AoPS colleague Sarah Trebat-Leder wrote a paper titled The 1729 K3 Surface which tied Ramanujan's observation about the number 1729 to some important curves and surfaces in the modern field of algebraic geometry.
wow!
Thanks David...And this appears prominently in black hole physics.
I searched it up and they found the 7th taxicab number a month ago at MIT
Really??? I didn't know that! I'll have to look it up afterwards.
Is there a proof that there are infinitely many taxicab numbers?
Indeed there is.
The buzz words are: K3 surfaces, 3d quantum gravity...
It's been proven that for any $n$, there is some number that can be written as a sum of two positive cubes in at least $n$ different ways. Hardy proved this in 1938 along with his colleague E. M. Wright.
One more fun note about taxicab numbers: How many people here are fans of the TV show Futurama?
It hasn't been on in a while, but it's still a favorite.
Apparently the creators of Futurama are big fans of taxicab numbers.
In particular, the number $1729$ is hidden in several episodes. It's the registration number of the Planet Express ship, for one thing.
Image from mathworld.wolfram.com
i never realized this
Woah!
And in one scene someone hails a taxi. Can you make out the number on the cab's roof?
Image from Gawker Media
87539319
87539319
the 3rd taxicab number
87539319
It's hard to make out, but it's the 3rd taxicab number $87539319$.
It may not surprise you to learn that Futurama's head writer did math contests in high school and has a degree in physics from Harvard!
One more bit of math tonight before we talk about the Spirit of Ramanujan program.
We showed you the Spirit of Ramanujan Project logo at the start of the Math Jam:
Does anybody know the significance of this logo?
ahh the nested radical looking thingy
Has lots of square roots
sqrt rt?
square roots?
idk infinite square root
That's right!
The logo was inspired by the following famous problem that was posed by Ramanujan:
Evaluate the following expression:
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
This is an example of a nested radical.
OH THAT ONE
Oh I've seen that before!
Ohh
Yes, it's a pretty famous item in Ramanujan's history.
He submitted this problem to the Journal of the Indian Mathematical Society as a challenge for others to solve.
After 6 months passed and no one had solved it(!), he provided the solution himself.
Any idea how we might solve it?
(If you've seen this before and know how it works, please don't spoil it for others!)
set one part as a variable
rewrite the things within the radicals
Good ideas...let's try a simpler problem first to see how these ideas work.
A similar problem is in the intro to algebra textbook
Absolutely! I think it may be this one in fact:
Suppose we want to evaluate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}$.
let x = sqrt(2 + sqrt2..)
Set that value to a variable, $x$.
Good idea. We can set it to be equal to a variable $x$, so that
$$x = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}.$$
Square both sides of the equation
square it
Square it
Sounds like a good thing to try.
We square both sides:
$$x^2 = 2 + \sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}.$$
But now what do you notice?
x^2 = 2 + x
let the other square root parts be x
We have $x$ again
get x again
$x^2=2+x$
This means that $x^2 = 2+x$
hey, the thing on the right looks like $x$
There's the same thing inside
right side is 2+x
it is just x^2=2+x
x^2 = 2 + x
the secopnd part is the same as the original
Aha! That big square root on the right side is just another copy of $x$!
So our equation becomes just $x^2 = 2 + x$.
Quadratic!
Turn it into a quadratic: $x^2-x-2=0$
quadratic!
it'S A QUADRATIC!
Then we have a quadratic
This becomes $x^2 - x + 2 = 0$, which factors as $(x-2)(x+1) = 0$.
Then solve the quadratic and use the positive root
x^2-x-2=(x-2)(x+1) x has to be positive so x=2!!
and x is positive
its equal to 2
2
answer is 2
Right: either $x=2$ or $x=-1$. And since square roots are positive, we must have $x=2$.
So we conclude that
$$2 = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}.$$
So now back to Ramanujan's problem:
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
Ramanujan did the same sort of clever substitution, but he had to be a little more clever.
we don't have complete repetition here though, it changes a little bit each time
Right. His nested radical doesn't exactly repeat, but there's definitely a pattern.
So he couldn't just use a single variable to represent a repeating part.
Instead he used a function to capture the pattern.
Specifically, he let $$f(x) = \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}.$$
Wow that was good thinking
Notice that if we plug in $x=2$ into this function, we get the value that we want! So our goal is to determine $f(2)$.
And so now what should we do?
What did we do in the easier problem at this stage?
squared
square
square both sides
square it
square it
Square it?
square it?
We squared it I think...
Yeah! Let's square it!
When we square this, we get $$(f(x))^2 = 1 + x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}.$$
And now, what is the big square root term?
f(x+1)
$f(x+1)$?
f(x+1)
f(x plus 1)
f(x+1)
f(x+1)
f(x+1)
$f(x+1)$
It's $f(x+1)$. Everywhere there was an $x$ in the original function, we replace it by $x+1$. So the old $x+1$ becomes $(x+1)+1$ or $x+2$, and so on.
And now we have the functional equation $$(f(x))^2 = 1 + xf(x+1).$$
And then because Ramanujan was very good with algebra, he was essentially able to "guess and check" what $f(x)$ must be. Can you?
x+1
Right! It's $f(x) = x+1$. Let's check.
If we use $f(x) = x+1$, then our functional equation become $$(x+1)^2 = 1 + x((x+1)+1).$$
Does this work?
yes
Yes?
YES!
Yes
yes
yes
yes
Yes
It does! Both sides are $x^2 + 2x + 1$. So it worked!
And how does this answer the original problem posed by Ramanujan?
f(2)=3
plug in x = 2
so f(2) = 3
x=2
since x=2, f(x)=3!
THE ANSWER IS 3!!! WOW THIS IS INGENIOUS!!!
$f(2) = 2+1 = 3$
$f(2) = 2+1 = 3$
plug in for x = 2
x+1 = f(2), so the answer is 3
Our original quantity is $f(2)$, so if $f(x) = x+1$, then $f(2) = 2+1 = \boxed{3}$.
Ramanujan was pretty clever, eh?
how do you show it is the only function
What if there are other functions?
How do you know that $x+1$ is the only solution to the functional equation?
Well...that's a really good point. There are certainly some icky details that I'm leaving out.
What Ramanujan did is something a little more general.
He proved that $$x+n+a = \sqrt{ax + (n+a)^2 + x\sqrt{a(x+n) + (n+a)^2 + (x+n)\sqrt{\cdots}}}$$ for any values of $x$, $n$, and $a$, using a very similar functional equation.
The original problem is the special case where $a=0$, $n=1$, and $x=2$.
It's believed that Ramanujan proved this formula when he was about 16 years old!
Wow!
cool
cool!
WOW! HOW?
Wow thats really young
So now that we've talked a little about the life and math of Ramanujan, let's transition into discussing the Spirit of Ramanujan initiative.
Let me start by citing an excerpt from an article called "Why Ramanujan Matters" written by Ken Ono and Robert Schneider:
Ramanujan matters because he represents endless curiosity and untapped potential, which we all have to believe in to proceed in the sciences. Science usually advances on the work of thousands, over generations, fine-tuning and extending the scope of understanding. But from time to time, creative fireballs like Ramanujan burst onto the scene propelling human thought forward. Yet what if Ramanujan had not reached out to, or been taken seriously by Hardy? The loss of scientific understanding is something our modern world could not absorb. He matters because science matters: curiosity and creativity drive scientific inquiry.
its true
Nice speech
That's what the Spirit of Ramanujan (which I'm going to abbreviate as SoR from now on) is all about: finding young people who have the potential, with the right resources and mentorship, to contribute to humankind's knowledge like Ramanujan did.
We are AoPS are pleased that Ken Ono asked us to be a part of this search. SoR's goals are also part of AoPS's goals: to create a broader community of aspiring mathematicians and problems solvers, to make connections between people who might not otherwise have the opportunity to contact, and to provide resources via our website to all corners of the country and of the world.
My pleasure.
Specifically, SoR is a grant program that provides two types of awards.
The Templeton-Ramanujan Development Awards will provide books and monetary grants to offset the costs of online resources.
The Templeton-Ramanujan Fellowships will provide monetary grants to offset the costs of enrolling in approved enrichment programs, such as summer camps.
This can include AoPS books and online programs.
By the way, the "Templeton" in these names is the Templeton World Charity Foundation, created by the late Sir John Templeton, and which is providing the funding for SoR.
Basically, SoR is hoping to play to role for future Ramanujans that G. H. Hardy did for the actual Ramanujan: access to resources and mentorship to nurture future mathematical and scientific talent.
2020-21 is the 5th year of the program. To date, the program has identified 46 award winners, with about half from the U.S. and half from elsewhere in the world.
how can one qualify for the awards?
To learn more about SoR, please visit the website spiritoframanujan.com.
The SoR website includes a link to the application page. You can also read brief bios of the previous winners on the website.
Please note that although awards will be made on a rolling basis, you should apply by March 1, 2021 in order to guarantee full consideration.
We have made several awards to the AoPs community, and we hope to do so again in 2021.
And with that, this concludes our Math Jam! Thanks for attending! Ken Ono and I will stay around for a little while to answer questions.
Great show David...
Lots of very talented folks online.
Thanks Ken! And very special thanks to Deven for jumping in while the gremlins attacked.
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