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The Spirit of Ramanujan

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AoPS 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

devenware 2020-11-10 19:30:08
Welcome to the 2020-21 Spirit of Ramanujan Math Jam!
devenware 2020-11-10 19:30:09
https://spiritoframanujan.com/wp-content/uploads/2018/09/Ramanujan_LOGO.png
Adul 2020-11-10 19:30:31
woohoo
Jianning 2020-11-10 19:30:31
whoo hoo! so excited
devenware 2020-11-10 19:30:37
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.
ryanzhao 2020-11-10 19:30:43
um hi
devenware 2020-11-10 19:30:45
Wait a second...
leafwhisker 2020-11-10 19:30:49
OMG HI DAVE
Adul 2020-11-10 19:31:00
DAVID PATRICK!!!!?
Countcountcount 2020-11-10 19:31:00
You're not Dave Patrick...
Mathbee2020 2020-11-10 19:31:00
rrllly
NehaSeban 2020-11-10 19:31:00
Ur not Dave Patrick tho
devenware 2020-11-10 19:31:04
I think they prepared this for someone else.
devenware 2020-11-10 19:31:21
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.
devenware 2020-11-10 19:31:29
Say hi, Ken!
KenOno 2020-11-10 19:31:47
Hi everyone! Pleased to be here.
vmr123 2020-11-10 19:31:56
Hi Ken
leafwhisker 2020-11-10 19:31:56
Hi Ken!
NehaSeban 2020-11-10 19:31:56
Hello Ken!
Pandayue 2020-11-10 19:31:56
Now, THAT's right! Hi!
Mathbee2020 2020-11-10 19:31:56
Hello Ken!
AOPqghj 2020-11-10 19:31:56
hi, Ken!
SmartGroot 2020-11-10 19:31:56
hello ken!
Andrew_Dou 2020-11-10 19:31:56
hi
nathanj 2020-11-10 19:31:56
Hi!
devenware 2020-11-10 19:32:05
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.
Anthony0512 2020-11-10 19:32:50
THANKS IN ADVANCE FOR ALL THE HARD WORK THE MODS ARE DOING
devenware 2020-11-10 19:32:57
Speaking of which, today we have a teaching assistant: Victoria Li (mag1c).
bsu1 2020-11-10 19:33:16
HI!
yesufsa 2020-11-10 19:33:16
Hi
Skyhigh1218 2020-11-10 19:33:16
Hi mag1c!
GXboy12345 2020-11-10 19:33:16
Hello!
Adul 2020-11-10 19:33:16
hello!!
Anthony0512 2020-11-10 19:33:16
hello!
vmr123 2020-11-10 19:33:16
Hi Victoria
interactivemath 2020-11-10 19:33:16
Hello mag1c!
yesufsa 2020-11-10 19:33:16
Hi!
bsu1 2020-11-10 19:33:16
Hi Victoria!
GXboy12345 2020-11-10 19:33:16
Hello Victoria!
Countcountcount 2020-11-10 19:33:16
Hi Victoria!
KenOno 2020-11-10 19:33:19
Hi Victoria.
mag1c 2020-11-10 19:33:29
Hi!
devenware 2020-11-10 19:33:34
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..."
devenware 2020-11-10 19:33:43
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.
Rishab-AOPS 2020-11-10 19:34:03
Wat is this?
Senguamar 2020-11-10 19:34:03
what will this be about If not relevant I gotta go
devenware 2020-11-10 19:34:04
Tonight, we're going to talk about the life and mathematics of Srinivasa Ramanujan.
devenware 2020-11-10 19:34:09
https://spiritoframanujan.com/wp-content/uploads/2018/09/postage-copy.jpg
GXboy12345 2020-11-10 19:34:28
Mathematics or Mathemagics?
Anthony0512 2020-11-10 19:34:28
so history?
devenware 2020-11-10 19:34:33
All of the above.
devenware 2020-11-10 19:34:46
But I'll let Dr. Patrick show you all that. I think his internet kicked back on.
KenOno 2020-11-10 19:35:14
I have a first day issue of that stamp.
DPatrick 2020-11-10 19:35:23
Hopefully I shooed all the gremlins away that were gnawing at my internet cable...
DPatrick 2020-11-10 19:35:51
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.
DPatrick 2020-11-10 19:36:12
Did anybody here see the movie?
ananyasharma 2020-11-10 19:36:34
I did!
SuperJJ 2020-11-10 19:36:34
Meee!
Askumar 2020-11-10 19:36:34
Yes
devenware 2020-11-10 19:36:34
That was a very entertaining movie! I saw it in theaters!
BlazingSun1200 2020-11-10 19:36:34
YES!
hwdaniel 2020-11-10 19:36:34
Me!
lilac613 2020-11-10 19:36:34
I saw it a week ago because of this math jam!
BrightMonkey 2020-11-10 19:36:34
yes!
SuperJJ 2020-11-10 19:36:34
I loved the movie!
NightFury101 2020-11-10 19:36:34
me!
DPatrick 2020-11-10 19:36:44
Our guest Ken Ono was one of the mathematical consultants on the film.
DPatrick 2020-11-10 19:36:57
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.
DPatrick 2020-11-10 19:37:14
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.
BlazingSun1200 2020-11-10 19:37:28
Srinivasa Ramanujan had his interest in mathematics unlocked by a book.
DPatrick 2020-11-10 19:37:38
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.
DPatrick 2020-11-10 19:38:02
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.
DPatrick 2020-11-10 19:38:25
(And don't worry if you don't know what "quartic equation" means -- I'll explain that in a little bit!)
DPatrick 2020-11-10 19:38:45
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.
GXboy12345 2020-11-10 19:39:15
Shoutout to G. H. Hardy for beginning Ramanujan's career! ;P
DPatrick 2020-11-10 19:39:29
Indeed, we're coming to that!
DPatrick 2020-11-10 19:39:38
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.
BlazingSun1200 2020-11-10 19:39:49
He sent multiple letters to Cambridge!
DPatrick 2020-11-10 19:40:11
A major turning point in Ramanujan's life was his writing a letter to the prominent British mathematician G. H. Hardy in January 1913.
Noam2007 2020-11-10 19:40:48
Was there a reply?
Countcountcount 2020-11-10 19:40:48
Did G. H. Hardy reply to him?
devenware 2020-11-10 19:41:28
Good question, I think Dave left you with a cliff hanger.
Skyhigh1218 2020-11-10 19:41:48
Noo, I hate those!
devenware 2020-11-10 19:41:54
Before telling you if he got a reply, here's how the letter began.
devenware 2020-11-10 19:42:00
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'.
devenware 2020-11-10 19:42:06
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.
nemoatl 2020-11-10 19:42:27
Did G.H Hardy listen?
Noam2007 2020-11-10 19:42:47
(drumroll, please!)
devenware 2020-11-10 19:42:48
Hardy took the letter seriously!
Noam2007 2020-11-10 19:43:04
Yay!
hipeeps1104 2020-11-10 19:43:04
yayayayay
ananyasharma 2020-11-10 19:43:04
Yay!!
NehaSeban 2020-11-10 19:43:04
YAAAY
devenware 2020-11-10 19:43:05
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.
devenware 2020-11-10 19:43:13
Finally, in 1914, Hardy arranged for Ramanujan to be able to travel and visit Trinity College at Cambridge University in England.
devenware 2020-11-10 19:43:16
Do you know how long it took Ramanujan to travel from India to London? Remember, this was slightly over 100 years ago!
danielWang816 2020-11-10 19:43:49
2 years
Peter0527 2020-11-10 19:43:49
1 year?
hipeeps1104 2020-11-10 19:43:49
4 years?
stevecabbage 2020-11-10 19:43:49
A year?
devenware 2020-11-10 19:43:55
I said 100 years ago, not 500.
vmr123 2020-11-10 19:44:21
a month???
MinervaPi 2020-11-10 19:44:21
months?
TheGeometer 2020-11-10 19:44:21
About a month?
amazingxin777 2020-11-10 19:44:21
one month
amazingxin777 2020-11-10 19:44:21
one month?
devenware 2020-11-10 19:44:24
27 days!
SmartGroot 2020-11-10 19:44:42
oh, wow
KLBBC 2020-11-10 19:44:42
wow
Yash1212 2020-11-10 19:44:42
wow?
BlazingSun1200 2020-11-10 19:44:42
Wow!
stevecabbage 2020-11-10 19:44:42
That's a lot faster than I expected
Morrigan_Black 2020-11-10 19:44:42
wow it would only take like 9 hours now
devenware 2020-11-10 19:44:45
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.
devenware 2020-11-10 19:44:54
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.
danielWang816 2020-11-10 19:45:07
yay!
devenware 2020-11-10 19:45:15
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.
amazingxin777 2020-11-10 19:47:34
c'mon devenware
devenware 2020-11-10 19:47:37
Apparently Dave's Gremlins came to my house too.
miya0829 2020-11-10 19:47:50
lol
Morrigan_Black 2020-11-10 19:47:50
Atoms789 hi
stevecabbage 2020-11-10 19:47:50
Huh?
Mathbee2020 2020-11-10 19:47:50
nuu
devenware 2020-11-10 19:47:52
I'm now on a mobile hotspot, we got this.
SmartGroot 2020-11-10 19:48:26
lol I thought I lost connection
devenware 2020-11-10 19:48:30
Nope, just me.
devenware 2020-11-10 19:48:55
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.
vmr123 2020-11-10 19:49:41
But sadly, he died.
BobDBuilder321 2020-11-10 19:49:41
hope...
Infinite_May 2020-11-10 19:49:41
he got sick again
devenware 2020-11-10 19:49:44
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.
SmartGroot 2020-11-10 19:50:31
oh.
micyang 2020-11-10 19:50:31
rjeagle2019 2020-11-10 19:50:31
ouyangc 2020-11-10 19:50:31
devenware 2020-11-10 19:50:35
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.
devenware 2020-11-10 19:50:47
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.
amazingxin777 2020-11-10 19:50:57
is that the whole story of his lfe?
devenware 2020-11-10 19:51:06
Well, we gave a brief summary! You can find much more detail on that page.
EpicAarn 2020-11-10 19:51:15
is this it? or is there more?
Anthony0512 2020-11-10 19:51:15
We still got math right?
devenware 2020-11-10 19:51:20
Now let's talk about some the mathematics that Ramanujan studied.
URcurious2 2020-11-10 19:51:33
Yay math!!!!
Mathbee2020 2020-11-10 19:51:33
yessss
rjeagle2019 2020-11-10 19:51:33
Yay Math
devenware 2020-11-10 19:51:38
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.
BlazingSun1200 2020-11-10 19:51:42
MATH IS AS INTERESTING AS IT IS BECAUSE OF PEOPLE LIKE HIM!!
devenware 2020-11-10 19:51:48
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.
devenware 2020-11-10 19:52:01
To start, many of you already know how to solve quadratic equations, which is an equation involving a polynomial of degree 2.
devenware 2020-11-10 19:52:03
In particular, how can we solve $ax^2 + bx + c = 0$?
Andrew_Dou 2020-11-10 19:53:03
we can use the quadratic formula
umikakv2013 2020-11-10 19:53:03
quardratic equation
Jianning 2020-11-10 19:53:03
factoring quadratic
Spaced_Out 2020-11-10 19:53:03
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
dolphin7 2020-11-10 19:53:03
quadratic formula
InsaneMoosen 2020-11-10 19:53:03
quadratic formula
devenware 2020-11-10 19:53:06
We use the well-known quadratic formula: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
devenware 2020-11-10 19:53:11
This is not too hard to prove using basic algebra. (We prove it in AoPS's Introduction to Algebra textbook and classes.)
BlazingSun1200 2020-11-10 19:53:14
complete the square
amazingxin777 2020-11-10 19:53:20
or complete the square
devenware 2020-11-10 19:53:24
Yeah, that's how you do it!
devenware 2020-11-10 19:53:28
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.
devenware 2020-11-10 19:53:45
If we set$$\begin{array}{rcl}


p &=& -\frac{b}{3a} \\


q &=& p^3 + \frac{bc-3ad}{6a^2} \\


r &=& \frac{c}{3a}


\end{array}$$
Anthony0512 2020-11-10 19:53:50
cubic formula?
devenware 2020-11-10 19:53:52
then


$$x = \sqrt[3]{q + \sqrt{q^2 + (r-p^2)^3}} + \sqrt[3]{q - \sqrt{q^2 + (r-p^2)^3}} + p.$$
Peter0527 2020-11-10 19:54:05
wow
rjeagle2019 2020-11-10 19:54:05
Wow that's big
Mathbee2020 2020-11-10 19:54:05
wow long formula
Countcountcount 2020-11-10 19:54:07
Wow. That's complex
devenware 2020-11-10 19:54:14
And we simplified it using our new variables...
asengo 2020-11-10 19:54:23
holy cow that's complicated
Skyhigh1218 2020-11-10 19:54:23
cOMPLICATED
CosmicIQ 2020-11-10 19:54:23
It's confusing!
rjeagle2019 2020-11-10 19:54:26
Oh it gets even worse
devenware 2020-11-10 19:54:33
Exactly!
devenware 2020-11-10 19:54:48
But, you know who loves cubics?
riverteng 2020-11-10 19:55:01
me
ezRobots 2020-11-10 19:55:01
me?
devenware 2020-11-10 19:55:08
Then you and DPatrick have something in common!
devenware 2020-11-10 19:55:16
The mention of cubics woke him up from his slumber.
DPatrick 2020-11-10 19:55:18
I do love cubics.
DPatrick 2020-11-10 19:55:34
And I hope the gremlins are gone for good this time.
DPatrick 2020-11-10 19:55:55
But...I don't love cubics enough to have memorized that formula for solving them.
DPatrick 2020-11-10 19:56:15
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!)
mathworm04 2020-11-10 19:56:24
can't there be up to 3 solutions?
DPatrick 2020-11-10 19:56:29
Indeed, good observation!
DPatrick 2020-11-10 19:56:38
That formula only give us one solution...
DPatrick 2020-11-10 19:56:40
...or so it seems.
DPatrick 2020-11-10 19:57:03
But there are three cube roots if you consider complex numbers!
TheGeometer 2020-11-10 19:57:11
Use polynomial division and the quadratic formula to find the other two.
Heavytoothpaste 2020-11-10 19:57:11
we can then use quadratic formula
DPatrick 2020-11-10 19:57:31
Or you can do that: once you have one root, you can divide it down to a quadratic.
DPatrick 2020-11-10 19:57:45
You'll learn about all this stuff in algebra classes down the line if you don't know it already.
DPatrick 2020-11-10 19:58:00
This formula, and other similar formulas for solving cubics, were discovered in the 1500s. Ramanujan would have learned this in his early teens.
amazingxin777 2020-11-10 19:58:20
is this mentioned because he worked with cubics?
DPatrick 2020-11-10 19:58:29
Yes---and he took it one step further!
rjeagle2019 2020-11-10 19:58:37
And he found the quartic?
Heavytoothpaste 2020-11-10 19:58:41
quartics
DPatrick 2020-11-10 19:58:43
What Ramanujan then undertook, and eventually successfully solved, was solving the quartic equation $ax^4 + bx^3 + cx^2 + dx + e = 0$.
DPatrick 2020-11-10 19:59:07
You saw how complicated the cubic formula was...imagine how complicated the quartic formula is!
DPatrick 2020-11-10 19:59:23
I'm not even going to write the formula down -- it would take way way too much space.
samrocksnature 2020-11-10 19:59:35
Can we see it?
yiz76 2020-11-10 19:59:35
$\boxed{\text{show me it}}$
os31415 2020-11-10 19:59:35
I wish I could see it
DPatrick 2020-11-10 19:59:39
Oh, OK...
DPatrick 2020-11-10 19:59:44
http://pjk.scripts.mit.edu/pkj/wp-content/uploads/2018/11/quartic_formula.jpg
amazingxin777 2020-11-10 20:00:02
yikes
A1K2C3 2020-11-10 20:00:02
aaaah
mathworm04 2020-11-10 20:00:02
!!
amazingxin777 2020-11-10 20:00:02
that is horrifying
Allen31415 2020-11-10 20:00:06
a wolfram alpha screenshot?
DPatrick 2020-11-10 20:00:15
It's from http://pjk.scripts.mit.edu/pkj/wp-content/uploads/2018/11/quartic_formula.jpg
Countcountcount 2020-11-10 20:00:28
that is soooo long
sturdycrab 2020-11-10 20:00:28
He derived that himself??
DPatrick 2020-11-10 20:00:49
Yes. It might now have been this equal formula, but it was something equivalent.
DPatrick 2020-11-10 20:00:56
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.
DPatrick 2020-11-10 20:01:11
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.
sanickode 2020-11-10 20:01:27
w==but you cant solve a plynomial to the 5thdegree right?
DPatrick 2020-11-10 20:01:37
Good question! Ramanujan attempted to solve the quintic equation:
$$ ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0 $$
DPatrick 2020-11-10 20:01:50
He did not have success. Does anyone know why?
rjeagle2019 2020-11-10 20:01:57
Yes Quintics are impossible
boing123 2020-11-10 20:02:10
IT'S IMPOSSIBLE
NCEE 2020-11-10 20:02:10
impossible?
RaymondOuyang 2020-11-10 20:02:10
Is it not possible?
Pandayue 2020-11-10 20:02:10
Cuz it's impossible?
ezRobots 2020-11-10 20:02:10
Quintics are impossible?
TheGeometer 2020-11-10 20:02:18
Abel-Ruffini I believe...
Afo 2020-11-10 20:02:18
Abel–Ruffini theorem
DPatrick 2020-11-10 20:02:25
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.)
UntamedMathBeast 2020-11-10 20:02:48
Wow
URcurious2 2020-11-10 20:02:48
Cool
Quacker88 2020-11-10 20:02:48
does that mean degree 5 and up are all impossible?
DPatrick 2020-11-10 20:03:00
That's right. Degree 5 and up are all impossible.
DPatrick 2020-11-10 20:03:07
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.
A1K2C3 2020-11-10 20:03:22
Ooh can we see the proof?
DPatrick 2020-11-10 20:03:44
I used to teach this proof! But it came at the year of a year-long abstract algebra class taught at university.
DPatrick 2020-11-10 20:03:48
So maybe we'll skip it today.
DPatrick 2020-11-10 20:04:10
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.
rama1728 2020-11-10 20:04:28
Also, sir what is ramanujan's biggest achievement
DPatrick 2020-11-10 20:04:39
There are so many to choose from! But the next one I'll talk about might be the most famous.
DPatrick 2020-11-10 20:04:52
One of the most famous stories regarding Ramanujan occurred when Hardy was visiting Ramanujan in the hospital during one of his many illnesses.
URcurious2 2020-11-10 20:05:03
1729?
BobDBuilder321 2020-11-10 20:05:03
1729!!!
hwdaniel 2020-11-10 20:05:03
Because of Ramanujan, 1729 is now my favourite number
rama1728 2020-11-10 20:05:09
oh you mean 1729?
audio-on 2020-11-10 20:05:09
$1729$
DPatrick 2020-11-10 20:05:15
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.
DPatrick 2020-11-10 20:05:17
Do you know why?
BobDBuilder321 2020-11-10 20:05:46
Sum of 2 cubes, 2 ways
aopsuser305 2020-11-10 20:05:46
smallest number that can be represented as the sum of 2 cubes in 2 ways
amazingxin777 2020-11-10 20:05:46
it is the least one that is the sum of two cubes in two different ways
audio-on 2020-11-10 20:05:46
$1729$ is the smallest positive integer than is the sum of $2$ positive pairs of $2$ positive cubes.
DPatrick 2020-11-10 20:05:57
It is the smallest number that can be written as a sum of two positive cubes in two different ways! Can you find them?
KenOno 2020-11-10 20:06:04
Nice...
URcurious2 2020-11-10 20:06:34
12^3 + 1^3 = 10^3 + 9^3 = 1729
LikMCAMC 2020-11-10 20:06:34
12^3+1=10^3+9^3
rama1728 2020-11-10 20:06:34
12^3+1^3=10^3+9^3
Aaryabhatta1 2020-11-10 20:06:34
${10}^3 + 9^3, 1^3 + {12}^3$
aopsuser305 2020-11-10 20:06:34
$12^3+1^3$ and $10^3+9^3$
os31415 2020-11-10 20:06:34
$1^3+12^3=9^3+10^3=1729$
audio-on 2020-11-10 20:06:34
$1^3+12^3$ and $9^3+10^3$
tennisplayer2007 2020-11-10 20:06:34
1^3 + 12^3 and 9^3 + 10^3?
BobDBuilder321 2020-11-10 20:06:34
$12^{3}+1^{3}$ and $10^{3}+9^{3}$
DPatrick 2020-11-10 20:06:41
$1729 = 1^3 + 12^3 = 1 + 1728$





$1729 = 9^3 + 10^3 = 729 + 1000$
DPatrick 2020-11-10 20:06:53
For this reason, $1729$ is called the 2nd taxicab number. It's also sometimes called the Hardy-Ramanujan number.
Bluedolphingogo 2020-11-10 20:07:04
is there a proof?
dolphin7 2020-11-10 20:07:20
bashing is the proof
DPatrick 2020-11-10 20:07:23
Right. The proof is essentially: you check that all of 1 through 1728 don't work.
pandapo 2020-11-10 20:07:31
why is it the second and not the first?
Champ1 2020-11-10 20:07:31
what is the first taxicab number?
DPatrick 2020-11-10 20:07:40
Great question with a slightly silly answer.
DPatrick 2020-11-10 20:07:53
The first is the smallest number that can be written as a sum of two positive cubes in at least one way.
DPatrick 2020-11-10 20:08:09
Which is...?
URcurious2 2020-11-10 20:08:26
2
NCEE 2020-11-10 20:08:26
wow so it is 2
pandapo 2020-11-10 20:08:26
2?
NCEE 2020-11-10 20:08:26
2
aopsuser305 2020-11-10 20:08:26
2
dolphin7 2020-11-10 20:08:26
2
asengo 2020-11-10 20:08:26
2
NCEE 2020-11-10 20:08:26
2?
BobDBuilder321 2020-11-10 20:08:26
2
aopsuser305 2020-11-10 20:08:26
$2$
URcurious2 2020-11-10 20:08:26
2?
KingJerryX 2020-11-10 20:08:26
2
DPatrick 2020-11-10 20:08:41
Right. The first taxicab number is $2 = 1^3 + 1^3$.
DPatrick 2020-11-10 20:08:56
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?)
atl3236 2020-11-10 20:09:22
8?
truffle 2020-11-10 20:09:22
8
DPatrick 2020-11-10 20:09:28
It's the 8-digit number $87{,}539{,}319$, which is $167^3 + 436^3 = 228^3 + 423^3 = 255^3 + 414^3$.
DPatrick 2020-11-10 20:09:39
It's likely that this was first discovered as recently as 1957.
DPatrick 2020-11-10 20:09:50
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.
tokumi 2020-11-10 20:10:04
what's the largest known taxicab number?
Countcountcount 2020-11-10 20:10:04
Have we discovered the 4th taxicab number yet?
DPatrick 2020-11-10 20:10:10
Only the first 6 taxicab numbers are known:


$2$


$1729$


$87539319$


$6963472309248$


$48988659276962496$


$24153319581254312065344$
DPatrick 2020-11-10 20:10:20
This is all pretty new! The last of these was discovered by computer search in 2008.
Arr0w 2020-11-10 20:10:28
What is the practicality of the the taxicab number though? Does it translate to an applied purpose?
DPatrick 2020-11-10 20:10:36
I am so glad you asked!
DPatrick 2020-11-10 20:10:46
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.
Countcountcount 2020-11-10 20:11:19
wow!
KenOno 2020-11-10 20:11:20
Thanks David...And this appears prominently in black hole physics.
sanickode 2020-11-10 20:11:29
I searched it up and they found the 7th taxicab number a month ago at MIT
DPatrick 2020-11-10 20:11:47
Really??? I didn't know that! I'll have to look it up afterwards.
yofro 2020-11-10 20:11:57
Is there a proof that there are infinitely many taxicab numbers?
DPatrick 2020-11-10 20:12:15
Indeed there is.
KenOno 2020-11-10 20:12:21
The buzz words are: K3 surfaces, 3d quantum gravity...
DPatrick 2020-11-10 20:12:28
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.
DPatrick 2020-11-10 20:12:54
One more fun note about taxicab numbers: How many people here are fans of the TV show Futurama?
DPatrick 2020-11-10 20:13:26
It hasn't been on in a while, but it's still a favorite.
DPatrick 2020-11-10 20:13:39
Apparently the creators of Futurama are big fans of taxicab numbers.
DPatrick 2020-11-10 20:13:50
In particular, the number $1729$ is hidden in several episodes. It's the registration number of the Planet Express ship, for one thing.
DPatrick 2020-11-10 20:13:54
http://mathworld.wolfram.com/images/gifs/1729Futurama.jpg
DPatrick 2020-11-10 20:14:01
Image from mathworld.wolfram.com
amazingxin777 2020-11-10 20:14:14
i never realized this
MinervaPi 2020-11-10 20:14:14
Woah!
DPatrick 2020-11-10 20:14:17
And in one scene someone hails a taxi. Can you make out the number on the cab's roof?
DPatrick 2020-11-10 20:14:21
https://i.kinja-img.com/gawker-media/image/upload/s--ya_coZqV--/c_fit,f_auto,fl_progressive,q_80,w_636/193a4a0f4oqtopng.png
DPatrick 2020-11-10 20:14:27
Image from Gawker Media
fasterthanlight 2020-11-10 20:14:33
87539319
A1K2C3 2020-11-10 20:14:40
87539319
happyhari 2020-11-10 20:14:40
the 3rd taxicab number
amazingxin777 2020-11-10 20:14:40
87539319
DPatrick 2020-11-10 20:14:44
It's hard to make out, but it's the 3rd taxicab number $87539319$.
DPatrick 2020-11-10 20:14:54
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!
DPatrick 2020-11-10 20:15:17
One more bit of math tonight before we talk about the Spirit of Ramanujan program.
DPatrick 2020-11-10 20:15:37
We showed you the Spirit of Ramanujan Project logo at the start of the Math Jam:
DPatrick 2020-11-10 20:15:42
https://spiritoframanujan.com/wp-content/uploads/2018/09/Ramanujan_LOGO.png
DPatrick 2020-11-10 20:15:55
Does anybody know the significance of this logo?
Bluedolphingogo 2020-11-10 20:16:14
ahh the nested radical looking thingy
Allen31415 2020-11-10 20:16:14
Has lots of square roots
swan11 2020-11-10 20:16:14
sqrt rt?
KingJerryX 2020-11-10 20:16:14
square roots?
williamxiao 2020-11-10 20:16:14
idk infinite square root
DPatrick 2020-11-10 20:16:18
That's right!
DPatrick 2020-11-10 20:16:28
The logo was inspired by the following famous problem that was posed by Ramanujan:
DPatrick 2020-11-10 20:16:33
Evaluate the following expression:
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
DPatrick 2020-11-10 20:16:46
This is an example of a nested radical.
samrocksnature 2020-11-10 20:16:56
OH THAT ONE
Arr0w 2020-11-10 20:16:56
Oh I've seen that before!
Skyhigh1218 2020-11-10 20:16:56
Ohh
DPatrick 2020-11-10 20:17:13
Yes, it's a pretty famous item in Ramanujan's history.
DPatrick 2020-11-10 20:17:21
He submitted this problem to the Journal of the Indian Mathematical Society as a challenge for others to solve.
DPatrick 2020-11-10 20:17:31
After 6 months passed and no one had solved it(!), he provided the solution himself.
DPatrick 2020-11-10 20:17:43
Any idea how we might solve it?
DPatrick 2020-11-10 20:17:52
(If you've seen this before and know how it works, please don't spoil it for others!)
KingJerryX 2020-11-10 20:18:12
set one part as a variable
Peter0527 2020-11-10 20:18:16
rewrite the things within the radicals
DPatrick 2020-11-10 20:18:47
Good ideas...let's try a simpler problem first to see how these ideas work.
gladIasked 2020-11-10 20:18:56
A similar problem is in the intro to algebra textbook
DPatrick 2020-11-10 20:19:05
Absolutely! I think it may be this one in fact:
DPatrick 2020-11-10 20:19:11
Suppose we want to evaluate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}$.
Bluedolphingogo 2020-11-10 20:19:31
let x = sqrt(2 + sqrt2..)
fasterthanlight 2020-11-10 20:19:31
Set that value to a variable, $x$.
DPatrick 2020-11-10 20:19:38
Good idea. We can set it to be equal to a variable $x$, so that


$$x = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}.$$
yesufsa 2020-11-10 20:19:53
Square both sides of the equation
mathwiz03 2020-11-10 20:19:53
square it
suvamkonar 2020-11-10 20:19:53
Square it
DPatrick 2020-11-10 20:19:59
Sounds like a good thing to try.
DPatrick 2020-11-10 20:20:03
We square both sides:


$$x^2 = 2 + \sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}.$$
DPatrick 2020-11-10 20:20:13
But now what do you notice?
Bluedolphingogo 2020-11-10 20:20:35
x^2 = 2 + x
mathwiz03 2020-11-10 20:20:35
let the other square root parts be x
mathchampion1 2020-11-10 20:20:35
We have $x$ again
PhysKid11 2020-11-10 20:20:35
get x again
fasterthanlight 2020-11-10 20:20:35
$x^2=2+x$
buckystark 2020-11-10 20:20:35
This means that $x^2 = 2+x$
boing123 2020-11-10 20:20:35
hey, the thing on the right looks like $x$
gladIasked 2020-11-10 20:20:35
There's the same thing inside
dolphin7 2020-11-10 20:20:35
right side is 2+x
LikMCAMC 2020-11-10 20:20:35
it is just x^2=2+x
URcurious2 2020-11-10 20:20:35
x^2 = 2 + x
happyhari 2020-11-10 20:20:35
the secopnd part is the same as the original
DPatrick 2020-11-10 20:20:43
Aha! That big square root on the right side is just another copy of $x$!
DPatrick 2020-11-10 20:20:52
So our equation becomes just $x^2 = 2 + x$.
audio-on 2020-11-10 20:21:07
Quadratic!
fasterthanlight 2020-11-10 20:21:07
Turn it into a quadratic: $x^2-x-2=0$
Bob.The.Builder 2020-11-10 20:21:07
quadratic!
happyhari 2020-11-10 20:21:16
it'S A QUADRATIC!
gladIasked 2020-11-10 20:21:16
Then we have a quadratic
DPatrick 2020-11-10 20:21:20
This becomes $x^2 - x + 2 = 0$, which factors as $(x-2)(x+1) = 0$.
tennisplayer2007 2020-11-10 20:21:27
Then solve the quadratic and use the positive root
LikMCAMC 2020-11-10 20:21:32
x^2-x-2=(x-2)(x+1) x has to be positive so x=2!!
boing123 2020-11-10 20:21:36
and x is positive
Bluedolphingogo 2020-11-10 20:21:36
its equal to 2
URcurious2 2020-11-10 20:21:36
2
KingJerryX 2020-11-10 20:21:36
answer is 2
DPatrick 2020-11-10 20:21:45
Right: either $x=2$ or $x=-1$. And since square roots are positive, we must have $x=2$.
DPatrick 2020-11-10 20:22:03
So we conclude that


$$2 = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}.$$
DPatrick 2020-11-10 20:22:20
So now back to Ramanujan's problem:


$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
DPatrick 2020-11-10 20:22:41
Ramanujan did the same sort of clever substitution, but he had to be a little more clever.
tokumi 2020-11-10 20:22:46
we don't have complete repetition here though, it changes a little bit each time
DPatrick 2020-11-10 20:22:54
Right. His nested radical doesn't exactly repeat, but there's definitely a pattern.
DPatrick 2020-11-10 20:23:16
So he couldn't just use a single variable to represent a repeating part.
DPatrick 2020-11-10 20:23:40
Instead he used a function to capture the pattern.
DPatrick 2020-11-10 20:23:46
Specifically, he let $$f(x) = \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}.$$
DARTHTATOR0112 2020-11-10 20:24:09
Wow that was good thinking
DPatrick 2020-11-10 20:24:12
Notice that if we plug in $x=2$ into this function, we get the value that we want! So our goal is to determine $f(2)$.
DPatrick 2020-11-10 20:24:19
And so now what should we do?
DPatrick 2020-11-10 20:24:48
What did we do in the easier problem at this stage?
tokumi 2020-11-10 20:25:07
squared
Zhaom 2020-11-10 20:25:07
square
happyhari 2020-11-10 20:25:07
square both sides
Countcountcount 2020-11-10 20:25:07
square it
connorlonnorl 2020-11-10 20:25:07
square it
mathcounting 2020-11-10 20:25:07
Square it?
rjeagle2019 2020-11-10 20:25:07
square it?
Arr0w 2020-11-10 20:25:07
We squared it I think...
DPatrick 2020-11-10 20:25:13
Yeah! Let's square it!
DPatrick 2020-11-10 20:25:19
When we square this, we get $$(f(x))^2 = 1 + x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}.$$
DPatrick 2020-11-10 20:25:27
And now, what is the big square root term?
happyhari 2020-11-10 20:25:49
f(x+1)
boing123 2020-11-10 20:25:49
$f(x+1)$?
fasterthanlight 2020-11-10 20:25:49
f(x+1)
Hamroldt 2020-11-10 20:25:49
f(x plus 1)
aopsuser305 2020-11-10 20:25:49
f(x+1)
Concerto-in-A-minor 2020-11-10 20:25:49
f(x+1)
First_Mathematician 2020-11-10 20:25:49
f(x+1)
fasterthanlight 2020-11-10 20:25:49
$f(x+1)$
DPatrick 2020-11-10 20:26:10
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.
DPatrick 2020-11-10 20:26:17
And now we have the functional equation $$(f(x))^2 = 1 + xf(x+1).$$
DPatrick 2020-11-10 20:26:29
And then because Ramanujan was very good with algebra, he was essentially able to "guess and check" what $f(x)$ must be. Can you?
hwdaniel 2020-11-10 20:26:52
x+1
DPatrick 2020-11-10 20:27:00
Right! It's $f(x) = x+1$. Let's check.
DPatrick 2020-11-10 20:27:09
If we use $f(x) = x+1$, then our functional equation become $$(x+1)^2 = 1 + x((x+1)+1).$$
DPatrick 2020-11-10 20:27:16
Does this work?
yofro 2020-11-10 20:27:30
yes
tonyme.2 2020-11-10 20:27:30
Yes?
Countcountcount 2020-11-10 20:27:30
YES!
yesufsa 2020-11-10 20:27:30
Yes
CrazyVideoGamez 2020-11-10 20:27:30
yes
jxc516 2020-11-10 20:27:30
yes
Peter0527 2020-11-10 20:27:30
yes
Allen31415 2020-11-10 20:27:30
Yes
DPatrick 2020-11-10 20:27:35
It does! Both sides are $x^2 + 2x + 1$. So it worked!
DPatrick 2020-11-10 20:27:41
And how does this answer the original problem posed by Ramanujan?
Zhaom 2020-11-10 20:28:18
f(2)=3
Bluedolphingogo 2020-11-10 20:28:18
plug in x = 2
KingJerryX 2020-11-10 20:28:18
so f(2) = 3
AOPqghj 2020-11-10 20:28:18
x=2
tokumi 2020-11-10 20:28:18
since x=2, f(x)=3!
PhysKid11 2020-11-10 20:28:18
THE ANSWER IS 3!!! WOW THIS IS INGENIOUS!!!
buckystark 2020-11-10 20:28:18
$f(2) = 2+1 = 3$
Aaryabhatta1 2020-11-10 20:28:18
$f(2) = 2+1 = 3$
AOPqghj 2020-11-10 20:28:18
plug in for x = 2
Wesey 2020-11-10 20:28:18
x+1 = f(2), so the answer is 3
DPatrick 2020-11-10 20:28:30
Our original quantity is $f(2)$, so if $f(x) = x+1$, then $f(2) = 2+1 = \boxed{3}$.
DPatrick 2020-11-10 20:28:44
Ramanujan was pretty clever, eh?
asbodke 2020-11-10 20:28:47
how do you show it is the only function
boing123 2020-11-10 20:28:47
What if there are other functions?
TheGeometer 2020-11-10 20:28:47
How do you know that $x+1$ is the only solution to the functional equation?
DPatrick 2020-11-10 20:29:05
Well...that's a really good point. There are certainly some icky details that I'm leaving out.
DPatrick 2020-11-10 20:29:19
What Ramanujan did is something a little more general.
DPatrick 2020-11-10 20:29:25
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.
DPatrick 2020-11-10 20:29:43
The original problem is the special case where $a=0$, $n=1$, and $x=2$.
DPatrick 2020-11-10 20:29:57
It's believed that Ramanujan proved this formula when he was about 16 years old!
fasterthanlight 2020-11-10 20:30:23
Wow!
S304564 2020-11-10 20:30:23
cool
Countcountcount 2020-11-10 20:30:23
cool!
tonyme.2 2020-11-10 20:30:23
WOW! HOW?
Awesomeness_in_a_bun 2020-11-10 20:30:23
Wow thats really young
DPatrick 2020-11-10 20:30:56
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.
DPatrick 2020-11-10 20:31:08
Let me start by citing an excerpt from an article called "Why Ramanujan Matters" written by Ken Ono and Robert Schneider:
DPatrick 2020-11-10 20:31:17
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.
PrinceRoyale 2020-11-10 20:31:49
its true
URcurious2 2020-11-10 20:31:49
Nice speech
DPatrick 2020-11-10 20:31:53
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.
DPatrick 2020-11-10 20:32:17
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.
KenOno 2020-11-10 20:32:38
My pleasure.
DPatrick 2020-11-10 20:32:46
Specifically, SoR is a grant program that provides two types of awards.
DPatrick 2020-11-10 20:32:57
The Templeton-Ramanujan Development Awards will provide books and monetary grants to offset the costs of online resources.
DPatrick 2020-11-10 20:33:09
The Templeton-Ramanujan Fellowships will provide monetary grants to offset the costs of enrolling in approved enrichment programs, such as summer camps.
KenOno 2020-11-10 20:33:18
This can include AoPS books and online programs.
DPatrick 2020-11-10 20:33:23
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.
DPatrick 2020-11-10 20:33:34
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.
DPatrick 2020-11-10 20:33:50
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.
Butterfly0707 2020-11-10 20:33:54
how can one qualify for the awards?
DPatrick 2020-11-10 20:34:01
To learn more about SoR, please visit the website spiritoframanujan.com.
DPatrick 2020-11-10 20:34:11
The SoR website includes a link to the application page. You can also read brief bios of the previous winners on the website.
DPatrick 2020-11-10 20:34:22
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.
KenOno 2020-11-10 20:35:10
We have made several awards to the AoPs community, and we hope to do so again in 2021.
DPatrick 2020-11-10 20:35:15
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.
KenOno 2020-11-10 20:35:54
Great show David...
KenOno 2020-11-10 20:36:06
Lots of very talented folks online.
DPatrick 2020-11-10 20:36:11
Thanks Ken! And very special thanks to Deven for jumping in while the gremlins attacked.

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