Mathematics desk
< November 5	<< Oct \| November \| Dec >>	November 7 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

November 6

Ramanujan, and Nobel Prize.

Saw a video on Srinivasa_Ramanujan last night, then looked at his article Srinivasa_Ramanujan. So he never won a Nobel prize in anything, as his works were not useful in the world of physics. But has any mathematician did math useful to the world of physics, or and won a Nobel prize in physics for it? Without being a physicist? Was Ramunujan considered the Einstein of math? Has there ever been talk on starting a Nobel prize in math? Thanks. 67.165.185.178 (talk) 18:18, 6 November 2022 (UTC).[reply]

See List of prizes known as the Nobel or the highest honors of a field § Applied mathematics and ~ § Mathematics. It is not obvious whether the former or the latter might be more relevant to the physical sciences, and other areas of mathematics such as statistics could also be significant. -- Verbarson ^talk_edits 18:27, 6 November 2022 (UTC)[reply]

Nobel laureate Roger Penrose is many things, but I think of him foremost as being a mathematician. Frank Wilczek is perhaps also about as much a mathematician as a theoretical physicist. --Lambiam 20:51, 6 November 2022 (UTC)[reply]

The Fields Medal is considered (by many at least) the Nobel Prize of mathematics, though the conditions for winning and the frequency of awards are somewhat different. Ramanujan died before the prize began, and even if he were alive he would have been over the 40 year age limit, but one has to wonder if he would have won if the prize had it started a few decades earlier. There's also the Abel prize, but that only started in 2001. See this page from the MAA for more on mathematicians who won the Nobel. --RDBury (talk) 21:57, 6 November 2022 (UTC)[reply]

Allvar Gullstrand got a Nobel prize in physiology for some optical mathematics. Rgdboer (talk) 05:11, 7 November 2022 (UTC)[reply]

John Forbes Nash Jr. and Lloyd Shapley each won the Nobel Memorial Prize in Economic Sciences in different years for their work in game theory, they're both mathematicians. Bertrand Russell was something of a polymath, but was a mathematician first and foremost, he won the 1950 Nobel Prize in Literature for his philosophical writings. John Pople straddles the lines between mathematics and chemistry. He was a computational chemist, and his own Ph.D. was in mathematics, he won the 1998 Nobel Prize in Chemistry. Max Born, similarly, had his Ph.D. in mathematics, and won the Nobel Prize in Physics in 1954, primarily for his work on the math behind quantum mechanics. Clive Granger is another mathematician who worked in economics, earning the Economics Nobel in 2003.--Jayron 32 16:54, 7 November 2022 (UTC)[reply]

Kenneth Arrow is another if you're interested in economics ones. NadVolum (talk) 14:01, 10 November 2022 (UTC)[reply]

Combinatorial identity

For a given n, start with the number of partitions of n with parts 1 and/or 2. For example for n=9 there are 5: 111111111, 21111111, 2211111, 222111, 22221. Add the number of partitions of n-1 with parts 2 and/or 3, in the example this is 2: 2222, 332. Add the number of partitions of n-2 with parts 3 and/or 4, in the example 1: 43, and continue this way until you the reach the partitions of 0. In the example there are only two more, a partition of 5 as 5, and the empty partition of 0. Show that the total reached is n+1. Is there a direct combinatorial proof of this? I think I'm converging on a proof with generating functions, though I haven't filled in the details yet. It amounts to showing:

{\frac {1}{(1-t)(1-t^{2})}}+{\frac {t}{(1-t^{2})(1-t^{3})}}+{\frac {t^{2}}{(1-t^{3})(1-t^{4})}}+{\frac {t^{3}}{(1-t^{4})(1-t^{5})}}+\dots ={\frac {1}{(1-t)^{2}}}.

This is just for fun btw; no one's thesis is riding on this :) --RDBury (talk) 22:34, 6 November 2022 (UTC)[reply]

If you split the fractions using

{\frac {t^{n}}{(1-t^{n+1})(1-t^{n+2})}}={\frac {1}{1-t}}\left({\frac {t^{n}}{1-t^{n+1}}}-{\frac {t^{n+1}}{1-t^{n+2}}}\right)

everything disapprears after the first term and you get your result. NadVolum (talk) 16:25, 7 November 2022 (UTC)[reply]

I haven't beem able to think of a direct combinatorial proof, in fact I'd never have thought of it except by working backwards from the generating function! NadVolum (talk) 17:01, 7 November 2022 (UTC)[reply]

The fraction split was the way I thinking the proof would go. More generally:

{\frac {1}{(1-t^{a})(1-t^{a+b})}}={\frac {1}{1-t^{b}}}\left({\frac {1}{1-t^{a}}}-{\frac {t^{b}}{1-t^{a+b}}}\right).

It's not hard to show combinatorially a more symmetric identity:

{\frac {1}{(1-t^{a})(1-t^{b})}}+{\frac {1}{1-t^{a+b}}}={\frac {1}{(1-t^{a})(1-t^{a+b})}}+{\frac {1}{(1-t^{b})(1-t^{a+b})}}.

From there it's a small step to the previous identity. It'd be a mess to untangle the steps to make a completely combinatorial proof of the original, but I'm satisfied it's there somewhere. As a bonus, the proof would generalize. Something along the lines of:

{\frac {1}{(1-t^{a})(1-t^{b})}}={\frac {1}{(1-t^{a})(1-t^{a+b})}}+{\frac {t^{b}}{(1-t^{a+b})(1-t^{a+2b})}}+{\frac {t^{2b}}{(1-t^{a+2b})(1-t^{a+3b})}}+{\frac {t^{3b}}{(1-t^{a+3b})(1-t^{a+4b})}}+\dots .

--RDBury (talk) 22:43, 7 November 2022 (UTC)[reply]

Or that can be written as

{\frac {1}{(1-x)(1-y)}}={\frac {1}{(1-x)(1-xy)}}+{\frac {y}{(1-xy)(1-xy^{2})}}+{\frac {y^{2}}{(1-xy^{2})(1-xy^{3})}}+{\frac {y^{3}}{(1-xy^{3})(1-xy^{4})}}+\dots .

A bit surprising looking. NadVolum (talk) 01:30, 10 November 2022 (UTC)[reply]

Nice. Combinatorial interpretation: For any pair of natural numbers a, b, exactly one of the following holds, 0≤b≤a, a+1≤b≤2a+1, 2a+2≤b≤3a+2, 3a+3≤b≤4a+3, ... . A combinatorial proof of this is straightforward, basically integer divide b by a+1. --RDBury (talk) 00:33, 12 November 2022 (UTC)[reply]