首页 > 其他分享 >The Steele Prize

The Steele Prize

时间:2024-05-12 17:08:02浏览次数:15  
标签:Mathematical Prize Steele Hill mathematics prize

Onday I found an interesting prize established by AMS, called The Steele Prize, see ams.org/prizes-awards.

''From 1970 to 1976 one or more prizes were awarded each year for outstanding published mathematical research; most favorable consideration was given to papers distinguished for their exposition and covering broad areas of mathematics. ''

Just take a look at the​ prize list, for example,

Lawrence C. Evans, a professor emeritus at the University of California, Berkeley, has received the 2023 Leroy P. Steele Prize for Mathematical Exposition for his book Partial Differential Equations published by the American Mathematical Society, Providence, RI, 1998 (first edition) and 2010 (second edition).

and also

Aise Johan de Jong, a professor of mathematics at Columbia University, will receive the 2022 AMS Leroy P. Steele Prize for Mathematical Exposition. De Jong is the originator and maintainer of The Stacks Project, which is at once a comprehensive online textbook on the foundations of contemporary algebraic geometry and a remarkable instance of a carefully structured large-scale collaborative writing process.

To Walter Rudin for his books Principles of Mathematical Analysis, McGraw-Hill (1953, 1964, and 1976); and Real and Complex Analysis, McGraw-Hill (1966, 1974, and1976).

I guess these names will be familiar to you if you have studied some college mathematics. Therefore, I believe the books on the prize list must be of high quality. From my viewpoint, Rudin's textbooks are impressive and phenomenal. Also, Evans' book is very helpful and beautiful.

标签:Mathematical,Prize,Steele,Hill,mathematics,prize
From: https://www.cnblogs.com/crossLH/p/18187963

相关文章

  • CodeTON Round 8 (Div. 1 + Div. 2, Rated, Prizes!) 补题记录(A~A)
    A猜测结论。发现当且仅当\(k=1\)或者\(n=k\)时有解,否则无解。对于\(k=1\)时构造序列\(1,2,3,\ldots,n\)满足条件。对于\(k=n\)时构造序列\(1,1,1,\ldots,1\)满足条件。时间复杂度为\(O(n)\)。#include<bits/stdc++.h>#defineintlonglongusingnamespaces......
  • [CF1942] CodeTON Round 8 (Div. 1 + Div. 2, Rated, Prizes! A~E 题解
    [CF1942]CodeTONRound8(Div.1+Div.2,Rated,Prizes!A~E题解A.FarmerJohn'sChallenge只有两种情况,一种是单调递增,这时\(k=1\),另一种是所有数都相同,这时\(k=n\)。B.BessieandMEX首位可以确定,然后从前往后增量构造\(p\)即可。voidwork(){cin>>......
  • CodeTON Round 8 (Div. 1 + Div. 2, Rated, Prizes!)
    目录写在前面ABC1C2DE写在最后写在前面比赛地址:https://codeforces.com/contest/1942。过了这么长时间才来补太唐了、、、赛时写D写了一坨呃呃,用刷表法总是不可避免地要多枚举一个\(O(n)\)比较+转移妈的,赛后一看填表法+堆就不用枚举了笑烂了A签到。完全相等的数列随便......
  • CodeTON Round 8 (Div. 1 + Div. 2, Rated, Prizes!)做题笔记
    A.FarmerJohn'sChallengeProblem-A-Codeforces题意:构造出满足条件的数组a,否则输出-1做法:判断k和n或者1的关系;k==1则输出1就行,k==n就从1输出到n;都不满足就输出-1;代码:#include<iostream>usingnamespacestd;intmain(){intt;cin>>t;while(t--){intn,k;cin......
  • CodeTON Round 8 (Div. 1 + Div. 2, Rated, Prizes!) D
    链接开始的时候看错题了。以为区间是可以我划分的,后面才发现是连着的区域是被强制合并的。导致我第一个写了给k短路。紫砂了。然后我的第二个思路是,从后往前和从前往后做两边dp,然后尝试枚举断点,看看有没有比最优稍微劣一点的解法。然后样例就是反例。正解是想到过的,但是因为......
  • 14 CodeTON Round 5 (Div. 1 + Div. 2, Rated, Prizes!)C. Tenzing and Balls(dp+前缀
    思路:dp还是挺明显的,思路可以参考最长上升子序列有点dp的感觉\(f[i]\)表示考虑前\(i\)个数,的最大值当前数有两种删或不删不删:\(f[i]=f[i-1]\);删:\(f[i]=max{f[j-1]+i-j+1}\)这个转移是\(O(n^2)\)的显然时间上来不及考虑优化,第一层循环一定是省不了的考虑优化掉第二层循环......
  • CodeTON Round 7 (Div. 1 + Div. 2, Rated, Prizes!)
    CodeTONRound7(Div.1+Div.2,Rated,Prizes!)比赛链接A.JaggedSwaps思路:考虑到题目要求,给定的排列第一位必须是1才会构造出可行性序列,如果不是就是没有办法Code#include<bits/stdc++.h>usingnamespacestd;#defineintlonglongvoidsolve(){ intn; cin......
  • CodeTON Round 7 (Div. 1 + Div. 2, Rated, Prizes!)
    Preface补题,经典不会F,看了会题解发现看不懂,索性直接开摆A.JaggedSwaps判断\(a_1\)是否为\(1\)即可#include<cstdio>#include<iostream>#include<utility>#include<vector>#include<cstring>#include<cmath>#include<cstdlib>#include<al......
  • CodeTON Round 7 (Div. 1 + Div. 2, Rated, Prizes!)
    看到B官方题解写了一堆,而如果能注意到一些性质,几行就写完了题意:给一个A,B构成的字符串,可以将“AB”翻转成"BA",问最多可以进行多少次翻转?实际上在手动模拟以后发现,由于题目限制了每个位置只能翻转一次,所以情况简单了不少。只要还没过最后一个B,那么最后一个B之前的所有A就会被反......
  • CodeTON Round 7 (Div. 1 + Div. 2, Rated, Prizes!)
    CodeTONRound7(Div.1+Div.2,Rated,Prizes!)A-JaggedSwaps思路:a2到an的数只要相邻为逆序都可以交换,只需要判断a1是否为1即可#include<bits/stdc++.h>usingnamespacestd;#defineintlonglong//#defineint__int128#definedoublelongdoubletypedefpai......