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Cuộc đời và sự nghiệp của Paul Erdos :)

Chủ đề trong 'Toán học' bởi username, 18/03/2003.

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  1. username

    username Thành viên rất tích cực

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    Cuộc đời và sự nghiệp của Paul Erdos :)

    Chắc các bác ai cũng từng nghe nói về nhà toán học "truyền thuyết" này. Tôi định viết/dịch một bài về cuộc đời bác này, các bác chờ đọc nha .
  2. nguoiiuaodai

    nguoiiuaodai Thành viên mới

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    bác username dịch lẹ lên cho em đọc đi ha, với lại bác giúp em về vấn đề ma phương dùm cái đi
  3. username

    username Thành viên rất tích cực

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    Chả phải là loại đánh trống bỏ dùi nhưng tôi không có máy nên không có điều kiện để lên mạng bi bô nhiều. Tôi định dịch bài "Paul Erdos - Life and Work" của Béla Bollobás trong tuyển tập The Mathematics of Paul Erdos nhưng hơi nhác, các bác đọc tạm bài này đã :
    Paul Erdös dies at 83
    Béla Bollobás
    Paul Erdös, mathematician, died on September 20, 1996, in Warsaw, aged 83. He was born on March 26, 1913, in Budapest.
    Paul Erdös was one of the most brilliant and probably the most remarkable of mathematicians of this century. Through his prodigious output and many collaborators, he greatly influenced many branches of mathematics and was the prime mover in the rapid growth of combinatorics. He never had a `proper' teaching job, but constantly travelled around the world, in search of new challenges. Considering material possessions a nuisance, he lived for over sixty years out of half-full suitcases, which he never learned to pack. His discarded suit was rejected by Oxfam. He was the quintessential mathematician: although he was interested in history, medicine and politics, he was totally dedicated to mathematics. He wrote about 1500 papers, about five times as many as other prolific mathematicians, and had close to 500 collaborators. His enormous output even inspired a limerick:
    A conjecture both deep and profound
    Is whether the circle is round.
    In a paper of Erdös,
    Written in Kurdish,
    A counterexample is found.
    According to a wit, on a long train journey he would write a joint paper with the conductor.
    Erdös was born into an intellectual Hungarian-Jewish family in Budapest amidst tragic circumstances: when his mother returned home from the hospital she found that her two daughters had died of scarlet fever. Soon after the outbreak of the First World War, Erdös's father was taken prisoner by the Russians and returned home from Siberia only six years later. The young Erdös was brought up by his mother, a teacher of mathematics like his father, and he remained devoted to her throughout his life.
    He was a child prodigy: as a small boy, he amused people by asking them how old they were and telling them how many seconds they had lived. Erdös was educated mostly at home, by his father, until 1930, when he entered the Péter Pázmány University in Budapest, where he was soon at the centre of a small group of outstanding young Jewish mathematicians. As a second year undergraduate, he practically completed his doctorate under Leopold Fejér. His main result was a simple proof of an extension of Bertrand's Postulate, first proved by the Russian mathematician Chebyshev, that there is always at least one prime number between any positive integer and its double. For Erdös, 1934 was a momentous year: not only did he graduate from the university, but he also received his doctorate, and got a fellowship to join the remarkable group of mathematicians that was brought together by Louis Mordell in Manchester. He also met Richard Rado and Harold Davenport, who became his great friends and collaborators.
    In 1938 Erdös sailed for the United States, where he was to stay for the next decade. During his first year, at the Institute for Advanced Study in Princeton, he wrote ground-breaking papers with Wintner and Kac, which founded probabilistic number theory, with Turán he proved great results in approximation theory, and he solved the then outstanding problem in dimension theory. When his Fellowship at the Institute was not renewed, he started his peregrinations, with longer stays at the University of Pennsylvania, Notre Dame, Purdue and Stanford. The great mathematical event of 1949 was an elementary proof of the Prime Number Theorem, given by Atle Selberg and Erdös. The result, which predicts the distribution of primes with some accuracy, was first proved in 1896 by sophisticated methods, and it had been thought that no elementary proof could be given.
    In 1954 he fell foul of the McCarthy era: despite being refused a reentry visa, he left the United States and, as a result, for the next nine years he was not allowed to return to America. Israel came to his aid with a job for three months at the Hebrew University of Jerusalem. Although officially he became a resident of Israel, he refused its citizenship and kept his Hungarian passport, claiming that he was a citizen of the world. Although in 1963 he was allowed to return to America, and from then on spent most of his time there, he could never forgive the American government. From 1964 on, his mother, who was then aged 84, accompanied him on his travels. This was a golden period for Erdös, who never recovered from her death in 1971.
    In over six decades of furious activity, he wrote fundamental papers on number theory, real analysis, geometry, probability theory, complex analysis, approximation theory, set theory and combinatorics, among other areas. His first great love was number theory, while in his later years he worked mostly in combinatorics. In 1966, with Selfridge, he solved a notorious problem in number theory that had been open for over 100 years, namely that the product of consecutive positive integers (like 4·5·6·7·8) is never an exact square, cube or any higher power. With Rado and Hajnal, he founded partition calculus, a branch of set theory, which is a detailed study of the relative sizes of large infinite sets. Nevertheless, he will be best remembered for his contributions to combinatorics, an area of mathematics fundamental to computer science. He founded extremal graph theory, his theorem with Stone being of prime importance, and with Rényi he started probabilistic graph theory. He advocated the use of elementary methods, in ad***ion to techniques requiring vast preparation, and decades before it became commonly accepted, he had showed the power of random methods in mathematics. He showed that simply stated problems often lead to exciting phenomena, and left behind hundreds of exciting problems whose solutions will influence combinatorics for many years to come.
    ***ual pleasure revolted him; even an accidental touch by anyone made him feel uncomfortable. He never married or had a family, although he was very good with children. As a truly unworldly figure, he lived for mathematics and relied on his friends to look after him; in his later years he particularly liked to be in Budapest, Memphis and Kalamazoo where, in ad***ion to his mathematical friends, he found good medical care. He hated to be alone, and almost never was; he loved to attend conferences and enjoyed the attention of mathematicians. His main aim in life was ``to do mathematics: to prove and conjecture''.
    A favourite saying of his was that ``every human activity, good or bad, must come to an end, except mathematics.'' He died as he wished to, before his powers were greatly diminished: while attending a conference, he was killed by a massive heart attack.
    Source: http://www.ams.org/new-in-math/erdosobit.html
  4. Thanhha

    Thanhha Thành viên quen thuộc

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    Bác Username chịu khó lúc nào dịch cho anh em cái, nhìn tiếng Anh trên mạng ngại đọc lắm . Giúp bác đoạn đầu cái .
    Paul Erdos mất ngày 20 tháng 9 năm 1996 ở Warsaw ở tuổi 83. Ông sinh ngày 26 tháng 3 năm 1913 tại Budapest.
    Paul Erdos là một trong những nhà toán học lỗi lạc nhất và có thể là xuất sắc nhất của thế kỷ 20. Những bài viết thiên tài cùng với rất nhiều người cộng tác của ông đã để lại rất nhiều ảnh hưởng cho nhiều lĩnh vực trong toán học và là động lực thúc đẩy lớn cho sự phát triển của tổ hợp. Ông chưa bao giờ dạy học một cách chính thức nhưng thường xuyên đi vòng quanh thế giới để tìm kiếm những thách thức mới. Trong suốt hơn 60 năm ông sống, sự sở hữu của cải luôn là những thứ tầm thường, ông chỉ có một vài chiếc vali mà ông cũng thậm chí chưa bao giờ học cách sắp xếp. Một trong những chiếc vali đó bị từ chối ngay cả bởi Oxfam (*). Erdos là một nhà toán học tinh tuý, mặc dù ông quan tâm nhiều đến lịch sử, y học và chính trị, ông hoàn toàn cống hiến cuộc đời mình cho toán học. Ông viết khoảng 1500 bài viết, khoảng 5 lần những nhà toán học sung mãn khác và có hơn 500 người cộng tác (**). Những kết quả khổng lồ đó cũng được nhắc đến trong bài thơ:
    Một giả thuyết vừa đẹp vừa sâu
    Là đường tròn có tròn hay không
    Trong một bài viết của Erdos,
    Viết ở Kurdish,
    Một phản ví dụ đã được tìm thấy.
    (*) : Oxfam là một cửa hiệu chuyên bán đồ cũ.
    (**) : Trong bài viết "Paul Erdos, nhà toán học thiên tài được yêu quý nhất đã qua đời" (Paul Erdos, The World's Most Beloved Mathematical Genius "Leaves") của A. Vázsonyi đăng trên Pure Math. Appl. có đoạn viết:
    Sự nổi tiếng và những bài viết của Erdos đã dẫn đến một cách phân loại các nhà toán học khá là tốt. Nhà toán học bậc cao nhất (ngoại trừ bản thân Erdos mang hệ số Erdos 0) có hệ số Erdos 1, nghĩa là người đó đã từng viết chung bài với Erdos. Nếu một người chưa bao giờ cộng tác với Erdos nhưng có viết chung với một người có hệ số Erdos 1 sẽ có hệ số Erdos 2. Cứ như vậy cho 3, 4, 5. Và như thế hầu như những nhà toán học thực thụ đều có hệ số Erdos nhỏ hơn hoặc bằng 5. Hệ số của Einstein là 2, thông qua Straus...

    Strawhero
  5. zuken21

    zuken21 Thành viên mới

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    Có ai biết mấy quyển sách về complex analysis tiếng việt có bán ở đâu không ? Help me.
    À quên complex analysis dịch sang tiếng việt là gì vậy các bác, chắc ko phải là Giải tích số phức chứ
    THanks
    HEHEHE
  6. thuanton

    thuanton Thành viên mới

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    Complex analysis là Giải tích phức ! Nhưng viết thế này thì chung chung quá! Chẳng biết bạn cần kiểu gì! Tạm nói về sách nhập môn nhé! Sách tiếng Việt thì có bộ Nhập môn GTP của B.V.Sabat (2 tập) đọc cũng được! Các bộ do tác giả VN viết thì có lẽ tạm được là quyển GTP của các tác giả Nguyễn Văn Khuê_Lê Mậu Hải! Hơ mà đang nói về Nhà Toán học Paul Erdos cơ mà! Cái này để thày Vũ Đình Hoà nói chắc cũng hay!
    Được thuanton sửa chữa / chuyển vào 04:07 ngày 08/07/2003
  7. heroes

    heroes Thành viên quen thuộc

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    Em xin phép được UP 1 số chứng minh của Erdös.
    Trước tiên là Bertrand''s postulate:
    (mọi người thông cảm, ngại dịch ra tiếng Việt quá)
    Bertrand''s postulate: For every n>=1, there is some prime number p with n<p<=2n
    The proof is taken from Erdös'' first published paper, which appeared in 1932, when Erdös was 19
    Proof. We will estimate the size of the binomial coefficient (2n,n) ( (m,n) = m!/(!(m-n)!)) carefully enough to see that if it didn''t have any prime factors in the range n<p<=2n, then it would be "too small". Our argument is in five steps.
    (1) We first prove Bertrand''s postulate for n<4000. For this one does not need to check 4000 cases: it suffices (this is "Landau''s trick") to check that
    2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001
    is a sequence of prime numbers, where each is smaller than twice the previous one. Hence every interval ( y: n< y <=2n), with n<= 4000, contains one of these 14 primes.
    (2) Next we prove that: product p (p<=x) <= 4^(x-1) for all real x>=2 (1)
    Where our notation - here and in the following - is meant to imply that the product is taken over all prime numbers p <= x. The proof that we present for this fact is not from Erdös'' original paper, but it is also due to Erdös. First we note that if q is the largest prime with q<=x then
    product p (p<=x) = product p (p<=q) and 4^(q-1)<= 4^(x-1)
    Thus it suffices to check (1) for the case where x=q is a prime number. For q=2, we get "2<=4" so we proceed to consider odd primes q=2m+1. For these we split the product and compute
    product p (p<=2m+1) = product p (p<=m+1) . product p (m+1<p<=2m+1) <= 4^m . ( 2m+1, m) <= 4^m. 2^(2m) = 4^2m
    All the pieces of this "one-line computation" are easy to see. In fact,
    product p ( p<=m+1) <= 4^m holds by induction. The inequality
    product p (m+1<p<=2m+1) <= (2m+1,m)
    follows from the observation that (2m+1,m)=(2m+1)!/m!(m+1)! is an integer when the primes that we consider all are factors of the numerator (2m+1)!, but not of the denominator m!(m+1)!. Finally,
    (2m+1,m) <= 2^(2n)
    since (2m+1,m) and (2m+1,m+1) are two equal! summands that appear in sigma (2m+1,k) (k=0..2m+1) = 2^(2m+1)
    (3) From Legendre''s theorem we get that (2n,n) = (2n)!/n!n! contains the prime factor p exactly
    sigma ( [2n/p^k] - 2[n/p^k]) (k>=1)
    times. Here each summand is at most 1, since it satisfies
    [2n/p^k] - 2[n/p^k] < 2n/p^k - 2(n/p^k - 1 ) = 2
    and is an integer. Furthermore, the summands vanish whenever p^k>2n
    Thus (2n,n) contains exactly sigma ( [2n/p^k] - 2[n/p^k]) (k>=1)
    <= max {r: p^r <= 2n} times. Hence the largest power of p that divides (2n,n) is not larger than 2n. In particular, primes p> sqrt(2n) appear at most once in (2n,n)
    Furthermore - and this, according to Erdös, is the key fact for his proof - primes p that satisfy (2/3)n < p <=n do not divide (2n,n) at all! Indeed 3p>2n implies (for n>=3 and hence p>=3) that p and 2p are the only multiples of p that appear as factors in the numerator of (2n)!/n!n! while we get two p-factor in the denominator.
    (4) Now we are ready to estimate (2n,n). For n>=3, using an estimate for the lower bound, we get
    4^n/2n <= (2n,n) <= product 2n (p<=sqrt(2n)) . product p (sqrt(2n)<p<=2/3n). product p (n<p<=2n)
    and thus, since there are not more than sqrt (2n) primes p <= sqrt(2n)
    4^n < = (2n)^(1+sqrt(2n). product p (sqrt(2n)<p<=2/3n). product p (n<p<=2n) for n>=3 (2)
    (5) Assume now that there is no prime p with n < p <=2n, so the seconde product in (2) is 1. Substituting (1) to (2) we get
    4^n <= (2n)^(1+sqrt(2n)).4^(2n/3)
    or 4^(n/3) <= (2n)^(1+sqrt(2n)) (3)
    which is false for n large enough! In fact, using a + 1 < 2^a (which holds for all a>=2, by induction) we get
    2n = ((2n)^(1/6))^6< ([(2n)^(1/6)]+1)^6<2^(6[(2n)^(1/6)]) <= 2^6((2n)^1/6) (4)
    and thus for n>=50 (and hence 18<2sqrt(2n)) we obtain from (3) and (4)
    2^(2n) <= (2n)^3(1+sqrt(2n)) < 2^((2n)^(1/6).(18+18sqrt(2n)) < 2^(20sqrt(2n).((2n)^1/6)) = 2^(20(2n)^(2/3))
    This implies (2n)^(1/3)<20 and thus n<4000
    Source: Proofs from the book, written by Martin Aigner and Gunter M. Ziegler
    Heroes

    Shinichi Kudo

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