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Show that the Bernoulli number B₁ satisfies B₁ = (-1)" ( 2 + 2d²+1) Σ (mod 1) for any positive integer n where d runs over all of the divisors of n such that 2d+1 is a prime. For example, 1-B₁ = 1/2+3/ 1+ B₂ 1 1/2 + 1/3 + 1 − B₁ = 1/2 + 1/3 + 1/7² 1 1 1+ B4 = 1- B5 = 2 3 2 3 11 1 + B6 = 2 + 3 + 5 +132-B₁ 7 This is known as the Staudt-Clausen theorem, which is found by Staudt (1840) and by Clausen (1840) independently. The latter was published in 'Astronomis- che Nachrichten', the oldest astronomical journal founded in 1821 by H. C. Schumacher (one of the friends and astronomical collaborators of Gauss), as a brief announcement of the result without proof. Later Schwering (1899) gave another proof using 1 1 (n-1)! + + + x + 1 2(x + 1)(x+2) n(x + 1)(x + 2)...(x + n) 1 1 B₁ B₂ = - - + 2x² x3 x³ + = + 17 15 + - lin + + - im B3 x7

Show that the Bernoulli number B₁ satisfies B₁ = (-1)" ( 2 + 2d²+1) Σ (mod 1) for any positive integer n where d runs over all of the divisors of n such that 2d+1 is a prime. For example, 1-B₁ = 1/2+3/ 1+ B₂ 1 1/2 + 1/3 + 1 − B₁ = 1/2 + 1/3 + 1/7² 1 1 1+ B4 = 1- B5 = 2 3 2 3 11 1 + B6 = 2 + 3 + 5 +132-B₁ 7 This is known as the Staudt-Clausen theorem, which is found by Staudt (1840) and by Clausen (1840) independently. The latter was published in 'Astronomis- che Nachrichten', the oldest astronomical journal founded in 1821 by H. C. Schumacher (one of the friends and astronomical collaborators of Gauss), as a brief announcement of the result without proof. Later Schwering (1899) gave another proof using 1 1 (n-1)! + + + x + 1 2(x + 1)(x+2) n(x + 1)(x + 2)...(x + n) 1 1 B₁ B₂ = - - + 2x² x3 x³ + = + 17 15 + - lin + + - im B3 x7

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Show that the Bernoulli number B₁ satisfies B₁ = (-1) ² ( ²2² +²24 ²+1) Σ (mod 1) 2d d for any positive integer n where d runs over all of the divisors of n such that 2d + 1 is a prime. For example, 1 1 1 1 1 1-B₁ = 1 + B₂ = 1- B3 = = 1/2+1/3+2/2+ 2 3 2 3 5' 7' 1 1 1 1 1 1 1+ B4 = + + 1 - B5 = + 2 3 5 2 3 11 1 1 1 1 + B6 = 1 1 + +-+ + 2-B₁ = 2 3 5 7 13' 2 3 This is known as the Staudt-Clausen theorem, which is found by Staudt (1840) and by Clausen (1840) independently. The latter was published in 'Astronomis- che Nachrichten', the oldest astronomical journal founded in 1821 by H. C. Schumacher (one of the friends and astronomical collaborators of Gauss), as a brief announcement of the result without proof. Later Schwering (1899) gave another proof using 1 1 (n-1)! + + + +... x+1 2(x + 1)(x + 2) n(x + 1)(x + 2)... (x + n) 1 1 B₁ B₂ B3 = + + X 2x² x3 x³ + 1 1 +-+ + +
Transcribed Image Text:Show that the Bernoulli number B₁ satisfies B₁ = (-1) ² ( ²2² +²24 ²+1) Σ (mod 1) 2d d for any positive integer n where d runs over all of the divisors of n such that 2d + 1 is a prime. For example, 1 1 1 1 1 1-B₁ = 1 + B₂ = 1- B3 = = 1/2+1/3+2/2+ 2 3 2 3 5' 7' 1 1 1 1 1 1 1+ B4 = + + 1 - B5 = + 2 3 5 2 3 11 1 1 1 1 + B6 = 1 1 + +-+ + 2-B₁ = 2 3 5 7 13' 2 3 This is known as the Staudt-Clausen theorem, which is found by Staudt (1840) and by Clausen (1840) independently. The latter was published in 'Astronomis- che Nachrichten', the oldest astronomical journal founded in 1821 by H. C. Schumacher (one of the friends and astronomical collaborators of Gauss), as a brief announcement of the result without proof. Later Schwering (1899) gave another proof using 1 1 (n-1)! + + + +... x+1 2(x + 1)(x + 2) n(x + 1)(x + 2)... (x + n) 1 1 B₁ B₂ B3 = + + X 2x² x3 x³ + 1 1 +-+ + +
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