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Prove that Σ prime p≤x p=3 (mod 10) 17 - Р 1 = log log x + A +0 +A+O log x for some constant A. (Hint: You may wish to use a method similar to the proof of Theorem 4.12.) Theorem 4.12 There is a constant A such that Hence Equation (31) can be written as follows: 1 1 (29) Σ log log x + A +0 for all x ≥ 2. PSxP log x PSX P Σlog log x + 1 - log log 2 + log R(t) dt+0 log x This proves the theorem with PROOF. Let R(t) log P A(x) = Σ A-1 - log log 2+ dt. log PSX P and let Then Σ = PSxP Therefore if we take f(t) 1 < 2, Σ = (30) a(n) - (1 if n is prime, 0 otherwise. a(n) п and A(x)=Σ a(n) log n. Sx n 1/login Theorem 4.2 we find, since A(t) = 0 for 1 Σ PSxP A(x) log x + ·S, 1 log² A(t) dt. From (27) we have A(x) log x + R(x), where R(x) = O(1). Using this on the right of (30) we find (31) Σ logt + R(t) 11x+(1) + PSX P log x =1+0 log x dt t log² t + dt 2 log log t + Slog² R(t) dt. Now dt t log t log log x log log 2 and R(t) t log² t - dt = R(t) tlog' dt - 00 R(t) dt, -S. 1 log² 1 the existence of the improper integral being assured by the condition R(t) O(1). But ၂ R(t) t log² dt - (S." 1103-1) = (103 x). (log²) dt O log =

Prove that Σ prime p≤x p=3 (mod 10) 17 - Р 1 = log log x + A +0 +A+O log x for some constant A. (Hint: You may wish to use a method similar to the proof of Theorem 4.12.) Theorem 4.12 There is a constant A such that Hence Equation (31) can be written as follows: 1 1 (29) Σ log log x + A +0 for all x ≥ 2. PSxP log x PSX P Σlog log x + 1 - log log 2 + log R(t) dt+0 log x This proves the theorem with PROOF. Let R(t) log P A(x) = Σ A-1 - log log 2+ dt. log PSX P and let Then Σ = PSxP Therefore if we take f(t) 1 < 2, Σ = (30) a(n) - (1 if n is prime, 0 otherwise. a(n) п and A(x)=Σ a(n) log n. Sx n 1/login Theorem 4.2 we find, since A(t) = 0 for 1 Σ PSxP A(x) log x + ·S, 1 log² A(t) dt. From (27) we have A(x) log x + R(x), where R(x) = O(1). Using this on the right of (30) we find (31) Σ logt + R(t) 11x+(1) + PSX P log x =1+0 log x dt t log² t + dt 2 log log t + Slog² R(t) dt. Now dt t log t log log x log log 2 and R(t) t log² t - dt = R(t) tlog' dt - 00 R(t) dt, -S. 1 log² 1 the existence of the improper integral being assured by the condition R(t) O(1). But ၂ R(t) t log² dt - (S." 1103-1) = (103 x). (log²) dt O log =

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 74E
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Prove that Σ prime p≤x p=3 (mod 10) 17 - Р 1 = log log x + A +0 +A+O log x for some constant A. (Hint: You may wish to use a method similar to the proof of Theorem 4.12.)
Transcribed Image Text:Prove that Σ prime p≤x p=3 (mod 10) 17 - Р 1 = log log x + A +0 +A+O log x for some constant A. (Hint: You may wish to use a method similar to the proof of Theorem 4.12.)
Theorem 4.12 There is a constant A such that Hence Equation (31) can be written as follows: 1 1 (29) Σ log log x + A +0 for all x ≥ 2. PSxP log x PSX P Σlog log x + 1 - log log 2 + log R(t) dt+0 log x This proves the theorem with PROOF. Let R(t) log P A(x) = Σ A-1 - log log 2+ dt. log PSX P and let Then Σ = PSxP Therefore if we take f(t) 1 < 2, Σ = (30) a(n) - (1 if n is prime, 0 otherwise. a(n) п and A(x)=Σ a(n) log n. Sx n 1/login Theorem 4.2 we find, since A(t) = 0 for 1 Σ PSxP A(x) log x + ·S, 1 log² A(t) dt. From (27) we have A(x) log x + R(x), where R(x) = O(1). Using this on the right of (30) we find (31) Σ logt + R(t) 11x+(1) + PSX P log x =1+0 log x dt t log² t + dt 2 log log t + Slog² R(t) dt. Now dt t log t log log x log log 2 and R(t) t log² t - dt = R(t) tlog' dt - 00 R(t) dt, -S. 1 log² 1 the existence of the improper integral being assured by the condition R(t) O(1). But ၂ R(t) t log² dt - (S." 1103-1) = (103 x). (log²) dt O log =
Transcribed Image Text:Theorem 4.12 There is a constant A such that Hence Equation (31) can be written as follows: 1 1 (29) Σ log log x + A +0 for all x ≥ 2. PSxP log x PSX P Σlog log x + 1 - log log 2 + log R(t) dt+0 log x This proves the theorem with PROOF. Let R(t) log P A(x) = Σ A-1 - log log 2+ dt. log PSX P and let Then Σ = PSxP Therefore if we take f(t) 1 < 2, Σ = (30) a(n) - (1 if n is prime, 0 otherwise. a(n) п and A(x)=Σ a(n) log n. Sx n 1/login Theorem 4.2 we find, since A(t) = 0 for 1 Σ PSxP A(x) log x + ·S, 1 log² A(t) dt. From (27) we have A(x) log x + R(x), where R(x) = O(1). Using this on the right of (30) we find (31) Σ logt + R(t) 11x+(1) + PSX P log x =1+0 log x dt t log² t + dt 2 log log t + Slog² R(t) dt. Now dt t log t log log x log log 2 and R(t) t log² t - dt = R(t) tlog' dt - 00 R(t) dt, -S. 1 log² 1 the existence of the improper integral being assured by the condition R(t) O(1). But ၂ R(t) t log² dt - (S." 1103-1) = (103 x). (log²) dt O log =
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