Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Let Σ 1 and g(x) = Σ logp. f(x) = prime p≤x p=3 (mod 10) prime p≤x p=3 (mod 10) g(x) = f(x) logx - Ր _☑ t¯¹ƒ(t) dt. Assuming that f(x) ~ 1½π(x), prove that g(x) ~ 1x. 米 (You may assume the Prime Number Theorem: 7(x) ~ x/log x.) *

Let Σ 1 and g(x) = Σ logp. f(x) = prime p≤x p=3 (mod 10) prime p≤x p=3 (mod 10) g(x) = f(x) logx - Ր _☑ t¯¹ƒ(t) dt. Assuming that f(x) ~ 1½π(x), prove that g(x) ~ 1x. 米 (You may assume the Prime Number Theorem: 7(x) ~ x/log x.) *

College Algebra
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter4: Exponential And Logarithmic Functions
Section: Chapter Questions
Problem 3CC: If xis large, which function grows faster, f(x)=2x or g(x)=x2?
See similar textbooks
icon
Related questions
Question
100%
Let Σ 1 and g(x) = Σ logp. f(x) = prime p≤x p=3 (mod 10) prime p≤x p=3 (mod 10) g(x) = f(x) logx - Ր _☑ t¯¹ƒ(t) dt. Assuming that f(x) ~ 1½π(x), prove that g(x) ~ 1x. 米 (You may assume the Prime Number Theorem: 7(x) ~ x/log x.) *
Transcribed Image Text:Let Σ 1 and g(x) = Σ logp. f(x) = prime p≤x p=3 (mod 10) prime p≤x p=3 (mod 10) g(x) = f(x) logx - Ր _☑ t¯¹ƒ(t) dt. Assuming that f(x) ~ 1½π(x), prove that g(x) ~ 1x. 米 (You may assume the Prime Number Theorem: 7(x) ~ x/log x.) *
Expert Solution
steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer