My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum


Thanks Tree1Thanks
  • 1 Post By mathbalarka
Reply
 
LinkBack Thread Tools Display Modes
June 7th, 2014, 05:11 AM   #1
Senior Member
 
raul21's Avatar
 
Joined: Apr 2014
From: zagreb, croatia

Posts: 234
Thanks: 33

Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology
Averages of arithmetical functions

Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum_{n<x} \sigma_a (n) = \frac_{\zeta(a+1)}{a+1}x^{a+1}+O(x^b)$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.

Last edited by raul21; June 7th, 2014 at 05:15 AM.
raul21 is offline  
 
June 7th, 2014, 05:16 AM   #2
Senior Member
 
raul21's Avatar
 
Joined: Apr 2014
From: zagreb, croatia

Posts: 234
Thanks: 33

Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology
Quote:
Originally Posted by raul21 View Post
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum_{n<x} \sigma_a (n) = \frac_{\zeta(a+1)}{a+1}x^{a+1}+O(x^b)$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum_{n<x} \sigma_a {n} = \frac_{\zeta{a+1}}{a+1}x^{a+1}+O{x^b}$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.

Last edited by raul21; June 7th, 2014 at 05:23 AM.
raul21 is offline  
June 7th, 2014, 05:23 AM   #3
Senior Member
 
raul21's Avatar
 
Joined: Apr 2014
From: zagreb, croatia

Posts: 234
Thanks: 33

Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology
Quote:
Originally Posted by raul21 View Post
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum_{n<x} \sigma_a {n} = \frac_{\zeta{a+1}}{a+1}x^{a+1}+O{x^b}$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum_{n<x} \sigma_a {n} = \frac_{\zeta{a+1}}{a+1}x^{a+1}+O{x^b}$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.
raul21 is offline  
June 7th, 2014, 07:03 AM   #4
Senior Member
 
raul21's Avatar
 
Joined: Apr 2014
From: zagreb, croatia

Posts: 234
Thanks: 33

Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology
Quote:
Originally Posted by raul21 View Post
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum_{n<x} \sigma_a {n} = \frac_{\zeta{a+1}}{a+1}x^{a+1}+O{x^b}$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum${n<x} $\sigma_a{n} = \frac_{\zeta{a+1}}{a+1}x^{a+1}+O{x^b}$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.
raul21 is offline  
June 7th, 2014, 09:43 AM   #5
Senior Member
 
raul21's Avatar
 
Joined: Apr 2014
From: zagreb, croatia

Posts: 234
Thanks: 33

Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology
Quote:
Originally Posted by raul21 View Post
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum${n<x} $\sigma_a{n} = \frac_{\zeta{a+1}}{a+1}x^{a+1}+O{x^b}$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum${n<=x} $\sigma_a (n) = \frac_{\zeta (a+1)}{a+1}x^{a+1}+O(x^b)$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.
raul21 is offline  
June 7th, 2014, 09:46 AM   #6
Senior Member
 
raul21's Avatar
 
Joined: Apr 2014
From: zagreb, croatia

Posts: 234
Thanks: 33

Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology
Quote:
Originally Posted by raul21 View Post
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum${n<=x} $\sigma_a (n) = \frac_{\zeta (a+1)}{a+1}x^{a+1}+O(x^b)$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum$ (n<=x) $\sigma_a (n) = \frac_{\zeta (a+1)} {a+1} x^(a+1) + O (x^b)$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.
raul21 is offline  
June 7th, 2014, 09:51 AM   #7
Senior Member
 
raul21's Avatar
 
Joined: Apr 2014
From: zagreb, croatia

Posts: 234
Thanks: 33

Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology
Quote:
Originally Posted by raul21 View Post
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum$ (n<=x) $\sigma_a (n) = \frac_{\zeta (a+1)} {a+1} x^(a+1) + O (x^b)$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.
Prove

If x is greater or equal to 1, and a > 0, a not equal to 1, we have

$\sum$ (n<=x) $\sigma_a (n)$ = $\frac{\zeta (a+1)} {a+1} x^(a+1) + O (x^b)$,

where b = max{1, a} and $\sigma_a (n)$ is divisor function.
raul21 is offline  
June 10th, 2014, 01:41 PM   #8
Math Team
 
mathbalarka's Avatar
 
Joined: Mar 2012
From: India, West Bengal

Posts: 3,871
Thanks: 86

Math Focus: Number Theory
Well, that's a pretty stinky estimate in general but if you want it ...

$$\sum_{n \leq x} \sigma_a (n) = \sigma_{n \leq x} \sum_{d|n} d^a = \sum_{k \leq x} \sum_{d \leq x/k} d^a$$

Now using this teensy little twiddling in the index, you can recognize the first sum as a geometric series and evaluating it gives something like $$\frac{x^a}{1+a}\sum_{k \leq x} \frac1{k^{1+a}}$$ with an error of $\ll x^a$. Approximating the partial-zeta for large enough $x$ gives you the desired.

Balarka
.
Thanks from raul21
mathbalarka is offline  
Reply

  My Math Forum > College Math Forum > Number Theory

Tags
arithmetical, averages, functions



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Updated asin arithmetical function PKSpark Computer Science 7 February 28th, 2014 02:16 AM
An arithmetical function to approximate inverse trigonometry PKSpark Computer Science 9 February 25th, 2014 02:59 PM
Unknown high-order arithmetical methods. mirror Applied Math 2 September 8th, 2013 08:04 AM
Help on these unknown arithmetical methods mirror Number Theory 3 September 7th, 2013 08:21 AM
Van der Wearden Theorem about arithmetical progression teodork Number Theory 1 December 10th, 2007 02:24 PM





Copyright © 2019 My Math Forum. All rights reserved.