# Math Help - average order

1. ## average order

How do you find the value of $\sum\limits_{n \leq x} \mu(n)\Bigl[\frac{x}{n}\Bigr]^{2}$

2. Originally Posted by Chandru1
How do you find the value of $\sum\limits_{n \leq x} \mu(n)\Bigl[\frac{x}{n}\Bigr]^{2}$
What do you mean by "find the value"? Do you mean a closed form expression without a $\sum$?

3. ## Average order

well

Prove that $\sum\limits_{n \leq x} \mu(n) \left[\frac{x}{n}\right]= \frac{1}{\zeta(2)} x^{2} + O(x \log{x})$

4. Originally Posted by Chandru1
well

Prove that $\sum\limits_{n \leq x} \mu(n) \left[\frac{x}{n}\right]= \frac{1}{\zeta(2)} x^{2} + O(x \log{x})$
Wait, is it $\left[\frac{x}{n}\right]$ or $\left[\frac{x}{n}\right]^2$?

5. I'm having trouble getting the desired error term, but I'm close to solving the problem. Maybe you could polish my method to get what you want...

$\sum\limits_{n \leq x} \mu(n)\Bigl[\frac{x}{n}\Bigr]^{2} = \sum\limits_{n \leq x} \mu(n)\left(\frac{x}{n}-\left\{\frac{x}{n}\right\}\right)^{2}$

$= \sum\limits_{n \leq x} \mu(n)\frac{x^2}{n^2}-2\sum\limits_{n \leq x} \mu(n)\frac{x}{n}\left\{\frac{x}{n}\right\}+\sum\l imits_{n \leq x} \mu(n)\left\{\frac{x}{n}\right\}^{2}$

$= x^2\sum\limits_{n \leq x}\frac{\mu(n)}{n^2}-2x\cdot O\left(\sum\limits_{n \leq x}\frac{\mu(n)}{n}\right)+O\left(\sum\limits_{n \leq x} \mu(n)\right)$

First sum: $x^2\sum\limits_{n \leq x}\frac{\mu(n)}{n^2} = x^2\sum\limits_{n=1}^{\infty} \frac{\mu(n)}{n^2} - x^2\sum\limits_{n > x}\frac{\mu(n)}{n^2} = \frac{x^2}{\zeta(2)} + O\left(x^2\sum_{n>x}\frac{1}{n^2}\right) \longleftarrow$ Here's where you need to improve the error term.

Second sum: $2x\cdot O\left(\sum\limits_{n \leq x}\frac{\mu(n)}{n}\right) = o(x)$ since $\sum\limits_{n \leq x}\frac{\mu(n)}{n} = o(1)$ (which is equivalent to the prime number theorem!).

Third sum: $O\left(\sum\limits_{n \leq x} \mu(n)\right) = O(x)$ since $|\mu(n)|\leq 1$.

Enumerating we get $\sum\limits_{n \leq x} \mu(n)\Bigl[\frac{x}{n}\Bigr]^{2} = \frac{x^2}{\zeta(2)} + O(f(x)) + o(x) + O(x) = \frac{x^2}{\zeta(2)} + O(f(x))$. Where $f(x)$ is the error term that you need to improve upon.

6. ## Mistake

$

$
$= \sum\limits_{n \leq x} \mu(n)\frac{x^2}{n^2}+2\sum\limits_{n
\leq x} \mu(n)\frac{x}{n}\left\{\frac{x}{n}\right\}+\sum\l imits_{n \leq
x} \mu(n)\left\{\frac{x}{n}\right\}^{2}

" alt="= \sum\limits_{n \leq x} \mu(n)\frac{x^2}{n^2}+2\sum\limits_{n
\leq x} \mu(n)\frac{x}{n}\left\{\frac{x}{n}\right\}+\sum\l imits_{n \leq
x} \mu(n)\left\{\frac{x}{n}\right\}^{2}
" />

before this a minus sign..

7. ## Done

Hi--

i think that i have found the solution, Here i upload it in a file. Just check it.

8. Originally Posted by Chandru1
Hi--

i think that i have found the solution, Here i upload it in a file. Just check it.
That looks good, could you show me where you learned $\sum_{n\leq x}\frac{1}{n^s} = \frac{x^{1-s}}{1-s}+\zeta(s)+O(x^{-s})$?

The formula looks familiar, but I can't find it anywhere for some reason.

9. ## Proof of the formula

Dear

Chiph, sorry for the delay in writing. I hope you are aware of the Eulers summation formula which says:

If $f$ has a continuous derivative on the interval $[y,x]$ then

$\sum\limits_{y

Please apply $f(n)= \frac{1}{n^s}$ in the Eulers Summation formula to get the desired result.

All of these can be found in Tom Apostol's Analytic Number Theory book.

10. Originally Posted by Chandru1
Dear

Chiph, sorry for the delay in writing. I hope you are aware of the Eulers summation formula which says:

If $f$ has a continuous derivative on the interval $[y,x]$ then

$\sum\limits_{y

Please apply $f(n)= \frac{1}{n^s}$ in the Eulers Summation formula to get the desired result.

All of these can be found in Tom Apostol's Analytic Number Theory book.
Of course! it just slipped my mind!

11. Just to point out a combinatorial interpretation of the sum $\sum\limits_{k \leq n} \mu(k)\left\lfloor\frac{n}{k}\right\rfloor^{2}$

It's equal to the cardinal of $\left\{(x,y)\in\left(\left[n\right]\right)^2\text{such that gcd}(x,y)=1\right\}
$
(1)

The explanation is simple:

Fix $n$, let $A_p$ be $\left\{(x,y)\in\left(\left[n\right]\right)^2\text{such that } p|x \wedge p|y\right\}$ (for each prime $p\leq n$ ).

Now (1) is the set of pairs of integers in $[n]=\left\{1,2,...,n\right\}$, such that they - the pairs- do not belong to the union $\bigcup_{p\leq n}A_p$.
Note that $A_{i_1}\cup...\cup A_{i_m}=\left\{(x,y)\in\left(\left[n\right]\right)^2\text{such that }(i_1...i_m)|x\wedge (i_1...i_m)|y\right\}$ -since we are working with primes- which cardinal is $\left\lfloor\frac{n}{i_1\cdot ...\cdot i_m}\right\rfloor^2$

The rest is inclusion-exclusion. Evidently you can substitute that $2$ for any $m\in\left\{2,3,...\right\}$