Is there an easy way of counting this?

Number Theory Level 5

For a positive integer n n , let r ( n ) r(n) denote the number of ways n n can be written as the sum of the squares of an ordered pair of integers. Find the value of

lim ⁡ k → ∞ 1 k ∑ n = 1 k r ( n ) \large \lim \limits_{k \to \infty} \dfrac{1}{k} \sum \limits_{n=1}^{k} r(n)


The answer is 3.1416.

Sharky Kesa by Sharky Kesa , 19, United Kingdom

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Patrick Corn
Dec 27, 2017

We have lim ⁡ k → ∞ 1 k ∑ n = 1 k r ( n ) = lim ⁡ k → ∞ 1 k ∣ S k ∣ , \lim_{k\to\infty} \frac1{k} \sum_{n=1}^k r(n) = \lim_{k\to\infty} \frac1{k} |S_k|, where S k = { ( x , y ) ∈ Z 2 : x 2 + y 2 ≤ k } . S_k = \{(x,y) \in {\mathbb Z}^2 : x^2 + y^2 \le k\}.

The number of lattice points inside a circle of radius r r centered at the origin is π r 2 + O ( r ) \pi r^2 + O(r) (this is a classical result; an easy proof uses Pick's theorem ). So we get 1 k ∣ S k ∣ = π + O ( 1 / k ) , \frac1{k} |S_k| = \pi + O(1/\sqrt{k}), which approaches π \pi as k → ∞ . k \to \infty.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...