Infinite Product of Sums of Squares

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1. The Start

The story begins with the expression $𝑥 2 - 𝑥 - 2$ , where $𝑥 > 0$ . By the difference of squares formula, we have

𝑥 2 - 𝑥 - 2 = (𝑥 + 𝑥 - 1) (𝑥 - 𝑥 - 1) = (𝑥 + 𝑥 - 1) (𝑥 12 + 𝑥 - 12) (𝑥 12 - 𝑥 - 12) = \dots

𝑥 2 - 𝑥 - 2 = (𝑥 + 𝑥 - 1) (𝑥 - 𝑥 - 1) = (𝑥 + 𝑥 - 1) (𝑥 12 + 𝑥 - 12) (𝑥 12 - 𝑥 - 12) = \dots

𝑥 2 - 𝑥 - 2 = (𝑥 + 𝑥 - 1) (𝑥 - 𝑥 - 1) = (𝑥 + 𝑥 - 1) (𝑥 12 + 𝑥 - 12) (𝑥 12 - 𝑥 - 12) = \dots

𝑥 2 - 𝑥 - 2 = (𝑥 + 𝑥 - 1) (𝑥 - 𝑥 - 1) = (𝑥 + 𝑥 - 1) (𝑥 12 + 𝑥 - 12) (𝑥 12 - 𝑥 - 12) = \dots

Hence, let $𝑎 𝑖 = 𝑥 2 - 𝑖 + 𝑥 - 2 - 𝑖$ , then $𝑎 2 𝑖 = 𝑥 2 - (𝑖 - 1) + 𝑥 - 2 - (𝑖 - 1) + 2 = 𝑎 𝑖 - 1 + 2$ , which is $𝑎 𝑖 = \sqrt 𝑎 𝑖 - 1 + 2$ . Let $𝑓 (𝑛) ≔ 𝑎 0 \cdot 𝑎 1 \cdot 𝑎 2 \cdot \dots \cdot 𝑎 𝑛$ , $𝑔 (𝑛) ≔ 𝑥 2 - 𝑛 - 𝑥 - 2 - 𝑛$ , then $𝑓 (𝑛) \cdot 𝑔 (𝑛) = 𝑥 2 - 𝑥 - 2$ , which immediately gives that $𝑓$ $𝑓$ converges. Now the problem is to find how fast $𝑓$ $𝑓$ increases and $𝑔$ $𝑔$ decreases. It is obvious that we would require some approximations here.

2. The Climax

First let $𝑡 = 𝑥 2 - 𝑥 - 2$ . Thus, $𝑓 (𝑛) \cdot 𝑔 (𝑛) = 𝑡$ . Since we are making approximations for $𝑓$ $𝑓$ and $𝑔$ $𝑔$ , it is a good idea to apply $𝜕 𝑛$ $𝜕 𝑛$ to both sides. Hence, we obtain $𝑓' (𝑛) 𝑔 (𝑛) + 𝑓 (𝑛) 𝑔' (𝑛) = 0$ , which is $𝑓' (𝑛) = - t g' (𝑛) 𝑔 (𝑛) 2$ .

Without loss of generality, set $𝑡 = 1$ and $𝑥 = 𝑒$ . Thus, $𝑔 (𝑛) = 𝑒 2 - 𝑛 - 𝑒 - 2 - 𝑛$ , $𝑓' (𝑛) = - 𝑔' (𝑛) 𝑔 (𝑛) 2$ . Let $𝑚 = 2 𝑛$ , $̂ 𝑔 (𝑚) = 𝑒 1 𝑚 - 𝑒 - 1 𝑚$ . Then $ˆ 𝑓 (𝑚) = - ̂ 𝑔' (𝑚) ̂ 𝑔 (𝑚) 2 = (1 𝑚 2) 𝑒 1 𝑚 + 1 𝑚 2 𝑒 - 1 𝑚 (𝑒 1 𝑚 - 𝑒 - 1 𝑚) 2$ . Now, it is clear how to best estimate how fast $𝑓$ $𝑓$ increases and $𝑔$ $𝑔$ decreases.

3. The End

The final result is here.

Theorem 3.1. $𝑓 (𝑛)$ $𝑓 (𝑛)$ increases and $𝑔 (𝑛)$ $𝑔 (𝑛)$ decreases like exponentials.

Proof. From the previous resut, we have

ˆ 𝑓 (𝑚) ​ = (1 𝑚 2) 𝑒 1 𝑚 + 1 𝑚 2 𝑒 - 1 𝑚 (𝑒 1 𝑚 - 𝑒 - 1 𝑚) 2

. When

𝑚

𝑚

is large enough,

𝑒 1 𝑚 ​ \approx 1 𝑚 ​ + 1

𝑒 - 1 𝑚 ​ \approx ​ - 1 𝑚 ​ + 1

∴ ˆ 𝑓' (𝑚) ​ \approx 2 𝑚 2 (2 𝑚) 2 ​ = 12

. and

̂ 𝑔' (𝑚) ​ = ​ - 1 𝑚 2 𝑒 1 𝑚 ​ - 1 𝑚 2 𝑒 - 1 𝑚 ​ \approx ​ - 2 𝑚 2

∴ ˆ 𝑓 (𝑚) ​ = 12 𝑚

and

̂ 𝑔 (𝑚) ​ = 2 𝑚

. Hence,

𝑓 (𝑛) ​ = 2 𝑛 - 1

and

𝑔 (𝑛) ​ = 2 1 - 𝑛

□

View Typst source

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#import "/themes/site/notes.typ": *
#show: web-equations
#import "/typ/shared.typ": *

#show: tylenotes.with(
  title: "Infinite Product of Sums of Squares",
  date: "2026-09-18",
  tags: ("real", "lean"),
  summary: "Here we discuss how quickly the infinite product of sums of squares increases.",
)

= The Start

The story begins with the expression $x^2 - x^(-2)$, where $x > 0$.
By the difference of squares formula, we have
$ x^2 - x^(-2) = (x+x^(-1))(x-x^(-1)) = (x+x^(-1))(x^(1/2)+x^(-1/2))(x^(1/2)-x^(-1/2))=... $
Hence, let $a_i = x^(2^(-i)) + x^(-2^(-i))$, then $a_i^2 = x^(2^(-(i-1)))+x^(-2^(-(i-1)))+2 = a_(i-1) + 2$,
which is $a_i = sqrt(a_(i-1)+2)$. Let $f(n) := a_0 dot a_1 dot a_2 dot ... dot a_n$,
$g(n) := x^(2^(-n)) - x^(-2^(-n))$, then $f(n) dot g(n) = x^2 - x^(-2)$,
which immediately gives that $f$ converges.
Now the problem is to find how fast $f$ increases and $g$ decreases.
It is obvious that we would require some approximations here.

= The Climax

First let $t = x^2 - x^(-2)$. Thus, $f(n) dot g(n) = t$.
Since we are making approximations for $f$ and $g$,
it is a good idea to apply $partial_n$ to both sides.
Hence, we obtain $f'(n)g(n) + f(n)g'(n)=0$,
which is $f'(n) = - (tg'(n))/(g(n)^2)$.

Without loss of generality, set $t = 1$ and $x = e$.
Thus, $g(n) = e^(2^(-n)) - e^(-2^(-n))$, $f'(n) = - (g'(n))/(g(n)^2)$.
Let $m = 2^n$, $hat(g)(m) = e^(1/m) - e^(-1/m)$.
Then $hat(f)(m) = - (hat(g)'(m))/(hat(g)(m)^2) = ((1/(m^2))e^(1/m) + 1/(m^2)e^(-1/m))/((e^(1/m)-e^(-1/m))^2)$.
Now, it is clear how to best estimate how fast $f$ increases and $g$ decreases.

= The End

The final result is here.

#theorem[
  $f(n)$ increases and $g(n)$ decreases like exponentials.
]

#proof[
 From the previous resut, we have
 $hat(f)(m) = ((1/(m^2))e^(1/m) + 1/(m^2)e^(-1/m))/((e^(1/m)-e^(-1/m))^2)$. 
 When $m$ is large enough, $e^(1/m) approx 1/m + 1$, $e^(-1/m) approx -1/m+1$.
 $therefore hat(f)'(m) approx (2/(m^2))/((2/m)^2)=1/2$.
 and $hat(g)'(m)=-1/(m^2)e^(1/m)-1/(m^2)e^(-1/m) approx -2/(m^2)$.
 $therefore hat(f)(m)=1/2 m$ and $hat(g)(m)=2/m$.
 Hence, $f(n)=2^(n-1)$ and $g(n)=2^(1-n)$.
]