Let $a_k
$$
l = \frac{b_1 + b_2 + b_3 + \cdots + b_k}{c_1 + c_2 + c_3 + \cdots + c_k}, \text{ } s = \frac{a_1 + a_2 + a_3 + \cdots + a_k}{c_1 + c_2 + c_3 + \cdots + c_k}
$$
Question: What is the limiting value of $l$ and $s$?
The difference between this question and the related question: Part 2: Does the arithmetic mean of sides right triangles to the mean of their hypotenuse converge? is that here the triangles are in sequenced in ascending order of the hypotenuse $c_k$ where as in the related question, they are sequenced in ascending order of $r$ and $s$, and depending on the choice of sequencing, the limiting value differs.
SageMath Code
c = 1
sa = 1
sb = 1
sc = 1
f = 0
sx = 0
while(c <= 10^20):
a = c - 1
b = 3
while(a > b):
b = (c^2 - a^2)^0.5
if(b%1 == 0):
if(b <= a):
if(gcd(a,b) == 1):
f = f + 1
sa = sa + a
sb = sb + b
sc = sc + c
sx = sx + 1/c.n()
print(f,c, sa/sc.n(),sb/sc.n(),sx)
else:
break
a = a - 1
c = c + 1
Answer
Quantities $x=a/c$ and $y=b/c$ are the legs of a pythagorean triangle having unit hypotenuse, hence they are the coordinates of points lying on a unit circle centred at the origin, and $x=\cos\theta$, $y=\sin\theta$ with $\pi/4<\theta<\pi/2$.
It is reasonable to think that, at least in the case of primitive triples, those points are spread evenly on that arc. In that case their average values are:
$$
\langle x\rangle={\int_{\pi/4}^{\pi/2}\cos\theta\,d\theta\over\int_{\pi/4}^{\pi/2}d\theta}=
{4-2\sqrt2\over\pi}\approx 0.372923,
$$
$$
\langle y\rangle={\int_{\pi/4}^{\pi/2}\sin\theta\,d\theta\over\int_{\pi/4}^{\pi/2}d\theta}={2\sqrt2\over\pi}\approx 0.900316.
$$
One should then justify that $\langle a\rangle/\langle c\rangle$ and $\langle a/c\rangle$ have the same limiting value, but that also seems very reasonable. I ran a simulation up to $k\approx800000$ and found the encouraging results:
$$
s_k\approx0.373,\quad l_k\approx0.900.
$$
No comments:
Post a Comment