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Moritz Firsching
Moritz Firsching
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Looking at Average distance of the mean of n random complex numbers in a unit disc, I tried to figure out  what is the expected absolute value $|\frac{z_1 + z_2}{2}|$ of two numbers $z_1, z_2\in\mathbb{C}$ drawn uniformly from the unit disc.  We can look at the quadruple integral $$\int_0^1r_1\int_0^{2\pi}\int_0^1r_2\int_0^{2\pi}|\frac{1}{2}(r_1e^{\phi_1} + r_2e^{\phi_2}) d\phi_2 dr_2 d\phi_1dr_1 $$$$\int_0^1r_1\int_0^{2\pi}\int_0^1r_2\int_0^{2\pi}|\frac{1}{2}(r_1e^{\phi_1} + r_2e^{\phi_2})| d\phi_2 dr_2 d\phi_1dr_1 $$ This can be reduced to two integrals, so that we can actually calculate the values with high precision (see the following code)

from mpmath import mp, sqrt, ellipe, quad, pimp.dps = 15
def ellipe_integrand(r, s):
    return  sqrt(r*r + 2*r*s + s*s)*(ellipe(+pi, 4*r*s/(r*r + 2*r*s + s*s)))
def integrand_2(s):
    return quad(lambda r: 2*r*ellipe_integrand(r, s), [0, s, 1])mp.dps=150quad(lambda s: s*integrand_2(s), [0,1])
# after ~40 minutes on my machine we get:
mpc(real='1.42222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222182628758433670366371075596783355537463385287071932198285251', imag='4.76485167346043265351378514322413241842941120327748264250820934786695559445396622361703586368896605436500687259315655829464952759337107787438513621789267e-246')

The result of that integration looks suspiciously like $\frac{64}{45}$. To get the expected absolute volume we still have to divide by the area of the disc:$$ \frac{64}{45\pi} \approx 0.45270739368361339952038048248181862978690743677285389866003155$$ (I haven't tried running numerical simulation of simply drawing two points repeatedly, but I would hope that one could recover the first few digits here..)

Let's denote the expected absolute value of the mean of n points by $\operatorname{exp\_abs}(n)$. It is easy to see that $\operatorname{exp\_abs}(1) = \frac{2}{3}$.

Question 1 Is $ \operatorname{exp\_abs}(2) = \frac{64}{45\pi}$, like the numerical evidence would suggest? How to prove that?

Question 2 Is there some reason to expect that $\operatorname{exp\_abs}(n)\in\mathbb{Q}(\pi)$?

Question 3 What is the value of  $\operatorname{exp\_abs}(3)$? Here I would be interested in numerical evidence if the exact value is not easily derived.

Looking at Average distance of the mean of n random complex numbers in a unit disc, I tried to figure out  what is the expected absolute value $|\frac{z_1 + z_2}{2}|$ of two numbers $z_1, z_2\in\mathbb{C}$ drawn uniformly from the unit disc.  We can look at the quadruple integral $$\int_0^1r_1\int_0^{2\pi}\int_0^1r_2\int_0^{2\pi}|\frac{1}{2}(r_1e^{\phi_1} + r_2e^{\phi_2}) d\phi_2 dr_2 d\phi_1dr_1 $$ This can be reduced to two integrals, so that we can actually calculate the values with high precision (see the following code)

from mpmath import mp, sqrt, ellipe, quad, pimp.dps = 15
def ellipe_integrand(r, s):
    return  sqrt(r*r + 2*r*s + s*s)*(ellipe(+pi, 4*r*s/(r*r + 2*r*s + s*s)))
def integrand_2(s):
    return quad(lambda r: 2*r*ellipe_integrand(r, s), [0, s, 1])mp.dps=150quad(lambda s: s*integrand_2(s), [0,1])
# after ~40 minutes on my machine we get:
mpc(real='1.42222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222182628758433670366371075596783355537463385287071932198285251', imag='4.76485167346043265351378514322413241842941120327748264250820934786695559445396622361703586368896605436500687259315655829464952759337107787438513621789267e-246')

The result of that integration looks suspiciously like $\frac{64}{45}$. To get the expected absolute volume we still have to divide by the area of the disc:$$ \frac{64}{45\pi} \approx 0.45270739368361339952038048248181862978690743677285389866003155$$ (I haven't tried running numerical simulation of simply drawing two points repeatedly, but I would hope that one could recover the first few digits here..)

Let's denote the expected absolute value of the mean of n points by $\operatorname{exp\_abs}(n)$. It is easy to see that $\operatorname{exp\_abs}(1) = \frac{2}{3}$.

Question 1 Is $ \operatorname{exp\_abs}(2) = \frac{64}{45\pi}$, like the numerical evidence would suggest? How to prove that?

Question 2 Is there some reason to expect that $\operatorname{exp\_abs}(n)\in\mathbb{Q}(\pi)$?

Question 3 What is the value of  $\operatorname{exp\_abs}(3)$? Here I would be interested in numerical evidence if the exact value is not easily derived.

Looking at Average distance of the mean of n random complex numbers in a unit disc, I tried to figure out  what is the expected absolute value $|\frac{z_1 + z_2}{2}|$ of two numbers $z_1, z_2\in\mathbb{C}$ drawn uniformly from the unit disc.  We can look at the quadruple integral $$\int_0^1r_1\int_0^{2\pi}\int_0^1r_2\int_0^{2\pi}|\frac{1}{2}(r_1e^{\phi_1} + r_2e^{\phi_2})| d\phi_2 dr_2 d\phi_1dr_1 $$ This can be reduced to two integrals, so that we can actually calculate the values with high precision (see the following code)

from mpmath import mp, sqrt, ellipe, quad, pimp.dps = 15
def ellipe_integrand(r, s):
    return  sqrt(r*r + 2*r*s + s*s)*(ellipe(+pi, 4*r*s/(r*r + 2*r*s + s*s)))
def integrand_2(s):
    return quad(lambda r: 2*r*ellipe_integrand(r, s), [0, s, 1])mp.dps=150quad(lambda s: s*integrand_2(s), [0,1])
# after ~40 minutes on my machine we get:
mpc(real='1.42222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222182628758433670366371075596783355537463385287071932198285251', imag='4.76485167346043265351378514322413241842941120327748264250820934786695559445396622361703586368896605436500687259315655829464952759337107787438513621789267e-246')

The result of that integration looks suspiciously like $\frac{64}{45}$. To get the expected absolute volume we still have to divide by the area of the disc:$$ \frac{64}{45\pi} \approx 0.45270739368361339952038048248181862978690743677285389866003155$$ (I haven't tried running numerical simulation of simply drawing two points repeatedly, but I would hope that one could recover the first few digits here..)

Let's denote the expected absolute value of the mean of n points by $\operatorname{exp\_abs}(n)$. It is easy to see that $\operatorname{exp\_abs}(1) = \frac{2}{3}$.

Question 1 Is $ \operatorname{exp\_abs}(2) = \frac{64}{45\pi}$, like the numerical evidence would suggest? How to prove that?

Question 2 Is there some reason to expect that $\operatorname{exp\_abs}(n)\in\mathbb{Q}(\pi)$?

Question 3 What is the value of  $\operatorname{exp\_abs}(3)$? Here I would be interested in numerical evidence if the exact value is not easily derived.

Source Link
Moritz Firsching
Moritz Firsching
  • 10.7k
  • 3
  • 63
  • 88

Expected absolute value of the average of two points from the disc

Looking at Average distance of the mean of n random complex numbers in a unit disc, I tried to figure out  what is the expected absolute value $|\frac{z_1 + z_2}{2}|$ of two numbers $z_1, z_2\in\mathbb{C}$ drawn uniformly from the unit disc.  We can look at the quadruple integral $$\int_0^1r_1\int_0^{2\pi}\int_0^1r_2\int_0^{2\pi}|\frac{1}{2}(r_1e^{\phi_1} + r_2e^{\phi_2}) d\phi_2 dr_2 d\phi_1dr_1 $$ This can be reduced to two integrals, so that we can actually calculate the values with high precision (see the following code)

from mpmath import mp, sqrt, ellipe, quad, pimp.dps = 15
def ellipe_integrand(r, s):
    return  sqrt(r*r + 2*r*s + s*s)*(ellipe(+pi, 4*r*s/(r*r + 2*r*s + s*s)))
def integrand_2(s):
    return quad(lambda r: 2*r*ellipe_integrand(r, s), [0, s, 1])mp.dps=150quad(lambda s: s*integrand_2(s), [0,1])
# after ~40 minutes on my machine we get:
mpc(real='1.42222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222182628758433670366371075596783355537463385287071932198285251', imag='4.76485167346043265351378514322413241842941120327748264250820934786695559445396622361703586368896605436500687259315655829464952759337107787438513621789267e-246')

The result of that integration looks suspiciously like $\frac{64}{45}$. To get the expected absolute volume we still have to divide by the area of the disc:$$ \frac{64}{45\pi} \approx 0.45270739368361339952038048248181862978690743677285389866003155$$ (I haven't tried running numerical simulation of simply drawing two points repeatedly, but I would hope that one could recover the first few digits here..)

Let's denote the expected absolute value of the mean of n points by $\operatorname{exp\_abs}(n)$. It is easy to see that $\operatorname{exp\_abs}(1) = \frac{2}{3}$.

Question 1 Is $ \operatorname{exp\_abs}(2) = \frac{64}{45\pi}$, like the numerical evidence would suggest? How to prove that?

Question 2 Is there some reason to expect that $\operatorname{exp\_abs}(n)\in\mathbb{Q}(\pi)$?

Question 3 What is the value of  $\operatorname{exp\_abs}(3)$? Here I would be interested in numerical evidence if the exact value is not easily derived.