So you want to create a universe

Thu 11 March 2021 by Moshe Zadka

A story about looking for a universe, and finding a pi(e)

This is fine. You need not feel shame. Many want to create a universe. But it is good you are being careful. A universe with sentient beings is a big moral responsibility.

It is good to start with something small. The smallest. Move up from there. So there is one point, that can move in one dimension. Just interesting enough to be exciting, but little chances of messing anything serious up.

It is good if the way the point moves can be described by a function. At each point in time, $t$, the point is at a place, $x$. This mapping is a function.

$$ f: \mathbb{R} \to \mathbb{R} $$

Right now there are no sentient beings in the universe. But inevitably, you will want to create a universe with such beings. Beings that want the universe to be predictable. Better start practicing now!

This means you want $f$ to have a simple mathematical description. The concept of death is morally complicated, but you want to allow for the potential of beings with a limited lifespan. This means that figuring out how $f$ changes should not rely on $t$.

One way to have a predictable function that does not depend on $t$ is to define $f$ with a position-independent differential equation (PIDE): an equation that involves $f$ and its derivatives.

You are just starting out, so why not have the simplest PIDE?

$$ f' = f $$

Any simpler and your universe will be constant! Solving differential equations is hard. A solution probably exists, right? Hopefully, one that is more interesting than the constant zero function.

$$ f = 0 $$

Yes, it definitely solves it, but that sounds like a really boring universe. If a solution that is not $0$ at $0$ exists, $f$, then

$$ f/f(0) $$

is also a solution since derivatives are linear. The function $f/f(0)$ is an interesting solution. Call it $e$.

For a constant $c$, $e(x + c)/e(c)$ solves the equation and is $1$ at $0$. Differential equations have a unique solution with the same starting condition, so

$$ e(x + c)/e(c) = e(x) $$

or, equivalently

$$ e(x + c) = e(x)e(c) $$

As a result, with a little induction,

$$ e(n/m) ^ m = e(1)^n $$

$$ e(n/m) = \sqrt[m]{e(1) ^ n} $$

If you use the convention that

$$ \sqrt[m](a) = a^{1/m} $$

, you get

$$ e(n/m) = e(1)^{n/m} $$

Your universe is one dimensional, but you cannot help thinking about two dimensions. Is there a way to get a second dimension for "free"? You decide to leave the details for later, and for now, just see if $e$ can be extended to the complex numbers. This is sure to come handy later.

Since $e' = e$, $e'' = e$ nd so on. In particular

$$ 1 = e(0) = e'(0) = e''(0) ... $$

This means that the Taylor series for $e$ looks like

$$ e(x) = \Sigma_{n=0}^{\infty} x^n / n! $$

This converges for every $x$. Because it converges absolutely everywhere, it can be extended to complex number with the same formula, and $e' = e$ over the complex numbers as well.

If $t$ is a real number,

$$ e(-it) = \overline {e (it) } $$

and so

$$ 1 = e(0) = e(it + (-it)) = e(it)e(-it)=e(it)\overline{(e(it))} = || e(it) || $$

Nifty, for real $t$, you get that $e(it)$ is on the unit circle. But where on the unit circle?

$$ \operatorname{Re} e(2i) = 1 - 2^2/2! + 2^4/4! - 2^6 / 6! + d $$

$$ = 1 - 2 + 16/24 - 64/720 + ... = -1 + 2/3 - 4/45 + d = -0.4\bar{2} + d $$

Where $d$ represents the rest of the series.

We can estimate $|d|$ as follows:

$$ |d| \leq \Sigma_{n=0}^{\infty} 2^{8 + n}/(8 + n)! \leq \Sigma_{n=0}^{\infty} ((2/315) 2^n / 4^n)) $$

$$ \leq 2/315 < 1/10 = 0.1 $$

and

$$ \operatorname{Re} e(2i) < -0.4\bar{2} + d < -0.4\bar{2} + 0.1 < -0.3 < 0 $$

Now

$$ \operatorname{Re} e(0i) = \operatorname{Re} e(0) = \operatorname{Re} 1 = 1 > 0 $$

and $t \to \operatorname{Re} e(ti)$ is a continuous function. This means there must be a minimal $t$ such that

$$ \operatorname{Re} e(ti) = 0 $$

This is fun! It is not obvious how this ties into the original goal of building universes, but it definitely cannot hurt! This is an interesting number. You decide to give it a name.

The minimal $t$ such that $\operatorname{Re} e(ti) = 0$ will be $\rho$. Since $||e(\rho i)|| = 1$, this means that

$$ e(\rho i) = \pm i $$

and so

$$ e(4 \rho i) = (\pm 1)^4 i^4 = 1 $$

Looks like $4 \rho$ is even more interesting than $\rho$, maybe it should have its own name. How about $\tau$? With this new symbol, we get

$$ e (\tau i + x) = e(\tau i)e(x) = 1e(x) = e(x) $$

So $e$, the tentative universe-evolution function, has a period of $\tau i$. You did not expect it. It is good you started with a simple universe, there are many things to learn before creating an interesting one.

You had a special name for $\rho$ and for $4 \rho$, it seems almost rude not to have a name for their geometric mean. All this universe creation is hungry work, though. It would be so much easier to think if you had a piece of...

Back to the topic at hand, you decide to call the geometric mean of $\rho$ and $\tau$, $\tau / 2$, $\pi$:

$$ \pi = \tau / 2 $$

Time to relax and eat a nice piece of pie. You definitely deserved it. Whether it is savory or sweet, a pie is delicious. Enjoy it. Savor it. The universe will be waiting for you, right here.

Satisfied and with a full tummy, you get back to the universe. You gave $\rho$, $\pi$, and $\tau$ names.

Any idea what their approximate value is?

You know that $0 < \rho < 2$, but this is a pretty wide gap of ignorance. Calculating $e(it)$, for $t$ in that range, seems to converge quickly.

Just a handful of terms gave you something accurate to within $2/315$. Time to leave abstract thinking, and crank up your universe simulation machine. You want to have a sense of the values involved here.

import math

def approximate_re_e_i(t):
    return sum((-1) ** n * t ** (2*n) / math.factorial(2 * n) for n in range(0, 10))

With a decent approximation of $ \operatorname{Re} e(it) $, you look for where the function is zero using binary search. It might not be the fastest way, but your universe simulator can handle it.

def find_zero():
    low, high = 0, 2
    while (high - low) > 0.001:
        midpoint = (high + low) / 2
        value = approximate_re_e_i(midpoint)
        if value < 0:
            high = midpoint
        else:
            low = midpoint
    return (high + low) / 2

Now it is time to activate the simulator, and find the values.

rho = find_zero()
tau = 4 * rho
pi = tau / 2

Wonderful progress on the universe for today. A great day. A wonderful day. A day you want to celebrate.

But when?

With $\rho <2$, you know that $\pi <8$. Sounds like the integer part of it could be the month, maybe?

month = int(pi)

All that is left is to choose a day. There are thirty days in a month, so hopefully two digits will do here.

rest = pi - month
day = int(rest * 100)

import datetime

year = datetime.date.today().year

celebration = datetime.date(year=year, month=month, day=day)

print("Celebrate on", celebration)

Celebrate on 2021-03-14

Enjoy your pie!

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misc