$\require{mathtools}
%
%%% GENERIC MATH %%%
%
% Environments
\newcommand{\al}[1]{\begin{align}#1\end{align}} % need this for \tag{} to work
\renewcommand{\r}{\mathrm}
%
% Greek
\newcommand{\eps}{\epsilon}
\newcommand{\veps}{\varepsilon}
\newcommand{\Om}{\Omega}
\newcommand{\om}{\omega}
\newcommand{\Th}{\Theta}
\let\fi\phi % because it looks like an f
\let\phi\varphi % because it looks like a p
%
% Miscellaneous shortcuts
% .. over and under
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\newcommand{\s}[2]{#1 /\mathopen{}#2}
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% .. relations
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\newcommand{\gs}{\gtrsim}
% .. miscer
\newcommand{\q}{\quad}
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\newcommand{\heart}{\heartsuit}
%
% Delimiters
% (I needed to create my own because the MathJax version of \DeclarePairedDelimiter doesn't have \mathopen{} and that messes up the spacing)
% .. one-part
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% .... (use phantom to force at least the standard height of double bars)
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%% .. two-part
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% ..... (use phantom to force at least the standard height of double bar)
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\newcommand{#6}{\mathrel{(\sdot{#1})}}
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}
\let\lt\undefined
\let\gt\undefined
% .. vanilla versions (is it within a constant?)
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% .. dotted versions (is it equal in the limit?)
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% .. log versions (is it equal up to log?)
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%
% Miscellaneous
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% .. operators
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\DeclareMathOperator{\negl}{negl}
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% .. functions
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\DeclareMathOperator{\ReLU}{ReLU}
% .. analysis
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% .. sets of numbers
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%
%%% SPECIALIZED MATH %%%
%
% Logic
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%
% Linear algebra
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% ... named tensors
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%
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% ... information theory
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%
%%% SPECIALIZED COMPUTER SCIENCE %%%
%
% Complexity classes
% .. classical
\newcommand{\Poly}{\mathsf{P}}
\newcommand{\NP}{\mathsf{NP}}
\newcommand{\PH}{\mathsf{PH}}
\newcommand{\PSPACE}{\mathsf{PSPACE}}
\renewcommand{\L}{\mathsf{L}}
% .. probabilistic
\newcommand{\formost}{\mathsf{Я}}
\newcommand{\RP}{\mathsf{RP}}
\newcommand{\BPP}{\mathsf{BPP}}
\newcommand{\MA}{\mathsf{MA}}
\newcommand{\AM}{\mathsf{AM}}
\newcommand{\IP}{\mathsf{IP}}
\newcommand{\RL}{\mathsf{RL}}
% .. circuits
\newcommand{\NC}{\mathsf{NC}}
\newcommand{\AC}{\mathsf{AC}}
\newcommand{\ACC}{\mathsf{ACC}}
\newcommand{\TC}{\mathsf{TC}}
\newcommand{\Ppoly}{\mathsf{P}/\poly}
\newcommand{\Lpoly}{\mathsf{L}/\poly}
% .. resources
\newcommand{\TIME}{\mathsf{TIME}}
\newcommand{\SPACE}{\mathsf{SPACE}}
\newcommand{\TISP}{\mathsf{TISP}}
\newcommand{\SIZE}{\mathsf{SIZE}}
% .. keywords
\newcommand{\co}{\mathsf{co}}
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%
% Boolean analysis
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% .. functions (small caps sadly doesn't work)
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%
% Dynamic optimality
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%
% Alignment
\DeclareMathOperator{\Amp}{\mathrm{Amp}}
%
%%% TYPESETTING %%%
%
% In text
\renewcommand{\th}{^{\mathrm{th}}}
\newcommand{\degree}{^\circ}
%
% Fonts
% .. bold
\newcommand{\BA}{\boldsymbol{A}}
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% .. calligraphic
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% .. typewriter
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\newcommand{\TZ}{\mathtt{Z}}$
Say some quantity $f(t)$ is growing with time $t$, and its relative derivative is a power of itself
\[
\frac{\ds{f}{t}}{f} = sf^{\alpha},
\]
for some constant “speed” $s>0$ and power $\alpha$. What does that tell you about how fast $f$ will grow to a certain value?
Intuition
Let’s build intuition by thinking about how long doubling times are in some special cases. Or, more precisely, we’ll think about how long it takes to multiply by $e$, which happens after the relative derivative has “accumulated to $1$”:
\[
\frac{f(t_2)}{f(t_1)} = e \Leftrightarrow \int_{t_1}^{t_2} \at{\frac{\ds{f}{t}}{f}}{t'} \d t' = 1
\]
Doubling slows down: polynomial growth
First, say $\frac{\ds{f}{t}}{f} = 1/f$, and we start at some value $f_0$. As $f$ grows from $f_0$ to $ef_0$, we have
$\frac{\ds{f}{t}}{f} \in \b{\frac{1}{ef_0}, \frac{1}{f_0}}$, so it should take somewhere between $f_0$ and $ef_0$ units of time to “double”. For the next doubling, though, $f$ will be betwen $ef_0$ and $e^2f_0$, so it will take between $ef_0$ and $e^2f_0$ units of time, and so on. Each doubling is a constant factor slower than the previous one.
This is precisely what happens when $f$ grows polynomially, and indeed $\frac{\ds{f}{t}}{f} = 1/f$ solves to
\[
f(t) = t-t_0,
\]
i.e. in this case, $f$ grows linearly.
Constant doubling: exponential growth
Now, say $\frac{\ds{f}{t}}{f} = 1$. In this case, takes exactly one unit of time to get from $f_0$ to $ef_0$, then one again to get from $ef_0$ to $e^2f_0$, etc. Each doubling takes the same amount of time.
This is what happens when $f$ grows exponentially, and indeed $\frac{\ds{f}{t}}{f} = 1$ solves to
\[
f(t) = e^{t-t_0}.
\]
Doubling speeds up: hyperbolic growth
Finally, suppose $\frac{\ds{f}{t}}{f} = f$, and we start at $f=f_0$. While $f$ grows from $f_0$ to $ef_0$, we have $\frac{\ds{f}{t}}{f} \ge f_0$, so it should take time at most $1/f_0$. But then as $f$ grows from $ef_0$ to $e^2 f_0$, we have $\frac{\ds{f}{t}}{f} \ge ef_0$, so it should take time at most $\frac1{ef_0}$, and so on. Each doubling is a constant factor faster than the previous one. This means that we will get through an infinite number of doublings ($f$ blows up to infinity) in finite time!
This is what happens when $f$ grows hyperbolically, and indeed $\frac{\ds{f}{t}}{f} = f$ solves to
\[
f(t) = \frac{1}{t_0 - t},
\]
where $t_0$ is the time at which $f$ blows up to infinity.
General case
Polynomial growth
Say we have
\[
\frac{\ds{f}{t}}{f} = \frac{s}{f^{\alpha}},
\]
for $\alpha >0$. Guessing that $f$ is a monomial of the form $f(t) = Ct^d$ for some constants $C,d>0$, then by the properties of relative derivatives we have
\[
f = \p{\frac{s}{\frac{\ds{f}{t}}{f}}}^{1/\alpha} = \p{\frac{s t}{d}}^{1/\alpha},
\]
and thus $d = 1/\alpha$, which gives $f(t) = (\alpha st)^{1/\alpha}$, and more generally
\[
f(t) = \p{\alpha s (t-t_0)}^{1/\alpha}.
\]
In particular, the time it takes to get from a small value $\eps$ to a large value $M$ is
\[
\frac{M^\alpha - \eps^\alpha}{\alpha s}.
\]
As long as $\alpha$ is not too small, this is $\approx M^{\alpha}/(\alpha s)$: since doubling gets harder as you grow, the time you need is mostly determined by where you want to go.
Exponential growth
Say we have
\[
\frac{\ds{f}{t}}{f} = s,
\]
then this solves to
\[
f(t) = e^{s(t-t_0)}.
\]
In particular, the time it takes to get from a small value $\eps$ to a large value $M$ is $\log(M/\eps)$, which is “equally determined” by where you start from and where you want to go.
Hyperbolic growth
Say we have
\[
\frac{\ds{f}{t}}{f} = sf^{\alpha},
\]
for $\alpha >0$. If we guess that $f$ is a monomial of the form $f(t) = (-C/t)^d$ for $t<0$ (for some constants $C,d>0$), then by the properties of relative derivatives we have
\[
f = \p{\frac{\frac{\ds{f}{t}}{f}}{s}}^{1/\alpha} = \p{-\frac{d}{st}}^{1/\alpha}.
\]
Thus $d = 1/\alpha$, which gives $f(t) = (-1/(\alpha st))^{1/\alpha}$, and more generally
\[
f(t) = \p{\frac{1}{\alpha s (t_0-t)}}^{1/\alpha}.
\]
In particular, the time it takes to get from a small value $\eps$ to a large value $M$ is
\[
\frac{\frac{1}{\eps^\alpha} - \frac{1}{M^\alpha}}{\alpha s}.
\]
As long as $M \gg \eps$ and $\alpha$ is not too small, this is $\approx \eps^{-\alpha}/(\alpha s)$: since doubling gets easier as you grow, the time you need is mostly determined by where you’re starting from.