[Link to Stackexchange post: Link]

Ref:

• “A Course of Pure Mathematics” by Hardy.

[Product to Sum]

Note that computing sums is easier than computing products.

Can we have a nice bijective function ${ f : (0, \infty) \to \mathbb{R} }$ such that equations “${ x y = z }$” in ${ (0, \infty) }$ give equations “${ f(x) + f(y) = f(z) }$” in ${ \mathbb{R} }$?

Note that equivalently, can we have a nice bijective function ${ f : (0, \infty) \to \mathbb{R} }$ such that ${ f(x) + f(y) = f(xy) }$?

Note that supposing such a function ${ f, f ^{-1} }$ are easy to compute, we can easily compute products. Note that for ${ x, y \in (0, \infty), }$ we can compute ${ x y }$ as ${ f ^{-1} (f(xy)) }$ ${ = f ^{-1} (f(x) + f(y)) . }$

Suppose ${ f }$ is such a function. Note that

\[{ f(xy) = f(x) + f(y) . }\]

Note that taking ${ \frac{\partial}{\partial x} }$ we have

\[{ y f ^{'} (xy) = f ^{'} (x) . }\]

Hence ${ f ^{‘} (x) }$ is pseudolinear in the sense of

\[{ f ^{'} (a x) = a ^{-1} f ^{'} (x) . }\]

Note that

\[{ f ^{'} (x) = \frac{1}{x} }\]

is a potential candidate for ${ f ^{‘} . }$

Hence consider the function

\[{ f(x) := \int _c ^x \frac{1}{t} \, dt . }\]

Note that ${ f (1) = 0 . }$

Hence consider the function

\[{ f(x) := \int _1 ^x \frac{1}{t} \, dt . }\]

Does this ${ f }$ satisfy ${ f(xy) = f(x) + f(y) }$?

It turns out yes.

Note that

\[{ {\begin{aligned} &\, f(xy) \\ = &\, \int _1 ^{xy} \frac{1}{t} \, dt \\ = &\, \int _1 ^{x} \frac{1}{t} \, dt + \int _{x} ^{xy} \frac{1}{t} \, dt \\ \stackrel{t = xu}{=} &\, \int _1 ^{x} \frac{1}{t} \, dt + \int _{1} ^{y} \frac{1}{xu} (x \, du ) \\ = &\, f(x) + f(y) \end{aligned}} }\]

as needed. ${ \blacksquare }$

Is ${ f }$ a bijection?

It turns out yes.

What is ${ f(+ \infty) }$?

Note that

\[{ {\begin{aligned} &\, f (2 ^n) \\ = &\, \int _1 ^{2 ^n} \frac{1}{t} \, dt \\ = &\, \int _1 ^{2} \frac{1}{t} \, dt + \int _{2} ^{2 ^2} \frac{1}{t} \, dt + \ldots + \int _{2 ^{n-1}} ^{2 ^n} \frac{1}{t} \, dt \\ = &\, n \int _{1} ^{2} \frac{1}{t} \, dt \longrightarrow \infty \quad \text{ as } n \longrightarrow \infty . \end{aligned}} }\]

Hence

\[{ f(+ \infty) = + \infty . }\]

What is ${ f(0 +) }$?

Note that

\[{ {\begin{aligned} &\, f(\varepsilon) \\ = &\, \int _1 ^{\varepsilon} \frac{1}{t} \, dt \\ \stackrel{t = 1/u}{=} &\, \int _{u = 1} ^{u = 1/\varepsilon} \frac{1}{1/u} \left( \frac{-1}{u ^2} \, du \right) \\ = &\, - f \left( \frac{1}{\varepsilon} \right) \longrightarrow - \infty \quad \text{ as } \varepsilon \longrightarrow 0 . \end{aligned}} }\]

Hence

\[{ f(0+) = - \infty . }\]

Hence ${ f(0+) = - \infty, }$ ${ f(+ \infty) = + \infty , }$ and ${ f ^{‘} (x) = 1/x > 0 . }$

Hence ${ f: (0, \infty) \to \mathbb{R} }$ is a smooth increasing bijection, as needed. ${ \blacksquare }$

Hence ${ \log : (0, \infty) \longrightarrow \mathbb{R} }$ given by

\[{ \log(x) := \int _1 ^x \frac{1}{t} \, dt }\]

is a smooth increasing bijection satisfying

\[{ \boxed{\log(xy) = \log(x) + \log(y)}. }\]

We call this the natural logarithm.

[Sum to Product]

Note that ${ \log : (0, \infty) \longrightarrow \mathbb{R} }$ given by

\[{ \log(x) := \int _1 ^x \frac{1}{t} \, dt }\]

is a smooth increasing bijection satisfying ${ \log(xy) = \log(x) + \log(y) . }$

Consider the inverse function

\[{ \exp = \log ^{-1} : \mathbb{R} \longrightarrow (0, \infty) . }\]

Note that

\[{ \log(xy) = \log(x) + \log(y) }\]

for ${ x, y \in (0, \infty) . }$

Hence

\[{ xy = \log ^{-1} (\underbrace{\log(x)} _{\overline{x}} + \underbrace{\log(y)} _{\overline{y}}) }\]

for ${ x, y \in (0, \infty) . }$

Hence

\[{ \log ^{-1} (\overline{x}) \log ^{-1} (\overline{y}) = \log ^{-1} (\overline{x} + \overline{y}) }\]

for ${ \overline{x}, \overline{y} \in \mathbb{R} . }$

Hence

\[{ \exp(x) \exp(y) = \exp(x + y) }\]

for ${ x, y \in \mathbb{R} . }$

Hence ${ \exp : \mathbb{R} \longrightarrow (0, \infty) }$ given by

\[{ \exp = \log ^{-1} }\]

is a smooth increasing bijection satisfying

\[{ \boxed{\exp(x + y) = \exp(x) \exp(y)} . }\]

We call this the exponential function.

Note that via the exponential function equations “${ x + y = z }$” in ${ \mathbb{R} }$ give equations “${ \exp(x) \exp(y) = \exp(z) }$” in ${ (0, \infty) . }$

[${ x ^a }$ for ${ x \in (0, \infty) }$]

Let ${ a \in \mathbb{R} . }$ Can we define ${ x ^a }$ for ${ x \in (0, \infty) }$?

Let ${ p, q \in \mathbb{Z}, }$ ${ q > 0 . }$ Let ${ x \in (0, \infty) . }$

Note that ${ x ^{p/q} }$ satisfies

\[{ \underbrace{x ^{p / q} \ldots x ^{p / q}} _{q \text{ times}} = x ^ p . }\]

Hence

\[{ \underbrace{\log(x ^{p / q}) + \ldots + \log(x ^{p/q})} _{q \text{ times}} = p \log(x) . }\]

Hence

\[{ \log(x ^{p/q}) = \frac{p}{q} \log(x) . }\]

Hence

\[{ x ^{p/q} = \exp \left( \frac{p}{q} \log(x) \right) . }\]

Let ${ a > 0 . }$ Let ${ x \in (0, \infty) . }$

Suppose there is a definition of ${ x ^a }$ which varies continuously with ${ a > 0 . }$

Note that under this assumption

\[{ {\begin{aligned} &\, x ^a \\ = &\, \lim _{n \to \infty} x ^{r _n} \\ = &\, \lim _{n \to \infty} \exp \left( r _n \log(x) \right) \\ = &\, \exp(a \log(x)) \end{aligned}} }\]

where ${ r _n }$ is a sequence of rationals ${ r _n \to a . }$

Note that this suggests defining

\[{ \boxed{x ^a = \exp(a \log(x)) } }\]

for ${ a \in \mathbb{R} }$ and ${ x \in (0, \infty) . }$

Note that the definition agrees with the usual definition of ${ x ^a }$ for ${ a }$ rational and ${ x \in (0, \infty) . }$