How to define the exponential function without calculus?

Question

For fun, I would like to define the complex exponential function from these two properties:

• $\exp(0) = 1$
• $\exp(z + w) = \exp(z) \exp(w)$

From here, I would like to find a way to compute values of $\exp(z)$, or at least to compute $\exp(1)$.

So far, I found only two ways:

1. Noting that $\exp'(z) = \exp(z)$ and solving the differential equation, which leads to $\int \frac{\exp'(z)}{\exp(z)} dz = \log(\exp(z)) + C = z$.

2. Noting that $\exp'(z) = \exp(z)$, computing its Taylor series and checking that what I get is an entire function.

The first approach is simply wrong because it involves logarithms, which I have not defined yet. The second approach looks much better. I haven't tried, but I guess I can find a way to manipulate the Taylor series to obtain the limit definition of $e$ and conclude that $\exp(1) = e$, which is my aim.

However, I'm struggling to find another way that does not involve differentiation or limits in general. I would be happy to find a way to say $\exp(1) = e$ without calculus. I think that the irrational nature of $e$ forces me to use limits -- am I right?

Answer 1

You won't be able to do derive

$exp(1)=e$

from your definition, since your definition works for the exponential function for any base $b$:

$b^0=1$

and

$b^{x+z}=b^x\cdot b^z$

are true for any $b$!

Answer 2

As Bram28 alludes, your definition is invariant under scalar multiplication. That is, if $u=cx$ for some $c$, then both of your conditions work just as well for $u$ as for $x$. For instance, $f(2(x+y)) = f(x+y)f(x+y) = f(x)f(y)f(x)f(y)$. And $f(2x)f(2y) = f(x)f(x)f(y)f(y)$. So if $f(x+y) = f(x)f(y)$, then $f(2(x+y)) = f(2x)f(2y)$. There is therefore no way to distinguish between non-zero numbers. If you have an argument for why $f(1) = e$, I can just multiply everything by two and everything will work the same as before, and I'll end up for an argument for why $f(2) = e$.

You have to take $f(1)$ as a constant, and then find $f(x)$ in terms of that constant. Then $f(n+1) = f(n)f(1)$, so by induction $f(n) = f(1)^n$ for natural number $n$, and a similar argument gets negative integers. You can then argue that $f(1) = f(.5+.5)=f(.5)f(.5)$, so $f(.5) =\sqrt{f(1)}$ (assuming $f(1)$ is positive). It's then not too difficult to get $f(x)$ defined for any rational number. For irrational numbers, though, you have to assume $f(x)$ is continuous.

Answer 3

$$\begin{align}\exp(x)&=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\\&=\sum_{i=0}^{\infty}\frac{x^i}{i!} \end{align}$$

Answer 4

You cannot really do this "without calculus". However, you can get close with a little bit of heuristic work.

Define $e^x$ by your two properties together with the property that $e^x \approx 1+x$ for small values of $x$. Making this precise involves Calculus, but I am not sure how much precision you need.

In any case, from this you can get

$e^1 = (e^{0.0001})^{10000} \approx (1+0.0001)^{10000} \approx 2.71814592682$

You can approximate other function values as well, like

$e^\pi \approx e^{3.1415} = (e^{0.0001})^{31415} \approx (1+0.0001)^{31415} \approx 23.13$

Finding the precise values using this idea does involve a limit.

Answer 5

Not sure how helpful this is, but you might compare the situation with $\pi$.

There is a geometric "definition" of $\pi$ that relies on the intuitive ideas of the diameter and the circumference of a circle.
Actually "calculating" $\pi$ requires a method of successive approximation.

If we are willing to accept an intuitive idea of the (directed) area between the graph of the reciprocal function and the horizontal axis "from" $1$ to $r$, then we can give a geometric "definition" of the real exponential function.
$\exp{x}$ will be the value of $r$ needed to get an area of $x$.
Actually "calculating" values of this function will require a method of successive approximation.

Answer 6

Of course this can't be done without assuming more of the function. Explicitly, suppose $F:R \rightarrow R $ is a non-identically 0 function such that $F(x+y) = F(x)F(y) \text{ for all } x,y \in R$.

Claim 1: F need not be continuous. To see this, let $L:R \rightarrow Q$ be a non zero $Q$-linear function, such that $L(1)=1$. $R$ is an infinite dimensional $Q$ vector space, and there are lots of linear functions like that (although you may need some weak form of the axiom of choice to prove that one exists). Then the function $F$ defined as $F(x) = exp(L(x)) \text{ for } x \in R$ satisfies the functional equation, and $F(1) = e$, but $F \ne exp$, since $F$ takes only values in a countable set.

If one assumes continuity of $F$ than it follows that $F(x) = F(1)^x$, but continuity is more than what one needs to show $F$ is an exponential, for example the following result holds.

Claim 2: If $F$ satisfies the functional equation and it is Borel-measurable then $F$ is an exponential.

Answer 7

You can define the (real) exponential as the inverse of the strictly monotonic function $$L(x) = \int_1^x \frac{1}{t} \, dt$$

(This may look strange but actually it's the first definition I've learned).

Answer 8

You need one more definition. $ \exp(1) = e $. We construct the $exp$ function from complex numbers, using trig functions. But however I carve it, I need a limit to define $e$,

Series: $$ e = \sum_{k=0}^{\infty} \frac{x^k}{k!} $$ As a Binomial expansion, I need limits to reduce to that. A Taylor Series assumes derivatives, which is Calculus. From cos and sin series, but those series need Calculus,

$$ e = \lim_{n \to \infty} (1 + \frac1{n})^n $$ Needs a limit.

Eulers Identity, $$ e^{i \pi} + 1 = 0 $$

assumes a definition of $e^{ix}$ as trig functions. These functions are ambivalent about the value of $e$. Defining $exp$ as Analytic gives the value, but that is Calculus, and it requires limits to get the value.

---

Using: $$ \exp(0) = 1 \wedge \exp(1) = e \wedge \exp(x + y) = \exp(x) exp(y) $$

Natural Numbers Any natural number is a sum of ones. So your definitions allow you to build up $exp$ for natural numbers. To get rational numbers you need to take roots.

**Rational Numbers **

Rational numbers are in the form $p/q$. These may be constructed by taking roots.

$$ \exp(x q) = \exp(p) \implies \exp(x) = \exp(p/q) = \sqrt[q]{\exp(p)}$$

Rational Numbers

For irrational numbers, you need limits on the rationals. That is the way they are defined, but this isn't specific to $\exp$

Imaginary Numbers

For imaginary numbers, use trigonometry. $$ \exp(ix) = \cos(x) + i \sin(x) $$

Which matches your definition because, $$ \exp(ix+iy) = \exp(ix) \exp(iy) $$ $$ \cos(x+y) + i \sin(x+y) = (\cos(x) + i \sin(x)) (\cos(y) + i \sin(y)) $$

And this result can be shown geometrically.

Complex numbers

Define: $$ \exp(x+iy) = \exp(x) (\cos(y) + i \sin(y)) $$

Even at this stage, having defined $\exp$ for complex numbers, the value of $e$ is not defined. There are many paths to define it but all that I know of require a limit.

Answer 9

If you're willing to accept that $\pi$ can be defined as the ratio of a circle's circumference to its diameter, then $\exp(i\pi) = -1$ should be the only extra condition you need to uniquely define $\exp$.

However, this is a rather non-constructive definition since it defines $\pi$ without helping you calculate it, so I suspect you won't find it satisfying.

But I'm not sure there is a more satisfying solution, so I thought I'd post this.

Answer 10

Let $g:\mathbb{R}\to\mathbb{R}^+$ a function such that $g(0)=1$ and $g(z+w)=g(z)g(w)$ for all $z,w\in\mathbb{R}$. (Note, incidentally, that the first one is superfluous because it can be derived from the second.)

First of all, all functions of the form $g(x)=a^x$, for any $a\in\mathbb{R}^+$, satisfy those properties. So no, you can't get the value of $\mathrm{e}$ out of those two properties only, simply because $\exp$ is not the only solution to your functional equation. Even worse: for each $a\in\mathbb{R}^+$, there is a solution with $g(1)=a$. So you need to plug in an extra normalization condition to identify that function uniquely. For instance, $g(1)=\mathrm{e}$ --- but of course this is the exact opposite of what you wanted to do...

But there is more... This problem is equivalent to asking that $f = \log g$ satisfies $f(0)=0$, $f(z+w)=f(z)+f(w)$. This is known as Cauchy's functional equation, and (as VictorZurkowski noted) it has lots of discontinuous solutions solutions in addition to the "obvious" ones of the form $f(x)=cx$ (which correspond to $g(x)=a^x$). So you are going to need to assume some other property if you wish to show that $g(x)=a^x$: for instance continuity, or monotonicity if you don't want to use the c-word, or even only the weaker assumption "the graph of this function is not a dense subset of $\mathbb{R}^2$". Without an additional hypothesis, you can only show that $g(x)=a^x$ for $x\in\mathbb{Q}$, as a others answers did.

Answer 11

We can actually circumvent the limit by using the supremum instead.

Let the base of the exponential function $f$ be $a \in \langle0, +\infty\rangle \setminus \{1\}$.

For $n \in \mathbb{N}$ define the exponential function recursively: $$a^1 = a \\ a^{n+1} = a\cdot a^n, \quad\forall n\geq 2$$

We can prove $a^{m+n} = a^ma^n, \forall m, n \in \mathbb{N}$ easily using induction.

This defines $f : \mathbb{N} \to \langle0, +\infty\rangle$.

$f$ can be expanded to $\mathbb{Z}$ by defining:

$$f(n) = \begin{cases} a^n, & \text{if $n > 0$} \\ \frac{1}{a^{-n}}, & \text{if $n < 0$} \\ 1, & \text{if $n = 0$} \end{cases}$$

$f$ can be expanded to $\mathbb{Q}$:

For $q \in \mathbb{Q}, q = \frac{m}{n}$ where $m \in \mathbb{Z}, n\in\mathbb{N}$ we define $$f(q) = \sqrt[m]{a^n}$$

Notice that $f(q + q') = f(q)f(q'), \forall q,q'\in\mathbb{Q}$:

\begin{align}f\left(q + q'\right) &= f\left(\frac{m}{n} + \frac{m'}{n'}\right)\\ &= f\left(\frac{mn'+m'n}{nn'}\right) \\ &= \sqrt[nn']{a^{mn'+m'n}} \\ &= \sqrt[nn']{a^{mn'}}\cdot\sqrt[nn']{a^{m'n}} \\ &= \sqrt[n]{a^{m}}\sqrt[n']{a^{m'}} \\ &= f\left(\frac{m}{n}\right)f\left(\frac{m'}{n'}\right) \end{align}

Now expand $f$ to $f : \mathbb{R} \to \langle0, +\infty\rangle$ by defining:

$$f(x) = \sup\{a^q : q \in \mathbb{Q}, q < x\}, \quad\forall x \in \mathbb{R}$$

To show that $f(x+y) = f(x)f(y), \forall x,y\in\mathbb{R}$, we shall use the following proposition:

Let $A, B \subseteq \langle0, +\infty\rangle$ be bounded from above. Then $AB = \{a\cdot b : a\in A, b\in B\}$ is also bounded from above and: $$\sup(AB) = \sup(A)\sup(B)$$

\begin{align}f(x)f(y) &= \sup\{a^q : q \in \mathbb{Q}, q < x\}\sup\{a^{q'} : q' \in \mathbb{Q}, q' < y\}\\ &= \sup\{a^q\cdot a^{q'} : q, q' \in \mathbb{Q}, q < x, q' < y\}\\ &= \sup\{a^{q+q'} : q, q' \in \mathbb{Q}, q + q' < x + y\}\\ &= f(x+y) \end{align}

Answer 12

(An echo of previous answers)

If $f(x+y) = f(x)f(y) $ and $f$ is differentiable, then $f(0) = 1$ (set $y=0$), so $\lim_{x \to 0} f(x) = 1 $.

Switching to standard calculus variables,

$\begin{array}\\ f(x+h)-f(x) &=f(x)f(h)-f(x)\\ &=f(x)(f(h)-1)\\ &=f(x)(f(h)-f(0))\\ \text{so}\\ \dfrac{f(x+h)-f(x)}{h} &=f(x)\dfrac{f(h)-f(0)}{h}\\ \end{array} $

Letting $h \to 0$, $f'(x) = f(x)f'(0)$.

$f(x)=a^x$ satisfies $f(x+y)=f(x)f(y)$. $a=e$ is special because that makes $f'(0) = 1$.

In general, $(a^x)' = a^x\ln(a) $.