Exponential

With the identity \( a^{x+y} = a^x a^y \), exponential bridges between addition and multiplication, and with the identity \( \d a^x = a^x\d x \cdot \ln(a) \), exponential plays an essential role in differential-integral calculus. There are various phenomena related to exponential: power law, scale invariance, exponential growth, exponential decay, etc. Here we start from the normal exponential on real number to see its role as the eigenvector of differential-integral calculus and derive its definition in the form of limit. Then using that limit-form definition, we show a deeper meaning of exponential as a mapping from tangent space to the background manifold in general.

The eigenvector of differential-integral calculus

For exponential function \( f(x) = a^x \) with the identity \( f(x+y) = a^{x+y} = a^x a^y \), we have its derivative
\[ f'(x) = \frac{\d(a^x)}{\d x} = \lim_{ε\to 0}{\frac{a^{x+ε}-a^{x}}{ε}} = \lim_{ε\to 0}{\frac{a^{x}a^{ε}-a^{x}}{ε}} \] \[ = a^{x}\cdot \left(\lim_{ε\to 0}{\frac {a^{ε}-1}{ε}}\right) = a^{x}\cdot \left(\lim_{ε\to 0}{\frac {a^{0+ε}-a^0}{ε}}\right) = a^{x}\cdot \left(\frac{\d(a^x)}{\d x}\Bigr|_{\substack{x=0}}\right) \] \[ = f(x)\cdot f'(0) \] where \( f'(0) \) is a constant only depending on the base a. That means the derivative of exponential function is the exponential function itself scaled by a constant, i.e. exponential is the eigenvector of differential-integral calculus. The scaling constant is the eigenvalue of that eigenvector and is called "rate constant" \( λ = f'(x)/f(x) = f'(0) \), which is the "growth constant" when positive, and the "decay constant" when negative.

The natural exponential \( \exp(x) = \e^x \) is defined to have unit rate constant \( \exp'(0) = 1 \), so that it's the unit eigenvector of differential-integral calculus, i.e. \( \exp'(x) = \exp(x) \). Its inverse function, natural logarithm \( \ln(x) := \exp^{-1}(x) \), gives meaning to the rate constant as the natural logarithm of the base a, i.e. \( λ = f'(0) = \ln(a) \), as shown in the following: From the definition of inverse function, we have \( a = \e^{\ln(a)} \), combined with the exponential identity \( (\e^α)^x = \e^{α\cdot x} \), we have the identity \( a^x = \e^{x\cdot \ln(a)} \) to convert arbitrary base-a exponential \( f(x) = a^x \) to natural exponential \( f(x) = \exp(x\cdot \ln(a)) \). With \( y := x\cdot \ln(a) \), we have \( f(x) = \exp(y) \) and

\( f'(x) = \)  ...  \[ \frac{\d(\exp(y))}{\d x} = \frac{\d(\exp(y))}{\d(x\cdot \ln(a))} \cdot \ln(a) = \frac{\d(\exp(y))}{\d(y)} \cdot \ln(a) \] \( = \exp'(y) \cdot \ln(a) = \exp(y) \cdot \ln(a) = f(x) \cdot \ln(a) \)

compared to the previous result \( f'(x) = f(x)\cdot f'(0) \), we have \( f'(0) = \ln(a) \). That gives us the differential identity:

\[ \d(a^x) = a^x\d x \cdot \ln(a) \] \[ \d(\e^x) = \e^x\d x \]

(Id.Eigen)

The limit of scaling operations

\[ \lim_{ε\to 0}{\frac {\e^{ε}-1}{ε}} = 1 \Leftrightarrow \lim_{ε\to 0}{\left(\frac {\e^{ε}-1}{ε} - 1\right)} = 0 \Leftrightarrow \lim_{n\to\infty}{n\cdot( \e^{1/n} - (1+1/n))} = 0 \] ... \[ \Leftrightarrow \lim_{n\to\infty}{\frac{n} {\sum_{k=0}^{n-1} \left(\e^{\frac{n-1-k}{n}} (1+\frac{1}{n})^k \right)} \cdot\left(\e - (1+\frac{1}{n})^n \right)} = 0 \]

\[ \Leftarrow \lim_{n\to\infty}{\left(\e - (1+\frac{1}{n})^n \right)} = 0 \]

, because

\[ n < \sum_{k=0}^{n-1} \left(\e^{\frac{n-1-k}{n}} (1+\frac{1}{n})^k \right) \]

\[ \Leftarrow \e = \lim_{n\to\infty}{\left(1+\frac{1}{n}\right)^n} \]

With this definition of the constant e (Euler's number), we have
\[ f'(0) = \ln(a) = \lim_{ε\to 0}{\frac {a^{ε}-1}{ε}} \] \[ \Leftrightarrow \lim_{n\to\infty}{n\cdot( a^{1/n} - (1+\ln(a)/n))} = 0 \] ... (similar to the above) \[ \Leftarrow a = \lim_{n\to\infty}{\left(1+\frac{\ln(a)}{n}\right)^n} = \e^α = \lim_{n\to\infty}{\left(1+\frac{α}{n}\right)^n} \]

Exponential map

The expression of \( \e^α \) via limit

\[ \e^α = \exp(α) = \lim_{n\to\infty}{\left(1+\frac{α}{n}\right)^n} \]

(Def.1)

shows a deeper meaning of exponential: For any ring \(A \ni α\) endowed with addition, multiplication and scalar multiplication, we can define a function \( \exp(α) \) satisfying exponential identities using Def.1 (see derivation steps below). Moreover, we should consider \(α\) and \(\e^α\) as parameters of some operators acting on some background object \(T_α(o) = α*o = o_t \) and \(X_α(o) = (\e^α)*o = o' \), just like the functions in programming, so \(A\) is a ring of operators. For the simplest case, the "background object" is just the normal number and the operator is the normal multiplication \(α*x = α\cdot x\). In exponential growth/decay, while \(T_α\) acts on intermediate values \(T_α(x_t) = α\cdot x_t\), \(X_α\) acts on the initial value \(x_0\) so that the value at time \(t\) is \(x_t = X_α(x_0) = (\e^{α_0 t})\cdot x_0\). And in physics, \(α\) is usually a matrix, a complex number, a quaternion, etc. In this sense, using the propery of scalar multiplication, we take derivative with respect to that scalar \(\frac{\partial α}{\partial t} = α_0\) \(\Rightarrow α = α_0 t\), so that we can divide the operator \(T_α\) into \(n\) differentials \(T_{\d α} = (α_0 t/n)*\), where \(\d α := α_0\d t = α_0 t/n = α/n\), and then "integrate" them up to get \(X_α = \e^α\).

To be general, we consider a background object in a manifold \(o\in M\) and the operator \(T_α\) as a tangent operator of that manifold, i.e. \( \forall o\in M, o + T_{\dα}(o)\in M \). Applying the differential \(T_{\d α}\) to the whole manifold, we have a vector field on \(M\). Then the displacement operator \(X_α(o) = o'\in M\) generated by that vector field, tracing a geodesic \( [o,o'] \subset M\), is expressed as the exponential \(\e^α\). That's because the infinitesimal displacement is \( X_{\dα}(o) = o + T_{\dα}(o) = (1+α_0\d t)*o \), and the whole displacement is the composition of infinitely many infinitesimal displacements \( X_α = X_{\dα}\circ X_{\dα}\circ X_{\dα}\cdots \) \(= (1+α_0\d t)\cdot(1+α_0\d t)\cdot(1+α_0\d t)\cdots \) \(= \e^{α_0 t} = \e^α \).

In this sense, the exponential is the fundamental building block of differential-integral calculus which is called "exponential map" (in Lie theory as well as in Riemannian geometry). In essence, any continuous operator mapping objects from one domain into the same domain, i.e. any flow \(X_α\) on \(M\), given its differential operator \(\frac{\partial}{\partial t} = T_{α_0} = α_0*\), can be expressed as an exponential operator \((\e^{α_0 t})*\) on \(M\).

\[ \forall M, \forall X_α:M\rightarrow M, F(t) := X_{α_0,M}(t) = X_α(M),\] \[ \frac{\d F(t)}{\d t} = α_0*F(t) \Rightarrow F(t) = (\e^{α_0 t})*F(0) = (\e^{α_0 t})*M \]

(Id.Flow)

Here, instead of acting on individual objects in the manifold, the operators \( α_0*\) and \( (\e^{α_0 t})*\) act on the whole manifold so that the result \(F(t)\) has only one variable \(t\), and the differential operator is total \(\frac{\d}{\d t}\) instead of partial \(\frac{\partial}{\partial t}\), as shown in these vizualizations for Lorenz attractor, Mandelbrot set. In physics, the manifold \(M\) is considered as (phase) space and the differential operator \(\frac{\partial}{\partial t} = T_{α_0}\) is usually considered as time derivative operator, as shown in this video clip. However in general, the differential operator can be with respect to any spatial coordinate or any other scalar variable. Note that exponential map in dynamical systems is just a mapping from complex plane to itself, which is somehow relevant but different from the exponential map here.

Identities

Here we derive the identities of exponential directly from the limit definition Def.1 and the expression of composition of operators above.

• \( \e^{0} = 1 \): When \(α=0\), there is no displacement, thus \(\e^{0} = X_0\) is an identity operator \(X_0(o) = o\).
• \( \e^{α+β} = \e^β\cdot \e^α \): The displacement of a sum \(α+β\) is the composition of component displacements \( \e^{α+β} = X_{α+β} = X_{β}\circ X_{α} = \e^{β}\cdot \e^{α} \), just like the parallelogram rule of vector addition (but may not be commutative).
  
• \( \e^{αs} = (\e^α)^s \): The displacement of a parameter \(α\) scaled by a real number \(s\) is \(X_α\) raised to the power of \(s\) in the sense of "\(s\) times" (more/less) infinitesimal displacements. \[ \e^{αs} = \lim_{n\to\infty}\left(1+\frac{αs}{n}\right)^n = \lim_{n\to\infty}\left(1+\frac{α}{n/s}\right)^{(n/s)s} = \left(\lim_{n'\to\infty}\left(1+\frac{α}{n'}\right)^{n'}\right)^s = (\e^{α})^s \]
• \( \d(\e^{α}) = \e^α\cdot \dα \): We deduce this from the limit definition, which is the converse of the induction in Id.Flow. From \(α = α_0 t\) and Def.1, we have \[ \frac{\d(\e^{α})}{\d t} = \lim_{n\to\infty}\frac{\d}{\d t}\left(1+\frac{α_0 t}{n}\right)^n = \lim_{n\to\infty}\left(\left(1+\frac{α}{n}\right)^{n-1}\cdot n\frac{α_0}{n}\right) \] \[= α_0 \lim_{n\to\infty}\left(1+\frac{α}{n}\right)^{n\frac{n-1}{n}} = α_0 \e^α \] \[ \Rightarrow \d(\e^{α}) = \e^α\cdot α_0\d t = \e^α\cdot \dα \]

Taylor series

We can also expand the binomial in Def.1 to retrieve its Taylor series
\[ \exp(α) = \lim_{n\to\infty}{\left(1+\frac{α}{n}\right)^n} = \lim_{n\to\infty} {\sum _{k=0}^{n}{\binom{n}{k} \left(\frac{α}{n}\right)^k}} \]  ...  \[ = \lim_{n\to\infty} {\sum _{k=0}^{n} {\frac{\prod_{j=0}^{k-1}{(n-j)}}{k!} \frac{α^k}{n^k}} } = \lim_{n\to\infty} {\sum _{k=0}^{n} {\left( \frac{α^k}{k!} \cdot \prod_{j=0}^{k-1}{\frac{n-j}{n}} \right)} } \] \[ = \lim_{n\to\infty} {\sum _{k=0}^{n}{\frac{α^{k}}{k!}}} = \sum _{k=0}^{\infty }{\frac{α^{k}}{k!}}=1+α+{\frac {α^{2}}{2}}+{\frac {α^{3}}{6}}+{\frac {α^{4}}{24}}+\cdots \]