Cauchy-Product of the Exponential Function

Cauchy Product of the Exponential Function

$\forall z, w \in C : exp (w + z) = exp (w) exp (z)$

The claim can be verified by computing the Cauchy product of the right hand side. We will compute the Cauchy product of the following two series:

$i = 0 \sum \infty \frac{w ^{i}}{i !}, j = 0 \sum \infty \frac{z ^{j}}{j !}$

Recall the Cauchy product and the series expansion of the exponential function $exp (z)$ :

💡 The Cauchy-Product of the two series

$i = 0 \sum \infty a_{i}, j = 0 \sum \infty b_{j}$

is the series

$n = 0 \sum \infty (j = 0 \sum n a_{n - j} b_{j}) = a_{0} b_{0} + (a_{0} b_{1} + a_{1} b_{0}) + (a_{0} b_{2} + a_{1} b_{1} + a_{2} b_{0}) + \dots$

💡 The exponential function

We define the exponential function $exp (z)$ as:

$exp (z) = 1 + z + \frac{z ^{2}}{2 !} + \frac{z ^{3}}{3 !} + \dots = i = 0 \sum \infty \frac{z ^{i}}{i !}$

(Note: this is a Taylor series expansion of $e^{z}$ )

The Cauchy product of the two is

$n = 0 \sum \infty (j = 0 \sum n \frac{w ^{n - j}}{( n - j )!} \frac{z ^{j}}{j !})$

This part in brackets evaluates to

$j = 0 \sum n \frac{w ^{n - j}}{( n - j )!} \frac{z ^{j}}{j !} = \frac{1}{n !} j = 0 \sum n (j n) w^{n - j} z^{j} = \frac{( w + z ) ^{n}}{n !}$

which by definition of $exp (w + z)$ completes the proof.