# MandysNotes

Friday, 13 May 2011 23:01

## Multiplication of Exponentials

By  Gideon
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Consider first the exponential of a constant, a:

$e^{a} = \lim_{n \to \infty} (1 + \frac{a}{n})^{n} = \sum_{k=0}^{\infty} \frac{a^{k}}{k!}.$

If b is another constant, what is $e^{a}e^{b} ?$

Let's go back to the definition of exponential:

$e^{a}e^{b} \ = \lim_{n \to \infty} f(n) .$

Where:

$f(n) = (1 + \frac{a}{n})^{n} (1 + \frac{b}{n})^{n} = ( 1 + \frac{ a + b}{n} + \frac{ab}{n^{2}})^{n}$

$= \sum_{k=0}^{n} \binom{n}{k} c^{k},$

for $c = \frac{ a + b}{n} + \frac{ab}{n^{2}}.$

Now, by the binomial theorem (again):

$c^{k} =$

$\sum_{i=0}^{k} \binom{k}{i} \left(\frac{ a + b}{n}\right)^{k} \left( \frac{ab}{n^{2}}^{n} \right)^{k-i}$

$= \left(\frac{ a + b}{n}\right)^{k} + {\bf O}\left(\frac{1}{n^{2}} \right) .$

Therefore:

$(1 + \frac{a}{n})^{n} (1 + \frac{b}{n})^{n}$

$= ( 1 + \frac{a +b}{n})^{n} + {\bf O}\left( \frac{1}{n^{2}}\right) .$

So that:

$\lim_{n \to \infty} (1 + \frac{a}{n})^{n} (1 + \frac{b}{n})^{n}$

$= \lim_{n \to \infty} ( 1 + \frac{a +b}{n})^{n} =$

$\sum_{k=0}^{\infty}\frac{(a + b)^{k}}{k!} = e^{(a + b)} .$

We have proved that:

$e^{a} e^{b} = e^{a + b} .$

(Important note: this formula is true whenever a and b commute; that is, whenever ab = ba. This is always true (by definition) when a and b are members of a field, such as the field of real numbers or the field of complex numbers. However, if a and b are members of a noncommutative ring, then the above formula does not hold and must be replaced by the Baker-Campbell-Hausdorff formula. For the purposes of this lesson, a and b will always be either real or imaginary numbers, and so exp(a)exp(b) = exp(a+b) will always hold. )