Using Maclaurin Series to define exponents

It is easy to define exponents in terms of positive integers.

xⁿ = x₁ * x₂ * ... * x_n

It is also straight forward to define exponents for 0, negative integers, and rational numbers. I wrote more about this in a previous blog.

But what about the complex numbers? What does it mean to have a value such as 5ⁱ. In fact, a very well known equation is Euler's Identity (see here):

e^iπ + 1 = 0

To handle this, we need a new definition for exponents that is consistent with their use for rational numbers but which can also handle the complex domain.

One way is to use the Maclaurin Series. In a previous blog, I showed that the Maclaurin series is:

f(x) = f(0) + (x/1!)f'(0) + (x²/2!)f''(0) + (x³/3!)f'''(0) + ... + (xⁿ/n!)fⁿ(0)


Now, if we define an exponent as a function so that e^x becomes:

f(x) = e^x

Applying derivatives, we get:

f(0) = e⁰ = 1

And likewise, all fⁿ(0) = 1

Substituting these results into the above equation gives us:

e^x = 1 + (x/1!) + (x²/2!) + ... + xⁿ/n!

Now a^y = e^y*ln(a) since:

(a) e^ln(a) = a [See here for details on ln if needed]

(b) ln(a^y) = y*ln(a) [See here for detalis on ln if needed]

So, we can use the equation for e^x to define a^y

a^y = e^y*ln(a) = 1 + (y*ln(a))/1! + (y*ln(a))²/2! + ... + (y*ln(a))ⁿ/n!

So for example,

1³ = 1 + (3*ln(1))/1! + (3*ln(2))²/2! + ... + =

= 1 + (0)/1! + (0)/2! + ...

2¹ = 1 + (1*ln(2))/1! + (1*ln(2))²/2! + ... + =

1 + ln(2) + ln(2)*ln(2)/2 + ln(2)*ln(2)*ln(3)/6 + ...

= 1 + 0.693... + 0.240... + 0.0555... + ... = 2

Finally, let's ask the question about aⁱ, which equals:

aⁱ = 1 + (i*ln(a)) + (i*ln(a))²/2! + ... + (i*ln(a))ⁿ/n!

= 1 + i*ln(a) - [ln(a)]²/2 -i*[ln(a)]³/3! + [ln(a)]⁴/4! + ...

= (1 - [ln(a)]²/2 + [ln(a)⁴/4! + ...) + i(ln(a) - [ln(a)³/3! + ln(a)⁵/5! + ...)