## Sunday, May 07, 2006

### Using Maclaurin Series to define exponents

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

xn = x1 * x2 * ... * xn

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 + 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 ex becomes:

f(x) = ex

Applying derivatives, we get:

f(0) = e0 = 1

And likewise, all fn(0) = 1

Substituting these results into the above equation gives us:

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

Now ay = ey*ln(a) since:

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

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

So, we can use the equation for ex to define ay

ay = ey*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

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! + ...)

The idea of radians is to define degrees in terms of pi. In a previous blog, I showed how Archimedes' proof for the area of a circle can be used to establish pi.

In other words radians = 360 degrees.

A straight line has π radians and a right angle has π/2 radians.

This is useful because it enables trigonometric functions to be expressed in terms of taking π as a value.

Here are some examples:

sin (π/2) = 1
sin (π) = 0
sin(2π) = 0

cos (π/2) = 0
cos(π) = -1
cos(2π) = 1

Lemma 1: sin(nπ) = 0 where n is any integer

Proof:

(1) sin(0) = 0 and sin(π)=0 [See Property 1, here]

(2) sin(x + &2pi;) = x [See Property 5, here]

(3) So, we can see that:

if n is even, then n = 2x and sin (nπ) = sin(x*2π) = sin(0 + x*2π) = sin(0) = 0

if n is odd, then n = 2x+1 and sin(nπ) = sin(x*2π + π) = sin(π) = 0

QED

Lemma 2: The area of a sector = (1/2)θr2

Proof:

(1) We know that the area of a circle = πr2 (see Corollary 2, here)

(2) An angle of a circle = (θ/2π) of a circle.

(3) So therefore, the area of a sector = (θ/2π)(πr2) = (1/2)θr2

QED