i was looking at euler's formula recently, and i decided i simply HAD to understand it. i also decided i had to blog about it.

preliminaries:

euler's formula (also referred to as euler's identity, but i found this most often referred to a special case, (e^πx) -1 = 0):

e^ix = sin(x) + i cos(x)

useful constants/functions:

i = sqrt(-1)

e = 1/0! + 1/1! + 1/2!...

e^x = 1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4!...

cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6!...

sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7!...

and also, keep in mind, multiplication distributes over addition, and this applies even to the elements of a function which can be represented by an infinite sum.

first, the taylor series expansion of e^ix...

e^ix = 1 + (ix^1)/1! + (ix^2)/2! + (ix^3)/3! + (ix^4)/4! + (ix^5)/5! + (ix^6)/6!...

as a result of raising i^n, we end up with some pairs of negative terms and positive terms (the second of each pair being complex, the first being real.)

= 1 + ix - (x^2)/2! - (i(x)^3)/3! + (x^4)/4!) + (i(x^5))/5! - (x^6)/6!...

then, if we separate out each term raised to an even number and put them on the left hand, and the remaining odds on the right hand, we end up with an equivalent of sin(x) + i cos(x):

= (1 - (x^2)/2! + (x^4)/4! - (x^6)/6!)... ) + i(x - (x^3)/3! + (x^5)/5! - (x^7)/7! ...)

(note that the series on the right is multiplied by i, which raises i to the respective powers of each term in the series, hence yielding some positive and some negative terms, and some real and some complex terms.)

interspersing the terms after distributing i over the right hand series, we end up with this

= 1 + ix - (x^2)/2! - (i(x^3))/3! + (x^4)/4! +(i(x^5)/5! - (x^6)/6!...

which you can see is also the expansion e^ix that we started with, hence proving (altho perhaps not with usual mathematical rigour...) euler's formula.

## 17.10.06

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