e

^{ix}= Cos(x)+iSin(x)

But this is not exactly intuitive why this is the case. The answer lies in a brilliant piece of maths devised by Brook Taylor, it is called the Taylor series. You can represent any function as the sum of an infinite series of polynomials. This is incredibly useful when it comes to Sin(x), Cos(x) and e

^{x}, and when you delve into the Taylor series of these you can begin to see where e

^{ix}= Cos(x)+iSin(x) comes from:

Now, it might not be immediately obvious how those are related, but all the right terms are there we just need to piece it together. If we begin to manipulate the Taylor series for e

^{x}and we replace 'x' with 'ix' we will begin to see our proof. For the purposes of this I must mention how i

^{x}works i

^{1}= i, i

^{2}= -1, i

^{3}= -i, i

^{4}= 1, i

^{5}= i,... and it continues in this fashion for all integer powers of i.

Now we have that e^ix = (1 - x^2/2!+ x^4/4!- ((x^6)/6!)+ ...) + i (x - x^3/3!+ (x^5/5!)- x^7/7!+ ...), and these look awfully familiar. In fact if I refer you back to the Taylor series' of Cos(x) and Sin(x):

You can see that these are apparent in what we have now discovered e

^{ix }to equal. This then means that:

And hence we have our proof of Euler's formula and that e

^{ix}= Cos(x)+iSin(x).

You may have also seen Euler's formula in action as Euler's identity which is often described as the most beautifully profound equation in maths. Euler's identity is e

^{iπ }+ 1 = 0, and it is so beautiful because it incorporates the five most important numbers in maths: π, e, 1, 0 and i.

Why does e

^{iπ }+ 1 = 0? Well if we look at our proof of Euler's formula, e

^{ix}= Cos(x)+iSin(x) and we input π we get: e

^{iπ }= Cos(π)+iSin(π). Sin(π) = 0, Cos(π) = -1. Therefore e

^{iπ }= -1, so e

^{iπ }+ 1 = 0.

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