Archie Howitt analyses Euler’s Identity

Euler’s Identity

Euler’s identity is an example of mathematical beauty, as it combines some of the most fundamental and irrational numbers to form a whole answer. It combines the additive identity (0) along with the multiplicative identity (1) with the most used irrational numbers of; π (the ratio of the circumference of a circle to its diameter), e (the base of the natural logarithm) and i (the imaginary unit for the complex numbers). The equation itself is eiπ + 1 = 0   the equation can be derived from Euler’s identity (eix = cos(x) + i*sin(x)) which only works when in radians, by having x = π it will derive Euler’s identity as cos(π) = -1 and sin(π) = 0. To prove this, we need to investigate the Taylor and MacLaurin series’, these two series allow us to generalise a function, for example we could generalise f(x) = cosx. How the Taylor series works is that it suggests that you wish to put f(x) in the form of c0 + c1(x-a) + c2(x-a)2 + c3(x-a)3 … etc. where a is a constant that the series is centred around. Next we say a=x as a result we get f(a) = c0 which we can then substitute into the starting form giving              f(x) = f(a) + c1(x-a) … etc. next if we differentiate the equation we shift the equation over giving                         f’(x) = c1 + 2c2(x-a) + 3c3(x-a)2 … etc. now if we say a=x again we can find the value of c1 as f’(a). By continuing this trend we can come up with the generalisation of Taylor’s law: f(x) = f(a) + f’(a)*(x-a) + f”(a)/2!*(x-a)2 + f’”(a)/3!*(x-a)3 … etc. The reason we must divide by the factorial of the degree of the (x-a) is because when differentiated that degree will keep coming down until it is gone multiplying the component by its factorial. Now we can give a a value and this value will be where the equation is centred around, the MacLaurin series is the Taylor series when a=0, these series are only valid for numbers within a given range around a, luckily for us π is within this range allowing us to use these series to derive Euler’s formula. First we find the MacLaurin series for cos(x), sin(x), and ez. Let’s start with cos(x), cos differentiates into -sin, and sin differentiates into cos, as a result we get a pattern of: f(x)=cos(x), f’(x)=-sin(x), f”(x)=-cos(x), f”’(x)=sin(x) then it repeats, now as    cos(0) = 1 and sin(0)=0 we can complete the MacLaurin series as follows, cos(x) = 1 – x2/2! + x4/4! – x6/6! … etc. next we can go onto the series of sin(x), f(x)=sin(x), f’(x)=cos(x), f”(x)=-sin(x), f”’(x)=-cos(x), repeat. Therefore f(a) = 0, f’(a) = 1, f”(a) = 0, f”’(a) = -1, with these values we can form the MacLaurin series of sin(x) as:                   sin(x) = x – x3/3! + x5/5! – x7/7!… etc. Finally the MacLaurin series of ez is shown as ez = 1 + z + z2/2! + z3/3! … etc. This is due to the fact that ez will differentiate to ez. Now we can substitute in ix = z, the series becomes             eix = 1 + ix + (ix)2/2! + (ix)3/3! … etc. We know i2 = -1 therefore we can simplify this expression to be:                              eix = 1 + ix – x2/2! – ix3/3! + x4/4! + ix5/5! … etc. which can be simplified to: 1 – x2/2! + x4/4! + …. + i(x – x3/3! + x5/5! – …. ), now we can form a simultaneous equation with these 3 series allowing us to form the expression   eix = cos(x) + i*sin(x), this is Euler’s formula which can easily form Euler’s identity as I did at the start by substituting in x=π. It is important to note that these series only work because sin and cos of x converge for all real x and ez converges for pure-imaginary z.

References:

https://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-ei-varphi-cos-varphi-i-sin-varphi