__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 e^{i}^{π} + 1 = 0 the equation can be derived from Euler’s identity (e^{ix} = 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 c_{0} + c_{1}(x-a) + c_{2}(x-a)^{2} + c_{3}(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) = c_{0} which we can then substitute into the starting form giving f(x) = f(a) + c_{1}(x-a) … etc. next if we differentiate the equation we shift the equation over giving f’(x) = c_{1} + 2c_{2}(x-a) + 3c_{3}(x-a)^{2} … etc. now if we say a=x again we can find the value of c_{1} 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 e^{z}. 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 – x^{2}/2! + x^{4}/4! – x^{6}/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 – x^{3}/3! + x^{5}/5! – x^{7}/7!… etc. Finally the MacLaurin series of e^{z} is shown as e^{z} = 1 + z + z^{2}/2! + z^{3}/3! … etc. This is due to the fact that e^{z} will differentiate to e^{z}. Now we can substitute in ix = z, the series becomes e^{ix} = 1 + ix + (ix)^{2}/2! + (ix)^{3}/3! … etc. We know i^{2} = -1 therefore we can simplify this expression to be: e^{ix} = 1 + ix – x^{2}/2! – ix^{3}/3! + x^{4}/4! + ix^{5}/5! … etc. which can be simplified to: 1 – x^{2}/2! + x^{4}/4! + …. + i(x – x^{3}/3! + x^{5}/5! – …. ), now we can form a simultaneous equation with these 3 series allowing us to form the expression e^{ix} = 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 e^{z} converges for pure-imaginary z.

**References:**

- https://en.wikipedia.org/wiki/Euler%27s_identity#:~:text=%CF%80%20is%20pi%2C%20the%20ratio,most%20fundamental%20numbers%20in%20mathematics.
- https://medium.com/swlh/eulers-identity-990b52f0d8fc
- https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:complex/x2ec2f6f830c9fb89:complex-plane/a/the-complex-plane
- https://www.dsprelated.com/freebooks/mdft/Proof_Euler_s_Identity.html
- https://ccrma.stanford.edu/~jos/mdft/Real_Exponents.html