The Euler’s Formula 𝑒^𝑖𝑥 can be expanded into a Taylor series:

• $e^{ix}=\frac{x^0}{0!}+\frac{i{x}^1}{1!}-\frac{x^2}{2!}-\frac{i{x}^3}{3!}+\frac{x^4}{4!}+\frac{i{x}^5}{5!}-\frac{x^6}{6!}...$

This is also a power series expansion of the Euler’s equation

Proof

Let’s consider the exponential function:

• 𝑒^𝑧

The power series expansion of 𝑒𝑧 is:

• $e^z=\frac{z^0}{0!}+\frac{z^1}{1!}+\frac{z^2}{2!}+\frac{z^3}{3!}+...$

Let 𝑧 = 𝑖𝑥:

• $e^{ix}=\frac{(ix{)}^0}{0!}+\frac{(ix{)}^1}{1!}+\frac{(ix{)}^2}{2!}+\frac{(ix{)}^3}{3!}+...$

Extracting the powers of 𝑖, we get:

• $e^{ix}=\frac{x^0}{0!}+\frac{i{x}^1}{1!}-\frac{x^2}{2!}-\frac{i{x}^3}{3!}+\frac{x^4}{4!}+\frac{i{x}^5}{5!}-\frac{x^6}{6!}...$