The exponential function 𝑒𝑥𝑝(𝑥) can be expanded into a Taylor series:

• $exp(x)=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$

This is also a power series expansion of the exponential function

Proof

We start with the definition of a Taylor series

• $f(x)=\frac{f(a)(x-a{)}^0}{0!}+\frac{f^{\prime }(a)(x-a{)}^1}{1!}+\frac{f^{\prime \prime }(a)(x-a{)}^2}{2!}+\frac{f^{\prime \prime \prime }(a)(x-a{)}^3}{3!}+\frac{f^{\prime \prime \prime \prime }(a)(x-a{)}^4}{4!}+...$
• $f(x)=\frac{f(0)x^0}{0!}+\frac{f^{\prime }(0)x^1}{1!}+\frac{f^{\prime \prime }(0)x^2}{2!}+\frac{f^{\prime \prime \prime }(0)x^3}{3!}+\frac{f^{\prime \prime \prime \prime }(0)x^4}{4!}+...\text{ let }a\text{ = 0}$
• $exp(x)=\frac{exp(0)x^0}{0!}+\frac{ex{p}^{\prime }(0)x^1}{1!}+\frac{ex{p}^{\prime \prime }(0)x^2}{2!}+\frac{ex{p}^{\prime \prime \prime }(0)x^3}{3!}+\frac{ex{p}^{\prime \prime \prime \prime }(0)x^4}{4!}+...\text{ let }f(x)=exp(x)$
• $exp(x)=\frac{exp(0)x^0}{0!}+\frac{exp(0)x^1}{1!}+\frac{exp(0)x^2}{2!}+\frac{exp(0)x^3}{3!}+\frac{exp(0)x^4}{4!}+...\text{ evaluate derivatives}$
• $exp(x)=\frac{1x^0}{0!}+\frac{1x^1}{1!}+\frac{1x^2}{2!}+\frac{1x^3}{3!}+\frac{1x^4}{4!}+...$
• $exp(x)=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$