Euler's identity

Many equations can be written as a series of terms added together. This is called a Taylor series.

The exponential function ${\displaystyle e^{x}}$ can be written as the Taylor series

${\displaystyle e^{x}=1+x+{x^{2} \over {2!}}+{x^{3} \over {3!}}+{x^{4} \over {4!}}\cdots =\sum _{k=0}^{\infty }{x^{n} \over n!}}$

As well, the sine function can be written as

${\displaystyle \sin {x}=x-{x^{3} \over 3!}+{x^{5} \over 5!}-{x^{7} \over 7!}\cdots =\sum _{k=0}^{\infty }{(-1)^{n} \over (2n+1)!}{x^{2n+1}}}$

and cosine as

${\displaystyle \cos {x}=1-{x^{2} \over 2!}+{x^{4} \over 4!}-{x^{6} \over 6!}\cdots =\sum _{k=0}^{\infty }{(-1)^{n} \over (2n)!}{x^{2n}}}$

Here, we see a pattern take form. ${\displaystyle e^{x}}$ seems to be a sum of sine and cosine's Taylor series, except with all of the signs changed to positive. The identity we are actually proving is ${\displaystyle e^{ix}=\cos(x)+i\sin(x)}$ .

So, on the left side is ${\displaystyle e^{ix}}$ , whose Taylor series is ${\displaystyle 1+ix-{x^{2} \over 2!}-{ix^{3} \over 3!}+{x^{4} \over 4!}+{ix^{5} \over 5!}\cdots }$

We can see a pattern here, that every second term is i times sine's terms, and that the other terms are cosine's terms.

On the right side is ${\displaystyle \cos(x)+i\sin(x)}$ , whose Taylor series is the Taylor series of cosine, plus i times the Taylor series of sine, which can be shown as:

${\displaystyle (1-{x^{2} \over 2!}+{x^{4} \over 4!}\cdots )+(ix-{ix^{3} \over 3!}+{ix^{5} \over 5!}\cdots )}$

if we add these together, we have

${\displaystyle 1+ix-{x^{2} \over 2!}-{ix^{3} \over 3!}+{x^{4} \over 4!}+{ix^{5} \over 5!}\cdots }$

Therefore,

${\displaystyle e^{ix}=\cos(x)+i\sin(x)}$

Now, if we replace x with ${\displaystyle \pi }$ , we have:

${\displaystyle e^{i\pi }=\cos(\pi )+i\sin(\pi )}$

Since we know that ${\displaystyle \cos(\pi )=-1}$ and ${\displaystyle \sin(\pi )=0}$ , we have:

• ${\displaystyle e^{i\pi }=-1}$
• ${\displaystyle e^{i\pi }+1=0}$

which is the statement of Euler's identity.

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• Euler's formula