# What is Euler's identity really saying?

1. May 4, 2012

So it is true that ei∏+1=0. But what does this mean? Why are all these numbers linked?

2. May 4, 2012

### DonAntonio

They are linked precisely by that equation, and since the equality $e^{i\theta}:=\cos\theta+i\sin\theta\,\,,\,\,\theta\in\mathbb{R}\,\,$ follows at once say from the definition

of the complex exponential function as power series (or as limit of a sequence), the above identity is really trivial.

DonAntonio

3. May 4, 2012

### HallsofIvy

Look at the MacLaurin series for those functions:
$$e^x= 1+ x+ x^2/2!+ x^3/3!+ \cdot\cdot\cdot+ x^n/n!$$
$$cos(x)= 1- x^2/2!+ x^4/4!- x^6/6!+ \cdot\cdot\cdot+ (-1)^nx^{2n}/(2n)!$$
$$sin(x)= x- x^3/3!+ x^5/5!- x^7/7!+ \cdot\cdot\cdot+ (-1)^nx^{2n+1}/(2n)!$$

If you replace x with the imaginary number ix (x is still real) that becomes
$$e^{ix}= 1+ ix+ (ix)^2/2!+ (ix)^3/3!+ \cdot\cdot\cdot+ (ix)^n/n!$$
$$e^{ix}= 1+ ix+ i^2x^2/2!+ i^3x^3/3!+ \cdot\cdot\cdot+ i^nx^n/n!$$

But it is easy to see that, since $i^2= -1$, $(i)^3= (i)^2(i)= -i$, $(i)^4= (i^3)(i)= -i(i)= -(-1)= 1$ so then it starts all over: $i^5= (i^5)i= i$, etc. That is, all even powers of i are 1 if the power is 0 mod 4 and -1 if it is 2 mod 4. All odd powers are i if the power is 1 mod 4 and -i if it is 3 mod 4.

$$e^{ix}= 1+ ix- x^2/2!- ix^3/3!+ \cdot\cdot\cdot$$

Separating into real and imaginary parts,
$$e^{ix}= (1- x^2/2!+ x^4/4!- x^6/6!+ \cdot\cdot\cdot)+ i(x- x^3/3!+ x^5/5!+ \cdot\cdot\cdot)$$
$$e^{ix}= cos(x)+ i sin(x)$$

Now, take $= \pi$ so that $cos(x)= cos(\pi)= -1$ and $sin(x)= sin(\pi)= 0$ and that becomes
$$e^{i\pi}= -1$$
or
$$e^{i\pi}+ 1= 0$$

I hope that is what you are looking for. Otherwise, what you are asking is uncomfortably close to "number mysticism".

Last edited by a moderator: May 4, 2012
4. May 5, 2012