Proving Euler's Formula

**anomaly** · November 7th 2011, 10:32 AM

e^(iΘ) = cos (Θ) + i sin (Θ)

Typically the proof involves using the summation e^x = Sum (x^n / n!) from 1 to infinity. Then you set x = iΘ, and separate the summation into two halves: cosine and sine.

The part I never understood is, how do you establish that the summation e^x = Sum (x^n / n!) holds for complex values of x? The proofs I've seen just assume that it does. Is there a way to develop summation definition for C?

**Amer** · November 7th 2011, 11:02 AM

the taylor series of any function f is

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} \cdot (x-a)^n$

you can check

Taylor series - Wikipedia, the free encyclopedia

**Siron** · November 7th 2011, 11:08 AM

You can also try to prove it with induction.

**anomaly** · November 8th 2011, 11:29 AM

How would I approach the induction proof?

I'm asking this because typically when people get introduced to Euler's formula proof, they've been dealing with real numbers and are used to working with series of all real-number quantities and variables. It looks questionable when you suddenly plug in complex numbers to the series and expect it to work.

**Amer** · November 8th 2011, 09:38 PM

The Taylor series of a real or complex function ƒ(x) that is infinitely differentiable in a neighborhood of a real or complex number a is the power series

$f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots.$

from WIKI

Taylor series for complex and real function dose not matter so

$e^{i\theta} = \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!}=1 + \frac{i\theta}{1} + \frac{-\theta ^2}{2!}+ + \frac{-i\theta ^3}{3!}+\frac{\theta ^4}{4!} + \frac{i\theta ^5}{5!} +...$

i terms together and without i together

$\sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} = i \sum_{n=0}^{\infty} \frac{\theta ^{2n+1}}{(2n+1)!} + \sum_{n=0}^{\infty} \frac{\theta ^{2n}}{(2n)!}$

it is clear that the term with i will be sin theta and without will be cosine

**Prove It** · November 8th 2011, 09:52 PM

Originally Posted by anomaly

e^(iΘ) = cos (Θ) + i sin (Θ)

Typically the proof involves using the summation e^x = Sum (x^n / n!) from 1 to infinity. Then you set x = iΘ, and separate the summation into two halves: cosine and sine.

The part I never understood is, how do you establish that the summation e^x = Sum (x^n / n!) holds for complex values of x? The proofs I've seen just assume that it does. Is there a way to develop summation definition for C?

Here's how I prove it.

Let $\displaystyle z = \cos{\theta} + i\sin{\theta}$ . Then

$\displaystyle \begin{align*} \frac{dz}{d\theta} &= -\sin{\theta} + i\cos{\theta} \\ \frac{dz}{d\theta} &= i^2\sin{\theta} + i\cos{\theta} \\ \frac{dz}{d\theta} &= i\left(\cos{\theta} + i\sin{\theta}\right) \\ \frac{dz}{d\theta} &= i\,z \\ \frac{1}{z}\,\frac{dz}{d\theta} &= i \\ \int{\frac{1}{z}\,\frac{dz}{d\theta}\,d\theta} &= \int{i\,d\theta} \\ \int{\frac{1}{z}\,dz} &= i\,\theta + C_1 \\ \log{z} + C_2 &= i\,\theta + C_1 \\ \log{z} &= i\,\theta + C \textrm{ where }C = C_1 - C_2 \\ z &= e^{i\,\theta + C} \\ z&= e^{i\,\theta}e^C \\ z &= A\,e^{i\,\theta} \textrm{ where }A = e^C \\ \cos{\theta} + i\sin{\theta} &= A\,e^{i\,\theta} \end{align*}$

Now if we let $\displaystyle \theta = 0$ we find

$\displaystyle \begin{align*} \cos{0} + i\sin{0} &= A\,e^{0i} \\ 1 &= A \end{align*}$

Therefore

$\displaystyle \begin{align*} \cos{\theta} + i\sin{\theta} &= e^{i\theta} \\ r\left(\cos{\theta} + i\sin{\theta}\right) &= r\,e^{i\theta} \end{align*}$

**anomaly** · November 8th 2011, 10:08 PM

But that, again, assumes that raising e to something imaginary is valid in the first place.

What if you've been just introduced to complex numbers and are trying to make sense out of this concept? Then the first thing you would probably prove is e^(iΘ) = cos (Θ) + i sin (Θ).

Originally Posted by Amer

Taylor series for complex and real function dose not matter so

$e^{i\theta} = \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!}=1 + \frac{i\theta}{1} + \frac{-\theta ^2}{2!}+ + \frac{-i\theta ^3}{3!}+\frac{\theta ^4}{4!} + \frac{i\theta ^5}{5!} +...$

i terms together and without i together

$\sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} = i \sum_{n=0}^{\infty} \frac{\theta ^{2n+1}}{(2n+1)!} + \sum_{n=0}^{\infty} \frac{\theta ^{2n}}{(2n)!}$

it is clear that the term with i will be sin theta and without will be cosine

I guess I would like to see how the concept of series can be developed in terms of complex numbers.

**Prove It** · November 8th 2011, 10:13 PM

Originally Posted by anomaly

But that, again, assumes that raising e to something imaginary is valid in the first place.

Which it is...

**Deveno** · November 9th 2011, 12:07 AM

Originally Posted by anomaly

e^(iΘ) = cos (Θ) + i sin (Θ)

Typically the proof involves using the summation e^x = Sum (x^n / n!) from 1 to infinity. Then you set x = iΘ, and separate the summation into two halves: cosine and sine.

The part I never understood is, how do you establish that the summation e^x = Sum (x^n / n!) holds for complex values of x? The proofs I've seen just assume that it does. Is there a way to develop summation definition for C?

one establishes convergence for complex power series in the same way that one does for real series:

a power series $\sum_{k=1}^{\infty}a_kz^k$ converges iff the sequence of partial sums $\{S_n = \sum_{k=1}^n a_kz^k\}$ converges as a sequence.

a complex sequence $\{S_n\}$ converges to a limit L iff for every $\epsilon > 0$ , there exists $N \in \mathbb{Z}^+$ such that :

$|S_n-L| < \epsilon$ for all n > N. the only difference here is that $|z| = |a+bi| = \sqrt{a^2+b^2}$ instead of $|x| = \sqrt{x^2}$ ,

we measure distance between complex numbers a little differently than distance between real numbers, instead of "intervals" we use "disks" (so intsead of having an interval of convergence, we have a radius of convergence).

for the particular power series:

$\sum_{k=1}^{\infty}\frac{z^k}{k!}$

it turns out that this converges for every complex number z (the radius of convergence is infinite), so we can assign the limit of this series to be the image of a function, f(z).

of course, on the real-axis, this obviously converges to $e^x$ , so it is natural to define this function f to be the complex exponential.

glossing over some technicalities, here, one can show that not only is this function f continuous, it is also infinitely-differentiable. one can (but i will not) show that for a convergent-everywhere complex power series, term-by-term differentiation is justified; doing so, we find that f'(z) = f(z), that is, not only does the extension of the real exponential to complex values behave well with respect to convergence, it also behaves the same under differentiation (the complex derivative is defined in exactly the same way as the real derivative, since complex numbers form a field, all the operations involved in the definition of the derivative still "make sense").

at this point, you can see that one can limit z to numbers of the form 0+iy, which then quickly gives euler's identity (since we wind up with two real series, one of which is multiplied by i, giving a complex series).

if you wish to prove this for yourself, a good first place to start, is by verifying that the triangle inequality still holds, when absolute value is replaced by the complex norm (or modulus). this will go a long way to convincing you many of the methods used in "epsilon-delta" proofs for real numbers, carry over to complex numbers with little or no modification.

**Siron** · November 9th 2011, 06:52 AM

Proof: $e^{i\theta}=\cos(\theta)+i\sin(\theta)$
By induction.
Firs step: take $\theta=0$ therefore:
$e^{0}=\cos(0)+i\sin(0) \Rightarrow 1=1$
So the statement is true for $\theta=0$

Induction step: Suppose the statement is true for $\theta=k$ , therefore we suppose this statement is true:
$e^{i\cdot k}=\cos(k)+i\sin(k)$

We have to proof it's true for $\theta=k+1$ :
$e^{i(k+1)}=\cos(k+1)+i\sin(k+1)$
We can write:
$e^{i(k+1)}=e^{i\cdot k+i}=e^{i\cdot k}\cdot e^{i}=[\cos(k)+i\sin(k)]\cdot [\cos(1)+i\sin(1)]=\cos(k)\cos(1)+i\sin(k)\cos(1)+\cos(k)[i\sin(1)]-\sin(k)\sin(1)=[\cos(k)\cos(1)-\sin(k)\sin(1)]+i[\sin(k)\cos(1)+\cos(k)\sin(1)]=\cos(k+1)+i\sin(k+1)$

Which we wanted to prove, therefore the statement is true.

I'm sure there're much better proves then this one, but I think it's an easy way to prove it.

**anomaly** · November 9th 2011, 08:25 AM

@Prove It: I mean that I want to prove that e^(imaginary) actually works before I go on any further. When students first learn complex variables, they don't know what e^(imaginary) means. They start with e^(iΘ) because it's the most accessible. But you have to show that it's meaningful.

A proof that uses log z would come after e^(imaginary). Ordinarily students first learn the concept of e^(imaginary) before learning about log z, unless there's another way to establish logs first.

@Deveno: Thank you. That's what I meant, a more rigorous development of series or whatever you would need for a proof. I'll read that and try to understand it.

@Siron: I guess that works, though I prefer a proof that doesn't rely on complex exponents until it's well-defined, which is why in the first post, I was asking for a way to build up series in C.

Essentially I want to assume that I don't know anything about complex numbers other than a+bi and z^n = z * z * ... * z.

Thread: Proving Euler's Formula

LinkBack

Thread Tools

Display

Proving Euler's Formula

Re: Proving Euler's Formula

Re: Proving Euler's Formula

Re: Proving Euler's Formula

Re: Proving Euler's Formula

Re: Proving Euler's Formula

Re: Proving Euler's Formula

Re: Proving Euler's Formula

Re: Proving Euler's Formula

Re: Proving Euler's Formula

Re: Proving Euler's Formula

Similar Math Help Forum Discussions

Euler's formula

Euler's Formula

Euler Formula..

Euler's Formula

Euler's formula

Search Tags