# Euler's Formula Proof

1. Mar 7, 2015

### Astudious

Can someone point me to a good, rigorous proof of Euler's formula (e^(ix) = cos(x) + i*sin(x)), starting from the definitions of sin(x) and cos(x) using the triangle (sin(x) = Opposite / Hypotenuse, cos(x) = Adjacent / Hypotenuse) and the definition of e^x as either Bernoulli's number (from the limit) or as the non-trivial self-differentiating function, and then proving everything from there?

Pythagoras' formula can be assumed. Basic common-knowledge trig identities can be assumed. But these are only because I already know how to derive them! I want to be able to write a rigorous proof.

Before someone proposes the Taylor series proof: fine. I can prove it using the Taylor series. But that is not how I wish to define sin(x) and cos(x) or e^x, so if you want to do it like that, you need to first give me a rigorous proof of Taylor's theorem from first-principles! I suspect there are easier ways of proving Euler's formula?

2. Mar 7, 2015

3. Mar 8, 2015

### HomogenousCow

What's wrong with that exactly? Ultimately, all proofs go back to the similarity in differentiation structure between the exponential function and the trig functions

4. Mar 27, 2015

### Astudious

1) My goal here is to try and follow a chain of reasoning which Euler might have used and understand how the theorem really arises from the proof. It would be both trivial and unhelpful to prove anything here with Taylor's theorem.

2) Is the easiest or most straightforward way to prove Euler's theorem from first-principles, really to prove Taylor's theorem first?

3) Let us say we can prove Taylor's theorem. How can we use it here? The power-series of e^x can be easily given from the starting points I suggested, but I cannot see how to prove anything about the differentiability of sin(x) or cos(x) from the definitions I have given them. Indeed, you need to define sin(x) = 1/(2i) * (e^(ix)-e^(-ix)) and cos(x) = 1/2 * (e^(ix) + e^(-ix)), formulae which themselves come only AFAIK from Euler's (and whose derivation trivialises Euler's), to prove that d/dx(sin(x)) = cos(x). This, of course, we cannot use given the set-up of the problem.

5. Mar 29, 2015

### Matternot

1) The chain of reasoning Euler used uses Taylor's theorem. Why would it be unhelpful?

2) yes.

3) It is easy to prove from first principles that the derivative of sin(x) is cos(x) and vice versa. This can be done by using geometrical methods to show sin(x)/x goes to 1 from small x. From this, and Taylor's theorem, It can be shown that the infinite power series of cos(x) and sin(x) converge and are equal to what you expect. This means the functions are EQUAL to their power series. They can be interchanged at free will. ex from the definition has derivative ex and therefore power series (1+x+x2/2 +...). From this, Euler's formula can be shown by comparing power series. This is fully rigorous and is considered the most simple way of showing it.

later, to make the function sin(x) and cos(x) more immediately usable, their definitions were replaced by -isinh(ix) and cosh(ix) respectively. These are now the definitions of sin and cos.

Last edited: Mar 29, 2015
6. Mar 29, 2015

### Astudious

Thank you.

For readers' reference, I satisfied myself by "using geometrical methods to show sin(x)/x goes to 1 from small x" using this page: http://www.themathpage.com/aCalc/sine.htm. Once d/dx(sin(x)) = cos(x) has been established and d/dx(cos(x)) = -sin(x), then Euler's formula follows readily from Taylor's theorem (for which there are proofs on Wikipedia).

7. Mar 29, 2015

### Matternot

8. Mar 29, 2015

### WWGD

I don't know if this is what you're looking for, but here is another approach: you can look at $e^{i\theta}$ as an instance of $e^z$, where z= $i\theta$. Here $e^z$ can be seen as the (local) inverse of the function logz. The function logz takes in a (non-zero) complex number , given in polar form as input and returns the (ln of) the radius and the angle with the origin ( once you choose a branch of logz, which we assume here , the angle $\theta$ is well-defined ). Then $e^z$ as the local inverse takes in an expression $ln(r)+ i\theta$ and returns the number in polar form that is represented by the pair $lnr, \theta$ . Then $e^{i\theta}$ is the number in polar form corresponding to the pair: (lnr=0 , \theta). Now, if lnr=0 , we know that r=1. Then $e^{i\theta}$ is the complex number given in polars with radius 1 and angle $\theta$ , and this is precisely the complex number given by $cos\theta + isin \theta$ .

9. Mar 29, 2015

### Matternot

Not quite sure how you're justifying ln(z) = ln(r)+iθ

10. Mar 29, 2015

### WWGD

I am starting with it as a definition. I need to start somewhere with definitions. Then I use that logz , for a chosen branch, is a local inverse of $e^z$.

11. Apr 1, 2015

### Astudious

I suggested the starting points for definitions in the OP. Assuming your definition is scarcely different from assuming Euler's formula, and seems to come from it in common practice rather than the other way round, if I understand right. Or is there some geometric way to arrive at that (from the starting points I suggested in the first post)?

Edit: The definition of a complex number should, I think, be taken as z = a + ib where i^2=-1 and a and b are real. Using a definition closer to Euler's feels like cheating.

12. Apr 1, 2015

### WWGD

No, I started by assuming the definition of the complex log, logz=ln(r)+#i\ theta#, then use that logz has a local inverse which coincides with #e^z# .But I don't see how to derive it geometrically.

logz is a function that assigns to a complex number the log of its radius and (one of its) argument(s), i.e., logz=lnr+i# \ theta # . And logz is locally- invertible (after a fixed choice of branch ) , and its inverse coincides with #e^z #; the inverse of a function that takes the lnr+ i# \ theta # is a function that takes # \ theta # (we are working with r=0, so we can ignore the ln(r) part ) and assigns to it a complex number whose argument is # \ theta # . This is precisely a function defined as #e^(x+iy)=e^x(cosx+isiny)#. So I only use that logz (fixed branch) is locally invertible, and this can be shown by, e.g., the inverse function theorem. And its inverse is a function that "undoes" what logz does, so it coincides with #e^z#.

Last edited: Apr 1, 2015
13. Apr 2, 2015

### HallsofIvy

You certainly can't because this is an insufficiently general definition! For one thing, x would have to be between 0 and 90 degrees (or 0 and $\pi/2$ radians while in Euler's formula x must be any number (as well as not having units such as "degrees" or "radians".

14. Jan 3, 2016

### matica

15. Jan 4, 2016

### Samy_A

16. Jan 4, 2016

### matica

Thanks Samy_A. Typo corrected. It now says
$\begin{equation*} \lim_{t\to 0}\frac{1-\cos t}t=0 \end{equation*}$
It had no effect on the calculations afterward. since only this line was mistyped, and the correct relation was used.

Last edited: Jan 4, 2016