Prove Euler Identity without using Euler Formula

[maze]

But Euler's formula is right there!

Yeah you're right it would take only epsilon more effort, at any of several steps here, to get the full Euler's formula. Your post 19, for one, but you could also trivially get the full formula by noting uniqueness of unit speed parameterization up to sign and initial position.

 

[yasiru89]

Perhaps the following might help, due to John Bernoulli which we might adapt for this case;

Consider the area in the first quadrant of a unit circle centred about the origin.

$$A = \int_{0}^{1} (1 - x^{2})^{1/2} dx$$

With the change of variable [itex]u = ix[/tex] the integral is now,<br /> <br /> $$A = -i \int_{0}^{i} (1 - u^{2})^{1/2} du$$<br /> <br /> We already know that [itex]A = \frac{\pi}{4}[/tex] and by evaluating the integral [itex]A = -\frac{1}{2} i \log{i}[/tex]<br /> <br /> Equating the expressions we have,<br /> <br /> $$\frac{\pi}{2} = \frac{1}{i} \log{i} = \log{i^{\frac{1}{i}}}$$<br /> <br /> Which in turn, on taking the exponent on both sides and raising to the [itex]i[/tex] th power yields;<br /> <br /> $$e^{\frac{i \pi}{2}} = i$$<br /> <br /> Take the square on both sides, et voila,<br /> $$e^{i \pi} = -1$$ , as required![/itex][/itex][/itex][/itex]

 

[morphism]

There are a lot of things to be said about that proof. For one, how are you defining the complex logarithm and complex exponentiation?

 

[yasiru89]

Throughout the logarithm is applied only to real quantities (like [itex]i^{i}[/tex]), so simply treating it as the real logarithm whenever the argument can at least be 'made' real would be justified (of course, for this we may require an alternate proof that [itex]i^{i}[/tex] is real, finding one that doesn't rely on Euler's identity shouldn't be too hard).<br /> <br /> Then the exponential is required only as the inverse of the said logarithm and we are done.[/itex][/itex]

 

[morphism]

You're missing my point. i^i doesn't even make sense unless you define what complex exponentiation is. Same comment applies to "1/i logi = log(i^i)": this is meaningless unless you already have a complex logarithm which you know behaves like this (I'm not even going to mention branches). Remedying this will almost certainly require Euler's formula.

 

[mathwonk]

if you integrate the form dlog(z) = dz/z around the upper half of the unit circle, and get ipi, it seems to me you have proved that e^(ipi) = -1.

i.e. here e^z is defined as the inverse of the integral of dz/z.

 

[morphism]

How are you evaluating that integral?