What's the simplest way to prove Euler's formula? Complex number retard here

What's the simplest way to prove Euler's formula? Complex number retard here

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literally just expand the taylor series you ingrate

If you expand exp(x) in a Taylor series, then plug in x = i*pi, then you get -1

Find the Euler's Formula page on Wikipedia. The Power Series proof under the "Proofs" section is as simple as it gets. You don't need to understand the Maclaurin series, just know that it works. Euler's identity is proved when you substitute pi for x.

Just give up. If it wasn't obvious to you from the start you'll never be a first class mathematician.

Its mostly true by definition.
Sin and cos are defined by using the exp() function you can just plug it in.

>Ingredients:

-Index laws
--e^a.e^b=e^(a+b) even if a and b are complex.
--e^0=1

-A function that commutes with complex conjugation is a generalisation of a real-valued function
--Any function f of complex numbers that obeys f(z*)=(f(z))* is a function that restricts to the real numbers; real number input means real number output.

-The distance of a complex number z from the origin is computable via zz*

-Some small amount of differentiation and integration
--Derivatives of trig functions
--Implicit differentiation
--Integration by separating variables


Cooking:

e^ix.e^-ix=e^(ix-ix)=e^0=1

Now since
(e^ix)*=e^(ix*)=e^(-ix)

we have calculated zz* for z=e^(ix). So z lies somewhere on the unit circle in the complex plane. This means for some value y,

e^ix=cosy+isiny

Now we differentiate this equation wrt x:
i.e^ix = dy/dx.(-siny) + dy/dx(icosy)

Factorise and divide by i:
e^ix = dy/dx (icosy-siny)/i

Times by i/i and cancel negatives:
e^ix = dy/dx (cosy+isiny)

We can now cancel on both sides to give
1=dy/dx

So dx=dy, which integrates to x=y+C. We know that when x=0, e^ix=1, so this forces C=0.

Hence

e^ix=cosx+isinx


Tasty result:

e^ipi=cospi+isinpi=-1.


Proofs with Taylor series in them are lazy, ugly, and stupid. I bet you're all engineers.

Mr. Gauss, thank you for joining us.

Another way to go about it without using taylor series is through simple differential equations as follows:

nice

you can get to the result from the third line. the rest is a bit of unnecessary wankery

For a more intuitive approach, observe that the derivative of y=e^(i*theta) is iy, which is just y rotated 90 degrees, and that the initial condition e^(i*0) is 1, so we can conclude that the function draws out a circle (because the tangent is perpendicular to the radius) with radius 1.
Then we can conclude that since |y| is always equal to 1, |dy/dx| is always equal to 1, since it's just y rotated.
Thus, as theta increases linearly, y draws out a circular arc with a constant rate of 1, so the distance y travels around this arc is equal to theta. Setting theta to pi, we see y travels a distance of pi around the unit circle in the complex plane, thus placing it at -1.

what's * ?

>Calling other proofs lazy when you abuse notation with "dy/dx = 1 implies dy = dx"
You can't make this shit up

but dy=(dy/dx)dx so if dy/dx=1 then it does imply that dy=dx

No, that's laziness and abuse of notation

You might be able to create a less nice geometric proof if you want to aviod using tayler-series. Not sure how basic you want this proof to be or what assumtions you would start with, but it should be easy enough to show that e^{i\pi} * complex conjugate (e^{-i\pi}) is a norm anb thus that |e^{i\pi}|=1 (so that e^{i\pi} lies on the unit circle around 0 in the complex plane). From there on ist just some geometry to define the real and the complex part of the triangle (with the line from 0 to e^{i\pi} as hypotenuse) as Sin and Cos or maybe just check that the real and complex part are in fact cos and sin...
Its kinda late right now so I cut it short: construc a geometric proof that e^{i\phi} = Cos(\phi) + i Sin(\phi) and then use that (Cos(\phi)=-1)
But maybe this doenst work this simple and I wasted the last 3 minutes... in any case: good night guys.

geometrically in the complex plane

>take out graphing calculator
>enter e^(I*π)+1
>result is 0

Why do mathfags over think everything? Serious question. Were you hoping for some kind of proof? Give me a real world scenario outside of a school assignment where writing a formal proof for that equation is of any use.

>abuse of notation
Well meme'd my friend

>No, that's laziness and abuse of notation

I've never seen that issue discussed in a math textbook, nor mentioned by any of my math professors. (I took all the standard engineering calc, up to diff.eq.)

Can you point to a web page that explains why some mathematicians feel that "dy/dx=1 implies dy=dx" (or similar shortcuts) is lazy notation and should be avoided?

because dx is not a real number

Yes it is. It's an infinitesimal number. And it's in the Real line.

I'm guessing he means complex conjugate. It's usually represented by a bar over the variable.
(a+bi)*=a-bi

...

> Infinitesimals are real numbers
> What is the Archimedean principle

Engineers shouldn't be allowed on Veeky Forums

No. It is real.

It's a unit pointer in the complex plane, starting at 1 and going counterclockwise. Pi is half a circle rotation, so from 1 to -1, thus it equals 0.

t. Engineer

Are you serious? fucking retard

dy/dx is an operator. Like +,-,÷,×.
Babbys first calculus shows the proof of using the operator correctly in succession, but it's a waste of space so we treat them as terms with context in practice.

youtube.com/watch?v=F_0yfvm0UoU

to thank me, i have a TREE(3) word essay on inductors due in 1 planck time and haven't even started yet, pic related

My pleasure. By the way, it's spelled GauB.

>I took all the standard engineering calc
Well, that explains it. It's not even the fact that engineers are shit at maths that bothers me. It's that they're not even aware that they're shit.

dear anons please help your threadly leech, thanks

>It's an infinitesimal number
You would think people this retarded wouldn't be attracted to a science/mathematics discussion board.

thanks

>engineering
there's your problem. an engineer, of all of people, should be careful of calling others lazy. of course, you really don't need to know what's going on in the background, but if you don't you have no right to claim superiority.
dx is not a real number, just like infinity is not a real number. moreover, dy/dx is a (linear) operator, not a fraction or a number.

I would be quite interested in your definition of "dx" and "dy", user.

There's a video on YouTube that explain it graphically. Can't remember the name, though

Let h denote the "infinitesimal" number, such that h is real and 0

[math] e^{ix}= lim_{n- \infty} { \left 1+ \frac{ix}{n} \right }^n [/math]

the argument of each term is
[math] n.arctan \left \frac{x}{n} \right [/math]
it tends to x as n tends to infinity
the modulus
[math] {1+x^2/n^2}^\frac{n}{2} [/math]
tends to 1
if I fucked up the latex imma off myself

>tfw engineer
>all of my math courses were very rigorous, with plenty of proofs
>still treated like a cum lover brainlet even though physicists where the ones doing bullshit like multiplying by dx and other garbage

Proof : its trivial and left as an exercise for the reader

On a more serious note, the Taylor is probably your best bet or alternatively if you have already proved exp(iθ)=cosθ + sinθ you can just plug in θ=π