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Euler's formula.

Anonymous Sat 25 Dec 03:42:19 2021 No.14007310 View Reply Original Report

Quoted By: >>14007315 >>14007355 >>14007398 >>14007491 >>14007550 >>14007569 >>14007952 >>14008020 >>14009345 >>14011353 >>14013386

How the fuck can I geometrically understand this formula?
Where does the $e$ come from and why does it relate to $/sqrt{-1} = i$ ?

Anonymous

Anonymous Sat 25 Dec 2021 03:44:14 No.14007315 Report

Quoted By:

>>14007310
I have no idea

Anonymous

Anonymous Sat 25 Dec 2021 03:59:05 No.14007355 Report

Quoted By: >>14007488

>>14007310
Your picture is exactly how to geometrically understand it
To algebraically understand it, Taylor series

Anonymous

Anonymous Sat 25 Dec 2021 04:21:51 No.14007398 Report

Quoted By: >>14007488

>>14007310
By reincarnating with more iq points

Anonymous

Anonymous Sat 25 Dec 2021 04:46:24 No.14007464 Report

Quoted By: >>14007488

If you plug ix into the infinite series for e you get the infinite series for cosine plus i times the infinite series for sine.

Anonymous

Anonymous Sat 25 Dec 2021 04:53:00 No.14007488 Report

Quoted By: >>14007502 >>14007503 >>14007614 >>14007620 >>14007634

>>14007355
>>14007398
>>14007464
you guys don't get it.
I can plug and chug all day long, but if you don't VISUALIZE it then you don't really understand it.
I wanna know why cosx + i*sinx = e^ix
I wanna geometrically visualize it.

Anonymous

Anonymous Sat 25 Dec 2021 04:54:46 No.14007491 Report

Quoted By: >>14007495 >>14007514 >>14007517

>>14007310
Come back after you've taken Calc 2 otherwise ifs hopeless to attempt to explain it to you

Anonymous

Anonymous Sat 25 Dec 2021 04:55:50 No.14007495 Report

Quoted By: >>14007518

>>14007491
i can visualize derivatives and integrals better than you

Anonymous

Anonymous Sat 25 Dec 2021 04:57:11 No.14007502 Report

Quoted By:

>>14007488
it's a circle in the complex plane

Anonymous

Anonymous Sat 25 Dec 2021 04:57:18 No.14007503 Report

Quoted By: >>14012639

>>14007488
Visualize a bunch of letters flying around and then gathering into that formula

Anonymous

Anonymous Sat 25 Dec 2021 04:59:30 No.14007514 Report

Quoted By: >>14007518

>>14007491
Cool so you don't need our help then. Google your question and read an article

Anonymous

Anonymous Sat 25 Dec 2021 05:00:40 No.14007517 Report

Quoted By:

>>14007491
this is the best answer. OP asked where e comes from and how it relates to i. the answer is algebrical, not geometrical. figure is 100% all you need for geometrical understanding.

Anonymous

Anonymous Sat 25 Dec 2021 05:00:42 No.14007518 Report

Quoted By:

>>14007514
Quoted myself, meant to quote
>>14007495

Anonymous

Anonymous Sat 25 Dec 2021 05:09:31 No.14007550 Report

Quoted By: >>14007554 >>14007568

>>14007310
$e^{x} = \sum_{n=0} ^{\infty}$
$e^{x} = \frac {x^{0}}{0!} + \frac {x^{1}}{1!} + \frac {x^{2}}{2!} + \frac {x^{3}}{3!} + \frac {x^{4}}{4!} + \cdots$
Substitute $ix$ for $x$ . You notice that $(ix)^{n} = (i^{n}) (x^{n})$ , which means that you can do the following:
$e^{ix} = \frac {i^{0} x^{0}}{0!} + \frac {i^{1} x^{1}}{1!} + \frac {i^{2} x^{2}}{2!} + \frac {i^{3} x^{3}}{3!} + \frac {i^{4} x^{4}}{4!} + \cdots$ .
Anyway, it turns out that this is equal to the power series for cosx + the power series for i*sinx. So, that's the simple explanation.
There are more advanced questions revealed by this, however:
Why does e^x have the power series that it does?
Why does sinx have the power series that it does?
Why does cosx have the power series that it does?

I've got a master's degree in math and I don't know the answer to those three questions, probably because I specialised in combinatorics and got a paying job.

Anonymous

Anonymous Sat 25 Dec 2021 05:11:18 No.14007554 Report

Quoted By: >>14007649

>>14007550
>>14007550
>>14007550
That whole post is garbage because of a huge typo I made, here's the fixed version:

$e^{x} = \sum_{n=0} ^{\infty} \frac {x^{n}{n!}$
$e^{x} = \frac {x^{0}}{0!} + \frac {x^{1}}{1!} + \frac {x^{2}}{2!} + \frac {x^{3}}{3!} + \frac {x^{4}}{4!} + \cdots$
Substitute $ix$ for $x$ . You notice that $(ix)^{n} = (i^{n}) (x^{n})$ , which means that you can do the following:
$e^{ix} = \frac {i^{0} x^{0}}{0!} + \frac {i^{1} x^{1}}{1!} + \frac {i^{2} x^{2}}{2!} + \frac {i^{3} x^{3}}{3!} + \frac {i^{4} x^{4}}{4!} + \cdots$ .
Anyway, it turns out that this is equal to the power series for cosx + the power series for i*sinx. So, that's the simple explanation.
There are more advanced questions revealed by this, however:
Why does e^x have the power series that it does?
Why does sinx have the power series that it does?
Why does cosx have the power series that it does?

I've got a master's degree in math and I don't know the answer to those three questions, probably because I specialised in combinatorics and got a paying job.

Anonymous

Anonymous Sat 25 Dec 2021 05:14:23 No.14007568 Report

Quoted By: >>14007595

>>14007550
>specialised in combinatorics and got a paying job.
what are the industry applications of coombinatorics?

Anonymous

Anonymous Sat 25 Dec 2021 05:14:38 No.14007569 Report

Quoted By:

>>14007310
Our guy Qiaochu knows
https://math.stackexchange.com/a/3973362

Anonymous

Anonymous Sat 25 Dec 2021 05:24:42 No.14007595 Report

Quoted By:

>>14007568
Basically just programming different variations of travelling salesman problem solutions for business efficiency. My main focus is in building programs that find the least inefficient layouts for conveyor belts in factories.

Anonymous

Anonymous Sat 25 Dec 2021 05:37:37 No.14007614 Report

Quoted By: >>14007630

>>14007488
How do you even visualize a complex function? You would need to be able to visualize four-dimensional space. You can trying coloring the Argand plane but good luck getting any good intuition with that. Just settle for the analytic proof using Taylor's series and move on with life.

Anonymous

View Same Google iqdb SauceNAO multbyi.gif, 9KiB, 344x356

Anonymous Sat 25 Dec 2021 05:42:23 No.14007620 Report

Quoted By: >>14007630

>>14007488
its just abusing the fact that multiplying by imaginary number have an effect of rotating the result around the real-imaginary plane, exactly like your pic suggests

Anonymous

Anonymous Sat 25 Dec 2021 05:50:21 No.14007630 Report

Quoted By: >>14007649

>>14007614
doesn't need to be 4D, we can draw a 2D plane like in the OP
>>14007620
woah that actually makes geometric sense.
now why does z=e here?

Anonymous

Anonymous Sat 25 Dec 2021 05:52:52 No.14007634 Report

Quoted By: >>14007969

>>14007488
Just go watch veritasium on imaginaries midwit

Anonymous

Anonymous Sat 25 Dec 2021 06:03:30 No.14007649 Report

Quoted By:

>>14007630
If you have a complex number z, you can usually write it in the form z = a + bi, however, you can also write it as a vector in the complex plane z = (a, b). Now, since you can write it as a vector, you can also write z as a function of an angle "t" and it's modulus |z|, so that: z = |z|cos(t) + i|z|sin(t). Now, just like >>14007554
pointed out, if you write their respective taylor series, it turns out that exp(it) = cos(t) + isin(t), with that, you can write z = |z|*exp(it)

Anonymous

Anonymous Sat 25 Dec 2021 08:44:18 No.14007912 Report

Quoted By:

$\displaystyle f(x) = e^{-ix}(\cos x + i \sin x) \\ f^{\prime}(x) = e^{-i x}(i \cos x - \sin x) - i e^{-i x}(\cos x + i \sin x) \\ f^{\prime}(x) = e^{-i x}(i \cos x - \sin x) - e^{-i x}(i \cos x + i^2 \sin x) \equiv 0 \\ f^{\prime}(x) = 0 \;\;\; \forall \; x \in \mathbb{R}\Rightarrow f(x) \text{ is a constant} \\ f(0) = e^{0}(\cos 0 + i \sin 0) = 1 \cdot(1+0) = 1 \Rightarrow f(x) = 1 \;\;\; \forall \; x \in \mathbb{R} \\ \\ 1 = e^{-ix}(\cos x + i \sin x) \Rightarrow e^{ix}=\cos x + i \sin x \;\;\; \forall \; x \in \mathbb{R} $

Anonymous

Anonymous Sat 25 Dec 2021 09:02:52 No.14007952 Report

Quoted By:

>>14007310
>Where does the e come from
taylor expand exp(x) and find out for yourself

Anonymous

Anonymous Sat 25 Dec 2021 09:15:08 No.14007969 Report

Quoted By: >>14011286

>>14007634
Someone not understanding complex numbers is not a midwit. They just don't understand something yet. But someone demeaning someone elses intelligence while recommending they watch veritasium is like as midwit as you can get.

Anonymous

Anonymous Sat 25 Dec 2021 09:44:15 No.14008020 Report

Quoted By:

>>14007310
Its not a geometric relationship though, how do you want to understand it geometrically? In fact what makes complex trig functions so wondrous is that they are no longer confined to the strict geometric notion of an angle and are instead rational transformations of the exponential function.
Best you can do is maybe graph e^ix and observe that the real part of the part of the graph equivalent to the cosine of the argument and vice versa with sine.

Anonymous

Anonymous Sat 25 Dec 2021 12:29:51 No.14009345 Report

Quoted By:

>>14007310
a function $\mathbb{R} \to \mathbb{C}$ is geometrically a parametrized curve in the complex plane. $t \mapsto e^{it}$ is a curve defined by the property that its time derivative is $i\cdot e^{it}$ and its value at $t = 0$ is $1$ . since multiplication by $i$ is counterclockwise rotation 90 degrees, the condition says that the tangent vector of the curve at any given point is the radius vector rotated counterclockwise by 90 degrees. the only curve which satisfies this is the unit circle centered at the origin. thus $e^{it}$ agrees with the known counterclockwise parametrization of the unit circle which is $t \mapsto \cos t + i\sin t$ .

Anonymous

Anonymous Sat 25 Dec 2021 16:53:49 No.14011286 Report

Quoted By:

>>14007969
youre right. not understanding complex numbers is simply an indicator of being dumb as shit/hopeless in math and science.

Anonymous

Anonymous Sat 25 Dec 2021 17:06:57 No.14011353 Report

Quoted By:

>>14007310
https://youtu.be/v0YEaeIClKY
As geometrical as you can get

Anonymous

Anonymous Sat 25 Dec 2021 22:25:39 No.14012639 Report

Quoted By: >>14012644

>>14007503
this is the funniest thing i've read all day. jej

Anonymous

Anonymous Sat 25 Dec 2021 22:27:36 No.14012644 Report

Quoted By:

>>14012639
so funny xD

Anonymous

Anonymous Sun 26 Dec 2021 02:05:37 No.14013386 Report

Quoted By:

>>14007310
If you sqrt 1 you and up with 1 on real axis.
If you sqrt -1 you get 1 on imaginary axis.
As people posted the euler’s derivation is from e^x series.
You just replace x with ix and have fun with powers of i. You’ll group the expressions with even and odd powers. Even will eliminate i. With odd you have to take i in front of the group so you’ll end up with even i powers. Minus flickers in both of them as square of sqrt -1 gives you -1 and if you square it once more it gives you 1.
The series you end up with are Maclaurin series representing trigonometric functions in series form. Real part happens to be cos and imaginary sin.
That’s my retard explanation

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