/sci/ - Science & Math

>>15807934
>08/21/20 New boards added: /vrpg/, /vmg/, /vst/ and /vm/
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>Euler's identity is often cited as an example of deep mathematical beauty.
>The identity also links five fundamental mathematical constants.
>The number 0.
Not a number. Could be considered a representation of infinity.
>The number 1.
A number, but there are arguments it doesn't have to be one. Also can be a representation of infintiy.
>The number π (π = 3.141...).
Pi is not a number. But could be a representation of infinity.
>The number e (e = 2.718...), which occurs widely in mathematical analysis.
'e' is not a number. But could be a representation of infinity.
>The number i, the imaginary unit of the complex numbers.
'i' is not a number. But could be a representation of infinity.

All the "equation" is saying is that infinity, to the power of infinity infinity, plus infinity, equals infinity. Come on, fuck this brainlet shit.

go half way around a circle and you point the other way

>>15807934
>Euler's identity is often cited as an example of deep mathematical beauty.
>The identity also links five fundamental mathematical constants.
>The number 0.
Not a number. Could be considered a representation of infinity.
>The number 1.
A number, but there are arguments it doesn't have to be one. Also can be a representation of infintiy.
>The number π (π = 3.141...).
Pi is not a number. But could be a representation of infinity.
>The number e (e = 2.718...), which occurs widely in mathematical analysis.
'e' is not a number. But could be a representation of infinity.
>The number i, the imaginary unit of the complex numbers.
'i' is not a number. But could be a representation of infinity.

All the "equation" is saying is that infinity, to the power of infinity infinity, plus infinity, equals infinity. Come on, fuck this brainlet shit.

>>15807966
oh cool so inf^(inf * inf) + inf = inf, got it.

use exp(ix)=cos(x)+i*sin(x), which makes perfect sense through the infinite series definitions of exp, sin, and cos.

>>15807934
Never had the makings of a varsity mathlete

>>15807934

Imaginary numbers were a mistake. Seeing this equation makes me think that mathematics took a wrong turn 200 years ago.

>>15807934
[math] \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}
[/math]

>>15808183
good post

>>15808183
This is an ad hoc "proof" that completely misses the point. It only works of you already know the result, it gives you zero insight and has zero learning potential. Also you are taking the derivatives of complex functions without even defining both the functions themselves and how the derivation works.
0/10

>>15808244
The proof is either correct or it's not. It doesn't matter if you pull the end result out of your ass. If you're looking for a physical interpretation try something like >>15807957

>>15808246
Way to misinterpret what I said, no wonder you thought that dog shit series of steps would help OP understand the identity any better you fucking sperg

>>15807966
I love this pasta

This guy is the Aphex Twin of math. Prove me wrong (protip: you can't)

>>15807966
Oh my god kill yourself please

>>15807957
this

>>15807966
You're a nigger.
Kill yourself.

>>15808330
A number line isn't actually multidimensional, though.

>>15808339
just detour around that problem

>>15808353
You would need multiple dimensions to have a detour, but lines are too linear.

>>15808244
>derivatives of complex functions
treating i as a constant is perfectly reasonable

>>15808330
the complex numbers can be seen as a 2D plane, where the number a+bi is the ordered pair (a,b)

>>15807934
3dumb1down makes it so easy an actual toddler could follow

watch his shit or give up

>>15808244
we're not your 3rd grade teachers anon. You ask for a proof, you get a proof

>>15807934
doing half a turn and then turning around is like doing nothing

>>15808339
yeah but a plane is... which is where this transformation is happening

>>15808456
Its called the number line, not the number plane.

>>15808466
its the complex plane. it's made of two number lines at right angles to each other.

>>15808470
If a complex plane actually exists, then there should be a complex space for each dimension of space that is just as easily measurable as the threes dimensions of space rather than just pulled out of your imagination.

>>15808477
its math all of it is almost entirely detached from physical reality sans natural numbers (excluding zero). and a complex number is nth dimensional where n > 1 and 2^d=n where d is the dimension you need to represent a transformation in, so there is a complex space for each dimension though commutativity breaks down

>>15808398
sorry meant to reply to this
>>15808339
>>15808466

>>15808486
>I can totally explain reality with math, but you will have to bear with me because it is almost entirely detached from reality.

>>15808493
your dreams are almost entirely detached from reality but that doesnt mean it doesnt have effects for you in consensus reality when you wake up sad from a nightmare. being detached from reality doesnt imply it doesn't affect it just that it can exist without its presence. and at this point its an argument of semantics and I can safely stop caring about responding

>>15808426
>You ask for a proof, you get a proof
Where did OP ask for a proof?
This is indeed not 3rd grade anon, you can't make such mistakes on understanding a simple sentence

>>15808499
>My best mathematical explanation of reality is that its just like a dream man.

>>15808502
i dont have to take that from someone bordering on telling me negative numbers are a jewish conspiracy

>>15808515
Oh yes you do especially while your pilpul devolves into ever increasingly mystical nonsense like detached realities and dreams as an accurate representation of reality.

>>15808183
>f′(x)=0∀x∈Rf(x)is a constant
someone interpret this math language for this line
f'(x) is 0 when x is real and f(x) is constant?

180 rotation on a complex plane is -1. What's so magical here?

>>15808536
you don't necessarily start from 1

>>15808536
The magical transmutation of a simple number line into a complex plane with imaginary dimensions.

>>15807934
First you take this definition of e
[math]
e = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n
[/math]
and build e^z
[math]
e^z = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^{nz} = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^{u} = \lim_{u \to \infty} \left(1+\frac{z}{u}\right)^{u}
[/math]
Then you do this
https://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function#Characterization_1_%E2%87%94_characterization_2
to get the power series.
Separate it into sin and cos series
https://en.wikipedia.org/wiki/Euler%27s_formula#Using_power_series
And you're basically done

>>15808391
>treating i as a constant is perfectly reasonable
Again, it's reasonable if you already know about complex analysis

>>15808540
[math]e^{i\pi}[/math] means you make an angle [math]\pi[/math] with the positive x-axis, so you start at 1 on the unit circle.
>>15808543
[math]i[/math] is the phase, [math]i\pi[/math] is the rotation. If you think that's magical, maybe you should go back to highschool retard

>>15807934
If you take 1 and rotate by pi, you end up at -1. The sum of these two numbers is 0.

>>15808339
Notice how in the equation there is a reference to i. So the universe of discourse in this context is a plane, more specifically the Argand plane.

>>15807934
rotate, that's it, if THAT filters you... best of luck with life

>>15808477
ah, i though you sounded familiar, you are that schizo that think that for complex numbers we need six dimensions, because you are to dumb to comprehend that i is the second dimension

>>15808522
math ain't reality, its just a game, touch grass faggot

I'm pretty sure the root of the connection is when you force the Pythagorean identity on improper right triangles, on right triangles where x > r, so that y has to be complex in order for the x^2 + y^2 = r^2 to hold. I'm looking for a clean way to show it with geometry/algebra, not sure if it's possible

>>15808528
basically, if f'(x) is 0 for all real numbers, then f(x) is a constant-valued function.
in other words, if da funcshun dun change, it's a constant

(i^0)=1
(i^1)=i
(i^2)=-1
(i^3)=-i
(i^4)=1 and so on…
(e^x)= 1+x+((x^2)/2!)+…+((x^n)/n!)
(e^ix)= 1+ix+((ix^2)/2!)+…+((ix^n)/n!)
i’s cancel per above rules
(e^ix)=(1-((x^2)/2!)+((x^4)/4!)-((x^6)/6!)+…)+i(x-((x^3)/3!)+((x^5)/5!)-((x^7)/7!)+…)
Therefore (e^ix)=cosx + isinx by definition.
(e^ipi)=cos(pi)+isin(pi)
(e^ipi)=-1+i0
(e^ipi)=-1

>>15807966
>>15808088
>>15808244
>>15808466
>>15808493
>>15808502
>>15808522
>>15808543

>>15808088
Oh sweet child

>>15808244
Can you understand the imaginary unit?

>>15809360
Can you get the point of a fucking english sentence?

>>15809467
Fuck that. I proclaim Icala, from the Xhosa language meaning side, will replace the derogatory term "imaginary".

I think I'm close, but I keep getting [math]e^{i \theta} = \sec{\theta} + i \tan{\theta} [/math]. Weird.

https://www.youtube.com/watch?v=B1J6Ou4q8vE

>>15809467
my bad.

Ignore the blue and green lines.

Start off with [math]i*\sin(x)[/math] and [math]\cos(x)[/math].

[math]i*\sin(x)+\cos(x) [/math] will be the intersection point but we don't know what it is yet.


To find the angle x in radians from the intersection point the complex logarithm is used

Why the complex logarithm? Well if you know common inverse functions such as cosine[math](0,\pi][/math] and sine [math][{-\pi \over 2},{\pi \over 2}][/math], it similarly maps lengths to angles so a complex logarithm takes complex numbers and gives an angle in [math](-i\pi,i\pi][/math].

The output is restricted for uniqueness but otherwise there are infinite values differing by [math]{2\pi}i[/math].

[math]\operatorname{Log} z = \ln r + i \theta = \ln |z| + i \operatorname{Arg} z[/math]

Because of the pythagorean theorem [math]\ln |z| = 1[/math] for all angles. what is left is [math]i \operatorname{Arg} z[/math]. [math]\operatorname{Arg} [/math] is basically [math]atan2[/math] function.

That is where we get to. Now we know the angle.
[math]\ln(i*\sin(x)+\cos(x)) = i\theta [/math]

If you inverse again, sort of like when you do sine after inverse sine, we can actually induce that
[math]i*\sin(x)+\cos(x) = e^{i\theta} [/math]

Put [math]\pi[/math] in for [math]\theta[/math] and voila you have.
[math]-1 = e^{i\pi} [/math] or most famously
[math]e^{i\pi} + 1 = 0 [/math]

>>15809587
>maps lengths to angles
ratios not lengths

some more hints

[math](\cos(x),\sin(y))[/math] in real unit circle is analogous to [math]i∗\sin(x)+\cos(x)[/math] in complex unit circle.

>>15807934
OP Look at this circle >>15809360
and read >>15809587 . Basically the original proof.

>>15807957
>> the universe is shaped exactly like the earth
>> if you go straight long enough you'll end up where you were

Half the people in this thread are trying desperately to explain the relationship between sine/cosine and the complex plane, and the other half is spitting and dribbling glue out their mouths to say "NUH UH" louder than they did before.
What a waste of time.

>>15809587
>Because of the pythagorean theorem ln|z|=1 for all angles

was this a typo? [math]\ln|z| = 0 [/math] for all angles. I think you mean [math]\mathbb{z} = 1 [/math].

>>15809740
that's been /sci/ for along time

Just Taylor expansion it bro by replace “pi” with “x,” or it’s the first thing prof do in dfq class or the end of calc2.

>>15809921
yeah it was nice catch

1. there is a function in real space that is it's own differential and is one when argument is zero
2. this function can be extended to imaginary space by replacing the taylor series real argument by complex variable
3. turns out the function's both real part and imaginary is a sinusoid of the argument times the real function of the real argument, comparing the taylor series'
4. inserting the desired value yields the imaginary part is zero and the real part is minus one.

>>15807966
>>15807998

>infinity^(infinity*infinity) + infinity = infinity
>infinity^(infinity*infinity) = infinity - infinity = 0
0 = infinity
>infinity^(infinity*infinity) = infinity

infroot(X) means X^1/infinity
>infroot( infinity^(infinity*infinity) ) = infroot(inf)

1 = infinity
>infinity^infinity = 1
>infinity^infinity = infinity
>infroot( infinity^infinity ) = 1
>infinity = 1
>infinity = infinity

Cool. Seems to check out.

>>15807934
[math]
e^(i \tau) = 1
[\math]

>>15810943
this

>>15810943
>>>/leddit/

>>15807934
i is the "sideways" number (with absolute value 1). Multiplying a "real" number by a positive or negative value takes you go to a new point to the right or left respectively along the number line. But multiplying by i takes you off the line, either above or below it. The whole collection of "real" and sideways numbers fill out a plane, rather than just a line. Note that multiplying a sideways number by another (ex: -5i * i = 5) puts you back on the "real" line. We can think of the operator *i as a 90° left turn on the plane.

Note that for any positive numbers p & q, p^x = q^(x * log(q,p) ) - ie: All such exponentials can be written in terms of each other.
Also note, p^0 = 1

The essence of exponentiation is captured in repeated multiplication. Whereas discretely multiplying by i gives you a 90° turn to the left, consider what it should mean to continuously vary x in the function p^(i*x) . The answer is a continuous turn to the left, rather than a discrete one, and this traces out a circular path. A perfect analog is is how a uniform magnetic field will makes a charged particle move in a circle, or a swinging ball on a tether.
Considering A*p^(i*x) scales the circle by a factor of A, it must be centered at the origin.

Recalling that exponentials can be written in terms of each other, the difference between p^(i*x) and q^(i*x) is simply the rate at which the function moves around the circle for a given x. The number e has many special properties, among which is that e^(i*2π) = e^(i*4π) = e^(i*6π) =...
Thus, e^(i*x) perfectly matches progression around a unit circle, where the input x corresponds to exactly that amount of radians.

It follows from { p^0 = 1 = e^(i*0) } that this function trace is a circle centered at the origin which:
-starts such that 0° => 1
-has a radius of 1

π radians corresponds to half the distance around a circle, and the point opposite the starting point (1) is -1.