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The "beauty" is in the fact that the constants are from very different branches of mathematics. 0, 1 and i are from algebra, e is from calculus/analysis, and π is from geometry.


And in combining those fundamental constants it uses each of the 4 fundamental math operations: Addition, multiplication, exponentiation and equality. All to arrive at a result that seems impossible. How can that be anything but beautiful? Now its your turn say something nice about it,chud
>>
[math] i^i = e^{-\pi/2}[/math]
>>
>>14472172
>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 dosent have to be one. Also can be a representation of infinity.

>the number pi
Pi is not a number. But could be a representation of infinity.

>the number e
e is not a number but could be a representation of infinity

>the number i
'i' is not a number. But could be a representation of infinity.

All this "equation" is saying is that infinity, to the power of infinity infinity plus infinity equals infinity. Come on, fuck this brainlet shit
>>
>>14472190
All i see is;
thing to the power of thingthing + value = no value
>>
>fact that the constants are from very different branches of mathematics.
e is the constant of proportion, for e^x:
dy/dx = y
If you imagine a particle travelling around a circle it's distance from center never changes. Since i was defined geometrically to move stuff around the graph, you can use it in many things by definition. You can redefine this parametrically without using i. I will say it is neat but I wouldn't say that it's beautiful, the equation is that way by definition.
>>
I find the Euler-Mascheroni constant much more beautiful than e or pi. It also pops up in derivatives of the (di)gamma functions or the zeta functions. Also the renormalized result, if not the actual result itself, of ζ(1) = 0.57721...

Now that is beautiful as hell but no one cares.
>>
It is amazing how i can be related to two other constants that seem unrelated like this. Considering how i is an abstraction and has no numerical value.
>>
>>14472213
It's because a sine wave is kind of like exponentiation, but it's as if saying "naw, I don't wanna diverge"
>>
How did Euler come up with it, considering you can't actually plug in a numeric value for i?
>>
>>14472190
can someone here come up with one but for

>addition

>multiplication

>substraction

>division


i have an idea but im not mathematically mature enough to post it
>>
>>14472190
>is not a number
You called?
>Could be considered a representation of infinity
How dare you?!
>>
Wait, what's this?
>>
>>14472224
>And in combining those fundamental constants it uses each of the 4 fundamental math operations

>Addition

>multiplication

>exponentiation

>equality

someone pls do a new version of this shitpost
>>
its nothing more than a convenient coming together for imaginary radial coordinates
e^(xπi) = -1, if x is an odd integer
e^(xπi) = 1, if x is an even integer
e^((x + 1/2)πi) = -i, if x is an odd integer
e^((x + 1/2)πi) = i, if x is an even integer
>>
>>14472219
i = -1. Anything to do with negative numbers is arbitrary.
>>
>>14472241
>And in combining those fundamental constants it uses each of the 4 fundamental math operations

>Addition
An operator, but there are arguments it dosent have to be one as it cant work with infinities (real numbers).

>multiplication
Not an operator because multiplication is repeated addition which may not be an operator as it cant work with infinites (real numbers)

>exponentiation
Exponentiation is not an operator as it is repeated multiplication which is not an operator as it is repeated addition which is not an operator as it cant work with infinites(real numbers)

>equality
Equality is not an operator as it can not check whether two infinities (real numbers) are the same or different
someone pls do a new version of this shitpost
>>
>>14472190
>1 can be a representation of infinity
>>
>>14472172
Hate that equation. Looks like a mishmash of meme numbers.
>>
>>14472172
cool story, bro. but
0=
is the trueest most beautiful formula. Mathematics is just art, an attempt to encapsule that beauty.
Prove me wrong
Pro tip: You can't
>>
>>14472172
>you WILL realize the beauty in this equation
but this equation is just a special case of an even better equation
>>
>>14472190

I am curious, what medications do you take?
You are clearly autistic in your endless struggle to show that these constants are "not a number" an absolutely insane statement, but clearly it must have strong meaning to you.
>>
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[eqn] \forall k\in\mathbb{Z}\quad\sqrt[i]{i}=e^{(2k+1)\pi/2},\quad k=0\implies\sqrt[i]{i}\approx4.8 [/eqn]
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>>14472190
>A number, but there are arguments it dosent have to be one. Also can be a representation of infinity.
Can I see these arguments?
>>
>>14473251
Number is plural
1 is not plural
>>
>>14472987
[math]z_1^{z_2} = e^{z_2 \log_{e}{z_1}}[/math]
never trust logarithms of complex numbers
>>
i=matrix[0-1;1 0]
e^x=1+x+x^2/2...
put x=i pi
and you will get matrix[-1 0;0 -1] from the infinite summation. Not so beautiful now, or is it more beautiful?
>>
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>>14472190
>the number 1
>A number, but there are arguments it dosent have to be one. Also can be a representation of infinity.
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>>14472213
isn't "i" sqrt(-1)?
>>
don't let doron zeilberger see this thread
>>
>>14473519
no number statisfy number*number=negative
>>
>>14473251
1=0.9999... is a representation of infinity. Infinity isn't a number.
>>
>>14472172
Euler's equation is really just a meme at this point.
>>
>>14473522
https://en.wikipedia.org/wiki/Imaginary_number
>>
>>14473312
>[math]\log_e[/math]
>>
>>14472190
kek, reminds me of the Elder Scrolls Stealth archer pasta. Good work anon
>>
>>14473434
I guess Ancient Greeks (Euclid, Archimedes, ...) "mistook" it so too?
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>>14473614
post it,im curious now
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>>14473562
based
>>
>>14472172
I don't really get why it's true despite many sites saying so.
>>
>>14473609
emphasize the word imaginary
as in imaginary friend
>>
>>14472281
He is technically not incorrect, as you can redefine 1 as to be a representation of pretty much anything. It is after all just a symbol.
>>
>>14473655
that's because you don't understand what e^x is (it's not "multiplying e together x times")
>>
>>14472172
okay, but what can i do with this?
>>
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>>14472172
>The "beauty" is in the fact that the constants are from very different branches of mathematics. 0, 1 and i are from algebra, e is from calculus/analysis, and π is from geometry.
>And in combining those fundamental constants it uses each of the 4 fundamental math operations: Addition, multiplication, exponentiation and equality. All to arrive at a result that seems impossible. How can that be anything but beautiful?

Well said and well explained.
OP you did good.
>>
>>14475115
solve literally any problem that involves vectors, but much easier than using regular vectors
>>
>>14473655
[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]
>>
>>14472226
>0 is a placeholder from a null space
No. 0 is the identity of a (additive) group. If 0 is a "placeholder" then 1 is also a placeholder because they're the exact same thing on different groups.
>>
>>14475188

this reads like a proof a high schooler would write up to seem extra smart
>>
>>14475059
it literally is
>>14473655
we define e^x as the function that is it's own derivative. we can represent it as a sum of an infinite number of monomial expressions called its series expansion
we define sinx and cosx similarly as functions whose second derivatives are themselves negated. They can also be defined as the sums of an infinite number of monomials
we know that cos(pi)=-1 and sin(pi)=0
we also have the imaginary unit i which we define as the square root of negative 1
the series expansion of e^(ix) is exactly the same as the series expansion of cosx + the series expansion of sinx multiplied by i. Plug in x = pi and you get e^ipi = -1
>>
>>14475750
>it literally is
>proceeds to elaborate how e^x is actually defined
>>
>>14472246
i^2 = -1
i is the square root of -1
>>
>>14475750
>it literally is
>we define e^x as the function that is it's own derivative
LMAO this is the post of the day
>>
>>14472172
make me onions guzzling space rotator
>>
>>14472172
Pi is disgusting.

e^iτ = 1
>>
>>14473519
Yes. Now tell me its numerical value.
>>
>>14472240
e to the complex power you faggot
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File: The_graph_y_=_√x(s).png (757 B, 310x196)
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>>14472240
What is this shape of e^x anyway? Looks like a distorted slice of an infinitely large circle.
The super-root is kinda weird too, pic related.
>>
>>14473312
das rite
>>
>>14473701
Why would I emphasize that word?
>>
That's not Euler's equation.. Euler never wrote about pi.. The enemies of the Pythagoreans attribute it to Euler to detract attention from his writing on geometry.

Now consider why you guys are arguing with each other..
>>
>>14478395
>Euler never wrote about pi
One of his most famous equations is
[eqn]\sum_{n=1}^\infty {1 \over n^2} = {\pi^2 \over 6} [/eqn]
>>
>>14472190
your bait, sir, is discretely finite and not even countably infinite.
>>
>>14473625
it's basically going over the strengths of playing each race and the punchline is all of the descriptions ends with "they make great stealth archers"
>>
>>14472172
what happens if it is not e but 2^ipi?
wouldn't it also rotate just like e^ipi?
>>
>>14480399
Since 2 = e^ln2, 2^ix = e^ixln2. So, it is pretty similar. The difference is that the period is longer because ln2 < 1.
Going the other way, e^ix = 2^(ix/ln2), so the equivalent of e^ipi would be 2^(ipi/ln2).
>>
>>14472176
not in my branch.
>>
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>>14481308
Ok, so essentially if the base e then when rotating around the circle the speed of rotation = the distance, since the derivative of e is e, or something like that? which the euler's formula confirms, i think. but if we change the base to 2 then
2^ix = cos (ln2 * x) + i sin (ln2 * y) ?
>>
>>14477827
exactly, it's just a unit circle. adding 1 to it so it becomes 0 is the opposite of elegant.
>>
The proof is easy: When you square [math]e^{i\pi}+1[/math] you have
[math](e^{i\pi}+1)^2 = e^{i2\pi}+2e^{i\pi}+1}=1+2e^{i\pi}+1 = 2(e^{i\pi}+1),[/math]
and this implies [math]e^{i\pi}+1=0,[/math] since [math]x=0[/math] is the solution of [math]x^2 = 2x.[/math]
>>
>>14482420
The derivative of e^ix is ie^ix due to the chain rule, so the derivative of a point on the circle is itself * i.
This makes sense visually if you imagine a circle and the direction the tangent is going (counterclockwise). For example, on the right side of the circle, at e^i0 = 1, the derivative is just i because it is upwards pointing (in positive i direction).
With base 2, the derivative is i*ln(2)*2^ix, which is consistent with e^ix 's period being ln2 of 2^ix 's.
Your equation at the end is correct, except you put a y instead of x in the sin().
>>14482593
Your proof already takes for granted e^i2pi = 1, so I don't think it's very good. An easier way (which does assume a lot still):
e^ix = cosx + isinx
e^ipi = cospi + isinpi
= -1 + i0
>>
>>14483105
>>14483105
yeah none of this explains the intuition behind e^ix = cosx + isinx. the taylor series based proof is good but not intuitive at all.
i would assume the derivative of e^ix being ie^ix is part of the intuition but i still don't get it.
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>>14483654
>muh intooooition
you will never ever succeed at math
>>
>>14483654
For [math]{d \over dx} e^{ix} = ie^{ix}[/math] to work, you need to prove it under complex differentiation. It does work, but you have the additional directionality condition to worry about.
Personally, I'm a lot more fond of the Taylor series "trick", even if it's done more as a purely formal algebraic move. I remember recently watching Youtube video demonstrating the complex exponential via involutes, which at least makes for a geometric interpretation.
>>
>>14483654
Well, since e^ix = cosx + isinx, then the derivative can also be seen as i * (cosx + isinx), which simplifies to icosx - sinx. When x increases with a higher real (cos) part than imaginary (sin), then the imaginary part (icosx) increases while the real part (sinx) decreases, making them sort of "balance out," forming a circle.
I hope this puts it into easily understandable terms. Maybe graphing out each of these real and imaginary parts with respect to x of both the function and the derivative will help.



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