/sci/ - Science & Math

>>10524404
think about the series expression for exp(x)
if you place (ix) in there then you should recognize the series for sine and cosine

from there its easier to understand what is going on

that result is one of euler"s greatest accomplishments, dont be disheartened if it doesnt come easy

>>10524404
what is that symbol mean sir
care to explain dear

>>10524410
what this mean sir
I know Pythagoras theorem tho

>>10524414
before you understand this result you must learn about series expansions for functions (taylor polynomials)
so basically this result comes after differential and integral calculus

>>10524410
Thanks. I'll give that a shot

>>10524419
that result is an application of the complex exponential
in particular exp(0+ipi)
the complex exponential is a function that expands its input by exp(real) and rotates the input by imag

>>10524416
uh oh ok
thanks for your kindness sir
I wont make it tho because im not as smart as you sir
anyway appreciate

>>10524410
Meh I like the explanation using the derivative of e^ix

f(0) = e^0i = 1
f’(x) = i * e^ix = i * f(x)
In other words starting at 1 the direction you’re traveling is tangental (multiplication by i is the same as a rotation by 90 degrees in the complex plane) to your postion vector which is a way of defining a circle specifically the unit circle.

I’m not sure how mathematically legal this is so fuck off rigorous fags

>>10524441
Oh forgot to mention you have speed 1 since its a unit circle and it takes two pi radii to go around the circle so thats where pi comes from

>>10524404
You should look into how these symbols work and how mathematicians came up with them.
eulers number,complex numbers just like natural ones are not special on themselves, there's an specific way to create eulers number, manipulating the expansion series from which we come up with the symbol e gives us this equation, it becomes pretty intuitive once you start manipulating the exponential of e in different ways, like, what would happen if you put an square in there? or what would happen if you put a polynomial in there? or what would happen if you put an complex number?
euler and pi comes into way more difficult equation than this and is not witchcraft, just mathematicians using the properties of the numbers they created in very cleaver ways.

>>10524404
0.999... = 1

>>10524441
Its rigorous.
But yeah, excess rigour can be a buttpain.

>>10524404
It's not completely unexpected because sin, cos, exp are all related by the same differential equation: y' = ky

>>10524471
what is sin and cos

>>10524482
You'll know once you finish middle school

>>10524404
https://www.youtube.com/watch?v=-dhHrg-KbJ0

>>10524485
I'm 19 sir

>>10524496
What's your IQ?

>>10524512
106
I will never make it
you guys are so lucky ...

>>10524404
Step one: What is i?

Step two: What is 2^i? (protip, use wolframalpha and check the position on the complex plane)

Step three: What is e?

Step four: What is e^i? (protip, use wolframalpha and check the alternative form)

Step five: What is e^i*pi?

>>10524516
How did you not go through trigonometry are you not from the US?

>>10524522
I did sir but im brainlet.
I never passed math
the only thing im good at is posting apus from my apu folder sir

>>10524404
euler identity proof
[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]

>>10524404
AAAAAAAAAAAAAAAAAAAAAA

>>10524404

>>10524579

this proof is horrible, because it only works if you know they are equal in the first place

>>10524404
Here's one I still don't get.

[math]\phi+\phi+1=\phi^3[/math]

>>10524404
[eqn]exp(i\pi) = \sum_{k = 0}^{\infty} \frac{(i\pi)^k}{k!}[/eqn]

>>10524404
If a logarithmic constant is raised by the power of an imaginary multiplication product of pi BEFORE you add an identity, the answer will always be 0.

Maybe I just don't get how stupid other mathematicians are?

>>10525566
[eqn]exp(i\pi) = \frac{(i\pi)^0}{0!} + \frac{(i\pi)^1}{1!}+\frac{(i\pi)^2}{2!}+\frac{(i\pi)^3}{3!}+\frac{(i\pi)^4}{4!}+...
[/eqn]
[eqn]exp(i\pi) = 1 + i\pi -\frac{\pi^2}{2}- \frac{i\pi^3}{6}+\frac{\pi^4}{24}+...
[/eqn]
[eqn]= (1 - \frac{\pi^2}{2}+ \frac{\pi^4}{24} ...) + (\pi - \frac{i\pi^3}{6} +\frac{i\pi^5}{120 } + ...)i[/eqn]
(insert taylor series magic here)...
[eqn]= cos ~\pi + sin~{i\pi} = -1[/eqn]

>>10524404
[math] e^u = \sum\limits_{n=0}^{\infty} \frac{u^n}{n!} [/math]
[math] \sin(u) = \sum\limits_{n=0}^{\infty} (-1)^n \frac{u^{2n+1}}{(2n+1)!}[/math]
[math] \cos(u) = \sum\limits_{n=0}^{\infty} (-1)^n \frac{u^{2n}}{(2n)!}[/math]

now let u = ix.
[math] e^{ix} = \sum\limits_{n=0}^{\infty} \frac{(ix)^n}{n!} = 1 + ix - \frac{x^2}{2!} - i \frac{x^3}{3!} + \cdots = \left [ 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \right ] + i \left [ x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \right ] = \cos(x) +i\sin(x)[/math]

>>10525610
cos = because
sin = signed
i = signature
pi = all potentially measurable distributions over a random field


sin cos pi i sin cos pi

Holy shit maths is boring. https://www.wolframalpha.com/input/?i=sin+cos+pi+i+sin+cos+pi

You guys just fuck with moon runes until magic falls out, right fellow number shamans?

>>10525610
Ah fuck I nigged the last half.
[eqn] + (\pi + \frac{\pi^3}{6} + \frac{\pi^5}{120} + ... ) i = cos~\pi + i(sin~\pi) = -1[/eqn]

>>10525619
cos i sin pi = 0

https://www.wolframalpha.com/input/?i=cos+i+sin+pi

Mathematicians only care about if 0 equalizes, not if it is attachable to an identity function.

>>10524482
sin is something God does not want you to do. cos is short for co-sin that is also something God does not want you to do.

>>10525498
>>10524404
It's easy to show that one though, it's the hat made my brain go "click" many years ago, before I even understood series expansions.
Just show that both sides of the euler identity, expressed as functions of x, satisfy the same differential equation:

f''(x)=-f(x) for f(x)=e^ix and
g''(x)=-g(x) for g(x)= cos(x)+i*sin(x)
If both equations are identical, so their solutions must be identical.

I know this is a very mess proof from a "strict" mathematical perspective, but it's very useful for beginners and helps to conceptualize the complex exponential function geometrically.

>>10526444
>it's the hat
the explanation that