/sci/ - Science & Math

>>11511146
Mathmagic
Just memorize it

It's an symbolic representation of special case of the Barnett integrable functions

this is as simple as it gets

>>11511146
https://en.wikipedia.org/wiki/Euler%27s_identity

>>11511157

>>11511146
test
[math]x &= y \\ &= z[/math]

>>11511174
test 2
[math]
\begin{align}
x &= y \\
&= z
\end{align}
[/math]

>>11511146
https://en.wikipedia.org/wiki/Euler%27s_formula#Using_power_series
Now substitute [math]x = \pi[/math].

>>11511157
>Growth>
>lasting for half a circle
>points you backwards
These make sense.
>pushing sideways
I don't get it. Could you explain this part?

>>11511183
>>11511158
>>11511157
>>11511160
I don't like these explanations. Does anyone have one with a more abstract algebra feel?

>>11511157
>>11511158
>>11511160
>>11511183
While these may technically be "correct," Euler's formula requires an axiom that on it's own, does not follow from anything foundation of the field R with exponentiation. It's really just convention akin to defining 0^0 or 0! to be 1. Just define -e^i(pi) to be 1. This is the reason that the study of complex analysis always involves the study of branch cuts and analytic continuation. Euler's formula is really just putting all the symbols we know and love together in an equation that is both reasonable and elegant enough that using >>11511183
doesn't break the math, so all mathematicians (except wildberger) accept it.

>>11511146
[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]

>>11511232
Those are some pretty bold claims. You have to prove that in any model, no codification of a structure supporting the reals, the root of -1, and taylor series will imply Euler's formula.

>>11511232
The fuck are you talking about. Euler's formula is a theorem not an axiom.

>>11511276
Here's an axiom: i is the unique root of -1 such that Euler's formula does not hold. Prove me wrong. (Protip: you can't)

>>11511146
[math]\begin{align*}
f(t)&=\cos(t)+i\sin(t)\qquad (\Rightarrow f(0)=1)\\
\Rightarrow f'(t)&=-\sin(t)+i\cos(t)\\
&=i(\cos(t)+i\sin(t))\\
&=if(t)\\
\Rightarrow\frac{f'(t)}{f(t)}&=i\\
\Rightarrow\int_{0}^{x}\frac{f'(t)}{f(t)}\,dt&=\int_{0}^{x}i\,dt\\
\Rightarrow\log(f(x))-\log(f(0))&=i(x-0)\\
\Rightarrow\log(f(x))-\log(1)&=ix\\
\Rightarrow\log(f(x))&=ix\\
\Rightarrow f(x)&=\exp(ix)\\
\Rightarrow \cos(x)+i\sin(x)&=\exp(ix)\\
\Rightarrow \cos(\pi)+i\sin(\pi)&=\exp(i\pi)\\
\Rightarrow -1&=\exp(i\pi)
\end{align*}[/math]

>>11511237
this is it cheef

>>11511329
woah.

[math] \displaystyle
e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots
\\ \\
e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots
\\ \\
sin(x)=\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots
\\ \\
cos(x)=\frac{x^0}{0!}-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots
\\ \\
cos(x)+sin(x)=1+x-\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}-\frac{x^6}{6!}-\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^9}{9!}-\cdots
\\ \\
e^{ix}=\frac{(ix)^0}{0!}+\frac{(ix)^1}{1!}+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\cdots
\\ \\
e^{ix}=1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}-\frac{x^6}{6!}-\frac{ix^7}{7!}+\frac{x^8}{8!}+\frac{ix^9}{9!}-\cdots
\\ \\
e^{ix}=\left ( 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots \right )
+i \left ( x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots \right )
\\ \\
e^{ix}=cos(x)+i \, sin(x)
[/math]

>>11511220
Define a homomorphism [math]exp:\mathbb{I}=i\mathbb{R}\to S_1[/math] with the properties [eqn]exp(0)=1,[/eqn] and [eqn]exp(a+b)=exp(a)exp(b).[/eqn] Now choose [math]exp^{-1}(-1)=\pi.[/math]
As with what >>11511232 said, there isn't any reason that it needs to be true other than convenience, and constructing it with algebra illustrates this.

>>11511385
Sorry, should be [math]exp^{-1}(-1)=i\pi.[/math]

>>11511385
what about the literal taylor series?

>>11511352
i like this one, ya little nigga.

>>11511388
How would it involve abstract algebra if I define exp using the literal Taylor series?

>>11511400
I think he's trying to say that taylor series must imply euler's formula. He's not the one that asked for an abstract algebra explanation.

>>11511403
Oh, okay, but that still only holds if you define the complex exponential function as being equal to that Taylor series. In either case, it requires an arbitrary definition.

its called learn some basic fucking calculus like a white man and don't be a retarded gorilla nigger who doesn't know taylor series stupid nigger.

>>11511152
>>11511157
>>11511160
not science or math
>>11511232
>>11511237
>>11511264
>>11511276
>>11511281
>>11511329
>>11511332
>>11511352
Is this what /sci/ has been reduced to? You guys suck.
>>11511214
>>11511220
>>11511146
>Fucking how
Fucking simple: The intuition is that [eqn] \frac{d}{dt} e^{it} = ie^{it} [/eqn]
If you think of the complex number as a 2-d vector, you can work out that multiplying by i is rotating by 90 degrees.

So this equation above tells you that the rate of change of this 2-d vector [math] e^{it} [/math] with [math] t [/math] is always 90 degrees perpendicular to the vector and of unit length. This is how you get something to go in a circle, the tangent to a circle is 90 degrees perpendicular to the radius. And the length is constant, so the speed is constant. So this vector is going around a circle at constant speed, so that [eqn] e^{it} = \cos{t} + i \sin{t} [/eqn] and the result follows.

>>11511409
How do you prove that the length is constant without presupposing [math]e^{it}=\cos t+i\sin t?[/math]

>>11511385
>here isn't any reason that it needs to be true other than convenience
Yes there is you little shit.
[math]\exp x = \sum_{n=0}^\infty \frac{x^k}{k!}[/math]
Now let [math]x = i \pi[/math] and you ALWAYS get -1.

>>11511404
>that still only holds if you define the complex exponential function the exact same way its defined for real numbers

[math]\begin{align}
\exp(it)&=x+iy\\
\frac{d}{dt}\exp(it)&=\frac{d}{dt}(x+iy)\\
i\exp(it)&=\dot{x}+i\dot{y}\\
&=i(x+iy)\\
\dot{x}+i\dot{y}&=-y+ix\\
\dot{x}=-y&,\,\dot{y}=x,x_0=\operatorname{Re}(\exp(0i))=1,\,y_0=\operatorname{Im}(\exp(0i))=0\\
x=\cos(t)&,\,y=\sin(t)\\
\exp(it)&=\cos(t)+i\sin(t)\\
\exp(i\pi)&=-1\quad \blacksquare
\end{align}[/math]

>>11511409
>intuition
Not science or math, pathetic larper

>>11511415
at t=0, it has unit length, and since the rate of change is perpendicular, the length never changes.

>>11511416
Yes, we define it that way over the complex numbers for convenience.
>>11511417
Yes, it's not obvious that we should. Consider e.g. logarithms or square roots.

>>11511409
Great response, I’m even finding it weird that it has been posted here
/thread

>>11511146
That right there? that my friend is proof of God.

>>11511443
>we define it that way ... for convenience
Ebin man, spending Saturday night trolling /sci/
Literally every function is defined as it is "for convenience", that is what a definition is, congratulations, you've figured out how logic works. What exactly is your point?

>>11511463
> WOAHHH LIKE SICENCE AND MATH AND SHITT YOO THATS SO COOL like I LOVE SICNEC AND SHIT MAN IM SUCH A NERD

Shut the fuck up you fucking retarded nigger gorilla and be a stupid retarded nigger gorilla elsewhere, fucking cuck fag and learn what a cotangent bundle is you stupid nigger.

>>11511474
>> WOAHHH LIKE SICENCE AND MATH AND SHITT YOO THATS SO COOL like I LOVE SICNEC AND SHIT MAN IM SUCH A NERD
>Shut the fuck up you fucking retarded nigger gorilla and be a stupid retarded nigger gorilla elsewhere, fucking cuck fag and learn what a cotangent bundle is you stupid nigger.

>>11511146
Easy, just throw in some imaginary voodoo and anything is possible

>>11511494
what is this supposed to prove how much of an inpure nigger you are and your lack of aryan blood

>>11511146
That actually makes sense to me because of I understand phasors

>>11511415
The complex conjugate of exp(it) is exp(-it), so length[exp(it)] = sqrt[exp(it)(exp(-it)] = sqrt[exp(0)] = 1.

>>11511618
>the complex conjugate of exp(it) is exp(-it)
Prove it without using the formula you're attempting to derive.

>>11511496
You being too stupid to understand something doesn't make it voodoo.

>>11511146
Carl gauss said if why this identity is true is not immediately apparent to you, then you should quit math, bye.

>>11511637
Shoot, I forgot that's where this identity came from. Thanks for the catch.

>>11511281
are you actually retarded?

>>11511400
how do you prove that such a homomorphism exists using only abstract algebra?

>>11511503
>aryan blood
Prove it.

Consider a uniform circular motion along the unit circle. Clearly this curve must satisfy that its tangent vector is its radius vector rotated by pi/2. Therefore the components satisfy the differential equation

x' = -y
y' = x

with x(0) = 1, y(0) = 0, assuming you start at (1,0).

Using complex numbers this exact consideration can be written as

z' = iz

with z(0) = 1.

Since initial problems of ODE are unique, z(t) = x(t) + i*y(t)

>>11511146
isnt this just a representation of -1 on a complex plane.
x-real number y- complex number
-1 lies on negative x so can be represented as e^i*pi.

>>11511409
this is the correct answer

>>11511817
It's literally not even an answer.

>>11511232
>Euler's formula requires an axiom
and that axiom is i^2 = -1

>>11511146
It's made up bullshit, just like all math.

>>11511847
It requires another axiom as well.

>>11511232
Neither 0^0 nor 0! are defined to be 1. The first is proven by noting that the empty function is the only map from the empty set to itself. The second is also easy to prove: n! = (n+1)!/(n+1), so 0! = 1!/1 = 1

>>11511899
>It requires another axiom as well.
just some definition of exp, sin and cos

>>11511473
I guess that the identity is a direct (and fairly obvious) consequence of the definition. I think most people who struggle to understand the identity think that it is something inherent to the numbers e, i, and pi, and while that might be somewhat true (I'm not actually sure) it is easier to understand if you think of it as a consequence of how the exponential function. A lot of people don't understand the difference between repeated addition and multiplication, and also the difference between repeated multiplication and exponentials, so I do think that knowing that we make an arbitrary choice to get to the identity is helpful to the understanding.

tl;dr: I'm trying to help someone who doesn't understand how the identity comes about, not someone who does.

>>11511682
Ramanujan once said: peepee poopoo peepoopeepoo

>>11511902
>proven
no no no no no, that's wrong. You have to look at mathematical history to see that factorials and exponentiation were defined on positive integers exclusively. In the case of factorials, you can't just say, "this result holds for its domain, thus we may extend the domain to prove the same result on the extended domain." You have to take it as an axiom that it is defined that way on the extended domain. As for the "empty function" and "maps on sets," that's completely irrelevant. You are talking about sets, I'm talking about numbers.

>>11511709
Probably using polar coordinates, noticing that elements in S_1 can be described by a single variable - angle - so we have [math]S_1\cong\mathbb{R}/\tau.[/math] define the identity as 0, and multiplication on angles as [math]\theta_1+\theta_2,[/math] i.e. regular addition. Since we have a homomorphism from the additive reals to any interval [0,t], with t:=0, and hence to S_1. Obviously, we choose [math]\tau=2\pi[/math] to get our identity.
I guess for continuity you could need topology..? I'm not really sure if the algebraic result implies continuity. Regardless, there does exist a homemorphism from the reals to S_1 using a similar map to the one I used - essentially the same map as the one I described above.

>>11511928
you haven't constructed anything
>S1≅R/τ.
this is what you want to prove

>>11511899
WHAT AXIOM

>>11511146
Tibees is comfy
https://www.youtube.com/watch?v=-AyE1Wpgo3Q

>>11511961
so now she's into asmr videos disguised as mathematics ?

>>11511146
e^i*pi might equal -1?

>>11511946
Euler's formula itself is the axiom.

>>11512017
lmao it's not

>>11512065
can you prove it?

>>11512138
can you fucking read this thread?
it's been proved 5 different ways at least

>>11512148
>>11512148
none of those are proof. Refer to >>11511264
Without taking euler's formula as an axiom, you can't prove it follows simply from the definition of i. there's more to it than that.

>>11512178
you need only definitions of sin, cos, and exp

>>11511942
>you haven't constructed anything
>>S1≅R/τ.
>this is what you want to prove
"Since we have a homomorphism from the additive reals to any interval [0,t], with t:=0, and hence to S_1."
I guess your problem is with the fact that I presupposed that angles behave the way they do - i.e. are additive? This is fairly straightforward to do over the rationals using a symmetry argument. To do it over the reals, I would guess you either use Lie groups (I have no idea about this one, I'm just guessing that you could), define an ordering on a proper subset of S_1 that either contains the element 1 as an endpoint or doesn't contain 1, or just use trig.

>>11512217
I think the ordering idea might need an appeal to trig as well, I'm not sure.

>>11511352
As a highschooler, this is the most understandable explanation to me. Thanks!

>>11511975
She should just do a gf amsr roleplay already

>>11512217
>"Since we have a homomorphism from the additive reals to any interval [0,t], with t:=0, and hence to S_1."
that's the thing, this homomorphism *is* the euler's formula. you need to prove that it exists somehow.

>>11512178
>t. didn't read a single proof

its a limit, not real math

>>11512233
Read >>11512217
I haven't used any complex numbers or exponentials anywhere, it just relies on the idea that angles are additive - which is a symmetry argument.

>>11512236
heh, I'm a mathematician. I've seen 'em all before, kiddo.

>>11511409
>If you think of the complex number as a 2-d vector
The undergrad brainlet reveals himself

>>11511922
>you can't just say, "this result holds for its domain, thus we may extend the domain to prove the same result on the extended domain."
Yes, you can.

>>11511922
>You are talking about sets, I'm talking about numbers.
And exponentiation on the naturals (and even transfinite cardinals) is defined in terms of the cardinality of the set of functions between sets with the given cardinalities.

>>11513184
so what are you saying exactly ? that e^it = cos(t) + i*sin(t) must be taken as an axiom ? that's not true, there are many definitions of exp(z) and there are many definitions of sin(x) and cos(x). no matter how you define these terms, euler formula will follow.

>>11513196
>>If you think of the complex number as a 2-d vector
and this is somehow wrong because ?

>>11511961
she's so fucking boring

>>11511823
>It's literally not even an answer.
Wrong. Read it again. Did you not notice the use of language, and what it implies?
>>11513196
The underage /sci/lon rears its tiny head

>>11513207
>exponentiation on the naturals is defined in terms of the cardinality of the set of functions between sets with the given cardinalities.
No it's not. That's merely a coincidental relationship.
>>11513204
Only if you redefine your original function to the extended domain, i.e take as an axiom. These functions have always been defined recursively on positive naturals.
>>11513208
Incorrect.

>>11513367
I understand your basic argument: the function [math]\exp:\mathbb{R}\to\mathbb{R}[/math] such that [math]\exp(x)=\sum_{k=0}^{\infty}x^k/k![/math] is defined only on the real numbers, so the function [math]f:\mathbb{C}\to\mathbb{C}[/math] such that [math]f(z)=\sum_{k=0}^{\infty}z^k/k![/math] is a different function because it has a different domain and codomain.
This is correct. Before all else, a function is defined by its domain and codomain. This is why an actual mathematical text will usually distinguish between the /real/ exponential function [math]\exp:\mathbb{R}\to\mathbb{R}[/math] and the /complex/ exponential function [math]\exp:\mathbb{C}\to\mathbb{C}[/math]. Where you are incorrect is your claim that the definition of a function is an "axiom". It is not, at least not in the sense that anybody except people looking to start arguments on /sci/ uses it as such.

>>11513367
>take as an axiom
2nd the other anon, what the fuck are you saying?

>>11511146

It's an elegant relation which reveals that, in fact, π = e = 3, whereas the quantities are usually introduced as being nonequal in earlier math education. It's because of this equality that the whole thing cancels out nicely.

>>11513367
>That's merely a coincidental relationship.
Nevermind, didn't know I was talking to an actual retard.

>>11513313
Handwaving is not an answer.
While you are correct that complex numbers are used to represent coordinates in two dimensions (hence Gauss's recommendation of "lateral numbers"), you haven't provided an adequate explanation of /why/ that is. Someone unfamiliar with Euler's formula is certainly someone unfamiliar with the complex plane, how it may represent 2D vectors, how multiplying by [math]i[/math] "rotates" by [math]\pi/2[/math], and so on; simply telling them these things are true without illuminating on why does not lead to any increased understanding.

>>11511146
theres a negative number involved

>>11511961
that's not what the word comfortable means. stop relying on falsehoods sheltered in "slang".
expand your vocabulary.

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

>>11511146
It makes sense if you just use the dark number

>>11513563
>everyone, from the ancient greeks to enlightenment age number theorists was talking about set theory, not iterated multiplication
>never mind the fact that set theory wasn't invented until the 1900s
Read any set theory text and you will see that the book defines exponentiation as iterated multiplication, with base case 0^0 = 1.
>>11513437
>>11513485
>In M, for all x in S, a subset of A (the set of all a, of which x is a type of, in M), for all P in T (the subset of all P-types such that the following), |= P(x).
>In M, for some y in A\S, for all P in T, there is a P1 in T1, such that |= P1(x) and P1(y) := P(y) and |= P1(y)
>therefore |= P(y)
This is essentially what you're argument boils down to, but it must require that A exists. However, A and T cannot exist, as they are defined as subsets of the set of all sets. Thus there can be no proof that a function applied to an element not in its domain is identically equal to a superset of that function.

>>11513932
Nah, it's as simple as the fact that there exists a unique analytic function that agrees with the real exponential function on the real line.
Plot twist: It's the complex exponential function.
Now fuck off, retard.

>>11513932
x^y is the cardinality of the set of functions from a y-element set to an x-element set. There is exactly 1 function from a 0-element set to a 0-element set. Therefore 0^0 = 1.

>>11513932
>completely misses the point
Yes, yes, [math]\exp:\mathbb{C}\to\mathbb{C}[/math] is a different function from [math]\exp:\mathbb{R}\to\mathbb{R}[/math]. Doesn't matter. All the characterizations of [math]\exp:\mathbb{R}\to\mathbb{R}[/math] (e.g., [math]\exp'=\exp[/math]; [math]\exp(0)=1[/math]; [math]\exp(a+b)=\exp(a)\exp(b)[/math] for all real numbers [math]a[/math] and [math]b[/math]; [math]\exp(x)=\lim_{n\to\infty}(1+x/n)^n[/math] for all real numbers [math]x[/math]; [math]\exp(x)=\sum_{k=0}^{\infty}x^k/k![/math] for all real numbers [math]x[/math]) are true of [math]\exp:\mathbb{C}\to\mathbb{C}[/math]. You have nothing to contribute besides nonsense.
Oh, and learn some fucking [math]\LaTeX[/math] if you want your shit to be legible shit.

>>11513943
>defining exponentiation by set cardinality
And now you can't extend your function to the reals. NEXT!
>>11513942
>complex exponential function
To say that the real exponential function and the complex exponential function are equivalent is an axiom
>>11513949
Heh, sorry, but I don't read sentences that don't have spaces.

>>11513960
Fucking /sci/ fucked up my LaTeX.
Anyway, "To say that the real exponential function and the complex exponential function are equivalent is an axiom" is objectively false and to even say something like that betrays your utter pseudery. Two functions cannot be equal (which is what I assume you meant by "equivalent") if they have different domains and codomains, so the real exponential function and the complex exponential function are not equal, and an "axiom" that said they were would lead to a contradiction. However, there is no need for an "axiom" to prove that the two have the same characterizing properties that are relevant to the problem at hand.

>>11513986
>to say that they have the same properties
So you are saying that the set of all "characterizing properties" of the real exponential function is equal to the set of all "characterizing properties" of the complex exponential function? And I don't need an axiom for that?

>>11513996
>the set of all "characterizing properties"
that are relevant to the problem at hand, bitch. Literally the only property needed to prove that [math]\exp(i\pi)=-1[/math] is [math]\exp'=\exp,\,\exp(0)=1[/math]. The complex exponential function is the unique complex-valued solution to that equation.

>>11514018
The issue is that in math you can't say "the set of all" anything, without following it with that are elements of this concrete set. Otherwise you are talking about the set of all sets.

>>11511352
based

>>11514049
I'm not the one fucking talking about sets, you are. And there is no "set of all sets" in ZFC, unless that is your point, but at this point it seems you have no point except that which might seem intelligent to your utterly unintelligent mind.

>>11514055
you are implying the existence of the set of all sets in ZFC, by claiming that equivalence of two functions is implied by equivalence of all their properties. Remember, every mathematical object must be codified in some way that doesn't lead to a known contradiction.

>>11514060
>claiming that equivalence of two functions is implied by equivalence of all their properties
You are the one claiming that I am making this claim. In fact, you are the only one who was even brought up "equivalence", which you have yet to define.

>>11514063
Yes, I claim you make that claim because you are. Equivalence is defined two ways: Identical equivalence- evaluation of the symbols will always return the exact same formula, and logical equivalence- under certain constraints, evaluation of symbols will result in identical equivalence. Your claim is that the two exponential functions are identically equivalent without requiring an axiom, because they are logically equivalent when restricted to the reals, and you can't find a contradiction in a model where you can prove the exponential function satisfies euler's formula by theorems already in the model.

>>11512221
underage and B&

what is the science behind threads like this going for days and containing nothing but shitflinging in them? what's the science behind the entire board being just that?
maybe that's the point of 4chan - people are too lazy to expend calories to post and they need an incentive to do so, with someone else triggering them into replying being the best incentive to achieve it as it rewards the ego and makes it invested so it cannot pull back. maybe the whole point of this place is to keep triggering each other, to incentivize ourselves to participate in a topic and to therefor reach the local minimum after 500 iterations of the exact same shitpost across 3 years. maybe 4chan is just a debate optimization algorithm

>1 rotated half a circle around 0 is -1

WOW SO FUCKIN SICK DUDE. BEAUTIFUL. WHOA. I HOPE AFTER RETAKING CALC 3 AND WATCHING ENOUGH NUMBERPHILE VIDS I CAN UNDERSTAND THIS SOME DAY!

I wish fucking retard niggers that never took a high level math class weren't allowed here. It's so bad. They have such a low iq they can't look at a simple calculation and think it's some kind of great magic. There are much more "beautiful" things in complex analysis than a simple calculation, like Picard's Great Theorem, Liouville Theorem, Riemann Mapping Theorem, and Weierstrass Factorization. Plus all the different theorems in geometric function theory like Schwarz-Christoffel.

Just goes to show you how bad at math the average retarded normie is, this includes engineering undergrads that think they're "good at math".

>>11514073
>Your claim is that the two exponential functions are identically equivalent without requiring an axiom, because they are logically equivalent when restricted to the reals, and you can't find a contradiction in a model where you can prove the exponential function satisfies euler's formula by theorems already in the model.
Who made this claim? I think you're arguing against yourself.

>>11514261
>sighs
>rolls eyes
>this guy again
Quit being so anal about not understanding math.

>>11514278
Lmao what did you accomplish by spending your day trolling /sci/ ebinly

>>11514282
in fact, I got a lot done today. I analyzed functions, wrote programs, had deep, intellectual thought, and worked from home. Pretty productive.

>>11511157
What's the best introduction to spinny-spiraly math?

>>11511447
Nothing weird. Some iq people occasionally visit /sci/.

>>11514049
Utter retard

Remember Eular's number represents the equation for a series that if you take the derivative or the integral, it does not change.

>>11513960
>And now you can't extend your function to the reals.
Sure you can. Once you've defined it on the naturals, you can prove that, for every real [math]x \geq 0[/math] and every natural [math]n[math], there is exactly one [math]y \geq 0[/math] such that [math]y^n = x[/math] and define [math]x^{1/n}[/math] as this [math]y[/math].

>>11513960
>To say that the real exponential function and the complex exponential function are equivalent is an axiom
Nobody said they were """equivalent""" (whatever that means) you dishonest, retarded fuck. I said there's only ONE analytic function that agrees with the real exponential function on the real line. Learn to fucking read you dumb monkey.

>>11514355
>he's never heard of the set of all sets paradox
>>11514363
>sets have irrational cardinalities
>>11514367
"equivalent"... equals. dummy

>>11514571
>irrational cardinalities
Where did I say "irrational cardinalities"?

>>11514571
No one here has brought up the set of all sets but you.
That aside and completely off-topic to this discussion, the set of all sets DOES exist in some set theories, showing once again that even in this unrelated topic you have no idea what you're talking about.

>>11514571
>"equivalent"... equals. dummy
NOBODY SAYS THEY'RE EQUAL YOU TREMENDOUS RETARD. I repeat, since you have trouble reading: There's only ONE analytic function that agrees with the real exponential function on the real line.

>>11513367
>Incorrect
what exactly is incorrent ? and provide an explanation

>>11514571
>"equivalent"... equals. dummy
I have NEVER seen it used in this sense

>>11514577
>define exponentiation on the reals
>define exponentiation by cardinalities of sets
Pick one
>>11514578
>the set of all sets exists
heh,
>>11514579
>>11514654
>can't even read english properly
You realize math iq is highly correlated with verbal iq, right?
>>11514641
You are now defining the exponential function as a sum of trig functions, and expect me not to take it as an axiom?

>>11514671
>You are now defining the exponential function as a sum of trig functions, and expect me not to take it as an axiom?
no, I didn't define exponential functions as a sum of trig functions. I've asked you, whether YOU think that this is the only possible definition.
it's not. there are many definitions. I prefer the one using ODE.

>>11514671
>Pick one
You start with the latter and extend it to the former in a way that is consistent with the latter.

>>11514671
>>the set of all sets exists
>heh,
It exists in positive set theory, new foundations, and other set theories, you ABSOLUTE RETARD. Not that this has anything to do with the topic like your schizo ass keeps claiming.

>>11514671
>>can't even read english properly
Yes, you can't. Good to see you have at least some self-awareness.

There's only 1 (ONE) analytic function that agrees with the real exponential function on the real line.

No amount of cope will change this fact. Sorry, idiot.

>>11514684
>implying irrational cardinalities make any sense at all
By all means, explain it to me so I can steal your work and win the fields metal
>>11514685
>ah yes, the well-known "other" set theory
So exactly what is the "set of all sets that does not contain itself"
>>11514683
no u
>>11514686

>>11514691
you still haven't answered >>11514683

>>11513748

>>11514694
>there are many possible definitions
that's right. Some of them don't agree with euler's formula. That's what i've been saying this whole time.

>>11514701
>Some of them don't agree with euler's formula
give one which doesn't

>>11514703
e^i(pi) (equals) undefined. Not e^i(pi) is undefined, which would imply that the exponential function is not defined on complex domain.

>>11514707
what ?
would you say that I need an extra axiom for [math]\sin^2 x + \cos^2x = 1[/math] ?. I can define [math]\sin x = 5[/math] and [math]\cos x = 3[/math] then it's not true.

>>11514698
It's almost always the smartest people who are humble enough to say they don't understand something, while still maintaining an arrogance about themselves..

>>11513748
>Someone unfamiliar with Euler's formula is certainly someone unfamiliar with the complex plane, how it may represent 2D vectors, how multiplying by ii "rotates" by π/2π/2, and so on;
this is certainly not the case in Europe. I've been taught complex numbers in high school, the Euler's formula had to wait until third or fourth semester I think.

>>11514777
I also learned about complex numbers in high school. But I would say that once you know the Taylor expansions of the exponential function and of the basic trigonometric functions, then the Euler formula becomes almost immediate.

>ITT: naive set theorist attempts to bombard a thread of analysts' civil debate and gets utterly BTFOed

>>11514796
Didn't they die out with Russell long time ago?

>>11514801
apparently they evolved into creatures believing you need axioms to restrict or extend functions

Oh hey
[math]\mathbf{r}=\exp(it)[/math]
[math]\mathbf{v}=i\exp(it)[/math]
[math]\mathbf{a}=-\exp(it)[/math]
[math]\mathbf{a}=\mathbf{v}^2/\mathbf{r}[/math]
Never noticed that, thank you to the analysts and Chads in this thread and not the algebra subhumans

>>11514809
Have yet to see such a creature. I did expect them to be rare nowadays.

>>11514809
I think you confuse the notion of "axiom" with the notion of "definition". Much the same way Trump confuses "hoax" with "virus".

>>11512206
and complex numbers.

>>11514859
maybe. so what exactly is the difference ?

>>11514865
Most /sci/-tards would be perfectly willing to explain why something doesn't make sense, even if they have no idea whatsoever what it means. Which makes definitions totally irrelevant. Axioms too. And logic, obviously.

>>11511146
it's a construct, dumbo

>>11511387
>exp^-1()
isn’t that just ln

>>11514833
>Never noticed an obvious fact
>Calls algebraists subhuman
Okayy, sub-troglodyte.

>>11514919
I was defining an arbitrary function - I just called it exp because that's what it is. The inverse does end up being the logarithm, but that is irrelevant to the point.
In fact, I just realized that what I wrote isn't quite accurate; I needed to restrict the domain of exp in order for it to have an inverse.

>>11514933
Anon was being sarcastic though.

>>11511329
this is a fucking brillant explanation

>>11511329
based

>>11514691
>>implying irrational cardinalities make any sense at all
>said no one

>>11514691
Defining a notion of irrational cardinalities won't win you a Fields Medal, you stupid dunce.

But it can be done: https://math.stackexchange.com/a/73506

>>11511146
Multiplying a base by itself an imaginary, irrational number of times: how do you write this out in an expanded form? If you took 2 to the third as an example, there would be 3 2s and 2 multiplication symbols: 2x2x2.

How do you write out i*pi e's separated by i*pi-1 multiplication symbols?

>>11515892
You have to be 18 to post here.

>>11515901
Don't know, then? THOUGHT SO. Pro tip: bow to the master.

>>11515892
multiplication as repeated addition works only for natural numbers

>>11515832
>>11515836
Hmm lets check the link
>There are no notion of a fraction of an element in ZF
learn to read retard
>>11514713
only if x is not real

>>11515952
ZF IS NOT THE ONLY SET THEORY YOU BRAINDEAD RETARD

>>11515952
>only if x is not real
so is this "axiom" anything else than a definition of exp(z) for a complex argument ?

>>11514691
>>11515952
What on earth does this have to do with the topic? Why do you keep jabbering about set theory?

>>11511146

There are many ways to how.

One way, is to start with the Taylor expansion of sin x. Multiply each term by i. Then, add in the Taylor expansion of cos x.

What you get, is exactly the same as the Taylor expansion of e.

So e^x = cos x + i sin x.

Then, just plug in i*pi for x in that last equation and solve, and you'll get e^(i*pi) = -1.

>>11516001

'Scuse me, I made a typo. Try e^(ix) = cos x + i sin x instead.

Now that everyone in the thread knows Euler's formula, use it to prove[eqn]\int_{\infty}^{\infty}\!e^{-x^2}\,dx=\sqrt{\pi}[/eqn]

>>11516035
fairly sure this integral is zero

>>11516035
Multiply by the same expression in y, and you can evaluate it as a double integral, then convert to polar coordinates using the Jacobian, solve the resulting double integral and then take the square root.

>>11516040
Fuck[eqn]\int_{-\infty}^{\infty}\!e^{-x^2}\,dx=\sqrt{\pi}[/eqn]
>>11516049
Sure, but can you solve it with Euler's formula?

>>11515984
Yep, haha. Math is all about the technicalities. Get used to it kiddo.
>>11515982
It's the only one us professional mathematicians use.
>>11515996
You are talking about defining something. That is technically an axiom.

>>11516058
>Yep, haha. Math is all about the technicalities. Get used to it kiddo.
my point isn't that you need an axiom. my point is that you keep on insisting that this axiom is the Euler's formula.

>>11511146

With the complex e function you have two parts which you can imagine as „2d-vector“ from the origin (—> the point (0,0)).
The „real number“ part is the length of the vector.
The „irrational part“ is the angle of the vector, going counterclockwise and with 2*pi being the same as 0 (a full circle).

>>11516065
Technically every definition is an axiom. And yes, you do need multiple axioms for this definition, otherwise I could say that i is some element that is the root of -1 and does not satisfy euler's formula. Yes, we would lose a bunch of properties, but that doesn't mean these axioms are unnecessary. I know it's confusing at first, but you'll get the hang of it.

>>11516056
That method implicitly uses it. Idk, is there a different method?

>>11516080
I'll say it once again. my point isn't that you need axioms or definitions. my point is that you keep insisting that one of these axioms needs to be the Euler's formula itself.

>>11516093
yes, it does.

>>11511409
>you can work out that multiplying by i is rotating by 90 degrees.
Gotta explain that
>And the length is constant
Why?
> so the speed is constant
Why?

Your communication skills need work

>>11516095
axiom/definition: exp is the unique analytic function which agrees with e^x for real arguments

axiom/definition: exp is the unique solution to the initial value problem exp' = exp and exp(0) = 1

axiom/definition: exp(z) is the sum of the power series z^k/k!

axiom/definition: exp(z) is the limit (1 + z/n)^n for n->inf

all of these give Euler's formula and neither of them is Euler's formula

>>11516058
>It's the only one us professional mathematicians use.
No it’s not, you LARPing retard. Other theories are routinely used not only in logic and foundations but in areas like category theory and algebraic geometry.

>>11516113
>>>>>>professional mathematicians
LMAOOOOOO SECOND YEAR UNDERGRAD DETECTED

>>11516058
>You are talking about defining something. That is technically an axiom.
You don’t need set theory to have axioms. So again I ask you: Why do you keep blabbering about set theory?

>>11514796
Please don’t take that LARPing retard to be representative of set theorists.

>>11516110
sure, but you needed more than one axiom. Consider this: none of the axioms you gave imply that i is the principal value of root -1. Thus you must also take that as an axiom.
>>11516113
>algebraic geometry
You mean the dead field
>category theory
>>11516116
Yes you do. You need a foundation. Otherwise you might as well be coming up with bullshit.

>>11516105

Not him, but..

>you can work out that multiplying by i is rotating by 90 degrees.
>Gotta explain that
It‘s pretty much the definition of the complex e function.


>And the length is constant
>Why?

The real part of the exponent is the length of the „vector“.

e^(i*pi) = e^(1 * (i*pi))

The real part (length of the vector) stays alwas 1, so by raising the i-part you are simply doing a circle arround the origin.

If you raise both parts, you are doing a spiral (change angle and length of the „vector“).

> so the speed is constant
>Why?

See above.

>>11516121
>You need a foundation.
You don’t need a *set-theoretic* foundation.

>>11516127
It's not really any different, so I will assume it is set theoretic.

>>11516132
Glad you admit you were wrong. Now let’s turn back to the topic at hand.

>>11516121
>sure
so you agree that >>11516095 is wrong
>but you needed more than one axiom. Consider this: none of the axioms you gave imply that i is the principal value of root -1. Thus you must also take that as an axiom.
irrelevant, because my point isn't how many axioms you need. my point is that you keep insisting that the Euler's formula needs to be one of them.

>>11511421
how do you go from [math]\dot{x}=-y&,\,\dot{y}=x[/math] to [math]x=\cos(t)&,\,y=\sin(t)\\[/math] ? don't you need to prove that its the unique solution to the first line?

>>11516135
it's not wrong if there's no difference in the end result. In fact you are wrong. see>>11516121
> none of the axioms you gave imply that i is the principal value of root -1

>>11516105
>you can work out that multiplying by i is rotating by 90 degrees.
i(a+bi)=-b+ai
(a+b)^2+(b-a)^2=a^2+b^2+a^2+b^2,
and it follows from the converse of Pythagoras's theorem.
>And the length is constant
See >>11511431
>so the speed is constant
|i|=1, and |xy|=|x||y|.

You should learn to figure things out yourself instead of requiring someone to spoon-feed you every little thing.

>>11516139
not that anon, but I think it's pretty obvious that sin and cos are the unique solution to the problem

>>11516120
Sorry, set theory is actually really interesting. I was just trying to think of an insult for the retard and "naive" fit almost too well.

>>11516137
This brings me to my next point. Since we want to minimize the amount of axioms in our theory, we take Euler's formula and princinple root of -1 is i.

>>11516125
Please don't come back here until you finish high school.

>>11516125

Oh sorry, I think it‘s supposed to be:

e ^ (i*pi) = e ^ (0 + i*pi).

Therefor the „length“ is e^0 = 1 and the angle is pi which translates to 180 degrees.


I could expand further using cos as adjacent side and sin as opposite side, but I am too lazy and my mobile phone sucks.

>>11516154

Dude, University.
Every moron can give a proof, OP wanted an intuitive explanation.

See my correction here:
>>11516155

>>11516153
>Since we want to minimize the amount of axioms in our theory, we take Euler's formula and princinple root of -1 is i.
my point isn't how many axioms you need. my point is that you keep insisting that the Euler's formula needs to be one of them.
also this has to be one of the most retarded posts I've read on sci holy shit

>>11516149
No no no, it's not obvious at all. Luckily, there is a wonderful theorem that tells that any initial value problem for a non-linear system of ODEs has a unique solution.

>>11516137
There’s no reasoning with the clown you’re replying to. He’ll keep backpedalling and changing his claims ad infinitum, or going around in circles. I’m inclined to believe he’s a troll at this point.

>>11516157
You weren't responding to OP, and that garbage you wrote relies on the assumption that e^(it)=cos(t)+isin(t).
Again, finish high school, then come back here. You are supposed to be 18 anyway.

>>11516166
>my point isn't amount of axioms
but mine is, sucka

>>11516169
>. I’m inclined to believe he’s a troll at this point.
see >>11514278

>>11516169
I'm inclined to believe he tried to appear smart in his first post but it backfired greatly
math is hard, don't give up

>>11516125
>It‘s pretty much the definition of the complex e function.
The whole point is to derive that, no? That's essentially begging the question.

>>11516139
>>11516149
>>11516167
I could have proven it, but it seemed outside the scope of the question. Plus, the proof depends on which of the (equivalent) definitions of the trig functions you choose: if you define [math]\sin[/math] and [math]\cos[/math] by the circle, then it's one proof, another if you define them by their Taylor series, and another if you just define them as the unique solutions to those diff eqs.

>>11516175
Liar >>11516095

>>11516196
>backfired
I am smart. You're the one's typing up big paragraphs and getting mad after all.
>>11516199
I can have multiple points.

>>11516203
>I am smart
we can see

>>11516175
Seriously, what are you even trying to accomplish at this point? Saving face? You know this is an anonymous forum, right?

>>11516205
I'm just trying to help you fellas on sci learn about math.

>>11516203
So you admit your claim that Euler’s formula needs to be an axiom (one of the stupidest things I’ve ever heard) is wrong, right?

>>11516211
nope ;^)

>>11516211
he was just trolling man

also this thread right now is sci at it's best, fucking nice

>>11516221
ITS

>>11516218
I'm no troll, but I definitely saw a lot of caps locks replies and calling me retard, heh. The logical fallacies came from the other guys tho.

>>11516146
Dude, I understand how it works, but if you're going to explain it to someone who doesn't, you can't just say "you can work out that". If they don't get e^pi*i, I'd presume they don't know about complex numbers, or i, or vectors.

>Is this what /sci/ has been reduced to? You guys suck.

Those guys gave way clearer demonstrations, because there wasn't any hand wavy bullshit.

>>11516146
Terrible communicator, 0/10, the differential equation approach is more understandable than yours

>>11517386
>You need differential equations to show that multiplying by i is a 90 degree rotation, preserving length.

>>11511220
Look at the group SO(2) of rotations in the plane. These are 2x2 real orthogonal matrices. This can be viewed as a circle, picking out (1,0) as the identity, and define multiplication of points on the circle by seeing how far you have to rotate from (1,0). So (0,1)^2 = (-1,0) or (0,1)(-1,0) = (0,-1) etc.

This forms a Lie group. A smooth manifold with group structure. We look at its Lie algebra, the tangent space at the identity. The exponential map exp(x) = e^x is interesting because it takes elements of the Lie algebra and maps them to elements of the Lie group.

So we have the unit circle in the complex plane to model SO(2). Then its Lie algebra is the vertical line at (1,0), the purely imaginary line of elements ix. Then the exponential takes ix to a point on the circle, given by exp(ix) = e^ix = cosx + i sinx. You can view this geometrically as folding the line segment from (1,0) to (1,x) on the vertical line over onto the circle. This also makes it clear why it's periodic in 2pi, since if your line segment is longer than 2pi it folds back over the circle.

If you don't like complex numbers. Look at the Lie algebra of SO(2) in matrix form. Matrices in SO(2) satisfy AA^T = I. Consider a path in the Lie group A(t) with A(0) = I. Then differentiate and evaluate at 0 AA^T = 1 to get A'(0) + A^T'(0) = 0. The Lie algebra is anti-symmetric 2x2 real matrices. Up to scalar multiples there is only one, J = ((0,-1),(1,0)). Note that J^2 = -I. The set of matrices of the form aI + bJ is isomorphic to the complex numbers a + ib.

In this matrix form the formula is e^(xJ) = cosx I + sinx J

>>11511409
This is what you get if you identify the Lie algebra elements with velocity.
>>11513748
Identify each point (a,b) with the matrix aI+bJ and that algebra is isomorphic to a+ib.
>>11511421
If you want to ignore intuition, just use this
d/dx[e^-ix(cosx+isinx)] = 0 so its constant
at 0 its 1
therefore e^ix = cosx+isinx
>>11516105
Try thinking

>>11511329
Okay now this is epic

>>11511329
I don't think I've ever seen it quite like this. Many sources seem to use the series definition/taylor expansion/whatever you want to call it. It's probably good to know both.

This proof isn't mine, but is this even a bit simpler for a proof? Is this correct? Better? Neither?

https://math.stackexchange.com/a/8612

>>11517506
Actually just saw that what you said there is probably equivalent to the Math Stack Exchange answer.

>>11511157
By that logic you would end up with -e instead of -1.

>>11517975
Holy shit that proof is excellent.

>>11518132
e^0=1, and since it is growth pushing sideways (probably bad terminology, perpendicular would be better), you get to -1.

>>11517975
>>11518169
Why is it obvious that there is only one function [math] f [/math] such that [math] e^{-it}f(t)=1? [/math]

>>11518178
it's not perpendicular, that's the point - if if was exactly perpendicular it would not change the vector's length.

>>11518190
What vector? What are you on about? Also, the length doesn't change, so yes, it is perpendicular.

>>11518184
I'll try to explain:

His first step is to construct f(t) which is the right side of the formula divided by the left side. This division will be important.

His second step is deriving the function to get its tangent which, as it turns out is zero -> f(t) is constant for all t, and that means you only have to solve f(t) once to get the result for all t. 0 is an easy candidate so he solves it for t=0 which yields 1 -> f(t) is 1 for all t.

Now we're coming back to why f(t) was constructed as the right side of the euler formula divided by the left side. Because f(t) = 1 for all t, this means the division is 1, and that proves that both sides of the euler formula are the same number for all t -> the formula is true for any input t.

>>11517407
THE SUMMONING THREAD WORKED!

>>11518217
>deriving the function to get its tangent
sorry, should have been "to get its tangent function"

>>11518203
In your posted image you show a vector getting rotated.
Anyways, I realize what you meant by "growth". It still doesn't explain why it ends at -1 and not at, say, -5 for example.

>>11518233
Not my image btw. Again, sideways is probably the wrong word. Perpendicular would be a better word, and would explain why it ends at -1.

>>11518217
>Rewriting the proof instead of answering my question.

>>11518265
Then what do you mean by "only one function f"? He took both sides of the euler formula, why would he take one side of the formula and the other side from somewhere else? I'm not really sure if I understand your question if it has nothing to do with you not getting the proof, to be honest.

>>11518283
>He took both sides of the euler formula...
You can't start a proof with what you are trying to prove. You have to prove it forwards.
>I'm not really sure if I understand your question
Basically, why cos(t)+isin(t) is the only function f such that f(t)e^(-it)=1. I think it basically comes down to showing that e^(-it) is well defined. It just doesn't seem immediately obvious that it should be true.

>>11518292
>You can't start a proof with what you are trying to prove. You have to prove it forwards.
No, he didn't start by saying the formula is correct, he wrote the formula as a division and then stated that if the formula was indeed correct, that division must always be 1 (which is easily grasped, if a/b = 1 that means a=b). Then he continued to prove that the division is indeed always 1, which made the conclusion porrible that the formula is correct.

>Basically, why cos(t)+isin(t) is the only function f such that f(t)e^(-it)=1.
But he didn't try to prove that cos(t)+isin(t) is the only possible f, he just proved that cos(t)+isin(t) is one possible f. There might be others, but then it wouldn't be the euler formula he'd be proving.

>>11518301
I'm not certain, but I think that if there were other possible functions, then e^(-it) would be ill defined.

>>11518307
I didn't find anything about the euler function being well-defined, so it might be true or not; I'm not that deep into the topic that I could state one or the other. The proof was just for the euler formula being true for any number t you throw at it.

>>11518313
e^(it) is well defined. It's just that I feel like the proof given doesn't really address it, because it doesn't really even define what e^(-it) is; just that its derivative has a certain property.

>>11518184
>>11518265
>>11518292
he proved that the function e^{-it}*(cost + isint) is constant and equal to one. therefore

1 = e^{-it} * (cost +isint)

is true for all t. now you just multiply by e^it which you can, because it's always nonzero.

there's nothing about uniqueness going on

>>11517975
That's an elegant proof.
You'll notice that all these proofs, when it comes down to it, fundamentally rely on the fact that [math]\exp'=\exp[/math]. The series argument relies on the Taylor series representations of the functions, which are defined by the functions' derivatives (check out Taylor series, they're pretty cool); my proof (the one you replied to) relies on the fact that [math]\log'(x)=1/x[/math] (another way of saying that [math]\exp'=\exp[/math], since they are inverses); and the one you linked also relies on the fact that [math]\exp'=\exp[/math].
It's not really a question of correctness, since all these proofs are correct and rely on the same information in the end. It's just personal preference, and most importantly what is most convincing for your audience. Someone more experienced in the field will probably appreciate a short and elegant proof, but a total novice may see it as "cheating" in some way and prefer a longer, more detailed proof.

http://us.metamath.org/mpeuni/eulerid.html
no ambiguity there, and it shows every axiom needed

>>11518917
Thank you for the link Anon, this website looks fantastic for rigor autists like myself.

>>11511146
have a look at this formula and substitute into it i*pi
what you'll notice is, because i^x cycles in powers of 4, you end up adding and taking away real numbers every second and fourth power and adding and taking away imaginary numbers every first and third power, they end up cancelling out and tending towawrds -1 .

>>11511214
I think "rotating sideways" sounds better

>>11511329
The integrating factor saves the day
/thread