[math]i^i=e^{-frac{pi}{2}}=e^{frac{7pi}{2}}=e^{frac{3pi}{2}}=e^{frac{-2pi n-2}{2}}[/math]
Therefor [math]-frac{pi}{2}=frac{7pi}{2}=frac{3pi}{2}=frac{-2pi n-2}{2}[/math]
uhhh...no it doesn't. Yeah, something's wrong here. Mathematicians are literally contradicting themselves with this shit; and this is simple substitution, a basic property of all logic so what's going on here is completely unacceptable if we want a logic system. I'm thinking we should tear all down and start all over again.
google what the words "injective" and "surjective" mean
Just did, what about it?
My penis inside your mother is isomorphic to your birth.
Wait, my penis inside your mother is isomorphic to your conception. There, that's better.
>my uncle is called Tim
>my brother is also called Tim
>mfw my uncle must be my brother
So what you're saying is [math]underbrace{ecdot ecdot ecdotldotscdot ecdot e}_{-frac{pi}{2}}=underbrace{ecdot ecdot ecdotldotscdot ecdot e}_{frac{7pi}{2}}[/math] right? Wrong!
[math]ab=aciff b=c[/math], therefor if [math]b=underbrace{ecdot ecdot ecdotldotscdot ecdot e}_{n}[/math] and [math]c=underbrace{ecdot ecdot ecdotldotscdot ecdot e}_{m}[/math], then [math]eb=eciff b=c[/math];
[math]b_1=underbrace{ecdot ecdot ecdotldotscdot ecdot e}_{n-1}[/math] and [math]c_1=underbrace{ecdot ecdot ecdotldotscdot ecdot e}_{m-1}[/math], so [math]b=eb_1[/math] and [math]c=ec_1[/math] and [math]b=c[/math] then [math]eb_1=ec_1iff b_1=c_1[/math]; Now repeat this process over and over and you'll find that:
if [math]nneq m[/math], then if [math]n>m[/math] (or vice versa, it doesn't matter for the argument, just change everything to make sense) like [math]frac{7pi}{2}>frac{3pi}{2}[/math] then you'll run out of [math]e[/math]s for one side faster than the other (which means both sides cannot be equal) hence the contradiction.
One and zero are different due to the very definition of multiplication, idiot (viz. [math]0a=0[/math] and [math]1cdot a=a[/math]). There is no such law for [math]e[/math] so it just acts like every other number. Try to calculate [math]e^{frac{7pi}{2}}[/math] and [math]e^{frac{3pi}{2}}[/math] THESE ARE DIFFERENT NUMBERS.
[math] sin(0) = 0 [/math]
[math] sin(pi) = 0[/math]
Calculate 0 and [math]pi[/math]. THEY ARE DIFFERENT NUMBERS
> The sine function
Non sequitur.
> sine 0 = sine pi
but e^7pi/2 != e^3pi/2. That 0 and pi are different numbers is irrelevant as the definition of sine does not work like exponentiation.
>the definition of sine does not work like exponentiation.
[math] sin(x) := operatorname{Im} {(e^{ix})} [/math]
This is not the definition of sine, this is something you must prove. Plus, you assume what I'm disproving rn so you can't use it to prove my disproof wrong.
>Because the operation you're working with is in C.
The operation I'm working with? What? e^x? that's an exponential of a real number so it cannot be complex.
It's a complex number with imaginary part zero. And no, this is NOT the same thing as a real number. For example, log(x) is valid only for x > 0 if you're in R, but log(x) is valid for all x in C.
>And no, this is NOT the same thing as a real number.
So 0i+2 is a real number but 0i+e somehow is a complex number with imaginary part zero? lol, this just comes off as massive cope. They literally are the same thing. e is a real number but you want to treat it as if it had an imaginary part, which it doesn't.
> For example, log(x) is valid only for x > 0 if you're in R, but log(x) is valid for all x in C.
This is dubious. if powers of imaginary numbers work like i^i then there doesn't seem to be a consistent definition of exponentiation in this domain.
Explain how e^7pi/2 and e^3pi/2, which are both clearly different numbers, can somehow be equal? it's not possible in mathematics.
They're not equal, dingus.
Exactly my point, dingus, and yet what does i^i equal?
>provided it's equivalent to the other ordinary definitions.
Yeah, moron, and that's exactly what you have to prove.
[math]i^i = e^{n + pi/2}[/math] for any integer n. Not sure why this is filtering you so hard. It's the same thing as saying
[math]x = pm 2[/math]. It doesn't mean that [math] 2 = -2[/math], dumbass.
honestly i was a little with OP(At least, I saw his point but could tell it was my flaw, not the math's), but only because im too moronic to understand e or complex numbers more than superficially. This changed my mind, makes perfect sense.
have a nice day
>This is not the definition of sine, this is something you must prove
Prescriptivist nonsense.
You can take anything as a definition of sine provided it's equivalent to the other ordinary definitions.
If you want to define sine analytically by its Taylor series, or by the differential equation y''=-y, y(0)=0, y'(0)=1, or as the imaginary part of the complex exponential, it doesn't matter. They're all valid.
>This is not the definition of sine
It's the definition of the complex exponential function, and there is a proof of their equality (second answer here is the standard one https://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-ei-varphi-cos-varphi-i-sin-varphi )
Regardless, what do you think is more likely, that you simply didn't know that the function wasn't injective in the complex plane, or that the entirety of mathematicians somehow missed this for hundreds of years, with some IQfy(nel) poster finally bringing this to light.
e^z being periodic is well known and taught in any complex analysis textbook.
>that's an exponential of a real number so it cannot be complex.
imagine getting filted at 17th century mathematics
>[math] ab=aciff b=c [/math]
Not true in [math]mathbb{C}[/math].
>[math] underbrace{ecdot ecdot ecdotldotscdot ecdot e}_{-frac{pi}{2}}=underbrace{ecdot ecdot ecdotldotscdot ecdot e}_{frac{7pi}{2}} [/math]
My dude. [math] pi/2[/math] is a not natural number.
> Not true in C.
My dude, e is not a complex number, so why does that matter?
> My dude. π/2 is a not natural number.
Irrelevant.
>so why does that matter?
Because the operation you're working with is in [math]mathbb{C}[/math].
The numbers 2 and 1 are Natural numbers. But the division operator doesn't exist for Natural numbers; it does however exist for rational numbers. So when you write [math]1/2[/math], you can be certain you're working with rational numbers and NOT Natural numbers.
His argument is moronic but that is true in C, it is in fact true in any field.
Youre assuming real and even compex powers are defined exactly the same way as integer powers which is fricking stupid.
The complex exponential function is defined as
[eqn]exp(z) = sum_{n=0}^{infty} frac{z^n}{n!}[/eqn]
Learn some math before making these dumbass claims dude
>Not true in C
That is true for any integral domain. What kind of drugs are you taking?
[math]f(a)=f(b)[/math] does not always imply [math]a=b[/math]
I never said otherwise.
imagine not even understanding how exponentiation is extended beyond the naturals LOL do you seriously think e^-1 is e multiplied by itself "-1 times"?
fricking moron just have a nice day
So your granddad had sex with your mom. No contradiction here.
Yes.
[math]x^2 = 1[/math]
[math] x = 1 [/math]
[math] x = -1[/math]
[math] 1 = -1 [/math]
guys i broke math
Somebody was asleep in freshman complex variable class lol. We try to define i^i = e^{i*log(i)}. The problem doesn't arise from taking real powers of e, but of taking logarithms of complex numbers. The function exp is not injective on the complex numbers (try it! clearly 1=e^0=e^{i*2pi} for example) so the log function, taken as the inverse the exp function, is not well defined. To make a complex log function for the complex exp function, we need to look at just one branch of the the log function (recall the way we define arcsine, this is similar). Call this branch Log or the principal logarithm, and let it be the logarithm whose imaginary part lies in (-pi, pi]. In general we use this as the inverse function to exp (note, just as how complex exp agrees with real exp when given a real input, so does complex Log with real log). We define w^z = e^{z*Log(w)}. This gives us i^i=e^{i*Log(i)}=e^{i*i*(pi/2)} =e^{-pi/2} (and is in particular not equal to e^{7pi/2} or e^{3pi/2}, which are different numbers).This may seem wrong somehow, but it is just like saying sqrt(4) = 2 or arcsin(1) = pi/2. Raising a complex number to a complex power isn't something we can intuit through (multiply a complex number by itself an imaginary number of times??) but is a function that must be carefully defined.
I don't remember this stuff that well lol, definitely worth a quick browse through Brown Churchill Complex Variables and Applications or a similar level complex variables book.
This only applies in a unit circle moron
the issue in OPs reasoning is that he linked them all as = to each other
i^i equals each of those individually, but you can't just chain them like that
after e^-pi/2 each answer should be separated by a comma not another =
if a=b and a = c, then b=c is basic logic. What are you on about?
f(a), f(b), f(c)
So you agree that [math]e^{-frac{pi}{2}}=e^{frac{7pi}{2}}=e^{frac{3pi}{2}}[/math]?
transitivity is only sometimes applicable to n-ary relations given a domain D and a situation-space S. If you want to use predicate logic then I'd recommend looking into it. There are plenty of intransitive and nontransitive relations that are permitted within a space. I'd recommend 'logic' by wilfred hodges if you are curious.
>exp(-pi/2)=exp(-pi/2 + 2pi)=exp(7pi/2)
This is precisely the critical mistake you made, you assumed that the real exponential function was periodic with period 2pi in particular for some reason, in reality it's monotonic and you adding 2pi and attempting to preserve the output is absolute nonsense.
nice troll
>what is a multi-valued function
>why does log(z) go bad at z = 0
>what happens when you wrap around a singularity in complex analysis
0*2 = 0*5 therefore 2 = 5.
> [math] sin(frac{pi}{2}) = sin(frac{7 pi}{2}) = sin(frac{3 pi}{2}) [/math]
> Therefore [math] frac{pi}{2} = frac{7 pi}{2} = frac{3 pi}{2} [/math]
[math] sin(frac{-pi}{2}) [/math] but you get the point
taking log of both sides is not a valid operation
Epic.
[math]i^{1} = i^{5}[/math]
[math]1 = 5[/math]
The definition of i is that it is equal to the sqrt of -1 so it is therefor necessarily cyclical; 0a=0; 1a=a; e is nothing like any of these numbers, moron. Nice try though.
The definition of i is [math]i^2 = -1[/math], not [math]i = sqrt{-1}[/math]
its a joke
No mathematician worth their shit will tell you that i^i = a single value. It’s only popsci that pushes the meme.
1^2. = 1^50 = 1
Does 50 equal 2?
how do we get rid of malevolent trolls like OP?
>how do we get rid of malevolent trolls like OP?
why would you want to destroy beauty
(-1)^2=(-1)^6 so 2=6
this is how fricking stupid you sound
Okay, I get that OP is trolling but can somebody explain this?
[eqn]e=lim_{ntoinfty}(1+frac{1}{n})^n[/eqn]
[eqn]e^x=lim_{ntoinfty}(1+frac{x}{n})^n[/eqn]
How does the [math]x[/math] get in there?
[eqn]
lim_{n rightarrow infty} left(1 + {x over n}right)^n =
lim_{n rightarrow infty} sum_{k=0}^n {n choose k}left({x over n}right)^k =
lim_{n rightarrow infty} sum_{k=0}^n {n! over i!(n - k)!}left({x over n}right)^k =
sum_{k=0}^infty {x^k over k!} left( lim_{n rightarrow infty} {n! over k!(n - k)!}left({1 over n}right)^k right)
\
{n! over (n - k)!} = overbrace{n(n-1)(n-2)...(n-k+1)}^{k text{ terms}} sim overbrace{n cdot n cdot ... cdot n}^{k text{ terms}} = n^k implies lim_{n rightarrow infty} {n! over (n-k)!} cdot {1 over n^k} = 1 \
sum_{k=0}^infty {x^k over k!} left( lim_{n rightarrow infty} {n! over k!(n - k)!}left({1 over n}right)^k right) = sum_{k=0}^infty {x^k over k!} = e^x
[/eqn]
Yes I'm not being careful with my limit of sums/sum of limits manipulations. No, I don't care.
stop embarrassing yourself
Isn't that his point? that they're different numbers? and yet they're all supposedly equal to i^i?