New result
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Give creddit to Robert Chuck Knigge
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If it was incorrect then you would be able to cite an equation that produced the wrong answer
Utter nonsense.
Firstly, infinity is not a number, but a limit, so the equation is not even wrong, but undefined.
Secondly, with $$n \in \R$$,
$$\lim_{n\to\infty} e^{in} + 1 = 2 \Rightarrow \lim_{n\to\infty} e^{in} = 1$$
$$\Rightarrow \lim_{n\to\infty} \left(e^{in}\right)^i = 1^i$$
$$\Rightarrow \lim_{n\to\infty} e^{iin} = 1$$
$$\Rightarrow \lim_{n\to\infty} e^{-n} = 1$$
But $$\lim_{n\to\infty} e^{-n} = 0$.
Thus $$\lim_{n\to\infty} e^{in} + 1 \neq 2$$
Shit, I thought 8ch had a LaTeX mode.
This comment is totally stupid because you are talking about the real numbers but the paper used the extended real numbers in which infinity is a number, not a limit. Since you don't even say what type of number you are talking about I can see that you are totally stupid, or else you are trying to deceive the uneducated by making a point in a completely different number system than that which appears in the paper.
In the paper it says that "This paper examines some familiar results from complex analysis in the framework of hypercomplex analysis." Hypercomplex numbers are constructed by adding the transfinite component of the hyperreals to the extended complex numbers which are, themselves, constructed by adding the imaginary component of the complex numbers to the extended reals.
You are stupid to try to extend you point from R to all number systems, and you are extra stupid to try to raise that point in the extended reals which were invented specifically to nullify that point that you raise, and only that point. Other than that, the extended reals are just like the reals but you are so stupid (or deceitful) that you the ONLY point from real analysis that doesn't carry over to extended real analysis.
Damn you are stupid, and you suck.
So... since you are so bold as to attempt to demonstrate your scholarliness, tell me why you have chosen to make a point of R relating to a paper that uses extended R specifically not to bump into that issue you raise? Don't be stupid, answer the question.
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How is it clearly false?
Stand aside, this little proof's gonna make me rich and famous.
If it did make me rich and famous, that would be really great. However, that's not what I'm complaining about. I'm complaining that it doesn't earn me enough to buy a meal or rent a room.
So the result is only relevant in the irrelevant little world that the author imagined for himself and the title is pure clickbait. Got it.
More like, a paper that extends R in just the right way to produce some wacky "unintuitive" results in order to make the author feel like he's done something interesting.
Also, 1 + 2 + 3 + 4 + 5 + ... = -1/12. Prove me wrong :^)
Math tends to degenerate into mental gymnastics very quickly as you ascend the complexity curve. Do note however that it's still technically correct and is perfectly valid. Shit like doubling a sphere. You can't produce two separated spheres that simultaneously use all points of the original sphere, obviously a fraction of these went into each of the new spheres. But with the uncountably infinite real numbers and ill-defined mess that is "metric", you do actually get two identical spheres. Or however many as you want actually.
Both of these point are wrong and stupid. Someone else extended R that way like 100 years ago. The extended real line isn't something I invented. Furthermore, whoever invented extended R did so precisely to get rid of your stupid "infinity isn't a number" claim. So please, acknowledge that I am not the inventor of the extended real line, and that you argument fails in the extended reals as much as it does in the little world I invented.
If I can paraphrase your argument, you're saying, "Analytic continuation isn't allowed," but it is allowed. If you think it isn't allowed then that you means you don't know shit and are probably retarded.
I have not ascended the complexity curve very high here. All I really do is say, "Consider the analytic continuation of R in stead of R itself," and that is a very low-complexity proposition.
Okay. 17+12 = 13
So you are saying that e^-∞ = 1 and not 0?
Because your results implies that it equals 1.
I don't see that. Please explain how my result regarding the complex exponential gives the result you wrote.
there's not even a smidge of math anywhere in the universe that didn't get made up in someone's head.
e^(i∞) +1 = 2 => e^(i∞) = 1
=> (e^(i∞))^i = 1^i
=> e^(ii∞) = 1
=> e^-∞ = 1
OP BTFO!
38. and 39. evaluates inf - inf to be zero when it's actually an indeterminate form so you can't evaluate it.
You see OP, nobody pays you for your mathematical "results" because they are wrong.
If they were new and correct, you could easily get a PHD out of them and find yourself a job as a university professor.
That is not true in the extended real numbers. In the extended reals inf - inf = 0. Please be less stupid.
If something was wrong then you would be able to point it out
Done already in this thread.
Now delete this thread and then delete yourself!
Extended real number set just adds +inf and -inf to the real number set. It sounds as if they just slapped them on there without much reasoning. I haven't read anything beyond that though.
Here's what you did with π/2 subbed for infinity:
exp( iπ/2 ) = i
[ exp( iπ/2 )]^i = i^i
exp( -π/2) = i^i
Does exp( π/2)=i^i? No, it doesn't. Pic related, there are a lot of things like this than can be shown, and yet we don't say
exp(ix) = 1
even though my pic clearly shows that it is. There is some nuance about the definition of exp as a function of a complex variable versus a function of a real variable but honestly I can't remember the detail.
Yes that's right. Someone got tired of hearing "but infinity is a limit not a number" so they invented the extended real line. That's all there is to it.
First of all, you removed the negative from the exponent you sneaky little shit.
exp( -π/2) = i^i is a TRUE statement you fucking IDIOT OP!
Even software can figure this out.
wolframalpha.com
I don't see why you need a number that no matter what you do to it, it will only return itself.
I'm starting to think OP isn't the actual author because he is just embarrassing himself after being completely proven wrong.
You have bigger and smaller infinities. The size of the set of integers is infinite but smaller than the size of the set of reals.
I'm a total mathlet, but
how did you get rid of the ^i on the right side of the equation?
1 to the power of i is equal to 1.
Stop bumping this retarded thread.
1 to any power is 1.
Just adding in infinities doesn't give you the permission to be reckless. For example: assuming inf - inf = 0
4 - 2 = 2
4(inf) - 2(inf) = 2(inf)
inf - inf = inf
0 = inf
We have reached a contradiction
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KDE krunner got it correct
that's pretty cool
One should use notation
"inf" --> \hat{Phi}
which suggests not to take the product of 4 and 2 with infinity because of the basis vector interpretation bestowed by the hat. If one uses the inf symbol then the rule must apply accordingly.
Well that is very interesting that it is possible to derive an identity which says i^i ~ 0.207. I didn't realize what that trick there was because I dropped the minus. That is a very interesting result. Therefore, your criticism deserves more scrutiny.
The identity that i^i = 0.207 is derived from a more general result that says it can take many values. Therefore, you have to choose it to be equal to to exp(-pi/2). Therefore, I do not see a hard refutation in your criticism.
Furthermore, I would note that the infinity in your contradiction
exp(-inf) = 1
is like the result that sine at infinity was like infinity, which was equation (27) in the paper. We needed to shift the coordinate from the hypercomplex piecewise coordinate back to the regular coordinate. Therefore, when we convert from y+/- back to y', with
y = inf - y^+
then we get
exp(-0)=1
which is the correct answer. Obviously I didn't write out all the steps but, to interpret this result in the regular way, we need to unconvert the hypercomplex numbers. Clearly the result
exp(-inf) = 1
is valid in the y+ coordinate. Therefore, I consider this a refutation of your criticism although truly one would write out all the steps. If you do it and show that I am wrong then I will gladly have a look. Thank you for you well formed, thoughtful, and constructive criticism.
actually, you wouldn't switch the coordinate. That was wrong. I would refer the y+ coordinate only being defined in (0,inf). If the exponent is negative, then there is problem due to pic related because the sign is not supposed to change. If it the sign changes, then it becomes like y- on the higher level of infinity. Your criticism is really good, and I will examine it more closely and post a response later. Thanks :)
that was the single best criticism I have ever received any point in my research. Huzzah!
infinity is both negative numbers and positive numbers
negative span + positive span = 0
you can get the same with
> e ^ ( i 0 ) + 1 = 2 (approximation)
combine all numbers = infinity
infinity includes itself (0)
0 = ... 3 + 2 + 1 + 0 + -1 - 2 -3 ...
the question at hand is infinity the the true infinity?
I have not yet completely resolved this.
That system is different from what OP is working in.
I'm not sure exponentiation is commutative with complex numbers. I know it's not with rational numbers.
Fucking nonstandard analysis fuckery
Alright, so the author didn't extend R himself, he just carefully cherrypicked an already existing special snowflake extension to get an "unintuitive" result in order to make himself feel smart. That's better.
And we can see from OP's paper where that leads. From bullshit axioms, bullshit results. Garbage in, garbage out.
If I can paraphrase your argument, you're saying, "I'm a faggot and I like to put words in people's mouths, please rape my face," and I agree.
Also,
e^i∞ + 1 = 2
e^i∞ = 1
e^i∞ = e^0
i∞ = 0
∞ = 0
∞ + 1 = 1
∞ = 1
Therefore, given that 0 = ∞ = ∞ = 1, we conclude that 0 = 1. This must be the birth of a mathematical revolution!
this criticism is merely an extension of an existing criticism in complex analysis. Since the hypercomplex numbers continue the complex numbers, it makes sense that you can derive the case of 0=2npi for n=infinity. Your argument totally fails because I claim to continue the notation, not cure its pathologies. The result 0=2npi is well-known and is not accepted to invalidate all of complex analysis so neither should it be taken to invalidate hypercomplex analysis. However, pic related, I have looked that i^i issue raised earlier and come up with a hard refutation, I think, where several of my previous arguments were not totally robust.
How about equation (18)? It says 0*infinity^i is zero. What is infinity to the i? We can probably choose this to be zero too.
That's not how exponentiation works. For example, e^2(pi)i = e^0, but 2(pi)i is not 0.
You crealy don't know what an injective function is, 2^2 = (-2)^2 but 2 != -2.
t. mathfag
The complex plane, the range of wavefunctions if not also their domain, goes onto the Riemann sphere. I wanted to continue the region beyond infinity onto another Riemann sphere tangent at the polar point. To include the connective polar point in the topology, I needed to make the continuation to the extended complex plane which is a superset of the extension to the extended real line. Therefore, I chose this snowflake extension because it was the one created to include the point at infinity. It is needed to smoothly go beyond infinity, as is the gist of the MCM transfinite schema.
that's a good one
arithmetic error
more edits.
I see I still even dropped an exponent "n"
I usually edit this stuff a lot before I post it. I am horribly out of practice with basic mathematics.
Consider that e^(n*pi*i) = 1 when n is even and -1 when n is odd
Also, a*∞ = ∞ where a is any real number.
Then, e^(∞*pi*i) = e^(∞*i) = 1 when ∞ is even and -1 when ∞ is odd.
Assume that ∞ is odd (it is a pretty odd "number", after all).
Therefore, e^(i*∞) + 1 = 0. Checkmate, atheists!
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There was nothing commutative in that proof.
For exponentiation to be commutative would require x^y = y^x.
Nowhere is such a thing done in the proof you are responding to! It is a correct proof!
I fucked up when I said commutativity. I think it's associativity. Here's what could be the error.
a^(bc) isn't always (a^b)^c
For example. (-1)^(2*1/2) is not the same as (-1^2)^(1/2).
I'm not sure how this would play out with complex numbers though. I haven't done much for complex analysis, but I know weird shit happens. For example, I think log becomes a multivalued function when complex numbers are involved.
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Implying he is not the author. Just look at the references.
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Hmm, I forgot that it is only associative for integers, and i isn't an integer.
The identity (e^(ix))^i = e^-x does hold for x < pi. Infinity is greater than pi though...
It seems I will need to come up with another proof to destroy OP.
I should note that the equality seems to hold empirically, but I haven't proved it yet.
You could just ignore it and let the homeless indigence take care of it for you.
You claim you're using non-standard analysis, so why am I seeing limits in your paper? Where are the standard parts? Also you say
lim(theta->inf)[cos(theta)] = 1
Is this some different definition of the limit?
Your questions are too stupid for me to take seriously.
GTFO you cancerous wannabe namefag
>>>/out/
It's OK, every true Zig Forumsnician can parse TeX in their head.
The level of mathematical retardation in this thread is amazing. Stick to computer programs, guys. Math is not your forte.
Alright, I'll ignore your stupid result then.
