Brain teaser

Anyone?

Come back with a better sketch and a written explanation of the problem.

>...
off yeself and your unrigorous teaser.

The problem is to find a numerical evaluation (or a set of such) of the expression, if there is one.

-i is under recursive negative roots. The roots themselves are to the root "recursive negative roots of -i". Those roots themselves are root "recursive negative roots of -i"... and so on, recursively.

Also, sorry, I missed a couple of minus signs before a couple of roots, but I don't think that affects the understanding of the concept too much.

Here's a simpler, clearer sketch. Apologies for the confusing nature of the first one.

Why are the indices on top of the radicals?

They are not indices.
The expressions written in smaller font "above" the radicals represent nth roots of the expressions under the radicals.

That's... that's what an index is?

...

Well, yeah, I could just as well have written the whole thing in terms like ((-i)^(1/-i^...^1/-i))^...^(-i^(1/-i^...^1/-i)), but that would be confusing.

...

S=sqrt-expression
=> S^S=-S
Now you can solve it.

...

Didn't you miss the "Sth root" part? It's not a sqrt, it's nth root, or rather in this case, "Sth root".

So:

S = (-S)^(1/S)

[math]S=e^{\frac{i \pi }{S}} S^{\frac{1}{S}}[/math]

Now we need to incorporate the -i term into that somehow.

[math][(z+1)^a=\frac{\int_{\gamma -i \infty }^{\gamma +i \infty } \frac{\Gamma (s) \Gamma (-a-s)}{z^s} \, ds}{(2 \pi i) \Gamma (-a)}\text{ : }0

I'm not sure, but I think these are two different answers.

Well, sqrt(-i) = +i

so [math]i[/math] is closer to 'transposition' or 'flip around the positive/negative'.

So, it is a - operator on a collective value. At least that's the interpretation that allows all the math I understand to function.

>sqrt(-i) = +i

really? I though (+i)^2 = -1 (not -i)

sqrt(-i) should be i*sqrt(i) = i^3

>sqrt(-i) should be i*sqrt(i) = i^3

sorry, I mean just i*sqrt(i).

i^3 = -i = (i*sqrt(i))^2

Ohhh, you want that lookup table thing. That's Integran knowledge.

True : [math]i^3=-i=\left(i \sqrt{i}\right)^2[/math]

[math]i==p+\frac{e}{p}==x-(i e)==\frac{y}{t}+yi==p+\frac{e}{p}==x-i e==\frac{y}{t}+y[/math]

Wait a sec.. how can i be both simultaneously larger and smaller than 0?

Oh, because 0/0 and 0/1 are the only two states where 'divide by zero' is a valid operant.

The 0/0 is just non-constructive (i.e. NaN)

0/1 is when 'you' are saying that you are the prime example in an equation.

You also know this as binary.

You are never 'outside' the circle, so you cannot reference a 2d-circle directly and that is the ONLY thing denied to you. Everything else is 'oki doki'

It's a left/right thing (aka male/female | fire/water | up/down | black/white)

Sorry, to expand

N/0 = FALSE
0/N = TRUE

N is zero in Roman Numerals, if you're curious.

It's why we use it for our notation of 'n[x]'

Godel called it 'symbolic shunting'.

Ah, upper-case and lower-case can also be a fucker :P

n = Integer
N = Set

n/0 = False
0/N! = True

N is also 'superset' or basically a value you can perform factorial on. Also used for Newtons (When maths becomes 'applied something').

If you still need a hand.

Ultimately we are constructing pyramids. Every argument must be balanced.

Equations are only for arguments unrelated to each other (why is this set and that set giving a fuck to talk to each other?) So we have n-recursion and n-levels. If you have x + y, then literally infinity can 'manipulate that however it wants'. It only ever needs the intersection of two events, in an 'n-circle'. Micro and Macro are just 'circles at scale'.

To sacrifice things it doesn't understand it just throws it in a universally understandable 'but life goes on' kinda way.

Communication only happens at the layers of existence that are 'unrelated on one axis, but are on this other axis', and it achieves the process via 'translation'. Literally.

Guys, how would I sketch all solutions to Log(-1), I got it to Log(e)iPi≈ 1.364i and obviously there can be many solutions for pi+2Pi*n, would it just be a circle of radius 1.364...

Thanks.
Not sure I understand much, but I kind of get a little bit of the gist.
I appreciate your help though. If I were a qt girl I'd be your gf.

You'd have to be my [math]P H E N O T Y P E[/math]

Oh, basically mathematics is 'counting all the linked two-ring sets'. It keeps forgetting that ITSELF is an isolation filter so can get a little lost in its moon-runery.

If I let myself suck my own dick for a second here, I'd attend the fucking lectures I gave.

Man... LSD needed to come to the maths department, like, WAY sooner in our species education cycle.

[eqn]Log(-1)=(2n+1)\pi i,\space n\in\mathbb{Z}[/eqn]

Can you explain to me why its 2n+1?

[math]n = 1/2 i (π + i)[/math]

[math]Log(-1)=x[/math]
[math]e^{x}=-1[/math]
[math]re^{i\theta}=-1[/math]
[math]r=1[/math]
[math]\theta=\pi + 2n\pi[/math]
[math]\theta=(2n+1)\pi[/math]
[math]e^{(2n+1)\pi i}=-1[/math]
[math]x=(2n+1)\pi i[/math]

Am I being a complete retard here or is my understanding fucked

e^(2n)pi*i = 1

(Euler's identity) - whenever n is Z, you just add a full revolution of 2pi on the complex plane and end up back at 1.

e^(2n+1)pi*i = -1

This is the same as the previous one: 2n gives us n full revolutions taking us to 1, but now we add another half-revolution (+1pi), taking us from 1 to -1.

take ln of both sides:

ln(e^(2n+1)pi*i) = ln(-1)

ln(-1) = (2n+1)pi*i

Seems okay to me so far.
log(-1) = 2 n + 1
i π = 2 n + 1
-2 n + i π - 1 = 0
n = 1/2 i (π + i)

oh shit yeah haha Im so retarded, so the 2n+1 keeps it an odd coefficient of Pi, thanks a bunch mate

[math]\frac{1}{2i(\pi+i)}[/math] is not in [math]\mathbb{Z}[/math]

-i/(2 (i + π)) is a transcendental number

[math]cos(\pi) = -1[/math]
[math]z = r(cos(\theta) + isin(\theta))[/math]

why are you setting n equal to a non-integer number

otherwise how can I fit it into [math]\frac{1}{2 i (\pi +i)}=-\frac{i}{2 \left(4 \sum _{k=0}^{\infty } \frac{(-1)^k}{2 k+1}+i\right)}[/math]

my last question is, if you calculate the exact value of Log(-1) you get 1.3644..i, but this value cannot be obtained by the form iPi(2n+1), whats going on here?

Math. You're on a math board. This is math.

dude im in my first year I know fuck all, I need you to enlighten me with your infinitesimal wisdom

> if you calculate the exact value of Log(-1) you get 1.3644..i

Where did you calculate this? Wolfram gives i(pi)

>infinitesimal wisdom

lol nice one

This is going to appear kitchy, but it is a pretty good example of 'LOOK AT HOW FUCKING UNIVERSAL MATHEMATICS IS!' Basically the number 7 is the 'computational cieling' for cubes (powers of 3).

[math]E D U C A T E[/math]
[math]A C D E^2 T U[/math]
[math]\int A C D E^2 T U \, dA=\frac{1}{2} A^2 C D E^2 T U+\text{constant}[/math]
[math]\int _{-L}^L\int _{-L}^L\int _{-L}^L\int _{-L}^L\int _{-L}^L\int _{-L}^LA C D E^2 T UdUdTdEdDdCdA=0[/math]

if you google Log(-1) it gives you 1.36437635 i , thank you for having some form of patience with me.

>if you google Log(-1)

google recognises the "log" function as log(base 10), not the log(base e).

You're getting the answer for log(base 10)(-1).

Log(base e) is usually written as "ln" anyway.

yeah, shit the bed, I was really sweating hard about this, always because of some some simple stupid shit, I was thinking why would it be base 10.

Probably helps to think of the left-most associated rule of order of operations/powers. The S identity.
[math]0

although the other question remaining is this:

if you try and calculate 10^(1.3644*i), you'll get about 23, and nowhere close to -1.
Which I can't explain.

> Γ(n+1)

>if you try and calculate 10^(1.3644*i), you'll get about 23, and nowhere close to -1.
Which I can't explain.

Oh wait, no. It works fine now.

[math]n!1\land n^n>n\land n!>n^n\land n \log (x+y)>n![/math]

Anyway, thanks for the assistance guys, much appreciated

No, I didn't miss it. I took it to the Sth power to get rid of it, then we are left with -root-thingy again. multiplying that by -1 makes it into S once more.

So does that mean it doesn't matter what the term under the root is?
It could be any x (instead of -i as in the OP), and the expression would yield the same answer?

Is this University maths? Because I'm in 5 high school ( european ) and I found this very easy ( did Z last year and I didn't do logarithms applied to Z )

You do complex analysis in high school?

High schools usually only go up to trigonometry and real analysis.

I live in Italy then school program is different from yours, in 4th year I did exponential, logarithms,
goniometry, trigonometry, complex numbers ( a+ib form, goniometric form, equations of few curves like Line, Circle, Archimedean spiral and cardioid, but we only introduced Euler's form not solving any log ), then we did a little of solid geometry and cartesian coordinates ( x, y, z ) and then we introduce limits. This year I asked my professor if we are gonna do Complex number in calculus but she said no

I answered to the wrong thread


I live in Italy then school program is different from yours, in 4th year I did exponential, logarithms,
goniometry, trigonometry, complex numbers ( a+ib form, goniometric form, equations of few curves like Line, Circle, Archimedean spiral and cardioid, but we only introduced Euler's form not solving any log ), then we did a little of solid geometry and cartesian coordinates ( x, y, z ) and then we introduce limits. This year I asked my professor if we are gonna do Complex number in calculus but she said no

Ok, so no complex analysis then. Which is what I said.

unless by "goniometric form" you mean "trigonometric/polar form" of complex numbers, in which case, yes, that's complex analysis.

What I don't understand is how you can introduce euler's formula without first introducing limits, since euler's number can't be understood without a basic understanding of a limit.

Yes I mean polar form
We demonstrate everything but Euler's form, on the books there is no demonstration it was just a notion to make us know that in Complex you can reassume algebric, trigonometry and exponential in one mathematical formula... and that Euler identity is considered "the most beautiful formula of maths" because of this reasons I said before
Now looking at wikipedia I understand the demonstration by derivative

Sorry I don't really know the differences between calculus and analysis

as soon as I finish to have dinner I'm gonna take a look to demonstration by limits

Analysis studies functions.
Calculus is a major part of analysis that focuses on rates of change.
Complex analysis studies functions applied to complex numbers.

Had to search "rate of change" but thanks

What is this abomination??

We just had a whole thread dedicated to answering that very question. You couldn't be bothered to scroll down?

you know how I can tell you didn't pass Calc II?

It seems like this is the case. However the answer will still probably be something complex.