/sci/ - Science & Math

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indeterminate

Indeterminate (±1)

A whole number.

if you treat replace the infinity symbol with x and take lim x-> infinity, then well, if you are just doing integers ( n=1,2,3,4,.....) then it's a non-convergent sequence alternating between -1 and 1.

If you want x to be continuous then it's still -1 and 1 for when x is an integer but if not then it goes off into the complex plane a little bit then comes back out....


undastand?

>>5256695
It could be -1 though.

>>5256705
FALSE

>>5256695
i think he meant to say |x| = 1

-1^(inf) = -(0^0)^inf = -(0^0(1/0) = -(0^(0/0)) = 0^(0/0) = 0^(0^1/0^1) = 0^(0^0) = 0^1 = 0

>>5256729
0^0 = 1

>>5256729

0^0 is undefined

<span class="quote deadlink">>>5256736[/spoiler]
it is, use an online calculator

-1^inf = - (1^inf) = - (1) = -1

>>5256748

NU

HELP

DONT WANT TO GET BANNED FOR HW THREAD

ITS A GRE PROBLEM

"A teacher is administering a test to her 3 remedial students, and found that the scores were consecutive. The average of the three consecutive real numbers is 36, what is the smallest of these numbers"

so, is it a + b + c / 3 = 36, then what? 3 random variables?

<span class="quote deadlink">>>5256758[/spoiler]

2

<span class="quote deadlink">>>5256758[/spoiler]

N
N + 1
N + 2
3N + 3 = 36
N + 1 = 12
11

<span class="quote deadlink">>>5256758[/spoiler]
Real numbers don't have any notion of "consecutive". Unlucky.

>>5256691
divergent

>>5256778

Is the set of integers in the set of real numbers?

Are integers consecutive?

Then do real numbers have a "notion of "consecutive""?

>>5256783
>Is the set of integers in the set of real numbers?
Yes
>Are integers consecutive?
Not all of them. Most combinations of two aren't
>Then do real numbers have a "notion of "consecutive""?
No.

Infinity is a multiple of 2, and thus an even number; -1 raised to the power of an even number is 1. Thus (-1)^inf is 1

>>5256786

Can integers be consecutive?
Yes. (Hint: Not all of them implies "yes")
Not to mention, this is a strawman of an answer: is there not a subset of numbers within the reals that is consecutive? Yes.

Do real numbers have a notion of consecutive?
Yes.

checkmate. Get out.

>>5256800
infinity != 2*x

x E R

>>5256809
>Is the set of integers in the set of real numbers?
No
1.000... =/= 1

fuck you, infinity.

>>5256809
>Is the set of integers in the set of real numbers?
No
1.000... =/= 1

fuck you infinity.

-1

well, -1*-1=1. if you take -1 to the power of infinity, you're multiplying -1 by -1 for a result of 1 then multiplying 1 by -1 for a result of -1, in sequence, forever. so it's +/-1 as far as we can determine.

Sn=(-1)^n=-1+2-2+2-2+...2, with n occurrences of 2.
-S=1-2 -2+2-2+2... is the limit of Sn
2-S=2+1-2 -2+2-2+2...=1 -2+2-2+2...=-1+2 -2+2-2+2...=S

2-S=S => S=1.
<span class="math">(-1)^{\infty}=1[/spoiler]
Thus (-1)^infinity=1

>>5256918
Ups, disregard the last line, I reformatted it in LaTeX and forgot to delete it afterward.

<span class="quote deadlink">>>5256758[/spoiler]
If they were integers it would be 35

>>5256783
>Then do real numbers have a "notion of "consecutive""?
Real numbers are uncountable, meaning you cannot enumerate them by saying "this is the first, then comes this one, then this one, etc" and end up counting each of them if you wait long enough. Integers are countable because for any integer, say for instance 142857, if you count long enough, you will reach it.

The negative is not in parenthesis so i'm going to say -1 is the answer.

>>5258302
>in parenthesis
a lone parenthesis does nothing

>>5258309

whoops, meant to say I viewed it as -(1^infinity) not as (-1)^infinity

>>5257038
What is the first integer? Can you say where the set of integers begins and ends? The only difference in this respect between integers and reals is that there is nothing between consecutive integers (such as 1 and 2), whereas in between every real are infinitely more reals.

>Integers are countable because for any integer, say for instance 142857, if you count long enough, you will reach it.
That's entirely dependent on where you start counting and in which direction.

>>5258313
Well you viewed it wrong. Unary operators have the highest precedence.

>>5258323
>Implying both operators in OP aren't unary.

i cant believe no one has gotten it so far. its +-i

+-i.

(-1)^inf = (-1)^(1+1+1+1+1+1+1+1+1+1+1+.....)
1+1+1+1+1+1... = -0.5
(-1)^-0.5 = 1/sqrt(1) = +-i

>>5258326
Do you agree that the definition of "unary operator" is an operator that applies to one operand? And do you agree that exponentiation applies to two operands (<span class="math">x^y[/spoiler] where <span class="math">x[/spoiler] and <span class="math">y[/math are operands)? Then you agree that exponentiation is not a unary operator.[/spoiler]

>>5258328
*sqrt(-1)

>>5258332
You are right. I have to admit that i derp'd.

>>5258321
You can construct a list that contains every integer (0, 1, -1, 2, -2, 3, -3...) in which every integer appears. This is no true of real numbers. What are you doing in a maththread if you don't even know about countable vs. uncountable infinity?

<div class="math">1^\infty=\lim_{n \to \infty}\left(1+\frac1n \right)^n = e</div>
<div class="math">-1^\infty=-e</div>

it's equal to the limit of the following as n -> infinity

sum(((-1)^n)*n)

Doesn't converge, therefore the limit doesn't exist.

>>5258321
You have no idea of what you're talking about, I suggest googling "cardinality real natural numbers" or something

It's whatever you want it to be. That's how mathematics works.

>>5258321
Oh man... oh man... I'm trying to give the intuition of what is countable and what is not. Obviously when I call an integer "the first integer", it's obviously because "X is countable" is defined as "there exist an injection phi from X to N", and when you "count" a countable set, you count in order of the images by phi. The set of real numbers is uncountable because there is no such injection, thus you cannot choose a sequence that lists all real numbers. You can for integers. You can for rational numbers, even if there is an infinite amount of rational numbers between any two rational numbers.

You seem to think that the fact that there is an infinite amount of real numbers between any two real numbers has anything to do with R being countable or not: this is plain wrong. Q shares this property and is trivially countable as it is in bijection with N^2.

Beside,
>>Integers are countable because for any integer, say for instance 142857, if you count long enough, you will reach it.
>That's entirely dependent on where you start counting and in which direction.
In my whole post, I meant "non-negative integers" when I said integers. People use different definitions, usually the context helps understand which one is used. But anyway, I here assumed that we would count from 0 and increase by one at every step, because that's the easiest way to count N. But if you want to count Z, the same property happens. Any of the infinite amount of bijections phi from Z to N is such that phi(142857) is finite, obviously.

>>5258661
<div class="math">1^\infty=\lim_{n \to \infty}\left(1+\frac1n \right)^n</div>
Yeah that is obviously right. What could go wrong here?

>>5258328 and >>5256918 are much better. I root for >>5256918 because the summation is direct instead of being done in the power first.

>>5258728
Saying integers when you mean naturals is just plain sloppy, and bad form.

{1, even infinity
-1, odd infinity}

>>5258753
well then, the true question is: is the limit of infinity as infinity approaches infinity odd or ever?

>>5258741
Maybe. I'm not a native English speaker. It's not like that makes my post wrong anyway.

Also, I googled a bit to see if indeed it was more or less universal that integers are Z and not N, and I found http://math.about.com/od/mathhelpandtutorials/a/Understanding-Classification-Of-Numbers.htm
>Rational numbers have integers AND fractions AND decimals.
>An example of a well known irrational number is pi which as we all know is 3.14 but if we look deeper at it, it is actually 3.14159265358979323846264338327950288419.....and this goes on for somewhere around 5 trillion digits!
>Real numbers include natural numbers, whole numbers, integers, rational numbers and irrational numbers. Real numbers also include fraction and decimnal numbers.
>I'll leave it that complex numbers are real and imaginary.

Reading this made me extremely sad.

>>5256918
This is the only reply ITT that has actually made any sense, beside the ones claiming that the result is inderterminate, which may or not be the case.

>>5258328
I think you may have misread or just completely forgotten what you were doing? Lol,

>1+1+1+1+1+1... = -0.5

You're confusing that with 1+1-1+1-1...

Also:
>>5258818
>>5258741
>>5258728
>>5258321
>>5257038
>>5256783
>>5256778
guys what the fuck is your conversation even accomplishing. None of you have even mentioned how it is relevant to the OP's question.

test

>>5258989
>You're confusing that with 1+1-1+1-1...
no, im using a summation method to get a value for a divergent sum.

http://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%C2%B7_%C2%B7_%C2%B7
1+2+3+4+... = -1/12 is a better example:
http://en.wikipedia.org/wiki/1_+_2_%2B_3_+_4_%2B_%C2%B7_%C2%B7_%C2%B7

-(1^infnity) = -(0.999...^infnity) = -0 = 0

is infinite even or negative or imaginary or zero?

Indeterminate, and please, stop trolling.

>>5259065
no, its +-i

>>5259065
>Indeterminate

That word only has a mathematical meaning when you're in high school.

if there were a solution, we could at least say it has to be on the unit circle on the complex plane

but every element of the empty set says fuck you

both of the statements above are right

>>5259225
i want a blojob from him.

>>5259203
Learn some analysis.

Indeterminate = undefined.

>>5258808
infinity is not a number that's an abstract notion that does not cover the odd/even concept. But if you *really* want to have this kind of shape in mind (but that's misleading ...), infinity is neither odd or even, or if you want infinity is odd and even at the same time.
But you can define it as a limit that *converges* to infinity. In this case:
lim x->infinity (-1)^infinity =
{ -1, when x%2 = 1 (x is even)
1, when x%2 = 0 (x is odd)}

>>5259262
sorry if mixed even/odd congruence,
{ -1, when x%2 = 1 (x is odd)
1, when x%2 = 0 (x is even)}

>>5259047
That isn't the arithmetic sum, though - how is it useful here? You can't take the zeta sum and just apply it to this problem, then you're solving a completely different problem. The zeta sum doesn't actually give you a value equal to the sum, it just assigns the term a useful non-infinite value, but I don't see how that value is useful here. Except in that it produces a defined answer. But that answer might as well be anything.

>>5259065
How could a simple math question possibly be considered trolling...?

Everybody in this thread is retarded:

According to the order of operations, we take 1^(infinity) first. This is 1. Then we multiply it by -1. The answer is -1.

>>5259356
well you cant just trough infinity into a problem, you need to define the limit or series leading to the infinite term, so i assumed it is 1+1+1+.... I guess you can formulate the problem as (-1)^any term which in the limit goes to infinity, then you find that it can be any value on the unit circle in the complex plane, so (-1)^infinity = e^i*theta for any theta. so its indeterminate.

>>5259371
we like discussing math more than simple notational tricks, so we are assuming OP means (-1)^inf

>>5259378
Fair point, fair point. It's better than just putting infinity in there. Still... it's a very specific case of the infinite series interpretation. I think in most contexts, it would be indeterminate, to be honest, as you said.

>>5259371
But... but... you are contradicting this guy >>5258323

>>5259451
Unless this anon got it right:
>>5256918
But I don't really understand where the extra -2 comes from at the beginning of -S... culd someone explain that proof a bit further, if anyone gets it?? I don't really understand how that -S term was found.

>>5259469
Wait, shit, I get it now. 1-2. I'm an idiot.

>>5259460
W-well this guy >>5258323 is retarded.

>>5259469
it shows 1 is a possible answer, just like >>5258328 showed +-i is. he assumed the additional structure of -1+2-2+... and took the limit of the series as n goes to infinity while the other guy made the assumption of the infinity being from the series 1+1+1+1+...

>>5259535
Yeah. I guess you could also sum up the -2+2 terms using a different set of brackets and leave an extra -2 at the end; the same ambiguity present in the Grandi series. So even with his method there are other answers

Fucking maths, man, why is the answer to all the interesting stuff always "it depends on your interpretation", lol.

0 cause I'm guessing.

>>5259573
Presumably because if it can only be interpreted in one way, it's too trivial to be interesting.

-1 because order of operations. I have no idea what <span class="math">(-1)^\infty[/spoiler] is though

>>5259670
holy shitballs kill yourself

±inf ?

>>5256691
no parentheses => -1

But with parentheses it gives wolfram-alpha a boner

<span class="math">1+1+1+... =-1/2[/spoiler]

<span class="math">a.a.a...=a^{1+1+1+...}=1/\sqrta[/spoiler]

Set <span class="math">a=-1[/spoiler]

<span class="math">(-1).(-1).(-1)...=-i[/spoiler]

<span class="math">1+1+1+... =-1/2[/spoiler]

<span class="math">a.a.a...=a^{1+1+1+...}=1/√a[/spoiler]

Set <span class="math">a=-1[/spoiler]

<span class="math">(-1).(-1).(-1)...=-i[/spoiler]

>>5259810

SHPADOINKLE

>>5259810
why does 1+1+1+1... = -1/2?

>>5259891

Oh, "everyone" knows why.

>Posts mathematically meaningless statement
>Not even ninth grade difficulty
>81 posts omitted
Oh /sci/, you so silly.

>>5259810
see>>5259378

>>5259957
That post betrays a severe misunderstanding of how mathematics works, on your part. The whole reason mathematics develops over time is because we come up with so called "meaningless" statements, like (0-1), sqrt(-1), etc.

To call it "ninth grade difficulty" is just dodging the question.

>>5259896
WP anon, you read the wikipedia page so well that you can now do "in-jokes" about spanish physics profs, that people who didn't take the time to teach themselves will understand as random trolling. WP.

infinity is divisible by 2 therefore it's an even number

therefore (-1)^∞ = 1

>>5259255
>Learn some analysis.

Coming from someone that doesn't understand the difference between "undefined" and "indeterminate"...

Tell me of a serious (as in, at least undergraduate level) analysis book that uses "indeterminate" in a properly defined way for something else than treating poor notations that usually result from a highschooler not being able to write limits properly.

<span class="math">\lim_{n \to\infty}(-1 )^n[/spoiler] isn't "indeterminate". Indeterminate assumes that there is a value, we just don't know which one it is. When you take the limit of a quotient whose numerator and denominator both go to 0, 0/0 is indeterminate because it doesn't give you enough information to figure out the limit, and you're a highschooler so you don't know how to solve the problem differently so you leave it there. If you're not a highschooler, you will find (if it exists) the limit, and lift the indetermination.

In OP's case, it is not indeterminate. The sequence just does not converge. If you pretend to know some analysis and you can't understand how fundamental the difference is, that's really pretty fucking sad. <span class="math">1^{\infty}[/spoiler] is one of this "highschool" indeterminate things, <span class="math">(-1)^{\infty}[/spoiler] doesn't exist (if we leave out summation methods, because they are above the level of anyone who doesn't think that writing "indeterminate" in a serious maths article automatically makes the article look like pig shit to the community).

>>5260855
What is infinity divided by 2?

>>5260855
But since (infinity + 7) = infinity and infinity is even, (infinity + 7) = 2k for some k. Therefore infinity = 2k-7, so (-1)^infinity = (-1)^2k * (-1)^7 = 1 * (-1) = -1.

Damn!

>>5260863
You better be trolling boy, or I fear for your future in mathematics, and scientific studies in general.

>>5260870
Just assume people like him are trolling. Either that or he's an engineering student

>>5260870
>>5260874
are you people retarded?

>>5260863
infinity / x|xER = infinity
x|xER / infinity = 0
infinity +- x|xER = infinity

>>5260878
No. You probably are, though.

±1

>>5256691
That's the same thing as asking "what is <span class="math">cos(\infty)[/spoiler]". It is a limit that clearly does not exist.

<span class="math"> (-1)^n = (e^{i \pi})^n=e^{i \pi n}<span class="math">

e^{i \pi n}= \mathrm{Cos}(\pi n)+i \mathrm{Sin}(\pi n) [/spoiler][/spoiler]

>>5261387
for got to make it <span class="math"> \frac{\pi}{2} <span class="math">

(-1)^n=e^{i \frac{\pi}{2} n}

e^{i \frac{\pi}{2} n}= \mathrm{Cos}(\frac{\pi}{2} n)+i \mathrm{Sin}(\frac{\pi}{2} n) [/spoiler][/spoiler]

>>5261393
Fuck, the first one was correc the second one is for <span class="math"> (i)^n <span class="math">[/spoiler][/spoiler]

>>5261387
>>5261393
what are you trying to do?

Indeterminate... Stop trolling please

"The divergent series are the invention of the devil, and it is a shame to base on them any demonstration whatsoever" -- N. H. Abel

Also, ,I've always wondered what is wrong with you people like OP asking this sort of questions? If the limits of your mathematical knowledge consist of understanding of the order of basic operations then either learn some calculus and the answer to (-1)^inf will be clear to you, or give up and go do something useful.

The beauty of math lies in the fact that there's no philosophizing about it. You cannot say anything about math that's not total bullshit unless you actually learn it.

>>5261885
Again, it's not indeterminate. You mean "it doesn't converge". 0/0 is indeterminate because it can represent the limit of various quotients, so that some of them converge (to different values), and some of them diverge (either to +/- infinity or to nothing at all). It is therefore indeterminate. However (-1)^infinity always has the same behaviour regardless of what -1 is the limit of and what infinity is the limit of. It never converges. It is not indeterminate.

You call people trolls and you can't even get your facts straight...

>>5262126

No. Indeterminate means you cannot assign any value to it (even in the extended reals) in any meaningful way.

Also, your explanation why "0/0 is indeterminate" is bullshit.

Pure math grad student here.

-1. Exponentiation has a higher bind than unary negation.

>>5262173
This has been mentioned multiple times in this thread already and the consensus is you are a pedantic faggot and it is obvious from the context what OP wanted to ask.

>>5262160
Quoting wikipedia. You don't like it, find a better source.

>In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form.

"0/0" in the context of limits is written when you have something like <span class="math">L=\lim_{x \to a} \frac{f(x)}{g(x)}[/spoiler] with <span class="math">\lim_{x \to a} f(x)= \lim_{x \to a} g(x)= 0[/spoiler] and a is in the closure of the metric space in which x lives. It is indeterminate because depending on what f and g are, L can exist or not and take different values. For instance, in high-school settings (on reals, etc) and for a=0:
If <span class="math">f(x)=g(x)=1/x[/spoiler], then L=1.
If <span class="math">f(x)=1/x \wedge g(x)=1/(2x)[/spoiler], then L=2.
If <span class="math">f(x)=1/x \wedge g(x)=1/x^2[/spoiler], then L=0.
If <span class="math">f(x)=1/x^4 \wedge g(x)=1/x^2[/spoiler], then <span class="math">L=\infty[/spoiler].
If <span class="math">f(x)=1/x \wedge g(x)=\exp(i/x) /x[/spoiler], then L does not exist because f/g does not converge in 0.

However in the case of <span class="math">(-1)^{\infty}[/spoiler], for any function pair (f,g) and any a such that the limit of f in a is (-1) and the limit of g in a is infinity, the limit of f^g will not exist. Therefore <span class="math">(-1)^{\infty}[/spoiler] is not indeterminate: it doesn't exist.

Now, either find yourself a source or keep making yourself a fool. If you're really a pure math grad student, I certainly hope that you are not studying maths in English, so that you have an excuse. Otherwise, as a guy that last studied maths in as a grad CS student, I feel bad for you, having more knowledge than you on such simple things that are never useful in my domain.

>>5262160
you are fucking retarded.

>>5262160
>>5262220
see>>5259378

its indeterminate and does converge to a value of the complex unit circle depending on the limit giving the infinity.

>>5262220
>for any function pair (f,g) and any a such that the limit of f in a is (-1) and the limit of g in a is infinity, the limit of f^g will not exist.
Proof it

>>5264333
there are already counterexamples to that claim in this thread.

-1^n = -(1^n) = -1

so, that means that -1^∞ = -(1^∞) = -1

>>5264370
FUUUUUUUUUCK YOOOOUUUUUU

>>5264370
1^∞ is not 1 see >>5258661

m = -1*1*-1*1...
-m = 1*-1*1*-1*1... = 1(-1*1*-1*1...) = -1*1*-1*1
m/(-m) = (-1*1*-1*1...)/(-1*1*-1*1...) = 1
m = -m
m = 0

Oh boy

<span class="quote deadlink">>>5264389[/spoiler]
you arnt handling the brackets correctly. you changed the amount of 1ns, which is important in infinite series. especially if they diverge in the strict sense so-that you need to use summation methods. using your method you would -1, which is a possible answer since it must me a complex number with absolute value 1, depending on the limit giving you the infinity.

>>5264401
I was trolling.

<span class="quote deadlink">>>5264406[/spoiler]
are you sure? its hard to tell in this thread.

>>5264407
I have done more derp in my process.
See
m/(-m) = 1
m = -m
m = 0

<span class="quote deadlink">>>5264411[/spoiler]
but thats correct.

>>5256691

1 to any power is one. Because the negative on the 1 is outside any parentheses, we can rewrite the equation to:

(-1)*(1^inf) = ?

From here, it can be deduced the answer is -1.

>>5264439
for fuck sakes.

BUMP

OK, this thread's kinda gone haywire - Imma try and sum up what I think is more or less the case so far; to see if anyone else can expend upon it or correct me if I'm wrong here:

Despite what
>>5262220
said, I'm pretty sure there /are/ multiple limits of functions that are equivalent to this expression, as at least two have been mentioned, one here:
>>5256918
and one here:
>>5258328

However the first one is vulnerable to the same ambiguity present in the Grandi series (1+1-1+1-1 ...) in that you can bracket together the 2s and -2s in such a way that either they all cancel with each other, or they all cancel with each other and one of them gets left over. So the 2-S term in that post could actually be equal to either 1 or -1.

The second attempt to substitute the infinite term for a series limit, uses a series that diverges (1+1+1+1+1 ...) and then takes the "Zeta function regularisation" of that, whose limit is apparently -0.5. I have no idea why that is but wiki seems to have a number of sources on it. There's no particular reason to arbitrarily pick /that/ series though, as the poster mentioned. So it could equally be (1+2+3+4 ...), which (again, apparently) has a different limit when substitued into the regularised version of the Riemann Zeta funtion. Apparently it's -1/12. For more info on how the fuck that works, see here: http://en.wikipedia.org/wiki/Riemann_zeta_function#Theta_functions

Part 2 coming up.

Part 2 of
>>5266582

OK. So, it's inderterminate as we have multiple cases in which the limit of some different functions approaches infinity such that the expression in the OP is defined when we substitute the functions into it. And it then gives us a bunch of different values, depending on which function was used. I've worded this quite badly as in some parts I confused the words "series" with "function", I think, but I'm pretty sure it still makes sense. Indeterminate.

Amirite?

>inb4 you just re-stated what everyone else said
Most people ITT seem to have no fucking clue what's going on here anyway, so I'm gathering this all together to see if we can once and for all put this shit to bed.

>>5266582
correction: I explained that Grandi thing wrong.

I mean the 2-S term could have a -2 left over, or none left over. Not that it could be equal to -1 or 1, rather that it contains an ambiguity in the form of a difference of 2.

There is no ambiguity, the answer is undefined.

However, the limit of -1^x as x approaches infinity is in fact -1.

>>5266619
Y'know it'd be nice if you could at least say why you think so -_-

*sigh*... doesn't really matter now anyway; this thread will presumably 404 soon.