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Poll:STOOPID question: Does (infinity + 1)>Infinity?

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Originally posted by: MangoTBG
Do you mind explaining why they aren't...Maybe contribute more than just "You're wrong".

Well I don't have time right now to look up what I learned back in Calculus. Quickly, I can say that X^X grows much faster than E^X as we approach infinity. Can anyone help us with a function that grows slower than ln(X)? I know it's out there, but my mind escapes me at the moment...

 
Sorry but you won't be able to find something slower than ln(x). And I forget why my calculus teacher ruled out x^x but there was some reason we did not consider in the functions we talked about. I'll give her a call and PM you my results. I'm not trying to prove I know more or anything, I just like knowing things 😉
 
Originally posted by: MangoTBG
Sorry but you won't be able to find something slower than ln(x). And I forget why my calculus teacher ruled out x^x but there was some reason we did not consider in the functions we talked about. I'll give her a call and PM you my results. I'm not trying to prove I know more or anything, I just like knowing things 😉

It's been a while - maybe ln(x) is the slowest growing. I can't see of any reason to discount X^X though - it's perfectly valid and I recall it coming up a few times in class. Let me know what you find out 😉

EDIT: Oh! I was just doing some work and thought of a function that grows slower than f(x)=ln(x). How about f(x)=5. It's a valid function, and it definitely grows slower than f(x)=ln(x).
 
As indicated by my PM...F(X)=5 is not a valid function for consideration as it does not grow at all. We'll see what my teacher says when she calls me back. 😀
 
Originally posted by: DeathByAnts
infinity + 1 = infinity + 1

same as x + 1 = x + 1


then again, I failed the same math class the last 2 semesters...

You can't treat infinity as a variable, since it is not a set value. If the above were possible, than infinity/infinity would always be 1, which is not true.

Take, for example, lim (ln x / x) as x goes to infinity. The above reasoning would state that the answer is 1, since both the numerator and the denominator approach infinity as x increases. But, they do so at different rates, so that x is always greater than ln x. Therefore, the answer is in fact 0.
 
I remember a play I did in 3rd grade where some witch(I think) challenged someone to a counting contest , and all she kept saying was plus one.. . .I'm sure you all care😀 anyway you just drug up that memory..
 
Originally posted by: MangoTBG
Sorry but you won't be able to find something slower than ln(x). And I forget why my calculus teacher ruled out x^x but there was some reason we did not consider in the functions we talked about. I'll give her a call and PM you my results. I'm not trying to prove I know more or anything, I just like knowing things 😉

If we are talking about asymptotic order, it is pretty trival to show that (ln(x))^(1/2) grows slower then ln(x). Or hell, f(x)=1 does as well 😀

-Chu
 
Originally posted by: Chu
Originally posted by: MangoTBG
Sorry but you won't be able to find something slower than ln(x). And I forget why my calculus teacher ruled out x^x but there was some reason we did not consider in the functions we talked about. I'll give her a call and PM you my results. I'm not trying to prove I know more or anything, I just like knowing things 😉

If we are talking about asymptotic order, it is pretty trival to show that (ln(x))^(1/2) grows slower then ln(x). Or hell, f(x)=1 does as well 😀

-Chu

or ln(ln(ln(ln(x)))) 😀
 
If you could add 1 to infinity you'd also be able to subtract 1 from infinity. Conclusion, you cannot add 1 to infinity.

 
http://www.amazon.com/exec/obidos/tg/detail/-/053439339X/qid=1062615602/sr=8-1/ref=sr_8_1/002-7559959-2613628?v=glance&s=books&n=507846

Stewart is a tool though.

I voted infinity + 1 = infinity, although i really wanted to vote "You're a dumbass!"

As it was said before infinity is a concept not a number, if it were a number you could add 1 to it. Infinity is a concept based on the the fact that you could always add 1 to any number.

As for the function argument that you started, i wish i could remember my calculus better. PSA Don't talke calc 1, 2, 3 and then take a year off before doing DiffEQ, man im so screwed.
 
Originally posted by: Bootprint
Infinity + Sqr(-1)

Complex analysis is an evil, EVIL thing. Believe it or not, this expression does make sense if your make certain assumptions.

-Chu
 
lets say you are talking about the amount of numbers in the set of Integers. Which is infinity.

And then say, you take that same set, and remove all the odd numbers.

So. You have this: one set, and another set that is equal to half the first set, right? WRONG!

There are degrees of infinity. So, an infinite set can be more or less infinite than another set. Here is how you figgir it out.

What you must do is show that all the elements of a set can be mapped in a one-one function to all the elements of another set, and then you prove they have the same degree of infinity. If you can't do that mapping, then one is a lesser degree of infinity.

For example.

the set of integer:

{ . . . , -1, 0, 1, 2, 3, 4, 5, . . . }
The set of evens:

{ . . . , -2, 0, 2, 4, 6, 8, 10, . . . }

Sets say you multiplied ever number in the set of integers by 2. You get this: { . . . -1(2), 0(2), 1(2), 3(2), 4(2), 5(2), . . . }. Simplified, this is { . . . , -2, 0, 2, 4, 6, 8, 10, . . . }

basically, you have shown that for every number in the set of integers, there is a corresponding number in the set of evens, so the two are of the same degree of infinity. There is actually much more to this, but I ask you just to look at it and accept it in principal, because I don't want to type the rest. The same trick can be done with the set of odds if you, instead of multiplying everything by 2, multiply by 2 and then add or subtract 1.

However, no such operation can be performed with the Reals or the Rationals onto the Integers, so the degree of infinity that the reals have is greater than the degree of infinity represented by the integers, the whole numbers (which are also equivalent to the integers. I challenge those who have not done this to try and figure it out), the evens, the odds, etc.


So, I hope I answered your question.

but you can't add one to infinity, so I kinda just answered the unspoken question: "can one infinity be greater than another."
 
Originally posted by: FallenHero
infinity is a concept, not a defined number
yepp.

Infinity stands for "all numbers" , therefore "all numbers +1" does not make sense, as you are adding numbers to a concept.
 
infinity + 1 isn't a legal operation.

Infinity is NOT a number, or a variable or anything. It's a concept. Infinity is thought of in terms of rate of growth...for example, the limit as x approaches infinity of x! (factorial) is a larger order of infinity than say, the limit as x approahces infinity of x. Though, lim x->oo (x!) is no bigger or smaller than lim x->oo (x! + 1), because for limiting (read: large) values of x, the one becomes insignificant.
 
Originally posted by: MustangSVT
hmm how about this?

kid A: SuperMegaZord is stronger than ultraman
Kid B: Ultraman is stronger than SuperMegaZord
kid A: SuperMegaZord is stronger infinity times!
Kid B: Ultraman is stronger infinity times PLUS 1!

who wins?

Oh, yeah? SuperMegaZord is stronger A(infinity, infinity) where A is the Ackerman function. So there.
 
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