1^Infinity (One raised to the infinite power)?

Question

Though the answer may look like it is, of course, "1"---and the limit to the answer is, of course, "1"--- in mathematics 'one' cannot simply 'assume' the answers to things such as this. Mathematics is based on fact, not presumption. Now then, if anyone can find something concrete that'll make the answer whatever "n" it may be, or even possibly whatever 'n' it may be, I'm interested in seeing it. =) So far I've read one person actually give an example on why it may even be 'e'!

Remember, 'infinity' is absolutely unreal. Though we can easily imagine the concept of it, it cannot and thus does not and will not exist. Please, unless you've multiplied one times itself infinity number of times, do not tell me that the answer is "1". -_-

I don't like to repeat myself. Honestly enough, I don't mind if you give me "1" so long as you use some pretty highly advanced mathematics. But if you use elementary-level techniques---then at least tell me you're giving the answer for some 'points'. =)

It's a shame I won't be here when people start giving other opinions. So far the best answers (for what I'm looking for) involve indeterminates and 'any integer'.

In any case, I, TOO BELIEVE THE MOST LOGICAL (and probably) THE ANSWER IS ONE---I just wanted a few opinions on the matter other than the most obvious and what everyone is thinking.

"So what is 1^Negative Infinity"? <-- If that threw you off don't even bother pressing "Submit"! X-D

Answer 1

It is indeterminate (undefined)

Proof:
log b to the base a = (log b)/(log a)

Example:
log 8 base 2 = 3 ---- 2^3 = 8
(log 8)/(log 2) = 3

log 8 base 1
= (log 8)/(log 1)
= (log 8)/0 = infinity
It follows that 1^infinity = 8

Similarly
log 9 base 1 = infinity
It follows that 1^infinity = 9

1^infinity can have any positive value greater than or equal to 1.

1^infinity >= 1
1^(-infinity) = 1/1^(infinity)
0 <= 1^(-infinity) <= 1

1^infinity is indeterminate (undefined).

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Copied the following text from Wikipedia:

Powers with infinity:
The expressions and arise in analysis for the same reason as 0^0, and they are undefined for the same reason. That is, it is true that lim f(x)^lim g(x) = lim f(x)^g(x) when f and g approach nonzero finite constants, but not when they approach 0 or infinity; then, the limit of the power can be anything, not predictable from the limits of f and g.

There is one exception. If f and g both approach infinity as x approaches a, then lim f(x)^g(x) does equal infinity. Thus, it makes sense to say the expression inf^inf is well defined and equals inf.

Answer 2

You have somewhat of a point, but you have to understand some of mathematics is about inference - especially when you are dealing with infinity. If you know 1^googolplex = 1, then you can infer that 1^infinity is 1. But I do see where you're coming from. I obviously can't name a single person who has every multiplied 1 by itself an infinite number of times - it is of course impossible to do this.

It basically comes down to the properties of the number 1. Since 1*1 = 1, then 1*1*1 = (1*1)*1 = 1*1 = 1 . It's a recurrence relation, and as you approach infinity, you are allowed to infer, by mathematical induction, that the answer will be 1.

edit:
Ok here's your advanced mathematics, though not very advanced because that isn't necessary for this problem:

Prove 1^n = 1 for any n ≥ 1

Base case:
n = 1
1^1 = 1

Suppose the statement holds for n=k s.t.
1^k = 1

Show that the statement holds for n=k+1
1^(k+1)
= 1^k * 1^1
1^k = 1 by the hypothesis
1^1 = 1 by the base case
So 1^k * 1^1 = 1 * 1 = 1

There. Proven by mathematical induction.

edit2:
So I looked this up, and I need to admit that I am wrong! 1^∞ is an indeterminate form. I do, however, believe that this can (and will one day) be proven untrue. It's possible that the definition of an indeterminate form needs to be altered a little bit. In fact, when you do this on Maple, the answer is 1! My proof doesn't work. Induction only proves for the natural numbers. Infinity is not a natural number - it's not a number at all! Man, I feel like an idiot.

Answer 3

Well, since infinity is not a number, all we can do is ask: what is the limit of the sequence of numbers 1^n, as n gets larger and larger, if such a limit exists. This sequence is:

1^1, 1^2, 1^3, ..., 1^n, ...

But each term in the sequence is 1, since 1 raised to any finite power is still 1 (by induction). So the sequence is:

1, 1, 1, 1, ... forever.

the limit of such a sequence is clearly 1 (an epsilon-delta proof would be too basic and unnecessary). So 1 is the answer.

EDIT:
The "proof" immediately below this is complete nonsense (nothing personal). Here is why: You cannot throw around logarithms and memorized properties of them and claim it makes sense. To the point: the function f(x) = log base b of x is a shorthand way of saying: the number that b must be raised to to equal x. (this is the inverse of the one-to-one, and thus invertible, function y = b^x). So if b = 1, the function f(x) = log base b of x makes no sense, since 1 raised to any number will not equal x, unless x is 1, but even then it doesnt work since 1 raised to any number is still 1, showing f(x) is NOT EVEN A FUNCTION when b=1 (as the input x =1 gives more than one possible output). So please ignore the proof of the next person, it is ludicrous.

In order for it to be indeterminate, both the base and the exponent need to be 0 or infinity, or a combination of the two. Here the base is neither, it is 1. The wikipedia entry referring to 1^inf being indeterminate has to do with a sequence of numbers in the base that converge to 1. This is different. Here, the base is always one, and this is not indeterminate.

Also, the problem with Dr D's proof (two answers down) is that if you look closely in the middle, there is a step where you get zero times infinity, which is definitely indeterminate, and thus invalidates the proof.

Answer 4

1 To The Power Of Infinity

Answer 5

Your are using the wrong logic to prove that. If you were to use the function y=-(x^2) you would have a parabola that has an upper limit of zero even if you went to infinity front and back for X. the same hold true of 1^X where x goes to infinity. Your limit will be one because of the IDENTITY property of one(you should look this part up to understand it) . One times it self no matter how many times will = 1. You don't need to prove it by multiplying it infinite number of times(because that would be impossible). But by using definitions and identities of one we can indeed prove that it is ...1.

Answer 6

Unless you have multiplied one times itself infinity number of times, then don't tell me that the answer is not "1".

As you said, math is not based on presumption, but rather on facts and patterns. It is an undeniable fact that if you start multiplying one times itself, your answer will repeatedly be "1". By following this pattern, there is no reason to believe that there is suddenly an anomaly that occurs at any given point.

Also, consider it this way:

Multiplication can be considered as repetitive grouping. For instance, 4 x 4 means create four groups each containing four elements. If you want to find 4^3, then you would make four groups of "four groups of four elements". Translate this to 1^any power. Begin with 1^2, which is one group of 1 element. Then 1^3 would be one group of one group of one element, still only 1 total element. No matter what power you raise 1 to, you will always be creating only 1 group of whatever amount came previously, which will always be one element. Unlike other exponents, there is no growth because you always have only one group.

Answer 7

I believe the answer is 1.

But here's how you can get e, although I do not necessarily agree with this.

Let y = (1 + x)^(1/x)
We want to consider the behavior of y as x --> 0
ln(y) = 1/x * ln(1+x)
= 1/x * (x - x^2/2 + x^3 / 3 + ...)
= 1 + O(x)

So lny --> 1
y --> e as x --> 0

The problem with this is that it's arbitrary.
If you considered (1+2x)^(1/x) you'd get e^2

Answer 8

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RE:
1^Infinity (One raised to the infinite power)?
Though the answer may look like it is, of course, "1"---and the limit to the answer is, of course, "1"--- in mathematics 'one' cannot simply 'assume' the answers to things such as this. Mathematics is based on fact, not presumption. Now then, if anyone can find...

Answer 9

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None. And, having this infinite power and knowledge, it would take me about half a second to figure out how to save them all without violating "free will." How about that. Heck, I can figure out how to do that right now (I already have, in fact) -- and I *don't* have any infinite power or knowledge...

Answer 10

We could try a proof by induction.

Base: 1^1 = 1
Assume: 1^k = 1
What's the value of 1 ^ (k+1)?
1 ^ (k+1) = (1 ^ k) * 1 = 1 * 1 = 1

I'm convinced the answer is 1.

> Though we can easily imagine the concept of it, it
> cannot and thus does not and will not exist.

Then again, we can say that about of things. Show me "one". Not one apple or one electron or one inch; just one. It's a concept you and I can easily imagine, but "one" as a concrete object doesn't exist.