What Are the Different Orders of Infinity and How Do They Compare?

[Unicyclist]

I wasn't sure where to post this, so I'll post it here. Depending on your answers, I may have a few more questions.

What's greater: $$\infty^2$$ or $$2^\infty$$?

Why?

 

[mjsd]

i would say $$2^\infty$$ is greater since an exponential beats a polynomial ...like saying for large x, $$2^x>x^2$$

 

[Unicyclist]

Good.

Now, then. Consider this: $$\infty^x$$ and $$x^\infty$$
If x = 2, then the first expression is greater than the second, right? And if x = 1, then the second is greater than the first. So, is there a point for which $$\infty^x$$ = $$x^\infty$$? How do you find it?

 

[VietDao29]

I wasn't sure where to post this, so I'll post it here. Depending on your answers, I may have a few more questions.

What's greater: $$\infty^2$$ or $$2^\infty$$?

Why?

Well, they are both infinity, we can not say which is greater. We don't compare infinity's.

If you give us the 2 functions, and a number that they tend to, say, one of then is of the form $$\infty ^ 2$$, and the other is $$2 ^ \infty$$, then we can compare the ratio of the two to know which one is greater.

We have:
$$\lim_{x \rightarrow + \infty} \frac{x ^ 2}{2 ^ x} = 0$$

So, as x tends to + infinity, we have x² < 2^x, x² is of the form $$\infty ^ 2$$, and 2^x is of the form $$2 ^ \infty$$.

$$\lim_{x \rightarrow + \infty} \frac{{(x ^ x)} ^ 2}{2 ^ x} = \lim_{x \rightarrow + \infty} \frac{{x ^ x} \times x ^ x}{2 ^ x} = \infty$$

So, when x tends to + infinity, we have: $$(x ^ x) ^ 2 > 2 ^ x$$

 

[matt grime]

As functions from the extended real line to the extended real line (or complex plane), the result is the same - infinity.

If you wish to talk about limiting behaviour that is a slightly different thing.

Of course in cardinal arithmetic $\alpha^2$ and $2^\alpha=|P(\alpha)|$ are (with one exception) different.

 

[Unicyclist]

VietDao29: Didn't get that, sorry. Why did the $$x^2$$ turn into $$(x^x)^2$$?

 

[VietDao29]

VietDao29: Didn't get that, sorry. Why did the $$x^2$$ turn into $$(x^x)^2$$?

I gave you 2 example, one of which is: $$\infty ^ 2 < 2 ^ \infty$$, and the other is $$\infty ^ 2 > 2 ^ \infty$$.

They are 2 separate examples.

 

[Unicyclist]

I gave you 2 example, one of which is: $$\infty ^ 2 < 2 ^ \infty$$, and the other is $$\infty ^ 2 > 2 ^ \infty$$.

They are 2 separate examples.

Right, sorry. Now I get it.

What about my 2nd post?

 

[VietDao29]

Right, sorry. Now I get it.

What about my 2nd post?

No, again, we don't compare infinity's. They are incomparable.
Say, we have: $$\infty , \ \infty ^ 2 , \ \infty ^ 3 , ..., \infty ^ n, ...$$. They are all infinity, however, we cannot compare them.

Now, then. Consider this: $$\infty^x$$. and $$x ^ \infty$$.
If x = 2, then the first expression is greater than the second, right?

Nope, this is not true.

And if x = 1, then the second is greater than the first. So, is there a point for which $$\infty^x$$ = $$x ^ \infty$$.? How do you find it?

Since infinity can not be compared with each other, we cannot solve for x like that.

 

[Unicyclist]

What if you use the limits method like you did before? I'm not familiar with it, that's why I'm asking.

 

[VietDao29]

What if you use the limits method like you did before? I'm not familiar with it, that's why I'm asking.

I just used limit to show you that, in some cases, we have $$\infty ^ 2 < 2 ^ \infty$$, and in some other cases, $$\infty ^ 2 > 2 ^ \infty$$. We may also have $$\infty ^ 2 = 2 ^ \infty$$.

So, in general, we cannot compare $$\infty ^ 2$$, and $$2 ^ \infty$$.

 

[Unicyclist]

I think you're too smart for me. But thanks anyway.

I still don't get how $$\infty ^ 2 > 2 ^ \infty$$ one works. It's pretty much a totally different expression that you used for proving it.

 

[VietDao29]

I think you're too smart for me. But thanks anyway.

I still don't get how $$\infty ^ 2 > 2 ^ \infty$$ one works. It's pretty much a totally different expression that you used for proving it.

Ok, I'll go more slowly for you to get it, as x tends to infinity, then x^x also tends to infinity, right?
So, when x tends to infinity, (x^x)² is in the form $$\infty ^ 2$$

When x tends to infinity, we have (x / 2) also tends to infinity. So (x / 2)^x also tends to infinity. That means x^x / 2^x tends to infinity. Right?

x^x itself tends to infinity, when x tends to infinity, so (x^x x^x) / 2^x ~~> infinity (since, infinity x infinity = infinity).

So we have:
$$\lim_{x \rightarrow + \infty} \frac{(x ^ x) ^ 2}{2 ^ x} = \infty$$. That means, for x large enough, we'll have: (x^x)² > 2^x.

 

[uart]

I think that the question that OP was really trying to get at was the one he wrote in his post #3.

Good.

Now, then. Consider this: $$\infty^x$$ and $$x^\infty$$
If x = 2, then the first expression is greater than the second, right? And if x = 1, then the second is greater than the first. So, is there a point for which $$\infty^x$$ = $$x^\infty$$? How do you find it?

It seems that what he was really interested in is this :

lim x->infty : a^x / x^a, for a given a.

For a=1 the above limit is zero. For a>1 the above limit does not exist (or if you like, it goes to infinity with x). I think OP was assuming that this limit would be a continuous function of a so that there would be a specific value of a for which the above limit is unity. This however is not the case.

Do you see how that works Unicyclist? Your argument about the limit attaining a specific intermediate value for some intermediate value of a is only valid if the limit is a continuous function of a. But since the limit jumps discontinuously from zero to unbounded then there is no intermediate value for which the limit is unity.

 

[HallsofIvy]

Of course, you never did define "infinity"- or say which "infinity" you are talking about! It is well known that 2 times aleph-null is aleph-null while 2^(aleph-null) is c which is larger than aleph-null.

 

[Unicyclist]

VietDao29: Okay, thank you. That clears it up a bit. The only problem with that is: the inequality is sort of left for you to define. You can kind of decide which side is bigger, based on your whim. Or does it depend on through which expressions you have arrived to the infinity inequality?

Uart: yes, you're very right. I was assuming it's a continuous function for this question.

Anyway, thank you for your help.

 

[VietDao29]

VietDao29: Okay, thank you. That clears it up a bit. The only problem with that is: the inequality is sort of left for you to define. You can kind of decide which side is bigger, based on your whim. Or does it depend on through which expressions you have arrived to the infinity inequality?

No, I don't define which sides is bigger.
If a / b < 1, then it means that a > b, right?
The first example is:
$$\lim_{x \rightarrow + \infty} \frac{x ^ 2}{2 ^ x} = 0 < 1$$, so for x large enough, we'll have x² < 2^x.
$$\lim_{x \rightarrow + \infty} \frac{{(x ^ x)} ^ 2}{2 ^ x} = \lim_{x \rightarrow + \infty} \frac{{x ^ x} \times x ^ x}{2 ^ x} = \lim_{x \rightarrow + \infty} \left( \frac{{x}}{2} \right) ^ x \times x ^ x = \infty > 1$$
So, for x large enough, we'll have: (x^x)² < 2^x.

In conclusion, $$\infty ^ 2$$, and $$2 ^ \infty$$ are both infinity, we don't compare them, or determine which is larger than which. Well, it's just not appropriate.

 

[Unicyclist]

Okay, thank you very much for your patience and help.

I think I understand the concepts of limits and infinity better now.

 

[robert Ihnot]

HallsofIvy: Of course, you never did define "infinity"- or say which "infinity" you are talking about! It is well known that 2 times aleph-null is aleph-null while 2^(aleph-null) is c which is larger than aleph-null.

Consider all numbers from 0 to 1 on the real line in base two and try counting them. (Remember .1111111... is the same as 1.) What is the total? You have two choices for each place, so the answer is 2^infinity.

Infinity (aleph-null) consists of, say, all the integers. Infinity squared is the set of integers on the real plane consisting of (n,m). This set can be shown to equal the cardinality of the integers themselves. For example start by listing:
(0,0), (1,0,) (0,1) (1,1), (2,0), (0,2),(2,1), (1,2), (2,2), (3,0),(0,3)...

 

[Werg22]

Unicyclist, Courant offers a moderately extensive treatment of "The Order of Magnitude of Functions", you might want to take a look for further insight.

 

[picklefeet]

I'm just wondering... wouldn't infinite squared and 2^infinite be the same?because you can't get higher than the concept of infinite? As stated earlier in the thread, infinite times infinite equals infinite. And 2 to the infinite power would be 2X2X2X2... with no end which means it is infinetly great. Isn't that the definition of infinite? On a different note, how do you put the symbols in the replies? I can't find it anywhere.

 

[VietDao29]

I'm just wondering... wouldn't infinite squared and 2^infinite be the same?

Well, this has been discussed earlier in the thread. :)

On a different note, how do you put the symbols in the replies? I can't find it anywhere.

You can give this thread a look, Introducing LaTeX Math Typesetting. It's a good place for one to start learning LaTeX. There are 3 PDF files there, in the first post, teaching all the common technique used in LaTeX.

 

[robert Ihnot]

picklefeet: I'm just wondering... wouldn't infinite squared and 2^infinite be the same?because you can't get higher than the concept of infinite?

Well if you are just talking limits in the calculus normally there is no distinction between orders of infinity. However, George Cantor was the one who proposed different orders of infinity. In this case, the matter turns on cardinality. That is, we set up one to one correspondences. I have already shown how this can be done with the set of integers, and the set of integers in two dimensions, (N,M) Since this comparison can be made, then infinity squared (that is alpha-null infinity) is shown to be simply equal to infinity.

There is also the Galileo paradox, where it was shown that the squares can be put into 1-1 correspondence with the integers. Very simply, send N to N^2, so there is as many squares as integers. Also the matter of: if we have an infinite hotel containing an infinite number of guests, we can make room for another infinity by moving every guest from room N to room 2N.

However, Cantor showed that the line itself, or the continuum is larger than the set of integers. Now the matter of 2^(infinity) represents all numbers from 0 to 1, as I have shown, and is an aspect of the continuum, and, happens to be, equal in cardinality to the whole continuum. It is larger than alpha-null.

The question was raised as to whether there is no other infinity between c that is, 2^(infinity), and alpha-null, the cardinality of the integers. This was referred to as the continuum hypothesis.