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Powers and Infinity

  1. Jun 12, 2007 #1
    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: [tex]\infty^2[/tex] or [tex]2^\infty[/tex]?

    Why?
     
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  3. Jun 12, 2007 #2

    mjsd

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    i would say [tex]2^\infty[/tex] is greater since an exponential beats a polynomial ...like saying for large x, [tex]2^x>x^2[/tex]
     
  4. Jun 12, 2007 #3
    Good.

    Now, then. Consider this: [tex]\infty^x[/tex] and [tex]x^\infty[/tex]
    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 [tex]\infty^x[/tex] = [tex]x^\infty[/tex]? How do you find it?
     
  5. Jun 12, 2007 #4

    VietDao29

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    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 [tex]\infty ^ 2[/tex], and the other is [tex]2 ^ \infty[/tex], then we can compare the ratio of the two to know which one is greater.

    We have:
    [tex]\lim_{x \rightarrow + \infty} \frac{x ^ 2}{2 ^ x} = 0[/tex]

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

    [tex]\lim_{x \rightarrow + \infty} \frac{{(x ^ x)} ^ 2}{2 ^ x} = \lim_{x \rightarrow + \infty} \frac{{x ^ x} \times x ^ x}{2 ^ x} = \infty[/tex]

    So, when x tends to + infinity, we have: [tex](x ^ x) ^ 2 > 2 ^ x[/tex]
     
  6. Jun 12, 2007 #5

    matt grime

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    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 [itex]\alpha^2[/itex] and [itex]2^\alpha=|P(\alpha)|[/itex] are (with one exception) different.
     
  7. Jun 12, 2007 #6
    VietDao29: Didn't get that, sorry. Why did the [tex]x^2[/tex] turn into [tex](x^x)^2[/tex]?
     
  8. Jun 12, 2007 #7

    VietDao29

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

    They are 2 separate examples.
     
  9. Jun 12, 2007 #8
    Right, sorry. Now I get it.

    What about my 2nd post?
     
  10. Jun 12, 2007 #9

    VietDao29

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

    Nope, this is not true.

    Since infinity can not be compared with each other, we cannot solve for x like that.
     
    Last edited: Jun 12, 2007
  11. Jun 12, 2007 #10
    What if you use the limits method like you did before? I'm not familiar with it, that's why I'm asking.
     
  12. Jun 12, 2007 #11

    VietDao29

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

    So, in general, we cannot compare [tex]\infty ^ 2[/tex], and [tex]2 ^ \infty[/tex].
     
  13. Jun 12, 2007 #12
    I think you're too smart for me. But thanks anyway.

    I still don't get how [tex]\infty ^ 2 > 2 ^ \infty[/tex] one works. It's pretty much a totally different expression that you used for proving it.
     
  14. Jun 12, 2007 #13

    VietDao29

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

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

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

    So we have:
    [tex]\lim_{x \rightarrow + \infty} \frac{(x ^ x) ^ 2}{2 ^ x} = \infty[/tex]. That means, for x large enough, we'll have: (xx)2 > 2x.
     
  15. Jun 12, 2007 #14

    uart

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

    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 continous 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.
     
    Last edited: Jun 12, 2007
  16. Jun 12, 2007 #15

    HallsofIvy

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    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.
     
  17. Jun 12, 2007 #16
    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.
     
  18. Jun 12, 2007 #17

    VietDao29

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

    In conclusion, [tex]\infty ^ 2[/tex], and [tex]2 ^ \infty[/tex] are both infinity, we don't compare them, or determine which is larger than which. Well, it's just not appropriate.
     
  19. Jun 13, 2007 #18
    Okay, thank you very much for your patience and help.

    I think I understand the concepts of limits and infinity better now.
     
  20. Jun 13, 2007 #19
    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).......
     
    Last edited: Jun 13, 2007
  21. Jun 13, 2007 #20
    Unicyclist, Courant offers a moderately extensive treatment of "The Order of Magnitude of Functions", you might want to take a look for further insight.
     
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