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How do you take limit of k-roots when the terms within the root contains powers of k?

  1. Sep 10, 2012 #1
    Me and a friend was contemplating the ratio of even to odd number in Pascals triangle. After some thought we arrived at a ratio like this when looking at a triangle of 2k-1 rows.

    [itex]\frac{2^{2k-3}+2^{k-2}}{3^{k-1}} - 1 [/itex]

    As expected, this ratio grows exponentially. (I.e. almost all numbers in Pascals triangle will be even) However, we then started to wonder with what base this will grow, I.e. we wondered about

    [itex] \lim_{k \rightarrow \infty} (\frac{2^{2k-3}+2^{k-2}}{3^{k-1}} - 1)^\frac{1}{k}[/itex]

    After some calculations we got it reduced to

    [itex] \lim_{k \rightarrow \infty} \frac{4}{3}(2^{\log_23-3}+2^{\log_23-2-k} - 2^{-k(2-\log_23)})^\frac{1}{k}[/itex]

    Now, we belive this to be (4/3) but we get stuck with the limit

    [itex]\lim_{k \rightarrow \infty} [\frac{3}{8} +\frac{3}{2^{k+2}} - \frac{3^k}{4^k}]^\frac{1}{k} [/itex]

    Our thinking is that the last two terms within the brackets will be small so that this is somewhat like taking

    [itex] \lim_{k\rightarrow \infty} (\frac{3}{8})^{\frac{1}{k}}=1 [/itex]?

    and hence that the base tends to (4/3)? Empirically, this seems to be the case. However, how to you take this limit with mathematical rigour?
     
    Last edited: Sep 10, 2012
  2. jcsd
  3. Sep 10, 2012 #2

    Borek

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    Staff: Mentor

    Re: How do you take limit of k-roots when the terms within the root contains powers o

    Not that it answers your problem, but it will make formatting much easier: LaTeX guide.
     
  4. Sep 10, 2012 #3
    Re: How do you take limit of k-roots when the terms within the root contains powers o

    Thanks, I didn't know it was available. I'm new to forums like these.
     
  5. Sep 11, 2012 #4

    chiro

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    Re: How do you take limit of k-roots when the terms within the root contains powers o

    Hey bolle and welcome to the forums.

    I'm pretty sure that since this function is continuous across the interval (since 3/8 > 0) you can basically treat it as an evaluation of f(1/k) where f(x) = a^x as k -> infinity which means it will just be a^0 = 1.

    The reason you can do this is because for a continuous function the limit is equal to the value of that function (this is actually how you define continuity rigorously in analysis).

    So as long as you have this property, it's basically a thing where you just evaluate the limit and plug that in for the function (but it has to be continuous).

    We know this is continuous and smooth since you can take the derivative of (3/8)^x and it is also a smooth function.

    Whenever you can't be sure when something is continuous or not, then you need to use the epsilon-delta definition of a limit to find out. You also need to consider when the limit gives a non-finite value (like infinity or negative infinity) or whether the limit doesn't even exist (the right hand limit is not equal to the left hand limit).
     
  6. Sep 12, 2012 #5

    Stephen Tashi

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    Re: How do you take limit of k-roots when the terms within the root contains powers o

    [itex] \frac {3}{8} - \frac{3^k}{4^k} \lt \frac{3}{8} +\frac{3}{2^{k+2}} - \frac{3^k}{4^k} \lt \frac{3}{8} + \frac{3}{2^{k+2}} [/itex]

    For [itex] k > 1, f(x) = x^{\frac{1}{k}} [/itex] are there inequalities comparing

    [itex] f(a + b) [/itex] to [itex] f(a) + f(b) [/itex] and [itex] f( p - q) [/itex] to [itex] f(p) - f(q) [/itex] when [itex] p - q > 0 [/itex] ?

    Thiis isn't a real hint. I'm just thinking out loud.
     
  7. Sep 14, 2012 #6
    Re: How do you take limit of k-roots when the terms within the root contains powers o

    You are of course correct. We can enclose the limit in this way. I seem to have completely forgotten my real analysis.
     
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