1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit of

  1. Dec 29, 2006 #1
    i've to show (as part of a bigger assignment) that

    a^n/p(n), where p is any polynomial and a>1, tends to infinity as n does. i've proved that:

    a^n/n^k

    does so, but i'm not sure how to extend this to a complete polynomial such as

    (c1)n+(c2)n^2+(c3)n^3...

    thanks for any help

    NB: edited
     
    Last edited: Dec 29, 2006
  2. jcsd
  3. Dec 29, 2006 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    You've used a for two different things; and forgotton a condition on the first a.

    Remember the triangle inequality:

    |x+y| <= |x|+|y|

    that will help.
     
  4. Dec 29, 2006 #3
    right...well all i can think of at the moment is changing (n^k) to (n+b)^k and prove that it works for that - however i don't think thats valid since (i don't think) every possible polynomial can be created by such an expansion.

    i've not a clue where to start with the triangle inequality
     
  5. Dec 29, 2006 #4

    StatusX

    User Avatar
    Homework Helper

    Try proving the reciprical tends to zero, then you can just add different powers of n in the numerator.
     
  6. Dec 29, 2006 #5
    proving the reciprocal goes to zero is exactly how i did a^n/n^k.....

    but i didn't think you could just add terms in various powers of n and expect the same result as a^n/n^k (or its reciprocal) without proving it?
     
    Last edited: Dec 29, 2006
  7. Dec 29, 2006 #6
    also, i think all that is needed is the ratio test for convergence (of the reciprocal) however if I use this when I replace n^k with (b+n)^k (to create the polynomial), i end up with series within series, and i'm not sure what to do with that.

    (The ratio test is also what i used for a^n/n^k)
     
  8. Dec 30, 2006 #7

    Gib Z

    User Avatar
    Homework Helper

    The nth root test, where you take the nth root as n approaches infinity.

    For the numerator, you get (a^n)^(1/n) as n approaches infinity. The powers equate leaving us with just a. The denominator, it becomes p(n)^(1/n) as n approaches infinity. so in the end we get a*p(n)^n. As n approaches infinity, the polynomial will as well. So the result of our nth root test is more than 1. That means it does not converge, and therefore diverges. As n approaches inifinity, both a^n and p(n)^n will be positive, meaning it diverges positively, to infinity.

    Note: I didnt use that triangle inequality that matt grime posted, which makes me think i may have done something wrong lol, matt grimes always right...
     
    Last edited: Dec 30, 2006
  9. Dec 30, 2006 #8

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    There is always more than one way to skin cat. However, you have asserted that

    a/p(n)^{1/n} = a*p(n)^n

    which is not correct. Your 'proof' also works for a=1, which is unfortunate, as 1/p(n) tends to zero as n tends to infinity.
     
  10. Dec 30, 2006 #9

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    The hint was to use the triangle inequality. I don't see anyone doing that. Try to use the triangle inequality to go from studying p(n) to a k*n^r, where k is a constant and r is the degree of p(n) (and no, k is not the coeffeicient of n^r in p(n)).
     
  11. Dec 30, 2006 #10

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    That is what the triangle inequality will let you do - You can replace

    [tex] | b_r n^r + b_{r-1}n^{r-1} + \ldots b_0|[/tex]

    with something larger. Notice that n^r > n^{r-1} for n>1, r>2.
     
  12. Dec 30, 2006 #11
    do you mean:

    [tex]| b_r n^r + b_{r-1}n^{r-1} + \ldots b_0| \leq |n^r + n^r + n^r + \ldots \mbox{(to the extent of k)}|[/tex]

    which only is true if

    [tex]b_{r-s} < n^s \mbox{ and } \ k=r[/tex] where s=1,2,3...

    but the coefficients of [tex]n^{r-s}[/tex] are going to be much smaller than any possible value of [tex]n^{r-s}[/tex] once n approaches infinity, such that the first condition can be ignored?

    nb: only just started with latex so bear with me...
     
    Last edited: Dec 30, 2006
  13. Dec 30, 2006 #12

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    What are you using k to mean? Not what I used it to mean.
     
  14. Dec 30, 2006 #13

    Gib Z

    User Avatar
    Homework Helper

    ahh damn i knew i must have stuffed something up, couldnt have been that simple that someone else didnt see...ahh o well
     
  15. Jan 2, 2007 #14
    You are talking about orders of magnitude. First, we prove that a^x/ x tends to infinity as x increases beyond all bounds. Consider the function f(x) = log [a^x/ x] = x log a - log x. Surly if we prove that if f(x) goes to infinity, we prove that a^x/ x goes to infinity as well. Now let us consider f'(x) = log a - 1/x. Now for x>c, we have f'(x)>1/2 log a. Let us define g'(x) = 1/2 log a. By the principles of integral,

    f(x) - f(c) > g(x) - g(c).


    f(x) - f(c) > (x-c)*1/2 log a

    f(x) > (x-c)*1/2 log a + f(c)

    Clearly the right expression goes to infinity as x increases. Thus we have proved that f(x) goes to infinity as well.

    Now we have a polynomial of power n and a x^n+1. If p(x) remains positive past a certain point, the expression x^n+1/p(x) goes to infinity - this can be proven using Hopital's law. Now

    a^x/p(x) = [a^x/x^n+1]*[x^n+1/p(x)].

    We need only to prove that a^x/x^n+1 has a positive limit or is unbounded in order to prove the statement that a^x/p(x) goes to infinity as x increases. We consider, the n+1 th root of the expression that is,

    a^x/n+1 / x

    This can be rewritten as

    1/n+1 * [a^x/n+1 / (x/n+1)]

    Putting x/n+1 = y, we have

    1/n+1 * [a^y / y]

    As was shown, a^y/y goes to infinity as y increases. Thus, the n+1 th root goes to infinity and a^x/x^n+1 goes to infinity as well. Finally, a^x/p(x) goes to infinity.
     
    Last edited: Jan 2, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Limit of
  1. Limit of this (Replies: 21)

  2. No limit (Replies: 3)

  3. Limit ? (Replies: 2)

  4. Limit of a limit? (Replies: 24)

Loading...