1. Limited time only! Sign up for a free 30min personal 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!

Formal limit proof

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove [itex] \lim_{n\rightarrow\infty} \frac{p(n)}{e^n} = 0[/itex] where [itex] p(x) = a_k x^k + ... + a_1 x + a_0 [/itex] (with real coefficients [itex] a_i [/itex] in [itex] \mathbb{R} ) [/itex]


    3. The attempt at a solution

    I thought about using series to try and prove this, but I couldn't get it to work out and I think there is an easier way.

    [itex] \frac{p(n)}{e^n} [/itex] = [itex] \frac{ \sum_{n=0}^\infty a_k n^k}{\sum_{n=0}^\infty \frac{n^k}{k!}} [/itex] = [itex] \frac{ \sum_{n=0}^\infty a_k}{\sum_{n=0}^\infty \frac{1}{k!}} [/itex]
     
  2. jcsd
  3. Sep 27, 2011 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Replace n by a continuous variable x and use L'Hospital's rule.
     
  4. Sep 27, 2011 #3
    I don't think I'm allowed to use L'Hospital's rule, we have not proven it in class yet.
     
  5. Sep 27, 2011 #4
    Could you show that [itex]a_kn^k/e^n[/itex] goes to 0 as n goes to infinity? Then each lower order term would go to 0, as well. What tests/methods are you allowed to use? Does this have to be strict [itex]\epsilon - \delta[/itex] from the defimition of limit?
     
  6. Sep 27, 2011 #5
    No we are proving things based on the definition of convergence, for instance, our proofs start off as "Let [itex] \epsilon > 0 [/itex] be arbitrary, and choose N > ... such that for all n > N ... etc. We haven't done epsilon delta proofs yet.
     
  7. Sep 27, 2011 #6
    Whoops! yes, [itex] \epsilon - N[/itex] is what I meant. But anyway, could you show that [itex]a_kn^k/e^n[/itex] goes to 0? For instance, could you do something like this:
    Let [itex]a = sup \{|a_i| : 0 \leq i \leq k \}[/itex] then [itex] a_kn^k/e^n \leq an^k/e^n [/itex] and then show that [itex] an^k/e^n [/itex] goes to 0? Then deduce that the lower order terms go to 0? Of course, you are going to have to take more care than I did because 1)you'll need to find an N and 2)you'll need to take care of the times when you have negative [itex]a_k[/itex]'s. Anyway, am I making sense?
     
  8. Sep 27, 2011 #7
    Actually, since [itex] a_kn^k/e^n \leq an^k/e^n [/itex] is the largest term of the polynomial, couldn't you just find a N that suits this expression? Because [itex] p(n)/e^n [/itex] can be written as [itex] \frac{a_0}{e^n} + \frac{a_1 n}{e^n} + ... + \frac{a_k n^k}{e^n} [/itex], and since you choose n > N always, then the N that you pick for the largest term will be true for all terms, which will satisfy the definition, correct?
     
  9. Sep 27, 2011 #8
    Correct; that's what I had in mind. Now, find that N. Is this an Analysis class? I had terrible trouble proving things like this because it is so freaking hard to know what you are allowed to use. For example, L'Hospital's is off the table, but what can you use? The obvious answer, at first, is that you can only use stuff that you have already learnt in the analysis class but I'd be willing to bet you haven't defined e^x, yet, is this correct? Do you find this a confusing as I did?
     
  10. Sep 27, 2011 #9
    Yes, an analysis class, but yeah it's a little awkward, especially since the professor used a derivative to prove some lemma or something, then goes and says "we shouldn't be using derivatives yet, but I'll use it anyways" :) And no we have not defined e^x yet, correct
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Formal limit proof
Loading...