Hey there Sign in to join this conversationNew here? Join for free

Analysis (Tripos Question) Help

Announcements Posted on
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    http://www.maths.cam.ac.uk/undergrad.../PaperIA_1.pdf

    Hi, I need help on 9F) iii) and the last part of 11E)

    For 9F) iii) I tried the ratio test, I can't think of any other test which would work and I think I for the last part of 11 E) I need an auxiliary function, I tried f(x) = g(x + 1/n ) - g(x) , it doesn't quite give me the result.

    Thanks for any help in advance.
    • 13 followers
    Offline

    ReputationRep:
    For 9F (iii): You should be able to see the terms behave like \dfrac{1}{n} for large n. So you're expecting it to diverge. Since you're expecting it to diverge, the way forwards is to show that for large n the numerator is > kn^3 for some constant k, and the denominator is < Kn^4 for some other constant N. Then it diverges by the comparison test.

    For 11F, that function should work. Show (by contradiction), that there must be an integer k with 0 <= k < n and f(k/n)f((k+1)/n) < 0.
    • 0 followers
    Offline

    ReputationRep:
    Alternatively for 11F, consider  \sum_{r=0}^{n-1} f(\frac{r}{n}) . Directly you can then see that f(r/n)=0 for each r or at least one is positive and at least one is negative.
    • 13 followers
    Offline

    ReputationRep:
    (Original post by IrrationalNumber)
    Alternatively for 11F, consider  \sum_{r=0}^{n-1} f(\frac{r}{n}) . Directly you can then see that f(r/n)=0 for each r or at least one is positive and at least one is negative.
    Yeah, that's what I meant.
    • 0 followers
    Offline

    ReputationRep:
    (Original post by DFranklin)
    Yeah, that's what I meant.
    Sorry, to be honest, I couldn't see how to use your hint (it looks like a product).
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    (Original post by DFranklin)
    For 9F (iii): You should be able to see the terms behave like \dfrac{1}{n} for large n. So you're expecting it to diverge. Since you're expecting it to diverge, the way forwards is to show that for large n the numerator is > kn^3 for some constant k, and the denominator is < Kn^4 for some other constant N. Then it diverges by the comparison test.

    For 11F, that function should work. Show (by contradiction), that there must be an integer k with 0 <= k < n and f(k/n)f((k+1)/n) < 0.

    (Original post by IrrationalNumber)
    Alternatively for 11F, consider  \sum_{r=0}^{n-1} f(\frac{r}{n}) . Directly you can then see that f(r/n)=0 for each r or at least one is positive and at least one is negative.
    Thanks that worked.
    • 13 followers
    Offline

    ReputationRep:
    (Original post by IrrationalNumber)
    Sorry, to be honest, I couldn't see how to use your hint (it looks like a product).
    f(a)f(b)<0 is just a "lazy" way of writing "one of f(a), f(b) is negative and one positive". I should have considered the equality case though.
    • 0 followers
    Offline

    ReputationRep:
    (Original post by IrrationalNumber)
    Alternatively for 11F, consider  \sum_{r=0}^{n-1} f(\frac{r}{n}) . Directly you can then see that f(r/n)=0 for each r or at least one is positive and at least one is negative.

    Very cute solution.
    • 0 followers
    Offline

    ReputationRep:
    (Original post by DFranklin)
    f(a)f(b)<0 is just a "lazy" way of writing "one of f(a), f(b) is negative and one positive". I should have considered the equality case though.
    For the first part of the question is one supposed to prove the Intermediate Value Theorem? Question 12 starts with "Prove Rolle's Theorem" which is clear but I am not sure what to do here.
    • 3 followers
    Offline

    (Original post by worriedParnt)
    For the first part of the question is one supposed to prove the Intermediate Value Theorem? Question 12 starts with "Prove Rolle's Theorem" which is clear but I am not sure what to do here.
    The first part of Q11 is essentially saying "prove the intermediate value theorem", except it's the special case where the intermediate value is 0 (which simplifies notation). So if you really want to then you could prove the full IVT, but it's not necessary.
    • 0 followers
    Offline

    ReputationRep:
    (Original post by nuodai)
    The first part of Q11 is essentially saying "prove the intermediate value theorem", except it's the special case where the intermediate value is 0 (which simplifies notation). So if you really want to then you could prove the full IVT, but it's not necessary.
    Thanks very much

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?

    this is what you'll be called on TSR

  2. this can't be left blank
    this email is already registered. Forgotten your password?

    never shared and never spammed

  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide the button to the right to create your account

    Slide to join now Processing…

Updated: April 3, 2012
New on TSR

Moving on from GCSEs

What advice would you give someone starting A-levels?

Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.