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Analysis (Tripos Question) Help

http://www.maths.cam.ac.uk/undergrad/pastpapers/2005/Part_IA/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.

Reply 1

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.

Reply 2

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.

Reply 3

DFranklin

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.

Reply 4

IrrationalNumber

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).

Reply 5

tsrguru

Original post by DFranklin

For 9F (iii): You should be able to see the terms behave like

\dfrac{1}{n}

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.

Reply 6

DFranklin

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.

Reply 7

worriedParnt

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.

(edited 12 years ago)

Reply 8

worriedParnt

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.

Reply 9

nuodai

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.