University of Warwick

Coventry

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Oh yeah. Still not too difficult.

I'm having some problems with h), show that there is only one prime triple (3 primes of the form n, n+2, n+4).

There's also problems with pretty much every other question after it. Why's this one so much harder than the more recent ones...

I'm having some problems with h), show that there is only one prime triple (3 primes of the form n, n+2, n+4).

There's also problems with pretty much every other question after it. Why's this one so much harder than the more recent ones...

University of Warwick

Coventry

benwellsday

Oh yeah. Still not too difficult.

I'm having some problems with h), show that there is only one prime triple (3 primes of the form n, n+2, n+4).

There's also problems with pretty much every other question after it. Why's this one so much harder than the more recent ones...

I'm having some problems with h), show that there is only one prime triple (3 primes of the form n, n+2, n+4).

There's also problems with pretty much every other question after it. Why's this one so much harder than the more recent ones...

that one is a doddle.

any natural number can be written in the form 3m 3m+1 or 3m+2

show that for each individual case one of the 3 n n+2 or n+4 must have be a multiple of 3 and hence not prime.

Ok here is my fail attempt at 2006 analysis

I may improve on this in the morning

I may improve on this in the morning

The first few answers for (1) don't seem to match the questions. Not sure what happened there. Actually, nothing matches 2006. I think this is 2007.

Here's anything I've got to say anyway

Here's anything I've got to say anyway

benwellsday

The first few answers for (1) don't seem to match the questions. Not sure what happened there. Actually, nothing matches 2006. I think this is 2007.

Tom told me to do 2006, but I was very drunk and didn't notice I was actually doing 2007.

benwellsday

benwellsday

benwellsday

benwellsday

benwellsday

benwellsday

benwellsday

benwellsday

And I didn't do (d) on paper, so no ideas there.

And I didn't do (d) on paper, so no ideas there.

No matter. Thanks for your help

hey guys,

how do you prove from the completeness axiom that an increasing sequence an which is bounded above is convergent.

do you start with saying its non-empty and bounded above so there's a least upper bound, say U.

then a(n) is less than or equal to U for all n.

then for all positive epilson there's N s.t

la(n) - Ul < epilson for all n>N

but it seems like i'm fudging it to get epilson into this. i've lost workbook 5, and i think it's from there.

thanks for any help, i just want a rigorous answer.

how do you prove from the completeness axiom that an increasing sequence an which is bounded above is convergent.

do you start with saying its non-empty and bounded above so there's a least upper bound, say U.

then a(n) is less than or equal to U for all n.

then for all positive epilson there's N s.t

la(n) - Ul < epilson for all n>N

but it seems like i'm fudging it to get epilson into this. i've lost workbook 5, and i think it's from there.

thanks for any help, i just want a rigorous answer.

Ok, the sequence isn't what's non empty. The set of the members of sequence is non empty, so it has a supremum because of the completeness axiom.

I've written an answer above per Q3 from the part which begins 'lemma'. Benswellsday points out where the completeness axiom is used in the comments related to the question

I've written an answer above per Q3 from the part which begins 'lemma'. Benswellsday points out where the completeness axiom is used in the comments related to the question

i'm doing 2005 question 4a. for analysis. can someone please quickly compare answers.

first, is convergent as less than 1/(n^2) and is absolutely convergent too.

2. is bigger than 1/n so diverges by comparison test to +infinity

3.converges by alternating series test, and converges conditionally ( as n^-.5 doesn't converge)

4.converges absolutely ( as equal to 2^-n)

Is what i'm writing what you would write in an exam or for this question would you have to show how you figured it out???

thanks in advance guys

first, is convergent as less than 1/(n^2) and is absolutely convergent too.

2. is bigger than 1/n so diverges by comparison test to +infinity

3.converges by alternating series test, and converges conditionally ( as n^-.5 doesn't converge)

4.converges absolutely ( as equal to 2^-n)

Is what i'm writing what you would write in an exam or for this question would you have to show how you figured it out???

thanks in advance guys

amo17

thanks, so you have to definietely prove the lemma as well for the proof or can you just use it?

Because it makes up most of the proof, I think you have to prove it.

amo17

i'm doing 2005 question 4a. for analysis. can someone please quickly compare answers.

first, is convergent as less than 1/(n^2) and is absolutely convergent too.

2. is bigger than 1/n so diverges by comparison test to +infinity

3.converges by alternating series test, and converges conditionally

4.converges absolutely

Is what i'm writing what you would write in an exam or for this question would you have to show how you figured it out???

thanks in advance guys

first, is convergent as less than 1/(n^2) and is absolutely convergent too.

2. is bigger than 1/n so diverges by comparison test to +infinity

3.converges by alternating series test, and converges conditionally

4.converges absolutely

Is what i'm writing what you would write in an exam or for this question would you have to show how you figured it out???

thanks in advance guys

I agree with those answers, though you might want to say what test you used in part 4. Also, you might want to say why 3 doesn't converge absolutely.

You might not want to write things like 'as n^-0.5 doesn't converge', since it's not true. The sum doesn't converge.

Also, I think your reasoning is wrong for iv). It's not n!/2(n!) it's n!/(2n)!.

I haven't looked at part C) and I'm going out in a minute. I'll have a crack when I get back if someone else hasn't already and I'm sober enough...

Also, I think your reasoning is wrong for iv). It's not n!/2(n!) it's n!/(2n)!.

I haven't looked at part C) and I'm going out in a minute. I'll have a crack when I get back if someone else hasn't already and I'm sober enough...

Well usually when it involves factorails in sums.... I would try the ratio test

k guys, i have another question but i've attached it cos i can't write it out clearly. is the answer N=1000?

cheers

cheers

amo17

k guys, i have another question but i've attached it cos i can't write it out clearly. is the answer N=1000?

cheers

cheers

When faced with a similar question in my real exam, I wrote N=420^456723287329 or something ridiculously high, because I couldn't be assed to do the algebra for two marks. They don't ask you to prove it after all, you just have to spot that (3n+1)/n converges to 3 . I amused myself by adding more digits onto the exponent whenever I got stuck elsewhere.

JohnnySPal

When faced with a similar question in my real exam, I wrote N=420^456723287329 or something ridiculously high, because I couldn't be assed to do the algebra for two marks. They don't ask you to prove it after all, you just have to spot that (3n+1)/n converges to 3 . I amused myself by adding more digits onto the exponent whenever I got stuck elsewhere.

thanks ok, i will take that advice. but seriously out of interest i want to know how to do the question. is it just a rearrangement?

thanks

amo17

oh ok, thanks so the ratio test goes to 1/2(2n+1), which converges to zero, so the sum converges to 0.

would they ever ask us to work out what the sums actually converge to or just if they converge, do you think?

would they ever ask us to work out what the sums actually converge to or just if they converge, do you think?

Varies depending on the sum in the last case, there are a number of neat tricks(summation by parts, finite calculus, generating functions, using known series) that can sometimes help you obtain the value of the sum

And you've misunderstood the ratio test, it's the fact that the ratio goes to 0 as n tends to infinity that shows you it converges, but not that the sum is equal to 0.

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