xZero's maths question

[xZero]

Prove for all integers m ≥ 3, there is a bound |B(m)| < 3.289868... x , where B(m) denotes the mth-Bernoulli number.

If I understood this question correctly, for any odd mth Bernoulli number greater or equal to 3, this equation stands ( B(mth odd term ≥3) = 0), but how do I go about proving the even mth Bernoulli number also satisfy this equation?  Any hints on this question will be appreciated


Another question, solve the recurrence relation , . I tried to convert it to a homogeneous equation of higher order but the 7^k doesnt disappear :S


Thanks in advance

[xZero]

I think I solved the second q, let's solve the homogeneous part of the equation (LHS).

, let





, since this is a first order there should only be 1 solution





Now we have to look for a particular solution, let













So the general solution is ,









Please correct me if I'm wrong

[xZero]

A quick question, to disprove a partial ordering can I use actual numbers as an example or do I have to stick with algebra? Thanks

[xZero]

Need help with this question, case 1 is trivial but I don't understand case 2. I worked out if a=b then a=c and a=d but how does that prove that a=c and b=d? thanks

question and solution attached below

[TrueTears]

A quick question, to disprove a partial ordering can I use actual numbers as an example or do I have to stick with algebra? Thanks

well to find a counter example you'd find an actual example

[TrueTears]

Need help with this question, case 1 is trivial but I don't understand case 2. I worked out if a=b then a=c and a=d but how does that prove that a=c and b=d? thanks

question and solution attached below

maybe u missed something obvious but your assumption was a = b, so if a = d then that implies b = d

[xZero]

Need help with this question, case 1 is trivial but I don't understand case 2. I worked out if a=b then a=c and a=d but how does that prove that a=c and b=d? thanks

question and solution attached below

maybe u missed something obvious but your assumption was a = b, so if a = d then that implies b = d

hmm but it also implies that b = c since a = c so i was a bit confused there. hopefully this type of question won't show up in the exam tmrw

Thanks for the help anyways

[xZero]

Damn the exam room was freezing, ended up leaving an hour early, anyways just a couple of questions.

1. Use induction to prove the AM and GM inequality.
2. Let P be the set of all prime numbers and O be all odd natural numbers, what is P-O?

[Mao]

question two: P-O = {2}

[TrueTears]

question two: P-O = {2}

That's right, just to expand on Mao's thinking, the reason for this is because all primes (cept for 2) are odd, but not the converse (see if you can prove this), hence the resulting set.

[Mao]

Question 1 is quite interesting. I am able to prove it in a method very similar to Polya's proof, however this is not induction.

For the induction proof, see: http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

[xZero]

mad! got question 2 right, i was shitting myself when i saw P-O thinking i missed something during my revision. And dammit for question 1, didn't realise it was a lead on from the previous parts, it's so obvious now!

@TrueTears, if you mean prove that all primes are odd except 2, i think i can do it

Prove by contradition, let's assume that there exist a prime except 2 that is even, let's call this prime x. If x is even, then it is divisible by 2 but by the definition of prime, it can not be divided by anything other than 1 and itself. Hence even numbers are not prime

Thanks guys for the help!

[xZero]

New question:

Taylor Series
Find an infinite series for the function

What I thought was if you find the taylor series of then you can integrate that to find the taylor series of s(x) but will that work? Does the interval in the integral do anything to the taylor expansion? Thanks

[moekamo]

yea your right, just find the Taylor series for



then integrate term by term. Note: you can only do this integrating term by term if the series converges absolutely in the radius of convergence, i.e. if converges for , then is absolutely convergent for

Basically the terminals on the integral(if this is what you meant about the interval on the integral?) will result in you replacing the u's with x's, since the lower terminal results in every term being 0:



and is this an MTH1030 question? we did this exact thing in lectures lol... except Leo said find the taylor series for so it just had that in front of it lol

[xZero]

omg can't believe i didn't find that in my lecture notes -.-' well that was 20 min well spent. Yup that question's from lab 8, trying to finish all the lab questions before moving on to exams

Also another question, whats the deal with convergence and divergence, it seems like the entire series and sequences base around it and I don't recall leo said why they are so special.

Thanks in advance

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Don't want to double post, got another question

let , find a power series for valid for

well i got

if you rearrange x you would get

heres the problem, i worked out so if you want this series to be valid, but if sub y=-2 in, this obviously breaks down but y=-2 is apart of the condition in the question.

The solution on blackboard is so im really confused :S any help?