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xZero's maths question

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moekamo:
do you mean whats so special about knowing whether a series diverges or converges?

i think that if you know that they converge, then you can try and find out what it converges to, obviously this is good because then we may be able to generalise it, like with the geometric series where S=1/(1-r) given |r|<1, there may be other reasons, im just making stuff up here :P

but is this what your asking? i cant really understand the question...lol

xZero:
Just incase if you missed it, i had an extra question in my last post :P

well you know how we're trying to see whether a series converges or not, why do we want to know that? I can understand it for taylor expansion because if it diverges then the approximated answer will go up to infinity but why are we trying to find it for like any other series?

xZero:
Ceebs starting a new thread, gonna revive my old question thread.

I have a question on real analysis (MTH3140)

Let be non-empty and bounded by below and define B={: b is a lower bound for A}

Prove that sup B = inf A

My tutor did it some complicated way, I was wondering if this way works as well

To prove that sup B= inf A, we must first prove the existence of sup B and inf A, since the set A is defined such that it is bounded by below and is non-empty, by the axiom of completeness, inf A exists. Similarly, since lower bound for A exists, b which is an element of B also exists, thus it is non empty. Let , sup B exists if such that , . Since B={b}, it can be said that x=k=b and , which satisfy the definition of sup B. Hence sup B exists and sup B = b.

By the definition of b, which is the lower bound for A, inf A = b, thus we can conclude that sup B = b = inf A

Cheers

kamil9876:
I can't quite follow, there are some typos in there? The following is definitely false however:


--- Quote from: xZero on March 21, 2013, 10:05:04 pm ---
Since B={b}

--- End quote ---

You are suggesting that B consists of only one element? This is certainly not the case, for e.g: if then

xZero:

--- Quote from: kamil9876 on March 21, 2013, 11:16:22 pm ---I can't quite follow, there are some typos in there? The following is definitely false however:

You are suggesting that B consists of only one element? This is certainly not the case, for e.g: if then

--- End quote ---
mm I guess my understanding of lower bound was wrong, so if b is a lower bound of A, it can be a range of numbers rather than just a single element?

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