xZero's maths question

[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

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

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:

Since B={b}

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

[xZero]

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

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?

[kamil9876]

A lower bound for a set is a number such that for all .

So in particular if is a lower bound, so is every number smaller than . Hence at least one lower bound implies infinitely many.

What you are confusing is "lower bound" and "greatest lower bound" a.k.a , this is unique (and exists if the set is bounded from below by basic properties of ). There can be many lower bounds but there is only (at most) one greatest lower bound.

[xZero]

Another question on real analysis,

Let f:[0,1] -> [0,1] be a non increasing, non-continuous function, i.e. whenever . Prove that there exists such that f(c)+c=1. Hint: Let and define c=inf A.

I'm really confused with this question, the first step would be proving that inf A exists, since the set A is both upper and lower bounded (0 and 1), by axiom of completeness, inf A and sup A also exists. Next would be proving that inf A is in the set A, since A is in a inclusive interval, inf A is also in A (not sure if that's enough of a proof).

Now I'm stuck at this point, I have no idea what to do from here on, any hints/examples will be highly appreciated. Thanks

[Mao]

I'm not quite sure about the proper notation (I've never paid any attention to this kind of rigor, because applied maths major), but I believe the logic will suffice.

To show that and , it is sufficient to show that and intersects in the domain

Since spans the entire range , and by definition, there must be an intersection.

[xZero]

Opps, sorry but I completely missed the part that f(x) is non-continuous, don't think the proof works for a non-continuous function. Thanks tho

[xZero]

turns out there are parts to this question on the other side of the sheet, which I completely missed, again -.- the questions are kinda confusing so if anyone can help me with some of them that would be awesome.

a)Explain why c exists.
Since X is bounded below, by axiom of completeness, inf X exists, since A is a subset of X, then inf A also exists.

b)Let be a sequence of elements of A that converges to c. Prove that, for all n,
By definition of infimum, for all , and from the non-increasing function, , since

*This is where I got confused, if c is inf A then the sequence of x_n should be decreasing, meaning that x_n > c, but in the later parts it assumes that c>x_n.

c)Deduce that
since x is within [0,1], a closed interval, min X exists, so min A also exists. In this case min A = inf A = c, so c is within A, thus satisfying the condition that

d)If c=0, prove that
Let c=0 since f:[0,1] ->[0,1], the value of f(c) must exist between 0 and 1, thus

e)If c>0, by considering , or otherwise, prove that

This is the most confusing part, from b I assumed the sequence to be decreasing, but in this part it says that x_n<c, so c is no longer inf A. Maybe I misunderstood the original question or something but I've been struggling for ages to figure this one out so can someone please give me some sort of hint?

Thanks

[kamil9876]

First of all let me make some comments: (I think) Mao was alluding to the use of the intermediate value theorem, this is a standard way of approaching this problem if we had assumed that f was continous (but clearly this theorem does not hold for functions which aren't continous). However since we are given that the function is not continous, we cannot use this theorem. Indeed, what if we have for and . Then you can check that there does there does not exist a such that . However this isn't a counterexample to the exercise since our example isn't non-increasing (because ). So the catch is that although this is easy with the intermediate value theorem (as Mao showed) we can't use it as we must show it is also true for discontinous non-increasing functions too and hence the more complicated argument below.

Now back to the actual problem:

a)Explain why c exists.
Since X is bounded below, by axiom of completeness, inf X exists, since A is a subset of X, then inf A also exists.

Of course, by "c" we are refering to the c from the hint, i.e where

What is X in your answer? You also need to verify that is non-empty(to use the completeness axiom) and not just bounded. To show that it is non-empty notice that since since the codomain of is

b)Let be a sequence of elements of A that converges to c. Prove that, for all n,
By definition of infimum, for all , and from the non-increasing function, , since

*This is where I got confused, if c is inf A then the sequence of x_n should be decreasing, meaning that x_n > c, but in the later parts it assumes that c>x_n.

is all that is used, so that you get the inequality .
.

c)Deduce that
since x is within [0,1], a closed interval, min X exists, so min A also exists. In this case min A = inf A = c, so c is within A, thus satisfying the condition that

I am confused? what is this set "X" that you are refering to? I don't follow your argument. The standard way is to see that you have already shown above that and since the we also have (a standard fact that limits preserve which you may want to prove yourself as an exercise).

d)If c=0, prove that
Let c=0 since f:[0,1] ->[0,1], the value of f(c) must exist between 0 and 1, thus

Good.

e)If c>0, by considering , or otherwise, prove that

This is the most confusing part, from b I assumed the sequence to be decreasing, but in this part it says that x_n<c, so c is no longer inf A. Maybe I misunderstood the original question or something but I've been struggling for ages to figure this one out so can someone please give me some sort of hint?

Thanks

I think the hint to consider is NOT saying that these 's are the same as the 's in the previous part of the question. Perhaps the author should give them another name to avoid this unfortunate confusion. So we will instead say

Hint: By definition of the and so (technically need to be large enough so that the otherwise does not make sense). Then argue similairly as in part b and c to show that .

[Deleted User]

How do I do these questions?

Find the coordinate vector of v with respect to the given basis B for the vector space V.

a) v= 2-5x, B = {x+1,x-1}, V=P1

b) v= [1 2 1; -1 1 2], B={E^(ij)|i=1,2; j=1,2,3}, V=M2,3

Thanks!

[xZero]

Yes the question is intermediate value problem, however in the footnote it said that since its non-continuous and we haven't covered ivp, we don't need to know. Sorry about that Mao, still appreciate the help tho <3

"What is X in your answer? You also need to verify that  is non-empty(to use the completeness axiom) and not just bounded. To show that it is non-empty notice that  since  since the codomain of is "
Sorry, my X should be X=[0,1], and x is an element of X.

"I am confused? what is this set "X" that you are refering to? I don't follow your argument. The standard way is to see that you have already shown above that  and since the  we also have  (a standard fact that limits preserve  which you may want to prove yourself as an exercise). "
mm, I didn't know that limits preserve greater or equal, I tried proving it but all I came up with is take a limit on both side, which doesn't break the inequality.

"I think the hint to consider  is NOT saying that these 's are the same as the 's in the previous part of the question. Perhaps the author should give them another name to avoid this unfortunate confusion. So we will instead say "
Your right, I just got an email back from the lecturer, x_n is different in different parts so I was confused there. If I apply what I did in part b and c, I will get , I'm having problem making the sign less or equal rather than just less. I guess a very loosely argument would be since c is in A, f(c)+c must be equal to 1 or greater, so the sign change to less or equal but I don't see that as a legit proof.

Anyways thanks Kamil for the help <3

[kamil9876]

If I apply what I did in part b and c, I will get f(c)+c <1

You actually can't get f(c)+c<1 because it isn't true! The point is that limits preserve but NOT . For example but after taking limits as n goes to infinity you get 0>0 which isn't true.

I'm having problem making the sign less or equal rather than just less.

But clearly IMPLIES so this puts us into essentially the same situation as in part b and c.

[xZero]

You actually can't get f(c)+c<1 because it isn't true! The point is that limits preserve but NOT . For example but after taking limits as n goes to infinity you get 0>0 which isn't true.

But clearly IMPLIES so this puts us into essentially the same situation as in part b and c.

oh my god, you're a genius! Thank you!!!