Mathematical Analysis Questions

Announcements Posted on
    • Thread Starter
    Online

    3
    ReputationRep:
    If anyone would be able to check my answers, and/or point me in the right direction on those which are wrong or I'm stuck on, I'd appreciate it.



    Claim 1: One such counterexample would be if we let \displaystyle a_n=1-\frac{1}{n} which is a strictly increasing sequence that is bounded above at the value of 1.

    Claim 2: a_n=(-2)^n is a counter-example \forall n \in \mathbb{N}

    Claim 3: Stuck on this as I wasn't sure if b_n must be a constant or not, but I settled on the decision that it cannot be. So if b_n=\frac{1}{n} then (a_nb_n) \not \rightarrow \infty as it would tend to 0 instead, I think.

    -----------------------------------------------------------------------------------



    I suppose this one leads off the previous part. I think this is not necessarily true but I'm unsure how to prove it, and I haven't made much progress here.
    Online

    3
    (Original post by RDKGames)
    If anyone would be able to check my answers, and/or point me in the right direction on those which are wrong or I'm stuck on, I'd appreciate it.



    Claim 1: One such counterexample would be if we let \displaystyle a_n=1-\frac{1}{n} which is a strictly increasing sequence that is bounded above at the value of 1.

    Claim 2: a_n=(-2)^n is a counter-example \forall n \in \mathbb{N}

    Claim 3: Stuck on this as I wasn't sure if b_n must be a constant or not, but I settled on the decision that it cannot be. So if b_n=\frac{1}{n} then (a_nb_n) \not \rightarrow \infty as it would tend to 0 instead, I think.

    -----------------------------------------------------------------------------------



    I suppose this one leads off the previous part. I think this is not necessarily true but I'm unsure how to prove it, and I haven't made much progress here.
    Your counterexample for claim 3 is incomplete. You need to specify an (a_n) sequence to go with that (b_n)

    If a_n=n^2, for example, then your sequence for b_n would still have (a_nb_n)\to\infty

    For the last bit, rather than look for a counterexample, try and prove it's true.
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by ghostwalker)
    Your counterexample for claim 3 is incomplete. You need to specify an (a_n) sequence to go with that (b_n)

    If a_n=n^2, for example, then your sequence for b_n would still have (a_nb_n)\to\infty
    So if I say that a_n=n then I assume my counter-example would hold as then (a_nb_n)={ 1,1,1,1,1,... }?

    For the last bit, rather than look for a counterexample, try and prove it's true.
    I'm not quite sure how to go about the proof here for the general case, but I gave it some sort of go.

    Since b_n>\frac{1}{2} we can say that b_n=\frac{1}{2}+\frac{1}{n}, \forall n \in \mathbb{N}

    Then we know that (a_n) \rightarrow \infty so we can say it is a_n=n

    Leaving us with (a_nb_n)=n(\frac{1}{2}+\frac{1}{  n})=\frac{1}{2}n+1 which indeed tends to infinity.

    Generalising exponents we can also get a_n=n^k, b_n=\frac{1}{2}+\frac{1}{n^l}, k \geq 0, l \geq 0 \Rightarrow (a_nb_n)=\frac{n^k}{2}+n^{k-l} which is always tending to infinity by the looks of it, but I do not feel like this is a good enough proof as it doesn't seem to branch out to different functions tending to infinity.
    Online

    3
    ReputationRep:
    (Original post by RDKGames)



    I suppose this one leads off the previous part. I think this is not necessarily true but I'm unsure how to prove it, and I haven't made much progress here.
    If a_n \to \infty then there is an N_+ after which it is always positive. So

    n > N_+ \Rightarrow a_n >0 \Rightarrow a_n b_n > ?
    Online

    3
    (Original post by RDKGames)
    So if I say that a_n=n then I assume my counter-example would hold as then (a_nb_n)={ 1,1,1,1,1,... }?
    Yes.

    I'm not quite sure how to go about the proof here for the general case, but I gave it some sort of go.

    Since b_n>\frac{1}{2} we can say that b_n=\frac{1}{2}+\frac{1}{n}, \forall n \in \mathbb{N}

    Then we know that (a_n) \rightarrow \infty so we can say it is a_n=n

    Leaving us with (a_nb_n)=n(\frac{1}{2}+\frac{1}{  n})=\frac{1}{2}n+1 which indeed tends to infinity.
    Well that's a specific example of two sequences that meet the criterion.

    Generalising exponents we can also get a_n=n^k, b_n=\frac{1}{2}+\frac{1}{n^l}, k \geq 0, l \geq 0 \Rightarrow (a_nb_n)=\frac{n^k}{2}+n^{k-l} which is always tending to infinity by the looks of it, but I do not feel like this is a good enough proof as it doesn't seem to branch out to different functions tending to infinity.
    Whilst a specific example can sometimes help to give you the idea of how to go about a general proof, I think it will take longer to analyse how/why that's happening here and we could get distracted by details in the specific sequences, than it would be to do it for the general case.

    (Note: A sequence isn't necessarily going to have a nice formula defining it.)

    We need to start with, what does it mean a sequence tends to infinity (which is what we've been given in the question for a_n)? What's the mathematical definition you've been given?

    Qualitatively, we can see that if a sequence goes off to infinity, and we multiply each element of it by at least half, it's still going to go off to infinity, only it's rate of growth may be slower.

    PS: Analysis proofs tend to be a shock to the system when first encountered. Can be a bit like you're sailing along nicely, then suddenly lolwtf!!
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by ghostwalker)
    Yes.



    Well that's a specific example of two sequences that meet the criterion.



    Whilst a specific example can sometimes help to give you the idea of how to go about a general proof, I think it will take longer to analyse how/why that's happening here and we could get distracted by details in the specific sequences, than it would be to do it for the general case.

    (Note: A sequence isn't necessarily going to have a nice formula defining it.)

    We need to start with, what does it mean a sequence tends to infinity (which is what we've been given in the question for a_n)? What's the mathematical definition you've been given?

    Qualitatively, we can see that if a sequence goes off to infinity, and we multiply each element of it by at least half, it's still going to go off to infinity, only it's rate of growth may be slower.

    PS: Analysis proofs tend to be a shock to the system when first encountered. Can be a bit like you're sailing along nicely, then suddenly lolwtf!!
    The definition for a sequence tending to infinity has been given as:

    (a_n) \rightarrow \infty if, and only if, \forall C >0, \exists N\in \mathbb{N} such that \forall n > N, a_n>C

    and this makes sense to me but I am struggling to apply it when introducing, within a proof, a different sequence that has to be strictly above \frac{1}{2}.

    Since (a_n) \rightarrow \infty \Rightarrow (a_n)>C, \forall C > 0 and if (b_n)>\frac{1}{2} then (a_nb_n) > \frac{1}{2}C which still tends to infinity \forall n > N, N \in \mathbb{N}
    Online

    3
    (Original post by RDKGames)
    The definition for a sequence tending to infinity has been given as:

    (a_n) \rightarrow \infty if, and only if, \forall C >0, \exists N\in \mathbb{N} such that \forall n > N, a_n>C
    OK, that's the standard definition, and what we need to work with.

    and this makes sense to me but I am struggling to apply it when introducing, within a proof, a different sequence that has to be strictly above \frac{1}{2}.
    It's not so much the b_n we need to worry about as we know its terms are always > 1/2. So we know a_nb_n> \frac{a_n}{2}, for a_n > 0.

    Since (a_n) \rightarrow \infty \Rightarrow (a_n)>C, \forall C > 0 and if (b_n)>\frac{1}{2} then (a_nb_n) > \frac{1}{2}C which still tends to infinity \forall n > N, N \in \mathbb{N}
    (a_n)>C, \forall C > 0 has no meaning as such.

    We know that for any value C>0 we choose, we can find N (dependent on C) such that \forall n > N,a_n > C

    We want to show (I use different letters to avoid confusion),
    for any value D>0 we choose, we can find M (dependent on D) such that \forall n > M,a_nb_n > D

    The problem becomes tying these two definitions together.

    Edit: Sorry, struggling to explain clearly and have to go out just now. Will get back to it again in a few hours.
    Offline

    3
    ReputationRep:
    (Original post by RDKGames)
    The definition for a sequence tending to infinity has been given as:

    (a_n) \rightarrow \infty if, and only if, \forall C >0, \exists N\in \mathbb{N} such that \forall n > N, a_n>C

    and this makes sense to me but I am struggling to apply it when introducing, within a proof, a different sequence that has to be strictly above \frac{1}{2}.
    Hope ghostwalker won't mind me taking a stab at clarifying.

    Although we usually interpret \forall as "for all", or "for every", it's often more helpful to think of it as "for any".

    Similarly, rather than thinking of \exists as "exists", think of it as "we can find".

    So your statement becomes "(a_n)\to \infty if for any C > 0, we can find  N\in \mathbb{N} such that for any n > N, a_n>C"

    Moving on, with a similar rephrasing, the result we actually want to prove (that (a_n b_n)\to\infty) is:

    "for any C > 0, we can find  N\in \mathbb{N} such that for any n > N, a_n b_n>C"

    In a way this is just playing with words, but hopefully it makes it a bit easier to understand that to prove the statement above, your answer is going to look like:

    Take C > 0; here's how we can find N\in \mathbb{N} such that for any n > N, a_n b_n>C,

    (where the bit I've underlined is where the "real work" of the proof is going to lie).

    To do the underlined bit, you're going to need to use the definiton for a_n \to \infty in order to find a suitable "N".

    Edit: regarding going from "for any C" to "Take C > 0". It basically just saves a bit of writing/mental effort. Rather than continually doing/thinking "here's how you do it for all values of C", it's easier to write/think "here's how you do it for a particular value of C" (this is particularly true when you want to use C as a value in another definition). But since the "particular value" was arbitrary, at the end you've still shown it for arbitary C.
    Online

    3
    (Original post by DFranklin)
    Hope ghostwalker won't mind me taking a stab at clarifying.
    Not in the slightest, and much appreciated. I have added notation/definitions for the sake of clarity of the explanation that wouldn't be part of a standard proof, but I can't see offhand how to avoid this. Please feel free to criticise - think I'm making a bit of a pig's ear of it.


    (Original post by RDKGames)
    ...
    OK, lets take a stab at a proof. Whilst the format below should be sound, it won't actually work, but hopefully you can see how it could be adapted.

    With C,D,M,N symbols I defined in my previous post.

    Showing (a_nb_n) is unbounded:

    We have an arbitrary D > 0.

    By the unboundedness of (a_n) and choosing C equal to D

    \exists N, s.t. \forall n>N, a_n > D

    But we're interested in a_nb_n, and since b_n> 0 we can say

    \exists N, \forall n>N, a_nb_n > Db_n

    and since b_n > 1/2

    \exists N, \forall n>N, a_nb_n > D/2

    We now choose our M to equal N.

    So, we've shown:

    \forall D>0, \exists M\in\mathbb{N}s.t.\forall n>M, a_nb_n>D/2

    Which is almost what we want, but not quite. There's a pesky "/2". So, we need to adjust this, somewhere in the point we either go into, or come out of our orginal sequence (a_n)
    Online

    3
    ReputationRep:
    (Original post by ghostwalker)
    Which is almost what we want, but not quite. There's a pesky "/2". So, we need to adjust this, somewhere in the point we either go into, or come out of our orginal sequence (a_n)

    Doesn't D being an arbitrary value not just imply the sequence tends to infinity anyway?
    Online

    3
    (Original post by B_9710)
    Doesn't D being an arbitrary value not just imply the sequence tends to infinity anyway?
    It does.

    BUT, it's more important that the OP gets a feel for how these proofs are done (even though I don't feel I've explained it very well) and their exactness, than look for the "shortcuts" (for want of a better term).
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by ghostwalker)
    Not in the slightest, and much appreciated. I have added notation/definitions for the sake of clarity of the explanation that wouldn't be part of a standard proof, but I can't see offhand how to avoid this. Please feel free to criticise - think I'm making a bit of a pig's ear of it.


    OK, lets take a stab at a proof. Whilst the format below should be sound, it won't actually work, but hopefully you can see how it could be adapted.

    With C,D,M,N symbols I defined in my previous post.

    Showing (a_nb_n) is unbounded:

    We have an arbitrary D > 0.

    By the unboundedness of (a_n) and choosing C equal to D

    \exists N, s.t. \forall n>N, a_n > D

    But we're interested in a_nb_n, and since b_n> 0 we can say

    \exists N, \forall n>N, a_nb_n > Db_n

    and since b_n > 1/2

    \exists N, \forall n>N, a_nb_n > D/2

    We now choose our M to equal N.

    So, we've shown:

    \forall D>0, \exists M\in\mathbb{N}s.t.\forall n>M, a_nb_n>D/2

    Which is almost what we want, but not quite. There's a pesky "/2". So, we need to adjust this, somewhere in the point we either go into, or come out of our orginal sequence (a_n)
    Ah that is a good example of how to approach these proofs, thank you! So for the D/2 at the end, would you just multiply both sides by 2 and show that 2(a_nb_n)>D \Rightarrow 2(a_nb_n) \rightarrow \infty \Rightarrow (a_nb_n) \rightarrow \infty ?? Other method of getting rid off that "/2", as you say, I think would be to let a different variable equal D/2.

    Would this be the correct approach?

    Analysis seems like out of this world so far as I am 2 weeks into it aha, trying to get used to all of it is proving to be a hurdle!
    Online

    3
    (Original post by RDKGames)
    Ah that is a good example of how to approach these proofs, thank you! So for the D/2 at the end, would you just multiply both sides by 2 and show that 2(a_nb_n)>D \Rightarrow 2(a_nb_n) \rightarrow \infty \Rightarrow (a_nb_n) \rightarrow \infty ?? Other method of getting rid off that "/2", as you say, I think would be to let a different variable equal D/2.

    Would this be the correct approach?

    Analysis seems like out of this world so far as I am 2 weeks into it aha, trying to get used to all of it is proving to be a hurdle!
    [I should emphaszie I'm being pedantic here.

    Saying 2(a_nb_2)\to\infty \Rightarrow (a_nb_n)\to\infty seems and is obvious, but prove it from the definitions. There are a several simple results that it is useful to use/have, but you really need to prove them first before using them.]

    "let a different variable equal D/2." may be heading in the right direction, but I'm not clear what you mean.

    I think it best to just say:
    We chose D to start with, and then we chose C based on D, but there is no requirement for them to be equal. So, suppose we let C=2D....
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by ghostwalker)
    [I should emphaszie I'm being pedantic here.

    Saying 2(a_nb_2)\to\infty \Rightarrow (a_nb_n)\to\infty seems and is obvious, but prove it from the definitions. There are a several simple results that it is useful to use/have, but you really need to prove them first before using them.]

    "let a different variable equal D/2." may be heading in the right direction, but I'm not clear what you mean.

    I think it best to just say:
    We chose D to start with, and then we chose C based on D, but there is no requirement for them to be equal. So, suppose we let C=2D....
    Thank you very much for the clear explanation!
    • Thread Starter
    Online

    3
    ReputationRep:
    ghostwalker DFranklin

    Practicing the ways of Analysis proofs I stumbled on this one though it seems as if it should be straightforward. I'm unsure;

    Name:  Capture.PNG
Views: 17
Size:  9.0 KB

    (a) (a_n) \rightarrow -\infty if, and only if, \forall C<0, \exists N \in \mathbb{N} : \forall n > N, a_n<C



    (b) We know that if (a_n) \rightarrow -\infty then \exists N \in \mathbb{N} : \forall n > N, a_n<C, \forall C<0

    We want the conclusion that  \forall \epsilon>0, \exists M \in \mathbb{N} : \forall n>M, \lvert \frac{1}{a_n} \lvert <\epsilon

    So if \epsilon >0 \Rightarrow \frac{1}{\epsilon}>0 and if 0>C \Rightarrow \frac{1}{\epsilon}>C

    Since \forall n > N, a_n < C \Rightarrow a_n < \frac{1}{\epsilon}

    \Rightarrow \epsilon < \frac{1}{a_n} but this doesn't make sense as \frac{1}{a_n} is negative and \epsilon is positive, so the negative cannot be greater than the positive. Not sure if taking the modulus of the RHS would resolve this.
    Offline

    3
    ReputationRep:
    (Original post by RDKGames)
    ,,
    TBH, I'm finding it very hard to follow your argument here. A piece of general advice: using \implies can be very confusing in these kinds of proofs, especially where you are sometimes using it in the sense of a formal definition (e.g. \exists N s.t. n > n \implies ...), and using it in terms of normal mathematical argument (e.g. x+3 > y \implies x > y-3).

    To give a concrete example of why it's confusing, look at your line

    So if \epsilon > 0 \implies \frac{1}{\epsilon} > 0 and if  0 > C \implies \frac{1}{\epsilon} > C
    I don't know if you mean "if epsilon > 0 then 1/epsilon > 0 and ...", or you mean "if it is true that 'epsilon > 0 implies 1/epsilon > 0' and ..."

    It is usually a lot clearer to simply write "so" or "then" if you are just using normal mathematical arguments.

    In this case I am guessing you meant something along the lines of

    "So, if epsilon > 0 then 1/epsilon > 0, and if 0 > C then 1/epsilon > C".

    As far as the actual proof here, I would focus on why (by intuition) you would expect the result to be true.

    That is: if a_n is consistently large and negative, then 1/a_n is consistently small and negative, so 1/|a_n| is small and positive.
    You want to show 1/|a_n| < epsilon. Working backwards shows you you will need 0 > 1/a_n > -epsilon (note use of > since a_n and - epsilon are both negative). So you need a_n < -1/epsilon.
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by DFranklin)

    I don't know if you mean "if epsilon > 0 then 1/epsilon > 0 and ...", or you mean "if it is true that 'epsilon > 0 implies 1/epsilon > 0' and ..."
    I began the proof by stating what we have, and what we need to achieve to make it clear before heading into the main bodywork of it.

    Here I simply worked from our wanted form that comes straight from the definition: \forall \epsilon&gt;0, \exists M \in \mathbb{N} : \forall n&gt;M, \lvert \frac{1}{a_n} \lvert &lt;\epsilon, and so this tells us that it must be true that \epsilon &gt; 0 before proceeding on with the argument.

    As far as the actual proof here, I would focus on why (by intuition) you would expect the result to be true.

    That is: if a_n is consistently large and negative, then 1/a_n is consistently small and negative, so 1/|a_n| is small and positive.
    You want to show 1/|a_n| < epsilon. Working backwards shows you you will need 0 > 1/a_n > -epsilon (note use of > since a_n and - epsilon are both negative). So you need a_n < -1/epsilon.
    I understand that but I'm having difficulties seeing where this inequality comes from exactly.

    \frac{1}{a_n} &lt; 0 \implies\lvert \frac{1}{a_n} \lvert &gt; 0 and \epsilon &gt; 0 but how do you get the inequality you've shown from this?
    Online

    3
    ReputationRep:
    if bn = 1/an that would be a counter example for c)
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by the bear)
    if bn = 1/an that would be a counter example for c)
    Thanks but I'm already done with that part, onto some new Analysis questions now
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: October 19, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
How are you feeling about doing A-levels?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.