Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    17
    ReputationRep:
    Hey guys, so I'm sort of iffy on whether or not my answer to this question is correct so I was just wondering if someone could quickly glance over it. In particular im not so sure on the part about letting  \varepsilon ' = \varepsilon (x^* x_n) but I couldnt really think of any other way to show it.

    Question and attempt are (hopefully) attached.

    Thanks
    Attached Images
      
    Offline

    17
    ReputationRep:
    (Original post by DylanJ42)
    ..
    This proof isn't valid. Your new choice for epsilon (i.e. \epsilon') must be a constant - it cannot depend on n.

    Hint: The result you show in brackets is going to be more helpful to you than "we have x_n having the same sign as x*".
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by DFranklin)
    This proof isn't valid. Your new choice for epsilon (i.e. \epsilon') must be a constant - it cannot depend on n.

    Hint: The result you show in brackets is going to be more helpful to you than "we have x_n having the same sign as x*".
    ah okay thank you, Ill give it another go now
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by DFranklin)
    This proof isn't valid. Your new choice for epsilon (i.e. \epsilon') must be a constant - it cannot depend on n.

    Hint: The result you show in brackets is going to be more helpful to you than "we have x_n having the same sign as x*".
    so I think I have something but it relies on epsilon being less than x*, also the negative and positive cases for x* had to be considered separately. Does this sound along the right lines or am I miles off?
    Offline

    22
    ReputationRep:
    (Original post by DylanJ42)
    so I think I have something but it relies on epsilon being less than x*, also the negative and positive cases for x* had to be considered separately. Does this sound along the right lines or am I miles off?
    This sounds too complicated to be correct, although it may well be.

    You have \displaystyle \left| \frac{1}{x_n} - \frac{1}{x^{\ast}} \right| = \frac{|x_n - x^{\ast} |}{|x_n x^{\ast}|}; you can control the size of the numerator as much as you want since you know that x_n - x^{\ast} \to 0, so it can be made arbitrarily small. All you need to do is show that the denominator is bounded. You can now either (as DFrank) mentions use something akin to what you have in your brackets or use a fairly elementary theorem about convergent sequences and boundedness. I'll leave you to it.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    x
    ok so I understand your second suggestion, although im still sort of confused on DFranklins hint,

    is the idea to show that for all n>= 1 there's an epsilon which satisfies |xn - x*|< E therefore the difference in value of xn and x* is finite therefore xn is bounded?
    Offline

    17
    ReputationRep:
    (Original post by DylanJ42)
    ok so I understand your second suggestion, although im still sort of confused on DFranklins hint,

    is the idea to show that for all n>= 1 there's an epsilon which satisfies |xn - x*|< E therefore the difference in value of xn and x* is finite therefore xn is bounded?
    No. The big issue you have in trying to show 1/x_n is bounded is that everything blows up if x_n can get arbitrarily close to 0. So part of what you want to show that there's a bound m such that |x_n| &gt; m for all sufficiently large n.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by DFranklin)
    No. The big issue you have in trying to show 1/x_n is bounded is that everything blows up if x_n can get arbitrarily close to 0. So part of what you want to show that there's a bound m such that |x_n| &gt; m for all sufficiently large n.
    something really isnt clicking here sorry

    i get that showing x_n > m for large enough n tells us that 1/x_n isnt going to go off to infinity, but yea im not really getting where you're going with that.

    Doesn't the question saying that x_n -> x* and x* =/= 0 just imply that x_n doesn't get arbitrarily close to 0 since it must be getting very very close to some non zero x*?
    Offline

    17
    ReputationRep:
    (Original post by DylanJ42)
    something really isnt clicking here sorry

    i get that showing x_n > m for large enough n tells us that 1/x_n isnt going to go off to infinity, but yea im not really getting where you're going with that.

    Doesn't the question saying that x_n -> x* and x* =/= 0 just imply that x_n doesn't get arbitrarily close to 0 since it must be getting very very close to some non zero x*?
    Yes, but you do need to actually put that into proper mathematical language.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by DFranklin)
    Yes, but you do need to actually put that into proper mathematical language.
    oh okay i see I didnt realise it had to be explicitly shown, ill try my best from there and hopefully get it.

    Thanks for your help
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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.