Hey there! Sign in to join this conversationNew here? Join for free

Maclaurins Series Expansions, FP3 Watch

Announcements
    • Thread Starter
    Offline

    2
    ReputationRep:
    I've been doing the miscallaneous exercises from the AQA FP3 textbook, and I'm really stuck on the last question about Maclaurin's series. It asks you to show that the series for 1/(sin 2x) is 1/(2x) + (1/3) x + (7/45) x^3, having already found sin (2x). I tried rewriting it as (1+(sin(2x) -1))^(-1) and using binomial series, but it was impossible to calculate as each term in x changed each time you added a new term, so it would change infinitely. Also, isn't f(0) for this function undefined? So you couldn't get a series anyway?
    Can somebody please explain this to me!
    Thanks
    p.s. It then asks for 1/(1-e^(-x)), having already found 1-e^(-x), which I can't do either.
    • Very Important Poster
    Offline

    21
    ReputationRep:
    (Original post by lambda98)
    I've been doing the miscallaneous exercises from the AQA FP3 textbook, and I'm really stuck on the last question about Maclaurin's series. It asks you to show that the series for 1/(sin 2x) is 1/(2x) + (1/3) x + (7/45) x^3, having already found sin (2x). I tried rewriting it as (1+(sin(2x) -1))^(-1) and using binomial series, but it was impossible to calculate as each term in x changed each time you added a new term, so it would change infinitely. Also, isn't f(0) for this function undefined? So you couldn't get a series anyway?
    Can somebody please explain this to me!
    Thanks
    p.s. It then asks for 1/(1-e^(-x)), having already found 1-e^(-x), which I can't do either.
    Perhaps this would be better off in the Maths forum.
    Offline

    5
    ReputationRep:
    (Original post by lambda98)
    I've been doing the miscallaneous exercises from the AQA FP3 textbook, and I'm really stuck on the last question about Maclaurin's series. It asks you to show that the series for 1/(sin 2x) is 1/(2x) + (1/3) x + (7/45) x^3, having already found sin (2x). I tried rewriting it as (1+(sin(2x) -1))^(-1) and using binomial series, but it was impossible to calculate as each term in x changed each time you added a new term, so it would change infinitely. Also, isn't f(0) for this function undefined? So you couldn't get a series anyway?
    Can somebody please explain this to me!
    Thanks
    p.s. It then asks for 1/(1-e^(-x)), having already found 1-e^(-x), which I can't do either.

    I feel like there is a trick here, finding the it normally I think is impossible since as you said f(0) would be undefined. The term that has gotten my interest is 1/2x which I think carries a lot of weight here since it is abnormal.
    Offline

    5
    ReputationRep:
    I think I have it, the 1/2x term should be a major clue however I could see why you don't immediately see it.

    Maclaurin expansions only exist for functions which are defined at f(0) and since 1/sin(2x) isn't defined at 0 a standard Maclaurin expansion doesn't exist for it, another approach would be finding the series of a similar function which does exist at x = 0 and then manipulating every term in the series to get the original function.

    Spoiler:
    Show
     f(x) = \dfrac{x}{\sin{2x}}

     \displaystyle\lim_{x\to 0} \bigg[ \dfrac{x}{\sin{2x}} \bigg] = \dfrac{1}{2} using that  \sin{\phi} \approx \phi for small x.

     \dfrac{1}{\sin{2x}}  = \dfrac{1}{x} \displaystyle\sum_{n=0}^{\infty} \dfrac{f^n(0) x^n}{n!}
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by SeanFM)
    Perhaps this would be better off in the Maths forum.
    Oh right yeah thanks!
    Offline

    20
    ReputationRep:
    i used the McLauren expansion for sin(2x)... then applied the binomial expansion to the first few terms of the reciprocal , having factorised out 1/2x to give 1 as the first term.
    Offline

    21
    ReputationRep:
    Moved to maths.
    Offline

    15
    ReputationRep:
    this expansion is nonsense about 0
    Offline

    2
    ReputationRep:
    This question is far simpler than you are thinking it is.

    Calculating the series directly is beyond A level - you need to calculate the Laurent series for that.

    BUT you have all the information you need from the series for Sin(2x) itself.

    Just take the reciprocal of that series :

    \displaystyle \frac{1}{\frac{4 x^5}{15}-\frac{4 x^3}{3}+2 x}

    factor out 2x

    \displaystyle \frac{1}{2 x \left(\frac{2 x^4}{15}-\frac{2 x^2}{3}+1\right)}

    and then expand the term in brackets as a binomial series. Then divide each term from that by the 2x you factored out at the start.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by kc_chiefs)
    This question is far simpler than you are thinking it is.

    Calculating the series directly is beyond A level - you need to calculate the Laurent series for that.

    BUT you have all the information you need from the series for Sin(2x) itself.

    Just take the reciprocal of that series :

    \displaystyle \frac{1}{\frac{4 x^5}{15}-\frac{4 x^3}{3}+2 x}

    factor out 2x

    \displaystyle \frac{1}{2 x \left(\frac{2 x^4}{15}-\frac{2 x^2}{3}+1\right)}

    and then expand the term in brackets as a binomial series. Then divide each term from that by the 2x you factored out at the start.
    Thank you!
 
 
 
  • 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
    Should Spain allow Catalonia to declare independence?
    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.