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Maths Problem: Maclaurin Series

  • 01-05-2009 7:03pm
    #1
    Des23
    Closed Accounts Posts: 287 ✭✭


    I'm not sure what I'm missing but it must be something.
    What I've done is get the maclaurin series for (i) which was fine.
    Then for (ii) am I meant to get the series from the beginning again?
    And am I meant to do all that for a third time for part (iii)?

    I presume for part (iv) once u get to then end you use the fact that ln(1/2)=-ln2 to get an approximation for ln2 by filling in x as 1.

    I have never seen a Q like this before I don't think or else I'm missing something obvious. Q8 is usually pretty straightforward.

    Any help would be greatly appreciated.

    It is off an old pre paper I think, so I couldnt find help anywhere.


Comments

  • #2
    PurpleFistMixer
    Registered Users, Registered Users 2 Posts: 5,851 ✭✭✭PurpleFistMixer


    I'm a bit rusty on the aul Maclaurin (actually on looking it up, I really shouldn't be, considering we do Taylor Series in college), but in part 2 can you not just substitute 3x for x where it appeared in part 1?
    Then for part three, ln(1+x/1+3x) is just ln(1+x) - ln(1+3x), and you've already worked out the series for each of them individually.


  • #3
    andyman
    Closed Accounts Posts: 3,641 ✭✭✭andyman


    That's easy enough (well, the first 3 parts anyway), just looks kind of tricky.

    (i) Expand it out, no bother's really

    (ii) Get the Maclaurin Series expansion that you got for (i) except this time just sub in 3x instead of x.

    (iii) Basic rules of logs here. ln(A/B) = lnA - lnB. So in this case just get your expansions from (i) and (ii) and you should be able to work it out from there

    (iv) Not 100% sure about this one. I think you just put 1 + x/1 + 3x = 2 and work out a value for x and sub it in. Could be wrong though.


  • #4
    Des23
    Closed Accounts Posts: 287 ✭✭Des23


    I thought about subbing x in which is obviously the easiest method, but thought that if you differentiate ln(1 + x) and sub in 3x you get a different answer to differentiating ln(1 + 3x), i.e. you would have to have the 3 outside the bracket... or something along those lines. Therefor the sieries would have been different?

    But apparantly not. Subbing in is probably what I would have resorted to in an exam, if I didn't really understand what I was doing. Thanks a lot for the help. It is very much appreciated.


  • #5
    JSK 252
    Registered Users, Registered Users 2 Posts: 1,257 ✭✭✭JSK 252


    Des23 wrote: »
    I thought about subbing x in which is obviously the easiest method, but thought that if you differentiate ln(1 + x) and sub in 3x you get a different answer to differentiating ln(1 + 3x), i.e. you would have to have the 3 outside the bracket... or something along those lines. Therefor the sieries would have been different?

    But apparantly not. Subbing in is probably what I would have resorted to in an exam, if I didn't really understand what I was doing. Thanks a lot for the help. It is very much appreciated.

    It shouldnt make a difference if you sub in 3x because the series has already been derived in (i).

    For part 4 andyman is correct. Cancel the logs on both sides, cross multiply, find x and fill back into the original maclaurin series.


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