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I tried to solve this again

  1. Jan 26, 2008 #1
    a question that i tried to solve it before

    my new solution in the link


    now i solved it like in the method of the last post

    i develeloped the ln series till the x^2 power
    the "e" series to the second power too

    when i multiplied the lan series by x i added 1 to the O member

    and i cut down in the end all the members which are bigger then the 4th power

    now i got the solution to this question and i cant see no subtitution
    in there
    he is not adding 1 to the "O" object when multiplying the series by x

    and he gets all the points

    i dont care if you too dont get any sense in it

    where did i go wrong???
  2. jcsd
  3. Jan 26, 2008 #2

    Gib Z

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    You only expanded some series to their 2nd term, how do you expect to get an error on the order of x^4?
  4. Jan 26, 2008 #3
    so your saying that i should expand each one of the serieses to the 4th term
    (even if the last term is zero)
    like the sinus function


    what about the multiplication by x of a series
    does it changes the "O" object ???
    Last edited: Jan 26, 2008
  5. Jan 26, 2008 #4
  6. Jan 26, 2008 #5
    Three errors.

    The coefficient of [itex]u^3[/itex] is 5/81.
    The coefficient of [itex]u^4[/itex] is -10/243.
    You forgot the term [tex]-\frac{4}{9}\,x^2[/tex] in the final answer
  7. Jan 26, 2008 #6
    As Gib said you should write [itex]\ln(1+x)[/itex] till 3rd power, in order to pick up a x4 term in [itex]x\,\ln(1+x)[/itex] expansion
    Last edited: Jan 26, 2008
  8. Jan 26, 2008 #7
    gib said that i should expand both of them till the 4th power
    in order to get the error of the 4th power

    i'll try to solve it again like that

    but do i need to add 1 to the "O" member from O(x^3) >>> O(x^4)
    Last edited: Jan 26, 2008
  9. Jan 26, 2008 #8
    No you don't. You write the series as it is.
  10. Jan 26, 2008 #9
    here is the solution of this question

    and he did it not by your rule of "developing the series till the 3rd power"

    and he got all the point


    what is this solution??
    he didnt do like you told me

    why ???
    is it a wrong solution??
    Last edited: Jan 26, 2008
  11. Jan 26, 2008 #10
    :confused: He developed the [itex]\ln(1+x)[/itex] till the 4th power and kept the terms in [itex]x\,\ln(1+x)[/itex] till the 4th power, i.e. the 3rd power of [itex]\ln(1+x)[/itex].

    What's the question about my post??? :confused:

    Develope till the 3rd power is enough. Isn't it clear?
  12. Jan 26, 2008 #11
    ahhhhh ok thanks ill try to solve it again
    and the other question too
  13. Jan 27, 2008 #12

    Gib Z

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    No offence but do you actually know what exactly Big -Oh notation is? Are you familiar with the mean value theorem? If you are, you should have no problem with these, which you are.
  14. Feb 21, 2009 #13
    whats the link of the mean value theorem with the O notation??
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