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Taylor Series Problem!

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data
    The first three terms of a Taylor Series centered about 1 for [itex]ln(x)[/itex] is given by:

    [itex]\frac{x^{3}}{3}[/itex] - [itex]\frac{3x^{2}}{2}[/itex] + [itex]3x[/itex] - [itex]\frac{11}{6}[/itex]

    and that

    [itex]\int{ln(x)dx}[/itex] = [itex]xlnx - x + c[/itex]

    Show that an approximation of [itex]ln(x)[/itex] is given by:

    [itex]\frac{x^3}{12}[/itex] - [itex]\frac{x^2}{2}[/itex] + [itex]\frac{3x}{2}[/itex] - [itex]\frac{5}{6}[/itex] - [itex]\frac{1}{4x}[/itex]


    2. The attempt at a solution
    I have tried this problem a few times, but it is becoming clear that I am missing some crucial step/idea. Basically, what I have tried is setting lnx equal to the Taylor Series, integrating both sides, and solving for lnx. However, when I do this, I manage to get all of the necessary terms EXCEPT for the 1/4x. Where does that come from, exactly? If someone could help, that'd be awesome! :D

    Thanks!
     
  2. jcsd
  3. Apr 23, 2012 #2

    micromass

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    I think you likely forgot the +C. C is a constant and should equal some number. What do you think that number is?
     
  4. Apr 23, 2012 #3
    Does c = 1 because the center is at 1?
     
  5. Apr 23, 2012 #4
    Even if that is the case, I'm not sure how I would get 1/4x from that.
     
  6. Apr 23, 2012 #5

    micromass

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    No.

    What do you get after you subsitute ln(x) with the polynomial and integrate it??
     
  7. Apr 23, 2012 #6
    Please don't judge my slowness :( I'm really not bad at math. I just haven't done any of my homework, lol.
     
  8. Apr 23, 2012 #7
    After you integrate the polynomial, you get:

    [itex]\frac{x^4}{12}-\frac{x^3}{2}+3x^2-\frac{11x}{6} + c[/itex]

    Which is what I did, and then set it equal to [itex]xln(x)-x+C[/itex]

    But every time I do that, I get the wrong thing :( (meaning, I don't get the 1/4x)

    I feel like I'm misunderstanding the nudges that you are giving me, lol.
     
  9. Apr 23, 2012 #8

    micromass

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    OK, that's good (although the middle term should be [itex]\frac{3x^2}{2}[/itex])

    What you should get is

    [tex]\frac{x^4}{12}-\frac{x^3}{2}+\frac{3x^2}{2}-\frac{11x}{6} + c = x ln(x)-x+C[/tex]

    Yes, the constant of integration c and the other constant C are distinct in general, so you can't eliminate them!! This is likely your mistake.

    The above is equivalent to:

    [tex]\frac{x^4}{12}-\frac{x^3}{2}+\frac{3x^2}{2}-\frac{11x}{6} = x ln(x)-x+(C-c)[/tex]

    Can you figure out what number C-c is?? (we're not interested in the specific values of C and c here, we just want a number instead of constants).
     
  10. Apr 23, 2012 #9
    That's what I actually got, I just mistyped it :/

    I know :D That's why I denoted one as c and the other as C.

    This is where I am confused..Like, I'm not trying to sound annoyingly difficult, but I honestly have no idea what to do with the C-c. :( I understand that it is just some number, but I have absolutely no clue what that number is or how to get it.

    My confidence in my math skills is quickly plummeting :(
     
  11. Apr 23, 2012 #10
    I mean, I assume it is 1/4, because that is what I am trying to prove. But what is the reasoning behind that?
     
  12. Apr 23, 2012 #11

    micromass

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    Maybe you can substitute in some value for x and see what you get??

    For example, if I want to determine a constant C such that

    [tex]\sin(x)=x^2+x+C[/tex]

    Then I can substitute in x=0 and get

    [itex]0=C[/itex].

    Can you do something like that to determine C-c?
     
  13. Apr 23, 2012 #12
    Wow..Oh my freaking gosh, I think I am going to go into a corner and cry. I am so stupid :(

    Thanks for the help :( I got it now :(
     
  14. Apr 23, 2012 #13

    micromass

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    Don't worry about it. Yes, it is a trivial problem, but it's one of these things you had to see before knowing you could do that.

    Instead of feeling depressed, you should feel happy because you found a new technique!! I bet that next time (on a test perhaps), you won't forget how to do this! :tongue2:
     
  15. Apr 23, 2012 #14
    That's not even a "new technique." That is basic, elementary calculus that everyone should know how to do. Plugging in a given value and solving for C is something that every Calc student learns the moment they learn integration. I am in calculus 2 and didn't even consider it.

    AKA, I think my feelings of idiocy are justified :P
     
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