1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

  2. jcsd
  3. Apr 23, 2012 #2
    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

    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
    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
    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


    Then I can substitute in x=0 and get


    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
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook