Results 1 to 11 of 11

Math Help - Taylor Polynomial Problem help

  1. #1
    Newbie DaveDammit's Avatar
    Joined
    Jan 2009
    From
    Tennessee
    Posts
    15

    Taylor Polynomial Problem help

    Very hard take home test. We're trying to figure out how to integrate:

    f(x) = \int_{1}^{x}\frac{lnt}{1+t^2}dt


    Any help would be much appreciated. Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by DaveDammit View Post
    Very hard take home test. We're trying to figure out how to integrate:

    f(x) = \int_{1}^{x}\frac{lnt}{1+t^2}dt


    Any help would be much appreciated. Thanks!
    Please post the entire question, complete with all instructions.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jan 2009
    Posts
    715
    sub  y = \frac{1}{t}

     dt = - \frac{1}{y^2}~dy

    the integral becomes

     - \int_{\frac{1}{x}}^1 \frac{ - \ln(y) }{1 + y^2 } ~dy

    Since  0 < \frac{1}{x} \leq y <  1

     \frac{1}{1 + y^2} can be changed to infinite series form .


     - \sum_{n=0}^{\infty} (-1)^n \int_{\frac{1}{x}}^1 \ln(y) y^{2n}~dy


    Buy by integration by parts

     \int \ln(y) y^{2n}~dy = \ln(y) \frac{y^{2n+1}}{2n+1} - \frac{ y^{2n+1}}{(2n+1)^2 }

    Finally , the integral becomes

     \sum_{n=0}^{\infty} \frac{ (-1)^n}{(2n+1)^2}  - \ln(x) \tan^{-1}(\frac{1}{x}) - \sum_{n=0}^{\infty} \frac{ (-1)^n }{ x^{2n+1} (2n+1)^2}


    note that  \tan^{-1}(\frac{1}{x}) can be changed to

     \frac{\pi}{2} - \tan^{-1}(x)


    If  x \to \infty

    the integral  \int_1^{\infty} \frac{\ln(x)}{1 + x^2 }~dx

     = \sum_{n=0}^{\infty} \frac{ (-1)^n}{(2n+1)^2}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    simplependulum, what you found is not a Taylor series. i think DaveDammit is probably looking for the Taylor series of f(x) at x = 1. hopefully some day he'll give us more details!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie DaveDammit's Avatar
    Joined
    Jan 2009
    From
    Tennessee
    Posts
    15
    Quote Originally Posted by mr fantastic View Post
    Please post the entire question, complete with all instructions.
    Sorry guys, was working late on the test last night and was very tired when I posted. Pretty cool problems, here's the complete question:

    Find the third Taylor polynomial of f(x) = \int_{1}^{x}\frac{lnt}{1+t^2}dt about a=1.

    Thanks for any help!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by DaveDammit View Post
    Sorry guys, was working late on the test last night and was very tired when I posted. Pretty cool problems, here's the complete question:

    Find the third Taylor polynomial of f(x) = \int_{1}^{x}\frac{lnt}{1+t^2}dt about a=1.

    Thanks for any help!
    It seems simplependulum already answered this by giving the general Taylor series for the integral. If you want just the first 3 terms you can write them out through his work.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2009
    Posts
    151
    You do know the definition that a third order Taylor polynomial of f(x) about a is:

    T_3(f(x),a)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3.

    So all you need to do is find the first 3 derivatives of f(x), and sub a=1.

    Hope that helps.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie DaveDammit's Avatar
    Joined
    Jan 2009
    From
    Tennessee
    Posts
    15
    Quote Originally Posted by hjortur View Post
    You do know the definition that a third order Taylor polynomial of f(x) about a is:

    T_3(f(x),a)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3.

    So all you need to do is find the first 3 derivatives of f(x), and sub a=1.

    Hope that helps.
    I do know the definition of a third order Taylor polynomial. That's not the problem I'm having.

    The problem I'm having is using one of the Fundemental Theorems of Calculus to integrate the function with respect to t in order to find my f(x) function. This specific function is fairly challenging for me to integrate. It is a beautiful problem and I will attempt to solve it.

    Thanks for your input.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by DaveDammit View Post
    I do know the definition of a third order Taylor polynomial. That's not the problem I'm having.

    The problem I'm having is using one of the Fundemental Theorems of Calculus to integrate the function with respect to t in order to find my f(x) function. This specific function is fairly challenging for me to integrate. It is a beautiful problem and I will attempt to solve it.

    Thanks for your input.
    I didn't see NCA's note that simplependulum didn't write a Taylor polynomial so my previous post isn't correct.

    I think the point of this problem is you don't need to actually integrate the function because you can use FTCII to find f(1). What is f(1) if the integral of f(t) has bounds from 1 to 1?

    From there, f'(a) is just f(t) and all further derivatives can easily be found. You don't need to find the integral of f(t) to write the Taylor polynomial of f(x).
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Sep 2009
    Posts
    151
    <br />
f(x) = \int_{1}^{x}\frac{lnt}{1+t^2}dt<br />

    Then according to the fundamental theorem of calculus:

    f'(x)=\frac{ln(x)}{1+x^2}

    Do you see why?

    You don't need to integrate at all.

    Can you finish this one now?
    Last edited by Jameson; November 29th 2009 at 10:04 AM. Reason: latex
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie DaveDammit's Avatar
    Joined
    Jan 2009
    From
    Tennessee
    Posts
    15
    Hey guys,

    Sorry I'm so late in responding to this but I just wanted to let you guys know that I completed the problem and I think I did very well on the test. I had to calculate a wicked third derivative of the function but I'm pretty sure I got the answer right.

    Thanks for all your help!

    Dave
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. taylor polynomial help!
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 21st 2011, 01:46 PM
  2. Taylor Polynomial, Please Help!
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 16th 2009, 08:52 PM
  3. Taylor polynomial #2
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 6th 2009, 02:09 AM
  4. taylor polynomial help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 13th 2009, 03:11 AM
  5. Replies: 9
    Last Post: April 3rd 2008, 05:50 PM

Search Tags


/mathhelpforum @mathhelpforum