Results 1 to 5 of 5

Math Help - Question on Taylor power series

  1. #1
    Junior Member
    Joined
    Nov 2007
    Posts
    53

    Question on Taylor power series

    Ok, so the instructions state to apply Taylor's Theorum to find the power series (centered at c) for the function, and find the radius of convergence.

    Problem: f(x)=x^{1/2} with the center at c=1

    I took the first through the fourth derivative and then evaluated them at the center to get an idea of the pattern, here's what I got:

    1st Derivative: f'(x)=\frac{1}{2x^{1/2}} Evaluation: f'(1)=\frac{1}{2}

    2nd Derivative: f''(x)=-\frac{1}{4x^{3/2}} Evaluation: f''(1)=-\frac{1}{4}

    3rd Derivative: f'''(x)=\frac{3}{8x^{5/2}} Evaluation: f'''(1)=\frac{3}{8}

    4th Derivative: f''''(x)=-\frac{15}{16x^{7/2}} Evaluation: f''''(1)=-\frac{15}{16}

    I can use the Taylor series formula easily enough, however, I'm not sure how they came up with the answer for f^n(1)=(-1)^{n-1}\frac{1*3*....*5*(2n-3)}{2^n} which holds true when n is equal to or greater than 2.

    Now, I can write up the power series and perform the ratio test to find the radius of convergence on my own easily enough, however, I'm unfamiliar with this notation. I understand the (-1)^{n-1} in the answer because that must be true for the signs to alternate back and forth. Due to the denominators in each evaluation, I can also see where 2^n comes into play since the denominator is just 2 raised to an exponent equal to n. What I don't see is where that frickin' 1*3*....*5*(2n-3) comes into play. Can anyone explain to me what that notation is, why it's part of the answer and how I might have to treat it when taking a limit? Thanks in advance guys.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    I'll ignore the minus signs because we can make them with (-1)^n later.

    f^2 = \frac{1}{2}\cdot \frac{1}{2}

    f^3 = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{3}{2}

    f^4 = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{3}{2} \cdot\frac{5}{2}
    ....

    Did you see the pattern?

    f^n = \frac{1}{2}\prod^{n-2}_{k=0}\left ( \frac{1}{2}+k \right )

    First find \prod^{n-2}_{k=0}\left ( \frac{1}{2}+k \right ).

    \prod^{n-2}_{k=0}\left ( \frac{1}{2}+k \right )

    \prod^{n-2}_{k=0}\left ( \frac{2k+1}{2} \right )

    \left ( \frac{1}{2} \right ) \cdot \left ( \frac{3}{2} \right ) \cdot \left ( \frac{5}{2} \right ) \cdot \left ( \frac{7}{2} \right ) \cdot \cdot \cdot \cdot \left ( \frac{2n-3}{2} \right ) \cdot

    And that makes,

    \frac{1 \cdot 3\cdot 5 \cdot 7 \cdot \cdot \cdot \cdot \left ( 2n-3\right ) }{2^{n-1}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Isnt that

    just (2n-3)!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by Mathstud28 View Post
    just (2n-3)!
    No, (2n-3)! = 1.2.3.4.5....(2n-3)

    But this is 1.3.5.7.....(2n-3)
    (Without the even numbers)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Forgive me I am stupid

    but then isnt it (2n-1)! since 2n-1 gives the series of odd numbers?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: January 26th 2010, 08:06 AM
  2. taylor series power question..
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 13th 2010, 01:54 PM
  3. Question - Power/taylor/MacLaurin series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 9th 2009, 10:25 PM
  4. Formal power series & Taylor series
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 19th 2009, 09:01 AM
  5. Taylor + Power Series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 1st 2008, 09:26 AM

Search Tags


/mathhelpforum @mathhelpforum