Results 1 to 8 of 8

Math Help - Taylor and Maclaurin series questions

  1. #1
    Junior Member
    Joined
    Feb 2008
    Posts
    48

    Taylor and Maclaurin series questions

    1. Find the Taylor series for f(x) = cos (3x) centered at a = (Pi/2). Find its radius of convergence.

    2. Let f(x) = (1+x)^(-1/2)
    a) Find the Maclaurin series for f.
    b) Use the series found in 2a to find the maclaurin series for (1+x^2)^(-1/2) and use that result to find a Maclaurin series for sinh^-1 (x).

    I cannot figure out how to do these. For 1, I substituted the 3x for x in the cos(x) series... what do I do from there? I know I have to put (x-a) at the end for the center.

    For 2, I have no idea what to do. It says to use the binomial series, which we didn't really cover. It's a bonus on a homework.
    Last edited by thegame189; May 4th 2008 at 04:51 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    For the binomial series of \frac{1}{\sqrt{1+x}}

    If we let m=\frac{-1}{2} in the binomial series,

    (1+x)^{m}=1+mx+\frac{m(m-1)}{2!}x^{2}+.......+\frac{m(m-1)...(m-k+1)}{k!}x^{k}, if |x|<1.

    \frac{1}{\sqrt{1+x}}=1-\frac{1}{2}x+\frac{(\frac{-1}{2})(\frac{-1}{2}-1)}{2!}x^{2} +\frac{(\frac{-1}{2})(\frac{-1}{2}-1)(\frac{-1}{2}-2)}{3!}x^{3}+.....+\frac{(\frac{-1}{2})(3/2)(5/2)...(-1/2-k+1)}{k!}x^{k}+...

    =1-\frac{1}{2}x+\frac{3}{2^{2}2!}x^{2}-\frac{15}{2^{3}3!}x^{3}+....+(-1)^{k}\frac{1*3*5....(2k-1)}{2^{k}k!}x^{k}+....

    1-\frac{x}{2}+\frac{3x^{2}}{8}-\frac{5}{16}x^{3}+..........

    (1+x)^{\frac{-1}{2}}=1+\sum_{k=1}^{\infty}\frac{(\frac{-1}{2})(\frac{-3}{2})....(\frac{-1}{2}-k+1)}{k!}x^{k}

    The derivative of sinh^{-1}(x)=\frac{1}{\sqrt{1+x^{2}}}
    Last edited by galactus; May 4th 2008 at 02:34 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2008
    Posts
    48
    Quote Originally Posted by galactus View Post
    =1-\frac{1}{2}x+\frac{3}{2^{2}2!}x^{2}-\frac{15}{2^{3}3!}x^{3}+....+(-1)^{k}\frac{1*3*5....(2k-1)}{2^{k}k!}x^{k}+....

    1-\frac{x^{2}}{2}+\frac{3x^{4}}{8}-\frac{5}{16}x^{6}+..........

    How did you go from x to x^2, from x^2 to x^4, from x^3 to x^6, etc.?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    That's a stupid typo. Here is what it should be.

    1-\frac{x}{2}+\frac{3x^{2}}{8}-\frac{5x^{3}}{16}+\frac{35x^{4}}{128}-\frac{63x^{5}}{256}+..............
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2008
    Posts
    48
    So how would I do part b? Would I substitute x^2 in for every x?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Yes.

    Here is something to go on to know what to look for:

    The series for sinh^{-1}(x)=x-\frac{x^{3}}{6}+\frac{3x^{5}}{40}-\frac{5x^{7}}{112}+\frac{35x^{9}}{1152}-................
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Feb 2008
    Posts
    48
    Quote Originally Posted by galactus View Post
    (1+x)^{\frac{-1}{2}}=1+\sum_{k=1}^{\infty}\frac{(\frac{-1}{2})(\frac{-3}{2})....(\frac{-1}{2}-k+1)}{k!}x^{k}
    I'm still pretty confused with everything. Is that the answer?


    And I'm very much lost with the (1+x^2)^(-1/2) series.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Feb 2008
    Posts
    48
    Can someone help with #1? I did a table of derivatives and evaluated them at Pi/2:

    derivative, evaluated at Pi/2

    f(x) = cos3x ,0
    f '(x) = -3sin3x ,3
    f ''(x) = -9cos3x , 0
    f '''(x) = 27sin3x ,-27
    f ''''(x) = 81cos3x ,0
    f^5 (x) = -243sin3x ,243

    The nth derivative would be (-1)^n * [3^(2n)], right? SO then I just used the nth derivative evaluated at (pi/2) / n! formula to get c sub n.

    Summation n=0 to infinity [(nth derivative evaluated at Pi/2) / n!] * (x-Pi/2)^n

    Then I substituted the actual nth derivative values to get:
    0 + ([3(x-Pi/2)] / 1!) - ([27(x-Pi/2)^3] / 3!) + 0 + ([243(x-Pi/2)^5] / 5!) +...

    So then I got the Taylor series of cos(3x) to be:

    (-1)^n * 3^(2n+1) * (x-Pi/2)^(2n+1)
    (2n+1)!

    I'm really confused with this stuff. It's been awhile since I've done it and it's final time.
    Last edited by thegame189; May 4th 2008 at 04:07 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: January 26th 2010, 09:06 AM
  2. Taylor/Maclaurin series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 11th 2009, 08:50 AM
  3. Using Series (MacLaurin/Taylor)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 5th 2009, 09:37 AM
  4. Taylor & MacLaurin series
    Posted in the Calculus Forum
    Replies: 15
    Last Post: May 11th 2008, 08:13 PM
  5. Taylor ans Maclaurin series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 7th 2008, 12:32 PM

Search Tags


/mathhelpforum @mathhelpforum