Results 1 to 15 of 15

Math Help - maclaurin series

  1. #1
    Junior Member
    Joined
    Jul 2007
    Posts
    72

    maclaurin series

    How would you find the maclaurin series for this
    sin(x^5)
    thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by davecs77 View Post
    How would you find the maclaurin series for this
    sin(x^5)
    thanks.
    recall that the Maclaurin series for sine is:


    \sin x = \sum_{n = 0}^{\infty} \frac {(-1)^n}{(2n + 1)!} x^{2n + 1} = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} +... for ALL x

    so for \sin \left( x^5 \right), just replace x with x^5 in the series above and simplify
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by davecs77 View Post
    How would you find the maclaurin series for this
    sin(x^5)
    thanks.
    A McLaurin series is a Taylor series for x = 0. So we have:
    f(x) \approx f(0) + \frac{1}{1!}f^{\prime}(0)x + <br />
\frac{1}{2!}f^{\prime \prime}(0)x^2 + ~ ...

    f(x) = sin(x^5)

    So
    f(0) = 0

    f^{\prime}(x) = cos(x^5) \cdot 4x^5 ==> f^{\prime}(0) = 0

    f^{\prime \prime}(x) = 20x^3 cos(x^5) - 25x^8sin(x^5) ==> f^{\prime \prime}(0) = 0

    f^{\prime \prime \prime}(x) = -125x^{12}cos(x^5) + 60x^2 cos(x^5) - 300x^7sin(x^5) ==> f^{\prime \prime \prime}(0) = 0

    etc.

    I'll give you a hint. After a lot of hard work on your part, you will find that f^{(6)}(0) is the first non-zero term in the series.

    (This one's just nasty to do by hand!)

    -Dan
    Last edited by topsquark; July 23rd 2007 at 05:14 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2007
    Posts
    72
    Quote Originally Posted by Jhevon View Post
    recall that the Maclaurin series for sine is:


    \sin x = \sum_{n = 0}^{\infty} \frac {(-1)^n}{(2n + 1)!} x^{2n + 1} = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} +... for ALL x

    so for \sin \left( x^5 \right), just replace x with x^5 in the series above and simplify
    So you are just memorizing that for sin(x) or Maclaurn series?
    Would that be the same for Taylor series?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by davecs77 View Post
    So you are just memorizing that for sin(x) or Maclaurn series?
    Would that be the same for Taylor series?
    I don't like telling people to just memorize stuff, but yes, you should have the Maclaurin series for functions like sine, cosine, e^x, 1/(1 - x) ... memorized.

    the Maclaurin series is just the Taylor series centered at zero. topsquark showed you how to derive it, but in this case, i think the memorization route is easiest. unless your professor requires you to derive it
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    recall that the Maclaurin series for sine is:


    \sin x = \sum_{n = 0}^{\infty} \frac {(-1)^n}{(2n + 1)!} x^{2n + 1} = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} +... for ALL x

    so for \sin \left( x^5 \right), just replace x with x^5 in the series above and simplify
    Or you can take Jhevon's pansy-butted approach and skip all the work. But why do that when you could do all that work?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jul 2007
    Posts
    72
    how would you find the Maclaurin series for this then:
    1/(1+x^5)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by topsquark View Post
    But why do that when you could do all that work?

    -Dan
    because mathematicians don't like hard work.

    there are some things you want to try once just to see if you can do them, and then afterwards you take the shortcut. finding Taylor/Maclaurin series expansions are such things.

    "pansy-butted"?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    "pansy-butted"?
    It was the best I could come up with on short notice. Hey, I'm a Mormon. I'm not supposed to swear so that limits my insulting remarks.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Jul 2007
    Posts
    72
    Quote Originally Posted by davecs77 View Post
    how would you find the Maclaurin series for this then:
    1/(1+x^5)

    Does anyone know? I don't think you can memorize this one. Thanks.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by davecs77 View Post
    how would you find the Maclaurin series for this then:
    1/(1+x^5)
    \frac{1}{1+x}=1-x+x^2-x^3+...

    You do the rest.

    Note: You need to remember the most important power series as well as you name.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by davecs77 View Post
    Does anyone know? I don't think you can memorize this one. Thanks.
    actually you can. this happens to be one of the easiest to memorize in fact. the appropriate one to remember is

    \frac {1}{1 - x} = \sum_{n = 0}^{\infty} x^n = 1 + x + x^2 + x^3 + ... for |x|<1
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Now me, I tend to forget that \sum_{i = 1}^n 1 = n. (I try not to memorize too much! )

    -Dan
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Jul 2007
    Posts
    72
    Quote Originally Posted by ThePerfectHacker View Post
    \frac{1}{1+x}=1-x+x^2-x^3+...

    You do the rest.

    Note: You need to remember the most important power series as well as you name.
    The answer in the book is summation from n = 1 to infinity of (-1)^n times x^(5n)

    I see the (-1)^n since the values are changing from positive to negative every time, but where does the x^(5n) come from? Also how did you find your results... 1 - x + x^2 - x^3 ... I don't quite understand. Thank you.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by davecs77 View Post
    The answer in the book is summation from n = 1 to infinity of (-1)^n times x^(5n)

    I see the (-1)^n since the values are changing from positive to negative every time, but where does the x^(5n) come from? Also how did you find your results... 1 - x + x^2 - x^3 ... I don't quite understand. Thank you.
    he treated \frac {1}{1 + x} as \frac {1}{1 - (-x)} and then used the formula i gave you.

    where did the x^{5n} come from? did you actually look at the series? \sum_{n = 0}^{\infty}(-1)^n\left( x^5 \right)^n = \sum_{n = 0}^{\infty}(-1)^nx^{5n}
    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. Replies: 2
    Last Post: September 16th 2009, 08:56 AM
  3. Binomial Series to find a Maclaurin Series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 21st 2009, 08:15 AM
  4. Multiplying power series - Maclaurin series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 8th 2009, 12:24 AM
  5. Replies: 1
    Last Post: May 5th 2008, 10:44 PM

Search Tags


/mathhelpforum @mathhelpforum