Results 1 to 7 of 7
Like Tree4Thanks
  • 1 Post By Prove It
  • 1 Post By SlipEternal
  • 2 Post By SlipEternal

Math Help - expanding f(x) into the maclaurin series

  1. #1
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    expanding f(x) into the maclaurin series

    Expand the function f(x) = sin^6(x) + cos^6(x) into Maclaurin series on the whole real line \mathbb{R}.

    So here is what I have done so far:

    From what I have been able to figure out, though I am not sure this is correct, the Maclaurin series is defined as

    f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f''''(0)x^4}{4!} + \frac{f^5(0)x^5}{5!} + ... .

    I solved for f(0) = sin^6(0) + cos^6(0) = 0 + 1 = 1.

    Similarly, f'(x) = 6cos(x)sin^5(x) - 6sin(x)cos^5(x), giving f'(0) = 0.

    Once again, f''(x) = 6[5sin^4(x)cos^2(x) - sin^6(x)] - 6[cos^6(x) - 5cos^4(x)sin^2(x)], giving f''(0) = -6.

    Is the right direction? Should I be finding some pattern here? What is the final answer supposed to look like?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,079
    Thanks
    375
    Awards
    1

    Re: expanding f(x) into the maclaurin series

    Quote Originally Posted by vidomagru View Post
    Expand the function f(x) = sin^6(x) + cos^6(x) into Maclaurin series on the whole real line \mathbb{R}.

    So here is what I have done so far:

    From what I have been able to figure out, though I am not sure this is correct, the Maclaurin series is defined as

    f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f''''(0)x^4}{4!} + \frac{f^5(0)x^5}{5!} + ... .

    I solved for f(0) = sin^6(0) + cos^6(0) = 0 + 1 = 1.

    Similarly, f'(x) = 6cos(x)sin^5(x) - 6sin(x)cos^5(x), giving f'(0) = 0.

    Once again, f''(x) = 6[5sin^4(x)cos^2(x) - sin^6(x)] - 6[cos^6(x) - 5cos^4(x)sin^2(x)], giving f''(0) = -6.

    Is the right direction? Should I be finding some pattern here? What is the final answer supposed to look like?
    I'm not seeing any particular pattern. Note though, that the odd n terms are 0.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,677
    Thanks
    1499

    Re: expanding f(x) into the maclaurin series

    Quote Originally Posted by vidomagru View Post
    Expand the function f(x) = sin^6(x) + cos^6(x) into Maclaurin series on the whole real line \mathbb{R}.

    So here is what I have done so far:

    From what I have been able to figure out, though I am not sure this is correct, the Maclaurin series is defined as

    f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f''''(0)x^4}{4!} + \frac{f^5(0)x^5}{5!} + ... .

    I solved for f(0) = sin^6(0) + cos^6(0) = 0 + 1 = 1.

    Similarly, f'(x) = 6cos(x)sin^5(x) - 6sin(x)cos^5(x), giving f'(0) = 0.

    Once again, f''(x) = 6[5sin^4(x)cos^2(x) - sin^6(x)] - 6[cos^6(x) - 5cos^4(x)sin^2(x)], giving f''(0) = -6.

    Is the right direction? Should I be finding some pattern here? What is the final answer supposed to look like?
    I'd start like this...

    \displaystyle \begin{align*} \sin^6{(x)} + \cos^6{(x)} &= \left[ \sin^2{(x)} \right] ^3 + \cos^6{(x)} \\ &= \left[ 1 - \cos^2{(x)} \right] ^3 + \cos^6{(x)} \\ &= 1 - 3\cos^2{(x)} + 3\cos^4{(x)} - \cos^6{(x)} + \cos^6{(x)} \\ &= 1 - 3\cos^2{(x)} + 3\cos^4{(x)} \\ &= 1 - 3 \left[ \frac{1}{2} + \frac{1}{2}\cos{(2x)} \right] + 3 \left[ \frac{1}{2} + \frac{1}{2}\cos{(2x)} \right] ^2 \\ &= 1 - \frac{3}{2} - \frac{3}{2}\cos{(2x)} + 3 \left[ \frac{1}{4} + \frac{1}{2}\cos{(2x)} + \frac{1}{4}\cos^2{(2x)} \right] \\ &= -\frac{1}{2} - \frac{3}{2}\cos{(2x)} + \frac{3}{4} + \frac{3}{2}\cos{(2x)} + \frac{3}{4}\cos^2{(2x)} \\ &= \frac{1}{4} + \frac{3}{4}\cos^2{(2x)} \\ &= \frac{1}{4} + \frac{3}{4} \left[ \frac{1}{2} + \frac{1}{2}\cos{(4x)} \right] \\ &= \frac{1}{4} + \frac{3}{8} + \frac{3}{8}\cos{(4x)} \\ &= \frac{5}{8} + \frac{3}{8}\cos{(4x)} \end{align*}

    Now if you recall the MacLaurin Series for \displaystyle \begin{align*} \cos{(t)} = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n)!}\,t^{2n} = 1 - \frac{t^2}{2} + \frac{t^4}{4!} - \frac{t^6}{6!} + \dots \end{align*} you should be able to manipulate it to get the series you are after...
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    Re: expanding f(x) into the maclaurin series

    Quote Originally Posted by Prove It View Post
    Now if you recall the MacLaurin Series for \displaystyle \begin{align*} \cos{(t)} = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n)!}\,t^{2n} = 1 - \frac{t^2}{2} + \frac{t^4}{4!} - \frac{t^6}{6!} + \dots \end{align*} you should be able to manipulate it to get the series you are after...
    I have 1 - \frac{6x^2}{2!} + \frac{96x^4}{4!} - \frac{1536x^6}{6!} + ... but I cannot figure out how to manipulate that into a series similar to the Maclaurin Series Expansion of cos(x).

    Thoughts?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,931
    Thanks
    782

    Re: expanding f(x) into the maclaurin series

    \dfrac{7}{8} + \dfrac{3}{8}\sum_{n\ge 0} \dfrac{(-1)^n}{(2n)!}(4x)^{2n}
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    Re: expanding f(x) into the maclaurin series

    Quote Originally Posted by SlipEternal View Post
    \dfrac{7}{8} + \dfrac{3}{8}\sum_{n\ge 0} \dfrac{(-1)^n}{(2n)!}(4x)^{2n}
    I don't think that evaluates out to 1 - \frac{6x^2}{2!} + \frac{96x^4}{4!} - \frac{1536x^6}{6!} + ....
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,931
    Thanks
    782

    Re: expanding f(x) into the maclaurin series

    That was supposed to be a 5/8, not a 7/8. Let's calculate the first few terms:

    \dfrac{5}{8} + \dfrac{3}{8} - \dfrac{3\cdot 16x^2}{8\cdot 2!} + \dfrac{3\cdot 256x^4}{8\cdot 4!} - \dfrac{3\cdot 1024x^6}{8\cdot 6!} \pm \cdots

    = 1 - \dfrac{6x^2}{2!} + \dfrac{96x^4}{4!} - \dfrac{1536x^6}{6!} \pm \cdots

    Looks the same to me.
    Thanks from Prove It and topsquark
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: December 9th 2012, 07:12 PM
  2. Expanding the product of a series
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 25th 2012, 01:23 PM
  3. Replies: 0
    Last Post: January 26th 2010, 08:06 AM
  4. Expanding Maclaurin Series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 11th 2009, 06:36 PM
  5. Replies: 1
    Last Post: May 5th 2008, 09:44 PM

Search Tags


/mathhelpforum @mathhelpforum