Results 1 to 4 of 4

Math Help - Series Manipulation

  1. #1
    Newbie Elvis's Avatar
    Joined
    Jan 2010
    Posts
    12

    Series Manipulation

    I wanted to find a Maclaurin series for 3\sin^2\dfrac{x}{2}. Here is what I did:

    Spoiler:
    Since \sin{x} = x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...

    Then 3\sin^2\dfrac{x}{2} = 3\bigg(\dfrac{x}{2}-\dfrac{\left(\dfrac{x}{2}\right)^3}{3!}+\dfrac{\le  ft(\dfrac{x}{2}\right)^5}{5!}-\dfrac{\left(\dfrac{x}{2}\right)^7}{7!}+...\bigg) = 3\bigg(\dfrac{x}{2}-\dfrac{\dfrac{x}{2^3}^3}{3!}+\dfrac{\dfrac{x}{2^5}  ^5}{5!}-\dfrac{\dfrac{x}{2^7}^7}{7!}+...\bigg)

     = 3\bigg(\dfrac{x}{2}-\dfrac{x^3}{{2^3}3!}+\dfrac{x^5}{2^55!}-\dfrac{{x}^7}{{2}^77!}+...\bigg) = \dfrac{3x}{2}-\dfrac{3x^3}{{2^3}3!}+\dfrac{3x^5}{2^55!}-\dfrac{{3x}^7}{{2}^77!}+...+ \dfrac{(-1)^{n}(3x^{2n+1})}{2^{2n+1}(2n+1)!}-...

    \Rightarrow 3\sin^2{\dfrac{x}{2}} = \sum_{n=0}^{\infty}\dfrac{(-1)^{n}(3x^{2n+1})}{2^{2n+1}(2n+1)!}.


    But since 3\sin^2\dfrac{x}{2} = 3\left(\sin\dfrac{x}{2}\right)^2, can't I directly use the formula \sin{x} = \sum_{n=0}^{\infty}\dfrac{(-1)^nx^{2n+1}}{(2n+1)!} without expanding at all? If so, could someone please show me how to do it. I have tried to compute 3\sum_{n=0}^{\infty}\bigg(\dfrac{(-1)^n\left(\dfrac{x}{2}\right)^{2n+1}}{(2n+1)!}\big  g)^2 but it doesn't appear to yield \sum_{n=0}^{\infty}\dfrac{(-1)^{n}(3x^{2n+1})}{2^{2n+1}(2n+1)!}.
    Last edited by Elvis; April 14th 2010 at 11:50 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    It looks to me like you forgot to take the square of the expansion... What you gave is an expansion for 3\sin\frac{x}{2}, isn't it?

    You should consider using the formula \sin^2\frac x2=\frac{1-\cos x}{2}.

    Or painfully expand the square \left(\sum_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{2^{2n+1}(2n+1)!}\right)^2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie Elvis's Avatar
    Joined
    Jan 2010
    Posts
    12
    Damn it! I see what happened. Thank you, Laurent.

    \left(\sum_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{2^{2n+1}(2n+1)!}\right)^2 = \sum_{n=0}^{\infty}\dfrac{(3x^{4n+2})}{2^{4n+2}(2n  +1)!(2n+1)!}. This is where I stuck before as well when I was trying to calculate 3\sum_{n=0}^{\infty}\bigg(\dfrac{(-1)^n\left(\dfrac{x}{2}\right)^{2n+1}}{(2n+1)!}\big  g)^2 = 3\sum_{n=0}^{\infty}\dfrac{\dfrac{(-1)^{2n}x^{4n+2}}{2^{4n+2}}}{(2n+1)!} = \sum_{n=0}^{\infty}\dfrac{(3x^{4n+2})}{2^{4n+2}(2n  +1)!(2n+1)!}

    Let me try it with \sin^2\frac x2=\frac{1-\cos x}{2}:


    We have 3\sin^2\frac x2= \dfrac{3}{2}\left({1-\cos{x}}\right) = -\dfrac{3}{2}\left(\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{2n}\right). Fabulous! I wouldn't really mind if someone showed me how to manipulate the above series into this one, though.
    Last edited by Elvis; April 14th 2010 at 01:07 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Elvis View Post
    \left(\sum_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{2^{2n+1}(2n+1)!}\right)^2 = \sum_{n=0}^{\infty}\dfrac{(3x^{4n+2})}{2^{4n+2}(2n  +1)!(2n+1)!}.
    Since when do you have (a+b)^2=a^2+b^2? You just wrote \left(\sum_{n=0}^\infty a_n\right)^2=\sum_{n=0}^\infty (a_n)^2, which is just the same mistake. There are many other terms in the square of a sum.

    The product of two series is sometimes called "Cauchy product". If you expand the product, you get \left(\sum_{n=0}^\infty a_n x^n\right)^2=\sum_{n=0}^\infty\left(\sum_{k=0}^n a_k a_{n-k}\right)x^n (look for "Cauchy product" on the web, if you've never seen this before). Substitute your value for a_k, and now you understand why I mentioned pain (well, probably not that much actually). I guess the binomial formula would lead you back to the result obtained by the "fabulous" method.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Set manipulation
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: January 25th 2011, 11:10 PM
  2. Set Manipulation
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: September 30th 2009, 03:17 PM
  3. Manipulation of Power Series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 20th 2009, 06:34 PM
  4. Power Series Manipulation #2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2009, 06:33 PM
  5. [SOLVED] Manipulation of power series
    Posted in the Calculus Forum
    Replies: 7
    Last Post: April 9th 2009, 10:15 PM

Search Tags


/mathhelpforum @mathhelpforum