Results 1 to 6 of 6

Math Help - Math problem, help needed.

  1. #1
    Junior Member
    Joined
    Aug 2006
    Posts
    36

    Math problem, help needed.

    Hi guys, another problem. If anybody could guide me through the solution that would be great!

    See screen shot on following link....

    CLICK HERE
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,054
    Thanks
    368
    Awards
    1
    Quote Originally Posted by c00ky
    Hi guys, another problem. If anybody could guide me through the solution that would be great!

    See screen shot on following link....

    CLICK HERE
    Well, the best thing to do would be to work out
    1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6
    for several values of x between 0 and 2 \pi and see what happens when you compare that value to cos(x) when x is in radians.

    What you will find is that this function only works well when x is "small." I don't know how well the approximation works for x near, say  \pi , but for x less than, say, 1 it's probably pretty decent.

    For the record, as a Taylor series about x = 0 we may approximate cos(x) as
    \sum_{r = 0}^{\infty}(-1)^r \frac{x^{2r}}{(2r)!} = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 + ...
    so this approximation should be valid.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2006
    Posts
    36
    You have lost me, i'm still a little confused.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Let me help.

    you have expression: (-1)^r\times\frac{x^{2r}}{(2r)!}

    expand to four terms:

    (-1)^r\times\frac{x^{2r}}{(2r)!}\quad+\quad(-1)^r\times \frac{x^{2r}}{(2r)!}\quad+\quad(-1)^r\times \frac{x^{2r}}{(2r)!}\quad+\quad(-1)^r \times\frac{x^{2r}}{(2r)!}

    now substitute 0,1,2 and 3 for "r"

    (-1)^0\times\frac{x^{2(0)}}{(2(0))!}\quad+\quad(-1)^1 \times\frac{x^{2(1)}}{(2(1))!}\quad+\quad(-1)^2 \times\frac{x^{2(2)}}{(2(2))!}\quad+\quad(-1)^3 \times\frac{x^{2(3)}}{(2(3))!}

    simplify:

    1\times\frac{x^{0}}{0!}\quad-\quad1 \times\frac{x^{2}}{2!}\quad+\quad1\times \frac{x^{4}}{4!}\quad-\quad1\times \frac{x^{6}}{6!}

    simplify:

    \frac{1}{1}\quad-\quad\frac{x^{2}}{2!}\quad+\quad\frac{x^{4}}{4!}<br />
\quad<br />
-\quad\frac{x^{6}}{6!}

    simplify:

    1 -\frac{1}{2!}x^{2} +\frac{1}{4!}x^{4} -\frac{1}{6!}x^{6}

    now substitute a small value for x, then compare it to cos(x) and you will see they're similar
    Last edited by Quick; August 13th 2006 at 09:54 AM. Reason: making it look better
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by c00ky
    Hi guys, another problem. If anybody could guide me through the solution that would be great!

    See screen shot on following link....

    CLICK HERE
    The Taylor Expansion for \cos x is given by,
    f(x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+... so, f(0)=1
    Note that, f(x) has radius of convergence of infinite. And,
    f'(x)=-\frac{x}{1!}+\frac{x^3}{3!}-\frac{x^5}{5!}+... so, f'(0)=0
    Also,
    f''(x)=-1+\frac{x^2}{2!}-\frac{x^4}{4!}+...=-f(x)
    Thus, you have that,
    f''(x)+f(x)=0 with, f(0)=1,f'(0)=0 by the existence theorem there is a unique function satisfing this equation. Thus, whatever is the solution must have that expansion stated above. Using various methods we find that the solution is,
    f(x)=\cos x
    ----
    Now, if you expand it to 4 terms you have the approximation for cosine,
    \cos x \approx 1- \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}
    To find the accurary of the approximation note that, the remainder term is,
    R_4=\frac{-\sin (z) x^{5}}{5!} where z is some number on the interval. If we take the interval [-\pi/4,\pi/4] the maximum error is,
    \left| \frac{-\sin z (\pi/4)^5}{120}} \right| \leq \left| \frac{ |1|\cdot|1|^5}{120}\right| \leq .008
    Which is accurate to 2 decimal places.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Aug 2006
    Posts
    36
    I see, excellent. Thank you very much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Math help needed ASAP please
    Posted in the Statistics Forum
    Replies: 24
    Last Post: February 19th 2010, 07:01 AM
  2. Replies: 2
    Last Post: February 15th 2010, 03:09 PM
  3. urgent math help needed on max and min
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 12th 2008, 12:59 AM
  4. Help with Simple Math Problem Needed
    Posted in the Algebra Forum
    Replies: 10
    Last Post: July 6th 2008, 10:22 PM
  5. Business Math Help needed
    Posted in the Business Math Forum
    Replies: 2
    Last Post: June 2nd 2008, 10:55 AM

Search Tags


/mathhelpforum @mathhelpforum