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Math Help - Series expansion?

  1. #1
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    Series expansion?

    My problem is....

    Use the series expansions for sin x and cos x to find the first two
    terms of a series expansion for tan x


    but which series do i use? Power, maclaurin?
    also how do I find tan x ( i know sinx/cosx=tanx) but how do i get there using series?


    many thanks
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Use polynomial division.

    Expand {x-x^3/3!}/{1-x^2/2}. The first two terms is your answer.
    Last edited by Also sprach Zarathustra; June 10th 2011 at 02:40 PM.
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  3. #3
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    im having some trouble using polynomial division. what is the method for having a fraction in the nominator/denominator.
    when i divide straight down as seen i get


    but i know the answer is


    if someone could help me with the method please.
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    Quote Originally Posted by decoy808 View Post
    im having some trouble using polynomial division. what is the method for having a fraction in the nominator/denominator.
    when i divide straight down as seen i get


    but i know the answer is


    if someone could help me with the method please.
    You should learn the polynomial long division method. Please refer, Polynomial Long Division. I hope you will find it simple and illustrative.
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  5. #5
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    Quote Originally Posted by decoy808 View Post
    My problem is....

    Use the series expansions for sin x and cos x to find the first two
    terms of a series expansion for tan x


    but which series do i use? Power, maclaurin?
    also how do I find tan x ( i know sinx/cosx=tanx) but how do i get there using series?


    many thanks
    sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-.....

    cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-....

    tanx=\frac{sinx}{cosx}\Rightarrow\ sinx=cosxtanx

    \Rightarrow\ x-\frac{x^3}{3!}+\frac{x^5}{5!}-....=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-....\right)tanx

    Therefore the first term of tanx is x and the 2nd term involves x^3 as an x^2 would give even powers of x.

    x-\frac{x^3}{3!}+\frac{x^5}{5!}-....=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-....\right)\left(x+\frac{x^3}{k}+...\right)

    Multiplying out and comparing terms to find k,

    x-\frac{x^3}{3!}+\frac{x^5}{5!}-...=x-\frac{x^3}{2!}+\frac{x^5}{4!}+...+\frac{x^3}{k}-\frac{x^5}{(k)2!}+\frac{x^7}{(k)4!}-...

    \Rightarrow\frac{ x^3}{k}-\frac{x^3}{2!}=-\frac{x^3}{3!}

    \Rightarrow\frac{2x^3-kx^3}{(k)2!}=-\frac{x^3}{3!}

    which gives k and therefore the second term in the expansion of tanx.
    Last edited by Archie Meade; June 12th 2011 at 07:19 AM.
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