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The Taylor series expansion for sin about z_0 = (pi/2)

  1. May 10, 2007 #1
    1. The problem statement, all variables and given/known data
    Expand cos z into a Taylor series about the point z_0 = (pi)/2

    With the aid of the identity

    cos(z) = -sin(z - pi/2)

    2. Relevant equations
    Taylor series expansion for sin

    sinu = \sum^{infty}_{n=0} (-1)^n * \frac{u^{2n+1}}{(2n+1)!}

    and the identity as given above

    3. The attempt at a solution

    I've subbed in -sin(z- pi/2) into the identity my first prob was how to deal with the negative in front of the sin so I've done

    u = z-pi/2

    -sin(z-(pi/2)) = -(z-(pi/2)) + 1/3! * (z-(pi/2))^3 - 1/5! * (z-(pi/2))^5

    So if that is even correct (because I'm not sure about where to put the negative signs... then what does it mean when it says "about the point" z_0 = (pi/2)

    How do I sub that in to my answer???

    Any suggestions will be much appreciated

  2. jcsd
  3. May 10, 2007 #2


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    The Taylor's series of a function 'about' [itex]z_0= \pi/2[/itex] is the series
    [tex]\sum_{n=0}^\infty \frac{d^n f}{d x^n}(\pi/2)(z- \pi/2)^n[/tex].
    That is, you have powers of [itex]z- \pi/2[/itex] as well as evaluating the derivatives at [itex]\pi/2[/itex]. One method of determining the derivative is to use that definition directly: all derivatives of cos(x) are cos(x), sin(x), -cos(x), -sin(x) and those have values of 0, 1, and -1 at [itex]\pi/2[/itex].

    However, you are completely correct that [itex]cos(z)= -sin(z-\pi/2)[/itex]. That means all you need to do is write the Taylor's series for sin(z), which is
    [tex]z- (1/3!)z^3+ \cdot\cdot\cdot+ ((-1)^n/(2n+1)!)z^{2n+1}+\cdot\cdot\dot[/tex]
    Multiply by -1 to get -sin(z):
    [tex]-z+ (1/3!)z^3+ \cdot\cdot\cdot+ ((-1)^{n+1}/(2n+1)!)z^{2n+1}+\cdot\cdot\dot[/tex]
    and finally replace z by [itex]z-\pi/2[/itex]:
    [tex]-(z-\pi/2)+ (1/3!)(z-\pi/2)^3+\cdot\cdot\cdot+ ((-1)^{n+1}/(2n+1)!)(z- \pi/2)^{2n+1}+\cdot\cdot\cdot[/tex]
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