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Derivation of Power Series

  1. Apr 1, 2005 #1
    How do I write a derivation of the cosine power series?

    (I understand and can derive it, but it takes much space and is disjointed! :grumpy: ; how do you write the shortest and fastest derivation for it--briefly and fluently??)

    Attached Files:

  2. jcsd
  3. Apr 1, 2005 #2
    Did you not attach the answer to your question?
  4. Apr 1, 2005 #3
    I believe he means whether there was a faster way to come up with a power series than equating derivatives.

    I do not believe there is...
  5. Apr 1, 2005 #4
    You can define it as the derivative of the power series for sin(x). Thats the shortest derivation of cosin, but then you'd have to define sin :)
  6. Apr 1, 2005 #5
    Of course, that would be much faster--->just differentiate the sum formula for sin(x).
    But suppose we can't do that!-->what would be another fast way to derive cos(x)?

    Currently I start with cos(x), and then explain that as a Taylor series, cos(x-a) in this case is really the McLaurin series for cos (x), where a=0.
    Then, I write cos(x) = [tex] c_0 + c_1 x + c_2 x^2 + c_3 x^3 +... [/tex]
    Next, I express each derivative of cos (x) as the same series, except I reduce the powers of x and subunits of the constants appropriately.
    Then, I substitute x=0 (b/c it is McLaurin) and express each derivative of cos(0) as the constant "c" with the appropriate subunits.
    Next, I show that the odd derivatives of cos(0) are zero, and explain why the sum formula includes [tex] (-1)^n [/tex] and [tex] x^{2n} [/tex]
    Then, I show why, as a power series, the factorial (2n)! is needed in the denominator so that the terms will match the derivatives when multiplied out.

    Finally, I write the formula (I wish I knew LaTex...better!) as [tex] cos(x) = \sum\limits_{n = 0}^\infty {\frac{{\left( {-1} \right)^n x^{2n} }}{{\left( {2n} \right)!}}} [/tex]

    My question is: HOW to constrict/shorten this procedure??
    Are there parts here I can connect or skip??
    Last edited: Apr 2, 2005
  7. Apr 2, 2005 #6
    maybe the squareroot algorithm
  8. Apr 2, 2005 #7
    Do you know Euler's identity,

    [tex] e^{i\theta} = \cos \theta + i \sin \theta[/tex]


    If so, the Taylor series for [itex]e^x[/itex] is obviously really easy to derive. Just substitute in [itex]x = i \theta[/itex], and note that

    [tex] \cos \theta = \mbox{Re} \left[ e^{i \theta} \right] = \frac{ e^{i \theta} + e^{-i \theta}}{2}[/tex]

    from which you can just take the series for each of those and add them.
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