Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook