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

How to calculate this function?

  1. Jul 2, 2009 #1
    I have a queation as below


    How to solve it? I just konw that the angle is an operator, but the function of x is out of the cosine function.
  2. jcsd
  3. Jul 2, 2009 #2


    User Avatar
    Homework Helper
    Gold Member

    They probably mean

    [tex] \cos ({d/dx}) = 1 - \frac{1}{2!}\frac{d^2}{dx^2} + \frac{1}{4!}\frac{d^4}{dx^4} - ... [/tex]
  4. Jul 2, 2009 #3


    User Avatar
    Science Advisor

    Just as I would interpret f(g)(x) to mean f(g(x)), I would interpret cos(d/dx)f(x) to mean cos(df/dx). That would, I believe, give the same thing as dx's Taylor's series interpretation.
  5. Jul 2, 2009 #4
    The two suggestions are different. Because the OP says the answer is an operator probably the interpretation of dx is the intended one.

    Let's see why they are different. Try the example [tex]f(x) = x^2[/tex] so that [tex]f'(x) = 2x[/tex] and [tex]f''(x) = 2[/tex], and all higher derivatives are zero.

    Then under the dx method, we get

    f(x) - \frac{1}{2!}\;f''(x) = x^2 - 1

    but under the HallsOfIvy method, we get

    \sin(f'(x)) = \sin(2x)

    Not the same.
  6. Jul 2, 2009 #5

    But, this term comes from the Mathieu’s equation as below


    If I use series, this equation would be so terrible
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook