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Does cos(sqrt(x)) have a valid Taylor series expansion at a=0?

  1. Apr 15, 2015 #1
    Find the Taylor series about [itex] a=0 [/itex] for the function [itex] F(x) = \cos(\sqrt{x}). [/itex]

    Taylor series expansion of a function [itex] f(x) [/itex] about [itex] a [/itex]
    [tex] \sum^{\infty}_0 \frac{f^{(n)}(a)}{n!}(x-a)^n [/tex]

    Taylor series of [itex] \cos{x} [/itex] about [itex] a=0 [/itex] [tex] 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \ldots [/tex]

    From these expansions I could easily find a solution by a simple substitution of [itex] \sqrt{x} [/itex] into the Taylor series of [itex] \cos{x} [/itex] about [itex] a=0 [/itex] which produces the series
    [tex] 1 - \frac{x}{2!} + \frac{x^2}{4!} - \frac{x^3}{6!} \ldots [/tex]
    I found (through the use of a graphics calculator and the equality cos(z) = cosh(iz)) that this series converges to [tex] \cos(\sqrt{x}) [/tex] for [tex] x \ge 0 [/tex] and [tex] \cosh(\sqrt{-x})
    [/tex] for [tex] x \le 0 [/tex]
    My question is, is this Taylor series about [itex] a [/itex] valid at [itex] a=0,\, [/itex] as for a Taylor series to exist about a point [itex] a [/itex] the function has to have the property that it is infinitely differentiable at [itex] a [/itex] and [itex] \cos(\sqrt{x}) [/itex] is not differentiable at all at 0 (at least if you don't extend the function [itex] \cos{x} [/itex] to take imaginary values or my understanding of differentiability of endpoints is incorrect).

    I know that a function's continuity is defined differently at endpoints (for example the function [itex] F(x) = \sqrt{x} [/itex] is not defined as [itex] x \to 0^- [/itex] but is continuous at [itex] x=0 [/itex]), is a function's differentiability also defined differently at endpoints?

    I think the main point of my question is, is the function [itex] F(x) = \cos(\sqrt{x}),\, f : \mathbb{R} \geq 0 \to \mathbb{R} [/itex] differentiable at [itex] x=0 [/itex]? Or in a more general form, are functions on closed intervals differentiable at their endpoints?
     
  2. jcsd
  3. Apr 15, 2015 #2

    Dick

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    I think this is a question for the lawyers. There is a concept of a one-sided derivative and don't see any reason why you can't Taylor expand based on that. The point about the form of the function for ##x \lt 0## is interesting, but I don't think that's very important. The point is that your series does converge for ##x \ge 0##. My say.
     
  4. Apr 15, 2015 #3
    Though you found the series by a smart and easy trick, i think it would be beneficial to prove that there is the n-th derivative of the function F(x) (as you defined it) at the point x=0 and it is equal to [itex]F^{n}(0)=\frac{(-1)^n}{(n+1)...(2n)}[/itex]

    Notice we care only for the right derivative as the function is only defined in the right of x=0.

    Functions on close intervals are not necessarily differentiable in the end points, but for a function f/[a,b] we care only for the right derivatives at a, in order to apply taylor's theorem around point a.
     
    Last edited: Apr 16, 2015
  5. Apr 16, 2015 #4

    PeroK

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    Using the substitution of ##\sqrt{x}## into the Taylor series for ##cos(x)## is perfectly valid and produces a series expansion for ##cos(\sqrt{x})## valid for ##x \ge 0##.

    The way to look at it is that the Taylor series for ##cos(x)## is valid for whatever numerical value you choose for ##x##. For example, you could also get a perfectly valid series expansion for ##cos(|x|)## or even ##cos(f(x))## where ##f(x)## is not continuous anywhere.

    In this case, as Delta2 points out, ##cos(\sqrt{x})## may be infinitely (right) differentiable at 0. Remember that a derivative is a limit, not a formula.

    In fact, you could use your valid series expansion for ##cos(\sqrt{x})## to see that this function is infinitely differentiable and calculate its nth derivatives at 0.
     
    Last edited: Apr 16, 2015
  6. Apr 16, 2015 #5
    There is a slight problem though, it seems to me its not so easy to find the generic form of the n-th derivative of the function [itex]f(x)=cos(\sqrt x)[/itex] at (0,∞) which we need in order to find [itex]f^n(0)[/itex] as the limit [itex]\lim_{x \to 0^{+}}\frac{f^{n-1}(x)-f^{n-1}(0)}{x-0}[/itex] which can be simplified using De Hospital rule to [itex]f^n(0)=\lim_{x \to 0^+}f^n(x)[/itex] since we know the n-th derivative exists at (0,∞).
     
    Last edited: Apr 16, 2015
  7. Apr 16, 2015 #6

    PeroK

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    It's easy if you use:

    ##cos(\sqrt{x}) = 1 - \frac{x}{2!} + \frac{x^2}{4!} - \frac{x^3}{6!} \ldots ##
     
  8. Apr 16, 2015 #7
    Oh well yes but thats the thing we want to prove (i mean that we ignore the proof that has been given though it is 100% correct) with an alternative method.
     
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