- #1

chickensandwich

- 1

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*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?