Can f(x)=cos(x) be written as the difference of two increasing functions on the open interval $(0,2\pi )$ ?
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6$\cos x =(2x+\cos x )-(2x)$. – Kavi Rama Murthy Apr 08 '22 at 07:37
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5Re Taylor series : $$\cos(x) = \left[1 + \frac{x^4}{4!} + \frac{x^8}{8!} + \cdots\right] - \left[\frac{x^2}{2!} + \frac{x^6}{6!} + \cdots\right].$$ – user2661923 Apr 08 '22 at 07:51
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As a freebie: "A real valued function is of bounded variation iff. it is the difference of two increasing functions" and any (almost everywhere) continuously differentiable function (e.g. cosine) is of bounded variation (since they are absolutely continuous) – FShrike Apr 08 '22 at 10:37
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@FShrike Skip “almost everywhere.” There is an easy example of a differentiable function whose derivative is continuous at all but one point, but that function does not have bounded variation. – B. S. Thomson Apr 08 '22 at 13:28
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@B.S.Thomson Ok, thank you for the note – FShrike Apr 08 '22 at 13:29