Monday, 25 April 2016

calculus - Proving the smoothness (indefinitely differentiability) of 2cos1(1x).

I am trying to prove that dnθdxn where x=1cosθ2 exists n>0. I found that dθdx=2sinθ2 and based on this answer I think I only need to prove that sinθ2 is smooth (notice is a function of θ and its firsts derivatives looks like fun).



I also found that cos1(x)=π2+k=0x1+2k(12)kk!+2kk! (at wolfram alpha). But I am not sure that the existence of its Taylor Series implies its smoothness. I thought that by linearity of differentiation and because all terms in the series are polynomials (smooth funcions) there could be a way to make it work.



Is there a simple (elegant) way to prove its smoothness? Until now my only approach is to find a pattern of the n-th derivative, however I am actively trying to avoid that because of this.



edit:
I am trying to prove that θ=2cos1(1x) is smooth (in other words, has infinitely many derivatives dnθdxnn>0)




I tried different approaches, but based on the Taylor Expansion of cos1 I found that θ=2cos1(1x) can be expressed by polynomials (which are smooth functions) and therefore its derivatives.



I have never worked with Taylor Series, so I am not really confident about this last statement.



Side note: My original expression was x=1cosθ2,1x1 (which implies θ=2cos1(1x)) and because of that I derived dθdx=2sinθ2 applying ddx to both sides.

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