The problem is: f(x) = [cos(3x^2)-1]/(x^2) at x=0 I know that the Mclaurin series for cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + .... That is where I'm stuck, I can't remember how to use the Mclaurin series with this function
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Originally Posted by pakman The problem is: f(x) = [cos(3x^2)-1]/(x^2) at x=0 I know that the Mclaurin series for cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + .... That is where I'm stuck, I can't remember how to use the Mclaurin series with this function Learn and watch. Thus, (composition of two analyic functions at a point is analytic), Thus, When you divide by , is the same as dividing term by term, thus,
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