find a taylor series for f(x)=ln(x) centered at x=1

Differentiate: f '(x) = 1/x; f ''(x) = - 1/x^2; f '''(x) = 2/x^3; ..., f^{(n)}(x) = (n - 1)!/x^n; ...

The Taylor coefficients are

a_{0} = f(1) = ln 1 = 0; a_{1} = f '(1) = 1/1 = 1; a2 = (1/2!)f ''(1) = -1/2; … a_{n} = (1/n!)f^{(n)}(1) = (-1)^{n-1}/n; …

The Taylor series is (x – 1) – (x – 1)^{2}/2 + (x – 1)^{3}/3 - … + (-1)^{n-1}(x – 1)^{n}/n + ... . It converges to the function ln x in the interval 0 < x <= 2.

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