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

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

Apr 19th, 2015

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 

a0 = f(1) = ln 1 = 0; a1 = f '(1) = 1/1 = 1; a2 = (1/2!)f ''(1) = -1/2; … an = (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.

Apr 19th, 2015

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Apr 19th, 2015
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Apr 19th, 2015
May 23rd, 2017
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