What is taylor ln(x),4 ?

The answer to taylor ln(x),4 is 2ln(2)+1/4 (x-4)-1/32 (x-4)^2+1/192 (x-4)^3-1/1024 (x-4)^4+\ldots

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taylor ln(x),4

taylor

ln (x), 4

2 ln (2) + \frac{1}{4} (x - 4) - \frac{1}{32} (x - 4)^{2} + \frac{1}{192} (x - 4)^{3} - \frac{1}{1024} (x - 4)^{4} + \dots

Solution steps

Apply the Taylor Formula

= 2 ln (2) + \frac{\frac{d}{d x} ( ln ( x ) ) ( 4 )}{1 !} (x - 4) + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( x ) ) ( 4 )}{2 !} (x - 4)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( x ) ) ( 4 )}{3 !} (x - 4)^{3} + \dots

Evaluate Derivatives

= 2 ln (2) + \frac{\frac{1}{4}}{1 !} (x - 4) + \frac{- \frac{1}{16}}{2 !} (x - 4)^{2} + \frac{\frac{1}{32}}{3 !} (x - 4)^{3} + \frac{- \frac{3}{128}}{4 !} (x - 4)^{4} + \dots

Refine

= 2 ln (2) + \frac{1}{4} (x - 4) - \frac{1}{32} (x - 4)^{2} + \frac{1}{192} (x - 4)^{3} - \frac{1}{1024} (x - 4)^{4} + \dots

g (x) = 1 - 2 x - x^{(2)}

\frac{d y}{d x} = - \frac{x e ^{- x^{2}}}{y}

\int \frac{1}{e ^{2 t}} d t

\int (\frac{2 x}{y} + x y^{2}) d x

f (x) = x^{2} \cdot e^{x^{2} - 1}, at x = - 1

What is taylor ln(x),4 ?
The answer to taylor ln(x),4 is 2ln(2)+1/4 (x-4)-1/32 (x-4)^2+1/192 (x-4)^3-1/1024 (x-4)^4+\ldots