What is taylor ln(1+x),5 ?

The answer to taylor ln(1+x),5 is ln(6)+1/6 (x-5)-1/72 (x-5)^2+1/648 (x-5)^3-1/5184 (x-5)^4+\ldots

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taylor ln(1+x),5

taylor

ln (1 + x), 5

ln (6) + \frac{1}{6} (x - 5) - \frac{1}{72} (x - 5)^{2} + \frac{1}{648} (x - 5)^{3} - \frac{1}{5184} (x - 5)^{4} + \dots

Solution steps

Apply the Taylor Formula

= ln (6) + \frac{\frac{d}{d x} ( ln ( 1 + x ) ) ( 5 )}{1 !} (x - 5) + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 1 + x ) ) ( 5 )}{2 !} (x - 5)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 1 + x ) ) ( 5 )}{3 !} (x - 5)^{3} + \dots

Evaluate Derivatives

= ln (6) + \frac{\frac{1}{6}}{1 !} (x - 5) + \frac{- \frac{1}{36}}{2 !} (x - 5)^{2} + \frac{\frac{1}{108}}{3 !} (x - 5)^{3} + \frac{- \frac{1}{216}}{4 !} (x - 5)^{4} + \dots

Refine

= ln (6) + \frac{1}{6} (x - 5) - \frac{1}{72} (x - 5)^{2} + \frac{1}{648} (x - 5)^{3} - \frac{1}{5184} (x - 5)^{4} + \dots

2 x^{2} - 4, x^{2}

\frac{d}{d x} (2 e^{x} (cos (x) - sin (x)))

f (x) = (x - x^{2})

(y) d x + (y - x) d y = 0

What is taylor ln(1+x),5 ?
The answer to taylor ln(1+x),5 is ln(6)+1/6 (x-5)-1/72 (x-5)^2+1/648 (x-5)^3-1/5184 (x-5)^4+\ldots