What is taylor ln(5-x) ?

The answer to taylor ln(5-x) is ln(5)-1/5 x-1/50 x^2-1/375 x^3-1/2500 x^4+\ldots

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taylor ln(5-x)

taylor

ln (5 - x)

ln (5) - \frac{1}{5} x - \frac{1}{50} x^{2} - \frac{1}{375} x^{3} - \frac{1}{2500} x^{4} + \dots

Solution steps

Apply the Taylor Formula

= ln (5) + \frac{\frac{d}{d x} ( ln ( 5 - x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 5 - x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 5 - x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= ln (5) + \frac{- \frac{1}{5}}{1 !} x + \frac{- \frac{1}{25}}{2 !} x^{2} + \frac{- \frac{2}{125}}{3 !} x^{3} + \frac{- \frac{6}{625}}{4 !} x^{4} + \dots

Refine

= ln (5) - \frac{1}{5} x - \frac{1}{50} x^{2} - \frac{1}{375} x^{3} - \frac{1}{2500} x^{4} + \dots

y^{^{''}} + 484 y = sec (22 x)

x \to \infty lim (7 + \frac{9}{x})

\frac{\partial}{\partial y} (sin (x yz))

\frac{d}{d x} (\frac{4}{5 x})

\frac{d y}{d x} + 3 x^{2} y = 0

What is taylor ln(5-x) ?
The answer to taylor ln(5-x) is ln(5)-1/5 x-1/50 x^2-1/375 x^3-1/2500 x^4+\ldots