What is taylor ln(x/2) ?

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

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taylor ln(x/2)

taylor

ln (\frac{x}{2})

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

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

\frac{d y}{d x} (x + 1) + y = ln (x), y (1) = 10

2 x^{2}, (3, 18)

x^{2} y^{^{''}} + 4 x y^{^{'}} + 2 y = \frac{1}{x}

3 \frac{d y}{d x} + \frac{8}{x y ^{2}} = 0

\frac{d}{d x} (\frac{2}{x + 5})

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