What is taylor ln(1+4x),0 ?

The answer to taylor ln(1+4x),0 is 4x-8x^2+64/3 x^3-64x^4+1024/5 x^5+\ldots

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taylor ln(1+4x),0

taylor

ln (1 + 4 x), 0

4 x - 8 x^{2} + \frac{64}{3} x^{3} - 64 x^{4} + \frac{1024}{5} x^{5} + \dots

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( ln ( 1 + 4 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 1 + 4 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 1 + 4 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{4}{1 !} x + \frac{- 16}{2 !} x^{2} + \frac{128}{3 !} x^{3} + \frac{- 1536}{4 !} x^{4} + \frac{24576}{5 !} x^{5} + \dots

Refine

= 4 x - 8 x^{2} + \frac{64}{3} x^{3} - 64 x^{4} + \frac{1024}{5} x^{5} + \dots

\frac{25}{s ^{2} + 6 s + 25}

\int arctan (8 t) d t

(3 x y + 3 y - 4) d x + (x + 1)^{2} d y = 0

n = 0 \sum \infty n e^{- 8 n}

\frac{\partial}{\partial x} (e^{2 x - 3 y})

What is taylor ln(1+4x),0 ?
The answer to taylor ln(1+4x),0 is 4x-8x^2+64/3 x^3-64x^4+1024/5 x^5+\ldots