What is taylor ln(1+8x^2) ?

The answer to taylor ln(1+8x^2) is 8x^2-32x^4+512/3 x^6-1024x^8+32768/5 x^{10}+\ldots

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taylor ln(1+8x^2)

taylor

ln (1 + 8 x^{2})

8 x^{2} - 32 x^{4} + \frac{512}{3} x^{6} - 1024 x^{8} + \frac{32768}{5} x^{10} + \dots

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( ln ( 1 + 8 x ^{2} ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 1 + 8 x ^{2} ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 1 + 8 x ^{2} ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{16}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{- 768}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{122880}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{- 41287680}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{23781703680}{10 !} x^{10} + \dots

Refine

= 8 x^{2} - 32 x^{4} + \frac{512}{3} x^{6} - 1024 x^{8} + \frac{32768}{5} x^{10} + \dots

cos (\frac{1 - e ^{9 x}}{1 + e ^{9 x}})

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

x \to + 0 lim (7)

x \to - 4 lim (f (x))

\frac{d}{d x} (csc (\frac{x}{2}))

What is taylor ln(1+8x^2) ?
The answer to taylor ln(1+8x^2) is 8x^2-32x^4+512/3 x^6-1024x^8+32768/5 x^{10}+\ldots