Taylor series?

[匿名]

2016-03-26 9:58 am

Find the Taylor expansion of
ln[(1+x)/(1-x)]=a0+a1x+a2x^2+a3x^3+R3

Use this formula to find the approximate values of ln(2) and ln(3)

回答 (1)

YA HOO！知識＋管理員

2016-03-26 6:19 pm

✔ 最佳答案

ln [(1 + x)/(1 - x)] = a₀ + a₁x + a₂x² + a₃x³ + R₃ ...... ①

Let f(x) = ln [(1 + x)/(1 - x)] = ln (1 + x) - ln (1 - x)
f'(x) = 1/(x + 1) + 1/(1 - x)
f''(x) = -1/(x + 1)² + 1/(1 - x)²
f'''(x) = 2/(x + 1)³ + 2/(1 - x)³

a₀ = 0 ...... [ Sub x = 0 into ① ]
a₁ = f'(0)/1! = 2
a₂ = f''(0)/2! = 0
a₃ = f'''(0)/3! = 2/3

∴ ln [(1 + x)/(1 - x)] = 2x + 2x³/3 + R₃

ln[(1 + x)/(1 - x)] = ln 2
1 + x = 2 - 2x
x = 1/3
∴ ln 2 ≈ 2(1/3) + 2(1/3)³/3 = 0.69

ln[(1 + x)/(1 - x)] = ln 3
1 + x = 3 - 3x
x = 1/2
∴ ln 3 ≈ 2(1/2) + 2(1/2)³/3 = 1.08

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