Taylor series?

[匿名]

2016-04-21 8:02 am

using Taylor's Theorem, approximate the function f(x)=x^(1/2) about x0=16 by a polynomial of degree 2 in (x-x0). Hence find the value of 17^(1/2) and estimate the error

回答 (1)

用戶2053

2016-04-21 4:17 pm

✔ 最佳答案

f(x) = √x , f '(x) = (1/2)x^(-1/2) , f "(x) = (-1/4)x^(-3/2) , f "'(x) = (3/8)x^(-5/2)
∴ f(16) = √16 = 4 , f '(16) = (1/2)16^(-1/2) = 1/8 , f "(16) = (-1/4)16^(-3/2) = - 1/256 ,
f "'(16) = (3/8)16^(-5/2) = 3/8192

Thus f(x) = x^(1/2) about x₀=16 by a polynomial of degree 2 in (x - x₀) is
f(16) + f '(16)(x-16) + f "(16)(x-16)² / 2!
= 4 + (1/8)(x-16) - (1/256)(x-16)² / 2!

Putting x = 17 , we have
√17 ≈ 4 + 1/8 - 1/512 = 2111/512 = 4.1230...
The error < f '"(16)(17-16)² / 3! = (3/8192) / 3! = 1/16384 ≈ 0.000061

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