Question

Consider the following function. f(x) = ln(1 + 2x), a = 2, n = 3, 1.8 ≤ x ≤ 2.2 (a) Approximate f by a Taylor polynomial with degree n at the number a. T3(x) = Correct: Your answer is correct. (b) Use Taylor's Inequality to estimate the accuracy of the approximation f(x) ≈ Tn(x) when x lies in the given interval. (Round your answer to six decimal places.) |R3(x)| ≤

Answer 1

Answer:

a) $T_{3}(x) = f(a) + (x-a)f'(a) + \frac{(x-a)^{2} }{2!} f''(a)+ \frac{(x-a)^{3} }{3!}f'''(a)$

b) $R_{3}(x) =0.000010$

Step-by-step explanation:

$f(x) = ln(1+2x)$, a = 2, n = 3

1.8 ≤ x ≤ 2.2

The Taylor's polynomial to the third degree is given by:

$T_{3}(x) = f(a) + (x-a)f'(a) + \frac{(x-a)^{2} }{2!} f''(a)+ \frac{(x-a)^{3} }{3!}f'''(a)$

Since a = 2

$T_{3}(x) = f(2) + (x-2)f'(2) + \frac{(x-2)^{2} }{2!} f''(2)+ \frac{(x-2)^{3} }{3!}f'''(2)$.......(1)

$f(x) = ln(1+2x)\\f(2) = ln5\\f'(x) = \frac{2}{1+ 2x}\\ f'(2) = 2/5\\f''(x) = \frac{-4}{(1+2x)^{2} } \\f''(2) = -4/25\\f'''(x) = \frac{16}{(1+2x)^{3} } \\f'''(2) = 16/125$

Substituting these values into equation (1)

$T_{3}(x) =ln5 + \frac{2}{5} (x-2) - \frac{2 }{25}(x-2)^{2}+ \frac{ 8}{375}(x-2)^{3}$

b) Taylor's inequality is given by the equation:

$R_{n}(x) = \frac{f^{n+1}(x) }{(n+1)!} (x-a)^{n+1}$

n = 3 ( third degree), a =2

$R_{3}(x) = \frac{f^{iv}(x) }{(4)!} (x-2)^{4}$

$f^{iv} (x) =- \frac{96}{(1+2x)^{4} } \\ f^{iv} (x)= -96/625$

$R_{3}(x) = \frac{(-96/625)}{24} (x-2)^{4}\\}$

$R_{3}(x) = (0.0064) (x-2)^{4}$

$R_{3}(x) = (0.0064) (1.8-2)^{4}$

$R_{3}(x) = 1.025 * 10^{-5} \\R_{3}(x) =0.00001025\\R_{3}(x) =0.000010$