Question

Use Taylor's Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error. Round your answers to three significant figures.)

cos(0.3)=1−(0.3)22!+(0.3)44!
R4≤ .
R4=.

Answer 1

Explanation:

According to Tylor theorem,

f(x)=f(a)+f′(a)/1!×(x−a)+f′′(a)/2!×(x−a)2+......+fn(a)/n!×(x−a)n

Tylor series for cosx is

1 - x^2/2! + x^4/4! + x^6/6! + ……

cos (0.3) = 1 - 〖(0.3)〗^2/2! + 〖(0.3)〗^4/4!

The R4 valur is fourth derivative of cosx

f’(x) = -sinx

f’’ (x) = -cosx

f’’’ (x) = sinx

f’’’’ (x) = cosx

Here, a = 0, n = 3, x = 0.3

max l f^4l = cos (0.3)

= 0.9999

lR_4 (x)l ≤ |0.3-0|^(4+1)/4! × 0.9999

≤ 0.00015186

= 1.51786 × 10^(-4)

R_n (x) = cos (0.3) – (1 - 〖(0.3)〗^2/2!-(0.3)^4/4!)

= 0.0446

The error of approximation R_4 (x) ≤ 1.2 × 10^(-2) and exact value of error R_4 = 0.0446