Compute the Taylor poly 𝑇5(𝑥) and use the Error Bound to find the maximum possible size of the error. 𝑓(𝑥)=cos(𝑥), 𝑎=0, 𝑥=0.2. Find the answer to 𝑇5(0.2)= and |𝑓(0.2)−𝑇5(0.2)|≤ _*10^-7

Answer 1

Hi Hanas,

First we compute the Taylor series using the derivatives of f(x) = cos(x)

f⁽¹⁾(x) = -sin(x)

f⁽²⁾(x) = -cos(x)

f⁽³⁾(x) = sin(x)

f⁽⁴⁾(x) = cox(x)

f⁽⁵⁾(x) = -sin(x)

and find that T₅(x) = 1 - (1/2)x² + (1/24)x⁴

so we have T₅(0.2) = 0.980067. Now, we use Taylor's approximation theorem to find that the bound on the remainder. The Taylor approximation theorem states that for any value of n, if the (n+1)^th derivative of f is continuous, and | f⁽ⁿ⁺¹⁾(t) | ≤ M for all t between 0 and x (for us, x = 0.2) then

| f(x) - T_n(x) | ≤ M/(n+1)! |x|ⁿ⁺¹

so in our case we have f⁽⁶⁾(x) = -cos(x). Notice that the largest value that |-cos(x)| takes between 0 and 0.2 is 1 (at x=0), so we have |f⁽⁶⁾(t)| ≤ 1 for all t between 0 and 0.2. Now,

| f(0.2) - T₅(.2) | ≤ 1/(5+1)! |0.2|⁵⁺¹

⁼ (1/720) (0.2)⁶

which gives the maximum error (approximately .88 × 10^-7).

I hope that helps!