Use the Maclaurin Series for the cosine. Graph f(x) = cos x and *P_2, P_4, P_6 o

Question

Use the Maclaurin Series for the cosine. Graph f(x) = cos x and *P_2, P_4, P_6 on the same set of axes. x [-pi, pi] y[-2, 2] using a graphing calculator. (Use different colors of pen.) Sec example 4 in 11.10 in your book. Use the first 3 terms of the series to approximate f(x) for x =.05, x = pi/6 and x = pi/2. Since this is an alternating series, the Error lessthanorequalto |the 1^st neglected term|. Find the error for each of the value of x. cos x = 1 - x^2/2! + x^4/4! - x^6/6! * P_2 = 1 - x^2/2, etc Explain why the series approximation has more error for the last value of x.

Explanation / Answer

cos 0.05 = 1- (0.05)^2/2+ (0.05)^4/24 - 0.05^6 / 720 = 0.99875026

error = actual value - calculated value = 0.99875026- P2 = 0.00000026

cos (pi/6) = 1- (pi/6)^2/2 +(pi/6)^4/24 - (pi/6)^6/720 = 0.866025264

error = actual value - P4 = 0.866025264- 0.866053883 = -2.86194*10^ (-5)

P4 = 1- x^2/2! + x^4/4!

cos (pi/2) = 1- (pi/2)^2/2! + (pi/2)^4/4! - (pi/2) ^6/6! =  -8.94523*10^(-4)

error = 8.94523*10^(-4) since cos(pi/2) = 0

here for last case error is more because neglected terms x^8/8! is very close to error value

if we add

cos (pi/2) = 1- (pi/2)^2/2! + (pi/2)^4/4! - (pi/2) ^6/6! + (pi/2)^8/8! - (pi/2)^10/10!+ (pi/2)^12/12!

we get proper answer and

if x value is more the error is more

here x= pi/2 >1 so error is more

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