Asked by jessy

Use Taylor series expansion (degree 0 to 6) about x’=pi/4 to find an approximation for f(x) =
cosx at point x=pi/3 and then compare the result with the exact data
Answered by Jennifer
The Taylor expansion abour x = pi/4 to sixth order is

f(x) = f(pi/4) + f'(pi/4)*(x-pi/4) + f''(pi/4)*(x-pi/4)^2/2! + f'''(pi/4)*(x-pi/4)^3/3! + f''''(pi/4)*(x-pi/4)^4/4! + f'''''(pi/4)*(x-pi/4)^5/5!

f(x) = cos(x)
f'(x) = -sin(x)
f''(x) = -cos(x)
f'''(x) = sin(x)
f''''(x) = cos(x)
f'''''(x) = -sin(x)

f(pi/4) = 0.707
f'(pi/4) = -0.707
f''(pi/4) = -0.707
f'''(pi/4) = 0.707
f''''(pi/4) = 0.707
f'''''(pi/4) = -0.707

pi/4-pi/3 = -pi/12

f(x) = 0.707 - 0.707*(-pi/12) - 0.707*(-pi/12)^2/2! + 0.707*(-pi/12)^3/3! + 0.707*(-pi/12)^4/4! - 0.707*(-pi/12)^5/5!

calculate this number and compare it with cos(pi/3)
Answered by Arash
The Taylor expansion abour x = pi/4 to sixth order is

f(x) = f(pi/4) + f'(pi/4)*(x-pi/4) + f''(pi/4)*(x-pi/4)^2/2! + f'''(pi/4)*(x-pi/4)^3/3! + f''''(pi/4)*(x-pi/4)^4/4! + f'''''(pi/4)*(x-pi/4)^5/5!

f(x) = cos(x)
f'(x) = -sin(x)
f''(x) = -cos(x)
f'''(x) = sin(x)
f''''(x) = cos(x)
f'''''(x) = -sin(x)

f(pi/4) = 0.707
f'(pi/4) = -0.707
f''(pi/4) = -0.707
f'''(pi/4) = 0.707
f''''(pi/4) = 0.707
f'''''(pi/4) = -0.707

pi/4-pi/3 = -pi/12

f(x) = 0.707 - 0.707*(-pi/12) - 0.707*(-pi/12)^2/2! + 0.707*(-pi/12)^3/3! + 0.707*(-pi/12)^4/4! - 0.707*(-pi/12)^5/5!

calculate this number and compare it with cos(pi/3)
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