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
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)
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)
To find an approximation for f(x) = cos(x) using Taylor series expansion, we can follow these steps:
Step 1: Find the Taylor series expansion of f(x) = cos(x) about x' = pi/4.
The Taylor series expansion for a function f(x) centered around x = a is given by:
f(x) = f(a) + f'(a)(x - a)/1! + f''(a)(x - a)^2/2! + f'''(a)(x - a)^3/3! + ...
For f(x) = cos(x), we need to calculate the derivatives of cos(x) up to the desired degree and evaluate them at x' = pi/4.
Let's calculate the derivatives:
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''''''(x) = -cos(x).
Evaluating them at x' = pi/4:
f(pi/4) = cos(pi/4) = sqrt(2)/2,
f'(pi/4) = -sin(pi/4) = -sqrt(2)/2,
f''(pi/4) = -cos(pi/4) = -sqrt(2)/2,
f'''(pi/4) = sin(pi/4) = sqrt(2)/2,
f''''(pi/4) = cos(pi/4) = sqrt(2)/2,
f'''''(pi/4) = -sin(pi/4) = -sqrt(2)/2.
Step 2: Utilize the Taylor series expansion formula to approximate f(x) at x = pi/3.
Using degree 0 to 6, the Taylor series expansion of f(x) = cos(x) about x' = pi/4 becomes:
f(x) ≈ f(pi/4) + f'(pi/4)(x - pi/4)/1! + 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''''''(pi/4)(x - pi/4)^6/6!
Substituting the values we calculated earlier and x = pi/3:
f(pi/3) ≈ sqrt(2)/2 + (-sqrt(2)/2)(pi/3 - pi/4) + (-sqrt(2)/2)(pi/3 - pi/4)^2/2 + (sqrt(2)/2)(pi/3 - pi/4)^3/3!
Simplifying the expression gives us the approximation for f(pi/3).
Step 3: Compare the approximation with the exact value.
To compare the approximation with the exact value, we need to calculate the exact value of f(pi/3) = cos(pi/3).
Using a calculator, we find that cos(pi/3) = 0.5√3.
Now, we can compare the approximation obtained in Step 2 with the exact value.
Evaluate f(pi/3) using the approximation derived earlier, and then compare it with the exact value.
By comparing these two values, you can assess the accuracy of the approximation and observe how it compares to the exact data.