In this article, we will guide you through solving the problem of finding the second-order approximation of the function f(x) = x^(1/3) at x = 1, and then using it to approximate the value of (1.1)^(1/3). Additionally, we will explain how to evaluate the error involved in this approximation. We will address common mistakes and provide the correct steps to follow in this type of problem.
Step 1: Find the Second-Order Approximation
To begin, let’s recall the formula for the second-order Taylor expansion of a function at a point x = a:
f(x) ≈ f(a) + f'(a)(x – a) + (f”(a)/2)(x – a)^2
For f(x) = x^(1/3), we need to find the function value, the first derivative, and the second derivative evaluated at x = 1. Let’s calculate each step:
- f(1) = 1^(1/3) = 1
- f'(x) = (1/3)x^(-2/3), so f'(1) = (1/3)(1^(-2/3)) = 1/3
- f”(x) = (-2/9)x^(-5/3), so f”(1) = (-2/9)(1^(-5/3)) = -2/9
Substitute these values into the second-order Taylor expansion to get the approximation:
f(x) ≈ 1 + (1/3)(x – 1) – (1/9)(x – 1)^2
This is the second-order approximation for f(x) at x = 1.
Step 2: Approximate f(1.1) Using the Second-Order Approximation
Now, we can use this second-order approximation to estimate the value of f(1.1), which is (1.1)^(1/3). Substitute x = 1.1 into the second-order approximation:
f(1.1) ≈ 1 + (1/3)(1.1 – 1) – (1/9)(1.1 – 1)^2
f(1.1) ≈ 1 + (1/3)(0.1) – (1/9)(0.1)^2
f(1.1) ≈ 1 + 0.0333 – 0.0011 ≈ 1.0344
This gives us the approximate value of (1.1)^(1/3) ≈ 1.0344.
Step 3: Error Estimation
To estimate the error, we need to calculate the remainder term in the Taylor series expansion. The remainder term for a second-order approximation is given by:
R(x) = (f”'(c)/6)(x – a)^3
for some value c between a and x. To compute the error, we need to find the third derivative of f(x):
- f”'(x) = (10/27)x^(-8/3), so f”'(1) = 10/27
Using the formula for the remainder term, we can estimate the error:
R(1.1) = (10/27)(1.1 – 1)^3 / 6 = (10/27)(0.1)^3 / 6 ≈ 0.00000617
Thus, the error in our approximation is approximately 0.00000617.
Step 4: Conclusion
To summarize, the second-order approximation for f(x) = x^(1/3) at x = 1 is:
f(x) ≈ 1 + (1/3)(x – 1) – (1/9)(x – 1)^2
Using this approximation, we estimated that (1.1)^(1/3) ≈ 1.0344. The error in this approximation is approximately 0.00000617. By following the steps outlined above, you can solve similar problems and evaluate the accuracy of your approximations.
コメント