Answer:
To find a bound for the error in approximating the quantity with a third-degree Taylor polynomial for the function f(x) = sin(x) about x = 0, we can use the Lagrange Error Bound for Pn(x).
The Lagrange Error Bound formula is given by:
|f(x) - Pn(x)| ≤ M * |x - a|^(n+1) / (n+1)!
Where:
- f(x) is the given function
- Pn(x) is the nth degree Taylor polynomial approximation of f(x)
- M is the maximum value of the (n+1)th derivative of f(x) on the interval [a, x]
- x is the value at which we want to approximate f(x)
- a is the center of the Taylor polynomial approximation
For this problem, we want to find the error in approximating sin(0.3) using a third-degree Taylor polynomial about x = 0.
Since we are using a third-degree Taylor polynomial, n = 3. The center of the approximation is a = 0.
To find the bound, we need to find the maximum value of the fourth derivative of f(x) = sin(x) on the interval [0, 0.3]. We can do this by finding the maximum value of the absolute value of the fourth derivative of sin(x) on that interval.
Taking the derivative of sin(x) four times, we get:
f''''(x) = -sin(x)
To find the maximum value of -sin(x) on the interval [0, 0.3], we can compare the values at the endpoints and any critical points in between.
At x = 0, -sin(x) = 0.
At x = 0.3, -sin(x) = -sin(0.3) ≈ -0.29552.
Therefore, the maximum value of -sin(x) on the interval [0, 0.3] is approximately 0.29552.
Now we can plug these values into the Lagrange Error Bound formula:
|f(0.3) - P3(0.3)| ≤ M * |0.3 - 0|^4 / (3+1)!
Since f(x) = sin(x) and P3(x) is the third-degree Taylor polynomial approximation of f(x) about x = 0, we have:
|sin(0.3) - P3(0.3)| ≤ 0.29552 * |0.3 - 0|^4 / 4!
Simplifying further:
|sin(0.3) - P3(0.3)| ≤ 0.29552 * 0.3^4 / 24
|sin(0.3) - P3(0.3)| ≤ 0.002087
Therefore, the bound for the error in approximating sin(0.3) with a third-degree Taylor polynomial about x = 0 is approximately 0.002087.
Remember to round your answer to six decimal places.