To estimate the maximum error when approximating a function with a Taylor polynomial, we use the Lagrange remainder formula:

$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1}$

Here:

• $f(x) = x^{2/3}$
• $n = 3$
• $a = 1$
• $0.8 \leq x \leq 1.2$

We need to find the maximum error on the interval $[0.8, 1.2]$.

Step 1: Calculate the 4th derivative ($f^{(4)}(x)$)

First, find the derivatives of $f(x)$:

1. $f(x) = x^{2/3}$ $f'(x) = \frac{2}{3}x^{-1/3}$
2. $f''(x) = \frac{-2}{9}x^{-4/3}$
3. $f^{(3)}(x) = \frac{8}{27}x^{-7/3}$
4. $f^{(4)}(x) = \frac{-56}{81}x^{-10/3}$

Step 2: Find the maximum of $|f^{(4)}(x)|$ on $[0.8, 1.2]$

On the interval $0.8 \leq x \leq 1.2$, $f^{(4)}(x) = \frac{-56}{81}x^{-10/3}$ is negative, so we use its absolute value: $|f^{(4)}(x)| = \frac{56}{81}x^{-10/3}$ The function $x^{-10/3}$ decreases as $x$ increases. Thus:

• Maximum occurs at $x = 0.8$: $\frac{56}{81}(0.8)^{-10/3}$
• Minimum occurs at $x = 1.2$: $\frac{56}{81}(1.2)^{-10/3}$

Step 3: Compute the remainder

The Lagrange remainder formula gives: $R_3(x) = \frac{f^{(4)}(c)}{4!} (x-1)^4$ For the maximum error, substitute $|f^{(4)}(c)|$ with its maximum value $\frac{56}{81}(0.8)^{-10/3}$:

1. Compute $|x-1|^4$ for $x$ in $[0.8, 1.2]$. The maximum is $(1.2-1)^4 = 0.2^4 = 0.0016$.
2. Compute $\frac{f^{(4)}(c)}{4!} = \frac{\frac{56}{81}(0.8)^{-10/3}}{24}$.

Step 4: Simplify numerical calculation

Let’s compute this explicitly. Numerical computation for maximum error estimation

from sympy import symbols, factorial, simplify

Define variables and constants

x = symbols('x', real=True, positive=True) a = 1 # Center of Taylor polynomial n = 3 # Degree of Taylor polynomial

Define f(x) and its 4th derivative

f_4 = (-56/81) * x**(-10/3)

Interval bounds

x_min, x_max = 0.8, 1.2

Calculate |f^(4)(c)| maximum on the interval

f_4_max = f_4.subs(x, x_min) # f_4 is largest at x_min f_4_max_abs = abs(f_4_max)

Compute the maximum value of (x-a)^4 on the interval

x_diff_max = (x_max - a)**4

Compute R_3(x) max error

remainder_max = f_4_max_abs / factorial(4) * x_diff_max remainder_maxThe maximum error when approximating $f(x) = x^{2/3}$ with the third-degree Taylor polynomial centered at $a = 1$ on the interval $[0.8, 1.2]$ is approximately:

$R_3(x) \leq 9.697 \times 10^{-5}$

Let me know if you'd like further explanation or steps.

Expanding Questions

1. How does the choice of the Taylor polynomial's degree $n$ affect the error estimate?
2. Why is the Lagrange remainder formula suitable for this approximation?
3. How would the error change if the interval was narrower (e.g., $[0.9, 1.1]$)?
4. Can you apply a similar process to find the error for $f(x) = \ln(x)$?
5. What role does the behavior of the $n+1$-th derivative play in determining the maximum error?

Tip

For better accuracy, always evaluate the highest derivative at the most extreme point of the interval to account for maximum deviation.