Math Problem Statement

Suppose we want to approximate the value of 63^(1/3) using a Taylor polynomial. (a) Determine an appropriate power function f(x). (b) What would be an appropriate basepoint a? (c) Find T_1(x). (d) Use T_1(x) to estimate the value of 63^(1/3).

Solution

Let's walk through this problem step-by-step!

Problem Summary

You are asked to approximate 631/363^{1/3} using a Taylor polynomial.

(a) Determine an appropriate power function f(x)f(x).

  • The function related to finding a cube root is f(x)=x1/3f(x) = x^{1/3}, since we are interested in 631/363^{1/3}.

(b) What would be an appropriate base point aa?

  • In Taylor approximations, we pick a point aa close to the value we want to approximate (63), and one that makes calculations easier.
  • A good choice is a=64a = 64, because 641/3=464^{1/3} = 4 is easy to compute. This gives us a simple starting point for the approximation.

(c) Find T1(x)T_1(x).

This part asks for the first-order Taylor polynomial. The Taylor expansion of a function f(x)f(x) at aa is: T1(x)=f(a)+f(a)(xa)T_1(x) = f(a) + f'(a) (x - a) For f(x)=x1/3f(x) = x^{1/3}:

  1. Evaluate f(a)f(a): f(64)=641/3=4f(64) = 64^{1/3} = 4
  2. Find the derivative f(x)f'(x): f(x)=13x2/3f'(x) = \frac{1}{3} x^{-2/3}
  3. Evaluate f(a)f'(a) at a=64a = 64: f(64)=13642/3=13116=148f'(64) = \frac{1}{3} \cdot 64^{-2/3} = \frac{1}{3} \cdot \frac{1}{16} = \frac{1}{48}
  4. Write the first-order Taylor polynomial: T1(x)=4+148(x64)T_1(x) = 4 + \frac{1}{48} (x - 64)

(d) Use T1(x)T_1(x) to estimate the value of 631/363^{1/3}.

Now, substitute x=63x = 63 into the Taylor polynomial T1(x)T_1(x): T1(63)=4+148(6364)=4+148(1)T_1(63) = 4 + \frac{1}{48} (63 - 64) = 4 + \frac{1}{48} (-1) T1(63)=4148=40.02083.9792T_1(63) = 4 - \frac{1}{48} = 4 - 0.0208 \approx 3.9792

Thus, the estimate for 631/363^{1/3} is approximately: 631/33.979263^{1/3} \approx 3.9792


Conclusion

  • The Taylor polynomial approximation suggests 631/33.979263^{1/3} \approx 3.9792. This is close to the actual value, which is about 3.979057. So our approximation is quite good!

Do you have any questions, or would you like further details?


5 Related Questions:

  1. What is a second-order Taylor polynomial, and how would it improve the approximation?
  2. How does the error bound behave for Taylor polynomials? Can we calculate the error for this approximation?
  3. Can Taylor approximations work for other roots, such as x1/5x^{1/5}?
  4. How can we derive a general formula for the nn-th derivative of f(x)=x1/3f(x) = x^{1/3}?
  5. How do higher-order polynomials compare in accuracy to first-order approximations?

Tip:

When choosing the base point aa, try to select values that simplify calculations (like perfect squares or cubes). This reduces computational errors and simplifies derivatives.

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Math Problem Analysis

Mathematical Concepts

Taylor Polynomials
Approximations
Derivatives
Cube Roots

Formulas

First-order Taylor polynomial: T_1(x) = f(a) + f'(a) * (x - a)
Derivative of f(x) = x^(1/3): f'(x) = 1/3 * x^(-2/3)

Theorems

Taylor Polynomial Theorem

Suitable Grade Level

Grades 11-12, Early College