Math Problem Statement

Consider the following function f(x) = e^(3x^2), 0 <= x <= 0.2 (a) Approximate f by a Taylor polynomial with degree n = 3 at a = 0. p_3(x) = ? (b) Use Taylor's remainder to estimate the accuracy of the approximation f(x) ≈ p_n(x) when x lies in the given interval.

Solution

Let's solve the problem step by step.

Part (a): Taylor Polynomial Approximation

The function is given as: f(x)=e3x2f(x) = e^{3x^2} We need to find the Taylor polynomial of degree n=3n = 3 at a=0a = 0.

The general formula for a Taylor series expansion of a function f(x)f(x) around x=ax = a is: f(x)pn(x)=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3+f(x) \approx p_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots

Here, a=0a = 0. Let's find the derivatives and evaluate them at x=0x = 0:

  1. f(x)=e3x2f(x) = e^{3x^2} f(0)=e0=1f(0) = e^{0} = 1

  2. First derivative f(x)f'(x) f(x)=ddx(e3x2)=6xe3x2f'(x) = \frac{d}{dx}(e^{3x^2}) = 6x e^{3x^2} f(0)=6(0)e0=0f'(0) = 6(0)e^{0} = 0

  3. Second derivative f(x)f''(x) f(x)=ddx(6xe3x2)=6e3x2+36x2e3x2f''(x) = \frac{d}{dx}(6x e^{3x^2}) = 6e^{3x^2} + 36x^2 e^{3x^2} f(0)=6e0=6f''(0) = 6e^{0} = 6

  4. Third derivative f(x)f'''(x) f(x)=ddx(6e3x2+36x2e3x2)=72xe3x2+216xe3x2=288xe3x2f'''(x) = \frac{d}{dx}(6e^{3x^2} + 36x^2 e^{3x^2}) = 72x e^{3x^2} + 216x e^{3x^2} = 288x e^{3x^2} f(0)=288(0)e0=0f'''(0) = 288(0)e^{0} = 0

Now, we can write the Taylor polynomial p3(x)p_3(x): p3(x)=f(0)+f(0)x+f(0)2!x2+f(0)3!x3p_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 p3(x)=1+0x+62x2+0x3p_3(x) = 1 + 0 \cdot x + \frac{6}{2}x^2 + 0 \cdot x^3 p3(x)=1+3x2p_3(x) = 1 + 3x^2

Part (b): Taylor's Remainder Estimation

To estimate the accuracy, we use Taylor's remainder formula: Rn(x)=f(n+1)(c)(n+1)!xn+1R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} where cc is some point in the interval [0,x][0, x], and n=3n = 3.

We need the fourth derivative f(4)(x)f^{(4)}(x): f(4)(x)=ddx(288xe3x2)=288e3x2+864x2e3x2f^{(4)}(x) = \frac{d}{dx}(288x e^{3x^2}) = 288e^{3x^2} + 864x^2 e^{3x^2}

The maximum value of f(4)(x)f^{(4)}(x) on the interval 0x0.20 \leq x \leq 0.2 occurs at x=0.2x = 0.2. Let's calculate it: f(4)(0.2)=288e3(0.2)2+864(0.2)2e3(0.2)2f^{(4)}(0.2) = 288e^{3(0.2)^2} + 864(0.2)^2 e^{3(0.2)^2} =288e0.12+864(0.04)e0.12= 288e^{0.12} + 864(0.04) e^{0.12} =(288+34.56)e0.12= (288 + 34.56) e^{0.12}

Let's approximate e0.12e^{0.12} up to six decimal places: e0.121.127496e^{0.12} \approx 1.127496 f(4)(0.2)(322.56)1.127496363.625f^{(4)}(0.2) \approx (322.56) \cdot 1.127496 \approx 363.625

Now, the remainder can be estimated by: R3(x)363.6254!(0.2)4R_3(x) \leq \frac{363.625}{4!} \cdot (0.2)^4 R3(x)363.625240.0016R_3(x) \leq \frac{363.625}{24} \cdot 0.0016 R3(x)0.024241R_3(x) \leq 0.024241

Final Results:

(a) p3(x)=1+3x2p_3(x) = 1 + 3x^2

(b) R3(x)0.024241R_3(x) \approx 0.024241

Would you like more details or have any questions about the solution?

Related Questions:

  1. How do you derive higher-order Taylor polynomials?
  2. What are the steps to compute the remainder for different degrees of approximation?
  3. Can the Taylor series be used for functions with more complex terms, such as trigonometric functions?
  4. How can you determine the interval of convergence for a Taylor series?
  5. What is the practical use of Taylor approximations in real-world applications?

Tip:

When estimating errors using Taylor's remainder, always check the behavior of higher-order derivatives over the interval to identify where they might be maximized.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Taylor Series
Polynomial Approximation
Exponential Functions
Error Estimation

Formulas

Taylor series formula: f(x) ≈ p_n(x) = f(a) + f'(a)(x-a) + f''(a)/2!(x-a)^2 + ...
Remainder estimation: R_n(x) = f^(n+1)(c)/(n+1)! * x^(n+1)

Theorems

Taylor's Remainder Theorem

Suitable Grade Level

Undergraduate Calculus