https://file.aiquickdraw.com/mathbot/file/1729136060668mt24hlhf.jpg

Let's solve the problem step by step.

### Part (a): Taylor Polynomial Approximation

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

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

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

1. **\( f(x) = e^{3x^2} \)**
   \[
   f(0) = e^{0} = 1
   \]

2. **First derivative \( f'(x) \)**
   \[
   f'(x) = \frac{d}{dx}(e^{3x^2}) = 6x e^{3x^2}
   \]
   \[
   f'(0) = 6(0)e^{0} = 0
   \]

3. **Second derivative \( f''(x) \)**
   \[
   f''(x) = \frac{d}{dx}(6x e^{3x^2}) = 6e^{3x^2} + 36x^2 e^{3x^2}
   \]
   \[
   f''(0) = 6e^{0} = 6
   \]

4. **Third derivative \( f'''(x) \)**
   \[
   f'''(x) = \frac{d}{dx}(6e^{3x^2} + 36x^2 e^{3x^2}) = 72x e^{3x^2} + 216x e^{3x^2} = 288x e^{3x^2}
   \]
   \[
   f'''(0) = 288(0)e^{0} = 0
   \]

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

### Part (b): Taylor's Remainder Estimation

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

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

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

Let's approximate \( e^{0.12} \) up to six decimal places:
\[
e^{0.12} \approx 1.127496
\]
\[
f^{(4)}(0.2) \approx (322.56) \cdot 1.127496 \approx 363.625
\]

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

### Final Results:
**(a)**
\[
p_3(x) = 1 + 3x^2
\]

**(b)**
\[
R_3(x) \approx 0.024241
\]

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.

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.

Learn how to approximate the function f(x) = e^(3x^2) using a Taylor polynomial of degree 3 at a = 0, and estimate the error using Taylor's remainder. This solution involves finding derivatives, constructing the polynomial p_3(x) = 1 + 3x^2, and evaluating the accuracy of the approximation. Suitable for undergraduate calculus students.

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