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solve all three questions
6. The function $f$ has derivatives of all orders for all real numbers. It is known that $f(0)=2, f^{\prime}(0)=3$, $f^{\prime \prime}(x)=-f\left(x^{2}\right)$, and $f^{\prime \prime \prime}(x)=-2 x \cdot f^{\prime}\left(x^{2}\right)$.
(a) Find $f^{(4)}(x)$, the fourth derivative of $f$ with respect to $x$. Write the fourth-degree Taylor polynomial for $f$ about $x=0$. Show the work that leads to your answer.
(b) The fourth-degree Taylor polynomial for $f$ about $x=0$ is used to approximate $f(0.1)$. Given that $\left|f^{(5)}(x)\right| \leq 15$ for $0 \leq x \leq 0.5$, use the Lagrange error bound to show that this approximation is within $\frac{1}{10^{5}}$ of the exact value of $f(0.1)$.
(c) Let $g$ be the function such that $g(0)=4$ and $g^{\prime}(x)=e^{x} f(x)$. Write the second-degree Taylor polynomial for $g$ about $x=0$.
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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

Given $f''(x) = -f(x^2)$, we need to find $f^{(4)}(x)$

step 2

Differentiate $f''(x)$ to find $f'''(x)$: $f'''(x) = -2x \cdot f'(x^2)$

step 3

Differentiate $f'''(x)$ to find $f^{(4)}(x)$: $f^{(4)}(x) = -2 \cdot f'(x^2) - 4x^2 \cdot f''(x^2)$

step 4

Evaluate at $x=0$: $f^{(4)}(0) = -2 \cdot f'(0) = -2 \cdot 3 = -6$

step 5

The fourth-degree Taylor polynomial for $f$ about $x=0$ is $P_4(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4$

step 6

Substitute the known values: $P_4(x) = 2 + 3x - \frac{1}{2}x^2 - \frac{1}{6}x^3 - \frac{1}{4}x^4$

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Part (b)
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step 1

The Lagrange error bound for the Taylor polynomial is given by $R_n(x) = \frac{M}{(n+1)!} |x|^{n+1}$, where $M$ is the maximum value of $|f^{(n+1)}(x)|$ on the interval

step 2

Given $|f^{(5)}(x)| \leq 15$ for $0 \leq x \leq 0.5$, we use $M = 15$

step 3

For $x = 0.1$ and $n = 4$, the error bound is $R_4(0.1) = \frac{15}{5!} (0.1)^5 = \frac{15}{120} (0.1)^5 = \frac{15}{120} \cdot 10^{-5} = \frac{1}{8} \cdot 10^{-5} = \frac{1}{10^5}$

step 4

Therefore, the approximation $P_4(0.1)$ is within $\frac{1}{10^5}$ of the exact value of $f(0.1)$

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Part (c)
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step 1

Given $g(0) = 4$ and $g'(x) = e^x f(x)$, we need to find the second-degree Taylor polynomial for $g$ about $x=0$

step 2

The Taylor polynomial is $P_2(x) = g(0) + g'(0)x + \frac{g''(0)}{2!}x^2$

step 3

Evaluate $g'(0) = e^0 f(0) = 1 \cdot 2 = 2$

step 4

Differentiate $g'(x)$ to find $g''(x)$: $g''(x) = e^x f(x) + e^x f'(x)$

step 5

Evaluate $g''(0) = e^0 f(0) + e^0 f'(0) = 1 \cdot 2 + 1 \cdot 3 = 5$

step 6

Substitute the known values: $P_2(x) = 4 + 2x + \frac{5}{2}x^2$

Answer

Part (a): $f^{(4)}(0) = -6$, Taylor polynomial: $P_4(x) = 2 + 3x - \frac{1}{2}x^2 - \frac{1}{6}x^3 - \frac{1}{4}x^4$

Part (b): The approximation $P_4(0.1)$ is within $\frac{1}{10^5}$ of the exact value of $f(0.1)$.

Part (c): Second-degree Taylor polynomial for $g$: $P_2(x) = 4 + 2x + \frac{5}{2}x^2$

Key Concept

Taylor Polynomial and Lagrange Error Bound

Explanation

The Taylor polynomial provides an approximation of a function around a point, and the Lagrange error bound gives an estimate of the error in this approximation

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