To estimate $\sqrt{7}$ using the second Taylor polynomial for $\sqrt{x}$ at $x = 6$, we perform the following steps:

Step 1: Taylor series expansion for $\sqrt{x}$

The Taylor series expansion of $\sqrt{x}$ around $x = a$ is: $\sqrt{x} \approx \sqrt{a} + \frac{1}{2\sqrt{a}}(x-a) - \frac{1}{4a^{3/2}}(x-a)^2 + \dots$

Here, $a = 6$, and we are considering the second-order polynomial.

Step 2: Calculate the polynomial terms

1. First term: $\sqrt{a} = \sqrt{6}$.
2. Second term: $\frac{1}{2\sqrt{a}}(x-a) = \frac{1}{2\sqrt{6}}(7-6) = \frac{1}{2\sqrt{6}}$.
3. Third term: $-\frac{1}{4a^{3/2}}(x-a)^2 = -\frac{1}{4(6^{3/2})}(7-6)^2 = -\frac{1}{4(6^{3/2})}$.

The second Taylor polynomial becomes: $P_2(7) = \sqrt{6} + \frac{1}{2\sqrt{6}} - \frac{1}{4(6^{3/2})}.$

Step 3: Compute $P_2(7)$

1. Compute $\sqrt{6} \approx 2.44949$.
2. Compute $\frac{1}{2\sqrt{6}} \approx \frac{1}{4.89898} \approx 0.20412$.
3. Compute $-\frac{1}{4(6^{3/2})} \approx -\frac{1}{4(14.69694)} \approx -0.01701$.

Thus, $P_2(7) \approx 2.44949 + 0.20412 - 0.01701 \approx 2.63660.$

Step 4: Error bound using Taylor’s theorem

The remainder term for the $n$-th Taylor polynomial is given by: $R_2(7) = \frac{f^{(3)}(c)}{3!}(x-a)^3,$ where $c$ is some value in $[a, x]$. For $\sqrt{x}$, the third derivative is: $f^{(3)}(x) = -\frac{3}{8}x^{-5/2}.$

Since $c \in [6, 7]$, the maximum value of $|f^{(3)}(c)|$ occurs at $c = 6$: $|f^{(3)}(6)| = \frac{3}{8(6^{5/2})} \approx \frac{3}{8(46.76537)} \approx 0.000802.$

The error bound is: $|R_2(7)| \leq \frac{|f^{(3)}(c)|}{6}(7-6)^3 \approx \frac{0.000802}{6}(1)^3 \approx 0.000134.$

Final Answer

The estimate of $\sqrt{7}$ using the second Taylor polynomial is approximately: $P_2(7) \approx 2.63660$ with an error bound: $|R_2(7)| \leq 0.000134.$

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Related Questions

1. How is Taylor's theorem applied for higher-order approximations?
2. What are practical applications of Taylor polynomials in numerical analysis?
3. How does the remainder term affect approximation accuracy?
4. What is the derivation of the error bound formula?
5. How does the choice of expansion point $a$ affect the Taylor approximation?

Tip

Always evaluate the derivatives carefully and choose the expansion point near the desired