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To solve this problem, we need to use the alternating series error bound for the Taylor series of \( \cos(x) \).

The Taylor series expansion of \( \cos(x) \) about 0 is:

\[ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \]

The error term \( E_n(x) \) for the Taylor polynomial of degree \( n \) for an alternating series can be given by the first omitted term:

\[ |E_n(x)| \leq \frac{|x|^{2n+2}}{(2n+2)!} \]

We are given that \( n = 4 \) and \( x = 1 \). The term we need is:

\[ |E_4(1)| \leq \frac{|1|^{2 \cdot 4 + 2}}{(2 \cdot 4 + 2)!} = \frac{1^{10}}{10!} \]

Now, calculating \( 10! \):

\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3628800 \]

Thus, the error bound is:

\[ |E_4(1)| \leq \frac{1}{3628800} \approx 2.75573 \times 10^{-7} \]

So, the smallest constant \( K \) given by the alternating series error bound is:

\[ K = 2.75573 \times 10^{-7} \]

Would you like more details or have any questions?

Here are some other questions you might want to ask:
1. Can you explain the Taylor series expansion for a different function?
2. How do you determine the radius of convergence for a Taylor series?
3. What are some practical applications of Taylor series in engineering?
4. How does the error bound change for different values of \( n \)?
5. Can you explain how to derive the alternating series error bound?

**Tip:** When working with Taylor series, always ensure to check the interval of convergence to understand where the series accurately represents the function.

Learn how to calculate the error bound for the Taylor series expansion of cos(x) using the alternating series error bound theorem. This problem involves advanced mathematical concepts and is suitable for advanced undergraduate students or anyone interested in numerical analysis.

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