Evaluate the Maclaurin series.
 
 
 
\[\displaystyle\pi -\frac{{{\pi }^{3}}}{3!}+\frac{{{\pi }^{5}}}{5!}-\unicode{0x2026}+{{\left( -1 \right)}^{n}}\frac{{{\pi }^{2n+1}}}{\left( 2n+1 \right)!}+\unicode{0x2026}\]
Choose 1 answer:
Choose 1 answer:
(Choice A)   
\[0\]
A
\[0\]
(Choice B)   
\[1\]
B
\[1\]
(Choice C)   
\[\cos ( \pi^2 )\]
C
\[\cos ( \pi^2 )\]
(Choice D)   
\[-\cos ( \pi^2 )\]
D
\[-\cos ( \pi^2 )\]

The given series is:

\[
\pi - \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \cdots + (-1)^n \frac{\pi^{2n+1}}{(2n+1)!} + \cdots
\]

This series is a Maclaurin series (Taylor series at \( x = 0 \)) for a known function. Specifically, it matches the Maclaurin series for the sine function:

\[
\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots + (-1)^n \frac{x^{2n+1}}{(2n+1)!} + \cdots
\]

By comparing the series, we can see that the expression given is the sine function evaluated at \( x = \pi \):

\[
\sin(\pi) = \pi - \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \cdots
\]

We know from trigonometry that:

\[
\sin(\pi) = 0
\]

Therefore, the correct answer is:

**(Choice A) 0**

Would you like more details or have any questions? Here are 5 related questions that could further your understanding:

1. What is the general formula for the Maclaurin series of a function?
2. How do you determine the radius of convergence for a Maclaurin series?
3. Can you derive the Maclaurin series for the cosine function?
4. What are the key differences between the Maclaurin and Taylor series?
5. How does the sine function's periodicity affect its Maclaurin series?

**Tip:** When evaluating a series, try to recognize if it matches a known function's series expansion. This can simplify the problem significantly.

Learn how to evaluate the Maclaurin series for pi using the sine function's series expansion. Understand the trigonometric identity involved and how to apply series expansion to calculate the value. Suitable for advanced high school students and anyone interested in mathematical series.

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