Math Problem Statement

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 )]

Solution

The given series is:

ππ33!+π55!+(1)nπ2n+1(2n+1)!+\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=0x = 0) for a known function. Specifically, it matches the Maclaurin series for the sine function:

sin(x)=xx33!+x55!+(1)nx2n+1(2n+1)!+\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=πx = \pi:

sin(π)=ππ33!+π55!\sin(\pi) = \pi - \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \cdots

We know from trigonometry that:

sin(π)=0\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.

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Math Problem Analysis

Mathematical Concepts

Maclaurin series
Taylor series
Trigonometry

Formulas

Maclaurin series formula
Taylor series formula
Sine function series expansion

Theorems

Trigonometric identities

Suitable Grade Level

Advanced High School