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Math  /  Calculus

QuestionThe Taylor series for sin⁡x\sin x about x=0x=0 is given by ∑n=1∞(−1)n+1x2n−1(2n−1)!\sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^{2 n-1}}{(2 n-1)!} and converges to sin⁡x\sin x for all xx. If the ninth-degree Taylor polynomial for sin⁡x\sin x about x=0x=0 is used to approximate sin⁡2\sin 2, what is the alternating series error bound? (A) 299!\frac{2^{9}}{9!} (B) 21010!\frac{2^{10}}{10!} (C) 21111!\frac{2^{11}}{11!} (D) 21019!\frac{2^{10}}{19!}

Studdy Solution

STEP 1

1. The Taylor series for sin⁡x\sin x is given by ∑n=1∞(−1)n+1x2n−1(2n−1)!\sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^{2n-1}}{(2n-1)!}.
2. We are using the ninth-degree Taylor polynomial to approximate sin⁡2\sin 2.
3. The alternating series error bound is given by the absolute value of the first omitted term.

STEP 2

1. Identify the first omitted term in the Taylor series.
2. Calculate the alternating series error bound using the first omitted term.

STEP 3

The ninth-degree Taylor polynomial for sin⁡x\sin x includes terms up to x9x^9. Therefore, the first omitted term is the term with x11x^{11}.
The general term in the Taylor series is:
(−1)n+1x2n−1(2n−1)! (-1)^{n+1} \frac{x^{2n-1}}{(2n-1)!}
For the first omitted term, we need 2n−1=112n-1 = 11.
Solving for nn, we have:
2n−1=11 2n-1 = 11 2n=12 2n = 12 n=6 n = 6
Thus, the first omitted term is:
(−1)6+1x1111!=−x1111! (-1)^{6+1} \frac{x^{11}}{11!} = -\frac{x^{11}}{11!}

STEP 4

To find the alternating series error bound when approximating sin⁡2\sin 2, we substitute x=2x = 2 into the first omitted term:
∣−21111!∣=21111! \left| -\frac{2^{11}}{11!} \right| = \frac{2^{11}}{11!}
This is the error bound for the approximation.
The alternating series error bound is 21111!\boxed{\frac{2^{11}}{11!}}.

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