Question

Math

Posted 5 months ago

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4. The Taylor series for a function $f$ about $x=0$ converges to $f$ for $-1 \leq x \leq 1$. The $n$ th-degree Taylor polynomial for $f$ about $x=0$ is given by $P_{n}(x)=\sum_{k=1}^{n}(-1)^{k} \frac{x^{k}}{k^{2}+k+1}$. Of the following, which is the smallest number $M$ for which the alternating series error bound guarantees that $\left|f(1)-P_{4}(1)\right| \leq M ?$
(A) $\frac{1}{5!} \cdot \frac{1}{31}$
(B) $\frac{1}{4!} \cdot \frac{1}{21}$
(C) $\frac{1}{31}$
(D) $\frac{1}{21}$
```

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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

To apply the alternating series error bound, we use the first term not included in the Taylor polynomial $P_4(x)$. This is the 5th term of the series

step 2

The 5th term of the Taylor series is given by $(-1)^5 \frac{x^5}{5^2+5+1}$ when $x = 1$

step 3

Simplifying the 5th term: $(-1)^5 \frac{1^5}{5^2+5+1} = -\frac{1}{31}$

step 4

The error bound is the absolute value of the 5th term, which is $\left| -\frac{1}{31} \right| = \frac{1}{31}$

Answer

$M = \frac{1}{31}$

Key Concept

Alternating Series Error Bound

Explanation

The error bound for an alternating series is the absolute value of the first omitted term.

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