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) $\frac{1}{21}$

Question

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) $\frac{1}{21}$

Accepted Answer

∻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 $(-1)^{5} \frac{x^{5}}{5^{2}+5+1}$. ‖ ‖ step 3 ⋮ Simplifying the 5th term at $x=1$: $(-1)^{5} \frac{1^{5}}{5^{2}+5+1} = -\frac{1}{31}$. The negative sign indicates the direction of the error, but the error bound is the absolute value. ‖ ‖ step 4 ⋮ The error bound is therefore the absolute value of the 5th term, which is $\frac{1}{31}$. ‖ ∻Answer∻ ⚹ (C) $\frac{1}{31}$ ⚹ ∻Key Concept∻ ⚹Alternating Series Error Bound⚹ ∻Explanation∻ ⚹The error bound for an alternating series is given by the absolute value of the first omitted term.⚹◊What is the formula for the alternating series error bound in a Taylor polynomial approximation?⍭ Generate me a similar question◊

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