$\begin{array}{l}f(x)=\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n+1}}{(2 n)!} \ f^{\prime \prime \prime}(0)=\square\end{array}$

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the third derivative of $ f(x) $ at $ x = 0 $, we need to differentiate the given series term by term three times and then evaluate at $ x = 0 $. ‖ ‖ step 2 ⋮ The general term of the series is $ (-1)^n \frac{x^{2n+1}}{(2n)!} $. The first derivative of this term with respect to $ x $ is $ (-1)^n \frac{(2n+1)x^{2n}}{(2n)!} $. ‖ ‖ step 3 ⋮ The second derivative of the general term with respect to $ x $ is $ (-1)^n \frac{(2n+1)(2n)x^{2n-1}}{(2n)!} $. ‖ ‖ step 4 ⋮ The third derivative of the general term with respect to $ x $ is $ (-1)^n \frac{(2n+1)(2n)(2n-1)x^{2n-2}}{(2n)!} $. ‖ ‖ step 5 ⋮ Evaluating the third derivative at $ x = 0 $, we see that all terms where $ n \geq 1 $ will be zero because they will have a factor of $ x $ raised to a positive power. ‖ ‖ step 6 ⋮ The only term that survives is the one with $ n = 0 $, which is $ (-1)^0 \frac{(2 \cdot 0 + 1)(2 \cdot 0)(2 \cdot 0 - 1)}{(2 \cdot 0)!} $. ‖ ‖ step 7 ⋮ Simplifying the term with $ n = 0 $, we get $ \frac{1 \cdot 0 \cdot (-1)}{1} = 0 $. ‖ ∻Answer∻ ⚹ $ f'''(0) = 0 $ ⚹ ∻Key Concept∻ ⚹Differentiation of power series term by term and evaluation at a point⚹ ∻Explanation∻ ⚹The third derivative of the function at $ x = 0 $ is found by differentiating the series term by term three times and then evaluating at $ x = 0 $. All terms with $ n \geq 1 $ vanish, leaving only the term with $ n = 0 $, which evaluates to zero.⚹◊What is the general form of the $n^{th}$ derivative of the function $f(x)$ at $x=0$?⍭ Generate me a similar question◊

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