Question

Math

Posted 5 months ago

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

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Answer from Sia

Posted 5 months ago

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.

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