Suppose that $f(x)$ has the following power series representation.
$$
\sum_{n=0}^{\infty} \frac{3^{\sqrt{n}}}{2^{n}}(x-5)^{n}
$$

Find $f^{(4)}(5)$.

Question

Suppose that $f(x)$ has the following power series representation.
$$
\sum_{n=0}^{\infty} \frac{3^{\sqrt{n}}}{2^{n}}(x-5)^{n}
$$

Find $f^{(4)}(5)$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Identify the term in the power series that will contribute to the fourth derivative at $x=5$. ‖ ‖ step 2 ⋮ The $n$th term of the power series is $\frac{3^{\sqrt{n}}}{2^n}(x-5)^n$. The fourth derivative of this term with respect to $x$ is $\frac{d^4}{dx^4}\left(\frac{3^{\sqrt{n}}}{2^n}(x-5)^n\right)$. ‖ ‖ step 3 ⋮ When $n=4$, the term becomes $\frac{3^{\sqrt{4}}}{2^4}(x-5)^4$. The fourth derivative at $x=5$ is non-zero only for $n=4$. ‖ ‖ step 4 ⋮ Calculate the fourth derivative of the term when $n=4$: $\frac{d^4}{dx^4}\left(\frac{3^{\sqrt{4}}}{2^4}(x-5)^4\right) = \frac{3^2}{2^4} \cdot 4!$ since the fourth derivative of $(x-5)^4$ is $4!$. ‖ ‖ step 5 ⋮ Evaluate the fourth derivative at $x=5$: $\frac{3^2}{2^4} \cdot 4! = \frac{9}{16} \cdot 24$. ‖ ‖ step 6 ⋮ Simplify the expression to find $f^{(4)}(5)$: $\frac{9}{16} \cdot 24 = \frac{9}{16} \cdot 24 = \frac{9 \cdot 24}{16} = 9 \cdot 1.5 = 13.5$. ‖ ∻Answer∻ ⚹ $f^{(4)}(5) = 13.5$ ⚹ ∻Key Concept∻ ⚹Power Series and Derivatives⚹ ∻Explanation∻ ⚹The fourth derivative of the function at $x=5$ is found by differentiating the term in the power series where $n=4$ four times and evaluating it at $x=5$. This term is the only one that contributes to the fourth derivative at $x=5$ because for all other terms, the derivative will include a factor of $(x-5)$ which becomes zero when $x=5$.⚹◊What is the formula for the nth derivative of $f(x)$ at $x=5$ in terms of the power series representation provided?⍭ Generate me a similar question◊

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