The Maclaurin series for $f(x)$ is given by $1+\frac{x}{2!}+\frac{x^{2}}{3!}+\frac{x^{3}}{4!}+\cdots+\frac{x^{n}}{(n+1)!}+\cdots$
(a) Find $f^{\prime}(0)$ and $f^{(17)}(0)$.
(b) For what values of $x$ does the given series converge? Show your reasoning.
(c) Let $g(x)=x f(x)$. Write the Maclaurin series for $g(x)$, showing the first three nonzero terms and the general term.
(d) Write $g(x)$ in terms of a familiar function without using series. Then, write $f(x)$ in terms of the same familiar function.

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Accepted Answer

∻Solution by Steps∻ ### Part (a) ‖ step 1 ⋮ The Maclaurin series for $f(x)$ is given by $f(x) = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \frac{x^3}{4!} + \cdots + \frac{x^n}{(n+1)!} + \cdots$. ‖ ‖ step 2 ⋮ To find $f'(0)$, we need the coefficient of $x$ in the series. The coefficient of $x$ is $\frac{1}{2!} = \frac{1}{2}$. Therefore, $f'(0) = \frac{1}{2}$. ‖ ‖ step 3 ⋮ To find $f^{(17)}(0)$, we need the coefficient of $x^{17}$ in the series. The coefficient of $x^{17}$ is $\frac{1}{18!}$. Therefore, $f^{(17)}(0) = \frac{1}{18!}$. ‖ ∻Answer∻ ⚹ $f'(0) = \frac{1}{2}$, $f^{(17)}(0) = \frac{1}{18!}$ ⚹ ### Part (b) ‖ step 1 ⋮ The given series is $\sum_{n=0}^{\infty} \frac{x^n}{(n+1)!}$. ‖ ‖ step 2 ⋮ To determine the interval of convergence, we use the ratio test. ‖ ‖ step 3 ⋮ Compute the ratio $\left| \frac{a_{n+1}}{a_n} ight| = \left| \frac{x^{n+1}}{(n+2)!} \cdot \frac{(n+1)!}{x^n} ight| = \left| \frac{x}{n+2} ight|$. ‖ ‖ step 4 ⋮ As $n o \infty$, $\left| \frac{x}{n+2} ight| o 0$ for all $x$. ‖ ‖ step 5 ⋮ Since the ratio test shows that the series converges for all $x$, the interval of convergence is $(-\infty, \infty)$. ‖ ∻Answer∻ ⚹ The series converges for all $x \in (-\infty, \infty)$. ⚹ ### Part (c) ‖ step 1 ⋮ Let $g(x) = x f(x)$. ‖ ‖ step 2 ⋮ Substitute the Maclaurin series for $f(x)$: $g(x) = x \left( 1 + \frac{x}{2!} + \frac{x^2}{3!} + \frac{x^3}{4!} + \cdots ight)$. ‖ ‖ step 3 ⋮ Distribute $x$: $g(x) = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$. ‖ ‖ step 4 ⋮ The first three nonzero terms are $x$, $\frac{x^2}{2!}$, and $\frac{x^3}{3!}$. ‖ ‖ step 5 ⋮ The general term is $\frac{x^{n+1}}{(n+1)!}$. ‖ ∻Answer∻ ⚹ The Maclaurin series for $g(x)$ is $x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{n+1}}{(n+1)!} + \cdots$. ⚹ ### Part (d) ‖ step 1 ⋮ We know $g(x) = x f(x)$. ‖ ‖ step 2 ⋮ From the series, $f(x) = e^x - 1$. ‖ ‖ step 3 ⋮ Therefore, $g(x) = x (e^x - 1) = x e^x - x$. ‖ ‖ step 4 ⋮ Thus, $f(x) = \frac{g(x) + x}{x} = e^x - 1$. ‖ ∻Answer∻ ⚹ $g(x) = x e^x - x$, $f(x) = e^x - 1$ ⚹ ∻Key Concept∻ ⚹Maclaurin series and interval of convergence⚹ ∻Explanation∻ ⚹The Maclaurin series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. The interval of convergence is the set of all $x$ values for which the series converges.⚹◊What is the general term of the Maclaurin series for the function $f(x)$?⍭ Generate me a similar question◊

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