attempt 1: 10 attempts remaining. find the maclaurin series for f(x)=e^3x using the definition of a…

Explanation:

Step1: Recall Maclaurin series formula

The Maclaurin series of a function $f(x)$ is given by $\sum_{k = 0}^{\infty}\frac{f^{(k)}(0)}{k!}x^{k}$, where $f^{(k)}(x)$ is the $k$-th derivative of $f(x)$.

Step2: Find derivatives of $f(x)=e^{3x}$

Let $f(x)=e^{3x}$. Then $f^{(0)}(x)=e^{3x}$, $f^{(1)}(x) = 3e^{3x}$, $f^{(2)}(x)=3^{2}e^{3x},\cdots,f^{(k)}(x)=3^{k}e^{3x}$.

Step3: Evaluate derivatives at $x = 0$

$f^{(k)}(0)=3^{k}e^{0}=3^{k}$ for $k = 0,1,2,\cdots$.

Step4: Write out the Maclaurin series

Substitute $f^{(k)}(0)$ into the Maclaurin - series formula: $\sum_{k = 0}^{\infty}\frac{f^{(k)}(0)}{k!}x^{k}=\sum_{k = 0}^{\infty}\frac{3^{k}}{k!}x^{k}$.

Step5: Find the interval of convergence using the ratio - test

Let $a_{k}=\frac{3^{k}}{k!}x^{k}$. Then $\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|=\lim_{k\rightarrow\infty}\left|\frac{\frac{3^{k+1}}{(k + 1)!}x^{k+1}}{\frac{3^{k}}{k!}x^{k}}\right|=\lim_{k\rightarrow\infty}\left|\frac{3x}{k + 1}\right| = 0$ for all $x\in(-\infty,\infty)$.

Answer:

The Maclaurin series is $\sum_{k = 0}^{\infty}\frac{3^{k}}{k!}x^{k}$. The interval of convergence is $(-\infty,\infty)$.