w 2025 ww 9: problem 2 (1 point) question fin...
w 2025 ww 9: problem 2 (1 point) question find the first five non - zero terms of the maclaurin series for e^(-3x) and then use that to write e^(-3x) as a maclaurin series using sigma notation. solution we have f(x)= f(0)= f(x)= f(0)= f(x)= f(0)= f(x)= f(0)= f^(4)(x)= f^(4)(0)= now, the first five non - zero terms of the maclaurin series for e^(-3x) are + + + + +⋯ and using sigma notation, we have
Answer
# Explanation:
## Step1: Recall Maclaurin series formula
The Maclaurin series of a function $f(x)$ is given by $f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4}+\cdots$
## Step2: Define the function and find $f(0)$
Let $f(x)=e^{-3x}$. Then $f(0)=e^{-3\times0}=1$
## Step3: Find the first - derivative and $f^{\prime}(0)$
Differentiate $f(x)=e^{-3x}$ using the chain - rule. $f^{\prime}(x)=- 3e^{-3x}$. So $f^{\prime}(0)=-3e^{-3\times0}=-3$
## Step4: Find the second - derivative and $f^{\prime\prime}(0)$
Differentiate $f^{\prime}(x)=-3e^{-3x}$ again. $f^{\prime\prime}(x)=(-3)\times(-3)e^{-3x}=9e^{-3x}$. Then $f^{\prime\prime}(0)=9e^{-3\times0}=9$
## Step5: Find the third - derivative and $f^{\prime\prime\prime}(0)$
Differentiate $f^{\prime\prime}(x)=9e^{-3x}$. $f^{\prime\prime\prime}(x)=9\times(-3)e^{-3x}=-27e^{-3x}$. So $f^{\prime\prime\prime}(0)=-27e^{-3\times0}=-27$
## Step6: Find the fourth - derivative and $f^{(4)}(0)$
Differentiate $f^{\prime\prime\prime}(x)=-27e^{-3x}$. $f^{(4)}(x)=(-27)\times(-3)e^{-3x}=81e^{-3x}$. Then $f^{(4)}(0)=81e^{-3\times0}=81$
## Step7: Write the first five non - zero terms
The first five non - zero terms of the Maclaurin series are $1-3x+\frac{9}{2}x^{2}-\frac{27}{6}x^{3}+\frac{81}{24}x^{4}=1 - 3x+\frac{9}{2}x^{2}-\frac{9}{2}x^{3}+\frac{27}{8}x^{4}$
## Step8: Write in sigma notation
The Maclaurin series of $e^{-3x}$ in sigma notation is $\sum_{n = 0}^{\infty}\frac{(-3)^{n}}{n!}x^{n}$
# Answer:
The first five non - zero terms: $1-3x+\frac{9}{2}x^{2}-\frac{9}{2}x^{3}+\frac{27}{8}x^{4}$; Sigma notation: $\sum_{n = 0}^{\infty}\frac{(-3)^{n}}{n!}x^{n}$