find the taylor polynomial pn(x) centered at x = 0 for the function f(x) = e^4x

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function $f(x)$ centered at $x = a$ is given by $f(x)=\sum_{k = 0}^{\infty}\frac{f^{(k)}(a)}{k!}(x - a)^k$. When $a = 0$, it is the Maclaurin series: $f(x)=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}x^k$, and the $n$-th - degree Taylor polynomial $P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(0)}{k!}x^k$.

Step2: Find the $k$-th derivative of $f(x)=e^{4x}$

The first - derivative of $y = e^{4x}$ using the chain rule: $y^\prime=\frac{d}{dx}(e^{4x})=4e^{4x}$. The second - derivative $y^{\prime\prime}=\frac{d}{dx}(4e^{4x}) = 4^2e^{4x}$. By induction, the $k$-th derivative $f^{(k)}(x)=4^k e^{4x}$.

Step3: Evaluate the $k$-th derivative at $x = 0$

Substitute $x = 0$ into $f^{(k)}(x)$. We get $f^{(k)}(0)=4^k e^{4\times0}=4^k$.

Step4: Construct the Taylor polynomial

Substitute $f^{(k)}(0)=4^k$ into the formula for $P_n(x)$. We have $P_n(x)=\sum_{k = 0}^{n}\frac{4^k}{k!}x^k=1 + 4x+\frac{4^2}{2!}x^2+\frac{4^3}{3!}x^3+\cdots+\frac{4^n}{n!}x^n$.

Answer:

$P_n(x)=\sum_{k = 0}^{n}\frac{4^k}{k!}x^k$