Question

The $n$th derivative of $f(x)=e^{c x}$ is $f^{(n)}(x)=c^{n} e^{c x}$ The Maclaurin series for $f(x)=e^{x}$ is $e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}$

Answer 1

The Maclaurin series for $f(x) = e^{cx}$ is $\sum_{n=0}^{\infty} \frac{(cx)^n}{n!}$.

Answer 2

The Maclaurin series for a function $f(x)$ is given by $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$. For $f(x) = e^{cx}$, the $n$th derivative is $f^{(n)}(x) = c^n e^{cx}$. Evaluating this derivative at $x=0$ gives $f^{(n)}(0) = c^n e^{c \cdot 0} = c^n$. Substituting $f^{(n)}(0) = c^n$ into the Maclaurin series formula yields: $e^{cx} = \sum_{n=0}^{\infty} \frac{c^n}{n!} x^n$ This can be rewritten as: $e^{cx} = \sum_{n=0}^{\infty} \frac{(cx)^n}{n!}$