Example 5.20. Expand the function $f(x)=e^{3 x}$ into a Taylor series.
Solution. Let's compute the values of the given function and its successive derivatives at $x=0$:

$$
\begin{array}{rr}
f(x)=e^{3 x}, & f(0)=1, \\
f^{\prime}(x)=3 e^{3 x}, & f^{\prime}(0)=3 \\
f^{\prime \prime}(x)=3^{2} e^{3 x}, & f^{\prime \prime}(0)=3^{2} \\
f^{\prime \prime \prime}(x)=3^{3} e^{3 x}, & f^{\prime \prime \prime}(0)=3^{3} \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
f^{(n)}(x)=3^{n} e^{3 x}, & f^{(n)}(0)=3^{n}
\end{array}
$$

Substituting the obtained values into the general expression for the Taylor series of an arbitrary function, we get

$$
e^{3 x}=1+\frac{3}{1!} x+\frac{3^{2}}{2!} x^{2}+\frac{3^{3}}{3!} x^{3}+\ldots+\frac{3^{n}}{n!} x^{n}+\ldots
$$

This is the Taylor series expansion for the function $f(x)=e^{3 x}$. The obtained series converges to the generating function $f(x)=e^{3 x}$ for any value of $x$ (see problem 5.19).

Note that the same expansion can be obtained from the Taylor series for the function $e^{3 x}$ by replacing $x$ with $3 x$.