The infinite series $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$ corresponds to

Understanding Infinite Series Expansions

The given infinite series is presented as: $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$

This type of series, where terms involve powers of $x$ and factorials, is a common form of a power series, specifically a Maclaurin series. A Maclaurin series helps us represent a function as an infinite sum of terms calculated from the function's derivatives at a single point, $x=0$.

Maclaurin Series Definition

A Maclaurin series is a special case of a Taylor series expansion of a function about $x=0$. For a function $f(x)$ that is infinitely differentiable at $x=0$, its Maclaurin series is given by the formula:

$f(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3+...+\frac{f^{(n)}(0)}{n!}{x}^n+...$

This can also be written in a more compact summation notation as:

$f(x)=\sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{x}^n$

**Deriving the Series for $e^x$ **

Let's consider the exponential function $f(x)=e^x$. To find its Maclaurin series, we need to calculate its derivatives and then evaluate them at $x=0$.

- For the function itself: $f(x)=e^x⟹f(0)=e^0=1$
- For the first derivative: $f^{\prime }(x)=\frac{d}{dx}(e^x)=e^x⟹f^{\prime }(0)=e^0=1$
- For the second derivative: $f^{\prime \prime }(x)=\frac{d^2}{d{x}^2}(e^x)=e^x⟹f^{\prime \prime }(0)=e^0=1$
- For the third derivative: $f^{\prime \prime \prime }(x)=\frac{d^3}{d{x}^3}(e^x)=e^x⟹f^{\prime \prime \prime }(0)=e^0=1$
- In general, for any positive integer $n$, the $n^{th}$ derivative of $e^x$ is always $e^x$. Therefore, $f^{(n)}(0)=e^0=1$ for all $n$.

Now, we substitute these values into the general Maclaurin series formula:

$e^x=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3+...$

Substituting the values we found:

$e^x=1+(1)x+\frac{1}{2!}{x}^2+\frac{1}{3!}{x}^3+...$

Which simplifies to:

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$

Comparing Series and Identifying the Function

When we compare the derived Maclaurin series for $e^x$ with the given infinite series:

Given Series: $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$

Series for $e^x$: $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$

We can clearly observe that they are identical.

For additional context, here are some other common Maclaurin series:Based on this comparison and derivation, it is evident that the given infinite series precisely corresponds to the exponential function $e^x$.

Therefore, the correct option is **e${}^x$ **.