The Maclaurin series expansion of e^{x} or the Taylor series expansion of e^{x} at x=0 is given by the following summation:

e^{x} = $\sum_{n=0}^\infty \dfrac{x^n}{n!}$

= $1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots$.

In this post, we will learn how to find the series expansion of e^{x}.

## Taylor Series Expansion of e^{x} at x=0

The Maclaurin series expansion of a function f(x), that is, the Taylor series of f(x) at x=0 is given as follows:

$f(x)=\sum_{n=0}^\infty \dfrac{f^{(n)}(0)}{n!} x^n$ **…(∗) **

where f^{(n)} denotes the n-th derivative of f(x). The 0-th derivative of f(x) is equal to f(0).

Put f(x)=e^{x}.

As the derivative of e^{x} is e^{x} itself, we have:

f(0) = e^{0} = 1.

f$’$(x) = e^{x} ⇒ f$’$(0) = e^{0} = 1

f$^{”}$(x) = e^{x} ⇒ f$^{”}$(0) = e^{0} = 1

f^{(3)}(x) = e^{x} ⇒ f^{(3)}(0) = e^{0} = 1

f^{(4)}(x) = e^{x} ⇒ f^{(4)}(0) = e^{0} = 1

ans so on.

Substituting the above values in **(∗) **, the Maclaurin series expansion for f(x)=e^{x} is equal to

$f(x)=f(0)+xf'(0)+\dfrac{x^2}{2!}f^{”}(0)$ $+\dfrac{x^3}{3!}f^{(3)}(0)$ $+\dfrac{x^4}{4!}f^{(4)}(0)+\cdots$

⇒ $e^x$ $=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots$.

In sigma notation, the Maclaurin series expansion of e^{x} will be equal to

e^{x} = $\sum_{n=0}^\infty \dfrac{x^n}{n!}$.

**Note** that the above series for e^{x} converges for all real values. So the radius of converges of e^{x} series expansion is the interval (-∞, ∞).

**Also Read:**

**Maclaurin series expansion of sinx**

**Maclaurin series expansion of cosx**

## FAQs

**Q1: What is the Taylor series expansion of e^x at x=0?**

**Answer:** The Taylor series expansion of e^{x} at x=0 is equal to 1+x+x^{2}/2! + x^{3}/3! +x^{4}/4!+ …

**Q2: What is the Maclaurin series expansion of e^x?**

**Answer:** The Maclaurin series expansion of e^{x} is equal to 1+x+x^{2}/2! + x^{3}/3! +x^{4}/4!+ …