The exponential function e^x has a Taylor series expansion given by the formula,

e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!}\ldots+ \frac{x^n}{n!}+\ldotsThe above formula is the taylor series for e^x around the point x=0. It is also known as the Maclaurin series and it expresses the exponential function in the form of a “polynomial of infinite degree”. The infinite series on the right hand side means that the value of the infinite series at a point converges to the value of the exponential function at that point.

This series is valid for all real numbers, that is, the series on the right hand side is convergent for all real values of x . In fact, this series is sometimes taken to be the definition of the exponential function.

### Example 1

Consider the function f(x) = e^{2x} . We can expand it in the form of a Taylor series using the above formula by substituting 2x instead of x . We have that,

e^{2x} = 1 + \frac{2x}{1!} + \frac{(2x)^2}{2!}\ldots+ \frac{(2x)^n}{n!}+\ldots e^{2x} = 1 + \frac{2x}{1!} + \frac{4x^2}{2!}\ldots+ \frac{{2^n}{x^n}}{n!}+\ldots### Derivation of e^x Taylor Series:

We now give a heuristic proof/derivation for the Taylor series formula. We begin by assuming that the exponential function can be expanded in the form of an infinite polynomial as shown below.

e^x = a_0 + a_1x + a_2x^2\ldots+ a_nx^n+\ldotsWe need to determine the coefficients a_i of the above “infinite degree polynomial”.

Putting x=0 on both sides in the above series we get that a_0=1

Let us assume that the above series can be differentiated term by term. Differentiating once on both sides we get,

e^x = a_1 + 2a_2x\ldots+ na_nx^{n-1}+\ldotsOnce again substituting x=0 on both sides, the higher order terms vanish and we get that a_1=1

Similarly on differentiating the series ‘n’ number of times we get,

e^x = n!a_n + (n+1)n..2 a_{n+1}x+ \ldotsOnce again substituting x=0 on both sides, the higher order terms vanish and we get,

a_n = \frac{1}{n!}This “proves” the formula for the Taylor series of e^x. The above proof is merely a heuristic proof and not rigorous because we have simply assumed that the function can be expanded as a polynomial of infinite degree. We also assumed that term by term differentiation of an infinite series is a valid operation.

### e^x Taylor Series about the point x=a:

We have seen above the formula for the taylor series of e^x about the point x=0. The formula for the Taylor series of e^x about the point x=a is given as,

e^x = e^a + \frac{e^a(x-a)}{1!} + \frac{e^a(x-a)^2}{2!}\ldots+ \frac{e^a(x-a)^n}{n!}+\ldots