# Derive the analytic expression of a function by its Taylor expansion

1. Mar 17, 2014

### kexanie

1. The problem statement, all variables and given/known data
Actually this is not from homework. It occurs in my brain this afternoon.

Is it possible to derive the analytic expression of a function by its Taylor series expansion?

For example, given the following expansion, how to derive the analytic expression of it?

f(x) = 1- x / (1!) + (x^2) / (2!) - … + (-1)^n * (x^n) / (n!)

2. Relevant equations

3. The attempt at a solution

2. Mar 17, 2014

### jbunniii

Not necessarily. Indeed, one definition of the exponential function is
$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$
so given the right hand side, you could not "derive" the fact that it is $e^x$. However, given that fact, it's relatively easy to recognize variations such as
$$e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!}$$
or
$$e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!}$$

3. Mar 17, 2014

### HallsofIvy

Staff Emeritus
Given the series, one method of determining the function the series sums to is to derive a differential equation. In this case, given that $f(x)= \sum_{n=0}^\infty x^n/n!$ we can see that this series is convergent for all x. In particular, then, it is uniformly convergent for any closed and bounded interval and so differentiable, term by term, for any x.

That is, for any x, $f'(x)= \sum_{n=0}^\infty nx^{n-1}/n!= \sum_{n=1}^\infty x^{n-1}/(n-1)!$
Letting j= n-1, that is $\sum_{j=0} x^j/j!= f(x)$. So f(x) satisfies the differential equation f'(x)= f(x) and it is easy to show that any solution of that differential equation is of the form $Ce^x$ for some constant C. And since $f(0)= 1+ 0+ 0+ ...= 1$, C= 1 and $\sum_{n=0}^\infty x^n/n!= e^x$

(By the way, what you wrote, $f(x)= 1+ x+ x^2/2!+ \cdot\cdot\cdot+ x^n/n!$ is NOT the infinite sum and is a polynomial.)

4. Mar 17, 2014

### vanceEE

For common taylor series it is possible, if you are fimliar with series. For example, if you were given the sum $\sum\limits_{n=0}^∞ 3(\frac{x}{2})^n = 3 + \frac{3x}{2} + \frac{3x^2}{4} + ...$
You could derive the "analytic expression" as you might recognize this to be a convergent power series on the interval (-2,2). After a bit of algebra you could represent the taylor series as a function: $f(x) = \frac{6}{2-x}$ where -2 < x < 2. But this can only be said for the special/common series; with that being said, many infinite series are better left off as an infinite series :yuck: