# What does the "a" means in the Taylor Series expansion?

Jul 24, 2015

The "$a$" is the number where the series is "centered". There are usually infinitely many different choices that can be made for $a$, though the most common one is $a = 0$.

#### Explanation:

In general, a power series of the form $\setminus {\sum}_{n = 0}^{\setminus \infty} {c}_{n} {\left(x - a\right)}^{n} = {c}_{0} + {c}_{1} \left(x - a\right) + {c}_{2} {\left(x - a\right)}^{2} + {c}_{3} {\left(x - a\right)}^{3} + \setminus \cdots$ is said to be "centered at $x = a$". The reason for this terminology is that a power series like this will converge on some interval centered at $x = a$ (though it might just converge at $x = a$ and it can also converge over the entire real line $\mathbb{R}$, which is not, technically, centered anywhere). In other words, it will converge on some interval with $a$ at the midpoint of the interval (again, excluding the cases where it only converges at $x = a$ or where it converges for all $x \setminus \in \mathbb{R}$).

For example, the real power series $\setminus {\sum}_{n = 0}^{\setminus \infty} \frac{1}{3} ^ \left\{n\right\} {\left(x - 7\right)}^{n} = 1 + \frac{1}{3} \left(x - 7\right) + \frac{1}{9} {\left(x - 7\right)}^{2} + \frac{1}{27} {\left(x - 7\right)}^{3} + \setminus \cdots$ is a geometric series and can be shown to converge for all $x$ in the open interval $\left(4 , 10\right)$. Note that this interval is centered at $a = 7$.

Furthermore, in this last example, you can actually take $x$ to be a complex number and the series converges for all $x \setminus \in \mathbb{C}$ with the property that $| x - 7 | < 3$ (the distance between $x$ and 7 is less than 3), which is an open "disk" (the inside of a circle) in the complex plane centered at 7.

Taylor series are special kinds of power series. Given a function $f$ that is infinitely differentiable over some open interval $\left(c , d\right)$ containing the number $x = a$, you can compute the corresponding Taylor series for $f$ centered at $x = a$ as the following expression:

\sum_{n=0}^{\infty} f^{(n)}(a)/(n!)(x-a)^{n}=f(a)+f'(a)(x-a)+(f''(a))/(2!)(x-a)^2+(f'''(a))/(3!) (x-a)^3+\cdots

As before, this will converge on some interval centered at $x = a$.

For the geometric series example above, it happens to be the Taylor series for $f \left(x\right) = \frac{3}{10 - x}$ centered at $a = 7$.