# Explanation of taylor series

I understand that for a Taylor series of a function $f(x)$, centered around the point a, the general expression can be written as: \begin{align} &f(x) \\ &= f(a) + f'(a) (x-a) + \frac{f''(a)}{2!} (x - a)^2 + \frac{f^{(3)}(a)}{3!} (x - a)^3 + \dots + \frac{f^{(n)}(a)}{n!} (x - a)^n + \cdots \end{align}￼

I can wrap my head around what the function and meaning of every part of this equation is except for the factorial. Why does each term have to be divided by that order factorial?

• Calculate $f'(x)$, for instance. – TokenToucan May 24 '15 at 3:30
• If you suppose the function $f$ has a power series representation about $a$, so that $f(x) = \sum_{n=0}^{\infty} a_n(x-a)^n$. You can differentiate the power series term by term and solve for $a_n$. – MathNewbie May 24 '15 at 3:43
• the right hand side of the equation is supposed to approximate the left hand side when $x$ is close to $a.$ now what would you want for the constant term to be? , etc. – abel May 24 '15 at 3:54

Remember that the property of a Taylor polynomial $P_n(x)$ is that the values of its first $n$ derivatives match those of $f$ at the point $x=a$.
For instance, taking the derivative of the third-degree Taylor polynomial for $f$, denoted $P_3(x)$,
\begin{align}P_3^\prime(x)&=f^\prime(a)+2\frac{f^{\prime\prime}(a)}{2!}(x-a)+3\frac{f^{\prime\prime\prime}}{3!}(x-a)^2 \\ &= f^\prime(a)+f^{\prime\prime}(a)(x-a)+\frac{f^{\prime\prime\prime}(a)}{2!}(x-a)^2\end{align}
When we evaluate $P_3^\prime(x)$ at $x=a$, we will get $f^\prime(a)$, which is the expected behavior. If we take the derivative a second time, the $(x-a)$ term will disappear and leave $f^{\prime\prime}(a)$, etc. The factorials are necessary to preserve this equality because the power rule causes each term to be multiplied by the power of $(x-a)$. After repeated differentiations, each factorial will disappear one by one and produce an original derivative value of $f$.