Why is there a factorial in the denominator of a Taylor series term?

Conditions

• The function is n-times differentiable at the center point a.
• n is a non-negative integer.

Reasoning, step by step

1. Isolate the nth term of the Taylor series: $\frac{f^{(n)}(a)}{n!} (x - a)^n$.
2. Apply the nth derivative operator $\frac{d^n}{dx^n}$ to the term.
3. Factor out the constant coefficient $\frac{f^{(n)}(a)}{n!}$ using the linearity of the derivative.
4. Repeatedly apply the power rule to differentiate $(x - a)^n$ exactly n times.
5. Observe that the resulting coefficient from the polynomial differentiation is the product $n(n-1)\cdots(1)$, which equals $n!$.
6. Multiply the factored-out constant by the new coefficient: $\frac{f^{(n)}(a)}{n!} \cdot n!$.
7. Conclude that the $n!$ terms cancel out, leaving exactly $f^{(n)}(a)$.

Example

The video shows the step-by-step derivation: $\frac{d^n}{dx^n} ((x - a)^n) = n!$, and then substitutes this back to get $\frac{f^{(n)}(a)}{n!} n! = f^{(n)}(a)$.

Common misconceptions

• Assuming the factorial is just a standard formatting convention rather than a necessary mathematical cancellation.
• Thinking that the nth derivative of $(x-a)^n$ is simply $(x-a)^0$ without the accompanying $n!$ coefficient.

Watch the explanation

Explore next

• What happens to the other terms in the Taylor series when taking the nth derivative at $x=a$?
• What is the core intuition behind how Taylor polynomials approximate functions?