Calculus I · Building a Function From Its Derivatives

Match the curvature: the quadratic approximation

Bending is what the second derivative measures. So let’s look for a parabola that matches $f$ not in two ways but in three — value, slope, and curvature at $a$. Write the unknown parabola in powers of $(x-a)$, which makes everything easy to evaluate at $a$:

$$P(x) = c_0 + c_1(x-a) + c_2(x-a)^2.$$

Now force the three matches, computing each derivative and plugging in $x = a$ (where every $(x-a)$ becomes $0$):

• Value: $P(a) = c_0$, so matching $f(a)$ forces $c_0 = f(a)$.
• Slope: $P'(x) = c_1 + 2c_2(x-a)$, so $P'(a) = c_1$, forcing $c_1 = f'(a)$.
• Curvature: $P''(x) = 2c_2$, so $P''(a) = 2c_2$, forcing $c_2 = \dfrac{f''(a)}{2}$.

So the best quadratic is

$$P(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2.$$

It contains the tangent line (the first two terms) and adds one bending term. It hugs the curve better, and over a wider window. (Figure 7.2: the same curve with its tangent and its matching parabola; the parabola tracks the curve far past where the line gave up.)

The pattern is already visible — keep matching higher derivatives, keep adding terms — so let’s find the general rule, and the factorial it hides.

The general pattern, and why the factorial is forced

Suppose we want a degree-$n$ polynomial matching $f$ and its first $n$ derivatives at $a$:

$$P(x) = \sum_{k=0}^{n} c_k\,(x-a)^k.$$

To find $c_k$, differentiate $P$ exactly $k$ times and evaluate at $a$. The key is what happens to a single power $(x-a)^j$ when you hit it with $k$ derivatives:

• If $j < k$: you differentiate it down to a constant and then past it — it becomes $0$.
• If $j > k$: after $k$ derivatives it still carries a factor of $(x-a)$, which is $0$ at $x = a$.
• If $j = k$: differentiating $(x-a)^k$ exactly $k$ times peels off $k, k-1, \ldots, 1$ in turn, leaving the constant $k! = k\cdot(k-1)\cdots 2\cdot 1$.

So when you compute $P^{(k)}(a)$, every term dies except the $k$-th, which leaves $c_k \cdot k!$. Setting that equal to $f^{(k)}(a)$:

$$P^{(k)}(a) = c_k\,k! = f^{(k)}(a) \qquad\Longrightarrow\qquad \boxed{\;c_k = \frac{f^{(k)}(a)}{k!}\;}$$

There’s your factorial, and now you know why it’s there: it’s the $k!$ that falls out of differentiating $(x-a)^k$ down to the ground. We give the result a name.

The Taylor polynomial of degree $n$ for $f$ centered at $a$ is $$T_n(x) = \sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n.$$ Centered at $a = 0$ it’s often called a Maclaurin polynomial. Let $n\to\infty$ and (where it’s valid — see the remainder, below) you get the Taylor series.

Toolkit: Building a Taylor polynomial

1. Compute $f(a), f'(a), f''(a), \ldots, f^{(n)}(a)$ — the derivatives at the center.
2. Divide the $k$-th one by $k!$.
3. Assemble: $\sum_k \frac{f^{(k)}(a)}{k!}(x-a)^k$. Look for a pattern in the coefficients — the best functions have one.

All 12 chapters of Calculus I · Foundations · All courses