Deriving Coefficients by Matching Derivatives

Why Derivatives Determine the Coefficients

A Taylor polynomial is designed to agree with a function near a chosen center \(a\). One way to impose that agreement is to require the polynomial and function to have the same value, slope, and higher derivatives at \(a\). These conditions do more than make the polynomial resemble the function: they determine each coefficient in turn.

The key is to write the polynomial using powers of \(x-a\). At the center, those powers have a useful property: differentiating \(k\) times makes the \(k\)th power contribute a nonzero value, while the other powers contribute zero. The size of that contribution is \(k!\), which explains the factorial in the Taylor coefficient formula.

Set Up the Matching Conditions

Let \(n\) be a nonnegative integer, let \(a\) be a real number, and suppose \(f\) has derivatives through order \(n\) at \(a\). A polynomial of degree at most \(n\), written in powers of \(x-a\), has the form

To match the function’s derivatives through order \(n\), require that \(P_n^{(k)}(a)=f^{(k)}(a)\) for each integer \(k\) from \(0\) to \(n\). The order-zero condition means \(P_n(a)=f(a)\). For positive \(k\), the condition matches the \(k\)th derivatives at the center.

The phrase “degree at most \(n\)” allows some of the highest coefficients to be zero. The derivative-matching setup requires only that the relevant derivatives of \(f\) exist at \(a\); it does not require that \(f\) equal this polynomial away from the center.

How the Shifted Powers Behave

Consider one term \((x-a)^j\) and differentiate it \(k\) times. If \(k\leq j\), repeated application of the power rule gives

If \(k>j\), the derivative is zero because differentiating a degree-\(j\) polynomial more than \(j\) times gives zero. Now evaluate at \(x=a\). When \(j>k\), the remaining power has positive exponent and vanishes. When \(j=k\), no power remains, and the product of factors is \(k!\). When \(j<k\), the derivative is already zero.

Here \(0!=1\), so the statement includes \(k=0\): the zeroth derivative is the original term, and \((x-a)^j\) evaluated at \(a\) is \(1\) if \(j=0\) and \(0\) otherwise.

Deriving the Coefficient Formula

Differentiate \(P_n(x)=\sum_{j=0}^{n}c_j(x-a)^j\) exactly \(k\) times and evaluate at \(a\). Differentiation and evaluation apply term by term because the sum is finite. By the behavior of shifted powers, every term disappears except the one whose exponent is \(k\). Thus \(P_n^{(k)}(a)=k!c_k\). The matching condition says this value must be \(f^{(k)}(a)\), so solving for \(c_k\) gives the result.

This argument also explains the role of each derivative: the value \(f^{(k)}(a)\) is not itself the coefficient in general. It equals \(k!\) times that coefficient because the \(k\)th derivative of \((x-a)^k\), evaluated at \(a\), is \(k!\). For \(k=0\), the formula reads \(c_0=f(a)\), as expected from matching the function’s value.

Worked Examples

A Common Factorial Mistake

A frequent error is to use \(f^{(k)}(a)\) directly as the coefficient of \((x-a)^k\). That ignores the factor produced by differentiating the power \(k\) times. For example, let \(f(x)=x^2\) at \(a=0\), and suppose we try \(P_2(x)=c_0+c_1x+c_2x^2\). Since \(f''(0)=2\), the incorrect choice \(c_2=2\) would give \(P_2(x)=2x^2\). But

The correction is to solve the matching equation \(P_2''(0)=2!c_2=f''(0)=2\). Thus \(c_2=2/2!=1\), giving \(P_2(x)=x^2\), whose second derivative at zero is indeed \(2\). The factorial is therefore required for the derivative-matching condition to hold.

Why the Derivation Matters

The coefficient formula is not an arbitrary convention to memorize. It follows from the way derivatives act on powers centered at \(a\). Expressing the polynomial in powers of \(x-a\) makes each matching condition isolate exactly one coefficient. This gives a systematic route from derivative information to a polynomial model.

It also provides a practical check on calculations: after forming a Taylor polynomial, differentiate it and evaluate at the center. Its derivatives of orders \(0\) through \(n\) should reproduce the corresponding derivative values of the function. A mismatch usually indicates an incorrect derivative, an omitted factorial, or a mistake in evaluating at the center.

Derivative matching describes how the polynomial behaves at the center; by itself, it does not state how accurate the polynomial is at points away from the center. Questions about approximation error require additional results. Here the conclusion is specifically about the coefficients needed to match the prescribed derivatives.