Taylor Coefficients from Derivatives

From Derivatives to Polynomial Coefficients

A Taylor polynomial records information about a function and its derivatives at one chosen input, called the center. The coefficient of each power of \(x-a\) comes from a particular derivative evaluated at that center. The derivative order determines which coefficient is used, and a factorial adjusts for the repeated differentiation of that power.

This tutorial focuses on calculating those coefficients. A Taylor polynomial can be formed from derivative values at a point, but that fact alone does not assert that the polynomial equals the function everywhere, or even that an infinite series represents the function. Here we work with finite polynomials and the derivatives needed to construct them.

Written term by term, the same definition is

The index \(k\) identifies both the derivative and the power: the \(k\)th derivative supplies the coefficient of \((x-a)^k\). Do not confuse the derivative value \(f^{(k)}(a)\) with the coefficient itself. The coefficient is that value divided by \(k!\).

Why the Factorial Appears

To see where the factorial comes from, consider a polynomial written in powers of \(x-a\): \(P(x)=c_0+c_1(x-a)+c_2(x-a)^2+\cdots+c_n(x-a)^n\). Differentiate a term \((x-a)^j\) exactly \(k\) times and then set \(x=a\). If \(j<k\), that term has already differentiated to zero. If \(j>k\), its \(k\)th derivative still contains a positive power of \(x-a\), so it is zero at \(x=a\). Only the term with \(j=k\) remains: its \(k\)th derivative is the constant \(k!\).

Consequently, for a polynomial written in powers of \(x-a\), its \(k\)th derivative at \(a\) is \(k!c_k\). Solving for the coefficient gives \(c_k=P^{(k)}(a)/k!\). The Taylor-polynomial formula uses the corresponding derivative values of \(f\) to set each coefficient.

A Reliable Calculation Process

Keep the center visible throughout the calculation. In particular, the powers are powers of \(x-a\), not automatically powers of \(x\). A useful process is to write down the center and degree, calculate the derivatives in order, evaluate each at the center, and only then divide by the appropriate factorial.

Worked Examples

Common Mistake: Forgetting the Factorial

A frequent error is to use \(f^{(k)}(a)\) itself as the coefficient of \((x-a)^k\). For example, in the rational-function example, \(f''(1)=1/4\). Writing \(1/4\) as the coefficient of \((x-1)^2\) is incorrect. The second-order coefficient is \(f''(1)/2!=1/8\).

The error is visible by differentiating the incorrect term. If the term were \(\frac14(x-1)^2\), its second derivative would be \(2(\frac14)=\frac12\), not the required value \(f''(1)=\frac14\). With the corrected coefficient, the second derivative is \(2(\frac18)=\frac14\), as required.

What These Coefficients Do—and Do Not—Tell You

The coefficient calculations encode the function’s value and successive derivative values at the center. In particular, the degree-zero term gives the function value there; the degree-one coefficient is the first derivative there; and higher coefficients require the corresponding factorial adjustment.

These facts describe how to construct the finite Taylor polynomial. They do not, by themselves, prove that the polynomial equals \(f(x)\) away from \(a\), or determine how close its values are to the function. Those are separate questions. For this calculation, the essential requirements are that the needed derivatives exist at the center and that each derivative is evaluated and divided by the correct factorial.