Taylor Polynomials and Remainder Estimates (BC)

Local Approximation from Derivatives

A function’s value and derivatives at one point describe its local behavior. A Taylor polynomial uses that information to build a polynomial that agrees with the function and several of its derivatives at a chosen center. Near that center, the polynomial can provide a useful approximation to the function.

An approximation is not complete without a way to assess its accuracy. Taylor’s theorem supplies a remainder formula, and bounding the relevant derivative throughout the interval between the center and the target gives a guaranteed error bound.

The polynomial’s coefficients come from derivative values at the center, not at the target value. For instance, the constant term is \(f(a)\), the coefficient of \(x-a\) is \(f'(a)\), and the coefficient of \((x-a)^2\) is \(f''(a)/2!\). The factorials ensure that the polynomial’s derivatives have the intended values at the center.

Why the Coefficients Have This Form

Suppose a polynomial centered at \(a\) is written as \(Q_n(x)=c_0+c_1(x-a)+\cdots+c_n(x-a)^n\). At \(x=a\), every positive power of \(x-a\) vanishes, so \(Q_n(a)=c_0\). Differentiating once and evaluating at \(a\) gives \(Q_n'(a)=c_1\). Differentiating twice gives \(Q_n''(a)=2!c_2\), and, in general, differentiating \(k\) times gives \(Q_n^{(k)}(a)=k!c_k\). Therefore, to make \(Q_n^{(k)}(a)=f^{(k)}(a)\), its coefficient must be \(c_k=f^{(k)}(a)/k!\). This establishes the stated formula and explains why the factorials appear.

Taylor’s Theorem and the Error Bound

The difference between \(f(x)\) and its degree-\(n\) Taylor polynomial is the remainder \(R_n(x)=f(x)-P_n(x)\). Taylor’s theorem describes that difference exactly using a derivative of order \(n+1\) at an intermediate point. The intermediate point is generally not known, but a bound on that derivative over the whole interval gives a usable estimate.

The hypothesis on the derivative matters: \(M\) must bound the absolute value of the derivative throughout the interval between the center and the target. It is not enough to evaluate the derivative at \(a\) alone unless that value is known to bound the derivative everywhere on the interval.

Why Taylor’s Remainder Formula Works

Here is a proof of the theorem’s remainder formula. Fix \(x\ne a\), and write \(K=R_n(x)/(x-a)^{n+1}\). Define \(\phi(t)=f(t)-P_n(t)-K(t-a)^{n+1}\) on the interval with endpoints \(a\) and \(x\). By the definition of \(K\), \(\phi(x)=0\). Also, \(\phi(a)=0\), and its derivatives of orders \(1\) through \(n\) vanish at \(a\): the polynomial \(P_n\) matches the derivatives of \(f\) there, while the added power \((t-a)^{n+1}\) and its first \(n\) derivatives vanish there.

Applying Rolle’s theorem repeatedly to these zeros gives a point \(\xi\) between \(a\) and \(x\) where \(\phi^{(n+1)}(\xi)=0\). Since \(P_n\) has degree \(n\), its derivative of order \(n+1\) is zero. The derivative of order \(n+1\) of \(K(t-a)^{n+1}\) is \(K(n+1)!\). Thus $$ 0=\phi^{(n+1)}(\xi)=f^{(n+1)}(\xi)-K(n+1)!. $$ It follows that \(K=f^{(n+1)}(\xi)/(n+1)!\). Substituting the definition of \(K\) gives the remainder formula. Taking absolute values and using \(\left|f^{(n+1)}(\xi)\right|\le M\) gives the error bound. If \(x=a\), the remainder is zero directly.

A Reliable Calculation Process

In an approximation problem, identify the center, degree, and target before calculating. Then find the needed derivatives and evaluate them at the center. After constructing the polynomial, use the target value in it and separately establish the derivative bound required by Taylor’s theorem.

Worked Examples

Common Pitfall: Using the Derivative at the Center as the Bound

A Taylor remainder bound requires a bound for the next derivative across the interval, not simply its value at the center. For example, in the approximation of \(e^{0.2}\), the fourth derivative at zero is \(1\). Substituting that value directly for \(M\) would give $$ |R_3(0.2)|\le\frac{1(0.2)^4}{24}=0.0000666\ldots . $$ That estimate is not justified: the fourth derivative is \(e^t\), which increases above \(1\) for \(t>0\). A valid choice is \(M=1.23\), since \(e^t<1.23\) throughout \([0,0.2]\). The resulting justified bound is \(0.000082\).

A second useful check is to keep the center visible in every term: powers are \((x-a)^k\), not generally \(x^k\). This is especially important when the Taylor polynomial is not centered at zero. The derivatives must also be evaluated at that same center.

Using the Result Effectively

The polynomial gives the requested approximation; the remainder estimate explains what the approximation guarantees. A small value of \(|x-a|\) helps because the error bound contains \(|x-a|^{n+1}\), but the derivative bound \(M\) and the factorial \((n+1)!\) also affect the result. Increasing the degree does not by itself justify a particular error bound: the next derivative still needs to be bounded on the relevant interval.

When stating a conclusion, distinguish the approximation from its accuracy statement. For example, report both \(P_n(x)\) and the inequality \(|f(x)-P_n(x)|\le\text{bound}\). In an AP free-response solution, showing the derivative bound and the interval on which it holds makes clear why the stated accuracy is guaranteed.