Choosing Taylor Degree From a Factorial Bound

From an Error Tolerance to a Degree

A Taylor polynomial approximates a function near a center \(a\). The degree controls how many terms are included, but a larger degree is useful only if it gives a sufficiently small error bound at the point of interest. A practical question is therefore: what is the least degree that guarantees an error no larger than a prescribed tolerance?

When the Lagrange remainder theorem applies and the relevant derivative is bounded by a known constant \(M\), the error after a degree-\(n\) Taylor polynomial is bounded by a factorial expression. Selecting a degree means finding the first nonnegative integer \(n\) for which that bound is at most the tolerance. The word “first” matters: a degree that works is not necessarily the smallest degree that works.

Here \(d\) is the distance from the Taylor center to the evaluation point. The factorial is \((n+1)! = 1\cdot 2\cdots(n+1)\). If \(M\) is positive and \(d=0\), the evaluation point is the center and every bound is zero, so the least allowed degree is \(0\). If \(M=0\), the bounds are also zero. The interesting selection problem has \(M>0\), \(d>0\), and \(\varepsilon>0\).

Why the Degree Test Works

The Lagrange remainder theorem says that, when its hypotheses hold on the interval joining \(a\) and \(x\), the error after the degree-\(n\) Taylor polynomial \(T_n\) satisfies $$ |f(x)-T_n(x)|\leq \frac{M|x-a|^{n+1}}{(n+1)!}. $$ Thus \(B_n\leq\varepsilon\) guarantees \(|f(x)-T_n(x)|\leq\varepsilon\). It is a sufficient guarantee: the actual error can be smaller than the bound, but the bound alone does not promise that it is.

For a fixed \(M\) and \(d\), successive bounds are easy to compute without evaluating a new power and factorial from scratch. Dividing consecutive expressions gives $$ \frac{B_{n+1}}{B_n} = \frac{M d^{n+2}}{(n+2)!} \cdot \frac{(n+1)!}{M d^{n+1}} = \frac{d}{n+2}, $$ when \(M,d>0\). Therefore, $$ B_{n+1}=B_n\frac{d}{n+2}. $$ This recurrence is the main computational tool.

It also explains an important qualification: the bounds need not decrease from the start. If \(d>n+2\), then \(B_{n+1}>B_n\); if \(d=n+2\), they are equal; and if \(d<n+2\), they decrease at that step. Since \(d/(n+2)\) gets smaller as \(n\) grows, the bounds eventually decrease. In fact, once \(n+2\geq 2d\), the next bound is at most half the current one. Consequently, for positive \(M,d,\varepsilon\), the bounds eventually fall below the tolerance.

This procedure is a discrete search, not an ordinary algebraic rearrangement. The unknown integer appears in both a power and a factorial, so there is generally no useful elementary formula that directly solves for \(n\). The recurrence makes the search efficient and keeps the minimality check visible.

A Short Justification of Minimality and Termination

Suppose the bounds are calculated in order \(B_0,B_1,B_2,\ldots\), and the first one at or below \(\varepsilon\) is \(B_N\). By the meaning of “first,” each value \(B_k\) for \(0\leq k<N\) was checked and found to satisfy \(B_k>\varepsilon\). Therefore \(N\) works, and every smaller allowed degree fails the stated bound test. No claim that the bounds decrease at every step is needed for this reasoning.

For completeness, the search does terminate in the nontrivial case \(M,d,\varepsilon>0\). For all sufficiently large \(n\), \(d/(n+2)\leq 1/2\). From that point onward the recurrence gives \(B_{n+1}\leq B_n/2\). Repeating this inequality makes the bound as small as desired, so eventually \(B_n\leq\varepsilon\). This proves that the step-by-step selection reaches a passing degree.

Using the Recurrence at a Larger Distance

When \(d\) is greater than one, early bounds can rise. It is especially important in this situation not to assume that testing one degree tells you whether the next degree will be better. The recurrence gives the exact comparison at each step.

This example illustrates why the search should be organized by integer degree rather than by an assumption that each new bound is smaller. Even when early bounds rise, trying the degrees in order correctly identifies the first one that satisfies the tolerance.

Verifying a Candidate Degree

It is useful to show both that the proposed degree passes and that the immediately preceding degree fails. When bounds are known to decrease throughout the relevant range, these two checks establish that no lower degree passes. If the sequence may have risen earlier, check those earlier degrees too, or use the recurrence to document all of them. A passing bound by itself establishes sufficiency, not minimality.

Common Pitfall: Confusing a Passing Degree With the Smallest Degree

A frequent error is to find some degree whose bound passes and then call it the minimum without checking smaller degrees. For example, in the previous calculation, degree \(6\) also passes: its bound is $$ B_6=B_5\frac{2}{8}=\frac4{45}\cdot\frac14=\frac1{45}\approx0.0222\leq0.1. $$ But this does not make degree \(6\) the smallest suitable degree. Degree \(5\) already passes, and its bound \(4/45\) is also at most \(0.1\). The correction is to begin at degree zero and stop at the first passing value, or otherwise verify every lower degree.

A related mistake is to assert that increasing the degree always decreases the Lagrange bound. The ratio formula disproves that claim whenever \(d>n+2\). For instance, with \(d=3\), \(B_1/B_0=3/2>1\). The safe method is to use the actual recurrence and test the degrees in order.

Conditions to Keep in View

The degree search guarantees an error tolerance only when the remainder bound itself is valid. The constant \(M\) must bound the absolute value of the required derivative throughout the interval between the Taylor center and evaluation point. The function must also meet the differentiability conditions needed for the Lagrange remainder theorem. A numerical result from the factorial expression cannot compensate for a derivative bound that does not apply.

Also check whether the same \(M\) is genuinely valid for all candidate degrees being compared. In this tutorial’s fixed-bound setup, it is. In a problem where the derivative bound changes with the order, the expression to test may instead be \(M_n d^{n+1}/(n+1)!\). Then calculate the appropriate bound for each degree; the fixed-\(M\) recurrence does not apply without modification.

Choosing a degree this way turns a qualitative instruction—“make the Taylor approximation accurate enough”—into a verifiable decision. The computation identifies a degree, the remainder theorem explains what the bound guarantees, and checking the lower degrees establishes that the chosen degree is the least one certified by this bound.