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Tightening Derivative Bounds For Taylor Errors

Use the tightest valid derivative bound on the target interval to improve a Lagrange estimate without changing the Taylor polynomial.

Advanced 10 min read

What You'll Learn

  • Identify the interval on which the Lagrange remainder’s derivative must be bounded.
  • Find the maximum or supremum of the absolute derivative on that interval.
  • Convert the sharper derivative bound into a uniform Taylor error estimate.
  • Verify derivative maxima at endpoints and at interior critical points.
  • Distinguish a maximum from a supremum when the interval is not closed.

Why the Derivative Bound Can Be Improved

A Lagrange error estimate depends on a bound for the derivative of order one greater than the Taylor polynomial’s degree. A common first estimate uses a convenient bound that is valid over a large interval. But if the approximation is needed only on a smaller target interval, the derivative may be substantially smaller there.

The improvement is systematic: restrict attention to the interval containing the center and all target points, then find the largest value of the absolute derivative on that interval. This may mean checking endpoints and critical points, rather than using a rough inequality that discards information about the function.

Key idea: The Lagrange remainder only uses values of the \((n+1)\)st derivative between the Taylor center and the point being approximated. A bound for that derivative on the relevant interval is enough; a bound on a larger region is valid but may be unnecessarily large.

The Sharpened Uniform Estimate

Let \(f\) have a continuous \((n+1)\)st derivative on a closed interval \(K\), and let \(a\in K\). Its Taylor polynomial of degree \(n\) centered at \(a\) is

$$ P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k. $$

The Lagrange remainder theorem says that, for each \(x\in K\), there is a point \(\xi_x\) between \(a\) and \(x\) such that

$$ f(x)-P_n(x)=\frac{f^{(n+1)}(\xi_x)}{(n+1)!}(x-a)^{n+1}. $$

The point \(\xi_x\) can depend on \(x\). To obtain a bound that holds for every target point, bound the derivative throughout the interval containing all these intermediate points.

Theorem: Suppose \(J\) is a bounded target interval, \(a\) is the Taylor center, and \(f\) has a continuous \((n+1)\)st derivative on the closed interval \(K\) spanning \(a\) and \(J\). Set \(M=\max_{t\in K}|f^{(n+1)}(t)|\) and \(D=\sup_{x\in J}|x-a|\). Then, for every \(x\in J\), \(|f(x)-P_n(x)|\leq M|x-a|^{n+1}/(n+1)!\leq MD^{n+1}/(n+1)!\). The maximum defining \(M\) exists because the derivative is continuous on the closed bounded interval \(K\). The supremum defining \(D\) is finite because \(J\) is bounded; it need not be attained.

Why the Sharper Estimate Works

The improvement is a direct consequence of the Lagrange remainder formula. Its proof also shows exactly where the derivative bound enters.

1
Start with the remainder formula.
For each fixed \(x\in J\), the Lagrange theorem supplies an intermediate point \(\xi_x\) between \(a\) and \(x\).
2
Apply the interval-specific derivative maximum.
Since \(\xi_x\in K\), we have \(|f^{(n+1)}(\xi_x)|\leq M\). Therefore \(|f(x)-P_n(x)|\leq M|x-a|^{n+1}/(n+1)!\).
3
Make the estimate uniform over the target.
By the definition of \(D\), \(|x-a|\leq D\) for every \(x\in J\). Raising both nonnegative sides to the power \(n+1\) gives \(|x-a|^{n+1}\leq D^{n+1}\), proving the stated uniform bound.

This argument does not require a single intermediate point to work for every \(x\). It only requires that every possible \(\xi_x\) lie in \(K\), where the same value \(M\) bounds the derivative.

If the target interval is open or half-open, its distance from the center might not attain a largest value. In that case, use the supremum \(D\), not a maximum. For example, on \(J=(0,1)\) with \(a=0\), every \(|x-a|\) is less than \(1\), but the supremum is \(1\). The estimate with \(D=1\) remains valid. If the derivative is continuous on the closure of the relevant interval, its maximum there can still be used for \(M\).

Worked Example: Exponential Approximation on a Short Interval

Worked Example: Exponential Approximation on a Short Interval

Estimate the Taylor remainder for \(e^x\) using the degree-two Taylor polynomial centered at \(0\), uniformly for \(x\in[0,\tfrac12]\).

First compute the derivatives: \(f(x)=e^x\), \(f'(x)=e^x\), \(f''(x)=e^x\), and \(f'''(x)=e^x\). At \(0\), all four values are \(1\), so

$$ P_2(x)=1+x+\frac{x^2}{2}. $$

Here \(n+1=3\). On the target interval, \(f'''(t)=e^t\) is increasing, since its derivative is \(e^t>0\). Thus its maximum is attained at \(t=\tfrac12\), and

$$ M=e^{1/2}=\sqrt e,\qquad D=\sup_{0\leq x\leq 1/2}|x|=\frac12. $$

The uniform Lagrange estimate is therefore

$$ |e^x-P_2(x)| \leq \frac{\sqrt e}{3!}\left(\frac12\right)^3 =\frac{\sqrt e}{6\cdot 8} =\frac{\sqrt e}{48} \qquad (0\leq x\leq\tfrac12). $$

As a check on the polynomial, \(P_2(0)=1\), \(P_2'(x)=1+x\), and \(P_2''(x)=1\), so \(P_2'(0)=P_2''(0)=1\), matching the derivatives of \(e^x\) at the center. The derivative maximum used in the bound is indeed \(e^{1/2}\), not the larger value \(e\) that would result from bounding \(e^t\) on \([0,1]\).

Finding the Maximum Carefully

For a differentiable derivative function, maximizing its absolute value generally requires considering endpoints and interior critical points. If \(g\) denotes the relevant derivative, a positive maximum of \(|g|\) at an interior point occurs where \(g\) has a critical point (\(g^{\prime}=0\)), or at an endpoint. Zeros of \(g\) itself are not maxima of \(|g|\) unless \(g\) is identically zero, so compare \(|g|\) at the endpoints and at the critical points of \(g\). When a simple monotonicity argument applies, it can make the check shorter.

Worked Example: A Decreasing Derivative Magnitude

Use the degree-two Taylor polynomial for \(f(x)=1/(1+x)\), centered at \(0\), to bound the remainder on \([0,\tfrac12]\).

Differentiate successively:

$$ f'(x)=-\frac{1}{(1+x)^2},\qquad f''(x)=\frac{2}{(1+x)^3},\qquad f'''(x)=-\frac{6}{(1+x)^4}. $$

At the center, \(f(0)=1\), \(f'(0)=-1\), and \(f''(0)=2\). Hence

$$ P_2(x)=1-x+\frac{2}{2!}x^2=1-x+x^2. $$

On \([0,\tfrac12]\), \(|f'''(t)|=6/(1+t)^4\). Its derivative is \(-24/(1+t)^5<0\), so this absolute derivative decreases throughout the target interval. Its maximum is at \(t=0\): \(M=6/(1+0)^4=6\). Also \(D=\tfrac12\). Therefore

$$ |f(x)-P_2(x)| \leq \frac{6}{3!}\left(\frac12\right)^3 =\frac{6}{6\cdot 8} =\frac18 \qquad (0\leq x\leq\tfrac12). $$

To check the Taylor polynomial, \(P_2'(x)=-1+2x\) and \(P_2''(x)=2\), so \(P_2(0)=1\), \(P_2'(0)=-1\), and \(P_2''(0)=2\), as required. The derivative bound uses the actual maximum on the target interval; a rough estimate that replaces \(1+t\) by a smaller lower bound would give a larger, less useful value.

An Interior Maximum Can Matter

The maximum need not occur at an endpoint. Checking only endpoints is not a valid general method: the absolute value of the relevant derivative can have a larger value at an interior critical point. The next example makes that maximum explicit.

Worked Example: Locating an Interior Derivative Maximum

Let \(f(x)=x^4/6-x^5/15\), and use its degree-two Taylor polynomial at \(0\) to bound the remainder on \([0,1]\).

Differentiate step by step:

$$ f'(x)=\frac{2}{3}x^3-\frac{1}{3}x^4,\qquad f''(x)=2x^2-\frac{4}{3}x^3,\qquad f'''(x)=4x-4x^2=4x(1-x). $$

At \(0\), \(f(0)=f'(0)=f''(0)=0\), so the degree-two Taylor polynomial is \(P_2(x)=0\). On \([0,1]\), \(f'''(x)=4x(1-x)\geq0\), so its absolute value equals \(4x(1-x)\). Its derivative is \(4-8x\), which vanishes at \(x=\tfrac12\). The endpoint values are \(f'''(0)=0\) and \(f'''(1)=0\); at the critical point, \(f'''(\tfrac12)=4(\tfrac12)(\tfrac12)=1\). Thus the maximum is \(M=1\), attained in the interior, and \(D=1\). The uniform estimate is

$$ |f(x)-P_2(x)|\leq\frac{1\cdot 1^3}{3!}=\frac16 \qquad (0\leq x\leq1). $$

As a check, the displayed derivatives follow by differentiating the given \(f\), and \(P_2=0\) has value and first two derivatives equal to \(0\) at the center, matching \(f\). The interior critical point is essential: evaluating \(|f'''|\) only at the two endpoints would give \(0\), which is not a valid derivative bound on the interval.

Common Pitfall: Using a Coarse Bound or an Unattained Maximum

One common mistake is to estimate the derivative on a larger interval than needed and then treat that estimate as if it were the best available. Another is to write “maximum” for a quantity on an interval where it may not be attained. For a bounded target interval \(J\), the safe general quantity for distance is \(\sup_{x\in J}|x-a|\). If \(J\) is closed and bounded, this supremum is a maximum; otherwise it may not be.

Worked Example: Correcting an Unattained Distance Maximum

Suppose the Taylor center is \(a=0\), the target is \(J=(0,1)\), and a valid derivative bound on the relevant interval is \(M=2\). The distance \(|x-a|=x\) has no maximum on \(J\): for every \(x<1\), a larger point in \(J\) has greater distance. Its supremum is \(D=1\).

For a degree-one Taylor polynomial, the uniform estimate is consequently

$$ |f(x)-P_1(x)| \leq \frac{2|x|^2}{2!} =x^2 <1 \qquad (0<x<1). $$

In particular, the non-strict uniform bound \(|f(x)-P_1(x)|\leq1\) holds throughout \(J\). The correction is to use the supremum \(D=1\), not to assert that \(x\) attains a maximum on the open interval. If the target were \([0,1]\), the same distance value would be a maximum instead.

A Practical Procedure

For a sharper Lagrange estimate, first make the region and the quantity being maximized explicit. The relevant derivative order is determined by the Taylor degree, while the target interval determines where its maximum must be found.

1
Set the target and center.
Identify the target interval \(J\), the Taylor center \(a\), and the closed interval \(K\) spanning them.
2
Differentiate to the required order.
For a degree-\(n\) polynomial, compute \(f^{(n+1)}\), then consider \(|f^{(n+1)}|\) on \(K\).
3
Find the largest derivative magnitude.
Check endpoints and interior critical points, or establish monotonicity. Use a supremum if the relevant set is not closed and attainment is not guaranteed.
4
Apply the remainder estimate.
Use \(M|x-a|^{n+1}/(n+1)!\) pointwise, or use \(MD^{n+1}/(n+1)!\) for a uniform bound, with \(D=\sup_{x\in J}|x-a|\).

The sharper derivative maximum makes the Lagrange estimate more informative while leaving the Taylor polynomial and the remainder theorem unchanged. The key is to optimize the bound on the interval that actually matters, and to distinguish carefully between a maximum that is attained and a supremum that is only approached.

Check Your Understanding

Use the definitions and estimates in this tutorial to answer the following questions.

  1. For a degree-\(n\) Taylor polynomial, which derivative must be bounded in the Lagrange remainder estimate?
  2. Why is it valid to bound the derivative only on the interval spanning the center and the target points?
  3. What should replace a maximum distance if the target interval is bounded but open?
  4. For \(f'''(x)=4x(1-x)\) on \([0,1]\), where is the maximum of \(|f'''(x)|\), and what is its value?
  5. Why can checking only the endpoints fail when finding the maximum of an absolute derivative?

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