When Should You Trust a Linear Estimate?
A linear estimate replaces a function near a chosen base point with the tangent-line expression. The estimate is designed for inputs near that point, but “near” by itself does not guarantee a useful answer. A small input displacement can still produce a substantial error if the function curves sharply in the interval.
To assess an estimate, examine three things together: the displacement from the base point, the function’s curvature throughout the interval, and the accuracy required for the task. A second-derivative bound turns these checks into a quantitative test.
The Estimate and Its Error
Suppose \(f\) is differentiable at a base point \(a\), and the target input is \(x\). The linear estimate of \(f(x)\) based at \(a\) is
The quantity \(h=x-a\) is the input displacement. The estimate uses the function’s value and slope at \(a\), so it captures the tangent-line behavior there. Its error at \(x\) is the difference between the true value and the estimate:
The absolute error is \(\lvert E(x)\rvert\). Whether that error matters depends on the purpose of the estimate. An error of \(0.1\) may be acceptable when estimating a quantity of \(10{,}000\), but unacceptable when the required precision is \(0.01\).
A Second-Derivative Error Bound
The second derivative measures how the slope changes. If the second derivative remains small over the interval from \(a\) to \(x\), the tangent line has limited opportunity to separate from the function. This intuition is made precise by the following result.
The bound depends on the square of the displacement and on a bound for the magnitude of the curvature over the whole segment. In particular, checking \(f''\) only at the base point does not establish the required bound.
Why the Bound Works
Here is a derivation. If \(x=a\), the estimate equals \(f(a)\), so the error is zero. Now suppose \(x\ne a\), and write \(h=x-a\). Set
Define an auxiliary function on the segment between \(a\) and \(x\):
By construction, \(\phi(a)=0\) and \(\phi(x)=0\). Also, differentiating gives \(\phi'(a)=0\). Rolle’s theorem applied to \(\phi\) gives a point between \(a\) and \(x\) where \(\phi'\) is zero. Together with \(\phi'(a)=0\), a second application of Rolle’s theorem to \(\phi'\) gives some \(\xi\) strictly between \(a\) and \(x\) for which \(\phi''(\xi)=0\). Since
we have \(K=f''(\xi)/2\). Substituting the definition of \(K\) and using \(\lvert f''(\xi)\rvert\le M\) yields
This proves the bound. It also explains why curvature matters: the error is the displacement squared multiplied by half a curvature value somewhere in the interval.
Worked Example: A Close Estimate with a Small Error Bound
Worked Example: Estimating a Square Root Near 100
Estimate \(\sqrt{102}\) by linearizing \(f(x)=\sqrt{x}\) at \(a=100\), then use a second-derivative bound to assess the estimate.
First, \(f(100)=10\), and \(f'(x)=1/(2\sqrt{x})\), so \(f'(100)=1/20=0.05\). The input displacement is \(h=102-100=2\). Thus
For \(x>0\), \(f''(x)=-1/(4x^{3/2})\). On the interval \([100,102]\), its magnitude is largest at \(100\), so we can take \(M=1/(4\cdot100^{3/2})=1/4000=0.00025\). The error bound is therefore
Indeed, \(\sqrt{102}\approx10.099504938\), so the actual absolute error is approximately \(\lvert10.099504938-10.1\rvert=0.000495062\), which is below \(0.0005\). As a direct check on the estimate’s position, \(10.1^2=102.01>102\), consistent with \(10.1\) being slightly greater than \(\sqrt{102}\). The bound confirms that the estimate is accurate to within \(0.0005\).
Worked Example: A Small Displacement Is Not Enough
Worked Example: Estimating Near a Region of High Curvature
Consider \(f(x)=1/(1-x)\), based at \(a=0.90\), and estimate its value at \(x=0.91\). Although the input changes by only \(0.01\), the function is steep and curved in this region.
At the base point, \(f(0.90)=1/0.10=10\). Since \(f'(x)=1/(1-x)^2\), we have \(f'(0.90)=100\). The displacement is \(h=0.91-0.90=0.01\), giving
The exact value is \(f(0.91)=1/0.09=100/9\). Therefore the actual error is
Now \(f''(x)=2/(1-x)^3\), which increases on \([0.90,0.91]\). Its maximum there is \(2/(0.09)^3=2{,}000{,}000/729\). Hence the theorem gives
The actual error \(1/9=81/729\) is below the bound \(100/729\), as required. The displacement is about \(1.11\%\) of the base point, but the actual error is about \(1\%\) of the true value \(100/9\). Whether that is acceptable depends on the task. The example shows why a seemingly small input change cannot, by itself, certify accuracy.
Worked Example: A Bound Must Cover the Whole Interval
Worked Example: Why Curvature at the Base Point Can Mislead
Estimate \(e^1\) using \(f(x)=e^x\) based at \(a=0\). Here \(f(0)=1\), \(f'(0)=1\), and \(h=1\), so
The exact value is \(e\approx2.718282\), so the actual absolute error is \(e-2\approx0.718282\). Because \(f''(x)=e^x\), the largest value of \(\lvert f''(x)\rvert\) on \([0,1]\) is \(e\). Taking \(M=e\) gives
The actual error is less than this bound. However, using only \(f''(0)=1\) would give the proposed bound \(0.5\), which is false because \(0.718282>0.5\). The correction is to bound \(\lvert f''(t)\rvert\) over the entire segment, not to substitute its base-point value without checking that it bounds the curvature elsewhere.
A Practical Reliability Check
An error bound is useful only when compared with a stated accuracy requirement. If the task allows an absolute error of at most \(\tau\), the estimate is certified to meet that requirement whenever the hypotheses hold and
If the bound is larger than \(\tau\), that does not prove the estimate fails; the bound may be conservative. It means only that this bound cannot certify the required accuracy. The actual error might be smaller, as the examples show.
A useful diagnostic sequence is:
Compute the actual displacement \(h=x-a\). The linear estimate is local to this point, not to an unspecified nearby region.
Identify the segment joining \(a\) and \(x\). Confirm that the function is twice differentiable there and that no point on the segment invalidates the curvature bound.
Find a value \(M\) that satisfies \(\lvert f''(t)\rvert\le M\) throughout the segment. A bound based only on \(f''(a)\) is not enough unless it is known to bound the whole interval.
Calculate \(M|h|^2/2\) and compare it with the task’s allowable error. If the result is too large, the estimate may be unsuitable or the bound may simply be inconclusive.
Common Pitfall: Treating “Near” as a Guarantee
A frequent mistake is to argue, “The input changed by only a small amount, so the linear estimate must be accurate.” This omits the function’s curvature and the scale of accuracy needed. The example involving \(1/(1-x)\) had a displacement of only \(0.01\), yet its error was about \(0.1111\).
The second common mistake is to evaluate the second derivative only at the base point and use that value as \(M\). The exponential example shows the problem: \(f''(0)=1\), but \(f''(t)\) grows to \(e\) on \([0,1]\). The incorrect bound \(0.5\) does not cover the actual error. A valid bound must hold over every point between the base point and the target input.
Check Your Understanding
Use the stated conditions in each question; distinguish a guaranteed conclusion from one that is only plausible.
- For an estimate based at \(a\), what is the input displacement, and which interval must a second-derivative bound cover?
- If \(\lvert f''(t)\rvert\le 6\) between \(a=2\) and \(x=2.1\), what error bound follows?
- Why does checking only the value \(f''(a)\) generally fail to establish the error bound?
- If the second-derivative error bound exceeds a required tolerance, has the estimate necessarily failed? Explain.
- Why can a small input displacement still lead to a noticeable error?
