From a Linear Estimate to an Error Guarantee
A linearization gives a tangent-line estimate near a base point \(a\). But how accurate is that estimate across a particular interval? A bound on the second derivative answers this by placing a guaranteed limit on the absolute error.
The important detail is that the bound must apply throughout the interval between the base point and the input being estimated. The second derivative at the base point alone does not generally provide that guarantee.
The Error Bound
Let \(f\) be a function with a continuous second derivative on an interval containing \(a\) and \(x\). Its linearization at \(a\) is
The absolute error at \(x\) is the distance between the function value and the linear estimate:
If the estimate is needed for every \(x\) in an interval extending at most \(h\) units from \(a\), and the same bound \(|f''(t)|\le M\) holds throughout that interval, then
Thus the maximum absolute error over the interval is no greater than \(\frac{M}{2}h^2\). This is a guaranteed upper bound, not necessarily the exact maximum error.
Why the Bound Works
The error-bound theorem follows from the integral form of the remainder. Define \(g(t)=f(t)-L_a(t)\). Since \(L_a(a)=f(a)\) and \(L_a'(a)=f'(a)\), the difference satisfies \(g(a)=0\) and \(g'(a)=0\). Also, \(g''(t)=f''(t)\).
For \(x\ge a\), applying the Fundamental Theorem of Calculus twice gives
The last equality changes the order of integration: for each \(t\) from \(a\) to \(x\), the variable \(u\) runs from \(t\) to \(x\), producing the factor \(x-t\). Taking absolute values and using \(|f''(t)|\le M\),
For \(x<a\), the same argument integrates from \(x\) to \(a\); the nonnegative weight in the error integral is \(t-x\), whose integral is \(\frac12(a-x)^2\). Since \((a-x)^2=(x-a)^2\), the same bound follows. This explains both the factor \(\frac12\) and why the distance from the base point is squared.
Worked Example: Bounding the Error for a Square Root
Worked Example: A Square-Root Estimate Near 4
Use the linearization of \(f(x)=\sqrt{x}\) at \(a=4\) to bound the absolute error for \(3\le x\le5\).
First, \(f(4)=2\) and \(f'(x)=\frac{1}{2\sqrt{x}}\), so \(f'(4)=\frac14\). Therefore,
Next, \(f''(x)=-\frac{1}{4x^{3/2}}\), so \(|f''(x)|=\frac{1}{4x^{3/2}}\). On \([3,5]\), this magnitude is largest at \(x=3\), giving
Every point in the interval is at most \(h=1\) unit from \(a=4\). The error bound is therefore
For a check at the endpoint \(x=5\), the linear estimate is \(L_4(5)=2+\frac14=\frac94\), while \(f(5)=\sqrt5\). The absolute error is \(\left|\sqrt5-\frac94\right|\approx0.0139\), which is less than \(0.0241\). The check is consistent with the guaranteed bound.
Comparing Step Sizes
When the same curvature bound \(M\) applies, the error bound varies with the square of the step size. If the step size is halved, the bound is multiplied by \((\frac12)^2=\frac14\). If the step size is doubled, the bound is multiplied by \(2^2=4\).
When the curvature bounds differ across the intervals being compared, compare the full quantities \(\frac12 M h^2\). A smaller step does not by itself guarantee a smaller calculated bound if the second-derivative bound is much larger on the smaller interval.
Worked Example: Comparing Two Step Sizes
For \(f(x)=x^2\), compare error bounds for linearization at \(a=2\) when estimating throughout \([1.6,2.4]\) and \([1.8,2.2]\).
Since \(f(2)=4\), \(f'(x)=2x\), and \(f'(2)=4\), the linearization is
Also, \(f''(x)=2\), so \(M=2\) works on both intervals. For the first interval, \(h=0.4\), and the bound is
For the second interval, \(h=0.2\), and the bound is
The step size was halved, and the bound fell to one quarter: \(0.04=0.16/4\). To check the estimates directly, \(f(x)-L_2(x)=x^2-[4+4(x-2)]=(x-2)^2\). Its largest value on the first interval is \(0.4^2=0.16\), and on the second it is \(0.2^2=0.04\). In this example, the bounds equal the actual maximum absolute errors.
Worked Example: A Curvature Bound for a Logarithm
Worked Example: Estimating \(\ln x\) Near 1
Use the linearization of \(f(x)=\ln x\) at \(a=1\) to bound the error for \(0.9\le x\le1.1\).
Because \(f(1)=0\) and \(f'(x)=1/x\), the derivative at the base point is \(f'(1)=1\). Thus
The second derivative is \(f''(x)=-1/x^2\), so \(|f''(x)|=1/x^2\). This is largest on \([0.9,1.1]\) at \(x=0.9\), and
The largest distance from the base point is \(h=0.1=\frac{1}{10}\). Hence
Check at \(x=0.9\): \(L_1(0.9)=-0.1\), and \(\ln(0.9)\approx-0.105361\). The absolute error is approximately \(0.005361\), below \(0.006173\). At \(x=1.1\), the estimate is \(0.1\) and \(\ln(1.1)\approx0.095310\), giving an absolute error of approximately \(0.004690\), also below the bound.
A Common Mistake: Using Only the Curvature at the Base Point
A tempting shortcut is to substitute \(|f''(a)|\) for \(M\) without checking the rest of the interval. That number describes curvature at one point; it need not bound the curvature everywhere the estimate is being used.
Worked Example: Why a Point Value May Fail
Consider \(f(x)=e^x\), linearized at \(a=0\), for \(0\le x\le1\). Since \(f(0)=1\) and \(f'(0)=1\), \(L_0(x)=1+x\). Also, \(f''(x)=e^x\).
The incorrect calculation uses \(f''(0)=1\) as though it were a bound on the whole interval. At \(x=1\), it would claim
But \(e-2\approx0.7183\), which is greater than \(0.5\); the claimed bound fails. The correction is to bound the second derivative throughout \([0,1]\). There, \(e^x\le e\), so \(M=e\) is valid and the theorem gives
At \(x=1\), the actual error \(e-2\approx0.7183\) is below this corrected bound. The upper bound is less tight, but it is guaranteed.
Using the Bound Carefully
A reliable estimate requires four choices: the base point, the interval of inputs, a valid upper bound \(M\) for \(|f''|\) on that interval, and the greatest distance \(h\) from the base point. The resulting value \(\frac12Mh^2\) is an error guarantee for all the inputs considered.
- Use the magnitude \(|f''|\) when finding \(M\); the bound concerns absolute error.
- Check the entire interval, including the part between the base point and each target input.
- Keep \(M\) and \(h\) paired with the interval on which they were determined.
- Interpret the result as an upper bound, not automatically as the actual error or exact maximum.
This estimate is useful when a linear approximation is convenient but its accuracy must be assessed. It converts information about how rapidly the derivative changes into a quantitative guarantee about how far the function can depart from its tangent-line estimate.
Check Your Understanding
Use the second-derivative bound and show the interval and curvature bound you use in each calculation.
- If \(|f''(x)|\le 6\) between \(a\) and \(a+0.1\), what upper bound does the theorem give for the absolute linearization error at \(a+0.1\)?
- If a valid curvature bound \(M\) stays the same and the step size changes from \(h\) to \(3h\), by what factor does the error bound change?
- For \(f(x)=x^2\) linearized at \(a=1\), use \(M=2\) to bound the error when \(|x-1|\le0.3\).
- Why is \(|f''(a)|\) not automatically a valid choice for \(M\) on an interval?
- If two intervals have different curvature bounds \(M_1,M_2\) and step sizes \(h_1,h_2\), what quantities should be compared?
