From an Approximation to a Guarantee
A linearization gives a convenient estimate of a function near a chosen point. But an estimate alone does not tell us how far the function value could be from that estimate. A bound on the second derivative supplies a quantitative guarantee: if the curvature stays controlled across the interval of interest, then the linearization error cannot exceed a specific amount.
The interval matters. It must include every point between the linearization point and the input being estimated. Checking the second derivative only at the linearization point does not, in general, establish a valid error bound.
The Linearization Error Bound
At a point \(a\), the linearization of \(f\) is the function \(L\) defined by \(L(x)=f(a)+f'(a)(x-a)\). At the input \(a+h\), its error is the difference between the actual function value and the linear estimate, \(f(a+h)-L(a+h)\). The absolute value measures the size of that error without regard to its sign.
The interval in the condition is \([\min(a,a+h),\max(a,a+h)]\). This formulation works whether \(h\) is positive or negative. The right-hand side uses \(h^2\), so the bound depends on the size of the input change, not its direction. If \(h=0\), the error is zero.
The number \(M\) must be an upper bound for \(|f''|\) throughout the interval, not merely a value of \(|f''|\) at one selected point. A larger valid \(M\) still gives a correct guarantee, though the resulting error bound may be less precise.
Why the Bound Works
The key fact is the second-order Taylor theorem with a remainder: for a nonzero \(h\), there is a point \(c\) strictly between \(a\) and \(a+h\) such that the difference between \(f(a+h)\) and its linearization equals \(\frac12 f''(c)h^2\). Here is a derivation under the theorem’s stated smoothness condition.
Set \(R=f(a+h)-f(a)-f'(a)h\). This is exactly \(f(a+h)-L(a+h)\).
For \(t\) between \(a\) and \(a+h\), define \(F(t)=f(t)-f(a)-f'(a)(t-a)-\frac{R}{h^2}(t-a)^2\). It satisfies \(F(a)=0\) and \(F(a+h)=0\).
Rolle’s theorem gives a point \(u\) strictly between the endpoints with \(F'(u)=0\). Since \(F'(a)=0\), applying Rolle’s theorem to \(F'\) between \(a\) and \(u\) gives a point \(c\) between them with \(F''(c)=0\).
Since \(F''(t)=f''(t)-2R/h^2\), the equation \(F''(c)=0\) gives \(R=\frac12 f''(c)h^2\).
Taking absolute values yields \(|R|=\frac12|f''(c)|h^2\). The point \(c\) lies inside the interval where \(|f''(x)|\leq M\), so \(|R|\leq \frac12 M h^2\), as claimed. This argument explains why the interval condition is essential: the curvature value determining the error is taken at an intermediate point, which is not known in advance.
Worked Examples
Worked Example: An Exact Bound for a Quadratic
Approximate \(f(3.2)\) using the linearization of \(f(x)=x^2\) at \(a=3\), and bound the error.
First, \(f(3)=9\) and \(f'(x)=2x\), so \(f'(3)=6\). The linearization is \(L(x)=9+6(x-3)\). With \(h=0.2\),
Next, \(f''(x)=2\) everywhere. Thus \(M=2\) is valid on the interval \([3,3.2]\). The error bound is
Check directly: \(f(3.2)=(3.2)^2=10.24\), so the actual error is \(|10.24-10.2|=0.04\). It equals the bound. The estimate \(10.2\) is therefore within \(0.04\) of the exact value.
Worked Example: Bounding an Exponential Approximation
Use the linearization of \(f(x)=e^x\) at \(a=0\) to estimate \(e^{0.2}\), and find a guaranteed error bound.
Since \(f(0)=1\) and \(f'(x)=e^x\), the linearization is \(L(x)=1+x\). Therefore \(L(0.2)=1.2\). The second derivative is \(f''(x)=e^x\). On \([0,0.2]\), \(e^x\leq e^{0.2}\), so take \(M=e^{0.2}\). The theorem gives
As a numerical check, \(e^{0.2}\approx 1.22140\), so the actual error is approximately \(1.22140-1.2=0.02140\), which is less than \(0.02443\). The decimal values here are rounded to five significant digits; the inequality itself follows from the exact bound \(0.02e^{0.2}\).
Worked Example: A Negative Input Change
Estimate \(1/1.8\) by linearizing \(f(x)=1/x\) at \(a=2\), and bound the error. The change is \(h=-0.2\), so the interval to check is \([1.8,2]\).
We have \(f(2)=1/2\), \(f'(x)=-1/x^2\), and \(f'(2)=-1/4\). Thus
The second derivative is \(f''(x)=2/x^3\). It decreases on \([1.8,2]\), so its largest value on that interval occurs at \(1.8=9/5\). Therefore
The error is bounded by
To check, \(f(1.8)=1/(9/5)=5/9\), and the actual error is \(\left|5/9-11/20\right|=|100/180-99/180|=1/180\approx 0.00556\). This is less than \(5/729\approx 0.00686\). The negative value of \(h\) causes no difficulty because the bound uses \(h^2\).
Choosing a Bound That Is Actually Valid
A common error is to evaluate the second derivative at \(a\), call that value \(M\), and apply the theorem without checking the rest of the interval. The reciprocal example shows why this can fail. At \(a=2\), \(|f''(2)|=1/4\), but at \(x=1.8\), \(|f''(1.8)|=250/729\), which is larger than \(1/4\). Thus \(1/4\) is not an upper bound on the required interval.
To apply the theorem reliably, first mark both endpoints \(a\) and \(a+h\). Then find an upper bound for \(|f''|\) over the entire interval between them. If the second derivative changes, its value at the linearization point alone may not reveal the largest magnitude. Once a valid \(M\) is established, substitute \(M\) and \(h\) into \(Mh^2/2\).
Using the Bound to Plan an Approximation
The inequality can also answer a practical question: how small must the input change be to guarantee an error at most \(\varepsilon\)? If a single bound \(M>0\) is valid throughout the interval under consideration, it is sufficient to require
This condition is useful only when \(M\) really applies to the interval containing the target input. If that interval changes as \(h\) changes, the suitable value of \(M\) may change too. When \(M=0\), the theorem says the linearization error is zero on the interval.
The estimate is a worst-case guarantee, not necessarily the actual error. It may be conservative because the largest allowed value \(M\) might occur at a different point from the intermediate point that determines the actual remainder. Still, it provides a rigorous way to certify accuracy using information about the second derivative, without first knowing the exact function value.
Check Your Understanding
For each question, pay close attention to the interval on which the second-derivative bound must hold.
- For a change \(h=-0.3\) from a linearization point \(a=4\), what closed interval must be checked?
- If \(|f''(x)|\leq 6\) throughout the interval from \(a\) to \(a+h\), what error bound does the theorem give?
- Why is knowing \(|f''(a)|\leq M\) alone insufficient to apply the theorem?
- For \(f(x)=x^2\), why does the bound at any input change equal the actual linearization error?
- If a valid constant \(M\) is \(8\), what condition on \(|h|\) guarantees an error no greater than \(0.01\)?
