Why a Linear Estimate Has an Error
A linearization uses the tangent line to approximate a function near a chosen base point. It matches the function’s value and slope at that point, but away from the base point the function may bend away from the line. The second derivative measures how the slope changes, so a bound on the second derivative can limit how far the function moves from its linearization.
The aim here is to derive a precise bound, not just an informal claim that the approximation is good when the step is small. The result will show exactly which interval needs to be checked and how the error depends on the step size.
Linearization and Error
Suppose \(f\) is differentiable at a base point \(a\). Its linearization at \(a\) is the tangent-line function
For an input \(a+h\), the change from the base point is \(h\). Substituting \(x=a+h\) into the linearization gives
The linearization error at that input is the difference between the actual function value and the estimate. We will use its absolute value so that the bound describes the size of the error regardless of its sign:
The step \(h\) may be positive or negative. The relevant interval is the entire segment with endpoints \(a\) and \(a+h\), not just the part of the number line to the right of \(a\).
The Error Bound and Its Conditions
For a nonzero step, the proof gives a more precise statement: for some point \(\xi\) strictly between \(a\) and \(a+h\), the error equals \(f''(\xi)h^2/2\). The bound follows because the second derivative at that particular point cannot have absolute value greater than \(M\).
Deriving the Second-Derivative Remainder
The Mean Value Theorem relates changes in a function to a derivative at an intermediate point. To obtain the factor \(1/2\) in this bound, we use Rolle’s Theorem, which follows from the Mean Value Theorem. In particular, if a function is continuous on a closed interval, differentiable inside it, and has equal values at the endpoints, then its derivative is zero somewhere inside.
Here is a complete derivation. Assume \(h\ne 0\), and set
Construct an auxiliary function by subtracting the linearization and a quadratic term from \(f\):
At \(t=a\), each term involving \(t-a\) is zero, so \(g(a)=0\). At the other endpoint, the definition of \(A\) gives
Also, differentiating \(g\) gives
so \(g'(a)=0\). Since \(g(a)=g(a+h)\), Rolle’s Theorem gives a point \(c\) strictly between \(a\) and \(a+h\) for which \(g'(c)=0\). Now \(g'\) has equal values at \(a\) and \(c\), because both are zero. Applying Rolle’s Theorem to \(g'\) on the segment between those points gives a point \(\xi\) strictly between \(a\) and \(c\) with \(g''(\xi)=0\). This also places \(\xi\) strictly between \(a\) and \(a+h\), whether \(h\) is positive or negative.
Twice differentiating the auxiliary function shows that
Thus \(g''(\xi)=0\) means \(f''(\xi)=2A\). Substituting the definition of \(A\) proves the exact remainder formula:
Taking absolute values and using the assumed bound at \(\xi\) gives
This completes the derivation. The squared step size appears because the first-order change has been removed by the linearization, leaving a second-order remainder. The factor \(1/2\) is part of the exact remainder formula, not an extra estimate.
Worked Examples
Worked Example: A Quadratic with an Exact Bound
Let \(f(x)=x^2\), with base point \(a=1\) and step \(h=\frac15\). Find the linearization error and check the bound.
First, calculate the value and derivative at the base point:
Therefore the linearization is
The new input is \(a+h=1+\frac15=\frac65\). Evaluating both the function and the linearization there gives
Thus the absolute error is
Since \(f''(x)=2\), the bound holds on the whole segment from \(1\) to \(\frac65\) with \(M=2\). The theorem gives
The error equals the bound. As a check, the linearization has derivative \(2\), matching \(f'(1)\), and the computed function value minus linear estimate is exactly \(\frac1{25}\).
Worked Example: A Cubic with a Positive Step
Let \(f(x)=x^3\), \(a=1\), and \(h=\frac12\). Find the error and a valid bound.
The value and derivative at the base point are
So the linearization is \(L(x)=1+3(x-1)\). The input is \(a+h=\frac32\), and
The error is therefore
To choose \(M\), check the second derivative across the full segment \(\left[1,\frac32\right]\). Since \(f''(x)=6x\), on that segment \(6\le f''(x)\le 9\), so \(\left|f''(x)\right|\le 9\). The error bound is
Indeed, \(\frac78\le\frac98\). The calculation checks the theorem’s conclusion. Also, differentiating \(L(x)=1+3(x-1)\) gives \(L'(x)=3=f'(1)\), as required for the tangent-line linearization.
Worked Example: A Negative Step
Let \(f(x)=\frac1x\), with base point \(a=1\) and step \(h=-\frac12\). The segment is \(\left[\frac12,1\right]\), so this example checks the bound when the input is less than the base point.
At \(a=1\),
The linearization is \(L(x)=1-(x-1)\). The new input is \(1-\frac12=\frac12\). Its function and linearization values are
The absolute error is \(\left|2-\frac32\right|=\frac12\). Next, \(f''(x)=\frac{2}{x^3}\). On \(\left[\frac12,1\right]\), this is positive and at most \(16\), so \(M=16\) is valid. The resulting bound is
The computed error satisfies \(\frac12\le 2\). The sign of \(h\) causes no problem: its absolute value is used in the bound, and its square appears in the exact remainder.
A Common Mistake: Checking Only at the Base Point
A bound on \(f''\) at a single point does not automatically bound \(f''\) throughout the interval. The theorem requires control over the entire segment from \(a\) to \(a+h\).
For a concrete example, return to \(f(x)=x^3\), \(a=1\), and \(h=\frac12\). A mistaken calculation might use only \(f''(1)=6\), set \(M=6\), and claim that the error is at most
But the actual error, computed above, is \(\frac78\), which is greater than \(\frac34\). The claim fails because \(\left|f''(x)\right|=6x\) increases from \(6\) to \(9\) on the segment \(\left[1,\frac32\right]\). The corrected bound uses \(M=9\), giving \(\frac98\), and \(\frac78\le\frac98\).
Why the Step Size Matters
The error bound contains \(|h|^2\). If the interval and the value of \(M\) stay fixed, halving the step size changes the bound by a factor of \(\left(\frac12\right)^2=\frac14\). This is a consequence of the second-order remainder, not a guarantee that a particular example’s error will equal the bound.
The estimate can be exact, as it was for the quadratic example, or it can be larger than the actual error. It is a guarantee that applies whenever its assumptions hold. It does not tell you the sign of the error, and it does not justify using a second-derivative value from outside the interval.
Check Your Understanding
For each question, pay attention to the entire interval between the base point and the input.
- State the linearization of \(f(x)=x^2\) at \(a=2\), and identify the input corresponding to the step \(h=-\frac14\).
- For \(f(x)=x^3\), \(a=0\), and \(h=\frac12\), what is the largest value of \(\left|f''(x)\right|\) on the segment? What bound does the theorem give?
- Why does the proof use Rolle’s Theorem twice to obtain a point where \(g''(\xi)=0\)?
- What condition must \(M\) satisfy on the segment between \(a\) and \(a+h\)?
- If \(h=0\), what is the linearization error, and why does the theorem’s bound agree?
