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Integrated Linearization And Error-Propagation Problems

Learn to turn uncertain measurements into a linearized output estimate, an uncertainty range, and a checked comparison with exact endpoint values.

Intermediate 11 min read

What You'll Learn

  • Build a mathematical model from measured inputs and an output quantity
  • Choose and state a linearization point with units
  • Propagate input uncertainties using partial derivatives
  • Combine contributions into a worst-case estimated output range
  • Check estimates against exact endpoint values when possible
  • Distinguish a local uncertainty estimate from a guaranteed bound

From Measured Inputs to an Output Range

A measurement problem often has several stages. You measure one or more inputs, use a model to calculate an output, and then ask how uncertainty in the measurements affects that output. The calculation is not complete until the model, the point of linearization, the units, and the uncertainty estimate are all clear.

Linearization gives a systematic way to estimate the effect of small input changes. It uses derivatives at a specified base point to approximate how the output changes nearby. The result is useful, but it is generally an estimate rather than an exact guarantee. When possible, comparing it with exact values at the ends of the input ranges provides an important check.

Key idea: Build the output model first, evaluate its derivatives at the measured base point, and use those derivatives to estimate how the stated input uncertainties affect the output.

The Linearization and Uncertainty Estimate

Suppose an output \(y\) is modeled as a differentiable function \(f\) of measured inputs \(x_1,\ldots,x_n\). Let the measured values be the base point \((a_1,\ldots,a_n)\), and let \(\Delta x_i\) denote a possible change in input \(x_i\). The first-order change in the output is approximated by

$$ \Delta y \approx \sum_{i=1}^{n} \frac{\partial f}{\partial x_i}(a_1,\ldots,a_n)\,\Delta x_i. $$

If input \(x_i\) is known to lie within \(a_i\pm \delta_i\), where \(\delta_i\geq 0\), the largest first-order change in magnitude is estimated by adding the absolute contributions:

$$ \delta_y \approx \sum_{i=1}^{n} \left| \frac{\partial f}{\partial x_i}(a_1,\ldots,a_n) \right|\delta_i. $$

This is a worst-case linear estimate: each input’s contribution is allowed to act in whichever direction increases the output’s deviation. It is not a statistical standard deviation, and it does not assume that measurement errors cancel one another.

Important condition: Linearization is a local approximation. It is most useful when the input changes are small enough that the function is well approximated by its tangent or tangent plane near the chosen base point. The resulting range is not automatically a guaranteed enclosure of all exact outputs.

Why the Propagation Formula Works

Differentiability at the base point means that, for a small input change \(\mathbf{h}=(h_1,\ldots,h_n)\), the output change can be written as a linear part plus a remainder that becomes small relative to the size of \(\mathbf{h}\). In symbols, with \(\mathbf{a}=(a_1,\ldots,a_n)\),

$$ f(\mathbf{a}+\mathbf{h})-f(\mathbf{a}) = \sum_{i=1}^{n} \frac{\partial f}{\partial x_i}(\mathbf{a})h_i +r(\mathbf{h}), \qquad \frac{r(\mathbf{h})}{\|\mathbf{h}\|}\longrightarrow 0 \quad\text{as }\mathbf{h}\longrightarrow \mathbf{0}. $$

The linear part is the part used for the estimate. If \(|h_i|\leq\delta_i\), the triangle inequality gives

$$ \left| \sum_{i=1}^{n} \frac{\partial f}{\partial x_i}(\mathbf{a})h_i \right| \leq \sum_{i=1}^{n} \left|\frac{\partial f}{\partial x_i}(\mathbf{a})\right||h_i| \leq \sum_{i=1}^{n} \left|\frac{\partial f}{\partial x_i}(\mathbf{a})\right|\delta_i. $$

This proves the stated worst-case bound for the linear part. The differentiability remainder is not included in that bound, which is why the propagated range is an estimate rather than a guaranteed range for the original nonlinear model.

A Practical Sequence of Steps

1
Write the output model.
Define the output as a function of the measured inputs, and identify the units of each quantity.
2
Record the base point and uncertainties.
State each measured value and its uncertainty in matching units.
3
Differentiate and evaluate.
Find the relevant derivative or partial derivatives, then evaluate them at the base point.
4
Propagate the uncertainties.
Multiply each derivative magnitude by its input uncertainty and add the contributions for a worst-case linear estimate.
5
Check the result.
Check units and, when feasible, compare with exact model values at the input-range endpoints.

Worked Problems

Worked Example: Area from Two Measured Sides

A rectangular plate has measured side lengths \(L=12.0\pm0.1\text{ cm}\) and \(W=8.0\pm0.2\text{ cm}\). Estimate its area and a worst-case linearized uncertainty range, then check the result using the exact endpoint values.

Step 1: Model and base point. The area is \(A(L,W)=LW\), measured in \(\text{cm}^2\). The base point is \((L,W)=(12.0\text{ cm},8.0\text{ cm})\), and the uncertainties are \(\delta_L=0.1\text{ cm}\) and \(\delta_W=0.2\text{ cm}\).

Step 2: Evaluate the model and partial derivatives.

$$ A(12.0,8.0)=(12.0)(8.0)=96.0\text{ cm}^2, \qquad \frac{\partial A}{\partial L}=W=8.0\text{ cm}, \qquad \frac{\partial A}{\partial W}=L=12.0\text{ cm}. $$

Step 3: Propagate the uncertainties. The side-length contribution is \((8.0\text{ cm})(0.1\text{ cm})=0.8\text{ cm}^2\). The width contribution is \((12.0\text{ cm})(0.2\text{ cm})=2.4\text{ cm}^2\). Thus \(\delta_A\approx0.8+2.4=3.2\text{ cm}^2\), giving the linearized range \([96.0-3.2,96.0+3.2]=[92.8,99.2]\text{ cm}^2\).

Step 4: Check exact endpoint values. Since both side lengths are positive and area increases with either side, the smallest and largest areas occur at the corresponding endpoints:

$$ A_{\min}=(11.9)(7.8)=92.82\text{ cm}^2, \qquad A_{\max}=(12.1)(8.2)=99.22\text{ cm}^2. $$

The exact range is \([92.82,99.22]\text{ cm}^2\), close to the linearized range \([92.8,99.2]\text{ cm}^2\). Each exact endpoint is \(0.02\text{ cm}^2\) above its corresponding linearized endpoint. The derivative units also check: centimetres times centimetres give square centimetres.

Worked Example: Volume from an Uncertain Radius

A sphere has measured radius \(r=3.0\pm0.1\text{ cm}\). Estimate its volume and uncertainty range by linearization, and compare with the exact volumes at the radius endpoints.

Step 1: Model and base point. The model is \(V(r)=\frac{4}{3}\pi r^3\), with volume in \(\text{cm}^3\). Use the base point \(r=3.0\text{ cm}\) and uncertainty \(\delta_r=0.1\text{ cm}\).

Step 2: Evaluate the model and derivative.

$$ V(3.0)=\frac{4}{3}\pi(3.0)^3=36\pi\text{ cm}^3\approx113.097\text{ cm}^3, \qquad V'(r)=4\pi r^2, \qquad V'(3.0)=36\pi\text{ cm}^2. $$

Step 3: Propagate the radius uncertainty. The estimated volume change is \((36\pi\text{ cm}^2)(0.1\text{ cm})=3.6\pi\text{ cm}^3\approx11.310\text{ cm}^3\). The linearized range is therefore

$$ [36\pi-3.6\pi,\;36\pi+3.6\pi]\text{ cm}^3 = [32.4\pi,\;39.6\pi]\text{ cm}^3 \approx[101.788,\;124.407]\text{ cm}^3. $$

Step 4: Check the exact endpoint values. The radius endpoints are \(2.9\text{ cm}\) and \(3.1\text{ cm}\). The model increases for positive \(r\), so these give the minimum and maximum volumes:

$$ V(2.9)=\frac{4}{3}\pi(2.9)^3 =\frac{4}{3}\pi(24.389) \approx102.160\text{ cm}^3, $$ $$ V(3.1)=\frac{4}{3}\pi(3.1)^3 =\frac{4}{3}\pi(29.791) \approx124.788\text{ cm}^3. $$

The exact range is approximately \([102.160,124.788]\text{ cm}^3\). Compared with the linearized endpoints, the exact lower endpoint is about \(0.373\text{ cm}^3\) higher and the exact upper endpoint is about \(0.381\text{ cm}^3\) higher. The range is shifted upward and is slightly wider. The units are consistent because \(V'(r)\) has units \(\text{cm}^2\), which become \(\text{cm}^3\) after multiplication by a radius change in centimetres.

Worked Example: A Logarithmic Calibration Model

A calibration model converts a dimensionless sensor reading \(x\) into a reported quantity \(q=2.5\ln(1+x)\text{ mg}\). If \(x=0.50\pm0.02\), estimate \(q\) and its uncertainty range, then check the exact endpoint values.

Step 1: Model and base point. Let \(q(x)=2.5\ln(1+x)\text{ mg}\), with dimensionless \(x\). The base point is \(x=0.50\), and \(\delta_x=0.02\).

Step 2: Evaluate the model and derivative.

$$ q(0.50)=2.5\ln(1.50)\text{ mg} \approx2.5(0.405465)\text{ mg} =1.013663\text{ mg}, $$ $$ q'(x)=\frac{2.5}{1+x}\text{ mg}, \qquad q'(0.50)=\frac{2.5}{1.50}\text{ mg per unit of }x =\frac{5}{3}\text{ mg per unit of }x. $$

Step 3: Propagate the input uncertainty. The estimated output change is \(\frac{5}{3}(0.02)\text{ mg}=0.033333\text{ mg}\). The linearized range is

$$ [1.013663-0.033333,\;1.013663+0.033333]\text{ mg} \approx[0.980329,\;1.046996]\text{ mg}. $$

Step 4: Check the exact endpoint values. The input endpoints are \(0.48\) and \(0.52\). Since \(q'(x)>0\) for \(x>-1\), the model is increasing throughout this interval. Thus

$$ q(0.48)=2.5\ln(1.48)\text{ mg}\approx0.980105\text{ mg}, \qquad q(0.52)=2.5\ln(1.52)\text{ mg}\approx1.046776\text{ mg}. $$

The exact range is approximately \([0.980105,1.046776]\text{ mg}\), slightly below the linearized range at both endpoints. The input is dimensionless, so the derivative has units of milligrams per unit of \(x\); multiplying by a change in \(x\) gives milligrams, as required.

Common Pitfall: Using Only the Product of the Input Changes

For the rectangle, a tempting but incorrect calculation is to treat the area change as the product of the two side changes: \((0.1\text{ cm})(0.2\text{ cm})=0.02\text{ cm}^2\). That is not the first-order area uncertainty. It accounts only for the small product term formed when both sides change, and misses the main contributions from each side changing against the other side’s full length.

$$ dA=W\,dL+L\,dW =(8.0\text{ cm})(0.1\text{ cm}) +(12.0\text{ cm})(0.2\text{ cm}) =3.2\text{ cm}^2. $$

The correction is to differentiate the complete output model before substituting the input uncertainties. The resulting terms have the right units and reflect the sensitivity of the output at the chosen base point.

Interpreting and Reporting the Result

A useful report gives the model, base point, output estimate, uncertainty estimate, and units. For example, a result can be stated as “the linearized area estimate is \(96.0\text{ cm}^2\), with a worst-case first-order uncertainty of \(3.2\text{ cm}^2\).” Writing the interval as well makes the estimated spread explicit.

When the model is monotone over the full input interval, exact endpoint evaluation can provide the actual output range for a single uncertain input. For multiple inputs, checking the appropriate corners can also work when the model’s behavior makes those corners the extrema. If monotonicity or the location of extrema is not established, endpoint or corner calculations alone do not guarantee the range. In all cases, keep the distinction clear: the derivative calculation supplies a local estimate; a guaranteed range needs a separate argument.

Takeaway: A dependable multi-step estimate starts with a correctly stated model and base point. It propagates uncertainty using derivatives with units, then checks the approximation without presenting a linearized interval as an exact bound.

Check Your Understanding

Use the model, base point, uncertainty, and exact-check ideas from this tutorial to answer the following questions.

  1. For a differentiable model with two measured inputs, what quantities are needed to calculate a first-order output uncertainty estimate?
  2. Why are the absolute values of the partial derivatives used in a worst-case uncertainty estimate?
  3. A model gives an output in square centimetres and has a derivative with respect to a length input in centimetres. What units must the corresponding output-change contribution have?
  4. Why is a propagated linearized range not automatically a guaranteed range for the exact model?
  5. For a monotone increasing model of one input, which input endpoints should be evaluated to find the exact output range?

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