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For seismic source inversion, collect calibrated, consistently filtered samples from all available stations and components into a data vector $\mathbf{d}$. For each trial source position and origin time, compute the corresponding elastodynamic Green tensor in an Earth model, including the instrument response used for the data. Write the symmetric seismic moment tensor in six independent symmetric basis tensors:Each column of $\mathbf{G}$ is the synthetic seismogram for one basis tensor. An off-diagonal basis tensor contains both symmetric entries, so the double contraction and its factors of two remain consistent.

A suitable objective iswith noise covariance ${\mathbf{C}}_d$; a justified penalty can be added for regularization. At fixed $\mathbf{q}$, solve for the six entries of $\mathbf{m}$ by weighted least squares, preferably using a QR decomposition or singular value decomposition rather than explicitly inverting the matrix of the normal equations. Then search over the four nonlinear parameters using a coarse location/time grid followed by local refinement. This variable projection reduces a ten-parameter search to a four-parameter nonlinear search with an inner six-parameter linear fit. Merely having more than ten data samples does not ensure identifiability: the design matrix must have adequate rank and the network must resolve the source's radiation patterns.

Potential errors include inaccurate velocity, density, anisotropy and attenuation in the assumed Earth model; neglected finite rupture and source-time effects; timing, instrument calibration or component-orientation errors; correlated noise; uneven station coverage; and tradeoffs between depth, origin time and the moment tensor. Imposing a double-couple seismic source mechanism without evidence can hide genuine non-double-couple components. Regularization also changes the recovered source and should be reported.

Inspect residuals, singular values, recovery of synthetic sources and uncertainty under resampling or noise perturbations. Linearized covariance or a posterior distribution can quantify uncertainty within the assumed forward model, but does not automatically account for errors in that model. A good waveform fit alone is insufficient evidence that all ten parameters are uniquely recovered.

A rotational spectral line assignment associates a measured spectral line with a molecule, an internal quantum state, and an initial and final set of quantum numbers. For the spin-free linear rigid rotor, absorption is labelled $J\to J+1$. A regularly spaced group suggests a candidate rotational constant, but the lowest line inside a scanned band need not be $0\to 1$. Several molecules can contribute to the same spectrum, and an apparent missing member can lie outside the band or below the detection threshold.

First determine line centers and their uncertainties from the instrument's readout, accounting for resolved Doppler shifts or hyperfine structure rather than treating every peak as a separate rotational transition. Then propose integer assignments, fit a rotational Hamiltonian, and predict additional spectral lines. For a weakly distorted linear molecule within one vibrational quantum state, the centrifugal distortion model gives the useful diagnosticHere $B,D,{\nu }_m$ are all frequencies in $\mathrm{Hz}$. Correct assignments make the second expression a straight line against $m^2$, with intercept $B$ and slope $-2D$. This follows by division of the transition formula; no additional model of molecular geometry is required. A direct weighted least squares fit of ${\nu }_m$ to $2Bm-4D{m}^3$ uses the measured uncertainties. If instead the transformed values are fitted, their uncertainties must also be divided by $2m$.^[ref]

For an original example, stipulate $B=5.000000\mathrm{GHz}$ and $D=2.000\mathrm{kHz}$, and suppose a scan covers $38$–$72\mathrm{GHz}$. The model places four consecutive transitions at $39.999488$, $49.999000$, $59.998272$, and $69.997256\mathrm{GHz}$. Their assignments are $m=4,5,6,7$, hence $J=3,4,5,6$. An extra stipulated line at $53.2\mathrm{GHz}$ is unrelated to this series. All these frequencies are calculated example inputs, not measured OCS data.

At $m=4$ and $m=7$, the transformed values are $4999.936$ and $4999.804\mathrm{MHz}$. Their straight line has slope $(4999.804-4999.936)/(49-16)=-0.004\mathrm{MHz}$, recovering $D=0.002\mathrm{MHz}$ and $B=5000\mathrm{MHz}$. Both intermediate transitions satisfy the same line exactly. Merely relabelling the first visible transition $0\to 1$ fails this diagnostic. The model predicts a withheld $7\to 8$ transition at $79.995904\mathrm{GHz}$: an additional measurement there would test the assignment rather than merely refit the original four lines.

An assignment earns confidence when several independently observed spectral lines, including lines outside the fitted interval, agree within uncertainty and expected strong accessible lines are present. Inspect the residual of an inverse problem for each line against $J$: systematic curvature can signal inadequate centrifugal distortion, whereas isolated discrepancies suggest a blend or wrong assignment. Comparing a known sample, a blank, and an isotope shift of a linear rotor supplies different checks. These are general inference controls; the actual reported controls of the NIST trace-gas FTMW validation experiment are described separately. One isolated match or one fitted rotational constant alone cannot certify molecular identity.

In variable projection, a prediction has the form $\mathbf{G}(\mathbf{q})\mathbf{m}$, nonlinear in $\mathbf{q}$ but linear in $\mathbf{m}$. For each trial $\mathbf{q}$, solve the inner weighted least squares problem for $\mathbf{m}$, then optimize the resulting residual over $\mathbf{q}$. This reduces the nonlinear dimension without pretending that the whole inverse problem is linear.

For a finite list of real numbers $x_i$ and nonnegative weights of positive total $W$, the weighted mean uniquely minimizes the weighted sum of squared deviationsExpanding around ${\bar{x}}_w$ and using ${\sum }_i{w}_i(x_i-{\bar{x}}_w)=0$ givesEvery displacement from the weighted mean increases the objective. This is the simplest one-parameter weighted least squares problem: represent all entries by one common value.

For the weighted values $90$ and $60$ with weights $2$ and $8$,Thus $S(66)=1440$, while $S(75)=2250$. The minimum is defined by this squared-error criterion; a different loss can select a different summary, such as a median under unweighted absolute deviations.