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An affine combination is a weighted point $P={\sum }_{i=0}^k{\lambda }_i{P}_i$ with ${\sum }_{i=0}^k{\lambda }_i=1$. The coefficients can be negative. The sum is meaningful in an affine space without selecting a preferred origin: relative to any point $O$, it means $P=O+{\sum }_i{\lambda }_i(P_i-O)$. Affine maps preserve these weighted points. ^[ref]

For an original example in the Euclidean plane, take $A=(0,0)$, $B=(2,0)$ and $C=(0,2)$. Then $P=\frac{1}{4}A+\frac{1}{2}B+\frac{1}{4}C=(1,\frac{1}{2})$ lies inside their triangle. Its coefficients are nonnegative, making it a convex combination. In contrast, $R=-\frac{1}{2}A+\frac{3}{2}B=(3,0)$ is beyond $B$: it is an affine combination but not a convex combination of these three points. Negative weights allow extrapolation, rather than only interpolation inside the convex hull.

For the affine map $T(x,y)=(2x+y+1,y-2)$, the same weights produce $T(P)=(\frac{7}{2},-\frac{3}{2})$ from $T(A),T(B),T(C)$, and produce $T(R)=(7,-2)$ from $T(A),T(B)$. Thus one can compute a weighted point before transforming the vertices or afterwards, obtaining exactly the same point.

This is useful whenever an object is built from weighted control points: its construction commutes with an affine map, so transforming the control points transforms the constructed object as well. The sum of the weights is the essential condition; no assumption about lengths or angles is needed.

Claude Brezinski works in numerical analysis, particularly extrapolation, Padé approximation, continued fractions, orthogonal polynomials and numerical linear algebra. His research also treats systems of nonlinear equations. These subjects connect different representations of an approximation with ways to accelerate or stabilize a numerical computation.^[ref]

Identity source: claude-brezinski.fr/

A current-sign certificate with a curvature bound uses a bound on the second derivative of an electric current to certify the sign of an extrapolated current. An affine ramp is predicted exactly by two correctly timed samples, so its uncertainty should depend on curvature rather than counting the entire ramp as error. Current direction matters physically because it selects recirculation paths during switching dead time.^[ref]

Let samples at $-T$ and $0$ produce measurements $y_{-}=i(-T)+e_{-}$ and $y_0=i(0)+e_0$, with $|e_{-}|\le {ϵ}_{-}$ and $|e_0|\le {ϵ}_0$. For a future time $t\ge 0$, use the linear extrapolationIf $i$ is twice continuously differentiable and $|i^{\prime \prime }(s)|\le M$ throughout $[-T,t]$, the polynomial interpolation error givesThe model term follows by applying the remainder to the two-node interpolant: $i(t)-[(1+t/T)i(0)-(t/T)i(-T)]=i^{\prime \prime }(\xi )t(t+T)/2$ for some $\xi \in (-T,t)$. The sample-error terms use absolute coefficients, as required by the signed-filter current-sign certificate. At $t=T$, the model contribution is $M{T}^2$ and equal sample bounds contribute $3ϵ$.

For an original example, take $T=6\mu \mathrm{s}$, $M=10^{8}A/{\mathrm{s}}^2$ and each sample error bounded by $0.002\mathrm{A}$. At $t=T$,If $y_0=0.030\mathrm{A}$ and $y_{-}=0.040\mathrm{A}$, the predicted current is $0.020\mathrm{A}$, certifying actual current at least $0.0104\mathrm{A}$ at that future instant. To certify the entire interval $0\le t\le T$, check $\hat{i}(t)-B(t)>0$ everywhere. In this example the prediction decreases linearly and $B$ increases, so the minimum margin occurs at $T$ and the same lower bound applies throughout.

The curvature limit is a physical assumption. A switching event, sensor clipping or a slope discontinuity can invalidate it even when a fitted ramp looks smooth at the sample times. A bounded first derivative alone does not imply any finite $M$. Timing uncertainty in a curvature-based current-sign certificate also matters: two accurate current values at incorrectly modeled times need not extrapolate an affine ramp correctly.

A proposed test supplies known smooth ramps and curved waveforms, records sample timestamps and checks the independently measured future waveform against the enclosure. Test transitions separately. The result is a deterministic sign guarantee only when the sample and curvature bounds are deterministic; substituting one standard uncertainty for an absolute error bound changes the claim.

The fitted simple linear regression isIts regression slope is a predicted decrease of approximately 0.006314 trust-score units per extra year, or 0.06314 units per decade. This is an estimated statistical association between age and the score. A cross-sectional study does not establish that the same person's trust decreases at this rate as they age.

The estimated standard error of the slope is 0.005427. Testing a zero regression slope gives $t\approx -1.163$ on 746 residual degrees of freedom and $p\approx 0.245$. An approximate 95% confidence interval isThere is insufficient evidence of a nonzero linear age association at the 5% level. The interval includes both small negative and small positive slopes; failure to reject does not establish an exactly zero relationship or rule out a nonlinear one.

The regression intercept 6.388920 predicts the score at age zero. That age is outside the observed 19–89 range, so interpreting it substantively would be extrapolation. Its very small reported p-value tests a zero intercept, rather than the age hypothesis. The 746 residual degrees of freedom correspond to 748 observations minus the intercept and slope.

The coefficient of determination is $R^2=0.001811$, so the fitted line explains approximately 0.1811% of the observed centered variation, rather than 18.11%. The adjusted coefficient of determination is approximately 0.0004732. The residual standard error is 2.602 score units, close to the reported response standard deviation; most variation remains unexplained by the fitted line. With one predictor and an intercept, the slope's F-test is the square of its Student's t-test statistic, up to printed rounding, so $F\approx 1.354$ and $p\approx 0.245$ agree with the slope test.

Output 6 is a residual-versus-fitted plot: its vertical coordinates should be regression residuals $e_i=y_i-{\hat{y}}_i$. The fitted values range over only approximately 5.83–6.27, as the weak slope predicts over ages 19–89. The printed scatter shows no obvious strong mean curvature or clear widening funnel, but it has marked horizontal concentrations near its extreme residual levels. An absence of obvious curvature does not prove a correct linear conditional mean, and the plot alone cannot establish homoscedasticity, independent errors or a normal distribution. A normal probability plot and residuals against observation order address different aspects. Large residuals also do not by themselves establish influential observations; regression leverage and Cook's distance matter.

There is an additional inconsistency in the original figure. The descriptive output reports observed responses between zero and ten, so residual constraints from a bounded response requireYet the printed graphic contains points around $(\hat{y},e)=(6.25,4.2)$, which would imply $y\approx 10.45$, above the stated observed maximum. Some lower-edge points likewise imply negative responses. This is present in the original PDF, rather than introduced by the transcription. The graphic therefore cannot be treated as a fully consistent diagnostic of the reported dataset. The observations about its visible shape above describe the supplied picture; reliable model checking would require the actual fitted values and residuals.

The printed code has separate typographical problems. The displayed model call supports the intended formula with ~, and the plot needs one closing parenthesis:

model1 <- lm(t_guard ~ age, data = exam)
plot(fitted(model1), residuals(model1),
     xlab = "Fitted values", ylab = "Residuals")
abline(h = 0)

These calls correct the syntax; they cannot be rerun on the original observations because the individual-level dataset is not supplied. The regression interpretation above uses the available numerical summaries and states the diagnostic limitation explicitly.

Conditional predictions for observed birds and years add their fitted random intercepts to the fixed part of the Gaussian linear mixed model:These are predicted expected values, so they need not be integers. The first is likely more reliable: age three is within the displayed age range, whereas age seven requires extrapolation beyond ages one to six. A quantitative comparison would need the prediction intervals, including uncertainty in both fixed coefficients and random effects; the displayed coefficients alone do not supply them.

Write ${\tilde{p}}_1$ and $\tilde{q}$ for the quantities computed from the perturbed second sample. The endpoint linear interpolation and its derivative areCompare them on their interpolation interval $0\le x\le h$, with fixed $a>0$. Put $x=th$, $0\le t\le 1$. The Taylor series of the sine gives, uniformly in $t$,Here the second equality compares with $f^{\prime }(0)=a$. More generally, the exact derivative is $f^{\prime }(th)=a\cos (ath)$ andThe input perturbation itself changes the value interpolant by exactly $aϵx/h$, and changes its derivative by exactly $aϵ/h$. ThusFor fixed nonzero $ϵ$, the value approximation has a bounded noise floor: its maximum error approaches $a|ϵ|$, and at $x=h$ that error is exactly $a|ϵ|$. It does not converge uniformly to the exact values as $h\to 0$. However, it does not amplify the absolute sample perturbation on $[0,h]$. The derivative error instead grows without bound like $a|ϵ|/h$.

With exact samples, $ϵ=0$, both approximations do converge on the shrinking interval: the value error is $O(h^3)$ for this particular sine, and the derivative error is $O(h^2)$. With step-dependent noise, value convergence requires $ϵ(h)\to 0$, whereas derivative convergence requires the stronger $ϵ(h)=o(h)$, where $o$ denotes little o notation.

This is noise amplification by differentiation of a linear interpolant. Value interpolation has an absolute perturbation bound independent of $h$ on the sampled interval. Numerical differentiation has a perturbation factor $1/h$, expressing the ill-posedness of numerical differentiation from values with fixed absolute uncertainty. The amplification follows from the mathematical data map even in exact arithmetic; it is distinct from additional roundoff error in a particular implementation. For fixed $x$ outside $[0,h]$, the operation becomes extrapolation, and the value perturbation $aϵx/h$ can also diverge.

Sensor self-heating in a ratiometric resistance measurement occurs when the sensing electric current changes the sensor's own temperature through Joule heating. A ratiometric measurement can remove the direct dependence of the electrical reading on that electric current while still reading the heated sensor correctly. For a resistance thermometer, that means the indicated temperature can differ from the unperturbed surroundings.

With actual sensor electrical resistance $R_T(I)$ and a stable reference resistor $R_{\mathrm{ref}}$ carrying the same electric current,The cancellation leaves the dependence of $R_T$ on $I$ intact. In a small, steady perturbation, a model with thermal resistance $R_{\mathrm{th}}$ gives $\Delta T\simeq R_{\mathrm{th}}{I}^2{R}_T(0)$ andwhere ${\alpha }_R$ is the temperature coefficient of resistance. These approximations require small heating and thermal conditions that remain comparable between readings. The BIPM guide distinguishes heat flow inside the probe from heat flow into its surroundings; changing the surrounding medium can change the effect.^[ref]

A practical diagnostic is a two-current extrapolation. Model $R_T(I)\simeq R_0+c{I}^2$ over the tested range. Measurements $R_1,R_2$ at distinct nonzero current magnitudes $I_1,I_2$ imply$R_0$ is an inferred zero-power electrical resistance, not an electrical reading taken with zero excitation. The guide describes this squared-current method and requires a stable surrounding temperature and settled readings.^[ref]

For an original numerical example, suppose a $1000\Omega$ reference resistor gives ratios $0.120006$ at $0.5\mathrm{mA}$ and $0.120024$ at $1.0\mathrm{mA}$. The inferred resistances are $120.006\Omega$ and $120.024\Omega$. Since $I_2^2=4I_1^2$, the estimate is $(4R_1-R_2)/3=120.000\Omega$. Both ratios canceled their excitation exactly, yet they disagree because the sensor changed.

A bath experiment should alternate the two excitation settings, allow thermal settling and monitor the bath temperature. Repeated low/high/low readings help reveal drift rather than attributing every difference to heating. Resistance thermometer self-heating gives the wider physical setting; reference-resistor error in ratiometric resistance measurement and mismatched filters are separate contributions.