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Write $p_1(x)=a+bx$. It suffices to impose orthogonality of the error to the basis $1,x$ of ${\Pi }_1$. With the printed weight $x^{-1/2}$ and $f=x^{3/2}$, the normal equations areHere, for example, $⟨f,1⟩={\int }_0^{1}xdx=1/2$ and $⟨f,x⟩={\int }_0^1{x}^{2}dx=1/3$. Solving gives $a=-1/16$, $b=15/16$. ThereforeThe two normal equations make the error orthogonal to every linear polynomial, and part (b) then proves optimality, rather than merely furnishing a candidate fit.

Minimize the residual sum of squares $\Vert Y-X\theta {\Vert }^2$. Differentiating gives the normal equations $X^{T}X\hat{\theta }=X^{T}Y$. Full column rank makes $X^{T}X$ positive definite and invertible, so the unique minimizer isSince $\mathbb{E}Y=X\theta$, it is unbiased. The independent common-variance errors give $\mathrm{Cov}(Y)={\sigma }^{2}I$, henceIndependent Gaussian errors with common variance are a sufficient condition for this estimator also to be the maximum likelihood estimator: their log-likelihood differs from $-\Vert Y-X\theta {\Vert }^2/(2{\sigma }^2)$ by terms independent of $\theta$. Profiling an unknown common variance gives the same minimizer. Mean and variance assumptions alone do not specify a likelihood.

For the seven design points, $\bar{x}=0$ and $\overline{x^2}=4$. In increasing time order the design matrix isIts columns are mutually orthogonal: the odd sums vanish and $\sum (x_i^2-4)=0$. Their squared lengths are $7,28,84$, respectively. ThereforeIn particular $\hat{\gamma }={\sum }_i(x_i^2-4)Y_i/84$. Testing for a quadratic contribution means testing $H_0:\gamma =0$ against $\gamma \ne 0$.

The cooling measurements are stated to have zero-mean errors and known common variance, but normality is not explicitly assumed for them. Under the independent-error linear regression model as stated, Chebyshev's inequality gives a valid conservative level-$\delta$ test: reject whenIts null rejection probability is at most $\delta$ because the estimator has mean zero and the calculated variance. If Gaussian errors are additionally assumed, the sharper exact test uses $Z=\sqrt{84}\hat{\gamma }/\sigma \sim N(0,1)$ under the null and rejects for $|Z|>z_{1-\delta /2}$. With only seven observations, exact normal calibration cannot be deduced from the printed moment assumptions alone.

Minimize $\Vert Y-X\theta {\Vert }^2$. Differentiation gives the normal equations $X^{T}X\hat{\theta }=X^{T}Y$. Full column rank makes $X^{T}X$ positive definite, soThe expectation uses the zero error mean; the covariance uses $\mathrm{Cov}(ϵ)={\sigma }^{2}I$. With independent Gaussian errors of common variance, maximizing likelihood minimizes exactly the same residual sum of squares, so this is also the maximum likelihood estimator, whether the common variance is known or estimated.

For the quadratic design, decompose each group's residual sum of squares into its within-group constant plus the group size times the squared residual of its sample mean. The three parameter combinations $\alpha$, $\alpha +\beta +\gamma$ and $\alpha -\beta +\gamma$ can fit all three group means exactly. ThereforeThe group means are independent normal variables, with variances ${\sigma }^2/m,{\sigma }^2/n,{\sigma }^2/n$. Hence their two stated linear combinations are jointly normal, andJoint normality therefore proves the two estimators are independent. Their variances are ${\sigma }^2/(2n)$ and ${\sigma }^2/(2n)+{\sigma }^2/m$, respectively. Equal replication at the two off-center temperatures is what cancels the covariance.

Define $s_i=\sin [2\pi (t_i-T)/P+\pi ]$ and use the independent quoted Gaussian errors $\sigma =10\mathrm{m}{\mathrm{s}}^{-1}$, with period and exoplanet transit epoch fixed. The model is linear in $K,V_0$ despite its sinusoidal time dependence. Minimize ${\chi }^2={\sum }_i(V_i-K{s}_i-V_0{)}^2/{\sigma }^2$. With $N=5$ the normal equations areUsing $\sum s_i=-0.741472$ and $\sum s_i^2=2.824633$, the solution isThe residuals in observation order are $(3.67,11.01,2.74,-7.90,-9.51)\mathrm{m}{\mathrm{s}}^{-1}$. Therefore ${\chi }_{\min }^2=2.950$ for $5-2=3$ degrees of freedom, giving reduced ${\chi }^2=0.983$. The upper-tail probability is about $0.40$: the fit is consistent with the stated errors, although five points do not establish a unique physical explanation.

For known errors the covariance matrix is ${\sigma }^2(X^{T}X{)}^{-1}$. Equivalently, profiling out the fitted offset givesThis is the $\Delta {\chi }^2=1$ one-parameter uncertainty, not the curvature with $V_0$ incorrectly held fixed. Propagating it through the supplied mass relation, and treating its coefficient and $P$ as fixed,The symbol $\pm nP$ in the exoplanet transit epoch denotes periodic copies of the same exoplanet transit phase, not an uncertainty distribution to propagate. No independent error for the empirical mass coefficient was supplied.

Additional independent observations retain positive weight even with twice the error: each has one quarter of the previous per-observation weight. Their information adds to the existing information matrix, so the formal mass uncertainty decreases for informative additional orbital phases and cannot increase under this fixed-error model. It can remain unchanged in a degenerate phase configuration that supplies no further amplitude information after fitting the offset. The conclusion assumes the same correct model and independent errors, not a new fitted error rescaling or unmodelled systematic effects.

Assume independent normally distributed angle errors with known standard deviation ${\sigma }_y=0.002$. The relevant chi-squared test uses the sum of squared standardized residuals,For the model through the origin, the predictions and residuals, expressed in milliradians, areConsequently ${\chi }_0^2=30.0280$. There are $4-1=3$ statistical degrees of freedom. The upper-tail P-value is approximately $1.36\times 10^{-6}$; equivalently the value greatly exceeds the $5\%$ critical value $7.815$. The zero-intercept model is strongly rejected under the stated errors.

For the alternative linear regression, minimize ${\sum }_i(y_i-c^{\prime }{x}_i-d{)}^2$. Setting both partial derivatives to zero gives the normal equations, whose solution isFor these data $\bar{x}=2508.83699{\mathrm{m}}^{-2}$, $\bar{y}=0.0475$, and $S_{xx}=4.830232\times 10^6{\mathrm{m}}^{-4}$. This yieldsThe new predictions are $(66.6548,55.4171,41.4315,26.4966)\mathrm{mrad}$, with residuals $(-0.6548,-0.4171,2.5685,-1.4966)\mathrm{mrad}$. ThereforeThe affine model is compatible with the data. Allowing for its additional fitted parameter, the decrease $\Delta {\chi }^2=27.668$ has a chi-squared distribution with one statistical degree of freedom under the nested zero-intercept hypothesis, corresponding to $p\simeq 1.44\times 10^{-7}$. Thus the model with a fitted intercept provides the better description, rather than merely lowering the residual by adding a parameter.

With the measurement variance treated as known, the coefficient uncertainties areHence $d=(13.11\pm 2.49)\mathrm{mrad}$, about $5.26$ standard deviations above zero. A positive constant systematic error in the angular zero, or an additional nearly constant torque, could account for it. Its presence is evidence against the ideal zero-intercept model; four points alone do not identify the physical cause uniquely. A calibration error in $\kappa$ only rescales the inferred gravitational constant and cannot remove this residual pattern.

Using the fitted slope after allowing for this offset would give $G^{\prime }=\kappa {\hat{c}}^{\prime }/(mML)=(7.27\pm 0.51)\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, with the period and slope errors combined in quadrature. This is conditional on the constant-offset model.

Minimizing the squared residual norm gives the normal equationsFull column rank makes $X^{T}X$ invertible. With columns corresponding to $1,x,x^2-4$, the design matrix isSymmetry gives zero cross products between the odd column and either even column, and $\sum (x^2-4)=28-28=0$ gives the remaining orthogonality. Their squared norms are seven, 28 and 84, so $X^{T}X=\mathrm{diag}(7,28,84)$. ThereforeSince the error covariance is ${\sigma }^{2}I$, least squares has covariance ${\sigma }^2(X^{T}X{)}^{-1}$. ThusThe estimators have zero covariances. independence would require an additional distributional assumption such as Gaussian errors; the stated mean and variance assumptions alone give only the covariance calculation.

Differentiating the least-squares residual norm gives the normal equationsIf $X$ has full column rank, $\hat{\theta }=(X^{T}X{)}^{-1}{X}^{T}Y$, and thereforeHere $\bar{x}=0$ and $\bar{z}=2$, so the design matrix and its Gram matrix areThe centered predictor columns are perpendicular, givingBoth estimates are linear combinations of the jointly normal errors, hence jointly normal. Zero covariance therefore also gives independence. The normality assumption is what permits this last step; orthogonality alone would only prove uncorrelatedness.

Substitute the QR decomposition into the normal equations:The diagonal of $R$ is nonzero, so $R^T$ is invertible. Cancel it algebraically to obtainCompute the three inner products in $Q^{T}b$, then use back substitution in this upper triangular matrix. There is no need to form an inverse or explicitly construct $A^{T}A$. The orthogonal projection of $b$ onto the column space is $Q{Q}^{T}b$; its orthogonal residual proves that this solution minimizes $\Vert Ay-b{\Vert }_2$.

For a positive-definite weighted inner product on a compact interval, a least-squares polynomial is the orthogonal projection of a function onto the finite-dimensional space ${\Pi }_n$ of polynomials of degree at most $n$. In a basis ${\varphi }_0,\dots ,{\varphi }_n$, its coefficients solve the normal equations $Gc=d$, where $G_{ij}=({\varphi }_i,{\varphi }_j)$ and $d_i=(u,{\varphi }_i)$. The Gram matrix is positive definite: $z^{T}Gz=\Vert \sum z_i{\varphi }_i{\Vert }^2>0$ for $z\ne 0$. Thus the coefficients exist and are unique. The residual is orthogonal to every candidate polynomial, giving $\Vert u-q{\Vert }^2=\Vert u-p_n{\Vert }^2+\Vert p_n-q{\Vert }^2$ and proving optimality. An orthogonal basis makes these equations diagonal, as in a least-squares polynomial in an orthogonal-polynomial basis.

An exposure-conditioned adaptation measure. With distress $D_i$, exposure $S_i$ and predicted distress $\hat{m}(S_i)$, the displayed convention makes higher scores mean less distress than predicted. The result depends on exposure measurement, reference population and model. A regression residual can include measurement error and omitted causes rather than a distinct protective capacity.

For ordinary least squares $D=X\hat{\beta }+e$, the normal equation gives $X^{T}e=0$. Thus $R=-e$ is also orthogonal to the fitted predictors; with an intercept its sample mean is zero. A zero in-sample linear association between exposure and this score can therefore be an algebraic consequence of its construction, not evidence that exposure has no effect. Independent validation is needed before assigning substantive meaning to the regression residual.

Let nonnegative partition-normalized B-splines $N_i$ form a basis, with ${\sum }_i{N}_i\le 1$ on the integration interval and $d_i=\int N_i>0$. The row-normalized mixed Gram matrix is $G_{ij}=\int N_i{N}_j/d_i$. The normal equations for the orthogonal projection onto their linear span become $Ga=b$, with $b_i=\int f{N}_i/d_i$ and $\Vert b{\Vert }_{{\ell }^{\infty }}\le \Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines gives $\Vert {\sum }_i{a}_i{N}_i{\Vert }_{\infty }\le \Vert a{\Vert }_{{\ell }^{\infty }}$ and proves the bound. The ordinary symmetric Gram matrix omits this essential row normalization. The projection maps continuous functions into the spline space, not onto all continuous functions.

In the weighted inner product $⟨f,g⟩={\int }_0^{1}xf(x)g(x)dx$, the best approximation to $x^2$ in the span of $1,x$ is the displayed polynomial. The normal equations are $1/4-A/2-B/3=0$ and $1/5-A/3-B/4=0$. Their solution gives $A=-3/10$, $B=6/5$. The residual is orthogonal to that span, so the Pythagorean theorem proves minimality and uniqueness. Its squared norm is $1/600$. A weight vanishing at a single endpoint still gives a positive-definite inner product on continuous functions: a nonzero continuous function remains nonzero on an interval where the weight is positive.

For the weight $w(x)=x$ on $[0,1]$, the normal equations for projecting $\sqrt{1-x^2}$ onto $\mathrm{span}\{1,x\}$ have Gram matrix $\left(\begin{array}{cc}1/2& 1/3\\ 1/3& 1/4\end{array}\right)$ and right side $(1/3,\pi /16{)}^T$. Its determinant is $1/72$, and inversion gives the displayed polynomial. orthogonality of the residual to both basis functions proves best approximation in the weighted norm.