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A stable multiplicative sensitivity factor $s_i$ at detector pixel $i$ changes both means to $N_i=s_i{N}_0$. The noise-free ratio is still one, so flat-field correction is not required merely to remove that fixed pattern from the mean ratio. This is the principal advantage of using two matched flats rather than the variance of a single image.

However, the photon shot noise is not cancelled. At high counts the conditional ratio variance is $2/(g{N}_i)$. Pooling detector pixels gives $E^2\simeq (2/g)⟨1/N_i⟩$, where brackets denote a spatial average. If the formula uses $\overline{N}=⟨N_i⟩$, it returnsThe inequality follows from the Jensen inequality for $1/x$ on positive signals. For small fractional response variation, the relative bias is approximately minus its squared coefficient of variation. A locally nearly uniform region or a fit using detector pixel-dependent variances avoids this bias from mixing the harmonic mean with the arithmetic mean. Stable response cancellation also presumes the conversion gain itself is uniform; gain variations, offset errors or changing sensitivity do not obey that simple cancellation. The fixed response pattern cancels in the ratio mean; unequal shot-noise levels can still bias a globally pooled gain estimate.

If spatial illumination is stable between the two sums, it acts just like the sensitivity factor: the mean ratio remains one but each detector pixel has a different shot-noise variance. A whole-image standard deviation is disproportionately influenced by faint detector pixels, so substituting a global arithmetic mean signal generally underestimates detector conversion gain. If the illumination pattern differs between the sums, their noise-free ratio is not constant; its structure adds apparent variance and causes a further systematic error.

Use locally uniform signal bins or subregions, remove offsets, mask defective or saturated detector pixels, and propagate the local shot and read variances. In the stable-pattern case a useful alternative to pooling raw ratios is the normalized differenceAt high signal and negligible read noise, treating the denominator at its mean gives $\mathrm{Var}(U_i)\simeq (2N_i/g)/(2N_i)=1/g$, independent of the illumination. Thus a variance fit or $g\simeq 1/\mathrm{Var}(U)$ can use a spatially varying flat without the arithmetic/harmonic mismatch. Measured noisy denominators cause higher-order corrections, so independent or smoothed signal estimates can be preferable. Fit and subtract a genuine changing illumination pattern before noise analysis, using sufficient smoothing not to fit away the random noise. Arbitrary flat-field normalization without propagating its noise does not by itself recover the original gain.

Equal variances make every weight $1/N$, so the inverse-variance weighted mean reduces to the arithmetic mean. Independence givesThus the variance scales as $N^{-1}$, whereas the uncertainty expressed as a standard deviation scales as $N^{-1/2}$.

Write $u_Q=|Q{|}^{-1}{\int }_{Q}u$. Since $u-u_Q$ has arithmetic mean zero,and the pairwise-difference identity givesFirst take $u$ smooth. Join $x$ to $y$ by changing one coordinate at a time and apply the Cauchy-Schwarz inequality:For a one-dimensional slice $v$, the fundamental theorem of calculus yieldsIntegrating the $i$th summand over $x,y\in Q$ therefore gives at most $L^{n+2}\Vert D_{i}u{\Vert }_{L^2(Q)}^2$. HenceThe density of smooth functions in a Sobolev space extends the estimate to every $u\in H^1({\mathbb{R}}^n)$. Thus

Among all depth-one regression tree splits, the best split separates the second observation from the first and third by cutting advert1 between $0.89$ and $1.13$. The test point with advert1 equal to zero reaches the leaf containing responses $60$ and $53$, so the output is their arithmetic mean,This decision stump makes a piecewise-constant prediction far outside the observed predictor range and cannot extrapolate the spending trend towards the origin. Its shallow structure and leaf averaging keep its variance of an estimator modest, while that extrapolation failure can produce substantial bias of an estimator.

A Pauling bond-energy excess subtracts a chosen like-atom baseline from the unlike-atom bond-dissociation energy. For an unequal-input arithmetic example, stipulate $D_{AA}=200$, $D_{BB}=450$ and $D_{AB}=421.485\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, all with one dissociation convention. The arithmetic mean baseline and excess areThe arithmetic-baseline Pauling scale model therefore gives $|{\chi }_{\mathrm{P}}(A)-{\chi }_{\mathrm{P}}(B)|=\sqrt{96.485/96.485}=1$. The bond-dissociation energy $421.485$ is not itself the excess: subtracting the baseline is essential. All three numbers are stipulated, not experimental values for named chemical elements.

Changing the baseline changes the model. For the same stipulated inputs, a geometric mean baseline would be $\sqrt{200\times 450}=300\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, yielding an excess of $121.485\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ and a difference of approximately $1.122$. It is a different construction, even though both start from the same bond-dissociation energies. State which baseline is used before comparing electronegativity differences.

The arithmetic-baseline Pauling scale also has a useful inverse: a difference of $0.5$ corresponds to an excess of $24.12125\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, and a difference of $2$ corresponds to $385.94\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Doubling the electronegativity difference quadruples the modeled excess. Interchanging the endpoint labels changes neither quantity; the square root alone cannot choose the direction of bond polarization.

In The Nature of the Chemical Bond. IV. The Energy of Single Bonds and the Relative Electronegativity of Atoms (printed pp. 3576–3578, Table II and equation (1)), Linus Pauling organized excess energies for chemical bonds into an approximately additive map of their square roots. He also identified a limitation: his H–F map separation of $2$ predicted an excess of $4.00\mathrm{eV}$, exceeding his table's value by $1.23\mathrm{eV}$. Those are historical values and a historical model discrepancy, not modern recommended inputs.^[ref]

Pauling electronegativity is a relative, dimensionless scale motivated by bond-dissociation energies. The interesting observation is that a chemical bond between unlike chemical atoms can be stronger than a simple baseline inferred from the corresponding like-atom chemical bonds. An empirical excess energy can then be associated with an electronegativity difference. The 1932 paper The Nature of the Chemical Bond. IV. The Energy of Single Bonds and the Relative Electronegativity of Atoms built such a map; its original coordinates used hydrogen as zero, so those coordinates should not be mistaken for the modern tabulated Pauling electronegativities.^[ref]

One arithmetic-baseline version of the Pauling scale relation isHere $D_{AB}$, $D_{AA}$ and $D_{BB}$ are consistently defined bond-dissociation energies per molecule, and the arithmetic mean is the chosen baseline. With molar energies instead, write $\sqrt{{\Delta }_D/(96.485\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1})}$, approximately $0.102\sqrt{{\Delta }_D/(1\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1})}$. The denominator makes the result dimensionless; the factor $0.102$ is not a unit-independent magic constant. This is an empirical relation with a specified baseline, not a definition valid for every chemical bond.^[ref]

For an explicitly synthetic example, suppose $D_{AA}=D_{BB}=300\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ and $D_{AB}=396.485\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ with the same temperature and dissociation convention. Then ${\Delta }_D=96.485\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ and the model gives an electronegativity difference of $1$. Replacing $D_{AB}$ by $300\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ gives zero; a negative excess would make this real square-root model inapplicable, not produce a negative electronegativity difference. These invented inputs illustrate the formula and are not measurements of named chemical substances.

Only differences appear, so adding the same constant to every Pauling electronegativity leaves this relation unchanged. A reference value must fix the origin. The bond-dissociation energies alone also do not tell which endpoint attracts electron density more strongly, because interchanging $A$ and $B$ leaves the formula unchanged. Chemical comparisons fix that direction. Thus Pauling electronegativity is useful as a compact ordering, while the underlying measured energies and approximations remain essential.

Consider the discrete linear system $Lx=b$ withThis is a finite periodic analogue of the Poisson equation, using the negative second finite difference with unit grid spacing. It is also a Graph Laplacian equation on a cycle graph; no continuum approximation is being claimed here.

Summing the four equations gives zero on the left, since each value occurs twice positively and twice negatively. A necessary condition is therefore ${\sum }_n{b}_n=0$. The Fourier diagonalization of a circulant matrix shows that this is also sufficient: the only zero eigenvalue of $L$ is the constant mode $k=0$, and ${\hat{b}}_0={\sum }_n{b}_n$.

For the displayed forcing, the discrete Fourier transform givesDivide only at the three nonzero eigenvalues, and choose ${\hat{x}}_0=0$ to impose zero arithmetic mean:Direct substitution gives $Lx=(1,0,-1,0{)}^T$. All other vectors satisfying the linear system are $x+a(1,1,1,1{)}^T$, because the kernel is the space of constant vectors. Thus balanced forcing determines the differences between values, while a reference level is an extra choice. This small model explains why a periodic Poisson equation needs both a compatibility condition and a normalization.

A sequence indexed from zero with an integer period $p\ge 1$ satisfying the displayed identity for all $n\ge 0$. Repeating a finite cycle is an example. The Cesàro mean article explains why its cumulative averages approach the one-cycle arithmetic mean.

Phase folding of a light curve maps timestamps onto a trial cycle of duration $P>0$. Given a reference time $t_0$, define the phase in cycles bywhere $\mathrm{frac}(x)=x-\lfloor x\rfloor$ is the fractional part, using the floor function also for negative $x$. Plot brightness against ${\varphi }_i$ rather than $t_i$; measurements separated by an integer multiple of $P$ occupy the same phase. For a periodic function $F(t+P)=F(t)$, this reveals the shared shape of many cycles. Folding changes coordinates; it neither averages samples nor chooses the correct $P$ by itself. Welsh et al.'s folded Kepler plot separates the exoplanet transit and orbital brightness modulation. ^[ref].

The second figure uses a completely synthetic trapezoidal dip: $P=3$ days, $t_0=1$ day, and the same $1\%$ depth and four-hour duration as the first figure. Different colours identify four successive cycles. Correct folding superimposes the dips; a wrong trial orbital period spreads them out. If $P^{\prime }=P+ϵ$, the centre of the $n$th repeated event shifts by approximately $-nϵ/P$ cycles when folded on $P^{\prime }$, for $|ϵ|\ll P$. With $P^{\prime }=3.03$ days, the fourth plotted event is displaced by $-3(0.03)/3.03\simeq -0.0297$ cycles. Transit-timing variations can also smear a constant-period fold, so retain the original timestamps and inspect individual events.

One may combine nearby phases with a weighted mean, but the uncertainty reduction requires assumptions. For $N$ independent measurements with common standard deviation $\sigma$ and effectively the same expected brightness within the bin, their arithmetic mean has standard deviation $\sigma /\sqrt{N}$. Thus sixteen independent $0.0006$-precision samples give $0.00015$ precision on that phase-bin mean. Shared calibration errors, changing starspots, and bins wide enough to mix ingress with out-of-transit brightness violate these assumptions. A folded plot can conceal changes between cycles. Gaussian process models can represent correlated variability and interpolate irregular samples, but their predictions and uncertainties depend on their assumed covariance function.

The bottom panels show the same model as relative brightness $f=F/F_0$ and as magnitude difference in one fixed photometric passband with a fixed magnitude zero point:A dip to $f=0.99$ gives $\Delta m\simeq +0.010912$ magnitudes: fainter means a larger apparent magnitude. The figure places larger magnitude differences lower on the page so the two dip shapes are easy to compare; reversing that plotting convention does not change the formula. For $f=1-\delta$ and small $\delta$, expansion of the logarithm gives $\Delta m\simeq (2.5/\ln 10)\delta \simeq 1.085736\delta$. The exact transform is nonlinear: transforming an exposure's mean brightness is generally different from averaging its instantaneous apparent magnitudes. The ordinary logarithm cannot be applied to a nonpositive background-subtracted measurement; keep such measurements and their uncertainties in brightness units for an appropriate statistical model.

A pollen tube germination assay makes tube emergence visible outside a flower. The published Practical Biology experiment, Observing the growth of pollen tubes (Practical Biology), Technical note 3, lists florist lilies among successful materials without specifying a species. Record the actual identification; success with one sample cannot establish a universal growth rate.^[ref]

Pearson's Core practical 4: Investigate the effect of sucrose concentration on pollen tube growth by Pearson Education Ltd, Teacher sheet, pp. 1–2, describes an inspectable experiment. Mix equal volumes of sucrose solution and mineral-salt medium on clean cavity slides. Dust in pollen from mature anthers with a mounted needle, record the addition time, and incubate without coverslips in covered dishes containing wet filter paper. The moist chamber reduces drying; the uncovered drops avoid oxygen limitation. Brief observations with an optical microscope at $100\times$, using an eyepiece graticule calibrated against a stage micrometer, reveal tube emergence and length. Return slides to the chambers and switch off the lamp between observations to reduce heating and drying. For comparisons of sucrose concentration, use the same flower across treatments to control maturity, then repeat with other flowers.^[ref]

An additional proposed experimental design check is a medium-only slide: it tests whether debris is being mistaken for pollen tubes. Compare matched pollen samples at different final sucrose concentrations, keeping temperature, elapsed time and observation rules fixed. A sample known to produce pollen tubes in a previously successful medium would test whether an apparent failure comes from handling or the tested condition. These controls are questions to test, not guaranteed outcomes. A negative result cannot distinguish dead pollen from an unsuitable medium without further comparisons.

Two original calculations clarify the readout. First, take $0.40\mathrm{mL}$ of $0.80\mathrm{mol}{\mathrm{L}}^{-1}$ sucrose solution and $0.40\mathrm{mL}$ of sucrose-free medium. Assuming additive volumes, conservation of the amount of substance givesThis dilution also halves the medium's original component concentrations. Label the final molar concentration, not the starting bottle.

Second, stipulate a toy field containing $40$ pollen grains, of which $12$ have emerged pollen tubes. Its emergence fraction is $12/40=0.30$. If those twelve pollen tubes have total measured length $960\mu \mathrm{m}$, their arithmetic mean length is $80\mu \mathrm{m}$. Assigning length zero to the other grains instead gives $960/40=24\mu \mathrm{m}$ per grain. These invented numbers show why emergence fraction and mean length among emerged tubes answer different questions. Specify the denominator and the emergence criterion before counting, and avoid selecting only conspicuous long pollen tubes.

Tube emergence in a drop demonstrates growth under that medium's conditions. It observes no stigma interaction, passage through a style, delivery into an ovule, gamete fusion, or seed development. Thus a positive pollen tube germination assay does not demonstrate successful fertilization in a living flower. Its beauty is making one otherwise hidden part of sexual reproduction directly observable.^[ref]

A prefix sum stores the accumulated sum through a specified position in a finite sequence or array. For inputs $a_1,\dots ,a_n$, include the empty sum $P_0=0$ and setFor the original input $(4,-2,5,-1,3,-4)$, the prefix array including $P_0$ is $(0,4,2,7,6,9,5)$. A negative entry makes the accumulated sum decrease; a prefix sum does not have to be monotone. Conversely, the original entry is recovered by subtraction: $a_m=P_m-P_{m-1}$.

The useful cancellation is that any contiguous block has sumBoth prefixes include the entries before $\ell$; subtracting cancels those entries. In the example, the sum from positions two through four is $-2+5-1=2$, recovered as $P_4-P_1=6-4$. The arithmetic mean of that block is $2/3$, obtained by dividing its sum by its length. Prefix sums therefore make many repeated questions about an unchanged array cheap: reuse accumulated totals instead of resumming every block.

The sequential recurrence builds all $n+1$ prefixes in $O(n)$ time complexity and storage; each block query needs two reads and one subtraction, hence $O(1)$ time complexity. These counts assume constant-cost arithmetic and access to array entries. If one input changes, all later stored prefixes change, so this simple representation is most useful when queries outnumber updates. For integer inputs, sufficient numeric capacity must be retained; with floating-point arithmetic, subtracting large nearby prefixes can lose relative accuracy for a small block total. The exact mathematical identity does not eliminate representation error.

The apparently sequential recurrence also motivates a prefix scan: associativity permits a tree of partial calculations that yields all accumulated results in parallel. Guy E. Blelloch's 1990 technical report systematically develops this operation and its applications; its authored chapter version explains the two tree sweeps used to obtain all prefixes. ^[ref]. ^[ref].

Process scheduling chooses which runnable operating-system process or software thread uses a CPU next, and when it gives up that CPU. Runnable means able to execute now; work waiting for an input-output operation is not eligible until that wait ends. A ready queue records eligible work. The policy chooses a recipient; a context switch supplies the mechanism for transferring execution. Sharing a single CPU lets a long calculation and an interactive program both make progress, although only one execution stream runs on that CPU at an instant.

Consider three single-software thread operating-system processes on one CPU, with abstract time units, zero context switch or scheduler overhead, and no input-output blocking. Their arrival times and CPU bursts are $(a_A,b_A)=(0,5)$, $(a_B,b_B)=(1,3)$ and $(a_C,b_C)=(2,1)$. Each CPU burst is all the execution that its task needs. With FCFS, the current task runs to completion: A occupies $[0,5)$, B $[5,8)$ and C $[8,9)$. With round-robin scheduling and scheduling quantum $q=2$, the intervals are A $[0,2)$, B $[2,4)$, C $[4,5)$, A $[5,7)$, B $[7,8)$ and A $[8,9)$. At a boundary, arrivals join the ready queue before the unfinished task whose scheduling quantum has expired; this fixes the otherwise ambiguous ordering at time 2.

The following triples always list A, B, C. Completion is an absolute time; scheduling turnaround time, scheduling waiting time and scheduling response time are durations measured from the relevant events.

For example, B first runs at time 2 under round-robin scheduling, so its scheduling response time is $2-1=1$. It finishes at time 8, giving scheduling turnaround time $8-1=7$ and scheduling waiting time $7-3=4$. Under FCFS, C waits $8-2=6$ before its sole execution interval. Under round-robin scheduling, A waits during $[2,5)$ and $[7,8)$, totaling $3+1=4$. Both timelines do exactly $5+3+1=9$ units of useful CPU work; changing the order changes who benefits first.

The appeal is that shared progress can improve responsiveness: B and C get their first CPU service sooner under round-robin scheduling, while A completes later. The arithmetic mean of the scheduling response times falls from $10/3$ to $1$; the arithmetic mean of the scheduling turnaround times is $19/3$ for both policies in this particular example. These are different goals, and this workload establishes no universal ranking. Giving each continuously runnable task repeated turns is a practical form of fairness. A smaller scheduling quantum can shorten waits between turns, but makes context switches more frequent when competitors remain runnable. If each full slice performs $q$ units of useful work and incurs a fixed switching cost $s$, the overhead fraction in that simplified repeated cycle is $s/(q+s)$. Real costs and workloads determine the choice. ^[ref]

Quantization maps numerical values to a specified discrete collection of representable values. In lossy block averaging, $Q(m)=\lfloor m+1/2\rfloor$ maps block arithmetic means in $[0,255]$ to eight-bit integers. Thus $Q(40.25)=40$ and $Q(40.5)=41$. Different means can share a stored value, making this step irreversible.

The sample median is a central value of an ordered statistical sample. The usual numerical convention selects the middle order statistic for odd sample size and the arithmetic mean of the two central order statistics for even sample size:NIST uses this convention. ^[ref]. For $(0.7,0.1,0.9,0.4,0.2)$, the sample median is $0.4$; for $(0.1,0.2,0.4,0.7)$, it is $(0.2+0.4)/2=0.3$, which need not be an observed value.

More generally, every value between the two central order statistics in an even statistical sample is a median of its empirical distribution function. The generalized-inverse sample quantile at $p=1/2$ selects the lower central order statistic in that case, so it can differ from the midpoint convention. Ties are allowed in all these definitions.

The sample median measures the center using sorted positions. Changing the largest observation from $0.9$ to $9$ in the five-value example leaves the sample median at $0.4$, while the sample mean rises from $0.46$ to $2.08$. Its sampling distribution still depends on the population probability distribution: the mean $1/2$ found for five observations from a uniform distribution is a special calculation, and a sample median's expected value need not equal the population median in general.

For a concrete counterexample, three independent and identically distributed random variables with rate-one exponential distribution have sample median $X_{(2)}$ with expected value $1/3+1/2=5/6$, by independent exponential order-statistic spacings. The population median instead solves $1-e^{-m}=1/2$, giving $m=\log 2\approx 0.693147$. Thus even the middle order statistic can have mean different from the population median.

Equivalent chemical bonds in an intact molecule need not have equal successive bond-dissociation enthalpies. Once one chemical bond is removed, the next reactant has a different electronic quantum state and may relax to a different molecular geometry. The successive steps compare different species. This distinction also explains why a tabulated average chemical bond strength is not necessarily the cost of the first cleavage.^[ref]

For an invented species $A{B}_2$, stipulate the following gas-phase steps, with all enthalpies at the same temperature and the same standard states:The symbols stand for hypothetical species with consistently chosen electronic states. They are not values for water or another measured molecule. By Hess's law, the total enthalpy of atomization for $A{B}_2(g)\to A(g)+2B(g)$ is $700\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; the arithmetic mean per broken chemical bond is $350\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Neither individual step equals that mean. Nor must the second step be cheaper: here it is deliberately more expensive. An exact sequence sums exactly; substituting averages taken from other molecules generally gives only an estimate.

For nonnegative real numbers $a_i$ and positive weights $w_i$ with ${\sum }_i{w}_i=1$, the weighted geometric mean does not exceed the weighted arithmetic mean:For positive $a_i$, this follows from Jensen's inequality and the concavity of the natural logarithm; a limit extends it to zero inputs. For two terms it gives $a^s{b}^{1-s}\le sa+(1-s)b$ when $0<s<1$.