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The log odds ratio is the natural logarithm of an odds ratio. For two independent binomial groups with positive event and non-event counts $(a,b)$ and $(c,d)$, its estimate is $\log (ad/(bc))$, with first-order variance $1/a+1/b+1/c+1/d$ by the delta method. Zero cells need a method that handles boundary data; the displayed normal approximation is not valid there.

The Mellin transform is an integral transform suited to changes of scale: replacing $f(x)$ by $f(ax)$ multiplies its transform by $a^{-s}$ for $a>0$. For a locally integrable function $f:(0,\infty )\to \mathbb{C}$, definewhere the improper integral converges. The complex number $x^{s-1}=\exp ((s-1)\log x)$ uses the real natural logarithm because $x>0$. Its absolute value is $x^{\sigma -1}$, so absolute convergence depends on the real part $\sigma$ rather than the imaginary part $\tau$. The Mellin fundamental strip records the open range of such real parts.^[ref]

A familiar example is the exponential function $f(x)=e^{-x}$:This is exactly the Gamma integral. Near zero $e^{-x}$ tends to one, so absolute convergence requires $\sigma >0$; at infinity $x^r{e}^{-x}\to 0$ for every real $r$, so there is no upper boundary. The Gamma function has a meromorphic continuation with simple poles at $0,-1,-2,\dots$, but a finite continued value elsewhere in $\sigma \le 0$ does not make the defining improper integral converge.^[ref]

The Mellin transform of an interval indicator gives an elementary example with absolute convergence for every complex number $s$. The Mellin transform in logarithmic coordinates explains why changes of scale become translations, connecting the scaling property of the Mellin transform to the Fourier transform.

For $0<b<c<\infty$, the indicator function $f(x)=1_{[b,c]}(x)$ has Mellin transformThe interval stays away from both zero and infinity, so the integral is absolutely convergent for every complex number $s$. Its Mellin fundamental strip is the whole complex plane. Evaluating the antiderivative $x^s/s$ proves the formula for $s\ne 0$; at $s=0$, integrating $1/x$ gives the real natural logarithm $\log (c/b)$. The apparent division by zero is a removable singularity, since $c^s-b^s=s(\log c-\log b)+O(s^2)$ by the Taylor series of the exponential function. The transform is an entire function.

For example, $1_{[1,4]}$ gives $(4^s-1)/s$ with value $\log 4$ at zero. On $s=i\tau$, the Mellin transform sums oscillations over the width $\log 4$ of the interval in natural logarithm coordinates; this is the same cancellation mechanism as in the Fourier transform of an interval indicator.

Membrane transport is the passage of substances across a cell membrane. For the small solutes considered here, distinguish the available route from its driving force: a lipid bilayer, ion channel, or carrier protein determines how passage can occur, while differences in chemical potential and membrane potential determine which direction is energetically downhill. A large driving force cannot make an unavailable route conduct. See Alberts et al.: principles of membrane transport.

For an ideal dilute chemical solution at uniform temperature and pressure, with the same reference molar chemical potential on both sides, the molar electrochemical potential change for transferring a solute from compartment A to B isHere $c_{\mathrm{A}},c_{\mathrm{B}}$ are concentrations, $T$ is absolute temperature, $R$ is the molar gas constant, $z$ is signed ion charge number, $F$ is the Faraday constant, and $V_{\mathrm{B}}-V_{\mathrm{A}}$ is the electric potential difference. The natural logarithm contains a dimensionless ratio. The electrical term vanishes for an uncharged solute. More generally, use thermodynamic activities in place of the ideal concentration ratio. This combines the ideal-solution chemical potential difference with electrical energy per mole; see IUPAC Gold Book: electrochemical potential. For a biological cell, taking B inside and A outside makes the voltage term use the existing membrane potential convention.

The method of types combines two facts: there are few possible empirical distributions, and strings within each type class are equiprobable under an independent and identically distributed source. Information entropy determines how many strings share a frequency profile; relative entropy determines how unlikely that profile is under a specified source law. The connection is useful both for lossless source coding and for understanding rare observations.

The following formulas use natural logarithms, so information entropy and relative entropy are measured in nats. DefineHere $p$ is the source probability mass function and $q$ is an attainable empirical distribution. Use $0\ln 0=0$; first assume every $p(a)>0$. Base-two logarithms give the equivalent formulas with $2^{n{H}_2(q)}$ and $2^{-n{D}_2(q\Vert p)}$.

A morphogen response threshold is a concentration level used to distinguish developmental responses to a morphogen. In an idealized threshold model, otherwise equivalent biological cells exposed to $c\ge \theta$ take one response and those exposed to $c<\theta$ take another. Two morphogen response thresholds can assign three bands: fate A for $c\ge 50$, fate B for $20\le c<50$, and fate C for $c<20$, all in the toy arbitrary concentration units. The A/B/C labels specify a response rule, not named cell types or a claim that every morphogen produces sharp boundaries. See the positional-information model in Gilbert's reference under morphogen.

For the decreasing morphogen gradient $c(x)=c_0{e}^{-x/\ell }$ and $0<\theta <c_0$, solve $c(x_{\theta })=\theta$ using the natural logarithm:With $c_0=100$ and diffusion length $\ell =100\mu \mathrm{m}$, the morphogen response thresholds $50$ and $20$ giveThus fate A occupies $0\le x\le x_{50}$, fate B occupies $x_{50}<x\le x_{20}$, and fate C occupies $x>x_{20}$ in this idealization. The equality convention belongs to the response rule. If $\theta =c_0$, the crossing is at the source; if $\theta >c_0$, that morphogen response threshold is never reached anywhere in the tissue.

Now double the prescribed source concentration from $c_0$ to $2c_0$, keeping $D$, $k$, diffusion length $\ell$ and a fixed morphogen response threshold $0<\theta <c_0$ unchanged. The whole morphogen gradient doubles, but the boundary moves by an additive distance:For $\theta =20$, the crossing moves from $160.9438\mu \mathrm{m}$ to $100\ln 10\mu \mathrm{m}\approx 230.2585\mu \mathrm{m}$, a shift of $69.3147\mu \mathrm{m}$. Every fixed morphogen response threshold below the original source concentration shifts by this same distance in the toy model. The example compares two Dirichlet boundary conditions, $c(0)=100$ and $c(0)=200$; doubling an experimentally supplied source flux is a different boundary specification.

These calculations are useful because they separate changes in source concentration, diffusion length and response rule. Their limits are equally concrete: growing tissues change the geometry, a transient response can precede steady state, molecular fluctuations can blur a boundary, and signaling feedback or altered transport can change the morphogen gradient or its interpretation. A local instantaneous morphogen response threshold need not capture exposure duration, signaling adaptation or the receiving biological cell's state. Experiments on cultured human biological cells found different responses to the level and rate of change of developmental signals, illustrating why signal dynamics can matter. ^[ref] Quantitative measurements of bicoid protein and its downstream gene expression also make precision a question for experiment, rather than something guaranteed by drawing threshold lines. ^[ref] Morphogen gradients provide one mechanism for spatial patterning in biological development, alongside other signals and interactions.

The unit for information content or information entropy computed with natural logarithms. Converting to bits divides the numerical value by $\log 2$, so $1\mathrm{nat}=(1/\log 2)\mathrm{bits}$.

The natural logarithm $\ln x$ is the logarithm with base Euler's number, and the inverse function of the real exponential function. A particularly geometric definition isFor $x>1$, it is the area under the hyperbola $y=1/t$ from $1$ to $x$. For $0<x<1$, the reversed integration limits give a negative value. The fundamental theorem of calculus gives $(\ln x{)}^{\prime }=1/x$, so the natural logarithm is increasing, while $(\ln x{)}^{\prime \prime }=-1/x^2$ gives concavity.

The change of variables $t=au$ shows ${\int }_a^{ab}dt/t={\int }_1^{b}du/u=\ln b$ for $a,b>0$. Thus the intervals from $1$ to $2$ and from $2$ to $4$ each contribute area $\ln 2$, explaining $\ln 4=2\ln 2$. Equal ratios give equal increments on a logarithmic scale, even though the ordinary differences grow. This makes the natural logarithm useful for comparing multiplicative changes and solving for time in exponential growth or exponential decay. The integral construction and inverse function properties are discussed in OpenStax, the natural logarithm as an integral.

In a power-form rate law $v=k[\mathrm{A}{]}^{\alpha }[\mathrm{B}{]}^{\beta }$, the partial orders of reaction are $\alpha$ and $\beta$, and the overall order of reaction is $n=\alpha +\beta$. For example, doubling $[\mathrm{A}]$ at fixed $[\mathrm{B}]$ multiplies $v$ by $2^{\alpha }$: factors of one, two, and four correspond to partial orders of reaction zero, one, and two. These exponents can also be fractional or negative. A single constant order of reaction need not describe a rate law that is not a power form. See IUPAC Gold Book: order of reaction.

For a simple irreversible conversion $\mathrm{A}⟶\mathrm{B}$ with rate equation $-d[\mathrm{A}]/dt=k[\mathrm{A}{]}^n$, initial concentration $A_0>0$, and fixed $k>0$, compare zero-order reactions, first-order reactions, and second-order reactions using the dimensionless quantitiesTheir integrated curves, starting at $X(0)=1$, areFor the zero-order reaction, the plotted extension $X=\max (1-\tau ,0)$ stops conversion at depletion. It does not continue to obey $dX/d\tau =-1$ once $X=0$. A negative concentration would be unphysical. The first-order reaction and second-order reaction approach zero without reaching it at finite time in these models.

The common $\tau$ axis compares curve shapes after each model's own time rescaling, rather than equal laboratory times for an identical numerical $k$. In the lower panels, $X$, $\ln X$, and $1/X$ respectively give straight lines with slopes $-1$, $-1$, and $+1$ for the matching order of reaction. The natural logarithm uses a dimensionless quantity. These transformed plots help test a proposed rate law; the zero-order reaction line applies only before depletion. See Flowers et al., Chemistry 2e, §12.4: integrated rate laws.

Taking the natural logarithm of the positive product constraint gives $h(x)={\sum }_i{a}_i\log x_i=0$. Its gradient has entries $a_i/x_i>0$, so the equality constraint qualification holds everywhere on the feasible set. Use the optimization LagrangianStationarity requires $a_i-\lambda a_i/x_i=0$, so every $x_i=\lambda$. Substitution into the constraint, using ${\sum }_i{a}_i=1$, gives $\lambda =1$. Thus the only candidate for a local minimum is $x=(1,\dots ,1)$.

To prove global minimality without assuming the inequality to be established, set $u_i=\log x_i$. The constraint becomes the affine hyperplane ${\sum }_i{a}_i{u}_i=0$, and the objective function becomes $F(u)={\sum }_i{a}_i{e}^{u_i}$. Its Hessian matrix $\mathrm{diag}(a_i{e}^{u_i})$ is positive-definite, so $F$ is strictly convex. The elementary tangent inequality $e^t\ge 1+t$, with equality only at zero, givesEquality forces every $u_i=0$. ThereforeThis point is the unique global minimizer and the only local minimizer. The logarithm change of variables also explains why a multiplicative constraint leads to a convex optimization argument.

On $(0,1)$, the integrand is bounded by one. For $x\to \infty$, $\alpha x^{\beta }-2\log x\to \infty$, since every positive power grows faster than the natural logarithm. Thus $e^{-\alpha x^{\beta }}\le x^{-2}$ eventually. The comparison test for integrals proves Lebesgue integrability.

To prove joint continuity, fix $({\alpha }_0,{\beta }_0)$ with both coordinates positive. In a sufficiently small parameter rectangle, $\alpha \ge a:={\alpha }_0/2>0$ and $\beta \ge b:={\beta }_0/2>0$. For every parameter pair in this rectangle,The first argument shows that $H$ is integrable. Along any parameter sequence tending to $({\alpha }_0,{\beta }_0)$, the integrands converge pointwise, and the same $H$ dominates their eventual tail. The dominated convergence theorem gives $g(\alpha ,\beta )\to g({\alpha }_0,{\beta }_0)$. The sequential criterion for continuity proves the required joint continuity throughout the positive quadrant.

For comparison, the change of variables $u=\alpha x^{\beta }$ identifies the value with a gamma function:The domination argument establishes continuity directly, without needing a separate theorem about parameter dependence of the gamma function.

Use a uniform independent DNA background, so each nucleotide has probability $1/4$. Conditional on length five, its sequence probability is $4^{-5}=1/1024$. The requested score, often called log-odds in sequence analysis, is a log-likelihood ratio against this background:Using natural logarithms instead gives $\ln (27/4)\approx 1.90954$. Identically distributed independent positions do not, by themselves, require equal base frequencies. If the intended background instead has specified frequencies $q_A,q_T,q_G,q_C$, the general answer isThe numerical value above uses the conventional equal-frequency random DNA model.

Taking a logarithm only after multiplying probabilities does not prevent floating-point underflow: the product may already have rounded to zero, and its logarithm is then $-\infty$. Accumulate logarithms before forming tiny products; where alternative paths must be added, use log-sum-exp in the forward algorithm or rescale its intermediate probabilities.

For the real-valued expression, first consider the one-sided limit $x\to 0^{+}$. Its natural logarithm is $x\ln (\ln (1+x))$. Since $\ln (1+x)/x\to 1$,using $x\ln x\to 0$. The continuity of the exponential function therefore gives right-hand limit $1$.

The printed two-sided notation needs qualification: $\ln (1+x)<0$ for $-1<x<0$, so raising it to an arbitrary real power does not define a real-valued function throughout a left neighbourhood of zero. There is consequently no ordinary two-sided real limit on an open punctured neighbourhood with the expression as written. The value $1$ is the natural right-hand interpretation.

Let $N=4700$ and let the daily distances be independent and identically distributed random variables with probabilities $p_1=p_3=0.4995$, $p_{100}=0.001$. The unconditioned mean daily distance is $2.098$, whereas the observed trip requires a daily average near $10$. Conditioning on that rare average changes the typical frequencies substantially. It does not determine an exact boost count.

Use the method of types: for a fixed finite alphabet, the probability of an empirical distribution $q$ iswhere $D(q\Vert p)={\sum }_i{q}_i\log (q_i/p_i)$ is relative entropy, using natural logarithms. This follows directly from Stirling's approximation. There are only polynomially many types, so the empirical distribution conditional on a mean constraint concentrates at its unique relative entropy minimizer, when admissible lattice types approximate that minimizer. Any fixed neighborhood of it has a strictly smaller minimum cost than its complement.

Minimize $D(q\Vert p)$ subject to ${\sum }_i{q}_i=1$ and $q_1+3q_3+100q_{100}=10$. The entropy minimization under a prescribed sample mean gives the exponential tiltingTo verify minimization without relying only on Lagrange multipliers, any feasible distribution $r$ satisfiesRelative entropy nonnegativity makes $q$ the unique minimizer. Its mean determines $\theta$ through $Z^{\prime }(\theta )/Z(\theta )=10$, orThe derivative of the tilted mean is the tilted variance, which is strictly positive. The mean rises from $2.098$ at $\theta =0$ to $100$ as $\theta \to \infty$, so this equation has one positive solution. Numerically,The predicted boost count is $N{q}_{100}=381.6597211$. Thus the conditional large-deviation estimate is about 382 boosts.

For exact arrival at total distance $47000$, let $k,m,l$ count days of distances $100,3,1$. The constraints giveConsequently $k$ can be any even integer from $340$ to $426$. In particular, a unique count cannot be inferred from the return time. The exact conditional weights are proportional toTheir most probable count is $k=382$, consistent with the exponential tilting calculation. This finite-sample inference is a conditional probability statement, not an assertion that all journeys with the stated duration have that count.

If returning home means first crossing the distance threshold on day $4700$, the final total instead lies between $47000$ and $47099$. The last increment must exceed the overshoot; for fixed counts this selects the corresponding fraction of possible last days. Near the tilted optimum that factor is bounded away from zero, so it does not alter the leading exponential type cost. The same approximate answer follows. Exact conditional enumeration of this first-arrival interpretation also has mode $382$.

scan reads the numerical prices, including the three missing-value markers, into a vector. The data must be read row by row for the subsequent factor construction to match them. The sample size available for analysis is therefore $37$, rather than $40$. The numerical summary describes the observed prices: the sample median is £43.54 and the sample mean is £46.94. The upper tail, particularly the boots, raises the sample mean above the sample median. These summaries mix very different products and are not adjusted country comparisons.

The two character scans create labels. gl(8,5,length=40,labels=item) makes a regression factor with each of eight item labels repeated five times. gl(5,1,length=40,labels=country) cycles through the five country labels once for each item. For a regression factor and numerical response, the two plot commands produce grouped box plots. The country box plots suggest lower US prices, while the item box plots show particularly expensive boots and inexpensive cardigans. They are descriptive displays of this small basket, with different item compositions where prices are missing; they cannot separate country effects from item effects.

The first lm fits an additive two-way analysis of variance:The constraints are R's usual treatment coding. The normal linear model used for the printed tests assumes independent errors with common variance and approximately a normal distribution, and no item-by-country interaction in the mean. Missing prices are omitted, leaving $37$ responses and $1+4+7=12$ independent mean coefficients. Thus the residual degrees of freedom are $37-12=25$. This deletion is defensible for the conditional price model when missingness does not depend on the unobserved error after conditioning on the two factors; the output itself does not establish such a missing-data mechanism.

The anova tables report sequential sums of squares, rather than order-independent adjusted effects. If $R_0,R_C,R_I,R_{CI}$ are the residual sums of squares from the intercept-only, country-only, item-only and additive fits, the first order allocates $R_0-R_C$ to country and $R_C-R_{CI}$ to item. The second order allocates $R_0-R_I$ to item and $R_I-R_{CI}$ to country. Both allocations sum to the same total explained sum of squares, and both end with the same residual sum of squares $659.4$. The three missing combinations destroy balanced factorial orthogonality, so the allocation changes with order. To test country after allowing for items, use the country row in the second table; to test item after allowing for countries, use the item row in the first table. The first-entered rows do not supply those adjusted tests.

Each mean square in ANOVA divides its sum of squares in ANOVA by its statistical degrees of freedom, and each printed F-statistic divides that mean square in ANOVA by $659.4/25=26.38$. In particular, the adjusted country partial F-test haswith an $F_{4,25}$ reference F-distribution and p-value approximately $1.86\times 10^{-6}$. The adjusted item partial F-test gives $F=91.58$ on $7,25$ statistical degrees of freedom. Both factors contribute strongly under the additive normal linear model. The larger item sum of squares in ANOVA measures the much greater differences between these products, but is not evidence that every pair of items differs significantly.

The intercept $50.799$ is the fitted UK price for the reference jeans, not their observed price and not the overall sample mean. Each country regression coefficient is its adjusted price difference from the UK, assumed common across items: Sweden is £5.224 lower, France £10.708 lower, Germany £7.676 lower and the US £21.328 lower. Each item regression coefficient is its adjusted difference from the reference jeans, assumed common across countries. Thus the fitted UK boots price is $50.799+52.854=103.653$, and the fitted US boots price is $103.653-21.328=82.325$.

The coefficient standard errors come from $s^2(X^{T}X{)}^{-1}$, where $s^2=659.4/25$ and $X$ is the design matrix. Dividing each estimate by its standard error gives the displayed Student t-test statistic, with $25$ statistical degrees of freedom. The stars summarize the corresponding two-sided p-values. The individual tests compare each indicated country or item with its reference level, conditional on all other additive terms; they are neither tests of every pairwise comparison nor a simultaneous confidence interval procedure. In particular, Sweden's p-value $0.0626$ provides weaker evidence than the other country contrasts, and does not establish equality with the UK. The intercept test has little substantive relevance, and interpreting many individual stars also raises multiple hypothesis testing issues.

The residual standard error $s=5.136$ measures unexplained prices in pounds. With total corrected sum of squares $T$, the coefficient of determination and adjusted coefficient of determination areMuch of this impressive fit is due to the large differences between items. The overall F-test compares the additive fit with a constant mean: $F=((T-659.4)/11)/(659.4/25)=62.12$ on $11,25$ statistical degrees of freedom. It tests that all eleven non-intercept coefficients vanish, not that the additive model is adequate. Regression diagnostics should examine regression residuals, changing price variability across items and possible interaction. Percentage differences may be more plausible than common pound differences, motivating a natural logarithm of price. No random sample of countries or products, and no independent replication of a price within a cell, is supplied; broad claims about all designer goods or the causes of country differences require more than these conditional tests.

Finally, Item*Country expands to both main effects and their full interaction. A cell-means parameterization has one separate fitted mean for every observed cell. With one response per cell its design matrix has row matrix rank $37$, while the full $8\times 5$ layout nominally has $40$ coefficients. Three coefficient combinations are aliased because their cells are missing. This is saturation of an unreplicated two-factor regression: every observed price is fitted exactly and the residual degrees of freedom are zero. The final fit has zero residuals, $R^2=1$ and no residual-based estimate of the error variance or ordinary interaction significance test. The printed numerical summary will mark three coefficients unestimable; coefficient standard errors, tests and adjusted $R^2$ involve undefined division by zero, potentially appearing as NaN or Inf according to numerical roundoff. Perfect interpolation here is a consequence of a saturated statistical model, not evidence for a superior predictive model.

The response errors have standard deviations proportional to $|{\mu }_i|$. Thus, when fitted means approximate the true means, increasing absolute fitted values tends to accompany increasing residual spread. A residual-versus-fitted plot shows a fan for positive means, or spread increasing away from zero for means of either sign. Standardization by one common residual standard deviation, even with the usual regression leverage correction, does not remove this systematic heteroscedasticity.

Under independent errors the exact covariance matrix after fitting iswhere $k>0$ is the common proportionality constant. Hence the simple fan description is a diagnostic tendency, not an assertion that each fitted residual has exactly variance $k{\mu }_i^2$.

For positive responses, the intended variance-stabilizing transformation is the natural logarithm. The delta method givesUse $\log Y_i$ for positive data with modest relative error, then refit and check the transformed mean and residual structure. In particular, if the original mean was $X_i^T\beta$, the transformed mean is approximately $\log (X_i^T\beta )-k/2$, not automatically a linear function of the original covariates.

There is a domain qualification in the literal printed model: a nondegenerate normal distribution has negative support. For positive ${\mu }_i$, the probability of $Y_i\le 0$ is $\Phi (-1/\sqrt{k})$, so a real $\log Y_i$ is not defined throughout the stated sampling model. When $k$ is small this is a negligible-tail approximation, not an exact transformation to normal errors. If signed values must be retained and all ${\mu }_i\ne 0$, write $Y_i={\mu }_i(1+\sqrt{k}{Z}_i)$ with standard normal $Z_i$. ThenThe second term has a common distribution with finite variance, so log absolute value stabilization of a normal scale family is exact, but loses the sign of the mean and does not produce normal errors. Zeros and a zero mean are excluded from this transform. Alternatively retain the response scale and fit the stated variance model with weighted least squares weights proportional to $1/{\hat{\mu }}_i^2$ where the means stay away from zero. No arbitrary addition of a constant before taking logarithms exactly solves the given variance relation.

The limit comparison test says that if $u_n,v_n>0$ and $u_n/v_n\to L$ with $0<L<\infty$, then their series either both converge or both diverge. This follows because, eventually, $(L/2)v_n\le u_n\le (3L/2)v_n$, and the comparison test for series applies in both directions.

Here comparison with the harmonic series givesFor completeness, the natural logarithm grows slowly enough because for $x\ge 1$,so $(\log n{)}^2/n\le 16/\sqrt{n}\to 0$. The harmonic series diverges: every block from $n=2^k$ to $2^{k+1}-1$ has $2^k$ terms at least $2^{-k-1}$ and therefore contributes at least $1/2$. Hence

Write $1(n)=1$ and let $\epsilon$ be the Identity for Dirichlet convolution. The Möbius function uses the empty-product convention $\mu (1)=1$; the Von Mangoldt function has $\Lambda (1)=0$. We first establish all the arithmetic function identities needed below. Every integer divisor of $n$ contributes one to $\tau (n)$, so $\tau =1\ast 1$. If $n>1$ has $r$ distinct prime factors, the squarefree divisors correspond to their subsets andAt $n=1$ the sum is one. Thus $\mu \ast 1=\epsilon$. Writing $n={\prod }_p{p}^{v_p(n)}$, the only divisors with nonzero Von Mangoldt function are $p,p^2,\dots ,p^{v_p(n)}$, and henceThe Dirichlet convolution is commutative by replacing $d$ with $n/d$, and associative because either bracketing sums $f(a)g(b)h(c)$ over $abc=n$. ConsequentlyThese are consequences of the definitions, rather than unproved invocations of Möbius inversion.

To identify the constant, integrate on each interval $[j,j+1)$:The tail is $O(1/J)$, since the integrand is at most $2/x^2$ for $x\ge 2$. Summation therefore gives $\gamma ={\lim }_{J\to \infty }(H_J-\log (J+1))$, where $H_J$ is the harmonic number, and $H_J=\log J+\gamma +O(1/J)$. Thus this is Euler's constant.

Put $r=\lfloor \sqrt{N}\rfloor$. Counting pairs of positive integers whose product is at most $N$, and subtracting the square counted twice, gives the Dirichlet hyperbola method identitySince $r=\sqrt{N}+O(1)$, the harmonic number estimate yieldsThe increasing natural logarithm also satisfies ${\sum }_{n\le N}\log n=N\log N-N+O(\log (2N))$ by comparison with its integral. Subtracting proves

For the second implication, denote the Mertens function by $M(x)$ and the Second Chebyshev function by $\psi (x)$. Apply the proved Dirichlet convolution identities to $b(n)=\tau (n)-\log n-2\gamma 1(n)$:Summing this identity over integers up to $N$ gives the exact relationFix an integer $K\ge 2$. The contribution to $E(N)$ from $d\le N/K$ has absolute value at mostFor $d>N/K$, group by $j=\lfloor N/d\rfloor$. This gives exactlyunder the assumed cancellation $M(x)=o(x)$; extending that assumption from integers to real $x$ uses $M(x)=M(\lfloor x\rfloor )$. Therefore ${\mathrm{limsup}}_{N\to \infty }|E(N)|/N\ll K^{-1/2}$. Letting $K\to \infty$ proves the prime number asymptotic from Mertens cancellation:

For each element $x$ of the product set $A\cdot B$, let $s_x$ count its representations as $ab$ with $(a,b)\in A\times B$. The ratio equality $a/b=c/d$ is equivalent to $ad=cb$. Swapping the two $B$ coordinates is a bijection between the ratio-equality quadruples and the equal-product quadruples. Therefore the multiplicative energy satisfiesThe Cauchy-Schwarz inequality gives the energy lower bound:For the upper bound, place the Cartesian product $P=A\times B$ in the strictly positive quadrant. For every occupied Euclidean ray from the origin let $P_{\lambda }$ be its points and $r_{\lambda }=|P_{\lambda }|$. The slope is $\lambda =b/a$, soTake $\log$ to be the natural logarithm and put $L=\lceil \log |B|\rceil \ge 1$, using the non-triviality assumption $|B|\ge 2$. Partition the possible occupancies into $L$ classes: $[e^j,e^{j+1})$ for $0\le j<L-1$, and $[e^{L-1},|B|]$ for the last class. This last closed endpoint ensures that the partition works even when the top endpoint is attained. Within each class the ratio of any two occupancies is at most $e$.

Fix one class and order its occupied Euclidean rays by increasing slope, with occupancies $r_1,\dots ,r_t$. Write $S=|A+A||B+B|$, the cardinality of $(A+A)\times (B+B)$. If $t=1$, then $r_1^2\le |A||B|\le S$, since adding any fixed element gives an injection of $A$ into its sumset $A+A$, and similarly for $B$.

If $t\ge 2$, the sums $P_i+P_{i+1}$ contain exactly $r_i{r}_{i+1}$ different points, by injectivity of sums on two distinct rays. Indeed, the two direction vectors are linearly independent, so the coefficients of a sum uniquely recover its two summands. Positivity puts every sum strictly inside the open planar sector between those two Euclidean rays. The sectors between successive selected Euclidean rays are disjoint, even if there are additional unselected rays between them. All these sums belong to $(A+A)\times (B+B)$, and henceFor neighbouring occupancies their ratio lies between $e^{-1}$ and $e$, soSumming covers every $r_i^2$ at least once, and gives ${\sum }_i{r}_i^2\le 4S$. The same bound holds for empty or singleton classes by the preceding observations. Adding over the $L$ classes proves the multiplicative energy sumset bound $E(A,B)\le 4LS$. Combining both bounds yields the sum-product conclusion:If instead $\log$ is interpreted as base two, use the same $L=\lceil {\log }_2|B|\rceil$ occupancy classes with $2$ in place of $e$; $2+2^{-1}<4$ proves that convention as well. Positivity is essential to the sector argument.

Since the natural logarithm is increasing, comparison with an integral givesThe lower bound is $n\log n-n+1$. Dividing all three expressions by $n\log n$ and applying the squeeze theorem gives