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Complete the square in the denominator and split the numerator into a multiple of its derivative plus a constant:ThusThe natural logarithm needs no absolute-value correction because its quadratic argument is strictly positive for every real $x$. The second term is an inverse tangent antiderivative.

On any interval where $\sin x>0$, the chain rule gives $y^{\prime }=\cos x/\sin x=\cot x$. Differentiating again,The domain condition is necessary because the real natural logarithm requires a positive argument.

Expand the natural logarithm and reciprocal separately:Multiplication gives the first three nonzero terms of the Maclaurin series:The local expansions converge for $|x|<2$.

For $u>0$, the principal inverse tangent satisfies $\arctan u+\arctan (1/u)=\pi /2$: the two positive acute angles are complementary. Taking $u=\pi /4$ therefore makes the argument of the natural logarithm equal to $(2/\pi )(\pi /2)=1$. The requested value is $\ln 1=0$.

For the natural logarithm, $f(1)=0$, $f^{\prime }(1)=1$, $f^{\prime \prime }(1)=-1$, and $f^{(3)}(x)=2/x^3$. The quadratic Taylor polynomial therefore givesThe Taylor remainder is $R_3=2(0.1{)}^3/(6{\xi }^3)$ with $1<\xi <1.1$. It is positive, so the approximation is an underestimate. Moreover,Thus $1/3000\approx 0.0003334$ is a valid upper bound on the absolute error.

Take $t=e^x>0$ in the substitution rule, so $dx=dt/t$. The transformed integral isThe two terms follow from the derivatives of the natural logarithm and the inverse tangent. Returning to $x$ gives the antiderivative

For the natural logarithm, $f^{\prime }(x)=1/x$. The chord slope is $(\log 2-\log 1)/(2-1)=\log 2$. Hence the mean value theorem point satisfies $1/\xi =\log 2$, givingIt lies strictly between one and two. For example, $\log 2={\int }_1^{2}du/u$ lies strictly between $1/2$ and one.

The chord through the endpoints is the straight line $y=(x-1)\log 2$. At $\xi =1/\log 2$, the tangent line isThese two straight lines have the same slope, as required by the mean value theorem. Since $f^{\prime \prime }(x)=-1/x^2<0$, the natural logarithm is a concave function: its graph lies above the chord and below this tangent line.

Cancellation of the factorials in the consecutive ratio givesSet $t=1/n$. The derivative of the natural logarithm at $1$ gives $\ln (1+t)/t\to 1$. ThereforeThus the necessary condition from the term test for divergence is satisfied. On its own it does not prove that $\sum u_n$ converges.

For $n>1$, the Taylor series of the natural logarithm converges absolutely at $1/n$:Multiply this power series by $n+1/2$ and subtract $1$. The constant term cancels the subtracted $1$, and the coefficient of $1/n$ is $-1/2+1/2=0$. For $k\ge 2$, the coefficient of $n^{-k}$ isHenceIn particular $u_n=1/(12n^2)-1/(12n^3)+O(n^{-4})$, in Big O notation.

For a piecewise continuous function of exponential order on $t\ge 0$, its unilateral Laplace transform isThe Bromwich inversion formula, at a point of continuity, isThe Bromwich contour is a straight vertical line, traversed upwards, with $\gamma$ to the right of the abscissa of convergence and hence of all singularities of the transform. Under the usual piecewise smoothness assumptions, at a jump the symmetric inversion limit gives the mean of the two one-sided values.

In the transform in the question, put $\lambda =e^{-t}$. Then $d\lambda =-e^{-t}dt$ andThus the restricted Mellin transform on the unit interval is the Laplace transform of $f(t)=h(e^{-t})$. Apply the Bromwich inversion formula with $t=-\log \lambda$, where $0<\lambda <1$, to obtainHere $\gamma$ lies in the convergence half-plane of $H$. At a discontinuity this formula has the same mean-value interpretation. The real natural logarithm defines ${\lambda }^{-s}$ unambiguously.

Take the natural logarithm of the exponential decay model. To keep the logarithm dimensionless, choose the reference rate $R_{\ast }=1\mathrm{count}\mathrm{da}{\mathrm{y}}^{-1}$ and defineThen $y=mx+c$. With time measured in picoseconds, the slope has units ${\mathrm{ps}}^{-1}$ and the intercept is dimensionless. A straight-line linear regression in these variables therefore estimates the inverse mean lifetime and logarithmic normalization.

Write $A=-{\sum }_i\log X_i>0$, with $\log$ the natural logarithm. The likelihood function and log-likelihood function areThe score function is ${\ell }^{\prime }(\theta )=n/\theta -A$ and the second derivative is ${\ell }^{\prime \prime }(\theta )=-n/{\theta }^2<0$. Thus the log-likelihood function is strictly concave, its stationary point is the unique global maximum, and the maximum-likelihood estimator isThe one-observation Fisher information is $I(\theta )=1/{\theta }^2$. The support of a probability distribution here is independent of $\theta >0$, so the regular asymptotic normality of a maximum likelihood estimator givesThere is also a direct check: $U_i=-\log X_i$ has an exponential distribution with rate $\theta$, since $\mathbb{P}(U_i>u)=\mathbb{P}(X_i<e^{-u})=e^{-\theta u}$ for $u>0$. Its mean and variance are $1/\theta$ and $1/{\theta }^2$. Apply the central limit theorem to $A/n$ and the delta method to $h(u)=1/u$ to obtain the same result.

A likelihood-ratio test rejects for small $\lambda$. The derivative of the natural logarithm of its variable factor isThus $t^n(1-t{)}^m$ increases up to $n/(n+m)$ and decreases thereafter, tending to zero at both endpoints. Every nontrivial lower level set of $\lambda$ therefore consists of two tails. For a prescribed test size $0<\alpha <1$, the exact critical region isWriting $F_{n,m}$ for the cumulative distribution function of the beta distribution with parameters $(n,m)$, the required two equations areThe first makes both cutoffs have the same likelihood ratio; the second calibrates the null hypothesis rejection probability. Equivalently, the second left-hand side isAs the common likelihood ratio threshold increases from zero to one, this probability increases continuously and strictly from zero to one; hence the equations determine unique cutoffs. The two tail probabilities need not each equal $\alpha /2$ when $n\ne m$. Exact calibration holds for every common rate because the given null distribution of $T$ is independent of that rate. Boundary equalities have zero probability.

Put $A=-{\sum }_i\log x_i>0$, $B=-{\sum }_j\log y_j>0$, and $N=n+m$. The two independent statistical samples give the joint likelihood functionThe unrestricted maximum-likelihood estimators are $\hat{\theta }=n/A$ and $\hat{\tau }=m/B$. Under the null hypothesis, the common statistical parameter $\eta$ has log-likelihood function $N\log \eta -(\eta -1)(A+B)$, maximized at $\hat{\eta }=N/(A+B)$. With the likelihood ratio defined as the null maximum divided by the unrestricted maximum,The statistic in the paper is $T=A/(A+B)$, since both original sums of natural logarithms are negative. HenceIt has maximum one at $T=n/N$, corresponding to equal fitted rates. This is the likelihood-ratio test for equality of beta shape parameters.

For functions $f,g:\mathbb{N}\to \mathbb{R}$, big O notation meansUsing the natural logarithm and exponential function, $3^{\log n}=n^{\log 3}$. Since $\log 3>1$, for every $n\ge 1$ we have $n\le n^{\log 3}$. The witnesses $C=1,N=1$ therefore prove

For natural logarithms and sufficiently large $n$, take the logarithm of the positive expression:Set $t=\log \log n$. Then $\log n=e^t$. The exponential function grows faster than $t^2$: for example, its power series gives $e^t\ge t^3/6$ for $t>0$, so $t^2/e^t\le 6/t\to 0$. In particular, eventually $(\log \log n{)}^2\le \log n$. Exponentiating yieldsFor logarithms to a fixed base $b>1$, the natural logarithm of the left side is $(\log \log n-\log \log b{)}^2/\log b$, still smaller than $\log n$ eventually. The big O notation conclusion is therefore independent of that base choice.

Use natural logarithms; the same argument works for any fixed base greater than $1$. The function $\log \log x$ is increasing for $x>1$, since its derivative is $1/(x\log x)>0$. With natural logs the sole negative summand is at $i=2$. Put $C_0=|\log \log 2|$.

For all sufficiently large $n$ with $\log \log n\ge 1$, monotonicity and the triangle inequality giveThese are fixed constants and an eventual bound, provingFor another fixed logarithm base there are still only finitely many negative initial terms, which are absorbed in $C_0$.

Both injective and surjective. The natural logarithm is strictly increasing on the positive real numbers, so it is injective. For every $y\in \mathbb{R}$, $x=e^y>0$ satisfies $\ln x=y$; hence the function is surjective, with the exponential function as its inverse function.

True. The logarithm identity givesFor $n\ge 1$, $e^n\ge n$, so $0\le n{e}^{-n}\le 1$. By monotonicity of the natural logarithm, the expression lies between $0$ and $\ln 2$. It is therefore $O(1)$ in Big O notation.