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Truncation error is introduced by replacing an exact operation with a finite approximation. It remains even when all arithmetic is exact. For $f(x)=x^2$, the forward finite difference for its derivative at $x=1$ givesThus the truncation error is exactly $h$: $h=0.1$ gives $2.1$ and $h=0.01$ gives $2.01$. In finite precision, however, the numerator subtracts nearly equal values, and their roundoff error can be amplified by division by $h$. For smooth data with bounded evaluation errors of scale $u$, a typical bound has competing terms $C_{1}h+C_{2}u/h$. Smaller $h$ reduces the first term and increases the second; with fixed positive constants the balancing scale is $h\sim \sqrt{u}$. This is why blindly refining a finite difference eventually stops improving it.

Let a portfolio hold $x\in [0,T]$ in a risk-free asset with net return $r$ and $T-x$ in an asset returning $g$ or $b$, with probabilities $p$ and $1-p$. Assume $T>0$, $-1<b<r<g$, $0<p<1$, and $pg+(1-p)b>r$. Its state wealths are $w_g=T(1+g)-(g-r)x$ and $w_b=T(1+b)+(r-b)x$. The expected utility $p\log w_g+(1-p)\log w_b$ has strictly negative second derivative. Setting its first derivative to zero gives the displayed unconstrained root; the derivative at $x=T$ is negative, so only the lower constraint can bind. Thus the displayed formula is the unique optimum. All equity is optimal exactly when $p\ge (r-b)(1+g)/[(g-b)(1+r)]$. The positivity of both state wealths is part of the logarithmic utility domain.

A proper Riemann integral on a compact interval requires a bounded integrand, but a derivative can be unbounded. For $f(x)=x^2\sin (1/x^2)$ away from zero and $f(0)=0$, the difference quotient gives $f^{\prime }(0)=0$, whereas $f^{\prime }(x)=2x\sin (1/x^2)-2\cos (1/x^2)/x$ for $x\ne 0$. At $x_n=(2\pi n{)}^{-1/2}$, this derivative is $-2/x_n$ and is unbounded. The fundamental theorem of calculus version that integrates a derivative therefore needs an explicit integrability hypothesis.

On a nondegenerate interval, point evaluation of the derivative is unbounded for the supremum norm on smooth functions. The functions $f_N(x)=\sin (N(x-x_0))$ have supremum norm at most one but $f_N^{\prime }(x_0)=N$. Adding the supremum norm of the derivative makes this evaluation a bounded linear functional.

Suppose a family of polynomial coordinate maps has a fixed bound on its degrees and coefficients depending continuously on a compact parameter set, and sends a fixed real compact set into a compact subset of the domain of a real analytic function. A common complex neighborhood can then be chosen so the compositions are holomorphic and uniformly bounded there. The multivariable Cauchy estimate gives uniform factorial derivative bounds. For the change $\tilde{x}=x+y^2-\epsilon$, the coefficient pullback uses its inverse: $f_{\epsilon }(\tilde{x},\tilde{y})=f(\tilde{x}-{\tilde{y}}^2+\epsilon ,\tilde{y})$.

For a linear real analytic non-characteristic Cauchy problem for a partial differential equation with zero data, the local real analytic radius may be chosen uniformly over polynomial forcing terms. Fix an analyticity exponent $R^{-1}$. Each polynomial, including a polynomial obtained by a polynomial coordinate change, satisfies the corresponding factorial derivative bound with some finite prefactor, uniformly over a compact parameter range. Divide the forcing by that prefactor to put all such data in one fixed majorant class. The Cauchy-Kovalevskaya theorem supplies a common radius, and linearity rescales the solution back without shrinking it. This argument does not bound the amplitudes of the resulting solutions uniformly.

If Riemann integrable functions $f_n:[a,b]\to \mathbb{R}$ converge by uniform convergence to $f$, then $f$ is Riemann integrable andThis concerns one fixed compact interval, with finite length.^[ref]

Here is a proof outline that also establishes Riemann integrability of the uniform limit. Assume $a<b$, and put $\ell =b-a$ and $E_N=\Vert f_N-f{\Vert }_{\infty }$. One bounded $f_N$ and finite $E_N$ show that $f$ is a bounded function. For any partition of an interval $P$, its upper Darboux sum and lower Darboux sum satisfyGiven $\eta >0$, choose $N$ with $E_N<\eta /(4\ell )$, then a partition of an interval with $U(f_N,P)-L(f_N,P)<\eta /2$. The displayed gap is below $\eta$, proving Riemann integrability. Finally, $|f_n-f|\le E_n$ and monotonicity of the Riemann integral give the stated error estimate. When $a=b$, every integral is zero.

For the polynomial functions $f_n(x)=x^n$ on $[0,0.8]$, direct integration and the uniform bound giveAt $n=21$, the exact integral is approximately $0.0003353953$, while the general error bound is approximately $0.0073786976$. The bound certifies the result without an exact antiderivative. On $[0,1]$, these same integrals still tend to zero, although there is no uniform convergence: sufficient conditions need not be necessary. On an unbounded interval, even uniform convergence can fail to control total area, as uniformly vanishing functions can retain a nonzero integral shows.

Uniform convergence alone also fails to preserve derivatives: $f_n(x)=\sin (n^{2}x)/n$ converges uniformly to zero on $[-1,1]$, while $f_n^{\prime }(0)=n$. The small amplitude does not control the slopes of the sine oscillations.

Let $f_n$ be continuously differentiable functions on an interval $I$. If $f_n^{\prime }\to g$ by uniform convergence on $I$ and $f_n(c)\to A$ at one fixed $c\in I$, then $f_n$ converge locally uniformly to a continuously differentiable function $f$ with $f^{\prime }=g$. If $I$ is bounded, the convergence is uniform on all of $I$.^[ref]

The anchor fixes the additive constant that a derivative cannot see. By the fundamental theorem of calculus, defineThe uniform limit theorem makes $g$ a continuous function, so the fundamental theorem of calculus gives $f^{\prime }=g$. The displayed estimate proves uniform convergence wherever $|x-c|$ has a common finite bound, and therefore proves locally uniform convergence on any interval.

For example, on $[0,2]$ take $f_n(x)=3+1/n+x^2/2+x/n$. Its derivative is $x+1/n$, with supremum error $1/n$ from $g(x)=x$, and $f_n(0)\to 3$. The reconstructed uniform limit is $f(x)=3+x^2/2$; the exact supremum error is $3/n$, equal to the bound above. Without an anchor, the constant functions $f_n(x)=n$ have zero derivatives but no finite pointwise limit. On the real line, $f_n(x)=x/n$ has convergent anchored values and uniformly convergent derivatives, but its supremum error from zero is infinite. The bounded-distance condition has a real role.

In the common PDE convention, a uniformly convex scalar flux is a $C^2$ flux satisfying $F^{\prime \prime }(z)\ge \kappa >0$ throughout the specified state interval, with one constant $\kappa$. Equivalently, $F(z)-\kappa z^2/2$ is a convex function. This guarantees strict inequality below nontrivial chords and strict ordering of the two endpoint derivatives around the chord slope. On the whole real line it is incompatible with a globally bounded first derivative.

For independent and identically distributed random variables $U_1,\dots ,U_n$ with the continuous uniform distribution on $(0,1)$, the $r$th order statistic has Beta distributionIts probability density function, mean, and variance areThe probability density function is zero outside $[0,1]$; values assigned to its endpoints do not change any probability. Peter Kempthorne's MIT lecture notes derive the uniform order statistic law and these moments. ^[ref].

For $n=5$, the minimum, sample median, and sample maximum correspond to $r=1,3,5$. Their probability density functions are $5(1-t{)}^4$, $30t^2(1-t{)}^2$, and $5t^4$, respectively. The sample median therefore has Beta distribution $\mathrm{Beta}(3,3)$, mean $1/2$, and variance $1/28\approx 0.0357143$. Its cumulative distribution function isFor instance, $\mathbb{P}(U_{(3)}\le 1/4)=53/512\approx 0.103516$. The mean of the minimum is $1/6$, and the mean of the sample maximum is $5/6$. The fixed observed sample median $0.4$ in the order statistic example is one possible outcome; $1/2$ is the average over repeated statistical samples from this uniform distribution.

Each curve has area one. The extreme order statistics favor the ends, while the sample median favors the center. The endpoint density height $5$ is legitimate: the probability density function gives area per unit length, rather than point probability. In particular, a continuous uniform distribution has zero probability of producing an endpoint.

For a short proof, the binomial distribution counting identity under order statistic gives $\mathbb{P}(U_{(r)}\le t)={\sum }_{k=r}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)t^k(1-t{)}^{n-k}$. Taking its derivative cancels consecutive terms, leaving the displayed probability density function. Recognize its normalizing factorial coefficient as that of the Beta distribution, then substitute $\alpha =r$, $\beta =n+1-r$ into the Beta distribution mean and variance formulas. Although the original observations are independent, their order statistics generally are dependent; see the positive covariance of two uniform order statistics.

For an increasing concave utility on positive wealth, its utility conjugate uses the indicated supremum and is a convex function of the positive price variable. If $F(x)=-U(x)$ on positive $x$, extended by infinity elsewhere, then $\hat{U}(y)=F^{\ast }(-y)$. Under the Inada conditions its optimizer is the inverse marginal utility, and its derivative is the negative of that optimizer. The sign convention is part of the definition.

A Van der Waals isotherm is the pressure–particle volume curve at a fixed temperature. Put $x=v/v_c$, $t=T/T_c$, and $q=p/p_c$, using the Van der Waals critical point values below. Then the existing Van der Waals law of corresponding states becomesAll curves in the figure are calculations in this reduced model.

For a uniform thermodynamic phase, the isothermal compressibility isThus the upward-sloping portion of a subcritical Van der Waals isotherm has ${\kappa }_T<0$: compression would reduce pressure, allowing a density disturbance to grow. It is mechanically unstable. The spinodal points bound that portion; setting the derivative $\partial q/\partial x$ to zero gives their curve $t=(3x-1{)}^2/(4x^3)$. Negative slope alone guarantees only local mechanical stability: portions between the spinodal and the phase coexistence endpoints have metastability. Equilibrium replaces the entire interval between those endpoints, including these metastable portions, by a Maxwell construction plateau.^[ref]

Use the Fourier transform convention $\hat{\psi }(\omega )=\int \psi (x)e^{-i\omega x}dx$. If $\psi$ and $x^r\psi$ for $r<q$ are integrable functions, moment differentiation of the Fourier transform yields a $C^{q-1}$ Fourier transform andThe dominated convergence theorem justifies these derivatives. Under these assumptions, $q$ vanishing moments are equivalent to vanishing derivatives through order $q-1$ at zero. For $q>1$ this gives a Peano zero $\hat{\psi }(\omega )=o(|\omega {|}^{q-1})$; for $q=1$ it gives $\hat{\psi }(\omega )=o(1)$. It does not by itself justify a $q$th derivative or an $O(|\omega {|}^q)$ estimate.

If also $\int |x{|}^q|\psi (x)|dx<\infty$, subtract the degree-$(q-1)$ Taylor polynomial of $e^{-i\omega x}$ inside the integral. Its Taylor remainder gives the stronger boundFor an integrable function with compact support, every weighted integral exists and its Fourier transform extends to an entire function. The language “zero of order at least $q$” is then literal analytic function terminology. If the next integral is nonzero, the order is exactly $q$. The smooth-mask vanishing-moment criterion relates these cancellations to a refinement mask under its additional hypotheses.^[ref]