The derivative is represented by the gradient, and the Hessian matrix contains the second partial derivatives:Equivalently the derivative as a linear map is $Df(x,y)(u,v)=(4x^3-4y)u+(4y-4x)v$.

Write the computed function values as $f(x\pm h)+e_{\pm }$, where $|e_{\pm }|\le \delta$. The computed central finite difference $\tilde{d}$ satisfiesThe two contributions are respectively evaluation absolute error and truncation error. For $\delta >0$, step-size choice for a noisy central derivative minimizes their sum: its derivative is $h/3-\delta /h^2$, negative before and positive afterIf $\delta =0$, the bound tends to zero as $h\downarrow 0$ and has no positive minimizing step.

Apply the preceding Taylor's theorem at $t=h$ and $t=-h$, with $h>0$. The even-order terms cancel in the central finite difference, givingBoth Lagrange remainders have magnitude at most $h^3/6$, so the triangle inequality gives the sharp central-secant derivative error

By Taylor's theorem with Lagrange remainder, for an increment $t$,where ${\xi }_t$ lies between $x$ and $x+t$. The given third derivative bound implies $|R_x(t)|\le |t{|}^3/6$.

Minimize the square of the Euclidean distance, using minimizers under strictly increasing transformations. For $s=x+y>0$,For fixed $s$ the unique minimum has $x=y=s/2$. Put $x=y=\alpha >0$ and $J(\alpha )=2{\alpha }^2+{\ln }^2(2\alpha )$. Its derivative isLet $h(\alpha )=2{\alpha }^2+\ln (2\alpha )$. Since $h^{\prime }(\alpha )=4\alpha +1/\alpha >0$, it is strictly increasing, with limits $-\infty$ and $+\infty$ at the two ends of $(0,\infty )$. It therefore has exactly one zero. The sign of $J^{\prime }$ changes there from negative to positive, proving the unique global minimum. The nearest point is $P=(\alpha ,\alpha ,\ln (2\alpha ))$ at that zero.^[ref]

The Newton root-finding iteration isStarting at $0.4$ gives $0.376376476$, $0.376544572$, and $0.376544582$. Moreover $h(0.3765)<0<h(0.3766)$ certifies the rounding. Thus $\alpha =0.377$ to three significant figures; more accurately $P\approx (0.3765445825,0.3765445825,-0.2835716452)$.

The map $u\mapsto u^2$ is strictly increasing on $[0,\infty )$: for $u,v\ge 0$, $u\le v$ if and only if $u^2\le v^2$. Therefore a point has the smallest $f$-value if and only if it has the smallest $f^2$-value, and the same applies to $g$. By minimizers under strictly increasing transformations, both asserted argmin equalities hold, including when the argmin sets are empty. The differentiability assumptions are unnecessary for this order argument.

Let $W_i=U_i^2$. Applying the same measurable function separately to independent and identically distributed random variables preserves those two properties, andIn particular, these bounded random variables are integrable random variables. The weak law of large numbers givesIf $c>1/3$, then $\{{\overline{W}}_n>c\}\subseteq \{|{\overline{W}}_n-1/3|⩾c-1/3\}$, whose probability tends to zero. If $0⩽c<1/3$, then $\{{\overline{W}}_n⩽c\}\subseteq \{|{\overline{W}}_n-1/3|⩾1/3-c\}$, whose probability also tends to zero. Taking a complement in the second case provesThe positive gap from $1/3$ is essential to this argument. Convergence in probability alone does not determine the answer at the excluded threshold $c=1/3$.

Write $Z=U^a$. Since $a>0$, the map is increasing on $[0,1]$, and for $0⩽z⩽1$,Differentiation givesand zero outside this interval. Since $B(1/a,1)=a$, this is exactly the beta distribution probability density function with parameters $(1/a,1)$. The power transform of a uniform random variable therefore satisfies $U^a\sim \mathrm{Beta}(1/a,1)$; the second parameter does not depend on $a$.

The same integral works with both powers shifted:The assumptions $s>-\alpha$ and $t>-\beta$ say precisely that the exponents at each endpoint are greater than $-1$, ensuring integrability there. Values of a negative power at the endpoints themselves can be assigned arbitrarily, because those endpoints have probability zero. This establishes the mixed power moments of a beta distribution without requiring $s$ and $t$ to be integers.^[ref]

Multiply the probability density function by $x$ and integrate over its support:The middle equality is the beta function integral with parameters $\alpha +1$ and $\beta$; the last uses the recurrence for the gamma function. Both parameters are positive, so the integral converges. Equivalently, multiplying the beta distribution probability density function by $x$ converts its kernel to that of a beta distribution probability density function with its first parameter increased by one.^[ref]

For a probability density function $f$, the finite expected value isThe absolute integrability condition makes the signed integral unambiguous. One can also allow an extended expectation when exactly one of the positive and negative parts has infinite integral, obtaining $+\infty$ or $-\infty$. If both parts have infinite integral, the expectation is undefined; subtracting two infinities is not a definition.