Research proof index

The stopped valued of the predictable part with respect to τₙ(c) is less than or equal to c.

The mesh step-extension of the discrete predictable part is strongly adapted.

If $(X_i)_{1 \leq i \leq m}$ is a $\tau$-minimizer, and $k := D[(X_i)_{1 \leq i \leq m}]$, then for any other tuples $(X'_i)_{1 \leq i \leq m}$ and $(Y_i)_{1 \leq i \leq m}$ with the $X'_i$ G$-valued, one has $k - D[(X'_i)_{1 \leq i \leq m} | (Y_i)_{1 \leq i \leq m}] \leq \eta \sum_{i=1}^m d[X_i; X'_i|Y_i].$

In the $\tau$-minimizer endgame, let $X_1',X_2'$ be independent copies of $X_1,X_2$, set $k=d[X_1;X_2]$, $U=X_1+X_2$, $V=X_1'+X_2$, $W=X_1'+X_1$, $S=X_1+X_2+X_1'+X_2'$, and $I_1=I[U:V\mid S]$. If $c[A\mid S\#A\mid S]=\sum_{i=1}^2\bigl(d[X_i^0;A\mid S]-d[X_i^0;X_i]\bigr)$, then $c[U\mid S\#U\mid S]+c[V\mid S\#V\mid S]+c[W\mid S\#W\mid S]\le(6-3\eta)k+3(2\eta k-I_1)$.

A first-order Euler–Maclaurin formula. For $0\le a\le b$ and a differentiable function $f$ whose derivative is continuous on $[a,b]$, the sum over integers $\lfloor a\rfloor<k\le\lfloor b\rfloor$ equals $f(a)B_1(a)-f(b)B_1(b)+\int_a^b f(t)\,dt+\int_a^b f'(t)B_1(t)\,dt,$ where $B_1(t)=t-\lfloor t\rfloor-\tfrac12$.

Single contraction. Contracting the last k of k+1 index pairs leaves one free pair σ, τ, with the factorial factor (4-1)(4-2)….