Centered Taylor models with derivative intervals enclose the function range

Fix $x\in X$. If $x=c$, the polynomial already equals $f(c)$ and the remainder is zero. Otherwise the segment from $c$ to $x$ lies in $X$, so [L1] places $f^{(n+1)}$ there in $D$ and gives $\mathrm{\mid }{f}^{(n+1)}\mathrm{\mid }\le M$; also $\mathrm{\mid }x-c\mathrm{\mid }\le R$. The Taylor integral remainder bound [L2] is therefore at most $M{R}^{n+1}\mathrm{/}(n+1)!$.

Step 1.1 proves the stated analytic bound independently of storage. Suppose now that every operation and endpoint conversion in the Taylor-model evaluation returns a finite interval. For every $x\in X$, [L1] puts each exact coefficient of $P_n(x)$ inside the corresponding $C_j$, so the polynomial interval evaluation contains $P_n(x)$ by [L3]. The remainder from step 1.1 lies in the exact symmetric interval, whose outward endpoint conversion contains it. Thus $f(x)$ lies in the exact sum, and outward evaluation preserves that inclusion by [L3]. Since $x$ was arbitrary, the whole range $f(X)$ is enclosed. For $n=0$, $P_0(x)=f(c)\in C_0$, so the exact set-sum inclusion follows from step 1.1 even if its machine evaluation fails. On a non-finite status [L3] supplies no finite enclosure.