S01 · implies
Before
n\ge0
After
n+1>0,\quad f_n:D\to\mathbb R
Conditions: none
Rule: algebra · Uses: G1
Leaf basis: algebra
The denominator is nonzero and all maps share the specified domain and codomain.
Pointwise tail certificates on a finite domain
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derivation-mko-pointwise-convergence-of-functions-01 · ocme-new142-reader-adoption-20261021
Goal and scope
Prove pointwise convergence of f_n(p)=1+2/(n+1), f_n(q)=-2-3/(n+1) to f.
The common finite domain {p,q}, real values and n≥0; exact rational-tolerance tail certificates are extended to every positive real epsilon using the Archimedean property, with no infinite-domain extrapolation.
Givens and symbols
algebraepsilondefinition_here
G1— D={p,q}, f(p)=1,f(q)=-2; the displayed full formulas define f_n for every n≥0.
D— common finite domainN_p,N_q— point-dependent tail indices
Rules and sources
mko_definition· cited — This slice uses a complete finite domain and the symbolic tail f_n(x)=f(x)+c_x/(n+s). Supply an exact epsilon witness for each x. A maximum of the finitely many witnesses gives uniform convergence for this finite model, not for an arbitrary infinite domain. A finite prefix is insufficient. · mko-pointwise-convergence-of-functionsalgebra· prerequisite — Use real-field identities; division requires a nonzero denominator and multiplication by a positive number preserves inequalities.epsilon· prerequisite — Convergence requires for every positive epsilon an integer N controlling every n≥N by a strict error bound; use the Archimedean property of the reals.definition_here· prerequisite — Pointwise convergence: for each x and positive epsilon there exists N(x,epsilon) making the error at that x smaller than epsilon for every n≥N.
Main derivation
S02 · equivalent
Before
f_n(p)-f(p)=2/(n+1)
After
|f_n(p)-f(p)|=2/(n+1)
Conditions: none
Rule: algebra · Uses: G1, S01
Leaf basis: algebra
Fix p first; positivity of numerator and denominator removes the absolute value.
S03 · implies
Before
\varepsilon>0,\quad N_p=\lfloor2/\varepsilon\rfloor+1
After
n\ge N_p\Longrightarrow n+1>2/\varepsilon
Conditions: \varepsilon\in\mathbb Q_{>0}
Rule: epsilon · Uses: S02
Leaf basis: epsilon
Choose an integer strictly greater than 2/epsilon; every later denominator is larger.
S04 · implies
Before
n+1>2/\varepsilon
After
|f_n(p)-f(p)|<\varepsilon\quad(n\ge N_p)
Conditions: none
Rule: algebra · Uses: S02, S03
Leaf basis: algebra
Reciprocating positive numbers reverses the inequality and yields the strict epsilon bound.
S05 · equivalent
Before
f_n(q)-f(q)=-3/(n+1)
After
|f_n(q)-f(q)|=3/(n+1)
Conditions: none
Rule: algebra · Uses: G1, S01
Leaf basis: algebra
At q the negative sign disappears under absolute value.
S06 · implies
Before
N_q=\lfloor3/\varepsilon\rfloor+1
After
n\ge N_q\Longrightarrow |f_n(q)-f(q)|<\varepsilon
Conditions: \varepsilon>0
Rule: epsilon · Uses: S05
Leaf basis: epsilon
The same denominator comparison supplies q's all-tail certificate.
S07 · definition
Before
\forall x\in D\ \forall\varepsilon>0\ \exists N_x\ \forall n\ge N_x
After
f_n(x)\to f(x)\text{ pointwise on }DConditions: none
Rule: definition_here · Uses: G1, S04, S06
Leaf basis: definition_here
Given any real epsilon>0, the declared Archimedean property supplies a positive integer k>1/epsilon. Set delta=1/k, so delta is rational and 0<delta<epsilon. Instantiate the preceding tail certificates at tolerance delta to obtain N_p=2k+1 and N_q=3k+1, and take N=3k+1. For every n≥N, both pointwise errors are at most 3/(n+1)<1/k<epsilon. Thus the pointwise quantifiers hold for every positive real tolerance; this two-point model additionally has a common tail.
S08 · implies
Before
N=\max(N_p,N_q)
After
n\ge N\Longrightarrow\forall x\in D\ |f_n(x)-f(x)|<\varepsilon
Conditions: none
Rule: epsilon · Uses: S03, S06
Leaf basis: epsilon
For any positive real epsilon, use the rational tolerance delta=1/k chosen in S07. Here N_p and N_q are the indices instantiated at delta; their finite maximum N=3k+1 controls both points for every n≥N with error smaller than epsilon. This is an additional finite-domain inference, not an extrapolation to infinite domains.
Substeps and lemmas
No substeps in this guide.
Reverse audits
S08· premise_audit · S03, S06 — The common N is an additional finiteness inference, not part of pointwise convergence alone.S07· premise_audit · G1, S04, S06 — The x and N quantifiers have not been interchanged in the definition.S06· premise_audit · S05 — Pointwise convergence permits different N_q and N_p.S05· inverse_equivalence · G1, S01 — Negative function values do not make the absolute error negative.S04· premise_audit · S02, S03 — A non-strict epsilon bound is not silently substituted for the strict one.S03· premise_audit · S02 — N_p depends only on p and epsilon, not the later index n.S02· inverse_equivalence · G1, S01 — The sign check uses the full formula, not a finite table.S01· premise_audit · G1 — The domain does not vary with n.
Error controls and negative controls
Strict epsilon
Use floor(C/epsilon)+1 to preserve strict inequality.
Complete domain
The two points exhaust this domain, not an infinite one.
- Checking n≤100 proves pointwise convergence. →
reject— Every n≥N remains an obligation. - The minus sign makes the q absolute error negative. →
reject— The error is 3/(n+1)>0. - Every infinite-domain pointwise limit permits a finite maximum N. →
reject— Infinitely many pointwise indices may have no common bound.
Result and limitations
Both points converge pointwise, and this finite instance additionally has common N=max(N_p,N_q).
Verification and source binding
2 declared checks for this guide; only the listed expressions are covered.
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b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f - adopted source SHA-256:
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SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW
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