derivation-mko-analytic-ordinary-generating-function-01 · ocme-new142-reader-adoption-20261021
Derive the OGF from a recurrence on a justified convergence domain.
An ordinary analytic OGF, not unrestricted evaluation of a formal identity.
workG1 — a0=0,a1=1 and a_n=a_(n-1)+a_(n-2) for all n≥2; A(x)=sum_(n=0)^infinity a_n x^n. Derive analytic identities only on|x|<1/2.A(x) — ordinary generating function of a_n[x^n]A — coefficient a_nmko_definition · cited — Coefficient extraction [x^n]A=a_n is the encoding obligation. A finite partial sum uses only its declared prefix. For a_n=r^n, A(x)=1/(1−r*x) holds analytically only when |r*x|<1. For a_n=a_(n−1)+a_(n−2) with explicit a0,a1, the OGF is (a0+(a1−a0)*x)/(1−x−x^2); initial boundary terms must not be omitted. A formal power-series identity is not automatically an analytic equality. Do not evaluate outside convergence, determine a whole OGF/radius from an unknown finite prefix or infer unproved termwise differentiation from the word analytic. · mko-analytic-ordinary-generating-functionwork · prerequisite — Induction, geometric convergence and comparison, index shifts and finite linear combinations of absolutely convergent series; neither the OGF answer nor termwise differentiation is assumed.Before
a_0=0\le2^0,\quad a_1=1\le2^1
After
a_{n-1}\le2^{n-1},\ a_{n-2}\le2^{n-2}\implies a_n\le3\cdot2^{n-2}\le2^nConditions: none
Rule: work · Uses: G1, work
Leaf basis: work
The initial values form the base and recurrence the step proving0≤a_n≤2^n by induction.
Before
|x|<1/2
After
\sum_{n=0}^{\infty}|a_nx^n|\le\sum_{n=0}^{\infty}(2|x|)^n<\inftyConditions: none
Rule: work · Uses: G1, S01, work
Leaf basis: work
Geometric comparison proves absolute convergence, permitting later shifts and finite sums.
Before
A(x)=a_0+a_1x+\sum_{n=2}^{\infty}a_nx^nAfter
A(x)=x+\sum_{n=2}^{\infty}a_{n-1}x^n+\sum_{n=2}^{\infty}a_{n-2}x^nConditions: none
Rule: work · Uses: G1, S02, work
Leaf basis: work
Retain a0 and a1x boundary terms before substituting the recurrence.
Before
\sum_{n=2}^{\infty}a_{n-1}x^nAfter
x\sum_{j=1}^{\infty}a_jx^j=x(A-a_0)=xAConditions: none
Rule: work · Uses: G1, S03, work
Leaf basis: work
With j=n-1 the lower limit is 1; explicitly restore the missing a0 before substituting0.
Before
\sum_{n=2}^{\infty}a_{n-2}x^nAfter
x^2\sum_{j=0}^{\infty}a_jx^j=x^2AConditions: none
Rule: work · Uses: G1, S04, work
Leaf basis: work
The second shift starts at 0 and gives the entire A.
Before
A=x+xA+x^2A
After
(1-x-x^2)A=x
Conditions: none
Rule: work · Uses: G1, S05, work
Leaf basis: work
Move terms and factor A.
Before
|x+x^2|\le|x|+|x|^2<3/4<1
After
1-x-x^2\ne0,\quad A(x)=\frac{x}{1-x-x^2}Conditions: none
Rule: work · Uses: G1, S06, work
Leaf basis: work
Prove the denominator nonzero before division.
Before
x=1/4
After
A(1/4)=\frac{1/4}{1-1/4-1/16}=4/11Conditions: none
Rule: work · Uses: G1, S07, work
Leaf basis: work
The point 1/4 lies in|x|<1/2, so exact evaluation is justified.
Before
b_n=2^n,\quad B(x)=\sum_{n=0}^{\infty}(2x)^nAfter
B(x)=1/(1-2x)\quad(|2x|<1)
Conditions: none
Rule: work · Uses: G1, S08, work
Leaf basis: work
For the geometric family state the analytic domain explicitly; at x=1 the original series diverges.
No substeps in this guide.
S09 · premise_audit · G1, S08, work — The rational value -1 at x=1 is not a sum of the divergent series.S08 · premise_audit · G1, S07, work — No finite partial sum is substituted for the full A value.S07 · premise_audit · G1, S06, work — A nonzero denominator alone does not give convergence outside the domain.S06 · premise_audit · G1, S05, work — The equation holds on the justified convergence domain.S05 · premise_audit · G1, S04, work — The two shifts have different lower limits.S04 · premise_audit · G1, S03, work — Handle the boundary before simplifying the shift.S03 · premise_audit · G1, S02, work — Do not write A=xA+x²A and omit x.S02 · premise_audit · G1, S01, work — The domain obligation2|x|<1 is explicit.S01 · premise_audit · G1, work — The bound covers all n rather than guessing a radius from a prefix.The shifts start at j=1 and j=0; retain the a0 difference.
First use the 2^n geometric comparison to prove absolute convergence on |x|<1/2, then evaluate at the point 1/4 inside that domain.
reject — It omits a1x=x and incorrectly yields A=0.reject — Here|2x|=2≥1 and the series terms do not tend to zero.reject — Changing a later coefficient preserves the first five, so an all-index recurrence is needed.For|x|<1/2, A=x/(1-x-x²) and A(1/4)=4/11; no maximal-domain claim is made.
2 declared checks for this guide; only the listed expressions are covered.
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