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ConsciousnessLearn · Explore · Joseph Anady

Fixed-point polynomial of the average recurrence

At a nonzero fixed point, multiplying z=F(z) by 7z yields 2z²−z−1=0. Factoring gives (2z+1)(z−1)=0. This algebra identifies all finite real fixed points of the displayed recurrence.

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eq-e5065fb34cf142mathematical expression
7z=5z+1+1/z⇔2z2−z−1=0⇔(2z+1)(z−1)=0.7z=5z+1+1/z⇔2{z}^{2}-z-1=0⇔\left(2z+1\right)\left(z-1\right)=0.

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At a nonzero fixed point, multiplying z=F(z) by 7z yields 2z²−z−1=0. Factoring gives (2z+1)(z−1)=0. This algebra identifies all finite real fixed points of the displayed recurrence.

Terms & definitions

TermMeaning in this source
z

Candidate fixed point

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Original context

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Averaging is an additional coupling rule. It is not implied by drawing seven operators around a center. It should not be called the unique consequence of the proposed geometry.

The fixed-point equation is

7z=5z+1+1/z⇔2z2−z−1=0⇔2z+1z−1=0.

Hence

zfixed,1=1,  zfixed,2=−12.

The derivative is
Exact original notation
7z=5z+1+1/z⇔2{z}^{2}-z-1=0⇔\left(2z+1\right)\left(z-1\right)=0.

Sources & provenance

01
Machine_Consciousness_Mathematical_Audit.docx
paragraph 161, math 1 · Sep 17, 2026
7z=5z+1+1/z⇔2{z}^{2}-z-1=0⇔\left(2z+1\right)\left(z-1\right)=0.
02
Machine_Consciousness_Mathematical_Audit.docx
paragraph 161, math 1 · Sep 17, 2026
7z=5z+1+1/z⇔2{z}^{2}-z-1=0⇔\left(2z+1\right)\left(z-1\right)=0.

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