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Explicit constants at a=100: the contraction factor, the fixed points r± and the constant-code solutions

Example

Take a=100 in The square-root recurrence Ts(x)n=sna+xn−1+xn+1 is a contraction of BR with a unique coded solution, so that

R=100+3=13,L=a−2R=74,a+2R=126.

Then 126<169=132, so 126<R=13; 74>1, so L>1; and the contraction constant of Ts on B13 is

1L=174<1.

The constant codes uˉ (u=±1) are solved by the constant sequences xn=ru with

11<r+=1+101<12,−10<r−=1−101<−9,

the two displayed bounds hold because 100<101<121; and both satisfy the exact identity ru2=100+2ru. Every coded sequence has ∣xn∣≥74, and its sign is the corresponding symbol.

Facts & Assumptions

Given: The parameter a=100, the constants R=13, L=74, and the numbers r±=1±101.

[L1]

Model data (The square-root recurrence Ts(x)n=sna+xn−1+xn+1 is a contraction of BR with a unique coded solution, The two-sign bilateral shift space, its coding map, and the ball of bounded bilateral real sequences): for a≥100, R=a+3, L=a−2R, the operator Ts is a contraction of BR with constant 1/L; its unique fixed point satisfies sign⁡xn(s)=sn, ∣xn(s)∣≥L and xn+1(s)+xn−1(s)=xn(s)2−a.

[L2]

Square roots and squares (Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}, Absolute value in an ordered field, Integer powers am, Basic properties of the absolute value): 132=169, (101)2=101, 74>1 because 74>1, and 126<13 because 126<169 and both sides are nonnegative.

Verification

technique · direct; evaluate the defining inequalities and identities at the explicit constants
1.1givenL2algebra

Endpoint inequality: a+2R=100+26=126 and R2=169, so a+2R<R2; by [L2] 126<13=R, which is the self-mapping inequality of The square-root recurrence Ts(x)n=sna+xn−1+xn+1 is a contraction of BR with a unique coded solution. Similarly a−2R=100−26=74>1, so L=74>1 and 1/L<1.

1.2givenL1L2algebra

The fixed points: for r±=1±101 one has, using [L2], r±2=1±2101+101=102±2101=100+2(1±101)=100+2r±. Hence (Tuˉx~)n=u100+ru+ru=u100+2ru=uru2=ru by [L2] and sign⁡ru=u, so the constant sequence x~n=ru is a fixed point of Tuˉ; by uniqueness it is x(uˉ).

2.1givenstep 1.2L1L2algebra∎

Bounds: ∣r±∣≥101−1>9>74=L and ∣r±∣<12<13=R, consistent with the general bounds of [L1]; the strict bounds 9>74 and 101<11 follow from 74<81 and 101<121.

Remarks

  • All verifications are exact algebraic identities involving 101, 74 and 126, with rational comparisons for the inequalities; no numerical integration, simulation or measurement is involved.

Depends on

Dependency tree · 2 levels

Sources