An interesting result occurs, if e<>1, thus if rescaling is allowed.

I tried to find one e, for which the resulting powerseries are linear scalings of each other, such that

g0(x)=c * g1(x)

This is possible.

Assume again, that, denoting the fixpoints p and q, such that

f(x) = x + (x - p) (x - q)

Denote also the substitution

z = x' = x/e - d

and the backsubstitution

x" = (x + d)*e

Recall, that for a fixpoint p, d and e must be selected that p = e*d

If I select

then the resulting G0 and G1-matrices provide the same powerseries with a linear scaling.

Example

The matrices and eigenmatrices for g0 and g1 are as follows:

The additionally interesting thing is, that we get by this column-scaled binomial-matrices for G0 and G1

X = G0 * dV(2) = G1* dV(2/3)

so maybe I have just introduced some triviality.

Anyway- this does not (yet?) provide a linear scaling for the half-powers. Perhaps one can find an x->x' substitution, which provides such a linear scaling - but this is only an idea, don't know, whether this is even possible.

Gottfried

I tried to find one e, for which the resulting powerseries are linear scalings of each other, such that

g0(x)=c * g1(x)

This is possible.

Assume again, that, denoting the fixpoints p and q, such that

f(x) = x + (x - p) (x - q)

Denote also the substitution

z = x' = x/e - d

and the backsubstitution

x" = (x + d)*e

Recall, that for a fixpoint p, d and e must be selected that p = e*d

If I select

Code:

`´ `

e0 = 1 + (p-q) d0=p/e0 // first fixpoint

e1 = 1 + (q-p) d1=q/e1 // second fixpoint

Example

Code:

`´`

using fixpoint p using fixpoint q

e0 = 1 + (p-q) e1 = 1 + (q-p)

z0 = x/e0 - d0 = (x-p)/e0 z1 = x/e1 - d1 = (x-q)/e1

g0(z) = e0 z^2 + (1+(p-q))z g1(z) = e1 z^2 + (1+(q-p))z

The matrices and eigenmatrices for g0 and g1 are as follows:

Code:

`´`

W0 W1

1 0 0 0 0 0 | 1 0 0 0 0 0

0 1 0 0 0 0 | 0 1 0 0 0 0

0 2 1 0 0 0 | 0 -2 1 0 0 0

0 8/3 4 1 0 0 | 0 24/5 -4 1 0 0

0 24/7 28/3 6 1 0 | 0 -1176/95 68/5 -6 1 0

0 416/105 368/21 20 8 1 | 0 204768/6175 -4176/95 132/5 -8 1

... ...

D0 D1

1 1/2 1/4 1/8 1/16 1/32 | 1 3/2 9/4 27/8 81/16 243/32

W0^-1 W1^-1

1 0 0 0 0 0 | 1 0 0 0 0 0

0 1 0 0 0 0 | 0 1 0 0 0 0

0 -2 1 0 0 0 | 0 2 1 0 0 0

0 16/3 -4 1 0 0 | 0 16/5 4 1 0 0

0 -352/21 44/3 -6 1 0 | 0 416/95 52/5 6 1 0

0 2048/35 -384/7 28 -8 1 | 0 32768/6175 2048/95 108/5 8 1

... ...

--------------------------------------- -------------------------------------------------

G0 G1

1 0 0 0 0 0 | 1 0 0 0 0 0

0 1/2 0 0 0 0 | 0 3/2 0 0 0 0

0 1/2 1/4 0 0 0 | 0 3/2 9/4 0 0 0

0 0 1/2 1/8 0 0 | 0 0 9/2 27/8 0 0

0 0 1/4 3/8 1/16 0 | 0 0 9/4 81/8 81/16 0

0 0 0 3/8 1/4 1/32 | 0 0 0 81/8 81/4 243/32

... ...

The additionally interesting thing is, that we get by this column-scaled binomial-matrices for G0 and G1

X = G0 * dV(2) = G1* dV(2/3)

Code:

`X=`

1 . . . . . . .

0 1 . . . . . .

0 1 1 . . . . .

0 0 2 1 . . . .

0 0 1 3 1 . . .

0 0 0 3 4 1 . .

0 0 0 1 6 5 1 .

0 0 0 0 4 10 6 1

Anyway- this does not (yet?) provide a linear scaling for the half-powers. Perhaps one can find an x->x' substitution, which provides such a linear scaling - but this is only an idea, don't know, whether this is even possible.

Gottfried

Gottfried Helms, Kassel