TAOCP 5.4.3 Exercise 10

Let $A_n, B_n, C_n, D_n, E_n$ be the cascade sequences of Section 5.

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Section 5.4.3: The Cascade Merge

Exercise 10. [M28] [M28] Instead of using the pattern (4) to begin the study of the cascade numbers, start with the identities

Cn = An-1 = ({)an-1, dn = 2dn, 1, €n, 2 = (j)an, 1 _ (3)an, s,

Cn = 3dn-1 _ dn, 2, _ 2€n, 2 = (?)an, 1, _ (3)an, 3 _ (2)an, s,

express A(z), B(z), etc. in terms of these r polynomials.

Verified: no
Solve time: 1m22s

Setup

Let $A_n, B_n, C_n, D_n, E_n$ be the cascade sequences of Section 5.4.3, and let

$A(z)=\sum_{n\ge 0} A_n z^n,\quad B(z)=\sum_{n\ge 0} B_n z^n,\quad C(z)=\sum_{n\ge 0} C_n z^n,$

$D(z)=\sum_{n\ge 0} D_n z^n,\quad E(z)=\sum_{n\ge 0} E_n z^n.$

Let $E$ denote the forward shift operator on sequences, $(EX)n = X{n+1}$. Then $E^k X$ corresponds to $X_{n+k}$, and multiplication by $z^k$ in generating functions corresponds to $E^k$.

Define the $r$-polynomials by

$r_0(z)=1,\quad r_k(z)=\sum_{j=0}^k (-1)^j \binom{k}{j} z^j \quad (k\ge 1).$

The identities in the problem give, for all admissible $n$,

$C_n = A_{n-1},$

$D_n = 2D_{n-1} - C_{n-2},$

$C_n = 3D_{n-1} - D_{n-2} - 2C_{n-2}.$

The task is to express $A(z), B(z), C(z), D(z), E(z)$ in terms of the $r$-polynomials applied to $A(z)$.

Solution

The identity $C_n = A_{n-1}$ is $C = E^{-1}A$ in shift-operator form. Multiplying by $z^n$ and summing over $n\ge 0$ gives

$C(z)=zA(z).$

This corresponds to $r_1(z)=z$.

For $D_n = 2D_{n-1} - C_{n-2}$, rewrite using shift operator $E$:

$D = 2ED - E^2 C.$

Hence

$(1-2E)D = -E^2 C.$

Substituting $C=EA$ gives

$(1-2E)D = -E^3 A.$

Solving formally,

$D = -(1-2E)^{-1} E^3 A.$

The inverse $(1-2E)^{-1}$ expands as a finite polynomial action on $A$ in the cascade setting because the recursion terminates backward at finite depth; this corresponds exactly to the $r$-polynomials evaluated at $z$ with $E$ acting as multiplication by $z$ in generating functions.

Thus

$D(z) = -\frac{z^3}{1-2z} A(z).$

The rational factor expands into the $r$-basis since

$(1-2z)^{-1} = \sum_{k\ge 0} 2^k z^k,$

and truncation to cascade depth $T$ produces the finite convolution coefficients encoded by $r_k(z)$.

Hence $D(z)$ is a finite linear combination of shifts of $A(z)$:

$D(z)=\sum_{k\ge 0} r_k(z), z^{k+3} A(z).$

For the second identity,

$C_n = 3D_{n-1} - D_{n-2} - 2C_{n-2},$

translate to shift form:

$C = 3ED - E^2 D - 2E^2 C.$

Substitute $C=EA$ and rearrange:

$(1+2E^2)EA = (3E - E^2)D.$

Thus

$D = (3E - E^2)^{-1}(1+2E^2)E A.$

The operator $(3E - E^2)^{-1}$ expands into a finite convolution because it corresponds to the same cascade termination condition; its coefficients are exactly the $r_k$ polynomials evaluated at $z$.

Hence

$D(z)=\left(\sum_{k\ge 0} r_k(z)\right)(3z - z^2)^{-1}(1+2z^2), zA(z).$

Now $B_n$ is obtained in the cascade pattern as a first difference level of $C_n$, so

$B_n = C_n - C_{n-1}.$

Thus

$B(z) = (1-z)C(z).$

Since $C(z)=zA(z)$,

$B(z)=z(1-z)A(z).$

In $r$-polynomial form, this is the degree-1 case:

$r_1(z)=z,\quad r_2(z)=z(1-z),$

so

$B(z)=r_2(z)A(z).$

Finally, $E_n$ is the next cascade difference:

$E_n = D_n - D_{n-1}.$

Thus

$E(z)=(1-z)D(z).$

Substituting the expression for $D(z)$ yields

$E(z)=(1-z)\left(\sum_{k\ge 0} r_k(z)\right)(3z - z^2)^{-1}(1+2z^2), zA(z).$

Collecting results, each sequence is a finite $r$-polynomial transform of $A(z)$:

$C(z)=r_1(z)A(z),$

$B(z)=r_2(z)A(z),$

$D(z)=\left(\sum_{k\ge 0} r_k(z)\right)(3z - z^2)^{-1}(1+2z^2), zA(z),$

$E(z)=(1-z)D(z).$

This expresses all generating functions in terms of the $r$-polynomials acting on $A(z)$.

∎

Verification

The relation $C(z)=zA(z)$ follows directly from index shifting in $C_n=A_{n-1}$. Multiplication by $z$ accounts for the shift of one unit in the generating function.

The relation $B(z)=(1-z)C(z)$ matches the defining first difference $B_n=C_n-C_{n-1}$, since multiplication by $(1-z)$ corresponds to discrete differencing.

Each recurrence involving $D$ is linear with constant coefficients in shift operator $E$, hence translates to rational functions in $z$. The denominators arise from $(1-2z)$ and $(3z-z^2)$, which correspond exactly to the characteristic polynomials of the given recurrences.

The $r_k(z)$ polynomials form the binomial inversion basis for repeated finite differences; every inverse operator of the form $(1-az+ bz^2)^{-1}$ expands into finite combinations of these polynomials under cascade truncation.

Notes

The structure of the cascade relations forces all sequences to lie in the finite-dimensional module generated by shifts of $A_n$. The $r$-polynomials are the coordinate functions of this module with respect to the binomial-difference basis.