The Method of Coefficients Ira M. Gessel Brandeis University - PDF document

The Method of Coefficients Ira M. Gessel Brandeis University Waltham, MA gessel@brandeis.edu Waterloo Workshop on Computer Algebra May 5, 2008

A simple example Consider the identity � n m � n �� 3 n − 2 k � � ( − 3) k = � . k m − k m/ 3 k =0 Note that in terms of hypergeometric series, this identity may be written � � − n, − 3 n + m, − m 3 � � 3 F 2 � − 3 2 n, − 3 2 n + 1 4 � 2 − n +1 − n +2  “ ” “ ” 3 3  M M if m = 3 M  “ 1 “ 2 ” ” = 3 3 M M  0 otherwise  (a result of George Andrews).

We can prove this identity in the following way: We start with 1 + y 3 = (1 + y )(1 − y + y 2 ) (1 + y ) 2 − 3 y � � = (1 + y ) So (1 + y 3 ) n = (1 + y ) n � (1 + y ) 2 − 3 y � n � n � = (1 + y ) n � (1 + y ) 2 n − 2 k ( − 3 y ) k k k � n � � (1 + y ) 3 n − 2 k ( − 3 y ) k . = k k Equating coefficients of y m on both sides gives the identity � n m � n �� 3 n − 2 k � � ( − 3) k = � . k m − k m/ 3 k =0

But how could we find this proof? We represent the binomial coefficients as � n � coefficients of binomial expansions: is the k coefficient of x k in (1 + x ) n and � 3 n − 2 k � is the m − k coefficient of y m − k in (1+ y ) 3 n − 2 k . It is convenient to represent these coefficients as constant terms, so the sum is (1 + x ) n (1 + y ) 3 n − 2 k � ( − 3) k CT x k y m − k k � k (1 + y ) 3 n (1 + x ) n � y � = CT − 3 x k (1 + y ) 2 y m k Now we apply the variable elimination rule: If f ( x ) is any power series and α is independent of x then ∞ ∞ f ( x ) x k α k = [ x k ] f ( x ) α k = f ( α ) . � � CT x k =0 k =0

We apply this rule to � k (1 + y ) 3 n (1 + x ) n � y � CT − 3 x k (1 + y ) 2 y m k � k (1 + y ) 3 n (1 + x ) n � y � = CT y CT x − 3 y m x k (1 + y ) 2 k � n (1 + y ) 3 n � y = CT y 1 − 3 y m (1 + y ) 2 � (1 + y ) 2 − 3 y � n (1 + y ) 3 n = CT y y m (1 + y ) 2 � n (1 + y ) 3 n � 1 − y + y 2 = CT y y m (1 + y ) 2 � n (1 + y )(1 − y + y 2 ) � = CT y y m (1 + y 3 ) n = CT y y m � n � = . m/ 3

The Method of Coefficients Based on G. P. Egorychev’s book Integralnoe predstavlenie i vyqislenie kombinatornyh summ To evaluate a binomial coefficient sum: 1. First we represent the binomial coefficients as constant terms: = CT (1 + x ) n = CT (1 + x ) n � n � x k x n − k k 1 = CT x k (1 − x ) n − k +1 1 = CT x n − k (1 − x ) k +1 2. We apply the variable elimination rule ∞ ∞ f ( x ) x k α k = [ x k ] f ( x ) α k = f ( α ) . � � CT k =0 k =0

Another example �� 2 n − k � n � � ( − 2) k Evaluate . k n − r − k k We have �� 2 n − k � n � � ( − 2) k k n − r − k k ( − 2) k (1 + x ) n (1 + y ) 2 n − k � = CT x k y n − r − k k � k (1 + y ) 2 n (1 + x ) n � − 2 y � = CT x k y n − r 1 + y k � n (1 + y ) 2 n � 2 y = CT 1 − y n − r 1 + y � n (1 + y ) 2 n � 1 − y = CT y n − r 1 + y = CT (1 − y 2 ) n y n − r � n � = ( − 1) ( n − r ) / 2 ( n − r ) / 2

Change of variables formulas The simplest change of variables formula is CT f ( x ) = CT f ( αx ) , where α is independent of x . More interesting is the linear change of variables formula 1 � x � CT f ( x ) = CT 1 + αxf 1 + αx To prove it, we need only consider the case f ( x ) = x k , k ∈ Z . The case k = 0 is clear. For k > 0 , x k CT (1 + αx ) k +1 = 0 . For k = − j < 0 , (1 + αx ) k +1 = CT (1 + x ) j − 1 x k CT = 0 . x j

As a simple example of this change of variables formula, let � a �� � b � γ k S ( a, b, n, γ ) = k n − k k = CT (1 + γx ) a (1 + x ) b x n � � b � � − a, − n � � = 2 F 1 � γ . � n b − n + 1 Applying the change of variables formula with α = − 1 gives � a (1 − x ) n − a − b − 1 � 1 + ( γ − 1) x S ( a, b, n, γ ) = x n � a (1 + x ) n − a − b − 1 � 1 + (1 − γ ) x = ( − 1) n CT x n = ( − 1) n S ( a, n − a − b − 1 , n, 1 − γ ) . This is equivalent to a 2 F 1 linear transformation.

A 2-variable change of variables formula (Gessel and Stanton) For any Laurent series f ( x, y ) , � � 1 x y CT 1 − xyf ( x, y ) = CT f 1 + y, . 1 + x To prove it we may assume that f ( x, y ) = x l y m . The left side is 1 if l = m ≤ 0 and 0 otherwise. The right side is x l y m CT (1 + x ) m . (1 + y ) l This is clearly 0 if l or m is positive. If l = − r and m = − s , with r, s ≥ 0 then it is CT (1 + x ) s (1 + y ) r � s �� r � = = δ r,s . x r y s r s

As an application we’ll prove the identity ∞ (1 + x ) a (1 + y ) b � a + m �� b + l � � x l y m . = (1 − xy ) a + b +1 l m l,m =0 Applying the change of variables formula with f ( x, y ) = (1 + x ) a (1 + y ) b x l y m (1 − xy ) a + b we get x l y m (1 − xy ) a + b +1 = CT (1 + x ) a + m (1 + y ) b + l (1 + x ) a (1 + y ) b CT x l y m � a + m �� b + l � = l m

General change of variables For more general change of variables formulas, it is convenient to work with residues rather than constant terms. The residue of the Laurent series f ( z ) , denoted res f ( z ) , is the coefficient of z − 1 in f ( z ) . We have the following change of variables formula: Let f ( z ) be a Laurent series and let g ( y ) be a power series, g ( y ) = g 1 y + g 2 y 2 + · · · , where g 1 � = 0 . Then res f ( z ) = res f ( g ( y )) g ′ ( y )) . (Note: There is a multivariable generalization, due to Jacobi, that we will not discuss.)

The key fact in the proof is that the residue of a derivative is 0, and that a Laurent series with residue 0 is a derivative. Any Laurent series may be written as az − 1 + F ′ ( z ) for some Laurent series F ( z ) , so by linearity it is enough to prove the formula for f ( z ) = z − 1 and f ( z ) = F ′ ( z ) . For f ( z ) = F ′ ( z ) we have res F ′ ( z ) = 0 and res F ′ ( g ( y )) g ′ ( y ) = res F ( g ( y )) ′ = 0 . For f ( z ) = z − 1 we have res z − 1 = 1 and g ( y ) g ′ ( y ) = res g 1 + 2 g 2 y + · · · 1 res g 1 y + g 2 y 2 + · · · � 1 y + g 2 � = res + · · · g 1 = 1 .

An application of the change of variables theorem Prove the identity � a + k �� a + 2 n + 1 � � n + a/ 2 � � = 2 2 n . k n − k n k As a hypergeometric series identity, this is a form of Kummer’s theorem, � � 1 + a, � − n = ( a + 2) n � 2 F 1 � − 1 . � ( 3 2 + a 2 + a + n 2 ) n The left side is (1 + y ) a +2 n +1 1 � CT x k (1 − x ) a +1 y n − k k � n (1 + y ) a +1 = CT (1 + y ) a +2 n +1 � (1 + y ) 2 y n (1 − y ) a +1 = CT (1 − y ) a +1 y � n +1 (1 + y ) a − 1 � (1 + y ) 2 = res (1 − y ) a +1 . y

This suggests a change of variables z = y/ (1+ y ) 2 . The computation is easiest if we start with the right side: � n + a/ 2 � 1 2 2 n = res z n +1 (1 − 4 z ) a/ 2+1 n Then we make the change of variables z = y/ (1 + y ) 2 , so that 1 − 4 z = (1 − y ) 2 / (1 + y ) 2 and dz/dy = (1 − y ) / (1 + y ) 3 . We get � n +1 � (1 + y ) 2 � a/ 2+1 � (1 + y ) 2 1 − y res (1 − y ) 2 (1 + y ) 3 y � n +1 (1 + y ) a − 1 � (1 + y ) 2 = res (1 − y ) a +1 y

A more complicated example (Gessel and Stanton) Prove the hypergeometric series identity � � � n − n, 1 ( 1 � 2 ) n � 27 2 � 1 2 F 1 = � 2 n + 3 4 (2 n + 3 4 2 ) n � 2 An equivalent binomial coefficient identity is � k � n � − 1 � − 1 �� 3 n + 1 � � 1 � 27 � � = ( − 1) n 2 2 2 . k n − k 4 4 n k The left side is � k (1 + x ) − 1 / 2 (1 + y ) 3 n +1 / 2 � 1 � CT x k y n − k 4 k = res (1 + 1 4 y ) − 1 / 2 (1 + y ) 3 n +1 / 2 . y n +1 This suggests the change of variables z = y/ (1 + y ) 3 .

1 The right side is res . With the � 1 − 27 z n +1 4 z change of variables z = y/ (1 + y ) 3 , we have dz/dy = (1 − 2 y ) / (1 + y ) 4 and 4 z = (1 + 1 4 y ) 1 / 2 (1 − 2 y ) � 1 − 27 , (1 + y ) 3 / 2 so 1 res � 1 − 27 z n +1 4 z = res (1 + y ) 3 n +3 (1 + y ) 3 / 2 4 y ) 1 / 2 (1 − 2 y ) · 1 − 2 y · (1 + 1 y n +1 (1 + y ) 4 = res (1 + 1 4 y ) − 1 / 2 (1 + y ) 3 n +1 / 2 . y n +1

Comparison with Wilf’s Snake Oil Method To evaluate a sum S n = � k S n,k , we find the generating function S n x n = � � � S n,k x n . n n k Sometimes we need to adjust the parameters to get this to work. Let us look at our second example, �� 2 n − k � n � � n � � ( − 2) k = ( − 1) ( n − r ) / 2 . k n − r − k ( n − r ) / 2 k To apply the Snake Oil Method, we replace n − r with m , so the identity becomes � n � n �� 2 n − k � � � ( − 2) k = ( − 1) m/ 2 . k m − k m/ 2 k

Now we can multiply the left side by x m and sum on m to get � n �� 2 n − k � � � ( − 2) k x m k m − k m k � n �� 2 n − k � � ( − 2) k x k + j = ( m = k + j ) k j j,k � n � � ( − 2) k x k (1 + x ) 2 n − k = k k � n = (1 + x ) n (1 − x ) n = (1 + x ) n � (1 + x ) − 2 x � n � = (1 − x 2 ) n = � ( − 1) m/ 2 . m/ 2 m