§ WZ (Wilf Zeilberger) pairs
- Suppose we want to find a closed form for ∑k=−∞∞F(n,k).
- If we can show that ∑k=−∞∞F(n,k)=∑k=−∞∞F(n+1,k), then we have that ∑k=−inftyinftyF(n,k)=∑k=−infty∞F(0,k). So this will be independent of n, and we can find a closed form that is independent of k, so we can get a concrete numeric value.
- Now, the question then becomes, how do we show that ∑k=−∞∞F(n,k)=∑k=−∞∞F(n+1,k)?
- We can rewrite it as ∑k=−∞∞(F(n+1,k)−F(n,k))=0
- Since we're doing wishful thinking, let's wish that the sum telescoes in k!
- So we ask that there's a function G(n,k), such that F(n+1,k)−F(n,k)=G(n,k+1)−G(n,k), and that limK→±∞G(K)=0, so G vanishes at the boundary. This forces the sum to be identically zero.
- This gives us the sum to telescope, as ∑k=−∞∞F(n+1,k)−F(n,k)=∑k=−∞∞G(n,k+1)−G(n,k).
- We write the RHS as limM→∞∑k=−MMG(n,k+1)−G(n,k). But that equals limM→∞(G(n,M+1)−G(n,−M)). But both these terms vanish at the limit, so we get 0−0=0.
- Thus, we see that we've checked that ∑kF(n+1,k)−F(n,k)=0 by showing that ∑kG(n,k+1)−G(n,k) vanishes.
- If F,G form a WZ pair, there is a rationl function R such that G(n,k)=R(n,k)F(n,k−1). R(n,k) is the certificate.
- See that we are computing an indefinite sum for ∑kF(n,k), like an indefinite integral.
§ Certificate for F,G.
- It will turn out that the G(n,k) will always be of the form G(n,k)=R(n,k)F(n,k−1), since Gosper's algorithm always finds such as G(n,k)! So, the R(n,k) forms a certificate, where we check that the corresponding G(n,k) obeys the two identities above.
§ References
- Generatingfunctionlogy
- A = B
- Concrete Mathematics