[math-fun] An identity by Wrench generalized

6 Apr 2014

      Knuth, TAOCP, vol.3, p.644, ex.27 gives
the following identity, attributed to Wrench:

  sum(k>=1, A(k)*x^k / (1-x^k) ) =
  sum(k>=1, x^(k*k) * (A(k) + sum(j>=1, (A(k) + A(k+j)) * x^(k*j)) ) )

This can be verified by writing the terms in

  sum(k>=1, sum(j>=1, A(k) * x^(k*j) ) )

into a rectangular table and taking sums starting
from the diagonal entries, simultaneously to the right
and downwards.

Just today I managed to find a generalization:
  sum(k>=1, A(k)*t^k / (1 - B(k)*x^k) ) =
  sum(k>=1, x^(k*(k-1)) * t^k *
    (A(k)*B(k)^(k-1) + sum(j>=1, ( A(k)*B(k)^(k+j-1)*x^j + A(k+j)*B(k+j)^(k-1)*t^j ) * x^(j*(k-1)) ) ) )
Setting t=x and B(k)=1 gives Wrench's identity.

This is of a kind of formulas I am extremely fond of.

Plus it is (at least computationally) even more general than
R. P. Agarwal's identity for sum(n>=0, t^n/(1-x*q^n))
(see http://arxiv.org/abs/1202.6525 )

Should I sent this one to Knuth?

Best,   jj

Pari code to verify:
-----------------------
N=33;  x='x+O('x^N);
A(k)=eval(Str("a" k));
B(k)=eval(Str("b" k));
t2 = sum(k=1,N, A(k)*t^k / (1 - B(k)*x^k) );
{ t6 = sum(k=1,N, x^(k*(k-1)) * t^k * (A(k)*B(k)^(k-1) + sum(j=1,N, ( A(k)*B(k)^(k+j-1)*x^j + A(k+j)*B(k+j)^(k-1)*t^j ) * x^(j*(k-1)) ) ) ); }
t2-t6  \\ == zero (up to orders x^N and t^N)
-----------------------

Joerg Arndt

Fred Lunnon

Marc LeBrun

Joerg Arndt

Fred Lunnon

Michael Kleber

Joerg Arndt

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