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2.5. A closed-form expression for f d 65Finally, the only tuples we must treat are the ones such that i > j = 1 and n = m. We writethe degree i ≠ 0 in first position even if it is not the greatest one. Then( )a d,nl(i,...,1) = (−1)n+1 b d,nl,ni, . . . , 1,( )(i+1,...,0) = l(−1)n+1 b d,nl,n−1i + 1, . . . , 0,a d,nso it suffices to show that b d,nl,n = bd,n l,n−1 .We use the same notation as in Proposition 2.5.2 except that S and h denote the quantitiesS = l + ∑ j∈I∪J,1≤j≤n−1 k j − n and h = d − n − j. For b d,nl,n, I must be empty:d−nb d,nl,n = ∑( ) d − njj=0d−n∑( ) d − n=jj=0∑k j≥0,j∈J,1≤j≤n⎛⎝ ∑ k≥12 k(h − k)!k j≥0,j∈J,1≤j≤n−1(S + k n )!l![ h − k∑ 1 ∏l!j∈J⎛∑(S + k n )!k n≥0S + k n] ⎞ ⎠ ∏ j∈J⎝ ∑ k≥1A kjk j !n−1∏j=1A kjk j !n∏j=1C kj−1|k j − 1|!C kj−1|k j − 1|!2 k [ ] ⎞ h − k ⎠ C k n−1(h − k)! S + k n |k n − 1|!;whereas for b d,nl,n−1, I can contain n:b d,nl,n−1 =1∑( d−n1 ∑( ) d − ni)jj=0⎛i=0⎝ ∑ k≥1d−n∑( ) d − n=jj=0⎡2 k(h + 1 − k − i)!∑k j≥0,j∈J,1≤j≤n−1⎛∑k j≥0,j∈I∪J,1≤j≤n−12 k⎣(S + 1)! ⎝ ∑ (h + 1 − k)!k≥1⎛+ ∑(S + k n + 1)! ⎝ ∑k n≥0k≥1[ h + 1 − k − iS + 11 ∏ A kjl! k j !j∈J(S + 1)!l!] ⎞ ⎠ ∏ j∈Jn−1∏j=1[ ] ⎞ h + 1 − k ⎠S + 12 k [(h − k)!A kjk j !C kj−1|k j − 1|!∏j∈Ih − kS + k n + 1A kj − 3 kj=0k j !n−1∏j=1⎤] ⎞ ⎠ A k n− 3 kn=0⎦ .|k n − 1|!C kj−1|k j − 1|!The sums on j and k j for j ∈ J and 1 ≤ j ≤ n − 1 are identical, so it is sufficient to show the