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2.5. A closed-form expression for f d 63and the monomials of non-zero degree only come from Ξ d X .Moreover, Ξ d X can be written as[] ) ( ∑Ξ d X = 13 j1 ∑k j≥0,{j|X j≠2}( j2−1∑k=02 −kk!∑k{j|X k j≠2} j{j|X j≠2} k j∏{j|X j≠2} k j!)!⎛⎝∏{j|X j=0}⎞β kj ⎠j 4−βj Π d X .So, to get a multinomial of multi-degree (i 1 , . . . , i n ), different choices can be made for the k j ’s:• If X j = 0, then we must take k j = i j . This happens for 1 ≤ j ≤ n.• If X j = 1, then we can take any k j ≥ min(i j − 1, 0) and take into account the correctcoefficient in Π d X . This happens for 1 ≤ j ≤ d.• If X j = 2, then there is no choice to make. This happens for m + 1 ≤ j ≤ d.In the following sum, we gathered the contributions of all X’s. We denote by I the set ofindices m + 1 ≤ j ≤ n such that X j = 0, 1 (the other ones are such that X j = 2) and by J theset of indices n + 1 ≤ j ≤ d such that X j = 1 (the other ones are such that X j = 2).The summation variables k j where j is in I ∪ J or [1, m] are to be understood as the degreewe choose in the above expression of Ξ d X . Following the above discussion on the choice of the k j’s:• If j ∈ J, then we can choose any positive degree k j and extract the constant coefficient A kj .• If j ∈ I, then we can choose any positive degree k j and we extract the constant coefficientA kj as above if k j > 0, and A 0 − 3 if k j = 0 (the −3 comes from the choice X j = 0 whichgives 1 = 3 · 1/3).• Finally, if 1 ≤ j ≤ m, then we have to choose k j ≥ i j − 1, and the corresponding coefficient1isk = 1 j+1 i jif k j = i j − 1, 5/6 − 3 = −13/6 if k j = i j (as above the −3 comes fromthe choice X j = 0) and ( k j) ( )i jA kj−i j+ B k j −i j +1k j−i j+1if k j > i j . We denote that coefficient byD kj,i j.We denote S and h the quantities S = ∑ j∈I∪J,1≤j≤m k j and h = d − m − #J − #I. Thena d,n(i 1,...,i n)can be expressed asa d,n(i 1,...,i n) = (−1)n+1 ∑I⊂{m+1,...,n}J⊂{n+1,...,d}∑k j≥0,j∈I∪Jk j≥i j−1,1≤j≤mS!∏j∈I∪J k j! ∏ mj=1 k j!⎛⎝ ∑ 2 k [ ] ⎞ h − k ⎠ ∏ ∏ ( ∏A(h − k)! Skj Akj − 3 kj=0) m D kj,i j.k≥1j∈J j∈Ij=1Extracting the binomial coefficient of D kj,i j, we can factor out the multinomial coefficient ((remember that l was defined as l = ∑ nj=1 i j):( )(i = l1,...,i n) (−1)n+1 i 1 , . . . , i na d,n∑I⊂{m+1,...,n}J⊂{n+1,...,d}∏j∈J∑k j≥0,j∈I∪Jk j≥i j−1,1≤j≤mA kjk j !∏j∈IS!l!A kj − 3 kj=0k j !⎛⎝ ∑ k≥1m∏j=12 k(h − k)!C kj−i j|k j − i j |![ ] ⎞ h − k ⎠S,)li 1,...,i n