Proof of a combination identity: ∑_(j=0)^n (−1)^j (n j)(1−j/n)^n=(n!)/(n^n)

Heergerneuu 2022-09-26 Answered
Proof of a combination identity: j = 0 n ( 1 ) j ( n j ) ( 1 j n ) n = n ! n n
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Answers (2)

Kaiden Stevens
Answered 2022-09-27 Author has 12 answers
Start with
j = 0 n ( 1 ) j ( n j ) ( 1 + x ) n j = x n
which follows readily from the binomial theorem.
Now differentiate, and multiply by ( 1 + x ) . Repeat this n times and set x = 0 .
Notice that the constant term of the resulting polynomial on the right side is n ! .
You can prove by induction that the lowest degree of x that appears on the right side after k steps 0 < k n is x n k and has the coefficient n ( n 1 ) ( n k + 1 ) .
This gives us the set of identities
j = 0 n ( 1 ) j ( n j ) ( n j ) k = 0 ,     0 k < n
j = 0 n ( 1 ) j ( n j ) ( n j ) n = n !
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furajat4h
Answered 2022-09-28 Author has 1 answers
Lemma: Let f ( j ) = k = 0 n f k j k be a degree n polynomial. Claim that j = 0 n ( 1 ) j ( n j ) f ( j ) = n ! f n .
Proof: For any polynomial g, define Δ ( g ) to be the polynomial Δ ( g ) Δ ( g ) ( j ) = g ( j ) g ( j + 1 ). Observe that, if g is a polynomial with leading term a x d + , then Δ ( g ) is a polynomial with leading term d a x d 1 + . In particular, if f is as in the statement of the lemma, then Δ n ( f ) is the constant ( 1 ) n n ! f n . The sum in question is Δ n ( f ) evaluated at 0. QED
Now, apply the lemma to f ( j ) = ( 1 j / n ) n .
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