Section 3

Partitions of Sets, Cycles in Permutations

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Problem 1.186

Tabulate the numbers SknS_k^n for n≤8n \leq 8, 1≤k≤n1 \leq k \leq n.

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Remark.
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List concretely the different possibilities for small nn, use problem 187 for all n≥3n \geq 3.

Problem 1.187

Obviously

S1n=Snn=1. S_1^n = S_n^n = 1 .

Show that

Skn+1=Sk−1n+kSkn. S_k^{n+1} = S_{k-1}^n + k S_k^n .
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Problem 1.188

Show that

Skkzk+1+Skk+1zk+2+⋯+Sknzn+1+⋯=1z(z−1)(z−2)⋯(z−k). \frac{S_k^k}{z^{k+1}} + \frac{S_k^{k+1}}{z^{k+2}} + \cdots + \frac{S_k^n}{z^{n+1}} + \cdots = \frac{1}{z(z-1)(z-2) \cdots (z-k)} .
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Problem 1.189

Show that, for n≥1n \geq 1,

Skn=1k![kn−(k1)(k−1)n+(k2)(k−2)n−⋯+(−1)k0n]. S_k^n = \frac{1}{k!} \left[ k^n - \binom {k}{1}(k-1)^n + \binom {k}{2}(k-2)^n - \cdots + (-1)^k 0^n \right] .
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Problem 1.190

Show that

∑n=k∞Sknznn!=(ez−1)kk!. \sum _{n=k}^{\infty } \frac{S_k^n z^n}{n!} = \frac{(e^z-1)^k}{k!} .
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Problem 1.191

Prove the identity in xx

xn=S1nx+S2nx(x−1)+⋯+Snnx(x−1)⋯(x−n+1).[Problem 189; see also Problem 221 of Part III.] x^n = S_1^n x + S_2^n x(x-1) + \cdots + S_n^n x(x-1) \cdots (x-n+1) . \qquad \text{[Problem 189; see also Problem 221 of Part III.]}
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Problem 1.192

Prove the identity of problem 191 independently of problem 189 and derive hence a new proof for problem 189.

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Remark.
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By a combinatorial argument.

Problem 1.193

Define T0=1T_0=1. Then

T0+T1z1!+T2z22!+⋯+Tnznn!+⋯=eez−1. T_0 + \frac{T_1 z}{1!} + \frac{T_2 z^2}{2!} + \cdots + \frac{T_n z^n}{n!} + \cdots = e^{e^z-1} .
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Problem 1.194

Show that

Tn+1=(n0)Tn+(n1)Tn−1+(n2)Tn−2+⋯+(nn)T0. T_{n+1} = \binom {n}{0} T_n + \binom {n}{1} T_{n-1} + \binom {n}{2} T_{n-2} + \cdots + \binom {n}{n} T_0 .
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Remark.
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By a combinatorial argument.

Problem 1.195

Use problem 194 to prove problem 193 [see also problem 34].

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Problem 1.196

Show that for n≥1n \geq 1

Tn=1e(1n1!+2n2!+3n3!+⋯ ). T_n = \frac{1}{e} \left( \frac{1^n}{1!} + \frac{2^n}{2!} + \frac{3^n}{3!} + \cdots \right) .
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Problem 1.197

Tabulate the numbers skns_k^n for n≤8n \leq 8, 1≤k≤n1 \leq k \leq n.

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Remark.
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List concretely the different possibilities for small nn, use problem 198 for all n≥3n \geq 3.

Problem 1.198

Obviously

s1n=(n−1)!,snn=1. s_1^n = (n-1)!, \qquad s_n^n = 1 .

Show that

skn+1=sk−1n+nskn. s_k^{n+1} = s_{k-1}^n + n s_k^n .
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Problem 1.199

Show that

x(x+1)(x+2)⋯(x+n−1)=s1nx+s2nx2+⋯+snnxn x(x+1)(x+2) \cdots (x+n-1) = s_1^n x + s_2^n x^2 + \cdots + s_n^n x^n

or, which is the same, that

x(x−1)(x−2)⋯(x−n+1)=(−1)n−1s1nx+⋯−sn−1nxn−1+snnxn. x(x-1)(x-2) \cdots (x-n+1) = (-1)^{n-1} s_1^n x + \cdots - s_{n-1}^n x^{n-1} + s_n^n x^n .
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Problem 1.200

Show that

∑n=k∞sknznn!=1k!(log⁡11−z)k. \sum _{n=k}^{\infty } \frac{s_k^n z^n}{n!} = \frac{1}{k!} \left( \log \frac{1}{1-z} \right)^k .
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Remark.
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Compare problem 200 with problem 190, problem 199 with problem 188, and again problem 199 with problem 191. See also Problem 54.2 of Part VII and Problem 247.1 of Part VIII.

Problem 1.201

Tabulate the numbers S~kn\widetilde{S}_k^n for n≤8n \leq 8, 1≤k≤n1 \leq k \leq n.

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Remark.
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List concretely the different possibilities for small nn, use problem 202 for all n≥4n \geq 4.

Problem 1.202

Obviously

S~1n=1 when n≥2,S~kn=0 when n<2k. \widetilde{S}_1^n = 1 \text{ when } n \geq 2, \qquad \widetilde{S}_k^n = 0 \text{ when } n<2k .

Show that

S~kn+1=nS~k−1n−1+kS~kn. \widetilde{S}_k^{n+1} = n \widetilde{S}_{k-1}^{n-1} + k \widetilde{S}_k^n .
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Problem 1.203

Show that

S~2n=2n−1−n−1,S~n2n=1⋅3⋅5⋯(2n−1). \widetilde{S}_2^n = 2^{n-1}-n-1, \qquad \widetilde{S}_n^{2n} = 1 \cdot 3 \cdot 5 \cdots (2n-1) .
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Problem 1.204

Returning to SknS_k^n, show that

Sn−1n=(n2),Sn−2n=(n3)3n−54,Sn−3n=(n4)(n−2)(n−3)2 S_{n-1}^n = \binom {n}{2}, \qquad S_{n-2}^n = \binom {n}{3} \frac{3n-5}{4}, \qquad S_{n-3}^n = \binom {n}{4} \frac{(n-2)(n-3)}{2}

and, generally, that Sn−anS_{n-a}^n is a polynomial in nn of degree 2a2a divisible by n(n−1)(n−2)⋯(n−a)n(n-1)(n-2) \cdots (n-a); of course, a≥1a \geq 1.

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Problem 1.205

Define T~0=1\widetilde{T}_0=1,

T~n=S~1n+S~2n+⋯+S~nn \widetilde{T}_n = \widetilde{S}_1^n + \widetilde{S}_2^n + \cdots + \widetilde{S}_n^n

for n≥1n \geq 1. Compute TnT_n and T~n\widetilde{T}_n for n≤8n \leq 8.

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Problem 1.206
T~n+1=(n1)T~n−1+(n2)T~n−2+⋯+(nn−1)T~1+(nn)T~0. \widetilde{T}_{n+1} = \binom {n}{1} \widetilde{T}_{n-1} + \binom {n}{2} \widetilde{T}_{n-2} + \cdots + \binom {n}{n-1} \widetilde{T}_1 + \binom {n}{n} \widetilde{T}_0 .
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Problem 1.207
T~0+T~1z1!+T~2z22!+⋯+T~nznn!+⋯=eez−1−z. \widetilde{T}_0 + \frac{\widetilde{T}_1 z}{1!} + \frac{\widetilde{T}_2 z^2}{2!} + \cdots + \frac{\widetilde{T}_n z^n}{n!} + \cdots = e^{e^z-1-z} .
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Problem 1.208
T~n=(n0)Tn−(n1)Tn−1+(n2)Tn−2−⋯+(−1)n(nn)T0. \widetilde{T}_n = \binom {n}{0} T_n - \binom {n}{1} T_{n-1} + \binom {n}{2} T_{n-2} - \cdots + (-1)^n \binom {n}{n} T_0 .
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Remark.
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T~n=ΔnT0\widetilde{T}_n = \Delta^n T_0, if we use the notation of the calculus of finite differences, see the introduction to Problem 220 of Part III.

Problem 1.209

If the function F(t)F(t) has an nn-th derivative

(ddx)nF(ex)=S1nF′(ex)ex+S2nF′′(ex)e2x+⋯+SnnF(n)(ex)enx. \left( \frac{d}{dx} \right)^n F(e^x) = S_1^n F'(e^x) e^x + S_2^n F''(e^x) e^{2x} + \cdots + S_n^n F^{(n)}(e^x) e^{nx} .
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Problem 1.210

Derive the identities in the variables zz and ww:

∑n=0∞∑k≤n(nk)wkznn!=ez(w+1),(1) \sum _{n=0}^{\infty } \sum _{k \leq n} \binom {n}{k} \frac{w^k z^n}{n!} = e^{z(w+1)} , \tag {1} ∑n=0∞∑k≤nSknwkznn!=e(ez−1)w,(2) \sum _{n=0}^{\infty } \sum _{k \leq n} S_k^n \frac{w^k z^n}{n!} = e^{(e^z-1)w} , \tag {2} ∑n=0∞∑k≤nsknwkznn!=(1−z)−w(3) \sum _{n=0}^{\infty } \sum _{k \leq n} s_k^n \frac{w^k z^n}{n!} = (1-z)^{-w} \tag {3}

either on the basis of the foregoing [problems 10, 190, 200] or independently of the foregoing.

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Remark.
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On the other hand, we could regard (1), (2) and (3) as defining (nk)\binom {n}{k}, SknS_k^n and skns_k^n, respectively, and then take them as basis and starting point for establishing the propositions, especially the combinatorial propositions, discussed in the section herewith concluded. For additional material on the subject treated in this section see Problem 62.1 of Part V, Problem 54.2 of Part VII, Problems 22.1, 22.2, 22.3, 58.3, 247.1 of Part VIII, and Problems 191.1, 191.2 of this Part.