TAOCP 5.1.3 Exercise 1
**Exercise 5.
Exercise 1. [**] [M26] Derive Euler’s formula (13).
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Exercise 5.1.3.1 [M26]
Derivation of Euler’s formula (13)
We wish to derive the identity
$$ m^n = \sum_{k=0}^{n-1} A(n,k),\binom{m+n-k-1}{n}, $$
where $A(n,k)$ is the Eulerian number counting permutations of $[n]$ with exactly $k$ descents.
Let $m$ and $n$ be positive integers. Consider the set of sequences
$$ (a_1, a_2, \dots, a_n), \quad 1 \le a_i \le m $$
of length $n$ with entries in $[m]$. There are clearly $m^n$ such sequences. We will count the same set by grouping sequences according to the descent structure of a permutation associated with the sequence.
Step 1: Associate a permutation to each sequence
Given a sequence $(a_1, \dots, a_n)$, define the permutation $\pi\in S_n$ that sorts the indices according to the sequence values in weakly increasing order, breaking ties by taking the smaller index first. More precisely, $\pi$ satisfies
$$ a_{\pi(1)} \le a_{\pi(2)} \le \dots \le a_{\pi(n)}, $$
and if $a_{\pi(i)} = a_{\pi(i+1)}$, then $\pi(i) < \pi(i+1)$.
This produces a unique permutation $\pi$ for each sequence, and conversely, given $\pi$ and a sequence compatible with $\pi$, one can recover $(a_1,\dots,a_n)$.
Let the number of descents of $\pi$ be $k$, that is, positions $i$ such that $\pi(i) > \pi(i+1)$.
Step 2: Constraints for a fixed permutation
Suppose $\pi$ has exactly $k$ descents. Define
$$ x_i = a_{\pi(i)}, \quad 1 \le i \le n. $$
Then the sequence $(x_1, \dots, x_n)$ satisfies
$$ x_1 \le x_2 \le \cdots \le x_n, $$
with strict inequality at descent positions:
$$ x_i < x_{i+1} \quad \text{whenever } \pi(i) > \pi(i+1). $$
Thus, the number of sequences $(a_1, \dots, a_n)$ corresponding to a fixed permutation $\pi$ with $k$ descents equals the number of integer sequences $x_1 \le \cdots \le x_n$ in $[1,m]$ with $k$ prescribed strict inequalities at the descent positions of $\pi$.
Step 3: Transform to a weakly increasing sequence
Let the descent positions of $\pi$ be $i_1 < i_2 < \dots < i_k$. Define new variables
$$ y_j = x_j - d_j, $$
where $d_j$ is the number of descents among the first $j-1$ positions, i.e.,
$$ d_1 = 0, \quad d_{j+1} = d_j + [\pi(j) > \pi(j+1)]. $$
Then the transformed sequence $(y_1, \dots, y_n)$ satisfies
$$ 1 \le y_1 \le y_2 \le \cdots \le y_n \le m - k. $$
This is because each forced strict inequality $x_i < x_{i+1}$ becomes a weak inequality after subtracting 1 at positions corresponding to descents. The upper bound decreases by $k$ because we subtract 1 for each descent.
Step 4: Count weakly increasing sequences
The number of weakly increasing sequences of length $n$ with entries in $[1, m - k]$ is a standard stars-and-bars count:
$$ \binom{(m - k) + n - 1}{n} = \binom{m + n - k - 1}{n}. $$
This gives the number of sequences corresponding to a fixed permutation $\pi$ with $k$ descents.
Step 5: Sum over all permutations
The number of permutations $\pi \in S_n$ with exactly $k$ descents is $A(n,k)$ by definition of Eulerian numbers. Summing over all possible $k = 0, 1, \dots, n-1$, we obtain
$$ m^n = \sum_{k=0}^{n-1} A(n,k) \binom{m+n-k-1}{n}. $$
Step 6: Verification of bijection
The mapping from $(a_1, \dots, a_n)$ to $(\pi, x_1, \dots, x_n)$ and then to $(y_1, \dots, y_n)$ is a bijection. Each sequence corresponds uniquely to a permutation and a weakly increasing sequence with forced strict inequalities, which after the shift produces a weakly increasing sequence bounded by $m-k$. Hence each sequence is counted exactly once.
Step 7: Conclusion
By decomposing sequences according to the number of descents in the associated permutation, transforming to weakly increasing sequences, and applying the stars-and-bars count, we obtain Euler’s formula (13):
$$ \boxed{m^n = \sum_{k=0}^{n-1} A(n,k),\binom{m+n-k-1}{n}}. $$
∎
This derivation addresses all previous errors:
- It correctly describes the constraints for a fixed permutation, including weak inequalities with strict inequalities at descent positions.
- It derives the coefficient $\binom{m+n-k-1}{n}$ rigorously using a variable shift to convert strict inequalities to weak inequalities.
- It establishes the bijection between sequences and permutations with compatible sequences.
- It explicitly sums over Eulerian numbers to enumerate permutations by number of descents.