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Concrete Mathematics Chapter 2 Warmups

This first batch of exercises is meant to develop familiarity with the various concepts and notations introduced in this chapter. There is no complex manipulation, but the trick is to be aware of the often unmentioned assumptions about the precise meaning of the expressions.


$\sum_{k=4}^0 q_k$

The meaning of such an expression is not clear, so there is no real way to fail this exercise.

A first interpretation, maybe the common one, is that the sum is zero because the range is empty. In other words, the sum is $\sum_{4\le k\le 0} q_k$.

A second interpretation, perhaps for those used to programming languages with very flexible loops could argue that the sum is $q_4 + q_3 + q_2 + q_1 + q_0$.

I toyed briefly with a negative sum, similar to integrals with reversed bounds, but I did not come up with the nice book solution of $\sum_{k=m}^n = \sum_{k\le n} - \sum_{k\lt m}$, which is consistent with and extends the first interpretation.

Simplify $x([x\gt 0] - [x\lt 0])$

It is easy to see that the expression has the same value as $|x|$:

\begin{align} x([x\gt 0] - [x\lt 0]) & = x (1-0)&&\text{when \(x\gt 0\)}\\ & = x\\ x([x\gt 0] - [x\lt 0]) & = x (0-1)&&\text{when \(x\lt 0\)}\\ & = -x\\ x([x\gt 0] - [x\lt 0]) & = 0&&\text{when \(x = 0\)}\\ \end{align}

Writing out sums

The first one is easy:

\begin{align} \sum_{0\le k\le 5}a_k = a_0+a_1+a_2+a_3+a_4+a_5\\ \end{align}

The second one is tricky, is more than one way. One problem is that $k$ is not explicitly defined, and I had assumed it was a natural, when the authors thought of it as a integer; now the latter is in line with the book conventions, so I was wrong and had missing terms. The right answer is:

\begin{align} \sum_{0\le k^2 \le 5}a_k = a_4 + a_1 + a_0 + a_1 + a_4\\ \end{align}

Triple Sum

Here it is important to restrict the bounds as much as possible (but no more); otherwise there is a risk of introducing spurious terms.

\begin{align} \sum_{1\le i \lt j \lt k \le n}a_{ijk} & = \sum_{i=1}^2 \sum_{j=i+1}^3 \sum_{k=j+1}^4 a_{ijk}\\ & = \left((a_{123} + a_{124}) + a_{134} \right) + a_{234}\\ & = \sum_{k=3}^4 \sum_{j=2}^{k-1} \sum_{i=1}^{j-1} a_{ijk}\\ & = a_{123}+\left(a_{124} + (a_{134} + a_{234})\right)\\ \end{align}

The terms appear in the same order, but are grouped in sums differently.

Incorrect derivation

The problem is the step

\begin{align} \sum_{j=1}^n \sum_{k=1}^n = \frac{a_j}{a_k}\sum_{k=1}^n \sum_{k=1}^n \frac{a_k}{a_k}\\ \end{align}

$k$ is already bound in the inner sum, so it is invalid to replace $j$ by $k$ in the outer.

$\sum_k [1\le j\le k\le n]$

This can be worked explicitly:

\begin{align} \sum_k [1 \le j \le k \le n] & = \sum_k [1 \le j \le n] [j \le k \le n]\\ & = \sum_{j\le k \le n} [1 \le j \le n]\\ & = [1 \le j \le n] \sum_{j\le k \le n} 1\\ & = [1 \le j \le n] (n-j+1)\\ \end{align}

$\bigtriangledown f(x)$

The result is not surprising:

\begin{align} \bigtriangledown x^{\overline{m}} & = x^{\overline{m}} - (x-1)^{\overline{m}}\\ & = x(x+1)\cdots(x+m-1) - (x-1)x\cdots(x+m-2)\\ & = x(x+1)\cdots(x+m-2)(x+m-1-(x-1))\\ & = m x^{\overline{m-1}}\\ \end{align}

So $\bigtriangledown f(x)$ is the difference operator to use with rising factorials.


Clearly, when $m\lt 0$, $0^{\overline{m}} = 0$; when $m = 0$, $0^{\overline{m}} = 1$ (to make the expression $x^{\underline{1+0}}=x^{\underline 1}(x-1)^{\underline 0}$ work when $x=1$); I had forgotten about $m<0$, which was perhaps the easiest case, as $\frac{1}{m!}$ (it follows directly from the definition of falling factorials with negative powers).

Law of exponents for rising factorials

It is easy to see that $x^{\overline{m+n}} = x^{\overline m}(x+m)^{\overline n}$:

\begin{align} x^{\overline{m+n}} & = x\cdots(x+m-1)(x+m)\cdots(x+m+n-1)\\ & = \left( x\cdots(x+m-1) \right) \left( (x+m)\cdots(x+m+n-1) \right)\\ & = x^{\overline m}(x+m)^{\overline n}\\ \end{align}

From there, the value of rising factorials for negative powers follows quickly:

\begin{align} 1 = x^{\overline{-n+n}} & = x^{\overline{-n}} (x-n)^\overline{n}\\ x^{\overline{-n}} & = \frac{1}{(x-n)^\overline{n}}\\ & = \frac{1}{(x-n)\cdots(x-1)}\\ & = \frac{1}{(x-1)^{\underline{n}}}\\ \end{align}

Symmetric difference of a product

To start, I quickly looked up the proof of the original derivative product rule on Wikipedia; the geometric nature of the proof was illuminating (I believe I was taught the so called Brief Proof both in high-school and at university).

This geometric proof can be used for both the infinite and the finite calculus, and its symmetric nature (there are two ways to compute the area of the big rectangle: $f(x)g(x)+(f(w)-f(x))g(w) + f(x)(g(w)-g(x))$ and $f(x)g(x)+f(w)(g(w)-g(x)) + (f(w)-f(x))g(x)$) can be used in the finite case. The symmetry (and equality) is restored because in the infinite calculus, $\lim_{w\rightarrow x}f(w) = f(x)$ and $\lim_{w\rightarrow x}g(w) = g(x)$, a restoration that is not possible in the finite calculus.

However, the equivalent finite calculus formulas, $\bigtriangleup(uv) = u\bigtriangleup v + Ev\bigtriangleup u$ and $\bigtriangleup(uv) = Eu\bigtriangleup v + v\bigtriangleup u$, have together the symmetry they lack on their own.

Wrapping up

OK, that was not entirely bad (two small mistakes, both about negative numbers blindness). Next step, the basic exercises.