Tag: Step Functions

Calculate the values of an integral of a step function

by RoRi

July 19, 2015

For $p \in \mathbb{Z}_{>0}$ define a step function on the interval $[0,p]$ by

$s(x) = \begin{cases} (-1)^n \cdot n & \text{if } n \leq x < n+1, \ n=0,1, \ldots, p-1 \\ 0 & \text{if } x =p. \end{cases}$

Then, define

$f(p) = \int_0^p s(x) \, dx.$

Calculate $f(3), f(4), f(f(3))$ .
Find all values of $p$ such that $|f(p)| = 7$ .

We calculuate:
$\begin{align*} f(3) &= \int_0^3 s(x) \, dx = (-1)^0 0 (1-0) + (-1)^1 (1) (2-1) + (-1)^2 (2)(3-2) = 0 - 1 + 2 &= \phantom{-}1.\\ f(4) &= \int_0^4 s(x) \, dx = \int_0^3 s(x) \, dx + \int_3^4 s(x) \, dx = 1 + (-1)^3 (3) (4-3) &= -2. \\ f(f(3)) &= f(1) = \int_0^1 s(x) \, dx = (-1)^0 0 (1-0) &= \phantom{-}0. \end{align*}$
$p = 14,15$ .

Note: There is an error in the book. The answers in the back of the book claim that $f(4) = -1$ , which is incorrect.

More integrals of step functions

by RoRi

July 18, 2015

Compute
$\int_0^9 \left[ \sqrt{t} \right] \, dt.$
For $n \in \mathbb{Z}_{>0}$ prove
$\int_0^n \left[ \sqrt{t} \right] \, dt = \frac{n(n-1)(4n+1)}{6}.$

We compute,
$\int_0^9 \left[ \sqrt{t} \right] \, dt = (0 \cdot 1) + (1 \cdot (4-1)) + (2 \cdot (9-4)) = 13.$
Proof. Let $P = \{ 0,1,4,9, \ldots, n^2 \}$ . Then $P$ is a partition of $[0,n^2]$ and $\left[ \sqrt{t} \right]$ is constant on the open subintervals of $P$ . Further, for $(k-1)^2 < t < k^2$ we have $\left[ \sqrt{t} \right] = (k-1)$ . Thus, we have (using this exercise and this exercise to evaluate some of the sums),
$\begin{align*} \int_0^{n^2} \left[ \sqrt{t} \right] \, dt &= \sum_{k=1}^n (k-1) \cdot (k^2 - (k-1)^2) \\ &= \sum_{k=1}^n (k-1) \cdot (2k-1) \\ &= \sum_{k=1}^n (2k^2 - 3k + 1 ) \\ &= 2 \sum_{k=1}^n k^2 - 3 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\ &= \frac{2n^3}{3} + n^2 + \frac{n}{3} - \frac{3n^2}{2} - \frac{3n}{2} + n \\ &= \frac{4n^3 - 3n^2 - n}{6} \\ &= \frac{n(n-1)(4n+1)}{6}. \qquad \blacksquare \end{align*}$

Find some formulas for the integral of the step function [t]^2

by RoRi

July 18, 2015

For $n \in \mathbb{Z}_{>0}$ , prove
$\int_0^n [t]^2 \, dt = \frac{n(n-1)(2n-1)}{6}.$
For $x \in \mathbb{R}$ , with $x \geq 0$ , define
$f(x) = \int_0^x [t]^2 \, dt.$

Draw the graph of $f$ on the interval $[0,3]$ .
Find all real $x > 0$ such that
$\int_0^x [t]^2 \, dt = 2(x-1).$

Proof. Let $P = \{ 0,1, \ldots, n \}$ . Then $P$ is a partition of $[0,n]$ and $[t]^2$ is constant on the open subintervals of $P$ . Further, $[t]^2 = (k-1)^2$ for $k-1 < t < k$ . So,
$\begin{align*} \int_0^n [t]^2 \, dt &= \sum_{k=1}^n (k-1)^2 \cdot (k-(k-1)) \\ &= \sum_{k=1}^n (k-1)^2 \\ &= \sum_{k=0}^{n-1} k^2 \\ &= \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} \\ &= \frac{n(n-1)(2n-1)}{6}. \end{align*}$

The second to last line follows from this exercise (I.4.7, #6) $. \qquad \blacksquare$
The graph is:
By inspection, we have, $x = 1, \frac{5}{2}$ .

Compute integrals of some more step functions

by RoRi

July 18, 2015

Prove
$\int_0^2 [t^2] \, dt = 5 - \sqrt{2} - \sqrt{3}.$
Compute
$\int_{-3}^3 [t^2] dt.$

Proof. Let $P = \{ 0, 1, \sqrt{2}, \sqrt{3}, 2 \}$ be a partition of $[0,2]$ . Then $[t^2]$ is constant on the open subintervals of $P$ , so,
$\begin{align*} \int_0^2 [t^2] \, dt &= \sum_{k=1}^5 s_k \cdot (x_k - x_{k-1}) \\ &= (0 \cdot 1) + (1)(\sqrt{2}-1) + (2)(\sqrt{3}- \sqrt{2}) + (3)(2 - \sqrt{3}) \\ &= 5 - \sqrt{2} - \sqrt{3}. \qquad \blacksquare \end{align*}$
We compute,
$\begin{align*} \int_{-3}^3 [t^2] \, dt &= 2 \cdot \int_0^3 [t^2] \, dt \\ &= 2 \cdot \left( \int_0^2 [t^2] \, dt + \int_2^3 [t^2] \, dt \right) \\ &= 2 \cdot \left( 5 - \sqrt{2} - \sqrt{3} + 4(\sqrt{5}-2) + 5(\sqrt{6} - \sqrt{5}) + 6(\sqrt{7}-\sqrt{6}) \right.\\ & \qquad \quad \left.+ 7(\sqrt{8} - \sqrt{7}) + 8(3 - \sqrt{8}) \right) \\ &= 2 (21 - 3\sqrt{2} - \sqrt{3} - \sqrt{5} - \sqrt{6} - \sqrt{7}). \end{align*}$

Compute some integrals of step functions

by RoRi

July 18, 2015

Let $n \in \mathbb{Z}_{>0}$ , prove
$\int_0^n [t] \, dt = \frac{n(n-1)}{2}.$
Let $x \in \mathbb{R}_{>0}$ , and define
$f(x) = \int_0^x [t] \, dt.$

Draw the graph of $f$ for $0 \leq x \leq 4$ .

Proof. Let $P = \{ 0,1,2, \ldots, n \}$ be a partition of $[0,n]$ . Then, by the definition of the greatest integer function, $[t]$ is constant on the open subintervals of $P$ , so
$\begin{align*} \int_0^n [t] \, dt &= \sum_{k=1}^n (k-1)(k-(k-1)) & (s_k = k-1 \text{ since } [t]=k-1 \text{ if } k-1 < t < k)\\ &= \sum_{k=1}^n (k-1) \\ &= \sum_{k=0}^{n-1} k \\ &= \frac{n(n-1)}{2}. \end{align*}$

The final equality follows from here (I.4.7, #5) $. \qquad \blacksquare$
The graph is:

Prove an identity for the sum of integrals of particular step functions

by RoRi

July 18, 2015

Prove

$\int_a^b [x] \, dx + \int_a^b [-x] \, dx = a-b.$

Proof. From this exercise (1.11 #4, part b) we know

$[-x] = \begin{cases} -[x] & \text{if } x \in \mathbb{Z} \\ -[x]-1 & \text{if } x \notin \mathbb{Z}. \end{cases}$

But, since $[x]$ is constant on the open subintervals of the partition

$P = \{ a, [a] + 1, \ldots, [a] + [b-a], b \}$

which contains every integer between $a$ and $b$ , we have $[-x] = -[x]-1$ on the open subintervals of $P$ (since there are no integers in the open subintervals). Hence, $\int_a^b [-x] \, dx = \int_a^b (-[x] -1)\, dx$ . Thus,

$\begin{align*} \int_a^b [x] \, dx + \int_a^b [-x] \, dx &= \int_a^b [x] \, dx + \int_a^b -[x]-1 \, dx \\ &= \int_a^b -1 \, dx \\ &= \int_b^a 1 \, dx & (\text{let } P = \{ a,b \})\\ &= a-b. \qquad \blacksquare \end{align*}$

Find a step function whose integral has specific properties

by RoRi

July 17, 2015

Find a step function $s$ such that

$\int_0^2 s(x) \, dx = 5, \qquad \text{and} \qquad \int_0^5 s(x) \, dx = 2.$

Let

$s(x) = \begin{cases} \phantom{-}\frac{5}{2} & \text{if } 0 \leq x < 2, \\ -1 & \text{if } -2 \leq x \leq 5. \end{cases}$

Then,

$\begin{align*} \int_0^2 s(x) \, dx &= 2 \cdot \frac{5}{2} = 5 \\ \int_0^5 s(x) \, dx &= 2 \cdot \frac{5}{2} + 3 \cdot -1 = 2 \end{align*}$

as requested.

Compute the integrals of some step functions

by RoRi

July 17, 2015

Compute the following integrals where $[x]$ is the greatest integer less than or equal to $x$ .

$\displaystyle{\int_{-1}^3 [x] \, dx}$ .
$\displaystyle{\int_{-1}^3 \left[ x + \frac{1}{2} \right] \, dx}$ .
$\displaystyle{\int_{-1}^3 \left( [x] + \left[ x + \frac{1}{2} \right] \right) \, dx}$ .
$\displaystyle{\int_{-1}^3 2[x] \, dx}$ .
$\displaystyle{\int_{-1}^3 [2x] \, dx}$ .
$\displaystyle{\int_{-1}^3 [-x] \, dx}$ .

$\displaystyle{ \int_{-1}^3 [x] \, dx = (-1 \cdot 1)+(0 \cdot 1) + (1 \cdot 1) + (2 \cdot 1) = 2}$ .
$\displaystyle{ \int_{-1}^3 \left[ x + \frac{1}{2} \right] \, dx = \left( -1 \cdot \frac{1}{2} \right) + (0 \cdot 1) + (1 \cdot 1) + (2 \cdot 1) + \left( 3 \cdot \frac{1}{2} \right) = 4}$ .
$\displaystyle{ \int_{-1}^3 \left( [x] + \left[ x + \frac{1}{2} \right] \right) \, dx = \int_{-1}^3 [x] \, dx + \int_{-1}^3 \left[ x + \frac{1}{2} \right] \, dx = 2 + 4 = 6}$ .
$\displaystyle{ \int_{-1}^3 2[x] \, dx = 2 \cdot \int_{-1}^3 [x] \, dx = 4}$ .
$\displaystyle{ \int_{-1}^3 [2x] \, dx = \int_{-1}^3 \left( [x] + \left[ x + \frac{1}{2} \right] \right) \, dx = 6}$ . (See, 1.11 #4 (d) here).
$\displaystyle{ \int_{-1}^3 [-x] \, dx = -\int_1^{-3} [x] \, dx = \int_{-3}^1 [x] \, dx = (-3 \cdot 1) + (-2 \cdot 1) + (-1 \cdot 1) + (0 \cdot 1) = -6}$ .

Every step function is a linear combination of characteristic functions

by RoRi

July 17, 2015

We define a characteristic function, $\chi_S$ , on a set $S$ of points on $\mathbb{R}$ by

$\chi_S(x) = \begin{cases} 1 & \text{if } x \in S \\ 0 & \text{if } x \notin S. \end{cases}$

Let $f$ be a step functions taking the (constant) value $c_k$ on the $k$ th open subinterval $I_k$ of a partition of $[a,b]$ . Prove that for each $x \in I_1 \cup \cdots \cup I_n$ , we have

$f(x) = \sum_{k=1}^n c_k \chi_{I_k} (x).$

Proof. First, we note that the open subintervals of some partition of $[a,b]$ are necessarily disjoint since $P = \{ x_0, x_1, \ldots, x_n \} \ \implies \ x_0 < x_1 < \cdots < x_n$ . Hence, if $x \in I_1 \cup \cdots \cup I_n$ then $x \in I_j$ for exactly one $j, \ 1 \leq j \leq n$ .
So, we have

$\chi_{I_k}(x) = \begin{cases} 1 & \text{if } j = k \\ 0 & \text{if } j \neq k \end{cases}$

for all $1 \leq k \leq n$ , and for any $x$ . Further, by definition of $f$ , we know $f(x) = c_k$ if $x \in I_k$ . So,

$\begin{align*} \sum_{k=1}^n c_k \chi_{I_k}(x) &= c_1 \cdot 0 + c_2 \cdot 0 + \cdots + c_k \cdot 1 + \cdots + c_n \cdot 0 \\ &= c_k \\ &= f(x) \end{align*}$