Tag: Periodic Functions

Prove some properties of even, periodic, integrable functions

by RoRi

August 24, 2015

Let $f$ be an integrable, even, periodic function with period 2. Define

$g(x) = \int_0^x f(t) \, dt, \qquad \text{and} \qquad A = g(1).$

Prove $g$ is odd and $g(x+2) - g(x) = g(2)$ .
Compute $g(2)$ and $g(5)$ using that $A = g(1)$ .
Find the value of $A$ that makes $g$ periodic with period 2.

Proof. Using the expansion/contraction property (Apostol, Thm 1.19 with $k = -1$ ) we compute,
$\begin{align*} g(-x) &= \int_0^{-x} f(t) \, dt \\ & = - \int_0^x f(-t) \, dt \\ & = - \int_0^x f(t) \, dt & (f \text{ is even}) \\ & = - g(x). \end{align*}$

Thus, $g$ is odd. Next,

$\begin{align*} g(x+2) - g(x) &= \int_0^{x+2} f(t) \, dt - \int_0^x f(t) \, dt \\ &= \int_0^2 f(t) \,dt + \int_2^{x+2} f(t) \, dt - \int_0^x f(t) \, dt \\ &= g(2) + \int_0^x f(t+2) \, dt - \int_0^x f(t) \, dt \\ & = g(2) + \int_0^x f(t) \, dt - \int_0^x f(t) \, dt & (f \text{ periodic}) \\ & = g(2). \qquad \blacksquare \end{align*}$
First, we compute $g(2)$ ,
$\begin{align*} g(2) &= g(3) - g(1) &(\text{part (a) with } x = 1) \\ &= \int_0^3 f(t) \, dt - \int_0^1 f(t) \, dt \\ &= \int_1^3 f(t) \, dt \\ & = \int_{-1}^1 f(t+2) \, dt &(\text{Translation prop, Thm 1.18})\\ & = 2 \int_0^1 f(t) \, dt \\ & = 2A, \end{align*}$

where the second to last line followed from the fact that $f$ is even and this exercise, so $\int_{-b}^b f(t)\, dt = 2 \int_0^b f(t) \, dt$ .

Since $g(1) = A$ by definition, the above also gives us $g(3) = g(2) + g(1) = 3A$ . So, again using part (a) (this time with $x = 3$ ) we have

$\begin{align*} g(5) &= g(2) + g(3) \\ &= 2A + 3A \\ &= 5A. \end{align*}$
Here we want to find a value of $A$ such that $g$ is periodic with period 2. For $g$ to be periodic with period 2, we must have
$g(x+2) = g(x) \quad \implies \quad g(x+2) - g(x) = 0.$

But, since $g(0) = \int_0^0 f(t) \, dt = 0$ , and the above must be true for all $x$ , we evaluate at $x=0$ and use part (b) that gave us $g(2) = 2A$ to compute,

$g(2) = 0 \quad \implies \quad 2A = 0 \quad \implies \quad A = 0.$

Prove some formulas for odd periodic functions

by RoRi

August 24, 2015

Let $f$ be an odd, periodic, integrable everywhere function with period 2. Then, define

$g(x) = \int_0^x f(t) \, dt.$

Prove that $g(2n) = 0$ for all $n \in \mathbb{Z}$ .
Prove that $g$ is an even, periodic function with period 2.

Proof. We can establish this by a direct computation.
$\begin{align*} g(2n) &= \int_0^{2n} f(t) \, dt \\ &= \int_{-2n}^0 f(t+2n) \, dt & (\text{translation property, Thm 1.18})\\ &= \int_{-2n}^0 f(t) \, dt & (f \text{ periodic, period 2}) \\ &= - \int_0^{-2n} f(t) \, dt \\ &= \int_0^{2n} f(-t) \, dt & (\text{Expansion/contraction property}) \\ &= - \int_0^{2n} f(t) \, dt & (f \text{ is odd so } f(-t) = -f(t)) \end{align*}$

But then we have

$\int_0^{2n} f(t) \, dt = - \int_0^{2n} f(t) \, dt \quad \implies \quad \int_0^{2n} f(t) \, dt = 0$

Since $g(2n) = \int_0^{2n} f(t) \, dt$ we then have the requested result,

$g(2n) = \int_0^{2n} f(t) \, dt = 0. \qquad blacksquare$
First, to show that $g$ is even we need to show $g(-x) = g(x)$ for all $x$ . We compute, using the fact that $f$ is odd and the expansion/contraction property of the integral (Theorem 1.19 in Apostol) with $k=-1$ .
$\begin{align*} g(-x) &= \int_0^{-x} f(t) \, dt \\ &= - \int_0^x f(-t) \, dt & (\text{Exp/Cont property}) \\ &= \int_0^x f(t) \, dt & (f \text{ is odd}) \\ &= g(x) & (\text{Def of } g). \end{align*}$

Next, to show that $g$ is periodic with period 2 we must show $g(x+2) = g(x)$ for all $x$ .

$\begin{align*} g(x+2) &= \int_0^{x+2} f(t) \, dt \\ &= \int_0^2 f(t) \, dt + \int_2^{x+2} f(t) \, dt \\ &= 0 + \int_0^x f(t+2) \, dt &(\text{part (a) and translation property}) \\ &= \int_0^x f(t) \, dt & (f \text{ is periodic, period 2}) \\ & = g(x). \qquad \blacksquare \end{align*}$

Prove an integral formula for periodic functions

by RoRi

August 6, 2015

Let $f$ be an integralable function on $[0,p]$ , and let $f$ be periodic with period $p > 0$ . Prove

$\int_0^p f(x) \, dx = \int_a^{a+p} f(x) \, dx$

for all $a \in \mathbb{R}$ .

Proof. For any $a, p \in \mathbb{R}$ with $p > 0$ we know from a previous exercise that there exists a unique $n \in \mathbb{Z}$ such that

$n \leq \frac{a}{p} < n + 1 \quad \implies \quad np \leq a < (n+1)p \leq a + p.$

Then, we start by splitting the integral into two pieces. (The goal here is to rearrange the integral so that it starts at $np$ and ends at $(n+1)p$ , and then use the fact that $f$ is periodic to conclude that this integral is the same as the one from 0 to $p$ .)

$\begin{align*} \int_a^{a+p} f(x) \, dx &= \int_a^{(n+1)p} f(x) \, dx + \int_{(n+1)p}^{a+p} f(x) \, dx \\ &= \int_{a-np}^{(n+1)p - np} f(x+np) \, dx + \int_{(n+1)p - (n+1)p}^{a+p-(n+1)p} f(x+(n+1)p) \, dx \\ &= \int_{a-np}^p f(x) \, dx + \int_0^{a - np} f(x) \, dx & (\text{using periodicity})\\ &= \int_0^p f(x) \, dx & (\text{since } 0 \leq a-np < p ). \end{align*}$

This completes the proof $. \qquad \blacksquare$