2
$\begingroup$

Let's say I've a function $y\colon [0,\infty]\to\mathbb{R}$ and it is periodic with $T$. If we take a look at the average value of this function over the hole region we can write:

$$\lim_{n\to\infty}\frac{1}{n}\int_0^nf(x)\,dx=\frac{1}{T}\int_0^Tf(x)\,dx$$

But how can prove that that is indeed true?

My work:

When $f(x)$ is periodic with T, the integral on the left hand side is infinite. So we can use lhopitals rule:

$$\lim_{n\to\infty}\frac{1}{n}\int_0^nf(x)\,dx=\lim_{n\to\infty}\frac{\frac{d}{dn}\left(\int_0^nf(x)\,dx\right)}{\frac{d}{dn}\left(n\right)}=\lim_{n\to\infty}\frac{f(n)}{1}=\lim_{n\to\infty}f(n)$$

So, we get:

$$\lim_{n\to\infty}f(n)=\frac{1}{T}\int_0^Tf(x)\,dx$$

But because $f(x)$ is periodic $f(\infty)$ does not have a 'value'. So this leads to noting.

$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

Suppose that $n=kT+n'$, where $0\le n'<T$. Then:

$$\frac{1}{n}\int_0^n f(x)dx = \frac{k\int_0^T f(x)dx + \int_{kT}^{kT+n'}f(x)dx}{n}=\frac{k}{kT+n'}\int_0^T f(x)dx+\frac{1}{n}\int_{kT}^{kT+n'}f(x)dx$$

As $n\to\infty$, we have $k\to \infty$, while $n'$ and $\int_{kT}^{kT+n'}f(x)dx$ remain bounded. Hence, in the limit, the first summand approaches $\frac{1}{T}\int_0^T f(x)dx$, while the second summand approaches $0$.

$\endgroup$
1
$\begingroup$

Let $m_n = \lfloor n/T \rfloor$.

\begin{align} \frac{1}{n} \int_0^n f(x) \, dx &= \frac{1}{n} \int_0^{m_n T} f(x) \, dx + \frac{1}{n} \int_{m_n T}^n f(x) \, dx \\ &= \frac{m_n T}{n} \cdot \left(\frac{1}{T} \int_0^T f(x) \, dx\right) + \frac{1}{n} \int_{m_n T}^n f(x) \, dx. \end{align}

The second term is bounded by $\frac{n-m_n T}{n} \sup_{x \in [0,T]} |f(x)| \le \frac{T}{n} \sup_{x \in [0,T]} |f(x)| \to 0$ as $n \to \infty$. The first term converges to $1$ the desired limit, since $1 - \frac{T}{n} \le \frac{m_n T}{n} \le 1$.

$\endgroup$
0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.