calculusconstantsfunctionsperiodic functionstrigonometry
I dont understand the meaning of this line in my book –
" $\sin^2x + \cos^2x$ is periodic but the fundamental period is not defined. "
Why is the period not defined? $F(x)$ is $1$ here so it is a constant function which should be periodic ?
Does this mean all constant functions are not periodic?
Please explain
A function $f(x)$ is "periodic with period $c$", with $c\gt 0$, if $f(x+c) = f(x)$ for all values of $x$. In particular, $f(x+mc) = f(x)$ for all integers $m$, by using induction.
A function is periodic if it is periodic with period $c$ for some $c\neq 0$.
Every periodic function has lots of periods: if $c$ is a period, then so is $mc$ for any positive integer $m$, at the very least; it may have others. So we may be interested in knowing what is the "quickest" that the function starts repeating. For instance, $\tan x$ certainly repeats every $2\pi$, since $\tan(x+2\pi) = \tan(x)$ for all $x$ (either both are defined and equal, or both are undefined); and by the comment above, it will repeat every $2m\pi$ for any integer $m$. But it actually repeats more often than every $2\pi$: we know that $\tan(x+\pi) = \tan(x)$ for all $x$. So altough it is true that $\tan(x)$ is periodic with period $2\pi$, it is also true that $\tan(x)$ is periodic with period $\pi$.
So we would like to have something more than just a "period"; we would like to be able to talk about the smallest possible period. That's what the "fundamental period" is trying to capture:
We say that a periodic function $f(x)$ has "fundamental period" $k$ if and only if $k$ is the smallest positive number such that $f(x)$ is periodic with period $k$. That means that we require:
So, in general, we would take the set of all positive periods of the function $f$, and look for its minimum (smallest element). The one possible problem is that not every set of positive real numbers has a smallest element, so it may be possible, at least in principle, that a function is periodic but has no fundamental period, because it does have periods, but it doesn't have one that is the smallest period.
The constant functions, $f(x) = a$ for all $x$, are an example of this pathology: they are periodic, because they satisfy the defintion of being periodic. In order to show that they satisfy the definition, we need to show that there is at least one number $c\gt 0$ such that $f(x)=f(x+c)$ for all $x$. Well, we can take $c=1$, because $f(x) = a = f(x+1)$ for all $x$. So the constant function is certainly periodic. But in fact, for every positive number $c$, $f(x+c) = a =f(x)$. That means that if you take any $c\gt 0$, then you can rightly say that "$f(x)$ is periodic with period $c$". So if we look at the set of all periods, $$P=\{ c \in \mathbb{R}\mid c\gt 0\text{ and $f(x)$ is periodic of period $c$}\}$$ then $P=(0,\infty)$. Since the "fundamental period" is supposed to be the smallest element of $P$, but $P$ does not have a smallest element, then the constant function does not have a fundamental period. But it is periodic (we just saw it satisfies the definition).
In the case of the greatest integer function, $f$ is not periodic at all. Again, "$f$ is periodic" means "there exists $c\gt 0$ such that $f(x+c)=f(x)$ for all $x$". So the negation of "$f$ is periodic" is "for every $c\gt 0$, there exists at least one $x$ such that $f(x)\neq f(x+c)$."
In order to show that the greatest integer function $f(x)=\lfloor x\rfloor$ is not periodic, we need to show that given any $c\gt 0$ (think of it as a candidate for a period), there is at least one point $x$ for which $f(x)\neq f(x+c)$.
So, let $c\gt 0$ be given. I'm going to come up with some point $x$ (which will depend on $c$) with the property that $\lfloor x\rfloor\neq \lfloor x+c\rfloor$. The idea is simply to take a value of $x$ that is before some integer, but once you add $c$ to it you go above that integer. Then the greatest integer function will give two different values.
The simplest example to take is $x=-\frac{c}{2}$; this number is negative, so $f(-\frac{c}{2}) = \lfloor -\frac{c}{2}\rfloor\lt 0$; on the other hand, $x+c = -\frac{c}{2}+c = \frac{c}{2}$ is positive, so $f(x+c) = \lfloor \frac{c}{2}\rfloor \geq 0$. Since $f(x)$ is strictly less than $0$ and $f(x+c)$ is at least $0$, $f(x)\neq f(x+c)$. So $f(x)$ is not periodic with period $c$.
But $c$ was an arbitrary positive number. That means that we cannot say "$f(x)$ is periodic with period $c$" for any positive number, and that means, by definition, that $f(x) = \lfloor x\rfloor$ is not periodic.
The usual definition of a periodic function presented in undergraduate calculus classes is something like
$f$ is periodic with period $T$ if for all $x \in \mathbb{R}$ we have $f(x+T) = f(x)$.
In particular, $f(x)$ must be defined for all real numbers. Hence every function that is periodic has, by definition, an unbounded domain. Implicit in the definition is that $f(x+kT) = f(x)$ for all integers $k$. As a pathological counterexample, consider the following pathological function:
For each $n\in\mathbb{N}$, choose (inductively) some $x_n \in (0,1)$ such that $x_i - x_j$ is irrational for all $i\ne j$. Define a function $f$ by setting $$ f(x_n + n + m) = 1, $$ where $m$ and $n$ range over the natural numbers. This function is pretty far from periodic, if you ask me (it doesn't really repeat itself in any meaningful way from one interval to the next), but if you drop the requirement that a periodic function be defined for all of $\mathbb{R}$, you could perhaps argue that $f$ is 1-periodic.
Another approach is to consider that a periodic function is really defined on an $n$-dimensional torus (a circle, in the case of functions that we want to think of as taking real inputs, but we can also talk about periodicity in $\mathbb{R}^n$). A periodic function on $\mathbb{R}$ then becomes some kind of lifting of a function on the circle via a covering map. Again, this would imply that the domain (thought of as a subset of $\mathbb{R}$) must be unbounded, though it needn't imply that the domain is all of $\mathbb{R}$.
Best Answer
A function $f:\mathbb{R} \to \mathbb{R}$ is periodic if there exists some number $t > 0$ such that $$ f(x) = f(x + t) $$ A constant function is periodic since you can take $t = 1, t = 2$, etc. (Hint: Hover over the tag "periodic-functions". What do you see?)
The fundamental period of $f$ is the smallest of such $t$'s. Since $t$ cannot be $0$, you are looking for the minimum of $(0,\infty)$, which does not exist.