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What is the period of the function, $x(t)=1+\cos(2\pi t)$

when, $t=0$, $x(0)=2$

when, $t=\frac{1}{2}$, $x(\frac{1}{2})=0$

when, $t=1$, $x(1)=2$

when, $t=-1$, $x(-1)=2$

when, $t=-\frac{1}{2}$, $x(-\frac{1}{2})=0$

when, $t=\frac{1}{4}$, $x(\frac{1}{4})=1$

when, $t=-\frac{1}{4}$, $x(-\frac{1}{4})=1$

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The period is the period of your cosine : 1. Draw the graph of the function.

A simple way to do it is solving the equation $$\begin{align} x(t) &= x(t+T) \\ 1+\cos(2\pi t) &= 1+\cos(2\pi(t+T)) \\ \cos(2\pi t) &= \cos(2\pi T + 2\pi t) \end{align}$$

As you see, the constant factor doesn't matter.

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  • Figure out (or, if you really have to, look up) the period of $\cos(t)$.
  • If a function $f(t)$ has the property that $f(t)=f(t+p)$ for any $t$, and a function $g$ is defined to be $g(t)=f(at)$ for some number $a\neq 0$, determine a $q$ such that $g(t)=g(t+q)$ for any $t$.
  • Adding a constant is irrelevant to periodicity.
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If $f(x)=1+\cos2\pi t$ and the $T$ be the period

$f(x+T)=f(x)\implies 1+\cos2\pi t=1+\cos2\pi(t+T)$

$\implies \cos2\pi t=\cos2\pi(t+T)$

$\implies 2\pi t=2n\pi \pm2\pi(t+T)$ where $n$ is any integer

$\implies t=n\pm(t+T)$

$'-'\implies t=n-t-T\implies T=n-2t$ which is not constant

$'+'\implies t=n+t+T\implies T=-n$

So, the fundamental/prime/primitive period, being the least positive will be $1$

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A period of a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is a set $A$ of values such that $\forall x\in A, \forall t\in\mathbb{R}, f(x+t)=f(t)$. Now let's consider your function: $f(t)= 1 + \cos(2 \pi x)$. The period is the same of the function $f(t)=\cos(2 \pi x)$ (why is that so? Hint: double inclusion on the two respective sets of periods). Now we can use (it isn't hard do prove anyhow) that the period of the function $f(x) = \cos(x)$ is $T=\lgroup 2k\pi: k\in \mathbb{Z}\rgroup $. Now you want to know when $\cos(2\pi (x+t))=\cos(2\pi t), \forall t \in \mathbb{R}$. This happens only if $2\pi x \in T, (\Rightarrow x\in \mathbb{Z})$ or else you would have that the period of $\cos(x)$ is not $2k\pi$. I leave to you this last proof, but if you want I would be happy to post it afterwards.

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