How do you draw the graph of $f(x)=e^{x-[x]}$ where $[x]$ denotes greatest integer function? I am trying to draw the graph of $f(x)=e^{x-[x]}$.
I have come to conclusion that:
$$f(x) = \begin{cases} e^x &\text{, if } 0\le x \lt 1\\
    e^{x-1} &\text{, if } -1\le x \lt 0\\
    e^{x+1} &\text{, if } 1 \le x \lt 2 \\ \vdots & \vdots \end{cases}$$
but when I am trying to plot the graph it does not corroborate.
 A: Rewrite as
$$f(x)=e^{x-[x]}=e^{\{x\}}$$
As @dezdichato stated in his comment, the function has a period of $1$, i.e $f(x)=f(x+1)$ since $\{x\} = \{x + 1\}$. It's well-known that $0 \le \{x\} < 1$ (the graph is attached), and therefore
$$e^0 = 1 \le f(x) < e^1$$
Knowing this, the period and the fact that $\{x\}=x, \forall x \in [0,1)$, you should be able to sketch the graph.
Basically, it will be the graph of $e^x$ for $x\in[0,1)$ extended according to its period.

Below is the graph of $\{x\}$:


Below is the segment of $e^x$ to be extended:

A: Any function, lets say $f:\mathbb R \rightarrow \mathbb C $ ,can be extended as a $\mathcal T -periodic$
function across $\mathbb R$ in a unique way with the formula: $f(x)=f(x-\mathcal T[\frac{x}{\mathcal T}])$.
Hence your function is $1-periodic$, so you only have to plot the function in $[L,L+1)$, for instance: $[0,1)$ its easy to notice that
the graph starts from the point $(0,1)$ and ends at $(1,e)$ (with the same curvature as $e^x)$, and so forth extended in $\mathbb R$.
