Manual Graph Sketching For $\lfloor(x)\rfloor+\sqrt{x-\lfloor(x)\rfloor}$

Question

How to prove $$\lfloor x\rfloor+\sqrt{x-\lfloor x \rfloor}$$ is continous for all $x$?And how to plot the nature of the graph manually?

I can prove that at integer points it is continous.But what about other points?Rather how to trace the graph manually?

Answer 1

Let $x=n+d$ be the decomposition in integer and fractional parts.

Then

$$f(x)=f(n+d)=n+\sqrt d=n+\sqrt{x-n}.$$

For $x$ between two integers, $n$ is constant and the square root function is known to be continuous.

To plot the function, first consider the range $0\le x < 1$ (you get an arc of a parabola) and replicate by translating the graph, using $f(n+x)=n+f(x)$.

Answer 2

Addressing each piece of this function and how it affects the whole makes it easy to understand:

$\sqrt{x−⌊x⌋}$ <--- You might recognize the function $x−⌊x⌋$ as the ramp function. In fact, this is a composite function of the ramp function and $x^{1/2}$. It'll look like a periodic of the square root of x, a repeating right-facing half-wishbone pattern.

$⌊x⌋ +$ <--- adding the previous function to a floor function will simply sift the periodic upward by 1 every iteration.

So, in summary, it's a periodic root of x function that is shifted up by 1 every iteration. No need to sketch.

EDIT: it's also continuous.

Answer 3

This hopping-jumping root function can be analyzed by a recursion as well. First, since $x \ge \lfloor x \rfloor$, the function $f(x) = \lfloor x\rfloor+\sqrt{x-\lfloor x \rfloor}$ is defined for every $x\in \mathbb{R}$.

Now start from a smaller and simpler interval, and take $d\in[0,1[$, where $\lfloor d \rfloor = 0$. Thus $f(d) = \sqrt{d}$, which can be drawn by hand like an arch of rotated parabola taking values in $[0,1[$, joigning $(0,0)$ and $(1_-,1_-)$. Since $f(1) = 1 + \sqrt{1-1} = \sqrt{1_-}$, $f$ is continuous on $[0,1]$ with $f(0)=0$ and $f(1)=1$.

Assume that $f$ is continuous on $[n,n+1]$ for some integer $n\ge0$, with $f(n)=n$ and $f(n+1)=n+1$. Take $x\in[n,n+1]$ and $y=x+1$. Then $f(y) = f(x)+1$ is continuous on $[n+1,n+2]$, and $f(n+1)=n+1$ and $f(n+2)=n+2$.

You can do the same, going negative. Graphically, you start from a pattern on $[0,1]$, and obtain your function on each interval with a translation of vector $(n,n)$.