Find the number of elements in the range$ f(x) =[x] + [2x] +[2x/3] +[3x] +[4x] +[5x]$ for $0\le x \le3$.

Question

Find the number of elements in the range $f(x) =[x] + [2x] +[2x/3] +[3x] +[4x] +[5x]$ for $0\le x \le3.$

I cant understand...It will go very long if i keep breaking them into small intervals .

Answer 1

$$f(x) =[x] + [2x] +[2x/3] +[3x] +[4x] +[5x]$$

Range is $0 \le x \le 3$.

• $[x]$ changes it's value every $\frac 11$ , or 3 times $\tag A$
• $[2x]$ changes it's value every $\frac 12$, or 6 times $\tag B$
• $[2x/3]$ changes it's value every $\frac 32$, or 2 times $\tag C$
• $[3x]$ changes it's value every $\frac 13$, or 9 times $\tag D$
• $[4x]$ changes it's value every $\frac 14$, or 12 times $\tag E$
• $[5x]$ changes it's value every $\frac 15$, or 15 times $\tag F$

So the number of changes is $3 + 6 + 2 + 9 + 12 + 15$ minus duplicates or something less than $47$.

Of the changes in $(F)$, the changes at $x = [1,2,3]$ will occur in one of $(A), (B), ... (E)$, so $(F)$ overcounts $3$.

Of the changes in $(E)$, the changes at $x = [1/2, 1, 3/2, 2, 5/2, 3]$ occur in $(A), (B), (C), (D)$, so $(E)$ overcounts $6$ times.

Of the changes in $(D)$, the changes at $x = [1,2,3]$ will occur in $(A), (B)$, or $(C)$, so $(D)$ overcounts $3$.

Of the changes in $(C)$, the changes at $x = [3/2, 3]$ will occur in $(A)$ or $(B)$, so it overcounts $2$.

Of the changes in $(B)$, the changes at $x = [1, 2, 3]$ will occur in $(A)$, so it overcounts $3$.

So the total number of changes is $47 - 3 - 6 - 3 - 2 - 3 = 30$. Counting the inital value of $0$, that leaves $31$ possible values.

---

Or for the short version:

$\begin{align} U_{[x]} &= \{0,\,1,\,2,\,3\} \\ U_{[2x]} &= \{0,\,1/2,\,1,\,3/2,\,2,\,5/2,\,3\} \\ U_{[2x/3]} &= \{0,\,3/2,\,3\} \\ U_{[3x]} &= \{0,\,1/3,\,2/3,\,1,\,4/3,\,5/3,\,2,\,7/3,\,8/3,\,3\} \\ U_{[4x]} &= \{0,\,1/4,\,1/2,\,3/4,\,1,\,5/4,\,3/2,\,7/4,\,2,\,9/4,\,5/2,\,11/4,\,3\} \\ U_{[5x]} &= \{0,\,1/5,\,2/5,\,3/5,\,4/5,\,1,\,6/5,\,7/5,\,8/5,\,9/5,\,2,\,11/5,\,12/5,\,13/5,\,14/5,\,3\} \\ \end{align}$

$U = U_{[x]} \cup U_{[2x]} \cup U_{[3x/2]} \cup U_{[3x]} \cup U_{[4x]} \cup U_{[5x]}$

$|U| = 31$

Answer 2

The lowest value is 0, the highest value is 3+6+2+9+12+15. Then you need to find the values where two or more of the terms increase at the same time, that is where two or more are integers. Five terms are integers at x=1, so the sum increases by 5, and four possible values for the sum are skipped over. [3x],[4x] and [5x] are only integers at the same time if x=integer. There are only a few places, other than x=integer, where two or more terms are integers.

Answer 3

Given, $f(x)=[x]+[2x]+[2x/3]+[3x]+[4x]+[5x]$.

Segment this function into smaller functions,

$f_1(x)=[x],f_2(x)=[2x],f_3(x)=[2x/3],f_4(x)=[3x],f_5(x)=[4x],f_6(x)=[5x]$

$ f_1(x) \begin{cases} 0& {x \in [0,1)}\\ 1& {x \in [1,2)}\\ 2& {x \in [2,3)}\\ 3& {x=3}\\ \end{cases}$

$ f_2(x) \begin{cases} 0& {x \in [0,1/2)}\\ 1& {x \in [1/2,1)}\\ 2& {x \in [1,3/2)}\\ 3& {x \in [3/2,2)}\\ 4& {x \in [2,5/2)}\\ 5& {x \in [5/2,3)}\\ 6& {x=3}\\ \end{cases}$

$ f_3(x) \begin{cases} 0& {x \in [0,3/2)}\\ 1& {x \in [3/2,3)}\\ 2& {x=3}\\ \end{cases}$

$ f_4(x) \begin{cases} 0& {x \in [0,1/3)}\\ 1& {x \in [1/3,2/3)}\\ 2& {x \in [2/3,1)}\\ 3& {x \in [1,4/3)}\\ 4& {x \in [4/3,5/3)}\\ 5& {x \in [5/3,2)}\\ 6& {x \in [2,7/3)}\\ 7& {x \in [7/3,8/3)}\\ 8& {x \in [8/3,3)}\\ 9& {x=3} \end{cases}$

$ f_5(x) \begin{cases} 0& {x \in [0,1/4)}\\ 1& {x \in [1/4,1/2)}\\ 2& {x \in [1/2,3/4)}\\ 3& {x \in [3/4,1)}\\ 4& {x \in [1,5/4)}\\ 5& {x \in [5/4,3/2)}\\ 6& {x \in [3/2,7/4)}\\ 7& {x \in [7/4,2)}\\ 8& {x \in [2,9/4)}\\ 9& {x \in [9/4,5/2)}\\ 10& {x \in [5/2,11/4)}\\ 11& {x \in [11/4,3)}\\ 12& {x=3} \end{cases}$

$ f_6(x) \begin{cases} 0& {x \in [0,1/5)}\\ 1& {x \in [1/5,2/5)}\\ 2& {x \in [2/5,3/5)}\\ 3& {x \in [3/5,4/5)}\\ 4& {x \in [4/5,1)}\\ 5& {x \in [1,6/5)}\\ 6& {x \in [6/5,7/5)}\\ 7& {x \in [7/5,8/5)}\\ 8& {x \in [8/5,9/5)}\\ 9& {x \in [9/5,2)}\\ 10& {x \in [2,11/5)}\\ 11& {x \in [11/5,12/5)}\\ 12& {x \in [12/5,13/5)}\\ 13& {x \in [13/5,14/5)}\\ 14& {x \in [14/5,3)}\\ 15& {x=3} \end{cases}$

Now, all you have to do, is find out the number of intervals, from above, where the main function changes its value and you will reach solution. I think you can take it from here.

Answer 4

Denote the function $f$ without the $\left\lfloor{2x\over3}\right\rfloor$ term by $g$. The "derivative" of $g$ is periodic with period $1$ and has spikes of weight $1$ at ${1\over5}$, ${1\over4}$, ${1\over3}$, ${2\over5}$, ${3\over5}$, ${2\over3}$, ${3\over4}$, ${4\over5}$, then a spike of weight $2$ at ${1\over2}$, and a spike of weight $5$ at $1$.

It follows that $g$ begins with $0$ at $x=0$, then increases by steps of $1$, but omits the value $5$, and jumps directly from $10$ to $15$ at $x=1$. This behavior is repeated in the intervals $[1,2]$ and $[2,3]$. $\langle$It follows that $20$ is omitted, and we have a jump from $25$ to $30$ at $x=2$, then $35$ is omitted, and we have a jump from $40$ to $45$.$\rangle$

Finally we have to take care of the $\left\lfloor{2x\over3}\right\rfloor$ term. It has unit jumps at $x={3\over2}$ and $x=3$. Therefore $f$ coincides with $g$ for $0\leq x<{3\over2}$, is one larger than $g$ for ${3\over2}\leq x<3$ and is $2$ larger than $g$ at $x=3$.

The sentence in $\langle\ldots\rangle$ above therefore has to be changed to: It follows that $20$ and $21$ are omitted, and we have a jump from $26$ to $31$ at $x=2$, then $36$ is omitted, and we have a jump from $41$ to $47$.

To sum it all up, one has $$\eqalign{f\bigl([0,3]\bigr)&=\{k\in{\mathbb Z}\>|\>0\leq k\leq 47\}\setminus\cr &\quad\{5,11,12,13,14,20,21,27,28,29,30,36,42,43,44,45,46\}\ ,\cr}$$ so the range in question has 31 elements. Compare with the following Mathematica output: