prove that $\quad n = E\left[ {\frac{{n - 1}}{2}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 4}}{4}} \right]$

Question

good evening I want to show that: $\forall n \in \mathbb {Z}\quad :\quad n = E\left[ {\frac{{n - 1}}{2}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 4}}{4}} \right]$

$\begin{array}{l} m = E\left[ {\frac{{n - 1}}{2}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 4}}{4}} \right] \\ \forall k \in \mathbb {Z}:E\left[ {\frac{k}{2}} \right] + E\left[ {\frac{k}{2}} \right] = k \\ E\left[ {\frac{{n - 1}}{4}} \right] + E\left[ {\frac{{n - 1}}{4}} \right] \le E\left[ {\frac{{n - 1}}{2}} \right] \\ \Rightarrow 2E\left[ {\frac{{n - 1}}{4}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 4}}{4}} \right] \le m \\ \Rightarrow 4\left( {\frac{{n - 1}}{4}} \right) \le m \\ \Rightarrow n - 1 \le m.....\left( 1 \right) \\ m = E\left[ {\frac{{2\left( {n - 1} \right)}}{4}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 2}}{4} + \frac{1}{2}} \right] \\ E\left[ {\frac{{n + 4}}{4}} \right] = \left\{ \begin{array}{l} E\left[ {\frac{{n + 2}}{4}} \right] \\ E\left[ {\frac{{n + 2}}{4}} \right] + 1 \\ \end{array} \right. \\ m = E\left[ {\frac{{2\left( {n - 1} \right)}}{4}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 4}}{4}} \right] \\ \Rightarrow m = \left\{ \begin{array}{l} E\left[ {\frac{{2\left( {n - 1} \right)}}{4}} \right] + 2E\left[ {\frac{{n + 2}}{4}} \right] \\ E\left[ {\frac{{2\left( {n - 1} \right)}}{4}} \right] + 2E\left[ {\frac{{n + 2}}{4}} \right] + 1 \\ \end{array} \right. \\ \Rightarrow m < \left\{ \begin{array}{l} \frac{{2\left( {n - 1} \right)}}{4} + 2\frac{{n + 2}}{4} + 1 = n + 1 \\ \frac{{2\left( {n - 1} \right)}}{4} + 2\frac{{n + 2}}{4} + 2 = n + 2 \\ \end{array} \right. \\ \Rightarrow m < n + 1...\left( 2 \right) \\ \end{array}$

$ \begin{array}{l} E\left[ x \right] \\ x \le E\left[ x \right] < x + 1 \\ \end{array} $

disigne integral part How can we exclude the case $n-1$ thank you in advance

Answer 1

First, the function you are denoting E[.] is more usually called the "floor" function these days, and is abbreviated as $\lfloor x \rfloor$ in latex.

The trick here is to consider the four cases n $\equiv$ 0,1,2,3 (mod 4) separately. For example, if $n \equiv$ 0 (mod 4) then $\lfloor (n-1)/2 \rfloor = n/2 - 1$. Similarly $\lfloor (n+2)/4 \rfloor = n/4$ in this case and $\lfloor (n+4)/4 \rfloor = n/4 + 1$. Adding these gives the result for the case n $\equiv$ 0 (mod 4). Now do the same for the other cases.

Answer 2

disigne integr part $\begin{array}{l} E\left[ x \right] \\ x \le E\left[ x \right] < x + 1 \\ \end{array}$

$ \left\{ \begin{array}{l} n \equiv 1\left[ 4 \right] \\ n \equiv 2\left[ 4 \right] \\ \end{array} \right. \Rightarrow \left\{ \begin{array}{l} E\left[ {\frac{{n - 1}}{2}} \right] = \frac{n}{2} - 1 \\ E\left[ {\frac{{n + 2}}{4}} \right] = \frac{n}{4} \\ E\left[ {\frac{{n + 4}}{4}} \right] = \frac{n}{4} + 1 \\ \end{array} \right. \Rightarrow E\left[ {\frac{{n - 1}}{2}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 4}}{4}} \right] = n $

Answer 3

$\left\{ \begin{array}{l} n \equiv 1\left[ 4 \right] \\ n \equiv 2\left[ 4 \right] \\ \end{array} \right. \Rightarrow \left\{ \begin{array}{l} E\left[ {\frac{{n - 1}}{2}} \right] = \frac{n}{2} - 1 \\ E\left[ {\frac{{n + 2}}{4}} \right] = \frac{n}{4} \\ E\left[ {\frac{{n + 4}}{4}} \right] = \frac{n}{4} + 1 \\ \end{array} \right. \Rightarrow E\left[ {\frac{{n - 1}}{2}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 4}}{4}} \right] = n$