Prove that
$$\lfloor{x}\rfloor=\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots$$ $x\geq{0}$
I asked this question here yesterday. Although I was able to come up with a proof, the second part was left unanswered.
I proved it using,
Subtract $\lfloor{x}\rfloor$ from both sides.
Then we need to prove that, $$-\lfloor{x}\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots=0$$
We know that, $\lfloor{x}\rfloor=\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor$
Therefore, $-\lfloor{x}\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor=-\bigg\lfloor{\frac{x}{2}}\bigg\rfloor$
Subsituting this,
$$-\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots$$
$$-\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{\frac{x}{2}+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots$$
Similarly, I reach,
$$\lim_{n\to\infty}-\bigg\lfloor{\frac{x}{2^n}}\bigg\rfloor+\bigg\lfloor{\frac{\frac{x}{2^n}+1}{2}}\bigg\rfloor$$
i.e. $$=0$$
My question is,as $\lfloor{x}\rfloor=\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor$ is true $\forall x\in{\mathbb{R}}$.
Why isn't $$\lfloor{x}\rfloor=\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots$$ true $\forall x\in{\mathbb{R}}$.
Also how would the expression change for $x\lt{0}$.
Thank you.
For $x=-1$ the lest side is $-1$ and the right side is $0$.