Find the value of :
$\displaystyle \lfloor\frac {x + 1}{2}\rfloor + \lfloor\frac {x + 2}{4}\rfloor + \lfloor\frac {x + 3}{6}\rfloor + \lfloor\frac {x + 4}{8}\rfloor $+...........upto $\displaystyle \infty$
Hello everyoneFor finite positive $\displaystyle x$, all terms after a certain point are zero, so it certainly has a finite sum.
What the answer is, though, I don't know. I'm thinking about it...but I think it may be to do with expressing [an integer] $\displaystyle x$ in binary form.
Grandad
Hello pankajSorry, I don't really understand what you mean here.
If $\displaystyle f(x) =\lfloor\frac {x + 1}{2}\rfloor + \lfloor\frac {x + 2}{4}\rfloor + \lfloor\frac {x + 3}{6}\rfloor + \lfloor\frac {x + 4}{8}\rfloor + ...$ then
$\displaystyle f(1) = 1,\, f(2) = 2,\, f(3) = 4,\, f(4) = 5,\, f(5) = 7,\, ...$
and the differences between values of $\displaystyle f(x)$ for consecutive integer values of $\displaystyle x$ don't form any obvious pattern.
However (using a spreadsheet to do the number-crunching) I have found that for all the values of $\displaystyle n$ I've tried so far (which is up to $\displaystyle n = 7$), $\displaystyle f(2^n) = f(2^{n-1}) + 1$. Which is what made me think it might be worth looking at powers of $\displaystyle 2$ or the binary representation of $\displaystyle x$.
Further thoughts: as $\displaystyle x$ increases through consecutive integer values:
- the value of the first term, $\displaystyle \lfloor\frac {x + 1}{2}\rfloor$ increases by $\displaystyle 1$ every second value of $\displaystyle x$; i.e. when $\displaystyle x =2, 4, 6, 8, ...$
- the value of the second term, $\displaystyle \lfloor\frac {x + 2}{4}\rfloor$ increases by $\displaystyle 1$ every 4th value of $\displaystyle x$; i.e. when $\displaystyle x = 2, 6, 10, 14, 18, ...$
- the value of the third term, $\displaystyle \lfloor\frac {x + 3}{6}\rfloor$ increases by $\displaystyle 1$ every 6th value of $\displaystyle x$; i.e. when $\displaystyle x = 3, 9, 15, 21, ...$
- ... and so on.
Grandad