calculate-S-n-x-x-1-2-x-2-4-x-4-8-x-2-n-2-n-1-

Question Number 49810 by Abdo msup. last updated on 10/Dec/18

c a l c u l a t e

S_{n} (x) = [\frac{x + 1}{2}] + [\frac{x + 2}{4}] + [\frac{x + 4}{8}] + \dots [\frac{x + 2^{n}}{2^{n + 1}}]

Commented by maxmathsup by imad last updated on 13/Jan/19

S_{n} (x) = \sum_{k = 1}^{n} [\frac{x}{2^{k}} + \frac{1}{2}] n o w l e t p r o v e t h a t [t + \frac{1}{2}] = [2 t] - [t]

l e t [t] = p w i t h p f r o m Z \Rightarrow p ⩽ t < p + 1 \Rightarrow 2 p ⩽ 2 t < 2 p + 2

p + \frac{1}{2} ⩽ t + \frac{1}{2} < p + \frac{3}{2} i f p + \frac{1}{2} ⩽ t + \frac{1}{2} < p + 1 \Rightarrow [t + \frac{1}{2}] = p a n d w e h a v e

2 p + 1 ⩽ 2 t + 1 < 2 p + 2 \Rightarrow 2 p ⩽ 2 t < 2 p + 1 \Rightarrow [2 t] = 2 p \Rightarrow [2 t] - [t] = p = [t + \frac{1}{2}]

i f p + 1 ⩽ t + \frac{1}{2} < p + \frac{3}{2} w e p r v e a l s o t h a t [2 t] - [t] = [t + \frac{1}{2}] \Rightarrow

S_{n} (x) = \sum_{k = 1}^{n} [\frac{2 x}{2^{k}}] - [\frac{x}{2^{k}}] = \sum_{k = 1}^{n} [\frac{x}{2^{k - 1}}] - [\frac{x}{2^{k}}]

= [x] - [\frac{x}{2}] + [\frac{x}{2}] - [\frac{x}{2^{2}}] + \dots . [\frac{x}{2^{n - 2}}] - [\frac{x}{2^{n - 1}}] + [\frac{x}{2^{n - 1}}] - [\frac{x}{2^{n}}]

= [x] - [\frac{x}{2^{n}}] \Rightarrow {l i m}_{n \to + \infty} S_{n} = [x] .

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18

T_{r} = [\frac{x + 2^{r - 1}}{2^{r}}] = [\frac{x}{2^{r}} + \frac{1}{2}] > [\frac{x}{2^{r}}] + [\frac{1}{2}]

[\frac{x}{2^{r}} + \frac{1}{2}] > [\frac{x}{2^{r}}] + 0

S_{n} (x) = \underset{r = 1}{\sum^{n}} [\frac{x}{2^{r}}]

v a l u e o f S_{n} (x) c a n b e d e t e r m i n e d w h e n w e k n o w

h o w x r e l a t e d t o 2

x = f (2)