Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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

$$calculate$$$$S_n\left(x\right)=\left[\frac{x+1}{2}\right]+\left[\frac{x+2}{4}\right]+\left[\frac{x+4}{8}\right]+...\left[\frac{x+2^n}{2^{n+1}}\right]$$

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

$$S_n\left(x\right)=\sum _{k=1}^n[\frac{x}{2^k}+\frac{1}{2}]nowletprovethat[t+\frac{1}{2}]=\left[2t\right]-\left[t\right]$$$$let\left[t\right]=pwithpfromZ\Rightarrow p⩽t<p+1\Rightarrow 2p⩽2t<2p+2$$$$p+\frac{1}{2}⩽t+\frac{1}{2}<p+\frac{3}{2}ifp+\frac{1}{2}⩽t+\frac{1}{2}<p+1\Rightarrow [t+\frac{1}{2}]=pandwehave$$$$2p+1⩽2t+1<2p+2\Rightarrow 2p⩽2t<2p+1\Rightarrow \left[2t\right]=2p\Rightarrow \left[2t\right]-\left[t\right]=p=[t+\frac{1}{2}]$$$$ifp+1⩽t+\frac{1}{2}<p+\frac{3}{2}weprvealsothat\left[2t\right]-\left[t\right]=[t+\frac{1}{2}]\Rightarrow$$$$S_n\left(x\right)=\sum _{k=1}^n\left[\frac{2x}{2^k}\right]-\left[\frac{x}{2^k}\right]=\sum _{k=1}^n\left[\frac{x}{2^{k-1}}\right]-\left[\frac{x}{2^k}\right]$$$$=\left[x\right]-\left[\frac{x}{2}\right]+\left[\frac{x}{2}\right]-\left[\frac{x}{2^2}\right]+....\left[\frac{x}{2^{n-2}}\right]-\left[\frac{x}{2^{n-1}}\right]+\left[\frac{x}{2^{n-1}}\right]-\left[\frac{x}{2^n}\right]$$$$=\left[x\right]-\left[\frac{x}{2^n}\right]\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n=\left[x\right].$$

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

$$T_r=\left[\frac{x+2^{r-1}}{2^r}\right]=[\frac{x}{2^r}+\frac{1}{2}]>\left[\frac{x}{2^r}\right]+\left[\frac{1}{2}\right]$$$$[\frac{x}{2^r}+\frac{1}{2}]>\left[\frac{x}{2^r}\right]+0$$$$S_n\left(x\right)=\underset{r=1}{\sum ^n}\left[\frac{x}{2^r}\right]$$$$valueof{S}_n\left(x\right)canbedeterminedwhenweknow$$$$howxrelatedto2$$$$x=f\left(2\right)$$$$$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com