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Question Number 49810 by Abdo msup. last updated on 10/Dec/18
![calculate S_n (x)=[((x+1)/2)] + [((x+2)/4)] +[((x+4)/8)]+...[((x+2^n )/2^(n+1) )]](Q49810.png)
$${calculate}\: \\ $$$${S}_{{n}} \left({x}\right)=\left[\frac{{x}+\mathrm{1}}{\mathrm{2}}\right]\:+\:\left[\frac{{x}+\mathrm{2}}{\mathrm{4}}\right]\:+\left[\frac{{x}+\mathrm{4}}{\mathrm{8}}\right]+...\left[\frac{{x}+\mathrm{2}^{{n}} }{\mathrm{2}^{{n}+\mathrm{1}} }\right] \\ $$
Commented by maxmathsup by imad last updated on 13/Jan/19
![S_n (x) =Σ_(k=1) ^n [(x/2^k ) +(1/2)] now let prove that [t+(1/2)]=[2t]−[t] let [t]=p with p from Z ⇒ p≤t<p+1 ⇒2p≤2t<2p+2 p+(1/2)≤t+(1/2)<p+(3/2) if p+(1/2)≤t+(1/2)<p+1 ⇒[t+(1/2)]=p and we have 2p+1≤2t+1<2p+2 ⇒2p≤2t<2p+1 ⇒[2t]=2p ⇒[2t]−[t]=p=[t+(1/2)] if p+1≤t+(1/2)<p+(3/2) we prve also that [2t]−[t]=[t+(1/2)] ⇒ S_n (x)=Σ_(k=1) ^n [((2x)/2^k )]−[(x/2^k )]=Σ_(k=1) ^n [(x/2^(k−1) )]−[(x/2^k )] =[x]−[(x/2)]+[(x/2)]−[(x/2^2 )] +....[(x/2^(n−2) )]−[(x/2^(n−1) )] +[(x/2^(n−1) )]−[(x/2^n )] =[x]−[(x/2^n )] ⇒lim_(n→+∞) S_n =[x] .](Q52839.png)
$${S}_{{n}} \left({x}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \left[\frac{{x}}{\mathrm{2}^{{k}} }\:+\frac{\mathrm{1}}{\mathrm{2}}\right]\:\:{now}\:{let}\:{prove}\:{that}\:\left[{t}+\frac{\mathrm{1}}{\mathrm{2}}\right]=\left[\mathrm{2}{t}\right]−\left[{t}\right] \\ $$$${let}\:\left[{t}\right]={p}\:\:{with}\:{p}\:{from}\:{Z}\:\Rightarrow\:{p}\leqslant{t}<{p}+\mathrm{1}\:\Rightarrow\mathrm{2}{p}\leqslant\mathrm{2}{t}<\mathrm{2}{p}+\mathrm{2} \\ $$$${p}+\frac{\mathrm{1}}{\mathrm{2}}\leqslant{t}+\frac{\mathrm{1}}{\mathrm{2}}<{p}+\frac{\mathrm{3}}{\mathrm{2}}\:\:\:{if}\:{p}+\frac{\mathrm{1}}{\mathrm{2}}\leqslant{t}+\frac{\mathrm{1}}{\mathrm{2}}<{p}+\mathrm{1}\:\Rightarrow\left[{t}+\frac{\mathrm{1}}{\mathrm{2}}\right]={p}\:{and}\:{we}\:{have} \\ $$$$\mathrm{2}{p}+\mathrm{1}\leqslant\mathrm{2}{t}+\mathrm{1}<\mathrm{2}{p}+\mathrm{2}\:\Rightarrow\mathrm{2}{p}\leqslant\mathrm{2}{t}<\mathrm{2}{p}+\mathrm{1}\:\Rightarrow\left[\mathrm{2}{t}\right]=\mathrm{2}{p}\:\Rightarrow\left[\mathrm{2}{t}\right]−\left[{t}\right]={p}=\left[{t}+\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$${if}\:\:{p}+\mathrm{1}\leqslant{t}+\frac{\mathrm{1}}{\mathrm{2}}<{p}+\frac{\mathrm{3}}{\mathrm{2}}\:{we}\:{prve}\:{also}\:{that}\:\left[\mathrm{2}{t}\right]−\left[{t}\right]=\left[{t}+\frac{\mathrm{1}}{\mathrm{2}}\right]\:\Rightarrow \\ $$$${S}_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \left[\frac{\mathrm{2}{x}}{\mathrm{2}^{{k}} }\right]−\left[\frac{{x}}{\mathrm{2}^{{k}} }\right]=\sum_{{k}=\mathrm{1}} ^{{n}} \:\left[\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }\right]−\left[\frac{{x}}{\mathrm{2}^{{k}} }\right] \\ $$$$=\left[{x}\right]−\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{{x}}{\mathrm{2}}\right]−\left[\frac{{x}}{\mathrm{2}^{\mathrm{2}} }\right]\:+....\left[\frac{{x}}{\mathrm{2}^{{n}−\mathrm{2}} }\right]−\left[\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }\right]\:+\left[\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }\right]−\left[\frac{{x}}{\mathrm{2}^{{n}} }\right] \\ $$$$=\left[{x}\right]−\left[\frac{{x}}{\mathrm{2}^{{n}} }\right]\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:\:{S}_{{n}} =\left[{x}\right]\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18
![T_r =[((x+2^(r−1) )/2^r )]=[(x/2^r )+(1/2)]>[(x/2^r )]+[(1/2)] [(x/2^r )+(1/2)]>[(x/2^r )]+0 S_n (x)=Σ_(r=1) ^n [(x/2^r )] value of S_n (x) can be determined when we know how x related to 2 x=f(2)](Q49812.png)
$${T}_{{r}} =\left[\frac{{x}+\mathrm{2}^{{r}−\mathrm{1}} }{\mathrm{2}^{{r}} }\right]=\left[\frac{{x}}{\mathrm{2}^{{r}} }+\frac{\mathrm{1}}{\mathrm{2}}\right]>\left[\frac{{x}}{\mathrm{2}^{{r}} }\right]+\left[\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$\left[\frac{{x}}{\mathrm{2}^{{r}} }+\frac{\mathrm{1}}{\mathrm{2}}\right]>\left[\frac{{x}}{\mathrm{2}^{{r}} }\right]+\mathrm{0} \\ $$$${S}_{{n}} \left({x}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\frac{{x}}{\mathrm{2}^{{r}} }\right] \\ $$$${value}\:{of}\:{S}_{{n}} \left({x}\right)\:{can}\:{be}\:{determined}\:{when}\:{we}\:{know} \\ $$$${how}\:{x}\:{related}\:{to}\:\mathrm{2} \\ $$$${x}={f}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$