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