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Question Number 197919 by universe last updated on 04/Oct/23

$${\mathrm{I}}_m={\int }_0^1\left(\frac{\lfloor 2^{m}x\rfloor }{3^m}\underset{n=m+1}{\sum ^{\mathrm{\infty }}}\frac{\lfloor 2^{n}x\rfloor }{3^n}\right)dx$$$$\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}$$$$\mathrm{I}=\underset{m=1}{\sum ^{\mathrm{\infty }}}{\mathrm{I}}_m=?$$

Answered by witcher3 last updated on 06/Oct/23

$$2^{\mathrm{m}}\mathrm{x}=\mathrm{t}$$$$\iff {\mathrm{I}}_{\mathrm{m}}=\frac{1}{2^{\mathrm{m}}}{\int }_0^{2^{\mathrm{m}}}\frac{\left[\mathrm{t}\right]}{3^{2\mathrm{m}}}\sum _{\mathrm{n}⩾\mathrm{m}+1}\frac{\left[2^{\mathrm{n}-\mathrm{m}}\mathrm{t}\right]}{3^{\mathrm{n}-\mathrm{m}}}$$$$\Rightarrow {18}^{\mathrm{m}}{\mathrm{I}}_{\mathrm{m}}={\int }_0^{2^{\mathrm{m}}}\left[\mathrm{t}\right]\sum _{\mathrm{n}⩾1}\frac{\left[2^{\mathrm{n}}\mathrm{t}\right]}{3^{\mathrm{n}}}\mathrm{dt}$$$$=\underset{\mathrm{k}=0}{\sum ^{2^{\mathrm{m}}-1}}{\int }_{\mathrm{k}}^{\mathrm{k}+1}\mathrm{k}.\sum _{\mathrm{n}⩾1}\frac{\left[2^{\mathrm{n}}\mathrm{t}\right]}{3^{\mathrm{n}}}\mathrm{dt},2^{\mathrm{n}}\mathrm{t}=\mathrm{y}$$$$=\sum _{\mathrm{k}}\sum _{\mathrm{n}}{\int }_{2^{\mathrm{n}}\mathrm{k}}^{(\mathrm{k}+1)2^{\mathrm{n}}}\frac{\mathrm{k}}{6^{\mathrm{n}}}\left[\mathrm{y}\right]\mathrm{dy}={18}^{\mathrm{m}}{\mathrm{I}}_{\mathrm{m}}$$$${\int }_{2^{\mathrm{n}}\mathrm{k}}^{(\mathrm{k}+1)2^{\mathrm{n}}}\left[\mathrm{y}\right]\mathrm{dy}=\underset{\mathrm{j}=0}{\sum ^{2^{\mathrm{n}}-1}}{\int }_{2^{\mathrm{n}}\mathrm{k}+\mathrm{j}}^{2^{\mathrm{n}}\mathrm{k}+1+\mathrm{j}}\left[\mathrm{y}\right]\mathrm{dy}$$$$=\underset{\mathrm{j}=0}{\sum ^{2^{\mathrm{n}}-1}}{2}^{\mathrm{n}}\mathrm{k}+\mathrm{jdy}$$$$=2^{2\mathrm{n}}\mathrm{k}+2^{\mathrm{n}-1}(2^{\mathrm{n}}-1)=$$$${18}^{\mathrm{m}}{\mathrm{I}}_{\mathrm{m}}=\sum _{\mathrm{k}}\sum _{\mathrm{n}⩾1}\frac{\mathrm{k}(2^{2\mathrm{n}}\mathrm{k}+2^{\mathrm{n}-1}(2^{\mathrm{n}}-1)}{6^{\mathrm{n}}}$$$$=\sum _{\mathrm{k}}\sum _{\mathrm{n}⩾1}{\mathrm{k}}^2{\left(\frac{2}{3}\right)}^{\mathrm{n}}+\frac{\mathrm{k}}{2}.{\left(\frac{2}{3}\right)}^{\mathrm{n}}-\frac{\mathrm{k}}{2}{\left(\frac{2}{6}\right)}^{\mathrm{n}}$$$$=\underset{\mathrm{k}=0}{\sum ^{2^{\mathrm{m}}-1}}.({2\mathrm{k}}^2+\frac{3}{4}\mathrm{k})=\frac{(2^{\mathrm{m}}-1)\left(2^{\mathrm{m}}\right)(2^{\mathrm{m}+1}-1)}{3}+\frac{3}{4}{2}^{\mathrm{m}-1}(2^{\mathrm{m}}-1)$$$${\mathrm{I}}_{\mathrm{m}}=\frac{2^{3\mathrm{m}+1}-3.2^{2\mathrm{m}}+2^{\mathrm{m}}}{3.{18}^{\mathrm{m}}}+\frac{3.2^{2\mathrm{m}-1}-3.2^{\mathrm{m}}}{4.{18}^{\mathrm{m}}}$$$$\mathrm{I}=\sum _{\mathrm{m}⩾1}\frac{2}{3}.{\left(\frac{2}{3}\right)}^{2\mathrm{m}}-{\left(\frac{2}{9}\right)}^{\mathrm{m}}+\frac{1}{3}.\frac{1}{9^{\mathrm{m}}}+\frac{3}{8}.{\left(\frac{2}{9}\right)}^{\mathrm{m}}-\frac{3}{4}.{\left(\frac{1}{9}\right)}^{\mathrm{m}}$$$$\mathrm{easy}\mathrm{from}\mathrm{here}$$$$\sum _{\mathrm{n}⩾1}{\mathrm{a}}^{\mathrm{n}}=\frac{\mathrm{a}}{1-\mathrm{a}},\mathrm{\forall }\mid \mathrm{a}\mid <1$$$$$$$$$$$$$$$$$$

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