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Question Number 115645 by ZiYangLee last updated on 27/Sep/20

Commented by Dwaipayan Shikari last updated on 27/Sep/20

$$\frac{(\mathrm{n}+1)\mathrm{p}+(\mathrm{n}-1)\mathrm{q}}{(\mathrm{n}-1)\mathrm{p}+(\mathrm{n}+1)\mathrm{q}}=\mathrm{C}$$$$\frac{2\mathrm{np}+2\mathrm{nq}}{(\mathrm{n}+1)\mathrm{p}-(\mathrm{n}-1)\mathrm{p}+(\mathrm{n}-1)\mathrm{q}-(\mathrm{n}+1)\mathrm{q}}=\frac{\mathrm{C}+1}{\mathrm{C}-1}$$$$2\mathrm{n}.\frac{\mathrm{p}+\mathrm{q}}{\mathrm{p}-\mathrm{q}}=\frac{\mathrm{C}+1}{\mathrm{C}-1}$$$$2\mathrm{n}\frac{\frac{\mathrm{p}}{\mathrm{q}}+1}{\frac{\mathrm{p}}{\mathrm{q}}-1}=\frac{{\left(\frac{\mathrm{p}}{\mathrm{q}}\right)}^{\mathrm{k}}+1}{{\left(\frac{\mathrm{p}}{\mathrm{q}}\right)}^{\mathrm{k}}-1}$$$$2\mathrm{n}\frac{\mathrm{a}+1}{\mathrm{a}-1}=\frac{{\mathrm{a}}^{\mathrm{k}}+1}{{\mathrm{a}}^{\mathrm{k}}-1}$$$$2\mathrm{n}({\mathrm{a}}^{\mathrm{k}+1}+{\mathrm{a}}^{\mathrm{k}}-\mathrm{a}-1)=({\mathrm{a}}^{\mathrm{k}+1}+\mathrm{a}-{\mathrm{a}}^{\mathrm{k}}-1)$$$$2\mathrm{n}=\frac{{\mathrm{a}}^{\mathrm{k}+1}+\mathrm{a}-{\mathrm{a}}^{\mathrm{k}}-1}{{\mathrm{a}}^{\mathrm{k}+1}+{\mathrm{a}}^{\mathrm{k}}-\mathrm{a}-1}$$$$\mathrm{n}>1$$$$\frac{{\mathrm{a}}^{\mathrm{k}+1}+\mathrm{a}-{\mathrm{a}}^{\mathrm{k}}-1}{{\mathrm{a}}^{\mathrm{k}+1}+{\mathrm{a}}^{\mathrm{k}}-\mathrm{a}-1}>2$$$$-{\mathrm{a}}^{\mathrm{k}+1}-{2\mathrm{a}}^{\mathrm{k}}+\mathrm{a}+1>0$$$${\mathrm{a}}^{\mathrm{k}}(\mathrm{a}+2)-1(\mathrm{a}+1)<0$$$${\mathrm{a}}^{\mathrm{k}}<\frac{\mathrm{a}+1}{\mathrm{a}+2}$$$$\mathrm{klog}\left(\mathrm{a}\right)<\log \left(\frac{\mathrm{a}+1}{\mathrm{a}+2}\right)$$$$\mathrm{k}<\log \left(\frac{\mathrm{a}+1}{\mathrm{a}+2}\right).\frac{1}{\log \left(\mathrm{a}\right)}$$$$\mathrm{k}<\log \left(\frac{\mathrm{p}+\mathrm{q}}{\mathrm{p}+2\mathrm{q}}\right).\frac{1}{\log \left(\frac{\mathrm{p}}{\mathrm{q}}\right)}$$

Commented by PRITHWISH SEN 2 last updated on 01/Oct/20

$$(\mathrm{n}+1)\mathrm{p}-(\mathrm{n}-1)\mathrm{p}+(\mathrm{n}-1)\mathrm{q}-(\mathrm{n}+1)\mathrm{q}$$$$=\mathrm{np}+\mathrm{p}-\mathrm{np}+\mathrm{p}+\mathbf{nq}-\mathbf{q}-\mathbf{np}-\mathbf{q}$$$$=2(\mathbf{p}-\mathbf{q})$$$$\therefore \frac{2\mathrm{nq}+2\mathrm{np}}{(\mathrm{n}+1)\mathrm{p}-(\mathrm{n}-1)\mathrm{p}+(\mathrm{n}-1)\mathrm{q}-(\mathrm{n}+1)\mathrm{q}}$$$$=\frac{2\mathrm{n}(\mathrm{p}+\mathrm{q})}{2(\mathrm{p}-\mathrm{q})}=\frac{\mathrm{n}(\mathrm{p}+\mathrm{q})}{(\mathrm{p}-\mathrm{q})}$$$$\mathbf{please}\mathbf{check}$$$$$$

Answered by PRITHWISH SEN 2 last updated on 27/Sep/20

$$\frac{\mathrm{n}(\mathrm{p}+\mathrm{q})+(\mathrm{p}-\mathrm{q})}{\mathrm{n}(\mathrm{p}+\mathrm{q})-(\mathrm{p}-\mathrm{q})}={\left(\frac{\mathrm{p}}{\mathrm{q}}\right)}^{\mathrm{k}}$$$$\because \mathrm{p}\to \mathrm{q}\Rightarrow (\mathrm{p}-\mathrm{q})\to 0$$$$\frac{\mathrm{n}(\mathrm{p}+\mathrm{q})}{\mathrm{n}(\mathrm{p}+\mathrm{q})}={\left(\frac{\mathrm{p}}{\mathrm{q}}\right)}^{\mathrm{k}}$$$$\mathrm{k}=0\mathrm{please}\mathrm{check}$$

Commented by ZiYangLee last updated on 27/Sep/20

$$Butpisjustapproximatelyequal$$$$toq,ithinkwecannotassumethat$$$$pisequaltoq......hmmm$$

Commented by PRITHWISH SEN 2 last updated on 27/Sep/20

$$\mathrm{that}\mathrm{is}\mathrm{why}\mathrm{I}\mathrm{have}\mathrm{written}$$$$\mathrm{p}-\mathrm{q}\to 0\mathrm{not}\mathrm{as}\mathrm{p}-\mathrm{q}=0$$

Commented by PRITHWISH SEN 2 last updated on 27/Sep/20

$$\mathrm{now}\mathrm{if}\mathrm{p}-\mathrm{q}=ϵϵ=\mathrm{a}\mathrm{very}\mathrm{small}\mathrm{positive}\mathrm{quantity}$$$$\mathrm{then}\mathrm{n}(\mathrm{p}+\mathrm{q})+ϵ\to \mathrm{n}(\mathrm{p}+\mathrm{q})-ϵ$$$$\therefore \frac{\mathrm{n}(\mathrm{p}+\mathrm{q})+ϵ}{\mathrm{n}(\mathrm{p}+\mathrm{q})-ϵ}\to 1\mathrm{I}\mathrm{think}$$

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