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Author: Tinku Tara

lets-two-polynimies-p-n-q-n-givwn-by-p-1-q-1-x-p-n-1-p-n-q-n-q-n-1-p-n-q-n-then-1-1-1-2-2-3-lets-W-u-v-determinant-u-v-u-v-is-true-that-W-p-n-q-n-0-n-gt-1-

Question Number 1844 by 123456 last updated on 12/Oct/15 $$\mathrm{lets}\:\mathrm{two}\:\mathrm{polynimies}\:{p}_{{n}} ,{q}_{{n}} \:\mathrm{givwn}\:\mathrm{by} \\ $$$${p}_{\mathrm{1}} ={q}_{\mathrm{1}} ={x} \\ $$$${p}_{{n}+\mathrm{1}} ={p}_{{n}} +{q}_{{n}} \\ $$$${q}_{{n}+\mathrm{1}} ={p}_{{n}} {q}_{{n}} \\…

f-x-y-f-x-y-xy-f-x-y-x-2-y-2-f-x-y-y-x-100-y-100-f-0-0-f-1-4-

Question Number 1842 by 123456 last updated on 12/Oct/15 $${f}\left({x},{y}\right)={f}\left({x}+{y},{xy}\right) \\ $$$${f}\left({x},{y}\right)={x},−\mathrm{2}\leqslant{y}\leqslant\mathrm{2} \\ $$$${f}\left({x},{y}\right)={y},\mid{x}\mid\geqslant\mathrm{100}\vee\mid{y}\mid\geqslant\mathrm{100} \\ $$$${f}\left(\mathrm{0},\mathrm{0}\right)=? \\ $$$${f}\left(\mathrm{1},\mathrm{4}\right)=? \\ $$ Commented by 123456 last updated…

find-1-1-x-2-arctan-1-1-x-dx-

Question Number 67374 by mathmax by abdo last updated on 26/Aug/19 $${find}\:\int\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctan}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){dx} \\ $$ Commented by Abdo msup. last updated on 27/Aug/19 $${let}\:{I}\:=\int\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arcran}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){dx}\:{by}\:{psrts}…

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Question Number 1838 by 112358 last updated on 11/Oct/15 $${Show}\:{that} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{sinz}}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{{z}+{n}\pi}+\frac{\mathrm{1}}{{z}−{n}\pi}\right) \\ $$$$\left(\mathrm{2}\right)\:{cotz}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{z}+{n}\pi}+\frac{\mathrm{1}}{{z}−{n}\pi}\right) \\ $$$${where}\:{z}\neq{m}\pi,\:{m}\in\mathbb{Z}\:,\:{given}\:{the}\:{fact}\:{that} \\ $$$${cosax}=\frac{\mathrm{2}{sina}\pi}{\pi}\left[\frac{\mathrm{1}}{\mathrm{2}{a}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}}…