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Question-118791

Question Number 118791 by Algoritm last updated on 19/Oct/20 Answered by Olaf last updated on 19/Oct/20 $${f}_{{p}} \left({a}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{{p}} {\sum}}\frac{\mathrm{sin}\left(\mathrm{2}{n}\ast{a}\right)}{\mathrm{2}^{{n}} } \\ $$$${f}_{{p}} \left({a}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{{p}}…

lim-x-0-x-sinx-x-3-lim-x-0-x-x-x-3-3-x-3-lim-x-0-1-3-1-6-

Question Number 118776 by obaidullah last updated on 19/Oct/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{sinx}}{{x}^{\mathrm{3}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}}{{x}^{\mathrm{3}} } \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{3}!}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$ Terms of Service Privacy Policy…

Question-118781

Question Number 118781 by Algoritm last updated on 19/Oct/20 Commented by MJS_new last updated on 19/Oct/20 $$\mathrm{minimum}\:\mathrm{of}\:{f}\left({x}\right)={x}^{{x}^{{x}^{{x}} } } \:\mathrm{is}\:\approx.\mathrm{59} \\ $$$$\mathrm{but}\:\frac{\mathrm{1}}{\mathrm{3}^{\sqrt{\mathrm{48}}} }\approx.\mathrm{00049} \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{solution}…

Question-118768

Question Number 118768 by mathdave last updated on 19/Oct/20 Answered by mindispower last updated on 23/Oct/20 $$\underset{{m}=\mathrm{0}} {\overset{{n}} {\sum}}{e}^{{imx}} =\frac{\mathrm{1}−\left({e}^{{ix}} \right)^{{n}+\mathrm{1}} }{\mathrm{1}−{e}^{{ix}} }=\frac{{e}^{{i}\frac{{nx}}{\mathrm{2}}} \left({e}^{−{i}\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}} −{e}^{{i}\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}}…

The-first-three-terms-in-the-binomial-expansion-p-q-m-in-ascending-order-of-q-are-denoted-by-a-b-and-c-respectively-Show-that-b-2-ac-2m-m-1-

Question Number 118759 by ZiYangLee last updated on 19/Oct/20 $$\mathrm{The}\:\mathrm{first}\:\mathrm{three}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{binomial}\:\mathrm{expansion} \\ $$$$\left({p}−{q}\right)^{{m}} \:,\:\mathrm{in}\:\mathrm{ascending}\:\mathrm{order}\:\mathrm{of}\:{q},\:\mathrm{are}\:\mathrm{denoted} \\ $$$$\mathrm{by}\:{a},{b}\:\mathrm{and}\:{c}\:\mathrm{respectively}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:\frac{{b}^{\mathrm{2}} }{{ac}}=\frac{\mathrm{2}{m}}{{m}−\mathrm{1}} \\ $$ Commented by PRITHWISH SEN 2…

Question-53205

Question Number 53205 by mondodotto@gmail.com last updated on 19/Jan/19 Commented by MJS last updated on 22/Jan/19 $$\mathrm{2}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$$${t}_{\mathrm{1}} \left({x}\right)=\frac{\mathrm{25}^{{x}} }{\mathrm{5}^{{x}} }=\left(\frac{\mathrm{25}}{\mathrm{5}}\right)^{{x}} =\mathrm{5}^{{x}} ;\:{t}_{\mathrm{1}} \left(\mathrm{2}\right)=\mathrm{25}…

Question-118733

Question Number 118733 by mohammad17 last updated on 19/Oct/20 Commented by bemath last updated on 19/Oct/20 $$\left(\mathrm{4}\right)\:\frac{{d}}{{dx}}\:\left[\:\int\:_{{x}^{\mathrm{3}} } ^{\mathrm{2}{x}^{\mathrm{2}} } \:\mathrm{cos}\:\left(\mathrm{2}{t}^{\mathrm{3}} +\mathrm{1}\right)\:{dt}\:\right]\:= \\ $$$$\mathrm{4}{x}\:\mathrm{cos}\:\left(\mathrm{2}.\left(\mathrm{8}{x}^{\mathrm{6}} \right)+\mathrm{1}\right)−\mathrm{3}{x}^{\mathrm{2}}…