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Category: Operation Research

Given-the-function-f-x-2-0-lt-x-lt-2-2-2-lt-x-lt-0-of-period-4-a-sketch-the-graph-of-y-f-x-for-6-lt-x-lt-6-b-Find-the-Fourier-coefficient-a-0-a-n-and-b-n-

Question Number 204469 by DeArtist last updated on 18/Feb/24 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)\:=\:\begin{cases}{\mathrm{2},\:\mathrm{0}<\:{x}\:<\mathrm{2}}\\{−\mathrm{2},\:−\mathrm{2}\:<{x}\:<\:\mathrm{0}}\end{cases} \\ $$$$\mathrm{of}\:\mathrm{period}\:\mathrm{4} \\ $$$$\left(\mathrm{a}\right)\:\:\mathrm{sketch}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of}\:{y}\:=\:{f}\left({x}\right)\:,\:\mathrm{for}\:−\mathrm{6}\:<\:{x}\:<\:\mathrm{6} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{Fourier}\:\mathrm{coefficient}\:{a}_{\mathrm{0}} ,\:{a}_{{n}} ,\:\mathrm{and}\:{b}_{{n}} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{write}\:\mathrm{down}\:\mathrm{the}\:\mathrm{Fourier}\:\mathrm{series}.\: \\ $$$$\left(\mathrm{d}\right)\:\mathrm{hence}\:\mathrm{show}\:\mathrm{that}\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}−\mathrm{1}}\:=\:\frac{\pi}{\mathrm{4}}…

x-1-2-D-2-x-1-D-1-y-4cos-ln-x-1-

Question Number 130827 by bemath last updated on 29/Jan/21 $$\left[\:\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}\:} \mathrm{D}^{\mathrm{2}} +\left(\mathrm{x}+\mathrm{1}\right)\mathrm{D}+\mathrm{1}\:\right]\mathrm{y}\:=\:\mathrm{4cos}\:\left(\mathrm{ln}\left(\:\mathrm{x}+\mathrm{1}\right)\right) \\ $$ Answered by EDWIN88 last updated on 29/Jan/21 $${let}\:\mathrm{ln}\:\left({x}+\mathrm{1}\right)={t}\:\Rightarrow{x}+\mathrm{1}\:=\:{e}^{{t}} \\ $$$$\begin{cases}{\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dt}}×\frac{{dt}}{{dx}}\:=\:\frac{\mathrm{1}}{{x}+\mathrm{1}}.\frac{{dy}}{{dt}}}\\{\frac{{d}^{\mathrm{2}} {y}}{{dx}}\:=\:\frac{{d}}{{dx}}\:\left[\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\frac{{dy}}{{dt}}\:\right]=\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}}…

Question-129102

Question Number 129102 by AbdulAzizabir last updated on 12/Jan/21 Answered by Olaf last updated on 12/Jan/21 $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \:=\:\left\{\mathrm{1},\:\mathrm{5},\:\mathrm{14},\:\mathrm{30},\:\mathrm{55},…\right\}\:=\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{20}}+\frac{\mathrm{1}}{\mathrm{56}}−\frac{\mathrm{1}}{\mathrm{120}}+\frac{\mathrm{1}}{\mathrm{220}}−…\right) \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{14}}−\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{55}}−…\right) \\…

A-particle-P-moves-on-the-curve-with-polar-equation-r-e-k-where-r-are-polar-coordinates-referred-to-a-fixed-pole-and-k-is-a-positive-constant-Given-that-the-radial-velocity-of-P-is-k-

Question Number 63298 by Rio Michael last updated on 02/Jul/19 $${A}\:{particle}\:{P},\:{moves}\:{on}\:{the}\:{curve}\:{with}\:{polar}\:{equation}\:\: \\ $$$${r}\:=\:{e}^{{k}\theta} \:,\:{where}\:\left({r},\theta\right)\:{are}\:{polar}\:{coordinates}\:{referred}\:{to}\:{a}\:{fixed} \\ $$$${pole}\:{and}\:{k}\:{is}\:{a}\:{positive}\:{constant}.\:{Given}\:{that}\:{the}\:{radial}\:{velocity} \\ $$$${of}\:{P}\:{is}\:\frac{{k}}{{r}}\:\:{show}\:{that}\:{the}\:{transverse}\:{acceleration}\:{of}\:{th}\:{particle} \\ $$$${is}\:{zero}. \\ $$$$ \\ $$ Commented…

W-n-k-1-2n-k-n-2-k-2-montrer-que-W-n-converge-et-calculer-la-valeur-de-W-n-

Question Number 123826 by pticantor last updated on 28/Nov/20 $$ \\ $$$$\boldsymbol{{W}}_{\boldsymbol{{n}}} =\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\mathrm{2}\boldsymbol{{n}}} {\sum}}\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} } \\ $$$$\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\boldsymbol{{W}}_{\boldsymbol{{n}}} \:\boldsymbol{{converge}} \\ $$$$\boldsymbol{{et}}\:\boldsymbol{{calculer}}\:\boldsymbol{{la}}\:\boldsymbol{{valeur}}\:\boldsymbol{{de}}\:\boldsymbol{{W}}_{\boldsymbol{{n}}} \\ $$ Commented…