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Question Number 68206 Answers: 1 Comments: 0

find S(θ)=Σ_(n=0) ^∞ ((sin^3 (nθ))/(n!))

$${find}\:{S}\left(\theta\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{sin}^{\mathrm{3}} \left({n}\theta\right)}{{n}!} \\ $$

Question Number 68203 Answers: 0 Comments: 2

Question Number 68191 Answers: 2 Comments: 0

Question Number 68189 Answers: 0 Comments: 4

Question Number 68188 Answers: 0 Comments: 2

Σ_(n=3) ^∝ 1/n(ln n)^2 is the function converg or diverg ? pleas help me

$$\sum_{{n}=\mathrm{3}} ^{\propto} \:\mathrm{1}/{n}\left({ln}\:{n}\right)^{\mathrm{2}} \:\:\:{is}\:{the}\:{function}\:{converg}\:{or}\:{diverg}\:?\:{pleas}\:{help}\:{me} \\ $$

Question Number 68183 Answers: 0 Comments: 1

Question Number 68178 Answers: 0 Comments: 2

lim_(x→0) (((cosx)^(sin2x) −1)/x^3 )=?

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\left({cosx}\right)^{{sin}\mathrm{2}{x}} −\mathrm{1}}{{x}^{\mathrm{3}} }=? \\ $$

Question Number 68171 Answers: 1 Comments: 2

Question Number 68161 Answers: 0 Comments: 7

In my textbook its written: In applying the nth−term test we can see that: Σ_(n=1) ^∞ (−1)^(n+1) diverges because lim_(n→∞) (−1)^(n+1) does not exist. But then why Σ_(n=1) ^∞ (−1)^(n+1) (1/n^2 ) , Σ_(n=1) ^∞ (−1)^(n+1) (1/(ln(n))) converges ?

$${In}\:{my}\:{textbook}\:{its}\:{written}: \\ $$$${In}\:{applying}\:{the}\:{nth}−{term}\:{test}\:{we}\: \\ $$$${can}\:{see}\:{that}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{diverges}\:{because}\: \\ $$$${lim}_{{n}\rightarrow\infty} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{does}\:{not}\:{exist}. \\ $$$${But}\:{then}\:{why}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:,\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{ln}\left({n}\right)} \\ $$$${converges}\:? \\ $$

Question Number 68151 Answers: 0 Comments: 1

If ∣a∣< 1 and ∣b∣< 1, then the sum of the series 1+(1+a)b+(1+a+a^2 )b^2 +(1+a+a^2 +a^3 )b^3 +... is

$$\mathrm{If}\:\mid{a}\mid<\:\mathrm{1}\:\mathrm{and}\:\mid{b}\mid<\:\mathrm{1},\:\mathrm{then}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series} \\ $$$$\mathrm{1}+\left(\mathrm{1}+{a}\right){b}+\left(\mathrm{1}+{a}+{a}^{\mathrm{2}} \right){b}^{\mathrm{2}} +\left(\mathrm{1}+{a}+{a}^{\mathrm{2}} +{a}^{\mathrm{3}} \right){b}^{\mathrm{3}} +... \\ $$$$\mathrm{is} \\ $$

Question Number 68150 Answers: 0 Comments: 0

Question Number 68149 Answers: 0 Comments: 2

Explicit f(a)=Σ_(n=1) ^∞ (((−1)^n )/(n(an+1)))

$$\:{Explicit}\:\:\:{f}\left({a}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left({an}+\mathrm{1}\right)}\:\:\:\: \\ $$

Question Number 68145 Answers: 0 Comments: 5

Find the arc length, given the curve x(t) = sin (πt), y(t) = t , 0 ≤ t ≤ 1

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{arc}\:\mathrm{length},\:\mathrm{given}\:\mathrm{the}\:\mathrm{curve} \\ $$$${x}\left({t}\right)\:=\:\mathrm{sin}\:\left(\pi{t}\right),\:\:{y}\left({t}\right)\:=\:{t}\:,\:\:\mathrm{0}\:\leqslant\:{t}\:\leqslant\:\mathrm{1} \\ $$

Question Number 68133 Answers: 2 Comments: 0

solve y′′=y′y

$${solve}\:{y}''={y}'{y} \\ $$

Question Number 68132 Answers: 0 Comments: 3

Question Number 68129 Answers: 0 Comments: 0

prove that πcotan(απ)=(1/α) +Σ_(n=1) ^∞ ((2α)/(α^2 −n^2 )) with α ∈R−Z . prove also that for t≠0 cotan(t) =(1/t) +Σ_(n=1) ^∞ ((2t)/(t^2 −n^2 π^2 ))

$${prove}\:{that}\:\pi{cotan}\left(\alpha\pi\right)=\frac{\mathrm{1}}{\alpha}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}\alpha}{\alpha^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$$${with}\:\alpha\:\in{R}−{Z}\:\:. \\ $$$${prove}\:{also}\:{that}\:\:\:{for}\:{t}\neq\mathrm{0} \\ $$$${cotan}\left({t}\right)\:=\frac{\mathrm{1}}{{t}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} −{n}^{\mathrm{2}} \pi^{\mathrm{2}} } \\ $$

Question Number 68141 Answers: 0 Comments: 1

the 2 formulas for solving ∫(dx/(x^3 +px+q)) with “nasty” solutions of x^3 +px+q=0 with p, q ∈R case 1 D=(p^3 /(27))+(q^2 /4)>0 ⇒ x^3 +px+q=0 has got 1 real and 2 conjugated complex solutions u=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3) ∧v=((−(q/2)−p(√((p^3 /(27))+(q^2 /4)))))^(1/3) x_1 =u+v x_2 =(−(1/2)+((√3)/2)i)u+(−(1/2)−((√3)/2)i)v x_3 =(−(1/2)−((√3)/2)i)u+(−(1/2)+((√3)/2)i)v α=u+v∧β=((√3)/2)(u−v) ⇔ u=(α/2)+(β/(√3))∧v=(α/2)−(β/(√3)) x_1 =α x_2 =−(α/2)+βi x_3 =−(α/2)−βi ∫(dx/(x^3 +px+q))=∫(dx/((x−α)(x^2 +αx+((α^2 +4β^2 )/4))))= =(1/(9α^2 +4β^2 ))(∫(dx/((x−α)))−∫((x+2α)/(x^2 +αx+((α^2 +4β^2 )/4)))dx)= =(1/(9α^2 +4β^2 ))(ln ∣x−α∣ −(1/2)ln ((2x+α)^2 +4β^2 ) −((3α)/(2β))arctan ((2x+α)/(2β))) +C ...now calculate the constants case 2 D=(p^3 /(27))+(q^2 /4)<0 ⇒ x^3 +px+q=0 has got 3 real solutions x_k =(2/3)(√(−3p)) sin (((2π)/3)k+(1/3)arcsin ((3(√3)q)/(2(√(−p^3 ))))) with k=0, 1, 2 let x_1 =α, x_2 =β, x_3 =γ ∫(dx/(x^3 +px+q))=∫(dx/((x−α)(x−β)(x−γ)))= =(1/((α−β)(α−γ)))∫(dx/(x−α))+(1/((β−α)(β−γ)))∫(dx/(x−β))+(1/((γ−α)(γ−β)))∫(dx/(x−γ))= =((ln ∣x−α∣)/((α−β)(α−γ)))+((ln ∣x−β∣)/((β−α)(β−γ)))+((ln ∣x−γ∣)/((γ−α)(γ−β)))+C ...now calculate the constants

$$\mathrm{the}\:\mathrm{2}\:\mathrm{formulas}\:\mathrm{for}\:\mathrm{solving}\:\int\frac{{dx}}{{x}^{\mathrm{3}} +{px}+{q}}\:\mathrm{with} \\ $$$$``\mathrm{nasty}''\:\mathrm{solutions}\:\mathrm{of}\:{x}^{\mathrm{3}} +{px}+{q}=\mathrm{0}\:\mathrm{with}\:{p},\:{q}\:\in\mathbb{R} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{1} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}>\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{3}} +{px}+{q}=\mathrm{0}\:\mathrm{has}\:\mathrm{got}\:\mathrm{1}\:\mathrm{real} \\ $$$$\mathrm{and}\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$${u}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}}\wedge{v}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}−{p}\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}} \\ $$$${x}_{\mathrm{1}} ={u}+{v} \\ $$$${x}_{\mathrm{2}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$${x}_{\mathrm{3}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$$\alpha={u}+{v}\wedge\beta=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({u}−{v}\right)\:\Leftrightarrow\:{u}=\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\sqrt{\mathrm{3}}}\wedge{v}=\frac{\alpha}{\mathrm{2}}−\frac{\beta}{\sqrt{\mathrm{3}}} \\ $$$${x}_{\mathrm{1}} =\alpha \\ $$$${x}_{\mathrm{2}} =−\frac{\alpha}{\mathrm{2}}+\beta\mathrm{i} \\ $$$${x}_{\mathrm{3}} =−\frac{\alpha}{\mathrm{2}}−\beta\mathrm{i} \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{3}} +{px}+{q}}=\int\frac{{dx}}{\left({x}−\alpha\right)\left({x}^{\mathrm{2}} +\alpha{x}+\frac{\alpha^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} }{\mathrm{4}}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}\alpha^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} }\left(\int\frac{{dx}}{\left({x}−\alpha\right)}−\int\frac{{x}+\mathrm{2}\alpha}{{x}^{\mathrm{2}} +\alpha{x}+\frac{\alpha^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} }{\mathrm{4}}}{dx}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}\alpha^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} }\left(\mathrm{ln}\:\mid{x}−\alpha\mid\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\left(\mathrm{2}{x}+\alpha\right)^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} \right)\:−\frac{\mathrm{3}\alpha}{\mathrm{2}\beta}\mathrm{arctan}\:\frac{\mathrm{2}{x}+\alpha}{\mathrm{2}\beta}\right)\:+{C} \\ $$$$...\mathrm{now}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{constants} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{2} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}<\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{3}} +{px}+{q}=\mathrm{0}\:\mathrm{has}\:\mathrm{got}\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions} \\ $$$${x}_{{k}} =\frac{\mathrm{2}}{\mathrm{3}}\sqrt{−\mathrm{3}{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}{k}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{3}\sqrt{\mathrm{3}}{q}}{\mathrm{2}\sqrt{−{p}^{\mathrm{3}} }}\right)\:\mathrm{with}\:{k}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$$\mathrm{let}\:{x}_{\mathrm{1}} =\alpha,\:{x}_{\mathrm{2}} =\beta,\:{x}_{\mathrm{3}} =\gamma \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{3}} +{px}+{q}}=\int\frac{{dx}}{\left({x}−\alpha\right)\left({x}−\beta\right)\left({x}−\gamma\right)}= \\ $$$$=\frac{\mathrm{1}}{\left(\alpha−\beta\right)\left(\alpha−\gamma\right)}\int\frac{{dx}}{{x}−\alpha}+\frac{\mathrm{1}}{\left(\beta−\alpha\right)\left(\beta−\gamma\right)}\int\frac{{dx}}{{x}−\beta}+\frac{\mathrm{1}}{\left(\gamma−\alpha\right)\left(\gamma−\beta\right)}\int\frac{{dx}}{{x}−\gamma}= \\ $$$$=\frac{\mathrm{ln}\:\mid{x}−\alpha\mid}{\left(\alpha−\beta\right)\left(\alpha−\gamma\right)}+\frac{\mathrm{ln}\:\mid{x}−\beta\mid}{\left(\beta−\alpha\right)\left(\beta−\gamma\right)}+\frac{\mathrm{ln}\:\mid{x}−\gamma\mid}{\left(\gamma−\alpha\right)\left(\gamma−\beta\right)}+{C} \\ $$$$...\mathrm{now}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{constants} \\ $$

Question Number 68122 Answers: 1 Comments: 1