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Question Number 220495    Answers: 0   Comments: 5

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please help me .The photos I upload becomes blurred HELP please

$${please}\:{help}\:{me}\:.{The}\:{photos}\:{I}\:{upload} \\ $$$${becomes}\:{blurred} \\ $$$${HELP}\:{please} \\ $$

Question Number 220007    Answers: 2   Comments: 1

Question Number 219988    Answers: 1   Comments: 1

let s>1 be a real number. for all continues function f:[0,1]→R such that ∫_( 0) ^( 1) f(x)=0, determind of the exist a positive constant K(s) statisfying: (∫_0 ^( 1) f(x)∙Li_s (x)dx)^2 ≥K(s)∫_( 0) ^( 1) (f(x))^2 ∙Li_(2s−1 ) where Li_s (x)=Σ_(k=1) ^∞ (x^k /k^s ) is the polylogarithm function. if such a constants exists, find the optimal value of K(s).

$$ \\ $$$$\:\:\:\:\mathrm{let}\:{s}>\mathrm{1}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}.\:\mathrm{for}\:\mathrm{all}\:\mathrm{continues}\:\mathrm{function}\:{f}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\mathbb{R} \\ $$$$\:\:\:\mathrm{such}\:\mathrm{that}\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} {f}\left({x}\right)=\mathrm{0},\:\mathrm{determind}\:\mathrm{of}\:\mathrm{the}\:\mathrm{exist}\:\mathrm{a} \\ $$$$\:\:\:\:\:\mathrm{positive}\:\mathrm{constant}\:{K}\left({s}\right)\:\mathrm{statisfying}: \\ $$$$\:\:\:\left(\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}\left({x}\right)\centerdot\mathrm{Li}_{{s}} \left({x}\right){dx}\right)^{\mathrm{2}} \geqslant{K}\left({s}\right)\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \left({f}\left({x}\right)\right)^{\mathrm{2}} \centerdot\mathrm{Li}_{\mathrm{2}{s}−\mathrm{1}\:} \\ $$$$\:\:\:\mathrm{where}\:\mathrm{Li}_{{s}} \left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{k}} }{{k}^{{s}} }\:\mathrm{is}\:\mathrm{the}\:\mathrm{polylogarithm}\:\mathrm{function}.\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\mathrm{if}\:\mathrm{such}\:\mathrm{a}\:\mathrm{constants}\:\mathrm{exists},\:\mathrm{find}\:\mathrm{the}\:\mathrm{optimal}\:\mathrm{value}\:\mathrm{of}\:{K}\left({s}\right).\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 219724    Answers: 1   Comments: 0

prove; Π_(n=1) ^∞ (((2n+1)^3 −3(2n+1)+2)/((2n+1)^3 )) = (π/6)

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}; \\ $$$$\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\:\frac{\pi}{\mathrm{6}} \\ $$$$ \\ $$

Question Number 219676    Answers: 0   Comments: 0

The latex converter is not converting some symbols. Any reason why?

$$\mathrm{The}\:\mathrm{latex}\:\mathrm{converter}\:\mathrm{is}\:\mathrm{not}\:\mathrm{converting}\:\mathrm{some}\:\mathrm{symbols}.\:\mathrm{Any}\:\mathrm{reason}\:\mathrm{why}?\: \\ $$

Question Number 219624    Answers: 1   Comments: 0

Question Number 219606    Answers: 2   Comments: 0

prove that for positive real numbers a,b,c, the following inequality holds; (a^2 /(b + c)) + (b^2 /(c + a)) + (c^2 /(a + b)) ≥ ((a + b + c)/2)

$$ \\ $$$$\:\mathrm{prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:{a},{b},{c},\:\:\: \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}; \\ $$$$\:\:\frac{{a}^{\mathrm{2}} }{{b}\:+\:{c}}\:+\:\frac{{b}^{\mathrm{2}} }{{c}\:+\:{a}}\:+\:\frac{{c}^{\mathrm{2}} }{{a}\:+\:{b}}\:\:\geqslant\:\frac{{a}\:+\:{b}\:+\:{c}}{\mathrm{2}} \\ $$$$ \\ $$

Question Number 219589    Answers: 0   Comments: 2

Evaluate; L(tan^(−1) (t−(1/t))) solution; ⇒F(s)= L(tan^(−1) (t−(1/t))) ⇔ sF(s)+(π/2)=L(((t^2 +1)/(t^4 −t^2 +1)))(s) ⇒ sF(s)+(π/2)=L(((1/2)/(t^2 −(√3) t +1))+((1/2)/(t^2 −(√3) t+1)))(s) ⇒ sF(s)+(π/2)=(1/2)L((1/((t−((√3)/2))^2 +(1/4))))(s)+(1/2)L((1/((t+((√3)/2))^2 +(1/4))))(s) ⇒ sF(s)+(π/2)=(1/2)e^(−((√3)/2) s) L((1/(t^2 +((1/2))^2 )))(s)+(1/2)e^(((√3)/2) s) L((1/(t^2 +((1/2))^2 )))(s) ⇒ sF(s)+(π/2)=(1/2)e^(−((√3)/2) s) (1/(1/2)) sin ((1/2) s)∗(1/s)+(1/2)e^(((√3)/2) s) (1/(1/2))sin((1/2) s)∗(1/s) ⇒ sF(s)+(π/2)=e^(−((√3)/2) s) L(sin((1/2) t))(s)∗(1/s)+e^(((√3)/2) s) L(sin((1/2) t))(s)∗(1/s) ⇒ sF(s)+(π/2)=e^(−((√3)/2) s) ((1/2)/(s^2 +1/4))∗(1/s)+e^(((√3)/(2 ))s) ((1/2)/(s^2 +1/4))∗(1/s) Final Answer; F(s)=(1/s)(e^(−((√3)/2)s) ((1/2)/(s^2 +1/4))∗(1/s)+e^(((√3)/2)s) ((1/2)/(s^2 +1/4))∗(1/s))−(π/(2s))

$${Evaluate};\:\mathscr{L}\left({tan}^{−\mathrm{1}} \left({t}−\frac{\mathrm{1}}{{t}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{solution}; \\ $$$$\:\Rightarrow{F}\left({s}\right)=\:\mathscr{L}\left({tan}^{−\mathrm{1}} \left({t}−\frac{\mathrm{1}}{{t}}\right)\right) \\ $$$$\Leftrightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\mathscr{L}\left(\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\mathrm{1}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\mathscr{L}\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:{t}\:+\mathrm{1}}+\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:{t}+\mathrm{1}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\mathscr{L}\left(\frac{\mathrm{1}}{\left({t}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\right)\left({s}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathscr{L}\left(\frac{\mathrm{1}}{\left({t}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)\left({s}\right)+\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:{s}\right)\ast\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{s}\right)\ast\frac{\mathrm{1}}{{s}} \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}={e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{t}\right)\right)\left({s}\right)\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{t}\right)\right)\left({s}\right)\ast\frac{\mathrm{1}}{{s}} \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}={e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}\:\:\:\:\: \\ $$$$\mathrm{Final}\:\mathrm{Answer}; \\ $$$$\:\:\:{F}\left({s}\right)=\frac{\mathrm{1}}{{s}}\left({e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}\right)−\frac{\pi}{\mathrm{2}{s}}\:\: \\ $$$$ \\ $$

Question Number 219587    Answers: 1   Comments: 1

Question Number 219521    Answers: 0   Comments: 1

Question Number 219520    Answers: 1   Comments: 0

find the laplace transform of f(t)=∫_0 ^t ((sint)/t)dt

$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{laplace}}\:\boldsymbol{{transform}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{t}}\right)=\int_{\mathrm{0}} ^{\boldsymbol{{t}}} \frac{\boldsymbol{{sint}}}{\boldsymbol{{t}}}\boldsymbol{{dt}} \\ $$

Question Number 219519    Answers: 1   Comments: 0

find the laplace transform of ∫_0 ^∞ te^(−2t) sintdt

$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{laplace}}\:\boldsymbol{{transform}}\:\boldsymbol{{of}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \boldsymbol{{te}}^{−\mathrm{2}\boldsymbol{{t}}} \boldsymbol{{sintdt}} \\ $$

Question Number 219491    Answers: 1   Comments: 0

if x′′−2x′+10x=e^(2t) , at t=0,x=0 and x′=1 find x(t) using laplace transform

$$\boldsymbol{{if}}\:\boldsymbol{{x}}''−\mathrm{2}\boldsymbol{{x}}'+\mathrm{10}\boldsymbol{{x}}=\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{t}}} ,\:\boldsymbol{{at}}\:\boldsymbol{{t}}=\mathrm{0},\boldsymbol{{x}}=\mathrm{0}\:\boldsymbol{{and}}\:\boldsymbol{{x}}'=\mathrm{1} \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)\:\boldsymbol{{using}}\:\boldsymbol{{laplace}}\:\boldsymbol{{transform}} \\ $$

Question Number 219488    Answers: 1   Comments: 0

solve the initial value problem y′−2e^(−t^2 ) +2ty=0 y(0)=1

$$\boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{initial}}\:\boldsymbol{{value}}\:\boldsymbol{{problem}}\: \\ $$$$\boldsymbol{{y}}'−\mathrm{2}\boldsymbol{{e}}^{−\boldsymbol{{t}}^{\mathrm{2}} } +\mathrm{2}\boldsymbol{{ty}}=\mathrm{0}\:\:\boldsymbol{{y}}\left(\mathrm{0}\right)=\mathrm{1} \\ $$

Question Number 219427    Answers: 0   Comments: 4

Question Number 219373    Answers: 0   Comments: 0

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Question Number 219370    Answers: 0   Comments: 0

Question Number 219289    Answers: 2   Comments: 1

if a,b,c ∈ Z , and a^2 + b^2 = c^2 , then 3∣(ab) = ?

$$ \\ $$$$\:\:\:\:{if}\:\:\:\:\:\:\:\:\:\:\:{a},{b},{c}\:\in\:\mathbb{Z}\:\:\:, \\ $$$$\:\:\:\:\:{and}\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:=\:\:{c}^{\mathrm{2}} \:\:\:, \\ $$$$\:\:\:\:\:\:\:\:{then}\:\:\:\:\:\:\mathrm{3}\mid\left({ab}\right)\:=\:? \\ $$$$ \\ $$

Question Number 219211    Answers: 2   Comments: 0

Question Number 218891    Answers: 0   Comments: 0

suppose y(x) = Σ_(n=0) ^∞ a_n x^n statisfies y′′y−(y′)^2 =e^y −1with y(0)=0 and y′(0)=1. determin a_4 .

$$ \\ $$$$\:\boldsymbol{{suppose}}\:\boldsymbol{{y}}\left(\boldsymbol{{x}}\right)\:=\:\:\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\boldsymbol{{a}}_{\boldsymbol{{n}}} \boldsymbol{{x}}^{\boldsymbol{{n}}} \boldsymbol{{statisfies}}\: \\ $$$$\:\:\:\boldsymbol{{y}}''\boldsymbol{{y}}−\left(\boldsymbol{{y}}'\right)^{\mathrm{2}} =\boldsymbol{{e}}^{\boldsymbol{{y}}} −\mathrm{1}\boldsymbol{{with}}\:\boldsymbol{{y}}\left(\mathrm{0}\right)=\mathrm{0}\:\boldsymbol{{and}}\:\boldsymbol{{y}}'\left(\mathrm{0}\right)=\mathrm{1}.\:\:\:\:\: \\ $$$$\:\:\:\:\boldsymbol{{determin}}\:\boldsymbol{{a}}_{\mathrm{4}} . \\ $$$$ \\ $$

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