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Operation ResearchQuestion and Answers: Page 2

Question Number 139152 Answers: 0 Comments: 0

wich program to run genetic algorithm?

$${wich}\:{program}\:{to}\:{run}\:\:{genetic}\:{algorithm}? \\ $$

Question Number 137575 Answers: 0 Comments: 0

Question Number 137574 Answers: 0 Comments: 0

Question Number 137347 Answers: 1 Comments: 0

(1/(1+(π^2 /(1+(((π+1)^2 )/(1+(((π+2)^2 )/(1+...))))))))+(1/(1+(((π+1)^2 )/(1+(((π+2)^2 )/(1+...))))))=(1/π) Prove or disprove

$$\frac{\mathrm{1}}{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}+...}}}}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\left(\pi+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}+...}}}=\frac{\mathrm{1}}{\pi}\:\: \\ $$$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{disprove}} \\ $$

Question Number 137059 Answers: 0 Comments: 0

((1+(1/2^2 )+(1/3^2 )+...)/4^2 )+((1+(1/2^3 )+(1/3^3 )+(1/4^3 )+..)/4^3 )+((1+(1/2^4 )+(1/3^4 )+...)/4^4 )+..=(3/4)log(2)−(π/8)

$$\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+...}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+..}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }+...}{\mathrm{4}^{\mathrm{4}} }+..=\frac{\mathrm{3}}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{8}} \\ $$

Question Number 136645 Answers: 0 Comments: 0

(1/(1+(η^2 /(1+(((η+1)^2 )/(1+(((η+2)^2 )/(1+(((η+3)^2 )/(1+..)) ))))))))=(1/2)ψ(((η+2)/2))−(1/2)ψ(((η+1)/2)) (η>0) Or K_(r=0) ^∞ (η+r)^2 =(2/(ψ((η/2)+1)−ψ(((η+1)/2))))

$$\frac{\mathrm{1}}{\mathrm{1}+\frac{\eta^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\eta+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\eta+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\eta+\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{1}+..}\:\:}}}}=\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{\eta+\mathrm{2}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{\eta+\mathrm{1}}{\mathrm{2}}\right)\:\:\left(\eta>\mathrm{0}\right) \\ $$$${Or}\:\:\underset{{r}=\mathrm{0}} {\overset{\infty} {\boldsymbol{\mathrm{K}}}}\left(\eta+{r}\right)^{\mathrm{2}} =\frac{\mathrm{2}}{\psi\left(\frac{\eta}{\mathrm{2}}+\mathrm{1}\right)−\psi\left(\frac{\eta+\mathrm{1}}{\mathrm{2}}\right)} \\ $$

Question Number 135647 Answers: 0 Comments: 0

...+(1/((1−2π^2 )^2 ))+(1/((1−π^2 )^2 ))+1+(1/((1+π^2 )^2 ))+(1/((1+2π^2 )^2 ))+...=((csc^2 ((1/π)))/π^2 )

$$...+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{2}\pi^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\pi^{\mathrm{2}} \right)^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{\left(\mathrm{1}+\pi^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{2}\pi^{\mathrm{2}} \right)^{\mathrm{2}} }+...=\frac{{csc}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\pi}\right)}{\pi^{\mathrm{2}} } \\ $$

Question Number 135127 Answers: 1 Comments: 0

∫_0 ^1 log^2 (Γ(x))dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\Gamma\left({x}\right)\right){dx} \\ $$

Question Number 130827 Answers: 1 Comments: 0

[ (x+1)^(2 ) D^2 +(x+1)D+1 ]y = 4cos (ln( x+1))

$$\left[\:\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}\:} \mathrm{D}^{\mathrm{2}} +\left(\mathrm{x}+\mathrm{1}\right)\mathrm{D}+\mathrm{1}\:\right]\mathrm{y}\:=\:\mathrm{4cos}\:\left(\mathrm{ln}\left(\:\mathrm{x}+\mathrm{1}\right)\right) \\ $$

Question Number 129131 Answers: 0 Comments: 5

Solve 134^(x+1) =16^x −768

$${Solve}\:\:\mathrm{134}^{{x}+\mathrm{1}} =\mathrm{16}^{{x}} −\mathrm{768} \\ $$

Question Number 129102 Answers: 1 Comments: 0

Question Number 124041 Answers: 1 Comments: 1

A_n =(1+(√5))^n −(1−(√5))^n A_n =????

$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\boldsymbol{{A}}_{{n}} =\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\boldsymbol{{n}}} −\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)^{\boldsymbol{{n}}} \\ $$$$ \\ $$$$\:\:\:\boldsymbol{{A}}_{\boldsymbol{{n}}} =???? \\ $$

Question Number 123826 Answers: 1 Comments: 1

W_n =Σ_(k=1) ^(2n) (k/(n^2 +k^2 )) montrer que W_n converge et calculer la valeur de W_n

$$ \\ $$$$\boldsymbol{{W}}_{\boldsymbol{{n}}} =\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\mathrm{2}\boldsymbol{{n}}} {\sum}}\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} } \\ $$$$\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\boldsymbol{{W}}_{\boldsymbol{{n}}} \:\boldsymbol{{converge}} \\ $$$$\boldsymbol{{et}}\:\boldsymbol{{calculer}}\:\boldsymbol{{la}}\:\boldsymbol{{valeur}}\:\boldsymbol{{de}}\:\boldsymbol{{W}}_{\boldsymbol{{n}}} \\ $$

Question Number 119150 Answers: 2 Comments: 0

Given that a, b and c are real numbers that stisfy the system of equation above a − (√(b^2 − (1/(16)) )) = (√(c^2 − (1/(16)) )) b − (√(c^2 − (1/(25)))) = (√(a^2 − (1/(25)) )) c − (√(a^2 − (1/(36)) )) = (√(b^2 − (1/(36)))) if a+ b + c = (x/( (√y))) where x, y are positive integers and y is square free, find the value of x + y !

$${Given}\:{that}\:{a},\:{b}\:{and}\:{c}\:{are}\:{real}\:{numbers}\: \\ $$$${that}\:{stisfy}\:{the}\:{system}\:{of}\:{equation}\:{above} \\ $$$$ \\ $$$${a}\:−\:\sqrt{{b}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{16}}\:}\:=\:\sqrt{{c}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{16}}\:}\: \\ $$$${b}\:−\:\sqrt{{c}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{25}}}\:=\:\sqrt{{a}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{25}}\:} \\ $$$${c}\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{36}}\:}\:=\:\sqrt{{b}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{36}}}\: \\ $$$$ \\ $$$${if}\:\:{a}+\:{b}\:+\:{c}\:=\:\frac{{x}}{\:\sqrt{{y}}}\:{where}\:{x},\:{y}\:{are}\:{positive}\:{integers} \\ $$$${and}\:{y}\:{is}\:{square}\:{free},\:{find}\:{the}\:{value}\:\:{of}\:{x}\:+\:{y}\:! \\ $$

Question Number 113901 Answers: 0 Comments: 4

when do I use ∣x∣ for (√x^2 )?

$${when}\:{do}\:{I}\:{use}\:\mid{x}\mid \\ $$$${for}\:\sqrt{{x}^{\mathrm{2}} }? \\ $$

Question Number 112813 Answers: 0 Comments: 2

A binary operation has the property a∗(b∗c) = (a∗b)•c and that a∗a=1 for all non−zero real numbers a,b and c. (′•′ here represent multiplication). The solution of the equation 2016∗(6∗x)=100 can be written as (p/q) where p and q are relatively prime positive integers. What is q−p?

$$\mathrm{A}\:\mathrm{binary}\:\mathrm{operation}\:\mathrm{has}\:\mathrm{the}\:\mathrm{property} \\ $$$$\mathrm{a}\ast\left(\mathrm{b}\ast\mathrm{c}\right)\:=\:\left(\mathrm{a}\ast\mathrm{b}\right)\bullet\mathrm{c}\:\mathrm{and}\:\mathrm{that}\:\mathrm{a}\ast\mathrm{a}=\mathrm{1}\:\mathrm{for} \\ $$$$\mathrm{all}\:\mathrm{non}−\mathrm{zero}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}. \\ $$$$\left('\bullet'\:\mathrm{here}\:\mathrm{represent}\:\mathrm{multiplication}\right). \\ $$$$\mathrm{The}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{2016}\ast\left(\mathrm{6}\ast\mathrm{x}\right)=\mathrm{100}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\:\frac{\mathrm{p}}{\mathrm{q}} \\ $$$$\mathrm{where}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime} \\ $$$$\mathrm{positive}\:\mathrm{integers}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{q}−\mathrm{p}? \\ $$

Question Number 111105 Answers: 1 Comments: 1

y′′+2y′+y=e^(−2x) +2x+3

$$\mathrm{y}''+\mathrm{2y}'+\mathrm{y}=\mathrm{e}^{−\mathrm{2x}} +\mathrm{2x}+\mathrm{3} \\ $$

Question Number 109888 Answers: 3 Comments: 0

2^(x+5) =(√8^x )

$$\mathrm{2}^{{x}+\mathrm{5}} =\sqrt{\mathrm{8}^{{x}} } \\ $$

Question Number 108476 Answers: 2 Comments: 0

S_n =Σ_(k=1 ) ^n k^2 (−1)^k C_n ^k =? please help

$$\boldsymbol{{S}}_{{n}} =\underset{{k}=\mathrm{1}\:} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \left(−\mathrm{1}\right)^{{k}} \boldsymbol{{C}}_{\boldsymbol{{n}}} ^{\boldsymbol{{k}}} =? \\ $$$$\:\:\boldsymbol{{ple}}{ase}\:{help} \\ $$

Question Number 108406 Answers: 1 Comments: 0

solve the differential equation y^′ −2e^x y=2e^x (√y)