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Question Number 143627 Answers: 2 Comments: 0

Question Number 143635 Answers: 1 Comments: 0

Question Number 143633 Answers: 1 Comments: 0

Question Number 143624 Answers: 1 Comments: 0

∫_0 ^( ∞) z^2 e^(1/z) dz

$$\int_{\mathrm{0}} ^{\:\infty} {z}^{\mathrm{2}} {e}^{\frac{\mathrm{1}}{{z}}} {dz} \\ $$

Question Number 143622 Answers: 1 Comments: 0

∫_0 ^∝ e^(2arctg(t^2 )) dt

$$\int_{\mathrm{0}} ^{\propto} {e}^{\mathrm{2}{arctg}\left({t}^{\mathrm{2}} \right)} {dt} \\ $$

Question Number 143794 Answers: 0 Comments: 3

Prove 3^(111) +1⋮223

$$\mathrm{Prove}\:\:\:\:\:\:\:\:\:\mathrm{3}^{\mathrm{111}} +\mathrm{1}\vdots\mathrm{223} \\ $$

Question Number 143609 Answers: 1 Comments: 0

Prove that:: Ω:=∫_0 ^( 1) ((ln^2 (1−x).ln(x))/x)dx=((−1)/2) ζ (4 ) Without using the “Beta function” m.n

$$\:\:\: \\ $$$$\:{Prove}\:{that}:: \\ $$$$\:\:\Omega:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right).{ln}\left({x}\right)}{{x}}{dx}=\frac{−\mathrm{1}}{\mathrm{2}}\:\zeta\:\left(\mathrm{4}\:\right) \\ $$$${Without}\:{using}\:{the}\:``{Beta}\:{function}'' \\ $$$$\:\:{m}.{n} \\ $$

Question Number 143606 Answers: 1 Comments: 0

Question Number 143603 Answers: 2 Comments: 0

.....Calculus..... Ω:=Σ_(n=1) ^∞ (1/(n^k (1+n))) (k≥ 2) ......

$$\:\:\:\:\:\:\:\:\:\:\:\:\:.....{Calculus}..... \\ $$$$\:\:\:\:\:\:\:\:\Omega:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{k}} \left(\mathrm{1}+{n}\right)}\:\:\:\left({k}\geqslant\:\mathrm{2}\right)\:...... \\ $$

Question Number 143602 Answers: 2 Comments: 1

Question Number 143598 Answers: 3 Comments: 2

Question Number 143597 Answers: 0 Comments: 0

s = ut + (1/2)at^2 ⇒ t^2 + 2(u/a)t − 2(s/a) = 0 by the use of quadratic formula t = ((−((2u)/a) ± (√(((4u^2 )/a^2 ) + 4s)))/2) t = −(u/a) ± (√((u^2 /a^2 ) + s)) Victor Francis

$${s}\:=\:{ut}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:\Rightarrow\:\:{t}^{\mathrm{2}} \:+\:\mathrm{2}\frac{{u}}{{a}}{t}\:−\:\mathrm{2}\frac{{s}}{{a}}\:=\:\mathrm{0} \\ $$$${by}\:{the}\:{use}\:{of}\:{quadratic}\:{formula} \\ $$$${t}\:=\:\frac{−\frac{\mathrm{2}{u}}{{a}}\:\pm\:\sqrt{\frac{\mathrm{4}{u}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:\mathrm{4}{s}}}{\mathrm{2}} \\ $$$${t}\:=\:−\frac{{u}}{{a}}\:\:\pm\:\:\sqrt{\frac{{u}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:{s}} \\ $$$${Victor}\:\:{Francis} \\ $$$$ \\ $$$$ \\ $$

Question Number 143596 Answers: 1 Comments: 0

Given that the series (x/3^ )+(y/3^2 )+(x/3^3 )+(y/3^4 )+…+(x/3^(n−1) )+(y/3^n )=8, x,y∈R. Find the value of 3x+y.

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{series} \\ $$$$\frac{{x}}{\mathrm{3}^{} }+\frac{{y}}{\mathrm{3}^{\mathrm{2}} }+\frac{{x}}{\mathrm{3}^{\mathrm{3}} }+\frac{{y}}{\mathrm{3}^{\mathrm{4}} }+\ldots+\frac{{x}}{\mathrm{3}^{{n}−\mathrm{1}} }+\frac{{y}}{\mathrm{3}^{{n}} }=\mathrm{8},\:{x},{y}\in\mathbb{R}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{3}{x}+{y}. \\ $$

Question Number 143595 Answers: 2 Comments: 0

If f(x)=((10^x −10^(−x) )/(10^x +10^(−x) )), find f^( −1) (x).

$$\mathrm{If}\:{f}\left({x}\right)=\frac{\mathrm{10}^{{x}} −\mathrm{10}^{−{x}} }{\mathrm{10}^{{x}} +\mathrm{10}^{−{x}} },\:\mathrm{find}\:{f}^{\:−\mathrm{1}} \left({x}\right). \\ $$

Question Number 143593 Answers: 0 Comments: 0

soit (X_i ),i∈{1,2,......,n} une suite de variable aleartoire independante (iid) suivant la loi binomiale B(n,p) montrer que X_n ^_ converge en loi vers E(X) 2− montrer que (1/n)Σ_(i=1) ^2 X_(i ) ^2 converge en loi vers E (X^2 ) 3−montrer que (1/n)Σ_(i=1) ^n X_i Y_i converge en loi vers E(XY)

$${soit}\:\left(\boldsymbol{{X}}_{{i}} \right),{i}\in\left\{\mathrm{1},\mathrm{2},......,{n}\right\}\:{une}\:{suite}\:{de}\:{variable}\:{a}\boldsymbol{{leartoire}}\:\boldsymbol{{independante}} \\ $$$$\left(\boldsymbol{{iid}}\right)\:\boldsymbol{{suivant}}\:\boldsymbol{{la}}\:\boldsymbol{{loi}}\:\boldsymbol{{binomiale}}\:\boldsymbol{{B}}\left(\boldsymbol{{n}},\boldsymbol{{p}}\right)\: \\ $$$$\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\overset{\_} {\boldsymbol{{X}}}_{\boldsymbol{{n}}} \boldsymbol{{converge}}\:\boldsymbol{{en}}\:\boldsymbol{{loi}}\:\boldsymbol{{vers}}\:\boldsymbol{{E}}\left(\boldsymbol{{X}}\right) \\ $$$$\mathrm{2}−\:\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\frac{\mathrm{1}}{\boldsymbol{{n}}}\underset{{i}=\mathrm{1}} {\overset{\mathrm{2}} {\sum}}\boldsymbol{{X}}_{\boldsymbol{{i}}\:\:} ^{\mathrm{2}} \boldsymbol{{converge}}\:\boldsymbol{{en}}\:\boldsymbol{{loi}}\:\:\boldsymbol{{vers}}\:\:\:\:\:\:\:\:\boldsymbol{{E}}\:\left(\boldsymbol{{X}}^{\mathrm{2}} \right) \\ $$$$\mathrm{3}−\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\frac{\mathrm{1}}{\boldsymbol{{n}}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\boldsymbol{{X}}_{{i}} \boldsymbol{{Y}}_{{i}} \:\boldsymbol{{converge}}\:\boldsymbol{{en}}\:\:\boldsymbol{{loi}}\:\boldsymbol{{vers}}\:\boldsymbol{{E}}\left(\boldsymbol{{XY}}\right) \\ $$

Question Number 143591 Answers: 1 Comments: 0

a;b;c>0 and a+b+c=k min((1/(1+a^2 )) + (1/(1+b^2 )) + (1/(1+c^2 )))=?

$${a};{b};{c}>\mathrm{0}\:\:{and}\:\:{a}+{b}+{c}={k} \\ $$$$\boldsymbol{{min}}\left(\frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}+{b}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}+{c}^{\mathrm{2}} }\right)=? \\ $$

Question Number 143588 Answers: 3 Comments: 0

Question Number 143582 Answers: 0 Comments: 0

x^3 −x−c=0 let x=t+h (t−s){t^3 +3ht^2 +(3h^2 −1)t +(h^3 −h−c)}=0 let t=p+q (p+q−s){p^3 +q^3 +3pq(p+q) +3h(p^2 +q^2 )+6hpq +(3h^2 −1)(p+q)+(h^3 −h−c)} = 0 p^4 +q^4 +pq(p^2 +q^2 )+3pq(p+q)^2 +3h(p^2 +q^2 )(p+q)+6hpq(p+q) +(3h^2 −1)(p+q)^2 +(h^3 −h−c)(p+q) −s(p^3 +q^3 )−3pqs(p+q) −3hs(p^2 +q^2 )−6hspq +(3h^2 −1)s(p+q) −s(h^3 −h−c)=0 let p^4 +q^4 =p^2 +q^2 (p^2 +q^2 )^2 −(p^2 +q^2 )−2p^2 q^2 =0 ⇒ p^2 +q^2 =(1/2)±(√((1/4)+2p^2 q^2 )) say z=(1/2)+(√((1/4)+2m^2 )) (p+q)^2 =z+2m ⇒z(1+m+3m+3h^2 −1−3hs) +(z+2m)^(1/2) {3hz+6mh +h^3 −h−c−3ms −s(z−m)+(3h^2 −1)s} +6m^2 +2(3h^2 −1)m−6hsm −s(h^3 −h−c)=0 let (3h−s)z+m(6h−2s) +h^3 −h−c+3sh^2 −s=0 & 4m+3h(h−s)=0 ⇒ 2(3h−s)z−3h(h−s)(3h−s) +3sh^2 −s+h^3 −h−c=0 let s=−3h ⇒ 12hz−80h^3 −h−c=0 & m=−3h^2 ⇒ 54h^4 −6h^2 (3h^2 −1) −54h^4 +3h(h^3 −h−c)=0 ⇒ 15h^4 −3h^2 +3ch=0 And if h≠0 ⇒ 5h^3 −h+c=0 D=(c^2 /(100))−((1/(15)))^3 ⇒ D≥0 if c≥ (2/(3(√(15)))) m=−3h^2 ; s=−3h; And 12hz−80h^3 −h−c=0 ⇒ z=((20h^2 )/3)+(1/(12))+(c/(12h)) p+q=(√(z+2m)) =(√(((2h^2 )/3)+((c+h)/(12h)))) x=t+h=p+q+h x=h+(√(((2h^2 )/3)+((c+h)/(12h)))) but (c/(12h))>−(((2h^2 )/3)+(1/(12))) ⇒ c<−8h^3 −h ⇒ 8h^3 +h+c<0 5h^3 −h+c=0 ⇒ 13h^3 +2c<0 ......

Question Number 143578 Answers: 0 Comments: 2