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Question Number 220081 Answers: 2 Comments: 3

Question Number 220074 Answers: 0 Comments: 0

Question Number 220072 Answers: 1 Comments: 0

If x,y∈(0,(π/2)) Then prove that: log_(sinx) ^2 (((sin2x)/(sinx + cosx))) + log_(cosx) ^2 (((sin2x)/(sinx + cosx))) ≥ 2

$$\mathrm{If}\:\:\:\mathrm{x},\mathrm{y}\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{log}_{\boldsymbol{\mathrm{sinx}}} ^{\mathrm{2}} \:\left(\frac{\mathrm{sin2x}}{\mathrm{sinx}\:+\:\mathrm{cosx}}\right)\:+\:\mathrm{log}_{\boldsymbol{\mathrm{cosx}}} ^{\mathrm{2}} \:\left(\frac{\mathrm{sin2x}}{\mathrm{sinx}\:+\:\mathrm{cosx}}\right)\:\geqslant\:\mathrm{2} \\ $$

Question Number 220069 Answers: 1 Comments: 0

Let be (H_n )_(n≥1) H_n = Σ_(k=1) ^n (1/k) Find: lim_(n→∞) e^(2H_n ) ((((n+1)!))^(1/(n+1)) − ((n!))^(1/n) ) sin (π/n^2 ) = ?

Question Number 220066 Answers: 0 Comments: 0

evaluate −((csc(πs))/(iπ))∫_( C) (−t)^(s−1) e^(−t) dt , path C;(−∞,∞) −((𝚪(1−s))/(2πi)) ∫_( C) (((−t)^(s−1) )/(e^t −1)) dt , path C;(−∞,∞)

$$\mathrm{evaluate} \\ $$$$−\frac{\mathrm{csc}\left(\pi{s}\right)}{\boldsymbol{{i}}\pi}\int_{\:\boldsymbol{\mathcal{C}}} \:\left(−{t}\right)^{{s}−\mathrm{1}} {e}^{−{t}} \:\mathrm{d}{t}\:,\:\mathrm{path}\:\boldsymbol{\mathcal{C}};\left(−\infty,\infty\right) \\ $$$$−\frac{\boldsymbol{\Gamma}\left(\mathrm{1}−{s}\right)}{\mathrm{2}\pi\boldsymbol{{i}}}\:\int_{\:\boldsymbol{\mathcal{C}}} \:\frac{\left(−{t}\right)^{{s}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}\:\mathrm{d}{t}\:,\:\mathrm{path}\:\boldsymbol{\mathcal{C}};\left(−\infty,\infty\right) \\ $$

Question Number 220065 Answers: 2 Comments: 0

α∈R ; ω∈R^+ I(α) = ∫_(−∞) ^( ∞) ((x^2 ln(1+x^2 ))/((x^4 +1)^α )) e^(−x^2 ) cos(ωx) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha\in\mathbb{R}\:\:\:;\:\:\:\:\omega\in\mathbb{R}^{+} \\ $$$$\:\:\:\:\:{I}\left(\alpha\right)\:=\:\int_{−\infty} ^{\:\infty} \:\frac{{x}^{\mathrm{2}} \:\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\alpha} }\:{e}^{−{x}^{\mathrm{2}} } \:\mathrm{cos}\left(\omega{x}\right)\:{dx} \\ $$$$ \\ $$

Question Number 220040 Answers: 1 Comments: 0

∫^( ∞) _0 (1/( (√(x^8 + 1)))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{0}} {\int}^{\:\infty} \:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{8}} \:+\:\mathrm{1}}}\:{dx} \\ $$$$ \\ $$

Question Number 220038 Answers: 1 Comments: 0

∫^( 1) _( 0) (1/( (√(x^8 + 1)))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{8}} \:+\:\mathrm{1}}}\:{dx} \\ $$$$\: \\ $$

Question Number 220037 Answers: 1 Comments: 0

∫ (1/( (√(x^8 + 1)))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{8}} \:+\:\mathrm{1}}}\:{dx} \\ $$$$ \\ $$

Question Number 220034 Answers: 5 Comments: 0

Question Number 220022 Answers: 1 Comments: 0

If 0 ≤ a ≤ b ≤ 1 Then prove that: ∫_a ^( b) ∫_a ^( b) ∫_a ^( b) ((dxdydz)/( (√(1 + xyz)))) ≥ (b−a)^2 ∫_a ^( b) (dx/( (√(1 + x^3 ))))

Question Number 220020 Answers: 2 Comments: 0

If f:[a,b]→[−1,∞) a,b∈R a ≤ b f-continuous Then prove that: (∫_a ^( b) (1+f(x))dx)^3 ≥ (b−a)^3 + 3(b−a)^2 ∫_a ^( b) f(x)dx

$$\mathrm{If}\:\:\:\mathrm{f}:\left[\mathrm{a},\mathrm{b}\right]\rightarrow\left[−\mathrm{1},\infty\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{a},\mathrm{b}\in\mathbb{R} \\ $$$$\:\:\:\:\:\:\:\mathrm{a}\:\leqslant\:\mathrm{b} \\ $$$$\:\:\:\:\:\:\:\mathrm{f}-\mathrm{continuous} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\left(\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\left(\mathrm{1}+\mathrm{f}\left(\mathrm{x}\right)\right)\mathrm{dx}\right)^{\mathrm{3}} \geqslant\:\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{3}} +\:\mathrm{3}\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} \:\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$

Question Number 220019 Answers: 0 Comments: 1

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

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

Question Number 220016 Answers: 1 Comments: 0

Question Number 220015 Answers: 1 Comments: 0

∫_0 ^( ∞) ∣∣J_ν (r)∣∣e^(−rt) dr=Σ_(h=1) ^∞ ∫_z_h ^( z_(h+1) ) J_ν (r)e^(−rt) dr z_j is point of J_ν (z)=0 , z_1 =0 Σ_(h=1) ^∞ [F(r,t)]_(r=z_h ) ^(r=z_(h+1) ) i can′t solve anymore

$$\int_{\mathrm{0}} ^{\:\infty} \:\mid\mid{J}_{\nu} \left({r}\right)\mid\mid{e}^{−{rt}} \:\mathrm{d}{r}=\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\int_{{z}_{{h}} } ^{\:{z}_{{h}+\mathrm{1}} } \:{J}_{\nu} \left({r}\right){e}^{−{rt}} \mathrm{d}{r} \\ $$$${z}_{{j}} \:\mathrm{is}\:\mathrm{point}\:\mathrm{of}\:\:{J}_{\nu} \left({z}\right)=\mathrm{0}\:,\:{z}_{\mathrm{1}} =\mathrm{0} \\ $$$$\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left[{F}\left({r},{t}\right)\right]_{{r}={z}_{{h}} } ^{{r}={z}_{{h}+\mathrm{1}} } \: \\ $$$$\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{anymore} \\ $$

Question Number 220007 Answers: 2 Comments: 1

Question Number 219998 Answers: 1 Comments: 0

Question Number 219996 Answers: 1 Comments: 0

Solve Equation ((dx(t))/dt)=2x(t)+y(t) ((dy(t))/dt)=−3y(t) (((x^((1)) (t))),((y^((1)) (t))) )= ((2,( 1)),(0,(−3)) ) (((x(t))),((y(t))) ) A= ((2,( 1)),(0,(−3)) ) det{A−𝛌E}=0 det{ ((2,( 1)),(0,(−3)) )− ((𝛌,0),(0,𝛌) )}=0 𝛌=2,−3 v_1 = (((−1)),(( 5)) ) v_2 = ((1),(0) ) ((x),(y) )=C_1 e^(λt) v_1 +C_2 e^(λt) v_2 = (((−1),1),(( 5),0) ) (((C_1 e^(2t) )),((C_2 e^(−3t) )) ) ((x),(y) )= (((−C_1 e^(2t) +C_2 e^(−3t) )),(( 5C_1 e^(2t) )) ) ((dx(t))/dt)=−2C_1 e^(2t) −3C_2 e^(−3t) ≠ 2x(t)+y(t)=−2C_1 e^(2t) −2C_2 e^(−3t) +5C_1 e^(2t) ((dy(t))/dt)=10e^(2t) ≠ −3y(t)=−15C_2 e^(2t) Wrong..... Help

$$\mathrm{Solve}\:\mathrm{Equation} \\ $$$$\frac{\mathrm{d}{x}\left({t}\right)}{\mathrm{d}{t}}=\mathrm{2}{x}\left({t}\right)+{y}\left({t}\right) \\ $$$$\frac{\mathrm{d}{y}\left({t}\right)}{\mathrm{d}{t}}=−\mathrm{3}{y}\left({t}\right) \\ $$$$\begin{pmatrix}{{x}^{\left(\mathrm{1}\right)} \left({t}\right)}\\{{y}^{\left(\mathrm{1}\right)} \left({t}\right)}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}\end{pmatrix}\begin{pmatrix}{{x}\left({t}\right)}\\{{y}\left({t}\right)}\end{pmatrix} \\ $$$$\mathrm{A}=\begin{pmatrix}{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}\end{pmatrix} \\ $$$$\mathrm{det}\left\{\mathrm{A}−\boldsymbol{\lambda}\mathrm{E}\right\}=\mathrm{0} \\ $$$$\mathrm{det}\left\{\begin{pmatrix}{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}\end{pmatrix}−\begin{pmatrix}{\boldsymbol{\lambda}}&{\mathrm{0}}\\{\mathrm{0}}&{\boldsymbol{\lambda}}\end{pmatrix}\right\}=\mathrm{0} \\ $$$$\boldsymbol{\lambda}=\mathrm{2},−\mathrm{3} \\ $$$$\boldsymbol{\mathrm{v}}_{\mathrm{1}} =\begin{pmatrix}{−\mathrm{1}}\\{\:\:\:\:\mathrm{5}}\end{pmatrix} \\ $$$$\boldsymbol{\mathrm{v}}_{\mathrm{2}} =\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}={C}_{\mathrm{1}} {e}^{\lambda{t}} \boldsymbol{\mathrm{v}}_{\mathrm{1}} +{C}_{\mathrm{2}} {e}^{\lambda{t}} \boldsymbol{\mathrm{v}}_{\mathrm{2}} =\begin{pmatrix}{−\mathrm{1}}&{\mathrm{1}}\\{\:\:\:\:\mathrm{5}}&{\mathrm{0}}\end{pmatrix}\begin{pmatrix}{{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} }\\{{C}_{\mathrm{2}} {e}^{−\mathrm{3}{t}} }\end{pmatrix} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{−{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} +{C}_{\mathrm{2}} {e}^{−\mathrm{3}{t}} }\\{\:\:\:\:\:\:\:\mathrm{5C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} }\end{pmatrix} \\ $$$$\frac{\mathrm{d}{x}\left({t}\right)}{\mathrm{d}{t}}=−\mathrm{2}{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} −\mathrm{3}{C}_{\mathrm{2}} {e}^{−\mathrm{3}{t}} \neq \\ $$$$\mathrm{2}{x}\left({t}\right)+{y}\left({t}\right)=−\mathrm{2}{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} −\mathrm{2}{C}_{\mathrm{2}} {e}^{−\mathrm{3}{t}} +\mathrm{5}{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} \\ $$$$\frac{\mathrm{d}{y}\left({t}\right)}{\mathrm{d}{t}}=\mathrm{10}{e}^{\mathrm{2}{t}} \neq \\ $$$$−\mathrm{3}{y}\left({t}\right)=−\mathrm{15}{C}_{\mathrm{2}} {e}^{\mathrm{2}{t}} \\ $$$$\:\mathrm{Wrong}.....\:\mathrm{Help} \\ $$

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 219970 Answers: 2 Comments: 0

∫ e^x^2 dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:{e}^{{x}^{\mathrm{2}} } \:{dx} \\ $$$$ \\ $$

Question Number 219957 Answers: 2 Comments: 0

if f:(0,∞)→(0,∞) f continuous and f((1/x)) + f((1/y)) = 2f((1/(x+y))) ∀ x,y > 0 then ∀ a,b > 0: ∫_a ^( b) ∫_a ^( b) ∫_a ^( b) f((1/(x+y+z)))dxdydz = (b−a)^2 ∫_a ^( b) f((1/x))dx

$$\mathrm{if}\:\:\:\mathrm{f}:\left(\mathrm{0},\infty\right)\rightarrow\left(\mathrm{0},\infty\right) \\ $$$$\:\:\:\:\:\:\mathrm{f}\:\:\mathrm{continuous} \\ $$$$\mathrm{and}\:\:\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:+\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{y}}\right)\:=\:\mathrm{2f}\left(\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}\right) \\ $$$$\forall\:\mathrm{x},\mathrm{y}\:>\:\mathrm{0}\:\:\:\mathrm{then}\:\:\:\forall\:\mathrm{a},\mathrm{b}\:>\:\mathrm{0}: \\ $$$$\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}+\mathrm{z}}\right)\mathrm{dxdydz}\:=\:\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{dx} \\ $$

Question Number 219956 Answers: 1 Comments: 0

let a,b,c,d > 1 f : [a , b] → [c , d] a continuous function for which ∃λ ∈ (a , b) such that a ∫_a ^( 𝛌) f(x) dx + b ∫_b ^( 𝛌) f(x) dx ≥ a + c then prove ∫_a ^( b) (x/(f(x))) dx ≤ ((1/a) + (1/b)) ((b^2 −a^2 −2)/2)

$$\mathrm{let}\:\:\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:>\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{f}\::\:\left[\mathrm{a}\:,\:\mathrm{b}\right]\:\rightarrow\:\left[\mathrm{c}\:,\:\mathrm{d}\right] \\ $$$$\mathrm{a}\:\mathrm{continuous}\:\mathrm{function} \\ $$$$\mathrm{for}\:\mathrm{which}\:\:\exists\lambda\:\in\:\left(\mathrm{a}\:,\:\mathrm{b}\right) \\ $$$$\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{a}\:\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\lambda}} \:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:+\:\mathrm{b}\:\int_{\boldsymbol{\mathrm{b}}} ^{\:\boldsymbol{\lambda}} \:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:\geqslant\:\mathrm{a}\:+\:\mathrm{c} \\ $$$$\mathrm{then}\:\mathrm{prove} \\ $$$$\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\frac{\mathrm{x}}{\mathrm{f}\left(\mathrm{x}\right)}\:\mathrm{dx}\:\leqslant\:\left(\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\right)\:\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}} \\ $$

Question Number 219952 Answers: 0 Comments: 4

solve the system differential x′=3x−y+z y′=2x+z z′=−2x+y with x,y,and z are the function of t.

$${solve}\:{the}\:{system}\:{differential} \\ $$$${x}'=\mathrm{3}{x}−{y}+{z} \\ $$$${y}'=\mathrm{2}{x}+{z} \\ $$$${z}'=−\mathrm{2}{x}+{y} \\ $$$${with}\:{x},{y},{and}\:{z}\:{are}\:{the}\:{function}\:{of}\: \\ $$$${t}. \\ $$$$ \\ $$

Question Number 219949 Answers: 3 Comments: 0

Question Number 219948 Answers: 1 Comments: 0

Question Number 219946 Answers: 2 Comments: 0

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