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AllQuestion and Answers: Page 6

Question Number 212821    Answers: 1   Comments: 0

Question Number 212820    Answers: 2   Comments: 0

Question Number 212816    Answers: 1   Comments: 0

Question Number 212829    Answers: 4   Comments: 1

Question Number 213376    Answers: 2   Comments: 1

∫(dx/( (√((4x^2 +1)^3 ))))

$$\int\frac{{dx}}{\:\sqrt{\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }} \\ $$

Question Number 212808    Answers: 0   Comments: 0

lim_(x→∞) [(x^(.1) −x^(.9) )^(1o) −x]

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\left({x}^{.\mathrm{1}} −{x}^{.\mathrm{9}} \right)^{\mathrm{1}{o}} −{x}\right] \\ $$

Question Number 212803    Answers: 3   Comments: 0

Question Number 212802    Answers: 1   Comments: 0

Question Number 212801    Answers: 2   Comments: 0

Question Number 212798    Answers: 1   Comments: 0

lim_(x→∞) (1+(1/x))^(x^2 /e^x ) =?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\frac{{x}^{\mathrm{2}} }{{e}^{{x}} }} =? \\ $$

Question Number 212790    Answers: 1   Comments: 0

Question Number 213374    Answers: 2   Comments: 4

Find: ((1 − 2 (5)^(1/4) + (√5))/(((√3) − (5)^(1/4) )^2 )) = ?

$$\mathrm{Find}:\:\:\:\frac{\mathrm{1}\:−\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\:+\:\sqrt{\mathrm{5}}}{\left(\sqrt{\mathrm{3}}\:−\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\right)^{\mathrm{2}} }\:=\:? \\ $$

Question Number 213371    Answers: 0   Comments: 1

Question Number 212784    Answers: 1   Comments: 0

Question Number 212783    Answers: 1   Comments: 1

Question Number 212779    Answers: 0   Comments: 3

excuse me???? am i invisible?? why isn′t anyone answering my qusestion

$$\mathrm{excuse}\:\mathrm{me}???? \\ $$$$\mathrm{am}\:\mathrm{i}\:\mathrm{invisible}??\: \\ $$$$\mathrm{why}\:\mathrm{isn}'\mathrm{t}\:\mathrm{anyone}\:\mathrm{answering}\:\mathrm{my}\:\mathrm{qusestion} \\ $$$$ \\ $$

Question Number 212777    Answers: 0   Comments: 0

question 212462 prove that ∫_0 ^∞ (dx/( (√(x^4 +25x^2 +160))))=∫_0 ^∞ (dx/( (√(x^4 −95x^2 +2560)))) my attempt (but is it a proof?): I_b ^a :=∫_0 ^∞ (dx/( (√(x^4 +ax^2 +b))))=(2/( π(b)^(1/4) agm (1, ((√(a+2(√b)))/(2(b)^(1/4) ))))) ∫_0 ^∞ (dx/( (√(x^4 +ax^2 +b))))= [t=2arctan (x/( (b)^(1/4) ))] =(1/( (b)^(1/4) ))∫_0 ^(π/2) (dt/( (√(1−((1/2)−(a/(4(√b))))sin^2 t))))= [K (k^2 ):=∫_0 ^(π/2) (dt/( (√(1−k^2 sin^2 t))))] =(1/( (b)^(1/4) ))K ((1/2)−(a/(4(√b)))) K (k^2 ) =(2/(πagm (1, (√(1−k^2 ))))) (1/( (b)^(1/4) ))K ((1/2)−(a/(4(√b)))) =(2/( π(b)^(1/4) agm (1, ((√(a+2(√b)))/(2(b)^(1/4) ))))) u_0 =p∧v_0 =q u_(n+1) =((u_n +v_n )/2)∧v_(n+1) =(√(u_n v_n )) agm (p, q) =lim_(n→∞) u_n =lim_(n→∞) v_n I_(160) ^(25) =(1/( π((10))^(1/4) agm (1, ((√(25+8(√(10))))/(4((10))^(1/4) ))))) I_(2560) ^(−95) =(1/(2π((10))^(1/4) agm (1, ((√(−95+32(√(10))))/(8((10))^(1/4) ))))) I_(160) ^(25) =I_(2560) ^(−95) ⇔ agm (1, ((√(25+8(√(10))))/(4((10))^(1/4) )))=2agm (1, ((√(−95+32(√(10))))/(8((10))^(1/4) ))) which is true

$$\mathrm{question}\:\mathrm{212462} \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} +\mathrm{25}{x}^{\mathrm{2}} +\mathrm{160}}}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{95}{x}^{\mathrm{2}} +\mathrm{2560}}} \\ $$$$ \\ $$$$\mathrm{my}\:\mathrm{attempt}\:\left(\mathrm{but}\:\mathrm{is}\:\mathrm{it}\:\mathrm{a}\:\mathrm{proof}?\right): \\ $$$$ \\ $$$${I}_{{b}} ^{{a}} :=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} +{ax}^{\mathrm{2}} +{b}}}=\frac{\mathrm{2}}{\:\pi\sqrt[{\mathrm{4}}]{{b}}\:\mathrm{agm}\:\left(\mathrm{1},\:\frac{\sqrt{{a}+\mathrm{2}\sqrt{{b}}}}{\mathrm{2}\sqrt[{\mathrm{4}}]{{b}}}\right)} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} +{ax}^{\mathrm{2}} +{b}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{t}=\mathrm{2arctan}\:\frac{{x}}{\:\sqrt[{\mathrm{4}}]{{b}}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{b}}}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{dt}}{\:\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}\sqrt{{b}}}\right)\mathrm{sin}^{\mathrm{2}} \:{t}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{K}\:\left({k}^{\mathrm{2}} \right):=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{dt}}{\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{t}}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{b}}}{K}\:\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}\sqrt{{b}}}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{K}\:\left({k}^{\mathrm{2}} \right)\:=\frac{\mathrm{2}}{\pi\mathrm{agm}\:\left(\mathrm{1},\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{b}}}{K}\:\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}\sqrt{{b}}}\right)\:=\frac{\mathrm{2}}{\:\pi\sqrt[{\mathrm{4}}]{{b}}\:\mathrm{agm}\:\left(\mathrm{1},\:\frac{\sqrt{{a}+\mathrm{2}\sqrt{{b}}}}{\mathrm{2}\sqrt[{\mathrm{4}}]{{b}}}\right)} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{u}_{\mathrm{0}} ={p}\wedge{v}_{\mathrm{0}} ={q} \\ $$$$\:\:\:\:\:\:\:\:\:\:{u}_{{n}+\mathrm{1}} =\frac{{u}_{{n}} +{v}_{{n}} }{\mathrm{2}}\wedge{v}_{{n}+\mathrm{1}} =\sqrt{{u}_{{n}} {v}_{{n}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{agm}\:\left({p},\:{q}\right)\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{u}_{{n}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}{v}_{{n}} \\ $$$$ \\ $$$${I}_{\mathrm{160}} ^{\mathrm{25}} =\frac{\mathrm{1}}{\:\pi\sqrt[{\mathrm{4}}]{\mathrm{10}}\:\mathrm{agm}\:\left(\mathrm{1},\:\frac{\sqrt{\mathrm{25}+\mathrm{8}\sqrt{\mathrm{10}}}}{\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{10}}}\right)} \\ $$$${I}_{\mathrm{2560}} ^{−\mathrm{95}} =\frac{\mathrm{1}}{\mathrm{2}\pi\sqrt[{\mathrm{4}}]{\mathrm{10}}\:\mathrm{agm}\:\left(\mathrm{1},\:\frac{\sqrt{−\mathrm{95}+\mathrm{32}\sqrt{\mathrm{10}}}}{\mathrm{8}\sqrt[{\mathrm{4}}]{\mathrm{10}}}\right)} \\ $$$$ \\ $$$$ \\ $$$${I}_{\mathrm{160}} ^{\mathrm{25}} ={I}_{\mathrm{2560}} ^{−\mathrm{95}} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{agm}\:\left(\mathrm{1},\:\frac{\sqrt{\mathrm{25}+\mathrm{8}\sqrt{\mathrm{10}}}}{\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{10}}}\right)=\mathrm{2agm}\:\left(\mathrm{1},\:\frac{\sqrt{−\mathrm{95}+\mathrm{32}\sqrt{\mathrm{10}}}}{\mathrm{8}\sqrt[{\mathrm{4}}]{\mathrm{10}}}\right) \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{true} \\ $$

Question Number 212775    Answers: 0   Comments: 1

vector×scalar=? vector or scalar?

$${vector}×{scalar}=? \\ $$$${vector}\:{or}\:{scalar}? \\ $$

Question Number 212899    Answers: 1   Comments: 0

Question Number 212765    Answers: 1   Comments: 0

i generalized Bessel function′s Laplace Transform L.TJ_ν (z)=(((s+(√(s^2 +1)))^(−ν) )/( (√(s^2 +1)))) , s∈[0,∞) , ν∈R L.T Y_ν (z)=((cot(πν)(s+(√(s^2 +1)))^(−ν) )/( (√(s^2 +1))))−((csc(πν)(s+(√(s^2 +1)))^ν )/( (√(s^2 +1)))) s∈[0,∞) , ν∈R^+ \{0,Z^+ } but.... i can′t explain why L.T Y_ν (z) is undefined when ν∈Z^+ \{0}..... Help me.....!!!

$$\mathrm{i}\:\:\mathrm{generalized}\:\boldsymbol{\mathrm{Bessel}}\:\boldsymbol{\mathrm{function}}'\mathrm{s} \\ $$$$\mathrm{Laplace}\:\mathrm{Transform} \\ $$$$\mathrm{L}.\mathrm{T}{J}_{\nu} \left({z}\right)=\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}\:,\:{s}\in\left[\mathrm{0},\infty\right)\:,\:\nu\in\mathbb{R} \\ $$$$\mathrm{L}.\mathrm{T}\:{Y}_{\nu} \left({z}\right)=\frac{\mathrm{cot}\left(\pi\nu\right)\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}−\frac{\mathrm{csc}\left(\pi\nu\right)\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${s}\in\left[\mathrm{0},\infty\right)\:,\:\nu\in\mathbb{R}^{+} \backslash\left\{\mathrm{0},\mathbb{Z}^{+} \right\} \\ $$$$\mathrm{but}....\:\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{explain}\:\mathrm{why}\:\mathrm{L}.\mathrm{T}\:{Y}_{\nu} \left({z}\right) \\ $$$$\mathrm{is}\:\mathrm{undefined}\:\mathrm{when}\:\nu\in\mathbb{Z}^{+} \backslash\left\{\mathrm{0}\right\}..... \\ $$$$\mathrm{Help}\:\mathrm{me}.....!!! \\ $$

Question Number 212762    Answers: 1   Comments: 1

Question Number 212760    Answers: 1   Comments: 0

Question Number 212761    Answers: 2   Comments: 1

Question Number 212752    Answers: 1   Comments: 1

Question Number 212745    Answers: 2   Comments: 0

Question Number 212741    Answers: 1   Comments: 0

a, b,c ∈ N^∗ Show that ab<c ⇒ a+b≤c

$${a},\:{b},{c}\:\in\:\mathbb{N}^{\ast} \\ $$$$ \\ $$$${Show}\:{that}\:{ab}<{c}\:\Rightarrow\:{a}+{b}\leqslant{c} \\ $$$$ \\ $$

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