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Question Number 214059    Answers: 2   Comments: 0

Question Number 214098    Answers: 1   Comments: 0

Let F be Field of characteristic 0 L_i (i=1,2) be two algebraic extension of F , and L_1 L_2 be a field in F^ (where F^ is the algebraic closure of F) defined by {l_1 l_2 ∣l_i ∈L_i (i=1,2)} 1. show that if L_1 and L_2 are galois over F then L_1 L_2 is also Galois over F 2. show that if G(L_1 /F^ ) and G(L_2 /F^ ) are Solvable , then Gal(L_1 L_2 /F^ ) is also Solvable

$$\mathrm{Let}\:{F}\:\mathrm{be}\:\:\mathrm{Field}\:\mathrm{of}\:\mathrm{characteristic}\:\mathrm{0} \\ $$$${L}_{{i}} \:\left({i}=\mathrm{1},\mathrm{2}\right)\:\mathrm{be}\:\mathrm{two}\:\mathrm{algebraic}\:\mathrm{extension} \\ $$$$\mathrm{of}\:{F}\:,\:\mathrm{and}\:{L}_{\mathrm{1}} {L}_{\mathrm{2}} \:\mathrm{be}\:\mathrm{a}\:\mathrm{field}\:\mathrm{in}\:\bar {{F}}\: \\ $$$$\left(\mathrm{where}\:\bar {{F}}\:\:\mathrm{is}\:\mathrm{the}\:\mathrm{algebraic}\:\mathrm{closure}\:\:\mathrm{of}\:{F}\right) \\ $$$$\mathrm{defined}\:\mathrm{by}\:\left\{{l}_{\mathrm{1}} {l}_{\mathrm{2}} \mid{l}_{{i}} \in{L}_{{i}} \:\left({i}=\mathrm{1},\mathrm{2}\right)\right\} \\ $$$$\mathrm{1}.\:\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:{L}_{\mathrm{1}} \:\mathrm{and}\:{L}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{galois}\:\mathrm{over}\:{F} \\ $$$$\mathrm{then}\:{L}_{\mathrm{1}} {L}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{also}\:\mathrm{Galois}\:\mathrm{over}\:{F} \\ $$$$\mathrm{2}.\:\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:{G}\left({L}_{\mathrm{1}} /{F}^{\:} \right)\:\mathrm{and}\:{G}\left({L}_{\mathrm{2}} /{F}^{\:} \right) \\ $$$$\mathrm{are}\:\mathrm{Solvable}\:,\:\mathrm{then}\:\mathrm{Gal}\left({L}_{\mathrm{1}} {L}_{\mathrm{2}} /{F}^{\:} \right)\:\mathrm{is}\:\mathrm{also} \\ $$$$\mathrm{Solvable} \\ $$

Question Number 214097    Answers: 1   Comments: 0

Question Number 214051    Answers: 1   Comments: 1

evaluate ∫_0 ^( π) e^(sin^2 (u)) du... i use Feynman′s trick to solve integral ∫_0 ^( π) e^(sin^2 (u)) du=I I(t)=∫_0 ^( π) e^(tsin^2 (u)) du I^((1)) (t)=∫_0 ^( π) sin^2 (u)e^(tsin^2 (u)) du ∫_0 ^( π) (1−cos^2 (u))e^(t∙sin^2 (u)) du ∫_0 ^( π) e^(t∙sin^2 (u)) du−∫_0 ^( π) cos^2 (u)e^(t∙sin^2 (u)) du ∫_0 ^( π) e^(t∙sin^2 (u)) du−∫_0 ^( π) sin^2 (u+(π/2))e^(t∙sin^2 (u)) du u+(π/2) =w → du=dw ∫_(π/2) ^((3π)/2) sin^2 (w)e^(t∙cos^2 (w)) dw= ∫_(π/2) ^( ((3π)/2)) sin^2 (w)e^(t(1−sin^2 (w))) dw ∫_(π/2) ^((3π)/2) sin^2 (w)e^t e^(−tsin^2 (w)) dw=e^t ∫_(π/2) ^((3π)/2) sin^2 (w)e^(−tsin^2 (w)) dw I^((1)) (t)=∫_0 ^( π) e^(t∙sin^2 (u)) du−e^t ∫_0 ^( π) sin^2 (u)e^(−t∙sin^2 (u)) du I can′t ...anymore...help

$$\mathrm{evaluate}\:\int_{\mathrm{0}} ^{\:\pi} \:{e}^{\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u}...\: \\ $$$$\mathrm{i}\:\mathrm{use}\:\mathrm{Feynman}'\mathrm{s}\:\mathrm{trick}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{integral} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \:{e}^{\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u}={I} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\pi} \:{e}^{{t}\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u} \\ $$$${I}^{\left(\mathrm{1}\right)} \left({t}\right)=\int_{\mathrm{0}} ^{\:\pi} \:\mathrm{sin}^{\mathrm{2}} \left({u}\right){e}^{{t}\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left({u}\right)\right){e}^{{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \:{e}^{{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u}−\int_{\mathrm{0}} ^{\:\pi} \:\mathrm{cos}^{\mathrm{2}} \left({u}\right){e}^{{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \:{e}^{{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u}−\int_{\mathrm{0}} ^{\:\pi} \:\mathrm{sin}^{\mathrm{2}} \left({u}+\frac{\pi}{\mathrm{2}}\right){e}^{{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u} \\ $$$${u}+\frac{\pi}{\mathrm{2}}\:={w}\:\:\rightarrow\:\:\mathrm{d}{u}=\mathrm{d}{w} \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{3}\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{2}} \left({w}\right){e}^{{t}\centerdot\mathrm{cos}^{\mathrm{2}} \left({w}\right)} \:\mathrm{d}{w}= \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\:\frac{\mathrm{3}\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{2}} \left({w}\right){e}^{{t}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left({w}\right)\right)} \mathrm{d}{w} \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{3}\pi}{\mathrm{2}}} \:\:\mathrm{sin}^{\mathrm{2}} \left({w}\right){e}^{{t}} {e}^{−{t}\mathrm{sin}^{\mathrm{2}} \left({w}\right)} \mathrm{d}{w}={e}^{{t}} \int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{3}\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{2}} \left({w}\right){e}^{−{t}\mathrm{sin}^{\mathrm{2}} \left({w}\right)} \mathrm{d}{w} \\ $$$${I}^{\left(\mathrm{1}\right)} \left({t}\right)=\int_{\mathrm{0}} ^{\:\pi} \:{e}^{{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u}−{e}^{{t}} \int_{\mathrm{0}} ^{\:\pi} \:\mathrm{sin}^{\mathrm{2}} \left({u}\right){e}^{−{t}\centerdot\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u} \\ $$$$\mathrm{I}\:\mathrm{can}'\mathrm{t}\:...\mathrm{anymore}...\mathrm{help} \\ $$

Question Number 214050    Answers: 0   Comments: 0

If A: a_1 , a_2 , ..., a_(62) and B: b_1 , b_2 , ..., b_(62) are two strictly increasing natural number sequences such that a_(62) ≤755 and b_(62) ≤755. Find the maximum of Σ_(i=1) ^(62) ∣a_i −b_i ∣−∣Σ_(i=1) ^(62) (a_i −b_i )∣.

$$\mathrm{If}\:{A}:\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:...,\:{a}_{\mathrm{62}} \:\mathrm{and}\:{B}:\:{b}_{\mathrm{1}} ,\:{b}_{\mathrm{2}} ,\:...,\:{b}_{\mathrm{62}} \:\mathrm{are}\:\mathrm{two} \\ $$$$\mathrm{strictly}\:\mathrm{increasing}\:\mathrm{natural}\:\mathrm{number}\:\mathrm{sequences} \\ $$$$\mathrm{such}\:\mathrm{that}\:{a}_{\mathrm{62}} \leqslant\mathrm{755}\:\mathrm{and}\:{b}_{\mathrm{62}} \leqslant\mathrm{755}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:\underset{{i}=\mathrm{1}} {\overset{\mathrm{62}} {\sum}}\mid{a}_{{i}} −{b}_{{i}} \mid−\mid\underset{{i}=\mathrm{1}} {\overset{\mathrm{62}} {\sum}}\left({a}_{{i}} −{b}_{{i}} \right)\mid. \\ $$

Question Number 214047    Answers: 1   Comments: 1

Question Number 214040    Answers: 1   Comments: 1

Question Number 214030    Answers: 1   Comments: 0

Question Number 214020    Answers: 2   Comments: 1

Question Number 214012    Answers: 1   Comments: 1

Question Number 214005    Answers: 1   Comments: 0

find all zero divisors of Z_(24)

$${find}\:{all}\:{zero}\:{divisors}\:{of}\:{Z}_{\mathrm{24}} \\ $$

Question Number 214002    Answers: 2   Comments: 0

find the integers x that satisfies a congruence 3x=4 (mod 11) .

$${find}\:{the}\:{integers}\:{x}\:{that}\:{satisfies}\:{a}\:{congruence}\:\mathrm{3}{x}=\mathrm{4}\:\left({mod}\:\mathrm{11}\right)\:. \\ $$

Question Number 214001    Answers: 2   Comments: 0

find all solutions of the equation x^2 =x in each of the rings Z_2 Z_3 and Z_6

$$\mathrm{find}\:\mathrm{all}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} ={x}\:\mathrm{in}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rings}\: \\ $$$${Z}_{\mathrm{2}} \: \\ $$$${Z}_{\mathrm{3}} \\ $$$$\mathrm{and}\:\mathrm{Z}_{\mathrm{6}} \\ $$

Question Number 214000    Answers: 0   Comments: 4

Let y(x) be the solution of diff eq. y ′= ((cos x+y)/(cos x)) , y(0)=0 Find y((π/6)).

$$\:\:\mathrm{Let}\:\mathrm{y}\left(\mathrm{x}\right)\:\mathrm{be}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{diff}\:\mathrm{eq}. \\ $$$$\:\:\mathrm{y}\:'=\:\frac{\mathrm{cos}\:\mathrm{x}+\mathrm{y}}{\mathrm{cos}\:\mathrm{x}}\:,\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\:\:\mathrm{Find}\:\mathrm{y}\left(\frac{\pi}{\mathrm{6}}\right). \\ $$

Question Number 213999    Answers: 1   Comments: 1

∫∫...∫_( D) e^(−(z_1 ^2 +z_2 ^2 ...+z_n ^2 )) da D=[0,∞)×[0,∞)......[0,∞)_(n times) ∫_0 ^( π) e^(−sin^2 (z)) dz help

$$\int\int...\int_{\:\mathcal{D}} \:\:{e}^{−\left({z}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} ...+{z}_{{n}} ^{\mathrm{2}} \right)} \mathrm{da} \\ $$$$\mathcal{D}=\underset{\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{times}}} {\left[\mathrm{0},\infty\right)×\left[\mathrm{0},\infty\right)......\left[\mathrm{0},\infty\right)} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \:{e}^{−\mathrm{sin}^{\mathrm{2}} \left({z}\right)} \mathrm{d}{z} \\ $$$$\mathrm{help} \\ $$

Question Number 213992    Answers: 0   Comments: 0

Question Number 213991    Answers: 0   Comments: 0

Question Number 213962    Answers: 1   Comments: 0

∫((x^4 −1)/(x(x^4 −5)(x^5 −5x+1)))dx

$$\int\frac{{x}^{\mathrm{4}} −\mathrm{1}}{{x}\left({x}^{\mathrm{4}} −\mathrm{5}\right)\left({x}^{\mathrm{5}} −\mathrm{5}{x}+\mathrm{1}\right)}{dx} \\ $$

Question Number 213960    Answers: 1   Comments: 1

Question Number 213956    Answers: 2   Comments: 0

Question Number 213953    Answers: 1   Comments: 0

Question Number 213948    Answers: 0   Comments: 0

evaluate. 1. (1/π)∫_0 ^( π) e^(−i(t−sin(t))) dt 2. ∫_0 ^( a) ∫_0 ^( a) (√(u^2 +v^2 −6u+9)) dudv 3. ∫_0 ^( π/2) e^(cos(t)) cos(2t+sin(t))dt 4. ∫_(−∞) ^( ∞) ((sin(3z))/(z^2 +2z+5)) dz 5.∫_0 ^( 2π) (1/(2+cos(θ))) dθ

$$\mathrm{evaluate}. \\ $$$$\mathrm{1}.\:\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\:\pi} \:\:{e}^{−\boldsymbol{{i}}\left({t}−\mathrm{sin}\left({t}\right)\right)} \mathrm{d}{t} \\ $$$$\mathrm{2}.\:\int_{\mathrm{0}} ^{\:\mathrm{a}} \int_{\mathrm{0}} ^{\:\mathrm{a}} \:\:\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} −\mathrm{6}{u}+\mathrm{9}}\:\mathrm{d}{u}\mathrm{d}{v} \\ $$$$\mathrm{3}.\:\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \:\:{e}^{\mathrm{cos}\left({t}\right)} \mathrm{cos}\left(\mathrm{2}{t}+\mathrm{sin}\left({t}\right)\right)\mathrm{d}{t} \\ $$$$\mathrm{4}.\:\int_{−\infty} ^{\:\infty} \:\frac{\mathrm{sin}\left(\mathrm{3}{z}\right)}{{z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{5}}\:\mathrm{d}{z} \\ $$$$\mathrm{5}.\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\:\frac{\mathrm{1}}{\mathrm{2}+\mathrm{cos}\left(\theta\right)}\:\mathrm{d}\theta \\ $$

Question Number 213945    Answers: 3   Comments: 0

Question Number 213944    Answers: 1   Comments: 0

Question Number 213939    Answers: 2   Comments: 1

Question Number 213923    Answers: 3   Comments: 0

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