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Question Number 152653 Answers: 1 Comments: 0

Ω :=∫_0 ^( ∞) (( e^( −x^( 3) ) . sin (x^( 3) ))/x)dx= ((ζ (2 ))/2) m.n...

$$ \\ $$$$\:\:\Omega\::=\int_{\mathrm{0}} ^{\:\infty} \frac{\:{e}^{\:−{x}^{\:\mathrm{3}} } .\:{sin}\:\left({x}^{\:\mathrm{3}} \:\right)}{{x}}{dx}=\:\frac{\zeta\:\left(\mathrm{2}\:\right)}{\mathrm{2}} \\ $$$$\:{m}.{n}... \\ $$$$ \\ $$

Question Number 152652 Answers: 0 Comments: 1

Question Number 152647 Answers: 1 Comments: 1

Question Number 152631 Answers: 1 Comments: 0

Question Number 152626 Answers: 1 Comments: 0

Solve for real numbers: (1/(x-1)) + (2/(x-2)) + (3/(x-3)) + (4/(x-4)) = 2x^2 -5x-4

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{1}}{\mathrm{x}-\mathrm{1}}\:+\:\frac{\mathrm{2}}{\mathrm{x}-\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{x}-\mathrm{3}}\:+\:\frac{\mathrm{4}}{\mathrm{x}-\mathrm{4}}\:=\:\mathrm{2x}^{\mathrm{2}} -\mathrm{5x}-\mathrm{4} \\ $$

Question Number 152625 Answers: 1 Comments: 2

x^4 +c_3 x^3 +c_2 x^2 +c_1 x+c_0 =0 for c_n ∈R this can have 4 unique zeros ∈R 2 unique zeros + 1 double zero ∈R 2 double zeros ∈R 1 triple + 1 unique zeros ∈R 1 fourfold zero ∈R 2 unique zeros ∈R + 1 pair of complex zeros 1 double zero ∈R + 1 pair of complex zeros 2 pairs of complex zeros 2 double imaginary zeros for given c_n ; can we decide which case we have without solving?

$${x}^{\mathrm{4}} +{c}_{\mathrm{3}} {x}^{\mathrm{3}} +{c}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{0}} =\mathrm{0} \\ $$$$\mathrm{for}\:{c}_{{n}} \in\mathbb{R}\:\mathrm{this}\:\mathrm{can}\:\mathrm{have} \\ $$$$\mathrm{4}\:\mathrm{unique}\:\mathrm{zeros}\:\in\mathbb{R} \\ $$$$\mathrm{2}\:\mathrm{unique}\:\mathrm{zeros}\:+\:\mathrm{1}\:\mathrm{double}\:\mathrm{zero}\:\in\mathbb{R} \\ $$$$\mathrm{2}\:\mathrm{double}\:\mathrm{zeros}\:\in\mathbb{R} \\ $$$$\mathrm{1}\:\mathrm{triple}\:+\:\mathrm{1}\:\mathrm{unique}\:\mathrm{zeros}\:\in\mathbb{R} \\ $$$$\mathrm{1}\:\mathrm{fourfold}\:\mathrm{zero}\:\in\mathbb{R} \\ $$$$\mathrm{2}\:\mathrm{unique}\:\mathrm{zeros}\:\in\mathbb{R}\:+\:\mathrm{1}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{zeros} \\ $$$$\mathrm{1}\:\mathrm{double}\:\mathrm{zero}\:\in\mathbb{R}\:+\:\mathrm{1}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{zeros} \\ $$$$\mathrm{2}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{zeros} \\ $$$$\mathrm{2}\:\mathrm{double}\:\mathrm{imaginary}\:\mathrm{zeros} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{given}\:{c}_{{n}} ;\:\mathrm{can}\:\mathrm{we}\:\mathrm{decide}\:\mathrm{which}\:\mathrm{case}\:\mathrm{we} \\ $$$$\mathrm{have}\:\mathrm{without}\:\mathrm{solving}? \\ $$

Question Number 152617 Answers: 0 Comments: 1