Question and Answers Forum

AllQuestion and Answers: Page 26

Question Number 219696 Answers: 2 Comments: 0

Question Number 219695 Answers: 1 Comments: 0

Question Number 219682 Answers: 0 Comments: 0

∫_0 ^( 1) ((log(1+x)∙Li_s (x))/(x∙ζ(s+1,x))) dx

$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{1}+{x}\right)\centerdot\mathrm{Li}_{{s}} \left({x}\right)}{{x}\centerdot\zeta\left({s}+\mathrm{1},{x}\right)}\:{dx} \\ $$$$ \\ $$

Question Number 219678 Answers: 2 Comments: 2

Question Number 219676 Answers: 0 Comments: 0

The latex converter is not converting some symbols. Any reason why?

$$\mathrm{The}\:\mathrm{latex}\:\mathrm{converter}\:\mathrm{is}\:\mathrm{not}\:\mathrm{converting}\:\mathrm{some}\:\mathrm{symbols}.\:\mathrm{Any}\:\mathrm{reason}\:\mathrm{why}?\: \\ $$

Question Number 219668 Answers: 2 Comments: 0

Question Number 219663 Answers: 0 Comments: 0

Question Number 219662 Answers: 0 Comments: 0

Question Number 219660 Answers: 1 Comments: 0

∫_( 0) ^( 2π) (1/(a + b cos (x))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{\mathrm{1}}{{a}\:+\:{b}\:{cos}\:\left({x}\right)}\:{dx} \\ $$$$ \\ $$

Question Number 219659 Answers: 1 Comments: 0

prove; cos (B+C−A)−cos(C+A−B)+cos(A+B−C)−cos(A+B+C) = 4sinAcosBsinC

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{prove}; \\ $$$$\:\:{cos}\:\left({B}+{C}−{A}\right)−{cos}\left({C}+{A}−{B}\right)+{cos}\left({A}+{B}−{C}\right)−{cos}\left({A}+{B}+{C}\right)\:=\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{sinAcosBsinC} \\ $$$$ \\ $$

Question Number 219658 Answers: 2 Comments: 0

prove Π_(n=2) ^∞ e(1−(1/n^2 ))^n^2 = (π/(e(√e)))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{prove} \\ $$$$\:\:\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}\:{e}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\overset{{n}^{\mathrm{2}} } {\right)}=\:\frac{\pi}{{e}\sqrt{{e}}} \\ $$$$ \\ $$

Question Number 219657 Answers: 1 Comments: 0

prove; Π_(n=1) ^∞ (((5n−2)(5n−3))/((5n−1)(5n−4))) = ϕ

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{prove}; \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\left(\mathrm{5}{n}−\mathrm{2}\right)\left(\mathrm{5}{n}−\mathrm{3}\right)}{\left(\mathrm{5}{n}−\mathrm{1}\right)\left(\mathrm{5}{n}−\mathrm{4}\right)}\:=\:\varphi \\ $$$$ \\ $$

Question Number 219651 Answers: 1 Comments: 0

Solve x^2 y^((2)) (x)+xy^((1)) (x)+(x^2 −ν^2 )y(x)=e^(−kx)

$$\mathrm{Solve} \\ $$$${x}^{\mathrm{2}} {y}^{\left(\mathrm{2}\right)} \left({x}\right)+{xy}^{\left(\mathrm{1}\right)} \left({x}\right)+\left({x}^{\mathrm{2}} −\nu^{\mathrm{2}} \right){y}\left({x}\right)={e}^{−{kx}} \\ $$

Question Number 219649 Answers: 0 Comments: 0

Solve y^((2)) (t)−t∙y(t)=0

$$\mathrm{Solve}\:{y}^{\left(\mathrm{2}\right)} \left({t}\right)−{t}\centerdot{y}\left({t}\right)=\mathrm{0} \\ $$

Question Number 219642 Answers: 1 Comments: 0

Question Number 219685 Answers: 2 Comments: 0

Question Number 219637 Answers: 0 Comments: 0

∫_0 ^( ∞) ((J_ν (s)e^(−μs) )/( (√(s^2 +R^2 ))))ds , (ν,μ∈R^+ , R∈R^+ \{0})

$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{{J}_{\nu} \left({s}\right){e}^{−\mu{s}} }{\:\sqrt{{s}^{\mathrm{2}} +{R}^{\mathrm{2}} }}\mathrm{d}{s}\:,\:\left(\nu,\mu\in\mathbb{R}^{+} \:,\:\mathrm{R}\in\mathbb{R}^{+} \backslash\left\{\mathrm{0}\right\}\right) \\ $$

Question Number 219634 Answers: 1 Comments: 0

Question Number 219624 Answers: 1 Comments: 0

Question Number 219625 Answers: 0 Comments: 0

Determine all real numbers x that statisfy the following inequality; (√(x^4 +1))+(√(x^4 +4x^2 +4)) ≤ (√((x^2 +1)(x^2 +4))) + ∣x∣(√(x^2 +3))

$$ \\ $$$$\:\mathrm{Determine}\:\mathrm{all}\:\mathrm{real}\:\mathrm{numbers}\:{x}\: \\ $$$$\:\:\:\mathrm{that}\:\mathrm{statisfy}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}; \\ $$$$\:\:\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}+\sqrt{{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}}\:\leqslant\:\sqrt{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:+\:\mid{x}\mid\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 219621 Answers: 2 Comments: 1

Question Number 219620 Answers: 1 Comments: 0

Prove; ∫_( 0) ^( ∞) ((50x^8 )/(x^(20) +2x^(10) +1)) dx = φπ

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{Prove};\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{50}{x}^{\mathrm{8}} }{{x}^{\mathrm{20}} +\mathrm{2}{x}^{\mathrm{10}} +\mathrm{1}}\:{dx}\:=\:\phi\pi \\ $$$$ \\ $$

Question Number 219619 Answers: 2 Comments: 0

∫_0 ^∞ (x^2 +1)^(−1/2) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−\mathrm{1}/\mathrm{2}} {dx} \\ $$$$ \\ $$

Question Number 219618 Answers: 0 Comments: 0

Prove; ∫_0 ^( 1) ((ln ln (1/x))/((1+x)^2 )) dx = (1/2)(ln((π/2))−γ)

$$ \\ $$$$\:\:\:\:\:\:\:\:{Prove}; \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{{ln}\:{ln}\:\frac{\mathrm{1}}{{x}}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\frac{\pi}{\mathrm{2}}\right)−\gamma\right) \\ $$$$ \\ $$

Question Number 219617 Answers: 0 Comments: 0

ok, let′s all answer questions from anywhere on the www using the given results from the sources or wolframalpha or any AI available. this promises great fun!

$$\mathrm{ok},\:\mathrm{let}'\mathrm{s}\:\mathrm{all}\:\mathrm{answer}\:\mathrm{questions}\:\mathrm{from}\:\mathrm{anywhere} \\ $$$$\mathrm{on}\:\mathrm{the}\:{www}\:\mathrm{using}\:\mathrm{the}\:\mathrm{given}\:\mathrm{results}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{sources}\:\mathrm{or}\:{wolframalpha}\:\mathrm{or}\:\mathrm{any}\:\mathrm{AI} \\ $$$$\mathrm{available}.\:\mathrm{this}\:\mathrm{promises}\:\mathrm{great}\:\mathrm{fun}! \\ $$

Question Number 219606 Answers: 2 Comments: 0

prove that for positive real numbers a,b,c, the following inequality holds; (a^2 /(b + c)) + (b^2 /(c + a)) + (c^2 /(a + b)) ≥ ((a + b + c)/2)

$$ \\ $$$$\:\mathrm{prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:{a},{b},{c},\:\:\: \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}; \\ $$$$\:\:\frac{{a}^{\mathrm{2}} }{{b}\:+\:{c}}\:+\:\frac{{b}^{\mathrm{2}} }{{c}\:+\:{a}}\:+\:\frac{{c}^{\mathrm{2}} }{{a}\:+\:{b}}\:\:\geqslant\:\frac{{a}\:+\:{b}\:+\:{c}}{\mathrm{2}} \\ $$$$ \\ $$

Pg 21 Pg 22 Pg 23 Pg 24 Pg 25 Pg 26 Pg 27 Pg 28 Pg 29 Pg 30

Contact: info@tinkutara.com