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Question Number 192118 Answers: 0 Comments: 0

given points (a,b) where a,b ∈R how to get the best fit parabola that go through the origin and open downward (coefficient of x^2 is negative)?

$$\mathrm{given}\:\mathrm{points}\:\left({a},{b}\right)\:\mathrm{where}\:{a},{b}\:\in\mathbb{R} \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{best}\:\mathrm{fit}\:\mathrm{parabola}\:\mathrm{that}\:\mathrm{go}\:\mathrm{through}\:\mathrm{the}\:\mathrm{origin} \\ $$$$\mathrm{and}\:\mathrm{open}\:\mathrm{downward}\:\left(\mathrm{coefficient}\:\mathrm{of}\:{x}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{negative}\right)? \\ $$

Question Number 173157 Answers: 1 Comments: 0

If (1/a) + (1/b) + (1/c) = (1/(a + b + c)) then prove that (1/a^3 ) + (1/b^3 ) + (1/c^3 ) = (1/((a + b + c)^3 )) .

$$\mathrm{If}\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:=\:\frac{\mathrm{1}}{{a}\:+\:{b}\:+\:{c}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{{b}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{{c}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\left({a}\:+\:{b}\:+\:{c}\right)^{\mathrm{3}} }\:. \\ $$

Question Number 173507 Answers: 0 Comments: 3

solve y−(x+1) y′ + ((2x(2x−1))/((x+1)^2 )) = 0 and x(x^2 +x+1) y′ +((x/2) +1) y = (x^2 +x+1)^(3/2)

$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${solve}\:\:\:{y}−\left({x}+\mathrm{1}\right)\:{y}'\:+\:\frac{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:\:\:\:\:\:{x}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:{y}'\:+\left(\frac{{x}}{\mathrm{2}}\:+\mathrm{1}\right)\:{y}\:=\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 173152 Answers: 2 Comments: 0

If x = 2 + 2^(2/3) + 2^(1/3) then prove that x^3 − 6x^2 + 6x − 2 = 0.

$$\mathrm{If}\:{x}\:=\:\mathrm{2}\:+\:\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} \:+\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$${x}^{\mathrm{3}} \:−\:\mathrm{6}{x}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:−\:\mathrm{2}\:=\:\mathrm{0}. \\ $$

Question Number 173150 Answers: 1 Comments: 0

let ∀n∈N: I_n (f(x))= the n^(th) antiderivate of f(x) with I_0 =f(x) find the formula for the constants a_n , b_n of I_n (ln x)=a_n x^n ln x +b_n x^n

$$\mathrm{let}\:\forall{n}\in\mathbb{N}:\:{I}_{{n}} \left({f}\left({x}\right)\right)=\:\mathrm{the}\:{n}^{\mathrm{th}} \:\mathrm{antiderivate} \\ $$$$\mathrm{of}\:{f}\left({x}\right)\:\mathrm{with}\:{I}_{\mathrm{0}} ={f}\left({x}\right) \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{the}\:\mathrm{constants}\:{a}_{{n}} ,\:{b}_{{n}} \:\mathrm{of} \\ $$$${I}_{{n}} \left(\mathrm{ln}\:{x}\right)={a}_{{n}} {x}^{{n}} \mathrm{ln}\:{x}\:+{b}_{{n}} {x}^{{n}} \\ $$

Question Number 173149 Answers: 0 Comments: 1

∫_0 ^( ∞) (( dx)/( (√(1+x)) .(2+2x +x^( 2) )))=(1/σ) (π−ln(3+2(√3) )) σ = ? −− solution −− Ω=^((√(1+x)) =t) 2∫_0 ^( ∞) (dt/( 1+ t^( 4) )) = 2∫_0 ^( 1) (dt/(1 + t^( 4) )) Ψ = ∫_0 ^( 1) (( dt)/(1+t^( 4) )) = (1/2)∫_0 ^( 1) (( 1+t^( 2) −(t^( −2) −1))/(1+t^( 4) ))dt = (1/2) ∫_0 ^( 1) (( 1+ t^( 2) )/(1+t^( 4) )) dt +(1/2) ∫_0 ^( 1) ((1−t^( 2) )/(1+ t^( 4) )) dt Φ=∫_0 ^( 1) ((1 +t^( 2) )/(1+t^( 4) ))dt =∫_0 ^( 1) (( t^( −2) +1)/(t^( −2) + t^( 2) )) dt =∫_0 ^( 1) (( 1+t^( −2) )/(( t^ − t^( −1) )^( 2) +2)) =^(sub) [ (1/( (√2))) tan^( −1) (t −t^( −1) )]_0 ^1 =(π/(2(√2))) ∗ 𝛗 = ∫_0 ^( 1) (( 1−t^( 2) )/(1+t^( 4) )) dt = ∫_0 ^( 1) ((t^( −2) −1)/(( t +t^( −1) )^( 2) −2))dt =^(t +(1/t) =u) −∫_2 ^( ∞) (du/(( u−(√2) )(u+ (√2) ))) = ((−1)/(2(√2))) [ln(((u −(√2))/(u+(√2))))]_2 ^( ∞) =(1/(2(√2))) ln(((2−(√2))/(2+(√2))) ) 𝛗 =−(1/(2(√2))) ln( 3 +2(√2) ) ∗∗ (∗) & (∗∗):: Ω=2Ψ= (Φ + 𝛗) =(1/(2(√2))) ( π −ln( 3 +2(√2) ) ...■m.n

$$ \\ $$$$\:\:\:\:\:\int_{\mathrm{0}} ^{\:\:\infty} \frac{\:{dx}}{\:\sqrt{\mathrm{1}+{x}}\:.\left(\mathrm{2}+\mathrm{2}{x}\:+{x}^{\:\mathrm{2}} \right)}=\frac{\mathrm{1}}{\sigma}\:\left(\pi−\mathrm{ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}\:\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sigma\:=\:? \\ $$$$\:\:\:\:\:\:\:\:−−\:\:\mathrm{solution}\:−− \\ $$$$\:\:\:\:\:\Omega\overset{\sqrt{\mathrm{1}+{x}}\:={t}} {=}\:\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{{dt}}{\:\mathrm{1}+\:{t}^{\:\mathrm{4}} }\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{dt}}{\mathrm{1}\:+\:{t}^{\:\mathrm{4}} } \\ $$$$\:\:\:\:\:\:\Psi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{dt}}{\mathrm{1}+{t}^{\:\mathrm{4}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{1}+{t}^{\:\mathrm{2}} −\left({t}^{\:−\mathrm{2}} −\mathrm{1}\right)}{\mathrm{1}+{t}^{\:\mathrm{4}} }{dt} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{1}+\:{t}^{\:\mathrm{2}} }{\mathrm{1}+{t}^{\:\mathrm{4}} }\:{dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{t}^{\:\mathrm{2}} }{\mathrm{1}+\:{t}^{\:\mathrm{4}} }\:{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Phi=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}\:+{t}^{\:\mathrm{2}} }{\mathrm{1}+{t}^{\:\mathrm{4}} }{dt}\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{t}^{\:−\mathrm{2}} +\mathrm{1}}{{t}^{\:−\mathrm{2}} +\:{t}^{\:\mathrm{2}} }\:{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\:\mathrm{1}+{t}^{\:−\mathrm{2}} }{\left(\:{t}^{\:} −\:{t}^{\:−\mathrm{1}} \right)^{\:\mathrm{2}} +\mathrm{2}}\:\overset{{sub}} {=}\:\left[\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{tan}^{\:−\mathrm{1}} \left({t}\:−{t}^{\:−\mathrm{1}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\:\:\:\:\ast \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{1}−{t}^{\:\mathrm{2}} }{\mathrm{1}+{t}^{\:\mathrm{4}} }\:{dt}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{t}^{\:−\mathrm{2}} −\mathrm{1}}{\left(\:{t}\:+{t}^{\:−\mathrm{1}} \right)^{\:\mathrm{2}} −\mathrm{2}}{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\overset{{t}\:+\frac{\mathrm{1}}{{t}}\:={u}} {=}−\int_{\mathrm{2}} ^{\:\infty} \frac{{du}}{\left(\:{u}−\sqrt{\mathrm{2}}\:\right)\left({u}+\:\sqrt{\mathrm{2}}\:\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\:\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\left[\mathrm{ln}\left(\frac{{u}\:−\sqrt{\mathrm{2}}}{{u}+\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{2}} ^{\:\infty} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{ln}\left(\:\mathrm{3}\:+\mathrm{2}\sqrt{\mathrm{2}}\:\right)\:\:\:\:\:\ast\ast \\ $$$$\:\:\:\left(\ast\right)\:\&\:\left(\ast\ast\right)::\:\:\:\:\:\:\:\Omega=\mathrm{2}\Psi=\:\:\left(\Phi\:+\:\boldsymbol{\phi}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\left(\:\pi\:−\mathrm{ln}\left(\:\mathrm{3}\:+\mathrm{2}\sqrt{\mathrm{2}}\:\right)\:\:...\blacksquare\mathrm{m}.\mathrm{n}\right. \\ $$$$ \\ $$

Question Number 173148 Answers: 1 Comments: 0

∫_0 ^1 ^n (√x) (arcsin x) dx

$$ \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:^{\mathrm{n}} \sqrt{\mathrm{x}}\:\left(\mathrm{arcsin}\:\mathrm{x}\right)\:\mathrm{dx} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 173132 Answers: 0 Comments: 5

U_n = ((((−4)^(n+1) −1)/(1−(−4)^n )))U_(n−1) with U_0 =1 find U_(n ) in terms of n

$${U}_{{n}} \:=\:\left(\frac{\left(−\mathrm{4}\right)^{{n}+\mathrm{1}} −\mathrm{1}}{\mathrm{1}−\left(−\mathrm{4}\right)^{{n}} }\right){U}_{{n}−\mathrm{1}} \:{with}\:{U}_{\mathrm{0}} =\mathrm{1} \\ $$$${find}\:{U}_{{n}\:} \:{in}\:{terms}\:{of}\:{n}\:\: \\ $$

Question Number 173129 Answers: 0 Comments: 0