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Question Number 1117 by 123456 last updated on 17/Jun/15
a_0 =1 a_n =((√a_(n−1) )/a_(n−1) )+1 lim_(n→+∞) a_n =?
$${a}_{\mathrm{0}} =\mathrm{1} \\ $$$${a}_{{n}} =\frac{\sqrt{{a}_{{n}−\mathrm{1}} }}{{a}_{{n}−\mathrm{1}} }+\mathrm{1} \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{a}_{{n}} =? \\ $$
Commented by 123456 last updated on 17/Jun/15
x>0 x=((√x)/x)+1 x^2 −x=(√x) x^4 −2x^3 +x^2 =x x^4 −2x^3 +x^2 −x=0 x(x^3 −2x^2 +x−1)=0 x=0∨x^3 −2x^2 +x−1=0 f(x)=x^3 −2x^2 +x−1 f(−2)=−19 f(+2)=1 ∃ξ∈(−2,+2),f(ξ)=0 f(x)=x^3 −2x^2 +x−1⇒?,?,? f′(x)=3x^2 −4x+1⇒(1/3),1 f′′(x)=6x−4⇒(2/3) f′′′(x)=6
$${x}>\mathrm{0} \\ $$$${x}=\frac{\sqrt{{x}}}{{x}}+\mathrm{1} \\ $$$${x}^{\mathrm{2}} −{x}=\sqrt{{x}} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} ={x} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}=\mathrm{0} \\ $$$${x}\left({x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\vee{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}=\mathrm{0} \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1} \\ $$$${f}\left(−\mathrm{2}\right)=−\mathrm{19} \\ $$$${f}\left(+\mathrm{2}\right)=\mathrm{1} \\ $$$$\exists\xi\in\left(−\mathrm{2},+\mathrm{2}\right),{f}\left(\xi\right)=\mathrm{0} \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}\Rightarrow?,?,? \\ $$$${f}'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\mathrm{3}},\mathrm{1} \\ $$$${f}''\left({x}\right)=\mathrm{6}{x}−\mathrm{4}\Rightarrow\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${f}'''\left({x}\right)=\mathrm{6} \\ $$
Commented by 123456 last updated on 17/Jun/15
f(x)=a(x−x_1 )(x−x_2 )(x−x_3 )⇒x_1 ,x_2 ,x_3 f′(x)=a[(x−x_2 )(x−x_3 )+(x−x_1 )(x−x_3 )+(x−x_1 )(x−x_2 )]⇒?,? f′′(x)=2a[3x−(x_1 +x_2 +x_3 )]⇒((x_1 +x_2 +x_3 )/3)
$${f}\left({x}\right)={a}\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)\left({x}−{x}_{\mathrm{3}} \right)\Rightarrow{x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{x}_{\mathrm{3}} \\ $$$${f}'\left({x}\right)={a}\left[\left({x}−{x}_{\mathrm{2}} \right)\left({x}−{x}_{\mathrm{3}} \right)+\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{3}} \right)+\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)\right]\Rightarrow?,? \\ $$$${f}''\left({x}\right)=\mathrm{2}{a}\left[\mathrm{3}{x}−\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} \right)\right]\Rightarrow\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} }{\mathrm{3}} \\ $$
Commented by 123456 last updated on 17/Jun/15
f=x^4 −2x^3 +x^2 −x f^((3)) =24x−12⇒(1/2) g=x^3 −2x^2 +x−1 g^((2)) =6x−4⇒(2/3) { ((a+b+c=2)),((ab+ac+bc=1)),((abc=1)) :}
$${f}={x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x} \\ $$$${f}^{\left(\mathrm{3}\right)} =\mathrm{24}{x}−\mathrm{12}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${g}={x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1} \\ $$$${g}^{\left(\mathrm{2}\right)} =\mathrm{6}{x}−\mathrm{4}\Rightarrow\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\begin{cases}{{a}+{b}+{c}=\mathrm{2}}\\{{ab}+{ac}+{bc}=\mathrm{1}}\\{{abc}=\mathrm{1}}\end{cases} \\ $$
Answered by prakash jain last updated on 17/Jun/15
x^3 −2x^2 +x−1=0 x=(1/6)(4+2^(2/3) ((25−3(√(69))))^(1/3) +2^(2/3) ((25+3(√(69))))^(1/3) ) From cubic formula for ax^3 +bx^2 +cx+d=0.
$${x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{4}+\mathrm{2}^{\mathrm{2}/\mathrm{3}} \sqrt[{\mathrm{3}}]{\mathrm{25}−\mathrm{3}\sqrt{\mathrm{69}}}+\mathrm{2}^{\mathrm{2}/\mathrm{3}} \:\sqrt[{\mathrm{3}}]{\mathrm{25}+\mathrm{3}\sqrt{\mathrm{69}}}\right) \\ $$$$\mathrm{From}\:\mathrm{cubic}\:\mathrm{formula}\:\mathrm{for}\:{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0}. \\ $$

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