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Question-4954




Question Number 4954 by Dnilka228 last updated on 26/Mar/16
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Commented by Dnilka228 last updated on 26/Mar/16
(√e^2_𝛒  )×(1/4)+f^(+n) −f(n)  ((f(n_1 +(n_2 +(n_3 +...))))/(x(y−(2a)^i )))=x(y±(2−x)^σ )  2a∝e
$$\sqrt{\boldsymbol{\mathrm{e}}^{\mathrm{2}_{\boldsymbol{\rho}} } }×\frac{\mathrm{1}}{\mathrm{4}}+{f}^{+{n}} −{f}\left({n}\right) \\ $$$$\frac{{f}\left({n}_{\mathrm{1}} +\left({n}_{\mathrm{2}} +\left({n}_{\mathrm{3}} +…\right)\right)\right)}{{x}\left({y}−\left(\mathrm{2}{a}\right)^{{i}} \right)}={x}\left({y}\pm\left(\mathrm{2}−{x}\right)^{\sigma} \right) \\ $$$$\mathrm{2}{a}\propto{e} \\ $$
Answered by 123456 last updated on 26/Mar/16
x_(n+1) =xx_n +αyy_n   y_(n+1) =x_n y+xy_n +βyy_n   x_0 =1  y_0 =0
$${x}_{{n}+\mathrm{1}} ={xx}_{{n}} +\alpha{yy}_{{n}} \\ $$$${y}_{{n}+\mathrm{1}} ={x}_{{n}} {y}+{xy}_{{n}} +\beta{yy}_{{n}} \\ $$$${x}_{\mathrm{0}} =\mathrm{1} \\ $$$${y}_{\mathrm{0}} =\mathrm{0} \\ $$
Answered by FilupSmith last updated on 27/Mar/16
((f(n_1 +(n_2 +(n_3 +...))))/(x(y−(2a)^i )))=x(y±(2−x)^σ )  ∴ ((f(Σ_(t=1) ^∞ n_t ))/(x(y−(2a)^i )))=x(y±(2−x)^σ )   f(Σ_(t=1) ^∞ n_t )=x^2 (y±(2−x)^σ )(y−(2a)^i )  x, y, σ, a ∉ f,   i^2 =−1  ∴ LHS=RHS=constant
$$\frac{{f}\left({n}_{\mathrm{1}} +\left({n}_{\mathrm{2}} +\left({n}_{\mathrm{3}} +…\right)\right)\right)}{{x}\left({y}−\left(\mathrm{2}{a}\right)^{{i}} \right)}={x}\left({y}\pm\left(\mathrm{2}−{x}\right)^{\sigma} \right) \\ $$$$\therefore\:\frac{{f}\left(\underset{{t}=\mathrm{1}} {\overset{\infty} {\sum}}{n}_{{t}} \right)}{{x}\left({y}−\left(\mathrm{2}{a}\right)^{{i}} \right)}={x}\left({y}\pm\left(\mathrm{2}−{x}\right)^{\sigma} \right) \\ $$$$\:{f}\left(\underset{{t}=\mathrm{1}} {\overset{\infty} {\sum}}{n}_{{t}} \right)={x}^{\mathrm{2}} \left({y}\pm\left(\mathrm{2}−{x}\right)^{\sigma} \right)\left({y}−\left(\mathrm{2}{a}\right)^{{i}} \right) \\ $$$${x},\:{y},\:\sigma,\:{a}\:\notin\:{f},\:\:\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\therefore\:{LHS}={RHS}={constant} \\ $$

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