How do you prove that the curves of $y = x^{x}$ $y=x^x$ and y = the Functional Continued Fraction (FCF) generated by $y = x^{x (1 + \frac{1}{y})}$ $y=x^(x(1+1/y))$ touch $y = x$ $y=x$ , at their common point ( 1, 1 )?

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How do you prove that the curves of $y = x^{x}$ $y=x^x$ and y = the Functional Continued Fraction (FCF) generated by $y = x^{x (1 + \frac{1}{y})}$ $y=x^(x(1+1/y))$ touch $y = x$ $y=x$ , at their common point ( 1, 1 )?

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Answer 1 · 2016-08-27T16:16:49

The curves intersect without osculating.

Explanation:

Considering

$y = x^{x (1 + \frac{1}{y})} \equiv y^{\frac{y}{y + 1}} - x^{x}$ $y=x^(x(1+1/y)) equiv y^(y/(y+1))-x^x$ we have

$f_{1} (x, y) = y^{\frac{y}{y + 1}} - x^{x} = 0$ $f_1(x,y)= y^(y/(y+1))-x^x = 0$ and
$f_{2} (x, y) = y - x^{x} = 0$ $f_2(x,y)=y-x^x=0$

Those functions intersect at clearly at ${x = 1, y = 1}$ ${x = 1,y=1}$ . This can be verified by simple substitution.

Now the tangency at this point obeys

${(\frac{d y}{d x})}_{1} = - \frac{{(f_{1})}_{x}}{{(f_{1})}_{y}} = \frac{{(1 + y)}^{2} (1 + {log}_{e} (x))}{1 + y + {log}_{e} (y)} = 2$ $((dy)/(dx))_1 = -(f_1)_x/((f_1)_y) = ( (1 + y)^2 (1 + Log_e(x)))/(1 + y + Log_e(y)) = 2$
${(\frac{d y}{d x})}_{2} = - \frac{{(f_{2})}_{x}}{{(f_{2})}_{y}} = x^{x} (1 + {log}_{e} (x)) = 1$ $((dy)/(dx))_2 = -(f_2)_x/((f_2)_y) =x^x (1 + Log_e(x)) = 1$

So they intersect without osculating.

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How do you prove that the curves of $y = x^{x}$ $y=x^x$ and y = the Functional Continued Fraction (FCF) generated by $y = x^{x (1 + \frac{1}{y})}$ $y=x^(x(1+1/y))$ touch $y = x$ $y=x$ , at their common point ( 1, 1 )?

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Explanation:

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How do you prove that the curves of y=xxy=x^x and y = the Functional Continued Fraction (FCF) generated by y=xx(1+1y)y=x^(x(1+1/y)) touch y=xy=x, at their common point ( 1, 1 )?

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How do you prove that the curves of $y = x^{x}$ $y=x^x$ and y = the Functional Continued Fraction (FCF) generated by $y = x^{x (1 + \frac{1}{y})}$ $y=x^(x(1+1/y))$ touch $y = x$ $y=x$ , at their common point ( 1, 1 )?