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

Question

Samantha Adkison · Accepted Answer

The curves intersect without osculating.

Taking into account #y=x^(x(1+1/y)) equiv y^(y/(y+1))-x^x# we have #f_1(x,y)= y^(y/(y+1))-x^x = 0# and #f_2(x,y)=y-x^x=0# Those functions intersect at clearly at #{x = 1,y=1}#. This can be verified by simple substitution. At this moment, the tangency complies with #((dy)/(dx))_1 = -(f_1)_x/((f_1)_y) = ( (1 + y)^2 (1 + Log_e(x)))/(1 + y + Log_e(y)) = 2# #((dy)/(dx))_2 = -(f_2)_x/((f_2)_y) =x^x (1 + Log_e(x)) = 1# As a result, they cross without osculating.

Aurora Alvarado · Answer

undefined To prove that the curves of y=x^x and y = the Functional Continued Fraction (FCF) generated by y=x^(x(1+1/y)) touch y=x at their common point (1, 1), we need to show that both curves pass through this point and have the same slope at that point.

First, let's evaluate the point (1, 1) on both curves:

For y=x^x:
When x=1, y=1^1=1. So, the point (1, 1) lies on the curve y=x^x.

For y = the FCF generated by y=x^(x(1+1/y)):
When x=1, y=1^(1(1+1/1))=1^2=1. So, the point (1, 1) also lies on this curve.

Now, let's find the slope of both curves at the point (1, 1):

For y=x^x:
To find the slope at (1, 1), we differentiate y=x^x with respect to x:
dy/dx = x^x * (1 + ln(x))

Substituting x=1, we get:
dy/dx = 1^1 * (1 + ln(1)) = 1 * (1 + 0) = 1

For y = the FCF generated by y=x^(x(1+1/y)):
To find the slope at (1, 1), we differentiate y=x^(x(1+1/y)) with respect to x:
dy/dx = x^(x(1+1/y)) * (1 + 1/y - ln(x)/y^2)

Substituting x=1 and y=1, we get:
dy/dx = 1^(1(1+1/1)) * (1 + 1/1 - ln(1)/1^2) = 1 * (1 + 1 - 0) = 2

Since both curves pass through the point (1, 1) and have the same slope of 1 at that point, we can conclude that the curves of y=x^x and y = the FCF generated by y=x^(x(1+1/y)) touch y=x at their common point (1, 1).