Given #f(x) = cosh(x+a/cosh(x+a/cosh( cdots)))# and #g(x)# its inverse, what is the minimum distance between then for #a > 0#?

Answer 1

Our goal is to find the shortest path between the function and

# f(x) = cosh(x+a/cosh(x+a/cosh( x+a/cosh(cdots))) ) #
and its inverse, for #a gt 0#
Let, #y=f(x)#, Then we can write this repeated recursive function definition concisely as:
# y = cosh(x+a/y ) = coshw#, say where #w=x+a/y#
Now, we attempt to find the inverse function, #f^(-1)(x)#. Using the definition of the hyperbolic cosine:
# cosh w = (e^w+e^-w)/2 #

so then:

# y = coshw = (e^w+e^-w)/2 #
# :. 2y = e^w+e^-w #
# :. 2ye^w = e^(2w)+1 #
# :. e^(2w) -2ye^w + 1 = 0 #
# :. (e^(w) -y)^2 - y^2+1= 0 #
# :. (e^(w) -y)^2 = y^2-1 #
# :. e^(w) -y = +-sqrt(y^2-1) #
# :. e^(w) = y +-sqrt(y^2-1) #
# :. w = ln(y +-sqrt(y^2-1)) #
# :. x+a/y = ln(y +-sqrt(y^2-1)) #
# :. x = ln(y +-sqrt(y^2-1)) -a/y #
Therefore although we cannot write an exact function for the original repeated recursive function, #f(x)#, we can write an exact form for its inverse, #f^(-1)(x)#
# :. f^(-1)(x) = ln(x +-sqrt(x^2-1)) -a/x #
In order to minimize the distance between any function and its inverse, which are reflection in the line #y=x#, we can reduce the problem to that of minimizing the distance, #D#, say, between either function and #y=x#, then the distance we seek is #2D#.
Now, the distance from the point #(p,q)# and the line #Ax+By+C=0# is given by:
# D = |Ap+Bq+C|/sqrt(A^2+B^2) #
So, we can find the distance between a generic coordinate #(x,f^(-1)(x))# on #f^(-1)(x)# and #y=x# (or #x-y=0#), using:
# D = |-x+ln(x +-sqrt(x^2-1)) -a/x|/sqrt(1^2+(-1)^2) #
# :. 2D^2 = ln(x +-sqrt(x^2-1)) -a/x - x #
Differentiating wrt #x#, we have:
# 4(dD)/(dx) = ( 1+- x/sqrt(x^2-1) )/(x +-sqrt(x^2-1))+1/x^2-1 #

We need the derivative to vanish at a certain point in order for one of the two possible outcomes to occur:

# ( 1+ x/sqrt(x^2-1) )/(x +sqrt(x^2-1))+1/x^2-1 = 0 #
# ( 1- x/sqrt(x^2-1) )/(x -sqrt(x^2-1))+1/x^2-1 = 0 #
Solving these equations (which are independent of #a#), we get approximate real solutions:
# x ~~ +- 1.7742 #
From which we can calculate the corresponding value of #D#.

If I have more time, I will go back to the solution.

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Answer 2

Assuming that y = g(x) ( in my view f(x) ) is the inverse, the distance is 0, everywhere.

Note on Disambiguation:

Writing a relation as y = f(x) is known as inversion.

x = inverse of y = #f^(-1)(y)#,
for an x-interval, in which the relation is bijective, #1 - 1#.
f is a function operator and #f^(-1)# is its inverse operator. So,
the successive operations # f^(-1) f# and #ff^(-1)# both
produce the multiplication unit operator 1. Also, #f^(-1) f^(-1)# is f.

Thus, if y equals f(x),

#x = f^(-1)f(x) = f^(-1)(f(x)) = f^(-1)(y)#

I believe it is past time to end the calling.

#y = f^(-1)(x)#,
obtained by the swapping #(x, y) to (y, x)#,

as the opposite.

The fact that the graphs for both are identical is significant.

This graph is rotated and altered by switching x and y.

Given that the two graphs are one and the distance is zero, I certify that

similar. For the shared graph, a = 1. It was necessary to define the inverse.

in two pieces, #x >=0# and #x <= 0#, as
#x = ln(y+(y^2-1)^0.5)# and #x = - ln(y+(y^2-1)^0.5)#,

correspondingly.

Y = cosh(x+1/y) = the provided FCF (Functional Continued Function) graph

The expansion of a fraction f(x) is as follows: graph{(x - ln(y+(y^2-1)^0.5))(x + ln(y+(y^2-1)^0.5))=0}

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Answer 3

The minimum distance between f(x) and g(x) for a > 0 is 2a.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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