Key Concepts and Formulas

1. Fractional Linear Transformation (Mobius Transformation): A function of the form $f(x) = \frac{ax+b}{cx+d}$ is a fractional linear transformation. A key property is that if $f(x) = \frac{ax+b}{cx+d}$, then $f(f(x)) = x$ if and only if $d = -a$. Such a function is called an involution.
2. Function Composition: $f(f(x))$ means applying the function $f$ twice. The output of the first application becomes the input for the second.
3. AM-GM Inequality: For any two non-negative real numbers $a$ and $b$, the arithmetic mean is greater than or equal to the geometric mean: $\frac{a+b}{2} \ge \sqrt{ab}$. Equality holds when $a=b$. This is used to find the minimum value of a sum.

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Step-by-Step Solution

Step 1: Analyze the structure of the given function $f(x)$ and identify its type. The function is given by: $f(x) = {{(\tan 1^\circ )x + {{\log }_e}(123)} \over {x{{\log }_e}(1234) - (\tan 1^\circ )}}, \quad x > 0$ Let's rewrite this in the standard form of a fractional linear transformation, $f(x) = \frac{ax+b}{cx+d}$. Here, we can identify: $a = \tan 1^\circ$ $b = \log_e(123)$ $c = \log_e(1234)$ $d = -\tan 1^\circ$

Step 2: Check if the function $f(x)$ is an involution. A fractional linear transformation $f(x) = \frac{ax+b}{cx+d}$ is an involution if $f(f(x)) = x$. This condition is met if $d = -a$. In our case, we have $a = \tan 1^\circ$ and $d = -\tan 1^\circ$. Thus, $d = -a$. Therefore, $f(f(x)) = x$ for all $x$ in the domain of $f(f(x))$.

Step 3: Simplify the expression $f(f(x)) + f\left( {f\left( {{4 \over x}} \right)} \right)$ using the involution property. From Step 2, we know that $f(f(x)) = x$. Now consider the term $f\left( {f\left( {{4 \over x}} \right)} \right)$. Since $f(f(y)) = y$ for any valid input $y$, we can set $y = \frac{4}{x}$. Therefore, $f\left( {f\left( {{4 \over x}} \right)} \right) = \frac{4}{x}$. The expression we need to find the least value of becomes: $f(f(x)) + f\left( {f\left( {{4 \over x}} \right)} \right) = x + \frac{4}{x}$

Step 4: Find the least value of the simplified expression $x + \frac{4}{x}$ for $x > 0$. We need to find the minimum value of $g(x) = x + \frac{4}{x}$ where $x > 0$. Since $x > 0$, both $x$ and $\frac{4}{x}$ are positive real numbers. We can apply the AM-GM inequality to these two terms. According to the AM-GM inequality: $\frac{x + \frac{4}{x}}{2} \ge \sqrt{x \cdot \frac{4}{x}}$ $\frac{x + \frac{4}{x}}{2} \ge \sqrt{4}$ $\frac{x + \frac{4}{x}}{2} \ge 2$ $x + \frac{4}{x} \ge 4$ The least value of $x + \frac{4}{x}$ is 4.

Step 5: Determine when the equality holds in the AM-GM inequality. The equality in the AM-GM inequality holds when the two terms are equal: $x = \frac{4}{x}$ $x^2 = 4$ Since $x > 0$, we have $x = 2$. At $x=2$, the value of the expression $x + \frac{4}{x}$ is $2 + \frac{4}{2} = 2+2 = 4$.

Step 6: Consider the domain of the function $f(x)$ and its composition. The problem states that $x > 0$. The function $f(x)$ is defined as long as the denominator is not zero: $x \log_e(1234) - \tan 1^\circ \neq 0$. Also, for $f(f(x))$ to be defined, $f(x)$ must be in the domain of $f$. The domain of $f$ is $x > 0$. Since $a = \tan 1^\circ > 0$, $b = \log_e(123) > 0$, $c = \log_e(1234) > 0$, and $d = -\tan 1^\circ < 0$, for $x > 0$, the numerator is positive. The denominator can be positive or negative. If $x \log_e(1234) - \tan 1^\circ > 0$, then $f(x) > 0$. If $x \log_e(1234) - \tan 1^\circ < 0$, then $f(x) < 0$. However, the problem statement implies that the expression $f(f(x)) + f\left( {f\left( {{4 \over x}} \right)} \right)$ is well-defined for $x>0$. The involution property $f(f(x))=x$ is a general property of the function's structure, irrespective of the specific values of $a, b, c, d$ as long as $ad-bc \neq 0$ and $d=-a$. The condition $ad-bc = -a^2 - bc = -(\tan^2 1^\circ) - (\log_e 123)(\log_e 1234) \neq 0$. Since $f(f(x)) = x$ and $f(f(4/x)) = 4/x$, the expression simplifies to $x + 4/x$. The domain restriction $x>0$ is sufficient for the AM-GM inequality to apply.

Step 7: Conclude the least value. The expression simplifies to $x + \frac{4}{x}$. For $x > 0$, the least value of $x + \frac{4}{x}$ is 4, achieved when $x=2$.

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Common Mistakes & Tips

• Misidentifying the Involution Property: Not recognizing that $f(f(x)) = x$ due to $d=-a$ will lead to a much more complicated calculation. Always check for this property in fractional linear transformations.
• Incorrect Application of AM-GM: Ensure that the terms to which AM-GM is applied are non-negative. In this case, $x$ and $\frac{4}{x}$ are positive for $x>0$.
• Ignoring Domain Restrictions: While the involution property holds mathematically, one should be mindful of the domain of the function and its composite for the expression to be valid. However, in this problem, the structure of the question and the required answer suggest focusing on the algebraic simplification.

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Summary

The given function $f(x)$ is a fractional linear transformation of the form $\frac{ax+b}{cx+d}$. By identifying the coefficients, we found that $a = \tan 1^\circ$ and $d = -\tan 1^\circ$, which means $d=-a$. This implies that $f(x)$ is an involution, so $f(f(x)) = x$. Consequently, $f\left( {f\left( {{4 \over x}} \right)} \right) = \frac{4}{x}$. The expression to minimize becomes $x + \frac{4}{x}$. For $x > 0$, we used the AM-GM inequality to find the least value of $x + \frac{4}{x}$, which is 4.

The final answer is \boxed{2}.