Let $f(n)$ be a real-valued sequence defined for $n \in \mathbb{N}$, with $f(n) > 0$ for all $n$. Define a new sequence: $$ g(n) = \log_b(f(n)) $$
I know that when $0 < b < 1$, the logarithmic function is strictly decreasing. Thus, if $x_1 < x_2$, then $$ \log_{1/2}(x_1) > \log_{1/2}(x_2) $$
Logically this should means that when $f(n)$ attains a minimum (maximum), then $g(n)$ attains a maximum (minimum)?
After if I have a $b>1$, if $f(n)$ have a minimum (maximum) then $g(n)$ have a maximum (minimum)?
For example: I want found the infimum and supremum of the following set, specifying whether they correspond to a maximum and/or minimum: $$ X = \left\{ \log_{\frac{1}{2}}(n^2 - 7n + 13) : n \in \mathbb{N} \right\} $$ I observe that the argument of the logarithm is the parabola: $$ f(n) = n^2 - 7n + 13, $$ which opens upwards (the coefficient of $n^2$ is positive), thus it has an absolute minimum at its vertex. Let's calculate it: $$ n_V = \frac{7}{2} = 3.5, \quad f(3.5) = (3.5)^2 - 7 \cdot 3.5 + 13 = 12.25 - 24.5 + 13 = 0.75. $$ Since $n \in \mathbb{N}$, consider the nearest integers: $$ f(3) = 9 - 21 + 13 = 1, \quad f(4) = 16 - 28 + 13 = 1. $$ Therefore, the minimum value taken by the parabola for $n \in \mathbb{N}$ is 1, reached at $n=3$ and $n=4$. Now consider the function: $$ g(n) = \log_{\frac{1}{2}}(f(n)) = \log_{\frac{1}{2}}(n^2 - 7n + 13). $$ Since the base of the logarithm is in the interval $]0,1[$, the logarithmic function is strictly decreasing. Hence, if $x_1 < x_2$ then $$ \log_{\frac{1}{2}}(x_1) > \log_{\frac{1}{2}}(x_2). $$
This means that when $f(n)$ is minimum, $g(n)$ is maximum?
$$ \begin{array}{c|c|c} n & f(n) = n^2 - 7n + 13 & g(n) = \log_{\frac{1}{2}}(f(n)) \\ \hline 1 & 7 & \approx -2.807 \\ 2 & 3 & \approx -1.585 \\ 3 & 1 & 0 \\ 4 & 1 & 0 \\ 5 & 3 & \approx -1.585 \\ 6 & 7 & \approx -2.807 \\ 7 & 13 & \approx -3.700 \\ 8 & 21 & \approx -4.393 \\ 9 & 31 & \approx -4.954 \\ 10 & 43 & \approx -5.426 \\ \end{array} $$
The supremum of $X$ is $\sup X = \max X = 0$, corresponding to the maximum value reached at $n=3$ and $n=4$.
The infimum of $X$ is $\inf X = -\infty$, because $f(n) \to +\infty$ as $n \to +\infty$ and thus $g(n) = \log_{\frac{1}{2}}(f(n)) \to -\infty$. There is no minimum because the infimum $-\infty$ is not attained by any element of the set.
Same question for the exponential function $$g(n) = b^{f(n)}.$$ for the exercise:
$$X=\left\{\left(\frac12\right)^{n^2-7n+4} \mid n\in \mathbb N\right\}$$
looking the exponent of $1/2$.
I have create table $$ \begin{array}{|c|c|c|c|} \hline \textbf{Function} & \textbf{Base interval} & \textbf{Monotonicity} & \textbf{Effect on min/max of } f(n) \\ \hline g(n)=\log_b(f(n)) & 0 < b < 1 & \text{decreasing } f & \text{min of } f \rightarrow \text{max of } g \\ g(n)=\log_b(f(n)) & 0 < b < 1 & \text{increasing } f& \text{max of } f \rightarrow \text{min of } g \\ g(n)=\log_b(f(n)) & b > 1 & \text{decreasing } f& \text{min of } f \rightarrow \text{min of } g \\ g(n)=\log_b(f(n)) & b > 1 & \text{increasing } f& \text{max of } f \rightarrow \text{max of } g \\ b^{f(n)} & 0 < b < 1 & \text{decreasing } f & \text{min of } f \rightarrow \text{max of } g \\ b^{f(n)} & 0 < b < 1 & \text{increasing } f& \text{max of } f \rightarrow \text{min of } g \\ b^{f(n)} & b > 1 & \text{decreasing } f& \text{min of } f \rightarrow \text{min of } g\\ b^{f(n)} & b > 1 & \text{increasing } f& \text{max of } f \rightarrow \text{max of } g \\ \hline \end{array} $$
to understand whether there exist theorems or proofs that logically establish the correctness of the assertions about how the monotonicity of $f$ affects $g$, without relying on derivatives, and if such theorems are available.
This concept for me is important to found the $\inf X (\min X)$, $\sup X (\max X)$ of a generic set $X$ that involves with exponentials and logarithms.
