Minimum of the Sum of Logarithms

In summary, the minimum of the given expression can be found by setting all the variables to be equal to each other and minimizing the function $\displaystyle f(x) = \log_{x} (x - \frac{1}{4})$. This can be done by finding the critical point of the function, which is achieved when $\displaystyle x = \frac{1}{2}$. Therefore, the minimum value of the given expression is achieved when all the variables are equal to $\frac{1}{2}$.
  • #1
anemone
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Find the minimum of $\large \log_{a_1}\left(a_2-\dfrac{1}{4}\right)+\log_{a_2}\left(a_3-\dfrac{1}{4}\right)+\cdots+\log_{a_n}\left(a_1-\dfrac{1}{4}\right)$ where $a_1,\,a_2,\cdots,a_n$ are real numbers in the interval $\left(\dfrac{1}{4},\,1\right)$.
 
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  • #2
anemone said:
Find the minimum of $\large \log_{a_1}\left(a_2-\dfrac{1}{4}\right)+\log_{a_2}\left(a_3-\dfrac{1}{4}\right)+\cdots+\log_{a_n}\left(a_1-\dfrac{1}{4}\right)$ where $a_1,\,a_2,\cdots,a_n$ are real numbers in the interval $\left(\dfrac{1}{4},\,1\right)$.

[sp]Attempt ... because changing the order of the variables the problem remains the same, the solution will be such that $ a_ {1} = a_ {2} = ... = a_ {n} = x $ and the function to be minimized is...

$\displaystyle f(x) = log_{x} (x - \frac{1}{4})\ (1)$

Proceeding in standard fashion is...

$\displaystyle f^{\ '} (x) = \frac{\frac{\ln x}{x - \frac{1}{4}} - \frac{\ln (x - \frac{1}{4})}{x}}{\ln ^{2} x}\ (2)$

... and the (2) vanishes for...

$\displaystyle x\ \ln x = (x - \frac{1}{4})\ \ln (x - \frac{1}{4}) \implies x = \frac{1}{2}\ (3)$[/sp]

Kind regards

$\chi$ $\sigma$
 
  • #3
chisigma said:
[sp]Attempt ... because changing the order of the variables the problem remains the same, the solution will be such that $ a_ {1} = a_ {2} = ... = a_ {n} = x $ and the function to be minimized is...

$\displaystyle f(x) = log_{x} (x - \frac{1}{4})\ (1)$

Proceeding in standard fashion is...

$\displaystyle f^{\ '} (x) = \frac{\frac{\ln x}{x - \frac{1}{4}} - \frac{\ln (x - \frac{1}{4})}{x}}{\ln ^{2} x}\ (2)$

... and the (2) vanishes for...

$\displaystyle x\ \ln x = (x - \frac{1}{4})\ \ln (x - \frac{1}{4}) \implies x = \frac{1}{2}\ (3)$[/sp]

Kind regards

$\chi$ $\sigma$

Thanks chisigma for participating in this challenge!

And in your method,the minimum of $\large \log_{a_1}\left(a_2-\dfrac{1}{4}\right)+\log_{a_2}\left(a_3-\dfrac{1}{4}\right)+\cdots+\log_{a_n}\left(a_1-\dfrac{1}{4}\right)$ where $a_1,\,a_2,\cdots,a_n$ are real numbers in the interval $\left(\dfrac{1}{4},\,1\right)$ would be $2n$.

Here is the solution of other that I wanted to share:

Since $\log_m a$ is a decreasing function of $a$ when $0<m<1$, and since $\left(a-\dfrac{1}{2}\right)^2\ge 0$ which implies $a^2\ge a-\dfrac{1}{4}$, we have

$\large \log_{a_k}\left(a_{k+1}-\dfrac{1}{4}\right)\ge \log_{a_k} a_{k+1}^2=2\log_{a_k} a_{k+1}=2\dfrac{\log a_{k+1}}{\log a_{k}}$

It follows that

$\begin{align*}\log_{a_1}\left(a_2-\dfrac{1}{4}\right)+\log_{a_2}\left(a_3-\dfrac{1}{4}\right)+\cdots+\log_{a_n}\left(a_1-\dfrac{1}{4}\right)&\ge 2\left(\dfrac{\log a_2}{\log a_1}+\dfrac{\log a_3}{\log a_2}+\cdots+\dfrac{\log a_n}{\log a_{n-1}}+\dfrac{\log a_1}{\log a_n}\right)\\&\ge 2n\,\,\,\text{by AM-GM inequality}\end{align*}$

Equalities hold iff $a_1=a_2=\cdots=a_n=\dfrac{1}{2}$.
 

Related to Minimum of the Sum of Logarithms

1. What is the "Minimum of the Sum of Logarithms" (MSL)?

The MSL is a mathematical concept that involves finding the minimum value of a sum of logarithms. It is often used in optimization problems and can be solved using techniques such as gradient descent.

2. How is MSL different from other optimization methods?

MSL is unique in that it involves taking the logarithm of the variables in the objective function, which can help to simplify and solve complex problems. It is also more robust and efficient compared to other methods, especially for non-linear and non-convex problems.

3. What are the applications of MSL?

MSL can be applied to various fields such as engineering, economics, and machine learning. It is commonly used in optimization problems, such as finding the best parameters for a machine learning model or maximizing profits in a business.

4. How do you solve for the minimum of the sum of logarithms?

To solve for the minimum of the sum of logarithms, you can use techniques such as gradient descent or Newton's method. These methods involve iteratively updating the values of the variables until the minimum is reached. Alternatively, you can also use mathematical software or programming languages to solve for the minimum.

5. Are there any limitations to using MSL?

While MSL is a powerful optimization method, it may not be suitable for all types of problems. It is most effective for problems with a single global minimum, and may not work well for problems with multiple local minima. In addition, it may require a significant amount of computation time for complex problems.

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