When is a * x * Ln(x) = b * (1-x) Ln(1-x) with 0<x<1?

[BJU]

TL : DR: What is the general expression for the null?

I have the function

f(x) = a * x * Ln(x) - b * (1-x) Ln(1-x) = 0

with 0 < x < 1 and a, b > 0.

I know for a = b = 1 this looks like this:



I can show that the general shape of the function (down-up-down, starting and ending at 0) will remain irrespective of a and b.

The first three derivatives are

f(x)' = a + b + a Ln[x]+ b Ln[1 - x]

f(x)'' = a/x - b/(1 - x)

f(x)''' = - a/x² - b/(1 - x)²

Since f(x)''' is always negative, f(x)' must be concave. Since lim f(x)' is negative infinity as x approaches 0 from the right and 1 from the left, it must be positive in-between since f(0) = f(1) = 0.

Since f(a/(a+b))'' = 0, the inflection point is always at x = a/(a+b).

So I do know the general shape.

But, if possible, I would like to get an expression for the null with a and b.

Obviously, increasing a and b will move the null further right and left, respectively, since the partial derivatives of f(x) are positive and negative, respectively, but I cannot think of a way to get a general expression for the null.

Does anyone have any suggestions?

 

[JeffM]

I am not completely sure what you mean by the null. If you mean the zeroes or roots, notice that the function is defined only if 0 < x < 1.

If f(p) = 0, then

[MATH]ap ln(p) - b(1 - p)ln(1 - p) = 0 \implies[/MATH]
[MATH]ap ln(p) = b(1 - p)ln(1 - p) \implies[/MATH]
[MATH]ln(p^{ap}) = ln\{(1 - p)^{(b-bp)}\} \implies[/MATH]
[MATH]p^{ap} = (1 - p)^{(b-bp)}.[/MATH]
I see no way to get a closed form solution in a and b. In fact, even if you know a and b, I think you will have to find the root by numerical methods.

 

[BJU]

Yes, I'm looking for f(x)=0. I thought null was the same as zero and root, sorry (I'm not a native speaker).

I was afraid that this would not be possible, but it's good to have confirmation, thank you!