Conjecture about the equality : $f(y)=y\ln(y)+\sum_{n=2}^{\infty}\pm\frac{\left(y\ln y\right)^n}{2^{a_n}}$

Question

I try here because I expect I cannot have any answer on MSE :

Problem :

Let :

$$f\left(x\right)=\frac{\left(1+\ln27\right)x!\ln x!}{x+1},g\left(x\right)=x\ln x$$

Then it seems $\exists y\in(0,1)$ and $a_n$ be an infinite set of positive integer such that :

$$f(y)=y\ln(y)+\sum_{n=2}^{\infty}\pm\frac{\left(y\ln y\right)^n}{2^{a_n}},g(x)=x\ln(x)+\sum_{n=2}^{\infty}\pm\frac{\left(x\ln x\right)^n}{2^{a_n}},f'(y)-g'(y)=0$$

For consolidate my conjecture see :

$$H\left(x\right)=x\ln x+\frac{x^{2}\ln^{2}\left(x\right)}{4}-\frac{x^{3}\left(\ln x\right)^{3}}{8}+\frac{x^{4}\left(\ln x\right)^{4}}{4\cdot8}-\frac{x^{5}\left(\ln x\right)^{5}}{4\cdot32}+\frac{x^{6}\left(\ln x\right)^{6}}{4\cdot16}-\frac{x^{7}\left(\ln x\right)^{7}}{4\cdot64}+\frac{x^{8}\left(\ln x\right)^{8}}{2048}+\frac{x^{9}\left(\ln x\right)^{9}}{2048\cdot4}-\frac{x^{10}\left(\ln x\right)^{10}}{2048\cdot2}-\frac{x^{11}\left(\ln x\right)^{11}}{2048\cdot2}-\frac{x^{12}\left(\ln x\right)^{12}}{2048\cdot8}-\frac{x^{12}\left(\ln x\right)^{12}}{2^{22}}-\frac{x^{13}\left(\ln x\right)^{13}}{2^{24}}+\frac{x^{14}\left(\ln x\right)^{14}}{2^{24}}-\frac{\left(1+\ln27\right)x!\ln\left(x!\right)}{x+1}$$

Then :

$$|H\left(0.35925959\right)|<3\cdot10^{-14},|H'\left(0.35925959\right)|<6\cdot10^{-8}$$

For this conjecture I have two question (if true):

How to show the convergence ?If yes : How to find the $a_n$ ?

Thanks everyone !

Ps: for $\pm$ it's not $(-1)^n$ as you can see for order $8,9$

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I cannot go further with Desmos Software but look at the following partial sum :

$$H\left(x\right)=x\ln x+\frac{x^{2}\ln^{2}\left(x\right)}{4}-\frac{x^{3}\left(\ln x\right)^{3}}{8}+\frac{x^{4}\left(\ln x\right)^{4}}{4\cdot8}-\frac{x^{5}\left(\ln x\right)^{5}}{4\cdot32}+\frac{x^{6}\left(\ln x\right)^{6}}{4\cdot16}-\frac{x^{7}\left(\ln x\right)^{7}}{4\cdot64}+\frac{x^{8}\left(\ln x\right)^{8}}{2048}+\frac{x^{9}\left(\ln x\right)^{9}}{2048\cdot4}-\frac{x^{10}\left(\ln x\right)^{10}}{2048\cdot2}-\frac{x^{11}\left(\ln x\right)^{11}}{2048\cdot2}-\frac{x^{12}\left(\ln x\right)^{12}}{2048\cdot8}-\frac{x^{12}\left(\ln x\right)^{12}}{2^{22}}-\frac{x^{13}\left(\ln x\right)^{13}}{2^{24}}+\frac{x^{14}\left(\ln x\right)^{14}}{2^{24}}+\frac{x^{15}\left(\ln x\right)^{15}}{2^{24}}-\frac{x^{16}\left(\ln x\right)^{16}}{2^{25}}+\frac{x^{17}\left(\ln x\right)^{17}}{2^{26}}+\frac{x^{18}\left(\ln x\right)^{18}}{2^{26}}+\frac{x^{19}\left(\ln x\right)^{19}}{2^{26}}-\frac{\left(1+\ln27\right)x!\ln\left(x!\right)}{x+1}$$

Where : $x=0.35925959\cdots$

Update 08/10/2022 :

We have a similar problem with the Binet formula slighlty modified see :

$$g\left(x\right)=\sqrt{2\pi x}\left(\frac{x}{e}\right)^{x}e^{\frac{47000000000}{116593560186976815022080000x^{13}}+\frac{4388413659}{9716130015581401251840000x^{12}}+\frac{310}{13494625021640835072000x^{11}}+\frac{390}{86504006548979712000x^{10}}+\frac{800}{514904800886784000x^{9}}+\frac{1}{86684309913600x^{8}}+\frac{2}{902961561600x^{7}}+\frac{7}{75246796800x^{6}}+\frac{1}{836075520x^{5}}+\frac{2}{24883200x^{4}}-\frac{1}{10368x^{3}}-\frac{1}{291x^{2}}+\frac{1}{12x}}-x!$$

And $g(0.285)$

Answer 1

Such as a $y$ might exist but its VERY specific to that choice of $y$ if we try to analyze the function generally we won't find any such magic set of coefficients.

So we start with $f(x) = \frac{\left( 1 + \ln(27) x! \ln(x!) \right) }{1+x}$

For the sake of our readers we will re write this to:

$$f(x) = \frac{\left( 1 + \ln(27) \Gamma(x+1) \ln(\Gamma(x+1)) \right) }{1+x}$$

We can now subtract off $ x \ln(x) $ to yield

$$ Q(x) = f(x) - x \ln(x) = \frac{\left( 1 + \ln(27) \Gamma(x+1) \ln(\Gamma(x+1)) \right) }{1+x} - x \ln(x) $$

Now we will consider the linear operator $O: \left( \mathbb{C} \rightarrow \mathbb{C} \right) \rightarrow \left( \mathbb{C} \rightarrow \mathbb{C} \right), O [f]= \frac{1}{1 + \ln(x)} \frac{df}{dx} $.

It's easy to see that $O[\left( x \ln(x) \right)^k] = k \left( x \ln(x) \right)^{k-1}$ , so its the right operator for our job. (In general the series h(x)^k is connected to the operator $\frac{1}{h'(x)} \frac{d}{dx} $).

We need to evaluate this series around either $x=0$ or $x=1$ since these are the points where $\left( x \ln(x) \right)^k = 0$ much like how $x^k = 0$ at $x=0$ in taylors theorem. We will start with looking at $x=1$ since I see some logarithms in the denominator in the near future.

$$ O[Q] = \frac{1 + 3 \ln(3) }{1+ \ln(x)} \frac{d}{dx} \left[ \frac{\Gamma(x+1) \ln(\Gamma(x+1)) }{1+x} \right] - 1 $$

$$ O[Q] = \frac{1 + 3 \ln(3) }{1+ \ln(x)} \frac{d}{dx} \left[ \frac{\Gamma(x+1) \ln(\Gamma(x+1)) }{1+x} \right] - 1 $$

At $x=1$ this is $ \frac{ 1 + 3 \ln(3)}{2} \left( 1 - \gamma \right) - 1 $ At $x=0$ this is $-1$.

However in your expansion that linear coeffient on $y \ln(y)$ is $1$.

So this will require much more specific methods relating to your choice of $y$.