Let $r>1$ be real and \begin{align*} f_1(x) &= 1,\\ f_2(x) &= (x+4)^r,\\ f_3(x) &=(x+4)^r(x+3)^r,\\ f_4(x) &= (x+4)^r(x+3)^r(x+2)^r,\\ f_5(x) &=(x+4)^r(x+3)^r(x+2)^r(x+1)^r. \end{align*} How can I show that these function form a T-system on $(0,\infty)$, i.e., $$\det[f_j(x_i)]\ne 0$$ for every mutually distinct $0<x_1,\ldots,x_5$?
I am looking for a good hint for a solution (not a full solution) to this, and its generalization.
A simple sufficient criteria is to verify that the system of functions is $M_{+}$ system, i.e., for each $k=1, \ldots, 4$ the determinants of the matrices $\{f_{i}^{(j-1)}(t)\}_{i,j=1}^{k+1}$ are positive on $(0, \infty)$ (see the discussion right before Theorem 1.1 here and the references therein https://arxiv.org/pdf/2201.12932 )
One could write a script to verify signs of these simple determinants. I used LLMs to do this computation. I checked it and it looks correct to me (but please also double check for yourself). Here are the Wronskians:
Let
$$
f_1(x)=1,\quad
f_2(x)=(x+4)^r,\quad
f_3(x)=(x+4)^r(x+3)^r,\quad
f_4(x)=(x+4)^r(x+3)^r(x+2)^r,\quad
f_5(x)=(x+4)^r(x+3)^r(x+2)^r(x+1)^r .
$$
We denote by
$
\Delta_k(x)=\det\big(f_i^{(j-1)}(x)\big)_{i,j=1}^{k+1}
$
the $(k+1)\times (k+1)$ Wronskian determinant.
Since $f_1\equiv1$, each $\Delta_k$ reduces to the usual Wronskian of $(f_2,\dots,f_{k+1})$.
Order $k=1$ $$ \Delta_1(x)=W(f_2)=f_2'(x)=r(x+4)^{r-1}>0. $$
Order $k=2$. Using the identity $W(u,ug)=u^2 g'$ with $u=(x+4)^r$, $g=(x+3)^r$, we get $$ \Delta_2(x)=W(f_2,f_3) =r(x+3)^{\,r-1}(x+4)^{\,2r}>0. $$
Order $k=3$. A direct computation yields $$ \Delta_3(x) =\frac{r^2\,(x+2)^{\,r}(x+3)^{\,2r}(x+4)^{\,3r}}{(x+2)^2(x+3)^2}\, \bigl(2r x+5r-1\bigr), $$ and since $r>1$ one has $2r x+5r-1>0$ for all $x>0$. Hence $\Delta_3(x)>0$.
Order $k=4$. A longer but straightforward calculation gives $$ \Delta_4(x) =\frac{r^3\,(x+1)^{\,r}(x+2)^{\,2r}(x+3)^{\,3r}(x+4)^{\,4r}} {(x+1)^3(x+2)^4(x+3)^3}\, Q_r(x), $$ where $$ \begin{aligned} Q_r(x)=&\;12r^3x^4+(96r^3-24r^2)x^3+(281r^3-144r^2+19r)x^2\\ &\quad+(356r^3-284r^2+76r-4)x+(165r^3-184r^2+75r-8). \end{aligned} $$ For every $r>1$ all coefficients of $Q_r(x)$ are positive, hence $Q_r(x)>0$ on $(0,\infty)$.
Therefore, for $r>1$ we have $\Delta_k(x)>0$ for $k=1,2,3,4$ and all $x>0$, so $(f_1,\dots,f_5)$ satisfies the sufficient Wronskian positivity condition for an $M_{+}$ (or $T_{+}$) system on $(0,\infty)$.
