How to solve $2(k-1)-2z-z\frac{U^\prime}{U}=0$?

Question

Consider the function

$$f_X(x)= \frac{(x^2)^{\frac{k-1}{2}}e^{-\frac{x^2}{2}}}{\sqrt{\pi}\Gamma(\frac{k}{2}) 2^{\frac{k}{2}}} U(\frac{1}{2} , \frac{k+1}{2},\frac{x^2}{2}), \quad k>0, \quad x\in R,$$

where $U(a,b,z)$ is the Trichome's function.

For different values of $k$ the figure of $f(x)$ is as follows:

for $k<2$ it is obvious it has one peak at $x=0$. But for $k\geq 3$ it has two peaks. The main problem is how to find these peaks(for k>=3)?!

I want to find these peaks by solving $$\frac{\partial }{\partial x} \log f(x)=0$$

By HypergeometricU we have

$$\frac{\partial }{\partial z} U(a,b,z)=-a U(a+1,b+1,z).$$

By defining $z=x^2$ and $\frac{\partial }{\partial x} \log f(x)=\frac{\partial }{\partial z}\frac{\partial z }{\partial x} \log f(x)$ so it is obvious in $x=0$, $f^\prime (x)=0$.

So it is enough to find $$\frac{\partial }{\partial z} \log f(z)=0,$$ where $$f(z)= \frac{(z)^{\frac{k-1}{2}}e^{-\frac{z}{2}}}{\sqrt{\pi}\Gamma(\frac{k}{2}) 2^{\frac{k}{2}}} U(\frac{1}{2} , \frac{k+1}{2},\frac{z}{2}).$$

By derivation we have

\begin{eqnarray*} \frac{\partial }{\partial z} \log f(z) &=&\frac{\partial }{\partial z} \log \left( \frac{(z)^{\frac{k-1}{2}}e^{-\frac{z}{2}}}{\sqrt{\pi}\Gamma(\frac{k}{2}) 2^{\frac{k}{2}}} U(\frac{1}{2} , \frac{k+1}{2},\frac{z}{2}) \right) \\ &= & \frac{k-1}{2} \frac{\partial }{\partial z} \log z +\frac{\partial }{\partial z} \left( -\frac{z}{2} \right) + \frac{\partial }{\partial z}\log U(\frac{1}{2} , \frac{k+1}{2},\frac{z}{2}) \\ &= & \frac{k-1}{2} \frac{1}{z} -\frac{1}{2}-\frac{1}{2} \times \frac{1}{2} \times \frac{U(\frac{1}{2}+1 , \frac{k+1}{2}+1,\frac{z}{2})}{U(\frac{1}{2} , \frac{k+1}{2},\frac{z}{2})} \\ &\propto & \frac{1}{4z} \left( 2(k-1)-2z-z \frac{U(\frac{3}{2} , \frac{k+3}{2},\frac{z}{2})}{U(\frac{1}{2} , \frac{k+1}{2},\frac{z}{2})} \right) \end{eqnarray*}

The problem is How to solve $2(k-1)-2z-z \frac{U(\frac{3}{2} , \frac{k+3}{2},\frac{z}{2})}{U(\frac{1}{2} , \frac{k+1}{2},\frac{z}{2})}=0$ ? Is there a way to solve the above equation?! Is there a better way to solve the main problem(finding the peak location)?

Answer 1

This being a purely numerical problem, I prefered to directly maximize the function $$f_k(x)= \frac{(x^2)^{\frac{k-1}{2}}e^{-\frac{x^2}{2}}}{\sqrt{\pi}\,\Gamma(\frac{k}{2})\, 2^{\frac{k}{2}}}\,\, U(\frac{1}{2} , \frac{k+1}{2},\frac{x^2}{2})$$

Once the results obtained for $3 \leq k \leq 50$, a quick and dirty regression leads to a cubic polynomial in $k^{1/3}$ (with $R^2 > 0.999999$)

$$x_{\text{max}}=a+ b\, k^{1/3}+ c\, k^{2/3}+ d\, k^{3/3}$$

$$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -2.97220 & 0.06143 & \{-3.09617,-2.84823\} \\ b & +2.71735 & 0.07473 & \{+2.56653,+2.86817\} \\ c & -0.20862 & 0.02925 & \{-0.26764,-0.14960\} \\ d & +0.04433 & 0.00370 & \{+0.03687,+0.05180\} \end{array}$$

Extroplated to $k=100$, this gives $9.57927$ while the exact solution is $9.37982$. This means that solving for the zero of $$-x \frac {d}{dx} \log[f(x)]=x^2 \left(1+\frac 12\frac{U\left(\frac{3}{2},\frac{k+3}{2},\frac{x^2}{2}\right)}{ U\left(\frac{1}{2},\frac{k+1}{2},\frac{x^2}{2}\right)}\right)-k+1$$ shoulfd not present any problem using Newton method with the guess given above.

Trying for $k=200$, Newton iterates are

$$\left( \begin{array}{cc} n & x_n \\ 0 & 14.6507 \\ 1 & 13.7594 \\ 2 & 13.5657 \\ 3 & 13.5456 \\ 4 & 13.5454 \end{array} \right)$$

Edit

Looking closer, it seems that $$x_k =-\frac 9 {14}+\sqrt k$$ could be a sufficient approximation.

For $k=100$ it gives $9.35714$ (the solution being $9.37982$.

For $k=200$ it gives $13.4993$ (the solution being $13.5454$.

For $k=500$, Newton iterates are $$\left( \begin{array}{cc} n & x_n \\ 0 & 21.7178 \\ 1 & 21.7875 \\ 2 & 21.7846 \end{array} \right)$$