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I want to find an expression for:

$$ (1) \quad \underset{x}{argmax} \Big \{ \frac{1}{2b} \Big(\frac{x}{Na}\Big)^{\frac{N-1}{2}} e^{- \frac{1}{2b} (x+Na)} I_{N-1}\Big( \frac{\sqrt{Nax}}{b} \Big) \Big \} $$

Where $a,b,x \in \mathbb{R_{++}}$, $N \in \mathbb{N}$, and $I_{N-1}(\cdot)$ is the modified Bessel function of the first kind of order $N-1$.

Since this function is concave in $x$, all I need to do is take the partial derivative w.r.t $x$, set it equal to zero, and solve for $x$. Unfortunately, this seems like a pretty intractable problem. Try computing the $x$ partial in Wolfram to see what I mean.

So, I am going to approximate the answer instead. I know from NIST that for large orders:

$$I_{N-1}(z) \sim \frac{1}{\sqrt{2 \pi (N-1)}} \Big( \frac{e z}{2 (N-1)} \Big)^{N-1}$$

So then I have:

$$ (2) \quad \underset{x}{argmax} \Big \{ \frac{1}{2b} \Big(\frac{x}{Na}\Big)^{\frac{N-1}{2}} e^{- \frac{1}{2b} (x+Na)} \frac{1}{\sqrt{2 \pi (N-1)}} \Big( \frac{\frac{e}{b} \sqrt{Nax} }{2 (N-1)} \Big)^{N-1} \Big \} $$

This problem is easy. After taking the logarithm, expanding terms, taking the derivative, and setting it to zero, I get:

$$ x^{\ast} = 2b (N-1) $$

This is exact for (2), but nowhere near the truth for (1), especially as $N$ gets large and $a,b \approx 1$. This can be checked numerically. I think this is because the NIST expansion doesn't account for the argument growing as well.

Questions:

  1. How can I solve (1) analytically? Can you solve it exactly via the derivative?
  2. Is there an expansion, similar to the one used in (2), for when both the order and argument of the Bessel function are growing?
  3. Any other ideas on how I can approach this problem? Maybe approximate the log of the Bessel function directly?

Thanks!

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We simplify slightly the expression by writing \begin{align} &\frac{1}{2b} \left(\frac{x}{Na}\right)^{\frac{N-1}{2}} e^{- \frac{1}{2b} (x+Na)} I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)=\frac{1}{2b}\left( Na \right)^{\frac{1-N}{2}}e^{-\frac{Na}{2b}}f(x)\\ &f(x)=x^{\frac{N-1}{2}}e^{-\frac{x}{2b}} I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right) \end{align} Then, we want to obtain the solutions of $f'(x)=0$. The derivative reads \begin{equation} f'(x)=\frac{N-1}{2}\frac{f(x)}{x}-\frac{1}{2b}f(x)+\frac{\sqrt{Na}}{2b\sqrt{x}}\frac{I'_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}{I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}f(x) \end{equation} The equation to solve is thus \begin{equation} \frac{N-1}{x}-\frac{1}{b}+\frac{\sqrt{Na}}{b\sqrt{x}}\frac{I'_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}{I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}=0 \end{equation} When both the order and the argument are large, a uniform asymptotic expansion can be used for $I_{N-1}$ and $I_{N-1}$: \begin{align} I_{N-1}\left((N-1) z\right)&\sim\frac{e^{(N-1)\eta}}{(2\pi(N-1))^{\frac{1}{2}}(1+z^{2% })^{\frac{1}{4}}}\sum_{k=0}^{\infty}\frac{U_{k}(p)}{(N-1)^{k}}\\ I_{N-1}'\left((N-1) z\right)&\sim\frac{(1+z^{2})^{\frac{1}{4}}e^{(N-1)\eta}}{(2\pi% (N-1))^{\frac{1}{2}}z}\sum_{k=0}^{\infty}\frac{V_{k}(p)}{(N-1)^{k}} \end{align} where \begin{equation} \eta=(1+z^{2})^{\frac{1}{2}}+\ln\frac{z}{1+(1+z^{2})^{\frac{1}{2}}} \end{equation} Here, we choose \begin{equation} z=\frac{\sqrt{Nax}}{b(N-1)} \end{equation} Keeping the term $k=0$ only, we find (with $U_0(p)=V_0(p)=1$) \begin{equation} \frac{I'_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}{I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}=\frac{\sqrt{1+z^2}}{z} \end{equation} The equation to be solved becomes \begin{equation} \frac{N-1}{x}-\frac{1}{b}+\frac{\sqrt{Na}}{b\sqrt{x}} \sqrt{1+\frac{b^2(N-1)^2}{Nax}}=0 \end{equation} or \begin{equation} N-1-\frac{x}{b}+ \sqrt{\frac{Nax}{b^2}+(N-1)^2}=0 \end{equation} which has a simple non-zero solution \begin{equation} x=Na+2(N-1)b \end{equation} which seems to be numerically correct. For $a=1,b=1,N=20$, the approximation gives $x=58$ while a numerical evaluation of the solution is $x=57.49$. For $a=1,b=1,N=50$ we find $148$ to be compared to $147.496$.

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  • $\begingroup$ Wow that is amazing. I wish I could do that. I have so much to learn... $\endgroup$ – The Dude Sep 20 '19 at 0:44
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    $\begingroup$ We all have so much to learn, in so many domains... $\endgroup$ – Paul Enta Sep 20 '19 at 6:53

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