0
$\begingroup$

Peace be upon you,

I had the following system of equations to be solved \begin{align*} \begin{cases} \psi(\alpha)-\psi(\alpha+\beta)=c_1\\ \psi(\beta)-\psi(\alpha+\beta)=c_2 \end{cases} \end{align*} where $c_1$ and $c_2$ are constants (I should solve the above, for miscellaneous constant couples). First of all, I wanted to solve it analytically so I tested some ideas and then asked for the solution in the question "A late-diverging "approximating solution" for a system of functional equations"; but after visiting no answer I proceeded to solve it numerically, using the popular approximation of the Digamma function \begin{align*} \psi(x)= \ln(x)-\frac{1}{2x}-\frac{1}{12x^2}+\frac{1}{120x^4}-\frac{1}{252x^6}+\frac{1}{240x^8}-\frac{5}{660x^{10}}+\frac{691}{32760x^{12}}-\frac{1}{12x^{14}}+O\left(\frac{1}{x^{16}}\right) \end{align*} as you can see from the approximation order, the error in 1 can even be 1! and for the values less that 1 the story is worse; while the answer of my system ($\alpha$ and $\beta$) maybe in this interval ($(0,1]$). I wonder if someone can present a better approximation of the Beta function which can cope my task for such values.

Any lighting up ideas?

$\endgroup$
1
$\begingroup$

An approximation close to $1$ (i.e. $x=1+\epsilon$ ) is given below (from WolframAlpha) :

enter image description here

For more accuracy, you can use the Taylor's expansion : $$\psi(1+\epsilon)=\Sigma_{n=0}^{\infty} \frac{\psi^{(n)}(1)}{n!}\epsilon^n$$ with numerical values of $\psi^{(n)}(1)$

$\endgroup$
  • 1
    $\begingroup$ Let's observe that $\;\psi^{(2)}(1)=-2\,\zeta(3)$ and $\;\psi^{(4)}(1)=-24\,\zeta(5)$. You obtained the classical generating function $\;\psi(1+\epsilon)\;$ for $\,\zeta(n)$. $\endgroup$ – Raymond Manzoni Nov 1 '14 at 10:28
  • $\begingroup$ Could you please link me the wolfram page from which you have copied this equation? $\endgroup$ – hossayni Nov 1 '14 at 10:31
  • 1
    $\begingroup$ The equation is the Taylor's expansion formula that I added to my preceeding answer. wolframalpha.com/input/?i=digamma%281%2Bx%29 $\endgroup$ – JJacquelin Nov 1 '14 at 10:35
  • $\begingroup$ @JJacquelin don't you have any idea on the analytic solution of the system which I pointed to, in the other question? $\endgroup$ – hossayni Nov 1 '14 at 12:27
  • $\begingroup$ I think that there is no hope to find sufficiently accurate formulas. We can expect a rough approximate, in order to start a numerical iterative process with more chance of convergence. $\endgroup$ – JJacquelin Nov 1 '14 at 15:44
1
$\begingroup$

I dare to post a second answer because it is important to clearly separate the answers to the first and the second question raised. The subject of one question was the analytical properties of the digamma function. The subject of the other question is how to compute approximate solutions of a system of equations.

Since the inverse function of the digamma is not related to a referenced special function, there is no hope to find the solution on a closed form. So, I think that numerical calculus is required : There are several methods for numerical solving of non-linear equations, based on more or less sophisticated iterative process.

These processes require a first estimate to start : Sometimes, this is a major difficulty on the practical viewpoint.

In the present case, I propose a method to compute a rough first estimate ( principle shown in attachment). Since the unknowns $\alpha$ and $\beta$ are in the range $0$ to $1$, the Taylor's expansions are made around $\frac{1}{2}$. Of course, if it was known that the expected values are closer to $1$ (for example) the formulas should be rewritten according to it.

enter image description here

$\endgroup$
  • $\begingroup$ Thanks so much for your idea; but I think that my sentence "while the answer of my system (α and β) maybe in $(0,1]$" made you make a mistake. I only meant that my answer probably is in $(0,1]$ and it threatens my accuracy. but I do not have any limitation in my results. Though it is not an analytic answer, since it is in closed form, you may like to edit & move it to the other question hope that if no analytic way proposed you can get the bounty. About the analytic solution though there is not a close inverse for Digamma, the system may be solved by integral equation techniques... $\endgroup$ – hossayni Nov 2 '14 at 9:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.