Mixing
Finding the inverse of an incomplete beta function
$begingroup$
Is there a rigorous way of inverting
$$rho(r)=frac{b_{0}}{1-q}B(1-(frac{b_{0}}{r})^{1-q},frac{1}{2},frac{1}{q-1})$$
where $B(1-(frac{b_{0}}{r})^{1-q},frac{1}{2},frac{1}{q-1})$ is an incomplete beta function, $b_{0}$ is some positive constant, $-infty<q<1$. Basically I want to obtain $r(rho)$. I am thinking of getting the asymptotic expansion of $rho(r)$ (since I am interested in the regime where $r$ is very large) and then invert using Lagrange inversion method. Maybe I only need the first three terms of the expansion. Am I allowed to do this? Or is there a more convenient way of inverting the given function containing an incomplete beta function?
asymptotics special-functions inverse inverse-function beta-function
asked Jan 26 at 14:13
user583893user583893
286
$endgroup$
add a comment |
$begingroup$
Is there a rigorous way of inverting
$$rho(r)=frac{b_{0}}{1-q}B(1-(frac{b_{0}}{r})^{1-q},frac{1}{2},frac{1}{q-1})$$
where $B(1-(frac{b_{0}}{r})^{1-q},frac{1}{2},frac{1}{q-1})$ is an incomplete beta function, $b_{0}$ is some positive constant, $-infty<q<1$. Basically I want to obtain $r(rho)$. I am thinking of getting the asymptotic expansion of $rho(r)$ (since I am interested in the regime where $r$ is very large) and then invert using Lagrange inversion method. Maybe I only need the first three terms of the expansion. Am I allowed to do this? Or is there a more convenient way of inverting the given function containing an incomplete beta function?
asymptotics special-functions inverse inverse-function beta-function
asked Jan 26 at 14:13
user583893user583893
286
$endgroup$
add a comment |
$begingroup$
Is there a rigorous way of inverting
$$rho(r)=frac{b_{0}}{1-q}B(1-(frac{b_{0}}{r})^{1-q},frac{1}{2},frac{1}{q-1})$$
where $B(1-(frac{b_{0}}{r})^{1-q},frac{1}{2},frac{1}{q-1})$ is an incomplete beta function, $b_{0}$ is some positive constant, $-infty<q<1$. Basically I want to obtain $r(rho)$. I am thinking of getting the asymptotic expansion of $rho(r)$ (since I am interested in the regime where $r$ is very large) and then invert using Lagrange inversion method. Maybe I only need the first three terms of the expansion. Am I allowed to do this? Or is there a more convenient way of inverting the given function containing an incomplete beta function?
asymptotics special-functions inverse inverse-function beta-function
asked Jan 26 at 14:13
user583893user583893
286
$endgroup$
Is there a rigorous way of inverting
$$rho(r)=frac{b_{0}}{1-q}B(1-(frac{b_{0}}{r})^{1-q},frac{1}{2},frac{1}{q-1})$$
where $B(1-(frac{b_{0}}{r})^{1-q},frac{1}{2},frac{1}{q-1})$ is an incomplete beta function, $b_{0}$ is some positive constant, $-infty<q<1$. Basically I want to obtain $r(rho)$. I am thinking of getting the asymptotic expansion of $rho(r)$ (since I am interested in the regime where $r$ is very large) and then invert using Lagrange inversion method. Maybe I only need the first three terms of the expansion. Am I allowed to do this? Or is there a more convenient way of inverting the given function containing an incomplete beta function?
asymptotics special-functions inverse inverse-function beta-function
asymptotics special-functions inverse inverse-function beta-function
asked Jan 26 at 14:13
user583893user583893
286
asked Jan 26 at 14:13
user583893user583893
286
asked Jan 26 at 14:13
user583893user583893
286
asked Jan 26 at 14:13
user583893user583893
286
asked Jan 26 at 14:13
user583893user583893
286
286
add a comment |
add a comment |
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