Mixing
How to find a specific inverse function.
$begingroup$
I am developing a new probability distribution. The probability density function is as follows:
$$f(x) , = ,
frac{x^{r+1}}{(r+1)^{x}}frac{ln(r+1)^{r+2}}{
(r+1)!}$$ with $0<x$ and $r = 1, , 2, , 3,, ... ,$ . The distribution function is as follows:
$$F(x) = 1 - sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
Define $G(x) = 1 - F(x)$:
$$G(x) = sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
I am searching for the inverse of $G(x)$. Does a solution exist?
functions inverse
asked Jan 28 at 11:33
Ad van der VenAd van der Ven
23
$endgroup$
add a comment |
$begingroup$
I am developing a new probability distribution. The probability density function is as follows:
$$f(x) , = ,
frac{x^{r+1}}{(r+1)^{x}}frac{ln(r+1)^{r+2}}{
(r+1)!}$$ with $0<x$ and $r = 1, , 2, , 3,, ... ,$ . The distribution function is as follows:
$$F(x) = 1 - sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
Define $G(x) = 1 - F(x)$:
$$G(x) = sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
I am searching for the inverse of $G(x)$. Does a solution exist?
functions inverse
asked Jan 28 at 11:33
Ad van der VenAd van der Ven
23
$endgroup$
add a comment |
$begingroup$
I am developing a new probability distribution. The probability density function is as follows:
$$f(x) , = ,
frac{x^{r+1}}{(r+1)^{x}}frac{ln(r+1)^{r+2}}{
(r+1)!}$$ with $0<x$ and $r = 1, , 2, , 3,, ... ,$ . The distribution function is as follows:
$$F(x) = 1 - sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
Define $G(x) = 1 - F(x)$:
$$G(x) = sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
I am searching for the inverse of $G(x)$. Does a solution exist?
functions inverse
asked Jan 28 at 11:33
Ad van der VenAd van der Ven
23
$endgroup$
I am developing a new probability distribution. The probability density function is as follows:
$$f(x) , = ,
frac{x^{r+1}}{(r+1)^{x}}frac{ln(r+1)^{r+2}}{
(r+1)!}$$ with $0<x$ and $r = 1, , 2, , 3,, ... ,$ . The distribution function is as follows:
$$F(x) = 1 - sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
Define $G(x) = 1 - F(x)$:
$$G(x) = sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
I am searching for the inverse of $G(x)$. Does a solution exist?
functions inverse
functions inverse
asked Jan 28 at 11:33
Ad van der VenAd van der Ven
23
asked Jan 28 at 11:33
Ad van der VenAd van der Ven
23
edited Jan 29 at 8:41
Ad van der Ven
asked Jan 28 at 11:33
Ad van der VenAd van der Ven
23
asked Jan 28 at 11:33
Ad van der VenAd van der Ven
23
asked Jan 28 at 11:33
Ad van der VenAd van der Ven
23
23
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For $r=1$,
$$G(x)= 2^{-x}left(1+ x ln2+ frac12(x ln2)^2right)$$
is of the form $2^{-x}P(x)$ where $P$ is a polynomial. Such functions are not invertible analytically. The same holds for $r>1$.
answered Jan 28 at 14:21
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Yves Daoust Thank you for your valuable answer.
$endgroup$
– Ad van der Ven
Jan 28 at 14:48
add a comment |
$begingroup$
May I start remarking that it looks for me like some Poisson distribution:
$$G(x) = sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
Now, to start set $a:= ln(r+1) >0 $ then
$$G(x) = sum_{k=0}^{r+1} e^{-ax} frac{(ax)^k}{k!}= e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} $$
This shows (which is needed later), that $G$ can be expressed as a product of two power series and thus can be written as a power series itself.
Now let us get the derivative of $G$ and find out if it is monotonic:
$$G'(x) = -a e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} + e^{-ax} sum_{k=1}^{r+1} afrac{(ax)^{k-1}}{(k-1)!} = \
-a e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} + ae^{-ax} sum_{k=0}^{r} frac{(ax)^k}{k!} =\
=-a e^{-ax}frac{(ax)^{r+1}}{(r+1)!}
$$
Then if $r>0$ we have $a =ln(r+1) > 0$ and hence the the last line gets negative for positive $x>0$ . Under this condition for $r$ and $x$ then $G'$ is negative on $mathbb{R}^+$, thus it is monotonic (decreasing) and therefore invertable.
Furthermore as has been noted previously that $G$ can be expressed as a power series, we could apply the Lagrange–Bürmann formula to find the coefficients of the inverse function's power series coefficients.
Addendum: Please note that
$$G(x) = e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} = frac{Gamma(r+2,ax)}{ Gamma(r+2)}=Q(r+2,ax)$$ (see here)where $Gamma(.,..)$ is the incomplete Gamma function, $Gamma(.)$ is the well known Gamma function, and $Q$ is the regulized Gamma function. All these functions are available by many math libraries, and their inverses are so as well. Just in case you need explicit expressions take a look here.
answered Jan 28 at 19:06
MaksimMaksim
94019
$endgroup$
$begingroup$
Maksim, you concluded that $$G(x) = e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} $$ The derivative of this function is $$ G'(x) = frac{ a^{2} e^{-ax} (ax)^{r} x}{(r+1)!} $$ This is not the same as the derivative you gave: $$G'(x) = (1-a)e^{-ax}sum_{k=0}^{r} frac{(ax)^k}{k!}-a frac{(ax)^{r+1}}{(r+1)!} $$ Did I make an error?
$endgroup$
– Ad van der Ven
Jan 29 at 10:01
$begingroup$
You didn't - but I did. My apologies. I corrected the calculation, and I think we now are in line. :-)
$endgroup$
– Maksim
Jan 29 at 10:37
$begingroup$
I intend to write an article about the probability distribution above and I need the inverse of the function $G$ to be able to simulate $x$ values. How many coefficients of the power series of the inverse function do you think I will need?
$endgroup$
– Ad van der Ven
Jan 29 at 12:31
$begingroup$
Pls. have a look at my addendum in the answer.
$endgroup$
– Maksim
Jan 29 at 20:46
$begingroup$
On [look here][1] I read at the end: The inverse of the regularized incomplete gamma function $Q^{-1}(a,z)$ satisfies the following ordinary nonlinear second-order differential equation: $$w(z) w(z)'' - w(z)'^{2} (w(z)-a+1) = 0$$ with $w(z) = Q^{-1}(a,z)$. You conluded that $$G(x) = frac{Gamma(r+2,ax)}{ Gamma(r+2)}=Q(r+2,a x)$$ My question is now: how to proceed further. I am aware that $a$ in the first equation is not the same as $a$ in the second equation. [1]: functions.wolfram.com/GammaBetaErf/InverseGammaRegularized3/…
$endgroup$
– Ad van der Ven
Jan 30 at 14:50
|
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
oldest
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$begingroup$
For $r=1$,
$$G(x)= 2^{-x}left(1+ x ln2+ frac12(x ln2)^2right)$$
is of the form $2^{-x}P(x)$ where $P$ is a polynomial. Such functions are not invertible analytically. The same holds for $r>1$.
answered Jan 28 at 14:21
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Yves Daoust Thank you for your valuable answer.
$endgroup$
– Ad van der Ven
Jan 28 at 14:48
add a comment |
$begingroup$
For $r=1$,
$$G(x)= 2^{-x}left(1+ x ln2+ frac12(x ln2)^2right)$$
is of the form $2^{-x}P(x)$ where $P$ is a polynomial. Such functions are not invertible analytically. The same holds for $r>1$.
answered Jan 28 at 14:21
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Yves Daoust Thank you for your valuable answer.
$endgroup$
– Ad van der Ven
Jan 28 at 14:48
add a comment |
$begingroup$
For $r=1$,
$$G(x)= 2^{-x}left(1+ x ln2+ frac12(x ln2)^2right)$$
is of the form $2^{-x}P(x)$ where $P$ is a polynomial. Such functions are not invertible analytically. The same holds for $r>1$.
answered Jan 28 at 14:21
Yves DaoustYves Daoust
131k676229
$endgroup$
For $r=1$,
$$G(x)= 2^{-x}left(1+ x ln2+ frac12(x ln2)^2right)$$
is of the form $2^{-x}P(x)$ where $P$ is a polynomial. Such functions are not invertible analytically. The same holds for $r>1$.
answered Jan 28 at 14:21
Yves DaoustYves Daoust
131k676229
answered Jan 28 at 14:21
Yves DaoustYves Daoust
131k676229
answered Jan 28 at 14:21
Yves DaoustYves Daoust
131k676229
answered Jan 28 at 14:21
Yves DaoustYves Daoust
131k676229
131k676229
$begingroup$
Yves Daoust Thank you for your valuable answer.
$endgroup$
– Ad van der Ven
Jan 28 at 14:48
add a comment |
$begingroup$
Yves Daoust Thank you for your valuable answer.
$endgroup$
– Ad van der Ven
Jan 28 at 14:48
$begingroup$
Yves Daoust Thank you for your valuable answer.
$endgroup$
– Ad van der Ven
Jan 28 at 14:48
$begingroup$
Yves Daoust Thank you for your valuable answer.
$endgroup$
– Ad van der Ven
Jan 28 at 14:48
add a comment |
$begingroup$
May I start remarking that it looks for me like some Poisson distribution:
$$G(x) = sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
Now, to start set $a:= ln(r+1) >0 $ then
$$G(x) = sum_{k=0}^{r+1} e^{-ax} frac{(ax)^k}{k!}= e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} $$
This shows (which is needed later), that $G$ can be expressed as a product of two power series and thus can be written as a power series itself.
Now let us get the derivative of $G$ and find out if it is monotonic:
$$G'(x) = -a e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} + e^{-ax} sum_{k=1}^{r+1} afrac{(ax)^{k-1}}{(k-1)!} = \
-a e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} + ae^{-ax} sum_{k=0}^{r} frac{(ax)^k}{k!} =\
=-a e^{-ax}frac{(ax)^{r+1}}{(r+1)!}
$$
Then if $r>0$ we have $a =ln(r+1) > 0$ and hence the the last line gets negative for positive $x>0$ . Under this condition for $r$ and $x$ then $G'$ is negative on $mathbb{R}^+$, thus it is monotonic (decreasing) and therefore invertable.
Furthermore as has been noted previously that $G$ can be expressed as a power series, we could apply the Lagrange–Bürmann formula to find the coefficients of the inverse function's power series coefficients.
Addendum: Please note that
$$G(x) = e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} = frac{Gamma(r+2,ax)}{ Gamma(r+2)}=Q(r+2,ax)$$ (see here)where $Gamma(.,..)$ is the incomplete Gamma function, $Gamma(.)$ is the well known Gamma function, and $Q$ is the regulized Gamma function. All these functions are available by many math libraries, and their inverses are so as well. Just in case you need explicit expressions take a look here.
answered Jan 28 at 19:06
MaksimMaksim
94019
$endgroup$
$begingroup$
Maksim, you concluded that $$G(x) = e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} $$ The derivative of this function is $$ G'(x) = frac{ a^{2} e^{-ax} (ax)^{r} x}{(r+1)!} $$ This is not the same as the derivative you gave: $$G'(x) = (1-a)e^{-ax}sum_{k=0}^{r} frac{(ax)^k}{k!}-a frac{(ax)^{r+1}}{(r+1)!} $$ Did I make an error?
$endgroup$
– Ad van der Ven
Jan 29 at 10:01
$begingroup$
You didn't - but I did. My apologies. I corrected the calculation, and I think we now are in line. :-)
$endgroup$
– Maksim
Jan 29 at 10:37
$begingroup$
I intend to write an article about the probability distribution above and I need the inverse of the function $G$ to be able to simulate $x$ values. How many coefficients of the power series of the inverse function do you think I will need?
$endgroup$
– Ad van der Ven
Jan 29 at 12:31
$begingroup$
Pls. have a look at my addendum in the answer.
$endgroup$
– Maksim
Jan 29 at 20:46
$begingroup$
On [look here][1] I read at the end: The inverse of the regularized incomplete gamma function $Q^{-1}(a,z)$ satisfies the following ordinary nonlinear second-order differential equation: $$w(z) w(z)'' - w(z)'^{2} (w(z)-a+1) = 0$$ with $w(z) = Q^{-1}(a,z)$. You conluded that $$G(x) = frac{Gamma(r+2,ax)}{ Gamma(r+2)}=Q(r+2,a x)$$ My question is now: how to proceed further. I am aware that $a$ in the first equation is not the same as $a$ in the second equation. [1]: functions.wolfram.com/GammaBetaErf/InverseGammaRegularized3/…
$endgroup$
– Ad van der Ven
Jan 30 at 14:50
|
show 3 more comments
$begingroup$
May I start remarking that it looks for me like some Poisson distribution:
$$G(x) = sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
Now, to start set $a:= ln(r+1) >0 $ then
$$G(x) = sum_{k=0}^{r+1} e^{-ax} frac{(ax)^k}{k!}= e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} $$
This shows (which is needed later), that $G$ can be expressed as a product of two power series and thus can be written as a power series itself.
Now let us get the derivative of $G$ and find out if it is monotonic:
$$G'(x) = -a e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} + e^{-ax} sum_{k=1}^{r+1} afrac{(ax)^{k-1}}{(k-1)!} = \
-a e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} + ae^{-ax} sum_{k=0}^{r} frac{(ax)^k}{k!} =\
=-a e^{-ax}frac{(ax)^{r+1}}{(r+1)!}
$$
Then if $r>0$ we have $a =ln(r+1) > 0$ and hence the the last line gets negative for positive $x>0$ . Under this condition for $r$ and $x$ then $G'$ is negative on $mathbb{R}^+$, thus it is monotonic (decreasing) and therefore invertable.
Furthermore as has been noted previously that $G$ can be expressed as a power series, we could apply the Lagrange–Bürmann formula to find the coefficients of the inverse function's power series coefficients.
Addendum: Please note that
$$G(x) = e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} = frac{Gamma(r+2,ax)}{ Gamma(r+2)}=Q(r+2,ax)$$ (see here)where $Gamma(.,..)$ is the incomplete Gamma function, $Gamma(.)$ is the well known Gamma function, and $Q$ is the regulized Gamma function. All these functions are available by many math libraries, and their inverses are so as well. Just in case you need explicit expressions take a look here.
answered Jan 28 at 19:06
MaksimMaksim
94019
$endgroup$
$begingroup$
Maksim, you concluded that $$G(x) = e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} $$ The derivative of this function is $$ G'(x) = frac{ a^{2} e^{-ax} (ax)^{r} x}{(r+1)!} $$ This is not the same as the derivative you gave: $$G'(x) = (1-a)e^{-ax}sum_{k=0}^{r} frac{(ax)^k}{k!}-a frac{(ax)^{r+1}}{(r+1)!} $$ Did I make an error?
$endgroup$
– Ad van der Ven
Jan 29 at 10:01
$begingroup$
You didn't - but I did. My apologies. I corrected the calculation, and I think we now are in line. :-)
$endgroup$
– Maksim
Jan 29 at 10:37
$begingroup$
I intend to write an article about the probability distribution above and I need the inverse of the function $G$ to be able to simulate $x$ values. How many coefficients of the power series of the inverse function do you think I will need?
$endgroup$
– Ad van der Ven
Jan 29 at 12:31
$begingroup$
Pls. have a look at my addendum in the answer.
$endgroup$
– Maksim
Jan 29 at 20:46
$begingroup$
On [look here][1] I read at the end: The inverse of the regularized incomplete gamma function $Q^{-1}(a,z)$ satisfies the following ordinary nonlinear second-order differential equation: $$w(z) w(z)'' - w(z)'^{2} (w(z)-a+1) = 0$$ with $w(z) = Q^{-1}(a,z)$. You conluded that $$G(x) = frac{Gamma(r+2,ax)}{ Gamma(r+2)}=Q(r+2,a x)$$ My question is now: how to proceed further. I am aware that $a$ in the first equation is not the same as $a$ in the second equation. [1]: functions.wolfram.com/GammaBetaErf/InverseGammaRegularized3/…
$endgroup$
– Ad van der Ven
Jan 30 at 14:50
|
show 3 more comments
$begingroup$
May I start remarking that it looks for me like some Poisson distribution:
$$G(x) = sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
Now, to start set $a:= ln(r+1) >0 $ then
$$G(x) = sum_{k=0}^{r+1} e^{-ax} frac{(ax)^k}{k!}= e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} $$
This shows (which is needed later), that $G$ can be expressed as a product of two power series and thus can be written as a power series itself.
Now let us get the derivative of $G$ and find out if it is monotonic:
$$G'(x) = -a e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} + e^{-ax} sum_{k=1}^{r+1} afrac{(ax)^{k-1}}{(k-1)!} = \
-a e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} + ae^{-ax} sum_{k=0}^{r} frac{(ax)^k}{k!} =\
=-a e^{-ax}frac{(ax)^{r+1}}{(r+1)!}
$$
Then if $r>0$ we have $a =ln(r+1) > 0$ and hence the the last line gets negative for positive $x>0$ . Under this condition for $r$ and $x$ then $G'$ is negative on $mathbb{R}^+$, thus it is monotonic (decreasing) and therefore invertable.
Furthermore as has been noted previously that $G$ can be expressed as a power series, we could apply the Lagrange–Bürmann formula to find the coefficients of the inverse function's power series coefficients.
Addendum: Please note that
$$G(x) = e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} = frac{Gamma(r+2,ax)}{ Gamma(r+2)}=Q(r+2,ax)$$ (see here)where $Gamma(.,..)$ is the incomplete Gamma function, $Gamma(.)$ is the well known Gamma function, and $Q$ is the regulized Gamma function. All these functions are available by many math libraries, and their inverses are so as well. Just in case you need explicit expressions take a look here.
answered Jan 28 at 19:06
MaksimMaksim
94019
$endgroup$
May I start remarking that it looks for me like some Poisson distribution:
$$G(x) = sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$
Now, to start set $a:= ln(r+1) >0 $ then
$$G(x) = sum_{k=0}^{r+1} e^{-ax} frac{(ax)^k}{k!}= e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} $$
This shows (which is needed later), that $G$ can be expressed as a product of two power series and thus can be written as a power series itself.
Now let us get the derivative of $G$ and find out if it is monotonic:
$$G'(x) = -a e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} + e^{-ax} sum_{k=1}^{r+1} afrac{(ax)^{k-1}}{(k-1)!} = \
-a e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} + ae^{-ax} sum_{k=0}^{r} frac{(ax)^k}{k!} =\
=-a e^{-ax}frac{(ax)^{r+1}}{(r+1)!}
$$
Then if $r>0$ we have $a =ln(r+1) > 0$ and hence the the last line gets negative for positive $x>0$ . Under this condition for $r$ and $x$ then $G'$ is negative on $mathbb{R}^+$, thus it is monotonic (decreasing) and therefore invertable.
Furthermore as has been noted previously that $G$ can be expressed as a power series, we could apply the Lagrange–Bürmann formula to find the coefficients of the inverse function's power series coefficients.
Addendum: Please note that
$$G(x) = e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} = frac{Gamma(r+2,ax)}{ Gamma(r+2)}=Q(r+2,ax)$$ (see here)where $Gamma(.,..)$ is the incomplete Gamma function, $Gamma(.)$ is the well known Gamma function, and $Q$ is the regulized Gamma function. All these functions are available by many math libraries, and their inverses are so as well. Just in case you need explicit expressions take a look here.
answered Jan 28 at 19:06
MaksimMaksim
94019
edited Jan 29 at 20:45
answered Jan 28 at 19:06
MaksimMaksim
94019
answered Jan 28 at 19:06
MaksimMaksim
94019
answered Jan 28 at 19:06
MaksimMaksim
94019
94019
$begingroup$
Maksim, you concluded that $$G(x) = e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} $$ The derivative of this function is $$ G'(x) = frac{ a^{2} e^{-ax} (ax)^{r} x}{(r+1)!} $$ This is not the same as the derivative you gave: $$G'(x) = (1-a)e^{-ax}sum_{k=0}^{r} frac{(ax)^k}{k!}-a frac{(ax)^{r+1}}{(r+1)!} $$ Did I make an error?
$endgroup$
– Ad van der Ven
Jan 29 at 10:01
$begingroup$
You didn't - but I did. My apologies. I corrected the calculation, and I think we now are in line. :-)
$endgroup$
– Maksim
Jan 29 at 10:37
$begingroup$
I intend to write an article about the probability distribution above and I need the inverse of the function $G$ to be able to simulate $x$ values. How many coefficients of the power series of the inverse function do you think I will need?
$endgroup$
– Ad van der Ven
Jan 29 at 12:31
$begingroup$
Pls. have a look at my addendum in the answer.
$endgroup$
– Maksim
Jan 29 at 20:46
$begingroup$
On [look here][1] I read at the end: The inverse of the regularized incomplete gamma function $Q^{-1}(a,z)$ satisfies the following ordinary nonlinear second-order differential equation: $$w(z) w(z)'' - w(z)'^{2} (w(z)-a+1) = 0$$ with $w(z) = Q^{-1}(a,z)$. You conluded that $$G(x) = frac{Gamma(r+2,ax)}{ Gamma(r+2)}=Q(r+2,a x)$$ My question is now: how to proceed further. I am aware that $a$ in the first equation is not the same as $a$ in the second equation. [1]: functions.wolfram.com/GammaBetaErf/InverseGammaRegularized3/…
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– Ad van der Ven
Jan 30 at 14:50
|
show 3 more comments
$begingroup$
Maksim, you concluded that $$G(x) = e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} $$ The derivative of this function is $$ G'(x) = frac{ a^{2} e^{-ax} (ax)^{r} x}{(r+1)!} $$ This is not the same as the derivative you gave: $$G'(x) = (1-a)e^{-ax}sum_{k=0}^{r} frac{(ax)^k}{k!}-a frac{(ax)^{r+1}}{(r+1)!} $$ Did I make an error?
$endgroup$
– Ad van der Ven
Jan 29 at 10:01
$begingroup$
You didn't - but I did. My apologies. I corrected the calculation, and I think we now are in line. :-)
$endgroup$
– Maksim
Jan 29 at 10:37
$begingroup$
I intend to write an article about the probability distribution above and I need the inverse of the function $G$ to be able to simulate $x$ values. How many coefficients of the power series of the inverse function do you think I will need?
$endgroup$
– Ad van der Ven
Jan 29 at 12:31
$begingroup$
Pls. have a look at my addendum in the answer.
$endgroup$
– Maksim
Jan 29 at 20:46
$begingroup$
On [look here][1] I read at the end: The inverse of the regularized incomplete gamma function $Q^{-1}(a,z)$ satisfies the following ordinary nonlinear second-order differential equation: $$w(z) w(z)'' - w(z)'^{2} (w(z)-a+1) = 0$$ with $w(z) = Q^{-1}(a,z)$. You conluded that $$G(x) = frac{Gamma(r+2,ax)}{ Gamma(r+2)}=Q(r+2,a x)$$ My question is now: how to proceed further. I am aware that $a$ in the first equation is not the same as $a$ in the second equation. [1]: functions.wolfram.com/GammaBetaErf/InverseGammaRegularized3/…
$endgroup$
– Ad van der Ven
Jan 30 at 14:50
$begingroup$
Maksim, you concluded that $$G(x) = e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} $$ The derivative of this function is $$ G'(x) = frac{ a^{2} e^{-ax} (ax)^{r} x}{(r+1)!} $$ This is not the same as the derivative you gave: $$G'(x) = (1-a)e^{-ax}sum_{k=0}^{r} frac{(ax)^k}{k!}-a frac{(ax)^{r+1}}{(r+1)!} $$ Did I make an error?
$endgroup$
– Ad van der Ven
Jan 29 at 10:01
$begingroup$
Maksim, you concluded that $$G(x) = e^{-ax} sum_{k=0}^{r+1} frac{(ax)^k}{k!} $$ The derivative of this function is $$ G'(x) = frac{ a^{2} e^{-ax} (ax)^{r} x}{(r+1)!} $$ This is not the same as the derivative you gave: $$G'(x) = (1-a)e^{-ax}sum_{k=0}^{r} frac{(ax)^k}{k!}-a frac{(ax)^{r+1}}{(r+1)!} $$ Did I make an error?
$endgroup$
– Ad van der Ven
Jan 29 at 10:01
$begingroup$
You didn't - but I did. My apologies. I corrected the calculation, and I think we now are in line. :-)
$endgroup$
– Maksim
Jan 29 at 10:37
$begingroup$
You didn't - but I did. My apologies. I corrected the calculation, and I think we now are in line. :-)
$endgroup$
– Maksim
Jan 29 at 10:37
$begingroup$
I intend to write an article about the probability distribution above and I need the inverse of the function $G$ to be able to simulate $x$ values. How many coefficients of the power series of the inverse function do you think I will need?
$endgroup$
– Ad van der Ven
Jan 29 at 12:31
$begingroup$
I intend to write an article about the probability distribution above and I need the inverse of the function $G$ to be able to simulate $x$ values. How many coefficients of the power series of the inverse function do you think I will need?
$endgroup$
– Ad van der Ven
Jan 29 at 12:31
$begingroup$
Pls. have a look at my addendum in the answer.
$endgroup$
– Maksim
Jan 29 at 20:46
$begingroup$
Pls. have a look at my addendum in the answer.
$endgroup$
– Maksim
Jan 29 at 20:46
$begingroup$
On [look here][1] I read at the end: The inverse of the regularized incomplete gamma function $Q^{-1}(a,z)$ satisfies the following ordinary nonlinear second-order differential equation: $$w(z) w(z)'' - w(z)'^{2} (w(z)-a+1) = 0$$ with $w(z) = Q^{-1}(a,z)$. You conluded that $$G(x) = frac{Gamma(r+2,ax)}{ Gamma(r+2)}=Q(r+2,a x)$$ My question is now: how to proceed further. I am aware that $a$ in the first equation is not the same as $a$ in the second equation. [1]: functions.wolfram.com/GammaBetaErf/InverseGammaRegularized3/…
$endgroup$
– Ad van der Ven
Jan 30 at 14:50
$begingroup$
On [look here][1] I read at the end: The inverse of the regularized incomplete gamma function $Q^{-1}(a,z)$ satisfies the following ordinary nonlinear second-order differential equation: $$w(z) w(z)'' - w(z)'^{2} (w(z)-a+1) = 0$$ with $w(z) = Q^{-1}(a,z)$. You conluded that $$G(x) = frac{Gamma(r+2,ax)}{ Gamma(r+2)}=Q(r+2,a x)$$ My question is now: how to proceed further. I am aware that $a$ in the first equation is not the same as $a$ in the second equation. [1]: functions.wolfram.com/GammaBetaErf/InverseGammaRegularized3/…
$endgroup$
– Ad van der Ven
Jan 30 at 14:50
|
show 3 more comments
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