Approximate a solution for $frac{e^{-a}(v+u)(a^x+e^{a}xGamma(x,a))}{Gamma(x+1)}-uapprox 0$
Is it possible to approximate (or even find) a solution for the following equation:
$$frac{e^{-a}(v+u)(a^x+e^{a}xGamma(x,a))}{Gamma(x+1)}-uapprox0,$$
where $xge 0$ and integer, and the parameters $0<a<1$ and $u>0$, $v>0$.
It should be solved for $x$, can it be expressed in terms of the parameters?
Numerical methods show that such solution exists. For example, for $x=1$, $a=0.7$, $v=96.5543$ and $u=523$ the term value is $0.0246$.
Note: $x$ can be taken as a real number and then use the floor or ceiling functions to define the final expression.
calculus real-analysis exponential-function roots approximation-theory
asked Nov 30 at 22:05
Y.L
597
|
show 5 more comments
Is it possible to approximate (or even find) a solution for the following equation:
$$frac{e^{-a}(v+u)(a^x+e^{a}xGamma(x,a))}{Gamma(x+1)}-uapprox0,$$
where $xge 0$ and integer, and the parameters $0<a<1$ and $u>0$, $v>0$.
It should be solved for $x$, can it be expressed in terms of the parameters?
Numerical methods show that such solution exists. For example, for $x=1$, $a=0.7$, $v=96.5543$ and $u=523$ the term value is $0.0246$.
Note: $x$ can be taken as a real number and then use the floor or ceiling functions to define the final expression.
calculus real-analysis exponential-function roots approximation-theory
asked Nov 30 at 22:05
Y.L
597
1
Within you equation there is a $p$ used but you did not defined it.
– mrtaurho
Nov 30 at 22:20
2
Can you give a numerical example? I'm not even sure which variable you're solving for. Also, what would be considered a good approximation?
– Maxim
Dec 1 at 5:11
1
Using $v=96.52509765$ while keeping all of the other values the same gets you much closer to zero. Also, all solutions for $x$ will only depend on $a$ and the ratio $r=v/u$. To see that divide both sides of the equation by $u$.
– JimB
Dec 1 at 23:54
1
Not sure there's much more than noting that the ratio $u/v$ can be written in terms of $x$ and $a$: $frac{-a^x+e^a Gamma (x+1)-e^a x Gamma (x,a)}{a^x+e^a x Gamma (x,a)}$. Plotting $a$ vs $r$ for various integer values of $x$ might be informative.
– JimB
Dec 2 at 2:31
1
If $a$ and $q = u/v$ are fixed and you're solving for $x$, it seems the lhs as a function of $x$ is monotonically increasing on $(-1, infty)$ and always has a root there, finding which should be straightforward (the root may be negative). If you're allowing $x < -1$, the function has multiple zeroes on $(-infty, -1)$, finding the zero closest to an integer will be more tricky.
– Maxim
Dec 2 at 3:41
|
show 5 more comments
Is it possible to approximate (or even find) a solution for the following equation:
$$frac{e^{-a}(v+u)(a^x+e^{a}xGamma(x,a))}{Gamma(x+1)}-uapprox0,$$
where $xge 0$ and integer, and the parameters $0<a<1$ and $u>0$, $v>0$.
It should be solved for $x$, can it be expressed in terms of the parameters?
Numerical methods show that such solution exists. For example, for $x=1$, $a=0.7$, $v=96.5543$ and $u=523$ the term value is $0.0246$.
Note: $x$ can be taken as a real number and then use the floor or ceiling functions to define the final expression.
calculus real-analysis exponential-function roots approximation-theory
asked Nov 30 at 22:05
Y.L
597
Is it possible to approximate (or even find) a solution for the following equation:
$$frac{e^{-a}(v+u)(a^x+e^{a}xGamma(x,a))}{Gamma(x+1)}-uapprox0,$$
where $xge 0$ and integer, and the parameters $0<a<1$ and $u>0$, $v>0$.
It should be solved for $x$, can it be expressed in terms of the parameters?
Numerical methods show that such solution exists. For example, for $x=1$, $a=0.7$, $v=96.5543$ and $u=523$ the term value is $0.0246$.
Note: $x$ can be taken as a real number and then use the floor or ceiling functions to define the final expression.
calculus real-analysis exponential-function roots approximation-theory
calculus real-analysis exponential-function roots approximation-theory
asked Nov 30 at 22:05
Y.L
597
asked Nov 30 at 22:05
Y.L
597
edited Dec 2 at 5:22
asked Nov 30 at 22:05
Y.L
597
asked Nov 30 at 22:05
Y.L
597
asked Nov 30 at 22:05
Y.L
597
597
1
Within you equation there is a $p$ used but you did not defined it.
– mrtaurho
Nov 30 at 22:20
2
Can you give a numerical example? I'm not even sure which variable you're solving for. Also, what would be considered a good approximation?
– Maxim
Dec 1 at 5:11
1
Using $v=96.52509765$ while keeping all of the other values the same gets you much closer to zero. Also, all solutions for $x$ will only depend on $a$ and the ratio $r=v/u$. To see that divide both sides of the equation by $u$.
– JimB
Dec 1 at 23:54
1
Not sure there's much more than noting that the ratio $u/v$ can be written in terms of $x$ and $a$: $frac{-a^x+e^a Gamma (x+1)-e^a x Gamma (x,a)}{a^x+e^a x Gamma (x,a)}$. Plotting $a$ vs $r$ for various integer values of $x$ might be informative.
– JimB
Dec 2 at 2:31
1
If $a$ and $q = u/v$ are fixed and you're solving for $x$, it seems the lhs as a function of $x$ is monotonically increasing on $(-1, infty)$ and always has a root there, finding which should be straightforward (the root may be negative). If you're allowing $x < -1$, the function has multiple zeroes on $(-infty, -1)$, finding the zero closest to an integer will be more tricky.
– Maxim
Dec 2 at 3:41
|
show 5 more comments
1
Within you equation there is a $p$ used but you did not defined it.
– mrtaurho
Nov 30 at 22:20
2
Can you give a numerical example? I'm not even sure which variable you're solving for. Also, what would be considered a good approximation?
– Maxim
Dec 1 at 5:11
1
Using $v=96.52509765$ while keeping all of the other values the same gets you much closer to zero. Also, all solutions for $x$ will only depend on $a$ and the ratio $r=v/u$. To see that divide both sides of the equation by $u$.
– JimB
Dec 1 at 23:54
1
Not sure there's much more than noting that the ratio $u/v$ can be written in terms of $x$ and $a$: $frac{-a^x+e^a Gamma (x+1)-e^a x Gamma (x,a)}{a^x+e^a x Gamma (x,a)}$. Plotting $a$ vs $r$ for various integer values of $x$ might be informative.
– JimB
Dec 2 at 2:31
1
If $a$ and $q = u/v$ are fixed and you're solving for $x$, it seems the lhs as a function of $x$ is monotonically increasing on $(-1, infty)$ and always has a root there, finding which should be straightforward (the root may be negative). If you're allowing $x < -1$, the function has multiple zeroes on $(-infty, -1)$, finding the zero closest to an integer will be more tricky.
– Maxim
Dec 2 at 3:41
1
1
Within you equation there is a $p$ used but you did not defined it.
– mrtaurho
Nov 30 at 22:20
Within you equation there is a $p$ used but you did not defined it.
– mrtaurho
Nov 30 at 22:20
2
2
Can you give a numerical example? I'm not even sure which variable you're solving for. Also, what would be considered a good approximation?
– Maxim
Dec 1 at 5:11
Can you give a numerical example? I'm not even sure which variable you're solving for. Also, what would be considered a good approximation?
– Maxim
Dec 1 at 5:11
1
1
Using $v=96.52509765$ while keeping all of the other values the same gets you much closer to zero. Also, all solutions for $x$ will only depend on $a$ and the ratio $r=v/u$. To see that divide both sides of the equation by $u$.
– JimB
Dec 1 at 23:54
Using $v=96.52509765$ while keeping all of the other values the same gets you much closer to zero. Also, all solutions for $x$ will only depend on $a$ and the ratio $r=v/u$. To see that divide both sides of the equation by $u$.
– JimB
Dec 1 at 23:54
1
1
Not sure there's much more than noting that the ratio $u/v$ can be written in terms of $x$ and $a$: $frac{-a^x+e^a Gamma (x+1)-e^a x Gamma (x,a)}{a^x+e^a x Gamma (x,a)}$. Plotting $a$ vs $r$ for various integer values of $x$ might be informative.
– JimB
Dec 2 at 2:31
Not sure there's much more than noting that the ratio $u/v$ can be written in terms of $x$ and $a$: $frac{-a^x+e^a Gamma (x+1)-e^a x Gamma (x,a)}{a^x+e^a x Gamma (x,a)}$. Plotting $a$ vs $r$ for various integer values of $x$ might be informative.
– JimB
Dec 2 at 2:31
1
1
If $a$ and $q = u/v$ are fixed and you're solving for $x$, it seems the lhs as a function of $x$ is monotonically increasing on $(-1, infty)$ and always has a root there, finding which should be straightforward (the root may be negative). If you're allowing $x < -1$, the function has multiple zeroes on $(-infty, -1)$, finding the zero closest to an integer will be more tricky.
– Maxim
Dec 2 at 3:41
If $a$ and $q = u/v$ are fixed and you're solving for $x$, it seems the lhs as a function of $x$ is monotonically increasing on $(-1, infty)$ and always has a root there, finding which should be straightforward (the root may be negative). If you're allowing $x < -1$, the function has multiple zeroes on $(-infty, -1)$, finding the zero closest to an integer will be more tricky.
– Maxim
Dec 2 at 3:41
|
show 5 more comments
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1
Within you equation there is a $p$ used but you did not defined it.
– mrtaurho
Nov 30 at 22:20
2
Can you give a numerical example? I'm not even sure which variable you're solving for. Also, what would be considered a good approximation?
– Maxim
Dec 1 at 5:11
1
Using $v=96.52509765$ while keeping all of the other values the same gets you much closer to zero. Also, all solutions for $x$ will only depend on $a$ and the ratio $r=v/u$. To see that divide both sides of the equation by $u$.
– JimB
Dec 1 at 23:54
1
Not sure there's much more than noting that the ratio $u/v$ can be written in terms of $x$ and $a$: $frac{-a^x+e^a Gamma (x+1)-e^a x Gamma (x,a)}{a^x+e^a x Gamma (x,a)}$. Plotting $a$ vs $r$ for various integer values of $x$ might be informative.
– JimB
Dec 2 at 2:31
1
If $a$ and $q = u/v$ are fixed and you're solving for $x$, it seems the lhs as a function of $x$ is monotonically increasing on $(-1, infty)$ and always has a root there, finding which should be straightforward (the root may be negative). If you're allowing $x < -1$, the function has multiple zeroes on $(-infty, -1)$, finding the zero closest to an integer will be more tricky.
– Maxim
Dec 2 at 3:41