Simplifying $sum_{n=1}^{x}ne^{-a}frac{a^{x-n}}{(x-n)!}$, where $x$ is an integer and $a<1$
$begingroup$
I would like to simplify the following expression,
$$sum_{n=1}^{x}ne^{-a}frac{a^{x-n}}{(x-n)!}$$
where $x$ is an integer and $a<1$.
Is it possible to lose the sum?
An approximation for the sum will be also helpful.
calculus real-analysis sequences-and-series taylor-expansion
asked Nov 29 '18 at 18:34
Y.LY.L
647
$endgroup$
add a comment |
$begingroup$
I would like to simplify the following expression,
$$sum_{n=1}^{x}ne^{-a}frac{a^{x-n}}{(x-n)!}$$
where $x$ is an integer and $a<1$.
Is it possible to lose the sum?
An approximation for the sum will be also helpful.
calculus real-analysis sequences-and-series taylor-expansion
asked Nov 29 '18 at 18:34
Y.LY.L
647
$endgroup$
add a comment |
$begingroup$
I would like to simplify the following expression,
$$sum_{n=1}^{x}ne^{-a}frac{a^{x-n}}{(x-n)!}$$
where $x$ is an integer and $a<1$.
Is it possible to lose the sum?
An approximation for the sum will be also helpful.
calculus real-analysis sequences-and-series taylor-expansion
asked Nov 29 '18 at 18:34
Y.LY.L
647
$endgroup$
I would like to simplify the following expression,
$$sum_{n=1}^{x}ne^{-a}frac{a^{x-n}}{(x-n)!}$$
where $x$ is an integer and $a<1$.
Is it possible to lose the sum?
An approximation for the sum will be also helpful.
calculus real-analysis sequences-and-series taylor-expansion
calculus real-analysis sequences-and-series taylor-expansion
asked Nov 29 '18 at 18:34
Y.LY.L
647
asked Nov 29 '18 at 18:34
Y.LY.L
647
asked Nov 29 '18 at 18:34
Y.LY.L
647
asked Nov 29 '18 at 18:34
Y.LY.L
647
asked Nov 29 '18 at 18:34
Y.LY.L
647
647
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
It can be written using the Incomplete Gamma function as
$$ {frac {{{rm e}^{-a}}{a}^{x}-Gamma left( x,a right) left( a-x
right) }{Gamma left( x right) }}
$$
answered Nov 29 '18 at 18:36
Robert IsraelRobert Israel
322k23212465
$endgroup$
$begingroup$
Thanks @Robert Israel, but I was thinking maybe there is a simpler option, without using the Gamma.
$endgroup$
– Y.L
Nov 29 '18 at 18:49
$begingroup$
can you please show how you get to the term?
$endgroup$
– Y.L
Nov 29 '18 at 19:24
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It can be written using the Incomplete Gamma function as
$$ {frac {{{rm e}^{-a}}{a}^{x}-Gamma left( x,a right) left( a-x
right) }{Gamma left( x right) }}
$$
answered Nov 29 '18 at 18:36
Robert IsraelRobert Israel
322k23212465
$endgroup$
$begingroup$
Thanks @Robert Israel, but I was thinking maybe there is a simpler option, without using the Gamma.
$endgroup$
– Y.L
Nov 29 '18 at 18:49
$begingroup$
can you please show how you get to the term?
$endgroup$
– Y.L
Nov 29 '18 at 19:24
add a comment |
$begingroup$
It can be written using the Incomplete Gamma function as
$$ {frac {{{rm e}^{-a}}{a}^{x}-Gamma left( x,a right) left( a-x
right) }{Gamma left( x right) }}
$$
answered Nov 29 '18 at 18:36
Robert IsraelRobert Israel
322k23212465
$endgroup$
$begingroup$
Thanks @Robert Israel, but I was thinking maybe there is a simpler option, without using the Gamma.
$endgroup$
– Y.L
Nov 29 '18 at 18:49
$begingroup$
can you please show how you get to the term?
$endgroup$
– Y.L
Nov 29 '18 at 19:24
add a comment |
$begingroup$
It can be written using the Incomplete Gamma function as
$$ {frac {{{rm e}^{-a}}{a}^{x}-Gamma left( x,a right) left( a-x
right) }{Gamma left( x right) }}
$$
answered Nov 29 '18 at 18:36
Robert IsraelRobert Israel
322k23212465
$endgroup$
It can be written using the Incomplete Gamma function as
$$ {frac {{{rm e}^{-a}}{a}^{x}-Gamma left( x,a right) left( a-x
right) }{Gamma left( x right) }}
$$
answered Nov 29 '18 at 18:36
Robert IsraelRobert Israel
322k23212465
answered Nov 29 '18 at 18:36
Robert IsraelRobert Israel
322k23212465
answered Nov 29 '18 at 18:36
Robert IsraelRobert Israel
322k23212465
answered Nov 29 '18 at 18:36
Robert IsraelRobert Israel
322k23212465
322k23212465
$begingroup$
Thanks @Robert Israel, but I was thinking maybe there is a simpler option, without using the Gamma.
$endgroup$
– Y.L
Nov 29 '18 at 18:49
$begingroup$
can you please show how you get to the term?
$endgroup$
– Y.L
Nov 29 '18 at 19:24
add a comment |
$begingroup$
Thanks @Robert Israel, but I was thinking maybe there is a simpler option, without using the Gamma.
$endgroup$
– Y.L
Nov 29 '18 at 18:49
$begingroup$
can you please show how you get to the term?
$endgroup$
– Y.L
Nov 29 '18 at 19:24
$begingroup$
Thanks @Robert Israel, but I was thinking maybe there is a simpler option, without using the Gamma.
$endgroup$
– Y.L
Nov 29 '18 at 18:49
$begingroup$
Thanks @Robert Israel, but I was thinking maybe there is a simpler option, without using the Gamma.
$endgroup$
– Y.L
Nov 29 '18 at 18:49
$begingroup$
can you please show how you get to the term?
$endgroup$
– Y.L
Nov 29 '18 at 19:24
$begingroup$
can you please show how you get to the term?
$endgroup$
– Y.L
Nov 29 '18 at 19:24
add a comment |
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