Mixing
An equal expression to $sum^{z}_{n=0}nfrac{x^{z-n}}{(z-n)!}$
$begingroup$
Is there a way to rewrite an equal expression to the following
$$sum^{z}_{n=0}nfrac{x^{z-n}}{(z-n)!}$$
but without the sum?
$z$ is integer and $xgeq 0$.
real-analysis calculus sequences-and-series summation
asked Jan 3 at 22:31
Y.LY.L
597
$endgroup$
add a comment |
$begingroup$
Is there a way to rewrite an equal expression to the following
$$sum^{z}_{n=0}nfrac{x^{z-n}}{(z-n)!}$$
but without the sum?
$z$ is integer and $xgeq 0$.
real-analysis calculus sequences-and-series summation
asked Jan 3 at 22:31
Y.LY.L
597
$endgroup$
add a comment |
$begingroup$
Is there a way to rewrite an equal expression to the following
$$sum^{z}_{n=0}nfrac{x^{z-n}}{(z-n)!}$$
but without the sum?
$z$ is integer and $xgeq 0$.
real-analysis calculus sequences-and-series summation
asked Jan 3 at 22:31
Y.LY.L
597
$endgroup$
Is there a way to rewrite an equal expression to the following
$$sum^{z}_{n=0}nfrac{x^{z-n}}{(z-n)!}$$
but without the sum?
$z$ is integer and $xgeq 0$.
real-analysis calculus sequences-and-series summation
real-analysis calculus sequences-and-series summation
asked Jan 3 at 22:31
Y.LY.L
597
asked Jan 3 at 22:31
Y.LY.L
597
asked Jan 3 at 22:31
Y.LY.L
597
asked Jan 3 at 22:31
Y.LY.L
597
asked Jan 3 at 22:31
Y.LY.L
597
597
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer is no. By making the change of avriable $k=z-n$ and splitting the sum into two parts you get $(z-x)sum_{k=1}^{z-1} frac {x^{k}} {k!} +frac {zx^{z}} {z!}$ from which you cannot get rid of the sum.
answered Jan 3 at 23:38
Kavi Rama MurthyKavi Rama Murthy
54.2k32055
$endgroup$
$begingroup$
Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
$endgroup$
– Y.L
Jan 4 at 0:28
add a comment |
$begingroup$
Sticking to a more standard notation ($z$ normally denotes complex, not integral, number)
begin{align}
sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
&= sum_{k=0}^{n} (n-k) frac{x^k}{k!}
= nsum_{k=0}^n frac{x^k}{k!} - sum_{k=0}^n xfrac{d}{dx} frac{x^k}{k!}\
&= nP_n(x) - xfrac{d}{dx}P_n(x)
end{align}
where $P_n(x) = sum_{k=0}^n frac{x^k}{k!}$ is the partial sum of the exponential's Taylor series. But as can be seen here, the incomplete gamma function $Gamma(n,x) = int_x^{infty} x^{n-1}e^{-x};dx$ may be expressed as
$$Gamma(n,x) = (n-1)! e^{-x} sum_{k=0}^{n-1}frac{x^k}{k!}$$
hence,
$$P_n(x) = frac{Gamma(n+1,x)}{n!} e^x$$
We may then compute
begin{align}
frac{d}{dx}P_n(x)
&= P_n(x) frac{d}{dx} ln P_n(x) = P_n(x) left[,frac{frac{partial}{partial{x}}Gamma(n+1,x)}{Gamma(n+1,x)} + 1,right]\
&= frac{Gamma(n+1,x)}{n!} e^x left[,frac{-x^ne^{-x}}{Gamma(n+1,x)} + 1,right]\
&= frac{Gamma(n+1,x)}{n!} e^x - frac{x^n}{n!}
end{align}
Solving and simplifying, we get
begin{align}
sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
&= nP_n(x) - xfrac{d}{dx}P_n(x)\
&= frac{1}{n!} left[,(n-x)Gamma(n+1,x)e^x + x^{n+1},right]
end{align}
answered Jan 4 at 1:18
adfriedmanadfriedman
3,161169
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is no. By making the change of avriable $k=z-n$ and splitting the sum into two parts you get $(z-x)sum_{k=1}^{z-1} frac {x^{k}} {k!} +frac {zx^{z}} {z!}$ from which you cannot get rid of the sum.
answered Jan 3 at 23:38
Kavi Rama MurthyKavi Rama Murthy
54.2k32055
$endgroup$
$begingroup$
Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
$endgroup$
– Y.L
Jan 4 at 0:28
add a comment |
$begingroup$
The answer is no. By making the change of avriable $k=z-n$ and splitting the sum into two parts you get $(z-x)sum_{k=1}^{z-1} frac {x^{k}} {k!} +frac {zx^{z}} {z!}$ from which you cannot get rid of the sum.
answered Jan 3 at 23:38
Kavi Rama MurthyKavi Rama Murthy
54.2k32055
$endgroup$
$begingroup$
Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
$endgroup$
– Y.L
Jan 4 at 0:28
add a comment |
$begingroup$
The answer is no. By making the change of avriable $k=z-n$ and splitting the sum into two parts you get $(z-x)sum_{k=1}^{z-1} frac {x^{k}} {k!} +frac {zx^{z}} {z!}$ from which you cannot get rid of the sum.
answered Jan 3 at 23:38
Kavi Rama MurthyKavi Rama Murthy
54.2k32055
$endgroup$
The answer is no. By making the change of avriable $k=z-n$ and splitting the sum into two parts you get $(z-x)sum_{k=1}^{z-1} frac {x^{k}} {k!} +frac {zx^{z}} {z!}$ from which you cannot get rid of the sum.
answered Jan 3 at 23:38
Kavi Rama MurthyKavi Rama Murthy
54.2k32055
answered Jan 3 at 23:38
Kavi Rama MurthyKavi Rama Murthy
54.2k32055
answered Jan 3 at 23:38
Kavi Rama MurthyKavi Rama Murthy
54.2k32055
answered Jan 3 at 23:38
Kavi Rama MurthyKavi Rama Murthy
54.2k32055
54.2k32055
$begingroup$
Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
$endgroup$
– Y.L
Jan 4 at 0:28
add a comment |
$begingroup$
Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
$endgroup$
– Y.L
Jan 4 at 0:28
$begingroup$
Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
$endgroup$
– Y.L
Jan 4 at 0:28
$begingroup$
Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
$endgroup$
– Y.L
Jan 4 at 0:28
add a comment |
$begingroup$
Sticking to a more standard notation ($z$ normally denotes complex, not integral, number)
begin{align}
sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
&= sum_{k=0}^{n} (n-k) frac{x^k}{k!}
= nsum_{k=0}^n frac{x^k}{k!} - sum_{k=0}^n xfrac{d}{dx} frac{x^k}{k!}\
&= nP_n(x) - xfrac{d}{dx}P_n(x)
end{align}
where $P_n(x) = sum_{k=0}^n frac{x^k}{k!}$ is the partial sum of the exponential's Taylor series. But as can be seen here, the incomplete gamma function $Gamma(n,x) = int_x^{infty} x^{n-1}e^{-x};dx$ may be expressed as
$$Gamma(n,x) = (n-1)! e^{-x} sum_{k=0}^{n-1}frac{x^k}{k!}$$
hence,
$$P_n(x) = frac{Gamma(n+1,x)}{n!} e^x$$
We may then compute
begin{align}
frac{d}{dx}P_n(x)
&= P_n(x) frac{d}{dx} ln P_n(x) = P_n(x) left[,frac{frac{partial}{partial{x}}Gamma(n+1,x)}{Gamma(n+1,x)} + 1,right]\
&= frac{Gamma(n+1,x)}{n!} e^x left[,frac{-x^ne^{-x}}{Gamma(n+1,x)} + 1,right]\
&= frac{Gamma(n+1,x)}{n!} e^x - frac{x^n}{n!}
end{align}
Solving and simplifying, we get
begin{align}
sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
&= nP_n(x) - xfrac{d}{dx}P_n(x)\
&= frac{1}{n!} left[,(n-x)Gamma(n+1,x)e^x + x^{n+1},right]
end{align}
answered Jan 4 at 1:18
adfriedmanadfriedman
3,161169
$endgroup$
add a comment |
$begingroup$
Sticking to a more standard notation ($z$ normally denotes complex, not integral, number)
begin{align}
sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
&= sum_{k=0}^{n} (n-k) frac{x^k}{k!}
= nsum_{k=0}^n frac{x^k}{k!} - sum_{k=0}^n xfrac{d}{dx} frac{x^k}{k!}\
&= nP_n(x) - xfrac{d}{dx}P_n(x)
end{align}
where $P_n(x) = sum_{k=0}^n frac{x^k}{k!}$ is the partial sum of the exponential's Taylor series. But as can be seen here, the incomplete gamma function $Gamma(n,x) = int_x^{infty} x^{n-1}e^{-x};dx$ may be expressed as
$$Gamma(n,x) = (n-1)! e^{-x} sum_{k=0}^{n-1}frac{x^k}{k!}$$
hence,
$$P_n(x) = frac{Gamma(n+1,x)}{n!} e^x$$
We may then compute
begin{align}
frac{d}{dx}P_n(x)
&= P_n(x) frac{d}{dx} ln P_n(x) = P_n(x) left[,frac{frac{partial}{partial{x}}Gamma(n+1,x)}{Gamma(n+1,x)} + 1,right]\
&= frac{Gamma(n+1,x)}{n!} e^x left[,frac{-x^ne^{-x}}{Gamma(n+1,x)} + 1,right]\
&= frac{Gamma(n+1,x)}{n!} e^x - frac{x^n}{n!}
end{align}
Solving and simplifying, we get
begin{align}
sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
&= nP_n(x) - xfrac{d}{dx}P_n(x)\
&= frac{1}{n!} left[,(n-x)Gamma(n+1,x)e^x + x^{n+1},right]
end{align}
answered Jan 4 at 1:18
adfriedmanadfriedman
3,161169
$endgroup$
add a comment |
$begingroup$
Sticking to a more standard notation ($z$ normally denotes complex, not integral, number)
begin{align}
sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
&= sum_{k=0}^{n} (n-k) frac{x^k}{k!}
= nsum_{k=0}^n frac{x^k}{k!} - sum_{k=0}^n xfrac{d}{dx} frac{x^k}{k!}\
&= nP_n(x) - xfrac{d}{dx}P_n(x)
end{align}
where $P_n(x) = sum_{k=0}^n frac{x^k}{k!}$ is the partial sum of the exponential's Taylor series. But as can be seen here, the incomplete gamma function $Gamma(n,x) = int_x^{infty} x^{n-1}e^{-x};dx$ may be expressed as
$$Gamma(n,x) = (n-1)! e^{-x} sum_{k=0}^{n-1}frac{x^k}{k!}$$
hence,
$$P_n(x) = frac{Gamma(n+1,x)}{n!} e^x$$
We may then compute
begin{align}
frac{d}{dx}P_n(x)
&= P_n(x) frac{d}{dx} ln P_n(x) = P_n(x) left[,frac{frac{partial}{partial{x}}Gamma(n+1,x)}{Gamma(n+1,x)} + 1,right]\
&= frac{Gamma(n+1,x)}{n!} e^x left[,frac{-x^ne^{-x}}{Gamma(n+1,x)} + 1,right]\
&= frac{Gamma(n+1,x)}{n!} e^x - frac{x^n}{n!}
end{align}
Solving and simplifying, we get
begin{align}
sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
&= nP_n(x) - xfrac{d}{dx}P_n(x)\
&= frac{1}{n!} left[,(n-x)Gamma(n+1,x)e^x + x^{n+1},right]
end{align}
answered Jan 4 at 1:18
adfriedmanadfriedman
3,161169
$endgroup$
Sticking to a more standard notation ($z$ normally denotes complex, not integral, number)
begin{align}
sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
&= sum_{k=0}^{n} (n-k) frac{x^k}{k!}
= nsum_{k=0}^n frac{x^k}{k!} - sum_{k=0}^n xfrac{d}{dx} frac{x^k}{k!}\
&= nP_n(x) - xfrac{d}{dx}P_n(x)
end{align}
where $P_n(x) = sum_{k=0}^n frac{x^k}{k!}$ is the partial sum of the exponential's Taylor series. But as can be seen here, the incomplete gamma function $Gamma(n,x) = int_x^{infty} x^{n-1}e^{-x};dx$ may be expressed as
$$Gamma(n,x) = (n-1)! e^{-x} sum_{k=0}^{n-1}frac{x^k}{k!}$$
hence,
$$P_n(x) = frac{Gamma(n+1,x)}{n!} e^x$$
We may then compute
begin{align}
frac{d}{dx}P_n(x)
&= P_n(x) frac{d}{dx} ln P_n(x) = P_n(x) left[,frac{frac{partial}{partial{x}}Gamma(n+1,x)}{Gamma(n+1,x)} + 1,right]\
&= frac{Gamma(n+1,x)}{n!} e^x left[,frac{-x^ne^{-x}}{Gamma(n+1,x)} + 1,right]\
&= frac{Gamma(n+1,x)}{n!} e^x - frac{x^n}{n!}
end{align}
Solving and simplifying, we get
begin{align}
sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
&= nP_n(x) - xfrac{d}{dx}P_n(x)\
&= frac{1}{n!} left[,(n-x)Gamma(n+1,x)e^x + x^{n+1},right]
end{align}
answered Jan 4 at 1:18
adfriedmanadfriedman
3,161169
answered Jan 4 at 1:18
adfriedmanadfriedman
3,161169
answered Jan 4 at 1:18
adfriedmanadfriedman
3,161169
answered Jan 4 at 1:18
adfriedmanadfriedman
3,161169
3,161169
add a comment |
add a comment |
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