$Gamma(s)= lim_{n to infty} frac{n^s n!}{s(s+1)…(s+n)}$ , the product formula of Gamma function .
$begingroup$
Prove that $$Gamma(s)= lim_{n to infty} frac{n^s n!}{s(s+1)...(s+n)}$$
Whenever $s neq 0,-1,-2,...$
My attempt :
By applying product formula for $frac{1}{Gamma}$ ,
$$Gamma(s)=lim_{nto infty} frac{n!}{s(s+1)...(s+n)}e^{s(1+frac12+...+frac1n-gamma)}$$
Notice that $1+frac12+...+frac1n-gammato log n$ , so it seems that we then get the desired conclusion . But whenever $Re(s)gt0$ , $n^s to infty$ , we can not have
$$Gamma(s)=lim_{nto infty} frac{n!}{s(s+1)...(s+n)}lim_{nto infty} e^{s(1+frac12+...+frac1n-gamma)}$$
Let $A(s)=lim_{nto infty} frac{n!}{s(s+1)...(s+n)}n^s$ , I want to prove this exercise by showing the following statement.
1) $A(s)$ defines a meromorphic function with simple poles at $0,-1,-2,...$ and nowhere else.
2)$A(s)=Gamma (s)$ whenever $s in (frac14,frac13)$ .
To show 2) , we need some estimate of $gamma$ ,
$$sum_{n=1}^N frac1n-log N =sum_{n=1}^{N-1}int_n^{n+1} frac1n-frac1x ,dx +frac1N$$ Then we may write $a_n=int_n^{n+1} frac1n-frac1x ,dx $ and $gamma=sum_1^{infty}a_n$ . Moreover , $a_n le frac{1}{n(n+1)}$
Notice that whenever $sin (frac14,frac13)$ , $frac{n!}{s(s+1)...(s+n)}$ is bounded by some fixed $M$ .
$$Gamma(s)-A(s)le Mlim_{nto infty} |e^{s(1+frac12+...+frac1n-gamma)}-e^{s log n}|$$
apple mean-value theorem we have $$|e^{s(1+frac12+...+frac1n-gamma)}-e^{s log n}|le s|1+frac12+...+frac1n-gamma-log n||e^{s*2 log n}|$$
$$le sn^{frac23} |sum_{k=n}^{infty} a_k+frac1n|le sn^{frac23} |sum_{k=n}^{infty} frac{1}{k(k+1)}+frac1n|le frac{2sn^{frac23}}{n} to 0$$
And we complete the proof of 2) . To prove 1) , we need to prove that for every compact subset $Omega$ of $C-{0,-1,-2,...}$ , $f_n=frac{n!}{s(s+1)...(s+n)}n^s$ converges uniformly . I have no idea how to deal with this .
complex-analysis gamma-function
edited Jan 18 at 13:04
José Carlos Santos
161k22127232
asked Jan 18 at 13:02
J.GuoJ.Guo
3759
$endgroup$
add a comment |
$begingroup$
Prove that $$Gamma(s)= lim_{n to infty} frac{n^s n!}{s(s+1)...(s+n)}$$
Whenever $s neq 0,-1,-2,...$
My attempt :
By applying product formula for $frac{1}{Gamma}$ ,
$$Gamma(s)=lim_{nto infty} frac{n!}{s(s+1)...(s+n)}e^{s(1+frac12+...+frac1n-gamma)}$$
Notice that $1+frac12+...+frac1n-gammato log n$ , so it seems that we then get the desired conclusion . But whenever $Re(s)gt0$ , $n^s to infty$ , we can not have
$$Gamma(s)=lim_{nto infty} frac{n!}{s(s+1)...(s+n)}lim_{nto infty} e^{s(1+frac12+...+frac1n-gamma)}$$
Let $A(s)=lim_{nto infty} frac{n!}{s(s+1)...(s+n)}n^s$ , I want to prove this exercise by showing the following statement.
1) $A(s)$ defines a meromorphic function with simple poles at $0,-1,-2,...$ and nowhere else.
2)$A(s)=Gamma (s)$ whenever $s in (frac14,frac13)$ .
To show 2) , we need some estimate of $gamma$ ,
$$sum_{n=1}^N frac1n-log N =sum_{n=1}^{N-1}int_n^{n+1} frac1n-frac1x ,dx +frac1N$$ Then we may write $a_n=int_n^{n+1} frac1n-frac1x ,dx $ and $gamma=sum_1^{infty}a_n$ . Moreover , $a_n le frac{1}{n(n+1)}$
Notice that whenever $sin (frac14,frac13)$ , $frac{n!}{s(s+1)...(s+n)}$ is bounded by some fixed $M$ .
$$Gamma(s)-A(s)le Mlim_{nto infty} |e^{s(1+frac12+...+frac1n-gamma)}-e^{s log n}|$$
apple mean-value theorem we have $$|e^{s(1+frac12+...+frac1n-gamma)}-e^{s log n}|le s|1+frac12+...+frac1n-gamma-log n||e^{s*2 log n}|$$
$$le sn^{frac23} |sum_{k=n}^{infty} a_k+frac1n|le sn^{frac23} |sum_{k=n}^{infty} frac{1}{k(k+1)}+frac1n|le frac{2sn^{frac23}}{n} to 0$$
And we complete the proof of 2) . To prove 1) , we need to prove that for every compact subset $Omega$ of $C-{0,-1,-2,...}$ , $f_n=frac{n!}{s(s+1)...(s+n)}n^s$ converges uniformly . I have no idea how to deal with this .
complex-analysis gamma-function
edited Jan 18 at 13:04
José Carlos Santos
161k22127232
asked Jan 18 at 13:02
J.GuoJ.Guo
3759
$endgroup$
$begingroup$
Definition of $Gamma$ you are using?
$endgroup$
– Simply Beautiful Art
Jan 18 at 15:13
$begingroup$
@ Simply Beautiful Art $Gamma_0(s)=int_0^{infty}e^{-t} t^{s-1} , dt$ whenever $Re(s)gt0$ and $Gamma(s)$ is the Analytic continuaton of $Gamma_0$ which defined a meromorphic function on $C$ .
$endgroup$
– J.Guo
Jan 18 at 15:33
add a comment |
$begingroup$
Prove that $$Gamma(s)= lim_{n to infty} frac{n^s n!}{s(s+1)...(s+n)}$$
Whenever $s neq 0,-1,-2,...$
My attempt :
By applying product formula for $frac{1}{Gamma}$ ,
$$Gamma(s)=lim_{nto infty} frac{n!}{s(s+1)...(s+n)}e^{s(1+frac12+...+frac1n-gamma)}$$
Notice that $1+frac12+...+frac1n-gammato log n$ , so it seems that we then get the desired conclusion . But whenever $Re(s)gt0$ , $n^s to infty$ , we can not have
$$Gamma(s)=lim_{nto infty} frac{n!}{s(s+1)...(s+n)}lim_{nto infty} e^{s(1+frac12+...+frac1n-gamma)}$$
Let $A(s)=lim_{nto infty} frac{n!}{s(s+1)...(s+n)}n^s$ , I want to prove this exercise by showing the following statement.
1) $A(s)$ defines a meromorphic function with simple poles at $0,-1,-2,...$ and nowhere else.
2)$A(s)=Gamma (s)$ whenever $s in (frac14,frac13)$ .
To show 2) , we need some estimate of $gamma$ ,
$$sum_{n=1}^N frac1n-log N =sum_{n=1}^{N-1}int_n^{n+1} frac1n-frac1x ,dx +frac1N$$ Then we may write $a_n=int_n^{n+1} frac1n-frac1x ,dx $ and $gamma=sum_1^{infty}a_n$ . Moreover , $a_n le frac{1}{n(n+1)}$
Notice that whenever $sin (frac14,frac13)$ , $frac{n!}{s(s+1)...(s+n)}$ is bounded by some fixed $M$ .
$$Gamma(s)-A(s)le Mlim_{nto infty} |e^{s(1+frac12+...+frac1n-gamma)}-e^{s log n}|$$
apple mean-value theorem we have $$|e^{s(1+frac12+...+frac1n-gamma)}-e^{s log n}|le s|1+frac12+...+frac1n-gamma-log n||e^{s*2 log n}|$$
$$le sn^{frac23} |sum_{k=n}^{infty} a_k+frac1n|le sn^{frac23} |sum_{k=n}^{infty} frac{1}{k(k+1)}+frac1n|le frac{2sn^{frac23}}{n} to 0$$
And we complete the proof of 2) . To prove 1) , we need to prove that for every compact subset $Omega$ of $C-{0,-1,-2,...}$ , $f_n=frac{n!}{s(s+1)...(s+n)}n^s$ converges uniformly . I have no idea how to deal with this .
complex-analysis gamma-function
edited Jan 18 at 13:04
José Carlos Santos
161k22127232
asked Jan 18 at 13:02
J.GuoJ.Guo
3759
$endgroup$
Prove that $$Gamma(s)= lim_{n to infty} frac{n^s n!}{s(s+1)...(s+n)}$$
Whenever $s neq 0,-1,-2,...$
My attempt :
By applying product formula for $frac{1}{Gamma}$ ,
$$Gamma(s)=lim_{nto infty} frac{n!}{s(s+1)...(s+n)}e^{s(1+frac12+...+frac1n-gamma)}$$
Notice that $1+frac12+...+frac1n-gammato log n$ , so it seems that we then get the desired conclusion . But whenever $Re(s)gt0$ , $n^s to infty$ , we can not have
$$Gamma(s)=lim_{nto infty} frac{n!}{s(s+1)...(s+n)}lim_{nto infty} e^{s(1+frac12+...+frac1n-gamma)}$$
Let $A(s)=lim_{nto infty} frac{n!}{s(s+1)...(s+n)}n^s$ , I want to prove this exercise by showing the following statement.
1) $A(s)$ defines a meromorphic function with simple poles at $0,-1,-2,...$ and nowhere else.
2)$A(s)=Gamma (s)$ whenever $s in (frac14,frac13)$ .
To show 2) , we need some estimate of $gamma$ ,
$$sum_{n=1}^N frac1n-log N =sum_{n=1}^{N-1}int_n^{n+1} frac1n-frac1x ,dx +frac1N$$ Then we may write $a_n=int_n^{n+1} frac1n-frac1x ,dx $ and $gamma=sum_1^{infty}a_n$ . Moreover , $a_n le frac{1}{n(n+1)}$
Notice that whenever $sin (frac14,frac13)$ , $frac{n!}{s(s+1)...(s+n)}$ is bounded by some fixed $M$ .
$$Gamma(s)-A(s)le Mlim_{nto infty} |e^{s(1+frac12+...+frac1n-gamma)}-e^{s log n}|$$
apple mean-value theorem we have $$|e^{s(1+frac12+...+frac1n-gamma)}-e^{s log n}|le s|1+frac12+...+frac1n-gamma-log n||e^{s*2 log n}|$$
$$le sn^{frac23} |sum_{k=n}^{infty} a_k+frac1n|le sn^{frac23} |sum_{k=n}^{infty} frac{1}{k(k+1)}+frac1n|le frac{2sn^{frac23}}{n} to 0$$
And we complete the proof of 2) . To prove 1) , we need to prove that for every compact subset $Omega$ of $C-{0,-1,-2,...}$ , $f_n=frac{n!}{s(s+1)...(s+n)}n^s$ converges uniformly . I have no idea how to deal with this .
complex-analysis gamma-function
complex-analysis gamma-function
edited Jan 18 at 13:04
José Carlos Santos
161k22127232
asked Jan 18 at 13:02
J.GuoJ.Guo
3759
edited Jan 18 at 13:04
José Carlos Santos
161k22127232
asked Jan 18 at 13:02
J.GuoJ.Guo
3759
edited Jan 18 at 13:04
José Carlos Santos
161k22127232
edited Jan 18 at 13:04
José Carlos Santos
161k22127232
edited Jan 18 at 13:04
José Carlos Santos
161k22127232
161k22127232
asked Jan 18 at 13:02
J.GuoJ.Guo
3759
asked Jan 18 at 13:02
J.GuoJ.Guo
3759
asked Jan 18 at 13:02
J.GuoJ.Guo
3759
3759
$begingroup$
Definition of $Gamma$ you are using?
$endgroup$
– Simply Beautiful Art
Jan 18 at 15:13
$begingroup$
@ Simply Beautiful Art $Gamma_0(s)=int_0^{infty}e^{-t} t^{s-1} , dt$ whenever $Re(s)gt0$ and $Gamma(s)$ is the Analytic continuaton of $Gamma_0$ which defined a meromorphic function on $C$ .
$endgroup$
– J.Guo
Jan 18 at 15:33
add a comment |
$begingroup$
Definition of $Gamma$ you are using?
$endgroup$
– Simply Beautiful Art
Jan 18 at 15:13
$begingroup$
@ Simply Beautiful Art $Gamma_0(s)=int_0^{infty}e^{-t} t^{s-1} , dt$ whenever $Re(s)gt0$ and $Gamma(s)$ is the Analytic continuaton of $Gamma_0$ which defined a meromorphic function on $C$ .
$endgroup$
– J.Guo
Jan 18 at 15:33
$begingroup$
Definition of $Gamma$ you are using?
$endgroup$
– Simply Beautiful Art
Jan 18 at 15:13
$begingroup$
Definition of $Gamma$ you are using?
$endgroup$
– Simply Beautiful Art
Jan 18 at 15:13
$begingroup$
@ Simply Beautiful Art $Gamma_0(s)=int_0^{infty}e^{-t} t^{s-1} , dt$ whenever $Re(s)gt0$ and $Gamma(s)$ is the Analytic continuaton of $Gamma_0$ which defined a meromorphic function on $C$ .
$endgroup$
– J.Guo
Jan 18 at 15:33
$begingroup$
@ Simply Beautiful Art $Gamma_0(s)=int_0^{infty}e^{-t} t^{s-1} , dt$ whenever $Re(s)gt0$ and $Gamma(s)$ is the Analytic continuaton of $Gamma_0$ which defined a meromorphic function on $C$ .
$endgroup$
– J.Guo
Jan 18 at 15:33
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A much more straightforward proof from the given definition for $Re(s)>0$:
$$f_n(t):=begin{cases}0,&t>n\left(1-frac tnright)^n,&0le tle nend{cases}$$
begin{align}Gamma(s):!&=int_0^infty t^{s-1}e^{-t}~mathrm dt\&=lim_{ntoinfty}int_0^infty t^{s-1}f_n(t)~mathrm dttag{DCT}\&=lim_{ntoinfty}int_0^nt^{s-1}f_n(t)~mathrm dt\&=lim_{ntoinfty}int_0^nt^{s-1}left(1-frac tnright)^n~mathrm dt\&=lim_{ntoinfty}n^sint_0^1u^{s-1}left(1-uright)^n~mathrm dutag{$t=nu$}\&=lim_{ntoinfty}n^sprod_{k=1}^nfrac k{k+s-1}tag{Induction over $n$}end{align}
The last expression being your limit in product form. One can then verify the last form to satisfy the recursive equation $Gamma(s+1)=sGamma(s)$ and hence extends to $Re(s)le0$ as the Gamma function.
answered Jan 19 at 2:23
Simply Beautiful ArtSimply Beautiful Art
50.5k578182
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add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{lim_{n to infty}{n^{s}, n! over spars{s + 1}ldotspars{s + n}}} =
lim_{n to infty}{n^{s}, n! over s^{overline{n + 1}}} =
lim_{n to infty}{n^{s}, n! over Gammapars{s + n + 1}/Gammapars{s}}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s}, n! over pars{s + n}!}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s}bracks{root{2pi}n^{n + 1/2}expo{-n}} over root{2pi}pars{n + s}^{n + s + 1/2}
expo{-pars{n + s}}}qquadpars{~Stirling Asymptotic~}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s + n + 1/2},expo{s} over
n^{n + s 1/2},pars{1 + s/n}^{n + s + 1/2}}
=
Gammapars{s}lim_{n to infty}{expo{s} over pars{1 + s/n}^{n}} \[5mm] = &
Gammapars{s},{expo{s} over expo{s}} = bbx{Gammapars{s}}
end{align}
answered Jan 19 at 4:20
Felix MarinFelix Marin
68k7107142
$endgroup$
add a comment |
$begingroup$
To prove uniform convergence of $f_n$ on a compact set $Omega in mathbb{C} setminus {0,-1,-2,ldots}$ note that
$$tag{*}f_n(s) = frac{n^s n!}{s(s+1)...(s+n)} = frac{n^s}{s} prod_{k=1}^n left(1 + frac{s}{k} right)^{-1} \ = left(frac{n}{n+1} right)^s frac{1}{s}prod_{k=1}^n left(1 + frac{1}{k} right)^{s}left(1 + frac{s}{k} right)^{-1} $$
where the last equality follows from
$$prod_{k=1}^n left(1 + frac{1}{k} right)^{s} = left[ prod_{k=1}^n left(frac{k+1}{k}right) right]^s = left(frac{(n+1)!}{n!}right)^s = (n+1)^s$$
Since $frac{1}{2} leqslant frac{n}{n+1} < 1$ it can be shown that $left(frac{n}{n+1} right)^s frac{1}{s}$ is uniformly bounded on $Omega$. Thus, uniform convergence of $f_n$ holds if the product on the RHS of (*) is uniformly convergent.
We can apply a general theorem for complex infinite products that states that the product $prod [1 + g_n(s)]$ is uniformly and absolutely convergent on a compact set if $sum |g_n(s)|$ is uniformly convergent. This holds here because for large $K$ we have
$$left(1 + frac{1}{k} right)^{s}left(1 + frac{s}{k} right)^{-1} = 1 + frac{s((s-1)}{2k^2} + mathcal{O}(k^{-3})$$
Thus, the product on the RHS of (*) and, consequently $f_n$ are uniformly convergent on the compact set $Omega$ that avoids singularities of the gamma function.
answered Jan 19 at 5:07
RRLRRL
51.2k42573
$endgroup$
$begingroup$
Here $s$ is a complex variable , so we can not have $log (ab) = log (a) + log (b) $ , then how to deduce that $log f_n(s) = -log s - underbrace{sleft(sum_{k=1}^nfrac{1}{k} - log nright)}_{A_n(s)} -underbrace{sum_{k=1}^nleft[log left(1 + frac{s}{k}right)- frac{s}{k} right]}_{B_n(s)}$
$endgroup$
– J.Guo
Jan 19 at 12:17
$begingroup$
@J.Guo: You are absolutely correct but that technicality can be avoided as shown in the revised answer proving uniform convergence. That should provide the piece you are missing.
$endgroup$
– RRL
Jan 20 at 2:47
add a comment |
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3 Answers
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3 Answers
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$begingroup$
A much more straightforward proof from the given definition for $Re(s)>0$:
$$f_n(t):=begin{cases}0,&t>n\left(1-frac tnright)^n,&0le tle nend{cases}$$
begin{align}Gamma(s):!&=int_0^infty t^{s-1}e^{-t}~mathrm dt\&=lim_{ntoinfty}int_0^infty t^{s-1}f_n(t)~mathrm dttag{DCT}\&=lim_{ntoinfty}int_0^nt^{s-1}f_n(t)~mathrm dt\&=lim_{ntoinfty}int_0^nt^{s-1}left(1-frac tnright)^n~mathrm dt\&=lim_{ntoinfty}n^sint_0^1u^{s-1}left(1-uright)^n~mathrm dutag{$t=nu$}\&=lim_{ntoinfty}n^sprod_{k=1}^nfrac k{k+s-1}tag{Induction over $n$}end{align}
The last expression being your limit in product form. One can then verify the last form to satisfy the recursive equation $Gamma(s+1)=sGamma(s)$ and hence extends to $Re(s)le0$ as the Gamma function.
answered Jan 19 at 2:23
Simply Beautiful ArtSimply Beautiful Art
50.5k578182
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add a comment |
$begingroup$
A much more straightforward proof from the given definition for $Re(s)>0$:
$$f_n(t):=begin{cases}0,&t>n\left(1-frac tnright)^n,&0le tle nend{cases}$$
begin{align}Gamma(s):!&=int_0^infty t^{s-1}e^{-t}~mathrm dt\&=lim_{ntoinfty}int_0^infty t^{s-1}f_n(t)~mathrm dttag{DCT}\&=lim_{ntoinfty}int_0^nt^{s-1}f_n(t)~mathrm dt\&=lim_{ntoinfty}int_0^nt^{s-1}left(1-frac tnright)^n~mathrm dt\&=lim_{ntoinfty}n^sint_0^1u^{s-1}left(1-uright)^n~mathrm dutag{$t=nu$}\&=lim_{ntoinfty}n^sprod_{k=1}^nfrac k{k+s-1}tag{Induction over $n$}end{align}
The last expression being your limit in product form. One can then verify the last form to satisfy the recursive equation $Gamma(s+1)=sGamma(s)$ and hence extends to $Re(s)le0$ as the Gamma function.
answered Jan 19 at 2:23
Simply Beautiful ArtSimply Beautiful Art
50.5k578182
$endgroup$
add a comment |
$begingroup$
A much more straightforward proof from the given definition for $Re(s)>0$:
$$f_n(t):=begin{cases}0,&t>n\left(1-frac tnright)^n,&0le tle nend{cases}$$
begin{align}Gamma(s):!&=int_0^infty t^{s-1}e^{-t}~mathrm dt\&=lim_{ntoinfty}int_0^infty t^{s-1}f_n(t)~mathrm dttag{DCT}\&=lim_{ntoinfty}int_0^nt^{s-1}f_n(t)~mathrm dt\&=lim_{ntoinfty}int_0^nt^{s-1}left(1-frac tnright)^n~mathrm dt\&=lim_{ntoinfty}n^sint_0^1u^{s-1}left(1-uright)^n~mathrm dutag{$t=nu$}\&=lim_{ntoinfty}n^sprod_{k=1}^nfrac k{k+s-1}tag{Induction over $n$}end{align}
The last expression being your limit in product form. One can then verify the last form to satisfy the recursive equation $Gamma(s+1)=sGamma(s)$ and hence extends to $Re(s)le0$ as the Gamma function.
answered Jan 19 at 2:23
Simply Beautiful ArtSimply Beautiful Art
50.5k578182
$endgroup$
A much more straightforward proof from the given definition for $Re(s)>0$:
$$f_n(t):=begin{cases}0,&t>n\left(1-frac tnright)^n,&0le tle nend{cases}$$
begin{align}Gamma(s):!&=int_0^infty t^{s-1}e^{-t}~mathrm dt\&=lim_{ntoinfty}int_0^infty t^{s-1}f_n(t)~mathrm dttag{DCT}\&=lim_{ntoinfty}int_0^nt^{s-1}f_n(t)~mathrm dt\&=lim_{ntoinfty}int_0^nt^{s-1}left(1-frac tnright)^n~mathrm dt\&=lim_{ntoinfty}n^sint_0^1u^{s-1}left(1-uright)^n~mathrm dutag{$t=nu$}\&=lim_{ntoinfty}n^sprod_{k=1}^nfrac k{k+s-1}tag{Induction over $n$}end{align}
The last expression being your limit in product form. One can then verify the last form to satisfy the recursive equation $Gamma(s+1)=sGamma(s)$ and hence extends to $Re(s)le0$ as the Gamma function.
answered Jan 19 at 2:23
Simply Beautiful ArtSimply Beautiful Art
50.5k578182
answered Jan 19 at 2:23
Simply Beautiful ArtSimply Beautiful Art
50.5k578182
answered Jan 19 at 2:23
Simply Beautiful ArtSimply Beautiful Art
50.5k578182
answered Jan 19 at 2:23
Simply Beautiful ArtSimply Beautiful Art
50.5k578182
50.5k578182
add a comment |
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{lim_{n to infty}{n^{s}, n! over spars{s + 1}ldotspars{s + n}}} =
lim_{n to infty}{n^{s}, n! over s^{overline{n + 1}}} =
lim_{n to infty}{n^{s}, n! over Gammapars{s + n + 1}/Gammapars{s}}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s}, n! over pars{s + n}!}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s}bracks{root{2pi}n^{n + 1/2}expo{-n}} over root{2pi}pars{n + s}^{n + s + 1/2}
expo{-pars{n + s}}}qquadpars{~Stirling Asymptotic~}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s + n + 1/2},expo{s} over
n^{n + s 1/2},pars{1 + s/n}^{n + s + 1/2}}
=
Gammapars{s}lim_{n to infty}{expo{s} over pars{1 + s/n}^{n}} \[5mm] = &
Gammapars{s},{expo{s} over expo{s}} = bbx{Gammapars{s}}
end{align}
answered Jan 19 at 4:20
Felix MarinFelix Marin
68k7107142
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{lim_{n to infty}{n^{s}, n! over spars{s + 1}ldotspars{s + n}}} =
lim_{n to infty}{n^{s}, n! over s^{overline{n + 1}}} =
lim_{n to infty}{n^{s}, n! over Gammapars{s + n + 1}/Gammapars{s}}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s}, n! over pars{s + n}!}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s}bracks{root{2pi}n^{n + 1/2}expo{-n}} over root{2pi}pars{n + s}^{n + s + 1/2}
expo{-pars{n + s}}}qquadpars{~Stirling Asymptotic~}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s + n + 1/2},expo{s} over
n^{n + s 1/2},pars{1 + s/n}^{n + s + 1/2}}
=
Gammapars{s}lim_{n to infty}{expo{s} over pars{1 + s/n}^{n}} \[5mm] = &
Gammapars{s},{expo{s} over expo{s}} = bbx{Gammapars{s}}
end{align}
answered Jan 19 at 4:20
Felix MarinFelix Marin
68k7107142
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{lim_{n to infty}{n^{s}, n! over spars{s + 1}ldotspars{s + n}}} =
lim_{n to infty}{n^{s}, n! over s^{overline{n + 1}}} =
lim_{n to infty}{n^{s}, n! over Gammapars{s + n + 1}/Gammapars{s}}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s}, n! over pars{s + n}!}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s}bracks{root{2pi}n^{n + 1/2}expo{-n}} over root{2pi}pars{n + s}^{n + s + 1/2}
expo{-pars{n + s}}}qquadpars{~Stirling Asymptotic~}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s + n + 1/2},expo{s} over
n^{n + s 1/2},pars{1 + s/n}^{n + s + 1/2}}
=
Gammapars{s}lim_{n to infty}{expo{s} over pars{1 + s/n}^{n}} \[5mm] = &
Gammapars{s},{expo{s} over expo{s}} = bbx{Gammapars{s}}
end{align}
answered Jan 19 at 4:20
Felix MarinFelix Marin
68k7107142
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{lim_{n to infty}{n^{s}, n! over spars{s + 1}ldotspars{s + n}}} =
lim_{n to infty}{n^{s}, n! over s^{overline{n + 1}}} =
lim_{n to infty}{n^{s}, n! over Gammapars{s + n + 1}/Gammapars{s}}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s}, n! over pars{s + n}!}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s}bracks{root{2pi}n^{n + 1/2}expo{-n}} over root{2pi}pars{n + s}^{n + s + 1/2}
expo{-pars{n + s}}}qquadpars{~Stirling Asymptotic~}
\[5mm] = &
Gammapars{s}lim_{n to infty}{n^{s + n + 1/2},expo{s} over
n^{n + s 1/2},pars{1 + s/n}^{n + s + 1/2}}
=
Gammapars{s}lim_{n to infty}{expo{s} over pars{1 + s/n}^{n}} \[5mm] = &
Gammapars{s},{expo{s} over expo{s}} = bbx{Gammapars{s}}
end{align}
answered Jan 19 at 4:20
Felix MarinFelix Marin
68k7107142
edited Jan 19 at 17:31
answered Jan 19 at 4:20
Felix MarinFelix Marin
68k7107142
answered Jan 19 at 4:20
Felix MarinFelix Marin
68k7107142
answered Jan 19 at 4:20
Felix MarinFelix Marin
68k7107142
68k7107142
add a comment |
add a comment |
$begingroup$
To prove uniform convergence of $f_n$ on a compact set $Omega in mathbb{C} setminus {0,-1,-2,ldots}$ note that
$$tag{*}f_n(s) = frac{n^s n!}{s(s+1)...(s+n)} = frac{n^s}{s} prod_{k=1}^n left(1 + frac{s}{k} right)^{-1} \ = left(frac{n}{n+1} right)^s frac{1}{s}prod_{k=1}^n left(1 + frac{1}{k} right)^{s}left(1 + frac{s}{k} right)^{-1} $$
where the last equality follows from
$$prod_{k=1}^n left(1 + frac{1}{k} right)^{s} = left[ prod_{k=1}^n left(frac{k+1}{k}right) right]^s = left(frac{(n+1)!}{n!}right)^s = (n+1)^s$$
Since $frac{1}{2} leqslant frac{n}{n+1} < 1$ it can be shown that $left(frac{n}{n+1} right)^s frac{1}{s}$ is uniformly bounded on $Omega$. Thus, uniform convergence of $f_n$ holds if the product on the RHS of (*) is uniformly convergent.
We can apply a general theorem for complex infinite products that states that the product $prod [1 + g_n(s)]$ is uniformly and absolutely convergent on a compact set if $sum |g_n(s)|$ is uniformly convergent. This holds here because for large $K$ we have
$$left(1 + frac{1}{k} right)^{s}left(1 + frac{s}{k} right)^{-1} = 1 + frac{s((s-1)}{2k^2} + mathcal{O}(k^{-3})$$
Thus, the product on the RHS of (*) and, consequently $f_n$ are uniformly convergent on the compact set $Omega$ that avoids singularities of the gamma function.
answered Jan 19 at 5:07
RRLRRL
51.2k42573
$endgroup$
$begingroup$
Here $s$ is a complex variable , so we can not have $log (ab) = log (a) + log (b) $ , then how to deduce that $log f_n(s) = -log s - underbrace{sleft(sum_{k=1}^nfrac{1}{k} - log nright)}_{A_n(s)} -underbrace{sum_{k=1}^nleft[log left(1 + frac{s}{k}right)- frac{s}{k} right]}_{B_n(s)}$
$endgroup$
– J.Guo
Jan 19 at 12:17
$begingroup$
@J.Guo: You are absolutely correct but that technicality can be avoided as shown in the revised answer proving uniform convergence. That should provide the piece you are missing.
$endgroup$
– RRL
Jan 20 at 2:47
add a comment |
$begingroup$
To prove uniform convergence of $f_n$ on a compact set $Omega in mathbb{C} setminus {0,-1,-2,ldots}$ note that
$$tag{*}f_n(s) = frac{n^s n!}{s(s+1)...(s+n)} = frac{n^s}{s} prod_{k=1}^n left(1 + frac{s}{k} right)^{-1} \ = left(frac{n}{n+1} right)^s frac{1}{s}prod_{k=1}^n left(1 + frac{1}{k} right)^{s}left(1 + frac{s}{k} right)^{-1} $$
where the last equality follows from
$$prod_{k=1}^n left(1 + frac{1}{k} right)^{s} = left[ prod_{k=1}^n left(frac{k+1}{k}right) right]^s = left(frac{(n+1)!}{n!}right)^s = (n+1)^s$$
Since $frac{1}{2} leqslant frac{n}{n+1} < 1$ it can be shown that $left(frac{n}{n+1} right)^s frac{1}{s}$ is uniformly bounded on $Omega$. Thus, uniform convergence of $f_n$ holds if the product on the RHS of (*) is uniformly convergent.
We can apply a general theorem for complex infinite products that states that the product $prod [1 + g_n(s)]$ is uniformly and absolutely convergent on a compact set if $sum |g_n(s)|$ is uniformly convergent. This holds here because for large $K$ we have
$$left(1 + frac{1}{k} right)^{s}left(1 + frac{s}{k} right)^{-1} = 1 + frac{s((s-1)}{2k^2} + mathcal{O}(k^{-3})$$
Thus, the product on the RHS of (*) and, consequently $f_n$ are uniformly convergent on the compact set $Omega$ that avoids singularities of the gamma function.
answered Jan 19 at 5:07
RRLRRL
51.2k42573
$endgroup$
$begingroup$
Here $s$ is a complex variable , so we can not have $log (ab) = log (a) + log (b) $ , then how to deduce that $log f_n(s) = -log s - underbrace{sleft(sum_{k=1}^nfrac{1}{k} - log nright)}_{A_n(s)} -underbrace{sum_{k=1}^nleft[log left(1 + frac{s}{k}right)- frac{s}{k} right]}_{B_n(s)}$
$endgroup$
– J.Guo
Jan 19 at 12:17
$begingroup$
@J.Guo: You are absolutely correct but that technicality can be avoided as shown in the revised answer proving uniform convergence. That should provide the piece you are missing.
$endgroup$
– RRL
Jan 20 at 2:47
add a comment |
$begingroup$
To prove uniform convergence of $f_n$ on a compact set $Omega in mathbb{C} setminus {0,-1,-2,ldots}$ note that
$$tag{*}f_n(s) = frac{n^s n!}{s(s+1)...(s+n)} = frac{n^s}{s} prod_{k=1}^n left(1 + frac{s}{k} right)^{-1} \ = left(frac{n}{n+1} right)^s frac{1}{s}prod_{k=1}^n left(1 + frac{1}{k} right)^{s}left(1 + frac{s}{k} right)^{-1} $$
where the last equality follows from
$$prod_{k=1}^n left(1 + frac{1}{k} right)^{s} = left[ prod_{k=1}^n left(frac{k+1}{k}right) right]^s = left(frac{(n+1)!}{n!}right)^s = (n+1)^s$$
Since $frac{1}{2} leqslant frac{n}{n+1} < 1$ it can be shown that $left(frac{n}{n+1} right)^s frac{1}{s}$ is uniformly bounded on $Omega$. Thus, uniform convergence of $f_n$ holds if the product on the RHS of (*) is uniformly convergent.
We can apply a general theorem for complex infinite products that states that the product $prod [1 + g_n(s)]$ is uniformly and absolutely convergent on a compact set if $sum |g_n(s)|$ is uniformly convergent. This holds here because for large $K$ we have
$$left(1 + frac{1}{k} right)^{s}left(1 + frac{s}{k} right)^{-1} = 1 + frac{s((s-1)}{2k^2} + mathcal{O}(k^{-3})$$
Thus, the product on the RHS of (*) and, consequently $f_n$ are uniformly convergent on the compact set $Omega$ that avoids singularities of the gamma function.
answered Jan 19 at 5:07
RRLRRL
51.2k42573
$endgroup$
To prove uniform convergence of $f_n$ on a compact set $Omega in mathbb{C} setminus {0,-1,-2,ldots}$ note that
$$tag{*}f_n(s) = frac{n^s n!}{s(s+1)...(s+n)} = frac{n^s}{s} prod_{k=1}^n left(1 + frac{s}{k} right)^{-1} \ = left(frac{n}{n+1} right)^s frac{1}{s}prod_{k=1}^n left(1 + frac{1}{k} right)^{s}left(1 + frac{s}{k} right)^{-1} $$
where the last equality follows from
$$prod_{k=1}^n left(1 + frac{1}{k} right)^{s} = left[ prod_{k=1}^n left(frac{k+1}{k}right) right]^s = left(frac{(n+1)!}{n!}right)^s = (n+1)^s$$
Since $frac{1}{2} leqslant frac{n}{n+1} < 1$ it can be shown that $left(frac{n}{n+1} right)^s frac{1}{s}$ is uniformly bounded on $Omega$. Thus, uniform convergence of $f_n$ holds if the product on the RHS of (*) is uniformly convergent.
We can apply a general theorem for complex infinite products that states that the product $prod [1 + g_n(s)]$ is uniformly and absolutely convergent on a compact set if $sum |g_n(s)|$ is uniformly convergent. This holds here because for large $K$ we have
$$left(1 + frac{1}{k} right)^{s}left(1 + frac{s}{k} right)^{-1} = 1 + frac{s((s-1)}{2k^2} + mathcal{O}(k^{-3})$$
Thus, the product on the RHS of (*) and, consequently $f_n$ are uniformly convergent on the compact set $Omega$ that avoids singularities of the gamma function.
answered Jan 19 at 5:07
RRLRRL
51.2k42573
edited Jan 20 at 2:45
answered Jan 19 at 5:07
RRLRRL
51.2k42573
answered Jan 19 at 5:07
RRLRRL
51.2k42573
answered Jan 19 at 5:07
RRLRRL
51.2k42573
51.2k42573
$begingroup$
Here $s$ is a complex variable , so we can not have $log (ab) = log (a) + log (b) $ , then how to deduce that $log f_n(s) = -log s - underbrace{sleft(sum_{k=1}^nfrac{1}{k} - log nright)}_{A_n(s)} -underbrace{sum_{k=1}^nleft[log left(1 + frac{s}{k}right)- frac{s}{k} right]}_{B_n(s)}$
$endgroup$
– J.Guo
Jan 19 at 12:17
$begingroup$
@J.Guo: You are absolutely correct but that technicality can be avoided as shown in the revised answer proving uniform convergence. That should provide the piece you are missing.
$endgroup$
– RRL
Jan 20 at 2:47
add a comment |
$begingroup$
Here $s$ is a complex variable , so we can not have $log (ab) = log (a) + log (b) $ , then how to deduce that $log f_n(s) = -log s - underbrace{sleft(sum_{k=1}^nfrac{1}{k} - log nright)}_{A_n(s)} -underbrace{sum_{k=1}^nleft[log left(1 + frac{s}{k}right)- frac{s}{k} right]}_{B_n(s)}$
$endgroup$
– J.Guo
Jan 19 at 12:17
$begingroup$
@J.Guo: You are absolutely correct but that technicality can be avoided as shown in the revised answer proving uniform convergence. That should provide the piece you are missing.
$endgroup$
– RRL
Jan 20 at 2:47
$begingroup$
Here $s$ is a complex variable , so we can not have $log (ab) = log (a) + log (b) $ , then how to deduce that $log f_n(s) = -log s - underbrace{sleft(sum_{k=1}^nfrac{1}{k} - log nright)}_{A_n(s)} -underbrace{sum_{k=1}^nleft[log left(1 + frac{s}{k}right)- frac{s}{k} right]}_{B_n(s)}$
$endgroup$
– J.Guo
Jan 19 at 12:17
$begingroup$
Here $s$ is a complex variable , so we can not have $log (ab) = log (a) + log (b) $ , then how to deduce that $log f_n(s) = -log s - underbrace{sleft(sum_{k=1}^nfrac{1}{k} - log nright)}_{A_n(s)} -underbrace{sum_{k=1}^nleft[log left(1 + frac{s}{k}right)- frac{s}{k} right]}_{B_n(s)}$
$endgroup$
– J.Guo
Jan 19 at 12:17
$begingroup$
@J.Guo: You are absolutely correct but that technicality can be avoided as shown in the revised answer proving uniform convergence. That should provide the piece you are missing.
$endgroup$
– RRL
Jan 20 at 2:47
$begingroup$
@J.Guo: You are absolutely correct but that technicality can be avoided as shown in the revised answer proving uniform convergence. That should provide the piece you are missing.
$endgroup$
– RRL
Jan 20 at 2:47
add a comment |
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$begingroup$
Definition of $Gamma$ you are using?
$endgroup$
– Simply Beautiful Art
Jan 18 at 15:13
$begingroup$
@ Simply Beautiful Art $Gamma_0(s)=int_0^{infty}e^{-t} t^{s-1} , dt$ whenever $Re(s)gt0$ and $Gamma(s)$ is the Analytic continuaton of $Gamma_0$ which defined a meromorphic function on $C$ .
$endgroup$
– J.Guo
Jan 18 at 15:33