Solving the limit of $lim_{xto 1-n} (exp(2 pi i x)-1)Gamma(x)$












0












$begingroup$


I am reading the book Introduction to Cyclotomic Fields. I found the following limit on page number 33. I have no idea how to obtain the following limit.



$$lim_{xto 1-n} (e^{(2 pi i x)}-1)Gamma(x)=frac{(2pi i)(-1)^{(n-1)}}{(n-1)!}.$$



Do we need to use L'Hospital's rule for this?
Is there any connection with the residue of Gamma function at negative values?



The residue at $z=k$ is given by
$operatorname{Re}s_{z=k} Gamma(z)=frac{(-1)^{k}}{k!}.$



Please help me to understand this.






complex-analysis number-theory limits gamma-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Should that be $lim_{xto1-n}cdots$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 18 '18 at 18:51










  • $begingroup$
    Yes, it should be $1-n$.
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 20:23










  • $begingroup$
    I think you mean the residue at $z=-k,$ not $z=k.$
    $endgroup$
    – Thomas Andrews
    Dec 18 '18 at 20:42
















0












$begingroup$


I am reading the book Introduction to Cyclotomic Fields. I found the following limit on page number 33. I have no idea how to obtain the following limit.



$$lim_{xto 1-n} (e^{(2 pi i x)}-1)Gamma(x)=frac{(2pi i)(-1)^{(n-1)}}{(n-1)!}.$$



Do we need to use L'Hospital's rule for this?
Is there any connection with the residue of Gamma function at negative values?



The residue at $z=k$ is given by
$operatorname{Re}s_{z=k} Gamma(z)=frac{(-1)^{k}}{k!}.$



Please help me to understand this.






complex-analysis number-theory limits gamma-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Should that be $lim_{xto1-n}cdots$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 18 '18 at 18:51










  • $begingroup$
    Yes, it should be $1-n$.
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 20:23










  • $begingroup$
    I think you mean the residue at $z=-k,$ not $z=k.$
    $endgroup$
    – Thomas Andrews
    Dec 18 '18 at 20:42














0












0








0





$begingroup$


I am reading the book Introduction to Cyclotomic Fields. I found the following limit on page number 33. I have no idea how to obtain the following limit.



$$lim_{xto 1-n} (e^{(2 pi i x)}-1)Gamma(x)=frac{(2pi i)(-1)^{(n-1)}}{(n-1)!}.$$



Do we need to use L'Hospital's rule for this?
Is there any connection with the residue of Gamma function at negative values?



The residue at $z=k$ is given by
$operatorname{Re}s_{z=k} Gamma(z)=frac{(-1)^{k}}{k!}.$



Please help me to understand this.






complex-analysis number-theory limits gamma-function







share|cite|improve this question











$endgroup$




I am reading the book Introduction to Cyclotomic Fields. I found the following limit on page number 33. I have no idea how to obtain the following limit.



$$lim_{xto 1-n} (e^{(2 pi i x)}-1)Gamma(x)=frac{(2pi i)(-1)^{(n-1)}}{(n-1)!}.$$



Do we need to use L'Hospital's rule for this?
Is there any connection with the residue of Gamma function at negative values?



The residue at $z=k$ is given by
$operatorname{Re}s_{z=k} Gamma(z)=frac{(-1)^{k}}{k!}.$



Please help me to understand this.






complex-analysis number-theory limits gamma-function




complex-analysis number-theory limits gamma-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 21:09









Thomas Andrews

130k11146297




130k11146297










asked Dec 18 '18 at 18:48









Soma WickSoma Wick

305




305








  • 1




    $begingroup$
    Should that be $lim_{xto1-n}cdots$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 18 '18 at 18:51










  • $begingroup$
    Yes, it should be $1-n$.
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 20:23










  • $begingroup$
    I think you mean the residue at $z=-k,$ not $z=k.$
    $endgroup$
    – Thomas Andrews
    Dec 18 '18 at 20:42














  • 1




    $begingroup$
    Should that be $lim_{xto1-n}cdots$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 18 '18 at 18:51










  • $begingroup$
    Yes, it should be $1-n$.
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 20:23










  • $begingroup$
    I think you mean the residue at $z=-k,$ not $z=k.$
    $endgroup$
    – Thomas Andrews
    Dec 18 '18 at 20:42








1




1




$begingroup$
Should that be $lim_{xto1-n}cdots$?
$endgroup$
– Lord Shark the Unknown
Dec 18 '18 at 18:51




$begingroup$
Should that be $lim_{xto1-n}cdots$?
$endgroup$
– Lord Shark the Unknown
Dec 18 '18 at 18:51












$begingroup$
Yes, it should be $1-n$.
$endgroup$
– Soma Wick
Dec 18 '18 at 20:23




$begingroup$
Yes, it should be $1-n$.
$endgroup$
– Soma Wick
Dec 18 '18 at 20:23












$begingroup$
I think you mean the residue at $z=-k,$ not $z=k.$
$endgroup$
– Thomas Andrews
Dec 18 '18 at 20:42




$begingroup$
I think you mean the residue at $z=-k,$ not $z=k.$
$endgroup$
– Thomas Andrews
Dec 18 '18 at 20:42










1 Answer
1






active

oldest

votes


















0












$begingroup$

Start:



Note that if $f(x)=e^{2pi i x}$ then $f(1-n)=1$ when $n$ is an integer, and you can write your limit is $$lim_{xto 1-n} frac{f(x)-f(1-n)}{x-(1-n)} cdot (x-(1-n)) Gamma(x)=f'(1-n) cdot lim_{nto 1-n}(x-(1-n))Gamma(x)$$



But $f'(1-n)=2pi i$, so you need to show:



$$lim_{xto 1-n} (x-(1-n))Gamma(x)=frac{(-1)^{n-1}}{(n-1)!}$$



This follows from your residue result (corrected to be at $z=-k$,) if you can show that the pole at $z=1-n$ is of degree $1.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much. But I don`t understand the last part. How can we get the result with the residue part? Is there a theory on that?
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 21:12












  • $begingroup$
    The residue for a function $f(x)=sum_{k=-N}^{infty} a_n(x-x_0)^k$ at $x=x_0$ is $a_{-1}$. If $N=-1,$ so the pole at $x_0$ is not of degree $2$ or more, then the reside at $x_0$ is $lim_{xto x_0} (x-x_0)f(x).$ @SomaWick
    $endgroup$
    – Thomas Andrews
    Dec 18 '18 at 21:15












  • $begingroup$
    I understood. Thank you
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 21:35











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Start:



Note that if $f(x)=e^{2pi i x}$ then $f(1-n)=1$ when $n$ is an integer, and you can write your limit is $$lim_{xto 1-n} frac{f(x)-f(1-n)}{x-(1-n)} cdot (x-(1-n)) Gamma(x)=f'(1-n) cdot lim_{nto 1-n}(x-(1-n))Gamma(x)$$



But $f'(1-n)=2pi i$, so you need to show:



$$lim_{xto 1-n} (x-(1-n))Gamma(x)=frac{(-1)^{n-1}}{(n-1)!}$$



This follows from your residue result (corrected to be at $z=-k$,) if you can show that the pole at $z=1-n$ is of degree $1.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much. But I don`t understand the last part. How can we get the result with the residue part? Is there a theory on that?
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 21:12












  • $begingroup$
    The residue for a function $f(x)=sum_{k=-N}^{infty} a_n(x-x_0)^k$ at $x=x_0$ is $a_{-1}$. If $N=-1,$ so the pole at $x_0$ is not of degree $2$ or more, then the reside at $x_0$ is $lim_{xto x_0} (x-x_0)f(x).$ @SomaWick
    $endgroup$
    – Thomas Andrews
    Dec 18 '18 at 21:15












  • $begingroup$
    I understood. Thank you
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 21:35
















0












$begingroup$

Start:



Note that if $f(x)=e^{2pi i x}$ then $f(1-n)=1$ when $n$ is an integer, and you can write your limit is $$lim_{xto 1-n} frac{f(x)-f(1-n)}{x-(1-n)} cdot (x-(1-n)) Gamma(x)=f'(1-n) cdot lim_{nto 1-n}(x-(1-n))Gamma(x)$$



But $f'(1-n)=2pi i$, so you need to show:



$$lim_{xto 1-n} (x-(1-n))Gamma(x)=frac{(-1)^{n-1}}{(n-1)!}$$



This follows from your residue result (corrected to be at $z=-k$,) if you can show that the pole at $z=1-n$ is of degree $1.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much. But I don`t understand the last part. How can we get the result with the residue part? Is there a theory on that?
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 21:12












  • $begingroup$
    The residue for a function $f(x)=sum_{k=-N}^{infty} a_n(x-x_0)^k$ at $x=x_0$ is $a_{-1}$. If $N=-1,$ so the pole at $x_0$ is not of degree $2$ or more, then the reside at $x_0$ is $lim_{xto x_0} (x-x_0)f(x).$ @SomaWick
    $endgroup$
    – Thomas Andrews
    Dec 18 '18 at 21:15












  • $begingroup$
    I understood. Thank you
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 21:35














0












0








0





$begingroup$

Start:



Note that if $f(x)=e^{2pi i x}$ then $f(1-n)=1$ when $n$ is an integer, and you can write your limit is $$lim_{xto 1-n} frac{f(x)-f(1-n)}{x-(1-n)} cdot (x-(1-n)) Gamma(x)=f'(1-n) cdot lim_{nto 1-n}(x-(1-n))Gamma(x)$$



But $f'(1-n)=2pi i$, so you need to show:



$$lim_{xto 1-n} (x-(1-n))Gamma(x)=frac{(-1)^{n-1}}{(n-1)!}$$



This follows from your residue result (corrected to be at $z=-k$,) if you can show that the pole at $z=1-n$ is of degree $1.$






share|cite|improve this answer











$endgroup$



Start:



Note that if $f(x)=e^{2pi i x}$ then $f(1-n)=1$ when $n$ is an integer, and you can write your limit is $$lim_{xto 1-n} frac{f(x)-f(1-n)}{x-(1-n)} cdot (x-(1-n)) Gamma(x)=f'(1-n) cdot lim_{nto 1-n}(x-(1-n))Gamma(x)$$



But $f'(1-n)=2pi i$, so you need to show:



$$lim_{xto 1-n} (x-(1-n))Gamma(x)=frac{(-1)^{n-1}}{(n-1)!}$$



This follows from your residue result (corrected to be at $z=-k$,) if you can show that the pole at $z=1-n$ is of degree $1.$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 20:43

























answered Dec 18 '18 at 20:36









Thomas AndrewsThomas Andrews

130k11146297




130k11146297












  • $begingroup$
    Thank you very much. But I don`t understand the last part. How can we get the result with the residue part? Is there a theory on that?
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 21:12












  • $begingroup$
    The residue for a function $f(x)=sum_{k=-N}^{infty} a_n(x-x_0)^k$ at $x=x_0$ is $a_{-1}$. If $N=-1,$ so the pole at $x_0$ is not of degree $2$ or more, then the reside at $x_0$ is $lim_{xto x_0} (x-x_0)f(x).$ @SomaWick
    $endgroup$
    – Thomas Andrews
    Dec 18 '18 at 21:15












  • $begingroup$
    I understood. Thank you
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 21:35


















  • $begingroup$
    Thank you very much. But I don`t understand the last part. How can we get the result with the residue part? Is there a theory on that?
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 21:12












  • $begingroup$
    The residue for a function $f(x)=sum_{k=-N}^{infty} a_n(x-x_0)^k$ at $x=x_0$ is $a_{-1}$. If $N=-1,$ so the pole at $x_0$ is not of degree $2$ or more, then the reside at $x_0$ is $lim_{xto x_0} (x-x_0)f(x).$ @SomaWick
    $endgroup$
    – Thomas Andrews
    Dec 18 '18 at 21:15












  • $begingroup$
    I understood. Thank you
    $endgroup$
    – Soma Wick
    Dec 18 '18 at 21:35
















$begingroup$
Thank you very much. But I don`t understand the last part. How can we get the result with the residue part? Is there a theory on that?
$endgroup$
– Soma Wick
Dec 18 '18 at 21:12






$begingroup$
Thank you very much. But I don`t understand the last part. How can we get the result with the residue part? Is there a theory on that?
$endgroup$
– Soma Wick
Dec 18 '18 at 21:12














$begingroup$
The residue for a function $f(x)=sum_{k=-N}^{infty} a_n(x-x_0)^k$ at $x=x_0$ is $a_{-1}$. If $N=-1,$ so the pole at $x_0$ is not of degree $2$ or more, then the reside at $x_0$ is $lim_{xto x_0} (x-x_0)f(x).$ @SomaWick
$endgroup$
– Thomas Andrews
Dec 18 '18 at 21:15






$begingroup$
The residue for a function $f(x)=sum_{k=-N}^{infty} a_n(x-x_0)^k$ at $x=x_0$ is $a_{-1}$. If $N=-1,$ so the pole at $x_0$ is not of degree $2$ or more, then the reside at $x_0$ is $lim_{xto x_0} (x-x_0)f(x).$ @SomaWick
$endgroup$
– Thomas Andrews
Dec 18 '18 at 21:15














$begingroup$
I understood. Thank you
$endgroup$
– Soma Wick
Dec 18 '18 at 21:35




$begingroup$
I understood. Thank you
$endgroup$
– Soma Wick
Dec 18 '18 at 21:35


















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