Solving the limit of $lim_{xto 1-n} (exp(2 pi i x)-1)Gamma(x)$
$begingroup$
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
edited Dec 18 '18 at 21:09
Thomas Andrews
130k11146297
asked Dec 18 '18 at 18:48
Soma WickSoma Wick
225
$endgroup$
add a comment |
$begingroup$
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
edited Dec 18 '18 at 21:09
Thomas Andrews
130k11146297
asked Dec 18 '18 at 18:48
Soma WickSoma Wick
225
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 18 '18 at 21:09
Thomas Andrews
130k11146297
asked Dec 18 '18 at 18:48
Soma WickSoma Wick
225
$endgroup$
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
edited Dec 18 '18 at 21:09
Thomas Andrews
130k11146297
asked Dec 18 '18 at 18:48
Soma WickSoma Wick
225
edited Dec 18 '18 at 21:09
Thomas Andrews
130k11146297
asked Dec 18 '18 at 18:48
Soma WickSoma Wick
225
edited Dec 18 '18 at 21:09
Thomas Andrews
130k11146297
edited Dec 18 '18 at 21:09
Thomas Andrews
130k11146297
edited Dec 18 '18 at 21:09
Thomas Andrews
130k11146297
130k11146297
asked Dec 18 '18 at 18:48
Soma WickSoma Wick
225
asked Dec 18 '18 at 18:48
Soma WickSoma Wick
225
asked Dec 18 '18 at 18:48
Soma WickSoma Wick
225
225
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $1.$
answered Dec 18 '18 at 20:36
Thomas AndrewsThomas Andrews
130k11146297
$endgroup$
$begingroup$
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045534%2fsolving-the-limit-of-lim-x-to-1-n-exp2-pi-i-x-1-gammax%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $1.$
answered Dec 18 '18 at 20:36
Thomas AndrewsThomas Andrews
130k11146297
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $1.$
answered Dec 18 '18 at 20:36
Thomas AndrewsThomas Andrews
130k11146297
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $1.$
answered Dec 18 '18 at 20:36
Thomas AndrewsThomas Andrews
130k11146297
$endgroup$
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 $1.$
answered Dec 18 '18 at 20:36
Thomas AndrewsThomas Andrews
130k11146297
edited Dec 18 '18 at 20:43
answered Dec 18 '18 at 20:36
Thomas AndrewsThomas Andrews
130k11146297
answered Dec 18 '18 at 20:36
Thomas AndrewsThomas Andrews
130k11146297
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
add a comment |
$begingroup$
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
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045534%2fsolving-the-limit-of-lim-x-to-1-n-exp2-pi-i-x-1-gammax%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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