Binomial series for $|z|=1$?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite
1












The binomial series is given by
$$(*) quad (1-z)^-alpha=sum_k=0^inftyfracGamma(k+alpha)Gamma(alpha), k! , z^k; ,, |z|<1,$$
where $alpha in mathbb C$ is an arbitrary complex number. My question is that the identity $(*)$ is true for $|z|=1$ ?



In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$, i.e, $(1-xi)^-1=...; xi in mathbb S^1$.




calculus sequences-and-series complex-analysis binomial-coefficients




share|cite|improve this question

















  • 2




    The convergence on the unit circle depends on what $alpha$ is, and what $z$ is.
    – Lord Shark the Unknown
    Aug 2 at 10:32










  • if $z in mathbfR$ and $|z| = 1$ then LHS is 0, while RHS is some positive sum.
    – pointguard0
    Aug 2 at 10:33






  • 3




    For example, the series for $(1-z)^-1$ diverges for $|z|=1$ but the series for $(1-z)^-1/2$ converges for $z=-1$.
    – GEdgar
    Aug 2 at 10:36











  • In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$.
    – Z. Alfata
    Aug 2 at 10:40











  • @Z.Alfata When $alpha=-1$ then $Gamma(alpha)$ doesn't exist...
    – Lord Shark the Unknown
    Aug 2 at 10:42














up vote
0
down vote

favorite
1












The binomial series is given by
$$(*) quad (1-z)^-alpha=sum_k=0^inftyfracGamma(k+alpha)Gamma(alpha), k! , z^k; ,, |z|<1,$$
where $alpha in mathbb C$ is an arbitrary complex number. My question is that the identity $(*)$ is true for $|z|=1$ ?



In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$, i.e, $(1-xi)^-1=...; xi in mathbb S^1$.




calculus sequences-and-series complex-analysis binomial-coefficients




share|cite|improve this question

















  • 2




    The convergence on the unit circle depends on what $alpha$ is, and what $z$ is.
    – Lord Shark the Unknown
    Aug 2 at 10:32










  • if $z in mathbfR$ and $|z| = 1$ then LHS is 0, while RHS is some positive sum.
    – pointguard0
    Aug 2 at 10:33






  • 3




    For example, the series for $(1-z)^-1$ diverges for $|z|=1$ but the series for $(1-z)^-1/2$ converges for $z=-1$.
    – GEdgar
    Aug 2 at 10:36











  • In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$.
    – Z. Alfata
    Aug 2 at 10:40











  • @Z.Alfata When $alpha=-1$ then $Gamma(alpha)$ doesn't exist...
    – Lord Shark the Unknown
    Aug 2 at 10:42












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





The binomial series is given by
$$(*) quad (1-z)^-alpha=sum_k=0^inftyfracGamma(k+alpha)Gamma(alpha), k! , z^k; ,, |z|<1,$$
where $alpha in mathbb C$ is an arbitrary complex number. My question is that the identity $(*)$ is true for $|z|=1$ ?



In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$, i.e, $(1-xi)^-1=...; xi in mathbb S^1$.




calculus sequences-and-series complex-analysis binomial-coefficients




share|cite|improve this question













The binomial series is given by
$$(*) quad (1-z)^-alpha=sum_k=0^inftyfracGamma(k+alpha)Gamma(alpha), k! , z^k; ,, |z|<1,$$
where $alpha in mathbb C$ is an arbitrary complex number. My question is that the identity $(*)$ is true for $|z|=1$ ?



In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$, i.e, $(1-xi)^-1=...; xi in mathbb S^1$.





calculus sequences-and-series complex-analysis binomial-coefficients





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 10:43
























asked Aug 2 at 10:30









Z. Alfata

864413




864413







  • 2




    The convergence on the unit circle depends on what $alpha$ is, and what $z$ is.
    – Lord Shark the Unknown
    Aug 2 at 10:32










  • if $z in mathbfR$ and $|z| = 1$ then LHS is 0, while RHS is some positive sum.
    – pointguard0
    Aug 2 at 10:33






  • 3




    For example, the series for $(1-z)^-1$ diverges for $|z|=1$ but the series for $(1-z)^-1/2$ converges for $z=-1$.
    – GEdgar
    Aug 2 at 10:36











  • In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$.
    – Z. Alfata
    Aug 2 at 10:40











  • @Z.Alfata When $alpha=-1$ then $Gamma(alpha)$ doesn't exist...
    – Lord Shark the Unknown
    Aug 2 at 10:42












  • 2




    The convergence on the unit circle depends on what $alpha$ is, and what $z$ is.
    – Lord Shark the Unknown
    Aug 2 at 10:32










  • if $z in mathbfR$ and $|z| = 1$ then LHS is 0, while RHS is some positive sum.
    – pointguard0
    Aug 2 at 10:33






  • 3




    For example, the series for $(1-z)^-1$ diverges for $|z|=1$ but the series for $(1-z)^-1/2$ converges for $z=-1$.
    – GEdgar
    Aug 2 at 10:36











  • In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$.
    – Z. Alfata
    Aug 2 at 10:40











  • @Z.Alfata When $alpha=-1$ then $Gamma(alpha)$ doesn't exist...
    – Lord Shark the Unknown
    Aug 2 at 10:42







2




2




The convergence on the unit circle depends on what $alpha$ is, and what $z$ is.
– Lord Shark the Unknown
Aug 2 at 10:32




The convergence on the unit circle depends on what $alpha$ is, and what $z$ is.
– Lord Shark the Unknown
Aug 2 at 10:32












if $z in mathbfR$ and $|z| = 1$ then LHS is 0, while RHS is some positive sum.
– pointguard0
Aug 2 at 10:33




if $z in mathbfR$ and $|z| = 1$ then LHS is 0, while RHS is some positive sum.
– pointguard0
Aug 2 at 10:33




3




3




For example, the series for $(1-z)^-1$ diverges for $|z|=1$ but the series for $(1-z)^-1/2$ converges for $z=-1$.
– GEdgar
Aug 2 at 10:36





For example, the series for $(1-z)^-1$ diverges for $|z|=1$ but the series for $(1-z)^-1/2$ converges for $z=-1$.
– GEdgar
Aug 2 at 10:36













In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$.
– Z. Alfata
Aug 2 at 10:40





In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$.
– Z. Alfata
Aug 2 at 10:40













@Z.Alfata When $alpha=-1$ then $Gamma(alpha)$ doesn't exist...
– Lord Shark the Unknown
Aug 2 at 10:42




@Z.Alfata When $alpha=-1$ then $Gamma(alpha)$ doesn't exist...
– Lord Shark the Unknown
Aug 2 at 10:42















active

oldest

votes











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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2869929%2fbinomial-series-for-z-1%23new-answer', 'question_page');

);

Post as a guest






















active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes










 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2869929%2fbinomial-series-for-z-1%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/… – Hector Blandin Nov 17 &
Read more

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an
Read more