Generalizing rising and falling factorial to complex arguments preserving zeroes











up vote
0
down vote

favorite












Falling factorials count injective functions. There are no injective functions $ A to B $ if $|A| gt |B|$ . Is there a way to extend the definition of the falling factorial, ideally to $mathbb{C} times mathbb{C} to mathbb{C}$, so that the zeroes are preserved? The most obvious way of informally generalizing the falling factorial involves taking ratios of the gamma function, but I'm not convinced that that works out.





When the right argument of the rising factorial is a natural number, it is defined in the following way:



$$ (x)^{overline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x + i) $$



$$ (x)^{overline{k}} = (x)(x+1)cdots(x+k-2)(x+k-1) $$



And the falling factorial is defined in the following way



$$ (x)^{underline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x - i) $$



$$ (x)^{underline{k}} = (x)(x-1)cdots (x-k+2)(x-k+1) $$



The falling factorial can be defined in terms of the rising factorial in two ways.



$$ (x)^{underline{k}} = (-1)^k (-x)^{overline{k}} $$



or



$$ (x)^{underline{k}} = (x-k+1) ^ {overline{k}} $$



The rising factorial satisfies the following identity:



$$ (x)^{overline{k}} = frac{(x+k-1)!}{(x-1)!} $$



Which suggests the generalization



$$ (x)^{overline{k}} = frac{Gamma(x+k)}{Gamma(x)} $$



and



$$ (x)^{underline{k}} = frac{Gamma(x + 1)}{Gamma(x-k + 1)} $$



This makes me nervous since $0 notin text{Im}(Gamma)$ and $Gamma$ has a pole when $ x le -1 $ .





Does a nice way of extending the rising/falling factorial to complex arguments exist?










share|cite|improve this question
























  • I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers
    – G Cab
    Nov 17 at 23:18






  • 1




    by the way, it is actually $$ x^{,underline {,k,} } = left( {x - k + 1} right)^{,overline {,k,} } = {{Gamma (x + 1)} over {Gamma (x - k + 1)}} $$
    – G Cab
    Nov 17 at 23:24















up vote
0
down vote

favorite












Falling factorials count injective functions. There are no injective functions $ A to B $ if $|A| gt |B|$ . Is there a way to extend the definition of the falling factorial, ideally to $mathbb{C} times mathbb{C} to mathbb{C}$, so that the zeroes are preserved? The most obvious way of informally generalizing the falling factorial involves taking ratios of the gamma function, but I'm not convinced that that works out.





When the right argument of the rising factorial is a natural number, it is defined in the following way:



$$ (x)^{overline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x + i) $$



$$ (x)^{overline{k}} = (x)(x+1)cdots(x+k-2)(x+k-1) $$



And the falling factorial is defined in the following way



$$ (x)^{underline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x - i) $$



$$ (x)^{underline{k}} = (x)(x-1)cdots (x-k+2)(x-k+1) $$



The falling factorial can be defined in terms of the rising factorial in two ways.



$$ (x)^{underline{k}} = (-1)^k (-x)^{overline{k}} $$



or



$$ (x)^{underline{k}} = (x-k+1) ^ {overline{k}} $$



The rising factorial satisfies the following identity:



$$ (x)^{overline{k}} = frac{(x+k-1)!}{(x-1)!} $$



Which suggests the generalization



$$ (x)^{overline{k}} = frac{Gamma(x+k)}{Gamma(x)} $$



and



$$ (x)^{underline{k}} = frac{Gamma(x + 1)}{Gamma(x-k + 1)} $$



This makes me nervous since $0 notin text{Im}(Gamma)$ and $Gamma$ has a pole when $ x le -1 $ .





Does a nice way of extending the rising/falling factorial to complex arguments exist?










share|cite|improve this question
























  • I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers
    – G Cab
    Nov 17 at 23:18






  • 1




    by the way, it is actually $$ x^{,underline {,k,} } = left( {x - k + 1} right)^{,overline {,k,} } = {{Gamma (x + 1)} over {Gamma (x - k + 1)}} $$
    – G Cab
    Nov 17 at 23:24













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Falling factorials count injective functions. There are no injective functions $ A to B $ if $|A| gt |B|$ . Is there a way to extend the definition of the falling factorial, ideally to $mathbb{C} times mathbb{C} to mathbb{C}$, so that the zeroes are preserved? The most obvious way of informally generalizing the falling factorial involves taking ratios of the gamma function, but I'm not convinced that that works out.





When the right argument of the rising factorial is a natural number, it is defined in the following way:



$$ (x)^{overline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x + i) $$



$$ (x)^{overline{k}} = (x)(x+1)cdots(x+k-2)(x+k-1) $$



And the falling factorial is defined in the following way



$$ (x)^{underline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x - i) $$



$$ (x)^{underline{k}} = (x)(x-1)cdots (x-k+2)(x-k+1) $$



The falling factorial can be defined in terms of the rising factorial in two ways.



$$ (x)^{underline{k}} = (-1)^k (-x)^{overline{k}} $$



or



$$ (x)^{underline{k}} = (x-k+1) ^ {overline{k}} $$



The rising factorial satisfies the following identity:



$$ (x)^{overline{k}} = frac{(x+k-1)!}{(x-1)!} $$



Which suggests the generalization



$$ (x)^{overline{k}} = frac{Gamma(x+k)}{Gamma(x)} $$



and



$$ (x)^{underline{k}} = frac{Gamma(x + 1)}{Gamma(x-k + 1)} $$



This makes me nervous since $0 notin text{Im}(Gamma)$ and $Gamma$ has a pole when $ x le -1 $ .





Does a nice way of extending the rising/falling factorial to complex arguments exist?










share|cite|improve this question















Falling factorials count injective functions. There are no injective functions $ A to B $ if $|A| gt |B|$ . Is there a way to extend the definition of the falling factorial, ideally to $mathbb{C} times mathbb{C} to mathbb{C}$, so that the zeroes are preserved? The most obvious way of informally generalizing the falling factorial involves taking ratios of the gamma function, but I'm not convinced that that works out.





When the right argument of the rising factorial is a natural number, it is defined in the following way:



$$ (x)^{overline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x + i) $$



$$ (x)^{overline{k}} = (x)(x+1)cdots(x+k-2)(x+k-1) $$



And the falling factorial is defined in the following way



$$ (x)^{underline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x - i) $$



$$ (x)^{underline{k}} = (x)(x-1)cdots (x-k+2)(x-k+1) $$



The falling factorial can be defined in terms of the rising factorial in two ways.



$$ (x)^{underline{k}} = (-1)^k (-x)^{overline{k}} $$



or



$$ (x)^{underline{k}} = (x-k+1) ^ {overline{k}} $$



The rising factorial satisfies the following identity:



$$ (x)^{overline{k}} = frac{(x+k-1)!}{(x-1)!} $$



Which suggests the generalization



$$ (x)^{overline{k}} = frac{Gamma(x+k)}{Gamma(x)} $$



and



$$ (x)^{underline{k}} = frac{Gamma(x + 1)}{Gamma(x-k + 1)} $$



This makes me nervous since $0 notin text{Im}(Gamma)$ and $Gamma$ has a pole when $ x le -1 $ .





Does a nice way of extending the rising/falling factorial to complex arguments exist?







factorial analytic-continuation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 17 at 23:50

























asked Nov 17 at 20:38









Gregory Nisbet

403311




403311












  • I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers
    – G Cab
    Nov 17 at 23:18






  • 1




    by the way, it is actually $$ x^{,underline {,k,} } = left( {x - k + 1} right)^{,overline {,k,} } = {{Gamma (x + 1)} over {Gamma (x - k + 1)}} $$
    – G Cab
    Nov 17 at 23:24


















  • I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers
    – G Cab
    Nov 17 at 23:18






  • 1




    by the way, it is actually $$ x^{,underline {,k,} } = left( {x - k + 1} right)^{,overline {,k,} } = {{Gamma (x + 1)} over {Gamma (x - k + 1)}} $$
    – G Cab
    Nov 17 at 23:24
















I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers
– G Cab
Nov 17 at 23:18




I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers
– G Cab
Nov 17 at 23:18




1




1




by the way, it is actually $$ x^{,underline {,k,} } = left( {x - k + 1} right)^{,overline {,k,} } = {{Gamma (x + 1)} over {Gamma (x - k + 1)}} $$
– G Cab
Nov 17 at 23:24




by the way, it is actually $$ x^{,underline {,k,} } = left( {x - k + 1} right)^{,overline {,k,} } = {{Gamma (x + 1)} over {Gamma (x - k + 1)}} $$
– G Cab
Nov 17 at 23:24















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: 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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002783%2fgeneralizing-rising-and-falling-factorial-to-complex-arguments-preserving-zeroes%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown






























active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002783%2fgeneralizing-rising-and-falling-factorial-to-complex-arguments-preserving-zeroes%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Plaza Victoria

In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...