Trying to Express A Factorial As A Polynomial
$begingroup$
I'd like to express the following as a polynomial.
$$(a-1)(a-2)(a-3) . . . (a-b)$$
where $b<a$
I'm currently working on it now, but wanted to see if anyone's already done it, or already know what the answer is.
polynomials factorial
asked May 5 '16 at 22:36
Ben MarconiBen Marconi
62
$endgroup$
add a comment |
$begingroup$
I'd like to express the following as a polynomial.
$$(a-1)(a-2)(a-3) . . . (a-b)$$
where $b<a$
I'm currently working on it now, but wanted to see if anyone's already done it, or already know what the answer is.
polynomials factorial
asked May 5 '16 at 22:36
Ben MarconiBen Marconi
62
$endgroup$
5
$begingroup$
That is already a polynomial :-)
$endgroup$
– parsiad
May 5 '16 at 22:41
$begingroup$
Expended form : $sum_{i=1}^b a^i (-1)^{b-i} (sum_{I subseteq[1,b], |I|=b-i+1} prod_{k in I} k) $. Do you need something clearer?
$endgroup$
– Vincent
May 5 '16 at 22:43
$begingroup$
This is often called the "falling factorial" and often written as $(a-1)_b$. The coefficients are messy. en.wikipedia.org/wiki/Falling_and_rising_factorials
$endgroup$
– Thomas Andrews
May 5 '16 at 23:14
add a comment |
$begingroup$
I'd like to express the following as a polynomial.
$$(a-1)(a-2)(a-3) . . . (a-b)$$
where $b<a$
I'm currently working on it now, but wanted to see if anyone's already done it, or already know what the answer is.
polynomials factorial
asked May 5 '16 at 22:36
Ben MarconiBen Marconi
62
$endgroup$
I'd like to express the following as a polynomial.
$$(a-1)(a-2)(a-3) . . . (a-b)$$
where $b<a$
I'm currently working on it now, but wanted to see if anyone's already done it, or already know what the answer is.
polynomials factorial
polynomials factorial
asked May 5 '16 at 22:36
Ben MarconiBen Marconi
62
asked May 5 '16 at 22:36
Ben MarconiBen Marconi
62
asked May 5 '16 at 22:36
Ben MarconiBen Marconi
62
asked May 5 '16 at 22:36
Ben MarconiBen Marconi
62
asked May 5 '16 at 22:36
Ben MarconiBen Marconi
62
62
5
$begingroup$
That is already a polynomial :-)
$endgroup$
– parsiad
May 5 '16 at 22:41
$begingroup$
Expended form : $sum_{i=1}^b a^i (-1)^{b-i} (sum_{I subseteq[1,b], |I|=b-i+1} prod_{k in I} k) $. Do you need something clearer?
$endgroup$
– Vincent
May 5 '16 at 22:43
$begingroup$
This is often called the "falling factorial" and often written as $(a-1)_b$. The coefficients are messy. en.wikipedia.org/wiki/Falling_and_rising_factorials
$endgroup$
– Thomas Andrews
May 5 '16 at 23:14
add a comment |
5
$begingroup$
That is already a polynomial :-)
$endgroup$
– parsiad
May 5 '16 at 22:41
$begingroup$
Expended form : $sum_{i=1}^b a^i (-1)^{b-i} (sum_{I subseteq[1,b], |I|=b-i+1} prod_{k in I} k) $. Do you need something clearer?
$endgroup$
– Vincent
May 5 '16 at 22:43
$begingroup$
This is often called the "falling factorial" and often written as $(a-1)_b$. The coefficients are messy. en.wikipedia.org/wiki/Falling_and_rising_factorials
$endgroup$
– Thomas Andrews
May 5 '16 at 23:14
5
5
$begingroup$
That is already a polynomial :-)
$endgroup$
– parsiad
May 5 '16 at 22:41
$begingroup$
That is already a polynomial :-)
$endgroup$
– parsiad
May 5 '16 at 22:41
$begingroup$
Expended form : $sum_{i=1}^b a^i (-1)^{b-i} (sum_{I subseteq[1,b], |I|=b-i+1} prod_{k in I} k) $. Do you need something clearer?
$endgroup$
– Vincent
May 5 '16 at 22:43
$begingroup$
Expended form : $sum_{i=1}^b a^i (-1)^{b-i} (sum_{I subseteq[1,b], |I|=b-i+1} prod_{k in I} k) $. Do you need something clearer?
$endgroup$
– Vincent
May 5 '16 at 22:43
$begingroup$
This is often called the "falling factorial" and often written as $(a-1)_b$. The coefficients are messy. en.wikipedia.org/wiki/Falling_and_rising_factorials
$endgroup$
– Thomas Andrews
May 5 '16 at 23:14
$begingroup$
This is often called the "falling factorial" and often written as $(a-1)_b$. The coefficients are messy. en.wikipedia.org/wiki/Falling_and_rising_factorials
$endgroup$
– Thomas Andrews
May 5 '16 at 23:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here are some asymptotic results of mine
that might be of some use
(although the product
goes in the wrong direction):
$$
lim_{n to infty}dfrac{(x+(n-1)/2)^n-x(x+1)...(x+n-1)}{(x+(n-1)/2)^{n-2}(n^3-n)/24}
=1
$$
and
$$
lim_{n to infty}
(x+(n-1)/2)dfrac{(x+(n-1)/2)-(x(x+1)...(x+n-1))^{1/n}}
{(n^2-1)/24}
=1
$$
An outline of the proofs is here:
Limit of $sqrt[n]{(x+1)...(x+n)} - x$ as $x to +infty$
answered May 5 '16 at 23:44
marty cohenmarty cohen
75.4k549130
$endgroup$
add a comment |
Your Answer
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%2f1773384%2ftrying-to-express-a-factorial-as-a-polynomial%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
$begingroup$
Here are some asymptotic results of mine
that might be of some use
(although the product
goes in the wrong direction):
$$
lim_{n to infty}dfrac{(x+(n-1)/2)^n-x(x+1)...(x+n-1)}{(x+(n-1)/2)^{n-2}(n^3-n)/24}
=1
$$
and
$$
lim_{n to infty}
(x+(n-1)/2)dfrac{(x+(n-1)/2)-(x(x+1)...(x+n-1))^{1/n}}
{(n^2-1)/24}
=1
$$
An outline of the proofs is here:
Limit of $sqrt[n]{(x+1)...(x+n)} - x$ as $x to +infty$
answered May 5 '16 at 23:44
marty cohenmarty cohen
75.4k549130
$endgroup$
add a comment |
$begingroup$
Here are some asymptotic results of mine
that might be of some use
(although the product
goes in the wrong direction):
$$
lim_{n to infty}dfrac{(x+(n-1)/2)^n-x(x+1)...(x+n-1)}{(x+(n-1)/2)^{n-2}(n^3-n)/24}
=1
$$
and
$$
lim_{n to infty}
(x+(n-1)/2)dfrac{(x+(n-1)/2)-(x(x+1)...(x+n-1))^{1/n}}
{(n^2-1)/24}
=1
$$
An outline of the proofs is here:
Limit of $sqrt[n]{(x+1)...(x+n)} - x$ as $x to +infty$
answered May 5 '16 at 23:44
marty cohenmarty cohen
75.4k549130
$endgroup$
add a comment |
$begingroup$
Here are some asymptotic results of mine
that might be of some use
(although the product
goes in the wrong direction):
$$
lim_{n to infty}dfrac{(x+(n-1)/2)^n-x(x+1)...(x+n-1)}{(x+(n-1)/2)^{n-2}(n^3-n)/24}
=1
$$
and
$$
lim_{n to infty}
(x+(n-1)/2)dfrac{(x+(n-1)/2)-(x(x+1)...(x+n-1))^{1/n}}
{(n^2-1)/24}
=1
$$
An outline of the proofs is here:
Limit of $sqrt[n]{(x+1)...(x+n)} - x$ as $x to +infty$
answered May 5 '16 at 23:44
marty cohenmarty cohen
75.4k549130
$endgroup$
Here are some asymptotic results of mine
that might be of some use
(although the product
goes in the wrong direction):
$$
lim_{n to infty}dfrac{(x+(n-1)/2)^n-x(x+1)...(x+n-1)}{(x+(n-1)/2)^{n-2}(n^3-n)/24}
=1
$$
and
$$
lim_{n to infty}
(x+(n-1)/2)dfrac{(x+(n-1)/2)-(x(x+1)...(x+n-1))^{1/n}}
{(n^2-1)/24}
=1
$$
An outline of the proofs is here:
Limit of $sqrt[n]{(x+1)...(x+n)} - x$ as $x to +infty$
answered May 5 '16 at 23:44
marty cohenmarty cohen
75.4k549130
answered May 5 '16 at 23:44
marty cohenmarty cohen
75.4k549130
answered May 5 '16 at 23:44
marty cohenmarty cohen
75.4k549130
answered May 5 '16 at 23:44
marty cohenmarty cohen
75.4k549130
75.4k549130
add a comment |
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%2f1773384%2ftrying-to-express-a-factorial-as-a-polynomial%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
5
$begingroup$
That is already a polynomial :-)
$endgroup$
– parsiad
May 5 '16 at 22:41
$begingroup$
Expended form : $sum_{i=1}^b a^i (-1)^{b-i} (sum_{I subseteq[1,b], |I|=b-i+1} prod_{k in I} k) $. Do you need something clearer?
$endgroup$
– Vincent
May 5 '16 at 22:43
$begingroup$
This is often called the "falling factorial" and often written as $(a-1)_b$. The coefficients are messy. en.wikipedia.org/wiki/Falling_and_rising_factorials
$endgroup$
– Thomas Andrews
May 5 '16 at 23:14