Is $f(x;a)=sum_{n=0}^{infty}cfrac{n!(1-x)^n}{(1-a)(2-a)cdots(n-a)}$ some kind of special function?
$begingroup$
In deriving the solution to a differential equation I arrived at the following series expression:
$f(x;a)=sum_{n=0}^{infty}cfrac{n!(1-x)^n}{(1-a)(2-a)cdots(n-a)};quad 0le x le 1$
where $a$ is a very small parameter of the equation, and is NOT an integer (otherwise the formula is not valid). After some brief analysis I found this series to be convergent, but I couldn't find a closed form expression.
Does it belong to any existing families of special functions?
sequences-and-series complex-analysis taylor-expansion special-functions
asked Jan 10 at 2:24
yixianshuiesuanyixianshuiesuan
54
$endgroup$
add a comment |
$begingroup$
In deriving the solution to a differential equation I arrived at the following series expression:
$f(x;a)=sum_{n=0}^{infty}cfrac{n!(1-x)^n}{(1-a)(2-a)cdots(n-a)};quad 0le x le 1$
where $a$ is a very small parameter of the equation, and is NOT an integer (otherwise the formula is not valid). After some brief analysis I found this series to be convergent, but I couldn't find a closed form expression.
Does it belong to any existing families of special functions?
sequences-and-series complex-analysis taylor-expansion special-functions
asked Jan 10 at 2:24
yixianshuiesuanyixianshuiesuan
54
$endgroup$
1
$begingroup$
the series sums to a hypergeometric function ${}_2F_1(1,1;1-a;1-x)$
$endgroup$
– achille hui
Jan 10 at 2:33
$begingroup$
Hypergeometric functions pop up a lot in solving differential equations, namely when using Frobenius series. Specifically, if $$f(x)=x^rsum_{k=0}^infty a_kx^k$$ and $frac{a_{k+1}}{a_k}$ forms a rational function in $k$, you have yourself a generalized hypergeometric function.
$endgroup$
– Simply Beautiful Art
Jan 10 at 3:06
$begingroup$
Thanks! My original differential equation arises from the forward-time Fokker-Planck equation regarding a derivative: $frac{1}{2}partial_x^2[x(1-x)partial_xf]+mpartial_x^2f=delta(x)$
$endgroup$
– yixianshuiesuan
Jan 10 at 3:09
add a comment |
$begingroup$
In deriving the solution to a differential equation I arrived at the following series expression:
$f(x;a)=sum_{n=0}^{infty}cfrac{n!(1-x)^n}{(1-a)(2-a)cdots(n-a)};quad 0le x le 1$
where $a$ is a very small parameter of the equation, and is NOT an integer (otherwise the formula is not valid). After some brief analysis I found this series to be convergent, but I couldn't find a closed form expression.
Does it belong to any existing families of special functions?
sequences-and-series complex-analysis taylor-expansion special-functions
asked Jan 10 at 2:24
yixianshuiesuanyixianshuiesuan
54
$endgroup$
In deriving the solution to a differential equation I arrived at the following series expression:
$f(x;a)=sum_{n=0}^{infty}cfrac{n!(1-x)^n}{(1-a)(2-a)cdots(n-a)};quad 0le x le 1$
where $a$ is a very small parameter of the equation, and is NOT an integer (otherwise the formula is not valid). After some brief analysis I found this series to be convergent, but I couldn't find a closed form expression.
Does it belong to any existing families of special functions?
sequences-and-series complex-analysis taylor-expansion special-functions
sequences-and-series complex-analysis taylor-expansion special-functions
asked Jan 10 at 2:24
yixianshuiesuanyixianshuiesuan
54
asked Jan 10 at 2:24
yixianshuiesuanyixianshuiesuan
54
asked Jan 10 at 2:24
yixianshuiesuanyixianshuiesuan
54
asked Jan 10 at 2:24
yixianshuiesuanyixianshuiesuan
54
asked Jan 10 at 2:24
yixianshuiesuanyixianshuiesuan
54
54
1
$begingroup$
the series sums to a hypergeometric function ${}_2F_1(1,1;1-a;1-x)$
$endgroup$
– achille hui
Jan 10 at 2:33
$begingroup$
Hypergeometric functions pop up a lot in solving differential equations, namely when using Frobenius series. Specifically, if $$f(x)=x^rsum_{k=0}^infty a_kx^k$$ and $frac{a_{k+1}}{a_k}$ forms a rational function in $k$, you have yourself a generalized hypergeometric function.
$endgroup$
– Simply Beautiful Art
Jan 10 at 3:06
$begingroup$
Thanks! My original differential equation arises from the forward-time Fokker-Planck equation regarding a derivative: $frac{1}{2}partial_x^2[x(1-x)partial_xf]+mpartial_x^2f=delta(x)$
$endgroup$
– yixianshuiesuan
Jan 10 at 3:09
add a comment |
1
$begingroup$
the series sums to a hypergeometric function ${}_2F_1(1,1;1-a;1-x)$
$endgroup$
– achille hui
Jan 10 at 2:33
$begingroup$
Hypergeometric functions pop up a lot in solving differential equations, namely when using Frobenius series. Specifically, if $$f(x)=x^rsum_{k=0}^infty a_kx^k$$ and $frac{a_{k+1}}{a_k}$ forms a rational function in $k$, you have yourself a generalized hypergeometric function.
$endgroup$
– Simply Beautiful Art
Jan 10 at 3:06
$begingroup$
Thanks! My original differential equation arises from the forward-time Fokker-Planck equation regarding a derivative: $frac{1}{2}partial_x^2[x(1-x)partial_xf]+mpartial_x^2f=delta(x)$
$endgroup$
– yixianshuiesuan
Jan 10 at 3:09
1
1
$begingroup$
the series sums to a hypergeometric function ${}_2F_1(1,1;1-a;1-x)$
$endgroup$
– achille hui
Jan 10 at 2:33
$begingroup$
the series sums to a hypergeometric function ${}_2F_1(1,1;1-a;1-x)$
$endgroup$
– achille hui
Jan 10 at 2:33
$begingroup$
Hypergeometric functions pop up a lot in solving differential equations, namely when using Frobenius series. Specifically, if $$f(x)=x^rsum_{k=0}^infty a_kx^k$$ and $frac{a_{k+1}}{a_k}$ forms a rational function in $k$, you have yourself a generalized hypergeometric function.
$endgroup$
– Simply Beautiful Art
Jan 10 at 3:06
$begingroup$
Hypergeometric functions pop up a lot in solving differential equations, namely when using Frobenius series. Specifically, if $$f(x)=x^rsum_{k=0}^infty a_kx^k$$ and $frac{a_{k+1}}{a_k}$ forms a rational function in $k$, you have yourself a generalized hypergeometric function.
$endgroup$
– Simply Beautiful Art
Jan 10 at 3:06
$begingroup$
Thanks! My original differential equation arises from the forward-time Fokker-Planck equation regarding a derivative: $frac{1}{2}partial_x^2[x(1-x)partial_xf]+mpartial_x^2f=delta(x)$
$endgroup$
– yixianshuiesuan
Jan 10 at 3:09
$begingroup$
Thanks! My original differential equation arises from the forward-time Fokker-Planck equation regarding a derivative: $frac{1}{2}partial_x^2[x(1-x)partial_xf]+mpartial_x^2f=delta(x)$
$endgroup$
– yixianshuiesuan
Jan 10 at 3:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Indeed it is - it's a kind of hypergeometric function. More specifically, we rewrite it through a series of steps as
$$begin{align}
f(x; a) &= sum_{n=0}^{infty} frac{n!}{(1 - a)(2 - a) cdots (n - a)} frac{n!}{n!} (1 - x)^n\
&= sum_{n=0}^{infty} frac{n! n!}{(1 - a)(2 - a) cdots (n - a)} frac{(1 - x)^n}{n!}\
&= sum_{n=0}^{infty} frac{n! n!}{(1 - a)([1 - a] + 1) cdots ([1 - a] + n - 1)} frac{(1 - x)^n}{n!}end{align}$$
and now converting to rising factorials, using $n! = 1^{(n)}$, we get
$$f(x; a) = sum_{n=0}^{infty} frac{1^{(n)} 1^{(n)}}{[1 - a]^{(n)}} frac{(1 - x)^n}{n!}$$
which we can read off now as
$$f(x; a) = _2 F_1left(begin{matrix}1, 1 \ 1 - aend{matrix} Big| 1 - xright)$$
answered Jan 10 at 2:37
The_SympathizerThe_Sympathizer
7,8002246
$endgroup$
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%2f3068161%2fis-fxa-sum-n-0-infty-cfracn1-xn1-a2-a-cdotsn-a-some-kin%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
$begingroup$
Indeed it is - it's a kind of hypergeometric function. More specifically, we rewrite it through a series of steps as
$$begin{align}
f(x; a) &= sum_{n=0}^{infty} frac{n!}{(1 - a)(2 - a) cdots (n - a)} frac{n!}{n!} (1 - x)^n\
&= sum_{n=0}^{infty} frac{n! n!}{(1 - a)(2 - a) cdots (n - a)} frac{(1 - x)^n}{n!}\
&= sum_{n=0}^{infty} frac{n! n!}{(1 - a)([1 - a] + 1) cdots ([1 - a] + n - 1)} frac{(1 - x)^n}{n!}end{align}$$
and now converting to rising factorials, using $n! = 1^{(n)}$, we get
$$f(x; a) = sum_{n=0}^{infty} frac{1^{(n)} 1^{(n)}}{[1 - a]^{(n)}} frac{(1 - x)^n}{n!}$$
which we can read off now as
$$f(x; a) = _2 F_1left(begin{matrix}1, 1 \ 1 - aend{matrix} Big| 1 - xright)$$
answered Jan 10 at 2:37
The_SympathizerThe_Sympathizer
7,8002246
$endgroup$
add a comment |
$begingroup$
Indeed it is - it's a kind of hypergeometric function. More specifically, we rewrite it through a series of steps as
$$begin{align}
f(x; a) &= sum_{n=0}^{infty} frac{n!}{(1 - a)(2 - a) cdots (n - a)} frac{n!}{n!} (1 - x)^n\
&= sum_{n=0}^{infty} frac{n! n!}{(1 - a)(2 - a) cdots (n - a)} frac{(1 - x)^n}{n!}\
&= sum_{n=0}^{infty} frac{n! n!}{(1 - a)([1 - a] + 1) cdots ([1 - a] + n - 1)} frac{(1 - x)^n}{n!}end{align}$$
and now converting to rising factorials, using $n! = 1^{(n)}$, we get
$$f(x; a) = sum_{n=0}^{infty} frac{1^{(n)} 1^{(n)}}{[1 - a]^{(n)}} frac{(1 - x)^n}{n!}$$
which we can read off now as
$$f(x; a) = _2 F_1left(begin{matrix}1, 1 \ 1 - aend{matrix} Big| 1 - xright)$$
answered Jan 10 at 2:37
The_SympathizerThe_Sympathizer
7,8002246
$endgroup$
add a comment |
$begingroup$
Indeed it is - it's a kind of hypergeometric function. More specifically, we rewrite it through a series of steps as
$$begin{align}
f(x; a) &= sum_{n=0}^{infty} frac{n!}{(1 - a)(2 - a) cdots (n - a)} frac{n!}{n!} (1 - x)^n\
&= sum_{n=0}^{infty} frac{n! n!}{(1 - a)(2 - a) cdots (n - a)} frac{(1 - x)^n}{n!}\
&= sum_{n=0}^{infty} frac{n! n!}{(1 - a)([1 - a] + 1) cdots ([1 - a] + n - 1)} frac{(1 - x)^n}{n!}end{align}$$
and now converting to rising factorials, using $n! = 1^{(n)}$, we get
$$f(x; a) = sum_{n=0}^{infty} frac{1^{(n)} 1^{(n)}}{[1 - a]^{(n)}} frac{(1 - x)^n}{n!}$$
which we can read off now as
$$f(x; a) = _2 F_1left(begin{matrix}1, 1 \ 1 - aend{matrix} Big| 1 - xright)$$
answered Jan 10 at 2:37
The_SympathizerThe_Sympathizer
7,8002246
$endgroup$
Indeed it is - it's a kind of hypergeometric function. More specifically, we rewrite it through a series of steps as
$$begin{align}
f(x; a) &= sum_{n=0}^{infty} frac{n!}{(1 - a)(2 - a) cdots (n - a)} frac{n!}{n!} (1 - x)^n\
&= sum_{n=0}^{infty} frac{n! n!}{(1 - a)(2 - a) cdots (n - a)} frac{(1 - x)^n}{n!}\
&= sum_{n=0}^{infty} frac{n! n!}{(1 - a)([1 - a] + 1) cdots ([1 - a] + n - 1)} frac{(1 - x)^n}{n!}end{align}$$
and now converting to rising factorials, using $n! = 1^{(n)}$, we get
$$f(x; a) = sum_{n=0}^{infty} frac{1^{(n)} 1^{(n)}}{[1 - a]^{(n)}} frac{(1 - x)^n}{n!}$$
which we can read off now as
$$f(x; a) = _2 F_1left(begin{matrix}1, 1 \ 1 - aend{matrix} Big| 1 - xright)$$
answered Jan 10 at 2:37
The_SympathizerThe_Sympathizer
7,8002246
answered Jan 10 at 2:37
The_SympathizerThe_Sympathizer
7,8002246
answered Jan 10 at 2:37
The_SympathizerThe_Sympathizer
7,8002246
answered Jan 10 at 2:37
The_SympathizerThe_Sympathizer
7,8002246
7,8002246
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%2f3068161%2fis-fxa-sum-n-0-infty-cfracn1-xn1-a2-a-cdotsn-a-some-kin%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
$begingroup$
the series sums to a hypergeometric function ${}_2F_1(1,1;1-a;1-x)$
$endgroup$
– achille hui
Jan 10 at 2:33
$begingroup$
Hypergeometric functions pop up a lot in solving differential equations, namely when using Frobenius series. Specifically, if $$f(x)=x^rsum_{k=0}^infty a_kx^k$$ and $frac{a_{k+1}}{a_k}$ forms a rational function in $k$, you have yourself a generalized hypergeometric function.
$endgroup$
– Simply Beautiful Art
Jan 10 at 3:06
$begingroup$
Thanks! My original differential equation arises from the forward-time Fokker-Planck equation regarding a derivative: $frac{1}{2}partial_x^2[x(1-x)partial_xf]+mpartial_x^2f=delta(x)$
$endgroup$
– yixianshuiesuan
Jan 10 at 3:09