How to find the $n$- th term of the generating function $frac1(1-x-x^2)^k$?
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I know how to find the $n$ -th term for $k = 1$, of $frac1left(1-x^2-xright)^k$ but I want to know if there's a general formula for $k > 1$.
generating-functions
edited Aug 31 at 4:14
dmtri
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asked Aug 31 at 4:03
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up vote
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I know how to find the $n$ -th term for $k = 1$, of $frac1left(1-x^2-xright)^k$ but I want to know if there's a general formula for $k > 1$.
generating-functions
edited Aug 31 at 4:14
dmtri
852317
asked Aug 31 at 4:03
Mercado
205
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know how to find the $n$ -th term for $k = 1$, of $frac1left(1-x^2-xright)^k$ but I want to know if there's a general formula for $k > 1$.
generating-functions
edited Aug 31 at 4:14
dmtri
852317
asked Aug 31 at 4:03
Mercado
205
I know how to find the $n$ -th term for $k = 1$, of $frac1left(1-x^2-xright)^k$ but I want to know if there's a general formula for $k > 1$.
generating-functions
generating-functions
edited Aug 31 at 4:14
dmtri
852317
asked Aug 31 at 4:03
Mercado
205
edited Aug 31 at 4:14
dmtri
852317
asked Aug 31 at 4:03
Mercado
205
edited Aug 31 at 4:14
dmtri
852317
edited Aug 31 at 4:14
dmtri
852317
edited Aug 31 at 4:14
dmtri
852317
852317
asked Aug 31 at 4:03
Mercado
205
asked Aug 31 at 4:03
Mercado
205
asked Aug 31 at 4:03
Mercado
205
205
add a comment |Â
add a comment |Â
1 Answer
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First, notice that the auxiliar function $R(s)=dfrac11-s$ satisfies the curious formula:
$$
dfrac1(1-s)^k=(-1)^kdfracR^(k)(s)k!
$$
Thus, one can simply find the $nth$ term of the EGF progression
$$
dfrac1(1-s)^k=sum_n=0^infty a_n,k s^n
$$
by simply compute the $kth$ derivative of the geometric series expansion
$$ dfrac11-s=sum_n=0^infty s^n.$$
That is,
$$
dfrac1(1-s)^k=(-1)^kdfracR^(k)(s)k!=sum_n=k^infty (-1)^kleft(beginarraycn \ kendarrayright)s^n-k=sum_n=0^infty (-1)^kleft(beginarraycn+k \ kendarrayright)s^n.
$$
Thus, $$a_n,k=(-1)^kleft(beginarraycn+k \ kendarrayright).$$
In order to conclude our computation, one needs to perform the substitution $s=x^2+x$ on the above series expansion.
A straightorward application of the binomial identity to $(x^2+x)^n$ will allows you to determine a close formula for the coefficients of the aforementioned EGF.
answered Aug 31 at 4:59
Nelson Faustino
1376
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
First, notice that the auxiliar function $R(s)=dfrac11-s$ satisfies the curious formula:
$$
dfrac1(1-s)^k=(-1)^kdfracR^(k)(s)k!
$$
Thus, one can simply find the $nth$ term of the EGF progression
$$
dfrac1(1-s)^k=sum_n=0^infty a_n,k s^n
$$
by simply compute the $kth$ derivative of the geometric series expansion
$$ dfrac11-s=sum_n=0^infty s^n.$$
That is,
$$
dfrac1(1-s)^k=(-1)^kdfracR^(k)(s)k!=sum_n=k^infty (-1)^kleft(beginarraycn \ kendarrayright)s^n-k=sum_n=0^infty (-1)^kleft(beginarraycn+k \ kendarrayright)s^n.
$$
Thus, $$a_n,k=(-1)^kleft(beginarraycn+k \ kendarrayright).$$
In order to conclude our computation, one needs to perform the substitution $s=x^2+x$ on the above series expansion.
A straightorward application of the binomial identity to $(x^2+x)^n$ will allows you to determine a close formula for the coefficients of the aforementioned EGF.
answered Aug 31 at 4:59
Nelson Faustino
1376
add a comment |Â
up vote
1
down vote
accepted
First, notice that the auxiliar function $R(s)=dfrac11-s$ satisfies the curious formula:
$$
dfrac1(1-s)^k=(-1)^kdfracR^(k)(s)k!
$$
Thus, one can simply find the $nth$ term of the EGF progression
$$
dfrac1(1-s)^k=sum_n=0^infty a_n,k s^n
$$
by simply compute the $kth$ derivative of the geometric series expansion
$$ dfrac11-s=sum_n=0^infty s^n.$$
That is,
$$
dfrac1(1-s)^k=(-1)^kdfracR^(k)(s)k!=sum_n=k^infty (-1)^kleft(beginarraycn \ kendarrayright)s^n-k=sum_n=0^infty (-1)^kleft(beginarraycn+k \ kendarrayright)s^n.
$$
Thus, $$a_n,k=(-1)^kleft(beginarraycn+k \ kendarrayright).$$
In order to conclude our computation, one needs to perform the substitution $s=x^2+x$ on the above series expansion.
A straightorward application of the binomial identity to $(x^2+x)^n$ will allows you to determine a close formula for the coefficients of the aforementioned EGF.
answered Aug 31 at 4:59
Nelson Faustino
1376
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First, notice that the auxiliar function $R(s)=dfrac11-s$ satisfies the curious formula:
$$
dfrac1(1-s)^k=(-1)^kdfracR^(k)(s)k!
$$
Thus, one can simply find the $nth$ term of the EGF progression
$$
dfrac1(1-s)^k=sum_n=0^infty a_n,k s^n
$$
by simply compute the $kth$ derivative of the geometric series expansion
$$ dfrac11-s=sum_n=0^infty s^n.$$
That is,
$$
dfrac1(1-s)^k=(-1)^kdfracR^(k)(s)k!=sum_n=k^infty (-1)^kleft(beginarraycn \ kendarrayright)s^n-k=sum_n=0^infty (-1)^kleft(beginarraycn+k \ kendarrayright)s^n.
$$
Thus, $$a_n,k=(-1)^kleft(beginarraycn+k \ kendarrayright).$$
In order to conclude our computation, one needs to perform the substitution $s=x^2+x$ on the above series expansion.
A straightorward application of the binomial identity to $(x^2+x)^n$ will allows you to determine a close formula for the coefficients of the aforementioned EGF.
answered Aug 31 at 4:59
Nelson Faustino
1376
First, notice that the auxiliar function $R(s)=dfrac11-s$ satisfies the curious formula:
$$
dfrac1(1-s)^k=(-1)^kdfracR^(k)(s)k!
$$
Thus, one can simply find the $nth$ term of the EGF progression
$$
dfrac1(1-s)^k=sum_n=0^infty a_n,k s^n
$$
by simply compute the $kth$ derivative of the geometric series expansion
$$ dfrac11-s=sum_n=0^infty s^n.$$
That is,
$$
dfrac1(1-s)^k=(-1)^kdfracR^(k)(s)k!=sum_n=k^infty (-1)^kleft(beginarraycn \ kendarrayright)s^n-k=sum_n=0^infty (-1)^kleft(beginarraycn+k \ kendarrayright)s^n.
$$
Thus, $$a_n,k=(-1)^kleft(beginarraycn+k \ kendarrayright).$$
In order to conclude our computation, one needs to perform the substitution $s=x^2+x$ on the above series expansion.
A straightorward application of the binomial identity to $(x^2+x)^n$ will allows you to determine a close formula for the coefficients of the aforementioned EGF.
answered Aug 31 at 4:59
Nelson Faustino
1376
answered Aug 31 at 4:59
Nelson Faustino
1376
answered Aug 31 at 4:59
Nelson Faustino
1376
answered Aug 31 at 4:59
Nelson Faustino
1376
1376
add a comment |Â
add a comment |Â
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