Find generating function for the following conditions:












0












$begingroup$


$a_n=$ $2^{n+1}$ for $n:3|n$



$a_n=$ $0$ for the rest



So theres what I tried to do:



$$sum_{n=0}^{infty}a_nz^{n}=2*z^0+0*z^1+0*z^2+2^4z^3+0*z^4+0*z^5+2^7z^6+...2^{k+1}z^k$$



$$2^{k+1}z^k=b^k$$



$$frac{b^{k+1}}{b^k}$$



$$frac{2^{k+2}z^{k+1}}{2^{k+1}z^{k}}=4z$$



$$q=4z$$
$$W_0=(4z)^2$$
$$A(z)=frac{4^{z}}{1+(4z)^2}$$



Is this the right answer?






discrete-mathematics generating-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    $3 mid 0$ is $displaystylecolor{red}{texttt{true}}$ !!!.
    $endgroup$
    – Felix Marin
    Jan 27 at 22:14


















0












$begingroup$


$a_n=$ $2^{n+1}$ for $n:3|n$



$a_n=$ $0$ for the rest



So theres what I tried to do:



$$sum_{n=0}^{infty}a_nz^{n}=2*z^0+0*z^1+0*z^2+2^4z^3+0*z^4+0*z^5+2^7z^6+...2^{k+1}z^k$$



$$2^{k+1}z^k=b^k$$



$$frac{b^{k+1}}{b^k}$$



$$frac{2^{k+2}z^{k+1}}{2^{k+1}z^{k}}=4z$$



$$q=4z$$
$$W_0=(4z)^2$$
$$A(z)=frac{4^{z}}{1+(4z)^2}$$



Is this the right answer?






discrete-mathematics generating-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    $3 mid 0$ is $displaystylecolor{red}{texttt{true}}$ !!!.
    $endgroup$
    – Felix Marin
    Jan 27 at 22:14
















0












0








0





$begingroup$


$a_n=$ $2^{n+1}$ for $n:3|n$



$a_n=$ $0$ for the rest



So theres what I tried to do:



$$sum_{n=0}^{infty}a_nz^{n}=2*z^0+0*z^1+0*z^2+2^4z^3+0*z^4+0*z^5+2^7z^6+...2^{k+1}z^k$$



$$2^{k+1}z^k=b^k$$



$$frac{b^{k+1}}{b^k}$$



$$frac{2^{k+2}z^{k+1}}{2^{k+1}z^{k}}=4z$$



$$q=4z$$
$$W_0=(4z)^2$$
$$A(z)=frac{4^{z}}{1+(4z)^2}$$



Is this the right answer?






discrete-mathematics generating-functions







share|cite|improve this question











$endgroup$




$a_n=$ $2^{n+1}$ for $n:3|n$



$a_n=$ $0$ for the rest



So theres what I tried to do:



$$sum_{n=0}^{infty}a_nz^{n}=2*z^0+0*z^1+0*z^2+2^4z^3+0*z^4+0*z^5+2^7z^6+...2^{k+1}z^k$$



$$2^{k+1}z^k=b^k$$



$$frac{b^{k+1}}{b^k}$$



$$frac{2^{k+2}z^{k+1}}{2^{k+1}z^{k}}=4z$$



$$q=4z$$
$$W_0=(4z)^2$$
$$A(z)=frac{4^{z}}{1+(4z)^2}$$



Is this the right answer?






discrete-mathematics generating-functions




discrete-mathematics generating-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 27 at 22:15







Arnolt Infern Kitler

















asked Jan 27 at 22:02









Arnolt Infern KitlerArnolt Infern Kitler

62




62












  • $begingroup$
    $3 mid 0$ is $displaystylecolor{red}{texttt{true}}$ !!!.
    $endgroup$
    – Felix Marin
    Jan 27 at 22:14




















  • $begingroup$
    $3 mid 0$ is $displaystylecolor{red}{texttt{true}}$ !!!.
    $endgroup$
    – Felix Marin
    Jan 27 at 22:14


















$begingroup$
$3 mid 0$ is $displaystylecolor{red}{texttt{true}}$ !!!.
$endgroup$
– Felix Marin
Jan 27 at 22:14






$begingroup$
$3 mid 0$ is $displaystylecolor{red}{texttt{true}}$ !!!.
$endgroup$
– Felix Marin
Jan 27 at 22:14












1 Answer
1






active

oldest

votes


















0












$begingroup$

With $displaystyleleftvert zrightvert < {1 over 2}$,
$displaystylequadsum_{n = 0}^{infty}a_{n}z^{n} = sum_{n = 0}^{infty}2^{3n + 1}, z^{3n} =
2sum_{n = 0}^{infty}left[left(2zright)^{3}right]^{n} =
bbox[#ffd,10px,border:1px groove navy]{2 over 1 - left(2zright)^{3}}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Or explained with less notation, the series is a geometric series: $2 + 2^4z^3 + 2^7z^6+cdots = 2+2(8z^3)+2(8z^3)^2+cdots$. The first term is $2$ and the common ratio is $8z^3$, so the sum is $displaystyle{ 2over{1-8z^3}}$.
    $endgroup$
    – Steve Kass
    Jan 27 at 22:27










  • $begingroup$
    @SteveKass Fine. Thanks.
    $endgroup$
    – Felix Marin
    Jan 27 at 22:55











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

With $displaystyleleftvert zrightvert < {1 over 2}$,
$displaystylequadsum_{n = 0}^{infty}a_{n}z^{n} = sum_{n = 0}^{infty}2^{3n + 1}, z^{3n} =
2sum_{n = 0}^{infty}left[left(2zright)^{3}right]^{n} =
bbox[#ffd,10px,border:1px groove navy]{2 over 1 - left(2zright)^{3}}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Or explained with less notation, the series is a geometric series: $2 + 2^4z^3 + 2^7z^6+cdots = 2+2(8z^3)+2(8z^3)^2+cdots$. The first term is $2$ and the common ratio is $8z^3$, so the sum is $displaystyle{ 2over{1-8z^3}}$.
    $endgroup$
    – Steve Kass
    Jan 27 at 22:27










  • $begingroup$
    @SteveKass Fine. Thanks.
    $endgroup$
    – Felix Marin
    Jan 27 at 22:55
















0












$begingroup$

With $displaystyleleftvert zrightvert < {1 over 2}$,
$displaystylequadsum_{n = 0}^{infty}a_{n}z^{n} = sum_{n = 0}^{infty}2^{3n + 1}, z^{3n} =
2sum_{n = 0}^{infty}left[left(2zright)^{3}right]^{n} =
bbox[#ffd,10px,border:1px groove navy]{2 over 1 - left(2zright)^{3}}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Or explained with less notation, the series is a geometric series: $2 + 2^4z^3 + 2^7z^6+cdots = 2+2(8z^3)+2(8z^3)^2+cdots$. The first term is $2$ and the common ratio is $8z^3$, so the sum is $displaystyle{ 2over{1-8z^3}}$.
    $endgroup$
    – Steve Kass
    Jan 27 at 22:27










  • $begingroup$
    @SteveKass Fine. Thanks.
    $endgroup$
    – Felix Marin
    Jan 27 at 22:55














0












0








0





$begingroup$

With $displaystyleleftvert zrightvert < {1 over 2}$,
$displaystylequadsum_{n = 0}^{infty}a_{n}z^{n} = sum_{n = 0}^{infty}2^{3n + 1}, z^{3n} =
2sum_{n = 0}^{infty}left[left(2zright)^{3}right]^{n} =
bbox[#ffd,10px,border:1px groove navy]{2 over 1 - left(2zright)^{3}}$






share|cite|improve this answer









$endgroup$



With $displaystyleleftvert zrightvert < {1 over 2}$,
$displaystylequadsum_{n = 0}^{infty}a_{n}z^{n} = sum_{n = 0}^{infty}2^{3n + 1}, z^{3n} =
2sum_{n = 0}^{infty}left[left(2zright)^{3}right]^{n} =
bbox[#ffd,10px,border:1px groove navy]{2 over 1 - left(2zright)^{3}}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 27 at 22:21









Felix MarinFelix Marin

68.8k7109146




68.8k7109146












  • $begingroup$
    Or explained with less notation, the series is a geometric series: $2 + 2^4z^3 + 2^7z^6+cdots = 2+2(8z^3)+2(8z^3)^2+cdots$. The first term is $2$ and the common ratio is $8z^3$, so the sum is $displaystyle{ 2over{1-8z^3}}$.
    $endgroup$
    – Steve Kass
    Jan 27 at 22:27










  • $begingroup$
    @SteveKass Fine. Thanks.
    $endgroup$
    – Felix Marin
    Jan 27 at 22:55


















  • $begingroup$
    Or explained with less notation, the series is a geometric series: $2 + 2^4z^3 + 2^7z^6+cdots = 2+2(8z^3)+2(8z^3)^2+cdots$. The first term is $2$ and the common ratio is $8z^3$, so the sum is $displaystyle{ 2over{1-8z^3}}$.
    $endgroup$
    – Steve Kass
    Jan 27 at 22:27










  • $begingroup$
    @SteveKass Fine. Thanks.
    $endgroup$
    – Felix Marin
    Jan 27 at 22:55
















$begingroup$
Or explained with less notation, the series is a geometric series: $2 + 2^4z^3 + 2^7z^6+cdots = 2+2(8z^3)+2(8z^3)^2+cdots$. The first term is $2$ and the common ratio is $8z^3$, so the sum is $displaystyle{ 2over{1-8z^3}}$.
$endgroup$
– Steve Kass
Jan 27 at 22:27




$begingroup$
Or explained with less notation, the series is a geometric series: $2 + 2^4z^3 + 2^7z^6+cdots = 2+2(8z^3)+2(8z^3)^2+cdots$. The first term is $2$ and the common ratio is $8z^3$, so the sum is $displaystyle{ 2over{1-8z^3}}$.
$endgroup$
– Steve Kass
Jan 27 at 22:27












$begingroup$
@SteveKass Fine. Thanks.
$endgroup$
– Felix Marin
Jan 27 at 22:55




$begingroup$
@SteveKass Fine. Thanks.
$endgroup$
– Felix Marin
Jan 27 at 22:55


















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