Mixing
Find generating function for the following conditions:
$begingroup$
$a_n=$ $2^{n+1}$ 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
asked Jan 27 at 22:02
Arnolt Infern KitlerArnolt Infern Kitler
62
$endgroup$
add a comment |
$begingroup$
$a_n=$ $2^{n+1}$ 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
asked Jan 27 at 22:02
Arnolt Infern KitlerArnolt Infern Kitler
62
$endgroup$
$begingroup$
$3 mid 0$ is $displaystylecolor{red}{texttt{true}}$ !!!.
$endgroup$
– Felix Marin
Jan 27 at 22:14
add a comment |
$begingroup$
$a_n=$ $2^{n+1}$ 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
asked Jan 27 at 22:02
Arnolt Infern KitlerArnolt Infern Kitler
62
$endgroup$
$a_n=$ $2^{n+1}$ 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
asked Jan 27 at 22:02
Arnolt Infern KitlerArnolt Infern Kitler
62
asked Jan 27 at 22:02
Arnolt Infern KitlerArnolt Infern Kitler
62
edited Jan 27 at 22:15
Arnolt Infern Kitler
asked Jan 27 at 22:02
Arnolt Infern KitlerArnolt Infern Kitler
62
asked Jan 27 at 22:02
Arnolt Infern KitlerArnolt Infern Kitler
62
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
add a comment |
$begingroup$
$3 mid 0$ 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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
With $displaystyleleftvert zrightvert < {1 over 2}$,
$displaystylequadsum_{n = 0}^{infty}a_{n}z^{n} = sum_{n = 0}^{infty}2^{3n + 1}, z^{3n} =
2sum_{n = 0}^{infty}left[left(2zright)^{3}right]^{n} =
bbox[#ffd,10px,border:1px groove navy]{2 over 1 - left(2zright)^{3}}$
answered Jan 27 at 22:21
Felix MarinFelix Marin
68.8k7109146
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With $displaystyleleftvert zrightvert < {1 over 2}$,
$displaystylequadsum_{n = 0}^{infty}a_{n}z^{n} = sum_{n = 0}^{infty}2^{3n + 1}, z^{3n} =
2sum_{n = 0}^{infty}left[left(2zright)^{3}right]^{n} =
bbox[#ffd,10px,border:1px groove navy]{2 over 1 - left(2zright)^{3}}$
answered Jan 27 at 22:21
Felix MarinFelix Marin
68.8k7109146
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
With $displaystyleleftvert zrightvert < {1 over 2}$,
$displaystylequadsum_{n = 0}^{infty}a_{n}z^{n} = sum_{n = 0}^{infty}2^{3n + 1}, z^{3n} =
2sum_{n = 0}^{infty}left[left(2zright)^{3}right]^{n} =
bbox[#ffd,10px,border:1px groove navy]{2 over 1 - left(2zright)^{3}}$
answered Jan 27 at 22:21
Felix MarinFelix Marin
68.8k7109146
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
With $displaystyleleftvert zrightvert < {1 over 2}$,
$displaystylequadsum_{n = 0}^{infty}a_{n}z^{n} = sum_{n = 0}^{infty}2^{3n + 1}, z^{3n} =
2sum_{n = 0}^{infty}left[left(2zright)^{3}right]^{n} =
bbox[#ffd,10px,border:1px groove navy]{2 over 1 - left(2zright)^{3}}$
answered Jan 27 at 22:21
Felix MarinFelix Marin
68.8k7109146
$endgroup$
With $displaystyleleftvert zrightvert < {1 over 2}$,
$displaystylequadsum_{n = 0}^{infty}a_{n}z^{n} = sum_{n = 0}^{infty}2^{3n + 1}, z^{3n} =
2sum_{n = 0}^{infty}left[left(2zright)^{3}right]^{n} =
bbox[#ffd,10px,border:1px groove navy]{2 over 1 - left(2zright)^{3}}$
answered Jan 27 at 22:21
Felix MarinFelix Marin
68.8k7109146
answered Jan 27 at 22:21
Felix MarinFelix Marin
68.8k7109146
answered Jan 27 at 22:21
Felix MarinFelix Marin
68.8k7109146
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
$3 mid 0$ is $displaystylecolor{red}{texttt{true}}$ !!!.
$endgroup$
– Felix Marin
Jan 27 at 22:14