Mixing
Find limit of $(a_n)$ if $a_n=frac{1}{k}sumlimits_{i=1}^{k}a_{n-i}$, $a_0=1$ and $a_{-n}=0$ for $n>0$
$begingroup$
We have a sequence $a_n$ which solves the recurrence relation
$$a_n=frac{1}{k}sumlimits_{i=1}^{k}a_{n-i}$$
for any $n>0$ with a given $k$. Let the initial values be $a_0=1$ and $a_{-n}=0$ for any $n>0$.
I wrote a program and found that the sequence should have a limit $frac{2}{k+1}$, but I can't figure out how to prove it.
sequences-and-series limits recurrence-relations
edited Jan 19 at 17:48
Did
248k23224463
asked Jan 19 at 15:12
Kangping YanKangping Yan
1817
$endgroup$
add a comment |
$begingroup$
We have a sequence $a_n$ which solves the recurrence relation
$$a_n=frac{1}{k}sumlimits_{i=1}^{k}a_{n-i}$$
for any $n>0$ with a given $k$. Let the initial values be $a_0=1$ and $a_{-n}=0$ for any $n>0$.
I wrote a program and found that the sequence should have a limit $frac{2}{k+1}$, but I can't figure out how to prove it.
sequences-and-series limits recurrence-relations
edited Jan 19 at 17:48
Did
248k23224463
asked Jan 19 at 15:12
Kangping YanKangping Yan
1817
$endgroup$
add a comment |
$begingroup$
We have a sequence $a_n$ which solves the recurrence relation
$$a_n=frac{1}{k}sumlimits_{i=1}^{k}a_{n-i}$$
for any $n>0$ with a given $k$. Let the initial values be $a_0=1$ and $a_{-n}=0$ for any $n>0$.
I wrote a program and found that the sequence should have a limit $frac{2}{k+1}$, but I can't figure out how to prove it.
sequences-and-series limits recurrence-relations
edited Jan 19 at 17:48
Did
248k23224463
asked Jan 19 at 15:12
Kangping YanKangping Yan
1817
$endgroup$
We have a sequence $a_n$ which solves the recurrence relation
$$a_n=frac{1}{k}sumlimits_{i=1}^{k}a_{n-i}$$
for any $n>0$ with a given $k$. Let the initial values be $a_0=1$ and $a_{-n}=0$ for any $n>0$.
I wrote a program and found that the sequence should have a limit $frac{2}{k+1}$, but I can't figure out how to prove it.
sequences-and-series limits recurrence-relations
sequences-and-series limits recurrence-relations
edited Jan 19 at 17:48
Did
248k23224463
asked Jan 19 at 15:12
Kangping YanKangping Yan
1817
edited Jan 19 at 17:48
Did
248k23224463
asked Jan 19 at 15:12
Kangping YanKangping Yan
1817
edited Jan 19 at 17:48
Did
248k23224463
edited Jan 19 at 17:48
Did
248k23224463
edited Jan 19 at 17:48
Did
248k23224463
248k23224463
asked Jan 19 at 15:12
Kangping YanKangping Yan
1817
asked Jan 19 at 15:12
Kangping YanKangping Yan
1817
asked Jan 19 at 15:12
Kangping YanKangping Yan
1817
1817
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your equation can be rewritten in the form of a renewal equation:
$$
a_n = b_n +(a*p)_n
$$ where, for every $nge0$, $$p_n =frac{1}{k}1_{1le nle k}qquad b_n=1_{n=0}$$ and
$$
(a*p)_n :=sum_{j=0}^n a_{n-j}p_j
$$ Here, ${b_j}$ is added since $a_n =frac{1}{k}sumlimits_{j=1}^k a_{n-j}$ does not hold for $n=0$.
Now, it is a result of the (discrete version of) Blackwell's renewal theorem that
$$
lim_{ntoinfty} a_n =frac{sumlimits_{n=0}^infty b_n}{sumlimits_{n=0}^infty np_n}=frac{1}{frac{1}{k}frac{k(k+1)}{2}}=frac{2}{k+1}
$$
edited Jan 19 at 16:28
Did
248k23224463
answered Jan 19 at 15:30
SongSong
16.1k1739
$endgroup$
add a comment |
$begingroup$
Could be of help to associate to the cute answer already given, the expression of the generating function (z-Transform).
We have
$$
a_{,n} = {1 over k}sumlimits_{1, le ,i, le ,k} {a_{,n, - ,i} } + left[ {n = 0} right]
= {1 over k}sumlimits_i {left[ {1, le ,i, le ,k} right]a_{,,n, - ,i} } + left[ {n = 0} right]
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
The generating function will then be
$$
eqalign{
& A(z) = sumlimits_{0, le ,n} {a_{,n} ,z^{,n} }
= {1 over k}sumlimits_{0, le ,n} {sumlimits_i {left( {left[ {1, le ,i, le ,k} right]z^{,i} } right)left( {a_{,,n, - ,i} ,z^{,n - i} } right)} } + 1 = cr
& = {1 over k}left( {sumlimits_{0, le ,n} {left[ {1, le ,n, le ,k} right]z^{,n} } } right)
sumlimits_{0, le ,n} {left( {a_{,,n,} ,z^{,n} } right)} + 1 = cr
& = {1 over k}left( {zsumlimits_{1, le ,n, le ,k} {z^{,n - 1} } } right)sumlimits_{0, le ,n} {left( {a_{,,n,} ,z^{,n} } right)} + 1 = cr
& = {1 over k}left( {z{{1 - z^{,k} } over {1 - z}}} right)A(z) + 1 cr}
$$
that is
$$
A(z) = {1 over {1 - {z over k}left( {{{1 - z^{,k} } over {1 - z}}} right)}} = {{1 - z} over {z^{,k} - left( {k + 1} right)z + k}}
$$
answered Jan 19 at 21:46
G CabG Cab
19.8k31339
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your equation can be rewritten in the form of a renewal equation:
$$
a_n = b_n +(a*p)_n
$$ where, for every $nge0$, $$p_n =frac{1}{k}1_{1le nle k}qquad b_n=1_{n=0}$$ and
$$
(a*p)_n :=sum_{j=0}^n a_{n-j}p_j
$$ Here, ${b_j}$ is added since $a_n =frac{1}{k}sumlimits_{j=1}^k a_{n-j}$ does not hold for $n=0$.
Now, it is a result of the (discrete version of) Blackwell's renewal theorem that
$$
lim_{ntoinfty} a_n =frac{sumlimits_{n=0}^infty b_n}{sumlimits_{n=0}^infty np_n}=frac{1}{frac{1}{k}frac{k(k+1)}{2}}=frac{2}{k+1}
$$
edited Jan 19 at 16:28
Did
248k23224463
answered Jan 19 at 15:30
SongSong
16.1k1739
$endgroup$
add a comment |
$begingroup$
Your equation can be rewritten in the form of a renewal equation:
$$
a_n = b_n +(a*p)_n
$$ where, for every $nge0$, $$p_n =frac{1}{k}1_{1le nle k}qquad b_n=1_{n=0}$$ and
$$
(a*p)_n :=sum_{j=0}^n a_{n-j}p_j
$$ Here, ${b_j}$ is added since $a_n =frac{1}{k}sumlimits_{j=1}^k a_{n-j}$ does not hold for $n=0$.
Now, it is a result of the (discrete version of) Blackwell's renewal theorem that
$$
lim_{ntoinfty} a_n =frac{sumlimits_{n=0}^infty b_n}{sumlimits_{n=0}^infty np_n}=frac{1}{frac{1}{k}frac{k(k+1)}{2}}=frac{2}{k+1}
$$
edited Jan 19 at 16:28
Did
248k23224463
answered Jan 19 at 15:30
SongSong
16.1k1739
$endgroup$
add a comment |
$begingroup$
Your equation can be rewritten in the form of a renewal equation:
$$
a_n = b_n +(a*p)_n
$$ where, for every $nge0$, $$p_n =frac{1}{k}1_{1le nle k}qquad b_n=1_{n=0}$$ and
$$
(a*p)_n :=sum_{j=0}^n a_{n-j}p_j
$$ Here, ${b_j}$ is added since $a_n =frac{1}{k}sumlimits_{j=1}^k a_{n-j}$ does not hold for $n=0$.
Now, it is a result of the (discrete version of) Blackwell's renewal theorem that
$$
lim_{ntoinfty} a_n =frac{sumlimits_{n=0}^infty b_n}{sumlimits_{n=0}^infty np_n}=frac{1}{frac{1}{k}frac{k(k+1)}{2}}=frac{2}{k+1}
$$
edited Jan 19 at 16:28
Did
248k23224463
answered Jan 19 at 15:30
SongSong
16.1k1739
$endgroup$
Your equation can be rewritten in the form of a renewal equation:
$$
a_n = b_n +(a*p)_n
$$ where, for every $nge0$, $$p_n =frac{1}{k}1_{1le nle k}qquad b_n=1_{n=0}$$ and
$$
(a*p)_n :=sum_{j=0}^n a_{n-j}p_j
$$ Here, ${b_j}$ is added since $a_n =frac{1}{k}sumlimits_{j=1}^k a_{n-j}$ does not hold for $n=0$.
Now, it is a result of the (discrete version of) Blackwell's renewal theorem that
$$
lim_{ntoinfty} a_n =frac{sumlimits_{n=0}^infty b_n}{sumlimits_{n=0}^infty np_n}=frac{1}{frac{1}{k}frac{k(k+1)}{2}}=frac{2}{k+1}
$$
edited Jan 19 at 16:28
Did
248k23224463
answered Jan 19 at 15:30
SongSong
16.1k1739
edited Jan 19 at 16:28
Did
248k23224463
edited Jan 19 at 16:28
Did
248k23224463
edited Jan 19 at 16:28
Did
248k23224463
248k23224463
answered Jan 19 at 15:30
SongSong
16.1k1739
answered Jan 19 at 15:30
SongSong
16.1k1739
answered Jan 19 at 15:30
SongSong
16.1k1739
16.1k1739
add a comment |
add a comment |
$begingroup$
Could be of help to associate to the cute answer already given, the expression of the generating function (z-Transform).
We have
$$
a_{,n} = {1 over k}sumlimits_{1, le ,i, le ,k} {a_{,n, - ,i} } + left[ {n = 0} right]
= {1 over k}sumlimits_i {left[ {1, le ,i, le ,k} right]a_{,,n, - ,i} } + left[ {n = 0} right]
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
The generating function will then be
$$
eqalign{
& A(z) = sumlimits_{0, le ,n} {a_{,n} ,z^{,n} }
= {1 over k}sumlimits_{0, le ,n} {sumlimits_i {left( {left[ {1, le ,i, le ,k} right]z^{,i} } right)left( {a_{,,n, - ,i} ,z^{,n - i} } right)} } + 1 = cr
& = {1 over k}left( {sumlimits_{0, le ,n} {left[ {1, le ,n, le ,k} right]z^{,n} } } right)
sumlimits_{0, le ,n} {left( {a_{,,n,} ,z^{,n} } right)} + 1 = cr
& = {1 over k}left( {zsumlimits_{1, le ,n, le ,k} {z^{,n - 1} } } right)sumlimits_{0, le ,n} {left( {a_{,,n,} ,z^{,n} } right)} + 1 = cr
& = {1 over k}left( {z{{1 - z^{,k} } over {1 - z}}} right)A(z) + 1 cr}
$$
that is
$$
A(z) = {1 over {1 - {z over k}left( {{{1 - z^{,k} } over {1 - z}}} right)}} = {{1 - z} over {z^{,k} - left( {k + 1} right)z + k}}
$$
answered Jan 19 at 21:46
G CabG Cab
19.8k31339
$endgroup$
add a comment |
$begingroup$
Could be of help to associate to the cute answer already given, the expression of the generating function (z-Transform).
We have
$$
a_{,n} = {1 over k}sumlimits_{1, le ,i, le ,k} {a_{,n, - ,i} } + left[ {n = 0} right]
= {1 over k}sumlimits_i {left[ {1, le ,i, le ,k} right]a_{,,n, - ,i} } + left[ {n = 0} right]
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
The generating function will then be
$$
eqalign{
& A(z) = sumlimits_{0, le ,n} {a_{,n} ,z^{,n} }
= {1 over k}sumlimits_{0, le ,n} {sumlimits_i {left( {left[ {1, le ,i, le ,k} right]z^{,i} } right)left( {a_{,,n, - ,i} ,z^{,n - i} } right)} } + 1 = cr
& = {1 over k}left( {sumlimits_{0, le ,n} {left[ {1, le ,n, le ,k} right]z^{,n} } } right)
sumlimits_{0, le ,n} {left( {a_{,,n,} ,z^{,n} } right)} + 1 = cr
& = {1 over k}left( {zsumlimits_{1, le ,n, le ,k} {z^{,n - 1} } } right)sumlimits_{0, le ,n} {left( {a_{,,n,} ,z^{,n} } right)} + 1 = cr
& = {1 over k}left( {z{{1 - z^{,k} } over {1 - z}}} right)A(z) + 1 cr}
$$
that is
$$
A(z) = {1 over {1 - {z over k}left( {{{1 - z^{,k} } over {1 - z}}} right)}} = {{1 - z} over {z^{,k} - left( {k + 1} right)z + k}}
$$
answered Jan 19 at 21:46
G CabG Cab
19.8k31339
$endgroup$
add a comment |
$begingroup$
Could be of help to associate to the cute answer already given, the expression of the generating function (z-Transform).
We have
$$
a_{,n} = {1 over k}sumlimits_{1, le ,i, le ,k} {a_{,n, - ,i} } + left[ {n = 0} right]
= {1 over k}sumlimits_i {left[ {1, le ,i, le ,k} right]a_{,,n, - ,i} } + left[ {n = 0} right]
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
The generating function will then be
$$
eqalign{
& A(z) = sumlimits_{0, le ,n} {a_{,n} ,z^{,n} }
= {1 over k}sumlimits_{0, le ,n} {sumlimits_i {left( {left[ {1, le ,i, le ,k} right]z^{,i} } right)left( {a_{,,n, - ,i} ,z^{,n - i} } right)} } + 1 = cr
& = {1 over k}left( {sumlimits_{0, le ,n} {left[ {1, le ,n, le ,k} right]z^{,n} } } right)
sumlimits_{0, le ,n} {left( {a_{,,n,} ,z^{,n} } right)} + 1 = cr
& = {1 over k}left( {zsumlimits_{1, le ,n, le ,k} {z^{,n - 1} } } right)sumlimits_{0, le ,n} {left( {a_{,,n,} ,z^{,n} } right)} + 1 = cr
& = {1 over k}left( {z{{1 - z^{,k} } over {1 - z}}} right)A(z) + 1 cr}
$$
that is
$$
A(z) = {1 over {1 - {z over k}left( {{{1 - z^{,k} } over {1 - z}}} right)}} = {{1 - z} over {z^{,k} - left( {k + 1} right)z + k}}
$$
answered Jan 19 at 21:46
G CabG Cab
19.8k31339
$endgroup$
Could be of help to associate to the cute answer already given, the expression of the generating function (z-Transform).
We have
$$
a_{,n} = {1 over k}sumlimits_{1, le ,i, le ,k} {a_{,n, - ,i} } + left[ {n = 0} right]
= {1 over k}sumlimits_i {left[ {1, le ,i, le ,k} right]a_{,,n, - ,i} } + left[ {n = 0} right]
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
The generating function will then be
$$
eqalign{
& A(z) = sumlimits_{0, le ,n} {a_{,n} ,z^{,n} }
= {1 over k}sumlimits_{0, le ,n} {sumlimits_i {left( {left[ {1, le ,i, le ,k} right]z^{,i} } right)left( {a_{,,n, - ,i} ,z^{,n - i} } right)} } + 1 = cr
& = {1 over k}left( {sumlimits_{0, le ,n} {left[ {1, le ,n, le ,k} right]z^{,n} } } right)
sumlimits_{0, le ,n} {left( {a_{,,n,} ,z^{,n} } right)} + 1 = cr
& = {1 over k}left( {zsumlimits_{1, le ,n, le ,k} {z^{,n - 1} } } right)sumlimits_{0, le ,n} {left( {a_{,,n,} ,z^{,n} } right)} + 1 = cr
& = {1 over k}left( {z{{1 - z^{,k} } over {1 - z}}} right)A(z) + 1 cr}
$$
that is
$$
A(z) = {1 over {1 - {z over k}left( {{{1 - z^{,k} } over {1 - z}}} right)}} = {{1 - z} over {z^{,k} - left( {k + 1} right)z + k}}
$$
answered Jan 19 at 21:46
G CabG Cab
19.8k31339
answered Jan 19 at 21:46
G CabG Cab
19.8k31339
answered Jan 19 at 21:46
G CabG Cab
19.8k31339
answered Jan 19 at 21:46
G CabG Cab
19.8k31339
19.8k31339
add a comment |
add a comment |
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