Mixing
A remarkable almost-identity
Clash Royale CLAN TAG#URR8PPP
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OEIS sequence A210247 gives the signs of $textli(-n,-1/3) = sum_k=1^infty (-1)^k k^n/3^k$, also the signs of the Maclaurin coefficients of $4/(3 + exp(4x))$.
Mikhail Kurkov noticed that it appeared that $a(n+28) = -a(n)$ for this sequence.
It's not quite true: the first $n$'s for which this is not true are $578, 1143,$ and $1736$. But still it's remarkably close to true.
A nice illustration of the Strong Law of Small Numbers? Is there any explanation for the almost-identity, or is it just coincidence? Is it still true for large $n$ that $a(n+28)$ is usually $-a(n)$?
nt.number-theory ca.classical-analysis-and-odes sequences-and-series integer-sequences polylogarithms
asked Sep 5 at 20:28
Robert Israel
40.3k46113
add a comment |Â
up vote
23
down vote
favorite
OEIS sequence A210247 gives the signs of $textli(-n,-1/3) = sum_k=1^infty (-1)^k k^n/3^k$, also the signs of the Maclaurin coefficients of $4/(3 + exp(4x))$.
Mikhail Kurkov noticed that it appeared that $a(n+28) = -a(n)$ for this sequence.
It's not quite true: the first $n$'s for which this is not true are $578, 1143,$ and $1736$. But still it's remarkably close to true.
A nice illustration of the Strong Law of Small Numbers? Is there any explanation for the almost-identity, or is it just coincidence? Is it still true for large $n$ that $a(n+28)$ is usually $-a(n)$?
nt.number-theory ca.classical-analysis-and-odes sequences-and-series integer-sequences polylogarithms
asked Sep 5 at 20:28
Robert Israel
40.3k46113
add a comment |Â
up vote
23
down vote
favorite
up vote
23
down vote
favorite
OEIS sequence A210247 gives the signs of $textli(-n,-1/3) = sum_k=1^infty (-1)^k k^n/3^k$, also the signs of the Maclaurin coefficients of $4/(3 + exp(4x))$.
Mikhail Kurkov noticed that it appeared that $a(n+28) = -a(n)$ for this sequence.
It's not quite true: the first $n$'s for which this is not true are $578, 1143,$ and $1736$. But still it's remarkably close to true.
A nice illustration of the Strong Law of Small Numbers? Is there any explanation for the almost-identity, or is it just coincidence? Is it still true for large $n$ that $a(n+28)$ is usually $-a(n)$?
nt.number-theory ca.classical-analysis-and-odes sequences-and-series integer-sequences polylogarithms
asked Sep 5 at 20:28
Robert Israel
40.3k46113
OEIS sequence A210247 gives the signs of $textli(-n,-1/3) = sum_k=1^infty (-1)^k k^n/3^k$, also the signs of the Maclaurin coefficients of $4/(3 + exp(4x))$.
Mikhail Kurkov noticed that it appeared that $a(n+28) = -a(n)$ for this sequence.
It's not quite true: the first $n$'s for which this is not true are $578, 1143,$ and $1736$. But still it's remarkably close to true.
A nice illustration of the Strong Law of Small Numbers? Is there any explanation for the almost-identity, or is it just coincidence? Is it still true for large $n$ that $a(n+28)$ is usually $-a(n)$?
nt.number-theory ca.classical-analysis-and-odes sequences-and-series integer-sequences polylogarithms
asked Sep 5 at 20:28
Robert Israel
40.3k46113
asked Sep 5 at 20:28
Robert Israel
40.3k46113
asked Sep 5 at 20:28
Robert Israel
40.3k46113
asked Sep 5 at 20:28
Robert Israel
40.3k46113
40.3k46113
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
38
down vote
accepted
Consider $F(z) = 4/(3+exp(4z))$ as a function of the complex variable $z$. It is meromorphic and has simple poles where the denominator vanishes. Namely when $4z = log 3 + (2k +1)pi i$ for integers $k$. The poles with the smallest magnitude of $z$ occur when $4z = log 3 pm pi i$. We can compute the Taylor series coefficients of $F$ by looking at
$$
frac12pi i int_z F(z) z^-n fracdzz,
$$
starting with $r$ suitably small. Now we can estimate the integral asymptotically by taking larger values of $r$, and accounting for poles that are encountered. As noted above, the smallest poles are at $(log 3 pm pi i)/4$ and these will account for the leading asymptotics of these coefficients. Now, the argument of $(log 3 pm pi i)/4$ is $pm 1.23438ldots $ which is very nearly $11 pi/28=1.23419ldots$. This accounts for the observed phenomenon.
answered Sep 5 at 21:02
Lucia
33.8k5146175
Out of curiosity: does it answer the question that for large $n$ that $a(n+28)$ is usually $-a(n)$?
– Andrew
9 hours ago
@Andrew: Yes, for large $n$ most of the time the signs will flip, but a small proportion of the time it won't. The answer roughly says that the signs look like the signs of $cos (1.23438 n)$ (not being too careful here), which will have the similar feature of flipping very often when $n$ is increased by $28$.
– Lucia
8 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
38
down vote
accepted
Consider $F(z) = 4/(3+exp(4z))$ as a function of the complex variable $z$. It is meromorphic and has simple poles where the denominator vanishes. Namely when $4z = log 3 + (2k +1)pi i$ for integers $k$. The poles with the smallest magnitude of $z$ occur when $4z = log 3 pm pi i$. We can compute the Taylor series coefficients of $F$ by looking at
$$
frac12pi i int_z F(z) z^-n fracdzz,
$$
starting with $r$ suitably small. Now we can estimate the integral asymptotically by taking larger values of $r$, and accounting for poles that are encountered. As noted above, the smallest poles are at $(log 3 pm pi i)/4$ and these will account for the leading asymptotics of these coefficients. Now, the argument of $(log 3 pm pi i)/4$ is $pm 1.23438ldots $ which is very nearly $11 pi/28=1.23419ldots$. This accounts for the observed phenomenon.
answered Sep 5 at 21:02
Lucia
33.8k5146175
Out of curiosity: does it answer the question that for large $n$ that $a(n+28)$ is usually $-a(n)$?
– Andrew
9 hours ago
@Andrew: Yes, for large $n$ most of the time the signs will flip, but a small proportion of the time it won't. The answer roughly says that the signs look like the signs of $cos (1.23438 n)$ (not being too careful here), which will have the similar feature of flipping very often when $n$ is increased by $28$.
– Lucia
8 hours ago
add a comment |Â
up vote
38
down vote
accepted
Consider $F(z) = 4/(3+exp(4z))$ as a function of the complex variable $z$. It is meromorphic and has simple poles where the denominator vanishes. Namely when $4z = log 3 + (2k +1)pi i$ for integers $k$. The poles with the smallest magnitude of $z$ occur when $4z = log 3 pm pi i$. We can compute the Taylor series coefficients of $F$ by looking at
$$
frac12pi i int_z F(z) z^-n fracdzz,
$$
starting with $r$ suitably small. Now we can estimate the integral asymptotically by taking larger values of $r$, and accounting for poles that are encountered. As noted above, the smallest poles are at $(log 3 pm pi i)/4$ and these will account for the leading asymptotics of these coefficients. Now, the argument of $(log 3 pm pi i)/4$ is $pm 1.23438ldots $ which is very nearly $11 pi/28=1.23419ldots$. This accounts for the observed phenomenon.
answered Sep 5 at 21:02
Lucia
33.8k5146175
Out of curiosity: does it answer the question that for large $n$ that $a(n+28)$ is usually $-a(n)$?
– Andrew
9 hours ago
@Andrew: Yes, for large $n$ most of the time the signs will flip, but a small proportion of the time it won't. The answer roughly says that the signs look like the signs of $cos (1.23438 n)$ (not being too careful here), which will have the similar feature of flipping very often when $n$ is increased by $28$.
– Lucia
8 hours ago
add a comment |Â
up vote
38
down vote
accepted
up vote
38
down vote
accepted
Consider $F(z) = 4/(3+exp(4z))$ as a function of the complex variable $z$. It is meromorphic and has simple poles where the denominator vanishes. Namely when $4z = log 3 + (2k +1)pi i$ for integers $k$. The poles with the smallest magnitude of $z$ occur when $4z = log 3 pm pi i$. We can compute the Taylor series coefficients of $F$ by looking at
$$
frac12pi i int_z F(z) z^-n fracdzz,
$$
starting with $r$ suitably small. Now we can estimate the integral asymptotically by taking larger values of $r$, and accounting for poles that are encountered. As noted above, the smallest poles are at $(log 3 pm pi i)/4$ and these will account for the leading asymptotics of these coefficients. Now, the argument of $(log 3 pm pi i)/4$ is $pm 1.23438ldots $ which is very nearly $11 pi/28=1.23419ldots$. This accounts for the observed phenomenon.
answered Sep 5 at 21:02
Lucia
33.8k5146175
Consider $F(z) = 4/(3+exp(4z))$ as a function of the complex variable $z$. It is meromorphic and has simple poles where the denominator vanishes. Namely when $4z = log 3 + (2k +1)pi i$ for integers $k$. The poles with the smallest magnitude of $z$ occur when $4z = log 3 pm pi i$. We can compute the Taylor series coefficients of $F$ by looking at
$$
frac12pi i int_z F(z) z^-n fracdzz,
$$
starting with $r$ suitably small. Now we can estimate the integral asymptotically by taking larger values of $r$, and accounting for poles that are encountered. As noted above, the smallest poles are at $(log 3 pm pi i)/4$ and these will account for the leading asymptotics of these coefficients. Now, the argument of $(log 3 pm pi i)/4$ is $pm 1.23438ldots $ which is very nearly $11 pi/28=1.23419ldots$. This accounts for the observed phenomenon.
answered Sep 5 at 21:02
Lucia
33.8k5146175
answered Sep 5 at 21:02
Lucia
33.8k5146175
answered Sep 5 at 21:02
Lucia
33.8k5146175
answered Sep 5 at 21:02
Lucia
33.8k5146175
33.8k5146175
Out of curiosity: does it answer the question that for large $n$ that $a(n+28)$ is usually $-a(n)$?
– Andrew
9 hours ago
@Andrew: Yes, for large $n$ most of the time the signs will flip, but a small proportion of the time it won't. The answer roughly says that the signs look like the signs of $cos (1.23438 n)$ (not being too careful here), which will have the similar feature of flipping very often when $n$ is increased by $28$.
– Lucia
8 hours ago
add a comment |Â
Out of curiosity: does it answer the question that for large $n$ that $a(n+28)$ is usually $-a(n)$?
– Andrew
9 hours ago
@Andrew: Yes, for large $n$ most of the time the signs will flip, but a small proportion of the time it won't. The answer roughly says that the signs look like the signs of $cos (1.23438 n)$ (not being too careful here), which will have the similar feature of flipping very often when $n$ is increased by $28$.
– Lucia
8 hours ago
Out of curiosity: does it answer the question that for large $n$ that $a(n+28)$ is usually $-a(n)$?
– Andrew
9 hours ago
Out of curiosity: does it answer the question that for large $n$ that $a(n+28)$ is usually $-a(n)$?
– Andrew
9 hours ago
@Andrew: Yes, for large $n$ most of the time the signs will flip, but a small proportion of the time it won't. The answer roughly says that the signs look like the signs of $cos (1.23438 n)$ (not being too careful here), which will have the similar feature of flipping very often when $n$ is increased by $28$.
– Lucia
8 hours ago
@Andrew: Yes, for large $n$ most of the time the signs will flip, but a small proportion of the time it won't. The answer roughly says that the signs look like the signs of $cos (1.23438 n)$ (not being too careful here), which will have the similar feature of flipping very often when $n$ is increased by $28$.
– Lucia
8 hours ago
add a comment |Â
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