Skip to main content
added 419 characters in body
Source Link
Aaron Meyerowitz
  • 30.1k
  • 1
  • 48
  • 104

I would guess with you that it isn't true, but would not be surprised if it is.REVISED

I wanted to ask how much you know and provide this, for what it is worth:No.

My calculations,Let which I would need to repeat to be confident of, are that there are$$A=(1+x^2+x^4+x^6+\ldots)(1+x^3+x^6+x^9+\ldots)\ldots$$ and call a few, but not that many, ways to pick the signs out pastpolynomial $\pm x^{N}$ for$f=1+\sum_1^da_ix^i$ $N$ aroundfeasible if all the $15$ so that$a_i \in \{{-1,1\}}$ and all the coefficients of the product are all $0,\pm1$ out$fA$ up to thethat of $x^{N}$ term. For example one must start$x^d$ are in $1-x-x^2$ and the next thing must be$\{{-1,0,1\}}.$ Finally, call $+x^{3}$ because otherwise the first term$f$ maximal if it is feasible but neither of $$1-x-x^2-x^3+x^4+x^5+x^6+x^7-x^8+x^9$$$$-x^{10}+x^{11}-x^{12}-x^{13}-x^{14}-x^{15}-x^{16}+cx^{17}-x^{18}$$ $$-cx^{19}-cx^{20}-cx^{21}+x^{22}+x^{23}+x^{24}+dx^{25}+\cdots$$ which leaves$f+x^{d+1}$ nor $$(c+d-4)x^{25}$$ in the product$f-x^{d+1}$ is feasible.

I did not as carefully check the continuationsThere are no feasible polynomials of degree 31. There are $1-x-x^2+x^3$ although things were not quite as constrained$40$ maximal polynomials.

Again, I am not confident thatThe largest degree maximal polynomials are

${x}^{30}+{x}^{29}+{x}^{28}-{x}^{27}-{x}^{26}+{x}^{25}-{x}^{24}-{x}^{23 }-{x}^{22}-{x}^{21}-{x}^{20}+{x}^{19}+{x}^{18}+{x}^{17}-{x}^{16}-{x}^{ 15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x}^{10}-{x}^{9}+{x}^{8}+{x}^{ 7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}-x+1 $

${x}^{30}+{x}^{29}-{x}^{28}-{x}^{27}-{x}^{26}+{x}^{25}-{x}^{24}+{x}^{23 }-{x}^{22}+{x}^{21}-{x}^{20}+{x}^{19}-{x}^{18}-{x}^{17}+{x}^{16}-{x}^{ 15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x}^{10}-{x}^{9}+{x}^{8}+{x}^{ 7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}-x+1 $ and $-{x}^{25}-{x}^{24}-{x}^{23}+{x}^{22}+{x}^{21}+{x}^{20}+{x}^{19}-{x}^{ 18}+{x}^{17}-{x}^{16}-{x}^{15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x} ^{10}-{x}^{9}+{x}^{8}+{x}^{7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}- x+1. $

the above is error free, butdegrees of the approach seems feasible. Is this in line with what you did and, if so, how far did you push it out?maximal polynomials are

$30, 30, 25, 24, 24, 24, 24, 24, 24, 24, 22, 21, 21, 20, 19, 18, 18, 18, 18, 18, 18, 17, 17, 16, 15, 15, 14, 14, 14, 14, 14, 13, 10, 10, 9, 8, 8, 7, 7, 6.$

I would guess with you that it isn't true, but would not be surprised if it is.

I wanted to ask how much you know and provide this, for what it is worth:

My calculations, which I would need to repeat to be confident of, are that there are a few, but not that many, ways to pick the signs out past $\pm x^{N}$ for $N$ around $15$ so that the coefficients of the product are all $0,\pm1$ out to the $x^{N}$ term. For example one must start $1-x-x^2$ and the next thing must be $+x^{3}$ because otherwise the first term is $$1-x-x^2-x^3+x^4+x^5+x^6+x^7-x^8+x^9$$$$-x^{10}+x^{11}-x^{12}-x^{13}-x^{14}-x^{15}-x^{16}+cx^{17}-x^{18}$$ $$-cx^{19}-cx^{20}-cx^{21}+x^{22}+x^{23}+x^{24}+dx^{25}+\cdots$$ which leaves $$(c+d-4)x^{25}$$ in the product.

I did not as carefully check the continuations of $1-x-x^2+x^3$ although things were not quite as constrained.

Again, I am not confident that the above is error free, but the approach seems feasible. Is this in line with what you did and, if so, how far did you push it out?

REVISED

No.

Let $$A=(1+x^2+x^4+x^6+\ldots)(1+x^3+x^6+x^9+\ldots)\ldots$$ and call a polynomial $f=1+\sum_1^da_ix^i$ feasible if all the $a_i \in \{{-1,1\}}$ and all the coefficients of $fA$ up to that of $x^d$ are in $\{{-1,0,1\}}.$ Finally, call $f$ maximal if it is feasible but neither of $f+x^{d+1}$ nor $f-x^{d+1}$ is feasible.

There are no feasible polynomials of degree 31. There are $40$ maximal polynomials.

The largest degree maximal polynomials are

${x}^{30}+{x}^{29}+{x}^{28}-{x}^{27}-{x}^{26}+{x}^{25}-{x}^{24}-{x}^{23 }-{x}^{22}-{x}^{21}-{x}^{20}+{x}^{19}+{x}^{18}+{x}^{17}-{x}^{16}-{x}^{ 15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x}^{10}-{x}^{9}+{x}^{8}+{x}^{ 7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}-x+1 $

${x}^{30}+{x}^{29}-{x}^{28}-{x}^{27}-{x}^{26}+{x}^{25}-{x}^{24}+{x}^{23 }-{x}^{22}+{x}^{21}-{x}^{20}+{x}^{19}-{x}^{18}-{x}^{17}+{x}^{16}-{x}^{ 15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x}^{10}-{x}^{9}+{x}^{8}+{x}^{ 7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}-x+1 $ and $-{x}^{25}-{x}^{24}-{x}^{23}+{x}^{22}+{x}^{21}+{x}^{20}+{x}^{19}-{x}^{ 18}+{x}^{17}-{x}^{16}-{x}^{15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x} ^{10}-{x}^{9}+{x}^{8}+{x}^{7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}- x+1. $

the degrees of the maximal polynomials are

$30, 30, 25, 24, 24, 24, 24, 24, 24, 24, 22, 21, 21, 20, 19, 18, 18, 18, 18, 18, 18, 17, 17, 16, 15, 15, 14, 14, 14, 14, 14, 13, 10, 10, 9, 8, 8, 7, 7, 6.$

Source Link
Aaron Meyerowitz
  • 30.1k
  • 1
  • 48
  • 104

I would guess with you that it isn't true, but would not be surprised if it is.

I wanted to ask how much you know and provide this, for what it is worth:

My calculations, which I would need to repeat to be confident of, are that there are a few, but not that many, ways to pick the signs out past $\pm x^{N}$ for $N$ around $15$ so that the coefficients of the product are all $0,\pm1$ out to the $x^{N}$ term. For example one must start $1-x-x^2$ and the next thing must be $+x^{3}$ because otherwise the first term is $$1-x-x^2-x^3+x^4+x^5+x^6+x^7-x^8+x^9$$$$-x^{10}+x^{11}-x^{12}-x^{13}-x^{14}-x^{15}-x^{16}+cx^{17}-x^{18}$$ $$-cx^{19}-cx^{20}-cx^{21}+x^{22}+x^{23}+x^{24}+dx^{25}+\cdots$$ which leaves $$(c+d-4)x^{25}$$ in the product.

I did not as carefully check the continuations of $1-x-x^2+x^3$ although things were not quite as constrained.

Again, I am not confident that the above is error free, but the approach seems feasible. Is this in line with what you did and, if so, how far did you push it out?