Rewriting a sum of harmonic powers as a minimal polynomial
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Revisiting one of my older questions, I've decided to try to tackle a simpler version of the problem, this time without the square root coefficients.
Let $x_0$ be a real number such that it satisfies the equation $$x_0+x_0^1/2+x_0^1/3+cdots+x_0^1/n=1$$ for a natural number $n$. What is the minimal polynomial in $mathbbZ[x]$?
Of course, this is possible by brute force: isolating the smallest power of $x$ then raising both sides by its reciprocal and repeating, but it becomes extremely tedious to do when $n$ is large. Also, this does not guarantee that the polynomial obtained is minimal.
This works fine for $n=1,2,3$. The minimal polynomials are, respectively, $$x-1,quad x^2-3x+1,quad x^5-8x^4+24x^3-21x^2+10x-1$$ and it may be interesting to note that the sign of the coefficients are alternating.
Is there an efficient way of doing this for the general case?
roots minimal-polynomials
asked Aug 27 at 18:45
TheSimpliFire
10.7k62054
add a comment |Â
up vote
3
down vote
favorite
Revisiting one of my older questions, I've decided to try to tackle a simpler version of the problem, this time without the square root coefficients.
Let $x_0$ be a real number such that it satisfies the equation $$x_0+x_0^1/2+x_0^1/3+cdots+x_0^1/n=1$$ for a natural number $n$. What is the minimal polynomial in $mathbbZ[x]$?
Of course, this is possible by brute force: isolating the smallest power of $x$ then raising both sides by its reciprocal and repeating, but it becomes extremely tedious to do when $n$ is large. Also, this does not guarantee that the polynomial obtained is minimal.
This works fine for $n=1,2,3$. The minimal polynomials are, respectively, $$x-1,quad x^2-3x+1,quad x^5-8x^4+24x^3-21x^2+10x-1$$ and it may be interesting to note that the sign of the coefficients are alternating.
Is there an efficient way of doing this for the general case?
roots minimal-polynomials
asked Aug 27 at 18:45
TheSimpliFire
10.7k62054
The "$=1$" seems arbitrary. If there is no sensible way which uses that it's the multiplicative identity (as in, say, any power of the LHS is also $=1$), then I would replece it with "$=c$", play around with that and see if there's a clearer pattern.
– Torsten Schoeneberg
Aug 28 at 10:50
Any thoughts on my answer, Fire?
– Gerry Myerson
Aug 29 at 13:34
@GerryMyerson sorry, I'm still trying to work something out from the equation you wrote in your answer. I'll get back to you asap
– TheSimpliFire
Aug 29 at 18:10
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Revisiting one of my older questions, I've decided to try to tackle a simpler version of the problem, this time without the square root coefficients.
Let $x_0$ be a real number such that it satisfies the equation $$x_0+x_0^1/2+x_0^1/3+cdots+x_0^1/n=1$$ for a natural number $n$. What is the minimal polynomial in $mathbbZ[x]$?
Of course, this is possible by brute force: isolating the smallest power of $x$ then raising both sides by its reciprocal and repeating, but it becomes extremely tedious to do when $n$ is large. Also, this does not guarantee that the polynomial obtained is minimal.
This works fine for $n=1,2,3$. The minimal polynomials are, respectively, $$x-1,quad x^2-3x+1,quad x^5-8x^4+24x^3-21x^2+10x-1$$ and it may be interesting to note that the sign of the coefficients are alternating.
Is there an efficient way of doing this for the general case?
roots minimal-polynomials
asked Aug 27 at 18:45
TheSimpliFire
10.7k62054
Revisiting one of my older questions, I've decided to try to tackle a simpler version of the problem, this time without the square root coefficients.
Let $x_0$ be a real number such that it satisfies the equation $$x_0+x_0^1/2+x_0^1/3+cdots+x_0^1/n=1$$ for a natural number $n$. What is the minimal polynomial in $mathbbZ[x]$?
Of course, this is possible by brute force: isolating the smallest power of $x$ then raising both sides by its reciprocal and repeating, but it becomes extremely tedious to do when $n$ is large. Also, this does not guarantee that the polynomial obtained is minimal.
This works fine for $n=1,2,3$. The minimal polynomials are, respectively, $$x-1,quad x^2-3x+1,quad x^5-8x^4+24x^3-21x^2+10x-1$$ and it may be interesting to note that the sign of the coefficients are alternating.
Is there an efficient way of doing this for the general case?
roots minimal-polynomials
asked Aug 27 at 18:45
TheSimpliFire
10.7k62054
edited Aug 28 at 7:02
asked Aug 27 at 18:45
TheSimpliFire
10.7k62054
asked Aug 27 at 18:45
TheSimpliFire
10.7k62054
asked Aug 27 at 18:45
TheSimpliFire
10.7k62054
10.7k62054
The "$=1$" seems arbitrary. If there is no sensible way which uses that it's the multiplicative identity (as in, say, any power of the LHS is also $=1$), then I would replece it with "$=c$", play around with that and see if there's a clearer pattern.
– Torsten Schoeneberg
Aug 28 at 10:50
Any thoughts on my answer, Fire?
– Gerry Myerson
Aug 29 at 13:34
@GerryMyerson sorry, I'm still trying to work something out from the equation you wrote in your answer. I'll get back to you asap
– TheSimpliFire
Aug 29 at 18:10
add a comment |Â
The "$=1$" seems arbitrary. If there is no sensible way which uses that it's the multiplicative identity (as in, say, any power of the LHS is also $=1$), then I would replece it with "$=c$", play around with that and see if there's a clearer pattern.
– Torsten Schoeneberg
Aug 28 at 10:50
Any thoughts on my answer, Fire?
– Gerry Myerson
Aug 29 at 13:34
@GerryMyerson sorry, I'm still trying to work something out from the equation you wrote in your answer. I'll get back to you asap
– TheSimpliFire
Aug 29 at 18:10
The "$=1$" seems arbitrary. If there is no sensible way which uses that it's the multiplicative identity (as in, say, any power of the LHS is also $=1$), then I would replece it with "$=c$", play around with that and see if there's a clearer pattern.
– Torsten Schoeneberg
Aug 28 at 10:50
The "$=1$" seems arbitrary. If there is no sensible way which uses that it's the multiplicative identity (as in, say, any power of the LHS is also $=1$), then I would replece it with "$=c$", play around with that and see if there's a clearer pattern.
– Torsten Schoeneberg
Aug 28 at 10:50
Any thoughts on my answer, Fire?
– Gerry Myerson
Aug 29 at 13:34
Any thoughts on my answer, Fire?
– Gerry Myerson
Aug 29 at 13:34
@GerryMyerson sorry, I'm still trying to work something out from the equation you wrote in your answer. I'll get back to you asap
– TheSimpliFire
Aug 29 at 18:10
@GerryMyerson sorry, I'm still trying to work something out from the equation you wrote in your answer. I'll get back to you asap
– TheSimpliFire
Aug 29 at 18:10
add a comment |Â
1 Answer
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1
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Let $x=y^L$, where $L$ is the least common multiple of $2,3,dots,n$. Then you have the two equations, $y^L-x=0,y^L+y^L/2+y^L/3+cdots+y^L/n-1=0$. The resultant of these two polynomials will be a polynomial satisfied by $x$. I suspect, but am not sure I could prove, that it's the minimal polynomial.
The resultant can be computed as a determinant. Unfortunately, it's the determinant of a $2Ltimes2L$ matrix, but at least it's a very sparse matrix. I have a hunch the problem is inherently tedious and that there's no efficient way to do it.
answered Aug 28 at 7:34
Gerry Myerson
143k8145295
It looks like the first equation is trivial and the second equation is just raising each term to a power of $textlcm$. However, this still avoids the method of finding a minimal polynomial.
– TheSimpliFire
Sep 1 at 13:21
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $x=y^L$, where $L$ is the least common multiple of $2,3,dots,n$. Then you have the two equations, $y^L-x=0,y^L+y^L/2+y^L/3+cdots+y^L/n-1=0$. The resultant of these two polynomials will be a polynomial satisfied by $x$. I suspect, but am not sure I could prove, that it's the minimal polynomial.
The resultant can be computed as a determinant. Unfortunately, it's the determinant of a $2Ltimes2L$ matrix, but at least it's a very sparse matrix. I have a hunch the problem is inherently tedious and that there's no efficient way to do it.
answered Aug 28 at 7:34
Gerry Myerson
143k8145295
It looks like the first equation is trivial and the second equation is just raising each term to a power of $textlcm$. However, this still avoids the method of finding a minimal polynomial.
– TheSimpliFire
Sep 1 at 13:21
add a comment |Â
up vote
1
down vote
Let $x=y^L$, where $L$ is the least common multiple of $2,3,dots,n$. Then you have the two equations, $y^L-x=0,y^L+y^L/2+y^L/3+cdots+y^L/n-1=0$. The resultant of these two polynomials will be a polynomial satisfied by $x$. I suspect, but am not sure I could prove, that it's the minimal polynomial.
The resultant can be computed as a determinant. Unfortunately, it's the determinant of a $2Ltimes2L$ matrix, but at least it's a very sparse matrix. I have a hunch the problem is inherently tedious and that there's no efficient way to do it.
answered Aug 28 at 7:34
Gerry Myerson
143k8145295
It looks like the first equation is trivial and the second equation is just raising each term to a power of $textlcm$. However, this still avoids the method of finding a minimal polynomial.
– TheSimpliFire
Sep 1 at 13:21
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $x=y^L$, where $L$ is the least common multiple of $2,3,dots,n$. Then you have the two equations, $y^L-x=0,y^L+y^L/2+y^L/3+cdots+y^L/n-1=0$. The resultant of these two polynomials will be a polynomial satisfied by $x$. I suspect, but am not sure I could prove, that it's the minimal polynomial.
The resultant can be computed as a determinant. Unfortunately, it's the determinant of a $2Ltimes2L$ matrix, but at least it's a very sparse matrix. I have a hunch the problem is inherently tedious and that there's no efficient way to do it.
answered Aug 28 at 7:34
Gerry Myerson
143k8145295
Let $x=y^L$, where $L$ is the least common multiple of $2,3,dots,n$. Then you have the two equations, $y^L-x=0,y^L+y^L/2+y^L/3+cdots+y^L/n-1=0$. The resultant of these two polynomials will be a polynomial satisfied by $x$. I suspect, but am not sure I could prove, that it's the minimal polynomial.
The resultant can be computed as a determinant. Unfortunately, it's the determinant of a $2Ltimes2L$ matrix, but at least it's a very sparse matrix. I have a hunch the problem is inherently tedious and that there's no efficient way to do it.
answered Aug 28 at 7:34
Gerry Myerson
143k8145295
answered Aug 28 at 7:34
Gerry Myerson
143k8145295
answered Aug 28 at 7:34
Gerry Myerson
143k8145295
answered Aug 28 at 7:34
Gerry Myerson
143k8145295
143k8145295
It looks like the first equation is trivial and the second equation is just raising each term to a power of $textlcm$. However, this still avoids the method of finding a minimal polynomial.
– TheSimpliFire
Sep 1 at 13:21
add a comment |Â
It looks like the first equation is trivial and the second equation is just raising each term to a power of $textlcm$. However, this still avoids the method of finding a minimal polynomial.
– TheSimpliFire
Sep 1 at 13:21
It looks like the first equation is trivial and the second equation is just raising each term to a power of $textlcm$. However, this still avoids the method of finding a minimal polynomial.
– TheSimpliFire
Sep 1 at 13:21
It looks like the first equation is trivial and the second equation is just raising each term to a power of $textlcm$. However, this still avoids the method of finding a minimal polynomial.
– TheSimpliFire
Sep 1 at 13:21
add a comment |Â
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The "$=1$" seems arbitrary. If there is no sensible way which uses that it's the multiplicative identity (as in, say, any power of the LHS is also $=1$), then I would replece it with "$=c$", play around with that and see if there's a clearer pattern.
– Torsten Schoeneberg
Aug 28 at 10:50
Any thoughts on my answer, Fire?
– Gerry Myerson
Aug 29 at 13:34
@GerryMyerson sorry, I'm still trying to work something out from the equation you wrote in your answer. I'll get back to you asap
– TheSimpliFire
Aug 29 at 18:10