Mixing
Methods for calculating the number of zeros of a polynomial with a specified real part?
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
Given a polynomial with real coefficients is there a method (e.g. from algebra or complex analysis) to calculate the number of complex zeros with a specified real part?
Background. This question is motivated by my tests related to this problem.
Let $p>3$ be a prime number. Let $G_p(x)=(x+1)^p-x^p-1$, and let
$$F_p(x)=frac(x+1)^p-x^p-1px(x+1)(x^2+x+1)^n_p$$
where the exponent $n_p$ is equal to $1$ (resp. $2$) when $pequiv-1pmod 6$ (resp. $pequiv1pmod 6$).
The answer by Lord Shark the Unknown (loc. linked) implies that $F_p(x)$ is a monic polynomial with integer coefficients. The degree of $F_p$ is equal to $6lfloor(p-3)/6rfloor$. I can show that the complex zeros of $F_p(x)$ come in groups of six. Each of the form $alpha,-alpha-1,1/alpha,-1/(alpha+1),-alpha/(alpha+1),-(alpha+1)/alpha.$ That is, orbits of a familiar group (isomorphic to $S_3$) of fractional linear transformations.
My conjecture. Exactly one third of the zeros of $F_p(x)$ have real part equal to $-1/2$.
I tested this with Mathematica for a few of the smallest primes and it seems to hold. Also, each sextet of zeros of the above form seems to be stable under complex conjugation, and seems to contain a complex conjugate pair of numbers with real part $=-1/2$.
Anyway, I am curious about the number of zeros $z=s+it$ of the polynomial $F_p(x)$ on the line $s=-1/2$.
Summary and thoughts.
- Any general method or formula is welcome, but I will be extra grateful if you want to test a method on the polynomial $G_p(x)$ or $F_p(x)$ :-)
- My first idea was to try the following: Given a polynomial $P(x)=prod_i(x-z_i)$ is there a way of getting $R(x):=prod_i(x-z_i-overlinez_i)$? If this can be done, then we get the answer by calculating the multiplicity of $-1$ as a zero of $R(x)$.
- May be a method for calculating the number of real zeros can be used with suitable substitution that maps the real axes to the line $s=-1/2$ (need to check on this)?
- Of course, if you can prove that $F_p(x)$ is irreducible it is better that you post the answer to the linked question. The previous bounty expired, but that can be fixed.
complex-analysis polynomials
asked Jul 16 at 14:18
Jyrki Lahtonen
105k12161355
add a comment |Â
up vote
5
down vote
favorite
Given a polynomial with real coefficients is there a method (e.g. from algebra or complex analysis) to calculate the number of complex zeros with a specified real part?
Background. This question is motivated by my tests related to this problem.
Let $p>3$ be a prime number. Let $G_p(x)=(x+1)^p-x^p-1$, and let
$$F_p(x)=frac(x+1)^p-x^p-1px(x+1)(x^2+x+1)^n_p$$
where the exponent $n_p$ is equal to $1$ (resp. $2$) when $pequiv-1pmod 6$ (resp. $pequiv1pmod 6$).
The answer by Lord Shark the Unknown (loc. linked) implies that $F_p(x)$ is a monic polynomial with integer coefficients. The degree of $F_p$ is equal to $6lfloor(p-3)/6rfloor$. I can show that the complex zeros of $F_p(x)$ come in groups of six. Each of the form $alpha,-alpha-1,1/alpha,-1/(alpha+1),-alpha/(alpha+1),-(alpha+1)/alpha.$ That is, orbits of a familiar group (isomorphic to $S_3$) of fractional linear transformations.
My conjecture. Exactly one third of the zeros of $F_p(x)$ have real part equal to $-1/2$.
I tested this with Mathematica for a few of the smallest primes and it seems to hold. Also, each sextet of zeros of the above form seems to be stable under complex conjugation, and seems to contain a complex conjugate pair of numbers with real part $=-1/2$.
Anyway, I am curious about the number of zeros $z=s+it$ of the polynomial $F_p(x)$ on the line $s=-1/2$.
Summary and thoughts.
- Any general method or formula is welcome, but I will be extra grateful if you want to test a method on the polynomial $G_p(x)$ or $F_p(x)$ :-)
- My first idea was to try the following: Given a polynomial $P(x)=prod_i(x-z_i)$ is there a way of getting $R(x):=prod_i(x-z_i-overlinez_i)$? If this can be done, then we get the answer by calculating the multiplicity of $-1$ as a zero of $R(x)$.
- May be a method for calculating the number of real zeros can be used with suitable substitution that maps the real axes to the line $s=-1/2$ (need to check on this)?
- Of course, if you can prove that $F_p(x)$ is irreducible it is better that you post the answer to the linked question. The previous bounty expired, but that can be fixed.
complex-analysis polynomials
asked Jul 16 at 14:18
Jyrki Lahtonen
105k12161355
At least a fairly simple way to quickly count the number of roots of $(x+1)^p-x^p-1$ on the “critical lineâ€Â: compute $$2^-frac1p cosleft(frac2 pi np-1right)^1-frac1p$$ for $n=0, ldots, lfloor(p-1)/4rfloor$. Every value in $(0, tfrac12]$ adds four roots and $0$ adds two roots.
– WimC
Jul 17 at 19:29
Sounds interesting @WimC. I'm afraid I don't see it. Can you please elaborate?
– Jyrki Lahtonen
Jul 17 at 19:51
1
See my answer below. It is based on the observation that solutions correspond to points $z$ on the ciritical line such that $z^p$ is also on the critical line. Then a bit of computation leads to the stated result.
– WimC
Jul 17 at 19:55
1
@JyrkiLahtonenMay be a method for calculating the number of real zeros can be usedLet $,x=-1/2 + i z,$, then $,P(x)=A(z) + i B(z),$ with $,A,B in mathbbR[textx],$ has roots with real part $,-1/2,$ iff $,gcd(A,B),$ has real roots. The computations would not be pretty, though.
– dxiv
Jul 17 at 20:55
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Given a polynomial with real coefficients is there a method (e.g. from algebra or complex analysis) to calculate the number of complex zeros with a specified real part?
Background. This question is motivated by my tests related to this problem.
Let $p>3$ be a prime number. Let $G_p(x)=(x+1)^p-x^p-1$, and let
$$F_p(x)=frac(x+1)^p-x^p-1px(x+1)(x^2+x+1)^n_p$$
where the exponent $n_p$ is equal to $1$ (resp. $2$) when $pequiv-1pmod 6$ (resp. $pequiv1pmod 6$).
The answer by Lord Shark the Unknown (loc. linked) implies that $F_p(x)$ is a monic polynomial with integer coefficients. The degree of $F_p$ is equal to $6lfloor(p-3)/6rfloor$. I can show that the complex zeros of $F_p(x)$ come in groups of six. Each of the form $alpha,-alpha-1,1/alpha,-1/(alpha+1),-alpha/(alpha+1),-(alpha+1)/alpha.$ That is, orbits of a familiar group (isomorphic to $S_3$) of fractional linear transformations.
My conjecture. Exactly one third of the zeros of $F_p(x)$ have real part equal to $-1/2$.
I tested this with Mathematica for a few of the smallest primes and it seems to hold. Also, each sextet of zeros of the above form seems to be stable under complex conjugation, and seems to contain a complex conjugate pair of numbers with real part $=-1/2$.
Anyway, I am curious about the number of zeros $z=s+it$ of the polynomial $F_p(x)$ on the line $s=-1/2$.
Summary and thoughts.
- Any general method or formula is welcome, but I will be extra grateful if you want to test a method on the polynomial $G_p(x)$ or $F_p(x)$ :-)
- My first idea was to try the following: Given a polynomial $P(x)=prod_i(x-z_i)$ is there a way of getting $R(x):=prod_i(x-z_i-overlinez_i)$? If this can be done, then we get the answer by calculating the multiplicity of $-1$ as a zero of $R(x)$.
- May be a method for calculating the number of real zeros can be used with suitable substitution that maps the real axes to the line $s=-1/2$ (need to check on this)?
- Of course, if you can prove that $F_p(x)$ is irreducible it is better that you post the answer to the linked question. The previous bounty expired, but that can be fixed.
complex-analysis polynomials
asked Jul 16 at 14:18
Jyrki Lahtonen
105k12161355
Given a polynomial with real coefficients is there a method (e.g. from algebra or complex analysis) to calculate the number of complex zeros with a specified real part?
Background. This question is motivated by my tests related to this problem.
Let $p>3$ be a prime number. Let $G_p(x)=(x+1)^p-x^p-1$, and let
$$F_p(x)=frac(x+1)^p-x^p-1px(x+1)(x^2+x+1)^n_p$$
where the exponent $n_p$ is equal to $1$ (resp. $2$) when $pequiv-1pmod 6$ (resp. $pequiv1pmod 6$).
The answer by Lord Shark the Unknown (loc. linked) implies that $F_p(x)$ is a monic polynomial with integer coefficients. The degree of $F_p$ is equal to $6lfloor(p-3)/6rfloor$. I can show that the complex zeros of $F_p(x)$ come in groups of six. Each of the form $alpha,-alpha-1,1/alpha,-1/(alpha+1),-alpha/(alpha+1),-(alpha+1)/alpha.$ That is, orbits of a familiar group (isomorphic to $S_3$) of fractional linear transformations.
My conjecture. Exactly one third of the zeros of $F_p(x)$ have real part equal to $-1/2$.
I tested this with Mathematica for a few of the smallest primes and it seems to hold. Also, each sextet of zeros of the above form seems to be stable under complex conjugation, and seems to contain a complex conjugate pair of numbers with real part $=-1/2$.
Anyway, I am curious about the number of zeros $z=s+it$ of the polynomial $F_p(x)$ on the line $s=-1/2$.
Summary and thoughts.
- Any general method or formula is welcome, but I will be extra grateful if you want to test a method on the polynomial $G_p(x)$ or $F_p(x)$ :-)
- My first idea was to try the following: Given a polynomial $P(x)=prod_i(x-z_i)$ is there a way of getting $R(x):=prod_i(x-z_i-overlinez_i)$? If this can be done, then we get the answer by calculating the multiplicity of $-1$ as a zero of $R(x)$.
- May be a method for calculating the number of real zeros can be used with suitable substitution that maps the real axes to the line $s=-1/2$ (need to check on this)?
- Of course, if you can prove that $F_p(x)$ is irreducible it is better that you post the answer to the linked question. The previous bounty expired, but that can be fixed.
complex-analysis polynomials
asked Jul 16 at 14:18
Jyrki Lahtonen
105k12161355
edited Jul 17 at 19:53
asked Jul 16 at 14:18
Jyrki Lahtonen
105k12161355
asked Jul 16 at 14:18
Jyrki Lahtonen
105k12161355
asked Jul 16 at 14:18
Jyrki Lahtonen
105k12161355
105k12161355
At least a fairly simple way to quickly count the number of roots of $(x+1)^p-x^p-1$ on the “critical lineâ€Â: compute $$2^-frac1p cosleft(frac2 pi np-1right)^1-frac1p$$ for $n=0, ldots, lfloor(p-1)/4rfloor$. Every value in $(0, tfrac12]$ adds four roots and $0$ adds two roots.
– WimC
Jul 17 at 19:29
Sounds interesting @WimC. I'm afraid I don't see it. Can you please elaborate?
– Jyrki Lahtonen
Jul 17 at 19:51
1
See my answer below. It is based on the observation that solutions correspond to points $z$ on the ciritical line such that $z^p$ is also on the critical line. Then a bit of computation leads to the stated result.
– WimC
Jul 17 at 19:55
1
@JyrkiLahtonenMay be a method for calculating the number of real zeros can be usedLet $,x=-1/2 + i z,$, then $,P(x)=A(z) + i B(z),$ with $,A,B in mathbbR[textx],$ has roots with real part $,-1/2,$ iff $,gcd(A,B),$ has real roots. The computations would not be pretty, though.
– dxiv
Jul 17 at 20:55
add a comment |Â
At least a fairly simple way to quickly count the number of roots of $(x+1)^p-x^p-1$ on the “critical lineâ€Â: compute $$2^-frac1p cosleft(frac2 pi np-1right)^1-frac1p$$ for $n=0, ldots, lfloor(p-1)/4rfloor$. Every value in $(0, tfrac12]$ adds four roots and $0$ adds two roots.
– WimC
Jul 17 at 19:29
Sounds interesting @WimC. I'm afraid I don't see it. Can you please elaborate?
– Jyrki Lahtonen
Jul 17 at 19:51
1
See my answer below. It is based on the observation that solutions correspond to points $z$ on the ciritical line such that $z^p$ is also on the critical line. Then a bit of computation leads to the stated result.
– WimC
Jul 17 at 19:55
1
@JyrkiLahtonenMay be a method for calculating the number of real zeros can be usedLet $,x=-1/2 + i z,$, then $,P(x)=A(z) + i B(z),$ with $,A,B in mathbbR[textx],$ has roots with real part $,-1/2,$ iff $,gcd(A,B),$ has real roots. The computations would not be pretty, though.
– dxiv
Jul 17 at 20:55
At least a fairly simple way to quickly count the number of roots of $(x+1)^p-x^p-1$ on the “critical lineâ€Â: compute $$2^-frac1p cosleft(frac2 pi np-1right)^1-frac1p$$ for $n=0, ldots, lfloor(p-1)/4rfloor$. Every value in $(0, tfrac12]$ adds four roots and $0$ adds two roots.
– WimC
Jul 17 at 19:29
At least a fairly simple way to quickly count the number of roots of $(x+1)^p-x^p-1$ on the “critical lineâ€Â: compute $$2^-frac1p cosleft(frac2 pi np-1right)^1-frac1p$$ for $n=0, ldots, lfloor(p-1)/4rfloor$. Every value in $(0, tfrac12]$ adds four roots and $0$ adds two roots.
– WimC
Jul 17 at 19:29
Sounds interesting @WimC. I'm afraid I don't see it. Can you please elaborate?
– Jyrki Lahtonen
Jul 17 at 19:51
Sounds interesting @WimC. I'm afraid I don't see it. Can you please elaborate?
– Jyrki Lahtonen
Jul 17 at 19:51
1
1
See my answer below. It is based on the observation that solutions correspond to points $z$ on the ciritical line such that $z^p$ is also on the critical line. Then a bit of computation leads to the stated result.
– WimC
Jul 17 at 19:55
See my answer below. It is based on the observation that solutions correspond to points $z$ on the ciritical line such that $z^p$ is also on the critical line. Then a bit of computation leads to the stated result.
– WimC
Jul 17 at 19:55
1
1
@JyrkiLahtonen
May be a method for calculating the number of real zeros can be used Let $,x=-1/2 + i z,$, then $,P(x)=A(z) + i B(z),$ with $,A,B in mathbbR[textx],$ has roots with real part $,-1/2,$ iff $,gcd(A,B),$ has real roots. The computations would not be pretty, though.– dxiv
Jul 17 at 20:55
@JyrkiLahtonen
May be a method for calculating the number of real zeros can be used Let $,x=-1/2 + i z,$, then $,P(x)=A(z) + i B(z),$ with $,A,B in mathbbR[textx],$ has roots with real part $,-1/2,$ iff $,gcd(A,B),$ has real roots. The computations would not be pretty, though.– dxiv
Jul 17 at 20:55
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Note The part below that states that $g_c$ can only lose roots on the critical line for increasing $c$ is not rigorous yet, even though it must be true given my other answer...
Another approach. Let $p geq 3$ be odd and consider $g_c(z) = (1-z)^p + z^p - c$ for real $c geq 0$. Note that the zeroes of $g_c$ are conjugate symmetric and symmetric in the critical line $operatornameRe(z) = tfrac12$. Now start at $c=0$ and track what happens with the roots on the critical line when $c$ is increased.
Note that $z$ is a root of $g_c$ on the critical line if and only if $2 operatornameRe(z^p) = c$. For $c=0$ the situation is easy: $g_0$ has all its $p-1$ roots on the critical line and all roots are simple. By symmetry, the number of roots on the critical line can only change if $g_c$ has a double root on the critical line. Either a new double root appears or two roots disappear from such $c$ upward.
So when does $g_c$ have a double root on the critical line? Exactly if $g_c$ and $g_c'$ have a common root there. Now $$p^-1(1-z), g_c'(z )+ g_c(z) = z^p-1 - c$$ so a double root is a positive multiple of a $(p-1)$-th root of unity. This shows that $g_c$ has a double root on the critical line precisely at $$frac12left(1 pm mathrmi,tanleft(frac2 pi mp-1right)right)$$ for some integer $m$ when $$c=frac12^p-1cos^p-1left(frac2 pi mp-1right).$$ For $c in(0,1)$ this happens for $m in [0, (p-1)/6)$ and $g_c$ loses two ($m=0$) or four ($m>0$) roots on the critical line every time $c$ passes such a point. Conclusion $g_1(z)$ has $$p+1 - 4lceilfracp-16rceil$$ roots on the critical line.
answered Jul 18 at 13:35
WimC
23.7k22860
add a comment |Â
up vote
4
down vote
I realized that my comment actually answers your question completely for $(x+1)^p-x^p-1$. After raising to the power $p/(p-1)$ the counting criterion simplifies quite a bit: Compute $$ cosleft(frac2 pi np-1right)$$ for $n=0, ldots, lfloor(p-1)/4rfloor$. Every value in $(0, tfrac12]$ adds four roots and $0$ adds two roots. In other words: every $ (p-1)/6 leq n < (p-1)/4$ counts for four roots and $n=(p-1)/4$ (if $pequiv 1 pmod 4$) counts for two roots.
Here is how I derived this root counting method. To avoid my sign mistakes I make the substitution $x leftarrow -x$ and investigate the roots of $(1-x)^p + x^p - 1$ for odd $p$ on the critical line $operatornameRe(z) = tfrac12$. Note that on the critical line $1-z=overlinez$ so $z$ is a root if and only if $operatornameRe(z^p) = tfrac12$. The strategy is now to investigate the image of the critical line under all branches of $z^1/p$ and see how often this image intersects the critical line.
Let $f_0(z)=z^1/p$ indicate the principle branch. The other branches are then $$f_m(z)=expleft(frac2 pi mathrmi,mpright)f_0(z)$$ for integral $m$. Here $m$ will be restricted to $[0, (p-1)/4]$, i.e. those $m$ for which the primitive $p$-th root lies in the upper right quadrant.
Now parameterise the critical line by $$z = tfrac12(1 + mathrmitan(alpha))$$ for $alpha in (-pi/2, pi/2)$. A straight forward computation shows that
$$N_m(z) = (operatornameRef_m(z))^p = fraccos^p((alpha + 2 pi m)/p)2 cos(alpha).$$ Another straight forward calculation shows that $N_m$ attains its extremal value at $$alpha = frac2 pi mp-1$$ with extremal value $$N_m(z) = tfrac12cos^p-1left(frac2 pi mp-1right).$$
Now the central observation is this: The image of the critical line under $f_m$ looks a bit like a hyperbola. See this picture for $p=5$ which shows all branches:
So if at this extremal angle $0 < N_m(z) leq 2^-p$ then $0 < operatornameRef_m(z) leq tfrac12$ and the image of $f_m$ will intersect the critical line in two places (counting multiplicity) since the branch is located at the right of the extremal value. By conjugate symmetry this $m$ accounts for four zeroes on the critical line.
for $m=(p-1)/4$ the situation is a bit different: the image now has the imaginary axis as one of its asymptotes. (As visible in the picture for $p=5$. The extremal value $N_m(z)$ is $0$ in this case.) This branch clearly intersects the critical line only at a single point, accounting for two zeroes on the critical line in total.
answered Jul 17 at 19:53
WimC
23.7k22860
1
Please add the details :-)
– Jyrki Lahtonen
Jul 18 at 6:30
1
Added my derivation of the root counting method. Things work out so nicely that one cannot help but wonder if there is not a much more natural approach to all this. But until then...
– WimC
Jul 18 at 10:34
it is a lovely idea, indeed, that $z$ from the critical line is a solution if and only if $z^p$ is also on the critical line. I don't doubt your details at all, but I need to set aside a little time to check evrything. Meanwhile, THANK YOU!
– Jyrki Lahtonen
Jul 18 at 10:39
add a comment |Â
up vote
1
down vote
How about Cauchy argument variation
$$int_C fracf'(z)f(z)dz=2pi i N$$
around a rectangular contour $a pm epsilon + bi$ where $a$ is specified and $b$ leaps $pm infty$?
If it doesn't help maybe for Rouche's theorem you can choose some dominant function.
answered Jul 17 at 20:07
mathreadler
13.6k71857
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note The part below that states that $g_c$ can only lose roots on the critical line for increasing $c$ is not rigorous yet, even though it must be true given my other answer...
Another approach. Let $p geq 3$ be odd and consider $g_c(z) = (1-z)^p + z^p - c$ for real $c geq 0$. Note that the zeroes of $g_c$ are conjugate symmetric and symmetric in the critical line $operatornameRe(z) = tfrac12$. Now start at $c=0$ and track what happens with the roots on the critical line when $c$ is increased.
Note that $z$ is a root of $g_c$ on the critical line if and only if $2 operatornameRe(z^p) = c$. For $c=0$ the situation is easy: $g_0$ has all its $p-1$ roots on the critical line and all roots are simple. By symmetry, the number of roots on the critical line can only change if $g_c$ has a double root on the critical line. Either a new double root appears or two roots disappear from such $c$ upward.
So when does $g_c$ have a double root on the critical line? Exactly if $g_c$ and $g_c'$ have a common root there. Now $$p^-1(1-z), g_c'(z )+ g_c(z) = z^p-1 - c$$ so a double root is a positive multiple of a $(p-1)$-th root of unity. This shows that $g_c$ has a double root on the critical line precisely at $$frac12left(1 pm mathrmi,tanleft(frac2 pi mp-1right)right)$$ for some integer $m$ when $$c=frac12^p-1cos^p-1left(frac2 pi mp-1right).$$ For $c in(0,1)$ this happens for $m in [0, (p-1)/6)$ and $g_c$ loses two ($m=0$) or four ($m>0$) roots on the critical line every time $c$ passes such a point. Conclusion $g_1(z)$ has $$p+1 - 4lceilfracp-16rceil$$ roots on the critical line.
answered Jul 18 at 13:35
WimC
23.7k22860
add a comment |Â
up vote
1
down vote
accepted
Note The part below that states that $g_c$ can only lose roots on the critical line for increasing $c$ is not rigorous yet, even though it must be true given my other answer...
Another approach. Let $p geq 3$ be odd and consider $g_c(z) = (1-z)^p + z^p - c$ for real $c geq 0$. Note that the zeroes of $g_c$ are conjugate symmetric and symmetric in the critical line $operatornameRe(z) = tfrac12$. Now start at $c=0$ and track what happens with the roots on the critical line when $c$ is increased.
Note that $z$ is a root of $g_c$ on the critical line if and only if $2 operatornameRe(z^p) = c$. For $c=0$ the situation is easy: $g_0$ has all its $p-1$ roots on the critical line and all roots are simple. By symmetry, the number of roots on the critical line can only change if $g_c$ has a double root on the critical line. Either a new double root appears or two roots disappear from such $c$ upward.
So when does $g_c$ have a double root on the critical line? Exactly if $g_c$ and $g_c'$ have a common root there. Now $$p^-1(1-z), g_c'(z )+ g_c(z) = z^p-1 - c$$ so a double root is a positive multiple of a $(p-1)$-th root of unity. This shows that $g_c$ has a double root on the critical line precisely at $$frac12left(1 pm mathrmi,tanleft(frac2 pi mp-1right)right)$$ for some integer $m$ when $$c=frac12^p-1cos^p-1left(frac2 pi mp-1right).$$ For $c in(0,1)$ this happens for $m in [0, (p-1)/6)$ and $g_c$ loses two ($m=0$) or four ($m>0$) roots on the critical line every time $c$ passes such a point. Conclusion $g_1(z)$ has $$p+1 - 4lceilfracp-16rceil$$ roots on the critical line.
answered Jul 18 at 13:35
WimC
23.7k22860
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note The part below that states that $g_c$ can only lose roots on the critical line for increasing $c$ is not rigorous yet, even though it must be true given my other answer...
Another approach. Let $p geq 3$ be odd and consider $g_c(z) = (1-z)^p + z^p - c$ for real $c geq 0$. Note that the zeroes of $g_c$ are conjugate symmetric and symmetric in the critical line $operatornameRe(z) = tfrac12$. Now start at $c=0$ and track what happens with the roots on the critical line when $c$ is increased.
Note that $z$ is a root of $g_c$ on the critical line if and only if $2 operatornameRe(z^p) = c$. For $c=0$ the situation is easy: $g_0$ has all its $p-1$ roots on the critical line and all roots are simple. By symmetry, the number of roots on the critical line can only change if $g_c$ has a double root on the critical line. Either a new double root appears or two roots disappear from such $c$ upward.
So when does $g_c$ have a double root on the critical line? Exactly if $g_c$ and $g_c'$ have a common root there. Now $$p^-1(1-z), g_c'(z )+ g_c(z) = z^p-1 - c$$ so a double root is a positive multiple of a $(p-1)$-th root of unity. This shows that $g_c$ has a double root on the critical line precisely at $$frac12left(1 pm mathrmi,tanleft(frac2 pi mp-1right)right)$$ for some integer $m$ when $$c=frac12^p-1cos^p-1left(frac2 pi mp-1right).$$ For $c in(0,1)$ this happens for $m in [0, (p-1)/6)$ and $g_c$ loses two ($m=0$) or four ($m>0$) roots on the critical line every time $c$ passes such a point. Conclusion $g_1(z)$ has $$p+1 - 4lceilfracp-16rceil$$ roots on the critical line.
answered Jul 18 at 13:35
WimC
23.7k22860
Note The part below that states that $g_c$ can only lose roots on the critical line for increasing $c$ is not rigorous yet, even though it must be true given my other answer...
Another approach. Let $p geq 3$ be odd and consider $g_c(z) = (1-z)^p + z^p - c$ for real $c geq 0$. Note that the zeroes of $g_c$ are conjugate symmetric and symmetric in the critical line $operatornameRe(z) = tfrac12$. Now start at $c=0$ and track what happens with the roots on the critical line when $c$ is increased.
Note that $z$ is a root of $g_c$ on the critical line if and only if $2 operatornameRe(z^p) = c$. For $c=0$ the situation is easy: $g_0$ has all its $p-1$ roots on the critical line and all roots are simple. By symmetry, the number of roots on the critical line can only change if $g_c$ has a double root on the critical line. Either a new double root appears or two roots disappear from such $c$ upward.
So when does $g_c$ have a double root on the critical line? Exactly if $g_c$ and $g_c'$ have a common root there. Now $$p^-1(1-z), g_c'(z )+ g_c(z) = z^p-1 - c$$ so a double root is a positive multiple of a $(p-1)$-th root of unity. This shows that $g_c$ has a double root on the critical line precisely at $$frac12left(1 pm mathrmi,tanleft(frac2 pi mp-1right)right)$$ for some integer $m$ when $$c=frac12^p-1cos^p-1left(frac2 pi mp-1right).$$ For $c in(0,1)$ this happens for $m in [0, (p-1)/6)$ and $g_c$ loses two ($m=0$) or four ($m>0$) roots on the critical line every time $c$ passes such a point. Conclusion $g_1(z)$ has $$p+1 - 4lceilfracp-16rceil$$ roots on the critical line.
answered Jul 18 at 13:35
WimC
23.7k22860
answered Jul 18 at 13:35
WimC
23.7k22860
answered Jul 18 at 13:35
WimC
23.7k22860
answered Jul 18 at 13:35
WimC
23.7k22860
23.7k22860
add a comment |Â
add a comment |Â
up vote
4
down vote
I realized that my comment actually answers your question completely for $(x+1)^p-x^p-1$. After raising to the power $p/(p-1)$ the counting criterion simplifies quite a bit: Compute $$ cosleft(frac2 pi np-1right)$$ for $n=0, ldots, lfloor(p-1)/4rfloor$. Every value in $(0, tfrac12]$ adds four roots and $0$ adds two roots. In other words: every $ (p-1)/6 leq n < (p-1)/4$ counts for four roots and $n=(p-1)/4$ (if $pequiv 1 pmod 4$) counts for two roots.
Here is how I derived this root counting method. To avoid my sign mistakes I make the substitution $x leftarrow -x$ and investigate the roots of $(1-x)^p + x^p - 1$ for odd $p$ on the critical line $operatornameRe(z) = tfrac12$. Note that on the critical line $1-z=overlinez$ so $z$ is a root if and only if $operatornameRe(z^p) = tfrac12$. The strategy is now to investigate the image of the critical line under all branches of $z^1/p$ and see how often this image intersects the critical line.
Let $f_0(z)=z^1/p$ indicate the principle branch. The other branches are then $$f_m(z)=expleft(frac2 pi mathrmi,mpright)f_0(z)$$ for integral $m$. Here $m$ will be restricted to $[0, (p-1)/4]$, i.e. those $m$ for which the primitive $p$-th root lies in the upper right quadrant.
Now parameterise the critical line by $$z = tfrac12(1 + mathrmitan(alpha))$$ for $alpha in (-pi/2, pi/2)$. A straight forward computation shows that
$$N_m(z) = (operatornameRef_m(z))^p = fraccos^p((alpha + 2 pi m)/p)2 cos(alpha).$$ Another straight forward calculation shows that $N_m$ attains its extremal value at $$alpha = frac2 pi mp-1$$ with extremal value $$N_m(z) = tfrac12cos^p-1left(frac2 pi mp-1right).$$
Now the central observation is this: The image of the critical line under $f_m$ looks a bit like a hyperbola. See this picture for $p=5$ which shows all branches:
So if at this extremal angle $0 < N_m(z) leq 2^-p$ then $0 < operatornameRef_m(z) leq tfrac12$ and the image of $f_m$ will intersect the critical line in two places (counting multiplicity) since the branch is located at the right of the extremal value. By conjugate symmetry this $m$ accounts for four zeroes on the critical line.
for $m=(p-1)/4$ the situation is a bit different: the image now has the imaginary axis as one of its asymptotes. (As visible in the picture for $p=5$. The extremal value $N_m(z)$ is $0$ in this case.) This branch clearly intersects the critical line only at a single point, accounting for two zeroes on the critical line in total.
answered Jul 17 at 19:53
WimC
23.7k22860
1
Please add the details :-)
– Jyrki Lahtonen
Jul 18 at 6:30
1
Added my derivation of the root counting method. Things work out so nicely that one cannot help but wonder if there is not a much more natural approach to all this. But until then...
– WimC
Jul 18 at 10:34
it is a lovely idea, indeed, that $z$ from the critical line is a solution if and only if $z^p$ is also on the critical line. I don't doubt your details at all, but I need to set aside a little time to check evrything. Meanwhile, THANK YOU!
– Jyrki Lahtonen
Jul 18 at 10:39
add a comment |Â
up vote
4
down vote
I realized that my comment actually answers your question completely for $(x+1)^p-x^p-1$. After raising to the power $p/(p-1)$ the counting criterion simplifies quite a bit: Compute $$ cosleft(frac2 pi np-1right)$$ for $n=0, ldots, lfloor(p-1)/4rfloor$. Every value in $(0, tfrac12]$ adds four roots and $0$ adds two roots. In other words: every $ (p-1)/6 leq n < (p-1)/4$ counts for four roots and $n=(p-1)/4$ (if $pequiv 1 pmod 4$) counts for two roots.
Here is how I derived this root counting method. To avoid my sign mistakes I make the substitution $x leftarrow -x$ and investigate the roots of $(1-x)^p + x^p - 1$ for odd $p$ on the critical line $operatornameRe(z) = tfrac12$. Note that on the critical line $1-z=overlinez$ so $z$ is a root if and only if $operatornameRe(z^p) = tfrac12$. The strategy is now to investigate the image of the critical line under all branches of $z^1/p$ and see how often this image intersects the critical line.
Let $f_0(z)=z^1/p$ indicate the principle branch. The other branches are then $$f_m(z)=expleft(frac2 pi mathrmi,mpright)f_0(z)$$ for integral $m$. Here $m$ will be restricted to $[0, (p-1)/4]$, i.e. those $m$ for which the primitive $p$-th root lies in the upper right quadrant.
Now parameterise the critical line by $$z = tfrac12(1 + mathrmitan(alpha))$$ for $alpha in (-pi/2, pi/2)$. A straight forward computation shows that
$$N_m(z) = (operatornameRef_m(z))^p = fraccos^p((alpha + 2 pi m)/p)2 cos(alpha).$$ Another straight forward calculation shows that $N_m$ attains its extremal value at $$alpha = frac2 pi mp-1$$ with extremal value $$N_m(z) = tfrac12cos^p-1left(frac2 pi mp-1right).$$
Now the central observation is this: The image of the critical line under $f_m$ looks a bit like a hyperbola. See this picture for $p=5$ which shows all branches:
So if at this extremal angle $0 < N_m(z) leq 2^-p$ then $0 < operatornameRef_m(z) leq tfrac12$ and the image of $f_m$ will intersect the critical line in two places (counting multiplicity) since the branch is located at the right of the extremal value. By conjugate symmetry this $m$ accounts for four zeroes on the critical line.
for $m=(p-1)/4$ the situation is a bit different: the image now has the imaginary axis as one of its asymptotes. (As visible in the picture for $p=5$. The extremal value $N_m(z)$ is $0$ in this case.) This branch clearly intersects the critical line only at a single point, accounting for two zeroes on the critical line in total.
answered Jul 17 at 19:53
WimC
23.7k22860
1
Please add the details :-)
– Jyrki Lahtonen
Jul 18 at 6:30
1
Added my derivation of the root counting method. Things work out so nicely that one cannot help but wonder if there is not a much more natural approach to all this. But until then...
– WimC
Jul 18 at 10:34
it is a lovely idea, indeed, that $z$ from the critical line is a solution if and only if $z^p$ is also on the critical line. I don't doubt your details at all, but I need to set aside a little time to check evrything. Meanwhile, THANK YOU!
– Jyrki Lahtonen
Jul 18 at 10:39
add a comment |Â
up vote
4
down vote
up vote
4
down vote
I realized that my comment actually answers your question completely for $(x+1)^p-x^p-1$. After raising to the power $p/(p-1)$ the counting criterion simplifies quite a bit: Compute $$ cosleft(frac2 pi np-1right)$$ for $n=0, ldots, lfloor(p-1)/4rfloor$. Every value in $(0, tfrac12]$ adds four roots and $0$ adds two roots. In other words: every $ (p-1)/6 leq n < (p-1)/4$ counts for four roots and $n=(p-1)/4$ (if $pequiv 1 pmod 4$) counts for two roots.
Here is how I derived this root counting method. To avoid my sign mistakes I make the substitution $x leftarrow -x$ and investigate the roots of $(1-x)^p + x^p - 1$ for odd $p$ on the critical line $operatornameRe(z) = tfrac12$. Note that on the critical line $1-z=overlinez$ so $z$ is a root if and only if $operatornameRe(z^p) = tfrac12$. The strategy is now to investigate the image of the critical line under all branches of $z^1/p$ and see how often this image intersects the critical line.
Let $f_0(z)=z^1/p$ indicate the principle branch. The other branches are then $$f_m(z)=expleft(frac2 pi mathrmi,mpright)f_0(z)$$ for integral $m$. Here $m$ will be restricted to $[0, (p-1)/4]$, i.e. those $m$ for which the primitive $p$-th root lies in the upper right quadrant.
Now parameterise the critical line by $$z = tfrac12(1 + mathrmitan(alpha))$$ for $alpha in (-pi/2, pi/2)$. A straight forward computation shows that
$$N_m(z) = (operatornameRef_m(z))^p = fraccos^p((alpha + 2 pi m)/p)2 cos(alpha).$$ Another straight forward calculation shows that $N_m$ attains its extremal value at $$alpha = frac2 pi mp-1$$ with extremal value $$N_m(z) = tfrac12cos^p-1left(frac2 pi mp-1right).$$
Now the central observation is this: The image of the critical line under $f_m$ looks a bit like a hyperbola. See this picture for $p=5$ which shows all branches:
So if at this extremal angle $0 < N_m(z) leq 2^-p$ then $0 < operatornameRef_m(z) leq tfrac12$ and the image of $f_m$ will intersect the critical line in two places (counting multiplicity) since the branch is located at the right of the extremal value. By conjugate symmetry this $m$ accounts for four zeroes on the critical line.
for $m=(p-1)/4$ the situation is a bit different: the image now has the imaginary axis as one of its asymptotes. (As visible in the picture for $p=5$. The extremal value $N_m(z)$ is $0$ in this case.) This branch clearly intersects the critical line only at a single point, accounting for two zeroes on the critical line in total.
answered Jul 17 at 19:53
WimC
23.7k22860
I realized that my comment actually answers your question completely for $(x+1)^p-x^p-1$. After raising to the power $p/(p-1)$ the counting criterion simplifies quite a bit: Compute $$ cosleft(frac2 pi np-1right)$$ for $n=0, ldots, lfloor(p-1)/4rfloor$. Every value in $(0, tfrac12]$ adds four roots and $0$ adds two roots. In other words: every $ (p-1)/6 leq n < (p-1)/4$ counts for four roots and $n=(p-1)/4$ (if $pequiv 1 pmod 4$) counts for two roots.
Here is how I derived this root counting method. To avoid my sign mistakes I make the substitution $x leftarrow -x$ and investigate the roots of $(1-x)^p + x^p - 1$ for odd $p$ on the critical line $operatornameRe(z) = tfrac12$. Note that on the critical line $1-z=overlinez$ so $z$ is a root if and only if $operatornameRe(z^p) = tfrac12$. The strategy is now to investigate the image of the critical line under all branches of $z^1/p$ and see how often this image intersects the critical line.
Let $f_0(z)=z^1/p$ indicate the principle branch. The other branches are then $$f_m(z)=expleft(frac2 pi mathrmi,mpright)f_0(z)$$ for integral $m$. Here $m$ will be restricted to $[0, (p-1)/4]$, i.e. those $m$ for which the primitive $p$-th root lies in the upper right quadrant.
Now parameterise the critical line by $$z = tfrac12(1 + mathrmitan(alpha))$$ for $alpha in (-pi/2, pi/2)$. A straight forward computation shows that
$$N_m(z) = (operatornameRef_m(z))^p = fraccos^p((alpha + 2 pi m)/p)2 cos(alpha).$$ Another straight forward calculation shows that $N_m$ attains its extremal value at $$alpha = frac2 pi mp-1$$ with extremal value $$N_m(z) = tfrac12cos^p-1left(frac2 pi mp-1right).$$
Now the central observation is this: The image of the critical line under $f_m$ looks a bit like a hyperbola. See this picture for $p=5$ which shows all branches:
So if at this extremal angle $0 < N_m(z) leq 2^-p$ then $0 < operatornameRef_m(z) leq tfrac12$ and the image of $f_m$ will intersect the critical line in two places (counting multiplicity) since the branch is located at the right of the extremal value. By conjugate symmetry this $m$ accounts for four zeroes on the critical line.
for $m=(p-1)/4$ the situation is a bit different: the image now has the imaginary axis as one of its asymptotes. (As visible in the picture for $p=5$. The extremal value $N_m(z)$ is $0$ in this case.) This branch clearly intersects the critical line only at a single point, accounting for two zeroes on the critical line in total.
answered Jul 17 at 19:53
WimC
23.7k22860
edited Jul 18 at 10:44
answered Jul 17 at 19:53
WimC
23.7k22860
answered Jul 17 at 19:53
WimC
23.7k22860
answered Jul 17 at 19:53
WimC
23.7k22860
23.7k22860
1
Please add the details :-)
– Jyrki Lahtonen
Jul 18 at 6:30
1
Added my derivation of the root counting method. Things work out so nicely that one cannot help but wonder if there is not a much more natural approach to all this. But until then...
– WimC
Jul 18 at 10:34
it is a lovely idea, indeed, that $z$ from the critical line is a solution if and only if $z^p$ is also on the critical line. I don't doubt your details at all, but I need to set aside a little time to check evrything. Meanwhile, THANK YOU!
– Jyrki Lahtonen
Jul 18 at 10:39
add a comment |Â
1
Please add the details :-)
– Jyrki Lahtonen
Jul 18 at 6:30
1
Added my derivation of the root counting method. Things work out so nicely that one cannot help but wonder if there is not a much more natural approach to all this. But until then...
– WimC
Jul 18 at 10:34
it is a lovely idea, indeed, that $z$ from the critical line is a solution if and only if $z^p$ is also on the critical line. I don't doubt your details at all, but I need to set aside a little time to check evrything. Meanwhile, THANK YOU!
– Jyrki Lahtonen
Jul 18 at 10:39
1
1
Please add the details :-)
– Jyrki Lahtonen
Jul 18 at 6:30
Please add the details :-)
– Jyrki Lahtonen
Jul 18 at 6:30
1
1
Added my derivation of the root counting method. Things work out so nicely that one cannot help but wonder if there is not a much more natural approach to all this. But until then...
– WimC
Jul 18 at 10:34
Added my derivation of the root counting method. Things work out so nicely that one cannot help but wonder if there is not a much more natural approach to all this. But until then...
– WimC
Jul 18 at 10:34
it is a lovely idea, indeed, that $z$ from the critical line is a solution if and only if $z^p$ is also on the critical line. I don't doubt your details at all, but I need to set aside a little time to check evrything. Meanwhile, THANK YOU!
– Jyrki Lahtonen
Jul 18 at 10:39
it is a lovely idea, indeed, that $z$ from the critical line is a solution if and only if $z^p$ is also on the critical line. I don't doubt your details at all, but I need to set aside a little time to check evrything. Meanwhile, THANK YOU!
– Jyrki Lahtonen
Jul 18 at 10:39
add a comment |Â
up vote
1
down vote
How about Cauchy argument variation
$$int_C fracf'(z)f(z)dz=2pi i N$$
around a rectangular contour $a pm epsilon + bi$ where $a$ is specified and $b$ leaps $pm infty$?
If it doesn't help maybe for Rouche's theorem you can choose some dominant function.
answered Jul 17 at 20:07
mathreadler
13.6k71857
add a comment |Â
up vote
1
down vote
How about Cauchy argument variation
$$int_C fracf'(z)f(z)dz=2pi i N$$
around a rectangular contour $a pm epsilon + bi$ where $a$ is specified and $b$ leaps $pm infty$?
If it doesn't help maybe for Rouche's theorem you can choose some dominant function.
answered Jul 17 at 20:07
mathreadler
13.6k71857
add a comment |Â
up vote
1
down vote
up vote
1
down vote
How about Cauchy argument variation
$$int_C fracf'(z)f(z)dz=2pi i N$$
around a rectangular contour $a pm epsilon + bi$ where $a$ is specified and $b$ leaps $pm infty$?
If it doesn't help maybe for Rouche's theorem you can choose some dominant function.
answered Jul 17 at 20:07
mathreadler
13.6k71857
How about Cauchy argument variation
$$int_C fracf'(z)f(z)dz=2pi i N$$
around a rectangular contour $a pm epsilon + bi$ where $a$ is specified and $b$ leaps $pm infty$?
If it doesn't help maybe for Rouche's theorem you can choose some dominant function.
answered Jul 17 at 20:07
mathreadler
13.6k71857
answered Jul 17 at 20:07
mathreadler
13.6k71857
answered Jul 17 at 20:07
mathreadler
13.6k71857
answered Jul 17 at 20:07
mathreadler
13.6k71857
13.6k71857
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853450%2fmethods-for-calculating-the-number-of-zeros-of-a-polynomial-with-a-specified-rea%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
At least a fairly simple way to quickly count the number of roots of $(x+1)^p-x^p-1$ on the “critical lineâ€Â: compute $$2^-frac1p cosleft(frac2 pi np-1right)^1-frac1p$$ for $n=0, ldots, lfloor(p-1)/4rfloor$. Every value in $(0, tfrac12]$ adds four roots and $0$ adds two roots.
– WimC
Jul 17 at 19:29
Sounds interesting @WimC. I'm afraid I don't see it. Can you please elaborate?
– Jyrki Lahtonen
Jul 17 at 19:51
1
See my answer below. It is based on the observation that solutions correspond to points $z$ on the ciritical line such that $z^p$ is also on the critical line. Then a bit of computation leads to the stated result.
– WimC
Jul 17 at 19:55
1
@JyrkiLahtonen
May be a method for calculating the number of real zeros can be usedLet $,x=-1/2 + i z,$, then $,P(x)=A(z) + i B(z),$ with $,A,B in mathbbR[textx],$ has roots with real part $,-1/2,$ iff $,gcd(A,B),$ has real roots. The computations would not be pretty, though.– dxiv
Jul 17 at 20:55