Sum of fifth roots of roots of cubic.
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$a,b,c$ are the (real) roots of $x^3-16x^2-57x+1=0$. Prove that
$sqrt[5]{a} + sqrt[5]{b} +sqrt[5]{c} = 1 $
................................................................................
edit : my answer to this question on another forum
Let $p=sqrt[5]{a},q=sqrt[5]{b}, r=sqrt[5]{c}, u=p+q+r, v=qr+rp+pq, w=pqr$
From given cubic p⁵+q⁵+r⁵=16, q⁵r⁵+r⁵p⁵+p⁵q⁵=-57, p⁵q⁵r⁵=-1 (so w=-1)
Thus results at end gives the following simultaneous equations for u,v
p⁵+q⁵+r⁵ = 16 = u⁵-5u³v+5uv²-5u²+5v
q⁵r⁵+r⁵p⁵+p⁵q⁵ = -57 = v⁵+5uv³+5u²v+5v²+5u
I solved these via Mathematica to give u=1, v=-2 (only real solution) so your answer is 1
====================
p,q,r are roots of z³-uz²+vz-w=0
Multiply by zⁿ and let s(n)=pⁿ+qⁿ+rⁿ
Set z=p,q,r in turn and sum for the recurrence s(n+3) = us(n+2)-vs(n+1)+ws(n)
Noting that s(0)=3, s(1)=u, s(2)=u²−2v generate s(3), s(4), s(5) as follows …
n=0 : s(3) = us(2)-vs(1)+ws(0) = u³-3uv+3w
n=1 : s(4) = us(3)-vs(2)+ws(1) = u⁴-4u²v+2v²+4uw
n=2 : s(5) = us(4)-vs(3)+ws(2) = u⁵-5u³v+5uv²+5u²w-5vw
p⁵+q⁵+r⁵ = u⁵-5u³v+5uv²+5u²w-5vw
Replace p by qr, q by rp, r by pq and get
q⁵r⁵+r⁵p⁵+p⁵q⁵ = v⁵-5uv³w+5u²vw²+5v²w²-5uw³
====================
algebra-precalculus
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up vote
10
down vote
favorite
$a,b,c$ are the (real) roots of $x^3-16x^2-57x+1=0$. Prove that
$sqrt[5]{a} + sqrt[5]{b} +sqrt[5]{c} = 1 $
................................................................................
edit : my answer to this question on another forum
Let $p=sqrt[5]{a},q=sqrt[5]{b}, r=sqrt[5]{c}, u=p+q+r, v=qr+rp+pq, w=pqr$
From given cubic p⁵+q⁵+r⁵=16, q⁵r⁵+r⁵p⁵+p⁵q⁵=-57, p⁵q⁵r⁵=-1 (so w=-1)
Thus results at end gives the following simultaneous equations for u,v
p⁵+q⁵+r⁵ = 16 = u⁵-5u³v+5uv²-5u²+5v
q⁵r⁵+r⁵p⁵+p⁵q⁵ = -57 = v⁵+5uv³+5u²v+5v²+5u
I solved these via Mathematica to give u=1, v=-2 (only real solution) so your answer is 1
====================
p,q,r are roots of z³-uz²+vz-w=0
Multiply by zⁿ and let s(n)=pⁿ+qⁿ+rⁿ
Set z=p,q,r in turn and sum for the recurrence s(n+3) = us(n+2)-vs(n+1)+ws(n)
Noting that s(0)=3, s(1)=u, s(2)=u²−2v generate s(3), s(4), s(5) as follows …
n=0 : s(3) = us(2)-vs(1)+ws(0) = u³-3uv+3w
n=1 : s(4) = us(3)-vs(2)+ws(1) = u⁴-4u²v+2v²+4uw
n=2 : s(5) = us(4)-vs(3)+ws(2) = u⁵-5u³v+5uv²+5u²w-5vw
p⁵+q⁵+r⁵ = u⁵-5u³v+5uv²+5u²w-5vw
Replace p by qr, q by rp, r by pq and get
q⁵r⁵+r⁵p⁵+p⁵q⁵ = v⁵-5uv³w+5u²vw²+5v²w²-5uw³
====================
algebra-precalculus
@molarmass OP made an edit.
– Jack Bauer
Aug 16 at 8:47
2
$x^{15}-16x^{10}-57x^5+1$ factors as Hint: $(x^3-x^2-2x+1)(x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1)$ and the sum of the roots of the first factor is $1$.
– Yves Daoust
Aug 16 at 9:05
Why is this closed ? The OP did show some non-trivial effort, even though presented via an uncommon medium.
– Yves Daoust
Aug 16 at 9:08
@YvesDaoust In that link I only see efforts that others have done to answer his question.
– Jaap Scherphuis
Aug 16 at 9:11
@JaapScherphuis: the first answer is by the OP himself. Check the signatures.
– Yves Daoust
Aug 16 at 9:12
|
show 4 more comments
up vote
10
down vote
favorite
up vote
10
down vote
favorite
$a,b,c$ are the (real) roots of $x^3-16x^2-57x+1=0$. Prove that
$sqrt[5]{a} + sqrt[5]{b} +sqrt[5]{c} = 1 $
................................................................................
edit : my answer to this question on another forum
Let $p=sqrt[5]{a},q=sqrt[5]{b}, r=sqrt[5]{c}, u=p+q+r, v=qr+rp+pq, w=pqr$
From given cubic p⁵+q⁵+r⁵=16, q⁵r⁵+r⁵p⁵+p⁵q⁵=-57, p⁵q⁵r⁵=-1 (so w=-1)
Thus results at end gives the following simultaneous equations for u,v
p⁵+q⁵+r⁵ = 16 = u⁵-5u³v+5uv²-5u²+5v
q⁵r⁵+r⁵p⁵+p⁵q⁵ = -57 = v⁵+5uv³+5u²v+5v²+5u
I solved these via Mathematica to give u=1, v=-2 (only real solution) so your answer is 1
====================
p,q,r are roots of z³-uz²+vz-w=0
Multiply by zⁿ and let s(n)=pⁿ+qⁿ+rⁿ
Set z=p,q,r in turn and sum for the recurrence s(n+3) = us(n+2)-vs(n+1)+ws(n)
Noting that s(0)=3, s(1)=u, s(2)=u²−2v generate s(3), s(4), s(5) as follows …
n=0 : s(3) = us(2)-vs(1)+ws(0) = u³-3uv+3w
n=1 : s(4) = us(3)-vs(2)+ws(1) = u⁴-4u²v+2v²+4uw
n=2 : s(5) = us(4)-vs(3)+ws(2) = u⁵-5u³v+5uv²+5u²w-5vw
p⁵+q⁵+r⁵ = u⁵-5u³v+5uv²+5u²w-5vw
Replace p by qr, q by rp, r by pq and get
q⁵r⁵+r⁵p⁵+p⁵q⁵ = v⁵-5uv³w+5u²vw²+5v²w²-5uw³
====================
algebra-precalculus
$a,b,c$ are the (real) roots of $x^3-16x^2-57x+1=0$. Prove that
$sqrt[5]{a} + sqrt[5]{b} +sqrt[5]{c} = 1 $
................................................................................
edit : my answer to this question on another forum
Let $p=sqrt[5]{a},q=sqrt[5]{b}, r=sqrt[5]{c}, u=p+q+r, v=qr+rp+pq, w=pqr$
From given cubic p⁵+q⁵+r⁵=16, q⁵r⁵+r⁵p⁵+p⁵q⁵=-57, p⁵q⁵r⁵=-1 (so w=-1)
Thus results at end gives the following simultaneous equations for u,v
p⁵+q⁵+r⁵ = 16 = u⁵-5u³v+5uv²-5u²+5v
q⁵r⁵+r⁵p⁵+p⁵q⁵ = -57 = v⁵+5uv³+5u²v+5v²+5u
I solved these via Mathematica to give u=1, v=-2 (only real solution) so your answer is 1
====================
p,q,r are roots of z³-uz²+vz-w=0
Multiply by zⁿ and let s(n)=pⁿ+qⁿ+rⁿ
Set z=p,q,r in turn and sum for the recurrence s(n+3) = us(n+2)-vs(n+1)+ws(n)
Noting that s(0)=3, s(1)=u, s(2)=u²−2v generate s(3), s(4), s(5) as follows …
n=0 : s(3) = us(2)-vs(1)+ws(0) = u³-3uv+3w
n=1 : s(4) = us(3)-vs(2)+ws(1) = u⁴-4u²v+2v²+4uw
n=2 : s(5) = us(4)-vs(3)+ws(2) = u⁵-5u³v+5uv²+5u²w-5vw
p⁵+q⁵+r⁵ = u⁵-5u³v+5uv²+5u²w-5vw
Replace p by qr, q by rp, r by pq and get
q⁵r⁵+r⁵p⁵+p⁵q⁵ = v⁵-5uv³w+5u²vw²+5v²w²-5uw³
====================
algebra-precalculus
algebra-precalculus
edited Aug 16 at 9:50
asked Aug 16 at 6:02
URCHIN
645
645
@molarmass OP made an edit.
– Jack Bauer
Aug 16 at 8:47
2
$x^{15}-16x^{10}-57x^5+1$ factors as Hint: $(x^3-x^2-2x+1)(x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1)$ and the sum of the roots of the first factor is $1$.
– Yves Daoust
Aug 16 at 9:05
Why is this closed ? The OP did show some non-trivial effort, even though presented via an uncommon medium.
– Yves Daoust
Aug 16 at 9:08
@YvesDaoust In that link I only see efforts that others have done to answer his question.
– Jaap Scherphuis
Aug 16 at 9:11
@JaapScherphuis: the first answer is by the OP himself. Check the signatures.
– Yves Daoust
Aug 16 at 9:12
|
show 4 more comments
@molarmass OP made an edit.
– Jack Bauer
Aug 16 at 8:47
2
$x^{15}-16x^{10}-57x^5+1$ factors as Hint: $(x^3-x^2-2x+1)(x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1)$ and the sum of the roots of the first factor is $1$.
– Yves Daoust
Aug 16 at 9:05
Why is this closed ? The OP did show some non-trivial effort, even though presented via an uncommon medium.
– Yves Daoust
Aug 16 at 9:08
@YvesDaoust In that link I only see efforts that others have done to answer his question.
– Jaap Scherphuis
Aug 16 at 9:11
@JaapScherphuis: the first answer is by the OP himself. Check the signatures.
– Yves Daoust
Aug 16 at 9:12
@molarmass OP made an edit.
– Jack Bauer
Aug 16 at 8:47
@molarmass OP made an edit.
– Jack Bauer
Aug 16 at 8:47
2
2
$x^{15}-16x^{10}-57x^5+1$ factors as Hint: $(x^3-x^2-2x+1)(x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1)$ and the sum of the roots of the first factor is $1$.
– Yves Daoust
Aug 16 at 9:05
$x^{15}-16x^{10}-57x^5+1$ factors as Hint: $(x^3-x^2-2x+1)(x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1)$ and the sum of the roots of the first factor is $1$.
– Yves Daoust
Aug 16 at 9:05
Why is this closed ? The OP did show some non-trivial effort, even though presented via an uncommon medium.
– Yves Daoust
Aug 16 at 9:08
Why is this closed ? The OP did show some non-trivial effort, even though presented via an uncommon medium.
– Yves Daoust
Aug 16 at 9:08
@YvesDaoust In that link I only see efforts that others have done to answer his question.
– Jaap Scherphuis
Aug 16 at 9:11
@YvesDaoust In that link I only see efforts that others have done to answer his question.
– Jaap Scherphuis
Aug 16 at 9:11
@JaapScherphuis: the first answer is by the OP himself. Check the signatures.
– Yves Daoust
Aug 16 at 9:12
@JaapScherphuis: the first answer is by the OP himself. Check the signatures.
– Yves Daoust
Aug 16 at 9:12
|
show 4 more comments
2 Answers
2
active
oldest
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up vote
3
down vote
Let $A,B,C$ be the roots of the cubic polynomial
$$
g(x) = x^3 - x^2 - 2x + 1 .
$$
So $A,B,C$ satisfy (Vieta):
$$
begin{aligned}
e_1 &=A + B + C &&=+1 ,\
e_2 &= AB + BC + CA &&=-2 ,\
e_3 &= A B C &&= -1 .
end{aligned}
$$
(Computer support motivating the above is postponed.)
These values are numerically:
sage: R.<x> = PolynomialRing(QQ)
sage: g = x^3 - x^2 - 2*x + 1
sage: g(x).roots(ring=QQbar, multiplicities=False)
[-1.246979603717467?, 0.4450418679126288?, 1.801937735804839?]
so they are real.
We compute the polynomial with roots $a'=A^5$, $b'=B^5$, $c'=C^3$.
(We "hope" that $a=a'$, $b=b'$, $c=c'$.)
For this we have to start computations involving symmetrical polynomials,
trying to get
$e_1(a',b',c')$,
$e_2(a',b',c')$,
$e_3(a',b',c')$, in terms of $e_1(A,B,C)$, $e_2(A,B,C)$, $e_3(A,B,C)$.
Here, $e_1$, $e_2$, $e_3$ are the first three symmetric polynomials.
(We have formally $e_4=e_5=dots=0$.)
One result is immediate:
$$
e_3(a',b',c')
=a'b'c'
=A^5B^5C^5=(ABC)^5=e_3(A,B,C)^5=-1 .
$$
We need now the Newton polynomial $p_5$ in Newton's identities of fifth degree...
$$
begin{aligned}
e_1(a',b',c')
&=
a'+b'+c'
\
&=A^5+B^5+C^5
\
&=p_5(A,B,C) ,\
&qquadtext{so we compute successively}\
p_1(A,B,C) &= e_1(A,B,C)=1 ,\
p_2(A,B,C) &= (e_1p_1-2e_2)(A,B,C)=1cdot 1-2cdot(-2)=5 ,\
p_3(A,B,C) &= (e_1p_2-e_2p_1+3e_2)(A,B,C)=1cdot 5-(-2)cdot1+3cdot(-1)=4 ,\
p_4(A,B,C) &= (e_1p_3-e_2p_2+e_3p_1)(A,B,C)=1cdot 4-(-2)cdot5+(-1)cdot1=13 ,\
e_1(a',b',c') &=p_5(A,B,C)
\&=(e_1p_4-e_2p_3+e_3p_2)(A,B,C)=1cdot13-(-2)cdot4+(-1)cdot5=16 .
end{aligned}
$$
We need now finally $a'b'+b'c'+c'a'$.
For this, we repeat the same procedure as above, but not for $A,B,C$, but for $s,t,u$, which are respectively $AB$, $BC$, $CA$, with
$$
begin{aligned}
e_1(s,t,u) &=s+t+u=AB + BC + CA &&=-2 ,\
e_2(s,t,u) &=st+tu+us= ABC(A+B+C) &&=-1 ,\
e_3(s,t,u) &=stu= A^2 B^2 C^2 &&= +1 .
end{aligned}
$$
So we have:
$$
begin{aligned}
p_1(s,t,u)&=e_1(s,t,u)=-2 ,\
p_2(s,t,u)&=(e_1p_1-2e_2)(s,t,u)=(-2)cdot (-2)-2cdot(-1)=6 ,\
p_3(s,t,u)&=(e_1p_2-e_2p_1+3e_2)(s,t,u)=(-2)cdot6-(-1)cdot(-2)+3cdot1=-11 ,\
p_4(s,t,u)&=(e_1p_3-e_2p_2+e_3p_1)(s,t,u)
=(-2)cdot(-11)-(-1)cdot6+1cdot(-2)=26 ,\
e_2(a',b',c')&=p_5(s,t,u)
\
&=(e_1p_4-e_2p_3+e_3p_2)(s,t,u)
=(-2)cdot26-(-1)cdot(-11)+1cdot6=-57 .
end{aligned}
$$
So $a',b',c'$ are the roots of the given polynomial
$$
x^3-E_1x^2+E_2x-E_3
=
x^3 - 16x^2 -57 x +1 .
$$
So $a',b',c'$ are (up to reordering) the "given" values $a,b,c$.
We conclude:
$$
a^{1/5} + b^{1/5} + c^{1/5}
=A+B+C=1 .
$$
$square$
Numerical support for the computations done so far, and the motivation for the abrupt start with the roots $A,B,C$ of the polynomial $x^3 - x^2 - 2x + 1$:
Sage code:
R.<x> = PolynomialRing(QQ)
f = x^3 - 16*x^2 - 57*x + 1
a, b, c = f.roots( ring=QQbar, multiplicities=False )
a, b, c
a^(1/5), b^(1/5), c^(1/5)
This gives:
(-3.015065490237851?, 0.01745839634379104?, 18.99760709389406?)
(1.008827691046369? + 0.7329562209746396?*I,
0.4450418679126288?,
1.801937735804839?)
OK, the computer needs human assitance to get the right fifth root of $a$ in $Bbb R$.
A, B, C = -(-a)^(1/5), b^(1/5), c^(1/5)
giving
sage: A, B, C
(-1.246979603717467?, 0.4450418679126288?, 1.801937735804839?)
this is good, three real numbers. Let us compute the elementary symmetric functions,
and the first Newton polynomials for them:
print "A + B + C = %s" % (A+B+C)
print "AB + BC + CA = %s" % (A*B + B*C + C*A)
print "A B C = %s" % (A*B*C)
for k in [1..5]:
print "A^%s + B^%s + C^%s = %s" % (k, k, k, A^k+B^k+C^k)
This gives so far:
A + B + C = 1.000000000000000?
AB + BC + CA = -2.000000000000000?
A B C = -1.000000000000000?
A^1 + B^1 + C^1 = 1.000000000000000?
A^2 + B^2 + C^2 = 5.000000000000000?
A^3 + B^3 + C^3 = 4.000000000000000?
A^4 + B^4 + C^4 = 13.00000000000000?
A^5 + B^5 + C^5 = 16.00000000000000?
Same for the values $s,t,u$:
s, t, u = A*B, B*C, C*A
print "s + t + u = %s" % (s+t+u)
print "st + tu + us = %s" % (s*t + t*u + u*s)
print "s t u = %s" % (s*t*u)
for k in [1..5]:
print "s^%s + t^%s + u^%s = %s" % (k, k, k, s^k+t^k+u^k)
This gives:
s + t + u = -2.000000000000000?
st + tu + us = -1.000000000000000?
s t u = 1.000000000000000?
s^1 + t^1 + u^1 = -2.000000000000000?
s^2 + t^2 + u^2 = 6.000000000000000?
s^3 + t^3 + u^3 = -11.00000000000000?
s^4 + t^4 + u^4 = 26.00000000000000?
s^5 + t^5 + u^5 = -57.00000000000000?
The auxiliary cubic polynomial is readily obtained as a factor of $p(x^5)$ where $p$ is the original cubic.
– Yves Daoust
Aug 16 at 9:22
Yes, as in the comments to the OP. My problem was to see its roots and the fact that they are real. Else one has to take the suitable branch of the fifth root. to match this one "beautiful factor". People wanted computations explicitly, i did it explicitly. On a piece of paper with own notations the calculus is done in some five minutes, typing all the stuff...
– dan_fulea
Aug 16 at 9:28
add a comment |
up vote
3
down vote
The given polynomial $p(x)$ has three real roots so $p(x^5)$ also has three real roots as all others are complex fifth roots of reals.
So we need to factor $p(x^5)=q(x)r(x)$ where $q$ is cubic with real roots and $r$ is an unimportant polynomial of degree twelve with complex ones.
With a simple numerical solver or with Cardano's formulas, we estimate the coefficients of $q$ using Vieta's formulas on the fifth roots of $p$, giving
$$q(x)approx (x-1.80194) (x-0.445042) (x+1.24698)\approx x^3-1.000002 x^2-2.000002633x+1.000001871.$$
So we can hypothetize $q(x)=x^3-x^2-2x+1$ and long division confirms this hypothesis exactly *.
Then the sum of the real roots of $q$ is $1$.
*$$dfrac{x^{15}-16x^{10}-57x^5+1}{x^3-x^2-2x+1}\=x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1.$$
Yes, a useful trick. Wolfram will also factor p(x⁵) directly, which is something |I should have tried.
– URCHIN
Aug 16 at 14:57
@URCHIN: the next challenge is to find a way to solve fully by hand.
– Yves Daoust
Aug 16 at 21:27
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Let $A,B,C$ be the roots of the cubic polynomial
$$
g(x) = x^3 - x^2 - 2x + 1 .
$$
So $A,B,C$ satisfy (Vieta):
$$
begin{aligned}
e_1 &=A + B + C &&=+1 ,\
e_2 &= AB + BC + CA &&=-2 ,\
e_3 &= A B C &&= -1 .
end{aligned}
$$
(Computer support motivating the above is postponed.)
These values are numerically:
sage: R.<x> = PolynomialRing(QQ)
sage: g = x^3 - x^2 - 2*x + 1
sage: g(x).roots(ring=QQbar, multiplicities=False)
[-1.246979603717467?, 0.4450418679126288?, 1.801937735804839?]
so they are real.
We compute the polynomial with roots $a'=A^5$, $b'=B^5$, $c'=C^3$.
(We "hope" that $a=a'$, $b=b'$, $c=c'$.)
For this we have to start computations involving symmetrical polynomials,
trying to get
$e_1(a',b',c')$,
$e_2(a',b',c')$,
$e_3(a',b',c')$, in terms of $e_1(A,B,C)$, $e_2(A,B,C)$, $e_3(A,B,C)$.
Here, $e_1$, $e_2$, $e_3$ are the first three symmetric polynomials.
(We have formally $e_4=e_5=dots=0$.)
One result is immediate:
$$
e_3(a',b',c')
=a'b'c'
=A^5B^5C^5=(ABC)^5=e_3(A,B,C)^5=-1 .
$$
We need now the Newton polynomial $p_5$ in Newton's identities of fifth degree...
$$
begin{aligned}
e_1(a',b',c')
&=
a'+b'+c'
\
&=A^5+B^5+C^5
\
&=p_5(A,B,C) ,\
&qquadtext{so we compute successively}\
p_1(A,B,C) &= e_1(A,B,C)=1 ,\
p_2(A,B,C) &= (e_1p_1-2e_2)(A,B,C)=1cdot 1-2cdot(-2)=5 ,\
p_3(A,B,C) &= (e_1p_2-e_2p_1+3e_2)(A,B,C)=1cdot 5-(-2)cdot1+3cdot(-1)=4 ,\
p_4(A,B,C) &= (e_1p_3-e_2p_2+e_3p_1)(A,B,C)=1cdot 4-(-2)cdot5+(-1)cdot1=13 ,\
e_1(a',b',c') &=p_5(A,B,C)
\&=(e_1p_4-e_2p_3+e_3p_2)(A,B,C)=1cdot13-(-2)cdot4+(-1)cdot5=16 .
end{aligned}
$$
We need now finally $a'b'+b'c'+c'a'$.
For this, we repeat the same procedure as above, but not for $A,B,C$, but for $s,t,u$, which are respectively $AB$, $BC$, $CA$, with
$$
begin{aligned}
e_1(s,t,u) &=s+t+u=AB + BC + CA &&=-2 ,\
e_2(s,t,u) &=st+tu+us= ABC(A+B+C) &&=-1 ,\
e_3(s,t,u) &=stu= A^2 B^2 C^2 &&= +1 .
end{aligned}
$$
So we have:
$$
begin{aligned}
p_1(s,t,u)&=e_1(s,t,u)=-2 ,\
p_2(s,t,u)&=(e_1p_1-2e_2)(s,t,u)=(-2)cdot (-2)-2cdot(-1)=6 ,\
p_3(s,t,u)&=(e_1p_2-e_2p_1+3e_2)(s,t,u)=(-2)cdot6-(-1)cdot(-2)+3cdot1=-11 ,\
p_4(s,t,u)&=(e_1p_3-e_2p_2+e_3p_1)(s,t,u)
=(-2)cdot(-11)-(-1)cdot6+1cdot(-2)=26 ,\
e_2(a',b',c')&=p_5(s,t,u)
\
&=(e_1p_4-e_2p_3+e_3p_2)(s,t,u)
=(-2)cdot26-(-1)cdot(-11)+1cdot6=-57 .
end{aligned}
$$
So $a',b',c'$ are the roots of the given polynomial
$$
x^3-E_1x^2+E_2x-E_3
=
x^3 - 16x^2 -57 x +1 .
$$
So $a',b',c'$ are (up to reordering) the "given" values $a,b,c$.
We conclude:
$$
a^{1/5} + b^{1/5} + c^{1/5}
=A+B+C=1 .
$$
$square$
Numerical support for the computations done so far, and the motivation for the abrupt start with the roots $A,B,C$ of the polynomial $x^3 - x^2 - 2x + 1$:
Sage code:
R.<x> = PolynomialRing(QQ)
f = x^3 - 16*x^2 - 57*x + 1
a, b, c = f.roots( ring=QQbar, multiplicities=False )
a, b, c
a^(1/5), b^(1/5), c^(1/5)
This gives:
(-3.015065490237851?, 0.01745839634379104?, 18.99760709389406?)
(1.008827691046369? + 0.7329562209746396?*I,
0.4450418679126288?,
1.801937735804839?)
OK, the computer needs human assitance to get the right fifth root of $a$ in $Bbb R$.
A, B, C = -(-a)^(1/5), b^(1/5), c^(1/5)
giving
sage: A, B, C
(-1.246979603717467?, 0.4450418679126288?, 1.801937735804839?)
this is good, three real numbers. Let us compute the elementary symmetric functions,
and the first Newton polynomials for them:
print "A + B + C = %s" % (A+B+C)
print "AB + BC + CA = %s" % (A*B + B*C + C*A)
print "A B C = %s" % (A*B*C)
for k in [1..5]:
print "A^%s + B^%s + C^%s = %s" % (k, k, k, A^k+B^k+C^k)
This gives so far:
A + B + C = 1.000000000000000?
AB + BC + CA = -2.000000000000000?
A B C = -1.000000000000000?
A^1 + B^1 + C^1 = 1.000000000000000?
A^2 + B^2 + C^2 = 5.000000000000000?
A^3 + B^3 + C^3 = 4.000000000000000?
A^4 + B^4 + C^4 = 13.00000000000000?
A^5 + B^5 + C^5 = 16.00000000000000?
Same for the values $s,t,u$:
s, t, u = A*B, B*C, C*A
print "s + t + u = %s" % (s+t+u)
print "st + tu + us = %s" % (s*t + t*u + u*s)
print "s t u = %s" % (s*t*u)
for k in [1..5]:
print "s^%s + t^%s + u^%s = %s" % (k, k, k, s^k+t^k+u^k)
This gives:
s + t + u = -2.000000000000000?
st + tu + us = -1.000000000000000?
s t u = 1.000000000000000?
s^1 + t^1 + u^1 = -2.000000000000000?
s^2 + t^2 + u^2 = 6.000000000000000?
s^3 + t^3 + u^3 = -11.00000000000000?
s^4 + t^4 + u^4 = 26.00000000000000?
s^5 + t^5 + u^5 = -57.00000000000000?
The auxiliary cubic polynomial is readily obtained as a factor of $p(x^5)$ where $p$ is the original cubic.
– Yves Daoust
Aug 16 at 9:22
Yes, as in the comments to the OP. My problem was to see its roots and the fact that they are real. Else one has to take the suitable branch of the fifth root. to match this one "beautiful factor". People wanted computations explicitly, i did it explicitly. On a piece of paper with own notations the calculus is done in some five minutes, typing all the stuff...
– dan_fulea
Aug 16 at 9:28
add a comment |
up vote
3
down vote
Let $A,B,C$ be the roots of the cubic polynomial
$$
g(x) = x^3 - x^2 - 2x + 1 .
$$
So $A,B,C$ satisfy (Vieta):
$$
begin{aligned}
e_1 &=A + B + C &&=+1 ,\
e_2 &= AB + BC + CA &&=-2 ,\
e_3 &= A B C &&= -1 .
end{aligned}
$$
(Computer support motivating the above is postponed.)
These values are numerically:
sage: R.<x> = PolynomialRing(QQ)
sage: g = x^3 - x^2 - 2*x + 1
sage: g(x).roots(ring=QQbar, multiplicities=False)
[-1.246979603717467?, 0.4450418679126288?, 1.801937735804839?]
so they are real.
We compute the polynomial with roots $a'=A^5$, $b'=B^5$, $c'=C^3$.
(We "hope" that $a=a'$, $b=b'$, $c=c'$.)
For this we have to start computations involving symmetrical polynomials,
trying to get
$e_1(a',b',c')$,
$e_2(a',b',c')$,
$e_3(a',b',c')$, in terms of $e_1(A,B,C)$, $e_2(A,B,C)$, $e_3(A,B,C)$.
Here, $e_1$, $e_2$, $e_3$ are the first three symmetric polynomials.
(We have formally $e_4=e_5=dots=0$.)
One result is immediate:
$$
e_3(a',b',c')
=a'b'c'
=A^5B^5C^5=(ABC)^5=e_3(A,B,C)^5=-1 .
$$
We need now the Newton polynomial $p_5$ in Newton's identities of fifth degree...
$$
begin{aligned}
e_1(a',b',c')
&=
a'+b'+c'
\
&=A^5+B^5+C^5
\
&=p_5(A,B,C) ,\
&qquadtext{so we compute successively}\
p_1(A,B,C) &= e_1(A,B,C)=1 ,\
p_2(A,B,C) &= (e_1p_1-2e_2)(A,B,C)=1cdot 1-2cdot(-2)=5 ,\
p_3(A,B,C) &= (e_1p_2-e_2p_1+3e_2)(A,B,C)=1cdot 5-(-2)cdot1+3cdot(-1)=4 ,\
p_4(A,B,C) &= (e_1p_3-e_2p_2+e_3p_1)(A,B,C)=1cdot 4-(-2)cdot5+(-1)cdot1=13 ,\
e_1(a',b',c') &=p_5(A,B,C)
\&=(e_1p_4-e_2p_3+e_3p_2)(A,B,C)=1cdot13-(-2)cdot4+(-1)cdot5=16 .
end{aligned}
$$
We need now finally $a'b'+b'c'+c'a'$.
For this, we repeat the same procedure as above, but not for $A,B,C$, but for $s,t,u$, which are respectively $AB$, $BC$, $CA$, with
$$
begin{aligned}
e_1(s,t,u) &=s+t+u=AB + BC + CA &&=-2 ,\
e_2(s,t,u) &=st+tu+us= ABC(A+B+C) &&=-1 ,\
e_3(s,t,u) &=stu= A^2 B^2 C^2 &&= +1 .
end{aligned}
$$
So we have:
$$
begin{aligned}
p_1(s,t,u)&=e_1(s,t,u)=-2 ,\
p_2(s,t,u)&=(e_1p_1-2e_2)(s,t,u)=(-2)cdot (-2)-2cdot(-1)=6 ,\
p_3(s,t,u)&=(e_1p_2-e_2p_1+3e_2)(s,t,u)=(-2)cdot6-(-1)cdot(-2)+3cdot1=-11 ,\
p_4(s,t,u)&=(e_1p_3-e_2p_2+e_3p_1)(s,t,u)
=(-2)cdot(-11)-(-1)cdot6+1cdot(-2)=26 ,\
e_2(a',b',c')&=p_5(s,t,u)
\
&=(e_1p_4-e_2p_3+e_3p_2)(s,t,u)
=(-2)cdot26-(-1)cdot(-11)+1cdot6=-57 .
end{aligned}
$$
So $a',b',c'$ are the roots of the given polynomial
$$
x^3-E_1x^2+E_2x-E_3
=
x^3 - 16x^2 -57 x +1 .
$$
So $a',b',c'$ are (up to reordering) the "given" values $a,b,c$.
We conclude:
$$
a^{1/5} + b^{1/5} + c^{1/5}
=A+B+C=1 .
$$
$square$
Numerical support for the computations done so far, and the motivation for the abrupt start with the roots $A,B,C$ of the polynomial $x^3 - x^2 - 2x + 1$:
Sage code:
R.<x> = PolynomialRing(QQ)
f = x^3 - 16*x^2 - 57*x + 1
a, b, c = f.roots( ring=QQbar, multiplicities=False )
a, b, c
a^(1/5), b^(1/5), c^(1/5)
This gives:
(-3.015065490237851?, 0.01745839634379104?, 18.99760709389406?)
(1.008827691046369? + 0.7329562209746396?*I,
0.4450418679126288?,
1.801937735804839?)
OK, the computer needs human assitance to get the right fifth root of $a$ in $Bbb R$.
A, B, C = -(-a)^(1/5), b^(1/5), c^(1/5)
giving
sage: A, B, C
(-1.246979603717467?, 0.4450418679126288?, 1.801937735804839?)
this is good, three real numbers. Let us compute the elementary symmetric functions,
and the first Newton polynomials for them:
print "A + B + C = %s" % (A+B+C)
print "AB + BC + CA = %s" % (A*B + B*C + C*A)
print "A B C = %s" % (A*B*C)
for k in [1..5]:
print "A^%s + B^%s + C^%s = %s" % (k, k, k, A^k+B^k+C^k)
This gives so far:
A + B + C = 1.000000000000000?
AB + BC + CA = -2.000000000000000?
A B C = -1.000000000000000?
A^1 + B^1 + C^1 = 1.000000000000000?
A^2 + B^2 + C^2 = 5.000000000000000?
A^3 + B^3 + C^3 = 4.000000000000000?
A^4 + B^4 + C^4 = 13.00000000000000?
A^5 + B^5 + C^5 = 16.00000000000000?
Same for the values $s,t,u$:
s, t, u = A*B, B*C, C*A
print "s + t + u = %s" % (s+t+u)
print "st + tu + us = %s" % (s*t + t*u + u*s)
print "s t u = %s" % (s*t*u)
for k in [1..5]:
print "s^%s + t^%s + u^%s = %s" % (k, k, k, s^k+t^k+u^k)
This gives:
s + t + u = -2.000000000000000?
st + tu + us = -1.000000000000000?
s t u = 1.000000000000000?
s^1 + t^1 + u^1 = -2.000000000000000?
s^2 + t^2 + u^2 = 6.000000000000000?
s^3 + t^3 + u^3 = -11.00000000000000?
s^4 + t^4 + u^4 = 26.00000000000000?
s^5 + t^5 + u^5 = -57.00000000000000?
The auxiliary cubic polynomial is readily obtained as a factor of $p(x^5)$ where $p$ is the original cubic.
– Yves Daoust
Aug 16 at 9:22
Yes, as in the comments to the OP. My problem was to see its roots and the fact that they are real. Else one has to take the suitable branch of the fifth root. to match this one "beautiful factor". People wanted computations explicitly, i did it explicitly. On a piece of paper with own notations the calculus is done in some five minutes, typing all the stuff...
– dan_fulea
Aug 16 at 9:28
add a comment |
up vote
3
down vote
up vote
3
down vote
Let $A,B,C$ be the roots of the cubic polynomial
$$
g(x) = x^3 - x^2 - 2x + 1 .
$$
So $A,B,C$ satisfy (Vieta):
$$
begin{aligned}
e_1 &=A + B + C &&=+1 ,\
e_2 &= AB + BC + CA &&=-2 ,\
e_3 &= A B C &&= -1 .
end{aligned}
$$
(Computer support motivating the above is postponed.)
These values are numerically:
sage: R.<x> = PolynomialRing(QQ)
sage: g = x^3 - x^2 - 2*x + 1
sage: g(x).roots(ring=QQbar, multiplicities=False)
[-1.246979603717467?, 0.4450418679126288?, 1.801937735804839?]
so they are real.
We compute the polynomial with roots $a'=A^5$, $b'=B^5$, $c'=C^3$.
(We "hope" that $a=a'$, $b=b'$, $c=c'$.)
For this we have to start computations involving symmetrical polynomials,
trying to get
$e_1(a',b',c')$,
$e_2(a',b',c')$,
$e_3(a',b',c')$, in terms of $e_1(A,B,C)$, $e_2(A,B,C)$, $e_3(A,B,C)$.
Here, $e_1$, $e_2$, $e_3$ are the first three symmetric polynomials.
(We have formally $e_4=e_5=dots=0$.)
One result is immediate:
$$
e_3(a',b',c')
=a'b'c'
=A^5B^5C^5=(ABC)^5=e_3(A,B,C)^5=-1 .
$$
We need now the Newton polynomial $p_5$ in Newton's identities of fifth degree...
$$
begin{aligned}
e_1(a',b',c')
&=
a'+b'+c'
\
&=A^5+B^5+C^5
\
&=p_5(A,B,C) ,\
&qquadtext{so we compute successively}\
p_1(A,B,C) &= e_1(A,B,C)=1 ,\
p_2(A,B,C) &= (e_1p_1-2e_2)(A,B,C)=1cdot 1-2cdot(-2)=5 ,\
p_3(A,B,C) &= (e_1p_2-e_2p_1+3e_2)(A,B,C)=1cdot 5-(-2)cdot1+3cdot(-1)=4 ,\
p_4(A,B,C) &= (e_1p_3-e_2p_2+e_3p_1)(A,B,C)=1cdot 4-(-2)cdot5+(-1)cdot1=13 ,\
e_1(a',b',c') &=p_5(A,B,C)
\&=(e_1p_4-e_2p_3+e_3p_2)(A,B,C)=1cdot13-(-2)cdot4+(-1)cdot5=16 .
end{aligned}
$$
We need now finally $a'b'+b'c'+c'a'$.
For this, we repeat the same procedure as above, but not for $A,B,C$, but for $s,t,u$, which are respectively $AB$, $BC$, $CA$, with
$$
begin{aligned}
e_1(s,t,u) &=s+t+u=AB + BC + CA &&=-2 ,\
e_2(s,t,u) &=st+tu+us= ABC(A+B+C) &&=-1 ,\
e_3(s,t,u) &=stu= A^2 B^2 C^2 &&= +1 .
end{aligned}
$$
So we have:
$$
begin{aligned}
p_1(s,t,u)&=e_1(s,t,u)=-2 ,\
p_2(s,t,u)&=(e_1p_1-2e_2)(s,t,u)=(-2)cdot (-2)-2cdot(-1)=6 ,\
p_3(s,t,u)&=(e_1p_2-e_2p_1+3e_2)(s,t,u)=(-2)cdot6-(-1)cdot(-2)+3cdot1=-11 ,\
p_4(s,t,u)&=(e_1p_3-e_2p_2+e_3p_1)(s,t,u)
=(-2)cdot(-11)-(-1)cdot6+1cdot(-2)=26 ,\
e_2(a',b',c')&=p_5(s,t,u)
\
&=(e_1p_4-e_2p_3+e_3p_2)(s,t,u)
=(-2)cdot26-(-1)cdot(-11)+1cdot6=-57 .
end{aligned}
$$
So $a',b',c'$ are the roots of the given polynomial
$$
x^3-E_1x^2+E_2x-E_3
=
x^3 - 16x^2 -57 x +1 .
$$
So $a',b',c'$ are (up to reordering) the "given" values $a,b,c$.
We conclude:
$$
a^{1/5} + b^{1/5} + c^{1/5}
=A+B+C=1 .
$$
$square$
Numerical support for the computations done so far, and the motivation for the abrupt start with the roots $A,B,C$ of the polynomial $x^3 - x^2 - 2x + 1$:
Sage code:
R.<x> = PolynomialRing(QQ)
f = x^3 - 16*x^2 - 57*x + 1
a, b, c = f.roots( ring=QQbar, multiplicities=False )
a, b, c
a^(1/5), b^(1/5), c^(1/5)
This gives:
(-3.015065490237851?, 0.01745839634379104?, 18.99760709389406?)
(1.008827691046369? + 0.7329562209746396?*I,
0.4450418679126288?,
1.801937735804839?)
OK, the computer needs human assitance to get the right fifth root of $a$ in $Bbb R$.
A, B, C = -(-a)^(1/5), b^(1/5), c^(1/5)
giving
sage: A, B, C
(-1.246979603717467?, 0.4450418679126288?, 1.801937735804839?)
this is good, three real numbers. Let us compute the elementary symmetric functions,
and the first Newton polynomials for them:
print "A + B + C = %s" % (A+B+C)
print "AB + BC + CA = %s" % (A*B + B*C + C*A)
print "A B C = %s" % (A*B*C)
for k in [1..5]:
print "A^%s + B^%s + C^%s = %s" % (k, k, k, A^k+B^k+C^k)
This gives so far:
A + B + C = 1.000000000000000?
AB + BC + CA = -2.000000000000000?
A B C = -1.000000000000000?
A^1 + B^1 + C^1 = 1.000000000000000?
A^2 + B^2 + C^2 = 5.000000000000000?
A^3 + B^3 + C^3 = 4.000000000000000?
A^4 + B^4 + C^4 = 13.00000000000000?
A^5 + B^5 + C^5 = 16.00000000000000?
Same for the values $s,t,u$:
s, t, u = A*B, B*C, C*A
print "s + t + u = %s" % (s+t+u)
print "st + tu + us = %s" % (s*t + t*u + u*s)
print "s t u = %s" % (s*t*u)
for k in [1..5]:
print "s^%s + t^%s + u^%s = %s" % (k, k, k, s^k+t^k+u^k)
This gives:
s + t + u = -2.000000000000000?
st + tu + us = -1.000000000000000?
s t u = 1.000000000000000?
s^1 + t^1 + u^1 = -2.000000000000000?
s^2 + t^2 + u^2 = 6.000000000000000?
s^3 + t^3 + u^3 = -11.00000000000000?
s^4 + t^4 + u^4 = 26.00000000000000?
s^5 + t^5 + u^5 = -57.00000000000000?
Let $A,B,C$ be the roots of the cubic polynomial
$$
g(x) = x^3 - x^2 - 2x + 1 .
$$
So $A,B,C$ satisfy (Vieta):
$$
begin{aligned}
e_1 &=A + B + C &&=+1 ,\
e_2 &= AB + BC + CA &&=-2 ,\
e_3 &= A B C &&= -1 .
end{aligned}
$$
(Computer support motivating the above is postponed.)
These values are numerically:
sage: R.<x> = PolynomialRing(QQ)
sage: g = x^3 - x^2 - 2*x + 1
sage: g(x).roots(ring=QQbar, multiplicities=False)
[-1.246979603717467?, 0.4450418679126288?, 1.801937735804839?]
so they are real.
We compute the polynomial with roots $a'=A^5$, $b'=B^5$, $c'=C^3$.
(We "hope" that $a=a'$, $b=b'$, $c=c'$.)
For this we have to start computations involving symmetrical polynomials,
trying to get
$e_1(a',b',c')$,
$e_2(a',b',c')$,
$e_3(a',b',c')$, in terms of $e_1(A,B,C)$, $e_2(A,B,C)$, $e_3(A,B,C)$.
Here, $e_1$, $e_2$, $e_3$ are the first three symmetric polynomials.
(We have formally $e_4=e_5=dots=0$.)
One result is immediate:
$$
e_3(a',b',c')
=a'b'c'
=A^5B^5C^5=(ABC)^5=e_3(A,B,C)^5=-1 .
$$
We need now the Newton polynomial $p_5$ in Newton's identities of fifth degree...
$$
begin{aligned}
e_1(a',b',c')
&=
a'+b'+c'
\
&=A^5+B^5+C^5
\
&=p_5(A,B,C) ,\
&qquadtext{so we compute successively}\
p_1(A,B,C) &= e_1(A,B,C)=1 ,\
p_2(A,B,C) &= (e_1p_1-2e_2)(A,B,C)=1cdot 1-2cdot(-2)=5 ,\
p_3(A,B,C) &= (e_1p_2-e_2p_1+3e_2)(A,B,C)=1cdot 5-(-2)cdot1+3cdot(-1)=4 ,\
p_4(A,B,C) &= (e_1p_3-e_2p_2+e_3p_1)(A,B,C)=1cdot 4-(-2)cdot5+(-1)cdot1=13 ,\
e_1(a',b',c') &=p_5(A,B,C)
\&=(e_1p_4-e_2p_3+e_3p_2)(A,B,C)=1cdot13-(-2)cdot4+(-1)cdot5=16 .
end{aligned}
$$
We need now finally $a'b'+b'c'+c'a'$.
For this, we repeat the same procedure as above, but not for $A,B,C$, but for $s,t,u$, which are respectively $AB$, $BC$, $CA$, with
$$
begin{aligned}
e_1(s,t,u) &=s+t+u=AB + BC + CA &&=-2 ,\
e_2(s,t,u) &=st+tu+us= ABC(A+B+C) &&=-1 ,\
e_3(s,t,u) &=stu= A^2 B^2 C^2 &&= +1 .
end{aligned}
$$
So we have:
$$
begin{aligned}
p_1(s,t,u)&=e_1(s,t,u)=-2 ,\
p_2(s,t,u)&=(e_1p_1-2e_2)(s,t,u)=(-2)cdot (-2)-2cdot(-1)=6 ,\
p_3(s,t,u)&=(e_1p_2-e_2p_1+3e_2)(s,t,u)=(-2)cdot6-(-1)cdot(-2)+3cdot1=-11 ,\
p_4(s,t,u)&=(e_1p_3-e_2p_2+e_3p_1)(s,t,u)
=(-2)cdot(-11)-(-1)cdot6+1cdot(-2)=26 ,\
e_2(a',b',c')&=p_5(s,t,u)
\
&=(e_1p_4-e_2p_3+e_3p_2)(s,t,u)
=(-2)cdot26-(-1)cdot(-11)+1cdot6=-57 .
end{aligned}
$$
So $a',b',c'$ are the roots of the given polynomial
$$
x^3-E_1x^2+E_2x-E_3
=
x^3 - 16x^2 -57 x +1 .
$$
So $a',b',c'$ are (up to reordering) the "given" values $a,b,c$.
We conclude:
$$
a^{1/5} + b^{1/5} + c^{1/5}
=A+B+C=1 .
$$
$square$
Numerical support for the computations done so far, and the motivation for the abrupt start with the roots $A,B,C$ of the polynomial $x^3 - x^2 - 2x + 1$:
Sage code:
R.<x> = PolynomialRing(QQ)
f = x^3 - 16*x^2 - 57*x + 1
a, b, c = f.roots( ring=QQbar, multiplicities=False )
a, b, c
a^(1/5), b^(1/5), c^(1/5)
This gives:
(-3.015065490237851?, 0.01745839634379104?, 18.99760709389406?)
(1.008827691046369? + 0.7329562209746396?*I,
0.4450418679126288?,
1.801937735804839?)
OK, the computer needs human assitance to get the right fifth root of $a$ in $Bbb R$.
A, B, C = -(-a)^(1/5), b^(1/5), c^(1/5)
giving
sage: A, B, C
(-1.246979603717467?, 0.4450418679126288?, 1.801937735804839?)
this is good, three real numbers. Let us compute the elementary symmetric functions,
and the first Newton polynomials for them:
print "A + B + C = %s" % (A+B+C)
print "AB + BC + CA = %s" % (A*B + B*C + C*A)
print "A B C = %s" % (A*B*C)
for k in [1..5]:
print "A^%s + B^%s + C^%s = %s" % (k, k, k, A^k+B^k+C^k)
This gives so far:
A + B + C = 1.000000000000000?
AB + BC + CA = -2.000000000000000?
A B C = -1.000000000000000?
A^1 + B^1 + C^1 = 1.000000000000000?
A^2 + B^2 + C^2 = 5.000000000000000?
A^3 + B^3 + C^3 = 4.000000000000000?
A^4 + B^4 + C^4 = 13.00000000000000?
A^5 + B^5 + C^5 = 16.00000000000000?
Same for the values $s,t,u$:
s, t, u = A*B, B*C, C*A
print "s + t + u = %s" % (s+t+u)
print "st + tu + us = %s" % (s*t + t*u + u*s)
print "s t u = %s" % (s*t*u)
for k in [1..5]:
print "s^%s + t^%s + u^%s = %s" % (k, k, k, s^k+t^k+u^k)
This gives:
s + t + u = -2.000000000000000?
st + tu + us = -1.000000000000000?
s t u = 1.000000000000000?
s^1 + t^1 + u^1 = -2.000000000000000?
s^2 + t^2 + u^2 = 6.000000000000000?
s^3 + t^3 + u^3 = -11.00000000000000?
s^4 + t^4 + u^4 = 26.00000000000000?
s^5 + t^5 + u^5 = -57.00000000000000?
edited Aug 16 at 9:24
answered Aug 16 at 9:19
dan_fulea
6,1051312
6,1051312
The auxiliary cubic polynomial is readily obtained as a factor of $p(x^5)$ where $p$ is the original cubic.
– Yves Daoust
Aug 16 at 9:22
Yes, as in the comments to the OP. My problem was to see its roots and the fact that they are real. Else one has to take the suitable branch of the fifth root. to match this one "beautiful factor". People wanted computations explicitly, i did it explicitly. On a piece of paper with own notations the calculus is done in some five minutes, typing all the stuff...
– dan_fulea
Aug 16 at 9:28
add a comment |
The auxiliary cubic polynomial is readily obtained as a factor of $p(x^5)$ where $p$ is the original cubic.
– Yves Daoust
Aug 16 at 9:22
Yes, as in the comments to the OP. My problem was to see its roots and the fact that they are real. Else one has to take the suitable branch of the fifth root. to match this one "beautiful factor". People wanted computations explicitly, i did it explicitly. On a piece of paper with own notations the calculus is done in some five minutes, typing all the stuff...
– dan_fulea
Aug 16 at 9:28
The auxiliary cubic polynomial is readily obtained as a factor of $p(x^5)$ where $p$ is the original cubic.
– Yves Daoust
Aug 16 at 9:22
The auxiliary cubic polynomial is readily obtained as a factor of $p(x^5)$ where $p$ is the original cubic.
– Yves Daoust
Aug 16 at 9:22
Yes, as in the comments to the OP. My problem was to see its roots and the fact that they are real. Else one has to take the suitable branch of the fifth root. to match this one "beautiful factor". People wanted computations explicitly, i did it explicitly. On a piece of paper with own notations the calculus is done in some five minutes, typing all the stuff...
– dan_fulea
Aug 16 at 9:28
Yes, as in the comments to the OP. My problem was to see its roots and the fact that they are real. Else one has to take the suitable branch of the fifth root. to match this one "beautiful factor". People wanted computations explicitly, i did it explicitly. On a piece of paper with own notations the calculus is done in some five minutes, typing all the stuff...
– dan_fulea
Aug 16 at 9:28
add a comment |
up vote
3
down vote
The given polynomial $p(x)$ has three real roots so $p(x^5)$ also has three real roots as all others are complex fifth roots of reals.
So we need to factor $p(x^5)=q(x)r(x)$ where $q$ is cubic with real roots and $r$ is an unimportant polynomial of degree twelve with complex ones.
With a simple numerical solver or with Cardano's formulas, we estimate the coefficients of $q$ using Vieta's formulas on the fifth roots of $p$, giving
$$q(x)approx (x-1.80194) (x-0.445042) (x+1.24698)\approx x^3-1.000002 x^2-2.000002633x+1.000001871.$$
So we can hypothetize $q(x)=x^3-x^2-2x+1$ and long division confirms this hypothesis exactly *.
Then the sum of the real roots of $q$ is $1$.
*$$dfrac{x^{15}-16x^{10}-57x^5+1}{x^3-x^2-2x+1}\=x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1.$$
Yes, a useful trick. Wolfram will also factor p(x⁵) directly, which is something |I should have tried.
– URCHIN
Aug 16 at 14:57
@URCHIN: the next challenge is to find a way to solve fully by hand.
– Yves Daoust
Aug 16 at 21:27
add a comment |
up vote
3
down vote
The given polynomial $p(x)$ has three real roots so $p(x^5)$ also has three real roots as all others are complex fifth roots of reals.
So we need to factor $p(x^5)=q(x)r(x)$ where $q$ is cubic with real roots and $r$ is an unimportant polynomial of degree twelve with complex ones.
With a simple numerical solver or with Cardano's formulas, we estimate the coefficients of $q$ using Vieta's formulas on the fifth roots of $p$, giving
$$q(x)approx (x-1.80194) (x-0.445042) (x+1.24698)\approx x^3-1.000002 x^2-2.000002633x+1.000001871.$$
So we can hypothetize $q(x)=x^3-x^2-2x+1$ and long division confirms this hypothesis exactly *.
Then the sum of the real roots of $q$ is $1$.
*$$dfrac{x^{15}-16x^{10}-57x^5+1}{x^3-x^2-2x+1}\=x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1.$$
Yes, a useful trick. Wolfram will also factor p(x⁵) directly, which is something |I should have tried.
– URCHIN
Aug 16 at 14:57
@URCHIN: the next challenge is to find a way to solve fully by hand.
– Yves Daoust
Aug 16 at 21:27
add a comment |
up vote
3
down vote
up vote
3
down vote
The given polynomial $p(x)$ has three real roots so $p(x^5)$ also has three real roots as all others are complex fifth roots of reals.
So we need to factor $p(x^5)=q(x)r(x)$ where $q$ is cubic with real roots and $r$ is an unimportant polynomial of degree twelve with complex ones.
With a simple numerical solver or with Cardano's formulas, we estimate the coefficients of $q$ using Vieta's formulas on the fifth roots of $p$, giving
$$q(x)approx (x-1.80194) (x-0.445042) (x+1.24698)\approx x^3-1.000002 x^2-2.000002633x+1.000001871.$$
So we can hypothetize $q(x)=x^3-x^2-2x+1$ and long division confirms this hypothesis exactly *.
Then the sum of the real roots of $q$ is $1$.
*$$dfrac{x^{15}-16x^{10}-57x^5+1}{x^3-x^2-2x+1}\=x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1.$$
The given polynomial $p(x)$ has three real roots so $p(x^5)$ also has three real roots as all others are complex fifth roots of reals.
So we need to factor $p(x^5)=q(x)r(x)$ where $q$ is cubic with real roots and $r$ is an unimportant polynomial of degree twelve with complex ones.
With a simple numerical solver or with Cardano's formulas, we estimate the coefficients of $q$ using Vieta's formulas on the fifth roots of $p$, giving
$$q(x)approx (x-1.80194) (x-0.445042) (x+1.24698)\approx x^3-1.000002 x^2-2.000002633x+1.000001871.$$
So we can hypothetize $q(x)=x^3-x^2-2x+1$ and long division confirms this hypothesis exactly *.
Then the sum of the real roots of $q$ is $1$.
*$$dfrac{x^{15}-16x^{10}-57x^5+1}{x^3-x^2-2x+1}\=x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1.$$
edited Aug 16 at 13:31
answered Aug 16 at 10:20
Yves Daoust
122k668218
122k668218
Yes, a useful trick. Wolfram will also factor p(x⁵) directly, which is something |I should have tried.
– URCHIN
Aug 16 at 14:57
@URCHIN: the next challenge is to find a way to solve fully by hand.
– Yves Daoust
Aug 16 at 21:27
add a comment |
Yes, a useful trick. Wolfram will also factor p(x⁵) directly, which is something |I should have tried.
– URCHIN
Aug 16 at 14:57
@URCHIN: the next challenge is to find a way to solve fully by hand.
– Yves Daoust
Aug 16 at 21:27
Yes, a useful trick. Wolfram will also factor p(x⁵) directly, which is something |I should have tried.
– URCHIN
Aug 16 at 14:57
Yes, a useful trick. Wolfram will also factor p(x⁵) directly, which is something |I should have tried.
– URCHIN
Aug 16 at 14:57
@URCHIN: the next challenge is to find a way to solve fully by hand.
– Yves Daoust
Aug 16 at 21:27
@URCHIN: the next challenge is to find a way to solve fully by hand.
– Yves Daoust
Aug 16 at 21:27
add a comment |
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@molarmass OP made an edit.
– Jack Bauer
Aug 16 at 8:47
2
$x^{15}-16x^{10}-57x^5+1$ factors as Hint: $(x^3-x^2-2x+1)(x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1)$ and the sum of the roots of the first factor is $1$.
– Yves Daoust
Aug 16 at 9:05
Why is this closed ? The OP did show some non-trivial effort, even though presented via an uncommon medium.
– Yves Daoust
Aug 16 at 9:08
@YvesDaoust In that link I only see efforts that others have done to answer his question.
– Jaap Scherphuis
Aug 16 at 9:11
@JaapScherphuis: the first answer is by the OP himself. Check the signatures.
– Yves Daoust
Aug 16 at 9:12