Mixing
Show that solution $(0,0)$ is unique
$begingroup$
Assume the system:
$$
begin{align*}
& 2x-5y-x(x^2+2y^2)^2=0 \
& 5x+2y-3y(2x^2+y^2)^2=0
end{align*}
$$
Obviously, $(0,0)$ is a solution. The thing I find hard doing is showing that it's the only solution.
Attempt:
Suppose that there exist another solution $(bar{x},bar{y}) neq (0,0)$. Then, for $x neq0$ and $y neq 0$:
$$
2x-5y=x(x^2+2y^2)^2 iff 2-5frac{y}{x}=(x^2+2y^2)^2=c(x,y)geq 0 tag{1}
$$
$$
5x+2y=3y(2x^2+y^2)^2 iff 5 frac{x}{y}+2=3(2x^2+y^2)^2=k(x,y) geq 0tag{2}
$$
From equations $(1)$ and $(2)$, we yield:
$$
2-5frac{5}{k-2}=ciff (2-c)(k-2)=25
$$
But I can't seem to spot a contradiction. Is there another (more straightforward) way to prove this?
algebra-precalculus systems-of-equations substitution
edited Jan 28 at 18:49
Michael Rozenberg
109k1896201
asked Jan 28 at 18:11
LoneBoneLoneBone
968
$endgroup$
add a comment |
$begingroup$
Assume the system:
$$
begin{align*}
& 2x-5y-x(x^2+2y^2)^2=0 \
& 5x+2y-3y(2x^2+y^2)^2=0
end{align*}
$$
Obviously, $(0,0)$ is a solution. The thing I find hard doing is showing that it's the only solution.
Attempt:
Suppose that there exist another solution $(bar{x},bar{y}) neq (0,0)$. Then, for $x neq0$ and $y neq 0$:
$$
2x-5y=x(x^2+2y^2)^2 iff 2-5frac{y}{x}=(x^2+2y^2)^2=c(x,y)geq 0 tag{1}
$$
$$
5x+2y=3y(2x^2+y^2)^2 iff 5 frac{x}{y}+2=3(2x^2+y^2)^2=k(x,y) geq 0tag{2}
$$
From equations $(1)$ and $(2)$, we yield:
$$
2-5frac{5}{k-2}=ciff (2-c)(k-2)=25
$$
But I can't seem to spot a contradiction. Is there another (more straightforward) way to prove this?
algebra-precalculus systems-of-equations substitution
edited Jan 28 at 18:49
Michael Rozenberg
109k1896201
asked Jan 28 at 18:11
LoneBoneLoneBone
968
$endgroup$
add a comment |
$begingroup$
Assume the system:
$$
begin{align*}
& 2x-5y-x(x^2+2y^2)^2=0 \
& 5x+2y-3y(2x^2+y^2)^2=0
end{align*}
$$
Obviously, $(0,0)$ is a solution. The thing I find hard doing is showing that it's the only solution.
Attempt:
Suppose that there exist another solution $(bar{x},bar{y}) neq (0,0)$. Then, for $x neq0$ and $y neq 0$:
$$
2x-5y=x(x^2+2y^2)^2 iff 2-5frac{y}{x}=(x^2+2y^2)^2=c(x,y)geq 0 tag{1}
$$
$$
5x+2y=3y(2x^2+y^2)^2 iff 5 frac{x}{y}+2=3(2x^2+y^2)^2=k(x,y) geq 0tag{2}
$$
From equations $(1)$ and $(2)$, we yield:
$$
2-5frac{5}{k-2}=ciff (2-c)(k-2)=25
$$
But I can't seem to spot a contradiction. Is there another (more straightforward) way to prove this?
algebra-precalculus systems-of-equations substitution
edited Jan 28 at 18:49
Michael Rozenberg
109k1896201
asked Jan 28 at 18:11
LoneBoneLoneBone
968
$endgroup$
Assume the system:
$$
begin{align*}
& 2x-5y-x(x^2+2y^2)^2=0 \
& 5x+2y-3y(2x^2+y^2)^2=0
end{align*}
$$
Obviously, $(0,0)$ is a solution. The thing I find hard doing is showing that it's the only solution.
Attempt:
Suppose that there exist another solution $(bar{x},bar{y}) neq (0,0)$. Then, for $x neq0$ and $y neq 0$:
$$
2x-5y=x(x^2+2y^2)^2 iff 2-5frac{y}{x}=(x^2+2y^2)^2=c(x,y)geq 0 tag{1}
$$
$$
5x+2y=3y(2x^2+y^2)^2 iff 5 frac{x}{y}+2=3(2x^2+y^2)^2=k(x,y) geq 0tag{2}
$$
From equations $(1)$ and $(2)$, we yield:
$$
2-5frac{5}{k-2}=ciff (2-c)(k-2)=25
$$
But I can't seem to spot a contradiction. Is there another (more straightforward) way to prove this?
algebra-precalculus systems-of-equations substitution
algebra-precalculus systems-of-equations substitution
edited Jan 28 at 18:49
Michael Rozenberg
109k1896201
asked Jan 28 at 18:11
LoneBoneLoneBone
968
edited Jan 28 at 18:49
Michael Rozenberg
109k1896201
asked Jan 28 at 18:11
LoneBoneLoneBone
968
edited Jan 28 at 18:49
Michael Rozenberg
109k1896201
edited Jan 28 at 18:49
Michael Rozenberg
109k1896201
edited Jan 28 at 18:49
Michael Rozenberg
109k1896201
109k1896201
asked Jan 28 at 18:11
LoneBoneLoneBone
968
asked Jan 28 at 18:11
LoneBoneLoneBone
968
asked Jan 28 at 18:11
LoneBoneLoneBone
968
968
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If $y=0$ so $x=0$ and we get $(0,0)$.
Let $yneq0$ and $x=ty$.
Thus, $$frac{2t-5}{5t+2}=frac{t(t^2+2)^2}{3(2t^2+1)^2}$$ or
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=0,$$ which is impossible for all real $t$.
Indeed,
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=$$
$$=(5t^6-22t^5+25t^4)+(55t^4-16t^3+40t^2)+(40t^2+2t+15)>0.$$
answered Jan 28 at 18:29
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
I like pictures; If we increase the coefficient of $y$ in the second equation from $3$ to $7.421911190758908573490159799$ we get the first nontrivial intersections (with tangency) for this problem. The $x$ value at tangency is very close to $1,$ it may or may not actually be $1$ itself. The last picture, with no such coefficient, has rotational symmetry.
? g
%7 = 10*x^6 + 50*x^5 - 45*x^4 - x^3 + 45*x^2 + 50*x - 10
? factor(g)
%8 =
[10*x^6 + 50*x^5 - 45*x^4 - x^3 + 45*x^2 + 50*x - 10 1]
? polroots(g)
%9 = [-5.759456718855120444516616926 + 0.E-28*I,
0.1736274875937921035052204217 + 0.E-28*I,
-0.6374628782453508481344867754 - 0.5280217041759846233761928738*I,
-0.6374628782453508481344867754 + 0.5280217041759846233761928738*I,
0.9303774938760150186401850276 - 0.7706480276869016371542943284*I,
0.9303774938760150186401850276 + 0.7706480276869016371542943284*I]~
?
x = 0.1736274875937921035052204217
?
? r = x * (2 + x^2)^2 * (2-5*x) / ( (1+2*x^2)^2 * (5+2*x) )
%12 = 0.1347361851008281269891649571
?
? a = 1/r
%13 = 7.421911190758908573490159799
?
answered Jan 28 at 19:45
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
The resultant of the two left sides with respect to $y$ is
$$ -x left( 531441,{x}^{24}-2165130,{x}^{20}+7884864,{x}^{16}-
11409984,{x}^{12}+12205512,{x}^{8}+504218,{x}^{4}+489375 right)
$$
and using Sturm's theorem one can show that $x=0$ is the only real root of that. There are of course complex roots.
EDIT: With the $3$ in $-3y$ replaced by $s$, the resultant becomes
$$ eqalign{ -x &left( 6561,{s}^{4}{x}^{24}-18954,{s}^{4}{x}^{20}-23328,{s}^{3}{
x}^{20}+21708,{s}^{4}{x}^{16}+173340,{s}^{3}{x}^{16}right.cr &+160704,{s}^{2}
{x}^{16}-23976,{s}^{4}{x}^{12}-191016,{s}^{3}{x}^{12}-396000,{s}^{2
}{x}^{12}cr &
-248832,s{x}^{12}+10692,{s}^{4}{x}^{8}+24540,{s}^{3}{x}^{8
}+597520,{s}^{2}{x}^{8}+1517568,s{x}^{8}-5832,{s}^{4}{x}^{4}
cr &left.+746496
,{x}^{8}-14938,{s}^{3}{x}^{4}+47328,{s}^{2}{x}^{4}+337792,s{x}^{4}
-59392,{x}^{4}+18125,{s}^{3} right)}
$$
The discriminant of this with respect to $x$ is
$$ 16618447044007316502648165488386203399239363151697251980300315466139733122351104000000000000000000000000000000000000
,{s}^{83} left( 295612416,{s}^{6}-946111104,{s}^{5}-7364205035,{s
}^{4}-13075015530,{s}^{3}-7364205035,{s}^{2}-946111104,s+295612416
right) ^{4} left( 2772210825,{s}^{9}-14150310144,{s}^{8}+
50430196864,{s}^{7}+1618020643840,{s}^{6}+5389314830336,{s}^{5}+
11451158167552,{s}^{4}+18821256052736,{s}^{3}+28395282890752,{s}^{2
}+12309929918464,s-11093095219200 right) ^{8}
$$
The roots of this polynomial should correspond to values at which the two curves are tangent. In particular, the factor $$295612416,{s}^{6}-946111104,{s}^{5}-7364205035,{s}^{4}-13075015530
,{s}^{3}-7364205035,{s}^{2}-946111104,s+295612416
$$ has a root at approximately $7.421911191$. Hmm: Maple indicates that the Galois group of this sextic polynomial is solvable, so the root should have an expression in radicals.
There's also a root at approximately $0.1347361851$ which makes the two curves tangent:
On the other hand, the root at $0.4013844674$ does not appear to produce tangent curves: presumably in this case the multiple root is complex.
answered Jan 28 at 18:31
Robert IsraelRobert Israel
330k23218473
$endgroup$
$begingroup$
I found the coefficient with which to replace the $3$ in $-3y$ to get the two curves tangent. Came out about 7.42. There is probably some elegant way to do that, and for that matter correctly find the point of tangency that has been created this way
$endgroup$
– Will Jagy
Jan 28 at 19:49
add a comment |
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3 Answers
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3 Answers
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$begingroup$
If $y=0$ so $x=0$ and we get $(0,0)$.
Let $yneq0$ and $x=ty$.
Thus, $$frac{2t-5}{5t+2}=frac{t(t^2+2)^2}{3(2t^2+1)^2}$$ or
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=0,$$ which is impossible for all real $t$.
Indeed,
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=$$
$$=(5t^6-22t^5+25t^4)+(55t^4-16t^3+40t^2)+(40t^2+2t+15)>0.$$
answered Jan 28 at 18:29
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
If $y=0$ so $x=0$ and we get $(0,0)$.
Let $yneq0$ and $x=ty$.
Thus, $$frac{2t-5}{5t+2}=frac{t(t^2+2)^2}{3(2t^2+1)^2}$$ or
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=0,$$ which is impossible for all real $t$.
Indeed,
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=$$
$$=(5t^6-22t^5+25t^4)+(55t^4-16t^3+40t^2)+(40t^2+2t+15)>0.$$
answered Jan 28 at 18:29
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
If $y=0$ so $x=0$ and we get $(0,0)$.
Let $yneq0$ and $x=ty$.
Thus, $$frac{2t-5}{5t+2}=frac{t(t^2+2)^2}{3(2t^2+1)^2}$$ or
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=0,$$ which is impossible for all real $t$.
Indeed,
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=$$
$$=(5t^6-22t^5+25t^4)+(55t^4-16t^3+40t^2)+(40t^2+2t+15)>0.$$
answered Jan 28 at 18:29
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
If $y=0$ so $x=0$ and we get $(0,0)$.
Let $yneq0$ and $x=ty$.
Thus, $$frac{2t-5}{5t+2}=frac{t(t^2+2)^2}{3(2t^2+1)^2}$$ or
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=0,$$ which is impossible for all real $t$.
Indeed,
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=$$
$$=(5t^6-22t^5+25t^4)+(55t^4-16t^3+40t^2)+(40t^2+2t+15)>0.$$
answered Jan 28 at 18:29
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 28 at 18:29
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 28 at 18:29
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 28 at 18:29
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
add a comment |
add a comment |
$begingroup$
I like pictures; If we increase the coefficient of $y$ in the second equation from $3$ to $7.421911190758908573490159799$ we get the first nontrivial intersections (with tangency) for this problem. The $x$ value at tangency is very close to $1,$ it may or may not actually be $1$ itself. The last picture, with no such coefficient, has rotational symmetry.
? g
%7 = 10*x^6 + 50*x^5 - 45*x^4 - x^3 + 45*x^2 + 50*x - 10
? factor(g)
%8 =
[10*x^6 + 50*x^5 - 45*x^4 - x^3 + 45*x^2 + 50*x - 10 1]
? polroots(g)
%9 = [-5.759456718855120444516616926 + 0.E-28*I,
0.1736274875937921035052204217 + 0.E-28*I,
-0.6374628782453508481344867754 - 0.5280217041759846233761928738*I,
-0.6374628782453508481344867754 + 0.5280217041759846233761928738*I,
0.9303774938760150186401850276 - 0.7706480276869016371542943284*I,
0.9303774938760150186401850276 + 0.7706480276869016371542943284*I]~
?
x = 0.1736274875937921035052204217
?
? r = x * (2 + x^2)^2 * (2-5*x) / ( (1+2*x^2)^2 * (5+2*x) )
%12 = 0.1347361851008281269891649571
?
? a = 1/r
%13 = 7.421911190758908573490159799
?
answered Jan 28 at 19:45
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
I like pictures; If we increase the coefficient of $y$ in the second equation from $3$ to $7.421911190758908573490159799$ we get the first nontrivial intersections (with tangency) for this problem. The $x$ value at tangency is very close to $1,$ it may or may not actually be $1$ itself. The last picture, with no such coefficient, has rotational symmetry.
? g
%7 = 10*x^6 + 50*x^5 - 45*x^4 - x^3 + 45*x^2 + 50*x - 10
? factor(g)
%8 =
[10*x^6 + 50*x^5 - 45*x^4 - x^3 + 45*x^2 + 50*x - 10 1]
? polroots(g)
%9 = [-5.759456718855120444516616926 + 0.E-28*I,
0.1736274875937921035052204217 + 0.E-28*I,
-0.6374628782453508481344867754 - 0.5280217041759846233761928738*I,
-0.6374628782453508481344867754 + 0.5280217041759846233761928738*I,
0.9303774938760150186401850276 - 0.7706480276869016371542943284*I,
0.9303774938760150186401850276 + 0.7706480276869016371542943284*I]~
?
x = 0.1736274875937921035052204217
?
? r = x * (2 + x^2)^2 * (2-5*x) / ( (1+2*x^2)^2 * (5+2*x) )
%12 = 0.1347361851008281269891649571
?
? a = 1/r
%13 = 7.421911190758908573490159799
?
answered Jan 28 at 19:45
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
I like pictures; If we increase the coefficient of $y$ in the second equation from $3$ to $7.421911190758908573490159799$ we get the first nontrivial intersections (with tangency) for this problem. The $x$ value at tangency is very close to $1,$ it may or may not actually be $1$ itself. The last picture, with no such coefficient, has rotational symmetry.
? g
%7 = 10*x^6 + 50*x^5 - 45*x^4 - x^3 + 45*x^2 + 50*x - 10
? factor(g)
%8 =
[10*x^6 + 50*x^5 - 45*x^4 - x^3 + 45*x^2 + 50*x - 10 1]
? polroots(g)
%9 = [-5.759456718855120444516616926 + 0.E-28*I,
0.1736274875937921035052204217 + 0.E-28*I,
-0.6374628782453508481344867754 - 0.5280217041759846233761928738*I,
-0.6374628782453508481344867754 + 0.5280217041759846233761928738*I,
0.9303774938760150186401850276 - 0.7706480276869016371542943284*I,
0.9303774938760150186401850276 + 0.7706480276869016371542943284*I]~
?
x = 0.1736274875937921035052204217
?
? r = x * (2 + x^2)^2 * (2-5*x) / ( (1+2*x^2)^2 * (5+2*x) )
%12 = 0.1347361851008281269891649571
?
? a = 1/r
%13 = 7.421911190758908573490159799
?
answered Jan 28 at 19:45
Will JagyWill Jagy
104k5102201
$endgroup$
I like pictures; If we increase the coefficient of $y$ in the second equation from $3$ to $7.421911190758908573490159799$ we get the first nontrivial intersections (with tangency) for this problem. The $x$ value at tangency is very close to $1,$ it may or may not actually be $1$ itself. The last picture, with no such coefficient, has rotational symmetry.
? g
%7 = 10*x^6 + 50*x^5 - 45*x^4 - x^3 + 45*x^2 + 50*x - 10
? factor(g)
%8 =
[10*x^6 + 50*x^5 - 45*x^4 - x^3 + 45*x^2 + 50*x - 10 1]
? polroots(g)
%9 = [-5.759456718855120444516616926 + 0.E-28*I,
0.1736274875937921035052204217 + 0.E-28*I,
-0.6374628782453508481344867754 - 0.5280217041759846233761928738*I,
-0.6374628782453508481344867754 + 0.5280217041759846233761928738*I,
0.9303774938760150186401850276 - 0.7706480276869016371542943284*I,
0.9303774938760150186401850276 + 0.7706480276869016371542943284*I]~
?
x = 0.1736274875937921035052204217
?
? r = x * (2 + x^2)^2 * (2-5*x) / ( (1+2*x^2)^2 * (5+2*x) )
%12 = 0.1347361851008281269891649571
?
? a = 1/r
%13 = 7.421911190758908573490159799
?
answered Jan 28 at 19:45
Will JagyWill Jagy
104k5102201
answered Jan 28 at 19:45
Will JagyWill Jagy
104k5102201
answered Jan 28 at 19:45
Will JagyWill Jagy
104k5102201
answered Jan 28 at 19:45
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
$begingroup$
The resultant of the two left sides with respect to $y$ is
$$ -x left( 531441,{x}^{24}-2165130,{x}^{20}+7884864,{x}^{16}-
11409984,{x}^{12}+12205512,{x}^{8}+504218,{x}^{4}+489375 right)
$$
and using Sturm's theorem one can show that $x=0$ is the only real root of that. There are of course complex roots.
EDIT: With the $3$ in $-3y$ replaced by $s$, the resultant becomes
$$ eqalign{ -x &left( 6561,{s}^{4}{x}^{24}-18954,{s}^{4}{x}^{20}-23328,{s}^{3}{
x}^{20}+21708,{s}^{4}{x}^{16}+173340,{s}^{3}{x}^{16}right.cr &+160704,{s}^{2}
{x}^{16}-23976,{s}^{4}{x}^{12}-191016,{s}^{3}{x}^{12}-396000,{s}^{2
}{x}^{12}cr &
-248832,s{x}^{12}+10692,{s}^{4}{x}^{8}+24540,{s}^{3}{x}^{8
}+597520,{s}^{2}{x}^{8}+1517568,s{x}^{8}-5832,{s}^{4}{x}^{4}
cr &left.+746496
,{x}^{8}-14938,{s}^{3}{x}^{4}+47328,{s}^{2}{x}^{4}+337792,s{x}^{4}
-59392,{x}^{4}+18125,{s}^{3} right)}
$$
The discriminant of this with respect to $x$ is
$$ 16618447044007316502648165488386203399239363151697251980300315466139733122351104000000000000000000000000000000000000
,{s}^{83} left( 295612416,{s}^{6}-946111104,{s}^{5}-7364205035,{s
}^{4}-13075015530,{s}^{3}-7364205035,{s}^{2}-946111104,s+295612416
right) ^{4} left( 2772210825,{s}^{9}-14150310144,{s}^{8}+
50430196864,{s}^{7}+1618020643840,{s}^{6}+5389314830336,{s}^{5}+
11451158167552,{s}^{4}+18821256052736,{s}^{3}+28395282890752,{s}^{2
}+12309929918464,s-11093095219200 right) ^{8}
$$
The roots of this polynomial should correspond to values at which the two curves are tangent. In particular, the factor $$295612416,{s}^{6}-946111104,{s}^{5}-7364205035,{s}^{4}-13075015530
,{s}^{3}-7364205035,{s}^{2}-946111104,s+295612416
$$ has a root at approximately $7.421911191$. Hmm: Maple indicates that the Galois group of this sextic polynomial is solvable, so the root should have an expression in radicals.
There's also a root at approximately $0.1347361851$ which makes the two curves tangent:
On the other hand, the root at $0.4013844674$ does not appear to produce tangent curves: presumably in this case the multiple root is complex.
answered Jan 28 at 18:31
Robert IsraelRobert Israel
330k23218473
$endgroup$
$begingroup$
I found the coefficient with which to replace the $3$ in $-3y$ to get the two curves tangent. Came out about 7.42. There is probably some elegant way to do that, and for that matter correctly find the point of tangency that has been created this way
$endgroup$
– Will Jagy
Jan 28 at 19:49
add a comment |
$begingroup$
The resultant of the two left sides with respect to $y$ is
$$ -x left( 531441,{x}^{24}-2165130,{x}^{20}+7884864,{x}^{16}-
11409984,{x}^{12}+12205512,{x}^{8}+504218,{x}^{4}+489375 right)
$$
and using Sturm's theorem one can show that $x=0$ is the only real root of that. There are of course complex roots.
EDIT: With the $3$ in $-3y$ replaced by $s$, the resultant becomes
$$ eqalign{ -x &left( 6561,{s}^{4}{x}^{24}-18954,{s}^{4}{x}^{20}-23328,{s}^{3}{
x}^{20}+21708,{s}^{4}{x}^{16}+173340,{s}^{3}{x}^{16}right.cr &+160704,{s}^{2}
{x}^{16}-23976,{s}^{4}{x}^{12}-191016,{s}^{3}{x}^{12}-396000,{s}^{2
}{x}^{12}cr &
-248832,s{x}^{12}+10692,{s}^{4}{x}^{8}+24540,{s}^{3}{x}^{8
}+597520,{s}^{2}{x}^{8}+1517568,s{x}^{8}-5832,{s}^{4}{x}^{4}
cr &left.+746496
,{x}^{8}-14938,{s}^{3}{x}^{4}+47328,{s}^{2}{x}^{4}+337792,s{x}^{4}
-59392,{x}^{4}+18125,{s}^{3} right)}
$$
The discriminant of this with respect to $x$ is
$$ 16618447044007316502648165488386203399239363151697251980300315466139733122351104000000000000000000000000000000000000
,{s}^{83} left( 295612416,{s}^{6}-946111104,{s}^{5}-7364205035,{s
}^{4}-13075015530,{s}^{3}-7364205035,{s}^{2}-946111104,s+295612416
right) ^{4} left( 2772210825,{s}^{9}-14150310144,{s}^{8}+
50430196864,{s}^{7}+1618020643840,{s}^{6}+5389314830336,{s}^{5}+
11451158167552,{s}^{4}+18821256052736,{s}^{3}+28395282890752,{s}^{2
}+12309929918464,s-11093095219200 right) ^{8}
$$
The roots of this polynomial should correspond to values at which the two curves are tangent. In particular, the factor $$295612416,{s}^{6}-946111104,{s}^{5}-7364205035,{s}^{4}-13075015530
,{s}^{3}-7364205035,{s}^{2}-946111104,s+295612416
$$ has a root at approximately $7.421911191$. Hmm: Maple indicates that the Galois group of this sextic polynomial is solvable, so the root should have an expression in radicals.
There's also a root at approximately $0.1347361851$ which makes the two curves tangent:
On the other hand, the root at $0.4013844674$ does not appear to produce tangent curves: presumably in this case the multiple root is complex.
answered Jan 28 at 18:31
Robert IsraelRobert Israel
330k23218473
$endgroup$
$begingroup$
I found the coefficient with which to replace the $3$ in $-3y$ to get the two curves tangent. Came out about 7.42. There is probably some elegant way to do that, and for that matter correctly find the point of tangency that has been created this way
$endgroup$
– Will Jagy
Jan 28 at 19:49
add a comment |
$begingroup$
The resultant of the two left sides with respect to $y$ is
$$ -x left( 531441,{x}^{24}-2165130,{x}^{20}+7884864,{x}^{16}-
11409984,{x}^{12}+12205512,{x}^{8}+504218,{x}^{4}+489375 right)
$$
and using Sturm's theorem one can show that $x=0$ is the only real root of that. There are of course complex roots.
EDIT: With the $3$ in $-3y$ replaced by $s$, the resultant becomes
$$ eqalign{ -x &left( 6561,{s}^{4}{x}^{24}-18954,{s}^{4}{x}^{20}-23328,{s}^{3}{
x}^{20}+21708,{s}^{4}{x}^{16}+173340,{s}^{3}{x}^{16}right.cr &+160704,{s}^{2}
{x}^{16}-23976,{s}^{4}{x}^{12}-191016,{s}^{3}{x}^{12}-396000,{s}^{2
}{x}^{12}cr &
-248832,s{x}^{12}+10692,{s}^{4}{x}^{8}+24540,{s}^{3}{x}^{8
}+597520,{s}^{2}{x}^{8}+1517568,s{x}^{8}-5832,{s}^{4}{x}^{4}
cr &left.+746496
,{x}^{8}-14938,{s}^{3}{x}^{4}+47328,{s}^{2}{x}^{4}+337792,s{x}^{4}
-59392,{x}^{4}+18125,{s}^{3} right)}
$$
The discriminant of this with respect to $x$ is
$$ 16618447044007316502648165488386203399239363151697251980300315466139733122351104000000000000000000000000000000000000
,{s}^{83} left( 295612416,{s}^{6}-946111104,{s}^{5}-7364205035,{s
}^{4}-13075015530,{s}^{3}-7364205035,{s}^{2}-946111104,s+295612416
right) ^{4} left( 2772210825,{s}^{9}-14150310144,{s}^{8}+
50430196864,{s}^{7}+1618020643840,{s}^{6}+5389314830336,{s}^{5}+
11451158167552,{s}^{4}+18821256052736,{s}^{3}+28395282890752,{s}^{2
}+12309929918464,s-11093095219200 right) ^{8}
$$
The roots of this polynomial should correspond to values at which the two curves are tangent. In particular, the factor $$295612416,{s}^{6}-946111104,{s}^{5}-7364205035,{s}^{4}-13075015530
,{s}^{3}-7364205035,{s}^{2}-946111104,s+295612416
$$ has a root at approximately $7.421911191$. Hmm: Maple indicates that the Galois group of this sextic polynomial is solvable, so the root should have an expression in radicals.
There's also a root at approximately $0.1347361851$ which makes the two curves tangent:
On the other hand, the root at $0.4013844674$ does not appear to produce tangent curves: presumably in this case the multiple root is complex.
answered Jan 28 at 18:31
Robert IsraelRobert Israel
330k23218473
$endgroup$
The resultant of the two left sides with respect to $y$ is
$$ -x left( 531441,{x}^{24}-2165130,{x}^{20}+7884864,{x}^{16}-
11409984,{x}^{12}+12205512,{x}^{8}+504218,{x}^{4}+489375 right)
$$
and using Sturm's theorem one can show that $x=0$ is the only real root of that. There are of course complex roots.
EDIT: With the $3$ in $-3y$ replaced by $s$, the resultant becomes
$$ eqalign{ -x &left( 6561,{s}^{4}{x}^{24}-18954,{s}^{4}{x}^{20}-23328,{s}^{3}{
x}^{20}+21708,{s}^{4}{x}^{16}+173340,{s}^{3}{x}^{16}right.cr &+160704,{s}^{2}
{x}^{16}-23976,{s}^{4}{x}^{12}-191016,{s}^{3}{x}^{12}-396000,{s}^{2
}{x}^{12}cr &
-248832,s{x}^{12}+10692,{s}^{4}{x}^{8}+24540,{s}^{3}{x}^{8
}+597520,{s}^{2}{x}^{8}+1517568,s{x}^{8}-5832,{s}^{4}{x}^{4}
cr &left.+746496
,{x}^{8}-14938,{s}^{3}{x}^{4}+47328,{s}^{2}{x}^{4}+337792,s{x}^{4}
-59392,{x}^{4}+18125,{s}^{3} right)}
$$
The discriminant of this with respect to $x$ is
$$ 16618447044007316502648165488386203399239363151697251980300315466139733122351104000000000000000000000000000000000000
,{s}^{83} left( 295612416,{s}^{6}-946111104,{s}^{5}-7364205035,{s
}^{4}-13075015530,{s}^{3}-7364205035,{s}^{2}-946111104,s+295612416
right) ^{4} left( 2772210825,{s}^{9}-14150310144,{s}^{8}+
50430196864,{s}^{7}+1618020643840,{s}^{6}+5389314830336,{s}^{5}+
11451158167552,{s}^{4}+18821256052736,{s}^{3}+28395282890752,{s}^{2
}+12309929918464,s-11093095219200 right) ^{8}
$$
The roots of this polynomial should correspond to values at which the two curves are tangent. In particular, the factor $$295612416,{s}^{6}-946111104,{s}^{5}-7364205035,{s}^{4}-13075015530
,{s}^{3}-7364205035,{s}^{2}-946111104,s+295612416
$$ has a root at approximately $7.421911191$. Hmm: Maple indicates that the Galois group of this sextic polynomial is solvable, so the root should have an expression in radicals.
There's also a root at approximately $0.1347361851$ which makes the two curves tangent:
On the other hand, the root at $0.4013844674$ does not appear to produce tangent curves: presumably in this case the multiple root is complex.
answered Jan 28 at 18:31
Robert IsraelRobert Israel
330k23218473
edited Jan 29 at 0:25
answered Jan 28 at 18:31
Robert IsraelRobert Israel
330k23218473
answered Jan 28 at 18:31
Robert IsraelRobert Israel
330k23218473
answered Jan 28 at 18:31
Robert IsraelRobert Israel
330k23218473
330k23218473
$begingroup$
I found the coefficient with which to replace the $3$ in $-3y$ to get the two curves tangent. Came out about 7.42. There is probably some elegant way to do that, and for that matter correctly find the point of tangency that has been created this way
$endgroup$
– Will Jagy
Jan 28 at 19:49
add a comment |
$begingroup$
I found the coefficient with which to replace the $3$ in $-3y$ to get the two curves tangent. Came out about 7.42. There is probably some elegant way to do that, and for that matter correctly find the point of tangency that has been created this way
$endgroup$
– Will Jagy
Jan 28 at 19:49
$begingroup$
I found the coefficient with which to replace the $3$ in $-3y$ to get the two curves tangent. Came out about 7.42. There is probably some elegant way to do that, and for that matter correctly find the point of tangency that has been created this way
$endgroup$
– Will Jagy
Jan 28 at 19:49
$begingroup$
I found the coefficient with which to replace the $3$ in $-3y$ to get the two curves tangent. Came out about 7.42. There is probably some elegant way to do that, and for that matter correctly find the point of tangency that has been created this way
$endgroup$
– Will Jagy
Jan 28 at 19:49
add a comment |
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