Why does this method for solving cubics seem to fail?What's so special about primes $x^2+27y^2 = 31,43, 109, 157,dots$ for cubics?I know there are three real roots for cubic however cubic formula is giving me non-real answer. What am I doing wrong?Is it possible to write $cos left( frac13arccos frac3764-fracpi 3 right)$ as a radical expression of real numberA cubic equation with at most one integer rootIntegral involving multiple variablesAlgebraic Method for Inverse MapsArctangent addition formula for any range of the argumentsSolving $8x^3 - 6x + 1$ using Cardano's method
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Why does this method for solving cubics seem to fail?
What's so special about primes $x^2+27y^2 = 31,43, 109, 157,dots$ for cubics?I know there are three real roots for cubic however cubic formula is giving me non-real answer. What am I doing wrong?Is it possible to write $cos left( frac13arccos frac3764-fracpi 3 right)$ as a radical expression of real numberA cubic equation with at most one integer rootIntegral involving multiple variablesAlgebraic Method for Inverse MapsArctangent addition formula for any range of the argumentsSolving $8x^3 - 6x + 1$ using Cardano's method
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margin-bottom:0;
.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;
$begingroup$
I was trying to see if I could apply the method below to solve a certain subclass of the class of cubic equations with only one real root, and I came up with a block whose cause I couldn't find.
OK, so let $$y=ax^3+bx^2+cx+d,$$ where $a,b,c,d$ are real and $ane 0.$ Then $y'=3ax^2+2bx+c.$ Now the plan is to consider those cubics as defined above for which $y'$ has the same sign for all $x.$ This would ensure that an inverse function $$x=f(y)$$ exists, and the plan is to get this function by integration. Let's continue.
Now, for $y'>0,$ without loss of generality, we must have $(2b)^2-4(3a)clt 0,$ or $$3ac-b^2>0.$$ Thus, with $k^2=3ac-b^2,$ we have $$y'=3aleft[left(x+fracb3aright)^2+frack^29a^2right],$$ by attempting to complete squares. It then follows, since $y'ne 0$ for all $x,$ that $$x=intfracmathrm d y3aleft[left(y+fracb3aright)^2+frack^29a^2right]$$ exists for all $y.$ This integral can be easily evaluated to give $$frac1k arctanleft(frac3ayk+fracbkright)+C.$$ To determine the constant $C,$ note that with the original equation, we have $y=d$ when $x=0,$ so that we have $$C=-frac1k arctanleft(frac3adk+fracbkright),$$ so that $$x=frac1k arctanleft(frac3ayk+fracbkright)-frac1k arctanleft(frac3adk+fracbkright).$$
It is now an easy matter to find the real root $x$ of the original equation, which is the value of $x$ when $y=0.$ This gives the number $$x=frac1k arctanleft(fracbkright)-frac1k arctanleft(frac3adk+fracbkright).$$
Now consider applying this procedure to the example $$2x^3+3x^2+2x+3=0.$$ Clearly $3^2-3(2)(2)<0,$ so it is of the required type. Also, it is easy to see that it has only one real root, namely $x=-3/2.$ However, applying the method above, we obtain $k=sqrt 3,$ so that the root as given by the arctangent should be $$x=frac 1sqrt 3 arctanleft(frac3sqrt 3right)-frac1sqrt 3 arctanleft(frac21sqrt 3right).$$ If this is so, then we must have $$frac1sqrt 3 arctanleft(frac3sqrt 3right)-frac1sqrt 3 arctanleft(frac21sqrt 3right)=-frac 3 2,$$ which gives $$arctan(7sqrt 3)-arctansqrt 3=frac3sqrt 32,$$ which is clearly false.
I have checked again and again, but have failed to see where I went wrong. Please help me spot the false step. Many thanks!
PS. This method works seamlessly well with equations of first order, as can be easily checked; so I kept wondering where the analogy breaks down.
proof-verification indefinite-integrals inverse-function cubic-equations
asked Jul 7 at 22:54
AllawonderAllawonder
7,7861 gold badge11 silver badges21 bronze badges
$endgroup$
|
show 14 more comments
$begingroup$
I was trying to see if I could apply the method below to solve a certain subclass of the class of cubic equations with only one real root, and I came up with a block whose cause I couldn't find.
OK, so let $$y=ax^3+bx^2+cx+d,$$ where $a,b,c,d$ are real and $ane 0.$ Then $y'=3ax^2+2bx+c.$ Now the plan is to consider those cubics as defined above for which $y'$ has the same sign for all $x.$ This would ensure that an inverse function $$x=f(y)$$ exists, and the plan is to get this function by integration. Let's continue.
Now, for $y'>0,$ without loss of generality, we must have $(2b)^2-4(3a)clt 0,$ or $$3ac-b^2>0.$$ Thus, with $k^2=3ac-b^2,$ we have $$y'=3aleft[left(x+fracb3aright)^2+frack^29a^2right],$$ by attempting to complete squares. It then follows, since $y'ne 0$ for all $x,$ that $$x=intfracmathrm d y3aleft[left(y+fracb3aright)^2+frack^29a^2right]$$ exists for all $y.$ This integral can be easily evaluated to give $$frac1k arctanleft(frac3ayk+fracbkright)+C.$$ To determine the constant $C,$ note that with the original equation, we have $y=d$ when $x=0,$ so that we have $$C=-frac1k arctanleft(frac3adk+fracbkright),$$ so that $$x=frac1k arctanleft(frac3ayk+fracbkright)-frac1k arctanleft(frac3adk+fracbkright).$$
It is now an easy matter to find the real root $x$ of the original equation, which is the value of $x$ when $y=0.$ This gives the number $$x=frac1k arctanleft(fracbkright)-frac1k arctanleft(frac3adk+fracbkright).$$
Now consider applying this procedure to the example $$2x^3+3x^2+2x+3=0.$$ Clearly $3^2-3(2)(2)<0,$ so it is of the required type. Also, it is easy to see that it has only one real root, namely $x=-3/2.$ However, applying the method above, we obtain $k=sqrt 3,$ so that the root as given by the arctangent should be $$x=frac 1sqrt 3 arctanleft(frac3sqrt 3right)-frac1sqrt 3 arctanleft(frac21sqrt 3right).$$ If this is so, then we must have $$frac1sqrt 3 arctanleft(frac3sqrt 3right)-frac1sqrt 3 arctanleft(frac21sqrt 3right)=-frac 3 2,$$ which gives $$arctan(7sqrt 3)-arctansqrt 3=frac3sqrt 32,$$ which is clearly false.
I have checked again and again, but have failed to see where I went wrong. Please help me spot the false step. Many thanks!
PS. This method works seamlessly well with equations of first order, as can be easily checked; so I kept wondering where the analogy breaks down.
proof-verification indefinite-integrals inverse-function cubic-equations
asked Jul 7 at 22:54
AllawonderAllawonder
7,7861 gold badge11 silver badges21 bronze badges
$endgroup$
$begingroup$
Interesting approach. Could it be applied to quadratic equations? Just wondering how was "c" calculated in the expression: y'=3a[...] near the top? Thanks.
$endgroup$
– NoChance
Jul 7 at 23:19
5
$begingroup$
Sorry if I'm missing something obvious: how did you obtain that $$x=intfracmathrm d y3aleft[left(y+fracb3aright)^2+frack^29a^2right]?$$
$endgroup$
– YiFan
Jul 7 at 23:33
$begingroup$
I guess this would be from the definition of the inverse function x=f(y).
$endgroup$
– NoChance
Jul 7 at 23:39
2
$begingroup$
Error. It should be $k^2 =3ac-b^2$.
$endgroup$
– steven gregory
Jul 8 at 0:19
1
$begingroup$
@stevengregory Oh, my. Thank you. That's what I'd meant to type -- I actually used the correct form in my work, as you can confirm. That is, this error is not the source of the discrepancy described. I will presently correct it.
$endgroup$
– Allawonder
Jul 8 at 0:22
|
show 14 more comments
$begingroup$
I was trying to see if I could apply the method below to solve a certain subclass of the class of cubic equations with only one real root, and I came up with a block whose cause I couldn't find.
OK, so let $$y=ax^3+bx^2+cx+d,$$ where $a,b,c,d$ are real and $ane 0.$ Then $y'=3ax^2+2bx+c.$ Now the plan is to consider those cubics as defined above for which $y'$ has the same sign for all $x.$ This would ensure that an inverse function $$x=f(y)$$ exists, and the plan is to get this function by integration. Let's continue.
Now, for $y'>0,$ without loss of generality, we must have $(2b)^2-4(3a)clt 0,$ or $$3ac-b^2>0.$$ Thus, with $k^2=3ac-b^2,$ we have $$y'=3aleft[left(x+fracb3aright)^2+frack^29a^2right],$$ by attempting to complete squares. It then follows, since $y'ne 0$ for all $x,$ that $$x=intfracmathrm d y3aleft[left(y+fracb3aright)^2+frack^29a^2right]$$ exists for all $y.$ This integral can be easily evaluated to give $$frac1k arctanleft(frac3ayk+fracbkright)+C.$$ To determine the constant $C,$ note that with the original equation, we have $y=d$ when $x=0,$ so that we have $$C=-frac1k arctanleft(frac3adk+fracbkright),$$ so that $$x=frac1k arctanleft(frac3ayk+fracbkright)-frac1k arctanleft(frac3adk+fracbkright).$$
It is now an easy matter to find the real root $x$ of the original equation, which is the value of $x$ when $y=0.$ This gives the number $$x=frac1k arctanleft(fracbkright)-frac1k arctanleft(frac3adk+fracbkright).$$
Now consider applying this procedure to the example $$2x^3+3x^2+2x+3=0.$$ Clearly $3^2-3(2)(2)<0,$ so it is of the required type. Also, it is easy to see that it has only one real root, namely $x=-3/2.$ However, applying the method above, we obtain $k=sqrt 3,$ so that the root as given by the arctangent should be $$x=frac 1sqrt 3 arctanleft(frac3sqrt 3right)-frac1sqrt 3 arctanleft(frac21sqrt 3right).$$ If this is so, then we must have $$frac1sqrt 3 arctanleft(frac3sqrt 3right)-frac1sqrt 3 arctanleft(frac21sqrt 3right)=-frac 3 2,$$ which gives $$arctan(7sqrt 3)-arctansqrt 3=frac3sqrt 32,$$ which is clearly false.
I have checked again and again, but have failed to see where I went wrong. Please help me spot the false step. Many thanks!
PS. This method works seamlessly well with equations of first order, as can be easily checked; so I kept wondering where the analogy breaks down.
proof-verification indefinite-integrals inverse-function cubic-equations
asked Jul 7 at 22:54
AllawonderAllawonder
7,7861 gold badge11 silver badges21 bronze badges
$endgroup$
I was trying to see if I could apply the method below to solve a certain subclass of the class of cubic equations with only one real root, and I came up with a block whose cause I couldn't find.
OK, so let $$y=ax^3+bx^2+cx+d,$$ where $a,b,c,d$ are real and $ane 0.$ Then $y'=3ax^2+2bx+c.$ Now the plan is to consider those cubics as defined above for which $y'$ has the same sign for all $x.$ This would ensure that an inverse function $$x=f(y)$$ exists, and the plan is to get this function by integration. Let's continue.
Now, for $y'>0,$ without loss of generality, we must have $(2b)^2-4(3a)clt 0,$ or $$3ac-b^2>0.$$ Thus, with $k^2=3ac-b^2,$ we have $$y'=3aleft[left(x+fracb3aright)^2+frack^29a^2right],$$ by attempting to complete squares. It then follows, since $y'ne 0$ for all $x,$ that $$x=intfracmathrm d y3aleft[left(y+fracb3aright)^2+frack^29a^2right]$$ exists for all $y.$ This integral can be easily evaluated to give $$frac1k arctanleft(frac3ayk+fracbkright)+C.$$ To determine the constant $C,$ note that with the original equation, we have $y=d$ when $x=0,$ so that we have $$C=-frac1k arctanleft(frac3adk+fracbkright),$$ so that $$x=frac1k arctanleft(frac3ayk+fracbkright)-frac1k arctanleft(frac3adk+fracbkright).$$
It is now an easy matter to find the real root $x$ of the original equation, which is the value of $x$ when $y=0.$ This gives the number $$x=frac1k arctanleft(fracbkright)-frac1k arctanleft(frac3adk+fracbkright).$$
Now consider applying this procedure to the example $$2x^3+3x^2+2x+3=0.$$ Clearly $3^2-3(2)(2)<0,$ so it is of the required type. Also, it is easy to see that it has only one real root, namely $x=-3/2.$ However, applying the method above, we obtain $k=sqrt 3,$ so that the root as given by the arctangent should be $$x=frac 1sqrt 3 arctanleft(frac3sqrt 3right)-frac1sqrt 3 arctanleft(frac21sqrt 3right).$$ If this is so, then we must have $$frac1sqrt 3 arctanleft(frac3sqrt 3right)-frac1sqrt 3 arctanleft(frac21sqrt 3right)=-frac 3 2,$$ which gives $$arctan(7sqrt 3)-arctansqrt 3=frac3sqrt 32,$$ which is clearly false.
I have checked again and again, but have failed to see where I went wrong. Please help me spot the false step. Many thanks!
PS. This method works seamlessly well with equations of first order, as can be easily checked; so I kept wondering where the analogy breaks down.
proof-verification indefinite-integrals inverse-function cubic-equations
proof-verification indefinite-integrals inverse-function cubic-equations
asked Jul 7 at 22:54
AllawonderAllawonder
7,7861 gold badge11 silver badges21 bronze badges
asked Jul 7 at 22:54
AllawonderAllawonder
7,7861 gold badge11 silver badges21 bronze badges
edited Jul 8 at 0:23
Allawonder
asked Jul 7 at 22:54
AllawonderAllawonder
7,7861 gold badge11 silver badges21 bronze badges
asked Jul 7 at 22:54
AllawonderAllawonder
7,7861 gold badge11 silver badges21 bronze badges
asked Jul 7 at 22:54
AllawonderAllawonder
7,7861 gold badge11 silver badges21 bronze badges
7,7861 gold badge11 silver badges21 bronze badges
$begingroup$
Interesting approach. Could it be applied to quadratic equations? Just wondering how was "c" calculated in the expression: y'=3a[...] near the top? Thanks.
$endgroup$
– NoChance
Jul 7 at 23:19
5
$begingroup$
Sorry if I'm missing something obvious: how did you obtain that $$x=intfracmathrm d y3aleft[left(y+fracb3aright)^2+frack^29a^2right]?$$
$endgroup$
– YiFan
Jul 7 at 23:33
$begingroup$
I guess this would be from the definition of the inverse function x=f(y).
$endgroup$
– NoChance
Jul 7 at 23:39
2
$begingroup$
Error. It should be $k^2 =3ac-b^2$.
$endgroup$
– steven gregory
Jul 8 at 0:19
1
$begingroup$
@stevengregory Oh, my. Thank you. That's what I'd meant to type -- I actually used the correct form in my work, as you can confirm. That is, this error is not the source of the discrepancy described. I will presently correct it.
$endgroup$
– Allawonder
Jul 8 at 0:22
|
show 14 more comments
$begingroup$
Interesting approach. Could it be applied to quadratic equations? Just wondering how was "c" calculated in the expression: y'=3a[...] near the top? Thanks.
$endgroup$
– NoChance
Jul 7 at 23:19
5
$begingroup$
Sorry if I'm missing something obvious: how did you obtain that $$x=intfracmathrm d y3aleft[left(y+fracb3aright)^2+frack^29a^2right]?$$
$endgroup$
– YiFan
Jul 7 at 23:33
$begingroup$
I guess this would be from the definition of the inverse function x=f(y).
$endgroup$
– NoChance
Jul 7 at 23:39
2
$begingroup$
Error. It should be $k^2 =3ac-b^2$.
$endgroup$
– steven gregory
Jul 8 at 0:19
1
$begingroup$
@stevengregory Oh, my. Thank you. That's what I'd meant to type -- I actually used the correct form in my work, as you can confirm. That is, this error is not the source of the discrepancy described. I will presently correct it.
$endgroup$
– Allawonder
Jul 8 at 0:22
$begingroup$
Interesting approach. Could it be applied to quadratic equations? Just wondering how was "c" calculated in the expression: y'=3a[...] near the top? Thanks.
$endgroup$
– NoChance
Jul 7 at 23:19
$begingroup$
Interesting approach. Could it be applied to quadratic equations? Just wondering how was "c" calculated in the expression: y'=3a[...] near the top? Thanks.
$endgroup$
– NoChance
Jul 7 at 23:19
5
5
$begingroup$
Sorry if I'm missing something obvious: how did you obtain that $$x=intfracmathrm d y3aleft[left(y+fracb3aright)^2+frack^29a^2right]?$$
$endgroup$
– YiFan
Jul 7 at 23:33
$begingroup$
Sorry if I'm missing something obvious: how did you obtain that $$x=intfracmathrm d y3aleft[left(y+fracb3aright)^2+frack^29a^2right]?$$
$endgroup$
– YiFan
Jul 7 at 23:33
$begingroup$
I guess this would be from the definition of the inverse function x=f(y).
$endgroup$
– NoChance
Jul 7 at 23:39
$begingroup$
I guess this would be from the definition of the inverse function x=f(y).
$endgroup$
– NoChance
Jul 7 at 23:39
2
2
$begingroup$
Error. It should be $k^2 =3ac-b^2$.
$endgroup$
– steven gregory
Jul 8 at 0:19
$begingroup$
Error. It should be $k^2 =3ac-b^2$.
$endgroup$
– steven gregory
Jul 8 at 0:19
1
1
$begingroup$
@stevengregory Oh, my. Thank you. That's what I'd meant to type -- I actually used the correct form in my work, as you can confirm. That is, this error is not the source of the discrepancy described. I will presently correct it.
$endgroup$
– Allawonder
Jul 8 at 0:22
$begingroup$
@stevengregory Oh, my. Thank you. That's what I'd meant to type -- I actually used the correct form in my work, as you can confirm. That is, this error is not the source of the discrepancy described. I will presently correct it.
$endgroup$
– Allawonder
Jul 8 at 0:22
|
show 14 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The observation of @YiFan is right.
You actually have
$$
eqalign
& y' = 3aleft( left( x + b over 3a right)^,2 + k^,2 over left( 3a right)^,2 right)quad Rightarrow cr
& Rightarrow quad dx = dy over 3aleft( left( x + b over 3a right)^,2 + k^,2 over left( 3a right)^,2 right) cr
$$
where you shall read the $x$ at the denominator as $x(y)$.
answered Jul 8 at 0:08
G CabG Cab
23.9k3 gold badges13 silver badges45 bronze badges
$endgroup$
$begingroup$
My thinking was this: I integrated $1/y'$ from some fixed point $p$ to a variable $y.$ That is, I treated the $x$ in the denominator as a dummy variable. That is equivalent to seeking a primitive and using FTC as displayed above, isn't it? But I appreciate your input. Will look into it.
$endgroup$
– Allawonder
Jul 8 at 0:15
2
$begingroup$
@Allawonder: You can't just wishfully turn $x$ into $y$. This is meaningless.
$endgroup$
– Ted Shifrin
Jul 8 at 0:27
6
$begingroup$
@Allawonder: " .. I treated the $x$ in the denominator as a dummy variable .." : you can't, it depends on $y$ !
$endgroup$
– G Cab
Jul 8 at 0:36
$begingroup$
Yes, I have convinced myself that this was indeed where my problem was. Turns out there's no easy way here too, beyond the linear equations, which is nothing doing. Alright, thank you.
$endgroup$
– Allawonder
Jul 8 at 1:32
$begingroup$
@Allawonder: sorry I had to pin down that, because otherwise your "plan of attack" was interesting.
$endgroup$
– G Cab
Jul 8 at 9:55
|
show 1 more comment
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$begingroup$
The observation of @YiFan is right.
You actually have
$$
eqalign
& y' = 3aleft( left( x + b over 3a right)^,2 + k^,2 over left( 3a right)^,2 right)quad Rightarrow cr
& Rightarrow quad dx = dy over 3aleft( left( x + b over 3a right)^,2 + k^,2 over left( 3a right)^,2 right) cr
$$
where you shall read the $x$ at the denominator as $x(y)$.
answered Jul 8 at 0:08
G CabG Cab
23.9k3 gold badges13 silver badges45 bronze badges
$endgroup$
$begingroup$
My thinking was this: I integrated $1/y'$ from some fixed point $p$ to a variable $y.$ That is, I treated the $x$ in the denominator as a dummy variable. That is equivalent to seeking a primitive and using FTC as displayed above, isn't it? But I appreciate your input. Will look into it.
$endgroup$
– Allawonder
Jul 8 at 0:15
2
$begingroup$
@Allawonder: You can't just wishfully turn $x$ into $y$. This is meaningless.
$endgroup$
– Ted Shifrin
Jul 8 at 0:27
6
$begingroup$
@Allawonder: " .. I treated the $x$ in the denominator as a dummy variable .." : you can't, it depends on $y$ !
$endgroup$
– G Cab
Jul 8 at 0:36
$begingroup$
Yes, I have convinced myself that this was indeed where my problem was. Turns out there's no easy way here too, beyond the linear equations, which is nothing doing. Alright, thank you.
$endgroup$
– Allawonder
Jul 8 at 1:32
$begingroup$
@Allawonder: sorry I had to pin down that, because otherwise your "plan of attack" was interesting.
$endgroup$
– G Cab
Jul 8 at 9:55
|
show 1 more comment
$begingroup$
The observation of @YiFan is right.
You actually have
$$
eqalign
& y' = 3aleft( left( x + b over 3a right)^,2 + k^,2 over left( 3a right)^,2 right)quad Rightarrow cr
& Rightarrow quad dx = dy over 3aleft( left( x + b over 3a right)^,2 + k^,2 over left( 3a right)^,2 right) cr
$$
where you shall read the $x$ at the denominator as $x(y)$.
answered Jul 8 at 0:08
G CabG Cab
23.9k3 gold badges13 silver badges45 bronze badges
$endgroup$
$begingroup$
My thinking was this: I integrated $1/y'$ from some fixed point $p$ to a variable $y.$ That is, I treated the $x$ in the denominator as a dummy variable. That is equivalent to seeking a primitive and using FTC as displayed above, isn't it? But I appreciate your input. Will look into it.
$endgroup$
– Allawonder
Jul 8 at 0:15
2
$begingroup$
@Allawonder: You can't just wishfully turn $x$ into $y$. This is meaningless.
$endgroup$
– Ted Shifrin
Jul 8 at 0:27
6
$begingroup$
@Allawonder: " .. I treated the $x$ in the denominator as a dummy variable .." : you can't, it depends on $y$ !
$endgroup$
– G Cab
Jul 8 at 0:36
$begingroup$
Yes, I have convinced myself that this was indeed where my problem was. Turns out there's no easy way here too, beyond the linear equations, which is nothing doing. Alright, thank you.
$endgroup$
– Allawonder
Jul 8 at 1:32
$begingroup$
@Allawonder: sorry I had to pin down that, because otherwise your "plan of attack" was interesting.
$endgroup$
– G Cab
Jul 8 at 9:55
|
show 1 more comment
$begingroup$
The observation of @YiFan is right.
You actually have
$$
eqalign
& y' = 3aleft( left( x + b over 3a right)^,2 + k^,2 over left( 3a right)^,2 right)quad Rightarrow cr
& Rightarrow quad dx = dy over 3aleft( left( x + b over 3a right)^,2 + k^,2 over left( 3a right)^,2 right) cr
$$
where you shall read the $x$ at the denominator as $x(y)$.
answered Jul 8 at 0:08
G CabG Cab
23.9k3 gold badges13 silver badges45 bronze badges
$endgroup$
The observation of @YiFan is right.
You actually have
$$
eqalign
& y' = 3aleft( left( x + b over 3a right)^,2 + k^,2 over left( 3a right)^,2 right)quad Rightarrow cr
& Rightarrow quad dx = dy over 3aleft( left( x + b over 3a right)^,2 + k^,2 over left( 3a right)^,2 right) cr
$$
where you shall read the $x$ at the denominator as $x(y)$.
answered Jul 8 at 0:08
G CabG Cab
23.9k3 gold badges13 silver badges45 bronze badges
answered Jul 8 at 0:08
G CabG Cab
23.9k3 gold badges13 silver badges45 bronze badges
answered Jul 8 at 0:08
G CabG Cab
23.9k3 gold badges13 silver badges45 bronze badges
answered Jul 8 at 0:08
G CabG Cab
23.9k3 gold badges13 silver badges45 bronze badges
23.9k3 gold badges13 silver badges45 bronze badges
$begingroup$
My thinking was this: I integrated $1/y'$ from some fixed point $p$ to a variable $y.$ That is, I treated the $x$ in the denominator as a dummy variable. That is equivalent to seeking a primitive and using FTC as displayed above, isn't it? But I appreciate your input. Will look into it.
$endgroup$
– Allawonder
Jul 8 at 0:15
2
$begingroup$
@Allawonder: You can't just wishfully turn $x$ into $y$. This is meaningless.
$endgroup$
– Ted Shifrin
Jul 8 at 0:27
6
$begingroup$
@Allawonder: " .. I treated the $x$ in the denominator as a dummy variable .." : you can't, it depends on $y$ !
$endgroup$
– G Cab
Jul 8 at 0:36
$begingroup$
Yes, I have convinced myself that this was indeed where my problem was. Turns out there's no easy way here too, beyond the linear equations, which is nothing doing. Alright, thank you.
$endgroup$
– Allawonder
Jul 8 at 1:32
$begingroup$
@Allawonder: sorry I had to pin down that, because otherwise your "plan of attack" was interesting.
$endgroup$
– G Cab
Jul 8 at 9:55
|
show 1 more comment
$begingroup$
My thinking was this: I integrated $1/y'$ from some fixed point $p$ to a variable $y.$ That is, I treated the $x$ in the denominator as a dummy variable. That is equivalent to seeking a primitive and using FTC as displayed above, isn't it? But I appreciate your input. Will look into it.
$endgroup$
– Allawonder
Jul 8 at 0:15
2
$begingroup$
@Allawonder: You can't just wishfully turn $x$ into $y$. This is meaningless.
$endgroup$
– Ted Shifrin
Jul 8 at 0:27
6
$begingroup$
@Allawonder: " .. I treated the $x$ in the denominator as a dummy variable .." : you can't, it depends on $y$ !
$endgroup$
– G Cab
Jul 8 at 0:36
$begingroup$
Yes, I have convinced myself that this was indeed where my problem was. Turns out there's no easy way here too, beyond the linear equations, which is nothing doing. Alright, thank you.
$endgroup$
– Allawonder
Jul 8 at 1:32
$begingroup$
@Allawonder: sorry I had to pin down that, because otherwise your "plan of attack" was interesting.
$endgroup$
– G Cab
Jul 8 at 9:55
$begingroup$
My thinking was this: I integrated $1/y'$ from some fixed point $p$ to a variable $y.$ That is, I treated the $x$ in the denominator as a dummy variable. That is equivalent to seeking a primitive and using FTC as displayed above, isn't it? But I appreciate your input. Will look into it.
$endgroup$
– Allawonder
Jul 8 at 0:15
$begingroup$
My thinking was this: I integrated $1/y'$ from some fixed point $p$ to a variable $y.$ That is, I treated the $x$ in the denominator as a dummy variable. That is equivalent to seeking a primitive and using FTC as displayed above, isn't it? But I appreciate your input. Will look into it.
$endgroup$
– Allawonder
Jul 8 at 0:15
2
2
$begingroup$
@Allawonder: You can't just wishfully turn $x$ into $y$. This is meaningless.
$endgroup$
– Ted Shifrin
Jul 8 at 0:27
$begingroup$
@Allawonder: You can't just wishfully turn $x$ into $y$. This is meaningless.
$endgroup$
– Ted Shifrin
Jul 8 at 0:27
6
6
$begingroup$
@Allawonder: " .. I treated the $x$ in the denominator as a dummy variable .." : you can't, it depends on $y$ !
$endgroup$
– G Cab
Jul 8 at 0:36
$begingroup$
@Allawonder: " .. I treated the $x$ in the denominator as a dummy variable .." : you can't, it depends on $y$ !
$endgroup$
– G Cab
Jul 8 at 0:36
$begingroup$
Yes, I have convinced myself that this was indeed where my problem was. Turns out there's no easy way here too, beyond the linear equations, which is nothing doing. Alright, thank you.
$endgroup$
– Allawonder
Jul 8 at 1:32
$begingroup$
Yes, I have convinced myself that this was indeed where my problem was. Turns out there's no easy way here too, beyond the linear equations, which is nothing doing. Alright, thank you.
$endgroup$
– Allawonder
Jul 8 at 1:32
$begingroup$
@Allawonder: sorry I had to pin down that, because otherwise your "plan of attack" was interesting.
$endgroup$
– G Cab
Jul 8 at 9:55
$begingroup$
@Allawonder: sorry I had to pin down that, because otherwise your "plan of attack" was interesting.
$endgroup$
– G Cab
Jul 8 at 9:55
|
show 1 more comment
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$begingroup$
Interesting approach. Could it be applied to quadratic equations? Just wondering how was "c" calculated in the expression: y'=3a[...] near the top? Thanks.
$endgroup$
– NoChance
Jul 7 at 23:19
5
$begingroup$
Sorry if I'm missing something obvious: how did you obtain that $$x=intfracmathrm d y3aleft[left(y+fracb3aright)^2+frack^29a^2right]?$$
$endgroup$
– YiFan
Jul 7 at 23:33
$begingroup$
I guess this would be from the definition of the inverse function x=f(y).
$endgroup$
– NoChance
Jul 7 at 23:39
2
$begingroup$
Error. It should be $k^2 =3ac-b^2$.
$endgroup$
– steven gregory
Jul 8 at 0:19
1
$begingroup$
@stevengregory Oh, my. Thank you. That's what I'd meant to type -- I actually used the correct form in my work, as you can confirm. That is, this error is not the source of the discrepancy described. I will presently correct it.
$endgroup$
– Allawonder
Jul 8 at 0:22