Prove that there is a solution that depends smoothly on the coefficients
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Consider the equation $x^4+a_0x^3+a_1x^2+2a_2x+a_3=0$. Prove that there is $delta > 0$ such that if $|a_i-1| < delta$ for $i=0,1,2,3$, then the above equation has a solution that depends smoothly on the $a_i$'s.
I assume this is an exercise on the Implicit function theorem, but I don't know how to apply it in this case. I thought about the function $mathbb R^4to mathbb R, (a_0,a_1,a_2,a_3)mapsto x^4+a_0x^3+a_1x^2+2a_2x+a_3$, but the theorem requires that the function vanishes at some point, whereas this function contains a formal variable and cannot vanish.
calculus real-analysis multivariable-calculus implicit-function-theorem
asked Aug 12 at 1:53
user557902851
407117
add a comment |Â
up vote
1
down vote
favorite
Consider the equation $x^4+a_0x^3+a_1x^2+2a_2x+a_3=0$. Prove that there is $delta > 0$ such that if $|a_i-1| < delta$ for $i=0,1,2,3$, then the above equation has a solution that depends smoothly on the $a_i$'s.
I assume this is an exercise on the Implicit function theorem, but I don't know how to apply it in this case. I thought about the function $mathbb R^4to mathbb R, (a_0,a_1,a_2,a_3)mapsto x^4+a_0x^3+a_1x^2+2a_2x+a_3$, but the theorem requires that the function vanishes at some point, whereas this function contains a formal variable and cannot vanish.
calculus real-analysis multivariable-calculus implicit-function-theorem
asked Aug 12 at 1:53
user557902851
407117
Note: If all $a_i$ are equal to $1$, then $x=-1$ is a root.
– quasi
Aug 12 at 1:58
1
The map you want is probably from the coefficients to the four roots, not to the polynomial. In any case the crux of the problem lies in showing that there are no repeated roots when all $a_i$ are approximately equal to one.
– hardmath
Aug 12 at 1:59
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the equation $x^4+a_0x^3+a_1x^2+2a_2x+a_3=0$. Prove that there is $delta > 0$ such that if $|a_i-1| < delta$ for $i=0,1,2,3$, then the above equation has a solution that depends smoothly on the $a_i$'s.
I assume this is an exercise on the Implicit function theorem, but I don't know how to apply it in this case. I thought about the function $mathbb R^4to mathbb R, (a_0,a_1,a_2,a_3)mapsto x^4+a_0x^3+a_1x^2+2a_2x+a_3$, but the theorem requires that the function vanishes at some point, whereas this function contains a formal variable and cannot vanish.
calculus real-analysis multivariable-calculus implicit-function-theorem
asked Aug 12 at 1:53
user557902851
407117
Consider the equation $x^4+a_0x^3+a_1x^2+2a_2x+a_3=0$. Prove that there is $delta > 0$ such that if $|a_i-1| < delta$ for $i=0,1,2,3$, then the above equation has a solution that depends smoothly on the $a_i$'s.
I assume this is an exercise on the Implicit function theorem, but I don't know how to apply it in this case. I thought about the function $mathbb R^4to mathbb R, (a_0,a_1,a_2,a_3)mapsto x^4+a_0x^3+a_1x^2+2a_2x+a_3$, but the theorem requires that the function vanishes at some point, whereas this function contains a formal variable and cannot vanish.
calculus real-analysis multivariable-calculus implicit-function-theorem
asked Aug 12 at 1:53
user557902851
407117
asked Aug 12 at 1:53
user557902851
407117
asked Aug 12 at 1:53
user557902851
407117
asked Aug 12 at 1:53
user557902851
407117
407117
Note: If all $a_i$ are equal to $1$, then $x=-1$ is a root.
– quasi
Aug 12 at 1:58
1
The map you want is probably from the coefficients to the four roots, not to the polynomial. In any case the crux of the problem lies in showing that there are no repeated roots when all $a_i$ are approximately equal to one.
– hardmath
Aug 12 at 1:59
add a comment |Â
Note: If all $a_i$ are equal to $1$, then $x=-1$ is a root.
– quasi
Aug 12 at 1:58
1
The map you want is probably from the coefficients to the four roots, not to the polynomial. In any case the crux of the problem lies in showing that there are no repeated roots when all $a_i$ are approximately equal to one.
– hardmath
Aug 12 at 1:59
Note: If all $a_i$ are equal to $1$, then $x=-1$ is a root.
– quasi
Aug 12 at 1:58
Note: If all $a_i$ are equal to $1$, then $x=-1$ is a root.
– quasi
Aug 12 at 1:58
1
1
The map you want is probably from the coefficients to the four roots, not to the polynomial. In any case the crux of the problem lies in showing that there are no repeated roots when all $a_i$ are approximately equal to one.
– hardmath
Aug 12 at 1:59
The map you want is probably from the coefficients to the four roots, not to the polynomial. In any case the crux of the problem lies in showing that there are no repeated roots when all $a_i$ are approximately equal to one.
– hardmath
Aug 12 at 1:59
add a comment |Â
1 Answer
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The function you have defined is not a map to $mathbbR$. It is a map to $mathbbR[x]$. Almost certainly you want a map $mathbbR^5 to mathbbR$ which is defined by $(a_0, a_1, a_2, a_3, x) mapsto x^4 + a_0x^3 + a_1x^2 + 2a_2x + a_3 $. In this case, if you think of this as a function $mathbbR^4+1 to mathbbR^1$, you'll find the implicit function theorem to be a little more friendly. In particular, if you take for example all the $a_i = 1$, you can find a solution given by $x = -1$. Now you just need to show that $x$ can be eliminated in terms of the $a_i$ by using the implicit function theorem, i.e. checking that the $x$ derivative of this function of $5$ variables has a non-zero derivative at the point in question (as it is clearly more than just $C^1$, it is $C^infty$ as a polynomial in these variables).
answered Aug 12 at 2:02
Alfred Yerger
9,7972044
Thanks! What point do you mean? I believe I can take $p=(1,1,1,1,-1)$. In this case, writing the function you defined as $F$, we have $F(p)=0$ and $F_x(p)ne 0$. Then by the Implicit function theorem, there is $delta > 0$ and a smooth $phi: V=(1-delta,1+delta)^4to mathbb R$ with $phi(1,1,1,1)=-1$ and $F(a_0,a_1,a_2,a_3,phi(a_0,a_1,a_2,a_3))=0$ for all $(a_0,a_1,a_2,a_3)in V$. So $phi(a_0,a_1,a_2,a_3)$ is the local solution we need.
– user557902851
Aug 12 at 2:38
1
Yes, that point. This is pretty much exactly what I had in mind.
– Alfred Yerger
Aug 12 at 2:44
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The function you have defined is not a map to $mathbbR$. It is a map to $mathbbR[x]$. Almost certainly you want a map $mathbbR^5 to mathbbR$ which is defined by $(a_0, a_1, a_2, a_3, x) mapsto x^4 + a_0x^3 + a_1x^2 + 2a_2x + a_3 $. In this case, if you think of this as a function $mathbbR^4+1 to mathbbR^1$, you'll find the implicit function theorem to be a little more friendly. In particular, if you take for example all the $a_i = 1$, you can find a solution given by $x = -1$. Now you just need to show that $x$ can be eliminated in terms of the $a_i$ by using the implicit function theorem, i.e. checking that the $x$ derivative of this function of $5$ variables has a non-zero derivative at the point in question (as it is clearly more than just $C^1$, it is $C^infty$ as a polynomial in these variables).
answered Aug 12 at 2:02
Alfred Yerger
9,7972044
Thanks! What point do you mean? I believe I can take $p=(1,1,1,1,-1)$. In this case, writing the function you defined as $F$, we have $F(p)=0$ and $F_x(p)ne 0$. Then by the Implicit function theorem, there is $delta > 0$ and a smooth $phi: V=(1-delta,1+delta)^4to mathbb R$ with $phi(1,1,1,1)=-1$ and $F(a_0,a_1,a_2,a_3,phi(a_0,a_1,a_2,a_3))=0$ for all $(a_0,a_1,a_2,a_3)in V$. So $phi(a_0,a_1,a_2,a_3)$ is the local solution we need.
– user557902851
Aug 12 at 2:38
1
Yes, that point. This is pretty much exactly what I had in mind.
– Alfred Yerger
Aug 12 at 2:44
add a comment |Â
up vote
2
down vote
accepted
The function you have defined is not a map to $mathbbR$. It is a map to $mathbbR[x]$. Almost certainly you want a map $mathbbR^5 to mathbbR$ which is defined by $(a_0, a_1, a_2, a_3, x) mapsto x^4 + a_0x^3 + a_1x^2 + 2a_2x + a_3 $. In this case, if you think of this as a function $mathbbR^4+1 to mathbbR^1$, you'll find the implicit function theorem to be a little more friendly. In particular, if you take for example all the $a_i = 1$, you can find a solution given by $x = -1$. Now you just need to show that $x$ can be eliminated in terms of the $a_i$ by using the implicit function theorem, i.e. checking that the $x$ derivative of this function of $5$ variables has a non-zero derivative at the point in question (as it is clearly more than just $C^1$, it is $C^infty$ as a polynomial in these variables).
answered Aug 12 at 2:02
Alfred Yerger
9,7972044
Thanks! What point do you mean? I believe I can take $p=(1,1,1,1,-1)$. In this case, writing the function you defined as $F$, we have $F(p)=0$ and $F_x(p)ne 0$. Then by the Implicit function theorem, there is $delta > 0$ and a smooth $phi: V=(1-delta,1+delta)^4to mathbb R$ with $phi(1,1,1,1)=-1$ and $F(a_0,a_1,a_2,a_3,phi(a_0,a_1,a_2,a_3))=0$ for all $(a_0,a_1,a_2,a_3)in V$. So $phi(a_0,a_1,a_2,a_3)$ is the local solution we need.
– user557902851
Aug 12 at 2:38
1
Yes, that point. This is pretty much exactly what I had in mind.
– Alfred Yerger
Aug 12 at 2:44
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The function you have defined is not a map to $mathbbR$. It is a map to $mathbbR[x]$. Almost certainly you want a map $mathbbR^5 to mathbbR$ which is defined by $(a_0, a_1, a_2, a_3, x) mapsto x^4 + a_0x^3 + a_1x^2 + 2a_2x + a_3 $. In this case, if you think of this as a function $mathbbR^4+1 to mathbbR^1$, you'll find the implicit function theorem to be a little more friendly. In particular, if you take for example all the $a_i = 1$, you can find a solution given by $x = -1$. Now you just need to show that $x$ can be eliminated in terms of the $a_i$ by using the implicit function theorem, i.e. checking that the $x$ derivative of this function of $5$ variables has a non-zero derivative at the point in question (as it is clearly more than just $C^1$, it is $C^infty$ as a polynomial in these variables).
answered Aug 12 at 2:02
Alfred Yerger
9,7972044
The function you have defined is not a map to $mathbbR$. It is a map to $mathbbR[x]$. Almost certainly you want a map $mathbbR^5 to mathbbR$ which is defined by $(a_0, a_1, a_2, a_3, x) mapsto x^4 + a_0x^3 + a_1x^2 + 2a_2x + a_3 $. In this case, if you think of this as a function $mathbbR^4+1 to mathbbR^1$, you'll find the implicit function theorem to be a little more friendly. In particular, if you take for example all the $a_i = 1$, you can find a solution given by $x = -1$. Now you just need to show that $x$ can be eliminated in terms of the $a_i$ by using the implicit function theorem, i.e. checking that the $x$ derivative of this function of $5$ variables has a non-zero derivative at the point in question (as it is clearly more than just $C^1$, it is $C^infty$ as a polynomial in these variables).
answered Aug 12 at 2:02
Alfred Yerger
9,7972044
answered Aug 12 at 2:02
Alfred Yerger
9,7972044
answered Aug 12 at 2:02
Alfred Yerger
9,7972044
answered Aug 12 at 2:02
Alfred Yerger
9,7972044
9,7972044
Thanks! What point do you mean? I believe I can take $p=(1,1,1,1,-1)$. In this case, writing the function you defined as $F$, we have $F(p)=0$ and $F_x(p)ne 0$. Then by the Implicit function theorem, there is $delta > 0$ and a smooth $phi: V=(1-delta,1+delta)^4to mathbb R$ with $phi(1,1,1,1)=-1$ and $F(a_0,a_1,a_2,a_3,phi(a_0,a_1,a_2,a_3))=0$ for all $(a_0,a_1,a_2,a_3)in V$. So $phi(a_0,a_1,a_2,a_3)$ is the local solution we need.
– user557902851
Aug 12 at 2:38
1
Yes, that point. This is pretty much exactly what I had in mind.
– Alfred Yerger
Aug 12 at 2:44
add a comment |Â
Thanks! What point do you mean? I believe I can take $p=(1,1,1,1,-1)$. In this case, writing the function you defined as $F$, we have $F(p)=0$ and $F_x(p)ne 0$. Then by the Implicit function theorem, there is $delta > 0$ and a smooth $phi: V=(1-delta,1+delta)^4to mathbb R$ with $phi(1,1,1,1)=-1$ and $F(a_0,a_1,a_2,a_3,phi(a_0,a_1,a_2,a_3))=0$ for all $(a_0,a_1,a_2,a_3)in V$. So $phi(a_0,a_1,a_2,a_3)$ is the local solution we need.
– user557902851
Aug 12 at 2:38
1
Yes, that point. This is pretty much exactly what I had in mind.
– Alfred Yerger
Aug 12 at 2:44
Thanks! What point do you mean? I believe I can take $p=(1,1,1,1,-1)$. In this case, writing the function you defined as $F$, we have $F(p)=0$ and $F_x(p)ne 0$. Then by the Implicit function theorem, there is $delta > 0$ and a smooth $phi: V=(1-delta,1+delta)^4to mathbb R$ with $phi(1,1,1,1)=-1$ and $F(a_0,a_1,a_2,a_3,phi(a_0,a_1,a_2,a_3))=0$ for all $(a_0,a_1,a_2,a_3)in V$. So $phi(a_0,a_1,a_2,a_3)$ is the local solution we need.
– user557902851
Aug 12 at 2:38
Thanks! What point do you mean? I believe I can take $p=(1,1,1,1,-1)$. In this case, writing the function you defined as $F$, we have $F(p)=0$ and $F_x(p)ne 0$. Then by the Implicit function theorem, there is $delta > 0$ and a smooth $phi: V=(1-delta,1+delta)^4to mathbb R$ with $phi(1,1,1,1)=-1$ and $F(a_0,a_1,a_2,a_3,phi(a_0,a_1,a_2,a_3))=0$ for all $(a_0,a_1,a_2,a_3)in V$. So $phi(a_0,a_1,a_2,a_3)$ is the local solution we need.
– user557902851
Aug 12 at 2:38
1
1
Yes, that point. This is pretty much exactly what I had in mind.
– Alfred Yerger
Aug 12 at 2:44
Yes, that point. This is pretty much exactly what I had in mind.
– Alfred Yerger
Aug 12 at 2:44
add a comment |Â
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Note: If all $a_i$ are equal to $1$, then $x=-1$ is a root.
– quasi
Aug 12 at 1:58
1
The map you want is probably from the coefficients to the four roots, not to the polynomial. In any case the crux of the problem lies in showing that there are no repeated roots when all $a_i$ are approximately equal to one.
– hardmath
Aug 12 at 1:59