When is the polynomial $P(|x+y|)$ total differentiable?
$begingroup$
If $P in mathbb{R}[x]$ is a polynomial, under which sufficient condition is the function: $$f: mathbb{R}^2 to mathbb{R}: f(x,y) = P(|x+y|)$$ total differentiable?
So for a function to be total differentiable, every partial derivative need to be continuous or the function needs to fulfill the definition of total differentiability:
$$lim_{x to a} frac{||f(x) - f(a) -A(x-a)||}{||x-a||} = 0$$
where A is the linear map with matrix $A = (df)(a).$
So the first condition I have is that $x+y neq 0 $. But I have the feeling that there needs be a more general or specific condition.
real-analysis derivatives absolute-value
asked Jan 22 at 14:35
Belgium_PhysicsBelgium_Physics
325110
$endgroup$
add a comment |
$begingroup$
If $P in mathbb{R}[x]$ is a polynomial, under which sufficient condition is the function: $$f: mathbb{R}^2 to mathbb{R}: f(x,y) = P(|x+y|)$$ total differentiable?
So for a function to be total differentiable, every partial derivative need to be continuous or the function needs to fulfill the definition of total differentiability:
$$lim_{x to a} frac{||f(x) - f(a) -A(x-a)||}{||x-a||} = 0$$
where A is the linear map with matrix $A = (df)(a).$
So the first condition I have is that $x+y neq 0 $. But I have the feeling that there needs be a more general or specific condition.
real-analysis derivatives absolute-value
asked Jan 22 at 14:35
Belgium_PhysicsBelgium_Physics
325110
$endgroup$
1
$begingroup$
The question as is, is not about the set of $(x,y)$ such that the function is differentiable at these points, but about the set of polynomials such that the function is differentiable everywhere.
$endgroup$
– Mindlack
Jan 22 at 14:49
add a comment |
$begingroup$
If $P in mathbb{R}[x]$ is a polynomial, under which sufficient condition is the function: $$f: mathbb{R}^2 to mathbb{R}: f(x,y) = P(|x+y|)$$ total differentiable?
So for a function to be total differentiable, every partial derivative need to be continuous or the function needs to fulfill the definition of total differentiability:
$$lim_{x to a} frac{||f(x) - f(a) -A(x-a)||}{||x-a||} = 0$$
where A is the linear map with matrix $A = (df)(a).$
So the first condition I have is that $x+y neq 0 $. But I have the feeling that there needs be a more general or specific condition.
real-analysis derivatives absolute-value
asked Jan 22 at 14:35
Belgium_PhysicsBelgium_Physics
325110
$endgroup$
If $P in mathbb{R}[x]$ is a polynomial, under which sufficient condition is the function: $$f: mathbb{R}^2 to mathbb{R}: f(x,y) = P(|x+y|)$$ total differentiable?
So for a function to be total differentiable, every partial derivative need to be continuous or the function needs to fulfill the definition of total differentiability:
$$lim_{x to a} frac{||f(x) - f(a) -A(x-a)||}{||x-a||} = 0$$
where A is the linear map with matrix $A = (df)(a).$
So the first condition I have is that $x+y neq 0 $. But I have the feeling that there needs be a more general or specific condition.
real-analysis derivatives absolute-value
real-analysis derivatives absolute-value
asked Jan 22 at 14:35
Belgium_PhysicsBelgium_Physics
325110
asked Jan 22 at 14:35
Belgium_PhysicsBelgium_Physics
325110
edited Jan 22 at 14:43
Belgium_Physics
asked Jan 22 at 14:35
Belgium_PhysicsBelgium_Physics
325110
asked Jan 22 at 14:35
Belgium_PhysicsBelgium_Physics
325110
asked Jan 22 at 14:35
Belgium_PhysicsBelgium_Physics
325110
325110
1
$begingroup$
The question as is, is not about the set of $(x,y)$ such that the function is differentiable at these points, but about the set of polynomials such that the function is differentiable everywhere.
$endgroup$
– Mindlack
Jan 22 at 14:49
add a comment |
1
$begingroup$
The question as is, is not about the set of $(x,y)$ such that the function is differentiable at these points, but about the set of polynomials such that the function is differentiable everywhere.
$endgroup$
– Mindlack
Jan 22 at 14:49
1
1
$begingroup$
The question as is, is not about the set of $(x,y)$ such that the function is differentiable at these points, but about the set of polynomials such that the function is differentiable everywhere.
$endgroup$
– Mindlack
Jan 22 at 14:49
$begingroup$
The question as is, is not about the set of $(x,y)$ such that the function is differentiable at these points, but about the set of polynomials such that the function is differentiable everywhere.
$endgroup$
– Mindlack
Jan 22 at 14:49
add a comment |
1 Answer
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$begingroup$
Clearly $f$ is differentiable at points where $x+yne 0$ since $|cdot|$ is differentiable on $mathbb{R}setminus {0}$.
If we write $P(t) = sum_{i=0}^na _i t^i$ then
$f$ is differentiable everywhere if and only if $a_1 = 0$.
Indeed, if $a_1 = 0$ then $f(x,y) = a_0+sum_{i=2}^n a_i|x+y|^i$ is differentiable because $|cdot|^alpha$ is differentiable for $alpha > 1$.
Conversely, if $f(x,y) = sum_{i=0}^n a_i|x+y|^i$ is differentiable, then the function $$g(x,y)= f(x,y) - underbrace{left( a_0+sum_{i=2}^n a_i|x+y|^iright)}_{text{this is differentiable}} = a_1|x+y|$$
is also differentiable.
However this implies $a_1 = 0$ since otherwise for $y = x$ we would get the function $g(x,x)=2a_1|x|$ which is not differentiable at $0$.
answered Jan 22 at 14:47
mechanodroidmechanodroid
27.8k62447
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add a comment |
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$begingroup$
Clearly $f$ is differentiable at points where $x+yne 0$ since $|cdot|$ is differentiable on $mathbb{R}setminus {0}$.
If we write $P(t) = sum_{i=0}^na _i t^i$ then
$f$ is differentiable everywhere if and only if $a_1 = 0$.
Indeed, if $a_1 = 0$ then $f(x,y) = a_0+sum_{i=2}^n a_i|x+y|^i$ is differentiable because $|cdot|^alpha$ is differentiable for $alpha > 1$.
Conversely, if $f(x,y) = sum_{i=0}^n a_i|x+y|^i$ is differentiable, then the function $$g(x,y)= f(x,y) - underbrace{left( a_0+sum_{i=2}^n a_i|x+y|^iright)}_{text{this is differentiable}} = a_1|x+y|$$
is also differentiable.
However this implies $a_1 = 0$ since otherwise for $y = x$ we would get the function $g(x,x)=2a_1|x|$ which is not differentiable at $0$.
answered Jan 22 at 14:47
mechanodroidmechanodroid
27.8k62447
$endgroup$
add a comment |
$begingroup$
Clearly $f$ is differentiable at points where $x+yne 0$ since $|cdot|$ is differentiable on $mathbb{R}setminus {0}$.
If we write $P(t) = sum_{i=0}^na _i t^i$ then
$f$ is differentiable everywhere if and only if $a_1 = 0$.
Indeed, if $a_1 = 0$ then $f(x,y) = a_0+sum_{i=2}^n a_i|x+y|^i$ is differentiable because $|cdot|^alpha$ is differentiable for $alpha > 1$.
Conversely, if $f(x,y) = sum_{i=0}^n a_i|x+y|^i$ is differentiable, then the function $$g(x,y)= f(x,y) - underbrace{left( a_0+sum_{i=2}^n a_i|x+y|^iright)}_{text{this is differentiable}} = a_1|x+y|$$
is also differentiable.
However this implies $a_1 = 0$ since otherwise for $y = x$ we would get the function $g(x,x)=2a_1|x|$ which is not differentiable at $0$.
answered Jan 22 at 14:47
mechanodroidmechanodroid
27.8k62447
$endgroup$
add a comment |
$begingroup$
Clearly $f$ is differentiable at points where $x+yne 0$ since $|cdot|$ is differentiable on $mathbb{R}setminus {0}$.
If we write $P(t) = sum_{i=0}^na _i t^i$ then
$f$ is differentiable everywhere if and only if $a_1 = 0$.
Indeed, if $a_1 = 0$ then $f(x,y) = a_0+sum_{i=2}^n a_i|x+y|^i$ is differentiable because $|cdot|^alpha$ is differentiable for $alpha > 1$.
Conversely, if $f(x,y) = sum_{i=0}^n a_i|x+y|^i$ is differentiable, then the function $$g(x,y)= f(x,y) - underbrace{left( a_0+sum_{i=2}^n a_i|x+y|^iright)}_{text{this is differentiable}} = a_1|x+y|$$
is also differentiable.
However this implies $a_1 = 0$ since otherwise for $y = x$ we would get the function $g(x,x)=2a_1|x|$ which is not differentiable at $0$.
answered Jan 22 at 14:47
mechanodroidmechanodroid
27.8k62447
$endgroup$
Clearly $f$ is differentiable at points where $x+yne 0$ since $|cdot|$ is differentiable on $mathbb{R}setminus {0}$.
If we write $P(t) = sum_{i=0}^na _i t^i$ then
$f$ is differentiable everywhere if and only if $a_1 = 0$.
Indeed, if $a_1 = 0$ then $f(x,y) = a_0+sum_{i=2}^n a_i|x+y|^i$ is differentiable because $|cdot|^alpha$ is differentiable for $alpha > 1$.
Conversely, if $f(x,y) = sum_{i=0}^n a_i|x+y|^i$ is differentiable, then the function $$g(x,y)= f(x,y) - underbrace{left( a_0+sum_{i=2}^n a_i|x+y|^iright)}_{text{this is differentiable}} = a_1|x+y|$$
is also differentiable.
However this implies $a_1 = 0$ since otherwise for $y = x$ we would get the function $g(x,x)=2a_1|x|$ which is not differentiable at $0$.
answered Jan 22 at 14:47
mechanodroidmechanodroid
27.8k62447
answered Jan 22 at 14:47
mechanodroidmechanodroid
27.8k62447
answered Jan 22 at 14:47
mechanodroidmechanodroid
27.8k62447
answered Jan 22 at 14:47
mechanodroidmechanodroid
27.8k62447
27.8k62447
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$begingroup$
The question as is, is not about the set of $(x,y)$ such that the function is differentiable at these points, but about the set of polynomials such that the function is differentiable everywhere.
$endgroup$
– Mindlack
Jan 22 at 14:49