Confusion about non-derivable continuous functions The 2019 Stack Overflow Developer Survey...
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Confusion about non-derivable continuous functions
The 2019 Stack Overflow Developer Survey Results Are InAre there any implicit, continuous, non-differentiable functions?Logical Relations Between Three Statements about Continuous FunctionsCombination of continuous and discontinuous functionsIs there only one continuous-everywhere non-differentiable function?Intuition behind uniformly continuous functionsWhy weren't continuous functions defined as Darboux functions?Examples of functions that do not belong to any Baire classFind all continuous functions that satisfy the Jensen inequality(?) $f(frac{x+y}{2})=frac{f(x)+f(y)}{2}$Confused About Limit Points and Closed SetsConfusion About Differentiability of Function
$begingroup$
I am reading a definition which claims that a function is continuous in point $p$ iff all its first derivations exist and are continuous in the point $p$.
And what confuses me are functions such as $f(x)=|x|$ which should be continuous by intuition, but is clearly not derivable in $x=0$.
I am almost certain I am getting something wrong here, but I can not even pin-point what.
real-analysis functions derivatives continuity
asked 5 hours ago
fazanfazan
537
$endgroup$
add a comment |
$begingroup$
I am reading a definition which claims that a function is continuous in point $p$ iff all its first derivations exist and are continuous in the point $p$.
And what confuses me are functions such as $f(x)=|x|$ which should be continuous by intuition, but is clearly not derivable in $x=0$.
I am almost certain I am getting something wrong here, but I can not even pin-point what.
real-analysis functions derivatives continuity
asked 5 hours ago
fazanfazan
537
$endgroup$
$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
5 hours ago
$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
5 hours ago
$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
5 hours ago
1
$begingroup$
@avs That is false.
$endgroup$
– zhw.
4 hours ago
2
$begingroup$
@avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous).
$endgroup$
– Robert Furber
1 hour ago
add a comment |
$begingroup$
I am reading a definition which claims that a function is continuous in point $p$ iff all its first derivations exist and are continuous in the point $p$.
And what confuses me are functions such as $f(x)=|x|$ which should be continuous by intuition, but is clearly not derivable in $x=0$.
I am almost certain I am getting something wrong here, but I can not even pin-point what.
real-analysis functions derivatives continuity
asked 5 hours ago
fazanfazan
537
$endgroup$
I am reading a definition which claims that a function is continuous in point $p$ iff all its first derivations exist and are continuous in the point $p$.
And what confuses me are functions such as $f(x)=|x|$ which should be continuous by intuition, but is clearly not derivable in $x=0$.
I am almost certain I am getting something wrong here, but I can not even pin-point what.
real-analysis functions derivatives continuity
real-analysis functions derivatives continuity
asked 5 hours ago
fazanfazan
537
asked 5 hours ago
fazanfazan
537
asked 5 hours ago
fazanfazan
537
asked 5 hours ago
fazanfazan
537
asked 5 hours ago
fazanfazan
537
537
$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
5 hours ago
$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
5 hours ago
$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
5 hours ago
1
$begingroup$
@avs That is false.
$endgroup$
– zhw.
4 hours ago
2
$begingroup$
@avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous).
$endgroup$
– Robert Furber
1 hour ago
add a comment |
$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
5 hours ago
$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
5 hours ago
$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
5 hours ago
1
$begingroup$
@avs That is false.
$endgroup$
– zhw.
4 hours ago
2
$begingroup$
@avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous).
$endgroup$
– Robert Furber
1 hour ago
$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
5 hours ago
$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
5 hours ago
$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
5 hours ago
$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
5 hours ago
$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
5 hours ago
$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
5 hours ago
1
1
$begingroup$
@avs That is false.
$endgroup$
– zhw.
4 hours ago
$begingroup$
@avs That is false.
$endgroup$
– zhw.
4 hours ago
2
2
$begingroup$
@avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous).
$endgroup$
– Robert Furber
1 hour ago
$begingroup$
@avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous).
$endgroup$
– Robert Furber
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
That "definition" is wrong. You are right, the function $|x|$ is continuous but is not differentiable at $x=0$. Continuity doesn't imply differentiability. However, differentiability does imply continuity.
The definition you stated looks to me as an attempt to define a smooth function, although it is not correct.
answered 5 hours ago
Haris GusicHaris Gusic
3,516627
$endgroup$
$begingroup$
It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
$endgroup$
– Robert Furber
1 hour ago
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
1 hour ago
add a comment |
$begingroup$
As has been pointed out this definition is incorrect, as it is inconsistent with the usual definitions of continuity and differentiability. Your example $|x|$ suffices to show this.
If you are encountering this in multivariable calculus then your professor might be trying to state the theorem mentioned by avs in the comments: that a function is differentiable at a point if all its first order partial derivatives exist in a neighbourhood of that point, and are continuous at that point. However the converse is not generally true: consider for example the function
$$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
at the origin. Thus this assumption might be completely false. It might be best to give a word for word reproduction of the statement and the paragraph before and after.
answered 5 hours ago
K.PowerK.Power
3,710926
$endgroup$
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
5 hours ago
1
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
5 hours ago
add a comment |
$begingroup$
Sounds like the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ like in your original post is continuous at every point $left(a,fleft(aright)right)$ wherever $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly)
answered 2 hours ago
ManRowManRow
24618
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That "definition" is wrong. You are right, the function $|x|$ is continuous but is not differentiable at $x=0$. Continuity doesn't imply differentiability. However, differentiability does imply continuity.
The definition you stated looks to me as an attempt to define a smooth function, although it is not correct.
answered 5 hours ago
Haris GusicHaris Gusic
3,516627
$endgroup$
$begingroup$
It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
$endgroup$
– Robert Furber
1 hour ago
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
1 hour ago
add a comment |
$begingroup$
That "definition" is wrong. You are right, the function $|x|$ is continuous but is not differentiable at $x=0$. Continuity doesn't imply differentiability. However, differentiability does imply continuity.
The definition you stated looks to me as an attempt to define a smooth function, although it is not correct.
answered 5 hours ago
Haris GusicHaris Gusic
3,516627
$endgroup$
$begingroup$
It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
$endgroup$
– Robert Furber
1 hour ago
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
1 hour ago
add a comment |
$begingroup$
That "definition" is wrong. You are right, the function $|x|$ is continuous but is not differentiable at $x=0$. Continuity doesn't imply differentiability. However, differentiability does imply continuity.
The definition you stated looks to me as an attempt to define a smooth function, although it is not correct.
answered 5 hours ago
Haris GusicHaris Gusic
3,516627
$endgroup$
That "definition" is wrong. You are right, the function $|x|$ is continuous but is not differentiable at $x=0$. Continuity doesn't imply differentiability. However, differentiability does imply continuity.
The definition you stated looks to me as an attempt to define a smooth function, although it is not correct.
answered 5 hours ago
Haris GusicHaris Gusic
3,516627
edited 5 hours ago
answered 5 hours ago
Haris GusicHaris Gusic
3,516627
answered 5 hours ago
Haris GusicHaris Gusic
3,516627
answered 5 hours ago
Haris GusicHaris Gusic
3,516627
3,516627
$begingroup$
It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
$endgroup$
– Robert Furber
1 hour ago
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
1 hour ago
add a comment |
$begingroup$
It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
$endgroup$
– Robert Furber
1 hour ago
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
1 hour ago
$begingroup$
It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
$endgroup$
– Robert Furber
1 hour ago
$begingroup$
It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not.
$endgroup$
– Robert Furber
1 hour ago
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
1 hour ago
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
1 hour ago
add a comment |
$begingroup$
As has been pointed out this definition is incorrect, as it is inconsistent with the usual definitions of continuity and differentiability. Your example $|x|$ suffices to show this.
If you are encountering this in multivariable calculus then your professor might be trying to state the theorem mentioned by avs in the comments: that a function is differentiable at a point if all its first order partial derivatives exist in a neighbourhood of that point, and are continuous at that point. However the converse is not generally true: consider for example the function
$$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
at the origin. Thus this assumption might be completely false. It might be best to give a word for word reproduction of the statement and the paragraph before and after.
answered 5 hours ago
K.PowerK.Power
3,710926
$endgroup$
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
5 hours ago
1
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
5 hours ago
add a comment |
$begingroup$
As has been pointed out this definition is incorrect, as it is inconsistent with the usual definitions of continuity and differentiability. Your example $|x|$ suffices to show this.
If you are encountering this in multivariable calculus then your professor might be trying to state the theorem mentioned by avs in the comments: that a function is differentiable at a point if all its first order partial derivatives exist in a neighbourhood of that point, and are continuous at that point. However the converse is not generally true: consider for example the function
$$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
at the origin. Thus this assumption might be completely false. It might be best to give a word for word reproduction of the statement and the paragraph before and after.
answered 5 hours ago
K.PowerK.Power
3,710926
$endgroup$
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
5 hours ago
1
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
5 hours ago
add a comment |
$begingroup$
As has been pointed out this definition is incorrect, as it is inconsistent with the usual definitions of continuity and differentiability. Your example $|x|$ suffices to show this.
If you are encountering this in multivariable calculus then your professor might be trying to state the theorem mentioned by avs in the comments: that a function is differentiable at a point if all its first order partial derivatives exist in a neighbourhood of that point, and are continuous at that point. However the converse is not generally true: consider for example the function
$$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
at the origin. Thus this assumption might be completely false. It might be best to give a word for word reproduction of the statement and the paragraph before and after.
answered 5 hours ago
K.PowerK.Power
3,710926
$endgroup$
As has been pointed out this definition is incorrect, as it is inconsistent with the usual definitions of continuity and differentiability. Your example $|x|$ suffices to show this.
If you are encountering this in multivariable calculus then your professor might be trying to state the theorem mentioned by avs in the comments: that a function is differentiable at a point if all its first order partial derivatives exist in a neighbourhood of that point, and are continuous at that point. However the converse is not generally true: consider for example the function
$$f(x,y)=begin{cases}(x^2+y^2)sin(frac{1}{sqrt{x^2+y^2}}) &(x,y)neq(0,0)\0&(x,y)=(0,0)end{cases}$$
at the origin. Thus this assumption might be completely false. It might be best to give a word for word reproduction of the statement and the paragraph before and after.
answered 5 hours ago
K.PowerK.Power
3,710926
edited 4 hours ago
answered 5 hours ago
K.PowerK.Power
3,710926
answered 5 hours ago
K.PowerK.Power
3,710926
answered 5 hours ago
K.PowerK.Power
3,710926
3,710926
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
5 hours ago
1
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
5 hours ago
add a comment |
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
5 hours ago
1
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
5 hours ago
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
5 hours ago
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
5 hours ago
1
1
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
5 hours ago
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
5 hours ago
add a comment |
$begingroup$
Sounds like the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ like in your original post is continuous at every point $left(a,fleft(aright)right)$ wherever $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly)
answered 2 hours ago
ManRowManRow
24618
$endgroup$
add a comment |
$begingroup$
Sounds like the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ like in your original post is continuous at every point $left(a,fleft(aright)right)$ wherever $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly)
answered 2 hours ago
ManRowManRow
24618
$endgroup$
add a comment |
$begingroup$
Sounds like the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ like in your original post is continuous at every point $left(a,fleft(aright)right)$ wherever $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly)
answered 2 hours ago
ManRowManRow
24618
$endgroup$
Sounds like the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ like in your original post is continuous at every point $left(a,fleft(aright)right)$ wherever $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly)
answered 2 hours ago
ManRowManRow
24618
answered 2 hours ago
ManRowManRow
24618
answered 2 hours ago
ManRowManRow
24618
answered 2 hours ago
ManRowManRow
24618
24618
add a comment |
add a comment |
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$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
5 hours ago
$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
5 hours ago
$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
5 hours ago
1
$begingroup$
@avs That is false.
$endgroup$
– zhw.
4 hours ago
2
$begingroup$
@avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous).
$endgroup$
– Robert Furber
1 hour ago