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 yesterday
fazanfazan
608
$endgroup$
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 yesterday
fazanfazan
608
$endgroup$
$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
yesterday
$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
yesterday
$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
yesterday
1
$begingroup$
@avs That is false.
$endgroup$
– zhw.
yesterday
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
yesterday
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 yesterday
fazanfazan
608
$endgroup$
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 yesterday
fazanfazan
608
asked yesterday
fazanfazan
608
asked yesterday
fazanfazan
608
asked yesterday
fazanfazan
608
asked yesterday
fazanfazan
608
608
$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
yesterday
$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
yesterday
$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
yesterday
1
$begingroup$
@avs That is false.
$endgroup$
– zhw.
yesterday
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
yesterday
add a comment |
$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
yesterday
$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
yesterday
$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
yesterday
1
$begingroup$
@avs That is false.
$endgroup$
– zhw.
yesterday
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
yesterday
$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
yesterday
$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
yesterday
$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
yesterday
$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
yesterday
$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
yesterday
$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
yesterday
1
1
$begingroup$
@avs That is false.
$endgroup$
– zhw.
yesterday
$begingroup$
@avs That is false.
$endgroup$
– zhw.
yesterday
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
yesterday
$begingroup$
@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
yesterday
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 yesterday
Haris GusicHaris Gusic
3,531627
$endgroup$
$begingroup$
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
yesterday
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
yesterday
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 yesterday
K.PowerK.Power
3,710926
$endgroup$
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
yesterday
1
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
yesterday
add a comment |
$begingroup$
Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ (like in your original post!) is continuous at any point $left(a,fleft(aright)right)$ for which $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly and provided both limits exist).
And finally, note that some functions can even be nowhere-continuous as well! Such as
$$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$
edited 22 hours ago
avs
4,137515
answered yesterday
ManRowManRow
25618
$endgroup$
add a comment |
Your Answer
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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 yesterday
Haris GusicHaris Gusic
3,531627
$endgroup$
$begingroup$
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
yesterday
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
yesterday
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 yesterday
Haris GusicHaris Gusic
3,531627
$endgroup$
$begingroup$
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
yesterday
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
yesterday
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 yesterday
Haris GusicHaris Gusic
3,531627
$endgroup$
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 yesterday
Haris GusicHaris Gusic
3,531627
edited yesterday
answered yesterday
Haris GusicHaris Gusic
3,531627
answered yesterday
Haris GusicHaris Gusic
3,531627
answered yesterday
Haris GusicHaris Gusic
3,531627
3,531627
$begingroup$
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
yesterday
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
yesterday
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
yesterday
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
yesterday
$begingroup$
Here is the relevant wikipedia page: en.wikipedia.org/wiki/…
$endgroup$
– Robert Furber
yesterday
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 yesterday
K.PowerK.Power
3,710926
$endgroup$
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
yesterday
1
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
yesterday
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 yesterday
K.PowerK.Power
3,710926
$endgroup$
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
yesterday
1
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
yesterday
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 yesterday
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 yesterday
K.PowerK.Power
3,710926
edited yesterday
answered yesterday
K.PowerK.Power
3,710926
answered yesterday
K.PowerK.Power
3,710926
answered yesterday
K.PowerK.Power
3,710926
3,710926
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
yesterday
1
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
yesterday
add a comment |
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
yesterday
1
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
yesterday
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
yesterday
$begingroup$
The partial derivatives need not be continuous for differentiability.
$endgroup$
– Haris Gusic
yesterday
1
1
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
yesterday
$begingroup$
@HarisGusic yes I realized as I posted. Fixed it
$endgroup$
– K.Power
yesterday
add a comment |
$begingroup$
Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ (like in your original post!) is continuous at any point $left(a,fleft(aright)right)$ for which $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly and provided both limits exist).
And finally, note that some functions can even be nowhere-continuous as well! Such as
$$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$
edited 22 hours ago
avs
4,137515
answered yesterday
ManRowManRow
25618
$endgroup$
add a comment |
$begingroup$
Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ (like in your original post!) is continuous at any point $left(a,fleft(aright)right)$ for which $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly and provided both limits exist).
And finally, note that some functions can even be nowhere-continuous as well! Such as
$$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$
edited 22 hours ago
avs
4,137515
answered yesterday
ManRowManRow
25618
$endgroup$
add a comment |
$begingroup$
Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ (like in your original post!) is continuous at any point $left(a,fleft(aright)right)$ for which $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly and provided both limits exist).
And finally, note that some functions can even be nowhere-continuous as well! Such as
$$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$
edited 22 hours ago
avs
4,137515
answered yesterday
ManRowManRow
25618
$endgroup$
Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:mathbb{R}tomathbb {R}$ (like in your original post!) is continuous at any point $left(a,fleft(aright)right)$ for which $$limlimits_{xto a^-}fleft(xright)=limlimits_{xto a^+}fleft(xright)$$ (denoting the left and right-hand limits accordingly and provided both limits exist).
And finally, note that some functions can even be nowhere-continuous as well! Such as
$$fleft(xright)=left{begin{matrix}1, xinmathbb{Q}\0,xnotinmathbb {Q}end{matrix}right.$$
edited 22 hours ago
avs
4,137515
answered yesterday
ManRowManRow
25618
edited 22 hours ago
avs
4,137515
edited 22 hours ago
avs
4,137515
edited 22 hours ago
avs
4,137515
4,137515
answered yesterday
ManRowManRow
25618
answered yesterday
ManRowManRow
25618
answered yesterday
ManRowManRow
25618
25618
add a comment |
add a comment |
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$begingroup$
For $|x|$ its derivative isn't continuous t zero.
$endgroup$
– coffeemath
yesterday
$begingroup$
Where did you read that erroneous definition?
$endgroup$
– bof
yesterday
$begingroup$
lecture notes by my prof. i might be mosreading them though
$endgroup$
– fazan
yesterday
1
$begingroup$
@avs That is false.
$endgroup$
– zhw.
yesterday
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
yesterday