Question about smooth functions?
0
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I know the definition of a smooth function. I understand why the function needs to be differentiable but why we need the derivative to be continuous?
calculus
asked Dec 24 '18 at 16:11
spyerspyer
1188
$endgroup$
|
show 2 more comments
0
$begingroup$
I know the definition of a smooth function. I understand why the function needs to be differentiable but why we need the derivative to be continuous?
calculus
asked Dec 24 '18 at 16:11
spyerspyer
1188
$endgroup$
1
$begingroup$
Smooth normally means infinitely differentiable: that is, that we can continue taking derivatives as many times as we like. In particular, to take second derivatives we require that the first derivative be differentiable and hence continuous..
$endgroup$
– ODF
Dec 24 '18 at 16:20
$begingroup$
OK. But let say we are talking only about one-time smooth function. Why is the condition for the first derivative to be continuous?
$endgroup$
– spyer
Dec 24 '18 at 16:21
2
$begingroup$
A differentiable function need not have continuous derivative (e.g. $f(x) = x^2 sin(frac{1}{x})$ for $x neq 0$ and $f(0) = 0$). Functions which are $C^1$ (which is what you mean by one time smooth?) are continuously differentiable - they are simply defined this way.
$endgroup$
– ODF
Dec 24 '18 at 16:25
$begingroup$
There is nothing like geometrical intuition behind the definition?
$endgroup$
– spyer
Dec 24 '18 at 17:25
2
$begingroup$
Derivatives can never have jump discontinuities : the only discontinuity a derivative can have is the sort of really horrible one that $f'(x)$ from above will have at 0. I suppose we require continuously differentiable functions precisely in order to rule out really horrible examples like that.
$endgroup$
– ODF
Dec 24 '18 at 17:51
|
show 2 more comments
0
0
0
$begingroup$
I know the definition of a smooth function. I understand why the function needs to be differentiable but why we need the derivative to be continuous?
calculus
asked Dec 24 '18 at 16:11
spyerspyer
1188
$endgroup$
I know the definition of a smooth function. I understand why the function needs to be differentiable but why we need the derivative to be continuous?
calculus
calculus
asked Dec 24 '18 at 16:11
spyerspyer
1188
asked Dec 24 '18 at 16:11
spyerspyer
1188
asked Dec 24 '18 at 16:11
spyerspyer
1188
asked Dec 24 '18 at 16:11
spyerspyer
1188
asked Dec 24 '18 at 16:11
spyerspyer
1188
1188
1
$begingroup$
Smooth normally means infinitely differentiable: that is, that we can continue taking derivatives as many times as we like. In particular, to take second derivatives we require that the first derivative be differentiable and hence continuous..
$endgroup$
– ODF
Dec 24 '18 at 16:20
$begingroup$
OK. But let say we are talking only about one-time smooth function. Why is the condition for the first derivative to be continuous?
$endgroup$
– spyer
Dec 24 '18 at 16:21
2
$begingroup$
A differentiable function need not have continuous derivative (e.g. $f(x) = x^2 sin(frac{1}{x})$ for $x neq 0$ and $f(0) = 0$). Functions which are $C^1$ (which is what you mean by one time smooth?) are continuously differentiable - they are simply defined this way.
$endgroup$
– ODF
Dec 24 '18 at 16:25
$begingroup$
There is nothing like geometrical intuition behind the definition?
$endgroup$
– spyer
Dec 24 '18 at 17:25
2
$begingroup$
Derivatives can never have jump discontinuities : the only discontinuity a derivative can have is the sort of really horrible one that $f'(x)$ from above will have at 0. I suppose we require continuously differentiable functions precisely in order to rule out really horrible examples like that.
$endgroup$
– ODF
Dec 24 '18 at 17:51
|
show 2 more comments
1
$begingroup$
Smooth normally means infinitely differentiable: that is, that we can continue taking derivatives as many times as we like. In particular, to take second derivatives we require that the first derivative be differentiable and hence continuous..
$endgroup$
– ODF
Dec 24 '18 at 16:20
$begingroup$
OK. But let say we are talking only about one-time smooth function. Why is the condition for the first derivative to be continuous?
$endgroup$
– spyer
Dec 24 '18 at 16:21
2
$begingroup$
A differentiable function need not have continuous derivative (e.g. $f(x) = x^2 sin(frac{1}{x})$ for $x neq 0$ and $f(0) = 0$). Functions which are $C^1$ (which is what you mean by one time smooth?) are continuously differentiable - they are simply defined this way.
$endgroup$
– ODF
Dec 24 '18 at 16:25
$begingroup$
There is nothing like geometrical intuition behind the definition?
$endgroup$
– spyer
Dec 24 '18 at 17:25
2
$begingroup$
Derivatives can never have jump discontinuities : the only discontinuity a derivative can have is the sort of really horrible one that $f'(x)$ from above will have at 0. I suppose we require continuously differentiable functions precisely in order to rule out really horrible examples like that.
$endgroup$
– ODF
Dec 24 '18 at 17:51
1
1
$begingroup$
Smooth normally means infinitely differentiable: that is, that we can continue taking derivatives as many times as we like. In particular, to take second derivatives we require that the first derivative be differentiable and hence continuous..
$endgroup$
– ODF
Dec 24 '18 at 16:20
$begingroup$
Smooth normally means infinitely differentiable: that is, that we can continue taking derivatives as many times as we like. In particular, to take second derivatives we require that the first derivative be differentiable and hence continuous..
$endgroup$
– ODF
Dec 24 '18 at 16:20
$begingroup$
OK. But let say we are talking only about one-time smooth function. Why is the condition for the first derivative to be continuous?
$endgroup$
– spyer
Dec 24 '18 at 16:21
$begingroup$
OK. But let say we are talking only about one-time smooth function. Why is the condition for the first derivative to be continuous?
$endgroup$
– spyer
Dec 24 '18 at 16:21
2
2
$begingroup$
A differentiable function need not have continuous derivative (e.g. $f(x) = x^2 sin(frac{1}{x})$ for $x neq 0$ and $f(0) = 0$). Functions which are $C^1$ (which is what you mean by one time smooth?) are continuously differentiable - they are simply defined this way.
$endgroup$
– ODF
Dec 24 '18 at 16:25
$begingroup$
A differentiable function need not have continuous derivative (e.g. $f(x) = x^2 sin(frac{1}{x})$ for $x neq 0$ and $f(0) = 0$). Functions which are $C^1$ (which is what you mean by one time smooth?) are continuously differentiable - they are simply defined this way.
$endgroup$
– ODF
Dec 24 '18 at 16:25
$begingroup$
There is nothing like geometrical intuition behind the definition?
$endgroup$
– spyer
Dec 24 '18 at 17:25
$begingroup$
There is nothing like geometrical intuition behind the definition?
$endgroup$
– spyer
Dec 24 '18 at 17:25
2
2
$begingroup$
Derivatives can never have jump discontinuities : the only discontinuity a derivative can have is the sort of really horrible one that $f'(x)$ from above will have at 0. I suppose we require continuously differentiable functions precisely in order to rule out really horrible examples like that.
$endgroup$
– ODF
Dec 24 '18 at 17:51
$begingroup$
Derivatives can never have jump discontinuities : the only discontinuity a derivative can have is the sort of really horrible one that $f'(x)$ from above will have at 0. I suppose we require continuously differentiable functions precisely in order to rule out really horrible examples like that.
$endgroup$
– ODF
Dec 24 '18 at 17:51
|
show 2 more comments
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1
$begingroup$
Smooth normally means infinitely differentiable: that is, that we can continue taking derivatives as many times as we like. In particular, to take second derivatives we require that the first derivative be differentiable and hence continuous..
$endgroup$
– ODF
Dec 24 '18 at 16:20
$begingroup$
OK. But let say we are talking only about one-time smooth function. Why is the condition for the first derivative to be continuous?
$endgroup$
– spyer
Dec 24 '18 at 16:21
2
$begingroup$
A differentiable function need not have continuous derivative (e.g. $f(x) = x^2 sin(frac{1}{x})$ for $x neq 0$ and $f(0) = 0$). Functions which are $C^1$ (which is what you mean by one time smooth?) are continuously differentiable - they are simply defined this way.
$endgroup$
– ODF
Dec 24 '18 at 16:25
$begingroup$
There is nothing like geometrical intuition behind the definition?
$endgroup$
– spyer
Dec 24 '18 at 17:25
2
$begingroup$
Derivatives can never have jump discontinuities : the only discontinuity a derivative can have is the sort of really horrible one that $f'(x)$ from above will have at 0. I suppose we require continuously differentiable functions precisely in order to rule out really horrible examples like that.
$endgroup$
– ODF
Dec 24 '18 at 17:51