Differentiability and continously differentiable definition/concepts.

[binbagsss]

Theorem: ctsly differentiable at a if the function is cts and its partial derivatives exist and are cts in a neighborhood of a. [1]

- so to be differentiable we can check whether this conditions holds, and if it does ctsly diff => diff.

- the definition of a scalar function being differentiable at the point a is f(a+h)-f(a)=h.v(a)+R(h)... [2] ; for a scalar function of f.

and lim $_{h \rightarrow 0}$ $\frac{R(h)}{ h }$ = 0 [3]

(sorry this should be modulus h . I can't get it to work ! )

- BUT, if this doesn't hold, then we can go back to the scalar differentiable definition and check if R(h) obeys condition [3]

Questions:

- when we deduce what R(h) is, what should we take ∇f as - should it be the value you get from the partial derivatives (limit definition), or from 'directly differentiating' f. (so this would assume that the partial derivative does exist , and failed on theorem [1] condition by the partials not being cts, being the reason I am looking back at definition [2]).

- The definition of cts, is, that the limit needs to exist in a neighborhood of point a, and not at the point a. So if condition [1] fails on the partial derivative not existing at a , this should not matter? we just need to check they are cts in a neighborhood of a? And ctsly differentiable is still a possibility? ( I ask because my solutions always seem to take the partial at the point a, or would this be more for ∇f?

- cts diff => diff. condition [1]. This does not work the other way around , so differentiability is still a possibility. Am I correct in thinking that a function can still be differentiable if:
a) its partial derivative does not exist at a
b) they are not cts in a neighborhood of a

- but regarding a) , if the partial derivatives do not exist, from [2] the only candidate for v(a) is ∇f , which is attained from the partial derivatives , so if these do not exist, as a limit, (the partial derivatives can not be cts) and we must get the partial derivatives from the function without the limit definition ?

(in the case that they are cts , this limit should equal the partial derivatives attained from the function evaluated at this point, so you take either for ∇f (as they are same ) - is this correct? (I Know you wouldn't need to go back to the definition in this case as theorem [1] conditions are met, but I'm checking my understanding..

Many Thanks for any assistance, greatly appreciated !

 

[SammyS]

Theorem: ctsly differentiable at a if the function is cts and its partial derivatives exist and are cts in a neighborhood of a. [1]

- so to be differentiable we can check whether this conditions holds, and if it does ctsly diff => diff.

- the definition of a scalar function being differentiable at the point a is f(a+h)-f(a)=h.v(a)+R(h)... [2] ; for a scalar function of f.

and lim $_{h \rightarrow 0}$ $\frac{R(h)}{ h }$ = 0 [3]

(sorry this should be modulus h . I can't get it to work ! )

- BUT, if this doesn't hold, then we can go back to the scalar differentiable definition and check if R(h) obeys condition [3]

Questions:

- when we deduce what R(h) is, what should we take ∇f as - should it be the value you get from the partial derivatives (limit definition), or from 'directly differentiating' f. (so this would assume that the partial derivative does exist , and failed on theorem [1] condition by the partials not being cts, being the reason I am looking back at definition [2]).

- The definition of cts, is, that the limit needs to exist in a neighborhood of point a, and not at the point a. So if condition [1] fails on the partial derivative not existing at a , this should not matter? we just need to check they are cts in a neighborhood of a? And ctsly differentiable is still a possibility? ( I ask because my solutions always seem to take the partial at the point a, or would this be more for ∇f?

- cts diff => diff. condition [1]. This does not work the other way around , so differentiability is still a possibility. Am I correct in thinking that a function can still be differentiable if:
a) its partial derivative does not exist at a
b) they are not cts in a neighborhood of a

- but regarding a) , if the partial derivatives do not exist, from [2] the only candidate for v(a) is ∇f , which is attained from the partial derivatives , so if these do not exist, as a limit, (the partial derivatives can not be cts) and we must get the partial derivatives from the function without the limit definition ?

(in the case that they are cts , this limit should equal the partial derivatives attained from the function evaluated at this point, so you take either for ∇f (as they are same ) - is this correct? (I Know you wouldn't need to go back to the definition in this case as theorem [1] conditions are met, but I'm checking my understanding..

Many Thanks for any assistance, greatly appreciated !

What are these abbreviations ?

 

[binbagsss]

Which ones? Ctsly = continuously, and cts= continuous. Any others?

 

[SammyS]

Which ones? Ctsly = continuously, and cts= continuous. Any others?

Those are the two in particular which encouraged me to not read through the Original Post .

I suspect that other helpers on the forums may have the same reaction. Those who generally chime in on such subjects haven't responded to this thread so far.

 

[micromass]

Theorem: ctsly differentiable at a if the function is cts and its partial derivatives exist and are cts in a neighborhood of a. [1]

- so to be differentiable we can check whether this conditions holds, and if it does ctsly diff => diff.

- the definition of a scalar function being differentiable at the point a is f(a+h)-f(a)=h.v(a)+R(h)... [2] ; for a scalar function of f.

and lim $_{h \rightarrow 0}$ $\frac{R(h)}{ h }$ = 0 [3]

(sorry this should be modulus h . I can't get it to work ! )

- BUT, if this doesn't hold, then we can go back to the scalar differentiable definition and check if R(h) obeys condition [3]

Questions:

- when we deduce what R(h) is, what should we take ∇f as - should it be the value you get from the partial derivatives (limit definition), or from 'directly differentiating' f. (so this would assume that the partial derivative does exist , and failed on theorem [1] condition by the partials not being cts, being the reason I am looking back at definition [2]).

The idea is to take the partial derivatives and use them to find the gradient of $f$. That is, we have

$$\nabla f(a) = (\frac{\partial f}{\partial x_1} (a), ..., \frac{\partial f}{\partial x_n}(a))$$

Use this to define $\nabla f$. Then you can check differentiability using this definition.
You know that if $f$ is differentiable, then $\nabla f$ has the previous form, so it can't be anything else.

- The definition of cts, is, that the limit needs to exist in a neighborhood of point a, and not at the point a. So if condition [1] fails on the partial derivative not existing at a , this should not matter? we just need to check they are cts in a neighborhood of a? And ctsly differentiable is still a possibility? ( I ask because my solutions always seem to take the partial at the point a, or would this be more for ∇f?

You need to find the partial derivatives at every point possible, not only at $a$. You will need the partial derivatives to exist at some neighborhood of $a$ and you will need them to be continuous there. Just checking continuity at one single point is not sufficient.

- cts diff => diff. condition [1]. This does not work the other way around , so differentiability is still a possibility. Am I correct in thinking that a function can still be differentiable if:
a) its partial derivative does not exist at a

No. If a function is differentiable, then its partial derivatives must exist.

b) they are not cts in a neighborhood of a

That can happen. It can happen that the function is differentiable but that the partials don't exist.

 

[binbagsss]

The idea is to take the partial derivatives and use them to find the gradient of $f$. That is, we have

$$\nabla f(a) = (\frac{\partial f}{\partial x_1} (a), ..., \frac{\partial f}{\partial x_n}(a))$$

Use this to define $\nabla f$. Then you can check differentiability using this definition.
You know that if $f$ is differentiable, then $\nabla f$ has the previous form, so it can't be anything else.

What if, for example, you have f = lx^2-y^2 l and are investigating differentiability at the origin. for
x^2-y^2 > 0 we get f= x^2-y^2 ,
for
x^2-y^2<0 we get f= -x^2+y^2.

So how do we 'read of/apply the gradient function' to the origin ? If we did the limit partial deritivate definition, then we can keep the mod signs , and don't need to decide? In this case isn't the only way to get ∇f, by partial deriviative definition?

Thanks for your help.

 

[micromass]

What if, for example, you have f = lx^2-y^2 l and are investigating differentiability at the origin. for
x^2-y^2 > 0 we get f= x^2-y^2 ,
for
x^2-y^2<0 we get f= -x^2+y^2.

So how do we 'read of/apply the gradient function' to the origin ? If we did the limit partial deritivate definition, then we can keep the mod signs , and don't need to decide? In this case isn't the only way to get ∇f, by partial deriviative definition?

First you'll need to find the partial derivatives at the origin. Then you can use that to make a candidate for the gradient $\nabla f$.

 

[binbagsss]

so you use the limit definition?

 

[micromass]

so you use the limit definition?

In this case, yes.