Proving Differentiability: Vector Calculus Homework

[Ted123]

Homework Statement

[PLAIN]http://img261.imageshack.us/img261/1228/vectorcalc.png

Homework Equations

The Attempt at a Solution

$f({\bf a+h})-f(\bf{a})={\bf c\times h} + \|{\bf a+h} \| ^2 {\bf c} - \|{\bf a}\| ^2 {\bf c}$

$f({\bf a+h})-f({\bf a}) = {\bf c} \times {\bf h} + {\bf c} [({\bf a}+ {\bf h}) \cdot ({\bf a} + {\bf h})] - {\bf c}({\bf a}\cdot {\bf a})$

$f({\bf a+h})-f({\bf a}) = {\bf c} \times {\bf h} + {\bf c} [({\bf a}\cdot {\bf a})+ 2({\bf a} \cdot {\bf h}) + ({\bf h}\cdot {\bf h})] - {\bf c}({\bf a}\cdot {\bf a})$

$f({\bf a+h})-f({\bf a}) = {\bf c} \times {\bf h} + {\bf c} ({\bf a}\cdot {\bf a}) + 2{\bf c}({\bf a} \cdot {\bf h}) + {\bf c}({\bf h}\cdot {\bf h}) - {\bf c}({\bf a}\cdot {\bf a})$

$f({\bf a+h})-f({\bf a}) = {\bf c} \times {\bf h} + 2 {\bf c} ({\bf a}\cdot {\bf h}) + {\bf c}({\bf h}\cdot {\bf h})$

Is this correct up to here?

From here can I do this?:

$f({\bf a+h})-f({\bf a}) = {\bf c} \times {\bf h} + {\bf c} (2{\bf a}\cdot {\bf h} + {\bf h}\cdot {\bf h})$

Can I say that $Df(a)(h) = (2{\bf a}\cdot {\bf c} + {\bf h}\cdot {\bf c})(h)$

$E_f ({\bf a, h}) = c \times h$

But $\frac{E_f ({\bf a, h})}{\|h\|} = \frac{c \times h}{\|h\|}$ doesn't tend to 0 as h tends to 0!

 

[micromass]

Checking differentiability is always a little involved. Certainly if you try to prove it from the defintion, which is quite hard sometimes. Luckily, there are theorems which make our life easier. I'm talking about the theorem which says that "if a function has continuous partial derivatives at a, then it is differentiable at a".

So, what I suggest is try to take the partial derivatives of f. But before we do this, we must try to write f in a suitable form. To do this, write $$c=(c_1,c_2,c_3)$$ and $$h=(h_1,h_2,h_3)$$. Then

$$f(h_1,h_2,h_3)=(c_1,c_2,c_3)\times (h_1,h_2,h_3)+\|(h_1,h_2,h_3)\|^2(c_1,c_2,c_3)$$

Try to calculate the vectorial product and the norm, then you will obtain a nice function.

The next step is to take the partial derivatives $$\frac{\partial f}{\partial h_i}$$ and to see if the result is continuous. If so, then our function is differentiable.

After doing this, if you really want to know the function Df(a)(h), then you can use the formula

$$Df(a)(h)=\sum_{i=1}^3{h_i\frac{\partial f}{\partial h_i}(a)}$$

Good luck!

 

[Ted123]

Checking differentiability is always a little involved. Certainly if you try to prove it from the defintion, which is quite hard sometimes. Luckily, there are theorems which make our life easier. I'm talking about the theorem which says that "if a function has continuous partial derivatives at a, then it is differentiable at a".

So, what I suggest is try to take the partial derivatives of f. But before we do this, we must try to write f in a suitable form. To do this, write $$c=(c_1,c_2,c_3)$$ and $$h=(h_1,h_2,h_3)$$. Then

$$f(h_1,h_2,h_3)=(c_1,c_2,c_3)\times (h_1,h_2,h_3)+\|(h_1,h_2,h_3)\|^2(c_1,c_2,c_3)$$

Try to calculate the vectorial product and the norm, then you will obtain a nice function.

The next step is to take the partial derivatives $$\frac{\partial f}{\partial h_i}$$ and to see if the result is continuous. If so, then our function is differentiable.

After doing this, if you really want to know the function Df(a)(h), then you can use the formula

$$Df(a)(h)=\sum_{i=1}^3{h_i\frac{\partial f}{\partial h_i}(a)}$$

Good luck!

Thanks but there is a very big problem with that. That is the question tells me to use the given definition!

 

[micromass]

Hmm. Thne I still suggest to write out the function with $$c=(c_1,c_2,c_3)$$ and $$h=(h_1,h_2,h_3)$$ and see where that gets us...

 

[Ted123]

Hmm. Thne I still suggest to write out the function with $$c=(c_1,c_2,c_3)$$ and $$h=(h_1,h_2,h_3)$$ and see where that gets us...

Well I've realized that, most fundamentally, Df(a) should be a linear map depending only on a (not h). So what I've written isn't even potentially correct.

On a secondary level, I've ignored the c x h term (and left it as part of the error). This term doesn't go to 0 as h does, so I have to account for it in the derivative rather than the error term.

Can you see how?

 

[micromass]

So you've gotten to the expression

$$f(a+h)-f(a)=c\times h+2c(a.h)+c(h.h)$$

Try taking $$Df(a)(h)=c\times h+ 2c(a.h)$$. That should do the trick...

 

[Ted123]

So you've gotten to the expression

$$f(a+h)-f(a)=c\times h+2c(a.h)+c(h.h)$$

Try taking $$Df(a)(h)=c\times h+ 2c(a.h)$$. That should do the trick...

But Df(a) is still not a linear map is it (as it contains h)?

 

[micromass]

I don't really see your problem? why can't it be linear?
Surely we have that $$Df(a)(\alpha h+\beta h^\prime)=\alpha Df(a)(h)+\beta Df(a)(h^\prime)$$

 

[Ted123]

I don't really see your problem? why can't it be linear?
Surely we have that $$Df(a)(\alpha h+\beta h^\prime)=\alpha Df(a)(h)+\beta Df(a)(h^\prime)$$

So from the expression

[itex]f({\bf a} + {\bf h})-f({\bf a})= {\bf c} \times {\bf h} + 2 {\bf c}({\bf a}\cdot {\bf h}) + {\bf c}({\bf h} \cdot {\bf h})[/tex]

can I take $Df({\bf a})({\bf h})= {\bf c} \times {\bf h} + 2{\bf c} ({\bf a} \cdot {\bf h})$ .

This is linear since we have that $Df({\bf a})(\alpha {\bf h} + \beta {\bf h^\prime})=\alpha Df({\bf a})({\bf h})+\beta Df({\bf a})({\bf h^\prime})$

And take $E_f ({\bf a, h}) = {\bf c(h\cdot h}) = {\bf c\|h\|}^2$

And $\displaystyle\lim_{{\bf h}\to {\bf 0}} \frac{E_f ({\bf a, h})}{\|h\|} = \lim_{{\bf h}\to {\bf 0}} \frac{{\bf c\| h\|}^2}{\|{\bf h}\|} = \lim_{{\bf h}\to {\bf 0}} {\bf c\|h\|} = {\bf 0}$

 

[micromass]

Yes, that's about it!

 

[Ted123]

Yes, that's about it!

But $Df(a)$ is dependent upon h as well as a isn't it? It should only be dependent on a.

 

[micromass]

Df(a) is only dependent of a. Df(a)(h) is dependent of a and h, as it should be...

 

[Ted123]

Df(a) is only dependent of a. Df(a)(h) is dependent of a and h, as it should be...

So what does $Df(a)$ equal?

 

[micromass]

Df(a) is a function with variable h. It only depends of a.