True or false: Differentiability with vectors

[Justhanging]

If all the first partial derivatives of f exist at $$\vec{x}$$, and if $$\lim_{\vec{h}\rightarrow\vec{0}}\frac {f(\vec{x})-(\nabla f(\vec{x}))\cdot\vec{h}}{||\vec{h}||} = 0$$

Then f is differentiable at $$\vec{x}$$

Note: Its the magnitude of h on the bottom.

First of all, I don't exactly understand what a function of a vector is like f($$\vec{x}$$). Does it mean that this function is evaluated at the terminal point of this vector?

 

[LCKurtz]

Just guessing trying to fix your TeX for you:

$$\lim_{\vec{h}\rightarrow\vec{0}}\frac {f(\vec{x})-(\nabla f(\vec{x}))\cdot\vec{h}}{|\vec{h}|} = 0$$

Click on it to see it.

 

[Justhanging]

Just guessing trying to fix your TeX for you:

$$\lim_{\vec{h}\rightarrow\vec{0}}\frac {f(\vec{x})-(\nabla f(\vec{x}))\cdot\vec{h}}{|\vec{h}|} = 0$$

Click on it to see it.

You sure know your latex, sorry its my first time and it was late. Everything is right except its the magnitude of h on the bottom not the absolute value. I don't if the notation is the same.

 

[Rasalhague]

Yes, say the vector x has components (1,-4,3). Then f(x) = f(1,-4,3). And yes, the absolute value notation, |h|, is often used instead of ||h|| to mean the magnitude of a vector.

 

[HallsofIvy]

First, do you understand what $\nabla\vec{f}$ means?

And what is your definition of "differentiable" for a vector valued function of 3 variables?

 

[Justhanging]

First, do you understand what $\nabla\vec{f}$ [means?

And what is your definition of "differentiable" for a vector valued function of 3 variables?

Yes it means the gradient of f evaluated at the terminal point of the vector.

My equation for differentiability taken from the book is:

$$\lim_{(\Delta x, \Delta y)\rightarrow\ (0,0)}\frac {\Delta f - f_x (x_0, y_0)\Delta x - f_y(x_0,y_0) \Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0$$

If $$\vec{h}$$ is taken to be $$(\Delta x, \Delta y)$$ than the dot product of h and the gradient evaluated at the vector x is:

$$f_x (\vec{x}) \Delta x + f_y(\vec{x}) \Delta y$$

Then

$$\lim_{(\Delta x, \Delta y)\rightarrow\ (0,0)}\frac {f(\vec{x}) - f_x (\vec{x})\Delta x - f_y(\vec{x}) \Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0$$

I distributed the minus from the original problem but am not exactly sure if I can. It looks very similar to the definition with the exception of deltaF being missing. Is this the only reason why the statement is false? I don't exactly understand the deltaF concept as being approximately $$f_x (\vec{x}) \Delta x + f_y(\vec{x}) \Delta y$$