Saturday, October 6, 2018

real analysis - Differentiability of a three variable function


Since I usually studied and checked differentiability and continuity of one variable, and only a few two variable functions, proceeding to the chapter of multivariables, I encountered a problem.



One of the problems for practice says: Check the differentiability (alongside continuity) of a piecewise function: $$f(x,y,z) = \begin{cases} (x^2+y^2+z^2)\sin(\frac{1}{x^2+y^2+z^2}), & (x,y,z)\neq(0,0,0) \\[2ex] 0, & (x,y,z)=(0,0,0) \end{cases}$$ We usually always checked continuity aswell even though it might suffice to prove differentiability only.


Continuity in one var. functions was proved by checking limits or by the epsilon-delta definition. We proved differentiability by checking if both left and right derivatives (by definition) are equal.


I'm a bit confused here on both continuity and differentiability, which method to use. I don't know how I'd go about continuity. And for differentiability I suppose if all partial derivatives exist, the function is differentiable. So I'd have to check the equality of all sides of partial derivatives with limits? How is this usually handled?


Answer



The function $f$ is differentiable at $(0,0,0)$ and $f'(0,0,0)$ is the null function. What this means is that$$\lim_{(x,y,z)\to(0,0,0)}\frac{(x^2+y^2+z^2)\sin\left(\frac1{x^2+y^2+z^2}\right)}{\|(x,y,z)\|}=0,$$which is true, since the previous limit is equal to$$\lim_{(x,y,z)\to(0,0,0)}\|(x,y,z)\|\sin\left(\frac1{x^2+y^2+z^2}\right)$$and the sine function is bounded.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...