0
$\begingroup$

Question from Calculus 3:

  1. Let $f(x,y)$ be a function that's defined in a neighborhood of $(x_0,y_0)$. Show that exist $a(x,y), b(x,y) $ that are continuous at $(x_0, y_0)$, such that:

$f(x,y)=f(x_0, y_0)+a(x,y)(x-x_0)+b(x,y)(y-y_0)$.

I was able to solve this, but am writing this to explain the next part of the question which I couldn't solve.

  1. Do there exist any points $(x_1, y_1)\neq (x_0 ,y_0)$, such that $b(x_1,y_1)=f_y(x_1,y_1)$. This I wasn't able to solve, though I found that $b(x,y)=f_y (x_0,y_0)$.

For more background, the way I calculated $b$, was through the definition of differentiability, when $f(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+r(x,y)$, when $r(x,y)=o(d((x,y), (x_0,y_0))$, and using a theorem, I wrote:

$r(x,y)=u(x,y)(x-x_0)+v(x,y)(y-y_0)$, when $lim_{(x,y)\rightarrow(x_0,y_0)}v(x,y)=0$, and I defined $b(x,y)=f_y(x_0,y_0)+v(x,y)$.

So to say that $b(x_1,y_1)=f_y(x_1,y_1)$ is to say that $f_y(x_1,y_1)=f_y(x_0,y_0)+v(x_1,y_1)$. Now does there exist $(x_1,y_1)\neq (x_0,y_0)$, such that this is true?

$\endgroup$
0
$\begingroup$

Consider e.g. $f(x,y) = y^2$ with $x_0 = 0$, $y_0 = 0$. Then you can take $a(x,y) = 0$ and $b(x,y) = y$. You never have $y_1 = f_y(x_1,y_21) = 2 y_1$ unless $y_1 = 0$.

EDIT: OK, here's a counterexample. Take $$\eqalign{f(x,y) &= x^2 y + {y}^{3}\cr a(x,y) &= x y\cr b(x,y) &= y^2 \cr x_0 = y_0 = 0\cr} $$ Then $f_y(x_1,y_1) = x^2 + 3 y^2 \ne b(x,y) = y^2$ unless $x=y=0$.

$\endgroup$
  • $\begingroup$ This definitely isn't a counter example since for all $\varepsilon >0$, we can take $|x|<\varepsilon$, such that $||(x,y)||<\varepsilon$, and therefore, in every neighborhood of $(0,0)$, exists $(x,y)$ in that neighborhood such that the theorem is true. Are you trying to say that the theorem is true? $\endgroup$ – NSA Mar 28 '16 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.