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Let $f(x,y)\in C^1$ in $\mathbb{R^2}$ and let $(x_0+\Delta x,y_0+\Delta y)$ and $(x_0,y_0)$ be points in $\mathbb{R^2}$.

Prove that $\exists\theta\in(0,1)$ such that: $$f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0)=\\\frac{\partial f}{\partial x}(x_0+\theta\Delta x,y_0+\theta\Delta y)\Delta x+\frac{\partial f}{\partial y}(x_0+\theta\Delta x,y_0+\theta\Delta y)\Delta y$$

At first glance, this seemed like the differentiability definition, but I tried to make the connection and unfortunately failed. I guess that MVT hides here, but I don't see how to rigorously reach it.

Thanks!

Note: I found a solution to this problem online (not here) but I couldn't understand it, so I'd really appreciate a somewhat detailed solution.

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Use $a$ and $b$ for the endpoints, for convenience. Define $g:[0,1]\to \mathbb R$ by $g(t)=f(bt+(1-t)a).$ Then, by the mean value theorem and the chain rule, there is a $\theta\in (0,1)$ such that

$g(1)-g(0)=g'(\theta)(1-0)=\nabla f(g(\theta)) \cdot (b-a)$.

I'll let you unravel this to obtain your result.

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  • $\begingroup$ First of all, thank you very much! Now - I thought that I could say, that there exist $t_1,t_2\in(0,1)$ such that $x*=x_0+t_1 h$ and $y*=y_0+t_2 k$, but how do I show that $t_1=t_2\triangleq\theta$? Also, even if I manage to prove that, I am not exactly left with that I need to prove $\endgroup$
    – Amit Zach
    Apr 18 '19 at 22:14
  • $\begingroup$ I misread your question sorry! I will repost. $\endgroup$ Apr 18 '19 at 22:22
  • $\begingroup$ Now I understand. Thank you very much! $\endgroup$
    – Amit Zach
    Apr 19 '19 at 7:52
  • $\begingroup$ You´re welcome! $\endgroup$ Apr 19 '19 at 13:50

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