2

How do you prove Euler's Theorem

$$du=\left({\partial u \over \partial x}\right)dx+\left({\partial u \over \partial y}\right)dy $$ if $u=f(x,y)$.

I also heard that Ramanujan developed another method can somebody know that?

Xoque55
  • 4,384
DebojyotiMukh
  • 915
  • Are you looking for a geometrically "heavy" proof? Such as "Let $\sigma$ be a curve in $\mathbb{R}^2$ such that $\dot{\sigma}(0) = v \in T_p \mathbb{R}^2$ where $\sigma(0) = p$. Then $du(v) = ...$ from which we conclude the result"? – Tom Nov 04 '13 at 18:17
  • I need all methods you guys would suggest. – DebojyotiMukh Nov 04 '13 at 18:26
  • If it's what you're talking about, it simply follows from the definition of the differentiation operator $\text d$ of differential forms. – pppqqq Nov 04 '13 at 18:41

1 Answers1

2

I would propose an "Analysis II based" approach to start with.

Let $p=(x,y)$ and $f:\Omega\subseteq \mathbb R^2\rightarrow \mathbb R$ be a function defined on the open set $\Omega\ni p$.

We say that $f$ is differentiable at $p$ if for every $h:=(dx,dy)$ s.t. $p+h\in\Omega$ there exists a vector $a\in\mathbb R^2$ s.t. $df(p):\mathbb R^2\rightarrow\mathbb R$ s.t.

$$f(p+h)-f(p)=\langle a,h\rangle+O(\|h\|^2).$$

The linear map $a\rightarrow \langle a,h\rangle$ from $\mathbb R^2$ to $\mathbb R$ is called differential of $f$ at $p$ and it is denoted by $df(p)$. The vector $h$ can be considered as a small increment around $p$.

If $f$ is differentiable it follows that (this is shown on every textbook of Analysis)

$$df(p)(a):=\langle a,h\rangle=\langle \nabla f(p),h\rangle,$$

where $\nabla f(p)$ is the gradient of $f$ at $p$. In summary, recalling that $p=(x,y)$ and $h=(dx,dy)$ we arrive at

$$df(p)=\langle \nabla f(p),h\rangle=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy, $$

with the partial derivatives computed at $p$.

Avitus
  • 14,018