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I was reading more on differentials and I found the phrasing "let there be a point $c$ in a neighborhood $D$ (...) $Df(c)$ is a one to one function". I understand that for a differential we also have a direction, but I fail to understand how a differential in a fixed point can be a function. I might have understanding problems with the definition of a differential and I would appreciate it if someone could explain it to me better. Thank you.

anne
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  • The differential at a point $x$ is a linear function $Df(x)(h)$, in the variable $h$, that satisfies that $\lim_{h\to0}\frac{|f(x+h)-f(x)-Df(x)(h)|}{|h|}=0$. – OscarRascal Jan 27 '20 at 11:22

2 Answers2

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Given a differentiable function $f:\Bbb R^m\to \Bbb R^n$ and $c\in \Bbb R^m$ a point. Then the differential of $f$ at $c$ is a linear map $Df(c):\Bbb R^m\to\Bbb R^n$ given by $$ f(c+x)\approx f(c)+Df(c)(x) $$ for a certain rigorously defined meaning of $\approx$.

As an example, if $f:\Bbb R\to\Bbb R$ is the function $f(x)=x^2$, you might be used to thinking of its derivative at $c$ as the number $2c$. In this framework, the differential of $f$ at $c$ is the linear function $x\mapsto 2c\cdot x$. In general, the differential is represented in matrix form by the Jacobian matrix.

Arthur
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  • in that $2cx$, $x$ is the direction of the function, say a real number? – anne Jan 27 '20 at 11:34
  • @anne In that differential, $x$ is a displacement from $c$, so yes, in this case it's a real number. Which is to say, we have $f(c+x)\approx c^2+2c\cdot x$. For small $x$ this is indeed a good approximation. – Arthur Jan 27 '20 at 11:35
  • so, $x$ is a real number in the neighborhood of $c$? – anne Jan 27 '20 at 11:36
  • @anne $x$ is any real number. It doesn't have to be close to $c$ for the differential to be defined on it. – OscarRascal Jan 27 '20 at 11:38
  • @anne $x$ can be any real number, but for the approximation to be good, it has to be small. Not close to $c$, but close to $0$. This makes $c+x$ close to $c$, which makes $f(c+x)$ close to $f(c)$, and $Df(c)(x)$ tells you how close. – Arthur Jan 27 '20 at 11:39
  • i think i understand now. but what is, then, the direction of a differential? is it this $x$ number? – anne Jan 27 '20 at 11:39
  • @anne No. $x$ is an input to the differential. The differential itself is the linear function "multiply by $2c$". – Arthur Jan 27 '20 at 11:40
  • i understand. thank you very much for the patience – anne Jan 27 '20 at 11:41
  • @anne The differential is a linear function. It doesn't have a direction. Now, in the case of one real variable, the differential is a linear functional. In coordinates it is the dot product by a vector. You can consider the direction of that vector, when it is non-zero. In your example, the vector $(1-2x, 1-3x^2)$ – OscarRascal Jan 27 '20 at 11:42
  • it clicked to me when i said it. thank you @OscarRascal – anne Jan 27 '20 at 11:43
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The differential of a function $f$ at a point $c$ is the best linear approximation $L$ of the values of the function increments in a neighbourhood of $c$, in a very precise sense: $$f(c+h)-f(c)=L\cdot h+o(h).$$

Bernard
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  • so let's say for the function $f(x)=(x-x^2, x-x^3)$ $Df(x)(u)=(1-2x,q-3x^2)u$ where $\forall u \in R$ and so this $u$ makes it a function even if we compute $Df(x)$ in a fixed point $c$? – anne Jan 27 '20 at 11:30
  • Yes: $Df(c)\cdot u=\bigl((1-2c)u , (1-3c^2)u\bigr)$. That's why the differential of a function in several variables is represented by its Jacobian matrix. – Bernard Jan 27 '20 at 11:50
  • i understand now. thank you so much. – anne Jan 27 '20 at 11:54
  • You're welcome. Always glad to help! – Bernard Jan 27 '20 at 11:55