Example of two theorems I have problems with:
Mean value theorem:
$U\subseteq\mathbb R^n$ open, $f:U\to\mathbb R^m$continuously differentiable, $x\in U$, $\xi\in\mathbb R^n$ such that $x+t\xi\in U$ for $t\in[0,1]$.
Then $f(x+\xi)-f(x) = \left(\int_0^1Df(x+t\xi)\text{dt}\right)\cdot\xi$
Taylor's theorem:
$U\subseteq\mathbb R^n$ open, $x\in U$, $\xi\in\mathbb R^n$ such that $x+t\xi\in U$ for $t\in[0,1]$, $f:U\to\mathbb R$ $k+1$-times continuously differentiable.
Then there exists $\theta\in[0,1]$ such that $ f(x+\xi) = \sum_{|\alpha|\le k}\frac{D^\alpha f(x)}{\alpha!}\xi^\alpha + \sum_{|\alpha|=k+1}\frac{D^\alpha f(x+\theta\xi)}{\alpha!}\xi^\alpha$.
I dont understand the arguments that are passed to the functions. I know that $x+t\xi$ is a line.
However we pass things like $x+\xi$ as parameters to both functions, and then use the line-representation on the RHS. Is the line on the LHS just with $t=1$ ?
Why do we even have to pass a line to a function, cant we just pick a vector from the open set?
Is there maybe a visual representation of what is happening there in low dimensions?
For the Mean Value Theorem, you need to regard $x+\xi$ and $x+t\xi$ as simply points in $\mathbb{R}^n$. Yes, as $t$ ranges from $0$ to $1$, $x+t\xi$ traces a line segment from $x$ to $x+\xi$, as you said, but for any given value of $t$, $x+t\xi$ is just a point on that line segment; i.e., a point in $\mathbb{R}^n$. Taylor's Theorem is similar.