Well, honestly, I wouldn't know where to start. I suppose I'd explore the problem by looking at powers of $a$ and $b$ and try to determine whether or not they're in the kernel of this homomorphism.
In the answer given to this question (math.stackexchange.com/a/1023152), why does the Cauchy Schwarz inequality yield the $n^6$ term? Shouldn’t it be $n^3$ if CS is applied to each sum with a vector of entries with only $1$s? Also, are $h_l, h_k, h_m$ the components of $\textbf{h}$? If so, how is the equality that follows justified - that is, the product of the sums of the individual components squared equals the norm of $\textbf{h}$ to the power of 6.
@TedShifrin I’m looking for an intuitive explanation for why the error term in a multivariable Taylor polynomial of degree $n$ is big O of the norm of $\textbf{h}$ raised to the power of $n+1$, where $\textbf{h}$ is the point of evaluation.
The last term will be a polynomial in the components of $\textbf{h}$, but can’t see how to derive the norm from the polynomial.
So you're starting with a formula for the error term?
That estimate is precisely what I told you this morning. All you need is $|h_i| \le \|h\|$ for each $i$.
Nothing fancy.
So any polynomial of degree $k$ is bounded by $N\|h\|^k$, where $N$ is the sum of the absolute values of the coefficients.
The best way to understand the Taylor error, I think, is in terms of the single-variable Taylor polynomial and something using the compactness of the unit sphere (more or less the set of directions in $\Bbb R^n$).
Theorem:
Let $(X,\tau)$ be a topological space and $A\subseteq X$. Let $\tau_{A}$ be the induced topology on $A$. Then
(1) $U$ is open in $A$ $\iff$ $\exists U'$(open in X) : $ U= U' \cap A$
(2)$F$ is closed $in$ $A$ $\iff$ $\exists F'$(closed in X) $:$ $F= F' \cap A$
My Proof:
(1) Follows f...
Alright, with regards to everything prior to "For the converse.." I was trying to show that if $F$ is closed in the subspace $A$ , which is the set with the induced topology then there exists a closed subset of , F, of $X$ ( the parent space) such that the intersection with A is F
@TedShifrin haha, yeah, ill try to stop using it, idk why I love to use that word
OK. I'm satisfied. This was about right (once things were cleared up) for learning the basics in a topology/analysis course at the beginning. Later on, you might make things less detailed.
Proposition 2:
Let $\textbf{f} : U \subseteq \mathbb{R}^n \rightarrow \mathbb{R}^m$ be defined on an open set U.
$\lim_{\textbf{x} \rightarrow \textbf{a}} \textbf{f(x)} = \textbf{L}$ if and only if $\lim_{\textbf{x} \rightarrow \textbf{a}} f_j(x) = L_{j}$ for each $j =1,2,3,…,m$.
What I want ...
Theorem:
Let $(X,\tau)$ be a topological space and $A\subseteq X$. Let $\tau_{A}$ be the induced topology on $A$. Then
(1) $U$ is open in $A$ $\iff$ $\exists U'$(open in X) : $ U= U' \cap A$
(2)$F$ is closed $in$ $A$ $\iff$ $\exists F'$(closed in X) $:$ $F= F' \cap A$
My Proof:
(1) Follows f...
