Well I set it up through $\|f(A+H) - f(A) \| = \|(A+H)^{-1} - A^{-1}\|$. I wanted to ask about the $A+H$ notion as well. But for the current moment I used that to represent a general invertible map in the space. So that whole set is less than $\|A^{-1}\| \frac{\epsilon}{1- \epsilon}$ which I have from part (b) of the question.
@Jakobian It's very unclear. But from their comments, I guess they want some way to evaluate the divergent Taylor expansion of $$\lim_{v\to c}\,1/\sqrt{1-v^2/c^2}$$ similar to 1+2+3+4+...=-1/12.
So another idea from the same part (b) which can also be applied is that if $\|H\| < \epsilon / \|A^{-1}\|$ then $\|(A+H)^{-1} - A^{-1}\|< \frac{\epsilon}{1- \epsilon}$ but this time $\epsilon = \epsilon / \|A^{-1}\|$
I used the relationship you allude to from section 5.1 when I did 6 (c): $\|A\| \leq (\sum_{i,j}a_{i,j}^2)^{1/2} \leq \sqrt{n} \|A\|$. And to your second question that's what I'm attempting. THe last part was just me playing around with what I know
well actually I didn't but I know that it was the hint...
but that doesn't exactly work with my inequality relationship that I have because what I want is $ \|(A+H)^{-1} - A^{-1}\| < \delta$, but what I have is: $ \|(A+H)^{-1} - A^{-1}\| < \|A^{-1}\| \frac{\epsilon}{1- \epsilon}$ as a relationship. But going back to the original inequalities if I can bound $\|H\|$ properly then I'll be in business
for all $\epsilon > 0 $, and for all $A \in X$ there exists a $\delta > 0$ such that for all $H \in X$ if $\| H - A \| < \delta$, the $\|f(H) - f(A)\| < \epsilon$. Here $X$ is just the set of invertible matrices
I do wonder if I should have used $A+H$ instead of just $H$ in the defintion.
I was just trying to see how the $\delta$ creates the implication in the whole idea. it is because we have $\|A^{-1}\|\|H\| < \epsilon$, but in this case $\epsilon$ is my $\epsilon^{'}$ so the relationship that I established in part (b) can be used.
