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00:22
ugh just took the test, as I expected my calc 1 knowledge has failed me. I had to take $\frac{df}{dt}$ of $f(x,y)=3xe^{x^2y}$ with $x(t)=\sin t$ and $y(t) = \cos t$ I set up the chain rule properly with $\frac{df}{dt}= \frac{df}{dx}\frac{dx}{dt}+\frac{df}{dy}\frac{dy}{dt}$ but I think I did the computation wrong.
like $\frac{df}{dx} = \frac{d(3x\cdot e^{x^2y})}{dx}$
i did the product rule but I think i differentiated $e^{x^2y}$ incorrectly
i forget what I did
I think i just brought down the whole exponent term
like $x^2ye^{x^2y}$
i know d/dx of e^x is e^x and e^ax is ae^ax
but i didn't know what to do with the x^2 term
oh.. i think i was meant to just bring down the y term then
no idea
ALSO big fail on $f(x,y) = 2xy - x^3 - y^2$ finding all critical points and relative extrema. I set up the 2nd partials test but was left with $d = 12x - 4$ from $d = F_{xx}F_{yy}-(F_{xy})^2$
everything else seemed okay, but I'm just mad because those two were worth 26 points and I hope I get partial credit for showing the setup and process.
Sounds like you should have been studying/practicing for weeks, not minutes.
00:39
if ted had his way, all we'd do is study. no time for fun, just study study study.
Something like that, yes.
40 years as a teacher ….
00:51
Can I have something to eat? No! And stop breathing and study!
No food. Only vodka.
01:17
Vodka is gross, but yeah I should have studied more but I think I did okay for the amount of time i had
Just wish I didn't take such a huge gap between calc 1/2 and calc 3 (6 years)
i remember being so motivated in high school proving theorems and studying calc all day ery day. i just assumed it was all up there in the noggin and would come out when the time was right
alas on opening night my brain performer choked and failed on stage
yeah, wow that is a gap.
For many years, we flagged students who’d taken more than a year after the preceding course. Hard to be successful with a big gap.
But you had weeks to do homework and practice. I hope you can be more diligent now. I’m not being mean — just trying to
be realistic.
P.S. i drink gin, not vodka.
i like vodka well enough, but i don't think i could win a debate where i had to argue that it is delicious.
Tasteless = delicious
01:58
i mean who drinks alcohol for the taste lets be real
Yeah starting now getting 100s baby, just gotta focus up.
I’m in your corner, Obliv. Happy to help … just not at the last minute! …. P.S. I am weird, but I do drink (good) gin for the taste.
Thank you Prof, Yes I do find I enjoy certain whiskeys/wines more than others but I don't know if I chalk it up to taste or what lol. Haven't actually tried gin so I can't say if you're weird or not yet.
oh, he's weird. gin's got nothing to do with that.
i like gin too but will admit it is an unconventional taste. again, couldn't win a debate about why it's delicious. it just is, for those who enjoy it.
02:15
Even the best gins are very, very different. Floral (nah), herbal (yes).
I think the grossest thing I've done was just drink plain tito's vodka with oj. Just terrible
but u gotta do what u gotta do to save money and get drunk lol
What were you expecting?
that's a fine mix, unless you use bargain basement OJ.
ROFL
“Orange drink”
Did Munchkin get mommy a nice bday present?
well.. i had a really nice vodka tonic at a bar and thought OJ couldn't be that bad. It really didn't mask the flavor like the tonic did mostly because I put way too much titos in it. Actually I think I blacked out that night which made me averse to OJ for months after.
you learn a lot in college
02:28
like anything from concentrate, yeah, that's going to suck.
funny how it's always the mixers that get the blame, or the food. a friend of mine in college swore off an entire restaurant due to some digestive upset he had after a session that i would have described as a marathon of drinking. but sure, blame the food.
Lol yeah that's true, I bet they get a lot of complaints. Maybe that's partially why they don't put a lot of alcohol in those drinks.
@TedShifrin Hey Ted
I need your help sir
Also to save money, but like I doubt they save that much.
are you free?
cheap vodka is generally also fine. you are more at risk from your orange juice.
i got e coli from fresh squeezed orange juice at a local bagel place about one month into the pandemic. my wife was so nervous that i had covid until she saw posts about e coli from other people who had gone to that place on social media.
02:33
that sucks, only ever had something like salmonella (didn't get it checked out) but your body just goes HAM on getting rid of something like that. High fever, in and out of bathroom constantly, whole body feels shot.
I don't think I've got covid yet, or I could have been asymptomatic, either way I rarely get sick. It was a bad batch of chicken for me.
yes, i spent about 72 hours in the bathroom.
@leslietownes How the hell did they manage that? No cleaning?
@Jack For a few minutes. No, I’m not going to proofread LaTeX.
I think people in the olden days died to these types of diseases out of pure dehydration (not sure).
ted: yes i think it was a general failure to wash the oranges, or the tank in which the oranges were placed.
@TedShifrin not the latex sir ! it is finished , just want your opinion on my proofs !
it is already done :D
02:35
obliv: absolutely. because often the water that you could get was contaminated with the same thing that made you sick. this is why cholera was so deadly. it's not that bad of a disease if you have clean water to get over it with.
can I email you ? since it is a homework , i dont want it to be on the internet
No, I’m not going to be a personal homework vetter.
okay :/
thanks anyway
Doesn't your body get resistant to certain bacteria over time though? I remember googling why humans can't drink from stagnant sources of water once because I was curious. Apparently we can it's just we need to get used to the bacteria like other animals would.
And our immunity systems are compromised to various extents.
02:37
obviously the process isn't fun as you would get dysentery or whatever it's called
there's definitely stuff like that. 'montezuma's revenge' being one example. i don't think anyone has developed resistance to cholera.
That's a hilarious name, probably coined by tourists.
Yeah our immune systems probably aren't as good as other animals, not sure though just a guess.
my cat can't stop chewing small pieces of plastic and then vomiting them up like 90 seconds later. this isn't an immune system thing, it's a brain power thing.
For continuous functions $f,g:\Bbb R\to\Bbb R$, $f(x) = O(g(x))$ does not imply $|f(x)|\leq C\cdot g(x)$ for all $x\in\Bbb R$.
The inequality holds only at the tail part $x\to\infty$.
Or as $x\to 0$, depending on context.
03:00
Oh but if $f$ is an even function then... no still not.
03:22
The context needs to be made explicit.
Yes, I know but I solved it.
It's about this. The second claim about $g$. Being $\hat{f},f$ even and exponential never vanishes imply the statement.
The context ($\infty$) is clear!
its a sort of uncertainty principle
03:48
I’m always uncertain.
04:03
certainly uncertain
ted: learned today that some day care staff refer to my daughter, semi affectionately, as "la gritona" (the shouter).
one of my principles is to always be uncertain.
the apple does not fall far from the tree
@copper.hat -heisenberg 1927 (said not really)
ah, the Heisenberg disaster
oh the uncertainty
04:06
fun fact heisenberg was awful at experimental physics because he didn't put any effort into it and felt so embarrassed because his theoretical physics prof had to vouch for him that he developed the uncertainty principle as a project
grossly paraphrased/simplified
i am sure some physicist opened a pub named hbar.
i am sure there were some great puns at the time
it's a golden opportunity. we have room 71 here
imagine the jokes if he was late for a lecture
if i had the money i'd open a very high end but casual bar called hbar and it'd be so cool.
where's heisenberg?
04:08
lol
he was observed in a fast moving stream, but they were not quite sure where he was
0
Q: Stein complex analysis exercise 4.12

one potato two potato Let $f$ be a function on $\Bbb R$ that satisfies $$f(x) = O(e^{-\pi x^2})\quad\text{and}\quad\hat{f}(\xi) = O(e^{-\pi\xi^2}).$$ Then if $f$ is even then $\hat{f}$ extends to an even entire function. Moreover, if $g(z) = \hat{f}(z^{1/2})$, then g satisfies $$|g(x)|\leq ce^{-\pi x}\quad\text{and}\...

ffs, maths?
As cool as physics is as a subject, one cannot deny one of the greatest appeals is the ability to wear a trench coat and talk about the universe with colleagues.
Like, old pics of heisenberg and others in the early 1900s they rocked trench coats..
I definitely prefer trench coats to lab coats.
reminds me of a comic a friend gave me, one of the strips (so to speak) was about fred filstrop the flasher
i think Phragman would be a great name for a Marvel character with super mathematical powers
or a porn movie star, not sure which
04:16
Lool, dang Fred Filstrop discrediting the swag of trench coats.
i had a trench coat in my late teens/early 20s
actually i stole my dad's coat
sounds a bit phallic for sure. also what in the world is $\xi$
bless you
do you pronounce that z-eye?
04:17
i guess part of learning maths is learning greek
@leslietownes semi-affectionate may be an exaggeration
sort of like kiss-see but one syllable
leslie, i have friends down there who could help with re-education
at what point do we stop using greek letters and a different language? I wonder how great a maths achievement must be for the world to let you choose the symbols.
a parent at one of my aunt's kids' school made a slightly negative comment about one of her kids. that night she went to the parent's house and cut every single flower off every plant in their rather large (for urban dublin) garden.
LOL
04:21
one wonders why it took so long to approximate peace in ireland
That's crazy
she is a little. some of those genes in all of the family.
thankfully seems to not have expressed in my own offspring.
Wait until they get older
If they haven't already absorbed your insanity that is
neither of them have been 'triggered' yet
sort of like a russian mole, its the irish gene
that's good. gives me hope that having kids is a good idea
04:24
i thin my kids think that having parents is a bad idea...
but i'm not irish so my craziness might not be safeguarded.
having parents is a great idea until you're a teenager then they're an awful idea.,
and then they're a great idea again once you DIE. mwahaha
strangely, mine were more accepting of me in their late teens
but it is thin ice
my daughter gets mad when i scream at other drivers. can't understand why
I do see that, and it always bewilders me, when people have functioning families as the kid is growing up. It's definitely possible.
i think we had a sort of functional sibling unit, there were 5 of us, so critical mass
you need > 2:1 ratio, i think
i'm only an armchair psychiatrist but to project my own experiences, sometimes when you see anger in an adult even if it's not directed to you if you're used to anger being directed to you you just have a negative aversion in general.
Could also be the critical mass ratio yes
04:27
have posted it before, but its still amusing youtube.com/watch?v=pzRhlwJ49Os
haha, reminds me of this youtube.com/watch?v=-iLKHMYwmFg
bonus points since it's scottish
omg, my kids will be a psychiatrist's twin prime conjecture should they ever go to one
that reminds me of all the stuff i need to do...
stuff you need to do? what stuff you need to do. blankets self into earth
@TedShifrin one of them really likes her, and another one of them definitely doesn't. unfortunately the former is leaving soon and being replaced with the latter.
hmm, staff should not leave kids with such a negative impression
04:38
@leslietownes you may have to teach her to behave better!
you can't, like, teach a kid to behave man, that's like, discipline. that would be like telling gene krupa not to go boom-boom-bah- [mimes drumming]
i was the problem in my kid's schools.
rolls $8^\pi + e^9$ eyes
woohoo, a convex problem, l8r
04:53
In general, if $\{(f_i,D_i)\}_{i=1}^n$ is a chain of function elements such that $D_i\cap D_{i+1}\neq\emptyset$ for each $i$, then it's not true in general that a function $f$ can be defined on $\bigcup_{i=1}^n D_i$ so that $f|_{D_i}\equiv f_i$ for each $i$. The example of the failure of such extension is the Log function. The problem is the well-definedness (also because an analytic continuation is not transitive).
Is this the only obstruction? I mean if it's well-defined by Monodromy theorem on a simply connected domain, then we can find the desired extension $f$?
And the extension $f$ is a trivial one right?
now i'm uncertain. what does 'trivial' mean in this context?
$f(x) := f_i(x)$ if $x\in D_i$.
i should also be more nuanced. if you only want to extend along a path, and you can locally find extensions, there will be no problem in gluing them together unless you begin wanting to do something like form a loop.
but some functions, for just whatever reason, may not be so extendible, outside of a domain.
like there's stuff analytic on the unit disc that can't be analytically continued to any point in the boundary. that kind of crap.
05:06
Say we have two geometric circles in $S^3\subseteq\Bbb R^4$
They each bound a (geometric) disk in $\Bbb R^4$
Is it true that those disks intersect (in a point) iff the circles are linked?
Since one circle intersects the other disk in a point, by the IVT don’t we expect the disks to intersect more?
I admit that putting this a dimension up confuses me.
If the circles link in $S^3$, why do they link in $\Bbb R^4$? They should not.
05:23
There is no linking in $\Bbb R^4$
Also, two 2-planes generically intersect in a single point
For example, the circles $(\cos\theta,\sin\theta,0,0)$ and $(0,0,\cos\theta,\sin\theta)$ bound parts of the xy- and zw-planes respectively, which intersect in the origin
I guess the point is that these disks can't intersect in a line that starts at one circle and ends at the other one, 'cause then you get four points in a line that should all be on the sphere
so if these disks intersect in more than one point, and therefore in a line, then their boundaries intersect too
which can't happen
Intersection being preserved by deformation in $S^3$ doesn’t imply it by deformation in $\Bbb R^4$.
So the circle needn’t intersect the other disk, I guess. I will need to ponder this.
Yes, but consider this
Intersect this all with shrinking spheres
Assuming neither disk contains the origin, eventually the sphere is small enough to hit neither disk
Within those spheres, it looks like the circles are shrinking to nothing
If they're linked they must intersect at some point.
(I suppose this argument doesn't require the disks to be geometrically flat, just to be fully contained inside the unit ball)
06:18
In this context, it then holds on all of $\Bbb C$ by analytic continuation means the identity theorem?
yes. 'on all of C' should maybe be 'on all of C \ integers.' which is totally fine as an input to the identity theorem, if you've proved it the right way.
Ok, thank you
06:48
An example of a metric space in which not every closed set is compact:-
Let $\mathbb N$ be endowed with discrete metric. Then $N$ is a closed but not compact.
infinite set with discrete topology?
yeah, there you go
Because $\mathbb N$ has an open cover $\{\{n\} \}_{n=1}^\infty$ without a finite subcover.
you can find other examples of course. e.g. the unit ball in any banach space will be closed in the norm topology, but it is not going to be compact if that space is infinite dimensional.
:-)
Disjoint compact subspaces A and B in a Hausdorff space can be separated by disjoint open sets.
Proof: For any b in B, there exist disjoint open sets U_b (containing A) and V_b (containing b). B is covered by V_b's and by compactness of B there are finitely many b' s such that B is contained in $\cup_{1\le i\le k} V_{b_i} $.
It can be seen now that the open sets U= $\cap_{1\le i\le k} U_{b_i} $ and $\cup_{1\le i\le k} V_{b_i} $ work.
QED
07:10
This is part of the proof of the estimate
$$\left|{1\over\Gamma(s)}\right|\leq c_1e^{c_2|s|\log |s|}.$$
During the proof, author divides the estimate of the term
$$\sum_{n=0}^\infty{(-1)^n\over n!(n+1-s)}{\sin\pi s\over\pi}$$
into two parts $|Im(s)|>1$ and $|Im(s)|\leq 1$. I don't know why this is required. $|\sin\pi s|\leq e^{\pi |s|}$ for all $s\in\Bbb C$ so the displayed sum is majorized by $ce^{\pi |s|}$ isn't it?
07:52
Does anyone know why the norm on C^n (space of continously differentiable functions) a sum of sups? |f|_C^n = sup|f| + sup|f'| + \dots + \sup|f^n|?
08:27
Anyone got p-rated version of William Wade Intro to analysis? I remember someone posting a link to that can't find it :(
Ignore me I found the global edition in lib ge*.
When I get rich I will pay. I promise.
Student loan is killing me.
Hey! Can you help me out please with a geometry question?
 
3 hours later…
11:31
1
Q: Stein complex analysis exercise 4.12

one potato two potato Let $f$ be a function on $\Bbb R$ that satisfies $$f(x) = O(e^{-\pi x^2})\quad\text{and}\quad\hat{f}(\xi) = O(e^{-\pi\xi^2}).$$ Then if $f$ is even then $\hat{f}$ extends to an even entire function. Moreover, if $g(z) = \hat{f}(z^{1/2})$, then g satisfies $$|g(x)|\leq ce^{-\pi x}\quad\text{and}\...

Oh, it's just a reminder. It's not answered yet.
@robjohn Could you take a look if you have spare time?
12:04
0
Q: Finding parallel vectors inside a ball

Rubi ShnolI am reading through a paper that uses a geometric construction not quite clear to me. Namely, let $B$ be a closed ball in $\mathbb R^n$ centered at $x$. Let $s \in B$. Next, let $w$ be an arbitrary vector. Then, for a sufficiently small $\varepsilon > 0$, there is $z \in \partial B$ such that $$...

12:42
@Lemon is your question why this is a norm or what the motivation for introducing this norm is?
 
3 hours later…
16:06
Suppose that [0,1] is with subspace topology from R. Let's call this topological space ([0,1], $T_1$). Let $T_2$ be any topology of [0,1]. The statement "If $T_1$ is proper subset of $T_2$, then $[0,1]$ is not compact in $T_2$." is true or false?
It is clear that the statement is true for some $T_2$'s (like discrete topology for example).
Intuitively, if $T_2\subset T_1$ (proper containment), then every open set in T_2 is open in T_1 and hence implies compactness in T_2.
So is the statement true for all $T_2$'s properly containing $T_1$?
16:37
@Koro think about the identity map from $T_2$ to $T_1$. If $T_2$ were compact, what would happen?
folland had a sequence of exercises about this. compact hausdorff spaces are 'just right'
weird way of spelling "compact smooth manifolds"
or "spaces with the homotopy type of a CW complex", depending on mood
16:55
i'm not enough of a pervert to remember what other kinds of spaces are out there.
@AlessandroCodenotti For any open set U in $T_1$, $i^{-1}(U)=U\in T_2$ so the identity map i is continuous. If $T_2$ is compact, then noting that 1) i is bijection, and 2) [0,1] with $T_1$ is Hausdorff, it follows that $i$ is a homeomorphism, which implies that $T_1= T_2$ contradiction!
@AlessandroCodenotti Thanks a lot. The identity map idea is so amazing :-).
@leslietownes I'll check them out. Thanks a lot. :-).
17:25
@Koro indeed. Note that the same argument works for any compact Hausdorff space, there is nothing special about $[0,1]$
@AlessandroCodenotti what alessandro said
@Lemon Because you want the norm to control all the derivatives. If the overall norm is $<\epsilon$, then you know that $\sup |f^{(k)}|<\epsilon$ for all $k=0,1,\dots,n$.
17:54
Any ideas for a bivariate function which is not coercive but $lim_{|x_1| \rightarrow \infty} f(x_1, ax_1) \rightarrow \infty$ and the same for $f(ax_2, x_2)$
It has to be continuous
and for all $a$, I was thinking first to take $\frac{x_1^2}{x_2}$
Because along $x_2 = x_1^2$ we do not explode as per coerciveness
But its not continuous :/
18:18
If I prove the existence of an $N$ such that if $n\gt N$ then $|a_n-L|\lt \varepsilon$ for all $\varepsilon \gt 0$, did I prove the sequence just converges or converges to $L$? I don’t have to choose the lowest possible $N$ do I?
@AlessandroCodenotti yes, noted.
cotton: the order of quantifiers is a little confusing there. if you can find such an N given any epsilon, you've not only proved that the sequence converges, but that it converges to L.
Now, I want to consider a similar statement but this time, I want to comment on Hausdorff property of $T_2$.
the cauchy criterion (google it) is often handy in proving that a sequence converges without explicit reference to what it converges to.
i.e., the statement-if $T_2$ is a proper subset of $T_1$ then [0,1] with $T_2$ is Hausdorff.
The statement is clearly false if I take $T_2\equiv $ indiscrete topology.
But is the statement true for any $T_2$ proper subset of $T_1$?
the identity map: $i$ from $[0,1]$ with T_1 to $T_2$ (notation abuse here) is continuous.
$([0,1], T_1)$ is compact so gets mapped to a compact set i.e., $([0,1], T_2)$ is compact.
this doesn't help.
Using the same argument as before: if $T_2$ were Hausdorff then the map i would be a homeomorphism resulting in $T_1=T_2$ a contradiction!
So the statement is false in general.
18:42
Understanding a question statement: Consider $(\mathbb Z, T)$, where $T$ is the topology generated by sets of the form $A_{m,n} = \{m + nk | k \in \mathbb Z\}$,for m, n in $\mathbb Z$ and $n \ne 0$.
"Topology generated by.." means that T={all A_{m,n}'s} $\cup \{\emptyset\}\cup \{\mathbb Z\}$.
is my understanding correct?
i don't think so. union up even numbers and multiples of 5. is that an arithmetic progression? i don't think so. but it's in T?
Ahh, $A_{m,n}$'s form a basis for $T$.
topology generated by means the smallest topology in which all of these are open
Z= {odd} union {even} so (Z,T) is not connected.
@Thorgott isn't this same as saying the topology generated by the sets as basis?
18:58
@Koro subbasis
sure?
any partition of set X will give rise to a subbasis.
I'm saying that in general you will only get a subbasis
I think basis is more appropriate and I think that it's same as what Thorgott said.
is there any connection between the domain of a map and a domain defined as a nonzero ring with no zero divisors
yes, this is called partition topology and it generalizes the discrete and indiscrete topologies
19:00
@Koro the "topology generated by <collection of subsets of a space>" always makes sense. the collection of subsets of a space may or may not be a basis of the resulting topology.
silly: no, just two uses of the same word
@SillyGoose I think that's a confident "no"
oh okay
@leslietownes oh ok, thank you
shrug
19:06
@Thorgott Does topology generated by {collection of subsets} make sense only when union of the subsets is the given space?
i.e., given $A_\alpha\subset X$ and $\cup _\alpha A=X$, only then we define topology generated by $A_\alpha$'s.
nvm, I understood now.
Topology generated by A (subset of X): Consider the set T={topologies on X containing A as element}. S is non empty as it contains discrete topology P(X).
Intersection of all elements of T is a topology which is called the topology generated by A.
yes, but $A$ is a subset of $P(X)$, not $X$
It doesn't matter?
it does, the definition doesn't make sense otherwise
hey I have question, how to find derivative of matrix function, for example f(A) = A^3
actually, you need to look at the topologies $\tau$ such that $A\subseteq\tau$, not $A\in\tau$
19:16
if I take A as subset of P(X), then yes.
Otherwise, why doesn't it make sense?
well, I guess it does make sense, but you're just defining the topology generated by a single subset $A$ of $X$ then
this is not a very interesting notion, I can also write down that topology by hand
{phi, X, A}
haha
yeah :)
right, one should define it for a collection of subsets of X.
thanks a lot :-).
can someone pls help me
for example if its 3 by 3 matrix, what would be derivative of f(A)=A^3
19:25
The production cost for $x$ products is described approximately by $f(x) = \sqrt{x}$.
The function $p$ describes the production cost of one product, when $x$ products are being produced.

So is this again one of those where it's technically $p(x) = \frac{f(x)}{x} = \frac{1}{\sqrt{x}}$ but $f'(x)$ is a good approximate for $p(x)$ and it's easier to calculate so you would pick that?
$f'(x) = \frac{1}{2\sqrt{x}}$
@smth think of the product rule
@Thorgott I am confused about how can I do such things with matrix? What exactly do you think? f'(A) = 3 A^2 A' ?
Or do I imagine this matrix as "transformation" so its actually f(A) = A^3 x
I am confused
(With "again one of those" I mean that we discussed something maybe similar with @XanderHenderson and @Jakobian recently which was a bit different though, it was about f(x + 1) - f(x) versus f'(x))
So is $p(x) = f'(x)$ the way to go here?
@smth: it will be a linear map from M_3 to M_3. I think that the derivative at A (i.e. Df(A))will satisfy $Df(A)(H)=A^2H+AHA+HA^2$ for all $H\in M_3$.
@Koro how you got that?
19:37
Use the relation between derivative and directional derivative.
keep in mind that matrix multiplication is not commutative
19:57
I still don't get it
Do you know the relation between derivative and directional derivative?
actually I am not sure, I am starter at this course, we have learnt that directional derivative is something like derivative in some direction of some vector, I know limit definition of that (h replace with h times vector v) but how can I use that here?
@Koro what is H in Df(A)(H)?
H is any 3 by 3 matrix.
(real)
so that derivative is in respect of matrix?
what is 'in respect of'? You are in several variables here.
20:09
should I imagine f(A) where A is 3 by 3 matrix as 9 variable function?
yes, you could do that.
But not required really.
@smth yes, that's what you need here. By what you said: $Df (A) (H)=\lim_{t\to 0}\frac{f(A+tH)-f(A)}{t}$. Simplify this to get the desired result.
how to simplify that? how to get rid of limit
$f(A)= A^3, f(A+tH)=(A+tH)^3$
bearing in mind that:
25 mins ago, by Thorgott
keep in mind that matrix multiplication is not commutative
just a second please
wait, I got $Df (A) (H)=\lim_{t\to 0}\frac{f(A+tH)-f(A)}{t} = \lim_{t\to 0} \frac{3A^2tH + 3At^2H^2 + t^3H^3}{t} = 3A^2H$. What happened?
Matrix multiplication is not commutative. You made some mistake while multiplying.
20:22
oh I get it, can't use formula for (a+b)^3, just a moment
okay so I get that $Df(A)(H)=A^2H+AHA+HA^2$, but I have question, why do we write Df(A)(H), why not just Df(A) what is point of H here?
Df(A) is a linear map from M_3 (set of all real 3 by 3 matrices) to M_3. You should tell how it's defined, i.e., what it does to any $H\in M_3$.
If you just say "Df_A$ is a linear map", then it doesn't give much details.
one more question, how to prove that $f \in C^1$ (that f is continuous and its first derivative is also a continuous function)?
is it okay to just say that it is composition of continuous functions?
talking about first derivative
also note that the above calculations are based on the fact that "f is differentiable". In case it is not given that f is differentiable, you should use definition of derivative to establish that Df(A) that you obtained earlier is indeed the derivative.
@Thorgott Yeah I was looking for the motivation behind it, but it turns out it is nothing more than iterating the construct on bounded and continuous functions space and that's why we are adding derivatives of f in the C^n norm and that's probably why I can't find many interesting analysis that talks about these C^n spaces because it is build from already known space, i.e. the C^0 space.
20:38
@smth So DF(A) is a linear map and hence continuous. For C1, apart from showing continuity of f, you also need to show that the map $G: M_3\to L(M_3,M_3)$ defined by $G(A):= DF(A)$ is continuous. L(M_3,M_3) is the space of all linear maps from M_3 to M_3.
how to prove the continuity of Df(A)? Is it okay to say that it is composition of linear mappings so it is continuous?
@Koro
20:57
why is it a composition of linear maps?
 
1 hour later…
21:59
yeah, the motivation is what Ted has also mentioned. in this norm, a sequence of functions $(f_k)_k$ converges to a function $f$ if and only if $f_k,f_k^{\prime},\dotsc,f_k^{(n)}$ all uniformly converge to $f,f^{\prime},\dotsc,f^{(n)}$ respectively. this means that this metric captures not only information about the function, but also its derivatives.
in particular, taking the $k$-th derivative yields a linear map $C^n\rightarrow C^{n-k}$ for $k=0,\dotsc,n$ and this maps are continuous with respect to these metrics. if you equipped all these spaces with the $C^0$-metric, the differentiation
other than the additive identity, is every element of a division ring a unit?
i think necessarily so
isn't that the definition?
i thought so, but D&F doesn't explicitly connect having an inverse to being a unit
well the definition of a division ring in D&F is a ring with identity 1 $\neq$ 0 in which every nonzero element has a multiplicative inverse
having a multiplicative inverse and being a unit is the same thing
 
1 hour later…
23:19
@Thorgott Can you elaborate on the last point a bit about this map $C^n\rightarrow C^{n-k}$? But yes, now I also see the further consequences of what Ted wrote.

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