Hey. I'm going to ask a very weird and may seem like I'm joking question but I'm really not. I don't know what is meant by last digit of a number. That would mean there is a first digit. Is it right to assume that the last digit in 123567 is 1 and the first digit is 7?
i think most of the time, when people say 'last digit,' they mean the last one you'd write when writing the decimal expansion of the number from left to right. so 7 in this case.
Thanks leslie. This language has always confused me. That makes sense. I think i was thinking of camparing place value. Earlier place values of the number are at the beginning of that number so they are first. I'm going to try to shake that thinking or that connection and try to use your thinking instead.
i googled that. apparently they do it the same way, although left-to-right is something of a red herring i guess.
'most significant' and 'least significant' are sometimes used to remove the ambiguity. as on that page on endianness.
a red herring in that it doesn't explain why we start with the biggest digit. if you were thinking of a number as a power series in 10 with digit coefficients it might seem more natural to go the other way.
on a properly populated chat, someone would have brought up the p-adics by now. i guess the party really is at copper's house.
Is there a way that I can back out a general form of a linear operator if it is on a vector space of functions?
So to be specific, I was asked to find the adjoint of $T(f) = f' + 3f$. So if I put in a general form of a vector in $P_1(\mathbb{R})$ such as $a + bx$ I'll get $(3a +b) + 3bx$
the inner product is $\langle f,g \rangle = \int_{-1}^{1}f(t)g(t)dt$
but I'm not even specifically talking about that. So I have these two expressions for the linear operators. Is there a way I could express $T^*(f)$ in the form $T^*(f) = f_{something} + c \cdot f_{something\ else}$
That's what I'm meaning by "backing out"
or whatever else would be added onto my $T^*(f)$. THat was just a suggestion of what I'm thinking about
do you understand what I'm explaining @copper.hat ?
If the tangent space is one-dimensional, and I look at the Ricci tensor Ric(x,x), is this just straight up equal to the scalar curvature since I am only "adding" 1 summand?
@Jakobian Here is a question you might find interesting that I've been thinking about lately. Let $X$ be a continuum and let $K(X)$ be its hyperspace of compact sets (with the Hausdorff metric/Vietoris topology). Must $K(X)$ contain a copy of the Hilbert cube (that is $[0,1]^\omega$) with nonempty interior?
dc3rd, pay attention to domains. if they haven't been given, work generally with the definition of the domain of the adjoint (hint it is not always everything) and try to find conditions that make it interesting
When a wire is full of electron even if electrons are accelerating I think current will remain same all over the wire. If a single electron acclerates then other electron will acclerate at same rate so I think current will be same right?
Can I ask a naive question from which I have no experience of doing yet? Can this be extended by analytic continuation? Because this is only valid at 0<x<2
Regarding my previous post:
A ‘quite’ elementary function $f(x)=\frac{x^{s-1}}{s^{x}-1}$ related to Riemann-Zeta function
About the relation of the function $f(x)$ in terms of the Zeta-Function, I now give a proof of this relation via the incomplete gamma function
Again let’s supposed the integra...
I'm thinking of Dummit Abstract Algebra, Chap 7, sectin 4, exercise 13b. I've tried to show that
$R/\varphi^{-1}\left( M\right) $ is division ring, but was not successful. Can anybody help me?
Let $\varphi:R\rightarrow S$ be a homomorphism of commutative rings.
(a) If $P$ is a prime ideal of...
Here, what is a counterexample to b) in case subjectivity condition is not given?
I tried thinking of inclusion homomorphism $f: Z_{12}\to Z$ and then (3) is a maximal ideal of Z but $f^{-1}((3))=\{0,3,6,9\}$ is also a maximal ideal in $Z_{12}$.
A non-vanishing $1-$dim. Killing field $X$ bounded by $[0,1]$ where $X$ flows from $(0,0)$ to $(1,0)$ s.t. we have a global flow on $M=(0,1).$ Does a construction exist?
@dc3rd Excuse my brain out last night. (Excuse the slight abuse of notation.) The general inner product is $\langle a+bx, c+dx \rangle = 2ac +{2 \over 3}bd$. Then $\langle a+bx, T(c+dx) \rangle = \langle a+bx, 3c+d+3dx) \rangle = 2a[3c+d] +{2 \over 3}b[3d] = 2[3a]c+ {2 \over 3}[3a+3b]d$. From this alone we get $T^*(a+bx) = 3a+(3a+3b)x$ (assuming I have made no mistakes, which is a biggie). In this case no need to deal with matrices, etc.
Ok that makes sense @copper.hat , but how do I determine what $g$ is from that expression? I get that all polynomials will be of that form once the operator is applied, but what is the application on the functions to make that form?
This might sound like a silly quesiton, but if ker f = ker (I - f)? They are the same space right? if Ker f = 0 where f is a Fredholm Operator (bounded linear map between Banach spaces), does that mean ker (I - f) = 0 also?
lemon: questions in front of things that appear to be hypotheses ("if ker f = ker(I - f)?") are a bit confusing. if f is fredholm, then ker(f) and ker(I-f) might not be the same, or even have the same dimension, even if you assume ker(f) = {0}. consider f = I
I m trying to resolve a simple issue right now in PDE. If ker L = 0 where L is the elliptic fredholm operator, i want to use the fredholm alternatives to argue L must also be surjective and so it is isomorphism.
hm, what is your definition of 'fredholm'? that conclusion should fall right out of either the definition, or a theorem that is fairly close to proximity to the definition in a textbook or set of course notes.
if it doesn't, there should be some examples. i don't know how important it is that L be 'elliptic' here. what is elliptic? does it have a non-PDE definition?
if there isn't one provided, you might need to do some work here. or worse, it might not be true.
OK, this might actually need some PDE, if it's true. i don't see how that definition easily boils down to a property that you could plug into the definition of 'fredholm.' i was hoping it might, but maybe it doesn't.
you can certainly have injective fredholm operators that aren't surjective. i was hoping there was some simple way of seeing, without detouring too much through PDE, that such operators could not be 'elliptic'. maybe there isn't.
i was thinking, if you're seeing this in a PDE book and there isn't a lot of general operator theory around it, then it ought to be provided to you from some PDE result. hopefully something about solvability of these things given certain data.
but it sounds like maybe that's not the case, or maybe this is something you're trying to prove on the way to getting solvability of those things given certain data.
what's your definition of 'fredholm'? there's only one notion of fredholm, but different books may introduce it in different, but equivalent ways. like invertibility, or other notions similar to fredholmness.
anyway, math.stackexchange.com/questions/3078469/… might help. i'm not sure of the PDE context if you are on a hilbert space or not. but dim ker L* is dim coker(L) in the hilbert space setting.
@dc3rd I don't understand what you are asking. I am just using the general form of an element of $P_1$. and matching coefficients. Just shortcutting the matrix formation, etc. So, for $ \langle g, Tf \rangle $ I have $g(x) = a+bx, f(x) = c+dx$. If I let $(T^* g)(x) = e +fx$, then the formula $\langle g, Tf \rangle = \langle T^*g, f \rangle$ gives $2ec +{2 \over 3}fd = 2[3a]c+ {2 \over 3}[3a+3b]d$ (for all $c,d$) from which we get $e=3a, f=3a+3b$ and so $(T^*g)(x) = 3a+(3a+3b)x$.
@copper.hat Yes I understand that Copper. So for $T(f)$ we had $T(f) = f' + 3f'$. Could I write something like that for $T^*(g)$? Just as an example $T^*(g) = g'' + 4g' + 2g$? That's what I was asking. Could I extract that sort of expression from the work you did above?
That is where I got hung up last night. You can, but (as far as I can determine) you end up with an awkward expression like $T^*g = 3(g-g'x)+ 3(g-g'x)x +3g'x$.
how did you even come up with that expression? Alright...I just preferred to respect the cloak of anonymity, but I suppose Joe is a regular enough name.... :P
point evaluation on P_1 is also representable by inner producting with elements of P_1. the calculational awkwardness arguably arises from the fact that there is no one function that does this for all P_n (because point evaluations do not extend to the completion of all polynomials under the norm)
i vaguely alluded to this with my remark about domains. we never got precise about the domain although i infer from subsequent conversation that everything is just P_1
i was worried it might be all C^1 functions, or something like that, where the domain of the adjoint is just {0} i think
I'm thinking of Dummit Abstract Algebra, Chap 7, sectin 4, exercise 13b. I've tried to show that
$R/\varphi^{-1}\left( M\right) $ is division ring, but was not successful. Can anybody help me?
Let $\varphi:R\rightarrow S$ be a homomorphism of commutative rings.
(a) If $P$ is a prime ideal of...
