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5:00 AM
In mathematics, a bump function is a function f : Rn → R on a Euclidean space Rn which is both smooth (in the sense of having continuous derivatives of all orders) and compactly supported. These are also called test functions [φ]. The space of all bump functions on Rn is denoted C 0 ∞ ( R n ) {\displaystyle C_{0}^{\infty }(\mathbf {R} ^{n})} or ...
Just a guess. I'm too tired to deny or confirm...
 
ok
 
eh nope I don't think it works...
wait so it has to be bounded below by 1...
 
why so ?
 
o wait, I mean it can't be bounded above by 1...
hmmm that doesn't really seem to get anywhere useful...
 
5:16 AM
actually such a function doesnot exist but I cannot visualize that ><
 
i feel like it shouldn't but I can't confirm... part of me wants to use an argument like: "It can't be bounded above by 1, so there is a portion that is greater than 1 and that will grow unbounded as $n \to \infty$"
I feel like there is some weird corner case tho...
 
hm
weird thinking - since x axis is from 0 to 1 so in order area to be 2
 
@Dair Not if f is continuous, I don't think. You'd need some interval > 1, and then that interval has arbitrarily large integral as n grows.
 
@Fargle You're saying it works if $f$ is continuous?
 
@Dair I would think so. I can't confirm that though, I'm no analyst.
(I'm no algebraist either. I'm just an undergrad. ;) )
 
5:24 AM
Like I want to use Reimann sums, do some algebra and find the answer, but I feel like if there is a counter example, it might be in the lebesgue integral...
 
yes
founs
a simi post
0
Q: Problem in solving a question related to real analysis.

Arnab Chatterjee.The question is : Does there exist a non-negative continuous function $f : [0,1] \longrightarrow \mathbb R$ such that $\int_{0}^{1} f^{n} (x)\ dx \rightarrow 2$ as $n \rightarrow \infty$? Is it true or not?If it isn't then how can I find example?If it is true then please give me a hint.Then I w...

 
looks like that is the answer.
 
how from monotone convergence we get the inequality?
answer by Robert Israel
Is it because measure of set A $\leq 1$ ?
 
@BAYMAX The inequality doesn't come from monotone convergence, I think. It just comes from $f$ nonnegative and $A \subset [0,1]$ by definition.
 
gotcha!
so $m(A) < 1$
where we used monotone convergence theorem here @Fargle
 
5:41 AM
Anybody have a link where I can check (or mind checking) that I have the right form for a quadratic equation of three variables? I think I have it right but am a math noob and want to be sure before I go off programming.
Here's what I have: w = (a1) + (a2)x + (a3)x^2 + (a4)y + (a5)y^2 + (a6)z + (a7)z^2 + (a8)xy + (a9)yz + (a10)xz + (a11)xyz
 
I think just with the limit of the integral over $A$ being the same as the integral of the limit, which is unbounded.
@mbrig Remove the xyz term and you're golden.
 
Number theory is sneaky
 
@Fargle ah, that makes sense. Thanks!
 
No problem!
@Daminark How so?
 
ok,next when $m(A) = 0$ then how the integral is less than 1 @Fargle
 
5:45 AM
So the problem is to find those primes $p$ such that $\mathbb{F}_p[i]$ is a field
 
@BAYMAX Because if you integrate anything over a measure 0 set, the integral is 0.
 
yes the integral is then $\geq 0$ but how we know it is less than $1$ ?
 
The part that I was able to do with someone else after much guidance was to come up with the conjecture that this was true if and only if $p \equiv_4 -1$, and prove the reverse direction holds
 
@BAYMAX Because any spot which could make it $> 1$ is part of a measure 0 set, so those parts don't factor in at all. Now $f^n(x)$ is guaranteed to converge either to $1$ or $0$ at all the other points, and the measure of that set of points is 1, so the interval can never be larger than 1.
 
$\int_{0}^{1}f^{n} dx \geq 0$ but how is $\int_{0}^{1}f^{n} dx \leq 1$
nice1,thanks @Fargle
 
5:49 AM
Now time to show the forward direction, which amounts to proving that $-1$ is a quadratic residue mod $p$ if $p\equiv_4 1$
 
@Daminark Ah yes, quadratic reciprocity.
 
Yea
Laci told me that there's a theorem to know in order to prove this, which for $p$ prime, there exists a primitive root mod $p$
 
@Daminark What's a primitive root? A number which generates the entire multiplicative group?
 
Yeah
So we know $p-1\equiv_4 0$ and that there exists some $k$ such that $r^k \equiv_p p-1$
 
Euuuuugh, my inexperience is becoming clear to me.
 
5:56 AM
As is mine
 
mine was clear to me quite some time ago.
 
Well we know that $r^{p-1} \equiv_p 1$
 
Wait no.
Apparently in my head, $pr^{p-1} = r^p$.
 
rip
You're thinking too much of analysis now
:P
 
I don't really have any knowledge of number theory besides canonical theorems etc.
 
6:02 AM
Well, let's figure something out
 
It's always been hard for me to approach and I haven't found a good introduction to it yet.
 
I don't know any number theory either
Maybe Chinese remainder theorem?
Oh wait no I just looked that up and it doesn't say what I thought it did
 
I thought that too, but it tells you just that $p - 1$ is the only class which is $0$ mod $4$ and $r^k$ mod $p$.
 
@Dami how many squares are there mod p?
 
Hmm
 
6:08 AM
Think about the map $x\mapsto x^2$
 
Hey @Ted
@Alessandro I'm not quite seeing how to look it as a map, though testing it on small cases it seems like half of the invertible elements in $\mathbb{Z}_p$ work?
 
I mean as a function $\Bbb F_p\to\Bbb F_p$
And that's a good guess (also do you know Fermat's little theorem?)
 
My point was more, I don't quite see the information gleaned by looking at it as a map
Yeah, $a^{p-1} \equiv_p 1$
 
...well now you know 1 is always a quadratic residue mod $p$...
>_>
 
Ok, nevermind that. Can you obtain something (hopefully) useful about squares rather than generic elements of $\Bbb F_p$ from Fermat's little theorem?
 
6:27 AM
I mean, if $a$ is a square than you're able to express it as $x^2$ so then $x^{2p-2} \equiv_p 1$, where and this exponent is now a multiple of $4$
 
Ok so if $a$ is a square $a^{\frac{p-1}{2}}=1$ in $\Bbb F_p$
 
Oh I think this makes sense
There are only two options when you're taking stuff to the power of $\frac{p-1}{2}$, $\pm 1$
 
(And 0 but only for 0)
Yeah what you have now is an upper bound on the number of squares since they satisfy the polynomial $X^{\frac{p-1}{2}}$, now as you guessed earlier you should prove that they are exactly $\frac{p-1}{2}$
And once that's done you also have a criteria to tell squares from non squares mod p
 
So what would make it true that this distribution is clean would be something like, if $a$ is sent to $\pm 1$, then $p-a$ is sent to $\mp 1$
 
@Daminark that's not true in general though
 
6:34 AM
Dammit
 
And if you write (-a) as (-1)(a) you'll see where the problem is
 
Hmm
Well, I think I'll sleep on it for now, it's kinda late (and also I suspect that there are probably some theorems floating around that would make this work...)
 
There's not much needed, ping me if you want to think about this problem again later
Good night!
 
7:02 AM
Hello, if $d(x,y)=\inf\{1.|x-y|\}$ how we find the open ball $B_{d}(x_0,r)$
please]
 
7:13 AM
@Vrouvrou What happens if $r > 1$?
 
i don't know
1<|x-y| ?
@Fargle
 
@Vrouvrou Right, and what does that tell you?
What is the maximum possible distance between two points under this distance function?
 
i don't know
 
Well, look at it like this: $\inf\{a,b\} \leq a$, right?
 
ok
 
7:19 AM
So do you have a bound on the distance function?
 
yes a
 
Which in this case is 1.
So you now know that all points are within 1 of each other; what does that tell you about balls with radius $> 1$?
 
sorry but i don't understand
 
Is this-ish question asked before here: Examples of deep mathematics behind common objects ? Otherwise I'm gonna ask.
 
@Vrouvrou If the radius is bigger than 1, what points are less than $r$ away from $x_0$?
Apologies if I'm not explaining myself well, I'm kinda tired D:
 
7:32 AM
all points because $\inf\{1,|x-y|\}\leq 1$
 
Right. And now you've classified half of the open balls.
(Well not literally half, but you've done one case)
And if $|x - y| \leq 1$, you have $d(x,y) = |x - y|$, so the other open balls are the normal ones.
 
@AlexK: That's a good question, and I think you should ask it on main; I can't find any similar thread after some preliminary searches.
 
Okay, so I'm adding the question text like this:
The question of usefulness of mathematics in everyday life is a cliche, and I am not asking that.

**What are some magic tricks/objects$^*$/algorithms/other curious stuff, which has very deep mathematical principle governing them ?**

Examples: (i) Rubik's cube (Lot's of stuff from group theory and combinatorics) (ii) The trick that you pick two number, and keep adding the largest to the second largest, and tell (after some iterations) only one, but from that you can derive the next one (Which is based on $\lim_{n \rightarrow \text{large numbers}} \frac{F_(n+1)}{F_n} \approx \phi$ )
Any suggestion to improve the wording ? (cc: @EricStucky)
 
@Fargle thank you
 
@Vrouvrou No problem!
 
7:38 AM
($^* \rightarrow$ objects means usually mechanical puzzles)
 
I feel like the magic trick question is a completely different thing. I suspect that if you allow for those, the vast majority of answers will be about them. But if you ask only for objects, you might get things that otherwise would be missed.
 
Yeah, I know magic from Gardner+Diaconis+Simon, so should I chuck that ?
 
'should' is a strong word; of course it depends on what you want... let's just leave it at 'I would', instead :P
 
@AlexKChen A pizza doesn't flop when you cup the crust, for the same reason that you can't make a perfect flat map of the Earth.
 
@Fargle I'm going to ask that question on main, so answer this there. (Yeah, good example). I am asking for question wording here.
Also, is the wording bit clunky ?
 
7:41 AM
Is it possible to have two distinct linear transformations $\mathbb{R}^2 \to \mathbb{R}^2$ with the same kernel but different ranges? I have been trying to come up with an example .. but in trying a bases for N(T) to be (0,0) or (1,0) or (0,1) or (1,0),(0,1), I always end up with something that spans R^2
 
@AlexKChen I think your wording is fine. Maybe "surprisingly deep" instead of "very". I just wanted to make sure it was the kind of thing you're looking for.
 
Yeah, MathWorld, it is possible. It's easy to convince yourself that the kernel of such a pair of maps has to be one-dimensional, so consider two vertical (not orthogonal!) projections onto different lines.
 
@EricStucky I like that example, that's elegant.
 
0
Q: Deep mathematics behind common and curious objects

Alex K ChenThe question of usefulness of mathematics in everyday life is a cliche, and I am not asking that. What are some objects$^*$/algorithms/other curious stuff/tricks, which has surprisingly deep mathematical principle governing them ? ($^* \rightarrow$ objects means concrete touchable stuff th...

Done.
Where I can find motivation behind polynomials ?
 
awat
 
7:50 AM
@EricStucky $?$
 
To me, this is like asking 'where can I find motivation for food'. Just try not eating for a while... I think you'll find the concept quite compelling.
 
Well, for a person who is beginning to eat for the first time, ..., your analogy fails. (Yeah, I'm using it for the first time)
 
But seriously, you already asked the question on MSE and it received a very good answer, plus a link to an MO thread which has even better answers.
 
I don't understand 95% of the terms in MO question, which is way too much over me.
 
ah, perhaps this is a cultural issue
 
7:57 AM
Cultural ?(!)
 
I also don't understand 95% of the terms in the MO question, but my response to that is just to ignore them
If you just go searching for the "reasonable-sounding but vague" phrases, you get things like:
In addition, polynomials sometimes serve as ''generalized integers'' in number theoretical contexts.
Polynomials can be used to extract information about finite sequences
 
I only have that two exact uses, not any others.
 
Polynomials are a generalization of number representation, replacing base 10 or base 2 by base x, while eliminating carry
Polynomials are, essentially by definition, precisely the operations one can write down starting from addition and multiplication.
Polynomials turn the idea of performing arithmetic operations (a dynamic, procedural notion: performing operations) into a static object of mathematics that can be considered in its own right
They are easy to compute and deal with.
 
@EricStucky What's that ?
 
Sometimes undeserved popularity: how many times have you seen a 15-term polynomial used in a regression equation?
what's what?
these are all direct quotes from the MO thread
 
8:00 AM
> Polynomials turn the idea of performing arithmetic operations (a dynamic, procedural notion: performing operations) into a static object of mathematics that can be considered in its own right
I don't understand this.
 
You don't understand what about this? You know what it means to perform arithmetic operations, presumably.
 
@AlexKChen Numbers are things that you can add and multiply with each other. Polynomials are functions; i.e. things that add and multiply their inputs.
 
Anyone knowledgeable about Fourier transform in here ?
 
I don't understand this (key) part:
> a static object of mathematics
 
ah
this is not really a precise concept: it's meant to invoke ideas like a number, or a set, or the cube group, or whatever
something that exists, as opposed to a process.
 
8:04 AM
Give me an concrete example. I have this arithmetic operation, say: $(stuff * 3)+1$, and how to turn this into a static object of mathematics ?
 
@AlexKChen I think what he's getting at is that a function is a process rather than an, which just is, but that you can think of functions as objects in their own right.
 
XD that corresponds to a polynomial, in fact
namely, "3x+1"
 
@AlexKChen You now have the function $f(x) = 3x + 1$, which you can now treat as its graph, for example. It's not just a process, now you have a line.
 
(Fargle is right; this is what I'm getting at)
 
Oh yeah, that's good. I will think about it. Thanks, @Fargle, and @EricStucky, too.
 
8:06 AM
It's this kind of thinking that's become really important even in really high mathematics in the last century or so.
 
@EricStucky wow thank you! I've found an example
 
The functions themselves become the objects of study, because the functions tell you more about the objects than an object-only view.
@AlexKChen No problem!
 
@Fargle What kind of thinking ? The process of static-ifying dynamic processes ?
 
Fargle, semi-related: have you read "When is one thing equal to some other thing?"? This is my favorite piece on why modern mathematics looks the way it does.
 
@AlexKChen Kind of, yeah. Linear algebra is a great example of this happening.
You turn a system of linear equations into a matrix problem, but that matrix also represents a linear transformation, i.e. a function that maps vectors to vectors in a nice way.
Specifically, $f(v + w) = f(v) + f(w)$ and $f(cw) = cf(w)$.
And then this brings up another example in calculus: the derivative operator.
You can think of the derivative as being an operator $D$ which takes in a differentiable function and spits out some other function.
 
8:15 AM
OK, nice, thanks.
There was a MO question on the use of concrete stuff in teaching, and there were some cool answers. I forgot the question ID, what's the link ?
 
 
1 hour later…
9:29 AM
@Fargle I am not a native English speaker, and I don't know what is pizza crust exactly, so what is "pizza flop" when you "cup the crust" ?
 
@AlexKChen The crust is on the outside of the circle.
 
The red or white portion ?
 
White.
 
What's cupping the crust ?
 
If you hold one piece of it by the white portion, but push the middle of the white part down and hold the sides up
You curve the pizza along that axis, basically.
It works with a sheet of paper too
 
9:35 AM
Like the cross section of a cone ?
 
@AlexKChen youtube.com/watch?v=gi-TBlh44gY This video explains the whole thing--the very beginning shows it.
 
Oh yeah, that's fantastic.
 
 
1 hour later…
10:59 AM
Hi chat
 
Hi $\forall$
 
11:14 AM
Is it just me or does MathJax / LaTeX not properly work in the chat?
 
Works for me
What do you mean by "properly" ?
 
It just displays the TeX command with dollar signs
Nevermind, on the left is a helpful link. Now it works
Right. Sorry.
 
np
 
11:31 AM
How to check whether two matrices are conjugate in $M_{2}(R)$ ?
 
Conjugate meaning ?
 
me too
?
 
What ?
 
1
Q: To show two matrices are conjugate to each other

tomb_raiderGiven two matrices A and B $$ A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 3 & 0 \\ 1 & 2 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 3 & 0 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ I have checked basic properties such as trace, charatersitic polynomial and are same . I also realise that i ne...

 
But what do you mean conjugate of each other ? Like there exists $P\in GL_n(\Bbb R), A = PAP^{-1}$ ?
 
11:45 AM
yup
 
Check their characteristic polynomial
 
Here's what my $(-1)↑↑x$ looks like for $x\in(-2,4)$
 
Whoa ! How's this proved ?
Nov 8 '14 at 9:18, by Chris's sis
Due to some personal research I proved that $$\sum_{n=1}^{\infty} \left(\frac{\displaystyle \psi(n)-\psi\left(n-\frac{1}{2}\right)}{\displaystyle \binom{4n-2}{2n-1}}+\frac{\displaystyle \psi\left(n+\frac{1}{2}\right)-\psi(n)}{\displaystyle \binom{4n}{2n}}\right) =$$

$$\frac{\pi}{3\sqrt{3}}+\frac{1}{75}\left(\log(1073741824)-3\sqrt{5} \pi^2+\pi\left(\Re \{{4 i \sqrt{5} \log(5- \sqrt{5} + i (\sqrt{15}- \sqrt{3}) ) -
2 i \sqrt{5} \log(5 +\sqrt{5}+ i (\sqrt{3} + \sqrt{15}))\}}+6\sqrt{5}\arctan\left(\sqrt{\frac{5}{3}}\right)-5\sqrt{3}\right)+\sqrt{5}\left(\log(79228162514264337593543950336) \l
$$\sum_{n=1}^{\infty} \left(\frac{\displaystyle \psi(n)-\psi\left(n-\frac{1}{2}\right)}{\displaystyle \binom{4n-2}{2n-1}}+\frac{\displaystyle \psi\left(n+\frac{1}{2}\right)-\psi(n)}{\displaystyle \binom{4n}{2n}}\right) =$$

$$\frac{\pi}{3\sqrt{3}}+\frac{1}{75}\left(\log(1073741824)-3\sqrt{5} \pi^2+\pi\left(\Re \{{4 i \sqrt{5} \log(5- \sqrt{5} + i (\sqrt{15}- \sqrt{3}) ) -
2 i \sqrt{5} \log(5 +\sqrt{5}+ i (\sqrt{3} + \sqrt{15}))\}}+6\sqrt{5}\arctan\left(\sqrt{\frac{5}{3}}\right)-5\sqrt{3}\right)+\sqrt{5}\left(\log(79228162514264337593543950336) \log\left(\frac{3-\sqrt{5}}{2}\right)
(Or is this a joke ?)
 
My Chat jax got hanged :)
 
Well, that certainly doesn't look appealing
 
11:52 AM
@SimplyBeautifulArt That looks crazy! Is that a function?
 
9
Q: A new interesting pattern to $i↑↑n$ that looks cool

Simply Beautiful ArtMany of you may recall "An obvious pattern to $i↑↑n$ that is eluding us all?", an old question of mine, and just recently, I saw this new question that poses a simple extension to tetration at non-integer values: $$a↑↑b=\begin{cases}a^b,&b\in[0,1]\\a^{a↑↑(b-1)},&b\in(1,+\infty)\\\log_a(a↑↑(b+1)),&...

 
@SimplyBeautifulArt Interesting. I will have a read.
 
when you jump seeing the question
 
11:56 AM
I was looking for the mathematics chat room in the long list of my favorite chat rooms, while I was already in it. Nice to know.
 
hey hi @nbro
 
@BAYMAX Hey! Did you solve your previous problem?
 
nope asked in the chatroom but still the mystery continues I hope!,Later I will post it as a question I think.@nbro
 
@AlexKChen It might be true, but in any case it's probably very tedious calculations or ... formal calculations programs
 
12:05 PM
What's the most accurate but, at the same time, general term to describe someone that proves a hypothesis (or conjecture), where the hypothesis may simply be a daily routine one? Should we call him/her/it a prover?
 
1:00 PM
is there a symbol for tautological equivalence?
*as opposed to regular equivalence*
 
$\equiv$
For instance $(p\implies q) \equiv ((\lnot p )\lor q)$
 
Oh never mind. I had it in my head that there was a difference between tautological equivalence and logical equivalence.

This was because there is a difference between tautological implication and regular implication. Trying out an example however, seems to show that the two equivalence relations are the same.
thanks for the help
 
I didn't see awesome user @ShaVuklia for a long while. I wonder what she's doing.
 
You haven't been there for a long while then :p
 
Seems so. :D
 
1:21 PM
Maybe you shouldn't wait too much...
 
Hmmmf. It seems I was wrong that tautological equivilant is the same as regular equivilance.

This is becuase regular equivilance is built with regular implication, while the other one requires tautological implication.

A tautologiucal implication must be a tautology. This means that tautological equivilance must be a tautology as well.
And that is the difference.
 
1:51 PM
I have a subset of the unitary matrices that is closed under inversion but does not contain the identity matrix. It couldn't be a group, since it wouldn't be closed under multiplication (right?). So it would be a semigroup? Is it anything beyond that? In general, if I throw in the identity element into such a subset, what is the result?
 
For instance take $O_n^-(\Bbb R)$
(orthogonal matrices with determinant -1)
 
@Astyx Oh, yes. Interesting. So, what if one throws in the identity matrix? What happens?
 
Nothing much, you get $O_n^-(\Bbb R) \cup \{I_n\}$
 
So, such a set is not a semigroup? Do semigroups have to be closed under multiplication?
 
Yes IIRC
(Well you get a semigroup for low $n$, but that's not interresting)
Any magma has to be closed under the operation
 
1:59 PM
@Astyx Why only for low $n$?
 
For instance for $n=1$, $O_1^-(\Bbb R) = \{[-1]\}$
So if you add the identity, you get $\{[1], [-1]\}$ which is a group
($\Bbb Z/2\Bbb Z$)
For $n\ge2$ this doesn't work
$$
\begin{pmatrix}
2 & 0 \\
0&-1/2\\
\end{pmatrix}$$
 
@Astyx Hmm. Interesting. Thanks for your help.
 
The square of this matrix is :
$$
\begin{pmatrix}
4 & 0 \\
0&1/4\\
\end{pmatrix}$$
The former is in $O_2^-(\Bbb R)$, the latter isn't in $O_2^-(\Bbb R)\cup \{I_2\}$
Wait no
I don't mean $O_2^-(\Bbb R)$, I mean $\det^{-1}(\{-1\})$
But something similar happens in $O_2^{-1}(\Bbb R)$ with rotation matrices (I don't have an example under my hand though)
Glad to help
 
So, I was wondering. I read this question on the forums:
https://math.stackexchange.com/questions/672825/invariant-subspace-of-the-inverse
I've seen that it doesn't hold for the infinite dimension case. But what if we also say that $V$ is a *closed* invariant subspace? Would it hold then?
 
2:17 PM
The shift operator works only if $V$ is not closed, doesn't it?
That's what got me wondering..
 
Unless I'm mistaken you can define a norm s.t. the set of sequences is closed
 
Hm
I'll try to find something..
 
The shift oeprator on sequences then is invertible and sends sequences to sequences starting with 0
So if you take sequences starting with 0, it's an invariant subspace for the shift operator
But it's not invariant for the inverse of the shift operator
Then again you might argue that it's not invertible because the domain is not the codomain
Depends on you notion of invertible I guess
 
Well it's not invertible in the regular sense of the definition we have, but perhaps that's a start
 
2:43 PM
10
Q: How to appreciate Riemannian geometry

abcdefI'm currently studying an introduction to Riemannian geometry i.e. connections, curvature and isometric immersions (the Gauss, Codazi and Ricci equations). I find the introduction to Riemannian geometry interesting, but whenever I look at some theorems beyond the introductory topics they seem qu...

 
@Astyx Actually, if $V$ is a subspace of a hilbert space, and it is closed, then it is a hilbert space in its own.
Being a hilbert space, we can show that $T(V)=V$. So $T^{-1}$ holds there!
 
3:09 PM
0
Q: Length of the projections of $v_1$ and $v_2$ onto respectively the plane orthogonal to $j_1$ and $j_2$ have the same length

nbroSuppose we have two unit vectors $j_1$ and $j_2$. Let's also define a two-dimensional plane orthogonal to each of them: let's call the first $P_1$ and the orthogonal plane to $j_2$ $P_2$. Now, suppose I have two other vectors $v_1$ and $v_2$. I want to show that the length of the projections of $...

 
Actually they're not certainly equal. Blah, I don't know.
 

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