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8:21 AM
@AkivaWeinberger Thanks!!
 
$\lim_{n\rightarrow \infty} \frac{sin(n+1)}{sin(n)} = 1$ is this correct?
 
@BAYMAX No
 
Intuitively I thought that as $n$ increases both numerator and denominator varies by almost same fashion! thus giving limit 1
ok
@TobiasKildetoft what is wrong in the intuition !
 
@BAYMAX they vary in the same way prediodically for all $n$, also for small values
Well, when we let $n$ be any real rather than just integers, but that doesn't matter here
 
how can i do this!
 
8:34 AM
Do what?
 
cannot i do $sin(n+1) \approx sin(n)$ for large n
 
@BAYMAX expand sin(n+1) with addition formulae and then simplify.
 
no, why would that be true?
 
like $sin(1003455.6) \approx sin(1003456.6)$
 
@BAYMAX Intuitively, adding one to the argument has essentially as large an effect for large $n$ as for small $n$
 
8:35 AM
@BAYMAX no because the sin function changes at the same rate irrespective of the argument
 
since the argument only matter mod $2\pi$
 
for example consider $\frac{d(sinx)}{dx} = cosx$. It doesn't matter how large x, the function still changes pretty fast
 
yeah I checked that
I think we could do that for linear functions only?
or not?
 
@BAYMAX You can use it whenever some kind of normalizing factor is involved. For example the small angle approximation $sin(x) \mapsto x$ is due to the fact that $\frac{sin(x)}{x}$ approaches 1 for small x.
 
so $\lim_{n \rightarrow \infty} \frac{\sin(n+1)}{\sin(n)} =\lim_{n \rightarrow \infty} \frac{\sin(n)\cos(1)+\sin(1)\cos(n)}{\sin(n)} $
$\lim_{n\rightarrow \infty} \cot(n) = ?$
 
8:41 AM
doesn't exist, and hence neither does the limit.
 
how can i show that limit doesnot exist?
 
@BAYMAX it's periodic
 
Plenty of ways
It's not defined on multiples of $\pi$
 
if its periodic then are we thinking that both left hand limit and right hand limit are not equal@LeakyNun
 
I am looking again at your answer about the composition of reflection and composition. You said:
"(reflect across x-axis + rotate about origin) + translation
the first two combine into like a reflection about some other line $\ell$
and if the translation is purely perpendicular to $\ell$, it's a reflection"

Why are the first two a reflection?
 
8:44 AM
@BAYMAX there is no right hand limit for infinity
 
@BAYMAX those terms do not make sense when we let $n$ go to infinity
 
it's periodic with two different values, so we can construct two subsequences that converge to different values
 
It's periodic with an irrational period
 
@Astyx the period doesn't have anything to do with it
 
$\lim \sin(\pi n)$ exists
 
8:45 AM
@Leaky it matters insofar as the period isn't $1$ or something
 
The sequence should keep hitting positive and negative values.
 
Constant functions are also periodic
 
oh, it's the discrete limit...
 
:^)
 
I thought it was the continuous limit
 
8:46 AM
What does continuous limit or discrete limit even mean
 
@BalarkaSen that we only consider the integers
limit of a sequence
 
@BalarkaSen n as a natural or real number
 
I just invented those two terms, so whatever
 
The limit of the sequence$(\cot(n))_n$
 
And as Steamy said constant functions are periodic
 
8:47 AM
@Astyx I did say "periodic with two different values" afterwards
 
So I should have said "periodic with an irrational minimal period"
 
I mean if the given sequence does not converge for integer values, then there is no limit at infinity of the given function.
 
(Since $x\mapsto 1$ also has an irrationnal period)
 
Which is to say, $\{f(n)\}_{n \in \Bbb N}$ does not converge => $\lim_{x \to \infty} f(x)$ does not exist.
 
gg @Balarka
 
8:48 AM
Yup...
but the other way around is not guaranteed
 
which is what leaky was talking about
 
$n$ usually denotes integers
It would be vicious to mean reals
 
what are the two different values the "cot" converges to ?
 
@BAYMAX everything
 
hi. this is a bit off topic tho, but, does any one have any personal page with his / her university name?
 
8:49 AM
"Let $n$ be a compact Hausdorff space"
8
 
Well, technically, you could pick any two real values :P
 
It has $\Bbb R$ as its set of limit points
 
@BAYMAX it only converges nicely for subsequences with irrational terms
 
But that would only make it ridiculously difficult :P
 
Let $\tilde{n}^*$ be a pointed cohomology theory
 
8:50 AM
@Daminark Well, I often use $k$ for an arbitraty field. But I might also use it as an index of summation
 
I also use $n$ as an element of the normal subgroup $N$ :P
 
"Let $B$ be BAYMAX !"
 
"Let $K$ be a set of indices and $(\Bbb k_k)_{k\in K}$ be a family of arbitrary fields"
 
@SteamyRoot I purposefully did not do that anywhere in my Masters thesis
 
"We denote by $(\underline{\sum_{k=1}^{\infty} x^k})$ the set of transcendental numbers"
 
8:52 AM
lmao
 
@Astyx now that's cheating, because you use the same $k$ :P
 
I am curious that $\{\cot(n)\}$ converges to every real number in $\Bbb{R}$ ?
 
Just put some mathfrak and mathcals in there
 
Better now ?
 
@BAYMAX Of course it does not converge to those
 
8:53 AM
Well
 
$\hat{\tilde{n}}_{n_n}$ be a modular form.
 
Kinda
 
@TobiasKildetoft Thanks to one of my professors, I usually use $\mathbb{F}$ for that :P
 
@BAYMAX subsequences do, the entire sequence doesn't converge at all
 
nice!
 
8:53 AM
what am i doing with my life
 
I use $\Bbb K$ for arbitrary fields
 
@Daminark I only use that $F$ for Galois fields
 
@BalarkaSen I was wondering the same thing
 
Life is an e-file
 
What's a Galois field?
 
8:54 AM
@Astyx you were wondering what i am doing with my life? that's nice of you
 
Finite field
 
Ah
 
Are there infinite fields ?
 
$\mathbb{Q}$
 
I hope so.
 
8:54 AM
@Astyx real nos
 
$\Bbb C$
only field ever
 
-Axler
 
@Astyx $\overline{k}$
 
"Norman Wildberger hates him ! Click here to find out why"
 
ayy lmao
 
8:56 AM
@TobiasKildetoft I'm inferring from topology but is this algebraic closure? I hadn't realized that algebraically closed fields were infinite
 
Yeah it's algebraic closure
 
@Daminark Indeed, all algebraically closed fields are infinite
 
Oh wait I can see it
 
$\lim_n \rightarrow \infty \frac{\log(n+1)}{\log(n)} = ?$
 
It's probably an Euler style argument
 
8:56 AM
Yup
 
How do I think of these?
 
Take the polynomial $1+ \prod_i (x-a_i)$
 
$1 + \prod (t-\lambda_i)$
Sniped...
 
get sniped you
 
@BAYMAX thats 1, try lhopitals
 
8:57 AM
Akiva sniped you all some weeks ago
 
Jul 5 at 13:48, by Balarka Sen
You will get this comment in the far future when you write "sniped"
 
M8 I think there's an expiration time on sniping. Like, after enough time has passed to reload it doesn't count anymore
 
So technically you are all sniped
 
Time to spam every answer to every question ever in this chat
So I can always snipe everyone who posts an answer here from now on :D
 
gotcha@tim
 
8:59 AM
Steamy proceeds to prove literally ALL of math, thus becoming the ultimate sniper
 
@SteamyRoot start with the millennium problems pls.
 
Haha :P
 
damn it dami
 
Nah, my purpose is only to snipe people.
 
i better not have been sniped
 
9:00 AM
So if I assume nobody will bother to answer questions whose answer require, say, more than 10000 characters
 
Use Draganov @SteamyRoot
 
shrugs in LaTeX
 
And using that the alphabet (including symbols) is finite
I can just get a bunch of monkeys with typewriters to write all those answers for me
 
Well you're in trouble though
In the time it takes them, we can write a bunch of answers
Plus there's everything that's been answered already
 
Darn it! My master plan has been ruined!
 
9:02 AM
any algebraic simplication for $a^{\ln p}$
?
 
$p^{ln a}$?
 
>simplification
 
@Daminark you got me there
 
Also is that true? I forgot a lot of this stuff
@Balarka all the precalc I've learned, I've forgotten
 
It's $e^{\ln a\ln p}$
 
9:04 AM
IM SHOCKED!!
 
@Astyx ah, clever
 
(intentional apostrophe error for Ted)
5
 
Better star it
 
SNIPED @Astyx
 
Not grammar
 
9:05 AM
alrighty then
time to tape over the apostrophe key on my laptop
 
@Daminark Technically you got sniped
 
He was proud when I finally figured out how the apostrophes worked on Stokes's's''' theorem
M8 I starred it before you said to do so
 
Not on my screen you didn't
 
Well you sniped your internet then
 
@Daminark MLG sound
 
9:07 AM
Funny thing is, I didn't even star the message
 
So you were sniped alright
 
Cause I don't like Balarka (and solely for that reason)
 
but you sure like to GET SNIPED
 
I do, it my passion in life
Along with claiming untrue math results
 
@Astyx FIGHT
:P
 
9:09 AM
And despairing over MLG memes you guys send me
 
COME AT ME BRO
 
Alright, alright I'll give you your star, please don't hit me :(
 
$FINISH \space HIM$
i have completely forgotten how to mathjax
 
Anyway, gotta go, seeya later
 
Yeah I was rolling under that assumption
 
9:12 AM
lol
history of the entire world is the dankest thing i have seen in a while
 
there was this video called, "the bork files"
and it was just gabe the dog barking to the tune of the x-files
 
@Daminark speaking of, I found some really dank videos from the instant regret list yesterday
 
$\lim_{x \rightarrow 0}\frac{\cos(x)}{x}$
doesnot exist?
 
Nice
@Baymax the limit is infinite
 
9:15 AM
@Balarka oh I've seen that
 
so doesnot exist @tim
 
it approaches infinities of different signs, so yes.
 
Oh yeah that's true
 
here also we can apply the convergent of subsequence thingy!
 
Lol extend it to complex $x$ and then think of the Riemann sphere
 
9:16 AM
@Daminark Have you seen this: youtube.com/…
 
like take a sequence of odd multiples of $\pi/2$
oh GOD!
 
if you mean that subsequences converge to different limits, then yes
 
@Daminark That's my favourite thing to do with weird limits :3
 
@BalarkaSen Truly the proper way
@SteamyRoot O lawd
 
what kind of subsequence you will take @tim
 
9:18 AM
@BAYMAX one with all positive terms and one with all negatives
thats effectively the same thing as taking the right hand and left hand limits separately
 
yes but can you explicitly state two
i thought of even and odd multiples of $\pi/2$
but htat wont work i think1
 
I dunno if that's gonna work
 
@BAYMAX $a_n = \pi/n$ and $b_n = -\pi/n$
 
Consider some sequence $a_n\to 0$ such that $a_n > 0$, then $-a_n$
 
its still gonna reach infinite limits
but opposite ones
 
9:20 AM
ok
 
Oh so I've discovered this result which is even more general than Liouville
So if $f$ is entire and $f(z) = O(|z|^k)$, then $f$ is a polynomial of degree at most $k$
Which is nifty
 
NIFTY
I see nifty in Sensex I think :)
 
What's Sensex?
Also @Benjamin we haven't had our battle to the death yet, haven't we?
 
leave that!
I meant Census
Save Census!
I am bla bla saying ")
gonna sleep
bye$\forall$
 
See you!
 
9:34 AM
Nightcore music are nice
I will not sleep now >< :)
 
user84215
10:12 AM
When I use "&" for the text part, then the command appears as the following:
$$\begin{align} v+w & =0 & \text{Given} \tag{1} \\ -w & =-w+0 & \text{additive identity} \tag{2} \\ -w+0 & =-w+(v+w) & \text{equations $(1)$ and $(2)$} \end{align}$$
and when I use "&&", it appears as:
$$\begin{align} v+w & =0 && \text{Given} \tag{1} \\ -w & =-w+0 && \text{additive identity} \tag{2} \\ -w+0 & =-w+(v+w) && \text{equations $(1)$ and $(2)$} \end{align}$$
Why does that happen?
 
10:55 AM
guys, what are the probability functions?
I know there are mass function, density function, distribution function, other than that is there any more?
 
Characteristic and moment generating?
 
user84215
11:12 AM
no idea?
 
11:40 AM
@aminliverpool Align works in pairs of rl-outlined math
So in your first example, it's rlr, in your second it's rlrl (but the third column is empty)
 
user84215
@SteamyRoot I do not understand.
 
first column outlines to the right, second to the left, third to the right, etc.
 
user84215
Could you explain more?
 
There's... nothing more to explain?
 
@SteamyRoot i need to ask you a question!
 
11:50 AM
Actually...
$$s=\int_{\Bbb{R}} \frac{a(x)}{10^x}dx,a(x)=\{0,1,2,3,4,5,6,7,8,9\}$$
I might have issues with measure zero sets for this one...
 
@Liad If it's a short question, I can help, but if it's an exercise I really don't have time I'm afraid.
 
@SteamyRoot not an exercise. and i can wait for latter if you cant now
 
user84215
12:06 PM
@SteamyRoot Now I understand it. Thanks.
 
user84215
12:46 PM
In the projective completion of an affine space, is the embedded affine space a hyperplane in the projective space?
 
does any one know about probability and conditional probability and pmfs?
 
Maybe, depends on the questions you have
 
1:08 PM
In general just ask, don't ask to ask, you'll find someone who can help more often than not
 
Consider the linear functional $f: \Bbb{N}^{\{0,...,9\}}\mapsto [0,1]$ where
$f = \left(\frac{1}{10^n}\right)_{n\in \Bbb{N}} \cdot$
 
ok , I know what pmf is, I know it is a function that the incomes are random variables, I know what conditional probability is, I know it's a change of sample space.
I know conditional pms and conditional probability are alike some how. but what if I have a sample space divided into independent events, and I want to calculate an event B inside them, using sum of the intersection of each event with the new event B. (here I should use conditionals).
now how can I do that for pmfs?
 
Let $s \in \Bbb{N}^{\{0,...,9\}}$. Then

$f(s)= \left(\frac{1}{10^n}\right)_{n\in \Bbb{N}} \cdot s = \sum_{k=0}^{\infty}\frac{s_k}{10^k}$
thus this gives the decimal expansion of all reals in $[0,1]$
 
@Secret you meant $s \in \{0,\cdots,9\}^\Bbb N$
 
Isn't given a set $A^B$, B is the domain and A is the image?
 
1:17 PM
yes
 
So to construct sequences such as (0,4,3,7,4,3,4,4,5,....) we should be mapping {0,...,9} to $\Bbb{N}$ slots?
 
no
what does 9 map to?
 
THe above is just one example of a sequence. Let me give a better example: (0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,...)
 
what does 9 map to?
 
Ah wait a sec..., I have {,,,,,,,,,,,,,} slots. Any function f is defined by {f,f,f,f,f,f,f,f,f,f}, and thus the image for each slot is taken from {0,...,9}
Ah ok, I understood now
So it is indeed $\{0,...,9\}^{\Bbb{N}}$
Ok, so let me rewrite everything:
Consider the linear functional $f: \{0,...,9\}^{\Bbb{N}}\mapsto [0,1]$ where $f = \left(\frac{1}{10^n}\right)_{n\in \Bbb{N}} \cdot$. Let $s \in \{0,...,9\}^{\Bbb{N}}$. Then $$f(s)= \left(\frac{1}{10^n}\right)_{n\in \Bbb{N}} \cdot s = \sum_{k=0}^{\infty}\frac{s_k}{10^k}$$
thus this gives the decimal expansion of all reals in $[0,1]$
 
1:33 PM
I want to show that for a rotation δ≠id a translation τ, δ∘τ is always a rotation. I have done the following:
A translation followed by a rotation of θ is still a rotation by θ. This is because if we take two points a and b which are mapped to a′ and b′, then θ is the angle between the vectors b−a and b′−a′, and even though the second one are translated.

Is this correct?
 
1:43 PM
Before we continue we review cardinal exponentiation:
$$\max (\kappa,2^{\mu}) \leq \kappa^{\mu} \leq \max (2^{\kappa},2^{\mu})$$
 
user84215
In the projective completion of an affine space, is the embedded affine space a hyperplane in the projective space?
 
Pretty sure the geometry guys have not wake up yet at this time
 
user84215
When will they wake up?
 
probably 3-4 hours later
 
user84215
At that time I may dream.
 
1:48 PM
Now the set $\{0,...,9\}^{\Bbb{N}}$ has carnality $10^{\aleph_0 }$. Thus
$$\max (10,2^{\aleph_0}) \leq 10^{\aleph_0} \leq \max (2^{10},2^{\aleph_0})\implies 10^{\aleph_0}=\aleph_1$$
More generally:
$$\max (10,2^{\aleph_{\alpha}}) \leq 10^{\aleph_{\alpha}} \leq \max (2^{10},2^{\aleph_{\alpha}})\implies 10^{\aleph_{\alpha}}=\aleph_{\alpha+1}$$
 
Is $\theta = \exp(\frac{1}{\log(x)})$ an element of $F = C(x)(t_1, t_2, ..., t_n)$ where $C$ is the constant field and $t_i$'s are monomials over $C(x)(t_1, ... t_{i - 1})$, for each $i$.
 
With these we are ready to analyse the set produced by the following function:
 
$F$ is overall a field here, made by building up the extension, similar to the tower of extensions.
 
2:10 PM
actually a slight corection to the above $s_0=0$ as otherwise we will be mapping to [0,10] instead
 
2:21 PM
Algebraists?
I think from what has been said thus far, it won't be hard to show the embedding, but what about the uniqueness up to isomorphism part?
 
2:51 PM
$g: \{0,...,9\}^{\Bbb{R}}\mapsto ?$ where $g = \left(\frac{1}{10^x}\right)_{x\in \Bbb{R_{\geq 0}}} \cdot$. Let $s \in \{0,...,9\}^{\Bbb{R}}$. Then $$g(s)= \left(\frac{1}{10^x}\right)_{x\in \Bbb{R_{\geq 0}}} \cdot s = \int_{0}^{\infty}\frac{s(x)}{10^x}dx$$

We can see that for all s such that $\lim_{x\to \infty}s(x)\neq 0$, the integral diverges. Therefore while the domain of $g$, the subset of integrable functions $X \subset\{0,...,9\}^{\Bbb{R}}$, has cardinality $\aleph_2$, its image is only $\Bbb{R_{\geq 0}}$. This means, compared to the countably infinite decimal expansion, the continuum
From this analysis, we found an interesting fact(?) about countably infinite vs continuum decimal expansions:
Actually typo, the integral always converges since the largest possible representation: $$\int_0^{\infty}\frac{9}{10^x}dx = \left[-\frac{9*10^{-x}}{\ln 10}\right]_0^{\infty} = \frac{1}{10^9\ln 10}$$ is finite
 
3:06 PM
I am trying to prove that $rank(cA)=rank(A)$, where $c$ is a nonzero scalar and $A \in M_{m \times n}(F)$. Couldn't I just note that $cA$ is equivalent to multiplying every row of $A$ by $c$, and therefore there exist $m$ elementary matrices implementing the multiplication; moreover, since elementary matrices are invertible, the rank is invariant under left multiplication by these elementary matrices? QED?
 
@user193319 what is the definition of rank?
 
The rank of a matrix is by definition $rank(L_A)$, where $L_A : F^n \to F^m$ with $L_A(x) = Ax$.
 
Normally I'd see that as a theorem but sure
 
But it can be shown to be equal to the maximum number of linearly independent columns of $A$.
In my proof I am appealing to the fact that $rank(PA)=rank(A)$ for any invertible $m \times m$ matrix $P$.
 
Lol so we're doing things backwards, I take the rank of a set $S$ to be the size of its largest linearly independent subset
That's potentially a way of doing it, but there's an easier way
So you have that the rank of a matrix is the largest number of linearly independent columns
 
3:14 PM
So, given the background information, does my proof appear correct?
 
Let's say the rank is $k$ and we have columns $v_1,\ldots,v_k$. What do we know about $cv_1,\ldots,cv_k$?
 
"So you have that the rank of a matrix is the largest number of linearly independent columns" Yes. This appears as a theorem in my book though, not a definition.
 
To each his own. And I mean yeah, your definition works
Also, your proof does, though there's an easier way of implementing your proof
 
"What do we know about $cv_1,\ldots,cv_k$?" I would say that they remain linearly independent under scalar multiplication, which would be an alternate proof to mine.
 
Yeah, and similarly, linear dependence is preserved
 
3:17 PM
@Daminark Thanks for the help!
 
But thing is, multiplying a matrix by $c$ is like multiplying it by $cI$ where $I$ is the identity. So that's invertible, and the rank is preserved!
If you prefer to do it using the theorem you cited
No problem!
 
Sorry, typos again (arrrrgggghhhh)
$$\int_0^{\infty}\frac{9}{10^x}dx = \left[-\frac{9*10^{-x}}{\ln 10}\right]_0^{\infty} = \frac{9}{\ln 10} \approx 3.90865$$
 
@user193319 @Daminark When you are just starting to learn linear algebra, it's of utmost importance that you actually show things like $cv_1,\ldots,cv_k$ remain linearly independent from the definition.
 
@PVAL 'tis true
 
Therefore one interesting fact about continuum sums (integral) and countably infinite sums involving terms $0<x<1$ is that, once you jump to the continuum from countably infinite, suddenly you have so much stuff that they accumulate to something larger than 1 even though each individual term is $0 < x < 1$
whereas you can never sum to anything larger than 1 for the countably infinite case if all your terms are $0 < x < 1$
 
3:31 PM
Hopefully he's either done this before or will do it
 
However, note that even with continuum number of things $$\lim_{x \to \infty} 0f(x) =0$$ as regardless of what function, the zero function is an absorber thus the result is still 0.
(For a more rigorous discussion, it is because the emptyset cartesian product with anything is zero as shown here math.stackexchange.com/questions/100820/…)
So to conclude, given $s_k \in \{0,...,b-1\}^{\aleph_{a}},a=\{0,1\}$

$$\text{Img}(f) =\text{Img} \left(\sum_{k=0}^{\infty} s_k b^{-k}\right)=[0,1]$$

but

$$\text{Img}(g) =\text{Img} \left(\int_{0}^{\infty} s(x) b^{-x}\right)=\left[0,\frac{b-1}{\ln b}\right]$$
Hence our conjecture:
1. Given the linear functional $h_{\alpha}: \{0,...,b-1\}^{\aleph_{\alpha}} \mapsto [0,u], u \in \Bbb{R}$, and $h_{\alpha} = \sum_{x \in S}s(x)b^{-x}$ Then

$$\text{Img}(h_{\alpha})<\text{Img}(h_{\alpha+1})$$
2. There exists $M = \alpha$ such that

$\text{Img}(h_{M}) = [0,\infty)$
 
3:59 PM
Btw, because all the about results (except the conjectures, which is to be checked) can be bijected back to all the reals, and that the base $b$ is arbitrary. We can then pick $b=e$ to get something familar to all of us:

$$F(1)=\int_0^{\infty}f(x)e^{-x}dx$$

which is the Laplace transform of a function evaluated at $s=1$
 
Is the composition of three reflections a reflection?
 
What about two first ? @MaryStar
 
Isn't the composition of two reflections a rotation? @Astyx
 
Yes, therefore ..?
 
So, we have a composition of a rotation and a reflection.
But what is the result of that composition?
 
4:10 PM
That's the question I'm asking you
 
Is it a reflection along a new line?
 
Oh wait, is a reflection only relative to a hyperplane, my bad
Take the rotation of angle $\theta$ and the reflection relative to the $x$-axis
Take a point with polar coordinates $(r, -\theta/2)$
Apply the rotation first, then the reflexion
You find that you have $1$ as eigenvalue
 
hmm.... I guess I need more topology skills in order to be able to do sums of the form $\sum_{k \in I} a_k$ where $|k| = \aleph_{\alpha}, \alpha > 1$. Will return to this investigation later...
 
Therefore you also have $-1$
So you have a reflection
 
I haven't understood the part with th eigenvalue. What exactly do you mean? @Astyx
 
4:18 PM
Sorry I have to go now
 
Ah ok
 
In dimension 3 this does not hold any more, take the reflections along the canonical basis
 
4:50 PM
@MaryStar In 2D, the only origin-preserving orientation-reversing isometries are reflections
 
So, for $\delta\circ\sigma$ we do the following:
First we reflect, then we rotate around the origin and the we translate.
After reflecting and rotating, is the center still the origin and that's why we have an origin-preserving isometry? @AkivaWeinberger
 
5:53 PM
That was my thinking, yeah
 

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