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QED
8:00 PM
some kind of grease monkey thing
 
There is no way to do it manually?
 
QED
maybe you can see it in the HTML source
 
Bleh.
 
8:01 PM
@MartinSleziak Display page source it is then.
 
Already voted, Srivatsan.
 
@Srivatsan Done.
 
@Matt The only way I know.
 
Thanks, guys! Only an executioner needed now.
 
@MartinSleziak How convenient.
Is there a way to do feature requests for SE sites?
 
8:03 PM
@Matt I guess if someone know better way, it will appear in that meta question. (Since I bumped it today.)
BTW if that link to meta.SE from Hendrik answers is tagged feature-request. meta.stackoverflow.com/questions/5436/direct-link-to-a-comment
 
There we go. Proved it.
 
And it is also tagged status-declined....
 
@MartinSleziak "status-declined"
 
Yes I've noticed it just now.
 
Me too.
Solution: open a new meta thread requesting the same feature, repeat until they can take it no longer and implement it.
 
QED
8:06 PM
ill upvote it
 
Or until you're banned.
 
QED
but like just install the greasemonkey script
 
I don't want to install anything.
 
QED
do you not use greasemonkey already?
 
No, what for?
 
QED
8:08 PM
you can change things about websites automatically
 
I have not encountered a situation where I wanted to do that.
 
@Srivatsan shouldn't one of those duplicates be closed as well?
 
I should really do something exercise-like, my neck has been bothering me quite a bit in the last few days. What's a sport that's not so "sporty"?
@MartinSleziak Unfortunately, I have that already I think. Because my finger slipped on one of my first days on SE. I removed the vote immediately but the badge stayed.
 
8:25 PM
@tb Thanks for the CW answer. Wouldn't the thread you linked me to be worth including too. (For his second question.)
 
Could have been nice if GEdgar would have answered this.
 
@MartinSleziak I didn't because of my comment here but maybe I should have for other users.
 
I see. I did not know you pointed out that thread to him before.
 
By the way: I hope I haven't misrepresented your notes.
 
Thanks for including them in the post. I hope that they will be useful to someone.
 
8:33 PM
There's a slight difference (< vs. $\le$), not sure if that is important. If that matters much, then we should close the other question leaving this one open.
 
The notes are basically typed theorems and definitions from that part; only the proof are missing.
 
@MartinSleziak what notes?
 
@Srivatsan My notes which t.b. linked to in this CW answer.
 
@Srivatsan I don't think it matters very much. It is very hard not to prove strict inequality.
 
@tb Yes; I'm just being safe after seeing Andre's comment in the other question. I think there'll always be some disagreement unless the post is an exact duplicate.
 
QED
8:47 PM
I don't know what to do i am bored
 
@Srivatsan Executioner left to be found :-)
 
@QED move to russia, drink vodka and solve problems?
 
Is there a way to tell Google to please not tell me about links from some websites by default? I'm having absolutely no use for mendeley.com or dict.cc or all of those wiki scraper sites, for example.
 
Cool, they will close (I think the biggest) water lock in NL.
 
Well, you could add this to your search engine settings in the browser.
 
QED
8:51 PM
you can use a greasemonkey script that hides them
 
Because the water level will rise above 2.95m NAP (that's the level of Amsterdam).
 
QED
I did that for the various wikipedia "mirrors" (i.e. search engine optimists)
 
@AsafKaragila what does that mean?
 
Hey guys.
 
QED
hi
 
8:56 PM
@tb Suppose you use the Firefox search bar, just change the setting of the Google query to automatically include the "not in site ..." as in the custom search.
 
Thanks to those of you who tried answering my bone-headed questions today. I appreciate it.
 
@tb Yes. When you do a search you can hover over the hit you want to block. Then "Block xyz.com" appears and you click on that. It will be stored to your search settings.
 
@KorganRivera I didn't answer any of your questions, but it would be nice if you marked some the answers you got as accepted if they were helpful: see here how to do it.
 
Hmmm... 30 more questions on AC, I just need to add answers to 4 existing ones, then I might hit silver specialist together with the Bronze one. That would be impressive.
 
9:06 PM
@Matt That doesn't seem to exist in my browser. Is this maybe Chrome specific?
 
@tb It doesn't even need the hovering bit: it's there already. Let me try with Safari.
 
@Matt Oh, maybe you need to be logged in into a Google account?
 
@tb Let me test with an incognito window.
@tb Yes. That's it.
 
Okay, thanks, good to know...
 
But if you don't want to have a Google account: Chrome is really good. Much faster than FF (I can't run FF on either of the laptops here) and more reliable (it restores your previous session if it crashes with 100% certainty).
And Chrome has a Personal Blocklist extension which does what you want.
: )
Of course you already have a Google account but Chrome is cool : )
 
9:11 PM
And what do you earn from them for this advertisement? :)
 
But that would mean change and man is change-resistant... : )
 
Chrome is very good. I resisted it for a long time in favour of FF. But FF started stalling every 20 mins after an update and I switched. I don't regret it.
 
@tb Nothing. It's my good deed for today. (as I have not answered any questions on SE)
 
Well, as Jonas likes to say: Google gives me the willies...
 
May I ask why?
 
9:14 PM
They are becoming too big for my taste. Admittedly, their search engine and their other tools are very good. But I don't really trust them.
 
: )
I left something in the other room for you. You don't have to look at it now. But just so you know : )
@tb So who do you trust then? Your bank every time you do online banking? Or do you go to the bank every time you have to transfer money? : )
 
@Matt Okay, I interpret that: look at it later. Thanks for the tips. I will consider them. But given my genetically imposed inertia...
 
QED
I tried to switch to a different search engine but I couldn't
didn't get good results
 
@QED Maybe you should try Bing...
 
9:19 PM
(They copy Google's results and present them as their own)
 
QED
bing just steals googles results
heh
 
Exactly : ) That's why I thought it was a good suggestion: different search engine but equally good search results.
 
@Matt well, I don't do that on a daily basis and I have some rights there. And they don't ask me to share my entire life with them because it's oh, so convenient.
 
QED
do you use facebook @tb
 
nope.
 
9:20 PM
: D
 
QED
I thought I was the only one
 
Guys, I feel a little defeated in trying to understand infinitesimals. I'm sure you all think this is hilarious. But if I can't understand this, then I'm yet again stalled. How did you guys come to terms with them, later in your studies?
 
QED
@KorganRivera, infinitesimals are a much more advanced topic than basic calculus
 
That's what I'm realising. So how can I move on from basic calculus knowing this?
 
QED
you can do calculus without any infinitesimals what so ever
 
9:24 PM
Even my 3rd coffee is saying no, I can't. :)
 
QED
What do you want to know? something about infinitesimals?
 
You know, I don't even know any more. O_O
I haven't slept well and, for all I know, there could be a perfect explanation already on my question page, that I can't 'see' right now.
 
QED
do you know the history? Calculus was invented based on the notion of infinitesimals. There were serious logical difficulties found in it, and a new theory developed based on limits. In modern times using some quite deep ideas from logic a new rigorous theory of infinitesimals was created.
 
Yes, I read about that today.
 
I can't get rid of the idea that the Abrahamson's stuff about infinitesimals is not really an improvement since it requires to learn a lot of logic.
 
QED
9:27 PM
In terms of pragmatics, everyone uses the limits theory of calculus - and it's very good. The infinitesimal theory is slightly useful but it's quite esoteric.
 
The limits idea makes more sense to me.
 
There are even people in my esoteric field that use esoteric stuff like non-standard analysis.
 
QED
did you see my answer about limits?
 
@JonasTeuwen Even if you never study non-standard analysis, logic is already a great improvement compared to that bore you do ;-)
 
I'm broken. Because of your words, Asaf.
 
9:29 PM
Hey Asaf, what's up? I'm hungry and fancy eating some of that beef stew you made yesterday.
 
Yes, but 1) your set notation intimidated me and, 2) I've rephrased my question since: math.stackexchange.com/questions/96420/…
 
Someone! Quick! Give me the title of some bad 80s song!
 
QED
@KorganRivera, I proved the most important theorem "uniqueness of limit".
 
@Matt You should come to Israel. I'll make some fancy dinner.
@JonasTeuwen Like a Virgin?
 
QED
@KorganRivera, you can't be intimidated by the symbols, that's what "lim" is
 
9:30 PM
@JonasTeuwen Wake Me Up (Before You Go-Go)
 
Oh no.
 
@QED I will spend more time on it.
 
That is too bad.
 
@AsafKaragila But the partner wants a beach holiday...
How to convince them otherwise?
 
QED
@KorganRivera, understanding the proof of the theorem is an important step in understanding the definition
 
9:30 PM
New Order with True Faith it will be!
 
@Matt We have beaches in Israel... just not great ones ;-)
 
Maybe we can do both. We've not been away in 1.5 years.
 
I'd take a different partner if mine wants a beach holiday 8-).
 
QED
@KorganRivera, if f is continous then lim_{x->a} f(x) = f(a)
 
@QED if set notation broke into your apartment and took your things, you'd be intimated too.
 
9:31 PM
@JonasTeuwen Ditto. I hate beaches, I hate the sea and I hate most lakes and whatnot.
 
There are two kinds of people: People with taste and people without taste.
 
@QED I understand that.
 
QED
@KorganRivera, I'm not sure why you're asking about the case where f is continous... since it's just f(a) which is easy
 
@JonasTeuwen Equivalently: There are two kinds of people, people with taste or people which disagree with me.
 
Exactly.
 
QED
9:33 PM
@KorganRivera, in the case of derivative, the function is g_x(h) = (f(x+h)-f(x))/h which is not continous at h=0
@KorganRivera, are you asking for a proof that if f is continous, then lim_{x -> a} f(x) = f(a)?
 
@JonasTeuwen @AsafKaragila You guys are harsh. I don't think I can find a better partner though.
 
@QED Hold on let me think :)
 
@Matt I know the feeling.
 
QED
@KorganRivera, please feel free to ask me about anything I say which doesn't make sense
 
That can be the case. Sometimes we have to be satisfied with less.
 
QED
9:38 PM
if your partner wants a beach holiday why don't you just give them it?
 
Let him/her go alone!
That's another option.
 
@QED No. This is my question as best as I can put it: I understand that lim_{x->a} f(x) = f(a), but then to say that the gradient of the tangent curve is some value, is like saying that when x=a, then f(x) = f(a). The whole point of the limit, I thought, was to say, instead, that we don't know what f(a) is, but we can say that it approaches some value.
 
Holidays should be relaxing, not frustrating.
 
But saying the gradient of the tangent is a value, is different that saying it approaches some value. So, I'm having a problem with understanding that the limit is the value.
@QED does that make sense?
 
QED
@KorganRivera, lim_{x->a} f(x) = f(a) is a theorem for continous f
@KorganRivera, the function you are taking a limit of, when you compute derivatives, is not continous
that function is not 'f', for example if f(x) = x, then to find the derivative of f at x=1... you need to compute lim_{h -> 0} (f(1+h)-f(1))/h.
so you're actually finding lim_{h -> 0}g(h) where g(h) = h/h. and that is not continous at h=0
@KorganRivera, Doesn't the theorem that if lim_{x->a} f(x) = A and lim_{x->a} f(x) = B then A = B answer that confusion about whether the thing really "is" or is not
 
9:45 PM
@Ok, let me think for a minute.
 
@JonasTeuwen I don't really remember what beach holiday is like. So I don't mind going.
 
@QED So if I'm working with function f, and f is continuous, my derivative dy/dx is by definition not continuous, since it is undefined at dx=0. Is that correct so far?
 
QED
@KorganRivera, What you just said didn't make sense
 
QED
@KorganRivera, let g(x) be the derivative of f(x) at x
 
9:48 PM
ok
 
QED
@KorganRivera, no need for any of this "dy/dx" nonsense that we've somehow been unable to get rid of even though it's been wrong since the 1600s
 
ok :)
 
QED
so for example if f(x) = x^3 then g(x) = 3x^2
 
test
 
QED
the reason people don't do it like I do, is there's nothing to get confused about
@Srivatsan, did you get kicked out the chat for not having enough rep?
 
9:51 PM
@QED I'm still following you.
 
QED
@KorganRivera, well do you have any more questions? I said what I wanted to say
 
@Matt Either way you should tell them that you'll go for a beach vacation if and only if they will come to Israel and we'll all have a nice dinner.
 
QED
maybe there should be a special chat room for this jury duty stuff
 
@QED I'm facing some tech problems with chat. I have to jump some hoops to chat, that's why I'm in and out.
 
QED
9:53 PM
that way you can leave a message there, and even if nobody is around - they can see it when they join (rather than here, where it gets scrolled away by the other chat)
 
@QED thanks, I'll see what I can do from here. /shotgun in mouth.
 
Thanks Matt. See you guys later.
 
guys the limit problem seems problematic
take a look
3
Q: Finding limit for the function

DavidI have problem with showing that the limit of the following function $$\frac{ \sqrt{\frac{3 \pi}{2n}} - \int_0^{\sqrt 6}( 1-\frac{x^2}{6} +\frac{x^4}{120})^ndx}{\frac{3}{20}\frac 1n \sqrt{\frac{3 \pi}{2n}}}$$ equal to $1$, with $n \to \infty$.

 
QED
where does a horrible big expression like that even come from?
 
exactly my question
and like input like output
integral is equally as ugly as its mother
 
9:59 PM
@QED When I said, "So if I'm working with function f, and f is continuous, my derivative dy/dx is by definition not continuous, since it is undefined at dx=0." I guess what I'm saying is that (f(x+h)-f(x))/h is not continuous since it's not defined at h=0.
 
QED
yeah, (f(x+h)-f(x))/h is not continous at h=0
 
@QED but when I say that with dy/dx notation you tell me it doesn't make sense. :)
 
@Srivatsan, @AsafKaragila, @Matt and the rest
please take a look at the limit question
 
QED
@KorganRivera There are lots of things wrong with that: dx=0 is wrong. dy/dx - what/s y? "dy/dx is by definition not continuous" it's not a function how can you ask whether or not it's continous, ... etc.
In general this stuff with 'dy/dx' is supposed to help as some kind of memory aid, but since there's no rigorous mathematics behind it - all it's going to do is confuse people
in fact there was a big controversy about it since using it in obvious ways suggested by the notation leads to wrong results
 
@QED. Ok, when I say "dy/dx" I'm not talking about a quotient. There is no y. I'm saying the change in f(x) over the change in x.
 
QED
10:03 PM
my recommendation is to forget about this dy/dx stuff but if it works for you fine
I think about calculus in terms of functions and limits, since that's what's actually happening
 
@QED Me too! I just write dy/dx to mean lim_{h->0} (f(x+h)-f(x))/h.
 
QED
you can use Df(x) or f'(x) instead ("D" being the differentiation operator)
 
@QED Just to make it clear, I'm not using δy/δx.
 
@NikhilBellarykar Please don't do that. Don't highlight everyone and ask them to look at the question that you already posted on the website.
 
No I havent posted the question. And I highlighted more than one because I thought there is something wrong with either the question or my approach.
 
10:09 PM
@QED I'll work on trying to understand that the gradient of the tangent is the limit, rather than the gradient of the tangent approaches the limit. I'll read your proof. Thanks for your help. I think I just need some sleep. O_O
 
QED
@KorganRivera, note that I never said anything about gradients or tangents.
@KorganRivera, you can do absolutely everything here without it: that stuff is just a geometric picture that's suppose to help people visualize what's going on
 
@NikhilBellarykar Either way, don't highlight everyone and ask them to check out some link. If you have a specific user which you think can say something in particular feel free to highlight them; you may also address "to all", but don't highlight several people like that.
 
QED
@KorganRivera, if you study the theorem "A=B" that I posted (after having slept). I think it will all become clear.
 
@QED Thx. Hope so. :)
 
@AsafKaragila I asked all in the beginning, its only after non-response that I highlighted specific guys. anyways can you please look at it now?
 
10:12 PM
I have no idea what I am looking at.
 
Ahoy thar!
 
QED
hi
 
@AsafKaragila shall I post the link once more?
 
Does there exist a number that is divisible by 7 and has a last digit of 9 ?
 
49 is the number
 
10:14 PM
@NikhilBellarykar No. I know what the link is. I have no idea why I am looking at it, what should I do about it, and frankly I have enough as it is. I use this chat to vent, not to exercise my better judgment.
 
Mmmm, so a number ending in 6 can only end in even digits, a number divisible by 7 can end in all digits, but a number divisible by 5 can only end in 0 and 5
 
@NikhilBellarykar 49 is such number. 539 is another.
 
I fail to see the pattern.
 
@AsafKaragila indeed. @N3buchadnezzar any number of the form 49+70*n where n is any natural number would do
 
QED
10:15 PM
@N3buchadnezzar, use modular arithmetic
 
$$ x \equiv 0 \pmod{3} $$

for an example ?
 
QED
no
$$x = 7^{-1} \cdot 9 \pmod {10}$$
 
See you later, @Srivatsan!
 
QED
then 7x ends in 9 and is divisible by 7
 
@AsafKaragila I really fancy that fancy dinner you offered : )
 
10:18 PM
@QED I think I just got it. And I think my answer was completely obvious. :)
 
QED
answer to what?
what was obvious?
 
@QED I think the answer to my question wasn't as difficult as I had thought, and I think my brain just found a way to understand it.
 
QED
oh good
 
@Matt I must warn you there are no romantic strings attached, but there will be some forcing and no choice ;-)
 
@AsafKaragila : D
 
10:21 PM
@QED just to make sure, would it be correct to think of the derivative of a function as the set of all limits of that function?
 
QED
no
 
@QED maybe I worded that wrong.
 
QED
also note that limits are unique
 
I understand that. What I'm trying to say is that I've realised that if I have two functions f(x) and g(x) and g(x) = (f(x+h)-f(x))/h, then lim_{h->0}g(x) = L = f'(x). Is that also wrong?
 
QED
how about g(h) = (f(x+h)-f(x))/h
then you can say lim_{h->0}g(h) = f'(x)
this is just the definition of the derivative
 
10:30 PM
ok guys, me is leaving. bye all
 
@Matt Even with my PhD salary I don't have enough money to go on a holiday :-/.
 
@QED So now it makes sense to me that the derivative is the limit. What I think I was doing in my head was saying to myself that g(x) isn't continuous at x=h so how can I evaluate g(h)? But that's not what's happening. The derivative is the limit, not g(h).
 
QED
"g(x) isn't continuous at x=h" doesn't make sense
do you mean g(h) isn't continous at h=0?
 
omfg
lol, yes.
 
QED
where g(h) = (f(x+h)-f(x))/h
 
10:34 PM
yes
 
QED
but yes,
 
@JonasTeuwen Brian Adams - Summer of '69? Is that 80ies enough? It's 1984.
 
QED
while for continous functions f, you can simply evaluate lim_{x -> a} f(x) = f(a)... For discontinous functions, the limit may or may not exist and computing it will require more work
 
@Matt It might be. Let me check it out again!
 
@QED Yes. That makes sense, I just wanted to straighten out my understanding of the continuous ones. I had a glitch in my code :)
 
QED
10:36 PM
@KorganRivera, in that case you'll need to be proving $\forall \varepsilon > 0,\,\,\,\, \exists \delta,\,\,\,\, \forall x,\,\,\,\, 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.$ by picking some correct L (somehow)
 
Does Jax not render in here?
 
QED
@KorganRivera, by the way, what's your definition of "continous"?
 
@JonasTeuwen What do you need it for?
 
To listen to. I feel sentimental.
 
Clash - Should I Stay Or Should I Go?
 
10:38 PM
That's better.
 
Phil Collins - Against All Odds.
 
QED
@KorganRivera, or do you not have one?
 
I think of a continuous function as a function that I can differentiate at all points. No breaks or jumps.
 
QED
@KorganRivera, thats a(n everywhere) differentiable function
 
10:39 PM
@Matt Ah! Gold!
 
I think the 80ies were a bad year for music.
And the 90ies.
 
But I'm from the 80s!
That's why I listen it.
Classical music or 80s.
 
@QED I'm not sure how to describe it, other than 'not discrete'.
 
QED
@KorganRivera, ok - but yeah basically it's a more advanced concept that isn't needed
you can do derivatives and stuff without it
 
@QED Isn't it necessary that a function be continuous to be differentiable, at least over the domain you need?
 
QED
10:44 PM
differentiable implies continuous, but not vice versa. As said this theory isn't needed
 
Ok, what's a simple example of a continuous function that is not differentiable?
 
f(x) = |x|.
Or the Weierstrass function.
$$f(x) = \sum_n \frac{\sin(3^n \pi x)}{2^n}$$
Or something like that.
 
ha, of course. :)
 
Yeah, just write down what it means for a function to not be differentiable. Then keep messing a function up in every point :-).
 
This is good. I've learned that a function that is differentiable at all points is continuous, but a continuous function is not necessarily differentiable.
 
What's a better, but simple, definition of a continuous function, other than 'not discrete'?
 
QED
@KorganRivera, 'not discrete' isn't a mathematical definition - that's just some kind of intuitive picture
 
Good night folks.
 
QED
night
 
night
 
10:53 PM
@Matt Babai!
 
@tb The "man" in that sentence was supposed to be gender neutral.
 
@Matt Nighty night.
 
Huy
Hey guys, I have a short question a friend of mine asked me which I cannot answer because I have not learnt about measure theory (or whatever is needed to answer the question) yet. He asks what is wrong with \int_0^{2 \pi} \frac{d}{dn} e^{inx} dx when he applies Lesbegue's dominated convergence theorem, because apparently, if he first integrates and then derives, the result is 0 but if he first derives and then integrates it's not 0. Does anyone know?
 
What are the conditions for that theorem? Check them.
Then come back!
 
QED
$$\int_0^{2 \pi} \frac{d}{dn} e^{inx} dx $$
 
Huy
10:54 PM
I don't understand any of them.
 
I appreciate you guys helping with my noob questions. I don't get to talk to anyone about this stuff, just study on my own from library books and online. So thanks :)
 
Then you shouldn't use that theorem, if you don't understand the hypothesis.
 
Huy
I'm not using that theorem.
 
QED
you shouldn't use that theorem at all
 
Huy
A friend of mine is preparing for an exam and asked me whether I knew the answer or whether I knew anyone who could help him out.
 
QED
10:56 PM
it doesn't say anything about commuting integral with derivative
 
Well, a quick corollary does.
 
QED
it's for commuting limits with integrals..
 
Yes, and how do you define a derivative?
 
Huy
So is a condition not met? Then I can tell him that and he at least has an idea where he went wrong.
 
Let him ask the question himself 8-).
 
Huy
10:58 PM
BTW, the function should be e^{-inx}
I don't see what the difference would be, but if that is what you suggest, then I will do it.
 
Okay. To apply the dominated convergence theorem we need to find integrable $g$ such that $|\exp(-i n x)| \leqslant g(x)$. So, $1$ would work.
Oh, right, we need a bound on the derivative, my bad.
 
Bye all, thanks for the help.
 
QED
bye
 
Huy
I believe g(x) = x should work.
 
I'm not sure why he wants to apply that theorem.
 
QED
11:06 PM
probably just playing around, found an anomaly and wants to understand what went wrong
 
There is no anomaly.
It just works.
I have tried it.
 
Huy
Well, integrating first and then deriving results into 0.
How is $\int_0^{2 \pi} -ix e^{-inx} dx = 0$?
 
Okay, let me check again.
It is not zero.
 
Huy
Both or only the latter?
 
I dislike people asking questions without verifying their claims.
Both.
 
Huy
11:11 PM
wolframalpha.com/input/?i=int+e^%28-inx%29+dx+from+0+to+2*pi
Hm, how do I insert the whole link?
 
I don't really understand why such a big point is made of exactness here.
 
Don't use Wolfram Alpha, to compute such a simple integral :-/. It wastes your mind.
 
Huy
Well, for all natural n the integral is zero.
Maybe he meant that, I don't know. :(
 
Doesn't the argument miss a $2\pi$ then?
Or the integral from $0$ to $1$.
 
Huy
What exactly do you mean?
 
11:15 PM
That the integral should run from $0$ to $1$ to obtain that or the integrand should say $\exp(2 \pi i n x)$.
 
Huy
$\int_0^{2 \pi} e^{-inx} dx = 0$ for natural n - is what I mean.
(and if I then derive, I again get 0)
 
But you shouldn't fill in $n$ yet, for the derivative that is a continuous parameter
Compute the integral and differentiate it.
 
Huy
I see what you mean.
So maybe that's where he went wrong.
 
Huy
11:30 PM
I just told him and yes, that's where he went wrong. Thank you for your help.
 
Good. Next time let he himself ask the question.
 
Huy
I will ;-)
 
11:49 PM
This thing will probably be closed: nl.wikipedia.org/wiki/Maeslantkering
If that thing doesn't function when it needs to it would be a waste of good whisky since mine would be underwater then :(.
 

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