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03:53
@user21820 u seem pretty good with analysis
can u help me with a quesiton I have?
@MoreAnonymous Okay, let's move to another room.
yay! mathematics?
4 messages moved from CRUDE
@MoreAnonymous This room. =)
I dont think the question is basic math
but k
The other room I have for maths stuff is the Logic room.
03:56
6
Q: A rough proof for infinitesimals?

More AnonymousI discovered the following relation for arbitrary $d_r$: $$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s}}_{\text{removable singularity}} \int_0^\infty...

ah ...
I guess I will rename this room afterward. =)
cool by me
its only heuristic proof ... But I;m very confident of the formula
Well have you tried using hard bounds?
Like asymptotic bounds up to some finite-order approximation.
(But I'm not that good in analysis, by the way.)
03:59
I dont think u can use finite order approimation of k going to infinity
because of the coeffiecients
oh ... :(
No that's not what I meant.
I meant you use asymptotic bounds on f.
And ζ.
hmmm ... didnt think of that
I was thinking f was something $e^{-x}$ and well we could sub in $\zeta$ with $s-1$
are you going through the proof? @user21820
I'm looking at it. But anyway I think you can try getting more eyes on your problem by posting it in the main chat-room.
I did do that
Oh. =)
04:04
but they said it was too long
:/
hi
is anyone know about the types of paradox?
were currently thinking about this paradox
6
Q: A rough proof for infinitesimals?

More AnonymousI discovered the following relation for arbitrary $d_r$: $$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s}}_{\text{removable singularity}} \int_0^\infty...

:P
But on a more serious note what do u mean?
Any thoughts @user21820 ?
@AdarshKumar If you are interested in a logical analysis of paradoxes, you can take a look at this:
8
A: Is Gödel's modified liar an illogical statement?

user21820Your question has two main facets. The first is that you did not grasp the way logic does not fall to the liar paradoxes. The second is that there are deeper reasons as to why we have such apparently innocuous sentences in natural language that seem to defy assimilation into formal logic systems....

But if you're just looking for common fallacies that people have when they don't use logical reasoning, Wikipedia has a long list hahaha..
@MoreAnonymous I'll get back to you later. I need to go off soon.
alright :/
04:30
@MoreAnonymous I just got back, but my guess is that ζ is actually unimportant except for its asymptotic behaviour up to the first few orders.
Okay ..
I think its zeroth order
according to my proof
Yea I won't be surprised. More like first-order.
In that case, it should be easy to use hard bounds.
Normally, the advantage of hard bounds is that you do not have to deal with infinite series.
yea i was seeing on wiki its basically $\zeta(s)(s-1) \sim 1 $
as s goes to 1
so $\zeta \sim 1/(s-1) $
Yea then what you do is you let t = s−1 and then it changes to t→0.
And then you just use first-order asymptotic approximation of 1/r^(1+t).
The integral doesn't even play a part on the right-hand side, since it's constant.
wait what?
04:40
It's constant, isn't it? You can take it out of the limit.
ah ...
didnt think of that
04:53
@user21820 sorry that was obvious was(am) a lil brain dead
No problem at all. It's still an interesting question. =)
05:29
Here is a nice curiosity from PIE (Principle of Inclusion and Exclusion):
> n! = n^n·C(n,n) − (n−1)^n·C(n,n−1) + (n−2)^n·C(n,n−2) − ... 0^n·C(n,0).
where C(n,k) denotes "n choose k", namely the k-th entry in the n-th row of Pascal's triangle, each counting from 0.
Jan
Jan
Hi, may I ask why $\frac{W_{ik}}{W_{ik}H_{kj}}$ (the denominator is not $ \sum_{k}{W_{ik}H_{kj}}$) is not equal to $\frac{1}{H_{kj}}$? Since it is not sum over $k$ yet?
$W$ and $H$ are matrices with dimension $ik$ and $kj$, respectively.
 
2 hours later…
07:02
@Jan I'm not sure what your question is. Does it boil down to whether x/(xy) = 1/y where x,y are reals?
Jan
Jan
07:46
@user21820 sry let me rewrite my question. Is $\frac{W_{ik}}{W_{ik}H_{kj}} = \frac{1}{H_{kj}}$ ? where $W$ and $H$ are matrices with only real number.
@Jan Well, is what I said correct?
Jan
Jan
@user21820 yes but x, y are real matrices
@Jan You wrote "W_{ik}". That is not a matrix.
Jan
Jan
and xy is not matrix multiplication. The row of $W_{ik}$ (x) is multiplied with the column of $H_{kj}$ (y) but without sum over $k$ dimension
oh what should I write for matrix named W with i row and k column?
08:36
@Jan That is not the point. If W is a real matrix, then W_{i,k} is an entry in W, which is a real number, so what you wrote has nothing to do with matrix multiplication. Just answer my question first: Is it true that x/(xy) = 1/y if x,y are reals?
Jan
Jan
yes
@user21820 yes except y is zero
@Jan Yes and that's all about it. What you wrote is also true only if W[i,k] and H[k,j] are both nonzero. Otherwise it can be considered meaningless (division by zero).
Whether you later sum over all k later or not is not relevant.
Jan
Jan
08:54
@user21820 ok, thank you so so much!
You're welcome!

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