« first day (2414 days earlier)      last day (2903 days later) » 

05:01
why if you calculate $d/dx (1 + x + x^2)^{-2} = -2(1 + x + x^2)^{-3}(1 + x)$ So here is an explicit formula
$({n +1})!(-1)^{2n + 1}(1 + x + x^2)(1 + x)^{n - 1}$
@MikeMiller write as $(1-2x+x^2)(1+2x^3+3x^6+4x^9+\cdots)$, find taylor coefficient
one edit
I think that is the explicit formula for it
(that's $(1-x)^2(1-x^3)^{-2})$ with $x^3$ plugged into the series for $(1-r)^{-2}$)
(which I got from noticing $1+x+x^2=(1-x^3)/(1-x)$)
Hello all, need help with my question here math.stackexchange.com/questions/2183428/…
@Adeek once you have the (1+x) in there you need to use a product rule
@arctictern clever
05:12
oh right haha I thought that always (1 + x) disappear since it is 1 but we have also the component coming from multiplication
05:25
It's easy to forget how calculus works after some time
:P jk
@Daminark lol
Are you TAing anything btw?
how are generating functions useful?
sure, if the generating function is something like P(x)/(x-a)^n then you can find its coefficients easily
Hi @DHMO
but with more terms in the denominator, you would need convolution, right?
@Learninguser hi
05:39
Did you check the sum
yesterday I have posted
can you post again?
@Daminark are you asking me ?
or mike ?
3
Q: Percentage based on Profit and Loss

Learning user What was the Percentage of Discount given? 23.5 % profit was earned by selling an almirah for rs 12,350. If there were no Discount,the earned profit would have been 30% The cost price of the almirah was rs 10,000 for these Question option are only I and II only I...

please check @DHMO
@Learninguser sorry, please find others
05:41
ok
Oh god^ I'm so happy that I don't have to do profit and loss problems any more :-D They creep me out :-P
@DHMO Good morning :)
@2017 namaste
So you learning generating functions?
not really
just exploring
Oh I see, you should see the method of snake oil, which heavily uses generating functions...they are pretty interesting!
05:43
?
(I don't know why it is named like that :-P)
05:55
@2017 oh thanks
np :)
now could you remind me what king rule and queen rule are
lol
f(a+b-x)=f(x) was the king
and the splitting (from 0 to 2a) one was the queen
I see
do you have jack rule?
Hello all, need help here. Im given a hint on a difficult question on midterm, it says "know how to construct an initial value problem for ODE using Picard Lindelof theorem".

Anyone has a clue what this means?
05:58
:D
There are many jack rules
ridiculous
Hey I'm back, and I did mean you @Adeek
@2017 evaluate $\displaystyle \sum_{n=0}^\infty \frac {x^{3n}} {\left({3n}\right)!}$
If a,b,c,d,p are different real numbers such that $$(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2) \leq 0$$ then $a,b,c,d$ are in A.P, G.P, H.P or A.G.P ? Try this @DHMO ! (Meanwhile I'm having a look at your sum)
@2017 sorry busy
06:02
oh sure :)
try it later then
$$\sum_{k\geq0}\frac{x^{3k}}{\left(3k\right)!}+\sum_{k\geq0}\frac{x^{3k+1}}{\lef‌​t(3k+1\right)!}+\sum_{k\geq0}\frac{x^{3k+2}}{\left(3k+2\right)!}=e^{x}$$
what is wrong with the latex!!
Anyhow, i think you'll get a second order differential eqn
then use the standard method to solve it
@DHMO ..
Hi, would anyone mind looking over my solution to this contest problem: mathb.in/134240 ? I think it might be correct, but I'm not sure.
Is anyone active in here? :)
06:28
@user50328 The final answer looks correct to me. (However, I don't know about lagrange multipliers)
Maximum ought to occur at symmetry
Hello! Why exactly do you say that the maximum ought to occur at symmetry? I don't have much intuition for the problem.
https://en.wikipedia.org/wiki/Kato_theorem
Quantum chemistry theorem which in order to understand the proof, requires knowledge in functional analysis and real analysis. I wonder if this guy is secretly a mathematician
@2017 oh that wasn't what I meant
I meant $\displaystyle \sum_{n=0}^\infty \frac 1 {\left({3n}\right)!}$
And try doing it without calculus
Okay, I'll try it a bit later..studying chem now :) I don't know a method without calculus...gotta think a bit
@Secret $\displaystyle \sum_{n=0}^\infty \frac 1 {\left({3n}\right)!}$
06:56
can someone suggest me some books on the theory of ordinals and in particular transfinite induction?
@DHMO hint?
@2017 do the variant with 2n first
I think it will be combination of e^x and sin(x)/cos(x)
@DHMO what is "variant" ?
Find $\displaystyle \sum_{n=0}^\infty \frac 1 {\left({2n}\right)!}$ instead
$(e+e^{-1})/2$
07:03
@DHMO am I allowed to use the binomial theorem and the expansion of just $e^1$?
@SoumyoB I don't know how you would use it
@2017 how?
@DHMO just square it?
ohhhhhh wait wait wait
Just series expansion
The alternate terms will cancel out
^
and for the $3n$ version we could just add $e^\omega$ where $\omega$'s are the cube roots of unity
@SoumyoB really?
07:07
am I making some mistake?
Some excess terms remain
and you won't get the answer like that..
did you divide it by 3 after adding $e+e^\omega+e^{\omega^2}$?
sorry, i need to go..
@SoumyoB that looks okay
@SoumyoB please continue
@DHMO that is my answer
07:18
please make it real
I know it's real
but please get rid of complex numbers
oh
well is the answer then $\frac{e+\sqrt{3}}{3}$?
why?
you can just write $e^\omega$ as $e^{i(-i\omega)}$ etc
and then?
write it as $\cos{\frac{\pi}{6}}+...$
07:23
so $-i\omega = \frac\pi6$?
yeah that's what I thought is the argument of $-i\omega$?
argument?
so multiplying something by $i$ gives it argument?
as in the complex number plotted on the argand plane?
what?
I solved this question $\int \frac{ 1 }{\displaystyle\sqrt {1 - 36x^2}} = \frac{ 1 }{6}arcsin(6x)$ but apparently he wants me to evaluate it using trig substitusion so this is not the right answer, but isn't this supposed to be trig sub?
07:25
I know what argument is
if you multiply it by $-i$ the argument becomes the argument of $-i$ plus the original argument
can you do it step by step?
argument of $-i$ is $-\frac{\pi}{2}$
continue
well sorry I gotta go, I'll talk about this later
07:26
So latex does not work in chat?
@someone see chat description
Now it's working
08:00
@someone How did you solve it?
My solution was 1/6sin(6x)
but he wanted me to solve it using trig sub
@DHMO Is it $\frac{e+e^{\frac{-1}{2}} \left( 2cos \left( \frac{ \sqrt{3} }{2} \right) \right)}{3}$?
@someone I agree with your solution, but if 'he' wanted you to use trig substitution, I'm going to assume you did it another way, which is?
hello
how to solve this equation
@ mike
Well, you could just invert either matrix since they both have determinant $1$
But, to give you a tip: the answer is already there in the question :P
@TimTheEnchanter I think I should be using $\displaystyle\sqrt {a^2 - x^2} \ {\text{let}}\ {x = asin(theta)} {\text{ and}} \displaystyle\sqrt {a^2 - x^2} = acos(theta)$
08:10
@steam
u answered to me
?ž
xn=1 ?
do not understand you entirely
@TimTheEnchanter and solve for theta then use a reference triangle and change to x ?
@someone That is trig substitution. What is 'his' problem?
His problem is I solved using the formula $\displaystyle \int \frac{ 1 }{\displaystyle\sqrt {a^2 - u^2}}\ du = arcsin(u/a)$
08:16
@someone Perhaps you ought to have shown the intermediates. It would help if we knew who 'he' was or what 'he' expects.
@ steam... how to start here
@TimTheEnchanter my math professor ?!
1. put into matrix?
@someone I don't know mate. Everything you've done is correct. The only possibility I can think of is that your prof expects you to explicitly write down the substitution $x=sin( \theta )$ and work from there.
@TimTheEnchanter If I did not use the above shortcut formula and used trig sub shouldn't one side of the triangle be $\displaystyle\sqrt {1 - 36x^2}$ ?
because if $x=sin(theta)$ then the adjacent side is $\displaystyle\sqrt {1 -x^2}$
08:30
@someone In this case you would have to use $6x = sin( \theta )$ but in principle what you've done is the same.
I would get the same answer, yet the grading system is not accepting $arcsin(6x)$ only $arcsin(x)$s :(
@Daminark I am doing TA work yeah
I am marking for multi-variable calculus and some other weird class
and I am doing help center to help people with math
08:48
@ADEEK
how to solve this
@someone That's rough man, I thought that kinda stuff only happens in standardised tests :(
It is very annoying but these questions are only for bonus credit so it's fine.
09:12
I've a quick topology question. Is $\langle a,b| ab\rangle$ a valid polygonal presentation of the closed disk $\overline{\mathbb{B}^2}$?
09:30
@ tobias
how to solve this
imgur.com/a/uiG0Z
09:50
@Algebra2015 Think about upper-triangular matrices.
how to write down the solution
tnx
The solution should be the construction of a matrix. If you write down the first and last couple of columns, the pattern should be clear enough to anyone who would be reading the answer.
What have you tried on that one?
i put into matrix
10:04
Try multiplying your matrix with a vector, and mess around with the values of your vector. How can you get the equalities you need?
@SteamyRoot Do they do this every April 1$^{st}$?
I think so
10:59
Hello :)
 
2 hours later…
12:59
Hello@SwapnilDas
13:14
$\pi $
Happy pi day!
Hi@Semiclassical
U free ?
Challenge: $\displaystyle \sum_{n=0}^\infty \frac 1 {\left({3n}\right)!}$ (please don't scroll up)
Whenever $\pi$ day comes up, I always think of this: smbc-comics.com/?id=1777
Yeah, that's easy once you know how to do root of unity filters.
13:18
@Semiclassical nice
I actually had a research problem that used that idea, though.
With, say, the 2nd root of unity filter used to figure out the probabilities of even/odd occupation numbers.
I wonder if $f:\mathscr P(\Bbb N) \to \Bbb R$, $ \displaystyle f: I \mapsto \sum_{n \in I} \frac 1 {n!}$ is bijective.
It seems plausible enough.
But I have no idea how one would show that two different subsets must give distinct results.
we can see that $f(\mathscr P(\Bbb N)) \subsetneq [0,e]$
13:22
@AkivaWeinberger tienes alguna idea?
And $(0,e)$ if you take $I$ to be a proper subset.
[0,e)
? The only way I can see is if $I$ is the empty set, which isn't a proper subset.
$\varnothing$ is a proper subset of $\Bbb N$
...am I forgetting definitions.
13:24
yes
Yes, you're right.
Well f({0}) = f({1}) but that's trivial
Hmm.
That probably muddles things a bit, but not much.
I think it may not be bijective
Maybe if you require positive integers?
In that case it'd be [0,e-1].
13:27
oh, it was a careless mistake in my calcuations, I think it is bijective
Well, it's not bijective as written (for the reason you just said). But if we take the domain to be N^+ then it again seems plausible.
I think I can prove it
I guess the question is whether or not $\sum_{k=1}^\infty c_k/k!$ can ever vanish when $|c_k|=1$.
what do you mean?
Let me write my proof though
I'm just writing $\sum_{n\in I_1}1/n!=\sum_{n\in I_2}1/n!$ in a different way.
13:30
@Balarka $\pi_1(\Sigma_2)=\langle a,b,\alpha,\beta\mid \alpha\beta\alpha^{-1}\beta^{-1}=aba^{-1}b^{-1}\rangle$, we did Seifert-Van Kampen's yesterday and I computed this, not sure whether it is correct (I also don't recognize it as a "nice" group)
But I'm wrong. Should have $c_k=-1,0,1$.
$\displaystyle \sum_{n=k+1}^\infty \frac 1 {n!}$
$\displaystyle < \sum_{n=0}^\infty \frac 1 {k! (k+1)^n}$
$\displaystyle = \frac 1 {k! \times k}$
$\displaystyle \le \frac 1 {k!}$
so we can compare head-to-head
I also got a similar expression (just longer) for $\Sigma_n$
because the else can never sum up to exceed the current
@Semiclassical do you understand me?
No, but that's just because I'm not awake.
I think I can see where you're going with this, and it sounds right.
But I'm not sufficiently awake to be a proof-checker right now :P
13:34
should I post in MSE to see other more elegant approaches?
Sure, it's a nice problem.
Plus it generalizes (the quesiton, not necessarily the result/solution). What if I take the sum to be $\sum_{n=1}^\infty g_n$ with $g_n>0$?
@DHMO There are some known results about representation numbers in the form $\sum\limits_{i\in I} a_i$. See here for a few references. You can probably find much more.
...aka my generalized problem, heh. Timing.
@MartinSleziak why must $\displaystyle \sum_{i \in I} f(i)$ diverge if $I$ is uncountable and $f(i)>0$?
I inch closer to 10k, but not quite there yet...
13:38
@DHMO Why do you always change the subject when I write a message to you?
However, this is ease enough.
@MartinSleziak I'm sorry for the sudden thoughts that pop up in my head
If you want the above sume to converge, then for each $n$ the set $\{i\in I; f(i)<\frac1n\}$ must be finite.
Union of countably many finite sets is countable.
I'm pretty sure there are a few posts about this on the main too.
@MartinSleziak why?
because then you'd have an infinite number of elements of at least finite size.
13:41
@MartinSleziak I would say that the sum isn't even defined
@DHMO My bad, I wrote the inequality in wrong direction.
because my current understanding is that $\displaystyle \sum_{k=0}^\infty a_k := \lim_{n \to\ infty} \sum_{k=0}^n a_k$
and I read it in the wrong direction, heh.
@MartinSleziak ic
@Semiclassical ... which is why I didn't understand your comment
13:42
So let us denote $I_n= \{i\in I; f(i)>\frac1n\}$. We have
$$|I_n| \cdot \frac 1n \le \sum_{i\in I_n} f(i) \le \sum_{i\in I} f(i) < +\infty$$
in that case @MartinSleziak thanks for your elegant proof
Detecting 1 error is easier than 2!
@MartinSleziak could you answer my question about the definition?
@DHMO You asked about uncountable sums. In this way you can define the sum only for $I=\mathbb N$. If you want to see a more general definition (i.e., sumation for arbitrary index sets) have a look at the links I posted above.
@DHMO I think I just did. (Even though I do not see any "question about definition".)
> $\displaystyle \sum_{s \in S} s = \sup \left \{ \sum_{s \in F} s\, :\, F \subseteq S \text{ finite} \right \}$
13:44
an uncountable sum is weird for me to think about, though I understand that it's a matter of definition.
(referring to the definition I just quoted) heh
@DHMO Yes, that is one of possible definitions. I guess it only works if you have non-negative summands.
> "When negative numbers are included it becomes impossible to define what you mean by the sum. "
of course, finding a uncountable sum which converges is presumably not too hard, since one can arrange for cancellation of positive/negative terms in the general case.
@MartinSleziak then can you give me a definition that works for all summands?
@Semiclassical you can't.
13:46
...huh.
@DHMO I already did.
There are several links above.
I'm browsing your links
struggling to find a definition
Now I'm confused. Is the point that such an uncountable sum is necessarily either ill-defined or divergent?
@Semiclassical or 0, if you let all entries be 0
or convergent, if all except a countable amount is 0
If you have a look at the link, there is a rather standard definition. You take sums over finite sets, and take a net defined on the directed set $[S]^{<\omega}=\{F\subseteq S; F\text{ is finite}\}$.
13:47
Hm, fair.
If this net converges to some limit, you call it a sum.
Okay, yeah.
It is given for example in Jonas Meyer's answer.
So a well-defined uncountable sum is either trivial (i.e. only a countable subset are nonzero) or divergent.
This answer gives a reformulation, which does not mention the notion of net.
13:49
Or: A well-defined uncountable sum can only be trivially convergent.
And Mariano's answer explains it using nets.
(trivial in the sense of actually being a countable sum in disguise)
@MartinSleziak "the sequence {1,-1/2,...} is convergent" is garbage
Aaanyways.
@DHMO Well, it is about the sum and not about sequence. And the answer linked says that it is not convergent.
13:51
The fact that this seems more interesting than the computations I shoudl be doing testifies to how little appetite I have for the them.
When did all interactions in this chat room became so frustrating?
@MartinSleziak you probably read the sentence before my quote
@MartinSleziak I'm sorry for the frustration caused, in part due to my stupid questions
Eh. This chat has always had its frustrating bits.
Have a nice day!
Perturbation theory, how I tire of thee.
13:55
@Semiclassical how many of the range of $f$ is transcendent?
Of what $f$?
35 mins ago, by DHMO
I wonder if $f:\mathscr P(\Bbb N) \to \Bbb R$, $ \displaystyle f: I \mapsto \sum_{n \in I} \frac 1 {n!}$ is bijective.
It's $\huge{\pi\text{ day!}}$
Well, any finite subset will give a rational result.
My intuition would be that an infinite subset would give a transcendental result, but I frankly have no idea.
I guess you can say for sure that the complement of any finite subset must give a transcendental.
13:57
@Semiclassical bingo
0
Q: Legitimacy of transformation using reciprocal of factorials

DHMONote: For the purpose of this question, $\Bbb N$ does not include $0$. I have a function $f:\mathscr P(\Bbb N) \to \Bbb R$ defined by: $$f(I) = \sum_{n \in I} \frac 1 {n!}$$ This is essentially a transformation from binary sequences of $\Bbb N$ to a number in $\Bbb R$. I would like to prove th...

Is the range of this function dense? @Semiclassical
You want a song on Pi day @heather here it is - youtube.com/watch?v=U2uVoDxZpaQ
@MartinSleziak welcome back
BTW this question is exactly about your question: The sum of an uncountable number of positive numbers
It starts: Claim:If $(x_\alpha)_{\alpha\in A}$ is a collection of real numbers $x_\alpha\in [0,\infty]$
such that $\sum_{\alpha\in A}x_\alpha<\infty$, then $x_\alpha=0$ for all but at most countably many $\alpha\in A$ (A need not be countable).
@DHMO Given the present context, you might appreciate the comments to an answer to an MO question I did a while back: mathoverflow.net/a/178561/55904
@MartinSleziak thanks
@Semiclassical are you thinking about my question?
14:04
Which? Oh.
No, not right now.
that video is hilarious @BAYMAX
I have a hard time thinking it's false, but shrug
@Semiclassical oh and the answer is no
Really? Hm.
it does have quite a lot of limit points
but all of its limit points are just themselves
14:06
Here's a hard generalization (though I'm not sure how hard). What if $f(x,I)=\sum_{n\in I}x^n/n!$ for $x\in(0,1], I\subset \mathbb{N}^+$?
so in a sense it is actually closed
@Semiclassical I should get back to work now
Same.
I suspect that this relates directly to the question I posed on MO since it'll necessarily be a lacunary series.
14:21
You like it ? @heather
14:36
@BAYMAX pretty funny, yeah
14:53
hello@AkivaWeinberger
ha ha :) @heather
Another one here -
@DHMO You know I suck at series, it will took me quite a deal of background reading before I can even have some idea on how to compute them (thus unlike all other problems you gave me before, where I at least have some idea what to do). Generating functions, however seemed to be a good start
Generating functions are fun.
Hi @Semiclassical I don't think $they$ have a solution

« first day (2414 days earlier)      last day (2903 days later) »