« first day (1443 days earlier)      last day (3579 days later) » 

9:00 PM
since log(log z) = log(1+log(1+(z/e-1)))
 
@BalarkaSen what's your point?
 
@blue there is a power series of $\log\log z$ in rational coeffs. that's precisely the point.
 
@BalarkaSen log(log z) is not in Q[[z]]
it's in Q[[z/e-1]]
 
@blue click the ping.
 
9:01 PM
I did click it
 
oh right. we are talking of $\log\log(1+z)$
 
even that's not in Q[[z]]
log(1+log(1+z)) is in Q[[z]]
 
i am not getting what you are saying. i think i need sleep.
 
loglog(1+z) is in the "coset" log(z)+Q[[z]]
 
log z is in Q[[z]]
 
9:04 PM
if it were, you could evaluate at 0
log(z) isn't even in Q((z))
 
heck. you are right.
log(1+z) is.
 
mmhmm
since loglog(z) = log(1+log(1+(z/e-1))), we know loglog(z) is in Q[[z/e-1]]
 
and that means log(1+log(1+z)) is in Q[[z]]
...which is just back to what you said
 
mmhmm
 
redo : can we determine if a function f(z) is in Q[[t]] or not, where t is C-linearly dependent over z?
 
9:08 PM
@BalarkaSen @blue I've compared the coefficients and I've got a few cases of the required summation. Is that sufficient to show that it's true?
 
@Shisui what do you mean " a few cases"?
 
I got many votes for 2 lhf today.
 
@Shisui just compare the coefficients of x^(n+m)
 
what blue said.
 
or wait no, of general l
forgetting what we were talking about
 
9:09 PM
@blue l, darn it.
the index is l.
 
@BalarkaSen I've got the coefficients for $l=0,1,2$
How would I find it for a general $l$?
 
@Shisui what is the coefficient of x^l on the LHS?
 
@Shisui multiply.
 
the powers look like x^(i+j), so you need to sum over all i+j=l
(exactly as it has in the desired answer!!!)
 
@blue the partition trick.
good ol' combinatorics.
 
9:12 PM
@BalarkaSen What's the partition trick?
 
@Shisui google partition function.
and euler's identity, i think it is called.
 
That's way beyond what I know! @BalarkaSen
 
@Shisui that's actually a very well-known trick. it has nothing to do about knowing.
 
I must've looked at the wrong link. I'm looking at this:
The partition function or configuration integral, as used in probability theory, information theory and dynamical systems, is a generalization of the definition of a partition function in statistical mechanics. It is a special case of a normalizing constant in probability theory, for the Boltzmann distribution. The partition function occurs in many problems of probability theory because, in situations where there is a natural symmetry, its associated probability measure, the Gibbs measure, has the Markov property. This means that the partition function occurs not only in physical systems w...
 
pfffft
hahahha
look at partition function number theory.
quantum fields just sprout out of nowhere.
poor partitions.
 
9:17 PM
Hmm, I'm still struggling with comparing the coefficients.
@BalarkaSen
 
@Shisui You're just adding stuffs up with the same degree.
(x + y)^2 has coefficient 2 for xy term since (x + y)(x+ y) = x^2 + xy + xy + y^2
So you're kind of collecting the coefficients from the same terms.
Do that for this one too.
Unfortunately, I have to leave.
Byes.
 
Thanks for your help so far!
I'll give it my best shot.
^_^
 
Where is the great anon?
 
@blue So far, I have: $$ \begin{aligned} & \binom{n}{0} \binom{m}{0} + \left[ \binom{n}{1} \binom{m}{0} + \binom{n}{0} \binom{m}{1} \right] x + \left[ \binom{n}{0} \binom{m}{2} + \binom{n}{1} \binom{m}{1} + \binom{n}{2} \binom{m}{0} \right] x^2 + \dots + \binom{n}{n} \binom{m}{n} x^{m+n} = \binom{m+n}{0} + \binom{m+n}{1} x + \binom{m+n}{2} + \dots + \binom{m+n}{m+n} x^{m+n} \end{aligned} $$
I'm guessing I'm not on the right lines ...
 
@JasperLoy He/she/it visits the chat mostly under the name of "blue" these days.
 
9:29 PM
@DanielFischer OMG!
 
Olive Melon Green, @Jasper?
 
@DanielFischer He renamed in honour of me.
 
@DanielFischer @JasperLoy Any ideas?
 
@Shisui I only know algebra-precalculus.
 
@JasperLoy Haven't you encountered the Binomial expansion yet?
 
9:34 PM
@Shisui Ah, yes, but I don't answer questions in chat, hehe.
 
@JasperLoy Cheeky!
 
@Shisui What do you want to show?
 
@DanielFischer $$ \sum_{k=0}^{l} \binom{n}{k} \binom{m}{l-k} = \binom{n+m}{l}$$ By using $$ \left( \sum_{i=0}^{n} \binom{n}{i} x^{i} \right) \left( \sum_{j=0}^{m} \binom{m}{j} x^{j} \right) = \sum_{l}^{n+m} \binom{n+m}{l} x^l $$
 
@Shisui $$(1 + x)^n\cdot (1+x)^m = (1+x)^{n+m}.$$
 
@DanielFischer I've got that so far and I've expanded it in the long post above.
 
9:38 PM
$$\sum_{i=0}^m\sum_{j=0}^n\binom{m}{i}x^i\binom{n}{j}x^j=\sum_{j=0}^{m+n}\sum_{i‌​=0}^j\binom{m}{j-i}x^{j-i}\binom{n}{i}x^i$$
 
I'm unsure of what to do. I was told to find the coefficient for a general $l$.
 
@Shisui Then you're done.
 
@DanielFischer So it's just enough to say that the coefficients follow the patter of the summation that I'm trying to show is true?
@robjohn I've not looked at double summations in a very long while, sorry!
 
@Shisui well, you can't really use this approach without using a double sum
 
@robjohn Would it be enough to say that the coefficients in the long line I've expanded follow the pattern I'm trying to prove?
(The question is in a section that comes way before double sums are introduced)
 
9:43 PM
@Shisui would it be enough? If you don't understand it, then it isn't enough.
 
@Shisui Well, "just saying it" is not enough. You need to show it. You have $$(1+x)^n(1+x)^m = \left(\sum_{k=0}^n \binom{n}{k}x^k\right)\left(\sum_{l=0}^m\binom{m}{l}x^l\right) = \sum_{k=0}^n\sum_{l=0}^m \binom{n}{k}\binom{m}{l}x^{k+l}.$$ Group according to constant $k+l$.
 
@Shisui that hint came before double sums are introduced?
 
@robjohn Yep. It's Spivak (3rd ed.) chapter 2.
 
@DanielFischer that is what I tried here
 
@Shisui $$\sum_{i=0}^m\sum_{j=0}^n\binom{m}{i}\binom{n}{j}x^{i+j}=\sum_{l\ge0}\left[\sum‌​_{i+j=l}\binom{m}{i}\binom{n}{j}\right]x^l $$
 
9:54 PM
Prof. @Ted - 90 in Multi. calc test, 95 final grade :) Thanks for your help! First in class, plus I feel I actually get the concepts I've studied.
 
@Studentmath Congrats!! ^_^
 
What is the part of a number called that comes before the comma ?
 
@salbeira Do you use a comma to separate the integer part from the fractional part or are you referring to a number like 1,999 where the comma is used?
 
Aaah ... integer part ... now I just need to know how that is called in german ...
We call the decimal places "Nachkommastelle" - "after comma place"
 
Hi
 
10:02 PM
It might have something to do with Einzelteil and Ganzzahl.
I don't know any German however!
@pourjour Are you French?
 
I think I can say Ganzzahl ... What I want to say is "If a number is $\in \mathbb{R}$ but is actually an integer (so it has no decimal places)"
And I for god's sake can not find the right wording in my own language because I know the right word in english
 
@salbeira You might be able to use the fractional part notation to just write it down in mathematical symbols if the German word eludes you!
All real numbers can be written in the form n + r where n is an integer (the integer part) and the remaining fractional part r is a nonnegative real number less than one. For a positive number written in decimal notation, the fractional part corresponds to the digits appearing after the decimal point. The fractional part of a real number x is x-\lfloor x\rfloor, where \lfloor\;\rfloor is the floor function. It is sometimes denoted \{x\}, \langle x \rangle or x\,\bmod\,1. If x is rational, then the fractional part of x can be expressed in the form p / q, where p and q are integers and 0 ...
 
@Shisui No, but I speak french
 
@pourjour Your username gave some French influence away! Je parle aussi un peut de Français!
 
Supposing $l_1 = (1,0, − 1), l_2 = (−2,1,1) et L = Vect(l_1 ,l_2 ).$
@Shisui Oui ce nom est composé de deux terme: pour 'la préposition' et jour 'c'est un nom'
How can I find an system of equation for the subspace L?
 
10:27 PM
I think that I've found a good solution here:
1
Q: System of equations for a subspace

wmnorthI realize this is a novice question, but I'm stuck here. Suppose you have an ordered basis $B = (v_1, v_2, v_3, v_4)$ of $R^4$. My problem is how to determine a system of equations that defines, relatively to the basis $B$, the linear subspace of $R^4$ generated by the vectors $v_1-v_2+v_4$ and $...

 
10:59 PM
@Martin Sleziak I saw the density tag, took care of it by retagging the posts. It should be auto-deleted soon, if not already.
 
In number theory, natural density (or asymptotic density or arithmetic density) is one of the possibilities to measure how large a subset of the set of natural numbers is. Intuitively, it is thought that there are more positive integers than perfect squares, since every perfect square is already positive, and many other positive integers exist besides. However, the set of positive integers is not in fact larger than the set of perfect squares: both sets are infinite and countable and can therefore be put in one-to-one correspondence. Nevertheless if one goes through the natural numbers, t...
1
Q: Density in the set of real numbers

birzhLet $S$ be the set of all rational numbers which are squares $x = p^2/q^2$ for some integers $p$ and $q$. How can I show that $S$ is dense in the set of non-negative real numbers?

 
I can't understand why $F^2 = U_1 \oplus W$
 
11:28 PM
@skullpatrol Yes, the word density means different things. Some of the questions tagged were about probability density functions. And those about topological density did not have the tag. Mo' tags, mo' tagging problems.
 
I see, keep up the great work @Thisismuchhealthier. :-)
 
11:47 PM
@Studentmath: Congratulations. I'm very proud of you.
 

« first day (1443 days earlier)      last day (3579 days later) »