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00:00 - 17:0017:00 - 00:00

00:44
back for a while before I go to the park
@robjohn How far do you walk per day?
@skullpatrol 1-2 miles
@robjohn Everyday?
@skullpatrol most days. we walk for about 40 minutes in the morning and some in the afternoon.
01:00
@robjohn Do you do any swimming? They say that's good for the upper body.
@robjohn Are you there?
@anon ?
01:27
Hi, Peter...probably not who you're looking for... ;-)
Now I'm going to develop an inferiority complex, @Peter!
@amWhy Heh, well, I have a problem.
A word problem!
In how many ways can you get a dollar using all coins available?
That is $1,5,10,25$ and $50$ cents.
@PeterTamaroff A word problem...! ooh...combinatorics...you can probably find that already asked on main...
I am saying that such number is the coefficient of $x^{100}$ in $$(x+x^5+x^{10}+x^{25}+x^{50})^{100}$$
Oops...I just turned on Chatjax...I see it now...
@amWhy Ah?
OK.
01:32
@PeterTamaroff Sorry...I was looking at unrendered text and thought you had a polynomial of degree $100$...
@PeterTamaroff I am back before getting dinner
@robjohn See above =)
@robjohn hi @rob by
@skull I'm waiting... ;-)
@robjohn I haven't eaten anything yet today, and I'm central time! Surely you can put off dinner to answer Peter's pressing word problem!
Ugh, I have a fever, I think.
Or am I thinking too hard? =)
@PeterTamaroff no dollar coins?
01:36
@robjohn Nope. The point is to "break up" the dollar.
@PeterTamaroff that coefficient is $1$
@robjohn So... I am doing something wrong.
Oh, wait.
I should add a $1$, right?
$$(1+x+x^5+x^{10}+x^{25}+x^{50})^{100}$$
Like that.
The answer is 292
@robjohn OK.
Think that that counts 95 pennies and one nickel in 96 different ways
01:44
A nickel is 25?
@PeterTamaroff 5
95 pennies and one nickel should only be counted once, right?
@robjohn Yes.
Guurddamn.
But $(x+x^5+x^{10}+x^{25}+x^{50})$ didn't work either.
Think of what $$\frac1{(1-x)(1-x^5)(1-x^{10})(1-x^{25})(1-x^{50})}$$ represents
01:47
@robjohn I know, yes.
So I need the coefficient of $100$ in that, yes?
Yes, which is 292
@robjohn How did you find it?
@PeterTamaroff Mathematica
Otherwise, it is just making up all the combinations.
@robjohn Heh...
This is meant to be obtained with the sweat of the brainz!
@PeterTamaroff really? Then that might be harder.
01:52
@robjohn =)
This is problem $1.$ in Polya and Szegö.
@PeterTamaroff and they don't want to use the generating function?
I don't know how I would approach that otherwise.
@robjohn Yes, sure.
But then the coefficients ought to be found using partial fractions? =O
@PeterTamaroff ah, I see same method up to the generating function, it's the Mathematica part that we need to replace
Section one is "Additive Number Theory, Combinatorial Problems and Applications"
Of course you've already solved ${}^*2$.
@PeterTamaroff the answer to the bottom question is the function I wrote above
01:59
@robjohn Yes.
2 mins ago, by Peter Tamaroff
Of course you've already solved ${}^*2$.
@PeterTamaroff :-) I missed the stuff below the image
This book is surely awesome.
@PeterTamaroff how would you set up number 1 to be worked by partial fractions?
@robjohn Roots of unity?
Heh, we would get $100+$ terms.
@PeterTamaroff I don't see how to easily get the coefficient of $x^{100}$
02:06
@Peter What book is that? (Sorry to interrupt).
@amWhy "Problems and Theorems in Analysis" by Pólya and Szegö.
@PeterTamaroff Thanks!
So you've found how to write
$$
\frac{A}{1-x}+\frac{B}{1-x^5}+\frac{C}{1-x^{10}}+\frac{D}{1-x^{25}}+\frac{E}{1-x^{50}}=\frac1{(1-x)(1-x^5)(1-x^{10})(1-x^{25})(1-x^{50})}
$$
then what?
There are three volumes, I think.
@robjohn We need more terms, as you note.
and I'm not sure that is a good partial fractions expansion
02:09
@robjohn Yes, we need more cancelling.
You might need to replace some of the numerators by polynomials
@robjohn Aha.
I stick with Mathematica for now :-) off to get dinner
bbl
@robjohn Byes.
02:45
@skull I just finished following up on your last comment. ;-)
@amWhy Yes, and I just finished reading it :D
@skullpatrol Good...I hope it's making sense :D
@amWhy You seem to prefer the first method...
@amWhy ...that is the method used in the book.
@skullpatrol Actually, yes...but the second method is fine.
@amWhy Me too :D but I was surprised to find the second method and in some ways it is better, no?
02:52
Yes, better in that it is easy to forget to check the case when $a = b$, in the second method, it jumps out as you as a case needing consideration! And it's a bit more creative...just an impression... Are you going to reward me for all my hard work? ;-)
No, seriously, accept the answer that made the most sense to you.
@amWhy Thank you for all your hard work. I want to sleep on it first.
@amWhy ;-)
@skullpatrol Of course! ;-)
Heh... obviously I missed the point of this question
 
1 hour later…
04:21
@Gmath Read number 3. Spam means that.
quick question: what are the elementary subsets of $\mathbb{R}^p$?
I presume any sort of interval, does $(-\infty , \infty)$ count then?
or is this just an arbitrary term im finding in the book im reading?
I think it means a finite union of open (p-dimensional) rectangles.
ok
so not with infity i presume then
because of the finiteness
Unless p=1, intervals (which are subsets of R) are not subsets of R^p btw
@DanZimm finiteness refers to the number of rectangles being union'd. I do not know if infinitely long/wide/etc rectangles count as rectangles in this definition.
ah ok didnt know that
and ok
04:34
@anon Can you make a quick check for me?
This. I just edited - Am I on the right path?
@BandeiraGustavo checking the n=0 and n=0+1 cases does not constitute an induction proof
the binomial theorem must be proven for all n, and proving two specific cases doesn't nearly cut it
@anon Then by doing for $n$ and $n+1$.
the idea behind induction is: (i) prove the very first case, (ii) prove that the truth of the nth case implies the (n+1)th case (do not assume any specific value for n)
so, prove that $\displaystyle(x+y)^n=\sum_{j=0}^n\binom{n}{j}x^{n-j}y^j\implies(x+y)^{n+1}=\sum_{j=0}^{n+1}\binom{n+1}{j}x^{n+1-j}y^j$
@anon Yeah, I was thinking something like that. Thanks.
I'll think a little more.
04:54
Hello
05:06
@BandeiraGustavo pardon of me I forgot []
I know
I added both the start and render to my bookmarks
but clicking on them doesnt render the latex
06:07
Hi @MarianoSuárez-Alvarez
@MarianoSuárez-Alvarez Would you like to comment on some algebraic low hanging fruit here?
06:20
@MarianoSuárez-Alvarez Specifically could you tell me what I'm missing in the thread with amWhy?
@DominicMichaelis huhu
@skullpatrol huhu
@dfeuer Nope, sorry. I'm not really knowledgeable regarding the use and necessity of AC in topology beyond the standard examples.
06:37
Is it true that all ODE's that cannot be put into Normal Form doesnt have solution ?
at least as we founded our topology it was elementary for everything. We make everything over filters and not nets
what is "Normal Form" ?
suppose the ODE is nth order
then it's in normal if its equated like this
ah you mean jordan normal form?
y' = f(y^(n), y(n-1),...,y,x)
where y is a function of x
it's a restriction upon the general form
f(y^(n), y^n(n-1), ..., y',y,x) = 0
What is about $$0=y'(x) \cdot y(x)$$
this one shouldn't have a normal form does it ?
06:42
no, general form
but it can be put in normal form
can it ?
dividing by y(x) where x doesnt make y , zero ?
if you only allow continuous differential solutions the only solution should be constant functions. In special $y(x)=0$ for all $x$ is a solution, so dividing could smallen your solution set
could, but doesn't :)
I'm a bit confused but i wanna understand
06:46
yeah I just try to learn what arithmetic is allowed to put it in normal form ^^
are you saying that ODEs that cant be put into normal form might have a solution ?
Of course, we can just take our favourite non-injective (hence non-invertible) function $f$, and pose $f(y') = f(y)$.
try (y'+c)y=0 with c nonzero for dividing by y to diminish the solution set
f = x²
then (y')² = y²
does it have a solution ?
y = e^x ?
and it can't be put in normal form
?
correct, not without getting a non-equivalent (logically weaker) DE (i.e. one with less solutions than the original)
06:49
it does have 2 solutions $c\cdot \exp(-x)$ and $c\cdot \exp(x)$
hm nice, i dont get how the non-invertible f requirement applies
why do we need him ?
if $f$ is not invertible you have no chance to get to a normal form
@nerdy If $f$ were invertible, we could infer from $f(y') = f(y)$ that $y' = y$.
you can't go from f(y')=blah to y'=blah without inverting a noninvertible function f...
@Lord_Farin oh I make to much category theory, I wanted to say we should say $f$ is not monic :D
06:51
can anyone give me a non-injective function witoith being powers or |x|
Another example is $\sin(y') = \sin(y)$, which has as solutions $y = c \exp(x) + 2\pi n$, $n \in \Bbb Z$ (I suspect that this may not necessarily be the full solution set).
?
ops
thanks farin
gonna think
07:08
hello every one
Dominic, you said "if you only allow continuous differential solutions the only solution should be constant functions".You said continuous differential solutions to represent functions that presents themselves as solutions in all of their domains ? I dont understand, if i allowed non-continous differential solutions you say i would allow functions that presents themselves as solutions not in all of their domain ?
@nerdy you missunterstood me. I was talking about allowing a more general set of functions as solutions. For example look here weak solutions those allows you functions as a solution for an ode which themself are not differentiable
hmm wierd, how would we check of they are a solution without differentiating them ?
since we wouldnt be able to
do you know what distributions are ?
07:22
yeah, random variables /
compare:
http://en.wikipedia.org/wiki/Distribution_(mathematics)
versus
http://en.wikipedia.org/wiki/Probability_distribution
I am very pleased to see that even wikipedia doesn't count stochastik to mathematics :D
oh, i see now
i often use the diract delta function
but didnt know it had anything to do with what you call distribution
bad guy
the dirac delta distribution is not a function
07:26
In mathematics, the Dirac delta function, or function, is (informally) a generalized function on the real number line that is zero everywhere except at zero, with an integral of one over the entire real line. The delta function is sometimes thought of as an infinitely high, infinitely thin spike at the origin, with total area one under the spike, and physically represents an idealized point mass or point charge. It was introduced by theoretical physicist Paul Dirac. Dirac explicitly spoke of infinitely great values of his integrand. In the context of signal processing it is often ref...
complain to wiki
"In mathematics, the Dirac delta function, or δ function, is (informally) a generalized function "
only phyisician call the delta distribution a function
"From a purely mathematical viewpoint, the Dirac delta is not strictly a function, because any extended-real function that is equal to zero everywhere but a single point must have total integral zero.[7] "
@DominicMichaelis That's physicist for you.
physician = medical doctor
Physicians don't call the $\delta$ distribution anything.
loool
07:32
@skullpatrol Nice teaming skull :).
@Lord_Farin oh thanks :D But remember some guys which study physics don't completly fail in math (maybe I am a counterexample)
you could ask me; I do pediatric math
@anon :D
I'm out, bye.
can someone help me please ?
07:43
@anon oh lol :D
thanks Lord_Farin
and anon
i goo too
vrou asking that often doesn't help that much :)
so you don't know the answer?
Knowing the answer doesn't make you a good teacher.
07:53
just an idea
let me rephrase it: When you ask two times for help in 10 minutes the motiviation for me to think about your problem is about 0
and possibly negative...
is it true that if : $\displaystyle k^2\leq \lim\inf_{|x|\rightarrow \infty} \frac{f(t,x)}{x}\leq \lim\sup_{|x|\rightarrow \infty}\frac{f(t,x)}{x} \leq (k+1)^2$
then
Gee what is wrong with you ?
I try to say that if nobody responds they might be busy, and you don't stop posting your problem and asking for help
so you can't help me no more you have no idea
just tell this
07:58
@anon you are longer here than I am. Is he kidding with me?
no, vrouvrou is known to do this from time to time
Hi @JayeshBadwaik
@skullpatrol Hello.....
@JayeshBadwaik how are you?
08:10
@skullpatrol Doing good..
May I ask to ask a very simple question?
askaway
May I ask a very simple question?
just ask
I was joking, in reference to the first pinned message.
08:15
so was I :D
Shoot first, ask questions later.
08:36
wish me luck I will try to test my ode system :D
good luck pal
 
4 hours later…
12:16
@user1 Can I answer that?
12:42
I like this formula:
$\text{li}(n) = \int_0^{\sqrt{\log (n)}} \left(\int_0^{\sqrt{\log (n)}} e^{a b} \, da\right) \, db+\log (\log (n))+\gamma$
@robjohn May I ask another question before replying to yours?
@user1 sure :-p
@robjohn Will you allow me the possibility to reply to your answer if I allow you to answer that?
@user1 Are you sure that is what you want?
@robjohn Why would you think that I would not want that?
13:00
@user1 Do you not want the unexpected once in a while?
13:16
@robjohn What if I expect everything I want in the first place?
13:26
Hello :)
In mathematics, a binary quadratic form is a quadratic form in two variables. More concretely, it is a homogeneous polynomial of degree 2 in two variables : q(x,y)=ax^2+bxy+cy^2, \, where a, b, c are the coefficients. Properties of binary quadratic forms depend in an essential way on the nature of the coefficients, which may be real numbers, rational numbers, or in the most delicate case, integers. Arithmetical aspects of the theory of binary quadratic forms are related to the arithmetic of quadratic fields and have been much studied, notably, by Gauss in Section V of Disquisitiones Ar...
what is the "set of numbers represented by a binary quadratic form"? Is it just the image of the associated map $\mathbb{Z}^2\to\mathbb{Z}$?
@Randal'Thor Yes.
@user1 Thanks! Is the mentioned equivalence relation on such binary forms such that the images of equivalent forms are the same then?
nevermind, I found it :)
13:56
Hello, :)
14:51
HEy all, just as an FYI; Springer Yellow is having a summer sale springer.com/mathematics/yellow+sale?SGWID=0-40050-0-0-0
15:12
@MarianoSuárez-Alvarez You around? I have a question about valuation rings
can someone give me an example of a separable ODE in differential form p(x)dx + h(y)dy = 0 that can't be put into normal form dx/dy = g(x)z(y) ?
15:33
Morning.
morning
Man, there have been a lot of shitty questions on main lately.
15:49
too simple questions ?
i'm a bit of a beginner ;d
Not too simple. Poorly asked. Poorly written. Expectations that the community will do homework for people.
16:17
hi guys.
lets say y(t) = x(t)*h(t), now is this valid? y(2t) = x(2t)*h(2t)? is it correct to make the assumption that we simply replace t by 2t?
@vvavepacket Yes, it is.
sure
unless t is a discrete variable
in which case it can yield a problem
its not a discrete, its continous
why nerdy
for general at
y(at) = x(at)*h(at) doesnt make sense for all a
if it's discrete
signals and systems ?:
:D
yeah signals and systems :))
16:24
h = imnpulse response ?
You nauled it nerdy
been through it
*nailed
electronics engineering hehe
ok im currently digesting your explanation..
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