Mathematics - 2014-04-05 (page 1 of 3)

Mike

00:33

hi @KarlKronenfeld

Karl Kronenfeld

@Mike ey yo

user127001

00:49

Hi

Mike

hey man

Pedro Tamaroff

hi

@KarlKronenfeld What's up?

How's Lee treating your brainz?

Karl Kronenfeld

@PedroTamaroff Not reading Lee right now, so his brainwashing was unsuccessful.

Pedro Tamaroff

@KarlKronenfeld Oh, look at that. =O

Karl Kronenfeld

I'll return soon enough. I'm just refreshing commutative algebra.

Pedro Tamaroff

00:53

@KarlKronenfeld Why is that?

Friend gave me that. Started the reading today.

@Mike @KarlKronenfeld

Karl Kronenfeld

@PedroTamaroff I missed a lot on the first go.

@PedroTamaroff fun stuff

Pedro Tamaroff

@KarlKronenfeld Oh. What book/resource are you using?

Mike

@PedroTamaroff I need to work more Lee... I've been distracted lately.

My last day and a half were on that problem I've been talking to Daniel about.

I could specify to compact spaces instead of locally compact but that would be terrible and unsatisfying :D

Karl Kronenfeld

@PedroTamaroff I am laying my (typed) notes in accordance with Matsumura's toc because I like how he organizes it. Matsumura also has some cool theorems that he proves early on. I will use various others (like A-M) to supplement. I'm also checking out an old textbook by Irving Kaplansky.

user127001

01:08

Which Lee book?

Mike

smooth manifolds

Pedro Tamaroff

01:25

@KarlKronenfeld ERMAGHERD Matsumura is typed.

Karl Kronenfeld

huh?

Pedro Tamaroff

The copy I have is typed. As in a typewriter.

Karl Kronenfeld

oh lol

Pedro Tamaroff

I have a book on K-theory that is typed too.

Mariano told me it is a great book.

Mike

lol

Jasper Loy

02:38

This chat is dead.

Anybody home?

user127001

02:58

Hi Jasper

Jasper Loy

@user127001 Hi, have we spoken before?

Sush

@JasperLoy, I too get confused with those User... s

@ccorn, i think the thing i learnt from you yesterday was generalization of Stars & Bars, right?

@JasperLoy, will you please help me finding coefficient of $x^2$, in $(1+x+x^2)^{10}$, without actually expanding it?

I think the fact $\dfrac{1-x^3}{1-x}=1+x+x^2$ may help.

But can't use it!

@user127001, can u please help me/

user127001

03:25

Jasper yes, very briefly

Sush I'm sorry I am very bad at math

Sush

@user127001, so i m!

@user127001, i think you should take a name, so that we can remember you easily. Though it is just my opinion.

user127001

What name should I take

ccorn

Hi @Sush

Sush

03:40

@ccorn, HI how are you sir!

I had the sum in dream which you taught me y.day!@ccorn

ccorn

You want to count partitions again?

Sush

@user127001HAHA! That you have to decide. You know remebering a number 127001 is very difficult, you know, when there are hundreds so!

@ccorn, that will be amazing!

@ccorn, i think the stars & bars is a particular example of what you taught yesterday, right? I checked several examples using Wolfram|Alpha, and that worked.

ccorn

Ok, for the coeff of $x^2$ in $(1+x+x^2)^{10}$, let me tell you about the multiply-and-square scheme. It's also useful for other things (when you have high exponents but need only the lowest parts of the result).

The idea is that we are expanding, but only the parts we need, and only in small steps, so that it does not get complicated.

Sush

Ok, so will you use $\mathcal O$?

Mike

I would suggest using the binomial theorem to expand $(1+(x+x^2))^{10}$, and disregarding the parts that don't matter :)

Sush

03:49

@Mike, ok :)

ccorn

@Mike: Actually that amounts to enumerating the partitions, which is what we do not want here :-)

Mike

It doesn't really.

Sush

Can $\dfrac{1-x^3}{1-x}=1+x+x^2$ fact help?

Mike

Unless you call finding the answer enumerating th partitions no matter what :P

ccorn

@Sush: It can, for $x^2$ it's probably even quickest. But it gets harder for higher exponents.

Sush

03:54

ok

ccorn

Here, it would mean: $(1-x^3)^{10} = 1 + O(x^3)$ (we are not interested in anything higher than $x^2$)

And $(1-x)^{-10} = 1 - \frac{-10}{1}x + \frac{(-10)(-11)}{(2)(1)}x^2 + O(x^3)$

Multiply both together and look for the coefficient of $x^2$, it's $1\cdot\frac{(-10)(-11)}{2} = 55$.

The expansions are examples of the binomial series

@Sush, still there?

Sush

@ccorn, YES, milkman came, so went, but now here.

Thank you SO MUCH!:-)

ccorn

I wanted to show you another computation strategy, but let's drop that. Another for the exercise: Distribute $6$ oranges among $10$ children, no child more than $2$. This means finding the coeff of $x^6$ in the product $(1-x^3)^{10}(1-x)^{-10}$. Can you do that?

(Bad thing to have only $6$ oranges here)

Sush

Trying!

ccorn

04:13

@Sush lab bhattacharjee has made that an answer to your question

Sush

@ccorn, yes, and that's the your way!

@ccorn, i think you were trying to teach me this

ccorn

Well, it's the way you have suggested

Sush

coefficient of $x^6$ is 2850, is that true answer?

ccorn

@Sush No that was more Mick's way, I suppose

@Sush yes! Correct.

@Sush Wow, Sasha has given answer too. He has done the most effort to not expand.

Sush

@ccorn, so whether oranges are less than chidren or not, doesn't make any difference in method, right?

ccorn

04:27

@Sush right.

Sush

SO, which answer should i accept? @ccorn, all are amazing!

ccorn

@Sush: difficult question :-)

Sush

I am accepting user125053's because he has least reputation but has worked too much to LaTeX the answer.

@ccorn

ccorn

@Sush Pls write a comment thanking all contributors cause there are lots of nice answers.

Sush

Yes. Shall i write it after every answer? or just beneath my question?

@ccorn

I think after every answer.

ccorn

04:33

@Sush does not matter, but include all usernames as in @Marc etc, so that they get to see the message

Sush

OK

@ccorn

Done.

I think comments beneath answers do not accept @Marc etc.

ccorn

@Sush: @name is automatically removed if that user is notified anyway (because it is below his own answer, for example)

It does not hurt if you write it though

Sush

@ccorn, ok. but, i had to write it manually!

ccorn

Very nice, all answers have your comments now

Sush

:)

ccorn

04:41

Thanks @Sush, you have attracted a lot of bigheads here it seems :-)

Sush

@ccorn, what is bigheads?

ccorn

@Sush I mean users known for giving good answers

Sush

ok!

ccorn

(I do not know whether bighead actually means such a thing in proper english. Probably not.)

Sush

it works for me! not native english speaker.

good bye, Had a nice day with your company . Am sure will have a nice day too! see you!

ccorn

04:45

good bye @Sush

Mike

04:56

so @ccorn

what's your story

ccorn

Nothing to tell @Mike

Mike

I don't believe that, but if you don't wanna, neither shall I ask further :)

ccorn

I mean: Problem done.

Amazing how attractive clear little problems can prove to be.

I originally wanted to demonstrate exponentiation mod $x^3$, but that tool would have been oversized.

Mike

05:30

I would like to retag this, because it's neither dimension theory nor low dimensional topology, and probably not homework, but I have no idea what the hell it actually is.

ccorn

@Mike Indeed strange.

Studentmath

05:51

Create a '42' tag. Or -1/12

Mike

that question looks more like a 420 tag

ccorn

@Studentmath Oh no, please, not that (2nd) number again. That's like entering Physics SE's h-bar and spelling nookular.

Studentmath

:D

I wonder what their reaction would be actually..

robjohn

06:38

@Chris'ssis Sorry, I was out with my son, getting his passport renewed.

@Chris'ssis We went over this very question in chat the other day, didn't we?

Mike

@robjohn I have to ask you what you think this question should actually be tagged.

robjohn

@Mike let me take a look...

Johan Larsson

does it make any sense to fit a fourier sine series to data and use it for prediction?

I'm prolly messing up the math lingo badly :)

robjohn

@Mike It is hard to tell... I guess the current tags look okay. Did you have some other suggestions?

@JohanLarsson is the function you are fitting odd? Sine series add up to an odd function. The cosine series picks up the even part.

Mike

Well, that's not really either low-dimensional topology (though I guess you could make the argument), or dimension theory. It doesn't look like homework either. I don't know what it should be is the problem.

ccorn

06:48

@JohanLarsson You'd use integer multiples of some base frequency, I suppose? Then your predicted series would necessarily be periodic, with the period you have given by means of the base frequency.

Mike

It frankly looks like a question from somebody who's misunderstanding some theory of physics (string theory or somesuch), but I don't want to tag it that unless it's the OP doing the tagging.

Johan Larsson

@robjohn I use sines + the phase part from the fft, pretty noob with this but think it results in sine + cos

@ccorn yea that is the idea, I've been playing with polynoms, moving averages and fft

don't remember much maths just playing around really

ccorn

@JohanLarsson problem is that you have to input the entire period, and what you get out is the thing itself, repeated. Therefore such things are used for interpolation, not prediction AFAIK.

Johan Larsson

the results from using the fourier stuff is not much better/worse than anything else, my question is more a check on the level of dumb for trying it :)

prolly gonna try to drop a neural network on it just for fun

@ccorn that makes sense

ccorn

@JohanLarsson Do you know how speech compression works?

At the core is a prediction method too.

Johan Larsson

06:55

I have a vague idea

ccorn

The idea is to find coefficients $c_1,\ldots,c_N$ with which to approximately predict the next sample value as a linear combination of the previous $N$ values.

One may take test data and use least-squares fitting to get the coefficients $c_1,\ldots,c_N$. Turns out that these are mostly computed like correlation coefficients.

Then one can try that set of coefficients on another data set.

Of course, this cannot make a stream of sample data more autocorrelated than it actually is, so typically there is a limited duration for which the predictions may be useful.

Johan Larsson

I have an ~easy~ prediction task, it is for controlling a physical property that is measured for each unit

I only need to predict the next item every time so I can adjust the process.

gonna google about what you wrote, it is fun to obsess with this :)

ccorn

Starting point. What I mentioned goes in the direction of AR, and you may have the tools to turn it to ARMA or ARIMA.

robjohn

07:12

@JohanLarsson okay, so you are talking about a full Fourier series. Fourier Series are periodic, so if you know the period, prediction (extrapolation) is actually interpolation. If your data is not periodic, then it doesn't make much sense to look at the sum of the series outside the domain of the data (prediction).

Johan Larsson

Is there a formal way to check if and how periodic a series is?

I don't know what the deviations are due to but it looks like sines in the data

ccorn

@JohanLarsson spectral analysis, for example maximum-entropy spectral analysis.

Johan Larsson

learning new words here :)

ccorn

Basically, you take the squared magnitude $|F(\omega)|^2$ of the fft of a (reasonably windowed) time series. That spectrum is then fitted to a rational polynomial (it shall decay to zero as $\omega\to\infty$, therefore use a reasonably high degree in the denominator poly.) From the result you can (more or less) read off predominant frequencies.

Johan Larsson

what about selecting window size, if I manage to select a window that is n*period I would imagine getting a perfect spike

the data is small, typically 1000 points so I can easily brute force it but still interesting to know if there is a correct way of doing it

ccorn

07:23

Yes, but if it's slightly off, you get a zoo of neighboring spikes. That is what the windowing functions shall minimize.

Johan Larsson

is the thing you described the windowing function?

ccorn

Software such as scilab, matlab, octave, FreeMat, R will offer windowing options.

Johan Larsson

I'm programming my own stuff in C# since I suck so hard at using all of the above :)

ccorn

The windowing function is a weight function applied to your timeseries window before the fft.

It must be zero at the begin and end of the window, and better vary smoothly.

Johan Larsson

do you have a preference between {scilab, octave, R} btw?

ccorn

07:27

I have a subset of those compiled. Every once in a while, some update fails to build, and I use something else from the set until I have managed to fix the darn codebase.

Johan Larsson

@ccorn I'm going to lag behind a couple of weeks on what you write, have some reading to do here. Huge ty for all the words to google for.

ccorn

Good luck @JohanLarsson

Johan Larsson

09:47

I think I got bored

Want to learn some scilab, find the transformation between two sets of points sounds fun and useful

Jasper Loy

@JohanLarsson I just woke up and I missed the postman who came with my books!

Johan Larsson

what does it mean? You have to go to the post office?

Jasper Loy

Yes, on Monday afternoon.

Johan Larsson

and how is your SVD?

Jasper Loy

Singular Value Decomposition?

Johan Larsson

09:49

yea

Jasper Loy

I don't know about it, lol.

Johan Larsson

#meh

Jasper Loy

Why do you need it?

Johan Larsson

I have a feeling SVD is a hammer turning the world to a place full of nails

do you know scilab well?

Jasper Loy

Not at all, lol.

Johan Larsson

09:52

that is not lol man :)

ccorn

10:20

@JohanLarsson: IIRC scilab has a mese function for maximum entropy spectral estimation. You might want to try whether you can get it to work for you.

Johan Larsson

I think I just managed to find the transformation between two sets of points without knowing much at all :D

@ccorn will check it out, ty sir

Jasper Loy

@JohanLarsson It might be a madam.

Johan Larsson

yea but I'm on a roll here :)

ccorn

@JasperLoy Might be, but isn't.

Johan Larsson

gist.github.com/JohanLarsson/9990092

was surprised to see the test pass

didn't even think about matrix dimensions just guessed :)

broke it now

ccorn

10:40

@JohanLarsson Floating-point issues?

NB: Does ToArray() return an IList<Point3D> or is there inheritance or an implicit conversion? (I don't know C#)

Johan Larsson

@ccorn nah bigger than that, think my dumb is in m[i, 3] = 1;

@ccorn yes, it enumerates the enumerable

hmm it looks pretty right

mirgee

10:59

Anyone here who knows about polarities and forms?

nerdy

suppose axo + byo = 1 and (xo,yo) are the linear combination which gives the least positive value ( namely 1 ). Then why is it true that if i multiply this equation by c, we get that acxo + bcyo = c and (xo,yo) still is the linear combination of ac and bc, which givs the least positive value ?

im trying to see why its true that if i know some pair ( say 8,7 ) is relatively prime, i can know that any k-multiple of the pair (8,7) has gcd = k
For example, gcd ( 6 . (8,7) ) = = gcd ( 48,42 ) = 6 and so forth

ccorn

@nerdy Simply multiply?

nerdy

what ?

What do you mean

ccorn

@nerdy Ah, you want to show that $\gcd(a,b)=1$ implies $\gcd(ca,cb)=c$?

Karl Kronenfeld

@nerdy Pick any linear combination of $ac, bc$: $acx+bcy=d$. Note that $c$ divides the left side of $acx+bcy=d$, therefore $c$ divides right side as well. That is, $c|d$.

ccorn

11:16

@nerdy? Silence

nerdy

a bit confused, i know that if c | a then c | acx + bcy = d

Pedro Tamaroff

@KarlKronenfeld Morning.

ccorn

@nerdy The GCD has two properties, do you recall them?

nerdy

why does gcd(ma,mb) = mgcd(a,b)

Karl Kronenfeld

@PedroTamaroff yo

ccorn

11:22

@nerdy Let $g=\gcd(a,b)$. First, you know, $g$ divides $a$ and $g$ divides $b$. Second, if any $h$ divides $a$ and $h$ divides $b$, then ...? (Please complete.)

nerdy

h divides ax + by

ccorn

True, but that does not relate $h$ to $g$. What's that relation?

($h$ is a common divisor of $a$ and $b$ by now. $g$ is the greatest common divisor. So?)

nerdy

h | g

h | gcd(a,b)

Pedro Tamaroff

@KarlKronenfeld This question might be a bit silly.

But.

ccorn

@nerdy Yes!

Pedro Tamaroff

11:30

Suppose $M$ is a module, and $S$ is a subset. Then we say $S$ is linearly independent if whenever we have a (finite) linear combination $\sum_{x\in S}a_x x=0$, each $a_x=0$. Here we do not allow something like $rx+r'x+r''x+r'''x=0$, or do we?

That is, repetition of the same element.

ccorn

@nerdy Now suppose $h$ divides $a$ and $h$ divides $b$, then $mh$ divides...?

Karl Kronenfeld

@PedroTamaroff No, the sum must be simplified.

Pedro Tamaroff

@KarlKronenfeld OK.

ccorn

@nerdy then $mh$ divides ... and ...?

Pedro Tamaroff

@KarlKronenfeld OK, that was a bit stoopid. We can always take $x+(-1)x=0$ =P

nerdy

11:35

well, a multiple of a common divisor between two numbers

ccorn

@PedroTamaroff Iteration over a set means: every member only once. So: not allowed.

nerdy

i don't know what a multiple of a common divisor between two numbers divide

:(

Pedro Tamaroff

@ccorn Ah, but if we consider indexed families this can happen.

ccorn

@nerdy Hmm... You know that if $h$ divides $g$, then $mh$ divides $mg$?

Pedro Tamaroff

Lang remarks it: the family $\{x\}$ is independent, but the family $\{x_i\}_{i\in I}$, $x_i=x$ is dependent for $|I|>1$.

nerdy

11:37

true

ccorn

@PedroTamaroff Yeah, but these then are not a set. A multiset at best. You said $S$ is a (sub)set.

Pedro Tamaroff

@ccorn Yes, I know.

ccorn

@PedroTamaroff Therefore, the wording disallows such multiplicities. For a family/sequence/multiset/..., you'd indeed have a problem with that def.

Pedro Tamaroff

@ccorn Agreed.

nerdy

man im progressing so slow in number theory because im trying to understand everything 100% and full intuition, i dont want just to be able to write a proof but understand it totally. so far i got really grasp of all profs for all basic properties of divisibility, euclidean division, euclidean algorithm, bezout identity, every common divisor divides gcd,

ccorn

11:40

@nerdy So: suppose h divides a and h divides b, then mh divides ma and mb.

nerdy

but the are some that im struggling

there are

agreed ccorn

ccorn

@nerdy Suppose also that $h$ divides $\gcd(a,b)$. Then $mh$ divides $m \gcd(a,b)$.

Karl Kronenfeld

@PedroTamaroff If we want to be pedantic, then the fact that the sum could be infinite would also be problematic.

nerdy

true

Pedro Tamaroff

@KarlKronenfeld Yeah, $a_x=0$ but finitely many $x$. I don't see the point of distinguishing families and sets, though. If we repeat one element in a family it is dependent, so we might as well see it as a set?

ccorn

11:44

@nerdy That is the second property. The first property: Show that $m \gcd(a,b)$ divides $ma$ and $mb$. Then you are done.

Karl Kronenfeld

@PedroTamaroff The best way to view it is to define a mapping from $A^S$ to $M$, by summing the coordinates as indicated.

Pedro Tamaroff

@KarlKronenfeld $A^{(S)}$?

=D

Right, right.

Karl Kronenfeld

Free module, since that is ambiguous, I just remembered.

Pedro Tamaroff

@KarlKronenfeld Ah, you mean the free module.

I thought you meant the $f:S\to A$.

Well....

Kinda the same.

Whatevs.

Karl Kronenfeld

No, the module of functions would be problematic.

nerdy

11:48

any m-multiple of a common divisor of a,b divides any m-multiple of gcd(a,b) . and any m-multiple of gcd(a,b) divides any m-multiple of a and any m-multiple of b

N3buchadnezzar

Hey all

Pedro Tamaroff

@KarlKronenfeld Well, functions with finite support.

That's how we build the free module after all.

nerdy

i dont grasp it now, but hopely after some hours i will

thanks man

i totally grasp the sequence i wrote now

but odnt know how to connect it with gcd(ma,mb) = mgcd(a,b)

skullpatrol

Heya @N3buchadnezzar

@N3buchadnezzar what's happening in the world of physics?

Johan Larsson

Noobing around with scilab

N3buchadnezzar

11:54

@skullpatrol -gonna judge ludøl in a couple of hours

People drinking about 24 beers in about 1.5 hours

skullpatrol

that is a physical challenge :D

N3buchadnezzar

If they win they have to drink the same amount a few hours later

Typhical norwegians

ccorn

@nerdy (It seems I need to clean up.) Let $g=\gcd(a,b)$. Since $g$ divides both $a$ and $b$, $mg$ divides both $ma$ and $mb$. Second, we know there exist Bezout coeffs $x$ and $y$ (integers) such that $xa+yb=g$. if some integer $k$ divides $ma$ and $mb$, then $k$ divides $m a x + m b y = m (ax+by) = mg$. All in all, this means that $mg = \gcd(ma,mb)$.

nerdy

are you using the fact that the only common divisor of a pair who can be divided by any other common divisor of the same pair, is the gcd of the pair

in the last part of the proof

ccorn

@nerdy Yes, that is the second property of the GCD.

No, actually. I show that. But I use that if $k$ divides $A$ and $B$, then $k$ divides $xA+yB$ for every integer pair $(x,y)$.

(Here $A=ma$ and $B=mb$)

(And for $(x,y)$ I use the Bezout coeffs that yield $xa+yb=g$)

Thus $k$ divides $mg$. This is the second property of the GCD that needed to be shown.

nerdy

12:08

hm o so if k divides ma and mb, it divides m.gcd(a,b) and if it divides m.gcd(a,b) that should mean that what it divides is the gcd(ma,mb).

ccorn

@nerdy if k divides ma and mb, it divides m.gcd(a,b) --- correct. That was the second part. The first part was: $m \gcd(a,b)$ divides $ma$ and $mb$ because $\gcd(a,b)$ divides $a$ and $b$.

Both together mean: $m \gcd(a,b)$ is the GCD of $ma$ and $mb$.

nerdy

I fully understand the first property : mgcd(a,b) divides ma and mb. i also understand ( the first second property i think ) that if h divides gcd(a,b) then mh divides mgcd(a,b). but i got confused with the k

ccorn

Argh. Let's start over.

nerdy

whats funny is that i understand by this view-point , but is it enough ? i still get so confused with number theory

First, if gcd(a,b) = k, that is , if the least positive linear combination if a,b is k and is obtained with some x,y then
ax + by = k.
So, that also means that the least positive linear combination of ma,mb is obtained with x,y and this least positive value is km :
max + mby = km

so gcd(ma,mb) = m.gcd(a,b).

of a,b*

ccorn

Let $g = \gcd(a,b) = xa+yb$. Set $A=ma$ and $B=mb$. (1) Show that $mg$ divides $A$ and $B$. (2) Show that if $k$ divides $A$ and $B$, then $k$ divides $mg$. (3) We conclude that $mg = \gcd(A,B)$.

Can you do (1) and (2) now?

nerdy

12:20

okay

hold on

mg = mxa + myb. if g divides a and b, then mg divides ma=A and mb=B

if K divides ma and mb, then it divides any lienar combination of it, including mg.

ccorn

@nerdy Since (not if) g divides a and b, then mg divides ma=A and mb=B --- yes, that's (1). That with K is (2), yes. You are done.

nerdy

we can only conclude that mg = gcd(ma,mb) because mg is the least positive linear combination of ma and mb , namely, max + mby ?

ccorn

@nerdy we can only conclude that mg = gcd(ma,mb) because we have shown mg = gcd(A,B). And A=ma, B=mb.

nerdy

i just got confused with the K

did the proof needed it ?

when you write mg = mxa + myb are you necessarily saying that gcd(ma,mb) = mg ? or we needed to assume more

ccorn

@nerdy You needed to begin with: If K divides A and B, ... (and we do not know whether K itself is divisible by m. Therefore yes, K is needed.)

@nerdy $x$ and $y$ are suitable Bezout coeffs. Holds only for those.

nerdy

12:31

when we go from ax + by = g to max + mby = mg , we are not necessarily saying that gcd(a,b) = g implies m.gcd(a,b) = mg ?

because the coeficientswhich are minimal for ax + by = g might not be minimal for (ma)x + (mb)y = mg ?

ccorn

@nerdy That's just the distributive law in reverse.

nerdy

ops i wrote it wrong, gcd(a,b) implies gcd(ma,mb) = mg

ccorn

max+mby=m(ax+by)

generally

nerdy

when we go from ax + by = g to max + mby = mg , we are not necessarily saying that gcd(a,b) = g implies gcd(ma,mb) = mg ?

because the coficients which are minimal for a and b, namely those x,y , might not be minimal for ma and mb

?

ccorn

@nerdy from ax + by = g to max + mby = mg --- this is just multiplication.

nerdy

12:34

because if it remains minimal, then its the definition of bezout theorem, that if x,y is minimal for ma and mb, then ma.x + mb.y is gcd(,ma,mb)

im just indoubt because if we knew that x,y would give rise to the least positive linear combination of ma,mb then we would be able to write x(ma) + y(mb) = gcd(ma,mb)

= m.g by first equation

ccorn

@nerdy not yet because you don't "know" yet that x and y are also Bezout coeffs for ma and mb.

You said above: if K divides ma and mb, then it divides any lienar combination of it, including mg. Go through that and note precisely what laws you use.

nerdy

but ccorn ... isnt it true that (ma)x + (mb)y = mg = gcd(ma,mb) ?

so x and y must be the bezout the coeficients

ccorn

@nerdy We are in the midst of a proof. Where d'ya know that?

nerdy

but we can't assume that, is thats ur saying

ccorn

No. $x$ and $y$ are Bezout coeffs for $a$ and $b$ because we already know that $g$ is the GCD of $a$ and $b$.

By hypothesis.

nerdy

12:42

maybe can we go in reverse ? gcd(ma,mb) = least positive value of max + mby = m [ least positive value of ax + by ] = m . gcd(a,b)

is that wrong too ?

ccorn

Ok, let's try it that way.

Mats Granvik

@meer2kat You got in me in a corner.

ccorn

You define $\gcd(A,B)$ as the least positive value of $xA+yB$ for integers $x,y$.

nerdy

yap

ccorn

OK. so if $g=\gcd(a,b)$, there are specific integers $x',y'$ such that $g=x'a+y'b$, and no other $xa+yb$ can yield a smaller positive value.

nerdy

12:47

yep

except with x,y = 0 i guess

but that doesnt count

ccorn

Now multiply the latter with the positive integer $m$. You get that $mg = x'ma+y'mb = x'A+y'B$ is the smallest positive integer of all $xA+yB$.

We are using here that if $0<s<t$ and $m>0$, then $0<ms<mt$.)

(and vice versa.)

nerdy

yeah ! so mg = gcd(A,B) = gcd(ma,mb)

skullpatrol

You may be interested in this @N3buchadnezzar

nerdy

ccorn thanks man :DD

ccorn

@nerdy Phew :-)

Mats Granvik

13:04

If the von Mangoldt function could be constructed entirely from complex numbers with real part 1/2, without knowing the zeta zeros, would it say something about the von Mangoldt function?

Ok, I see now the zeta function is not having real part equal to 1/2 when evaluated. But the argument is having real part 1/2 for the zeta function used in the product.

ccorn

I am a sinner. I have used an ordering relation other than ideal inclusion in a divisibility problem.

Sigh.

Mats Granvik

13:34

@ccorn What is an ordering relation?

ccorn

@MatsGranvik $<$

(for example)

Mats Granvik

What $<$ what?

ccorn

@MatsGranvik Wikipedia article My word bag is almost exhausted currently.

Feynman and the word bag

Mats Granvik

13:53

Hmm... when you have studied a subject long if enough you can get tired of describing it. Is that what you mean? Happened to me with the mobius function.

ccorn

@MatsGranvik No no. Tired yes. It's weekend, I have chosen to spend a sleepless night yesterday.

Mats Granvik

@ccorn ok, I see.

MickLH

I feel that @Chris'ssis and @robjohn have formulated some grand conspiracy to starve me of highly interesting integrals

Mats Granvik

Crazy, you don't need the whole zeta function in the fourier transform to get the von Mangoldt function spectrum. A partial sum of size equal to number of distinct mobius function terms will do.

Mats Granvik

14:21

processing...

Result: Not a known function.

New result: 0

Mats Granvik

14:33

Some Fourier transforms are simply not known.

Mats Granvik

15:01

Is FourierTransform( f(x) + g(x) ) = FourierTransform( f(x) ) + FourierTransform( g(x) ) ?

Studentmath

11

Q: How is the Fourier transform "linear"?

A "linear" function usually means one who's graph is a straight line, or that involves no powers higher than 1. And yet, many sources will tell you that the Fourier transform is a "linear transform". Both the discrete and continuous Fourier transforms fundamentally involve the sine and cosine fu...