This page contains selected channel logs of #math on the undernet. They are selected for various reasons for instance things as illustrating a mathematical point quite well, providing an exemplary picture of channel discussions or being outright funny.

**2007-11-11 Some Linear Algebra**

<kmh> what*s the problem ?

<Atm0s> Anyone that good, that he/she can see what’s wronge about this formula ( http://compsci.ca/v3/viewtopic.php?t=14897 ) It works, but not when the two circles have a different size.

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<Atm0s> See listing 2.

<leba> I have the matrix A = [4 0 1;-2 1 0;-2 0 1] and I need to find a base for the auto-space associated with the auto-value(lambda 1, 2 and 3).

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<kmh> not sure what the auto thing is

<leba> do you know..

<kmh> do you mean eigenvalue and egenspace by any chance ?

<leba> For example (for autovalue 1) : Ax = 1x ==> Ax – 1x = 0 ==> (A – 1I)x = 0

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<kmh> that is eigenvalue i guess

<leba> yes, I believe it is eigenvalue and eigenspace in english.

<kmh> ok

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<kmh> well if the eigenvalue m is known already

<leba> Then refraising I need to find a base for the eigen value (1 , 2, and 3) associated with with each eigenvector.

<kmh> then you can solve the equation Ax=mx to get the associated eigenvectors

<kmh> however the base is not for the eigenvalue or eigenvector but the eigenspace

<leba> yes, the eigenvalues are (1, 2 and, 3)… I need to find the base of the matrix A = [4 0 1;-2 1 0;-2 0 1] with those eigenvalues given.

<kmh> for every eigenvalue m the set of eigenvectors belonging to m forms a vector space called eigenspace

<kmh> and for that eigenspace you can compute a base

<kmh> in addition you can try to find a base for the original vector space that consists of eigenvectors only

<leba> ok, that is what I am trying

<kmh> which one ?

<leba> nono, I need to find a base for this eigenspace

<kmh> ?

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<leba> “kmh : and for that eigenspace you can compute a base”

<kmh> ok

<kmh> so v=(x,y,z) is a vector in R^3

<leba> ok, I have A = [4 0 1;-2 1 0;-2 0 1] and lambda 1 for start.

<leba> ok, I will shut up and listen you from now on.

<kmh> then you need to solve Av=mv <=> (A-m*I)v=0

<leba> yes, v={x,y,z} is a vector in R^3

<kmh> where I is the identity matrix

<kmh> the solution space of that equation is your eigenspace

<leba> I ended up with [1 0 0 0;0 0 1 0;0 0 0 0]

<kmh> not sure what you did there

<leba> I can explaine step by step.

<leba> A = [4 0 1;-2 1 0;-2 0 1]

<leba> [4 0 1;-2 1 0;-2 0 1] – [1 0 0;0 1 0;0 0 1] = [3 0 1;-2 0 0;-2 0 0]

<leba> [3 0 1 0;-2 0 0 0;-2 0 0 0] ~ [1 0 0 0;0 0 1 0;0 0 0 0]

<kmh> this is for eigenvalue 1 ?

<leba> yep

<leba> now I think I have to find the general solution (or the null space) but not sure..

<kmh> yes

<kmh> null space = solution space = eigenspace

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<leba> my book says lambda 1: [1 0 1], where did it take it from ?

<kmh> however i’m not quite sure what your last step was good for [3 0 1 0;-2 0 0 0;-2 0 0 0] ~ [1 0 0 0;0 0 1 0;0 0 0 0]

<leba> Sorry, the oposite: lambda 1: [0 1 0]

<leba> yes I did it by hand and then I applyed MatLab rref(A) on it.

<leba> applied*

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<DArcMatter> rawr

<leba> Hey DArcMatter, did you eat already? Now kmh and I are trying to solve it. You can either help us or watch it 😉 (welcome back)

<DArcMatter> I got a breakfast burger

<leba> great!

<leba> kmh, do you know what to do from now on ?

<kmh> in principle yes but i haven’t done it for a while

<leba> I have a similar problem easier to understand wich is in my book and I could copy it here.. I think it could hel us understand it.

<leba> I couldn’t solve for it because of the last matrix.

<kmh> your base vector is (0,1,0) i believe

<leba> yes, but “based” in what ?

<kmh> no that is the base vector

<kmh> and the eigenspace is <(0,1,0)>

<leba> look, A = [4 -1 6;2 1 6;2 -1 8]

<kmh> that*s a different matrix

<kmh> now let’s continue from your row echelon form

<leba> A – 2I = [4 -1 6;2 1 6;2 -1 8] – [2 0 0;0 2 0;0 0 2] = [2 -1 6;2 -1 6;2 -1 6]

<kmh> you need to solve (A-m*I)v=0

<leba> (A – 2I)x = 0 now

<kmh> with m = 1

<leba> Now a different example just to help.

<kmh> leave that alone you need to understand how it works

<kmh> if you’ve understood it for 1 you can do it for the other yourself

<kmh> so you start with (A-m*I)v=0 and m=1

<leba> I understand how it works with this example I am trying to show because of the last step.

<kmh> ?

<kmh> why do you ask if you now how it works ?

<leba> in the end I finally find x2[1/2 1 0] + x3[-3 0 1] wich helps me find the base of it.

<kmh> i’m trying to explain you the whole thing including the step after your rref command

<leba> ok. I will listen to you.

<kmh> so you start with (A-m*I)v=0 and m=1

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<leba> ok

<kmh> so solved that you computed the the row echelon form of the extemded matrix

<leba> You mean I reduce it right?

<kmh> which was [3 0 1 0;-2 0 0 0;-2 0 0 0] in your case (the extended matrix)

<leba> ok

<kmh> now you can row echelon rref on that extended matrix to get [1 0 0 0;0 0 1 0;0 0 0 0]

<leba> ok

<kmh> and now the important part is what does that result mean ?

<leba> yes! 🙂

<kmh> it means [1 0 0 ;0 0 1 ;0 0 0 ]*(x,y,z)^t=(0,0,0)^t

<kmh> [1 0 0 ;0 0 1 ;0 0 0 ]*(x,y,z)^t=(x,z,0)^t

<kmh> so you have (x,z,0)^t=(0,0,0)^t => x=0,z=0, y=anything

<leba> It is nor x=0, z= anything, y=0 ?

<kmh> so every vector of the form (0,y,0) is a solution of (A-m*I)v=0 and m=1

<kmh> no it is not

<kmh> you have (x,z,0)^t=(0,0,0)^t

<kmh> just compare the components

<kmh> so that the equality is true

<leba> yes, it moved the z from third to second vector.

<kmh> x=0,z=0, y=anything

<kmh> yes and the y value does not matter

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<kmh> so again, you have now that every vector of the form (0,y,0) is a solution of (A-m*I)v=0 and m=1

<leba> ok

<kmh> and all these vectors of the form are the eigenspace for m=1

<kmh> you you need a base vectors that create all vectors of the form (0,y,0)

<leba> This eigenspace forms a streight line. Isn’t it a base for a null space space ?

<kmh> and since y*(0,1,0)=(0,y,0)

<kmh> you can use use (0,1,0) as base vector to generate them all

<kmh> yes

<leba> So I have x1=0, x2 free and, x3=0 (0,x2,0)

<leba> x2(0,1,0)

<leba> man, it’s complicated!

<leba> 🙂

<kmh> as i said above “eigenspace” = “solution space of (A-m*I)v=0” = “null space of A-m*I”

<leba> And what is a base ?

<kmh> yes

<kmh> (0,1,0) is the base

<leba> What is it?

<paultwang> that is what it is

<leba> Just to confirm my thought.

<kmh> hmmm… maybe you need to look up base again to understand what that means

<leba> yes, I think it passed streight into my brains.

<kmh> base of a vector space = set of linearly independent vectors that generate the whole vector space

<leba> ok, thank you kmh!

<kmh> alright 🙂

<leba> you are great! 🙂

<kmh> np – youare welcome

<kmh> however i’m off now

**2007-11-8 #math does the fancy physics dance
**<Karlo_> According to the figures provided in the pilot episode, the Klingon homeworld is closer than Alpha Centauri.

<FatherPi> BRITISH!!!

<FatherPi> Ohayo, gozaimasu, Anjin-san.

<Cmd_Data> Karlo_ : i don’t get that either

<Cmd_Data> hi Sais

<FatherPi> aha….kingons on uranus…i thought so. hehe

<Cmd_Data> well in my recollection it is nowhere indicated in the episode

<Cmd_Data> but that is probably another case of different perception

<Karlo_> And since the ORIGINAL reason for introducing the transporter to ST:TOS was because of a budget constraint preventing them from using the shuttlecraft — why didn’t they just leave out the transporter entirely for the prequel?

<Karlo_> They didn’t explicitly give the distance, but they did give the speed and time, which is sufficient to compute it.

<Cmd_Data> hehe good point – maybe they still had budget constrains ?

<psilo-> did the shuttle already land?

<FatherPi> aha…the show was cancelled. nearly escaping the axe last year.

<Cmd_Data> Karlo_ : compute to based on what ?

<Karlo_> I can’t parse that

<FatherPi> velocity and time

<FatherPi> to compute distance

<Cmd_Data> exactly

<Karlo_> x <= v_max t

<Cmd_Data> switch the order of compute and to

<Cmd_Data> Karlo_ : that’s warp physics ?

<FatherPi> if i specify a max achievable V and give you the total t, and it aint enuf to get you 4 light years from SOL

<FatherPi> and you arrive at the Klingon home world…then something is amiss.

<FatherPi> when i first heard of the idea of a new prequel ST show, i had high hopes it would be a big budget, space oriented, scientific forum

<Karlo_> Even when it’s “warp physics”, it’s always been specified in terms of speed as if through normal space.

<FatherPi> so, when i tuned in to see Vulcans and other bs, i was very displeased

<FatherPi> WARP SPEED: hasn’t it always been equivalent to x ly’s per some unit time?

<Karlo_> warp x = x^3 * c.

<Cmd_Data> Karlo_ : well what exactly is stated in episode ?

<Karlo_> That’s never been strictly canonical, but they did confirm that warp 5 is about 125 c in the first ENT episode, when they stated how long it would take to reach Jupiter at that terrific new speed.

<FatherPi> Excellent wiki article: http://en.wikipedia.org/wiki/Warp_speed#Warp_velocities

<Cmd_Data> haha

<Cmd_Data> wiki rules

<Cmd_Data> actually if you google these days more and more you have wiki links in the top results

<Karlo_> So, using only the information in ENT ep 1, we can compute a maximum distance to the Klingon homeworld as a multiple of the distance to Jupiter. 🙂

<FatherPi> How long did they say to get to Jupiter?

<FatherPi> hey, do you have to accelerate to W5? and then de-accelerate?

<Cmd_Data> Karlo_ : that includes a lot of assumptions

<FatherPi> We need Mulder…the Master of Unprovable Assumptions! hehe

<psilo-> you could theoretically go faster than light

<Cmd_Data> he’s out there seeking the truth

<psilo-> if you could find enough exotic negative energy

<Cmd_Data> antimatter 🙂

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<FatherPi> antimatter is not at all exotic

<Cmd_Data> the TOS used antimatter in some episodes

<psilo-> antimatter won’t get you going faster than light

<FatherPi> it has the same gravitational properties as ‘normal’ matter

<Cmd_Data> FatherPi : well it is for real afaik, but not its application in TOS

<psilo-> i mean you have to bend spacetime in very funny ways to go faster than light

<psilo-> and that takes lots of exotic negative energy

<Mulder_> it’s friday!

<Cmd_Data> thank god ?

<FatherPi> it’s friday?

<FatherPi> it’s just barely Thursday here in Seattle

<Sais> Not quite.

<Cmd_Data> psilo- : event horizon – to throw in another SF movie

<Mulder_> -Mulder_- TIME Fri Nov 09 00:02:4

<FatherPi> antimatter is a principle player in the sun putting out 200 trillion trillion watts

<Karlo_> The problem with going faster than light — no matter what technology you use to do so — is that if special relativity is correct about having no preferred reference frame, then you can create a closed time-like loop by using two FTL communication devices at different relative speeds.

<FatherPi> university physics students routinely create antimatter in lab.

<Mulder_> it’s cold. 13’C

<FatherPi> it’s 0-1 C here too many nights these days

<Sais> Cover it up, then

<Karlo_> I find it amusing that Trek and other SF will say “We can go faster than c, but then the new limit is 1000 c” (or whatever).

<Mulder_> heh

<FatherPi> right…at Warp 10, the angle of your phallus is at maximum.

<Karlo_> Once you figure out how to get past c at all, in some reference frame, a finite adjustment to that gives you infinite speed (instantaneous travel) or even beyond (i.e., arriving before you left).

<Mulder_> my unprovable hunches work well most of the time

<Karlo_> I like to say that Star Trek physics goes BEYOND Einstein… Past Newton… All the way to Aristotle.

<FatherPi> LOL

<Karlo_> “If you turn off the engines, the ship will come to a stop!”

<Mulder_> i like douglas adams physics

<FatherPi> the improbability drive, brillint

<FatherPi> brilliant

<Karlo_> As far as FTL explanations go, I rather like the Kelly Drive.

<Karlo_> In some Harry Harrison story…

<Cmd_Data> FTL=?

<Mulder_> always down some Pan Galactic Gargle Blaster before travelling long haul economy

<FatherPi> i have THHGTTG on mp3’s btw

<Karlo_> FTL = Faster Than Light

<Cmd_Data> oh

<Mulder_> i own the 5 book trilogy

<Cmd_Data> Mulder_ : lol

<psilo-> karlo special relativity only holds locally

<Mulder_> 5th one was really horrible to read though. very unpolished

<Karlo_> Harrison’s story had some mention of using the Kelly drive to get somewhere at 7 times the speed of light. Then there’s a footnote.

<psilo-> if you could get a lot of exotic negative energy, you would be bending spacetime quite a lot

<FatherPi> Karlo…let’s say you achieve V of 2 c

<Cmd_Data> Mulder_ : well the is some truth to it that it is easier to handle the improbabilty drive while being drunk

<psilo-> so that the local special relativistic stuff would no longer apply to your system

<FatherPi> how would you set up a time loop?

<FatherPi> also, how would you arrive somewhere before you left?

<Mulder_> Cmd_Data, yeah except i wouldnt actually drink on plane. you get drunk faster and the hangover is apparently 100x worse

<FatherPi> that poor whale…

<Karlo_> (Footnote) When asked how his drive could reach 7 times lightspeed, when Einstein says that nothing can travel faster than light, Patrick Kelly replied: “Well, sure and I guess Einstein was wrong.”

<psilo-> special relativity is like the local approximation to general relativity

<psilo-> kind of in the way that the slope is a local approximation to a general differentiable function

<psilo-> if you stray far things break down and the simple stuff doesn’t apply anymore

<FatherPi> the trick is you can not travel through space-time faster than c.

<psilo-> it’s still totally consistent with einstein’s theory, just not with special relativity

<FatherPi> so, you warp space-time.

<FatherPi> there is no exclusion to that in GR

<psilo-> this is how you can travel faster than c within relativity

<psilo-> the problem is to bend spacetime in such ways requires a lot of exotic energy, which may be impossible to do

<psilo-> i’m not sure, but i think the verdict is still out on that one

<FatherPi> if i have an ‘engine’ which allow me to fold the fabric of space-time to create a virtual wormhole thru a section of space.

<FatherPi> i have not per se violated Relativity

<psilo-> pi, yeah it’s something like that

<psilo-> you squish the space in front of you and expand the space behind you

<psilo-> and you zoom forward

<Cmd_Data> FatherPi : a new doomsday machine – create a blackhole next to earth 🙂

<FatherPi> that’s the only chance we have at such an engine.

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<Cmd_Data> that’ll teach us

<burg> hello

<Karlo_> I still don’t believe it can work.

<psilo-> you’re not tearing the fabric of space

<FatherPi> cause, you are NOT gunna accelerate any mass to c.

<psilo-> and stitching faraway places next to each other

<psilo-> it’s totally smooth

<Mulder_> to travel quickly through space-time, we need to make space-time look like a wrinkled old man so that we can hope over the wrinkles

<psilo-> it doesn’t require singularities or any stuff beyond relativity

<Mulder_> hope/hop

<psilo-> you don’t hop over stuff

<psilo-> you don’t tear space

<psilo-> just push and pull it in the right way

<Mulder_> tearing is so violent. mine’s more PG

<Cmd_Data> hmmm

<psilo-> think of the big bang and inflation

<psilo-> you create inflation behind you, and contraction in front of you

<FatherPi> space-time isn’t smooth now.

<psilo-> this can propel you way faster than c

<psilo-> and you wouldn’t even feel acceleration if you do it right

* Cmd_Data bangs hi head big time

<Karlo_> When most people talk about wormholes connecting two points in spacetime, they focus on the space aspect. So what stops you from using the time aspect of that wormhole to send a message to your past self?

<FatherPi> it is full of quantum jiggles and changing from moment to moment due to proximity with mass.

<FatherPi> time is a scalar.

<FatherPi> there is no going to past time.

<FatherPi> you think the past is just sitting there, en toto, awaiting our return?

<Mulder_> unless you’re hiro nakamura

<Karlo_> But if a wormhole connects two points in spacetime — and allows travel in both directions — then even if they’re “different places, but the same time” in *our* reference frame, they’ll be different times in some *other* reference frame. So, one direction or the other is going into the past.

<Mulder_> good night

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<Karlo_> Hah! He couldn’t deal with my flawless logic. 🙂

<FatherPi> So, some dude in Andomeda is looking at Earth and sees dinosaurs roaming the planet, right?

<FatherPi> but that is just a phenomena of c

<Karlo_> I don’t know the distance to Andromeda offhand, but I’m willing to grant the basic premise so far.

<psilo-> yes

<psilo-> although to say “right now a dude in andromeda”

<FatherPi> he is not really looking in to the past.

<psilo-> is a dangerous thing

<psilo-> since “right now” is a very relative and nearly meaningless statement for something that far away

<Karlo_> Right — there is no simultaneity.

<FatherPi> right now, means, in my reference frame

<Karlo_> Okay fine

<psilo-> yeah you could do that

<FatherPi> my frame is as valid as any other…and we are on earth

<psilo-> although because those things are so far away

<psilo-> your reference frame is hard to be extended that far

<psilo-> curvature acts in between and warps things in bizzare ways

<psilo-> you can’t just use special relativity willy nilly and get sensible results necessarily

* Karlo_ makes a note to finish up his relativity computations from a few months ago.

<FatherPi> if i find a wormhole that takes me in a flash from here to Andromeda

<FatherPi> and then i return here…

<FatherPi> i have in neither case travelled into the past

<FatherPi> time travel is real. we are all of us, time travellers. it’s just a one-way trip.

<Karlo_> Suppose there are two planets, several light years apart, both moving at a velocity of c/2 northwards relative to us, and *those* people find a wormhole that takes them from planet A to planet B instantaneously (according to their reference frame).

<Karlo_> Then it becomes possible for us to leave a message at A when it passes us, have them send it to B instantaneously (in their frame), which is in the past according to *our* frame.

<Karlo_> Then B passes the other end of our wormhole, and passes on the message to our station there, which sends it through the wormhole instantaneously (in our frame) to get back to Earth, its starting point, before it was sent.

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<Karlo_> (Summary: You can create a chronophone out of two ansibles.)

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<FatherPi> i don’t quite see your scenario.

<Karlo_> It’s easier with a space-time graph…

<FatherPi> but, in my ignorance, i would say, the message is generated at time t0 in OUR frame, no wormhole or series of wormhole hopping…

<Karlo_> Oh, on a related topic,

<FatherPi> will get the message back to us in our frame prior to t0

<Karlo_> restricting attention to two space dimensions, for simplicity…

<FatherPi> google phone software…a killer winner??/

<FatherPi> (for potential investors…warning, stock usually moves on rumor, not on fact)

<Karlo_> I realized a while back that the geometry of velocities — i.e., a “point” corresponds to a velocity, and a “line” is a set of velocities reachable from each other by a common acceleration vector — is a hyperbolic plane with curvature -1/c^2.

<Karlo_> In this model, lightspeed velocities are points at infinity.

<Karlo_> But it’s possible to extend the hyperbolic plane to include virtual points (representing the intersection of lines that share a common perpendicular), which would correspond to speeds faster than light.

<FatherPi> karlo…forgive me for interrupting. my day starts early, and i see it’s approaching 4 am, here. i must go.

<FatherPi> i will review what you write, btw.

<FatherPi> be well, young man!

<Karlo_> Now, what I thought was an interesting result is that the extended model is a projective plane, and the slower-than-light points are a hyperbolic plane, which has a disk model…

* FatherPi is now known as PiAway

<Karlo_> Which means that the faster-than-light points are what’s left when you remove a disk from a projective plane — namely, it’s a Moebius strip.

<Karlo_> This had some rather mind-boggling implications when I tried to decide what it actually meant. Some things behave like angles when you want them to behave like distances, and vice versa.

**2007-11-6 Women in #math, , Limits **<FatherPi> i just thought of something rather interesting…

<FatherPi> heya, herbling !!!

<herbling> oh no!!!!

* herbling runs from pi.

<FatherPi> ltns. hehe

<herbling> hey

<kmh> hi herbling

<herbling> hey

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<FatherPi> get a load of this…

* zurt_work cuddles zurt`

<kmh> FatherPi : what is it then ?

<FatherPi> we have 3 reg ladies (1 is an old-timer who shows up now only once in a while)

<FatherPi> in #math

<Carve> kmh i have a new limit task: x->0 e^x-1/sqrt(x)

<Neuroner> Pi, did you se my problem for before

<FatherPi> Kim, HoneyCat, Koxi

<FatherPi> what do they have in common?

<zurt_work> potatoes?

<FatherPi> besides being wonderful ladies.

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<herbling> they are all number theroists?

<kmh> Carve : l*hostpital ?

<zurt_work> they have body parts that we don’t?

<Carve> well

<Carve> ofc but i doesnt grant me an answer

<Carve> other than that it doesnt exist

<zurt_work> oooh ooh I know… they’re all from Venus

<kmh> ofc ?

<slider142> number theorist parties?

<Carve> of course

<Karlo_> Carve: l’H should work.

<kmh> FatherPi : you missed emmy and zonia

<FatherPi> all 3 of them have PhDs.

<herbling> they are the nude ladies on the cover of “Time Enough For Love”!!!

<zurt_work> what about people who want to be women?

<herbling> eeep

<slider142>

<herbling> phd!

<FatherPi> kmh….emmy is new, zonia has just started coming back.

<kmh> zurt_work : do we have them ?

<Carve> ah okay

<FatherPi> slider! HEYA!

<Carve> sorry for the bother

<slider142> Pi! 😀

<kmh> actually zonia is around quite often but mostly idling

<herbling> zonia hides from me :(((

<zurt_work> kmh: ya never know until you know someone…

<emmy> I’m just working through math on my own when I have free time.

<Karlo_> I don’t know enough about those three to make a guess, unless the property in question is just a function of their nicknames, in which case you probably wouldn’t have abbreviated them in the question.

<FatherPi> HERBLING…silly. hehe

<Carve> if i have that x->0 lim e^x-1 / x = 1. how can i transform x->0 e^x-1/sqrt(x) so that i get to use the first answer?

<emmy> I dont have any special qualifications or anything like that

<FatherPi> Karlo…you appear lagged.

<FatherPi> kid was here ealier asking about AI stuff…in particular, kohonen

<Karlo_> Carve: You could write e^x-1 / sqrt(x) as the product of the thing you know, and something simpler.

<emmy> But I did make a channel called #not-math so that if anyone wants to talk about casual things we can leave this channel for focused math discussions.

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<slider142> Carve, ie., multiply by unity

<FatherPi> <relik> but in all lecture about the kohonen network they use the sigma as width of the neighbourhood function

<FatherPi> <relik> I try to understand the kohonen algorithm

* herbling yikes! and runs away from PI…strange man! 😛 🙂

<FatherPi> welcome, Rev_Null !!!!!!!!!!!!

* FatherPi bows deeply and respectfully to antizeus.

<Carve> i see, should i use substitution for this?

<FatherPi> emmy…no pluggin your own channels in here, please.

<FatherPi> hehe j/k

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<herbling> so whats new pi?

<Karlo_> Carve: Did you follow what I suggested?

<FatherPi> (she’s been helpful and nice in here so i’ll cut her some slack…and, she’s a girl!!!)

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<Carve> ye i think so

<herbling> male…female…whatever!

<Carve> sqrt(x) ?

* emmy squints

<Carve> was that something simpler

<Karlo_> Yes

<FatherPi> herbling, pi never changes. pi is pi. it’s a constant.

<FatherPi> new pi = old pi. 😛

<herbling> ah okay

<Carve> well i actually thought of a constant but ye =)

<Karlo_> And make use of the fact that lim{f g} = lim{f} lim{g} when both of those factors exist.

<Karlo_> Actually, I once read an article about pi as a function of time.

<Karlo_> “The ancient Egyptians used 3 1/8”, etc.

<slider142> Was it because of the fine structure constant?

<slider142> ah

<kmh> (e^x-1)/sqrt(x)= (e^x-1)/sqrt(x)*sqrt(x)/sqrt(x)=(e^x-1)/x*sqrt(x)

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<slider142> spoilsport :p

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<Karlo_> It was mildly amusing.

<kmh> oops lagging

<Carve> thanks =)

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<Carve> that way i can solve it without l’hopitals

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<kmh> yes

<slideaway> yep. most non-trigonometric indeterminates fall to factoring or adding a factor

<slideaway> at least the ones in textbooks 😀

<slideaway> bbl

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<kmh> rationalizing the denumerator/muktiplying with a conjugate or such often helps to get a solution based on known limits and algebra manuilations

<kmh> rationalizing the denumerator/muktiplying with a conjugate or such often helps to get a solution based on known limits and algebra manipulations only

<kmh> darn keyboard

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<herbling> *groans*

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<FatherPi> Karlo, 3 1/8 would be way too little–i am sure the Babylonians and Egyptians did a lot better.

<FatherPi> okay, here is one source (google):

<FatherPi> The ancient Egyptians seem to have sometimes used a value of 22/7 (3.142857 . . .) for pi. There is also evidence that they estimated the area of a circle with a square with a side that is 8/9 the size of the circle’s diameter. This gives a value of pi of 3.16049382716

<Karlo_> I seem to remember that they preferred 3 1/8 to 3 1/7 for some reason — possibly because it’s easy to divide by halving.

<FatherPi> well, that circle diameter thing is almost as bad as 3 1/8

<OC_80> i’m actually getting a pi tatto on my arm anybody else in here have one

<FatherPi> i have one, but i can not say where.

<OC_80> i’m passionate about pi radians

<Karlo_> Some people claim that the pyramids encode the value of pi, but I don’t believe that, myself.

<FatherPi> pimaster’s cough sounds suspiciously like (“BS”), hehe

<FatherPi> cough, cough

<kmh> http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Pi_through_the_ages.html

<Karlo_> Last time I did a search for this, or for something related, I found one statement that claimed that the pyramid measurements encode pi “to 15 decimal places” — which would be sub-atomic level.

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<kmh> i guess some phi claims are just like some phi claims, being more numerology than math (in nature)

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<kmh> which reminds of a quote in the pi movie: “if you you abandon mathematical rigour, you become a numerologist max”

<FatherPi> phi claims = phi claims? duh?

<FatherPi> j/k, i know you meant pi

<kmh> pi and phi

<FatherPi> good quote.

<kmh> i habe a rather annoying habit of lousy typing

<FatherPi> a little harsh perhaps.

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<FatherPi> kmh…not even your primary language, i’d say you were darn near perfect.

<Karlo_> I think the pyramid shape is more likely to encode phi than pi, but even that doesn’t seem exceptionally likely.

<FatherPi> much better than the americans i work with. hehe

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<herbling> darn americans! lazy bastards!!!!

<herbling> 🙂

<kmh> found it : Sol Robeson: As soon as you discard scientific rigor, you’re no longer a mathematician, you’re a numerologist.

<kmh> http://www.imdb.com/title/tt0138704/quotes <– at the bottom

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<herbling> I’m a pettie sophisticate!

* herbling shakes his booty

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<quaziman> hello 😀

<kmh> booty has a lot of meanings

<kmh> which somewhat irritates me at times 🙂

<kmh> howdy quaziman

<herbling> internet, define booty

<quaziman> im doing some work with t-distributions. I hava a question. Suppose we are given a sample mean, s and n. t=mean-mu / s/sqrt(n)

<kmh> wahhhaaaaa

* kmh hides

<kmh> bbl

<Karlo_> That pi article mentions the Shanks calculation to 707 digits, with a mistake after 527, which was discovered because of a statistical anomaly in the remaining digits.

<quaziman> what is that mu ??

<Karlo_> There’s something implausible about that.

<kmh> Karlo_ : i think van ceulen had an error as well

<FatherPi> Karlo…you have a point.

<Karlo_> If you make a random mistake in calculation, I would expect the resulting digits to still be as random as the real ones.

<quaziman> mu should be the population mean, shouldnt it?

<FatherPi> i never thought about it before! it’s prolly bs

<kmh> to sell the pattern in pi stuff

<zurt_work> no, no it’s plausible. the statistical anomaly they’re talking about is digit 604 which came up mysteriously as ‘A’ 😛

<kmh> however maybe you apply benford’s law ?

<Karlo_> This makes me wonder if Shanks got tired of calculating and just started filling in digits from his head… And we know that humans do not make good random number generators.

<FatherPi> zurt….hehe

<FatherPi> (i can’t figure out if i should laugh at zurt or perm ban him, hehe)

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<FatherPi> Karlo!!!!!!!!! no. he busted his butt till he nearly died to calculate out to 707 digits!

<Karlo_> Maybe he thought that nobody would ever bother to reproduce his work, and so he was safe. 🙂

<FatherPi> the poor guy…thank goodness he didn’t live to know about his error.

<kmh> those were the days when proper math fakes still worked 🙂

<FatherPi> Karlo…actually, very funny. hehe

<herbling> math fakes??!!?

<herbling> LOL

<herbling> :)))

<FatherPi> right…the proof’s too large for the margin (yeah, sure).

<kmh> did you read about those french physiscist and their scandal?

<herbling> oooooh

<herbling> what did they do????

<Karlo_> N-rays? Or something recent?

<kmh> well there’s reason to believe that their PhD thesis and jounrnal publicatons might be total nonsense

<kmh> however due to the very abstract nature/formulation nobody is really able to tell or willing to invest time to figure it out for sure

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<FatherPi> be well, ppl.

<FatherPi> ciao, tutti

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<herbling> wahhh!!!

<Karlo_> On the other hand, maybe Shanks really did make a mistake in which, even though the resulting digits were effectively random, they just happened to contain this statistical anomaly that caused the work to be reviewed — much the same as Pluto was accidentally discovered using measurements that turned out to be flawed.

<herbling> bye pi 😦

* Karlo_ starts dinner

**2007-11-6 Some Geometry**

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<evilmasterer> hey , can you guys help me with this problem now:P? : i got this triangle: http://img126.imageshack.us/my.php?image=wdh3vraag5fj1.png , line AB is called , vector b and line DB is called h , and the angle of BAD is called alpha , now i have to proof this right: h² = |vector b|² – |vector b|² * cos² alpha (and |vector b| simply is length of vector b :P)

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<evilmasterer> any1?

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<emmy> Are you familiar with the Pythagorean theorem?

<evilmasterer> you mean A² + B² = C² ?

<Carve> If i shall minimalize the perimeter based on the area of a circlesector, how do i go forward?

<emmy> evilmasterer, yes that’s it.

<evilmasterer> but i don’t have any values for the triangle , and i have to proof that its true

<emmy> What do you get if you solve a^2 + b^2 = c^2 for a^2 ?

<evilmasterer> it has something to do with (not sure if translated correctly) “inproduct” ( vector a * vector b = |vector a| * |vector b| * cos alpha

<emmy> evilmasterer, did you see my question?

<evilmasterer> uhm , you can’t say what you get cause you they’re all variable?

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<emmy> No we aren’t looking for any numerical solution.

<evilmasterer> oh , but isn’t a² + b² = c² sort of the fastest way of writing it?

<emmy> So you have a^2 + b^2 = c^2 can you express that as an equation that says what a^2 equals?

<evilmasterer> uhm , c² – b² = a² ?

<emmy> yes indeed.

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<emmy> so you have a^2 = c^2 – b^2

<emmy> now compare the form of this equation with the form of the equation you wish to prove.

<emmy> According to your diagram, what is the hypotenuse of the triangle we are referring to?

<evilmasterer> ok i see it , in that way you mean , a² = h² , c² = |vector b|² and b² = |vector b|² * cos² alpha?

<emmy> Yes for sure.

<evilmasterer> that should be , vector b , line AB

<emmy> The hypetenuse of the triangle in your diagram is the line segment AB. What you are calling vector b.

<emmy> so this must correspond to c in the pythagorean theorem. That is, c = ||b||

<emmy> ||b|| means the norm of b, the length of vector b.

<emmy> Unfortunately you are calling this vector b, and it is a different b than the b in the pythagorean theorem and that may cause some confusion.

<evilmasterer> but , we use length of vector b 😛 ( we write it with single line on both side)

<emmy> do you understand why the one of the legs of your right triangle has a length that is |vector b| cos alpha ?

<evilmasterer> i guess that has something to do with the “inproduct” if that is translated well :P?

<emmy> It has to do with the definition of cos alpha.

<emmy> For a right triangle cos alpha = adjacent / hypotenuse.

<emmy> Do you recall ever seeing that?

<evilmasterer> yea

<emmy> So according to the diagram which side is adjacent to alpha?

<evilmasterer> AD

<emmy> right and it would be very nice to know the length of his side.

<emmy> and by the diagram what are we calling the hypotenuse?

<evilmasterer> AB

<emmy> Yes, also known as ‘vector b’ apparently.

<evilmasterer> yea , we’re busy with 3 chapters about vectors , so the book uses that alot 😛

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<emmy> so we have cos alpha = AD / |vector b|

<emmy> could you write this an equation that says what AD is equal to?

<evilmasterer> cos alpha * |vector b| = AD

<emmy> Right. Now let’s use the pythagorean theorem. So we have AD^2 + h^2 = |vector b|^2 . But AD^2 = cos^2 (alpha)* |vector b|^2.

<emmy> So we have cos^2(alpha)*|vector b|^2 + h^2 = |vector b|^2

<emmy> write that as an equation that says what h^2 is equal to.

<evilmasterer> btw , how do i do cos² alpha? is that cos of alpha and that squared?

<emmy> cos^2(alpha) is just a conventional way of writing (cos(alpha))^2

<evilmasterer> ah , and that equatation would be , h² = |vector b|² – |vector b|² * cos² alpha :P?

<emmy> Yes 🙂

<evilmasterer> ok , thx ^^:)

<emmy> evilmasterer, to see if you understand these concepts you could try to show that (DC)^2 = (BC)^2 – |vector b|^2 * sin^2(alpha)

<evilmasterer> hmm , my guess would be something similar will be covered in one of the next questions 😛

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<sdnndi> how to differentiate -e^ (-y^2) ?

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<sdnndi> … = +y^2 * e^(-y^2-1) ?

<sdnndi> plz help me

<evilmasterer> yea … use that: vector DA = vector a ( line AC) – vector b cos alpha and vector BA = vector a – vector b to show that in triangle DBC there is: h² = |vector a – vector b|² – (|vector a| – |vector b| * cos alpha)² , i think i can use that one as a good way to practice 😛

<emmy> sdnndi, you use the product rule and the chain rule.

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<sdnndi> where should i apply the chain rule?

<Asriel> chain rule: d/dx g(f(x)) = f'(x) g'(f(x))

<Asriel> pattern match with your function. What’s your g? What’s your f?

**2007-11-6 A Conversation Stopper **

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<c64> I’m wondering how to properly define a complex unit hypercube in C^n, centered at the origin, and spanned by the orthonormal vectors e1,…,en

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<PiAway> c64–boy, you really shutdown the chat. hehe

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<PiAway> was c64 your first puter?

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<emmy> c64, I guess you put constraints on the coordinates so that |z_k| < .5 where z_k is the kth coordinate of z in C^n

<c64> well, you can’t use the abs. value, cause, I want the subspaces to be small hypercubes as well

<c64> yeah c64 was my first computer 🙂

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<c64> I’m gonna use this hypercube as a codebook

<c64> so I’ll have something like B = {\sum_{i=1}^n a_i e_i s.t. a_i \in [-1/2, 1/2], i=1, …, n}

<c64> but I’m wondering if this a_i should actually be complex

<c64> and not real

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<c64> should I put the constraint s.t. a_i makes sure that both the real and imaginary parts are in [-1/2, 1/2]?

<c64> I’m just really wondering if a_i should be real or complex

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<c64> any ideas?

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<emmy> I think that sounds right.

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<emmy> For these purposes, you can probably think of C^n as R^2n, and in R^2n it seems like the thing to do is make sure that (x_1,x_2,x_3….x_2n) is such that -.5 <= x_k <.5 for k = 1,2,3, .., 2n.

<c64> yeah, but isn’t it a bit dangerous to think of the extension to R^2n? The hypercube in R^2n would have 2^2n vertices, wouldn’t it?

<c64> a very different hypercube

<kmh> hey c64

<kmh> commodore rules

<c64> so I should define a_i as being in [-1/2, 1/2] for each (real and imaginary) dimension?

<c64> sure does : )

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<Neuroner> Any luck on the problem I typed in earlier?

<Neuroner> Abou the four guys and the brigde?

<kmh> c64 : i’m not sure whether a complex hypercube gives you a a nice geometrical interpretation still

<c64> yeah, that’s what I’m having problems with, the geometric interpretation

<c64> it’s fine for the real case

<c64> say, R^2

<emmy> I thought that C = R^2 so CxC = C^2 = R^2 x R^2 = R^4 etc. As far as metric properties go.

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<c64> I’m also wondering. In the real case, I need to divide up each side of B into smaller segments, making a mesh of subhypercubes. So, for instance, the ith side would be divided into sigma_i segments. How would this translate to the complex case? Would I divide each side into sub-spaces?

<kmh> c64 : unless you reread c^n as R^(2n) as emmy suggested i think you loose he geometric intuition

<c64> ok, that makes sense..

<c64> but any idea how would formulate the partition of the hypercube into smaller hypercubes?

<c64> would that be in terms of sub-spaces?

<c64> … instead of segments, which would be the case for R^n

<c64> each side is partitioned corresponding to some eigenvalue

<c64> so each side is not partitioned into the exact same number of segments/subspaces

<kmh> c64 : one problem is in R^2 for each interval/dimension part you have an order, which imho is probably essential to geometric intuition and you loose that with C

<c64> yeah

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