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Comments:
<0> @plot sin(x) + sin(pi*x)
<1> ihope: http://nanographer.nanosouffle.net/?func=sin%28x%29%20%2B%20sin%28pi*x%29%0A
<0> Fun, I'm sure.
<2> Hi just a quick, boring differential calc question. A bucket is attached to a pulley suspended 13.5m above the ground with a rope attached to the bucket and pulled through dangling loose to the ground. The rope is held at 1.5m above the ground and a worker walks away from the bucket raising it up. I need to find the rate of change of bucket's rising when the worker is 9m from where he started next to the bucket.
<2> Im just wondering how to set it up, the computation is fine. But the whole 1.5m above the ground seems to confound me.
<2> Do I just use pythagorean theorem and differentiate implicitly?
<2> Also, sorry for the bad nickname.
<3> i think you need to draw a picture
<4> .
<5> Well, I've got a picture but its not clear to me.
<6> _llll_: My professor looked at that modules question I had and said that my work looked correct and wasn't sure what there was to say about it. "You have until Monday, keep working on it."
<3> heh
<6> Other people seem to think that D^2(r) = r isn't strong enough. They think the ring has to be commutative.
<5> Like whatsay I go (13.5-y)^2 = x^2 - r^2 and differentiate implicitly?
<4> hey amy`
<3> it all depends exactly what the question is asking
<3> i can see the rgument for commutative, but it involves comparing two different structures on N**
<6> Yeah.
<6> I'm just supposed to say what the condition on D is that makes them the same though.
<6> Nothing seems to make them the same.
<7> chucktaylor: I may have miscalculated something, so don't trust me too much, but I get that it's 6/10'ths the rate at which the worker is moving away. I think your idea is basically the way to go, the problem is simpler if you only plug in the numbers at the end.
<3> oh, i thoight we ended up with D^2=id (somehow) before?
<6> That was what my professor said to me in an e-mail.
<7> Steve|Office: modules question?
<8> Hi, I'm having a bit of trouble with modelling the brick edges of a 3D tower in 2D, which probably requires a mathematical function of some sort. if [ ] are the edge of the tower, and | are the brick edges, [|| | | | | | | | ||]. Something like that anyway.
<3> but im sure you managed to get to that as well
<5> Cale, I was wondering about the value of r, the hypotenuse. None is given, can I basically ignore it and just implicitly differentiate the x and y values?
<7> chucktaylor: well, it's *something*.
<3> it just involved choosing "the original structure on N**" to be something a bit random
<6> I ended up with that by just guessing (r,g) |--> (f|-->D(D(r))g(f))
<3> yeah, quite :)
<6> Cale: Yeah.
<7> chucktaylor: and it varies negatively with the rate of change of the length of the hanging rope
<5> yeah but I can basically deduce the value of r from the final conditions.
<9> Hiyas kmh
<5> I guess I should just plug it in at the end.
<3> the problem is "what two structures are we being asked to compare", that one seems a reasonable one
<6> Cale: http://allenknutsonsothercl***.blogspot.com/ Problem 4.
<6> The issue was that f|->D(D(r))g(f) isn't R-linear.
<6> Well, it's not an element of Hom_R(N*,R) where N* and R are both _left_ R-modules.
<3> well, maybe it has the D^2 on the other side
<3> one way definitely worked!
<6> D^2 on the other side would fail ***ociativity.
<3> why isnt it R-linear? cos D is, g is
<6> And then we'd be comparing f|->g(f)D(D(r)) and g|->rg(f).
<3> well, you could choose f=1 and still get D^2=1 i guess
<6> Well, if we look at (rg)(sf) /= s(rg)(f).
<3> err, or g=1. hmm im confused already
<6> (rg)(sf) = D(D(r))g(sf) = D(D(r))sg(f), but then what do you do?
<6> You'd need D(D(r)) and s to commute.
<3> D is R-linear, so it does (i think, im a bit unsure what lvies where at this point)
<6> D isn't R-linear though.
<6> D(rs) = D(s)D(r).
<3> hmm, i tohught D was a map of modules
<6> D : R --> R^op.
<3> but are those R's rings or R-modules?
<6> It's the ring.
<6> N is a left R-module.
<5> thanks cale.
<6> D is an anti-automorphism of R.
<3> so D is just a map of rings
<6> Right.
<3> ok, so that just tells us the D's need to be on the other side
<10> is it reasonably possible to find an explicit mathematical solution for, say, the equation of motion of a point charge approaching another point charge? (one of them is fixed).
<3> so we should be looking at r.g being the map f|--> g(f)D(D(r))
<3> cos that is r-linear
<6> That is R-linear for sure.
<6> But it fails ***ociativity.
<10> I guess it just comes down to whether or not you can solve the tricky nonlinear second order differential equation
<3> so s.(r.g) is the map sending f to (r.g)(f)D(D(r))
<6> (rs.g)(f) = g(f)D(D(rs) = g(f)D(r)D(s).
<6> = (s.r.g)(f).
<10> anyone happen to know offhand?
<3> so s.(r.g) is the map sending f to (r.g)(f)D(D(s)) = g(f)D(D(r))D(D(s))
<6> er, = g(f)D(D(r))D(D(s)).
<3> i think it works, doesnt it?
<3> D^2 goes from R to R, should be non-anti again
<6> Did I make a mistake?
<3> no idea :)
<3> too many symbols for me
<10> a 'no' will suffice as I'm really curious and no response doesn't give me an insight into if the answer is no or if it is just being ignored and not given any thought
<6> So, what you said was (s.(r.g))(f) = (r.g)(f)D(D(s)) = g(f)D(D(r))D(D(s)). Okay.
<6> Now combine D(D(r))D(D(s)) = D(D(s)D(r) = D(D(rs)).
<6> So g(f)D(D(rs)) = (rs.g)(f) /= (s.r.g)(f).
<3> hmm, so it ends up being (rs).g instead
<6> Yeah. =\
<3> so it would be equal if R was commutative
<6> Right.
<3> or, it would be ok if we didnt have D(D(r)) but just one D
<6> Right. One D is what I have.
<6> r.g is the map sending f to g(f)D(r).
<6> That's ***ociative and R-linear.
<3> oh, right. that was what the first part was asking to show
<6> Yeah.
<11> hahaha. http://xkcd.com/c88.html
<3> so the "other structure on N**" was "r*g is the map sending f to r g(f)"
<6> So I've got two left R-module structures on N**. One of them as morphisms of left modules and the normal one as morphisms of right modules.
<6> Yes.
<3> so you'd need r g(f) = g(f)D(r) for all r,g,f. so f=1 gives r=D(r) and R would then be commutative
<3> but the answer is meant to have a D^2 in it...
<3> hmm
<6> Okay, I'm just putting down the commutative condition.
<6> Actually, that looks like part c. "Show the converse: if this new left R-module structure on N** agrees with the old one for all N, then D satisfies that condition.
<3> it seems kind of an odd answer. so im guesing it sint what is intended. but what the q actually is asking i dont really see
<6> Yeah, I have no idea.
<0> % Solve[x*y==x+y,x]
<1> ihope: {{x -> y/(-1 + y)}}
<0> y/(y-1), in other words?
<3> are you asking if -1+y equals y-1 ?
<6> And of course, problem 5 asks me to use problem 4.
<0> > (sqrt(.001*2), (.001+2)/2)
<1> The operator `.' [infixr 9] of a section
<1> must have lower precede...
<0> _llll_: that question was rhetorical.
<0> Or something.
<0> > (sqrt(0.001*2), (0.001+2)/2)
<3> :)
<1> (4.4721359549995794e-2,1.0005)
<3> Steve|Office: well a real vector space is a module for a commutative ring, so i guess that's a good sign...
<12> Hey _llll_ would you mind looking at and commenting on a mathmatical proof I wrote?
<3> maybe...
<3> luckily im not the only one in the channel
<0> __mikem: what proof?
<0> If it's the e^(ix) one, it's pretty nice. :-)
<12> it is
<12> But I want someone to check it for flaws.
<12> http://mikemiller.dyndns.org/proof.pdf
<13> It looks fine, but I don't know why you included the first part
<12> thermoplyae, because I wanted to include a reason why someone would state that e^(xi) = cos x + i sin x in the first place.
<14> __mikem: only suggestion I can think of is spelling of "derivative" ;)
<15> hey
<15> didja know that the average of all primes up to 100 is approximately 42?
<10> that's useful information
<15> yep
<15> :D
<10> EdBo1 answer my earlier question if you are bored please
<15> I just got in here
<15> so could you please restate it? :)
<10> let me paste
<10> is it reasonably possible to find an explicit mathematical solution for, say, the equation of motion of a point charge approaching another point charge? (one of them is fixed).
<16> ...uhhhhhhhhh
<16> I understood like, the first 10 words of that
<16> so, yeah, not able to answer that
<10> basically it's solving a
<10> second order nonlinear differential equation
<10> without using numerical methods
<16> ...riiight
<10> you don't understand that either?
<3> i imagine it depends a lot on the equation
<10> _llll_, true
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