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Comments:
<0> and it would allow for multiple pieces of data to be acted on at the same time (that is, which would otherwise be in separate iterations of the loop)
<0> which gives more freedom for moving things around
<1> neat.
<0> and it tried to maximise the amount of the processor in use for each cycle
<1> you do embedded stuff?
<0> nope
<1> pservers? macintoshes?
<0> This was for a project to produce a high-level language (and compiler) for doing signal and image processing tasks.
<1> *pseries
<0> Basically using ordinary (PowerPC at the time, I think they're targetting Cell now) equipment to do supercomputer-like tasks.
<1> so it is embedded? :)
<0> davidmccabe: In what sense?
<0> davidmccabe: These are ordinary server-like computers we're working with
<1> ah.
<1> I was under the (false, apparently) impression that PPC and especially Cell were designed primarily for embedded use.
<1> wikipedia correcteth me.
<2> my school blocks wikipedia with their proxy or w/e.. can you beleive that ?
<2> wikipedia is like the answer to highschool
<3> now you have to make up your facts on your own.
<3> instead of relying on other people to make them up.
<2> ahaha
<2> i like wikipedia
<3> statistically, somebody has to.
<4> ChrOnX: add .nyud:8090 to the url
<4> it'll give you the coral cache
<2> ?
<2> www.wikipedia.org.nyud:8090 ?
<4> ya
<2> hmm
<2> i'll try it
<4> nyud.net:8090
<4> forgot the .net :p
<2> hmm
<5> anyone care to tell me why the following isnt adding up (3/10 + 5/28) = (3*28+5*10)/(10*28) = (24+50)/280 = 74/280 = 37/140
<5> 3/10 + 5/28 != 37/140 :(
<6> 3*28=24 is news to me.
<4> lol
<5> doh
<5> lol
<2> haha
<5> wtf
<6> must be a new branch of math ;)
<0> He's just working modulo 60
<6> or any divisor thereof, yes
<6> everything becomes trivial when done mod 1.
<0> Some of those divisions seem questionable mod 60 though, seeing as some of those aren't units.
<5> still doing something wrong
<0> a/b + c/d = (ad + bc)/(bd)
<6> it's find mod 3 though.
<6> *fine
<0> So 3/10 + 5/28 = (3 * 28 + 10 * 5) / (10 * 28) = (84 + 50) / (280) = 134/280
<0> which is 67/140
<6> 3/10 + 5/28 = 42/140 + 25/140.
<0> (cancelling out the common 2)
<5> 67/140 is what i got too humm
<0> % 3/10 + 5/28
<7> Cale: 67/140
<5> Cale, what's the formula for subtraction?
<5> i know you subtract the nominators instead but forgot if there is anythingels
<0> well, you can easily derive it by knowing that a/b - c/d = a/b + (-c)/d
<0> = (ad + b(-c))/(bd)
<0> = (ad - bc)/(bd)
<2> everyone should turn their keyboard upside down and shake.. and watch carefully for a surprise
<0> Eyelashes and crumbs?
<8> ChrOnX: the vaseline prevents things from falling out
<2> ahahahahahaha
<2> i swear judging by the amount of cat hair in my keyboard that my cat sleeps on my keyboard
<6> she's probably chatting on IRC all night without you realizing it.
<4> lol
<2> haha
<9> cat on the prowl
<4> is there an easier way to show that two groups are isomorphic than going through and checking that phi(ab) = phi(a)phi(b) for every single element a,b?
<5> Cale so i can add the following to my notes a/b - c/d = a/b + (-c)/(bd)
<5> is that fair enough
<0> no, that's not true
<5> eh?
<6> that implies b=1 or c=0. I think.
<6> c*(b-1)=0. yes. :)
<5> what's the exact formula then?
<5> sorry been years since i've done this stuff
<0> a/b - c/d = a/b + (-c)/d is what I wrote
<0> note here that / has precedence over +
<0> so it happens first
<5> nod
<0> In general, a - b = a + (-b)
<0> and -(a/b) = (-a)/b = a/(-b)
<5> what is that term called again inverse operation?
<0> Additive inverse, or negation
<5> [01:11] <0>and -(a/b) = (-a)/b = a/(-b)
<5> ok cool thanks
<0> -a is read "negative a"
<5> ok i almost have it but if you can plug in the values into that formula for me it would clear up so much
<5> 4/5 - 2/3
<3> hardcore
<5> gotta start some where :-)
<3> in your case, efnet
<3> but now you're here, too
<5> I mastered addition
<5> just need help with subtracting.
<5> HiLander, people talk less here than efnet dont feel like wasting everyones time trying to learn basic algebra lol
<3> algebra is a cruel mistress
<5> just so much to cover which is annoying
<5> books make no attempt to explain anything
<0> hehe
<5> My book says "To subtract fractions, the same rules as in adding fraction applies (find the LCD), except you subtract numerators."
<0> Well, at least it would explain things -- might be a little hard on you though.
<5> then goes over to more complex fractions
<5> with no real explinations
<0> Well, in some sense, that's all there is to it
<0> If you're taking the approach of finding a common denominator and then adding
<0> It's really the distributive property, do you know what that is?
<0> For any numbers a,b,c: (a + b) c = (ac + bc)
<0> so, if we have:
<5> i was using that method at first but i found a/b + c/d = (a/d + b/c)/(bd) to be easier and faster
<0> (a/d + b/d)
<0> = (a (1/d) + b (1/d))
<0> = (a + b) (1/d)
<0> = (a + b)/d
<0> Of course, which way these proofs go tend to depend on what your axioms, or basic ***umptions about numbers are.
<5> can you fix that equation i think you forgot c
<5> and it confused me
<0> No, there never was a c
<0> in this case
<0> they were both d's
<5> oh humm..
<0> so what this means is that adding quotients with the same denominator is easy
<0> (easier than if they are different)
<5> yeah that i know already
<5> just add straight across 1/5 + 2/5 = 3/5
<0> same goes for subtracting them
<0> (a/d - b/d) = (a - b)/d
<0> Which is what the sentence from your book says
<0> So if you know how to find common denominators, you can also subtract
<6> and then (a/c + b/d) = ((ad)/(cd) + (cb)/(cd)) = (ad + bc)/cd
<5> so lets work though this problem
<5> 4/5 - 2/3
<5> first i should find the lcd?
<0> you could
<5> or any common dominator?
<0> Or you could just apply the formula
<6> . o O ( = 1/3 - 1/5 )
<5> Cale this one a/b + c/d = (a/d + b/c)/(bd)
<5> ?
<8> rodrickbrown: any common denominator will do, but then you'll have to reduce your final answer if its not the LCD (although you may ahve to reduce it anyway)
<0> rodrickbrown: that's not what the formula is
<0> a/b + c/d = (ad + bc)/(bd)
<8> rodrickbrown: if the denominators have no common factor, then simply multiplying them together will give you the LCD. 3 and 5 have no common factor.
<2> if there were 2 balls in 2d space.. and 1 ball was stationary .. and 1 hit the other what would each of their velocities look like ?
<0> reduction to lowest terms is overrated
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