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<doomie> lol
<lolnoyuo> haha
<xerox> Woot :D
<Cale> okay, I should loosen that ban a bit :)
<lolnoyuo> math is a very precise form of philosophy which can be applied to real world situations
<doomie> Mhh, its the first ban ive seen on freenode at all. :)
<c1re> in latex is there a way i can specify when i want a letter to represent a vector and it will make it bold and put an arrow on it?
<Cale> er
<xerox> hah.
<delta_> ouch
<toad-> c1re: \vec{foo}
<Cale> heh
<Cale> just that the other syntax I'm used to seems to be having no effect
<Cale> there
<c1re> ahh thanks a lot toad-
<Cale> for some reason my client is too smart and figured that since a more general ban was in place, it shouldn't set a more specific one :)
<doomie> isnt that the ircd? :)
<Cale> oh, possibly
<toad-> c1re: you might have to redefine it if it doesn't do exactly what you want
<Cale> if it is, then that's really dumb
<Cale> you want to be able to safely tune a ban without having to leave a gap where the user is completely unbanned
<Cale> @math hello
<mbot> Cale: "Good day to you!"
<Cale> sorry for not reacting sooner, I was up from the computer :)
<doomie> well, on uircd, its like if you ban a more general ban, it automatically unbans the more specific one and sets the more general one. so you dont have this gap.. :) dunno really about this ircd.
<Cale> that's kind of bad too, but it's more reasonable than this :)
<doomie> yes :)
<Cale> here it's just silently ignored
<doomie> its weird for a client somehow.
<Cale> I should probably limit @math requests
<doomie> Mhh, only x per minute for "unknows" users perhaps.
<doomie> unknown even.
<Cale> that requires per-user state, which is irritating
<xerox> Indeed.
<doomie> mh, yeah
<Anil> it's ok, he won't be here ever again!
<Cale> he was here before, iirc
<Cale> once
<doomie> and there will be others i guess
<sphex> hi. hey I got a question about boolean optimization: is there a faster way to get the canonical sum of products for a given expression than to test all the value combinations (which really gets tedious after ~20 terms)?
<sphex> c'mon...
<delta_> Prof_Vince, wanadoo is rebooting its servers?
<pmdboi> sphex: have you done anything with k-maps?
<pmdboi> and i take it to mean, if A and B are the two boolean variables in use, and A' indicates the inverse of A, then the canonical form of A + B is A'B + AB + AB'... right?
<pmdboi> what you can do is compute the boolean expression completely in terms of these sums of products
<sphex> pmdboi, yeah, and then I use Quine-McCluskey instead of Karnaugh to optimize
<pmdboi> so, for instance, if you want to figure out the canonical form of A + 'B
<pmdboi> you know that A = AB + A'B
<pmdboi> er... A = AB + AB', rather
<pmdboi> B = AB + A'B
<pmdboi> so A + B is the union of those two sums
<pmdboi> which is AB + AB' + A'B
<pmdboi> something like thiat
<pmdboi> that, even
<pmdboi> the sums could get really long for more variables, but that's one way to do it.
<sphex> hmm
<pmdboi> as another example, (A + B)C = ((ABC + AB'C + ABC' + AB'C') + (ABC + A'BC + ABC' + A'BC'))(ABC + A'BC + AB'C + A'B'C)
<pmdboi> = (ABC + AB'C + ABC' + AB'C' + A'BC + A'BC')(ABC + A'BC + AB'C + A'B'C)
<pmdboi> = (ABC + AB'C + A'BC)
<pmdboi> (that doesn't seem right... hmm)
<sphex> do you do that in your head?
<pmdboi> just think of it like this: the canonical form of a single symbol A is the sum of all the products that have A in the true state, and B, C, ... in the true or false states
<pmdboi> the sum of two canonical forms is the sum of the union of the elements in those forms...
<pmdboi> the product is the intersection of those elements
<pmdboi> and the boolean inverse is the complement.
<pmdboi> so if we're working with variables A and B...
<pmdboi> if foo = A'B + AB', then ~foo is AB + A'B'
<pmdboi> where ~ is boolean negation
<pmdboi> i'm not really sure that it's the best way, though
<pmdboi> but, yeah
<pmdboi> i hope that helps.
<sphex> yeah thanks a lot. guess I'm.. going to think about that. :p
<pmdboi> haha
<pmdboi> is there a specific example that you want to work through?
<pmdboi> gah, crap, i have to head out
<pmdboi> anyway, i hope that helped
<sphex> d'oh!
<c1re> how can i manually number an equation in latex?
<rbauld> waaaaaaaazaaaaa
<sphex> WAAAAAAaaaaaaaaaZZZZZAAAAAaaa?!
<rbauld> linux is sweet....
<sphex> newsflash!!
<rbauld> that jabberwalkike guy is cool....like really
<rbauld> :P
<rbauld> not that were one and the same...
<sphex> dam I wish I wasn't such a code monkey and knew some math
<teferi> so start learning
<rbauld> wishing never did ****
<sphex> yeah but I get so bored every time I see a greek letter
<lolnoyuo> use roman letters
<wig> cos x (-8/17), pi/2 < x < pi i have to find sin 2x, cos 2x, and tan 2x. the only problem is, sin 2x comes out to be negative?
<lolnoyuo> do you mean cos(x) = (-8/17) ?
<_llll1> @math Sin[2*(-8/17)] //N
<mbot> _llll1: -0.8082514096461221
<_llll_> not much of a problem ;-)
<de1337> Dsentinel: <<NumberTheory`NumberTheoryFunctions`
<Dsentinel> Null
<de1337> Dsentinel: PrimitiveRoot[2]
<Dsentinel> 1
<de1337> _llll_: about yesterday there you go, its 1
<_llll_> clearly primitive root does not mean what i thought it did. so what is it?
<de1337> PrimitiveRoot[n] gives a cyclic generator of the multiplicative group (mod n), when n is a prime power or twice a prime power
<_llll_> oh, so C_2 is {0,1} generated by the non-identity. i thought it was to do with field extensions
<de1337> Dsentinel: <<DiscreteMath`Permutations`
<Dsentinel> Null
<_llll_> so we now have 3 mathbots in here...
<teferi> perhaps that's a bit excessive
<teferi> de1337: is that yours? i hope you disabled all of the local filesystem access stuff and such
<de1337> teferi: yes all thats needed, and no you arent supposed to use this one, im mostly testing it and in difference to mbot it supports loading of modules/packages
<de1337> @mbot: PrimitiveRoot[2]
<mbot> de1337: PrimitiveRoot[2]
<teferi> de1337: there's a reason mbot doesn't persist the environment across expressions
<de1337> tereri: it dont allow loading of them just because it starts a new instance for every evaluation so it wouldnt be saved
<teferi> it doesn't persist the environment so some jerk can't ruin it for everyone else by redefining useful functions
<de1337> yep
<de1337> Dsentinel: <<NumberTheory`NumberTheoryFunctions`
<Dsentinel> Null
<u> irc://efnet
<Astinus> c/win 35
<Astinus> Ugh, sorry
<Manyfold> cool i burned myself :)
<Adam-> har
<Adam-> holy **** there is a maths room
<slava> if A and B are two linear maps, i want to find v such that <Av,Bv>=0 (<.,.> is the inner product)
<slava> is there any technique for doing this?
<Adam-> without the 2 it would be like 1/3 e^3x
<Adam-> ...
<Adam-> differentiate 2e^3x what do i do with the 2?
<slava> hang on... <Av,Bv>=0 iff <v,A*Bv> = 0
<slava> does that help?
<Adam-> i can solve if it doesnt have the 2
<Adam-> but not sure if it does
<Adam-> :(
<slava> well, <Tv,v>=0 iff a certain quadratic form is zero
<slava> hmm
<codejunky> @math 1+1
<mbot> codejunky: 2
<ichor> Adam-, you just keep the 2 there. (2 F(x))' = 2 F'(x)
<ichor> @math Derivative[2*Exp[3*x]]
<mbot> ichor: Derivative[2*E^(3*x)]
<ichor> @math help
<mbot> ichor: "See http://documents.wolfram.com/v5/ for detailed Mathematica help."
<ichor> ichor: Derivative[2*E^(3*x), x]
<ichor> @math Derivative[2*Exp[3*x],x]
<mbot> ichor: Derivative[2*E^(3*x), x]
<Adam-> :/
<ichor> oh, well. The aswer should 6e^(3x) by the chain rule.
<Adam-> cheers
<cjohnson> @math (2 * E^(3 * #)&)'[x]
<mbot> cjohnson: 6*E^(3*x)
<Adam-> :)
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