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Comments:
<0> Catfive: 0.1111111111111111
<1> no
<1> see catfive
<1> 9*0.111111....=0.999999...=1
<1> nicely consistent
<1> so the implication is 0.11111...=1/9
<2> given that it's completely useless for mathematics, the easiest solution is just to drop decimal representation of reals. Baby Rudin mentions it once in a 15 line paragraph IIRC and doesn't need it further.
<1> lieven : i don't think dropping the decimal representation is a good idea
<3> yes, they're entirely overrated, but the non-mathematical seize on them as a 'concrete' way of understanding real numbers, and the rest is misguided history.
<2> doing mathematics for the non-mathematical sounds about as right an idea as doing orchestra theory for the deaf
<3> of course it shouldn't be dropped - it's quite useful for some specific purposes. it just shouldn't be taken as the *primary* means for understanding what a real number is.
<1> lieven : i disagree with that as well
<4> but cale just because i cant write it with a sqrt sign doest mean i shud say rational and irrational it still is qrt(2) ;)
<1> though you don't need it for for doing (theoretical analysis) staying connected to "real" world isn't as mistake
<1> and non mathematicians do "math" all the time for a good reason
<2> kmh: a case could be made that all the real world need are rationals. measurements always end up with a finite precision as a rational number
<1> lieven : yes
<1> lieven : however you can read the ecimal representation as rationals
<2> kmh: that's where most of the confusion begins with
<1> i.e. it it more convenient to notate the decimal representation than the explicit fraction
<1> and that convenience imho is precisely one of the reason we have the decimal representation
<2> and it should have as much impact on *real* mathematics as the 1 1/2 conventions of notation and other stuff practically adopted
<1> Catfive : yes with i agree, it should not be the primary means to understand the reals
<3> you can as easily use the decimal representation for rationals, though (and in fact that's always what you're doing, unless your expansion is infinite)
<1> well the big plus about the decimal representation is they prove an immediate orientation and order comparison
<1> which doesn't matter for most theoretic aspects, but is quite handy for applications/computations
<3> @arrrr
<0> Eat maggoty hardtack, ye unkempt, jenny frequentin', son of a gun.
<5> arrr
<5> hmm quiet
<3> ... too quiet
<5> indeed
<3> did you get determinant with all the hints?
<5> nah, i gave up half way through
<3> aww.
<5> it got quite unappealing fast, to be honest
<5> it's hard to find the motivation at times
<6> actually starting is the biggest obstacle
<5> starting is nothing
<5> GAP is such a pain in the ***
<5> it works, but it's awful
<7> I have a 5 by 1 matrix. I want to complete it to have an orthogonal matrix, is there any command in Mathematica that can do that? (Using Gram-Schmidt or whatever)
<5> well, Gram-Schmidt is in LinearAlgebra`Orthogonalization`
<5> there's a GramSchmidt function
<7> But I think I have to give it a basis. If so, is there a way to complete from one vector to a basis automatically?
<5> i don't quite get what you're asking
<5> you need to consider a space somewhere before you can think about bases
<5> a nonzero vector can be a basis for a 1-dimensional space, for example
<3> is, in fact
<5> catfive: well, {(-1, 1)} is not a basis for the 1-d space spanned by (1, 1)
<7> Let me re-ask.
<3> one ***umes the nonzero vector is in the space, of course
<5> catfive: one has to make that ***umption, of course ;)
<3> not many vectors are parts of bases for spaces that don't contain them =)
<5> IIRC, there is a standard procedure of completing a set of vectors to a basis for a given space
<7> I'm writing an example of Schur decomposition. I have a 5 by 1 matrix, I get an eigenvector of it. Now I need to find an orthogonal basis for R^5 that includes this eigenvector.
<5> then that's easy, IIRC
<7> Yes, but I can not find the commands to do that.
<5> you augment your set with the standard basis vectors for R^5, gaussian eliminate, choose the vectors that correspond to pivot elements in the reduced matrix
<5> i'm rusty on this stuff, but i think that's the right way of doing it
<7> After that I need to orthogonalize(?) the basis.
<5> you have a basis then, use GramSchmidt like i said earlier
<7> That's what I want. I'm looking for a method to do it all automatically using Mathematica. GramSchmidt is great, now I need a way to complete a set to a basis automatically.
<5> i told you the method, would you like me to write the code for you?
<7> Nope. I guess that there's already a command that can do it and none of us has to write any code.
<5> no, there isn't, because the procedure is simple
<7> Ok.
<5> just for kicks
<5> is it not the case that the vector with the first four standard basis vectors for R^5 form a basis?
<8> hi
<8> I'm using chi square to ***ess whether the data I've collected deviates from the expected distribution
<8> but
<8> I want to ***ess whether or not it's deviated in a particular way
<8> I have a number of cl***es of events, and I want to check that one particular cl*** has significantly increased in frequency
<7> dysprosia nope. It's a random eigenvector.
<5> then one doesn't need to do much calculation; it becomes more complex with more than one vector you want to complete
<8> as opposed to saying "there is a difference"
<8> any ideas?
<5> kmd: yeah, that random eigenvector plus the first four standard basis vectors
<7> What if the random eigenvector is {1,0,0,0,0}? it won't form a basis with the first four standard basis vectors.
<5> you're right, you'll need a procedure
<5> i'm half way there anyway
<9> hey
<10> kmd: are you kmh in disguise?
<9> if I had Sum[Sum[P{N=k},k,i,infinity],i,1,infinity] how would I change the order of summation?
<7> Manyfold nope.
<11> Sum[Sum[P{N=k},i,1,k],k,1,Infinity]
<5> the tricky part is selecting the vectors corresponding to the pivot columns
<9> Olathe...it can't be that simple....I'd hate myself if it were that simple
<9> :(
<5> and i want to write this "properly" too
<9> oh...wait...I guess...
<9> ok
<9> thanks
<11> longodj : k >= i is the only thing stopping them from being independent.
<12> hi all!
<12> does "Van der pol" (in my case mu=1) equation have analitic solution?
<5> i've almost finished, except that something isn't working right :(
<7> :_(
<5> ok, i think i've got it
<5> % Take[ Sort[ (Transpose@{Transpose@RowReduce@Transpose@Prepend[IdentityMatrix[5], {1,3,0,0,1}], Prepend[IdentityMatrix[5], {1,3,0,0,1}]}), OrderedQ[{#1[[1]], #2[[1]]}]&, 5]
<0> dysprosia: $Failed
<5> aaaaw
<5> ****y
<5> the final version:
<5> completeBasisR5[v_] := Transpose[Take[Sort[(Transpose@{Transpose@RowReduce@Transpose@Prepend[IdentityMatrix[5], v],Prepend[IdentityMatrix[5], v]}), OrderedQ[{#1[[1]], #2[[1]]}] &], 5]][[1]]
<13> $failed
<5> kmd: that *should* work
<13> not just failed
<5> yeah, because it gives you a long list
<13> failed and you have to pay a fine, i prefer
<5> lol
<13> for wasting mbot's time
<5> oh crap, i don't think that's right
<5> >:(
<13> now the mood.pl face is sad
<3> @keal
<0> 99% of my book has been erased by faulty hdd's
<5> what does keal mean anyway
<13> think happy thoughts
<14> haha
<3> @help keal
<0> keal. Talk like Keal
<13> whoever keal is
<5> oh, it's a someone
<5> >:((
<5> if only Select seemed to work properly, i'd have this
<5> i'll do this the ugly way, i think
<5> the problem is picking those elements corresponding to pivot columns
<5> that's it
<5> i don't know why Select isn't working the way i want
<5> aha, progress
<5> aha!
<7> YEAH!
<5> ok!
<5> this was a fun exercise
<5> completeBasisR5[v_] := Table[First[Select[(Transpose@{Transpose@RowReduce@Transpose@Prepend[IdentityMatrix[5], v],Prepend[IdentityMatrix[5], v]}),#[[1]] == IdentityMatrix[5][[n]] &]][[2]], {n, 1, 5}]
<7> Nice.
<5> it does work?
<7> Apparently :_D
<5> oh thank god
<5> that is an audacious piece of mathematica programming
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