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Comments:
<0> hey i never got into like higher alebgra or trig or whatever but how can i solve a square root?
<1> newtons method
<0> newtons method?
<0> i was horrible at math >.<
<1> http://en.wikipedia.org/wiki/Newton's_method
<0> >.<
<1> http://en.wikipedia.org/wiki/Methods_of_computing_square_roots
<0> ill never learn that stuff :( why did they have to make math so complicated
<2> so it works
<3> haha
<3> that was possibly the best response ever.
<0> haha
<3> virusdotnet the only thing that makes mathematics complicated is the notation at lower levels, which you seem to be at. just get a good grasp of the symbolism and you'll be fine.
<0> why didn't i just stay in school and payed attention >.<
<4> contrariwise, it's notation that makes mathematics manageable
<3> oh absolutely
<4> try to solve even a simple problem with words only
<3> otherwise there would be no way to express such complex ideas in such concise ways
<4> Cajori's history of mathematical notation is a wonderful book showing this process
<3> eh, i don't know, i do a lot of proofs word-only.
<3> but proofs are sort of expository by nature, so it lends itself to word-only processes.
<0> lemme ask a simple question. if i have a 45% of $145.00 how can I do that? do i divide or subtract?
<3> what does "percent" mean?
<4> 145*45/100
<3> yeah, 45 per 100 is what it means
<0> .45
<3> right.
<4> is mbot alive ?
<3> mbot was rendered obsolete apparently from what i've heard.
<1> :O
<4> % 145*0.45//N
<4> poor bot! Alas, I knew him well!
<3> yeah i know
<3> it took me like 3 hours of trying to get it to solve a PDE before someone told me it was dead
<3> man can PDEs be a bitch.
<0> PDE?
<1> partial differential equation
<3> partial differential equation
<4> there isn't much systematic theory on how to solve them
<3> well, depends on the pde
<3> a lot of the time, a surprisingly large amount of the time actually, you can just factor the operators.
<3> some are insane though.
<3> i've been doing a bit of research on embedded solitons, and some insane PDEs come into play with that
<3> without ansatzen i'd be totally lost.
<3> i am sort of considering abandoning this research though since i've come across a really interesting doubly stochastic measure project one of my professors is doing
<3> measure theory is insanely beautiful
<5> hi
<5> what is the best latex editor this days?
<6> texshell
<5> texshell is better than winedt or texnic?
<7> vi is good. or emacs, if you're twisted
<6> yes it is better than everything
<6> simplicity is good
<5> that's a good point
<7> emacs is evil, but that's for another flamewar...er i mean discussion
<8> texniccenter if you're using windows
<5> thank you
<5> I will install texniccenter
<8> it's only an editor though. you need to install a latex thingy first. but then texniccenter will configure it for you
<1> in the set X of all 2x2 matrices with determinant -1, will the determinant of the product of any s in S also be -1?
<1> s/X/S/
<6> im serious tho
<7> det AB = det A det B iirc
<7> so no
<1> ok, thankyou very much
<5> I installed miktex
<8> that works
<6> everytime i try the fancy editors they wind up way more of a pain in the *** than they're worth
<5> I used to latex a few years ago so I still remember that you have to do it
<5> I used winedt then
<5> but I guess that there are better programs today
<1> can you consider 0 a polynomial if you're considering constant polynomials with arbitrary integer coefficients?
<9> What is a constant polynomial?
<1> i'm ***uming it's a constant
<1> any integer
<1> it looks like the answer is yes, but 0 as a polynomial sounds strange
<9> The polynomial with all cofficients equal to zero is 0! Isn't it?
<1> ok then
<1> thanks
<9> :)
<10> % 0!
<10> mbot gone :(
<1> r.i.p
<10> how could he left me?
<1> maybe your standards were to high
<11> 0 is a polynomial
<11> 0! = 1
<1> i gathered the exclamation mark was more to do with the silly nature of my question
<11> (while 0 is a polynomial, most constant polynomials are said to have degree 0, while the 0 polynomial traditionally either has no degree or its degree is -infinity)
<10> HiLander: one permutation of the empty set?
<11> yes, one function from it into itself.
<11> 0! and 0^0 are both 1.
<11> it saves a lot of time
<7> the polynomials wouldn't be a ring if 0 weren't a polynomial
<1> 0^0 is undefined isn't it?
<11> no, it's usually defined as 1
<1> i thought the only definition of 0^0 was x^x as x->0
<11> (0! is *always* defined as 1)
<1> ya, i know about the factorial
<7> hilander: it is undefined, we commonly define it as 1
<11> dysprosia: i hope you realize how stupid that sounds.
<1> lol
<11> 0^0 = 1 is a very natural definition that shows up in a lot of places
<7> that's just it
<7> it's a *definition*
<11> if it has a definition
<11> then it's *defined*
<7> you can just as easily define it as 0
<11> yes, and you can define 7^0 as 0, too
<7> it has no ``natural'' definition
<11> but that just screws other calculations up
<11> 0^0 = 1 is a very natural result when considering the entropy of random variables
<7> i'm not disagreeing that it's a natural definition
<11> yes you are: <7> it has no ``natural'' definition
<7> yeah words can have different connotations you know
<11> one of us is very confused, and it's not I.
<7> who are you talking about? it's not me either
<11> you're arguing that 0^0 = 1 is not defined and a definition at the same time.
<11> superpositional logic doesn't really work in the non-quantum world.
<7> hilander: no, no i'm not
<10> HiLander: unfortunatly our universe is a quantum world :)
<7> 0^0 does not have an immediately obvious definition unless you make a definition one way or the other
<11> i'd disagree about the "immediately obvious" part, but your statement is so general as to be almost useless
<9> http://www.google.com/search?q=0%5E0
<9> Google says! ;-)
<11> the 0th power of 0 makes no more or less sense than defining the 0th power of 7 or of pi.
<7> hilander: like iodine said, the natural way of determining such a value by limits doesn't work
<11> i never claimed that it was a unique way to approach it: x^x tends to 1 while 0^x tends to 0 as x->0 from the right. but which one actually comes up in applications? the limit of x^x, not of 0^x.
<7> hilander: you seemed to be saying that 0^0 is and must be 1 always, which isn't the case. one can take a stand one way or the other and say 0^0 := 1 for our purposes, but it's a consistent definition as any other
<11> sorry, i never claimed that there was a unique, unambiguous limit that people of the form 0^0. that's why it's an indeterminate form when it does show up. that doesn't mean that 0^0 doesn't equal 1.
<7> that's what i've been getting at
<7> i think we've been confusing our equals and our is defined as
<11> my use of an equals sign instead of an ***ignment operator is not to suggest that it is not a definition, if that's been the problem.
<11> but since such a tiny minority of mathematicians don't subscribe to the 0^0 = 1 definition, it seems fair to call it an accepted definition.
<7> sure
<7> i've never disagreed with that
<1> a vector is like a 1 x n matrix, and you can only take the inverse of an n x n matrix, therefore there is no inverse of a vector (unless it's size is 1 of course) ?
<12> i think 0^0 = 0
<11> you think wrong.
<1> i'm not sure how to define the inverse of x in R^3
<13> iodine: I'm not sure what you're asking.
<1> how do you compute the inverse of some variable x in the set R^3 ?
<1> or is there no such thing
<12> any x in R^3 has an additive inverse -x, is that what you are asking about?
<11> i like to divide by vectors.
<10> xx^-1 = 1
<12> you have to explain what you mean by inverse, there is no obvious multiplication
<12> of course, additive inverse makes sense, as i already saida
<1> hm
<1> ok, i think you may be right
<1> well to prove that R^3 is a group i'm required to show that for every g in G there is an element g^-1 in G such that g * g^-1 = g^-1 * g = 1
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