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Comments:

<0> TRWBW, I'm trying to split this: (x+1)/(x^2 + x)(x^2 -x - 1)
<1> I'm a little worried, this problem took me a LONG time. But I suppose practice makes perfect :X
<2> jin: okay, so doing it this way, you need to work back to get the ap+bq=1
<0> yea
<0> but when I do the reverse thing, I don't get the right answer
<2> jin: good luck, your method is sound, but the devil is in not making mistakes
<0> I got -1 = (1+2x)(x^2 - x - 1) + (3 - 2x)(x^2 + x)
<0> any way to verify this with mbot ?
<0> mbot: Simplify[(1+2x)(x^2 - x - 1) + (3 - 2x)(x^2 + x)]
<3> heh
<0> mbot: %Simplify[(1+2x)(x^2 - x - 1) + (3 - 2x)(x^2 + x)]
<3> Prepend it with a '% ' instead of 'mbot: '
<0> mbot, not responding
<3> Be sure there's a space
<3> % 1+1
<4> :P



<0> mbot: % Simplify[(1+2x)(x^2 - x - 1) + (3 - 2x)(x^2 + x)]
<3> Or maybe it's dead
<3> One way or another, it won't respond with an 'mbot: ' in front
<0> can any one verify it in mathematica?
<5> I get -1 in maple
<5> jin:
<0> what command did u use?
<5> simplify((1+2*x)*(x^2 - x - 1) + (3 - 2*x)*(x^2 + x));
<0> so P is (1+2x) en Q is (3 - 2x)
<0> now multiply with (x+1)
<1> Another question: If I have an integral with the form ?(c/x)dx where c is a constant, should I immediately do c?(1/x)dx?
<6> Squall`: yes
<6> I usually directly go c*ln|x| + C
<1> excelblue: yeah, but I'm just starting Calc 2 so I want to be extra verbose in all my steps :)
<6> but in a multi-step problem, moving the constants to the front is almost always the thing to do
<6> if you want to be extra verbose, that is a step to take
<1> haha, thanks
<6> though it probably won't be the best thing to do in this case: integral(2x dx)
<6> I usually do it when the constants make the problem look more complicated than it actually is
<0> somiaj, can you simplify this? -(1+3*x+2*x^2)(x^2 - x - 1) -(3+x-2*x^2)(x^2 + x)
<0> -(1+3*x+2*x^2)(x^2 - x - 1) -(3+x-2*x^2)(x^2 + x)
<5> jin: x+1
<7> lmao
<0> ah, so my answer is correct
<0> damnit, I have spent more than 2 hours on this
<0> wasted 2 hours
<7> jin: Which cl***?
<5> time is ment to be wasted, (:
<0> the answer **** says -1-x and x-1 are the answer
<0> so there are more than 1 answer posible
<7> WTF?
<7> No.
<0> Capso, Intro to some basic math
<7> Simplification leads to only one possible answer.
<7> Evaluation may lead to more.
<0> N,Z,Q,R ring and group etc
<0> since P and Q are in Q[x], my answer is right, I think, it satisfy the equation (x+1)/(x^2 + x)(x^2 -x - 1) = P/(x^2 + x) + q/(x^2 - x - 1)
<0> what a waste :(
<2> jin: once you've done it a few times, it's better to write a program to do it
<7> TRWBW: lmao
<7> Is "subduplicate" widely accepted mathematical term?
<2> Capso: hell if i know
<0> TRWBW, I wish I can use that program during exam
<1> Damn, I thought there was some free trial thing for mathematica
<0> Squall`, get it from uni
<1> University?
<0> some how maple 10 is not simplifying the polynomial I entered
<0> yea
<1> okay
<2> jin: of course you could have just cancelled out the (x+1) term. ;) ;)
<7> jin: Ahem... writing the program would be a good exercise.
<0> Capso, you mean a program in C++ alike?
<2> jin: sure, but no need to get fancy. C++, perl, whatever
<0> it's a good idea, I need some more c+= training anyway
<2> jin: perl i actually kinda nice for things like this, it's got a lot of builtin array stuff so you can get started fast
<0> TRWBW, nah, C++ can do it alot faster and I don't know perl atm
<0> I have not used it for years now
<2> jin: no doubt, c++ code would run faster. if i was coding it, i'd probably do it in ansi c actually, no objects.
<0> TRWBW, yea, but I do rather chose a language I already know , at least the basic of it
<2> jin: but after you did the work, do you see that (x+1)/(x^2 + x)(x^2 -x - 1)=1/x*(x^2-x-1)=-1/x+(x-1)/(x^2-x-1)
<7> jin: C++ if you need many functions to accomplish your task... C if you can do it in a small amount of functions -- this has been my basic rule, though often abused in bias of C. ;P



<2> jin, just wondering if you see the easy way to do this particular problem
<0> TRWBW, what can I do what that?
<2> jin, what was the original problem again
<0> I have to split it using GCD
<0> otherwise I lose points
<2> jin, fine, you can fix that up later
<2> jin, once you get the answer you can turn it into gcd form if you like
<0> (x+1)/(x^2 + x)(x^2 -x - 1) = P/(x^2 + x) + q/(x^2 - x - 1)
<2> jin, okay, so factor out the x+1
<0> why should P and Q have degree < 2?
<2> jin: you can work bacwards from the euclidean algorithm taking a degree off on each step, so n steps, then working back adding a degree
<7> jin: I've never encountered anything stating it involved ring theory...
<7> jin: Though it seems like basic algebra.
<0> Capso, I said the course is, not this problem
<0> it's one of the problems we got,
<2> jin, just trying to show you a trick for your exam, if not interested, i admire your pluck
<7> Is it just a term I've never heard before? Or is there more to it?
<0> TRWBW, that's exactly what I did
<8> hey guys, this is a HW question, but it's just methodology. when i am drawing direction fields, if y' is a fuction of t and y, do I need to draw a different slope at each point on my grid? e.g., when it's just y it seems pretty easy
<2> jin: in this case you don't need to use the gcd agorithm, you can solve it more easily and use that solution to reconstruct the ap+bq=1.
<0> I get ya
<0> but I still don't understand why P and Q must have degree < 2
<2> jin: when you divide you knock off a degree at each step, then when you go backwards you add one, look at the pattern
<9> Does anyone know if maxima is available on windows?
<9> nvm found a copy
<10> if I wanted to study cryptography. would I major in math or math and computer science?
<11> math
<11> with extra work in CS
<10> okay
<10> the problem is: I want to major in philosophy too
<11> Why?
<10> I enjoy it
<11> Books are cheap, courses are not
<12> double major
<11> fol, If you enjoy it and you think that you would enjoy majoring in philosophy, then by all means go ahead
<10> hmm
<10> I can get a double major in cs/phil and do pure math papers
<11> If you have any doubts, keep in mind that one philosophy course costs more than 20 books on philosophy
<10> but I have to drop out of multivariate calculus and linear algebra
<10> hmm, good point
<11> Personally, I am just taking cl***es that will help me get my degree. I still study all sorts of things outside of coursework, and it has the advantage that it is pretty much free, at my own leisure, and I get to pick the topics :)
<12> me too, kilimanjaro
<10> couldn't I learn all the mathematics from books too though?
<13> hi
<11> You can learn pretty much anything from books. But some things are easier :)
<10> hey Snootr
<11> I am trying to teach myself math, and that's why I ask so many stupid questions in here.
<10> okay
<10> what are you majoring in?
<11> Math :)
<13> does anybody knows stewart's book on calculus?
<11> Snootr, yes, I took a course from it
<10> well my plan originally was to major in math but it seems computer science has better career options
<13> I've been seeing it on Amazon and it seems to be a very popular book in the U.S.
<12> i have stewarts, 5th d
<13> Here (in Spain) nobody knows that book
<12> it is popular at my school
<12> snootr
<13> Is a good book?
<11> Snootr, it's not bad, but it's not really good either. If you can get it cheap, then by all means go ahead
<12> yes. (but it's the only calculus book i've used so far, so don't take my word for it)
<13> abasinisvacant, did you used it at school? At high-school?
<12> my university
<13> So isn't it a college level book?
<11> yea
<13> Ah, OK.
<12> i suppose so
<14> In sweden you are paid to study. And the courses are free :)
<11> It's designed for advanced highschool, intro university
<12> really? wow
<13> MizardX`, I read that on a magazine. That's fine.
<12> i'd love to get paid to study
<15> except for the fact that everyone pays for it their whole life
<5> Snootr: its a great reference book but I wouldn't suggest it for indvidual learnring, there are far better books to acually learn from (from a point of view of a pure mathematician)


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