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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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