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Strangers studying maths at uni level watch

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    WHAT THE HELL ARE YOU TALKING ABOUT?! confussinggggggggg

    seriously?! a stranger is defined as somebody you don't know! do you mean how do people who have no or little experience in maths find studying maths at uni?!
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    (Original post by generalebriety)
    For what it's worth (and I'm a third year undergrad), I still don't know the answers to questions like that. I always suspected that questions like this required some very specialised theory to answer on anything more than a superficial "look, substitute it in, it works" kind of way.
    Hi generalebriety

    I do tend to do this proof with students, and it really requires no more than a knowledge of the roots of a quadratic and the idea of using substitutions to reduce a 2nd order DE to a first order DE and then using an Integrating factor. Such approaches can add a lot to the topic and add a little depth to the topic. But I kid you not when I say the proof of the nature of all roots of 2nd order DE's are really very straightforwards.
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    (Original post by generalebriety)
    For what it's worth (and I'm a third year undergrad), I still don't know the answers to questions like that. I always suspected that questions like this required some very specialised theory to answer on anything more than a superficial "look, substitute it in, it works" kind of way.
    That depends on what you are willing to accept as an answer I suppose. If we're talking a moral* answer (as to, say, why it's linear), then I can't really offer any sensible comment. However, "detuning" and "variation of parameters" can offer some insight here:

    Suppose our equation is:

    y'' -2y'+y = 0 (The first DE I could think of)

    Doing the usual: Try e^{\lambda x} only gives us one linearly independent solution, so consider:

    y''-2y'+(1-\epsilon^2)=0

    Then, trying e^{\lambda x} gives:

    \lambda = 1+\epsilon, 1-\epsilon

    So y = e^{x} ( Ae^{\epsilon x} - B e^{-\epsilon x})

    Then using a Taylor expansion, we may write:

    y = e^x((A+B) + \epsilon x(A-B) + O(A\epsilon^2, B \epsilon^2))

    Choosing A and B appropriately, we can make the limit as \epsilon \to 0 such that A+B \to a, \epsilon(A-B) \to b so that we obtain the result we were looking for.


    * by which I mean the analytical equivalent to abstract nonsense
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    (Original post by generalebriety)
    Not bad at all, but all you've done there is the superficial work. Can you even tell me what a limit is (rigorously)? Can you tell me why taking the limit in two different ways (e.g. along two different sequences converging to the same point) doesn't give you two different answers (and hence why writing "u'(a)" even makes sense)? Can you use all of your machinery above to prove that the derivative of e^x is e^x?

    (I'm not necessarily doubting you can, by the way, but I'd be very surprised if you could, since it takes most university students a full lecture course in analysis to even get these basic facts down. If you can, and you're a typical baccalauréat student, the English education system has a lot to learn from you.)

    Haha well at least thank you for the compliment. Yes, I think I can do what you mentioned here, even though it might be less rigorous or not as good as you would do it. If you really doubt of my capacities (lol it is a good thing to do though), then maybe I should tell you that I was 10th of a national maths competition here in France, I was 14th in the French National Olympiads, and I was selected to participate in the International Olympiads (about 6 kids per country...however I did not get a place, but I was part of the 25 last)

    So if you really insist, I willl show you how I do these demonstrations. It is not extremely difficult, most of it we do it in class.

    In addition, I am not saying that the French system is better. However, I know that many people (the last one was the Mathematics Tutor of Jesus College, Oxford) think that the French are some of the best maths students (you may know the dreadful "prépas" we have here...and Polytechnique?)

    Thanks for your answer though, I appreciate.
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    (Original post by anandv05)
    WHAT THE HELL ARE YOU TALKING ABOUT?! confussinggggggggg

    seriously?! a stranger is defined as somebody you don't know! do you mean how do people who have no or little experience in maths find studying maths at uni?!
    Yeah sorry I was not very clear. By "strangers", I mean people that did not studied in the UK (pre-univ level) Please do not see a political/racial dimension to this, it is just too long and confusing to write "people who did not studied in the UK before uni, and now studying maths at uni in the UK"

    I'm sure you understand now
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    (Original post by paronomase)
    Yeah sorry I was not very clear. By "strangers", I mean people that did not studied in the UK (pre-univ level) Please do not see a political/racial dimension to this, it is just too long and confusing to write "people who did not studied in the UK before uni, and now studying maths at uni in the UK"

    I'm sure you understand now
    The word you were looking for is 'foreigners'.
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    (Original post by Kolya)
    The word you were looking for is 'foreigners'.
    Or "international students" or "overseas students" would probably be more accurate
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    (Original post by Kolya)
    The word you were looking for is 'foreigners'.
    Excuse me, I am myself a "foreigner", and do not speak a perfect English. Yet I am sure that you can understand the purpose of my thread, even though it mentioned "strangers" instead of "foreigners".
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    (Original post by SimonM)
    That depends on what you are willing to accept as an answer I suppose. If we're talking a moral* answer (as to, say, why it's linear), then I can't really offer any sensible comment. However, "detuning" and "variation of parameters" can offer some insight here:

    Suppose our equation is:

    y'' -2y'+y = 0 (The first DE I could think of)

    Doing the usual: Try e^{\lambda x} only gives us one linearly independent solution, so consider:

    y''-2y'+(1-\epsilon^2)=0

    Then, trying e^{\lambda x} gives:

    \lambda = 1+\epsilon, 1-\epsilon

    So y = e^{x} ( Ae^{\epsilon x} - B e^{\epsilon x})

    Then using a Taylor expansion, we may write:

    y = e^x((A+B) + \epsilon x(A-B) + O(A\epsilon^2, B \epsilon^2))

    Choosing A and B appropriately, we can make the limit as \epsilon \to 0 such that A+B \to a, \epsilon(A-B) \to b so that we obtain the result we were looking for.


    * by which I mean the analytical equivalent to abstract nonsense
    That provides insight, yes, but I don't like it. Why should the solutions behave in a nice way, so that you can actually argue that talking about the limit is sensible?
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    (Original post by Oharad)
    I hear that you'd much rather leave .75 as 3/4 as well.
    i think most people would leave .75 as 3/4 . as 3/4 is more accurate.
    only in final answers would i consider gettind rid of fractions
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    (Original post by SimonM)
    That depends on what you are willing to accept as an answer I suppose. If we're talking a moral* answer (as to, say, why it's linear), then I can't really offer any sensible comment. However, "detuning" and "variation of parameters" can offer some insight here:

    Suppose our equation is:

    y'' -2y'+y = 0 (The first DE I could think of)

    Doing the usual: Try e^{\lambda x} only gives us one linearly independent solution, so consider:

    y''-2y'+(1-\epsilon^2)=0

    Then, trying e^{\lambda x} gives:

    \lambda = 1+\epsilon, 1-\epsilon

    So y = e^{x} ( Ae^{\epsilon x} - B e^{\epsilon x})

    Then using a Taylor expansion, we may write:

    y = e^x((A+B) + \epsilon x(A-B) + O(A\epsilon^2, B \epsilon^2))

    Choosing A and B appropriately, we can make the limit as \epsilon \to 0 such that A+B \to a, \epsilon(A-B) \to b so that we obtain the result we were looking for.


    * by which I mean the analytical equivalent to abstract nonsense
    Unfortunately you are assuming the nature of the root rather than deriving it.
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    (Original post by Mrm.)
    Unfortunately you are assuming the nature of the root rather than deriving it.
    I don't follow I'm afraid
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    (Original post by SimonM)
    I don't follow I'm afraid

    I mean that the solution works because the assumption is made that e^{\lambda x}

    As opposed to an approach from first principles.
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    (Original post by Mrm.)
    I mean that the solution works because the assumption is made that e^{\lambda x}

    As opposed to an approach from first principles.
    You mean "why should a solution of the form" e^{\lambda x} exist?

    I suppose the best reason for that is that e^{\lambda x} is an eigenfunction for \displaystyle \frac{d}{dx}. If you want the reason why that should be true, then we have to start thinking about how we define e^x. The power series definition of e^x is equivalent to defining it in terms of solutions to a certain differential equation (which makes it's appearance as an eigenfunction "obvious".
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    (Original post by SimonM)
    You mean "why should a solution of the form" e^{\lambda x} exist?

    I suppose the best reason for that is that e^{\lambda x} is an eigenfunction for \displaystyle \frac{d}{dx}
    No, that isn't what I mean.
    What I mean is that if you approach the problem from 1st principles then the fact that the said exponential function leads to the roots comes directly from the working in much the same way as how one would derive the general integrating factor.
    No need for eigenfunctions here.
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    (Original post by Mrm.)
    No, that isn't what I mean.
    What I mean is that if you approach the problem from 1st principles then the fact that the said exponential function leads to the roots comes directly from the working in much the same way as how one would derive the general integrating factor.
    No need for eigenfunctions here.
    I have absolutely no idea about what you're talking about now. Would you like to give your own answer, rather than leaving us all to guess at it
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    (Original post by SimonM)
    I have absolutely no idea about what you're talking about now. Would you like to give your own answer, rather than leaving us all to guess at it
    Well what I meant was solving a DE without making assumptions as to the nature of the solutions.
    As for my solution, well TBH my Latex skills are not so good, so it would take me rather a long time to type up. I should add that the point I made on this DE topic was just as an example as the unfortunate lack of depth in modern day texts.
    Time permitting I will make a .pdf on the way I would approach this later.
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    (Original post by anandv05)
    a stranger is defined as somebody you don't know!
    The OP is French. With no knowledge of French, you can conclude that this was probably a linguistic mistake. With a small knowledge of French, you would realise that the word for "stranger" and "foreigner" in French is the same.
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    (Original post by paronomase)
    Haha well at least thank you for the compliment. Yes, I think I can do what you mentioned here, even though it might be less rigorous or not as good as you would do it. If you really doubt of my capacities (lol it is a good thing to do though), then maybe I should tell you that I was 10th of a national maths competition here in France, I was 14th in the French National Olympiads, and I was selected to participate in the International Olympiads (about 6 kids per country...however I did not get a place, but I was part of the 25 last)

    So if you really insist, I willl show you how I do these demonstrations. It is not extremely difficult, most of it we do it in class.
    No, I don't really want you to show me this, I believe you if you say you can - particularly given that you seem to be a very good student.

    (Original post by paronomase)
    In addition, I am not saying that the French system is better. However, I know that many people (the last one was the Mathematics Tutor of Jesus College, Oxford) think that the French are some of the best maths students (you may know the dreadful "prépas" we have here...and Polytechnique?)
    I do vaguely know about these, yes. It makes sense. The English education system isn't particularly good, so I wouldn't be offended if you said the French system was better - I'd probably agree...

    (Original post by paronomase)
    Excuse me, I am myself a "foreigner", and do not speak a perfect English. Yet I am sure that you can understand the purpose of my thread, even though it mentioned "strangers" instead of "foreigners".
    As a linguistic aside: there is no word in English that means both "foreign" and "strange", so there is no relation between the two words in our minds, and without knowing a small amount of French, there is no more reason to assume that the word "stranger" actually meant "foreigner" than that it actually meant "potato". The idea of a word that means both is very foreign (and, independently, very strange ) to us. But yeah, this isn't really relevant, just something you might like to know.
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    (Original post by didgeridoo12uk)
    i think most people would leave .75 as 3/4 as 3/4 is more accurate.
    I think you will find it is equally accurate but I agree that there is no advantage to be gained in converting \frac{3}{4} to 0.75. I wonder why Oharad (the person who raised this issue) believes a decimal is preferable?

    (Original post by Oharad)
    ...
    Your view Oharad?
 
 
 
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