Strangers studying maths at uni level

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.

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.

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.)

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

The word you were looking for is 'foreigners'.

The word you were looking for is 'foreigners'.

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

I hear that you'd much rather leave .75 as 3/4 as well.

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.

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.

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.

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

a stranger is defined as somebody you don't know!

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?)

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".

i think most people would leave .75 as 3/4 as 3/4 is more accurate.

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