Why would you use substitution in such an easy equation? It might be lazy but in our exams we have a limited amount of time so we have to choose the fastest way. However yes, the french baccalaureate is very rigorous compared to Alevel or IB but doesn't matter, in the UK you will probably get a good education (similar to the one you'll get in France probably). Maybe if you wanna study hardcore maths you should apply somewhere like Oxbridge or maybe to a politechnique in France. Perhaps you would study more there... Who knows(Original post by paronomase)
Hey,
I consider studying maths at uni, and I have few questions about the courses. Basically, I heard that in maths, at uni, you did not demonstrate anything or prove anything in particular during your course. Here, I mean specific demonstrations, not just calculations or reasoning. I'll try to make myself clearer by giving you an example in a second.
I also have a French friend at Imperial, and he says the same. I wonder how do strangers in general feel about it, or at least people who, during their preuni teaching, used to be very rigorous in maths.
I do not mean that Alevels students are not rigorous, but look at this example.
Q1: Differentiate f(x)=(3x)/[ln(x)]
So, according to what I saw, and what I've been told, here is the Alevels method;
f'(x)=[3.ln(x)3x/x]/(ln(x)²)
= 3(ln(x)1)/(lnx)²
Now, here is how we have to do it in France (and other countries)
f(x) is a quotient of two functions u(x)=3x and v(x)=ln(x)
u(x)=3x, Du=R and v(x)=ln(x), Dv=R+* (Du is the domain of u)
Now, we need v(x)=/=0 because it is the denominator, therefore ln(x)=/=0 so x=/=1
Therefore, Df=R+*/{1}
Now, u(x) is derivable on R as linear function, and v(x) is derivable for all x of Dv, that is for all x>0
By quotient, f(x) is derivable for all x>0 and x different of 1
Now, for all x of ]0,1[u]1,+inf[, f is derivable, and we have
f'(x)=[3ln(x)3x/x]/(lnx)²
=3(lnx1)/(lnx)²
Of course, I exagerated a bit, but still in France we have to solve it this way. therefore, I wondered how uni was in terms of maths demonstrations.
Another example would be the demonstration that e^(X+Y)=(e^X)(e^Y) that I could post the French way if you wanted...
x
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andreavivan
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 30012010 18:07
Last edited by andreavivan; 30012010 at 18:12. 
DFranklin
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 30012010 19:06
(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.
Writing D for d/dx, we wish to solve (D^2 +bD + c)y = 0.
Factorise in the usual way: (Ds)(Dt)y = 0 (where s,t are the roots)
Write g for (Dt)y: then (Ds)g = 0, and so g = Ae^sx.
So now we want to solve (Dt)y = g. That is, (Dt)y = Ae^sx.
We can rewrite as D(e^t y) = A e^(st)x (basically using the IF e^t).
And then integrating gives two solutions, including the (A+Bx)e^s solution if t = s.
The similarities with eigenvalues, and in particular Jordan normal form for repeated eigenvalues are pretty glaring, in my opinion. 
DFranklin
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 30012010 19:11
(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?) 
Simplicity
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 30012010 19:39
(Original post by paronomase)
Hey,
I consider studying maths at uni, and I have few questions about the courses. Basically, I heard that in maths, at uni, you did not demonstrate anything or prove anything in particular during your course. Here, I mean specific demonstrations, not just calculations or reasoning. I'll try to make myself clearer by giving you an example in a second.
I also have a French friend at Imperial, and he says the same. I wonder how do strangers in general feel about it, or at least people who, during their preuni teaching, used to be very rigorous in maths.
I do not mean that Alevels students are not rigorous, but look at this example.
Q1: Differentiate f(x)=(3x)/[ln(x)]
So, according to what I saw, and what I've been told, here is the Alevels method;
f'(x)=[3.ln(x)3x/x]/(ln(x)²)
= 3(ln(x)1)/(lnx)²
Now, here is how we have to do it in France (and other countries)
f(x) is a quotient of two functions u(x)=3x and v(x)=ln(x)
u(x)=3x, Du=R and v(x)=ln(x), Dv=R+* (Du is the domain of u)
Now, we need v(x)=/=0 because it is the denominator, therefore ln(x)=/=0 so x=/=1
Therefore, Df=R+*/{1}
Now, u(x) is derivable on R as linear function, and v(x) is derivable for all x of Dv, that is for all x>0
By quotient, f(x) is derivable for all x>0 and x different of 1
Now, for all x of ]0,1[u]1,+inf[, f is derivable, and we have
f'(x)=[3ln(x)3x/x]/(lnx)²
=3(lnx1)/(lnx)²
Of course, I exagerated a bit, but still in France we have to solve it this way. therefore, I wondered how uni was in terms of maths demonstrations.
Another example would be the demonstration that e^(X+Y)=(e^X)(e^Y) that I could post the French way if you wanted...
The point is if you did everything so called French way then it would be such a pain, certainly I doubt you would be able to do half the paper in the link if you did it that way(I highly doubt you would get more marks as most of the above seems like waffle). As lol that seems worse that A levels.
P.S. Not, like that at all. Certainly, in a proof in pure maths you assume stuff and you prove what is asked, not go off at some wild tangent.
P.P.S. e^(x+y)=e^x e^y, is a definition. 
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 30012010 19:46
(Original post by maxfire)
They are both sound methods. The A Level method is fairly objective, and lets you solve the problem relatively easily. Tbh, that's what maths is  and objective subject, and product differentiation should really be a fairly trivial process in respect to other parts of maths.
The French method requires you to have a really good understanding of why you get that answer, though it is fairly longwinded and unnecessary, but I can see why it is taught.
OP, your post is quite confusing... in A level, there isn't just one way to integrate/differentiate... there are couple and at the end of the day, maths is the same in france/uk/rest of the world. 
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 30012010 20:35
(Original post by Simplicity)
P.P.S. e^(x+y)=e^x e^y, is a definition.
In any sensible course in real analysis, you define e^x as exp(x) where the latter is defined either directly in terms of the power series or as a solution to the differential equation df/dx = f . The identity e^(x+y)=e^x e^y then follows from results about multiplication of power series. 
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 30012010 21:03
In mathematics you don't understand the things, you just get used to them. <3

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 30012010 21:10
(Original post by SomeStudent)
Your sig is disturbing, I've seen it about five times today
OP, your post is quite confusing... in A level, there isn't just one way to integrate/differentiate... there are couple and at the end of the day, maths is the same in france/uk/rest of the world. 
Simplicity
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 30012010 21:16
(Original post by Y__)
Just no.
In any sensible course in real analysis, you define e^x as exp(x) where the latter is defined either directly in terms of the power series or as a solution to the differential equation df/dx = f . The identity e^(x+y)=e^x e^y then follows from results about multiplication of power series.
For any real a that is non zero, then a^(x+y)=a^x a^y. Again, then let a=e.
Hmm, try to be bourbaki i.e. e^(x+y)=e^x e^y is just a special case of the more general definition.
P.S. Note, he said not defining e so why the hell would you off on a tangent to define e, when the question itself is not to define e. 
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 30012010 21:21
(Original post by Simplicity)
I didn't mean e itself, note e isn't special in this case.
For any real a that is non zero, then a^(x+y)=a^x a^y. Again, then let a=e.
Hmm, try to be bourbaki i.e. e^(x+y)=e^x e^y is just a special case of the more general definition.
P.S. Note, he said not defining e so why the hell would you off on a tangent to define e, when the question itself is not to define e.
Let me just stress once more for everyone who might read this thread that the bit in bold is not a definition but can be deduced as soon as you know how you define a^x for any real x and nonnegative real a. 
DFranklin
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 30012010 21:25
What would we do without Simplicity to defend the honour of the UK maths education system.

Oh I Really Don't Care
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 30012010 22:05
love ya simp

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 31012010 02:10
In my experience, the amount of proof that is expected of you at ALevel depends on your teacher.
Last year, my maths teacher would never prove or derive anything, she simply showed us how to use it and expected that we learnt this method.
She's not left, and my new teacher is a pure maths PhD who seems to think it's a crime against humanity to teach the method rather than the derivation, which i get on with very well. There are occasions when he has to omit something that would take up too much time to prove from first principles, but generally stuff gets proved. 
paronomase
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 31012010 10:52
(Original post by DFranklin)
Don't forget that the students who wish to go abroad to study tend not to be exactly typical. *Most* of the students I knew from the continent were on their country's IMO squads. It's not surprising they'll tend to be some of the better students.
(Original post by Simplicity)
Why the hell would you give the point at which it doesn't work...e^x+y=e^xe^y is a definition
Therefore, you are doing exactly what I've done, except that you do not apply it when doing a simple calculus problem. What is the point of learning what is "range" and what is "derivability" if you do not apply them?
furthermore, if (e^x+y)=(e^x)(e^y) is a definition...then maths is just made of definition you have to learn by heart? I don't think so...The proof is simple (using calculus for ex) but requires thought and reflection, what maths is really about in the first place.
And for a^(x+y)=(a^x)(a^y)..., would you know 1) how to prove it, 2) why it doesn't work for a=0 ? Do you know why it works for x=0 and y=0, provided a is non zero (or should I say could you prove it) ??
(Original post by generalebriety)
... 
DFranklin
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 31012010 11:14
(Original post by paronomase)
Yeah that's true...even though I'm not sure it really makes a difference. Because a good and a bad student, if they come from the same country, they will be equally representative of this country's system.
I don't even see the point of your remark.Thanks for your post, it does help You're one of the few honest guys in this thread lol 
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 31012010 11:25
(Original post by DFranklin)
But being "one of the best students" is reasonably orthogonal to how you were taught preuniversity  you are going to learn so much more maths at university that what you were taught before doesn't matter that much.
Note that Simplicity and GE are both representatives of the UK system...
And in my opinion what you learned before does matter...well this thread is just a proof of it. People have different approaches of maths, different ways of solving problems (as Somestudent said) and therefore many "foreigners" will have to adapt to the UK system if they did not learn maths the same way during their preuni teaching.
At least I'm sure that if Alevels come to study in France (for example in prépas, or at Polytechnique), they will have a lot of problems of adaptation, even for basic problems. 
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 31012010 11:40
(Original post by paronomase)
Haha well you tricked me there...well almost. Indeed, as you said, Simplicity defends the UK system, whereas GE admits that it is maybe not as excellent as it seems.
And in my opinion what you learned before does matter...well this thread is just a proof of it.
When I went to Cambridge, you'd certainly find people from other countries (or 'highend' private schools) who had covered far more material than the average undergrad. And in the first year, that could make quite a difference. By the 3rd year, not so much.
It's a cliched example, but Ramanujan is probably the "best" student the UK has seen over the last 100 years, and he his mathematical education left a *lot* to be desired (particularly if you're a stickler for rigour). 
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 31012010 11:56
(Original post by DFranklin)
But your disagreement with Simplicity's posting is not that he's defending the UK system, it's the actual mathematics he's posted.
For the rest of your post, I agree that since you go there to learn, it does not really matter what you've already learned.
Ramanujan is not really a good example though, he learned maths with books (no teachers) and did not go to uni (correct me if I'm wrong) and he's more a genius than anything else really... 
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 31012010 12:00
Ok, he's a first year at Manchester...but it doesn't change much.

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 31012010 12:13
(Original post by paronomase)
...then maths is just made of definition you have to learn by heart? I don't think so...
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