I think he's talking a load of nonsense. And I think GE is talking a load of sense. And they are both products of the UK system.(Original post by paronomase)
Yes, of course, but the mathematics he posted are based on what he learned. This is why I already said that I did not criticise the UK system. But what is YOUR opinion on this? For example, on the fact that e^(x+y)=(e^x)(e^y) is a definition...cuz I do not know what is his level of studies, and if you went to Cambridge you may have some detachment.
Ramanujan went to Cambridge (not as an undergrad, and you could say calling him a student is a cheat  he certainly wasn't a student in the usual sense) and was eventually elected a fellow. And yes, he was a genius. But that's the point: geniuses (and neargeniuses) tend to be the most rewarding people to teach. I don't think his prior education had much to do with that.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...
Incidentally, here's a quote from Littlewood on Ramanujan:
"He had no interest in rigour, which for that matter is not of firstrate importance in analysis beyond the undergraduate stage. [...] The cleancut ideal of what is meant by a proof [...] he perhaps did not possess at all"
Which brings us back to the start of the thread. Being incredibly 'rigourous' when everyone can see the points concerned are true doesn't actually show better ability or understanding.

DFranklin
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 31012010 12:16

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 31012010 12:19
(Original post by Y__)
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?
One of the questions I have about ODEs is about the existence and uniqueness of solutions. It's long been drummed into our heads that an nthorder ODE (with no initial conditions) has n linearly independent solutions  but surely there are some conditions here. What happens when the conditions are not satisfied? Is it possible to have more than n? I know we have the PicardLindelöf theorem which tells us that for nice enough 1storder ODEs with initial conditions, there is a unique solution; it follows that nice nthorder ODEs with enough initial conditions have a unique solution. 
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 31012010 12:28
Zhen: Are you talking about linear ODE's? If you're talking about nonlinear ones, I don't think it's true.
E.g. (dy/dx)^2 = y^2 sin^2 x. At each x=n \pi, you get to "choose" whether you will take dy/dx = y sin x or dy/dx = y sin x for the interval [n\pi, (n+1)\pi] , so there are an infinite number of lin. indep. solutions. (I think). 
Trevor 12345
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 31012010 16:40
(Original post by DFranklin)
Well, I'm fairly sure I know what you're getting at, and even if you don't explicitly use eigenfunctions, it's pretty closely linked.
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.
I quite simply stated that a first principle approach was there without the need to reference eigenfunctions.
Needless to say, most Alevel students will not know about Jordan normal form, but they will know enough (fp2 students) to approach the problem without making assumptions as to the nature of the solution. The point I made was that in my opinion it is a shame that such approaches are ommitted from many school tests, and replaced with a sytem of jumping through hoops.
I have no particular desire to argue with you, I simply find you too arrogant to bother with,
thanks though. 
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 31012010 17:09
(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.
(a^x)(a^y)=(e^xlog(a))(e^ylog(a) )=e^((x+y)log(a))=a^(x+y)
This requires that e^(x+y)=(e^x)(e^y), which can be proved from the definition of e as the evaluation of a power series. (Clearly it also requires that we define a log function and prove some facts about it.) 
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 31012010 19:28
(Original post by SimonM)
Well, actually, yes, it is made up of definitions. You don't necessarily have to learn them by heart. But there are accepted axioms and definitions of mathematical objects that we can't prove. It may not be just made up of definitions, but if Simplicity had given a correct example of something which is a definition, then.... 
Simplicity
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 31012010 20:01
(Original post by harr)
I would prove the statement a^(x+y)=(a^x)(a^y) for a=/=0 as follows:
(a^x)(a^y)=(e^xlog(a))(e^ylog(a) )=e^((x+y)log(a))=a^(x+y)
This requires that e^(x+y)=(e^x)(e^y), which can be proved from the definition of e as the evaluation of a power series. (Clearly it also requires that we define a log function and prove some facts about it.)
Even, then I can't see a big problem with taking it as a definition. Its well defined.
Hmm, on defining log functions,that would be just moving stuff back a couple of steps.
(Original post by DFranklin)
I think he's talking a load of nonsense. And I think GE is talking a load of sense. And they are both products of the UK system.
(Original post by Scallym)
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.
To be fair, the proofs in A level are stupid and so its sort of pointless learning it. Even, then truth tables and what not can't really be taught.
(Original post by paronomase)
don't even see the point of your remark. I've looked at your link, and...A b) "What are the domain and range"...isn't it what I've done? wait...c) "Provide an approximate sketch" Here, if you do not mention that the function is not defined or/and not derivable at some point, then you will sketch the wrong graph.
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?
(Original post by paronomase)
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 20:43
(Original post by Simplicity)
Hmm, I was under the impression that a^(x=y)=(a^x)(a^y) is a definition, from a book on Abstract algebra from Rotman and the fact that in Peter Eccles book that I have been using all term it has all the exponential laws defined inductively. Okay, I stand corrected.
Even, then I can't see a big problem with taking it as a definition. Its well defined.
Hmm, on defining log functions,that would be just moving stuff back a couple of steps.
Outside of an Analysis course it seems to me to make sense to just give it as a definition, but it is important that we know that the definition works for noninteger exponents (for integer exponents the result can follow from defining a^n as the product of n lots of a). 
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 31012010 21:13
(Original post by Simplicity)
I'm at a disadvantage as I haven't studied analysis at all, but lol. 
DFranklin
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 31012010 21:27
(Original post by harr)
It's possible that there are other ways to show that it makes sense, but I don't know what they'd be. You need to know how to define a^x, and the only way I've seen that done is by defining e^x as a power series, defining a log function and then letting a^x=e^alog(x). 
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 31012010 21:27
(Original post by Simplicity)
I doubt it. Certainly, finding points at which there is singularity and we know about domain and co domain, however not what the object actually is i.e. a set. In a way, we would only say let u=blah and dv=blah, if the choice matters as in integration by parts it does.
I can promise that they would have trouble studying in France. Being less verbose is easy, being more verbose is quite hard (I did not use the term "rigorous", otherwise all of you would have jumped at me but u see what I mean)
But then the debate will never end, cuz I'm afraid that there are very few people on TSR that went to study in prépas after doing their Alevels...
Concerning the e^(x+y)=(e^x)(e^y), you wouldn't use the fact that a^(x+y)=(a^x)(a^y)
In fact, you would use the first to prove the second, and you would use the definition of exp to show the first (that is exp'=exp and exp(0)=1)
Plus you're in a first year undergrad in Maths, so you should normally be more advanced... I would be interested to see how "rigorously" you could prove basic maths problems, for example that lim e^x/(x^n) = ? in as x tends to + inf, or that e is irrational, or, even more interesting, if you could prove me that the method of solving by induction works.
But then it's not the purpose of the thread.... 
Simplicity
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 31012010 21:41
(Original post by generalebriety)
Then, with all due respect, may I recommend that you stop talking about anything analytical until you have studied it?
(Original post by paronomase)
Yeah of course, you can doubt it. My mistake was to exagerate the example, and I agree that when you're at uni, you wouldn't bother to write all this (though in my opinion you would still say that you calculate f'(x) for x>0, and x=/=1, which would still need a little bit of work beforehand)
I can promise that they would have trouble studying in France. Being less verbose is easy, being more verbose is quite hard (I did not use the term "rigorous", otherwise all of you would have jumped at me but u see what I mean)
But then the debate will never end, cuz I'm afraid that there are very few people on TSR that went to study in prépas after doing their Alevels...
Concerning the e^(x+y)=(e^x)(e^y), you wouldn't use the fact that a^(x+y)=(a^x)(a^y)
In fact, you would use the first to prove the second, and you would use the definition of exp to show the first (that is exp'=exp and exp(0)=1)
Plus you're in a first year undergrad in Maths, so you should normally be more advanced... I would be interested to see how "rigorously" you could prove basic maths problems, for example that lim e^x/(x^n) = ? in as x tends to + inf, or that e is irrational, or, even more interesting, if you could prove me that the method of solving by induction works.
But then it's not the purpose of the thread....
I doubt it again.
I haven't done any analysis. Hmm, but I have done sets, numbers and functions. You can't prove induction its more of a definition you get from the construction of the integers. Hmm, in Rotmans book he deduces induction from the fact that a set of postive integers has a smallest element, so thats basically where induction comes from. 
Trevor 12345
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 31012010 21:51
The second order method, accessible to Alevel students is in the attached pdf ..... I have left a few gaps because I have done enough typing for one day.

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 31012010 21:56
(Original post by paronomase)
I can promise that they would have trouble studying in France. Being less verbose is easy, being more verbose is quite hard (I did not use the term "rigorous", otherwise all of you would have jumped at me but u see what I mean)
First bit in bold: Who are you referring to?
Second bit in bold: Sorry, but that's just not true. Being more rigorous is hard (eg when you're calculating limits you would sometimes skip more involved arguments and details during A levels whereas you won't be allowed to in a serious uni degree), being more verbose is certainly not: The example you give in your first post differs from your socalled A levels method only by mentioning that you have to work on the positive reals without 1 so as to avoid a zero denominator and have a sensible definition of ln  you just expand it into multiple lines of arguably unnecessary waffle to make it look like a lot.
In terms of personal experience, I can only talk about the Cambridge course, but I can assure you that the pure maths courses tend to do everything rigorously. It might be slightly less detailed than you're used to, but don't be tricked into thinking that your method is necessarily better (I'm having this kind of discussion once a week with a friend who gets mad at lecturers for omitting some detail that one could easily prove on one's own)  today's British mathematicians are no better or worse than today's French or German mathematicians even though the courses might have a different style. 
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 31012010 22:14
(Original post by Y__)
First bit in bold: Who are you referring to?
Second bit in bold: Sorry, but that's just not true. Being more rigorous is hard (eg when you're calculating limits you would sometimes skip more involved arguments and details during A levels whereas you won't be allowed to in a serious uni degree), being more verbose is certainly not:
Oh, and I was referring to Alevels student that study in French prépas (I do not know the level in maths in other countries such as Russia, Germany, Japan, China...)
For Simplicity:
Yes, there is a proof of the method of induction. And the fact that the set N has a smallest element is not enough...Consider a set that has a smallest element, but that has also a biggest element. (do you say largest element?)
Let "n" be this biggest element, P the proposition that you are trying to prove, and E the set for which you are trying to prove P. P(n) is true, because n belongs to E, however P(n+1) is not true, because n+1 does not belong to E. Does that mean that P is not true for all x of E? No...As you can see, it's not as intuitive and easy
Yeah ok you're right, you could derive on R, on C, on Z, on N...on any set for which it works. It's just that USUALLY you tend to do it for real numbers...especially if you're doing applied maths. 
Simplicity
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 31012010 22:46
(Original post by paronomase)
For Simplicity:
Yes, there is a proof of the method of induction. And the fact that the set N has a smallest element is not enough...Consider a set that has a smallest element, but that has also a biggest element. (do you say largest element?)
Let "n" be this biggest element, P the proposition that you are trying to prove, and E the set for which you are trying to prove P. P(n) is true, because n belongs to E, however P(n+1) is not true, because n+1 does not belong to E. Does that mean that P is not true for all x of E? No...As you can see, it's not as intuitive and easy
Yeah ok you're right, you could derive on R, on C, on Z, on N...on any set for which it works. It's just that USUALLY you tend to do it for real numbers...especially if you're doing applied maths.
Actually, in applied maths complex numbers are improtant. So lol, it would be wrong to assume that you work in reals because of applied maths, as that is wrong. Hmm, note Penrose actually said that real numbers don't appear in nature that much its actually complex numbers that are more useful.Last edited by Simplicity; 31012010 at 22:52. 
generalebriety
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 31012010 22:49
(Original post by paronomase)
Yes, there is a proof of the method of induction. And the fact that the set N has a smallest element is not enough...Consider a set that has a smallest element, but that has also a biggest element. (do you say largest element?)
And indeed, induction works on any wellordered set, regardless of whether it has a greatest element or not, with the added stipulation "don't go outside the set". This doesn't include Q, R or C (but could include Z for a good choice of ordering). 
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 31012010 22:54
(Original post by generalebriety)
Actually, we say "least element" and "greatest element". "Small" and "large" normally refer to absolute value (e.g. 500000 is larger than 5).
And indeed, induction works on any wellordered set, regardless of whether it has a greatest element or not, with the added stipulation "don't go outside the set". This doesn't include Q, R or C (but could include Z for a good choice of ordering). 
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 31012010 23:00
(Original post by Simplicity)
Hmm, I was under the impression that its best to see induction as saying it has a smallest element in the set and the set is ordered. 
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 31012010 23:14
(Original post by Simplicity)
How would I logically know if something is analytical or a definition if I haven't studied anything analytical?
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