Strangers studying maths at uni level

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generalebriety

Well, perhaps I'm being unfair by expecting first year maths undergraduates to have some idea that the study of functions like e^x is analytical in nature, but...

leave my brother alone..in maths he will kick your ass any day

Reply 81

Zhen Lin

DFranklin

Zhen: Are you talking about linear ODE's? If you're talking about non-linear 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).

Yeah, I guess I mean linear ODEs.

Reply 82

paronomase

Simplicity

Lol, wrong. Hmm, I can't be bothered to go into the full details. http://en.wikipedia.org/wiki/Well-ordering_principle is logically the same as induction. So all you need to do to show induction is that its well ordered i.e. contains a smallest element http://en.wikipedia.org/wiki/Mathematical_induction#Proof_or_reformulation_of_mathematical_induction

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.

Yeah, giving me the link to wikipedia doesn't really help. You would know that the N set does not have a greatest element...that is why you can assume that n+1 exists for all n...Or maybe it's just that you never saw inductions on sets that were not N (or Z)..and in this case I cannot do anything for you. Oh, but wait, it's maybe a definition, and then it would be useless to try to question it.

For what Penrose said, well...ok...good for you :wink:

But I hope that you do not think that I'm going to be convinced because one man said something, and that you vaguely quoted it...and if you prefer to work with complex numbers, then just do it, I never said that there was any problem with doing so. On my side, I very rarely see problems of physics (such as mechanics, magnetism, electricity, etc.) with complex solutions but then you could say that I do not know much about it (and I could answer "how the **** can you be first year undergrad and know nothing about analysis & stuff...)

You are also quite arrogant, and I would not be if I were you. You cannot really boast about anything here.

Reply 83

Y__

paronomase

This. As is said, I consider it to be a problem of rigour, not of being verbose or not (I used the term verbose to refer to what some people here think my reasoning is about). Therefore, as you said, being more rigorous is hard. And in many cases (exclude this one), A-levels student are not rigorous enough in my opinion.

Oh, and I was referring to A-levels student that study in French prépas (I do not know the level in maths in other countries such as Russia, Germany, Japan, China...)

Hmmm. How many A levels students do you know that are studying maths in prepas? It would seem a strange move for a student wishing to do maths to move to France after A levels for academic reasons. (I'm not saying French universities are bad, but they are not considerably better than UK ones as such - and there's always the language barrier!) Hence I would suggest that your sample is not representative.

I would also argue that those A level students who are genuinely interested in maths and in pursuing a maths degree at university are more than capable to close the gaps in the material they're presented and make up for the lack of rigour, so it doesn't really matter in the end.

Finally, please don't think that the average aspiring UK mathematician is in any way like Simplicity... they're not.

Reply 84

Oh I Really Don't Care

DFranklin

You can also define a^x for rational x in a fairly algebraic way (you probably need a bit of analysis as well); if you want a^x to be continuous then defining it on the rationals is enough to define it on the irrationals as well (if x is irrational, take a rational sequence x_n tending to x). There are quite a few nasty details along the way I believe - I've never seen it actually done.

Rudin highlights a method that isn't that bad;

If a = m/n = p/q show (a^m)^1/n = (a^p)^1/q and then corresponding results about addition.

Define B(x) = {b^t | t <= x} show when r is rational b^r = supB(r) and then define a^x = SupB(x) for all x.

Reply 85

DFranklin

Dean: I don't think that's too different from what I was saying: the main nasty detail I see is showing that your B(x) is continuous.

Reply 86

Simplicity

Y__

Finally, please don't think that the average aspiring UK mathematician is in any way like Simplicity... they're not.

Hmm, lol I'm not that bad at maths, honest. To be fair, I'm getting good marks so far.

Reply 87

Simplicity

generalebriety

Well, perhaps I'm being unfair by expecting first year maths undergraduates to have some idea that the study of functions like e^x is analytical in nature, but...

I only know what I have been taught. I will know in a couple of weeks.

Reply 88

Simplicity

paronomase

Actually, I have seen induction done on a lot of things. Okay, it needs to be well ordered and you can deduce it from that. Actually, you can take induction as an axiom, hence the wikipedia page. But, yeah I don't see the problem with taking induction as a basic axiom instead of deriving it. Seems sort of pointless.

Lol, Penrose. Do you know who he is? Anyway, I agree with that generally. Hmm, QM uses complex numbers and it wouldn't make no sense without it. http://answers.yahoo.com/question/index?qid=20080614195653AAlB6sX Generally first anwser. But, yeah QM, look it up.

Actually, I'm not arrogant. Certainly, if I'm wrong I will admit I'm wrong. Like I did on that definition which wasen't a definition after all. Its just I don't usually agree with anyone

Reply 89

generalebriety

Simplicity

I only know what I have been taught.

Clearly the ethos at Manchester is different to what I had expected. Alright.

Simplicity

But, yeah I don't see the problem with taking induction as a basic axiom instead of deriving it.

Because you are exactly the sort of person who'd be in favour of taking everything as an axiom, and never seeing how they connected up. If you took induction as a basic axiom, you'd have to prove the well-ordering principle from it, which (to me) is much more obvious than induction.

Reply 90

Scallym

generalebriety

Clearly the ethos at Manchester is different to what I had expected. Alright.

Don't say that, that's my first choice. :frown:

(On a serious note, am I misguided in thinking manchester is reasonably respected for maths?)

Anyway, would this not class as a proof that (e^x)(e^y)= e^(x+y):

Defining

e^{x}

such that

\frac{d}{dx}(e^{x})=e^{x}

Assuming the chain rule to be true.

Let

z=e^{x}e^{y}

\frac{dz}{dx} = e^{x}e^{y}\frac{dy}{dx} + e^{x}e^{y}

\frac{dz}{dx} = e^{x}e^{y}(\frac{dy}{dx}+1)

\frac{dz}{dx} = z(\frac{dy}{dx}+1)

z = e^{\int \frac{dy}{dx}+1 dx}

z = Ae^{x+y}

e^{x}e^{y} = Ae^{x+y}

Let

x=0

and

y=0

e^{0}e^{0}=Ae^{0}

1=A

Therefore,

e^{x}e^{y}=e^{x+y}

(May be talking ********, corrections if that's not valid would be appreciated. :smile:

)

Reply 91

Simplicity

generalebriety

Clearly the ethos at Manchester is different to what I had expected. Alright.

Because you are exactly the sort of person who'd be in favour of taking everything as an axiom, and never seeing how they connected up. If you took induction as a basic axiom, you'd have to prove the well-ordering principle from it, which (to me) is much more obvious than induction.

Hmm, the uni doesn't focus on analysis in the first year intill the second term. Is that so strange?

Hmmm, but still generally it is taken as an axiom. Certainly, the book I use takes it as an axiom. Isn't that the axiom of choice? well ordering principle?

P.S. I don't see the problem with taking some stuff as definitions or axioms. But, lol. In a pedantic way you could break axioms up into other things to deduce them, then you can do the same ad infinity. It would be pointless if the thing is obvious, or equivelent to something that is obvious i.e. well ordering.

Reply 92

DFranklin

Scallym

Don't say that, that's my first choice. :frown:

(On a serious note, am I misguided in thinking manchester is reasonably respected for maths?)

Anyway, would this not class as a proof that (e^x)(e^y)= e^(x+y):

Defining

e^{x}

such that

\frac{d}{dx}(e^{x})=e^{x}

This allows something like e^x = 3 exp(x) (where exp(x) is the function we all know and love).

(That objection is easy to understand but also easy to work around. A bigger question: why do you think such a function

e^{x}

exists? Why do you think it is unique?)

Incidentally, if you ignore those issues, it is somewhat cleaner to show that

\displaystyle \frac{d}{dx} \frac{e^x e^y}{e^{x+y}} = 0

Reply 93

Kolya

Simplicity

Hmmm, but still generally it is taken as an axiom. Certainly, the book I use takes it as an axiom. Isn't that the axiom of choice? well ordering principle?

I don't know this stuff so I could be wrong, but I think the confusion is coming from the use of different foundations. If we use PA, then induction is an axiom and the WOP follows. If we use ZFC, both induction and the WOT are theorems (with the WOT following from the AoC).

Reply 94

paronomase

Simplicity

Lol, Penrose. Do you know who he is? Anyway, I agree with that generally. Hmm, QM uses complex numbers and it wouldn't make no sense without it. http://answers.yahoo.com/question/index?qid=20080614195653AAlB6sX Generally first anwser. But, yeah QM, look it up.

Yes I know who he is, I read a lot of his pubblications (mainly the ones with S. Hawking cuz I particularly like their theory about black holes) but then does it make him more legimitate? If I didn't know him, should I agree with you? I can't really understand the difference btw knowing him and not knowing him...

And I know him because, unlike you, I do not only know what I've been teached..."

Simplicity

I only know what I have been taught. I will know in a couple of weeks.

If everyone was like you, then science wouldn't have evolved since the Babylonians...how sad.

And again, your proof of e^(x+y) may be valid, but a proof always depends on what you take for granted at the beginning (that is exp'=exp, or exp is the reciprocal bijection of ln(x), etc.)

If you are curious about knowing my method (which I really doubt...) then here it is
Let h(x) = (e^(x+y))/(e^y), Dh=R because e^x>0 for all x of R
Therefore h is differentiable on R, and h'(x)=[(e^x+y)(e^y)]/[(e^y)²]
= [(e^x+y)]/(e^y)
= h(x)

In addition, h(0)=e^y/e^y = 1

Since h'=h and h(0)=1, h=e^x
therefore, e^x=(e^(x+y))/e^y

However, you could still say that my proof is incorrect, or at least partially correct. I just chose to take for granted that e^x is the only function so that f'=f and f(0)=1

hhmmmmm

paronomase

However, you could still say that my proof is incorrect, or at least partially correct. I just chose to take for granted that e^x is the only function so that f'=f and f(0)=1

Which is why arguing about rigour at a pre-university level is largely pointless.

At university you will prove this without having to make assumptions. (You will typically do some other slightly different dodgy things, but still...)

Reply 97

Scallym

DFranklin

e^{x}

exists? Why do you think it is unique?)

Incidentally, if you ignore those issues, it is somewhat cleaner to show that

\displaystyle \frac{d}{dx} \frac{e^x e^y}{e^{x+y}} = 0

Eeesh, silly me. :ashamed:

So defining e as the real number such that

\frac{d}{dx}(e^{x}) = e^{x}

would make that valid? (Since the function is then unique?)
I'm not doing wonders for the rigour of the UK system here. :p:

(Just ran through the method you posted, is somewhat nicer.)

Reply 98

DFranklin

Scallym (and others):

The way this is normally done at university is essentially: define a function exp(x) (usually as a power series, though there are other methods). Show that exp'(x) = exp(x) and exp(0) = 1. From there you can differentiate exp(x)/exp(x+y) and so on.

Note that we *don't* say that exp(x) = e^x. One big reason for this is that it's actually very hard to define what a^x is if x is not rational, let alone if x is complex.

So we have this function exp(x) that behaves very much how we think e^x should behave, and then we define e^x = exp(x), and more generally a^x = exp(x log a). Of course, what we need to do at this point is justify that this really does make sense and do what we think it should, and it's this that most university courses gloss over a bit. (Partly because it's not obvious how heavily you should justify it).

Now, from some of what paranomose was saying, I was taking it that he (and by extension people replying to him) were going in the same direction as the university route, but being a little sloppy in talking about e^x instead of exp(x). That is, e^x is "some function" with nice derivative properties, but NOT necessarily the "raise e to the power of x" function. (Because if we knew it was, then it's automatic that e^x e^y = e^(x+y), pretty much by definition of the power function. (Some faffing with passing from x,y rational to irrational in the limit if you want to be hugely pedantic)).

In which case any multiple y=A exp(x) satisfies dy/dx = y, so you don't know that exp(0) = 1 unless you prove it, which was my objection.

Note this has nothing to do with the value of 'e' itself. If you have y = A a^x and you want dy/dx = y, you're forced to have a = e, there's no room for choosing something else. (But A can be anything).