• Mathematician Chat v. 3.999...
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Never thought I'd be doing another master's degree but just received an offer to do Part III at Cambridge. Will be an interesting experience to say the least.
[QUOTE=agentalexandre;50129984]Never thought I'd be doing another master's degree but just received an offer to do Part III at Cambridge. Will be an interesting experience to say the least.[/QUOTE] Fucking rad. Funded? [editline]14th April 2016[/editline] Not to diminish what is an awesome accomplishment, but I've heard practically anyone can go to Cambridge if they've got piles of cash.
[QUOTE=JohnnyMo1;50130552]Fucking rad. Funded? [editline]14th April 2016[/editline] Not to diminish what is an awesome accomplishment, but I've heard practically anyone can go to Cambridge if they've got piles of cash.[/QUOTE] I applied well after deadlines for funding unfortunately. In fact, my application went in after the actual deadline for the course, it was really a last minute decision - very glad I went through with it though. I would have liked to have been funded but I think I will be able to manage without.
[QUOTE=agentalexandre;50129984]Never thought I'd be doing another master's degree but just received an offer to do Part III at Cambridge. Will be an interesting experience to say the least.[/QUOTE] Nice! Part III was hard work, but I really enjoyed it. [QUOTE=JohnnyMo1;50130552] Not to diminish what is an awesome accomplishment, but I've heard practically anyone can go to Cambridge if they've got piles of cash.[/QUOTE] Where did you hear that? It's certainly not the case for people already at Cambridge (the route I took) - I can't really say how hard it is to get in from the outside though.
[QUOTE=Joey90;50132093]Nice! Part III was hard work, but I really enjoyed it. Where did you hear that? It's certainly not the case for people already at Cambridge (the route I took) - I can't really say how hard it is to get in from the outside though.[/QUOTE] Both online and my undergrad research advisor (he went to Imperial College London). I doubt any of them were talking about from within, though. Typically it's outsiders asking how hard it is to get into Cambridge graduate programs. [editline]14th April 2016[/editline] Now that I think about it, I think that was in reference to international students. Since they pay significantly more it's a big source of money for the university. Didn't notice agentalexandre is in the UK until now so doesn't really apply anyway.
[QUOTE=Joey90;50132093]Nice! Part III was hard work, but I really enjoyed it.[/QUOTE] What kind of courses did you concentrate on? I'm mostly interest in algebraic number theory so I guess I'll be focusing on number theoretic/algebraic courses myself. Also, did you do an essay? I've seen some examples of ones and I'm seriously overwhelmed by the level that's expected in them. [editline]16th April 2016[/editline] [QUOTE=JohnnyMo1;50132582]Both online and my undergrad research advisor (he went to Imperial College London). I doubt any of them were talking about from within, though. Typically it's outsiders asking how hard it is to get into Cambridge graduate programs. [editline]14th April 2016[/editline] Now that I think about it, I think that was in reference to international students. Since they pay significantly more it's a big source of money for the university. Didn't notice agentalexandre is in the UK until now so doesn't really apply anyway.[/QUOTE] From what I've heard from other people who have been to Cambridge, Part III is a huge cash cow for the uni in terms of people applying from the outside. I don't think they lower their standards but I do believe they are a lot more lenient with how many people they let in - for example, I applied after the deadline (and my references were in 2 weeks after the deadline!) and I still got in.
[QUOTE=NixNax123;50128708]calc 3 is definitely easier than calc 2 all around at least until you get to stuff like line integrals which might be challenging[/QUOTE] Line integrals are easy as piss. Even if inside vector field, it's just Integral plus scalar product.
What manifold would a shirt be (neglecting the thickness of the fabric)? I was wondering if there is a mathematical analysis of ironing a shirt and if it is possible to get out all the creases given certain restrictions :v: Couldn't find anything though. Don't know enough topology either... [Editline]16th March[/editline] Found [URL="http://math.stackexchange.com/a/2761/48460"]this[/URL]...
[QUOTE=Number-41;50142793]What manifold would a shirt be (neglecting the thickness of the fabric)? I was wondering if there is a mathematical analysis of ironing a shirt and if it is possible to get out all the creases given certain restrictions :v: Couldn't find anything though. Don't know enough topology either... [Editline]16th March[/editline] Found [URL="http://math.stackexchange.com/a/2761/48460"]this[/URL]...[/QUOTE] It's homeomorphic to a sphere with four punctures. Here's a related clothing-based geometry: [url]https://en.wikipedia.org/wiki/Pair_of_pants_(mathematics)[/url] They're definitely homeomorphic and hence essentially equivalent as topological manifolds and they're almost certainly diffeomorphic as well.
[QUOTE=agentalexandre;50139840]What kind of courses did you concentrate on? I'm mostly interest in algebraic number theory so I guess I'll be focusing on number theoretic/algebraic courses myself. Also, did you do an essay? I've seen some examples of ones and I'm seriously overwhelmed by the level that's expected in them. [/quote] I did all pure stuff as well - favourites were probably Algebraic Topology and Category Theory. Also did things like algebraic number theory, rep theory, groups, elliptic curves, lie algebras. I did do the essay - so long as you get a decent supervisor you're fairly sure to get marks. My title was 'Factorisation Systems' (Category Theory). Mainly fun because it had a whole pile of commutative diagrams [img]https://dl.dropboxusercontent.com/u/4081470/3D%21.png[/img] [QUOTE=agentalexandre;50139840] From what I've heard from other people who have been to Cambridge, Part III is a huge cash cow for the uni in terms of people applying from the outside. I don't think they lower their standards but I do believe they are a lot more lenient with how many people they let in - for example, I applied after the deadline (and my references were in 2 weeks after the deadline!) and I still got in.[/QUOTE] Possibly, I still had lots of friends in College so I didn't hang out in the CMS much - meant I didn't get to know many of the MASt people. I think the MMaths tend to fare slightly better, but perhaps because we're used to the system...
[IMG]http://i.imgur.com/828tVQA.png[/IMG] made a post asking how to figure this out, but i solved it. i was using chebyshev's inequality wrong, forgetting to subtract the mean from both sides of the inner inequality (I was trying to do P(X - 110 >= 140) instead of P(X - 110 >= 30)) [editline]26th April 2016[/editline] but i thought this was a cool problem because now i can impress my friends with worthless stats
Anyone can help me answer a few of these practice problems for my final? [editline]8th May 2016[/editline] [T]http://i.imgur.com/V8NJdsI.jpg[/T]
[QUOTE=cody8295;50283783]Anyone can help me answer a few of these practice problems for my final? [editline]8th May 2016[/editline] [T]http://i.imgur.com/V8NJdsI.jpg[/T][/QUOTE] I suggest you show your work/attempts and ask more specific questions.
[QUOTE=cody8295;50283783]Anyone can help me answer a few of these practice problems for my final? [editline]8th May 2016[/editline] [T]http://i.imgur.com/V8NJdsI.jpg[/T][/QUOTE] (a) must be convergent, because Sum ( n^2/n^4 ) = Sum (1/n^2), which converges. [code] Proof: Sum 0 -> inf (n^2 + 1)/((n^4 + 2) = Sum 2 -> inf (n^2 - 1)/((n^4) = Sum 2 -> inf (n^2)/((n^4) + Sum 2 -> inf (-1)/((n^4) = Sum 2 -> inf (1)/((n^2) + Sum 2 -> inf (-1)/((n^4) = ^ ^ | | Both sums converge [/code] (b) is convergent, because the factorial n! rises faster than 2^n. Not good argument, but it's a start. Too lazy to write a proof
Gonna sit a Dynamics & Vector Calculus exam tomorrow, decides wether I'm in uni next year or out. I can get the majority of vector calc but there's one or two things in these question sheets I don't really understand. for example, this bugger: [t]https://latex.codecogs.com/png.latex?\dpi{120} (\vec{v}\cdot\nabla)\vec{v}\quad\textrm{where}\quad \vec{v}%3D\vec{b}\times\vec{r}[/t] B is a constant vector; R the position most of these fairly small VC questions don't require needing to go too deep into the vector components, but I can't see a way to evalute or try to evalute that dot product otherwise. Is using tensor notation a good idea to approach this question?
snip
[QUOTE=Number-41;50292294]snip[/QUOTE] Surely it's not that simple, are you implying? [t]https://latex.codecogs.com/png.latex?\dpi{150} (\vec{v}\cdot\nabla)\vec{v} %3D \vec{v}\cdot(\nabla\vec{v})\quad\textrm{or}\quad(\nabla\cdot\vec{v})\vec{v}[/t] [editline]just[/editline] ahh, saw that it was a snip. it does look similar to one of the identities when expanded [t]https://latex.codecogs.com/png.latex?%5Cdpi%7B150%7D%20%28%28%5Cvec%7Bb%7D%5Ctimes%5Cvec%7Br%7D%29%5Ccdot%5Cvec%7B%5Cnabla%7D%29%28%5Cvec%7Bb%7D%5Ctimes%5Cvec%7Br%7D%29[/t] but it's missing another dot product for that to be nice. It seems it would be (lol) easy enough to just swap around the internal dot product but I'm pretty sure that the position of the nabla in it implies it acts as an operator on the next value ( placement of d/dx ); so it's position has to be static
[QUOTE=Fourier;50292225](b) is divergent, because the factorial n! rises faster than 2^n. Not good argument, but it's a start. Too lazy to write a proof[/QUOTE] I'm really confused by this post, both because that should imply that it [I]converges[/I] rather than diverging, but also because if that reasoning were good, it would imply that the harmonic series converges. Anyway, for that one, the ratio test should work.
[QUOTE=JohnnyMo1;50293833]I'm really confused by this post, both because that should imply that it [I]converges[/I] rather than diverging, but also because if that reasoning were good, it would imply that the harmonic series converges. Anyway, for that one, the ratio test should work.[/QUOTE] Brain fart sorry, should have written convergent.
[QUOTE=Instant Mix;50292314]Surely it's not that simple, are you implying? [t]https://latex.codecogs.com/png.latex?\dpi{150} (\vec{v}\cdot\nabla)\vec{v} %3D \vec{v}\cdot(\nabla\vec{v})\quad\textrm{or}\quad(\nabla\cdot\vec{v})\vec{v}[/t] [editline]just[/editline] ahh, saw that it was a snip. it does look similar to one of the identities when expanded [t]https://latex.codecogs.com/png.latex?%5Cdpi%7B150%7D%20%28%28%5Cvec%7Bb%7D%5Ctimes%5Cvec%7Br%7D%29%5Ccdot%5Cvec%7B%5Cnabla%7D%29%28%5Cvec%7Bb%7D%5Ctimes%5Cvec%7Br%7D%29[/t] but it's missing another dot product for that to be nice. It seems it would be (lol) easy enough to just swap around the internal dot product but I'm pretty sure that the position of the nabla in it implies it acts as an operator on the next value ( placement of d/dx ); so it's position has to be static[/QUOTE] My guess was that nabla is a linear operator, as well as the dot product so you could just switch around the parentheses but I'm not entirely sure. Can't be assed to work it out with components because that sort of defeats my original intentions of not using components :v:
Nabla is linear operator, since it consists out of derivatives :).
Took the calc2 exam today, wasnt awful
I'm trying to learn limits, and it's been a pretty heavy task, I'm even having trouble proving sequence limits by definition. How do I even prove that the limit of (2n-5)/(2n-7) is equal to 1 using the definition? It seems simple, but I haven't been able to find an example like this on the internet. In most of them, the absolute value can be trivially removed, but you just can't do that here. any help? same wiht the limit of (4(n^4)+2)/(5n^5-6n+1) is equal to 0. sorry if this looks like some kid asking for the answer of his homework, but I've been stuck on these things all day and google isn't helping at all.
So for the first one, you want to show that for all e>0, there exists N such that for n greater than or equal to N, we have: |(2n-5)/(2n-7) - 1| < e. Rewrite this last inequality as: |(2n-5)/(2n-7) - (2n-7)/(2n-7)| < e. Now just simplify: |-12/(2n-7)| < e. Can you see from this form why it satisfies the definition?
The arquimedian propierty, right? I have another question, if there is a convergent sequence (Un), with a limit equal to l, such that there exists an N such that for every e>0, n>n0 -> |Un-l|<e. How can I prove that there exists only a finite number of different terms on the sequence? Isn't this just the definition of limit?
[QUOTE=Cosa8888;50370619]The arquimedian propierty, right? I have another question, if there is a convergent sequence (Un), with a limit equal to l, such that there exists an N such that for every e>0, n>n0 -> |Un-l|<e. How can I prove that there exists only a finite number of different terms on the sequence? Isn't this just the definition of limit?[/QUOTE] Suppose the sequence had infinitely many distinct terms. Then such terms must occur arbitrarily far out in the sequence (else there would only be finitely many). But this is in direct contradiction with the convergence of the sequence since a convergent sequence necessarily converges to its limit point (in this case l) and stays put at that limit point (else it wouldn't be convergent!).
got a semester of real analysis after I finish (hopefully) this calc 3 course, apparently its mega hard, have no idea what it contains
[QUOTE=agentalexandre;50370783]Suppose the sequence had infinitely many distinct terms. Then such terms must occur arbitrarily far out in the sequence (else there would only be finitely many). But this is in direct contradiction with the convergence of the sequence since a convergent sequence necessarily converges to its limit point (in this case l) and stays put at that limit point (else it wouldn't be convergent!).[/QUOTE] Not really, what about 1/n? It converges to 0, but all it's terms are different, no matter how high n is. Is this ok? Define N such that for every e>0, n>N -> |Un-l|<e Then |Un-l| is always less that inf(e>0)=0, because it is a lower bound. Then, |Un-l|<=0. But the term it's always positive or equal to zero, so |Un-l|=0, and that clearly implies that Un=l. With this, I'm saying that there exists an N such that for all n>N, Un=l, and that implies that there exists a finite number of different terms. But why can't I apply this to the normal definition of limit? thereust be something wrong with this proof, in that case.
I'm taking 'Calc 1 + Analytical Geometry' but I haven't taken prereqs such as Precalc or Trigonometry. However, my SAT scores allowed me to hop right in to Calc 1. So far we've gotten into Limits and it's pretty dandy and algebraic, which I understand for the most part. But, I am not sure how difficult things will get since I haven't taken prereqs. Highest math I have taken is College Algebra. Anyone got any advice/warnings for me? [editline]22nd May 2016[/editline] I do feel like I ought to brush up on algebra, cause simplifying fractions has always been hell for me
[QUOTE=Vilous;50372912]I'm taking 'Calc 1 + Analytical Geometry' but I haven't taken prereqs such as Precalc or Trigonometry. However, my SAT scores allowed me to hop right in to Calc 1. So far we've gotten into Limits and it's pretty dandy and algebraic, which I understand for the most part. But, I am not sure how difficult things will get since I haven't taken prereqs. Highest math I have taken is College Algebra. Anyone got any advice/warnings for me? [editline]22nd May 2016[/editline] I do feel like I ought to brush up on algebra, cause simplifying fractions has always been hell for me[/QUOTE] You'll need to know your trig, learn your unit circle also try and get a grasp on various trig identities.
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