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A Summer of Maths (ASoM) 2016

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Avatar for Zacken
Zacken
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Original post by Gregorius
Oh what an idiot :colondollar:

Edited. I blame senility.


Was half-hoping that there was some deep theorem that would make everything into integers. :lol:
Original post by Gregorius
If zwˉ=zˉw=zwˉ‾ z \bar{w} = \bar{z} w = \overline{z \bar{w}} then zwˉ∈R z \bar{w} \in \mathbb{R}. Hence z=k/wˉ z = k/\bar{w} for a real k. Can you finish it off?


ahhh yes thank you
Original post by Zacken
Was half-hoping that there was some deep theorem that would make everything into integers. :lol:


It would make life so much simpler. Even better if there were only finitely many of them too...
Original post by Zacken
Z\mathbb{Z}? :curious:


Well R,Yeh cause Im=0
Y ?Is it wrong?


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Anyone have any videos to recommend for learning Green's theorem/Stoke's theorem?
Original post by Student403
Anyone have any videos to recommend for learning Green's theorem/Stoke's theorem?


MIT YouTube channel might have something on it?


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Original post by Student403
Anyone have any videos to recommend for learning Green's theorem/Stoke's theorem?


There are lectures on the MIT OpenCourseware for a large body of multivar and vector calculus concepts which are very good, and explain each concept in a lucid intuitive terms.
Original post by Student403
Anyone have any videos to recommend for learning Green's theorem/Stoke's theorem?


Take a look at section 3.6 of this chapter of the Feynman lectures, where he explains Stoke's theorem in a nice intuitive way:

http://www.feynmanlectures.caltech.edu/II_03.html

You would probably benefit from reading through chapters 2 and 3, where Feynman covers a lot of vector calculus from scratch.
Original post by Student403
Anyone have any videos to recommend for learning Green's theorem/Stoke's theorem?


Khanacademy explains the intuition of Green'/Stokes really well in my opinion. There is no rigour/proofs or anything but it served as an excellent introduction imo.
Any feedback on my argument for there being infinitely many rationals in the interval [a,b] [a,b] ? (a,b are any reals, b>a)
Let ϵ=1b−a \epsilon = \dfrac{1}{b-a} Then ∃N∈Z:1N<ϵ2 \exists N \in \mathbb{Z} : \frac{1}{N}< \frac{ \epsilon}{2} .
Let k k be the largest integer such that kN<a \frac{k}{N} <a . Then a<k+1N<k+2N<b a< \dfrac{k+1}{N} < \dfrac{k+2}{N} <b (There are therefore atleast 2 rationals between any 2 reals).
But take these rationals and find their mean, 2k+32N \dfrac{2k+3}{2N} . It can be easily seen that this is distinct from the other 2 rationals. Also can show the mean of any 2 rationals lies in the interval between them if needed. As we have not made any assumptions on the values of the 2 starting rationals, we can take any 2 distinct rationals in our interval and calculate the mean to find a new rational infinite times therefore infinite rationals in interval. (also its clear the mean is a rational as the products of and sums of any integers is an integer).
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Avatar for Zacken
Zacken
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Original post by EnglishMuon
Any feedback on my argument for there being infinitely many rationals in the interval [a,b] [a,b] ? (a,b are any reals, b>a)
Let ϵ=1b−a \epsilon = \dfrac{1}{b-a} Then ∃N∈Z:1N<ϵ2 \exists N \in \mathbb{Z} : \frac{1}{N}< \frac{ \epsilon}{2} .
Let k k be the largest integer such that kN<a \frac{k}{N} <a . Then a<k+1N<k+2N<b a< \dfrac{k+1}{N} < \dfrac{k+2}{N} <b


Seems okay (although there is a rather slightly shorter shorter proof), but how do you know that k+2N<b\frac{k+2}{N} < b? (I might just be missing something obvious).

I'd say that k+2N<aN+2N=a+2N<a+ϵ\frac{k+2}{N} < \frac{aN + 2}{N} = a + \frac{2}{N} < a + \epsilon where, if ϵ=b−a\epsilon = b-a, then this inequality would generate a+(b−a)=ba + (b-a) = b and we'd get k+2N<b\frac{k+2}{N} < b as expected, but that's not what ϵ\epsilon has been defined as.
Original post by Zacken
Seems okay (although there is a rather slightly shorter shorter proof), but how do you know that k+2N<b\frac{k+2}{N} < b? (I might just be missing something obvious).

I'd say that k+2N<aN+2N=a+2N<a+ϵ\frac{k+2}{N} < \frac{aN + 2}{N} = a + \frac{2}{N} < a + \epsilon where, if ϵ=b−a\epsilon = b-a, then this inequality would generate a+(b−a)=ba + (b-a) = b and we'd get k+2N<b\frac{k+2}{N} < b as expected, but that's not what ϵ\epsilon has been defined as.


yep that makes sense :smile: and yea sorry I meant to write my epsilon the other way up as 1/2(b−a) 1/2 (b-a) then it all works out I think, i just wrote upside down :tongue:
Not sure where else to post this:ImageUploadedByStudent Room1467936421.429411.jpg


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Definitely want to get involved in this! I'm nowhere near the level you guys are at, but I'm starting Maths and Economics at LSE through distance learning after summer, so want to bridge the gap between what I know and what I'll need to know.
While going through N&S I had a question. Are natural numbers positive integers and whole numbers non-negative integers? Or are natural numbers non-negative integers – in which case what are whole numbers? Or are both conventions followed by different people?
Original post by krishdesai7
While going through N&S I had a question. Are natural numbers positive integers and whole numbers non-negative integers? Or are natural numbers non-negative integers – in which case what are whole numbers? Or are both conventions followed by different people?


Natural numbers are the counting numbers, ( the non negative and non-zero integers ) whole numbers are just integers AFAIK


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Original post by drandy76
Natural numbers are the counting numbers, ( the non negative and non-zero integers ) whole numbers are just integers AFAIK


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The natural numbers are the positive integers. Some people include 0, others don't. Whole numbers are the same thing. I don't think you can include negatives in the whole numbers so that's where the whole numbers differ from the integers.
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Zacken
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Original post by krishdesai7
Or are both conventions followed by different people?


This. Some people think that whole numbers are numbers without a fractional part, hence making them equivalent to the integers. Others think they are non-negative integers. Others think they are positive integers. Mathematicians tend not to use the word "whole numbers" for that very precise reason, and instead use phrases like "non-negative integers", "non-positive integers", "integers", "positive integers", etc...

Some of the notation/terminology in maths just makes me shake my head, like the whole domain/co-domain/range thing and the whole f−1(A)f^{-1}(A) being the pre-image, but using inverse notation. Fun tymes.
(edited 7 years ago)
I propose that mathmos must carry a pen and pad everywhere, so that any verbally ambiguous/ awkward terms can be expressed using notation( thereby abandoning, whole/natural numbers etc)


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Original post by Zacken


Some of the notation/terminology in maths just makes me shake my head, like the whole domain/co-domain/range thing and the whole f−1(A)f^{-1}(A) being the pre-image, but using inverse notation. Fun tymes.


Yeah lol. I spend so much time trying to figure out what exactly the difference between co-domain and range is before I finally gave up.

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