A Summer of Maths (ASoM) 2016

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    (Original post by Zacken)
    \bigcup: \displaystyle \bigcup_{1 \leq i \leq m, 1 \leq j \leq n}
    Thanks
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    (Original post by EnglishMuon)
    Ok. whats wrong with the first sentence? We know that any two right cosets in a group are either equal or disjoint. So if we are looking at the intersection of 1 right coset  Hx_{i} and another  Ky_{i} and this set is not empty, they must both equal some  Hg for some  g \in G . There was a typo actually in the 2nd line as I meant to say  Hx_{i} \cap Ky_{j} = Hg \cap Kg = (H \cap K) g but I don't see why the rest of argument doesn't work.
    With the (Z, +) example, subgroups {3n} and {4n}, the intersection of the cosets {3n+1} and {4n+1} are certainly non empty, but this does not imply that Hx = Ky as you've written.


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    (Original post by Ecasx)
    With the (Z, +) example, subgroups {3n} and {4n}, the intersection of the cosets {3n+1} and {4n+1} are certainly non empty, but this does not imply that Hx = Ky as you've written.


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    lol yea that seems true (e.g. 13 occurs in both 3n+1, 4n+1) but I swear
    "any two left cosets of H in G are either identical or disjoint. In other words every element of G belongs to one and only one left coset and so the left cosets form a partition of G.[3] Corresponding statements are true for right cosets." must hold?? what on earth am i doing wrong here XD
    Actually yeah its cus they are different subgroups
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    (Original post by EnglishMuon)
    lol yea that seems true (e.g. 13 occurs in both 3n+1, 4n+1) but I swear
    "any two left cosets of H in G are either identical or disjoint. In other words every element of G belongs to one and only one left coset and so the left cosets form a partition of G.[3] Corresponding statements are true for right cosets." must hold?? what on earth am i doing wrong here XD
    Actually yeah its cus they are different subgroups
    Yeah they have to be cosets of the same subgroup.


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    (Original post by Ecasx)
    Yeah they have to be cosets of the same subgroup.


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    Aggh I was getting confused with typos:

    So let  g \in Hx_{i} \cap Ky_{j} . Then  g \in Hx_{i} and  g \in Ky_{j} \Rightarrow g=hx_{i}=ky_{j} for some  h \in H, k \in K .

    So  Hg=H(hx_{i})=Hx_{i} and  Kg=K(ky_{j})=Ky_{j}
    Therefore  Hx_{i} \cap Ky_{j}= Hg \cap Kg . My original typo here was that I said  Ky_{j}=Hg also instead of Kg.
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    (Original post by EnglishMuon)
    Aggh I was getting confused with typos:

    So let  g \in Hx_{i} \cap Ky_{j} . Then  g \in Hx_{i} and  g \in Ky_{j} \Rightarrow g=hx_{i}=ky_{j} for some  h \in H, k \in K .

    So  Hg=H(hx_{i})=Hx_{i} and  Kg=K(ky_{j})=Ky_{j}
    Therefore  Hx_{i} \cap Ky_{j}= Hg \cap Kg . My original typo here was that I said  Ky_{j}=Hg also instead of Kg.
    What are you trying to do here? I think you're saying that there exists x, y in G such that Hx and Ky both contain G, but this is an obvious statement, since x=y=g works (both H and K contain the identity). Therefore it is true that g belongs to Hg ^ Kg = (H ^ K)g, a coset of (H ^ K), as you have said. But this was already known to us, since we know that H ^ K is a subgroup, and that the cosets of a subgroup will partition the group.


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    (Original post by Ecasx)
    What are you trying to do here? I think you're saying that there exists x, y in G such that Hx and Ky both contain G, but this is an obvious statement, since x=y=g works (both H and K contain the identity). Therefore it is true that g belongs to Hg ^ Kg = (H ^ K)g, a coset of (H ^ K), as you have said. But this was already known to us, since we know that H ^ K is a subgroup, and that the cosets of a subgroup will partition the group.


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    Im showing that is true for all g in Hx and Ky, hence why I did that proof rather than just noting each contains the identity (there would no use in just saying each contains just 1 element the same). So this shows  Hg \cap Kg = Hx_{i} \cap Ky_{j} for any of our xi yj. So we can then say G is just  \displaystyle \bigcup_{1 \leq i \leq m, 1 \leq j \leq n} (Hx_{i} \cap Ky_{j} )= \displaystyle \bigcup_{i=1}^{nm} (H \cap K)g_{i}. i.e. H intersection K is of finite index.
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    Any one got any ideas on this?

    Question: A ball bearing rests on a ramp fixed to the top of a car which isaccelerating horizontally. The position of the ball bearing relative to theramp is used as a measure of the acceleration of the car. Show that ifthe acceleration is to be proportional to the horizontal distance moved bythe ball (measured relative to the ramp), then the ramp must be curvedupwards in the shape of a parabola. ++
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    (Original post by AsifHossain)
    Any one got any ideas on this?

    Question: A ball bearing rests on a ramp fixed to the top of a car which isaccelerating horizontally. The position of the ball bearing relative to theramp is used as a measure of the acceleration of the car. Show that ifthe acceleration is to be proportional to the horizontal distance moved bythe ball (measured relative to the ramp), then the ramp must be curvedupwards in the shape of a parabola. ++
    This is my solution (give or take a couple of small constants from scribbling it down quickly) Name:  ImageUploadedByStudent Room1469288276.436569.jpg
Views: 184
Size:  140.2 KB


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    (Original post by AsifHossain)
    Any one got any ideas on this?

    Question: A ball bearing rests on a ramp fixed to the top of a car which isaccelerating horizontally. The position of the ball bearing relative to theramp is used as a measure of the acceleration of the car. Show that ifthe acceleration is to be proportional to the horizontal distance moved bythe ball (measured relative to the ramp), then the ramp must be curvedupwards in the shape of a parabola. ++
    Is this from Upgrade Your Physics? In which case excellent book, I recommend it wholeheartedly.

    Your approach will want to be along the lines of showing that the gradient - or, equivalently, cot(theta) with theta to the horizontal if my mental diagram is vorrect - is proportional to x, and so a parabola must result. Any ideas on how to establish something about this angle?

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    (Original post by Krollo)
    Is this from Upgrade Your Physics? In which case excellent book, I recommend it wholeheartedly.

    Your approach will want to be along the lines of showing that the gradient - or, equivalently, cot(theta) with theta to the horizontal if my mental diagram is vorrect - is proportional to x, and so a parabola must result. Any ideas on how to establish something about this angle?

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    Yep it's off that book, glad to know it's useful haha.

    I'm unsure if we can say that the ball is in equilibrium and maximum displacement (i.e. Rcos(theta) = ma and Rsin(theta) = mg from which the result would follow) since surely the ball would some have velocity at equilibrium so would continue rising up the ramp and therefore the value of x would increase?
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    (Original post by AsifHossain)
    Yep it's off that book, glad to know it's useful haha.

    I'm unsure if we can say that the ball is in equilibrium and maximum displacement (i.e. Rcos(theta) = ma and Rsin(theta) = mg from which the result would follow) since surely the ball would some have velocity at equilibrium so would continue rising up the ramp and therefore the value of x would increase?
    Are you familiar with the idea of a fictitious force?


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    (Original post by Krollo)
    Are you familiar with the idea of a fictitious force?


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    Yes, we have an inertial force acting in the positive x direction equal to ma right?
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    (Original post by AsifHossain)
    Yes, we have an inertial force acting in the positive x direction equal to ma right?
    Yep. By equilibrium, it is in equilibrium relative to the ramp if that makes sense.
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    Zacken Are there any resources for GT on MIT OCW? I must be blind since I can only find "Introduction to Lie Groups".
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    (Original post by EnglishMuon)
    Zacken Are there any resources for GT on MIT OCW? I must be blind since I can only find "Introduction to Lie Groups".
    Nope, them americanos are weird.
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    (Original post by EnglishMuon)
    Zacken Are there any resources for GT on MIT OCW? I must be blind since I can only find "Introduction to Lie Groups".
    They'll be listed under 'algebra' or 'abstract algebra'.

    Groups:
    http://ocw.mit.edu/courses/mathemati...a-i-fall-2010/
    Rings:
    http://ocw.mit.edu/courses/mathemati...2011/index.htm
    Bit of both:
    http://ocw.mit.edu/courses/mathemati...2013/Syllabus/

    Lie groups is pretty hard stuff, you'll need a lot of prerequisites for that.
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    (Original post by Alex:)
    They'll be listed under 'algebra' or 'abstract algebra'.

    Groups:
    http://ocw.mit.edu/courses/mathemati...a-i-fall-2010/
    Rings:
    http://ocw.mit.edu/courses/mathemati...2011/index.htm
    Bit of both:
    http://ocw.mit.edu/courses/mathemati...2013/Syllabus/

    Lie groups is pretty hard stuff, you'll need a lot of prerequisites for that.
    Thanks. and yea haha I wasn't planning on starting the lie groups stuff. a couple of years away hopefully
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    (Original post by EnglishMuon)
    Thanks. and yea haha I wasn't planning on starting the lie groups stuff. a couple of years away hopefully
    You planning years early.
    Cmon mate.



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    (Original post by EnglishMuon)
    Thanks. and yea haha I wasn't planning on starting the lie groups stuff. a couple of years away hopefully
    It's good that you're interested in group theory though. I found groups to not be interesting early on, since I'm more of an analysis person. Groups pop up everywhere, which is why we learn about them.. For instance, Lie groups I quite like, as they are groups and differentiable manifolds, so it has all group properties and you can do analysis on them.
 
 
 
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