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    (Original post by Zacken)
    Okay, I've gotten that the identity element is \mathbf{X}_0. I'm unclear as to how you got each element to be its self-inverse though, why exactly is \mathbf{X}_i \star \mathbf{X}_i = \mathbf{X}_0?
    ah that was when I misread the words/ read them the opposite way round. Originally I originally thought we were swapping Xi for X0 but after rereading it is the opposite way round. I just reread it and I still not sure of the wording. please can you explain what it wants us to be 'swapping'?
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    (Original post by EnglishMuon)
    ah that was when I misread the words/ read them the opposite way round. Originally I originally thought we were swapping Xi for X0 but after rereading it is the opposite way round. I just reread it and I still not sure of the wording. please can you explain what it wants us to be 'swapping'?
    So, when we have \mathbf{X}_i \star \mathbf{X}_j, we take whatever is on the left and write it in terms of P and Q.

    i.e: \mathbf{X}_i = M^i X_0 + Q\sum_{n=0}^{i-1} M^n and replace the \mathbf{X}_0 in that expression with whatever is on the right, i.e: \mathbf{X}_j so that we get: X_i = M^i X_j + Q\sum_{n=0}^{i-1} M^n.

    Or at least, that's what I understood of it. Does that make any sense? There aren't any solutions for this, are there? :lol:
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    (Original post by Zacken)
    So, when we have \mathbf{X}_i \star \mathbf{X}_j, we take whatever is on the left and write it in terms of P and Q.

    i.e: \mathbf{X}_i = M^i X_0 + Q\sum_{n=0}^{i-1} M^n and replace the \mathbf{X}_0 in that expression with whatever is on the right, i.e: \mathbf{X}_j so that we get: X_i = M^i X_j + Q\sum_{n=0}^{i-1} M^n.

    Or at least, that's what I understood of it. Does that make any sense? There aren't any solutions for this, are there? :lol:
    Haha exactly! Im really confused, each time I read it I think we are subbing each term the opposite way round. The wording seems really ambiguous to me :lol: But yeah as above in my other workings I showed it doesnt form a group by replacing xi with x0 so I guess it must be this way round
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    (Original post by EnglishMuon)
    Haha exactly! Im really confused, each time I read it I think we are subbing each term the opposite way round. The wording seems really ambiguous to me :lol: But yeah as above in my other workings I showed it doesnt form a group by replacing xi with x0 so I guess it must be this way round
    But, yeah. Even with this definition, I'm not sure how to prove the existence for the inverse?
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    (Original post by Zacken)
    But, yeah. Even with this definition, I'm not sure how to prove the existence for the inverse?
    Me neither . Maybe they want us to substitute both X0 for Xj and Xj for X0 at the same time? (But once again, I cant see how that gets us anything).
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    (Original post by EnglishMuon)
    Me neither . Maybe they want us to substitute both X0 for Xj and Xj for X0 at the same time? (But once again, I cant see how that gets us anything).
    Oh... I think I can see, we need to substitute in some X_j such that M^{i}X_J + Q \sum_{r=0}^{i-1} M^n equals X_i by making X_j depend on i, \mathbf{X_0} and m (which means it'll need to depend on Q(?) in some way such that we can use the previous fact that \sum_{r=0}^{m-1} \mathbf{M}^{r} = \mathbf{O} and \mathbf{M}^m = \mathbf{I}. I may be completely wrong though, this sounds a bit too complicated.

    Probably best to just set \mathbf{X}_i \star \mathbf{X}_j = \mathbf{X}_0 and solve for \mathbf{X}_j.
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    (Original post by Zacken)
    Oh... I think I can see, we need to substitute in some X_j such that M^{i}X_J + Q \sum_{r=0}^{i-1} M^n equals X_i by making X_j depend on i, \mathbf{X_0} and m (which means it'll need to depend on Q(?) in some way such that we can use the previous fact that \sum_{r=0}^{m-1} \mathbf{M}^{r} = \mathbf{O} and \mathbf{M}^m = \mathbf{I}. I may be completely wrong though, this sounds a bit too complicated.

    Probably best to just set \mathbf{X}_i \star \mathbf{X}_j = \mathbf{X}_0 and solve for \mathbf{X}_j.
    Yeah that does sound a little complicated, but I think I get what you mean! Its the only thing that I can see that may be what they want so we'll have a go
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    Zacken You've probably long done this q by now but here were my workings for the solution that actually works. Also I was extremely dumb when I said "there must be more to do with complex numbers here" as if I literally just said  z= \mathrm{cos} \theta + i \mathrm{sin} \theta it's clear the sins and cos's cancel Attachment 526309 Name:  ImageUploadedByStudent Room1461749239.228165.jpg
Views: 42
Size:  146.9 KB


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    (Original post by EnglishMuon)
    Zacken You've probably long done this q by now but here were my workings for the solution that actually works. Also I was extremely dumb when I said "there must be more to do with complex numbers here" as if I literally just said  z= \mathrm{cos} \theta + i \mathrm{sin} \theta it's clear the sins and cos's cancel
    Oh, that is brilliant. Good going.

    By the way, you probably shouldn't use j as a dummy variable if you're going to use X_j as a non-dummy variable after that. :lol: I'm confused though, has everything attached properly? 'cause I don't see any mentions of complex numbers anywhere in your picture?

    P.S, you can do \cos and \sin to get \cos, \sin instead of \mathrm{cos}, \mathrm{sin}
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    (Original post by Zacken)
    Oh, that is brilliant. Good going.

    By the way, you probably shouldn't use j as a dummy variable if you're going to use X_j as a non-dummy variable after that. :lol: I'm confused though, has everything attached properly? 'cause I don't see any mentions of complex numbers anywhere in your picture?

    P.S, you can do \cos and \sin to get \cos, \sin instead of \mathrm{cos}, \mathrm{sin}
    Thanks Yeah sorry about that widespread use of j, forgot it came up again. I attached 2 images- the first one just shows as "attachment..." whereas the second actually shows so I think it should come up if you click on that.
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    (Original post by EnglishMuon)
    Thanks Yeah sorry about that widespread use of j, forgot it came up again. I attached 2 images- the first one just shows as "attachment..." whereas the second actually shows so I think it should come up if you click on that.
    It just says "invalid attachment" when I click it?
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    (Original post by Zacken)
    It just says "invalid attachment" when I click it?

    Hmm try this one Name:  ImageUploadedByStudent Room1461751969.305171.jpg
Views: 40
Size:  147.2 KB


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    (Original post by EnglishMuon)
    Hmm try this one


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    Yep, that works. - looks good! Neat little question and solution :yep:
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    (Original post by Zacken)
    Yep, that works. - looks good! Neat little question and solution :yep:
    Yep! Thanks for your help!
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    (Original post by EnglishMuon)
    Yep! Thanks for your help!
    I wasn't helping, it was more of a joint effort thing.

    (I really liked this, I don't often get to do maths with others 'cause solo-teaching on an island and shiz) - I've got something by the way (I haven't gotten around to doing it myself yet)):

    A Ford circle is a circle tangent to the x-axis at a point \left(\frac{p}{q}, 0\right) for coprime integers p, q with radius \frac{1}{2q^2}. Show that, whilst some Ford circles touch, none of them intersect.
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    (Original post by Zacken)
    I wasn't helping, it was more of a joint effort thing.

    (I really liked this, I don't often get to do maths with others 'cause solo-teaching on an island and shiz) - I've got something by the way (I haven't gotten around to doing it myself yet)):

    A Ford circle is a circle tangent to the x-axis at a point \left(\frac{p}{q}, 0\right) for coprime integers p, q with radius \frac{1}{2q^2}. Show that, whilst some Ford circles touch, none of them intersect.
    Me too Nice, ill have a think about that question.
    Out of interest, are there any other math Gs like yourself you know on Mauritius? Id guess there weren't many (hence why you're hanging out with simpletons like myself on tsr ) but idk
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    (Original post by EnglishMuon)
    Me too Nice, ill have a think about that question.
    Out of interest, are there any other math Gs like yourself you know on Mauritius? Id guess there weren't many (hence why you're hanging out with simpletons like myself on tsr ) but idk
    I don't think there's anybody that interested in maths here. :lol:

    BTW, the reason I hang with you on TSR is to have your cleverness rub off on me in some way so I can scrape my way into King's where I'll repeat the process till the end of the degree. :rofl:
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    (Original post by Zacken)
    I don't think there's anybody that interested in maths here. :lol:

    BTW, the reason I hang with you on TSR is to have your cleverness rub off on me in some way so I can scrape my way into King's where I'll repeat the process till the end of the degree. :rofl:
    Go to a local unis lectures or somehong. The lecturer must like maths a little bit 👀


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    (Original post by Zacken)
    I don't think there's anybody that interested in maths here. :lol:

    BTW, the reason I hang with you on TSR is to have your cleverness rub off on me in some way so I can scrape my way into King's where I'll repeat the process till the end of the degree. :rofl:
    lol same reason for me, but I think thats more true from my side :lol: I can just imagine you at the end of August: "Phew that was a close call, I just scraped my way into King's with 3 Ss and only average of 98 ums in my a levels"
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    (Original post by physicsmaths)
    Go to a local unis lectures or somehong. The lecturer must like maths a little bit 👀


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    There's only one uni and I'm not sure they even do maths :erm:
 
 
 
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