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    Test 5, Question 2 (Maths and Physics, Paper 1)

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    Look at squares modulo 7 and split it into cases for different values of c^2

    Solution:
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    x^2 \equiv 0, 1, 2, 4 \pmod{7} (we neglect 0 for obvious reasons in this question)

    Since a^2 + b^2 = c^2 we then have a^2 + b^2 \equiv 1, 2, 4 \pmod{7}

    Case 1: c^2 \equiv 1 \pmod{7} yields a^2 \equiv b^2 \equiv 4 \pmod{7}
    Case 2: c^2 \equiv 2 \pmod{7} yields a^2 \equiv b^2 \equiv 1 \pmod{7}
    Case 3: c^2 \equiv 4 \pmod{7} yields a^2 \equiv b^2 \equiv 2 \pmod{7}

    In all cases, we have a^2 \equiv b^2 \pmod{7} \Rightarrow a^2 - b^2 \equiv 0 \pmod{7} as required.
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    (Original post by Zacken)
    Test 5, Question 2 (Maths and Physics, Paper 1)

    Hints:
    Spoiler:
    Show
    Look at squares modulo 7 and split it into cases for different values of c^2
    Solution:
    Spoiler:
    Show

    x^2 \equiv 0, 1, 2, 4 \pmod{7} (we neglect 0 for obvious reasons in this question)

    Since a^2 + b^2 = c^2 we then have a^2 + b^2 \equiv 1, 2, 4 \pmod{7}

    Case 1: c^2 \equiv 1 \pmod{7} yields a^2 \equiv b^2 \equiv 4 \pmod{7}
    Case 2: c^2 \equiv 2 \pmod{7} yields a^2 \equiv b^2 \equiv 1 \pmod{7}
    Case 3: c^2 \equiv 4 \pmod{7} yields a^2 \equiv b^2 \equiv 2 \pmod{7}

    In all cases, we have a^2 \equiv b^2 \pmod{7} \Rightarrow a^2 - b^2 \equiv 0 \pmod{7} as required.
    Can you link me to it.
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    Trinity Test 4, Question 9

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    Consider using dtheta and dT, and try resolving radially and tangentially.


    Solution
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    Diagram attached. I've replaced some symbols for each of typing.
    Name:  Screen Shot 2015-12-08 at 18.29.56.png
Views: 137
Size:  14.7 KB

    Resolving radially,

    

R = Tsin(0.5dA) + (T+dT)sin(0.5dA)

    and since 0.5dA is small, we can approximate thus:

    

R = T * 0.5dA + T*0.5dA + dT*0.5dA

R = TdA

    since dT*dA is negligibly small.

    Resolving tangentially:

    

F + Tcos(0.5dA) = (T+dT)cos(0.5dA)

F = dTcos(0.5dA)

0.25TdA = dT
    since dA is small, and mu is 0.25. Letting dA, dT ->0, this gives us a differential equation which may be solved by separating the variables:
    0.25A +C = \ln{T}
    And since A=0 -> T=1:
    0.25A = \ln{T}
    So T = 10^6 -> A = 4*13.6 = 54.4
    As this is the angle in radians, M = 54.4/2pi = 9 turns or so.
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    (Original post by Krollo)
    Trinity Test 4, Question 9

    Hint
    Spoiler:
    Show

    Consider using dtheta and dT, and try resolving radially and tangentially.


    Solution
    Spoiler:
    Show

    Diagram attached. I've replaced some symbols for each of typing.
    Name:  Screen Shot 2015-12-08 at 18.29.56.png
Views: 137
Size:  14.7 KB

    Resolving radially,

    

R = Tsin(0.5dA) + (T+dT)sin(0.5dA)

    and since 0.5dA is small, we can approximate thus:

    

R = T * 0.5dA + T*0.5dA + dT*0.5dA

R = TdA

    since dT*dA is negligibly small.

    Resolving tangentially:

    

F + Tcos(0.5dA) = (T+dT)cos(0.5dA)

F = dTcos(0.5dA)

0.25TdA = dT
    since dA is small, and mu is 0.25. Letting dA, dT ->0, this gives us a differential equation which may be solved by separating the variables:
    0.25A +C = \ln{T}
    And since A=0 -> T=1:
    0.25A = \ln{T}
    So T = 10^6 -> A = 4*13.6 = 54.4
    As this is the angle in radians, M = 54.4/2pi = 9 turns or so.
    Damn my mechanics is bad, don't even understand this mate.


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    (Original post by physicsmaths)
    Damn my mechanics is bad, don't even understand this mate.


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    Neither do I

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    (Original post by Krollo)
    Trinity Test 4, Question 9

    Hint
    Spoiler:
    Show

    Consider using dtheta and dT, and try resolving radially and tangentially.

    Solution
    Spoiler:
    Show

    Diagram attached. I've replaced some symbols for each of typing.


    Resolving radially,

    

R = Tsin(0.5dA) + (T+dT)sin(0.5dA)

    and since 0.5dA is small, we can approximate thus:

    

R = T * 0.5dA + T*0.5dA + dT*0.5dA

R = TdA

    since dT*dA is negligibly small.

    Resolving tangentially:

    

F + Tcos(0.5dA) = (T+dT)cos(0.5dA)

F = dTcos(0.5dA)

0.25TdA = dT
    since dA is small, and mu is 0.25. Letting dA, dT ->0, this gives us a differential equation which may be solved by separating the variables:
    0.25A +C = \ln{T}
    And since A=0 -> T=1:
    0.25A = \ln{T}
    So T = 10^6 -> A = 4*13.6 = 54.4
    As this is the angle in radians, M = 54.4/2pi = 9 turns or so.
    I'll take your word for it

    I just scribbled the method I would use to solve it, didn't have the courage to try it myself
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    (Original post by Krollo)
    Neither do I

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    When is your interview? If i have no mechanics or stats i will bang it, but no mechanics and stats is unlikely


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    (Original post by physicsmaths)
    When is your interview? If i have no mechanics or stats i will bang it, but no mechanics and stats is unlikely


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    14th, I think you're a bit later? I'm sure you'll be absolutely fine. I wouldn't mind applied myself, to be honest, but each to their own. :-)
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    (Original post by Krollo)
    14th, I think you're a bit later? I'm sure you'll be absolutely fine. I wouldn't mind applied myself, to be honest, but each to their own. :-)
    I would not mind pure geometry 😃! But yh mechanics i can work with sometiems and the same with probability but the topics are iffy for me. Yeh I am on the 18th!


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    15rootpi/8 for the maths and CompSci Q1 Test 1?


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    (Original post by physicsmaths)
    15rootpi/8 for the maths and CompSci Q1 Test 1?


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    Yep (I think this was solved a page or so back)
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    (Original post by Krollo)
    Yep (I think this was solved a page or so back)
    Ah so those solutions are up aswell, Q2 is so quick.


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    Anybody know how to do Q3 without using jensens equality?
    https://www1.maths.leeds.ac.uk/~read/TQC1.pdf
    This is how I have done it,



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    Name:  ImageUploadedByStudent Room1449610482.335483.jpg
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    (Original post by physicsmaths)
    Anybody know how to do Q3 without using jensens equality?
    https://www1.maths.leeds.ac.uk/~read/TQC1.pdf
    This is how I have done it,



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    Yeah, I had to use Jensen's. Although, Jensen's in the case of two is *fairly* intuitive. Btw, the next part is actually Holder's inequality.
    Btw, I think you overcomplicated it, using weights of 1/p and 1/q, you dont have to do much manipulation.
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    (Original post by Renzhi10122)
    Yeah, I had to use Jensen's. Although, Jensen's in the case of two is *fairly* intuitive. Btw, the next part is actually Holder's inequality.
    Btw, I think you overcomplicated it, using weights of 1/p and 1/q, you dont have to do much manipulation.
    Yh I figured it out after that I was using the p,q weirdly but I used p,q for the sake of the powers haha, , Yh holders is the one with 3 sequences right and the sum of powers on the least side right? And when it is two sequences it is just Cauchy right?


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    (Original post by Renzhi10122)
    Yeah, I had to use Jensen's. Although, Jensen's in the case of two is *fairly* intuitive. Btw, the next part is actually Holder's inequality.
    Btw, I think you overcomplicated it, using weights of 1/p and 1/q, you dont have to do much manipulation.
    Would they really expect everyone to know jensens inequality? That is trinity for you


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    (Original post by physicsmaths)
    Yh I figured it out after that I was using the p,q weirdly but I used p,q for the sake of the powers haha, , Yh holders is the one with 3 sequences right and the sum of powers on the least side right? And when it is two sequences it is just Cauchy right?


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    Nah, more sequences is just more cauchy, i think. Holder's is where the powers arent 1/2 anymore (square root cauchy, you get square roots on the larger side), they are 1/p and 1/q where 1/p+1/q=1
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    (Original post by physicsmaths)
    Would they really expect everyone to know jensens inequality? That is trinity for you


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    Honestly, probably not. Only about max 20 or so people applying would know it, though some might spot the trick for the 2 variable case. I guess it would be like the new material kind of question, where they tell you the inequality in interview, and ask you to use it.
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    (Original post by Renzhi10122)
    Nah, more sequences is just more cauchy, i think. Holder's is where the powers arent 1/2 anymore (square root cauchy, you get square roots on the larger side), they are 1/p and 1/q where 1/p+1/q=1
    Yh that is what i meant, when the powers are 1/2 and it is 2 sequences holders becomes cauchy schwarz.
    I would cautious of such questions if they ask me some crazy indepth questions about Holders inequality ill be like 'sorry i only know the basic proofs mate'. Atleast I am fairly certain I won't need this in my 30 minute test 😂.


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