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    (Original post by JakeThomasLee)
    Specimen Paper III, Question 8:
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    In a grid of 8 dots, we have an area of 7 x 7 = 49 unit squares

    It follows that there are 49 unit squares (1x1 squares)

    There are 6 double squares on each row (2x2 squares) and hence 6x6 = 36 double squares in the whole grid

    There are 5 triple squares on each row (3x3 squares) and hence 5x5 = 25 triple squares in the whole grid

    There are 4 quadruple squares on each row (4x4 squares) and hence 4x4=16 quadruple squares in the whole grid

    There are 3 quintuple squares on each row (5x5 squares) and hence 3x3=9 quintuple squares in the whole grid

    There are 2 sixfold squares on each row (6x6 squares) and hence 2x2=4 sixfold squares in the whole grid

    There is 1 sevenfold square in the grid only

    Hence there are 49+36+25+16+9+4+1 = 140 squares in the grid who have sides parallel to the dots in the grid

    It's clear that this may be applicable to any grid of n dots such that the number of squares with sides parallel to the dots is equal to the sum of all square numbers less than n

    Assuming a square must have dots for all four of it's vertices:

    Considering diagonal squares:

    When I say a 1x1 diagonal square I mean a \sqrt{2}x\sqrt{2} square but have written it this way for simplicity.

    1x1 diagonal squares take up a 2x2 parallel area in the grid. Hence there are the same number of 1x1 diagonal squares as 2x2 parallel squares in the grid = 36

    2x2 diagonal squares take up a 4x4 parallel area in the grid. Hence there are the same number of 2x2 diagonal squares in the grid as 4x4 parallel squares = 16

    3x3 diagonal squares take up a 6x6 parallel area in the grid. Hence there are the same number of 3x3 diagonal squares in the grid as 6x6 parallel squares = 4

    There can be no 4x4 diagonal squares contained in the grid.

    Therefore there are 140 + 36 + 16 + 4 = 196 total squares
    I disagree with your answer. Squares do not have to be diagonal, they can be just slightly slanted.
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    (Original post by Renzhi10122)
    I disagree with your answer. Squares do not have to be diagonal, they can be just slightly slanted.
    Ah, yes you're right
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    (Original post by JakeThomasLee)
    I currently have this integral, not too sure how I'm going to evaluate it, although it may just be completely obvious and I can't see it

    \RIghtarrow I=\displaystyle\int_0^{\frac{\pi  }{2}} \dfrac{\cos(\theta)}{\sin(\theta  )+\cos(\theta)} \ d\theta
    Use the same sub as for the other one. . .
    I hope the others made it clear what I did
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    (Original post by JakeThomasLee)
    I currently have this integral, not too sure how I'm going to evaluate it, although it may just be completely obvious and I can't see it

    \RIghtarrow I=\displaystyle\int_0^{\frac{\pi  }{2}} \dfrac{\cos(\theta)}{\sin(\theta  )+\cos(\theta)} \ d\theta
    Consider \displaystyle J = \int_0^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta + \cos \theta} \, \mathrm{d}x then look at I + J and I-J.
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    (Original post by JakeThomasLee)
    I currently have this integral, not too sure how I'm going to evaluate it, although it may just be completely obvious and I can't see it

    \RIghtarrow I=\displaystyle\int_0^{\frac{\pi  }{2}} \dfrac{\cos(\theta)}{\sin(\theta  )+\cos(\theta)} \ d\theta
    Can't you just divide top and bottom by cos (theta)

    Giving 1/(1+tan)

    Let u = tan

    And using sec ^2 = tan ^2 +1

    Gives,

    1/(1+u)(1+u^2) which can be done by partial fractions?

    Not the most elegant though


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    (Original post by Zacken)
    Consider \displaystyle J = \int_0^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta + \cos \theta} \, \mathrm{d}x then look at I + J and I-J.
    Deja vu, Siklos somewhere!


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    (Original post by I am Ace)
    Deja vu, Siklos somewhere!


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    I will forever remember that STEP question.
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    (Original post by I am Ace)
    Deja vu, Siklos somewhere!


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    I better get reading... :o:
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    (Original post by physicsmaths)
    i don't agree with part 1, surely it is P(heads/ 3 heads) which is an 1/8?
    then again if he is right can someone explain since i am **** at probability.
    got it now my division of numbers was wrong
    The probability of t-2, t-1 and t all being heads is 1 though, by definition t. So the probability that s=t+1 is just the probability the coin is flipped heads at t+1.
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    (Original post by MadChickenMan)
    The probability of t-2, t-1 and t all being heads is 1 though, by definition t. So the probability that s=t+1 is just the probability the coin is flipped heads at t+1.
    p( h|3 heads)=1/2 I had an error with numbers causing me to get 1/8


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    (Original post by joostan)
    Specimen Paper 2, Question 9:
    Caveat - my probability is a bit weak and I'm not convinced about the last part.
    Also succinct notation was a pain.
    Spoiler:
    Show
    Notational stuff:
    Let N_x(n,r) be the number of cards of type x i.e x=Q , K.
    Then, assuming n \leq 52:
    P(N_K(n,r)=i)=\dfrac{{4 \choose i}{n -4 \choose r-i}}{{n\choose r}}

    a) P(N_K(52,13)=1)=\dfrac{{4 \choose 1}{48 \choose 12}}{{52 \choose 13}}

    b) P(N_Q(52,13) \geq 2)=1-P(N_Q(52,13) < 2)=1-\dfrac{{4 \choose 1}{48 \choose 12}+{4 \choose 0}{48 \choose 13}}{{52 \choose 13}}

    c) I claim (perhaps erroneously) that:
    P\left( N_K(52,13)=N_Q(52,13) \right) = \displaystyle\sum_{i=0}^4 P(N_Q(52,13)=i)P(N_K(48,13-i)=i)
    i.e I'm saying that we need i Queens from 52 cards and i Kings from 48 cards - the full pack without the Queens, and that these are independent.
    \Rightarrow P\left( N_K(52,13)=N_Q(52,13) \right) =\displaystyle\sum_{i=0}^4 \left( \dfrac{{4 \choose i}{48 \choose 13-i}}{{52 \choose 13}} \cdot \dfrac{{4 \choose i}{44 \choose 13-2i}}{{48 \choose 13-i}} \right)
    Simplifying a little we obtain:
    P\left( N_K(52,13)=N_Q(52,13) \right)=\displaystyle\sum_{i=0}^  4 \left(\dfrac{{4 \choose i}^2{44 \choose 13-2i}}{{52 \choose 13}} \right).
    for part c why is it 48 and not 52-i for the number of cards in which to draw a king?
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    Is there a thread for the Natural Sciences solutions?
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    (Original post by ak33m98)
    Is there a thread for the Natural Sciences solutions?
    Don't think so. U might aswell make one then


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    (Original post by mattyd375)
    for part c why is it 48 and not 52-i for the number of cards in which to draw a king?
    Because you want to pull a King from the pack, without pulling a Queen as this would change i though as I said - I'm not entirely certain I'm right.
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    (Original post by ak33m98)
    Is there a thread for the Natural Sciences solutions?
    NatSci's aren't as cool as Mathmos, it seems.

    jk, I don't think they have a thread - you might want to make one yourself, but it seems a little late for this years interview session.
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    Test 4, Question 5

    Hint
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    You have two unknowns - so you need two equations. Use the trigonometry the question suggests to find them.


    Solution
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    I attach a diagram. I've used A and B instead of theta and phi because latex is effort.

    Name:  Screen Shot 2015-12-05 at 21.14.52.png
Views: 191
Size:  6.7 KB

    Consider triangle XX'P (where P is the location of the deer), we can write down

    



(X'-X)tanA  = Y'

    Similarly from YY'P,

    



X'tanB = (Y'-Y)

    Substituting in the values of Y' from the first equation into the second, we obtain (after messy algebra)

    



X' = \dfrac{XtanA + Y}{tanA-tanB}

    and substituting the second equation into the first:

    



Y' = \dfrac{XtanAtanB + YtanA}{tanA-tanB}
    Someone should probably check my algebra.
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    Test 5, Question 1 (Maths and Physics, Paper 1)

    Hint:
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    A nice sub x = u^2 /2 then Integration by parts.
    Solution:
    Spoiler:
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    Using the sub x \mapsto \frac{x^2}{2} we have \displaystyle I_6 = \frac{1}{4 \sqrt{2}} \int_{0}^{\infty} x^6 e^{-\frac{x^2}{2}} \, \mathrm{d}x.

    Integration by parts with u=x^5 and \displaystyle \mathrm{d}v = ue^{-\frac{u^2}{2}} gives us \displaystyle I_6 = \frac{5}{4\sqrt{2}} \int_0^{\infty} u^4 e^{-\frac{u^2}{2}} \, \mathrm{d}u = \frac{5}{4\sqrt{2}}I_4.

    Repeated integration by parts yields I_4 = 3I_2 = 3I_0 (which is the integral we were given, with some fiddling with limits) so, finally:

    \displaystyle I_6 = \frac{1}{4\sqrt{2}} \cdot 5 \cdot 3 \cdot \frac{I_0}{2} = \frac{15\sqrt{\pi}}{8}

    N.B: \displaystyle I_0 = \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}} \, \mathrm{d}x \Rightarrow \frac{I_0}{2} = \int_{0}^{\infty} e^{-\frac{x^2}{2}} \mathrm{d}x = \frac{\sqrt{2\pi}}{2} = \sqrt{\frac{\pi}{2}}.
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    Test 5, Question 5 (Maths and Physics, Paper 1)

    Thoroughly enjoyed this question.

    Hint:
    Spoiler:
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    \displaystyle \frac{x}{e^x - 1} = \sum_{i=1}^{\infty} xe^{-ix}
    Solution:
    (a)
    Spoiler:
    Show

    This part is fairly trivial:
    \displaystyle a \sum_{i=1}^{\infty} \frac{1}{b^i} = \frac{a}{b-1}
    (b)
    Spoiler:
    Show

    \displaystyle \int_{0}^{\infty}\frac{x}{e^x - 1} \, \mathrm{d}x = \int_{0}^{\infty} \sum_{i=1}^{\infty} xe^{-ix} \, \mathrm{d}x

    (Without discussing convergence, as per the question, we then have):

    \displaystyle \int_{0}^{\infty} \sum_{i=1}^{\infty} xe^{-ix} \, \mathrm{d}x = \sum_{i=1}^{\infty} \int_0^{\infty} xe^{-ix} \, \mathrm{d}x

    Hence:

    \displaystyle \sum_{i=1}^{\infty} \int_0^{\infty} xe^{-ix} \, \mathrm{d}x = \sum_{i=1}^{\infty} \left(\left[x \cdot \frac{1}{i}e^{-ix}\right]_0^{\infty} + \frac{1}{i}\int_0^{\infty} e^{-ix} \, \mathrm{d}x\right)

    Where we used integration by parts with u = x and \mathrm{d}v = e^{-ix}.

    \displaystyle \sum_{i=1}^{\infty} \left(\left[x \cdot \frac{1}{i}e^{-ix}\right]_0^{\infty} + \frac{1}{i}\int_0^{\infty} e^{-ix} \, \mathrm{d}x\right) = \sum_{i=1}^{\infty}\left(\frac{1  }{i} \left[-\frac{1}{i}e^{-x}\right]_0^{\infty}\right) = \sum_{i=1}^{\infty} \frac{1}{i^2} = \frac{\pi^2}{6} as required.
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    (Original post by Zacken)
    Test 5, Question 5 (Maths and Physics, Paper 1)

    Thoroughly enjoyed this question.

    Hint:
    Spoiler:
    Show
    \displaystyle \frac{x}{e^x - 1} = \sum_{i=1}^{\infty} xe^{-ix}
    Solution:
    (a)
    Spoiler:
    Show

    This part is fairly trivial:
    \displaystyle a \sum_{i=1}^{\infty} \frac{1}{b^i} = \frac{a}{b-1}
    (b)
    Spoiler:
    Show

    \displaystyle \int_{0}^{\infty}\frac{x}{e^x - 1} \, \mathrm{d}x = \int_{0}^{\infty} \sum_{i=1}^{\infty} xe^{-ix} \, \mathrm{d}x

    (Without discussing convergence, as per the question, we then have):

    \displaystyle \int_{0}^{\infty} \sum_{i=1}^{\infty} xe^{-ix} \, \mathrm{d}x = \sum_{i=1}^{\infty} \int_0^{\infty} xe^{-ix} \, \mathrm{d}x

    Hence:

    \displaystyle \sum_{i=1}^{\infty} \int_0^{\infty} xe^{-ix} \, \mathrm{d}x = \sum_{i=1}^{\infty} \left(\left[x \cdot \frac{1}{i}e^{-ix}\right]_0^{\infty} + \frac{1}{i}\int_0^{\infty} e^{-ix} \, \mathrm{d}x\right)

    Where we used integration by parts with u = x and \mathrm{d}v = e^{-ix}.

    \displaystyle \sum_{i=1}^{\infty} \left(\left[x \cdot \frac{1}{i}e^{-ix}\right]_0^{\infty} + \frac{1}{i}\int_0^{\infty} e^{-ix} \, \mathrm{d}x\right) = \sum_{i=1}^{\infty}\left(\frac{1  }{i} \left[-\frac{1}{i}e^{-x}\right]_0^{\infty}\right) = \sum_{i=1}^{\infty} \frac{1}{i^2} = \frac{\pi^2}{6} as required.
    How did you manage to get Test 5 out of 4 numbered tests?
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    (Original post by FutureOxford3)
    How did you manage to get Test 5 out of 4 numbered tests?
    (Original post by Zacken)
    (Maths and Physics, Paper 1)
    ...
 
 
 
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