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    is paper 3 getting done?


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    (Original post by physicsmaths)
    is paper 3 getting done?


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    At some point. I'll start now in fact...

    Paper 3, Q5

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    Consider parity (even/odd). What can we apply it to?


    Solution
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    Initially, the sum of all the cards is 5050 (from arithmetic series formula), which is even.

    Suppose the two cards you choose are even. Then their sum is even, and their difference is even. The sum of all cards thus remains even.

    Suppose they are both odd. Then their sum is even, and their difference is even. The sum of all cards thus remains even.

    Suppose one is even and one is odd. Then their sum is odd, and their difference is odd. The sum of all cards thus remains even.

    Thus the sum of all cards will always remain even, which remains true when a single card remains. That card must hence be even.


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    Paper 3 Q3 Solution
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     \frac{e^{x} + e^{-x}}{2} = cosh(x) \hspace{1mm} , \hspace{1mm} \frac{e^{x} - e^{-x}}{2} = sinh(x)
    These are the hyperbolic trig functions, and it is simple to show that
     cosh^{2}(x) - sinh^{2}(x) = 1
    We will use this fact below.

     \sqrt{y^2 - 1} = sinh(x) \hspace{1mm}, \hspace{2mm} y^2 = sinh^{2}(x) + 1 = cosh^{2}(x) \hspace{3mm} y = \pm cosh(x)

     \sqrt{y^2 + 1} = cosh(x) \hspace{1mm}, \hspace{2mm} y^2 = cosh^{2}(x) - 1 = sinh^2(x) \hspace{3mm} y = \pm sinh(x)

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    (Original post by atsruser)
    Some further thoughts (I don't have a complete solution):

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    1. I think that the no-motion condition should be |x| < \frac{F}{k}

    2. If we write down the equations of motion we get:

    \dot{x} > 0: m\ddot{x} = -kx -F \Rightarrow \ddot{x}+\frac{k}{m}x = -\frac{F}{m}

    \dot{x} < 0: m\ddot{x} = -kx +F \Rightarrow \ddot{x}+\frac{k}{m}x = \frac{F}{m}

    with corresponding solutions:

    \dot{x} > 0: x =A\cos(\omega t-\phi)-\frac{F}{k}

    \dot{x} < 0: x =B\cos(\omega t-\phi)+\frac{F}{k}

    with \omega^2=\frac{k}{m}

    ~ snip ~

    So I don't know what to make of this at all. The solutions to the equations of motion don't seem to match what we "know" must be happening; the only explanation that I can see at the moment is that intuitively I'm wrong, and that over every half cycle we get SHM, but somehow we can make an argument the the "constants" A,B must change at the end of every half-cycle, or something like that.
    You're pretty much there,

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    The part of your answer I've emphasized is exactly what's going on here. Basically when the particle is moving in the "+ve direction" it's performing SHM about the point x = -F/k, when it's moving in the -ve direction it's performing SHM about the point x = +F/k. At the points where it's still, you need to switch between the two, and so both "versions" of the motion need to agree when \dot{x} = 0. Which means that the amplitude must decrease by 2F/k at each reversal of direction (until you get to where the block stops moving).
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    Paper 3 Question 1:
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    Proof by picture seems to be the best way to go, without any fancy expressions for the \zeta function.
    A quick sketch should show that:
    \displaystyle\int_1^{\infty} \dfrac{1}{x^s} \ dx \leq \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^s}.
    We can compute the integral:
    \displaystyle\int_1^{\infty} \dfrac{1}{x^s} \ dx = -\dfrac{1}{s-1} \left[ \dfrac{1}{x^{s-1}} \right] _1^{\infty}=\dfrac{1}{s-1}
    Also from the sketch we see that:
    \displaystyle\sum_{n=2}^{\infty} \dfrac{1}{n^s} \leq \displaystyle\int_2^{\infty} \dfrac{1}{(x-1)^s} \ dx
    Hence:
    \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^s} \leq 1+\displaystyle\int_1^{\infty} \dfrac{1}{x^s} \ dx
    Observing that 1+\dfrac{1}{s-1}=\dfrac{s}{s-1} gives the result:
    \dfrac{1}{s-1} \leq 1+2^{-s}+3^{-s}+... \leq\dfrac{s}{s-1}

    Paper 3 Question 2:
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    I=\displaystyle\int_0^1 \dfrac{dx}{x+\sqrt{1-x^2}} \ dx
    Let x=\sin(\theta)
    \Rightarrow I = \displaystyle\int_0^{\frac{\pi}{  2}} \dfrac{\sin(\theta)}{\sin(\theta  )+\cos(\theta)} \ d\theta

\theta \mapsto \dfrac{\pi}{2}-\theta

\RIghtarrow I=\displaystyle\int_0^{\frac{\pi  }{2}} \dfrac{\cos(\theta)}{\sin(\theta  )+\cos(\theta)} \ d\theta 

\Rightarrow 2I=\displaystyle\int_0^{\frac{ \pi}{2}} \ d\theta  = \dfrac{\pi}{2}

\therefore I=\dfrac{\pi}{4}.

    Paper 3 Question 6:
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    False.
    In decimal expansion, an n digit number N may be written as:
    N=\displaystyle\sum_{r=0}^n a_r 10^r \ , \ a_r \in \{0,1, . . . ,9 \}
    Let x=10^n-1 for some n \in \mathbb{N}.
    Then:
    x^2=(10^n-1)^2=10^{2n}-2(10^n)+1= \left( \displaystyle\sum_{r=n+1}^{2n} 9(10)^r \right) + 8(10^n)+1.
    Certainly then, we have n digits a_r \not= 0,1.
    Hence if n>1000 we have a square number with more than 1000 digits a_r \not= 0,1.
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    Test 3, Question 7:
    Thanks to Krollo for this solution.


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    Multiply by e^x and ponder over the similarity of both sides of the inequality.
    Solution:
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    By multiplying both sides by e^x > 0 we get e^x (a + \sin x) \leq e^y (a + \sin y).

    We are also given that x\leq y, so labelling f(x) = e^x (a + \sin x) our inequality reduces to x\leq y \Rightarrow f(x) \leq f(y).

    So what we really need to find is a such that f is increasing. This is easy, enough, we need only find a such that f'(x) = e^x(a + \sin x +\cos x) \geq 0, which is easy enough, we can divide through by e^x > 0 again and re-write \sin x + \cos x as \sqrt{2} \sin \left(x - \frac{\pi}{4}\right) which yields:

    a + \sqrt{2}\sin \left(x + \frac{\pi}{4}\right) \geq 0 \Rightarrow a \geq \sqrt{2}, so \min a = \sqrt{2}.

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    (Original post by DFranklin)
    You're pretty much there,

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    The part of your answer I've emphasized is exactly what's going on here. Basically when the particle is moving in the "+ve direction" it's performing SHM about the point x = -F/k, when it's moving in the -ve direction it's performing SHM about the point x = +F/k. At the points where it's still, you need to switch between the two, and so both "versions" of the motion need to agree when \dot{x} = 0. Which means that the amplitude must decrease by 2F/k at each reversal of direction (until you get to where the block stops moving).
    Yes, thanks for this. I've gone through this in detail now, and I've got a complete solution, I think. This is a pretty interesting result, and I can see why I was baffled last night (I didn't take the equations seriously enough and tried to rely on my intuition - my intuition was wrong).

    So, in full:

    Spoiler:
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    1. The no-motion condition is |x| < \frac{F}{k}

    2. The equations of motion are:

    \dot{x} > 0: m\ddot{x} = -kx -F \Rightarrow \ddot{x}+\frac{k}{m}x = -\frac{F}{m}

    \dot{x} < 0: m\ddot{x} = -kx +F \Rightarrow \ddot{x}+\frac{k}{m}x = \frac{F}{m}

    with corresponding solutions:

    \dot{x} > 0: x =A\cos(\omega t-\phi_A)-\frac{F}{k}

    \dot{x} < 0: x =B\cos(\omega t-\phi_B)+\frac{F}{k}

    with \omega^2=\frac{k}{m}. These solutions can be found by an easy CF+PI approach.

    Now let \dot{x} < 0 so that the particle is travelling to the left, and let x=x_0 > 0, \dot{x} = 0 at t=0

    So \dot{x} = -\omega B \sin(\omega t-\phi_B) \Rightarrow \dot{x}(0) = \omega B \sin\phi_B =0 \Rightarrow \phi_B=0

    (and a similar argument will give us that  \phi_A=0 later on)

    So we have x =B\cos(\omega t)+\frac{F}{k} and x(0)=B+\frac{F}{k} =x_0 \Rightarrow B=x_0-\frac{F}{k}

    So for \dot{x} < 0, we have x =(x_0-\frac{F}{k})\cos\omega t+\frac{F}{k}

    At the end of the first half-cycle:

    t=\frac{T}{2} = \frac{\pi}{\omega} \Rightarrow x(\frac{T}{2}) = (x_0-\frac{F}{k})\cos\pi +\frac{F}{k} =\frac{2F}{k}-x_0 = -(x_0-\frac{2F}{k})

    so the negative excursion is \frac{2F}{k} less than the original amplitude, x_0

    When the velocity instantaneously becomes +ve at t=\frac{T}{2}, we have x =A\cos\omega t-\frac{F}{k}, and by continuity of x(t), we have:

    \frac{2F}{k}-x_0 = A\cos\pi -\frac{F}{k} = -A-\frac{F}{k} \Rightarrow A=x_0-\frac{3F}{k}

    so for the second half-cycle, where \dot{x} > 0 , we have x =(x_0-\frac{3F}{k})\cos\omega t-\frac{F}{k}

    When t=T, x(T)=(x_0-\frac{3F}{k})\cos 2\pi-\frac{F}{k} = x_0-\frac{4F}{k} = (x_0-\frac{2F}{k})- \frac{2F}{k}

    so in the 2nd half-cycle, the positive excursion is \frac{2F}{k} less than the previous negative excursion, and by setting a new x'_0=x_0-\frac{4F}{k}, we can see that this pattern over the two half-cycles will repeat.

    3. Discussion

    a) Since \omega^2=\frac{k}{m}, the frequency of the motion is the same as that of the SHM that would arise without the frictional force.

    b) The total motion differs from that of damped SHM (with friction proportional to velocity) in that the amplitudes decrease by a constant amount on every half-cycle, as opposed to exponentially. However, the motion over each half-cycle is SHM, which we can see for the first half-cycle by setting u=x_-\frac{F}{k}, which gives us:

    x =(x_0-\frac{F}{k})\cos\omega t+\frac{F}{k} \Rightarrow x -\frac{F}{k}=(x_0-\frac{F}{k})\cos\omega t \Rightarrow u=u_0\cos\omega t

    with u_0=x_0-\frac{F}{k}. This is SHM with amplitude u_0 about the point \frac{F}{k} as measured on the x-axis. Note that the loss of energy due to friction manifests itself as a reduction in the amplitude as compared to the SHM without friction (which would be a constant x_0), but does not affect the frequency of the motion.

    c) At some point, the initial displacement at the end of movement to the left or right will fall below \frac{F}{k}, and the motion will cease.

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    (Original post by atsruser)
    Yes, thanks for this. I've gone through this in detail now, and I've got a complete solution, I think. This is a pretty interesting result, and I can see why I was baffled last night (I didn't take the equations seriously enough and tried to rely on my intuition - my intuition was wrong).

    So, in full:

    Spoiler:
    Show

    1. The no-motion condition is |x| < \frac{F}{k}

    2. The equations of motion are:

    \dot{x} > 0: m\ddot{x} = -kx -F \Rightarrow \ddot{x}+\frac{k}{m}x = -\frac{F}{m}

    \dot{x} < 0: m\ddot{x} = -kx +F \Rightarrow \ddot{x}+\frac{k}{m}x = \frac{F}{m}

    with corresponding solutions:

    \dot{x} > 0: x =A\cos(\omega t-\phi_A)-\frac{F}{k}

    \dot{x} < 0: x =B\cos(\omega t-\phi_B)+\frac{F}{k}

    with \omega^2=\frac{k}{m}. These solutions can be found by an easy CF+PI approach.

    Now let \dot{x} < 0 so that the particle is travelling to the left, and let x=x_0 > 0, \dot{x} = 0 at t=0

    So \dot{x} = -\omega B \sin(\omega t-\phi_B) \Rightarrow \dot{x}(0) = \omega B \sin\phi_B =0 \Rightarrow \phi_B=0

    (and a similar argument will give us that  \phi_B=0)

    So we have x =B\cos(\omega t)+\frac{F}{k} and x(0)=B+\frac{F}{k} =x_0 \Rightarrow B=x_0-\frac{F}{k}

    So for \dot{x} < 0, we have x =(x_0-\frac{F}{k})\cos\omega t+\frac{F}{k}

    At the end of the first half-cycle:

    t=\frac{T}{2} = \frac{\pi}{\omega} \Rightarrow x(\frac{T}{2}) = (x_0-\frac{F}{k})\cos\pi +\frac{F}{k} =\frac{2F}{k}-x_0 = -(x_0-\frac{2F}{k})

    so the negative excursion is \frac{2F}{k} less than the original amplitude, x_0

    When the velocity instantaneously becomes +ve at t=\frac{T}{2}, we have x =A\cos\omega t-\frac{F}{k}, and by continuity of x(t), we have:

    \frac{2F}{k}-x_0 = A\cos\pi -\frac{F}{k} = -A-\frac{F}{k} \Rightarrow A=x_0-\frac{3F}{k}

    so for the second half-cycle, where \dot{x} > 0 , we have x =(x_0-\frac{3F}{k})\cos\omega t-\frac{F}{k}

    When t=T, x(T)=(x_0-\frac{3F}{k})\cos 2\pi-\frac{F}{k} = x_0-\frac{4F}{k} = (x_0-\frac{2F}{k})- \frac{2F}{k}

    so in the 2nd half-cycle, the positive excursion is \frac{2F}{k} less than the previous negative excursion, and by setting a new x'_0=x_0-\frac{4F}{k}, we can see that this pattern over the two half-cycles will repeat.

    3. Discussion

    a) Since \omega^2=\frac{k}{m}, the frequency of the motion is the same as that of the SHM that would arise without the frictional force.

    b) The total motion differs from that of damped SHM (with friction proportional to velocity) in that the amplitudes decrease by a constant amount on every half-cycle, as opposed to exponentially. However, the motion over each half-cycle is SHM, which we can see for the first half-cycle by setting u=x_-\frac{F}{k}, which gives us:

    x =(x_0-\frac{F}{k})\cos\omega t+\frac{F}{k} \Rightarrow x -\frac{F}{k}=(x_0-\frac{F}{k})\cos\omega t \Rightarrow u=u_0\cos\omega t

    with u_0=x_0-\frac{F}{k}. This is SHM with amplitude u_0 about the point \frac{F}{k} as measured on the x-axis. Note that the loss of energy due to friction manifests itself as a reduction in the amplitude as compared to the SHM without friction (which would be a constant x_0), but does not affect the frequency of the motion.

    c) At some point, the initial displacement at the end of movement to the left or right will fall below \frac{F}{k}, and the motion will cease.

    Yes all looks good.

    One thing I'll add:

    If you're doing mechanics and have something like:

    \ddot{x} + kx = A, if you write u = x-A/k, you find

    \ddot{u} + ku = 0.

    That is, looking at u immediately shows we have SHM about u = 0.

    I don't know that it's objectively "better" than solving via CF+PI, but to me it shows more clearly what's going on.
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    (Original post by DFranklin)
    Yes all looks good.

    One thing I'll add:

    If you're doing mechanics and have something like:

    \ddot{x} + kx = A, if you write u = x-A/k, you find

    \ddot{u} + ku = 0.

    That is, looking at u immediately shows we have SHM about u = 0.

    I don't know that it's objectively "better" than solving via CF+PI, but to me it shows more clearly what's going on.
    Yes, I think that tends to crop up in questions where you have a weight hung from an spring vertically and you extend the system from its equilibrium point. It's objectively *quicker* than CF+PI, for sure, but it didn't occur to me immediately last night when I first wrote it up - still, it's just as well to demonstrate a variety of approaches, I guess.
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    Paper 3, Question 10

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    Momentum is conserved, but energy isn't.


    Solution
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    Suppose we model the dust cloud as a number of stationary particles, each of mass m.

    Initially the ship has speed U and mass M. Over its journey, its total momentum remains at MU by conservation of momentum.

    So as it hits the nth dust particle, the momentum conservation equation is

    MU = (M+nm)V.

    So when the speed of the ship is halved, V = U/2. It follows that nm = M.

    So the total mass of the dust that the ship collides with is M, and hence M=p*d*A where d is the distance travelled.

    So d = M/pa.

    (As mentioned earlier in the thread, we can also approach this as a single collision with a dust cloud of mass m, which I believe gives the same answer. I have no idea which, if either, is more correct...)

    Latex to come soon, as always.


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    Why aren't people touching the probability ones? What about paper 4 (from Krollo's interview prep thread?)
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    (Original post by shamika)
    Why aren't people touching the probability ones? What about paper 4 (from Krollo's interview prep thread?)
    :dontknow:

    I can't do any of them unfortunately and I think some people, especially those who can do them easily, have no need to come to this thread so they just aren't being done. If you have any you would like to do please go ahead it would be a big help

    I'll add Paper 4 now.
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    (Original post by Jordan\)
    :dontknow:

    I can't do any of them unfortunately and I think some people, especially those who can do them easily, have no need to come to this thread so they just aren't being done. If you have any you would like to do please go ahead it would be a big help

    I'll add Paper 4 now.
    Shotgun all the non probability ones

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    (Original post by Krollo)
    Shotgun all the non probability ones

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    Go ahead mate :lol: If I could do any of the others I would have them done already
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    Test 4 question 1

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    When you need to approximate a probability p think about how you might approximate firstly lnp. Can you work back from there?


    Solution
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    If the probability of one bond winning is 1/14000, then the probability of it not winning is 13999/14000. So the probability of at least one winning is 1-(13999/14000)^14000, evidently not 1.

    We need to approximate (13999/14000)^14000. Take natural log of it, giving 14000ln(1-1/14000) req 14000 * -1/14000 req -1 by the taylor expansion.

    So the probability required is about 1 - e^-1 req 0.63.



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    I've got Question 2 if you haven't done it already actually :lol:
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    (Original post by Jordan\)
    I've got Question 2 if you haven't done it already actually :lol:
    Go for it :-)

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    Test 4 Question 2

    @Jordan\ Whoops sorry I had not seen your above post as I was typing this.

    Hint:
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    I do not know how to hint this without giving it away, other than by saying that the Cayley-Hamilton will come in handy, and to follow the hint in the question.



    Solution:
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    The characteristic equation for this matrix is  \lambda^2 - 2\lambda + 1 = 0 and hence by the Cayley-Hamilton Theorem:
     A^2 = 2A - I (*) where I is the 2x2 identity matrix.

    Playing around with a few small numerical values of k suggests the relationship:
    A^k = kA - (k-1)I which can be seen to be true for k = 2 in (*).

    Assuming true for k = r we get that A^r = rA - (r-1)I .
    Multiplying by A:
     A^{r+1} = rA^2 - (r-1)A = r(2A - I) - (r-1)A = (r+1)A - rI
    So if it is true for k = r it is true for k = r+1. Hence it is true for all  k \geq 2 by induction.

    Therefore

    A^{1000} = 1000A - 999I =\begin{pmatrix} -5999 & 4000 \\ -9000 & 6001  \end{pmatrix}


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    (Original post by 16Characters....)
    Test 4 Question 2

    Jordan Whoops sorry I had not seen your above post as I was typing this.

    Hint:

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    I do not know how to hint this without giving it away, other than by saying that the Cayley-Hamilton will come in handy, and to follow the hint in the question.




    Solution:
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    The characteristic equation for this matrix is  \lambda^2 - \lambda + 1 = 0 and hence by the Cayley-Hamilton Theorem:
     A^2 = 2A - I (*) where I is the 2x2 identity matrix.

    Playing around with a few small numerical values of k suggests the relationship:
    A^k = kA - (k-1)I which can be seen to be true for k = 2 in (*).

    Assuming true for k = r we get that A^r = rA - (r-1)I . Multiplying by A:
     A^{r+1} = rA^2 - (r-1)A = r(2A - I) - (r-1)A = (r+1)A - rI so if it is true for k = r it is true for k = r+1. Hence it is true for all  k \geq 2 by induction.

    Therefore

    A^{1000} = 1000A - 999I =\begin{pmatrix} -5999 & 4000 \\ -9000 & 6001  \end{pmatrix}


    (Original post by Jordan\)
    Paper 4 Question 2

    Hint
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    Do the first few powers of A and notice a pattern.

    Solution
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     A = \begin{pmatrix} -5 & 4 \\-9 & 7 \\\end{pmatrix}



\Rightarrow A^{2} = \begin{pmatrix} -11 & 8 \\ -18 & 13 \\ \end{pmatrix}



\Rightarrow A^3 = \begin{pmatrix} -17 & 12 \\ -27 & 19 \\ \end{pmatrix}

    Hence  A^n = \begin{pmatrix} -5- 6(n-1) & 4n \\ -9n & 7 + 6(n-1) \\ \end{pmatrix}

     \therefore A^{1000} = \begin{pmatrix} -5- 6(1000-1) & 4(1000) \\ -9(1000) & 7 + 6(1000-1) \\ \end{pmatrix}



\Rightarrow A^{1000} = \begin{pmatrix} -5999 & 4000 \\ -9000 & 6001\\ \end{pmatrix}
    Two pleasingly different solutions. What maths is all about, perhaps

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    (Original post by Krollo)
    Two pleasingly different solutions. What maths is all about, perhaps

    Posted from TSR Mobile
    I agree.

    Definitely keep both solutions anyway, it will be good to have some alternatives.
 
 
 
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