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STEP Maths I,II,III 1987 Solutions

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    (Original post by coffeym)
    STEP I Q5

    Let
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle \psi=\int^{\beta}_{\alpha}{\frac {1}{\sqrt{(x-\alpha)(\beta-x)}}\,dx


    Now if we let \displaystyle x=\alpha\cos^2{\theta}+\beta\sin  ^2{\theta} then we can say:

    \displaystyle (x-\alpha)=(\beta-\alpha)\sin^2{\theta} and

    \displaystyle (\beta-x)=(\beta-\alpha)cos^2{\theta}

    Also, \displaystyle dx=2(\beta-\alpha)\cos{\theta}\sin{\theta} (best not to write this in a simpler form)

    Substituting all of these results into the integral, and changing limits as required, we reach the result:

    \displaystyle \psi=\int^{\frac{\pi}{2}}_0{\fra  c{2(\beta-\alpha)\sin{\theta}\cos{\theta}}  {\sqrt{(\beta-\alpha)^2\sin^2{\theta}\cos^2{\t  heta}}}}\,d\theta

    \displaystyle \implies \psi=2\int^{\frac{\pi}{2}}_0{}\,  d\theta=\pi as required by the first part.

    If \displaystyle \alpha>\beta, then simply \psi=-\pi

    For the next part, let

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle \omega=\int^{\beta}_{\alpha}{\fr ac{1}{x\sqrt{(x-\alpha)(\beta-x)}}\,dx


    and make the substitution \displaystyle x=\frac{1}{y}\implies dx=-\frac{1}{y^2}dy

    Hence, after making the substitution, and swapping limits with help from the minus sign, we reach:

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle \omega=\int^{\frac{1}{\alpha}}_{ \frac{1}{\beta}}{{\frac{1}{y\sqr t{(\frac{1}{y}-\alpha)(\beta-\frac{1}{y})}}\,dx


    Multiplying the lone y inside the square root, we find that

    \displaystyle \omega=\int^{\frac{1}{\alpha}}_{  \frac{1}{\beta}}{{\frac{1}{\sqrt  {(1-\alpha y)(\beta y-1)}}}}\,dy

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle =\int^{\frac{1}{\alpha}}_{\frac{ 1}{\beta}}{{\frac{1}{\sqrt{\alph a\beta (\frac{1}{\alpha}-y)(y-\frac{1}{\beta})}}}\,dy


    Now note that, taking out the factor of \displaystyle \frac{1}{\sqrt{\alpha\beta}}, that the integral is simply the same as the one we evaluated in the first part, with certain symbols being inverted. This does not change the value of the integral, as everything is consistent. Hence

    \displaystyle \omega=\frac{1}{\sqrt{\alpha\bet  a}}\psi

    \displaystyle \implies \omega=\frac{\pi}{\sqrt{\alpha\b  eta}} as required.

    This completes the question.
    I heve retyped this in npost no. 235
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    1987 STEP II question 14

    Revised solution

     \text{Let }y \text{ be the distance of an element of the band from the vertex of the cone. See diagram }
     \text{component of force on element along slant face of cone is }T\delta\theta \sin \alpha
     \text{ and }T=\dfrac{\lambda (x-l)}{l} \text{ so force }=\dfrac{\lambda(x-l)\sin \alpha}{l}\delta \theta
    \text{Hence, ignoring gravitational effects, equation of motion is }
     - \dfrac{m \delta\theta}{2\pi}\dfrac{d^2y}{  dt^2}=\dfrac{\lambda(x-l)\sin\alpha}{l}\delta\theta
    y=\dfrac{r}{\sin\alpha}=\dfrac{x  }{2\pi\sin\alpha} \text{ so }\dfrac{d^2y}{dt^2}=\dfrac{1}{2 \pi \sin\alpha}\dfrac{d^2x}{dt^2}
    \text{Hence, }-\dfrac{m}{4\pi^2\sin\alpha} \dfrac{d^2x}{dt^2}= \dfrac{\lambda(x-l)\sin\alpha}{l} \Rightarrow \dfrac{d^2x}{dt^2}+\dfrac{4\pi^2  \lambda(x-l)\sin^2\alpha}{ml}=0
     \text{writing this as } \dfrac{d^2x}{dt^2}+ \dfrac{4 \pi^2 \lambda x \sin^2\alpha}{ml}= \dfrac{4\pi^2 \lambda l \sin^2 \alpha}{ml} \text{ we see that it is a standard differential equation }
    \text{ with C.F. }A\cos{kt}+B\sin{kt} \text{ where }k=2\pi sin \alpha\sqrt \dfrac{\lambda}{ml}
    x=l \text{ is an obvious P.I. so general solution is }x=A \cos{kt}+B \sin{kt}+l
     \dfrac{dx}{dt}=0 \text{ when }t=0 \Rightarrow B=0 \text{ so } x=A\cos{kt}
    \dfrac{d^2x}{dt^2}=0 \text{ when }x=l \Rightarrow A=- \dfrac{1}{k^2} \text{ so }x=-\dfrac{1}{k^2}\cos{kt}+l
     \text{Band will become slack when }x=l \Rightarrow \cos{kt}=0 \Rightarrow t_0= \dfrac{ \pi}{2k}= \dfrac{1}{4 \sin \alpha } \sqrt{ \dfrac{ml}{ \lambda}}
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    1987 STEP II question 16

    Revised solution

    \text{Let }X =[\text{There is a body in the fridge}],H=[\text{ P says there is}] \text{ and} K=[\text{Q says there is}]
    \text{Then we require }Pr(X|H \cup K)= \dfrac{Pr(X \cup H \cup K)}{Pr(H \cup K)}= \dfrac{\frac{1}{2}pq}{\frac{1}2p  q+\frac{1}{2}(1-p)(1-q)}=\dfrac{pq}{1-p-q+2pq}

     \text{Now let }S=[\text{A is in the fridge}] \text{ and }T=[\text{B is in the fridge}], L=[\text{P says A is}],M=[\text{Q says A is}]
    \text{Then we require }Pr(S \cup T | L \cup M)= \dfrac{\frac{1}{3}pq}{_\frac{2}{  3}pq+\frac{1}{3}(1-p)(1-q)}=\dfrac{pq}{1-p-q+3pq}}
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    (Original post by SimonM)
    STEP III, Question 8

    Spoiler:
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    Since P_0(x) = 1

    From \displaystyle \frac{dP_1}{dx} = 1P_0

    P_1(x) = x +c

    \displaystyle  \int_0^1 (x+c) \, dx = \frac{1}{2} +c

    So \displaystyle P_1 (x) = x - \frac{1}{2}

    Base case:

    P_1(x+1)-P_1(x) = x+1-\frac{1}{2} - \left  (x+\frac{1}{2} \right) = 1 = 1x^0

    So we're done.

    Inductive Case:

    Let f_n ( x) = P_n(x+1)-P_n(x)

    Therefore \dfrac{df_n}{dx} = n(n-1)x^{n-2}

    Integrating, we get f_n = nx^{n-1}+C

    Integrating P_{n-1} we get

    \displaystyle \int_0^1 nP_{n-1} (x) \, dx = P_n(1)-P_n(0)=0

    So f_n(0)=0

    Therefore f_n(x)=nx^{n-1} for all integers n \ge 1

    \displaystyle n\sum_{m=0}^k m^{n-1} = \sum_{m=0}^k \left ( P_n (m+1)-P_n(m)\right) = P_n(k+1)-P_n(0)

    P_1(x)=x-\frac{1}{2}

    Therefore

    P_2 (x) = x^2 -x +c

    \displaystyle \int_0^1 \left (  x^2 -\frac{1}{2}x +c \right) \, dx = \left [ \frac{x^3}{3} - \frac{x^2}{4} + cx \right ]_0^1 = \frac{1}{3} - \frac{1}{2} + c

    Therefore P_2(x) = x^2-x+\frac{1}{6}

    Repeating a couple of times we get

    P_4(x) = x^4-2x^3+x^2-1/30

    Therefore

    \displaystyle 4\sum_{m=0}^{1000} m^3 = P_4(1001)-P_4(0)

    Therefore

    \displaystyle \sum_{m=0}^{1000} m^3 = \frac{t^4-2t^3+t^2}{4} = \left ( \frac{t(t-1)}{2} \right)^2 = \frac{(1001)(1000)}{2}^2 = (500500)^2
    This is actually question 10 NOT 8.
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    1987 STEP III question 8

     \text{Consider the parallelogram }ABCD \text{ where the angle between the diagonals is }\theta
     \text{Area of }ABCD=2 \times \dfrac{1}{2} \left(AD \times\dfrac{BD}{2} \sin \theta \right) =\dfrac{1}{2}AC.BD\sin \theta
    \text{which is clearly a maximum when }\sin \theta=1 \Rightarrow \text{ diagonals perpendicular}

    \text {if }AC \geq BD \text{ then } \dfrac{1}{2}AC.BD \sin \theta \leq AC^2 \sin \theta \leq \dfrac{1}{2}AC^2

     A \text{ is the parallelogram with sides defined by }
     a_1x+b_1y=\pm \delta \text{ and }a_2x+b_2y+\pm \delta
    \text{and is the inverse image under the linear map by the matrix }M=\begin{pmatrix} a_1\ b_1\\a_2\ b_2 \end{pmatrix}
    \text{of the square with vertices }(\pm\delta,\pm\delta) \text{ and hence of area }4\delta^2 \text{ so }A\text{ has area }\dfrac{4\delta^2}{|a_1b_2-a_2b_1|} ]
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    (Original post by ben-smith)
    (I've posted some solutions in the 1992 thread if you care)
    STEP III Q11

    Call the position of the mother A, the point where she enters the river B, and the position of the child C.
    AB=\sqrt{a^2+(b-x)^2}

BC=\sqrt{x^2+c^2}
    So, total time T=\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{x^2+c^2}}{  v}
    Differentiating, T'=\frac{x-b}{u\sqrt{a^2+(b-x)^2}}+\frac{x}{v\sqrt{x^2+c^2}}
    set the derivative to zero to find the minimum time. x must satisfy that.
    For the next part, AB is the same but the velocity in BC has an extra v in it's horizontal component. Think of it's velocity in terms of her swimming vector incllned at an angle \theta and then add on the horizontal component due to the river at the end. Because she can only swim in straight lines:
    \frac{x}{c}=\frac{vcos\theta +v}{vsin\theta}. Using the identity:
    1+Tan^2\theta =sec^2\theata and rearranging we get:
    (c^2-x^2)sec^2\theta+2c^2sec\theta+c^  2+x^2=0 which is a quadratic in sec\theta, so, using the quadratic formula:
    sec\theta=\frac{-2c^2\pm\sqrt{4c^4-4(c^2-x^2)(c^2+x^2)}}{2(c^2-x^2)}=\frac{-c^2 \pm x^2}{c^2-x^2}. We can discard the 'plus' root as it gives theta to be \pi which makes no sense because the mother would never reach the child if she had no vertical component in her velocity.So:
     sec\theta=\frac{-c^2 - x^2}{c^2-x^2} \Rightarrow cos\theta=\frac{c^2-x^2}{-c^2-x^2}
    To find the speed for BC we use pythagoras:
    speed_{BC}=\sqrt{v^2sin^2\theta+  (vcos\theta+v)^2}=\sqrt{v^2(2cos \theta+2)}, substituting in and simplifying:
    speed_{BC}=\frac{2vx}{\sqrt{c^2+  x^2}}
    so, just like in the previous part:
    T=\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{c^2+x^2}}{  \frac{2vx}{\sqrt{c^2+x^2}}}= \frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{c^2+x^2}{2vx} and differentiating we get:
    T'= \frac{x-b}{u\sqrt{a^2+(b-x)^2}}+\frac{4vx^2-2vx^2-2vc^2}{4v^2x^2}= \frac{x-b}{u\sqrt{a^2+(b-x)^2}}+\frac{x^2-c^2}{2vx^2}
    letting T'=0 to find minima and rearranging:
    \frac{b-x}{u\sqrt{a^2+(b-x)^2}}= \frac{x^2-c^2}{2vx^2} \Rightarrow 2vx^2(b-x)=u(x^2-c^2)[a^2+(b-x)^2]^{1/2} as required.
    Can't really upload the sketches so if anyone wants to do it be my guest
    Hey i have a much easier way in showing the second equation in x. Think of the relative velocity of the river to the position of the girl in the river.
    (x-vt) is the vertical distance as she enters the river at time t
    c is the horizontal distance still.
    Then by pythagoras as before (where t is the journy time in the river):
    t=\frac{\sqrt{c^2+(x-vt)^2}}{v}
    t^2v^2=c^2+x^2-2vtx+v^2t^2
    Rearrange for t (the square bits in t cancel)
    t=\frac{x^2+c^2}{2vx}
    as desired
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    (Original post by themaths)
    Hey i have a much easier way in showing the second equation in x. Think of the relative velocity of the river to the position of the girl in the river.
    (x-vt) is the vertical distance as she enters the river at time t
    c is the horizontal distance still.
    Then by pythagoras as before (where t is the journy time in the river):
    t=\frac{\sqrt{c^2+(x-vt)^2}}{v}
    t^2v^2=c^2+x^2-2vtx+v^2t^2
    Rearrange for t (the square bits in t cancel)
    t=\frac{x^2+c^2}{2vx}
    as desired
    Sorry, I don't follow. how is that last result the same as mine? Have you skipped over the calculus? I'm baffled as to how you could find the x that results in the shortest possible total time time without calculus.
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    (Original post by ben-smith)
    Sorry, I don't follow. how is that last result the same as mine? Have you skipped over the calculus? I'm baffled as to how you could find the x that results in the shortest possible total time time without calculus.
    Oh no i did the calculus part exactly the same as you to find the minimum time you obviously need to take dT/dx as 0 to find the minimum.
    I just meant for the part where you were using theta and omitting roots of quadratics and things to find her journey time in the river, when i did it i just used relative velocities to find the time she was in the river. Everything else after that i did the same as you (assuming the diagrams were the same also but they were quite standard after getting the result).
    Sorry i didn't really make it clear reading my comment back
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    (Original post by themaths)
    Oh no i did the calculus part exactly the same as you to find the minimum time you obviously need to take dT/dx as 0 to find the minimum.
    I just meant for the part where you were using theta and omitting roots of quadratics and things to find her journey time in the river, when i did it i just used relative velocities to find the time she was in the river. Everything else after that i did the same as you (assuming the diagrams were the same also but they were quite standard after getting the result).
    Sorry i didn't really make it clear reading my comment back
    Oh, I see. Well, your way is definitely simpler, it's good to show both methods I guess.
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    (Original post by brianeverit)
    1987 STEP Fma numbers 12 -16
    For question 12, aren't polar coordinates supposed to be taken anticlockwise from the +ve x axis? It probably doesn't matter, it just took me a while to realise why my answer differed from yours.


    (Original post by SimonM)
    ...
    Alternative solution to the first bit of question 12 STEP II.
    Brian did this by using an energy argument. I have a different way.
    let \mathbf{\gamma},\mathbf{T} denote the unit tangent vector and the tension in the string respectively.
    If the bead can be in static equilibrium fo all positions then:
    m \mathbf{g}. \mathbf{\gamma}+ \mathbf{T}.\mathbf{\gamma}=0

\Rightarrow (m \mathbf{g}+\mathbf{t}).\mathbf{ \gamma}=0

\Rightarrow-mg\begin{pmatrix} cos\theta \\ 1+sin\theta\end{pmatrix}.\frac{ \mathbf{ \dot{r}}}{|\mathbf{\dot{r}}|}=0 \Rightarrow  -mg\begin{pmatrix} cos\theta \\ 1+sin\theta\end{pmatrix}. \begin{pmatrix} \dot{r}cos\theta-rsin\theta \\ \dot{r} sin\theta+rcos\theta \end{pmatrix} \frac{1}{\sqrt{\dot{r}^2+r^2}}=0

\Rightarrow \dot{r}cos^2\theta-rsin\thetacos\theta+\dot{r} sin\theta+rcos\theta+\dot{r}sin^  2\theta+rsin\thetacos\theta=0

\Rightarrow \dot{r}(1+sin\theta)+rcos\theta=  0

\Rightarrow \dot{r}+r\frac{cos\theta}{1+sin\  theta}=0
    Multiplying through by an integrating factor of e^{ln(1+sin\theta)} we get:
    \frac{d}{d \theta}[re^{ln(1+sin\theta)}]=0 \Rightarrow r=\frac{C}{sin\theta+1} where C is a constant.
    The minimum obviously occurs when sin\theta=1 \Rightarrow r=\frac{2d}{sin\theta+1}
    Note that this is the same answer as Brian's. I am taking angles about the x axis whereas he has taken them about the y.
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    (Original post by Swayum)
    STEP I, question 4.

    Let S = \mathrm{log}_2 e - \mathrm{log}_4 e + \mathrm{log}_{16} e...

    Take each term to logarithm base 2 (change the base rule)

    \displaystyle S = \frac{\mathrm{log}_2 e}{1} - \frac{\mathrm{log}_2 e}{2} +  \frac{\mathrm{log}_2 e}{4}...

    S = (\mathrm{log}_2 e})(1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32...)

    Split the second brackets into two parts.

    P = 1 + 1/4 + 1/16...
    N = -1/2 - 1/8 - 1/32...

    So

    S = (\mathrm{log}_2 e})(P + N)

    P is a geometric series with first term 1 and common ratio 1/4. Its sum to infinity is:

    P = 1/(1-1/4) = 4/3

    N is a geometric series with first term -1/2 and common ratio 1/4. Its sum to infinity is:

    N = (-1/2)/(1-1/4) = -2/3

    S = (\mathrm{log}_2 e})(P + N) = (\mathrm{log}_2 e})(4/3 - 2/3)

    S = 2/3(\mathrm{log}_2 e})

    Change the base to e, so we have the natural logarithm.

    \displaystyle S = 2/3(\mathrm{log}_2 e}) = 2/3(\frac{\mathrm{log}_e e}{\mathrm{log}_e 2}) = 2/3(\frac{1}{\mathrm{log}_e 2})
    \displaystyle S = \frac{2}{3\mathrm{ln} 2}

    Using the power rule

    \displaystyle S = \frac{2}{3\mathrm{ln} 2} = \frac{2}{\mathrm{ln} 8}


    \displaystyle S = \frac{2*\frac{1}{2}}{\frac{1}{2}  \mathrm{ln} 8} (multiplying top and bottom by 1/2)

    \displaystyle S = \frac{1}{\frac{1}{2}\mathrm{ln} 8} = \frac{1}{\mathrm{ln} \sqrt 8}

    \displaystyle S = \frac{1}{\mathrm{ln} 2\sqrt 2}

    WWWWW.
    Slightly more succinct and using the given hint.....
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    (Original post by brianeverit)
    1987 Paper 2 no.11
    Your vector product seems to be wrong, it should be sin(theta) in the first term rather than cos(theta), though it get's the right answer at the end.
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    (Original post by brianeverit)
    1987 Paper 2 no.11
    Follow up post. I'm now not sure you have the right answer. Suppose theta is zero. Then there doesn't need to be any frictional force to keep rod in place. Hence coefficient of friction can be zero, so "mu" = tan(alpha)sin(theta).
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    (Original post by nota bene)
    I'll post up my attempt to a solution to I/7, using the approach with taking the imaginary parts...
    Spoiler:
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    \sin(\frac{2\pi}{23})+\sin(\frac  {6\pi}{23})+...+ \sin(\frac{42\pi}{23})=\displays  tyle\sum_{k=0}^{10} \sin(\frac{(2+4k)\pi}{23})=\Im\d  isplaystyle\sum_{k=0}^{10}e^{i\d  isplaystyle\frac{(2+4k)\pi}{23}}

    \displaystyle\sum_{k=0}^{10}e^{i  \displaystyle\frac{(2+4k)\pi}{23  }}=e^{i\displaystyle\frac{2\pi}{  23}}\displaystyle\sum_{k=0}^{10}  e^{i\displaystyle\frac{4k\pi}{23  }}
    Note that the later is a G.P., 1+x+x^2+...+x^n where x=e^{\frac{4\pi}{23}}.
    Using the sum for a G.P. and u=\frac{2\pi}{23} we have:
    e^{iu}\times\displaystyle\frac{1-e^{22iu}}{1-e^{2iu}}=e^{iu}\times\frac{e^{11  iu}(e^{-11iu}-e^{11iu})}{e^{iu}(e^{-iu}-e^{iu})}=\frac{e^{11iu}(e^{-11iu}-e^{11iu}}{e^{-iu}-e^{iu}}
    Now, note that e^{-a}-e^a=-2\sin(a) [is shown by writing it as cos(-a)+isin(-a)-cos(a)-isin(a)=-isin(a)-isin(a)].

    Then we have e^{11iu}\frac{\sin(11u)}{\sin(u)  }=\frac{\sin(11u)}{\sin(u)}(\cos  (11u)+i\sin(11u))

    Now, taking the imaginary part \Im\frac{\sin(11u)}{\sin(u)}(\co  s(11u)+i\sin(11u))=\frac{\sin^2(  11u)}{\sin(u)}
    And substituting back for u:
    \displaystyle\frac{\sin^2(\frac{  22\pi}{23})}{\sin(\frac{2\pi}{23  })}=\displaystyle\sum_{k=0}^{10} \sin(\frac{(2+4k)\pi}{23})
    Note that\sin(\frac{\pi}{23})=\sin(\frac{  22\pi}{23}), so \frac{\sin^2(\frac{\pi}{23})}{2\  sin(\frac{\pi}{23})\cos(\frac{\p  i}{23})}=\frac{sin(\frac{\pi}{23  })}{2\cos (\frac{\pi}{23})}=\frac{1}{2}\ta  n(\frac{\pi}{23})

    To the second part, where the series concerned can be written as \displaystyle\sum_{k=0}^{10}(-1)^k\sin(\frac{(2+4k)\pi}{23})=\  Im \displaystyle\sum_{k=0}^{10}(-1)^ke^{i\frac{(2+4k)\pi}{23}}
    \displaystyle\sum_{k=0}^{10}(-1)^ke^{i\frac{(2+4k)\pi}{23}}=e^  {i\frac{2\pi}{23}}\displaystyle\  sum_{k=0}^{10}(-e^{i\frac{4\pi}{23}})^k
    The latter is now a G.P.
    Again, introducing u=\frac{2\pi}{23} will simplify the latexing a bit...
    e^{iu}\frac{1+e^{2iu(k+1)}}{1+e^  {2iu}}=e^{iu}\frac{e^{iu(k+1)}(e  ^{-iu(k+1)}+e^{iu(k+1)})}{e^{iu}(e^  {-iu}+e^{iu})}=\frac{e^{iu(k+1)}(e  ^{-iu(k+1)}+e^{iu(k+1)})}{e^{-iu}+e^{iu}}
    Now, use the fact that e^{-a}+e^a=2\cos(a)[show it by writing cos(-a)+isin(-a)+cos(a)+isin(a)=cos(a)+cos(a)]

    Therefore \frac{e^{iu(k+1)}(e^{-iu(k+1)}+e^{iu(k+1)})}{e^{-iu}+e^{iu}}=\frac{\cos(u(k+1))}{  \cos(u)}(\cos(u(k+1))+i\sin(u(k+  1))
    Now, taking the imaginary part (and substituting back for u):
    \Im\frac{\cos(u(k+1))}{\cos(u)}(  \cos(u(k+1))+i\sin(u(k+1))=\frac  {\cos(\frac{2\pi}{23}(k+1))\sin( \frac{2\pi}{23}(k+1))}{\cos(\fra  c{2\pi}{23})}=\sin(\frac{2\pi}{2  3})-\sin(\frac{6\pi}{23})+...-\sin(\frac{38\pi}{23})+\sin(\fra  c{42\pi}{23})
    By earlier argument  \sin(\frac{22}{23}\pi)=\sin(\fra  c{1}{23}\pi) and by symmetry follows that \sin(\frac{44}{23}\pi)=- \sin(\frac{2}{23}\pi)
    Putting k=10 into my above general formula we have \displaystyle\frac{\cos( \frac{22}{23} \pi) \sin( \frac{22}{23}\pi)}{ \cos( \frac{2}{23} \pi)} =  \frac{ \sin( \frac{44}{23} \pi)}{2 \cos( \frac{2}{23} \pi)}[using the product formula for cos(a)sin(b)] And the top of that fraction is equal to - \sin(\frac{2}{23}\pi) and thus the required answer in -\frac{1}{2}\tan(\frac{2}{23}\pi)


    Tell me if I've made some silly mistakes, quite likely...

    edit: The solution should now be correct.

    I have compressed things slightly and I think you have the wrong sign for the last part...see document....
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    (Original post by mikelbird)
    I have compressed things slightly and I think you have the wrong sign for the last part...see document....
    corrected document
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    (Original post by Seernb)
    Follow up post. I'm now not sure you have the right answer. Suppose theta is zero. Then there doesn't need to be any frictional force to keep rod in place. Hence coefficient of friction can be zero, so "mu" = tan(alpha)sin(theta).
    Thanks for pointing out the error. I agree with your answer.
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    (Original post by mikelbird)
    Slightly more succinct and using the given hint.....
    Sorry for being a bit daft, but could you please tell me how you got from:

    \sum_{r=0}^{\infty}(-1)^{r}\frac{1}{2^{r}}

    to:

    \frac{1}{1.5}

    I have already done this question, but I can't see the link between the two without having to write a couple terms out.
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    (Original post by Blazy)
    Sorry for being a bit daft, but could you please tell me how you got from:

     \sum_{n=0}^{\infty}\frac{1}{2^{r  }}(-1)^{r}

    to:

     \frac{1}{1.5}

    I have already done this question, but I can't see the link between the two without having to write a couple terms out.
    GP? (-1)^r \frac{1}{2^r} = \left(-\frac{1}{2}\right)^r. The good old formula should work, then think about where this infinity symbol goes.
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    (Original post by gff)
    GP? (-1)^r \frac{1}{2^r} = \left(-\frac{1}{2}\right)^r. The good old formula should work, then think about where this infinity symbol goes.
    Yeah I got there eventually...D:. Thanks though

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