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    (Original post by Felix Felicis)
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    Very good but it would be perfect if you fixed the typos for clarity!
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    (Original post by und)
    Very good but it would be perfect if you fixed the typos for clarity!
    Damn, only just noticed how many I made Hope it's all done now
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    Problem 37*/**

    Let S_{r} for r = 1, 2, 3, ..., 100 denote the sum of the infinite geometric series whose first term is \dfrac{r-1}{r!} with common ratio \dfrac{1}{r}.

    Evaluate \dfrac{100^{2}}{100!} + \displaystyle\sum_{r=1}^{100} |(r^{2} - 3r + 1) S_{r} |
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    Solution 37

    Clearly S_r= \dfrac{1}{(r-1)!} with S_1=0 hence:

    \big| (r^2-3r+1)S_r\big|=\bigg| \dfrac{r-1}{(r-2)!}-\dfrac{r}{(r-1)!}\bigg|=\dfrac{r-1}{(r-2)!}-\dfrac{r}{(r-1)!} when r>2

    and \cdots =1 for r=2

    Hence it all telescopes and the desired sum is 3

    Really Felix?
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    (Original post by Lord of the Flies)
    Really Felix?
    http://images.wikia.com/walkingdead/...Did-There..png
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    Time for some polynomial banter.

    Problem 38*

    Show that there is no polynomial p with integer coefficients such that p(a)=b,\;p(b)=c,\;p(c)=a for distinct integers a,b,c

    Problem 39*

    Show that if p is a polynomial satisfying p(x)+p'''(x)\geq p'(x)+p''(x) for all x then p(x)\geq 0
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    Solution to Problem 38:

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    p(p(p(a))) = p(p(b)) = p(c) = a

    p(p(p(b) = p(p(c)) = p(a) = b

    p(p(p(c) = p(p(a)) = p(b) = c

    Apart from f(x) = x, no such function exists where f(f(f(...f(x)...))) = x (I think)?
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    (Original post by ThatRandomGuy)
    Solution to Problem 38:

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    p(p(p(a))) = p(p(b)) = p(c) = a

    p(p(p(b) = p(p(c)) = p(a) = b

    p(p(p(c) = p(p(a)) = p(b) = c

    Apart from f(x) = x, no such function exists where f(f(f(...f(x)...))) = x (I think)?
    Spoiler:
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    Those equalities do not imply f(f(f(x))) = x for all x, only for three fixed integers a,b,c - so there is nothing preventing such a function from existing!
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    Problem 40*

    Find all functions f:\mathbb{R}\to \mathbb{R} which satisfy f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2  )-f(xy)) for all real numbers x and y.
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    (Original post by Noble.)
    Here's an additional problem:

    Problem 27 **/*** (the ** rating very loosely)

    Let a,b,c \in \mathbb{R}. Determine

    I =  \displaystyle\quad \iiint\limits_{x^2+y^2+z^2 \leq 1} \cos(ax+by+cz) \ \, \mathrm{d} x\,\mathrm{d} y\,\mathrm{d}z

    Also, show that your result is consistent with the fact the volume of the unit sphere is 4 \pi/3
    Could you give me a hint of how to go about evaluating this integral, because whichever coordinate system I use, I eventually run into something nasty that I don't know what to do with. It's been driving me mad...

    Part 1:

    If I try cartesian/cylindrical coordinates, then:
    Spoiler:
    Show
    The region in \mathbb{R} ^3 that we are integrating over is the volume enclosed by the unit sphere x^2+y^2+z^2= 1 , and so -1\leq x \leq 1,\ -1\leq y\leq 1,\ -\sqrt{1-x^2-y^2} \leq z\leq \sqrt{1-x^2-y^2}

    \displaystyle I= \int^1_{-1} \int^1_{-1} \int^{\sqrt{1-x^2-y^2} }_{-\sqrt{1-x^2-y^2} } \cos (ax+by+cz) \ dz\ dy\ dx

    \displaystyle = \int^1_{-1} \int^1_{-1} \frac{1}{c} \left[ \sin (ax+by+cz) \right]_{-\sqrt{1-x^2-y^2} }^{\sqrt{1-x^2-y^2} } \ dy\ dx

    \displaystyle = \int^1_{-1} \int^1_{-1} \sin (ax+by+c\sqrt{1-x^2-y^2} ) - \sin (ax+by+c-\sqrt{1-x^2-y^2} ) \ dy\ dx and thus I reach a dead end because of the square root term in the brackets.

    If I then try a similar approach with cylindrical coordinates and define the region \mathbb{R} ^3 that we are integrating over as 0\leq r\leq 1,\ 0\leq \theta \leq 2\pi ,\ -\sqrt{1-r^2} \leq z \leq \sqrt{1-r^2} then a very similar dead end occurs in which I have to then evaluate

    \displaystyle I= \frac{1}{c} \int^{2\pi }_{0} \int^1_0 r\sin (c\sqrt{1-r^2} +a\cos \theta r+b\sin \theta r) -r\sin (-c\sqrt{1-r^2} +a\cos \theta r+b\sin \theta r) \ dr\ d\theta

    and again the square root term in the brackets is a problem.


    So then I tried spherical coordinates, and:

    Spoiler:
    Show
    The region in \mathbb{R} ^3 that we are integrating over is 0\leq \rho \leq 1,\ 0\leq \theta \leq 2\pi ,\ 0\leq \phi \leq \pi and so

    \displaystyle I= \int^{\pi }_0 \int^{2\pi }_0 \int^1_0 \rho ^2 \sin \phi \cos (a\rho \sin \phi \cos \theta +b\rho \sin \phi \sin \theta +c\rho \cos \theta ) \ d\rho \ d\theta \ d\phi

    Let e = f(\theta ,\phi )= a\sin \phi \cos \theta +b\sin \phi \sin \theta +c\cos \theta just to make things more manageable and so:

    \displaystyle I= \int^{\pi }_0 \int^{2\pi }_0 \int^1_0 \rho ^2 \sin \phi \cos e\rho \ d\rho \ d\theta \ d\phi

    \displaystyle I= \int^{\pi }_0 \sin \phi \int^{2\pi }_0 \left[ \int^1_0 \rho ^2 \cos e\rho \ d\rho \right] \ d\theta \ d\phi

    Two iterations of IBP on the integral in square brackets and simplifying yields:

    \displaystyle I= \int^{\pi }_0 \int^{2\pi }_0 \left( \frac{\sin e}{e} +\frac{2\cos e}{e^2} - \frac{2\sin e}{e^3} \right) \ d\theta \ d\phi which again is a nightmare of a dead end.


    Where am I going wrong? :confused:

    Part 2:

    Spoiler:
    Show
    Since the region in \mathbb{R} ^3 that we are integrating over is enclosed by the unit sphere, then if we call its volume V, then

    \displaystyle V = \quad \iiint\limits_{x^2+y^2+z^2 \leq 1} 1 \ dx\ dy\ dz and comparing with

    I =  \displaystyle\quad \iiint\limits_{x^2+y^2+z^2 \leq 1} \cos(ax+by+cz) \ \, \mathrm{d} x\,\mathrm{d} y\,\mathrm{d}z ,

    it's obvious that we require \cos (ax+by+cz) =1 \Leftrightarrow ax+by+cz = 2k\pi , k\in \mathbb{Z} and so if I was able to get a result for the first part, I'd substitute this wherever it appears and it should hopefully evaluate to \dfrac{4\pi }{3} .
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    (Original post by Star-girl)
    Can you give me a hint of how to go about evaluating this integral, because whichever coordinate system I use, I eventually run into something nasty that I don't know what to do with. It's been driving me mad...

    Part 1:

    If I try cartesian/cylindrical coordinates, then:
    Spoiler:
    Show
    The region in \mathbb{R} ^3 that we are integrating over is the region enclosed by the unit sphere x^2+y^2+z^2\leq 1 , and so -1\leq x \leq 1,\ -1\leq y\leq 1,\ -\sqrt{1-x^2-y^2} \leq z\leq \sqrt{1-x^2-y^2}

    \displaystyle I= \int^1_{-1} \int^1_{-1} \int^{\sqrt{1-x^2-y^2} }_{-\sqrt{1-x^2-y^2} } \cos (ax+by+cz) \ dz\ dy\ dx

    \displaystyle = \int^1_{-1} \int^1_{-1} \frac{1}{c} \left[ \sin (ax+by+cz) \right]_{-\sqrt{1-x^2-y^2} }^{\sqrt{1-x^2-y^2} } \ dy\ dx

    \displaystyle = \int^1_{-1} \int^1_{-1} \sin (ax+by+c\sqrt{1-x^2-y^2} ) - \sin (ax+by+c-\sqrt{1-x^2-y^2} ) \ dy\ dx and thus I reach a dead end because of the square root term in the brackets.

    If I then try a similar approach with cylindrical coordinates and define the region \mathbb{R} ^3 that we are integrating over as 0\leq r\leq 1,\ 0\leq \theta \leq 2\pi ,\ -\sqrt{1-r^2} \leq z \leq \sqrt{1-r^2} then a very similar dead end occurs in which I have to then evaluate

    \displaystyle \frac{1}{c} \int^{2\pi }_{0} \int^1_0 r\sin (c\sqrt{1-r^2} +a\cos \theta r+b\sin \theta r) -r\sin (c-\sqrt{1-r^2} +a\cos \theta r+b\sin \theta r) \ dr\ d\theta and again the square root term in the brackets is a problem.


    So then I tried spherical coordinates, and:

    Spoiler:
    Show
    The region in \mathbb{R} ^3 that we are integrating over is 0\leq \rho \leq 1,\ 0\leq \theta \leq 2\pi ,\ 0\leq \phi \leq \pi and so

    \displaystyle I= \int^{\pi }_0 \int^{2\pi }_0 \int^1_0 \rho ^2 \sin \phi \cos (a\rho \sin \phi \cos \theta +b\rho \sin \phi \sin \theta +c\rho \cos \theta ) \ d\rho \ d\theta \ d\phi

    Let e = f(\theta ,\phi )= a\sin \phi \cos \theta +b\sin \phi \sin \theta +c\cos \theta just to make things more manageable and so:

    \displaystyle I= \int^{\pi }_0 \int^{2\pi }_0 \int^1_0 \rho ^2 \sin \phi \cos e\rho \ d\rho \ d\theta \ d\phi

    \displaystyle I= \int^{\pi }_0 \sin \phi \int^{2\pi }_0 \left[ \int^1_0 \rho ^2 \cos e\rho \ d\rho \left] \ d\theta \ d\phi

    Two iterations of IBP on the integral in square brackets and simplifying yields:

    \displaystyle I= \int^{\pi }_0 \int^{2\pi }_0 \left( \frac{\sin e}{e} +\frac{2\cos e}{e^2} - \frac{2\sin e}{e^3} \right) \ d\theta \ d\phi which again is a nightmare of a dead end.


    Where am I going wrong? :confused:

    Part 2:

    Spoiler:
    Show
    Since the region in \mathbb{R} ^3 that we are integrating over is enclosed by the unit sphere, then if we call its volume V, then

    \displaystyle V = \quad \iiint\limits_{x^2+y^2+z^2 \leq 1} 1 \ dx\ dy\ dz and comparing with

    I =  \displaystyle\quad \iiint\limits_{x^2+y^2+z^2 \leq 1} \cos(ax+by+cz) \ \, \mathrm{d} x\,\mathrm{d} y\,\mathrm{d}z ,

    it's obvious that we require \cos (ax+by+cz) =1 \Leftrightarrow ax+by+cz = 2k\pi , k\in \mathbb{Z} and so if I was able to get a result for the first part, I'd substitute this wherever it appears and it should evaluate to \dfrac{4\pi }{3} .
    Haha, it is a bit of a horrible question!

    Hint:

    Spoiler:
    Show
    You can rewrite the integral as I =  \displaystyle\quad \iiint\limits_{x^2+y^2+z^2 \leq 1} \cos(\textbf{a} \cdot \textbf{r}) \ \, \mathrm{d} x\,\mathrm{d} y\,\mathrm{d}z

    Where \textbf{a} = (a,b,c)

    Then, the vector

    \textbf{e}_3 = \dfrac{\textbf{a}}{|\textbf{a}|} = \dfrac{(a,b,c)}{\sqrt{a^2+b^2+c^  2}}

    is of unit length so can be extended to an orthonormal basis \textbf{e}_1, \textbf{e}_2, \textbf{e}_3 with associated coordinates X,Y,Z, where:

    \textbf{a} \cdot \textbf{r} = (0,0,|\textbf{a}|) \cdot (X,Y,Z) = |\textbf{a}|Z

    The unit sphere x^2+y^2+z^2 \leq 1 is still given as X^2+Y^2+Z^2 \leq 1 as \textbf{e}_1, \textbf{e}_2, \textbf{e}_3 are orthonormal and dV = dX \ dY \ dZ so:

    I =  \displaystyle\quad \iiint\limits_{X^2+Y^2+Z^2 \leq 1} \cos(\textbf{a} Z) \ \, \mathrm{d} X\,\mathrm{d} Y\,\mathrm{d}Z

    Which you can now change to spherical polar cooridinates.
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    (Original post by DJMayes)
    Problem 36: */**

    A particle is projected from the top of a plane inclined at an angle  \phi to the horizontal. It is projected down the plane. Prove that; if the particle is to attain it's maximum range, the angle of projection  \theta must satisfy:

     \theta = \dfrac{\pi}{4} - \dfrac{\phi}{2}
    Is \theta measured from the horizontal or from the plane?
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    (Original post by Felix Felicis)
    Is \theta measured from the horizontal or from the plane?
    The horizontal. Will edit that back in to make it clear.
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    (Original post by Noble.)
    Haha, it is a bit of a horrible question!

    Hint:

    Spoiler:
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    You can rewrite the integral as I =  \displaystyle\quad \iiint\limits_{x^2+y^2+z^2 \leq 1} \cos(\textbf{a} \cdot \textbf{r}) \ \, \mathrm{d} x\,\mathrm{d} y\,\mathrm{d}z

    Where \textbf{a} = (a,b,c)

    Then, the vector

    \textbf{e}_3 = \dfrac{\textbf{a}}{|\textbf{a}|} = \dfrac{(a,b,c)}{\sqrt{a^2+b^2+c^  2}}

    is of unit length so can be extended to an orthonormal basis \textbf{e}_1, \textbf{e}_2, \textbf{e}_3 with associated coordinates X,Y,Z, where:

    \textbf{a} \cdot \textbf{r} = (0,0,|\textbf{a}|) \cdot (X,Y,Z) = |\textbf{a}|Z

    The unit sphere x^2+y^2+z^2 \leq 1 is still given as X^2+Y^2+Z^2 \leq 1 as \textbf{e}_1, \textbf{e}_2, \textbf{e}_3 are orthonormal and dV = dX \ dY \ dZ so:

    I =  \displaystyle\quad \iiint\limits_{X^2+Y^2+Z^2 \leq 1} \cos(\textbf{a} Z) \ \, \mathrm{d} X\,\mathrm{d} Y\,\mathrm{d}Z

    Which you can now change to spherical polar cooridinates.
    A bit horrible is an understatement...

    Oooooh - thanks. Well I could have sat here and scratched my head but I would not have thought of that! :lol:

    EDIT: I'll write up the full solution another time.
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    (Original post by und)
    Problem 40*

    Find all functions f:\mathbb{R}\to \mathbb{R} which satisfy f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2  )-f(xy)) for all real numbers x and y.
    Spoiler:
    Show

    x=y=0 gives f(0)=0
    x=y gives f(x^3)=xf(x^2)
    x=-y gives f(x^3)=-f(-x^3)
    ..I'm sure I only need one or two more to show it can only be f(x)=x (or -x), but I can't think what. Blast!
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    Solution 40

    y=0\Rightarrow f(x^3)=xf(x^2)\;(\star) and set xz (x)=f(x).

    (\star)\Rightarrow x^3z(x^3)=x^3z(x^2)\Rightarrow z(x^3)=z(x^2)\Rightarrow z (x)=\mathcal{C} hence f(x)=\mathcal{C}x where \mathcal{C}=f(1)
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    My thoughts on Q39:

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    Firstly, notice that if P is a polynomial of odd degree, then P will tend towards negative infinity for either sufficiently large positive or negative x, faster than the other polynomials, so the statement cannot hold true. This means P is a polynomial of even degree with the first term having a positive co-efficient, so that it doesn't tend towards negative infinity, and must have a minimum turning point. At this turning point, P' = 0, P'' is greater than or equal to 0 as it is a minimum. Ignoring the third derivative, it's easy to see this imples P is not less than zero. However, I can't see how to bring the third derivative into this argument.

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    (Original post by DJMayes)
    Problem 36: */**

    A particle is projected from the top of a plane inclined at an angle  \phi to the horizontal. It is projected down the plane. Prove that; if the particle is to attain it's maximum range, the angle of projection  \theta from the horizontal must satisfy:

     \theta = \dfrac{\pi}{4} - \dfrac{\phi}{2}
    Solution 36
    Here you go beautiful

    Solution 36

    Let s_{h} denote the horizontal displacement and the projectile is launched with intial velocity v.

    R( \swarrow) s_{h} = ut + \dfrac{1}{2} at^{2}
    R( \swarrow) u = v \cos \theta, a = g \sin \phi, t = \dfrac{-2v \sin \theta}{g \cos \phi}

    \Rightarrow s_{h} = v \cos \theta \cdot \dfrac{-2v \sin \theta}{g \cos \phi} + \dfrac{1}{2} g \sin \phi \left( \dfrac{-2v \sin \theta}{g \cos \phi} \right)^{2}

    = \dfrac{-2v^{2} \sin \theta \cos \theta}{g \cos \phi} + \dfrac{2v \sin^{2} \theta \sin \phi}{g \cos^{2} \phi}

    =\dfrac{2v^{2}}{g \cos^{2} \phi} (- \sin \theta \cos (\theta + \phi))

    The maximum occurs when \dfrac{d s_{h}}{d \theta} = 0

    = sin \theta \sin (\theta + \phi ) - \cos (\theta) \cos (\theta + \phi) = 0

    = \cos (\theta + \phi + \theta) = 0 \Rightarrow 2 \theta + \phi = \dfrac{\pi}{2} \Rightarrow \theta = \dfrac{\pi}{4} - \dfrac{ \phi}{2} as required
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    Problem 41***


    A (countably) infinite class of Cambridge students who believe in the Axiom of Choice are taking their finals for the Mathematical Tripos. At the start of the exam, the examiner places either a white or a black hat on each student at random. There is only one question in the exam: "What colour is your hat?" If only a finite number of students answers incorrectly then everyone becomes a Wrangler (very good), otherwise the Wooden Spoon is given to the whole class (very bad). Everyone can see the hats of everyone else besides their own, and since it's exam conditions they are not allowed to communicate. Students may not remove their hats or try to look at it in any way.

    What strategy could the students devise together so to avoid failure and the impending doom of unemployment? Being clever clogs from Cambridge, memorising infinitely large amounts of data is no problem to the students.

    Note: A student suggested that each person just guess at random. However this has already been ruled out by the cleverer students who realised that if they were really unlucky, infinite of them could guess wrong.
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    Solution 38

    WLOG a<b<c. Now b \equiv a \pmod {c-a}, c \equiv b \equiv a \pmod {b-a} and b \equiv c \equiv a \pmod {c-b}.
    Hence b-a=c-a and c-b=b-a. Consequently, c-a=c+b-2a, a=b=c, which is a contradiction.
 
 
 
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