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    STEP III, Question 12:

    (i)
    Since X_1 \sim B(k, 1/2) then E(X_1) = \frac{k}{2} and it follows easily that E(X_2) = \frac{1}{2}E(X_1) = \frac{k}{4}.

    In general, we have E(X_n) = \frac{1}{2} E(X_{n-1}) and hence E(X_n) = \frac{k}{2^{n-1}}.

    This gives us that

    \displaystyle

\begin{equation*}  \sum_{i=1}^{\infty} E(X_i) = \sum_{i=1}^{\infty} k \frac{1}{2^{i-1}} = k \times \frac{1/2}{1 - 1/2} = k\end{equation*}

    by sum of a G.P.


    (ii)
    Note that \mathbb{P}(Y_i = r) = \left(\frac{1}{2}\right)^{r+1} by a geometric distribution argument.

    Hence, it follows immediately that the probability generating function for each Y_i is:

    \displaystyle 

\begin{equation*}G_{Y_i}(t) = \frac{1}{2} + \frac{1}{4}t+ \frac{1}{8}t^2 + \cdots = \frac{1/2}{1 - \frac{t}{2}} = \frac{1}{2-t}\end{equation*}

    So that their sum is given by the product of each p.g.f, aka: G_Y (t) = (2-t)^{-k}.

    Taking the first derivative gives us G'_{Y}(1) = k and the second derivative is similarly G''_Y(1) = k(k+1).

    We then have E(Y) = k and \text{Var}(Y) = k(k+1) + k - k^2 = 2k.

    Taking the nth derivative, we see that:

    \displaystyle 

\begin{equation*}G^(r)_Y = \frac{k(k+1)\cdots (k+r-1)}{(2-t)^{k+r}} \Rightarrow \mathbb{P}(Y = r) = \frac{k(k+1)\cdots (k + r - 1)}{2^{k+r} r! } = \frac{1}{2^{k+r}} ^{k+r-1}\mathrm{C}_{r}\end{equation*}
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    STEP III, Question 3:

    (i)
    Since \displaystyle f(t) = \frac{t}{e^t -1} = \frac{t}{1 + t + o(t^2) - 1} = \frac{1}{1 + o(t)} then we have that, as t \to 0 \Rightarrow f(t) \to \frac{1}{1 + 0} = 1.

    Furthermore, \displaystyle f'(t) = \frac{e^t - 1 - te^t}{(e^t - 1)^2} = \frac{t + t^2/2 - t -t^2 - t^3/2 + o(t^4)}{t^2 + o(t^3)} = \frac{1 - \frac{1}{2!} + o(t)}{1 + o(t)} so that, as t \to 0 we have f'(t) \to -\frac{1}{2}


    (ii)
    Let:

    \displaystyle

\begin{equation*}g(t) = f(t) + \frac{t}{2} = \frac{t(e^t + 1)}{2(e^t - 1)}

    Then:

    \displaystyle 

\begin{equation*}g(-t) \equiv \frac{-t(e^{-t} + 1)}{2(e^{-t} -1 )} \times \frac{e^t}{e^{t}} \equiv \frac{t(e^t + 1)}{2(e^t) -1} \equiv g(t)\end{equation*}

    So g(t) is even, as required.


    (iii)
    The sketch is easy enough, note that it has a maximum at t=0, asymptotic to the t-axis as t \to -\infty and diverges to negative infinity as t \to \infty gives us our nice little graph:



    Since t \neq 0 then e^t(1-t) \neq 1 and hence the numerator of our derivative never vanishes, so the entire derivative never vanishes. This is instrumental in the next sketch.


    sketch


    Which follows pretty much elementarily from our above results.
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    Doing III Q8 right now .
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    STEP III, Question 8

    (i)
    Let t=-\ln x. Then 0=\displaystyle\lim_{x \to \infty}e^{-mt}t^n=\displaystyle\lim_{x \to 0}e^{m\ln x}(-\ln x)^n=\displaystyle\lim_{x \to 0} x^m(\ln x)^n

    as required (taking m=n=1 in first result).

    \displaystyle\lim_{x \to 0}x^x=\displaystyle\lim_{x \to 0}e^{x\ln x}=e^0=1

    as required.


    (ii)

    I_{n+1}=\displaystyle\int^1_0 x^m(\ln x)^{n+1}\ dx

=\displaystyle\lim_{\delta \to 0^+}\displaystyle\int^1_\delta x^m(\ln x)^{n+1}\ dx

=\displaystyle\lim_{\delta \to 0^+} \left[ \dfrac{x^{m+1}}{m+1}(\ln x)^{n+1}\right]_{x=\delta}^1-\displaystyle\int^1_0 \dfrac{x^{m+1}}{m+1}(n+1)(\ln x)^n(\dfrac{1}{x})\ dx

=\dfrac{1}{m+1}\displaystyle\lim  _{\delta \to 0^+}(-\delta^{m+1}(\ln\delta)^{n+1})-\dfrac{n+1}{m+1} \displaystyle\int^1_0 x^m(\ln x)^n\ dx

= -\dfrac{n+1}{m+1}I_n

    as required.

    Therefore

    I_n=\left(-\dfrac{n}{m+1}\right)I_{n-1}=\left(\dfrac{-n}{m+1}\right)\left(-\dfrac{n-1}{m+1}\right)I_{n-2}=\ldots=\left(\dfrac{-n}{m+1}\right)\left(-\dfrac{n-1}{m+1}\right)\left(-\dfrac{n-2}{m+1}\right) \ldots \left(-\dfrac{1}{m+1}\right)I_0

=\dfrac{(-1)^nn!}{(m+1)^n} \displaystyle \int^1_0 x^m\ dx=\dfrac{(-1)^nn!}{(m+1)^n} \left[\dfrac{x^{m+1}}{m+1}\right]_{x=0}^1=\dfrac{(-1)^nn!}{(m+1)^n}.


    (iii)
    \displaystyle\int^1_0 x^x\ dx=\displaystyle\int^1_0 e^{x\ln x}\ dx=\displaystyle\int^1_0 \left(1+x\ln x+\dfrac{x^2(\ln x)^2}{2!}+\dfrac{x^3(\ln x)^3}{3!}+\ldots \right)\ dx



=1+\dfrac{(-1)^11!}{2^2}+\dfrac{(-1)^22!}{3^3(2!)}+\dfrac{(-1)^33!}{4^4(3!)}+\ldots



=1 -- \left(\dfrac{1}{2}\right)^2+ \left(\dfrac{1}{3}\right)^3 -\left(\dfrac{1}{4}\right)^4+ \ldots

    as required.
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    Done Q7 I can try and type it up now.
    II
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    STEP II Question 7
    (i)
    Spoiler:
    Show
    If  \displaystyle y=(x-a)^ne^{bx}\sqrt{1+x^2} , using product rule in  (x-a)^ne^{bx} factor first and then using product rule in conjunction with the  \sqrt{1+x^2} , we obtain

     \displaystyle \frac{dy}{dx}=\frac{ \left (x-a)^{n-1}(bx^3+(n-ab+1)x^2+(b-a)x+n-ab \right ) }{\sqrt{1+x^2}} .

    So we could write  q(x) = bx^3+(n-ab+1)x^2+(b-a)x+n-ab .

    (i)
    Let \displaystyle I =\int \frac{(x-4)^{14}e^{4x}(4x^3-1)}{\sqrt{1+x^2}} dx , notice that this integrated is in the exam form of the pervious result with n=15, a=4 and b=4.
    Plugging these into the expression given for y at the start of the question yields
     \boxed{ \displaystyle I=(x-4)^{15}e^{4x}\sqrt{1+x^2}+c} .
    (ii)
    Spoiler:
    Show
    If we let  \displaystyle J = \int \frac{(x-1)^{21}e^{12x}(12x^4-x^2-11)}{\sqrt{1+x^2}} dx .
    Noticing that at the moment there is a quartic expression instead of a cubic expression, we factorise and it is clear that (x-1) is a factor of the quartic.
    Factorising  \displaystyle 12x^4-x^2-11 \equiv (x-1)(12x^3+12x^2+11x+11) and we see that this means that integral,  J , becomes

     \displaystyle \int \frac{(x-1)^{22}e^{12x}(12x^3+12x^2+11x+1  1)}{\sqrt{1+x^2}} dx .
    Plugging in n=23, a=1 and b=12 we see that again the integral takes the exact form of dy/dx found in the first part of the question.

    Therefore we have  \displaystyle \boxed{ J= \displaystyle (x-1)^{23}e^{12x}\sqrt{1+x^2} + c } .
    (iii)
    Spoiler:
    Show
    If we let  \displaystyle I = \int \frac{(x-6)^6e^{4x}(4x^4+x^3-2)}{\sqrt{1+x^2}} dx .
    We immediately run into the problem that not only is the polynomial expression a quartic, it can also not be factorised (at least not in any useful way to this problem).
    Realising that it may be useful to split up the quartic expression into a different form, maybe one with (x-2) in it somewhere we find that
     \displaystyle 4x^4+x^3-2 \equiv (x-2)(4x^3+x^2+2x)+2(4x^3+2x-1) we can rewrite the integral as

     \displaystyle I = \int \frac{(x-2)^7e^{4x}(4x^3+x^2+2x)}{\sqrt{1  +x^2}} dx + 2\int \frac{(x-2)^6e^{4x}(4x^3+2x-1)}{\sqrt{1+x^2}} dx .

    Now both of these are in the exact form found in the first part of the question, the first with n=8, a=2 and b=4 and for the second we have n=7, a=2 and b=4 which gives us
     \displaystyle I= (x-2)^8e^{4x}\sqrt{1+x^2} +2(x-2)^7e^{4x}\sqrt{1+x^2} +c .

    Cleaning this up gives
     \boxed{ \displaystyle I = x(x-2)^7e^{4x}\sqrt{1+x^2} + c} .
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    (Original post by B_9710)
    STEP III Question 7
    (i)
    Spoiler:
    Show
    If  \displaystyle y=(x-a)^ne^{bx}\sqrt{1+x^2} , using product rule in  (x-a)^ne^{bx} factor first and then using product rule in conjunction with the  \sqrt{1+x^2} , we obtain

     \displaystyle \frac{dy}{dx}=\frac{ \left (x-a)^{n-1}(bx^3+(n-ab+1)x^2+(b-a)x+n-ab \right ) }{\sqrt{1+x^2}} .

    So we could write  q(x) = bx^3+(n-ab+1)x^2+(b-a)x+n-ab .

    (i)
    Let \displaystyle I =\int \frac{(x-4)^{14}e^{4x}(4x^3-1)}{\sqrt{1+x^2}} dx , notice that this integrated is in the exam form of the pervious result with n=15, a=4 and b=4.
    Plugging these into the expression given for y at the start of the question yields
     \displaystyle I=(x-4)^{15}e^{4x}\sqrt{1+x^2}+c .
    (ii)
    Spoiler:
    Show
    If we let  \displaystyle J = \int \frac{(x-1)^21e^{12x}(12x^4-x^2-11)}{\sqrt{1+x^2}} dx .
    Noticing that at the moment there is a quartic expression instead of a cubic expression, we factorise and it is clear that (x-1) is a factor of the quartic.
    Factorising  \displaystyle 12x^4-x^2-11 \equiv (x-1)(12x^3+12x^2+11x+11) and we see that this means that integral,  J , becomes

     \displaystyle \int \frac{(x-1)^{22}e^{12x}(12x^3+12x^2+11x+1  1)}{\sqrt{1+x^2}} dx .
    Plugging in n=23, a=1 and b=12 we see that again the integral takes the exact form of dy/dx found in the first part of the question.

    Therefore we have  \displaystyle (x-1)^{23}e^{12x}\sqrt{1+x^2} + c .
    (iii)
    Spoiler:
    Show
    If we let  \displaystyle I = \int \frac{(x-6)^6e^{4x}(4x^4+x^3-2)}{\sqrt{1+x^2}} dx .
    We immediately run into the problem that not only is the polynomial expression a quartic, it can also not be factorised (at least not in any useful way to this problem).
    Realising that it may be useful to split up the quartic expression into a different form, maybe one with (x-2) in it somewhere we find that
     \displaystyle 4x^4+x^3-2 \equiv (x-2)(4x^3+x^2+2x)+2(4x^3+2x-1) we can rewrite the integral as

     \displaystyle I = \int \frac{(x-2)^7e^{4x}(4x^3+x^2+2x)}{\sqrt{1  +x^2}} dx + 2\int \frac{(x-2)^6e^{4x}(4x^3+2x-1)}{\sqrt{1+x^2}} dx .

    Now both of these are in the exact form found in the first part of the question, the first with n=8, a=2 and b=4 and for the second we have n=7, a=2 and b=4 which gives us
     \displaystyle I= (x-2)^8e^{4x}\sqrt{1+x^2} +2(x-2)^7e^{4x}\sqrt{1+x^2} +c .

    Cleaning this up gives
     \displaystyle I = x(x-2)^7e^{4x}\sqrt{1+x^2} + c .
    Probably better to use the fact that (fgh)'=f'gh+fg'h+fgh' for the differentiation at the start.
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    (Original post by B_9710)
    STEP II Question 7
    (i)
    Spoiler:
    Show
    If  \displaystyle y=(x-a)^ne^{bx}\sqrt{1+x^2} , using product rule in  (x-a)^ne^{bx} factor first and then using product rule in conjunction with the  \sqrt{1+x^2} , we obtain

     \displaystyle \frac{dy}{dx}=\frac{ \left (x-a)^{n-1}(bx^3+(n-ab+1)x^2+(b-a)x+n-ab \right ) }{\sqrt{1+x^2}} .

    So we could write  q(x) = bx^3+(n-ab+1)x^2+(b-a)x+n-ab .

    (i)
    Let \displaystyle I =\int \frac{(x-4)^{14}e^{4x}(4x^3-1)}{\sqrt{1+x^2}} dx , notice that this integrated is in the exam form of the pervious result with n=15, a=4 and b=4.
    Plugging these into the expression given for y at the start of the question yields
     \displaystyle I=(x-4)^{15}e^{4x}\sqrt{1+x^2}+c .
    As an alternative to the above, you can also take logs to get:

    \log y = n \log (x-a) + bx + \frac{1}{2} \log (1+x^2) then differentiate implicitly to get \frac{y'}{y} = \frac{n}{x-a} + b + \frac{x}{1 + x^2}
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    (Original post by Zacken)
    As an alternative to the above, you can also take logs to get:

    \log y = n \log (x-a) + bx + \frac{1}{2} \log (1+x^2) then differentiate implicitly to get \frac{y'}{y} = \frac{n}{x-a} + b + \frac{x}{1 + x^2}
    I tried that to begin with but it didn't seem to make it any less painless.
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    (Original post by IrrationalRoot)
    Probably better to use the fact that (fgh)'=f'gh+fg'h+fgh' for the differentiation at the start.
    You're probably right.
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    (Original post by B_9710)
    I tried that to begin with but it didn't seem to make it any less painless.
    It worked out okay if you pulled out the denominator and \frac{1}{x-a} right away. Either way is fine, anyways. :-)
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    Is there a thread for STEP I 2009 solutions already? I can't seem to find one.
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    (Original post by IrrationalRoot)
    Is there a thread for STEP I 2009 solutions already? I can't seem to find one.
    Nopes, it's just that the official examiner solutions are already beautifully laid out and intricately detailed. If there'd demand, we can always just add in a STEP I section after we've completed II and III.
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    (Original post by Zacken)
    Nopes, it's just that the official examiner solutions are already beautifully laid out and intricately detailed. If there'd demand, we can always just add in a STEP I section after we've completed II and III.
    Oh yeah the I solutions are lovely. I don't get why they don't do that for II and III, would be so nice.
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    (Original post by IrrationalRoot)
    Oh yeah the I solutions are lovely. I don't get why they don't do that for II and III, would be so nice.
    Different examiners - the examiner for STEP I is a nice dude, or so I've heard. :lol:
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    (Original post by Zacken)
    Different examiners - the examiner for STEP I is a nice dude, or so I've heard. :lol:
    Oh wow, so the II and III examiners are just lazy lol.
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    (Original post by IrrationalRoot)
    Oh wow, so the II and III examiners are just lazy lol.
    I think that tends to be the norm with maths examiners, unfortunately xD
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    STEP III Q10

    Solution
    Consider motion along the axis of the spring, let the spring have spring constant k and suppose P is of mass m. By the equilibrium condition of the particle at rest, note mg = kd \Rightarrow k = \dfrac{mg}{d}.

    Moreover, by conservation of energy from the time the ball is dropped until time t \leq T after it first makes contact with the spring:

    mgh = -mgx + \dfrac{1}{2}m\dot{x}^2 + \dfrac{1}{2}kx^2 \ (1)

    Differentiating (1) w.r.t. t and simplifying then yields

    -g\dot{x} + \dot{x}\ddot{x} + \dfrac{g}{d}x\dot{x} =0 \iff \boxed{\ddot{x} + \dfrac{g}{d}x = g} \ (2);

    A second order ODE that describes the motion, as desired (alternatively, use Newton's 2nd law if you prefer).

    Now consider a solution of the given form, then

    \begin{array}{lcl}

x(0) = 0 &\Rightarrow& B=-A \\

\dot{x}(0)= \sqrt{2gh} &\Rightarrow& C = \dfrac{1}{\omega}\sqrt{2gh} \\

(2) &\Rightarrow& \left(-\omega^2 + \dfrac{g}{d}\right) B\cos \omega t + \left(-\omega^2 + \dfrac{g}{d}\right) C\sin \omega t + \dfrac{g}{d}A = g \Rightarrow \omega =\sqrt{\dfrac{g}{d}}, A= d

\end{array}

    Using these results, we obtain the solution \boxed{x= d - d\cos \sqrt{\dfrac{g}{d}}t + \sqrt{2dh}\sin \sqrt{\dfrac{g}{d}}t} \ (3), as required.

    Note that the particle will first lose contact with the spring when the spring next returns to it's natural length. So seeking T s.t. x(T)=0, T\not= 0:

    \begin{array}{rcll}

(3)& \Rightarrow & \cos \sqrt{\dfrac{g}{d}}T - \sqrt{\dfrac{2h}{d}}\sin \sqrt{\dfrac{g}{d}}T = 1 \\

&\iff & \sqrt{\left(1+\dfrac{2h}{d} \right) } \cos\left(\sqrt{\dfrac{g}{d}}T + \arctan \sqrt{\dfrac{2h}{d}} \right) = 1\\

& \iff & \sqrt{\dfrac{g}{d}}T + \arctan \sqrt{\dfrac{2h}{d}} = \arccos \left( \dfrac{1}{\sqrt{1+(\sqrt{2h/d})^2}} \right) \text{or } \ 2\pi - \arccos \left( \dfrac{1}{\sqrt{1+(\sqrt{2h/d})^2}} \right)\\

& \iff &  \sqrt{\dfrac{g}{d}}T + \arctan \sqrt{\dfrac{2h}{d}} = \arctan \sqrt{\dfrac{2h}{d}}  \ \text{or } \ 2\pi - \arctan \sqrt{\dfrac{2h}{d}} \ (*)\\

\end{array}

    Where the line (*) follows from the identity \arccos\left(\dfrac{1}{\sqrt{1+x  ^2}}\right) \equiv \arctan x for x\geq 0, which is easily derived by considering a right angled triangle defined s.t. one it's other angles are \arctan x, with corresponding opposite and adjacent sides of length x and 1 respectively.

    Note that the former possibility in (*) represents when the particle first makes contact with the spring, leaving us with the latter by the T\not=0 condition. That is, T satisfies:

    \sqrt{\dfrac{g}{d}}T + \arctan \sqrt{\dfrac{2h}{d}} = 2\pi - \arctan \sqrt{\dfrac{2h}{d}} \iff \boxed{T = \sqrt{\dfrac{d}{g}}\left(2\pi -2\arctan \sqrt{\dfrac{2h}{d}} \right)}, as was to be shown.
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    STEP III Q6

    Solution
    Let A,B and M be the points in the complex plane represented by e^{i\alpha}, e^{i\beta} and the midpoint of AB respectively. By considering the right angled \triangle OAM and observing that A,B lie on the unit circle centred at O, note:

    \sin \left(\angle AOM \right) = \dfrac{AM}{OA}= \dfrac{AB}{2\cdot OA} \iff \sin \dfrac{1}{2}(\beta - \alpha) = \dfrac{|e^{i\beta}-e^{i\alpha}|}{2} (1)

    As desired.

    For ease of LaTeX, translate Greek to English. Also, recall 2\sin A \sin B = \cos(A-B) - \cos(A+B) (2).

    Then, taking care with the fact that moduli are non-negative, by (1):

    \frac{1}{2}|e^{ia}-e^{ib}||e^{ic}-e^{id}| + \frac{1}{2}|e^{ib}-e^{ic}||e^{ia}-e^{id}| = 2\sin \frac{1}{2}(b-a) \sin \frac{1}{2}(d-c) + 2\sin \frac{1}{2}(c-b) \sin \frac{1}{2} (d-a)

    Applying (2):

    = \cos \left[\frac{1}{2}(b-a) - \frac{1}{2}(d-c)\right] - \cos \left[\frac{1}{2}(b-a) + \frac{1}{2}(d-c)\right] + \cos \left[\frac{1}{2}(c-b) - \frac{1}{2}(d-a)\right] - \cos \left[\frac{1}{2}(c-b) + \frac{1}{2}(d-a)\right]

    Rearranging the arguments slightly, that is:

    = \cos \left[\frac{1}{2}(c-a) - \frac{1}{2}(d-b)\right] - \cos \left[\frac{1}{2}(b-a) + \frac{1}{2}(d-c)\right] + \cos \left[-(\frac{1}{2}(b-a) + \frac{1}{2}(d-c))\right] - \cos \left[\frac{1}{2}(c-a) + \frac{1}{2}(d-b)\right]

    Observing that \cos(-x)=\cos x, the middle two terms cancel, allowing us to again apply (2):

     = 2\sin \frac{1}{2}(c-a) \sin \frac{1}{2}(d-b)

    And finally (1):

    = \frac{1}{2}|e^{ia}-e^{ic}||e^{ib}-e^{id}|

    It follows that

    \boxed{|e^{i\alpha}-e^{i\beta}||e^{i\gamma}-e^{i\delta}| + |e^{i\beta}-e^{i\gamma}||e^{i\alpha}-e^{i\delta}| = |e^{i\alpha}-e^{i\gamma}||e^{i\beta}-e^{i\delta}|}

    As was to be shown.

    Defining A as the point in the complex plane represented by the complex number e^{i\alpha} with B,C,D defined similarly, and noting that these points define a cyclic quadrilateral ABCD, this result states that:

    AB\cdot CD + BC\cdot AD = AC\cdot BD

    i.e. the sum of the product of the lengths of each pair of opposing sides in a cyclic quadrilateral is given by the product of the lengths of it's pair of diagonals. See Ptolemy's Theorem.
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    STEP III Q7

    Solution

    (i)
    Note first that f_1(x) = -\dfrac{2x}{(1+x^2)^2} and f_2(x) = \dfrac{-2(1+x^2)^2 +8x^2(1+x^2)}{(1+x^2)^4} = \dfrac{2(3x^2-1)}{(1+x^2)^3}, then we have:

    (1+x^2)f_2(x)+2(1+1)xf_1(x) +1(1+1)f_0(x) = \dfrac{2(3x^2-1)}{(1+x^2)^2} -\dfrac{8x^2}{(1+x^2)^2}+2\dfrac{  1+x^2}{(1+x^2)^2}=0

    Furthermore, suppose \exists k\in \mathbb{N} s.t. (1+x^2)f_{k+1} +2(k+1)xf_k + k(k+1)f_{k-1}=0 (\star) . Then:

    \dfrac{d}{dx}(\star): (1+x^2)f_{k+2}+ [2x+2(k+1)x]f_{k+1} + [2(k+1)+k(k+1)]f_k=0

    \iff (1+x^2)f_{k+2} + 2(k+2)xf_{k+1} + (k+1)(k+2)f_k = 0

    By induction, it follows that \boxed{(1+x^2)f_{n+1}(x) +2(n+1)xf_n(x) + n(n+1)f_{n-1}(x)=0 \ \forall n\geq 1} \square

    (ii)
    P_0(x) = 1, P_1(x) = -2x, P_2(x) = 2(3x^2-1)

    Note that:

    \begin{array}{lcl}

(1+x^2)\dfrac{dP_n}{dx} - 2(n+1)xP_n &=& (1+x^2) [(1+x^2)^{n+1}f_{n+1} +2(n+1)x(1+x^2)^nf_n]-2(n+1)x(1+x^2)^{n+1}f_n\\

&=& (1+x^2)^{n+2}f_{n+1}(x)\\

&=& P_{n+1}(x)

\end{array}

    \iff \boxed{P_{n+1}(x) -(1+x^2)\dfrac{dP_n(x)}{dx} + 2(n+1)xP_n(x) =0 \ \forall n\geq 0} (\star \star), as required.

    Moreover, suppose \deg P_k(x) = k with leading coefficient a\not=0 for some k \geq 0. Then, by (\star \star):

    \deg P_{k+1}(x) \leq \max\{ \deg(1+x^2) + \deg \dfrac{dP_k(x)}{dx}\ , \  \deg (2(n+1)x) + \deg P_k(x) \}  = k+1

    Seeking the leading coefficient of P_{k+1}(x) using (\star \star):

    x^{k+1}:  1\cdot ak - 2(k+1)\cdot a =-(k+2)a \not= 0

    So \deg P_{k+1}(x) = k+1. Hence, observing that \deg P_0(x)=0, it follows by induction that P_n(x) is a polynomial of degree n for all n\geq 0 \square

 
 
 
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