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    (Original post by Number Nine)
    bagsy q1

    \displastyle

\begin{equation*} I_n = \int^{\infty}_{-\infty}\frac{1}{(x^2 + 2ax + b)^n} \mathrm{d}x\end{equation*}

    Using the substitution  x + a = \sqrt{b-a^2} \tan u we have \frac{\mathrm{d}x}{\mathrm{d}u} = \sqrt{b-a^2} \sec^2 u

    So that:

    \displaystyle 

\begin{align*}I_n = \int^{\infty}_{-\infty}\frac{1}{((x+a)^2 + b - a^2)^n} \mathrm{d}x & = \int^{\pi/2}_{-\pi/2}\frac{\sqrt{b-a^2}\sec ^2 u}{((b-a^2)\tan ^2 u + (b-a^2))^n } \mathrm{d}u \\ & = \int^{\pi/2}_{-\pi/2}\frac{\sqrt{b-a^2}\sec ^2 u}{((b-a^2)\sec ^2 u)^n } du \\ & =  \int^{\pi/2}_{-\pi/2}\frac{1}{(b-a^2)^{n-\frac{1}{2}} \sec ^{2n-2} u } du \end{align*}

    So:

    i) Find I_1

    \displaystyle 

\begin{align*}I_1 = \int^{\pi/2}_{-\pi/2}\frac{1}{(b-a^2)^{\frac{1}{2}}} \mathrm{d}u & = \left[ \frac{x}{(b-a^2)^\frac{1}{2}} \right]_{-\pi/2}^{\pi/2} \\& = \frac{\pi}{\sqrt{b-a^2}} \end{align*}

    as required
    Cleaning up the LaTeX a little for you.
    ..
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    (Original post by zacken)
    cleaning up the latex a little for you.
    ..
    yeah can
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    Anyone got the pdf? Willing to have a bash at any unpopular questions
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    wow everyone here is so clever. have fun at oxbridge you guys
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    (Original post by fefssdf)
    wow everyone here is so clever. have fun at oxbridge you guys
    We don't talk about Oxford round these parts


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    (Original post by drandy76)
    We don't talk about Oxford round these parts


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    ok im backing off
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    (Original post by Llewellyn)
    Anyone got the pdf? Willing to have a bash at any unpopular questions
    http://www.thestudentroom.co.uk/show...9#post66077019
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    Can someone write an answer for Q3 please
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    (Original post by Zacken)
    Cleaning up the LaTeX a little for you.
    ..
    http://www.thestudentroom.co.uk/show...9#post66074739
    Please can you make this of adequate size again?
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    (Original post by student0042)
    ..
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    (Original post by Number Nine)
    bagsy q1

     I_n = \int^{\infty}_{-\infty}\frac{1}{(x^2 + 2ax + b)^n} dx

       Using the substitution:

     x + a = \sqrt{b-a^2} \tan u

     \Rightarrow \frac{dx}{du} = \sqrt{b-a^2} \sec ^2 u

      \Rightarrow I_n = \int^{\infty}_{-\infty}\frac{1}{((x+a)^2 + b - a^2)^n} dx

     \Rightarrow I_n = \int^{\pi/2}_{-\pi/2}\frac{\sqrt{b-a^2}\sec ^2 u}{((b-a^2)\tan ^2 u + (b-a^2))^n } du

     \tan ^2 u + 1 = \sec ^2 u

     I_n = \int^{\pi/2}_{-\pi/2}\frac{\sqrt{b-a^2}\sec ^2 u}{((b-a^2)\sec ^2 u)^n } du

     I_n = \int^{\pi/2}_{-\pi/2}\frac{1}{(b-a^2)^{n-\frac{1}{2}} \sec ^{2n-2} u } du

     i) Find\  I_1

     I_1 = \int^{\pi/2}_{-\pi/2}\frac{1}{(b-a^2)^{\frac{1}{2}}} du

     I_1 = \left[ \frac{x}{(b-a^2)^\frac{1}{2}} \right]_{-\pi/2}^{\pi/2}

     I_1 = \frac{\pi}{\sqrt{b-a^2}}\ as\ required

    apparently my latex is ugly
    How many marks do you reckon the first part will be? 5-6?
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    (Original post by Shrek1234)
    How many marks do you reckon the first part will be? 5-6?
    4-5 id say Zacken opinion
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    (Original post by Shrek1234)
    How many marks do you reckon the first part will be? 5-6?
    (Original post by Number Nine)
    4-5 id say Zacken opinion
    5 sounds about right, yeah.
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    For the f(x), g(x), h(x) would an acceptable answer to the final part be the observation that it can be arranged to have (h(x)+x)+(h(1/1-x)+(1/1-x))=1 and so each of these pairs of terms are 'something' +the function of that 'something' and so it is clear that h(x)=1/2-x would be the solution as the -x part would cancel the 'something' and then the two +1/2 would add to give the one. Note: the 'something' is a function of x
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    (Original post by fefssdf)
    wow everyone here is so clever. have fun at oxbridge you guys
    thanks fam ill try ot enjoy it
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    (Original post by Ewanclementson)
    For the f(x), g(x), h(x) would an acceptable answer to the final part be the observation that it can be arranged to have (h(x)+x)+(h(1/1-x)+(1/1-x))=1 and so each of these pairs of terms are 'something' +the function of that 'something' and so it is clear that h(x)=1/2-x would be the solution as the -x part would cancel the 'something' and then the two +1/2 would add to give the one. Note: the 'something' is a function of x
    Nice! Yes I think it would be.
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    (Original post by Zacken)
    STEP III: Question 4

    Sum to N
    Note that \frac{1}{1+x^r} - \frac{1}{1+x^{r+1}} = \frac{x^r(x-1)}{(1+x^r)(1+x^{r+1}}

    So, it follows, by re-writing and telescoping, we have:

    \displaystyle 

\begin{equation*}\sum_{r=1}^{N} \frac{x^r}{(1+x^r)(1+x^{r+1})} = \frac{1}{x-1}\sum_{r=1}^N \frac{1}{1+x^r} - \frac{1}{1+x^{r+1}} =\frac{1}{x-1}\left( \frac{1}{1+x} - \frac{1}{1+x^N} \right)\end{equation*}



    Sum to infinity


    Now, take N \to \infty, so that for |x|<1 we have x^N \to 0 meaning that our sum is:

    \displaystyle

\begin{align*}\sum_{r=1}^{\infty  } \frac{x^r}{(1+x^r)(1+x^{r+1})} &= \lim_{N \to \infty} \frac{1}{x-1}\left(\frac{1}{1+x} - \frac{1}{1+x^{N+1}}\right) \\ &= \frac{1}{x-1}\left( \frac{1}{1+x} - 1 \right) \\&= \frac{1}{x-1} \times \frac{-x}{1+x} \\&= \frac{x}{1-x^2}\end{align*}


    Hyperbolic sum

    Write the hyperbolic sum in exponential form:

    \displaystyle \sum_{r=1}^{\infty} \frac{4}{(e^{ry} + e^{-ry})(e^{y(r+1)} + e^{-y(r+1)})} = e^{-y}\sum_{r=1}^{\infty} \frac{4e^{-2ry}}{(1 + e^{-2ry})(1 + e^{-2y(r+1)})}

    So take x = e^{-2y} and use the previous result, noting that \forall y \in \mathbb{R}, |x| < 1 we have:

    \displaystyle 

\begin{align*}\frac{4e^{-y}e^{-2y}}{1-e^{-4y}} = \frac{4e^{-y}}{e^{2y} - e^{-2y}} = 2e^{-y}\text{cosech} \, y\end{align*}

    Now, for the final result, it's simply double the sum and adding on 2\text{sech} \, y to account for the r=0 term. This cleans up to 2\text{cosech} \, y
    OMG OMG OMG OMG I Totally forgot that I had done most of this question too!! I think I'm gonna get 13 marks out of it because I used x=e^2y and didn't do the lastest bit. You know what i might actually get a 1 out of that messy exam.
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    (Original post by Hext)
    You would not be deducted more than 4 marks.
    Try quote people


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    Quite liked the simple harmonic question (q9) so I thought I'd write up and post a solution:
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    (Original post by L-Tyrosine)
    Quite liked the simple harmonic question (q9) so I thought I'd write up and post a solution:
    How many marks do you think I'd get for getting all the way to the F=ma part and then stoping because I was too scared of the algebra hahaha ?
 
 
 
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