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    so grade boundary predictions?
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    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
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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
    Yes so, I did the right thing for last part but didnt simplify and ofcourse i used x=e^2y. Lose like 5?


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    I'll post a solution to 8 if you'd like?
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    (Original post by student0042)
    I'll post a solution to 8 if you'd like?
    Please do
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    Q2 is my *****. Had a beauty to that, it was eazy. Need to go pick up my little brother first.


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    EDIT:
    someone else do q7, Latex ****ing trolled me hard
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    Q12 is mine guys

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    Anyone got a copy of the paper yet?
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    (Original post by Mathemagicien)
    2 late
    😥😥 I cri
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    (Original post by Shrek1234)
    Please do
    Alright Q8. I don't have access to a paper, so I may have remembered the question incorrectly, but I got the same answers as yesterday, so...
    Name:  STEP III 2016 Q8 1.jpg
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    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
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    (Original post by Mathemagicien)
    ...
    Q 12 was beautiful, probably the easiest STEP III stats question ever.
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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
    how much for all that but not rewriting blabla+2sechy as 2cosechy?
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    (Original post by gasfxekl)
    how much for all that but not rewriting blabla+2sechy as 2cosechy?
    19
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    (Original post by Zacken)
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    Zacin. I used x=e^2y but then in last prt i subtracted r=1 term cause i was rushin. I still doubled and added r=0. How many lost


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    (Original post by physicsmaths)
    Zacin. I used x=e^2y but then in last prt i subtracted r=1 term cause i was rushin. I still doubled and added r=0. How many lost


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    Probs lose 5 or 6 marks.
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    (Original post by Zacken)
    Probs lose 5 or 6 marks.

    Decent ttntntntnt
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    (Original post by sweeneyrod)
    Q 12 was beautiful, probably the easiest STEP III stats question ever.
    Ikr! I began it right at the end so ran out time 😥😥 Could've been an easy 20

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    (Original post by Mathemagicien)
    I know, I felt bad doing it, it really feels like cheating
    How do you think the marks willbe allocated between each part?

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