A Summer of Maths (ASoM) 2016

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    (Original post by physicsmaths)
    Or parts on xsinx and secx


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    How so? Neither one of those has a nice anti-derivative.
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    (Original post by Zacken)
    How so? Neither one of those has a nice anti-derivative.
    Duno just running some **** through my head.


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    In a moment of weakness nearly ran it through wolfram alpha, almost


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    (Original post by Zacken)
    How so? Neither one of those has a nice anti-derivative.
    Easy zain. Took few minutes once i started writing. Pi/2(ln2)
    Stop distracting me from d2 revision.
    Need to pass otherwise im gna B in AFM.


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    Groups looks so bloody interesting *o*
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    (Original post by drandy76)
    Unless I've goofed you end up were you started


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    Considering it's been posted on this thread by that person, I wouldn't expect anything less
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    (Original post by Student403)
    Considering it's been posted on this thread by that person, I wouldn't expect anything less
    I have two ideas and I'm too lazy to attempt either


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    (Original post by drandy76)
    I have two ideas and I'm too lazy to attempt either


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    Na the obvious parts direction is roght. Just carry on going and note some symmetry.
    A graph is symmetric about a if
    F(2a-x)=F(x) or some simple symmetry of sin which is the same lol.


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    Will post sol if any mone wants it. Just dnt wana ruin it.



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    (Original post by Snasher350)
    Substituting U=Cos x ? Should end up with a by parts integral with u and arccosU
    Doesn't sound like it'll work.

    (Original post by physicsmaths)
    Will post sol if any mone wants it. Just dnt wana ruin it.

    Did you basically end up finding integral of ln (sin x) between 0 and pi/2? If so, that wasn't really the intended method.
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    (Original post by physicsmaths)
    Na the obvious parts direction is roght. Just carry on going and note some symmetry.
    A graph is symmetric about a if
    F(2a-x)=F(x) or some simple symmetry of sin which is the same lol.
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    ah i see, was considering taylor expansions lol
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    (Original post by Zacken)
    Doesn't sound like it'll work.




    Did you basically end up finding integral of ln (sin x) between 0 and pi/2? If so, that wasn't really the intended method.
    Yeh. Not indefinite tho. Just added some **** and pi/2-x and u-pi/2 subs.
    Here it is don't look anyone who hasn't solved yet, it is a very nice problem.
    Name:  ImageUploadedByStudent Room1467150054.370639.jpg
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    Was gna do a tanx method etc. Seems easier thenz


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    FWIW, my approach:

    \displaystyle 

\begin{align*} I(\alpha) = \int_0^{\frac{\pi}{2}} \frac{\arctan (\alpha \tan x)}{\tan x} \, \mathrm{d}x & \Rightarrow I'(\alpha) = \int_0^{\frac{\pi}{2}} \frac{\mathrm{d}x}{(\alpha \tan x)^2 + 1} \\ & \Rightarrow I'(\alpha) = \frac{\pi}{2(\alpha +1)} \\ & \Rightarrow I(\alpha) = \frac{\pi}{2}\ln (\alpha + 1) + \mathcal{C} \end{align*}

    Now taking I(0) = 0 gives us \mathcal{C} and \alpha = 1 recovers \int_0^{\frac{\pi}{2}} x \cot x \, \mathrm{d}x = \frac{\pi}{2} \log 2.
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    (Original post by Insight314)
    So many people got the Beardon, and I don't. I am furious!

    I will have it in two days though.
    I just had a look at the contents page on Amazon.

    *goes to order the book*
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    (Original post by Zacken)
    FWIW, my approach:

    \displaystyle 

\begin{align*} I(\alpha) = \int_0^{\frac{\pi}{2}} \frac{\arctan (\alpha \tan x)}{\tan x} \, \mathrm{d}x & \Rightarrow I'(\alpha) = \int_0^{\frac{\pi}{2}} \frac{\mathrm{d}x}{(\alpha \tan x)^2 + 1} \\ & \Rightarrow I'(\alpha) = \frac{\pi}{2(\alpha +1)} \\ & \Rightarrow I(\alpha) = \frac{\pi}{2}\ln (\alpha + 1) + \mathcal{C} \end{align*}

    Now taking I(0) = 0 gives us \mathcal{C} and \alpha = 1 recovers \int_0^{\frac{\pi}{2}} x \cot x \, \mathrm{d}x = \frac{\pi}{2} \log 2.
    Ah ok. Very nyc.
    Shudda thought this was undergrad type ****.



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    Pesky people:

    \displaystyle 

\begin{equation*} \int_0^1 \frac{x^2 - 1}{\ln x} \, \mathrm{d}x \end{equation*}
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    (Original post by Zacken)
    Pesky people:

    \displaystyle 

\begin{equation*} \int_0^1 \frac{x^2 - 1}{\ln x} \, \mathrm{d}x \end{equation*}
    I got ln3 by considering I(a) = integral from 0 to 1 of e^(alnx)/lnx
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    (Original post by wlog54)
    I got ln3 by considering I(a) = integral from 0 to 1 of e^(alnx)/lnx
    The answer is correct, but I'm not entirely sure about your parametrisation. What a do you need to recover the original integral? The only way I can see you move on from your parametrisation is by integrating the power series term by term, which isn't exactly savoury.

    I'd use something like \displaystyle I(\alpha) = \int_0^1 \frac{x^{\alpha} - 1}{\ln x}.
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    (Original post by Zacken)
    The answer is correct, but I'm not entirely sure about your parametrisation. What a do you need to recover the original integral? The only way I can see you move on from your parametrisation is by integrating the power series term by term, which isn't exactly savoury.

    I'd use something like \displaystyle I(\alpha) = \int_0^1 \frac{x^{\alpha} - 1}{\ln x}.
    The parametrisation is fine. The required integral is  I(2) - I(0) .
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    Are any improvements on Dijkstra's Algorithm taught in Dx modules for x more or equal to 2?
 
 
 
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