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    (Original post by bananarama2)
    Since when have the compscis become such realists?
    When they had to work within the confines of real hardware? (seriously)
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    (Original post by ukdragon37)
    When they had to work within the confines of real hardware? (seriously)
    Tell bloody games designers that, designing games that run on computers that cost 32 billions pounds :nothing: :rant:
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    (Original post by bananarama2)
    Tell bloody games designers that, designing games that run on computers that cost 32 billions pounds :nothing: :rant:
    If you fancy a girl who only dates handsome guys, and you aren't up to the standard, who do you blame exactly?

    Clearly those people make enough money catering to those who do have the adequate hardware than to be bothered with you.
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    (Original post by ukdragon37)
    If you fancy a girl who only dates handsome guys, and you aren't up to the standard, who do you blame exactly?

    Clearly those people make enough money catering to those who do have the adequate hardware than to be bothered with you.
    Well I think there's a market for a game which runs on just not quite the best computer.

    (I have an i7 and some games still run like it's a DOS machine.)
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    (Original post by bananarama2)
    Well I think there's a market for a game which runs on just not quite the best computer.

    (I have an i7 and some games still run like it's a DOS machine.)
    I love DOS games thank you very much...
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    (Original post by Slumpy)
    I love DOS games thank you very much...
    I don't disagree Lemmings and Legacy FTW.
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    (Original post by bananarama2)
    Well I think there's a market for a game which runs on just not quite the best computer.

    (I have an i7 and some games still run like it's a DOS machine.)
    There are plenty! And many of them are great too! E.g. Minecraft, many of the games from the Humble Indie Bundles....

    Also for something most definitely new you can try looking at Kickstarter and invest in the start-up games at an amount that promises an eventual copy when it is released. Most of them are not hardware-intensive so as to appeal to the broadest market. Plus you get a nice surprise a few months down the line when the game you invested but forgot about sends you something.
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    (Original post by ukdragon37)
    There are plenty! And many of them are great too! E.g. Minecraft, many of the games from the Humble Indie Bundles....

    Also for something most definitely new you can try looking at Kickstarter and invest in the start-up games at an amount that promises an eventual copy when it is released. Most of them are not hardware-intensive so as to appeal to the broadest market.
    I guess, but they are all of a certain style. Minecraft is good, although surprisingly memory intensive

    I'll have a look
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    And here I am, thinking, if I didn't have an Internet connection that flickered 20 times a minute, I'd be blessed.
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    (Original post by MW24595)
    And here I am, thinking, if I didn't have an Internet connection that flickered 20 times a minute, I'd be blessed.
    Admittedly it's a spoiled kid problem I have.
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    Cool stuff!

    Solution 163

    Define t_i=a+\dfrac{i(b-a)}{n}

    \displaystyle\begin{aligned} n\left\{\int_a^b f(x)\,dx-\frac{b-a}{n}\sum_{i=1}^n f(t_i) \right\} &=n\sum_{i=1}^n \left \{ \int_{t_{i-1}}^{t_i} f(x)\,dx-\frac{b-a}{n} f(t_i) \right\} \\&=n\sum_{i=1}^n \left \{ \int_{t_{i-1}}^{t_i} f(x)-f(t_i)\,dx \right \} \\ &=n\sum_{i=1}^n \left \{ t_{i-1}\Big[ f(t_i) -f(t_{i-1}) \Big]-\int_{t_{i-1}}^{t_i} xf'(x)\,dx\right\} \\ &=n\sum_{i=1}^n \left \{ t_{i-1}\int_{t_{i-1}}^{t_i}f'(x)\,dx-\int_{t_{i-1}}^{t_i} xf'(x)\,dx\right\} \\&=\sum_{i=1}^n\int_{t_{i-1}}^{t_i} n(t_{i-1}-x)f'(x)\,dx \\& = \int_{0}^{b-a}-x\sum_{i=1}^n\left[\frac{1}{n}\, f'\left(\frac{x}{n}+t_{i-1}\right)\right]\,dx\;\;(n(x-t_{i-1})\to x) \\&=\int_0^{b-a} -\frac{x}{b-a} \int_a^b f'(z)\,dz\,dx\;\;(n\to\infty) \\&= \frac{b-a}{2}\Big( f(a)-f(b)\Big) \end{aligned}

    Solution 164

    \begin{aligned} \displaystyle\int_0^{ \infty}  \frac{\arctan x}{x^{\alpha}} \,dx=\frac{1}{\alpha-1}\int_0^{ \infty} \frac{dx}{x^{\alpha-1}(1+x^2)}

    Set \displaystyle t=(1+x^2)^{-1}:

    \displaystyle \begin{aligned} \int_0^{ \infty} \frac{dx}{x^{\alpha-1}(1+x^2)} \,dx &= \frac{1}{2} \int_0^{1} \frac{1}{t} \left( \frac{1}{t}-1 \right)^{-\frac{\alpha}{2}} \,dt \\ &= \frac{1}{2} \int_0^{1} t^{\frac{ \alpha}{2}-1} \left(1-t \right)^{-\frac{\alpha}{2}} \,dt\\ &=\frac{1}{2}\,\text{B}\left( \frac{ \alpha}{2},\, 1-\frac{\alpha}{2} \right) \\& = \frac{1}{2}\, \Gamma \left(\frac{\alpha}{2}\right) \Gamma \left(1-\frac{\alpha}{2}\right) \\&= \frac{\pi}{2} \csc \left(\frac{\pi \alpha}{2} \right)\;\;(\text{r.f.})\end{ali  gned}

    Hence \displaystyle\int_0^{ \infty}  \frac{\arctan x}{x^{\alpha}} \,dx= \frac{\pi}{2(\alpha-1)} \csc\left(\frac{ \pi \alpha}{2}\right)

    Time for bed. Here is a nice twist on a classic integral:

    Problem 168*

    \displaystyle\int_0^{\frac{\pi}{  2}} \ln \sin x\ln \cos x\,dx
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    (Original post by Lord of the Flies)
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    Cool stuff!

    Solution 163

    Define t_i=a+\dfrac{i(b-a)}{n}

    \displaystyle\begin{aligned} n\left\{\int_a^b f(x)\,dx-\frac{b-a}{n}\sum_{i=1}^n f(t_i) \right\} &=n\sum_{i=1}^n \left \{ \int_{t_{i-1}}^{t_i} f(x)\,dx-\frac{1}{n} f(t_i) \right\} \\&=n\sum_{i=1}^n \left \{ \int_{t_{i-1}}^{t_i} f(x)-f(t_i)\,dx \right \} \\ &=n\sum_{i=1}^n \left \{ t_{i-1}\Big[ f(t_i) -f(t_{i-1}) \Big]-\int_{t_{i-1}}^{t_i} xf'(x)\,dx\right\} \\ &=n\sum_{i=1}^n \left \{ t_{i-1}\int_{t_{i-1}}^{t_i}f'(x)\,dx-\int_{t_{i-1}}^{t_i} xf'(x)\,dx\right\} \\&=\sum_{i=1}^n\int_{t_{i-1}}^{t_i} n(t_{i-1}-x)f'(x)\,dx \\& = \int_{0}^{b-a}-x\sum_{i=1}^n\left[\frac{1}{n}\, f'\left(\frac{x}{n}+t_{i-1}\right)\right]\,dx\;\;(n(x-t_{i-1})\to x) \\&=\int_0^{b-a} -\frac{x}{b-a} \int_a^b f'(z)\,dz\,dx\;\;(n\to\infty) \\&= \frac{b-a}{2}\Big( f(a)-f(b)\Big) \end{aligned}

    Solution 164

    \begin{aligned} \displaystyle\int_0^{ \infty}  \frac{\arctan x}{x^{\alpha}} \,dx=\frac{1}{\alpha-1}\int_0^{ \infty} \frac{dx}{x^{\alpha-1}(1+x^2)} \,dx

    Set \displaystyle t=(1+x^2)^{-1}:

    \displaystyle \begin{aligned} \int_0^{ \infty} \frac{dx}{x^{\alpha-1}(1+x^2)} \,dx &= \frac{1}{2} \int_0^{1} \frac{1}{t} \left( \frac{1}{t}-1 \right)^{-\frac{\alpha}{2}} \,dt \\ &= \frac{1}{2} \int_0^{1} t^{\frac{ \alpha}{2}-1} \left(1-t \right)^{-\frac{\alpha}{2}} \,dt\\ &=\frac{1}{2}\,\text{B}\left( \frac{ \alpha}{2},\, 1-\frac{\alpha}{2} \right) \\& = \frac{1}{2}\, \Gamma \left(\frac{\alpha}{2}\right) \Gamma \left(1-\frac{\alpha}{2}\right) \\&= \frac{\pi}{2} \csc \left(\frac{\pi \alpha}{2} \right)\;\;(\text{r.f.})\end{ali  gned}

    Hence \displaystyle\int_0^{ \infty}  \frac{\arctan x}{x^{\alpha}} \,dx= \frac{\pi}{2(\alpha-1)} \csc\left(\frac{ \pi \alpha}{2}\right)
    Well done.

    (Original post by jack.hadamard)
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    I like this. Below is a hint; a question that I had posted in the inappropriate place at the inappropriate time.

    Spoiler:
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    The Riemann zeta function \zeta(s) is defined as

    \displaystyle \zeta(s)\ :=\ \sum_{n \geq 1} n^{-s}\ =\ \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots
    and has a finite value for all real numbers s with s > 1. It is given that \displaystyle \zeta(2) = \frac{\pi^2}{6}


    i) Let \displaystyle f(m, n) = \frac{1}{mn^3} + \frac{1}{2m^2n^2} + \frac{1}{m^3n} for positive integers m and n.

    Show that \displaystyle f(m, n) - f(m + n, n) - f(m, m + n) = \frac{1}{m^2n^2}.

    Hence, by summing over all m and n, show that

    \displaystyle \zeta(2)^2 = \left(\sum_{m,n \geq 1} - \sum_{m > n \geq 1} - \sum_{n > m \geq 1} \right) f(m, n) = \frac{5}{2}\zeta(4).


    ii) It is further given that \displaystyle \sum_{1 \leq n < k/2} \zeta(2n)\zeta(k - 2n) = \frac{k + 1}{2} \zeta(k) for all even k \geq 4.

    Show that \displaystyle \zeta(6) = \frac{\pi^6}{945} and express \zeta(8) in terms of \zeta(2), \zeta(4) and \zeta(6).


    iii)* Prove by induction that \zeta(k)\ =\ r \pi^k for all even positive integers k, where r is a rational number.

    EDIT: Nevermind, read it the right way.
    By the way, the second question is much more difficult. I can't think of anything simpler than L-functions.


    Solution 168

    Let \displaystyle I = \int_{0}^{\frac{\pi}{2}} \ln \sin x \ln \cos x \, dx; \displaystyle J = \int_{0}^{\frac{\pi}{2}} \ln^{2} \tan x \, dx; \displaystyle K = \int_{0}^{\frac{\pi}{2}} \ln^{2} \sin x \, dx

    It is clear that \displaystyle K = \int_{0}^{\frac{\pi}{2}} \ln^{2} \sin x \, dx \mathop =^{x \mapsto \frac{\pi}{2}-x} \int_{0}^{\frac{\pi}{2}} \ln^{2} \cos x \, dx
    Furthermore, \displaystyle \ln^{2} \sin 2x \, dx \mathop =^{x \mapsto \frac{x}{2}} \int_{0}^{\frac{\pi}{2}} \ln^{2} \sin x \, dx = K

    Now, \displaystyle 2I+2K = \int_{0}^{\frac{\pi}{2}} \left( \ln \cos x + \ln \sin x \right)^{2} \, dx = K - \frac{3\pi}{2} \ln^{2} 2.

    We also have \displaystyle 2K - 2I =  \int_{0}^{\frac{\pi}{2}} \left( \ln \sin x - \ln \cos x \right)^{2} \, dx = J. Thus \displaystyle I = \frac{\pi}{2} \ln^{2} 2 - \frac{1}{6}J

    Here we shall be done if we manage to evaluate J.

    We first observe that \displaystyle J \mathop =^{\tan x \mapsto x} \int_{0}^{\infty} \frac{1}{1+x^{2}} \ln^{2} x \, dx.

    Secondly, we consider \displaystyle f(z) = \frac{\ln^{3} z}{1+z^{2}}. This function has simple poles at z = \pm i.


    \begin{aligned} \displaystyle \int_{\gamma_{R, \epsilon}} \frac{\ln^{3} z}{1+z^{2}} \, dz &= \int_{\epsilon}^{R} \frac{\ln^{3} x}{1+x^{2}} \, dx - \int_{\epsilon}^{R} \frac{( \ln x + 2i \pi)^{3}}{1+x^{2}} \, dx + \int_{C_{R}} + \int_{C_{\epsilon}} \\& = 2i\pi \left(Res(f(z),i)+Res(f(z),-i) \right) \end{aligned}


    Clearly, as R \to \infty and \epsilon \to 0, we have \displaystyle \int_{C_{R}} \to 0 and \displaystyle \int_{C_{\epsilon}} \to 0


    Separating the real and imaginary parts, we get
    \begin{aligned} \displaystyle -6\pi i J + 8\pi^{3} i \int_{0}^{\infty} \frac{1}{1+x^{2}} \, dx = 2\pi i \left(Res(f(z),i)+Res(f(z),-i)) \end{aligned}


    We easily find \displaystyle Res(f(z),i) = -\frac{\pi^{3}}{16} and \displaystyle Res(f(z),-i) = \frac{27\pi^{3}}{16}.

    Thereupon, \displaystyle -3J + 2\pi^{3} = \frac{13\pi^{3}}{8}, which gives \displaystyle J = \frac{1}{8}\pi^{3}.

    Therefore, \displaystyle I = \frac{\pi}{2} \ln^{2} 2 - \frac{1}{48} \pi^{3}.
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    (Original post by bananarama2)
    Admittedly it's a spoiled kid problem I have.

    A potato generated internet connection would be more stable than his connection.
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    I think we need an amendment to the first post. **** now means LOTF and Mladenov difficulty.
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    (Original post by bananarama2)
    I think we need an amendment to the first post. **** now means LOTF and Mladenov difficulty.
    :lol:
    Or rather, * isn't really A-Level knowledge :cool:
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    (Original post by Lord of the Flies)
    \begin{aligned} \displaystyle\int_0^{ \infty}  \frac{\arctan x}{x^{\alpha}} \,dx=\frac{1}{\alpha-1}\int_0^{ \infty} \frac{dx}{x^{\alpha-1}(1+x^2)} \,dx
    How did you do this step? :eek:
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    (Original post by Jkn)
    How did you do this step? :eek:
    By parts...?
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    (Original post by Jkn)
    How did you do this step? :eek:
    As above, by parts and evaluating the limits around uv.
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    (Original post by Lord of the Flies)
    Solution 160

    \displaystyle f(t)=\int_0^1 \frac{\ln (1+tx)}{1+x^2}\,dx
    Is this differentiation under the integral sign? Or if not what is this technique, is it useful?
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    (Original post by bananarama2)
    I think we need an amendment to the first post. **** now means LOTF and Mladenov difficulty.
    I'm feeling increasingly disturbed that they are just about to (and most certainly will) encroach beyond techniques I understand soon. :laugh:

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    But then I was never a continuous maths person.
 
 
 
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