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    (Original post by henpen)
    Pretty much. A more direct way would be to differentiate the answer to problem 22, which of course is implicitly what you did.
    Ah true.. my instinct was to try and get rid of the logs before I start doing any potentially grizzly DUTIS (though I thought I better get typing before someone else steals it ).

    I can't think of any particularly useful/interesting generalisations tbh, can you? The substitution unzipped the problem very quickly so it is isn't really anything new... hmm.. :dontknow:
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    Problem 294**/***

    Prove
    \displaystyle \frac{\zeta^2(2)}{2}=\sum_{m=1}^  {\infty}\sum_{n=1}^{\infty}\frac  {1}{m^4+m^2n^2}

    I still can't really do any of Jkn's integrals, but I'll keep trying.

    Spoiler:
    Show
    Not necessarily a hint, but Problemette 294: Prove
    \displaystyle \frac{\pi}{m}\frac{1}{e^{2 \pi m}-1}+\frac{\pi}{2 m}-\frac{1}{2m^2}=\sum_{n=1}^{ \infty }\frac{1}{m^2+n^2}
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    (Original post by henpen)
    Problem 294**/***

    Prove
    \displaystyle \frac{\zeta^2(2)}{2}=\sum_{m=1}^  {\infty}\sum_{n=1}^{\infty}\frac  {1}{m^4+m^2n^2}
    Oh how I love waking up to random problems like this! I'm wondering if you found the same nice 'trick' that I did (the method is reminiscent of Fubini's Theorem for integrals).

    Solution 294


    Changing the order of the series, changing that bases WLOG, adding and then separating:

    \displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{m^2(  m^2+n^2)}= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{n^2(  m^2+n^2)} \Rightarrow \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{m^4+  m^2n^2}
    \displaystyle =\frac{1}{2} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{m^2+  n^2} \left( \frac{1}{m^2}+\frac{1}{n^2} \right) = \frac{1}{2} \left( \sum_{m=1}^{\infty}\frac{1}{m^2} \right) \left( \sum_{n=1}^{\infty}\frac{1}{n^2} \right) = \frac{\zeta^2 (2)}{2} \ \square
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    Problem 295***

    Suppose we have a closed contour \gamma in the complex plane. Then it defines an enclosed area. We shall define

     S = \dfrac{1}{2i}\displaystyle\int_{  \gamma}\bar z\ dz

    To be the signed area enclosed by \gamma

    Show that S is real, and that it recovers \pi r^2 for a circle, show both signs can occur, and explain the significance of the sign.
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    (Original post by FireGarden)
    Problem 295***

    Suppose we have a closed contour \gamma in the complex plane. Then it defines an enclosed area. We shall define

     S = \dfrac{1}{2i}\displaystyle\int_{  \gamma}\bar z\ dz

    To be the signed area enclosed by \gamma

    Show that S is real, and that it recovers \pi r^2 for a circle, show both signs can occur, and explain the significance of the sign.
    Consider the parametrization \gamma(t) = r e^{it} for t \in [0, 2 \pi] which is a circle of radius r about the original. Using Cauchy's integral theorem, we know that this contour will give a general result for \gamma.

    Integrating in the anticlockwise direction:

    Hence \displaystyle S = \frac{1}{2i} \int_{\gamma}\bar z\ dz = \frac{1}{2i} \int_0^{2 \pi} (r e^{-it})(ri e^{it}) \ dt = \frac{r^2}{2} \left[ t \right]_0^{2 \pi} = \pi r^2 \ \square

    Integration clockwise along this contour would return the value of  - \pi r^2.
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    (Original post by Jkn)
    Solution 295

    Consider the parametrization \gamma(t) = r e^{it} for t \in [0, 2 \pi] which is a circle of radius r about the original. Using Cauchy's integral theorem, we know that this contour will give a general result for \gamma.

    Integrating in the anticlockwise direction:

    Hence \displaystyle S = \frac{1}{2i} \int_{\gamma}\bar z\ dz = \frac{1}{2i} \int_0^{2 \pi} (r e^{-it})(ri e^{it}) \ dt = \frac{r^2}{2} \left[ t \right]_0^{2 \pi} = \pi r^2 \ \square

    Integration clockwise along this contour would return the value of  - \pi r^2.
    You didn't show S was real..? (I did mean for any smooth gamma!)
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    (Original post by FireGarden)
    You didn't show S was real..? (I did mean for any smooth gamma!)
    CIT implies the contour we choose does not affect our result so long as the turning number is the same, is that right? And an 'enclosed area' sort of implies either a simple smooth loop either clockwise or anticlockwise (real in both cases) .I'm new to Contour integration, so sorry if that's completely wrong . Also, there are no poles so it's quite a simple case, isn't it?
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    (Original post by Jkn)
    CIT implies the contour we choose does not affect our result so long as the turning number is the same, is that right?
    CIT requires the contour to be within a domain where the function is holomorphic (ie, differentiable). f(z)=\bar z is only differentiable at the origin! Though CIT isn't really even applicable here, anyway. It would only tell us (if the function were holomorphic) that its integral would simply be zero. With the ideas of poles within the region bound by the contours, we would then look to the residue theorem; somewhat of a generalisation of Cauchy's integral theorem.

    And an 'enclosed area' sort of implies either a simple smooth loop either clockwise or anticlockwise (real in both cases) .
    The curve may be as pathological and windy as desired, as the only requirement is that the contour is smooth. Of course, the interpretation of "area" you get from such a curve isn't as straightforward as a simple curve, but then, the idea of "negative area" isn't new really - the same issue occurs for real integrals.
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    Solution 271

    We have \displaystyle \Gamma(z)= \int_{0}^{\infty} t^{z-1}e^{-t}dt, so \displaystyle \int_{0}^{\infty} e^{-t} \ln tdt= \Gamma(1)'=\digamma(1). Now, using the fact that \displaystyle \frac{1}{\Gamma(z)}= \exp(\gamma z)z \prod \left(1+\frac{z}{n} \right)e^{-\frac{z}{n}}. We get \displaystyle \digamma(z)= \lim_{n \to \infty} (\ln n - \sum_{k=0}^{n} \frac{1}{k+z}), which gives the result.

    Solution 272

    From the above result, we obtain \displaystyle \gamma = -4\int_{0}^{\infty} e^{-x^{2}}x \ln x dx. Again, if we let \displaystyle x \mapsto \ln \frac{1}{x}, we get \displaystyle \gamma =- \int_{0}^{1} \ln \ln \frac{1}{x} dx.
    We also have \displaystyle \digamma(z) = \int_{0}^{\infty} \left(\frac{1}{e^{x}x}- \frac{1}{e^{zx}(1-e^{-x})} \right) dx (follows directly from the definition), which gives \displaystyle \gamma = \int_{0}^{\infty} \left( \frac{1}{xe^{x}}-\frac{1}{e^{x}-1} \right) dx. Letting x \mapsto -\ln x, we have the fourth integral. Considering \displaystyle \int_{0}^{\infty} \left( \frac{1}{e^{x}-1}-\frac{1}{x(1+x^{k})} \right) dx=0, we get the fifth representation.

    Solution 273

    \begin{aligned} \displaystyle \int_{0}^{1}\int_{0}^{1} \frac{1-x}{(1-xy)(-\ln xy)^{-z}}dx dy   & \overset{y \mapsto \frac{y}{x}}=  \int_{0}^{1}\int_{0}^{x} \frac{1-x}{x(1-y)(-\ln y)^{-z}}dydx  \\&=  \int_{0}^{1} \frac{y- \ln y -1}{(1-y)(-\ln y)^{-z}}dy  \\&=  \Gamma(z+2) \zeta(z+2) -\Gamma(z+1) \end{aligned}.
    Now, letting z \to -1, we get \displaystyle \int_{0}^{1}\int_{0}^{1} \frac{x-1}{(1-xy)\ln xy}dxdy= -\gamma.

    Solution 275

    Some complex analysis shows that \displaystyle \int_{0}^{1} \frac{1}{(1-xy)x(\ln^{2}(\frac{1-x}{x})+\pi^{2})}dx = \frac{1}{\ln(1-y)}+\frac{1}{y}. Hence, using \displaystyle \gamma = \int_{0}^{1} \left( \frac{1}{\ln x}+ \frac{1}{1-x} \right)dx, we get \displaystyle \gamma = \int_{-\infty}^{\infty} \frac{\ln(1+e^{-x})e^{x}}{x^{2}+\pi^{2}}dx \overset{e^{-x} \mapsto x}= \int_{0}^{\infty} \frac{\ln(1+x)}{x^{2}(\ln^{2}x+\  pi^{2}}dx.



    Let me propose a problem, which is very dear to my heart.

    Problem 296***

    Consider a smooth manifold X with an open cover n< \infty of sets \{X_{k}\}_{1}^{n}, which are contractible. Assume also that \pi_{0}(X_{k} \cap X_{l}) \le m, for all k,l. Find an upper bound to the first Betti number of X.

    (Original post by Smaug123)
    anyone can join this coven, but we insist that it remain finite…
    At least, I have picked a nice substitute.
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    (Original post by Mladenov)
    Consider a smooth manifold X with an open coven…
    anyone can join this coven, but we insist that it remain finite…
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    (Original post by Mladenov)
    Solution 271

    Now, using the fact that \displaystyle \frac{1}{\Gamma(z)}= \exp(\gamma z)z \prod \left(1+\frac{z}{n} \right)e^{-\frac{z}{n}}
    This is incomplete. The point of this question is to prove (or not assume) any non-*/** theorems. 'Felix Felicis' produced a similar proof but deleted it when he realized that he had to prove the theorems.
    Solution 272

    From the above result, we obtain \displaystyle \gamma = -4\int_{0}^{\infty} e^{-x^{2}}x \ln x dx.
    We also have \displaystyle \digamma(z) = \int_{0}^{\infty} \left(\frac{1}{e^{x}x}- \frac{1}{e^{zx}(1-e^{-x})} \right) dx (follows directly from the definition)
    which gives \displaystyle \gamma = \int_{0}^{\infty} \left( \frac{1}{xe^{x}}-\frac{1}{e^{x}-1} \right) dx.
    Considering \displaystyle \int_{0}^{\infty} \left( \frac{1}{e^{x}-1}-\frac{1}{x(1+x^{k})} \right) dx=0, we get the fifth representation.
    None of these verbal statements are sufficient for mathematical proof.
    Solution 273

    \begin{aligned} \displaystyle \int_{0}^{1}\int_{0}^{1} \frac{1-x}{(1-xy)(-\ln xy)^{-z}}dx dy   & \overset{y \mapsto \frac{y}{x}}=  \int_{0}^{1}\int_{0}^{x} \frac{1-x}{x(1-y)(-\ln y)^{-z}}dydx  \\&=  \int_{0}^{1} \frac{y- \ln y -1}{(1-y)(-\ln y)^{-z}}dy  \\&=  \Gamma(z+2) \zeta(z+2) -\Gamma(z+1) \end{aligned}.
    Now, letting z \to -1, we get \displaystyle \int_{0}^{1}\int_{0}^{1} \frac{x-1}{(1-xy)\ln xy}dxdy= -\gamma.
    Looks right, but you're going to need to include some more details.. I mean, who in their right mind could follow that without pen and paper?

    The integral actually began as a conjecture in 2004 (though it was immediately solved since it's not that hard) but the mathematician who came up with it was unable to prove it!

    Also, it is not true for all complex z so either you have made a mistake or that's another crucial point you have left out.
    Solution 275

    Some complex analysis shows that \displaystyle \int_{0}^{1} \frac{1}{(1-xy)x(\ln^{2}(\frac{1-x}{x})+\pi^{2})}dx = \frac{1}{\ln(1-y)}+\frac{1}{y}[/latex]
    Some Galois Representations shows that \displaystyle x^n+y^n \not= z^n \forall n >2 \land x,y,z,n \in \mathbb{N} \land x, y, z \not=0... you're going to either need to show your working or give the name of the theorem you are using. All you are essentially doing is telling us you can do it?

    Sorry to sound like such a **** man, but if anyone besides you posted those solutions, people would probably think they didn't actually do it given the lack of detail.. also the satisfaction is reading someone's solution is in understanding something, or seeing how someone else did something.
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    (Original post by Jkn)
    Oh how I love waking up to random problems like this! I'm wondering if you found the same nice 'trick' that I did (the method is reminiscent of Fubini's Theorem for integrals).

    Solution 294


    Changing the order of the series, changing that bases WLOG, adding and then separating:

    \displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{m^2(  m^2+n^2)}= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{n^2(  m^2+n^2)} \Rightarrow \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{m^4+  m^2n^2}
    \displaystyle =\frac{1}{2} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{m^2+  n^2} \left( \frac{1}{m^2}+\frac{1}{n^2} \right) = \frac{1}{2} \left( \sum_{m=1}^{\infty}\frac{1}{m^2} \right) \left( \sum_{n=1}^{\infty}\frac{1}{n^2} \right) = \frac{\zeta^2 (2)}{2} \ \square
    My solution was much more clunky, it involved doing the first summation (via the infinite series for cot), then the second after a closed form for the first as a function of m is reached.
    I'm not familiar with the content of Fubini's theorem or its discrete analogue. Why were you able to swap summation order? It makes sense intuitively but I'd like to formalise it.
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    (Original post by Jkn)
    This is incomplete. The point of this question is to prove (or not assume) any non-*/** theorems. 'Felix Felicis' produced a similar proof but deleted it when he realized that he had to prove the theorems.
    Well, we actually do not need this representation; I just came up with something else - integration by parts shows that \Gamma(x)=(x-1)\Gamma(x-1), and so, \displaystyle \ln \Gamma(x) = \ln \Gamma(x+n) - \sum_{k=0}^{n-1} \ln (x+k). It remains to be shown that \displaystyle (\ln \Gamma(x))' = \ln x + O(\frac{1}{x}).
    To this end, we note that (\ln \Gamma(x))' is monotonically increasing and by the mean value theorem, we have \ln (x-1) \le (\ln \Gamma(x))' \le \ln x; the conclusion follows.


    (Original post by Jkn)
    None of these verbal statements are sufficient for mathematical proof.
    Apart from the integral representation of digamma, everything is pretty obvious, I claim. If needed, I shall compare the definitions.

    (Original post by Jkn)
    Looks right, but you're going to need to include some more details.. I mean, who in their right mind could follow that without pen and paper?

    The integral actually began as a conjecture in 2004 (though it was immediately solved since it's not that hard) but the mathematician who came up with it was unable to prove it!

    Also, it is not true for all complex z so either you have made a mistake or that's another crucial point you have left out.
    I missed those steps, in which we change the order of integration, since they are easily justifiable. Also, to make it more clear, we can use a substitution in order to show the last line.
    Yes, the result is true for \text{Re}(z)>-1; that's why we need to take the limit, and not evaluate the integral directly, though there is a way of doing it. I used a standard result from the theory of the Riemann zeta function to evaluate the limit.

    And, just by the way, if you think that you will always be able to read mathematical text without a pen and paper, then, I dare say, you are completely wrong. Moreover, when the author omits the details, that one who benefits is the reader.
    (Original post by Jkn)
    Some Galois Representations shows that \displaystyle x^n+y^n \not= z^n \forall n >2 \land x,y,z,n \in \mathbb{N} \land x, y, z \not=0... you're going to either need to show your working or give the name of the theorem you are using. All you are essentially doing is telling us you can do it?

    Sorry to sound like such a **** man, but if anyone besides you posted those solutions, people would probably think they didn't actually do it given the lack of detail.. also the satisfaction is reading someone's solution is in understanding something, or seeing how someone else did something.
    Don't be that strict; I just thought, I could eschew some \LaTeX typing.
    I am going the have a dinner, and then I shall type the solution, which uses nothing more than some theory of Markov functions (The Theory of Analytic Functions by Markushevich is one of the best books on complex analysis).

    Apropos, the book I am currently reading goes like this: Theorem 4.6 Proof: Exercise 6; and so on. The author leaves even the crucial theorems as exercises..
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    (Original post by Jkn)
    Sorry to sound like such a **** man, but if anyone besides you posted those solutions, people would probably think they didn't actually do it given the lack of detail.. also the satisfaction is reading someone's solution is in understanding something, or seeing how someone else did something.
    (Original post by Mladenov)
    And, just by the way, if you think that you will always be able to read mathematical text without a pen and paper, then, I dare say, you are completely wrong. Moreover, when the author omits the details, that one who benefits is the reader.
    This is true for a textbook, or a paper, but not for a forum devoted to this kind of exercise, for which people would like to see working in order to follow a solution more easily. Especially since the ability and level of mathematical exposure varies wildly across the users of TSR.
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    (Original post by Jkn)

    Solution 294-5
    Hot stuff, man.
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    Don't want to clog up the thread:

    Spoiler:
    Show
    (Original post by FireGarden)
    CIT requires the contour to be within a domain where the function is holomorphic (ie, differentiable). f(z)=\bar z is only differentiable at the origin! Though CIT isn't really even applicable here, anyway. It would only tell us (if the function were holomorphic) that its integral would simply be zero. With the ideas of poles within the region bound by the contours, we would then look to the residue theorem; somewhat of a generalisation of Cauchy's integral theorem.

    The curve may be as pathological and windy as desired, as the only requirement is that the contour is smooth. Of course, the interpretation of "area" you get from such a curve isn't as straightforward as a simple curve, but then, the idea of "negative area" isn't new really - the same issue occurs for real integrals.
    Oh well it sounds as though it involves concepts I am yet to come to terms with (my approach to Contour Integration has been narrowly through the need for evaluation of real integrals - though it has seemed rather dissatisfying so far). I will leave it for someone else to do (though if no-one solves it I'll do some more studying and have anther go perhaps ).
    (Original post by henpen)
    My solution was much more clunky, it involved doing the first summation (via the infinite series for cot), then the second after a closed form for the first as a function of m is reached.
    I'm not familiar with the content of Fubini's theorem or its discrete analogue. Why were you able to swap summation order? It makes sense intuitively but I'd like to formalise it.
    Ah I thought you might have done that (based on seeing your spoiler). It's a really good thing to learn! In many double/triple/etc.. integrals/summations/products, the key often lies in being able to separate the solutions. Initially we could not separate the two sums but after using the nice symmetrical limit-swap, we were then able to put it in a separable form (in case it wasn't clear: I first swapped the summations and then swapped the symbols 'n' and 'm' as their only property is that they are different). Give it a google! It's one of those useful things, like DUTIS, that people often mindlessly use without even thinking it could be classed as a 'theorem'!
    (Original post by Mladenov)
    Well, we actually do not need this representation; I just came up with something else - integration by parts shows that \Gamma(x)=(x-1)\Gamma(x-1), and so, \displaystyle \ln \Gamma(x) = \ln \Gamma(x+n) - \sum_{k=0}^{n-1} \ln (x+k). It remains to be shown that \displaystyle (\ln \Gamma(x))' = \ln x + O(\frac{1}{x}).
    To this end, we note that (\ln \Gamma(x))' is monotonically increasing and by the mean value theorem, we have \ln (x-1) \le (\ln \Gamma(x))' \le \ln x; the conclusion follows.
    Hmm that's nice. Looks as though it would be rather grizzly to directly link with the limit definition of the E-M constant though!

    Ah but now you have to prove that theorem too..
    I missed those steps, in which we change the order of integration, since they are easily justifiable. Also, to make it more clear, we can use a substitution in order to show the last line.
    Yes, the result is true for \text{Re}(z)>-1; that's why we need to take the limit, and not evaluate the integral directly, though there is a way of doing it. I used a standard result from the theory of the Riemann zeta function to evaluate the limit.

    And, just by the way, if you think that you will always be able to read mathematical text without a pen and paper, then, I dare say, you are completely wrong. Moreover, when the author omits the details, that one who benefits is the reader.
    Yes but I am not talking about a mathematical text! Yes, that is often the case, but I'd like to think this thread can serve to teach people and help those to improve. It certainly has done for me! If you emit multiple uses of Fubini's, exponential substitutions, integral representations of the zeta function combined with alternate unstated representations of the E-M constant in a single line though, there is no way people are ever going to learn!
    Don't be that strict; I just thought, I could eschew some \LaTeX typing.
    I am going the have a dinner, and then I shall type the solution, which uses nothing more than some theory of Markov functions (The Theory of Analytic Functions by Markushevich is one of the best books on complex analysis).

    Apropos, the book I am currently reading goes like this: Theorem 4.6 Proof: Exercise 6; and so on. The author leaves even the crucial theorems as exercises..
    No idea what that is but, so long as you explain what it is (or offer some sort of learning resource), I look forward to it..

    Sounds pretty hardcore! You can only really do things like that once you've 'broken in' to a topic though. For me, I have found group/rings/fields resources that do that rather annoying as I am fairly new to it! Thankfully I have found some resource that have some worked examples which are helping me to start thinking about easing my way towards some non-trivial problems!
    (Original post by Smaug123)
    This is true for a textbook, or a paper, but not for a forum devoted to this kind of exercise, for which people would like to see working in order to follow a solution more easily. Especially since the ability and level of mathematical exposure varies wildly across the users of TSR.
    This^

    Pretty much sums up why I made ASOM - for a more 'supportive'/'learning-directed' side to these community threads!
    (Original post by MW24595)
    Hot stuff, man.
    Cheers bro!

    Oh and thanks for those group theory resources you posted (somewhere?) a while ago! (I refer to them above^) Really need to update my resource library...
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    (Original post by Jkn)
    Don't want to clog up the thread:
    Hmm, a page with comments is actually a really bad design for a forum. It should be a directed graph - usually it would be a tree, although you could draw branches together in a single reply. You'd then be able to reply to a comment by creating a child node to it, and quote people by making the original post a parent. Then a single conversation could be tracked really easily - just follow the arrows, the first-created arrow first (that is, weight the arcs by when they were formed).
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    Pretty much agree with what Smaug has said, I can't follow some of the solutions on here and I would say I have more maths experience (but not talent!) than most on TSR...
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    (Original post by Smaug123)
    Hmm, a page with comments is actually a really bad design for a forum. It should be a directed graph - usually it would be a tree, although you could draw branches together in a single reply. You'd then be able to reply to a comment by creating a child node to it, and quote people by making the original post a parent. Then a single conversation could be tracked really easily - just follow the arrows, the first-created arrow first (that is, weight the arcs by when they were formed).
    It sounds a lot like you want reddit:p:
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    (Original post by Slumpy)
    It sounds a lot like you want reddit:p:
    Kind of, but more flexible - I'm visualising something like the roots of a tree, stretching down, but sometimes meeting up and joining. Not just a straight binary tree, no meeting-and-joining.
 
 
 
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