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    (Original post by joostan)
    Stole this from the old thread courtesy of Felix :lol:
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    Show that the sum of the infinite series:

    is
    Don't know what e and ln is :s I know that I have to use: \frac{a}{1-r}

    (Original post by Felix Felicis)
    It doesn't take that long! And it's peculiar But fine, fine...

    How about this...

    If \displaystyle\prod_{r=1}^{n} f(r) denotes f(1) \cdot f(2) \cdot f(3) \cdots f(n) then evaluate

    \displaystyle\prod_{r=1}^{n} \left( \dfrac{r+1}{r} \right)
    Ahh...I see...
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    :nah:
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    (Original post by joostan)
    I mean exactly what I said. . . And whole number has no real meaning in this context. :ahee:
    You troll you
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    No one ever wants to do my questions. :emo:
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    (Original post by Felix Felicis)
    No one ever wants to do my questions. :emo:
    Wasn't that part of a STEP question? I remember the look of it
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    (Original post by MathsNerd1)
    Wasn't that part of a STEP question? I remember the look of it
    Haha yeah but come on, admittedly it's pretty straightforward xD
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    (Original post by tigerz)
    Don't know what e and ln is :s I know that I have to use: \frac{a}{1-r}
    e is defined such that:
    \dfrac{d}{dx}(e^x) = e^x
    In series form it's.
    e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + . . . + \frac{x^n}{n!}+ . . .
    or more formally.
    e^x = \displaystyle\sum_{n=0}^{\infty} \dfrac{x^n}{n!}

    Alternatively this is also:
    e = \lim{n \to \infty} (1+\frac{1}{n})^n

    \ln(x) = \log_e(x)

    (Original post by tigerz)
    Ahh...I see...
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    :nah:
    There's a link in one of my previous posts if you want to have a look.
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    (Original post by Felix Felicis)
    No one ever wants to do my questions. :emo:
    I would...If i could :ashamed:
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    (Original post by Felix Felicis)
    Haha yeah but come on, admittedly it's pretty straightforward xD
    Yeah it's rather simple once you think about it but the sight of it would probably scare some people off, that was a nice question in general
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    (Original post by Felix Felicis)
    No one ever wants to do my questions. :emo:
    They are doing your question, reubenkinara did your other one.
    And I'd already done the product one
    (Original post by Felix Felicis)
    You troll you
    Hehe -
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    It's how my teacher taught my class complex numbers, except without the ages :lol:
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    I just broke maths. Come at me bros.

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    \displaystyle\lim_{n \to \infty} \displaystyle\sum_{r=0}^{n} 2^{r} = \infty

    \displaystyle\sum_{r=0}^{\infty} 2^{r} = (2-1) \displaystyle\sum_{r=0}^{\infty} 2^{r} = 2 \displaystyle\sum_{r=0}^{\infty} 2^{r}  - \displaystyle\sum_{r=0}^{\infty} 2^{r}

    = \left(2+4+8+16+32+...) - (1+2+4+8+16+32+...)

    =-1

    \therefore \infty = -1

    :eek:
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    (Original post by joostan)
    e is defined such that:
    \dfrac{d}{dx}(e^x) = e^x
    In series form it's.
    e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + . . . + \frac{x^n}{n!}
    or more formally. e^x = \displaystyle\sum_{n=0}^\ifnty \dfrac{x^n}{n!}

    Alternatively this is also:
    e = \lim{n \to \ifnty} (1+\frac{1}{n}

    \ln(x) = \log_e(x)



    There's a link in one of my previous posts if you want to have a look.
    Ahh, my target is to solve one your questions, and one from felix felices, after the 10th of June though LOOOL
    For now I don't think I can solve that :'(
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    (Original post by tigerz)
    I would...If i could :ashamed:
    You can. xD It's honestly not as daunting as it seems and is entirely possible with C2 knowledge.
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    (Original post by Felix Felicis)
    You can. xD It's honestly not as daunting as it seems and is entirely possible with C2 knowledge.
    Ahh, I do understand it but I think it'll take me a while, maybe i'll try this one tomorrow
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    (Original post by tigerz)
    Ahh, my target is to solve one your questions, and one from felix felices, after the 10th of June though LOOOL
    For now I don't think I can solve that :'(
    lol, I tidied up the latex in my post - hopefully it's clearer now.
    Felix's one's not too bad
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    (Original post by tigerz)
    Ahh, I do understand it but I think it'll take me a while, maybe i'll try this one tomorrow
    If you work through his question it should only take about 5 minutes, once you 'see' it.
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    (Original post by Felix Felicis)
    I just broke maths. Come at me bros.

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    \displaystyle\lim_{n \to \infty} \displaystyle\sum_{r=0}^{n} 2^{r} = \infty

    \displaystyle\sum_{r=0}^{\infty} 2^{r} = (2-1) \displaystyle\sum_{r=0}^{\infty} 2^{r} = 2 \displaystyle\sum_{r=0}^{\infty} 2^{r}  - \displaystyle\sum_{r=0}^{\infty} 2^{r}

    = \left(2+4+8+16+32+...) - (1+2+4+8+16+32+...)

    =-1

    \therefore \infty = -1

    :eek:
    This implies that:
    \infty - \infty = -1
    Which doesn't really mean much :lol: just like:
    \dfrac{\infty}{\infty} = ?
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    (Original post by MathsNerd1)
    If you work through his question it should only take about 5 minutes, once you 'see' it.
    Okays Thank youu

    (Original post by joostan)
    lol, I tidied up the latex in my post - hopefully it's clearer now.
    Felix's one's not too bad
    Ahh thanks, for now i'll just do the simple ones

    I don't even know how many conversations i'm having anymore :s
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    (Original post by joostan)
    This implies that:
    \infty - \infty = -1
    Which doesn't really mean much :lol: just like:
    \dfrac{\infty}{\infty} = ?
    Yeah, I know. xD
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    (Original post by joostan)
    Personally? No, I just dive straight into it
    Hint: - If you want one.
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    I told tigerz that it's a Geometric series so it's only fair.
    I give up. What I have is a stupid mess. I'll post in spoilers
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    S_{\infty}=\dfrac{(log_2 e)^2}{log_2 \frac{e^2}{2}}
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    (Original post by tigerz)
    Okays Thank youu
    Good good, I'll now be back on this forum a lot more to test myself with the challenging questions and to help out others who need it as this morning I finished my exam which was taking such a great deal of my time up
 
 
 
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