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    (Original post by joostan)
    How pretty are you expecting the answer to be?
    Spoiler:
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    At present I have the rather unpleasant expression you have, x\ln(x) \mapsto -\frac{1}{e}, and with the x^x=e^{x\ln(x)}=e^{-\frac{1}{e}}, and I don't much fancy seeing if it simplifies.
    Oh effing eff - I meant abscissa - that'll teach me to use those fancy college kid words. I'll edit.
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    (Original post by atsruser)
    Oh effing eff - I meant abscissa - that'll teach me to use those fancy college kid words. I'll edit. Maybe you could spoiler your reply though?
    It is spoilered. . .?
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    (Original post by joostan)
    It is spoilered. . .?
    Ignore me. I'm tired and heading off to bed. I look forward to your complete argument in the morning.
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    (Original post by atsruser)
    Ignore me. I'm tired and heading off to bed. I look forward to your complete argument in the morning.
    Using the standard notation here to save TeXing all this junk.
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    Note that:
    f(x)=x\ln(x)+\displaystyle\sum_{  k=1}^n(^ke)^{e^{x\ln(x)}}.

\implies f'(x)=(1+\ln(x))\left(1+ \displaystyle \sum_{k=1}^n\left(\displaystyle \prod_{j=1}^k (^je)^{x\ln(x)}\right) \right).
    f'(x)=0 \iff 1+\ln(x)=0 or g(x)=\left(1+ \displaystyle \sum_{k=1}^n\left(\displaystyle \prod_{j=1}^k (^je)^{x\ln(x)}\right) \right)=0.
    Noting that all the terms of g(x) are necessarily positive, we deduce that we must have: 1+\ln(x)=0.
    Thus we have a single stationary point, which is in fact a minimum, at x=\dfrac{1}{e}.
    To see that this is a minimum, simply note that g(x)>0 and so the derivative changes sign from negative to positive as x increases at x=\dfrac{1}{e}.
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    (Original post by joostan)
    ...
    Minor typo, x can't be negative - but lovely solution! This notation fuzzles it up enough for me not to understand it. :laugh:
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    (Original post by Zacken)
    Minor typo, x can't be negative - but lovely solution! This notation fuzzles it up enough for me not to understand it. :laugh:
    whupsy daisy -that was x\ln(x) rather than x, will edit.
    It is a little confusing, but if you write it out it's simple enough, I was being lazy.
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    (Original post by joostan)
    Using the standard notation here to save TeXing all this junk.
    Spoiler:
    Show
    Note that:
    f(x)=x\ln(x)+\displaystyle\sum_{  k=1}^n(^ke)^{e^{x\ln(x)}}.

\implies f'(x)=(1+\ln(x))\left(1+ \displaystyle \sum_{k=1}^n\left(\displaystyle \prod_{j=1}^k (^je)^{x\ln(x)}\right) \right).
    f'(x)=0 \iff 1+\ln(x)=0 or g(x)=\left(1+ \displaystyle \sum_{k=1}^n\left(\displaystyle \prod_{j=1}^k (^je)^{x\ln(x)}\right) \right)=0.
    Noting that all the terms of g(x) are necessarily positive, we deduce that we must have: 1+\ln(x)=0.
    Thus we have a single stationary point, which is in fact a minimum, at x=\dfrac{1}{e}.
    To see that this is a minimum, simply note that g(x)>0 and so the derivative changes sign from negative to positive as x increases at x=\dfrac{1}{e}.
    That looks fine, but I think that there's a much simpler argument:

    Spoiler:
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    Note that f(x)=g(x) + e^{g(x)} + e^{e^{g(x)}} + \cdots +e^{e^{.^{.^{e^{g(x)}}}}}

    Now let y=e^{h(x)} \Rightarrow y' = h'(x)e^{h(x)} = 0 \Rightarrow h'(x) = 0 since e^{h(x)} > 0 for all x. So h(x), e^{h(x)} have the same turning points, and so do their sum.

    Thus we need only consider the first term in the series for f(x), which trivially gives a minimum of 1/e
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    (Original post by atsruser)
    That looks fine, but I think that there's a much simpler argument:
    Spoiler:
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    Note that f(x)=g(x) + e^{g(x)} + e^{e^{g(x)}} + \cdots +e^{e^{.^{.^{e^{g(x)}}}}}

    Now let y=e^{h(x)} \Rightarrow y' = h'(x)e^{h(x)} = 0 \Rightarrow h'(x) = 0 since e^{h(x)} > 0 for all x. So h(x), e^{h(x)} have the same turning points, and so do their sum.

    Thus we need only consider the first term in the series for f(x), which trivially gives a minimum of 1/e
    It is a little smoother I guess, but I calculated the ordinates first.
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    Since I took a problem, here's one I made up back, it's not very hard.
    Problem 572:*
    Compute the exact value of:
    x=13+\dfrac{1}{2+\dfrac{1}{26+ \dfrac{1}{2+\frac{1}{26+\frac{1}  {2+...}}}}}
    Where this fraction continues in this recurring way ad infinitum.
    Find an expression of this type (known as a continued fraction) for x=\dfrac{1+\sqrt{5}}{2}, and x=\sqrt{2}.
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    (Original post by joostan)
    Since I took a problem, here's one I made up back, it's not very hard.
    Problem 572:*
    Yaay! Iterative stuff!


    Compute the exact value of:

    x=13+\dfrac{1}{2+\dfrac{1}{26+ \dfrac{1}{2+\frac{1}{26+\frac{1}  {2+...}}}}}

    Where this fraction continues in this recurring way ad infinitum.
    \displaystyle 

\begin{equation*}x -13 = \frac{1}{2 + \frac{1}{26 + x-13}} \Rightarrow \frac{1}{x-13} = 2 + \frac{1}{x+13} \Rightarrow \frac{1}{x-13} - \frac{1}{x+13} = 2 \end{equation*}

    So that x^2 = \sqrt{182} \Rightarrow x = \sqrt{182} where we pick the positive root for obvious reasons.

    Find an expression of this type (known as a continued fraction) for x=\dfrac{1+\sqrt{5}}{2}
    We have:

    \displaystyle 

\begin{equation*}2x = 1 + \sqrt{5} \Rightarrow (2x-1)^2 = 5 \Rightarrow 4x^2 - 4x -4 = 0 \iff x^2 - x - 1 = 0\end{equation*}

    So:

    \displaystyle

\begin{equation*} x^2 = 1 + x \Rightarrow x = 1 + \frac{1}{x} = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \cdots}}}\end{equation*}

    and x=\sqrt{2}.
    We have:

    \displaystyle 

\begin{equation*}\sqrt{2} = 1 + \sqrt{2} - 1 = 1 + \frac{1}{1 + \sqrt{2}} \Rightarrow \sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2+ \cdots}}}
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    Problem 401:

    Determine whether

    1/1^5 + 1/2^5 + 1/3^5...

    is irrational.
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    (Original post by Krollo)
    Problem 401:

    Determine whether

    1/1^5 + 1/2^5 + 1/3^5...

    is irrational.
    1^5 is irrational so the sum of an irrational number and some other numbers are irrational, hence the sum is irrational. QED.
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    (Original post by Zacken)
    1^5 is irrational so the sum of an irrational number and some other numbers are irrational, hence the sum is irrational. QED.
    ^ me trying to learn analysis

    Posted from TSR Mobile
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    (Original post by Zacken)
    1^5 is irrational so the sum of an irrational number and some other numbers are irrational, hence the sum is irrational. QED.
    Such proof. . .
    Many brilliance. . .
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    Wow.
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    (Original post by Zacken)
    Yaay! Iterative stuff!




    \displaystyle 

\begin{equation*}x -13 = \frac{1}{2 + \frac{1}{26 + x-13}} \Rightarrow \frac{1}{x-13} = 2 + \frac{1}{x+13} \Rightarrow \frac{1}{x-13} - \frac{1}{x+13} = 2 \end{equation*}

    So that x^2 = \sqrt{182} \Rightarrow x = \sqrt{182} where we pick the positive root for obvious reasons.



    We have:

    \displaystyle 

\begin{equation*}2x = 1 + \sqrt{5} \Rightarrow (2x-1)^2 = 5 \Rightarrow 4x^2 - 4x -4 = 0 \iff x^2 - x - 1 = 0\end{equation*}

    So:

    \displaystyle

\begin{equation*} x^2 = 1 + x \Rightarrow x = 1 + \frac{1}{x} = \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \cdots}}}\end{equation*}



    We have:

    \displaystyle 

\begin{equation*}\sqrt{2} = 1 + \sqrt{2} - 1 = 1 + \frac{1}{1 + \sqrt{2}} \Rightarrow \sqrt{2} = \frac{1}{2 + \frac{1}{2 + \frac{1}{2+ \cdots}}}
    gah, just as I finished my solution I refresh the page and you've got there first! Atleast I got the same answers
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    (Original post by EnglishMuon)
    gah, just as I finished my solution I refresh the page and you've got there first! Atleast I got the same answers
    Ah, sorry - that's always frustrating!
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    (Original post by Zacken)
    x
    Having looked more closely I now notice you seem to have made a typo.
    In both cases you're missing the preceding 1+.
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    (Original post by joostan)
    Having looked more closely I now notice you seem to have made a typo.
    In both cases you're missing the preceding 1+.
    Urgh, thanks! Fixed. :-)
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    (Original post by Zacken)
    Ah, sorry - that's always frustrating!
    Lol don't be sorry, I'll have to be faster next time
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    (Original post by EnglishMuon)
    Lol don't be sorry, I'll have to be faster next time
    And I'll have to make less mistakes.
 
 
 
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