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    Problem 191*


    Determine all real numbers  z which satisfy this equation, and remarking as to why some roots are disregarded.


    \sqrt{\sqrt{3-z}-\sqrt{z+1}}>\frac{1}{2}
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    Solution 176:
    \displaystyle\int^1_0 x(1-x)^{99} \ dx = \displaystyle\int^1_0 (1-u)u^{99} \ du = \dfrac{1}{10100}
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    (Original post by LewisMead)
    \int^\frac{1}{2}_0 (arcsinx)^2 dx = \frac{\pi^2}{72}-\int^\frac{1}{2}_0\frac{2xarcsin  x}{\sqrt{1-x^2}} dx = \frac{\pi^2}{72}+\frac{\pi\sqrt{  3}}{6} - 1

    God typing LaTeX on a mobile is tedious.
    Yay, this is how I did it! Much more satisfying!
    (Original post by Lord of the Flies)
    Solution 187
    I should really get more confident with the Laplace Transform! I got as far as the last integral but it looked too messy to turn into something I could apply the inverse too :lol: Nicely done!
    (Original post by joostan)
    \Rightarrow I = 4\displaystyle \int \dfrac{u}{(u+1)^{10}} \ dx
    Nice job! Though, given how straightforward a lot of these integrals are, I believe the aim is to come up with with most elegant/quick method. I believe that the only chance someone is going to have to complete one of these questions every 50 seconds is to find such elegant methods! Heres mine

    Solution 187 (2)

    Let u=\sqrt[4]{x},

    \displaystyle \Rightarrow \int \dfrac{1}{\sqrt{x} (\sqrt[4]{x}+1)^{10}} dx = 4 \int \dfrac{(u+1)-1}{(u+1)^{10}} \ dx = 4 \int \dfrac{1}{(u+1)^9} -\dfrac{1}{(u+1)^{10}} du 

\displaystyle =-\frac{1}{2} (u+1)^{-8} +\frac{4}{9} (u+1)^{-9} +\mathcal{C} = - \dfrac{1}{18}\left( \dfrac{9\sqrt[4]{x}+1}{(\sqrt[4]{x}+1)^{9}}\right)+ \mathcal{C}
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    (Original post by Jkn)
    Nice job! Though, given how straightforward a lot of these integrals are, I believe the aim is to come up with with most elegant/quick method. I believe that the only chance someone is going to have to complete one of these questions every 50 seconds is to find such elegant methods! Heres mine

    Solution 187 (2)

    Let u=\sqrt[4]{x},

    \displaystyle \Rightarrow \int \dfrac{1}{\sqrt{x} (\sqrt[4]{x}+1)^{10}} dx = \int \dfrac{(u+1)-1}{(u+1)^{10}} \ dx = \int \dfrac{1}{(u+1)^9} -\dfrac{1}{(u+1)^{10}} du 

\displaystyle =-\frac{1}{8} (u+1)^{-8} +\frac{1}{9} (u+1)^{-9} +\mathcal{C} = - \dfrac{1}{18}\left( \dfrac{9\sqrt[4]{x}+1}{(\sqrt[4]{x}+1)^{9}}\right)+ \mathcal{C}
    The difference is marginal, I just showed more working, though, you're right, that would have been slightly neater.
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    (Original post by Jkn)
    [

    Problem 184*

    Evaluate \displaystyle \int \frac{x^4}{1-x^2} dx
    Solution 184

    Let  x = tanhy

     dx = sech^2y \ dy

     \Rightarrow I = \int \dfrac{tanh^4ysech^2y}{sech^2y} \ dy

     = \int tanh^4y \ dy

     = \int (1-sech^2y)tanh^2y \ dy

     = \int 1-sech^2y - sech^2ytanh^2y \ dy

     = y - tanhy - \frac{1}{3} tanh^3y + c

     = artanhx - x -\frac{1}{3}x^3 + c
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    (Original post by Zakee)
    Problem 191*


    Determine all real numbers  z which satisfy this equation, and remarking as to why some roots are disregarded.


    \sqrt{\sqrt{3-z}-\sqrt{z+1}}>\frac{1}{2}
    Solution 191

    Note \sqrt{ \sqrt{3-z} - \sqrt{z+1}} is only defined for -1 \leq z < 1

    \displaystyle \begin{aligned} \sqrt{ \sqrt{3-z} - \sqrt{z + 1}} > \dfrac{1}{2} & \Rightarrow \sqrt{3-z} \cdot \sqrt{z+1} > \dfrac{63}{32} \\ & \Rightarrow -1 \leq z < 1 - \dfrac{\sqrt{127}}{32} \end{aligned}

    Is there a more concise solution than my page and half of arithmetic? :lol:
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    (Original post by DJMayes)
    Solution 184

    Let  x = tanhy

     dx = sech^2y \ dy

     \Rightarrow I = \int \dfrac{tanh^4ysech^2y}{sech^2y} \ dy

     = \int tanh^4y \ dy

     = \int (1-sech^2y)tanh^2y \ dy

     = \int 1-sech^2y - sech^2ytanh^2y \ dy

     = y - tanhy - \frac{1}{3} tanh^3y + c

     = artanhx - x -\frac{1}{3}x^3 + c
    You love hyperbolic functions
    Solution 184 (alt)
    \int \frac{x^4}{1-x^2}\dx = \int -x^2 -1 +\frac{1}{1-x^2}\dx
    =\frac{-1}{3}x^3-x+arctanhx+c
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    (Original post by joostan)
    The difference is marginal, I just showed more working, though, you're right, that would have been slightly neater.
    The differences add up I do believe! And being clever with your algebra is a useful trick to know (see below)
    (Original post by DJMayes)
    Solution 184

    Let  x = tanhy

     dx = sech^2y \ dy

     \Rightarrow I = \int \dfrac{tanh^4ysech^2y}{sech^2y} \ dy

     = \int tanh^4y \ dy

     = \int (1-sech^2y)tanh^2y \ dy

     = \int 1-sech^2y - sech^2ytanh^2y \ dy

     = y - tanhy - \frac{1}{3} tanh^3y + c

     = artanhx - x -\frac{1}{3}x^3 + c
    -(snip)-

    Because jokes can be taken the wrong way :rolleyes:
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    (Original post by DJMayes)
    Solution 184

    Let  x = tanhy

     dx = sech^2y \ dy

     \Rightarrow I = \int \dfrac{tanh^4ysech^2y}{sech^2y} \ dy

     = \int tanh^4y \ dy

     = \int (1-sech^2y)tanh^2y \ dy

     = \int 1-sech^2y - sech^2ytanh^2y \ dy

     = y - tanhy - \frac{1}{3} tanh^3y + c

     = artanhx - x -\frac{1}{3}x^3 + c
    You love your hyperbolic functions!
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    Here is a truly sensuous result.

    Problem 192*

    f is a twice-differentiable function with continuous derivatives, and satisfies the following conditions over (a,b):

    (\text{i})\;f(x)>0\qquad (\text{ii})\; f''(x)+f(x)>0

    Additionally,

    (\text{iii})\; f(a)=f(b)=0

    Show that b-a>\pi
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    (Original post by Felix Felicis)
    Note \sqrt{ \sqrt{3-z} - \sqrt{z+1}} is only defined for -1 \leq z < 1

    \Rightarrow 1 \leq z < 1 + \dfrac{\sqrt{127}}{32}
    :pierre: *coughcough*
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    (Original post by Jkn)
    :pierre: *coughcough*
    Typo typo typo
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    (Original post by Felix Felicis)
    Solution 191

    Note \sqrt{ \sqrt{3-z} - \sqrt{z+1}} is only defined for -1 \leq z < 1

    \sqrt{ \sqrt{3-z} - \sqrt{z + 1}} > \dfrac{1}{2} \Rightarrow \sqrt{3-z} \cdot \sqrt{z+1} > \dfrac{63}{32}

    \Rightarrow 1 \leq z < 1 - \dfrac{\sqrt{127}}{32}

    Is there a more concise solution than my page and half of arithmetic? :lol:

    You defined the limits of the equation to be -1 \leq z < 1

    So I'm just curious as to where  f(-1) has gone? Can't it be...

    -1 \leq z < 1 - \dfrac{\sqrt{127}}{32}

    Oh, possibly. There may be a more elegant solution (mine is only 8-9 steps long of working), unless your writing is brobdingnagian.
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    (Original post by Lord of the Flies)

    (\text{i})\;f(x)>0\qquad  (\text{iii})\; f(a)=f(b)=0
    Really?
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    (Original post by Felix Felicis)
    Typo typo typo
    Of course. I bet you're just one of those reckless *******s who loves introducing additional solutions into equations. Something like this:

    http://i.qkme.me/3otjtl.jpg
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    (Original post by bananarama2)
    Really?
    Apologies, amended.
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    (Original post by Zakee;42894413

    So I'm just curious as to where [tex
    )

    f(-1) [/tex] has gone? Can't it be...
    Typo

    (Original post by Zakee)
    Of course. I bet you're just one of those reckless *******s who loves introducing additional solutions into equations. Something like this:

    http://i.qkme.me/3otjtl.jpg
    I like to live dangerously :pierre:

    #CARPEDIEM
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    (Original post by Mladenov)
    By the way, the second question is much more difficult. I can't think of anything simpler than L-functions.
    The second part can be done in three lines by using only value of Hurwitz's zeta function.

    Problem 193 / ****

    Prove that x^2 + y^2 + z^4 = p^2 has no integer solutions, with xyz \not= 0, for a prime p \equiv 7\ (8).
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    (Original post by Felix Felicis)
    Typo


    I like to live dangerously :pierre:

    #CARPEDIEM

    Don't you mean...

    http://i887.photobucket.com/albums/a...ikarpediem.jpg


    http://cache.ohinternet.com/images/a/a1/Awww_yeah.jpg
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    (Original post by bananarama2)
    You love your hyperbolic functions!
    I do - so much better than trig!

    (Original post by Jkn)

    A correct and perfectly acceptable solution. However, this was in the head-to-head and there is no chance you could have done that in less than 30 seconds let alone doing most of the working in your head!

    My approach on the other hand... :rolleyes: (hopefully now my point will be made)

    Solution 184 (2)

    \displaystyle \int \frac{x^4}{1-x^2} dx =\int \frac{1-(1-x^2)-x^2(1-x^2)}{1-x^2} dx = artanh(x)-x-\frac{1}{3} x^3 +\mathcal{C}
    What point are you trying to make? That you're obnoxious? That is all that has been proven here tonight and scarcely needed more rigorous justification anyway.
 
 
 
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