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April 4th, 2014, 12:10 PM   #11
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Yes , me too , but... the area of the volume element at (1 , 1 , 0) is 0 but the height of the volume element at (1 ,1 , 0) is infinity , how can we be sure the total volume (represented by the double integral) will converge nicely to $ \dfrac{ \pi^2}{6}$ ?

Would this concern be nullified if we remove (1 , 1 , f(1 , 1)) from the domain and range?

$$ [0 , t) × [0 , t) , t \to 1^- $$

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April 4th, 2014, 12:22 PM   #12
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Integrals aren't concerned by that kind of discontinuity at endpoints. They really only care about the open interval (min, max), even in well-behaved sequences. (Of course you need it to use the FTA...!)
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April 4th, 2014, 01:00 PM   #13
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Indeed a function can be discontinuous at finitely many points in an interval and still be integrable over the interval.
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April 4th, 2014, 06:39 PM   #14
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I relinquish control to Olinguito , you get to ask the next question!

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April 5th, 2014, 01:20 AM   #15
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Thanks Agentredlum.

$\displaystyle f(x)$ is a monic cubic polynomial with distinct real roots $\displaystyle \alpha,\beta,\gamma$, two of which are negative, such that
$$\alpha\,=\,\gamma^2-2 \\ \beta\,=\,\alpha^2-2 \\ \gamma\,=\,\beta^2-2$$
Show that $\displaystyle f(x)\,=\,x^3+x^2-2x-1$.
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April 5th, 2014, 10:54 PM   #16
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If no one can answer in 24 hours please post the full solution.

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April 6th, 2014, 01:28 AM   #17
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Well, I suppose the 24 hours are up. Anyone not wanting to see the solution yet, look away now!


$\displaystyle \beta\,=\,\alpha^2-2\,=\,(\gamma^2-2)^2-2\,=\,\gamma^4-4\gamma^2+2$.

$\displaystyle \gamma\,=\,\beta^2-2\,=\,(\gamma^4-4\gamma^2+2)^2-2\,=\,\gamma^8-8\gamma^6+20\gamma^4-16\gamma^2+2$.

$\displaystyle \therefore\ 0\,=\,\gamma^8-8\gamma^6+20\gamma^4-16\gamma^2-\gamma+2\,=\,(\gamma+1)(\gamma-2)(\gamma^3-3\gamma+1)(\gamma^3+\gamma^2-2\gamma-1)$.

$\gamma=-1$ or $\gamma=2$ would give rise to repeated roots. It is easy to check that the polynomial $x^3-3x+1$ has two positive real roots (one in the interval $(0,\,1)$ and the other in the interval $(1,\,2)$). Hence we must have $f(x)=x^3+x^2-2x-1$.
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Last edited by Olinguito; April 6th, 2014 at 01:31 AM.
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April 6th, 2014, 03:14 PM   #18
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This is a brilliant analysis , thank you for sharing.

I got the 8th degree polynomial but then didn't try factoring it because i couldn't see what good that would do.

Now i see.

Go ahead and ask another question please , the control is still yours.

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April 7th, 2014, 01:07 AM   #19
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I release control to MarkFL.
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April 8th, 2014, 08:39 AM   #20
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On second thoughts (and since Mark hasn’t posted) I will take back control and post another question.


Find all integers $n$ such that $\displaystyle \left(n^{2014}+2013\right)^{2012}+1$ is divisible by $2011$.
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