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 April 4th, 2014, 12:10 PM #11 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,260 Thanks: 198 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? $$[0 , t) × [0 , t) , t \to 1^-$$ Thanks from MarkFL
 April 4th, 2014, 12:22 PM #12 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 932 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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...!) Thanks from MarkFL
 April 4th, 2014, 01:00 PM #13 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra Indeed a function can be discontinuous at finitely many points in an interval and still be integrable over the interval. Thanks from CRGreathouse and MarkFL
 April 4th, 2014, 06:39 PM #14 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,260 Thanks: 198 I relinquish control to Olinguito , you get to ask the next question! Thanks from MarkFL and Olinguito
 April 5th, 2014, 01:20 AM #15 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra 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 $\displaystyle f(x)\,=\,x^3+x^2-2x-1$. Thanks from agentredlum
 April 5th, 2014, 10:54 PM #16 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,260 Thanks: 198 If no one can answer in 24 hours please post the full solution.
 April 6th, 2014, 01:28 AM #17 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra 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 $f(x)=x^3+x^2-2x-1$. Thanks from MarkFL and agentredlum Last edited by Olinguito; April 6th, 2014 at 01:31 AM.
 April 6th, 2014, 03:14 PM #18 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,260 Thanks: 198 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. Thanks from Olinguito
 April 7th, 2014, 01:07 AM #19 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra I release control to MarkFL.
 April 8th, 2014, 08:39 AM #20 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra 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 $2011$. Thanks from MarkFL and agentredlum

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