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? $$ [0 , t) × [0 , t) , t \to 1^ $$ 
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 wellbehaved sequences. (Of course you need it to use the FTA...!)

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. 
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! 
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 $$\alpha\,=\,\gamma^22 \\ \beta\,=\,\alpha^22 \\ \gamma\,=\,\beta^22$$Show that $\displaystyle f(x)\,=\,x^3+x^22x1$. 
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! $\displaystyle \beta\,=\,\alpha^22\,=\,(\gamma^22)^22\,=\,\gamma^44\gamma^2+2$. $\displaystyle \gamma\,=\,\beta^22\,=\,(\gamma^44\gamma^2+2)^22\,=\,\gamma^88\gamma^6+20\gamma^416\gamma^2+2$. $\displaystyle \therefore\ 0\,=\,\gamma^88\gamma^6+20\gamma^416\gamma^2\gamma+2\,=\,(\gamma+1)(\gamma2)(\gamma^33\gamma+1)(\gamma^3+\gamma^22\gamma1)$. $\gamma=1$ or $\gamma=2$ would give rise to repeated roots. It is easy to check that the polynomial $x^33x+1$ has two positive real roots (one in the interval $(0,\,1)$ and the other in the interval $(1,\,2)$). Hence we must have $f(x)=x^3+x^22x1$. 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. 
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 $2011$. 

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