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May 26, 2014 at 20:53 comment added username It's a question of notations. If $\partial_x^2 a =(\partial_x a)^2$ my counter example works. If $\partial_x^2 a = \partial_{xx} a$ the answer below gives the counter example.
May 26, 2014 at 0:30 review Reopen votes
May 26, 2014 at 9:30
May 26, 2014 at 0:14 comment added Wang Ming Thanks for your answer. For your example, the right hand side is 0, which does not give a contradiction.
May 26, 2014 at 0:13 history edited Wang Ming CC BY-SA 3.0
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May 25, 2014 at 9:38 history closed José Figueroa-O'Farrill
Stefan Kohl
Chris Godsil
Ben Webster
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May 24, 2014 at 22:16 comment added username counterexample: $\phi(x)=\cos(x)$ $a(x)= 2+\sin(Nx)$ gives $$ 2\pi > N^2\frac{\pi}{2}.$$
May 24, 2014 at 19:15 review Close votes
May 25, 2014 at 9:38
May 24, 2014 at 18:56 comment added José Figueroa-O'Farrill This reads like homework. Voted to close.
May 24, 2014 at 9:35 answer added marcoromito timeline score: 2
May 24, 2014 at 3:29 history edited Wang Ming CC BY-SA 3.0
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May 24, 2014 at 2:59 history asked Wang Ming CC BY-SA 3.0