20,668 reputation
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bio website math.ntnu.no/~hanche
location Trondheim, Norway
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visits member for 2 years, 11 months
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4h
awarded  Caucus
1d
answered Given matrix P such that $P^{102 } =0 $ , to show that $P^{2} = 0$.
Dec
17
comment Intuitively, why is the Gaussian the Fourier transform of itself?
The link you gave at the end results in a “request denied” error. Do you know of any other source?
Dec
17
answered Inequality $a^2b^2+2(a+b)\geq 4ab+1$
Dec
17
comment Inequality $a^2b^2+2(a+b)\geq 4ab+1$
The slightly (but only slightly) flippant answer is: Don't do that, then. To be less flippant, what you have discovered is that the inequality you used is to rough an estimate to work. So you'll need to abandon that approach and try something different.
Dec
17
comment Interchange of derivatives
(2) should be okay if you just add a term $\partial L/\partial t$ on the left. For (1), it is not even clear what $\partial\dot L/\partial\dot q$ should mean. I suppose $\dot L$ is a total derivative of $L$, but then that is only defined along some curve, and you can't differentiate it in the $\dot q$ direction. To elaborate a bit, $d/dt$ is a total derivative along a curve. It doesn't mix well with partial derivatives. A proper answer requires more time than I have available right now.
Dec
17
comment Prove inequalities with induction
Exactly. As an exercise in induction, the problem is unfortunately worthless. (On a side note, be careful in writing stuff like $P(2):2+\sqrt{2}>2 \Rightarrow true$ like you did above, for any statement, true or false, implies any true statement. It's one of the oddities of mathematical logic as opposed to everyday use of language. Perhaps you meant the arrow to mean “evaluates to” or something like that, but then you should have chosen a different style arrow to avoid confusion.)
Dec
17
comment Prove inequalities with induction
Sure, what @Henry says is true. Moreover, it shows that no induction is needed, since it immediately proves the requested inequality for $n\ge3$. I suspect whoever posed the problem did not think it through … that can happen from time to time, to the best of us.
Dec
16
comment Solve $x''(t)-\frac{x^2(t)}{\sin t}=\frac{\sin\left( (t-1)^2\right)}{\sin t}$.
I second your doubt!
Dec
16
comment Solve $x''(t)-\frac{x^2(t)}{\sin t}=\frac{\sin\left( (t-1)^2\right)}{\sin t}$.
@Dr.SonnhardGraubner Try reloading the page.
Dec
16
comment Solve $x''(t)-\frac{x^2(t)}{\sin t}=\frac{\sin\left( (t-1)^2\right)}{\sin t}$.
Is the numerator on the right $\sin\bigl((t-1)^2\bigr)$, or what is more commonly written $\sin^2(t-1)$?
Dec
16
comment Closed subspace. A Hahn–Banach theorem consequence
I was just going to say what @IvoTerek said. (I am too lazy myself, and don't need more rep anyhow.)
Dec
16
comment Closed subspace. A Hahn–Banach theorem consequence
By looking carefully at the meaning of $\bigcap\{\ker(\phi):\phi|_{M} = 0 \}$. Does $x$ belong, or does it not?
Dec
16
comment Well-foundedness is not a first order property.
Don't write $\langle$ and $\rangle$ using < and >. Use \langle and \rangle instead (I fixed it for you). That said, I am not sure I understand the question. Doesn't the same argument in the linked question apply to PA and ZF(C) as well?
Dec
16
revised Well-foundedness is not a first order property.
\langle, \rangle
Dec
16
comment Closed subspace. A Hahn–Banach theorem consequence
Consider a point $x\notin\overline M$, try to build some functional $\phi$ … ?
Dec
16
revised Closed subspace. A Hahn–Banach theorem consequence
edited title
Dec
16
comment Splitting the plane to fit convexes
@AlexRavsky Absolutely. However, the Hahn–Banach separation theorem is much easier in Hilbert spaces! The proof outlined by me and elaborated by Mike Miller works equally well in that more general setting.
Dec
15
answered Splitting the plane to fit convexes
Dec
15
comment Splitting the plane to fit convexes
Are the sets assumed open or closed? If not, you have to be careful: Consider the example where $K$ consists of points with $x^2+y^2<1$ and either $y>0$ or $y=0$ and $x>0$. Then let $L=-K$ (i.e., the reflection of $K$ through the origin).