Tag Info

1 vote

Proving that the function $f(x) := |x|^{\frac{\lambda - n}{p}} (1- \psi(x))$ satisfies two specific properties related with limits and supremums.

Your approach for property 1 seems sound. Let's proceed to address property 2. For property 2, we need to show that the expression tends to zero as $\xi$ approaches zero. To do this, let's expand the ...
• 355
Accepted

• 23
1 vote
Accepted

• 30.7k
Accepted

• 10.6k

Folium of Descartes - what is this point P?

Yes, that is the correct idea. At the point $P$, the derivative $dy/dx$ is undefined; or equivalently, the derivative $dx/dy = 0$. The function $f(x,y)$ satisfies reflection symmetry about $y = x$; ...
• 137k

• 631

In which points is $f: \mathbb{R}P^2 \rightarrow \mathbb{R}^3$ an inmersion.

Your chart $\phi_1$ is the one you would use if working with the projective space as the quotient $\Bbb{R}^3\setminus\{0\}\to\Bbb{RP}^2$, not $S^2\to\Bbb{RP}^2$. Thus your whole calculation is wrong. ...
• 56k

Finding a Fixed Point for nested Squareroots

With square roots you can use the binomial theorem $$|\sqrt{3+2y}-\sqrt{3+2x}|=\frac{2|y-x|}{\sqrt{3+2y}+\sqrt{3+2x}} \le \frac{|y-x|}{\sqrt3}.$$ This is sufficient for the fixed-point theorem. You ...
• 127k

Given $d(\vec{x},A):=\text{inf}\{\lVert\vec{x}-\vec{y}\rVert:\vec{y}\in A\}$, show that $f^{-1}(\{0\})=\overline{A}$

For $\overline{A} \subseteq f^{-1}(\{0\})$ suppose $d(x,A)>0$. Take $2\varepsilon = d(x,A)$. Suffices to note that $B(x,\varepsilon)\cap A = \emptyset$. If not, then there exists $y\in A$ such that ...
• 8,699
1 vote

Finding a Fixed Point for nested Squareroots

This works with a little bit more caution: $\phi'(x)<1$ is not sufficient, but we need more strongly $\phi'(x)\le k$ where $k$ is a constant that is strictly less than $1$. Surely this is easy in ...
• 15.4k
1 vote

• 7,577
1 vote

How do I check if this number is transcendental?

It looks as if the zeroes in the decimal expansion of your number may occur in long enough chunks to make it a Liouville number and thus transcendental. If not, you can probably modify your ...
• 95.7k

Generalization of $\max$ and $\min$ of $Ax^2+Bxy+Cy^2$ with given $x^2+y^2=k$.

Eliminating $x$ in $$\cases{ z = A x^2+B x y +C y^2\\ x^2+y^2=k }$$ we obtain $$p(y) = A^2 k^2-2 A^2 k y^2+A^2 y^4+2 A C k y^2-2 A C y^4-2 A k z+2 A y^2 z-B^2 k y^2+B^2 y^4+C^2 y^4-2 C y^2 z+z^2$$ ...
• 33.5k

Generalization of $\max$ and $\min$ of $Ax^2+Bxy+Cy^2$ with given $x^2+y^2=k$.

The linear algebra way to do this is to evaluate the objective as. $x^T\pmatrix {A & \frac {B}{2}\\\frac {B}{2} & C} x^T$ And the maximum will be associated with the largest eigenvalue. The ...
• 11.3k

• 1,835