1,193 reputation
212
bio website
location Copenhagen, Denmark
age 40
visits member for 1 year, 8 months
seen 11 hours ago

11h
comment A continuous function from the open ball to itself?
Of course one can also choose a homeomorphism between the open disk and the plane. Then the plane allows translations (additions of a constant vector) which clearly are continuous and without fixed points.
2d
comment For cardinals, if $\mathfrak{a}\ne\mathfrak{b}$ then $2^\mathfrak{a}\ne 2^\mathfrak{b}$
OK, I see now that really many people asked this question already. Will only link to Sets question, without Zorn's lemma (duplicate) also. But follow linked threads' linked threads. Maybe this question can simply be closed...
2d
comment For cardinals, if $\mathfrak{a}\ne\mathfrak{b}$ then $2^\mathfrak{a}\ne 2^\mathfrak{b}$
Great link. That gave me something to search for (I was not aware that $\mathfrak{a}\mapsto 2^\mathfrak{a}$ was called the continuum function). This highly related thread also turned up: Why continuum function isn't strictly increasing?.
2d
asked For cardinals, if $\mathfrak{a}\ne\mathfrak{b}$ then $2^\mathfrak{a}\ne 2^\mathfrak{b}$
2d
comment What properties are true for “almost all real numbers”?
The set of computable numbers is only countable, so already when considering the cardinality of the set, we see that there are "few" computables. In contrast, the set of non-normal numbers has the same cardinality, $2^{\aleph_0}$, as the set of normal numbers. So they are equally "potent" as sets. When we check the Lebesgue measure we see the difference: The set of non-normal numbers has measure ("length") $0$; the set of normal numbers has "full" measure.
2d
comment What properties are true for “almost all real numbers”?
For almost all $x$, with $a_i$ as above, we have $$\frac{a_0 + a_1 + \dots + a_n}{n} \to \infty$$ as $n\to\infty$.
2d
comment What properties are true for “almost all real numbers”?
You can construct an integer sequence $\{ a_n \}$ such that its geometric mean as above converges to $\kappa$, and then you can define a number $x$ from $x=[a_0; a_1, a_2, \ldots]$. But that is kind of cheating.
2d
comment What properties are true for “almost all real numbers”?
In each of the examples below, it is interesting to consider if the null set of exceptions is (1) finite, (2) countable, or (3) uncountable, possibly with the full cardinality $2^{\aleph_0}$.
Nov
21
revised What properties are true for “almost all real numbers”?
added 192 characters in body
Nov
20
comment What properties are true for “almost all real numbers”?
So $\frac92$ is to decimal expansions (and arithmetic means) what Khinchin's constant $2.685452001\ldots$ (see MJD's answer) is to continued fractions (and geometric means).
Nov
19
comment Find the smallest value of n for which the sum of the first n terms of the series exceeds 100. Divergent
To explain what @ganeshie8 said: If this series is specified by saying that (the first term is $1.2$ and) each term is 50 percent greater than the previous term, then this is a so-called geometric series (also known as quotient series). There is a well-known closed formula for the sums of finite sections of the series (partial sums) then. And the link gives the solution in that case. -- Ah, I see ganeshie8 improved his first comment a bit now.
Nov
18
comment Solving $ (x+1)^{x^2-4x+3} = 1 $ for x
@labbhattacharjee I like that. Now solve instead $(x+1)^{x^2-4x+2}=1$.
Nov
18
comment Sum of eigenvalues of a symmetric matrix
In a way it depends on whether the underlying field (or ring!) of scalars is an algebraically closed field. For example the matrix $$\begin{pmatrix} 1 & 1 \\ -1 & 1 \\ \end{pmatrix}$$ has a trace of $2$, clearly, but taken as a matrix over $\mathbb{R}$ it has no real eigenvalues. So in a sense its trace is not the sum of its eigenvalues. However, if we take it as a matrix over e.g. $\mathbb{C}$, it has eigenvalues, and their sum is "correct". Note: Any real symmetric matrix has only real eigenvalues.
Nov
14
comment How do you find the imaginary roots of a fourth degree polynomial that cannot be simplified?
Maybe they meant $f(x)=16x^4-1$ since the heading mentions "fourth degree". In that case the substitution $\xi=2x$ leads to $\xi^4=1$, so $\xi\in\{ 1,i,-1,-i \}$. So the solution, with decimal points, would be $x\in\{ 0.5,0.5i,-0.5,-0.5i \}$.
Nov
11
comment Why is an ellipse, hyperbola, and circle not a function?
To state the same in another way. The graph of the function $f(x,y)=\frac{x^2}9 - \frac{y^2}4$ would be a surface in space. Every vertical line in space (that is a line parallel to the $z$ axis) intersects the surface exactly once. The hyperbola mentioned is the inverse image $f^{-1}( \{ 0 \} )$. That is the intersection of the graph with the $xy$ plane (given by $z=0$).
Nov
8
comment Can the composite of two projections really fail to be a projection?
Further projecting $(1,1)$ onto $U$ gives $(1,0)$. We see that $(P_V \circ P_U) \circ (P_V \circ P_U) \ne (P_V \circ P_U)$ (just evaluate at $(2,0)$), so the composition is not idempotent. It also shows that $P_V \circ P_U \ne P_U \circ P_V$. The matrix representations can be found from $P_U(x,y)=(x,0)$ and $P_V(x,y)=((x+y)/2,(x+y)/2)$. Edit: Ah, after your edits you address this as well. Will upvote.
Nov
8
comment Can the composite of two projections really fail to be a projection?
What you call $P_U$ and $P_V$ are not the orthogonal projections onto subspaces $U$ and $V$. They do not respect the Hilbert space structure on $\mathbb{R}^2$. In the question the notation with $P$ was reserved for orthogonal projections. But with the same spaces $U$ and $V$ as above, one is a horizontal line through the origin (the $x$ axis), and the other one is a line at an angle of $\pi /4$ to the first one. Take the point $(x_0,y_0)=(2,0)$. If you project first onto $U$, you get $(2,0)$ again. If you project that to $V$, you get $(1,1)$. That is outside $U\cap V$ (which is $\{ 0\}$).
Nov
5
comment $(n-7)(n+7)=$ some perfect square
@Doeser It is much easier to use Erick's immediate hint, in his first comment to the question. In how many ways can 49 be written as a product of two integers?
Nov
5
comment $(n-7)(n+7)=$ some perfect square
@ErickWong Transforming $(n-7)(n+7)=a^2$ into $(n-a)(n+a)=7^2$ is a beautiful hint.
Nov
5
comment Yet another question about continuity in $\mathbb{R}$
And @HagenvonEitzen's example includes the easy example where $B$ is a singleton set, or even the empty set. So the most difficult part is coming up with a nowhere continuous function.