Jeppe Stig Nielsen
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 8h answered why do remainders show cyclic pattern? 9h comment Must all Lebesgue integrable functions really be invertible? You are absolutely right. While all functions you will think of (unless you have very advanced thoughts/phantasies for someone who is seeing the definition of measurable functions for the first time!) will be measurable, it is really easy for you to come up with a method $f:\mathbb{R}\to\mathbb{R}$ which does not have an inverse $f^{-1}:\mathbb{R}\to\mathbb{R}$. Jul20 comment Function that looks a lot like exponential, but isn't Of course if you add $g(x)=\sin (2\pi x)$ or a similar function to your solution, that would be a new "solution" (at least in the sense that it is continuous (even analytic) and has the right values at the natural numbers). Jul20 comment a natural number that is both a perfect square and a perfect cube is a perfect sixth power? It may be helpful to look at the first examples. The number $64$ is both a square and a cube because $64=8^2$ and $64=4^3$. And for $729$ we have $729=27^2$ and $729=9^3$. The next two are $4096=64^2, 4096=16^3$ and $15625=125^2, 15625=25^3$. Can you see where you confused yourself? Jul17 comment Do the matrices with maximum determinant always have integral values See also the comments in OEIS A003432 entry. Jul16 comment Is every noninvertible matrix a zero divisor? And it is also easy to see that a left-factor $C$ with $CA=O$ exists exactly if this is the case. Jul14 comment How to prove that the Fibonacci sequence is periodic mod 5 without using induction? It is a bit interesting that only 25 such pairs $(a,b)$ exist, and your cycle accounts for the 20 of them. The trivial sequence $(0,0) \to (0,0)$, a cycle of length one (fixpoint), accounts for 1 more. The last 4 pairs are in a cycle $(1,3) \to (3,4) \to (4,2) \to (2,1) \to (1,3)$. Jul8 comment Groups where all elements are order 3 Then if we consider the Heisenberg group over $\mathbb{Z}/2\mathbb{Z}$, we still have a non-abelian group, but this time there are elements whose orders do not divide $2$ (therefore no counter-example to the exercise result mentioned by the asker). Jul7 comment Is there an algorithm to compute the degree of a polynomial? @Travis Yes, that is cool. That algorithm would just be the usual Lagrange interpolation given the $\# \mathbb{K}$ pairs $(x_i,y_i)$ of arguments an corresponding values. Jul7 comment Is there an algorithm to compute the degree of a polynomial? The same is true for other infinite fields (no matter if the characteristic is positive): The values (i.e. the "evaluation mapping" for $f$) does specify the polynomial (including its degree), but a finite subset of values is clearly not enough. But the polynomial can be reconstructed from its values at $\aleph_0$ distinct points. Jul7 comment Is there an algorithm to compute the degree of a polynomial? If the field (is a field and) has characteristic 0, you can follow the infinite algorithm of choosing at each step the unique polynomial $p_n$ of degree at most $n$ which agrees with $f$ on $\{0,1,2,\ldots,n\}$. At each step that is elementary interpolation. So this gives an infinte sequence of polynomials $p_0, p_1, p_2, \ldots$. It is clear that this sequence will be constant from some index $N$, so all $p_i=f$ for $i\ge N$. However, it is clear (from your answer as well) that we have no way of knowing when we are past that $N$. So we can never stop the process "safely". Jul3 comment What's a group whose group of automorphisms is non-abelian? A link on the less intuitive fact you mention: MathOverflow: when is Aut(G) abelian Jul3 comment What is the $1469^\text{th}$ derivative of $x^{532}-5x^{37}-4$? Do you know what the degree of a (non-zero) polynomial is? Have you ever thought about what happens to the degree of a polynomial when you take the derivative? Jul2 comment Is $\pi(n)$ a Rational Function? Yes, that was what I thought when I wrote that comment. I hope we are not overlooking something. However this proof uses theorems that were not known before Zhāng's 2013 result. Maybe one can avoid that. Jul2 comment Is $\pi(n)$ a Rational Function? I suspect the growth of $\pi$ is too irregular for this to be possible. For example it was recently shown that for some constant $K$, the quantity $\pi(x+K)-\pi(x)$ is infinitely often over (say) $10$ (and of course is infinitely often zero). Can polynomials $P$ and $Q$ as above be chosen to get this behavoir? Jun30 comment How to define a bijection between $(0,1)$ and $(0,1]$? Here is another way to say the same thing. I reverse the half-open intervals. I want to show that the intervals $A=\left[ 0,\infty \right)$ and $B= \left( -\infty,\infty \right)$ are in bijective correspondence. By chopping up in half-open pieces (of length one) like $\ldots, \left[ 3,4 \right) , \left[ 4,5 \right) , \left[ 5,6 \right) ,\ldots$, the interval $A$ is naturally in correspondence $A \approx \left[ 0,1 \right) \times\mathbb{N}$, while for $B$ that is $B \approx \left[ 0,1 \right) \times\mathbb{Z}$. But since it is known that $\mathbb{N}$ and $\mathbb{Z}$ are "equal", we are done. Jun30 comment Why is the commutator defined differently for groups and rings? OK, so suppose we do consider a division ring. The usual example is $\mathbb{H}$, the quaternions. What is the commutator of $i$ and $j$ in $\mathbb{H}$? If we use the ring structure, we get $[i,j] = ij - ji = k + k = 2k$. If instead we use only the multiplicative group structure, we have $[i,j] = i^{-1}j^{-1}ij = (-i)(-j)ij = (ij)(ij) = k^2 = -1$. And of course if we take quaternions $a,b$ that actually commute, then $[a,b]$ in the ring sense is zero, while in the multiplicative sense the commutator is one. Jun22 comment Distributive property on fractions While $\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}$ is clearly not an identity which holds for all $a, b$, if you use complex numbers, you can find specific pairs $(a,b)$ for which this holds. If for example you fix one $a\in\mathbb{C}$ you can find an associated $b$ by solving a simple quadratic equation. Jun15 revised A number $n$ which is the sum of all numbers $k$ with $\sigma(k)=n$? added 217 characters in body Jun11 comment Measuring angles in a prism (cont.) For each interior angle has $\frac{\pi}{4}$ or $45^\circ$ on each side of the "fold" (cube edge). So the sum of interior angles for the triangle is $\frac{3\pi}{2}$ or $270^\circ$. Compared to a Euclidean triangle, the angle excess is $\frac{\pi}{2}$ or $90^\circ$. Also see the image truncated hexahedron where all eight corners have been cut off like this (and not yet glued back).