im not sure if this is what you're looking for, but from $f-Y$ vanishing on the two lines $Y=\pm 1$ (they vanish on $x\ne 0$, but because the vanishing locus is closed it follows that they also vanish on $(0,1)$ and $(0,-1)$), you can deduce that $f = Y + c(1-Y^2)$ in $k[X,Y]/(X(1-Y^2))$ for some $c\in k[x,y]/(X(1-Y^2))$ and I guess you can argue that this must vanish, by e.g. putting $x=0$ in
A few readers have Emailed me asking how I manage the time to read. But it should be noted that the books I have reviewed I have read over the span of about 35 years. The reviews are posted at my leisure, and it should not be concluded that a book review that is posted 3 days after a prior one implies that the book was read in three days. Most of the books I have reviewed I read long before Amazon was even conceived. The books were therefore not read in the sequential order in which their reviews are posted.
You know, I'd be delighted to buy that for Dr. Carson, but he may freakout by a 'stalker' sending him items from his wishlist and stop writing the amazing reviews. xD
Occupation: Mathematician/physicist. Education Level: PhD (Physics, 1989, University of Texas At Austin) Income Level: Live very comfortably Location: Belcamp (suburb of Baltimore) Race: White Ethnicity: Swedish/British hybrid Religion: Atheist Sexual Orientation: Heterosexual Research Projects: 'Quantization of Mathematics'(in progress). What would happen if mathematical structures were viewed from a strictly 'quantum' point of view?
This research is thus an attempt to view mathematics from such a 'quantum strategy'. My statement on Amazon reviews: All the books I review I have read completely or if not have indicated as such in the review itself.
In other words, if you choose $n$ numbers from $[-1,1]$ independently at random, the odds that the sum of their squares is 1 is much less than the odds that they're all positive.
In a recent Math Stack Exchange question I asked about the function $$f(z)=\sum_{n=0}^\infty z^{2^n},$$ and was informed of its status is a canonical example of a lacunary series with natural boundary at $|z|=1$. A phenomenon observed by the accepted answer was that this function has a multitude ...
More generally, such sums are examples of lacunary series since you've got larger and larger gaps between subsequent powers (i.e. most of the coefficients are zero)
@Abr001am I suspect there is no formula (closed form) for it. But finding it numerically (i.e., finding the first few digits of it for specific values of $x$ and $k$) should be pretty easy
"In addition to the unique real zero at z=−0.658626…, Mahler in a 1982 paper determined, to within eight decimal places, eight complex conjugate pairs of zeros of f(z). This gives a total of 17 fairly precisely located zeros. At the end of that paper he conjectures that there are infinitely many; in fact, in an earlier paper...he says that he expects every point on the unit circle to be a limit point of zeros of f. I do not know if his conjecture has since been proved or disproved."
I am interested in finding a proof of the following property that does not make reference to bases, and ideally doesn't use facts about determinants that depend on the block structure of a matrix.
Let $T \in L(V,V)$ be a linear operator on a finite-dimensional space $V$. Suppose $W \preccurl...
I remember I knew a nice to get that special case.
We have $\displaystyle \prod_{0 \le k \le n}\left(1-x^{2^{k}}\right) = \frac{\left(1-x\right)}{\left(1-x^{2^{n+1}}\right)}\prod_{1 \le k \le n+1}\left(1-x^{2^{k}}\right) = \frac{\left(1-x\right) }{\left(1-x^{2^{n+1}}\right)}\prod_{0 \le k \le n}\left(1-x^{2^{k+1}}\right). $
But $\displaystyle \left(1-x^{2^{k+1}}\right) = \left(1-x^{2^{k}}\right)\left(1+x^{2^{k}}\right)$, thus $\displaystyle \prod_{0 \le k \le n}\left(1-x^{2^{k+1}}\right) =\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)\prod_{0 \le k \le n}\left(1+x^{2^{k}}\right). $
We have the following cases for the number of subsets of size $k$ chosen from a set of $n$ distinct elements:
replacement and ordered, "permutation with repetition" $$n^k$$
no replacement and ordered, "k-permutations of n" $$\frac{n!}{(n-k)!}$$
no replacement and unordered, "combinations" $$\b...
@MatheinBoulomenos I don't think that's correct, $g\circ f$ with $g$ continuous and $f$ Lebesgue measurable is Lebesgue measurable, but that can fail if $g$ is Lebesgue measurable rather than continuous
Let $M$ be a smooth manifold. Consider $\Lambda ^k(M)=\bigcup_{p\in M}\Lambda ^k(T_{p}M)$ with the natural smooth structure. With this structure I showed that the $\pi :\Lambda ^k(M)\rightarrow M$ projection is smooth.
Show that a $k$-form $\omega$ is smooth iff it is smooth as a section $\omega : M\rightarrow \Lambda ^k(M)$ of $\pi$.
It is not clear to me the condition: "smooth as a section $\omega :M\rightarrow \Lambda ^k(M)$ of $\pi$".
@tatan Lol, I didn't have this in mind when I created it.
For a bounded real-valued function $f(x,y)$ (not necessarily continuous!) Fubini's theorem says that one can compute the double integral as iterated integral first over $y$ and then over $x$ for almost all $x$. Not necessarily for all $x$ since the inner integral over $y$ must not exist for every $x$. This means the problematic points $x$ are a Lebesgue null-set.
I was wondering if this poses any problem in real life when computing iterated integrals. Is it true, that we can just ignore the problematic points since the integral over a null set is zero anyways?
I have it as an exercise to prove that the axiom of replacement implies the axiom of specification. So the latter of these states, if $A$ is a set and $Q(x)$ is a statement pertaining to $x\in A$, that $z\in\{ y\mid\exists x\in A \ \text{s.t.} \ Q(x)\}\iff \exists x\in A \ \text{s.t.} Q(x)$. From here, how can get that $z\in \{x\in A\mid Q(x)\} \iff z\in A \land Q(z)$.
For a bounded real-valued function $f(x,y)$ (not necessarily continuous!) Fubini's theorem says that one can compute the double integral as iterated integral first over $y$ and then over $x$ for almost all $x$. Not necessarily for all $x$ since the inner integral over $y$ must not exist for every $x$. This means the problematic points $x$ are a Lebesgue null-set.
I was wondering if this poses any problem in real life when computing iterated integrals. Is it true, that we can just ignore the problematic points since the integral over a null set is zero anyways?
I am interested in finding a proof of the following property that does not make reference to bases, and ideally doesn't use facts about determinants that depend on the block structure of a matrix.
Let $T \in L(V,V)$ be a linear operator on a finite-dimensional space $V$. Suppose $W \preccurl...
We have the following cases for the number of subsets of size $k$ chosen from a set of $n$ distinct elements:
replacement and ordered, "permutation with repetition" $$n^k$$
no replacement and ordered, "k-permutations of n" $$\frac{n!}{(n-k)!}$$
no replacement and unordered, "combinations" $$\b...
