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1500 questions
107
votes
13 answers
References for multivariable calculus
Due to my ignorance, I find that most of the references for mathematical analysis (real analysis or advanced calculus) I have read do not talk much about the "multivariate calculus". After dealing with the single variable calculus theoretically, it…
user9464
107
votes
1 answer
All polynomials with no natural roots and integer coefficients such that $\phi(n)|\phi(P(n))$
Let $P(x)$ be a polynomial with integer coefficients such that the equation $P(x)=0$ has no positive integer solutions. Find all polynomials $P(x)$ such that for all positive integers $n$ we have $\phi(n) \mid \phi(P(n))$. It is conjectured there…
Amir Parvardi
- 4,896
107
votes
6 answers
Continuous bijection from $(0,1)$ to $[0,1]$
Does there exist a continuous bijection from $(0,1)$ to $[0,1]$? Of course the map should not be a proper map.
Alex
- 1,317
107
votes
8 answers
Is Bayes' Theorem really that interesting?
I have trouble understanding the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science.
From the purely mathematical point of view, I think it would be uncontroversial to say that Bayes'…
user78270
- 4,000
107
votes
3 answers
Direct proof that the wedge product preserves integral cohomology classes?
Let $H^k(M,\mathbb R)$ be the De Rham cohomology of a manifold $M$.
There is a canonical map $H^k(M;\mathbb Z) \to H^k(M;\mathbb R)$ from the integral cohomology to the cohomology with coefficients in $\mathbb R$, which is isomorphic to the De Rham…
Greg Graviton
- 5,184
107
votes
1 answer
Lebesgue measure theory vs differential forms?
I am currently reading various differential geometry books. From what I understand differential forms allow us to generalize calculus to manifolds and thus perform integration on manifolds. I gather that it is, in general, completely distinct from…
sonicboom
- 9,921
- 13
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- 84
107
votes
2 answers
A semigroup $X$ is a group iff for every $g\in X$, $\exists! x\in X$ such that $gxg = g$
The following could have shown up as an exercise in a basic Abstract Algebra text, and if anyone can give me a reference, I will be most grateful.
Consider a set $X$ with an associative law of composition, not known to have an identity or inverses.…
Lubin
- 62,818
107
votes
6 answers
How to learn from proofs?
Recently I finished my 4-year undergraduate studies in mathematics. During the four years, I met all kinds of proofs. Some of them are friendly: they either show you a basic skill in one field or give you a better understanding of concepts and…
Roun
- 3,017
107
votes
13 answers
Why is "the set of all sets" a paradox, in layman's terms?
I've heard of some other paradoxes involving sets (i.e., "the set of all sets that do not contain themselves") and I understand how paradoxes arise from them. But this one I do not understand.
Why is "the set of all sets" a paradox? It seems like…
Justin L.
- 14,532
107
votes
8 answers
What makes elementary functions elementary?
Is there a mathematical reason (or possibly a historical one) that the "elementary" functions are what they are? As I'm learning calculus, I seem to focus most of my attention on trigonometric, logarithmic, exponential, and $n$th roots, and solving…
user23784
107
votes
13 answers
Why would I want to multiply two polynomials?
I'm hoping that this isn't such a basic question that it gets completely laughed off the site, but why would I want to multiply two polynomials together?
I flipped through some algebra books and have googled around a bit, and whenever they…
user3818
- 1,079
106
votes
15 answers
Comparing $\pi^e$ and $e^\pi$ without calculating them
How can I
compare (without calculator or similar device) the values of $\pi^e$ and $e^\pi$ ?
Mirzodaler
- 1,317
106
votes
1 answer
Geometric interpretation of the Riemann-Roch for curves
Let $X$ be a smooth projective curve of genus $g\geq2$ over an algebraically closed field $k$ and denote by $K$ a canonical divisor.
I have some clues about the geometrical interpretation of the Riemann-Roch Theorem for smooth algebraic curves, but…
Abramo
- 6,917
106
votes
4 answers
Defining a manifold without reference to the reals
The standard definition I've seen for a manifold is basically that it's something that's locally the same as $\mathbb{R}^n$, without the metric structure normally associated with $\mathbb{R}^n$. Aesthetically, this seems kind of ugly to me. The real…
user13618
106
votes
2 answers
If $f_k \to f$ a.e. and the $L^p$ norms converge, then $f_k \to f$ in $L^p$
Let $1\leq p < \infty$. Suppose that
$\{f_k, f\} \subset L^p$ (the domain here does not necessarily have to be finite),
$f_k \to f$ almost everywhere, and
$\|f_k\|_{L^p} \to \|f\|_{L^p}$.
Why is it the case that $$\|f_k - f\|_{L^p} \to 0?$$
A…
user1736
- 8,573