Hot answers tagged intuition
14
The number $e$ is the number such that the area enclosed by the region bounded by $x=1$ on the left-hand side, the $X$ axis from below, $y=1/x$ from above and $x=e$ on the right-hand side is $1$. Hence, $e$ can be defined as the length (you need to add $1$, since you measure the length from $1$) you need to move along $X$ axis such that the area of the above ...
9
A different take on the 'classical' limit that I think is my favorite way of thinking about $e$ recreationally (and a remarkably useful approximation for many games): "I take a six-sided die and roll it six times. What are the odds I never roll '1' in those six rolls? Okay, now I take a twenty-sided die and roll it twenty times. What are the odds I never ...
8
Claim: (The above.)
Proof: First, notice that
\begin{eqnarray}
xy + yz + zx - 2xyz &=& xy(1 - z) + yz(1-x) + zx(1-y) + xyz\\
&=& xy(x+y) + yz(y+z) + zx(z+x) + xyz
\end{eqnarray}
since $x + y + z = 1$; since we can re-write our original expression as a sum of all positive terms, we plainly have
$$0 \le xy + yz + zx - 2xyz$$
which was ...
7
Many modern representation theorists are interested in "geometric representation theory". One of the goals in this field is to realize a representation (e.g. a representation of a Lie algebra) geometrically. What this means is to realize the underlying vector space as the (co)homology of some algebraic variety and the action (e.g. the action of the Lie ...
6
It turns out there are two separate issues to consider.
In functional notation, derivatives are things that are applied to functions, not variables. The derivative of a univariate function (i.e. a function with one argument) is always the derivative of the value of the function with respect to the argument of the function.
i.e. if $f$ is the function ...
5
The notion of period, which is introduced by Kontsevich and Zagier, would partially give a negative answer to your question. According to this article, it is now known whether $e$ is a period or not, though it is conjecturedly not a peroid. In particular, $e$ seems not to arise as an area or a length of a geometric figure defined by an algebraic equation.
...
4
There is a standard "trick" for evaluating this. First, note that
$$
\int_{\mathbb{R}}e^{-\pi x^2}\,\mathrm{d}x=1\tag{1}
$$
Multiplying $(1)$ together $n$ times, we get
$$
\int_{\mathbb{R}^n}e^{-\pi x^2}\,\mathrm{d}x=1\tag{2}
$$
Converting $(2)$ to polar coordinates yields
$$
\int_0^\infty\omega_{n-1}e^{-\pi r^2}r^{n-1}\,\mathrm{d}r=1\tag{3}
$$
where ...
4
Note that when we define the natural numbers we have a good sense of addition and multiplication (ordinal arithmetics), and from those we can define the operations on $\Bbb Z$ and $\Bbb Q$ and then by using Dedekind cuts construction we can extend these to $\Bbb R$ as well.
So we have that $\Bbb R$ has the operations $+,\cdot$ and they all satisfy all the ...
4
I wans't planning to write a "rant" but it turns out that my beer is pretty good.
Mr. Greinecker talked about a scenario (CH) where it is impossible to measure all sets of reals (this was shown by the Banach and Kuratowski a long time ago late 1920s maybe). Let me add that if you do not require translation invariance then it is conceivable that all sets of ...
3
In fact a much stronger $e$-related statement is true: let $X_i$ denote the number of $i$-cycles in a random permutation on $n$ elements. Then for fixed $k$, as $n \to \infty$ the random variables $X_1, X_2, ... X_k$ are asymptotically independently Poisson with rates $1, \frac{1}{2}, ... \frac{1}{k}$. This observation about derangements is a special case ...
3
$$e\approx 2.71828182846$$
Consider the equation:
$$f(n)=(1+\frac{1}{n})^n$$
As $n$ gets larger and larger, notice what the result approaches.
$$f(1)=2$$
$$f(2)=2.25$$
$$f(3)\approx2.3703703$$
$$...$$
$$f(100)\approx2.7048138$$
$$...$$
3
It all really comes down to HOW the functions are constructed.
Consider the following problem:
A man is 1.5m tall walking towards a street light that is 10m high. The man is walking
at a speed of 2.2m/s. How fast is his shadow shrinking when the man is 10m away from the street light?
Let y = the length of the man's shadow
First of all let's just sort ...
3
Maybe the following might interest you: Let $X$ be a smooth projective curve over $\mathbb{C}$ of genus $\geq 2$. It is a classical theorem by Narasimhan and Seshadri that there is an equivalence of categories between the category of stable vector bundles of degree $0$ on $X$ and the category of irreducible unitary finite dimensional complex representations ...
3
From a logician's point of view, logic is far more than just a tool!! But that's probably a topic for another time...
Certainly, I think people would (or should) agree that logic is at least at very useful tool. The first answer to the question you linked to has a fair point: for most mathematicians, the logical abilities are not as important as creativity. ...
2
The book $N$ contains much less information than the book $C$: it tells you how many times each letter appeared but not what order they appeared in (that's where the information is).
A simpler model for understanding what's going on here is to consider binary strings, so words on the alphabet $\{ 0, 1 \}$. There are $2^n$ possible binary strings of length ...
2
I use the picture of the rectangle in my own teaching (without the differential notation) and show it to grad students who are starting their teaching careers. It is far superior to the usual tricky addition-of-$0$ argument found in most textbooks.
Here is the argument in greater detail:
\begin{align*}
\frac{\Delta(uv)}{\Delta x} &= \frac{(u+\Delta ...
2
The complex plane is a two-dimensional generalization of the real numbers. Complex numbers (a, b) are added just like you'd expect: (a, b) + (c, d) = (a + c, b + d). They multiply a bit differently, though: (a, b)(c, d) = (ac - bd, ad + bc). You can get ordinary real numbers back by setting the second number of the pair to 0.
Usually (a, b) is written ...
2
Perhaps their homepage helps here...
One of the nice formulas involving them is the one for sum of powers discovered by Bernoulli (and from there they take their name):
$$
\sum_{0 \le k \le n - 1} k^m
= \frac{1}{m + 1} \sum_{0 \le k \le m} \binom{m + 1}{k} B_{m - k} n^k
$$
2
The big theorem about weak homotopy equivalences is Whitehead's Theorem. If $X$ and $Y$ are pointed CW-complexes, and there exists a pointed weak homotopy equivalence $f\colon X\rightarrow Y$, then $f$ is a homotopy equivalence between $X$ and $Y$. Probably the simplest corollary of this theorem is that every CW complex with trivial homotopy groups at all ...
2
There is a generalization of metric space, called continuity space, where the codomain of the metric is replaced by a quantale (a certain partially ordered set with some extra structure, see Flagg's "quantales and continuity spaces"). It it then shown that every topological space is metrizable as long as one interprets metrizable to mean with respect to a ...
2
The topologies for infinite dimensional Banach spaces can be a bit tricky. One nice "measure" of their weird behavior with respect to compactness is looking at the unit ball (vectors of norm $\leq 1$). The unit ball in a Banach space is compact (with respect to the norm topology) if and only if the Banach space is finite dimensional. In some sense, there are ...
2
I can think of lots of ways of visualizing this, so here are a couple that spring to mind. They are all essentially the same.
A crude way Imagine you colour even numbers white and odd numbers black. Now get a big tub, and lots of white and black paint. Then count through the integers, adding a dash of the paint of the corresponding colour, mixing it up as ...
2
This one is not visual or graphical, but may be easiest to understand for a non-mathematician.
Suppose you have $\$1000$ and you want to put it in a bank account. You have picked a bank that, besides giving you an absurd interest over your money, gives you a choice between several interest schemes:
An annual interest of $100\%$.
An interest of $50\%$, but ...
2
Let me try this $$(\pi)_{\mathbb{R}}=\left\{x\in \mathbb{Q}:\;\exists k\in\mathbb{N}.\;x< \sum\limits_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}\right\}$$ This would be the Dedekind construction of the reals, and note that there is no union here.
Or, I would complete the rationals first by taking the quotient of ring of fundamental sequences with maximal ...
2
Here's a calculus solution.
Let's begin with your substitution $x=p+\frac 13, y=q+\frac 13, z=-p-q+\frac 13$. The problem conditions impose that $p \ge -\frac 13, q \ge -\frac 13, -p-q \ge -\frac 13$. This is a triangular area shown here:
The polynomial simplifies to $f(p,q)=2p^2q+2pq^2-\frac{1}{3}(p^2+pq+q^2)+\frac{7}{27}$, and the problem reduces to ...
2
Here's another approach. Let $f(x,y,z) = xy+yz+zx-2xyz$, and suppose its maximum value for $x,y,$ and $z$ non-negative reals with sum 1 is $f(a,b,c)$. Because $f$ is symmetric in its arguments, assume without loss of generality that $a \le b \le c$.
A bit of algebra shows that $f(\frac{a+c}{2},b,\frac{a+c}{2}) - f(a,b,c)$ = $\frac{1}{4}(a-c)^2(2a+2c-1)$, ...
2
First, under the continuum hypothesis, every probability measure on the powerset of $[0,1]$ is concentrated on a countable set of numbers, so it would give a highly asymmetrical notion of "at random".
Second, $\mathbb{Q}\cap[0,1]$ is countable and any measure that does not put positive mass on some numbers will put zero measure on countable sets. Every ...
1
I assume the result of the 1d Fourier transform wouldn't be the results of the 2d Fourier transform in row-major order, but I can't quite convince myself of it.
Your question is not a silly one. Suppose we take a slice of the 2D Fourier transform of the image. This is equivalent to projecting/accumulating the 2D image onto the line parallel to the line ...
1
Apart from doubling $p$ at the end, your answer is correct: your denominator is actually equal to $1$. It can be rewritten as
$$\frac1{2^{2n}}\sum_{i=0}^n\binom{2n-i}n2^i=\frac1{2^{2n}}\sum_{m=n}^{2n}\binom{m}n2^{2n-m}=\frac1{2^{2n}}\sum_{i=0}^n\binom{n+i}n2^{n-i}\;,$$
and
$$\begin{align*}
...
1
In bringing the matrix to Jordan canonical form (see http://en.wikipedia.org/wiki/Jordan_canonical_form) it is crucial to have the multiplicities you denoted $m_i$ so as to describe the blocks into which the matrix decomposes (up to conjugation).
On the other hand, computationally speaking the coincidence of eigenvalues is not stable, and from this point of ...
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