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H: What's the recurrence relation to this problem?
A machine can perform $3$ types of operation $A$, $B$ and $C$. The memory is initially $0$. A Program $P$ is a series of these operations. If the machine does $A$, it will add $1$ to the memory's current value. If it does $B$, it will subtract $1$ from the memory. If it does $C$, it will add $2$ to the memory (It is kind of like a Turing machine). How many programs are there if they are of length $n$ and set the memory to zero after execution? And how can I derive a recurrence relation to solve this problem recursively?
AI: You are looking for words of length $n$ on A, B, and C. If there are $a$ A's, $b$ B's, and $c$ C's, the constraints are $a+b+c=n,\ \ a-b+2c=0$ With two equations in three unknowns, you have one free parameter. We can take $c$ as the free parameter. Then $a+b=n-c, \ \ a-b=-2c, a=\frac {n-3c}2,\ \ b=\frac {n+c}2$, so $c$ must have the same parity as $n$. The number of words is $$\sum_{i=0}^{\lfloor \frac n3 \rfloor}\binom n{\frac{n-3i}2}\binom{n-\frac{n-3i}2}{\frac{n+i}2}$$
I don't see a simple recurrence relation for this. |
H: Calculating the following definite integral
How may I approach and compute the following integral? Could polylogarithm functions and complex numbers be avoided?
$$ \int_0^{\frac{\pi}{2}} \frac{x}{\tan x} dx$$
AI: Integrate by parts with $u=x$ and $\displaystyle dv=\frac{dx}{\tan x}$ so $$ \int^{\pi/2}_0 \frac{x}{\tan x} dx = x \log (\sin x) \biggr|^{\pi/2}_0 - \int^{\pi/2}_0 \log (\sin x) dx=- \int^{\pi/2}_0 \log (\sin x) dx. $$
Let $x=\pi/2 - u$ into $\displaystyle I= \int^{\pi/2}_0 \log(\sin x) dx $ to get $\displaystyle I=\int^{\pi/2}_0 \log(\cos x) dx.$ Adding gives $$2I =\int^{\pi/2}_0 \log( \sin x \cos x) dx = \int^{\pi/2}_0 \log(\sin 2x) dx - \frac{\pi \log 2}{2}.$$
Now by $u=2x$ we have $$\int^{\pi/2}_0 \log (\sin 2x) dx = \int^{\pi}_0 \log(\sin u) \frac{du}{2} = I$$
so $$\int^{\pi/2}_0 \frac{x}{\tan x} dx = \frac{\pi \log 2}{2}.$$ |
H: Solving an equation for B
I am having an issue solving this. I wrote this equation to find the with of a bar for a bar chart that is comparative.
\begin{equation}
D-P*(N+1) = ((N-1)*M)*B*0.75+N*B
\end{equation}
AI: No guarantee that the result will be an integer, but just distribute the $B$ out of the terms on the right and divide.
$\begin{equation}
D-P*(N+1) = ((N-1)*M)*B*0.75+N*B
\end{equation}$
$\begin{equation}
D-P*(N+1) =( ((N-1)*M)*0.75+N)*B
\end{equation}$
$\begin{equation}
(D-P*(N+1))/( ((N-1)*M)*0.75+N)=B
\end{equation}$ |
H: Parallelogram With non-90 angles
I realized that in-case of a parallelogram with a non-90 degree angles the two diagonals would never be equal. Is my assumption correct ?
AI: The diagonals of a parallelogram bisect each other, so if the diagonals have equal length, then they cut the parallelogram into four isosceles triangles: a "top and bottom" pair with base angles (say) $\alpha$ and a "left and right" pair with base angles $\beta$. Each corner of the parallelogram has size $\alpha+\beta$, so they're all congruent, which makes them necessarily right angles.
Contrapositively, if the angles aren't right angles, then the diagonals have unequal length. |
H: $10+10\times 0$ equals $0$ or $10$
I thought $10+10\times 0$ equals $0$ because:
$$10+10 = 20$$
And
$$20\times 0 = 0$$
I know about BEDMAS and came up with conclusion it should be $$0$$ not $$10$$
But as per this, answer is $10$, are they right?
AI: There is a "precedence of operations": some operations should be done before other operations in the absence of indications to the contrary. This is a convention used, among other things, to make writing things like polynomials simpler and to require fewer parentheses.
Multiplication has higher precedence than addition. That means that to perform $10+10\times 0$, you should first do the product, then do the sum. So first you do $10\times 0$, then you add the answer of that to $10$. The answer is $10$. |
H: Determinant of an $n\times n$ complex matrix as an $2n\times 2n$ real determinant
If $A$ is an $n\times n$ complex matrix. Is it possible to write $\vert \det A\vert^2$ as a $2n\times 2n$ matrix with blocks containing the real and imaginary parts of $A$?
I remember seeing such a formula, but can not remember where. Any details, (and possibly references) for such a result would be greatly appreciated.
AI: Write $A=A_1+iA_2$ where $A_1$ and $A_2$ are real matrices. Let
$$B:=\pmatrix{A_1&iA_2\\iA_2&A_1}.$$
We have
$$\det B=\det\pmatrix{A_1+iA_2&iA_2\\A_1+iA_2&A_1}=\det\pmatrix{I&iA_2\\I&A_1}\cdot\det\pmatrix{A_1+iA_2&0\\0&I},$$
and
$$\det\pmatrix{I&iA_2\\I&A_1}=\det\pmatrix{I&iA_2\\0&A_1-iA_2}$$
hence
$$\det B=\det(A_1-iA_2)\det(A_1+iA_2)=\det A\cdot\det\bar A=|\det A|^2.$$ |
H: If each matrix representing $T$ contains a $0$, then $T$ is a scalar
Let $V$ be a finite-dimensional vector space over $\mathbb F$ ($\text{char}\ \mathbb F =0$) and $T$ is a linear transformation such that for any basis the matrix of linear transformation has at least one zero.
I want to show that there exists $c\in \mathbb F$ $\,$such that $T(x)=cx$ for any $x \in V.$
Any suggestion?
AI: It certainly won't work in general. Over the field $GF(2)$, the only matrix with no zero entries is all $1$'s, and there are lots of $n \times n$ matrices that are not similar to this besides $0$ and $I$.
EDIT: with the added condition $\text{char } {\mathbb F} = 0$, it works. Let $u$ be a vector such that $u$ and $Tu$ are linearly independent. There is a basis $b_1, b_2, \ldots, b_n$ with $b_1 = u$ and $b_2 = Tu$. Let $B$ be the matrix with these basis vectors as columns,
so $(B^{-1} T B)_{2,1} \ne 0$. Consider $C = \sum_{P} c_P B P$ where the sum is over all $n \times n$ permutation matrices and $c_P$ are indeterminates. $\det(C)$ is a polynomial in the $c_P$, and not identically $0$ (note that $\det(C) = \pm \det(B) \ne 0$ if any one of the $c_P$ is $1$ and the others are $0$). Similarly, any matrix element of $(\text{Adj}(C) T C)_{ij}$ (where $\text{Adj}$ is the adjugate or classical adjoint matrix) is a polynomial in the $c_P$, and not identically $0$: if we take one particular $c_P$ for permutation $\pi$ to be $1$ and the others $0$, $(\text{Adj}(C) T C)_{ij} = \det(BP) (P^{-1} B^{-1} T B P)_{ij}$, and if $\pi(j) = 1$ and $\pi(i)=2$ this is $\det(BP)$. There will therefore be some set of rational values of the $c_P$ for which all
those polynomials are nonzero, and the columns of the corresponding $C$ are a basis in which all matrix elements of $T$ are nonzero. |
H: General method to find inf, sup, maxs and mins of a function
Could someone explain how to find inf, sup, max and min values of a function (real-valued functions of real variable, generally continuous/differentiable, with some possible points of discontinuity)?
Some examples:
$$y = \frac{x}{2}\sqrt{\frac{\log{x} + 1}{\log{x} - 1}}$$
$$y = \frac{x^2}{1+\log|x|}$$
$$y = e^{x-|x^2-x|}$$
$$y = e^{\arctan{\frac{1}{x-1}}}$$
$$y = \left(\frac{1}{2}\right)^{\frac{|x|}{1+x}}$$
$$y = x\log{\frac{1}{|x|}}$$
AI: If you are dealing with real-valued functions of real-variable, then generally you want to:
Determine the domain of the function.
Determine where the function is continuous.
For intervals of continuity, any local extreme will occur at either an endpoint or a critical point (points where the derivative either does not exist, or is equal to $0$). Local extremes may also occur at points of discontinuity. You can also use the derivative to determine intervals on which the function is increasing or decreasing, which will of course help you locate extreme points.
So generally, you want to determine those intervals of continuity, and work each of them using the derivative and the sundry tests that exist for checking critical points to determine if they are local extremes or not.
Example. Let's take your last function:
$$y = x\log{\frac{1}{|x|}}.$$
First, the domain of this function is all $x\neq 0$. The function is continuous at every point of its domain, so we will want to look at what happens on $(0,\infty)$ and on $(-\infty,0)$. In addition, we want to understand what happens as the function approaches the "endpoints" of those intervals ($0$ on either side, $\infty$, and $-\infty$); this is done using limits.
First, as $x\to \infty$, we have the function
$$y = x\log\frac{1}{x} = -x\log x.$$
As $x\to\infty$, $\log x\to\infty$; multiplied by a function that goes to $-\infty$, the product will go to $-\infty$. That is,
$$\lim_{x\to\infty}x\log\frac{1}{|x|} = -\infty.$$
Moreover, if we look at the derivative,
$$y' = -\log x - x\left(\frac{1}{x}\right) = -\log x- 1.$$
This will be positive if $\log x \lt -1$, if and only if $x\lt \frac{1}{e}$; and negative if $x\gt \frac{1}{e}$. We have a critical point at $x=\frac{1}{e}$.
Since $y$ is increasing on $(0,\frac{1}{e})$ and decreasing on $(\frac{1}{e},\infty)$, the function has a local maximum at $\frac{1}{e}$ and no other local extremes on $(0,\infty)$.
We can also verify that
$$\lim_{x\to 0^+}-x\log x = 0,$$
e.g., using L'Hopital's Rule:
$$\begin{align*}
\lim_{x\to 0^+}-x\log x &= \lim_{x\to 0^+}\frac{-\log x}{\frac{1}{x}}\\
&= \lim_{x\to 0^+}\frac{-(1/x)}{-(1/x^2)}\\
&= \lim_{x\to 0^+} x\\
&= 0.
\end{align*}$$
Thus, there are no minimum or local minimum values.
We can do the same analysis on $(-\infty,0)$, or we can note that the function is odd (exploit symmetry whenever possible):
$$y(-x) = (-x)\log\frac{1}{|-x|} = -x\log\frac{1}{|x|} = -y(x).$$
So we can just reflect about the origin to see what happens on $(-\infty,0)$. The function approaches $0$ from below as $x\to 0^-$; the function is increasing on $(-\frac{1}{e},0)$ and decreasing on $(-\infty,-\frac{1}{e})$; there is a local minimum at $x=-\frac{1}{e}$, and
$$\lim_{x\to-\infty}x\log\frac{1}{|x|} = \infty.$$ |
H: Why should coordinate transformations be reversible?
Intuitively I understand why coordinate transformation should be reversible. New coordinates should cover the same area covered by the initial coordinates, i.e. there should be one-to-one mapping.
But still, are reversible transformations used only because it is convenient or is there any theoretical background?
AI: In a "coordinate transformation" (change-of-coordinates transformation), if you start with a basis $\beta=[\mathbf{v}_1,\ldots,\mathbf{v}_n]$, then because the transformation is one-to-one, the image of $\beta$ must be linearly independent, and hence a basis. Thus, the image of the transformation contains a basis, and so will necessarily be onto. Thus, by virtue of being one-to-one on a finite dimensional space, it must necessarily be onto as well, and thus will be invertible.
Or you can view a "change-of-coordinates" transformation as a way of expressing vectors in one basis, $\beta$, in terms of vectors in another basis, $\gamma$; if you then write out what it would mean to express the vectors in $\gamma$ in terms of $\beta$, and you compose the resulting two transformations, you get the elements of $\beta$ expressed in terms of $\beta$ (and composing the other way, the elements of $\gamma$ in terms of the vectors of $\gamma$). Because each vector can be expressed in a unique way in terms of the vectors of a basis, the only way to express the vectors of $\beta$ in terms of the vectors of $\beta$ is by the identity transformation:
$$\begin{align*}
\mathbf{v}_1 &= 1\mathbf{v}_1 + 0\mathbf{v}_2 + \cdots + 0\mathbf{v}_n\\
\mathbf{v}_2 &= 0\mathbf{v}_1 + 1\mathbf{v}_2 + \cdots + 0\mathbf{v}_n\\
&\vdots\\
\mathbf{v}_n &= 0\mathbf{v}_1 + 0\mathbf{v}_2 + \cdots + 1\mathbf{v}_n.
\end{align*}$$
so that the composition is the identity. Composing them the other way also gives the identity, so that shows the original change-of-coordinates transformation must be invertible. |
H: Cauchy's Residue Theorem on a Singularity Outside a Contour
I recently ran into the following exercise:
Evaluate
$$\oint_\Gamma\frac{\cos z}{(z-\pi)^2}dz,$$where $\Gamma$ is a complete circuit of the circle $|z|=1$.
Clearly, the singularity lies outside the contour:
However, recall Cauchy's Residue Theorem:
If $\Gamma$ is a simple closed positively oriented contour and $f$ is analytic inside and on $\Gamma$ except at the points $z_1$, $z_2$, ..., $z_n$, then
$$\int_\Gamma f(z)dz=2\pi i\sum_{j=1}^{n}\text{Res}(z_j),$$
where
$$\text{Res}(f;z_0)=\lim_{z\to z_0}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)].$$
If I use this to evaluate the integral, I will have that
$$\text{Res}(f;\pi)=\lim_{z\to\pi}\frac{d}{dz}[\cos z]=0,$$and hence the value of the integral will evaluate to $0$, which is the correct answer.
Is this purely coincidental? Because, as I understand it, for Cauchy's theorem to hold, the singularities must lie within the contour (or perhaps not?). Thanks in advance!
AI: You don't have to do anything to evaluate your integral, because $\cos(z)/(z-\pi)^2$ is holomorphic on the unit disc. So by Cauchy's theorem for a disk (or whichever more general variant you'd prefer), the integral is zero. You can also evaluate this using the residue theorem, but none of the residues are inside the disk so once again you get zero.
The function does indeed have a pole of order $2$ at $\pi$ and the residue is $0$. This does not mean that the function has a removable singularity at the pole, it just means that the function has an analytic antiderivative in some punctured disk about $\pi$. |
H: Number of elements in $\mathbb{Z}_p[x]/ \langle f \rangle$
I want to determine the number of elements in $\mathbb{Z}_p[x]/ \langle f \rangle$ where $f \in \mathbb{Z}_p[x]$ is an irreducible polynomial with $k$ degree bigger than 2. Is the number of elements always $p^k$? And are the equivalence classes $[f_1],[f_2], \dots, [f_{p^k}]$ always given be the polynomials in $\mathbb{Z}_p$ with degree smaller than $k$?
AI: Yes, you're on the right track of reasoning.
Use the fact that $\mathbb{F}[x]$ has a division algorithm to justify that all possible remainders (=all polynomials of degree strictly less than $k$) form a complete set of distinct residues.
To show that every polynomial is equivalent to a remainder, remember that modding out by $(f)$ is basically saying that $g\equiv h\iff f|(g-h)$. To show that the remainders are distinct, start by writing out what it would mean if two are equivalent, and show it's impossible.
After that, it's easy to count how many remainders! You have $p$ choices for each of the $k$ coefficients of a remainder. |
H: Recurrence Relation for the nth Cantor Set
Wikipedia gives a recurrence relation for the $nth$ Cantor Set: $C_n = \frac{C_{n-1}}{3} \cup(\frac{2}{3}+\frac{C_{n-1}}{3})$.
http://en.wikipedia.org/wiki/Cantor_Set#Construction_and_Formula_of_the_ternary_set
How does one verify such a relation? I considered using induction, but I couldn't do much more than the base case with this one. Can someone provide on how to use induction with a recurrence relation?
AI: $C_n$ isn’t a Cantor set; it’s the $n$-th approximation to the middle-thirds Cantor set, the $n$-th stage in the construction of the middle-thirds Cantor set. There really isn’t anything to verify: $$C_n=\frac{C_{n-1}}3\cup\left(\frac23+\frac{C_{n-1}}3\right)\tag{1}$$ is the recursion step of a definition whose initial step is $C_0=[0,1]$. The middle-thirds Cantor set is then defined to be $$C=\bigcap_{n\in\Bbb N}C_n\;,$$ and it’s at this point that there’s something to prove $-$ namely, that this $C$ has the various properties of the Cantor set.
For instance, in order to prove that $C$ is closed in $\Bbb R$, it suffices to show that each $C_n$ is closed. $C_0$ is certainly closed, so it suffices to show that the construction in $(1)$ preserves closedness. This isn’t hard: the map $\Bbb R\to\Bbb R:x\mapsto\frac{x}3$ is a homeomorphism, so if $C_{n-1}$ is closed, so is $\frac13C_{n-1}$. Similarly, translation by $\frac23$ is a homeomorphism, so $\frac23+\frac12C_{n-1}$ is closed. Finally, the union of two closed sets is closed, so $C_n$ is closed.
Similarly, one can use induction to show that $C_n\subseteq C_{n-1}$ for each $n\in\Bbb N$. It’s easy to verify directly that $C_1\subseteq C_0$. Now assume that $C_n\subseteq C_{n-1}$ for some $n\in\Bbb N$: $$C_n=\frac{C_{n-1}}3\cup\left(\frac23+\frac{C_{n-1}}3\right)\subseteq C_{n-1}\;.$$ Then
$$\begin{align*}
C_{n+1}&=\frac{C_n}3\cup\left(\frac23+\frac{C_n}3\right)\\
&=\frac13\left(\frac{C_{n-1}}3\cup\left(\frac23+\frac{C_{n-1}}3\right)\right)\cup\left(\frac23+\frac13\left(\frac{C_{n-1}}3\cup\left(\frac23+\frac{C_{n-1}}3\right)\right)\right)\\
&\subseteq\frac13C_{n-1}\cup\left(\frac23+\frac13C_{n-1}\right)\\
&=C_n\;,
\end{align*}$$
as desired.
With more work one can use this definition to show that $C$ contains no non-trivial interval.
Added: To see that this construction not only gives a Cantor set but gives the familiar middle-thirds Cantor set, look at the numbers in $[0,1]$ in terms of their ternary expansions. First, $$C_0=\left\{\sum_{n\ge 1}\frac{a_n}{3^n}:a_n\in\{0,1,2\}\text{ for }n\ge 1\right\}\;.$$ For convenience let me write $0.a_1a_2a_3\dots$ instead of $\sum_{n\ge 1}\frac{a_n}{3^n}$. Then $C_0$ contains all of the ternary expansions $0.a_1a_2a_3\dots$. Multiplying by $\frac13$ just moves every digit one place to the right, changing $0.a_1a_2a_3\dots$ to $0.0a_1a_2a_3\dots$, and adding $\frac23$ to each of these numbers gives us all of the ternary expansions of the form $0.2a_1a_2a_3\dots$. In other words, $C_1$ consists of the numbers in $[0,1]$ with ternary expansions beginning with $0$ or $2$: we’ve stripped out those numbers that have only ternary expansions beginning with $1$, i.e., those in $\left(\frac13,\frac23\right)$.
At each stage the recursion in $(1)$ takes the numbers in $C_{n-1}$, shifts their ternary expansions one place to the right, and inserts a leading $0$ or $2$. Thus, the numbers in $C_2$ are those with ternary expansions whose first two digits are either $0$ or $2$, and in general $C_n$ contains those numbers in $[0,1]$ with ternary expansions whose first $n$ digits are all either $0$ or $2$.
This is exactly the same thing that happens when we remove the open middle thirds in the other recursive construction. |
H: What do you call a graph of value vs rank of the value?
What do you call a graph of value vs rank of the value?
For example, say I make a census of the 2010 net income for a bunch of people.
The graph of net income vs age gives me a: scatterplot
The graph of net income vs location (latitude, longitude) gives me a: height map
After sorting the net income values so the value in rank 1 is highest income, rank 2 is the next-highest, etc., then the graph of net income vs rank gives me a: ???
(This may be a question more about information graphics than about math -- is there some other site, perhaps Graphic Design, that is a better place for this kind of question than Math.StackExchange?)
My understanding is that Zipf's law often applies to such graphs.
Ideally I'd like a name for all such graphs, whether or not Zipf's law applies to them.
What do you call such a graph of value vs rank of the value?
AI: I've seen them called simply "rank plots" or "rank–value plots" (or e.g. "rank–income plots", if income is what you're measuring). Apparently, the term "Zipf plot" is also sometimes used. |
H: Optimization, solving for the 'error' coefficient
Given a modified regression equation:
$\hat Y = \exp(\beta_0 + \sum\beta_ix_i + \varepsilon)*F$
where:
$\hat Y = 11353$
$\beta_0 = 8.693021$
$\sum\beta_ix_i = 5.95487177696$
$F = 0.21829$
what is:
$\varepsilon =$
AI: $$ \hat Y = \exp(\beta_0 + \sum\beta_ix_i + \varepsilon)*F \\
\hat Y / F = \exp(\beta_0 + \sum\beta_ix_i + \varepsilon) \\
\ln (\hat Y / F) = \beta_0 + \sum\beta_ix_i + \varepsilon \\
\ln (\hat Y / F) - \beta_0 - \sum\beta_ix_i = \varepsilon $$ |
H: Sequence of Uniformly Bounded functions
Consider a sequence $\{ f_k \}_{k=1}^{\infty}$ of locally-bounded functions $f_k: \mathbb{R}^n \rightarrow \mathbb{R}_{\geq 0}$.
Assume the following.
For any sequence $\{X_k\}_{k=1}^{\infty}$ of compact sets $X_k \subset \mathbb{R}^n$ such that $X_k \subseteq X_{k+1}$ and $X_k \rightarrow \mathbb{R}^n$, there exist (a uniform) $M \in \mathbb{R}_{>0}$ such that
$$ \sup_{x \in X_k} f_k(x) \leq M$$
Say if the following claim holds (or find a counterexample).
There exists $K \in \mathbb{Z}_{\geq 1}$ such that
$$ \sup_{x \in \mathbb{R}^n} f_K(x) < \infty$$
Note: can we use this argument?
AI: If it's not the case, then for all $k\geq 1$, we have $\sup_{x\in\Bbb R^n}f_k(x)=+\infty$.
There is $x_k\in\Bbb R^n$ such that $f_k(x_k)\geq k$. We define $X_1=\{x_1\}$ and by induction $X_k:=\{x_k\}\cup X_{k-1}\cup \{x,\lVert x\rVert\leq k\}$. Then $\{X_k\}$ is an increasing sequence of compact sets, and for all $k$,
$$k\leq \sup_{x\in X_k}f_k(x)\leq M$$
which is not possible. |
H: Lines in the plane and recurrence relation
I am trying to solve the following problem from Cohen's Basic Techniques of Combinatorial Theory:
A collection of $n$ lines in the plane are are said to be in general position if no two are parallel and no three are concurrent. Let $a_n$ be the number of regions into which $n$ lines in general position divide the plane. How big is $a_n$?
The recurrence relationship for $a_n$ is immediate, namely
$$a_{n+1}=a_n+n.$$
I have been trying to apply generating functions to solve this recurrence, but I am making a mistake somewhere. My attempt is as follows:
Let $G(X)=\sum_{k\geq 1} a_k X^k$. Multiplying both sides of the recurrence by $X^k$ and summing over all valid $k$ gives
$$\sum_{k\geq 1}a_{k+1}X^k=\sum_{k\geq 1}a_kX^k+\sum_{k\geq 1}kX^k.$$
Using the fact that $\frac{1}{(1-x)^2}=\sum_{k\geq 1} kX^k$, I get
$$\sum_{k\geq 1}a_{k+1}X^k=G(X)+\frac{1}{(1-x)^2}.$$
The left hand side is equal to
$$\frac{G(X)-a_1}{X}=\frac{G(X)-2}{X}$$
since $a_1=2$. Solving for $G(X)$, I get
$$G(X)=\frac{2-4X+3x^2}{(1-x)^3}.$$
Expanding with partial fractions gives
$$G(X)=\frac{3}{1-x}-\frac{2}{(1-x)^2}+\frac{1}{(1-x)^3}.$$
Transforming this into power series gives
\begin{align}
G(X)&=3\sum_{k\geq 0}X^k-2\sum_{k\geq 0} (k+1)X^k+\sum_{k\geq 0}(k+1)(k+2)X^k\\
&=\sum_{k\geq 0} (3-2(k+1)+(k+1)(k+2))X^k,
\end{align}
which is not correct. I have tried starting over, but I seems to be making a pretty serious mistake. Any thoughts?
Solution:
The real recurrence relation is $a_{n+1}=a_n+(n+1)$. Let $G(X)=\sum_{k\geq 1} a_k X^k$. Multiplying both sides of the recurrence by $X^k$ and summing over all valid $k$ gives
$$\sum_{k\geq 1}a_{k+1}X^k=\sum_{k\geq 1}a_kX^k+\sum_{k\geq 1}(k+1)X^k.$$
Solving in terms of $G(X)$ gives
$$G(X)=\frac{x(2-2x+x^2)}{(1-x)^3}.$$
By expanding with partial fractions we see
$$G(X)=\frac{x(2-2x+x^2)}{(1-x)^3}=\frac{x}{1-x}+\frac{x}{(1-x)^3}.$$
Using the power series identities $\frac{1}{2}\sum_{k\geq 1}k(k-1)X^{k-1}=\frac{x}{(1-x)^3}$ and $\sum_{k\geq 1}X^k=\frac{x}{1-x}$. Substituting these power series in gives
\begin{align}
G(X)&=\frac{1}{2}\sum_{k\geq 1}k(k-1)X^{k-1}+\sum_{k\geq 1}X^k\\
&=\frac{1}{2}\sum_{k\geq 1}k(k+1)X^{k}+\sum_{k\geq 1}X^k \quad \dagger\\
&=\sum_{n\geq 1} (\binom{k+1}{2}+1)X^k.
\end{align}
Therefore, by equating coefficients, we have $$a_n=\binom{n+1}{2}+1.$$
The step $\dagger$ follows from the fact that the first coefficient in the power series is $0$.
Note: This is NOT homework, I am merely working through exercises during summer vacation. I'm sure there are other ways to approach the problem, but I am trying to practice using generating functions.
AI: Your recurrence is off by $1$: it should be $a_n=a_{n-1}+n$ for $n>0$, and you could make life a little easier by setting $a_0=1$. Then if you set $a_n=0$ for $n<0$, you can write simply $$a_n=a_{n-1}+n+[n=0]\;,$$ where the last term is an Iverson bracket. Now you get
$$\begin{align*}
G(x)&=\sum_na_nx^n\\
&=\sum_na_{n-1}x^n+\sum_nnx^n+1\\
&=xG(x)+\frac{x}{(1-x)^2}+1\;,
\end{align*}$$
so
$$\begin{align*}
G(x)&=\frac{x}{(1-x)^3}+\frac1{1-x}\\
&=\sum_n\binom{n+2}2x^{n+1}+\sum_nx^n\\
&=\sum_n\binom{n+1}2x^n+\sum_nx^n\;,
\end{align*}$$
and $$a_n=\binom{n+1}2+1\;.$$
Note that $\sum_kkx^k=\frac{x}{(1-x)^2}$, not $\frac1{(1-x)^2}$. |
H: Computing the value of $(\frac{p-1}{2})!$ modulo $p$.
I want to prove that for $p \geq 3$, and for $a=(\frac{p-1}{2})!$, if $p \equiv1\pmod 4$, then $a^2\equiv -1 \pmod p$, and if $p \equiv 3\pmod4$, then $a \equiv \pm 1 \pmod p$.
For the first part, I used Wilson's theorem which says that for prime $p$, $(p-1)!=-1 \pmod p$.
so $a^2=((\frac{p-1}{2})!)^2=(1\cdot2\cdot3\cdot\cdot\cdot(\frac{p-1}{2}))^2$ and since $p-k \equiv -k \pmod p$ we get that $a^2=(1\cdot2\cdot3\cdot\cdot\cdot(\frac{p-1}{2}))(1\cdot2\cdot3\cdot\cdot\cdot(\frac{p-1}{2}))=(1\cdot2\cdot3\cdot\cdot\cdot(\frac{p-1}{2}))((p-1)(p-2)(p-3)\cdot\cdot\cdot(-1)^{(p-1)/2}).$
so since $p \equiv 1\pmod 4$ and using Wilson I get the what I need. Is this correct? How should I prove the second part?
Thank you.
AI: You got the hard part. To summarize, we have $a^2\equiv (-1)^{(p+1)/2}\pmod{p}$.
If $p$ is of the form $4k+1$, then $(p+1)/2$ is odd, and therefore $a^2\equiv -1\pmod{p}$.
If $p$ is of the form $4k+3$, then $(p+1)/2$ is even, and therefore $a^2\equiv 1\pmod{p}$. It follows that $a\equiv 1\pmod{p}$ or $a\equiv -1\pmod{p}$.
This is because from $p$ divides $a^2-1$, we see that $p$ divides $(a-1)(a+1)$. Therefore either $p$ divides $a-1$, in which case $a\equiv 1\pmod{p}$ or $p$ divides $a+1$, in which case $a\equiv -1\pmod{p}$.
Remark: Note that for example when $p=3$, then $a=1$, so $a\equiv 1\pmod{p}$. It is also the case that when $p=23$, we have $a\equiv 1\pmod{p}$. But, for example, when $p=7$, and when $p=11$, we have $a=\equiv -1\pmod{p}$. So for $p$ of the form $4k+3$, both $a\equiv 1\pmod{p}$ and $a\equiv -1\pmod{p}$ can occur. Whether $a$ in this case is congruent to $1$ or $-1$ turns out to be connected with the solvability of the congruence $x^2\equiv 2\pmod{p}$. |
H: Proving $\mathrm e <3$
Well I am just asking myself if there's a more elegant way of proving $$2<\exp(1)=\mathrm e<3$$ than doing it by induction and using the fact of $\lim\limits_{n\rightarrow\infty}\left(1+\frac1n\right)^n=\mathrm e$, is there one (or some) alternative way(s)?
AI: What answer you find most elegant may depend on what definition of $e$ you're starting with, as Dylan suggests, but I find this argument quite short and sweet:
$$\begin{align}
&\quad 1 + 1 &= 2\\
&< 1 + 1 + \frac12 + \frac1{2\cdot3} + \frac1{2\cdot3\cdot4} + \cdots &= e \\
&< 1 + 1 + \frac12 + \frac1{2\cdot2} + \frac1{2\cdot2\cdot2} + \cdots &= 3
\end{align}$$ |
H: Does this intuition for "calculus-ish" continuity generalize to topological continuity?
In the past, I've always motivated continuity of a function from (some subset of) $\mathbb R$ to $\mathbb R$ based on the (incomplete) definition $\lim_{x \to c} f(x) = f(c)$; continuity at isolated points was never really too hard to explain. The main tool behind my explanation is a "using your common sense, what should the value here be?" prompt.
My usual example is this: You have a video camera, you're taking a video of the projectile motion of a ball or something. At precisely 3 minutes, your battery dies. Your camera is so awesome that it records stuff up to 2.999...[however many (finitely many) 9's] minutes. Common sense, i.e. "how the real world works", should tell you that this is enough for you to find out where the ball is at 3 minutes, even without any footage at that specific instant. If it were just a little bit higher/lower than "the right answer", then the ball would have to magically teleport from one point to another in [snaps fingers], which doesn't make any sense.
Then I get to the punchline - that continuity is required for us to "predict" things within "reasonable expectations". Things don't magically jump/teleport from one point to another, neither does temperature, brightness, etc. At isolated points, there's not enough "surrounding data" for us to make predictions, so any value can potentially be "reasonable", because if you claim that my prediction is "too high/low", I can always argue that it's conceivable that there might be a spike/dip in the surrounding data, so who's to say my prediction is definitely wrong?
(If this is a bad explanation, please let me know how I can improve it.)
Fast-forward to a rather different scenario, now I'm trying to explain that the product topology is the coarsest topology that makes all projections continuous. I got countered with "but why should all projections be continuous?" I was stumped. I'd never really thought about the intuition behind continuity outside of $\mathbb R$. So my question is, can I generalize this "prediction" thing to motivate why projections of a product space onto its factor spaces should always be continuous? If so, how? If not, why not?
I have read the answers here, but I think/hope my question is a little different.
Also, feel free to yell at me if I've gotten continuity wrong for the last few years (and poisoned the minds of many others). But I would appreciate it if you could follow up with suggestions. Thanks.
AI: It’s essentially the same intuition: you want to define the topology on $X\times Y$ in such a way that as you move around continuously in $X\times Y$, your ‘shadows’ on the $X$ and $Y$ ‘walls’ move continuously as well. More specifically, you want to know that as you approach a point $\langle x,y\rangle\in X\times Y$, your $X$ and $Y$ projections approach $x$ and $y$ respectively. Of course you may have some trouble getting across the idea that approach depends on the topology chosen; the connection is really intuitive only for metric and linearly ordered spaces. Perhaps it’s better to avoid the idea of approaching a point dynamically and look at limit points. You want to choose the topology of $X\times Y$ so that if $\langle x,y\rangle$ is so close to $A$ that it can’t be separated from $A$ by an open set $-$ i.e., if it’s in $\operatorname{cl}_{X\times Y}A$ $-$ then $x$ and $y$ are so close to the ‘shadows’ $\pi_X[A]$ and $\pi_Y[A]$, respectively, that they can’t be separated from them. |
H: Can ≈ be used to express an arbitrarily rounded rational number?
Is it a formally acceptable use of ≈ to express a rational number rounded to an arbitrary number of significant digits?
For example, $\frac{4}{7}\approx0.57.$
If formally acceptable, is it expected?
AI: I see $\approx$ to mean close enough for the purpose at hand. Rounding rationals is one of many cases, $\sqrt 2 \approx 1.414$ is another, also earth population $ \approx 7$ billion. Some people will use $=$ for anything that comes out of a calculator, but $\sqrt 2 = 1.41421$ is discouraged here. |
H: HCF/LCM problem
Find the greatest number that will divide $x$, $y$ and $z$ leaving the same remainder in each case.
Now the solution for this is obtained by finding the HCF of $(x – y)$, $(y – z)$ and $(z – x)$.
Can you tell me why is that?
AI: HINT: $x,y$ have the same reminder when divided by $d$ if and only if $d$ divides $x-y$.
Once you prove this, the claim follows immediately observing that $x,y,z$ have the same reminder when divided by $d$ if and only if all three pair, $x,y$; $x,z$ and $y,z$ have the same remainder when divided by $d$; if and only if $d$ divides $x-y, x-z, y-z$. |
H: Example of a function $f: [a, b] \to \mathbb{R}$ that is unbounded.
Can anyone give me an example of a function $f\colon[a,b]\to\mathbb{R}$ that is unbounded?
AI: There can be no continuous function that has that property (since a continuous function defined on a finite closed interval must achieve a maximum and a minimum value, hence must be bounded; or, to use higher power results, the image of a compact set under a continuous function is compact). However, you can define discontinuous functions that do rather easily. E.g,
$$f(x) = \left\{\begin{array}{ll}
\frac{1}{x-a} - \frac{1}{b-x}&\text{if }x\neq a\text{ and }x\neq b\\
0 &\text{if }x=a\text{ or }x=b.
\end{array}\right.$$
This function has $\lim\limits_{x\to a^+}f(x) = \infty$ and $\lim\limits_{x\to b^{-}}f(x)=-\infty$. |
H: Normal subgroups of infinite symmetric group
I recently took a course on group theory, which mentioned that the following proposition is equivalent to the continuum hypothesis: "The infinite symmetric group (i.e. the group of permutations on the set $\mathbb{N}$) has exactly 4 normal subgroups." Does anyone have any references or explanation for this?
AI: You are inquiring about the Schreier-Ulam theorem. This old MO post contains an answer of mine with the statement of the result; here is a link to the original paper (thanks to t.b.). I would be happy to supplement this and/or that answer with a link to a free, electronically available English language proof, if anyone knows one. |
H: Not every metric is induced from a norm
I have studied that every normed space $(V, \lVert\cdot \lVert)$ is a metric space with respect to distance function
$d(u,v) = \lVert u - v \rVert$, $u,v \in V$.
My question is whether every metric on a linear space can be induced by norm? I know answer is no but I need proper justification.
Edit: Is there any method to check whether a given metric space is induced by norm ?
Thanks for help
AI: Let $V$ be a vector space over the field $\mathbb{F}$. A norm
$$\| \cdot \|: V \longrightarrow \mathbb{F}$$
on $V$ satisfies the homogeneity condition
$$\|ax\| = |a| \cdot \|x\|$$
for all $a \in \mathbb{F}$ and $x \in V$. So the metric
$$d: V \times V \longrightarrow \mathbb{F},$$
$$d(x,y) = \|x - y\|$$
defined by the norm is such that
$$d(ax,ay) = \|ax - ay\| = |a| \cdot \|x - y\| = |a| d(x,y)$$
for all $a \in \mathbb{F}$ and $x,y \in V$. This property is not satisfied by general metrics. For example, let $\delta$ be the discrete metric
$$\delta(x,y) = \begin{cases} 1, & x \neq y, \\ 0, & x = y. \end{cases}$$
Then $\delta$ clearly does not satisfy the homogeneity property of the a metric induced by a norm.
To answer your edit, call a metric
$$d: V \times V \longrightarrow \mathbb{F}$$
homogeneous if
$$d(ax, ay) = |a| d(x,y)$$
for all $a \in \mathbb{F}$ and $x,y \in V$, and translation invariant if
$$d(x + z, y + z) = d(x,y)$$
for all $x, y, z \in V$. Then a homogeneous, translation invariant metric $d$ can be used to define a norm $\| \cdot \|$ by
$$\|x\| = d(x,0)$$
for all $x \in V$. |
H: F distribution: $P\{S_1\geq 3S_2\}$ with $n_1 = 7, n_2 = 13$
Using the $F$ distribution table:
I think it should be $ P\{\frac{S_1}{S_2} \geq3\} = 1 - P\{\frac{S_1}{S_2} \leq 3\} = 1 - 0.05 = 0.95$. But my notes say it should be $0.05$?
AI: Let $X = \frac{S_1}{S_2}$ be F-distributed random variable. Then
$$
f_X(x) = \frac{1}{B\left(\frac{n_1}{2}, \frac{n_2}{2}\right)} \frac{n_1}{n_2} \left( \frac{n_1 x}{n_2} \right)^{\frac{n_1}{2} -1} \left( 1 + \frac{n_1 x}{n_2} \right)^{-\frac{n_1+n_2}{2}}
$$
and
$$
\mathbb{P}\left( X \geqslant 3 \right) = \int_3^\infty f_X(x) \mathrm{d} x
$$
The plot of $f_X(x)$ for given parameters looks as follows:
Clearly the probability $\mathbb{P}(X \geqslant 3)$ is small (it is the orange area in the plot). Using quadratures it equals $0.04156$. |
H: How to show that $0 \times 2 = 0$?
Here is the suggested proof:
$0 \cdot 2=2 \cdot 0 =(1+1) \cdot 0 = 1\cdot 0 + 1\cdot 0 = 0 + 0$.
But my question lies in this step.
Here is the definition of Zero: $0+a = a$ (for any number $a$)
therefore: $0+5=5$ or $0+1639=1639$,
but can we say $0+0 =0$ ? I mean, that "any" includes "Zero". But we have not defined zero yet.
In other words, we are using "something" before we defined that "something".
These days I'm struggling with these fundamental doubts. Especially after reading some quotes from Bertrand Russell saying:"
Mathematicians don't know what they're doing nor know their work is true( or something like this...)
and John von Neumann saying: "When we are doing mathematics we don't know whether our works are true or not. We just get used to them"
Why Russell wrote about 300 pages to show what is "$1$"?
Can anyone please introduce me a book which proves math's certainty (if it exists) ?
AI: If you want to ask in detail how something is proved, you need to specify what set of axioms and what rules of inference are allowed.
However, the usual situation is you have a set of axioms. Zero is a distinguished element, that is, one that is named in the axioms. It doesn't need a definition. As you say, one axiom (in PA, for example) is $\forall a: a+0=a$ This is not a definition of $0$, it is part of a definition of $+$. You can certainly substitute $0$ for $a$ to get $0+0=0$ Then you also have an axiom that $\forall a:a \cdot 0=0$. You also need the commutative law for multiplication (a theorem of PA), the distributive law (an axiom), the definition of $2$ (usually $SS0$ where $S$ is a unary function intended as the successor), and $1+1=2$ (a theorem of PA once you define $1$ as $S0$).
The most famous attempts to prove the certainty of mathematics were Hilbert's program and Russell and Whitehead's Principia Mathematica. This is an enormous area of study. |
H: Homeomorphism between a normed space and its open unit ball
I have studied that every normed space $X$ is homeomorphic to its open unit ball $B$. I want to know what conclusion can we draw from this statement? Does it mean that every normed space is open ball? I am confused with this statement. Could anyone clear my doubt?
Thanks for help
AI: It is true that every normed linear space $X$ is homeomorphic to its open unit ball $B$, for example via the homeomorphism $f:B\to X$ defined by $f(x)=\tan(\frac{\pi}{2}\|x\|)\cdot x$. This means that, as topological spaces, $X$ and $B$ are essentially identical. However, a normed linear space carries more information than a topological space, namely a norm. Thus the appropriate notion of "essentially identical" for normed linear space is the existence of an isometric isomorphism, a homeomorphism which preserves the norm. Clearly $X$ is not isometrically isomorphic to $B$, since $B$ is not even a normed linear space unless $X=\{0\}$. To see this, take any $x\neq 0$. Then $x/2\|x\|\in B$ but $2\cdot x/2\|x\|\notin B$. |
H: Sard's Theorem For Constant functions
It states:
Let $g:A \to R^n$ be continuously differentiable, where $A \subset R^n$ is open, and let $B=${${x \in A: \det g'(x)=0}$}. Thne $g(B)$ has measure $0$.
Okay.... obviously this theorem is right... but why don't constant functions violate this? After all, the derivative of a constant function is $0$ EVERYWHERE.... so that can't possibly be measure zero!
AI: It is $g(B)$, not $B$, that has measure $0$. If $g$ is constant, then g(B) contains a single point (the constant). |
H: trying to understand how to solve this kind of limits.
Ok Right now Im doing some excersises... and now I get stuck on this one..
$$\lim_{x\to2} \frac{2^{x+1}-8}{4-2x}$$
tried to do this
$$\lim_{x\to2} \frac{2^{x}\cdot 2-8}{4-2x}$$
but it's the same thing...
AI: After factoring $-2$ from the denominator, and $2$ from the numerator, we have:
$$\lim_{x\to 2}\frac{2^{x+1}-8}{4-2x} = \lim_{x\to 2}\frac{2(2^x - 4)}{-2(x-2)} = -\lim_{x\to 2}\frac{2^x - 4}{x-2}.$$
So we can concentrate on $$\lim\limits_{x\to 2}\frac{2^x-4}{x-2}.$$
We can rewrite this a bit more as
$$\lim_{x\to 2}\frac{2^x - 2^2}{x-2} = 4\lim_{x\to 2}\frac{2^{x-2}-1}{x-2}.$$
Now let $u=x-2$. Then $u\to 0$ as $x\to 2$, so this becomes
$$4\lim_{x\to 2}\frac{2^{x-2}-1}{x-2} = 4\lim_{u\to 0}\frac{2^u-1}{u}.$$
So we are reduced to finding
$$\lim_{u\to 0}\frac{2^u-1}{u}.$$
Rewriting $2^u = e^{\ln(2)u}$ and setting $h = \ln(2)u$, we have that $h\to 0$ as $u\to 0$, so
$$\lim_{u\to 0}\frac{2^u-1}{u} = \lim_{h\to 0}\frac{e^h-1}{h/\ln(2)} = \ln(2)\lim_{h\to 0}\frac{e^h-1}{h}.$$
So we are reduced to finding
$$\lim_{h\to 0}\frac{e^h-1}{h}.$$
This may be a limit you already know how to do, or else you can find several answers in this site, for example, here. The limit is equal to $1$. Putting it all together, we have:
$$\begin{align*}
\lim_{x\to 2}\frac{2^{x+1}-8}{4-2x} &= -\lim_{x\to 2}\frac{2^x-4}{x-2}\\
&= -4\lim_{x\to 2}\frac{2^{x-2}-1}{x-2}\\
&= -4\lim_{u\to 0}\frac{2^u-1}{u}\\
&= -4\ln(2)\lim_{h\to 0}\frac{e^h-1}{h}\\
&= -4\ln(2).
\end{align*}$$
Although I suspect that you already know how to do some of the limits we found along the way; e.g., if this is an exercise in your book before you know about derivatives, chances are you already know that
$$\lim_{h\to 0}\frac{a^h-1}{h} = \ln(a)$$
when $a\gt 0$. But, since you don't give us enough information, it's impossible to tell. |
H: A property equivalent to $K$ being a finite-dimensional, normal extension field of $F$
Prove that a finite-dimensional extension field $K$ of $F$ is normal if and only if it has this property: Whenever $L$ is an extension field of $K$ and $\sigma :K\rightarrow L$ an injective homomorphism such that $\sigma(c) = c$ for every $c\in F$, then $\sigma(K)\subseteq K$.
First off, one thing I don't understand is why do they put $\sigma(K)\subseteq K$ instead of $\sigma(K) = K$? Unless I'm missing something $\sigma(K)\subset K$ (strict inclusion) is impossible if $\sigma$ is injective.
As to the main part of my question, I've proved this tautology in the forward direction (after significant effort), but I've been unsuccessful in proving the other direction.
By contrapositive seems like the most likely to succeed, so I assumed $K$ wasn't a normal extension of $F$ and thus there exists some irreducible polynomial $f(x)\in F[x]$ which has some but not all its roots in $K$. What I need to do now is find some field extension $L$ of $K$ and some injective hom $\sigma : K\rightarrow L$ which fixes $F$ but sends an element of $K-F$ to an element of $L-K$. This is tough, what makes the most sense to me is choosing $L$ to be the splitting field of $f(x)$, so the part of $f(x)$ which factors in $K[x]$ has to have at least one of its roots sent outside $K$, I'm thinking maybe I should send them to roots of $f(x)$ which only exist in $L$, but I just can't figure out how to construct a homomorphism from this.
I tried looking at the example of $x^3 - 2$ with $K = \mathbb{Q}(\sqrt[3]{2})$ and $L = \mathbb{Q}(\sqrt[3]{2},\sqrt[3]{2}\omega,\sqrt[3]{2}\bar{\omega}$) to get some intuition but it seems too trivial, is there a better example to help my understanding? Anyways hopefully someone can help me with this, thanks.
Addition: In the other direction I made extensive use of an injective hom $\sigma:K\rightarrow L$ implying the existence of an injective hom $\phi:K[x] \rightarrow L[x]$.
AI: I think you're on the right track. What is written below mostly fleshes out and gives references for the plan given by Prof Magidin in the comments.
Let $\alpha$ be a root of $f$ in $K$, and view $K$ as being embedded in an algebraic closure $F^a$ of $F$. There is some root $\beta \in F^a$ of $f$ which is not contained in $K$, and there is an isomorphism $F(\alpha) \to F(\beta)$ over $F$ [As both of these fields are isomorphic to $F[X]/(f)$; see Proposition 2.1a of Milne for more discussion.]. Moreover, since $K/F$ is algebraic this extends to an embedding $\sigma\colon K \to F^a$ [See Corollary 2.8b of Milne for a sharper version of this claim.], and we've constructed this such that $\sigma(\alpha) \notin K$.
Regarding $\sigma(K) \subset K$ versus $\sigma(K) = K$: the answers to this question show that if $K/F$ is algebraic and $\sigma\colon K \to K$ is the identity on $F$ then $\sigma$ is surjective, so there is no difference. |
H: Curry-Howard correspondence
I read that the Curry-Howard correspondence introduces an isomorphism between typed functions and logical statements. For example, supposedly the function
$$\begin{array}{l}
I : \forall a. a \to a\\
I = \lambda x. x
\end{array}$$
can be interpreted as:
$$p \implies p \text{ for any proposition } p$$
Is that correct? How might I interpret a function $K : \forall ab. a \to (b \to a)$? What about $\text{succ} : \mathbb{N} \to \mathbb{N}$, where $\text{succ} = \lambda x. x + 1$?
If the scope of this question is too large, can someone recommend an introductory book that would cover this? A proof of the isomorphism would probably be enlightening, too: can someone point me to one?
AI: Have a look at Girard, Lafont and Taylor, Proofs and Types. It's available online in its entirety.
We can state the Curry–Howard correspondence in general terms as a correspondence between proof systems and type theories. Two ways of stating it as that proofs are programs and propositions are types. Your question looks like it's focused on the connection which Howard drew between natural deduction and the simply-typed $\lambda$-calculus.
You ask how to interpret a function $K : \forall{a, b} . a \rightarrow (b \rightarrow a)$, which is not precisely the right question to ask, because it misses the intensional aspect of the correspondence. Let $t$ be a term of type $\forall{a, b} . a \rightarrow (b \rightarrow a)$ whose extension is the function $K$. (In general there will be many such $t$ for any given $K$.) What does $t$ correspond to, then? A proof of the proposition which its type corresponds to: the proposition that for arbitrary propositions $a$ and $b$, $a \rightarrow (b \rightarrow a)$.
What might $t$ look like? We can figure this out to some extent because of its type. Start with $r : b \rightarrow a$. Let $v : a$ and $x : b$. Then let $r := \lambda x. v$. Then $t := \lambda y. (\lambda x. v)$.
Of course this isn't terribly informative: to figure out anything substantial we'd need to know what $v$ is. There are lots of things it might be: here's one which fits. Let $v := y$. Then $t := \lambda y. (\lambda x. y)$. In other words, provide terms $u : a$ and $s: b$ (which, remember, are proofs of $a$ and $b$ respectively), so then $(t[u / y])[s / x]$ reduces to just $u$, which is a proof of $b$. |
H: How to solve infinite repeating exponents
How do you approach a problem like (solve for $x$):
$$x^{x^{x^{x^{...}}}}=2$$
Also, I have no idea what to tag this as.
Thanks for any help.
AI: I'm just going to give you a HUGE hint. and you'll get it right way. Let $f(x)$ be the left hand expression. Clearly, we have that the left hand side is equal to $x^{f(x)}$. Now, see what you can do with it. |
H: Maximum numbers of underdetermined solution of $Ax=b$
Given $m \times n$ real matrix $A$, where $m<n$, we know that the nullity of $A$ is the dimension of the kernel $W=\{w| Aw=0\}$. Also all solutions of linear equation $Ax=b$ for $b\neq 0$ can be described as
$$\mathcal{A}(v):=\{v+w | Aw=0\}=v+W$$
where $v$ satisfies $Av=b$.
Does the maximum size of linearly independent set in $\mathcal{A}(v)$ always
equal to $\dim W$?
in other words:
Does the maximum numbers of independent solutions of the underdetermined
system $Ax=b$ always equal to the nullity of $A$?
I can only seeing this by plotting the solutions.
My idea seems naive:
Suppose that $\dim W=r$, and we have $r+1$-linearly independent vectors in $v+W$ say
$$v+x_1,v+x_2,\cdots,v+x_{r+1}$$
Can we prove that $x_1,x_2,\cdots,x_{r+1}$ are also linearly independent? (in order to get a contradiction)
AI: This is an answer to your question "Does the maximum size of linearly independent set in $\mathcal{A}(v)$ always
equal to $\dim W$?".
Basically no. Take $A=\begin{bmatrix}1 & 0 & 0\\ 0 & 0 & 1\end{bmatrix}$ and $b=e_1$. Then $\ker A = \mathbb{sp} \{ e_2 \} $, and since $A e_1 = b$, we may take $v = e_1$. However $v+\ker A$ contains $e_1+e_2$ and $e_1-e_2$ which are linearly independent, whereas $\ker A$ has dimension $1$.
(Note however, that the affine dimension of $\mathcal{A}(v)$ always equals that of $\ker A$, as it is just a translate.) |
H: Multiplication of Matrix Properties - bi-symmetric, symmetric and anti-symmetric
Once again, can the great maths minds here please give a hand to explain or solve this problem. Much appreciated!
How can you prove that a bi-symmetric matrix multiplied by symmetrical vector will give me a symmetrical vector?
Vice versa, a bi-symmetric matrix multiplied by anti-symmetrical vector will give me an anti-symmetrical vector?
Is there a way to solve it without using symbolic examples?
AI: Let $J$ be the (as wikipedia calls it) exchange matrix
\[
J = \begin{pmatrix} 0 & 0 & \cdots & 1\\\
0 & 0 & \dots & 0\\\
& & \vdots &\\\
0 & 1 & \cdots & 0\\\
1 & 0 & \cdots & 0
\end{pmatrix}
\]
I assume that you call $v \in K^n$ symmetric if $Jv = v$ and anti-symmetric if $Jv = -v$, right?
Now let $A \in \mathrm{Mat}(n, K)$ bisymmetric, that is $JA = AJ$ and $A^T = A$. If $v \in K^n$ is symmetic, then
\[
J(Av) = JAv = AJv = Av
\]
so $Av$ is symmetric, if $Jv = -v$, then
\[
J(Av) = AJv = A(-v) = -Av.
\]
Note, that we didn't use $A^T = A$ (i. e. $A$ being symmetric). |
H: Probability of someone being born the same day as you in a group of N people.
I got into a discussion with a friend about this simple question:
Provided that I went to a class with $N$ people, where everyone was born the same year in the same city where there are $K$ hospitals, what's the probability that someone was born in the same hospital and in the same day as me?
My approach would be as follows:
$$\left( 1- \left(\frac{364}{365} \right)^N \right) \cdot \left(1-\left(\frac{K-1}{K} \right)^N \right)$$
But I'm not really sure that this is the correct answer :)
AI: Under appropriate independence assumptions, and assumptions about equality of hospitals, the probability that someone chosen at random was born on the same day as you and in the same hospital is
$$\frac{1}{365}\cdot\frac{1}{K}=\frac{1}{365K}.$$
So the probability the person was born on a different day or a different hospital than you is
$$1-\frac{1}{365K}.$$
The probability that all $N$ were born on a different day or different hospital than you is
$$\left(1-\frac{1}{365K}\right)^N.$$
So the probability that at least one person "double matches" with you is
$$1-\left(1-\frac{1}{365K}\right)^N.$$
Remark: This is exactly the same reasoning as the pure days problem, except that we now have $365K$ abstract "days." |
H: Next generation numbers
1: Discovering of negative numbers.
Assume a and b are positive integers
$x+a=b$ ----> if $b>a$ then $x$ is positive integer
$x+a=b$ ----> if $b=a$ then $x=0$
$x+a=b$ ----> if $b<a$ then $x=b-a$ is negative integer
2: Discovering of rational numbers.
$x+x+....+x=a.x=b$ ----> $b \equiv k\pmod a$ if $k$ is not zer0, $x=\frac{b}{a}$ is not integer .
3: Discovering of irrational numbers
$x+x+x......+x=x.x=x^2=2$ ----> $x=\sqrt{2}$ is not rational number
4: Discovering of complex numbers
$x^2=-1$ ----> $x$ is not irrational number
I wonder what 5th step can be for next generation numbers. Is there any known operator or equation to find next generation numbers?
or in other words, Are the complex numbers end of story for numbers to be found as an equation via an operator?
Could you please tell me your ideas and share your knowledge about this subject?
NOTE: I know the quaternions that are a number system that extends the complex numbers. Actually I wonder if possible or not to define next generation numbers via known operators or new operator such as previous numbers (negative numbers,rational numbers,irrational numbers, complex numbers) were defined as equation of $x$.
Thank you very much for answers and links.
AI: Several extensions can be named. One of them is the set of "hypercomplex" numbers, see mathworld. They are somewhat like the quaternions in that you have additional elementary numbers $j$ and $k$, except the relations are chosen such that all numbers commute.
There are also "hyperreals", which is like a refining of real numbers. If you've heard about infinitesimals and infinities, this is how they're made rigorous. I'd recommend you read up on wikipedia. I'm confident the hyperreals can be extended to a 2D version, thus yielding a different "hypercomplex": this is not Davenport's algebra, but a 2D version which is, in a sense, more "dense" than the regular reals.
Though, as noted in the comments to the question, the real problem is Hurwitz' theorem, which says we cannot find very many algebras with interesting properties. |
H: Duality of a finitely generated projective modules
Let $M$ and $N$ be a finitely generated projective module over a ring $R$. Suppose we have a non degenerate bilinear pairing $\langle \ \cdot \ ,\ \cdot\ \rangle: M \times N \to R$.
I want to show $M$ is isomorphic to the dual $N^*$ of $N$.
The injectivity of $M$ into $N^*$ follows from the non degeneracy of the pairing by definiting a map $x \mapsto \langle x, \cdot\rangle$. What I cannot prove is surjectivity.
If I impose a condition that $R$ is an injective module then I think surjectivity also follows. But I want to prove it without any condition on $R$
Any help is appriciated.
AI: If by non-degenerate you mean that $\langle x , - \rangle = 0 \Leftrightarrow x = 0$ (or $\langle - , y \rangle \Leftrightarrow y = 0$), your question is equivalent to: Is every injective homomorphism between finitely generated projective $R$-modules an isomorphism? The answer is no, even for maps $R \to R$.
If by non-degenerate you mean that $\langle x , - \rangle = 0 \Leftrightarrow x = 0$ and $\langle - , y \rangle \Leftrightarrow y = 0$, then your question is equivalent to: If $M \to N$ is an injective homomorphism of f.g. proj. modules such that $N^* \to M^*$ is also injective, then $M \to N$ is an isomorphism. But again this is false for maps $R \to R$.
The "correct" definition of a perfect pairing $M \times N \to R$ is the following: Both maps $M \to N^*$ and $N \to M^*$ are isomorphisms. |
H: How to invert a polynomial function such as $f(x)=x^{\beta}\left((x-1)^6+1\right)$?
I'm trying to replicate a simulation study in a paper. For that I would need the inverse of this function:
$f(x)=x^{\beta}\left((x-1)^6+1\right), x\in[0,1]$
Plugging this unto Maxima returns:
solve(y=x^(beta)*((x-1)^6+1),x);
beta beta + 3
[x = y - x ]
which doesn't help me one bit. The fact that
it's used in this paper leads me to think that
the inverse of this function is probably well known.
It's just not known to me.
Alternatively, is there a general numerical strategy to find the inverse of a function for a grid of values of x?
AI: As noted in the comments, there is no closed form for the inverse. There are numerical techniques for finding very good approximations. One of the simplest and best is called Newton's Method - you will have no trouble finding it. |
H: How to handle group schemes by points?
I find it is very inconvenient to handle group schemes by its defination(i.e. everything is defined by morphism). And I have noticed that for group varieties, one can treat them as actual groups(i.e. talking about elements). Moreover, in the book Geometric Invariant Theory by Mumford, though he gave the defination of group scheme by relations of morphism, what he really used was "point"! Not the closed point, but R-point(i.e. morphism Spec(R) $\to G$, where R is a ring ).
I don't know anything about thoery on R-point, and I guess this might be the way of handling group scheme by "point". Does anyone know the corresponding theory(or reference)? And how could them hook together?
Moreover, I am also curious about how does one think of group schemes. What could go wrong if one think of group scheme by element?
This may not be a well-asked question, because of my ignorance of group schemes, and the words above is just my own feeling while reading GIT. Any comments would be great appreciated!
AI: The main point is that a group scheme does not have enough points :-)
For example, given any number field $K$ (=finite-dimensional extension of $\mathbb Q$), Mazur-Rubin have recently proved that there exists an elliptic curve $E$ defined over $K$ such that $E(K)$ (=points with coordinates in $K$) is the trivial group with one element: this is absurdly insufficient for studying $E$.
More importantly the possibility to vary the field (or even scheme) and look at points of the group-scheme with coordinates in that field, i.e. the morphism point of view, is incredibly important in arithmetic geometry.
For example, if $E$ is an elliptic curve defined over $\mathbb F_p$ it has a zeta function
$$Z(E)= \operatorname {exp} (\sum\operatorname {card}E(\mathbb F_{p^n})\cdot\frac{T^n}{n})
$$
which is, amazingly, a rational function in $T$ and has been the prototype of the arithmetic study of group-schemes.
That said, over an algebraically closed field $k$ , the classical study of a group-scheme just as a group $G(k)$ is quite interesting and fruitful: it is called the geometric study of $G$, as opposed to the arithmetic study evoked above.
Edit: points of a scheme
Given a scheme $X$ , a point with values in a another scheme $T$ is a morphism $T\to X$. This looks rather opaque (especially if you also consider that all schemes are over another scheme $S$ !) so let me try and show that this notion is not so unreasonable as it looks.
Consider the field (or even ring) $k$, the algebra $A=k[T_1,...,T_n]/I$ ( where $I\subset k[T_1,...,T_n]$ is an ideal) and the affine $k$-scheme $X=Spec(A)$.
Now, what is a $T$-point of $X$ if you take for $T$ the affine scheme $T=Spec(R)$ corresponding to some $k$-algebra $R$ ?
Answer: it just corresponds to $V(R)\subset R^n$, the set of solutions $(r_1,...,r_n)\in R^n$ of the system of equations $$P(r_1,...,r_n)=0 \quad \text{for all} \;P(T_1,...,T_n)\in I$$ which is the naïve interpretation of "points of $X$ with values in $R$".
And this is easy: a $k$-morphism $Spec(R)\to X=Spec(A)$ corresponds to a morphism of $k$-algebras $\phi: A=k[T_1,...,T_n]/I\to R$ to which you just associate the naïve $n$-tuple $(r_1,...,r_n)=(\phi(\bar T_1),...,\phi(\bar T_1))\in V(R)$ |
H: 3 consecutive numbers
I was playing with some numbers and just realized that:
For any 3 consecutive numbers X, Y and Z: $Y^2$ = (X*Z) + 1
For eg: Consider numbers 171, 172 and 173
$172^2$ = 29584
and
171*173 = 29583
Can anyone tell me if there is any proof for this and what it is known as?
AI: Let the middle number be $x$; the other two are $x-1$ and $x+1$. Basic algebra tells us that $(x-1)(x+1)=x^2-1$, and therefore $x^2=(x-1)(x+1)+1$. (This is true even if $x$ is not an integer.) |
H: How to use the Malgrange-Ehrenpreis-Theorem
In the Wikipedia article of this theorem
http://en.wikipedia.org/wiki/Malgrange%E2%80%93Ehrenpreis_theorem
it is said that i could be used to prove that
$P(\partial/\partial x_i)u(x)=f(x)$
has a solution for any distribution $f$, how this prove works?
AI: It is not something that I (or I suspect anyone) will want to re-derive in this limited space here.
Letting $E$ be the fundamental solution: $P(D)E = \delta$.
In the case $f$ is $C^\infty_0$, we obviously have a distributional solution by taking a simple convolution $u = E * f$.
In the case $f$ is a distribution with compact support, the same holds true: the distributional convolution $E*f$ is also well defined, and that it is a theorem (see equation (4.2.5)' in ALPDO v.1) $P(D) E*f = (P(D)E)*f = \delta * f = f$.
For the general case where $f$ is a distribution (an element of $(C^\infty_0)'$), one has to be more careful, and the discussion is nontrivial. One cannot just simplemindedly take the convolution, since convolutions of two arbitrary distributions may not be well-defined (and may not satisfy the expected properties). Sections 10.4 through 10.8 of ALDPO v. 2 is devoted to demonstrating that indeed, the equation $P(D)u = f$ can be solved for $f \in C^\infty_0(X)'$ when $X\subset\mathbb{R}^n$ is an open subset that is $P$-convex for singular support. (In particular, any convex open set is $P$-convex for singular support.)
The proof given in ALPDO is non-constructive: it passes through the Hahn-Banach theorem to show that $P(D)$ induces a mapping $C_0^\infty(X)' \to C_0^\infty(X)' / C^\infty(X)$ that is surjective. Essentially one "solves" the "singular part" of the equation first, and leaving a smooth part. The $P$-convexity property is used to solve this singular part. The smooth part can be corrected for using the linearity of the operator and the fact that $P(D)$ can be inverted over smooth functions (another nontrivial theorem proven earlier; its proof depends on knowing the regularity properties of the fundamental solution $E$).
To summarise: while it is true that Malgrange-Ehrenpreis can be used to derive the solvability of $P(D)u = f$, there are quite a lot more quantitative information one needs to get (from various sharper forms of the Malgrange-Ehrenpreis theorem) before one can deduce the claim that $P(D)u = f$ is solvable for any distribution $f$. If you really want to see how the proof works, and assuming you have a working knowledge of distribution theory, you should read Chapters 7 and 10 of ALDPO.
ALPDO stands for Analysis of Linear Partial Differential Operators by Lars Hörmander. |
H: $K\subset Y\subset X$
Here is a theorem in Rudin's Principles of Mathematical Analysis:
$K\subset Y\subset X$.Then $K$ is compact relative to $X$ if and only
if $K$ is relative to $Y$.
I read the proof in the book but I tried to construct a different proof:Here it is:
"Proof": If $K$ is compact relative to $Y$ and $K\subset Y\subset X$,there exists a finite collection of subsets of $\{Y_{\alpha_i}\}$ of $Y$ such that $$K\subset \cup^{n}_{i=1}\{Y_{\alpha_i}\} $$.As $Y\subset X$,$\exists \{Y_{\alpha_{n+1}}\}\subset X$ such that $X= \cup^{n+1}_{i=1}\{Y_{\alpha_i}\}$,and so $K\subset \cup^{n+1}_{i=1}\{Y_{\alpha_i}\}=X$ which forms an open cover of $K$ ,so $K$ is compact relative to $X$.
Again, if $K\subset Y\subset X$ and $K$ is compact relative to $X$, then let $\cup^{n}_{i=1}\{X_{\alpha_i}\}=X$ be the open cover relative to $K$.As $Y\subset X$ $\exists X_{a_j}$ where $1\leq j\leq n$ such that $$Y=X-\{X_{a_j}\}$$.As $K\subset Y$,$K$ is compact relative to $Y$.
I feel the proof is erroneous but I cannot find the mistake. Thank you.
AI: When we say $Y$ is compact relative to $X,$ this means that any collection of open sets in $X$ whose union contains $Y$ (an open cover) can be restricted to a finite collection whose union still contains $Y$ (the finite subcover.) It's important to bear in mind that this is quantifying over all open covers of $Y$ relative to $X.$ In your proof, you haven't mentioned that your finite subcovers are restricted from arbitrary open covers.
Second, it is important to note that openness is a relative condition. For instance, $(0,1)$ is open in $\mathbb{R}$ but not in $\mathbb{R}^2.$ In your notation, we don't necessarily know that the $Y_{\alpha }$ are open in $X.$ The proper way to handle this side of the implication is to start with an open cover of $K$ rekative to $X,$ transform it to an open cover of $K$ relative to $Y$ (by intersecting each of its members with $Y$), restrict to a finite subcover, and pull back to the cover that is open relative to $X.$ |
H: Wrong proof of convergence almost everywhere
Can you tell me where the mistake is?
If $(f_n) \in L^1$ is a sequence of functions such that $\sum_n \|f_n\|_1 < \infty$ I can prove that $f(x) = \sum_{n=1}^\infty f_n(x) < \infty$ for all $x \in X$.
To this end, let $n \in \mathbb N$, $x \in X$. Then
$$ \sum_{k=1}^n f_k (x) \leq \sum_{k=1}^n |f_k (x) | \leq \sum_{k=1}^n \int_X |f_k (x) | d \mu$$
Hence $$ f(x) = \lim_{n \to \infty} \sum_{k=1}^n f_k (x) \leq \lim_{n \to \infty} \sum_{k=1}^n \int_X |f_k (x) | d \mu < \infty$$
Since $x$ was arbitrary we have $f(x) < \infty$ for all $x \in X$.
This proof must be wrong since we should have $f(x) < \infty$ almost everywhere.
AI: In general it is not true that
$$
|f(x)|\leq\Vert f\Vert_1=\int\limits_{X}|f(x)|d\mu
$$
For counterexample consider function
$$
f(x)=
\begin{cases}
1&\quad\text{ if }\quad x=0\\
0&\quad\text{ if }\quad x\neq 0
\end{cases}
$$
then you get
$$
|f(0)|=1>0=\int\limits_{\mathbb{R}} |f(x)|d\mu
$$ |
H: Drying blood - an algorithm for calculating the geometry of blood stains
Motivation
A bucket full of blood gets spilled over the floor. Question: What shape will the dried blood stains have?
Abstraction
The blood is modeled by a set of interacting particles (e.g. SPH). As time goes to infinity, the blood particles come to rest on the floor (we do not care about fluid-dynamics).
Therefore: A cloud of particles is in 3D space. The particles are clustered in multiple, different sized puddles.
My Solution
Cluster the particles
Generate a surface over each cluster
Project the created surfaces onto the floor (a 2D plane)
The result is the shapes of the dried blood.
Finally the Question :-)
How would you solve the problem? Are there better/different ways? And what algorithms would you use for the different steps? (performance/ precision does not matter, I'm interested in general, what types of different algorithms exist :-)
Algorithm for:
Clustering the particles
Generating the surfaces (surface tension should be considered)
Projecting the objects
(please reference your answers)
PS: Any hints for related work/problems would also be appreciated
AI: Your three step procedure is good. But I am afraid that it gives infinitely many possibilities. These are the following things that are to be considered.
The flow of the liquid ( I don't prefer using the word blood , which looks ugly ) , on to the floor that decides the forces acting on each module. Sometimes the slightest change in the input can cause a great variation in the output. See the butterfly effect.
You must consider the fluid dynamics of the liquid that play a major role. Buoyancy forces, resistance, surface tension etc., all these entities have a connection with the formation of the resultant shape. For example, if you pour a glass of water , you get another shape, if you pour a glass of honey, you get some other shape.
But anyway as per your request , I am suggestion you some algorithm that I read in my Database Administration Book. Its called Cluster Analysis . I hope that it will completely help you in clustering the particles.
I am sure that following your three way step will not yield you to a particular solution. It takes you to a infinite set of possibilities ( as shapes can be any in number ). But if you combine the Cluster Analysis with Fluid Dynamics and finally garnish the total thing with Chaotic dynamics and properties of fluid, you get the perfect and exact shape.
For that you need to consider the surface tension of the medium on which the liquid is poured. The rate of flow of liquid, initial injection velocity, viscosity and the buoyancy force offered by the medium.
Then you need to simulate the model of fluid flow, by taking the initial condition and then pop it into the fluid flow equation and then approximating the condition of the fluid after time $t$ ( with some negligible error ) . So after performing interpolation and curve fitting you get a function that approximately gives you the position of the cluster at time $t$. Then after doing that, you need to simulate the resultant positions of all clusters using the vector analysis and cancel the internal forces that are opposite.
Thats how you generate the entire model perfectly.
I hope this is helps you to some extent. Ask me for more mathematical details in case if you want.
Thank you. |
H: Calculate $\int f(x) f''(x)dx$
I have a curiosity. If
$\int f(x) f'(x)dx=\int f(x) df(x)=\frac{\left(f(x)\right)^{2}}{2}+C$
what is the result of:
$\int f(x) f''(x)dx$
AI: $$
f(x)f'(x)-\int f'(x)^2 \, dx \ ?
$$ |
H: Matrix equation involving Kronecker product
I would like to know if exists an algorithm to find the solutions (if any) of the following matrix equation:
$$A\otimes B=B\otimes A$$
in which the $N \times N$ matrix $B$ is given and the $A$ ($A\neq0$) is the unknown $N\times N$ matrix. The conditions on B are:
$$B\neq 1, B\neq 0$$...
Thanks.
AI: $A=cB$ is a solution for any scalar $c$, but these are all the solutions as long as $B\ne0$.
As relevant here, the Kronecker product is just an outer product (since matrix multiplication is not relevant), so you can model the equation as $A (B^T) = B (A^T)$ which again boils down to a number of equations $a_ib_j=a_jb_i$ for all $i$ and $j$. From there it is easy to see that if there are nonzero entries anywhere, the two inputs must be related by a common proportionality constant. |
H: Some approximations for $\arccos(1/(1+x))$
I was trying to calculate the maximum ground distance you can see on mountains, with your elvation given.
After some simple geometry, I was able to come up with the following formula:
Let $h$
be your elevation, $d(h)$
be the maximal distance you can see, then
$$d(h)=2\pi R\arccos\frac{R}{R+h}$$
where R is the radius of earth. We take $R=6378100m$
as its value.
But when I plot it in excel, here's what I got:
The unit for the vertical axis is km while the unit for the horizontal axis is m.
Amazingly, for $d\in(0,20000m)$ (essentially the maximum elevation you can achieve without paying millions to board a spaceship)
, $d(h)$
can be approximated by
$$d(h)\approx22345\sqrt{h}$$
here
with a $r^{2}$
value of 1!.
Only when $h>5\cdot10^{5}m$
dose the the approximation begin to deviate away significantly.
Does any one have a explanation of this from a numerical prespective?
AI: The Taylor series for the cosine is $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dotsb.$$
Truncating this series after the $x^2$ term gives the rather good approximation $$\cos x \approx 1 - \frac12 x^2,$$ from which, by substituting $\sqrt{2y}$ for $x$, we get $$\cos \sqrt{2y} \approx 1 - y$$ and thus $$\arccos (1-y) \approx \sqrt{2y}.$$
Since, for small values of $y$, $$\dfrac{1}{1+y} = 1 - \dfrac{y}{1+y} \approx 1 - y,$$ it follows that $\sqrt{2y}$ is also a good approximation for $\arccos \dfrac{1}{1+y}$ when $y$ is small.
To verify this approximation, consider the original equation
$$d(h)
= 2\pi R \cdot \arccos\frac{R}{R+h}
= 2\pi R \cdot \arccos\frac{1}{1+\frac{h}{R}}.
$$
Since $h/R$ is small in this case,
$$d(h)
\approx 2\pi R\sqrt{2\tfrac{h}{R}}
= \pi\sqrt{8R} \cdot \sqrt{h}
\approx 22441\sqrt{h},
$$
which only differs from Excel's calculation of the coefficient by $0.42\%$. |
H: Solving non linear differential equation.
$\frac{d}{dt}\left(\frac{x'(t)}{x(t)}\right)=x(t)-x^2(t)$ where $x'(t)=\frac{d}{dt}x(t)$
What reasoning (if it exists) I can apply to solve this differential equation?
Thanks.
AI: Let $x=e^u$; then $u$ satisfies de differential equation
$$
u''=e^u-e^{2u}.
$$
Multiply by $u´$ and integrate once to get
$$
\frac12(u')^2=e^u-\frac12\,e^{2u}+C_1.
$$
This is a differential equation in separeted variables, whose solution is
$$
\int\frac{du}{\sqrt{2\,e^u-e^{2u}+2\,C_1}}=\pm t+C_2.
$$ |
H: Does "regular" implies collectionwise hausdorff?
Does "regular" implies collectionwise hausdorff?
A topological space is said to be collectionwise Hausdorff if given any closed discrete collection of points in the topological space, there are pairwise disjoint open sets containing the points.
I believe it and try to proof it, however I'm not sure. Thanks in advance:)
AI: No, it doesn't. The Sorgenfrey plane is regular, as the Sorgenfrey line is normal, but it isn't collectionwise Hausdorff as $\{(-x,x)\mid x \in \mathbb R\}$ is discrete and closed, but there aren't pairwise disjoint open neighbourhoods.
To prove this, let $U_x = [-x,-x+\epsilon_x) \times [x,x+\epsilon_x)$ be a neighbourhood of $(-x,x)$ in the Sorgenfrey plane, then there is some $\epsilon > 0$ such that $E_\epsilon= \{x \in \mathbb R\mid \epsilon_x > \epsilon \}$ is uncountable (as $\mathbb R = \bigcup_n E_{1/n}$). Then $E_\epsilon$ has an accumulation point in $\mathbb R$, choose $x_1, x_2 \in E_\epsilon$ with $|x_1 - x_2| < \frac \epsilon 4$, then $U_{x_1} \cap U_{x_2}\ne \emptyset$.
To answer your addional question in the comment below: Let $X$ be regular and $A =\{a_n \mid n \in \mathbb N\}$ closed, discrete and countable. For each $n \in \mathbb N$ there are by regularity and closedness of all subsets of $A$ in $X$ disjoint open sets $U_n \ni x_n$, $V_n \supseteq A \setminus \{x_n\}$. Now let $W_n = U_n \cap \bigcap_{i < n} V_i$, then $W_n$ is an open set containing $x_n$ (as finite intersection of such). Now let $n < m$, then $W_n \subseteq U_n$, $W_m \subseteq V_n$, so $W_n \cap W_m = \emptyset$. Hence $X$ is (don't know if this is a common term, but IMHO it describes the property) countable collectionwise Hausdorff. |
H: Equivalent definitions on SVD property
Please have a look at the property of SVD - section "Singular values, singular vectors, and their relation to the SVD" from http://en.wikipedia.org/wiki/Singular_value_decomposition.
Could it be state that "and" between
$$Mv = \sigma\cdot u \quad\text{ and }\quad M^Tu = \sigma\cdot v$$
is actually equivalence, and can be safely omitted?
AI: No. It is not an equivalence. We need both the properties to be satisfied.
Given any matrix $M$, and any vector $v$ and a scalar $\sigma$, we can get a $u$, if we just want to satisfy $$Mv = \sigma u.$$
This need not satisfy $$M^*u = \sigma v$$
For instance, let $M = \begin{bmatrix} 2 & 9\\ 3 & 6 \end{bmatrix}$, $v = \begin{bmatrix}1 \\ 0 \end{bmatrix}$ and $\sigma = 2$. This gives us $Mv = \begin{bmatrix} 2\\ 3 \end{bmatrix}$ and hence $u = \begin{bmatrix} 1 \\ 3/2 \end{bmatrix}$. However, $$\dfrac1{\sigma} M^*u = \dfrac12 \begin{bmatrix} 2 & 3\\ 9 & 6 \end{bmatrix} \begin{bmatrix} 1 \\ 3/2 \end{bmatrix} = \dfrac12 \begin{bmatrix} 2 + 9/2 \\ 9 + 9 \end{bmatrix} = \dfrac12 \begin{bmatrix}11/2 \\ 18 \end{bmatrix} = \begin{bmatrix}11/4 \\ 9 \end{bmatrix} \neq v$$
In fact, we see that we need to choose $v$ to be the eigenvector of $M^*M$ and choose $u$ to be the eigenvector of $MM^*$ to get what we want. |
H: find inverse function in given domain
Suppose that we have function $y=\sin(x)$ we need to find its inverse function, assuming that $D(f)=[-\pi/4.\pi/4]$
I know that inverse of $\sin(x)$ is $\arcsin(x)$, it would be answer of a given function too, but why do I need $D(f)=[-\pi/4.\pi/4]$? I don't know, should I introduce some variable $c$, so that $y=\sin(x)$ will look like $y=\sin(x-c)$ or $y=\sin(x)+c$? Please give me a hint. In case I meet similar problem, like "find inverse of function $y=f(x)$ where $D(f)=[a,b]$" what should I do? As I know domain of given function and range of inverse function are the same, so it means that the range of $\arcsin(x)$ is $[-\pi/4.\pi/4]$, but how to proceed?
AI: The typical range of $\arcsin(x)$ is $[-\pi/2,\pi/2]$, with domain $[-1,1]$. This is because we restrict $\sin(x)$ to the interval $[-\pi/2,\pi/2]$--a maximal interval on which $\sin(x)$ is one-to-one--and then take the inverse.
However, we didn't have to have a maximal interval, just some interval on which it's one-to-one, so the same principle will apply here, and you'll simply have $f^{-1}(x)$ as the restriction of $\arcsin(x)$ to the interval $[-1/\sqrt{2},1/\sqrt{2}]$. |
H: About the number of functions passing for N points
Given one point $P\equiv (x_0,y_0)$ on a two-dimensional space $\Re^2$, there are infinite functions passing for that point. The question is: given $2$ points $P_0\equiv(x_0,y_0)$ and $P_1\equiv(x_1,y_1)$ is it still correct to say they are infinite? More in general, given $N$ points $P_k\equiv(x_k,y_k)$ with $x_k\neq x_j$, how many functions passing for the $N$ points can be constructed? Thanks.
AI: This answer deals with polynomials rather than all functions.
Given any (finite number) of points in the plane, there are always an infinite number of polynomial functions of two variables with zeros at those points. For each point $(a_i,b_i)$, take the polynomial $(x-a_i)(y-b_i)$, then the product of these polynomials will be zero at all the points, as will any higher degree polynomial that it divides, leading to the infinite number.
A more interesting question is to ask how many polynomials of a specified degree have zeros at a collection of points. If the points are in general position, and there are $3d-1$ of them for some $d$, then there are no polynomials of degree less than $d$ zero at all of them, infinitely many of degree more than $d$, and a non-zero finite number of degree $d$. |
H: How do I check the continuity of this function?
I am doing basic calculus, can someone tell me how to find the continuity here at $x=1$
$f(x)=\begin{cases}5x-4 & 0<x \leq 1\\4x^2 -3x & 1 < x < 2.\end{cases}$
AI: Suppose $a,b\in\Bbb R$ with $a<b$, let $I=(a,b)$, $c\in I$, $f:I\to\Bbb R$. Then we say that $f$ is continuous at $c$ iff $$f(c)=\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x),$$ that is, iff the left and right limits exist, and are both equal to $f(c)$.
In this case, coming toward $c=1$ from the left, what does $f(x)$ look like--that is, how is $f(x)$ defined for $x<1$? What about from the right? |
H: When does non-negativity of the integral of a function imply that the function itself is non-negative?
Let $(\Omega,\Sigma)$ be a measurable space and $(\omega_k)_{k\in\mathbb{N}}$ a sequence of elements of $\Omega$. Let
$$
\mathcal{M}:=\left\{\sum_{k=1}^\infty a_k\cdot\delta_{\omega_k}: \quad(a_k)_{k\in\mathbb{N}}\subset(0,+\infty),\quad \sum_{k=1}^\infty a_k=1\right\},
$$ where $\delta_{\omega_k}$ denotes the point mass at $\omega_k$. Let $f:\Omega\rightarrow\mathbb{R}$ be a measurable function and denote by $\mathcal{N}$ the set of measures $\mu\in\mathcal{M}$ with respect to which $f$ is $\mu$-integrable. Assume $\mathcal{N}\neq\emptyset$. If $\vert f\vert$ is bounded, then we have the following implication:
$$
\forall \mu\in\mathcal{N}\qquad\int_{\Omega} fd\mu\geq 0\Rightarrow \forall \omega\in\Omega:f(\omega)\geq 0.$$ But is it also necessary to have $\vert f\vert$ bounded, or can one weaken this condition?
AI: If $|f|$ is not bounded, then $\int_{\Omega} f d \mu$ might not be well defined. Keep in mind that if $\mu= \sum_{k=1}^\infty a_k\cdot\delta_{\omega_k}$ then
$$\int_{\Omega} f d \mu= \sum_{k=1}^\infty a_k\cdot f(\omega_k) \,.$$
Too see that you cannot drop the bounedndess, if $f$ is unbounded, for each $k$ pick some $\omega_k$ so that $|f(\omega_k)| > 2^k$, and set $\mu = \sum_{k=1}^\infty \frac{1}{2^{k+1}}\cdot\delta_{\omega_k}$.
Then $\int_{\Omega} f d \mu$ is an infinite series, each term being in absolute value at least one, and they can be positive and negative....
*P.S. * The above idea actually can be used to show that $f \in L^1(\mu) ; \forall \mu \in {\mathcal M} \Leftrightarrow |f|$ is bounded. Also, keep in mind that when you calculate the sum $ \sum_{k=1}^\infty a_k\cdot f(\omega_k) $ you must have absolute convergence, or "absolute" divergence to infinity, you cannot speak of conditional convergence because for the measure $\sum_{k=1}^\infty a_k\cdot\delta_{\omega_k}$, the order in which you list the terms doesn't matter.
The only way you could make sense of this by dropping the boundedness condition is by asking that for all $\mu \in {\mathcal M}$ you have either $f \in L^1(\mu)$ and $\int f d \mu \geq 0$ or $\int f d \mu = + \infty$. But this condition is equivalent to the negative part $f^-$ of $f$ is bounded.... |
H: Which do you recommend for learning how to write proofs — How to Prove it by Velleman, or How to Solve it by Polya?
Which of these two books is suited for a student looking to learn how to write proofs?
I have a working knowledge of calculus and linear algebra, but I'm not good at writing proofs. My intention is to learn proofs in general, not necessarily for the calculus and linear algebra.
How to Prove it by Daniel Velleman
How to Solve It by George Polya
I ask because the latter is suggested on a highly voted question here, but the former has a more apt name. The reviews aren't helping. You can suggest other books.
AI: Velleman's How to Prove it is quite a bit more relevant to your needs. It is organized like a conventional text, and pays a lot of attention to proof writing.
Polya's book focuses on problem-solving. One can view it as a better book, certainly a historically far more important book. But it focuses on how one finds the idea that will crack a concrete problem.
There is quite a bit of material in Velleman that is useful for writing proofs in linear algebra, in particular on how to proceed from definitions. There is none of that in Polya. There is also essentially nothing in Polya on basic analysis. Polya beautifully accomplishes his aims: they just happen to be different from what you said you wanted. |
H: Radius of Convergence of power series, with n! and $n^n$
During revision, I came across this problem:
The set of real numbers $x$ for which the series $$\sum_{n=1}^{\infty}{\frac{n!x^{2n}}{n^n(1+x^{2n})}}$$ converges is __.
I tried using the ratio test, but got stuck in the process of simplification.
(The answer is the series converges for $\mathbb{R}$.)
Sincere thanks for any help.
AI: For every $x \in \mathbb{R}$ and $n \in \mathbb{N}$ we have
$$
\frac{n!x^{2n}}{n^n(1+x^{2n})} \le a_n:=\frac{n!}{n^n},
$$
and
$$
\frac{a_{n+1}}{a_n}=\left(\frac{n}{n+1}\right)^n=\left(1+\frac{1}{n}\right)^{-n} \to 1/e.
$$
Therefore the series converges for every $x \in \mathbb{R}$. |
H: Some equivalences for convex sets
For a subset $A\subset V$ of a vector space over $\mathbb{R}$, let
$\mbox{conv}(A) := \left\{ \sum_{i=1}^n a_i x_i\, \middle| \,x_i\in A, a_i \ge 0\text{ with } \sum_{i=1}^n a_i = 1 \right\}$.
I want to show that if for all $x,y\in A$ and $\alpha\in[0,1]$ the vector $(1-\alpha)x + \alpha y $ is an element of $A$, then $A = \mbox{conv}(A)$.
Is there an elegant solution? Thank you for your help!
AI: The inclusion $A \subset conv(A)$ is obvious.
Now, if A has that property, you can prove that $conv(A) \subset A$ by induction by $n$. Note that
$$\sum_{i=1}^{n+1} a_i x_i = (\sum_{i=1}^n a_i x_i ) + a_{n+1}x_{n+1}=(1-a_{n+1})(\sum_{i=1}^n \frac{a_i}{1-a_{n+1}} x_i ) + a_{n+1}x_{n+1}$$
and
$$\sum_{i=1}^n \frac{a_i}{1-a_{n+1}}=1 \,.$$ |
H: How to find the equation of a line tangent to a function
How it's possible to find the equation to a line tangent to a function in a point where the derivative of the function is an indeterminate form?
I'm analyzing this function:
$$y = \frac{x^2}{1+\log|x|}$$
And the first derivative is:
$$y\,' = \frac{x(1+2\log|x|)}{(1+\log|x|)^2}$$
I have to find the line tangent in $x = 0$.
..but I'm looking for a method applicable to all kind of real-valued functions of real variable (or at least for a wide range of this).
Thanks in advance!
AI: Our function is undefined at $0$. However, it is clear that it approaches $0$ as $x$ approaches $0$. So we have a removable discontinuity at $x=0$. Remove it! Define our function to be $0$ at $x=0$.
Let our modified function be $f(x)$. We want to calculate $f'(0)$. Go back to the definition of derivative. Since $f(0)=0$, we want
$$\lim_{h\to 0} \frac{h^2}{h(1+\log|h|)}.$$
This limit is easily seen to be $0$.
Remark: Since our original function is not defined at $0$, I think the proper conclusion is that the derivative does not exist at $0$. In principle, what we found is the derivative at $0$ of a different function. However, from the point of view of the geometry, there is really no problem at $0$. |
H: Which function can be a particular solution for $y''+a(t)y'+b(t)y=0$?
It's seems like a really simple question, but I can't understand how to solve it.
I am requested to decide which function $y=t^2$, or $y=t^2+1$ can be used for a particular solution of an order two equation: $y''+a(t)y'+b(t)y=0$ with the continues coefficients in all $\mathbb R$. I need to give the equation as well.
What should I do? usually I am asked to decide between two pairs, so I put each pair in it's Wronskian matrix and check if it's determinant can't be $o$ in all the given range. here $y=t^2+1$ is not $0$, but shouldn't I check its derive? and then usually I put both the pair and $y$ in a Wronskian matrix and find the equation, what should I do here?
Thanks!
AI: Plug $y=t^2$ and $y=t^2+1$ into the equation and set $t=0$. What do you get? |
H: Are there problems that are optimally solved by guess and check?
For example, let's say the problem is: What is the square root of 3 (to x bits of precision)?
One way to solve this is to choose a random real number less than 3 and square it.
1.40245^2 = 1.9668660025
2.69362^2 = 7.2555887044
...
Of course, this is a very slow process. Newton-Raphson gives the solution much more quickly. My question is: Is there a problem for which this process is the optimal way to arrive at its solution?
I should point out that information used in each guess cannot be used in future guesses. In the square root example, the next guess could be biased by the knowledge of whether the square of the number being checked was less than or greater than 3.
AI: There are certainly problems where a brute force search is quicker than trying to remember (or figure out) a smarter approach. Example: Does 5 have a cube root modulo 11?
An example of a slightly different nature is this recent question where an exhaustive search of the (very small) solution space saves a lot of grief and uncertainty compared to attempting to perfect a "forward" argument.
A third example: NIST is currently running a competition to design a next-generation cryptographic hash function. One among several requirements for such a function is that it should be practically impossible to find two inputs that map to the same output (a "collision"), so if anyone finding any collision by any method automatically disqualifies a proposal. One of the entries built on cellular automata, and its submitter no doubt thought it would be a good idea because there is no nice known way to run a general cellular automaton backwards. The submission, however, fell within days to (what I think must have been) a simple guess-and-check attack -- it turned out that there were two different one-byte inputs that hashed to the same value. Attempting to construct a complete theory that would allow one to derive a collision in an understanding-based way would have been much more difficult than just seeing where some initial aimless guessing takes you. |
H: Roots in a finite field
Given a finite field $|K|=q$ and an irreducible $f \in K[x]$ with $\deg(f)=n$ with $\alpha$ as a root. My candidates for the roots are $\alpha, \dots , \alpha^{q^{n-1}}$.
Assuming $\alpha^{q^i} = \alpha^{q^j}$ I want to conclude that $\alpha = \alpha^{q^{j-i}}$ or more precisly I want to conclude that $f \mid x^{q^{j-i}}-x$.
Its a basic algebra course and we had not had Galois theory yet. So assume we have to lead $\alpha^{q^i} = \alpha^{q^j}$ to a contradiction by hand.
AI: Judging from your comment to Dilip's answer to your previous question you have not fully absorbed one aspect of finite field arithmetic. Namely the fact that if $\mathrm{char} K=p$ (or equivalently
$q=p^m$ for some positive integer $m$) implies the rules
$$
(a+b)^p=a^p+b^p,\qquad (a-b)^p=a^p-b^p.
$$
I am willing to bet that this really is somewhere in your lecture notes, in which case this is just a refresher. Going from particular to general, let us first study the case $q=2^m$. Then $K$ has $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$ as a subfield, so in the field $K$ we have $2=1+1=0$, because that's the way it goes in that subfield. This implies that
$$
(a+b)^2=a^2+2ab+b^2=a^2+0ab+b^2=a^2+b^2
$$
as claimed. Similarly, if $|K|=3^m$ it has $\mathbb{F}_3=\mathbb{Z}/3\mathbb{Z}$ as a subfield, and consequently $3=1+1+1=0$. This then implies
$$
(a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+0a^2b+0ab^2+b^3=a^3+b^3.
$$
The general case $|K|=p^m, p>3$ follows from a similar calculation with the binomial formula, after we first observe that the binomial coefficient
$$
{p\choose k}=\frac{p!}{(p-k)!k!}
$$
is divisible by $p$ (the numerator is manifestly divisible by $p$, but the denominator is not, because $p$ is a prime and all the factors in the factorials in the denominator are smaller than $p$). The claim for the differences then follows from this rule by replacing $b$ with $-b$.
As a consequence of this fact we also get that in any extension field of $K$, e.g. in $K(\alpha)$ we have
$$
(a-b)^q=(a-b)^{p^m}=\left((a-b)^p\right)^{p^{m-1}}=(a^p-b^p)^{p^{m-1}}=\cdots a^q-b^q
$$
proving the formula you had trouble with. |
H: What is the quantified definition for proper subset
We're thinking its:
$ A \subset B \leftrightarrow \forall x [x \in A \rightarrow x \in B] \land \exists x [x \notin A \land x \in B] $
is this OK?
Thanks,
z.
AI: As Brian says: Yes, that works. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}$ |
H: Difference between supremum and upper bounds and between infimum and lower bounds
I'm having some difficulties catching the difference between upper bound and supremum and, similarly, between lower bound and infimum.
Let's take a look at this set:
$A=\{x\in \mathbb Q | 0<x\leq\sqrt2\}$
So the supremum is $\sup(A)=\sqrt2$, but I don't think an infimum exists at all so
$\nexists\inf(A)$. Isn't by definition $\sup(A)$ also an upper bound? Are there any other upper bounds? By consequence there aren't any lower bounds, are they?
$B=\{x\in\mathbb Q|x^2\leq4\}$
Do the upper bound and supremum coincide? (Similarly do the lower bound and infimum coincide?)
$\inf(B)=-2$, $\sup(B)=2$
AI: Let $P$ be a (partially) ordered set, and let $A$ be a subset of $P$.
We say $x\in P$ is an upper bound for $A$ if and only if $a\leq x$ for all $a\in A$.
We say $s\in P$ is the supremum of $A$ if and only if two things happen:
$s$ is an upper bound for $A$; that is, $a\leq s$ for all $a\in A$; and
$s$ is the smallest upper bound for $A$; that is, if $a\leq x$ for all $a\in A$, then $s\leq x$.
The definitions for lower bound and infimum are dual:
We say $y\in P$ is a lower bound for $A$ if and only if $y\leq a$ for all $a\in A$.
We say $t\in P$ is the infimum of $A$ if and only if two things happen:
$t$ is a lower bound for $A$; that is, $t\leq a$ for all $a\in A$; and
$t$ is the largest lower bound for $A$; that is, if $y\leq a$ for all $a\in A$, then $y\leq t$.
Note that neither the supremum nor the infinmum is required to be in $A$. If they are, then they are also called the maximum and the minimum, respectively. The difference between the supremum and the maximum is that the maximum must be in the set, and the supremum does not have to. (In the real numbers, every set that has upper bound must have a supremum, but it does not have to have a maximum.)
You should be able to verify that you are correct that $\sqrt{2}$ is the supremum of your set $A$. And that any number larger than $\sqrt{2}$ is also an upper bound for $A$.
Note also that $0$ is the infimum of $A$, so you are incorrect in saying that $A$ does not have an infimum.
For $B$, the supremum is $2$, and the infimum is $-2$. Check that.
Now, in the case of the real numbers there are other ways to express the condition of being the smallest upper bound for $A$. It's possible you have seen it defined a different way. Here is a typical one:
Proposition. Let $A\subseteq \mathbb{R}$. Then $s$ is the supremum of $A$ if and only if:
for all $a\in A$, we have $a\leq s$; and
For all $\epsilon\gt 0$, there exists $a\in A$ (which may depend on $\epsilon$) such that $s-\epsilon\lt a\leq s$.
Proof. Suppose that $s$ is the supremum of $A$. Then certainly $a\leq s$ for all $a\in A$. Now let $\epsilon\gt 0$. Then, using the contrapositive of the second part of the definition of supremum, we see that since $s-\epsilon\lt s$, then $s-\epsilon$ cannot be an upper bound for $A$, so therefore there exists $a\in A$ such that $s-\epsilon\lt a$. This gives the condition in the proposition.
Conversely, suppose that $s$ satisfies the conditions in the theorem. To show that it is the supremum of $A$, we must show it is the smallest upper bound for $A$. Indeed, it is an upper bound by the first condition. Now, if $x\lt s$, then there exists $\epsilon\gt 0$ such that $x=s-\epsilon$. By the second condition of the proposition, there exists $a\in A$ such that $x\lt a\leq s$. Thus $x$ is not an upper bound for $A$. So we have shown: if $x$ is smaller than $s$, then $x$ is not an upper bound for $A$. The contrapositive of this is: if $x$ is an upper bound for $A$, then $x$ is not smaller than $s$ (i.e.,, $s\leq x$). This proves $s$ is the supremum. $\Box$ |
H: Number of natural and real numbers
Possible Duplicate:
The simplest way of proving that $|\mathcal{P}(\mathbb{N})| = |\mathbb{R}| = c$
I was reading Rubin and came across the fact that $2^{\aleph_0}$ is the cardinality of reals.
I follow the proof but I cant seem to understand physically what this attempts to say. Is there a nice intuitive explanation for this property?
AI: $2$ raised to the cardinality of the natural numbers, means all functions from the natural numbers to a set of two members (lets say $\{0,1\}$) which practically means all sequences composed of $0$'s and $1$'s. Think of every number in its binary representation. If the number is rational, the sequence will just have an infinite number of zeros from a certain point (we'll avoid representations of numbers similar to $0.999\ldots$ in decimal base). If the number is irrational, it will be any (non constant from a certain point) infinite sequence of zeros and ones. So, any real number can be represented as an infinite sequence of zeros and ones, and that's exactly what the above equality of cardinals asserts. |
H: Which manifolds have a circle as their boundary?
The boundary of a disk or of a Möbius band is a circle.
Which other manifolds share that property?
AI: For the compact case, I believe the answer is, as you said in the comments above, any closed surface with a single puncture, i.e. a disk removed. I claim that this completely classifies compact surfaces with boundary $S^1$. This is because you can glue a disk to the surface along its boundary to obtain a closed surface, and there is a unique way to do this (see, for example, Example 4.1.4(c) in Gompf and Stipsicz's 4-Manifolds and Kirby Calculus). So by the classification of surfaces, there should then be a unique surface with boundary $S^1$ corresponding to each closed surface. |
H: Why is this prime factor not counted
I came across a question:
If j is divisible by 12 and 10, is it divisible by 24 ?
The example draws the following factor tree which I agree with
But then it states that
There are only two 2's that are definitely in the prime factorization of j. because the 2 in the prime factorization in 10 maybe redundant.That is it may be the same 2 as one of the 2's in the prime factorization of 12.
My question is what does
" That is it may be the same 2 as one of the 2's in the prime factorization of 12. " mean ?
Edit:
Can we consider it this way
"Since the 2s have already been accounted for in 12 of the first figure there is no need to consider the 2s from the second figure in the final figure - in short the final figure is a union of both"
AI: You are trying to compute the least common multiple of $12$ and $10$. The idea is that to compute the least common multiple, you prime factor each of the two numbers. Then in the lcm, the exponent of $p$ is the max of the exponent of $p$ in the two numbers. So if $p = 2$, $2^2$ appears in $12$ and $2^1$ appears in $10$. Hence the lcm must have $2^2$. |
H: The image of an ideal under a homomorphism may not be an ideal
This is an elementary question about ideals. Consider a ring homomorphism
$$
f: \mathbb{Z} \rightarrow \mathbb{Z}[x],
$$
and consider the ideal $\left< 2\right>$ in $\mathbb{Z}$. When why is it that $f(\left< 2\right>)$ is not an ideal?
Some websites say that $f(\left< 2\right>)$ is not an ideal because it
does not contain nonconstant polynomials. That still doesn't make sense on why it is not an ideal.
Thank you all.
AI: An ideal $I$ of a ring $R$ is, by definition, "multiplicatively absorptive"; that is, for any $s\in I$ and $r\in R$, we must have $rs\in I$.
Under the usual definition of ring homomorphism, given rings $A$ and $B$, a ring homomorphism $f:A\to B$ must satisfy $f(1_A)=1_B$, so that there is in fact a unique ring homomorphism $f:\mathbb{Z}\to R$ for any ring $R$. In the case of $R=\mathbb{Z}[x]$, this is exactly what you would expect; since $f(1)=1$, we have $f(n)=n$ for all $n\in\mathbb{Z}$.
The set $S=f(\langle 2\rangle)$ in $\mathbb{Z}[x]$ is of course just $$S=\{g\in\mathbb{Z}[x]\mid g=2k\text{ for some }k\in\mathbb{Z}\},$$ i.e. the even integers. For example, $2\in S$. But $x\in \mathbb{Z}[x]$, and $2x\notin S$, so $S$ is not an ideal of $\mathbb{Z}[x]$. |
H: Proving $\inf\limits_{f\in\Gamma} \{ F(y(t))(f(t))\}= -\|F(y(t))\| $
I would like to proof the next claim:
Let $X$ a Banach space, $F\colon X\to X^*$ a linear continuous function,
$$
\Gamma:=\{f\in (\mathcal{C}([0,1],X)\,:\, f(0)=f(1)=0\mbox{ and }\|f\|\leq 1\}
$$ and fix $t\in [0,1]$, $y\in \Gamma$. Then $F(y(t))\in X^*$. How I can prove that
$$
\inf\limits_{f\in\Gamma} \{ F(y(t))(f(t))\}= -\|F(y(t))\|
$$
? I tried it using Hahn-Banach, but maybe there is another easier way. Thanks in advance.
Edit: sorry, I forgot to put that F es linear... Thanks for the answers.
AI: You need to assume $F(0) = 0$ for this to hold as the left hand side is $0$ for $t\in \{0,1\}$, hence the right hand side should also be zero, i. e. $F(0) = 0$. As $F$ and $y$ are evaluated at exactly one point, let's write $x^* = F\bigl(y(t)\bigr)$, if $t \in (0,1)$, then $\Gamma(t) = \{f(t)\mid f\in \Gamma\} = B_X$, so you have to prove $\inf_{x\in B_X} x^*(x) = -\|x^*\|$, which holds as
\[
\|x^*\| = \sup_{x \in B_X} x^*(x) = -\inf_{x\in B_X} x^*(-x) = -\inf_{x\in B_X} x^*(x).
\] |
H: Dimension of a cyclic submodule of a finite group representation
Let $\rho: G \to GL(V)$ be a finite (complex) group representation. What is the maximum dimension of $span\{\rho_gv | g \in G\}$ over all $v \in V$? This quantity is not necessarily the degree of $V$: for example, if $V$ is the direct sum of two isomorphic 1-dimensional subrepresentations, then the maximum dimension is 1.
I think this can be rephrased as: given a $\mathbb{C}G$-module, what is the largest (vector space) dimension over $\mathbb{C}$ of a cyclic submodule?
What I have so far: by the canonical decomposition, $V \cong W_1 \oplus W_2 \oplus ... \oplus W_r$ where each $W_i$ is isomorphic to $n_i$ copies of an irreducible submodule $U_i$ with dimension $d_i$. For $v \in V$ define $\Phi = \sum_{g \in G}{\rho_gvv^*\rho_g^*}$ (* denotes adjoint). Then the dimension quantity in question is equal to $rank(\Phi)$. $\Phi$ is a $\mathbb{C}G$-module homomorphism of $V$ into itself. Therefore $\Phi$ maps each $W_i$ to itself. Hence if $v = w_1 + w_2 + ... + w_r$ is the direct sum decomposition of $v$, then the dimension of $span\{\rho_g{v} | g \in G\} = \sum_{i = 1}^r{dim (span\{\rho_gw_i | g \in G\})}$, thus reducing the problem to the situation where the representation in question is isomorphic to copies of an irreducible module.
If this irreducible module has degree 1, then the maximum dimension is 1. But for degree > 1, I'm not sure. I haven't been formally taught representation theory, so I'd appreciate references as well as a clue to this problem.
AI: Your question is about the dimension of the cyclic $\mathbb{C}G$-module generated by $v$.
Such a module is obviously isomorphic to a quotient of the left regular representation $M=\mathbb{C}G$ (= the free cyclic module generated by $1$), as there is a unique homomorphism $f:M\rightarrow \langle v\rangle$ determined by $f(1)=v$. Conversely any such quotient is a cyclic module as a homomorphic image of $M$ is generated by the image of $1$.
Let us denote by $m(V,U_i)$ the multiplicity of an irreducible representation $U_i$ as a composition factor of $V$.
For each $U_i$ we know that $m(M,U_i)=\dim U_i$.
Therefore the maximum multiplicity of $U_i$ as a composition factor of a cyclic module is also equal to $\dim U_i$.
Conversely (using Maschke's theorem), if a module $V$ satisfies the inequality $m(V,U_i)\le \dim U_i$ for all irreducible representations $U_i$, it can be written as a quotient of $M$, i.e. is a cyclic module.
Therefore the answer is that the maximum dimension of a cyclic submodule of a given module $V$ is
$$
\sum_i\min\{\dim U_i, m(V,U_i)\}\cdot\dim U_i.
$$
The summation ranges over the irreducible representations of $G$. |
H: Counter-example that $I\cup J$ in a ring $R$ may not be an ideal
I've been doing some reading about ideals and here is another question (to which I couldn't yet find or construct a counterexample).
Let $I, J$ be ideals in a ring $R$. Then $I\cup J$ is contained in $I+J$ but it may not be an ideal since it may not be closed under addition.
Can you give me a counterexample of ideals $I$ and $J$ in $R$ so that $I\cup J$ is not an ideal?
Note: $I\cup J\subseteq I+J$ since we can write $i \in I$ as $i+0\in I+J$ and similarly, we can write $j\in J$ as $0+j\subseteq I+ J$.
$$
$$
After reading so many great responses from this post,
would this be a counter-example? Take $I=\left< 2\right>$ and $J=\left< x\right>$ in $R=\mathbb{Z}[x]$. Then $2\in I, x\in J$, but $2+x$ is not in either $I$ or $J$? After thinking about this counter-example, I don't think this is a good one: it only shows that $I\cup J$ is properly contained in $I+J$.
Thanks again for your time.
AI: Let $R = \mathbb Z$ and $I = 2 \mathbb Z$ and $J = 5 \mathbb Z$. Then $2,5 \in I \cup J$ but $2 + 5 = 7 \notin I \cup J$. |
H: divide triangle into ratio 1:2
I have a problem related to triangles. Please give me some hint to progress.
Suppose we have the following coordinates $A(-2,3)$,$B(1,-1)$,$C(-1,-1)$.
From point $A$, draw a line which divides the area of triangle $ABC$ in the ratio $1:2$.
I have calculated the length of the sides:
$[AB]=5$,$[AC]=\sqrt{17}$ and $[BC]=2$.
I have searched online for what this line should be, like medians, altitudes or etc and found this,
where is written:
If one median of a triangle is drawn, the second median to be drawn will divide the areas of the two triangles formed by the first median in the ratio 1:2.
But is it what I need? I can write the equation of lines which go through $AC$,$BC$,$AB$, but how do I find the equation of the required line?
AI: Let $D$ be a point on $\overline{BC}$ between $B$ and $C$. Observe that $\triangle BDA$ and $\triangle DCA$ have the same height, $3-(-1)=4$. Let $x=|BD|$ and $y=|DC|$. The area of $\triangle BDA$ is $2x$, the area of $\triangle DCA$ is $2y$, and $x+y=|BC|=2$, so $y=2-x$. Now all you have to do is find $x$ so that $2(2-x)=2(2x)$. |
H: Prove or disprove an inequality
Let's consider the following equation where $m,n$ are real numbers:
$$ x^3+mx+n=0 $$
I need to prove/disprove without calculus that for any real root of the above equation we have that:
$$ m^2-4 x_1 n \ge 0$$
AI: Suppose that $x_1$ is a real root of the cubic, and consider the quadratic equation $x_1x^2+mx+n=0$. This must have $x_1$ as a real solution, so ... ? |
H: How can we take a power series and multiply each term, i.e. $c_n x^n$ by $y^n$?
In other words, given a power series $f(x)$, is there an alternative to taking $\lim_{x\to{x y}}f(x)$? I ask this because I thought that there may be a way to replace the limit by integration, or some other operation that I know how to do repeatedly.
To attempt to clarify slightly, I plan on taking this limit repeatedly in a "loop", along with a few other operations. I have seen repeated integration done in a book, e.g. Keith B. Oldham's and Jerome Spanier's The Fractional Calculus, published by Dover Publications, Inc.
For an example, suppose I have a generating function:
$$f(x) = 15x^3 + 27x^5 + 300x^9$$
I want
$$f(xy) = 15x^3y^3 + 27x^5y^5 + 300x^9y^9$$
Again, I'm hoping for one or more answers that show some generalized methods to go from a general function $f(x)$ to $f(xy)$.
AI: I don't see what this has to do with limits. The notation $\lim_{x\to xy}$ is misleading. The process of substitution, replacing $x$ by $xy$, is the algebraic operation of composing two functions. It can be done repeatedly if you want: in the second line, you can replace $x$ with $xy$ again, or replace $x$ with $xz$, or both $x$ and $y$ with $xz$ and $yz$, etc - depends on what you want to get out of this process. |
H: How to get the characteristic equation?
In my book, this succession defined by recurrence is presented:
$$U_n=3U_{n-1}-U_{n-3}$$
And it says that the characteristic equation of such is:
$$x^3=3x^2-1$$
Honestly, I don't understand how. How do I get the characteristic equation given a succession?
AI: Here’s a rote rule for doing so. Start with the recurrence:
$$U_n=3U_{n-1}-U_{n-3}$$
Convert each subscript to an exponent:
$$U^n=3U^{n-1}-U^{n-3}$$
Change the variable to the one that you want to use in the characteristic equation:
$$x^n=3x^{n-1}-x^{n-3}$$
Divide through by the smallest exponent, in this case $n-3$:
$$x^{n-(n-3)}=3x^{(n-1)-(n-3)}-1\;,$$
which simplifies to $$x^3=3x^2-1\;.$$
With a little practice you can do the conversion in one go. For instance, the recurrence $$a_n=4a_{n-2}-6a_{n-3}+3a_{n-4}$$ has characteristic equation $$x^4=4x^2-6x+3\;,$$ as you can check by following through the steps given above. |
H: Proving that an integer is even if and only if it is not odd
There is this question, but the definition of "even" and "odd" that I am using uses integers instead of just natural numbers; i.e.,
An integer $n$ is even iff there is some integer $k$ such that $n=2k$.
An integer $n$ is odd iff there is some integer $k$ such that $n=2k+1$.
Here is what I have so far:
First we show that an integer $n$ is even or odd.
We first use induction on the positive integers.
For the base case, $1=2\cdot0+1$ so we are done.
Now suppose inductively that $n$ is even or odd.
If $n$ is even, then $n=2k$ for some $k$ so that $n+1=2k+1$ (odd).
If $n$ is odd, then $n=2k+1$ for some $k$ so that $n+1=2(k+1)$ (even).
This closes the induction, so every $n\in\mathbf{Z}^+$ is even or odd.
Now we show every $n\in\mathbf{Z}^-$ is even or odd.
Let $n\in\mathbf{Z}^-$.
Then $n=-k$ for some $k\in\mathbf{Z}$ (I think this follows immediately from most definitions of the integers.).
Suppose $k$ is even.
Then $k=2j$ for some $j$ so that $n=-k=-2j=2(-j)$ (even).
Now suppose $k$ is odd.
Then $k=2j+1$ for some $j$ so that $n=-k=-(2j+1)=-2j-1=-2j-1+1-1=-2j-2+1=2(-j-1)+1$ (odd).
For $0$, note that $0=2\cdot0$ (even).
Now we show that $n\in\mathbf{Z}$ cannot be both even and odd.
Suppose for the sake of contradiction that $n\in\mathbf{Z}$ is both even and odd.
Then there are integers $k,j$ such that $n=2k=2j+1$.
This implies that $2(k-j)=1$ (like in the referenced question).
So we must show that $1$ cannot be even in order to complete the proof.
This is where I am having trouble.
I know that if I let $f:\mathbf{Z}\to\mathbf{Z};x\mapsto2x$ be a function, then $f$ is increasing so since $f(0)=0$ and $f(1)=2$ and $0<1<2$, there is no integer $m$ such that $f(m)=1$.
But this seems complicated so I was wondering if there was an easier way to do this.
So my real question is: how can I show that $1$ is not even?
(This is not homework.)
AI: To show that $1$ is not even:
I assume you can prove or accept that $a \cdot 0 = 0$ for all $a \in \mathbb{Z}$, the product of a positive and negative number is negative, and $2a > a$ when $a >0$.
If $1$ is even, then there must exists $a < 1$ such that $2a = 1$. However, the only $a < 1$, which is an integer, is $0$ and clearly $2 \cdot 0 = 0$. So $1$ can not be be written as $2 \cdot a$ for any $a \in \mathbb{Z}$. $1$ is odd. |
H: Bidding Item Problem
You are bidding on an item that has an unknown value uniformly
distributed between 0 and 1. You do not know the true value of the
item, but you know that if you end up winning the bid for the item,
the item will increase its value to 2x its original value. Your bid
can only go through if its at least as large as the original value of
the item. How do you bid to maximize expected payoff.
Here's what I have:
Let V be the true value of the item
Let B be the bid you make
Let f(V) represent the profit you make given V as the true original value
$$f(V) =
\begin{cases}
2V - B & B \geq V\\
0 & B< V
\end{cases}$$
Where I get confused is when I need to start applying integrals to calculate how to maximize the expected value.
Thanks for any help.
EDIT: Here is the solution from the book I'm working off of. I do not understand how they are doing the calculus.
Let B be your bid. Let S be the true value of the item. The density function of S equals unity for $0 \leq S \leq 1$, and 0 otherwise.
Your payoff P is
$$P(S) =
\begin{cases}
2S - B & B \geq S\\
0 & \text{otherwise}
\end{cases}$$
The maximum post bid item value is 2, so you should be no more than 2. You want to maximize $E[P(S)]$ with respect to choice of B in the interval [0, 2]. Your expected payoff is:
$$\begin{aligned}
E[P(S)] &= \int_{S=0}^{S=1} P(S)*1*\,\mathrm{d}S \\
&= \int_{S=0}^{S=\min(B,1)} (2S-B)\,\mathrm{d} \\
&= \left.(S^2-BS)\right|_{S=0}^{S=\min(B,1)} \\
&=
\begin{cases}
0 & B\leq1\\
1 - B & B>1
\end{cases}
\end{aligned}$$
so you should bid less than or equal to 1 and expect to break even.
AI: As I (and others) noted in the comments, the exercise seems to be really confusingly phrased (and not at all like any real auction or other transaction I've ever heard of), and the given solution doesn't look much better. So instead of trying to explain the book solution, let me rephrase the exercise in a (hopefully) slightly less confusing manner, and then show how I would solve it.
Exercise: You are bidding on an item that has an unknown nominal value $V$ uniformly distributed between $0$ and $1$. You do not know the nominal value of the item, but you do know that the value of the item to you is twice its nominal value. You know that, if you bid more than the nominal value of the item, you will win the item and have to pay your bid; otherwise you don't get the item and don't have to pay anything. How much should you bid to maximize your expected gain?
Solution:
Let $V \sim U(0,1)$ be a random variable denoting the nominal price of the item. The probability density function of $V$ is $$f(V) = \begin{cases} 1 & \text{if }0 < V < 1 \\ 0 & \text{otherwise.} \end{cases}$$
If you bid an amount $B$, your gain will be $$g(B,V) = \begin{cases} 2V -B & \text{if }V < B \\ 0 & \text{otherwise.} \end{cases}$$
Thus, your expected gain from bidding $B$ is $$\begin{aligned}
\mathbb E_V[g(B,V)] &= \int_{-\infty}^\infty g(B,V)\: f(V)\: \mathrm dV \\
&= \int_0^1 g(B,V)\: \mathrm dV \\
&= \int_0^{\min(1,B)} (2V-B)\: \mathrm dV \\
&= \int_0^{\min(1,B)} 2V\: \mathrm dV - \int_0^{\min(1,B)} B\: \mathrm dV \\
&= V^2 \bigg|_{V=0}^{V=\min(1,B)} - BV \bigg|_{V=0}^{V=\min(1,B)} \\
&= (\min(1,B)^2 - 0^2) - (B\min(1,B) - B\cdot0) \\
&= \min(1,B)^2 - B\min(1,B) \\
&= \begin{cases} 1-B & \text{if }B > 1 \\ 0 & \text{if }B \le 1. \end{cases}
\end{aligned}$$
(Since you said you had trouble following this part in the book solution, I included quite many intermediate steps. Let me know if there's still something you don't follow.)
As $1-B < 0$ whenever $B > 1$, you should never bid more than $1$. Instead, any bid of $1$ or less will result in an expected gain of $0$, so any such bid is as good as not bidding at all, which is the optimal strategy.
Actually, this came out looking more like that book solution after all, though hopefully a bit clearer. What I would probably do, if asked that in an exam or something, would be to start by noting that the chance of winning the item equals $1$ for any bid $B \ge 1$. Thus, any bid $B > 1$ is clearly suboptimal, as it increases the cost without changing the probability of winning. That out of the way, I'd the just calculate $$\mathbb E[g(B)]
= \int_0^1 g(B,V)\: \mathrm dV
= \int_0^B (2V-B)\: \mathrm dV
= B^2 - B^2 = 0$$
for all $B \le 1$. (Or I might note that the integrand $2V-B$ has odd symmetry around the midpoint $V = B/2$ of the integration interval, so the integral has to be zero by symmetry considerations alone.) |
H: Number of elements in a finite field extension for finite fields
Given an arbitrary finite field $K$ (not necessarily $\mathbb{F}_p$ with $p \in \mathbb{P}$) with $|K| = q$ and an irreducible polynomial $f$ with $\alpha$ as root and degree of $n$. Is $|K(\alpha)| = q^n$ and why? Its clear to me for $K$ isomorphic to $\mathbb{Z}_p$ with $p$ is a prime number.
AI: The division algorithm holds in $k[x]$ for any field $k$. In fact, it holds in $R[x]$ for any commutative ring $R$ provided that the divisor has leading term a unit.
Theorem. Let $R$ be a commutative ring, and let $a(x)$ be a polynomial in $R[x]$ whose leading coefficient is a unit in $R$. Then for every $b(x)\in R[x]$ there exist unique $q(x),r(x)\in R[x]$ such that $b(x) = q(x)a(x)+r(x)$, where $r(x)=0$ or $\deg(r)\lt\deg(a)$.
The proof is exactly the same as the proof in any field, and can be done by induction on $\deg(b)$.
With the division algorithm in hand, it is immediate that if $f(x)$ is of degree $n$, then $k[x]/\langle f\rangle$ is a vector space over $k$ of dimension $n$, with basis $1+\langle f\rangle$, $x+\langle f\rangle,\ldots,x^{n-1}+\langle f\rangle$.
In particular, for $k=\mathbb{F}_q$, the resulting quotient is dimension $n$ over $\mathbb{F}_q$, and so has $q^n$ elements. This holds whether or not $f(x)$ is irreducible, though we need $f(x)$ irreducible in order to know that the quotient is a field. |
H: Variable Substitution
I need to show that next equation stands:
$$\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = \frac{1}{r}\frac{\partial}{\partial r} (r\frac{\partial u}{\partial r} ) +\frac{1}{r^2}(\frac{\partial^2 u}{\partial \theta^2} )$$
where
$u=f(x,y)$ , $x = r\cos\theta$ , $y = r\sin\theta$ , $r = \sqrt{x^2+y^2}$ , $\theta = \arctan(\frac{y}{x})$
I got lost where I came around $$du=dx(\frac{\partial u}{\partial r}\frac{x}{r} - \frac{\partial u}{\partial \theta}\frac{y}{r^2})+dy(\frac{\partial u}{\partial r}\frac{y}{r} + \frac{\partial u}{\partial \theta}\frac{x}{r^2})\\\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{x}{r} - \frac{\partial u}{\partial \theta}\frac{y}{r^2},\ \ \ \ \ \ \ \frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\frac{y}{r} + \frac{\partial u}{\partial \theta}\frac{x}{r^2}$$
I just can't figure out, what is the next step and how do I go on with this.
Could you, please, help me?
AI: Some general points:
I prefer subscript notation for such computations, it simplifies both writing and typing.
It is important to choose the direction of transformation: left to right or right to left.
For example, it is easy to differentiate $x,y$ in terms of $r,\theta$. It's not as easy the other way around (and besides, the formula $\theta=\arctan\frac{y}{x}$ is not 100% correct). So it makes sense to use the chain rule in the form $u_r=u_x x_r+u_y y_r$ and $u_\theta=u_x x_\theta+u_y y_\theta$, which is the opposite direction to what you tried.
For example, we see that $u_r=u_x \cos\theta+u_y \sin\theta$. This formula holds for any smooth function, so we can use it again with $ru_r$ in place of $u$. Although not strictly necessary, it helps to notice that $ru_r=xu_x+yu_y$.
$(ru_r)_r=(ru_r)_x \cos\theta+(ru_r)_y \sin\theta = (xu_x+yu_y)_x \cos\theta+(xu_x+yu_y)_y \sin\theta$, and so on... |
H: Series convergence, finding values that cause convergence.
I am trying to find the $x$ values that make this series converge:
$$\sum_{n = 1}^\infty (x+2)^n.$$
To me it seems like $x = -2$ would make the series converge but that is a wrong answer, I am not sure why either.
AI: Recall the geometric progression $$\sum_{n=1}^{N} a^n = a \left(\dfrac{1-a^{N}}{1-a} \right)$$ and hence the geometric series $\displaystyle \sum_{n=1}^{\infty} a^n$ converges if and only if $\vert a \vert < 1$.
In your problem, $a = x+2$. |
H: set in $\mathbb{R}$ which is not a Borel-set
Possible Duplicate:
Lebesgue measurable but not Borel measurable
Constructing a subset not in $\mathcal{B}(\mathbb{R})$ explicitly
if i start from the topology of $\mathbb{R}$, i.e. all open sets, and then build the closure under countable union and complement i get the so called Borel-$\sigma$-Algebra, the smallest set which contains all open sets (i.e the topology) of $\mathbb{R}$ and is a $\sigma$-algebra? do you know any sets on $\mathbb{R}$ which are not cointained in this Borel-$\sigma$-algebra?
AI: There are $2^{2^{\aleph_0}}$ subset of $\mathbb{R}$. There are $2^{\aleph_0}$ Borel subsets. Hence there certainly exists subsets of $\mathbb{R}$ that are not Borel.
However, there are even some very nicely defined sets that are not Borel. For example, there is a Lebesgue Measureable set that is not Borel. The cantor set has measure zero and is uncountable. Hence every subset of the Cantor set is Lebesgue Measureable and by a cardinality argument, there exists one which is not Borel. Analytic sets can be defined to be continuous images of the real line. There even exists analytic sets which are not Borel. |
H: Proof that the Irrationals are Countable
Proof: Between any two irrationals lies a rational, by the Density of the rationals in the real number system. There are only countably many rationals; therefore, there are only countably many pairs of irrationals. Therefore the number of irrationals is countable since the cardinality of $2\mathbf{N}$ is $\mathbf{N}$.
I don't know why I came across this logic since I know the irrationals are uncountably infinite, but I don't see the hole in my logic.
AI: The claim that between two rationals there is only one irrational is false. In fact between two rationals there are many irrationals, so you will have to map a lot of irrationals to the same pair (for most pairs too).
Therefore your proof does not constitute of a bijection, or even a well-defined function. This is a common mishap with infinite objects, though. They tend to get very confusing! |
H: Primality and repeated digits
I recently worked on problem 51 through project euler, I solved it essentially through brute-force but afterwards I viewed the forum and there were some more clever solutions.
For those unfamiliar with project euler, the problem reads:
By replacing the 1st digit of *3, it turns out that six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.
By replacing the 3rd and 4th digits of 56**3 with the same digit, this 5-digit number is the first example having seven primes among the ten generated numbers, yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. Consequently 56003, being the first member of this family, is the smallest prime with this property.
Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime value family.
People were able to drastically cut down the run times of their program with some shortcuts. First, they only replaced repeated digits in the prime numbers, where as I replaced all digits. They also determined that they need not consider primes with 2 or 4 repeated digits, as these would produce composite numbers at some point. I am curious of the mathematical reasoning behind these shortcuts, and any theories to back it up. The only one I could glean from the thread was that if the sum of the digits in the number is divisible by 3, the number itself will also be divisible by 3, but I don't see how that relates to the shortcuts they used.
AI: I recommend you take a look at this Wikipedia article on modular arithmetic, which is the basis for most of these shortcuts. Here are some shortcuts with reasons:
To find prime value families you only need to replace digits in primes, since replacing digits in any member of a prime valued family will give you the whole family.
If the remainder of the sum of the digits of $n$ after dividing by $3$ is $x$, and you replace the digit $d$ with $d'$, then the new remainder will differ from $x+d-d'$ by a multiple of $3$. If our new number is a prime, then it cannot be divisible by $3$ so $x+d'-d$ cannot be a multiple of $3$. But for at least $3$ out of $10$ possible choices of $d'$, $x+d'-d$ will be divisible by $3$. Thus replacing just one digit cannot give you an eight digit family, so you need only check primes with repeated digits.
More eloquently, at least one of the eight choices for $d'-d$ falls in the inverse equivalence class modulo $3$ as the equivalence class of $n$, so the result is divisible by $3$.
In fact, the same argument shows that you can't be replacing $s$ digits unless $s$ is a multiple of $3$, since otherwise for at least $3$ out of $10$ possible choices of $d'$, $x+s(d'-d)$ will be divisible by $3$. Thus you can't replace $2$ or $4$ digits. |
H: Generators of von Neumann algebras
Let us suppose that M is a von Neumann algebra on some hilbert space $H$ such that $M = A''$ for some $C^*$-algebra $A\subseteq B(H)$.
I am wondering when $A$ will be dense in $M$ and by which topology. I also want to know that if I have two functional $S,T \in M^*$ such that $S|_A=T|_A$, can I say that $T=S$?
Thank you very much.
AI: Let $K\subseteq H$ be the closed subspace generated by $AH$. Then von Neumann's double commutant theorem says that if $K=H$, then $M$ is the closure of $A$ in the weak operator, strong operator, ultraweak, or ultrastrong topologies. If $K$ is a proper subspace of $H$, then $A=A|_K\times \{0\}|_{K^\perp}$, $A''=(A|_K)''\times \mathbb C I_{K^\perp}$, and $A$ is weakly dense in $(A|_K)''\times \{0\}$ but not $M$. The condition $K=H$ is also known as $A$ acting nondegenerately on $H$. In particular, it holds if $A$ contains the identity operator on $H$.
So assuming that $A$ acts nondegenerately, $A''=M$ is equivalent to density of $A$ in $M$ in various topologies that are coarser than the norm topology. If $M^*$ is the set of norm-continuous linear functionals on $M$, then it is possible in general for distinct functionals in $M^*$ to agree on $A$, if at least one of them is not ultraweakly continuous. $A$ is a norm-closed subspace of $M$, so if $A\neq M$ then there are nonzero continuous functionals on the Banach space $M/A$, hence by composing with the quotient map, there are nonzero continuous functionals on $M$ that vanish on $A$. |
H: Help with binomial theorem related proof
I'm currently working through Spivak on my own. I'm stuck on this proof, and the answer key is extremely vague on this problem. I think I'm missing a manipulation involving sums.
Prove that $\displaystyle\sum_{k=0}^{l}\dbinom{n}{k}\dbinom{m}{l-k}= \dbinom{n+m}{l}$.
As a hint, he gives "Apply the binomial theorem to $\displaystyle(1+x)^n(1+x)^m$"
Following the hint, I get:
$\displaystyle(1+x)^n(1+x)^m=(1+x)^{n+m}$
Applying the binomial theorem, we get:
$\displaystyle\sum_{j=0}^n\dbinom{n}{j}x^j\cdot\sum_{k=0}^m\dbinom{n}{k}x^k=\sum_{l=0}^{n+m}\dbinom{n+m}{l}x^l$
Here's where I get stuck. How do I manipulate this into looking like the above?
As an aside, this is not homework. I'm working through the book for my own benefit. I'm usually reticent about consulting the answer key, but I've been stuck on this one for about a day.
AI: Starting where you got stuck, how do you get a term in $x^{\ell}$ from the product on the right side? You get it when you multiply ${n\choose j}x^j$ from the first term with ${m\choose\ell-j}x^{\ell-j}$ from the second term. [Note that you have a typo, an $n$ in the second sum where you wanted to have $m$] So, the coefficient of $x^{\ell}$ is the sum of all the terms ${n\choose j}{m\choose\ell-j}$, as desired. |
H: Complex series: $\frac{z}{(z-1)(z-3)} = -3 \sum\limits_{n=0}^\infty \frac {(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$ for $0 < |z-1| < 2$
Show that when $0 < |z-1| < 2$,
$$\frac{z}{(z-1)(z-3)} = -3 \sum_{n=0}^\infty \frac {(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$$
I thought to attack this using a partial fraction decomposition and then breaking the partial fractions into Maclaurin series. I got the $$-\frac{1}{2(z-1)}$$ from that, but the other part has me a bit stumped.
I have:
$$
\frac{3}{2(z-3)}$$
for the other partial fraction. And I factor out the 3/2 and then have:
$$\frac{1}{z-3}$$
So I get:
$$\frac{1}{1 - (-(z-2))}$$
but this does not give me what I need.
AI: $$\dfrac{z}{(z-1)(z-3)} = \dfrac1{(z-1)} + \dfrac3{(z-1)(z-3)}$$
$$\dfrac1{z-3} = \dfrac1{(z-1) -2} = -\dfrac12 \left( \dfrac1{1-(z-1)/2}\right) = -\dfrac12 \left( \sum_{k=0}^{\infty} \left( \dfrac{z-1}2\right)^k\right)$$
Hence, $$\dfrac3{(z-1)(z-3)} = - \dfrac32 \left( \dfrac1{z-1} + \sum_{k=0}^{\infty} \left( \dfrac{(z-1)^k}{2^{k+1}}\right)\right)$$
Hence,
\begin{align}
\dfrac{z}{(z-1)(z-3)} & = \dfrac1{(z-1)} + \dfrac3{(z-1)(z-3)}\\
& = \dfrac1{z-1} - \dfrac32 \left( \dfrac1{z-1} + \sum_{k=0}^{\infty} \left( \dfrac{(z-1)^k}{2^{k+1}}\right)\right)\\
& = - \dfrac12 \dfrac1{z-1}- 3 \left( \sum_{k=0}^{\infty} \left( \dfrac{(z-1)^k}{2^{k+2}}\right)\right)
\end{align} |
H: How smooth is the distribution function of a convex polynomial?
Here is a prototype of the problem I have in mind: Let $P:\mathbb{R}^2\rightarrow\mathbb{R}$ be a strictly convex, nonnegative polynomial such that $P(0,0)=0$. Let $\alpha\geq 0$, and consider the following version of its distribution function (in the sense of, say, harmonic analysis)
$$\lambda_P(\alpha)=|\{(x,y)\in\mathbb{R}^2: P(x,y)<\alpha\}|;$$
here by $|\cdot|$ I mean Lebesgue measure on the plane.
Question. Is $\lambda=\lambda_P$ a smooth ($C^\infty$) function of $\alpha$, for $\alpha>0$?
Thank you.
AI: I think it is indeed $C^\infty$ smooth. Let $r,\theta$ be polar coordinates. For each fixed $\theta$ the function $p_\theta(r)=P(r,\theta)$ is $C^\infty$ smooth on $(0,\infty)$ and has strictly positive derivative. Therefore, the inverse $R_\theta : (0,\infty)\to(0,\infty)$ is also $C^\infty$ smooth. We have $\lambda_P(\alpha)=\frac{1}{2}\int_0^{2\pi}R_\theta^2(\alpha)\,d\theta$, and this can be differentiated with respect to $\alpha$ as many times as we wish. |
H: Simplifying to a certain expression structure
I have this expression:
$$4n^2-n+(8(n+1)-5)$$
And I know it is equivalent to this:
$$4(n+1)^2-(n+1)$$
I need to simplify my expression to get the same structure as that one. However, no matter what I try, I don't end up with that structure.
My question is, well, how to reach that. But truly, the real problem is that I seem to lack this sense of "knowing which method to use to simplify to get the structure I want". Is there some kind of rule or guide about this? Or is it all about practice and experience?
Edit
If anyone was curious, I ended up with this:
$$4n^2+7n+3$$
Yes, I did multiplications and additions. Couldn't find common factor (as far I can tell) or any other trick.
AI: To factor $\rm\:f = 4n^2+7n+3\:$ note $\rm\,a\,x^2 + (a\!+\!c)\,x + c = ax\,(x+1) + c\,(x+1) = (ax+c)\,(x+1).\,$ Alternatively, as in your prior question, apply the AC-method as follows:
$$\begin{eqnarray}\rm 4f\, &=&\rm\ \ 16n^2 +\ 7\cdot 4n\, +\, 4\cdot 3 \\
&=&\rm\ (4n)^2 + 7\,(4n)\, + 4\cdot 3 \\
&=&\rm\ \ N^2\ +\ 7\, N\ +\ 12\quad for\quad N = 4n \\
&=&\rm\ (N\ +\ 4)\,(N\ +\ 3) \\
&=&\rm\ (4n\, +\, 4)\,(4n\, +\, 3) \\
\rm f\, &=&\rm\ (\ n\ +\ 1)\,(4n\, +\, 3) \end{eqnarray}$$ |
H: Differentiation under the integral sign.
I'm working through an integral suggested for practice at the end of the Wikipedia article on differentiation under the integral sign, and I'm stuck.
I am attempting to evaluate this integral:
$$\int_0^{\pi/2} \frac{x}{\tan x} \ dx.$$
The article suggests the following parameterization:
$$F(a)=\int_0^{\pi/2} \frac{\tan^{-1}(a\tan (x))}{\tan x} \ dx.$$
Differentiating with respect to $a$, we get
$$F'(a)=\int_0^{\pi/2} \frac{1}{1+a^2\tan^2 x} \ dx.$$
I can't find a way to evaluate this, and neither can Wolfram Alpha. The special values $a=0,1$ are easy, but I fail to see how they help.
How can I finish evaluating this integral?
Edit: I think it's just substitution, maybe. I'll update the post accordingly soon.
Edit 2:
Indeed, the substitution $u=\tan x$ and identity $\sec^2 = 1 + \tan^2$ transform the above integral into
$$F'(a) = \int_0^{\pi/2} \frac{1}{(1+u^2)(1+a^2u^2)}.$$
This can be solved with partial fractions.
AI: You have that
$$\frac{1}{{\left( {1 + {u^2}} \right)\left( {1 + {a^2}{u^2}} \right)}} = \left( {\frac{A}{{1 + {u^2}}} + \frac{B}{{1 + {a^2}{u^2}}}} \right)$$
Thus you want (after cross mult.)
$$1 = A + A{a^2}{u^2} + B + B{u^2}$$
This is
$$\eqalign{
& A + B = 1 \cr
& A{a^2} + B = 0 \cr} $$
Which gives
$$A = \frac{1}{{1 - {a^2}}}$$
and in turn
$$B = 1 - A = \frac{{{a^2}}}{{{a^2} - 1}}$$
which means
$$\frac{1}{{\left( {1 + {u^2}} \right)\left( {1 + {a^2}{u^2}} \right)}} = \frac{1}{{{a^2} - 1}}\left( {\frac{{{a^2}}}{{1 + {a^2}{u^2}}} - \frac{1}{{1 + {u^2}}}} \right)$$
Can you move on? |
H: Prove/Show that a number is square if and only if its prime decomposition contains only even exponents.
Prove/Show that a number is square if and only if its prime decomposition contains only even exponents.
How would you write a formal proof for this.
AI: Suppose $a = p_1^{r_1}...p_n^{r_n}$. If $r_i$ are all even, then let $b = p_1^{\frac{r_1}{2}} ... p_n^{\frac{r_n}{2}}$. Then $b^2 = a$.
Suppose $a = b^2$. Let $q_1^{s_1} ... q_m^{s_m}$ be the unique prime factorization of $b$. Then $a = q_1^{2s_1} ... q_m^{2s_m}$. So by unique factorization all primes of $a$ have even powers. |
H: Calculating Variance from Math Problem
I was reading this post here when one particular problem caught my attention for a while -
The banker shuffles a standard deck of 52 cards and slowly deals them
face up. The dealt cards are left in full view where they can be
inspected at any time by the player. Whenever the player wants, he may
say "Red." If the next card is red, he wins the game, otherwise he
loses. He must call red before the deal ends, even if he waits to call
on the last card. What odds should the banker give to make it a fair
game, assuming that the player adopts his best strategy on the basis
of feedback form the dealt cards? The player must announce the size of
his bet before each game begins.
Surprisingly from first glance, the EV is even for both banker and player, regardless of whatever strategy is used.
This got me thinking about using this game in a hypothetical casino, which charges say a 5% rake/commission per game. The casino would then have an expected value of +5% due to the rake.
Now, suppose the player uses the following strategy: Every time there are more black cards than red cards (even just one), the player calls "red". Obviously in the event there are more black cards than red cards, such a play would be the best and yield a positive EV for the player in that game (not for subsequent games obviously).
My question is twofold.
1) What is the likelihood of a deal never presenting an opportunity for the player to make this +EV play? (ie: how many deals will never at any time contain more black cards than red cards whilst being dealt).
Since answer 1 must be below 50%, this would indicate that the advantage gained from the strategy would be offset by the disadvantage incurred when there are only ever more red cards dealt before black cards.
2) Therefore, as a casino, what would your variance look like over say 100 deals? How would you calculate the standard deviation here from the long-term EV which would be +5% (since that is the rake we charge for play)? What would make a sufficiently large sample of plays for the casino to near the +5% EV (considering the law of large numbers)?
AI: The number of paths from $(0,0)$ to $(n,n)$ which never go above the diagonal is the Catalan number $\frac{1}{n+1}\binom{2n}{n}$. For the probability that Black is never ahead, let $n=26$, and divide the Catalan number by $\binom{2n}{n}$. So the probability that Black is never ahead is $\frac{1}{27}$.
For the variance, we have here a Bernoulli trial with probability $p$ of success equal to $\frac{1}{27}$. The variance for any one trial is $p(1-p)$, and the variance over $n$ trials is $np(1-p)$.
This class of problems (often called Ballot Problems, a subclass of Random Walks) has been extensively studied, so there is a great deal of information available. |
H: Intersection between a cylinder and an axis-aligned bounding box
Given a 3D axis-aligned bounding box (represented as its minimum point and maximum point) and a 3D cylinder of infinite length, what's the best way to test for intersection?
AI: Suppose you are given a box defined by
$$x_{min}\leq x\leq x_{max},y_{min}\leq y\leq y_{max},z_{min}\leq z\leq z_{max}$$
and a cylinder of radius $r$ along the vector $\vec v=(v_1,v_2,v_3)^T$ with its center passing through $\vec p=(p_1,p_2,p_3)^T$. First construct the projection matrix $P$ which maps vectors onto their projection in the plane perpendicular to $\vec v$. I assume $v_1\neq 0$, otherwise the matrix will have to be constructed slightly differently. We have
$$P=\begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
\end{pmatrix}\begin{pmatrix}
v_1 & 0 & 0\\
v_2 & 1 & 0\\
v_3 & 0 & 1\\
\end{pmatrix}^{-1}=\begin{pmatrix}
\frac{-v_2}{v_1} & 1 & 0\\
\frac{-v_3}{v_1} & 0 & 1\\
\end{pmatrix}$$
and thus a vector $\vec x$ lies in the cylinder iff $\|P(\vec x-\vec p)\|\leq r$. Note that the projection of the box is a convex set and is equal to the convex hull of the images of its vertices, and that either the image of center of the cylinder is contained in the image of the box or the boundary of the image of the box intersects the image of the cylinder.
In the first case, we have that the line $\vec p + t\vec v$ intersects one of the three planes $x=x_{min},y=y_{min},z=z_{min}$ within the boundary of the box. The points of intersection are given by $$\begin{pmatrix}
x_{min}\\
p_2+\frac{x_{min}-p_1}{v_1}v_2\\
p_3+\frac{x_{min}-p_1}{v_1}v_3\\
\end{pmatrix}, \begin{pmatrix}
p_1+\frac{y_{min}-p_2}{v_2}v_1\\
y_{min}\\
p_3+\frac{y_{min}-p_2}{v_2}v_3\\
\end{pmatrix}, \begin{pmatrix}
p_1+\frac{z_{min}-p_3}{v_3}v_1\\
p_2+\frac{z_{min}-p_3}{v_3}v_2\\
z_{min}\\
\end{pmatrix}$$
and are nonexistant if $v_1,v_2$ or $v_3$ respectively are $0$. It is easy to check whether these are contained in the box. If any of them are, the intersection is nonempty.
In the second case, let $\vec x_1+t\vec y_1,\ldots,\vec x_{12}+t\vec y_{12}$ be the lines which form the edges of the box (and are exactly the edges when $0\leq t\leq 1$), which are easy to determine from the vertices. We want to know whether there is a solution to $\|P(\vec x_i+t\vec y_i-\vec p)\|\leq r$ for any $i$ with $0\leq t\leq 1$. Observe that $\|P(\vec x_i+t\vec y_i-\vec p)\|^2-r^2$ is a quadratic equation in $t$, so we can differentiate it to get a linear equation and solve for $0$ to get a value $t_i$. We then check for $i=1,\ldots,12$ whether $\|P(\vec x_i+t_i\vec y_i-\vec p)\|\leq r$ and $0\leq t_i\leq 1$. If this holds for some $i$, then the intersection is nonempty. Otherwise we test the endpoints $t=0$ and $t=1$ for each $i$ to see whether any of them satisfy $\|P(\vec x_i+t\vec y_i-\vec p)\|\leq r$. If so, the intersection is nonempty.
Finally, if all the above tests have failed then the intersection is empty.
In general, problems like this are studied in the field of Real Semialgebraic Geometry. More complicated questions of this form are approached by a method known as Cylindrical Algebraic Decomposition or CAD, which is similar to the method I employed but would have reduced the number of needed calculations by determining the six lines which bound the image rather than using all twelve lines that bound the box, but is more difficult to grasp. |
H: Find a single-valued analytic branch of $\sqrt{z^2-1}$ in $\mathbb{C} \backslash [-1,1]$.
I have the following question:
Show there is a single-valued analytic branch $f(z)$ for $\sqrt{z^2-1}$ in $\mathbb{C}\backslash [-1,1]$ such that $f(x) < 0$ for $x>1$. Here $[-1,1]$ denotes a closed interval in $\mathbb{R}$.
The following solution mimics a derivation from Moore and Hadlock's text.
Note that $z^2-1 = (z-1)(z+1)$. Consider the following four functions:
\begin{align*}
w_1(z) &= \sqrt{|z-1|} e^{(i\operatorname{arg_1}(z-1))/2} &\text{ where } \operatorname{arg_1}(z-1) \in [0,2\pi) \\
w_2(z) &= \sqrt{|z+1|} e^{(i\operatorname{arg_1}(z+1))/2} &\text{ where } \operatorname{arg_1}(z+1) \in [0, 2\pi) \\
w_3(z) &= \sqrt{|z-1|} e^{(i\operatorname{arg_2}(z-1))/2} &\text{ where } \operatorname{arg_2}(z-1) \in [-\pi,\pi) \\
w_4(z) &= \sqrt{|z+1|} e^{(i\operatorname{arg_2}(z+1))/2} &\text{ where } \operatorname{arg_2}(z+1) \in [-\pi, \pi)
\end{align*}
Note that
$w_1$ is analytic on the set $\{ z \in \mathbb{C}: \operatorname{arg}(z-1) \neq 0\} = \mathbb{C} \backslash [1, \infty)$;
$w_2$ is analytic on the set $\{ z \in \mathbb{C}: \operatorname{arg}(z+1) \neq 0\} = \mathbb{C} \backslash [-1, \infty)$;
$w_3$ is analytic on the set $\{ z \in \mathbb{C}: \operatorname{arg}(z-1) \neq \pi\} = \mathbb{C} \backslash (-\infty, 1]$; and
$w_4$ is analytic on the set $\{ z \in \mathbb{C}: \operatorname{arg}(z+1) \neq \pi \} = \mathbb{C} \backslash (-\infty, -1]$.
Therefore $w_1w_2$ is analytic on the set $\mathbb{C} \backslash [-1, \infty)$ and $w_3w_4$ is analytic on the set $\mathbb{C} \backslash (-\infty, 1]$.
Furthermore, on the set $\mathbb{C} \backslash (-\infty, 1]$, we have $w_1w_2 = w_3w_4$, as we see below:
If $x > 1$, then $\operatorname{arg_1}(x \pm 1) = \operatorname{arg_2}(x \pm 1) = 0$, and thus
$$w_1w_2(x)= w_3w_4(x) = \sqrt{x^2-1} \cdot e^{i \cdot 0} = \sqrt{x^2-1}. $$
If $\operatorname{Im}(z) > 0$, then $w_1w_2(z) = w_3w_4(z)$ since $\operatorname{arg}(z-1), \operatorname{arg}(z+1) \in (0, \pi)$.
If $\operatorname{Im}(z) < 0$, then $\operatorname{arg_1}(z\pm 1) = 2\pi + \operatorname{arg_2}(z\pm 1)$. Then
$$w_1w_2(z) = \sqrt{|z^2-1|} e^{i(\operatorname{arg_1}(z-1) +\operatorname{arg_1}(z-1))/2} = \sqrt{|z^2-1|} e^{i(\operatorname{arg_2}(z-1) +\operatorname{arg_2}(z-1))/2 + 2\pi i } $$
$$ = \sqrt{|z^2-1|} e^{i(\operatorname{arg_2}(z-1) +\operatorname{arg_2}(z-1))/2} = w_3w_4(z).$$
This implies that $w_1w_2$ is analytic not only on $\mathbb{C} \backslash [-1, \infty)$, but on the larger set $\mathbb{C} \backslash [-1, 1]$. This is a single-valued analytic branch of $\sqrt{z^2-1}$. However, for $x >1$, we have $w_1w_2(x) >0$.
To address this, multiply $w_1w_2$ by $e^{i\pi}$. The resulting function is a single-valued analytic branch with the desired property.
I understand the computations here, but I'm trying to grasp the motivation behind the definitions, and see how it might generalize. Is there a way to attack general problems of this type, or a simple explanation of why these particular $w$ functions were chosen?
AI: EDIT:
A more straightforward solution is to note that $z^2 - 1$ is in $(-\infty,0]$ exactly when $z$ is in the interval $[-1,1]$ or the imaginary axis. If we use the principal branch of $\sqrt{}$, $\sqrt{z^2-1}$ will be analytic everywhere else. This would be positive on $[1,\infty)$, so instead we take $-\sqrt{z^2-1}$. To avoid having a branch cut on the imaginary axis, we switch to $+\sqrt{z^2-1}$ in the left half plane. Thus
$$f(z) = \cases{-\sqrt{z^2-1} & for $\text{Re}(z) \ge 0$\cr
+\sqrt{z^2-1} & for $\text{Re}(z) < 0$\cr}$$ |
H: Prove if $n$ has a primitive root, then it has exactly $\phi(\phi(n))$ of them
Prove if $n$ has a primitive root, then it has exactly $\phi(\phi(n))$ of them.
Let $a$ be the primitive root then I know other primitive roots will be among $\{a,a^2,a^3 \cdots\cdots a^{\phi(n)} \}$ because any other number will be congruent modulo $n$ to one of these.
Then I figured as the answer is $\phi(\phi(n))$ so for $a^k$ to be primitive root, $k$ must be coprime with $\phi(n)$.But I don't know the reason.Am I missing any silly point?
AI: Hint: If $G$ is a cyclic group of order $n$ generated by an element $g$, then the order of the element $g^m$ is $n/\gcd(m,n)$ (you should have seen this result shortly after cyclic groups were introduced). In particular $g^m$ is of order $n$, if and only if $\gcd(m,n)=1$. |
H: Is '$10$' a magical number or I am missing something?
It's a hilarious witty joke that points out how every base is '$10$' in its base. Like,
\begin{align}
2 &= 10\ \text{(base 2)} \\
8 &= 10\ \text{(base 8)}
\end{align}
My question is if whoever invented the decimal system had chosen $9$ numbers or $11$, or whatever, would this still be applicable? I am confused - Is $10$ a special number which we had chosen several centuries ago or am I missing a point?
AI: Short answer: your confusion about whether ten is special may come from reading aloud "Every base is base 10" as "Every base is base ten" — this is wrong; not every base is base ten, only base ten is base ten. It is a joke that works better in writing. If you want to read it aloud, you should read it as "Every base is base one-zero".
You must distinguish between numbers and representations. A pile of rocks has some number of rocks; this number does not depend on what base you use. A representation is a string of symbols, like "10", and depends on the base. There are "four" rocks in the cartoon, whatever the base may be. (Well, the word "four" may vary with language, but the number is the same.) But the representation of this number "four" may be "4" or "10" or "11" or "100" depending on what base is used.
The number "ten" — the number of dots in ".........." — is not mathematically special. In different bases it has different representations: in base ten it is "10", in base six it is "14", etc.
The representation "10" (one-zero) is special: whatever your base is, this representation denotes that number. For base $b$, the representation "10" means $1\times b + 0 = b$.
When we consider the base ten that we normally use, then "ten" is by definition the base for this particular representation, so it is in that sense "special" for this representation. But this is only an artefact of the base ten representation. If we were using the base six representation, then the representation "10" would correspond to the number six, so six would be special in that sense, for that representation. |
H: Centroid for a rectangular section inclined at an angle theta
How to find Centroid for a rectangular section inclined at an angle theta? Is there any general formula available?
AI: The location of the centroid of a body doesn't change with rotation . If you are working in a coordinate system, then only the coordinates of the centroid change (not with respect to the body but with respect to axes). Thus,in a rectangle centroid $(x,y)$ is intersection of two diagonals and if the rectangle is rotated by $\theta$ then new coordinates$(x',y')$ are given by system of equations $$x'=x\cos\theta-y\sin\theta$$ and $$y'=x\sin\theta+y\cos\theta$$ |
H: Is there a standard operation to "rotate rings on matrices"?
Is there some standard operation to "rotate rings on matrices"? Look at the image below:
The numbers around the four empty squares are what I'm calling ring, In the second matrix, this ring has been rotated counterclockwise. I'm aware that ring may not be the right name.
AI: Use the vector representation $\mathrm{vec}$ of your matrix $X$, by stacking all columns on top of each other:
$$
\mathrm{vec}\:
\pmatrix{
1 &4&2&6\\
7&\square&\square&2 \\
7&\square&\square&4 \\
8&2&5&6\\
}
=\pmatrix{
1 &4&2&6&
7&\square&\square&2 &
7&\square&\square&4 &
8&2&5&6\\
}^T.
$$
Now apply $\pi_{\text{rot.ring}}$, a permutation (matrix $M$, with dimension $4^2$) on $\mathrm{vec}\;X$, such that $\left(\mathrm{vec}\;X\right)_1$ is sent to $\left(\mathrm{vec}\;X\right)_5$, $\left(\mathrm{vec}\;X\right)_2$ to $\left(\mathrm{vec}\;X\right)_1$,...
So $\pi_{\text{rot.ring}}=\left(5,1,2,3,4,8,12,16,15,14,13,9\right)$ and $M_{j,j+1}=\left(\pi_{\text{rot.ring}}\right)_{j,j+1}$ and $M_{jj}=1$ if $j\in\{6,7,10,11\}$.
Undo the $\mathrm{vec}$ operation and you'll get a matrix with your ring rotated. |
H: Unstable linear inverse problem: which "dampening" Tikhonov matrix should I use?
A linear inverse problem is given by:
$\ \mathbf{d}=\mathbf{A}\mathbf{m}+\mathbf{e}$
where d: observed data, A: theory operator, m: unknown model and e: error.
The Least Square Error (LSE) model estimate is given by:
$\ \mathbf{\tilde{m}}=(\mathbf{A^\top A})^{-1}\mathbf{A^\top d}$
Typically the linear equation is ill-conditioned and as a consequence the model estimate tends to oscillate.
The LSE solution to the linear equation is equivalent to minimizing the following cost function:
$\ C=\|\mathbf{d-Am}\|^2$
In order to stabilize the solution we may introduce a "dampening" term:
$\ C=\|\mathbf{d-Am}\|^2 + \|\mathbf{\Gamma m}\|^2$
where $\ \Gamma$ is called the Tikhonov matrix.
The Tikhonov matrix should penalize unwanted aspect of a solution.
The stabilized LSE model estimate then becomes:
$\ \mathbf{\tilde{m}}=(\mathbf{A^\top A+\Gamma^\top \Gamma})^{-1}\mathbf{A^\top d}$
As an example problem say I have the model $\ m(x)=x^2, 0 \leq x \leq 1 $.
Instead of observing the model directly I am observing the derivative of the model + some noise (n):
$\ d(x)=\frac{dm}{dx}+e$
From measurements of d(x) I want to estimate m(x). Say I know that m(0)=0.
My linear inverse problem is now:
$\ \mathbf{d}=\mathbf{A}\mathbf{m}+\mathbf{e}, \mathbf{A=D_x}$
where $\ \mathbf{D_x}$ is the differential operator (matrix) for which I use central differences.
The LSE solution to this problem oscillates a lot:
Therefore I want to dampen it by using an appropriate Tikhonov matrix.
Which Tikhonov matrix should I use?
AI: Try $\Gamma = \alpha \Delta$, where $\Delta$ is a discretized version of the Laplace operator (ie, a second difference operator), and $\alpha$ is a scalar regularization parameter. The action of the Laplacean is to amplify high frequency modes, so such regularization will smooth your solution. |
H: solving the congruence: $5^n\equiv3^n+2 \pmod{11}.$
I'd really like your help with solving the following congruence:
$$5^n\equiv3^n+2 \pmod{11}.$$
I don't know with what to start.
Any help?
Thanks
AI: Just note that 11 is prime so you only need to consider n modulo 10 here. Just list $5^n$ and $3^n$ modulo 11 for n going from 0 to 9:
$n\,\;\quad 0\quad 1\quad 2\quad 3\quad 4\quad 5\quad 6\quad 7\quad 8\quad 9$
$5^n\quad 1\quad 5\quad 3\quad 4\quad 9\quad 1\quad 5\quad 3\quad 4\quad 9$
$3^n\quad 1\quad 3\quad 9\quad 5\quad 4\quad 1\quad 3\quad 9\quad 5\quad 4$
Easy to see that this only happens at $n$ being $1$ or $6$ modulo $10$. So in other words we want $n$ to be $1$ modulo $5$. |
H: How do I have to interpret $\mathrm{nm}^2$?
My short question: How many $\mathrm{m}^2$ are $\mathrm{nm}^2$? Do I have to interpret it as $\mathrm{nm}^2=(\mathrm{nm})^2=(10^{-9}\mathrm{m})^2=10^{-18}\mathrm{m}^2$ or shall it be $\mathrm{nm}^2=\mathrm{n}(\mathrm{m}^2) = 10^{-9} \mathrm{m}^2$? What is the right convention?
AI: $\rm nm^2=(nm)^2\neq n(m^2)$, therefore, $\rm nm^2=10^{-18}m^2$. Here, $\rm nm$ represents nanometer and $\rm m$ meter. |
H: Proving that an expression divides a number
How do you prove that
$$n(n+1)(n+2)$$
is divisible by 6 by using the method of mathematical induction?
According to my book
$$\begin{aligned}
(n+1)(n+2)(n+3) &= n(n+1)(n+2)+3(n+1)(n+2)\\
&= 6k + 3*2k'\\
&= 6(k+k')\\
&=6k''
\end{aligned}$$
But I wonder, where does that k come from anyway?
AI: The $k$ comes from the induction hypothesis: you are supposed to prove that $(n+1)(n+2)(n+3)$ is divisible by 6 assuming that $n(n+1)(n+2)$ is divisible by 6.
That's from where you get an integer $k$ for which $$n(n+1)(n+2) = 6k.$$
The $k'$ comes from the observation that $(n+1)(n+2)$ is the product of two subsequent numbers. One of which has to be even, so the product is even. |
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