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H: Approximating Coins Flips Problem
Approximate the probability of getting 500 heads out of a 1000 coin
flip of unbiased coins to be within 5% of its true value (without the use of a calculator).
I know that an exact probability is $$\binom{1000}{500}(.5)^{1000} = .02522...$$
I am unsure how one could simplify this problem through estimation to get an approximate answer however.
Thanks for any help.
AI: To avoid a calculator, you certainly need Stirling's approximation for the factorials. So $P=\binom {1000}{500}2^{-1000} \approx \frac {1000^{1000}\exp(500)\exp(500)}{500^{500}500^{500}\exp(1000)}\frac {\sqrt{2\pi 1000}}{\sqrt{2\pi 500}\sqrt{2 \pi 500}}2^{-1000}=\frac 1{\sqrt{\pi 500}}\approx \frac 1{\sqrt{1550}} \approx \frac 1{40}=0.025$
Stirling's approximation is within a factor $\frac 1{12n}$, so the error is negligible. Using $\pi \approx 3.1$ is within 2%. The last we were within 4% under the square root sign, so the root is within 2%, which means the calculation is within 4%. In fact this is within 1% of your exact value. |
H: Change to standard basis
Let $A=\{a_1, a_2, a_3\}$ be a basis, of which each vector is aligned to a cartesian axis. Given a vector $v_{\langle A\rangle}$, how can I get transform it to the standard, canonical basis so it becomes $v_{\langle I\rangle}$?
I must apologize if this question is silly, I've completely forgotten my linear algebra classes (which makes me want to take them again).
AI: First express the basis vectors $\{a_1 , a_2, a_3\}$ in the standard basis:
$$ a_{1, \langle I \rangle} = \pmatrix{a_{1,1} \\ a_{1,2} \\ a_{1,3} } \\
a_{2, \langle I \rangle} = \pmatrix{a_{2,1} \\ a_{2,2} \\ a_{2,3} } \\
a_{3, \langle I \rangle} = \pmatrix{a_{3,1} \\ a_{3,2} \\ a_{3,3} } \\$$
Form the matrix
$$ A = \pmatrix{a_{1,1} & a_{2,1} & a_{3,1} \\ a_{1,2} & a_{2,2} & a_{3,2} \\ a_{1,3} & a_{2,3} & a_{3,3}} $$
Now, given any $v_{\langle A \rangle} = (v_1 , v_2, v_3),$ we have
$$v_{\langle I \rangle} = A \pmatrix{v_1 \\ v_2 \\v_3}. $$
TL;DR $v_{\langle I \rangle}= (v_1, v_2, v_3)_{\langle A \rangle} = v_1 a_{1, \langle I \rangle} + v_2 a_{2, \langle I \rangle} + v_3 a_{3, \langle I \rangle}.$ |
H: Some doubt on Linear Diophantine equation
We know $ax+by=c$ is solvable iff $(a,b)|c$ where $a,b,c,x,y$ are integers.
If $x=2$, $a=\dfrac{k(k+5)}{2}$, $y=k$, $b=k+3$ and $c=2k$, where $k$ is any integer, then
$$2 \frac{k(k+5)}{2} - k(k+3) = 2k.$$
So, $\left(\dfrac{k(k+5)}{2}, (k+3)\right) | 2$ for all integral value of $k$.
But $k+3$ cannot be even unless $k$ is odd.
Where I am going wrong?
AI: $2k(k+5)/2 - k(k+3)=2k$ with $a=k(k+5)/2$, $x=k$, $y=k$, would mean $b=k+3$, not $k(k+3)$.
The conclusion is that $(k(k+5)/2,k+3) | 2k$, not $2$. For example, with $k=3$, $(24,6)=6$. |
H: Question about inner product and Cauchy Schwarz inequality
I'm wondering where the following equality came from: $$ \langle x , y \rangle = \|x \|Â \| y \| \cos \theta$$
where the thing on the LHS is the inner product and $\|\cdot\|$ is the norm induced by $\langle \cdot, \cdot \rangle$. Do we need the Cauchy Schwarz inequality to prove this? I'm asking because I'm reading my notes and there is a proof of the C. S. - inequality. It's fairly short but longer than the following:
Claim: $|\langle x,y \rangle |Â \leq \|x \| \|y \|$
Proof: Since $\langle x , y \rangle = \|x \|Â \| y \| \cos \theta$ we have $-\|x \|Â \| y \| \leq \langle x , y \rangle \leq \|x \|Â \| y \|$.
Thanks.
AI: My favorite interpretation of the identity
$$
\langle x,y \rangle = \|x\| \|y\| \cos \theta \tag{1}
$$
is that (1) defines the angle $\theta \in [-\pi/2,\pi/2]$ between the vectors $x$ and $y$. Since
$$
\langle x,y \rangle =\sum_{j=1}^n x_j y_j
$$
is the standard inner product in $\mathbb{R}^n$, the elementary Cauchy-Schwarz inequality implies that
$$
-1 \leq \frac{\langle x,y \rangle}{\|x\| \|y\|} \leq 1,
$$
and therefore there exists a unique angle $\theta \in \left[ -\frac{\pi}{2},\frac{\pi}{2} \right]$ such that (1) holds.
However, once you have fixed $x$ and $y$, the question becomes two-dimensional in the plane spanned by $x$ and $y$. There you can apply stadard elementary geometry to show that the cosine of the angle between the two vectors is given by (1). |
H: Conditions for two normal R.Vs to satisfy bivariate normal distribution
I have to Normally distributed Random Variables X and Y which are correlated. What conditions should they satisfy so that their joint distribution is a bivariate Normal distribution?
AI: A necessary and sufficient condition is that, for every $(a,b)$, the linear combination $aX+bY$ is normal. (Recall that, by convention, every Dirac distribution is normal, with variance zero.) |
H: Germs of continuous functions
Let $R$ be the ring of germs of continuous functions $\mathbb R \rightarrow \mathbb R$.
It is clear that this is a local ring with maximal ideal $\mathcal m$ consisting of those germs $f$ with $f(0)=0$.
What is not clear to me is why $(\mathcal m)^2 =\mathcal m$ should hold.
Perhaps someone can give me a small hint how to show this.
AI: For $f\in \mathfrak m$ we have $f=\sqrt [3]f\cdot (\sqrt [3]f)^2$ and $\sqrt [3]f, (\sqrt [3]f)^2\in \mathfrak m$, so that indeed $\mathfrak m\subset \mathfrak m^2$.
The other inclusion is trivial and we have proved that $$\mathfrak m^2 = \mathfrak m $$ Edit: warning !
The analogue of the above result is false if one replaces the ring of germs of continuous functions by the ring of germs $\mathcal C_0^k\; (1\leq k\leq \infty)$ of $k$ times continuously differentiable functions at the origin of $\mathbb R$.
Indeed if $f\in \mathfrak m^2$, we can write $f=\sum g_kh_k\; (g_k, h_k\in \mathfrak m)$ and differentiating yields $f'(0)=\sum g_k'(0)h_k(0)+\sum g_k'(0)h_k(0)=0$, so that elements $f\in \mathfrak m^2$ satisfy $f'(0)=0$ .
In particular $x\notin \mathfrak m^2$, although $x\in \mathfrak m$.
Hence $$\mathfrak m^2 \subsetneq \mathfrak m$$ |
H: A trivial problem in calculus
Try to compute $$\int\frac{dx}{x\ln x}$$
I compute it this way: first we have $x>0$.
\begin{align*}
\int\frac{dx}{x\ln x}
&=\int\frac{d(\ln x)}{\ln x}\\
&=\ln|\ln x|+C
\end{align*}
But the answer to the problem is $\ln\ln x+C$. Which one is right? Thanks!
Source Григорий Михайлович Фихтенгольц
AI: Both and none. Your solution is a primitive both on $(0,1)$ and on $(1,+\infty)$. The answer of the book is a primitive only on $(1,+\infty)$.
Therefore, I'd say that your answer is better and more complete. |
H: Direct limit and exact sequences of abelian groups
Suppose having a set of direct systems of abelian groups
$\ldots\{G_{\alpha}\}_{\alpha\in A}$, $\{G_{\beta}\}_{\beta\in B}$, $\{G_{\gamma}\}_{\gamma\in \Gamma}\ldots$
If there is a (long) exact sequence for certain indexes:
$$\cdots\longrightarrow G_\alpha\longrightarrow G_\beta\longrightarrow G_{\gamma}\longrightarrow\cdots\ $$
Then, is the following sequence exact?
$$\cdots\longrightarrow \varinjlim G_\alpha\longrightarrow \varinjlim G_\beta\longrightarrow\varinjlim G_{\gamma}\longrightarrow\cdots\ $$
AI: You want this long exact sequence to be compatible with the direct systems; but this cannot be formulated since the index sets are different. In fact, there is no way to define a map $\varinjlim G_{\alpha} \to \varinjlim G_{\beta}$ in your setting.
Probably you would like to know the correct statement: If $I$ is a directed poset and for every $n \in \mathbb{N}$ we have a diagram $G^n : I \to \mathsf{Ab}$, and morphisms of diagrams $G^n \to G^{n+1}$ for all $n$ (i.e. natural transformations), such that for all $i$ and $n$ the induced sequence $G^{n-1}_i \to G^n_i \to G^{n+1}_i$ is exact. Then the induced sequence $\dotsc \to \varinjlim G^n \to \varinjlim G^{n+1} \to \dotsc$ is exact. |
H: Prove that a set defined by concave functions on $R^n$ is convex
I've been trying to prove this statement the whole weekend...
Let $g_1,\dots,g_m$ be concave functions on $\mathbb{R}^n$. Prove that the set $S=\{x:g_i(x)\geq{0},\ i=1,\dots,m\}$ is convex.
AI: Take $x,y\in S$ and $a\in [0,1]$ and $i\in\{1,\dots,m\}$. Then by concavity of $g_i$,
$$g_i(ax+(1-a)y)\geq ag_i(x)+(1-a)g_i(y).$$
This quantity is non-negative, because so are $a$, $1-a$, $g_i(x)$ and $g_i(y)$.
We conclude that $ax+(1-a)y\in S$, hence $S$ is convex. |
H: How to prove $f(x,y)=\sqrt {xy}$ is concave?
How can I prove (preferably elegantly) that $f(x,y)=\sqrt {xy}$ where $x≥0$ and $y≥0$ is concave in both $x$ and $y$?
AI: Take $ (x_0, y_0)$ and $ (x_1,y_1)$ and then you need to show
$ f( t( x_0 , y_0) + (1-t) (x_1 , y_1) ) \geq t f(( x_0 , y_0)) +(1-t) f((x_1 , y_1) )$
which by squaring it becomes
$$ (t x_0 +(1-t)x_1)( t y_0 +(1-t)y_1) \geq ( t \sqrt{ x_0 y_0} + (1-t) \sqrt{x_1 y_1 })^2$$
which hold trivially by Cauchy-Schwarz. |
H: Complex Numbers as a group under multiplication
While revising I came across this question:
Let $G$ be the group of complex numbers $\{1, i, -1, -i\}$ under multiplication.
Is it true that for every such homomorphism, there is an integer $k$ such that the homomorphism has the form $z\mapsto z^k$ ?
The answer is true, but how does one prove it?
I thought of considering four cases of $f(i)=1,i,-1,-i$, since $i$ is a generator.
Sincere thanks for any help.
AI: I'm assuming the question means: let $f:G\to G$ be an endomorphism. Prove that $f(z)=z^k$ for some $k\in\mathbb{Z}$.
This is trivial. Since $G$ is a cyclic group (with $i$ as a generator, for example), the homomorphism is completely defined by what the generator is sent to (since all elements are some power of the generator). But whatever it is sent to, it is sent to an element of $w\in G$, but all of those can be written as some power of $i$ (that is, $w=i^k$ for some $k\in\mathbb{Z}$). This directly implies that $f(z)=z^k$: let $z\in G$, then $z=i^n$ for some $n\in\mathbb{Z}$, and then $f(z)=f(i^n)=f(i)^n$ since $f$ is a homomorphism, and then $f(i)^n=w^n=i^{kn}=z^k$ by definition.
If $f$ were required to be an automorphism (ie. bijective), then necessarily we would have $k$ odd. |
H: Percentage increase in ratios..
The question is:
Seats for Math,Physics and biology in a school are in ratio 5:7:8. There is a proposal to increase these seats by 40%,50% and 75% respectively.What will be ratio of increased seats?
Apparently I am currently increasing 5 by 40% and 7 by 50% and 8 by 75% . But this is not giving the correct answer. Any suggestions what should be done here ??
AI: Let number of seats be $5x,7x$ and $8x$. Number of seats after increment =$7x,21x/2,14x$. Therefore the new ratio is $14:21:28=2:3:4$.isn't this the answer? |
H: How do I solve this pde (using finite differences)?
How do I solve the pde:
$\ -s_x(x,t) -p(x,t)s_t(x,t)=p(x,t)$
for s(x,t) when p(x,t)=2x,
subject to the condition s(0,t)=0?
Generally p(x,t) may not be analytical so I would like to use finite differences to transform this into a linear algebra problem:
$\ (-D_x-pD_t)s=p $
But how do I incorporate my "boundary" condition s(0,t)=0 into this?
Thanks in advance for any answers!
AI: To find the solution you will also need an initial condition like $s(x,0)=\phi(x)$, $x\ge0$. Choose $\Delta x>0$ and $\Delta t>0$, and let $s_{j,k}\approx s(j\,\Delta x,k\,\Delta t)$, $p_{j,k}=p(j\,\Delta x,k\,\Delta t)$ for $j,k\ge0$. The initial condition is then
$$s_{j,0}=\phi(j\,\Delta x),\quad j\ge0,$$
and the boundary condition
$$s_{0,k}=0,\quad k\ge0.$$
The simplest finite difference approximation is
$$
-\frac{s_{j+1,k}-s_{j,k}}{\Delta x}-p_{j,k}\,\frac{s_{j,k+1}-s_{j,k}}{\Delta t}=p_{j,k},\quad j,k\ge0.
$$
If $p_{j,k}\ne0$for all $j,k\ge0$, then
$$
s_{j,k+1}=\frac{1}{p_{j,k}}\frac{\Delta t}{\Delta x}(s_{j,k}-s_{j+1,k})-\Delta t,\quad j,k\ge0.
$$ |
H: Proof that convex open sets in $\mathbb{R}^n$ are homeomorphic?
This is an exercise from Kelley's book. Could someone help to show me a proof?
It seems very natural, and it is easy to prove by utilizing the arctan function in $\mathbb{R}^1$.
Thanks a lot.
AI: 3rd edit: I have now typed things up in a slightly more streamlined way:
http://relaunch.hcm.uni-bonn.de/fileadmin/geschke/papers/ConvexOpen.pdf
Old answer:
How about this: Call a set $U\subseteq\mathbb R^n$ star-shaped
if there is a point $x\in U$ such that for all lines $g$ through
$x$, $g\cap U$ is a connected open line segment.
Every convex set is star-shaped.
Translating $U$ if necessary we may assume that $x=0$.
Now use arctan to map $U$ homeomorphically to a bounded set.
This set is still star-shaped around $x=0$.
Now scale each ray from $0$ to the boundary of the open
set appropriately, obtaining a homeomorphism from the open set onto
the open unit ball.
Edit: I am very sorry that I didn't reply to the comments earlier.
It seems that my answer was too sketchy. I think we agree that we can assume that the origin is an element of the original open set $U$.
As Henning Makholm points out, we use arctan applied to the radius in Polar coordinates to map our potentially unbounded set $U$ to a bounded open set $V$.
Now we show that in any direction the distance of the boundary of $V$
from the origin depends continuously on that direction.
Let $v$ be a direction, i.e., let $v$ be a vector in $\mathbb R^n$ of length $1$.
First case: the ray from the origin in the direction of $v$ is contained in $U$.
In this case $U$ is unbounded in the direction of $v$.
Since $U$ is open, there is $\varepsilon>0$ such that the $\varepsilon$-ball around the origin is still contained in $U$. Since $U$ is convex, the convex hull of the union of the ray in the direction of $v$ and the open $\varepsilon$-ball is also contained in $U$. Let us call this convex open set $C$.
For every $N>0$ the set of elements of $C$ that have distance at least $N$ from the origin is open. It follows that the set of directions $v'$ such that the
ray in direction $v'$ has elements of $C$ of distance at least $N$ from the origin is open.
But this implies that the map assigning to each direction $v$ the distance of the boundary of the transformed set $V$ to the origin in that direction is actually continuous in all the directions in which the original set $U$ is unbounded.
Second case: the ray from the origin in the direction $v$ is not contained in
$U$, i.e., $U$ is bounded in the direction of $v$.
We have to show that the distance of the boundary of the transformed set
$V$ in some direction $v'$ is continuous in $v$.
But since arctan is continuous, it is enough to show the same statement for
the distance to the boundary of the original set $U$.
So, let $d$ be the distance of the boundary of $U$ from the origin in the direction of $v$.
Let $\varepsilon>0$. Choose $x\in U$ in the direction $v$ such that
$d-|x|<\varepsilon/2$. For some $\delta>0$, the $\delta$-ball around $x$ is contained in $U$. The set of directions $v'$ such that the ray in that direction passed through the open ball of radius $\delta$ around $x$ is an open
set of directions containing $v$. It follows that on an open set of directions
containing $v$ the distance of the boundary of $U$ is at least $d-\varepsilon$.
Now let $x$ be a point on the ray in the direction of $v$ with $|x|>d$.
If we can show that $x$ is not in the closure of $U$, then there is an open ball around $x$ that is disjoint from $U$ and we see that there an open set of directions that contains $v$ such that for all directions $v'$ in that set
the distance of the boundary in direction $v'$ from the origin is at most
$|x|$. This shows that the map assigning the distance of the boundary of $U$ to
the direction is continuous in $v$ and we are done.
So it remains to show that $x$ is not in the closure of $U$.
We assume it is.
Let $y$ be the point on the ray in direction $v$ that satisfies $|y|=d$.
$y$ is on the boundary of $U$. Then $|y|<|x|$.
Let $B$ be an open ball around the origin contained in $U$.
We may assume that $x\not in B$.
Now I am waving my hands a bit, but I think this should be clear:
There is some $\varepsilon>0$ such that when $|x-z|<\varepsilon$,
then $y$ is in the convex hull $C_z$ of the union of $B$ and $\{z\}$.
Now choose $z\in U$ such that $|x-z|<\varepsilon$.
Now $y\in C_z$ and by the convexity of $U$, $C_z\subseteq U$.
But $C_z\setminus\{z\}$ is an open neighborhood of $y$.
This shows that $y$ is not on the boundary of $U$, a contradiction.
This finishes the proof of "continuity of scaling". I hope this was comprehensible.
2nd edit: It doesn't seem clear why the map $d$ assigning to each direction $v$ the distance of the boundary of $V$ from the origin in direction $v$ is well defined. For $v\in\mathbb R^n$ of length $1$ (i.e., $v$ a direction) let
$$d(v)=\inf\{|x|:x\in\mathbb R^n\setminus V\wedge x/|x|=v\}.$$
Since $V$ is bounded, this is well defined. I take that as the definition of
"the distance from the origin of the boundary of $V$ in the direction of $v$". In the "so it remains to show"-paragraph above it is shown that no point $x$ in the direction of $v$ with $d(v)<|x|$ is in the closure of $V$.
(Strictly speaking, I prove this for $U$ instead of $V$, and only in the case that the ray in direction $v$ is not contained in $U$. But if the ray is not contained in $U$, the arctan transformation preserves the property that we have shown for $U$. If the ray is contained in $U$, then $d(v)$ is $\pi/2$ and
the point $\pi v/2$ is the unique point on the boundary of $V$ in the direction of $v$.
The problem with $U$ is that it is unbounded and hence we might be taking the infimum over the empty set. In this case the distance would be $\infty$.)
It follows that the unique point $y$ in the direction of $v$ with $|y|=d(v)$ is in the boundary of $V$. Hence every ray through the origin intersects the boundary of $V$ in exactly one point. |
H: How to prove with Galois theory that a cube root is not geometrically constructible?
I could not find any proof on the Internet. I am looking for a formal proof with an explanation for the uninitiated (my knowledge of Galois theory is very basic).
With geometrically constructible I mean with compass and straightedge.
AI: Since the minimal algebraic polynomial for the $2^{1/3}$ is $x^3-2=0$ and for $2^{1/3}$ to be constructible , the degree of this polynomial needs to be a power of $2$, but since $3$ is not a power of $2$, hence $2^{1/3}$ is not constructible. Read this http://www2.math.uu.se/~svante/papers/sjN8.pdf |
H: Restoring a point after transformation
I am given a point $
\begin{bmatrix}
u & v
\end{bmatrix}^T
$ which I know is in form $\begin{bmatrix}
\frac{x}{f(r)} & \frac{y}{f(r)}
\end{bmatrix}^T$ where $f(r)$ is polynomial function, $f(r)=a_nr^n + \cdots + a_1r + a_0$ and $r=\sqrt{x^2+y^2}$.
I want to restore $x$ and $y$ given $u$ and $v$. What I have done so far is below.
$$
u = \frac{x}{f(r)}\\
v = \frac{y}{f(r)}\\
$$
If we square both equations and sum them we get:
$$
\tag{1}
f(r) = \frac1{\sqrt{u^2+v^2}}r
$$
Which is:
$$
a_nr^n + \cdots + \left(a_1 - \frac1{\sqrt{u^2+v^2}}\right) r + a_0 = 0
$$
I can find the roots of a polynomial using roots() in MATLAB. Then using Equation 1 I can find the value of $f$ and then $u$ and $v$.
When I try this with a numerical example with known polynomial coefficients I cannot restore $x$ and $y$ given $u$ and $v$, the results I am getting do not match. Am I missing something here?
I can provide the coefficients and the numbers I am using if needed.
Update:
Here are my values which are not working. First polynomial coefficients $a_0, a_1, a_2$ are $-174.4486, 0, 0.0026$ respectively. Lets start with an original point $p = \begin{bmatrix} \frac{50}{-161.4486} & \frac{50}{-161.4486}\end{bmatrix}^T$. Observe that $p$ is in form described above where $f(r) = -161.4486$, you can calculate if you don't believe me :). Now my given point becomes $\begin{bmatrix}-0.3097 & -0.3097\end{bmatrix}^T$, using only this I want to find $x=50$ and $y=50$. Lets start. We calculate:
$$\frac1{\sqrt{u^2+v^2}} = 2.2832$$
We put it into the polynomial and find the roots which are $r_1= 948.875$ and $r_2=-70.7107$. Since we know $r$ is non negative we choose $r_1$, plug it into the Equation 1 and we get $f(r)=2166.5$ put it into the equations we get $-671$ for $x$ and $y$. Before calculating those values we can understand something is wrong just looking at $r$, it is not the correct value which is $70.7107$.
AI: (Edited)
The values of $u$ and $v$, and therefore $u^2+v^2$, are given to you, and you want to find $x$, $y$ such that
$$(u,v)=\Bigl({x\over f(r)}, {y\over f(r)}\Bigr)\ ,\qquad r:=\sqrt{x^2+y^2}\ .\qquad(2)$$
It follows that we necessarily have $$x=f(r)u\ , \quad y=f(r)v\qquad(3)$$ and therefore
$$r^2=(u^2+v^2)f^2(r)\ .\qquad(4)$$
This equation only involves given data and the unknown $r$. Solving it produces a list of values $r_k>0$ (and maybe some other solutions), and to each of these $r_k$ by $(3)$ correspond values
$$x_k= u f(r_k)\ ,\quad y_k=v f(r_k)\ .\qquad(5)$$
Now $(5)$ is just a necessary condition that solutions of the original equation $(2)$ would have to fulfill, and we have to prove that such pairs $(x_k,y_k)$ are in fact solutions of $(2)$, i.e., satisfy
$$(u,v)=\Bigl({x_k\over f\bigl(\sqrt{x_k^2+y_k^2}\bigr)},{y_k\over f\bigl(\sqrt{x_k^2+y_k^2}\bigr)}\Bigr)\ .$$ To this end we argue as follows: If $x_k$ and $y_k$ are given by $(5)$, where $r_k>0$ is a solution of $(4)$, then $$x_k^2+y_k^2=(u^2+v^2)f^2(r_k)=r_k^2\ .$$
As $r_k>0$ it follows that $\sqrt{x_k^2+y_k^2}=r_k$ and therefore
$${x_k\over f\bigl(\sqrt{x_k^2+y_k^2}\bigr)}={x_k\over f(r_k)}=u\ ,$$
and similarly for $y$ resp. $v$. |
H: Set Operations with Empty sets
I have a question regarding performing set operations on empty sets.
For example let A = ∅, Let B = {A, ∅}, Let C = {A, B}.
Would B = {∅, ∅} and C = {∅, {∅,∅}}? or would B = {∅} and C = {∅, {∅}} since {∅, ∅} reduces to {∅}?
Now if I wanted to do $A \cup C$ would the answer be $∅ \cup C$ -> C = {∅, {∅}}?
Now what if I wanted to do D = C - B, would the result be {{∅}}?
One last question. If P = {∅}, Q = {P}, R = {∅,P}, S = {∅, P, Q, R};
What would X = {$ x : (x ∈ R) \wedge ( x ⊆ S)$}?
My thoughts are that Q = {{∅}}, R = {∅, {{∅}}} and S = { ∅, {∅}, {{∅}}, {∅, {{∅}}} }
x ∈ R would mean x = ∅ and x = {{∅}}
x ⊆ S would mean x is a set containing any combination of ∅, {∅},
{{∅}}, {∅, {{∅}}} as members
Therefore X = {{{∅}}} as {{∅}} is the only member that fits both those conditions?
I've tried looking at some lectures online on set theory but I want to make sure my understanding is correct so far.
On a more general note, I am also wondering if lets say D = {1, 2} and E = {D, 5}. Would E = {1,2,5} or E = {{1,2},5}?
Thanks
AI: $B$ contains a single element which is an empty set (or, equivalently, is equal to $A$): $B = \{\emptyset\} = \{A\}$
$C$ contains two elements, one of which is an empty set (or, equivalently, is equal to $A$), and another is $B$: $C = \{\emptyset, \{\emptyset\}\}$.
$A \cup C$ contains all elements $A$ contains (there are no such elements, as $A$ is empty) and all elements $C$ contains: $A \cup C = C = \{\emptyset, \{\emptyset\}\}$.
$C = \{A\} \sqcup \{B\}$, and $B = \{A\}$, hence $C - B = (\{A\} \sqcup \{B\}) - \{A\} = \{B\} = \{\{A\}\} = \{\{\emptyset\}\}$.
As for the last question, $x \in R$ is equal to $x = \emptyset \vee x = P$ per $R$ definition. Empty set is obviously a subset of any set (including $S$); and $P \subseteq S$ because all elements $P$ contains (there is only one of them - an empty set) are also elements of $S$. So $X = R$.
On a more general note, I am also wondering if lets say D = {1, 2} and E = {D, 5}. Would E = {1,2,5} or E = {{1,2},5}?
$E$ is a two-element set (as long as $D$ is not equal to $5$): one element is $5$ and another is $D$. Thus, $E = \{\{1, 2\}, 5\}$. |
H: maximum possible value of $d$
We are given three integers $a , b,$ and $c$ such that $a , b, c,
a + b − c, a + c − b, b + c − a ,$ and $a + b + c$ are seven distinct primes. Let
$d$ be the difference between the largest and smallest of these seven primes.
Suppose that $800$ is an element in the set {$a + b, b + c, c + a $}. Determine
the maximum possible value of $d$ . What I did: $a,b,c\geq 3$ (odd primes, if $a$ is even,then $a+b-c$ is even too and hence are not distinct ) $\implies a+b+c$ is the largest prime and WLOG, let $a$ be the smallest one.Then $b+c\geq a+b$ and $a+c$ implying $b+c\geq 800$.Since, $b+c-a\geq 0 \implies a\leq (b+c\geq 800)$.Here I am stuck, how to bound $a$(i don't know if that's necessary) and get the upper bound for $d$. Am i on the right track??
AI: It is clear that without loss of generality we can choose $a+b=800$, and again, as you suggest, we can try to maximize $b+c-a$ while minimizing $a$.
For the well definition of the problem we need $a\geq |b-c|$ and $c\leq a+b=800$. The two biggest primes smaller than 800 are (you can check on this list) $787$ and $797$, and their difference is $10$. So we need $a$ to be at least $10$ and such that its sum with $787$ or $797$ is $800$. This clearly suggests the following choice:
$$a=13, \qquad b=787, \qquad c=797.$$
Which leads to:
$$a+b-c=3, \qquad a+c-b=23, \qquad b+c-a=1571.$$
Again using the list above you can check that we have been lucky and those are all distinct primes. Then we have actually computed (not only bounded) the maximum vale $d$, which is:
$d = b+c-a -a = 1571-13= 1558.$
The maximality follows by construction. |
H: $X = \log_{12} 18$ and $Y= \log_{24} 54$. Find $XY + 5(X - Y)$
$X = \log_{12} 18$ and $Y= \log_{24} 54$. Find $XY + 5(X - Y)$
I changed the bases to 10, then performed manual addition/multiplication but it didn't yield me any result except for long terms. Please show me the way.
All I'm getting is $$\frac{\lg 18\lg54 + 5 \lg18\lg24 - 5\lg54\lg12}{\lg12\lg24} $$
AI: Let $I=\dfrac{\log 18}{\log 12}\cdot\dfrac{\log 54}{\log 24}+5 \left( \dfrac{\log 18}{\log 12}-\dfrac{\log 54}{\log 24} \right)$. Also, let $\log 3=x $ and $\log 2=y$.
Then,
$$I= \frac{\log 3^2\cdot2}{\log 2^2\cdot 3}.\frac{\log 3^3 \cdot 2}{\log 2^3\cdot 3}+5\left(\frac{\log 3^2\cdot2}{\log 2^2\cdot3}-\frac{\log 3^3\cdot2}{\log 2^3\cdot3}\right)= \frac{2x+y}{2y+x}\cdot\frac{3x+y}{3y+x}+5 \left( \frac{2x+y}{2y+x}-\frac{3x+y}{3y+x} \right)$$
$$=\frac{6x^2+5xy+y^2+10x^2+35xy+15y^2-15x^2-35xy-10y^2}{(2y+x)(3y+x)}=\frac{x^2+5xy+6y^2}{x^2+5xy+6y^2}=1$$ |
H: While proving that every vector space has a basis, why are only finite linear combinations used in the proof?
Statement: Every vector space has a basis
Standard Proof:It is observed that a maximal linearly independent set is a basis.
Let $\mathscr{Y}$ be a chain of linearly independent subsets of a vector space $\mathscr{V}$. The union of such a set can serve as an upper bound for it.To apply Zorn's lemma,we have to check whether the union is linearly independent? Well, if $t_1,\dots,t_n$ belong to the union, then each $t_i$ belongs to some linearly independent set $L_i\in \mathscr{Y}$. Because $\mathscr{Y}$ is a chain, one of these sets $L_i$ contains all the others. If that is $L_j$, then the linear independence of $L_j$ implies that no non-trivial linear combination of $t_1,\dots,t_n$ can be zero, which proves that the union of the sets in $\mathscr{Y}$ is linearly independent. Therefore, by Zorn’s lemma, there is a maximal linearly independent set and hence a basis.
My question: Why are we using only finite linear combinations to show that the union is linearly independent. Surely, if the union is infinite,then there do exist many infinite linear combinations of elements of the union, which cannot be proven to be linearly independent by the same reasoning. I suspect, that perhaps we are not concerned with infinite linear combinations due to issues of convergence, but I'm not sure.
Clear answers will be appreciated.
AI: The definition of linear independence is that any finite linear relation is trivial.
Vector spaces in general do not have any concept of an infinite sum at all. For those vector space where the usual concept of an infinite sum of reals can be generalized, one may speak of a different kind of span/basis where one allows infinite linear combinations in addition to finite ones. That gives rise to a separate concept, different from the usual kind of linear-combinations basis.
When one needs to distinguish between the different notions of basis, an ordinary basis that works by finite linear combinations is called a "Hamel basis" or "algebraic basis", and one that needs infinite linear combinations to span everything is called a "Schauder basis" (though strictly speaking the latter name implies some additional conditions). |
H: Convergence of the sequence $\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots$
I recently came across this problem
Q1 Show that $\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times }} = 3$
After trying it I looked at the solution from that book which was very ingenious but it was incomplete because it assumed that the limit already exists.
So my question is
Q2 Prove that the sequence$$\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots,\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}$$ converges.
Though I only need solution for Q2, if you happen to know any complete solution for Q1 it would be a great help .
If the solution from that book is required I can post it but it is not complete as I mentioned.
Edit: I see that a similar question was asked before on this site but it was not proved that limit should exist.
AI: Let $f_n(0)=\sqrt{1+n}$ and $f_n(k)=\sqrt{1+(n-k)f_n(k-1)}$. Then $0<f_n(0)<n+1$ when $n>0$. Assume that $f_n(k)<n+1-k$ and we can show by induction that
$$
f_n(k+1) < \sqrt{1+(n-k-1)(n-k+1)} = \sqrt{1+(n-k)^2-1} = n+1-(k+1)
$$
for all k. Your expression is $f_n(n-2)$ which is increasing in $n$ and bounded above by $3$, so converges. |
H: What do you call a generalised Fourier-like transform?
The Fourier series is a decomposition of an arbitrary function into a superposition of sinusoidal functions. Some time back, I asked a question about whether it is possible to decompose functions using other families of functions and indeed, it is.
Today's question is much simpler: What do you call such an expansion and the associated transform? Is there a standard name for "a transform whose counterpart for finite domains decomposes a function into the sum of several other functions"?
AI: There's a concept called an integral transform that generalizes the Fourier transform -- but I don't think that's exactly what you're looking for.
Alternatively, you could be looking at a Schauder basis for the function space in question -- but that name just corresponds to an assertion that a series form for every function exists and doesn't say anything about how to find it. |
H: How can " $\small {n \over \varphi(n) } \text{ is integer only if } n=2^r \cdot 3^s $ " simply be shown?
I've tried to answer the question concerning $\small { 3^n-2^n\over n } $ and wanted to show this simply by referring to the property that
$$ 3^{\varphi (n)}-2^{\varphi (n)}\equiv 0 \pmod n $$
and
$$ 3^{\varphi (n) \cdot m}-2^{\varphi (n) \cdot m}\equiv 0 \pmod n $$
and thus that it is required that $$ n = \varphi(n) \cdot m $$ and then $$ {n \over \varphi(n)}=m$$ is integer. After a short consideration I got aware that I could not really prove easily, that this is in general impossible. The wikipedia-page does not have a short statement about this and a short heuristic indicates, that this is possible only for $ n=2^r 3^s \gt 1 $ where then $ m \in \{ 2 , 3\}$.
Q: What is a short and simple way to prove this?
(Remark: in the referred question there is an answer like "apply Fermat's little theorem repeatedly" - here I'd like an answer which is less abstract)
AI: $\frac{n}{\phi(n)}$ is multiplicative, so if $n=\prod p_i^{a_i}$ is the prime factorization of $n$, then
$$
\frac{n}{\phi(n)} = \prod \frac{p_i^{a_i}}{p_i^{a_i-1}(p_i-1)} = \prod \frac{p_i}{p_i-1}
$$
If $n$ is divisible by two odd primes then the denominator is divisible by $4$ but the numerator cannot be so the result is not an integer. If $n$ is divisible by one odd prime $p$ then the denominator is divisible by $2$ and we must have $n=2^rp^s$ and $n/\phi(n) = 2p/(p-1)$. If $p=3$ this is an integer, otherwise $1<p/(p-1)<3/2$ and it cannot be. If $n$ is not divisible by an odd prime then $n=2^r$. |
H: Combinatorics thinking
I saw this question in a book I've been reading: in a group of four mathematicians and five physicians, how many groups of four people can be created if at least two people are mathematicians?
The solution is obtained by ${4 \choose 2}{5 \choose 2} + {4 \choose 3}{5 \choose 1} + {4 \choose 4}{5 \choose 0} = 81$. But I thought of the present (wrong) solution:
Step 1. Choose two mathematicians. It gives ${4 \choose 2}$ different ways of choosing.
Step 2. Choose two people from the seven people left. It gives ${7 \choose 2}$ ways of choosing.
Step 3. Multiply. ${4 \choose 2}{7 \choose 2} = 126 = {9 \choose 4}$. It is equivalent of choosing four people in one step. Clearly wrong.
I really don't know what's wrong in my "solution". It seems like I am counting some cases twice, but I wasn't able to find the error. What am I missing here?
PS. It is not homework.
AI: If your group is all four mathematicians, you will count it six times. Each unique pair could be the two you pick first and the other two will be the two you pick second. Similarly, all groups of three mathematicians and one physicist will be counted three times. Since there are 20 groups of 3+1, your overcount is $20*2+5=45=126-81$ |
H: proving there exists an integer $ k$ such that $ak^2+bk+c\equiv 0 \pmod{2^n}$ by induction
$n$ is a given positive integer.
how to prove by using mathematical induction that there exists an integer $ k$ such that $ak^2+bk+c\equiv 0 \pmod{2^n}$, where $b$ is odd and at least one of $a, c$ is even
AI: First check directly that the result holds for $n=1$.
We do the induction step. Let $x$ be an integer such that
$$ax^2+bx+c\equiv 0\pmod{2^{n-1}}.$$
We would like to show that there is an integer $y$ such that $ay^2+by+c\equiv 0\pmod{2^n}$.
If $2^n$ already divides $ax^2+bx+c$, we are finished. So we need only worry about the case where $ax^2+bx+c\equiv 2^{n-1}\pmod{2^n}$.
Try for a $y$ of the shape $x+t$. So we want
$$a(x+t)^2+b(x+t)+c\equiv 0\pmod{2^n}.$$
But $ax^2+bx+c\equiv 2^{n-1}\pmod{p}$. So we want the difference
$$[a(x+t)^2+b(x+t)+c]-[ax^2+bx+c]$$
to be congruent to $2^{n-1}$ modulo $2^n$. That is, we want to choose $t$ so that
$$a(2xt+t^2)+bt\equiv 2^{n-1}\pmod{2^n}.$$
Note that $t=2^{n-1}$ works, since $b$ is odd and $2t$ and $t^2$ are each divisible by $2^n$. |
H: Computing large powers $\!\bmod n$ when $n$ has multiple prime factors.
How will the congruence modulo works for large exponents? What theorem/s may be used? For example to show that $7^{82}$ is congruent to $9 \pmod {40}$.
AI: Besides the methods mentioned by Andre, it's worth mentioning another more powerful technique that often proves crucial. Namely, instead of Euler's theorem, based on his $\phi$ (totient) function, one should use an improvement due to Carmichael, based on his $\lambda$ function (universal exponent).
The big gain in power arises from employing $\rm\: \ell = \color{#C00}{lcm}\:$ vs. product to combine exponents:
$$\begin{eqnarray}\rm a^j&\equiv&\rm 1\pmod m\\ \rm a^k&\equiv&\rm 1\pmod n\end{eqnarray}\Bigg\}\ \Rightarrow\rm\ a^{\color{#C00}{lcm}(j,k)}\equiv 1\pmod{lcm(m,n)}$$
Proof $\,\rm\bmod m\!:\ a^j\equiv 1,\,\ j\mid \ell\:\Rightarrow\: a^{\ell}\equiv 1\,$ by mod order reduction. Similarly $\rm\bmod n\!:\ a^{\ell}\equiv 1,\,$ so combining: $\rm a^{\ell}\equiv 1\bmod m\ \&\ n\Rightarrow $ $\rm\,a^{\ell}\equiv 1\pmod{\!{\rm lcm}(m,n)}\,$ by CCRT.
For example, if $\rm\:n\:$ is coprime to $2,3,5$ then Euler's theorem implies that $\rm\:n^{64}\equiv 1\pmod{240},\:$ versus Carmichael's theorem, which yields the much stronger result that $\rm\:n^4\equiv 1\pmod{240}.\,$ This allows one to immediately solve the problem there, viz. $\rm\,240\:|\:p^4\!-\!1\,$ for all primes $\rm\,p > 5.$ Compare below the computation of the Euler vs. Carmichael function:
$\qquad\qquad\rm\phi(240) = \phi(2^4\cdot 3\cdot 5) = \ \phi(2^4)\ \phi(3)\ \phi(5) = 8\cdot2\cdot 4 = 64$
$\qquad\qquad\rm\lambda(240) = \lambda(2^4\cdot 3\cdot 5) = \color{#C00}{lcm}(2^{2},\phi(3),\phi(5)) = \color{#C00}{lcm}(4,2,4) = 4$
Thus Carmichael generally yields a much lower bound on the order of elements, which greatly simplifies problems like yours and that above. For more examples see my posts here. |
H: Ratios and Mixture
I can't seem to to find a way to solve this problem:
In a mixture $60$ litres, the ratio of alcohol and water is $2:1$. If this ratio is to be $1:2$ then the quantity of water further added is? Ans $60$ litres
Any suggestions?
AI: In the original 60 liter mixture, the ratio of alcohol to water was $2:1$ which means that the original mixture was two-thirds alcohol and one-third water. This means $\frac{2}{3}\cdot 60 = 40$ liters of alcohol and $\frac{1}{3}\cdot 60 = 20$ liters of water.
Now you want to change the ratio to $1:2$ by adding water. a $1:2$ ratio means one-third alcohol and two-thirds water. Since we are adding only water, the amount of alcohol in the final mixture is still going to be $40$ liters (same as in your original mixture).
Now we know that $40$ liters of alcohol is one-third of the new mixture, which means there has to be $\dfrac{2/3}{1/3} \cdot 40 = 2\cdot 40 = 80$ liters of water in the new mixture. Since we already had $20$ liters originally, we have to add $80-20 = 60$ liters. |
H: Something weaker than the Riesz basis
I have some function $f$, real valued and continuous.
I formed functions $\{f_{m,k}, k \in \mathbb{Z}, m>0\}$ such that that $\mathrm{span}\{f_{m,k}, k \in \mathbb{Z}, m>0\}$ is dense in $L_p(R)$.
Now I would like to constract a Riesz basis from $\{f_{m,k}, k \in \mathbb{Z}, m>0\}$, but unfortunately, my $\{f_{m,k}\}$ does not generate a Riesz basis.
I am wondering if there is some weaker/stronger structure than the Riesz basis, such that $\{f_{m,k}, k \in \mathbb{Z}, m>0\}$ can possibly generate?
Thank you.
AI: The term you're looking for is probably Schauder basis, altough for that you also need some independence properties (as usual with bases). |
H: Calculating the norm of an element in a field extension.
Given a number field $\mathbb{Q}[\beta]$, where the minimal polynomial of $\beta$ in $\mathbb[Z][x]$ has degree $n$, I would like to calculate the norm of the general element $$a_0+a_1\beta+\cdots+a_{n-1}\beta^{n-1}.$$
In particular, here is my attempt when $\alpha=2^\frac{1}{3}$:
Let $K=Q[2^\frac{1}{3}]$. Using the definition of the norm, it is the determinant of the linear transformation. Consider $\alpha =a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ acting by multiplication on the element $d+2^{\frac{1}{3}}e+2^{\frac{2}{3}}f$. Since $$\left(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c\right)\left(d+2^{\frac{1}{3}}e+2^{\frac{2}{3}}f\right)=ad+2bf+2ce+2^{\frac{1}{3}}\left(ae+bd+2cf\right)+2^{\frac{2}{3}}\left(af+dc+be\right) $$ in the basis $[1,2^{\frac{1}{3}},2^{\frac{2}{3}}$ we may view multiplication by $\alpha$ as a linear transform $$\alpha\left[\begin{array}{c}
d\\
e\\
f
\end{array}\right]=\left[\begin{array}{c}
ad+2bf+2ce\\
ae+bd+2cf\\
af+dc+be
\end{array}\right].$$ Using the above, we see that $$\alpha=\left[\begin{array}{ccc}
a & 2c & 2b\\
b & a & 2c\\
c & b & a
\end{array}\right] $$ in this basis. Taking the determinant we find $$\det \left[\begin{array}{ccc}
a & 2c & 2b\\
b & a & 2c\\
c & b & a
\end{array}\right] =a\left(a^{2}-2bc\right)-2c\left(ba-2c^{2}\right)+2b(b^{2}-ac) $$ $$=a^{3}+2b^{3}+4c^{3}-6abc. $$
This means we have shown that $N_K(\alpha)=a^{3}+2b^{3}+4c^{3}-6abc$ for $\alpha=a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$
Questions:
(1) Was the above calculation correct? Can we conclude that the norm of a general element in that space is $a^{3}+2b^{3}+4c^{3}-6abc$?
(2) Is there a better way to do this computation? What about if the extension is Galois?
AI: 1) Yes, the calculation was correct. A decent way to check you haven't made any arithmetic errors is to try some small integers for $a,b,c,d,e,f$ and check the norm is multiplicative.
2) This is probably the best way to do the computation. If we have a Galois extension, then the norm of an element is the product of all of its Galois conjugates. There are two ways I can think of that we can use this fact:
Produce the elements of $G(K/\mathbb{Q}).$ Sometimes one has to do this anyway, so in those cases that could be a short cut.
Assume for now your general element has the relevant coefficient non-zero so that its degree is the degree of the extension. Then since the list of distinct Galois conjugates being $\alpha_1, \alpha_2, \cdots \alpha_n$ implies the minimal polynomial is $m(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n),$ we can read the norm of $\alpha$ off the constant term of the minimal polynomial. Then the norm for lower degree elements follows by a limit argument. |
H: How to prove $I + t X$ is invertible for small enough $ | t | ?$
Let $X \in \text{GL}_n(\mathbb{R})$ be an arbitrary real $n\times n$ matrix. How can we prove rigorously:
$$ \underset{b>0} {\exists} : \underset{|t|\le b} {\forall} : \det (I + t X) \neq 0 $$
If necessary, we could also assume that $t \ge 0.$
AI: Note that for $t\neq 0$
$$
\det(I+tX) = t^n\det(\frac{1}{t}I+X) = 0
$$
only if $\frac{-1}{t}$ is an eigenvalue of $X$. If $ t=0$ the statement is trivial so we exclude this case.
Since $X$ has only finitely many eigenvalues, its set of eigenvalues is bounded in magnitude. For all $t$ sufficiently small $|\frac{-1}{t}|$ will be larger than this magnitude bound, so $\frac{-1}{t}$ will not be an eigenvalue of $X$, and hence we will have
$$
\det(I+tX) \neq 0.
$$ |
H: Questions concerning a proof that $\mathcal{D}$ is dense in $\mathcal{S}$.
I am currently working through
this lecture notes
and on page 164, there it is said
The space of $\mathcal{D}(\mathbb{R}^n)$ of smooth complex-valued functions with compact support is contained in the Schwartz space $\mathcal{S}(\mathbb{R}^n)$. If $f_k \to f$ in $\mathcal{D}$, then $f_k \to f$ in $\mathcal{S}$, so $\mathcal{D}$ is continuously embedded in $\mathcal{S}$. Furthermore, if $f\in \mathcal{S}$, and $\eta \in C_c^{\infty}(\mathbb{R}^n)$ is a cutoff function with $\eta_k(x) = \eta(x/k)$, then $\eta_k f \to f$ in $\mathcal{S}$ as $k \to \infty$, so $\mathcal{D}$ is dense in $\mathcal{S}$.
I don't understand the arguments in this paragraph, for a subset $\mathcal D$ of $\mathcal S$ to be dense in $\mathcal S$ for every element $s$ of $\mathcal S$ I need to find a sequence in $\mathcal D$ which converges to $s$, but there just stands that $\eta_k f \to f$ in $\mathcal{S}$, but what i need is a sequence in $\mathcal{D}$ not in $\mathcal{S}$, so why does it follow from this that $\mathcal{D}$ is dense in $\mathcal{S}$?
AI: The sequence $\{\eta_k\}$ is contained in $\mathcal D(\Bbb R^n)$ and we can check that the product of a function in the Schwartz space with a test function is a test function (it's smooth because the product of two smooth functions is smooth, and it has a compact support because it's contained in the support of the test function).
We have $\mathcal D(\Bbb R^n)\subset \mathcal S(\Bbb R^n)$, and what we want to see is that it's a dense subset for the topology of $\mathcal S(\Bbb R^n)$. So what we have to show is that
$$\sup_{x\in \Bbb R^n}|x^p\partial^{\alpha}(\eta_kf-f)(x)|\to 0$$
for all integer $p$ and all $\alpha\in\Bbb N^n$. |
H: Covering $R^2$ by open half planes
Let $\{s_\iota, \iota\in I\}$ be any collection of vectors in $\mathbb{R}^2$ and let $U_\iota=\{\pi\in \mathbb{R}^2;\pi\cdot s_\iota >0\}$. Then, $U_\iota$ is either empty or an open half plane in $\mathbb{R}^2$ for any $\iota\in I$. Is it true, that we can find countably many $\iota_k\in I$, $k\in\mathbb{N}$ such that $\bigcup_{\iota\in I} U_\iota=\bigcup_{k=1}^\infty U_{\iota_k}$?
AI: It is not always true, but if they cover $\mathbb R^2 \setminus \{(0,0)\}$, then it is true there exists a finite subcover.
There is a quotient map $\mathbb R^2 \setminus \{(0,0)\} \to S^1$ by identifying vectors on a common half-line ($s \sim t $ if there is $\alpha > 0$ such that $s = \alpha.t$). It turns out that your open half-planes are compatible with $\sim$ (if $\pi$ and $\pi'$ are congruent, the sign of their product with $s_\iota$ is the same, so one of them is in the half plane if and only if the other is too), and they correspond to open intervals of $S^1$ (open semi-circles, even). But $S^1$ is compact, so if you have an open cover of $S^1$, you can extract from it a finite cover.
However, if they don't cover the whole circle, there may not be a finite covering.
For an example, pick $s_\iota = (1,1/\iota)$ for $\iota \in I = (0;1]$. Any finite subcover will be included in $\cup_{\iota> \epsilon} U_\iota$ for some $\epsilon > 0$, which is strictly smaller than $\cup_{\iota \in I} U_\iota$ : the vector $(\epsilon,-1)$ is in the latter but not in the former. |
H: Needs alternative algorithm for a graph problem
There is a network of connected electric bulbs each with a unique id. Each bulb has a switch. When the switch is clicked, the colour of the bulb changes and also the colours of connected bulbs change. The order in which the colour-change occurs is green-yellow-red-green. The network of bulbs is a connected graph (but not necesarily a complete graph). I want to find a way by which I can convert all bulbs to green. This could be represnted by sequence of ids of bulbs whose switches can be clicked to get the final outcome.
I tried to solve this by converting the problem to a graph. Each node of the graph representing a state of the network so that two states are different if there is at least one unique bulb which has different colour in those two states. Then by using Dijkstra's shortest path algorithm I tried to find a path to the final state where all bulbs have green colour. (Although in my problem I don't need a shortest path. But this is also one valid solution.) This worked fine for small networks (number of nodes less than 12). But as the number of nodes increased the number of states increased so much that my program was running forever.
AI: If the bulbs initially have different colors, and the only possible colors are green, yellow and red, there can be at most 3 nodes in the graph. There are only 4 possible connected graphs with at most 3 nodes, do it seems to be feasible simply to consult a precomputed table of solutions for each possible instance of the problem. (Especially since the answer is always "no" for three of the four possible graphs).
On the other hand, if some of the bulbs are allowed to have the same color initially, then the problem becomes more interesting. The key to improving your search algorithm is to notice that the order in which the buttons are pressed does not matter; the final state of each bulb depends only on its initial color and the number of times some button connected to it was pressed.
Given this (and the lucky coincidence that the number of colors in the cycle is prime), the simplest way to approach the problem is linear algebra: Represent each color an element of $\mathbb F_3$ (the field of integers modulo 3), and a state of the entire network as a vector with entries in $\mathbb F_3$. The actions of each button is another fixed vector (containing only 0's and 1's) that is added to the state vector, and the problem is then to find whether the difference between the initial state and the all-green state is a sum of multiples of button vectors. That is simply a system of $n$ linear equations in $n$ unknowns, where the unknowns are the number of times each button is pressed. This can be solved in $O(n^3)$ time using standard techniques such as Gaussian elimination. |
H: Is it possible to combine two integers in such a way that you can always take them apart later?
Given two integers $n$ and $m$ (assuming for both $0 < n < 1000000$) is there some function $f$ so that if I know $f(n, m) = x$ I can determine $n$ and $m$, given $x$? Order is important, so $f(n, m) \not= f(m,n)$ (unless $n=m$).
AI: If there are no bounds on your integers, use the Cantor pairing function. It is pleasantly easy to compute, as are its two "inverses."
For the case where your integers are bounded by say $10^6$, you can simply concatenate the decimal expansions, padding with initial $0$'s as appropriate. Or do something similar with binary expansions. Dirt cheap to combine and uncombine, an easy string manipulation even when we allow bounds much larger than $10^6$. |
H: Under what conditions does $(\frac{3}{p})(\frac{-1}{p})=1?$ Two ways, different results.
I have quite a problem, two methods, different results. something's wrong.
I'm trying to find under what conditions the Legendre symbol for $(\frac{3}{p})(\frac{-1}{p})=1$.
First Way: $(\frac{3}{p})(\frac{-1}{p})=(\frac{3}{p})\cdot(-1)^{(p-1)/2}$.
For $p\equiv1\pmod4$, I get that $(\frac{3}{p})=(\frac{p}{3})$ and it is $1$ iff $p\equiv1\pmod3$ and $p\equiv1\pmod4$. Otherwise I get that $(\frac{3}{p})=-(\frac{p}{3})$ so it has to be that $p\equiv1\pmod3$, in order to get $-1\cdot -1 \cdot 1=1$. The Chinese remainder theorem tells me that the $p\equiv1,7 \pmod{12}$.
Second way (I believe that it is the problematic one but yet I can't see what's wrong):
$$\left(\frac{3}{p}\right)\left(\frac{-1}{p}\right)=\left(\frac{3}{p}\right)\cdot(-1)^{(p-1)/2}=\left(\frac{p}{3}\right)(-1)^{((p-1)/2)\cdot((3-1)/2)}\cdot(-1)^{(p-1)/2}=\left(\frac{p}{3}\right),$$ this equals $1$ only for $p \equiv 1 \pmod 3$.
What's wrong with the second way? I believe it's all legal.
Thanks a lot!
AI: Nothing's wrong. The answer is that for a prime $p > 3$, $\left( \frac{3}{p} \right) \left( \frac{-1}{p} \right) = 1 \iff p \equiv 1 \pmod 3$. Note that the possible residue classes modulo $12$ for a prime $p > 3$ are the ones which are coprime to $12$, i.e., $1,5,7,11$. Of these the ones congruent to $1$ modulo $3$ are $1$ and $7$.
Let me give you yet a third way to solve this problem. For an odd prime $p$, set $p^* = (-1)^{\frac{p-1}{2}} p$. Then an equivalent statement of quadratic reciprocity is that for distinct odd primes $p,q$,
$\left( \frac{p}{q} \right) = \left( \frac{q^*}{p} \right)$.
(Admittedly, verifying this is similar in nature to what you're doing! The point though is that this is a formula worth proving and then remembering.)
Applying this with $q = 3$, we get $\left( \frac{3}{p} \right) \left( \frac{-1}{p} \right) = \left( \frac{-3}{p} \right) = \left(\frac{p}{3} \right)$, which is $1$ iff $p \equiv 1 \pmod{3}$. |
H: Area of polar coordinate $r = e^{- \theta/4}$
$$r = e^{- \theta/4}$$
$$\pi /2 \leq \theta \leq \pi$$
I know the formula is
$$\int_a^b \frac{1}{2} r^2 d\theta$$
$$\int_{\pi/2}^{\pi} \frac{1}{2} (e^{- \theta/4})^2 d\theta$$
From here I cannot figure out an easy way to integrate this.
AI: Note that $(e^{-\theta/4})^2=e^{-\theta/2}$. It is a special case of the general exponential rule $(a^b)^c=a^{bc}$. |
H: Probability of the events
For the events of $A$ and $B$, probabilites are $\Bbb P(A) = 3/11$ and $\Bbb P(B) = 4/11$. Define the $\Bbb P(A \cap B )$ if:
a) $\Bbb P(A \cup B) = 6/11$.
b) events are indenpendent
I have done a task, and it's following
$$\Bbb P(A \cup B) = \Bbb P(A) + \Bbb P(B) - \Bbb P(A \cap B ) = 1/11\;.$$
However I don't know how to solve b task. I need your help.
Thanks in advance.
AI: To say that $A$ and $B$ are independent is to say that $\Bbb P(A\cap B)=\Bbb P(A)\Bbb P(B)$.
By the way, it’s not correct to write
$\Bbb P(A\cup B)=\Bbb P(A)+\Bbb P(B)-\Bbb P(A\cap B)=1/11$;
what you mean is that $\Bbb P(A\cup B)=\Bbb P(A)+\Bbb P(B)-\Bbb P(A\cap B)$, so $\Bbb P(A\cap B)=1/11$. |
H: If $\int_0^1 f(x)x^n \ dx=0$ for every $n$, then $f=0$.
Possible Duplicates:
Nonzero $f \in C([0, 1])$ for which $\int_0^1 f(x)x^n dx = 0$ for all $n$
Slight generalization of an exercise in (blue) Rudin
What can we say about $f$ if $\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$?
I found a nice problem I would like to share.
Problem: If $f$ is continuous on $[0,1]$, and if
$$\int_0^1 f(x)x^n \ dx =0$$
for every non-negative integer $n$, prove that $f(x)=0$ on $[0,1]$.
Source: W. Rudin, Principles of Mathematical Analysis, Chapter 7, Exercise 20.
I have posted a proposed solution in the answers.
AI: We will show that the integral of $f^2$ over $[0,1]$ is 0. This will show that $f$ is zero, because if $f$ were not identically equal to zero, $f^2$ would be positive on some interval (by continuity) and have a nonzero integral.
Using the Stone-Weierstrass theorem, we can approximate $f$ uniformly by polynomials $P_n$ so that $||P_n-f||<1/n$ in the $\sup$ norm. The given condition obviously implies that the integral of $f(x)P(x)$ is zero for any polynomial $P$, by linearity. Note that
$$\left|\int_0^1 f(x)f(x) \ dx \right|=\left|\int_0^1 f(x)f(x) \ dx - \int_0^1 f(x)P_n(x)\right|\le \int_0^1 |f(x)| |f(x)-P_n(x)|\ dx \\\le \frac{1}{n}\int_0^1 |f(x)| \ dx.$$
The last integral is constant, so taking $n$ arbitrarily large completes the proof. |
H: Finding an imprimitive group on $12$ letters
By definition a permutation group $G$ acting on a set $\Omega$ is called primitive if $G$ acts transitively on $\Omega$ and $G$ preserves no nontrivial blocks of $\Omega$. Otherwise, if the group does preserve a nontrivial block then $G$ is called imprimitive.
Here I am asked to find an imprimitive permutation group $\Omega$ acting on $\Omega$ with $|\Omega|=12$ such that $|G|$ be of maximum possible order.
It would be difficult and unprofessionally finding a group which has a block for example with two elements. At least I cannot do that right now. :). Clearly, our $G$ is a proper subgroup of $S_{12}$ but would not be $A_{12}$.
I am wondered how can it be shown that any group I found is of maximum order. Thanks for any help.
AI: Hint: given a particular block $B$, consider the group $H_B$ of all permutations that preserve $B$. |
H: Probability game of independent events
In one game, probability that a gamer scores win is 1/4. If X is number of tries( games ) that gamer has done in order to achieve score:
(NOTE!: every try is independent from the other)
a) Define the probability that gamer scored a win in 3 tries.
b) Define the probability that gamer will have at least three failures before he scores 2 wins.
I have tried to do it this way, but I am not sure if it is OK.
a) P(win in 3 tries) = 1/4*1/4*1/4 = 1/64.
b) for this task I have no idea from where to start.( I need great help here )
AI: HINTS:
(a) The games tries are independent, so the probability that he scores a win on the third try is the same as the probability that he scores a win on the first try. What is that probability? (What you calculated is the probability that he wins all of his first three tries.)
(b) Think of it as a race: which comes first, his second win or his third failure? You want the probability that the third failure comes before the second win, but you might find it easier to calculate the probability that he scores $2$ wins before he has three failures and subtract that from $1$. How can he score $2$ wins before his third loss? Using W for a win and L for a loss, some possibilities are WW, WLW, LWW, WLLW, LWLW, and LLWW.
Are there any others?
Calculate the probability of each of the possibilies. For instance, $\Bbb P(\text{WLW})=\frac14\cdot\frac34\cdot\frac14$.
What do you have to do with these probabilities to get the total probability of scoring $2$ wins before his third loss? |
H: Spectrum of a real number
Whilst reading Concrete Mathematics, the authors mention something which they refer to as the "spectrum" of a real number (pg. 77):
We define the spectrum of a real number $\alpha$ to be an infinite multiset of integers, $$\operatorname{Spec}{(\alpha)}=\{\left\lfloor\alpha\right\rfloor, \left\lfloor2\alpha\right\rfloor,\left\lfloor3\alpha\right\rfloor,\cdots\}.$$
It then goes on to describe interesting properties of these "spectra", such as the inequality of any two spectra, i.e. given some $\alpha\in\mathbb{R},\space\not\exists\beta\in\mathbb{R}\setminus\{\alpha\}:\operatorname{Spec}{(\alpha)}=\operatorname{Spec}{(\beta)}$.
It also gives a proof regarding the partitioning of the set $\mathbb{N}$ into two spectra, $\operatorname{Spec}{(\sqrt{2})}$ and $\operatorname{Spec}{(\sqrt{2}+2)}$.
These properties intrigued me, so I wondered if there were any other properties of these multisets, however; a google search for "spectrum of a real number" doesn't appear to yield any relevant results, so I wondered if there was any research into these objects, and if there is whether "spectrum" is a non-standard name (if it is, I'd appreciate the common name for these objects).
Thanks in advance!
AI: This is non-standard (and does not agree well with any of the other meanings of "spectrum" in mathematics so I would not use it). The standard name is Beatty sequence (at least when $\alpha$ is irrational). |
H: Conditional probability
A girl goes to school by bus every day. If it doesn't rain, probability that she will be late for bus is 1/5. If it rains probability that she will be late is 2/3. Probability that it is raining is 1/4.
Girl forgot to pick up a bus. Define the probability that it was raining.
In this task I have done following using my logic, but I am not sure is it correct:
P(late for bus and rained that day) = P(rains)/P(late for school and is raining )= 1/4/2/3 = 3/8.
I am not sure if it is correct, so I plead for your help.
AI: This is an application of the Bayes' theorem:
$$
\mathbb{P}(\text{rain}| \text{on time}) = \frac{\mathbb{P}(\text{rain} \land \text{on time} )}{\mathbb{P}(\text{on time} )} =
\frac{\mathbb{P}( \text{on time} | \text{rain} ) \mathbb{P}(\text{rain})}{\mathbb{P}(\text{on time} | \text{rain} ) \mathbb{P}(\text{rain}) + \mathbb{P}(\text{on time} | \text{no rain} ) \mathbb{P}(\text{no rain})} =
\frac{\frac{1}{3} \cdot \frac{1}{4} }{ \frac{1}{3} \cdot \frac{1}{4} + \frac{4}{5} \cdot \frac{3}{4} } = \frac{5}{41}
$$ |
H: Computing determinant of a specific matrix.
How to calculate the determinant of
$$ A=(a_{i,j})_{n \times n}=\left(
\begin{array}{ccccc}
a&b&b& \cdots & b\\
b& a& b& \cdots& b\\
\vdots& \vdots& \vdots& \ddots&\cdots\\
b&b&b & \cdots&a
\end{array}
\right)? $$
AI: We note that the sum of the elements of each column is $a+(n-1)b$ for each column, hence
$$\det A=(a+(n-1)b)\det\pmatrix{1&1&\dots& 1&1\\
b&a&\dots&b&b\\
\vdots&\ddots&\ddots&\vdots&\vdots\\
b&b&\dots&b&a}.$$
To each row of index $\geq 2$, we do $R_j\leftarrow bR_1$ to get
$$\det A=(a+(n-1)b)\det\pmatrix{1&1&\dots& 1&1\\
0&a-b&\dots&0&0\\
\vdots&\ddots&\ddots&\vdots&\vdots\\
0&0&\dots&0&a-b}.$$
Finally, we obtain
$$\det A=(a+(n-1)b)(a-b)^{n-1}.$$ |
H: Unconditional and independent probability
Two archers, independently, target a mark firing one arrow at a time. Probability that one archer will hit a mark is 0.8 and another is 0.4. After contest it is determined that one hit got in a mark. Find a probability that first archer scored it.
I have no idea from where to start on this. It would be logical that the result is 0.8, 'cause that is the probability to hit a mark of the first archer, but that's not the result wanted here. :(
AI: Per your statement, only one arrow hit the mark. That is possible when either archer A hit, and archer B missed, or the other way around.
$$
p_1 = \mathbb{P}\left(\text{A hit} \land \text{B miss}\right) = \underline{\quad\quad\quad}?
$$
$$
p_2 = \mathbb{P}\left(\text{B hit} \land \text{A miss}\right) = \underline{\quad\quad\quad}?
$$
Since these events are exclusive (both can not happen at the same time), $\mathbb{P}\left(\text{only one hit}\right) = p_1 + p_2$. Now to answering the question
$$
\mathbb{P}\left( \text{A hit} | \text{only one hit}\right) = \frac{\mathbb{P}\left(\text{A hit} \land \text{B miss}\right)}{\mathbb{P}\left(\text{only one hit}\right)} = \frac{p_1}{p_1+p_2}
$$ |
H: Help in self-studying mathematics.
Is this a reasonable list for who seek to learn mathematics by "self learning" program?
and is it a well sorted list to follow?
http://www.math.niu.edu/~rusin/known-math/index/index.html
AI: Our very own Pete Clark, now of the University of Georgia faculty, was once upon a time a very talented undergraduate student at the University of Chicago. Back then, he and several of his fellow U of C honor students compiled a list of their recommended textbooks at various levels and posted it online. It does need to be updated, but I think you'll find it quite helpful. In fact,I've been asking Pete to revise it for a few months now. I hope he'll eventually find the time to do so.
Here it is:
http://www.ocf.berkeley.edu/~abhishek/chicmath.htm |
H: A good Open Source book on Analytic Geometry?
Hi my course specifically talks about :
Cartesian and Polar Coordinates in 3 Dim, second Degree eqns in 3 vars, reduction to canonical forms, straight lines, shortest distance between 2 skew lines, Plane, sphere, cone, cylinder, paraboloid, ellipsoid,hyperboloid of one and two sheets and their properties.
This is merely indicative, and I would really like if the book talks about these (atleast) and some more. Would appreciate if it contained solved examples and exercise problems as well
Help Much Appreciated,
Soham
AI: There are a huge number of older classics freely available on the internet, such as (google "last name" and "title" together to find them):
Maxime Bocher, Plane Analytic Geometry (1915)
Maria M. Roberts and Julia T. Colpitts, Analytic Geometry (1918)
William F. Osgood and William C. Graustein, Plane and Solid Analytic Geometry (1922)
Lewis Parker Siceloff, George Wentworth, and David Eugene Smith, Analytic Geometry (1922)
The following School Mathematics Study Group texts from the 1960's may also be of use:
Analytic Geometry, Student Text, Part 1. Revised Edition
Analytic Geometry, Student Text, Part 3. Revised Edition
Since you seem especially interested in 3-dimensional analytic geometry, you'll also want to google (in google-books) for books with "solid analytic geometry" in their titles.
(added next day) This morning I looked through the books I have at home and came up with the following additional suggestions (definitely not all the relevant books that I have, as I have a large number of old math books):
N. J. Lennes and A. S. Merrill, Plane Analytic Geometry (1929)
Virgil Snyder and C. H. Sisam, Analytic Geometry of Space (1914)
V. A. Ilyin and E. G. Poznyak, Analytic Geometry (1984, MIR Publishers)
Joseph H. Kindle, Schaum's Outline of Theory and Problems of Plane and Solid Analytic Geometry (1950)
John M. H. Olmsted, Prelude to Calculus and Linear Algebra (1968)
William H. McCrea, Analytical Geometry of Three Dimensions (1960; Dover edition 2006)
Barry Spain, Analytical Conics (1957; Dover edition 2007)
Of all the books I've listed, my guess is that Kindle's book and Olmsted's book would be best for you. These two books should fit together well for self-study, since Kindle is more old-fashioned in style (with a huge number of problems) and Olmsted is more modern in style (with fewer problems, but much more careful text writing). ["Modern" means heavy use of set notation, functions as certain sets of ordered pairs, etc.] |
H: Closed linear span of a frame in a Hilbert space $\mathcal{H}$ coincide with $\mathcal{H}$
Definition of the problem
Let $\mathcal{H}$ be a separable Hilbert space and $J\subset\mathbb{N}$
an index set. Let $\Phi:=\left(\varphi_{j}\right)_{j\in J}\subset\mathcal{H}$
be a frame for $\mathcal{H}$, i.e. $\exists A,B>0\,\forall x\in\mathcal{H},\, A\left\Vert x\right\Vert ^{2}\leq\sum_{j\in J}\left|\left\langle x,\varphi_{j}\right\rangle \right|^{2}\leq B\left\Vert x\right\Vert ^{2}$.
Show that the closed linear span $cls\left\{ \varphi_{j}:j\in J\right\} :=\overline{span\left\{ \varphi_{j}:j\in J\right\} }$
coincides with $\mathcal{H}$. Furthermore, show that $\left(\left\Vert \varphi_{j}\right\Vert \right)_{j\in J}$
is always bounded.
My efforts
For the first part, I have tried to prove that the synthesis operator
$T_{\Phi}^{\star}:\ell_{2}\left(J\right)\rightarrow\mathcal{H}$ is
surjective. We know that the analysis operator $T_{\Phi}:\mathcal{H}\rightarrow\ell_{2}\left(J\right)$ of $\Phi$ is injective. Since the synthesis operator is
the adjoint operator of the analysis operator, we have that
$$
(\star)\ \forall x\in\mathcal{H},\, y\in\ell_{2}\left(J\right):\quad\left\langle T_{\Phi}x,y\right\rangle _{\ell_{2}\left(J\right)}=\left\langle x,T_{\Phi}^{\star}y\right\rangle _{\mathcal{H}}.
$$
From this, I derive that $T_{\Phi}^{\star}$ has the form $T_{\Phi}^{\star}\left(c_{j}\right)=\sum_{j\in J}c_{j}\varphi_{j},\,\left(c_{j}\right)\in\ell_{2}\left(J\right)$.
Since $\left(\star\right)$ must hold, we have that $\left(\star\star\right)\ \ker T_{\Phi}=\left(ran\, T_{\Phi}^{\star}\right)^{\perp}.$
And since $T_{\Phi}$ is injective, we have that its kernel is the
trivial kernel, i.e. $\ker T_{\Phi}=\left\{ 0\right\} $. From this
point, I do not see how to use this to prove $\left(\star\star\right)$.
After having shown that the synthesis operator $T_{\Phi}^{\star}:\ell_{2}\left(J\right)\rightarrow\mathcal{H}$
is surjective, and has the form $T_{\Phi}^{\star}\left(c_{j}\right)=\sum_{j\in J}c_{j}\varphi_{j},\,\left(c_{j}\right)\in\ell_{2}\left(J\right)$,
we know that all points in codomain $\mathcal{H}$ are covered by
$T_{\Phi}^{\star}$. It seems to me that the synthesis operator has
kind of the form of a linear combination of the frame-vectors of $\Phi$.
My question
How can I complete the proof of the surjectivity of the synthesis
operator, and how can I properly interpert that to show that the closed
linear span of $\Phi$ coincide with $\mathcal{H}$.
Thank you, Franck!
AI: If a closed subspace $V\subset \mathcal H$ does not coincide with $\mathcal H$, then there exists a nonzero vector $x$ that is orthogonal to $V$. What happens with the lower frame bound when you plug in such $x$?
To show that $\|\varphi_j\|$ are uniformly bounded, plug $x=\varphi_j$ into the upper frame bound. |
H: What should I call this commutative monoid of order three?
I'm looking for a name for the monoid given by the following table:
$$
\begin{array}{c|ccc}&1&a&b\\ \hline
1&1&a&b\\
a&a&1&b\\
b&b&b&b
\end{array}
$$
Is there a name that would be understandable to an undergraduate student who hasn't read anything about semigroups but has had a first course in algebra and knows what a semigroup/monoid is? What name would it be good to go under in a list of order-three semigroups?
AI: Call it : $(\mathbb{Z}/3\mathbb{Z},*)$ |
H: some references for regularity theory
I am gathering material that contains results of regularity for solutions of the equation
\begin{equation}
\mbox{div}(A(x) \nabla u) = f
\end{equation}
where the coefficints $a_{i,j}$ of $A$ are only measurable and $f \in L^p$ for some $p \in \mathbb{N}$ and
\begin{equation}
\lambda |\xi|^{2} \le \langle A(x)\xi , \xi \rangle \le |\xi|^{2} \quad \forall x,\xi
\end{equation}. Can you give me some references? Assume others hypothesis if you want. Thank you.
AI: Ladyzhenskaya & Uraltseva, "Linear and quasilinear elliptic equations"
Gilbarg & Trudinger, "Elliptic PDE of second order"
Maly & Ziemer, "Fine regularity of solutions of elliptic PDE"
Heinonen, Kilpelainen, Martio "Nonlinear potential theory of degenerate elliptic equations"
Listed in approximate order of relevance to the question. The books 3 & 4 deal more with p-Laplace type equations, but you can still get something by setting $p=2$. |
H: Area computed using polar coordinates
I have $r = \sqrt{\theta}$
http://www.wolframalpha.com/input/?i=cartesian+r+%3D+%5Csqrt%7Btheta%7D+
The graph given in the book ends at the first time it approaches to right side of the x axis (or 2pi). I attempted to set up an integral that cut each section of the graph so I have 4 section to compute that integral in and it did not even give close to the correct answer ($\pi^2$).
I tried to do
$$\int_0^ {\pi/2} \sqrt {\theta} d \theta$$
$$\int_{\pi/2}^ {\pi} \sqrt {\theta} d \theta$$
$$\int_\pi^ {3\pi/2} \sqrt {\theta} d \theta$$
$$\int_{3\pi/2}^ {2\pi} \sqrt {\theta} d \theta$$
This basically gives me just the integral of $\frac{1}{4} \theta^{2}$ evaluated $2\pi$ since everything else cancels out.
AI: You have no need to divide the integral from $0$ to $2\pi$ into a sum of other integrals.
$$r=f(\theta)=\sqrt{\theta}$$
The integral from $0$ to $2\pi$ of $f(\theta)$ is given by:
$$\frac{1}{2}\int_{a}^{b}(f(\theta))^2\, d\theta=\frac{1}{2}\int_{0}^{2\pi}\sqrt{\theta}^2\, d\theta=\frac{1}{2}\int_{0}^{2\pi}\theta\, d\theta=\frac{1}{2}\left[\frac{1}{2}\theta^2\right]_0^{2\pi}=\frac{1}{4}((2\pi)^2-0^2)=\frac{4\pi^2}{4}=\pi^2$$ |
H: How to reciprocal this imaginary exponent?
Assume $x\gt 0$, how does one simplify $$e^{(-x^2t)/i}\ ?$$
I don't understand how we could change the i under to the top so I could use Euler's formula
AI: Remember that $-i^2 = 1$ implies that $-i = \frac{1}{i}.$ Then apply Euler's formula as you mention. |
H: What will be the rate of inflation after n months?
Let's say I have a utility rate of 0.15 which inflates by 0.5% per year. I then want to convert that annual rate to a monthly rate and determine what the rate would be for month $n$.
The way I'm doing it now (which just feels incorrect) is:
$$\mathrm{utilityRate}\Bigl( (1.0 + \text{rate of inflation})^{n/12}\Bigr)$$
Is this correct, or am I completely off base?
AI: If the rate of inflation is constant every month then $(1 + \text{annual rate of inflation})^{n/12}$ is indeed what happens to the price level starting at $1$. So your formula looks sensible.
So in your example with $0.5\%$ annual inflation, this becomes $1.005^{n/12}$. The utilities I know would only change rates once a year if inflation was so low. |
H: Prove diagonalizability of a continuous transform
We've been going over diagonalization in my linear algebra class, but we've only been dealing with matrices—nothing too complicated. All of a sudden this problem came along and blindsided me:
Let V be the vector space of continuous functions with basis {$e^t, e^{-t}$}. Let $L:V\rightarrow V$ be defined by $L(g(t)) = g'(t)$ for $g(t)$ in $V$. Show that $L$ is diagonalizable.
I have the feeling that this will somehow involve a matrix similar to
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} because of the basis given, but I don't really know what to do with the derivative or how it begins to fit in. Any ideas? Thanks in advance!
AI: Well, consider the what this linear transformation does to the basis given:
$$L(e^t) = \frac{d}{dt}e^t = e^t$$
and similarly
$$L(e^{-t}) = \frac{d}{dt}e^{-t} = -e^{-t}$$
In other words, taking coordinates from the ordered basis $\beta = \{e^{t}, e^{-t}\}$ we have
$$L(1,0) = (1,0)$$
and
$$L(0,1) = -(0,1)$$
Hence, by definition, the matrix of $L$ with respect to the ordered basis $\beta$ is given by
$$[L]_\beta = \begin{pmatrix}1 & 0\\ 0 & -1\\ \end{pmatrix}$$
which is diagonal. |
H: lebesgue measure of $\{ (x,y,z) \in \mathbb R ^3 : x \in \mathbb R, 0 \leq y \leq 10, z \in \mathbb Z \} $
Find the lebesgue measure of the set:
$$ \Bigl\{ (x,y,z) \in \mathbb R ^3 : x \in \mathbb R, \quad 0 \leq y \leq 10, \quad z \in \mathbb Z \Bigr\} $$
I think is a null set but for some reason I have stuck and I can't write down a complete solution.
Any help?
Thanks in advance!
AI: $$
M:=\{(x,y,z) \in \mathbb{R}^3:\ x \in \mathbb{R}, \ 0 \le y \le 10, \ z \in \mathbb{Z}\}=
\bigcup_{i,j \in \mathbb{Z}}M_{i,j},
$$
where each $M_{ij}:=[i,i+1]\times[0,10]\times\{k\}$ has measure $0$ since for every $\epsilon>0$ one has $M_{ij} \subset M_{ij}^\epsilon:=[i,i+1]\times[0,10]\times[-\epsilon/20,\epsilon/20]$ and $|M_{ij}^\epsilon|=\epsilon$. Therefore $M$ is a countable union of null sets, and thus $M$ is a set of measure $0$. |
H: Corollary of Cauchy's theorem about finite groups.
Cauchy's Theorem: Let $G$ be a finite group and let $p$ divide the order of $G$. Then $G$ has an element of order $p$ and consequently a subgroup of order $p$ (of course the cyclic subgroup generated by the aforementioned element of order $p$).
Corollary: Let $G$ be a finite group, and $p$ be a prime. Then $G$ is a $p$-group if and only if the order of $G$ is a power of $p$.
Relevant Definition: $G$ is a $p$-group (for a prime $p$) if every element of $G$ has order $p^m$ for some $m\geq 1$.
The proof of the theorem was a beautiful application of using group actions to count. Now I'm trying to use the theorem to prove the corollary which is stated in my text leaving the proof as an exercise, but I am stuck because there doesn't seem to be a natural way to apply the theorem. Please accept my 'request for a hint'.
Thanks very much!
AI: If the order $n$ of the group is not a power of $p$, there is a prime $q\ne p$ that divides $n$. But then $\dots$. |
H: A question regarding sphere packing
The question of how many smaller spheres can be fit into a larger sphere is fascinating and has been examined extensively. I was curious, though, about the scenario of packing spheres of different radii into a larger sphere. For example: What is the maximum number of rigid spheres of radius 1 and radius 2 which can be packed into a rigid sphere of radius 10, given that there must be an equal number of radius 1 and radius 2 spheres? I'm not sure what packing strategy would produce an optimal result, so any ideas would be appreciated. Thanks in advance.
AI: A recent paper is Marshall and Hudson, Dense binary sphere packings, Beiträge Algebra Geom. 51 (2010), no. 2, 337–344, MR2682460 (2011g:52030). It deals particularly with the case where the two radii are 1 and $\gamma$, where $.444\lt\gamma\lt.482$, whereas you want $\gamma=1/2$, but it also contains an overview of the problem and an extensive list of references. |
H: Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$
Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$.
Ok so originally I messed around with $x^3 + x +1$ for a bit looking for an easy way to factor it and eventually decided that the factors are probably made up of really messy nested roots. So then I tried looking at the quotient field $\mathbb{Z}_2[x]/x^3 + x + 1$ to see if I would get lucky and it would contain all three roots but it doesn't. Is there a clever way to easily find this splitting field besides using the cubic formula to find the roots and then just directly adjoining them to $\mathbb{Z}_2$?
Edit: Ok it turns out I miscalculated in my quotient field, $\mathbb{Z}_2[x]/x^3 + x + 1$ does contain all three roots.
AI: The splitting field is either degree $3$ or degree $6$ over $\mathbb{Z}_2$, hence it is either $\mathbb{F}_8$ or $\mathbb{F}_{64}$.
Let $\alpha$ be a root, so that $\mathbb{F}_8 = \mathbb{F}(\alpha)$. The elements are of the form $a+b\alpha+c\alpha^2$, with $\alpha^3=\alpha+1$.
Now, the question is whether any of these elements besides $\alpha$ is a root of the original polynomial $x^3+x+1$.
Note that $(\alpha^2)^3 = (\alpha^3)^2 = (\alpha+1)^2 = \alpha^2+1$, and so if we plug in $\alpha^2$ into the polynomial we have
$$\alpha^6 + \alpha^2 + 1 = \alpha^2+1+\alpha^2+1= 0.$$
Thus, $\alpha^2$ is also a root. So the polynomial has at least two roots in $\mathbb{F}_8$, and so splits there. |
H: Is vector space a field? Or more than that?
As you all know, vector space is closed under scalar multiplication, scalar product, vector product and addition. If I take scalar product, vector space is a field, but if i take vector product, vector space is a group. Is there any term designating this kind of set? Plus scalar multiplication is a binary operation from two same sets to different set. What kind of operation is this?
AI: Let $K$ be a field. A vector space over $K$ is a set $V$, together with an operation $+\colon V\times V\to V$, and a function $\cdot\colon K\times V\to V$ (or alternatively, a family of functions $\lambda_c\colon V\to V$, indexed by elements $c\in K$) subject to certain identities and conditions.
The set $V$ does not normally contain $K$, so it makes no sense to talk about the vector space $V$ as being "closed under scalar multiplication". In addition, there is in general no "product" of vectors, so it does not make sense to talk about "vector multiplication."
Even in the case where we have a cross product of vectors (e.g., $\mathbb{R}^3$), this operation does not make $V$ into a group (the product is not associative).
Scalar multiplication is not an operation; an operation is always a function from a cartesian power of a set to the set. However, you can use currying to view the scalar multiplication as a family of unary operations on $V$, as indicated above, where for each $c\in K$ and each $\mathbf{v}\in V$, we define $\lambda_c(\mathbf{v})=c\mathbf{v}$.
(Added. Once you take into account the vector space axioms that relate to the scalar multiplication, it turns out that the scalar multiplication $K\times V\to V$ is actually an action of the field $K$ on the additive group $(V,+)$. But to truly call it an "action", you need the function to satisfy certain properties; just having a function $K\times V\to V$ does not make it an action, whereas any function $S\times S\to S$ is a binary operation on $S$, regardless of its properties.)
So pretty much all of your questions are based on misstatements of fact. Seems hard to answer them accurately.
Now, there is a situation in which some of what you say might make sense: if $F$ and $K$ are fields, and $F\subseteq K$, then we can view $K$ as a vector space over $F$ by "forgetting" about multiplication inside of $K$ when neither factor is in $F$. But here you really are going the other way: you already have a field, and you are obtaining a vector space by restricting the multiplication $\cdot\colon K\times K\to K$ to $F\times K$.
There is another situation in which you may have a product between "vectors": when you have an algebra. If $F$ is a field, a $K$ algebra is a ring with (a copy of) $F$ in the center of $K$. But again, what you have is a richer structure that, by "forgetting" part of the structure, yields a vector space. Much like you can have a ring and, by forgetting the product, obtain a group. |
H: Simplifying to a desired expression structure
My book has this expression:
\begin{align}
((n(n+1)(2n+7))/6)+(n+1)(n+3)
\end{align}
And then the book simplified it, and ended up with the desired expression:
\begin{align}
((n+1)(n+2)(2n+9))/6
\end{align}
I tried to do such simplification. But I ended up with this:
\begin{align}
(2n^3+15n^2+31n+18)/6
\end{align}
Even though the "value" in my expression is the same as the desired result's, I need the structure of my expression to be the same as well. But I don't understand how to do such.
Here are the full steps my book shows. Of course, I understand that the steps make sense, but I don't understand why did my book take that approach? Is that the only approach possible to reach the desired expression? How was I supposed to know that? I mean, I know how to simplify, but clearly my book is using a different "style" or "path"
\begin{align}
((n(n+1)(2n+7))/6)+(n+1)(n+3)\\
(n(n+1)(2n+7)+6(n+1)(n+3))/6\\
((n+1)[n(2n+7)+6(n+3)])/6\\
((n+1)(2n^2 +7n + 6n + 18))/6\\
((n+1)(n+2)(2n+9))/6
\end{align}
So yes, how should I simplify to get a desired expression structure?
AI: Below I explain in detail the solution given in your book.
$\begin{eqnarray}
&&\rm\ \ \,n(n\!+\!1)(2n\!+\!7)/\color{#C00}6+(n\!+\!1)(n\!+\!3)\\
\rm put\ all\ over\ a\ common\ denominator = \color{#C00}6:\quad &= &\rm\ (n(\color{#0A0}{n\!+\!1})(2n\!+\!7)\,+\,\color{#C00}6\,(\color{#0A0}{n\!+\!1})(n\!+\!3))/\color{#C00}6\\
\rm pull\ out\ the\ common\ factor\ \color{#0A0}{n\!+\!1}:\quad &=&\rm\ (\color{#0A0}{n\!+\!1})\,(n(2n\!+\!7)+6(n\!+\!3))/6\\
\rm apply\ the\ distributive\ law:\quad &=&\rm\ (n\!+\!1)\,(2n^2 + 13n + 18)/6\\
\rm factor\ the\ quadratic,\ see\ below:\quad &=&\rm\ (n\!+\!1)\,(n\!+\!2)\,(2n\!+\!9)/6
\end{eqnarray}$
In order to factor the quadratic $\rm\:f = 2n^2+13n+18\:$ one can apply the AC-method as follows:
$$\begin{eqnarray}\rm 2f\, &=&\rm\ \ 4n^2\ +\ 13\cdot 2n\, +\, 2\cdot 18 \\
&=&\rm\ (2n)^2 + 13\,(2n) + 2\cdot 18 \\
&=&\rm\ \ N^2\ +\ 13\, N\ +\ 36\quad for\quad N = 2n \\
&=&\rm\ (N\ +\ 4)\,(N\ +\ 9) \\
&=&\rm\ (2n\, +\, 4)\,(2n\, +\, 9) \\
\rm f\, &=&\rm\ (\ n\ +\ 2)\,(2n\ +\ 9) \end{eqnarray}$$ |
H: Is there a closed form to this expression?
Consider
$$2^{n-1} + 2^{n-2}\dfrac{(n - 1)!}{1!(n - 2)!} + 2^{n-3}\dfrac{(n - 2)!}{2!(n - 2 \cdot 2)!} +
2^{n-4}\dfrac{(n - 3)!}{3!(n - 2 \cdot 3)!} + 2^{n-5} \dfrac{(n - 4)!}{4!(n - 2 \cdot 4)!} + \ldots $$
Until $(n - 2 \cdot k)$ equals to either $0$ or $1$ (even/odd). In other words: $$k = \lfloor \dfrac{n}{2} \rfloor$$
This expression comes from a programming puzzle, and it runs extremely slow when $n$ is large, so I try to find its closed form. Any idea or suggestion would be greatly appreciated.
AI: Hint: If you compute five or six terms you can find at simpler expression at OEIS. |
H: Prove or provide a counterexample: Finite simple group with $|G|\geq 6$ cannot have an index-2 or index-3 subgroup.
Let $G$ be a finite simple group with $|G|\geq 6$. For each of the following statements, prove or provide a counterexample.
$G$ cannot have an index-$2$ subgroup.
$G$ cannot have an index-$3$ subgroup.
$G$ cannot have a proper subgroup of index less than $6$.
AI: Let $G$ be a group. If $H$ is a subgroup of index $n$, then $H$ contains a normal subgroup of $G$ of index at most $n!$.
This is a common problem; consider the action of $G$ on the left cosets of $H$ by left multiplication, $g\cdot xH \longmapsto gxH$. This induces a homomorphism $G\to S_n$, and the kernel is contained in $H$. Since $G/K$ is isomorphic to a subgroup of $S_n$, the index of $K$ is at most $n!$.
If $G$ is assumed to be simple, then $K$ must be trivial.
This will tell you the only possible counterexamples for index 2 or 3. Try them and see if they meet all your hypothesis.
Added. Assume that $G$ is a simple group, and $H$ is a subgroup of index $2$ in $G$. Then $H$ contains a subgroup $K$ that is normal in $G$, and has index at most $2!$. Since $K\subseteq H$ and $[G:H]=2$, then $[G:K] = [G:H][H:K] = 2[H:K]\geq 2$, it follows that $[G:K]=2$, so $H=K$. Since $G$ is simple and $H$ is normal, it follows that $H=\{e\}$. But since $[G:H]=2$, it follows that $|G|=2$. That is, $G$ is the cyclic group of order $2$. In summary: the only simple group that has a subgroup of index $2$ is the cyclic group of order $2$.
Thus, if $G$ is a simple group and $|G|\geq 6$, then $G$ cannot have a subgroup of index $2$. Thus, 1 is true.
Assume $G$ is a simple group and $H$ is a subgroup of index $3$ in $G$. By the result above, $H$ contains a normal subgroup $K$ of $G$ that has index at most $3!$ in $G$. Thus, $K=\{e\}$ by the same argument as above, so $|G|= [G:K] \leq 6$. The only simple groups of order at most $6$ are $C_2$, $C_3$, and $C_5$; the only one that has a subgroup of index $3$ is $C_3$. So $G$ will be cyclic of order $3$.
Thus, if $G$ is a simple group of order at least $6$, then $G$ cannot have a subgroup of index $3$. Thus, 2 is true.
For subgroups of index $4$, we get as above that $G$ contains a normal subgroup of index at most $4! = 24$; this must be trivial, so $|G|\leq 24$. The only simple groups of order at most $24$ are $C_p$ with $p$ a prime, $p\leq 24$, and none of them has a subgroup of index $4$. So no simple group has a subgroup of index $4$.
Subgroups of index $5$ yield a normal subgroup of index at most $5!= 120$. Now we have some possibilities, since there are nonabelian groups of order less than $120$. For instance, we have $A_5$, of order $60$. Does $A_5$ have a subgroup of index $5$? That would be a subgroup of order $12$. And in fact, $A_5$ does contain subgroups of order $12$, for example, the stabilizer of a point is isomorphic to $A_4$, of order $12$. Thus, $A_5$ is a simple group of order at least $6$ that has a subgroup of index $5$. So 3 is false (though "barely", as it were, since a simple group as given does not have subgroups of indices 2, 3, or 4).
$S_4$ does you no good, since $S_4$ has lots of normal subgroups: $A_4$, and also a normal subgroup of order $4$. |
H: A Question about limit of function on real
First of all, assuming ONLY the knowledge of sequential characterization of limit and also the epsilon-delta formulation of limit, why is the following limit undefined?
$$\lim_{x\to 0} \sqrt{x}$$
This question is from Schroder's "Mathematical Analysis". My argument is following which somehow shows that for some epsilon I can always find a delta which justifies the guess that the above limit is zero:
Want: $|x|<\delta$ such that $|\sqrt{x}-0|<\epsilon$.
Argument:
$|\sqrt{x}|<\epsilon$
$\leftrightarrow$ $|x|<\epsilon^2$ by squaring both sides,
Take $\delta=\epsilon^2$ for any given epsilon. Therefore the limit holds.
Anyone is welcome to point out the mistake.
AI: A common convention is to require the function to be defined in a punctured neighborhood of the point in question (but see below). The limit is undefined because the function $\sqrt{x}$ is not defined in a punctured neighborhood of $0$ (it is not defined at any negative numbers); so the limit cannot exist.
Under this convention, the definition would look as follows:
Let $a$ be a point such that $f$ is defined on a punctured neighborhood of $a$. The limit of $f(x)$ as $x\to a$ is $L$ if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that
$$\text{if }0\lt |x-a|\lt\delta,\text{ then }|f(x)-L|\lt\epsilon.$$
Here, taking $x$ negative, and satisfying $|x|\lt\epsilon^2$, the premise is satisfied by the consequent is not satisfied because $f(x)$ does not even make sense.
That said: there is another convention which only requires the point to be an accumulation point of the domain. In that case, the definition would look something like:
Let $a$ be an accumulation point of the domain $D$ of $f(x)$. The limit of $f(x)$ at $a$ is $L$ if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that
$$\text{if }x\in D\text{ and }0\lt|x-a|\lt\delta,\text{ then }|f(x)-L|\lt\epsilon.$$
Under that convention, your argument would be generally correct, once you add the necessary caveats that you are only considering points in the domain. But I suspect this is not the convention followed by the book you are using. |
H: Equivalence of continuity of a functional on a locally convex space
Let $X$ be a locally convex space whose topology is defined by a family of seminorms $\mathcal{P}$. Let $f$ be a linear functional on $X$. Then, I am trying to prove that the following statements are equivalent.
$f$ is continuous.
There are seminorms $p_1,...,p_n\in \mathcal{P}$ and positive scalars $\alpha_1,...,\alpha_n$ such that $$|f(x)|\leq \sum_{k=1}^n\alpha_kp_k(x).$$
I am stuck in proving $1\Rightarrow 2$. Applying the definition, and that $f$ is continuous at $0$, I get that there exist seminorms $p_1,...,p_n$ and $\epsilon_1,...,\epsilon_n>0$, such that whenever $$x\in \bigcap_{k=1}^n\{y:p_k(y)<\epsilon_k\}\Rightarrow |f(x)|<1.$$ But I am not able to conjure up positive $\alpha_1,..,\alpha_n$ such that above inequality holds. A hint will be very appreciated. Thanks.
AI: Since $f$ is continuous at $0$, the preimage $U=f^{-1}[(-1,1)]$ is open. By linearity, $0\in U$, so $U$ contains an open neighborhood of $0$. Since sets of the form
$$V=\{x\in X: p_1(x),\ldots,p_n(x)<\epsilon\}$$
form a local basis for $0$, we have some such set $V\subseteq U$. Thus we have some $\epsilon>0$ and $p_1,\ldots,p_n\in \mathcal P$ such that
$$\sum_{i=1}^n p_i(x)<\epsilon\implies p_1(x),\ldots,p_n(x)<\epsilon\implies |f(x)|<1$$
and for any $x\in X$ we have
$$\sum_{i=1}^n p_i\left(\frac{\epsilon}{2\sum_{i=1}^n p_i(x)}x\right)=\frac{\epsilon}{2\sum_{i=1}^n p_i(x)}\sum_{i=1}^n p_i\left(x\right)=\epsilon/2$$
thus
$$\frac{\epsilon}{2\sum_{i=1}^n p_i(x)}|f(x)|=\left|f\left(\frac{\epsilon}{2\sum_{i=1}^n p_i(x)}x\right)\right|<1$$
which gives us
$$f(x)\leq \frac{2\sum_{i=1}^n p_i(x)}{\epsilon}=\sum_{i=1}^n\frac{2}{\epsilon}p_i(x).$$
Edit: If $\sum\limits_{i=1}^n p_i(x)=0$ for some $x\in V$, then for any $\lambda\in\mathbb R$ we have $$\sum\limits_{i=1}^n p_i(\lambda x)=|\lambda| \sum\limits_{i=1}^n p_i(x)=0$$
thus $\lambda x\in V$, so $|\lambda||f(x)|=|f(\lambda x)|<1$. Since $\lambda$ is arbitrary, this means $|f(x)|=0$. |
H: Prove this argument is valid: (~N v (~B*D), ~C --> ~D therefore ~(~C*N))
Prove the following argument is valid (and provide reasons):
~N v (~B*D)
~C --> ~D therefore ~(~C*N)
Our work (so far):
~N v (~B*D)
~C --> ~D therefore ~(~C*N)
D-->C (contrapositive of 2)
~N v (~B*C) (substitution 3 into 1)
~N v ~(B v ~C) (Demorgan's on 4)
(~N v ~B)*(~N v C) (distribute 4)
~(N*B)*(C v ~N) (demorgan's and commutive on 6)
...
C v ~N
~(~C*N) (Demorgan's on the previous statement, which is being numbered 1 again even though I gave it number 99)
Line 7 has C v ~N in it, but I can't show that ~(N*B) is true, or that (~Nv~B) in line 6 is true.
This is where I've been for the past 6 hours. Help me out if you can.
Thanks.
AI: As you observed, $\sim(\sim C\land N)$ is equivalent to $C\lor\sim N$, by de Morgan's law. So if we prove $C\lor\sim N$, we can use de Morgan's law in reverse to get what we want.
(1) says $\sim N \lor (\sim B\land D)$. So we have two cases to consider: either $\sim N$ is true, or $\sim B\land D$ is true.
In the former case, we have $\sim N$, so we have $C\lor \sim N$ by the addition law, which says that $x$ implies $y\lor x$ for any $y$.
In the latter case, we have $\sim B\land D$, so we have $D$. (2) says $\sim C\to\sim D$, and by the contrapositive law this is equivalent to $D\to C$. Since we have $D$, we can conclude $C$ by detachment, and then $C\lor \sim N$ by addition.
In either case we have $C\lor \sim N$ and we can use de Morgan's law to get $\sim(\sim C \land \sim\sim N)$ and then eliminate the double negative to get $\sim(\sim C \land N)$ , which was what we wanted. |
H: if $f(n)$ is multiplicative prove that $f(n)/n$ is also multiplicative.
The question asks that if $f(n)$ is multiplicative to prove that $f(n)/n\qquad$ is also multiplicative.
This is what I have:
So, $f(n)\quad$ is multiplicative means that if $p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}\qquad$ is the prime-power decomposition of $n$, then
$f(n)=f(p_1^{e_1})f(p_2^{e_2})\cdots f(p_k^{e_k})$
Now I say, let $g(n)=f(n)/n\qquad $
then $g(n)=f(p_1^{e_1})f(p_2^{e_2})\cdots f(p_k^{e_k})/p_1^{e_1}p_2^{e_2}\cdots p_{k}^{e_{k}}\quad$. Which is by definition multiplicative since $gcd(p_1^{e_1},p_2^{e_2},\ldots,p_k^{e_k})=1\qquad$ and $g(p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}) = f(p_1^{e_1}) f(p_2^{e_2}) \cdots f(p_k^{e_k})/p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$.
I can't think of any other thing to do. Thank you in advance!
AI: You have it! Just break up the fraction to recognize it as the product of values of $g$ at the prime powers.
Of course, you don't have to go all the way to prime factorizations. Let $g(n) = f(n)/n$. We need to show that if $\gcd(a,b)=1$, then $g(ab) = g(a)g(b)$. We know that $f(ab)=f(a)f(b)$. So
$$g(ab) = \frac{f(ab)}{ab} = \frac{f(a)f(b)}{ab} = \frac{f(a)}{a}\frac{f(b)}{b} = g(a)g(b).$$ |
H: Messenger Riddle
A column of troops one km long is moving along a straight road at a
uniform pace. A messenger is sent from the head of the column,
delivers a message at the rear of the column and returns. He also
moves at a uniform pace and arrives back at the head of the column
when it has just covered its own length. How far did the messenger
travel?
I can't get any ideas on how to start.
Thanks for any help.
AI: Let us assume that the speed of the column is $1$ km per unit of time. For convenience, call that unit an hour. The column took $1$ hour to cover its own length.
Let $v$ be the speed of the messenger. When she is travelling to the back, the combined speed of approach of the messenger and the column rear is $v+1$, so the time it takes is $\frac{1}{v+1}$. Going the other way, the speed at which the messenger gains on the head is $v-1$, so the time it takes to gain the whole $1$ km is $\frac{1}{v-1}$. The whole task took $1$ hour, and therefore
$$\frac{1}{v+1}+\frac{1}{v-1}=1.$$
This gives $v=1+\sqrt{2}$. The time taken is an hour, so the distance travelled is $1+\sqrt{2}$. |
H: What is the geometric meaning of singular matrix
Could anyone help explain what is the geometric meaning of singular matrix ? What's the difference between singular and non-singular matrix ? I know the definition, but couldn't understand it very well.
Thanks.
AI: A matrix can be thought of as a linear function from a vector space $V$ to a vector space $W$. Typically, one is concerned with $n\times n$ real matrices, which are linear functions from $\mathbb R^n$ to $\mathbb R^n$. An $n\times n$ real matrix is non-singular if its image as a function is all of $\mathbb R^n$ and singular otherwise. More intuitively, it is singular if it misses some point in $n$-dimensional space and non-singular if it doesn't. |
H: If $\phi^{-1}(0)$ in $\phi:X\rightarrow Spec\; \mathbb{C}[t]$ is a complete intersection, then is $\phi$ flat?
This is a simple question so I am hoping the answer is quite simple as well.
Suppose $\phi:X\rightarrow Spec\; \mathbb{C}[t]$ is a map such that the algebraic variety $\phi^{-1}(0)$ is a complete intersection. Does that mean $\phi$ is flat?
Note that I am not imposing any conditions on $X$.
$$
$$
AI: No.
If you take for $X$ the subvariety $X\subset \mathbb A^2(\mathbb C)$ defined by $xy=0$, the first projection $\phi: X\to \mathbb A^1_\mathbb C=Spec(\mathbb C[t])$ is not flat because the dimension of the fibers $\phi^{-1}(t)$ jumps at $t=0$, nevertheless $\phi^{-1}(0)\cong \mathbb A^1(\mathbb C)$ is a complete intersection.
Edit
It is easy to see purely algebraically that $\phi$ is not flat.
Indeed, the dual map $\phi^*:\mathbb C[t]\to \mathbb C[x,y]/(xy):t\mapsto \bar x$ is not flat because over a PID (like $\mathbb C[t]$) the flat modules are exactly the torsion-free modules and here the element $\bar y\in \mathbb C[x,y]$ is a torsion element: $t\cdot \bar y=\bar x \bar y=0$ |
H: sequence of open intervals
Let $ \displaystyle{ \{ (a_n , b_n) : n \in \mathbb N \} }$ a sequence of open intervals on $\mathbb R$ such that $ \displaystyle{[0,15] \subset \bigcup_{n=1}^{n_0} (a_n ,b_n) }$ for some $ n_0 \in \mathbb N$
Prove that $ \displaystyle{ \sum_{n=1}^{n_0} \mu(a_n ,b_n) = \sum_{n=1}^{n_0} (b_n -a_n) >15 } \quad$ where $ \mu$ is the Lebesgue measure on $ \mathbb R$
I need to prove this using induction but I can't see how.
AI: Use monotonicity and countable sub-additivity of $\mu$
$$15=\mu ([0,15])<\mu (\bigcup_{n=1}^{n_0}(a_n,b_n))\leq\sum_{n=1}^{n_0}\mu(a_n,b_n)$$
The first inequality is strict because $[0,15]$ is closed and $\bigcup_{n=1}^{n_0}(a_n,b_n)$ is open and the containment is strict. |
H: Likelihood of continuing a discrete series
If I have a series like "aaabbbabaabaabbaa" and I'd like to know whether next element in series is "a" or "b", would a Poisson distribution be best solution for this?
If so, checking Poisson for "3" wouldn't be right since it can be >3 in the future. Would the the proper solution be 1-(P(1)+P(2))?
Thanks
AI: That really depends on how you want to model the series. I would not use a Poisson distribution here because it basically corresponds to random coin flipping which means that you cannot predict "a" or "b" from the previous events.
If you can assume that the string distribution is stationary, which means that the probability of occurence of "a" or "b" does not depend on the absolute position in your string, I would use a nth-order Markov model. This means that you model the probability $P(s[i]|s[i-1],...,s[i-n])$. You can simply estimate it via relative frequency.
For example, if you use a first order Markov model, you get
$$P(s[i]=a|s[i-1]=a) = \frac{\#(s[i]=a \mbox{ and } s[i-1]=a)}{\#(s[i-1]=a)}$$
and similarly for $P(s[i]=a|s[i-1]=b)$, $P(s[i]=b|s[i-1]=a)$, and $P(s[i]=b|s[i-1]=b)$.
If you want to predict whether what the next letter will be, you simply take the one which has larger conditional probability. |
H: smooth maps between smooth manifolds - Jacobian coordinate independence
I have question about comment in Lee's Introduction to Smooth Manifolds - page 51.
Given smooth map $F:M\to N$ between smooth manifolds $M$ and $N$ we say that the total derivative of $F$ at $p\in M$ (given chart $(U,\varphi)$ around $p$ and $(V,\psi)$ around $F(p)$) is given by $D(\psi\circ F\circ \varphi^{-1})(\varphi(p))$. The comment in the book is that total derivative is chart independent. If there is another chart $(U',\varphi')$ around $p$ then we should have
$$D(\psi\circ F\circ\varphi'^{-1})(\varphi'(p))=D(\psi\circ F\circ \varphi^{-1}\circ \varphi \circ \varphi'^{-1})(\varphi'(p))=$$
$$D(\psi\circ F\circ\varphi^{-1})(\varphi(p))\cdot D(\varphi \circ\varphi'^{-1})(\varphi (p)).$$
If total derivative is chart independent we should have $D(\varphi\circ \varphi'^{-1})(\varphi'(p))=\mathbb{Id}$, which doesn't have to be the case.
Whats wrong in my reasoning?
AI: "Chart independent" does not mean what you think it means.
A smooth vector field $v$ on a manifold $M$ is chart independent: treating it as a derivation on smooth functions, $v(f)$ is a scalar independent of which vector field in which you compute it. But the expression of the vector field $v$ in a coordinate basis $v = \sum v^i \partial_i$ is clearly dependent on the choice of charts. In fact, any nontrivial element $k\in T_pM$, when written in a coordinate basis, will depend on the choice of charts. This is simply the fact that different coordinate charts give rise to possibly different coordinate bases of $T_pM$, and the only element $k$ in a vector space whose coordinate representation is the same in all bases is the 0 element.
In other words, "chart independent" should be taken to mean that the object is naturally co/contravariant under changes of bases in (co)tangent spaces, or equivalently changes of charts.
As you can easily verify: if $k\in T_pM$, then $DF(k) \in T_{F(p)}N$ is independent of choice of charts, since the change of representations of $k$ as a vector under the chart $(U,\varphi)$ and under the chart $(U',\varphi')$ precisely cancels the factor $D(\varphi\circ\varphi'^{-1})$ you get. |
H: Ball from platform with specific vertex
On earth in a vacuum. You throw a platon from a platform height
$h$ and want it to land at point $d$ distant. Note, h is absolutely fixed and d is absolutely fixed. It "must land" at point d, no matter what. You throw it with velocity expressed using $Vx$ and $Vy$.
Now,
(Problem AA)
you want the vertex HEIGHT to be at a percentage $H$ >100 of $h$ (say $H=120 \% $).
-- ~~ ~~ OR ~~ ~~ --
(Situation BB)
you want the vertex DISTANCE to be at a percentage $D$ <50 of $d$ (say $D=25 \% $).
NOTE: the two gentlemen below have generously explained that you CANNOT choose BOTH H and D. Thank you very much for this insight and proof!
So! For each of ProblemAA and ProblemBB, how to calculate $Vx$ and $Vy$ ?
If this is possible - thank you!
{Aside: I assume there's only one $Vx$ / $Vy$ solution for a given value in either ProblmeAA or ProblemBB - but could there be two, or more??}
AI: Instead of initial speed $v$ and angle $\alpha$, it's easier to work in Cartesian coordinates, with horizontal and vertical components of velocity $v_x$ and $v_y$.
The height of the vertex is $h + v_y^2/2g$, where $g$ is the acceleration due to gravity. To have a height of $h\cdot\frac H{100}$, you need
$$v_y = \sqrt{2gh\left(\frac H{100}-1\right)}.$$
As the horizontal velocity is constant, the ratio of distance in the horizontal direction is the same as the ratio of the corresponding times. The vertical displacement of the projectile is $h + v_yt - \frac12gt^2$. The time to reach the vertex is $v_y/g$, while the total time to drop to height $0$ is $\big(v_y + \sqrt{v_y^2+2gh}\big)/g$. You want their ratio to be $\frac D{100}$, which implies that
$$v_y = \sqrt{2gh\left(\frac{(D/100)^2}{1-2(D/100)}\right)}.$$
If the initial $h$ is fixed, you can't choose both $H$ and $D$ independently; there will be no solution that has both the $H$ you want and the $D$ you want, unless by coincidence they both give the same value of $v_y$.
The horizontal velocity of the projectile doesn't come into this at all. If you pick one of the equations, say you specify $H$ and decide you don't care about $D$, then after you have the solution for $v_y$ you can always choose $v_x$ to make the projectile land at distance $d$. In fact, since the time of flight is $\big(v_y + \sqrt{v_y^2+2gh}\big)/g$, and the projectile must travel a distance $d$ in that time, you can just set
$$v_x = \frac{dg}{v_y + \sqrt{v_y^2+2gh}}.$$ |
H: Behaviour of $\sum_{k=1}^n\left(\left(\frac{3}{2}\right)^k\ (\mathrm{mod}\ 1)\right)$
Using Mathematica I found that the relation
$$\sum_{k=1}^n\left(\left(\frac{3}{2}\right)^k\ (\mathrm{mod}\ 1)\right)\approx\frac{n}{2}$$
seems to hold. Actually, every fraction of the form $\frac{b}{a}$, with $b>a$ and $\mathrm{gcd}(a,b)=1$, seems to behave similarly.
Example, $\frac{3}{2}$:
So, can we prove some asymptotic formula or somehow show that this behavior is constant?
AI: You might be interested in this: http://mathworld.wolfram.com/PowerFractionalParts.html |
H: All integer solutions for $x^4-y^4=15$
I'm trying to find all the integer solutions for $x^4-y^4=15$.
I know that the options are $x^2-y^2=5, x^2+y^2=3$, or $x^2-y^2=1, x^2+y^2=15$, or $x^2-y^2=15, x^2+y^2=1$, and the last one $x^2-y^2=3, x^2+y^2=5$.
Only the last one is valid. $x^2+y^2=15$ is not solvable since the primes which have residue $3$ modulo $4$ is not of an equal power.
One particular solution for $x^2-y^2=3, x^2+y^2=5$, is $x_0=2, y_o=1$. How do I get to an expression of a general solution for this system?
Thanks!
AI: $$(x^2-y^2)+(x^2+y^2)=2x^2=3+5=8\Rightarrow x^2=4\Rightarrow x=2,-2$$
$$(x^2+y^2)-(x^2-y^2)=2y^2=5-3=2\Rightarrow y^2=1\Rightarrow y=1,-1$$
Substitute the values to check that these are indeed the soloutions. |
H: Changing domain of PDE (should be easy)
Suppose I have a PDE
$$u_t(x,t) + f(x,t)\cdot \nabla u(x,t) = 0 \quad \text{on $\gamma(t)$}$$
where $\gamma(t)$ is a curve for each fixed t in $\mathbb{R}^2$ and $f$ is given.
We have that $\gamma(t)$ is parametrised by the function $X(s,t)$ for each $t$. So $X(s,t) = x \in \gamma(t)$ for some $s$. So $x \in \mathbb{R}^2$, $s \in S^1$ or some interval in the real line.
I want to write the PDE as a function of $(s,t)$ instead of $(x,t).$ So basically I want to solve a PDE for the unknown function $v(p,t) = u(X(p,t), t)$. How can I convert my PDE to this domain? I can't find any expressions for $u_t$ and $\nabla u$ in terms of $v$..
AI: Start with
$$ v(s,t) = u(X(s,t),t) $$
By the chain rule you get
$$ \begin{cases} v_t(s,t) &= X_t(s,t) \cdot \nabla u(X(s,t),t) + u_t(X(s,t),t) \\
v_s(s,t) &= X_s(s,t) \cdot \nabla u(X(s,t),t) \end{cases}$$
To solve for $u_t$ and $\nabla u$, you need to solve for three scalar functions. But you only have two equations here. Clearly the system is underdetermined, so in general one cannot solve for $u_t$ and $\nabla u$ interms of $v$.
Plugging in the above into the equation you get
$$ v_t - X_t\cdot \nabla (u\circ X) + f \cdot \nabla (u\circ X) = 0 $$
So if $f - X_t \propto X_s$, that is, if $f - X_t = \alpha(t,s) X_s$, we can plug in
$$ v_t + \alpha X_s \nabla (u\circ X) = 0 \implies v_t + \alpha v_s = 0$$
The condition that $f - X_t \propto X_s$ above is a integrability condition. Given $\gamma(t)$ it specifies a two-dimensional surface is $\mathbb{R}^3 = \{t\}\times \mathbb{R}^2$. Your equation can be regarded as a transport equation in $\mathbb{R}^3$, that a certain vector field $V = (1,f)$ acts on $u$ to give zero. Generically one may expect the vector field $V$ to be transverse to the surface $\gamma(t)$: in that case your equation is underdetermined: values on the surface $\gamma(t)$ depends on values of $u$ off the surface, and in that case it is impossible to write the equation in terms of $v$, which is essentially just the value of $u$ on the surface $\gamma(t)$.
The condition $f - X_t \propto X_s$ ensures that the vector $(1,f)$ is in fact tangential to the surface $\gamma(t)$. This way your vector field $(1,f)$ restricts to a tangent vector field of $\gamma(t)$ and it makes sense to study the equation restricted to the surface. |
H: Finding last digit of numbers raised to large powers
This question came in a competitive exam I took recently.
The last digit of LCM of $3^{2003} - 1$ and $3^{2003} + 1$ is
is there any strategy by which we can quickly determine the answer? I am hoping there is, since the question came in a timed competitive test.
AI: The GCD of $x$ and $y$ divides $x-y$, hence the GCD $3^{2003}-1$ and $3^{2003}+1$ is either $1$ or $2$; it is easy to check that both numbers are even, so this GCD of these is 2.
The LCM of $x$ and $y$ multiplied by the GCD of $x$ and $y$ equals $x \times y$. So the LCM in question equals $\frac{(3^{2003}-1) \times (3^{2003}+1)}{2} = \frac{3^{4006}-1}{2}$.
$3^n \mod 20$ gives us the following sequence: $1, 3, 9, 7, 1, 3, 9, 7, \cdots$; hence $3^{4006} \mod 20 = 3^2 \mod 20 = 9$, and the answer is $\frac{9-1}{2} = 4$. |
H: What is the minimum $ \sigma$-algebra that contains open intervals with rational endpoints
What are the minimum $\sigma$-ring and $\sigma$-algebra on $\mathbb R$ which contain the open intervals with rational endpoints?
Is there a relation between this $\sigma$-algebra and Borels?
AI: Let $\mathcal C:=\{(a,b),a,b\in\Bbb Q\}$.
Denote $\mathcal R$ the $\sigma$-ring generated by $\mathcal C$ and $\mathcal A$ the $\sigma$-algebra generated by $\mathcal C$.
Let $a<b$ two real number, $\{a_n\}$ and $\{b_n\}$ two sequence of rational numbers, $a_n\downarrow a$ and $b_n\uparrow b$. Then
$$(a,b)=\bigcup_{n\geq 1}(a_n,b_n)$$
hence $(a,b)\in\mathcal R$ and $(a,b)\in\mathcal A$. Since we can write each open set as a countable union of intervals, $\mathcal A$ contains the Borel $\sigma$-algebra, and since it's the smallest $\sigma$-algebra containing $\mathcal C$, it's actually the Borel $\sigma$-algebra.
We have $\mathbb R=\bigcup_{n\geq 1}\underbrace{(-n,n)}_{\in\mathcal C}$ hence $\Bbb R\in\mathcal R$. Hence $\mathcal R$ is stable by countable union, complementation and contains the whole real line: it's a $\sigma$-algebra. By the same argument, $\mathcal R=\mathcal B(\Bbb R)$. |
H: proving that $2^{m-1}$ has reminder $1$ when divided by $m$
Let $$m = \frac{4^p - 1}{3}$$
Where $p$ is a prime number exceeding $3$. how to prove that $2^{m-1}$ has reminder $1$ when divided by $m$
AI: $2^{2p}=4^p=3m+1\equiv 1 \pmod m$ so the result follows if $2p\mid m-1$.
Since $m$ is odd $2\mid m-1$, and by Fermat's Little Theorem $p\mid 4^p-4=3(m-1)$. Since $p>3$ is prime we must have $p\mid m-1$. |
H: A question regarding analysis.
Let $d\in\mathbb N$ and
$$
P_d := \left\{p : K \to \mathbb{R} :\quad p(x) = \sum\limits_{i=1}^d p_i x^i\quad\text{ where }\quad \{p_i\}_{i=1}^d \subset \mathbb{R}\right\},
$$
the set of all polynomials of degree at most $d$, where $K$ is a compact set. We have to show that $P_d$ isn't dense in the space of all continuous functions $C(K,\mathbb{R})$.
Our teacher gave us a hint: Consider a continuous function $f$ which has $d + 1$ zeros and show that all polynomials
which approximate $f$ at its zeros with accuracy $\varepsilon > 0$ are uniformly bounded (Lagrange interpolation). Use this observation to show that there are continuous functions $f$ which cannot be approximated by $P_d$ with given accuracy $\varepsilon > 0$.
I can't figure out how to use this hint to solve the problem. Please help me out. Thanks.
AI: Assume $\{p_n\}_n$ is a bounded sequence in $P_d$. We can write
$$
p_n(x)=a_n (x-x_{1,n})(x-x_{1,n})\cdots (x-x_{d,n})
$$
for suitable sequences $\{a_n\}$, $\{x_{i,n}\}_n$, $i=1,\ldots ,d$. Since $x \in K$, a compact set, up to subsequences we can assume
$$
a_n \to a_\infty, \quad x_{i,n} \to x_{i,\infty}
$$
as $n \to +\infty$.
But then $\{p_n\}$ converges uniformly on $K$ to the polynomial
$$
p_\infty (x) = a_\infty (x-x_{1,\infty})\cdots (x-x_{d,\infty}).
$$
In particular, $\deg p_\infty \leq d$.
Remark. This is a rephrasing of Davide Giraudo's comment. |
H: Determine whether $\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$ is convergent or divergent. If convergent, find the sum.
$$\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$$
It's geometric, since the common ratio $r$ appears to be $\frac{-3}{4}$, but this is where I get stuck. I think I need to do this: let $f(x) = \frac{(-3)^{x-1}}{4^x}$.
$$\lim\limits_{x \to \infty}\frac{(-3)^{x-1}}{4^x}$$
Is this how I handle this exercise? I still cannot seem to get the answer $\frac{1}{7}$
AI: A geometric series is convergent if the $|r|<1$ where $r$ is the common ratio.
Let $S_n=\sum_{i=0}^n (-3/4)^i$ then $$S_n=\frac{(-3/4)^{n+1}-1}{(-3/4)-1}$$
Now take $n\rightarrow \infty$ then $$S_n\rightarrow \frac{0-1}{(-3/4)-1}=4/7$$
because $|-3/4|<1$ and so $(-3/4)^n\rightarrow 0$.
Now note that your sum is $$\mbox{lim }\sum_{i=1}^{n+1}\frac{(-3)^{i-1}}{4^{i}}=\mbox{lim }\frac{1}{4}\sum_{i=1}^{n+1}\frac{(-3)^{i-1}}{4^{i-1}}=1/4.\mbox{lim }S_n=1/7$$ |
H: Fraction of two binomial coefficients
In an exercise I was asked to simplify a term containing the following fraction: $${\binom{m}{k}\over\binom{n}{k}}$$
The solution does assume the following is true in the first step, without explaining why. I unfortunately cannot reconstruct the step:
$${\binom{m}{k}\over\binom{n}{k}} = {\binom{n-k}{m-k}\over\binom{n}{m}}$$
Has anyone a good explanation why this transformation is possible?
AI: I don't know if it should be a memorized formula necessarily, but I can justify it combinatorially:
$$\binom{n}{m}\binom{m}{k} $$
Counts the number of ways of selecting $m$ items out a pool of $n$ and placing them in bin A, and then selecting $k$ items out of bin A and putting them in bin B. You end up with $m-k$ items in the first bin and $k$ items in the second. We might have well as done it in reverse order though: select $k$ items out of $n$ to put into bin B, and then select $m-k$ items out of $n-k$ to put into bin A. Hence:
$$=\binom{n}{k}\binom{n-k}{m-k}.$$
Equating these two yields the given equality of ratios after rearrangement. |
H: infinite number of irreducible polynomials in $\mathbb{Z}/2{\mathbb Z}[X]$
For $A= \mathbb{Z}/2{\mathbb Z}[X]$ ring of polynomials with coefficient in the field $\mathbb{Z}/2{\mathbb Z},$ I need to show that there are infinite number of irreducible polynomials in $A.$
How do I show that? I didn't come to any conclusion. I though of series of polynomials but since it it modulo $2$ they were not suitable.
Any direction?
(And: does any one have a link to a web where I can choose Latex symbols and see how they are written? I had one, but I lost it, and can't find it in Google)
AI: A variant of Euclid's proof should work fine. Assume there are only finitely many irreducible polynomials $p_1,...,p_n$ in $A$, and consider the irreducible factors of $\prod_{i=1}^n p_i + 1$. |
H: Probability question (two draws)
If I draw 2 balls from a bag which contains 2 pink, 3 blue and 4 orange balls, what's the probability that the first ball would be pink and the second blue - as a decimal? Thanks. :)
AI: Hint: what is the chance that you get a pink ball on the first draw? Given that one pink ball is removed (I presume you are drawing without replacement), what is the chance that the second draw is blue? The chance of both is the product. Now you have a fraction to express in decimal. What fraction is it? |
H: Function derivative in the vector direction
Let there be a function $z=x^3+xy+y^2$. Is there a vector $\vec U$ that in the point $(1,1)$ the function derivative in the vector direction is equal to 6?
Now I know that the $|\nabla F(1,1)|$ is $\sqrt{4^2+5^2} = \sqrt{41} = 6.4 > 6$.
So there is a vector $\vec U$ that equal to 6.
How can I find it?
Thank!
AI: For every unit vector $\mathbf{u}=(u_1,u_2)$, you know that
$$
\frac{\partial z}{\partial \mathbf{u}} = \nabla z \cdot \mathbf{u} = \frac{\partial z}{\partial x}u_1 + \frac{\partial z}{\partial y}u_2.
$$
You can now evaluate all partial derivatives at $(1,1)$ and solve the equation
$$
\frac{\partial z}{\partial x}u_1 + \frac{\partial z}{\partial y}u_2=6.
$$
Since
$$
\frac{\partial z}{\partial x} = 2x+2y
$$
and
$$
\frac{\partial z}{\partial y} = 2x+3y^2
$$
you must solve
$$
4u_1 +5 u_2 = 6
$$
under the constraint $u_1^2+u_2^2=1$. |
H: Uniqueness of extension of zero measure
Let $(\Omega,\mathscr F)$ be a measurable space with two probability measures $\mu, \nu: \mathscr F\to[0,1]$ defined over it. Suppose that $\mathscr C\subset\mathscr F$ is some class of sets and
$$
(\mu-\nu)|_\mathscr C = 0.
$$
Which necessary and sufficient conditions are known on $\mathscr C$ in order to assure that $\mu = \nu$?
For example, it is sufficient for $\mathscr C$ to be a ring of sets generating $\mathscr F$, as it follows from Caratheodory's extension theorem. However, is that condition also necessary?
In particular, if $\sigma(\mathscr C) = \mathscr F$ can it be the case that $\mu\neq \nu$?
AI: A sufficient condition is that $\mathcal{C}$ is a set of generators of $\mathcal{F}$ closed under pairwise intersections (a $\pi$-system). The proof goes via Dynkins lemma and can be found in "Probability with Martingales" by Williams.
Edit: The first version was not quite right. This works: Let $\Omega=\{1,2,3,4\}$ and $\mathcal{F}=2^\Omega$. Let $$\mathcal{C}=\big\{\{1,2\},\{3,4\},\{1,3\},\{2,4\}\big\}.$$ Clearly $\mathcal{F}=\sigma(\mathcal{C})$. Define $\mu$ by $\mu\{1\}=\mu\{4\}=1/6$ and $\mu\{2\}=\mu\{3\}=1/3$. Define $\nu$ by $\nu\{1\}=\nu\{4\}=1/3$ and $\nu\{2\}=\nu\{3\}=1/6$. Then $\nu$ and $\mu$ agree on $\mathcal{C}$, but are different probability measures. The counterexample is from "Counterexamples in Probability" by Stoyanov. |
H: Equivalences of $S^n$ vs. $\Omega^nS^n$
Let $H(n)$ be the group of self-homotopy-equivalences of $S^n$ preserving the basepoint. I read that $H(n)$ may be identified with ''two components of $\Omega^nS^n$''.
What does this mean and how can I see it?
AI: Maps $[S^n,S^n]$ are classified by degree. For each $z\in\mathbb{Z}$ there is a component of $\Omega^nS^n$. The only maps that are homotopy equivalences have degree $1$ and $-1$.
EDIT: In answer to the comment below from JasonDevito: For any finite CW complex $X$, $[S^1,X]$ is the same thing as $[S^0,\Omega X]$, since $\Omega$ and $\Sigma$ (reduced suspension) are adjoints. But, based maps from $S^0$ to $Y$ is the same as $Y$ for any finite CW complex. Thus
$$
[S^1,X]=[S^0,\Omega X]=\Omega X.
$$
Iterate this and you get
$$
[S^n,S^n]=[\Sigma S^{n-1},S^n]=[S^{n-1},\Omega S^n]=\cdots =\Omega^n S^n.
$$ |
H: $\limsup$ and cluster points
Let $(x_n)$ be a sequence in $\mathbb R$. We call $y$ a cluster point of $(x_n)$ iff for every neighborhood $N$ of $y$ there are infinitely many $n$ such that $x_n \in N$.
Let $C$ denote the set of cluster points of $(x_n)$.
By definition, $\limsup_{n \to \infty} x_n = \inf_n \sup_{k \geq n} x_k$.
Can you show me how to prove $\limsup_n x_n = \sup C$? Thanks.
AI: Put $L = \limsup_n x_n$ and $s_n = \sup_{k \ge n} x_k$. Notice that $s_n$ is a decreasing sequence in $n$ and that $s_n \downarrow L$.
Let $\lambda > L$. then for some $n$, $\sup_{k\ge n} x_k <\lambda$. Hence there can be no limit point of the sequence in $(\lambda, \infty)$.
Since $\lambda > L$ was chosen arbitrarily, all limit points of the sequence must lie in $(-\infty, L]$.
We will be done if we can show that the limsup is a limit point of the $x_n$. We choose a subsequence as follows. Choose $n$ so that $\sup_{k\ge n} x_k < L - 1/2$; now fix $n_1$ so $\sup_{k\ge n_1} x_k > L - 1/2$. Now suppose $n_1 < n_2 < \cdots < n_k$ are chosen. Since $s_n\downarrow L$ there is $N$ so that $s_N > L - 1/2^n$. Since $S$ is decreasing, we can choose $N > n_k$. Put $N = n_{k+1}$.
The sequence $\{x_{n_k}\}$ converges to $L$, so $L$ is a limit point of the $x_n$. |
H: Normal distribution of the dosage
I am having an issue with the following task:
For the dose of the MemPro medicine it is known that has normal distribution. Sample of 30 parameters has been taken into consideration and based on the obtained data, we know that mean of a dose of this medicine is 5grams. With possibility that alpha error is 5%, test the hypothesis that mathematical expectation is 6grams.
Could you please elaborate this, I have no idea how to solve this matter. Thanks in advance.
AI: This is a test of a null hypothesis that the mean is 6 grams. Since you have 30 observations from a normal idstribution the appropriate test is a t test. The test considers the pivotal (sample mean - 6)/(sample std dev/√30) in your case since n=30 and the hypothesized mean = 6. Since sample mean=5 this pivotal quantity is -√30/sample std dev. This value is to be compared to a t distribution with 29 degrees of freedom. You need to calculate the sample standard deviation from your sample of 30 observation to arrive at this number. You did not specify whether this is a one-sided or two-sided test. For the moment I will assume it is one-sided at level alpha=0.05. Then you look up the 5th percentile for a t distribution with 29 degrees of freedom and reject the null hypothesis that the true mean parameter is 6 or more if the test statistic is below the 5th percentile. From the table I see that the critical value (5th percentile) is -1.699. So compare -30/sample std dev to -1.699 to determine whether or not to reject. |
H: Confidence interval of the car sample
In the sample of 25 cars of the VW Golf 5 recently sold in Augsburg, average cost was 9.000 euros with standard devation of 300 euros. Assume that the cost of all the cars of this type of vehicle are normally distributed. Construct 90% of confidence interval for the average cost of all the cars of this type of vehicle.
I need help with this task. I have tried many ways but none of them seems to be correct.
AI: To construct the confidence interval you again use a t statistic. In this case the pivotal quantity is (9000 -m)/(300/√25) where m is the true unknown mean. This simplifies to
(9000-m)/60 To construct a two-sided 90% confidence interval we need m to satisfy the condition
t(0.05)< (9000-m)/60 < t(0.95) where t(0.05) is the 5th percentile of the t distribution with n-1 = 24 degrees of freedom and t(0.95) is the 95th percentile for a t distribution with 24 degrees of freedom. Because the t distribution is symmetric about 0 t(0.05)=-t(0.95).
So you take the inequality -t(0.95) < (9000-m)/60 < t(0.95) and rearrange it to be an inequality for m as 9000-60 t(0.95) < m < 9000+60 t(0.95).
The interval [9000-60 t(0.95), 9000+60 t(0.95)] is your answer. From the t distribution with 24 degrees of freedom you will see that t(0.95)=1.711. |
H: Question on meaning of a symbol: long thin C
I don't know how to write it in $\LaTeX.$ It is a tall skinny bold C. This is the context: A set is defined by:
where $\complement\atop{\smash \scriptstyle i}$ is the thing I don't understand. The $i$ is actually directly underneath the weird $C$ in this case. Can anyone explain what this means?
AI: It's the coefficient operator. It extracts the ith coefficient of the Taylor expansion. This is used a lot in combinatorics with generating functions. |
H: Seeking formula for a custom game mechanic
We are game developers currently working on a new game. We are looking to solve the following problem:
We start with the following variables:
$$x = 0, h = 1;$$
The variable x should always increase over time depending on $h$—the larger the $h$ the larger the rate that $x$ increases. On the other hand $h$ decreases over time with a steady velocity. $h$ ranges from 0 to 1.
After time $t$ the variable $x$ will become:
$$x := x + \sum( C_1 \cdot ( h + V \cdot n ) ) \mathrm{\ for\ } n = [0, t]. $$
$C_1$ is a constant.
$V$ is the change rate of $h$.
Whenever we calculate $x$ we need to calculate the new value of $h$ like so:
$$h := h + V \cdot t$$
So what happens in order is this:
$x := x + \sum( C_1 \cdot ( h + V \cdot n ) ) \mathrm{\ for\ } n = [0,t]$
$h := h + V \cdot t$
Both of these formulas could be wrong of course, this is simply what we are trying to achieve.
The variable t is essentially time in seconds. The problem is that if say $t = 2$ then $x$ will have a certain value if we calculate it like so:
$x := x + \sum( C_1 \cdot ( h + V \cdot n ) ) \mathrm{\ for\ } n = [0,2]$
$h := h + V \cdot 2$
and a different value if we calculate it like so:
$x := x + \sum( C_1 \cdot ( h + V \cdot n ) ) \mathrm{\ for\ } n = [0,1]$
$h := h + V \cdot 1$
$x := x + \sum( C_1 \cdot ( h + V \cdot n ) ) \mathrm{\ for\ } n = [0,1]$
$h := h + V \cdot 1$
Essentially the values are different if we break up the calculation into more steps.
We need the values to be the same in both cases. We cannot do this with these formulas so I was wondering if there is another way to do this. I understand that my explanation might be a little difficult to understand but it's not very easy for me to explain this!
Thank you
AI: You can use the differential equations to get an exact solution. You have
$x'=C_1h$ where the prime is the derivative with respect to time
$h'=-C_2t$ Then
$h=h(0)-C_2t$
$x=x(0)+C_1h(0)t-\frac 12 C_2 t^2$
In your case $x(0)=0,\ h(0)=1$, so
$h=1-C_2t$
$x=C_1t-\frac 12 C_2 t^2$ |
H: A question concerning on the axiom of choice and Cauchy functional equation
The Cauchy functional equation:
$$f(x+y)=f(x)+f(y)$$
has solutions called 'additive functions'. If no conditions are imposed to $f$, there are infinitely many functions that satisfy the equation, called 'Hamel' functions. This is considered valid if and only if the Zermelo's axiom of choice is accepted as valid.
My question is: suppose we don't consider valid the axiom of choice, this means that we have a finite number of solutions? Or maybe the 'Hamel' functions are still valid?
Thanks for any hints ore answer.
AI: What you wrote is not true at all. The argument is not valid "if and only if the axiom of choice holds".
Note that there are always continuous functions of this form, all look like $f(x)=ax$, for some real number $a$. There are infinitely many of those.
The axiom of choice implies that there are discontinuous functions like this, furthermore a very very weak form of the axiom of choice implies this. In fact there is very little "choice" which can be inferred from the existence of discontinuous functions like this, namely the existence of non-measurable sets.
Even if the axiom of choice is false, it can still hold for the real numbers (i.e. the real numbers can be well-ordered even if the axiom of choice fails badly in the general universe). However even if the axiom of choice fails at the real numbers it need not imply that there are no such functions in the universe.
We know that there are models in which all functions which have this property must be continuous, for example models in which all sets of real numbers have the Baire property. There are models of ZF in which all sets of reals have the Baire property, but there are non-measurable sets. So we cannot even infer the existence of discontinuous solutions from the existence of non-measurable sets.
Observe that if there is one non-discontinuous then there are many different, since if $f,g$ are two additive functions then $f\circ g$ and $g\circ f$ are also additive functions. The correct question is to ask whether or not the algebra of additive functions is finitely generated over $\mathbb R$, but to this I do not know the answer (and I'm not sure if it is known at all).
More:
Is there a non-trivial example of a $\mathbb Q-$endomorphism of $\mathbb R$? |
H: Finding the diagonal of a rhombus
I am trying to show that the diagonals of a RHOMBUS intersect each other at 90 degrees
However I need to find the length of the diagonal without using Trig. ratios. Any ideas how i could find that ?
The length of each side in the figure is 6.
AI: We give a traditional "angle-chasing" argument. Draw lines $AC$ and $AD$. Label their intersection point $I$.
Note that by the definition of rhombus, $\triangle ABC$ is isosceles, so $\angle BAC=\angle ACB$. Since $\triangle ABC$ and $\triangle ADC$ are congruent, $\angle DAC=\angle DCA=\angle BAC=\angle ACB$.
Work now with the other diagonal. The same argument as in the preceding paragraph shows that all the angles it forms with the sides are equal.
Now look at $\triangle AIB$ and $\triangle CIB$. Since $\angle IAB=\angle ICB$, and $\angle IBA=\angle IBC$, their remaining angles must be equal.
So $\angle AIB=\angle CIB$. But these two angles add up to a "straight angle" ($180^\circ$), so each of them must be $90^\circ$.
The lengths of the diagonals are inextricably tied to trig functions of $80^\circ$ or relatives. These trig functions are not at all "nice," so there is no way to sneak around them. |
H: Probability that TV is defective
It is known that every 100th TV produced is defective. If we choose sample of 10 TVs from the tracks, find:
a) probability that one TV is defective.
b) probability that at most two TVs are defective.
I don't know where to start from. Should I use Poisson distribution here or not?
p = 0.01
a) $f(1,10,p) = (10 1) p(1-p)^{10} = 10p(1-p)^{10}$
b) $f(2,10,p) = 10p(1-p)^{10} + (10 2) p^2(1-p)^8 = 10p(1-p)^{10} + 45 p^2(1-p)^8$
AI: No you probably should not use the Poisson distribution as the Poisson is nothing more than the limiting case of the Binomial where n is large and p is small. So you usually shouldn't use Poisson when $np$ or $nq$ (that is, $n(1-p)$) is less than five.
As a clue:
If a mean or average probability of an event happening per unit
time/per page/per mile cycled etc., is given, and you are asked to
calculate a probability of n events happening in a given time/number
of pages/number of miles cycled, then the Poisson Distribution is
used.
If, on the other hand, an exact probability of an event
happening is given, or implied, in the question, and you are asked to
caclulate the probability of this event happening k times out of n,
then the Binomial Distribution must be used.
in your question
Here we are given a definite probability, in this case, of defective components, $p = 0.01$ and hence $q = 0.99$ = Prob. not defective. Hence, Binomial, with $n = 10$. Expand $(q + p)^{10}$
to get
$q^{10} + 10 q^9 p + 10(10-1)/2! q^8 p^2 + \dots$
So $P(1) = 10 q^9 p = 0.09135$. |
H: Studying this function: $y = \frac{x^2}{1+\log|x|}$
I'm still studying this function:
$$y = \frac{x^2}{1+\log|x|}$$
And now I'm dealing with the study of the monotony. So I got the first derivative, this:
$$y\,' = \frac{x(1+\log{x^2})}{(1+\log|x|)^2}$$
Then I put the first derivative $> 0$:
$$\frac{x(1+\log{x^2})}{(1+\log|x|)^2} > 0$$
And to solve it I set those conditions:
$x > 0$
$1+\log{x^2} > 0$
$(1+\log|x|)^2 > 0$
That give me back these results:
$x > 0$
$x < -\frac{1}{\sqrt{e}}\vee x > \frac{1}{\sqrt{e}}$
$x < -\frac{1}{e} \vee x > \frac{1}{e}$
The problem in these results don't match with Wolfram Alpha's results, so I tried to solve the last condition in another way: not doing te square root of both members, but splitting it in two factors, in this way:
$$(1+\log|x|)(1+\log|x|) > 0$$
In this way it gives me two times the same result, but since these results have to be put on different lines of the positivity graphic, it matches with Wolfram Alpha..
My question now is: why the first way I adopted is wrong? For which mathematical principle?
AI: You have:
$$\frac{x(1+\log{x^2})}{(1+\log|x|)^2} > 0$$
The denominator is greater than zero because it's a square. So you need $x(1+\log{x^2})$ to be greater than zero as well to have a positive fraction. Therefore:
$x>0$
$1+\log{x^2}>0$
Since $\log x$ is defined for all positive real numbers, you only need to worry about $1+\log{x^2}$. |
H: Is this mathematical definition iterative? If not, what does an iterative function look like?
I was debating with someone about iterative vs recursive in programming. I was defending the iterative side. He then said me that the true definition of Fibonacci number is this:
$$f(n) = f(n-1) + f(n-2);\space n > 2$$
with $f(0) = 0$ and $f(1) =1$.
Then I replied with the factorial example, because I had to admit that, for this one, this was right, I don't see any other definition. Factorial is defined by the following recursive mathematical definition:
$$\operatorname{fact}(n)=\begin{cases}1 & \text{if } n=0 \\ n\cdot\operatorname{fact}(n-1) & \text{if } n > 0\end{cases}$$
But, in the same page, they also say that factorial can be implemented iteratively, and the mathematical definition, with the pseudocode that goes with it, is:
function factorial is:
input: integer n such that n >= 0
output: [n × (n-1) × (n-2) × … × 1]
1. create new variable called running_total with a value of 1
2. begin loop
1. if n is 0, exit loop
2. set running_total to (running_total × n)
3. decrement n
4. repeat loop
3. return running_total
end factorial
$$\operatorname{fact}(n) = \operatorname{fact_{acc}}(n, 1)$$
$$\operatorname{fact_{acc}}(n, t)=\begin{cases}t & \text{if } n=0 \\ \operatorname{fact_{acc}}(n-1, n\cdot{t}) & \text{if } n > 0\end{cases}$$
Then he replied to me that it's not an iterative mathematical definition, but another recursive mathematical definition. But, if you check the source (I know that Wikipedia is not the best source, but...), it clearly states that it's the representation of the iterative pseudocode.
The way I read it is that, as long as $n>0$, repeat the content of the function and decrease $n$. $t$ is used as an accumulator, and when $n$ reaches 0, we simply keep the result of what $t$ contains.
Is the above a mathematical definition of a recursive or of an iterative function? If its not an iterative definition, how a iterative mathematical definition for that factorial look like? (Not the code to implement it, the mathematical definition). If there is no way to represent that function with a iterative mathematical definition, can you show some example of very easy to understand iterative mathematical definition, and explain me how to read such definition?
Don't point me out to this: https://math.stackexchange.com/questions/97820/factorial-function-recursive-and-iterative-implementation. It's only the implementation. It doesn't really talk about the mathematical definition. I'm only interested in the mathematical definition, nothing else.
Edit: I want to thank Shaktal who edited my post, and added some formatting. From the formatting he did to the 2 first mathematical definition of the function, I was able to format the rest.
I want to thanks everyone for the clarification about mathematical definition and the implementation. Thanks for your comprehension, I will only see those thing in math next scholar year, but already saw recursion and iterative long time ago in programming.
AI: The usual mathematical definition of factorial is $n! = \prod_{j=1}^n j$. I would say this is neither "iterative" nor "recursive" in your sense: the distinction between those is a matter of implementation rather than mathematics. You might also like $n! = \int_0^\infty x^n e^{-x}\ dx$. Or an even more purely "declarative", combinatorial definition: $n!$ is the number of permutations of $n$ objects.
As for the Fibonacci numbers, you might like the following declarative definition:
$F_n$ is the number of subsets of $\{1,2,\ldots,n-2\}$ that don't contain any two consecutive integers.
There is a notion of "recursive function" in mathematical logic, but that's something quite different. |
H: Give an example of a measure which is not complete
Give an example of a measure which is not complete ?
A measure is complete if its domain contains the null sets.
AI: The canonical example is the Borel measure on the Borel $\sigma$ algebra (the $\sigma$ algebra generated by the open intervals) on $\mathbb{R}.$ This example is often used as a motivation for the construction of the Lebesgue measure. |
H: Compact subsets of the real numbers
Let $C\subset \mathbb{R}$ be compact. I am wondering if
$$C=\bigcup_{i=1}^n[a_i,b_i]$$
then for some $a_i,b_i\in\mathbb{R}$, $a_1\le b_1 < a_2 \le b_2 \dots < a_n \le b_n$. By Heine-Borel, $C$ does indeed lie in some interval $[a,b]$, but is it the finite disjoint union of such intervals?
Could not find a proof for myself yet. Perhaps this is even wrong?
Thank you in advance :)
AI: Note that the interval $[x,y]$ has an interior, that is an open set contained in it, whenever $x<y$.
We know that there are compact sets whose interior is empty and cannot be written as such unions.
One example is $\{0\}\cup\{\frac1n\mid n\in\mathbb N\}$. It is bounded in $[0,1]$ and closed since it is really just a convergent sequence and its limit point. |
H: Some kind of relation between classical heat equation and Laplace .
If we have $k(x,t)= \frac {1}{(4t)^{\frac{n}{2}}} \exp\left(\frac{-|x|^2}{4t}\right)$ is the fundamental solution of heat equation. If we consider $n \ge 3 $, I would like to show that $\int_0^\infty k(x,t) dt$ is the fundamental solution of lLaplace equation.
I would like some hints.
I thought of integrating but don't know how to approach.
Thank you
ie , i need to arrive to a form like $\frac{1}{B} \frac{1}{|x|^{n-1}}$ $B $ is a constant depending on the measure of the space.
AI: We use the substitution $s=\frac t{|x|^2}$ (then $dt=|x|^2ds$) to get
\begin{align}
\int_0^{+\infty}k(x,t)dt&=\int_0^{+\infty}\frac 1{(4t)^{n/2}}\exp\left(-\frac{|x|^2}{4t}\right)dt\\
&=\frac 1{4^{n/2}}\int_0^{+\infty}\frac 1{(s|x|^2)^{n/2}}\exp\left(-\frac 1{4s}\right)|x|^2ds\\
&=\frac 1{4^{n/2}}|x|^{2-n}\int_0^{+\infty}\frac 1{s^{n/2}}\exp\left(-\frac 1{4s}\right)ds\\
&=\frac 1{4^{n/2}}|x|^{2-n}\int_0^{+\infty}y^{n/2-2}\exp(-y/4)dy\\
&=c_n|x|^{2-n}.
\end{align}
We have to show that $f\colon x\mapsto |x|^{2-n}=\left(\sum_{k=1}^nx_k^2\right)^{1-n/2}$ is harmonic. Let $j\in\{1,\dots,n\}$. We have
$$\partial_jf(x)=\left(\sum_{k=1}^nx_k^2\right)^{-n/2}2x_j\left(1-\frac n2\right)$$
and
$$\partial_{jj}f(x)=2\left(1-\frac n2\right)\left(\sum_{k=1}^nx_k^2\right)^{-n/2}-\left(1-\frac n2\right)nx_j\left(\sum_{k=1}^nx_k^2\right)^{-n/2-1}(2x_j).$$
Summing that, we get
\begin{align}
\Delta f(x)&=\sum_{j=1}^n\partial_{jj}f(x)\\
&=\left(1-\frac n2\right)\left(\sum_{k=1}^nx_k^2\right)^{-n/2-1}\left(2n|x|^2-n\cdot 2\cdot |x|^2\right)\\
&=0.
\end{align} |
H: Is a matrix with characteristic polynomial $t^2 +1$ invertible?
Given that $A$ is a square matrix with characteristic polynomial $t^2+1$, is $A$ invertible?
I'm not sure, but this question seems to depend on whether $A$ is over $\mathbb{R}$ or over $\mathbb{C}$. My reasoning is that if $A$ is over $\mathbb{C}$ then $A$ has two distinct eigenvalues $-i$ and $i$ and is diagonalizable. Since it's diagonalization is invertible, $A$ is also invertible.
However if $A$ is over $\mathbb{R}$ then $A$ has no eigenvalues and therefore... I don't know where to go from there.
AI: The eigenvalues of the matrix are all roots of the characteristic polynomial.
A square matrix is invertible if and only if $0$ is not an eigenvalue of the matrix.
Therefore, a square matrix is invertible if and only the constant term of its characteristic polynomial is <fill in the blank> |
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