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H: Set of numbers that are solutions for equations such as $ x = x + 1 $
Is there any number set that is used in that kind of equations without solution on the complex numbers?
AI: Extended real numbers. We add two numbers to the real numbers namely $-\infty,\infty$ with $c\infty=\infty$ if $c>0$ and $-\infty$ if $c<0$. $\infty+\infty=\infty$ , $\infty-\infty$ undefined.
Check
https://en.m.wikipedia.org/wiki/Extended_real_number_line
Clearly if you work with in $G$ for some set and you want to check the solution of and you each element have an inverse w.r.t $+$ then
$\begin{array}{ccc}x&=&x+1 \\ x-x&=&-x+x+1\\
0&=&1\end{array}$
hence in our $G$ , we have $1=0$ i.e. $1$ is the addition inverse.
But am sure you are not looking for something like this right? |
H: If $x$ is a root of $x^2 + x + 1$ over $GF(4)$, then so is $x + 1$
It's clear to me how $GF(4)$ can be gotten as the quotient ring of $GF(2)/X^2+X+1$. This leads to the four elements $0, 1, x, 1+x$. I'm struggling more as seen these ones as extending $GF(2)$ with the roots of the irreducible polynomial $x^2+x+1$ in $GF(2)$.
Everywhere I've inspected better on existing answer here on that, I saw the claim:
if $x$ is a root, then $x+1$ is a root as well
It's not clear here what root means, here. I mean we can extend $GF(2)$ by one of the cube root of unity, so let's call it x. But then $x+1$ is NOT yet another root of $x^2+x+1$.
May you help better in understanding that, please?
Thanks in advance
AI: You ask:
It's not clear here what root means, here
A number $a$ is a root of some polynomial $P$ if and only if $P(a) = 0$. Sometimes we say that $a$ is a βzeroβ of $P$, to emphasize this. The two terms mean the same thing.
Supposing that $a$ is a root of $x^2+x+1$, so that $a^2+a+1=0$, then $a+1$ is in fact the other root. Let's substitute $a+1$ for $x$ in $x^2+x+1$ and see what we get:
$$\begin{align}
(a+1)^2 + (a+1) + 1 & = \\
(a^2 + 2a + 1) + (a + 1) + 1 & = \\
a^2 + 3a + 3 & = & \text{(because $GF(2)$)} \\
a^2 + a + 1 & = \\
0
\end{align}
$$
Another to see this is to explicitly multiply $(x-a)(x-(a+1))$ and see what you get. The $a$s all cancel out and leave you with $x^2+x+1$ as you ought.
Here's a third way to see this. Suppose $a$ is a cube root of unity. Straightforward calculations (which I leave to you) show that $a+1$ must also be a cube root of unity because $a+1 = a^2$.
(I wrote a blog article about extension fields of $GF(2)$ that examines these points in great detail; maybe it will help. Especially see the part that begins βwe will [introduce] a new number, called $b$, which has the property that $b^2 + b + 1 = 0$. What are the consequences of this?β) |
H: If $f:X\to X$, $f(f(x))=x$, is $f$ onto?
I have been trying following question and was unable to solve.
Let $f: X \to X$ such that $f(f(x)) = x$ for all $x\in X. \space$ Then:
Is $f$ 1-1?
Is $f$ onto?
Clearly $f$ is 1-1 . But I am unable to deduce why $f$ must be onto or not.
AI: If $f$ were not onto, since $f:X\to X$, $f(X)\subset X$ and this containment is proper. Then $f(f(X))\subset f(X)\subset X$, but this contradicts $f(f(x))=x$ for every $x\in X$. |
H: Double sum that grows at sublinear rate
Is there an example of a non-zero function $f: \mathbb{N} \to \mathbb{R}^+$ such that for any $n \in \mathbb{N}$, the following term is sublinear (or $o(n)$)?
$$\sum_{j=1}^n \sum_{i=1}^j f(i)$$
AI: Work the other way around: You want:
$\begin{align*}
\sum_{1 \le j \le n} s(j)
= o(n)
\end{align*}$
Here $s(j)$ is your inner sum. This means $s(j) = o(1)$. This requires in turn:
$\begin{align*}
\sum_{1 \le i \le j} f(i)
= o(1)
\end{align*}$
But this is impossible, as the last sum is at least it's first term, it can't go down as $j$ increases. |
H: Proof of the Central Limit Theorem
I am reading the wikipedia article that proves the central limit theorem and had a question about one of the steps they take in the proof.(See the article here, the context is not really important to understand the question)
They prove that as $n$ approaches infinity, the characteristic function $\varphi: \mathbb{R} \to \mathbb{R}_{\geq 0}$ becomes:
$$ \varphi(t) = 1 - \dfrac{t^2}{2n} + o\left(\dfrac{t^2}{n}\right) $$
Where $o$ is the little-o notation. Then they claim that:
$$ (\varphi(t))^n = \left[1-\dfrac{t^2}{2n} + o\left(\dfrac{t^2}{n}\right)\right]^n \to e^{-\dfrac{t^2}{2}}, n \to \infty $$
I know that $\lim_{n\to\infty} (1+t/n)^n = e^t$, but how can they prove the above equation where we have $o(t^2 / n)$ inside the brackets? Also, I had never seen the small-o notation in use before; if that term were, say, $O(t^3/n\sqrt{n})$ with the big-O notation, would the proof still be valid?
AI: Replacing $t$ with $-t^2/2$ in the result you're familiar with gives $\lim_{n\to\infty}\left(1-\frac{t^2}{2n}\right)^n=e^{-t^2/2}$. But for the CLT, we need a slightly different result. In terms of the little-$o$ notation you asked about, $1-\frac{t^2}{2n}+o\left(\frac{t^2}{2n}\right)$ denotes a certain set of functions, or an arbitrary element thereof. What the CLT's proof notes is that, considered as a function of $n$ at fixed $t$, $\varphi$ is a function in $1-\frac{t^2}{2n}+o\left(\frac{t^2}{2n}\right)$, and each such function satisfies $\lim_{n\to\infty}\varphi^n=e^{-t^2/2}$. In fact, we can use little-$o$ notation on both sides, viz.$$\left[1-\frac{t^2}{2n}+o\left(\frac{t^2}{2n}\right)\right]^n=e^{-t^2/2n+o(t^2/2n)}.$$The advantage of each use of $o(t^2/2n)$ over $O(t^3/n^{3/2})$ is that we don't need any assumptions in the CLT about moments beyond the variance. |
H: The orbit of a non zero vector $v$ in $\mathbb{R}^n$ is $\mathbb{R}^n \setminus \{0\}$
Prove that given any non zero vectors $v$ and $w$ in $\mathbb{R}^n$, there exists an invertible $n\times n$ matrix $A$ such that $Av=w$.
(I don't know where to start. Any hints will be appreciated)
AI: If $w$ is a scalar multiple of $v$, you could just let $A$ be a scalar matrix. That is, if $w=\lambda v$, then take $A=\lambda I$.
Otherwise $v$ and $w$ are linearly independent, and take $\{v,w,b_3,\ldots,b_n\}$ to be some basis of $\mathbb{R}^n$. Let $P$ be the $n\times n$ matrix $[v\ w\ b_3\ \ldots\ b_n]$. Then take $$A=P\left[\begin{array}{cc|c}0&1&\\1&0&\\\hline&&I\end{array}\right]P^{-1}$$ |
H: INMO $2020$ P1: Prove that $PQ$ is the perpendicular bisector of the line segment $O_1O_2$.
Let $\Gamma_1$ and $\Gamma_2$ be two circles of unequal radii, with centres $O_1$ and $O_2$ respectively, intersecting in two distinct points $A$ and $B$. Assume that the centre of each circle is outside the other circle. The tangent to $\Gamma_1$ at $B$ intersects $\Gamma_2$ again in $C$, different from $B$; the tangent to $\Gamma_2$ at $B$ intersects $\Gamma_1$ again at $D$, different from $B$. The bisectors of $\angle DAB$ and $\angle CAB$ meet $\Gamma_1$ and $\Gamma_2$ again in $X$ and $Y$, respectively. Let $P$ and $Q$ be the circumcentres of triangles $ACD$ and $XAY$, respectively. Prove that $PQ$ is the perpendicular bisector of the line segment $O_1O_2$.
My progress: This problem is really intimidating to me !
I observed that XBY is collinear, which can be proved by angle chase. Just note that $\angle BDA = \angle CBA$ and $\angle ACB = \angle ABD $ . Then $\Delta ABD \sim \Delta ACB$ . By cyclic quads, we get XBY collinear .
Then I was able to show $PO_1=PO_2$ by noticing that $\angle PO_1O_2 = 180- \angle DAB$ and $\angle O_1O_2P = 180-\angle BAC$ .
Then I am stuck. I also observed that $O_1,P,O_2,Q$ is cyclic but was not able to prove.
Here is a diagram:
I am also thinking of using spiral symmetry but I don't have any idea on how to use it ?
Please if possible send hints rather than solution. It helps me a lot . Thanks in advance.
AI: Hints:
Since angle bisector meets $\Gamma_1$ at $Y$ we see that $YB = YD$ and similary $XB = XC$. (Further more, since $\angle DYB = \angle DBC = \angle BXC$ (tangent chord) we have $\Delta BDY\sim \Delta CBX$. You don't need this.)
You can prove $QO_1= Q_2O$ with spiral similarity around $A$ which takes $D$ to $O_1$ and $B$ to $O_2$. It takes $Y$ to $Q$ and since $YB = YD$ we have also $QO_1= Q_2O$.
Let $\angle ABD = x$ and $\angle ABC = y$. Prove that $\angle O_2O_1P = \angle O_1O_2P = x+y$ and you are done. |
H: Calculate $x$ from a given ratio calculation with a known answer
I have the below ratio calculation, and I need to find the value of $x$:
$ \dfrac{181.5 + 16.5x}{181.5 + 11x} = 1.251 $
Can I use this to find the value of $x$?
Let me know if I need to provide more info.
Cheers
AI: Assuming you meant $(181.5+16.5X)/(181.5+11X)=1.251$, we have
$181.5+16.5X=1.251(181.5+11X)$;
$(16.5-1.251\times11)X=2.739X=181.5(0.251)$;
$X\approx 16.6325$. |
H: What is the probability function of the sum of $N$ categorical distribution experiments
Let's say I have a categorical distribution defined by
$$P(X)=0.5\delta(X+1)+0.25\delta(X)+0.25\delta(X-1).$$
Suppose I want to repeat the experiment $N$ times and sum all the $X$ values
The resulting value would fit another categorical distribution with $2N+1$ coefficients.
Is there any analytical way to calculate them? Brute forcing through code is quite trivial, but that is not the approach I'm looking for.
AI: The easiest way to get what I think you want is to use the probability generating function of the individual distribution:
$$
F(z) = \sum_k p_k z^k = \frac12 z^{-1} + \frac14 + \frac14 z
$$
Then the probability generating function for the sum of $n$ such i.i.d. variables is
$$
G_n(z) = [F(z)]^n
$$
For instance, with $n = 3$ variables, the sum is distributed as
\begin{align}
G_3(z) & = [F(z)]^3 \\
& = \left( \frac12 z^{-1} + \frac14 + \frac14 z \right)^3 \\
& = \frac18 z^{-3} + \frac{3}{16} z^{-2} + \frac{9}{32} z^{-1}
+ \frac{13}{64} + \frac{9}{64} z + \frac{3}{64} z^2 + \frac{1}{64} z^3
\end{align}
meaning that the sum has the distribution
$$
\text{sum} = \begin{cases}
-3 & \text{with probability $1/8$} \\
-2 & \text{with probability $3/16$} \\
-1 & \text{with probability $9/32$} \\
0 & \text{with probability $13/64$} \\
1 & \text{with probability $9/64$} \\
2 & \text{with probability $3/64$} \\
3 & \text{with probability $1/64$}
\end{cases}
$$
ETA: This formulation doesn't give you the specific probabilities in any magical way beyond an exhaustive case analysis, but it does permit one to organize those cases in an effective way, especially if code is being written. (You could use FFT.) It can give things like expressions for the various moments. |
H: If $S$ is the unit square with opposite corners $(0, 0)$ and $(1, 1)$, show $\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}=\log(1+\sqrt{2})$
Let $S$ be the unit square with opposite corners $(0, 0)$ and $(1, 1)$. Show $$\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}=\log(1+\sqrt{2}).$$
What I tried:
Since $\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}$ converges in $S$. So the integral is convergent independent of choice of ${D_n}$, the let $D_n$ be square with opposite corners $(1/n, 1/n)$ and $(1, 1)$.
Then
\begin{align}
\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}&=\int^1_{1/n}\int^1_{1/n}\frac{y^2}{\sqrt{(x^2+y^2)^3}}
\\&=\int^1_{1/n}y^2dy\int^1_{1/n}\frac{1}{(x^2+y^2)^{2/3}}dx
\\&=\int^1_{1/n}y^2\frac{x}{y^2\sqrt{y^2+x^2}}|^1_{1/n}dy
\\&=\log(y+\sqrt{1+y^2}|^1_{1/n}-\frac{1}{n}\log(y+\sqrt{y^2+(1/n)^2})|^1_{1/n}
\\&=\log(1+\sqrt{2})-0-0+\frac{1}{n}\log\left(\frac{1}{n}+\sqrt{\frac{1}{n}^2+\frac{1}{n}^2}\right)
\end{align}
But $\frac{1}{n}\log(\frac{1}{n}+\sqrt{\frac{1}{n}^2+\frac{1}{n}^2})$ seem to go to infinity.
AI: Under the variable interchange $x\leftrightarrow y$ we have that
$$I = \iint\limits_{[0,1]^2} \frac{y^2}{(x^2+y^2)^{\frac{3}{2}}}dA = \iint\limits_{[0,1]^2} \frac{x^2}{(x^2+y^2)^{\frac{3}{2}}}dA$$
and adding the two integrals gets us
$$2I = \iint\limits_{[0,1]^2} \frac{x^2+y^2}{(x^2+y^2)^{\frac{3}{2}}}dA \implies I = \frac{1}{2} \iint\limits_{[0,1]^2} \frac{1}{\sqrt{x^2+y^2}}dA$$
By symmetry we can do the integral in polar coordinates
$$I = 2\cdot\frac{1}{2}\int_0^1 \int_0^x \frac{1}{\sqrt{x^2+y^2}}dA = \int_0^{\frac{\pi}{4}}\int_0^{\sec\theta}\:dr\:d\theta$$
$$ = \log|\sec\theta+\tan\theta|\Biggr|_0^{\frac{\pi}{4}} = \log(1+\sqrt{2})$$ |
H: Need help in understanding parametric equation of a circle in 3D
$ = \cos _1 + \sin W_1$
Where V1 is a unit vector on the circle and W1 is a unit vector perpendicular to V1
I am currently working on finding an equation for a unit circle in 3D. I have come across the above equation many times but I have no idea where it actually comes from. Could someone please explain the equation to me and how it works?
AI: In two dimensions, a circle with radius, r, and center $C=(c_1, c_2)$ can be expressed as
$$f(t) = C + r\cos(t)\vec u + r\sin(t)\vec v$$
where $\vec u$ and $\vec v$ are orthogonal unit vectors, for example$\vec u = (1,0)$ and $\vec v = (0,1)$.
To define a circle in $n$-dimensional space, the equation stays the same, except now $\vec u$ and $\vec v$ are orthogonal $n$-dimensional vectors. For example, in $3$-dimensional space,
$$\text{$\vec u = \frac 17(2,3,6)$ and $\vec v = \dfrac 17(3,-6,2)$}$$
will work.
Why does this work?
If you know what the scalar product of two vectors is, then you can argue as follows
Let $P = C + r\cos(t)\vec u + r\sin(t)\vec v$ Then
\begin{align}
\| P - C \| &= \| r(\cos(t)\vec u + \sin(t)\vec v) \| \\
&= r^2(\cos(t)\vec u + \sin(t)\vec v)\circ(\cos(t)\vec u + \sin(t)\vec v) \\
&= r^2(\cos^2(t) (\vec u \circ \vec u)
+ 2 \cos(t) \sin(t) (\vec u \circ \vec v)
+ r^2(\sin^2(t) (\vec v \circ \vec v) \\
&= r^2(\cos^2(t) \cdot 1 + 2 \cos(t) \sin(t) \cdot 0
+ r^2 \cdot \sin^2 (t) \\
&= r^2(\cos^2 t + \sin^2 t) \\
&= r^2
\end{align}
In words, the distance between $P$ and $C$ is $r$.
This is only true when $\vec u$ and $\vec v$ have unit lengths and are perpendicular to each other. |
H: True or false: infinite sequence in a compact topological group is dense.
This is from an exercise in Bredon's Topology and Geometry:
Let $G$ be a compact topological group (assumed to be Hausdorff). Let $g\in G$ and define $A=\{g^n:n=0,1,2...\}$. Then show that the closure $\bar{A}$ is a topological subgroup.
Note if the assumption of compactness is dropped, then the statement is false. A counterexample is $\mathbb{N}\subset \mathbb{R}$ as additive sets.
Note: if we add the assumption of $G$ being 1st countable, this question is easy to answer.
So, if the set $A$ is finite, then it is a cyclic group and it is already closed. In the infinite case, I can only think of the example of an infinite subgroup of the circle group, which is dense in compact group. I am not sure how to proceed with the proof of this general fact. Any hint will be appreciated.
AI: Here is a proof which assumes that $G$ is first countable. Since $G$ is compact, there is a subsequence $(g^{n_k})_{k\in\Bbb Z_+}$ of $(g^n)_{n\in\Bbb Z_+}$ which converges to some $a\in G$. Clearly, $a$ is in fact an element of $A$. But then the sequence $(g^{n_k-1}a)_{k\in\Bbb Z_+}$ is a sequence of elements of $A$ which converges to $g^{-1}a$. So, $g^{-1}a\in A$. By the same argument, $(\forall N\in\Bbb N):g^{-N}a\in A$. But then $(g^{-n_k-1}a)_{k\in\Bbb Z_+}$ is a sequence of elements of $A$ which converges to $g^{-1}$ and this proves that $g^{-1}\in A$. By the same argument, $(\forall N\in\Bbb N):g^{-N}\in A$. Can you take it from here?
If you drop the assumption that $G$ is first countable, you can still get the conclusion that you want to get, using nets in the proof, as suggested by Eric Wofsey in the comments. |
H: c.l.u.b. set and type of $T$
In Shelah's book Proper and Improper forcing , what is the type (i.e. what is the underlying set of $T$ and $T_\alpha$, respectively) and relationship of $T$ and $T_\alpha$
for $\alpha\in \omega_1$ in the following snippet ? I'd know though, what is $\Gamma$, if I knew
what is $T_\alpha$.
AI: $T_\alpha$ is the $\alpha$-th level of the tree $T$. Itβs in the little Notation section just before Theorem $\bf{6.1}$:
For an $\aleph_1$-tree $T$, $T_i$ is the $i$-th level, $T\upharpoonright i=\bigcup_{j<i}T_j$, and for $x\in T_\beta$, $\alpha\le\beta$, $x\upharpoonright\alpha$ is the unique $y\in T_\alpha$, $y\le x$.
Here $T$ is any Aronszajn tree. |
H: If $\overline B\subseteq\overset{Β°}{A}$ then is possible that $\partial (A\setminus B)=\partial A\cup\partial B$?
Lemma
If $X$ is a topological space then
$$
\partial(A\cap B)\subseteq[\overline{A}\cap\partial B]\cup[\partial A\cap\overline{B}]
$$
for any $A,B\subseteq X$.
Corollary
If $X$ is a topological space then
$$
\partial(A\setminus B)\subseteq\partial A\cup\partial B
$$
for any $A,B\subseteq X$.
Proof. By the first lemma we know that
$$
\partial(A\setminus B)=\partial\big(A\cap(X\setminus B)\big)\subseteq[\overline A\cap\partial(X\setminus B)]\cup[\partial A\cap\overline{X\setminus B}]=[\overline A\cap\partial B]\cup[\partial A\cap\overline{X\setminus B}]\subseteq\partial A\cup\partial B.
$$
for any $A,B\subseteq X$.
So I ask if generally is $\partial (A\setminus B)=\partial A\cup\partial B$ when $B\subseteq A$ and if not I ask if with some additional hypotheses about $X$ (hausdorff separability, connectedness, etc...) or $A$ and $B$ it could be true. For example if the closure of B is contained in the interior of A then does the equality hold? So could someone help me, please?
AI: Suppose that $B \subseteq A$. Then by using that $\overline{A_1\cup A_2} = \overline{A_1}\cup\overline{A_2}$ for any sets $A_1$ and $A_2$,
$$
\partial A\cup\partial B
\subseteq \overline{X\setminus A}\cup\overline{B}
= \overline{(X\setminus A)\cup B}
= \overline{X\setminus(A\setminus B)}. \tag{1} $$
Now we also assume that $\overline{B}\subseteq\mathring{A}$. Then
$$ \overline{X\setminus A} \cap \overline{B}
\subseteq \overline{X\setminus A} \cap \mathring{A}
= \varnothing. $$
So
\begin{align*}
\partial B
&= \overline{X\setminus B} \cap \overline{B} \\
&= (\overline{A\setminus B} \cup \overline{X\setminus A}) \cap \overline{B} \\
&= (\overline{A\setminus B} \cap \overline{B}) \cup (\overline{X\setminus A} \cap \overline{B}) \\
&= \overline{A\setminus B} \cap \overline{B}
\end{align*}
and this shows that $\partial B \subseteq \overline{A\setminus B}$. Simiarly,
\begin{align*}
\partial A
&= \overline{X\setminus A} \cap \overline{A} \\
&= \overline{X\setminus A} \cap (\overline{A \setminus B} \cup \overline{B}) \\
&= (\overline{X\setminus A} \cap \overline{A \setminus B}) \cup (\overline{X\setminus A} \cap \overline{B}) \\
&= \overline{X\setminus A} \cap \overline{A \setminus B}
\end{align*}
shows that $\partial A \subseteq \overline{A\setminus B}$. Consequently
$$ \partial A \cup \partial B \subseteq \overline{A\setminus B} \tag{2} $$
and combining $\text{(1)}$ and $\text{(2)}$ proves the inclusion $\partial A \cup \partial B \subseteq \partial (A\setminus B)$. |
H: Theorem on GCD of polynomials
Let
$F$ be a field and suppose that
$d(x)$ is a greatest common divisor of two polynomials $p(x)$ and $q(x)$ in $F[x]$. Then there exist polynomials $r(x)$ and $s(x)$ such that $d(x)=r(x)p(x)+s(x)q(x).$
Furthermore, the greatest common divisor of two polynomials is unique.
Let
$d(x)$ be the monic polynomial of smallest degree in the set
$S
=\{f(x)p(x)+g(x)q(x):f(x),g(x)βF\}$.
We can write
$d(x)=r(x)p(x)+s(x)q(x)$ for two polynomials
$r(x)$ and $s(x)$ in $F[x]$.
We need to show that
$d(x)$ divides both $p(x)$ and $q(x)$. We shall first show that
$d(x)$ divides $p(x)$. By the division algorithm, there exist polynomials
$a(x)$ and $b(x)$ such that $p(x)=a(x)d(x)+b(x)$, where
$b(x)$ is either the zero polynomial or
deg $b(x)$ < deg $d(x)$. Therefore,
$b(x)$
$=p(x)βa(x)d(x)$
$=p(x)βa(x)(r(x)p(x)+s(x)q(x))$
$=p(x)βa(x)r(x)p(x)βa(x)s(x)q(x)$
$=p(x)(1βa(x)r(x))+q(x)(βa(x)s(x))$
is a linear combination of
$p(x)$ and $q(x)$ and therefore must be in
$S$
.
However,
$b(x)$ must be the zero polynomial, since $d(x)$ was chosen to be of smallest degree; consequently,
$d(x)$ divides $p(x)$. A symmetric argument shows that $d(x)$ must also divide $q(x)$; hence, $d(x)$ is a common divisor of
$p(x)$ and $q(x)$.
My questions are
Why does $d(x)$ need to be a monic polynomial?
The theorem says that gcd is monic but that is not the case with all polynomials!?
AI: If $d(x)$ were not constrained to be monic, then the gcd of $p(x),q(x)\in F[x]$ would not be unique; we could multiply $d(x)$ by any non-zero element of $F$, and it would still be a greatest common divisor of $p(x)$ and $g(x)$. On the other hand, given any greatest common divisor of $p(x)$ and $g(x)$, we can multiply it by the inverse of the leading coefficient -- note that all non-zero elements of a field are invertible -- to get a monic greatest common divisor. For this reason, the monic polynomial g.c.d. is taken as the greatest common divisor, just as for integers the positive greatest common divisor is taken as the g.c.d., even though its opposite also divides both integers and all common divisors of both integers also divide its opposite. |
H: closed subspaces in $\ell^p$
Given the operator: $T: \ell^1 \rightarrow \ell^2$ with $Tx = x$ which of the following subsets of $\ell^1 \times \ell^2$ are closed?
$U = \ell^1 \times \{0\}$
$V = \Gamma_T$ the Graph of T
$U + V$
Defining the operator $A: \ell^1 \times \ell^2$ and $Ax = 0$ we have $U = \Gamma_A$.
$A$ and $T$ are both continuous and linear. by the closed graph theorem we have that $U$ and $V$ are both closed. is this correct so far?
I don't know how to prove or disprove that $U + V$ is closed. Thanks for any help.
AI: Consider a sequence $$x_n =\left( 1,\frac{1}{2} , \frac{1}{3} ,...,\frac{1}{n} , 0,0,...\right)$$
Tnen $(-x_n ,0)\in U$ and $(x_n , x_n ) \in V$ but $$(-x_n ,0) + (x_n , x_n )\to (0, x)$$
where $x=\left(\frac{1}{k}\right)_{k\in\mathbb{N}}$ and the element $(0,x)\notin U+V$ so the set $U+V$ is not closed. |
H: Arc length of function - confused about particular u-substitution in integral
I am trying to teach myself how to compute the arc length of a function through a textbook, and I am stuck on how they did a particular integral substitution in a worked example.
I'll provide the specific line of working I am stuck to provide relevant information quickly. I'll follow with my working, and then provide the context of the question below if necessary.
Where I am stuck
I am unsure how to get from:
$$\int_1^3\sqrt{1+4t^2}\,dt$$
to, as the book puts it exactly:
$$\int{\frac{1}{2}\cosh^2u \,du}$$
They say they are
putting $2t$ = $\sinh u$.
I am unsure of how to get to this line using that substitution. (I am also unsure why the bounds of integration are gone, I originally assumed out of convenience but different bounds due to the substitution are being used later in the example.)
My attempt
$\int_1^3\sqrt{1+4t^2}\,dt$
Using $2t$ = $\sinh u\Rightarrow2\,dt = \cosh u \Rightarrow dt = \frac{\cosh u}{2}\,du$.
$\int_1^3\frac{1}{2}\sqrt{1+\sinh^2 u} \cosh u \,du$
Using $\sqrt{1+\sinh^2 u} \equiv \cosh^2 u$:
$\int_1^3\frac{1}{2}\cosh^2u \cosh u \,du$
$\frac{1}{2}\int_1^3\cosh^3u \,du$
Clearly there is an issue here, as I have arrived at $\cosh^3u$ instead of $\cosh^2u$ (and potentially different bounds of integration.) I'm not sure if there's an error in my working and/or if there's some different integration technique that I need to apply here.
More context
The example relates to finding an arc length of the function $y = x^2$ from $(1, 1)$ to $(3, 9)$. This length $s$is being worked out using: $$\frac{ds}{dt} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$$
The function is parameterised: $x = t$, y = $t^2$. I am then confident in using this to find that:
$$s = \int_1^3\sqrt{1+4t^2} \,dt$$
If there is any more information I should provide please do not hesitate to answer. (Also, I've not posted here much so if there is any error in my MathJax formatting please correct it as well as telling me what I should have done instead. Thank you.)
AI: Note the correction: $$\sqrt{1+\sinh^2(u)} \equiv \cosh^{\color{red}{1}} (u)$$as $\cosh^2(u)-\sinh^2(u)=1$. |
H: Relation between $f(x)$ and $f(\sqrt{x})$
This might be silly, but if $f(\sqrt{x})=\frac{0.1}{a}x$, is $f(x)=\frac{0.1}{a}x^2$?
AI: The argument of a function is a dummy variable, meaning that it is a placeholder for a number. Consider $g(x)=x^2$; we could just as equally have written $g(y)=y^2$. The function is about the operation performed on the argument, not the argument itself. It might help to think of $f$ as
$$
f(\text{number})=(\text{number})^2\times\frac{0.1}{a}
$$
This makes it clearer what is going on: you take a number, you square it, and you multiply it by $\frac{0.1}{a}$. It is this kind of thinking that leads us directly to the answer:
$$
f(x)=x^2
\times\frac{0.1}{a}=\frac{x^2}{10a}$$
So remember, it is the rule that is the defining property of a function. The dummy variable is merely there to illustrate what happens to a number when it is inputted into the function. With this in mind, $\sqrt{x}$ seems like an odd choice. It is clearer, and less likely to cause confusion, to use something like $x$ or $y$ instead.
NB as Sangchul Lee has pointed out, the domain of the function is important: this answer assumes that $x\geq0$, as I presume you only care about real-valued functions. |
H: Let $A$ be a square matrix and $A=BC$ be its rank factorization. Show that rank $(A)$=rank $(A^2)$ if and only if $CB$ is non-singular.
Let $A$ be a square matrix and $A=BC$ be its rank factorization. Show that rank $(A)$=rank $(A^2)$ if and only if $CB$ is non-singular.
please give a hint.
My approach : Suppose that rank $(A)$=rank $(A^2)$. This implies range $A$=range $A^2$. Now from this how to proceed.
AI: Hint: What can we say about $\operatorname{range}(B) \cap \ker(C)$?
Alternatively, note that $A^2 = B[CB]C$.
One approach is as follows. Let $A$ be $n \times n$ with rank $r$, so that $B$ is $n \times r$ and $C$ is $r \times n$.
Suppose that $\operatorname{rank}(CB) < r$ (i.e. $CB$ is singular). It follows that
$$
\operatorname{rank}(A^2) = \operatorname{rank}(B(CB)C) \leq \operatorname{rank}(CB) < r = \operatorname{rank}(A).
$$
On the other hand, suppose that $CB$ is non-singular. It follows that
$$
\operatorname{rank}((CB)C) = \operatorname{rank}(C) = r.
$$
Now, using the fact that that $B$ has a trivial kernel, conclude that
$$
\operatorname{rank}(A^2) = \operatorname{rank}(B[CBC]) =
\operatorname{rank}(CBC) = r.
$$ |
H: Confused by an example of generalization of a point
Suppose $A=\operatorname{Spec}\mathbb{C}[x,y]$. We say that a point $P\in\operatorname{Spec}A$ is a generalization of a point $Q\in\operatorname{Spec}A$ if $Q\in\overline{\{P\}}$. So let $P=[(y-x^2)]$ and $Q=[(x-2,y-4)]$. The claim is that $P$ is a generalization of $Q$, but I'm having trouble seeing this.
I know that $(y-x^2)\in (x-2,y-4)$, which is equivalent to the ideal $(y-x^2)$ vanishing at the closed point $(x-2,y-4)$ (which we can think of as $(2,4)$). But this doesn't help me see that $(x-2,y-4)\in\overline{(y-x^2)}$ (where the closure is with respect to the Zariski topology).
Now, certainly, $[(x-2,y-4)]$ is a closed point, but why is it in the intersection of all closed subsets containing $(y-x^2)$? Is there a better way to see this? How would one go about finding the closure of $(y-x^2)$?
AI: To prove that $(x - 2, y - 4)$ is in the closure of $\{ (y - x^2 ) \}$, you have to prove that every closed set that contains $(y - x^2 )$ also contains $(x - 2, y - 4)$.
By the definition of the Zariski topology, the closed sets in ${\rm Spec \ }A$ are sets of the form $V(I) = \{ \mathfrak p { \rm \ prime \ } \subset A : I \subseteq \mathfrak p \}$.
But if $I \subseteq A$ is any ideal such that $V(I)$ contains $(y - x^2)$, then $I \subseteq (y - x^2) \subset (x - 2, y - 4)$ too, hence $V(I)$ contains $(x - 2, y - 4)$. Thus $(x - 2, y - 4)$ is in the closure of $\{ (y - x^2 ) \}$.
[By the way, I used the fact that $(y - x^2 ) \subset (x - 2, y - 4)$.] |
H: Orthocenter, Circumcenter, and Circumradius
In triangle $ABC,$ let $a = BC,$ $b = AC,$ and $c = AB$ be the sides of the triangle. Let $H$ be the orthocenter, and let $O$ and $R$ denote the circumcenter and circumradius, respectively. Express $HO^2$ in terms of $a,$ $b,$ $c,$ and $R.$
I know what orthocenter, circumcenter, and circumradius are, however, I am having trouble expressing it in terms of a, b, c, and R.
AI: This is a quite famous formula actually.
$HO^2 = 9R^2 - (a^2 + b^2 + c^2)$
For more details see here:
https://www.cut-the-knot.org/arithmetic/algebra/DistanceOH.shtml
Depending on your level (say if you are in 7th or 8th grade e.g.),
it may be quite a difficult formula to prove. |
H: Largest Component Interval Proof
I want to check whether the proof for the following is correct:
Every point in an open set $S\subset \mathbb{R}$ belongs to one and only one component interval of $S$, where an open interval $I$ is a component interval of $S$ if and only if $I\subset S$ and there exists no open interval $J$ such that $I\subset J\subset S$.
Here's my attempt:
Claim: The interval $I_{x}=(f(x),g(x))$, where $f(x)=\inf\left\{a:(a,x)\subset S\right\}$ and $g(x)=\sup\left\{b:(x,b)\subset S\right\}$, is the desired component interval (and the largest).
Proof: By the definition of $I_{x}$, there exists no open interval $J$ such that $I_{x}\subset J\subset S$. Then $I_{x}$ is the largest component interval of $S$.
If $J_{x}$ is another component interval of $S$, since $I_{x}$ is the largest, $J_{x}\subset I_{x}\subset S$. This forms a contradiction from the definition of component intervals, and so $I_{x}=J_{x}$.
Is the last paragraph logically sound? In Apostal, he uses unions but I don't quite understand why $J_{x}\cup I_{x}=J_{x}$. Is it because of the above arguments I was using?
AI: I like your approach!
First, a minor remark: It may be worth emphasising that $\inf \{ a : (a, x) \subset S \}$ and $\sup \{ b : (x, b) \subset S \}$ exist. This follows from the fact that $S$ is open.
To prove that the component interval of $S$ containing $x$ is unique, suppose that $I_x$ and $I'_x$ are two component intervals of $S$ containing $x$. Consider $J := I_x \cup I'_x$. Then $J$ is an open interval (since $I_x$ and $I'_x$ share a point in common, namely $x$). Furthermore, $I_x \subseteq J \subseteq S$ and $I'_x \subseteq J \subseteq S$. Since $I_x$ and $I'_x$ are components intervals of $S$, we must have $I_x = J$ and $I'_x = J$. Hence $I_x = I'_x$. This shows that the component interval of $S$ containing $x$ is unique.
In your last paragraph, the statement $J_x \subseteq I_x \subseteq S$ doesn't immediately follow what we know at that point in the argument. At that point in the argument, we know that $I_x$ is maximal amongst intervals of $S$ (in the sense that there is no interval of $S$ strictly bigger than $I_x$). But we haven't yet established that all intervals of $S$ containing $x$ are contained in $I_x$, which is a different statement. I think you conflated the two statements by using the word "largest" to mean one thing in your second paragraph and another thing in your third paragraph. |
H: Question about the proof of convergence in probability implies convergence in distribution
I am following the notes for my master's program, trying to prove conv.in probability implies conv. in distribution. The proof goes:
Let $F_n$ be the distribution function of $X$. Fix any $x$ s.t. $F$ is continuous at $x$, and fix any $\epsilon>0$. Observe that if $X_n \leq x$ then either $X\leq x+\epsilon$ or $|X_n-X|>\epsilon$.
Here I am confused about the observation. I tried drawing this and I feel like these two cases are not mutually exclusive. I understand the rest of the proof once I assume this part and I think I understand this part too. However, I think my problem is related to the use of "either ... or". When mathematicians use "or" do they use it like it is used in logic (where the statement is correct even when both predicates are correct) or daily use(which implies the two options are mutually exclusive)
AI: \begin{align}
\{X_n\le x\}&=\{X_n\le x,|X-X_n|\le \epsilon\}\cup\{X_n\le x,|X-X_n|>\epsilon\} \\
&\subseteq \{X\le x+\epsilon\}\cup\{|X-X_n|>\epsilon\},
\end{align}
which means that if $\omega\in\{X_n\le x\}$, then $\omega$ belongs to at least one of the sets in the union on the RHS. |
H: Question regarding algebraic proof for Pascal's identity.
I was looking at the proof for
$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$, where $n$ and $k$ are each $\ge 1$.
According to the proof, expressing the right-hand side in terms of factorials, we get
$$\frac{(n-1)!}{k!(n-k-1)!}+\frac{(n-1)!}{(k-1)!(n-k)!}.$$
However, I don't understand how we got $(n-k)!$ instead of $(n-k-2)!$ for the denominator in 2nd term.
The equality used is the following:
$\binom{n}{k}=\frac{n!}{k!(n-k)!}$
So shouldn't we have $(k-1)!(n-k-2)!$ in the denominator for the second term?
I assume I'm missing something really simple, but any help would be appreciated.
AI: No, it shouldn't be $n-k-2$, and, yes, you're missing something simple.
When you subtract $k-1$ from $n-1$, don't forget to distribute the $-1$ factor properly:
$(n-1)\color{blue}-(k\color{blue}-1)=n-1\color{blue}-k\color{blue}+1=n-k$. |
H: Multinomial coefficient of a sequence in specific form
I came across a question which asked me to find the coefficient of $x^{2n}$ in the following polynomial:
$$(\sum\limits_{i=0}^{n-1} x^i )^{2n+1}$$
My approach was to isolate every term, i.e. if we choose $x^2$ , n times and 1 n times again, we get a part of the coefficient of $x^{2n}$. Doing the same for $x^4$, n/2 times and iterating this process over and over would take a lot of time and the answer will be in the form of a sigma which maybe reduced as well.
However, the answer given was rather simple
Required form of answer
${2n+1}\choose{2}$-${{2n+1}\choose{1}}{{3n}\choose{n}}$+${4n}\choose{2n}$
My questions
(if possible)How should I proceed with my method so that I reach the same results?
Any other method to precisely find the number of solutions for the equation $$(\sum\limits_{i=1}^{2n+1} x_i )=2n$$ where all $0\leq x_i\leq n-1$
AI: You actually want the number of solutions in non-negative integers to the equation
$$\sum_{i=1}^{2n+1}x_i=2n\;.\tag{1}$$
This can be found by a combination of stars and bars calculations and an inclusion-exclusion calculation.
Without the upper limit of $n-1$ this has $\binom{2n+(2n+1)-1}{(2n+1)-1}=\binom{4n}{2n}$ solutions in non-negative integers by the usual stars and bars calculation. For each $k=1,\ldots,2n+1$ we need to subtract the number of solutions in which $x_k\ge n$. These are in bijective correspondence with solutions to
$$\sum_{i=1}^{2n+1}x_i=2n-n=n\;,$$
and by a similar calculation there are $\binom{3n}{2n}=\binom{3n}n$ of them for each $k$. Thus, we need to subtract $(2n+1)\binom{3n}n=\binom{2n+1}1\binom{3n}n$ to correct for overcounting. However, any solution to $(1)$ in which two of the $x_i$ exceed the upper bound have now been subtracted twice and need to be added back in. For each pair of indices $k$ and $\ell$ there is just one solution to $(1)$ in which $x_k\ge n$ and $x_\ell\ge n$, so we must add $\binom{2n+1}2$ to correct the excessive subtraction in the first correction. And now weβre done, since $(1)$ has no solutions in which more than two of the $x_i$ exceed the upper bound: the final result is
$$\binom{4n}{2n}-\binom{2n+1}1\binom{3n}n+\binom{2n+1}2\;.$$ |
H: Prove that $b^2-4ac$ can not be a perfect square
Given $a$,$b$,$c$ are odd integers Prove that $b^2-4ac$ can not be a perfect square.
My try:Let $a=2k_1+1,b=2n+1,c=2k_2+1;n,k_1,k_2 \in I$
$b^2-4ac=(2n+1)^2-4(2k_1+1)(2k_2+1)$
$\implies b^2-4ac=4n^2+4n+1-16k_1k_2-8k_2-8k_1-4 $
AI: Assume $b^2-4ac=d^2$.
Then $d$ is odd and $(b-d)(b+d)=4ac$.
Consequently, there exists odd $u,v$ such that $b-d=2u$ and $b+d=2v$.
This leads to the contradiction $b=u+v$ since $u+v$ is even. |
H: Prove that $T$ is normal if and only if $\|Tv\| = \|T^*v\|$ for all $v$
I'm going through the proof of the above theorem in Linear Algebra Done Right (3rd ed). It goes as follows:
\begin{align}
T \text{ is normal } &\iff T^*T - TT^* = 0\\
&\iff \langle (T^*T - TT^*)v, v \rangle = 0, \text{ for all } v \in V\\
&\iff \langle T^*Tv, v \rangle = \langle TT^*v, v \rangle, \text{ for all } v \in V\\
&\iff ||Tv||^2 = ||T^*v||^2 , \text{ for all } v \in V\\
\end{align}
Axler then goes on to mention that the second equivalence follows by (7.16) which states that if $T = T^*$ and $\langle Tv, v \rangle$ = $0$ for all $v$, then $T$ = $0$.
Now to my question:
I haven't managed to convince myself the above stated justification for the second equivalence is needed. More specifically, the first equivalence establishes that $T^*T - TT^*$ is the zero map, and therefore maps all $v \in V$ to $0$. Consequently, $(T^*T - TT^*)v = 0$. We also know that $\langle 0, v \rangle = 0$ for all $v \in V$. Hence, the second equivalence follows directly without having to invoke (7.16). In fact, isn't invoking (7.16) in this case not just redundant but purely wrong, since the implication follows only in one direction?
AI: It is necessary.
Claim. $T^*T - TT^* = 0$ if and only if $\langle (T^*T - TT^*)v,v \rangle = 0$ for any $v \in V$.
Proof. $(\Rightarrow)$ Suppose $T^*T - TT^* = 0$. Then, for any $v \in V$,
$$\langle (T^*T - TT^*)v,v \rangle = \langle 0,v \rangle = 0.$$
$(\Leftarrow)$ Suppose $\langle (T^*T - TT^*)v,v \rangle = 0$ for any $v \in V$. Define $U := T^*T - TT^*$ and note that $U^*=U$. Then $U^*=U$ and $\langle Uv,v \rangle = 0$ for all $v \in V$ implies that $U=0$ thanks to $(7.16)$. |
H: Show that $\sum_x \Big\lfloor\sqrt[m]{\frac{n}{x}}\Big\rfloor=\lfloor n\rfloor$
This seems to be a trivial fact however, I can't find a satisfactory proof for the following statement:
For any integer $n>0$, and $m>0$,
$$
\sum_x\left\lfloor\sqrt[m]{\frac{n}{x}}\right\rfloor=\lfloor n\rfloor
$$
where the sum is over all positive integers $x$ that are not divisible by a power $m$ of an integer larger than one, and $p\mapsto[p]$ stands for the floor function.
I tried to use the fact that every positive integer $k$ can be expressed uniquely as
$$k=x s^m$$
where $x$ is not divisible by the power $m$ of any integer larger than $1$.
AI: Each term in the left-hand side counts the number of positive integers less than or equal to $n$ of the form $xs^m$ for some $s$; i.e. let
$$A_x = \{k\in \mathbb{N}: \exists s\in \mathbb{N}\text{ such that }k =xs^m\text{ and }k\leq n\}.$$
Then it suffices to show that $\{1, \ldots, n\}$ is a disjoint union of $\bigcup_x A_x$, since this would give
$$n = \left|\bigcup_xA_x\right|= \sum_x|A_x|,$$
which is what we want.
The fact that we take $x$ to be the positive integers that are not divisible by powers if $n$ ensures that the sets are disjoint. The fact that you put in the problem description tells us that we have included all the needed elements. |
H: Let $R$ be the Ring of all functions in $\mathscr{C}^{0}[0, 1]$
$\bullet~$ Problem: Let $R$ be the Ring of all functions in $\mathscr{C}^{0}[0, 1]$. Prove that the map $\varphi : R \to \mathbb{R}$ defined by
$$ \varphi(f) = \int_0^1 f(t) dt \quad \text{for any } f \in \mathscr{C}^{0}([0, 1])$$
is a Homomorphism for additive groups but not a Ring Homomorphism.
$\bullet~$ My Solution: We are given the ring $R$ of all continuous real-valued functions on the closed interval $[0, 1]$.
According to question let's consider the map
\begin{align*}
\varphi : &~ R \rightarrow \mathbb{R}\\
& f(t) \mapsto \int_{0}^{1} f (t) dt
\end{align*}
$\circ~$ Now we'll check if it's an Additive Group Homomorphism.
\begin{align*}
\varphi(f + g)(t) & = \varphi( f (t) + g(t) ) = \int_{0}^{1} ( f (t) + g (t) )dt\\
& = \int_{0}^{1} f(t)dt + \int_{0}^{1} g (t)dt\\
& = \varphi(f) + \varphi(g)
\end{align*}
Hence it is an additive group homomorphism.
$\circ~$ Edit: Consider $f(t) = t \in \mathscr{C}^0[0, 1]$ and $g(t) = t^2 \in \mathscr{C}^{0}[0, 1]$ then $(f\cdot g)(t) = t^3 \in \mathscr{C}^0[0, 1] $ .
Now we'll check for the multiplicative one.
\begin{align*}
\varphi( fg )(t) & = \varphi( f(t) g(t))\\
& = \int_{0}^{1} f (t) g (t) dt\\
& = \int_0^1 t^3 dt\\
& = \frac{1}{4} \neq \frac{1}{6}\\
& = \int_0^1 t dt \cdot \int_0^1 t^2 dt\\
& = \bigg( \int_{0}^{1} f (t) dt \bigg) \bigg( \int_{0}^{1} g (t) dt \bigg) \\
& = \varphi( f) \cdot \varphi( g )
\end{align*}
And hence it's clear that, our map is not a ring homomorphism.
Please check the solution for glitches and give new ideas :)
AI: The fact that $\phi(|f||g|) \leq \phi(|f|)\phi(|g|)$ is perfectly consistent with the claim that $\phi(|f||g|) = \phi(|f|)\phi(|g|)$. You need to come up with a specific example where equality is violated. |
H: Defining a topology on $\mathbf{Q}$
We may define a topology on the set $\mathbf{Q}$ of rational numbers
by taking for opens sets all unions of bounded open intervals.
I am not sure I understand this. Do they mean that we take
$$\mathfrak{D}:=\left\{A\ \middle|\ (\exists x)(\exists y)\left(x\in\mathbf{Q}^{\mathbf{N}}\land y\in\mathbf{Q}^{\mathbf{N}}\land A=\bigcup_{n\in\mathbf{N}}\,]x_n,y_n[\right)\right\}$$
for the open sets of this topology?
AI: There are two reasonable interpretations of bounded open interval in $\Bbb Q$, but they yield the same topology. If by interval they mean order-convex set, then the set of intervals in question is
$$\mathscr{B}=\{(x,y)\cap\Bbb Q:x,y\in\Bbb R\text{ and }x<y\}\;;$$
if they also want the endpoints of the intervals to be rational, itβs
$$\mathscr{B}=\{(x,y)\cap\Bbb Q:x,y\in\Bbb Q\text{ and }x<y\}\;.$$
In either case the topology that theyβre defining is
$$\tau=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{B}\right\}\;,$$
the set of unions of arbitrary subcollections of $\mathscr{B}$. |
H: Question regarding Triangle Inequality.
Working on the book: Lang, Serge & Murrow, Gene. "Geometry - Second Edition" (p. 23)
If two sides of a triangle are 12 cm and 20 cm, the third side must be larger than __ cm, and smaller than __ cm.
The answer given by the author is:
$20 - 12 < x < 12 + 20$, where x is the length of the third side.
Triangle Inequality.
Let $P, Q, M$ be points. Then $d(P, M) < d(P, Q) + d(Q, M)$.
What calculations did the author perform in order to achieve the subtraction on the left side of $20 - 12 < x < 12 + 20$ ?
AI: It's from $20\lt x+12$; subtract $12$ from both sides. |
H: Expected number of coin flip
I need to calculate the expected number of coin flips needed to get two consecutive heads.
Below is my approach -
Expected number 1. Probability is 0 (because I need atleast 2 tosses)
Expected number 2. It is HH. Probability is .5^2
Expected number 3. It is THH. Probability is .5^3
So on.
So my total expected number will be infinite sum of 2*.5^2 + 3*.5^3 +...
Is my approach correct? If not, where am I doing wrong?
Is there any finite value of above sum?
Your pointer will be highly appreciated.
AI: This is a Markov Chain problem. Suppose we stop flipping once we get two consecutive heads We have the following states:
The last flip was tails.
The last flip was heads
Let $a, b$ denote the expected value of the number of steps it takes to stop flipping from each state. Here
$$a = 1 + \frac{1}{2}a + \frac{1}{2}b$$
$$b = 1 + \frac{1}{2}a$$
and solving gives $a=6$ and $b=4$.
From the starting state, this is equivalent to if the last flip was tails, so the expected value is $\boxed{6}$. |
H: Probability in Poisson distribution.
Q: The location of trees in a park follows a Poisson distribution with rate $1$ tree per $300$ square meters. Suppose that a parachutist lands in the park at a random location. What is the probability that the nearest tree is more than $4$ meters away from where he lands?
Is this asking me to find $P(\min X_i>4)$ where $X_i$ represent the distance of tree $i$ from the parachutist?
AI: The set of points within $4$ meters of him has area $\pi\cdot(\text{4 meters})^2.$
Thus the expected number of trees within that disk is $\dfrac{\pi\cdot 4^2}{300}\approx 0.1675516.$
That number of trees has a Poisson distribution. You're looking for the probability that that number is $0.$ |
H: How should I notate this function?
My apologies for the vague, non-descript title, I couldn't come up with a concise way to describe what I mean.
Basically, I have a sequence $A$ such that $\forall \;x \in A: x \in \{0, 1, 2, ..., n\}$, where $n \in \mathbb{N}$. Let's take $n = 6$ as an example. $A$ could look like this, for example:
βββββ¦ββββ¦ββββ¦ββββ¦ββββ¦ββββ¦ββββ¦ββββ¦ββββ¦ββββ
β 3 β 1 β 4 β 4 β 5 β 1 β 6 β 2 β 2 β 3 β
βββββ©ββββ©ββββ©ββββ©ββββ©ββββ©ββββ©ββββ©ββββ©ββββ
Now, I want to have a function that takes in a sequence $X$ and an index $i$ and returns the number of times that $X_{i}$ has appeared in $X$ up until $i$. An example would probably help here.
Let's say that our index is $4$ and we're using sequence $A$. We look at what's in the fourth place in $A$; it's the number 4. Then, starting from the beginning of the sequence, we check every item in the sequence to see if it equals 4. However, once we reach the $i$-th element, we stop.
βββββ¦ββββ¦ββββ¦ββββ¦ββββ¦ββββ¦ββββ¦ββββ¦ββββ¦ββββ
β 3 β 1 β 4 β 4 β 5 β 1 β 6 β 2 β 2 β 3 β
βββββ©ββββ©ββββ©ββββ©ββββ©ββββ©ββββ©ββββ©ββββ©ββββ
^ ^ we don't even bother
| | checking all of these
In this case, the function would return 2.
Is there any way that I could notate this? It doesn't even have to be compact or particularly legible, I just need a way to describe what I'm talking about mathematically.
Thanks.
AI: $$ X_i=\text{ cardinality of }\{A_k=A_i \vert k\le i\}$$ |
H: Uniform convergence of sequence on interval $[-b,0]$
I am aiming to prove that there exists a continous function $\exp:\mathbb{R}\rightarrow\mathbb{R}$ using the sequence $$e_n^x=\sum_{k=0}^{n}\frac{x^k}{k!}$$
So far: defining the sequence $e_n^x=\sum_{k=0}^{n}\frac{x^k}{k!}$ given that this is a polynomial, I have shown that this is cauchy and therefore uniformly convergent on the reals, and taking $b\in\mathbb{Q}$ the sequence is also cauchy and thus uniformly convergent. Then $e_n^x$ is uniformly convergent for $x\in[0,b]$, I then showed that the limit function is continuous, then it must be that $\exp\colon[0,b] \rightarrow\mathbb{R}$. Now assuming that $e_n^x$ is uniformly convergent in $[0,x]$.
Now I am aiming to show that $\{e_n^{-x}\}$ is uniformly convergent on $[-b,0]$, here I am aiming to use the triangle inequality somehow so that if $exp$ is converging in $[0,b]$ then $\exp$ is converging on the union set of $\mathbb{Q}$ and $[0,b]$. After doing this it has to be that if $\exp:[0,b]\rightarrow\mathbb{R}$ then $\exp:[0,\infty]\rightarrow\mathbb{R}$. From here I can take a real number $r\in\mathbb{R}$ so that for every element $q\in\mathbb{Q}$ where $r<q$ then $\exp:[0,q]\rightarrow\mathbb{R}$ and we have that exp is continous at $r$ and this must hold for all reals.So then it must be that there exists $\exp:\mathbb{R}\rightarrow\mathbb{R}$.
I would like to know if me general idea is correct? Further I am having some trouble with showing that $\{e_n^{-x}\}$ is uniformly convergent on $[-b,0]$ how would we go about showing this?
AI: Showing that the sequence of functions $e_n(x) := \sum_{k = 0}^n \frac{x^k}{k!}$ is uniformly convergent on $[- b, 0]$ requires just as much work as showing that it is uniformly convergent on $[0, b]$.
In fact, I would say that the thing to aim for is showing that $e_n(x)$ is uniformly convergent on $[-b, b]$, for any $b \in [0, \infty)$. This suffices for proving that $e^x$ is continuous everywhere.
To get there, we can appeal to this standard result:
Theorem: If the sequence $\lim_{k \to \infty} \left|\frac{a_k }{ a_{k+1}} \right|$ converges, then $r := \lim_{k \to \infty} \left|\frac{a_k }{ a_{k+1}} \right|$ is the radius of convergence for the power series $\sum_{k = 0}^\infty a_k x^k$. (This is true even if $r = \infty$.)
Furthermore, if $0 \leq b < r$, then the sequence of partial sums $x \mapsto \sum_{k = 0}^n a_k x^k$ is uniformly convergent on $[-b, b]$.
In our case, $a_k = 1 / k!$ and $\lim_{k \to \infty} \left|\frac{a_k }{ a_{k+1}} \right| = \lim_{k \to \infty} | k + 1 | = \infty$, so the radius of convergence is $r = \infty$. Hence for any $b \in [0, \infty)$, $e_n(x)$ is uniformly convergent on $[-b, b]$. |
H: Notation for a coordinate grid
Consider the coordinate grid of points between $(-5, -5)$ and $(5, 5)$, inclusive, spaced 0.3 units apart in both directions. Is there a more compact way of notating this without writing something long, such as $\{(-5, -5), (-5, -4.7), \dots, (-5, 5), (-4.7, -5), \dots, (5, 5)\}$?
AI: You could use
$(0.3\mathbb{Z}-5\cap[-5,5])^2$
Explanation: $\mathbb{Z}$ are the integers, and $0.3\mathbb{Z}$ are just values which are $0.3$ units apart. The $-5$ just moves the origin of the lattice around. Intersecting the lattice with $[-5,5]$ restricts it to the intended bounds. Finally, you take the Cartesian product with itself, aka square it, to get two dimensional.
PS: not both, 5 and -5 can be part of the solution, since $10/0.3=33$ remainder $0.1$. |
H: To what degree does a centered random walk $S_n$ behave asymptotically similar to $-S_n$?
Let $S_n$ be a centered random walk, i.e. increment mean $EX = 0$. For convenience let us say that $EX^2 < \infty$ and it only takes two values $\{a, -b\}$, $a,b > 0$. Considering objects such as $$ E ( e^{S_n}) \quad \text{ or } \quad E\left( \sum_{k = 1}^n e^{S_k}\right)$$
(or more generally $E(f(S_1, \dots, S_n))$, do these expectation always behave asymptotically the same as $$ E(e^{-S_n}), E\left( \sum_{k=1}^n e^{-S_k} \right), E (f(-S_1, \dots, -S_n)) \quad ?$$
More precisely, does $$ E (e^{S_n}) / E (e^{-S_n}) \to 1 $$
hold etc.?
I hope it is clear what my question is. Because the random walk is centered, it should not (so I suspected) make a difference whether we consider $S_n$ or $-S_n$ for large $n$, as the drift is normally the main criteria for the asymptotics of $S$. Is this true in general? And, can one show this rigorously?
AI: It is true that all random walks with a second moment looks the same "from a distance", namely that they converge to a Brownian motion after proper rescaling. But you are asking for much more. Take the simple example of
$$
\mathbb{P}(X_i = -1) = \frac23, \quad \mathbb{P}(X_i = 2) = \frac13.
$$
Then
$$
\mathbb{E}(e^{S_n}) = \left ( \frac23 e^{-1} + \frac13 e^2 \right )^n
$$
while
$$
\mathbb{E}(e^{-S_n}) = \left ( \frac23 e + \frac13 e^{-2} \right )^n,
$$
so the ratio of the two is
$$
\frac{\mathbb{E}(e^{S_n})}{\mathbb{E}(e^{-S_n})} = \left ( \frac{2e^{-1}+e^2}{2 e + e^{-2}} \right )^n,
$$
which tends to $+ \infty$ exponentially fast. |
H: Fourier transform of $e^{-at}u(t-b)$
I have this signal $ s(t) = e^{- \frac{t}{RC} } [ u(t) - u(t - T_c) ] $ and I have to calculate the Fourier transform. I obtained $$ S(f) = \frac{ RC( 1 - e^{-i2 \pi f T_c } )}{1 + i 2 \pi f RC } $$ but I should obtain $$ S(f) = \frac{ RC( 1 - e^{-i2 \pi f T_c } e^{\frac{-T_c}{RC}}) }{1 + i 2 \pi f RC } $$
The Fourier transform of the first part gave me $ \frac{1}{ \frac{1}{RC} + iw } $
and the Fourier transform of the second part of the subtraction gave me $ \frac{1}{ \frac{1}{RC} + iw } e^{-i 2 \pi f T_c } $. I this the error is in the second part but I donβt know how to solve it.
AI: Basically, we just need a repeated application of this Fourier transform:
$$\bbox[5px, border: 1pt solid blue]{{\cal{F}}\left\{e^{-at}u(t-b)\right\} = \int_b^\infty e^{-at} e^{-2\pi i f t}\,dt=\frac{e^{-b(a+2\pi i f)}}{a+2\pi i f}.}$$
Applying it twice:
$$\bbox[5px, border: 1pt solid green]{\begin{aligned}{\cal{F}}\left\{e^{-at}[u(t)-u(t-b)]\right\} &= \int_0^\infty e^{-at} e^{-2\pi i f t}\,dt-\int_b^\infty e^{-at} e^{-2\pi i f t}\,dt \\ &=\frac{1}{a+2\pi i f}-\frac{e^{-b(a+2\pi i f)}}{a+2\pi i f}.\end{aligned}}$$
Equivalently, this can be written as:
$$\bbox[5px, border: 1pt solid purple]{\begin{aligned}{\cal{F}}\left\{e^{-at}[u(t)-u(t-b)]\right\}
&= {\cal{F}}\left\{e^{-at}I_{[0,b]}\right\} \\\
&= \int_0^b e^{-at} e^{-2\pi i f t}\,dt\\ &=\frac{1-e^{-b(a+2\pi i f)}}{a+2\pi i f}.\end{aligned}}$$
We have $a=\frac{1}{RC}$ and $b=T_c$:
$$\bbox[5px, border: 1pt solid green]{
\begin{aligned}{\cal{F}}\left\{e^{-\frac{t}{RC}}[u(t)-u(t-T_c)]\right\} &= \int_0^\infty e^{-\frac{t}{RC}} e^{-2\pi i f t}\,dt-\int_{T_c}^\infty e^{-\frac{t}{RC}} e^{-2\pi i f t}\,dt\\ &=\frac{1}{\frac{1}{RC}+2\pi i f}-\frac{e^{-T_c\left(\frac{1}{RC}+2\pi i f\right)}}{\frac{1}{RC}+2\pi i f}.\end{aligned}}$$ |
H: Prove $x^n-p$ is irreducible over $Z[i]$ where $p$ is an odd prime.
Prove $x^n-p$ is irreducible over $Z[i]$ where $p$ is an odd prime.
By gausses lemma this is equivalent to irreducability over $\mathbb{Q}(i)$. Using field extensions this is easy. $[\mathbb{Q}(i,\sqrt[n]{p}):\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}]=[\mathbb{Q}(i,\sqrt[n]{p}):\mathbb{Q}(\sqrt[n]{p})][\mathbb{Q}(\sqrt[n]{p}):\mathbb{Q}]=2n$ Thus $[\mathbb{Q}(i,\sqrt[n]{p}):\mathbb{Q}(i)]=n$ and so $x^n-p$ must be the minimal polynomial, and so it is irreducible. However, the book says you can solve this problem using Eisenstein criterion. That is easy when $x^2+1$ is irreducible mod $p$ as $(p)$ is then prime. What do you do in the other cases?
AI: To prove this via Eisenstein's criterion, use the fact that $\mathbb Z[i]$ is a principal ideal domain. In fact, it is Euclidean. Also, for odd primes $p$ in $\mathbb Z$, $p$ remains prime in $\mathbb Z[i]$ for $p=3\mod 4$, and $p$ factors as a product of two distinct primes $p=p_1p_2$ in $\mathbb Z[i]$ for $p=1\mod 4$ (see this post).
Thus, in either case, there is a prime in $\mathbb Z[i]$ dividing the constant term (and all others except the leading term, since those others are $0$), and whose square does not divide the constant term. So we can apply Eisenstein.
Appropriate proof(s) about the remaining-prime and/or factoring into distinct factors depend on your context... but the explanation may give you some motivation to look at otherwise-technical points. |
H: Show that $y'(t)=y^{2/3}(t) \text{ with } y(0)=0$ has infinitely many solutions
Show that the problem
$$y'(t)=y^{2/3}(t) \text{ with } y(0)=0$$
has infinitely many solutions.
Explain why the existence and uniqueness theorem does not apply here
My attempt
By solving the differential equation by the variable separation method, We get:
$\int\frac{1}{y^{2/3}}dy=\int dt$
$\frac{y^{1/3}}{1/3}=t+c$
And by substituting the initial condition $y(0)=0$ we can get $c=0$
Thus $$y(t)=\frac{t^3}{3^3}$$
But from here how should I prove that there are infinitely many solutions?...
And for the second part (Uniqueness theorem) isn't it because for the solution of $y'=f(y)$ to be unique, we need $f$ to have a continuous first derivative. But here in this specific example, $$\frac{d}{dy}f=\frac{2}{3}y^{-1/3}$$ which is not continuous at zero.
AI: There is only one "separable solution", as you say: it is $y=t^3/3^3$. But to perform separation of variables, you needed to assume that $y \neq 0$ almost everywhere on the domain where you did the integration (so that it is justified to divide by $y^{2/3}$). And this isn't necessarily true, because certainly $y=0$ is a solution.
To get even more solutions, because the equation is autonomous, you can just stay at zero for however long you want, say up to some $t_1>0$, and then switch over to $(t-t_1)^3/3$. You can do the same backward in time. So all of these functions are solutions to the ODE:
$$y(t;t_0,t_1)=\begin{cases} (t-t_0)^3/3^3 & t < t_0 \\
0 & t_0 \leq t \leq t_1 \\
(t-t_1)^3/3^3 & t>t_1 \end{cases}$$
whenever $-\infty \leq t_0 \leq t_1 \leq \infty$. Of course you must assume that $t_0 \leq 0 \leq t_1$ in order to have a solution to the IVP as well. |
H: Convergence/Divergence of Complex Series $\sum\limits_{n=1}^{\infty} \frac{n(2+i)^n}{2^n}$
$$\sum\limits_{n=1}^{\infty} \frac{n(2+i)^n}{2^n}$$
My Attempt: I am new to analyzing complex series, so please forgive me in advance. I apply the ratio test:
$$\lim_{n \to \infty}\frac{|a_{n+1}|}{|a_n|} = \lim_{n \to \infty}\frac{|(n+1)(2+i)^{n+1}2^n|}{|2^{n+1}n \ (2+i)^n|} = \lim_{n \to \infty} |\frac{n+1}{2n}(2+i)| = \frac{1}{2} \lim_{n \to \infty} |2+i|$$
I know that $|z| = |a + bi|$ can be expressed as $\sqrt{a^2+b^2}$, hence:
$$\frac{1}{2}
\lim_{n \to \infty} \sqrt{5} > 1$$ By the ratio test, this makes the series diverging series. Is this approach correct?
AI: Your approach is good. You can alternatively solve it through the root test. One has that
\begin{align*}
|a_{n}| = \left|\frac{n(2+i)^{n}}{2^{n}}\right| = n\left(\frac{\sqrt{5}}{2}\right)^{n} \Longrightarrow \limsup_{n\to\infty}|a_{n}|^{1/n} = \limsup_{n\to\infty}n^{1/n}\left(\frac{\sqrt{5}}{2}\right) = \frac{\sqrt{5}}{2} > 1
\end{align*}
Thus the given series diverges.
Hopefully this helps. |
H: Derivative Greater Than 0 Implies One-To-One Function In Neighborhood
Let $f: \textrm{dom}(f) \rightarrow \mathbb{R}.$
Let $x_0 \in \mathbb{R}.$
Assume $f'(x_0) > 0$.
i.e. $~ \displaystyle\lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h} > 0$
i.e. $~ \exists l > 0 \textrm{ s.t. } \forall \varepsilon_1 > 0, \exists \delta_1 > 0 \textrm{ s.t. } \forall h \in \mathbb{R}, 0 < |h| < \delta_1 \Rightarrow \Bigg| \displaystyle\frac{f(x_0 + h) - f(x_0)}{h} - l \Bigg| < \varepsilon_1$
This implies that $f$ is continuous at $x_0$, as I have proven before.
i.e. $~ \forall \varepsilon_2 > 0, \exists \delta_2 > 0 \textrm{ s.t. } \forall x \in \mathbb{R}, |x - x_0| < \delta_2 \Rightarrow |f(x) - f(x_0)| < \varepsilon_2$
Also,
$\exists \delta_3 > 0 \textrm{ s.t. } \forall x \in \mathbb{R}, 0 < |x - x_0| < \delta_3 \Rightarrow \displaystyle\frac{f(x) - f(x_0)}{x - x_0} > 0$
I would like to prove there exists an open interval containing $x_0$ where $f(x)$ does not have a value of $f(x_0)$ for any $x$ in that interval apart from $x_0$.
i.e. $~ \exists a, b \in \mathbb{R} \textrm{ s.t. } a < x_0 < b \wedge \big( \forall x \in (a, b), x \neq x_0 \Rightarrow f(x) \neq f(x_0) \big)$
Unfortunately, I cannot say anything about double differentiability of $f$.
Because of this, I cannot mention continuity of $f$ in a neighborhood of $x_0$ (or can I?)
Perhaps there is a counterexample and I shouldn't try to prove this statement.
Help needed.
AI: It is not true that if $f'(x_0)>0$ then $f$ is one-to-one in some open neighborhood of $x_0.$
But it is true that if $f'(x_0)>0$ then there is some open neighborhood of $x_0$ within which $f$ takes the value $f(x_0)$ only at $x_0$ and nowhere else.
Let $\displaystyle f(x) = \begin{cases} f(x_0) & \text{if }x=x_0, \\[8pt] f(x_0) + (x-x_0) + (x-x_0)^2 \sin(1/(x-x_0)) & \text{if } x\ne x_0. \end{cases}$
Then $f'(x_0)=1,$ but in every open neighborhood of $x_0$ there are values of $x$ for which $f'(x)$ is positive and others for which it is negative. Thus $f$ is not one-to-one in that neighborhood.
However, there is some open neighborhood of $x_0$ within which every value of $x$ other than $x_0$ satisfies
$$
\frac{f(x) - f(x_0)}{x-x_0} > \frac 1 2.
$$
That implies $f(x)-f(x_0)>0$ if $x>x_0$ and $f(x)-f(x_0)<0$ if $x<x_0.$ |
H: Classifying the cyclic extensions of a fixed degree
This is the following result from Milne's Fields and Galois Theory (page 72):
Milne only showed the "only if" part but I admit that I cannot give the reason why the "if" part is true.
My attempt: In case we have $a= b^r c^n$, it is $a^{1/n}/c$ a root of $X^n - b^r$. Now I think it might be a good idea to use the fact that $r$ is coprime to $n$. However, here is the part where I don't know how to use the fact and continue.
Could you please help me elaborating on this part? Thank you!
AI: The idea is that the stated condition is, when $r$ is relatively prime to $n$, actually symmetric in $a$ and $b$, even though it appears otherwise.
In this it is clear that under those conditions, $a^{1/n}$ is in $F(b^{1/n})$, so you only need the other direction.
There are integers $x,y$ such that $rx + ny = 1$. Then take the $x$th power of both sides:
$$a^x = b^{rx} c^n.$$
Multipy both by $ny$ and combine the exponent on the RHS $b$ into a $1$:
$$a^x b^{ny} = b^{rx+ny} c^n = b^{1} c^n.$$
Then divide by $c^n$ and group terms on the left:
$$a^x \left(\frac{b^y}{c}\right)^n = b$$
And so $b^{1/n}$ is in $F(a^{1/n})$, giving the other inclusion. |
H: Rudin's Functional Analysis Theorem 7.23
At the end of (b) of Theorem 7.23 in Rudin's Functional Analysis, it says
"Since (1) holds for $z \in \mathbb{R}^n$ (by the choice of $u$), Lemma 7.21 completes the proof of (b)."
Here, (1) is $\mathbf{f(z) = u(e_{-z})}$, where $u \in \mathscr{D}'(\mathbb{R}^n)$. The latter part of the statement is clear, so the only thing I have to understand is why (1) holds for $z \in \mathbb{R}^n$.
The condition for $f$ is that $\vert f(x) \vert \le \gamma(1 + \vert x \vert)^N $ and the book goes on to say
"The restriction of $f$ to $\mathbb{R}^n$ is therefore in $\mathscr{S}_n'$ and is the Fourier transform of some tempered distribution." $(*)$
However, I really don't know why (1) holds for $z \in \mathbb{R}^n$ only based on the above statement $(*)$, especially provided that $e_{-z}(t) = e^{-iz \cdot t}$ is not in $\mathscr{S}_n'$ and $f$ is not in $L^1(\mathbb{R}^n)$. Maybe the approximation works, but I am still struggling to show (1).
Any help would be appreciated!
AI: Hint: If $\phi \in \mathcal S_n$ then $\hat {\phi} (y)=\int \phi (x) e_{-y} (x) dx$. Apply $u$ to get $\int f \phi=\int u(e_{-y}) \phi$ for all $\phi \in \mathcal S_n$. |
H: Properties of distribution function
Let $(\Omega, \mathcal{F}, P)$ be a probability space, $X$ a random variable and $F(x) = P(X^{-1}(]-\infty, x])$. The statement I am trying to prove is
The distribution function $F$ of a random variable $X$ is right continuous, non-decreasing and satisfies $\lim_{x \to \infty}F(x) = 1$, $\lim_{x \to -\infty} F(x) = 0$.
As $F(x + \delta) = F(x) + P(]x, x + \delta])$, we have that $F$ is non-decreasing, but is the measure of an interval bounded by its length? In that case we would have right continuity as well.
For the limits, we have $F(x) + P(X^{-1}(]x, \infty]) = P(\Omega) = 1$, so $F(x) = 1 - P(X^{-1}(]x, \infty])$, so it suffices for $P(X^{-1}(]x, \infty])$ to get small as $x$ gets large and to get large as $x$ gets small. This is not true for general measures, take the Lebesgue measure for example, but maybe because we need $P(X^{-1}(\mathbb{R}))$ to be $1$?
AI: It is a basic fact that for any finite measure $\mu$ the condition $A_n$ decreasing to $A$ implies that $\mu (A_n) \to \mu (A)$. [Lebesgue measure is an infinite measure and this property fails for Lebesgue measure]. This follow from the fact that $\mu(A_n^{c}) \to \mu(A^{c})$ since $A_n^{c}$ increases to $A$ and $\mu (E^{c})=\mu (\Omega)-\mu (E)$. With this result in hand it should be easy for you to complete your arguments.
Note that $(x,x+\delta]$ decreases to empty set as $\delta$ decreases to $0$ and $(x, \infty)$ decreases to empty set as as $x$ increases to $\infty$. |
H: Combinatorics of a Tournament
$8$ people participate in a tournament, so that each person plays with all the other people once. If a person wins against another, then the winner gets $2$ points, while the losing team will get none. If they tie, they each get $1$ point, respectively. When finished, the people are ranked depending on how many points they have in total. How many points does a person need to have, to secure their spot in the best four players.
I know that the maximum possible amount of points one can achieve is $14.$ I also know a score of $12$ gurantees you to be in the top $5.$ I'm not really sure how to continue with this information, how would I do this problem?
AI: Note that there are a total of $56 = 2\binom{8}{2}$ total points given out in the tournament. If anyone gets at least $11$ points, then they are guaranteed to be in the top 4. Suppose for contradtiction otherwise, that there are at least $5$ people who earn $11$ points. Then that would mean there's only $56 - 5\times 11 = 1$ point left over that could have been won by one of the 3 remaining players. This is impossible though because these players also play against each other, and so within these $3$ players there must have been $6 = 2\binom{3}{2}$ points allocated from these games.
Now we show that getting $10$ points isn't enough to be in the top 4 (assuming arbitrary tie-breaking). We show that there is a scenario where there are five players who get $10$ points each. Label the players 1 through 8.
Player 1 beats 2, 3, 6, 7, 8
Player 2 beats 3, 4, 6, 7, 8
Player 3 beats 4, 5, 6, 7, 8
Player 4 beats 5, 1, 6, 7, 8
Player 5 beats 1, 2, 6, 7, 8
Player 6 beats 7
Player 7 beats 8
Player 8 beats 6
This scenario works. |
H: Are there any elementary functions $\beta(x)$ that follows this integral $\int_{y-1}^{y} \beta(x) dx =\cos(y)$
Are there any simple functions $\beta(x)$ that follows this integral $$\int_{y-1}^{y} \beta(x) dx =\cos(y)$$
I think there is an infinite amount of solutions that are continuous everywhere but how can I find one that only uses elementary functions?
AI: Try this:
$$
\beta(x) = \frac{\sin(x+1)}{2} - \frac{\sin(1)\;\cos(x+1)}{2\cos(1)-2}
$$
How to find this? If $N$ is a positive integer, then
$$
\int_0^N \beta(x)\;dx = \sum_{n=1}^N\int_{n-1}^n\beta(x)\;dx
=\sum_{n=1}^N \cos n
=-\frac{\cos(N+1)}{2}-\frac{\sin(2)\;\sin(N+1)}{2(\cos(1)-1)} - \frac{1}{2}
$$
Now make a guess that the same formula will also work when we
replace $N$ with $y$ which is not
a positive integer. Differentiate to get $\beta(y)$.
Then check that it works.
Check:
\begin{align}
\beta(x) &= \frac{\sin(x+1)}{2} - \frac{\sin(1)\;\cos(x+1)}{2\cos(1)-2}
\\
\int\beta(x)\;dx &=
\frac{-\cos(x+1)}{2} - \frac{\sin(1)\;\sin(x+1)}{2\cos(1)-2}
\\
\int_{y-1}^y\beta(x)\;dx &=
\frac{-\cos(y+1)+\cos(y)}{2}
+ \frac{\sin(1)\;(-\sin(y+1)+\sin(y))}{2\cos(1)-2}
\\ &=
\frac{-\cos(1)\cos(y)+\sin(1)\sin(y)+\cos(y)}{2}
+ \frac{\sin(1)\;(-\cos(1)\sin(y)-\sin(1)\cos(y)+\sin(y))}{2\cos(1)-2}
\\ &=
\left[\frac{-\cos(1)+1}{2}
+\frac{-\sin^2(1)}{2(\cos(1)-1)}\right]\cos(y)
+\left[\frac{\sin(1)}{2}
+\frac{-\cos(1)\sin(1)-\sin(1)}{2(\cos(1)-1)}\right]\sin(y)
\\ &=
\frac{-\cos^2(1)+2\cos(1)-1-\sin^2(1)}{2(\cos(1)-1)}\cos(y)
+\frac{\cos(1)\sin(1)-\sin(1)-\sin(1)\cos(1)+\sin(1)}{2(\cos(1)-1)}\sin(y)
\\ &=
\frac{2\cos(1)-2}{2(\cos(1)-1)}\cos(y)
+\frac{\cos(1)\sin(1)-\sin(1)-\sin(1)\cos(1)+\sin(1)}{2(\cos(1)-1)}\sin(y)
\\ &=
\cos(y)
\end{align} |
H: Integral of modulus is the modulus of the integral iff arguments are constant.
I'm sure this has been asked before but I can't find it and it's bugging me. I'm reading Voison's Hodge theory book and I ran into this elementary inequality that I realize I don't understand as well as I should: if $f:\mathbb C\to \mathbb C$ is continuous (or as differentiable as you like, it doesn't matter) then integrating along a circle centered at $z_0\in \mathbb C$ and parametrizing in the usual way, the following inequality holds: $$\left|\int_0^1 f(z_0+e^{2\pi it})\mathrm dt\right|\le \int_0^1|f(z_0+e^{2\pi it})|\mathrm dt.$$
The claim is that equality holds if and only if the argument of $f(z_0+e^{2\pi it})$ is constant. This is of course sufficient for equality, but for the life of me I can't prove necessity. I think I'm just having a bad day but I need to know why.
AI: Assume that we have equality. By multiplying $f$ with the right phase, we may wlog assume that the integral on the LHS is nonnegative even without absolut value. Thus, one gets in this case
$$ \int_0^1 \ \operatorname{Re}( f(z_0 + e^{2\pi i t})) dt = \int_0^1 \ \sqrt{ \operatorname{Re}(f(z_0 + e^{2\pi i t}))^2 + \operatorname{Im}(f(z_0 + e^{2\pi i t}))^2} dt.$$
This can only be an equality if the imaginary part vanishes identically and the real part is nonnegative. I.e. exactly that the argument is constant. |
H: Dual image map restricts to open sets?
A book I'm reading on category theory says that if $A$ and $B$ are topological spaces and $f:A\to B$ is continuous, then the "dual image" map
$$f_*(U)=\{\,b\in B\mid f^{-1}(b)\subseteq U\,\}$$
restricts to open sets; that is, $f_*:\mathcal{O}(A)\to\mathcal{O}(B)$. (So then it's right adjoint to $f^{-1}:\mathcal{O}(B)\to\mathcal{O}(A)$.)
This seems wrong, since it would imply for example (taking $U=\varnothing$) that the image of a continuous function is always closed.
Are there natural conditions under which it does make sense to restrict to open sets?
AI: Youβre right that thereβs a problem here.
Let $A=(0,1)\times(0,1)$ and $B=(0,1)$, each with the usual topology, and let $$f:A\to B:\langle x,y\rangle\mapsto x$$ be the projection to the $x$-axis; this is a continuous, open map. Let $$U=\{\langle x,y\rangle\in A:y>2x-1\}\;.$$ $U$ is open in $A$, but
$$f_*(U)=\left(0,\frac12\right]\;,$$
which is not open in $B$.
You do get the result if $f$ is closed and continuous. In that case let $U$ be open in $A$, and let $F=A\setminus U$. Suppose that $b\in B$; then $f^{-1}[\{b\}]\subseteq U$ iff $b\notin f[F]$, i.e., iff $b\in B\setminus f[F]$, so $f_*(U)=B\setminus f[F]$, which is open in $B$. |
H: How to find the intersection of the graphs of $y= x^2$ and $y = 6 - |x|$?
I was trying to solve this question in preparation for the Math subject GRE exam:
The region bounded by the graphs of $y=x^{2}$ and $y=6-|x|$ is revolved around the $y$ -axis. What is the volume of the generated solid?
(A) $\frac{32}{3} \pi$
(B) $9 \pi$
(C) $8 \pi$
(D) $\frac{20}{3} \pi$
(E) $\frac{16}{3} \pi$
So first I was trying to find the intersection of the graphs of $y= x^2$ and $y = 6 - |x|.$ I did that by equating $ x^2 = 6 - |x|.$ Then I ended up having 2 equations which are $ x^2 - 6 + x = 0 $ and $ x^2 - 6 - x = 0 $, and then I got four values of $x$, which are $\pm 3, \pm 2.$ But I found in the answer for the question that they consider only the two values $\pm 2.$
Could anyone show me what is wrong in my solution or in the book solution (the book just gave me the values $\pm 2$ without explaining how and why), please?
AI: The solutions should be ($x^2-6+x=0$ and $x\ge0$) or ($x^2-6-x=0$ and $x\le0)$.
Can you take it from here? |
H: Why don't these two different methods of counting give the same result?
We have 4 bananas, 5 apples, 6 oranges. How many ways can we choose 7 fruits with at least 4 oranges?
The straight forward method is to divide this into cases with 4, 5, or 6 oranges and then picking from bananas and apples, so that we have chosen 7 fruits. If we compute each case and add them up, we get: $$\binom 64\binom 93+\binom 65\binom 92+\binom 66\binom 91=1485$$ But there is a simpler way; first pick 4 oranges from those 6 oranges (so that we have picked at least 4 oranges), and then choose 3 fruits from the remaining fruits (2 oranges, 5 apples and 4 bananas which sum to 11 fruits). This way we do have picked 7 fruits and at least 4 of them are oranges but the result is: $$\binom 64\binom {11}3=2475$$ which is different from the actual answer.
How can I rigorously check which method works without having to list all the combinations?
AI: Your second method counts each selection that has $5$ oranges $\binom54=5$ times, and each selection that has $6$ oranges $\binom64=15$ times. In each case you count the selection once for each set of $4$ oranges contained in it: any one of those sets could be the set of $4$ that you preselected. |
H: Constructing A Truth Table
How do I construct a truth table with a formula that has 3 logical operators that lack a parentheses?
$$P \lor Q \land \neg(R \lor \neg S)$$
AI: Generally
$$P \lor Q \land \neg(R \lor \neg S)$$ is same as $$P \lor (Q \land (\neg(R \lor (\neg S))))$$
Can you proceed now? |
H: List all the pairs $(x,y)$ s.t. $x^2 - y^2 = 2020$
List all of the pairs $(x,y) \in \mathbb{Z}^2$ s.t. $^2 - ^2 = 2020$.
The prime factorization of $2020$ is $2^2 \cdot 5 \cdot 101$. I used the fact that there exists a solution $(x,y) \in \mathbb{Z}^2$ to the Diophantine equation $x^2 - y^2 = n$ if and only if $n$ is odd or $n$ is a multiple of $4$. Since $n$ is a multiple of $4$, there exist a solution.
I know there is a solution but I am neither able to obtain the number of solution nor able to get the list all the pairs $(x,y)$.
AI: Hint
$x+y,x-y$ have the same parity as $x+y+x-y$ or $x+y-(x-y)$ is even
Now as $2020$ is even, both must be even
$$\implies\dfrac{x+y}2\cdot\dfrac{x-y}2=\dfrac{2020}4$$
Now if $x,y>0$ $$x+y>x-y$$
As $505=1\cdot505=5\cdot101,$
$\dfrac{x-y}2=1$ or $5$ |
H: Konig's lemma proof
Konig's lemma
In this proof of Konig's lemma taken from Kaye's book "The mathematics of logic: A guide to completeness theorems and their applications", I am having a lot of trouble understanding the "We are going to find a sequence of $s(n)$ of elements of $T$ such that..." portion. We are trying to find a sequence $s(n)$ such that it satisfies the given properties. Drawing attention to the first property, "$s(n)$ has length $n$", am I correct in saying these are not the same $n$ in that one is a dummy variable and the other is of course the length of the sequence hence $s(n) = (s(0),s(1),...,s(n-1))$? If this is the case, how is it that we can talk about $s(n+1)$ where $n+1$ is supposedly the next element of the sequence consisting of only $n$ terms as I have described? Note that the half-arrow in the subsequent property is a restriction operator taking the first $n$ elements of the sequence $s(n+1)$ as is shown in the proof.
AI: The $n$ in $s(n)$ and the length of the sequence most definitely are the same $n$. For each $n\in\omega$, $s(n)$ is both an element of $T$ and a sequence of zeroes and ones of length $n$. In particular, for any given $n$, $s(n+1)$ is an element of $T$ that is a sequence of $n+1$ zeroes and ones. If $s(n+1)=\langle b_0,b_1,\ldots,b_n\rangle$, then $s(n+1)\upharpoonright n=\langle b_0,b_1,\ldots,b_{n-1}\rangle$.
It might have been clearer if the author had written this:
We are going to find a sequence $\langle s(n):n\in\omega\rangle$ of elements of $T$ such that for each $n\in\omega$,
$s(n)$ has length $n$,
$s(n)=s(n+1)\upharpoonright n$, and
the tree $T_{s(n)}$ below $s(n)$ is infinite. |
H: ISL 2006 G3:Prove that the line $AP$ bisects the side $CD$.
Let $ABCDE$ be a convex pentagon such that
$$ \angle BAC = \angle CAD = \angle DAE \qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.$$ The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.
My Proof: Note that by $AAA$, we get $\Delta ABC \sim \Delta ADE$ . Hence $A$ is the spiral center of the spiral similarity that sends $CB$ to $ED$ . Hence, $EDPA$ is cyclic and $PACB$ is cyclic .
Now, note that, since $\angle AED= \angle ADC$ and $\angle ABC=\angle ACD$ , we have $DC$ as the common tangent of $(EDPA)$ and $(APCB)$ .
Let $AP\cap DC= M$. Note that, since $AP$ is the radical axis of $(EDPA)$ and $(APCB)$, and $DC$ is the common tangent of $(EDPA)$ and $(APCB)$ . We have $AP$ bisecting $DC$.
AI: Your proof is correct. An alternative proof is as follows:
By the angle conditions, we require $\triangle ABC \sim \triangle ADE \sim \triangle ACD$, so $ABCD$ is similar to $ACDE$. Ceva's Theorem now implies
$$\frac{AF}{FC}\frac{CX}{XD}\frac{DG}{FA} = 1,$$
which gives $\frac{CX}{XD} = 1$ and implies the desired result. |
H: Finding a closed formula for a sum
I wrote this sum (out of the blue) and wondered if it has a closed form:
$$\sum_{k=1}^{\infty} L^{\frac{1}{k}} \cdot(-1)^{k+1}$$ where $L \in \mathbb{N}$
I thought of a sum that would use "$\text{k-root}$" but with alternating sign ($+$ to $-$ etc..)
I couldn't find a way to do so, the only thing I did is write a program that calculates it, so I post here for help. Thank you so much!! :-)
AI: It doesnβt converge unless $L=0$. For $n\ge 1$ let $s_n=\sum_{k=1}^nL^{1/k}(-1)^{k+1}$. If $L\ge 1$, then $L^{1/k}\ge 1$ for all $k\ge 1$, so $|s_{n+1}-s_n|\ge 1$ for all $n\ge 1$. Thus, the sequence of partial sums is not Cauchy and cannot converge. |
H: Understand a derivation
I am learning formal logic through the book An Exposition of Symbolic Logic; in chapter 1, section 10, I am asked to derivate the following argument:
(PβQ) β S
S β T
~T β Q
β΄ T
I couldn't work out the solution, so I saw the answer the book tells:
Show T
~T ass id
Q 2 pr3 mp
Show PβQ
Q 3 r cd
S 4 pr1 mp
T 6 pr2 mp 2 id
When I saw the derivation on line 4 to 5, I said to myself "this is nonsense!". So I ask: is it wrong? If it is right, could you explain it?
AI: I don't want to spend too much time deciphering this book's idiosyncratic and obscure system of abbreviations. But I gather that the idea is:
We have already shown $Q$ in step 3.
We can hypothesize $P$ in step 5 and then claim that $P\to Q$ because $Q$ has already been shown.
This is sometimes called the rule of reiteration. (I suppose this is why the author chose the notation r.) Symbolically, we can write it as $$A \to (B \to A)$$ If we already know $A$, we can conclude that $B$ implies $A$, for any $B$. In your derivation, because we already know $Q$, we can conclude $P\to Q$.
This is indeed counterintuitive. It's one of the notorious paradoxes of material implication. The confusion arises because the formal logical meaning of $A\to B$ is so different from the conventional notion of "ifβ¦ thenβ¦β.
In formal logic, $A\to B$ is a very weak claim. It does not say that $A$ causes $B$, or that $A$ and $B$ are related in any way. All it says is that whenever $A$ is true, $B$ is also true.
But if we interpret $\to$ in this way, $A\to (B\to A)$ is always true. It says is that whenever $A$ is true, $B\to A$ is also true. And this is correct! Because if $A$ is true then, whenever $B$ is true, $A$ is also true.
Perhaps a more straightforward way to prove the theorem would be: Assume $\sim T$. From premise 2 conclude $\sim S$ by modus tollens. From $\sim S$ and premise 1, conclude $\sim(P \to Q)$ by modus tollens. $\sim (P\to Q)$ means $P\land \sim Q$ by definition, so extract $\sim Q$. From premise 3, conclude $Q$ by modus ponens, a contradiction, and the original assumption $\sim T$ is false.
But I make no guarantees that all these steps are valid in this author's unpleasant logical system. My advice would be to pick up a secondary text that explains how to do proofs via analytic tableaux, and learn logic from that. Once you understand it, writing the proofs in a more natural way, and then translating them into your book's weird formalism, will probably be easier than trying to construct the proofs directly in the weird formalism. |
H: Expressing a vector space over a finite field as a finite union of proper subspaces.
Am trying to solve the following exercise which appeared in an abstract algebra textbook :
Assume that $V$ is an $n$-dimensional vector space over a finite field ${\bf F}_{q}$ which consists of $q$ elements. If $V$ is a finite set theoretic union of $m$ proper linear subspaces ${W_1},\cdots {W_m},$ then it must be the case that $$m\geq{\frac{{q^n}-1}{q-1}}.$$ Prove that there exist ${\frac{{q^n}-1}{q-1}}$ subspaces whose union is $V$.
It is clear that each of the proper subspaces must consist of $q^k$ elements where $k<n$. Not sure how to approach this question!
AI: Despite the comments, I believe this is true. There are $q^n-1$ nonzero elements $v$, and each such $v$ is in the one-dimensional subspace $W_v=\{fv|f\in F_q\}.$ Each of these subspaces has $q-1$ nonzero elements, so there are $\frac{q^n-1}{q-1}$ distinct $W_v$ whose union is $V$. |
H: Is $f(x)=\frac{x^{2}-1}{x-1}$ continuous at $x=1$?
Given $f(x)=\frac{x^{2}-1}{x-1}$. The function is said to be discontinuous at $x=1$ but since we can simplify it and rewrite $f(x)=x+1$, this removes the discontinuity. So is the function continuous or discontinuous at $x=1$
How do the two forms of $f(x)$ differ as both expressions are equal to each other. What stops us from simplifying the earlier expression and saying the function is continuous ?There was a similar question here but it didn't address my latter point.
AI: $\lim_{x\rightarrow 1} f(x) = 2$, so the singularity is removable.
Let $$g(x) = \left\{ \begin{array}{cc} f(x), & x \ne 1\\ 2, & x=1 \end{array} \right.$$ The function $g(x)$ is continuous.
So, yes, $f(x)$ is discontinous but this discontinuity is easily repaired. If you were to graph $y=f(x)$, it would be the straight line $y=x+1$ with the point $(1,2)$ removed. |
H: Moore-Penrose Pseudoinverse: Because $(AA^+)^T = AA^+$, we have that $A^T(Ax-b) = ((AA^+)A)^Tb - A^Tb = 0$
I am studying this answer by user "Etienne dM". They claim that, because $(AA^+)^T = AA^+$, we have that $A^T(Ax-b) = ((AA^+)A)^Tb - A^Tb = 0$. However, I do not understand how $A^T(Ax-b) = ((AA^+)A)^Tb - A^Tb = 0$ is possible. In particular, it seems like an extra $AA^+$ term comes out of nowhere. I would greatly appreciate it if people would please take the time to explain what's going on here.
AI: The whole point of the MP inverse is that $AA^+A=A$ (and a few other properties). So
$$
A^TAx=A^TAA^+b=A^T(AA^+)^Tb=(AA^+A)^Tb=Ab.
$$ |
H: Solving Laplace Equation using Separation of Variables inside an annulus
Question: find the bounded function $u(x,y)$ that satisfies the following conditions
$$\nabla^2u(x,y)=0, \qquad 4<x^2+y^2<16$$
$$u(x,y)=x, \qquad x^2+y^2=4$$
$$u(x,y)=y, \qquad x^2+y^2=16$$
First of all I transform this problem to polar coordinates, obtaining
$$u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^2}u_{\theta\theta}=0, \qquad 3<r<4$$
$$u(3,\theta)=3\cos(\theta), \qquad -\pi\leq\theta\leq\pi$$
$$u(4,\theta)=4\sin(\theta), \qquad -\pi\leq\theta\leq\pi$$
I'll jump to the expression for the general solution to a Laplace Equation in polar coordinates, since where I'm getting stuck at is when evaluating the boundary conditions. A general expression would be the following:
$$u(r,\theta)=A+B\ln(r)+\sum_{n=1}^\infty (C_n\,r^n+D_n\,r^{-n})(E_n\sin n\theta+F_n\cos n\theta)$$
Since the boundary conditions are just a sine and a cosine, I thought it would be almost trivial to find expressions for the coefficients. But I've been stuck for quite some time now, how would one solve this system of equations to find said coefficients?
AI: There is no reason why $\sin n\theta$ and $\cos n\theta$ has to share the same function of $r$, which is probably why you are having troubles
Quicker way: We know the solution has to be of the form
$$
u(r,\theta)=f(r)\sin\theta+g(r)\cos\theta.
$$
So substituting, we want to solve
$$
f''(r)+\frac1rf'(r)-f(r)=0,\quad f(3)=0,\quad f(4)=4
$$
and similarly $g$. That gives you a solution in terms of the modified Bessel functions $I_0,K_0$. |
H: Projection matrix.
Suppose $P=X(X'X)^{-1}X'$ and $X$ can be descomposed as $X= [X_1 X_2]$, where $X$ is a matrix. Then is true that $PX_1=X_1$.
My proof:
$X'X=[X_1^2+X_2^2]$ Then $[X_1^2+X_2^2]^{-1}=1/(X_1^2+X_2^2)$. Thus $(X_1^2+X_2^2)X_1/(X_1^2+X_2^2)=X_1$
Is this proof correct?
AI: ok, So $X$ is tall skinny matrix, typically with many many more rows than columns.
Suppose, for example that $X$ is a $100\times5$ matrix. Then $X^\top X$ is a $5\times5$ matrix. If $X_1$ is a $100\times3$ matrix and $X_2$ is $100\times2,$ then what is meant by $X_1^2+X_2^2,$ let alone by its reciprocal?
If $x$ is any member of the column space of $X$, then $Px=x.$
This is proved as follows: $x = Xu$ for some suitable column vector $u$.
Then $Px = \Big(X(X^\top X)^{-1} X^\top\Big) (Xu) = X(X^\top X)^{-1}(X^\top X) u = Xu = x.$
(Similarly if $x$ is orthogonal to the column space of $X$, then $Px=0.$ The proof of that is much simpler.)
Now observe that the columns of $X_1$ are in the column space of $X.$ |
H: Prove $\mathcal{R}:=\left\{\left(a,b\right)\mid a\le b\right\}$ is not symmetric
Given a homogeneous binary relation $\mathcal{R}$ over a set $A$,and is defied as:
$$\mathcal{R}:=\left\{\left(a,b\right)\mid a\le b\right\}$$
Prove $\mathcal R$ is not symmetric.
I don't understand why $\mathcal R$ is not symmetric,indeed it's symmetric if :
$$\forall a,b \in A:a\le b \implies b\le a $$
Or in other words:
$$\forall a,b \in A$$ one of these are true:
$$a < b \implies b< a $$
$$a < b \implies b= a $$
$$a = b \implies b< a $$
$$a = b \implies b= a $$
As it's seen,the last one is true,and that's why I think the relation is symmetric.
AI: $$\forall a,b \in A:a\le b \implies b\le a $$
Or in other words:
$$\forall a,b \in A$$ one of these are true:
$$a < b \implies b< a $$
$$a < b \implies b= a $$
$$a = b \implies b< a $$
$$a = b \implies b= a $$
You are wrong.
Your second statement is not equivalent to the first. The first statemtent says that for every pair $a,b\in $A, the statement $a\leq b\implies b\leq a$ is true. The statement $a\leq b\implies b\leq a$ is not the same as
$$(a < b \implies b< a) \lor (a < b \implies b= a)\lor (a = b \implies b< a) \lor(a = b \implies b= a) $$
and you can quite clearly show this by taking $a=1$, $b=2$.
More generally, your argument is based on assuming that, for any collection of logical statements $A, B, C$, the statement $$(A\lor C)\implies(B\lor C)$$ is equivalent to the statement
$$(A\implies B)\lor(A\implies C)\lor(C\implies B)\lor (C\implies C)$$
but this is not the case.
(for clarity, in your argument, $A$ was the statement $a<b$, $B$ was the statement $b<a$, and $C$ was the statement $a=b$)
In fact, if you wanted to write out the logical implication (not that I recommend it), then what you would need to consider is that $$(P\lor Q)\implies (R)$$ is equivalent to:
$$\begin{align}(P\lor Q)\implies R&\sim \neg (P\lor Q)\lor R\\&\sim (\neg P \land \neg Q)\lor R \equiv (\neg P\lor R)\land(\neg q \lor R)\\&\sim (P\implies R)\color{red}{\land}(Q\implies R)\end{align}$$
Note the highlighted $\land$ appearing in the statement which is nowhere in your argument. |
H: Show that $X \in L^p$
Let $X$ and $Y$ be independent variables and $p \geq 1$. Show that $X + Y \in L^p$ $\implies$ $X, Y \in L^p$. I tried using the inequalities
$$|X + Y|^p \leq 2^{p}(|X|^p + |Y|^p),$$
and
$$|X|^p \leq 2^{p}(|X + Y|^p + |Y|^p),$$
but I didn't get anything useful.
AI: This is proved using Fubini's Theorem. We have $\int(E|x+Y|^{p})dF_X(x) <\infty$ and this implies that $E|x+Y|^{p} <\infty$ for at least one $x$. ( In fact for almost all $x$ w.r.t. $F_X$!). Now you can see that $E|Y|^{p} <\infty$? from the inequality $|Y|^{p} \leq 2^{p} (|x+Y|^{p}+|x|^{p})$. |
H: Don't understand the difference between $P(a \le X \le b)$ and $P(X\in B)$
By def, $P(a \le X \le b)=\int_a^b f_X(x) dx$ and $P(X\in B)=\int_B f_X(x) dx$.
I think I understand $P(a \le X \le b)$ well enough: it tells you that if $X$ is the random variable that maps the height of someone to $x$, then $P(a \le X \le b)$ asks what's the probability of someone's height, or $x$, being between $a$ and $b$.
$P(X\in B)$ is a lot more confusing. My textbook, Tsitsiklis' Intro to Probability and Statistics, said this is 'the PDF of $X$ for every subset $B$ of the real line'.
I have no idea what that is supposed to mean. Is it just the probability of $x$ being a real number?
The integral also doesn't make much sense to me. I can understand what $\int_a^b$ means - the area under the function between $x=a$ and $x=b$. But $\int_B$ only gives the lower bound...I am sure it doesn't mean the area from $B$ to infinite...so what does it mean?
AI: $B$ in the statement $P(X\in B)$ is a subset of $\mathbb R$.
In fact, you can rewrite $P(a\leq X\leq B)$ as $$P(X\in [a, b])$$ in which case you can use the second definition to get, $$P(X\in [a, b]) = \int_{[a,b]}f_X(x)dx$$
and, "luckily", measure theory also tells you that
$$\int_{[a,b]}f_X(x)dx = \int_a^bf_X(x)dx$$
so the two definitions are compatible. |
H: Compute $\int xy dx +(x+y)dy$ over the curve $Ξ$, $Ξ$ is the arc $AB$ in the 1st quadrant of the unit circle $x^2+y^2=1$ from $A(1,0)$ to $B(0,1)$.
Compute $\int xydx+(x+y)dy$ over the curve $Ξ$, where $Ξ$ is the arc $AB$ in the first quadrant of the unit circle $x^2+y^2=1$ from $A(1,0)$ to $B(0,1)$.
I solved this problem with the help of Green's theorem. My result is $\frac{\pi}{4}-\frac{1}{3}$. But the book I'm following, shows, the result is $\frac{\pi}{4}+\frac{1}{6}$. Since I'm a learner, I don't understand where have I done the mistake.
Can anyone please give the solution?
AI: Your book is correct. First of all, your application of Green's theorem here is incorrect, as indicated in the comment. You can't apply Green's theorem for a curve that's not simple closed. See https://en.wikipedia.org/wiki/Green%27s_theorem. So the correct answer is through the line integral evaluation as follows
$$\int_\Gamma xydx+(x+y)dy=\int_{AB} xydx+(x+y)dy$$ $$=\int_{y=0}^1\bigg[y\sqrt{1-y^2}d\bigg(\sqrt{1-y^2}\bigg)+\bigg(\sqrt{1-y^2}+y\bigg)dy\bigg]$$ $$=-\int_0^1y^2dy+\int_0^1\sqrt{1-y^2}dy+\int_0^1ydy=-\frac{1}{3}+\frac{\pi}{4}+\frac{1}{2}=\frac{\pi}{4}+\frac{1}{6}$$ |
H: Show that $(A\wedge Bβ) \vee B β‘ A \vee B$.
How do I prove this using laws of logical equivalency?
$$
(A\wedge Bβ) \vee B β‘ A \vee B
$$
What I tried. $(A \vee B') \wedge (A \vee A)$ I started by applying the distributive laws. Is this the right path?
AI: $%
\require{begingroup}
\begingroup
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\rightarrow}
\newcommand{\when}{\leftarrow}
\newcommand{\fa}[2]{\forall #1 \left( #2 \right) }
\newcommand{\ex}[2]{\exists #1 \left( #2 \right) }
\newcommand{\exun}[2]{\exists ! #1 \left( #2 \right) }
\newcommand{\F}{\mathcal F}
\newcommand{\equiv}{\leftrightarrow}
%$
$$\calc
(A\land B') \lor B
\op\equiv\hint{distributive law}
(A\lor B)\land(B'\lor B)
\op\equiv\hint{tautology}
(A\lor B)\land \text{tautology}
\op\equiv\hint{tautology law}
A\lor B
\endcalc$$ |
H: Result of $\int_0^1 Tr(e^{-a(1-t)T}A^{β}e^{-atT}B)) dt , a>0$
$T$ is a positive selfadjoint densely defined on $H$ (Hilbert), only has point spectrum $\sigma (T)=\{\lambda_k \}_{k\in\mathbb{N}}.\lambda_k < \lambda_{k+1}\forall k$ and $\{u_k\}_{k\in \mathbb{N}}$ is the respective orthonormal base of eigenvectors of $T$.
$Tr(A)=\sum_n <u_n,Au_n>$ for $A$ a positive operator on $H$. If $Tr(A)<\infty$ then $A$ is trace class.
Assume $e^{-aT}$ is trace class where $a>0$.
$A,B$ are positive bounded operators on $H$.
I wanna show $\int_0^1 Tr(e^{-a(1-t)T}A^{β}e^{-atT}B) dt=\sum_{m,n;m\neq n}<u_m,A^{β}u_n><u_m,Bu_n>\frac{e^{-a\lambda_m}-e^{-a\lambda_n}}{-a(\lambda_m-\lambda_n)}+\sum_n<u_n,A^{β}u_n><u_n,Bu_n>e^{-a\lambda_n}$
I know that the trace is independent of the base, its ciclicity, but I can't get a double summatory nor the product of inner products.
I got $...=\int_0^1\sum_n e^{-a(1-t)\lambda_n}<Au_n,e^{-atT}Bu_n>dt$
Any help is regarded.
AI: Observe
\begin{align}
\operatorname{Tr}(e^{-a(1-t)T}A^\ast e^{-atT}B) =&\ \sum_n \langle Ae^{-a(1-t)T}u_n, e^{-atT}Bu_n\rangle\\
=&\ \sum_{n, m}\langle Ae^{-a(1-t)T}u_n, u_m\rangle \langle u_m, e^{-atT}Bu_n\rangle\\
=&\ \sum_{n, m}e^{-a(1-t)\lambda_n}\langle Au_n, u_m\rangle \langle u_m, e^{-atT}Bu_n\rangle.
\end{align}
Summing up the diagonal terms yields
\begin{align}
\sum_n e^{-a(1-t)\lambda_n}\langle Au_n, u_n\rangle \langle u_n, e^{-atT}Bu_n\rangle =&\ \sum_n e^{-a(1-t)\lambda_n}\langle Au_n, u_n\rangle \langle e^{-atT}u_n, Bu_n\rangle\\
=&\ \sum_n e^{-a\lambda_n}\langle u_n, A^\ast u_n\rangle \langle u_n, Bu_n\rangle.
\end{align}
For the off-diagonal terms, we have
\begin{align}
\sum_{n\ne m}e^{-a(1-t)\lambda_n}\langle Au_n, u_m\rangle \langle u_m, e^{-atT}Bu_n\rangle =&\ \sum_{n\ne m}e^{-a(1-t)\lambda_n-at\lambda_m}\langle u_n, A^\ast u_m\rangle \langle u_m, Bu_n\rangle.
\end{align}
I think the rest should be self explanatory. |
H: Probability of being sick after three positive tests
Question: The probability of getting sick is 5%. The probability of correct detection after getting sick is 80%. The probability of detection error in healthy people is 1%. Three consecutive tests are all getting sick. What is the actual probability of getting sick?
So this problem is confusing me because it seems to be a Bayes' law rule (as follows), but I don't know if the question itself is a typo. Why are they asking us of the "actual" probability of getting sick, when they give us the probability of getting sick (5%)? Here's what I'm thinking:
$$\Pr(\text{getting sick | 3 test results are positive}) = \tfrac{\Pr(\text{getting sick AND 3 are sick})}{\Pr(\text{3 test results are positive})}$$
$$\Pr(\text{3 test results are "sick"}) = \Pr(\text{3 are actually sick; 3 tests are correct}) + \Pr(\text{2 are actually sick, 1 isn't sick; 2 tests are correct, 1 is incorrect}) + \Pr(\text{1 is actually sick, 2 aren't sick; 1 test is correct, 2 are incorrect}) + \Pr(\text{0 are actually sick, 3 aren't sick; 3 tests are incorrect})$$
So I know how to find this probability, but I'm still confused as to how to find the "actual" probability. Is the actual probability no longer 5%, now that we are given this info regarding the 3 people?
AI: I think that the question is about the same person. What the question asks is: without prior knowledge, the probability of someone being sick is 5%. A person is chosen at random and three tests made to him are positive, what is the probability of him being sick?
(All test results are independent given the person is sick or healthy)
Answer:
$$\Pr(sick\vert 3 tests + )=\tfrac{\Pr(3 tests + \vert sick ) \Pr (sick)}{\Pr (3 tests +)}=\tfrac{0.8^3 \cdot 0.05}{0.8^3 \cdot 0.05 + 0.01^3 \cdot 0.95}=0.99996$$ |
H: MLE of $(\theta_1,\theta_2)$ in a piecewise PDF
I am trying to find the MLE of $\theta=(\theta_1,\theta_2)$ in a random sample $\{X\}_{i=1}^n$ with the following pdf
$$f(x\mid\theta)= \begin{cases}
(\theta_1+\theta_2)^{-1}\exp\left(\frac{-x}{\theta_1}\right) &, x>0\\
(\theta_1+\theta_2)^{-1}\exp\left(\frac{x}{\theta_2}\right) &, x\le0\\
\end{cases}
$$
If I let $\bar{X}_1$ be the average of the $n_1$ values where $X_1>0$ and $\bar{X}_2$ the average of $n_2$ values where $X_i\le 0$ and $n_1+n_2=n$
Then the likelihood function is:
$$L(\theta\mid X)=\left(\frac 1 {\theta_1+\theta_2}\right)^n\exp\left(\frac{-n_1\bar{X}_1}{\theta_1}+\frac{n_2\bar{X}_2}{\theta_2}\right)$$
but I am having trouble maximizing this function.
AI: Working with the log-likelihood is easier. We write $$\ell(\theta_1, \theta_2 \mid n_1, n_2, \bar x_1, \bar x_2) = -(n_1 + n_2) \log (\theta_1 + \theta_2) - \frac{n_1 \bar x_1}{\theta_1} + \frac{n_2 \bar x_2}{\theta_2}.$$ Note that this function is subject to the restrictions $$\theta_1, \theta_2 > 0, \quad n_1, n_2 \in \mathbb Z^+, \quad \bar x_1 > 0, \quad \bar x_2 \le 0.$$ Taking the partial derivatives with respect to $\theta_1$, $\theta_2$ and equating these to $0$ yield respectively $$\frac{\partial \ell}{\partial \theta_1} = -\frac{n_1 + n_2}{\theta_1 + \theta_2} + \frac{n_1 \bar x_1}{\theta_1^2} = 0, \\ \frac{\partial \ell}{\partial \theta_2} = -\frac{n_1 + n_2}{\theta_1 + \theta_2} - \frac{n_2 \bar x_2}{\theta_2^2} = 0.$$ I leave it as an exercise for you to solve this simultaneous system (it is not difficult) and show that the unique critical point is $$(\theta_1, \theta_2) = \left(\frac{n_1 \bar x_1 + \sqrt{-n_1 \bar x_1 n_2 \bar x_2}}{n_1 + n_2}, \frac{-n_2 x_2 + \sqrt{-n_1 \bar x_1 n_2 \bar x_2}}{n_1 + n_2}\right),$$ which would suggest that it is better to use the sufficient statistics $$T_1 = \sum_{i=1}^n X_i \mathbb 1(X_i > 0), \quad T_2 = - \sum_{i=1}^n X_i \mathbb 1(X_i \le 0);$$ that is to say, $T_1$ is the sample total of positive observations, and $T_2$ is the negative of the sample total of negative or zero observations (thus is negative or zero). Then we may rewrite the joint MLE as $$(\hat \theta_1, \hat \theta_2) = \left(\frac{T_1 + \sqrt{T_1 T_2}}{n}, \frac{T_2 + \sqrt{T_1 T_2}}{n} \right),$$ which makes the symmetry apparent and does away with the auxiliary variables $n_1, n_2$. |
H: Name and proof of the general form of ${a_1}{b_1} + {a_2}{b_2} = \left( {{a_1} - {a_2}} \right){b_1} + {a_2}\left( {{b_1} + {b_2}} \right)$?
I was running into a strange identity that is
Given ${x_1},{x_1},...,{x_n}$ and ${y_1},{y_1},...,{y_n}$ are all real number.
Denote ${c_k} = {y_1} + {y_2} + {y_3} + ... + {y_k}$ where $1 \le k \le n$
Proof that
${x_1}{y_1} + {x_2}{y_2} + ...{x_n}{y_n} = \left( {{x_1} - {x_2}} \right){c_1} + \left( {{x_2} - {x_3}} \right){c_2} + ... + \left( {{x_{n - 1}} - {x_n}} \right){c_n} + {x_n}{c_n}$
By plug in some number, I was able to come up with some case but I am not sure how to proof this identity for the general case
For $n=2$, we have:
${a_1}{b_1} + {a_2}{b_2} = \left( {{a_1} - {a_2}} \right){b_1} + {a_2}\left( {{b_1} + {b_2}} \right)$
For $n=3$, we have:
${a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3} = \left( {{a_1} - {a_2}} \right){b_1} + \left( {{a_2} - {a_3}} \right)\left( {{b_1} + {b_2}} \right) + {a_3}\left( {{b_1} + {b_2} + {b_3}} \right)$
My country call this identity as Abel's expansion but I was unable to determine whether if this naming is correct or not.
Edit: I have finally found the name, this process is called Abel transformation
https://en.wikipedia.org/wiki/Summation_by_parts
AI: We could prove this by induction.
Givens: $(x_1,x_2,...,x_n)$, $(y_1,y_2,...,y_n)$, and $c_k=y_1+y_2+...+y_k$ for $1\leq k\leq n$. Show that $$\sum_{k=1}^n x_ky_k=c_{n}x_{n+1}+\sum_{k=1}^{n} (x_k-x_{k+1})c_k $$ Base case $n=1$: $$\begin{align*}x_1y_1 &= c_1x_2 + (x_1-x_2)c_1\\ &= (y_1)(x_2)+(x_1-x_2)(y_1)\\&=x_2y_1+x_1y_1-x_2y_1\\&=x_1y_1\end{align*}$$
Induction Hypothesis: Assume $$\sum_{k=1}^{j} x_ky_k=c_{j}x_{j+1}+\sum_{k=1}^{j} (x_k-x_{k+1})c_k $$ is true for some integer $j$. We need to show that it is true for $j+1$.
$$ \begin{align*}\sum_{k=1}^{j+1} x_ky_k &=x_{j+1}y_{j+1}+\sum_{k=1}^{j} x_ky_k\\&= x_{j+1}y_{j+1}+c_jx_{j+1}+\sum_{k=1}^{j}(x_k-x_{k+1})c_k\\&= x_{j+1}(c_j+y_{j+1})+ \sum_{k=1}^{j}(x_k-x_{k+1})c_k \\&= x_{j+1}c_{j+1}+\sum_{k=1}^{j}(x_k-x_{k+1})c_k\\&= (x_{j+2} +x_{j+1}-x_{j+2})c_{j+1} +\sum_{k=1}^{j}(x_k-x_{k+1})c_k\\&= x_{j+2}c_{j+1} +(x_{j+1}-x_{j+2})c_{j+1} + \sum_{k=1}^{j}(x_k-x_{k+1})c_k\\&= x_{j+2}c_{j+1} + \sum_{k=1}^{j+1}(x_k-x_{k+1})c_k\end{align*}$$
By induction, we are done. As for the name, I believe it was given in the comments. |
H: Better methods to approximate $2^{2\over 3}$
Recently while solving a problem on thermodynamics I ended up with $2^{2\over 3}$ .
Now the problem was on a test where no calculators were allowed and answer was required upto $2$ decimal digits.
I then resorted to binomial theorem for help (for $x\lt 1$) $$\left. \begin{array} { l } { ( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) x ^ { 2 } } { 2 ! } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } + \ldots \ldots + \frac { n ( n - 1 ) \ldots \ldots ( n - r + 1 ) } { r ! } x ^ { r } \ldots } \\ \end{array} \right.\text{upto}\, \, \infty$$
So the original problem can be written as :
$$2^{2\over 3}=4^{1\over 3}=(8-4)^{1\over 3}=2\left(1-\frac{1}{2}\right)^{1\over 3}$$
Now after evaluation first $3$ terms I ended up with $\left(2-\frac{1}{3}-\frac{1}{18}\right) \approx 1.61$ but the correct answer was $1.59$.
Also the average time you got per question was around $2$ minutes and I had already used more than half of it so I did not think of calculating more terms.
Now I am looking for a method which can help me evaluate $2^{2\over 3}$ faster and more precisely.
Can anyone please help me with this ?
AI: If you are around powers of 2 a lot (as computer scientists are, for example), then you can recognize $4.096=2^{12}/10^3=(2^4/10)^3=1.6^3$, and then $4^{1/3} = 1.6(1-\frac{0.096}{4.096})^{1/3}$ is much better approximated by its power series. Indeed, the approximation $1.6(1-\frac13\frac{0.096}{4.096}) \approx 1.6(1-\frac13\frac{0.1}4) = 1.6 - \frac1{75}$ is already within $0.05\%$ of the true answer $\approx 1.5874$. |
H: Concrete Mathematics: Chapter 1: Generalised Josephus Recurrence: Understanding Radix 3 to 10 digit-by-digit replacement
Summary
When converting $(201)_3$, specifically, converting the $2$, I am not entirely sure how they select $5$ (from $f(2) = 5$) and not $8$ (from $f(3n+2) =
10f(n)+8$) when doing the radix replacement operation for the recurrence following equation 1.18. My thinking is that
it is due to the start of the replacement being done with $\alpha_j$ which, in this case, is $5$. See below for more
context.
Details
The following radix independent, generalised recurrence is given as follows (1.17 in book)
$$
\begin{align}
f(j) &= \alpha_j,\ \ \text{for}\ 1 \leq j < d; \\
f(dn + j) &= cf(n) + \beta_j,\ \ \text{for}\ 0 \leq j < d\ \text{and}\ \ n \geq 1,
\end{align}
$$
The above recurrence can then start with numbers in radix $d$ and produce values in radix $c$. So it has the radix
changing solution (1.18 in book)
$$
f((b_m b_{m-1} ... b_1 b_0)_d) = (\alpha \beta_m\beta_{b_{m-1}} \beta_{b_{m-2}} ... \beta_{b_1} \beta_{b_0})_c
$$
Then, as the book says, by some stroke of luck we're given the recurrence
$$
\begin{align}
f(1) &= 34, \\
f(2) &= 5, \\
f(3n) &= 10f(n) + 76,\ \ \text{for}\ \ n \ge 1, \\
f(3n+1) &= 10f(n) - 2,\ \ \text{for}\ \ n \ge 1, \\
f(3n+2) &= 10f(n) + 8,\ \ \text{for}\ \ n \ge 1, \\
\end{align}
$$
Then the book proposes computing $f(19)$ where $d = 3$ and $c = 10$ (from 1.17). So $19 = (201)_3$ and the
radix-changing solution has us doing a digit-by-digit replacement of 201 so
$2_3$ becomes $5_{10}$ - this is the bit I'm unsure about. Why is it not $8$? I think because it is the first digit to
replace and thus applies to $\alpha_j$. If it where in the middle of the number (e.g. $120$) it may well have been $8$?
$0_3$ becomes $76$
$1_3$ becomes $-2$
So tying it all together as radix $10$ we get
$$
f(19) = ((201)_3) = (5 76 -2)_{10} = 500 + 760 -2 = 1258
$$
Being verbose: the final addition is because $500$ is in the radix $10$ "hundreds" column (multiply by $100$), $76$ in the
"tens" column (multiply by $10$), and $-2$
in the "ones" column.
AI: Your suspicion is correct: the function $f$ takes $$(b_mb_{m-1}\ldots b_1b_0)_d$$ to $$(\alpha_{b_m}\beta_{b_{m-1}}\beta_{b_{m-2}}\ldots\beta_{b_1}\beta_{b_0})_c\,.$$ (Note that you didnβt copy (1.18) quite correctly.) This is a digit-by-digit replacement: $b_m$ is replaced by $\alpha_{b_m}$, and $b_k$ is replaced by $\beta_{b_k}$ for $0\le k\le m-1$. In the example you have $c=10$, $d=3$, and the values
$$\left\{\begin{align*}
\alpha_1&=34\\
\alpha_2&=5\\
\beta_0&=76\\
\beta_1&=-2\\
\beta_2&=8\,.
\end{align*}\right.$$
If your input is $19=(201)_3$, then $m=2$, $b_2=2$, $b_1=0$, and $b_0=1$, so your output from $f$ is
$$\begin{align*}
(\alpha_2\beta_0\beta_1)_{10}&=(5\;76\;-2)_{10}\\
&=5\cdot 10^2+76\cdot 10^1-2\cdot 10^0\\
&=500+760-2\\
&=1258\,.
\end{align*}$$
Had the input been $46=(1201)_3$, the output would have been
$$\begin{align*}
(\alpha_1\beta_2\beta_0\beta_1)_{10}&=(34\;8\;76\;-2)_{10}\\
&=34\cdot 10^3+8\cdot 10^2+76\cdot 10^1-2\cdot 10^0\\
&=34000+800+760-2\\
&=35558\,.
\end{align*}$$ |
H: Why the improper integral $\int_{e}^{\infty} \frac{d x}{x(\log x)^{n}}$ converges iff $ - n + 1 < 0$?
I was trying to answer this problem in preparation for GRE exam:
And the answer was given below:
But I do not understand why the improper integral converges iff $ - n + 1 < 0,$ could anyone explain this for me, please?
AI: In $-\frac{1}{n+1}(\log x)^{-n+1}$ we're essentially dealing with a power of $\log(x)$ (the term before it is a constant), which is (as said) a function that increases without bound as $x \to \infty$. Recall that $f(x)^{-a} = \frac{1}{f(x)^a}$ for all $a>0$. So the only way a power of log can tend to $0$ (as it should for convergence) is that the exponent is negative so that it appears in the denominator of a fraction, and in general $\frac{1}{y} \to 0$ as $y \to \infty$.
So we need $-n+1 < 0$ which is equivalent to $n>1$. |
H: Is there a basis $\beta$ for $V$ such that $\langle \mathbf{v}, \mathbf{w}\rangle=\langle [\mathbf{v}]_{\beta}, [\mathbf{w}]_{\beta}\rangle$.
Given an inner product product $\langle \cdot, \cdot\rangle$ on a finite dimensional vector space $V$ over $F$, $F=\mathbb{R}$ or $F=\mathbb{C}$.
My Questions
Is there a basis $\beta$ for $V$ such that $\langle \mathbf{v}, \mathbf{w}\rangle=\langle [\mathbf{v}]_{\beta}, [\mathbf{w}]_{\beta}\rangle$ for every $\mathbf{v}, \mathbf{w}\in V$, where the second inner product is the standard inner product on $F^n$.
Which books I can find this theorem? (I believe it is true.)
AI: You need $\beta$ to be an orthonormal basis. You can always obtain such basis by using the Gram-Schmidt process. |
H: Proving the equation of an ellipse
If $(c, 0)$ and $(-c, 0)$ are the foci of an ellipse, and the sum of the distance of any point on the ellipse with the foci is $2a$ I am asked to prove thath the equation of the ellipse is:
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
where $b^2=a^2-c^2$.
I tried to first write the definition of the ellipse in mathematical terms:
$$
\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a
$$
But when I tried to proceed from here I didn't know where should I go.
AI: A slightly faster way
Hint:
Move $\sqrt{(x+c)^2+y^2}$ to other side of equality to give $\sqrt{(x-c)^2+y^2}=2a-\sqrt{(x+c)^2+y^2}$
Square both sides.
The calculation should simplify a lot as a lot of terms cancel after first step ,compared to simply squaring original equation.
Then move terms within roots to another side and square again.
The problem should be solved. |
H: When is it necessary to use proof by induction?
Suppose $T \in \mathcal{L} (V)$ and $U$ is a subspace of $V$ invariant
under $T$. Prove that $U$ is invariant under $p(T)$ for every
polynomial $p \in \mathcal{P} (\mathbb{F})$.
$\underline{\textrm{My attempt at the solution:}}$
Suppose $p (z) \in \mathcal{P}(\mathbb{F})$. Write $p (z) = \sum_{j=0}^{m} a_{j} z^{j}$ where $z \in \mathbb{F}$.
By definition, $p (T) = \sum_{j=0}^{m} a_{j} T^{j}$.
Then, for any $u \in U$, we have
\begin{align*}
p (T) (u) = \left( \sum_{j=0}^{m} a_{j} T^{j} \right) (u) = \sum_{j=0}^{m} a_{j} (T^{j} u)
\end{align*}
$\underline{Claim}:$ $T^{n} u \in U$ for any $n \in \mathbb{N}^{+}$.
Prove by induction:
Base case: $T^{0} u = I u = u \in U$.
Induction hypothesis: $T^{n} u \in U$.
Induction proof: Note $T^{n+1} u = T (T^{n} u)$. Because $U$ is invariant under $T$, we have $T(T^{n} u) \in U$.
Therefore, $T^{n+1} u \in U$ for $n \in \mathbb{N}$.
Hence the result.
By the inductive result, we have $T^{j} u \in U$ for all $j$. Since $U$ is a subspace of $V$, it is closed under addition and scalar multiplication. Therefore, $\sum_{j=0}^{m} a_{j} (T^{j} u) \in U$. By definition, $U$ is invariant under $p (T)$ for any $p \in \mathcal{P} (\mathbb{F})$.
$\underline{\textrm{Question:}}$
Would it be necessary to also prove by induction that this result holds for $p \in \mathcal{P} (\mathbb{F})$ of any degree? I thought it wouldn't be necessary since the choice of $m$ was arbitrary.
AI: No, you use a summation formula (these are defined by recursion, but you can assume those are well-defined, because you have to start somewhere, right?) and so the argument is valid and general as it stands.
There is a background (previously done?) argument hidden in that summation, namely that a subspace is closed under all finite sums, so that indeed from $\alpha_i \in U$ for $i=1,\ldots,m$ we can conclude $\sum_{i=1}^m \alpha_i \in U$. This is a completely standard induction argument based on the recursive definition of the summation $\sum_{i=1}^m \alpha_i$. I would assume that as a fact and not reprove it every time. The fact that $T^m(u) \in U$ for all natural $m$ and $u \in U$ is a special case of the full polynomial result and so IMO it merits a place as a minor lemma here. |
H: How to find quantity of subsets generated from subsets of a set?
What is the maximum number of subsets that can be formed from $n$ subsets of a fixed set using intersection, join, and complement operations? Answer is $2^{2^n}$, but can you explain why?
AI: Suppose the fixed set is $A$ with subsets $B_1,B_2,\ldots,B_n$
Then you can find minimal subsets which are of the form $C_k=B_1^c \cap B_2 \cap \cdots \cap B_n^c$ or similar, by intersecting each original subset or its complement. The $C_k$ are mutually exclusive, their union is $A$ and if there are $m$ distinct minimal subsets then $m \le 2^n$
Any obtainable subset of $A$ is then of the form $D_i=C_1 \cup C_4 \cup \cdots \cup C_{j}$ or similar, by taking the union of a number of $C_k$. The $D_i$ form a sigma-algebra and if there are $s$ distinct obtainable subsets then $s \le 2^m$. So $s \le 2^{2^n}$ |
H: Combinations with restrictions in a set
Q1. Alex and Erin are two of the eight students trying out for a certain five-person chess team. If Alex and Erin must make it to the chess team, in how many different ways can the chess team be populated?
To solve for the above we can assume that two players Erin and Alex are already part of the chess team leaving us with 4 players to choose from 6 players. Hence 6C3
Q2. In a treasure chest, there are five different flawed diamonds and three different perfect diamonds. If a pirate selects four diamonds from the chest, in how many ways can four diamonds be selected, if exactly two of them must be perfect diamonds?
In Q2 if I try to apply the same logic as of Q1 i.e. 2 diamonds must be perfect(I can assume them to be already selected) which would leave me to select 2 diamonds from five flawed diamonds which would be 5C2, but that's not the solution. What am I missing in here?
AI: In both questions you choose $a$ out of $A$ objects from one group and $b$ out of $B$ objects from the second group. In both cases, the answer is ${A \choose a}{B \choose b}$.
In Q1 $A=a=2$ so you don't see any term in the answer. In $Q2$ you see since $A=3$ and $a=2$. |
H: Tangent to an ellipse
I want to find the tangent on point $(x_0, y_0)$ to an ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
We assume $y_0$ is positive. We can derive $y=\frac b a \sqrt{a^2-x^2}$.
In order to find the slope I use the limit
$$
\lim_{x \to x_0}\frac b a \frac{\sqrt{a^2-x^2}-\sqrt{a^2-x_0^2}}{x-x_0}=\lim_{x \to x_0}\frac b a \frac{a^2-x^2-a^2+x_0^2}{(x-x_0)(\sqrt{a^2-x^2}+\sqrt{a^2-x_0^2})}
$$
The amount of this limit is obviously zero. Where's the problem?
AI: In the RHS of that equality, the numerator should be $(a^2-x^2)-(a^2-x_0^{\,2})$, which is equal to $-x^2+x_0^{\,2}$. This, in turn, is equal to $-(x-x_0)(x+x_0)$. So, your limit becomes$$-\lim_{x\to x_0}\frac ba\frac{x+x_0}{\sqrt{a^2-x^2}+\sqrt{a^2-x_0^{\,2}}}.$$Can you take it from here? |
H: Relation between diagonal entries of $A^{-1}$ and inverse values of $a_{ii}$ for positive definite $A$.
I'd like to expand upon this question. Namely, it says that if $A$, $A=A^T$, is a positive definite matrix, then it holds that \begin{equation}\tag{*}(A^{-1})_{ii}\ge \frac1{A_{ii}}.\end{equation}
Can we prove the converse, i.e.,
if (*) holds for all $1\le i\le n$, then $A$ is positive definite?
OK, as suggested by @Klaus and @Jan I accept this answer and continue here.
AI: No, that's even wrong for numbers, e.g. $A = -1$. |
H: There is only one disc with center i0 for which exists a holomorphic bijection from a given domain
Let D be a domain which is not the entire complex plane.
and let f a holomorphic bijection to a disc with center at 0, that satisfies $f(z_0)=0;f'(z_0)=1$ for some $z_0$ on a disc.
Prove that there is only one such disc; meaning there is only one radius for which this f exists.
EDIT: solve it without Riemann mapping theorem.
AI: Let $f_1: D \to B(0, r_1)$ and $f_2: D \to B(0, r_2)$ both be bijective with $f_j(z_0) = 0$ and $f_j'(z_0) = 1$. Then consider
$$
g(z) = \frac{1}{r_2} f_2(f_1^{-1}(r_1z)))
$$
which is a bijective mapping from the unit disk onto itself with $g(0) = 0$ and $g'(0) = r_1/r_2$. Use the Schwarz Lemma to conclude that $r_1 = r_2$ and then that $g$ is the identity function. |
H: Coset Enumeration: Defining Cosets
I have a problem with understanding the initial step in the Todd-Coxeter coset enumeration algorithm. One needs to define a few cosets when you start off, but I'm not sure how to define them.
As an example, I found the following example: For the presentation $\left\langle {x,y\;\left| {{x^3} = {y^3} = {{\left( {xy} \right)}^2} = 1} \right.} \right\rangle$ and subgroup $H = \left\langle x \right\rangle$, I do understand you firstly define $H: = 1$, and hence $1x=1$, but how does one know to define $1y=2$, $2y=3$, $3y=1$, $2x=3$, etc.? I know for example $1y=2$ follows from $Hy=2$ and $2y=3$ follows from $Hy^2=3$, but how does one know in which order to enumerate them? Why was $Hy^2$ not defined to be '$Hy^2=4$' for example?
Another example I found is the presentation $\left\langle {x,y\;\left| {{x^2} = {y^2} = {{\left( {xy} \right)}^3} = 1} \right.} \right\rangle$ and subgroup $H = \left\langle x \right\rangle$. Here, how does one know to initially define $2x=3$, $3y=4$ and $4x=5$?
AI: The simple answer is that you don't know. You have to use your skill and judgement to choose which new coset to define at any stage and, with experience, you get better at it, in the sense that you are more likely to choose definitions that lead to more rapid completion.
Of course, when programming it on a computer you have to choose some strategy (which could include a randomized component). There are two basic strategies that have been used extensively, often in combination with each other.
The first, often called "Felsch" is that you order the generators and their inverses in some way, and then find the smallest coset number $i$ for which there is an undefined entry, and define $ig_j$ where $j$ is minimal with $ig_j$ undefined. You make all possible deductions from this definition before making a new definition.
The second, called "HLT", does something similar, but works through the relations step by step, making definitions to complete the relator tables.
For hand calculations, the first of these, combined with personal experience, is usually preferable. The second one typically results in more unnecessary definitions, but is a little easier to program and runs fast on straightforward examples. As you probably know, making more definitions means that some of the cosets defined turn out to be equal, and then you have to perform a "coincidence" procedure, which is very awkward and tedious to do by hand, but relatively easy for a computer.
Unfortunately, it has been observed that, for every strategy, there exist examples on which that strategy performs badly, and others perform better. So a good coset enumeration implementation (like the ACE system) will allow flexibility and experimentation. (I suspect that the underlying reason for this is that the general question of whether the index $|G:H|$ is finite is theoretically undecidable, whereas a uniformly good strategy for coset enumeration would suggest otherwise - but that's just idle speculation!)
You could google "coset enumeration strategies" for further details. |
H: Expectation $E[e^{\lambda B_{T}}]$ where $T$ is a stopping time w.r.t. Brownian Motion
Consider a $1$-dimensional Brownian motion started from $0$. Compute $E[e^{\lambda B_{T}}]$, where $\lambda>0$ and $T$ was the first time $t$ for which $B_t=1$.
If this were $E[e^{\lambda {T}}]$, then I would know how to compute it. However, I am a bit confused with how one can compute this expectation with $B_T$ involved. I assume this should be quite straightforward; one just needs to find a suitable martingale and then apply the optional stopping theorem and dominated convergence. However, I am not seeing it.
Could I use that $e^{\lambda B_t-\lambda^2t/2}$ is a martingale?
AI: $B_T$ is just the constant random variable $1$ so $Ee^{\lambda B_T}=e^{\lambda}$. |
H: Proving that $\int_{-1}^{1} \frac{\{x^3\}(x^4+1)}{(x^6+1)} dx=\frac{\pi}{3}$, where $\{.\}$ is positive fractional part
Here, $\{-3.4\}=0.6$.
The said integral can be solved using $\{z\}+\{-z\}=1$, if $z$ is a non-zero real number;
after using the property that $$\int_{-a}^{a} f(x) dx= \int_{0}^{a} [ f(x)+f(-x)] dx$$
So here $$I=\int_{-1}^{1} \frac{\{x^3\}(x^4+1)}{(x^6+1)} dx=\int_{0}^{1} \frac{[\{x^3\}+\{-x^3\}](x^4+1)}{(x^6+1)} dx =\int_{0}^{1} \frac{(x^4+1)}{(x^6+1)} dx.$$
$$\implies I= \int_{0}^{1} \frac{(1+x^2)^2-2x^2}{(x^6+1)}dx=\int_{0}^{1}\frac{(1+x^2) dx}{x^4-x^2+1}-\int_{0}^{1} \frac{2x^2 dx}{x^6+1}=I_1-I_2.$$
In $I_1$, divide up and down by $x^2$ and use $x-1/x=u$, then
$$I_1=\int_{-\infty}^{0} \frac{du}{1+u^2}=\frac{\pi}{2}$$
Next, use $x^3=v$ $$I_2=\int_{0}^{1} \frac{2x^2 dx}{x^6+1}=\frac{2}{3} \int_{0}^{1} \frac{dv}{1+v^2}=\frac{\pi}{6}$$
Finally $$I=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$$
It will be interesting to see other approaches/methods of proving this integral.
AI: $$I=\int\dfrac{x^4+1}{x^6+1}\,dx=\int\dfrac{x^4+1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\,dx=\int\left(\dfrac{x^2+1}{3\left(x^4-x^2+1\right)}+\dfrac{2}{3\left(x^2+1\right)}\right)dx$$
$$\int\dfrac{x^2+1}{x^4-x^2+1}\,dx=\int\dfrac{x^2+1}{\left(x^2-\sqrt{3}x+1\right)\left(x^2+\sqrt{3}x+1\right)}\,dx=\int\left(\dfrac{1}{2\left(x^2+\sqrt{3}x+1\right)}+\dfrac{1}{2\left(x^2-\sqrt{3}x+1\right)}\right)dx$$ Complete the squares and finish to get
$$I=\dfrac{\arctan\left(2x+\sqrt{3}\right)+\arctan\left(2x-\sqrt{3}\right)+2\arctan\left(x\right)}{3}$$ Combine the arctangents to end with
$$I=\frac{1}{3} \left(\tan ^{-1}\left(\frac{x}{1-x^2}\right)+2 \tan ^{-1}(x)\right)$$ |
H: Measurability of composition of random vector w.r.t. $\sigma$(X)
Here is a problem from Resnick - Probability Path (3.3) :
Let $f:\mathbb{R}^k \rightarrow \mathbb{R}$, and $f \in \mathscr{B}(\mathbb{R}^k) / \mathscr{B}(\mathbb{R}) $.
Let also $X_1,...,X_k$ be random variables on $(\Omega,\mathscr{B}) $ .
Show that $f(X_1,...,X_k) \in \sigma(X_1,...,X_k)$ (which I guess it is as saying that $f$ is measurable w.r.t. the sigma algebra generated by the random vector X).
I think that it would be enough to show that $\sigma(f(X_1,...,X_k)) \subset \sigma(X_1,...,X_k)$, (with $\sigma(f(X)) =${$X^{-1}(f^{-1}(B)):B\in \mathscr{B}(\mathbb{R}) $} , if I'm not wrong); but I don't know how to prove it.
Any help would be appreciated, many thanks.
AI: Let $B\in \mathcal B(\mathbb R)$. Then $f^{-1}(B)\in \mathcal B(\mathbb R^k)$, and thus $$f(X_1,...,X_k)^{-1}(B)=(X_1,...,X_k)^{-1}(f(B))\in \mathcal B(\mathbb R^k).$$
Done. |
H: Values of $p$ for an improper integral to converge
I am trying to find values of $p \in \mathbb{R}$ such that $\displaystyle\int_0^{+\infty} x^p\sin(e^x)$ converges. All I have managed doing is using reduction formulas but I couldn't reach a result. Any ideas?
AI: Near $+\infty$ one has by partial integration
\begin{equation}
\int x^p \sin(e^x) d x = -x^p e^{-x} \cos(e^x) + \int(p x^{p-1}-x^p)e^{-x}\cos(e^x) d x
\end{equation}
which clearly converges for all $p\in {\mathbb R}$
Near $0$, we have $x^p \sin(e^x) \sim x^p \sin(1)$ which converges iff $p> -1$ |
H: Let $M$ be a non-empty set whose elements are sets. What are $F=\{AΓ\{A\} : AβM, Aβ β
\}$ and $βF$?
I think it's not so difficult but I struggling a little to figure it out, I want to make sure I'm correct, is $F$ a set of the form:
$F=\{ \{(a, A), (b, A), β¦\}, \{(Ξ±, B), (Ξ², B), β¦\}, ...\}$ for all $a,b,...βA$ and $Ξ±,Ξ²,...βB$, where $A,B,...βM$?
and $βF$ a set of the form:
$βF=\{(a, A), (b, A), β¦,(Ξ±, B), (Ξ², B), β¦\}$?
Thank you!
AI: $y$ is an element of $F$ iff it has the shape $y=A\times\{A\}$ where $A\subseteq M$ and $A\neq\varnothing$.
The equality in your title already states this and cannot essentially be improved, so you do not have to bother about $F$ itself.
$x$ is an element of $\bigcup F$ iff $x\in y$ for some $y\in F$.
Referring to the first line we get:
$x$ is an element of $\bigcup F$ iff $x\in A\times\{A\}$ for some $A$ with $A\subseteq M$ and $A\neq\varnothing$.
That means that $x$ must have the shape $x=(a,A)$ where $a\in A$.
Proved is now that:$$\bigcup F=\{(a,A)\mid a\in A\subseteq M\}$$
The condition $A\neq\varnothing$ does not need to be mentioned here because $a\in A$ implies that $A\neq\varnothing$. |
H: Taylor/Maclaurin series of $\arctan(e^x - 1)$
Right, so this keeps bugging me, and I'm probably stuck in some tunnel trying the same thing over and over again.
Give the Maclaurin series of the function $\arctan(e^x - 1)$. up to terms of degree three. Since I try to be as lazy as possible and differentiating this thing sounds like a lot of work, I want to cheat!
Knowing that the Maclaurin series for $\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} + ....$ and the Maclaurin series for $e^x - 1 = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + ..... $
So yeah, substition to the resque, $\arctan(e^x - 1)$, use the known series and substite in the known series for $e^x - 1$:
$$[1 + x + \frac{1}{2}x^2 + \frac{1}{3!}x^3] - \frac{[1 + x]}{3}^3 + \frac{[1]}{5!}^5$$
Something along the lines of:
$$[1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3] - (\frac{1 + 3x + 3x^2 + x^3}{3})$$
Okay, lets all get 6's:
$$[\frac{6}{6} + \frac{x}{6} + \frac{3}{6}x^2 + \frac{1}{6}x^3] - (\frac{2}{6} + \frac{18x}{6} + \frac{18x^2}{6} + \frac{2x^3}{6})$$
Great... Lets do some algebruh
$$\frac{6}{6} + \frac{x}{6} + \frac{3}{6}x^2 + \frac{1}{6}x^3 - \frac{18x}{6} - \frac{18x^2}{6} - + \frac{1}{6}x^3$$
Join some like terms
$$\frac{6}{6} - \frac{2}{6} + \frac{x}{6} - \frac{18x}{6} + \frac{3}{6}x^2 - \frac{18x^2}{6} + \frac{1}{6}x^3 - \frac{2}{6}x^3$$
This is so far removed from any of the possible answers and I've spent lots of time on this one, so it's time to get some real help.
Please help me :)
AI: The correct expansion should be $e^x-1=\sum_{i=1}^\infty \frac{x^i}{i!}$.
Hence substituting it, we get
\begin{align}
\arctan (e^x - 1) &\approx (x+\frac{x^2}{2}+\frac{x^3}{6}) - \frac{(x+\frac{x^2}{2}+\frac{x^3}{6})^3}{3} \\
&\approx (x+\frac{x^2}{2}+\frac{x^3}{6}) - \frac{x^3}{3}\\
&= x+\frac{x^2}2 - \frac{x^3}6
\end{align} |
H: Simplifying expectation of square of sum
I am working on a problem (not for homework) where one step involves simplifying an expectation. The solution looks like this:
Call $p_\mu(X)$ the PDF of the variable distributed as $N(\mu, \sigma^2I)$. Let $E_0$ denote the expectation under $N(0, \sigma^2I)$.
Then $E_0\left[\left(1+q\left(\frac{p_\mu(X) - p_0(X)}{p_0(X)}\right)\right)^2\right] = 1+q^2E_0\left[\left(\frac{p_\mu(X) - p_0(X)}{p_0(X)}\right)^2\right]$.
I'm not able to derive this step; I tried expanding the square and am left with a middle term that does not seem to equal zero. I also tried using $E[X^2] = Var(X) + E[X]^2$ which also left me with extra terms. Am I missing an identity here?
For context, the full solution is 6.2b here: http://web.stanford.edu/class/stats311/Exercises/2019-solutions.pdf
AI: The reason for this is that the linear part looks like
$$2q E_{0} \left[ \frac{p_{\mu}(X) - p_{0}(X)}{p_{0}(X)}\right] = \int_{X} \frac{p_{\mu}(X) - p_{0}(X)}{p_{0}(X)} p_{0}(X) = \int_{X} p_{\mu}(X) - \int_{X} p_{0}(X)$$
Then since this is a difference in total probability of these two distributions it must be zero (i.e. 1-1 = 0). |
H: Why Can a Plane Not Be Defined Solely With a Vector
Why can a vector not be used to define a plane, why does it have to be a vector and a point. Couldn't you just take a vector and draw a plane at the tip which is perpendicular to the "stem" of the vector in all directions?
AI: A plane through the origin can always be described this way, but a description like this cannot distinguish between a plane through the origin and a parallel translate of the same plane.
I like to emphasize that in the setting of multivariable calculus, a vector that is thought of as an arrow doesn't really have a base point; it doesn't 'start somewhere'. It's really only a direction and a magnitude |
H: Commutative von Neumann regular ring with more than two prime ideals.
Is there an example of commutative von Neumann regular ring which has more than 2 prime ideals?
Matrix rings are not commutative so I couldn't find any, please help.
AI: It is well-known that the quotient of a commutative ring by is nilradical is absolutely flat if and only if every prime ideal is maximal(Bourbaki, Commutative Algebra, Ch. II Localisation, Β§4, exercise 16 d) β in other words, a reduced ring is absolutely flat if and only if it has Krull dimension $0$.
Therefore a product of more than $2$ copies of a field provides such an example. |
H: Is it possible for a sequence of rational numbers to have a non rational limit?
I'm solving an exercise, and at some point I have to prove that any inteval $(a,b)$, with $a,b \in \mathbb R$, is the union of intervals $[\alpha,\beta)$, with $\alpha \in \mathbb R$ and $\beta \in \mathbb Q$. The values of $\alpha$ are easy to find:
$$(a,b) = \bigcup _{n = k}^{\infty} [a + \frac{1}{n},\text{____})$$
For a value of $k$ sufficiently large such that $a + 1/k \in (a,b)$. But the problem is that I don't know what to put in the place of $\text{_____}$, because we have to put a sequence of rational numbers, but the limit of the sequence has to be the real number $b$. Is it even possible for a sequence of rational numbers to have a limit that is not rational? Because a sequence of real numbers never has a limit that is complex for example, so if $(a_n)_n \in \mathbb Q$ it makes scene that $\lim_n a_n \in \mathbb Q$, but I can be wrong. Is this possible? If so what should be the sequence in _____?
AI: The sequence $b_n=10^{-n}\left\lfloor 10^nb\right\rfloor$ will do. You can see that $0\le 10^nb-\lfloor 10^nb\rfloor<1$ by definition, and therefore $0\le b-b_n<10^{-n}$.
You might end up having $[a+1/n, b_n)=\emptyset$ for a while, but it's no biggie. Unless, of course, you are a person who says that $[3,2)=[2,3)$ because he likes $[a,b]$ to denote the segment with endpoints $a$ and $b$. In that case you have to make a caveat of considering only sufficiently large $n$-s.
Addendum: The sequence $b_n$ is very similar to the everyday decimal digit expansion.
For positive real numbers it corresponds to the digit expansion rounded below: say, for $b=\sqrt3$ it's $$1,\, 1.7,\, 1.73,\, 1.732,\, 1.732\text{ (again)},\, 1.73205,\, 1.73205\text{ (again)},\, 1.7320508,\cdots$$
For negative real numbers it is the the digit expansion "rounded above" (which would still technically be "below", because the number is negative and larger absolute value corresponds to smaller number): say, for $b=-\sqrt3$ it's $$-2,\, -1.8,\, -1.74,\, -1.733,\, -1.7321,\, -1.73206,\,-1.732051,\,-1.7320509,\cdots$$ and for $b=-1.401$ it's $$-2,\, -1.5,\, -1.41,\, -1.401,\, -1.401,\, -1.401,\,-1.401,\, -1.401,\cdots$$ |
H: Integral of $\int\limits_0^{2\pi } {\operatorname{erfc}\left( {\cos \left( {a + \theta } \right)} \right)d\theta } $?
I am sorry if it does not fit here. I found some of the integral for the complementary error function e.g.
So far I did not find any integral regarding,
$\int\limits_0^{2\pi } {\operatorname{erfc}\left( {\cos \left( {a + \theta } \right)} \right)d\theta } $
Or, $\int\limits_0^{2\pi } {\operatorname{erfc}\left( {\cos \left( {a + \theta } \right)} \right)\operatorname{erfc}\left( {\sin \left( {a + \theta } \right)} \right)d\theta } $
Is it not possible to find the closed form of the integral of complimentary error function with trigonometric function inside.
Can anyone share me any integral of complimentary error function that has trigonometric function inside as argument?
AI: If $g(x)$ is periodic with a period $T$, then $$\int_{k}^{T+k} g(x) dx=\int_{0}^{T} g(x) dx ~~~~~(1)$$
$$I=\int_{0}^{2\pi} \text{Erfc}[\cos(a+t)] dt=\int_{a}^{2\pi+a} \text{Erfc}[\cos x] dx=\int_{0}^{2\pi} \text{Erfc}(\cos x) dx~~~~(2)$$
Next note the property that $$\int_{0}^{2a} f(x) dx=\int_{0}^{a}[ f(x)x+ f(2a-x)] dx~~~~(3).$$
As $\cos(2\pi-x)=\cos x,$ we get
$$I=2\int_{0}^{\pi} \text{Erf}(\cos x) dx~~~~(4)$$
Again using (3), we get
$$I=2[\int_{0}^{\pi/2}[\text{Erfc}(\cos x)+ \text{Erfc} (-\cos x)]dx=2\pi,~~~~(5)$$
as $\text{Erfc}(z)+\text{Erfc}(-z)=2$.
Edit: Now we take up the pther integral
$$J=\int_{0}^{2\pi} [\text{Erfc}(a+\sin x) ~\text{Erfc}(a+\cos x)] dx$$
Due to the property (1) again, we get $J$ independent of $a$
$$J=\int_{0}^{2\pi} [\text{Erfc}(\sin x) ~\text{Erfc}(\cos x)] dx$$
Using (3) again, we get
$$J=\int_{0}^{\pi} [[\text{Erfc}(\sin x) ~\text{Erfc}(\cos x)+[\text{Erfc}(-\sin x) ~\text{Erfc}(\cos x)] dx$$
Next, use $\text{Erf}(-z)=2-\text{Erf}(z)$,
to write from (4) and (5)
$$J=2\int_{0}^{\pi} \text{Erf}(\cos x) dx= I=2\pi$$ |
H: Triangle inside an open disk
Given an open disk $D=\{x \in \mathbb{R}^2 \mid |x-x_0|<r\}$ and fixed three points $x_1,x_2,x_3 \in D$, how can I show that the triangle (of vertices $x_1,x_2,x_3$) $\Delta=\{t_1x_1+t_2x_2+t_3x_3 \in \mathbb{R}^2 \mid t_1,t_2,t_3 \ge 0 \quad \land \quad t_1+t_2+t_3=1\}$ is all inside $D$, namely $\Delta \subseteq D$.
I know that I have to show that $|x-x_0|<r$ for all $x=t_1x_1+t_2x_2+t_3x_3 \in \Delta$ using the triangular inequality and the fact that $|x_1-x_0|<r,|x_2-x_0|<r,|x_3-x_0|<r$.
However after one hour I can't find any valid approach.
Any kind of hint would be very appreciated. Thank you!
AI: Use the fact that\begin{align}|t_1x_1+t_2x_2+t_3x_3-x_0|&=|t_1x_1+t_2x_2+t_3x_3-(t_1+t_2+t_3)x_0|\\&\leqslant t_1|x_1-x_0|+t_2|x_2-x_0|+t_3|x_3-x_0|\\&<(t_1+t_2+t_3)r\\&=r.\end{align} |
H: Maximum angle of separation between $n$ vectors in $m$ dimensions
My question is exactly as above. If there are $n$-vectors having the same origin in an $m$-dimensional Euclidean space then what is the maximum angle of separation that you can achieve.
Example:
In $2D$ Given $2$ vectors the max angle of separation is $180Β°$. For $3$ vectors it is $120Β°$; for $4$ it is $90Β°$; and so on (here we simply divide $360$ by $n$).
In $3D$ for $4$ vectors it $109.5Β°$ (I know this from chemistry, the arrangement of carbon compounds) and for $6$ vectors it is $90Β°$.
So if we are to generalize this to $n$ vectors in $m$ dimension what's the result?
Can you provide at least an approximate solution to this problem or some resources that pursue it?
AI: This is a very difficult question in general. I will assume unit vectors throughout.
In high dimensions, it's natural to talk about inner products and restrict them.
Let $\epsilon>0,$ then given dimension $m$ one might ask what is the maximum number of unit vectors $v_1,\ldots,v_n\in \mathbb{R}^m,$ so that the mutual inner product between any two distinct vectors is no larger than $\epsilon.$
It turns out, even if you restrict yourself to vectors of the form
$$(\pm 1,\pm 1,\ldots,\pm 1)/\sqrt{m},$$ you can have as many as $n=c 2^{\epsilon m}$ for large $m.$
In small dimensions there are some constructions, but general solutions attaining optima are unknown.
Look up Welch lower bounds for more information about the inner product formulation. Another keyword is "spherical codes". For example, the points at the centres of the pentagons and hexagons of a soccer ball form a spherical code in 3 dimensions. How many pentagons on a soccer ball?
See https://www.quantamagazine.org/a-new-path-to-equal-angle-lines-20170411/ for some information about the case that you restrict all pairwise angles to be the same. |
H: Combinations of towers
You have 25 red blocks and 25 blue blocks.
You can stack the blocks in any order into 5 towers with 5 blocks maximum in each tower.
You do not have to use all the blocks and blocks cannot float in mid-air.
How many combinations of towers can you make?
The base being 0 0 0 0 0 as 1 of the combinations.
So you could have in stack 1: ABBBA 2:0 3: BBA 4: BABAB 5:AAAAA.
How do you calculate how many options there are?
AI: Since there are 25 blocks of each color, there is no limitation on the use of block, which means there won't the case where you don't have enough block of either color.
Then, since each tower are independent from each other, the total amount options is just how many options you have for each tower and raise it to the power of 5. i.e.$\text{Total Options}=(\text{options of each tower})^5$
For each tower, you can have 0 block, 1 block, 2, 3, 4, or 5 blocks. Therefore, the options you have for each tower is $2^0+2^1+2^2+2^3+2^4+2^5=2^6-1=63$
So, $\text{Total Options}=63^5=992436543$
Hope this is helpful. |
H: Given $n \in \mathbb{N}$, find the number of odd numbers among ${n}\choose{0}$,${n}\choose{1}$,${n}\choose{2}$, $...,$ ${n}\choose{n}$ .
So here is the Question :-
Given $n \in \mathbb{N}$, find the number of odd numbers among ${n}\choose{0}$,${n}\choose{1}$,${n}\choose{2}$, $...,$${n}\choose{n}$ .
What I Tried :- I don't know but I guess maybe Lucas's Theorem can help doing it . It states that :-
${m}\choose{n}$ = $\prod_{i = 0}^{k}$${m_i}\choose{n_i}$ (mod p) for all:-
$m = m_kp^k + m_{k - 1}p^{k - 1} + ... + m_1p + m_0 , n = n_kp^k + n_{k - 1}p_{k - 1} + ... + n_1p + n_0$
I can use Lucas's Theorem by substituting $p = 2$ , but I am not sure how to do it, neither I am not sure whether this theorem will help or not. Can anyone help ?
AI: Write $n=(n_kn_{k-1}\cdots n_0)_2$ in base two notation, and similarly for $n$.
The Lucas theorem implies that $\binom n m$ is odd iff every $\binom{n_i}{m_i}$
is odd, and the only way that fails is if $n_i=0$ and $m_i=1$ for some $i$.
For oddness of $\binom nm$ then $m_i$ must be zero when $n_i=0$
but $m_i$ may be zero or one when $n_i=1$. So we have two valid
choices for each $i$ with $n_i=1$, so $2^k$ valid choices overall
where $k$ is the number of ones in the base two expansion of $n$. |
H: Suppose that $f(x) \to\ell$ as $x\to a$ and $g(y) \to k$ as $y \to\ell$. Does it follow that $g(f(x)) \to k$ as $x \to a$?
If $f$ and $g$ are continuous the limit of the composition is the composition of the limit, so the implication follows.
But what if $f$ and $g$ are not continuous? There exist counter examples to the implication? Or does it follow also in this case?
AI: No, not necessarily: we can abuse the fact that the definition of limit ignores the value the function may take there.
For example, take $a = l = k = 0$ and let $f,g:\mathbb R \to \mathbb R$ be as follows.
$$f(x) := 0,\qquad\text{ and}$$
$$g(x) := \begin{cases} 1 & x = 0 \\ 0 & x \neq 0\end{cases}.$$
Then:
$$\lim_{x\to 0} g(x) = \lim_{x\to 0} f(x) = 0,$$
but $g(f(x)) = 1$ for all $x \in \mathbb R$, so $\displaystyle\lim_{x \to 0} g(f(x)) = 1$, not $0$.
Claim The statement in the title holds if and only if one of the following hold:
$g$ is continuous at $l$, or
There is some neighbourhood of $a$ (excluding $a$) on which $f$ does not
obtain the value $l$.
Proof
Case 1. Assume $g$ is continuous at $l$.
Let $\epsilon > 0$. Since $\lim_{y \to l}g(y) = k$ and $g(y) = k$ (by continuity), there exists $\delta>0$ such that
$$ 0 \boldsymbol{\leq} |y - l| < \delta \implies |g(y)-k| < \epsilon. \tag{1}$$
(Note that we can only use "$\leq$" at $\ast$ because it is continuous!)
Since $\lim_{x \to a}f(x) = l$, there exists $\eta>0$ such that
$$ 0 < |y - l| < \eta \implies |f(x)-l| < \delta. \tag{2}$$
which, combined with one, gives you the limit you want.
Case 2. Assume there is some neighbourhood of $a$ (excluding $a$) on which $f$ does not obtain $l$ at all: i.e., there exists $\tau>0$ such that
$$0<|x - a|<\tau \implies f(x) \neq l. \tag{3}$$
Then, repeating the argument for the first case:
Given $\epsilon > 0$, since $\lim_{y \to l}g(y) = k$, there exists $\delta>0$ such that
$$ 0 < |y - l| < \delta \implies |g(y)-k| < \epsilon. \tag{4}$$
(And without continuity, that's all we get.)
Since $\lim_{x \to a}f(x) = l$, there exists $\eta>0$ such that
$$ 0 < |y - l| < \eta \implies |f(x)-l| < \delta. \tag{5}.$$
But moreover, combining $(5)$ with $(3)$ gives
$$ 0 < |y - l| < \min(\eta,\tau) \implies 0 < |f(x)-l| < \delta, \tag{6}$$
which you can combine with $(4)$ to get the desired limit.
Case 3. Assume $g$ is discontinuous at $l$, and that $f$ attains the value $l$ on any (arbitrarily small) neighbourhood of $a$.
We'll use a sequence. You can prove (by contradiction β with the second assumption) that there must be a sequence of points $$(x_n)_{n=1}^β \to a,$$ with $x_n \neq a$ for any $n$, such that, for every $n \in \mathbb N$,
$$f(x_n) = l.$$
Consequently,
$$g(f(x_n)) = g(l),$$
and by the discontinuity of $g$ at $l$, $k \neq g(l)$.
For contradiction, assume that $fg(x) \to k$ as $x \to a$. Then (by the equivalent sequential definition of continuous limit), every sequence $(k_n)_{n=1}^\infty\to a$ with $k_n \neq a$ satisfies
$$g(f(k_n)) \to k.$$
But aha! $(x_n)_n$ is exactly an example of a sequence for which $g(f(k_n)) \not\to k$ (because it already converges to $g(l)\neq k$). |
H: How to prove that $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$ in a simpler way.
EDIT: Preferably a LHS = RHS proof, where you work on one side only then yield the other side.
My way is as follows:
Prove: $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$
I use the fact that $\cos(2x)=2\cos^2(x)-1, \sin(2x)=2\sin(x)\cos(x)$
(1) LHS = $\frac{\cos(x)-2\cos^2(x)+1}{\sin(x)(1+2\cos(x))}$
(2) Thus it would suffice to simply prove that $\frac{\cos(x)-2\cos^2(x)+1}{1+2\cos(x)}=1-\cos(x)$
(3) Then I just used simple algebra by letting $u = \cos(x)$ then factorising and simplifying.
(4) Since that equals $1-\cos(x)$ then the LHS = $\frac{1-\cos(x)}{\sin(x)} = $ RHS.
Firstly, on the practice exam, we pretty much only had maximum 2-2.5 minutes to prove this, and this took me some trial and error figuring out which double angle formula to use for cos(2x).
This probably took me 5 minutes just experimenting, and on the final exam there is no way I can spend that long.
What is the better way to do this?
EDIT: I also proved it by multiplying the numerator and denominator by $1-\cos(x)$, since I saw it on the RHS. This worked a lot better, but is that a legitimate proof?
AI: $$\frac{\cos x-\cos2x}{\sin x+\sin 2x } = \frac{1-\cos x }{\sin x}\iff \sin x\cos x-\sin x\cos2x=\sin x-\sin x\cos x+$$
$$+\sin2x-\sin2x\cos x\iff \color{red}{\sin x\cos 2x}+\sin x+\sin2x-\color{red}{\sin2x\cos x}-2\sin x\cos x=0\iff$$
$$\color{red}{\sin(-x)}+\sin x=0$$
and we're done by the double implications all through (and assuming the first expression is well defined, of course)
Check all the cancellations are correct and check all the trigonometric identities used above.
Another way: We begin with the left side, again: assuming it is well defined
$$\frac{\cos x-\cos2x}{\sin x+\sin2x}\stackrel{\cos2x=2\cos^2x-1\\\sin2x=2\sin x\cos x}=\frac{\cos x-2\cos^2x+1}{\sin x(1+2\cos x)}\stackrel{-2t^2+t+1=-(2t+1)(t-1)}=$$
$$=\require{cancel}-\frac{\cancel{(2\cos x+1)}(\cos x-1)}{\sin x\cancel{(1+2\cos x)}}\stackrel{\cdot\frac{\cos x+1}{\cos x+1}}=-\frac{\overbrace{(\cos^2x-1)}^{=-\sin^2x}}{\sin x(\cos x+1)}=$$
$$=-\frac{(-\sin x)}{(\cos x+1)}=\frac{\sin x}{\cos x+1}$$
Finally, we show that last right side equals the right side of the original equation:
$$\frac{\sin x}{\cos x+1}\cdot\frac{\cos x-1}{\cos x-1}=\frac{\sin x(\cos x-1)}{\underbrace{\cos^2x-1}_{=-\sin^2x}}=-\frac{\cos x-1}{\sin x}=\frac{1-\cos x}{\sin x}$$ |
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