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H: Proof by Induction - Triangles
Given n non-parallel lines such that no three intersect at a point, there are n choose 3 triangles formed.
So far what I come up with is by using proof by induction:
Base Case: For every three lines a triangle is formed.
Inductive Hypothesis: Assume true for k line.
Inductive Step: Suppose that it holds for k lines, show true for k+1 lines.
But how do I show true for k+1 lines? Did I make a mistake in what I have in my induction proof so far? Any help is appreciated, and I don't need a direct solution, just how to begin will suffice.
AI: If this was a homework assignment in a class trying to teach induction, you should complain. You wrote already the main point: For every three lines, there is a triangle formed. The converse, of course, is also true: for every triangle formed, there are three lines. This establishes a bijection between $\{$triangles$\}$ and $\{$unordered triples of lines$\}$. But the number of unordered triples of elements from an $n$-element set (in this case the set of lines) is precisely $\binom n 3$ — this is the definition "choose".
The reason I could imagine this being assigned in a class on induction is that it's not uncommon to use stupid definitions (they are not stupid, but rather important and deep, as facts, but should not be taken as the definition) of binomial coefficients. Two facts, sometimes taken as definitions, are that $\binom n 3 = \frac16 n^3 - \frac12 n^2 + \frac13 n$, and that $\binom{n+1}3 = \binom n 3 + \binom n 2$. Although both of these can be proved by induction, the most natural proofs are not inductive. In any case, I could imagine an instructor looking for an argument like: by induction, there are $\binom n 3$ triangles formed from $n$ lines; adding another line creates a triangle for each of intersection among the original lines; there are $\binom n 2$ such intersections (why? by induction there there are $\binom n 2$ intersections among $n$ lines, and adding a line creates $n$ new intersections, and $\binom n 2 + n = \binom {n+1}2$); therefore there are $\binom n 3 + \binom n 2 = \binom{n+1}3$ triangles formed from $n+1$ lines.
But that inductive argument is stupid compared to the one-line argument that triangles correspond to triples. |
H: Why is $H^0(C,\mathcal O(D))$ a vector space?
Given a divisor $D$ on a smooth curve, one can define the sheaf $\mathcal O(D)$ by the prescription $\Gamma(U,\mathcal O(D) :=$ $\{$meromorphic functions on $U$ that satisfy $(f) + D \ge 0\}$. Then, one can define a line bundle as $\mathcal L(D) = H^0(C,\mathcal O(D))$.
So, line bundles are supposed to be rank one $\mathcal O_X$-modules. But how does this definition above guarantee that you can add two global sections (that's what I take the $H^0$ to mean) of $\mathcal O(D)$ and still get a section in $\mathcal O(D)$?
It seems to me that one can get no control over the divisor $(f + g)$, for $f,G \in \mathcal O(D)$, other than that its degree is zero (since the sum of two meromorphic functions is meromorphic).
AI: Let $D_+$, $D_-$ denote respectively the positive and negative parts of a divisor.
Just notice that $f + g$ has poles exactly at the points either $f$ or $g$ has poles and with order equal to the maximum of the order of the two. Hence $(f + g)_-$ plus $D_+$ is still $\ge 0$. On the other hand, $D_-$ dictates where $f$ and $g$ must have zeros (and to what order). Since each one of $f$ and $g$ have zeros at those points to at least the right order, so will $(f+g)_+ + D_- \ge 0$. |
H: Finding minimum value of multi-variable expression without partial derivatives
Minimize where $a$ and $b$ are positive real numbers
$\sqrt{a^{2}\; +\; 4}\; +\; \sqrt{\left( 3-a \right)^{2}\; +\; \left( b-2 \right)^{2}}\; +\; \sqrt{25\; +\; \left( 6-b \right)^{2}}$
I could take the partial derivatives, equate them to 0, and solve the system of equations, but that leads to some messy equations. Is there a more elegant way to do this?
AI: Hint: What is the distance from $(0,0)$ to $(a,2)$ to $(3,b)$ to $(8,6)$? |
H: Posterior distribution as a distribution for a new random variable?
So in Bayesian framework one uses observed data $X=\{x_1,\dots,x_n\}$ to update the prior $p(\theta)$. My question is it justified mathematically to say that $p(\theta\mid x_1,\dots,x_n)$ corresponds to a new random variable itself?
AI: If you believe that the prior can be represented by a random variable, then the assertion follows by Bayes' Theorem. The difficult and controversial issue with the Bayesian framework is whether you believe that there is a prior distribution. I happen to not believe in it. |
H: Remainder when $123456789101112\ldots$ is divided by $75$
How would you find the remainder when you divide
$$1234567891011121314151617\ldots201120122013$$
(The number formed by combining the numbers from $1$ to $2013$)
by $75$?
AI: The Chinese Remainder Theorem tells you that to calculate a remainder mod $75$ is the same as calculating mod $25$ and mod $3$. For the former, note that $25$ divides $100$; therefore the remainder for your number is $13$.
For the latter, note that $10$ gives remainder $1$ when divided by $3$, and so the remainder of any decimal number mod $3$ is the remainder of the sum of its digits. More generally, $10^k$ gives remainder $1$ upon division by $3$, and so $n \times 10^k$ gives the same remainder as $n$ upon division by $3$. It follows that your number gives the same mod-$3$ remainder as does $\sum_{i=1}^{2013} i = \frac12 \times 2013 \times 2012 = 2013 \times 1006$. But $2013$ is divisible by $3$ (e.g. by summing the digits) and so your number is too.
Thus the remainder is the number between $0$ and $75$ which is divisible by $3$ and which is $13$ more than a multiple of $25$. This number is $63$. |
H: Complicated multivariable implicit differentiation problem
Given that the surface $x^{6}y^{5}+y^{4}z^{5}+z^{9}x^{7}+4xyz=7$ has the equation $z=f(x,y)$ in a neighborhood of the point (1, 1, 1) with f(x,y) differentiable, find:
$\displaystyle\frac{\partial^{2} f}{\partial x^2}(1,1)$
I have already found an intermediate expression:
$\displaystyle\frac{\partial f}{\partial x}=\frac{-6x^{5}y^{5}-7z^{9}x^{6}-4yz}{5y^{4}z^{4}+9z^{8}x^{7}+4xy}$
At the point (1, 1), we have that it's value is -17/18. However, I have tried finding the second partial derivative in two different ways, differentiating both sides of the expression and then differentiating the complete quotient, but i get different results and my answer is still wrong. Could someone guide me in the right direction or have an easier method to solve this exercise? Thank you.
AI: If $$x^{6}y^{5}+y^{4}z^{5}+z^{9}x^{7}+4xyz=7,~~~z=f(x,y)$$ then $$6x^5y^5+\color{red}{\underline{5z^4z'y^5}}+\color{blue}{\underline{9z^8z'x^7}}+\color{green}{\underline{7z^9x^6}}+4yz+4xyz'=0$$ and so $$30x^5y^5+\color{red}{\underline{20z^3z'^2y^5+5z^4z''y^5}}+\color{blue}{\underline{72z^7z'^2x^7+9z^8z''x^7+63z^8z'x^6}}+\color{green}{\underline{63z^8z'x^6+42z^9x^6}}+4yz'+4yz'+4xyz''=0$$ |
H: Product of rotation and translation is a rotation
I have a homework question that I'm not sure how to answer.
Given rotation R and translation T (neither of which are the identity), show that T(R) must be a rotation.
My guess is that we can draw a triangle, then rotate and translate it, and then find some sort of intersection by extending lines from the triangles that is the center of the overall rotation, but I don't think that works.
AI: I assume you are working on the Euclidean plane. We can then choose coordinates so that $R$ is rotation by angle $\theta$ (in the counterclockwise direction; measured in radians) around the origin, and $T$ is translation vector $\vec v = (v_1,v_2)$. A good representation for the plane is as the complex numbers. Then $R$ is multiplication by $a = \exp(i\theta)$ and $T$ is addition by $v = v_1 + i v_2$. Then $T\circ R$ is the function that takes $z$ to $v + az$.
For $a \neq 1$, this transformation has a fixed point $p$, which can be found by solving $p = v + ap$, from which we see that $p = \frac v{1-a}$. It is well known that a rigid transformation which fixes a point is a rotation; in this case, it is not difficult to check that $T\circ R$ is rotation by angle $\theta$ around $p$.
When $a = 1$, so that $\theta$ is an integer multiple of $2\pi$, the composition $T\circ R$ is not a rotation, but a translation, and the problem as stated is false. |
H: If $B \subseteq A$ and $f:A \to B$ is 1-1, it must be onto
Let $B \subseteq A$ and $f: A \to B$ be a 1-1 function, then $f$ must be onto.
I understand that $f$ is onto if and only if every element of $B$ is in the image of $f$...
I believe this statement is true then?
AI: It is true by the Pigeonhole Principle if $B$ is a finite set, but need not be true, otherwise. In fact, if $B$ is finite, $B\subseteq A$ and $f:A\to B$ is one-to-one, then $f$ is onto, and in fact $A=B.$
To see why it need not be true when $B$ is infinite, let $A$ be the set of positive integers, and $B$ the set of positive integers not equal to $1$. Then $B$ is a proper subset of $A,$ but while the map $f:A\to B$ given by $f(n)=2n$ is one-to-one, it is not onto. |
H: The sum of the discount is the discount of the sum
Suppose I have a till. I have a special that I give a 10 percent discount on all dinner meals. I want to know that if I calculate the discount on each dinner item and round up the discount and apply it to each meal, I will not lose money than if I first added up the dinner items and then applied the 10 percent discount and then rounded up the answer.
AI: The consumer benefits from roundup of individual meals because $$\lceil .1(x_1+x_2+\cdots +x_n)\rceil \le \lceil .1 x_1\rceil + \lceil .1 x_2\rceil + \cdots + \lceil .1 x_n\rceil$$ where $\lceil\quad\rceil $ is the ceiling function. So if you're the restauranteur you'd prefer to round up after totaling the meals. |
H: definition clarification in graph theory
I was studying about Almost Self-Centered Graphs (ASC). ASC graphs are
introduced as the graphs with exactly two non-central vertices. Of course, the remaining two vertices are diametrical. My doubt is that if we have two non-central vertices x and y, does it mean that d(x,y) = ecc(x) = ecc(y). I am quite confused. Tried few examples and I found it to be true. If I am wrong please rectify me. thanks a lot for help :)
AI: This is true.
Suppose $r$ is the radius of the graph. Then any vertex except for $x$ and $y$ has eccentricity $r$, and $x$ and $y$ have eccentricity greater than $r$.
There must be some vertex $z$ with distance $d(x,z) = \operatorname{ecc}(x)$, by the definition of eccentricity. However, $z$ must in fact be $y$, because, were $z$ not $y$, $\operatorname{ecc}(z) < \operatorname{ecc}(x) = d(x,z)$, a contradiction. Therefore, $d(x,y) = \operatorname{ecc}(x)$.
The same logic holds for $y$, proving the theorem. |
H: On irreducible polynomial over normal extension
Let $L/K$ be a normal extension and a irreducible polynomial $f(X) \in K[X]$. Prove that, if $f$ is reducible over $L$ then $f$ is factored into product of irreducible factors with same degree. Furthermore, if $f$ has roots in $L$ then $f$ splits over $L$.
Help me a hint.
Thank for any insight.
AI: First, note that the first part of the question implies the second part: i.e. suppose $f$ has roots in $L$. Then, $f$ factors into product of irreducible factors with same degree, one of which has degree 1 (since $f$ has a root in $L$). Therefore, $f$ splits over $L$.
For the first part, take $\bar{K}$ to be the algebraic closure of $K$. And consider $\sigma\in \textrm{Gal}(\bar{K},K)$ where $\sigma$ permutes the roots of $f$. Now, suppose $f$ factors as the product of $f_1\cdots f_n$ over $L$. Can you use the fact that $L/K$ is normal to show $\sigma|_L$ sends each $f_i$ to some $f_j$? (we are not ruling out the case $i=j$)
After that, you can choose appropriate $\sigma$ to show that you can actually send $f_1$ to $f_j$ for any $j$.
I hope this helps! |
H: The number of ways in which four cards be selected from a pack of 52 cards such that there is exactly one pair ( pair has same number or alphabet)
Problem :
The number of ways in which four cards be selected from a pack of 52 cards such that there is exactly one pair ( pair has same number or alphabet)
What I tried :
Pairs may be : (2,2),(3,3) ....(K,K) etc.
$^{13}C_1 \times ^{13}C_1 \times ^{50}C_2$
But not getting the right answer.. please suggest..
AI: There are a few problems with your solution.
For one, ${}^{13}C_1$ gives the number of ways to choose a card denomination without specifying a suit, but we have to specify a suit and a denomination to specify a card. To choose two cards of the same denomination, then, we must choose a denomination and must choose $2$ of the $4$ possible suits.
Now, to pick $2$ cards that are not a pair, and do not have the same denomination as the pair we've chosen, we must pick $2$ of the remaining cards in the deck, to be sure. However, ${}^{50}C_2$ gives the number of ways to choose any $2$ of the remaining cards, which allows for $4$ of a kind, $3$ of a kind, and $2$ pairs. We don't want any of these things to happen! Instead, we should pick $2$ of the denominations we haven't already picked from (How many such denominations are there? How many ways can we pick two of them?) and then pick a suit for each denomination.
Can you put it all together? |
H: integral of $\,x/\ln(x)$
I'm tried to integrate this integral without any success using integration by parts and substitution. $$\int_0^1\int_{e^y}^e \frac{x}{\ln x} dx dy$$
Does this integral require some other technique?
AI: Let $R=\{(x,y):0\le y\le 1,\, e^y\le x\le e\}$. Then you can check that $R=\{(x,y):1\le x\le e,\, 0\le y\le\ln x\}$. So we get
$$\int_0^1 \int_{e^y}^e \frac{x}{\ln x} dxdy=\iint_R \frac{x}{\ln x}dA=\int_1^e\int_0^{\ln x} \frac{x}{\ln x} dydx=\int _1^e xdx=\frac{e^2-1}{2}.$$ |
H: Random variable $V$ has moment generating function $M(t)=e^{3e^{t}-1}$. What is $E(V)$? Note: you do not have to derive the mean.
Random variable $V$ has moment generating function $M(t)=e^{3e^{t}-1}$. What is $E(V)$? Note: you do not have to derive the mean.
I have tried deriving the function , which i got $2e^2$ after subbing in $t=0$
That was no where close to the answer $3$ .
AI: I don't think $M(t)=e^{3e^t-1}$ is a valid moment generating function:
$M(t) = E[e^{tV}]$ so
$M(0) = E[e^0]=E[1]=1$ necessarily, yet $e^{3e^0-1}=e^2$.
I think you meant $M(t)=e^{3(e^t-1)}$, in which case we recognize $V\sim Poisson(3)$. Thus $E[V]=3$.
Alternatively we could find $E[V]$ by evaluating the derivative of $M(t)$ at $0$. |
H: Trouble with simple consequence of the polarization identity
The polarization identity is
$[u,v] = \frac{1}{4} \sum_{k=0}^3 i^k \|u + i^k v\|^2$,
where $[u,v]$ is any sesquilinear hermitian form. My instructor claims as a simple consequence that:
if $[v,v] = 0$ for all $v\in V$, then $[u,v] = 0$ for all $u,v\in V$.
I don't see where this is coming from and he just repeats that its a simple consequence. The only thing I can figure is that $[v,v] = 0$ for all $v$ implies that $\|v\|=0$ by equating real and imaginary parts from the polarization identity to zero. I'm not sure how this implies that $[u,v]=0$ for all $u,v\in V$ as well.
AI: $\| u\|^2$ indicates $[u,u]$, right?
Then $$[u,v] = \frac{1}{4}\sum_{k=0}^3i^k [u + i^kv,u+i^kv] = \frac{1}{4}\sum_{k=0}^3i^k[w_k,w_k] = 0$$
(writing out the last step isn't strictly necessary; it's just to help you see how this follows) |
H: Non-trivial finite/infinite subgroups of infinite groups
Does an infinite group whose every non-trivial subgroup is also infinite exist? If yes, what can be an example of such a group?
also,
Does an infinite group whose every non-trivial subgroup is finite exist? If yes, what can be an example of such a group?
AI: Any group without non-trivial elements of finite order has this property (and the converse also holds). One example is $\mathbb{Z}$.
For your second question, one example is the group $\{x \in \mathbb{C} : \exists n \in \mathbb{N}\: \: x^{2^n} = 1\}$ under multiplication. |
H: If $ p \equiv 3 \mod 4$ and $r$ primitive root, then $\mathrm{ord}_p(-r) = (p-1)/2$
I've been looking at a bunch of number theory problems lately and I need help with a few. One of them is as follows:
Let $p$ be a prime number with $p \equiv 3 \mod 4$ and let $r$ be a primitive root modulo $p$. Prove that $\mathrm{ord}_p(-r) = (p-1)/2$.
First we note that $p=4k+3$ and $(p-1)/2=2k+1$. My idea so far has been to use that $r$ is a primitive root, so $$r^{p-1}\equiv 1 \mod p$$ or $$r^{4k+2} = r^{2(2k+1)} = \left( r^{2k+1}\right)^{2} \equiv 1 \mod p.$$ Hence $r^{2k+1}\equiv \pm 1 \mod p$, but since $r$ is a primitive root we must have $r^{2k+1} \equiv -1 \mod p$. Hence $-(r)^{2k+1} \equiv 1$, or, since $2k+1$ is odd, $$(-r)^{2k+1} \equiv 1 \mod{p}.$$ Is this the correct way of going about it?
AI: The argument is good, but it has not been shown that $-r$ has order $\frac{p-1}{2}$. What has been shown is that the order of $-r$ divides $\frac{p-1}{2}$.
But the ideas you used lead quickly to the finish. If $-r$ has order $q\lt 2k+1$, where $p=4k+3$, then since $q$ is odd we have $r^q\equiv -1\pmod{p}$, and therefore $r^{2q}\equiv 1\pmod{p}$. Since $2q\lt p-1$, this contradicts the fact that $r$ is a primitive root of $p$. |
H: Prove that if $x | (3x + 20)$ then the only positive $x$ for which the statement is true are $1,2,4,5,10,20$.
How can I give a valid proof to this problem?
AI: $x$ divides $3x$ since $x$ divides $x$. Now $3x$ and $20$ are in addition.
For $3x+20$ to be divisible by $x$, $(3x+20)/x$ should give an integer value (Problem Statement).
Now let us assume, for a moment, that $x$ doesn't divide $20$. This means that $20/x$ would give a non-integral value. Say this value is $p$. Now $(3x+20)/x = (3x)/x + 20/x = 3 + p$. From our assumption, since p is non-integral, $3+p$ should also be non-integral. But this is a contradiction from the statement of the question. Hence, our assumption was wrong and therefore $p$ is integral. Hence $20$ should be divisible by $x$ and therefore possible values of $x$ are $1,2,4,5,10,20$. |
H: Very elementary set theory inquiry
Suppose $A = \bigcup_{i=1}^{N} E_i$ and $B = \bigcup_{j=1}^{N} F_j $. Does it follow that
$$ A \cap B = \bigcup_{i=1}^{N} \bigcup_{j=1}^{N} (E_i \cap F_j) = \bigcup_{j=1}^{N} \bigcup_{i=1}^{N} (E_i \cap F_j)$$
?????
AI: Do you know the distributive property? Namely, that $(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$? If, so all you need to do is expand the expression with some careful algebra. I will do it for collections of $3$ sets and maybe you can try and do it for $N$ sets. You could either do this directly, or using induction (induction may give you less of a headache).
So say $A = E_1 \cup E_2 \cup E_3$ and $B = F_1 \cup F_2 \cup F_3$ then
$$A \cap B = (E_1 \cup E_2 \cup E_3) \cap (F_1 \cup F_2 \cup F_3)$$
$$=((E_1 \cup E_2 \cup E_3) \cap F_1) \cup ((E_1 \cup E_2 \cup E_3) \cap F_2) \cup ((E_1 \cup E_2 \cup E_3) \cap F_3) $$
$$=((E_1 \cap F_1) \cup (E_2 \cap F_1) \cup (E_3 \cap F_1)) \cup ((E_1 \cap F_2) \cup (E_2 \cap F_2) \cup (E_3 \cap F_2)) \cup ((E_1 \cap F_3) \cup (E_2 \cap F_3) \cup (E_3 \cap F_3))$$
$$=\bigcup_{i=1}^3 (E_i \cap F_1) \cup \bigcup_{i=1}^3 (E_i \cap F_2) \cup \bigcup_{i=1}^3 (E_i \cap F_3)$$
And then we get:
$$ A \cap B = \bigcup_{j=1}^3 \bigcup_{i=1}^3 (E_i \cap F_j)$$
Similarly, by distributing the other way you get
$$A \cap B = \bigcup_{i=1}^3 \bigcup_{j=1}^3 (E_i \cap F_j)$$ |
H: Quadratic residues modulo $p$ are congruent to the even powers of $r$ modulo $p$
This is another number theory problem I've been tackling:
Let $p$ be an odd prime number and let $r$ be a primitive root modulo
$p$. Prove that the quadratic residues modulo $p$ are congruent to the
even powers of $r$ modulo $p$ and the quadratic nonresidues modulo $p$
are congruent to the odd powers of $r$ modulo $p$.
We know that quadratic residues $a$ satisfy $x^2 \equiv a \mod p$. We must show that $a \equiv r^{2k} \mod p$, i.e. that $r^{2k}$ is congruent to a square modulo $p$. Letting $x=r^k$ we see that the equality holds. I'm not sure how to proceed from here!
AI: All quadratic residues modulo $p$ are of the form $(r^n)^2 = r^{2n}$, so all odd powers of $r$ must be nonresidues. |
H: Find norm of a linear functional
I have a normed space:
$$l_p = \{ (x_n)_{n = 1}^{\infty}: \sum\limits_{n = 1}^{\infty}|x_n|^{p} < \infty \}$$
With norm:
$$||x|| = (\sum\limits_{n = 1}^{\infty} |x_n|^p)^{1/p}$$ where $p = \dfrac{5}{4}$
And I want to find norm of following functional:
$$f(x) = -x_{1} + x_{100} + \sum\limits_{n = 1}^\infty \dfrac{x_{2n}}{n!}$$
I was try apply Hölder's inequality, but not shure it was right:
$$|f(x)| \le ||x||_p ||g||_q, q = 5$$ and $g$ looks like:
$$g(x) = (-1,\dfrac{x_2}{1!}, 0, \dfrac{x_4}{2!}, 0, \ldots, \dfrac{(50! + 1) x_{100}}{50!}, 0, \dfrac{x_{101}}{51!}, \ldots)$$
But I have no idea how to move forward.
AI: Mimicing proof from this answer you can show that the norm of functional $f$ equals to the $q$-norm of the sequence
$$
a=\left(-1,\frac{1}{1!},0,\frac{1}{2!},0,\ldots,0,\frac{1}{49!},0,1,0,\frac{1}{51!},0,\ldots,0,\frac{1}{n!},0\ldots\right)
$$
In other words
$$
\Vert f\Vert=\left(|-1|^5 + \left|\frac{1}{1!}\right|^5+\left|\frac{1}{2!}\right|^5+\ldots+\left|\frac{1}{49!}\right|^5+1^5+\left|\frac{1}{51!}\right|^5+\ldots\right)^{1/5}\\
=\left(2-\frac{1}{(50!)^5}+\sum\limits_{n=1}^\infty\frac{1}{(n!)^5}\right)^{1/5}
$$ |
H: Is there an expression using the main constants of mathematics as result of the following infinite sum:
$$\sum_{k=0}^\infty {{\pi^{k\over 2}}\over {\Gamma({k\over 2} +1)}}$$
I've found, that
$\sum_{k=0}^\infty {{\pi^{k\over 2}}\over {\Gamma({k\over 2} +1)}}$ = $e^{\pi} +
2\sum_{k=0}^\infty {{({4\pi})^{k}k!}\over {(2k+1)!}}$
But does that help?
AI: You can have a closed form in terms of the $\rm {erf}$ function
$$ \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2}\,\mathrm dt, $$
as
$$ {{\rm e}^{\pi }} \left( 1+{\rm erf} \left( \sqrt {
\pi } \right) \right) .$$ |
H: $p=2^n+1$. Prove that every quadratic nonresidue modulo $p$ is a primitive root modulo $p$
This is another one of the number theory problems I've been struggling with as of late (hopefully I'm not posting too many questions at once!).
Let $n$ be a positive integer and let $p=2^n+1$ be a prime number.
Prove that every quadratic nonresidue modulo $p$ is a primitive root
modulo $p$.
Suppose $a$ is a quadratic nonresidue and $r$ a primitive root muodulo $p$. Then there exists no $x$ such that $x^2 \equiv a \mod p$. We must determine the order of $a$. We know that $a$ must be congruent to some power of $r$, say $m$, since $\{r^1,r^2,\ldots,r^{2^n}\}$ is a reduced residue system modulo $p$. Suppose that $\mathrm{ord}_p a=b$. Then $a^b \equiv r^{mb} \equiv 1 \mod p$. We must show that $b=2^n$ since $\phi(p)=2n$. But $r$ is a primitive root and so $mb \equiv 0 \mod{2^n}$ (or $\equiv 2^n \mod{2^n}$). Hence $mb = 2^n k$. How can I finish it?
AI: We will use standard results that you may be familiar with. There are $2^{n-1}$ quadratic non-residues of $p$.
There are $\varphi(\varphi(p))$ primitive roots of $p$. Thus there are $2^{n-1}$ primitive roots of $p$.
It follows that every quadratic non-residue is a primitive root. |
H: Need help with a certain integrating technique
How do I integrate this?
$$\frac{\text{d}P}{\text{d}t} = \frac{(r(t) - B)}{z} \cdot P(t) + c\cdot w$$
The t is just the top limit in the integral.
Let me be more specific. I have this:
$$\frac{dP}{dt} = g(t)P(t) + k$$ where $k$ is a constant and $g(t)$ is any function of $t$, is to put all $P(t)$ on the same side and multiply as such:
\begin{eqnarray*}
\frac{dP}{dt} - g(t)P(t) &=& K\\
e^{\int^t-g(s)ds}\frac{dP}{dt} - g(t)e^{\int^t-g(s)ds}P(t) &=& Ke^{\int^t-g(s)ds}\\
\frac{d}{dt}\Big(e^{-\int^t g(s)ds}P(t)\Big) &=& Ke^{\int^t-g(s)ds}
\end{eqnarray*}
AI: (Assuming you just want to solve the differential equation that you stated).
Note: it was stated in the comments that it may be assumed that $r(t)=at$
$g(t)=\frac{at-B}{z}$
$\dfrac{dP}{dt} = g(t)P(t) + k \iff \dfrac{dP}{dt}-gP=k$
Use $e^{-\int g dt}=e^{-\int\frac{at-B}{z}dt}=e^{\frac{Bt-\frac{a}{2}t^2}{z}}$ as your integrating factor.
Then $e^{\frac{Bt-\frac{a}{2}t^2}{z}}\dfrac{dP}{dt} -\frac{at-B}{z}e^{\frac{Bt-\frac{a}{2}t^2}{z}} P=ke^{\frac{Bt-\frac{a}{2}t^2}{z}}$
Or equivalently, $\dfrac{d}{dt}\left(e^{\frac{Bt-\frac{a}{2}t^2}{z}}P \right)=ke^{\frac{Bt-\frac{a}{2}t^2}{z}}$,
From here you would need to integrate on both sides, but that doesn't seem like a walk in the park (the expression on the right hand side could in fact only be expressed in terms of error functions). |
H: Poisson Distribution Lambda, Probability, and Looking for Exactly k
Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with a rate of $ \lambda $ = 10 per hour. Suppose that with probability 0.5 an arriving vehicle will have no equipment violations.
What is the probability that exactly 5 have no violations?
I wanted to know if this process was correct?
And I'm a bit confused as to how the probability comes into play here.
$ p_X(k) = e^{-\lambda} \cdot \frac{\lambda^k}{k!} $
Then, $~ p_X(5) = e^{-10} \cdot \frac{10^5}{5!} = 0.0378 = 3.78\%$
AI: If $X$ has Poisson distribution with parameter $\lambda$, and $Y$ has binomial distribution with the number of trials equal to the random variable $X$, and $p$ any fixed probability $\ne 0$, then the number of "successes" has Poisson distribution with parameter $\lambda p$.
This has been proved repeatedly on MSE, at least twice by me. Here is a link to a proof.
In our case $\lambda=10$ and $p=0.5$, so the required probability is $e^{-5}\frac{5^5}{5!}$. (It is not the number you obtained.) |
H: Find the sum of $\sum_{n=1}^\infty (-1)^{n+1} (2n-1)x^{2n-1}$
Find the sum of $S(x) = \sum_{n=1}^\infty (-1)^{n+1} (2n-1)x^{2n-1}$.
I know convergence radius is $1$ because $\frac{1}{\sqrt[n] {(2n-1)}} = \frac{1}{1} = 1.$
Then:
$$x^{-1} S(x) = \sum_{n=1}^\infty (-1)^{n+1}(2n-1)x^{2n+2}. $$
$$\int x^{-1}S(x) = \sum_{n=1}^{\infty} (-1)^{n+1}(2n-1) \frac{x^{2n-1}}{2n-1} = $$
$$ \sum_{n=1}^\infty (-1)^{n+1}x^{2n+1} = $$
$$ -\frac{x}{1-x} = \frac{x}{-1+x}. $$
Now getting back our original series:
$$x^{-1} S(x) = (\frac{x}{-1+x})' = \frac{(-1+x)-x}{(-1+x)^2} = \frac{-1}{1-2x+x^2}$$
$$xx^{-1}S(x) = S(x) = \sum_{n=1}^\infty (-1)^{n+1} (2n-1)x^{2n-1} = \frac{-x}{1-2x+x^2} = \frac{1}{-\frac{1}{x} + 2 - x}.$$
What is my problem?
wolframalpha.com says the answer is wrong (it doesn't specify sum of series with $x$).
Thanks in advance!
AI: $$S=\sum_{n=0}^\infty (-1)^n (2n+1)x^{2n+1}$$
Thus, $$\begin{align}
S&=\sum_{n=0}^\infty (-1)^n(2n)x^{2n+1}+\sum_{n=0}^\infty (-1)^nx^{2n+1}\\
&=2x\sum_{n=0}^{\infty}n(-x^2)^{n}+\frac{x}{1+x^2}\\
&=2x\left(\sum_{n=0}^\infty(n+1)(-x^2)^n-\sum_{n=0}^\infty (-x^2)^n\right)+\frac{x}{1+x^2}\\
&=2x\left(\frac{1}{(1+x^2)^2}-\frac{1}{1+x^2}\right)+\frac{x}{1+x^2}\\
&=\frac{2x}{(1+x^2)^2}-\frac{x}{1+x^2}=\frac{x-x^3}{(1+x^2)^2}
\end{align}$$
Well, that's my way. In reference to your calculus,
$$\begin{align}
S&=x\frac{\mathrm d}{\mathrm dx}\left(\sum_{n=0}^\infty (-1)^nx^{2n+1}\right) \\
&=x\frac{\mathrm d}{\mathrm dx}\left(x\sum_{n=0}^\infty(-x^2)^n\right) \\
&=x\frac{\mathrm d}{\mathrm dx}\left(\frac{x}{1+x^2}\right) \\
&=x\left(\frac{1(1+x^2)-x(2x)}{(1+x^2)^2}\right)=\frac{x(1-x^2)}{(1+x^2)^2}
\end{align}$$ |
H: How to evaulate $\int \cos x \sqrt{5 + \cos^2 x} dx$?
How do I evaluate
$$
\int \cos x \sqrt{5 + \cos^2 x} dx?
$$
AI: HINT:
$$\int\cos x\sqrt{5+\cos^2x}dx=\int\cos x\sqrt{6-\sin^2x}dx$$
Now put $\sin x=u$ and use Point $\#8$ of this or this |
H: $L^1$ function is bounded almost everywhere
Suppose $f\in L^1(\mathbb{R})$. Is it true that $f$ is bounded (except for a set of measure zero)?
I think it should be true, but can't show it formally. If $f$ is unbounded, why would we have $\int_\mathbb{R}|f|dx=\infty$?
AI: Every measurable function is "approximately continuous" almost everywhere. See https://www.encyclopediaofmath.org/index.php/Approximate_continuity. In view of the examples you are given in the comment, this is the best you can hope for. |
H: Composite of two algebraic extensions is algebraic.
Let $L,F$ be extensions of the field $K$ and are contained in a common field. Prove that, if $L$ and $F$ are algebraic extensions over $K$ then $LF$ is also a algebraic extension over $K$.
Help me a hint.
Thank for any insight.
AI: Suppose $\alpha$ is in $LF$. Then it can be expressed in terms of some elements $\beta_1,\dots,\beta_r$ of $L$ and $\gamma_1,\dots,\gamma_s$ of $F$. So it's in the extension of $K$ generated by $\beta_1,\dots,\beta_r,\gamma_1,\dots,\gamma_s$. That's a finite extension of $K$, so $\alpha$ is algebraic over $K$. |
H: Validate or invalidate the propositional argument
Validate or invalidate the following arguments
$ p\to t$
$ p \to \lnot r$
$q \to p$
$\lnot t \lor r$
$r \to t$
$\therefore \lnot p \land \lnot q \land (r \iff t) $
I could only see why it is $(r \iff t)$
AI: Hint: You know $r \iff t$. What if $p$ were true? |
H: Solve for: $2\log_3\left(x^2-4\right)+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4$
Solve for: $$2\log_3\left(x^2-4\right)+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4$$
My try:
$2\log_3\left(x^2-4\right)+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarrow \log_3\left(x^2-4\right)^2+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarrow \log_3\left[\left(x-2\right)^2\times\left(x+2\right)^2\right]+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarrow \log_3\left(x-2\right)^2+\log_3\left(x+2\right)^2+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarrow \log_3\left(x+2\right)^2+3\sqrt{\log_3\left(x+2\right)^2}-4\leq0\,\,\,(*)$
Put: $t=\sqrt{\log_3\left(x+2\right)^2}\Rightarrow (*)\Leftrightarrow t^2+3t-4\leq0$
But I don't know Conditions defined for this math? Could help me?
AI: The domain of the first inequation is $x>3$ or $x\leq 3$, so the range of your $t$ is $t\geq 0$. Then you can use $t\geq 0$ as the domain to solve $t^2+3t−4\leq0$, which gets $0\leq t \leq 1$. At last, you just go back to solve $0 \leq t \leq 1$. |
H: Solve for: $8\log_4\sqrt{x^2-9}+3\sqrt{2\log_4\left(x+3\right)^2}=10+\log_2\left(x-3\right)^2$
Solve for: $$8\log_4\sqrt{x^2-9}+3\sqrt{2\log_4\left(x+3\right)^2}=10+\log_2\left(x-3\right)^2$$
My try:
$8\log_4\sqrt{x^2-9}+3\sqrt{2\log_4\left(x+3\right)^2}=10+\log_2\left(x-3\right)^2\\\Leftrightarrow \log_2\left(x^2-9\right)^2+3\sqrt{\log_2\left(x+3\right)^2}=10+\log_2\left(x-3\right)^2\\\Leftrightarrow \log_2\left(x-3\right)^2+\log_2\left(x+3\right)^2+3\sqrt{\log_2\left(x+3\right)^2}-10-\log_2\left(x-3\right)^2=0\\\Leftrightarrow \log_2\left(x+3\right)^2+3\sqrt{\log_2\left(x+3\right)^2}-10=0$
But I don't know Conditions defined for this math? Could you help me please?
AI: $$\log_2(x+3)=y$$
$$y^2+3y-10=0$$
$$y_{1,2}=\frac{-3\pm7}{2},y_1=2,y_2=-5$$
$$\log_2(x+3)=2,x+3=4,x=1$$
$$\log_2(x+3)=-5,x+3=2^{-5},x=2^{-5}-3$$ |
H: Let $L,F$ be extensions over the field $K$ and $L,F$ are contained in a common field.
Let $L,F$ be extensions over the field $K$ and $L,F$ are contained in a common field. Prove that if $L=K(S)$, with $S$ is a nonempty subset of $L$ then $LF=F(S)$.
Thank for any insight.
AI: Ok,
$S \subset L \subset LF$ and $F \subset LF$ and so $F(S) \subset LF$
Conversely, $F\subset F(S)$ and $L = K(S) \subset F(S)$. Since $LF$ is the smallest field containing both $L$ and $F$, it follows that $LF \subset F(S)$ |
H: Finding Polar Area Problem
So I've been staring at this problem for about half an hour because I cannot for the life of me understand what exactly I did wrong. I took a picture because with my stress level it would take me about 25 minutes of angry typing to type up everything using the html tags and all of that.
AI: The problem is with your bounds. Consider the graph of $r^2 = 11 \cos(2\theta)$. We start at the point $(0,\sqrt{11})$. Now for what (smallest positive) value of $\theta$ will $r=0$ so that the curve reaches the origin? Well since $\cos(\pi/2)=0$, this will occur at the point $(\pi/4, 0)$. So the integral you are looking for is:
$$
4 \cdot \left[ \frac{1}{2}\int_0^{\pi/4} 11\cos(2\theta) ~d\theta \right] = 22\left[ \frac{1}{2}\sin{2\theta} \right]_0^{\pi/4} = 11[\sin\tfrac{\pi}{2}- \sin 0] = 11
$$ |
H: Proving the Frobenius map is an endomorphism
I have prime $p$, and $K$ a field such that $p \cdot 1 = 1+1+\cdots+1 = 0$. I am asked to prove that $F: K \rightarrow K$, $a \mapsto a^p$ is a ring homomorphism.
I can prove this for everything except addition, $F(a+b)=F(a)+F(b)$. For addition, it seems to me that $F(a+b)=(a+b)^p=(a+b)^0=1$ but $F(a)+F(b)=a^p+b^p=a^0+b^0=1+1=2$. Can someone point out where I'm going wrong?
AI: If we multiply by $p$ we get zero (this is known as "the ring $K$ has characteristic $p$"). But if we raise to the $p$-th power we may not necessarily get $1$. Then you have to convince yourself that $F(a+b)=(a+b)^p=a^p+b^p$. If you write out $(a+b)^p$ for specific primes $p$ you will notice that all terms but $a^p$ and $b^p$ are multiples of $p$ - thus zero.
For instance for $p=2$ we have
$$
(a+b)^2=a^2+2ab+b^2
$$
where $2ab=0$ in characteristic $2$. Similarly for $p=3$ we get
$$
(a+b)^3=a^3+3a^2b+3ab^2+b^3
$$
where $3a^2b+3ab^2$ vanishes since they are multiples of $3$ hence equal to zero in characteristic $3$. |
H: Projection of ellipsoid
Find the projections of the ellipsoid
$$ x^2 + y^2 + z^2 -xy -1 = 0$$
on the cordinates plan
I have no idea how to do this. I couldn't find much on google to help me with it too.
Thanks in advance!
AI: By coordinate planes, I assume you mean the $xy$-plane, $xz$-plane, and $yz$-plane? If so, the projections can be found as follows:
$xy$-plane: Let $z=0$.
$xz$-plane and $yz$-plane: At first try, you may want to use $y=0$ or $z=0$ respectively; however, this will leave you with the cross section with the $xz$-plane, not the projection onto the $xz$-plane. To get the proper projections, we need to analyze the graph of the projection in the $xy$-plane; in particular, find the equations of the horizontal and vertical tangent lines.
In the $xy$-plane, the equation simplifies to $x^2+y^2-xy-1=0$. Implicitly differentiating with respect to $x$ gives us
$$2x+2y\frac{dy}{dx} - y - x\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{y-2x}{2y-x}.$$
Now, you get horizontal tangent lines when the numerator is zero (i.e. when $dy/dx = 0$), which implies that $y-2x=0\implies y=2x$. Substituting this back into the implicit equation gives us
$$x^2+4x^2-2x^2 -1 = 0 \implies 3x^2=1 \implies x=\pm \frac{\sqrt{3}}{3}$$
and thus $y = \pm\dfrac{2\sqrt{3}}{3}$ are the equations of the horizontal tangent lines.
Similarly, we find the vertical tangent lines when the denominator is zero (i.e. when $dy/dx$ is undefined), which implies that $2y-x=0\implies x=2y$. In a similar fashion, we get that $y=\pm\dfrac{\sqrt{3}}{3}$ and hence $x=\pm\dfrac{2\sqrt{3}}{3}$ are the equations of the vertical tangent lines.
Now why did we go through all of this? Well, the equations of the tangent lines tells us how far along the $x$ and $y$ axis the ellipse extends; in particular, the vertical tangents give us a bound on $x$ and $y$ for the ellipse (i.e. $-2\sqrt{3}/3 \leq x,y \leq 2\sqrt{3}/3$). These bounds play the role of the major axes for the $xz$ and $yz$ projections of the ellipsoid onto the $y=0$ and $x=0$ planes respectively. The minor axis will be along the $z$ direction; in particular, along the ellipsoid, $-1\leq z\leq 1$. With that said, the corresponding projection of the ellipsoid onto the $yz$-plane is
$$\frac{3y^2}{4} + z^2 = 1$$
and the corresponding projection onto the $xz$-plane is
$$\frac{3x^2}{4} + z^2 = 1$$
Note that the last two figures are cylinders of the projections as seen in three space; the purpose of visualizing them this way was to show you that they completely enclose the ellipsoid of interest. |
H: Cardinality of subbasis for topological space
Claim: If $(X,\tau)$ is a topological space, $\mathcal B$ a base for $\tau$ and $\mathcal U$ an open cover of $X$ then there is a subcover $\mathcal V \subset \mathcal U$ whose cardinality is not larger than that of $\mathcal B$.
I appreciate some hints, since since is homework. The idea is to remove redundand open sets in $\mathcal U$. My first idea was to remove all $U \in \mathcal U$ which are subeset of the union of other elements of $\mathcal U$, but I failed to show that the remaining subset of $\mathcal U$ still covers $X$.
AI: Hint: Let $\mathcal{B}^\prime = \{ W \in \mathcal{B} : ( \exists U \in \mathcal{U} ) ( W \subseteq U ) \}$. Now associate to each $W \in \mathcal{B}^\prime$ one element of $\mathcal{U}$ in an appropriate fashion. |
H: Indefinite Integral $\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx$
How can I evaluate this integral?
$$\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx$$
AI: Write $$3\sin x+2\cos x=A(2\sin x+3\cos x)+B\frac{d(2\sin x+3\cos x)}{dx}$$
$$\implies 3\sin x+2\cos x=A(2\sin x+3\cos x)+B(2\cos x-3\sin x)$$
$$\implies 3\sin x+2\cos x=(2A-3B)\sin x+(3A+2B)\cos x$$
Solve for $A,B$ equating the coefficients of $\cos x,\sin x$
So, $$\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx=A+B\int\frac{d(2\sin x+3\cos x)}{2\sin x+3\cos x}$$ |
H: What is the period
How to find the period of the function \begin{array}{cc} e^{z}\end{array} where "z" is a complex number?? Does this function really have any period??
AI: We say that a function $f(z)$ has period $C$ for some constant $C$ if and only if:
$$ f(z+C) = f(z) $$
for all $z$. In this case the function $e^z$ is defined in the complex plane, so it makes sense to consider complex values for $C$. In particular, since $e^{2\pi i} = 1$:
$$ e^{z+2\pi i} = e^z e^{2\pi i} = e^z $$
So the exponential function is periodic with purely imaginary period $C = 2\pi i$. |
H: Morse index and degeneracy
The function is as follows: $$f(x,y,z) = e^x(xy-y^2-z^2)$$ I have found the critical points to be $(0,0,0)$ and $(-2,1,0)$. The question asks to determine the morse index of the points and the degeneracy. How am I supposed to determine the morse index? Also, what exactly is a degenerate critical point? How do I determine whether it is a local maximum or local minimum?
AI: At a critical point of a function $f$ (that is, points where $\nabla f = 0$), its Hessian is a well-defined bilinear form and is independent of coordinate changes. By Sylvester's law of inertia, to each bilinear form on a vector space we can associate three numbers: the dimensions of the maximum subspaces on which the bilinear form is positive definite, zero, or negative definite.
The Morse index of a critical point is the negative index of inertia. In other words, it is the dimension of the maximum subspace on which the Hessian is negative definite.
The degeneracy refers to the maximum dimension of the zero subspace. If this dimension is nonzero, the critical point is said to be degenerate; and it is nondegenerate otherwise.
Therefore, to answer all your questions, you need to compute the Hessian matrix of $f$ at the two critical points. For each of the two matrices, you have to find its indices of inertia (see the Wikipedia page above). Based on that information you can read off the Morse index and degeneracy.
For a nondegenerate critical point, it is a local maximum if and only if the positive index of inertia is zero. Alternatively, for a nondegenerate critical point, it is a local minimum if and only if the Morse index is zero. If the critical point is degenerate, then based on Hessian alone you cannot determine whether the critical point is a local maximum or minimum (but you can tell when it is not: if both the positive and negative indices are nonzero, then you know the critical point must be a saddle point). |
H: Determine the interpolating polynomial
Determine the polynomial of $ deg \le 6 $ interpolating function $$ f(x) = x^3 + 2x^2 + x + 1 $$ at the points : $ -3, -2, -1, 0, 1, 2, 3 $.
My first idea it was to use Lagrange's formula, but it's too long.
Maybe you have other, faster method? Because I guess, that I should give only polynomial and maybe Lagrange's formula is not needed.
Thanks for answers.
AI: f does interpolate f in $ -3, -2, -1, 0, 1, 2, 3 $ and his degree is lower than 6. |
H: Parametric Curves Existence of Tangent
If $\frac{dy}{dt}$ and $\frac{dx}{dt}$ exist, then does $\frac{dy}{dx}$ always exist when $\frac{dx}{dt} \not=0$?
Indeed, this is a very simple question. Sorry but I'm just a beginner for Calculus so I don't know that much in depth. But here is the reason for asking.
During my studying the parametric curves, we learn that
$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
given that $\frac{dx}{dt} \not=0$. But from the Chain Rule, when $\frac{dy}{dt}$ and $\frac{dt}{dx}$ exist,
$$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$$
But as far as I know, for the given information, we cannot say that even though $\frac{dy}{dt}$ and $\frac{dt}{dx}$ exist, $\frac{dy}{dx}$ might not be the same as $\frac{dy/dt}{dx/dt}$ since
$$\frac{dt}{dx}=\frac{1}{\frac{dx}{dt}}$$
if the inverse of $t(x)$ exists (as far as I know... I'm not sure). So for my short knowledge, I'm guessing there might be a case even though $\frac{dy}{dt}$ and $\frac{dx}{dt}$ exist, $\frac{dy}{dx}$ might not exist(or have different value comapared to the suggested one up there). That might happen when $t$ is not one-to-one function, thus having no inverse function against $x$!
Thanks.
AI: Counter example:
$$x=\sin t$$
$$y=\cos t$$
$$\frac {dy}{dx}=-\tan t$$
$\frac {dy}{dx}$ does not exist when t=$\frac {\pi}{2}$ but $\frac{dx}{dt}$ exist at x=$\pi/2$ |
H: Inverse of function arcsin
I'm having trouble finding the solution of the inverse of the function
${\rm f}\left(y\right) = \arcsin\left(\,3 - x^2\,\right)$
Isn't $\arcsin$ the inverse of $\sin$ ?. This is what I have now as inverse:
$\sin\left(\,3 - x^2\,\right)$.
AI: $$y(x)=\arcsin(3-x^2)\iff x(y)=\sqrt{3-\sin(y)}$$. |
H: A Question Regarding Reverse Mathematics and V=L
Can all of "ordinary mathematics" ("ordinary mathematics" as understood by practitioners of reverse mathematics) be formulated in ZF[0]+V=L (ZF[0]is simply ZF without the Power Set Axiom)? It is my understanding that second order arithmetic interprets ZF[0]+V=L (this from Colin Mc'Larty's paper 'Interpreting Set theory in Higher Order Arithmetic", ArXiv:1207.6357 [math.LO] section 3.2).
AI: Apart from a few exceptions, all of the theorems that we study in reverse mathematics are provable in ZF+V=L. Indeed they are provable in "full second order arithmetic", $\mathsf{Z}_2$. So ZF+V=L is much stronger than is needed to prove the theorems that are usually considered.
When Simpson says "ordinary mathematics" in his book Subsystems of second-order arithmetic, he refers to his provisional definition on page 1:
We identify as ordinary or non-set-theoretic that body of mathematics which is prior to or independent of the introduction of abstract set-theoretic concepts. We have in mind such branches as geometry, number theory, calculus, differential equations, real and complex analysis, countable algebra, the topology of complete separable metric spaces, mathematical logic, and computability theory.
This type of theorem will have a "countable version" in which we restrict discrete objects to be countable and continuous objects (e.g. complete metric spaces) to be separable.
The hierarchy of subsystems comes by looking at which countable versions of these theorems imply which (countable versions of) others relative to a much weaker base system.
Regarding "ordinary", there are really three aspects of these theorems, from my perspective:
"ordinary" theorems do not depend on abstract set theory for their statements or proofs, but can be proved in the "usual" way. For example, a result in dynamics that uses the technique of enveloping semigroups is "extraordinary" in this sense. (Theorems that are ordinary in this sense would likely survive the discovery of an inconsistency in ZFC.)
"ordinary" theorems tend to relate to countable or separable objects. For example, Ramsey's theorem for $\mathbb{N}$ is an ordinary theorem, but the Erdős-Rado theorem that generalizes it to arbitrary cardinalities is "extraordinary" in this sense.
"ordinary" results tend to be found in undergraduate texts, and form the well-known core of mathematics. More esoteric theorems that are only found at the graduate level, or which are cooked up only to serve as examples in mathematical logic, are "extraordinary" in this sense.
Theorems are are "ordinary" in all three senses are particularly important for foundations of mathematics, and they are the theorems that traditional Reverse Mathematics is most interested in studying. |
H: Why is the expected value (mean) of a variable written using square brackets?
My question is told in a few words: Why do you write $E[X]$ in square brackets instead of something like $E(X)$? Probably it is not a "function". How would you call it then? This question also applies for $Var[X]$.
AI: Mathematicians usually adopt square bracket such that $f[g]$, instead of $f(g)$ to indicate $f$ is a functional. Obviously, random variable $X=X(\omega)$ is a function, which makes expectation, variance and so on functionals. But this is only an unestablished convention, not necessary. |
H: Recursive formula for the probability that the starting player wins [DBertsekas & JTsitsiklis P57, 1.21]
Two players take turns removing a ball from a jar that initially contains
$w$ white and $b$ black balls. The first player to remove a white ball wins. Develop a
recursive formula that allows the convenient computation of the probability that the
starting player wins.
Solution: Let $P(b, w)$ be the probability that the starting player wins when the jar initially contains $w$ white and $b$ black balls. We have, using the total probability theorem,
$\Large{\color{red}{[}}$ $P(b, w) = \frac{w}{b + w} + \frac{b}{b + w}[ \, 1 - P(b - 1, w) \,] = 1- \frac{b}{b + w}P(b - 1, w) \, \Large{\color{red}{]}}$.
The probabilities $P(1,w), P(2,w), ..., P(n,w)$ can be calculated sequentially using ths formula, starting with the initial condition $P(0,w) = 1$.
How do you derive the expression inside the red brackets? Since the question postulates the starting player to win, I was thinking: $P(b, w) = P($a white ball is chosen on the 1st turn, or on the 3rd turn, or on the 5th or ...).
Source: P57, 1.21, An Intro to Pr, 2nd Ed, D Bertsekas & J Tsitsiklis
AI: This formula says:
$I.$ either the starting player (call him $F$) wins in first move $(\frac{w}{b+w})$
$II.$ or $F$ draws a black ball $\frac{b}{b+w}$
AND the second player (label him $S$) has to lose.
In other words, $P(b,w) =$ Pr($F$ picks a white ball) + Pr($F$ picks a black ball)$\times$Pr($S$ loses)
Now, need Pr($S$ loses). Since the question provides only a probability for the winner, and imparts no other information on the loser, thus consider Pr($S$ loses) = $1 - Pr(S$ wins). What's $Pr(S$ wins)?
In case II, $F$ went first and drew a black ball.
Subsequently, at $S$'s turn, $b - 1$ & $w$ remain.
In other words, $S$ (and NOT $F$) is now the starting player in a game that begins with $b-1$ black balls and $w$ white ones !
Thus, $Pr(S$ wins) is exactly $P(b-1, w)$,
because $S$ can now be considered as the starting player in a game that begins with $b-1$ black balls and $w$ white ones.
Thus, Pr($S$ loses) = $1 - P(b-1, w)$. |
H: Characteristic of a field $F$ is prime
If Char$F$ $\neq 0$, then Char$F$ must be prime number.
MY try: If Char$F$$ = nk $ for integers $n$ and $k$, then by definition, $nk = 0 \implies n = 0$ or $k = 0$ which implies Char$F=0$ which is a contradiction.
Is this correct?
AI: We need to claim $F$ is also an integral domain
Complete proof
As $F$ is a field then $F$ is obviously an integral domain. Assume a contradiction that $\text{Char}(F)$ is not a prime then $\text{Char}(F)=a.b$ where $a,b$ is different from $0$.
So $0=\text{Char}(F).1=(ab).1=a.b$ which implies that $a=0$, or $b=0$ since $F$ is a integral domain. (a contradiction)
Hence, $\text{Char}(F)$ is a prime number. |
H: When is a polynomial irreducible over a field with characteristic $\neq2$?
When is a polynomial irreducible over a field with characteristic $\neq2$?
Help me a hint. I have no idea.
Thanks a lot.
AI: We can, in turn, check if this polynomial has roots on this field or not.
If $charF$ is not too large, it is not terribly difficult to check this. |
H: Volume 3 of Johnstone's "Sketches of an Elephant"
Recently, I read the Chapter 8 of Johnstone's "Topos theory" and got interested in the homotopy and cohomology theory of Grothendieck toposes. So I'm looking for the textbooks expanding these subjects, and it seems that it will be treated in the Volume 3 of "Sketches of an Elephant".
However, unfortunately, Volume 3 has not yet been published for 10 years. So, my question is
Volume 3 will not be published? Did he give up writing?
Is a part of Volume 3 available? Is there a preprint?
Does there exist a textbook which include those subjects?
(Homotopy theory of toposes, Model structures for sheaves, Torsors and non-abelian cohomology, etc..)
AI: Volume 3 is still in preparation. Some sections of Part F have been completed; you can try to ask the author for a copy if you're interested. I do not know if much progress has been made on Part E, but here are some alternatives:
The original reference for the homotopy theory of toposes is [Artin and Mazur, Étale homotopy], and some aspects are discussed in [Moerdijk, Classifying spaces and classifying topoi].
Model structures on presheaf toposes are the subject of Cisinski's thesis, [Les préfaisceaux comme modèles des types d'homotopie].
There doesn't seem to be a coherent reference for model structures for sheaves of simplicial sets at this time.
The original reference for non-abelian cohomology is probably [Giraud, Cohomologie non abélienne]. |
H: Defining iteration of function in set theory
I was trying to prove the axiom of dependant choices from AC, and got confronted with the following problem:
Given $x\in X$ and $f : X\rightarrow X$, can we define the sequence $(f^i(x))_{i\in\mathbb N}$?
Let $\mathcal R$ be a relation on $X\times X$ such that for every $x\in X$, there exists $y\in X$ such that $x\mathcal R y$. My goal is to build a chain $(x_n)_{n\in\mathbb N}$ such that $x_n\mathcal R x_{n+1}$. For $x\in X$, I define $N(x) = \{y\in X : x\mathcal R y\}$. By AC, $\prod_{x\in X} N(x)$ is not empty, hence I have this function that maps every $x$ to some $y$ which is in relation with $x$. Iterating $f$ would prove the axiom of dependant choices, but I can't define the aforementioned sequence using axioms of $ZF$. Could you provide me a hint on how to do this?
AI: You already have everything you need. The result you asked for in your first sentenced is called Dedekind's Recursion Theorem:
Given a set X, an element x in X, and a function f:X \to X, there exists a unique sequence (x_i) such that x_0=x, and x_{n+1}=f(x_n) for all n. |
H: Confused about taking absolute value after integrating reciprocal
I often seem to get caught out when integrating $1/x$ to $\log x$, or similar. Here's an example -- solve
$$ \frac{\mathrm{d}z}{\mathrm{d}x} + \frac{1}{2}z = \frac{1}{2}$$
My first attempt was using separation of variables, which proceeds as follows:
$$ \frac{\mathrm{d}z}{\mathrm{d}x} = - \frac{1}{2} (z-1)$$
$$ \int \frac{1}{z-1}\mathrm{d}x = - \frac{1}{2} $$
$$ \log |z-1| = - \frac{1}{2}x + c_1$$
$$ |z-1| = c_2 e^{- x/2}$$
but the text I'm following has $z = c_2 e^{- \frac{1}{2}x} + 1$. This is found fairly easily through integration by parts, using an integrating factor of $p = \exp(\int \mathrm{d}x/2) = \exp(x/2)$.
$$ \frac{\mathrm{d}z}{\mathrm{d}x}e^{x/2} + \frac{1}{2}ze^{x/2} = \frac{1}{2}e^{x/2}$$
$$ \frac{\mathrm{d}}{\mathrm{d}x} (ze^{x/2}) = \frac{1}{2}e^{x/2} $$
$$ ze^{x/2} = e^{x/2} + c_2 $$
$$ z = c_2 e^{-x/2} + 1 $$
Obviously, if I ignore the modulus signs in the first case, the first and second answers are exactly the same. Alternatively if I proceed using the integrating factor approach the issue doesn't arise. Neither approach seem very satisfactory though. What am I missing, and any advice from experience as to how to minimize the pain in addressing these kinds of problems?
AI: You have $|z-1| = c_2 e^{-x/2}$ where $c_2$ is an arbitrary constant. Thus either $z-1 = c_2 e^{-x/2}$ which simplifies to $z = 1 + c_2 e^{-x/2}$ or $1-z = c_2 e^{-x/2}$ which simplifies to $z = 1 - c_2 e^{-x/2}$. Since $c_2$ is arbitrary, both $c_2$ and $-c_2$ represent an unspecified constant. The two solutions are identical. |
H: Is there a means of analytically proving the following identity?
Okay, so before I begin, my background is more in the world of applied, rather than pure, mathematics, so this question is motivated by a physics problem I'm looking at just now. Mathematically, it boils down to looking for a minimum of a real valued function of a single, positive, real-valued variable, $u$. The part of differentiating the function and finding the condition for a stationary point is straightforward. I can, somewhat heuristically, convince myself that the function must be a minimum, but this is speaking from a physical standpoint. The condition for a minimum rests on the truth (or otherwise) of the following inequality:
$
\cosh^2(u) \geq \frac{u^2}{8}.
$
Now, I can plot this on a graph, and it clearly holds up for the range of values over which the plot is carried out (and would appear to be true in general). I can then say that it's true for all sensible values of the physical parameter $u$, which is simply a combination of a bunch of physical constants and a variable. But obviously I cannot draw an infinite graph, and would rather like a concrete proof to show that this is true for all positive, real-valued $u$. Is there a method that is recommended for dealing with such a problem?
I don't expect a full solution, as I realise it's quite elementary-looking, and once pointed in the right direction I could no doubt take care of it myself. I'm just curious to know what the best analytic method would (in your opinion) be to deal with it so the proof looks a wee bit neater and more rigorous anyway.
Thanks in advance.
AI: Expanding in a Taylor series, $cosh^2(x)=1+x^2+$ (higher order even terms) $\geq 1+x^2 \geq \frac{x^{2}}{8}$. |
H: Finding the nth derivative of $y=e^{ax+b}$
How to find the $n$-th derivative of $y=e^{ax+b}$
Please provide an explanation of the steps.
Thanks.
AI: Recall that for $y = e^{f(x)},\; y' = f'(x)e^{f(x)}$, where $f(x)$ is a function of $x$. This is making use of the chain rule, as it relates to the exponential function.
$\;y' = a \,e^{ax + b}$
$\,y'' = a^2 e^{ax + b}$
$y''' = a^3 e^{ax + b}$
$\vdots$
$y^{(n)} = a^n e^{ax + b}$. |
H: Dividing one equation by another equation
This is from Higher Algebra by Hall and Knight,
$u+v+\sqrt{uv}=39$...(1)
$u^2+v^2+uv=741$...(2)
we obtain by division
$u+v-\sqrt{uv}=19$
I don't know how do you divide one equation by another equation, can someone pls explain.
AI: It's just like any other operation: if you have an equation that is true, and you do the same thing to both sides, you get another equation that is true.
For example, if we know $a=b$ and $c=d$ are both true, then we know $a+c = b+d$ is true: since $a$ and $b$ are the same thing, we've done the same thing to both sides of the equation $c+d$.
The only thing that differentiates division in this respect is that you have to pay a little more attention that the thing you do is not nonsense. i.e. dividing by $x$ is nonsense if you don't know $x$ is nonzero. |
H: Prove that $\mathbb{R}_{\neq0}$ is not a real vector space
Assume a set of $\mathbb{R}_{\neq0}=\{a \in \mathbb{R} \mid a \neq 0\}$, where addition of elements in $\mathbb{R}_{\neq0}$ is the product in scalar $ab$. Prove that this is not a real vector space.
I have made the assumption that the scalar product for the elements is the power of it, $a^k$, $a \in \mathbb{R}_{\neq0}$, $k \in \mathbb{R}$.
And then I went through all the definitions for vector space and this vector space fulfills it. I've seen a rule somewhere that says if $ka=kb$, where $a,b\in\mathbb{R}_{\neq0}$ and $k\in\mathbb{R}$, then $a=b$. If I put $k=2$ and $a=1$, $b=-1$ then it doesn't satisfy the rule, but is that rule even correct?
AI: To make $V=\mathbb R_{\ne0}$ a vector space, we need two operations: An addition $\oplus\colon V\times V\to V$, which we are given by the problem statement: $v\oplus w=v\cdot w$; and a scalar multiplication $\odot\colon \mathbb R\times V\to V$, which we are not given. However, we can see that no matter how we try to define $\odot$, we run into trouble. Let us first study $\oplus$ further: $V$ should be an abelian group und $\oplus$ and we easily check that it indeed is. The neutral element is $1$ and hence we need $x\odot 1=1$ for all $x\in\mathbb R$. Apart from that we have $2\odot(-1)=(1+1)\odot(-1)=(1\odot(-1))\oplus(1\odot(-1))=(-1)\oplus(-1)=1$, but then $-1=1\odot(- 1)=(\frac12\cdot 2)\odot(-1)=\frac12\odot(2\odot(-1))=\frac12\odot 1=1$, contradiction. |
H: Use induction on $n$ to prove that $2n+1<2^n$ for all integers $n≥3$.
Use induction on n to prove that $2n+1<2^n$ for all integers $n\geq 3$.
My attempt: Let $P(n)$ be the statement $2n+1<2^n$.
Base case: Prove that $P(3)$ is true. $LS = 2(3)+1=7$ and $RS=2^3=8$. Since $LS<RS$, $P(3)$ is true.
Inductive Hypothesis: Suppose $P(k)$ is true for some arbitrary integer $k\geq3$, i.e., $2k+1<2^k$ $\Leftrightarrow 2^k>2k+1$ .
Inductive Step: We now show that $P(k+1)$ is true.
So $2^{k+1} = 2^k.2 > (2k+1).2$ by IH
Then where do I go from here?
AI: Nice work, so far. To finish off:
It is enough for you to simply show that on the right-hand side,
$$2(2k+1) = 4k + 2\; >\; 2k + 3 = 2(k+1)+1$$
because $4k+ 2 > 2k + 3, \;\forall k \geq 1$.
So, putting it all together, we have:
$$2^{k+1} = \;2(2^k) \overset{\text{I.H.}}{\;>\;} 2(2k+1) \;> \;2(k +1) + 1$$ |
H: Prove that if $f$ is integrable on $[a,b]$ then so is $|f|$?
Prove that if $f$ is integrable on $[a,b]$ then so is $|f|$.
I can prove the converse of this is false, I also try using the definition of integrable function $f$, but I don't know what to do after that
AI: Integrability means that certain sums involving value differences $|f(x)-f(x')|$ are small. Given that $\bigl||y|-|y'|\bigr|\leq|y-y'|$ the same sums with $|f|$ instead of $f$ tend to be even smaller. |
H: Prove that $SO(3)$ acts transitively on the unit sphere $S^2$ of $\Bbb R^3$
Prove that $SO(3)$ acts transitively on the unit sphere $S^2$ of $\Bbb R^3$.
I think $S^2$ means a 2-D sphere and $SO(3)$ is the usual $SO(3)$ group. I'm unsure how to prove that $SO(3)$ acts transitively.
My guess is to show that like $SO(2)$, $SO(3)$ preserves distance, but I'm unsure how that can guarantee that it will act transitively on the sphere. Any ideas if I'm thinking about this right?
AI: It suffices to show all of $S^2$ is the orbit of a single point, say $e_1$. The equation $Ae_1=x$ says that the first column of $A$ is $x\in S^2$. To create the rest of the matrix $A$, simply take any two vectors on our sphere that are orthogonal to each other and $x$. To do this, compute $e_1\times x$ and then normalize to obtain $y$ for example, then compute $x\times y$ and normalize to obtain $z$. Observe $\det[x~y~z]=\pm1$, so we can permute $y$ and $z$ as necessary to get either $A=[x~y~z]$ or $[x~z~y]$. |
H: let $L_f, L_g, L_{f+g}$ be the lower integral of $f, g, and f+g$. Prove that $L_{f+g}\ge L_f+L_g$
let $f,g$ be two bounded function on $[a,b]$, let $L_f, L_g, L_{f+g}$ be the lower integral of $f, g, and f+g$. Prove that $L_{f+g}\ge L_f+L_g$
I don't know how to start
AI: Here is how you should attack any problem like this (i.e. a problem so basic that looking at pure definitions is a productive route): Going from the definition of lower integrals, we are led to look at lower sums (because the lower integral is defined as a limit of lower sums with finer and finer partitions). Since lower sums sum over a finite number of intervals, we are first led to look at what happens on a single interval. Then we work our way back. This is what happens:
On an interval $[i, j]$, let $l_{f}, l_g, l_{f+g}$ be the infimum of $f, g$ and $f+g$ respectively. We have that $l_f + l_g \leq l_{f+g}$ (show this).
With this, you can show that the lower sums $L'_f, L'_g, L'_{f+g}$ for any finite partition of $[a, b]$ satisfy $L'_f + L'_g \leq L'_{f+g}$ (show this too).
Now take the limit as the partitions get finer and finer. The inequality still stands (show this as well).
Now you're done. |
H: Projection on a convex set
If I have a convex set $ S$ and if I project an $ x$ onto $S$. Is it true that $x $ would project onto a unique element of $S$. Why? What would be considered different if the set $S$ was non-convex?
AI: First of all, you need $S$ to be closed. For example, $S=(0,1)$ is convex but $2$ has no closest point in $S$.
The most general type of space where the closest point property (i.e. the projection uniquely exists) holds is a Hilbert space - look in any functional analysis text for the proof. It doesn't hold generally for any normed space even if $S$ is convex - see here.
To see why convexity is essential, take $S=[0,1]\cup[3,4]$ and think about $x=2$. |
H: Show that $n! \mid (p^n-1)(p^n-p) \cdots (p^n-p^{n-1})$ where $p$ is prime and $n \geq 1$.
So, I'm preparing for an exam and in one of the problems it asks us to find the number of distinct bases that we can have for an $n$ dimensional vector space over a finite field of $p$ elements ($p$ is a prime number, of course).
So, I started to reason this way: First I decided to solve it by simplifying the problem with putting an order. For the first vector, I can choose $p^n -1$ elements in $F^n$ because $0$ can't be chosen. For the second vector, I can choose $p^n - p$ elements because the second vector shouldn't lie in the span of the first vector. For the third vector I can choose $p^n - p^2$ vectors that don't lie in the span of the previously chosen vectors. Continuing with this reasoning I will have $p^n - p^{n-1}$ choices for the $n$-th vector in my basis set. Then since order does not matter in the problem, I have to divide by $n!$ to find the number of distinct bases without considering the order. It seems like I have solved the problem, but then a question popped up in my mind.
Does $n!$ really divide $(p^n-1)(p^n-p)(p^n-p^2)\cdots(p^n-p^{n-1})$ for $n \geq 1$ where $p$ is a prime number?
This counting formula shows that it must be true, but I'm looking for a number theoretic proof if possible or some argument that is independent from this combinatorial proof that I have.
AI: Your counting proof is a fine proof.
Alternatively, given any other prime $q$, find the highest power of $q$ that divides $n!$. Show that it is less than or equal to the highest power of $q$ that divides $(p-1)(p^2-1)\cdots(p^n-1)$.
Then you just need to deal with the case of $q=p$.
That's not a particularly direct proof, however.
This proof will work for $p$ not a prime, then just break the cases into $q\mid p$ and $q\not\mid p$.
Specifically, the number of time $q$ goes into $n!$ is:
$$\sum_{k=1}^\infty \left\lfloor\frac n {q^k}\right\rfloor$$
But we can show, by similar reason, that, if $q\not\mid p$, then $q$ goes into $(p-1)(p^2-1)\cdots(p^n-1)$ at least:
$$\sum_{k=1}^\infty \left\lfloor\frac n {\phi(q^k)}\right\rfloor$$
times. This is because, if $\phi(q^k)\mid m$ then $q^k\mid p^m-1$.
Since $\phi(q^k)<q^k$, we see that $ \left\lfloor\frac n {q^k}\right\rfloor\leq \left\lfloor\frac n {\phi(q^k)}\right\rfloor$ for each $k$. So $q$ goes into $n!$ no more times than $q$ goes into $(p-1)(p^2-1)\cdots(p^n-1)$.
I'm wondering if there is a combinatorial proof for the general $p$ - the counting proof in the question only generalizes for $p$ a prime power. |
H: Relation between a null space and a line.
I have a matrix $T$ and a line $x=y=z$. Lets say I find the null space of $T$ to be $\{\begin{bmatrix}-1\\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}-1\\0\\1\end{bmatrix}\}$. Is it correct to say that the nullspace of $T$ is the plane of $x=y=z$?
AI: I presume that you mean the null space of $T$ is the space spanned by those two vectors. What you'll find is that the null space of $T$ is the plane determined by the lines $y=-x,z=0$ and $z=-x,y=0$.
It is worth noting that this plane is perpendicular to the line $x=y=z$ (and this is fairly readily proven, as it is perpendicular to the two lines mentioned above), so if that's what you meant by "the plane of the line", then you're correct, but you should be precise. |
H: Necessary and Sufficient Condition for Vector Space
Problem Assume a finite set $F$, write the necessary and sufficient condition in terms of the number of elements of $F$, such that $F$ is a real vector space. (Assuming that the vector addition and scalar multiplication can be defined)
Attempt at Solution
Since the necessary conditions for being a vector space is to satisfy the 8 axioms, I thought that $F$ will require at least 3 elements, an identity element, an arbitrary element and the inverse of it.
How could I go from here?
AI: Since a real vector space of positive dimension necessarily contains infinitely many elements, $F$ must be the zero vector space. This suggests that it is necessary and sufficient that $F$ be a singleton set.
Depending on the intention of the problem, there's probably still a couple things to be proven regarding the operations of addition and scalar multiplication, namely that there is only one binary operation on a singleton set (which gives addition), and there is only one possible map $\mathbb R \times F \to F$ giving scalar multiplication. |
H: Is there any point in a logician studying $\infty$-categories?
My primary areas of interest lately have been set theory, logic, and category theory, so naturally topos theory has been a large part of what I'm learning (in between getting caught up on some other things I should really know). I've recently been introduced to the idea of weak Kan complexes and the idea of an "$\infty$-topos". They seem interesting in the sense that I find the formalism pretty, but I don't have a really solid idea for them yet.
As a logician, should I bother to learn more about $\infty$-categories? I gather it's more closely related to the topological end of category theory, but I can't gather from what I can Google whether a logician might have any reason to delve into them.
AI: I'm not an expert but I'd say that you could really interested in learning about $\infty$-topos theory because of homotopy type theory.
As I said I'm not an expert, I've just started learning this stuff so what follows maybe incorrect, I apologize in advance and thank anyone who will point out any mistaken.
For what I've got so far Homotopy type theory should be the internal language of a $\infty$-topos, pretty much like higher order logic is the internal language of a topos, or simply typed type theory should be the internal language of a cartesian closed category, etc....
Homotopy type theory is a sort of dependent (Martin Löf) type theory with some additional axioms, like the univalence axiom. This kind of type theory is intended to be a type theory in which types should behave like homotopy types of homotopy theory. For this reason such kind of type theory can be interpreted in $\bf SSet$ (the category of simplicial sets) and any other category with enough structure to allow to do homotopy theory.
Anyway as I've said homotopy type theory is not just the language of $\bf SSet$ but the language of $\infty$-topos, so just like we are not limited to interpret a first order language in $\bf Set$, but we can use any topos instead, we can use any $\infty$-topos to interpret homotopy type theory.
So as much one is interested in studying topos theory since they are the environment in which one can interpret HOL, one should also be interested in $\infty$-toposes since they are the categories in which interpret homotopy type theory.
If you interested in more on homotopy type theory you can start reading
the book or try to have a look at the homotopy type theory page. |
H: A relation between sine and cosine
I cannot figure out how this relation:
$$\cos(\omega t)+ \frac{\zeta}{\sqrt{1-\zeta^2}}\sin(\omega t) $$
is equal to:
$$\frac{1}{\sqrt{1-\zeta^2}}\sin\left(\omega t + \tan^{-1}\frac{\sqrt{1-\zeta^2}}{\zeta}\right)$$
I only found that this is not true for $\zeta<0$, and specifically it must be $0\leqslant\zeta<1$.
AI: The key thing to notice is that a right angled triangle with sides of length $\sqrt{1-\zeta^2}$ and $\zeta$ has hypotenuse of length $1$ by Pythagoras' Theorem. So
$\sin(\tan^{-1}(\frac{\sqrt{1-\zeta^2}}{\zeta}) = \sqrt{1-\zeta^2}$ and
$\cos(\tan^{-1}(\frac{\sqrt{1-\zeta^2}}{\zeta}) = \zeta$
Then use the rule for $\sin(a+b)$ and it should come out. |
H: Convert time duration a to decimal figure
I am in needs of a formula for converting a duration of a work, already expressed in hour and seconds to a decimal representation. I mean the clock is divided in four parts, of 25 units each.
For example, these are the entries and the desired results:
8h 00m = 8,00
8h 15m = 8,25
8h 30m = 6,50
8h 45m = 8,75
The recording system doesn't allow entries outside a quarter, like 8h 10m, 6h 35m etc... everything is constrained within the corresponding quarter, so the formula could take this into account.
I am not a math guru, so any correction to this question regarding tags, title and syntax is welcome.
AI: Time in decimal is equal to hours + minutes / 60.
So $$8 \textrm{hr}, 45 \textrm{min} = 8+\frac{45}{60} = 8.75 \textrm{hr}$$ |
H: $f(x)=x$ if $x$ is rational and $f(x)=1-x$ if $x$ is irrational
Show that $f$ assumes every value between $0$ and $1$ if
$$f(x) =
\begin{cases}
x, & x\ \text{is rational},\\
1-x, & x\ \text{is irrational}
\end{cases}$$
for all $x\in[0,1]$.
Is it enough to say that since the rationals and the irrationals are dense in the reals, every $x\in[0,1]$ will be assumed by $f$?
AI: It's not sufficient, no. A set can be dense without containing every value, e.g. $\mathbb{Q} \cap [0,1]$.
The answer is much simpler. Take any $y \in [0,1]$. If $y$ is rational, then $f(y) = y$ is a value assumed by $f$. On the other hand, if $y$ is irrational, $1-y$ is irrational too, and $f(1-y) = 1-(1-y) = y$ is also a value assumed by $f$. So every element of $[0,1]$ is assumed by $f$. |
H: Limited function.
Let $f:[0,+\infty) \rightarrow \mathbb{R}$ be a function bounded on each bounded interval. Prove that if $\lim_{x \rightarrow +\infty } [f(x+1)-f(x)] = L$, then $\lim_{x \rightarrow +\infty} \frac{f(x)}{x} = L$.
AI: Assume $|f(x+1)-f(x)-L|<\epsilon$ for $x \ge M$.
Use the triangle inequality as follows, for $x \ge M$:
\begin{align*}
|f(x+n)- f(x)-nL| &= \left\rvert \sum_{i = 1}^n f(x+i)-f(x+i-1)-L \right\rvert \\
&\le \sum_{i=1}^n |f(x+i)-f(x+i-1)-L| \\
& < \sum_{i=1}^n \epsilon \\
& = n\epsilon
\end{align*}
This gives you $|f(x+n)-f(x)-nL|<n\epsilon$. Now divide by $x+n$ and see what you get in the limit as $n \to \infty$, $x$ fixed. You should see that the left hand side of the inequality apporaches $\left\lvert\frac{f(x+n)}{x+n}-L\right\rvert$, while the right hand side approaches $\epsilon$.
Exercise: Suppose $|f(x)|<B$. In terms of $x$, $B$, and $L$, find a natural number $N$ large enough so that, for $n>N$, $\left\lvert\frac{f(x+n)}{x+n}-L\right\rvert<2\epsilon$. To do this, note that by the triangle inequality: $$\left\lvert\frac{f(x+n)}{x+n}-L\right\rvert \le \left\lvert\left(\frac{f(x+n)}{x+n}-L\right) - \left(\frac{f(x+n)-f(x)-nL}{x+n}\right)\right\rvert + \left\lvert\frac{f(x+n)-f(x)-nL}{x+n}\right\rvert.$$ The quantity $\left\lvert\frac{f(x+n)-f(x)-nL}{x+n}\right\rvert$ is already bounded by $\epsilon$; you just need to simplify $\left\lvert\left(\frac{f(x+n)}{x+n}-L\right) - \left(\frac{f(x+n)-f(x)-nL}{x+n}\right)\right\rvert$ and show that $N$ exists so that it is bounded by $\epsilon$ as well.
Now consider $x \in [M,M+1]$. Use the fact that $f$ is bounded on $[M,M+1]$, say by $B$, and the fact that $M \le x \le M+1$, and your work above, to get the bound $\left\lvert\frac{f(x+n)}{x+n}-L\right\rvert<2\epsilon$ for $n>N$, for some choice of $N$. By substitution, we get the result that $\left\lvert \frac{f(x)}{x} - L \right\rvert < 2\epsilon$ for $x>M+N$. Now take $\epsilon$ to zero. |
H: Proving a constant function $f(x) = c$ is Riemann integrable
Prove that a constant function $f(x) = c$, where $c$ is in the Real Numbers, is Riemann integrable on any interval $[a, b]$ and $\int_a^bf(x) dx = c(b-a)$.
By looking at the definition, it looks like I am going to explain that it's bounded (which would be obvious since the function is constant?) Additionally, it would appear that $\inf(f)$ and $\sup(f)$ also obviously exist since $f$ is constant. The parts I am having trouble understanding involve explaining that $\sup\{L(P,f)\} = \inf\{U(P,f)\}$, as well as proving that $\int_a^bf(x) dx = c(b-a)$. Let me know if what I have so far is okay, and please give me some guidance on the rest. Thanks!
AI: You're on the right track so far. For the sup=inf part, you're in a really nice position: if you take any partition and compute the max of f on each subinterval, you'll get c. So $U(P, f) = \sum_i c \cdot (t_{i+1} - t_i) = c \cdot \sum_i t_{i+1} - t_i = c(b-a)$. That means that EVERY SINGLE Upper sum turns out to be $c(b-a)$. I'll bet that you can compute the inf of a set that contains only a single number, right? Then you're on your way. |
H: Number of elements of group with specific order
Consider the group $(G,\cdot)$ where $$G=\left\{\left(\begin{matrix}1&a\\0&b\end{matrix}\right):a,b\in\mathbb{R}, b\neq0\right\}.$$ How many members of $G$ have order 2?
My Attemt
A member $M$ of $G$ will have order two iff $M^2=I$. I.e. $$\left(\begin{matrix}1&a\\0&b\end{matrix}\right)^2=\left(\begin{matrix}1&a+ab\\0&b^2\end{matrix}\right)=I.$$This is true for $b=1$ and $a=0$ in which case $M=I$ or in the case $b=-1$ and $a$ is any real. Hence, there is an infinite number of elements with order 2 in $G$.
Can somebody please verify that this is true?
AI: Very well argued. You are entirely correct.
On the light side: You can stamp your answer Verified @MSE. |
H: In characteristic $p$, the field extension $k(X,Y)$ over $k(X^p,Y^p)$ is not simple
Let $k$ be a field with characteristic $p>0$, $L=k(X,Y)$ be the field of rational fractions of two variables over $k$. Let $K=k(X^p,Y^p)$. Then $[L:K]=p^2$ (for a proof see In characteristic $p$, the field extension $k(X,Y)$ over $k(X^p,Y^p)$ has degree $p^2$). Show that $L^{p}\subseteq K$ and $L$ is not a simple extension over $K$.
Thank in advance.
AI: Obviously $k^p \subseteq k\subseteq K$ and furthermore clearly $X^p\in K$ and $Y^p\in K$. Since the characteristic of our fields is $p$, the Frobenius mapping $x\mapsto x^p$ is a field homomorphism. This implies $L^p \subseteq K$.
Assume that $L = K(a)$ with $a\in L$.
Since $L^p\subseteq K$, in particular $a^p\in K$. So $a$ is a zero of $f = T^p - a^p\in K[T]$.
Thus $[L : K] \leq \deg(f) = p$. Contradiction. |
H: $A$ is similar to $B$ if $A\oplus A$ is similar to $B\oplus B$
Question:
If the matrix $\begin{pmatrix} A & 0 \\ 0& A \end{pmatrix}$ is similar to $\begin{pmatrix} B & 0 \\ 0 & B \end{pmatrix}$
show that:
the matrix $A$ is similar the matrix $B$
My try:
since the matrix diag $(A,A)$ is similar matrix diag $(B,B)$;
and I want use this elementary divisor,But there is not in The plural number field
then I can't.Thank you someone help me,Thank you very much!
AI: This solution uses the notion of elementary divisors from the theory of canonical forms: see e.g. $\S$ 5 of these notes for the definition and properties. I would be interested to know if there is a more elementary argument.
Use the fact that the elementary divisors of $\operatorname{diag}(A,B)$ are obtained by combining the elementary divisors of $A$ and $B$ (i.e., add the multiplicities) together with the fact that two matrices are similar if and only if they have the same elementary divisors (Theorem 7.2 of loc. cit.). |
H: Finding the limit of $|x|^{1/x}$ as $x \rightarrow 0$.
I have to find the limit of $|x|^{1/x}$ as $x\rightarrow 0$
I can't use de l'Hopital's method and derivatives.
Now the limit itself does not exist but it have to be divided into two side the left and right. Then it have to be calculated separately. The right side is 0, the problem is in the left side. Someone managed to solve the limit giving the correct result that is -infinity but I can't figure it out. Basing on my calculus I've arrived to +infinity and if you got to see in wolfram alpha putting lim (-x)^(1/x) as x->0- (the left side) you can see that the limit goes to +infinity. In my opinion the book's solution is wrong, however can someone show me that I'm right (to have a confirm of my solution?)
AI: By definition, $|x|^{1/x}=\exp\frac{\ln|x|}x$. Since $\ln|x|\to-\infty$ and $\frac1x\to\pm\infty$ as $x\to 0^\pm$, the limit as $x\to 0$ does not exist: We have $$ \lim_{x\to0^+}|x|^{1/x}=0,\qquad \lim_{x\to0^-}|x|^{1/x}=+\infty.$$ |
H: Deriving time and distance
The distance an aircraft travels along a runway before takeoff is given by $D=(10/9)t^2$, where $D$ is measured in meters form the starting point and $t$ is measured in seconds from the time the brakes are released.
The aircraft will become airborne when its speed reaches $400\;\text{km/h}$.
$\text{(1)}$ How long will it take to become airborne, and
$\text{(2)}$ What distance will it travel in that time?
My try:
Finding the derivative of $D$ will give the velocity function $v(t)$.
Equate $v(t)$ with $400\;\text{km/h}$ to find $\text{(1)}$:
$\frac{\text{d}}{\text{dt}}D=\frac{\text{d}}{\text{dt}}(10/9)t^2 = v(t)=(20/9)t$
$400=(20/9)t$
$(\frac{9}{20})(400)=t$
$180\; \text{sec}=t$
This is wrong though. The correct answer is $50\; \text{sec}=t$ for $\text{(1)}$. Thank you.
AI: Notice that $t$ is measured in seconds and $D$ in metres meaning that $\tfrac{d}{dt}D$ will be is m/s.
The take-off speed if given in km/h. So that's where your problem comes from.
Notice that 1 km = 1,000 m and 1 h = 3600 s. It follow that
$$1 \, \text{km/h} = \frac{1 \, \text{km}}{1 \, \text{h}} = \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{5}{18} \, \text{m/s}$$ |
H: Probability Uniform Distribution
If $A$ is uniformly distributed over $[-25, 30]$, what is the probability that the roots of the equation
$$x^2 + Ax + A + 80 = 0$$
are both real?
I kept getting weird answers which was wrong. $$A (1 + x)+ x^2+ 80 = 0$$
$$A (1 + x) = 0$$
$$A =0 $$ $$x=-1$$
$$ x^2+80 = 0 $$
$$ x=-4i √(5) $$
$$ x= 4i √(5) $$
I am so sure this approach to the question is incorrect. because it shouldn't have any sort of imaginary answers.
AI: Hint: The roots are real if and only if $$A^2 - 4(A+80) \geq 0$$ by the quadratic formula. |
H: $U^*\otimes V$ versus $L(U,V)$ for infinite dimensional spaces
It's well known that $U^*\otimes V\cong L(U,V)$ for finite dimensional spaces. However, why people say that $U^*\otimes V$ is not isomorphic to $L(U,V)$ for infinite dimensional spaces and many books don't say anything related to this?
AI: Suppose we are over an infinite field $\Bbbk$, and let $\dim(U) = \kappa$, $\dim(V) = \lambda$.
This great answer of Arturo Magidin shows that $|U^*| = |\Bbbk|^\kappa$. Basically $U^*$ is in bijection with maps from a basis of $U$ to $\Bbbk$.
When the spaces are infinite, the cardinality of the tensor product is the product of the cardinalities.
There is a surjection $\bigsqcup_{n \geq 0} (X \times Y)^n$ to $X \otimes Y$ given by linear combinations.
$|X \otimes Y|$ is obviously at least $|X| |Y|$: there is an injection given by $x \otimes y$.
Therefore $|U^* \otimes V| = |U^*| |V| = |\Bbbk|^\kappa \lambda$
On the other hand, $L(U,V)$ is in bijection with maps from a basis of $U$ to $V$. Therefore $|L(U,V)| = |V|^\kappa = |\Bbbk|^\kappa \lambda^\kappa$.
So if $|\Bbbk|^\kappa \lambda < |\Bbbk|^\kappa \lambda^\kappa$ (equality can happen), then the two spaces don't have the same dimension. So they can't be isomorphic. For example, assume the continuum hypothesis and the axiom of choice (sorry set theorists) and take $|\Bbbk| = \aleph_0$ (e.g. $\Bbbk = \mathbb{Q}$), $\lambda = \aleph_\omega$ and $\kappa = \operatorname{cof}(\lambda) = \aleph_0$. Then by König's theorem $\lambda^{\operatorname{cof}(\lambda)} > \lambda$, so:
$$|\Bbbk|^\kappa \lambda = \aleph_0^{\aleph_0} \aleph_\omega = 2^{\aleph_0} \aleph_\omega \overset{\mathsf{CH}}{=} \aleph_1 \aleph_\omega = \aleph_\omega = \lambda < \lambda^\kappa = |\Bbbk|^\kappa \lambda^\kappa.$$
In general though, the spaces can be isomorphic. For example is both spaces have countably infinite dimension over $\mathbb{R}$, then both $U^* \otimes V$ and $L(U,V)$ have dimensions $2^{\aleph_0}$ and so they are isomorphic, though one would be hard-pressed to describe an explicit isomorphism.
A trivial example: if $V = \Bbbk$, then you're wondering whether $U^*$ is isomorphic to $L(U, \Bbbk)$...
If $U$ is finite dimensional, then the usual map is an isomorphism (see below).
But if both $U$, $V$ are infinite dimensional, they are not naturally isomorphic, at least through the only obvious natural map.
In the finite dimensional case, the natural isomorphism looks like this: $\varphi \otimes v \mapsto (u \mapsto \varphi(u)v)$. When the spaces are infinite dimensional, this is not an isomorphism! Indeed any element of $L(U,V)$ in the image has finite dimensional range. But there are elements in $L(U,V)$ with infinite dimensional range, obviously. |
H: Math behind Keynesian Expenditure Multiplier
Take a look at this page: http://wiki.ubc.ca/Keynesian_Multiplier
Why can you find out the sum of the geometric series just by dividing the mps by 1?
AI: Consider the series
$$
1+r+r^2+r^3+\cdots=\sum_{k=0}^\infty r^k
$$
If $|r|<1$, then this is a geometric series whose sum is the well-known
$$
\sum_{k=0}^\infty r^k=\frac{1}{1-r}.
$$
So if you have $MPC$ instead of $r$ and $MPS=1-MPC$, then this is
$$
\sum_{k=0}^\infty MPC^k=\frac{1}{1-MPC}=\frac{1}{MPS}.
$$ |
H: Proof related with prime numbers and congruence
How to (dis)prove this
$ (n-2)! \equiv 1 \mod n$
If n is said to be a prime number. I guess we'll have to use FERMAT’S LITTLE THEOREM, and I just don't know where to start from. Thanks in advance
AI: If $\;n=p\;$ is a prime, then by Wilson's theorem
$$\color{red}{-1}=(p-1)!=(p-2)!(p-1)=\color{red}{-(p-2)!\pmod p}\implies 1= (p-2)!\pmod p$$ |
H: Prime notation in integral?
Recall the definition of potential energy:
$$
U_x-U_{x_0} = -\int^x_{x_0}F_x(x')dx'
$$
I've seen the integral definition of work, but not this - the thing I'm specifically interested in is the notation. I've never seen differential and prime notation intermingled like this. What does it mean? Would it have been possible to simply express it as:
$$
U_x-U_{x_0} = -\int^x_{x_0}F_x(x)dx
$$
or is there an important reason for the given notation?
AI: $x'$ is a so-call dummy variable. The integral could just as well have been
$$
U_x-U_{x_0} = -\int^x_{x_0}F_x(\tau)d\tau
$$
without changing the result.
The variable $\tau$ is "integrated out" and the result is a function of $x$ with $x_0$ as a parameter.
See, for example, the Wolfram article for Dummy Variable:
A variable that appears in a calculation only as a placeholder and
which disappears completely in the final result. |
H: Multiplication modulo proving a set is a group conform conditions
Given is the following explanation.
A group is a set, together with a binary operation $\odot$, such that the following conditions hold:
For all a, b $\in$ S it holds that a $\odot$ b $\in$ S
Now, i have to show that the Set = {1, 2 ... 16, 17} together with multiplication modulo 18 is not a group.
AI: Multiplication modulo $14$ means the value of $a\cdot b \pmod {14}$. So for $a = 1, b = 2$, $a\odot b = 1 \cdot 2 \pmod {14} = 2.$ This is an associative operation on the integers, because multiplication is associative.
HINT for the first: The identity element here is $1$. Show that $2$ has no multiplicative inverse $a'$ (no element in the group such that $2\cdot a' = a'\cdot 2 = 1$. Indeed, you can simply show that your set is not closed under the binary operation $\odot$: Calculate for example, $2 \odot 7 = 2\cdot 7 \pmod{14}$.
HINT for the second: Note that $13$ is prime, and any group $\mathbb Z^*_{p}$ where p is prime has $p - 1$ elements, and is equal to $\{1, 2, \cdots, p-1\}$.
$1$ again is your identity element: this is easy enough to confirm. You can also verify that the inverse $a'$ of every element $a\in \mathbb Z^*{13} \in \mathbb Z^*_{13}$. You can also confirm that for any two elements $a, b \in \mathbb Z^*_{13}$, $a \odot b = a\cdot b \pmod {13} \in \mathbb Z^*_{13}.$ |
H: Prove a matrix maps to a point that make up the plane perpendicular to a line.
I'm having difficulty understanding what the following question is asking and was hoping someone could explain it to me.
$T = \begin{bmatrix}1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \end{bmatrix}$
Given a point on the line x = y = z, prove that the set of points in $\mathbb{R}^3$ that $T$ maps to that point make up the plane through that point that is perpendicular to the line $x = y = z$.
According to my instructor, I am suppose to solve for $Tx = \begin{bmatrix}a\\a\\a\end{bmatrix}$. Doing so I get
$\begin{bmatrix} 1/3 && 1/3 && 1/3 && a \\1/3 && 1/3 && 1/3 && a \\1/3 && 1/3 && 1/3 && a \end{bmatrix}$
Reducing:
$\begin{bmatrix} 1 && 1 && 1 && 3a \\ 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 \end{bmatrix}$
$ x_1 + x_2 + x_3 = 3a $
$ x_1 = 3a - x_2 - x_3$
So
$x = \begin{bmatrix} x_1\\ x_2\\x_3 \end{bmatrix} = x_1\begin{bmatrix} 3a \\ 0 \\ 0 \end{bmatrix} + x_2\begin{bmatrix} -1\\1\\0\end{bmatrix} + x_3\begin{bmatrix} -1\\0\\1 \end{bmatrix} $
Since I don't understand what the question is really asking, I'm not sure what to do with this answer.
AI: Well, check that the set of points you denoted as $\;x\;$ indeed passes through the point on $\;x=y=z\;$ and is perpendicular to that line:
$$\begin{pmatrix}a\\a\\a\end{pmatrix}=\begin{pmatrix}1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\frac13\begin{pmatrix}x_1+x_2+x_3\\x_1+x_2+x_3\\x_1+x_2+x_3\end{pmatrix}\iff$$
$$x_1+x_2+x_3=3a$$
Now, check the vector $\;u=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\;$ defined by the above relation actually passes throught the point $\;A:=\begin{pmatrix}a\\a\\a\end{pmatrix}\,$ (very easy) ,and that the vector $\;A-u\;$ is perpendicular to the given line there (very easy, too). |
H: Does there exist a semigroup such that every element factorizes in this way, which nonetheless lacks a left identity?
If a semigroup $S$ has a left identity-element, then for any $y \in S$ we can write $y = xy$ for some $x \in S$. Just take $x$ to be any of the left identities, of which there is at least one, by hypothesis.
Does there exist a semigroup $S$ such that every $y \in S$ factorizes in this way, which nonetheless lacks a left identity-element?
AI: Every idempotent semigroup has this property, because $y=yy$, but not all have left identities. |
H: If derivative of a function is non-zero then it is monotone. Since function is monotone, variable can be substituted in integration
I came across this in the text Differential Equations-An Introduction With Applications, by Lothar Collatz:
$y'(x)=\frac{dy}{dx}=\frac{f(x)}{g(y)}$
Suppose $f(x)$ is continuous in $[a,b]$ and $g(y)$ is continuous in $[c,d]$. We integrate both sides of the equation
$g(y)y'=f(x)$
with respect to $x$ from a chosen point $x_0$ up to a variable point $x$ (both points are to lie in $[a,b]$), and on the left we introduce $y$ as a new variable instead of $x$ (this is possible if $y'(x)\neq 0$, since then $y(x)$ is monotone in $[x_0,x]$)
Could somebody please explain the last statement?
EDIT: I had not included thought about this earlier but what does monotonicity have anything to do with substitution of a variable in integration.
AI: If $f$ is differentiable and $f'$ is continuous and $f'(x)\neq 0$ for all $x\in [a,b]$ then (by the intermediate values theorem) either $f'(x)>0$ for all $x\in [a,b]$ or $f'(x)<0$ for all $x\in [a,b]$. As a consequence $f$ is monotone. |
H: Tangent line to the curve $x^3+xy^2+x^3y^5=3$
Does the tangent line to the curve $x^3+xy^2+x^3y^5=3$ at the point $(1,1)$ pass through the point $(-2,3)$? (using implicit differentiation)
I got the implicit differentiation as $\frac{dy}{dx}=\frac{-3x^2-y^2-3x^2y^5}{2xy+5x^3y^4}$. Then I got stuck?
Explanation would be appreciated.
AI: You've got one too many $-y^2$ terms in your numerator: you should arrive at $$\frac{dy}{dx}=\frac{-3x^2-y^2-3x^2y^5}{2xy+5x^3y^4}$$
With the correct derivative, evaluate $\dfrac {dy}{dx}$ at the point $(1, 1)$. That will give you the slope $m$ of the line tangent to your curve at $(1, 1)$. Then using $m$ and the point $(x_0, y_0) = (1, 1)$, you can write the equation of your line in point-slope form:
$$y - y_0 = m(x - x_0)\tag{1}$$
Once you've obtained your equation, then plug in the coordinates of $(-2, 3)$ into your equation of the tangent line $(1)$ to see if the equation, given $x = -2, y = 3$ is true. If so, then $(-2, 3)$ is on the line $(1)$ tangent to your curve at $(1, 1)$. If not, then it is not on that line $(1)$. |
H: Finding the local minimum of $e^{3x} + e^{-x}$
$${\begin{array}{l} f(x) \; = \; e^{3x} + e^{-x} \\ f'(x) \; = \; 3e^{3x} - e^{-x}\end{array}}$$
The interval on which $f$ is increasing is $\left( \frac{1}{4}\ln \left(\frac{1}{3}\right), \; \infty \right).$
The interval on which $f$ is decreasing is $\left(-\infty, \; \frac{1}{4}\ln \left(\frac{1}{3}\right) \right).$
When trying to find the local minimum, I used $\left( \frac{1}{4}\ln \left(\frac{1}{3}\right), \; \infty \right).$
The interval on which $f$ is decreasing is $\frac{1}{4}\ln \left(\frac{1}{3}\right)$ for $x$ in $f(x)$ and set it equal to zero. I ended up getting $.71$ for the answer, but it's not correct according to WebAssign. I checked my math to see if I did something wrong, but I keep getting the same answer.
AI: Another way, $f(x) = e^{3x}+\frac13 e^{-x}+\frac13 e^{-x}+\frac13 e^{-x} \ge \dfrac4{\sqrt[4]{27}}$ by AM-GM, with equality only when $3e^{4x}=1$. |
H: Step in Proof of Cardinality of Product of two Groups
Let $A,B < G$ be two subgroups of some group $G$, then I have a question on the proof of the following:
$$
|AB| = \frac{|A||B|}{|A \cap B|}.
$$
Proof: Let $D = A \cap B$, arrange $A$ and $B$ in left cosets and right cosets regarding $D$ (which is also a subgroup)
$$
A = D \cup s_2 D \cup s_3 D \cup \ldots \cup s_n D, \quad
B = D \cup D t_2 \cup D t_3 \cup \ldots \cup D t_m
$$
Now the fact that $\frac{|A||B|}{|D|}$ are distinct results from the following equations:
If
$$
s_{\alpha}D D t_{\beta} = s_{\alpha_1} D D t_{\beta_1}
$$
then
$$
s_{\alpha_1}^{-1} s_{\alpha} D = D t_{\beta} t_{\beta}^{-1}
$$
As the first member of the last equation represents an element of $A$ while the second member represents an element of $B$, it results that the last equation implies $\alpha_1 = \alpha$ and $\beta_1 = \beta$.
The last step is not clear to me, why does $\alpha_1 = \alpha$ and $\beta_1 = \beta$ follows? Is it the case that if $aD = Db$ for $a \in A$ and $b \in B$ that $a=b=1$ follows? I don't believe this and cannot prove it either? Could someone please explain the last step?
AI: It is not true in general that $aD = Db$, $a \in A$, $b \in B$, implies $a = b$. In fact $aD = Db$ is true for any choice of elements $a, b \in D$. What's being used here is the fact that the $s_\alpha$ and $t_\beta$ are a set of representatives for the cosets. This means if $s_\alpha \neq s_{\alpha_1}$ then $s_\alpha D \neq s_{\alpha_1}D$. Turning this around, if $s_\alpha D = s_{\alpha_1}D$ then we must have $s_\alpha = s_{\alpha_1}$.
Now, you know that $s_{\alpha_1}, s_\alpha \in A$ and $D \subseteq A$ so $s_{\alpha_1}^{-1}s_\alpha D \subseteq A$. Similarly, $Dt_\beta t_{\beta_1}^{-1} \subseteq B$. As these are equal you get $s_{\alpha_1}^{-1}s_\alpha D \subseteq A \cap B = D$ and hence $s_{\alpha_1}^{-1}s_\alpha D = D$. But this can be rewritten as $s_{\alpha_1}D = s_\alpha D$, therefore $s_{\alpha_1} = s_\alpha$.
A similar argument works for the $t_\beta$. |
H: Trying to understand matrix image
On my linear-algebra lecture, we were given the definition of the image of a matrix $A \in \mathbb{K}^{m,n}$ as follows:
$$ \mathrm{im} A = \{\vec y \in \mathbb{K} ^{m}: \exists \vec x \in \mathbb{K}^n . \vec y = A\vec x\}$$
I just can't wrap my head around this one. Could someone please tell me a less formal definition of an image, what exactly does it do, and how can I find it's base for a given matrix? Also, just by the way, how does matrix' image compare to a function's image? Maybe this will help me understand...
Thanks!
AI: Each matrix $A\in\Bbb K^{m,n}$ defines a function (specifically, a linear transformation)
$$T_A:\Bbb K^n\to\Bbb K^m:\vec x\mapsto A\vec x\;;$$
$\operatorname{im}A$ is just the range of this function $T_A$. You could just as well describe it as
$$\operatorname{im}A=\{A\vec x\in\Bbb K^m:\vec x\in\Bbb K^n\}\;,$$
just as you define the range of any function $f:X\to Y$ to be $\{f(x)\in Y:x\in X\}$. The image of $A$ is the column space of $A$, i.e., the subspace of $\Bbb K^n$ spanned by the columns of $A$; to find a basis for it, just find a maximal linearly independent subset of the columns of $A$. |
H: If $f:[0,1] \to \mathbb{R}$ is continuous and $\int^{x}_{0} f = \int^{1}_xf,$ then $f(x) = 0, \forall x\in [0,1].$
If $f:[0,1] \to \mathbb{R}$ is continuous and $\int^{x}_{0} f = \int^{1}_xf,$ then $f(x) = 0, \forall x\in [0,1].$
May I verify if my proof is valid? Thank you:)
Proof: $\int^{c}_{0} f = \int^{1}_cf \implies\int^{1}_{0}f=2\int^{c}_{0}f=2\int^{d}_{0}f \implies \int^{d}_{c}f=0, \forall c,d \in [0,1]. $
Suppose $\exists c \in (0,1)$ such that $f(c)>0.$ Since $f$ is continuous at $c,$ given $\epsilon = \frac{f(c)}{2},$ $\exists \delta >0$ such that $f(x)>\frac{f(c)}{2}, \forall x\in (c-\delta,c+\delta)\subseteq [0,1].$
It follows that $\int^{c+\frac{\delta}{2}}_{c}f \geq \frac{f(c)}{4}\delta>0$ (Contradiction).
If $f(0)>0, $ given $\epsilon = \frac{f(0)}{2},$ $\exists \delta >0$ such that $f(x)>\frac{f(c)}{2}, \forall x\in [0,0+\delta)\subseteq [0,1].$
It follows that $\int^{0+\frac{\delta}{2}}_{0}f \geq \frac{f(0)}{4}\delta>0$ (Contradiction). Similary, $f(1)=0$ results in contradiction.
AI: I think you can be satisfied with the conclusion that $\int_\alpha^\beta f(x)\, dx=0$ for all $\alpha$ and $\beta$. As you noticed, it is standard to deduce that $f=0$ everywhere.
Another proof goes as follows. Let $F$ be a primitive of $f$ (which exists since $f$ is continuous). Then $F(x)-F(0)=\int_0^x f(t)\, dt$ and $F(1)-F(x)=\int_x^1 f(t)\, dt$. Hence $F'(x)=-F'(x)$ for every $x$, and therefore $2f(x)=0$ for every $x$. |
H: PSD matrices properties
If I have a matrix $X \in R^{n \times n} $ and an index set $ I \subseteq \{1,\dots,n\} $,
Is $X_I$ also positive-semidefinite $\forall \ \ I $? Why ?
$X_I $ is the submatrix that is formed by choosing all rows and columns from index-set $I $
Edit : How would you prove that the determinant is product of eigenvalues for $X$ ?
AI: (I presume your question has a statement error). $\mathbf{X}_I$ should be the Matrix formed by removing the rows and columns from index-set $I$. Yes, the new matrix $\mathbf{X}_I$ is also positive semi-definite. Proof: (already shown in comments). |
H: Are these isomorphic $\mathbb{Z}_{2}\times\mathbb{Z}_{3}$ and $\mathbb{Z}_{9}^{*}$
Is $\mathbb{Z}_{2}\times\mathbb{Z}_{3}$ isomorphic to $\mathbb{Z}_{9}^{*}$
both have orders 6
both have elements with orders 1,2,3,6 (1 element of order 1, 2 elements of order 3, 1 element of order 2 and 2 elements of order 6)
Both are cyclic thus Abelian
Can I assume then that they are isomorphic? Or a specific isomorphism must be constructed?
AI: Yes, the two groups are isomorphic. And you are almost there in proving this.
Note that $\mathbb Z_2\times \mathbb Z_3 = \mathbb Z_6$, since $\gcd(2, 3) = 1$.
And since the order of $\mathbb Z^*_9 = 6$ and is cyclic, we know that $\mathbb Z^*_9 \cong \mathbb Z_6$.
There is no need to construct an explicit isomorphism here, though you need to (implicitly or explicitly) invoke the following standard fact about finite cyclic groups:
Every finite cyclic group of order $n$ is isomorphic to $\mathbb Z_n$, the group of integers under addition, modulo $n$. |
H: Let $E \subset R$ and let $f$ be a real-valued function on $E$ that is continuous at $p \in E$
Let $E \subset R$ and let $f$ be a real-valued function on $E$ that is continuous at $p$ in $E$. If $f(p) > 0$, prove that there exists an $\alpha > 0$ and a $\delta > 0$ such that $f(x) \ge \alpha$ for all $x \in N_\delta(p) \cap E$
AI: Let $A:=f(p)>0$. $f$ is continuous at $p$ so for all $\epsilon>0,$ there exists a $\delta>0$ such that $|f(x)-f(p)|<\epsilon$ whenever $x\in N_{\delta}(p)\cap E$. In particular, pick $\epsilon=f(p)-\alpha$, where $\alpha$ is any number that satisfies $0<\alpha<f(p)$ (such a number exists by the Archimedean property of the real numbers). For example, let $\alpha=f(p)/2$. Then there exists a $\delta$ as before such that
$|f(x)-f(p)|<f(p)-\alpha$, whenever $x\in N_{\delta}(p)\cap E$.
The inequality resolves to $\alpha-f(p)<f(x)-f(p)<f(p)-\alpha$. In particular, the first inequality gives you $\alpha<f(x)$. |
H: Terminology: Groups, rings, fields, etc.
Groups, semigroups, fields, rings, integral domains, vector spaces, R-modules... these are all approximately the same sort of "stuff", but each one refers to a slightly different combination of required properties. Is there a general term that collectively refers to these types of "stuff"? Also, which branch of mathematics do such objects live in (broadly speaking)?
AI: All of the objects you listed could be collectively called "algebraic objects" and their theories individually are usually considered to be under the purview of abstract algebra. But of course nearly every other field uses them too!
The study of algebraic objects in the abstract is called universal algebra. The generic name for algebraic objects in universal algebra is that they are all "algebras" of different types. (This is not to be confused with the other very widespread term for an associative ring with a scalar operation over a field.)
They go so far as to define $n$-ary operations on sets. Many different theorems (for example, the basic isomorphism theorems) which are usually proven separately for all those objects are proven all at once for large classes of algebraic objects in the eyes of universal algebra. |
H: Let $m$ and $n$ be integers in the ring of integers. Show that $m\mathbb Z$ contains $n\mathbb Z$ if and only if $m$ divides $n$
I am working on the problem:
Let $m$ and $n$ be integers in the ring of integers. Show that $m\mathbb Z$ contains $n\mathbb Z$ if and only if $m$ divides $n$.
It's an if and only if proof so two directions to show. Start with the fact $n\mathbb Z$ is a subset of $m\mathbb Z$ and want to show $n=mx$ for some $x \in \mathbb Z$. Help to proceed?
AI: If $n\mathbb{Z}\subset m\mathbb{Z}$ then $n\in m\mathbb{Z}$ so $n=md$
for some $d\in\mathbb{Z}$.
If $n=md$ for some $d\in\mathbb{Z}$ and $k\in n\mathbb{Z}$ then
$k=ne=mde\in m\mathbb{Z}$ for some $e\in\mathbb{Z}$, so $n\mathbb{Z}\subset m\mathbb{Z}$. |
H: Rings whose elements are partitioned between units and zero-divisors.
In $\mathbb{Z}_n$ the elements are fully partitioned between the units and the zero-divisors. I believe this is the case, am I correct?
Now, I take it this does not hold true in general, there may be rings with elements that are neither units nor zero divisors?
AI: You're correct in this case, and more generally elements in Artinian rings are either units or are zero divisors. It's not hard to prove: basically you can show that if $x$ isn't a zero divisor, then then chain $xR\supseteq x^2R\supseteq\dots$ has to stabilize, whence there will be an $r$ such that $x^n=x^{n+1}r$. Rewriting that, you get $x^n(xr-1)=0$. If $x$ isn't a zero divisor, then the $x^n$ can be cancelled, resulting in $xr=1$, so that $x$ is a unit.
Any commutative domain which isn't a field has LOTS of nonunits which aren't zero divisors. So for example $\Bbb Z$ has two units $\{\pm1\}$, zero, and the rest of the elements are not zero divisors. |
H: How much Category theory one must learn?
I have learnt very basic category theory (up to Yoneda lemma from Hungerford's Algebra text). My question is how much category theory should every Mathematics student who is not planning to specialize in that area learn ?
I am not sure which area of Mathematics I would like to specialize in, though I guess it would have to be one of Topology or Geometry or Mathematical Physics. Is it a good idea for me to read Categories for the Working Mathematician by Saunders Mac Lane from cover to cover or would so much of category theory be useful only to specialists in Algebra ?
I have some knowledge of basic algebra (Groups, Rings , Fields, Galois Theory, Commutative algebra & basic Homological algebra) from the book by Lang, Manifolds & Differential Geometry (differential forms, de Rham cohomology, connections & curvature on principal bundles) though almost no knowledge of Algebraic topology. Is it a good time to learn more category theory ?
AI: My personal opinion is that category theory is like set theory; it's a language, everyone should know the basics, and everything in the "basics" is essentially trivial. Here "basics" for set theory means subsets, products, power sets, and identities like $f^{-1}(\bigcap A) = \bigcap f^{-1}(A)$. For category theory, I think "basics" means:
categories, functors, natural transformations;
duality;
basic constructions like product categories, comma categories (at least over- and under-categories), and functor categories;
universal properties, representable functors, and the Yoneda lemma/embedding;
limits and colimits;
adjunctions.
Basically the first 4 chapters of Mac Lane (ignoring the stuff about graphs and foundations). One could probably add "abelian categories" to that list, but I think a homological algebra text is a better place to learn that. |
H: approximating a measurable function using simple functions
I would like to understand the proof that shows that for any measurable function, we can approximate it with a sequence of simple functions. The proof is presented in (among others):
https://www.math.ucdavis.edu/~hunter/measure_theory/measure_notes_ch3.pdf
However, I am having some difficulty understanding why $\phi_n \to f$ in the ntoation of that pdf. I understand that since $f$ is measurable, we can take its pre-image: $E_{n,k} = f^{-1}(I_{n,k})$ and define indicator functions on that pre-image, $1_{E_{n,k}}$. I don't see, however, how $\phi_n =\sum_{k=0}^{2^n-1}\frac{k}{2^n}1_{E_{n,k}} \to f$.
This is not homework; just attempting to understand this.
AI: Fix $x\in X$. If $f(x)=\infty$, then $\phi_n(x)=2^n$ for all $n$, hence there is convergence.
Assume that $f(x)$ is finite. Then $f(x)\leqslant 2^n$ for $n$ large enough, hence for these $n$, we have $x\in E_{n,k_n}$ for some $0\leqslant k_n\leqslant 2^{2n}-1$. This means that $f(x)\in (k_n2^{—n},(k_n+1)2^{—n})$. Since $f_n(x)=k_n2^{-n}$, we have $0\leqslant f(x)-f_n(x)\leqslant 2^{—n}$. |
H: linear dependence on $\mathbb R^n$
$S_1$ and $S_2$ finite sets on $\mathbb{R}^n$, $S_1$ is a subset of $S_2$, $(S_1\neq S_2)$.
If $S_2$ is linearly dependent, so:
$S_1$ could be linearly dependent?
$S_1$ Could be linearly independent?
How can i answer this, and show some examples?
AI: You can have both of them. For example suppose $S_{2}=\lbrace e_{1} , \ldots, e_{n} , e_{1} + e_{2} \rbrace$. If you have $S_{1}= \lbrace e_{1} , \ldots , e_{n} \rbrace$, it is linear independent. On the other hand you can have $S_{1} = \lbrace e_{1} , e_{2} , e_{1} + e_{2} \rbrace$. So there is no generic answer. |
H: Stuck on equivalence relation question
I have been stuck on this question for a while. I was wondering for a set $A={1,2,3,4,5,6}$, given that its distinct equivalence classes are $\{1,4,5\},\{2,6\},\{3\}$, what is the equivalence relation R on A?
I have tried everything, such as powers, modulo, parity arguments, sum, difference, product, etc.
Any help would be appreciated.
AI: What you need to do is simply define $R$ by explicitly stating which ordered pairs of numbers belong to $R$. There's no need to discover some clever operation(s) on elements that define of a binary relation between elements.
All we need to know is:
Every element in an equivalence class is related to itself and all other elements in that class, and is NOT related to any element outside the class.
Because $R$ is an equivalence relation, we need to also ensure that we include pairs in $R$ satisfying:
reflexivity $(a, a) \forall a\in A$,
symmetry: $(a, b) \in R \implies (b, a) \in R$,
transitivity: $(a, b), (b, c)\in R \implies (a, c)\in R$
We simply list those ordered pairs:
$$R = \{\underbrace{(1, 1), (1, 4), (1, 5), (4, 1), (4, 4), (4, 5), (5, 1), (5, 4), (5, 5),}_{\text{first class}} \\\underbrace{(2, 2), (2, 6), (6, 2), (6, 6),}_{\text{second class}} \underbrace{(3, 3)}_{3^\text{rd}\text{class}}\}$$ |
H: Prove that the $x$-axis in $\Bbb R^2$ with the Euclidean metric is closed
I want to show that the $x$-axis is closed.
Below is my attempt - I would appreciate any tips on to improve my proof or corrections:
Let $(X,d)$ be a metric space with the usual metric.
Want to Show: $\{(x,y) | x ∈ \Bbb R, y = 0\}$ is closed
Claim: $\{(x,y) | x ∈ \Bbb R, y ≠ 0\}$ is open
Proof: Let $\{(x,y) | x ∈ \Bbb R, y ≠ 0\} = C$.
Let $z$ be an arbitrary $(a,b) ∈ C$ and let $$ε = \min\{d(z,(0,y)), d(z,(x,0))\}.$$
Then for any $p ∈ $B_ε$(z)$, $B(p) ∈ \{(x,y) | x ∈ \Bbb R, y ≠ 0\}$ with radius $ε/2$.
AI: Your definition of $\epsilon$ has problems: what are $x$ and $y$? They’ve not been defined up to this point, so $\epsilon$ isn’t defined. The only specific coordinates that are available at this point for you to work with are $a$ and $b$ (and specific real numbers like $0$).
HINT: If $z=\langle a,b\rangle\in C$, try letting $\epsilon=|b|$. |
H: Choosing a congress
Five scientists each have to choose a venue to attend out of 8 possible venues. Their choice is independent and each venue has the same chance of being chosen.
What is the probability that each of the 5 scientists choose a different venue?
What is the probability that 4 out of 8 venues will not be chosen?
For the first one, it seems:
$$
\text{n} = \frac{8!}{3!}=8*7*6*5*4=6720\\
\text{N} = 8^5 = 32768\\
P(\text{choosing different venues}) = \frac{6720}{32768} =0.20507... = 20.5 \%
$$
The second one seems trickier. They are essentially saying: 4 venues are taken by 4 scientists. And the last guy should pick one of those. What is the probability of this?
A = 4 scientists pick 4 different venues
B = remaining scientist picks one of those 4
$P(A\cup B) = P(A) + P(B)$ because of the independence. So
$$
\frac{\dfrac{8!}{4!}+\dfrac{4!}{1!}}{32768}
$$
My gut feeling says that I’m off on the second question. Any hints or tips?
AI: Let us suppose we decide to use the sample space of $8^5$ that you used for the first part. Then for the numerator we want to find the number of choices in which exactly $4$ venues are chosen.
Which $4$? They can be chosen in $\binom{8}{4}$ ways. For every choice of $4$ venues, there are $\binom{4}{1}$ ways to choose the one that will be attended by $2$ scientists. Which $2$ scientists go to this popular venue? They can be chosen in $\binom{5}{2}$ ways. And once this is done, the remaining $3$ scientists can be distributed among the $3$ remaining chosen venues in $3!$ ways, for a total of $\binom{8}{4}\binom{4}{1}\binom{5}{2}3!$.
Or else we could choose the popular venue first, $\binom{8}{1}$ ways, and the scientists that go there, $\binom{5}{2}$ ways. That leaves $7$ venues, of which we choose $3$, and then multiply by $3!$. Same number. |
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