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H: Proving that UB is transitive follows from B is transitive
I'm really stuck as to how to prove that if B is a transitive set, it must be the case that the union of B is transitive as well. I guess I'm just a bit shaky on the concepts of proofs overall, but even though I can intuit that it should be the case, I'm not sure how I would go about proving it. I think maybe I've just looked at it for too long. Any help would be appreciated.
AI: HINT: If $B$ is transitive, then $x\in B\implies x\subseteq B$. Suppose that $y\in\bigcup B$. Then by definition $y\in x$ for some $x\in B$. Use the transitivity of $B$ to conclude that $y\subseteq\bigcup B$. |
H: What does the symbol $\subset\subset$ mean?
In some texts (mainly complex analysis or harmonic analysis) I sometimes see the following double subset symbol $\subset\subset$ for inclusion relation of two regions, e.g., $\Omega$ and $\Omega'$ are two regions in $\mathbb{C}$ such that $\Omega \subset\subset \Omega'$. I never figured out what it means exactly; I always interpreted it as the closure $\overline{\Omega}$ is contained in $\Omega'$ (so that some nasty boundary effects can be avoided). Is that right? Or does $\subset\subset$ mean some other kind of inclusion? Thanks.
AI: It usually (the only meaning I've come across yet) means that $\Omega$ is relatively compact in $\Omega'$, so the closure of $\Omega$ is compact and contained in $\Omega'$. |
H: Logic/Quantifiers and Proofs/counterexample
How do I negate the following statement?
Also please help me with this exercise:
AI: First question :
$$\neg(c): (\exists x,y \in \Bbb R)(\exists z \in \Bbb Z)(\forall a \in \Bbb R) \quad a \geq \frac{x+y}{2} \vee a > z $$
Second question:
Let $x \in U$ where $U$ is the universal set indicated above. Then:
$$x \in (A\backslash B) \backslash C \iff x \in A \wedge x \notin B \wedge x \notin C $$ Let us call $(1)$ the right proposition of this equivalence.
$$x \in A\backslash(B \backslash C) \iff x \in A \wedge x \notin B \backslash C$$ Let us call $(2)$ the right proposition of this equivalence.
if $x \in A$ and $x \notin B $ and $x \in C$ then $(2)$ holds but not $(1)$
This allow us to construct a counter-example : let $A=C=\{1\}$ and $B=\{2\}$
Then $(A\backslash B) \backslash C= \emptyset$ and $A \backslash (B \backslash C)=\{1\}$ |
H: Harmonic Oscillator Homework, Require Verification
A particle of unit mass movies on a straight line under a force having potential energy $$V(x)=\frac{bx^3}{x^4 + a^4}$$ where a,b are positive constants. Find the period of small oscillations about the position of stable equilibrium.
So, I differentiated $V(x)$ once, found the equilibrium points which were $x_1=0, x_2=a\sqrt[4]{3}, x_3=-a\sqrt[4]{3}$. By finding $V''(x)$, I verified that the STABLE equilibrium point was $c=x_3=-a\sqrt[4]{3}$ since $V''(c)=\frac{12b}{a^3}>0$ With the value for $V''(c)$ obtained I can finally wok out the period of small oscillations about the position of stable equilibrium using the formula $$2\pi \sqrt[2]\frac{m}{V''(c)}$$, where m is the mass of the particle and c is the abscissa of the stable equilibrium point. So I ended up with the period being $\frac{\pi}{3} \sqrt[2]{3a^3}$
Can someone check if this is correct? Didn't' have the patientce to write out $V'(x)$ fully. Does anyone know a good programme for that as wolfram doesn't let me insert constants like a,b in only actually numbers.
AI: $V'(x) = -\frac{b\,{x}^{2}\,\left( {x}^{4}-3\,{a}^{4}\right) }{{\left( {x}^{4}+{a}^{4}\right) }^{2}}$. The stable equilibrium is at $\hat{x}=-\sqrt[4]{3}a$, and
$V''(x) = \frac{2\,b\,x\,\left( {x}^{8}-12\,{a}^{4}\,{x}^{4}+3\,{a}^{8}\right) }{{\left( {x}^{4}+{a}^{4}\right) }^{3}}$, with
$V''(\hat{x}) = \frac{3 \sqrt[4]{3} b}{ 4 a^3}$.
The linearized system around $\hat{x}$ becomes
$\ddot{\delta} = - V''(\hat{x}) \delta$, hence the angular frequency is $\omega_0 = \sqrt{V''(\hat{x})} = \frac{3^\frac{5}{8}}{2}\sqrt{\frac{b}{a^3}}$, and so the period is $T = \frac{2 \pi}{\omega_0} = \frac{4 \pi}{3^\frac{5}{8}} \sqrt{\frac{a^3}{b}}$.
(Assuming I have made no mistakes, which is a big presumption.) |
H: How to write product of three sums
I know that by the binomial theorem,
$\displaystyle \left(\sum_{n=0}^\infty a_nx^n \right)\left(\sum_{n=0}^\infty b_nx^n \right)= \sum_{n=0}^\infty \left(\sum_{k=0}^n a_kb_{n-k} x^n\right)$. How do I write the product of three sums
$\displaystyle \left(\sum_{n=0}^\infty a_nx^n \right)\left(\sum_{n=0}^\infty b_nx^n \right)\left(\sum_{n=0}^\infty c_nx^n \right)$?
Thank you!
AI: The first one isn’t the binomial theorem; it’s just an instance of the Cauchy product.
$$\large\left(\sum_{n\ge 0}a_nx^n\right)\left(\sum_{n\ge 0}b_nx^n\right)\left(\sum_{n\ge 0}c_nx^n\right)=\sum_{n\ge 0}\sum_{{i+j+k=n}\atop{0\le i,j,k\le n}}a_ib_jc_kx^n$$ |
H: How to verify the set of two vectors is a basis for the plane?
Verify that B={(3,2,0),(0,2,3)} is a basis for the plane 2x[1]-3x[2]+2x[3]=0 in R3.
How to solve out this question?
And I have a question about this question. I thought dim(Rn)=n.So in this question the basis should have 3 vectors rather than two. So my question is how two vectors can be a basis in R3?
Thank you very much!
AI: The plane is 2 dimensional sitting inside of $\mathbb{R}^3$ so you only need two vectors for a basis for the plane. To prove it is a basis you must show that it is a spanning set and it is linearly independent.
To show it is linearly independent suppose:
$$a(3,2,0) + b(0,2,3) = (0,0,0)$$
Then comparing the entries we see that $3a + 0b = 0$ so $a = 0$ and similarly $0a + 3b = 0$ so $b=0.$ Thus the set is linearly independent. Since the vectors lie in the plan (easy check) and they are linearly independent, and since the plane is 2-dimensional it follows that they are a basis. |
H: System $\dot x=Ax-|x|^2x$ has one limit ccle
I have a system in $\mathbb R^2$, namely $\dot x=Ax-|x|^2x$ where $A$ is constant real matrix with complex eigenvalues $p+-iq (q>0)$.
I want to show that there exists at least one limit cycle for $p>0$ and none for $q<0$ using Dulacs Theorem and Poincare-Bendixson teorem, as seen here for exmaple http://www2.warwick.ac.uk/fac/sci/maths/people/staff/colin_sparrow/qtode/kitsonnotes.pdf
I cannot find an argument that uses this two theorems to conclude that such cycle must exist, may you can help me with that?
AI: Hint: Take $V(x_1,x_2)=x_1^2+x_2^2$ and show that the infinity is always a repeller here. If $p>0$ conclude that the origin is unstable. Hence show that there is a positive invariant set that does not contain equilibria. Hence, by Poincare-Bendixson theorem... |
H: Is $G=\{A\in M_2(\mathbb{R}): A^2=I_2\}$ a group?
I am investigating whether $G=\{A\in M_2(\mathbb{R}): A^2=I_2\}$ is a group. It is clear that associativity and existence of identity and inverse are satisfied. So it only remains to determine wether $G$ is closed under matrix multiplication. Then, we consider $A,B\in G$ so that $$(AB)^2=ABAB.$$ After this point I couldn't think of a way to prove (or disprove) that $ABAB$ is $I_2$, even though I tried to use to the facts that $A=A^{-1}$, $B=B^{-1}$ and $B^{-1}A^{-1}=(AB)^{-1}.$
I am mostly looking for hints.
Thanks in advance.
AI: $G$ is NOT a group. In fact we have
$$
A=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right), \quad B=\left(\begin{array}{cc}0&1\\1&0\end{array}\right) \in G
$$
since
\begin{eqnarray}
A&=&\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)=I_2,\\
B&=&\left(\begin{array}{cc}0&1\\1&0\end{array}\right)\left(\begin{array}{cc}0&1\\1&0\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)=I_2.
\end{eqnarray}
But
$$
P:=AB=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\left(\begin{array}{cc}0&1\\1&0\end{array}\right)=\left(\begin{array}{cc}0&1\\-1&0\end{array}\right) \notin G,
$$
because we have
$$
P^2=\left(\begin{array}{cc}0&1\\-1&0\end{array}\right)\left(\begin{array}{cc}0&1\\-1&0\end{array}\right)=\left(\begin{array}{cc}-1&0\\0&-1\end{array}\right)=-I_2.
$$ |
H: Probability of packages
Suppose mail is delivered 6 days each week. Someone sent you ten packages, all of which are scheduled to arrive this week. But you don't know what day any of the packages will arrive. What is the expected number of packages that you receive on a particular weekday?
AI: Assuming the packages arrive at random, i.e., if any of the packages is equally-likely to arrive any of the 6 days then any package has a probability $P=1/6$ of arriving any given day. This means that , for $X$ =number of packages arriving per day, we have: $$E(X)=\frac{1}{6}(1)+\frac{1}{6}(2)+...+\frac{1}{6}(10)=\frac {1}{6}=(55/6)=8+1/6 $$ packages.
If the packages do not arrive randomly, then we must have a distribution function that tells us the probability of package #k ; $k=1,2,..,10$ arriving in day j $; j=1,2,..6.$ , we do a sum of a similar form as above, but substituting $\frac{1}{6}$ for $P_{kj}$ , i.e., the probability of pacage #k arriving in day j, for all pairs $(k,j)$ |
H: Continuous functions on Closed/Bounded sets
Ok so I know that if K is closed and bounded, f(K) is also closed and bounded, I reckon both statements are false but not entirely sure, I just cant seem to come up with counter examples. Thanks
AI: A counterexample for closed: consider the set $S:=(x,1/x)$ in $\mathbb R^2$ , and consider $f(x):=\pi_2(x)$ which is the projection to the second component (into $\mathbb R$).
Then $\pi_2(S)=\mathbb R-{0}$ , which is open, since it is the complement in $\mathbb R$ of the closed set {$0$}.
For the second case, consider the function $f(x)=1/x$ on the bounded set $(0,1)$. $f$ is continuous there, but grows without bound.
And, for completeness, here is an example of how the continuous image of an open set is not necessarily open: take the open interval $(-1,1)$ ( take it, please!) , and the continuous function $f(x)=x^2$. Then {$f(-1,1)$}=$[0,1)$ , which is not open. |
H: Solving for $y$ in $y= 14x + 1000 = y= 16x + 800$
Given $y= 14x + 1000 = y= 16x + 800$, solve for $y$.
I think I have it:
$x= 100$. Subbing that in would leave me with:
$y = 14x + 1000$
$y = 14 \cdot 100 + 1000$
$y = 2400$
$y = 16x + 800$
$y = 16 \cdot 100 + 800$
$y = 2400$
the lines break even at $y= 2400$ and $x= 100$
Correct me if i am wrong
AI: Hint: You know that $y=14x+1000$ and you know that $y=16x+800.$ You want to either use elimination or substitution to solve for either $x$ or for $y$, and then back substitute to find the other variable.
An example is for two equations: $y=5x+2$ and $y=2x-1$. I'll proceed by the substitution method. Since we know both equations are equal to $y$, instead of writing $y$, we can write the other equation instead, since their values are equal to $y$. So we get:
$5x+2=2x-1\\\implies 3x=-3\\\implies x=-1$.
So now we know the $x$ value. We can substitute this into the equation (either equation). I'm going to choose to substitute $x=-1$ into the equation $y=2x-1$. This gives us $y=2(-1)-1=-2-1=-3$. So in my example, $x=-1$ and $y=-3$. The process for your question is similar. |
H: Exponential distribution problem
Suppose that waiting times for M103 buses are independent and exponentially distributed with the same parameter. On average, there are 4 buses per hour.
(a) What is the probability that there are no buses for 30 minutes?
This is clearly an exponential random variable with $\lambda$ = 4 per hr.
$P(X > .5) = 1 - P(X \leq .5) = 1 - F(X) = 1 - (1-e^{(-4) * .5}) = e^{-2} = .135$
(b) What is the probability that in 2 hours you see more than 6 buses?
This I'm not sure about. Does lambda vary here?
$P(Y > 6) = 1 -P(Y \leq 6) = 1 - \sum_{i=1}^6 (1-e^{- (\lambda/i) * 2})=1 - \sum_{i=1}^6 (1-e^{(- 4/i) * 2})$
(c) What is the probability that the fifth bus appears not earlier than in 1.5 hours?
$\lambda = 4/5 $ ; Every success (i.e. 5 buses) occurs 4/5 per hour.
$P(X>1.5)=1-P(X \leq 1.5)= 1 - (1-e^{(-4/5) * .1.5}) =e^{-1.2} = .301$
AI: Let $X$ be the number of buses during time period $t$. Since the waiting time is exponential with parameter $\lambda$, then $X$ has a Poisson distribution with parameter $\lambda t$:
$$
{\rm P}\{X=k\}=\frac{(\lambda t)^k}{k!}e^{-\lambda t}.
$$
$t=0.5,k=0$;
$t=2,k>6$;
$t=1.5,k<5$. |
H: Examples of groups of mapping over a set which is not a subset of the symmetric group.
We can find examples of sets $X$ for which there exists a group $G$ (with |G| > 1) under function composition which is a subset of $X^X$ but not a subset of the group $Sym(X)$. The catch is that the identity of $G$ may not be the identity function, i.e. identity of $Sym(X)$. One such example is as follows:
Let $H$ be a group with identity element $e$. Take $X = H \times H$ and $G = \{(x, y) \mapsto (ax, e) : a \in H\}$ with the map $(x, y) \mapsto (x, e)$ as the identity of $G$. Clearly $G$ is a group.
I have two questions now:
Are there other such examples?
Can we classify the sets $X$ for which such groups $G$ exist?
Edit: This is inspired from problem $1.4$ in Algebra by I. Martin Isaacs
AI: You gave a special case of the "only" example: let $X$ be any set and $X = X' \cup Y$ a decomposition of $X$ into disjoint, non-empty subsets. Pick a subgroup $K\leq \mathop{\mathrm{Sym}}(X')$, a function $a:Y\to X'$, and a subgroup $H\leq \mathop{\mathrm{Sym}}(a(Y))$ Then
$$\{f:X\to X\mid f|_{X'}\in K \text{ and, for some $h\in H$, } f|_Y = ha \text{ and } f|_{a(Y)} = h\}$$
is a group.
To see why there are no other examples, suppose $G\subseteq X^X$ is a group under function composition. Set $X' = e(X)$ and $Y = X - X'$ for $e$ the identity element of $G$. The equation $e = e\cdot e$, shows that $e(x) = x$ for all $x\in X'$. Define $a:Y\to X'$ by $a = e|_{Y}$. Using $g^{-1}$ to denote group-theoretic inverse for $g\in G$ (i.e., not inverse image), the identity $g^{-1}(g(x)) = g(g^{-1}(x)) = e(x) = x$ for $x\in X'$ shows that $g|_{X'}\in \mathop{\mathrm{Sym}}(X')$. Further, the identity $g(x) = e(g(x))$ for $x\in X$ shows that $g(X)\subseteq X'$. The same identity restricted to $Y$ shows that $g(Y) \subseteq a(Y)$. The subgroup $K\leq \mathop{\mathrm{Sym}}(X')$ is the image of the group homomorphism $G\to \mathrm{Sym}(X'):g\mapsto g|_{X'}$.
It remains to construct the subgroup $H\leq \mathop{\mathrm{Sym}}(a(Y))$, which we realize as the image of a homomorphism $\phi:G\to \mathop{\mathrm{Sym}}(a(Y))$. For $g\in G$, define $\phi(g): a(Y)\to a(Y)$ to be the permutation sending $a(y)$ to $g(y)$ for $y\in Y$ – this is well-defined because $g(y) = g(y')$ if and only if $e(y) = g^{-1}(g(y)) = g^{-1}(g(y')) = e(y')$. |
H: Use partial fractions to find the integral.
Find the integral using partial factions.
$$\int\frac{(2x^2+5x+3)}{(x-1)(x^2+4)}\,dx$$
So do I do...$$\frac{2x^2+5x+3}{(x-1)(x^2+4)}=\frac{A}{x-1} + \frac{Bx+C}{x^2+4},$$
then get
\begin{align*}
2x^2+5x+3 &= A(x^2+4)+(Bx+C)(x-1) \\
2x^2+5x+3 &= Ax^2+4A+Bx^2-Bx+Cx-C?
\end{align*}
AI: Note that we can simply add zero: $2x^2 = 2(x^2 + 4) - 8 = 2x^2 + 8 - 8$. Then
$$\frac{2x^2 + 5x + 3}{(x - 1)(x^2 + 4)} = \frac{2(x^2 + 4) - 8 + 5x + 3}{(x - 1)(x^2 + 4)} = \frac{2}{x - 1} + \frac{5x - 5}{(x - 1)(x^2 + 4)}.$$
Do you think you can integrate
$$\frac{2x^2 + 5x + 3}{(x - 1)(x^2 + 4)}$$
now?
Hint:
$$\frac{5x - 5}{(x - 1)(x^2 + 4)} = \frac{5(x - 1)}{(x - 1)(x^2 + 4)} = \frac{5}{x^2 + 4}$$
and
$$\int \frac{1}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C, \quad a \ne 0.$$ |
H: Reversing a conditional probability
If I'm given a P(X|Y) table and P(Y), how can I find P(Y|X)?
I understand that $P(Y|X)=\frac{P(X|Y)P(Y)}{P(X)}$ but how do i find P(X)?
Furthermore, If i'm told that random variable X is given a value, does this affect P(Y)?
AI: Assume that you are given a decomposition $\mathscr D$ of your probability space $\Omega$, $\mathscr D=\{D_1,\ldots,D_k\}$, i.e., $D_i\cap D_j=\emptyset$ for $i\neq j$ and $D_1\cup\ldots\cup D_k=\Omega$. Then the formula for total probability for the event $A$ is given by
$$
{\rm P}\{A\}=\sum_i{\rm P}\{A\mid D_i\}{\rm P}\{D_i\}.
$$
To find ${\rm P}\{D_i\mid A\}$ you just go
$$
{\rm P}\{D_i\mid A\}=\frac{{\rm P}\{A\mid D_i\}{\rm P}\{D_i\}}{{\rm P}\{A\}}=\frac{{\rm P}\{A\mid D_i\}{\rm P}\{D_i\}}{\sum_j{\rm P}\{A\mid D_j\}{\rm P}\{D_j\}},
$$
which is called the Bayes formula.
The first is often referred as a priory probability (before the experiment), the second as a posteriori probability (after the experiment, you fix your probabilities of $D_j$ based on the information of the event $A$). |
H: Frobenius Norm Triangle Inequality
How can I go about proving the triangle inequality holds for the Frobenius norm?
I worked through $\|A+B\|_F \le \|A\|_F + \|B\|_F$ and was not able to make it work =/.
AI: You can use the fact that it's a norm from inner product :
$$\langle A,B \rangle = \text{Trace}(A^TB)$$
Then:
$$\|A+B\|^2=\|A\|^2+\|B\|^2+2\langle A,B\rangle$$
using Cauchy-Schwarz inequalitie we have : $$\langle A,B\rangle \leq \|A\| \|B\|$$
and this gives:
$$\|A+B\|^2 \leq \|A\|^2+\|B\|^2+2 \|A\| \|B\|= (\|A\| + \|B\|)^2$$ |
H: From 1-1000, choosing at random, what is the probability that number is prime or composite with a prime factor p $\leq$ 29?
An integer $k \in \{1,2, \dots, 999, 1000\}$ is selected at random. What is the probability that $k$ is a prime number or a composite number with a prime factor $p\leq29$?
AI: It is much easier to count the numbers that are not in our set. These are $1$ and non-primes whose prime factors are all $\gt 29$. There is only $1$ such number, namely $31^2$. So the reuired probability is $\frac{998}{1000}$.
Remark: Presumably the person who designed the problem expected us to forget about $1$, and to decide that the answer is $\frac{999}{1000}$. And then he/she would have the pleasure of marking our answer as wrong. |
H: A subgroup $H$ is normal in $G$ if and only if...
Prove that $H$ is a normal subgroup of $G$ if and only if $\{g^{-1}h^{-1}gh|g \in G, h\in H\}$ is a subgroup of $H$.
Attempt:
For the necessary condition, assume $H$ is normal in $G$. Then $g^{-1}h^{-1}g \in H$ for all $h \in H$. Then elements of the form $g^{-1}h^{-1}gh$ make a subgroup of $H$ by closure.
AI: Let $H$ be a normal subgroup of $G$. Then $g^{-1}h^{-1}gh=(g^{-1}h^{-1}g)h$ is the product of two elements of $H$ and thus is contained in $H$. Therefore, the set $\{g^{-1}h^{-1}gh|g\in G, h\in H\}$ is contained in $H$. We want to show that the subgroup generated by this set is a subgroup of $H$. To do this, it suffices to consider the product of two elements of the form $g^{-1}h^{-1}gh$, for $g\in G$ and $h\in H$: $$\begin{eqnarray*}
g_1^{-1}h_1^{-1}g_1h_1 g_2^{-1}h_2^{-1}g_2h_2&=&
(g_1^{-1}h_1^{-1}g_1)h_1 (g_2^{-1}h_2^{-1}g_2)h_2\\
\end{eqnarray*}$$
We have that $g_1^{-1}h_1^{-1}g_1$ and $g_2^{-1}h_2^{-1}g_2$ are elements of $H$ by normality, and of course $h_1$ and $h_2$ are elements of $H$, so $(g_1^{-1}h_1^{-1}g_1)h_1 (g_2^{-1}h_2^{-1}g_2)h_2\in H$. Therefore, $\langle g^{-1}h^{-1}gh|g\in G, h\in H\rangle$ is contained in $H$, so it is a subgroup of $H$.
Conversely, if the subgroup generated by the set $\{g^{-1}h^{-1}gh|g\in G, h\in H\}$ is a subgroup of $H$, then $g^{-1}h^{-1}g\in H$ by closure for every $g\in G, h\in H$, so $H$ is normal in $G$. |
H: Exactly $\frac{p-1}{2} - \phi(p-1)$ incongruent quadratic nonresidues mod p that are not primitive roots mod p
Prove that there are exactly $$\frac{p-1}{2} - \phi(p-1)$$ incongruent quadratic nonresidues modulo $p$ that are not primitive roots modulo $p$.
I have been looking at this problem for quite some time, but have not been able to make any headway. Can you assist?
AI: HINT:
First of all, if $a$ is a Quadratic Residue $\pmod p,$ there exists $x$ such that $x^2\equiv a\pmod p\implies a^{\frac{p-1}2}=x^{p-1}\equiv1\pmod p$
$\implies$ ord $_pa|\frac{p-1}2\implies a$ can not be a primitive roots i.e., primitive root must be quadratic nonresidue.
Now use this and this |
H: Prove that the limit is zero given integral is zero.
Let $f(t)$ be a real-valued function that is continuous, positive, and increasing on the real interval $(0,T)$. If
$$\int_0^T {\dfrac{f(s)}{s} ds} < \infty, $$
prove that $\lim_{t \to 0} f(t)= 0.$
I only have an intuitive idea of why this is true: if the conclusion does not hold, $f(s)/s$ will go to infinity as $s$ approaches zero. In this case, the integral cannot converge.
I would like a more rigorous solution.
AI: Your intuition is slightly off. Consider the integral $$\int_0^1 \frac{1}{s^p} ds$$ for $0<p<\infty$. It is well known that this integral diverges if $p\geq 1$, and converges if $p<1$. Yet $1/s^p$ is a function going to $\infty$ as $s\to 0$.
There is, however, a rigorous reason why the claim holds. Suppose, aiming for contradiction, that $\lim_{t\to 0} f(t) = c > 0$. Since $f$ is increasing on $(0,T)$, $f(t)\geq c$ for all $t\in(0,T)$. This implies that
$$ \infty > \int_0^T \frac{f(s)}{s}ds \geq \int_0^T \frac{c}{s}ds = c\int_0^T \frac{1}{s}ds = \infty.$$
This is a contradiction.
As an easy exercise, you should determine where I used the continuity of $f$. |
H: Help in a calculus question
I'm trying to solve this question:
In item (a)
I know $P'(200)=1/50$ and $P'(250)=8/625$, but how can I know the average rate of $P$ from $200\ in^3$ to $250\ in^3$? Sum and divide by two?
In item (b)
I think it's simple, because $V'=800/P^2$
How can I finish item (a)? the item (b) is correct?
Thanks
AI: (To summarize the comments...)
The average rate of change of a function $f$, from $x = a$ to $x = b$, is $$\dfrac{f(b)-f(a)}{b-a}$$
For part (b), you want $- \dfrac{800}{P^2}$. |
H: Norm of a principal ideal
I am trying to prove $N(\sigma(x)\mathcal O_k) = $$N(x\mathcal O_k) $ where $N(I)$ is the number $|\frac{\mathcal O_k}{I}|$ and $ x\mathcal O_k $ is the principal ideal generated by $x$. $\sigma \in Gal(K/Q)$
As $ \sigma(\mathcal O_k)= \mathcal O_k $.
Can I say $|\frac{\mathcal O_k}{x\mathcal O_k}| = |\frac{\sigma (\mathcal O_k)}{\sigma(x\mathcal O_k)}|$? and hence $N(\sigma(x)\mathcal O_k) = $$N(x\mathcal O_k) $.
I want to prove this with out considering $\mathbb Z$ basis.
AI: You probably want the reasoning to be a bit more explicit.
Can you find an explicit isomorphism ${\cal O}_K/x{\cal O}_K\cong{\cal O}_K/\sigma(x){\cal O}_K$?
Inspiration: already on hand you have an isomorphism $\sigma:K\to K$. |
H: Complex Integration poles real axis
In class my professor said that
$$
\int_{-\infty}^{\infty}\frac{e^{iax}}{x^2 - b^2}dx = -\frac{2\pi}{b}\sin(ab)
$$
where $a,b > 0$.
However, since the poles are on the real axis, isn't the integral equal to
$$
\pi i\sum_{\text{real axis}}\text{Res}(f(z); z_j)\mbox{?}
$$
If that is the case, the integral is
$$
-\frac{\pi}{b}\sin(ab).
$$
AI: Define
$$f(z):=\frac{e^{iaz}}{z^2-b^2}\;,\;a,b\in\Bbb R^+\;,$$
$$C_R:=[-R,-b-\epsilon]\cup\gamma_{-b,\epsilon}\cup[-b+\epsilon,b-\epsilon]\cup\gamma_{b\epsilon}\cup[b+\epsilon,R]\cup\Gamma_R$$
with
$$\gamma_{r,s}:=\{r+se^{it}\;;\;0\le t\le \pi\}\;,\;r,s\in\Bbb R^+\;,\;\Gamma_R:=\{Re^{it}\;;\;0\le t\le \pi\}\;,\;\;R\in\Bbb R^+$$
Since $\;f(z)\;$ analytic on and within $\;C_R\;$ , we get
$$\oint\limits_{C_R}f(z)\,dz=0$$
By the corollary to the lemma in the second answer here we get
$$\begin{align*}\lim_{\epsilon\to 0}\int\limits_{\gamma_{-b,\epsilon}}f(z)dz&=i\pi\,\text{Res}\,(f)_{z=-b}=i\pi\frac{e^{-iab}}{-2b}=-\frac{\pi i}{2b}e^{-iab}\\
\lim_{\epsilon\to 0}\int\limits_{\gamma_{b,\epsilon}}f(z)dz&=i\pi\,\text{Res}\,(f)_{z=b}=i\pi\frac{e^{iab}}{2b}=\frac{\pi i}{2b}e^{iab}\end{align*}$$
And applying Jordan's Lemma the integral on $\;\Gamma_R\;$ goes to zero when $\;R\to \infty\;$ , so in the end
$$0=\lim_{R\to\infty\,,\,\epsilon\to 0}\oint\limits_{C_R}f(z)dz=\int\limits_{-\infty}^\infty f(x)dx-\frac{\pi i}{2b}\left(e^{iab}-e^{-iab}\right)\implies$$
$$\implies \int\limits_{-\infty}^\infty f(x)dx=-\frac\pi b\frac{e^{iab}-e^{-iab}}{2i}=-\frac\pi b\sin(ab)$$
and I get the same as you did... |
H: proving of inequality $(n!)^2 \leq n^n\cdot n!\leq (2n!)$, where $n\in \mathbb{N}$
How can we prove the inequality $(n!)^2 \leq n^n\cdot n!\leq (2n!)$, where $n\in \mathbb{N}$
$\bf{My \; Try}$::
$1\leq n$
$2\leq n$
$3\leq n$
....
....
$n\leq n$
So $1\cdot 2 \cdot 3 ..n \leq n^n$
So $n!\leq n^n\Rightarrow (n!)^2 \leq n^n\cdot n!$
But did not understand how can we prove $n^n\cdot n! \leq (2n!)$
Help me
Thanks
AI: $$(2n)!=2n\cdot(2n-1)\cdot...\cdot(n+1)\cdot n!$$,
so $n\le 2n$, $n \le 2n-1$,... |
H: How find this inequality minimum $\sum\limits_{cyc}\sqrt{a^2+b^2+ab-2a-b+1}$
Let $0<a,b,c<1$, find this follow minimum
$$\sqrt{(a+b)^2-(a+1)(b+2)+3}+\sqrt{(b+c)^2-(b+1)(c+2)+3}+\sqrt{(c+a)^2-(c+1)(a+2)+3}$$
My try:
since
$$(a+b)^2-(a+1)(b+2)+3=a^2+b^2+2ab-ab-2a-b-2+3=a^2+b^2+ab-2a-b+1$$
so we only find this follow minimum
$$\sum_{cyc}\sqrt{a^2+b^2+ab-2a-b+1}$$
But I can't.Thank you
AI: $a^2+b^2+ab-2a-b+1=(b+\dfrac{a-1}{2})^2+\dfrac{3}{4}(a-1)^2$
$\sum\limits_{cyc}\sqrt{a^2+b^2+ab-2a-b+1} \ge \sqrt{(\sum\limits_{cyc} (b+\dfrac{a-1}{2})^2+\dfrac{3}{4}(\sum\limits_{cyc}(a-1))^2}=\sqrt{(\dfrac{3}{2}(\sum\limits_{cyc} a-1)^2+\dfrac{3}{4}(\sum\limits_{cyc} a-3)^2}=\sqrt{3((\sum\limits_{cyc} a)^2-3\sum\limits_{cyc} a+3)}\ge \dfrac{3}{2}$
when $\sum\limits_{cyc} a=\dfrac{3}{2}$ and $ a=b=c=\dfrac{1}{2}$ get min. |
H: Problems with axioms
I am currently exploring the idea of axiomatic truths. As of now, I have looked into axioms dealing with euclidean geometry and they are said to be self-evident truths. Each axiom in euclidean geometry seems very intuitive and easy to apply to geometry in many respects.
Do these axioms exist in disciplines such as Calculus and more advanced mathematics? I am curious to learn where axioms play roles in other areas of math.
I am mostly curious about the problems associated with these axioms. If there are such axioms that we base whole disciplines on, such as Calculus, does this imply that Calculus gives us truth. (of course with the assumption that the axioms are true).
Many subjects other than math can't offer truth the way math seems to offer it.
AI: As a direct answer, calculus has no axioms inherent to itself. Theorems of calculus derive from the axioms of the real, rational, integer, and natural number systems, as well as set theory.
Most disciplines of modern mathematics exhibit this sort of behavior, in which the discipline has no axioms inherent to itself. Modern disciplines of mathematics typically work under a unified axiomatic system, the most common one being Zermelo-Fraenkel set theory with the Axiom of Choice (ZFC). ZFC is powerful enough to encode our most frequently encountered structures, including our various number systems.
What, then, do the various disciplines of mathematics working under the axioms of ZFC study? The answer is the various structures that can be constructed using the axioms of ZFC. For example, single-variable calculus can be (very) broadly characterized as the study of real-valued functions with real domain. Such functions can be defined under ZFC, because the real number system can be axiomatized in ZFC.
Similarly, group theorists study algebraic structures called groups, which are defined in ZFC as a set (the building blocks of Zermelo-Fraenkel set theory) and a binary operation on the set, which is a function (which is also defined in ZFC using only sets) obeying certain properties. In a sense, one could interpret these properties as axioms for group theory, but they are actually merely definitions - the underlying axioms are those of ZFC.
The question of truth is a bit trickier to grasp, and there are many differing viewpoints. Complicating the question is that one could adopt an axiomatic system other than ZFC. If you take the perspective that true propositions are those that can be proved in your favorite axiomatic system, then it could easily be that a mathematician that believes in ZFC will disagree on the truth of a proposition with a mathematician that adopts a different system. And different axiomatic systems are not strange or unnatural - indeed, they constitute one of the major objects of study in mathematical logic.
One might also adopt the viewpoint that axioms are not self-evident truths, and therefore one should not believe or disbelieve in any particular axiomatic system. For many of these people, a collection of axioms is a list of meaningless rules to follow, and mathematics is the game of manipulation of symbols under these meaningless rules. (This is Hilbert's formalist perspective.) Truth, then, is a meaningless concept - all propositions are outcomes of some game with some particular rules.
There are other perspectives on truth, many of which I am not familiar with. But this should convince you that the concept of "mathematical truth" is more nuanced than you are probably aware, and really varies from person to person.
As for my personal opinion: I lean toward formalism, so to me it is meaningless to talk about whether calculus "gives me truth." Being inclined toward analysis and topology, I view calculus as an incredibly important and useful tool that agrees with my intuitive understanding of the world, but I view the underlying axioms as just a list of rules I am allowed to play with, without regard to whether I believe in their "truth." |
H: Transcendence of $\sqrt{\pi}$
So it is known that $\pi$ is transcendental. With a little thought I was able to prove that $k\pi$ and $\pi^{k}$ for all $k\in\mathbb{Z}$ was transcendental. After that I thought about $\pi^{b}$ for any rational number $b$ thinking this result wouldn't be to difficult but I got stumped.
Are there any results that tell whether or not numbers like $\pi^{b}$ are transcendental for algebraic $b$? If that is too broad start with numbers like $\pi^{1/2}$. The simple methods used to treat the integer cases failed miserably here in the rational case.
Ok the statement for $\pi^{1/2}$ seemed to be clear. What about $\pi^{1/3}$ or $\pi^{1/n}$
I was wondering if this was a result anyone already knew or something that someone had thought of before? Any ideas?
Thanks
AI: The square of an algebraic is also algebraic, hence, if $\sqrt\pi$ would be as such, then $\sqrt\pi^2=\pi$ would be so as well. Contradiction.
It is known that algebraics ($\mathbb{A}$), just like naturals ($\mathbb{N}$), integers ($\mathbb{Z}$), rationals ($\mathbb{Q}$), reals ($\mathbb{R}$), and complex ($\mathbb{C}$), form a group with multiplication, and a ring with both addition and multiplication. For more information on this topic, see here and here. |
H: Can $\mathbb{R}$ be written as an ascending union of proper additive subgroups?
Can the group $\mathbb{R}$ be written as countable ascending union of proper subgroups? (i.e. does there exists a series of proper subgroups $H_1\leq H_2\leq \cdots $ such that $\cup {H_i}=\mathbb{R}$?)
AI: The real numbers $\mathbb{R}$ is a vector space over $\mathbb{Q}$. By axiom of choice there is a basis $B$. Let $S_i$ be the set of all rational numbers $a/b$ where the prime factors of $b$ are among the $i$ first primes. The set of real numbers with coordinates (for the basis $B$) from $S_i$ is a subgroup $G_i$. It is clear from the definition that $G_i\subset G_{i+1}$. Now $B$ is a basis and hence for every real number there is some $G_i$ containing it. Or am I misunderstanding the question? |
H: How do I know if my answer satisfies Rolle's theorem?
Given a function on $[0,2]$,
$$f(x) = x^3 - x ^2 -2x +2$$
I know the answer has to be between $[0,2]$, but for some reason, my answer isn't being accepted. I derived the function and got the following:
$$f'(x) = 3x^2 - 2x -2 \ .$$
I then set it equal to zero and found $x$ using the quadratic formula. This is what I got for $x$.
$$x = \frac{2 \pm \sqrt{28}}{6}$$
AI: You should know the solution of a quadratic equation
$$ax^2+bx+c=0\implies x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$$
The solution is
$$ \frac{1}{3}+\frac{\sqrt {7}}{3},\, \frac{1}{3}-\frac{\sqrt {7}}{3}. $$
Now, you need to choose the right answer from the above |
H: Center of $D_6$ is $\mathbb{Z}_2$
The center of $D_6$ is isomorphic to $\mathbb{Z}_2$.
I have that
$$D_6=\left< a,b \mid a^6=b^2=e,\, ba=a^{-1}b\right>$$
$$\Rightarrow D_6=\{e,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b\}.$$
My method for trying to do this has been just checking elements that could be candidates. I've widdled it down to that the only elements that commute with all of $D_6$ must be $\{e,a^3\}$ but I got there by finding a pair of elements that didn't commute for all other elements and I still haven't even shown that $a^3$ commutes with everything. For example, I have been trying to show now that
$$a^3b=ba^3$$
and haven't gotten too far yet but if I had to answer a question like this on the exam, I feel it would be difficult, is there any kind of trick or hints other than brute force using the relations to get that $a^3$ commutes with everything?
For the solution once I have that the center of $D_6$ is what I think then as there is only one group of order $2$ up to isomorphism, it must be isomorphic to $\mathbb{Z}_2$.
Ideally a way that doesn't appeal to $D_6$ as symmetries of the hexagon if that seems possible.
AI: I wrote up a general classification for the centers of $D_n$, (the dihedral group of order $2n$, not $n$) just the other week. Perhaps it will be useful to read:
If $n=1,2$, then $D_n$ is of order $2$ or $4$, hence abelian, and $Z(D_n)=D_n$. Suppose $n\geq 3$. We have the presentation
$$
D_n=\langle x,y:x^2=y^n=1,\; xyx=y^{-1}\rangle.
$$
Then $yx=xy^{-1}$ implies the reduction $y^kx=xy^{-k}$. An element is in the center iff it commutes with $x$ and $y$, since $x$ and $y$ generate $D_n$. Let $z=x^iy^j$ be in the center. From $zy=yz$ we see
$$
x^iy^{j+1}=yx^iy^j\implies x^iy=yx^i.
$$
But $i\neq 1$, else we have $xy=yx=xy^{-1}$, so $y^2=1$, a contradiction since $n\geq 3$. So $i=0$, and $z=y^j$. Then from the equation $zx=xz$, we have
$$
y^jx=xy^j=xy^{-j}
$$
which implies $y^{2j}=1$. Thus $j=0$ or $j=n/2$. If $n$ is odd, we must necessarily have $j=0$, and $z=1$. If $n$ is even, either possibility works. But $y^{n/2}$ is indeed in the center as it clearly commutes with $y$, as well as with $x$ since $y^{n/2}x=xy^{-n/2}=x(y^{n/2})^{-1}=xy^{n/2}$. Summarizing, we have, for $n\geq 3$,
$$
Z(D_n)=\begin{cases}
\{1,y^{n/2}\} & \text{if }n\equiv 0\pmod{2},\\
\{1\} & \text{if }n\equiv 1\pmod{2}.
\end{cases}
$$ |
H: Is $\sum_{n=1}^\infty \frac{e^{\delta n^d}}{\sqrt{n}}\exp\left[-\frac{e^{\delta n^d}}{\sqrt{n}}\right]<\infty$?
Let $\delta>0$ and $d>0$. I am pretty sure that
$$\sum_{n=1}^\infty \frac{e^{\delta n^d}}{\sqrt{n}}\exp\left[-\frac{e^{\delta n^d}}{\sqrt{n}}\right]<\infty$$
as it seems that the series in the form $\sum_{n=1}^\infty f(n)\exp[-f(n)]$ ought to converge for increasing $f(n)$.
However, I am having trouble formally proving this, though the proof should be well-known if it exists. Maybe it's obvious and I am overlooking something. Perhaps I am wrong and my series doesn't converge. Can someone give a hint or pointer?
AI: All you need is that
$\frac{x^m}{e^x}
\to 0$
as $x \to \infty$
for any positive real $m$.
(This follows from
$e^x > \frac{x^{m+1}}{(m+1)!}$.)
This gives you
$\frac{x}{e^x}
< \frac{c}{x^m}
$
for any positive $c$
for large enough $x$.
Setting
$x = \frac{e^{\delta n^d}}{\sqrt{n}}$,
this gives
$\frac{x}{e^x}
<\frac{c}{x^m}$.
Using the inequality again,
$\frac{\delta n^d}{e^{\delta n^d}}
< \frac{c}{(\delta n^d)^m}
$
and the sum of these converges
if $md > 1$. |
H: Find all numbers $p$ such that both $p$ and $p^2+14$ are primes.
Find all numbers $p$ such that both $p, \ p^2+14$ are primes.
I believe only $ p= 3 $ works.
How do I prove this using the complete set of residues Modulo 3. Would something like this work.
Assume $\mathbb F = \{p, \ p+1, \ p+2 \} $ be the set of complete residues in $\mathbb Z/3 \mathbb Z$, then one of the elements of $\mathbb F$ is a zero. Meaning it is a multiple of $3$.
This cannot be $p+1$ because then $p$ would equal $2$ and $4+14$ is a composite integer. Same for $p+2$ as $p = 1$ is not a prime. Hence $p = 3$.
I am still unsatisfied with this proof.
AI: If the number $k$ is not a multiple of $3$, then $k=3n\pm 1$ for some $n$, and $$k^2=9n^2\pm6n+1=3(3n^2\pm2n)+1=3m+1$$ for some $m$, that is, $k^2\equiv1\pmod 3$ if $3$ does nor divide $k$, so $k^2+14\equiv 0\pmod 3$. In particular, if $3$ does not divide $p$, then $3$ divides $p^2+14>3$, so $p^2+14$ is not a prime.
On the other hand, if $p=3$, then $p^2+14=23$ is prime as well, so $p=3$ is the only solution. |
H: The number of $j$-dimensional subspaces which contain a given $i$-dimensional subspace.
Let $V=F_q^n$ be a vector space of dimension $n$ over the $q$-element field $F_q$ ($q$ is a prime number). Given a $i$-dimensional subspace of $V$. What is the number of the $j$-dimensional subspaces of $V$ which contain the given $i$-dimensional subspace ($j>i$). In the case $n=4$, $i=1$, $j=3$, the number is ${3 \brack 1}_q = \frac{q^3-1}{q-1}$. What is this number in general? Thank you very much.
AI: Hint (lattice correspondence): if $W\le V$ then
$$\{A:W\le A\le V\}\cong\{B:0\le B\le V/W\}.$$
Make this correspondence more precise by partitioning each set according to dimension.
This reduces the problem to counting fixed dimension subspaces of a given finite vector space. |
H: Find a bijection between the two sets
I know the function to find the number of bijections is $f(n) = f(n-1) * n$ where $f(n)$ is the number of bijections of $n$, and $n = |A|$, i.e the number of elements in the set (in this case, matching parentheses). But I have no idea where to go from here, I know it's not what the question is asking, can anyone help?
AI: One way to obtain a parntheses sequence from a triangulated polygon (with one edge marked as "base") is as follows:
The base belongs to a triangle (except for the case of a 2-gon, which we represent by the empty parentehses string ""). Remove the triangle and you obtain two smaller triangulated polygons, where you view the edge they share with the removed triangle as base. Now write down the parentheses sequence for the left polygon, followed by "(", the parentheses seqeunce for the right polygon, and ")".
In fact this method matches the examples in the picture (with the obvious choice of base) and it is instructive to verify this.
Can you find the inverse map, i.e. how to construct a polygon by "parsing" aparentheses string? And show that these maps are indeed inverse of each other? |
H: Modular quadratic equation (solve for 3-digit natural numbers)
$n^2 + 6n - 88$ is divisible by 97. Solve for all n if n is a 3-digit natural number.
Here's my progress so far
$$n^2 + 6n - 88 \equiv 0\pmod {97}$$
$$n^2 + 6n - 88 + 97 \equiv 0\pmod {97}$$
$$n^2 + 6n + 9 \equiv 0\pmod {97}$$
$$(n+3)^2 \equiv 0\pmod {97}$$
I know $97$ is prime, but can't figure out how to use this fact. I can't just take the sqrt, can I?
Any hints, tips or anything else highly appreciated.
AI: You are almost there. Your last step implies that (n+3) is divisible by 97 (since if a2 is divisible by b, a should also be divisible by b. Just list all three digit factors of 97, and subtract 3 from each to get to the answer. So your answer is 191, 288, 385, 482, 579, 676, 773, 870, 967. |
H: Defining sets using pairs, check if definition satisfies the pair correctness property - Kuratowski ordered pair
I know that (a,b) = (c,d) if a = c and b = d, but I have no idea what to do here. I assume I'm supposed to show that {{a},{a,b}} = {{c},{c,d}} if a = c and b = d, but how do I verify that using the sets?
AI: The first definition doesn't work. Just take any example with $x\neq y$ to generate two different ordered pairs that equal $\{x,y\}$.
Let $a,b,c,d$ be such that $\{\{a\},\{a,b\}\}=\{\{c\},\{c,d\}\}$. Either $a=b$ or $a\neq b$.
If $a=b$, then $\{\{a\}\}=\{\{a\},\{a,b\}\}=\{\{c\},\{c,d\}\}$. Thus, due to the cardinality of the sets, $\{a\}=\{c\}$ and $\{a\}=\{c,d\}$, it follows that $a=b=c=d$.
If $a\neq b$, then because $\{\{a\},\{a,b\}\}=\{\{c\},\{c,d\}\}$, again due to cardinality, it follows that $\{a\}=\{c\}$ and $\{a,b\}=\{c,d\}$, (this needs to be justified a little bit better, consider all possibilities). So $\color{blue}{a=c}$ and $b\in \{a,b\}=\{c,d\}=\{a,d\}$, hence $\color{blue}{b=d}$.
The other direction is trivial. |
H: Committee selection with no two consecutive people.
Assume that $10$ people are sitting around a round table. Determine the number of ways to choose a committee, where the committee is made up of two people who are NOT sitting next to each other.
Assume that $10$ people are sitting around a round table. Determine the number of ways to choose a committee, where the committee is made up of three people, Person X and two other people, such that NO two of the three people are sitting next to each other.
AI: Possible committees - Committees disallowed = $^{10}C_2$ - 10
Assuming Person X is fixed, we can have to pick 2 members from the 7 that are neither X nor sitting beside X. This gives $^7C_2 - 6$ since we have to discount the committees made up of neighbours |
H: Convergence of $\frac{1}{n}\sum_{i=1}^{n}\left[\frac{\left(\log\left(1+i/n\right)\right)^2}{1+i/n}\right]$
Sequence of real numbers
$$S_n=\frac{1}{n}\sum_{i=1}^{n}\left[\frac{\left(\log\left(1+\frac{i}{n}\right)\right)^2}{1+\frac{i}{n}}\right]$$
Does $\lim\limits_{n \to \infty} S_n$ exist? If so, compute the value.
My Solution:
A)
According to Cauchy's first theorem on limits.
This limit will be same as $$\lim_{n \to \infty}\left[\frac{(\log(1+\frac{n}{n}))^2}{1+\frac{n}{n}}\right] =\frac{(\log(2))^2}{2}$$
B) $S_n$ can also be written as $$\int_{0}^{1}\frac{(\log(1+x))^2}{1+x}dx=\frac{(\log(2))^3}{3}$$
Two answers by two methods. What am I missing here? Is this sequence not convergent?
AI: Cauchy's first theorem doesn't apply. The sequence of values you are summing together changes with $n$. |
H: How to restrict the output values of a continued fraction?
I understand that a continued fraction of the form: $g(n_1,n_2,n_3,n_4,n_5,\ldots)= n_1 + \cfrac{1}{n_2 + \cfrac{1}{n_3 + \cfrac{1}{n_4 + \cfrac{1}{n_5+\cdots} } } }$
gives a unique irrational number for every sequence of natural numbers $(n_1, n_2,n_3,n_4,n_5,...)$. I wish to however, restrict the output values of this continued fraction to say, an interval $(a,b) \in \mathbb R$. Is it possible to do so by somehow tweaking the fraction?
AI: Yes. First of all, you need $\lfloor a\rfloor\le n_1\le \lfloor b\rfloor$.
Then if $\lfloor a\rfloor< n_1< \lfloor b\rfloor$, no more restrictions apply.
If $\lfloor a\rfloor= n_1< \lfloor b\rfloor$, recursively restrict $g(n_1,n_3,\ldots)$ to $(1,\frac1{a-n_1})$, if $\lfloor a\rfloor< n_1= \lfloor b\rfloor$ to $(\frac1{b-n_1},\infty)$ and if $\lfloor a\rfloor= n_1= \lfloor b\rfloor$ to $(\frac1{b-n_1},\frac1{a-n_1})$.
In principle, you have to watch out for the boundary points and I didn't describe how, but that does not matter not for "practical" purposes, especially, if $a,b$ are rational and you want to produce irrationals. |
H: Sylow subgroup of a subgroup of a finite nilpotent group is normal
Let $G$ be a finite nilpotent group and $H \le G$. Let $P$ be a Sylow subgroup of $H$. I'd like to either prove or disprove that, under these conditions, $P$ is normal in $H$.
Any hints how to get started?
(This is part of a larger proof that any subgroup of a nilpotent group is nilpotent.)
AI: Proof: let $P$ be element of $Syl_p(H)$ then $P$ is an $p-$subgroup of $G$. Then it must be contained in a sylow-$p$ subgroup of $G$ .
Let say this is $Q$. We have $P≤Q\implies$ $P=H\cap{P}≤Q\cap{H}$ thus $Q\cap{H}$ is a $p-$subgroup of $H$ containing $P$.
By maximality of $P$ in $H$, $P=Q\cap{H}$. Now since $G$ is nilpotent then $Q$ is uniqe so is $P$.
Since $P$ is uniqe in $H$ then it is normal. By the way from that point, it is immediate that $H$ is nilpotent. |
H: How find this $\sum_{i=1}^{5}\tan^4{\frac{i\pi}{11}}$
show that:$$\tan^4{\dfrac{\pi}{11}}+\tan^4{\dfrac{2\pi}{11}}+\tan^4{\dfrac{3\pi}{11}}+\tan^4{\dfrac{4\pi}{11}}+\tan^4{\dfrac{5\pi}{11}}=2365$$
my try:
I think first we can find this value：$$x=\tan{\dfrac{\pi}{11}}+\tan{\dfrac{2\pi}{11}}+\tan{\dfrac{3\pi}{11}}+\tan{\dfrac{4\pi}{11}}+\tan{\dfrac{5\pi}{11}}$$
But I can't.Thank you for you help.
AI: Using this for odd $n=2m+1$ and setting $\tan(2m+1)x=0\implies x=\frac{r\pi}{2m+1}$ where $0\le r\le 2m$
So, the roots of $$\tan^{2m+1}x-\binom{2m+1}2\tan^{2m-1}x+\binom{2m+1}4\tan^{2m-3}x-\cdots=0$$ are $\tan\frac{r\pi}{2m+1}$ where $0\le r\le 2m$
Discarding $\tan 0=0,$ the roots of $$\tan^{2m}x-\binom{2m+1}2\tan^{2m-2}x+\binom{2m+1}4\tan^{2m-4}x-\cdots=0$$ are $\tan\frac{r\pi}{2m+1}$ where $1\le r\le 2m$
Now observe that $\displaystyle \tan\left(\frac{(2m+1-u)\pi}{2m+1}\right)=\tan\left(\pi-\frac{u\pi}{2m+1}\right)=-\tan\left(\frac{u\pi}{2m+1}\right)$
$\displaystyle\implies\tan^2\left(\frac{(2m+1-u)\pi}{2m+1}\right)=\tan^2\left(\frac{u\pi}{2m+1}\right)$
So, $\displaystyle\tan^2\frac{r\pi}{2m+1}$ where $1\le r\le m$ or $m+1\le r\le2m$ (more generally we can replace $u,1\le u\le m$ with $2m+1-u$) are the $m$ roots of $$t^mx-\binom{2m+1}2t^{m-1}x+\binom{2m+1}4t^{m-2}x-\cdots=0$$
$\displaystyle\sum_{1\le r\le m}\tan^4\frac{r\pi}{2m+1}= \left(\sum_{1\le r\le m}\tan^2\frac{r\pi}{2m+1}\right)^2-2\sum_{1\le r\le m}\tan^2\frac{r_i\pi}{2m+1}\tan^2\frac{r_j\pi}{2m+1}$ where $1\le i<j\le m$
Now using Vieta's formula,
$\displaystyle\sum_{1\le r\le m}\tan^2\frac{r\pi}{2m+1}=\binom{2m+1}2$
and $\displaystyle\sum_{1\le r\le m}\tan^2\frac{r_i\pi}{2m+1}\tan^2\frac{r_j\pi}{2m+1}=\binom{2m+1}4$ where $1\le i<j\le m$
In this problem $m=5$ |
H: Solve the following complex number: $\frac{1 + i\tan \theta}{1 - i\tan\theta}$
How do I solve the following complex number?
$$\frac{1 + i\tan \theta}{1 - i\tan\theta}$$
I know how to solve arithmetic problems with complex numbers, but this is the first time I have a function and variable inside.
Thanks in advance!
AI: Using: $\cos\theta+i\sin\theta=e^{i\theta}$ (De-moivre's formula)
$$\frac{1+i\tan\theta}{1-i\tan\theta}=\frac{\cos\theta+i\sin\theta}{\cos\theta-i\sin\theta}=\frac{e^{i\theta}}{e^{-i\theta}}=e^{i2\theta}=\cos2\theta+i\sin2\theta$$ |
H: Cubic Poynomial : In the equation $x^3 +3Hx +G=0$ if G and H are real and $G^2 +4H^3 >0$ then roots of the.........
Question:
In the equation $x^3 +3Hx +G=0$ if G and H are real and $G^2 +4H^3 >0$ then roots of the equation are
(a) all real and equal
(b) all real and distinct
(c) one real and two imaginary
(d) all real
What I did :
Let the cubic polynomial is $ax^3+bx^2+cx+d=0$ then let p,q,r are roots of equation then $ax^3 - a(p+q+r)x^2 + a(pq+pr+qr)x - a(pqr)$
How do I relate this with the given equation.. thanks..
AI: The answer is (c): The equation has one real root and two nonreal complex conjugate roots.
Proof: This Wiki article classifies the nature of the roots.
The discriminant of a cubic equation $ax^3+bx^2+cx+d=0\;$ is
$$\Delta = 18 a b c d - 4b^3d + b^2 c^2 -4ac^3 - 27 a^2 d^2.$$
If $\Delta > 0,$ then the equation has three distinct real roots.
If $\Delta = 0,$ then the equation has a multiple root and all its roots are real.
If $\Delta < 0,$ then the equation has one real root and two nonreal complex conjugate roots.
With your coefficients we get (since $b=0$)
$$\Delta = -4ac^3 - 27a^2d^2 = -108H^3-27G^2 = -27(4H^3 + G^2) < 0.$$
And therefore, because $\Delta < 0,$ the equation has one real root and two nonreal complex conjugate roots.
Note: If the question is a kind of multiple choice, the third roots of unity (1, and two conjugate-complex) would also give a hint for case (c). |
H: How to prove the two angles are equal?
It is from Young Double slit experiment. But How to prove the the two $\theta$ are equal, I meant, how $\angle EAD= \angle PEC$? I see from the both triagle have $90^0$ but what about others.
If we think θ→0 we get valid results. we can see the image from any article whoch is YOUNG's double slit experiment. citycollegiate.com/interference2.htm
AI: I believe both the angles can be equal only when L is infinity. Here is why---
If ∠PEC = θ then ∠AEP = 90 - θ
So in small triangle AEO where O is the point of intersection of EP and AD :--
∠AOE = 180 - (θ + 90 - θ) = 90
Now because ∠AOE = ∠ADB = 90 so it follows that EP and BP are parallel. It is only possible when L is infinity.
Does the answer make sense to you...?? |
H: find minimum of a function: $X^2 + X + 1$?
How do you find the minimum value of $X^2 + X + 1$?
I know it's $3/4$ from intuition. How do you prove it?
AI: For real $x,$ $$x^2+x+1=\left(x+\frac12\right)^2+1-\frac14\ge \frac34,$$ the equality occurs when $x+\frac12=0$
Alternatively, we can use Second derivative test |
H: Laplace Transformation spring question
Here is the question:
https://i.stack.imgur.com/yJyCO.jpg
I can't seem to get the answer.
Are those values in the writing like 1N/m even relevant?
Can someone give me some direction? Thanks!
AI: From that image, it just seems to me that they want you to use Laplace Transforms to solve the system
$$\left\{\begin{aligned} y_1^{\prime\prime}(t) &= y_2(t) - 2y_1(t) \\ y_2^{\prime\prime}(t) &= y_1(t) - 2y_2(t)\end{aligned}\right.$$
subject to the initial conditions $y_1(0) = 1$, $y_1^{\prime}(0) = \sqrt{3}$, $y_2(0)= 1$, $y_2^{\prime}(0) = -\sqrt{3}$. All that information at the beginning was more or less giving you the background information that was necessary to construct that system of equations.
Now, let $\mathcal{L}\{y_1(t)\}=X(s)$ and $\mathcal{L}\{y_2(t)\} = Y(s)$. Then, applying Laplace Transforms to the equations in your system leaves us with
$$\left\{\begin{aligned} s^2X(s) - sy_1(0) - y_1^{\prime}(0) &= Y(s) - 2X(s)\\ s^2Y(s) - sy_2(0) -y_2^{\prime}(0) &= X(s) - 2Y(s)\end{aligned}\right. \implies \left\{\begin{aligned} (s^2+2)X(s) -s-\sqrt{3} &= Y(s) \\ (s^2+2)Y(s) -s+\sqrt{3} &= X(s)\end{aligned}\right.$$
Let's substitute the first equation into the second. Then it follows that
$$(s^2+2)\left[(s^2+2)X(s) - s - \sqrt{3}\right] -s +\sqrt{3} = X(s)\\ \implies (s^2+2)^2X(s) -s(s^2+2) -\sqrt{3}(s^2+2) - s +\sqrt{3} = X(s) \\ \implies ((s^2+1)^2-1)X(s) = s(s^2+3)+\sqrt{3}(s^2+1)\\ \implies X(s) = \frac{s(s^2+3)}{(s^2+2)^2-1} + \frac{\sqrt{3}(s^2+1)}{(s^2+2)^2-1}$$
Now noting that $(s^2+2)^2-1 = (s^2+3)(s^2+1)$ by difference of squares, we see that
$$X(s) = \frac{s}{s^2+1} + \frac{\sqrt{3}}{s^2+3}$$
and hence $$y_1(t) = \mathcal{L}^{-1}\{X(s)\} = \cos t + \sin(\sqrt{3}t).$$
Now, to find $y_2(t)$, we substitute our solution into the first equation and solve for $y_2(t)$; this leaves you with
$$-\cos(t) -3\sin(\sqrt{3}t) = y_2(t) - 2(\cos t + \sin(\sqrt{3}t))\implies y_2(t) = \cos(t) -\sin(\sqrt{3}t)$$
(Note: You could have also shown this by using what we computed for $X(s)$ by plugging it into the second equation and solve for $Y(s)$; it's easy to see that
$$\begin{aligned}Y(s) &= \frac{s}{(s^2+2)(s^2+1)}+\frac{\sqrt{3}}{(s^2+2)(s^2+3)} +\frac{s}{s^2+2} - \frac{\sqrt{3}}{s^2+2} \\ &= \frac{s}{s^2+1} -\frac{\sqrt{3}}{s^2+3}\end{aligned}$$
and thus $y_2(t) = \mathcal{L}^{-1}\{Y(s)\} = \cos t - \sin(\sqrt{3}t)$.) |
H: $\lim\limits_{(x,y)\to(0,0)} \frac{x^2 \sin^2 (y)}{x^2 + 2y^2}$ how to prove this limit doesn't exist
Question is in the title.
I tried approaching from $x$ axis, $y$ axis, $y=x$, $y=x^2$, $y=x^3$... they all go to 0 as $(x,y)\to 0$. But wolframalpha says it doesn't exist.. how??? http://www.wolframalpha.com/input/?i=lim+%28x%2Cy%29+-%3E+%280%2C0%29+x%5E2sin%5E2%28y%29%2F%28x%5E2%2B2y%5E2%29
AI: So, consider the absolute value of this expression in the neighbourhood of origin:
$$ \left\vert \frac{x^2 \sin^2y}{x^2+2y^2} \right\vert \leqslant
\frac{x^2y^2}{x^2 + 2y^2}$$
Apply the following inequality: $x^2 + 2y^2 \geqslant 2\sqrt{ x^2 \cdot 2y^2}$ (Cauchy-Bunyakovsky really).
Then you get:
$$ \frac{x^2y^2}{x^2 + 2y^2} \leqslant \frac{x^2y^2}{2\sqrt{2} \vert x \vert \vert y \vert} = \frac{1}{2\sqrt{2}} \vert x \vert \vert y \vert$$
For me it seems that limit exists and is zero :)
NOTE:
Strictly speaking, I used inequality of form $a^2 + b^2 \geqslant 2ab$, which is just consequence of $(a-b)^2 \geqslant 0$. Also, it's just "rephrasing" of arithmetic-geometric mean inequality which is one of particular cases for mentioned Cauchy-Bunyakovsky. |
H: $\displaystyle 3^x+4^x=5^x$
Show that the equation $\displaystyle 3^x+4^x=5^x$ has exactly one root
I proved it graphically but I am in need of an analytic solution
AI: $3^x+4^x=5^x\iff\left(\frac35\right)^x+\left(\frac45\right)^x=1$. Since the exponential function on the left hand side is strictly decreasing, it follows that the equation, were it to ever be true for a certain x, then that x is unique. And it is indeed true for $x=2$, making it its only solution. |
H: why in $ C^*$-algebra generated by $ x$ that denote by $ A[x]$, $ A[x]$ is commutative?
suppose $ A$ is a $ C^*$-algebra and $ x$ is a normal element in $ A $. $ C^*$-algebra generated by $ x$ denote by $ A[x]$. then
1) $ A[x]$ is commutative.
2) $ A[x]$ is the clouser of polynomials of two variable $ x$ and $ x^*$.
in every book that I read, (1) and (2) is clear from author's point of view. But I dont understand why.
Please help me about (1),(2)and say what is the form of elaments of $ A[x]$.
thanks
AI: Let's do this backwards :
Let $D$ denote the closure of all polynomials in $x$ and $x^{\ast}$. Since $xx^{\ast} = x^{\ast}x$, any two polynomials commute. Hence, $D$ is commutative.
Clearly, $D$ is a C* algebra and contains $x$. Any other C* algebra $B$ that contains $x$, must contain $x^{\ast}$, and hence must contain all polynomials in $x$ and $x^{\ast}$. Thus, $D\subset B$.
Hence, $D = A[x]$, and so (1) and (2) hold. |
H: $f$ not measurable, but $\lvert f\rvert$ measurable
Do you know an example of a function $f\colon\mathbb{R}\to\mathbb{R}$ which is not $\mathcal{B}$-measurable but $\lvert f\rvert$ is $\mathcal{B}$-measurable?
AI: Pick any non-measurable set $A$. Define:
$$
f(x) = \begin{cases}
1 &: x \in A \\
-1 &: x \not\in A
\end{cases}
$$
$|f| = 1$ is a constant function, hence measurable. $f$ is easily seen to be non-measurable. |
H: Why is this is the derivative?
We are using the Euler equation to calculate the minimum:
Euler equation: $-\frac{d}{dt}\hat{L}_{x'}(t) + \hat{L}_x(t) = 0$
We have the following $L = 12tx + x'^2$ ($x$ is a function of $t$)
Now calculating these derivatives my book says it equals $-2x'' + 12t = 0$
Can anybody please explain how they came to this answer?
Thanks in advance.
AI: Hints. Rewrite as follows:
$$-\frac{d}{dt} \frac{\partial L}{\partial x'} + \frac{\partial L}{\partial x} = 0
\qquad ; \qquad L = 12tx + (x')^2$$ What is $\frac{\partial L}{\partial x'}$ , what is $-\frac{d}{dt} \frac{\partial L}{\partial x'}$ then, and, at last, what is $\frac{\partial L}{\partial x}$ ? If you can assemble these, then you're finished. |
H: Proof of binomial coefficient formula.
How can we prove that the number of ways choosing $k$ elements among $n$ is $\frac{n!}{k!(n-k)!} = \binom{n}{k}$ with $k\leq n$?
This is an accepted fact in every book but i couldn't find a proof.
AI: Well, let's first make sequences (ordered lists of distinct elements) of length $k$ from the $n$ element set.There are $n$ ways of picking the first element, $n-1$ ways of picking the second, ..., $n-k+1$ ways of picking the $k$th element. So there are $$n(n-1)\cdots (n-k+1) = \frac{n!}{(n-k)!}$$ such sequences.
But we want to count unordered lists, i.e., for each sequence, we want to identify the $k!$ permutations of its elements (each of which was counted separately above) as the same list. Hence the number of unordered lists is $$\frac{\frac{n!}{(n-k)!}}{k!} = \frac{n!}{k!(n-k)!}.$$ |
H: why from $ \int |f_n -f| \to 0$ we can conclude that $ \int f_n \to \int f$?
Let $(Ω,A,μ)$ be a measure space and $(f_n)$ a sequence of μ-integrable functions, which converges uniformly to $f:Ω→{\mathbb{R}}$. why from $ \int |f_n -f| \to 0$ we can conclude that $ \int f_n \to \int f$? thanks
AI: Because $$|\int f_n-\int f|=|\int (f_n-f)|\le \int |f_n-f|$$ which follows from the linearity of the integral, and the general inequality $$|\int g|\le |\int g|.$$ |
H: Principle of inclusion and exclusion problem
I've got 5 bottles of rum, 4 bottles of vodka and 3 bottles of whisky. How many ways are to arrange them into a line when all the bottles of the same kind can't stand next to each other(the bottles of one kind are indistinguishable).
My answer was $\frac{(5+4+3)!}{5!4!3!}-\frac{(5+4+1)!}{5!4!}-\frac{(5+1+3)!}{5!3!}-\frac{(1+4+3)!}{4!3!}$, because i subtract the ways of arrangement when all the bottles of one kind stand next to each other. I was told that it's false, and I'm to use PIE. It's really hard for me to just accept that without an explanation and use a formula I don't really understand. Could someone walk me through this problem and explain what I did wrong and how should I use the principle?
AI: Consider the arrangement RVVVVRRWWWRR: all of the bottles of vodka are together, so you counted this in your $\frac{(5+1+3)!}{5!3!}$ term, and all of the bottles of whisky are together, so you also counted it in your $\frac{(5+4+1)!}{5!4!}$ term. Thus, you’ve subtracted this arrangement twice from the original $\frac{(5+4+3)!}{5!4!3!}$, so you’ve counted it a total of $-1$ times. You want to count it $0$ times, so you need to add it back in. The same is true of all arrangements that have all of the vodka together and all of the whisky together, and there are $\frac{(5+1+1)!}{5!}$ such arrangements, so you need to add $\frac{(5+1+1)!}{5!}$ to your answer.
There is a similar problem with arrangements that have all of the vodka together and all of the rum together, and with the arrangements that have all of the rum together and all of the whisky together, and you have to make similar corrections for them, giving you
$$\begin{align*}\frac{(5+4+3)!}{5!4!3!}&-\frac{(5+4+1)!}{5!4!}-\frac{(5+1+3)!}{5!3!}-\frac{(1+4+3)!}{4!3!}\\\\
&+\frac{(5+1+1)!}{5!}+\frac{(1+4+1)!}{4!}+\frac{(1+1+3)!}{3!}\;.
\end{align*}$$
You’re still not quite finished, however: each arrangement that has all of the vodka bottles together, all of the rum bottles together, and all of the whisky bottles together has been counted once in the first term, $-3$ times in the next three terms, and $3$ times in the last three terms, for a net total of one time. We don’t want to include these arrangements, so we need to subtract them:
$$\begin{align*}\frac{(5+4+3)!}{5!4!3!}&-\frac{(5+4+1)!}{5!4!}-\frac{(5+1+3)!}{5!3!}-\frac{(1+4+3)!}{4!3!}\\\\
&+\frac{(5+1+1)!}{5!}+\frac{(1+4+1)!}{4!}+\frac{(1+1+3)!}{3!}\\\\
&-3!\;.
\end{align*}$$
Your answer was the beginning of the inclusion-exclusion calculation, but only the beginning. |
H: Value if term equals previous term count plus value
How would I calculate the value of the term in a range like this:
1:0
2:1
3:3
4:6
5:10
...
As you can see the value of any given term is equal to the value of the previous term plus the identifier of the previous term, but I have no idea what the formula would be for calculating the value of any given term.
AI: At identifier $n$ the value is $n(n-1)/2$. You can show this by induction from $f(n+1)=n+f(n).$ |
H: On principal non-maximal ideal
Let $R$ be an integral domain but $R$ is not a field. Prove that, in $R[x]$, $\langle x\rangle$ is maximal principal ideal (that is, maximal among principal ideals) but not a maximal ideal.
AI: Since $R$ is not a field, there is nonunit element $a\in R$. Consider the ideal
$$
\langle a,x\rangle =\{ra+sx:r,s\in R[x]\}.
$$
Add. To prove $(x)\subsetneq(a,x)$, check $a\notin (x)$. |
H: Evaluate $\lim_{x\rightarrow 0} \frac{\sin (6x)}{\sin(2x)}$ without L'Hopital
I am trying to evaluate the following limit without L'Hopital's:
$$\lim_{x\rightarrow 0} \frac{\sin (6x)}{\sin(2x)}$$
I know I have to use the fact that $\frac{\sin x}{x} = 1$ but I don't know how to get the limit from the above to $\frac{\sin x}{x}$ or even a portion of it to that. I know how to evaluate limits like the following
$$\lim_{x \rightarrow 0}\frac{\sin 3x}{x} = 3$$
if that's any help.
Any hints would be appreciated
Thanks!
AI: hint :
$$\frac{\sin6x}{\sin 2x}= \frac{\frac{\sin6x}{6x}}{\frac{\sin 2x}{2x}}\cdot 3 $$ |
H: Mean and Variance of probability distributions
I know how to calculate mean and variance of some given numbers but I have trouble computing them for probability distributions especially when it is a continuous probability distribution.
For example, can you show me how to calculate mean and variance of Gaussian distribution?
AI: $$E[X]=\int xf_X(x)\mathrm dx\qquad E[X^2]=\int x^2f_X(x)\mathrm dx$$ $$\mathrm{var}(X)=E[X^2]-E[X]^2$$ |
H: Divisibility by 37 proof
$\overline {abc}$ is divisible by $37$. Prove that $\overline {bca}$ and $\overline {cab}$ are also divisible by $37$.
$$\overline {abc} = 100a + 10b + c$$
$$\overline {bca} = 100b + 10c + a$$
$$\overline {cab} = 100c + 10a + b$$
When you add them:
$$\overline {abc} + \overline {bca} + \overline {cab} = 111a + 111b + 111c$$
$$\overline {abc} + \overline {bca} + \overline {cab} = 111(a + b + c)$$
Since $111$ is divisible by $37$, the whole sum is divisible by $37$, but how can i prove that $\overline {abc}$, $\overline {bca}$, $\overline {cab}$ separately are divisible by $37$?
Any tips or hints appreciated.
AI: $$\overline{bca}=10\cdot\overline{abc}-1000a+a = 10\cdot\overline{abc}-27\cdot 37\cdot a$$ |
H: Question on continuous functions on intervals
Problem Statement: Let $I = [a,b]$ and let $f: I \rightarrow \mathbb{R}$ and $g: I \rightarrow \mathbb{R}$ be continuous functions on $I$. Show that the set $E = \{x \in I: f(x) = g(x)\}$ has the property that if $(x_n) \subseteq E$ and $x_n \rightarrow x_0$, then $x_0 \in E$.
Here's what I have so far, I think it's complete, however I'd like to see a different proof if possible. Maybe one using the difference function, $lim[(f-g)(x_n)]$.
$Proof:$ Since $(x_n) \subseteq E$ and $f, g$ are continuous on $I$, then $limf(x_n) = f(x_0)$ and $limg(x_n)=g(x_0).$ Therefore $\forall \varepsilon >0.\ \exists \delta_1 >0$ such that $$|f(x_n) - f(x_0)| < \varepsilon /2$$ and similarly $\exists \delta_2 >0$ such that $$|g(x_n) - g(x_0)| < \varepsilon /2.$$ Choose $\delta = inf \{\delta_1, \delta_2 \}. \forall x \in E \cap V_\delta(x_0)$
\begin{align*}
|f(x_0) - g(x_0)| & = |f(x_0) - f(x_n) + g(x_n) - g(x_0)| \ & \ [\text{Since}\ f(x) = g(x)\ \forall x \in E]\\
& \le |f(x_n) - f(x_0)| + |g(x_n) - g(x_0)| \\
& < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*}
Since $\varepsilon$ is arbitrarily small $f(x_0) = g(x_0)$, therefore $x_0 \in E.
\square$
AI: Since each $x_n$ lies in the closed set $[a,b]$ and $x_n\to x_0,$ then $x_0\in [a,b],$ and so $h(x_0)$ is defined. (You left this out of your proof.)
Now, let $h=f-g.$ Note that $h$ is a continuous function such that $h(x_n)=0$ for all $n,$ so since $x_n\to x_0,$ then.... |
H: Fourier Series (Even and Odd Functions)
Let $f \in E$ (where $E$ is a linear space of complex-valued piecewise continuous functions defined on the interval $[-\pi,\pi]$) and $$f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n}\cos{nx}+b_{n}\sin{nx}\right]$$
denote its Fourier series. Define $$g(x)=\frac{f(x)+f(-x)}{2}, \quad \quad h(x)=\frac{f(x)-f(-x)}{2}$$
Find the Fourier series of $g$ and of $h$.
Is it just as simple as to write out the definition of $f$ and then use the properties of the trigonometric functions and series, as done below?
$$h(x)=\frac{\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n}\cos{nx}+b_{n}\sin{nx}\right]-\left(\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n}\cos{-nx}+b_{n}\sin{-nx}\right]\right)}{2}$$ $$=\frac{\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n}\cos{nx}+b_{n}\sin{nx}\right]-\left(\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n}\cos{nx}-b_{n}\sin{nx}\right]\right)}{2}$$
$$=\frac{\sum_{n=1}^{\infty}\left[\left(a_{n}\cos{nx}-a_{n}\cos{nx} \right)+\left(b_{n}\sin{nx}+b_{n}\sin{nx} \right) \right]}{2}=\sum_{n=1}^{\infty}b_{n}\sin{nx}$$
If so, then one easily see that $g(x)=\sum_{n=1}^{\infty}a_{n}\cos{nx}$, which is to say it is an even function, whereas $h(x)$ is odd.
AI: Adding function series works fine when you know they converge. It is however not a trivial matter to decide if a function series, or Fourier series, converge.
What you should do instead is to consider the definition of Fourier series. Thus, you should try to calculate $a_n(g)$ and $b_n(g)$ (and $a_n(h)$ and $b_n(h)$). |
H: Is $\tan\theta\cos\theta=\sin\theta$ an identity?
A friend of mine, who is a high school teacher, called me today and asked the question above in the title. In an abstract setting, this boils down to asking whether an expression like "$f=g$" is regarded as an "identity" when one of their domains is a proper subset of the other, and the two functions coincide on the smaller domain. Examples of such equalities are abundant in high school mathematics exercises, e.g. $\frac{x^2-1}{x-1}=x+1,\ e^{\log x}=x$ etc.. Often, the domains are not specified in the exercises.
A somewhat similar but subtly different case is when both functions are defined on the same domain but whether they are equal depends on the exact domain. For instance, $e^{x+y}\equiv e^xe^y$ for real or complex numbers but not for quaternions. Yet, for the purpose of discussion, let us focus on the aforementioned case of $f$ and $g$. For pedagogical purposes:
Do you consider $f=g$ an "identity"? What does an identity mean?
How to convince high school students that your definition is a good one?
AI: I've typically seen the term "identity" used to refer to an equation that is true wherever both sides are defined. This would include $\sin x=\tan x\cos x,$ to be sure, but it would also include $x=2x-x.$ Put another way, we say that $f=g$ is an identity iff $f$ and $g$ are identical wherever they are both defined. |
H: Do the non-units in a commutative ring form an ideal?
Do the non-units in a commutative ring form an ideal?
The following are my thoughts on this. Have I made any incorrect assumptions?
Let $R$ be a commutative ring. Let $a, b \in N$ with $N$ being the set of non-units in $R$. We must show the following to prove $N$ is an ideal -
$0 \in N$
$a + b \in N$
$-a \in N$
$ar, ra \in N \ \forall r \in R$
1. $0 \in N$
I.e. $0$ a non-unit. This is true as $\nexists \ 0^{-1}$ such that $0 \cdot 0^{-1} = 1$
2. $a + b \in N$
Assume $a + b$ is a unit. Then $\exists \ g \in R$, $g \neq 0$ such that
$(a + b)g = 1$ $\implies$ $ag + bg = 1$
For this to be true either $a$ or $b$ must be $0$. Consider the case when $a = 0$. Then we have $bg = 1$. But this is a contradiction as $b$ is a non-unit. Hence $\nexists \ g \in R$ such that $(a + b)g = 1$. Similarly for when $b = 0$.
Therefore, $a + b$ is a non-unit.
3. $-a \in N$
I.e. Does there exist $(-a) \in N$ such that $a + (-a) = 0$?
Assume $-a$ is a unit. Then $\exists \ g \in R, g \neq$ 0, such that $(-a)g = 1$
Consider $a + (-a) = 0$
Multiplying both sides by $g$ we get
$(a + (-a))g = 0 \cdot g$
$ag + (-a)g = 0$
$ag + 1 = 0$
$-ag = 1$
$a(-g) = 1$
But this is a contradiction as $a$ is a non-unit. Hence $(-a)$ is a non-unit.
4. $ar, ra \in N \ \forall r \in R$
Assume $ar$ is a unit. Then $\exists \ g \in R, g \neq$ 0, such that $(ar)g = 1$
I.e. $a(gr) = 1$
But this is a contradiction as $a$ is a non-unit. Hence $ar$ is a non-unit.
So, to conclude, the non-units in a commutative ring do form an ideal. Are my workings correct?
AI: In general, the set $N$ of non-units in a commutative ring $R$ doesn't form an ideal. The first point, $0\in N$, fails if and only if $R = \{0\}$ is the trivial ring, in which $0$ is a unit. Points 3. and 4. hold in any commutative ring, and your argument is correct.
However, in general, the sum of non-units need not be a non-unit. In $\mathbb{Z}$ for example, we have $3 + (-2) = 1 \notin N$. A commutative ring in which $N$ is an ideal is a local ring, a ring with a unique maximal ideal - that ideal is then the set of non-units. An example of a local ring is the ring of formal power series over a field, $K[[X]]$, in which the unique maximal ideal is the ideal generated by $X$. |
H: Following positive semi-definitness from matrix norm
How can I follow the following?
$$||A||_2 \le \sigma > 0$$
$$\Leftrightarrow A -\sigma I \mbox{ is positive semi-definit}$$
I always get it the other way around, i.e. that $\sigma I - A$ is positive semi-definit.
AI: I think you got it right. Obvious counterexample $A=0.5 I$ with $\sigma=1$ |
H: Proof by induction that fibonacci sequence are coprime
I have a bit difficulty to proofe that two consecutive numbers are coprime. I have the following
The property $P(n)$ is the equation $(F_{n+1},F_n)=1$ where F_i the sequence of fibonacci is and $n \in \mathbb {N}$.
Induction hypothesis: $gcd(F_{n+1},F_n)=1$
$P(0)$ is true because $F_1=1, F_0=1$ and $gcd(1,1)=1$
To show: $gcd(F_{n+2},F_{n+1})=1$
Proof:
Fibonacci formula says
$F_{n+2}=F_{n+1}+F_n$
This is as faar as I got. How can I now get to $gcd(F_{n+2},F_{n+1})=gcd(F_{n+1},F_n)$ which would then proofe it?
AI: $\gcd(F_{n+1},F_{n+2}) = \gcd(F_{n+1},F_{n+1}+F_n) = \gcd(F_{n+1},F_n)$
By the induction hypothesis $\gcd(F_{n+1},F_n)=1$ so $\gcd(F_{n+1},F_{n+2})=1$
To prove $\gcd(a,b) = \gcd(a,b-a)$ use the fact that if $c\vert a$ and $c\vert b$ then $c\vert na+mb$ |
H: $T$ be the operator from $C[0,1]$ to $C[0,1]$ defined by $Tf = f'+f''$. Show that the operator $T$ is unbounded.
$f \in C[0,1]$, the space of all continuous, complex-valued functions on $[0,1]$ with supremum norm.
$\|f\|=\sup_{x\in[0,1]}|f(x)|$.
Let $D$ be the set of $f \in C[0,1]$ such that the first derivative and second derivative are defined and continuous on $[0,1]$.
Let $T$ be the operator from $C[0,1]$ to $C[0,1]$ defined by $$Tf = f'+f''$$ on the domain $D$.
Show that the operator $T$ is unbounded by constructing a sequence $f_n$ such that $\|f_n\| \le M$ for some constant $M$ but $\|Tf_n\| \rightarrow \infty$ as $n \rightarrow \infty$
AI: Take $f_n(x)=e^{-nx}$ so $T(f_n)=-nf_n+n^2f_n=n(n-1)f_n$. Can you verify now that $T$ is unbounded? |
H: Equivalence of condition for Lebesgue-measurability
We had a proof in lecture, that showed a bunch of equivalences for Lebesgue-measurability. I have a problem understanding the following implication, where the professor said the reasoning was "trivial".
Let $A\subseteq\mathbb{R^n}$. If there exists $B\in F_\sigma$ such that $\lambda_n^*((A\setminus B)\cup(B\setminus A))=0$ then $A$ is Lebesgue-measurable. $F_\sigma$ denotes sets that are a countable union of closed sets, $\lambda_n^*$ is the $n$-dimensional Lebesgue outer measure.
Can anyone jump start me as to why this is "trivial"?
AI: $$
A=(B\setminus(B\setminus A))\cup(A\setminus B).
$$
The rest should be more trivial. |
H: Norm of an operator induced by $L^2$-kernel is bounded by $L^2$- norm of the kernel
I am currently studying Hilbert Schmidt operators on my own using the book "Functional Analysis" (Vol 1) by Reed and Simon.
There it is stated that a function $K \in L^2(M \times M, d\mu \otimes d\mu)$ induces an operator
$$
A_Kf(x) = \int_M K(x,y) f(y) \,d\mu(y)
$$
and the operator norm is dominated by the $L^2$-norm of $K$, that is $\|A\| \le \|K\|$.
I am trying to verify this norm estimate but got stuck. Here is what I did:
We have
\begin{align}
\|A_K\|^2 &= \sup_{\|f\| = 1} \int\left|\int K(x,y) f(y) \,d\mu(y)\right|^2\,d\mu(x)\\
&= \sup_{\|f\| = 1} \int\int K(x,y) f(y)\,d\mu(y) \overline{\int K(x,z)f(z)\,d\mu(z)} \,d\mu(x)\\
& \le \sup_{\|f\| = 1} \int\int |K(x,y) f(y)|\,d\mu(y) \int |K(x,z)f(z)|\,d\mu(z) \,d\mu(x) \\
& \le \sup_{\|f\| = 1} \int(\|K_x\|\|f\|)(\|K_x\|\|f\|) \,d\mu(x)\qquad \text{(Cauchy Schwarz)}\\
&= \int \|K_x\|^2 \,d\mu(x)
\end{align}
where $\|K_x\|$ refers to the norm of the function $K(x, \cdot) \in L^2(M,d\mu)$. But how can I finish this argument, and is it correct to start with? I don't have a lot of experience with product measures, this must be a pretty elementary fact that I am missing most likely about the relationship of $d\mu \otimes d\mu$ to its component measures ..
Thanks for your help!
AI: You are in fact done, since
$$\lVert K_x\rVert^2=\int K(x,y)^2\mathrm d\mu(y),$$
and we get what we want integrating with respect to $x$.
Notice that we can use Cauchy-Schwarz directly in the inner integral of the RHS of the first line. |
H: ZF Natural Even Numbers
Regarding $\Bbb N$ as constructed using ZF ($0=\emptyset, n+1=n^+=n\cup\{n\}$), how is the property "divisible by 2" expressed (using sets and logic)?
AI: We can define addition and multiplication on the natural numbers as defined. $m+n$ is defined to be the unique $k$ such that there is a bijection between $k$ and $m\times\{0\}\cup n\times\{1\}$. Similarly $m\cdot n=k$ when $k$ is the unique natural number such that $m\times n$ has a bijection with $k$.
Now we can say that $\exists k(k+k=n)$. |
H: does $(\overline{E})^{'}= E^{'} \cup ( E^{'})^{'}$ holds?
My question is as follows:
Suppose $E$ is a set in metric space $X$, let $\overline{E}$ denote the closure of E, let $E^{'}$ be the set of all the limit points of $E$. We all know that $\overline{E}=E\cup E^{'} $ Then my question is: Does the following equality hold?
$(\overline{E})^{'}= E^{'} \cup ( E^{'})^{'}$
if not, can you give me an exception in which the equality does not hold?
Thanks so much!
AI: We have more. In a Hausdorff space (even in a $T_1$ space, if you already know what that is), $x$ is a limit point of $A$ if and only if every neighbourhood of $x$ contains infinitely many points of $A$. Thus in such spaces, we have
$$(\overline{E})' = E'.$$
Since evidently $(E')' \subset (\overline{E})'$, the equality holds in Hausdorff (or $T_1$, nore generally) spaces, in particular in metric spaces. |
H: Equivalence of positive semi-definite matrices
Why can I follow
$$\sigma I - RB^{-1}R^\top \mbox{ is pos. semi-definite} \Leftrightarrow B- \frac{1}{\sigma}R^\top R \mbox{ is pos. semi-definite} $$
having $ \sigma > 0$ and $R$ being an upper triangle matrix?
EDIT: $B$ is positive definit and symmetric. $R$ originates from QR factorization ($R\neq 0$).
AI: The implication is false. Take $R=1$, $B=-1$, $\sigma=1$. The first equation writes $1+1=2>0$, yet the second one is $-1-1=-2<0$.
Another example is to take $R=0$ and $B$ negative definite.
Edit some thoughts on the second version of the question.
Assume that $\det R\ne 0$. Being positive semidefinite is equivalent to say $\forall u$
$$\sigma\|u\|^2-(B^{-1}R^Tu,R^Tu)\ge 0.$$
One can now take $v=R^Tu$ and say that we have $\forall v$
$$\sigma\|R^{-T}v\|^2-(B^{-1}v,v)\ge 0.$$
$B$ is positive definite, so it admits a $B^{-1/2}$ in the class of positive definte matrices, let's take $w=B^{-1/2}v$. Hence $\forall w$
$$\sigma\|R^{-T}B^{1/2}w\|^2-\|w\|^2\ge 0.$$
Now we can say that $w=Rz$ and write $\forall z$
$$\|R^{-T}B^{1/2}Rz\|^2-\frac 1\sigma\|Rw\|^2\ge 0,$$
which is equivalent to say that
$$R^TB^{1/2}R^{-1}R^{-T}B^{1/2}R -\frac 1\sigma R^TR $$is positive semidefinite. |
H: uniform convergence of series $\sum_{k=0}^{\infty} \frac{1}{1+k^2x} $
I know that the series
$\sum_{k=0}^{\infty}\frac{1}{\vphantom{\Large A}1\ +\ k^{2}\,x}$
is converges uniformly on $\left(a,\infty\right)$ for $a > 0$, but how can I show that it does not converge uniformly on $\left(0,\infty\right)$?
AI: If the series is uniformly convergent on $(0,+\infty)$ then
$$\lim_{x\to0} \sum_{k=0}^{\infty} \frac{1}{1+k^2x}= \sum_{k=0}^{\infty} 1\, \text{is finite}$$
which is obviously wrong. Conclude.
Added
By the definition the series is uniformly convergent if
$$0=\lim_{n\to\infty}\sup_{x\geq0}\sum_{k=n+1}^\infty \frac{1}{1+k^2x}\geq\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac{1}{1+k^2\times0}=\lim_{n\to\infty}n=\infty$$
which's wrong |
H: Proving that total variation is equal to $\int_{a}^{x}|g|$
Question:Suppose $g$ is continuous on $[a,b]$. Let f(x)=$\int_{a}^{x}g$ where $x∊[a,b]$. Show that $\int_{a}^{x}|g|$ gives the total variation of $f$ on $[a,x]$.
I managed to prove that $V_{f}(a,x)≤\int_{a}^{x}|g|$. But I still could not find a way to prove that $V_{f}(a,x)≥\int_{a}^{x}|g|$. I would much appreciate if someone could provide me a hint. Thanks
AI: Hint: The integral $\int_a^x |g|dx$ is defined to be the supremum of all Riemann sums of $|g|$ over $[a,x]$. Given a Riemann sum, find a partition $P$ for which the variation of $|g|$ over $P$ is greater than this sum. From there, we may conclude that the supremum of Riemann sums is less than the supremum of variations over partitions, yielding the necessary conclusion. |
H: Density with irrational number and trig function
We may assume the following theorem:
Theorem: A real number $\lambda$ is irrational iff the set $\{m+\lambda n\mid m,n\in\mathbb{Z}\}$ is a dense subset of $\mathbb{R}$.
Consider the points $$\gamma(t)=(a\cos t+b\sin t, a\sin t-b\cos t, c\cos(\lambda t)+d\sin(\lambda t), -c\sin(\lambda t)+d\cos(\lambda t))$$ for all $t\in\mathbb{R}$, for some constants $a,b,c,d$ where $a^2+b^2=c^2+d^2=1$.
Let $X=\{(x_1,x_2,x_3,x_4)\mid x_1^2+x_2^2=x_3^2+x_4^2=1\}$.
Why must the points $\gamma(t)$ be dense in $X$?
AI: HINT: If $a=\cos\theta$ and $b=\sin\theta$, then the first two coordinates are $\cos(t-\theta)$ and $\sin(t-\theta)$
Then for any particular value of $t$, you add multiples of $2\pi$ and the third and fourth coordinates are dense in that circle. |
H: A simple Inequality: $\frac{a+b}{\max\{a',b'\}}\leq\frac{a}{a'}+\frac{b}{b'}$?
We know $a \geq a' \geq 0$ and $b\geq b' \geq 0$ . How we can prove:
$\frac{a+b}{\max\{a',b'\}}\leq\frac{a}{a'}+\frac{b}{b'}$
AI: Hint: Assuming that $a',b'>0$, use the fact that $\max\{a',b'\} \ge a',\ \max\{a',b'\}\ge b'$. |
H: What is the probability space for Poisson process?
I'm studying discrete stochastic processes using notes and lectures of prof. Gallager [1], and I was wondering if somebody could intuitively explain what the probability space (usually $\Omega$) for Poisson processes is. Or how to understand that?
Particularly, in the text they say that these events are equal:
$$\{S_n > t\} = \{N(t) < n\}$$
$S_n$ is the time of n-th arrival
$N(t)$ is number of arrivals at time $t$
Assuming that an event is a set, I should be able to see that the set that the LHS generates is the same as the set that RHS generates. But I can't see how I can find these sets.
Could somebody help me with that please?
[1] Page 71 here: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-262-discrete-stochastic-processes-spring-2011/course-notes/MIT6_262S11_chap02.pdf
AI: One does not need to know what the probability space is to understand anything about Poisson processes.
In the case at hand, for every $t\geqslant0$, $N(t)=\max\{n\geqslant0\mid S_n\leqslant t\}$ with the convention that $S_0=0$. Equivalently, $N(t)=\sum\limits_{k=1}^\infty\mathbf 1_{S_k\leqslant t}$. Hence the event $[N(t)\geqslant n]$ corresponds to at least $n$ indexes $k\geqslant1$ being such that $S_k\leqslant t$, that is, to the fact that $S_k\leqslant t$ for every $1\leqslant k\leqslant n$, or, equivalently (since the sequence $(S_k)$ is almost surely nondecreasing), to the fact that $S_n\leqslant t$.
Thus, $[N(t)\geqslant n]=[S_n\leqslant t]$. Can you conclude? |
H: Expectation of $\frac{1}{1+X}$ for Gamma
I am trying to evaluate the following integral:
$$ \int_{0}^{\infty} \frac{1}{c+x} \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} dx $$
I have tried simple transformation and by-parts and nothing worked. Then I found this simple solution on internet at http://www.physicsforums.com/showthread.php?t=655393 :
$$ \int_{0}^{\infty} \frac{1}{c+x} \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} dx = \beta^{\alpha} c^{\alpha-1} e^{\beta c}\Gamma(1-\alpha, \beta c) $$
It would be great if someone could verify if this is indeed the answer. I don't know Maple or any other software that can evaluate integrals and hence there is no way to verify the correctness. Any reference (paper,book) would be great too.
I am a Bayesian Statistician and I need this for evaluating a posterior quantity.
AI: Use $\displaystyle\frac1{c+x}=\int_0^\infty\mathrm e^{-(c+x)t}\mathrm dt$ and interchange the order of the integrals. This yields that the integral you wish to compute is
$$
I=\int_0^\infty\mathrm e^{-ct}\int_{0}^{\infty}\frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1}\mathrm e^{-(\beta+t) x} \mathrm dx\mathrm dt=\int_0^\infty\mathrm e^{-ct}\frac{\beta^{\alpha}}{(\beta+t)^{\alpha}}\mathrm dt.
$$
The change of variable $s=c(\beta+t)$ yields
$$
I=\mathrm e^{c\beta}\beta^{\alpha}c^{1-\alpha}\int_{c\beta}^\infty s^{-\alpha}\mathrm e^{-s}\mathrm ds=\mathrm e^{c\beta}\beta^{\alpha}c^{1-\alpha}\Gamma(1-\alpha, \beta c). $$ |
H: Explicit computation $\operatorname{Tor}(M,N)$
Let $R=\mathbb{C}[t]/t^2$ the ring of dual numbers. Using the homomorphism $\phi:R \to \mathbb{C}=R/(t)$ we have that $\mathbb{C}$ is a $R$-module, infact we have
$$\psi: \mathbb{C} \times \mathbb{C}[t]/t^2 \to \mathbb{C} $$
taking $\psi(a,b)=a\phi(b)$. So we have that $\mathbb{C}$ is a $R$-module and it is obvious that $R$ is a $R$-module. I'd like to compute $\operatorname{Tor}_i(R,\mathbb{C})$ for all $i \ge 1$. In order to do this I have to find a free resolution of $R$ ($\cdots \to P_1 \to P_0 \to R \to 0$). I think that we can take $P_i=R$:
$$ \cdots \to R \to R\to R \to 0 .$$
If I had taken the correct resolution we have to consider the tensor product $- \otimes_{\mathbb{C}[t]/(t^2)} \mathbb{C}$ and have
$$ R \otimes_{R} \mathbb{C} .$$
So I have to calculate the homology of the complex having $\operatorname{Tor}_i(R,\mathbb{C})=\mathbb{C}$ for all $i \ge 0$. Is it correct? Thanks!
AI: For modules over any ring $R$ we have
$$
\operatorname{Tor}_i^R(M,N)=0 \quad(i>0)
$$
whenever either of $M$ and $N$ is flat, in particular projective (or free).
If $M$ is projective, it's quite easy to show: a projective resolution of $M$ is $$\dots\to0\to0\to M\to M\to 0,$$ so removing the last $M$ and tensoring with $N$ gives the complex
$$
\dots\to 0\to 0 \to M\otimes_R N\to 0.
$$
Therefore the homology is zero on all degrees $>0$.
Thus you need nothing else to show that $\operatorname{Tor}_i^R(R,\mathbb{C})=0$ for $i>0$. |
H: Is entailment biconditional or conditional?
When we say a KB entails Q it means that it is never the case that KB is true and Q is false. Does this mean entailment is similar to the conditional statement KB -> Q? I'm confused because our textbook keeps using "if and only if".
AI: Entailment is closest to the material conditional: To say "$KB$ implies $Q,$" or "If $KB$, then $Q$," is to assert $$KB \rightarrow Q$$
To say "$KB \;\text{ if and only if } \;Q$" is to say: $$KB \rightarrow Q\quad\text{ AND }\quad Q \rightarrow KB$$
In this case, in order for the biconditional to be true, $KB$ and $Q$ must both be true, or must both be false. If their truth-values differ, then exactly one or the other of the two conditionals must be false, and hence, the biconditional is false. |
H: First steps in algebraic geometry
I'm reading some introductory material in algebraic geometry. I'm trying to closely follow
this http://people.fas.harvard.edu/~amathew/287y.pdf. I keep getting confused with some basic
notions. I'd appreciate if someone could help me clarifying some points.
Let $X$ be a projective curve and $D$ be a divisor on $X$. I am comfortable with the classical definition of the Riemann-Roch space $\mathcal{L}(D)$. Here are few questions I have:
(1) I read that $\mathcal{O}_X$ stands for the sheaf of regular functions on $X$. But
I learned that every non-constant function has a pole. So is $\mathcal{O}_X$ just the set of constant functions?
(2) What is the meaning of the sheaf $\mathcal{O}_X(D)?$ Is it the same as $\mathcal{L}(D)$ or something more general? What are the line bundles $\mathcal{O}_X(D)?$ which are not of type
$\mathcal{L}(D)$?
(3) What should I understand by "global sections" of line bundles?
I apologize for the (perhaps) silly questions. I'd appreciate any help.
AI: The global sections of $\mathcal{O}_X$ (i.e. $\mathcal{O}_X(X)$) are constant functions, if $X$ is a projective variety. However take for example $\mathbb{P}^1$ over $k = \mathbb{C}$. Then $\mathcal{O}_X(\mathbb{P}^1 - \{\infty\}) = \mathcal{O}_X(\mathbb{A}^1)$ is given by $\mathbb{C}[x]$ (because something like $x$ or $x^3 - 3x + 1$ only has a pole at infinity).
$\mathcal{O}_X(D)$ and $\mathcal{L}(D)$ are both used for the line bundle associated to a Cartier divisor. Wikipedia has a decent description of them. However in the notes you linked to, it seems the convention is $\mathcal{L}(D) = \mathcal{L}(D)(X)$, i.e. the global sections.
As mentioned above, sections of a line bundle are with respect to some open subset $U$ of the whole space $X$. A global section is when you choose the open subset $U = X$. |
H: How to prove that $\lim_{x \to 0} \sin(x) = 0$ using the epsilon-delta definition?
How do I prove that
$\lim_{x \to 0} \sin(x) = 0$
using the episilon-delta definition of a limit?
Do I have to divide the domain of $x$ into 4 cases for each quadrant?
Update:
Based on the input from @vadim123,
For any $\epsilon>0$
if there exists some $\delta>0$ satisfying $|x-0|<\delta$, then
we have to show that $|\sin\ x-0|<\epsilon$
Choose $\delta=\epsilon$.
By transitivity of the $<$ and $\le$ relations,
$|\sin\ x |\le|x|<\epsilon$.
we have now shown that $|\sin\ x| < \epsilon$.
AI: Hint: $|\sin x|<|x|$, for all $x$. |
H: Models of the full theory of a structure
I'm reading Model theory: an introduction, by David Marker. I'm at page 14, where it says:
...one way to get a theory is to take $\operatorname{Th}(\mathcal{M})$, the full theory of an $\mathcal{L}$-structure $\mathcal{M}$. In this case, the elementary class of models of $\operatorname{Th}(\mathcal{M})$ is exactly the class of $\mathcal{L}$-structures elementarily equivalent to $\mathcal{M}$....
I have some problem understanding these few lines. In fact, call $\mathcal{K}$ the class of models of $\operatorname{Th}(\mathcal{M})$. Let $\mathcal{A}\in\mathcal{K}$. Let $\phi$ be a sentence which is true in $\mathcal{M}$. Then $\phi$ is also true in $\mathcal{A}$. But why should the converse also hold? In order to have $\mathcal{A}\equiv\mathcal{M}$ (i.e. $\mathcal{A}$ and $\mathcal{M}$ elementary equivalent) each sentence which is true in $\mathcal{A}$ must be true also in $\mathcal{M}$. By construction we have: true in $\mathcal{M}$ implies true in $\mathcal{A}$, but the opposite implication sounds strange to me. For instance, let $\mathcal{M}$ be a group, and $\mathcal{A}$ a ring. Then every sentence $\phi$ in $\operatorname{Th}(\mathcal{M})$ is true in $\mathcal{A}$, but why do every sentence which is true for a ring is also true for a group?
AI: Suppose that $\mathcal{A}\vDash\varphi$. If $\mathcal{M}\not\vDash\varphi$, then $\mathcal{M}\vDash\neg\varphi$, so $\neg\varphi\in\operatorname{Th}(\mathcal{M})$, and $\mathcal{A}\vDash\neg\varphi$. Oops! |
H: Regularity of Lebesgue measure
Let $A\subseteq \mathbb{R}$ be a Lebesgue measurable set of measure $m(A)=p>0$. Then for all $0<q<p$, show that there is a subset $B\subseteq A$ with $m(B)=q$.
Which Theorem do i have to use here, regularity or density of Lebesgue measure or something else ?
AI: Consider $f(x) = \int_A 1_{(-\infty, x]}$. Then $\lim_{x \to -\infty} f(x) = 0$, $\lim_{x \to +\infty} f(x) = p$ and $f$ is continuous.
Then, use the intermediate value theorem. |
H: Same eigenvectors for A and A+rI: why?
I read [Lambiotte 2010, "Multi-scale modularity in complex networks"] that the eigenvectors of an (adjacency) matrix A and the matrix A + rI, with r a scalar and I the identity matrix, are the same. Why is this so? Thanks!
AI: This is a special case of a more general result. Let $v$ be an eigenvector of $A$, then there exists $\lambda$ such that $Av = \lambda v$. Now let's look at $A+rI$.
$$(A+rI)v = Av+rIv = \lambda v + rv = (\lambda +r)v.$$
Thus if $v$ is an eigenvector of $A$, it is an eigenvector of $A+rI$. |
H: Is $y(x)=\frac{1}{2}M\left[1-\cos\left(\frac{\pi}{M}x\right)\right]$ an integer
Let
${\rm y}\left(x\right) = \frac{1}{2}M\left[1-\cos\left(\frac{\pi}{M}x\right)\right]$. Is ${\rm y}\left(x\right)$ an integer for each $x = 1,2,\ldots,M$ when
$M\to\infty$ ?.
AI: $\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
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\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert #1 \right\vert}%
\newcommand{\yy}{\Longleftrightarrow}$
For fixed $x$,
$$
\mbox{when}\ M \to \infty\,,\quad
{1 \over 2}\,M\bracks{1 - \cos\pars{{\pi \over M}x}}
\sim
{1 \over 2}\,\pars{{1 \over 2}\,{\pi^{2}x^{2} \over M^{2}}} = {\pi^{2}x^{2} \over 4}\,{1 \over M^{2}} \to 0
$$ |
H: Matrix with Functions as Entries
What do we call a matrix with functions as entries?
$$\textbf{f(x)}=\begin{bmatrix}
f_{11}(x) & f_{12}(x) \\
f_{21}(x) & f_{22}(x)
\end{bmatrix} $$
AI: Recall that you denote by $M_{2\times 2}(\mathbb{C})$ the set of matrices with entries in the complex numbers.
You can define matrices over other sets and depending on the structure of those sets these matrix algebras may or may not be interesting to you.
It is not clear what type of functions the $f_{ij}$ are here but because you might want to add and multiply the entries (why?), they will usually form a ring (briefly a set with addition and multiplication). So perhaps your functions come from the ring of continuous functions on the real line. You might denote this set by $\mathcal{C}(\mathbb{R})$.
Then your matrix above would be an element of the set of matrices over $\mathcal{C}(\mathbb{R})$ which we would denote by $M_{2\times 2}(\mathcal{C}(\mathbb{R}))$ and we could write
$$\mathbf{f}=\left(\begin{array}{cc}f_{11} & f_{12}\\ f_{21} & f_{22}\end{array}\right).$$
Your object above then seems to do more. It seems to take as an input $x\in\mathbb{R}$ and output a 2$\times$2 matrix:
$$\mathbf{f}:x\mapsto \left(\begin{array}{cc}f_{11}(x) & f_{12}(x)\\ f_{21}(x) & f_{22}(x)\end{array}\right),$$
so you have a function
$$\mathbf{f}:\mathbb{R}\rightarrow M_{2\times 2}(\mathbb{R}).$$
EDIT: i.e. what copperhat said.
Now you might begin to ask what kind of properties does this map have? |
H: Question about sheaves on projective varieties.
I am new to algebraic geometry, and really can't get idea of this:
For any product $X_{1} \times X_{2}$ of a projective varieties, with projections $p:X_{1}\times X_{2} \rightarrow X_{1}, q:X_{1}\times X_{2}\rightarrow X_{2}$ and let $L_{1},L_{2}$ be sheaves on $X_{1},X_{2}$.
We set $L_{1}\boxtimes L_{2}:= p^{*}L_1 \otimes q^{*}L_2$
i found this in a paper "Betti numbers of graded modules and cohomolgy of vector bundles", page $25$ http://arxiv.org/pdf/0712.1843v3.pdf
can anyone tell me what are $p^{*}$ and $q^{*}$?
AI: The usual notation is $L_1 \boxtimes L_2$. It is called the external tensor product of $L_1$ with $L_2$.
To your question: Every morphism of ringed spaces $f : X \to Y$ induces a pullback functor $f^* : \mathsf{Mod}(Y) \to \mathsf{Mod}(X)$. It is (defined to be) the left adjoint functor to the direct image functor $f_* : \mathsf{Mod}(X) \to \mathsf{Mod}(X)$ (with $(f_* F)(U)=F(f^{-1}(U))$ for $U \subseteq Y$). |
H: Inequality of scalar-product and norm
Why does the following inequality hold, given $A$ is symmetric and $\lambda_{\min} (A)$ is the smallest Eigenvalue of $A$?
$$v^\top A v \ge \lambda_{\min} (A) \; ||v||^2$$
AI: Assuming that $A$ is real and symmetric, it can be orthogonally diagonalized, that is, for some orthogonal $U$ (meaning $U^TU=I$) we have $U^T A U = \Lambda$, where $\Lambda$ is a diagonal whose entries are the eigenvalues of $A$. Since $A$ is real and symmetric, its eigenvalues are all real.
Note that since $U$ is orthogonal, we have $\|Ux\|= \|x\|$.
Then $\langle v , A v \rangle = \langle v , U^T \Lambda U v \rangle = \langle U v , \Lambda U v \rangle = \sum \limits_{k} \lambda_k [Uv]_k^2 \ge \lambda_\min \sum \limits_{k} [Uv]_k^2 = \lambda_\min \|U v\|^2 = \lambda_\min \|u\|^2$. |
H: Set intersection problem.
Is $ A \cap B' = A - B$ where $A \cup B$ is the universal set?
I am an absolute beginner at Sets, So please dont vote down my question because it might be too easy for you. $ B'$ refers to the complement of set $B$.
AI: It doesn’t matter whether $A\cup B$ is the universal set or not. If $x\in A\cap B\,'$, then by definition $x\in A$ and $x\in B\,'$, so $x\in A$ and $x\notin B$. But that means by definition that $x\in A\setminus B$. This shows that $A\cap B\,'\subseteq A\setminus B$.
Conversely, if $x\in A\setminus B$, then $x\in A$ and $x\notin B$, so $x\in A$ and $x\in B\,'$, and therefore $x\in A\cap B\,'$. This shows that $A\setminus B\subseteq A\cap B\,'$. Put the two together, and you have $A\cap B\,'=A\setminus B$.
Both simply describe the set of things that are in $A$ but not in $B$. |
H: Any finite set in $k^n$ is an algebraic set.
I'm trying to show that given a field $k$, and a finite set of points $\{a^i: i = 1\dots n\} \subset k^n$ is an algebraic set or equivalently is the set of common zeros of some set of polynomials $S \subset k[x_1, \dots, x_n]$.
For the case $n = 1$, we have $Z(x_1 - a_1^1, \dots, x_n - a_n^1)$ = $\{a^i\}$. For the case $n = 2$, we have two points $V = \{a^1, a^2 \}$. Let $f_i = (x_i - a_i^1)(x_i - a_i^2)$ , then clearly $V \subset Z(f_i : i = 1\dots n)$. Let $b \in Z(f_i)$. Then now $b$ can be a mixture of $a^1, a^2$, for example $b = (a^1_1, a^2_2, a^1_3, \dots)$. So that construction doesn't work!
AI: Hint:$V(I)\cup V(J)=V(I\cap J)$. Then take the intersection of those maximal ideal. |
H: Group isomorphism: $\mathbb{R}/\mathbb{Z}\cong S^1$
Let $\mathbb R$, $\mathbb{Z}$ be the groups of real numbers and integers respectively under addition, and $S^1$ denote the group of complex with modulus $1$ under multiplication.
Then show that $\mathbb{R}/\mathbb Z\cong S^1$.
My idea is to build a homomorphism with kernel $\mathbb Z$. I tried mapping $x\to \left(\{x\},\sqrt{1-\{x\}^2}\right)$, where $\{\cdot \}$ is the fractional part.
But its ineffective because my construction only maps to the complex number in the first quadrant.
AI: Note that $\mathbb{R}/\mathbb{Z}=\lbrace [r] : r\in \mathbb{R}\rbrace$, where $[r]=r+\mathbb{Z}$.
As Prahlad Vaidyanathan suggests above, consider the map $$f:\mathbb{R}/\mathbb{Z}\to S^1, \\ \;\;\;\;\;\;\;\;\;\;[r]\mapsto e^{2\pi i r}$$
This is a group homomorphism, since $$f([r]+[s])=f((r+s)+\mathbb{Z})=e^{2\pi i (r+s)}=e^{2\pi i r}e^{2\pi i s}=f([r])f([s]).$$
It only remains to prove that the map above is a bijection, or, in other words, that $f$ is both an injection and a surjection, which is not hard. |
H: Is there a name for a block-diagonal matrix with blocks of the form $\begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix}$?
Is there a name for a real square matrix of the form
$$\begin{pmatrix}
0 & a_1 \\ -a_1 & 0 \\
& & 0 & a_2 \\ & & -a_2 & 0 \\
& & & & \ddots \\
& & & & & 0 & a_k \\
& & & & & -a_k & 0
\end{pmatrix}$$
optionally with one or more additional rows/columns of 0?
For example, analogous to the fact that a real symmetric matrix is orthogonally equivalent to a real diagonal matrix, we can show that a real skew-symmetric matrix is orthogonally equivalent to a matrix of the form above.
AI: The above is a quasi-diagonal skew-symmetric matrix. Obviously, I don't need to explain the skew-symmetric part.
The term "quasi-diagonal" means "block-diagonal with the diagonal blocks of order at most $2$". They are studied, for example, here. The name is chosen with respect to quasi-triangular matrices which are often used in real Schur and similar decompositions (see here).
The term itself is mentioned here. |
H: prime number related proof
I want to prove if following is true for every integer a,b and c
$$a^2 - b^2 = cp $$
then p|(a+b) or p|(a-b) where p is a prime number. Any suggestion, help would be highly appreciated. Thanks in advance
AI: $$a^2-b^2=cp\implies a^2\equiv b^2\pmod p$$
If $p|a, b^2\equiv0\pmod p\implies b\equiv0$
Else $(ab,p)=1\implies \left(\frac ab\right)^2\equiv1\pmod p$
Now use this |
H: How to divide by 12 quickly?
Let $n\in\mathbb N$ be divisible by 12 and $n/12<100$. Is there a way of computing $n/12$ rather quickly using mental arithmetic (e.g. for 972/12, 1044/12, etc.)?
For example, the number 11 seems to have a nice property. When we consider $836/11=d$ then
770<836, but 880>836, so the first digit of $d$ must be 7. And since the last digit of $836$ is $6$, so is the last digit of $d$. This gives us $d=76$.
Or consider $693/11=d$. Then $660<693$ (and 770>693), so the first digit of $d$ must be $6$ and the second is $3$ since this is the last digit of $693$. This gives us $d=63$.
Now, is there another (possibly similar) approach for dividing by 12 (or even 13, 14, etc.)? But I am mostly interested in a "trick" for the number 12 (just using mental arithmetic).
AI: One can memorize multiples of $12$ that are less than $100$. Then if $n/12=10a+b$ you can guess $a$ quite fast. And there are two options for $b$. Now if $n=100a+(2a+b)10+2b$ then you can decide about $b$ by comparing $2a+b$ and the second least significant digit of $n$. |
H: differentiability and local Holder continuity
My analysis is really rusty, so apologies if this is a stupid question.
If $f\in C^1$ in a compact set $\Omega$, does this mean $f$ is Holder continuous for any $\alpha$ in $\Omega$?
I have tried googling but I couldn't find this result,
I have tried to do this myself. take an arbitrary $x$ then we have, for any $\epsilon >0$, we can choose a $\delta\leq 1$
$f'(x) -\epsilon < \dfrac{f(x)-f(y)}{x-y}<f'(x)+\epsilon$
so we have $|f(x)-f(y)|\leq K|x-y|\leq K|x-y|^\alpha$
but this only shows $f$ is Holder at every point inside $K$? Can we show $f$ is Holder on K?
Also are there weaker assumptions than $C^1$, I could have assumed here?
I assume $f$ is $\alpha$ Holder means it is $f$ is $\beta$ Holder for all $\beta<\alpha$?
AI: This is a good question.
In particular, you can conclude that $f$ is Lipschitz continuous (Holder continuous with $\alpha = 1$).
Because the derivative is continuous on a compact set, it is uniformly bounded, i.e. $|f'(x)| < K$ for all $x\in\Omega$.
Thus, by the mean value theorem, for any $x,y$, $\exists c$ such that
$$
|f(x) - f(y)| \leq |f'(c)||x-y| < K |x-y|.
$$ |
H: Frobenius Norm Unitary Operators
For something I'm working on, I have a matrix $A$ with other matrices $U$ and $V$ which are unitary ($U^*U = I$ and $V^*V = I$), and I'm trying to show that, for the Frobenius norm, $\|A\| =\|UA\| =\|AV\| = \|UAV\|$. Now, I solve out the first portion, but everything else is giving me trouble (the 3rd and 4th parts of the equation). Do I have to make use of singular decomposition somehow?
AI: If $U$ are unitary, then $\|U x\| = \|x\|$.
Then we have $\|A\|_F^2 = \sum_k \|A e_k\|_2^2 = \sum_k \|UA e_k\|_2^2 = \|UA\|_F^2$
We also have $\|A\|_F^2 = \|A^*\|_F^2$. The above result gives
$\|A\|_F^2 = \|A^*\|_F^2 = \|V^*A^*\|_F^2 = \|AV\|_F^2$.
Combining gives the desired result. |
H: Prove $f_n(x)=(1-x/n)^n$ converges uniformly on non-negative reals
I need to prove $f_n(x)=(1-x/n)^n$ converges on non-negative reals. I can easily prove it converges to $f(x)=e^{-x}$, but it is unclear to me whether this convergence is pointwise or uniform.
I have attempted the proof by taking the Taylor expansion of $f$ and do the binomial expansion of $f_n$, and then subtract.
\begin{equation}
|f(x)-f_n(x)|=|\sum_{k=0}^{\infty}(-1)^k\frac{x^k}{k!}-\sum_{k=0}^{n}(-1)^k\binom{n}{k}(\frac{x}{n})^k|\\
=|\sum_{k=0}^{n}(-1)^k\frac{x^k}{k!}-(-1)^k\binom{n}{k}(\frac{x}{n})^k+O(\frac{1}{n^{k+1}})|
\end{equation}
Toss out the big O term for now, since it is just part of the Taylor expansion of $e^{-x}$, with $k>n$. I tried manipulate the terms.
\begin{equation}
=|\sum_{k=0}^{n}(-1)^k\frac{x^k}{k!}(1-\frac{n(n-1)...(n-k+1)}{n^k})|\\
=|\sum_{k=0}^{n}(-1)^k\frac{x^k}{k!}\frac{n^k-n(n-1)...(n-k+1)}{n^k}|\\
\end{equation}
For any non-negative real $x$, each term in the summation clearly converges as we take $n$ large enough. But given a fixed $\epsilon>0$, I can't seem to find $N\in \mathbb{Z^+}$ such that such that $\forall n>N$, $|f(x)-f_n(x)|<\epsilon$. In other word, I have trouble proving the speed of the convergence does not depend on $x$.
Any help would be greatly appreciated, as I really would like to understand this. Thank you in advance!
AI: The convergence is not uniform: if it was, then for all $\varepsilon>0$ there exists an $n_0\in\mathbb N$ such that, if $n\geq n_0$, then $|f_n(x)-e^{-x}|<\varepsilon$ for all $x\geq 0$, so $|f_n(2n)-e^{-2n}|<\varepsilon$ for all $n\geq n_0$. Therefore, the sequence $(f_n(2n)-e^{-2n})$ converges to $0$. But, $$f_n(2n)-e^{-2n}=\left(1-\frac{2n}{n}\right)^n-e^{-2n}=(-1)^n-e^{-2n},$$ which does not converge to $0$, and this is a contradiction. |
H: Algebraic solution to: Do the functions $y=\frac{1}{x}$ and $y=x^3$ ever have the same slope?
The exercise doesn't specify how it must be answered, so I chose a graphical proof because I couldn't come up with an algebraic one. Sketching the graphs of $y=\frac{1}{x}$ and $y=x^3$, I noticed that $y=x^3$ always has a nonnegative slope, whereas $y=\frac{1}{x}$ always has a negative slope. Therefore these two functions never have the same slope.
However, I'm wondering if there's a algebraic way of showing this. I thought of differentiating each function and setting the values equal, but I think this would only prove that the two functions don't have the same slope at any particular x, and not that they don't have the same slopes anywhere over their domains.
AI: You can prove the exact observation you took from your sketch (note that sketch $\ne $ proof), and indeed you are on the right track with differentiation - no wonder if the problem is about slopes.
The derivative of $x\mapsto \frac 1x$ is $-\frac 1{x^2}$ (where $x\ne 0$), hence $<0$ as the square of a nonzero number is strictly positive.
The derivative of $x\mapsto x^3$ is $3x^2$, hence $\ge 0$ (again because squares are nonnegative).
A number $<0$ can never equal a number $\ge 0$. |
H: polynomial division, gcd, question
We are asked to show that there are polynomials $p,q \in Q[t]$ such that:
$p(t)*(t^4+2t^2+1)+q(t)*(t^4-3t^2-4) = t^2+1$
Is the answer the same for $t+5$ instead of $t^2+1$?
What I tried doing:
I don't really know why, but I thought maybe finding the gcd of $(t^4+2t^2+1)$ and $(t^4-3t^2-4)$ would help, maybe its $t^2+1$, but it wasnt.
I don't have an idea how to solve this.
AI: Note that
$$
t^4 + 2t^2 + 1 = (t^2+1)^2, \text{ and } t^4-3t^2-4 = (t^2+1)(t^2-4)
$$
So $t^2 + 1$ is the gcd.
You are in $\mathbb{Q}[t]$, which is a Euclidean domain - in which the gcd of two elements is a "linear combination" of those two elements (You should find a proof of this in pretty much any reasonable book on ring theory). Hence, there exist polynomials $p(t), q(t)$ as you require. |
H: How many automorphisms a countable field has?
Let $\mathbb{B}$ a countable algebraically closed field (car=0) of infinite transcendence degree. How many automorphisms $\mathbb{B}$ has? Are they $2^\omega$?
AI: Your field is isomorphic to the algebraic closure of the field $\mathbb{Q}(x_1,x_2,\dots,x_n,\dots)$, where the $x_i$ are "indeterminates." Any of the $2^\omega$ one to one onto maps from $\{x_1,x_2,\dots\}$ to itself induces an automorphism of $A(x_1,x_2,\dots)$. There can be no more than $2^\omega$ automorphisms, for there are only $2^\omega$ functions from our field to itself. |
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