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H: What will be expected value of smallest element of chosen set
We are given a set $X = \{x_1,x_2,\ldots,x_n\}$ where $x_i = 2^i$. A sample $S$ (which is a subset of $X$) drawn by selecting each $x_i$ independently with probability $p_i = \frac{1}{2^i}$. What will be the expected value of the smallest number in sample $S$?
AI: The answer $E_n$ depends on $n$, for $n=0$ it is clearly $E_0=0$.
For $n\ge 1$, we have the recursion
$$E_n=1 + \frac12 E_{n-1}$$
because with probability $\frac12$ we have $\min S=2$ and with probability $\frac12$ the result is the same as from taking a sample from $\{x_2,\ldots ,x_n\}=2\cdot\{x_1,\ldots,x_{n-1}\}$ and rejecting it with probabiliy $\frac12$. By induction, $E_n=2-2^{1-n}$. |
H: What's the limit?
Suppose $a>0$. Let $x_1=\sqrt{a}$, and define $x_{n+1}=\sqrt{a+x_n}$ for $n\ge 1$. I've already used induction to show that $x_n<1+\sqrt{a}$ for all $n$ and that $\{x_n\}$ is an increasing sequence. I know that it is bounded and must converge by the monotone convergence theorem. I'm just not sure how to find the limit.
Any suggestions?
AI: The trick is that if it converges, you can get at the limit $x = \sqrt{a + x}$. Now just solve for $x$.
Edit: In particular $x^2 - x - a = 0$, so that $x = \frac{1 \pm \sqrt{1 + 4a}}{2}$. Notice that $x > 0$, so we can discard the negative solution. |
H: Proof about some recursion
Let $f(n)=\overset{n}{\underset{k=1}{\Sigma}}\lceil log_{2}k\rceil$. Prove that
$$f(n)=n-1+f(\lceil\frac{n}{2}\rceil)+f(\lfloor\frac{n}{2}\rfloor)$$
for all $n\geq1$.
Hint i've got for this: divide $\overset{n}{\underset{k=1}{\Sigma}}\lceil log_{2}k\rceil$ for sums for odd and even $k$'s.
I have tried induction for this, but I can't hit on an idea how to use assumption there since I see only sums.
Thank you in advance for any hints for that.
AI: Let's use the hint, if $k$ is even, say $k = 2\kappa$, then
$$ \def\fl#1{\left\lfloor #1\right\rfloor}\def\cl#1{\left\lceil#1\right\rceil}
\cl{\log_2 k} = \cl{\log_2 2\kappa} = \cl{1 + \log_2\kappa} = 1 +\cl{\log_2 \kappa} $$
For odd $k$, say $k = 2\kappa - 1$, we get for $\kappa \ge 2$ noting that 1 is the only odd power of 2, that is, $\log_2 (2\kappa - 1)$ is never an integer for $\kappa \ge 2$, hence
$$ \cl{\log_2 k} = \cl{\log_2 (2\kappa - 1)} = \cl{\log 2\kappa} = 1 + \cl{\log_2 \kappa} $$
Now consider the cases $n$ odd or even. |
H: Finding the change in radius that induces an $11\%$ drop in gravitational force between two bodies.
How come differentials only estimate the answer and don't give an exact answer that you might get when you calculate the real figure using other methods?
Example: Let $F = Gm_1m_2/r^2$
Let $r=8$. What is the change in $r$ that would correspond to an 11% decrease in $F$?
My calculations is that $r$ changes by .44 but this isn't what I get when I calculate this by hand.
AI: Let $\rho$ be the new radius. Then you want to have $$0.89=\cfrac{\left(\cfrac{Gm_1m_2}{\rho^2}\right)}{\left(\cfrac{Gm_1m_2}{8^2}\right)}=\frac{64}{\rho^2},$$ so $$\rho^2=\frac{64}{0.89},$$ and so $\rho=\sqrt{\dfrac{64}{0.89}}.$ The difference, then, is $\sqrt{\dfrac{64}{0.89}}-8.$ An approximate answer is $0.48.$ I'm not sure what happened in your calculations, but you must have made a mistake somewhere.
Now I understand what you did. The principle behind differential approximation is that $$\Delta f\approx\frac{df}{dx}\cdot\Delta x,$$ and that this approximation can be made as close as we like, simply by making $\Delta x$ small enough. The problem is that "small enough" will vary from function to function. Here, you were able to get an approximation within about $0.04,$ which isn't bad, but to get a better approximation to $\Delta F$, we'd want a less substantial $\Delta r$ |
H: Find another recursive algorithm that is equal to a series
I have the following sequence:
$$
y_n = \int_0^1 \frac{x^n}{x+5}\,dx, n = 0,1,\dots
$$
Now I have the following recursive algorithm which is equal to the sequence:
$$
y_0 = \log{6} - \log{5}
$$
$$
y_n = \frac{1}{n} - 5y_{n-1}, n = 1,...
$$
I have to find another recursive algorithm equal to the sequence. I got the tip to write the recursion backwards e.g start with $y_9 \approx 0$
I don't get it, can you help me, thanks.
AI: You already have the equation $y_n = 1/n - 5 y_{n-1}$. But instead of using this to calculate $y_n$ from $y_{n-1}$, you can use it to calculate $y_{n-1}$ from $y_n$. As a starting point take $y_N \approx 0$ where $N$ is fairly large.
The fact that this "backwards" recursion is very stable (error is divided by $5$ at each step) means that even if the approximation for $y_N$ is not very good, you will get very good approximations after a few steps. |
H: Evaluating a sum with binomial coefficients: $\sum_{k=1}^n {n \choose k} \frac{1}{k^r} a^k b^{n-k}$
I have come across the following sum evoking the binomial theorem:
$$\sum_{k=1}^n {n \choose k} \frac{1}{k^r} a^k b^{n-k},$$
where $r > 0$ is a positive real constant and $a,b \in \mathbb{R}$ are arbitrary real numbers.
Could anyone suggest an effective method how to approach this sum in order to obtain a closed-form solution or at least an asymptotic estimate? Thank you in advance.
AI: This can be partially evaluated using convolutions. Let $c_k:=\frac{a^k}{k^r}$ and $d_k:=\binom{n}{k}b^k$, with the convention that $d_k=0$ for $k>n$. Notice that $\binom{n}{k}:=\binom{n}{n-k}$.
Then if $C(z):=\sum_{k=0}^\infty c_kz^k$ and $D(z)=\sum_{k=0}^\infty d_k z^k$, your sum will be the $z^n$ coefficient of $C(z)D(z)$. It's easy to see that $D(z)=(1+bz)^n$ and a little harder that $C(z)=L_r(az)$ where $L_r(z)$ is the polylogorithm function. Thus, to find your sum, you need to differentiate $(1+bz)^nL_r(az)$ exactly $n$ times and divide by $n!$. |
H: Calculating the number of triangles
I am trying to calculate how many triangles that can be found in an equilateral triangle with $2n$ lines starting at the bottom angles and ending at the opposite side, such that equally many lines start/end of either side.
This is rather hard to explain, so I drew that first 5 terms:
The 1st term has, of course, 1 triangle.
The 2nd term has 8 triangles.
The 3rd term has 27 triangles.
This problem is really killing me, so any help would be greatly appreciated.
AI: Well, the way I see it; break down the lines into the two groups of $n$ lines on the left and right, and add the "left" and "right" side of the original triangle to make two groups of $n+1$ lines. There are then two cases I see:
One side of the triangle is from a line on the "left", one from the "right", and the bottom side of the original triangle. There are $(n+1)^2$ such triangles.
Two of the sides are from lines in one group, while the third side is from a line in the other. This gets us $2(n+1)\binom{n+1}{2}$ more.
That gets me $(n+1)^2 + 2(n+1)\binom{n+1}{2} = (n+1)^3$. |
H: Find the Laplace transform?
Find the Laplace transform of $f(t) = 1 + (1 - t)u_1(t) + (t-2)u_3(t)$.
Obviously each term of the function must be of $f(t - c)u_c(t)$ or be clearly transformable. Thus we have for our first term simply $\frac{1}{s}$. Our second term has been shifted by $f(t) = -t \to f(t - 1) = 1 - t$. Our third term has been shifted by $f(t) = t + 1$ and so $f(t - 3) = t - 2$.
Now I have to transform it. I get $$G(t) = \frac{1}{s} + e^{-s}(t - \frac{t^2}{2}) + e^{-3s}(\frac{t^2}{2} - 2t)$$
I don't know if this is correct. I used transform #28 from the following table : http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
AI: $f(t) = 1+u_1(t) - t u_1(t) + t u_3(t) - 2 u_3(t)$.
You need to take care with multiplication by $t$. The rule is ${\cal L} (t \mapsto t f(t))(s) = - \frac{d \hat{f}(s)}{ds} $.
Then (abusing notation slightly)
\begin{eqnarray}
\hat{f}(s) &=& \frac{1}{s} + \frac{1}{s} e^{-s} - (- \frac{d}{ds} (\frac{1}{s} e^{-s})) + (- \frac{d}{ds} (\frac{1}{s} e^{-3s}))-2\frac{1}{s} e^{-3s} \\
&=& \frac{1}{s} + \frac{1}{s} e^{-s} - \frac{(s+1)}{s^2} e^{-s} + \frac{3s+1}{s^2} e^{-3s} -2\frac{1}{s} e^{-3s} \\
&=& \frac{1}{s} -\frac{1}{s^2}e^{-s} + \frac{(s+1)}{s^2} e^{-3s}
\end{eqnarray}
When you apply the transform, all '$t$'s should disappear. |
H: Negating "If no one is absent, then if the weather permits, we will study outside"
I am a beginner; please help solve this.
Write the negation of the statement:
"If no one is absent, then if the weather permits, we will study
outside."
AI: First, write the statement as a logical statement (with appropriate notation).
A: "No one is absent."
W: "Weather permits"
S: "We will study outside."
Then we have the proposition, symbolically:
$$A\rightarrow(W \rightarrow S)$$
Then, since we want to negate the proposition, we work with:
$$\begin{align} \lnot\Big(A\rightarrow(W \rightarrow S)\Big)& \equiv \lnot\Big(\lnot A \lor (\lnot W \lor S)\Big)\tag{$p\rightarrow q \equiv \lnot p \lor q$}\\ \\ & \equiv A \land W \land \lnot S\tag{DeMorgan's}\end{align}$$
So the negation of the statement translated back to English gives us:
"No one is absent and weather is permitting, but (and) we won't study outside." |
H: Linear Algebra problem, old Berkeley exam.
I came across this problem:
Let $G$ be the group of $2 \times 2$ matrices with determinant $1$ over the four-element field $\bf F$. Let $S$ be the set of lines through the origin in $\bf F^2$. Show that $G$ acts faithfully on $S$. (The action is faithful if the only element of $G$ wich fixes every element of $S$ is the identity.)
I solved it (not really sure about the correctness of the proof) without using any abstract algebra.The only "official" solution i found to this problem uses some abstract algebra and is more elegant.
Nonetheless I tried my best and would like to have some feeback (is my proof correct?) on my proof. I'm really thankful for your help. Here is my solution:
Solution: Suppose $G$ acts not faithfully on $S$.Then $\exists g \neq id \in G : gs=s , \ \forall s\in S$. We saw the multip. table of $\bf F$ in the lecture. I want to identify only $2$ elements of $S$. Those will suffice for our proof. Each line in $ \bf F$ is determinated by it's slope wich can be $0,1,a,b,$ or $\infty$. We consider the lines $S_1$ and $S_a$ with slopes $1$ and $a$ respectivly.
Define $g$ as $g:=\begin{pmatrix} g_1 & g_2 \\g_3 & g_4 \end{pmatrix}$ with $g_{1,2,3,4} \in \bf F$. It's easy to verify that
$$
S_a=\left \{(0,0),(1,a)\right \} \text{ and } S_1=\left \{ (0,0),(1,1),(a,a),(b,b) \right \}.
$$
Now the idea is to apply $g$ on $(1,1)$ and $(1,a)$ and try to find $g_{1,2,3,4}$ wich fit with the conditions on $g$. Applying $g$ on $(1,1)$ we get the condition:
$$g_1+g_2=1.\tag{1}$$
While applying $g$ on $(1,a)$ we get:
\begin{align}
g_1+g_2a&=1,\tag{2}\\
g_3+g_4a&=a.\tag{3}
\end{align}
Combining $(1)$ and $(2)$, we get $g_2a=g_2$, hence we have two possibilities. Either $a=1$ (wich cannot be, checking the mult. table) or $g_2=0$. But we know another important thing about $g_{1,2,3,4}$, namely $g_1g_4-g_2g_3=1$ since the determinant of $g$ is $1$. Knowing that $g_2=0$ this equation becomes $g_1g_4=1$. Now plugging $g_2=0$ in $(1)$ we get $g_1=1$ hence from the last equation $g_4=1$. Now,finally, considering $(3)$ and the fact that $g_4=1$ we get $g_3+a=a \Rightarrow g_3=0$ hence $g=id$, contradiction!
Hope the works and the proof is not too confusing.
Besides the feedback (wich is actually what i need), alternative solutions are of course also welcome!
AI: The problem with your proof is that you assumed that
$$g (1,a) = (1,a)$$
and $$g(1,1) = (1,1)$$
But the matrices act on lines, so when you have, for example, the line $s=\{ (1,1) ,(0,0) , (a,a) , (b,b) \}$ , then $gs=s$ only means that $\{ g(1,1) ,g(0,0) , g(a,a) , g(b,b) \} = \{ (1,1) ,(0,0) , (a,a) , (b,b) \}$. i.e. $g(1,1)$ could be any one of the four elements of $s$. That will make everything a bit more complicated and maybe you'd have to consider several cases, but if you're willing to solve even more linear equations, something similar to your attempt could work to solve the problem. |
H: Is this polynomial solvable? $\sum_{i=0}^{k} p_{i}x^{k-i}$
I'm given $p_{0},...,p_{k} \in \mathbb{R}$ and I only know, that $p_{0}=-1$. We create a polynomial using these values like this :
$\sum_{i=0}^{k} p_{i}x^{k-i}$
Can I tell for sure, that no roots exist? I got class notes that said so, yet I don't see why not. Or do they mean, that we don't know for sure, that these roots exist?
AI: I will assume that you are looking for roots over the reals. If $k$ is odd, then you polynomial will have a root. This follows from the intermediate value theorem. In the general case, you can use Sturm's Theorem to count the number of real roots. See here: http://en.wikipedia.org/wiki/Sturm%27s_theorem#Applications |
H: How to prove that for a sheaf functor F, F(empty set) = terminal object?
The empty set is covered by an empty family of sets. Therefore any two sections of F(empty set) are vacuously equal, so the F(empty set) can be at most a singleton. Now, why does this mean that it is in fact the terminal object, as Vakil claims in his Algebraic geometry notes? In Sets, this is true, but why in general? In particular, what happens if we are talking about a sheaf of rings with unity. Then I think that we actually have a problem because rings with unity have at least two elements.
AI: For any index set $I$ we can consider the covering of $U=\emptyset$ by the sets $U_i=\emptyset$ for $i\in I$. This gives us that $F(U)$ in
$$ \tag1F(U)\to\prod_{i\in I}F(U_i){{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}} \prod_{i,j\in I}F(U_i\cap U_j)$$
is an equalizer. For $U=\emptyset$ we may in fact pick $I=\emptyset$ and this still works. But the empty product of whatever is terminal, so $(1)$ becomes
$$ F(\emptyset)\to T{{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}}T$$
where $T$ is terminal and the two arrows on the right are $\operatorname{id}_T$.
The equalizer condition says that for any object $X$ together with a morphism $X\to T$ such that in $X\to T{{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}}T$ both paths to the right $T$ give the same morphism, there exists a unique morphism $X\to F(\emptyset)$ that makes
$$\tag2 \begin{matrix}F(\emptyset)&\longrightarrow&T\\\uparrow&\nearrow\\X\end{matrix}$$
commute. But the condition about the morphism $X\to T$ is not really a condition: There exists a unique morphism from $X$ to $T$ any way and the equality of the two paths is always given (again by uniqueness of morphism to $T$). On the other hand, any morphism $X\to F(\emptyset)$ make $(2)$ commutative. Therefore $F(\emptyset)$ is simply an object such for any $X$ there exists a unique morphism $X\to F(\emptyset)$. - In other words, $F(\emptyset)$ is terminal.
Now that $(1)$ can be used to define the notion of sheaf only for a category with products including the empty product, hence also having terminal objects. The category Ring of rings with unity has products only for the nonempty case, so $(1)$ does not apply. More precisely, we can use $(1)$ for Ring valued sheaves only if $I\ne \emptyset$. Still, the formulation of locality with sections also postulates that $F(\emptyset)$ has only one element, which is impossible. Everything makes sense again if we augment Ring by adding the zero ring artificially as terminal object. |
H: Equivalence relations, Cosets
Let G be a group and for elements a,b (elements of)G let a R b mean that there exists an element x(element of)G such that a=xbx^(-1). Show that R is an equivalence relation on G.
Not really sure how to go about this one.
AI: You need to check the three conditions for an equivalence relation:
Check that $x \sim x$ under your relation $R$. So is there an element $a$ in every group $G$ such that $x=a x a^{-1}$?
Now check that if $x \sim y$, that is if $x=aya^{-1}$ that $y=bxb^{-1}$ for some $b \in G$. For this try solving for $y$ and then relabeling to find your $b$'s in terms of your $a$'s. Then have you found elements that make this work?
Now check that if $x \sim y$, that is if $x=aya^{-1}$ and $y \sim z$, that is if $y=bzb^{-1}$, that $x \sim z$. That means that $x=czc^{-1}$ for some $c \in G$. Try solving for $z$ in terms of $b$ and $y$ and then substituting that in your equation for $x \sim y$. Once you have that equation, can you find an element $c$ in terms of $a,b$ so that you have $x \sim z$? |
H: Question About Dual Vector Spaces and the Adjoint Map
I'm having trouble understanding some things about this problem:
Suppose that $U \subset V$ is a subspace. Let $I : U \rightarrow V$ be the inclusion map. The adjoint map $I^* : V^* \rightarrow U^*$ has a kernel: $$U^{\bot} := \mathrm{ ker}\; I^* \subset V^*$$ Show that $U^* \cong V^*/U^{\bot}$ ($U$ dual is isomorphic to $V$ dual mod $U^{\bot}$) by showing that elements of $V^*/U^{\bot}$ act as linear functionals $U \rightarrow \mathbb{R}$. Conclude from this that subspaces of $V$ are in one-to-one correspondence with quotients of $V^*$ and vice-versa.
We know that if $\eta$ is in the kernel of $I^*$, then $I^*\eta = 0 \in U^*$, so then $\forall u \in U, (I^*\eta)u = 0 = \eta(Iu) = \eta(u)$. And then $\{ \eta : U \subset \mathrm{ker} \;\eta\} = U^{\bot}$.
I don't understand how $U$ perp is a set of functionals. Isn't U perp just an ordinary vector space? Or are we using a notion of U perp that is different from the conventional usage?
Thank you.
AI: In this context, the annihilator $U^\perp$ of $U$ is defined as the kernel of $I^\ast : V^\ast \to U^\ast$, and hence, in particular, is a subspace of $V^\ast$, making it a set of functionals on $V$.
Now, what you're thinking of is the orthogonal complement of $U$ in $V$, in the case that $V$ is an inner product space; the point is that the annihilator $U^\perp \subset V^\ast$ is the generalisation of the orthogonal complement $U^\perp \subset V$ in the absence of an inner product on $V$. So, in the finite dimensional case, for simplicity, an inner product $\langle \cdot,\cdot \rangle$ defines an isomorphism $R : V \to V^\ast$, conjugate-linear if you're working over $\mathbb{C}$, by $v \mapsto \langle v,\cdot \rangle$. You can then check that $R$ restricts to an isomorphism
$$
\text{orthogonal complement of $U$ in $V$} \to \text{annihilator of $U$ in $V^\ast$}.
$$ |
H: Derivatives of $\frac{\csc x}{e^{-x}}$ and $\ln\left(\frac{3x^2}{\sqrt{3+x^2}}\right)$
I have tried to mainly ask thoughtful conceptual questions here, but now I am reduced to asking for help on a specific problem that I have been wrestling with for over an hour.
Disclaimer: I am not a lazy student trying to get free internet homework help. I am an adult who is learning Calculus from a textbook. I am deeply grateful to the members of this community for their time.
$$f(x)= \frac{\csc(x)}{e^{-x}}$$
$$f'(x)= ?$$
Answer key says choice (1): $$e^x\csc2x(1-2\cot2x)$$
2) My answer does not match the answer key. Is there a typo?
Any guidance is enjoyed...it's driving me nuts!
$$\frac{d}{dx}\ln\left(\frac{3x^2}{\sqrt{3+x^2}}\right)$$
Answer key says choice (2): $$\frac{x^2+6}{x^3+3x}$$
AI: For 1, I would recommend re-writing it as:
$f(x)=\frac{e^x}{\sin{x}}$.
Then using the quotient rule with $u=e^x, v=\sin{x},$
we obtain:
$f'(x)=\frac{e^x\sin{x}-e^x\cos{x}}{\sin^2{x}}=\frac{e^x(\sin{x}-\cos{x})}{\sin^2{x}}=\frac{e^x}{\sin{x}}\frac{\sin{x}-\cos{x}}{\sin{x}}=e^x\csc{x}(1-\cot{x}).$
I do not think this is equivalent to what was given in the textbook, although Wolfram Alpha agrees with my answer so I am inclined to think it is the correct one.
As for 2, it makes it considerably simpler to use log laws first, to show that:
$\ln{\frac{3x^2}{\sqrt{3+x^2}}}=\ln{3}+2\ln{x}-\frac{1}{2}\ln{(x^2+3)}=g(x)$, say.
Then using the standard derivative rules for logs we get:
$g'(x)=\frac{2}{x}-\frac{x}{x^2+3}=\frac{x^2+6}{x^3+3x}$, as required. |
H: For all integers a, b, c, if a | b and b | c then a | c.
Is this T or F? and most importantly, why?
I'll be using any answers for a basis or completely my other questions, since my understanding is still a little poor.
AI: HINT: This is just a matter of applying the definitions, which is always the first thing that you should try. What does $a\mid b$ mean? It means that there is an integer $m$ such that $b=am$. Similarly, $b\mid c$ means that there is an integer $n$ such that $c=bn$. You now have the equations
$$\left\{\begin{align*}
&b=am\\
&c=bn\;,
\end{align*}\right.\tag{1}$$
and you want to show that $a\mid c$. Once again you should turn to the definition: this means that there is an integer $k$ such that $c=ak$. Can you combine the equations in $(1)$ to find such an integer $k$? |
H: Proving that the Union of Two Compact Sets is Compact
Prove if $S_1,S_2$ are compact, then their union $S_1\cup S_2$ is compact as well.
The attempt at a proof:
Since $S_1$ and $S_2$ are compact, every open cover contains a finite subcover. Let the open cover of $S_1$ and $S_2$ be $\mathscr{F}_1$ and $\mathscr{F}_2$, and let the finite subcover of $\mathscr{F}_1$ and $\mathscr{F}_2$ be $\mathscr{G}_1$ and $\mathscr{G}_2$, respectively. If I can show that $S_1\cup S_2$ contains a finite subcover for every open cover, then I will have showed that the union is indeed compact. I note that $\mathscr{G}_1\subset\mathscr{F}_1$ and that $\mathscr{G}_2\subset\mathscr{F}_2$ (by definition of an open subcover). Then, I note that $\mathscr{G}_1\subset\mathscr{F}_1\cup\mathscr{F}_2$ and that $\mathscr{G}_2\subset\mathscr{F}_1\cup\mathscr{F}_2.$
I'm not sure how to proceed from this point. I think I am on the right track though. Any suggestions to proceed would be appreciated. Thanks in advance.
AI: HINT: You’re starting in the wrong place. In order to show that $S_1\cup S_2$ is compact, you should start with an arbitrary open cover $\mathscr{U}$ of $S_1\cup S_2$ and show that it has a finite subcover. The hypothesis that $\mathscr{U}$ covers $S_1\cup S_2$ simply means that $S_1\cup S_2\subseteq\bigcup\mathscr{U}$. Clearly this implies that $S_1\subseteq\bigcup\mathscr{U}$ and $S_2\subseteq\bigcup\mathscr{U}$. Thus, $\mathscr{U}$ is an open cover of $S_1$ and also an open cover of $S_2$. Now use the compactness of $S_1$ and $S_2$ to produce a finite subset of $\mathscr{U}$ that covers $S_1\cup S_2$. |
H: Counting the Elements of a Finite Group
Let $G = \mathbb Z_6 \times \mathbb Z_4$, and find $[G:H]$ for:
a) $H = \{0\} \times \mathbb Z_4$
b) $H = \langle 2\rangle \times \langle 2\rangle$
AI: Determine the order of $G: \,|G|$, and the orders of $H$ in each case; then you can easily calculate the index of $H$ in $G$: $$[G:H] = \dfrac{|G|}{|H|}$$
Now, since the order of a group is the number of elements it contains, and the order of the direct product of two sets is given by the product of the orders: $$|G| = |\mathbb Z_6\times \mathbb Z_4| = |\{0, 1, 2, 3, 4, 5\}| \times |\{0, 1, 2, 3\}| = 6 \times 4 = 24$$ and so there are $24$ ordered pairs in $G$.
As for the number of elements of $H$, given
a. $H = \{0\} \times Z_4 = \{(0, 0), (0, 1), (0, 2), (0, 3)\}$. So the order of $H$ in this case is equal to $4 = 1\times 4$.
b. $H = \langle 2\rangle \times \langle 2\rangle = \{(0, 0), (0, 1), (1, 0), (1, 1)\}$. So the order of $H$ in this case is also equal to $4$. |
H: Percentage of Amounts
I'm studying and I'm not that sure how to answer this question.
Is $97.1%$ $=$ $650,000,000$?
I was going to find $2.9%$ of $650,000,000$ however this would be wrong as I would finding our the annum of the $2002$ population, if you know what I mean. I know this looks quite easy but thanks.
AI: Let's think about this. Africa has a population of $650,000,000$ and is increasing every year by $2.9\%$. So next year they will have all the $650$ million and another $2.9\%$, this means they will have $$650+2.9\%(650)=650+0.029(650)=668,850,000$$ million people next year. But notice that both the terms above have a $650$ in common! So we can factor that number out. This gives us $$650(1+2.9\%)=650(1+0.029)=650(1.029)$$ So the number we could have multiplied by was $1.029$.
What did we learn from this? If we want to find something if it increases by some percentage $x\%$ after a certain time, we can multiply it by $$1+d$$
where $d$ is the percentage written as a decimal! Of course, if the number is DECREASING by a certain percentage each year, instead we could multiply by $$1-d$$ where $d$ is the decimal representation of our percentage.
Now for the second part. Think after 1 year, the year 2003, the population in Africa was $$650(1+0.029)$$ The next year, it grows by another $2.9\%$. That means the number we just got grows by another $2.9\%$, but from what we just learned that means it's $$\big(650(1+0.029)\big)(1+0.029)=650(1+0.029)^2$$ We can continue this process again and again until we reach the desired year of 2020. This will give us a number of the form $$650(1+0.029)^t$$ I'll let you figure out what the number $t$ will be! |
H: Sample Variance question
The random variable
$$ S^2=\sum_{i=1}^n \frac{(X_i- \overline X)^2}{(n-1)} $$
is called the sample variance.
A) Show that $ (n-1)S^2=\left[\sum_{i-1}^n ((X_i-\mu)^2 \right]-n(\overline X-\mu)^2 $
(Hint: Start with $(n-1)S^2=\sum_{i-1}^n ((X_i-\mu)+(\mu-\overline X))^2 $)
B) Use result from A) to show that $ E[S^2]=\sigma^2 $
A)
So far I used the hint to obtain
$$ \sum_{i=1}^n [(X_i-\mu)^2+2[(X_i-\mu)(\mu-\overline X)]+(\mu-\overline X)^2] $$
$$ =\sum_{i=1}^n (X_i-\mu)^2+\sum_{i=1}^n2[(X_i-\mu)(\mu-\overline X)]+\sum_{i=1}^n(\mu-\overline X)^2 $$
$$ =\sum_{i=1}^n (X_i-\mu)^2+\sum_{i=1}^n2[(X_i-\mu)(\mu-\overline X)]+n(\mu-\overline X)^2 $$
My next step was to take the sum of the middle part but it got messy and didn't look like it would help me much. I know need to get rid of that middle part and also negate the third part - Is it possible to just negate the n and then switch the $\mu\ \text{and the} \ \overline X$?
Help! I'm confused
AI: For the middle portion, note that: $\mu-\overline{X}$ does not depend on the index $i$ in the sum and thus can be taken out of the sum.
In other words, $\sum_{i=1}^n2[(X_i-\mu)(\mu-\overline X)]$ can be re-written as:
$$2(\mu-\overline X)\sum_{i=1}^n(X_i-\mu)$$
Now, use the fact that $n \overline{X} =\sum_{i=1}^nX_i$ and simplify. |
H: Vector Calculus: to get the parametric equations of the tangent line, why can't you just use the derivative of the position vector?
As you can see in the example, the author generates the parametric equations of the tangent line through the use of r(3) and r'(3).
Why couldn't the author just use the derivative of the position vector as highlighted?
$$x=2t$$
$$y=2t-1$$
$$z=-7$$
AI: The derivative $r'(t)$ is positioned with its tail (or starting point) at the origin $(0,0)$. So, you need to translate it to the point in question $(9,6,-21)$. |
H: Example of two norms and ONE linear operator that is bounded and unbounded in a norm.
I am looking for an example of a linear operator that is bounded as well as unbounded depending on which norm you take. Since I do not have much experience with Functional Analysis, I do not know many examples.
AI: Assume $\varphi : (E, \| \cdot \|_{E}) \to (F, \| \cdot \|_F)$ is an unbounded linear operator, i.e.
$$
\sup_{x \neq 0} \frac{\|\varphi(x)\|_F}{\|x\|_E} = \infty.
$$
Then define
$$
\forall x \in E, \qquad \|x\| \overset{def}= \|x\|_E + \|\varphi(x)\|_F.
$$
You can check that this gives a well defined norm on $E$ under which $\varphi$ is bounded. You don't need $\varphi$ to be unbounded for that to work, but if it is unbounded, then you have an example. The reason why $\varphi$ becomes bounded is because
$$
\sup_{x \neq 0} \frac{\|\varphi(x)\|_F}{\|x\|} = \sup_{x \neq 0} \frac{\|\varphi(x)\|_F}{\|x\|_E + \|\varphi(x)\|_F} \le 1.
$$
This is an example that happens in practice, it is not just a counter-example ; for instance, if $E = C^1[a,b]$ and $F = C[a,b]$, if we equip both of them with the supremum norm $\| \cdot \|_{\infty}$, then $\varphi : E \to F$ defined by $\varphi(f) = f'$ is not a bounded linear operator. Thus we define the new norm on $E$, namely $\|f\| = \|f\|_{\infty} + \|f'\|_{\infty}$, so that differentiation becomes continuous.
Hope that helps, |
H: How to sample points on a triangle surface in 3D?
To take random uniform points inside a triangle Triangle Point Picking method is used. But this is for 2D points, how can I take random points from a triangle that is defined by 3 arbitrary 3D points?
In other words, let's say I have 3 points in 3D space, and these points define a plane, how can I take random points on the plane such that my points are uniformly sampled inside the triangle that is defined by these 3 points?
Thanks in advance..
AI: The same exact algorithm should work in $3D$ too:
$$\vec{x} = \vec{v_1} + a (\vec{v_2} -\vec{v_1}) + b (\vec{v_3} -\vec{v_1})$$
Where $a,b$ are uniformly distributed in $[0,1]$ and the $\vec{v}$'s are the triangle vertices in $3D$. |
H: What is this type of fixed point called?
After numerically solving the following differential equations:
$$v'(t)=\frac{-\frac{1}{3} v(t)^3+v(t)-\omega (t)}{\tau } $$ $$ \omega '(t)= \frac{a+v(t)}{\tau}$$
at $a=1,\tau=0.2$ and taking $t$ from 0 to 10, I get this phase space plot:
The fixed point $(-1,-\frac23)$ is similar to an attracting focus, in that all nearby trajectories get attracted to it in a spiral. Note that the above plot is of a finite set of time steps, and there are no limit cycles here.
However, as one gets closer and closer to the fixed point, it starts behaving more and more like a center in the sense that trajectories seem to form almost closed loops. Basically, the trajectories are reaching the fixed point, but asymptotically.
As the nature of a fixed point is determined by the behavior of the trajectories in its immediate neighborhood, is this fixed point a center? Or is it a stable focus? Or is it called something else entirely? (I've been calling it a "limit point")
AI: Stability analysis can enlighten you here. Your system is determined by the a system of equation
\begin{align}
\dot{\nu} = f(\nu,\omega)\\
\dot{\omega} = g(\nu,\omega)
\end{align}
with a jacobian
\begin{align}
J = \begin{bmatrix} \frac{1-\nu^2}{\tau} & -\frac{1}{\tau}\\ \frac{1}{\tau} & 0\end{bmatrix}
\end{align}
Near the fixed point $(\nu^*,\omega^*)=(-1,-2/3)$, it evaluates to
\begin{align}
J^* = \begin{bmatrix}0 & -\frac{1}{\tau}\\ \frac{1}{\tau} & 0\end{bmatrix}
\end{align}
with eigenvalues $\lambda=\pm i/\tau$. So, what can be said about this fixed point? It is a marginal case: eigenvalues have null real parts. In the case of interest the fixed point is called a center [Strogatz 2000, Nonlinear dynamics and chaos]. |
H: Volume of ellipsoid
How do I calculate the volume of the intersection of ellipsoid $x^2/36+y^2/49+z^2/49\leq 1$ and subspace $x/6+y/7+z/7\leq1$ ?
AI: If the 36 were replaced with 49, and the 6 were replaced with 7, you have a plane cutting a sphere, and the volume of the resulting pieces of the sphere can be found in standard ways that I could elaborate on if you'd like.
To get the volume you are looking for, all the components of this sphere will have been contracted by $6/7$ in one dimension to obtain that ellipse. Such a contraction has the effect of scaling volume by $6/7$.
Added: Consider a sphere with radius 7, centered at the origin. Also there is a horizontal plane at $z=7-a$, which slices the sphere $a$ units below the sphere's north pole. Let's find the volume of the "lens" that lies above. Using the method of discs,
$$V_{\text{lens}}=\int_{z=7-a}^{z=7}\pi(\sqrt{49-z^2})^2\,dz$$
This is an easy integral. From here, the volume of the rest of the sphere is $\frac{4}{3}\pi(7^3)-V_{\text{lens}}$.
For our purposes, $a$ should be the distance between $\left(\frac{7}{\sqrt{3}},\frac{7}{\sqrt{3}},\frac{7}{\sqrt{3}}\right)$ and $\left(\frac{7}{{3}},\frac{7}{{3}},\frac{7}{{3}}\right)$. (Imagine a lens cap that is sliced off from the sphere using the plane $x/7+y/7+z/7=1$; this distance would be its "altitude".) This distance should be easy to compute.
So now you have $\frac{4}{3}\pi(7^3)-V_{\text{lens}}$, the volume of the complement of a lens cut off from a sphere. Your actual problem asked for a version of this object that has been compressed by a factor of $6/7$ in the $x$-direction. So the volume will be compressed in the same way, and the result is $\frac{6}{7}\cdot\left(\frac{4}{3}\pi(7^3)-V_{\text{lens}}\right)$. |
H: Solving $x^x=\frac{1}{\sqrt 2}$
The equation $$x^x=\frac{1}{\sqrt 2},x\in \mathbb R$$
has two obvious solutions $0.5$ and $0.25$
One can easily prove they are the only ones using differential calculus.
Is there any natural algebraic manipulation that would lead to finding these solutions ?
AI: One natural thing to try is taking logarithms. This gives
$$x\log x = \log\left({1\over\sqrt2}\right)={1\over2}\log\left({1\over2}\right)$$
from which the solution $x=1/2$ stands out. One would likewise find the solution $x=1/3$ for the equation
$$x^x = {1\over\sqrt[3]3}$$
The other solution, $x=1/4$, however, can be traced to the fact that $2^2=2\times2$, so that
$${1\over4}\log\left({1\over4}\right)={1\over4}\log\left({1\over2^2}\right)={2\over4}\log\left({1\over2}\right)={1\over2}\log\left({1\over2}\right)$$
There is no correspondingly nice relationship when the $2$'s are replaced by $3$'s. |
H: Free product of the trivial group with another group
I'm new to the idea of a free product..
Basically I was wondering if G is an arbitrary group and 1 is the trivial group then is $1\star G \cong G$. If not.. what whould it look like?
AI: Yes, $1*G\cong G$. A proof would depend on which definition or construction of free products you know or like. Using the definition as the coproduct in the category of groups, the proof consists of observing that, for every group $H$, there's a bijection between the set of homomorphisms $G\to H$ and the set of pairs of homomorphisms, the first $1\to H$ and the second $G\to H$, because there's exactly one homomorphism $1\to H$. (More generally, a coproduct of an initial object and an arbitrary object $X$ is isomorphic to $X$.) |
H: Does $\cos(x+y)=\cos x + \cos y$?
Find the value using a calculator: $\cos 75°$
At first I thought all I need is to separate the simpler known values like this:
$\cos 75^\circ = \cos 30°+\cos45° = {\sqrt3}/{2} + {\sqrt2}/{2} $
$= {(\sqrt3+\sqrt2)}/{2} $ This is my answer which translates to= $1.5731$ by calculator
However, when I used the calculator directly on $\cos 75°$, I get $0.2588$.
Where I am going wrong?
AI: You used a formula $\cos(x+y) = \cos(x) + \cos(y)$ which is false. The correct formula is:
$$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$ |
H: Uniform convergence of $\sum_{n=0}^{\infty} \frac{(-1)^n 2i}{(2n+1)z^{2n+1}}$ on squarewith vertices $\pm6\pm6i$?
Can someone explain me how I can check whether the convergence of $\sum_{n=0}^{\infty} \frac{(-1)^n 2i}{(2n+1)z^{2n+1}}$ is uniform on the (boundary of) square $A$ with vertices $\pm6\pm6i$?
AI: When we have a series
$$\sum_{n=0}^\infty \frac{(-1)^n 2i}{(2n+1)z^{2n+1}},$$
we can bound the size of the terms from above when we can bound the absolute modulus of $z$ from below.
On the square (boundary/contour) with vertices $\pm 6 \pm 6i$, we know that $\lvert z\rvert \geqslant 6$. Thus we can majorise the series by
$$\sum_{n=0}^\infty \left\lvert \frac{(-1)^n 2i}{(2n+1)z^{2n+1}} \right\rvert \leqslant \sum_{n=0}^\infty \frac{2}{(2n+1)6^{2n+1}} < \frac13\sum_{n=0}^\infty \frac{1}{36^n}$$
and see that the series converges uniformly on the contour. More, we can see that it converges uniformly on $\{ z \in\mathbb{c} : \lvert z\rvert \geqslant r\}$ for every $r > 1$ since the analogous computation yields a majorisation by a geometric series with ratio $r^{-2} < 1$. |
H: A sequence $( a_n )$ is defined by $(a_1)=1$, and $a_{n+1}$=$\sqrt {a_n + 1}$, for $n \in \mathbb{N}$
How to prove that $( a_n )$ is an increasing sequence by induction?
$n\ge 1$
AI: *(for $n=1$) ${{a}_{2}}=\sqrt{2}>1={{a}_{1}}$ so $a_2>a_1$.
*now assume ${{a}_{n+1}}>{{a}_{n}}.$ we know $\sqrt{{{a}_{n}}+1}={{a}_{n+1}}>{{a}_{n}}$
so using
*${a}_{n+1}>{{a}_{n}}$ and
**$\sqrt{{{a}_{n}}+1}={{a}_{n+1}}$ we get
$${{a}_{n+2}}=\sqrt{{{a}_{n+1}}+1}>\sqrt{{{a}_{n}}+1}={{a}_{n+1}}$$ then $${{a}_{n+2}}>{{a}_{n+1}}$$ |
H: proving the lim inf and the lim sup for any listing of rational in the interval (a,b)
Let $r_n$ be any listing of the rational numbers in the interval $(a,b)$. Establish with proof $\liminf_{n\to \infty} r_n$ and $\limsup_{n\to\infty} r_n$
Please forgive, I have no idea how to write the actual mathematical language on here
We attempted a proof by contradiction but got stuck, we know that the lim inf and the lim sup are guaranteed by the Completeness Axiom but we aren't sure where to go from here. Please help
AI: Hint: There are infinitely many rational numbers in the intervals $(a, a+\varepsilon)$ and $(b-\varepsilon, b)$ for any $\varepsilon>0$. |
H: Intuitive understanding of work.
I am working on a problem that has to do with work.
I am also assuming that the acceleration due to gravity is $10m/s^2$.
A 15 kg crate is moved along a horizontal floor by a warehouse worker who's pulling on it with a rope that makes a 30 degree angle with the horizontal. The tension in the rope is 69 N and the crate slides a distance of 10 m. How much work is done on the crate by the worker ? (kinetic friction = 0.4)
I found out that the horizontal force exerted by the rope is about 60N and the force exerted by the friction is about 60N in the opposite direction. So, I cannot see how this object was able to move 10m in the first place.
I understand that the net force = 0 doesn't mean that it is at rest, but I don't quite understand the fact that the problem tells you that it moved 10m.
If I could have answers for the following it would really help.
1), Are we assuming that the crate was already moving ?
2), I calculated the work done by the force by the rope to be 600N and that of the friction to be -600N. Intuitively I want to say that the total work done was 0. But if the object moved, then some work must have been done. What am I thinking wrong ?
Thank you.
AI: Looking at the vertical forces on the crate, you must have an equation like $T \sin 30^{\circ} + F = mg$, where $T$ is the tension on the rope and $F$ is the normal reaction from the floor. This gives $F = 15\times 10 - 69 \sin 30^{\circ} = 115.5\,$N.
Now looking at the horizontal forces, you have $T \cos 30^{\circ} \approx 60$ in one direction and frictional force of $\mu F = 0.4 \times 115.5 \approx 46 $ in the other direction. So, there is a net unbalanced force of about $60-46 = 14\,$N which will accelerate the crate at $\frac{14}{15}$ $m/s^2$.
Total work done is then work done against friction $+$ kinetic energy gained by the acceleration, but this will be equal to force in the horizontal direction times displacement anyway, so $60*10 = 600\,$J. |
H: Complex contour integral with residue theory
I need to calculate the following contour integral using residue theory.
$z \in \mathbb{C}$
$f(z)=\exp(-1/z) \sin(1/z)$
$\oint_C f(z) dz$
$C: \left | z \right |=1$
The difficult points I detected include only z=0.
But i'm a bit stuck at calculating the residue of f at z=0. (I know the answer is -1)
I know that the residue is the coefficient of the 1/z term in the laurent series. However, I don't know how to create the laurent series of this function quickly,
I've been playing around with a substitution: $z=1/\xi$ but that didn't help me so far.
Its probably really easy but I'm new to the subject. So any tips are appreciated and the one that helps me to find the solution myself will be accepted as an answer.
AI: $$e^{-1/z} \sin \frac1z = \left(1 - \frac1z + \ldots\right)\left(\frac1z+\ldots\right).$$ |
H: Geometric meaning of minors
This is a bit silly question I found on another discussion forum. I know that determinants can be used to compute volumes of parallelepiped. I also know that determinants can be computed by linear combination of its minors. Is there any geometric meaning of minors or some proof/explanation why there is no such geometric meaning.
AI: If $A$ is an $n\times n$ matrix, then consider $P_iAQ_j$, where $P_i$ is the $(n-1)\times n$ matrix obtained from removing the $i$th row from the identity matrix and $Q_j$ is the $n\times (n-1)$ matrix obtained from removing the $j$th column from the identity matrix.
Then $P_iAQ_j$ is the $ij$ minor of $A$. So it may help you to think about the geometry related to $P_i:\mathbb{R}^n\to\mathbb{R}^{n-1}$ and $Q_j:\mathbb{R}^{n-1}\to\mathbb{R}^{n}$. |
H: Exponential distribution: Finding the parameter
Please help me solve the following problem
Time of production of one electronic component is given with
exponential distribution with parameter λ. If the process lasts less than 3 hours, the component is working, otherwise it is defective. Events that component is working or that is defective have equally probability. Find:
Parameter λ
Probability that the process will last less than 1 hour.
AI: Let $X$ be the random variable that measures length of life, in hours. We are told that $X$ has density function $\lambda e^{-\lambda x}$ (for $x\gt 0)$.
We are also told that $\Pr(X\lt 3)=\Pr(X\ge 3)$. Each is therefore $\frac{1}{2}$.
But $\Pr(X\gt x)$, by integration, is equal to $e^{-\lambda x}$. It follows that
$$e^{-3\lambda}=\frac{1}{2}.$$
Now we can solve for $\lambda$, by taking logarithm to the base $e$ of both sides.
And now that we have $\lambda$, we can find the probability the thing lasts less than $1$ hour, since $\Pr(X\le x)=1-e^{-\lambda x}$ for $x\gt 0$.
Remarks: $1.$ We were asked to find $\lambda$, so we did. But note that the second problem can be solved without finding $\lambda$ explicitly. For we had $e^{-3\lambda}=\frac{1}{2}$. Taking cube roots, we find that $e^{-\lambda}=2^{-1/3}$, and therefore $\Pr(X\lt 1)=1-2^{-1/3}$.
$2.$ The calculations of the problem are closely related to half-life calculations that you have undoubtedly done several times in the past. Individual atoms of a radioactive isotope have a lifetime that is exponentially distributed. The median lifetime is called the half-life of the substance. |
H: Subset of sets exrecise in discrete structures course
$\{\{b\}\}$ is a subset of $X=\{a,b,\varphi,\{a,b\},\{a,\{b\}\},\{c,\varphi\}\}$: true or false? Can someone explain why is it true/false?
AI: Is every element of $\{\{b\}\}$ an element of $X$?
That is: Is $\{b\}$ an element of $X$?
Well, we do have $b\in X$ and $\{a,b\}\in X$ and $\{b\}\in\{a,\{b\}\}\in X$ and these are all near misses, but we dont't have $\{b\}\in X$.
Unless it happens that $a$ equals $\{b\}$, that is. |
H: What does "under divisibility" mean?
I'm currently working on this problem, and the term "under divisibility" is kind of ambiguous to me. I googled the term "under divisibility" and I couldn't find any definitions.
The solution also doesn't make sense to me either. It states that all numbers divide 0, but isn't it NOT possible to divide by zero (when talking about natural numbers). Can anyone explain this question/solution in more explicit terms?
This is from this problem set.
AI: If $R$ denotes a weak partial ordering, for example $\subseteq$ ("is a subset of") or $\mid$ ("divides") then the statement "$x$ is minimal under $R$" means that the only $y$ satisfying the statement "$y \mathbin{R} x$" is $x$ itself.
To answer your second question, for any $x$ we have $x \mid 0$ ("$x$ divides zero") because there is a $y$ such that $xy = 0$, namely $y=0$. In other words, $0/x$ is a natural number $y = 0$. So we are not dividing anything by zero, but rather dividing zero by things. |
H: If a function is continuous and converges finitely, then it is bounded
Suppose that $f$ is a continuous function on $[a,+\infty)$ and that there exists a finite limit $\lim \limits_{x \to +\infty} f(x)$, then $f$ is bounded.
I know that from the assumption I can conclude that $f$ is uniformly continuous on that interval, but how can I use this to show that $f$ is bounded.
AI: Let $C := \lim\limits_{x \to +\infty} f(x)$. Then there is $b$ such that $|f(x)-C|<1$ for $x \ge b$. Now consider $f$ on $[a,b]$ and $[b,+\infty)$. |
H: Second Borel-Cantelli without independence assumption
The second Borel-Cantelli lemma says
Let $(X,F,\mu)$ be a probability space. Let $A_1,A_2,\ldots\in F$ be independent, and let $A=\bigcap_{i=1}^\infty \bigcup_{j=i}^\infty A_j$ (i.e. the probability that infinitely many $A_i$'s occur). If $\sum_{i=1}^{\infty}\mu(A_i)=\infty$, then $\mu(A)=1$.
One easy example to show that if the independence condition is omitted, then the theorem is not true, is to take $A_1=A_2=\ldots$. Is there an easy example where the sets $A_i$ are mutually distinct?
AI: One example would be to take $\Omega=[0,1]$ with the Lebesgue measure, and $A_i = [0,\frac{1}{i}]$. |
H: Bound of Standard Normal Integral
Consider the Standard Normal Integral given by:
$$ I=\int_{-\infty}^{\infty} \frac{1} { \sqrt{2\pi} } e^{ \left( -z^2 /2 \right)} dz $$
In order to prove that it exists we note that the integrand is a positive continuous function which is bounded by an integrable function; that is:
$$ 0<e^{-z^2/2}<e^{- \left|z \right| +1}\ \text{for}\ -\infty<z<\infty $$
and $$\int_{-\infty}^{\infty} e^{ \left|z \right| +1} dz =2e $$
I understand the essence of the comparison test and I can show that the integral of the last function is $ 2e $, my question is how can we show that the relation $e^{-z^2/2} <e^{-\left| z \right| +1}\ \text{for all z} $ holds? Thanks!
AI: Well, $\mathrm e^{-|z|^2/2}\lt\mathrm e^{-|z|+1}$ if and only if $-\frac12|z|^2\lt-|z|+1$ if and only if $|z|^2-2|z|+2\gt0$. And $|z|^2-2|z|+2=(|z|-1)^2+1$. Hence... |
H: Linear Algebra Projections and orthogonality
I am currently studying linear algebra and am getting very confused on orthogonality and projections. I am reviewing questions to better understand it but can not get a certain one. Here is the question:
$$
\vec{b} = \begin{bmatrix} 1 \\ 2 \\ 2\end{bmatrix} \quad\text{and}\quad\vec{a} = \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}.
$$
First thing I need to do is Project b onto the line through a. But how would I check that $\vec e = \vec b - \vec p$ is orthogonal to a? I am unclear on what that exactly means.
The questions asks me to find the projection matrix $P = \frac{\vec a\vec a^T}{\vec a^T\vec a}$ onto the line through $\vec a.$ I also need to verify $P^2 = P.$ Lastly, I need to multiply $P\vec b$ to compute the projection $\vec p.$
AI: The definition of projection is
$$pr_\vec{a}(\vec{b}) = \frac{(\vec{a},\vec{b})}{(\vec{a},\vec{a})}\vec{a}$$
where $(\vec{a},\vec{b})$ is an inner product between $\vec{a}$ and $\vec{b}$. (In this case, we have the dot product).
You can conclude with a geometric intuitive argument that $\vec{b}-pr_\vec{a}(\vec{b})$ is orthogonal to $\vec{a}$. Just draw a right triangle with $\vec{b}$ as it's hypotenuse and cathetus $pr_\vec{a}(\vec{b})$ (on the line through $\vec{a}$). $\vec{b}-pr_\vec{a}(\vec{b})$ will be the other cathetus, hence orthogonal to $\vec{a}$.
To prove orthogonality, just show that $(\vec{b}-pr_\vec{a}(\vec{b}),\vec{a})=0$.
EDIT:
To find the transformation matrix, just compute the transformation of the canonical basis and put these vectors as columns on $P$, and you can prove why it works really easy. For exaple, if you apply the transformation $P$ on the vectors $e_1=(1,0,0)$,$e_2=(0,1,0)$ and $e_3=(0,0,1)$ you can obtain, say, $Pe_1=v_1,Pe_2=v_2,Pe_3=v_3$. All you have to do is build the matrix:
$$P = \begin{bmatrix}v_1 v_2 v_3 \end{bmatrix} $$
that is, $v_1$ is the first column, $v_2$ is the second column and $v_3$ the third column.
Well, obviously you can't use the matrix $P$ to compute $Pe_1$ and so on. These transformations you must compute by hand, using the projection and orthogonality concepts introduced before. |
H: Disjoint union with limsup
For any sets $A_n,n\in\mathbb{N}$ consider
$$
A^+:=\limsup_{n\to\infty}A_n:=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k,~~~~~E_m:=\bigcup_{n\geq m}A_n.
$$
Show that the sets $E_m, m\geq 1$ can be written as a disjoint union
$$
E_m=A^+\uplus\biguplus_{n\geq m}(E_n\setminus E_{n+1}).
$$
I do not have a working idea. I started with writing $E_m$ as a disjoint union, i.e.
$$
E_m=A_m\uplus\biguplus_{i=m+1}^{\infty}A_i\setminus\bigcup_{j=m}^{i-1}A_j=A_m\uplus\biguplus_{i=m+1}^{\infty}\bigcap_{j=m}^{i-1}A_i\setminus A_j
$$
and additionally I see that
$$
A^+=\bigcap_{n=1}^{\infty}E_n.
$$
But I do not know if this is helpful...
Would be great to get a help resp. answer.
With kind regards,
math12
AI: Hint: For $x \in E_m$ consider
$$\sup_{A_k \ni x} k.$$
Show that if this supremum is infinite, then $x \in A^{+}$. Otherwise, denote this supremum by $n$ and show that $x \in E_n \setminus E_{n+1}$. |
H: Radius ratio for four packed circles
Suppose we are given four circles $A,B,C,D$ in the Euclidean plane having radii $r_A,r_B,r_C,r_D$ such that $r_A=r_C,r_B=r_D$ and circles $A,C$ are tangent to each other and to $B,D$ but $B,D$ are only tangent to $A,C$. Suppose further that a given bounding square $E$ in the plane is tangent to each circle exactly twice. What is the value of the ratio
$$r_A\over r_B$$
AI: Using alternate diagonals, it is possible to identify the following:
The center of $A(C)$ is $\sqrt{2r_A^2}$ distant from the closest corner of $E$, similarly for $B(D)$, and the distance from the center of $A$ to $C$ is simply $r_A+r_C=2r_A$, so we have diagonal length $d=2\sqrt2r_A+2r_A$. It is possible to arrive at a similar expression for the other diagonal, except that we must get a good value for the distance between the centers of $B$ and $D$. Since the four circle centers form a rhombus, the alternate centers cross perpendicularly, so we have $\sqrt{(r_A+r_B)^2-r_A^2}=\sqrt{r_B^2+2r_Ar_B}$ as the distance between centers of $B$ and $D$. Then we have alternate diagonal length
$$d=2\sqrt2r_B+2\sqrt{r_B^2+2r_Ar_B}=2\sqrt2r_A+2r_A$$
Rearranging, squaring and dividing by $r_A^2$ yields
$${r_B^2\over r_A^2}-2(\sqrt2+3){r_B\over r_A}+(1+\sqrt 2)^2=0$$
This quadratic yields
$${r_B\over r_A}=\sqrt2+3\pm\sqrt{4\sqrt2+8}\approx 0.718695432327948\dots$$
It is certainly the case that the negative branch is appropriate since $r_B\le r_A$ under the conditions and picture shown. Then the original requested ratio has value
$${r_A\over r_B}=\frac 1{\sqrt2+3-2\sqrt{\sqrt2+2}}$$
which does not appear to look any better upon conjugate multiplication and simplification. |
H: Line Integration over a broken line
I have taught myself Line Integration and when doing some practice questions I came across the following:
Evalute the line integral $$\int_{C}(y^2\:dx+xy\:dy+zx\:dz)$$ Where $C$ is the broken line from $A(0,0,0)$ to $B(1,1,1)$ connecting $(0,0,0)$, $(0,0,1)$, $(0,1,1)$ and $(1,1,1)$.
The first step I took was to split $C$ into 3 smaller curves: $C=C_{1}+C_{2}+C_{3}$, such that:
$C_{1}$ can be parametrized by $0\leq t\leq1$ with $x=y=0, z=t \implies dx=dy=0, \:dz=dt$
$C_{2}$ can be parametrized by $0\leq t \leq 1$ with $x=0, y=t, z=1 \implies dx=dz=0,\: dy=dt$
$C_{3}$ can be parametrized by $0 \leq t \leq 1$ with $x=t, y=z=1 \implies dx=dt,\:dy=dz=0$
I then split my integral up as follows:
$$\begin{align*}\int_{C}(y^2\:dx+xy\:dy+zx\:dz)&=\int_{C_{1}}(y^2\:dx+xy\:dy+zx\:dz) \\ &+\int_{C_{2}}(y^2\:dx+xy\:dy+zx\:dz)\\&+\int_{C_{3}}(y^2\:dx+xy\:dy+zx\:dz)\end{align*}$$
Evaluating these we get:
$$\int_{C}(y^2\:dx+xy\:dy+zx\:dz)=\int_{0}^{1}0\:dt+\int_{0}^{1}0\:dt+\int_{0}^{1}\:dt=1$$
Is this the correct way to go about evaluating such an integral?
AI: I agree with Mhenni. Strangely enough, I first thought that the answer could not be that simple :-), but reading along, I realized that the function you are integrating along this curve, and the curve itself, seem to have been designed to cancel all interesting contributions and make computation easy. |
H: Dense subset in Hilbert space: a basic question
I am trying to solve the following problem:
Let $\bar{H}$ be a Hilbert space, $H$ - its dense linear subspace, $z_0\in \bar{H}-H$. Consider the following subspace $M\subset H$ defined as $\{x\in H:<x,z_0>=0\}$. Is is true that $Cl(M)=Span(z_0)^{\perp}$ ($Cl(M)$ is the closure of $M$ in the topology of $\bar{H}$, $<,>$ is the given inner product).
It is easy to verify that $Cl(M)\subset Span(z_0)^{\perp}$ (really, if $x\in Cl(M)$ then there exists a sequence $\{x_n\}\subset M$ such that $x=\lim x_n$; we see that an inner product is continuous, hence $<x,z_0>=<\lim x_n,z_0>=\lim <x_n,z_0>=0$). But what about the converse $Span(z_0)^{\perp}\subset Cl(M)$?
I will be grateful for any help!
PS
$Span(z_0)^{\perp}$ is the orthogonal completion for one-dimensional $Span(z_0)$
AI: Yes.
Suppose $z \in \operatorname{Span}(z_0)^\perp$, i.e. $\langle z, z_0 \rangle = 0$. There is a sequence $x_n \in H$ with $x_n \to z$. Let $u$ be any vector in $H$ with $\langle u, z_0 \rangle \ne 0$ (this $u$ must exist because $\operatorname{Span}(z_0)^\perp$ is not dense, so it cannot contain $H$). By rescaling, assume $\langle u, z_0 \rangle = 1$. Now set $y_n = x_n - \langle x_n, z_0 \rangle u$; clearly $y_n \in H$. By continuity, $\langle x_n, z_0 \rangle \to \langle z, z_0 \rangle = 0$, so $y_n \to z$. And $\langle y_n, z_0 \rangle = \langle x_n, z_0 \rangle - \langle x_n, z_0 \rangle \langle u, z_0 \rangle = 0$, so $y_n \in \operatorname{Span}(z_0)^\perp$. |
H: Expectation/ independence of random variables
Let $X,Y$ be two correlated variables and $Z\sim N(0,1)$ independent of $X,Y$. Consider the expectation:
$$E[f(X,Y)Z].$$
If $f(X,Y)$ and $Z$ are independent then clearly $E[f(X,Y)Z]=E[f(X,Y)]E[Z]=0$ but I guess this is not in general true. Nevertheless, I can argue as follows, by conditioning on $X$ and $Z$.
\begin{align*}
E[f(X,Y)Z]=& E[E[f(x,y)Z| X=x,Y=x]]\\
=&E\left[f(x,y)\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty ze^{-z^2/2}dz \bigg|_{X=x,Y=y}\right]\\
=& 0.
\end{align*}
I really wonder where the error is! Could anyone help me? :)
Thank you very much!
AI: There is no error in your derivation. If Z is independent of (X,Y) then it is also independent of f(X,Y). Please, also see Are functions of independent variables also independent?. |
H: There are 6 cards with letters a, c, e, i, m, n in a box. Somebody picks cards in a random order.What is the probability of getting the word “cinema”?
This is a probability question.
There are 6 cards with letters a, c, e, i, m, n in a box. Somebody picks cards in a
random order. What is the probability of getting the
word “cinema”?
Don't solve it for me. I just need a hint or two. If you know the theory (theorem) required to solve the question please write it to me. Thank you.
Regards
John
AI: First consider the total number of ways to pick these letters - how many ways can you pick the first? then the second? and so on.
Now if the order of the final 6 letters are to spell the word "cinema", how many choices do you have for the first letter? the second? continue this way and you have the total number of ways you can pick the 6 letters such that they spell "cinema".
And so what is the required probability? |
H: Unit length vectors that sum to zero
Let's say we have a collection of $n$ vectors in $\mathbb{R}^2$ where $n$ is odd. Suppose each vector has unit length and that the sum of the vectors is zero. Is it necessarily true that the vectors will correspond to the vertices of a regular $n$-gon?
AI: Hint: Take an equilateral triangle, with centre at the origin, vertices at unit distance from the origin, and one of the vertices at $(1,0)$. Rotate it counterclockwise about the origin a little. We have $3$ vectors whose sum is $0$. Add to this collection the vectors $(1,0)$ and $(-1,0)$. |
H: Diffusion-advection equation with time-variable coefficients
Is the fundamental solution (Green's function) of the 1D advection-diffusion equation
$$\frac{\partial{\phi}}{\partial{t}} = D(t)\frac{\partial{^{2}\phi}}{\partial{x^{2}}} - c(t)\frac{\partial{\phi}}{\partial{x}}-k(t)\phi$$
(where the diffusivity $D(t)$, drift speed $c(t)$ and decay coefficient $k(t)$ are functions of $t$ only and not $x$) given by
$$
\phi(x,t)
=
\frac{1}{\sqrt{4\pi\int_{t_0}^{t}D\left(t'\right)dt'}}\,
\exp\left({-\frac{\left\lbrack x-\int_{t_{0}}^{t}c(t')\,{\rm d}t'\right\rbrack^{2}}{4\int_{t_0}^{t}D\left(t'\right){\rm d}t'}-\int_{t_0}^{t}k\left(t'\right)\,{\rm d}t'}\right)
$$
provided that $\int{D dt}$, $\int{c dt}$ and $\int{k dt}$ exist?
I believe that this solution is correct, I can't find an error in my derivation. I'll post more details if this result is incorrect, in order to identify the error. But is it correct?
Thanks!
AI: Yes, it is correct assuming your initial condition is $\phi(x,t_0)=\delta(x)$ and the domain is $x\in\mathbb{R}$, $t>t_0$.
First verify that it solves the equation for $t>t_0$.
input:
$$\frac{1}{\sqrt{4\pi \int _0^td[s] ds}}\text{Exp}\left[\frac{-\left(x-\text{x0}-\int _0^tc[s] ds\right){}^2}{4\int _0^td[s] ds}-\int _0^tk[s] ds\right]\text{//}\{-1,d[t],-c[t],-k[t]\}.\{D[\#,t],D[\#,x,x],D[\#,x],\#\}\&\text{//}\text{Simplify}$$
output: $0$
To verify that it satisfies that initial condition, compare it with the solution for constant coefficients, and note that $\int_0^t f(s)\,ds = t f(0) + O(t)$ for continuous functions $f$.
I can't resist making a general note that you prove an expression to be correct by proving it to be correct, rather than comparing with an expression (in a textbook maybe) that you suspect is correct. Otherwise, you seem to have got it right yourself. |
H: How do i calculate the maxima in a Polynomial between two roots.
I'm writing a program that plots compaction curves for soil density tests.
I have very little math background and I'm using Extreme Optimization's Math Library to curve fit my data(linear least squares) and plotting it as either a 2nd or 3rd degree polynomial. Finding max's in 2nd degree was straight forward but I'm having difficulty figuring out how to get the max for 3rd degree polys.
Here's a sample of what i have:
$$P(x) =-0.00399730972447669x^3-0.270240772455719x^2+6.44622423832016x+91.4780393989453$$
And roots at $x_1 =-83.6183853039161$, $x_2 = -10.3725525112563$, $x_3 =26.3852752364245$
how do i find the max that lives between the $-10$ and the $26$?
Thanks
AI: The (relative) maximum of a cubic polynomial $ax^3+bx^2+cx+d$ is attained at one of the roots of the derivative $3ax^2+2bx+c$. So, if you still remember how to solve the quadratic, just do it and see which root is in the required range. |
H: How to tell an ideal of integer polynomial ring is principal?
I understand that uni-variate polynomial rings with coefficients in a field only have principal ideals. For example, $\mathbb{C}[x]$. But how can I tell if an ideal of integer polynomial ring is principal, please? For example, a textbook claims that "the kernel of the map $\mathbb{Z[x]} \rightarrow \mathbb{Z[i]}$ sending $x \mapsto i$ is the principal ideal of $\mathbb{Z}[x]$ generated by $f=x^2+1$" without any justification. How to show this is true, please?
AI: In this case, it is fairly easy. Since the polynomial $f$ is monic, you can write every $p \in \mathbb{Z}[x]$ as
$$p = q\cdot f + r$$
with $r \in \mathbb{Z}[x]$ of degree at most one. Since - easily verified - $f$ maps to $0$, the kernel of $\pi \colon g \mapsto g(i)$ surely contains the principal ideal $(f)$ generated by $f$, and to show that it contains nothing more, by the above it suffices to show that no polynomial of degree $\leqslant 1$ except the zero polynomial is mapped to $0$. But $a + bx \mapsto a + bi$, so $a+bx \in \ker \pi \iff a = b = 0$. |
H: $NP$ problems not known to be in $P$ and not known to be $NP$-complete
I've read that solving Pell's equation is neither known to be in $P$ nor known to be $NP$-complete. What are other natural and important examples of such problems?
AI: They are called NP-intermediate problems (under the assumption of $P\neq NP$, otherwise the definition makes no sense).
By now, some natural/real-world problems are suspected to be in this class, yet no one has been able to prove that, since that would be a proof of $P\neq NP$.
Graph isomorphism problem, that is, to determine whether two graphs are isomorphic.
Integer factorization and the related problem of computing discrete logarithm
are among the most well-known ones. |
H: Prove that $\{f_n\cdot g\}\rightarrow f\cdot g$ in measure.
From Royden's Analysis book (4th Edition) Chapter 5 Section 2 Problem 7:
Let $E$ have finite measure, $\{f_n\}\rightarrow f$ in measure on $E$ and $g$ be a measurable function on $E$ that is finite a.e. on $E$. Prove that $\{f_n \cdot g\} \rightarrow f\cdot g$ in measure.
I've attempted many different approaches, but to no avail. If anyone could give a starting hint (not necessarily a solution), I would be extremely grateful.
AI: Idea: Try looking at what happens when you restrict yourself to $A_k = \{ x : |g(x)| \in [0,k+1)\}$ for each $k\in \{0,1,\ldots\}$. What can you say about $\lim_k \mu(A_k^c)$ ?
Note: $A_k^c$ stands for the complement of $A_k$ |
H: is the vector space $\mathbb{R}^\mathbb{N}$ locally compact?
is the vector space $\mathbb{R}^\mathbb{N}$ locally compact?
for example, let $x=(x_1,x_2,....)$ any point of $\mathbb{R}^\mathbb{N}$ and let $V=[x_1-\epsilon,x_1+\epsilon] \times [x_2-\epsilon,x_2+\epsilon]\times...$
is $V$ a compact neighborhood of $x$? or is there a mistake?
AI: A neighborhood of $0\in\mathbb{R}^{\mathbb{N}}$ must contain a subset of the form
$$
U_1\times U_2\times \dots \times U_n \times \dotsm
$$
where all but a finite number of the $U_i$ are equal to $\mathbb{R}$ and the others are open intervals containing $0\in\mathbb{R}$.
If this subset is contained in a compact subset of $\mathbb{R}^{\mathbb{N}}$, then also
$$
C_1\times C_2\times \dots \times C_n \times \dotsm
$$
would be compact, where $C_i$ is the closure of $U_i$. Obviously this is a contradiction.
The product topology is the least topology making all projections into continuous functions. Let's talk generally for $X=\prod_{i\in I}X_i$ where $X_i$ is a topological space (a metric space, if you prefer; or just $\mathbb{R}$). We can define, for every $i\in I$, the projection map $\pi_i\colon X\to X_i$.
Our choice for the topology on $X$ means that, in order to check whether a map $f\colon Z\to X$ is continuous, we just need to check that $\pi_i\circ f\colon Z\to X_i$ is continuous for all $i\in I$.
Now, what subsets of $X$ are neighborhoods of $x\in X$? Only those that need to be as a consequence of the definition. Remember that the family of neighborhoods of $x$ is closed under finite intersections and you'll know why, in my argument above, only a finite number of the subsets $U_i$ can be a proper subset of $\mathbb{R}$.
What you had in mind is the box topology; but this is, in general, strictly finer than the product topology when considering the product of an infinite family. You can try your hand at showing when the two topologies are equal.
Why isn't the box topology used? Well, an infinite product of compact spaces is not compact when given the box topology (again, apart for trivial cases). Tykhonov's theorem, one of the pillars of topology, tells us that the product of any family of compact spaces is compact.
Local compactness (which is preserved for the box topology) is not as relevant as compactness; moreover there are many other reasons for using the product topology rather than the box topology: basically, it is a “functorial” construction. |
H: Abstract Linear Algebra for Statistics
Is abstract linear algebra required for a deep understanding of statistics? I'm a computer science major deciding between a linear algebra for applications class versus a very theoretical proof based linear algebra class (where they don't cover applied linear algebra).
I'm very interested in the field of statistics and would like a deep understanding of how the statistical tools are developed and used. That being said I'm not very keen on mathematical proofs and would rather not take this class if I don't have to.
AI: I think an applied linear algebra class is perfect, since doing statistics you will probably use finite vectors space over the real numbers field.
If someday you need a more specific background in other vector spaces, it won't be hard to expand some concepts and theorems. If it is usefull to statistics that you learn finite vector spaces, you will see it in the applied linear algebra course and learn why it's important.
There is no problem in doing an abstract course, but having constant contact with applications is really good. In my case, I did an abstract course and later, an applied extension course. The first one was really helpfull to understand concepts, but the second one was the most significant to my projects (mainly because it had some programming activities with linear algebra too). |
H: Set in the Complex Plane
How can I describe the set:
$$ \left\vert z - {\rm i}\,\right\vert = 3\left\vert z\right\vert $$
It does appear quite unfamiliar.
Attempt: $$ \left\vert\frac{z-i}{z}\right\vert = 3 $$ so,
$$ \left\vert 1 - {\rm i}\,\frac{1}{z}\right\vert = 3 $$
But this seems to be even more difficult to visualize.
AI: Hint: Square both sides, and write $z = x + iy$:
$$9(x^2 + y^2) = |x + i (y - 1)|^2 = x^2 + (y - 1)^2$$
Rearranging,
$$8x^2 + 8y^2 + 2y - 1 = 0$$
Divide through by $8$ and find
$$x^2 + y^2 + \frac{1}{4} y = \frac{1}{8}$$
or even better,
$$x^2 + y^2 + \frac{1}{4} y + \frac{1}{8} = \frac{1}{4}$$
Try rewriting the left side as the equation of a circle. |
H: Solving for x - Trig
Someone mind helping on this? I think have done the question correct but the system isnt accepting my answer.
AI: Using $\sin{2x} = 2 \sin{x} \cos{x}$, we see that your equation is equivalent to
$$2 \sin{x} \cos{x} = \cos{x}$$
or alternatively
$$\cos{x} \left(2 \sin{x} - 1\right) = 0$$
Now $\cos{x} = 0$ has solutions at $\pi/2$ and $3\pi/2$, while $\sin{x} = \frac{1}{2}$ has solutions at $x = \pi/6$ and $x = 5\pi/6$.
The only error in your answer is the final $5\pi/2$, where I think you have a typo. |
H: if the improper integral $\int^\infty_a f(x)\,dx$ converges, then $\lim_{x→∞}f(x)=0$
I need to prove that:
$$\lim_{x→∞}f(x)=0$$
if
$$\displaystyle∫^∞_af(x)\,dx$$ converges.
I need a proof or an specific, and if possible simple, counterexample. Would really appreciate your help! Thank you in advance.
AI: This is false; here is a counterexample:
$$
\begin{align}
\int_0^\infty \sin(x^2)\, dx
&=\int_0^\infty \frac{\sin x}{2\sqrt x}\,dx\\
&= \frac{1}{2} \sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi} \frac{\sin x}{\sqrt x}\,dx\\
&= \frac{1}{2} \sum_{n=0}^\infty (-1)^n \left|\int_{n\pi}^{(n+1)\pi} \frac{\sin x}{\sqrt x}\,dx\right|
\end{align}
$$
converges by the alternating series test (in fact, it equals $\sqrt{\frac{\pi}{8}}$).
However,
$$\lim_{x\to\infty}\sin(x^2)$$
does not exist. |
H: Prove: $\liminf_{n \to \infty} s_n \le \limsup_{n \to \infty} s_n$
I am looking over examples and the definitions for this section but I am still not familiar with all the tricks. I appreciate any help with proving this (from hints to maybe a solution. It is only a review so I just need something to look at and make sure I am doing it right). Thank you.
Prove $$\liminf_{n \to \infty} s_n \le \limsup_{n \to \infty} s_n$$
AI: Hint: for every $n$ we have $$\inf_{m \ge n}s_m \le \sup_{m \ge n}s_m$$
Edit: if the sequence $\{s_n\}$ is bounded then just take the limit and you are done.
If the sequence is not bounded make sure you are defining things in the following way:
set $\alpha_n = \inf_{m \ge n}s_m$ then define $$\liminf_{n \to \infty}s_n = \begin{cases}-\infty & \text{if}\ \alpha_n = -\infty\ \text{for all}\ n\\
\lim_{n \to \infty}\alpha_n & \text{otherwise.}\ \end{cases}$$
Do the same for the limit superior and finally everything works. (this prevents you from meaningless things like taking the limit of $-\infty$) :D |
H: How to tell an ideal of integer polynomial ring is principal?
I understand that uni-variate polynomial rings with coefficients in a field only have principal ideals. For example, $\mathbb{C}[x]$. But how can I tell if an ideal of integer polynomial ring is principal, please? For example, a textbook claims that "the kernel of the map $\mathbb{Z[x]} \rightarrow \mathbb{Z[i]}$ sending $x \mapsto i$ is the principal ideal of $\mathbb{Z}[x]$ generated by $f=x^2+1$" without any justification. How to show this is true, please?
AI: The quote tells you the element $f\in Z[x]$ that generates the ideal. Suppose now another $g\in Z[x]$ is in that ideal. Since $f$ is monic you can use Eucledian algorithm to find $h,r\in Z[x]$ such that $g=fh+r$, and $deg(r)\leq 1$. Since degree 0 or 1 polynomial are in said kernel conclude that $r\equiv 0$, therefore all elements $g$ of the ideal have form $fh$ for some $h$, therefore the ideal is generated by $f$. |
H: Prove that $|a| < b \iff -b
I know I have to prove both sides here so:
$\implies$:
Suppose $|a| < b$. Since $|a|$ is a positive number, we know $b$ is a positive number greater than $|a|$:
If $a$ is positive, $a < b$
If $a$ is negative, $a < b$
But for $|a| < b$, it has to be that $a > -b$.
$a < b$ and $a > -b$, therefore $–b < a < b$ is true.
$\impliedby$:
Suppose $-b < a < b$.
If $a$ is positive, then $b$ is a greater positive number, and $-b$ would be less than $a$
If $a$ is negative, then $b$ can only be a greater positive number for $-b < a$ to hold
So, $b$ is a greater positive integer than $a$. Therefore, it must be that $|a| < b$.
This was what I came up with but it seems convoluted and kind of forced with the cases. Any suggestions?
AI: If $|a|<b$ then: if $a\geq 0$ then $a=|a|<b$; if $a<0$ then $-a=|a|<b$ implies that $-b<a$.
Can you prove the other part now, using less text and more math? |
H: Why use Euclidean distance in linear model?
The method of least squares is the most basic method in statistical linear models. For the simplest linear model$$Y_i=\beta_0+X_i\beta_1+\epsilon_i$$we are looking for $\beta_0$ and $\beta_1$ that minimizes the Euclidean distance $\sum\limits_{i=1}^n|Y_i-\beta_0-X_i\beta_1|^2$. I learned statistics and an engineering student asked me why not use $\sum\limits_{i=1}^n|Y_i-\beta_0-X_i\beta_1|$ or things like $\sum\limits_{i=1}^n|Y_i-\beta_0-X_i\beta_1|^4$, I got stuck...... Is there any convincing answer for this question?
AI: One of the most frequently given rationales is that if one assumes the errors are independent and distributed as $N(\mu,\sigma^2)$ then the least-squares estimators of $\beta_0$ and $\beta_1$ coincide with the maximum-likelihood estimators.
Another rationale is the Gauss--Markov theorem. In that theorem it is not assumed that anything has a normal (or "Gaussian") distribution and it is not even assumed that the errors all have the same distribution. The assumptions are
The errors $\varepsilon_i$ have expected value $0$;
The errors $\varepsilon_i$ all have the same (finite) variance, which let us denote $\sigma^2$;
The errors $\varepsilon_i$ are uncorrelated (but not necessarily independent).
The conclusion is that the least-squares estimators $\hat\beta_0$ and $\hat\beta_1$ are the "best linear unbiased estimators" of $\beta_0$ and $\beta_1$. That means that among all linear unbiased estimators of $\beta_0$ and $\beta_1$, they have the smallest mean squared error $\mathbb E\left(\left(\hat\beta_i - \beta_i\right)^2\right)$.
Let us be careful about what "linear" means. (I know of an instance where someone said they were "affine" but not "linear"; that person was confused.) "Linear" means that the mapping
$$
\begin{bmatrix} Y_1 \\ \vdots \\ Y_n\end{bmatrix} \mapsto \begin{bmatrix} \hat\beta_0 \\ \hat\beta_1 \end{bmatrix} \text{ (with $\begin{bmatrix} X_1 \\ \vdots \\ X_n\end{bmatrix}$ fixed)}
$$
is linear, i.e. it is additive and 1st-degree homogeneous.
The central idea of the proof is a lemma saying that the least-squares estimators are uncorrelated with every linear unbiased estimator of $0$. A linear unbiased estimator of $0$ is a linear combination of $Y_1,\ldots,Y_n$ with coefficients that may depend on the (observable) $X$ values but not on unobservables such as $\beta_0,\beta_1,\sigma^2$, and whose expected value remains $0$ regardless of the values of those unobservables.
Another rationale is that least-squares is easy to work with. For example, suppose you want a $90\%$ confidence interval for $\beta_1$. You get (assuming normality and independence) something based on a $t$-distribution with $n-1$ degrees of freedom, and making use of $\hat\beta_1$ and $X_1,\ldots,X_n$ (the dependence on the $X$s is non-linear, and the non-linearity can be isolated in the inversion of a $2\times 2$ matrix that depends on the $X$s). The hypothesis test corresponding to that confidence interval can be shown to be precisely the likelihood-ratio test. |
H: Domain of this function
The function is: $$y=\frac{x+6}{15-\sqrt{{x^2}-64}}$$
The domain is: $(-\infty,-17) \cup (-17,-8] \cup [8,17) \cup (17,\infty) $
The value inside the square root can't be negative, so I set up:
$$x^2-64 \geq0$$
$$(x-8)(x+8) \geq0$$
Then I found the roots, $x=8$ and $x=-8$, which are the only places where the function may change sign. Now I'm looking for the other part of the inequality, the $x>0$ part, so I then set up a number line and tested numbers in three different regions: $$x<-8$$ $$-8\leq x \leq 8$$ $$x>8.$$
I know that in the $-8\leq x \leq 8$ region, (interval notation: $[-8,8]$),
$x^2-64<0$, so that area is not part of the domain and I understand that part of the answer.
I don't understand where the $-17,17$, and $\infty$ part of the domain comes from.
Please explain, thanks.
Also, what would the range be?
AI: I know that in the $-8\leq x \leq 8$ region, (interval notation: $[-8,8]$), $x^2-64<0$, so that area is not part of the domain and I understand that part of the answer.
This isn’t quite right: it’s in the open interval $(-8,8)$ that $x^2-64<0$, not the closed interval $[-8,8]$, so it’s $(-8,8)$ that is ruled out by the square root. That leaves $(-\infty,-8]\cup[8,\infty)$. However, the denominator of the fraction cannot be $0$, so $\sqrt{x^2-64}$ cannot be $15$, and $x^2-64$ cannot be $15^2=225$. Thus, $x^2$ cannot be $225+64=289=17^2$, and $x$ cannot be $\pm 17$. Removing $-17$ and $17$ from the set $(-\infty,-8]\cup[8,\infty)$ leaves
$$(-\infty,-17)\cup(-17,-8]\cup[8,17)\cup(17,\infty)\;.$$ |
H: Finding the limit $ \lim_{x\to a^+} \frac{\cos(x)\ln(x-a)}{\ln(e^x-e^a)} $
I have been working on this question for a while now:
$$
\lim_{x\to a^+} \frac{\cos(x)\ln(x-a)}{\ln(e^x-e^a)}
$$
I know the answer is $\cos a$, that's what the solutions say. But I don't understand how it became to that. I've tried solving it using L'Hopital rule but I've gone nowhere because the expressions just keep getting messier and messier. What's a sensible way to go about doing this? Thank you in advance for your answer.
AI: Note that we can rewrite this as
$$
\lim_{x\to a^+} \cos(x) \cdot \lim_{x\to a^+} \frac{\ln(x-a)}{\ln(e^{x} - e^a)} =
\cos(a)\lim_{x\to a^+} \frac{\ln(x-a)}{\ln(e^{x} - e^a)}
$$
In order to show that the remaining limit becomes $1$, use L'Hôpital's rule:
$$
\begin{align}
\lim_{x\to a^+} \frac{\ln(x-a)}{\ln(e^{x} - e^a)} &=
\lim_{x\to a^+} \frac{1/(x-a)}{1/(e^{x} - e^a)\cdot e^x}\\
&=\lim_{x\to a^+} e^{-x}
\lim_{x\to a^+} \frac{e^x - e^a}{x-a}\\
&= e^{-a} \lim_{x\to a^+} \frac{e^x - e^a}{x-a}
\end{align}
$$
Using L'Hôpital's rule once again gives us
$$
\begin{align}
\lim_{x\to a^+} \frac{\ln(x-a)}{\ln(e^{x} - e^a)} &=
e^{-a} \lim_{x\to a^+} \frac{e^x}{1} = e^{-a}e^a = 1
\end{align}
$$
Thus, the limit of our original function is $\cos(a)$. |
H: Show the closure of $\mathbb{Q}$ in $\mathbb{R}$ is $\mathbb{R}$.
I know this question has been asked on here before, but the answers did not contain a formal proof which I am trying to do.
To show $\operatorname{cl}\mathbb{Q}$ = $\mathbb{R}$, I want to do it by showing two inclusions. Clearly we know $\operatorname{cl}\mathbb{Q} \subset \mathbb{R}$ but I am having trouble showing $\mathbb{R}$ $\subset\operatorname{cl}\mathbb{Q}$. We can write $\operatorname{cl}\mathbb{Q} = \bigcap_{I \in J}A_I$ where each $A_I$ is a closed set in $\mathbb{R}$ containing $\mathbb{Q}$. However I don't know how this helps for me to show the inclusion I am stuck on. Also limit points have not been introduced yet and shouldn't be needed for this problem.
AI: Let $A$ be a closed set containing $\mathbb{Q}$. Then $A^c$ is open. Suppose there is an $a\in A^c$. Then there is an open interval $(a-\epsilon,a+\epsilon)\in A^c$. So there is an open interval with no elements in $A$ and this implies there is an open interval with no elements in $\mathbb{Q}$, which is impossible since the rationals are dense. Thus $A^c=\emptyset$, so $A=\mathbb{R}$. So the only closed set containing $\mathbb{Q}$ is $\mathbb{R}$, so cl$(\mathbb{Q})=\mathbb{R}$. |
H: Evaluate the limit $\lim_{n\to\infty}\frac{n}{\ln n}\left(\frac{\sqrt[n]{n!}}{n}-\frac{1}{e}\right)$
Evaluate
$$\lim_{n\to\infty}\frac{n}{\ln n}\left(\frac{\sqrt[n]{n!}}{n}-\frac{1}{e}\right).$$
This sequence looks extremely horrible and it makes me crazy. How can we evaluate this?
AI: Use Stirling's approximation:
$$n!\sim\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}=\left(\frac{n}{e}\right)^{n}e^{\frac12 \ln 2\pi n}$$
It transforms your limit into
$$\lim_{n\rightarrow\infty}\frac{n}{\ln n}\frac{e^{\frac{\ln n}{2n}}-1}{e}=\lim_{x\rightarrow 0}\frac{e^{\frac{x}{2}}-1}{ex}=\frac{1}{2e}.$$
When obtaining the first expression, we neglect $\frac{\ln 2\pi}{2n}$ in the exponential (it is $o\left(\frac{\ln n}{2n}\right)$), and then we make the change of variables $x=\frac{\ln n}{n}$. |
H: How many 7-character passwords can be generated with...Stipulations?
Characters can only be (a-z), (A-Z), and (0-9) w/stipulation that first character must be a lower case letter, the last character must be an upper case letter, and of the 5-characters in the middle there must be at-least ones digit (0-9).
How many 7-character passwords can be generated with the above stipulations?
AI: Hint:
The number of ways to choose such a password is the number of ways to choose the first character, times the number of ways to choose the last character, times the number of ways to choose the 5 characters in the middle. So, you must count these.
Counting the number of possibilities for the first and last digits should be no sweat. As for the five in the middle: my hint to you is to remember that the number of ways to choose five characters, at least one of which is a digit, is given by
$$
(\text{total number of 5-character strings}) - (\text{number of 5-character strings with no digits}).
$$ |
H: Prove that $Tu(x)$ is a contraction. $Tu(x) = -\lambda\int_0^1g(x,y)\sin(u(y))\,dy$
I want to show that $Tu(x)$ is a contraction where
$$Tu(x) = -\lambda\int_0^1g(x,y)\sin(u(y))\,dy$$ and
$$g(x,y) = \begin{cases} x(1-y) & 0\leq x\leq y\leq 1, \\
y(1-x) & 0\leq y \leq x \leq 1. \end{cases}$$
I have
$$\begin{align*} \|Tu_1-Tu_2\|_\infty &= |\lambda|\left\|\int_0^1 g(x,y)\sin(u_1(y))-sin(u_2(y))\,dy\right\|_\infty \\
&\leq |\lambda|\| \sin(u_1(y))-\sin(u_2(y))\|_{\infty}\left\|\int_0^1 g(x,y)\,dy\right\|_{\infty}
\end{align*}$$
I know im supposed to get that this last inequality is $\leq |\lambda|\|u_1-u_2\|\frac{1}{8}$ but I'm not sure how to show that.
Any help and comments would be greatly appreciated. Thank you!
AI: Its just grind...
\begin{eqnarray}
|Tu(x)-Tv(x)| &\le & |\lambda| \int_0^1 |g(x,y)| |\sin u(y) - \sin v(y)| dy \\
&\le& |\lambda| \int_0^1 |g(x,y) | | u(y) - v(y)| dy \\
&\le& |\lambda| \|u-v\|_\infty \int_0^1 |g(x,y) | dy \\
\end{eqnarray}
Hence $ \|Tu-Tv\|_\infty \le |\lambda| \|u-v\|_\infty \sup_{x \in [0,1]}\int_0^1 |g(x,y) | dy $.
Now compute $\int_0^1 |g(x,y) | dy = \int_0^x y(1-x)dy + \int_x^1 x(1-y)dy = \frac{1}{2}x(1-x)$, and this is maximized at $x=\frac{1}{2}$, and so $\int_0^1 |g(x,y) | dy \le \frac{1}{8}$, which gives
$$\|Tu-Tv\|_\infty \le \frac{1}{8}|\lambda| \|u-v\|_\infty $$
(Of course, this only shows $T$ to be a contraction if $|\lambda| < 8$.) |
H: Proof that the gamma function is an extension of the factorial function
I've already proved that $$\Gamma (n)= (n-1)!$$ but I don´t really know what else to do to verify that $\Gamma$ is an extension of the factorial function for real numbers (positive) Thank you! And I´m sorry for my language, I am Spanish, so thank you again for trying to understand me.
AI: Consider $$\Gamma (z)=\int_0^\infty e^{-t}t^{z}{dt\over t}.$$ We can rewrite this as $$\Gamma (z)=\int_0^1 e^{-t}t^{z}{dt\over t}+\int_1^\infty e^{-t}t^{z}{dt\over t}.$$ In the first term of this sum we see that the power series representation of $e^{-t}$ converges uniformly which implies that the series can be integrated term by term. So, $$\Gamma (z)=\int_0^1\sum_{n=0}^\infty {(-1)^n\over n!}t^{z+n}{dt\over t}+\int_1^\infty e^{-t}t^{z}{dt\over t}=\sum_{n=0}^\infty {(-1)^n\over n!(z+n)}+\int_1^\infty e^{-t}t^{z}{dt\over t}.$$ We can see that the series converges for $z\neq 0, -1, -2,...$ which is a meromorphic function. Its poles are simple poles at the non-positive integers. The residue at $-n$ is $(-1)^n\over n!$. The last integral extends as an entire function of z.
Thus $\Gamma (z)$ has been analytically continued to the entire complex plane except for $z\neq0,1,2,...$. |
H: Continuous functions from one topological space to another topological space.
Question: Let $X=\lbrace 1,2,3,4,5 \rbrace$ with topology $\lbrace \emptyset,X,\lbrace 1 \rbrace,\lbrace 3,4 \rbrace,\lbrace 1,3,4 \rbrace\rbrace,$ and let $Y=\lbrace A,B\rbrace$ with topology $\lbrace \emptyset,Y,\lbrace A\rbrace\rbrace.$ Find all continuous functions from $X\to Y$.
I know that given any open subset U in Y, f^{-1}(U) must be open in X. Now, how do I list and find these functions for the above spaces?
AI: You need $f^{-1} V \in \tau_X$ for all $V \in \tau_Y$. The only member of $\tau_Y$ that is relevant is $\{A\}$. To be continuous, we must have $f^{-1}\{A\} \in \tau_X$. There are only $5$ possibilities, and $f$ must take the value $B$ on the complement.
The functions are straightforward to list. For example, if $f^{-1}\{A\} = \emptyset$, then $f(x) = B$ for all $x$. Another, if $f^{-1}\{A\} = \{1\}$, then $f(1) = A$ and $f(x) = B$ for all $x \neq 1$. |
H: Linear Algebra - Determine if the following 2 subspaces add up to $\mathbb R^3$
$W_1=\{ (0,-b,b)|\ b\in \mathbb R$}
$W_2=\{(a_1, a_1+a_2,a_2)\;|\ a_1,a_2\in \mathbb R\}$
Is $\mathbb R^3=W_1+W_2$?
I know that if it is true, then $\mathbb R^3\subset W_1+W_2$. How do I determine whether the 2 subspaces span $\mathbb R^3$?
AI: Note that $W_1$ has dimension $1$ while $W_2$ has dimension $2$. Thus, they will sum up to $\Bbb R^3\iff W_1\cap W_2=\{(0,0,0)\}$. |
H: Is there an easy way to factor polynomials with ugly numbers?
Say for example I have the polynomial $x^2-8x-153$. It is not easy to immediately see the factors of this are $9$ and $17$. Is there a method to find these factors easily when the terms in the polynomials are this big? The reason I ask this is because I don't want to spend a lot of time on a test guessing for the factors. Thanks in advance!
AI: Every quadratic equation can be solved by completing the square. Noting that $8/2 = 4$, we have
$$x^2 - 8x - 153 = x^2 - 8x + (4^2) - 4^2 - 153$$
or rearranging,
$$(x - 4)^2 = 153 + 4^2 = 169 = 13^2$$
Thus, $x - 4 = \pm 13$, leading to the two solutions $x = -9$ and $x = 17$. Then the polynomial factors as
$$(x + 9)(x - 17)$$
So this technique only requires knowing about square roots. |
H: Finding coordinates of closest approach
Given two lines $l_1=\mathbf E_1+k\mathbf E'_1$ and $l_2=\mathbf E_2+\mu\mathbf E'_2$ in 3D, there exists a shortest distance between the two lines. How does one find the coordinates of the points $P$ on $l_1$ and $Q$ on $l_2$, such that $P$ and $Q$ are the points where the distance between the two lines is the shortest?
AI: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert #1 \right\vert}%
\newcommand{\yy}{\Longleftrightarrow}$
${\bf d} = {\bf l}_{1} - {\bf l}_{2}$
$$
{\bf d}^{2}
=
\pars{{\bf E}_{1} + k{\bf E}_{1}' - {\bf E}_{2} - \mu{\bf E}_{2}'}^{2}
=
\pars{{\bf E} + k{\bf E}_{1}' - \mu{\bf E}_{2}'}^{2}\,,
\quad
{\bf E} \equiv {\bf E}_{1} - {\bf E}_{2}
$$
$$
{\bf d}^{2}
=
{\bf E}^{2} + {\bf E}_{1}'^{2}k^{2} + {\bf E}_{2}'^{2}\mu^{2}
+
2{\bf E}\cdot{\bf E}_{1}'k
-
2{\bf E}\cdot{\bf E}_{2}'\mu
-
2{\bf E}_{1}'\cdot{\bf E}_{2}'k\mu
$$
Minimize ${\bf d}^{2}$ respect of $k$ and $\mu$:
\begin{align}
\partiald{{\bf d}^{2}}{k}
=
0
&\imp\quad
2{\bf E}_{1}'^{2}k
+
2{\bf E}\cdot{\bf E}_{1}'
-
2{\bf E}_{1}'\cdot{\bf E}_{2}'\mu
=
0
\\[3mm]
\partiald{{\bf d}^{2}}{\mu}
=
0
&\imp\quad
2{\bf E}_{2}'^{2}\mu
-
2{\bf E}\cdot{\bf E}_{2}'
-
2{\bf E}_{1}'\cdot{\bf E}_{2}'k
=
0
\end{align}
Now, we have two equations for $k$ and $\mu$:
$$
\left\lbrace%
\begin{array}{rcrcl}
{\bf E}_{1}'^{2}\,k
& - &
{\bf E}_{1}'\cdot{\bf E}_{2}'\,\mu
& = &
-{\bf E}\cdot{\bf E}_{1}'
\\
{\bf E}_{1}'\cdot{\bf E}_{2}'\,k
& - &
{\bf E}_{2}'^{2}\,\mu
& = &
-{\bf E}\cdot{\bf E}_{2}'
\end{array}\right.
$$
Find the values of $k$ and $\mu$. They determine the points we are looking for on the lines ${\bf l}_{1}$ and ${\bf l}_{2}$, respectively. |
H: Sending each basis element $a,b,c,d$ to $y-x$.
Hatcher P99 last paragraph:
Define a homomorphism $\partial: C_1 \to C_0$ by sending each basis element $a,b,c,d$ to $y-x$, the vertex at the head of the edge minus the vertex at the tail.
So I am confused: $x,y$ are just vertices, how can they subtract?
http://www.math.cornell.edu/~hatcher/AT/ATch2.pdf
AI: $C_0$ is defined to be the free abelian group generated by the vertices; that is,
$$C_0 = \{n x + m y : n, m \in \mathbb{Z}\}$$
under the natural definitions of $+$ and $-$ for group elements of this form. So $x - y$ is simply the element corresponding to $n = 1$ and $m = -1$ in this group. |
H: Find the remainder when $2^{47}$ is divided by $47$
So I am trying to work out how to find remainders in several different way.
I have a few very similar question,
1.) Find the remainder when $2^{47}$ is divided by 47
So i have a solution that says
$$2^{24} \equiv 2$$
$$2^{48} \equiv 4$$
$$2^{47} \equiv 2$$ Since $(2,47)=1$
and
2.)Find the remainder when $2^{32}$ is divided by $47$:
We have $$2^6 = 64 \equiv 17$$
$$2^{12} \equiv 17^2 = 289 \equiv 7$$
$$2^{24} \equiv 7^2=49 \equiv 2$$
$$2^8 \equiv 4*17=68 \equiv 21$$
$$2^{32}=2^{8}*2^{24} \equiv21*2 \equiv42$$
Okay so although i have some solutions here, Im not 100 percent sure how they derived this. For example on the second question, why have they started with $2^{6}$ before finding the remainder? Is there some steps missing or am I just missing the point?
Is this called something in particular? There are not many problems similar to and i do not know if it has a special name to search for.
P.S Please do not use fermat's little theorem, i want to understand this method, thanks :)
AI: 1) Find the remainder when $2^{47}$ is divided by $47$.
$2^1\equiv2\pmod{47}$
$2^2\equiv4\pmod{47}$
$2^4\equiv16\pmod{47}$
$2^8=(2^4)^2\equiv(16)^2=256\equiv21\pmod{47}$
$2^{16}=(2^8)^2\equiv(21)^2=441\equiv18\pmod{47}$
$2^{24}=2^8\cdot2^{16}\equiv21\cdot18=378\equiv2\pmod{47}$
$2^{48}\equiv2^2=4\pmod{47}$
$2*2^{47}\equiv2*2\pmod{47}$
Since $ca\equiv cb\pmod m$ and $(c,m)=1$ implies $a\equiv b\pmod m$, and since $(2,47)=1$, we can cancel the $2$:
$2^{47}\equiv2\pmod{47}$
2) Find the remainder when $2^{32}$ is divided by $47$.
From the previous solution:
$2^{24}\equiv2\pmod{47}$
$2^8\equiv21\pmod{47}$
$2^{32}=2^{24+8}=2^{24}\cdot2^8\equiv2\cdot21=42\pmod{47}$ |
H: If $xy$ and $x+y$ are both even integers (with $x,y$ integers), then $x$ and $y$ are both even integers
The title statement can be proven using the contrapositive, note that $x$ odd or $y$ odd means that at least one of $x\cdot y,x+y$ is odd. Is there a way to prove the statement directly?
To generalize on this statement (in two integer variables), show that for every square free integer $n$, there exists a function $f:\mathbb Z^2\to\mathbb Z$ where
$$f(x,y)\equiv 0\pmod n\iff x\equiv y\equiv 0\pmod n$$
This generalization is formalized in this question: If $n$ is squarefree, $k\ge 2$, then $\exists f\in\Bbb Z[x_1,\dots,x_k] : f(\overline x)\equiv 0\pmod n\iff \overline x\equiv \overline 0\pmod n$
AI: Observe that $\displaystyle (x+1)(y+1)=xy+x+y+1$ is odd as $xy,x+y$ are even
If $x$ is odd $\iff x+1$ is even |
H: Continuity of an identity linear transformation between two spaces.
$V$ is a vector space of all continuous complex valued function on $ J = [a,b]$. Let $X_1 = (V,\|\|_{\infty})$ where $\|x\|_{\infty} = \max_{t \in J}\{x(t)\}$ and $X_2 = (V,\|\|_2)$ where $\|x\|_2 = <x,x>^{\frac{1}{2}}$. The the identity mapping $x \rightarrow x$ of $x_1$ onto $x_2$ is continuous. I want to prove it.
here $X_1$ is a normed linear space and $X_2$ is an inner product space. I have tried to use definition of the continuous function. Ordinary $\epsilon, \delta$ definition, definition of continuity using sequence and open set definition of continuity. But I can not match them with the data. Maybe some very easy point I am missing.
Thank you for your suggestion.
AI: You have $\| Id(x)\|_2 = \sqrt{\int_a^b x(t)^2 dt} \le \sqrt{\int_a^b \|x\|_\infty^2 dt} = \sqrt{b-a} \|x\|_\infty$, and so $\|Id\| \le \sqrt{b-a}$. |
H: How many ways are there to distribute $18$ different toys among $4$ children,(a) without restriction?(b) if $2$ children get $7$ toys each and $2$…
How many ways are there to distribute $18$ different toys among $4$ children?
a) without restrictions
b) if $2$ children get $7$ toys each and $2$ children get $2$ toys each.
For $1$ since toys are different, then there are $4^{18}$ ways to distribute .
for $2$, Im kind of stuck. I know I have two do it in $2$ cases. Can someone help me?
AI: Choose the $2$ children who will get $7$ toys. Then choose the $7$ toys the older of these will get. Then choose the $7$ for the younger one. Then choose the $2$ toys for the older of the remaining children. That gives a total of $\binom{4}{2}\binom{18}{7}\binom{11}{7}\binom{4}{2}$. |
H: Counter example of Zorn's lemma when we only take countable chains
I was learning Zorn's lemma yesterday and I couldn't find any example, where Zorn's lemma fails when we only require that every countable chain in P has a maximal element in P. Does anyone know an example?
AI: Consider the collection of countable subsets of $\Bbb R$, ordered by inclusion.
Every countable chain has an upper bound, since the countable union of countable sets is countable; but there is no "maximal countable subset".
(It is consistent for the axiom of choice to fail and $\Bbb R$ to be the countable union of countable sets, in which case this example is not a counterexample; and the same goes for Brian's example. But if the axiom of choice fails that bad, then there are other counterexamples to your question.) |
H: Greatest Lower Bound in $\mathbb{R}$ as a corollary of the LUB?
I can assume as fact that $\mathbb{R}$ is an ordered field in which every non-empty subset that is bounded above has a least upper bound.
My question is whether I can also assume as fact that every non-empty subset that is bounded below also has a greatest lower bound. I'm trying to show that if $A,B \subseteq \mathbb{R}$, and $A,B \neq \emptyset$, then there exists a least upper bound for $A$ (I've proven this), and a greatest lower bound for $B$ (this is what I'm currently concerned with).
Any tips, or helpful definitions?
AI: HINT: If $A\subseteq\Bbb R$ is bounded below, then $-A=\{-a:a\in A\}$ is bounded above.
Alternative HINT: Let $B$ be the set of lower bounds for $A$. $B$ is bounded above, and its supremum is ... ? |
H: Topology on cartesian product and product topology.
Let X and Y be sets. Does it have every topology on cartesian product X$\times$Y must be product topology ?
AI: It goes a bit further: one can show that for virtually any pair $X,Y$ of sets there are topologies on $X \times Y$ which cannot be expressed as the product topology on topologies on $X$ and $Y$, respectively.
For one example, consider the topology on $\mathbb{Z} \times \mathbb{Z}$ where the basic open sets are the "diagonals" $$U_k = \{ \langle n , n + k \rangle : n \in \mathbb{Z} \}$$ for each $k \in \mathbb{Z}$. (This means that $U \subseteq \mathbb{Z} \times \mathbb{Z}$ is open iff for each $\langle m,\ell \rangle \in U$ all pairs of the form $\langle n , n+(\ell-m) \rangle$ is also in $U$.)
To show that there is no topologies $\mathcal{O}_1 , \mathcal{O}_2$ on $\mathbb{Z}$ such that the topology above is the product topology of $\langle \mathbb{Z} , \mathcal{O}_1 \rangle \times \langle \mathbb{Z} , \mathcal{O}_2 \rangle$, consider the following:
Given $\langle m,\ell \rangle \in \mathbb{Z} \times \mathbb{Z}$, consider the open set $U_{\ell-m} \ni \langle m,\ell\rangle$. Note that as the only "rectangles" which are subsets of $U_{\ell-m}$ are the singletons $\{ \langle n , n + (\ell-m) \rangle \} = \{ n \} \times \{ n + (\ell-m) \}$, it follows that if $A , B \subseteq \mathbb{Z}$ are such that $m \in A$, $\ell \in B$ and $A \times B \subseteq U_{\ell-m}$, then $A = \{ m \}$ and $B = \{ \ell \}$. It thus follows that if the topology above were the product topology of two topologies on $\mathbb{Z}$, then these two topologies must be discrete. But note that the product of two discrete spaces is discrete, and the topology above is clearly not discrete.
Another (similar/simpler) example, let $X = \{ 0,1 \}$, and consider the topology $$\mathcal{O} = \{ \varnothing , \{ \langle 0,0 \rangle \} , \{ \langle 1,1 \rangle \} , \{ \langle 0,0 \rangle , \langle 1,1 \rangle \} , X \times X \}$$ on $X \times X$.
If $\mathcal{O}_1$ and $\mathcal{O}_2$ were topologies on $X$ such that $\mathcal{O}$ is the product of these two topologies, then as $\{ \langle 0 , 0 \rangle \}$ is open, it follows that the singleton $\{ 0 \}$ is open in both $\mathcal{O}_1$ and $\mathcal{O}_2$. Similarly, $\{ 1 \}$ is open in both $\mathcal{O}_1$ and $\mathcal{O}_2$. Note that by definition of the product topology it would follow that $\{ \langle 0,1 \rangle \} = \{ 0 \} \times \{ 1 \}$ is open in $\mathcal{O}_1 \times \mathcal{O}_2$, but it is clearly not open in $\mathcal{O}$.
This second example can be extended for any pair of sets $X , Y$ which contain at least two points. (It should be noted that if either $|X| = 1$ or $|Y|=1$ then any topology on $X \times Y$ is the product of topologies on $X$, $Y$.) |
H: construct x =ab by using compass alone, if a and b are given segments.
I found the problem in the book "What is mathematics?".
The following is a description of Mohr's constructions.(Macheroni problem)
9) Find $x = ab$, if $a$ and $b$ are given segments.
I found the question didn't give a segment denotes 1.I've no idea how to construct the x.
Can anyone give me a hint?
AI: The question makes no sense unless you have a $1$ given: $a$ and $b$ are lengths, so $ab$ is an area; the required $x$ is the length of a rectangle of equal area to $ab$ and of width $1$.
One construction is to plot axes, and mark $(0,a)$ on the $y$-axis and $(-1,0)$ on the $x$-axis. The line between them has gradient $a$, so prolong that to where it meets the (vertical) line perpendicular to the $x$-axis drawn through $(b-1,0)$. The intersection is at $(b-1, ab)$. |
H: Generating function of recurrence sequence
I have the following recurrence sequence
$$
a_{1} = 0\,,\quad
a_{2} = 1\,,
\qquad\qquad
a_{n}
=
{2 + 2\left(n - 2\right)\, a_{n - 2}
+
\left(n - 2\right)\left(n - 1\right)\, a_{n - 1}
\over
n\left(n - 1\right)}
$$
It starts form
$\displaystyle{%
\left\lbrace%
0,\ 1,\ {2 \over 3},\ {5 \over 6},\ {4 \over 5},\ {37 \over 45},\ {52 \over 63},\ {349 \over 420},\ {338 \over 405},\ {11873 \over 14175}
\right\rbrace
}$.
Please help me to find generating function
AI: Hint: Consider $A(x)=\sum\limits_{n=1}^\infty a_nx^n$ the generating function of $(a_n)$. The recursion you are interested in is that, for every $n\geqslant2$,
$$
n(n-1)a_n=2+2(n-2)a_{n-2}+(n-2)(n-1)a_{n-1}.
$$
Can you identify the generating function of $(b_n)$ when $b_n=n(n-1)a_n$? When $b_n=2$? When $b_n=2(n-2)a_{n-2}$? And when $b_n=(n-2)(n-1)a_{n-1}$? Then see what comes out. |
H: Definition of semi-ring homomorphism
I need the definition of semi-ring homomorphism. Thanks in advance!
AI: The definition should depend on your purposes, but I think that this should be the best definition:
$$f(a+b) = f(a) + f(b)$$
$$f(ab) = f(a)f(b)$$
$$f(0) = 0$$
$$f(1) = 1$$
Unlike with rings, you cannot deduce $f(0) = 0$ from the other axioms.
If you want to work with non-unital semirings, you might drop the last axiom. But as far as I can tell, semirings are always defined to be additive monoids, in which case you definitely should not drop $f(0) = 0$. |
H: Can't solve this problem $lim_{x \to 0}\frac{3^{5x}-2^{7x}}{\arcsin\left(2x\right) - x}$(Without using L'Hospital's rule)
I don't see the way to solve this limit.
$$\lim_{x \to 0}\frac{3^{5x}-2^{7x}}{\arcsin\left(2x\right) - x} $$
My attempt is
1) Divide the numerator by $3^{5x}$
$$\lim_{x \to 0}\frac{3^{5x}-2^{7x}}{\arcsin\left(2x\right) - x} = \lim_{x \to 0} \frac {1- \left(2/3 \right)^{5x}\,\,2^{2x}}{\arcsin\left(2x\right)-x} $$
And I am unable to do next step. So how this limit can be solved?
AI: Use the fact that
$$3^{5 x}-2^{7 x} = e^{(5 \log{3}) x} - e^{(7 \log{2}) x}$$
and that, for small $y$
$$e^{y} = 1 + y + \cdots$$
and further, for small $z$
$$\arcsin{z} = z + \cdots$$
so that, for $x$ near $0$:
$$\frac{3^{5 x}-2^{7 x}}{\arcsin{(2 x)}-x} \sim \frac{(5 \log{3}-7 \log{2}) x}{2 x-x} = 5 \log{3}-7 \log{2}$$ |
H: Proof of Continuous compounding formula
Following is the formula to calculate continuous compounding
A = P e^(RT)
Continuous Compound Interest Formula
where, P = principal amount (initial investment)
r = annual interest rate (as a decimal)
t = number of years
A = amount after time t
The above is specific to continuous compounding. The general compounding formula is
$$A=P\left(1+\frac{r}{n}\right)^{nt}$$
I want to understand how continuous compounding formula is derived from general compounding formula, given t=1, n=INFINITY.
AI: One of the more common definitions of the constant $e$ is that:
$$
e = \lim_{m \to \infty} \left(1 + \frac{1}{m}\right)^m
$$
Thus we have, with a change of variables $n = mr$, that
$$
\lim_{n \to \infty} P\left(1 + \frac{r}{n}\right)^{nt}\\
= \lim_{m \to \infty} P\left(1 + \frac{1}{m}\right)^{mrt}\\
= P\left(\lim_{m \to \infty}\left(1 + \frac{1}{m}\right)^m\right)^{rt}\\
= Pe^{rt}
$$
and you have your continuous compounding formula. |
H: Can this expression be reduced to a difference quotient?
Setting up an equation I've come into this factor:
$\displaystyle \lim_{h\rightarrow0}\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h}; \quad f\in \mathcal{C}^\infty$
To me this looks more or less like a derivative, but I've not been able to reduce it to a common difference quotient.
Is that actually $df/dx$ or there's more?
AI: Since $f\in C^2$, integration by parts gives Taylor's Theorem with remainder:
$$
\begin{align}
f(x)
&=f(a)+(x-a)f'(a)+\int_a^x(x-t)f''(t)\,\mathrm{d}t\\
&=f(a)+(x-a)f'(a)+\frac12(x-a)^2f''(\xi)\tag{1}
\end{align}
$$
for some $\xi$ between $a$ and $x$.
Using $(1)$ to represent $f(x+h)$ and $f(x-h)$, we get
$$
\begin{align}
\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h^2}
&=-\frac{f(x+h)-2f(x)+f(x-h)}{2f(x)h^2}\\
&=-\frac{\frac12h^2f''(\xi_+)+\frac12h^2f''(\xi_-)}{2f(x)h^2}\\[4pt]
&=-\frac{f''(\xi)}{2f(x)}\tag{2}
\end{align}
$$
where $\xi_+\in(x,x+h)$ and $\xi_-\in(x-h,x)$ and $\xi\in(\xi_-,\xi_+)\subset(x-h,x+h)$.
Therefore,
$$
\lim_{h\to0}\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h^2}=-\frac{f''(x)}{2f(x)}\tag{3}
$$
Of course, $(3)$ implies
$$
\begin{align}
\lim_{h\to0}\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h}
&=\lim_{h\to0}h\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h^2}\\
&=\lim_{h\to0}h\lim_{h\to0}\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h^2}\\[4pt]
&=0\cdot-\frac{f''(x)}{2f(x)}\\[12pt]
&=0\tag{4}
\end{align}
$$ |
H: Topology Questions
So I am just wanting to make sure I had these right. For the most part I think I have a solid understanding of both problems 1 and 2. Both the last 2 problems, I think I understand them, but I am not 100% sure.
Problem 1: Let $A= [0,8)$ be a subspace of $(\mathbb{R}, U)$. Which of the following sets are $U_A$-open? For each set $V$ that is $U_A$-open, find a $U$-open set $W$ for which $V = W \cap A$. Note: $U$ is the usual topology for $\mathbb{R}$.
(a) $[7,8)$ No.
(b) $(7,8)$ Yes, Let $W = (7,8)$.
(c) $[0,8)$ Yes, Let $W = (-2,10)$.
Problem 2: Let $A= (0,8]$ be a subspace of $(\mathbb{R}, H)$. Which of the following sets are $H_A$-open? For each set $V$ that is $H_A$-open, find a $H$-open set W for which $V = W \cap A$. Note: $H$ is the topology for $\mathbb{R}$ which has a base $B$ of half open intervals $[a,b)$.
(a) $(0,1]$ No
(b) $[2,4]$ No
(c) $[2,4)$ Yes, Let $W= [2,4).
Problem 3: Let $A= (-3,0] \cup [1,3)$ be a subspace of $(\mathbb{R}, C)$. Which of the following sets are $C_A$-open? For each set $V$ that is $C_A$-open, find a $C$-open set $W$ for which $V = W \cap A$. Note: $C$ is the topology for $\mathbb{R}$ where $C= \{ (a,\infty) : a ∈ \mathbb{R} \} \cup\{\mathbb{R},\varnothing\}$.
(a) $(-2,0] \cup [1,3)$ Yes, Let $W=(-2, \infty)$
(b) $(-2,0]\cup [2,3)$ No , the only $C$-open set would be $(-2, \infty)$ and this would not have $[2,3)$ in $V$.
(c) $(1,3)$ No, same as b more or less.
Problem 4: Let X = {a,b,c,d} and τ = {X, Ø, {a}, {b,c}, {a,b,c}}. List the sets in each of the Following topologies:
(a) $\tau_{a,d} = \{\varnothing, \{a\}\}$.
(b) $\tau_{b,d} = \{\varnothing\}$.
(c.) $\tau_{a,b,c} = \{\varnothing, \{a\}, \{b,c\}, \{a,b,c\}\}$.
(d) $\tau_{d} = \{\varnothing\}$.
AI: Problems $1$ and $2$ are perfect.
Your answer to $3($c$)$ is not correct; the situation is not the same as in part (b). You might want to look at part (b) more carefully, though. Your answer is correct but I'm not sure about your reasoning, since $[2,3)$ is a subset of $(-2,\infty)\cap A$.
For Problem $4$, the image you link (and the notation in the rest of the question) suggests that $\tau_A$ is the topology $\tau$ restricted to the set $A$. In particular, it cannot be that any set in $\tau_A$ contains any point not in $A$, so none of them should contain $X$. There are some other mistakes which I'll leave to you to figure out now that you understand the problem statement. |
H: Show $\mathcal D_D$ is Dynkin system
Let $\mathcal D\subset\mathcal P(\Omega)$ be a Dynkin system and $D\in\mathcal D$. Then $\mathcal D_D=\{A\subset\Omega|A\cap D\in\mathcal D\}$ is a Dynkin system.
It's clear that $\Omega\in\mathcal D_D$. But I'm not sure how to show that
i) for $A\in\mathcal D_D$ we also have $A^C\in\mathcal D_D$
ii) for $A_1,...\in\mathcal D_D$ with $A_i\cap A_j=\emptyset$ for $i\neq j$ we also have $\sum_{i=1}^\infty A_i\in\mathcal D_D$
For ii) I was thinking that if $(A_i)$ is a sequence in $\mathcal D_D$ we have $A_i\cap D\in\mathcal D$ for all $i$. So we have $\left(\sum_{i=1}^\infty A_i\right)\cap D = \sum_{i=1}^\infty \underbrace{A_i\cap D}_{\in \mathcal D}$. But can I argue that this union is also in $\mathcal D$?
And how do I show i)?
Thanks in advance for any help.
AI: For (ii): Yes, you can argue like this, note that
$$(A_i \cap D) \cap (A_j \cap D) = (A_i \cap A_j) \cap D = \emptyset ,\quad i \ne j $$
by assumption on the $A_i$. Now use that $\mathcal D$ is a Dynkin system.
For (i): Suppose $A \cap D \in \mathcal D$, we have
$$ A^c \cap D = (A \cup D^c)^c = \bigl( (A\cap D) \cup D^c \bigr)^c $$
Now use that $D \in \mathcal D$, that $A\cap D$ and $D^c$ are disjoint and $\mathcal D$ is a Dynkin system. |
H: 2 regular graphs and permutations
I have found this question on MSE before but I didn't find the answer satisfactory and it is so old I doubt anyone is still following it.
Let $f_{n}$ be the number of permutations on $[n]$ with no fixed points or two cycles. Let $g_{n}$ be the number of simple, labeled two regular graphs on $n$ vertices. Let $F(x)$ be the EGF for $f_{n}$ and $G(x)$ be the EGF for $g_{n}$. I've shown that $F(x)=G(x)^{2}$. So by composition of exponential genereating functions
$$f_{n}=\sum_{k\geq 0}^{n}\binom{n}{k}g_{k}g_{n-k}$$.
I am tasked to prove $F(x)=G(x)^{2}$ via a bijective proof. I know that given a graph on $k$ vertices we can decompose it into cycles. Then for each cycle we can orient the cycle two ways to make a permutation on $n$. But I think this count is too large.
Any hint would help thanks.
AI: HINT: Each permutation of $[n]$ with no fixed points and no $2$-cycles corresponds in an obvious way to a unique labelled two-regular directed graph in which each cyclic component is a directed cycle. Given such a graph $G$, we can split the components into two sets, $\mathscr{C}_I(G)$ and $\mathscr{C}_D(G)$, as follows:
Suppose that $C$ is a component whose lowest-numbered vertex is $v$; there are vertices $u$ and $w$ in $C$ such that $u\to v$ and $v\to w$ are directed edges of $C$. If $u<w$, put $C$ into $\mathscr{C}_I(G)$, and if $u>w$, put $C$ into $\mathscr{C}_D(G)$.
Let $G_I=\bigcup\mathscr{C}_I$ and $G_D=\bigcup\mathscr{C}_D$. Now show that for a fixed set of $k$ labels from $[n]$ there are $g_k$ possible graphs $G_I$ and $g_{n-k}$ possible graphs $G_D$. |
H: I want to prove that $m||v||_1\le ||v||$
Let $(V,||\cdot ||)$ a finite-dimensional real normed space, and $\{v_1,...,v_n\}$ a basis. We define $||v||_1=\sqrt{x_1^2+...+x_n^2}$, where $x_1,...,x_n$ are the coordinates of $v$.
I already proved that there exists $w\in S:=\{v\in V:||v||_1=1\}$ such that $$||w||\le ||v||$$ for every $v\in S$.
Now I want to prove that $$m||v||_1\le ||v||$$ for every $v\in V.$, but I can't see why we have that.
Any hint? Thanks.
Edit: I forgot... $m:=||w||$
AI: Hint: First note, that $w \ne 0$, let $m := \|w\| >0$. Now let $v \in V$. Then $v':= v/\|v\|_1 \in S$, hence $m = \|w\| \le \|v'\|$. Now use the defition of $v'$ and homogenity of $\|\cdot \|$. |
H: Ramification of a Galois extension
I understand that an extension of number field $L/K$ is unramified if every non-zero prime ideal of $\mathcal{O}_K$ is unramified in $L$ (where a prime ideal $\mathfrak{p}$ of $\mathcal{O}_K$ is ramified if it has $e_i > 1$ for some $i$ when you write it as the decomposition of primes of $\mathcal{O}_L$).
However I am not sure of the situation when $L/K$ is Galois. For example, if the inertia group is trivial does it imply that the whole extension is unramified or just one of its primes ?
Thanks for your help !
AI: The simplification that occurs when $L/K$ is Galois is that for any fixed prime $\mathfrak{p}$ of $\mathcal{O}_K$, the Galois group acts transitively on the primes $\mathfrak{P}_i$ of $\mathcal{O}_F$ above $\mathfrak{p}$. So the decomposition groups $D_{\mathfrak{P}_i/\mathfrak{p}}$ and the inertia groups $I_{\mathfrak{P}_i/\mathfrak{p}}$ for different $i$ are all conjugate, and so the ramification indices $e_{\mathfrak{P}_i/\mathfrak{p}}$ are all the same, as are the residue field degrees.
This was for a fixed prime downstairs. The story for another fixed prime $\mathfrak{p}'$ of $\mathcal{O}_K$ is (more or less) independent of the story for $\mathfrak{p}$. |
H: Integrals of trignometric functions
Question is to Prove that :
$$\int_0 ^{2\pi} \frac{d\theta}{a+b\sin \theta}=\frac{2\pi}{\sqrt{a^2-b^2}} \text{for}\ a>b>0$$
using residue theory.
What i have done so far is :
I transformed functio of $\theta$ as function of complex entity $z$ with $z=e^{i\theta}$.
Then, we have : $dz=i.e^{i\theta}.d\theta\Rightarrow dz=i.z.d\theta \Rightarrow d\theta =\frac{dz}{iz}$
$$\Rightarrow \int_0 ^{2\pi} \frac{d\theta}{a+b\sin \theta}=\int_{|z|\leq 1} \frac{1}{a+b.\frac{1}{2i}.(\frac{z^2-1}{z})}.\frac{dz}{iz}$$
$$\int_{|z|\leq 1} \frac{1}{a+b.\frac{1}{2i}.(\frac{z^2-1}{z})}.\frac{dz}{iz}= 2 \int_{|z|\leq 1} \frac{1}{bz^2+(2ai)z-b} dz$$
So, now the actual problem comes :
I have to find poles for this and use residue theorem to evaluate the integrals :
$bz^2+(2ai)z-b=0 \Rightarrow z=\frac{-2ai \pm \sqrt{-4a^2+4b^2}}{2b}$. But, $a>b$ so, i would like to see this as
$\frac{-2ai \pm 2i\sqrt{a^2-b^2}}{2b}=\frac{-ai \pm i \sqrt{a^2-b^2}}{b}$
Now, i am not able to decide which of $\frac{-ai + i \sqrt{a^2-b^2}}{b}$ and $\frac{-ai - i \sqrt{a^2-b^2}}{b}$ is in $\{ z\in \mathbb{C} : |z|\leq 1\}$
I tried considering their modulus..
Please help me to understand this problem.
As this is just my third problem in "solving integrals using residue theory", I do not have much experience to tackle this problem...
Once i know what are the poles in $\{ z\in \mathbb{C} : |z|\leq 1\}$ then i know how to use residue theorem and conclude required result.
Thank You.
EDIT :
we need to consider $|\frac{-ai - i \sqrt{a^2-b^2}}{b}|=\big|-i(\frac{a+\sqrt{a^2-b^2}}{b})\big|=\big|\frac{a+\sqrt{a^2-b^2}}{b}\big| \leq |\frac{a}{b}|+|\sqrt{(\frac{a}{b})^2-1}|$
As $a>b$, we have $\frac{a}{b}>1$ SO, I do not have specific upper bound for $\frac{a}{b}$ .. I am confused..
even if i have $|\frac{-ai + i \sqrt{a^2-b^2}}{b}|$ then also i can not see..
AI: Consider your roots:
$$z_{\pm} = -i \left [ \frac{a}{b} \pm \sqrt{\left ( \frac{a}{b}\right)^2-1}\right]$$
Note that, when $a>b$, $a/b > 1$. Therefore $|z_+| > 1$. Note also that
$$|z_-| = \frac{a}{b} - \sqrt{\left ( \frac{a}{b}\right)^2-1} = \frac{1}{\frac{a}{b} + \sqrt{\left ( \frac{a}{b}\right)^2-1}} = \frac{1}{|z_+|} < 1$$
The residue at the pole $z=z_-$ is
$$\frac{2}{2 b z_-+i 2 a} = \frac{i}{a-\sqrt{a^2-b^2}-a} = \frac{-i}{\sqrt{a^2-b^2}}$$
The value of the integral is then $2 \pi/\sqrt{a^2-b^2}$, as was to be demonstrated. |
H: How can $|x|= -x$, when $x<0$?
Why is the following true?
$$
|x|=
\begin{cases}
x,&x\ge 0\\
-x,&x<0
\end{cases}
$$
Isn't the modulus of a number always positive? According to the above formula $|-4|=-4$ because $4<0$, which is incorrect.
Please explain this to me. Thank you.
AI: No, the formula says that $|-4|=-(-4)=4$. The $x$ in this case is $-4$, and $-(-4)=4$, not $-4$. When you get the value $-4$, you’re not applying the definition correctly: you’re taking the negative of $4$, not of $-4$, but the number inside the absolute value is not $4$. |
H: How to show that all roots of $(11+v)q^3-18q^2+9q-2$ have their absolute value less than 1.
The equation is $(11+v)q^3-18q^2+9q-2=0$, where $v>0$
I need to show that either absolute value of all the roots is not greater than one or there exists a root $q: |q|>1$.
Using Weierstrass theorem I showed that there is a real root $q\in[0;1]$.
Using derivative I showed that there are no real roots in $(-\infty,0)\cup(1;+\infty)$.
As Cardano's theorem states, there are three possibilities:
All three roots are distinct and real. If I have this situation, I'm done because all $q\in[0;1]$
Two distinct roots (three real roots of which one is counted twice). The same situation is above.
One real root and two complex roots.
In the third situation is there possibility that $|q|>1$ for some complex root of the equation in question?
AI: $\textbf{Hint:}$ First show that the real root lies in the interval $(\frac{2}{11+v},1)$. Then think about the product of the three roots. |
H: Isolated Points [Confusion Regarding Size of Neighborhood]
I'm learning (general?) topology and am having trouble understanding the definition for Isolated Points. The definition from Wikipedia says:
"A point $x$ of a topological space X is called an isolated point of a subset $S$ of $X$ if $x$ belongs to $S$ and there exists in $X$ a neighborhood of $x$ not containing other points of $S$"
Now, my main confusion is with "...a neighborhood of $x$ not containing other points of $S$". How exactly large is this neighborhood? For example, for the set $S$ = ${0} \cup$ { $1$, $\frac{1}{2}$, $\frac{1}{3}$, $\cdots$} why is $0$ not an isolated point?
AI: In the usual topology on $\Bbb R$, a set $N$ is a nbhd of $0$ if and only if there is an $\epsilon>0$ such that $(-\epsilon,\epsilon)\subseteq N$. No matter how small $\epsilon$ is, we can always find an integer $n>\frac1\epsilon$, so that $\frac1n<\epsilon$, and then we have
$$\frac1n\in(-\epsilon,\epsilon)\cap S\;.$$
Thus, every nbhd of $0$ contains other points of $S$.
This is not the case with the other points of $S$. Consider the point $\frac15$, for instance. $N\subseteq\Bbb R$ is a nbhd of $\frac15$ if and only there is an $\epsilon>0$ such that $$\left(\frac15-\epsilon,\frac15+\epsilon\right)\subseteq N\;.$$ Let $\epsilon=\frac15-\frac16=\frac1{30}$; since $\frac14-\frac15=\frac1{20}>\frac1{30}$, it’s clear that
$$\left(\frac15-\epsilon,\frac15+\epsilon\right)\cap S=\left(\frac15-\frac1{30},\frac15+\frac1{30}\right)\cap S=\left\{\frac15\right\}\;,$$
so $\frac15$ is an isolated point of $S$. A similar calculation can be carried out for any other element of $S\setminus\{0\}$ to show that $0$ is the only non-isolated point of $S$.
However, all of this depends on the fact that we’re talking about the usual topology on $\Bbb R$, the one whose open sets are the unions of open intervals. There are many other topologies on $\Bbb R$, and in some of them $0$ is an isolated point of the set $S$. The simplest of these is the discrete topology on $\Bbb R$, in which every subset of $\Bbb R$ is open. In this topology every point of $S$ is an isolated point of $S$: if $x\in S$, then no matter what $x$ and $S$ are, $\{x\}$ is a nbhd of $x$ that contains no other point of $S$. |
H: Did I solve this limit problem correctly? $\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}$
Need to solve this limit
$$\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}$$
When I put $x$ in this expression I have indeterminate form $1^{\infty }.$
So I choose the way which was described on this page (2nd method)
http://www.vitutor.com/calculus/limits/one_infinity.html
$$e^{\lim_{x \to 3 }{\tan \frac{\pi x}{6}(\frac{6-x}{3}-1)}}=e^{\lim_{x \to 3 }{\tan \frac{\pi x}{6}(\frac{6-x}{3}-\frac{3}{3})}} = e^{\lim_{x \to 3 }{\tan \frac{\pi x}{6}(\frac{3-x}{3})}}$$
and now I just put $x$ in this expression and I got $$e^{\infty \cdot 0} = e^0 =1$$
P.S. If you don't see my formulas, then
Right click on the math, go to Math Settings > Scale All Math... and enter your favorite number bigger than 100.
AI: With your 2nd method the limit is
$$ A := e^{\lim_{x \to 3 }{\tan \frac{\pi x}{6}(\frac{3-x}{3})}}
=e^{\lim_{x \to 3 }\frac{\frac{3-x}{3}}{\cot \frac{\pi x}{6}}}
$$
Now use l'Hôpital's rule
$$
A = e^{\lim_{x \to 3 }\frac{\frac{3-x}{3}}{\cot \frac{\pi x}{6}}}
= e^{\frac{-\frac{1}{3}}{-\frac{\pi}{6}}} = e^{\frac{2}{\pi}}
$$ |
H: Disjoints dense sets
How can we define three disjoints dense sets? Then I have to define four and five disjoints dense sets and finally - infinity disjoints dense sets. Could you give me some hints? Thanks in advance!
AI: HINT: Consider sets of the form $x+\Bbb Q=\{x+q:q\in\Bbb Q\}$ for some $x\in\Bbb R$. |
H: What is the order of $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ and is it cyclic?
Evidently, $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ has order $4$, but I think it's infinite.
The four cosets are listed as $(0,0) + \langle (2,2) \rangle$, $(0,1) + \langle (2,2) \rangle$, $(1,0)+ \langle (2,2) \rangle$ and $(1,1) + \langle (2,2) \rangle$. However, $(2,0)$ doesn't appear to be in any of these cosets. Maybe the answer I'm being told is wrong.
AI: You are right, $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ is infinite. You can embed $\mathbb{Z}$ via $k\mapsto (k,0)$ (and in other ways) into it. The quotient is not cyclic, because it contains elements of finite order, $(1,1)$ for example.
Probably it was meant to be $(\mathbb{Z} \oplus \mathbb{Z})/ (\langle 2\rangle\oplus \langle 2 \rangle)$ which indeed is a group of order $4$ (a Klein $4$-group). |
H: Is every Boolean algebra a separative partial order?
A partially ordered set $\langle P,\leq\rangle$ is separative iff it satisfies the following condition:
\[
\neg x\leq y\Rightarrow\exists z(z\leq x\wedge z\bot y)
\]
where:
\[
x\bot y\iff\neg\exists z(z\leq x\wedge z\leq y).
\]
If $\langle B,\leq\rangle$ is a complete Boolean lattice (every subset of $B$ has supremum), then $B^+=B\setminus\{0\}$ is separative.
My question is: what if $B$ is not complete? Is it separative or does not have to be separative?
AI: Recall that $\lnot x\leq y$ is the same as saying that $x\cdot y\neq x$. Let $z=x\cdot\bar y$ (where $\bar y$ is the complement of $y$).
Then $z\leq x$, and of course $z\leq\bar y$. But also $z\cdot y=x\cdot\bar y\cdot y=x\cdot 0=0$. Therefore $z\perp y$ as wanted. |
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