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H: Laplace's equation is solved when the functional $E[u] = \int_{\Omega}|\nabla u|^2 $ is minimized
My professor mentioned something like "Laplace's equation is solved when the functional $E[u] = \int_{\Omega}|\nabla u|^2 $ is minimized."
I've been trying to understand this statement. If I say that $E[u+tv]$ has a minimum at $t=0$, I get the condition
$$\int_{\partial \Omega}v \frac{\partial u}{\partial \nu} = \int_\Omega v \Delta u.$$
But what does this show? Can someone help me to flesh this out? Must $\Omega$ be bounded?
AI: No, $\Omega$ need not be bounded. Rather, the support of the test function $v$ is bounded (more precisely, a compact subset of $\Omega$). This ensures that $u+tv$ has the same boundary values as $u$. By the way, the quoted statement by the professor implicitly continues with "... among $W^{1,2}$ functions with given boundary values".
If $\int_\Omega v g=0$ for all test function $v$, then $g$ is zero a.e. Indeed, otherwise $g$ would have a Lebesgue point $p$ with nonzero value. We would pick a small neighborhood $N$ of $p$ and let $g$ be a smooth approximation of the characteristic function of $N$. This would lead to $\int vg\ne 0$, a contradiction. |
H: $\mathbb{R}$ - algebras in topological spaces
I´m reading an introduction to $\mathbb{R}$-algebras and in the text there is an observation that says:
If $X$ is a topological space, then the set of functions $f : X \to \mathbb{R}$ are a $\mathbb{R}$ - algebra and that the continuous functions are a subalgebra of the last space.
How can I prove this?.
AI: You just have to show that $Hom(X,\mathbb{R})$ satisfies all the conditions to be an $\mathbb{R}$-algebra. The addition of two maps $f,g$ is given by $(f+g)(x)=f(x)+g(x)$ for all $x\in X$, multiplication is given by $(f.g)(x)=f(x).g(x)$, and scalar multiplication is given by $(r.f)(x)=r.(f(x))$. From this it's easy to see that the zero function is given by $0(x)=0$ for all $x\in X$ and the unit function is given by $1(x)=1$ for all $x\in X$.
Your job is to show that this is all well-defined and satisfies the necessary axioms of an associative $\mathbb{R}$-algebra. |
H: Rationals with subspace topology from the reals
Suppose $\mathbb{Q}$ is endowed with the subspace topology of$\mathbb{R}$ Does it follow that $\mathbb{Q}$ is connected?
MY attempt: We can sue fact that $\mathbb{Q}$ is countable and so $\mathbb{Q} = \bigcup \{x\} $. And singletons are connected. Can we conclude that the union is connected?
AI: In the relative topology, $\mathbb{Q}$ is totally
disconnected. |
H: Counting elements of $y^2 - y = x^3$ in finite fields
The problem I have to solve is the following:
Let $p$ be a prime number with $p \equiv 2$ mod $3$. Let $E$ be the elliptic curve given by $y^2 - y = x^3$. Show that $\#E(\mathbb{F}_p) = p+1$ and $\#E(\mathbb{F}_{p^2}) = (p+1)^2$.
I have solved the first part in the following way: $\mathbb{F}_p^*$ has order $p-1$ which is not divisible by $3$, and so every element in $\mathbb{F}_p$ has a unique cube root. So the elements of $E$ in $\mathbb{F}_p$ are the $p$ elements $(\sqrt[3]{y^2-y},y)$ and the point at infinity, which gives a total of $p+1$ elements.
In the second part this trick no longer works, as cube roots are no longer unique. I don't know where to start to show that $\#E(\mathbb{F}_{p^2}) = (p+1)^2$. Any help is appreciated.
AI: Use Lemma 4.2 of the lecture notes. |
H: Number Theory Contest Problem
Given that $x, y$ are positive integers with $x(x + 1)\mid y(y + 1)$, but neither $x$ nor $x + 1$ divides either of $y$ or $y + 1$, and $x^2+ y^2$ as small as possible, find $x^2+ y^2$.
I have tried looking at the values, and it seems that neither $x$ or $x+1$ or the $y$'s are prime.
AI: Yes, we don't want $x$ or $x+1$ to be a prime power. The first candidate is $x=14$. Then $y=20$ works.
Added: Suppose we find consecutive composites integers $x$ and $x+1$ neither of which is a prime power. Let $x=ab$ and $y=cd$, where $a$ and $b$ are relaively prime, as are $c$ and $d$, and none is equal to $1$. Consider the system of congruences $y\equiv 0\pmod{ac}$, $y=\equiv -1\pmod{bd}$. By the Chinese Remainder Theorem, this has a solution. Note that $x$ does not divide $y$, for $b$ divides $y+1$, so is relatively prime to $y$. The other required "non-divisibilities" can be verified in a similar way. But $x(x+1)$ divides $y(y+1)$. |
H: Show $g(z+1) = zg(z)$
This is for homework, and I am in need of a hint. Given the product
$$ g(z) = \prod_{k=1}^{\infty} \frac{k}{z+k}\left( 1 + \frac{1}{k} \right)^z, $$
I am trying to show that $g(z+1) = zg(z)$. Here is what I have so far.
I tried to get a better sense of $g$, and wrote
$$ g(z) = \frac{1}{z+1}\left( 1 + \frac{1}{1} \right)^z\frac{2}{z+2}\left( 1 + \frac{1}{2} \right)^z\frac{3}{z+3}\left( 1 + \frac{1}{3} \right)^z\dotsb. $$
Then, after some manipulation, I found that
\begin{align*}
g(z+1) &= \frac{1}{(z+1)+1}\left( 1 + \frac{1}{1} \right)^{z+1}\frac{2}{(z+1)+2}\left( 1 + \frac{1}{2} \right)^{z+1}\frac{3}{(z+1)+3}\left( 1 + \frac{1}{3} \right)^{z+1}\dotsb \\
&= \frac{2}{z+2}\left( 1 + \frac{1}{1} \right)^z\frac{3}{z+3}\left( 1 + \frac{1}{2} \right)^z\frac{4}{z+4}\left( 1 + \frac{1}{3} \right)^z\dotsb.
\end{align*}
Then I tried to calculate $zg(z)$ and see if they match. I found
\begin{align*}
zg(z) &= \frac{z}{z+1}\left( 1 + \frac{1}{1} \right)^z\frac{2}{z+2}\left( 1 + \frac{1}{2} \right)^z\frac{3}{z+3}\left( 1 + \frac{1}{3} \right)^z\dotsb \\
&= \frac{z}{z+1}\frac{2}{z+2}\left( 1 + \frac{1}{1} \right)^z\frac{3}{z+3}\left( 1 + \frac{1}{2} \right)^z\frac{4}{z+4}\left( 1 + \frac{1}{3} \right)^z\dotsb,
\end{align*}
where I am doing a rearrangement in the second line. However, now it looks like $g(z+1) = zg(z)$ if and only if $\frac{z}{z+1} = 1$. What am I doing wrong here?
AI: Notice that $g(z+1)$ is the product for $g(z)$ "shifted" by one: that is,
$$ g(z+1) = \prod_{k=2}^\infty \frac{k}{z+k}\left(1+\frac{1}{k-1}\right)^z.$$
We can try to rewrite this in terms of the product $\prod\frac{1}{z+k}(1+\frac{1}{k})^z$:
$$\prod_{k=2}^\infty \frac{k}{z+k}\left(1+\frac{1}{k-1}\right)^z = \frac{z+1}{1}\prod_{k=1}^\infty \frac{k}{z+k}\left(1+\frac{1}{k}\right)^z$$
where I have multiplied and divided the expression by $\frac{1}{z+1}$ to obtain the right-hand side. Rewriting, $g(z+1) = (z+1)g(z)$. |
H: For what prime $p$ is $x^2=-1\pmod{p}$ solvable?
This is essentially the same as the following question:
When $F_p[x]/(x^2+1)$ is a field?
I don't know much about number theory. I came up with such question when I doing the following exercise:
When
$$
R_p=\left\{\left(\begin{matrix}
a&b\\
-b&a
\end{matrix}\right)\mid a,b\in F_p\right\}
$$
is a field?
A necessary condition for the statement in the title I found is that
$$
p=1\pmod{4},
$$ since
$$
a^2=-1
$$
implies $a^4=1$ and thus $o(a)=4$ in $F_p^\times$. And by the Lagrange's theorem,
$$
4\mid(p-1)
$$
and thus
$$
p=1\pmod{4}.
$$
How should I go on?
AI: Argue the converse: if $p\equiv1~(4)$ then there is an element of order $4$ (hence $x^2\equiv-1$ is solvable..)
Hint: finite groups of units of a field are cyclic. |
H: I was trying to compute $\sum_{j=0}^{m} 3^j {m \choose j}$, but don't know where to start
compute $\sum_{j=0}^{m} 3^j {m \choose j}$. Then, use the binomial theorem to verify the result.
AI: $$4^m=(3+1)^m=\sum_{j=0}^m{m\choose j}\cdot3^j\cdot1^{m-j}=\sum_{j=0}^m{m\choose j}\cdot3^j$$ or $$\sum_{j=0}^m{m\choose j}\cdot3^j=\sum_{j=0}^m{m\choose j}\cdot3^j\cdot1^{m-j}=(3+1)^m=4^m$$ |
H: Given $f(x) = x\log_{2}x$, how do I compute $f^{-1}(10)$?
Let $f(x) = x\log_2 x$. Compute $f^{-1}(10)$ to at least three decimal places of accuracy. Explain how you did this.
Note: for a function $f:A\rightarrow B$ for which there is exactly one point $a$ that maps to each point $b \in B$, $f^{-1}(y)$ means: the unique x such that $f(x) = y$.
AI: You wish to solve $x \lg x = 10$ for $x$ numerically. This is equivalent to finding a root of $g$, where $g$ is the function
$$g(x) = x \lg x - 10$$
Note that $g'(x)$ can be computed as
$$g'(x) = \lg{x} + x \left(\frac{1}{x \ln{2}}\right) = \lg{x} + \frac{1}{\ln{2}}$$
Using Newton's method, our iteration is given by
$$x_{n + 1} = x_n - \frac{g(x_n)}{g'(x_n)} = x_n - \frac{x \lg x - 10}{\lg{x} + \frac{1}{\ln{2}}}$$
If we begin with, say, $x_1 = 1$, then our iteration gives
\begin{align*}
x_1 &= 1 \\
x_2 &= 7.9315 \\
x_3 &= 4.8400 \\
x_4 &= 4.5680 \\
x_5 &= 4.5650 \\
x_6 &= 4.5650
\end{align*}
So, to an accuracy of three decimal places, we'd take $x = 4.565$.
Note that $g(x) \approx 2.7 \cdot 10^{-14}$, so this worked well. |
H: Finding the order of a finite group
Let $x\in\mathbb{Z}/42$, and suppose that x has order $n\in\mathbb{Z^+}$. Without listing all of the subgroups of $\mathbb{Z}42$, determine all of the possible values that $n$ could be.
I'm having a hard time understanding the concept of the order of a group and don't know where to begin.
AI: By Thanking Lagrange theorem we see that :
order of subgroup generated by $x$ should divide order of $\mathbb{Z}_{42}$..
please see that "order of an element is same as order of subgroup generated by that element"...
order of $\mathbb{Z}_{42}$ is $42$ So, divisors of $42$ are ______________
So, order of group generated by $x$ should be ______________
So, possible orders of $x$ should be ______________ |
H: Uniform Continuity Proof
$$f(x) = \frac{\sin x^3}{x+1}$$
If $f(x)$ is defined for $x \in [0,\infty)$, I can see that its derivative is bounded in the interval so it is just the matter of proving it. Im going for an $e-s$ proof, but I'm stumbling in terms of finding an appropriate $s.$ Also, would be nice if someone could give an explanation if possible what it means for a function to be uniformly continuous graphically - such that I can tell almost instantly if a function is uniform continuous (if possible).
Thanks.
AI: Take $\epsilon > 0$. Find $k > 1$ large enough that $1/x < \epsilon/4$ whenever $x \geq k$. Now, since $f(x)$ is continuous and $[0,k+1]$ is compact, then $f$ restricted to $[0,k+1]$ is uniformly continuous. Let $\delta_1 > 0$ be small enough that if $x, y \in [0,k+1]$ then $|x-y|< \delta_1$ implies $|f(x)-f(y)|<\epsilon$. Now, suppose that $x \in (k+1,\infty)$ and $|x-y|<1$. In particular this implies that $k<x$ and $k<y$. Then
$$
|f(x)-f(y)| = \left| \frac{(y+1) \sin(x^3) - (x+1)\sin(y^3)}{(x+1)(y+1)}\right| \\ \leq \left| \frac{y\sin(x+1) - x\sin(y+1)}{(x+1)(y+1)}\right| + \left|\frac{\sin(x+1)-\sin(y+1)}{(x+1)(y+1)} \right| \\ \leq \frac{2 \max(x,y)}{(x+1)(y+1)}+\frac{2}{(x+1)(y+1)} \\ < \frac{2}{\min(x,y)}+\frac{2}{(x+1)(y+1)} \\ < 2\frac{1}{k}+2\frac{1}{k} \\ < 4(\epsilon/4) = \epsilon.
$$
The very last inequality came with our choice of $k$.
So, taking $\delta = \min(\delta_1, 1)$ will do the trick!
Graphically, you might think of uniform continuity as saying that $f$ can't move vertically too steeply at any points (otherwise any small changes in the domain [controlled by $\delta$] can still result in large changes in the range [controlled by $\epsilon$]). For example, $f(x) = 1/x$ is not uniformly continuous on $(0,1)$ since near $0$ it grows nearly vertically. Similarly, for any $n > 1$, $f(x) = \pm x^n$ is not uniformly continuous on $(0, \infty)$ because as $x$ approaches $\infty$ the function $f$ moves nearly vertically (up if $+x^n$ down if $-x^n$). |
H: can not find the proof that logarithms are the inverse of exponentials
I have been taught by the most powerful magic of mathematics - Hand waving - that the logarithms are inverses of the exponential. I have seen the graphs where each one is graph, showing they are reflective over the line $y=x$. I do not doubt they are inverses of each other; however, when I learned about the Fundamental Theorem of Calculates, I saw a proof. When I learned about the Mean Value Theorem, I saw a proof. When is learned about Riemann sums, I saw a proof. When I learned about derivatives, I saw a proof.
Where is the proof that logarithms are the inverse of exponential. I am just given a relationship $\log_b (a) = b^a$. Where's the proof?
I know there are many wise PHD here that can unleash the power of $\epsilon$ and $\delta$, crush conjectures with the squeeze theorem, follow the pigeon hole principle, make portals between fields, and even transform me into a coffee cup.
What is the proof of $\log_b(a) = b^a$?
AI: This proof might interest you:
Suppose we define $e^x$ to be the function $f(x)$ satisfying $f'(x) = f(x)$ with $f(0) = 1$. With implicit differentiation, we note that
$$
f(f^{-1}(x)) = x \implies f'(f^{-1}(x))(f^{-1})'(x) = 1
$$
Rearranging yields
$$
(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} = \frac{1}{f(f^{-1}(x))} = \frac 1x
$$
Now, define $f^{-1}$ to be this inverse of $f$. We deduce from the above that $f^{-1}$ has derivative $\frac 1x$. We note that since $f(0) = 1$, $f^{-1}(1) = 0$. $f^{-1}$, that is, $\ln x$, is the unique function satisfying the conditions $(f^{-1})'(x) = 1/x$ and $f^{-1}(0) = 1$. Thus, $f^{-1}$ must have all the properties of $\ln x$
We could go through the same process in reverse. The point is, once you know that $e^x$ has one of its defining properties, you may conclude that its inverse has all of the defining properties of $\ln x$. Similarly, once you know that $\ln x$ has one of its defining properties, you may conclude that its inverse has all of the defining properties of $e^x$.
I hope you find this satisfactory. |
H: A problem about class equation
Let $k$ be a finite field, where $|k|=q$ and $\operatorname{char}k\neq2$, and let
$$D=\{A\in\operatorname{SL}(2,k)\mid A \text{ is diagonalizable}\}.$$
Prove that $$|D|=2+\tfrac{1}{2}\cdot(q+1)\cdot q\cdot(q-3).$$
Can anyone help me? Thanks!
AI: We know that $|GL(2,k)| = (q^2-1)(q^2-q)$, and $|k^{\ast}| = (q-1)$, so
$$
|SL(2,k)| = (q+1)(q^2-q)
$$
Now, consider
$$
\hat{D} := \left\lbrace A_{\lambda} := \begin{pmatrix}
\lambda & 0 \\
0 & \lambda^{-1}
\end{pmatrix} : \lambda \in k^{\ast} \right\rbrace
$$
and consider the conjugacy class $C(A_{\lambda})$ of $A_{\lambda}$.
a) If $\lambda =1$ or $\lambda = -1$ (which are different since $2\neq 0$), $A_{\lambda}$ is in the center of $SL(2,k)$, and so
$$
|C(A_1)| = 1 = |C(A_{-1})|
$$
b) For any other $\lambda \in k^{\ast}$, the centralizer of $A_{\lambda}$ is precisely $\hat{D}$, which has cardinality $|k^{\ast}| = (q-1)$. By the Orbit-Stabilizer theorem,
$$
|C(A_{\lambda})| = \frac{|SL(2,k)|}{|\hat{D}|} = q(q+1)
$$
Each such $\lambda$ occurs in pairs $\{\lambda, \lambda^{-1}\}$, and there are
$$
\frac{1}{2}(q-3)
$$
such pairs (everything in $k$ excluding $\{0,1,-1\}$). Hence,
$$
|D| = 2 + \frac{1}{2}(q-3)q(q+1)
$$ |
H: Determinant of complex block matrix
Let $A$ be an $n\times n$ invertible matrix. Let $a \in \Bbb C$, let $\alpha$ be a row $n$-tuple of complex numbers and let
$\beta$ be a column $n$-tuple of complex numbers. Show that
$$(\det(A))^{-1}\, \det\left(\begin{bmatrix}a & \alpha\\
\beta & A
\end{bmatrix}\right)=\det\left(a-\alpha A^{-1}\beta\right).$$
Can anyone show me how to prove this?
AI: Hint:
$$
\begin{align}
(\det A)^{-1}
\det
\begin{bmatrix}
a & \alpha\\
\beta & A
\end{bmatrix}
&=
\det\left(
\begin{bmatrix}
1 & \left(\vec 0\right)^T\\
\vec 0 & A^{-1}
\end{bmatrix}
\right)\det\left(
\begin{bmatrix}
a & \alpha\\
\beta & A
\end{bmatrix}
\right)
\\ &=
\det\left(
\begin{bmatrix}
1 & \left(\vec 0\right)^T\\
\vec 0 & A^{-1}
\end{bmatrix}
\begin{bmatrix}
a & \alpha\\
\beta & A
\end{bmatrix}
\right)
\\
&= \det
\begin{bmatrix}
a & \alpha \\
A^{-1}\beta & I
\end{bmatrix}
\end{align}
$$
Where $\vec 0$ is the column $n$-tuple of zeros. |
H: What is the order of the sum of log x?
Let
$$f(n)=\sum_{x=1}^n\log(x)$$
What is $O(f(n))$?
I know how to deal with sums of powers of $x$. But how to solve for a sum of logs?
AI: Using Stirling's formula we have, for $n$ sufficiently large
$$
f(n)=\sum_{k=1}^n\log k=\log(n!)\simeq\log(\sqrt{2\pi}e^{-n}n^{n+1/2}).
$$
Hence
$$
O(f(n))=\log(\sqrt{2\pi}e^{-n}n^{n+1/2}).
$$ |
H: Expressing logarithms as ratios of natural logarithms
$$\frac{\log_2 x}{\log_3 x}=\frac{\ln x}{\ln 2} \div \frac{\ln x}{\ln3}$$
Why can logarithms be written as ratios of natural logarithms?
Can you explain it abstractly, please?
Example of an abstract explanation: the logarithm function is an isomorphism from the group of positive real numbers under multiplication to the group of real numbers under addition, represented as a function.
The teacher doesn't go into abstractions in class, so I would really like to understand it in an abstract sense. Thank you.
AI: Consider $y=\log_b x$. Then, by definition, $b^y=x$ and so $y \ln b = \ln x$. Thus,
$$\log_b x=\frac{\ln x}{\ln b}$$
A more sophisticated argument is the following. Consider a continuous function $L:\mathbb R^+ \to \mathbb R$ such that $L(xy)=L(x)+L(y)$. Then $F(x)=L(e^x)$ satisfies $F(x+y)=F(x)+F(y)$ and is thus a continuous automorphism of the additive group $\mathbb R$. It is easy to see that $F$ must be a scaling: $F(x)=ax$ for some $a$. Of course, $a=F(1)=L(e)$. When $L(x)=\log_b x$, we have $a=\log_b e=\dfrac1{\ln b}$. Since $L(x)=F(\ln x)$, we have, as before:
$$\log_b x=\frac{\ln x}{\ln b}$$ |
H: Prove that if a series converges to $a$, then that series to the power $k$ converges to $a^k$
Let $a_n$ be a sequence in the real numbers. Prove that:
$a_n\rightarrow a \implies (a_n)^k \rightarrow a^k$ $\forall k \in N$
I think I need to do this by induction. The base case is simple. When $k=1$:
$(a_n)^k = (a_n)^1=a_n \rightarrow a = a^1=a^k$.
Now assume $(a_n)^i \rightarrow a^i$. Then there exists $N>0$ such that if $n>N$, then $\mid (a_n)^i- a^i \mid < \epsilon$. I need to show that $(a_n)^{i+1} \rightarrow a^{i+1}$. So I think I need to manipulate $n$ and $N$, right?
Any ideas on how to approach the problem?
Thanks,
K
AI: When using induction, you are probably encouraged to use that if $a_n\to a$ and $b_n\to b$ then $a_nb_n\to ab$. Thus, assuming $a_n^{k-1}\to a^{k-1}$, you get $$\lim\limits_{n\to\infty} a_n^k=\lim\limits_{n\to\infty}a_n^{k-1}\lim\limits_{n\to\infty}a_n=a^{k-1}a=a^k$$ using the inductive hypothesis in the second equality. |
H: The limit of $(s_j)$ when said $(s_j)$ is finite
Suppose that we have a sequence $(s_j)$ that is finite (where $j$ is an element of the natural numbers); it has $j$ terms. Can we say that $(s_j)$ has a limit? Also what about convergence? Can we say that $(s_j)$ converges to some number? I would think no on both cases. What about subsequences? Can we say that they have limits or that they converge if they are finite?
AI: The answer is yes, but doing so is not really useful. A finite set of integers trivially forms a "directed set", meaning that it can be used as the domain of a "net", a sort of generalized sequence. The reason this is not useful is that every such net converges, and converges to its last value. Depending on what space you're working on, it may or may not converge to other values, but it does not, under any circumstances I can think of, have any useful or interesting properties. |
H: A form of the Gaussian integral
Consider the function $f(x,y)=ye^{-(1+x^{2})y^{2}}$ if $x\geq0$ and $y\geq0$ and $0$ otherwise. Integrate this function over $\mathbb{R}\times\mathbb{R}$ to show that $\int_{0}^{\infty}e^{-x^{2}}\,dx=\sqrt{\pi}/2$.
Here is what I have done:
\begin{align*}
\int_{0}^{\infty}\int_{0}^{\infty}f(x,y)\,dy\,dx&=\int_{0}^{\infty}\int_{0}^{\infty}ye^{-(1+x^{2})y^{2}}\,dy\,dx\\
&=\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{2}e^{-(1+x^{2})u}\,du\,dx\qquad(\text{set}\ u=y^{2},\ du=2y\,dy)\\
&=\int_{0}^{\infty}\left[-\frac{1}{2(1+x^{2})}e^{-(1+x^{2})u}\right]_{0}^{\infty}\,dy\\
&=\int_{0}^{\infty}\frac{1}{2(1+x^{2})}\,dx\\
&=\frac{1}{2}\int_{0}^{\infty}\frac{1}{1+x^{2}}\,dx\\
&=\frac{1}{2}[\arctan x]_{0}^{\infty}\\
&=\frac{\pi}{4}
\end{align*}
Now clearly the square root of $\pi/4$ will give the solution, but where does the $e^{-x^{2}}$ come in? Is there a way to break up the original integrand? Thanks.
AI: Write
$$
\int_0^{\infty} \int_0^{\infty} f(x,y)dydx = \int_0^{\infty}\int_0^{\infty} ye^{-y^2}e^{-x^2y^2}dydx
$$
By Fubini-Tonelli,
$$
= \int_0^{\infty} \int_0^{\infty} e^{-x^2y^2}ye^{-y^2}dxdy
$$
and substitute $u = xy$ to get
$$
\int_0^{\infty}\int_0^{\infty}e^{-u^2}e^{-y^2}dudy = \left( \int_0^{\infty} e^{-z^2}\right)^2
$$ |
H: Calculate time needed to solve problem
I have this question in an assignment and I was wondering if I could get help verifying whether my approach to this question is correct... The question is as follow:
Suppose that an algorithm uses 5n^2 + 3^n bit operations to solve a
problem of size n. Suppose that your machine can perform one bit
operation in 10^-9 seconds. How long does it take your algorithm to
solve a problem of the sizes given below. Note, if your algorithm
takes more than 60 seconds, answer in minutes. For more than 60
minutes, answer in hours. For more than 24 hours, answer in days. For
more than 365 days, answer in years. For more than 100 years, answer
in centuries!
Let's assume n is 10...
Would I,
1)Solve for 5(10)^2 + 3^(10)
2) Divide the number of bit operations by 10^-9 to find the answer represented in seconds
3)Divide by minutes or hours or days and so on as needed to find proper representation?
Thanks in advance for any input
AI: You would
1) calculate $5\cdot 10^2+3^{10}$-you solve equations but calculate expressions
2) multiply by $10^{-9}\frac {\text {seconds}}{\text { bit operation}}$ to get seconds-dividing gives an error of $10^{18}$ If you write out the units you will get it right
3) divide by the appropriate conversion factor-you don't divide by minutes. You are actually multiplying by $1$ in the form $\frac {1 \text { minute}}{60 \text { seconds}}$ or whichever is appropriate
Your basic approach is correct, but the writeup is not |
H: System of two ODEs with reverse sign
Consider the ordinary differential equations $$\dfrac{d}{dt}x_1(t)=x_2(t)$$ $$\dfrac{d}{dt}x_2(t)=-x_1(t)$$
for $t\in \mathbb{R}$. What are the solutions?
We have $\dfrac{d^2}{dt^2}x_1(t)=\dfrac{d}{dt} x_2(t)=-x_1(t)$ and also $\dfrac{d^2}{dt^2}x_2(t)=-\dfrac{d}{dt} x_1(t)=-x_2(t)$. So $x_1(t)$ and $x_2(t)$ satisfy the equation $\dfrac{d^2}{dt^2}u(t)+u(t)=0$. They could be cosine and sine. What else could they be?
AI: Hint:
What if you wrote the system as a $2x2$ matrix and found the eigenvalues?
The system would be:
$$x' = Ax = \begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix}$$
Spoiler - Do Not Peek
$x_1(t) = c_1 \cos t + c_2 \sin t$, $x_2(t) = -c_1 \sin t + c_2 \cos t$
Of course, we could have also solved the second order system you wrote:
$$u'' + u = 0 \rightarrow m^2 + 1 = 0 \rightarrow m_{1,2} = \pm ~i$$
This gives a solutions with sine and cosine terms.
We could also plot a phase portrait of the system as: |
H: Why is conjugation by an odd permutation in $S_n$ not an inner automorphism on $A_n$?
I was reading about the outer automorphism group on wikipedia, and it mentions that conjugation by an odd permutation is an outer automorphism on the alternating group $A_n$. This suggests the automorphism defined on $A_n$ by
$$
\varphi: A_n\to A_n:\varphi(\tau)=\sigma\tau\sigma^{-1}
$$
where $\sigma\in S_n$ is odd it not an inner automorphism of $A_n$. Is there a way to see explicitly that $\varphi$ isn't just $\tau\mapsto\rho\tau\rho^{-1}$ for $\rho\in A_n$ in disguise? To avoid trivial cases I guess we can assume $n>2$. Thanks.
AI: Note that if $\sigma$ is an odd permutation then $(12)\sigma$ is an even permutation, thus $\tau\mapsto (12)\sigma\tau\sigma^{-1}(12)$ is an inner automorphism, and so conjugation by $\sigma$ is an inner automorphism iff conjugation by $(12)$ is, as conjugation by $\sigma$ is the composition of conjugation by $(12)$ with conjugation by $(12)\sigma$ and inner automorphisms form a subgroup.
Suppose conjugation by $(12)$ is equivalent to conjugation by some $\pi\in S_n$. Let
$$m=\begin{cases} n&\text{if $n$ is odd}\\
n-1&\text{if $n$ is even}\end{cases}$$
and note that $(12)(123\cdots m)(12)=(213\cdots m)$, thus
$\pi(123\cdots m)\pi^{-1}=(213\cdots m)$. Since $m$-cycles can be written uniquely up to cyclic permutation, we have $\pi=\pi'(12)$ for some $m$-cycle $\pi'$, and thus $\pi\notin A_n$ so conjugation by $(12)$ is outer. |
H: How to graph absolute value equations that are not functions
Graph $|x + 1| + |y - 2| = 1 $
The graph should be a parallelogram, so it is not a function. How do you graph this?
AI: If we isolate $|y-2|$, then we get
$$|x+1|+|y-2|=1 \Rightarrow |y-2|=1-|x+1|.$$
Now we can apply the definition of absolute value,
$$
|y-2|=
\begin{cases}
\,\,\,\,(y-2) \quad \operatorname{if} \,\,(y-2) \geq 0 \\
-(y-2) \quad\operatorname{if} \,\,(y-2) <0.
\end{cases}
$$
So this gives rise to two equations for you to solve for $y$, namely,
$$
\begin{align*}
(y-2)&=1-|x+1| \\
-(y-2)&=1-|x+1|.
\end{align*}
$$
All that is left for you to do is solve both of these absolute value equations for $y$, and then plot them.
There is a domain restriction due to the absolute value definition. For the first equation, since $(y-2) \geq 0$,
$$1-|x+1|\geq 0 \Rightarrow |x+1|\leq 1 \Rightarrow -2 \leq x \leq 0.$$
For the second equation, since $(y-2)<0$, $-(y-2)>0$, thus
$$1-|x+1|>0 \Rightarrow |x+1|<1 \Rightarrow -2<x<0.$$
It is this domain restriction that makes your parallelogram pop out of the picture.
I suspect you will be ok at this point, but let us know if your path is still unclear. |
H: Allocation / Weighted Average question
I'm hoping someone can help with my problem. Forgive me as I don't even know what title to give this. I'm trying to come up with an allocation method for transportation expenses for multiple stops along a truck route based on a combination of mileage and cargo weight. I'm not sure if this is possible based on the known variables/constants.
I'll lay out an example:
3 stops all forming a straight line from a shipping location, the farthest point being 50 miles from the shipping location.
Stop 1: 10 miles from starting point, 10,000 lbs
Stop 2: 40 miles from starting point (30 miles from stop 1), 30,000 lbs
Stop 3: 50 miles from starting point (10 miles from stop 2), 20,000 lbs
Rules and/or known items, if you will:
Total miles to allocate = 50 (maximum value/farthest point of all of the stops)
Total weight = 60,000 lbs of cargo.
Each stop's allocation amount cannot be greater than the mileage for that stop (i.e. stop # 2 cannot have more than 40 miles allocated.)
You can't look at the mileage between the stops because that's not a fair allocation method. Stop 3 shouldn't only be charged for 10 miles when it made up of 33% of the weight.
Something like the following is easy to figure out based on simple logic:
Stop 1: 25 miles 10,000 lbs
Stop 2: 25 miles 20,000 lbs (drop off point is the same location as stop 1, the only variable is weight)
Solution:
Stop 1: 25 * (10,000/30,000 total weight) = 8.33 miles
Stop 2: 25 * (20,000/30,000 total weight) = 16.67 miles
Any ideas are greatly appreciated. I'm assuming I would at least have to determine how to weight the mileage vs. the cargo weight at the very least. Let's say that cargo weight is 2x as important as the mileage.
AI: There are many formulas that will give an answer. You need to specify what you are looking for. One simpleminded approach is to bill each stop for its share of the load for each segment that it is part of. So stop $1$ had $\frac 16$ of the load for $10$ miles, so pays $\frac {10}6$. Stop $2$ had $\frac 12$ of the load for $10$ miles and $\frac 35$ of the load for $30$ miles, so pays $\frac {10}2+ \frac {3 \cdot 30}5=23$ Stop $3$ has $\frac 13$ of $10$ miles, $\frac 25$ of $30$ and all of $10$, so pays $\frac {10}3+\frac {2\cdot 30}5+10=25\frac 13$ |
H: Let $p,q$ be 2 complex numbers with $|p|<|q|$.
I am stuck on the following problem that says:
Let $p,q$ be 2 complex numbers with $|p|<|q|$. Let $$f(z)=\sum\{3p^n-5q^n\}z^n$$ Then the radius of convergence of $f(z)$ is :
$|q|$
$|p|$
At least $\frac{1}{|q|}$
At most $\frac{1}{|q|}$
My Attempt: $f(z)=\sum(3p^n-5q^n)z^n=\sum\{3p^n\}z^n-\sum\{5q^n\}z^n=3\sum(pz)^n-5\sum(qz)^n$. Now for convergence,we must have $|pz|<1 \implies |z|<\frac{1}{|p|}$ and $|qz|<1 \implies |z|<\frac{1}{|q|}$. Also we are given that $|p|<|q| \implies \frac{1}{|q|} <\frac{1}{|p|}$.
Now,I am bit confused. Can someone help? Thanks in advance for your time.
AI: You are basically done: just think about the following.
If you want the series for $f$ to converge, then you'd better get the two series you got by splitting $f$ to both converge. If $|z|<1/|p|$, then the $p^nz^n$ series will converge, but is this enough to get the $q^nz^n$ series to converge? If $|z|<1/|q|$, will the $p^nz^n$ series converge? |
H: Median for Continuous Probability Distribution
Consider a continuous random variable X with probability density function given by:
$f(x)=4x(1-x^2)$ for $0 \le x \le 1$
Find the median.
So to calculate the median, I calculated the CDF and then set that equal to 0.5 and solve for x:
$F(x)=2x^2-x^4$
$0.5=2x^2-x^4\tag{1}$
So now we just have to solve equation (1) for x. We can do this by quadratic formula by setting $y=x^2$.
$y^2-2y+0.5=0\tag{2}$
$\implies y = \cfrac{2 \pm \sqrt{2}}{2}$
$\implies y= 1.71, y=0.293$
The answer in my book is $x_{0.5}=\cfrac{2 - \sqrt{2}}{2}$. Don't we have to solve for x by taking the sqrt of y to get the final answer? In other words, shouldn't the answer be $\sqrt{.293}$? We eliminate $\sqrt{1.71}$ because it's not in the domain...
Thanks in advance.
AI: I believe you are 100% correct. |
H: Solution to Hamilton-Jacobi differential equations
Let $H(x,y)$ be a $C^2$ function on $\mathbb{R}^2$ and let $(x(t),y(t))$ be a solution of the Hamilton-Jacobi equations $$\frac{dx}{dt}=\frac{\partial}{\partial y}H(x(t),y(t))$$$$\frac{dy}{dt}=-\frac{\partial}{\partial x}H(x(t),y(t))$$
Show that the function $H$ is constant along the integral curves of the equations above.
I'm not so sure what "$H$ constant along the integral curves" means. Does it mean $H(x(t),y(t))$ is a constant value for any $t$? If so, how to proceed in proving that?
AI: By integral curve they mean a solution to the system.
Suppose $t \mapsto (x(t),y(t))$ is a solution, then let $\phi(t) = H(x(t),y(t))$.
Then
\begin{eqnarray}
\dot{\phi}(t) &=& \frac{\partial H(x(t),y(t))}{\partial x} \dot{x}(t) + \frac{\partial H(x(t),y(t))}{\partial y} \dot{y}(t) \\
&=& (-\dot{y}(t)) \dot{x}(t) + (\dot{x}(t)) \dot{y}(t) \\
&=& 0
\end{eqnarray}
It follows that $\phi$ is constant. |
H: Well definedness of Lebesgue inner measure
This is for homework: if $A,A'$ are two elementary sets containing $E$, bounded set in $\mathbf{R}^d$, then $m(A)-m^*(A \backslash E)$ is equal to $m(A')-m^*(A \backslash E)$
So far my goal has been to show that they're both equal to the expression for $A \cap A'$ which is another elementary set.
I've done this work: $m(A)-m^*(A\backslash E) = m(A \cap A') + m(A \backslash A') - m^*((A \backslash A') \cup (A \cap A') \backslash E)$
Countable additivity of disjoint sets would be very nice here, but I can't think of why countable additivity would apply in this case (one set is elementary, the other not). So I don't know if I'm on the right track or not. I could try a regularity argument but I can't wrap my head around the strategy there.
AI: One idea is to show that the inner measure is in fact
$$\sup \left\{ \lambda(K) : K \subseteq E, \text{ Measurable } \right\} $$
I think an elementary set $A$ is bounded by definiton (perhaps we should check this, if not you may be able to work around, either way here is the case for if $A$ is bounded.)
$$ \lambda(A) - \lambda^{\star}(A\setminus E) = \lambda(A) - \inf\left\{ \lambda(B) : A\supseteq B \supset(A \setminus E), \text{ Measurable } \right\}$$
Now $B$ is measurable, so $\lambda(B) + \lambda( A \setminus B) = \lambda(A)$ hence since $A$ is bounded, we can rewrite $\lambda(B) = \lambda(A) - \lambda(A\setminus B)$, hence we have
$$\lambda(A) - \lambda^{\star}(A\setminus E) = \lambda(A) - \inf\left\{ \lambda(A) - \lambda(C) : C\subseteq E \right\}$$
But this is nothing else but what we want.
So we see the definition does not depend on $A$. |
H: Epsilon-Delta Continuity proof
Let $f_1, f_2$ be two functions from $\mathbb R\to\mathbb R$. Suppose that $f_1$ and $f_2$ are both continuous at $x_o∈ \mathbb R$. Let $g(x)=\min(f_1(x),f_2(x))$. Prove $g(x)$ is continuous at $x_o$. Has to be proven using the epsilon-delta definition of continuity: $∀ ε>0 ∃ δ>0$ such that if $x∈I$ and $|x−x_o|<δ$ then $|f(x)−f(x_o)|<ε$.
AI: Not sure who down voted because I think its a fair question.
Since both $f_1$ and $f_2$ are continuous at $x_0$, we have the following relations.
$\forall\epsilon_1>0$, there exists a $\delta_1$ such that $|x-x_0|<\delta_1\implies |f_{1}(x)- f_{1}(x_{0})|<\epsilon_1$ (Eq1).
Similarly we have that for $\forall\epsilon_2>0$, there exists a $\delta_2$ such that $|x-x_0|<\delta_2\implies |f_{2}(x)-f_{2}(x_0)|<\epsilon_2$ (Eq2).
Let $g(x)=min(f_{1}(x),f_{2}(x))$. Without loss of generality suppose that $f_{1}(x_0)<f_{2}(x_{0})$.
Then at $x_0$ we have that $g(x_0) = f_{1}(x_{0})$. To show that $g(x)$ is continuous we need to show that $\forall\epsilon>0$ there exists $\delta$ such that $|x-x_0|<\delta\implies |g(x)-g(x_0)|<\epsilon$.
Now as we have assumed $f_{1}(x_0)<f_2(x_0)$ we need $|g(x)-f_{1}(x_0)|<\epsilon|$.
If $g(x)$ is always equal to $f_1(x)$ (that is $f_1<f_2$ for all $x$, then we are done, since $f_1$ is continuous). If not, note that
$|f_2(x)-f_1(x_0)|< |f_2(x)-f_2(x_0)|<\epsilon_2$ by (Eq2). This tells us how to choose a $\delta$. That is for each $\epsilon >0$ choose $\delta = min\{\epsilon_1,\epsilon_2\}$. |
H: Easy question concerning notation(Abstract Algebra)
I have a very easy question concerning some notation I have been coming across in Abstract Algebra (Dummit & Foote)
Context:
The relation between actions and homomorphisms may be reversed. Namely, given any nonempty set A and any homomorphism $\phi$ of the group $G$ into $S_A$ we obtain an action of G on A by defining $g \cdot a = \phi(g)(a)$
What, in simplest terms, is the meaning of $g \cdot a = \phi(g)(a)$?
I understand the meaning of $g \cdot a$ but am unclear about $\phi(g)(a)$
AI: Since $\phi : G \to S_A$, $\phi(g)$ is a permutation on the set $A$. That is, $\phi(g)$ is a bijection from $A$ to $A$, and so $\phi(g) a$ is the element that the permutation $\phi(g)$ sends $a$ to. |
H: Residues of Complex Functions
I need to find the residues of $f$ at the isolated singular points, namely $z=1,z=0$.
Where $f(z)=\dfrac{2z+1}{z(z+1)}$.
I already have that the residue at $z=0$ is $1$, and I know I need to do some slight of hand to get res at $z=1$.
I tried, $f(z)=\dfrac{2z+1}{z(z+1)} = (\dfrac{2z+1}{z})(\dfrac{1}{2+(z-1)}) = (2+d\dfrac{1}{z})*\sum_{n=0}^{\infty}(-1)^n\dfrac{(z-1)^n}{2^{n+1}}$, but we never get a $\dfrac{1}{z-1}$ term, so can we conclude that the coefficient of $\dfrac{1}{z-1} = res =0?$
I would need this to conclude that, $\int_C f(z) dz = 2i\pi(1+0)=2i\pi$. Where C is the circle of radius 2.
Yes, I know I can use partial fractions to do this, but our professor wants us to use the Residue Theorem to evaluate the integral.
AI: Hint:
$$\frac{2z+1}{z(z+1)}=\frac1z+\frac1{z+1}=-\frac1{1-(z+1)}+\frac1{z+1}=$$
$$-1-(z+1)-(z+1)^2-\ldots+\frac1{z+1}\;\;\implies\;\ldots$$ |
H: Prove that if an integral is 0, the function is 0 across that interval (for $f(x) \geq 0$)
Assume $f:[a,b] \Rightarrow \mathbb{R}$ is continuous and $f(x) \geq 0$ for all $x\in[a,b]$. Prove that if $\int_a^b f dx = 0$, then $f(x) = 0$ for all x $\in [a,b]$.
My attempt at a proof a little obvious, but at the same time, pretty wordy. I think it can be done more easily/briefly; please recommend ways I could do that.
"Theorem 5.10" which I employ here states that for $a < c <b$, the integral from a to b is equal to the sum of the integrals from $a$ to $c$ and $a$ to $b$.
Suppose $f:[a,b]$ $\Rightarrow \mathbb{R}$ is continuous and $f(x) \geq 0$ for all $x\in[a,b]$. By theorem 5.10, interval $[a,b]$ can be cut into some large finite number of subintervals, and the sum of their integrals equal the integral from $a$ to $b$. No integral of any subinterval of $[a,b]$ can be less than 0, by definition. It follows that none can be greater than 0, since that would result in a positive value for the integral of $[a,b]$. Therefore, $f(x) = 0$ for all $[a,b]$.
AI: Hint Suppose $f(x)\ne 0\forall x$ so $\exists c$ such that $f(c)>0$ Then by sign preserving property of continuous function there exist $\delta>0$ such that $f(x)>0\forall x\in (c-\delta,c+\delta)$ |
H: Lie algebra homomorphism
I'm sure I'm missing something really obvious here. This seems too stupid.
On page 47 of Erdmann & Wildon's Introduction to Lie Algebras, we have the following set up. Let $L$ be a Lie subalgebra of $\mathfrak{gl}(V)$ of dimension $\geq 1$, and let $A \leq L$ be a maximal Lie subalgebra and let $\overline{L} = L/A$. Then they define $\varphi: A \to \mathfrak{gl}(\overline{L})$ by $\varphi(a)(x+A) := [a,x] + A$.
My question is: why do we not have $\varphi = 0$? Or do we have this? I mean, if $a \in A, x+A \in \overline{L}$, then $\varphi(a)(x+A) = [a,x] + A = [a+A,x+A]=[0,x+A]=0$, right?
Maybe I'm not understanding how the braket is defined on the quotient, or something.
AI: Let's have an example. Let $L = span_{\mathbb F} \{x, y\}$, and $[x, y] = y$ (Note that every two dimensional Lie algebra are of this form). Let $A = span\{x\}$. Then the map
$$\phi: A \to gl(\bar L)$$
is $\phi(cx)(y + A) = [cx, y] + A = cy+ A$. This is not a zero map. |
H: non prime generated cyclic numbers?
I recently watched a numberphile video on youtube talking about cyclic numbers. I was wondering if there was a number $1/a$, where $a$ wasn't prime, but $1/a$ turned out to be a cyclic number.
Or must $a$ always be prime?
AI: Indeed, if you want a number $\frac{1}{x}$, where $x$ is an integer, to be a cyclic number you need $x$ to be prime. In fact, it can't be just any prime number. The primes which generate these cyclic numbers actually have their own special names and are called full reptend primes. These cyclic numbers are heavily related to what are called repunits (where you have an integer consisting of only one digit which repeats) which are related to special primes called permutation primes (or absolute primes) as well as cyclic primes.
While it is necessary that the number in the denominator be primes, it is still unknown whether there exist an infinite amount of these cyclic numbers. But the question is related to another deeper problem in Mathematics called Artin's constant. |
H: equality for a measure $\mu(F\backslash E)= \mu(F)-\mu(E)$
Studying for Real Analysis I encountered this exercise and I am a bit confused about it.
Let $\mu$ be a measure on $(X,M)$, where $M$ is a $\sigma$-algebra on $X$. Show that if $E \subseteq F$ and $\mu(E)< \infty$ then $\mu(F\backslash E)= \mu(F)-\mu(E)$ (*). A statement like this means that if one side is finite then so is he other and they are equal.
() is pretty clear using countable additivity on $F\backslash E$ and $E$ since they are disjoint and their union gives $F$. I do not understand though why do we require that $\mu(E)< \infty$ ! Is it only not to have cases like $\infty - \infty$ in (*)?
AI: You are right: $\infty - \infty$ is meaningless and you want to avoid the situation when something meaningful, i.e. $\mu(F\setminus E)$, equals something meaningless.
You encounter the same situation when you define the integral of measurable function as $\int_X f\ d\mu := \int_X f^+\ d\mu - \int_X f^-\ d\mu$, provided at least one of the two is finite
Also, when you define a signed measure you want to ask that $\mu \colon \Sigma \to [-\infty,\infty)$ or $(-\infty,\infty]$. |
H: The closure of a product is the product of closures?
If $\{X_j:j\in J\}$ is a family of topological spaces and $A_j\subseteq X_j$, is it true that $\displaystyle\overline{\Pi_{j\in J}A_j}=\displaystyle{\Pi_{j\in J}\overline{A_j}}$? Is there an easy way to prove this?
Of course, we are considering in $\displaystyle{\Pi_{j\in J}X_j}$ the product topology.
Thanks.
AI: Hint: try to prove both inclusions. I am sure that the following characterization turns out to be very useful: $x \in \overline{A}$ if and only if for every open set $U$ containing $x$ we have $U \cap A \neq \emptyset$. Recall also that you need to consider only basic open set. |
H: Proof by cases, inequality
I have the following exercise:
For all real numbers $x$, if $x^2 - 5x + 4 \ge 0$, then either $x \leq 1$ or $x \geq 4$.
I need you to help me to identify the cases and explain to me how to resolve that. Don't resolve it for me please.
AI: HINT:
If $(x-a)(x-b)\ge0$
Now the product of two terms is $\ge0$
So, either both $\ge0$ or both $\le0$
Now in either case, find the intersection of the ranges of $x$ |
H: How to prove that $q^r$ and $q^s-1$ are relatively prime?
How to prove that $q^r$ and $q^s-1$ are relatively prime? ($q$ is prime or prime power)
AI: We do not need to assume anything about $q$, but we do need to assume that $s\gt 0$.
Let $d\gt 1$ be a divisor of $q^r$. Then some prime $p$ divides $d$, and hence $q$. Then $p$ divides $q^s$. Hence it cannot divide $q^s-1$, else it would divide $1$. |
H: Simple linear regression - understanding given
The question is to fill out the missing numbers (A-L) of a simple linear regression model.
I am having problems with converting and interpreting the given table in terms of variables. Would it be possible for someone to confirm and clarify things for me.
The first table represents regression statistics
True model
$$
Y_t = \beta_o + \beta_1X_t + \mu_t
$$
Estimated model
$$
\hat Y_t = \hat\beta_0 + \hat\beta_1x_t
$$
This is what I am confused about
Does the first standard error ($12.8478$) mean $\sum\hat\mu_t^2$ ?
Does the standard error for the intercept in last table ($14.6208$) mean $\sum\mu_t^2$ ?
Does $3.8508$ equal $\hat\beta_1$ ?
In order to calculate RSS (for J) I need $\sum \hat\mu_t^2$ does this confirm that my first two points are incorrect
I know $G=\hat\beta_1^2\sum x_t^2$, how do I find $\sum x_t^2$
If I am wrong, would it be possible to know what those numbers mean in terms of variables
AI: Let us first calculate all the unknowns:
D = 1 (as it has only one variable)
E = n-1-1 = 13
F = 15-1 = 14
MSE = (Standard Error of Estimate)^2 = 12.8478^2 = 165.06
So J = 165.06
I/J = F-statistic = 24.15
I = 24.15*165.06
= 3986.34
G = I*k = 3986.34
H = J*(n-k-1) = 165.06*13 = 2145.78
K/14.6208 = 1.3081
K = 19.125
3.8508/L = 4.9146
L = 3.8508/4.9146 = .7835
R-Squared = B = 3986.34/6132.85 = .649
Adj R-Squared =
1-[(n-1)/(n-k-1)*(1-R^2)] = .622
To give explanation
Sum of (Mu-Hat)^2 = Sum of Squared Errors
Standard Esimate of Error = SQRT(MSE)
Mean Squared Error = SSE/(n-k-1)
14.4208 means the Estimated Standard Error of Beta1-hat |
H: calculate the loss to shop owner
A customer purchases clothes worth 200 Rupees from a shop. customer gives 1000 Rupee note. Since shop owner does not have change, he collects change (100 * 10 notes) from the neighbor shop and gives 800 rupees back to the customer.
After few hours neighbor shop keeper tells the shop owner that 1000 rupee note is fake and collects his 1000 rupees.
So now calculate the total loss to the shop owner.
Assume that Shop owner is selling clothes without any profit.
AI: Customer gives fake 1000 note: 0
He collects 1000 from neighbour: 1000
Gives 800 back to customer: 200
Gives 1000 to neighbour: -800
End loss is 800 rupees and the clothes. |
H: Is $(-\infty,\infty)$ a closed **interval**?
Note that we are working in the reals, not the extended reals. Now consider two opposing claims:
$(-\infty,\infty)$ is a closed interval, because a closed
interval is an interval that is a closed set.
$(-\infty,\infty)$ is not a closed interval, because a closed
interval is an interval that includes both its endpoints.
Recently, I was very surprised to be informed by several mathematicians that $(-\infty,\infty)$ ought to be called a closed interval, not an open interval. Perhaps I could get some consensus from this community, if possible.
p.s. To be very clear, I am not asking if $(-\infty,\infty)$ is a closed set—it sure is.
Update
For what it's worth, the ISO 80000-2:2009 document explicitly calls $(a,b)$ an open interval and $[a,b]$ a closed interval.
AI: Amazing degree of consensus on the wrong answer. I won't bother giving my opinion (which I sort of hinted at in the previous sentence) because I am nobody and my opinion means nothing. Instead I will suggest a method you can use to get an answer, which may be slightly more valid than polling a handful of random internet addicts.
Go to the library and find the shelf with all the introductory analysis books. Look up the statements of theorems like "a continuous function on a finite closed interval attains its maximum" and "a continuous function on a finite closed interval is uniformly continuous". Count how many books include the word "finite" and how many omit it. I haven't done this myself, so I have no idea what result you will get. |
H: Interpreting inequalities
Two conditions are met for $x$:
$x \geq 4$ and $x \geq 1$
What's the final value of $x$? I believe it is $x \geq 4$, by simple logic if $x$ is said to be greater or equal than $1$ and greater or equal than $4$, then it means it is the greatest one..
However, what's the justification behind this? I cannot just put 'by looking at it' on my notebook.
AI: If $x \geq 4$ and $4 \geq 1$, then it follows by transitivity that $x \geq 1$.
So when we say that $x \geq 4$, it is implicit that $x \geq 1$.
So when you write $(x \geq 4$ and $x\geq1)$, that's exactly the same as writing $x \geq 4$. So you might as well just leave it at that. |
H: $X$ a topological space. If $A$ lies inside a closed set. Does it follow that the closure of $A$ also lies inside this closed set?
PROBLEM: $X$ a topological space. If $A$ lies inside a closed set. Does it follow that the closure of $A$ also lies inside this closed set?
MY TRY:
Suppose $A \subseteq F$ where $F$ is closed set. I want to show $\overline{A} \subseteq F $. Take aribtrary $x \in \overline{A} $. So $x$ is limit point of $A$. In particular, can find a nghbd $N$ of $x$ such that $N \cap A \neq \varnothing $.
IF $ x \in A$, then we are done since then we would have $x \in F$. Suppose $x \notin A $. Then either $x \in F \setminus A $ or $x \in X \setminus F$.
If $x \in F \setminus A \implies X \in F $
If $ x \in X \setminus F \implies x $ cannot be a limit point of $A$
Therefore, $\overline{A} \subseteq F $.
Is this a correct approach?
AI: Take it for what it is worth, I would argue something along the lines of:
Say $A \subseteq C$ some closed set.
$$\overline{A} = \bigcap_{A\subseteq F} F, \text{ $F$ closed in $X$.}$$
If $x \in \overline{A}$ then
$$x\in\bigcap_{A\subseteq F} F, \text{ $F$ closed in $X$}$$
and in particular, $C$ is a closed set of $X$ that contains $A$, and so as $x$ in all such sets, $x \in C$ and so $\overline{A}\subseteq C$. |
H: finding examples for a non negative and continuous function for which the infinite integral is finite but the limit at infinity doesn't exist
Question:
a. Find an example for a non-negative and continuous function s.t. $\int _0^\infty f(x)dx$ is finite but the following limit doesn't exist: $\lim_{x\to \infty} f(x)$.
b. Is it possible that $\int _0^\infty f(x)dx$ is bounded but $f(x)$ is not bounded?
What we did with A:
We suggested the function $|sin(x^2)|$ which has periods that get smaller and smaller until they no longer imply on the sum. But since it's a trig function it doesn't have a determinate limit. Wolfram didn't have the integral value for that func, and we were wondering if it really converges and if it is really a good example.
What we did with B:
We thought about the function: $f(x)= { x\in \Bbb N },{e^{-x} \notin \Bbb N }$ and we had a disagreement whether f(x) is integrable. I said no because similarly to Dirichlet function, one sum's limit will be 0 while the other one's will be infinite. My partner said that if I find a $\delta<1$ then the definition of Riemann's integral does hold and so this integral is equal to that of $f(x)=e^{-x}$ She was trying to say this functions integral will be 0, but still it won't be bounded. I disagreed. Who's right?
AI: For (b), and therefore (a), let $f(x)=0$ with the following exceptions. For every positive integer $n$, $f(x)$ climbs linearly from $f(x)=0$ at $x=n-2^{-2n}$ to $f(x)=2^n$ at $x=n$, then falls linearly to $0$ at $x=n+2^{-2n}$.
The area of the triangle "at" $n$ is $(2^{-2n})(2^n)$, that is, $2^{-n}$, and the sum of the areas of these triangles is $1$. |
H: Density of a set
Suppose $X$ is a topological space. We know by definition $A$ is dense in $X$ if $ \overline{A} = X $.
My question is. IS it enough that $\overline{A} \subseteq X$ to say that $A$ is dense in $X$ ??
AI: No. Take $X=[0,2]$ and $A=[0,1]$. Then $A= \overline{A} \subset X$, but clearly $A$ is not dense in $X$ (for example, $(1,2) \subset X\setminus A$). |
H: Proving the diameter is two times the radius
I am stuck on the following question:
Prove that each diameter is twice as long as each radius.
I drew a circle, with center O and diameter AB. Is there a theorem that could help me say that congruent segment AO and BO add up to form segment AB?
Or is there some other way to prove this?
I would really like it if anyone could give me a hint about this.
Thank you.
AI: Hint:
A circle is (in part) defined by having an equal distance from its center to its edge for all points on its edge, i.e. it has a constant radius. So then what's the distance from the edge to the center to the edge again? (And does that sound related to your definition of a diameter at all?)
You could also prove this pretty easily by contradiction: "Suppose $d \neq 2r$. Then..." I'll leave that to you. |
H: Tricky Puzzle!! Please help.
I stumbled upon a puzzle I can't crack. It goes like this:
In a certain Code language:
7321=6
5342=3
8645=15
Then 9312=?
The Answer is 9. But I can't seem to find the logic behind it??
AI: Cool Puzzle! Here is the pattern I found:
$7321:$ Take $7 \cdot 3 \cdot 2 \cdot 1 = 42$. Then $4+2 = 6$.
$5432:$ Take $5 \cdot 4 \cdot 3 \cdot 2 = 120$. Then $1+2+0 = 3$.
$8645:$ Take $8 \cdot 6 \cdot 4 \cdot 5 = 960$. Then $9+6+0 = 15$.
Et Cetera... :-) |
H: Let $f$ be a continuous function on $[a, b]$ such that $f(x)$ is rational for every $x$. What can be said about $f$?
can someone help me out with this question? I have been stuck on it for a while.
Suppose that $f$ is a continuous function on the closed interval $[a, b]$, and that $f(x)$ is rational for every $x$ in the set $[a, b]$. What can be said about $f$?
AI: $f$ must be constant.
Since $f$ is continuous and $[a,b]$ is connected and compact, then $f([a,b])$ is connected and compact. Hence $f([a,b])$ is an interval, that is $f([a,b])=[c,d]$ for some $c,d$. If $c<d$, then that means $f([a,b])$ contains some irrational, which is a contradiction. Hence $c=d$, and $f(x) = c$ for all $x \in [a,b]$. |
H: Proving a random variable
Here is another question from the book of V. Rohatgi and A. Saleh. I would like to ask help again. Here it goes:
Let $\mathcal{A}$ be a class of subsets of $\mathbb{R}$ which generates $\mathcal{B}$. Show that $X$ is an RV on $\Omega\;$ if and only if $X^{-1}(A)$ $\in \mathbb{R}$ for all $A\in \mathcal{A}$.
I actually do not know how/where to start. I hope someone can help. Thanks.
AI: Let $(\Omega,\mathcal{F},P)$ be a probability space and let $X:\Omega\to\mathbb{R}$ be a mapping from $\Omega$ to $\mathbb{R}$. By definition, $X$ is a random variable if $X^{-1}(A)\in\mathcal{F}$ for all $A\in\mathcal{B}$, where $\mathcal{B}$ denotes the Borel sets on $\mathbb{R}$.
Clearly, if $X$ is a random variable, then $X^{-1}(A)\in\mathcal{F}$ for all $A\in\mathcal{A}$ (why?).
To show the other direction, we can check that
$\Sigma:=\{A\subseteq\mathbb{R}\mid X^{-1}(A)\in \mathcal{F}\}$ is a sigma-algebra on $\mathbb{R}$, and
since $\mathcal{A}\subseteq\Sigma$ we can conclude that.. |
H: Proving the base of a gramian matrix with a defined form
Be $V$ n-dimensional in a $ \mathbb R $ vector space und be $q$ a nondegenerate quadratic form on $V$. To prove: It exists a base $B$ of $V$ in $ \mathbb R $, in which the gramian matrix of $q$ is defined as
$\begin{bmatrix}1 & & & & & \\ & ... & & & & \\ & & 1 & & & \\ & & & -1 & & \\ & & & & .. & \\ & & & & & -1\end{bmatrix}$
I don't know how to start proving this.
AI: Use induction on $n$.
The case $n=0$ is trivial. For $n>0$, there exists a vector $v$ such that $q(v)\neq0$, since the form is non-degenerate. After scaling $v$ by $1/\sqrt{|q(v)|}$ one will have obtained $q(v)\in\{-1,1\}$. Take this $v$ as first or last basis vector, depending on the sign of $q(v)$. Now $v^\perp$ is a a complementary hyperplane to $\langle v\rangle$, and the restriction of $q$ to$~v^\perp$ is easily seen to be non-degenerate. Induction applied to $v^\perp$ completes the basis. |
H: Why does the method to find out log and cube roots work?
To find cube roots of any number with a simple calculator, the following method was given to us by our teacher, which is accurate to atleast one-tenths.
1)Take the number $X$, whose cube root needs to be found out, and take its square root 13 times (or 10 times) i.e. $\sqrt{\sqrt{\sqrt{\sqrt{....X}}}}$
2)next, subtract $1$, divide by $3$ (for cube root, and any number $n$ for $n$th root), add $1$.
3) Then square the resultant number (say $c$) 13times (or 10 times if you had taken out root 10 times) i.e. $c^{2^{2^{....2}}}=c^{2^{13}}$. This yields the answer.
I am not sure whether taking the square root and the squares is limited to 10/13 times, but what I know is this method does yield answers accurate to atleast one-tenths.
For finding the log, the method is similar:-
1)Take 13 times square root of the number, subtract 1, and multiply by $3558$. This yield s the answer.
Why do these methods work? What is the underlying principle behind
this?
AI: Let's use these classical formulae :
$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
$$\ln\,x=\lim_{n\to\infty}n\left(x^{1/n}-1\right)$$
to get (replacing the limit by a large enough value of $n$ : $N=2^{13}$) :
\begin{align}
\sqrt[3]{x}=e^{\left(\ln x/3\right)}&\approx \left(1+\frac {\ln x/3}N\right)^N\\
&\approx \left(1+\frac {N\left(x^{1/N}-1\right)}{3\,N}\right)^N\\
&\approx \left(1+\frac {\left(x^{1/N}-1\right)}3\right)^N\\
\end{align}
Concerning the decimal logarithm we have :
$$\log_{10}\,x=\frac{\ln\,x}{\ln\,10}\approx \frac N{\ln\,10}\left(x^{1/N}-1\right)$$
For $N=2^{13}$ we may (as indicated by peterwhy) approximate the fraction with $$ \frac N{\ln\,10}=\frac {2^{13}}{\ln\,10}\approx 0.4343\times 8192\approx 3558$$
Hoping this clarified things, |
H: ${{p-1}\choose{j}}\equiv(-1)^j \pmod p$ for prime $p$
Can anyone share a link to proof of this?
$${{p-1}\choose{j}}\equiv(-1)^j(\text{mod}\ p)$$ for prime $p$.
AI: $$\binom {p-1}j=\prod_{1\le r\le j}\frac{p-r}r$$
Now, $\displaystyle p-r\equiv -r\pmod p\implies \frac{p-r}r\equiv-1\pmod p$ |
H: Derive a formula of a specific curve
I have this curve
And I know that the first point is
$$A(0,5)$$
and the last point is
$$C(1650,9.5)$$
The point almost at the center where the curve changes (if you look close, you can see a green dot) is
$$B(1000,8.5)$$
There is a way to derive a generic f(x) so that I can plug in a value of x and find f(x)?
AI: We know four properties: $f(0)=5$, $f(1000)=8.5$, $f(1650)=9.5$ and $f'(1000)=0$. To fit a polynomial with these constraints we need a degree of at least three:
$$f(x) = ax^3+bx^2+cx+d$$
$$f'(x) = 3ax^2+2bx+c$$
Plugging in what we know gives the following equation system.
$$\begin{cases}
a+b+c+d=5 \\ 10^9a+10^6b+10^3c+d=8.5 \\ 1650^3a+1650^2b+1650c+d=9.5 \\ 3000000a+2000b+c=0 \\
\end{cases}$$
Solving this and plotting the resulting $f(x)$, gives the following curve. Does this look good enough?
The solution:
$$\begin{cases}
a=3.562082728528 \cdot 10^{-9} \\
b=-1.063473805380 \cdot 10^{-5} \\
c=0.010583227922\\
d=4.989427403254\\
\end{cases}$$ |
H: Show there is a closed interval $[a, b]$ such that the function $f(x) = |x|^{\frac1{2}}$ is continuous but not Lipschitz on on $[a, b]$.
Hi guys I was given this as an "exercise" in my calculus class and we weren't told what a Lipschitz is so i really need some help, heres the question again:
Show there is a closed interval $[a, b]$ such that the function $f(x) = |x|^{\frac1{2}}$ is continuous but not Lipschitz on on $[a, b]$.
AI: Consider the interval $[0,1]$, then clearly $f(x) = x^{1/2}$ is continuous.
Can you show that $f$ is not Lipschitz on $[0,1]$? (Hint: Use the Mean-Value theorem on a closed sub-interval of $(0,1/n]$) |
H: If given a regular language, how can we prove that a sub-language is regular?
This question has been quite confusing me.
$\sum = \{a,b,c\}, L \text{ is a regular language}$ and we have to prove that $L^{'} = \{w \in L : w\text{ containts at least one c} \}$ is regular.
What are the steps of the proof? What methods can we use? I am new to automata and I have suggested the following which I am sure would not work:
$L$ is regular, then it has a regular expression $r \rightarrow L(r) = L $. Now if we write the expression $(r \cdot c \cdot c^* \cdot r^*)$ then it is a regular expression of the language $L^{'}$ where we would have a word of the language $L$ and also the word would contain at least one $c$.
I can deeply sense that there is a big flaw in that, what did I do wrong? and can you please direct me into a good solution?
Thank you!
AI: Consider the regular language $K = \Sigma^*c\Sigma^*$. Then your language $L'$ is equal to $L \cap K$. You should have learnt that regular languages are closed under intersection, so $L'$ regular. |
H: Help with a property of a convex function
I'm studying linear and nonlinear programming and on my book I bumped into the following statement:
$$\lim_{\alpha \to 0} \displaystyle \frac{f(\textbf{x}+\alpha (\textbf{y}-\textbf{x}))-f(\textbf{x})}{\alpha} = \nabla f(\textbf{x})(\textbf{y}-\textbf{x})$$
Could someone show me, why is the statement above true?...I get confused again with this for some reason :-)
Thank you for any help :-)
AI: Well, the equation is saying nothing more that the directional derivative of a differentiable function is equal to the inner product of the direction and the gradient of that function:
Your left hand side is the directional derivative of $f$ at $x$ in direction $(y-x)$ and your right hand side is the inner product of the gradient of $f$ at $x$ with the direction $(y-x)$. |
H: $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$
Problem:
Suppose $f$ and $g$ are two continuous functions such that $f: X \to Y
$ and $g : X \to Y $. $Y$ is a a Hausdorff space. Suppose $f(x) = g(x)
$ for all $x \in A \subseteq X $ where $A$ is dense in $X$, then $f(x)
= g(x) $ for all $x \in X $.
Attempt at a solution:
Put $h(x) = f - g $. Therefore, $h: X \to Y $ is continuous and $Y$ is Hausdorff by hypothesis. Also we know $h(x) = 0 $ for all $x \in A $ such that $A$ is dense in $X$. I want to show that $h(x)$ vanishes everywhere in $X$. We can show $h(x) = 0 $ for all $x \in X \setminus A $. Suppose $h(x) > 0 $ on $X \setminus A$. Pick points $y_1,y_2 \in Y $. Since $Y$ is Hausdorff, can find open set $O_1, O_2 \subseteq Y $ which are disjoint such that $y_1 \in O_1$ and $y_2 \in O_2$. By continuity, $f^{-1}(O_1), f^{-1}(O_2)$ are open in $X$.
I know that if I can show that one of the $f^{-1}(O_i)$ lies in $X \setminus A $, then we would have a contradiction since we have non-empty open set in $X \setminus A$ and this implies $A$ cannot be dense in $X$. But this is the part I am stuck. Any help would greatly be appreciated.
Also, Would be be possible to prove this without using the Hausdorff condition on $Y$?
AI: Suppose $f(x_0) \neq g(x_0)$, then since $Y$ is Hausdorff, there are open sets $U,V \subset Y$ such that
$$
f(x_0) \in U, g(x_0) \in V, \text{ and } U\cap V = \emptyset
$$
Now
$$
x_0\in f^{-1}(U)\cap g^{-1}(V) =: W
$$
and $W$ is open, and hence $\exists a\in A\cap W$, whence
$$
f(a) = g(a) \in U\cap V \Rightarrow U\cap V \neq \emptyset
$$
This contradiction proves the result. |
H: Finding the median of a probability distribution
A gambler makes a long sequence of bets against a rich friend. The gambler has initial capital C. On each round, a coin is tossed; if the coin comes up tails, he loses 30% of his current capital, but if the coin comes up heads he instead wins 35% of his current capital. Let $C_n$ be the gambler's capital after n rounds and write $C_n$ as a product $C*Y_1*Y_2*...*Y_n$.
Find the median of the distribution of $C_{(10)}$ and compare it to the expectation of $C_{(10)}$.
I worked out that the expectation of $C_n$ is $1.025^nC$ so I can calculate expectations but I'm a bit confused about finding the median.
AI: By symmetry the median has half heads and half tails. Each head/tail combination leads to a multiplication of capital by $(1.35)(0.70)=0.945$. |
H: Multiple disjunctions with a Tableaux proof system
I am using the Tableaux proof system, and have a question about branching and disjunctions.
Normally the example on how to use the Tableaux proof system is to get the formula to CNF, and then start branching it. It can look like this:
$$
(A \lor B), \lnot B,\lnot A
$$
And then you'd get the branches:
$A\lnot B,\lnot A$ and $B, \lnot B,\lnot A$
But sometimes I get a formula with multiple disjunctions and then I don't know if what I am doing is actually correct:
$$
(A \lor B \lor C), \lnot B,\lnot A, \lnot C
$$
branches:
$A\lnot B,\lnot A, \lnot C$ and $(B \lor C), \lnot B,\lnot A, \lnot C$
first branch is closed, so branch second branch: $B, \lnot B,\lnot A, \lnot C$ and $C, \lnot B,\lnot A, \lnot C$
Am I doing it correctly?
AI: Short answer: yes, but note that you generally don't have CNF in tableaux. You might be thinking of resolution (where you need to do a conversion to CNF before starting out).
Ask yourself how you got to the described step. An example:
$\{(A \lor B) \land \lnot B \land \lnot A\}$ (a set of one formula) is transformed to
$\{(A \lor B), \quad \lnot B \land \lnot A\}$ (two formulae) to
$\{(A \lor B) ,\quad \lnot B ,\quad \lnot A\}$ (three)
Here you repeatedly use the $\alpha$ rule to pick apart a single formula at a time.
Similarly, you can use $\beta$ rule to split $(A\lor B\lor C)$ one step at a time.
Note that you need to be careful (in the sense of thinking about operator priorities) if you encounter a formula like $A\lor B \land C$. |
H: Why are two definitions of ellipses equivalent?
In classical geometry an ellipse is usually defined as the locus of points in the plane such that the distances from each point to the two foci have a given sum.
When we speak of an ellipse analytically, we usually describe it as a circle that has been squashed in one direction, i.e. something similar to the curve $x^2+(y/b)^2 = 1$.
"Everyone knows" that these two definitions yield the same family of shapes. But how can that be proved?
AI: Suppose we have a classical ellipse with its two foci and major axis given. We want to prove that it's a squashed circle.
Select a coordinate system with its origin in the center if the ellipse, the $x$-axis passing through the foci, and a scale such that the common sum-of-distances-to-the-foci is $2$. The foci have coordinates $(\pm c,0)$ for some $c\in[0,1)$.
By considering the nodes at the end of the minor axis we find that the semiminor axis of the ellipse must be $b=\sqrt{1-c^2}$. We must then prove that the equations
$$ \tag{1} x^2 + \left(\frac{y}{\sqrt{1-c^2}}\right)^2 = 1 $$
$$ \tag{2} \sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 $$
are equivalent.
Rearranging (1) gives
$$\tag{1'} y^2 = (1-c^2)(1-x^2) $$
and therefore $(x+c)^2+y^2 = (1+xc)^2$ by multiplying out each side and collecting terms. Similarly, $(x-c)^2+y^2=(1-xc)^2$. So for points where (1) holds, (2) reduces to $(1+xc)+(1-xc)=2$, which is of course true.
Thus every solution of (1) is a solution of (2). On the other hand, for every fixed $x\in(-1,1)$, it is clear that the left-hand-side of (2) increases monotonically with $|y|$ so it can have at most two solutions, which must then be exactly the two solutions for $y$ we get from (1'). (And neither equation has solutions with $|x|>1$). So the equations are indeed equivalent.
Conversely, if we have a squashed circle with major and minor axes $2a$ and $2b$, we can find the focal distance $2c$ by $c^2+b^2=a^2$ and locate the foci on the major axis. Then the above argument shows that the classical ellipse with these foci coincides with our squashed circle. |
H: Optimisation of a rectangles area under a function curve
I have a questions asking for the dimensions of the rectangle with the largest area that has two bottom corners on the x axis and two top corners on the curve $y=12-x^2$.
I have plotted the curve and found it is a symmetrical parabola with a vertex of $x=0, y=12$.
It intersects the $x$ axis at $-2\sqrt3$ and $2\sqrt3$.
My thinking is that if I find when the derivative of the (area under the curve, minus the area inside the square) = 0, then I can determine what values make it a minimum.
I also thought that I could half the parabola and work with one side since it is symmetrical, then double those values at the end.
So the area under the curve in the positive x axis = $∫_0^{2\sqrt3}12-x^2 dx$
My problem is that I can't define area of the rectangle, or the sides. Can anyone give me any pointers?
AI: As you say, let us take into account the symetry. Let us put the two bottom corner at (a,0) and (-a,0). The top corners will be at (-a,b) and (a,b). But the top corners are also on the parabola; this means that b=12-a^2. Then, the area is 2ab = 2 a (12 - a^2) and you want this area be maximum. Are you able to continue with this ? If not, just post a message to me. |
H: Regular Language Operation
I need to show that the given regular language is closed under the following operation.
For example:
AllSuffixes(L) = {v : uv in L for some u in (0+1)* }
I do not have any idea about this question. I only know that regular languages are closed under + , . and * operations.
Do you have any idea(a solution way or a clue)?
AI: One approach is to start with a DFA that recognizes $L$ and has no inaccessible states. Now add empty transitions from the initial state to each other state and show that the resulting NFA does what you want.
You can also approach it via regular grammars. Start with a regular grammar for $L$. If $S$ is the initial symbol, add a production $S\to X$ for each accessible non-terminal symbol $X$ other than $S$. This is basically exactly the same idea. In both cases the idea is to let the machine/grammar start at any point that the original machine/grammar could have reached from the initial state/symbol. |
H: Probability on divisibility
Let S be the set of all 12-digit positive integers each of whose
digits is either 1 or 4 or 7 (for example, 477411171747 is a member
of S). What is the probability that a randomly picked member
of S is divisible by 12 ?
AI: HINT:
For a number to be divisible by twelve it must be divisible by $4$ and $3$.
Also the divisibility rule for a number to be divisible by $4$ is that its last two digits must be divisible by four.
Also for three is that its sum of digits is a multiple of $3$.
Can you take it from here? |
H: Questioning a Basis for $\mathbb{Q}[\sqrt[3]{2}]$ over $\mathbb{Q}$
Let $\omega = e^{2 \pi i /3}$ and $\alpha = \sqrt[3]{2}$. I'm seeing it claimed that $\mathcal{B} = \{\alpha, \alpha^2, \omega \alpha, \omega \alpha^2, \omega^2 \alpha, \omega^2 \alpha^2\}$ forms a basis for the vector space $\mathbb{Q}[\alpha, \omega]$ over $\mathbb{Q}$.
But the vector space $\mathbb{Q}[\alpha, \omega]$ over $\mathbb{Q}$ should have all of the elements of $\mathbb{Q}$ as members, and it's not clear to me how one could express some rational $q$ as a $\mathbb{Q}$ linear combination of the elements of $\mathcal{B}$. This is because all of the members of $\mathcal{B}$ are either strictly irrational or strictly complex so that any linear combination of them will result again in a strictly irrational or strictly complex number (and hence not a member of $\mathbb{Q}$).
Am I missing something, or is $\mathcal{B}$ in fact not a basis for $\mathbb{Q}[\omega, \alpha]$?
EDIT: I miswrote the claimed basis $\mathcal{B}$. I know that adding $1$ would make it a true basis. But as now written, the supposed basis is $\mathcal{B} = \{\alpha, \alpha^2, \omega \alpha, \omega \alpha^2, \omega^2 \alpha, \omega^2 \alpha^2\}$.
AI: You need to add 1 to your basis. First of all, since $[\mathbb{Q}(\alpha,\omega):\mathbb{Q}]=[\mathbb{Q}(\alpha,\omega):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]$, then this dimension can't be 5 since each of the factors is non-trivial. Therefore your intuition is right and you have to to add 1. The dimension is actually 6.
EDIT Due to the edit I will say what the comments already say above: the new set is not linearly independent due to the fact that $\alpha+\alpha\omega+\alpha\omega^2=0$. |
H: $(A\cap B)\cup C = A \cap (B\cup C)$ if and only if $C \subset A$
I have a set identity: $(A \cap B) \cup C = A \cap (B \cup C)$ if and only if $C \subset A$.
I started with Venn diagrams and here is the result:
It is evident that set identity is correct. So I started to prove it algebraic:
1) According to distributive law: $(A \cap B) \cup C = (A \cup C) \cap (B \cup C)$
2) ...
I stuck a little. Because $C$ is a subset of $A$. I thought of pulling out: $(B \cup C)$ but it seems wrong step to me.
How to prove this identity having in mind that $C \subset A$?
Updated
Venn diagram for $C ⊈ A$
AI: It’s a little easier to go the other way: $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$, and you’d like to show that this equals $(A\cap B)\cup C$ if and only if $C\subseteq A$.
Suppose first that $C\subseteq A$; how can you simplify $A\cap C$?
Now suppose that $C\nsubseteq A$; then there is some $c\in C\setminus A$. Show that this $c$ is an element of $(A\cap B)\cup C$ but not of $(A\cap B)\cup(A\cap C)$, so that these two sets cannot be equal.
Your Venn diagrams show what happens when $C\subseteq A$, so they’re useful in proving one direction of the desired result: if $C\subseteq A$, then the two sets are equal. To see how you might prove the other direction, i.e., that if $C\nsubseteq A$, then the two sets are not equal, you’d do better to look at a Venn diagram showing $C\nsubseteq A$. |
H: Convergence of a series of random elements
Given the normally distribuited random variable $\nu(t)$ with $\mu=0$ and variance $\sigma$, I have to find if the series:
$$G(\sigma)=\sum_{k=1}^{\infty}\frac{1}{\exp\left(\nu(k)\right)}$$ where $\nu(k)$ is the random value of the $\nu(t)$ for $t=1,2,...$, is convergent and if it is, how can be evaluated the result of the sum. Thanks
AI: As long as the parameters of the distribution of $\nu(t)$ do not depend on $t$, we can simply refer to this variable as $\nu$. Thus, $\frac{1}{\exp(\nu)}$ does not converge to anything, so the infinite sum of it goes to infinity. |
H: Construct context-free grammar for $\{a^ib^jc^k : i\le j+k\}$
I'm looking through several of old exam sets in order to prepare for the exam and now I'm stuck on this exercise, where we have to construct a context-free grammar for the language:
$$L = \{a^ib^jc^k: i \le j+k \}$$
The best solution I've obtained so far is the following that accepts $\{a^ib^jc^k: i = j+k \}$:
$$\begin{align*}
&S\to aSc \mid D\\
&D\to aDb \mid \epsilon
\end{align*}$$
This context-free grammar will not create any strings that are not in the language but it will however not create all the possible strings we get with the condition $i \le j+k$.
How do I proceed to solve this?
AI: Just allow $S$ to produce $Sc$ and $D$ to produce $Db$, as well as what you already have. |
H: Extensions of probability measures on fields
I am trying to solve exercise 3.3 in Billingsley's "Probability and measure" and I am not sure I correctly understood the text. I am not going to copy all the text, but I will ask directly some questions.
Some background:
Let P be a probability measure on the field $F_0$ and $P^*$ be the outer measure defined for each subset A of $\Omega$ as $P^*(A)=inf \sum_n P(A_n)$, where $A \subset \cup_n A_n$.
A theorem (3.1 in Billingsley) guarantees that $P^*$ is the unique probability measure on the $\sigma$-field $\sigma(F_0)$ such that $P^*(B)=P(B)$ for each $B \in F_0$.
Now the questions:
1) If $\Omega$ is uncountable and $F_0$ is the field of finite and cofinite sets, and P(A)=0 if A is finite and P(A)=1 if A is cofinite, do $P^*$ (defined as above) and $P$ agree on $F_0$?
My answer: P is a probability measure on $F_0$ (it is possible to check that it is countably additive) and then theorem 3.1 holds and P and $P^*$ agree on $F_0$.
My doubt comes from the fact that the text of the exercise seems to suggest that this is not the case and to explain why...
2) If \Omega is uncountable and F_0 is the field of countable and cocountable sets, and P(A)=0 if A is countable and P(A)=1 if A is cocountable, do $P^*$ (defined as above) and P agree on $F_0$?
My answer would be the same than before:
P is a probability measure on $F_0$ (it is possible to check that it is countably additive) and then theorem 3.1 holds and P^* agree on $F_0$.
3) Let $P(A)=I_A(\omega_0)$ for $A \in F_0$, and assume that ${\omega_0} \in \sigma(F_0)$, do $P^*$ (defined as above) and P agree on $F_0$?
My answer would be the same than before:
P is a probability measure on $F_0$ (it is possible to check that it is countably additive) and then theorem 3.1 holds and $P^*$ agree on $F_0$.
In this case I do not understand also why the text specifies that ${\omega_0} \in \sigma(F_0)$.
AI: The answer to 1 and 2 will be very similar since an uncountable set is never the union of a countable number of finite or countable sets. In this case, if $A$ is finite/countable, then $A \subset \cup A_n$ where for every $n$, $A_n = A$. Therefore $P^*(A) \leq \sum 0 = 0$. Since $P^*(A) \geq 0$ then you see that $P^*(A) = P(A) = 0$. On the other hand, if $B$ is a cofinite/cocountable set, then because $\Omega$ is uncountable, it must be that $B$ is uncountable. So if $A_n \in F_0$ such that $B \subset \cup A_n$ then for at least one of the $n$, $A_n$ must be uncountable (as remarked in the first sentence) which means that for this $n$, $P(A_n) = 1$ (since $A_n$ must in fact then be cofinite/cocountalbe). Therefore you will find that $P^*(B) \geq 1$. Choosing $A_1 = B$ and $A_n = \varnothing$ for $n >1$, you get $\sum P(A_n) = 1 \geq P^*(B)$, so $P^*(B) = 1 = P(B)$.
If the assumption on the first problem was that $\Omega$ is countable, then $P^*$ and $P$ won't agree on $F_0$. This happens since if $A \in F_0$ is cofinite, then $A$ is countable and we can denumerate the elements of $A=\{a_1, a_2, ...\}$. Now, for each $i$, $\{a_i\} \in F_0$ and $\cup\{a_i\} = A$. But then $0 \leq P^*(A) \leq \sum P(\{a_i\}) = 0$, so $P^*(A)=0$ while $P(A)=1$. What you should notice here is that since $A$ is the countable disjoint union of the $\{a_i\}$, you find that $1 = P(A) = P(\cup\{a_i\}) \neq \sum P(\{a_i\}) = 0$. So, $P$ is not a very nice candidate for a premeasure.
For question 3, we have $P(A)=1$ if $\omega_0 \in A$, otherwise $P(A) = 0$. If $A \subset \cup A_n$ and $\omega_0 \notin A$ then we can choose $A_n = A$ for every $n$ to find $P^*(A) = P(A) = 0$ using similar arguments as before. On the other hand, if $\omega_0 \in A$ and $A \subset \cup A_n$, then there is some $n$ such that $\omega_0 \in A_n$. Therefore you can infer that $P^*(A) \geq 1$. Since you can choose $A_1 = A$ and $A_n = \varnothing$ for every $n > 1$, you find that $P^*(A) \leq 1$ implying that $P^*(A) = P(A) =1$. |
H: Each player throws two dice
probability problem:
players a and b throws two dice and
a player wins if the sum for the first throw is 11 or 7 and Each player loses at once if it is 3, 2. For other
case,throw two dice is repeated.the probability that player a wins at the $k$th throw?
my attempt: $ p(\text{a win at the $k$th throw})= ( \frac{25}{36} )^{k-1}(\frac{8}{36})$
can someone help me?
thanks.
AI: From the description it appears that Player B can be completely ignored: his actions do not affect Player A’s wins and losses. If that is correct, then so is your analysis. On each roll the probability of rolling something other than $2,3,7$, or $11$ is $\frac{25}{36}$, so the probability of getting one of these non-decisive numbers $k-1$ times in a row is $\left(\frac{25}{36}\right)^{k-1}$. On each roll the probability of getting a winning number is $\frac8{36}$, so the probability of getting $k-1$ non-decisive numbers followed by a winning number, i.e., of winning on the $k$-th roll, is indeed $\left(\frac{25}{36}\right)^{k-1}\left(\frac8{36}\right)$. |
H: Is the sum of positive definite matrices still positive definite?
I have two symmetric positive definite (SPD) matrices. I would like to prove that the sum of these two matrices is still SPD. Symmetry is obvious, but what about PD-ness? Any clues, please?
AI: A real matrix $M$ is positive-definite if and only if it is symmetric and $u^TMu > 0$ for all nonzero vectors $u$.
Now if $A$ and $B$ are positive-definite, $u^T(A+B)u = \ldots?$ |
H: If $R,S$ are reflexive relations, so are $R \oplus S$ and $R \setminus S$?
Suppose $R$ and $S$ are reflexive relations on a set $A$. Prove or disprove each of these statements.
a) $R\oplus S$ is reflexive.
b) $R\setminus S$ is reflexive.
I think both of a) and b) are false, but I'm having trouble with coming up with counterexamples.
AI: Hint: What does it mean to say $(x,x) \in R \oplus S$, respectively $(x,x) \in R \setminus S$? Can both $R$ and $S$ be reflexive if this is the case?
The above amounts to a proof by contradiction. But we can avoid this; for example, by the following argument:
Let $x \in A$. We know that $(x,x)\in R$ and $(x,x) \in S$. Therefore, by definition of $R \oplus S$ (resp. $R \setminus S$), $(x,x)\notin R \oplus S$. Hence $R \oplus S$ is not reflexive. |
H: Is $\mathbb Z/p\mathbb Z$ a subfield of every finite field?
I translate this from a German book: "For every finite field
$K$ there exists a prime number $p$ such that $\mathbb Z/p\mathbb Z$ is a subfield of $K$"
But how is this possible? For example the field $K = \{0,1\}$ contains integers but $\mathbb Z/p\mathbb Z$ contains equivalence classes. To be a subfield it would also have to be a subset of $K$.
AI: Every field contains a subfield isomorphic to either $\mathbb{Z}/p\mathbb{Z}$ (for some prime $p$) or $\mathbb{Q}$.
This follows from basic laws of additive exponents.
Let $\mathbb{K}$ be a field with multiplicative identity $1_\mathbb{K}$. Then consider the map $f:\mathbb{Z}\to\mathbb{K}$ defined by $n \mapsto n1_\mathbb{K}=\underbrace{1_\mathbb{K}+1_\mathbb{K}+\cdots+1_\mathbb{K}}_{n-\mbox{times}}$.
Basic laws of (additive) exponents tell us that $f(n+m)=(n+m)1_\mathbb{K} = n1_\mathbb{K}+m1_\mathbb{K}=f(n)+f(m)$ and $f(nm)=(nm)1_\mathbb{K} = (n1_\mathbb{K})(m1_\mathbb{K})=f(n)f(m)$ so $f$ is a ring homomorphism. Thus by the first isomorphism theorem $\mathbb{Z}/\mathrm{Ker}(f) \cong \mathrm{Im}(f)$.
The kernel of $f$ is either $\{0\}$ (this means $\mathbb{K}$ has characteristic $0$) and so $\mathbb{Z} \cong \mathrm{Im}(f)$. Thus $\mathbb{K}$ has a subring isomorphic to the integers and so (since it's a field) must have (multiplicative) inverses for these elements and so has a subfield isomorphic to $\mathbb{Q}$.
Otherwise the kernel of $f$ is a nonzero ideal of $\mathbb{Z}$ and so has the form $p\mathbb{Z}$ where $p$ must be prime (otherwise $\mathbb{K}$ would a contain a subring which has zero divisors). Thus $\mathbb{K}$ has a subring isomorphic to $\mathbb{Z}/p\mathbb{Z}$ (for some prime $p$).
If we begin with the assumption that $\mathbb{K}$ is finite, this rules out characteristic zero (there's not enough room to fit the infinitely large copies of $\mathbb{Z}$ or $\mathbb{Q}$). So finite fields must contain a subfield isomorphic to $\mathbb{Z}/p\mathbb{Z}$ for some prime $p$.
By the way, these subfields (the subfield generated by $1_\mathbb{K}$) are called prime subfields.
For your particular example, $K=\{0,1\} \cong \mathbb{Z}/2\mathbb{Z}$ (this field is of characteristic 2 and is equal to its prime subfield). |
H: Can one prove that a language is regular without having a regular expression?
I was wondering if one could prove that a language is regular without showing a DFA/NFA or a regular expression that expresses it.
For example: $L = \{w \in \Sigma^* : w \text{ has at least two identical letters} \}$
AI: In this case it’s not hard to design a DFA that accepts the $L$.
For each $S\subseteq\Sigma$ give your DFA a state $q(S)$, and have one other state $q$ as well. Make $q(\varnothing)$ the initial state, and make $q$ the only acceptor state. For each $s\in\Sigma$ and $S\subseteq\Sigma$, the $s$-transition from $q(S)$ goes to $q(S\cup\{s\})$ if $s\notin S$ and to $q$ otherwise. All transition from $q$ are to $q$. It’s not hard to see that if you’re in state $q(S)$, you’ve seen each character in $S$ once; the moment you see a character a second time, you go to state $q$, and the word is accepted.
In general, however, the answer to your question is yes: there are ways to show that a language is regular without constructing a DFA, NFA, regular expression, or regular grammar for it, though only after one has proved some results that do use these methods. For example, if we know that $L_1$ and $L_2$ are regular, we can conclude that $L_1\cap L_2$ is regular, because the class of regular languages is closed under intersection. However, the proof of this fact uses one of the explicit characterizations. |
H: $-ia(1\pm \sqrt{1-1/a^2})$, $a>0$ inside unit circle?
Given $a>0$ I would like to know whether:
$\alpha=-ia(1+ \sqrt{1-1/a^2})$ and
$\beta =-ia(1- \sqrt{1-1/a^2})$ are inside the unit circle.
How can I check that?
AI: You have
$$\alpha\cdot\beta = (-i)^2a^2\left(1 - \sqrt{1-1/a^2}^2\right) = -a^2(1/a^2) = -1,$$
so either both lie on the unit circle, or one lies inside the unit disk and the other outside.
For $0 < a \leqslant 1$, the radicand $1-1/a^2$ is a non-positive real number, hence the square root purely imaginary, and then $\lvert\alpha\rvert = \lvert\beta\rvert = 1$.
For $a > 1$, the radicand $1-1/a^2$ is positive, and if you take the branch of the square root that is positive on the positive real axis, you have $\lvert\beta\rvert < \lvert\alpha\rvert$, and hence then $\beta$ lies inside the unit disk, $\alpha$ outside. |
H: Median for continuous distribution
Consider a continuous random variable X with probability density function given by $f(x)=cx$ for $1 \le x \le 5$, zero otherwise. Find the median.
First I calculate the CDF: $F(x)=cx^2/2$ for $1 \le x \le 5$, zero otherwise.
Now we have to solve for constant c by using the definition of PDF, namely:
$$\int\limits_{-\infty}^{\infty}f(x)dx=1 \implies \frac{c}{2}x^2\Big{|}_1^5=1 \implies c=\frac{1}{12} $$
Then to calculate the median, we set the CDF = 0.5:
$$\frac{1}{2}=\frac{1}{12}\cdot \frac{1}{2} \cdot x^2 \implies x=\sqrt{12}$$
But the book solution is $\sqrt{13}$. Can someone tell me what I am doing wrong?
Thank you.
AI: Recall that there is an integration constant when finding the CDF $F(x)$.
Also recall that the CDF should take on the value ZERO when $x$ is from minus infinity to $x=1$ and it must take on the value ONE from $x=5$ to plus infinity. (i.e. the CDF is a non-decreasing function on the support of the density $f(x)$).
If you counter verify, you will see that the above paragraph does not hold for the CDF you found above, in your question.
Reworking on the problem, you should find an appropriate CDF. Simply put: the CDF should be
$$
F(x) = \frac{x^2}{24} - \frac{1}{24}
$$
We see that $F(1) = 0$ and that $F(5) = 1$ indeed.
Finally,
$$
0.5 = \frac{x^2}{24} - \frac{1}{24}
$$
Solving for $x$ yields that the median equals $\sqrt{13}$ |
H: Painting $\mathbb R^+$ with two colors which sum of two same color numbers be the same.
Can any one paint $\mathbb R^+$ with two colors which sum of two numbers with the same color has the same color.
Additional condition: Both colors should be used.
I tried use Cauchy functions like ($f(x+y)=f(x)+f(y)$). But there was no result.
AI: Pick any $\Bbb Q$-automorphism $\sigma$ of $\Bbb R$ (as a $\Bbb Q$-vector field) (preferrably not of the form $\sigma(x)= \lambda x$ with $\lambda \in \Bbb R^*$), color $x$ blue if $\sigma(x)>0$, and color $x$ red if $\sigma(x)<0$. |
H: Number of non negative Integral solutions
Need to find the non negative integral solutions for the equation
$x+y+xy=x^{3}+y^{3}$
I have tried various methods for simplifying the RHS and LHS but could not arrive at the solution, so any help will be appreciated
AI: First get a bound for possible solutions. Suppose without loss of generality that $x \leqslant y$. Then you have
$$(y+1)^2 = y^2 + y + y + 1 > xy + y + x = y^3 + x^3 \geqslant y^3.$$
That gives you $y \leqslant 2$. But for $y = 2$, we have $(2+1)^2 = 9$ and $2^3 = 8$, so the equality could only hold if $x = y$, or the first inequality would lose too much, and $x = 0$, or the second inequality would lose too much. So actually $y \leqslant 1$. Then it is a simple exhaustive search that yields the solutions $(0,0),\, (0,1),\, (1,0)$. |
H: Does there exist a complement of a subgroup in a abelian group.
Let $G$ be an abelian group and $H$ subgroup of $G$.
Suppose that:
(i) $H$ has a complement in $G$.
(ii) $K$ is a subgroup of $G$ and K is isomorphic to $H$
Is there a complement of $K$ in $G$? If yes, what is the relation of complements of $H$ and complements of $K$?
Thanks in advance.
AI: Not true. Let $G$ be the group $\mathbb Z/4 \mathbb Z \oplus \mathbb Z/2\mathbb Z$. Let $H$ be the second factor in the direct sum, which is isomorphic to $\mathbb Z/2 \mathbb Z$. Let $K$ be the order $2$-subgroup of $\mathbb Z/4 \mathbb Z$, considered as a subgroup of the first factor of the aforementioned direct sum. |
H: Question regarding simple limit
Why is it, that:
$\lim_{x \to \infty} [x (1-\sqrt{1-\frac{c}{x}})] = \frac{c}{2}$
Link: Wolframalpha
and not $0$?
My (obviously incorrect) reasoning:
Since $c$ is an arbitrary constant, and as $x$ goes to infinity $\frac{c}{x}$ will practically equal $0$. thus $\sqrt{1-0} = \sqrt{1} = 1$.
Therefore,
$\lim_{x \to \infty} [x (1-1)] = \lim_{x \to \infty} [x*0] = 0$
Where did I go wrong?
AI: The reason your approach doesn't work is that you are multiplying $0$ by infinity, which does not have a well defined result. I would probably first substitute $a = \frac{1}{x}$, which gives the equivalent but (in my opinion) simpler
$\lim_{a\downarrow 0} \frac{1}{a}\left( 1 - \sqrt{1 - ca}\right)$
Then substituting the Taylor expansion $\sqrt{1-ca} = 1 - \frac{c}{2}a - \frac{c^2}{8}a^2 + O(a^3)$ one finds
$\lim_{a\downarrow 0} \frac{1}{a}\left(\frac{c}{2}a + \frac{c^2}{8}a^2 + O(a^3)\right) = \lim_{a\downarrow 0} \left(\frac{c}{2} + \frac{1}{8}c^2a + O(a^2)\right) = \frac{c}{2}$
The main thing to keep in mind is that you always want to avoid expressions looking like $0\cdot\infty$ because their value is undefined. |
H: How to document undergrad math knowledge?
If you did a degree which is low on math (read economics, psychology), but want to proceed to a more mathematically loaded master, how would you document your knowledge?
Are there standarized examinations that are widely recognized?
That would also be a way of discovering how weak one is in math, not only to prove towards others.
AI: If you are applying for math graduate school in the US many schools would require the Math subject GRE. See http://www.ets.org/gre/subject/about/content/mathematics for what is covered. It's a multiple choice test but it's pretty challenging (and requires you to work fast). |
H: Pullbacks of monic morphisms.
I'm trying to prove that pullbacks of monics are monic.
Let
$\require{AMScd}$
\begin{CD}
X_1 @>f>> X_2\\
@V m' V V @VV m V\\
X_3 @>>g > X_4
\end{CD}
be a pullback square with $m'$ monic.
Let $h, k$ be parallel such that $m\circ h=m\circ k$. Let $x_{0}$ be the domain of $h, k$.
Suppose there exists $\phi : x_{0}\to x_{3}$ such that $g\circ\phi =m\circ h=m\circ k$. Then, since the above is a pullback square, there exist unique $u, v: x_{0}\to x_{1}$ such that
$$h=f\circ u, \phi =m'\circ u$$ and
$$k=f\circ v, \phi =m'\circ v.$$
Since $u, v$ are parallel and $m'$ is monic, $u=v$. Thus $h=f\circ u=f\circ v=k$. But what if $\phi$ doesn't exist?
AI: What you're trying to prove is wrong, if I read you correctly. You're trying to prove that if $m'$ is monic then $m$ is if the diagram
is a pullback diagram. This is false. Consider the category of sets where $x_3 = x_1 = \emptyset$, $x_2$ is an arbitrary set with at least two elements and $x_4 = 1$. Then the diagram is clearly a pullback diagram but $m$ is not injective (not a mono).
On the other hand, if $m$ is monic then $m'$ will always be monic and this is the statement that is usually meant by "pullbacks of monics are monics".
To show that suppose you're given two parallel arrows $h, k$ such that $m' \circ k = m' \circ h$. Then consider the diagram
It should be easy to show that it commutes and that $f \circ h = f \circ k$ by using the fact that $m$ is a mono and that $m' \circ h = m' \circ k$. This should allow you to conclude that $h = k$. |
H: How to find the domain of $f\left(g\left(y\right)\right)=\sqrt{\left(\frac{x+1}{x-1}\right)^3-27}$
Please help me find the domain of the following equation.
\begin{eqnarray}
\\f\left(x\right)=\sqrt{x^3-27},\space \space \space g\left(y\right)=\frac{x+1}{x-1},\space \space \space find \space f。g\\
\end{eqnarray}
The following is my calculation process.
\begin{eqnarray}
\\f\left(g\left(y\right)\right)&=&\sqrt{\left(\frac{x+1}{x-1}\right)^3-27}\\
\\&=&\sqrt{\frac{\left(x+1\right)^3}{\left(x-1\right)^3}-3^3}\\
\\&=&\sqrt{\frac{\left(x+1\right)^3-\left[3\left(x-1\right)\right]^3}{\left(x-1\right)^3}}\\
\end{eqnarray}
Now, $x-1$ is not equal to 0 and found that when $x=1$, $f(x)$ does not exist.
\begin{eqnarray}
\\Using \space a^3-b^3&=&\left(a-b\right)\left(a^2+ab+b^2\right)\\
\\(4-2x)((x+1)^2+3(x+1)(x-1)+9(x-1)^2)&\ge&0\\
...
\\4-2x\ge/\le0 \space \space &and& \space \space 13x^2-16x+7\ge/\le0\\
...
\end{eqnarray}
Please give me some hints.
Thank your for your attention
The ans in book is (1, -4](exactly without mistyping)
AI: Don Antonio's answer is correct. If I may, I'd like to give a detailed explanation as to why this is so, just in case Don Antonio's answer is too difficult to follow for some who are still learning this stuff.
The domain of a function is the set of all allowed values you can put into the function, i.e. all allowed values of $x$. In this case, there are two parts of the function which may make a value of $x$ not allowed:
First, there is the fraction with denominator $x - 1$. Dividing by zero is not allowed, so this means $x = 1$ cannot be part of the domain.
Second, there is the square root. It is not allowed to put negative values in the square root (zero is okay), so whatever is under the square root needs to be larger than or equal to zero, so what is left to do is to figure out for which values of $x$ is
$\left(\frac{x+1}{x−1}\right)^3−27 \ge 0$
This is true for $\frac{x+1}{x−1} \ge 3$.
When solving an inequality, you should first check when the two sides are equal, ie solve $\frac{x+1}{x−1} = 3$. Multiplying by $x-1$ on both sides we get $x+1 = 3(x-1)$ which, as you may check is true if $x = 2$. Note that special care should be taken at $x = 1$ as well, since the fraction changes sign there.
The last step in solving the inequality is checking for which values of $x$ (smaller than 1, in between 1 and 2, larger than 2) the inequality $\frac{x+1}{x−1} \ge 3$ holds. This can be done by choosing values for $x$ in the respective intervals. If you do this, you'll see that $\frac{x+1}{x−1} \ge 3$ for $x \in (1, 2]$,
Hence, as Don Antonio said, the domain of the function is $(1, 2]$. |
H: Is $2^k = 2013...$ for some $k$?
I'm wondering if some power of $2$ can be written in base $10$ as $2013$ followed by other digits.
Formally, does there exist $k,q,r \in \mathbb N$ such that $$2^k=2013 \cdot 10^q+r \,\,\,; \,\,\,r<10^q $$
I'm not sure if it's true or not. I would go for a 'no', but I can't prove it.
Thanks for your help.
AI: Since $\dfrac{\log 2}{\log 10}$ is irrational, there are such $k$. We want
$$2.013 \cdot 10^m \leqslant 2^k < 2.014\cdot 10^m,$$
and taking the logarithm to the base $10$ that becomes
$$m + \frac{\log 2.013}{\log 10} \leqslant k\cdot\frac{\log 2}{\log 10} < m + \frac{\log 2.014}{\log 10}$$
or
$$\frac{\log 2.013}{\log 10} \leqslant \left\lbrace k\frac{\log 2}{\log 10}\right\rbrace < \frac{\log 2.014}{\log 10},$$
where $\{x\} = x - \lfloor x\rfloor$ is the fractional part. It is a well-known fact that for irrational $\alpha$ the sequence of $\{ k\alpha\}$ is dense in $[0,1]$.
Prelude> [n | n <- [1 .. 100000], take 4 (show (2^n)) == "2013"]
[1363,3499,5635,14664,16800,18936,30101,32237,34373,43402,45538
,47674,56703,58839,60975,63111,72140,74276,76412,85441,87577,89713] |
H: sum of square roots
I was wondering what the estimate for the value
$$
S= \sum_{j=1}^N \sqrt{j}
$$
is? Is there a way (or a formula) to estimate it well?
I am sure it is close to $\int_1^N \sqrt{t} dt$, but I guess I
was interested in estimating it better.
Thanks!
AI: Euler-Maclaurin series.
According to Maple, as $N \to \infty$
$$S = \frac{2}{3}\,{N}^{3/2}+\frac{1}{2}\,{N}^{1/2}+\zeta \left( -1/2 \right) +\frac{1}{24}\,N^{-1/2}-{\frac {1}{1920}}\,{N}^{-5/2}+{\frac {1}{9216}}
\,{N}^{-9/2}+O \left( {N}^{-13/2} \right)
$$ |
H: Negating statements / Finding $(A \cap B)',A \oplus B$ if $A=\{x \in\Bbb R \mid -3\le x\le0\}$ and $B=\{x \in \Bbb R\mid -1 < x < 2\}$
I am a bit new on this field and I am trying to solve some questions. I don't really think they are hard but there are some key points that I don't get it or I am stuck.
Lets see.
1) Write the negation of the sentences:
1a: "If the teacher is missing, some students will not do their homework."
My approach: Let $p=$ The teacher is missing; $q=$ some students; $r=$ do their h/w. That would be $(p \to (q \land \neg r) )$. I searched for "De Morgan's law", but didn't find anything about what to do with the negation of "$\to$".
1b. Find the negation of: $(\forall x \in D)(x+4 \le 8)$.
I really have no idea, but the obvious would be $(\not\exists x \notin D)(x+4 \le 8)$.
Also the negation of $\exists$ would be $\not\exists$, right?
2) $A=\{x \in\Bbb R \mid -3\le x\le0\}$ and $B=\{x \in \Bbb R\mid -1 < x < 2\}$
2a) Find $(A \cap B)'$.
I think this is simple, so the answer would be all the numbers $\in \Bbb R$, except the numbers $-1 < x \le 0$ (That would be their common numbers). That makes sense to me, I mean a lot of sense.
2b) $A \oplus B$.
No I cant even visualize this. I know the truth table of xor at least, but this cant really help me to continue.
I would really appreciate some input, at least for my answers if possible.
Thanks in advance!
AI: Since Peter already addressed question 1, let me comment on 2.
Your idea for 2a. is correct. So let us try to be a little more specific. We currently have:
$$(A\cap B)' = \{x \in \Bbb R\mid \text{not}(-1 < x \le 0)\}$$
The statement $\text{not}(-1<x\le 0)$ can be simplified: it must be that either $-1 \not <x$, i.e. $-1 \ge x$, or $x \not\le 0$, i.e. $x > 0$. In conclusion:
$$(A\cap B)' = \{x \in \Bbb R \mid x \le -1 \text{ or } x > 0\}$$
Now for 2b. By definition of $\oplus$, we have that: $$A \oplus B = \{x \mid x \in A \text{ or } x \in B, \text{ but not both}\}$$
which can be written as $$A \oplus B = (A\cup B)\setminus (A \cap B)$$
with $\setminus$ denoting set difference, $C \setminus D = \{x \in C: x \notin D\}$.
We had already determined $A \cap B$. Can you also do $A \cup B$ and after that, $A \oplus B$? I advise you to stay close to the definitions, write out everything meticulously, and simplify your answer as much as possible. That's the best way to get to grips with it. |
H: Number of digits of $2^{1000}$
A friend asks me to find the number of digits of $2^{1000}$. I tried to look for a pattern by calculating the first powers of $2$ but I didn't find it. How should I proceed? Thanks.
AI: Hint: Solve for $x$ using logs:
$$10^x = 2^{1000}$$
Spoiler
$x \approx 301$ digits, so round up and say $302$ digits. |
H: ${{p^k}\choose{j}}\equiv 0\pmod{p}$ for $0 < j < p^k$
$${{p^k}\choose{j}}\equiv 0\pmod{p}.\ \ \ \text{for $0 < j < p^k$ and p is prime}$$
I can show this for $k=1$ using the fact that in denominator all numbers are less than $p$. I need hint for proving this for $k>1$.
AI: $$\binom{p^k}j=p^k\frac{(p^k-1)(p^k-2)\cdot\ldots\cdot(p^k-(j-1))}{1\cdot 2\cdot\ldots\cdot j}$$
Now just check that $\;p\mid n\iff p\mid (p^k-n)\;$ so that all possible powers of $\;p\;$ in the fraction's denominator above cancel out with the corresponding powers of $\;p\;$ in the same fraction's numerator... |
H: The calculation of roots of complex numbers.
How to calculate the roots of $x^6+64=0$?
Or how to calculate the roots of $1+x^{2n}=0$?
Give its easy and understanble solution method. Thank you. In general, the results of "exp" are obtained.
AI: $$x^{2n}=-1=\cos\pi+i\sin\pi=\cos(2k+1)\pi+i\sin(2k+1)\pi$$ where $k$ is any integer
Using De Moivre's formula,
$$\implies x=\cos\left(\frac{(2k+1)\pi}{2n}\right)+i\sin\left(\frac{(2k+1)\pi}{2n}\right)$$ where $k=0,1,2\cdots, 2n-1$ as $2n$ degree equation has exactly $2n$ roots
In fact, we can make some further generalization on the set of values of $k,$ see here |
H: Differentiabilty of this function
I want to show that $(x^2+y^2)^{\alpha}$ is not differentiable for $\alpha\in(0,1)$. All other cases are pretty straightfoward.
AI: The non exstence of partial derivatives are enough to show that the function is not differentiable at $(0, 0)$. Namely,
$f_x(0, 0)=\displaystyle\lim_{h\rightarrow 0}\frac{f(h, 0)-f(0, 0)}{h}=\displaystyle\lim_{h\rightarrow 0}\frac{h^\alpha}{h}=\displaystyle\lim_{h\rightarrow 0}\frac{1}{h^{1-\alpha}}$ does not exist. Similarly, $f_y(0, 0)$ d.n.e. Of course, instead of $\alpha$ we should have $\alpha/2$ or $\alpha\in(1/2, 1)$. |
H: Is a set without limit points necessarily closed?
According to the definition on Rudin' Principles of Mathematicial Analysis, closed set is defined as:
$E$ is closed if every limit point of $E$ is a point of $E$.
Then I have a question: if a set has no limit point, is it necessarily closed? I think this idea doesn't contradict the definition of closed sets. Am I right?
And another question is #10 in Chap.2 of Rudin's:
Let $X$ be an infinite set. For $p\in X$ and $q\in X$, define
$d(p,q)=0$ when $p=q$. Otherwises,$d(p,q)=1$.
I have known that every set in the metric space is closed. To prove it, we can show that the complement of any set is open, which is quite trivial. My question is: how can we prove that every set is closed just by checking according to the definition of closed set, i.e. how can we show that every limit point of any arbitrary set $E$ in the metric space is a point of $E$?
Thanks in advanced!
AI: You are absolutely correct that a set $E$ without limit points is closed, vacuously, since the empty set of limit points of $E$ is necessarily a subset of $E$. In fact, this gives us an alternate way to show that every set in a discrete space (which is the sort of space $X$ you describe) is closed. Take any subset $E$ of $X$ and any point $x\in X$. The open $d$-ball centered at $x$ of radius $\frac12$ is simply $\{x\},$ which contains at most one point of $E$, so $x$ is not a limit point of $E$. Since $x$ was arbitrary, then $E$ is closed. |
H: The ring of integers in $\mathbb{Q}[\zeta]$ is $\mathbb{Z}[\zeta]$
I am working on a proof in the lecture of Milne "Proposition 6.2 b)" but there is a step I don't get:
We have an inclusion $\mathbb{Z}\hookrightarrow \mathcal{O}_K$ that induces the following isomorphism $\mathbb{Z}/(p)\to \mathcal{O}_K/(1-\zeta)$.
This means $\mathcal{O}_K=\mathbb{Z}+(1-\zeta)\mathcal{O}_K$.
I guess this should be easy to see, but where do we get this equation from?
The rest of the proof is clear.
I'd be happy if someone could help me with this equation!
All the best, Luca
AI: If you agree that $\mathbb Z/(p)\to\mathcal O_K/(1-\zeta)$ is an isomorphism, then you must agree that $\mathcal O_K=\{0,1,\cdots,p-1\}+(1-\zeta)\mathcal O_K$. This last is contained in $\mathbb Z+(1-\zeta)\mathcal O_K$. |
H: Is the intersection of a chain of covers also a cover?
Let $\mathcal{F}$ be a family of sets covering some set $X$, so $\bigcup_{F \in \mathcal{F}} F=X$.
Assume we take an infinite chain $\mathcal{X}$ of nested families $\mathcal{F}''\subset \mathcal{F}'\subset \mathcal{F}$ so for every $\mathcal{F}_1, \mathcal{F}_2 \in \mathcal{X}$ either $\mathcal{F}_1 \subset \mathcal{F}_2$ or the other way round, where inclusion is strict, such that we always have $\bigcup_{F \in \mathcal{F}'} F=X$ for all $\mathcal{F}' \in \mathcal{X}$.
Does it necessarily follows that $\bigcup_{F \in \bigcap_{\mathcal{F}' \in \mathcal{X}}} F=X$ or there is a counter example?
AI: As an explicit counterexample, consider:
$$X = \Bbb N,\quad \mathcal F_n = \{S \subseteq \Bbb N: S \text{ finite}, |S| \ge n\}$$
Then $\bigcap_n F_n = \varnothing$. |
H: Check if a given limit of a sequence is valid
Using the definition of the limit with vicinities (rough translation from my native language) , prove that:
$$ \lim_{n\to \infty} \frac {2^n+3}{2^n+4^n}=0 $$
For convenience I will take $a_n=\frac {2^n+3}{2^n+4^n}$
For this to be true I need to prove that inside any vicinity of the number 0 there are an infinity of terms from the sequence $(a_n)$.
In other words I need to prove that there exists a $n(\epsilon) \in \mathbb{N^*}$ such that $0-\epsilon < a_n < 0+\epsilon \ \forall \ \epsilon>0,n>n(\epsilon) $.
The first thing I observed is that $a_n$ si strictly positive so it will always be greater than $-\epsilon$.
What remains to be proven is this:$a_n < \epsilon \ \forall \ \epsilon>0,n>n(\epsilon) $
Here is where I got stuck, how should I prove this?
NOTE: I use the word vicinity as a rough translation of a term in my native language, I am not sure that is the right term in English. I would be grateful if someone could clarify this as I am not able to find anything related to this on Google.
AI: Note that for all $n\ge 2,$ we have $$\frac{2^n+3}{2^n+4^n}=\frac{2^n+3}{2^n(1+2^n)}\le\frac{2\cdot 2^n}{2^n(1+2^n)}\le\frac{2\cdot 2^n}{2^n\cdot 2^n}=\frac{2}{2^n}.$$ Can you take it from there? |
H: Show for a diagonalisable matrix that $\det A = \prod_{n=1}^n\lambda_i $
I am currently trying to solve this exercise:
Let A be a diagonalisable matrix with eigenvalues $\lambda_1,\lambda_2...\lambda_n.$ Show that $\det A = \prod_{n=1}^n\lambda_i $
I tried solved it for a $3x3$ matrix:
A diagonalisable matrix can be written as:
$$
\begin{pmatrix}
\lambda_1 & 0 & 0 \\
0 & \lambda_2 & 0 \\
0 & 0 & \lambda_3
\end{pmatrix}
$$
If we compute the determinant by the Rule of Sarrus
we get
$$\lambda_1\times\lambda_2\times\lambda_3 + 0 + 0 - 0 - 0 - 0$$
Therefore we can write:
$A = \prod_{n=1}^3\lambda_i $
My problem is how to show this generally?
AI: If $A$ is diagonalisable (i.e. it has distinct eigenvalues), then we can write:
\begin{align*}
Av_1 &= \lambda_1v_1 \\
Av_2 &= \lambda_2v_2 \\
& \vdots \\
Av_n &= \lambda_nv_n
\end{align*}
for distinct eigenvalues $\lambda_1,...,\lambda_n$ and linearly independent eigenvectors $v_1,...,v_n$.
Define a matrix $P = \bigg[v_1|...|v_n\bigg]$, where the columns of $P$ are the eigenvectors. Then
\begin{align*}
AP = A\bigg[v_1|...|v_n\bigg] = \bigg[\lambda_1 v_1|...|\lambda_nv_n\bigg] = \bigg[v_1|...|v_n\bigg]\left[\begin{matrix} \lambda_1 & & \\ & \ddots & \\ && \lambda_n\end{matrix}\right]
\end{align*}
i.e.
\begin{align*}
AP = PD
\end{align*}
where $D$ is a diagonal matrix with eigenvalues on its diagonal. Then $\det{A}\det{P} = \det{P}\det{D} \Rightarrow \det{A} = \det{D}$ |
H: What is the LCM of $3^{2001}-1$ and $3^{2001}+1$?
What is the LCM of $3^{2001}-1$ and $3^{2001}+1$?
I can not get whether the GCF is $2$ or more than that.
AI: Note that $b=3a+4$, where $a=3^{2001}-1$ and $b=3^{2002}+1$. So the greatest common divisor of $a$ and $b$ must divide $4$. So it is either $2$ or $4$. It is quite easy to see that $a$ isn't a multiple of $4$, so the GCD equals $2$. |
H: How to compute $\frac{t}{t+1}$ to the form $1-\frac{1}{t+1}$?
How to compute $\frac{t}{t+1}$ to the form $1-\frac{1}{t+1}$?
What else? Well.
Well can you use long division?
AI: HUGE EDIT IN RESPONSE TO OP CORRECTION:
$$\dfrac{t}{t+1} = \dfrac{t + 1 - 1}{t+1} = \dfrac{t + 1}{t+1} - \dfrac{1}{t+1} = 1 - \dfrac{1}{t+1}$$
When in doubt, remember your identities: try adding a weird form of zero (in this case 1-1), try multiplying by a weird form of 1, and in more advanced math, use your trig identities. |
H: A closed form for $\sum_{i=1}^{n} \prod_{k=1}^{i+2} (3k+2)$
I need to calculate the following expression. Is there any explanation to convert this expression into normal expression without those letters for sum and the product? Just normal expression.
$$ Z = \displaystyle\sum_{i=1}^{n} \displaystyle\prod_{k=1}^{i+2} (3k+2). $$
AI: As I said in my comment, this is a "normal expression". This notation comes up a lot in any type of math and so you should get used to it. The summation notation sums the enclosed expression from the lower bound to the upper bound. The product notation does the same but for products. So, for example, $$ \displaystyle\prod_{k=1}^{i+2} (3k+2) $$ multiplies $3k+2$ for $k=1$, $k=2$, etc., but to $i+2$. This yields an expression in terms of the upper bound $i$, which then gets summed by the $ \displaystyle\sum $. |
H: Comparing determinants of an odd-dimension matrix
If $\mathbf{A}$ is a square matrix with odd dimensions, can $|\mathbf{A}|=|-\mathbf{A}|$?
This is true if $\mathbf{A}$ is over the field $\mathbb{Z}_2$, but are there any other situations?
AI: If $A$ is $k\times k$ with $k$ odd, then $\mid -A\mid = (-1)^k\mid A\mid = -\mid A\mid$, so unless $-1 = 1$ in the field, this is possible only if $\mid A\mid = 0$. |
H: Linear operator $T$ on a finite dimensional vector space
Question:
A linear operator $T$ on a finite dimensional vector space is
diagonalizable if and only if the multiplicity of each eigenvalue
$\lambda$ equals the dimension of $E_\lambda$.
According to me the answer should be $true$, but my book says otherwise, and so, I know that I am probably wrong, but I can't figure out, what am I missing ?
Any help will be appreciated.
AI: I think I just understood what the author(s) meant.
If you consider a linear operator $T\colon V \to V$ where $V$ is a finite-dimensional vector space over a field $k$ that is not algebraically closed, then not all zeros of the characteristic polynomial $\chi_T$ of $T$ need to lie in $k$.
If only the zeros of $\chi_T$ that lie in $k$ are considered eigenvalues, then indeed $T$ need not be diagonalisable if the dimension of $E_\lambda$ equals the multiplicity of $\lambda$ as a zero of $\chi_T$, for example consider $T \colon \mathbb{R}^3 \to \mathbb{R}^3$ given by multiplication with
$$\begin{pmatrix}0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{pmatrix}.$$
The characteristic polynomial is
$$\chi_T(X) = X^3 - X^2 + X - 1 = (X-1)(X^2+1) = (X-1)(X-i)(X+i),$$
and the only real eigenvalue is $1$, the dimension of the corresponding eigenspace is $1$, as is the multiplicity of the eigenvalue. But $T$ is not diagonalisable over $\mathbb{R}$ (since it has non-real eigenvalues).
If you consider every zero of the characteristic polynomial (in an algebraic closure of $k$) as an eigenvalue, then you have the simple characterisation that $T$ is diagonalisable if and only if $\dim E_\lambda$ equals the multiplicity of $X-\lambda$ in the characteristic polynomial. (And things like this are why a lot of people consider every zero of the characteristic polynomial an eigenvalue; the theory becomes much simpler then.) |
H: Is $f: \mathbb{R}_l \to \mathbb{R}, f(x) = 1$ for $x\geq 0$ and $f(x) = -1$ for $x < 0$ continuous?
Let $\mathbb{R}$ be the set of real numbers with standard topology.
Let $\mathbb{R}_l$ is the set of real numbers with lower limit topology.
Is $f: \mathbb{R}_l$ $\to \mathbb{R} $ given by
$$
f(x) =
\begin{cases}
1, & \text{if $x$ $\geq$ 0} \\
-1, & \text{if $x$ $\lt$ 0} \\
\end{cases}
$$
continuous?
I know that $f$ is continuous if and only if $f^{-1}((a,b))$ is an open set in $\mathbb{R}_l$ for every open interval $(a,b)$.
It seems that if $(a,b)$ does not contain $1$ or $-1$, then $f^{-1}((a,b))$ is the empty set.
If $(a,b)$ contains $1$, then $f^{-1}((a,b))$ is all non-negative real numbers.
If $(a,b)$ contains $-1$, then $f^{-1}((a,b))$ is all negative numbers.
$(a,b)$ contains both, then $f^{-1}((a,b))$ is all real numbers.
In all cases, $f^{-1}((a,b))$ is in lower limit topo, so $f$ is continuous.
Is this right?
AI: Yes, this is correct. The easiest way to see it clearly is to regard $f$ as a function from $\Bbb R_\ell$ to $\{-1,1\}$ with the topology that $\{-1,1\}$ inherits from $\Bbb R$. This is the discrete topology, so the open sets in $\{-1,1\}$ are simply $\varnothing,\{-1\},\{1\}$, and of course $\{-1,1\}$ itself. The inverse images of these are $\varnothing$, $(\leftarrow,0)$, $[0,\to)$, and $\Bbb R_\ell$, respectively, all of which are open in $\Bbb R_\ell$, so $f$ is continuous.
By looking just at the range of $f$ (in its relative topology, of course) you avoid having to look at all of the open sets in $\Bbb R$: all you care about is their intersections with $\{-1,1\}$ anyway, since it’s those intersections that determine their inverse images under $f$. |
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