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H: antiderivative of $\sum _{ n=0 }^{ \infty }{ (n+1){ x }^{ 2n+2 } } $
I've proven that the radius of convergence of $\sum _{ n=0 }^{ \infty }{ (n+1){ x }^{ 2n+2 } } $ is $R=1$, and that it doesn't converge at the edges.
Now, I was told that this is the derivative of a function $f(x)$, which holds $f(0)=0$.
My next step is to find this function in simple terms, and here I get lost. My attempt:
$f(x)=\sum _{ n=0 }^{ \infty }{ \frac { n+1 }{ 2n+3 } { x }^{ 2n+3 } } $
and this doesn't seem to help.
I'd like to use the fact that $|x|<1$ so I'll get a nice sum based on the sum of a geometric series but I have those irritating coefficients.
Any tips?
AI: First, consider
$$
g(w)=\sum_{n=0}^{\infty}(n+1)w^n.
$$
Integrating term-by-term, we find that the antiderivative $G(w)$ for $g(w)$ is
$$
G(w):=\int g(w)\,dw=C+\sum_{n=0}^{\infty}w^{n+1}
$$
where $C$ is an arbitrary constant. To make $g(0)=0$, we take $C=0$; then
$$
G(w)=\sum_{n=0}^{\infty}w^{n+1}=\sum_{n=1}^{\infty}w^n=\frac{w}{1-w}\qquad\text{if}\qquad\lvert w\rvert<1.
$$
(Here, we've used that this last is a geometric series with first term $w$ and common ratio $w$.) So, we find
$$
g(w)=G'(w)=\frac{1}{(1-w)^2},\qquad \lvert w\rvert<1.
$$
Now, how does this relate to the problem at hand? Note that
$$
\sum_{n=0}^{\infty}(n+1)x^{2n+2}=x^2\sum_{n=0}^{\infty}(n+1)(x^2)^n=x^2g(x^2)=\frac{x^2}{(1-x^2)^2}
$$
as long as $\lvert x^2\vert<1$, or equivalently $\lvert x\rvert <1$.
From here, you can finish your exercise by integrating this last function with respect to $x$, and choosing the constant of integration that makes its graph pass through $(0,0)$.
|
H: why does the h in Torricelli's law (the form that relates height to time) go to zero rapidly?
https://class.coursera.org/calcsing-002/lecture/320
In the the above linked lecture at 5:44, we are trying to find how fast liquid leaks from a cone shaped tank. I understand the derivation but at the end it mentions that $h$, which is in $\mathcal O ( t_\mathrm{end}-t )^{ \frac{2}{5} }$, goes to zero rapidly, which I don't understand.
The derivation goes like this:
If the slope of the side of the cone is y=mx, the radius is x, the height is $y$, and the area is $$A(y)= \pi x^2 = \pi (y/m)^2$$
If this is a non-circular shape, we can just use whatever the area is as a function of height $$A(h) = \mathcal O(h^2)$$
$$\frac{dh}{dt} = -\frac{\kappa}{ h^2}\sqrt{h} = - \kappa h^{-\frac{3}{2}}$$
We can separate the variables.
$$h^{\frac{2}{3}} \, dh = - \kappa \, dt $$
$$\int h^{\frac{2}{3}} \, dh = \int - \kappa \, dt $$
$$\frac{2}{5}h^{\frac{5}{2}} \, dh = - \kappa t + c $$
The 2/5ths is absorbed into the constants
$$h = ( - \kappa t + c )^{\frac{2}{5}}$$
Here's my question:
From what I understand, a function $f(x)$ is in $O( g(x) )$ if $f(x)$ approaches zero as fast or faster than $g(x)$.
For example, when $x$ is close to zero, $f(x)=x^2$ is in $O(x^{\frac{1}{2}})$ because for every value of $x$, $f(x)=x^2 \le g(x)=x^{ \frac{1}{2} }$
Using a test value of $x=.01$, $x^2$ is smaller than $x^{\frac{2}{5}}$
$.01^2 - (.01)^{2/5}$
It seems like $h= ( x^{\frac{2}{5}})$ approaches zero slowly.
Initially, I thought $h= ( x_f-x)^{\frac{2}{5}}$ would be in $\mathcal O (x^{\frac{2}{5}})$
However, $ (x_f-x)^{\frac{2}{5}}$ can be written in binomial form and then expanded with a Taylor series. The leading term is ${x_f}^{\frac{2}{5}}$, which still doesn't seem to approach zero very fast. Why does the lecture say that $\mathcal O ( t_\mathrm{end}-t )^{ \frac{2}{5} }$ goes to zero quickly?
AI: Speed is the derivative w.r.t. time. If $h=K(t_e- t)^{2/5}$, then
$$
h'(t)=\frac{dh}{dt}=-(2K/5)(t_e-t)^{-3/5}.
$$
So when $t\to t_e$ from below, we have $h'(t)<0$ in accordance with the observation that the surface of the water is decreasing. Note that
$$
\lim_{t\to t_e-}h'(t)=-\lim_{t\to t_e-}\frac{2K}{5(t_e-t)^{3/5}}=-\infty.
$$
So the rate of descent of the surface grows without an upper bound as $t\to t_e$. That's fast in my book :-)
If the depth of the remaining liquid behaved like $Kt^2$, then the depth would, indeed, reach a small value earlier (in comparison to $Kt^{2/5}$), but this means that there is very little liquid remaining towards the end, and the depth is close to constant towards the end, i.e. not changing at all. I guess it is a matter of taste, which you would call fast. Whoever wrote that book at least appears to compare the speeds of change of depth.
|
H: Number of cusps of an modular curve $X_0(N)$
Let $X_0(N) = \Gamma_0(N) / (\mathbb{H} \cup \mathbb{P}^1(\mathbb{Q}))$. A lecture note (p. 2) lists the following (very easy!) formula for the corresponding cusps:
$$\nu_\infty = \sum_{d\mid N} \varphi(\gcd(d,N/d)).$$
Unfortunately, there is no source or derivation given for this formula. I know what a cusp is and that the set of cusps is finite for all subcongruence groups of $\operatorname{SL}_2(\mathbb{Z})$. However, I do not grasp why this formula for $\nu_\infty$ is correct.
AI: See Section 3.8 of Diamond and Shurman's "A first course in modular forms", page 103.
Alternatively, see Prop. 1.43 in page 24 of Shimura's "Introduction to the arithmetic theory of automorphic functions".
|
H: Normal bundle geometry
I am wondering what a normal bundle looks like given this definition for Riemannian manifolds:
Say $M = S^2$. $T_pM$ is the tangent plane ($\mathbb R^2$) at $p \in S^2$. Lets denote the north pole by $0$ and let $S$ be a ball of some radius around $0$.
Naturally, I expect the normal vector at $0$ to stick up perpendicularly to $T_0M$. But given the definition we're talking of normal to all vectors in $T_0S$. I don't see how this is possible. Could you explain to me what these normal vectors look like on $S^2$?
AI: Perhaps you are confusing a few notions: The normal bundle of a submanifold only makes sense for points in that submanifold, so if you are taking $S$ to be an open geodesic ball around the north pole, then the normal bundle at $0$ is trivial. Normals are taken to be with respect to the manifold the submanifold lives in; if you take a $1$-dimensional submanifold of $S^2$, you'll get a $1$-dimensional normal bundle (here's a crappy drawing of what the normal space at a point might look like), while if you take a $2$-dimensional submanifold, all directions will be tangent directions so the only possibility left for a normal direction is the zero one.
To get the case you are talking about, where the normal points upwards (or downwards), you could view $S^2$ as a submanifold of $\mathbb{R}^3$, embedded in the usual way. Then indeed the normal space $N_0 S^2 \subseteq T_0\mathbb{R}^3 \cong \mathbb{R}^3$ consists of the vectors in the vertical direction.
|
H: Is the quotients of a group of triangular distributed numbers still following a triangular distribution?
I have a group of numbers (about 10000 numbers) between 0.8 and 1.0 which follows simple triangular distribution (for example, lower limit: 0.8, upper limit: 1.0, mode: 0.9).
If I divide 2 by each number from this group, I got the quotients forming the second group of numbers which follows another distribution (i.e. lower limit: 2.0, upper limit: 2.5, mode: 20/9).
Here is my question: is the second distribution still a triangular distribution? If not, then how can I know quantitatively what kind of distribution it is?
I will appreciate very much for your kind answers!
AI: We work with your numbers $0.8$ and $1$. So the base of the triangle has length $0.2$. To make the area $1$, the height of the triangle needs to be $10$.
We are dealing with a random variable $X$ whose density function we could readily find. In order to follow what is coming, you need to draw a picture of that triangle.
Now we look at the random variable $Y=\dfrac{2}{X}$, and we want a description of the cumulative distribution function of $Y$, or maybe its density function, you did not specify. We go after the cumulative distribution function $F_Y(y)$. You can then differentiate if you need the density function. The distribution will not be triangular, though it it looks roughly triangular.
The effective range of $Y$ is, as you pointed out, $2$ to $2.5$. Let $F_Y(y)$ be the cumulative distribution function of $Y$, that is, $\Pr(Y\le y)$. It is clear that $F_Y(y)=1$ if $y\ge 2.5$, and $F_Y(y)=0$ for $y\le 2$. We now need to do some work to find $F_Y(y)$ for $y$ in the interesting interval, $2$ to $2.5$. We have
$$F_Y(y)=\Pr(Y\le y)=\Pr\left(
\frac{2}{X} \le y\right))=\Pr\left(X\ge \frac{2}{y}\right).$$
First deal with the case $y \ge \frac{2}{0.9}$. Then $\frac{2}{y}\le 0.9$. We want the probability that $X\ge \frac{2}{y}$. This is the area of the part of the triangle that is to the right of $\frac{2}{y}$.
The area of the part of the triangle that is to the left of $\frac{2}{y}$ is easier to find directly. This is a triangle with base $\frac{2}{y}-0.8$. The height of the triangle is $100(\frac{2}{y}-0.8)$. So the area of the triangle is $50(\frac{2}{y}-0.8)^2$, and therefore for $y \ge \frac{2}{0.9}$ (but $\le 2.5$) we have
$$F_y(y)= 1-50\left(\frac{2}{y}-0.8\right)^2.$$
Of course I don't trust my calculation. So let's check what happens at $y=\frac{2}{0.9}$, where the answer should be $0.5$. It is! The answer should be $1$ at $y=2.5$. Yup.
Now we need to deal with $y$ between $2$ and $\frac{2}{0.9}$. The probability that $Y\le y$ is the area of the part of our original triangle which is to the right of $\frac{2}{y}$. The base of this triangle is $1-\frac{2}{y}$. The height turns out to be $100$ times that, so for $2\lt y\le \frac{2}{0.9}$ we have
$$F_Y(y)=50\left(1-\frac{2}{y}\right)^2.$$
We can differentiate to find the density function $f_Y(y)$.
If you graph the resulting density, the distribution of $Y$ will probably look triangular to the casual eye. It isn't, we have the exact distribution above. But triangular is likely a good enough approximation for your purposes. If you went say $0.15$ to $1$ instead of $0.8$ to $1$, the departure from triangularity would be significant. Here it may not be.
Added: Explicitly, by differentiating the expressions for the cdf above, we find that $f_Y(y)$, the density function of $Y$, is in the interval $(2,2.5)$ given by:
$f_Y(y)=\dfrac{200}{y^3}(y-2)$ for $2\le y \le \frac{2}{0.9}$, and
$f_Y(y)=\dfrac{200}{y^3}(2-0.8y)$ for $\frac{2}{0.9} \lt y\lt 2.5$.
|
H: Is the number of irreducibles in any number field infinite?
Are there infinitely many irreducibles in the ring of integers of any algebraic number field ?
I tried to follow the same argument as we usually do for integers. Suppose there are finitely many irreducibles, say $p_1,\ldots ,p_n$ and let $\alpha :=1+ p_1\cdots p_n$. Now if $\alpha $ is not a unit then it must have an irreducible $p$ such that $p|\alpha$ but then $p$ can not be any of the $p_i$'s and we have a contradiction. $\textit{What if $\alpha$ is an unit ? Is it possible for $\alpha$ to be an unit ?}$
Of course, one can replace $p_1\cdots p_n$ by $p_1^{k_1}\cdots p_n^{k_n}$ for any $k_1,\ldots ,k_n\in\mathbb{N}$ and the same argument would go through.
AI: $\alpha$ can be a unit as the element $1+\sqrt{2}$ in the ring of integers $\mathbb{Z}[\sqrt{2}]$ shows. Here's a slight variation of your argument: Let $p_1,\ldots,p_n$ be all the irreducibles. Since $x=p_1\cdot\ldots\cdot p_n$ is integral over $\mathbb{Z}$ it satisfies:
$$
0\neq-a_0=x^n+\ldots+a_1x=x(x^{n-1}+a_{n-1}x^{n-2}+\ldots+a_1)=x\cdot y\in\mathbb{Z}.
$$
Taking the negative of $y$ if necessary wlog we assume $xy>0$. Then $1+xy$ is not a unit of $\mathbb{Z}$ (and so also not a unit of the ring of integers $R$). Now you can conclude with your argument.
|
H: Are the number of terms in an infinite series even or odd?
This question arose after I saw a youtube-vid where Grandi's series was discussed.It seems that the sum of the series will be 0 for an even, and 1 for an odd number of terms, where a term is defined as (-1)n, n indicating the n'th term.It seems that even when s is derived that the above tacit assumptions of even/odd is used.Is it valid to make these assumptions?
(My lay-opinion is that these assumptions are wrong, and therefor also the derived value of s (= 1/2), and that the series has no sum as stated in the wikipedia entry)
AI: You are exactly right. An "infinite number of terms" is neither even nor odd. With Grandi's series, you can only discuss the values of partial sums of the series - because the series itself does not converge, and therefore does not have any value.
A series is defined as the limit of its sequence of partial sums, if that limit exists:
$$
\sum_{n=0}^{\infty}a_n=\lim_{N\rightarrow\infty}\sum_{n=0}^{N}a_n
$$
For Grandi's series, the sequence of partial sums alternates between 0 and 1, and does not converge; hence the series itself is undefined.
|
H: Ambiguous notation for squared matrix
What does $A^{2}$ mean for square $A$? Is it $AA$ or $AA^{T}$? Sometimes, the result may differ.
Or there is no uniform approach?
AI: There's no ambiguity, $A^2=AA$, period.
|
H: Indefinite integral of $(2x+9)e^x$
What is the indefinite integral $\displaystyle\int (2x+9)e^x\,\mathrm dx$?
Attempt:
Integration by parts seems obvious.
$u = 2x + 9, \mathrm du = 2$
$\mathrm dv = e^x, v = e^x$
$uv - \int v\,\mathrm du$
$(2x+9)e^x - \int 2e^x$
$(2x+9)e^x - 2e^x$
This is wrong but I don't see why.
AI: As the comment mention, the result is not wrong; we can verify this by differentiation (using the product rule):
$$\frac{\mathrm d}{\mathrm dx}(2x+7)e^x = (2x+7)e^x + 2e^x = (2x+9)e^x$$
(Note that $(2x+9)e^x - 2e^x = ((2x+9)-2))e^x = (2x+7)e^x$.)
However, it is important to stress that for a function $u = u(x)$, one has:
$$\mathrm d u = u'(x) \,\mathrm dx$$
as opposed to simply $u'(x)$. When you get further in your mathematical studies, you will get to know that:
$\mathrm dx$ is a thing called a $1$-form;
For functions $u(x_1,x_2,\ldots,x_n)$, we have $\mathrm du = \dfrac{\partial u}{\partial x_1} \,\mathrm dx_1 +\ldots+\dfrac{\partial u}{\partial x_n}\,\mathrm dx_n$;
I hope that instigates some curiosity as to what these mysterious $\mathrm dx$s are, and what the intuition behind them should be.
|
H: Convergence of a complex power series
Let $a,b,c \in \mathbb C$ with $c \in \mathbb N$. Then I have to calculate the radius of convergence of the following power series:
$$
1+ \frac{ab}{c \cdot 1!} z + \frac{a (a+1)b(b+1)}{c(c+1)2!} z^2+ \frac{a(a+1)(a+2)b(b+1)(b+2)}{c(c+1)(c+2)3!}z^3 + \cdots
$$
Using the ratio-test I get that
$$
\left | \frac {a_{n+1}}{a_n} \right | = \left | \frac{(a+n)(b+n)}{(c+n)(n+1)}z \right |
$$
How can I proceed ? This is an exam-question and a given hint says to consider whether $a$ or $b$ are in $\mathbb N$ or not. I have no idea how to use the hint.
Thanks in advance.
AI: It would seem to me that
$$
\lim_{n\to\infty} \left| \frac{(a+b)(b+n)}{(c+n)(n+1)}z \right| = 0
$$
Regardless of $z$ because you're dividing a polynomial of degree $1$ by a polynomial of degree $2$. So, if you went through the ratio test correctly, the radius of convergence should just be $\infty$.
EDIT:
As per your comment, we now have
$$
\lim_{n\to\infty} \left| \frac{(a+n)(b+n)}{(c+n)(n+1)}z \right| =
\lim_{n\to\infty} \left| \frac{n^2+O(n)}{n^2+O(n)}\right| \cdot\left|z\right|=\left|z\right|
$$
Which means that although the ratio test fails when $\left|z\right|=1$, our radius of convergence is just $1$ (since the series converges for $|z|<1$).
I have no idea how to use the hint either; it seems you may have outsmarted the exam.
SECOND EDIT: I now understand the exam hint! The ratio test assumes $(a+n)$ and $(b+n)$ are always non-zero. In the event that $a$ or $b$ is a negative integer, the radius of convergence changes to $\infty$ since we will always have finitely many non-zero terms.
|
H: Find volume on a shape with base of an ellipse
I have an ellipse with area $\pi ab$ $a = 6$, $b = 4$ these are the axis lengths. I am suppose to compute the volume of a cone of height 12.
I tried many solutions but none of them worked and I don't know why. I would type them up but I doubt much could be learned. Basically I have been trying to use the fact that $\pi ab$ is the area that I can find $a$ or $b$ and then that ratio is my radius at any point. This doesn't really work though and I don't know why.
How do I do this?
AI: Assume the cone is vertical, its axis is the $z$-axis and its base is an
ellipse on the $xy$-plane, centered at $(x,y,z)=(0,0,0)$, with semi-major
axis $a$ and semi-minor axis $b$. The equation of the base is thus
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad z=0.$$
Let $h$ be the height of the cone. The
cone cross section at height $z$ is an ellipse with semi-major axis $$x_1=a\left( 1-\frac{z}{h}\right) $$ and semi-minor axis $$x_2=b\left( 1-\frac{z}{h}\right) $$ by similarity of (right) triangles, as shown in the following sketch:
The equation of the boundary of this cross section is
$$\frac{x^2}{x_1^2}+\frac{y^2}{x_2^2}=1,\qquad z=z.$$
Basically I have been trying to use the fact that $\pi ab$ is the area
The area $A(z)$ of this cross section is thus
$$
\begin{equation*}
A(z)=\pi x_1 x_2=\pi ab\left( 1-\frac{z}{h}\right) ^{2}.
\end{equation*}
$$
The volume is the integral of the area $A(z)$ from $z=0$ to $z=h$
$$
\begin{eqnarray*}
V &=&\int_{0}^{h}A(z)\, dz=\int_{0}^{h}\pi ab\left( 1-\frac{z}{h}\right) ^{2}dz
\\
&=&\pi ab\int_{0}^{h}\left( 1-2\frac{z}{h}+\frac{z^{2}}{h^{2}}\right) dz
\\
&=&\pi ab\left( h-h+\frac{h}{3}\right) =\frac{h}{3}\pi ab,
\end{eqnarray*}
$$
i.e.
$$V=\frac{1}{3}A_{\text{base}}\times \text{height},$$
as expected. For $a=6,b=4,h=12$, we have:
$$
\begin{equation*}
V=\frac{h}{3}\pi ab=96\pi.
\end{equation*}
$$
|
H: Proving that $f(x) = \frac{x^2}{1+\sin^2(1/x)},f(0)=0$ is continuous at $0$
I'm trying to prove that
$$f\left(x\right)=\begin{cases}
\frac{x^{2}}{1+\sin^{2}\left(\frac{1}{x}\right)}, & x\neq0\\
0, & x=0
\end{cases}$$
is continuous at 0.
I know that I should show that
$$\lim_{x\rightarrow0}\frac{x^{2}}{1+\sin^{2}\left(\frac{1}{x}\right)}=0$$
But I don't understand what is
$$\lim_{x\rightarrow0}\sin^{2}\left(\frac{1}{x}\right)=?$$
When $x$ goes to $0$, $\dfrac{1}{x}$ goes to infinity, but sine oscillates between $-1$ and $1$, hence I don't know what limit would be.
Could you, please, help me?
AI: You should apply squeeze theorem here. We have that for $x\in\mathbb{R}\setminus \{0\}$, $-1 \le \sin\left(\frac{1}{x}\right) \le 1$. So we have that $0 \le \sin^2\left(\frac{1}{x}\right) \le 1$. Let's make use of this property.
$$0 \le \lim_{x\rightarrow 0} \frac{x^2}{1+\sin^2\left(\frac{1}{x}\right)} \le \lim_{x\rightarrow 0} \frac{x^2}{1+0}.$$
Applying the limit on the right, we have that the overall limit is $0$.
|
H: If H, N are normal subgroups of G, then do all the commutators lie in the intersection?
Okay, I know that this is elementary, but, ah, well.
How do I show that if N and H are normal subgroups of a finite group G with coprime orders, then,
$xyx^{-1}y^{-1} \in H\cap N$ for all $x \in H, y \in N$?
I figured that, if all $xyx^{-1}y^{-1} \in H$, then, $HN/H$ is abelian. And further, that $HN/N$ is isomorphic to $N/ H\cap N$.
Thank you.
Edit: Yes, I also know that $H \cap N = (1),$ the point is eventually to use this fact to show that $xy=yx$.
AI: If $N$ is normal in $G$, and $y\in N$, then $xyx^{-1} \in xNx^{-1} = N$, so $xyx^{-1}y^{-1} \in Ny^{-1} = N$. (That is, if $N$ is normal in $G$, then $[N, G] \subseteq N$.) Apply similar reasoning to $H$. (If you like, $[N, H] \subseteq [N, G] \subseteq N$, but also $[N, H] \subseteq [G, H] \subseteq H$.)
|
H: If $a$ and $b$ are the roots of $0=3x^2+4x+9$, $(1+a)(1+b)$ can be expressed in the form $\large \frac uv$?
$u$ and $v$ are co prime positive integers. What is the value of $u+v$? This seems like an easy problem, but I can't figure out what I'm doing wrong.
$f(x)=3x^2+4x+9$
$f(x)=(x-a)(x-b)$
$f(-x)=(x+a)(x+b)$
$f(-1)=(1+a)(1+b)$
$f(-1)=3-4+9=8$
$\large \frac 81 =\large \frac uv$
$u+v=9$
The answer is supposed to be $11$. What am I doing wrong? Thanks.
AI: The formula $f(x)=(x-a)(x-b)$ is not correct. It should be $f(x)=3(x-a)(x-b)$. Then it works. ;)
|
H: Why don't the roots of this characteristic equation correspond to the given solution of this 2nd order ODE?
I am asked to solve
$$ y'' + 9y = 6\mathrm{sin}(3x) $$
using the method of undetermined coefficients.
The characteristic equation of this 2nd order ODE is
$$ \lambda^2 + 9\lambda = 0 $$
and its roots are $0$ and $-9$. These are both real and different, so the general solution of the homogenous equation ought to be
$$ y = c_1e^{0x} + c_2e^{-9x} $$
which is the same as
$$ y = c_1 + c_2e^{-9x} $$
In my table of reasonable guesses for the particular solution of the inhomogenous equation, it says that for output functions of the form $\textrm{sin}(\beta x)$, $\textrm{cos}(\beta x)$, the particular solution will be of the form
$$ y_p = A\cdot\textrm{sin}(\beta x) + B\cdot\textrm{cos}(\beta x) $$
if $i \beta$ is not a root of the characteristic equation, and
$$ y_p = x\left[ A\cdot\sin(\beta x) + B\cdot\textrm{cos}(\beta x) \right] $$
if $i \beta$ is a root of the characteristic equation.
So, given our output function $6\sin(3x) $, $\beta = 3$, and if we test it, substituting $3i$ in the characteristic equation, we get
$$(3i)^2 + 9(3i) = -9 + 27i \neq 0 $$
so it seems clear that $i \beta$ is not a root, and so I expect that the particular solution would be of the form
$$ y_p = A\cdot\textrm{sin}(3x) + B\cdot\textrm{cos}(3x) $$
Except that it's not, and it wasn't until someone told me to try substituting the output function in the ODE that I realized this. Only $ y_p = x\left[ A\cdot\sin(\beta x) + B\cdot\textrm{cos}(\beta x) \right] $ works. Is my formulary wrong, or have I misunderstood something here?
Using $ y_p = x\left[ A\cdot\sin(\beta x) + B\cdot\textrm{cos}(\beta x) \right] $ as the reasonable guess, the particular solution turns out to be $-x\cos 3x$.
So the general solution should be
$$ y = c_1 + c_2e^{-9x} -x\cos 3x $$
but when I try solving this ODE with various tools, I get
$$ y = c_1\cos(3x) + c_2\sin(3x) -x\cos(3x) $$
which I do not understand. The characteristic equation has, as near as I can tell, real (and distinct) roots, so why are there trigonometric identities in the solution to the homogeneous equation? More generally, where have I gone wrong here?
AI: You went wrong with the characteristic equation. Remember that $y$ is the $0^{th}$ derivative of $y$, so you should have
$$
\lambda^2 + 9 = 0
$$
|
H: Understanding summation decreasing index
I'me following some summation examples and I came to this situation
$$|4-4| + \sum_{n=1}^{\infty} |4\cdot0.1^n| = -4+4\sum_{n=0}^{\infty} 0.1^n$$
How do they get to the last result?
I thought that $|-4+4|=0$ and decreasing the index should become $\sum_{n=0}^{\infty} |4\cdot0.1^{n+1}|$
What am I doing wrong?
AI: $|4-4| = -4 + 4 = -4 + 4\cdot (0.1)^0$.
So $$|4-4|+\sum_{n=1}^{\infty} 4\cdot |0.1|^n = -4 + 4\cdot (0.1)^0 + 4\sum_{n=1}^\infty (0.1)^n = -4+4\sum_{n=0}^\infty (0.1)^n$$
|
H: Limit as $x\to 0$ of $x\sin(1/x)$
How to find limit as $x \to 0$ of $x\sin(1/x)$?
For $x^2\sin(1/x)$, I know it's $0$ since by the Squeeze theorem, $-x^2 \le x^2\sin(1/x) \le x^2$, but for $x\sin(1/x)$, I run into some problems when applying Squeeze theorem.
AI: To use the Squeeze Theorem, we do know that $0\leq|x \sin(1/x)|\leq|x|,\;$ so by the squeeze theorem $$\;\lim_{x\to 0} |x \sin(1/x)|=0,\;\implies \; \lim_{x\to 0} x \sin(1/x)=0.$$
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H: What is the relationship between the lengths of the binary and decimal representations of a number?
If a is 1024 bits, then how many digits will its decimal representation have?
AI: Bits are binary numerals. If you have an 8-bit numeral, say 10110001, that is exactly an 8-bit binary numeral. In fact "bit" is an abbreviation of "binary digit".
When converted to decimal, an $n$-bit binary numeral will have at most $kn$ decimal digits, rounded up, where $k = \log_{10} 2 \approx 0.301$. So an 8-bit binary numeral, say 10110001, will have around $8\cdot 0.301 = 2.408$ decimal digits—round up to 3. In this example, $10110001_2 = 177_{10}$, and 177 has three decimal digits.
A 1024-bit binary numeral therefore has about 309 decimal digits.
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H: Winding number of image curve
How many turns does $f(z) = z^{40} + 4$ make about the origin in the complex plane after one circuit of $|z| = 2$?
AI: First, note that your curve is homotopic to the image of $|z|=2$ under $g(z)=z^{40}$ in $\mathbb C\backslash 0$. (Draw a picture if you aren't convinced.) Because the curves are homotopic, their winding numbers are the same, so you can just compute the winding number of the image of $|z|=2$ under $g(z)$ to get the answer. This is easy.
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H: Does $\Gamma$ intersect $SL(2, \mathbb{R})$ transversely at $I$?
Identify the space of all $2 \times 2$ real matrices with $\mathbb{R}^4$ so that the matrix
$\left( \begin{array}{cc}
a & b\\
c & d\end{array} \right)$
corresponds to $(a, b, c, d)$.
Let $\Gamma$ denote the hyperplane in $\mathbb{R}^
4$ with equation $x_1 +x_2 +x_3 - x_4 = 0$.
Does $\Gamma$ intersect $SL(2, \mathbb{R})$ transversely at $I$?�
AI: At $\begin{pmatrix}a&b\\c&d\end{pmatrix}=I$, we have $$\begin{align}\frac{\partial}{\partial a}\det\begin{pmatrix}a&b\\c&d\end{pmatrix}&=d=1,\\\frac{\partial}{\partial b}\det\begin{pmatrix}a&b\\c&d\end{pmatrix}&=-c=0,
\\\frac{\partial}{\partial c}\det\begin{pmatrix}a&b\\c&d\end{pmatrix}&=-b=0,\\\frac{\partial}{\partial d}\det\begin{pmatrix}a&b\\c&d\end{pmatrix}&=a=1.\end{align}$$
hence the tangent space to $SL$ at $I$ is the orthogonal space to $(1,0,0,1)$.
On the other hand $\Gamma$ is th eorthogonal space to $(1,1,1,-1)$, hence these two tangent sapces span all of $\mathbb R^4$.
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H: Possibility of a number line that has variable density.
In my real analysis class, I have been informed that if you have a continuous line say $[0,1]$, and you do a mapping say $A\to B$ such that any element in $A$ is equal to $B^2$, where $A$ is every element in $[0,1]$ you would ultimately get a line that is denser toward the $0$ and less dense toward the $1$.
I have two question regarding this result,
Can we note that every element in $A$, say variable $a$ has both its square and square root already existing in the line $[0, 1]$, therefore the mapping is only an exchange of the position of points,
for example we know that if we have point $0.5$
both $0.5^2$ and $\sqrt{0.5}$ exist in $[0,1]$ therefore if we do the mapping, $0.5 \mapsto 0.25$ doesn't the other point $\sqrt{0.5}$ replaces $0.5$'s original position ?
How is it possible that you can have a denser part in a continuous line, does this not ultimately suggest that there will be a gap around the $1$'s side?
Any comment would be much appreciated
AI: All that is meant by the statement that the line is "more dense" towards 0 is this: say that you pick some small number $\epsilon>0$. Then the interval $[0,\sqrt{\epsilon}]$ all gets mapped (by $x\mapsto x^2$) in to the interval $[0,\epsilon]$; so, the preimage of $[0,\epsilon]$ under this map has total length $\sqrt{\epsilon}$.
On the other side, the numbers $[\sqrt{1-\epsilon},1]$ are mapped by $x\mapsto x^2$ to the interval $[1-\epsilon,1]$; so, in this case, the preimage of $[1-\epsilon,1]$ has length $1-\sqrt{1-\epsilon}$.
The interesting thing here: even though the intervals $[0,\epsilon]$ and $[1-\epsilon,1]$ have the same length... their preimages do not. In fact, for every $\epsilon\in(0,1)$, it turns out that $\sqrt{\epsilon}$ (the "size" of the preimage of $[0,\epsilon]$) is larger than $1-\sqrt{1-\epsilon}$ (the "size" of the preimage of $[1-\epsilon,1]$).
So, all that they really meant was that different intervals of $[0,1]$ get more or less "spread out" by the mapping $x\mapsto x^2$.
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H: The index of nilpotency of a nilpotent matrix
Let $A$ a matrix in $\mathcal{M}_5(\mathbb C)$ such that $A^5=0$ and $\mathrm{rank}(A^2)=2$, how prove that $A$ is nilpotent with index of nilpotency $4$? Thanks in advance.
AI: In the followings I'll use only words, but you should visualize the matrices I mention.
Jordan normal form tells you that $A=P^{-1}JP$, for some Jordan matrix $J$ and some invertible matrix $P$.
As you know, $\text{rank}(A)=\text{rank}(J)$, therefore $\text{rank}(J^2)=2$.
This tells you that the index of the only eigenvalue (which is $0$) is either $3,4$ or $5$.
Recall that JNF also tells you that the largest Jordan block on $J$ has size equal to the
index of $0$.
Now consider the possibilities for $J$ if the index is $5$. Square it and find a contradiction. Similarly if the index is $3$.
Conclude.
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H: First derivative of holomorphic function
I want to prove that $ |f'(z)| \le \frac{1}{1-|z|}$ where $f:B(0,1) \rightarrow B(0,1)$ is a holomorphic function. My idea was to use Cauchy's integral formula. The fact that $||f||\le 1$ might be helpful too. But I don't see how to get this $1-|z|$?
AI: It is rather easy to prove a stronger result. First, we prove the special case $z = 0$ using the Cauchy integral formula:
$$\lvert f'(0)\rvert = \Biggl\lvert\frac{1}{2\pi i} \int\limits_{\partial D_r} \frac{f(\zeta)}{\zeta^2}\, d\zeta\Biggr\rvert \leqslant \frac{1}{2\pi} \int\limits_{\partial D_r} \frac{1}{\lvert\zeta\rvert^2}\,\lvert\zeta\rvert\, d\varphi = \frac{1}{r}$$
for all $0 < r < 1$. Taking the limit $r \to 1$ proves the case $z = 0$.
Next, to handle the general case, consider the Möbius transformation
$$T \colon w \mapsto \frac{w + z}{1 + \overline{z}w}.$$
$T \in \operatorname{Aut}(\mathbb{D})$ is easy to verify, and obviously $T(0) = z$. Now consider $g = f \circ T \colon \mathbb{D} \to \mathbb{D}.$
By the special case we proved above, $\lvert g'(0)\rvert \leqslant 1$.
On the other hand,
$$g'(0) = f'(T(0)) \cdot T'(0) = f'(z)\cdot T'(0) \iff f'(z) = \frac{g'(0)}{T'(0)}.$$
Now,
$$T'(w) = \frac{d}{dw} \frac{w+z}{1+\overline{z}w} = \frac{(1+\overline{z}w) - \overline{z}(w+z)}{(1+\overline{z}w)^2} = \frac{1 - \lvert z\rvert^2}{(1+\overline{z}w)^2}$$
and hence $T'(0) = 1 - \lvert z\rvert^2$, and
$$\lvert f'(z)\rvert = \frac{\lvert g'(0)\rvert}{1 - \lvert z\rvert^2} \leqslant \frac{1}{1 - \lvert z\rvert^2} = \frac{1}{1 + \lvert z\rvert} \cdot \frac{1}{1 - \lvert z \rvert}.$$
A stronger result, and less easy to come by is the Schwarz-Pick lemma:
For every holomorphic $f \colon \mathbb{D} \to \mathbb{D}$, we have the inequality
$$\biggl\lvert\frac{f(z) - f(w)}{1 - \overline{f(w)}f(z)} \biggr\rvert \leqslant \biggl\lvert\frac{z-w}{1-\overline{w}z} \biggr\rvert$$
and the infinitesimal version (divide by $z-w$ and let $w \to z$)
$$\frac{\lvert f'(z)\rvert}{1 - \lvert f(z)\rvert^2} \leqslant \frac{1}{1 - \lvert z\rvert^2}, \quad z \in \mathbb{D}.$$
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H: Taylor or Maclaurin series for the factorial function?
I am new to Taylor/Maclaurin series and want to know if there is a series representation for the factorial function?
AI: If you mean the standard factorial function $!:\mathbb{N}\rightarrow\mathbb{N}$, a Taylor series cannot exist because $\mathbb{N}$ has no accumulation points so you cannot take derivatives of it. To see this, you could try evaluating the following difference quotient:
$$\lim_{x\rightarrow n}\frac{x!-n!}{x-n}.$$
Immediately you can see this limit is not well-posed because the factorial only takes integer values. You can't have $1.00001!$ for instance so you can't get arbitrary close to $n$.
You need some way to extend the idea of a factorial to the real numbers in order to take derivatives. One such generalization of the factorial to (almost all) real numbers is the Gamma function. For natural numbers, we have that $\Gamma(n+1) = n!$ and you can show this pretty easily. In this case, derivatives of all orders exist and you can have a Taylor series by evaluating around a certain point (that is not $0$ or a negative integer since the Gamma function is not defined for them).
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H: If F is a finite field, then $F^*$ is cyclic and $F=\Bbb{Z}_p(\alpha)$ for some $\alpha$.
From Galois Theory (Rotman):
If F is a finite field, then $F^*$ [which is the multiplicative group] is cyclic and $F=\Bbb{Z}_p(\alpha)$ for some $\alpha$.
Proof If $|F|=q$, take $\alpha$ to be a primitive (q-1)st root of unity.
I find this proof a bit confusing...
1) How do we know that F has a (q-1)st primitive root of unity, or any root of unity for that matter?
2) How do we know that $F=\Bbb{Z}_p(\alpha)$? I know that every finite field must have an isomorphic copy of $\Bbb{Z}_p$, but I don't understand how a finite field is equal $\Bbb{Z}_p(\alpha)$.
Thank you in advance
AI: Recall that every element of $\mathbb{F}_q$ is fixed by the Frobenius map $x\mapsto x^q$. But this means every element satisfies the polynomial equation $x^q-x=0$. If we throw out $0$, consider $x(x^{q-1}-1)=0$ and we get that every non-zero element is a $(q-1)$st root of unity. Conversely, since there are $q-1$ distinct non-zero elements we see that the groups are the same.
So $\mathbb{F}_q^*$ is the group of $(q-1)$st roots of unity which is cyclic and generated by a primitive root of unity. Depending on the author, $\mathbb{F}_q$ is the splitting field of $x^q-x$ over $\mathbb{F}_p$. By the previous part of the argument, if you adjoin a generator for $\mathbb{F}_q^*$, then you get all roots by taking powers. Thus $\mathbb{F}_p(\alpha)=\mathbb{F}_q$.
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H: How to max $f(D)$ over the space where matrix $D$ is diagonal?
I want to maximize some function $f(D).$
Obviously if there is no constraints, I can just form matrix $G$ by $G(D)_{ij} = \frac{\partial f(D)}{\partial D_{ij}}$ and solve $G(D) = 0$ for D.
However, if D is subject to constraints, for example D has {p,q} entries of D are fixed to zero. How should I solve this problem ? Can I even use $\frac{\partial f(D)}{\partial D_{i,j}}=0 (i\neq p, j \neq q) $ at all ?
I guess my question is , should the gradient be affected by the constraint ?
Thnaks a lot.
AI: You need to create the matrix $G(D)$ where $[G(D)]_{ij} = \frac{\partial f(D)}{\partial [D]_{ij}}$. If this is an unconstrained problem, then look for solutions to $G(D) = 0$.
Addendum: The domain in question is the set of diagonal matrices. Suppose $f$ is differentiable and $\Delta$ is diagonal, then $f(D+ \Delta) = f(D)+ \sum_i \frac{\partial f(D)}{\partial [D]_{ii}} [\Delta]_{ii} + o(\Delta)$. It follows that the derivative is $DF(D)(\Delta) = \sum_i \frac{\partial f(D)}{\partial [D]_{ii}} [\Delta]_{ii}$, which could be written as $DF(D)(\Delta) = G(D) \Delta$, where $G(D) = \operatorname{diag} (\frac{\partial f(D)}{\partial [D]_{11}},...,\frac{\partial f(D)}{\partial [D]_{nn}})$.
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H: Normal Distribution and a Discrete Amount
From what I understand about normal distribution is that you make a discrete number continuous by adding .5 which every way the question asks for.
What if you were to have a discrete number with a set amount you cannot have half of.
Would you keep it discrete?
Ex) say there were 30 prizes, would you calculate the z-score as 30 or 30.5?
If a game was attempted 100 times and the chance of winning is 30%, would 30 prizes be enough?
mean= np = 100*.2=20
standard deviation sqrt(100*0.2*0.8) = 16
Here is my dilemma:
z(30)
=(30-20)/16
=10/16
=0.63
0.63=0.7357
z(30.5)
=(30.5-20)/16
=10.5/16
=0.66
0.66=0.7454
SO, the chances of having enough prizes for 100 attempts is either 73.57 or 74.54 depending on that 0.5
AI: Suppose we have a binomially distributed random variable $X$, and we want to use the normal approximation. If "$n$" is quite big, say $1500$, the continuity correction makes no big difference, so it is of no great practical importance whether or not you use it. (But if an exercise wants you to use it, you have no choice.)
For middling $n$, such as $n=50$, the continuity correction often yields a more accurate approximation. Here is how I would apply it.
Method: Let $k$ be an integer. Then $\Pr(X\le k)$ gets approximated by the probability that the normal with the same mean $np$ and same variance $np(1-p)$ as $X$ is $\le k+0.5$. This latter probability is the same as the probability that $Z\le \frac{k+0.5-np}{\sqrt{np(1-p)}}$.
A similar continuity correction is made when we approximate a Poisson with not too small $\lambda$ by a normal. But the main place where you will be using the continuity correction is in the normal approximation to the Binomial.
A Caution: We do not automatically "add $0.5$." Here is an example. Use the normal approximation with continuity correction to approximate the probability that $X\lt k$, where $k$ is a given integer.
At first sight this looks like the approximation problem described under Method. But there is a difference, here we have $\lt k$, not $\le k$. So here is what we do. The probability that $X\lt k$ is exactly the same as the probability that $X\le k-1$. Now this looks exactly like the kind of problem discussed in Method, we have a $\le$.
So the required probability is approximated by the probability that a certain normal is $\le k-1+0.5$, in other words by the probability the normal is $\le k-0.5$.
So when you are using the continuity correction with the binomial, I would advise making sure to convert the problem into one that involves $\le$.
As another example, suppose we want to approximate the probability that $X\ge k$. Be careful in applying the continuity correction. We want $1$ minus the probability that $X\lt k)$. The $X\lt k$ problem has been just discussed.
I hope this short description has been helpful.
Remark: For the places where continuity correction can be helpful, it is less important than it used to be. For using software we can calculate binomial probabilities exactly.
About the prizes question, I can answer it properly only if I get the full exact statement of the problem.
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H: Small question regarding subspaces of order topology
When looking at $Y:=\left(0,1\right)\cup\left\{ 2\right\}
$ with the subspace topology induced by the order topology on $\mathbb{R}$ one immediately sees that $2$ is an isolated point of $Y$. Contrary to that, according to another thread I saw here when you look at $Y$ with the order topology induced by restricting the order on $\mathbb{R}$ to $Y$ then $2$ is not an isolated point. But to me it seems that in said topology since $2$ is the maximal element of $Y$ one gets that $\left\{ 2\right\} =\left(1,2\right]$ is an open set and thus $2$ should be isolated.
I have a feeling I'm missing something very silly and I apologize if the question might seem trivial to some.
AI: $2$ is not isolated in the order topology. To see this, let $U$ be a neighborhood of $2$ in the order topology. Recall that the open sets in the order topology are unions of open intervals and open rays, so $U$ must contain either an open interval or an open ray containing $2$. Clearly no open interval in $Y$ contains $2$, nor does any open ray in the negative direction, so $U$ must contain some open ray of the form $(a,\infty)$ which contains $2$. Note that $a$ must be in $Y$. Thus since $a<2$ we have $a<1$, so $\frac{a+1}{2}\in U$. Since every neighborhood of $2$ contains a point other than $2$, $2$ is not isolated.
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H: Power set and set of all mappings
I'm working with the Terence Tao's Analysis book. And I have a question in the part of set theory.
As power set axiom, Tao use the set of all function: "If X, and Y be sets. Then there exist a set which consist of all the functions from X to Y."
Using that axiom and the replacement axiom, I need to prove:
"Let X be a set. Then $$ \left \{ A : A \subseteq X \right \} \ $$ is a set"
I have worked in the next way, but I'm not satisfy with the result:
$$ Let\: \; P(X) : = \left \{ A : A\subseteq X \right \} $$
$$\ g: \left \{ 0,1 \right \}^{X}\rightarrow P(X)$$
$$ \forall f \left ( f \in \left \{ 0,1 \right \}^{X} \wedge g(f) := f^{-1} [\, \left \{ 1 \right \} \, ] \right ) $$
Then, using the axiom of replacement and the axiom of power set (as the book use it) I get the next:
$$ G: = \left \{ g(f) : f \in \left \{ 0,1 \right \}^{X} \right \} $$
And I supposed that I need: G = P(X)
Therefore
$$ A \in G \leftrightarrow f[A] = \left \{ 1 \right \} $$
And to some B be in P(X), we have: $$ B\in P(X) \leftrightarrow B \subseteq X $$ and as the images conserves the inclusion $$ f [B] \subseteq f[X] $$
I thought that only I needed to show it $$ f [A] \subseteq f[X] $$ but I'm stack here, so my approach was the next:
$$\bigcup_{f\in \left \{ 0,1 \right \}^{X} } f [X] = \left \{ 0,1 \right \} $$
$$ f[A] \subseteq \bigcup_{f\in \left \{ 0,1 \right \}^{X} } f [X] $$
and in other exercise I proved that the image conserves the union, so "I can conclude something a little odd (haha)"
$$ A\subseteq \bigcup X = X $$
I really don't feel comfortable with the result and also I don't know how to show :
$$ P(X)\subseteq G $$
(sorry for my mistakes, the English is not my mother language).
My question is indeed how can I prove that?
AI: You cannot begin by letting $\wp(X)=\{A:A\subseteq X\}$, because you don’t yet know that this object exists: that’s what you’re trying to prove. You do, however, know that $\{0,1\}^X$ exists. Let $\varphi(x,y)$ be the following formula:
$$\left(x\in\{0,1\}^X\land y=x^{-1}[\{1\}]\right)\lor\left(x\notin\{0,1\}^X\land y=0\right)$$
Then $\varphi(x,y)$ is functional: $\forall x\exists!y\,\varphi(x,y)$. Now you can apply replacement to conclude that there is a set $G$ such that
$$y\in G\leftrightarrow\exists x\in\{0,1\}^X\,\varphi(x,y)\;,$$
i.e., $$y\in G\leftrightarrow\exists f\in\{0,1\}^X\left(y=f^{-1}\big[\{1\}\big]\right)\;.\tag{1}$$
It remains to show that $\forall y(y\in G\leftrightarrow y\subseteq X)$, i.e., that this set $G$ really is the power set of $X$.
It’s straightforward to see that $\forall y(y\in G\to y\subseteq X)$. For the other implication, suppose that $y\subseteq X$, and define a function $f:X\to\{0,1\}$ that demonstrates (using $(1)$) that $y\in G$.
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H: Bijective map preserving inner products is linear
The question comes from Kaplansky's book Linear Algebra and Geometry on page 96 exercise 2
Let $V$ be a non-singular inner product space of characteristic $\neq2$. Let $T$ be a one-to-one map of $V$ onto itself, sending $0$ to $0$ and satisfying $(x-y, x-y) = (Tx - Ty, Tx - Ty)$ for all $x,y \in V$. Prove that $T$ is orthogonal.
That $T$ preserves inner products is easy, but I don't know how to prove that it is linear. Any help would be appreciated.
AI: Let $a, b \in F$, a field of char $\ne 2$ and $V$ be a $F$-vector space.
Noting that $T$ preserves inner products, we see
$0=\langle ax+by,z \rangle - a \langle x, z \rangle - b \langle y, z \rangle
=\langle T(ax+by)-aTx -bTy, Tz \rangle$ for any $x ,y, z \in V$.
As the inner product is non-singular and $T$ is onto, we have that $T(ax+by) -aTx-bTy = 0 \forall x,y \in V$ and $ a,b \in F$. This proves the linearity of $T$.
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H: Laplace-Beltrami operator on sphere.
Suppose that we have solution of
$$\delta d f = g$$
on sphere. Where $\delta d$ is Laplace-de Rham operator for functions, $f,g$ are scalar functions and $g$ has support on north hemisphere and it is non-negative there.
Than by Stokes theorem we have(I think that I have signs wrong but that does not solve the problem)
$$
\int_{\text{equator}}*df = \int_{\text{north hemisphere}} d*df = - \int_{\text{south hemisphere}} d*df$$
But
$$
\int_{\text{north hemisphere}} d*df = \int_{\text{north hemisphere}} *g \neq 0 $$
$$
\int_{\text{south hemisphere}} d*df = \int_{\text{south hemisphere}} *g = 0$$
What have I done wrong?? I can't really see it.
AI: There is no solution. By Stokes's Theorem, since the sphere $S$ is compact,
$$\pm\int_S \star g = \int_S d{\star}df = \int_{\partial S} \star df = 0\,,$$
since $\partial S = \emptyset$.
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H: A Covering Map $\mathbb{R}P^2\longrightarrow X$ is a homeomorphism
I came across the following problem: Any covering map $\mathbb{R}P^2\longrightarrow X$ is a homeomorphism. To solve the problem you can look at the composition of covering maps
$$
S^2\longrightarrow \mathbb{R}P^2\longrightarrow X
$$
and examine the deck transformations to show that the covering $S^2\longrightarrow X$ only has the identity and antipodal maps as deck transformations.
I've seen these types of problems solved by showing that the covering is one-sheeted. Is there a solution to the problem along those lines?
EDIT: Even if there isn't a way to do it by showing it is one-sheeted, are there other ways?
AI: What about using Euler characteristic? Euler characteristic is multiplicative for a covering map: If $E\to B$ is an $n$-sheeted covering space and $E$ is compact, then $\chi(E)=n\chi(B)$. Since $\chi(\mathbb RP^2)=1$, we're done.
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H: Check that $df_x(v) = (v,v).$
Here is a proof that I am totally different from my classmates'. So I am requesting for expert reference here. Thank you. :-)
Let $f: X \rightarrow X \times X$ be the mapping $f(x) = (x,x).$ Check that $df_x(v) = (v,v).$
$\begin{eqnarray*}
df_x(v) &=& \lim_{t \rightarrow 0} \frac{f(x + tv) - f(x)}{t}\\
& = & \lim_{t \rightarrow 0} \frac{(x+tv, x+tv) - (x,x)}{t}\\
& = & \lim_{t \rightarrow 0} \frac{(tv,tv)}{t}\\
& = & \lim_{t \rightarrow 0} (v,v)\\
& = & (v,v).
\end{eqnarray*}$
AI: If you replace $x + tv$ by a curve through $x$, $\gamma$ such that $\gamma(0)=x$ and $\gamma '(0)=v$, you would get the standard proof of the result.
In an arbitrary manifold $X$, that is the only way to make sense of the expression $x+tv$.
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H: When a 0-1-matrix with exactly two 1’s on each column and on each row is non-degenerated? [1]
Let $A$ be an $n\times n$ matrix with entries in the set $\{0,1\}$ which has exactly two ones in each column and two ones in each row. Give necessary and sufficient conditions for the rank of $A$ to be $n$.
AI: Let $G$ be an undirected bipartite graph with $2n$ nodes $p_1,\ldots,p_n,q_1,\ldots,q_n$, where $p_i$ is connected to $q_j$ if and only if $a_{ij}=1$. Since $A$ has exactly two $1$s on each column and each row, each node $p_i$ is connected to exactly two $q_j$s and vice versa. Therefore $A$ is a bipartite graph and it can be decomposed into disjoint union of cycles of the form
$$
p_{i_1}\to q_{j_1}\to p_{i_2}\to q_{j_2}\to \ldots p_{i_k}\to q_{j_k}\to p_{i_1},
$$
where $i_1,\ldots,i_k$ are distinct and $j_1,\ldots,j_k$ are distinct. In other words, by relabeling the rows and columns, $A$ is permutationally equivalent to a block-diagonal matrix whose diagonal blocks are of the form
$$
B_k = \pmatrix{1&&&1\\ 1&\ddots\\ &\ddots&\ddots\\ &&1&1}\tag{1}
$$
where $B_k$ is $k\times k$ with $k>1$. It is easy to see that
$$
\det(B_k)=\begin{cases}
0 &\text{ when } k \text{ is even},\\
2 &\text{ when } k\neq1 \text{ is odd }
\end{cases}.
$$
Therefore $A$ has full rank if and only if $A=P(B_{k_1}\oplus\ldots\oplus B_{k_m})Q$, where each block $B_{k_r}$ is of the form $(1)$ with an odd $k_r$ and $P,Q$ are some permutation matrices. Put it another way, $A$ has full rank iff its corresponding graph $G$ can be decomposed into a disjoint union of cycles of length $2$ modulo $4$.
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H: Primes in binary
Let $$S_n(k)=\{1\leq m\leq n: m\ \mbox{has $k$ ones in its binary representation and $m$ is prime}\}\ \\ \forall \ n\geq 2^k-1,\ k\geq1.$$ Let $\pi(x)$ be the prime number function. Then what can be said about $$f(k)=\lim_{n\rightarrow \infty} \frac{|S_n(k)|}{\pi(n)}? $$
AI: $\pi(n)$ is roughly $\frac{n}{\log n}$. We need at most $\log n$ bits to represent $n$ (or any number smaller than $n$). There are roughly $\binom{\log n}{k} \simeq (\log n)^k$ "$m$"s that have $k$ binary digits equal to $1$. Hence, $f(k) = 0,\,\forall k$. We do not need to count the primes.
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H: How do you solve this word problem?
The river is flowing from point A to B at a rate of 15 miles per hour. A boat moves on still water at 45 miles per hour. If it takes David 1 hour and 15 minutes to ride the boat on the river from A to B, how long does it take him to make the return trip from B to A?
Please explain the steps :) thankyou
AI: First, use what you are given about the time taken to travel from $A$ to $B$, and the given rates, to compute the distance $d$ between $A$ and $B$:
$$d = 1.25\;\text{hours}\;\cdot \left(\frac {15\;\text{miles}}{1\;\text{hour}} + \frac{45\;\text{miles}}{1\;\text{hour}}\right)$$
Above, we add the rate at which the river is moving to the rate at which the boat moves on still water, to get the overall rate at which the boat is traveling from point A to point B. (It's traveling with the current).
Then, try setting up the equation you need to solve for the time needed to travel distance $d$ from point $B$ to point $A$ at a rate of $(45\; \text{mph}\;- 15\;\text{mph})$. Here, we subtract the rate of the current, which is going in the opposite direction than the boat is, from the rate of movement of the boat in still water.
Try to set up the equation and then solving for the unknown, but desired time needed to travel from point B to point A. I'll be happy to check your progress, if you follow up in the comments below.
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H: When does $(\lim f_n)'=\lim f'_n$?
After defining
$BC^1(\mathbb R,\mathbb R):=\{f: \mathbb R \to \mathbb R \mid f \in C^1, \ \ \lVert f \rVert_\infty + \lVert f' \rVert_\infty < \infty \}$
and proving that it's complete, our lecturer made the following comment:
From the completeness of $BC^1$ immediately follows that if both $f_n$ and $f'_n$ converge uniformly then $(\lim f_n)'=\lim f'_n$.
I would highly appreciate an explanation on why should it follow.
AI: If you have a sequence $f_n \in \mathscr{C}^1(\mathbb{R})$ such that both $f_n$ and $f_n'$ converge uniformly, there exists an $n_0 \in \mathbb{N}$ such that
$$\bigl(\forall n, m \geqslant n_0\bigr) \bigl(\lVert f_n - f_m\rVert_\infty \leqslant 1 \land \lVert f_n' - f_m'\rVert_\infty \leqslant 1\bigr).$$
Then $(g_k)_{k \in \mathbb{N}}$ defined by $g_k = f_{n_0 + k} - f_{n_0}$ is a Cauchy sequence in $BC^1$.
By the completeness of $BC^1$, there exists a limit $g \in BC^1$ of $(g_k)$, that means $g_k \to g$ uniformly, and $g_k' \to g'$ uniformly.
But of course $g_k$ converges uniformly to $\lim f_n - f_{n_0}$, and $g_k'$ converges uniformly to $\lim f_n' - f_{n_0}'$, hence $\lim f_n' = (g' + f_{n_0}') = (g + f_{n_0})' = \bigl(\lim f_n\bigr)'$.
(Note: Neither the $f_n$ nor the $f_n'$ need to be bounded, we subtracted $f_{n_0}$ to obtain a sequence in $BC^1$.)
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H: A simple example of Lindelöf space.
Somebody can to give me a simple example of Lindelöf space?
Note. Lindelöf space is a topological space in which every open cover has a countable subcover.
AI: The natural numbers with the discrete topology.
Given an open cover, $U_i$ let $U_n$ be some open set such that $n\in U_n$, then $\{U_n\mid n\in\Bbb N\}$ is a countable subcover.
Although simpler example, perhaps, would be any compact set. I still think that you may benefit from a non-compact example.
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H: Cardinality of tautologies for propositional logic
I'm wondering how many tautologies there are in propositional logic. I'm thinking that it must be at least countable, since ($P_{1} \wedge P_{2} \wedge \cdots P_{n}) \models P_{i}$ should be a tautology for any natural number $n$ and any $i \in \{1,2,\cdots,n\}$, where each $P_{k}$ is a sentence symbol for $k \in \mathbb{N}$.
But there aren't uncountably any, are there? How would one explain/show that? I'm sure this has been asked before somewhere - any references would be appreciated. Thanks!
Sincerely,
Vien
AI: I am assuming that your language only has a countable number of propositional variables. In this case, you can easily show that the number of propositions to begin with is countable (hint: the set of finite strings over a countable language is countable).
Therefore there cannot be more than a countable set of tautologies. In the other direction, it's simpler to note that if $p_i$ is a propositional variable then $p_i\to p_i$ is a tautology. Therefore there are exactly $\aleph_0$ of them.
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H: Check the convergence of the series of matrices
$$ \sum_{k=1}^{\infty} ( 1/ k^2 ) A ^k $$ where
A =\begin{bmatrix}-1 & 1 \\ 0 & -1 \end{bmatrix}
AI: Hints.
Let $J=\pmatrix{0&1\\ 0&0}$. Then $A=-I+J$. Since $J^2=0$, in the binomial expansion of $A^k = (-I+J)^k$, only two terms remain.
The value of $\sum_{k=1}^\infty \frac{(-1)^k}{k^2}$ is known.
The value of $\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}$ is also known.
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H: Reference for the Law of the Unconscious Statistician?
Does anyone know of a reference (a book or journal article) for the Law of the Unconscious Statistician?
AI: Robert Israel gives a reference here. It is Ross' book "Introduction to Probability Models", but it is only in Ed. 1-3.
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H: Is it possible to have some number sequences that have no formula to solve them?
I'm by no means advanced at mathematics, but I'm trying to figure out a formula to get the nth value of the following sequence: $1,4,10,20,35,56,84$.
I'm using 'difference' tables to try and come up with a formula and I'm currently at the $n$-th term:
$$n^3-5n^3+25n^3-125n^3+625n^3+3750n^3+22500n^3+135000n^3+810000n^3+4860000n^3$$
I'm not sure if I'm using a bad method, or if I've gone wrong somewhere but it just seems like the number I'm multiplying by n is increasing with no sign of levelling out.
If I continue to use a difference table will I eventually reach a formula or is it possible this number will just continue to increase infinitely?
AI: Difference table looks like
\begin{array}{|c|c|c|c|}
1 &\\
& 3 &\\
4 & & 3\\
& 6 & & 1\\
10 & & 4 \\
& 10 & & 1\\
20 & & 5\\
& 15 & & 1\\
35 & & 6 \\
& 21 & & 1\\
56 & & 7\\
& 28 \\
84 \\
\end{array}
$1$-st column: given numbers.
$2$-nd column: differences: $3 = 4-1, \ \ 6 = 10-4, \ \ 10 = 20 - 10, \ \ldots$.
$3$-rd column: differences: $3 = 6-3, \ \ 4 = 10-6, \ \ 5 = 15-10, \ \ldots$.
$4$-th column is constant column, so it must be formula
$$
a_n = c_0 + c_1 n + c_2 n^2 + c_3 n^3.
$$
To find next value, we'll continue table:
\begin{array}{|c|c|c|c|}
1 &\\
& 3 &\\
4 & & 3\\
& 6 & & 1\\
10 & & 4 \\
& 10 & & 1\\
20 & & 5\\
& 15 & & 1\\
35 & & 6 \\
& 21 & & 1\\
56 & & 7\\
& 28 & & \color{red}{1}\\
84 & & \color{red}{8}\\
& \color{red}{36}\\
\color{red}{120}
\\
\end{array}
$\color{red}{1}$ $-$ because $4$-th column is constant;
$\color{red}{8} = 7+\color{red}{1}$;
$\color{red}{36} = 28+\color{red}{8}$;
$\color{red}{120} = 84+\color{red}{36}$;
If you want to find out formula, then you can create system:
$$
\left\{
\begin{array}{r}
c_0 + c_1+c_2+c_3 = 1; \\
c_0 + 2 c_1+4c_2+8c_3 = 4; \\
c_0 + 3 c_1+9c_2+27c_3 = 10; \\
c_0 + 4 c_1+16c_2+64c_3 = 20. \\
\end{array}
\right.
$$
System with Vandermonde matrix.
$$
\left\{
\begin{array}{r}
c_0 + c_1+c_2+c_3 = 1; \\
c_1+3c_2+7c_3 = 3; \\
2 c_1+8c_2+26c_3 = 9; \\
3 c_1+15c_2+63c_3 = 19. \\
\end{array}
\right.
$$
$$
\left\{
\begin{array}{r}
c_0 + c_1+c_2+c_3 = 1; \\
c_1+3c_2+7c_3 = 3; \\
2c_2+12c_3 = 3; \\
6c_2+42c_3 = 10. \\
\end{array}
\right.
$$
$$
\left\{
\begin{array}{r}
c_0 + c_1+c_2+c_3 = 1; \\
c_1+3c_2+7c_3 = 3; \\
2c_2+12c_3 = 3; \\
6c_3 = 1. \\
\end{array}
\right.
$$
Then we get:
$$
\left\{
\begin{array}{l}
c_3 = \dfrac{1}{6}; \\
c_2= \dfrac{1}{2}; \\
c_1 = \dfrac{1}{3}; \\
c_0 = 0. \\
\end{array}
\right.
$$
So,
$$a_n = \dfrac{n}{3} + \dfrac{n^2}{2}+\dfrac{n^3}{6} = \dfrac{2n+3n^2+n^3}{6}.$$
You can test this formula:
$a_1 = \dfrac{2+3+1}{6}=1$,
$a_2 = \dfrac{4+12+8}{6}=4$,
$a_3 = \dfrac{6+27+27}{6}=10$,
$\ldots \ldots$.
|
H: Show the given space is uncountable.
Let $X$ be a compact Hausdorff space without any isolated point. Show that $X$ is uncountable.
As $X$ is compact Hausdorff, it is normal. then for any two distinct points $x$ and $y$, we have a continuous map $f$ from $X$ to $[0,1]$ such that $f(x)=0$ and $f(y)=1.$
As $X$ is compact, $f(X)$ is closed and bounded in $[0,1]$. Now I want to show that $f(X)$ contains an interval by using the fact that $X$ has no isolated point. Some one Help me.
AI: One way to do this is via a version of the Baire Category Theorem, which states that every locally compact Hausdorff space is a Baire space, i.e. it is not the union of a countable set of nowhere dense subsets. Recall that compact implies locally compact, so $X$ is not the union of a countable set of nowhere dense subsets. But
$$X=\bigcup_{x\in X}\{x\}$$
and each $\{x\}$ is nowhere dense since $x$ is not an isolated point, thus $X$ must be uncountable.
|
H: What is the inverse of the function $x \mapsto \frac{ax}{\sqrt{a^2 - |x|^2}}$?
Let $B_a$ be the open ball{$x: |x|^2<a$} in $\mathbb{R}^k, |x|^2 = \sum x_i^2$.
What is the inverse of the function $x \mapsto \frac{ax}{\sqrt{a^2 - |x|^2}}$?
Here I want some justification to equate $|x|^2$ with $x^2$, I don't know to justify; but otherwise I don't know how to solve it.
AI: Try to think a minute about what this function does; as Rahul Narain stated, it multiplies a vector by a "scalar" which depends on the norm of the vector.
Second, remark the spherical invariance of this scalar; any vector with norm $r$ (origin to any point on the sphere of radius $r$) will have an associated scalar $\frac{a}{\sqrt{a^2 - r^2}}$.
Third, note that this scalar tends to $1$ when the radius tends to $0$, and that it tends to infinity when the radius tends to $a$. In other words, it "maps" the open $a$-ball to $\mathbb{R}^n$.
Now the only inverse you have to find is that of that last sentence... :) (Well, you do have to understand that mapping as well ;).
Ok, I gave you some time to think about it, now let's see the answer. As we said it's a mapping from $\mathbb{R}^n$ to the open $a$-ball, the question that drives us crazy is which one?
Take a look at the norm of the resulting vector. For any vector of $\mathbb{R}^n$ (this includes any vector in any open ball), we can write $\vec{x} = r\vec{u}$, where $\vec{u}$ is some unit vector, and $r$ is the norm of $\vec{x}$.
The norm of the "image" of $\vec{x}$ by our function (say $\vec{y}$) is $\frac{ar}{\sqrt{a^2 - r^2}}$. Because our application preserves the orientation, we can also write $\vec{y} = s\vec{u}$ (where $\vec{u}$ is the same than before), and we have $|y|=s=\frac{ar}{\sqrt{a^2 - r^2}}$.
If we simply inverse this last relation (see below) we obtain the inverse function, that is
$
\frac{a s}{\sqrt{a^2 + s^2}} \vec{u} = \frac{a y}{\sqrt{a^2 + |y|^2}}
$.
Details about the inversion:
$$
s^2 (a^2 - r^2) = a^2r^2 \iff a^2s^2 = r^2(s^2+a^2) \iff r = \frac{as}{\sqrt{a^2+s^2}}
$$
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H: Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$
Problem statement:
Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $.
, $n\in \mathbb{N}$
My progress
LHS is equivalent to $\frac{(2n-1)!}{(2n)!}=\frac{(2n-1)(2n-2)(2n-3)\cdot ....}{(2n)(2n-1)(2n-2)\cdot ....}=\frac{1}{2n}$ So we can rewrite our inequality as:
$\frac{1}{2n}\leq \frac{1}{\sqrt{2n+1}}$ Let's use induction:
For $n=1$ it is obviously true. Assume $n=k$ is correct and show that $n=k+1$ holds.
$\frac{1}{2k+2}\leq \frac{1}{\sqrt{2k+3}}\Leftrightarrow 2k+2\geq\sqrt{2k+3}\Leftrightarrow 4(k+\frac{3}{4})^2-\frac{5}{4}$ after squaring and completing the square. And this does not hold for all $n$
About convergence: Is it not enough to check that $\lim_{n \to \infty}\frac{1}{2n}=\infty$ and conclude that it does not converge?
AI: First, note that $2\cdot 4 \cdot 6\cdots (2n)=2^n(n!)$. Next, note that if we multiplied $1\cdot 3\cdot 5\cdots (2n-1)$ by $2\cdot 4\cdot 6\cdots (2n)$, that would exactly fill the gaps and produce $(2n)!$. Hence, the denominator of the LHS is $2^nn!$, while the numerator of the LHS is $\frac{(2n)!}{2^nn!}$ Combining, the LHS equals $$\frac{1}{2^{2n}}\frac{(2n)!}{n!n!}=2^{-2n}{2n\choose n}$$
This is a central binomial coefficient, which are well-studied. For example, one bound given is that ${2n \choose n}\le \frac{4^n}{\sqrt{3n+1}}$; applying it in this case gives $$LHS=4^{-n}{2n\choose n}\le \frac{1}{\sqrt{3n+1}}\le \frac{1}{\sqrt{2n+1}}$$
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H: Number of connected sets on $\mathbb{R}^n$
Let $\mathcal{A}_n$ denote the family of all connected sets over $\mathbb{R}^n$. If $n=1$ we have that the cardinality of $\mathcal{A}_1$ is the cardinality of the continuum. But if $n>1$ I've only shown that $\mathcal{A}_2$ is at least the cardinality of the continuum. If $n> 1$ does the cardinality of $\mathcal{A}_n$ is $2^{\mathbb{c}}$? or does it exist an $n$ such that $|\mathcal{A}_n| = 2^{\mathbb{c}}$?
AI: You are right for $n=1$, and the same is true for path connected in place of connected. For $n>1$, the cardinality of the connected spaces, and even the path connected spaces, is always $2^c$ because one can easily see that $\leq$ holds. For $\geq$ one needs to construct a $2^c $ family of path connected subsets. Consider the set of subsets of $\mathbb{R}^2$ of the form
$\{(x, y)|x=0\}\cup\bigcup_{r \in S} \{(x, y)|y=r\}$ for any subset $S$ of the reals.
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H: Proof Help: In a group $G$, there exists a $g$ such that $g^2 = e$
I'm working through an Abstract Algebra book to teach myself, and came across the problem:
Prove: If $G$ is a finite group of even order, then there exists a $g\in G$ such that $g^2 = e$ and $g \ne e$.
(In this book, $e$ is used as the identity element. I don't know if that's standard or not...)
I have a proof outline, but don't really know how to write it in a formal way. My idea is as follows:
First, note that this problem is equivalent to saying that "there exists a $g\in G$ such that $g = g^{-1}$. Also note that the identity element is its own inverse.
Since the identity element is its own inverse, we have an odd number of elements remaining in the group that need their inverse "assigned." Assign all but one of the remaining elements an inverse so that none satisfy $g = g^{-1}$.
You have one remaining element left; the rest of the elements already have inverses. As inverses are unique, this element must be its own inverse.
My question is twofold:
Am I even close to being on the right track as far as a proof outline goes?
If so, what can I do to make this proof rigorous, and not just an outline?
AI: You've done the hard part (of recognizing what's happening). The linked questions will help you nail it, though. The key idea is noting the existence of a partition of finite $G$ into two groups: those elements which are their own inverses, and those which are not.
Indeed, what we can conclude is more general: that a finite group $G$ of even order must have an odd number of elements $g\in G$ such that $g^{2} = e, \;g\neq e$. (Put differently, a finite group of even order must have an even number of elements such that $g^2 = e$, and one of those elements of course includes $e$).
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H: Proving that a graph is NOT bipartite
For any $n,k\in \mathbb{N}$ where $n\geq k$ let $G_{n,k}$ be the graph where $V(G_{n,k})$ is the set of all $k$-subsets of $[n]$, and two subsets $S,T$ are adjacent iff $|S\cap T|=1$.
Prove that when $n\geq 3k-3 > 0$, $G_{n,k}$ is not bipartite.
This is homework, so I don't want a full solution, just some hints. I have already derived that the number of vertices is ${n \choose k}$ and the number of edges, when there are edges, is $\frac{k^2\binom{n-k}{k}}{2}$. Is this correct? How can I proceed?
AI: Hint: Find a triangle in $G_{n,k}$; this proves it's not bipartite.
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H: Why is normal distribution more accurate than binomial distribution?
I'm having a tough time understanding this.
This is what I am told about comparing the two:
The probability that Saredo is late for school is 0.6.
What is the probability that in one month she is late 9 times?
Remember that one month would include 20 school days.
Using the binomial distribution
n = 20 and p = 0.6
So p(9) = 20C9 0.69 0.411 = 0.0710 = 7.10%
Using the normal distribution
n = 20 and p = 0.6
μ = np = 20 × 0.6 = 12
σ = √np(p − 1) = √20 × 0.6 × 0.4 = 2.19
The boundaries we will use to find the probability of 9 will be 8.5 and 9.5.
z(9.5) = 9.5 − 12 = -1.14
2.19
z(8.5) = 8.5 − 12 = -1.60
2.19
Using the table, -1.14 gives us a probability of 0.1271
-1.60 gives us a probability of 0.0548
0.1271 − 0.0548 = 0.0723 = 7.23%
The Comparison
The binomial distribution gave us a probability of 7.1% (to one decimal place).
The normal distribution gave us a probability of 7.2%
This shows us that the normal distribution can give us accurate approximations. However, in this case, it clearly involved more work.
I do not understand why it is more accurate, can anyone shed some light on this?
AI: The reason is because the normal distribution is actually a pretty good approximation for the binomial distribution.
If you look at the shape of a binomial distribution, as your $n$ gets large, it begins to look more and more like a normal distribution. Although it seems like more work, when $n$ is very large, it's actually much easier to use the normal approximation.
The mathematical reasons behind this coincidence are described by the de Moivre-Laplace Theorem, and it is basically a special case of the central limit theorem -- which turns out to be immensely important in statistics!
Now, because the normal distribution is a continuous distribution, you will probably compute an answer to arbitrarily many decimal places. But this does not mean the result is more accurate. In fact, the answer is always less accurate, because the binomial distribution gives us the exact result.
The issues of computing the exact result numerically in the binomial distribution, due to the large factorials, can induce a source of numerical error if you are not careful, but if you imagine yourself with a magical computer with infinite precision and memory, the binomial distribution will always return the exact error for problems of this type.
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H: Is it solvable? Six venn diagram problem. Very Complicated
It classifies 10000 people as
young or old
male or female
married or single
Of these 10000 people, 3000 are young, 4600 are male, 7000 are married
1320 = young and male, 3010 = married and male, 1400 = young and married
600 = young and married and male.
Y = young, O = old, M = male, F = female, W = married, S = single
Find how many are young and female and single.
Then I get
Y = 3000, O = 7000, M = 4600, F = 5400, W = 7000, S = 3000,
M and Y = 1320, M and W = 3010, Y and W = 1400,
F or O = 8680, F or S = 6990, O or F = 8600,
M and Y and W = 600, F or O or S = 9400
Y and F and S = ? (what we are looking for)
I need to find the intersection of Y and F and S
Y and F and S = 1000 - (O or M or W)
O or M or W = O + M + W - O and M - M and W - O and W + O and M and W
but I failed to get any numbers related with O and the other.
How would I get it? This is solvable? I am stuck on this for several hours but can't think of any solution.
Please help me~
AI: First, to find how many young females there are: there are 3000 young people, and 1320 young men. So, there must be 1680 young women.
Now, there are 1400 people who are both young and married, and 600 that are young, married, and male; hence there are 800 people who are young, married, and female.
Finally, the number of young, female, single people is the number of young female people minus the number of young, female, married people; so, there must be $1680-800=880$ people who are young, female, and single.
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H: Interval iff image is interval
Show that a nonempty set $E$ of real numbers is an interval if and only if every continuous real-valued function on $E$ has an interval as its image.
Suppose $E$ is an interval. For any $a,b\in E$ with $a<b$, we have that any $c\in(a,b)$ belongs to $E$. Let $f$ be a continuous real-valued function on $E$. Take two elements $f(a)$ and $f(b)$ in its image, and let $x\in(f(a),f(b))$. By the intermediate value theorem, there exists $c\in E$ such that $f(c)=x$. So the image is an interval.
What about the converse?
AI: $(\Rightarrow)$ Continuity preserves connectedness. Connected subsets of $\mathbb{R}$ are intervals.
$(\Leftarrow)$ The image $f(E)$ is an interval for every continuous $f$. In particular, $i(E) = E$ is an interval where $i$ is the inclusion.
|
H: Does $S(x) = x+1$ always hold in Peano arithmetic?
In some books, they seem to implicitly say that $S(x) = x+1$ holds always in Peano arithmetic. But does it really hold in all cases, even in non-standard ones? The standard model of course satisfies this, but non-standard ones don't seem obvious.
Also, in Peano arithmetic, some books also say that $0<1$ all the times. Is this true for all non-standard models of Peano arithmetic?
If all these are true, is this solely due to axiom schema of induction?
AI: Note that $1$ is a shorthand for $S(0)$. Our language only contains the symbols $0,S,+,\cdot$ and no symbol for $1$.
The axioms for addition tell us that $x+0=x$ and $S(x+y)=x+S(y)$. Therefore we have:
$$S(x)=S(x+0)=x+S(0)=x+1.$$
As you can see I didn't use induction, at all.
|
H: Solve for the domain of x
Solve for the domain of x
$$y=\sqrt{\text{Cos}\left[x^2\right]}\tag1$$
my answer:where is wrong?
$\text{Cos}\left[x^2\right]\geq 0$, so $x^2\in [2k \pi -\pi /2,2k \pi +\pi /2]\Rightarrow x\in \left[\sqrt{2k \pi -\pi /2},\sqrt{2k \pi +\pi /2}\right]$
The answer in my book
a little strange for me
$\text{Cos}\left[x^2\right]\geq 0$, so $0\leq x^2\leq \pi /2 \&\&(4k-1)\pi /2\leq x^2 \le (4k+1)\pi /2\quad (k=1,2,\dots)$
so, the domain is $|x|\leq \sqrt{\frac{\pi }{2}}\&\&(4k-1)\pi /2\leq |x|\leq \sqrt{(4k+1)\frac{\pi }{2}}\quad (k=1,2,\dots)$
AI: Your reasoning and results regarding $x^2$ are fine, and you conclusions about the domain of $x$ are understandable, but we want to cover the case of negative values of $x$, which we can't do using $k\lt 0$, because the radicand of the square-root function must be non-negative. But we can cover both negative and positive $x$-values to include in the domain by expressing the domain of $|x|$ as the union of all your intervals, but restricting $k$ the range of $k$ to the positive integers: ($k \leq 0$ gives one or more undefined intervals).
$$|x| \in\left[0, \sqrt{\pi/2}\right] \cup \bigcup_{k \in \mathbb Z^+} \left[\sqrt{2k \pi -\pi /2},\sqrt{2k \pi +\pi /2}\right]$$
I've explicitly written the interval we need to join, through union, to the "big" union, since it's not covered by any interval for which $k \geq 1$.
|
H: Is there a way to standardize the Poisson distribution?
For example, a variable of Normal distribution, $T$, with mean $\mu$ and variance $\sigma^2$ can be standardized into $S$ like this:
$$
S=\frac{T-\mu}{\sigma}\;\Longrightarrow\;F(x)=\Phi\left(\frac{x-\mu}{\sigma}\right)
$$
My question is, for the Poisson distribution with probability function
$$
f(k;\lambda)=\Pr(X=k)=\frac{\lambda^k e^{-\lambda}}{k!}.
$$
Is there a way to standardize the $X$ if I define the standard Poisson Distribution as the distribution that $\lambda=1$?
AI: Yes, there is a standard Poisson, the one with parameter $1$. Recall that if $X$ counts the number of "accidents" in a unit time interval, then under suitable conditions $X$ has Poisson distribution with parameter the mean number of accidents per unit time.
If that parameter is $\lambda$, then the number of accidents in a time interval $t$ is Poisson with parameter $\lambda t$.
We adjust the interval of time over which we count, until we get a time that gives mean count $1$. Of course $t=\frac{1}{\lambda}$ is that time interval.
Like in the standardization of the normal, a change of scale is involved. But in the case of the Poisson it is not the random variable that is being scaled.
|
H: Word problem (food for thought)
I thought of this question today as I was coming home from work in my car (probably because of my parents' anniversary).
This problem assumes the parents of everyone in the world got married and everyone in the future gets married at some point in time (a.k.a. it's hypothetical). If every couple planned their wedding day to occur exactly halfway between the dates of the anniversary of their parents' weddings, would everyone eventually have the same wedding day?
To illustrate this problem more clearly:
$P_m=$ date of wedding for parents of the male
$P_f =$ date of wedding for parents of the female
$C = \frac {(P_m + P_f)}{2} =$ date of future wedding for couple
AI: Not necessarily. Suppose everyone married a person of the same nationality as themselves. You could have all marriages in England occurring on 5 June, in Australia on 12 November etc.
Adding on edit:
Create a model in which every month has two days, and the first round of weddings consist of a single wedding on the first of each month. Each couple has two children. The first child born to the March couple marries a February child, the second March child marries an April child. The weddings all happen on the second of the month, and they all have two children. If the Dec/Jan wedding happens on 2 Dec (see Billy's comment) then you have a system where the weddings in each generation swap between the first and the second of the month.
If the Jan/Dec wedding happened on 2 June, there could be no future weddings on 1 Jan or 2 Dec. Depending what happens to $C$ when $P_m+P_f$ is odd, you can see that whenever a person who belongs to the earliest date or the latest date marries someone who belongs to a date at least two days around, (and depending on the relative birth rates for couples belonging to the different dates) the proportion of the population attached to the extreme dates will tend to fall, and the range of dates on which marriages take place will narrow.
But for a fully specified problem you need to pin down the answer to Billy's question, how to deal with dividing odd numbers by 2 (it would be possible on some scenarios - eg resolving in the direction of the man - to have a stable system with marriages taking place on the same two adjacent dates in each generation) and also to say something about mixing relative to birthrate.
Even in a system where people are mixing, if the mixing rate is low and the birth rate at extreme dates is high, there can be a persistence of weddings at the earliest and latest date rather than a narrowing of the date range.
So that illustrates a few of the possibilities and issues.
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H: What is an effective means to get senior high school students to write their complete working out as part of their answer.
In Australia and in the International Baccalaureate (2 systems I have worked in), for better or worse, mathematics is assessed by criteria. This increases the importance of students to express their mathematical processing and problem solving strategies in their working out - as this is explicitly assessed as part of the criteria.
This is also a source of an infrequent, yet prevalent pedagogical-headache, which is the basis of my question - what strategies are there to effectively train senior high school students to compose legible, coherent and logical mathematical sequences leading to their solution?
AI: This is a very common issue with students. There are several main concerns :
Students do not like writing up their solutions.
Students do not benefit from writing up their solutions.
Students hate to be wrong.
What you can do, is to change the motivations for them to write up their answers.
On Brilliant, I get a 75% submission rate when asking students who got a numerically correct answer to explain their full solutions. Note that this is not 'homework', nor are they required to submit answers in order to proceed on the site. When compared to math circles where this percentage is often close to 0, it is remarkable. [When I was training the singapore IMO team, we had to force the younger students to submit at least 1 solution as an entrance ticket to next week's session. That was the only way we were getting write-ups from them.]
The student solutions I receive on Brilliant are often well crafted, and heavily thought through. Several even start to learn Latex (though it is not required), in order to improve their presentation. Their solutions are then graded on a scale of incorrect, incomplete, correct, excellent. The best solution (up to my interpretation) is presented the following week for others to learn from. What I believe we are doing right, is to influence the motivations for students to write their answers.
It should not be seen as a chore, but rather a privilege. This can arise by saying "If you can solve the problem, there is a random chance that you will be asked to submit your solution."
Students want to know how much they are improving, and in what areas they can improve. Give them an easy way to see what a good solution is and why. Don't simply point out "You made this mistake", but instead say that "Several of the solutions made the following mistakes".
Students want to seek validation from their peers. The fact that their solutions are presented to the community to read, provides strong motivation for them to learn how to improve their answer. I often deal with the lament of "Why isn't my solution chosen to be featured?", to why I'd point out the good aspects of the solution (be it choice of argument, style of writing, , etc.)
I would suggest that you try adopting this method, and see if it works for your class. I would love to hear feedback about it.
Of course, we also offer them points to submit solutions. However, if bribery is so effective, I'd suggest that you simply bring cookies into class.
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H: Given a matrix C, calculate $e^C e^{C^T}$ and $e^{-C}$.
Given:
$$\mathcal C=\pmatrix{-5&3&-3\\2&-4&1\\2&-2&-1}$$
Calculate: $e^C e^{C^T}$
Calculate: $e^{-C}$
I am using the formula $\mathcal f(C)=\mathcal S f(\lambda)\mathcal S^{-1}$, therefore I used Matlab to calculate $det(\lambda I-\mathcal C)=0$ to get my eigenvalues:
$\lambda_1=-5$, $\lambda_2=-3$ and $\lambda_3=-2$.
For $\lambda_1=-5$:
$$U_1=\pmatrix{-2&-1&-1\\0&-3&3\\0&0&0}$$
For $\lambda_2=-3$:
$$U_2=\pmatrix{2&-3&3\\0&-2&2\\0&0&0}$$
But for $\lambda_3=-2$:
$$U_3=\pmatrix{3&-3&3\\0&0&1\\0&0&0}$$
Therefore:
$$\Lambda=\pmatrix{-5&0&0\\0&-3&0\\0&0&-2}$$
And $$S=\pmatrix{-1&0&0\\1&1&0\\1&1&0}$$
Which gives a nonexistent $\mathcal S^{-1}$ !! Could someone please tell me what went wrong? I should be able to calculate the exponentials this way.....or is there another way to deal with this problem? Any help is greatly appreciated as always. :)
AI: Note that, you need to compute only $e^{C}$ in full details, then you can use relations between the eigenvalues of a matrix and its transpose. Here is the final answer you need to reach
$$e^{C} =\left[ \begin {array}{ccc} {{\rm e}^{-5}}&-{{\rm e}^{-5}}+{{\rm e}^{-
2}}&-{{\rm e}^{-2}}+{{\rm e}^{-5}}\\{{\rm e}^{-3}}-
{{\rm e}^{-5}}&{{\rm e}^{-5}}+{{\rm e}^{-2}}-{{\rm e}^{-3}}&-{{\rm e}^
{-2}}+2\,{{\rm e}^{-3}}-{{\rm e}^{-5}}\\{{\rm e}^{-
3}}-{{\rm e}^{-5}}&-{{\rm e}^{-3}}+{{\rm e}^{-5}}&2\,{{\rm e}^{-3}}-{
{\rm e}^{-5}}\end {array} \right] .$$
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H: Vector spaces and direct sums.
What is the relationship between $S \oplus T$ and $T \oplus S$? Is the direct sum operation commutative? Formulate and prove a similar statement concerning associativity. Is there an "identitiy" for direct sum? What about "negatives"?
$S \oplus T = T \oplus S$ because elements of $S \oplus T$ are of the form $s+t$ where $s \in S$ and $t \in T$, and we know that $s + t = t + s$. Associativity is valid with the same argument. The identity is $\{0\}$. However, we cannot have negatives, because a negative subspace doesn't make sense, since if $s \in S$ then $-s \in S$.
Let V be a finite-dimensional vector space over an infinite field F. Prove that if $S_1, ... ,S_k$ are subspaces of $V$ of equal dimension, then there is a subspace $T$ of $V$ for which $V = S_i \oplus T$ for all $I = 1, ... ,k$. In other words, $T$ is a common complement of the subspaces $S_i$.
We know that every subspace has a complement. So let's say that the complement of $S_m$ where $1 \leq m \leq k$ is $T$. Then $$\dim(V) = \dim(S_m) + \dim(T).$$
For any $S_n$ where $1 \leq n \leq k$, we know that $$\dim(S_n) + \dim(T) = \dim(S_n + T) + \dim(S \cap T).$$
Since $\dim(S_n) = \dim(S_m)$,
$$(\dim(V) - \dim(T)) + \dim(T) = \dim(S_n + T) + \dim(S \cap T)$$
$$\implies \dim(V) = \dim(S_n + T) + \dim(S \cap T) = \dim(S_n) + \dim(T)$$
So $T$ must also be a complement of $S_n$ for all $1 \leq n \leq k$.
Do you think my answer is correct?
Thank you in advance
AI: $$\implies \dim(V) = \dim(S_n + T) + \dim(S \cap T) = \dim(S_n) + \dim(T)$$
So $T$ must also be a complement of $S_n$ for all $1 \leq n \leq k$.
This conclusion is incorrect. For example, consider $V=M_2(\mathbb{R})$, two-by-two matrices with real coefficients. Set $$S_1=\left\{\left(\begin{smallmatrix}a&b\\c&0\end{smallmatrix}\right):a,b,c\in\mathbb{R}\right\}$$
$$S_2=\left\{\left(\begin{smallmatrix}a&b\\0&c\end{smallmatrix}\right):a,b,c\in\mathbb{R}\right\}$$
$$S_3=\left\{\left(\begin{smallmatrix}a&0\\b&c\end{smallmatrix}\right):a,b,c\in\mathbb{R}\right\}$$
$$S_4=\left\{\left(\begin{smallmatrix}0&a\\b&c\end{smallmatrix}\right):a,b,c\in\mathbb{R}\right\}$$
Then $T=\left\{\left(\begin{smallmatrix}0&0\\0&d\end{smallmatrix}\right):d\in\mathbb{R}\right\}$ has $S_1\oplus T=V$, but $S_2\oplus T\neq V$.
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H: find the power representation of $x^2 \arctan(x^3)$
Wondering what im doing wrong in this problem im ask to find the power series representation of
$x^2 \arctan(x^3)$
now i know that arctan's power series representation is this
$$\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} $$
i could have sworn that for solving for this i could just have use that formula and then distribute the $x^2$ but i'm getting the wrong answer here is the steps im taking.
1- plug in $x^3$ for $x$
$$x^2\sum_{n=0}^\infty (-1)^n \frac{(x^3)^{2n+1}}{2n+1} $$
2- distribute the $2n+1$ to $x^3$
$$x^2\sum_{n=0}^\infty (-1)^n \frac{x^{6n+3}}{2n+1} $$
3- distribute the $x^2$
$$\sum_{n=0}^\infty (-1)^n \frac{x^{6n+5}}{2n+1} $$
however this is wrong according to my book the answer should be
$$\sum_{n=0}^\infty (-1)^n \frac{x^{3n+2}}{n} $$
can someone please point out my mistake. Please forgive me in advance for any mathematical blunders that i post. I'm truly sorry.
Thanks
Miguel
AI: The book's answer would be right if it said
$$
\sum_{\text{odd }n\ge 0} (-1)^{(n-1)/2)} \frac{x^{3n+2}}{n}.
$$
That would be the same as your answer, i.e.
$$
\sum_{\text{odd }n\ge 0} (-1)^{(n-1)/2)} \frac{x^{3n+2}}{n} = \sum_{n\ge0} (-1)^n\frac{x^{6n+5}}{2n+1}.
$$
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H: Filling in 'x' in a log function
if $3^5=x$ (exponential equation) converts to log form gives $log_3x=5$
that makes sense.
$$
3^5 = 243 \Rightarrow x=243
$$
So if I take the log form again: $log_3x=5$ and replace $x$ with $243$. I then take the log of $243$, expecting to get $5$?? But instead, I get $2.3856$??
Can anyone explain?
AI: The log button on a calculator takes the base-$10$ log of the number. You want the base-$3$ log, so you either have to use the change of base formula or use an fancier or online calculator like Wolfram Alpha.
To input a log of a certain base in Wolfram alpha, use the underscore _. So, typing: log_3(243) gives the result 5. (Or, as pointed out below, the underscore isn't really needed. I still like to use it for readability purposes.)
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H: Is it true that $cl_{\beta X} Z(f) = Z(\hat f)$?
Suppose $f\in C^* (X)$ and let $\hat f$ be its continuous extension to $\beta X$. It is clear that
$cl_{\beta X} Z(f)=cl_{\beta X} f^{-1}(${$0$}$)\subseteq \hat f ^{-1}(${$0$}$)= Z(\hat f)$, but I am having some difficulty showing the reverse inclusion.
AI: It is not necessarily true. Consider, for instance, $X=(0,1)$, and $f(x) = x$. Then $Z(f) = \emptyset$, but $Z(\hat{f})$ is not empty; for instance, the set $\{\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots\}$ has a limit point $y \in \beta X$ by compactness, and by continuity we must have $\hat{f}(y) = 0$.
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H: Proof of a Property of Vertical Asymptotes
I'm trying to understand a proof in my Calculus textbook of the following theorem:
Let the functions $f$ and $g$ be continuous on an interval containing
$c$. If $f(c) \neq 0$, $g(c) = 0$, and there is an open interval
containing $c$ such that $g(x) \neq 0$ for all $x \neq c$ in the
interval, then the graph of the function $h(x) = \frac{f(x)}{g(x)}$
has a vertical asymptote at $x = c$.
The proof the textbook provides is the following:
Consider the case for which $f(c) > 0$, and there exists $b > c :$
$c < x < b \implies g(x) > 0$. For $M > 0$, choose $\delta_{1}$
and $\delta_{2}$ such that:
$0<x-c<\delta_1 \implies \frac{f(c)}{2} <f(x)<\frac{3f(c)}{2}$
$0<x-c<\delta_2 \implies 0 < g(x) < \frac{f(c)}{2M}$
Set $\delta = min\{\delta_1, \delta_2\}$. Then $0 < x-c < \delta \implies \frac{f(x)}{g(x)} > \frac{f(c)}{2} \cdot \frac{2M}{f(c)}=
> M$, which implies that $\lim_{x \to c^+}\frac{f(x)}{g(x)} = \infty$,
so $x = c$ is a vertical asymptote of the graph of $h(x)$.
What I'm wondering is what the point is of
Saying that there exists $b > c : c < x < b \implies g(x) > 0$
Restricting $f(x) < \frac{3f(x)}{2}$
The way I understand the proof, $\frac{f(c)}{2}$ and $\frac{f(c)}{2M}$ were simply chosen because they cancel nicely when multiplied, correct? But what advantage comes from the other two restrictions I listed?
Any help or explanation would be appreciated, thanks.
AI: There is a significant typo in the greyed region, second displayed line. We want $0\lt g(x)\lt \frac{f(c)}{2M}$. But it looks as if that is not the reason for the concern about the proof.
By assumption there is an interval about $c$ for which $g(x)\ne 0$ except at $c$. The reason for that is that if as $x$ approached $c$, say from the right, $g(x)$ changed sign infinitely often (which can happen), then $\frac{f(x)}{g(x)}$ would oscillate wildly between very large positive and very large negative values, and moreover would be undefined at values of $x$ arbitrarily close to $c$ but not equal to $c$.
So if we are close enough to $c$ but on the right of $c$, then $\frac{f(x)}{g(x)}$ does not change sign. Similar considerations apply to the left of $c$ but close to $c$. However, the sign of $g(x)$ could be different for $x$ close to $c$ and bigger than $c$, and for $x$ close to $c$ and less than $c$,
The author chose to make $g(x)$ positive close enough to $c$ but to the right of $c$. That was an arbitrary choice, negative would have been OK too. Then a minor modification of the argument would show that $\frac{f(x)}{g(x)}\to -\infty$ as $x\to 0^+$.
The reason a $b$ was chosen so that in $(c,b)$ we have $g(x)\gt 0$ is to make sure that we are working in an interval in which the ratio $\frac{f(x)}{g(x)}$ is defined.
Now we look at the first restriction, $\frac{f(c)}{2}\lt f(x)\lt 3\frac{f(c)}{2}$. Only half of that restriction is necessary.
We know that $f(c)\ne 0$. The intuition is that we want to make $\frac{f(x)}{g(x)}$ huge. That will happen since $g(x)$ is approaching $0$, unless $f(x)$ decides to be very close to $0$ also. In that case, we would know nothing about the ratio. So we want to work in an interval such that in that interval $f(x)$ never gets close to $0$. That's why the condition $\frac{f(c)}{2}\lt f(x)$ is needed, to make sure $f(x)$ doesn't get too small.
The inequality $f(x)\lt \frac{3f(c)}{2}$ is not needed (and not used) in the proof. So you are quite right in wondering why it was mentioned. It gets produced automatically when we take care of the $\frac{f(c)}{2}$ part, but still need not have been mentioned.
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H: Slightly confused about the definition of upper limits and lower limits.
I'm reading "The way of Analysis" by Strichartz, and the following is the definition of an upperlimit.
The upper limit (limsup) of a sequence $\{x_j\}$ is the extended real number
$$\limsup _{k \to \infty} x_k = \lim_{k \to \infty }\sup_{j>k} x_j$$.
I am confused about the left side of the equality.
My understanding is, a limit point $x$ of a sequence $\{x_j\}$ is a point where for any given $\epsilon >0$ there exists infinitely many terms $x_j$ such that
$$|x-x_j|< \epsilon$$.
Another thing that I think I understand is that if $y_k = \sup_{j>k}x_j$, $\{y_k\}$ becomes a monotonic sequence where if we take $k \to \infty$ $\{y_k\} \to y$ where $y$ is not only a limit point, but the supremum of the set of limitpoints of $\{x_j\}$.
So I am thinking, the right side of the equation represents the "limit of the supremum of the subsequences" which is intuitively clear to me, but I don't get the left side. What does it mean ?
Since I am not too good at analysis, I will appreciate your help a lot if you could dumb down your explanation.
AI: The upper limit just means taking the "limit" of all supremums of a sequence. So if you have $$1,2,3,4,299,199,5,6,....198,198,198,....$$then $299$ would be included in the supremum of all subsequence start from $a_{i},i\le 4$; $199$ would be included in the supremum of the subsequence start from $a_{5}$; and once we pass $a_{5}$ the limit become $198$, which is the "true" upper limit of the whole sequence when we ignore the lower order terms.
Write in symbols, $c$ is the upper limit means for any $\epsilon$, there exist some $N$ such that for any $n\ge N$, we have the supremum of the subsequence $\langle a_{n}\rangle$ (define it as $b_{n}$) satisfying $|b_{n}-c|<\epsilon$. So $a_{n}$ may still be slightly higher than $c$, but as $n\to \infty$ we would observe $a_{n}$ goes "down" to "under" $c$.
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H: recurrence relation question
How can I build a recurrence equation if there isn't an $n$-variable? Example: $a_n = 3$. Also, how would I start making a recurrence equation for $a_n = 2n + 3$?
AI: For the first one, you could use $a_0=3$ and $a_n=a_{n-1}$. For the second, put $a_0=3$ and $a_n=2+a_{n-1}$.
Ultimately, we need to determine what the relationship between subsequent terms is. In the first case, there is no difference at all (hence the $a_n=a_{n-1}$ bit). For the second, we note that each term will be $2$ more than the previous one (hence the $a_n=2+a_{n-1}$ bit). But that isn't enough. We also need to figure out what (at least) one $a_n$ is for these cases, or our recurrence relation suggests an infinite family of sequences, rather than a specific closed form. Fortunately, this part is easily done ("plug in" whatever $n$ you like).
What do you notice about the relationship between the subsequent terms of $a_n=3^n\cdot 5$? What about $a_n=1+n^2$? (The latter is tricker.)
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H: Proving $(xyz)' = x'+y'+z'$
I'm trying to prove that $(xyz)' = x'+y'+z'$ using theorems/axioms.
I tried this but I just want to make sure if its the correct route or if I've done something "illegal"/wrong.
(xyz)' = [(xy)z]' by associativity
= [(x*y)'+z'] by DeMorgan's Law
= [(x'+y') + z'] by DeMorgan's Law
= [(x'+z')+(y'+z')] by Distribution
= x'+y'+z' by simplifying redundant z' terms.
Is this the correct method?
AI: I wouldn't say you did anything "illegal", but distribution is usually used in the following manner:
$\quad (x + y)z = xz + yz\quad $ or $\quad(xy)+ z = (x+z)(y+ z)$
or the "flip side"
$\quad x(y+z)=xy + xz \quad $ or $\quad x +(yz)=(x+ y)(x+z)$
Your work was done, essentially, when you after your second application of DeMorgan's. Then, we simply use associativity again:
$$[(x'+y') + z'] = x' + y' + z'\tag{by associativity}$$
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H: question about Graph Theory notation
I'm just starting to learn graph theory. I have two questions about notation:
1). For a graph $G$ we denote the vertex set $V$ and the edge set $E$ by $G=(V,E)$.
So we have a graph $G=$ ({$v_{1},v_{2},...,v_n$}, {$e_{1}e_{2},...e_{n}$}).
My textbook presents the edge set as $E=${($v_1,v_2$), ($v_1,v_3$),...,($v_i,v_n$)}.
If I remember correctly, set theory rules tell us that ( ) are used to indicate order, and { } are used when the order doesn't matter. Why are ( ) used in this case when denoting the vertices in the edge set?
2). I'm used to seeing this formula, $E\subseteq V\times V$, to indicate a relation on $V$ denoting the elements of $E\subseteq V\times V$ as edges. In another set of notes that I stumbled upon, this notation is used: $E\subseteq V\bigotimes V$ as edges. What is the $\bigotimes$, I'm confused about that.
thanks!
AI: Order of vertices in an edge is important if we consider directed edges.
The notation as pairs of vertices (which does not allow multiple edges, by the way!) fits well with viewing $E$ as subset of $V\times V$. The notation $\otimes$ is rather known from linear algebra where it denoites the tensor product, while here it apparently stands for the set of unordered pairs ov elements of $V$ (as needed for undirected graphs). Cf. also the notations $V^2$ for $V\times V$ versus $[V]^2$ for $V\otimes V$.
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H: Conversion between numeral systems.
I would like to know if there is a "standard" for converting a number of base N to a number of base N. For example, 117 decimal to 728 of "213 base". Any random base to any random base, not only decimal to binary, etc. I would like to know if there is some obscure law/rule/trick to convert a number in a base N to a number in a base N, whatever the base is.
I thank you very much for your attention!
AI: It's not very clear what OP is asking, since he is converting from base N to base N. The example does not make it clear either.
The usual process is that a number in a base, that any number is $a_0 = b a_1 + r_0$, and that one repeats this batching of $a_n$ until it becomes exhausted, eg
120 ) 1728 120) 16777216
120 ) 14. r 48 120) 139810. r16
0. r 14 120) 1165. r10
120) 9. r85
1728 d = 14.48 lh.
In essence, a number like $2^{24}$, is batched into divisions of the long hundred (120), leaving a remainder here of 16. The number of batches is then rebatched into 120's to get 1165 second-batches, and 10 first-batches. One repeats until exhaustion: until there is just a series of remainders. The outcome is then 9 third-order, 85 second-order, 10 first-order and 16 units.
For the fractions, one can do much the same, but by multiplication. For example 0.125 * 120 gives 15. So decimal 0.125 = 0:15, base 120.
Some times, you may want to convert between formats of the same base. An example of this is to switch between the canonical form of base 3 (digits 0,1,2), and the balanced form (-1=M,0,1). The trick here is to add and subtract the same number, the steps done in different formats. Here is the sum done for 20
202 dec 20 = 202
11111 add a string of 1's
------
12020
11111 subtract the strings of 1's digit-wise (no carry)
-----
01M1M 20 in the symmetric threes.
The process reverses completely.
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H: how come this summation after produkt?
I am really stuck in this step. I hope, the context does not matter here, so i didnot provide what this is about. I am trying to get ML-Estimator. but the problem is, as i see in my textbook, how they changed the produkt to summation, why it became $\sum$
$$L(a;X)=\prod_{i=1}^{n}\frac{1}{2}a \cdot exp(-a\cdot|x_i|) = \frac{1}{2^n}a^n\cdot exp(-a\sum_{i=1}^{n}|x_i|)$$
THANKS for help!
AI: This is just a property of the exponential function. Indeed $\exp(a+b)=\exp(a)\cdot\exp(b)$ and so induction shows that
$$
\prod_{i=1}^n\exp(a_i)=\exp\left(\sum_{i=1}^n a_i\right).
$$
Use this with $a_i=-\vartheta|x_i|$.
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H: Is every Tichonov space necessarily homeomorphic to a subset of a compact Hausdorff space?
My textbook says "A Tichonov space is homeomorphic to a subset of a compact Hausdorff space."
Doesn't the subset also have to be compact Hausdorff?
Motivation:- For the subset of the compact Hausdorff space to be homeomorphic to a Tichonov space, it will have to be regular and normal. We know that any compact Hausdorff space is both normal and regular. Is this necessarily true for every subset also?
Thanks for your time!
AI: No, the subset need not be compact Hausdorff. For a very simple example, note that the Tikhonov space $(0,1)$ is a non-compact subspace of the compact Hausdorff space $[0,1]$.
A subspace of a compact Hausdorff space is compact iff it is closed, so in general you can find lots of non-compact subspaces of a compact Hausdorff space. It is also entirely possible to find non-normal subspaces of compact Hausdorff spaces. For example, let $X=\omega_1+1$ with the order topology, and let $Y=\omega+1$ with the order topology. Both spaces are compact and Hausdorff, so their product is as well. But $(X\times Y)\setminus\{\langle\omega_1,\omega\rangle\}$ is not normal: the closed sets $\omega_1\times\{\omega\}$ and $\{\omega_1\}\times\omega$ cannot be separated by disjoint open sets. For that matter $(X\times X)\setminus\{\langle\omega_1,\omega_1\rangle\}$ is another example: the closed sets $\omega_1\times\{\omega_1\}$ and $\{\langle\alpha,\alpha\rangle:\alpha\in\omega_1\}$ cannot be separated by disjoint open sets.
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H: Confusing about sequence
when define a function, for example
$f(n)=1^2+2^2+\text{...}+(n-1)^2$
then what is $f(2n)$ ?
simply substitute $2n-1$ for $n-1$?
or $f(2n)=2^2+4^2+\text{...}(2n-2)^2$
or other?
What's the relation between
$1^2+2^2+\text{...}(n-1)^2$,
$1^2+2^2+\text{...}(2n)^2$,
$1^2+3^2+\text{...}(2n-1)^2$ and
$2^2+4^2+\text{...}(2n)^2$
AI: $f(2n)=1^2+2^2+\cdots+(2n-1)^2$
EDIT: I'm really not completely sure about what you mean by relation in second part of the question, but, in terms of $f(n)$,
$1^2+2^2+\cdots+(n-1)^2=f(n)$
$1^2+2^2+\cdots+(2n)^2=f(2n+1)$
For $4^{th}$ series,
$2^2+4^2+\cdots+(2n)^2=2^2(1^2+2^2+\cdots+(n)^2)=4f(n+1)$
For $3^{rd}$ series,
$3^{rd}$ series $=(1^2+2^2+\cdots+(2n)^2)-4^{th}$ series $= f(2n+1)-4f(n+1)$
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H: How can I show that the conditional expectation $E(X\mid X)=X$?
I tried to show that $E(X\mid X=x)=x$, which would lead me to get $E(X\mid X)=X$ but I am having trouble doing so. I know that the definition of conditional expectation (continuous case) is: $$E(X\mid Y=y)=\int_{-\infty}^{\infty}x f_{X\mid Y}(x\mid y)\,dx$$
This led me to write:
$$E(X\mid X=x)=\int_{-\infty}^{\infty}x f_{X\mid X}(x\mid x)\,dx$$ but I don't know what $f_{X\mid X}(x\mid x)$ is (if that's even a valid notation...)
I do note though that I don't know anything about $\sigma$-fields or Borel sets, as someone had tried to explain me once. Is it possible to prove this in an "elementary" way?
AI: It is actually easier to define $E(X\mid Y)$ than $E(X\mid Y=y)$ hence let us recall how this is done. By definition, $E(X\mid Y)$ is any random variable $Z$ such that:
$\qquad$ (1.) $Z$ is a measurable function of $Y$.
$\qquad$ (2.) For every bounded random variable $v(Y)$, $E(Zv(Y))=E(Xv(Y))$.
It happens that when $X$ is integrable, such a random variable $Z$ exists and is almost surely unique, in the sense that if $Z$ and $Z'$ both satisfy conditions (1.) and (2.), then $P(Z=Z')=1$.
When $X=Y$, things are quite simple since $Z=X$ obviously satisfies conditions (1.) and (2.), thus, $E(X\mid X)=X$.
Note finally that $Z=X$ always satisfies (2.) but does not satisfy (1.) in general, and, likewise, that $Z=E(X)$ always satisfies (1.) but does not satisfy (2.) in general.
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H: Is $R \setminus P$ a multiplicative subset?
Let $S$ be a subset of the ring $R$; we say that $S$ is multiplicative if
(a) $0 \notin S$,
(b) $1 \in S$, and
(c) whenever $a,b\in S$, we have $ab \in S$.
We can merge (b) and (c) by stating that $S$ is closed under multiplication, if we regard $1$ as the empty product.
Here are some standard examples of multiplicative sets.
The set of all nonzero elements of an integral domain.
The set of all nonzero elements of a commutative ring $R$ that are not zero divisors.
$R \setminus P$, where $P$ is a prime ideal of the commutative ring $R$.
I think 3 is not a correct example as $R \setminus P$ is an integral domain and has zero element $P$. It should be $R \setminus P$ except $P$. Is it right?
AI: The notation $\,R\backslash P\,$ means exactly "all the elements in the set $\;R\;$ except those belonging to $\;P\;$" .
I think you confused this with the quotient ring $\;R/P\;$ ...
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H: Prove that $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})^2\ge \frac{1}{a^2}+\frac{4}{a^2+b^2}+\frac{12}{a^2+b^2+c^2}+\frac{18}{a^2+b^2+c^2+d^2}$
Let $a,b,c,d$ be positive numbers. Show that
$$\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)^2\ge \dfrac{1}{a^2}+\dfrac{4}{a^2+b^2}+\dfrac{12}{a^2+b^2+c^2}+\dfrac{18}{a^2+b^2+c^2+d^2}$$
I have seen this Similar Problem
$$\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)^2\ge \dfrac{1}{a^2}+\dfrac{4}{a^2+b^2}+\dfrac{9}{a^2+b^2+c^2}+\dfrac{16}{a^2+b^2+c^2+d^2}$$
This problem pf:
\begin{align}
LHS &=\sum\dfrac{1}{a^2}+\sum\dfrac{2}{ab}\\
&=\dfrac{1}{a^2}+\dfrac{2}{ab}+(\dfrac{1}{b^2}+\dfrac{2}{ac}+(\dfrac{2}{ad}+\dfrac{2}{bc})++(\dfrac{1}{c^2}+\dfrac{1}{d^2}+\dfrac{2}{ab}+\dfrac{2}{bd}+\dfrac{2}{cd})\\
&\ge \dfrac{1}{a^2}+\dfrac{4}{a^2+b^2}+(\dfrac{1}{b^2}+\dfrac{4}{a^2+c^2})+(\dfrac{4}{a^2+d^2}+\dfrac{4}{b^2+c^2})+(\dfrac{1}{c^2}+\dfrac{1}{d^2}+\dfrac{2}{ab}+\dfrac{2}{bd}+\dfrac{2}{cd})\\
&\ge\dfrac{1}{a^2}+\dfrac{4}{a^2+b^2}+\dfrac{9}{a^2+b^2+c^2}+\dfrac{16}{a^2+b^2+c^2+d^2}
\end{align}
AI: There is the solution in the document you provided indeed - the proof is exactly the same. Note that on the top of p.25, there are four inequalities. Replace the third one by
$$\frac{1}{b^2} + \frac{1}{c^2} \ge \frac{4}{b^2+c^2} \ge \frac{4}{a^2+b^2+c^2}$$
The fourth one by
$$\frac{18}{ad+bd+cd} \ge \frac{18}{a^2+b^2+c^2+d^2}$$
and you get the result.
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H: Can the ceiling function be used to prove the Archimedean property?
Recall the following definition of the Archimedean property:
For each $x \in \Bbb{R}$, there exists some $n \in \Bbb{N}$ such that $n>x$.
My textbook proves this by invoking the completeness axiom. My first instinct however was to use the ceiling function:
Proof: Choose any $x \in \Bbb{R}$. Then consider $n=1+\lceil x \rceil$. By definition, $n>x$, as desired.
This seems like cheating though. Does the ceiling function implicitly depend on the completeness axiom? Is my proof circular?
AI: The ceiling function does not depend on completeness: its restriction to $\Bbb Q$ is well-defined, and the rationals are not complete. It does not, strictly speaking, depend on the Archimedean property either: it is perfectly possible to define $\lceil x\rceil$ to be $\min\{n\in\Bbb Z:x\le n\}$ when that minimum exists, and then to ask what the domain of this ceiling function is. It’s that domain that depends on the Archimedean property: the Archimedean property is equivalent to the statement that the domain of the ceiling function is $\Bbb R$.
Since your argument depends on the fact that the domain of the ceiling function is all of $\Bbb R$, and that fact is equivalent to the fact that $\Bbb R$ has the Archimedean property, your argument is indeed circular.
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H: The final number after $999$ operations.
I wanted to know, let the numbers $1,\frac12,\frac13,\dots,\frac1{1000}$ be written on a blackboard. One may delete two arbitrary numbers $a$ and $b$ and write $a+b+ab$ instead. After $999$ such operations only one number is left. What is this final number.
I tried, let $*$ be an operation $a*b = (1+a)(1+b) - 1$, and $(a*b)*c = (1+a)(1+b)(1+c) -1$. by induction we can show that for $N$ such no.s we have $(1+a)(1+b)(1+c)\dots(1+d) - 1$, where $n\{a,b,c,\dots,d\} = N$. In the question we have numbers $1,\frac12,\frac13,\dots,\frac1{1000}$. plugging in we get the final number $1000$, but the answer given is $100$.
What am i doing wrong?
Any help appreciated.
Thanks.
AI: Your answer is correct. We have
$$
1*(1/2)*(1/3)*\cdots*(1/1000)=\frac21\cdot\frac32\cdots\frac{1001}{1000}-1=1001-1=1000.
$$
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H: What's definition $(\ker \varphi)_P$ the stalk of kernel presheaf
If $\varphi : \mathcal{F} \to \mathcal{G}$ a morphism of sheaves then what definition of $(\ker \varphi)_P$? I digested that an element is $\langle U,r \rangle $ where $r \in \mathcal{F}(U)$ and $\varphi(U)(r) = 0$ and pairs are identified as in usual identifications.
AI: If $\mathscr{F}$ is a sheaf on any topological space $X$ (which I assume is what you're interested in here), the stalk of $\mathscr{F}$ at a point $x\in X$ is $\mathscr{F}_x:=\varinjlim_{x\in U}\mathscr{F}(U)$, the direct limit over the sets $\mathscr{F}(U)$ over all open sets $U\subseteq X$ containing $x$. So, an element of this set is an equivalence class of a pair $(s,U)$ where $x\in U\subseteq X$ is open and $s\in\mathscr{F}(U)$, and this equals the class represented by $(s^\prime,V)$ if and only if there is an open $W\subseteq U\cap V$ and a section $t\in\mathscr{F}(W)$ such that $s\vert_W=t=s^\prime\vert_W$.
In particular, for a morphism $\varphi:\mathscr{F}\rightarrow\mathscr{G}$, the sheaf kernel $\ker\varphi$ is defined by $(\ker\varphi)(U)=\ker(\varphi(U))$, where $\varphi(U):\mathscr{F}(U)\rightarrow\mathscr{G}(U)$ is the map of sections over $U$. So, as you say, an element of $(\ker\varphi)_x$ is specified by an open set $x\in U\subseteq X$ and a section $s\in\ker(\varphi(U))$.
EDIT: The answer to the question posed in the comments is yes. Even though the image presheaf $U\mapsto\mathrm{im}(\varphi(U))$ is not generally a sheaf, the natural map from this presheaf to its sheafification (which is by definition the image sheaf) induces isomorphisms on stalks, so the stalk of the image sheaf is the same as the stalk of the image presheaf, which is $\varinjlim_{x\in U}\mathrm{im}(\varphi(U))$, and so an element of this stalk is represented by a pair $(t,U)$ where $t=\varphi(U)(s)$ for some $s\in\mathscr{F}(U)$.
Moreover, one has $\mathrm{im}(\varphi)_x=\mathrm{im}(\varphi_x)$ via the stalk at $x$ of the canonical map $\mathrm{im}(\varphi)\rightarrow\mathscr{G}$, and similarly $\ker(\varphi)_x=\ker(\varphi_x)$ under the stalk at $x$ of the canonical inclusion $\ker(\varphi)\rightarrow\mathscr{F}$.
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H: Is $L^2(0,T;H_n)$ compactly embedded in $L^2(0,T;H)$?
Let $H$ be a separable Hilbert space with basis $h_i.$ Let
$$H_n := \text{span}\{h_1,...,h_n\}.$$
Questions:
1) Is $L^2(0,T;H_n)$ compactly embedded in $L^2(0,T;H)$?
2) Is $L^2(0,T;H_n^*)$ compactly embedded in $L^2(0,T;H^*)$?
I have no idea how to even begin. Any hints would be appreciated.
AI: No.
$\mathbb{R}$ is a separable Hilbert space with basis $\{h_1\} = \{1\}$; $L^2(0,T;\mathbb{R})$ clearly does not compactly embed in itself.
For any $\mathbb{R}$-Hilbert space $H$ we have that $H = H \otimes_{\mathbb{R}} \mathbb{R}$ and a bounded sequence $f_k$ in $L^2(0,T;\mathbb{R})$ with no converging subsequences gives a counterexample in $L^2(0,T;H)$ if you consider the elements $h_1 \otimes f_k$.
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H: Image of a normal Hall Subgroup under an automorphism
Let $G$ be a group such that $|G|= n = md$, where $\gcd(m,d)=1$.
Let $N$ be a normal subgroup of $G$ with order $m$.
Further, let us define a subgroup $H$ of order $d$. I managed to prove that, $H \cap N= [e] \implies HN = G$, and $HN \cong H \times N$.
Now, let $f$ be an automorphism of $G$.
Then, $|f(N)|= m$.
Consider that for some $n_i \in N$,
$f(n_i^m)= f(n_i)^m \implies f(e) = e = (f(n_i))^m$
So, the order of $f(n_i)$ divides $m$.
As $f(n_i) \in G$, $f(n_i) \in h_i N$, for some $h_i \in H$. So, let, $f(n_j) = h_i n_i \implies (h_in_i)^m = e$.
Now, consider a bijective map $g: HN \to H \times N$, such that,
$f(h_i n_i) = (h_i, n_i) \implies f((h_i n_i)^m)= f(e)= (h_i, n_i)^m= (h_i^m, n_i^m)$
$\implies (h_i^m, n_i^m)= (e, e) \implies h_i^m= e$ which is simply not possible given that $h_i \in H$ and that $gcd(m,d)=1$, unless, $h_i = e$.
So, for all $n_i \in N, f(n_i) \in N \implies f(N) = N$.
Okay, well, can somebody please just look through this reasoning and see if this is acceptable? Also, it seemed somewhat unnecessarily wordy. How could I prove it with greater brevity?
Thank you!
AI: There are several ways to do it. One is to note that
$$
\lvert N f(N) \rvert = \frac{\lvert N \rvert \cdot \lvert f(N) \rvert}{\lvert N \cap f(N) \rvert}.
$$
So this number is a multiple of $m$, a divisor of $n = m d$, and of $\lvert N \rvert \cdot \lvert f(N) \rvert = m^{2}$, and thus coprime to $d$. The only choice is
$$
\lvert N f(N) \rvert = m = \lvert N \rvert,
$$
so $f(N) \subseteq N$, and thus $f(N) = N$.
Alternatively, I think you are proving that
$$\tag{eq}
N = \{ x \in G : x^{m} = 1 \},
$$
which immediately implies that $N$ is invariant under automorphisms (even endomorphisms).
(eq) follows from the fact that if $x^m = 1$, then $(x N)^m = N$ in $G/N$, but since the order of $G/N$ is $d$, you also have $(xN)^d = N$, and since $(m, d) = 1$, we have $x N = N$, that is, $x \in N$.
PS The second way of doing it proves the statement that @TobiasKildetoft made in his comment. The first way works too, just take any subgroup $M$ of order dividing $m$, and consider $\lvert N M \rvert$, to obtain $M \le N$.
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H: Convergence of $\sum_{n=0}^\infty z^{2^n}$
Let formally $f(z) := \sum_{n=0}^\infty z^{2^n}$. What is the raduis of convergence of this series ?
AI: There is no need to think of how to apply any special test or theorem to this. Just think about how the terms grow, and how it compares to the most basic series you've seen before e.g. geometric series. If $|z|<1$ then the size of the terms drops off very quickly. Prove the series converges absolutely (and hence converges) if $|z|<1.$ Now see what happens where $z=1$ and this will determine the radius of convergence.
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H: Evaluating $\int{{x^2 -1}\over{x^3 \sqrt{2x^4-2x^2+1}}} \mathrm dx$
How to evaluate:
$$\int{{x^2 -1}\over{x^3 \sqrt{2x^4-2x^2+1}}} \mathrm dx$$
AI: HINT:
First put $x^2=y$ in
$$\int{{x^2 -1}\over{x^3 \sqrt{2x^4-2x^2+1}}} \mathrm dx=\int{{x^2 -1}\over{2x^4 \sqrt{2x^4-2x^2+1}}} \mathrm 2xdx$$
to get $$\int \frac{y-1}{2y^2\sqrt{2y^2-2y+1}}dy$$
Now $2y^2-2y+1=\frac{4y^2-4y+2}2=\frac{(2y-1)^2+1}2$
Put $2y-1=\tan\theta$
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H: The General Validity of $ab=cd$ Implies $ba=dc$.
I am doing some research of my own and I have a brief question and I don't recall studying this particular thing. Is it generally true that if a set coupled with a binary operation is closed under inverses and has an identity then $ab=cd$ implies $ba=dc$ for all $a,b,c,d$ in the set? If not, would this statement hold true for groups?
AI: As long as your operation has an identity, the property
$$ab = cd \Rightarrow ba=dc$$
is equivalent to commutativity: if the operation is commutative, the implication is obvious, for the other direction, put $c=1$ and $d=ab$ to get $ab = ba$.
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H: Is there a way to prove a boolean operator isn't universal?
In boolean algebra, I could prove an operator is universal by implementing a NAND or NOR gate with it. But is there a way to prove a boolean operator isn't universal?
I would like to know a general method that should work for every (or almost every) incomplete operator.
Right now, I want to prove incompleteness of this operator:
$$ T(w,x,y,z) = (\neg w \lor \neg x \lor \neg y) \oplus (xyz)$$
After simplification I could write it like one of the following:
$$(\neg(wx y)) \oplus (xyz) =wxyz \lor \neg w\neg x \lor \neg w\neg y \lor \neg w\neg z \lor \neg x\neg y \lor \neg x\neg z \lor \neg y\neg z$$
$\oplus$ is the eXclusive OR operator, $\lor$ is OR, and I'm using implicit AND for simple reading
AI: Yes, though the details generally depend considerably on the specific operator or operators. This answer is an example of the kind of argument that can be used. For more information you might read about functional completeness.
Added: If $T(w,x,y,z)=\neg(wxy)\oplus(xyz)$, then $T(x,x,x,x)=x\lor\neg x=\top$. (I use $\bot$ for false and $\top$ for true.) Suppose that each of the terms $t_1,t_2,t_3$, and $t_4$ is equivalent either to $x$ or to $\top$. Then $t_1t_2t_3$ is equivalent either to $x$ or to $\top$, as is $t_2t_3t_4$, and $T(t_1,t_2,t_3,t_4)$ is equivalent to one of $\neg x\oplus x=\top$, $\neg x\oplus\top=x$, $\bot\oplus x=x$, and $\bot\oplus\top=\top$. It follows by induction that every term built up from a single variable $x$ using $T$ is equivalent either to $x$ or to $\top$. In particular, $T$ cannot generate $\neg x$ and therefore is not universal.
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H: How to define a incomplete metric on $\mathbb{S}^1$?
Let $\mathbb{S}^1=\{x\in\mathbb{R}^2:\ \|x\|_2=1\}$.
My question is: Is it possible to define a incomplete metric on $\mathbb{S}^1$, i.e. a metric such that $\mathbb{S}^1$ is not complete.
Thank you.
AI: Let $f \colon (0,1) \to \mathbb S^1$ be a bijection. Define $d(x,y) = |f^{-1}(x) - f^{-1}(y)|$, then $f$ is an isometry, hence $(\mathbb S^1, d)$ is not complete.
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H: Confusion about Spec of quotient ring
Consider the ring $A := \dfrac{\mathbb C[x]}{(x(x-1)(x-2))}$. According to some sources (cf. Vakil) $sp(Spec (A))$ should be just the three points $\{0,1,2\}.$ It seems right, because $A$ is the ring of regular functions on these three points. However, I have the impression that there are many more prime ideals in $A$. Take for example $(x-0.1).$ Why is it not in the spec?
AI: To go with your example, if we consider $x-0.1$ in $\mathbb C [x]$ then we can multiply it by $x^2-2.9x+1.71$, which will give $x^3-3x^2+2x-0.171$, which is just $-0.171$ in our ring, since $x(x-1)(x-2)=x^3-3x^2+2x$. Hence it is invertible, since 0.171 is clearly invertible, and so is any other element.
I think that if you check that $(x-a)$ is the only prime ideal in $\frac{\mathbb C [x]}{(x-a)}$ then it will be simpler to compute and still give you the idea (indeed, if $a=0$ it is trivial that the only non-invertible ideal is $(x)$).
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H: Product of two compact spaces is compact
I read the proof that uses tube lemma and I do not have any problem with it but I cannot see what is wrong with the proof that first came to my mind:
Let $X$ and $Y$ be compact spaces. Let $\mathcal{A}$ be an open covering of $X\times Y$. Then, $\bigcup_{U\in \mathcal{A}}{U}=X\times Y$. And $\pi_1(\bigcup_{U\in\mathcal{A}}{U})=\bigcup_{U\in\mathcal{A}}{\pi_1(U)}=X$, similarly $\bigcup_{U\in\mathcal{A}}{\pi_2(U)}=Y$. Since $X$ is compact, there are finitely many $\pi_1(U)$'s and $\pi_2(U)$ that cover $X$ and $Y$. So the product of these finitely many sets covers $X\times Y$.
I presume that there is a mistake related to basic set theory.
AI: Yes, there is a mistake in this proof. The problem is that when you have your finite covers of $X$ and $Y$, it is true that the product of those open sets cover $X \times Y$ ; what is not true is that those products were in your original open cover!
Hope that helps,
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H: Questions about the composition of two dominant rational maps.
I am reading the lecture notes. On page 4, I have some difficulty in understanding the proof of the fact that $g \circ f$ exists for two dominant rational maps $f, g$.
Let $f: V \to W$ and $g: W \to U$ be dominant rational maps. It is said that "if we pick any point $Q \in W$ and any open set $S$ containing $Q$, then $S$ will also contain an element of $f(\operatorname{dom} f)$. Using the fact that rational maps are continuous we get that $f^{-1}(\operatorname{dom} g) \subset \operatorname{dom} f$ is a dense open set for any rational map $g : W \to U$, so $g \circ f$ is defined on a dense open set".
My questions are
(1) Why $S$ will also contain an element of $f(\operatorname{dom} f)$?
(2) Why $f^{-1}(\operatorname{dom} g) \subset \operatorname{dom} f$ and why $f^{-1}(\operatorname{dom} g)$ is a dense open subset of $V$?
(3) If $f$ is dominant but $g$ is not dominant, does $g \circ f$ always exist?
I think that since $f(\operatorname{dom} f)$ is dense in $W$, $$\overline{f(\operatorname{dom} f)} = \bigcap_{f(\operatorname{dom} f) \subseteq M, M \text{ is closed}} M = W. $$
Thank you very much.
AI: To answer point 1, suppose that $S$ contained no points in $f({\rm dom} f)$. Then $W\backslash S$ is closed, and contains $f({\rm dom} f)$. Thus the closure of $f({\rm dom} f)$ is contained in $W\backslash S$, and cannot be equal to $W$, contradicting it being dense.
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H: How to teach the division algorithm?
What is the best way to introduce the division algorithm? Are there real life examples of an application of this algorithm. At present I state and prove the division algorithm and then do some numerical examples but most of the students find this approach pretty dry and boring. I would like to bring this topic to live but how?
AI: [Cue sci-fi sequence.]
A number of clones in an underground cloning lab escape from their enclosure. Most of the clones are from the same batch, and they all weigh 150 pounds. There is also a clone from a later batch--not as fully developed, only weighing 50 pounds. They manage to reach the elevator out of the facility, and through [plot contrivance] are able to make it operational for one trip up, only. The elevator has a 2000 pound carrying capacity. How many of the grown clones will escape the facility (assuming that the volume of the elevator is not a restriction)? Will the elevator be able to carry the youngster?
Tune in next time, for Division Algorithm!
[Cut to black.]
Okay, that isn't a real life example, but it certainly isn't dull.
You have a pocket full of quarters (39 in all) and a serious gumball craving. The supermarket gumball machine charges a dollar per gumball. (The money-grubbers!) How many gumballs can you get, and how many quarters will you have left?
That isn't quite as exciting, but it's more "real."
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H: Qustions about the maps $\mathcal{O}(Y) \to \mathcal{O}_P \to K(Y)$.
I am reading the book Algebraic Geometry by Robin Hartshorne. I am trying to understand the maps $\mathcal{O}(Y) \to \mathcal{O}_P \to K(Y)$ on Page 16.
Are there some functions in $K(Y)$ but not in $\mathcal{O}_P$? Are there some functions in $\mathcal{O}_P$ but not in $\mathcal{O}(Y)$? Thank you very much.
AI: Yes. Take $Y=\mathbb{A}^2(k), P=(0,0)$ for example. Then $K(Y)$ is $k(x,y),$ the field of rational functions in two variables. $\mathcal{O}_P$ is the subring of $K(Y)$ which consists of rational functions defined at P. And $\mathcal{O}(Y)$ is the subring of $K(Y)$ which consists of rational functions defined at every point in $Y.$
So $\dfrac{1}{xy}$ is in $K(Y)$ but not in $\mathcal{O}_P$ since it is not defined at the origin. And $\dfrac{1}{x-1}$ is in $\mathcal{O}_P,$ but not in $\mathcal{O}(Y)$ since it is not defined everywhere in $Y$ - it is not defined at $(1,0)$ for instance. One can show (using the Nullstellensatz) that in this case, $\mathcal{O}(Y)=k[x,y]$ i.e. that the only rational functions in two variables defined everywhere on the affine plane are the polynomials.
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H: Meromorphic function on Riemann surface
I've got this exercise I can't solve. May someone help me? Thank you.
Let $X, Y, Z$ be homogeneous coordinates on complex projective plan and let $C=\{[X:Y:Z] |X^{4}+XY^{3}+Z^{4}=0\}$. Consider the meromorphic function $f=\frac{X}{Y}$ defined on $C$.
1) Calculate zeroes and poles of $f$ with their orders;
2) Calculate ramification points of $f$ with their indexes and the genus of $C$;
3) Find on $C$ three linearily independent holomorphic differentials
AI: To answer point 1)
The zeroes occur when $X$ is zero - in this case we are forced by the defining equation to have $Z=0$ and hence the only point at which $f$ has a zero is $[0:1:0]$.
Similarly we have a pole if $Y=0$, and in this case we are forced to have $X^4 = -Z^4$, which has various solutions over $\mathbb C$. Find these and you have your poles.
Finally, note that we cannot have $X=Y=0$, since then we would also have $Z=0$, and $[0:0:0]$ is not a valid point.
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H: If $a,b$ are roots for $x^2+3x+1=0$.Calculating $(\frac{a}{b+1})^2 +(\frac{b}{a+1})^2$
If $a,b$ are roots for the equation $x^2+3x+1=0$.How to calculate $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$
AI: Because $x^2+3x+1=0$, we have $x^2=-3x-1$ and also $x^2+2x+1=-x$, for $x=a,b$. Hence $$\left(\frac{a}{b+1}\right)^2=\frac{a^2}{(b+1)^2}=\frac{-3a-1}{-b}=\frac{3a+1}{b}$$
By symmetry, the desired expression is $$\frac{3a+1}{b}+\frac{3b+1}{a}=\frac{3a^2+a}{ab}+\frac{3b^2+b}{ab}=\frac{3a^2+a+3b^2+b}{ab}=\frac{3(-3a-1)+a+3(-3b-1)+b}{ab}=\frac{-8(a+b)-6}{ab}$$
Lastly, because $a,b$ are roots of $x^2+3x+1$, we know that $ab=1$ and $a+b=-3$. Plugging this into our final expression gives $$\frac{-8(-3)-6}{1}=18$$
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H: convergence of a series $a_1 + a_1 a_2 + a_1 a_2 a_3 +\cdots$
Suppose all $a_n$ are real numbers and $\lim_{n\to\infty} a_n$ exists.
What is the condition for the convergence( or divergence ) of the series
$$ a_1 + a_1 a_2 + a_1 a_2 a_3 +\cdots $$
I can prove that $ \lim_{n\to\infty} |a_n| < 1 $ ( or > 1 ) guarantees
absolute convergence ( or divergence ).
What if $ \lim_{n\to\infty} a_n = 1 \text{ and } a_n < 1 \text{ for all } n $ ?
AI: What if $\lim_{n\to\infty}a_n=1$ and $a_n<1$ for all $n$?
Then the series may or may not converge. A necessary criterion for the convergence of the series is that the sequence of products
$$p_n = \prod_{k = 1}^n a_k$$
converges to $0$.
If the $a_n$ converge to $1$ fast enough, say $a_n = 1 - \frac{1}{2^n}$ ($\sum \lvert 1 - a_n\rvert < \infty$ is sufficient, if no $a_n = 0$), the product converges to a nonzero value, and hence the series diverges.
If the convergence of $a_n \to 1$ is slow enough ($a_n = 1 - \frac{1}{\sqrt{n+1}}$ is slow enough), the product converges to $0$ fast enough for the series to converge.
Let $a_n = 1 - u_n$, with $0 < u_n < 1$ and $u_n \to 0$. Without loss of generality, assume $u_n < \frac14$.
Then $\log p_n = \sum\limits_{k = 1}^n \log (1 - u_k)$. Since for $0 < x < \frac14$, we have $-\frac32 x < \log (1-x) < -x$, we find
$$ -\frac32 \sum_{k=1}^n u_k < \log p_n < -\sum_{k=1}^n u_k,$$
and thus can deduce that if $\sum u_k < \infty$, then $\lim\limits_{n\to\infty}p_n > 0$, so the series does not converge.
On the other hand, if $\exists C > 1$ with $\sum\limits_{k = 1}^n u_k \geqslant C\cdot \log n$, then $p_n < \exp (-C \cdot\log n) = \frac{1}{n^C}$, and the series converges.
|
H: Continuity proof via one-sided derivatives
$f\left(x\right)$ is defined on $[a,b]$, differentiable on $\left(a,b\right)$ and has one-sided derivatives at points a and b. How to prove that $f\left(x\right)$ is continuous on $[a,b]$?
The function is continuous on $\left(a,b\right)$ due to necessary condition of differentiability. But I've no idea how to use one-sided derivatives to show continuity at a and b.
AI: The one-sided derivative at $a$ is $$\lim_{x\rightarrow a+}\frac{f(x)-f(a)}{x-a}$$ which must exists by hypothesis. If $\lim_{x\rightarrow a+}f(x)-f(a)$ did not exist and equal zero, that would provide a contradiction.
|
H: finding value of formula.
I am little bit confusing how to calculate $δβ/δρ$ value if I have set of values like this.
I have the values of $β$ and $ρ$ like this.
$$ \begin{array}{l|l}
β & ρ\\ \hline
0,324 & 0,687\\
0,322 & 0,695\\
0,319 & 0,721\\
0,317 & 0,759\\
0,316 & 0,798
\end{array} $$
from the above set of values i need to calculate this "$δβ/δρ$" $=$ ?
Can anyone explain me how to calculate it in microsoft excel or normal method?
AI: $\frac {\delta \beta}{\delta \rho}$ is the change in $\beta$ between two lines divided by the difference in $\rho$ in the same two lines. So for the first two, you would have $\frac {\delta \beta}{\delta \rho}=\frac{0.324-0.322}{0.687-0.695}=\frac{0.002}{-0.008}=\frac {-1}4$ If these are measured values, the error is quite large because of the subtraction.
|
H: Linear transformation $f$
I am tring to solve the following task:
Linear transformation $f: \mathbb R^2 \rightarrow \mathbb R^2$ is given by $f(\begin{bmatrix} x_1\\ x_2 \end{bmatrix}) =
\begin{bmatrix} 2x_1-x_2\\ x_1+x_2 \end{bmatrix}$.
Answer true or false to the following questions:
a) in some basis of transformation $ \mathbb R^2$ transformation matrix of $f$ is
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
b) $f$ is a bijection
c) transformation matrix of $f$ in basis $([1,0]^T, [1,1]^T)$ is
$\begin{bmatrix} 1 & 1 \\ -1 & 2 \end{bmatrix}$
Could you kindly give me any HINTS how to start it (not solution)?
AI: Hints (assuming some familiarity with various results of linear algebra):
a) If $A$ and $B$ are transformation matrices of $f$ with respect to different bases, then there is some matrix $X$ such that $XAX^{-1} = B$. What does this say if $A$ is the identity matrix?
b) A linear transformation is bijective if and only if for any (and thus all) basis, its transformation matrix is invertible.
c) This is a calculation.
|
H: Archimedes' derivation of the spherical cap area formula
Archimedes derived a formula for the area of a spherical cap.
so Archimedes says that the curved surface area of a spherical cap is equal to the area of a circle with radius equal to the distance between the vertex at the curved surface and the base of the spherical cap.
$$A = \pi(h^2+a^2)$$
I want to know how Archimedes derived this formula. I have searched on the net and only found solutions using integration. Is there a method to do this without using integration?
AI: Enclose the sphere inside a cylinder of radius $r$ and height $2r$ just touching at a great circle. The projection of the sphere onto the cylinder preserves area.
That is the way Archimedes derived that the area of the sphere is same as lateral surface area of the cylinder which is $= (2 \pi r) (2r)=4\pi r^2$. The projection of the cap on the cylinder has area $(2 \pi r)h$. And since $a^2=r^2-(r-h)^2=2rh-h^2 \Rightarrow 2rh=h^2+a^2$, the area of the cap is $\pi (2rh) = \pi (h^2+a^2).$
Edit: corrected grammar
|
H: Are $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ and $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ isomorphic as $\mathbb{R}$-vector spaces?
Are $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ and $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ isomorphic as $\mathbb{R}$-vector spaces?
I am having a very hard time at digesting tensor products and I do not know how to "compare" tensor products over different rings.
My hunch is that they are not isomorphic. It is easy to see that $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ is an $\mathbb{R}$-vector space of dimension $4$. I suspect that $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ is also an $\mathbb{R}$-vector space but of lower dimension, but I have no idea how to show this or if indeed this intuition is correct.
Thank you.
AI: No, they are not. Your intuition is correct.
Given a ground field $k$, and two finite-dimensional $k$-vector spaces $V,W$, then $\dim_k(V \otimes_k W) = (\dim_k V)(\dim_k W)$. So:
$\dim_{\mathbb C}(\mathbb C \otimes_{\mathbb C} \mathbb C) = (\dim_{\mathbb C} \mathbb C)^2= 1$, therefore $\dim_{\mathbb R} (\mathbb C \otimes_{\mathbb C} \mathbb C) = 2$.
$\dim_{\mathbb R}(\mathbb C \otimes_{\mathbb R} \mathbb C) = (\dim_{\mathbb R} \mathbb C)^2= 2^2 = 4$.
So the two are not isomorphic. In fact, $\mathbb C \otimes_{\mathbb C} \mathbb C \simeq \mathbb C$ canonically (and more generally $V \otimes_k k \simeq V$), while $\mathbb C \otimes_{\mathbb R} \mathbb C \simeq \mathbb C \oplus \mathbb C$.
|
H: $f_k$ differentiable$ \implies f(x)=\sum_{k=1}^{n}f_k(x_k)$ is differentiable
Let $f_k:(a,b)\rightarrow\mathbb{R}$ differentiable functions by $1\leq k\leq n$. Let $f:(a,b)^n\rightarrow\mathbb{R}$ defined by $$f(x)=\sum_{k=1}^{n}f_k(x_k)$$ Prove that $f$ is differentiable and calculate its derivative at any point of its domain
AI: Look at the functions:
$$(x_1,\ldots,x_n)\mapsto \sum_{k=1}^n x_k$$
and
$$(x_1,\ldots,x_n)\mapsto x_k\mapsto f_k(x_k)$$
Use these to build your function through composition. Now apply the chain rule.
|
H: How to calculate the inverse of a point with respect to a circle?
The theory said:
The inverse of a point $P$, with respect to a circle centered at $O$ and has a radius $r$, is the point $P'$ such that
The three points $O$, $P$ and $P'$ are colinear.
$OP \times OP'=r^2$
But I don't figure out how to calculate it. For example, given a point $P=(x,y)$, what is the formula to calculate its inverse $P'=(x',y')$ with respect to the circle centered at $O=(0,0)$ and has the radius $r=1$.
Solution
Thanks to @Cameron Buie 's hint the solution is $x'=\alpha x$ and $y'=\alpha y$ where $\alpha = \frac{r^2}{x^2 + y^2}$.
And for the more general case with the circle of inversion centered at any point $O=(h,k)$ rather than only at the origin, the solution becomes $x'=\alpha (x-h) + h$ and $y'=\alpha (y-k) + k$ where $\alpha = \frac{r^2}{(x-h)^2 + (y-k)^2}$
AI: Hint: Since you want the points collinear and your circle is centered at the origin, then you need $x'=\alpha x,y'=\alpha y$ for some positive $\alpha$. Can you use requirement number $2$ to solve for $\alpha$ in terms of $x,y,r$?
|
H: Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number?
$\large S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$
Multiplying by conjugate:
$\large S=\dfrac {-3}{\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}}$
From the original:
$\large S-2\sqrt[3]{5-2 \sqrt {13}}
=\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}$
Substituting:
$\large S=\dfrac{-3}{S-2\sqrt[3]{5-2 \sqrt {13}}}$
This leads to a quadratic equation in $\large S$ which I checked in wolframalpha and I got imaginary solutions. Why does this happen? I am not looking for an answer telling me how to solve this problem, I just want to know why this is wrong. Thanks.
AI: You are not right: the conjugate to $S$ is
$$
\left[\sqrt[3] {5+2 \sqrt {13}}\right]^2-\sqrt[3]{5+2 \sqrt {13}}\sqrt[3] {5-2 \sqrt {13}}+\left[\sqrt[3]{5-2 \sqrt {13}}\right]^2.
$$
|
H: Show that $(x_n)$ converge to $l$.
Let $(x_n)$ be a sequence of reals. Show that if every subsequence $(x_{n_k})$ of $(x_n)$ has a further subsequence $(x_{n_{k_r}})$ that converge to $l$, then $(x_n)$ converge to $l$.
I know the fact that subsequence of $(x_n)$ converge to the limit same as $(x_n)$ does, but I'm not sure if I can apply this.
Thank you.
AI: You are correct with your doubts as that argument applies only if you know that the sequence converges in the first place.
Now for a proof, assume the contrary, that is: there exists $\epsilon>0$ such that for all $N\in\mathbb N$ there exists $n>N$ with $|x_n-l|\ge\epsilon$.
For $N\in\mathbb N$ let $f(N)$ denote one such $n$.
Now consider the following subsequence $(x_{n_k})$ of $(x_n)$: Let $n_1=f(1)$ and then recursively $n_{k+1}=f(n_k)$. Since $(n_k)$ is a strictly increasing infinite subsequence of the naturals, this gives us a subsqeunece $(x_{n_k})$. By construction, $|x_{n_k}-l|\ge \epsilon$ for all $k$.
On the other hand, by the given condition, there exists a subsubsequence converging to $l$, so especially some (in fact almost all) terms of the subsubsequence fulfill $|x_{n_{k_r}}-l|<\epsilon$ - which is absurd. Therefore, the assumption in the beginning was wrong. Instead, $x_n\to l$.
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