text stringlengths 83 79.5k |
|---|
H: Can my MSE reputation be any positive integer?
As far as I know there are five kinds of vote
$+2$ for an edit
$-2$ for a downvote
$+10$ for an answer
$+15$ for an accepted answer
$+5$ for a question
Suppose that this is true. Can a MSE reputation take any positive integer value? How do you prove it?
AI: Yes. Just note that one upvote on a question, and then two downvotes equal $1$ point in reputation.
(Also note that $+2$ for edits only hold below a certain limit; and you can lose a point by downvoting an answer.) |
H: How to prove this Inverse Property of Group
I am given a Group $G$ with Projection $*:G\times G \Rightarrow G$ and with these properties:
$a*(b*c) = (a*b)*c$
$e*a=a$
$b*a=e$, $b$ is invers Element
$a*b=b*a$
I want to prove $(a*b)^{-1}=b^{-1}*a^{-1}$.
I am stuck not knowing how to interpret $^{-1}$, because it is not defined in given properties.
how do i prove this?
AI: The inverse of $a$, denoted as $a^{-1}$, is a unique element in $G$ which satisfies $a^{-1} * a = a * a^{-1} = e$.
The uniqueness can be proven; if $a^{-1} * a =b^{-1} * a = e$, then $a^{-1} = e * a^{-1} = (b^{-1} * a) * a^{-1}$, so... (try proving yourself)
Therefore, to prove that $(a*b)^{-1} = b^{-1} * a^{-1}$, you only need to prove that $b^{-1} * a^{-1}$ is an inverse (= the inverse) of $a*b$. What happens when you just multiply those two? |
H: is this operating procedure an Abelian Group?
I have to show if the following procedure gives a (Abelian) Group (G, *).
$G = \{ \textrm{true}, \textrm{false} \}$
$a*b := ( a \leftrightarrow b)$ (which means that $a$ is $\textrm{true}$ if and only if $b$ is $\textrm{true}$)
1.) Closure
For all $a,b \in G$, the result of the operation, $a * b$, is also in $G$.
This is NOT given, since $a,b$ are not in $G$.
However, the result of the operation $a*b$ is in $G$.
Well, do I have show "Associativity, ..." for this procedure, when $a,b$ are not in $G$?
None of the group axioms are working.
Or am I misunderstanding this?
Thank you very much :)
AI: Let's look at the operation. It seems to be XNOR, giving $$\begin{array}{ccc}
* & \textrm{true} & \textrm{false} \\
\textrm{true} & \textrm{true} & \textrm{false} \\
\textrm{false} & \textrm{false} & \textrm{true} \\
\end{array}$$
By inspection of the table you have closure
What's the identity?
Does each element have an inverse?
How can you show that $(a \leftrightarrow b) \leftrightarrow c = a \leftrightarrow (b \leftrightarrow c)$? (I see a few options. The most straightforward and least elegant would be to draw up two 8-row truth tables) |
H: what am I doing wrong in the following
Question:
Calculate the equation of tangent at the point $(4,2)$ if $y=\sqrt{X}$
My answer:
$(x_0,y_0) = (4,2)$,
$f'(X) = y^{1/2} y'= 1/2X^{-1/2}$
$f'(4) = (\frac{1}{2})(4)^{-1/2} = 1$
$y-y_0=f'(X)(x-X_0)$
$y-2=1 (X-4)$
$y= x-2$
Correct answer:
$y=\frac{1}{4}x +1$
what am I doing wrong? thanks
AI: $\frac{\mathrm{d}y}{\mathrm{d}x} \sqrt{x} = \frac{1}{2\sqrt{x}}$
When $x = 4$, $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{4}$.
So the equation of the line is
$y-2=\frac{1}{4}(x-4)$
$y=\frac{1}{4}x +1$ |
H: Calculation of integers $b,c,d,e,f,g$ such that $\frac{5}{7} = \frac{b}{2!}+\frac{c}{3!}+\frac{d}{4!}+\frac{e}{5!}+\frac{f}{6!}+\frac{g}{7!}$
There are unique integers $b,c,d,e,f,g$ such that $\displaystyle \frac{5}{7} = \frac{b}{2!}+\frac{c}{3!}+\frac{d}{4!}+\frac{e}{5!}+\frac{f}{6!}+\frac{g}{7!}$
Where $0\leq b,c,d,e,f,g <i$ for $i=2,3,4,5,6,7$. Then the value of $b+c+d+e+f+g = $
$\bf{My\; Try}::$ $\displaystyle \frac{5}{7} = \frac{2520\cdot b+840\cdot c+210\cdot d+42\cdot e+7\cdot f+g}{7\times 720}$
$\displaystyle 2520\cdot b+840\cdot c+210\cdot d+42\cdot e+7\cdot f+g = 720\times 5 = 3600$
Now I did not understand how can I solve after that.
Help Required
Thanks
AI: Look at this equality that you've got: $$ 2520 \cdot b + 840 \cdot c + 210 \cdot d + 42 \cdot e + 7 \cdot f + g = 3600.$$
Note that if you consider everything modulo $7$, then most of the summands disappear, because $2520,840,210,42$ and $7$ are all multiples of $7$. So, taking remainders modulo $7$, we get $g \equiv 2 \pmod 7$. Since $0 \leq g < 7$, it follows that $g = 2$. Now substitute $2$ for $g$ in your equality, subtract $2$ from both sides and divide everything by $7$. You get
$$
360 \cdot b + 120 \cdot c + 30 \cdot d + 6 \cdot e + f = 514.
$$
Now consider both sides modulo $6$, and go on in a similar fashion. You will eventually find the values for all variables. |
H: Find equation of the circular cross section of a unit sphere
I have a unit sphere in Cartesian coordinates:
$x^2 + y^2 + z^2 = 1$
or in spherical coordinates:
$x = \rho \sin(\phi) \cos(\theta)\\
y = \rho \sin(\phi) \sin(\theta)\\
z = \rho \cos(\phi)$
I select a point $P$ on the surface of the sphere, in the coordinate system that is more convenient. I want to identify the points, on the surface of the sphere, on the circle of radius $r$ centred in $P$, where $r$ is the distance of the shortest arc on the surface of the sphere from each point to $P$ (i.e. $r$ is the great arc between two points on a unit sphere).
I expect this problem to be pretty simple but I haven't found any clear resource on the web. Any idea? Essentially what I need is the equation in 3D of a "circular section" of a sphere, possibly as a function of one angle.
AI: I'm not entirely sure if you're asking for a closed expression that gives the points, or just a description of these points and am going to assume the latter.
The first step in solving this problem is to change it into an easier one. We may rotate the sphere in such a way the point $P$ becomes the north pole $N=(0,0,1)$. If we call this rotation (which may be expressed as a $3\times3$ matrix) $R$, a point $Q$ is on the circle with radius $r$ around $P$ iff $R(Q)$ is on the circle with radius $r$ around $N$. This is due to the fact that a rotation of the sphere does not change distances on the sphere.
The next step is to consider these particular circles of radius $r$ around $N$. The best way to see this is to draw a sketch of the intersection of the sphere with the $yz$-plane, i.e. a circle of radius 1 around the origin. We can also see to points $P_1, P_2$ on the circle at distance $r$ from $N$. (I'll assume $r$ is small enough for this to make sense) These points should be such that $P_1$ is obtained from $P_2$ by reflecting around the $z$-axis. The spherical coordinate $\theta$ corresponding to these points can now easily be obtained: this angle is given as an angle wrt the $z$-axis, while $r$ is precisely the arc length corresponding to this angle on a circle with radius 1. This leads to the conclusion that $r=\theta$ in this special case.
The next thing to notice, is that all points at distance $r$ from $N$ have the same $z$-coordinate. By projecting either $P_1$ or $P_2$ at this $z$-coordinate (which again can be done insightful in the same picture), one finds this height to be $z_0=\cos(\theta)=\cos(r)$. Moreover, yet again from considering the same picture, the required circle is a circle parallel to the $xy$-plane with center $(0,0,z_0)$ and radius $\sin(\theta)=\sin(r)$.
We may thus conclude that the points at distance $r$ from $N$ are of the form $(\sin(r)\cos(\phi),\sin(r)\sin(\phi),cos(r))$ with $\phi\in[0,2\pi)$. Using the rotation matrix $R$ from before, we thus get the set of points satisfying your original question:
$$\{R^{-1}(\sin(r)\cos(\phi),\sin(r)\sin(\phi),cos(r))\mid \phi\in[0,2\pi)\}.$$ |
H: Choosing Dense sets
More than one option can be correct.
Which of the following sets are dense in $R^2$ with respect to the usual topology.
(a) $\{(x,y)\in R^2:x\in N\}$
(b) $\{(x,y)\in R^2:x+y$ is a rational number$\}$
(c) $\{(x,y)\in R^2:x^2+y^2=5\}$
(d) $\{(x,y)\in R^2:xy\ne0\}$
Natural numbers are not limit points of real numbers. So, (a) is rejected.
Rational numbers can be limit points to irrational numbers, so (b) is selected.
We can draw a circle of radius $\sqrt5$ anywhere in the plane. So, I assume every point can be covered by this circle. So I would say it is also a dense set. If it is not then which set of points is it not covering?
$xy=c$ is a hyperbola with coordinate axes as its asymptotes. So, it will not cover all the points of the plane. So, (d) is rejected.
But answer is given as (b) and (d).
AI: Your answers to (a) and (b) are correct, but your reasons aren’t really sufficient, especially for (b). I’ll come back to those.
The set in (c) is specifically the circle of radius $\sqrt5$ centred at the origin: it’s a closed set in $\Bbb R^2$, and it’s not all of $\Bbb R^2$, so it cannot be dense in $\Bbb R^2$.
The set in (d) contains every point $\langle x,y\rangle$ such that $x\ne 0$ and $y\ne 0$: that’s the whole plane except the coordinate axes. Every point on the coordinate axes is the limit of a sequence from this set, so the closure of this set is $\Bbb R^2$, and this set is therefore dense in $\Bbb R^2$.
The points of (b) are the ones that lie on lines with slope $-1$ and rational $x$- and $y$-intercepts. This set is dense in $\Bbb R^2$, but in order to show that you must show that its closure is all of $\Bbb R^2$; it’s not enough to say that the rationals are dense in the reals.
Finally, the set in (a) is $\Bbb N\times\Bbb R$; since $\Bbb N$ is closed in $\Bbb R$, this set is closed in $\Bbb R^2$. It clearly isn’t all of $\Bbb R^2$, so it cannot be dense in $\Bbb R^2$. |
H: Prove the following language is not regular
The set of strings of 0's and 1's, beginning with a 1, such that when interpreted as an integer, that integer is prime.
I'm assuming the best way to move forward is to use the pumping lemma. I'm having difficulty developing a contradiction in this case because typically the membership criteria of the language involves some characteristic of the length of the members (e.g. the members are of length $n$, where $n$ is a perfect square), not their numerical value. Can someone help me apply the pumping lemma in this case?
AI: Pumping lemma is the key. If the language were regular, thered be strings $u,v,w$ such that $uv^*w\subseteq L$ and $|v|>0$ and wlog $|u|>0$, i.e. $u\in 1\{0,1\}^*$
If $U,V,W$ are the numbers represented by $u,v,w$ (where $v,w$ may have leading zeroes), then the number represented by $uv^kw$ is $U\cdot 2^{|w|+k|v|} + 2^{|w|}\cdot \frac{2^{k|v|}-1}{2^{|v|}-1}\cdot V+W$, where $U\ge 1$. Show that these cannot all be prime. |
H: Total number of divisors is a prime
Which numbers have prime number of divisors?
For example, $16$ has $1$, $2$, $4$, $8$, $16$, a total of $5$ divisors, $5$ being a prime.
I found that primes and the power of primes such that $p^{q-1}$, where $p$ and $q$ are prime numbers, all have prime number of divisors. Is this property
limited to only these numbers?
AI: If $n=p_1^{k_1}p_2^{k_2}\cdots p_h^{k_h}$ for ($p_i$ prime), then the number of divisors will be $(k_1+1)(k_2+1)\cdots(k_h+1)$.
So you are almost right. You need only one prime $p_1$ in the factorization and its exponent $k_1$ must be a prime minus one. |
H: Linearization by freezing the coefficients of the main part of the PDE
Let $\Omega\subset C^0$ a bounded domian in $\mathbb{R}^2$. Let $u\in C^2(\Omega)\cap C(\overline{\Omega})$ be a non negative classical solution of
$$
(1+x^2)u_{xx}-2xu_{xy}+(1+u)u_{yy}-(1+u^2)u_x+(1+u_x)u_y-u=1\text{ in }\Omega,\\u(x,y)=\frac{\sin^2(x)}{1+y^2}\text{ on }\partial\Omega.
$$
Show that $(0\leq) y\leq 1$ for $(x,y)\in\Omega$.
I already asked this here: Do I have to use a maximum principle?
But this time my question concerns a special method for solving this, I think it is called something like linearization by freezing the coefficients.
I do not know more about that, but I heard that it works by inserting the assumed solution $u$ in the coefficients of the main part of the PDE and the aim is to get a semi-linear (or linear?) PDE on which one can apply maximum principle.
Do you know something about that method?
AI: The idea behind the "freezing" technique is to introduce a new "source term" $v$ and replace enough appearances of $u$ by $v$ such that the resulting equation is linear - for example in this case we could try
$$F_v(u)=(1+x^2)u_{xx} - 2xu_{xy}+(1+v)u_{yy} - (1+v^2)u_x + (1-v_x) u_y - u - 1= 0.$$(For certain purposes it may be more fruitful to do something else with the first-order term - it's not immediately obvious what the "correct" choice is to me. For the purpose of establishing the maximum principle it doesn't matter.)
A function $u$ is then a solution of the original PDE if and only if $F_u(u) = 0$; i.e. iff it is a solution of the linear PDE $F_u (\cdot)= 0$. In some cases this could give a way to find solutions: if the linear PDE $F_v(u) = 0$ has unique solution given by $u = G(v)$ then solving the non-linear PDE is just solving the fixed-point problem $G(u)=u$.
Whether we establish the existence of solutions in this manner or another, the most useful application comes in analysing these solutions - since $u$ is a solution of the linear PDE $F_u(\cdot)=0$, you can apply all the same a priori estimates you're familiar with from the linear elliptic theory. All you have to do is control the properties of the solution enough for the operator $F_u$ to be elliptic.
In this case, you are given that $u$ is non-negative, so you should be able to verify that $F_u$ is indeed elliptic. Thus the maximum principle holds for $u$. |
H: How can a single integral equal a triple integral?
Here is part of a discussion about the gravitational potential of a sphere:
Let $dx$ $dy$ $dz$ represent an infinitesimal volume containing matter of density $\rho$ and mass $dm$. Then the potential at distance $R$ from the element will be
$$ dV = -\frac{G\rho}{R}dx\ dy\ dz. $$
We may express the total potential as an integral of the volume element over the spherical shell:
$$ \int{dV} = V = -G\rho \int\!\!\!\! \int\!\!\!\! \int{\frac{dx\ dy\ dz}{R}} = -G\rho \int^{r+dr}_{r}\!\!\!\int^{\pi}_{\theta=0}\int^{2\pi}_{\phi=0}{\frac{r^2\sin{\theta} \ dr\ d\theta \ d\phi}{R}}$$
where the volume element has been expressed in polar coordinates.
I do not understand how the author integrated both sides and ended up with a single integral on the left and a triple integral on the right. Is it true that (where $v$ is volume): $$\int{dv} = v = \int\!\!\!\! \int\!\!\!\! \int{dx\ dy\ dz}$$ because I previously thought that $$\int\!\!\!\! \int\!\!\!\! \int{dv} = \int\!\!\!\! \int\!\!\!\! \int{dx\ dy\ dz}$$ i.e., that you must have the same number of nested integrals on either side of an equation.
AI: Many students start calculus by learning to integrate over intervals on the real line. Then they integrate functions over 2D and 3D domains, adding more integral signs to match the number of dimensions. I suppose this is nice notation in the typical cases where you can do your integral dimension-by-dimension, e.g.
$$ \iiint\limits_V f(x, y, z) \,\mathrm{d}V = \int_{x_0}^{x_1} \left(\int_{y_0(x)}^{y_1(x)} \left(\int_{z_0(x,y)}^{z_1(x,y)} f(x, y, z) \,\mathrm{d}z \right) \,\mathrm{d}y \right) \,\mathrm{d}x. $$
The three integrals on the left remind you that you can reduce this multidimensional problem to nested single-dimensional problems. If you are really lucky, the whole thing separates and you have
$$ \iiint\limits_V f(x) g(y) h(z) \,\mathrm{d}V = \left(\int_{x_0}^{x_1} f(x) \,\mathrm{d}x\right) \left(\int_{y_0}^{y_1} g(y) \,\mathrm{d}y\right) \left(\int_{z_0}^{z_1} h(z) \,\mathrm{d}z\right). $$
But that triple integral was just a single symbol. What did it mean? Basically two things: (1) sum up the contributions over some domain of some integrand to follow, and (2) this domain is a subset of $\mathbb{R}^3$. Now point (2) is already conveyed in the other symbol $\mathrm{d}V$, which can be written $\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$ or $r^2 \sin(\theta) \,\mathrm{d}r \,\mathrm{d}\theta \,\mathrm{d}\phi$ or any number of other ways. In fact, these expanded forms are even more informative, conveying some information about the interpretation of the variables. As a result, all we really need the integrals for is to convey point (1), and a single symbol does this just as well as three.
For this reason, it is just as correct to write
$$ \int\limits_V f \,\mathrm{d}V $$
as it is to write
$$ \iiint\limits_V f \,\mathrm{d}V. $$
They are perfectly equivalent.
Having this mindset is also more useful in more general/advanced branches of analysis, where integration is closely tied to measure theory, and one can put measures on $\mathbb{R}^3$ just as easily as on $\mathbb{R}$, so there isn't much point in distinguishing.
In summary, feel free to use whichever notation you like, though it's usually a good idea to match the notation of your texts/instructors, at least until you are familiar enough to decide for yourself which is most reasonable. |
H: The cheapest offcut carpet to buy for two rooms word problem
Correct answer: C
This might come of as a slightly dumby question, but I've been googling for a while and have gone nowhere.
What I presumed the question was asking was to add up the two areas of the rooms, and choose the carpet area that is as close to that one, but bigger as possible. My answer was D ( the closest to 14.88, the combined area of the two rooms).
Am I not understanding the question?
AI: Can you arrange a $2.6m\times 4.4m$ rectangle, and a $1.8m\times 2.2m$ rectangle, on a $2.6m\times 5.8m$ rectangle without overlap? If you can, then you can cut out the two pieces from the big rectangle to carpet your office. |
H: How prove this limit $\lim_{n\to\infty}\frac{n^2}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}=0$
Assume that a positive term series $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges,
show that
$$\lim_{n\to\infty}\dfrac{n^2}{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}=0.$$
My try: since $a_{n}>0$. so
$$\dfrac{n^2}{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}>0.$$
But I can't find this right limit is zero!.Thank you.
AI: Let $\varepsilon >0$.
There is an index $r$ such that $\sum_{k=r}^{\infty}a_k \leq \frac{4}{9}\varepsilon$.
Let $n\geq r$. Consider the $n-r+1$ numbers $a_r,a_{r+1}, \ldots,a_n$. The
generalized means inequality
implies that
$$
M_{-1}(a_r,a_{r+1}, \ldots,a_n) \leq M_{0}(a_r,a_{r+1}, \ldots,a_n) \tag{1}
$$
i.e.
$$
\frac{n-r+1}{\sum_{k=r}^{n} \frac{1}{a_k}} \leq \frac{\sum_{k=r}^n a_k}{n-r+1} \tag{2}
$$
This implies that
$$
\frac{n^2}{\sum_{k=1}^{n} \frac{1}{a_k}} \leq \frac{n^2}{\sum_{k=r}^{n} \frac{1}{a_k}} \leq
\big(\frac{n}{n-r+1}\big)^2 \big(\sum_{k=r}^n a_k \big) \leq
\big(\frac{n}{n-r+1}\big)^2 \frac{4}{9}\varepsilon
$$
Now, if we take $n\geq 3(r-1)$ we will have $\frac{n}{n-r+1} \leq \frac{3}{2}$
so $\big(\frac{n}{n-r+1}\big)^2 \leq \frac{9}{4}$ and hence
$$
\frac{n^2}{\sum_{k=1}^{n} \frac{1}{a_k}} \leq \varepsilon
$$
which concludes the proof. |
H: If a sequnce $(a_n)_n \to L$, $(\sqrt{a_n})_n \to \sqrt L$
How do you prove the following:
If a sequnce $(a_n)_n \to L$, $(\sqrt{a_n})_n \to \sqrt L$
AI: $$|\sqrt{a_n}-\sqrt{L}|= |\frac {(\sqrt{a_n}-\sqrt{L})(\sqrt{a_n}+\sqrt{L})} {\sqrt{a_n}+\sqrt{L}}|=|\frac{a_n-L}{\sqrt{a_n}+\sqrt{L}}|$$
Now, $\sqrt{x} \geq 0$, so, $\sqrt{a_n}+\sqrt{L}\geq\sqrt{L}$ hence, $\frac{1}{\sqrt{a_n}+\sqrt{L}}\leq \frac{1}{\sqrt{L}}$
So, $$|\frac{a_n-L}{\sqrt{a_n}+\sqrt{L}}| \leq |\frac{a_n-L}{\sqrt{L}}|$$ Which goes to $0$ since $a_n$ goes to L. |
H: What condition satisfies the equation.
$f(x) = x^2 - 2x + \sin^2 \alpha$ = 0, if
$-1 \leq x \leq 1$
$\phantom{-}0 \leq x \leq 2$
$\phantom{-}0 \leq x \leq 4$
$-2 \leq x \leq 2$
How do I solve it?
AI: We complete the square, and express the equation as a difference of squares:
$$\begin{align} x^2 - 2x + \sin^2 \alpha = 0& \iff x^2 - 2x + 1 - 1 + \sin^2\alpha = 0\\ \\ &\iff(x-1)^2 - \underbrace{(1 - \sin^2 \alpha)}_{\large =\;\cos^2 \alpha} = 0 \\ \\ &\iff (x - 1)^2 - \cos^2 \alpha = 0\\ \\
& \iff \Big((x - 1) - \cos \alpha\Big)\Big((x - 1) + \cos \alpha\Big) = 0\end{align}$$
$$x = 1 + \cos \alpha$$
$$x = 1 - \cos \alpha$$
Where $$-1 \leq \cos \alpha \leq 1\quad \text{for all }\;\alpha \in \mathbb R$$ |
H: How to calculate area of this shape?
I was trying to solve a complicated problem then I came accros to this complicated problem. I believe that there is enough information to calculate the area. Can you help me to find a general formula for the area of this shape, in terms of $x,\alpha,\beta$?
I forgot to write on the figure: $|AB|$ is tilted $45^\circ$ w.r.t. "ground", $\beta<\alpha$ and $|AB|$ is not parallel to $|DC|$.
$|CB|=|DA|=1$ unit and $|AB|=x$.
AI: Set up a coordinate system with its original at $B$. Then, in this coordinate system, we have:
$$
A = (x_0, y_0) = \frac{1}{\sqrt2}(x,x)
$$
$$
B = (x_1, y_1) = (0,0)
$$
$$
C = (x_2,y_2) = (-\cos\beta, \sin\beta)
$$
$$
D = (x_3,y_3) = \frac{1}{\sqrt2}(x,x) + (-\cos\alpha, \sin\alpha)
$$
Then apply the area formula from here:
$$
\text{Area} = \frac12\sum_{i=0}^{n-1}(x_iy_{i+1} -x_{i+1}y_i)
$$ |
H: Solving for an $x$ in matrices, with condition $AB=BA$
I'm just starting to learn about matrices, and during one exercise I got a question to which I have no answer; Due to the fact that I haven't learned it yet...
The question is as follows:
Let $A = \left[\begin{matrix}1&x\\2&3\end{matrix}\right]$ and $B = \left[\begin{matrix}1&1\\1&2\end{matrix}\right]$
If $AB = BA$ what is the value of $x$?
I have absolutely no clue of how to solve this effectively.
One way I managed to get a (correct) result was through series of inequalities, which both looked, and was ugly to handle...
If anyone could concisely explain how one would go about solving this, and similar problems, it would be greatly appreciated!
AI: To solve this question, you just need to know how to multiply square matrices, and you need to know what it means for matrices to be equal.
Here is an effective way to solve the problem:
Compute $AB$.
Compute $BA$.
Compare the matrices for equality (means their corresponding entries must match.) |
H: If $a$ and $b$ are the zeroes of $x^2 + ax + b = 0$, then how many pairs of $(a,b)$ exist?
If $a$ and $b$ are the zeroes of $x^2 + ax + b = 0$, then how many pairs of $(a,b)$ exist?
One
Two
Three
Infinitely many
Also, what are these pairs?
AI: Since $a, b$ are zeros, they each satisfy the equation:
$$x=a: \quad a^2 + a\cdot a + b = 0\iff 2a^2 + b = 0 \iff b = -2a^2$$
$$x = b:\quad b^2 + ab + b = 0$$
Substituting $b = -2a^2$ into the second equation: $$(-2a^2)^2 + a(-2a^2) + -2a^2 = 0 \iff 4a^4 -2a^3 - 2a^2 = 2a^2(2a^2 -a - 1) = 0$$
The right-hand factor itself factors nicely, giving you 3 solutions to $a$, and then you can find the corresponding value for $b$ by using $b = -2a^2$. |
H: Showing the expectation of the third moment of a sum = the sum of the expectation of the third moment
Let $X_{1},\cdots,X_{n}$ be independent, each with mean 0, and each with finite third moments. Show that $E\left\{\left( \sum_{i=1}^{n}X_{i}\right)^{3}\right\} = \sum_{i=1}^{n}E\left\{ X_{i}^{3} \right \}$.
Thanks in advance!
AI: In the approach with characteristic functions, we have to use the fact that $\varphi_{S_n}(t)=\prod_{j=1}^n\varphi_{X_j}(t)$.
here is an other approach: we have, noting by $[n]$ the set $\{1,\dots,n\}$,
$$\mathbb E\left(\sum_{i=1}^nX_i\right)^3=\sum_{(i,j,k)\in [n]^3}\mathbb E(X_iX_jX_k).$$
The set $[n]^3$ can be divided into the $(i,j,k)$ such that:
$i=j=k$;
two indexes coincide, but not the third one;
all the indexes are different.
The sum of the indexes in 2. is $0$ as the terms are of the form $\mathbb E(X_i^2X_j)$, which is equal by independence to $\mathbb E(X_i^2)\mathbb EX_j=0$, and the same remark applies to 3.
It's actually the phenomenon which will appear when you will compute the third derivative of a product (either there is a factor differentiated three times, or two times an an other factor one time, or three different factors). |
H: Problem with semisimple ring theorem
Proposition: For a ring $R$ the following statements are equivalent:
(a) $R$ has a simple left generator;
(b) $R$ is simple left artinian;
(c) For some simple $_RT, _RR \cong T^{(n)}$ for some $n$;
(d) $R$ is simple and $_RR$ is semisimple.
My try to understand above proof:
$(a) \Longleftrightarrow (c)$: OK
$(a) \Longrightarrow (d)$: Please explain why $_RT$ is faithful. If i understand it, clearly $I \leq l_R(R/L)=l_R(T)=0$
$(d) \Longrightarrow (b)$: Since $_RR$ is semisimple and finite generate so $_RR= \oplus_{i=1} ^n T_i, T_i$ is simple. So, $_RR$ has composition series of length n:
$0=T_0\leq T_1\leq T_1\oplus T_2 \leq ... \leq \oplus_{i=1} ^n T_i = _RR$
Then $R$ is simple left artinian.
$(b) \Longrightarrow (a)$
First: If $R$ is left artinian, then $R$ has a minimal non-zero left ideal $T$: OK.
Please explain: $Tr_R(T) \neq 0$
Thanks for all!
AI: $(a)\implies (d)$ Generators have to be faithful. There exists $I$ and a surjection of $R$ modules $\oplus_I T\twoheadrightarrow R$. What would be a consequence if $T$ wasn't faithful?
$(d)\implies (b)$ No question?
$(b)\implies (a)$ The trace of $T$ in $R$ is the largest submodule of $R$ generated by $T$. You just got done saying that there is a copy of $T$ in $R$, so the trace of $T$ in $R$ is certainly larger than that... |
H: Spectrum of the multiplication operator
Let $B[0,1]$ be the Banach space of bounded complex functions on $[0,1]$ endowed with the supremum norm.
I've have to show that the spectrum of the multiplication operator $T_q: B[0,1] \rightarrow B[0,1]$
$$
(T_q f)(t) : = q(t)f(t), \,\,\, t \in [0,1]
$$
is
$
\sigma(T_q)=\overline{\{q(t):t\in[0,1]\}}, \,\,. \text{for each}\,\, q \in B[0,1].
$
My question is why do we need the closure?
AI: If $T$ is a linear operator on a (complex) Banach space $X$, then the resolvent set $\rho(T)$ is the collection of all complex numbers $\lambda$ such that $\lambda - T$ is a bounded invertible map $X \to \operatorname{Dom}(T)$. An important aspect of linear operators on Banach spaces is that their resolvent set is an open subset of $\mathbb{C}$. In fact, you learn that for $\lambda \in \rho(T)$, then the resolvent map $\lambda \mapsto (\lambda - T)^{-1}$ is analytic in $\lambda$ (that is, can be expressed as a power series within a sufficiently small ball around $\lambda$). This applies to your problem since the resolvent $\rho(T)$ is, in fact, the compliment of the spectrum $\sigma(T)$, implying that $\sigma(T)$ is closed. In your case, if $\lambda \in \{ q(s) : s \in [0,1]\}$ then suppose $q(t) = \lambda$. Now, let $f \in B[0,1]$ and $g \in B[0,1]$ be related by $f(s) = g(s)$ for every $s \neq t$ and $g(t) \neq f(t)$. Then for every $s \in [0,1]$ you have $(\lambda -q(s))f(s) = (\lambda-q(s))g(s)$. This means that $\lambda \in \sigma(T_q)$ since $\lambda - T_q$ is not invertible. Since $\sigma(T_q) = \overline{(\sigma(T_q))}$ (since it is closed) this implies that $\overline{\{q(t):t \in [0,1]\}} \subset \sigma(T_q)$. However, if $\lambda \notin \overline{\{q(t):t \in [0,1]\}}$ then $\lambda - q(s) \neq 0$ for any $s \in [0,1]$. In fact, there is some $\delta > 0$ such that $\sup_{s\in [0,1]} \| \lambda - q(s) \| > \delta$. It is easy enough now to see that $(\lambda - T_q)^{-1}f(s) = (\lambda - q(s))^{-1} f(s)$ and that $\| (\lambda - T_q)^{-1}\| \leq \delta^{-1} < \infty$. This means that if $\lambda \notin \overline{\{q(t):t \in [0,1]\}}$ then $\lambda \in \rho(T_q)$ and you have $\sigma(T_q) \subset \overline{\{q(t):t \in [0,1]\}}$ from which your question follows. |
H: How do I evaluate left and right limits?
I have this assignment:
$$\lim_{x \to 1} \frac{x^2 - 1}{|1 - x^3|}$$
I do not understand how I should do to separate this into two problems (one for $x \to 1^-$ and one for $x \to 1^+$ and then get rid of the absolute value. How do I do that?
AI: If $x \gt 1$, then $|1-x^3|=x^3-1$, whereas if $x \lt 1$, then $|1-x^3|=1-x^3$. You can always separate a limit into $\lim_{x \to 1^+}$, which means you are just considering values of $x$ that are greater than $1$ and $\lim_{x \to 1^-}$, considering values of $x$ that are less than $1$. It is not always useful, but when you have absolute value signs around it can be. To have a two-sided limit, both these have to exist and they have to agree. So you would write $$\frac {x^2-1}{|1-x^3|}=\begin {cases} \frac {x^2-1}{x^3-1}&x \gt 0 \\ \frac {x^2-1}{1-x^3} & -1 \lt x \lt 0 \end {cases}$$ and take the right side limit of the first, the left side limit of the second. |
H: How to integrate this integral,
$\int_{\mathbb{R}^ n} || \mathbf{x} - \mathbf{y} ||^{-k} d\mathbf{x}$? Here $k>0$.
AI: Hint:
$$
\int_{\mathbb R^n} f(\|\mathbf x\|)\,\mathrm d\mathbf x = \omega_n\int_0^\infty f(r)\,r^{n-1}\,\mathrm dr
$$
Where $\omega_n$ is the surface area of the $(n-1)$-dimensional unit sphere. |
H: Help with proof of boundedness of the union of bounded sets
I am trying to prove that the union of a finite number of bounded sets is bounded. A set $A$ is bounded if diameter, $d(A)$ is finite, and $d(A)=\sup_{x,y\in A}\rho(x,y)$ where $\rho(.)$ is metric.
My idea is to use induction. So, let $n=2$ and $A_1$ & $A_2$ be both bounded sets. Take $x,y\in A$ where $A=A_1\cup A_2$. In particular $x\in A_1$ and $y\in A_2$ Wlog $A_1\cap A_2=\emptyset$, otherwise consider $A_1\cup(A_2-A_1)$.
$$
\rho(x,y)\leq\rho(x,x_1)+\rho(x_1,y_1)+\rho(y_1,y)\leq d(A_1)+\rho(x_1,y_1)+d(A_2)
$$
where $x_1\in A_1, y_1\in A_2$. I can't find a way to show $\rho(x_1,y_1)$ is finite i.e., the distance between a point of $A_1$ and a point of $A_2$ is finite. Many thanks for any help!
AI: You are actually done because you managed to bound $\rho(x,y)$ by something independent of $x$ and $y$. You cannot avoid the $x_1$ and $y_1$ because the diameter of the union depends on the gap between the sets.
Notice that there is no need to assume $A$ and $B$ disjoint. |
H: How prove this inequality $\sum_{i=1}^{n}\frac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{i}}{i^2}\le\sqrt{2n-1}$
show that
$$\sum_{i=1}^{n}\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{i}}{i^2}\le\sqrt{2n-1}$$
My try:
$$x\in(n-1,n)\Longrightarrow \sqrt{x}>\sqrt{n-1}$$
so
$$\int_{n-1}^{n}\sqrt{x}dx>\sqrt{n-1}$$
so
$$\dfrac{2}{3}n^{\frac{3}{2}}=\int_{0}^{n}\sqrt{x}dx>\sqrt{n-1}+\sqrt{n-2}+\cdots+1$$
so
$$1+\sqrt{2}+\cdots+\sqrt{n}<\sqrt{n}+\dfrac{2}{3}n^{\frac{3}{2}}$$
But this not use.Thank you
AI: Try induction. $P(1)$ is obvious.
$P(n) \Rightarrow P(n_+1)$.
$$\sum_{i=1}^{n+1}\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{i}}{i^2}\le\sqrt{2n-1} +\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1}}{(n+1)^2}$$
To complete the proof we need to show
$$\sqrt{2n-1} +\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1}}{(n+1)^2} \leq \sqrt{2n+1} \Leftrightarrow \\
\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1}}{(n+1)^2} \leq \sqrt{2n+1}-\sqrt{2n-1} \Leftrightarrow \\
\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1} \leq (n+1)^2[ \sqrt{2n+1}-\sqrt{2n-1}] = \frac{2(n+1)^2}{\sqrt{2n-1}+\sqrt{2n+1}}
$$
Now, by your inequality we have
$$\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1} \leq \sqrt{n+1}+\frac{2}{3}(n+1)^{\frac{3}{2}}$$
and the inequality
$$\sqrt{n+1}+\frac{2}{3}(n+1)^{\frac{3}{2}} \leq \frac{2(n+1)^2}{2\sqrt{2n+1}} \leq \frac{2(n+1)^2}{\sqrt{2n-1}+\sqrt{2n+1}} $$
is easy to prove and completes the solution. |
H: Pythagoras numbers and fermats last theorem
I am reading "What Is Mathematics? An Elementary Approach to Ideas and Methods"
And I am stuck here, I don't get it. I have posted a screen shot underlining what my doubt is..
I dont get it when the author says while the pythagoras theorem is : $a^2 + b^2 = c^2$
and then he says $x=a/c$ and $y=b/c$
and then the equation should be according to me , $ax+by=c$.. right??
but the author writes y^2 = (1-x)(1+x)
Which I think may have came from something like
$x^2 + y^2 = 1$
$y^2= 1^2 + x^2$
$y^2= (1+x)(1-x)$ (since $a^2 + b^2 = (a+b)(a-b)$ )
but I dont get it where did that x^2 + y^2 = 1 came from ??
Is it that author assumed that x=a/c and y=b/c and then stoped talking about pythagoras theorem and started talking on x^2 + y^2 = 1???
and then further he introduces a number t.. i don't get it how it got converted into y=t(1+x) and (1-x)=ty ???
Can please someone help?
also , while i was writing the question it clicked me that if:
x=a/c (i.e. opposite upon hypotenuse means x is sin )
y=b/c (i.e. adjacent upon hypotenuse means y is cos)
therefore x^2 + y^2 = 1 (sin^2 + cos^2 = 1)
but then if i assume i am right about the sin cos thing then how come the last step is derived ?? (the one in blue color underline and box)
also that as per me x is sin , but the formula in that book is of cos2x? when t=tanx??
so m lil confused .. if you want any more clean way of me asking my doubt then do tell me i will rephrase the entire question .. :)
AI: For the first part of your question: the author divides both sides of the Pythagorean equation by $c^2$. This yields: $\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1$. Now, he defines $x = a/c$ and $y = b/c$. The equation can than be rewritten to: $x^2 + y^2 = 1$. Substracting $x^2$ from both sides yields $y^2 = 1 - x^2$, or equivalent: $y^2 = (1-x)(1+x)$.
He then devides both sides by $y$ and $(1+x)$. This gives $y/(1+x) = (1-x)/y$. His next step is to say that the expression is actually $t = t$, where $t$ is a rational. So, $ y/(1+x) = t \rightarrow y =t(1+x)$ and, in the same way, $(1-x)/y = t \rightarrow (1-x) = ty$.
Working this out gives the two equations $tx - y = -t$ and $x + ty = 1$. Now you have two unknowns in two equations.
This can be easily solved:
$tx - y = -t \rightarrow x = \frac{y-t}{t}$. Substitute this into the second equation. That gives us$\frac{y-t}{t} + ty = 1 \rightarrow (y-t) + t^2y = t$. Now, solve for y:
$y + t^2y = 2t \rightarrow y(1+t^2) = 2t \rightarrow y = \frac{2t}{1 + t^2}$.
Substitute this value for y back into the first to get the expression for $x$. |
H: How do I prove that $f(x) = x^2 : (0, 1/2) \to (0, 1/2)$ is not a contraction mapping?
How do I prove that $f(x) = x^2 : (0; 1/2) \to (0; 1/2)$ is not contraction mapping?
I'd like to prove this in $\mathbb{R}$ with the Euclidean metric.
AI: If it was a contraction map, then there would exist some $\lambda<1$ such that $|f(x)-f(y)| \le \lambda|x-y|$ for all $x,y \in I = (0,\frac{1}{2})$. If $f$ is differentiable, then this implies $\lim_{x \to x_0} \left| \frac{f(x)-f(x_0)}{x-x_0} \right| \le \lambda < 1$.
Note that $f'(x) = 2x$, so we see that $f'(x) \to 1$ as $x \to \frac{1}{2}$. In particular, we can choose $x_0$ so that $f'(x_0) > \lambda$.
Then
$\lim_{x \uparrow x_0} \frac{f(x)-f(x_0)}{x-x_0} = f'(x_0)$, and so $\lim_{x \uparrow x_0} |\frac{f(x)-f(x_0)}{x-x_0}| = f'(x_0) > \lambda$, which means that $f$ is not a contraction. |
H: Could anybody check this integral?
in a lengthy calculation by hand I got that
$$ \frac{2}{\pi \sigma_k} \int_{-\infty}^{\infty} \frac{sin^2(\frac{\sigma_k}{2}(v_gt-x))}{(v_gt-x)^2} dx =1$$
Now I was wondering whether there is anybody who could check this ( with a CAS or by hand )? If you decide to do the last option, please note that:
$$ \int_{-\infty}^{\infty} \frac{sin^2(x)}{x^2} dx = \pi$$
Thanks in advance
AI: Let's try together. Assume $\sigma > 0$. Substitute $u = \frac{\sigma}{2} (vt-x)$ to get $du = -\sigma dx/2$ and $[-\infty,\infty] \to [\infty,-\infty]$ we have
$$
\int_{-\infty}^\infty \frac{\sin^2 \left(\frac{\sigma}{2} (vt-x) \right)}{(vt-x)^2} dx
= \int_\infty^{-\infty} \frac{\sin^2 u}{\frac{4}{\sigma^2}u^2} \frac{-2du}{\sigma}
= \frac{-2}{\sigma} \frac{\sigma^2}{4} (-\pi)
= \frac{\sigma \pi}{2},
$$
so sounds like you did it correctly. |
H: show $\Bbb Z_6 \cong \Bbb Z_3 \times \Bbb Z_2 $.
I am trying to determine if $\Bbb Z_6 \cong \Bbb Z_3 \times \Bbb Z_2 $. I noticed that $\Bbb Z_6$ has a generator $1$ and $\Bbb Z_3 \times \Bbb Z_2$ has generator $(1,1)$. Now I set up the bijection $f :\Bbb Z_6 \to \Bbb Z_3 \times \Bbb Z_2 $ where
\begin{eqnarray}
f(1)&=&(1,1)\\
f(1+1)&=&(1,1)+(1,1) = (2,0)\\
f(2+1)&=&(2,0)+(1,1) = (0,1)\\
f(3+1)&=&(0,1)+(1,1) = (1,0)\\
f(4+1)&=&(1,0)+(1,1) = (2,1)\\
f(0)&=&(1,1)+(2,1) = (0,0)\\
\end{eqnarray}
So basically $f(1^n) = (1,1)^n$ where $1^n$ means we add the $1$ to itself $n$ times and take the result mod 6, so as per usual $1^n$ means applying the group operation on 1 to itself for $n$ times. So this is a bijection but I am stuck on showing that $f(ab)=f(a)f(b)$ for all $a,b \in \Bbb Z_6$. I could check all possibilities but there must be a better way to show this. I was thinking something like $f(1^n1^m) = f(1^{n+m})=(1,1)^{n+m}=(1,1)^n(1,1)^m = f(a)f(b) $ because every element in $\Bbb Z_6$ can be expressed as some power of $1$.
Is this right? And is it enough to show $\Bbb Z_6 \cong \Bbb Z_3 \times \Bbb Z_2 $?
AI: Usually the way to prove an isomorphism of this type is not to define a map for every element and then try to show that the map respects the group structure, but rather to define a homomorphism and then use the appropriate isomorphism theorem to prove that the resulting homomorphism is what you want. So here you would define the map on the generator by setting $f(1)=(1,1)$, then say "extend to a homomorphism" between the groups, and then prove that it's onto (or injective) which is enough to show isomorphism. |
H: Is holomorphic functions on (0, 1) (vanishing at endpoint) dense in $C_0((0, 1))$?
Here is my argument, please let me know if it works or not.
By Stone-Weierstrass Theorem (Complex Version), functions in $C_0((0, 1))$ can be uniformly approximated by polynomials in z and $\bar{z}$ which vanishes at 0 and 1. But on (0, 1), $z=\bar{z}$, so it suffices to consider only polynomials in $z$.
Larger Question:
I am trying to define $f(a)$ for $a \in A$, $a$ is an element in Banach algebra with spectrum $\sigma(a)\subset [0,1]$, and $f \in C_0(0,1)$. I know that $f(a)$ can be defined using Holomorphic Functional Calculus1 for $f$ holomoprhic on a neighborhood of [0,1], and if $f_n \to f$ on compact subsets of an open neighborhood of $\sigma(a)$, then
$f_n(a)$ converges. If the above were true, it seems that I can define $f(a)$ for general $f \in C_0((0,1))$ by using approximation with $f_n(a)$ with $f_n$ polynomials vanishing at 0 and 1.
AI: There's good news and bad news. The good news is that you can approximate every function in $C_0((0,1))$ uniformly by polynomials vanishing in $0$ and $1$ on $[0,1]$ per the Weierstraß approximation theorem. The bad news is that if such an approximating sequence of polynomials converges locally uniformly in a neighbourhood of the spectrum, then its limit function is holomorphic in that neighbourhood. So you may not gain anything, unfortunately. If the function you want to approximate is the restriction of a holomorphic function to the spectrum, you can directly use the holomorphic function calculus, and if it isn't, the approximation cannot be uniform in a neighbourhood of the spectrum, and it is in general not possible to deduce the convergence of $f_n(a)$. |
H: Largest Difference of Two Numbers in an Array
Given an integer sequence $s_1,s_2,...,s_n$, I want to find $\max(s_j-s_i)$, where $j>i$.
I wrote the following algorithm (s is a $1$-based array with n elements):
max = 0
for i from 1 to n - 1
for j from i + 1 to n
if s[j] - s[i] > max
max = s[j] - s[i]
It works, but it runs in $O(n^2)$ time. Is there an $O(n)$ algorithm? Is this a disguised instance of the maximum subarray problem?
AI: EDIT: Realized my prior algorithm was flawed. This new one should be right.
EDIT 2: And a second flaw was pointed out years later (despite its simplicity). Thanks @gfppoy
There is an $O(n)$ algorithm broken down into two cases. (Note that figuring out which of the two cases we're in boils down to figuring out whether an array is strictly decreasing, which is itself $O(n)$)
Case 1: The array is strictly decreasing. In this scenario, we simply compare pairs of adjacent numbers and return the value corresponding to the "closest" two.
Case 2: Everything else
Loop through the array, keeping track of the maximum element. The catch is, now we "reset" the maximum every time we find a new minimum element. Something like this in pseudocode:
max_difference = 0
min = s[0]
max = s[0]
for i = 1 to s.length - 1
if s[i] < min
min = s[i]
max = s[i]
else
max = max(s[i], max)
max_difference = max(max - min, max_difference)
That way, for each new "maximum" we find, we associate it with the smallest element to its left, and for that minimum element, we are only considering maxima to its right (and if we don't compare it with a maximum to its right, it will be because we found a smaller element that is still to the left of that maximum).
NB: It's possible to modify the pseudocode to make it cover both cases and still be $O(n)$. I've not done so since reasoning out the correctness of that code becomes unnecessarily complicated; the algorithm presented here is just easier to read and reason about. |
H: Interpreting results concerning the global sections ring being finitely generated
Let $A$ be a ring and $X$ be an $A$-scheme.
(Hartshorne exercise II.2.17) Suppose there exist $f_0,\dots,f_n\in \mathcal{O}_X(X)$ such that
a) $X_{f_i}:=\{x\in X: f_i(x)\not=0\}\subset X$ is an affine open for all $i$,
b) $(f_0,\dots,f_n)=\mathcal{O}_X(X)$.
Then $X$ is an affine $A$-scheme.
How can I get some intuition for this result? We did the proof on the course I'm taking, but I did not find it illuminating.
Then, I ran into the following exercise:
(Liu exercise 3.8) Suppose there exist $f_0,\dots,f_n\in \mathcal{O}_X(X)$ such that $(f_0,\dots,f_n)=\mathcal{O}_X(X)$.
Then there is a morphism $f:X\to P^n_A=Proj \ A[T_0,\dots,T_n]$ such that $f|_{X_{f_i}}:X_{f_i}\to D_+(T_i)$ is induced by the ring homomorphism $A[T_0/T_i,\dots,T_n/T_i] \to \mathcal{O}_X(X_{f_i}), \frac{T_j}{T_i}\mapsto \frac{f_j}{f_i}$.
How should I understand the conclusion? It must be related to the other exercise, too, but I can only see the superficial similarities. Perhaps it is relevant to say that I'm not very much familiar with classical algebraic geometry.
Also, is it alright to understand the hypothesis $(f_0,\dots,f_n)=\mathcal{O}_X(X)$ as a sort of "there exists a partition of unity"?
AI: Yes, $(f_0,\dotsc,f_n)=(1)$ could be imagined as a "partition of unity".
For the exercise in Hartshorne, one uses the canonical morphism $X \to \mathrm{Spec}(\mathcal{O}_X(X))$ (which exists for every locally ringed space $X$, see EGA I, §1.6). The assumptions say that it is an isomorphism locally on the base, hence an isomorphism.
I don't think that this has a connection to the exercise in Liu. More generally, if $f_0,\dotsc,f_n$ are global generators of an invertible sheaf $\mathcal{L}$ on $X$ (the exercise restricts to $\mathcal{L}=\mathcal{O}_X$, don't know why), then this datum corresponds 1:1 to a morphism of $A$-schemes $f : X \to \mathbb{P}^n_A$ which pulls back the Serre twist $\mathcal{O}(1)$ to $\mathcal{L}$ and correspondingly the canonical global generators $T_0,\dotsc,T_n$ of $\mathcal{O}(1)$ to $f_0,\dotsc,f_n$. This is the well-known description of $\mathbb{P}^n_A$ using its functor of points and actually can be seen as the definition of $\mathbb{P}^n_A$ (EGA I, §9).
If you want to get some intuition for the morphism $f$, look at $k$-valued points for a field $k$ (over $A$) and let's restrict to $\mathcal{L}=\mathcal{O}_X$ as in the exercise (which happens to be the case locally anyway). Then we have $n+1$ global sections of $X$, giving rise to a morphism $X \to \mathbb{A}^{n+1}$. Since they vanish nowhere, this factors over $\mathbb{A}^{n+1} \setminus \{0\}$. Now you just compose with the projection to projective space. Thus, $f(x)=[f_0(x):\dotsc:f_n(x)]$. |
H: Conjecture on integer solutions to the equation $ (ab + 1) \mid (a^{2}+b^{2})$
Inspired by the egregious problem in IMO 1988, I simulated the integer solutions to the equation
$$ (ab + 1) \mid (a^{2} + b^{2}) \tag{*}$$
for $1 \leq a, b \leq 3000$ and conjectured that every solution arises as an adjacent pair of numbers in the following sequence
$$ a_{0} = 0, \quad a_{1} = m, \quad \text{and} \quad a_{n+2} = ma_{n+1} - a_{n} $$
for some $m \in \Bbb{Z}$. Indeed, you can check that any pair $(a, b) = (a_{n}, a_{n+1})$ gives rise to a solution to $\text{(*)}$ with
$$ \frac{a_{n+1}^{2} + a_{n}^{2}}{a_{n+1}a_{n} + 1} = m^{2}. $$
I was unable to prove this conjecture as I'm ham-handed at number theory. Can you help me prove or disprove this?
AI: This problem has solve it six years ago in china,can you konw chinese Chinese language?
and this problem full solution on this book(page 457):you can download :
http://ishare.iask.sina.com.cn/f/22755033.html?sudaref=www.baidu.com&retcode=0
because this problem solution is very ugly,so I only post This problem results
we have $(a,b)=(a_{k},b_{k})$,and
$$a_{k}=\dfrac{d}{\sqrt{d^4-4}}\left(\left(\dfrac{d^2+\sqrt{d^4-4}}{2}\right)^k- \left(\dfrac{d^2-\sqrt{d^4-4}}{2}\right)^k\right)$$
and
$$b_{k}=a_{k+1},k=1,2,\cdots,b_{k}\ge a_{k}$$
where $d^2=m=gcd(a,b)$ |
H: Finding a formula for a repeating sequence of 1's and -1's
Is there a simple formula for the sequence $(a_n)$ given by $(1,1,-1,1,1,-1,1,1,-1,\cdots)$
(with the repeating pattern 1,1,-1), starting with $n=1$?
AI: $${ \left( -1 \right) }^{ (n+1)\mod 3 }$$ where $n$ starts from $1$.
Or
$${ \left( -1 \right) }^{ (n+2 ){ \mod 3 } }$$ where $n$ starts from $0$. |
H: Evaluate a Variable Defined in Terms of its Function
I have a variable x which is defined as follows:
x = 150 / (7 + f(x)) where f(x) = 0.005 * x if x > 200, or 100 otherwise.
This is actually a simplified version of a real world problem.
How do I evaluate x?
AI: You have 2 cases. If $x \leq 200$ then $f(x) = 100$ and your definition says
$$
x = \frac{150}{7+f(x)} = \frac{150}{107} < 200
$$
which seems to be one of the solutions.
Otherwise, $x > 200$ and $f(x) = 0.005x$ and your equation defines $x = \frac{150}{7+0.005x}$, which is equivalent to
$$
0.005x^2 + 7x - 150 = 0,
$$
which implies either $x = 100 \left(2\sqrt{13}-7\right) \approx 21.11 < 200$ (which is impossible since we are in the case $x > 200$) or $x = -100\left(7+2\sqrt{13}\right) < 0$ which is impossible for the same reason.
This leaves only one solution possible, $x = \frac{150}{107}$. |
H: On deductively closed theories
Definition A theory $T$ is a set of sentences. A structure $\mathcal{A}$ is a model of $T$ if $\mathcal{A}\vDash T$.
To better understand the situation, let us recall the classical Galois connection machinery used in these situations. Fix a language $\mathcal{L}$; with a set of formulae $T$ one can associate the class of $\mathcal{L}$-structures $\operatorname{Mod}(T)=\{\mathcal{A}: \mathcal{A}\vDash T\}$ and conversely with a class of $\mathcal{L}$-structures $\mathcal{C}$ one can associate a set of formulae $\operatorname{Th}(\mathcal{C})=\{B: \textrm{for all }\mathcal{A}\in\mathcal{C},\mathcal{A}\vDash B\}$. It is quite immediate to show that we have
$$T\subseteq \operatorname{Th}(\mathcal{C})\Leftrightarrow\mathcal{C}\subseteq\operatorname{Mod}(T)$$
for all $T,\mathcal{C}$.
Definition A theory $T$ is said to be deductively closed if $T=\operatorname{Th}(\operatorname{Mod}(T))$.
What I can't understand is what follows: a theory $T$ is deductively closed if and only if it is of the kind $\operatorname{Th}(\mathcal{C})$ for some $\mathcal{C}$.
One implication is trivial (take $\mathcal{C}:=\operatorname{Mod}(T))$. For the other one I'd need some help.
AI: Note that for any $T$, $T \subseteq \mbox{Th}(\mbox{Mod}(T))$ (which just says all the sentences in $T$ in hold in a model of $T$). Now assume $T = \mbox{Th}({\cal C})$. Then ${\cal C} \subseteq \mbox{Mod}(T)$ (which just says that every member of $\cal C$ is a model of $T$). So as the Galois connection reverses inclusions, we have $\mbox{Th}({\cal C}) \supseteq \mbox{Th}(\mbox{Mod}(T))$, so that:
$$T = \mbox{Th}({\cal C}) \supseteq \mbox{Th}(\mbox{Mod}(T)) \supseteq T.$$ |
H: Show that there are infinitely many bijections $f:\mathbb{Z_+} \rightarrow \mathbb{Z_+}$
Show that there are infinitely many bijections $f:\mathbb{Z_+} \rightarrow \mathbb{Z_+}$
Attempt: I dont really know where to start here. A little help is appreciated.
AI: There are lots and lots of ways to do this.
We know that $\mathbb{Z}_+ := \{1,2,3,4,\ldots\}$. For each $k \in \mathbb{Z}_+$ we can define a bijection $\operatorname{f}_k : \mathbb{Z}_+ \to \mathbb{Z}_+$ with the property that $\operatorname{f}_k(1)=k$, $\operatorname{f}_k(k)=1$ and $\operatorname{f}_k(n)=n$ for all $n \neq 1,k$.
In words, you exchange $1$ and $k$, and leave everything else unchanged.
These are all distinct and there are as many of the $\operatorname{f}_k$ as there are $k \in \mathbb{Z}_+$. |
H: Functions between Sets of different Cardinality
Let $\alpha : A \to B$ be a function between finite sets. Show that if $|A| > |B|$, then $α$ cannot be injective, and if $|A| < |B|$, then $α$ cannot be surjective .
AI: Suppose we have $\alpha$ injective with $|A|>|B|$. Then, $\forall b \in B$, $\alpha^{-1}(b)$ must have $0$ or $1$ element. Then, $|A|=|\alpha^{-1}(B)|\leq|B|$, what is a contradiction.
Suppose we have $\alpha$ surjective with $|A|<|B|$. Then $\forall b \in B$, $\alpha^{-1}(b)$ must have at least $1$ element. Then, $|A|=|\alpha^{-1}(B)|\geq|B|$, what is a contradiction. |
H: Showing that $\vec{a}\times\vec{c}=\vec{b}\times\vec{c} \implies \vec{c}\cdot\vec{a}-\vec{c}\cdot\vec{b}=\pm\|\vec{c}\|\cdot\|\vec{a}-\vec{b}\|$
For vectors $\vec{a},\vec{b},\vec{c}\in\mathbb{R}^{3}$, how do I show that:
If $\vec{a}\times\vec{c}=\vec{b}\times\vec{c} \implies \vec{c}\cdot\vec{a}-\vec{c}\cdot\vec{b}=\pm\|\vec{c}\|\cdot\|\vec{a}-\vec{b}\|$
I began by using the geometric definitions of the cross and dot products:
$$\vec{v}\times\vec{w}=\|\vec{v}\|\|\vec{w}\|\sin{\theta}\:\hat{n} \\ \vec{v}\cdot\vec{w}=\|\vec{v}\|\|\vec{w}\|\cos{\theta}$$
Therefore giving us:
If $\vec{a}\times\vec{c}=\vec{b}\times\vec{c}$, then $$\|\vec{a}\|\|\vec{c}\|\sin{\theta_{1}}\:\hat{n}_{1}=\|\vec{b}\|\|\vec{c}\|\sin{\theta_{2}}\:\hat{n}_{2} \implies \hat{n}_{1}=\pm\hat{n}_{2}$$
Therefore we can write:
$$\|\vec{a}\|\sin{\theta_{1}}=\pm\|\vec{b}\|\sin{\theta_{2}}$$
Using the geometric definition for the dot product we therefore have:
$$\vec{c}\cdot\vec{a}-\vec{c}\cdot\vec{b}=\|\vec{a}\|\|\vec{c}\|\cos{\theta_{1}}-\|\vec{b}\|\|\vec{c}\|\cos{\theta_{2}}=\|\vec{c}\|(\|\vec{a}\|\cos{\theta_{1}}-\|\vec{b}\|\cos{\theta_{2}})$$
However, I'm unsure of how to proceed further
AI: Here's how I would do it.
First, $(a-b)\times c = 0$, so
$$\|(a-b)\times c\|^2 = 0.$$
Now I will use the identity
$$\|x \times y\|^2 = \|x\|^2\|y\|^2-(x\cdot y)^2.$$
(You can verify this identity by expanding both sides using your angle expressions for dot and cross product.)
In this case, $x = a-b$ and $y=c$, so
$$0 = \|a-b\|^2\|c\|^2 - [(a-b)\cdot c]^2$$
or
$$[(a-b)\cdot c]^2 = \|c\|^2\|a-b\|^2.$$
Taking the square root of both sides gives me your expression. |
H: $Y^\varnothing$ has one element; if $X \ne \varnothing$, then $\varnothing^X = \varnothing$
From Section 8 of Halmos' Naive Set Theory.
(i) Show that $Y^\emptyset$ has exactly one element, namely $\emptyset$.
(ii) Show that if $X\neq\emptyset$, then $\emptyset^X=\emptyset$.
I'm not completely sure I understand what this means nor how to prove it.
For (i), we want to show that the set of functions $f:\emptyset\to X$ is exactly $\{\emptyset\}$. To me, this means that there is one "map", the "empty map", which associates nothing to something. I'm not really sure how to show this. We know that $Y^\emptyset\subseteq\mathcal P(\emptyset \times Y)=\{\emptyset\}$. Then we can show that $\{\emptyset\}\subseteq Y^\emptyset$ to obtain the equality. So this is vacuously true?
For (ii), since $X\neq\emptyset$, to me this exercise means that we're showing that there are no maps which associate something to nothing. But how to show that there is no function $X\to\emptyset$? Is this just immediate from the definition of function?
AI: The second part is the easier part, perhaps. Note that a function $f\colon A\to B$ must have an ordered pair $(a,f(a))$ for every $a\in A$. If $A$ is non-empty and $f\subseteq A\times\varnothing$, can $f$ be a function whose domain is $A$?
For the first part, yes. You have to show that $\{\varnothing\}=Y^\varnothing$. By showing that $\varnothing$ itself is a function from $\varnothing$ into $Y$ you will obtain this equality. But this is not vacuously true, you have to show that the definition of a function holds for $\varnothing$, when the domain is taken to be $\varnothing$. |
H: How can rate of change be with respect to time if you're not differentiating with respect to time?
A high school Calculus textbook asks: Determine the instantaneous rate of change in the surface area of a spherical balloon (as it is inflated) at the point in time when the radius reaches 10 cm.
My solution: The function relating surface area of a sphere to its radius is $A=4\pi r^2$. So, instantaneous rate of change of surface area with respect to radius is $A'=8\pi r$. When $r=10$, $A'=80\pi$. This answer is in agreement with the text, but what's confusing me are the units. The textbook gives the units as $\mathrm{cm}^2$/unit of time. However, I would think that the units would be $\mathrm{cm}^2/\mathrm{cm}$ because I'm finding rate of change of surface area with respect to the radius, not time. On the other hand, it does seem plausible that the rate of change of surface area would depend on time (as in, how quickly the balloon is being blown up). Also, the question uses the phrase "at the point in time" This is really confusing me.
Is the textbook mistaken, or am I? Any help is appreciated.
AI: You are correct. The units are cm.
The phrasing of the question is quite alright though "at the point in time" where the balloon is 10 cm, the rate of change is 80 pi, regardless of how fast you blew it full.
However, they are not correct in stating that the units are cm2/unit of time, unless the balloon is increasing in size by precisely 1 cm per unit of time. |
H: Pushforward commutes with external tensor product?
Let $f:X\rightarrow X'$ and $g:Y\rightarrow Y'$ be morphisms of varieties. Let $\mathcal F$ be a coherent sheaf on $X$ and $\mathcal G$ be a coherent sheaf on $Y$.
Is it true that
$$(f\times g)_* (\mathcal F \boxtimes \mathcal G)=
(f_* \mathcal F) \boxtimes (g_* \mathcal G)$$?
Here I used the notation
$$\mathcal F \boxtimes \mathcal G:= \pi_X^* \mathcal F \otimes \pi_Y^* \mathcal G $$
where the $\pi$'s stand for the projections $X \leftarrow X\times Y \rightarrow Y$.
Bonus points are awarded for an answer in the relative situation, where $f,g$ are maps of varieties flat over a base variety $B$ and $\boxtimes_B$ is defined using the projections $X \leftarrow X\times_B Y \rightarrow Y$
$$(f\times g)_* (\mathcal F \boxtimes_B \mathcal G)=
(f_* \mathcal F) \boxtimes_B (g_* \mathcal G)$$
AI: Let $S$ be a base scheme and $f : X \to X'$ and $g : Y \to Y'$ morphisms of $S$-schemes. If $F,G \in \mathsf{Qcoh}(X),\mathsf{Qcoh}(Y)$, we have the external tensor product $F \boxtimes G \in \mathsf{Qcoh}(X \times_S Y)$. We have a canonical homomorphism (induced by the usual adjunctions)
$$f_* F \boxtimes g_* G \to (f \times g)_* (F \boxtimes G)$$
in $\mathsf{Qcoh}(X' \times_S Y')$. We ask when it is an isomorphism. Obviously, we may assume that $S$ is affine. Also, we may assume that $X'$ and $Y'$ are affine. Let us write $S=\mathrm{Spec}(R)$, $X'=\mathrm{Spec}(A)$, $Y'=\mathrm{Spec}(B)$. Then $A,B$ are $R$-algebras and $X$ (resp. $Y$) is an $A$-(resp. $B$-) scheme, and the homomorphism above becomes
$$ \Gamma(X,F) \otimes_R \Gamma(Y,G) \to \Gamma(X \times_R Y,F \boxtimes G) ~~~~~ (\star)$$
in $\mathsf{Mod}(A \otimes_R B)$. The actions of $A,B$ don't matter anymore. Clearly $(\star)$ is an isomorphism when $X$ and $Y$ are affine. If $X$ is affine and $Y$ is separated and quasi-compact, then we set up the usual commutative diagram with exact columns to reduce to the affine case, provided that $\Gamma(X,F)$ is flat over $R$. As usual, the same proof can be used for generalizing to the case that $Y$ is quasi-compact and quasi-separated, and then also to the case that $X$ is quasi-compact and quasi-separated, provided that $\Gamma(Y,G)$ is flat.
Summary: When $f: X \to X'$ and $g : Y \to Y'$ are quasi-separated quasi-compact morphisms over some base $S$ and $F \in \mathsf{Qcoh}(X)$ is flat over $S$ and $G \in \mathsf{Qcoh}(Y)$ is flat over $S$, then
$$f_* F \boxtimes g_* G \cong (f \times g)_* (F \boxtimes G).$$
PS: I really would like to have a reference in the literature for this fact! |
H: Integral from zero to infinity of $\int_0^{\infty}\frac{(1-e^{-\lambda z})}{\lambda^{a+1}} d \lambda$
I know that the value of the integral is as follows
$$\int_0^{\infty}\frac{(1-e^{-\lambda z})}{\lambda^{a+1}} d \lambda =z^a \frac{\Gamma(1-a)}{a}$$
However, how exactly the integral is calculated? How one proves that the integral even exists? Why does the integral only exist for $0<a<1$?
AI: Existence: at $\lambda=\infty$, $a>0$ guarantees that the integrand is below $1/\lambda{1+a}$ which is integrable. At $\lambda=0$,
$$
\int_0^\delta\frac{(1-e^{-\lambda z})}{\lambda^{a+1}}\,d\lambda=\int_0^\delta\frac{\lambda z+O(\lambda^2)}{\lambda^{a+1}}\,d\lambda=\int_0^\delta\frac{z}{\lambda^a}+O(\lambda^{1-a})\,d\lambda.
$$
The first term requirest $a<1$ for convergence, and the second one $a<2$. So $0<a<1$ are precisely the values of $a$ for which your integral exists.
Regarding the calculation, I wouldn't know how to find the formula, but here is a way to justify it. It suffices to use differentiation under the integral sign: we have
$$
\frac d{dz}\,\int_0^\infty\frac{(1-e^{-\lambda z})}{\lambda^{a+1}}\,d\lambda=\int_0^\infty\frac{\lambda\,e^{-\lambda z}}{\lambda^{a+1}}\,d\lambda=\int_0^\infty\lambda^{-a}\,e^{-\lambda z}\,d\lambda=\int_0^\infty z^{a-1}\,t^{-a}\,e^{-t}\,dt=z^{a-1}\Gamma(1-a)=\frac d{dz}\,\left(\frac{z^a}a\,\Gamma(1-a)\right).
$$
So the two sides of your equality differ by a constant $c$. When $z=0$ both sides equal to $0$, so $c=0$. |
H: Translate a regular grammar to a regular expression
I want to translate the following grammar into a regular expression:
Set of variables
V := {S,T}
Set of terminals
Σ := {a,b}
Set of relations
S → ""
S → aS
S → bT
T → aT
T → bS
Start variable
S
For the regular expression I can only use the following operations:
concatenation, union and star (no difference)
and I can use the empty word and sigma (set of all words in the alphabet).
I do know how to "formulate" the expression in words... but I have problems to express it in a regular expression.
AI: HINTS:
Look at the grammar, find some stuff that you can produce, find some stuff that you can't produce. For instance, I can produce the strings $a, aa, aaa, aaaa, ...$. I can also produce the strings $bb, bab, baab, ...$. I cannot produce the strings $ab, aba, abaa, abaaa...$
Now, this is written in a way that makes it really easy to convert it into a DFA. Our two states can be labelled $S$ and $T$, and the transitions are (hopefully) fairly obvious. Look at the DFA and try and express as simply as possible what it will and will not accept before turning it into a regular expression. From there, just use the tools you have available.
SPOILERS:
The grammar can be represented by a DFA with state space $\{S,T\}$, accepting state $S$ and starting state $S$. On reading an $a$ we remain in our current state. On reading a $b$, we flip to the other state. The recognised language is the set of all strings containing an even number of $b$s.
The regular expression $a^*ba^*ba^*$ will recognise all strings with exactly two $b$s. The regex $(a^*ba^*ba^*)^*$ will recognise all words where the number of $b$s is a positive multiple of $2$. We can either unite or prepend this regex with $a^*$ to get the language of all words with an even number of $b$s. |
H: Inner Product Space (Continuously Differentiable Functions)
Let $V=C^{2}[-\pi,\pi]$ be the space of real valued twice continuously differentiable functions defined on the interval $[-\pi, \pi]$. Set $$\langle f,g \rangle=f(-\pi)g(-\pi)+\int_{-\pi}^{\pi}f''(x)g''(x)dx$$
Is this an inner product on $V$?
AI: Hint: the function
$$f(x)=x+\pi$$
belongs to $C^2([-\pi,\pi])$ and satisfies $f(-\pi)=0$, $f^{''}(x)=0$ for all $x\in\left[-\pi,\pi\right]$. This implies $\langle f,f\rangle=0$. |
H: Show that if $x>0$, then $\ln(x)\geq 1-\frac{1}{x} $
Show that if $x>0$, then
$$ \ln(x)\geq 1-\dfrac{1}{x}. $$
I tried a few things but so far nothing has worked, I could use a hint.
AI: You can use mean value theorem for $\ln$.
$x = 1$, then equality holds.
$0 < x < 1$:
$$\ln x \ge \frac{x-1}{x} \\
\frac{\ln x - \ln 1}{x- 1} \le \frac{1}{x} \\
\frac{1}{c} \le \frac{1}{x} \\
c \ge x$$
Which indeed is true, because $c \in (x,1)$. Change from $\ge$ to $\le$ is due to division by $(x-1)$ which is smaller than $0$.
$x>1$ similarly as in 2.:
$$\ln x \ge \frac{x-1}{x} \\
\frac{\ln x - \ln 1}{x- 1} \ge \frac{1}{x} \\
\frac{1}{c} \ge \frac{1}{x} \\
c \le x$$
here $c \in (1, x)$ so the inequality is true as well. |
H: What is the relationship between saying "a Taylor series converges for all $x$" and "a Taylor series converges to a function, f(x)"
Given the following Taylor series:
$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}- \dots$
We know that:
It converges for all of $x$
It converges to the function $\cos x$
The Taylor series converge for all of $x$ if for a fixed value of $x$, the partial sums converge to a limit, $L$.
The Taylor series converge to $\cos x$ if its error term is $0$ as $n$ (the number of terms in the Taylor series) goes to infinity.
My question is:
Are these two concepts related?
I thought point 1 would be useful when proving point 2. But when doing the proof of 2, I don't see any connection to point 1. If the two concepts are not related, then why is it useful to know the interval of convergence of a Taylor series (or any series for that matter)?
AI: Clearly they are related: 2. implies 1. But you are correct that using Taylor's Theorem with Remainder, it is sometimes possible to show 2. directly without first establishing the interval of convergence of the power series.
As for 1. Well, for instance what if the function is defined as a power series? E.g. this is perhaps the cleanest definition for the transcendental elementary functions like $e^x$, $\sin x$, etc.
Also, in practice it is usually much easier to determine the radius of convergence of the Taylor series than to apply Taylor's Theorem with Remainder. If one knows for some other reason that a function is analytic -- e.g. if it satisfies a suitably nice differential equation -- then one gets 2. from 1. more or less for free.
Added: Although I feel that the other answers may be missing a bit of the richness of the question (it's a very good question), they do make one good point: the interval of convergence of the Taylor series is certainly an upper bound for the interval on which the Taylor series could converge to $f$. When this interval is not simply all of $\mathbb{R}$ it is useful to know what it is. Consider for instance the case of the Binomial Series: one has to work a bit to get the full interval of convergence out of Taylor's Theorem with Remainder (although it can be done). If we didn't know that the radius of convergence is $1$, how would we know what we are trying to show? You can see the details worked out in $\S$ 12.4.3 of these notes. |
H: Show that F(t) is an immersion
I've got here an exercise that says: "Show that the map $F:\mathbb{R}\rightarrow \mathbb{R^2}$ defined by $F(t)=(\cos t, \sin t)$ is an immersion".
$F$ is an immersion if $dF_x:T_x\mathbb{R}\rightarrow T_{F(x)}\mathbb{R^2}$ is injective. Now $dF_x$ is $(-\sin x, \cos x)^t$ (am I wrong?) and it isn't injective, so it isn't an immersion.
Is it correct?
AI: Note that for all $x \in \mathbb R$ $$\text{rank } (-\sin x,\cos x)^T=1 $$since $\sin x$ and $\cos x$ don't have common zeros. |
H: What is zero choose one?
Algebraically it comes out to be undefined- but if I have zero elements, and I'm asked to pull elements from it, this should just be zero, right?
AI: Combinatorically speaking, you are absolutely right. There is no way to choose $1$ element from an empty set, so $\binom01=0$. It's a good idea to keep in mind that we do not always have $\binom{n}{k}=\frac{n!}{k!(n-k)!}.$ That formula only holds for integers $n,k$ such that $0\le k\le n$. If $n,k$ do not meet all those conditions, then $\binom{n}{k}=0.$ |
H: Homogeneous Equations and Such
"Consider the linear system $\begin{bmatrix} 1 & -2 & 3\\2 & 1 & 4\\1 & -7 & 5\end{bmatrix} * \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$ and find a general solution of the homogeneous system."
So a homogeneous system is just a system of equations that are equal to $0$. Through elementary row operations I find that the system of equations is:
$x_1 - 2x_2 + 3x_3 = 0$
$5x_2 - 2x_3 = 0$
$0 = 0$
Then I think the answer they want is $\underline{v} = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix} +$$\begin{pmatrix}-2.6t \\ \frac{2}{5}t \\ t\end{pmatrix}$ where $x_3 = t$. Is this what they want?
AI: That's the form that I suspect the answer should be; however, your calculation of $x_1$ is a tad off, given your equations: $$x_1 = -3t + \dfrac 45 t = -\dfrac{11}5 t= -2.2 t$$
After edit: The form you wrote in the first post is sufficient. There is no need to add the zero vector to it. What you can do is multiply the vector coefficient vector by $t$, but that too is unnecessary. $$\vec v = \begin{pmatrix}-2.2t \\ \frac{2}{5}t \\ t\end{pmatrix} = \begin{pmatrix}-2.2 \\ \frac{2}{5} \\ 1\end{pmatrix} t$$ |
H: Let R be a relation on set A. Prove that $R^2 \subseteq R <=>$ R is transitive $<=> R^i \subseteq R ,\forall i \geq 1$
this is my first question here. I'm still relatively new to more advanced
mathematics and don't have much experience with proofs yet. I'm self-studying
at the moment and therefore have no one to check whether my proofs
are valid. I hope that math.stackexchange can help me become better
at writing proofs.
I'm currently reading 'How to prove it' by Velleman and have been working through the section on relations.
Now I have found a statement somewhere that I want to prove, but I'm
not sure whether what I have come up with is reasonable and I also
have some questions on the logic used in these type of proofs.
The theorem is: Let $R$ be a relation on set $A$. Then it holds that:
$R^{2}\subseteq R\iff R\text{ is transitive}\iff R^{i}\subseteq R,\forall i\geq1$.
I'll start with $R^{2}\subseteq R\iff R\text{ is transitive}$:
$\Rightarrow$ Assume that $R^{2}\subseteq R$. Here $xR^{2}z$ means
that $xRy\wedge yRz$ for some $y\in A$. The goal is to show that
$R$ is transitive, which is basically a conditional statement: if
$xRy$ and $yRz$ hold, then I can infer that $xRz$. Therefore I
can assume $xRy$ and $yRz$, and now should prove that $xRz$. But
because $R^{2}\subseteq R$, I know that $xRy$ and $yRz$ is true
for an arbitrary $(x,z)\in A$. It follows that $(x,z)\in R$.
$\Leftarrow$ Assume that $R$ is transitive. Therefore if for any
$(x,z)\in A$ it is true that there is a $y\in A$ such that $xRy$
and $yRz$, I know that $xRz$. Now since $R^{2}\subseteq R$ is also
a conditional statement of the form $\forall(x,y)\in A\times A\;(x,y)\in R^{2}\Rightarrow(x,y)\in R$.
I can assume that there is an ordered pair, call it $(x,z)\in R^{2}$.
Since we assumed that $R$ is transitive, it follows by modus ponens
that $(x,z)\in R$.
Now I'll try to prove: $R\text{ is transitive}\iff R^{i}\subseteq R\forall i\geq1$.
$\Rightarrow$Assume $R$ is transitive. To prove that $R^{i}\subseteq R,\forall i\geq1$
I can use induction I guess?!
Base case: $i=1\Rightarrow R\subseteq R$ obviously holds.
Now I'm not really sure whether I actually have to use induction here.
I think what needs to be shown here is pretty straightforward. If
there is a $y\in A$ such that $x1Ry$ and $yRx2$ hold, then because
$R$ is transitive it follows that $x1Rx2$. And then you can continue
this way with $x2$ and $x3$ and in general any value $i$. But I
don't really know how to write this in a formally correct way. Can
anybody help me with that? And please give some feedback on the first
part of the proof.
AI: It looks fairly good, but I'd suggest a few refinements.
In your proof of the first implication, suppose that $\langle x,y\rangle,\langle y,z\rangle\in R.$ By definition, $\langle x,z\rangle\in R^2,$ and so by assumption, $\langle x,z\rangle\in R.$ Showing the converse is similarly done, much more briefly than your approach.
At this point, it would actually be easier to show that $$R^i\subseteq R\:\forall i\ge 1\iff R^2\subseteq R.$$ One of these implications is trivial. For the other, the $i=1$ case is readily true for any relation, and supposing that $R^i\subseteq R$ for some $i\ge 1,$ we can show readily that $$R^{i+1}=R^i\circ R\subseteq R\circ R=R^2,$$ and so $R^{i+1}\subseteq R$ by definition, finishing the induction step. |
H: When does "positive expected value" imply "positive conditional expectation with positive probability"?
Let $(\Omega,\mathcal G,\mathbb P)$ be a probability space, $\mathcal F\subseteq\mathcal G$ a sub-$\sigma$-algebra and $X:\Omega\rightarrow\mathbb R$ $\mathcal G$-measurable map. Assume that $\mathbb E_\mathbb P[X]>0.$ Do we always have $\mathbb P(\mathbb E_\mathbb P[X\mid\mathcal F]>0)>0$? Is there a reasonable condition for this to hold?
AI: By definition $\mathbb{E}[\mathbb{E}[X|\mathcal{F}]]=\mathbb{E}[X]>0$. It follows that $P(\mathbb{E}[X|\mathcal{F}]>0)>0$ by considering the set on which $\mathbb{E}[X|\mathcal{F}]>0$. This is just a restatement of the fact that if $\int_\Omega f> 0$ then $f$ must be positive on a subset of nonzero measure. |
H: Without using any Sylow theorem, if every element is a $p$-element then $G$ is a $p$-group
How can we prove the following theorem without using any Sylow theorem?
Let $p$ be a prime. In a finite group $G$, if every element is a $p$-element then $G$ is a $p$-group.
Or is it possible to generalize the following theorem to three or more subgroups?
If $H_{1},H_{2} \leq G$ then $\lvert H_{1}H_{2}\rvert = \frac{\lvert H_{1}\rvert\lvert H_{2}\rvert}{\lvert H_{1}\cap H_{2}\rvert}$
Thanks!
AI: If you'd like to avoid using Cauchy's theorem, you can use center equation and induction:
Let $G$ be a smallest counterexample. Let $q \neq p$ be a prime divisor of $|G|$. Then, $G$ is simple(why?). Use center equation to get the contradiction $q\mid \lvert Z(G)\rvert$. |
H: Equation of a line through a point on a plane
Find the line that passes through the point $(2, 5, 3)$ and is perpendicular to the plane $2x - 3y + 4z + 7 = 0$
My only real problem with this is how to shift the line
My first step is to find the norm of the plane which is $\vec{n}^{\ } = (2,-3,4)$
Is the answer as simple as $(x, y, z) = (2, 5, 3) + t(2, -3, 4)$
The question counts as 5 marks in the exam and this answer seems very simplistic.
AI: That answer looks correct to me. I think I'd write something like
$L = \{ (2, 5, 3) + t(2, -3, 4) : t \in \mathbb R \}$
to be clear that you were giving a description of the line as a set. I'd call this a "parametric description", since there's a parameter, t, in the answer. If you want an "implicit" description like $ax + by + cz + d = 0$ (i.e., a condition that each point $(x,y,z)$ satisfies if and only if it's on the line, you've got a problem: you'll really need to have TWO equations, since each defines a plane, and therefore the two of them define the line of intersection of the planes (assuming they intersect nicely). But I'll bet that's not what you're supposed to do. Why? (1) It's hard to grade. There are a million correct answers, and it's hard to check that a student got the right one, (2) it's pretty useless in practice, and (3) it's a question that almost no one ever asks, in part because of answer #2. :) |
H: Is $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ a normal extension?
I want to reason as to whether $K=\mathbb{Q}(\sqrt{2}, \sqrt{3})$ is a normal extension over $\mathbb{Q}$. There are competing definitions and so my question is re-phrased as:
Is it true that for every irreducible polynomial $f \in \mathbb{Q}[X]$ which has (at least) one root in $K$, $f$ splits completely in $K$?
Any help that can be offered is appreciated.
AI: Yes, it is. A basic theorem in Galois theory says that a finite extension $E \supseteq F$ is normal and separable if and only if $E$ is the splitting field of a separable polynomial $f(X) \in F[X]$.
It suffices to consider $f(X) = (X^2-2)(X^2-3)$. Its splitting field must contain $\sqrt{2}$ and $\sqrt{3}$, and since $K$ is the smallest field containing both of them, $K$ must in fact be the splitting field of $f(X)$. Therefore $K$ is normal over $\mathbb{Q}$. |
H: Why is it important to define that a logarithm and exponential function is one-to-one?
I'm currently studying the properties of logarithm in an open source pre-calculus textbook that can be found here (Page 438). Before the text goes on to the Algebraic properties of exponential and logarithmic functions it defines the "one-to-one" properties of exponential function and logarithmic functions (Theorem 6.4). As you can see:
And on the next page it describes the algebraic properties of logarithms...
Is there any particular reason for stating the one-to-one property of logarithmic and exponential function beforehand? I know the logarithmic function is the inverse of an exponential function, and as such it can only exists for a one-to-one function, but is there any other reason why this was mentioned? I have the feeling that this distinction is extremely important with regards to the algebraic properties of logarithms that come after (i.e. product rule, quotient rule, power rule) but I cannot see why.
I'm sorry if this was long winded, and thank you for taking the time to read this.
For who ever is so kind as to answer could you bear in mind that I'm studying at a pre-calculus level. Thanks.
AI: No, there's no connection between the injectivity (what you call "one-to-one") property and the various algebraic properties. The theorems just happen to be stated one after another in this particular book.
Knowing that the exponential and logarithmic functions are injective is just a useful fact to know. Really, knowing that any function $f$ is injective is useful if you plan on using $f$ a lot, because it allows you to cancel $f$, in other words, if in your calculations you should end up with:
$$f(x)=f(y)$$
You can immediately conclude that:
$$x=y$$
Which is only true if $f$ is injective, of course. For instance, $x^2$ is not injective, which is why $x^2=y^2$ does not imply $x=y$. |
H: Expected value proof : E(cx)=?
Let $E(x)$ be the expected value of random variable $x$. $c$ is a constant, then what will be the $E(cx)$ in simplest form and why?
AI: For the discrete case:
$$
\mathbf{E} \varphi(X)=\sum_{k=0}^{\infty}c k p_k= c \sum_{k=0}^{\infty}k p_k = c \mathbf{E}X
$$
that's if I understand your question correctly, of course. |
H: trigonometric equation opening
Solve:
$$ \sin x + \sin 3x + \sin 5x = 0 . $$
Attempt at a solution: applying formulas for summation of sine we get after a series of operations: $ \sin x(8 \cos x \cos 2x \cos x + 1) = 0$ equaling sine to $0$ we get one solution $180k$. comparing the other factor we eventually get $\cos^2 x = 0.25 $ this in turn produces the solutions plus/minus $120 + 180 k$ and plus/minus $60 + 180 k$ , giving us $5$ solutions in total. ( $K \in Z $ )
Is this ok?
AI: Yes, expressed in degrees, you've covered all the bases.
(Nit pick: why not specify that $k \in \mathbb Z$?)
Nice work on solving the equation. |
H: Uniform convergence
How can I prove the sequence of functions $x(1-x), x^2(1-x), ...$ converges uniformly on $[0,1]$?
I know from my earlier question here that I can prove it by taking the derivative. But how can I prove it without taking the derivative since we still haven't defined "derivative" yet in my class? I know I can divide the regions into two pieces.
AI: Let $f_n(x) = x^n(1-x)$.
Note that $f_n(0) = f_n(1) = 0$ and $f_n (x) \ge 0$ for $x \in [0,1]$. Hence $f$ has a maximum in $(0,1)$.
Setting $f_n'(x) = 0 $ gives $x = \frac{n}{n+1} $, hence we have $f_n(x) \le \frac{\left( \frac{n}{n+1} \right)^n}{n+1} \le \frac{1}{1+n}$.
Hence $f_n \to 0$ uniformly on $[0,1]$.
Alternatively: Let $\epsilon>0$ and choose $\theta \in (0,1)$ such that $1-\theta < \epsilon$. Now choose $N$ such that $\theta^N < \epsilon$. Then for all $n \ge N$, we have $|f_n(x)| < \epsilon.$
To see this, if $x \in [0,\theta]$, we have $|f_n(x)| \le x^n \le \theta^n \le \theta^N < \epsilon$. If $x \in (\theta,1]$, $|f_n(x)| \le 1-x < 1 - \theta < \epsilon$. |
H: Showing that if $x_k \rightarrow x \implies f(x_k) \rightarrow f(x)$, then $f^{-1}(C)$ is closed for any closed set
As part of proofs on continuity, I should show that
(i) $\forall x \in \mathbb{R}^n,$ if $x_k \rightarrow x \implies f(x_k) \rightarrow f(x)$
implies
(ii) $f^{-1}(C)$ is closed for any closed set $C \in \mathbb{R}^n$
As usual, my approach is to write down clearly what this actually means.
For (i), that is:
$$ (1)\ \forall \epsilon' > 0\ \exists n_0' \in \mathbb{N}\ s.t. ||x - x_n|| < \epsilon' \ \forall n > n_0'$$
$$ \implies $$
$$ (2)\ \forall \epsilon'' > 0\ \exists n_0'' \in \mathbb{N}\ s.t. ||f(x) - f(x_n)|| < \epsilon'' \ \forall n > n_0''$$
For (ii):
$$ f^{-1}(C) = \{ x \in X: f(x) \in C\}\ closed\Leftrightarrow X \setminus \{ x \in X: f(x) \in C\}\ open = \{ x \in X: f(x) \in C^c\} = S$$
Now, for similar proofs in the past I went for a proof by contradiction, which I feel looks also promising here.
So I start by supposing that $S$ is not open, i.e. $ \exists y \in S$ s.t. $\forall r > 0 , B(y,r) \notin S$, i.e. $\exists a \in B(y,r)$ s.t. $a \notin S$.
Then I understand that I should continue that this somehow contradicts with $(1) \implies (2)$, but I can't quite imagine how to establish that.
Thus, I would be very grateful for any hints about how to continue from there.
Thanks in advance!
AI: Hint: Try to use the condition $\exists a \in B(y,r)$ s.t. $a \notin S$ to make a sequence $\{a_k\}$ that converges to $y \in S$. Note that $f(y_k) \in C$ for all $y_k$. Now apply (1). The fact that closed sets in $\mathbb{R}^n$ have the property that if $C$ is closed and $\{a_k\}$ is a sequence in $C$ that converges to $c$, then $c \in C$, will be handy. |
H: In this diagram, why does the limit exist on this interval?
Does the $\lim_{x \to x_0} f(x)$ exist at every point $x_0$ in $(-1,1)?$
I answered False, but the correct answer is True. Why?
My thoughts:
$f(x)$ is not the same number as $x \rightarrow 1$ from the right and from the left, and as $x \rightarrow -1$, only the righthand limit exists. $x=0$ is the only place where the righthand and lefthand limits exist.
AI: Continuity is a pointwise condition for a function. You have to speak of a function $f$ being continuous at $x = x_0$, which as you know, means
$$
\lim_{x \to x_0} f(x) = f(x_0).
$$
The interval notation with the round brackets denotes an open interval (one that does not include its endpoints):
$$
(-1, 1) = \left\{ x \in \mathbb{R} \;\big|\; -1 < x < 1 \right\}.
$$
The fact that $f$ is not continuous at $x = 1$ (you're right) has no bearing on whether the function is continuous at every point in the interval $(-1, 1)$. (It is.) |
H: Isomorphism functorially
I was reading the lecture notes of Pierre Schapira
http://www.math.jussieu.fr/~schapira/lectnotes/AlTo.pdf
I am not able to understand one thing. Please help.
In page 75, theorem 4.6.1, the author says that 'quasi-isomorphism from X to $\lambda(X)$ is functorial in C. What does it mean. What does isomorphism functorially mean in general?
Please help.
AI: It means that given a morphism $f : X \rightarrow Y$, the diagram
$$\require{AMScd}
\begin{CD}
X @>{f}>> Y\\
@VVV @VVV \\
\lambda(X) @>{\lambda(f)}>> \lambda(Y)
\end{CD}
$$
is commutative, where the vertical arrows are the said quasi-isomorphisms. |
H: Necessary and sufficient conditions for a polynomial $f(x) \in \mathbb{Z}[x]$ to be irreducible.
This is not for homework, and I am not totally convinced I understand the question entirely. Also, I'm not allowed to use the fact that $\mathbb{Z}[x]$ is a UFD. The question asks
Show that $f(x) \in \mathbb{Z}[x]$ is irreducible if and only if $f(x)$ is either a prime integer or an irreducible polynomial in $\mathbb{Q}[x]$ such that the $\gcd$ in $\mathbb{Z}$ of the coefficients of $f(x)$ is $1$.
The part of the question I am finding confusing is the phrasing "...or an irreducible polynomial in $\mathbb{Q}[x]$ such that the $\gcd$ in $\mathbb{Z}$ of the coefficients...". What is the $\gcd$ in $\mathbb{Z}$ of the coefficients of a polynomial in $\mathbb{Q}[x]$? Does the author mean to multiply $f(x)$ by an integer $c$ such that the coefficients of $cf(x)$ are integers, and then look at the $\gcd$ of those coefficients? Under this tentative assumption of what the question is asking, here is what I have so far.
If $f(x)$ is a prime integer or an irreducible polynomial in $\mathbb{Q}[x]$ such that the $\gcd$ in $\mathbb{Z}$ of the coefficients of $f(x)$ is $1$, then $f(x)$ is irreducible in $\mathbb{Z}[x]$ (this follows from a theorem that states: "a polynomial $f(x)$ factors as a product of polynomials of degree $m$ and $n$ in $\mathbb{Q}[x]$ if and only if $f(x)$ factors as a product of polynomials of degree $m$ and $n$ in $\mathbb{Z}[x]$").
Conversely, suppose $f(x)$ is irreducible in $\mathbb{Z}[x]$. If $\deg (f(x)) = 0$, then it is not hard to see that $f(x)$ must be a prime integer. If $\deg (f(x)) > 0$, then I can again use the result quoted in the last paragraph to say that $f(x)$ cannot be written as the product of polynomials of lower degree in $\mathbb{Q}[x]$ (for if it could, then $f(x)$ would be reducible in $\mathbb{Z}[x]$). At this point, I'm not sure how to show that the $\gcd$ in $\mathbb{Z}$ of the coefficients of $f(x)$ is $1$. I'm also a bit uneasy about the approach in general. Critiques and hints are very welcome!
AI: The question asks about irreducibility in $\mathbb{Q}[x]$, but the polynomial itself is in $\mathbb{Z}[x]$. So there is always a g.c.d., without anything special happening.
As for your last question: how about a proof by contradiction? What would happen if the g.c.d. of the coefficients were not $1$? |
H: Proof of the limit of a sequence
The sequence is:
$a_n$ = $\frac 1n$ [$(\frac 1n)^2 + (\frac 2n)^2 + (\frac 3n)^2...(\frac nn)^2$]
The objective is a proof of the limit from 1 to infinity. Just from toying around with a few examples, I speculate the limit is (1/3), but I'm not sure. This is how far I am into the proof (basically just the skeleton).
Let $\epsilon$ > 0. Choose N> _. Let n$\in \mathbb{N}$, n>N. We need |$\frac 1n$ [$(\frac 1n)^2 + (\frac 2n)^2 + (\frac 3n)^2...(\frac nn)^2$] - $\frac 13$| < $\epsilon$.
I don't really know how to manipulate this to get something useful. Any advice?
AI: Hint: $$1^2+2^2+\ldots+n^2 = \dfrac{n(n+1)(2n+1)}{6}.$$ |
H: $\sum_{k=0}^{n-1}\sin(k\frac{\pi}{n}+\theta)$
I'm trying to find the closed form of the above formula. This link shows the solution of tan version. Sum of tangent functions where arguments are in specific arithmetic series
Though I'm trying to find the way to apply the argument used in the link to this case, it becomes quite complicated. Could you show another method to get the formula or modify my method? Do you think your formula is similar to the closed formula of sum of cos's?
My method:
Using De Moivre's formula,
$$\sin rx=\binom r1\cos^{r-1}x\sin x-\binom r1\cos^{r-3}x\sin^3x+\binom r5\cos^{r-5}x\sin^5x-\cdots$$
$$=\cos^nx\left(\binom r1 \tan x-\binom r1\tan^3x+\binom r5\tan^5x-\cdots\right)$$ and
If $r$ is even, $=2m$(say)
$$\sin2m\theta=\sin2mx=\cos^n{x}\left(\binom {2m}1\tan x-\binom {2m}3\tan^3x+\cdots+\binom {2m}{2m-1}(-1)^{m-1}\tan^{2m-1}x\right)$$
If $n$ is even, I can convert $$\cos^nx=\frac{1}{1+\tan^{n/2}x}$$
and if I investigate each case of the parity of $m$, I can get the formula of the case in which $r$ is even. But otherwise, I can't. Also, I have to investigate the case in which $n$ is odd.
AI: For sums of sines (and cosines) where the arguments are in an arithmetic progression, we have a much simpler way using Euler's formula:
$$\begin{align}
\sum_{k=0}^{n-1} \sin \left(k\frac\pi n + \theta\right)
&= \operatorname{Im}\sum_{k=0}^{n-1} \exp \left(i\left(k\frac\pi n+\theta\right) \right)\\
&= \operatorname{Im} e^{i\theta} \sum_{k=0}^{n-1} \exp\left(ik\frac\pi n\right)\\
&= \operatorname{Im} e^{i\theta} \frac{1-e^{i\pi}}{1-e^{i\pi/n}}\\
&= \operatorname{Im} \frac{2e^{i\theta}e^{-i\pi/(2n)}}{e^{-i\pi/(2n)}-e^{i\pi/(2n)}}\\
&= \operatorname{Im} \frac{i\exp\left(i\left(\theta-\frac{\pi}{2n}\right)\right)}{\sin \left(\frac{\pi}{2n}\right)}\\
&= \frac{\cos\left(\theta-\frac{\pi}{2n}\right)}{\sin \left(\frac{\pi}{2n}\right)}.
\end{align}$$ |
H: Prove that class of regular languages is closed on operation
Let's have an operation $$\odot(L)=\{w\in L \; | \; |w|=2k \land k>0\}$$Show that result of this operation will be regular.
PS: It's not homework, it's from last year's exam.
AI: The intersection of two regular languages is regular.
Let $E = \{ w : |w| \text{ even}, |w| > 0\}$. It's easy to show that $E$ is regular. It follows that $\odot(L) = E \cap L$ is regular too. |
H: Product of two random variables
How can one show that the product $X \cdot Y$ of two real-valued random variables $X,Y$ is again a random variable?
We can fix some set generating the Borel sigma algebra on the real line, then take for instance an arbitrary open interval, and consider $(X \cdot Y)^{-1}((a,b))$. We need to show it belongs to the sigma algebra on the underlying space $\Omega$.
We could take any $c$ in the interval, write $c = f \cdot \frac{c}{f}$, and consider $X^{-1}(f) \cap Y^{-1}(c/f)$, then take a union over $f$, and then again over $c$. But these are uncountable unions, so the argument doesn't work.
AI: Assuming that you already know that sums and constant multiples of random variables are again random variables, then all we really need to know is that the square of a random variable is a random variable, using the fact that $$XY=\frac14\bigl((X+Y)^2-(X-Y)^2\bigr).$$
As for how to prove that, suppose that $X$ is a random variable on a set $S$. For $\alpha\in\Bbb R,$ we have $$(X^2)^{-1}\bigl((\alpha,+\infty)\bigr)=\begin{cases}S & \text{if }\alpha<0\\X^{-1}\bigl((\sqrt\alpha,+\infty)\bigr)\cup(-X)^{-1}\bigl((\sqrt\alpha,+\infty)\bigr) & \text{if }\alpha\ge 0.\end{cases}$$ Then $(X^2)^{-1}\bigl((\alpha,+\infty)\bigr)$ is measurable for all $\alpha\in\Bbb R,$ and so $X^2$ is a random variable on $S$ whenever $X$ is. |
H: Linear density variation on a 2D plane
Here's my problem:
on a 2-dimensionnal plane, I know the $x$, $y$ coordinates of 3 points $A, B, C$
each point comes with an associated density $T$, that can vary from infinite minus to positive infinite
$T$ values vary in a linear fashion depending on the location of the points,
a linear gradient determines the local value of $T$
But,
I don't know the gradient's direction, nor its "rate" of variation.
What I'd like:
I have a list:
Ax // horizontal coordinate of point A
Ay // vertical coordinate of point A
AT // value of T at point A
Bx // etc.
By
BT
Cx
Cy
CT
From this data I'd like to calculate for any point $E$ $(x,y)$ the value of $T$ at location $E$.
AI: Pick one point at the reference point, and describe the other two as steps from it. Like
$$ (x_A,\,y_A) = (x_C+\Delta x_A,\,y_C + \Delta y_A)
\\ (x_B,\,y_B) = (x_C+\Delta x_B,\,y_C + \Delta y_B) $$
the find the determinant of the system with $D = \Delta x_A \Delta y_B - \Delta x_B \Delta y_A$ and make sure it is $D\neq 0$, or the points are co-linear.
Find the slopes in the $(x,y)$ directions
$$ e_x = \frac{ (T_A-T_C) \Delta y_B - (T_B-T_C) \Delta y_A}{D}
\\ e_y = \frac{ (T_B-T_C) \Delta x_A - (T_A-T_C) \Delta x_B}{D} $$
(edit: e_y should start (T_B-T_C) )
and now your value is
$$ \boxed{ T(x,y) = T_C + e_x (x-x_C) + e_y (y-y_C) }$$
Example
A = (8, 6) T_A = 11.3
B = (12, 3) T_B = -8.2
C = (2, -1) T_C = 0.7
$$ \Delta x_A = 8-2 = 6
\\ \Delta y_A = 6-(-1) = 7
\\ \Delta x_B = 12-2 = 10
\\ \Delta y_B = 3-(-1) = 4
$$
Check $D= (6)(4)-(10)(7) = -46 \neq 0 $
$$ e_x = \frac{(11.3-0.7)\,4 - (-8.2-0.7)\, 7}{-46} = -2.2761
\\ e_y = \frac{(-8.2-0.7)\,6-(11.3-0.7)\,10}{-46} = 4.4652
$$
So now $$T(x,y) = 0.7 -2.2761 (x-2) + 4.4652 (x+1)$$ |
H: Are Base Ten Logarithms Relics?
Just interested in your thoughts regarding the contention that
the pre-eminence of base ten logarithms is a relic from
pre-calculator days.
Firstly I understand that finding the (base-10) logarithm of positive real numbers without a calculator can be reduced to finding the (base-10) logarithm of numbers (strictly) between 1 and 10 via scientific notation
$$\log_{10}(x)=\log_{10}(a\times 10^n)=\log_{10}(a)+\log_{10}(10^n)=\log_{10}(a)+n,\,\,\,(*)$$
and that we could compile (approximate) logarithm tables for $1<a<10$ and hence we can calculate logs base 10. However this can now be done with a calculator... and why would you want to calculate $\log_{10}(x)$ in the first place?
The next reason that we might need $\log_{10}(x)$ is to solve equations like
$$b^x=n.$$
Now we know that $x=\log_bn$ but we can use the change of base "formula" to express this in terms of log base 10. Of course the change of base "formula" comes from a calculation like
$$\begin{align}
\log_{10}(b^x)&=\log_{10}n
\\\Rightarrow x\log_{10}(b)&=\log_{10}n
\\ \Rightarrow x&=\frac{\log_{10}n}{\log_{10}b}.
\end{align}$$
However the new modern calculators can calculate $\log_bn$ in the first place.
Then you could say what about solving
$$b^{f(x)}=c^{g(x)}.$$
Well you don't need to take a base-10 log: we have the perfectly good base $e$ natural log!
In my presently narrow view, it seems to me that it is only stuff like the Richter Scale and sound intensity and similar derived quantities and scales that really use base-10 logs and that while base $e$ logs are clearly useful, that the pre-eminence of base 10 logs is due only to the the by-hand-calculation (*).
To ask a specific question... base $e$ is clearly special:
Are base 10 logs 'special' only because of the "ease" of calculating (or
should I say approximating) logs base 10?
Or am I missing something else? The reason I am looking at this is I have a section of (precalculus) maths notes that is headed "Two Distinguished Bases" and I am thinking of throwing out base 10.
AI: Yes, $10$ is not mathematically significant as a base like $e$ is. Using base 10 logs is strictly for the benefit of non-computer calculation and estimation (which, note, can include such things as simply reading a graph with a scale in dB), and consistency with previously established conventions. This may not be of interest to mathematicians, but I doubt engineers would want to give it up.
For these purposes, $10$ does have at least one useful feature beyond being the base of our number system: $\log_{10} 2 = 0.301 ≈ 0.3$. This is a very common approximation that $3$ dB corresponds to a doubling or halving of the quantity of interest. We could get similar simplicity by using $\log_{2}$, but $\log_2 10 = 3.321…$ which is not nearly as convenient for estimation in decimal numbers.
Choosing base $10$ produces nice nearly-tenth-of-an-integer results for numbers of the form $10^x2^y$ (for integer $x$ and small integer $y$), whereas an arbitrary base $b$ is only guaranteed to be nice for $b^x$.
This suggests further investigation: evaluating bases other than $10$, $2$, and $e$ for having similar almost-integer approximations. I tried writing a program to measure/plot how many good approximations there were for various bases, but it turned out that defining the goodness of an approximation and whether it's good enough to count involves a few too many parameters and I didn't get around to refining it to a result worth sharing. |
H: On the definition of an exact sequence in an abelian category
I am slightly confused about the notion of exactness in a general abelian category (I want to stay clear of anything related to the Mitchell embedding theorem). Here are two definitions that I have seen:
A sequence $A \xrightarrow{f} B \xrightarrow{g} C$ is exact at $B$ if
$\operatorname{Im} f = \operatorname{Ker} g$, in the sense that they are isomorphic as subobjects;
or
$gf = 0$ and the canonical morphism $\operatorname{Im} f \rightarrow \operatorname{Ker} g$ is an isomorphism.
Could someone please shed some light on the equivalence between these two notions? A priori, one seems more general/stronger than the other.
Thanks for your time.
AI: $\operatorname{im}f$ is the kernel of the cokernel, hence is naturally equipped with a morphism $\to B$ and this is a monomorphism (as is always the case for kernels).
If we assume (1), then by the isomorphism of subobjects, $\operatorname{im}f$ factors over $\ker g$ (and vice versa), hence $\operatorname{im}f\to C$ is the zero morphism and finally $g\circ f=0$. To put it differently, the given isomorphism $\operatorname{im}f\to\ker g$ is the canonical morphism obtained from the fact the $\operatorname{im}f\to C$ is zero and so this is an isomorphism.
If we assume (2), then the canonical $\operatorname{im}f\to\ker g$ obtained from $g\circ f=0$ by definition is a factorization of $\operatorname{im} g\to B$ over $\ker g\to B$ and similar for its inverse, so we have $(1)$ |
H: What exactly is a stationary point?
I am asked to find the stationary points of the function $y=5+24x-9x^2-2x^3$. When I looked on the wikipedia page for the definition of stationary points, I read that a stationary point is a point where the derivative equals zero.
However, when I looked at the same article in a different languages, points of inflection were included too, and as far as I know points of inflection are points where the second derivative equals zero. So which one is correct?
AI: $y=x^3$ has a stationary point at $x=0$, even though this is also a point of inflection.
$y=x^3-x$ has two stationary points. It also has a point of inflection, bu that is not one of the stationary points. |
H: What is the least element of this set
What is the least element of the set
$$A=\{a\in \mathbb{R} \,\,| \,\,1<a<2\} $$?
Is incorrect to write $$1,000...1 $$?
AI: The open interval $(1,2)$ has no least element. For every element $a \in (1,2)$ there is an element $b \in (1,2)$ with $b< a$; for example, the point $b = (1+a)/2$ that is halfway between $1$ and $a$.
It is meaningless to denote a real number via its decimal expansion as $1.000\ldots 1$. A decimal expansion has a first digit, a second digit, and more generally an $n$-th digit for every natural number $n$, but that's all. It doesn't have any $\infty^{\text{th}}$ digit, whatever that might mean. |
H: Prove that $(|u-s|+|x-y|)^2\leq 2|u-s|^2+2|x-y|^2$.
Prove that $(|u-s|+|x-y|)^2\leq 2|x-y|^2+2|u-s|^2$.
My professor used this inequality for a proof last week. How would one prove this? I thought about using the Cauchy-Swartz inequality.
This is not a homework question. I am interested in the proof just for self-learning.
AI: $$(a+b)^2\le (a+b)^2+(a-b)^2 = (a^2+2ab+b^2)+(a^2-2ab+b^2)= 2a^2+2b^2.$$ |
H: If two submodules are isomorphic, so is their quotient… conditions on the ring!
In this question Isomorphic quotients by isomorphic normal subgroups it is shown that if we have isomorphic normal subgroups, their quotients need not be isomorphic.
Now, what if we take finite dimensional vector spaces, instead of groups, we do have that if two subspaces are isomorphic, then their quotients are isomorphic (simply these quotients have the same dimension, so their are isomorphic).
Now considering an arbitrary module $M$ over some ring $R$, what conditions do we have to impose on $R$ and $M$ for the above statement to be true? Do modules with this property have a name?
Thanks
AI: The statement holds for finitely generated modules over semisimple ring. More generally, the statement holds fo semisimlple finitely generated modules over arbitrary ring.
Consider a semisimple ring $R$. Then any right $R$-module $M$ is semisimple, which means that it is a direct sum of its simple submodules. If you further take $M$ to be finitely generated, then $M$ is finite direct sum of its simple submodules. (In the case of a general ring, just consider a finitely generated semisimple module $M$. The rest works in both cases.)
Suppose you have $K_1, K_2$ two isomorphic submodules of $M$. Then $M=K_1\oplus C_1=K_2 \oplus C_2$ for suitable submodules $C_1,C_2$ of $M$ (by alternative definition of semisimple module). So you need to show that $C_1 \simeq C_2$. However, consider the decompositions of $K_1,K_2,C_1,C_2$ into direct sums of simple submodules, i. e. say
$$K_1=\bigoplus_{i=1}^{k}S_i, \; K_2=\bigoplus_{i=1}^{k}S'_i, \; C_1=\bigoplus_{j=1}^{m}T_j,\; C_2=\bigoplus_{j=1}^{n}T'_j, $$
where $S_i,S'_i, T_i, T'_i$ are simple submodules of $M$.
Then
$$\bigoplus_{i=1}^{k}S_i\oplus\bigoplus_{j=1}^{m}T_j = M = \bigoplus_{i=1}^{k}S'_i\oplus\bigoplus_{j=1}^{n}T'_j, $$
are two decompositions of $M$ into sum of simple submodules, from which follows (after using the fact, that $K_1 \simeq K_2$ and, basically, Jordan-Holder theorem) that $m=n$ and the summands $T_j, T'_j$ are (in some suitable bijection of indices) isomorphic. |
H: Building a function to display of the sum of matrices at different powers in Matlab
I'm trying to write a for loop for the sum A+A^2+A^3+..+A^n.
Here is my code:
function [ x ] = Untitled2( A , n )
for k=1:n,
x =sum(A^k)
end
The problem I'm having with this is this function is listing the matrix A to each power without adding them. What can I do to add these together? Thank you in advance
AI: Matlab's sum returns the sum along the first dimension not equal to one. If you want to add to x you would be best to use:
function[x]=Untitled(A,n)
x=zeros(size(A));
for k=1:n
x = x + A^k
end |
H: angles of polynomials
Here is an improved question that was asked before.
Let $V$ be the space of real polynomials in one variable $t$ of degree less than or equal to three. Let our inner product be defined by:
$$
\langle p,q\rangle = p(1)q(1)+p'(1)q'(1)+p''(1)q''(1)+p'''(1)q'''(1).
$$
Find the angle between the polynomials given by $p(t)=-4+9t-6t^2+t^3$ and $q(t)=3-6t+3t^2$.
Can someone please help me to solve this? I have been trying to solve this for days and can not come up with an answer. I really need someone to help me with this answer please.
AI: The angle $\theta$ is given by $\cos\theta=\frac{\langle p,q\rangle}{\langle p,p\rangle^{1/2}\langle q,q\rangle^{1/2}}$ |
H: Find height of a triangle given length of three sides?
How can I find the height of a triangle given the length of all three sides?
The only solution I could find was to use Heron's formula to find area then $A=\frac{1}{2}bh$ to find height. Is there an easier way to do this?
I'm going to be using this in a piece software I'm developing, so a simple solution makes things much easier.
AI: Besides the solution you stated using Heron's formula, you can use the Law of Cosines to find one of the angles. The height is then easily obtained by dropping the appropriate perpendicular and using the definition of cosine. |
H: Solve the equation $\tan \theta = 2\sin \theta$.
Solve the equation $\tan \theta = 2\sin \theta$.
What I did was rewrite it to the form $$\sin \theta = 2 \sin \theta \cos \theta$$ You'll get $$\sin \theta = \sin\ 2 \theta.$$
How am I supposed to solve this when I have $\sin$ on both sides? My main problem with these types of 'solve' equations are that I don't know which forms I should rewrite them too. Usually just to the form $\sin \theta = n$, but I wonder if having a $\sin$ on both sides can result in an answer too.
AI: HINT:
$$\tan\theta = 2\sin\theta \iff \sin \theta = 2 \sin \theta \cos \theta \iff \sin\theta(1 - 2\cos\theta)=0$$
ADDED: I am rewriting this in a form where you can "read off" solutions. $$ab = 0\; \iff \;a = 0 \;\text{ or }\; b = 0$$ So in the case at hand, we have that
$$\sin\theta(1 - 2\cos\theta)=0 $$ $$\iff\sin \theta = 0,\;\text{ or } \; 1 - 2\cos \theta = 0 $$ $$\iff \sin \theta = 0 \;\text{ or }\; \cos\theta = \frac 12$$ |
H: If there exist open, disjoint sets $A, B$ in $X, d$, then is $\bar A \cap B = A \cap \bar B = \emptyset$?
Clearly $A, B$ are separated in $X, d$ if $A, B$ are closed and disjoint since $(A \cup A') \cap B = \emptyset$ and vice versa. For disjoint open sets, I cannot come to a conclusion. I tried constructing a set = {the limit points of $A\} = C$ but then what? Any hints?
AI: Since $B$ is open and disjoint from $A$, then $X\setminus B$ is a closed superset of $A.$ Since $\overline A$ is the smallest closed superset of $A$, then $\overline A\subseteq X\setminus B,$ and so $\overline A\cap B=\emptyset.$ Likewise, $A\cap\overline B=\emptyset.$ |
H: Help with this integral? I can't figure out the substitution.
I've been struggling to figure out this integral.
$$\int \frac {1}{x\sqrt{5-x^2}}$$
I'm almost certain it has something to do with this fact:
$$\int \frac 1{\sqrt{1-x^2}} = \sin^{-1}(x) + C$$
But I can't figure out how to use that to my advantage. No obvious substitution is jumping out at me either. I'm guessing there's something simple I'm missing... any help will be much appreciated!
Edit (11/13/13): An update on this, just for the sake of finishing what I started - I had gotten a bit stuck on this problem and could not seem to figure it out until I saw it worked out in this thread on reddit. Where I was stuck turned out to be a very simple algebra manipulation. Just thought I'd add that link for the sake of any Googlers or what have you who might find this post and want to see it worked out in more detail. Many thanks to all involved in helping me understand this problem!
AI: A natural substitution is $u=\sqrt{5-x^2}$, though I prefer the version $u^2=5-x^2$. Then $2u\,du=-2x\,dx$.
Rewrite our integral as
$$\int \frac{x\,dx}{x^2\sqrt{5-x^2}}.$$
After we make the substitution, we end up with
$$\int \frac{du}{u^2-5}.$$
Factor $u^2-5$ as $(u-\sqrt{5})(u+\sqrt{5})$ and use partial fractions. |
H: Closed unit disk homeomorphic to $\mathbb{R}^2$?
I've already shown the existence of a homeomorphism between the open unit disc and $\mathbb{R}^2$ and now I'm trying to work out whether the closed unit disc is homeomorphic to $\mathbb{R}^2$ or not. Clearly the only real difference is the fact it includes the points actually on the circle, but because the continuous, invertible, bijective map I defined for the open disc was $\dfrac{x}{1-|x|}$ this clearly is not defined for the points on the line.
Intuitively I'm thinking it's not possible to form a homeomorphism with the closed disk precisely because of these points on the circle but can't really form a decent argument. I was thinking along the lines of showing that the non-open, closed subset $\bar{D}(0,1) \backslash D(0,1)$ (which is just the set of points on the actual circle i.e the boundary/frontier of the closed disk) can't possibly bijectively map to a closed subset of $\mathbb{R}^2$ - which clearly would have to be the case in a homeomorphism.
Any advice would be greatly appreciated.
AI: Perhaps the most immediate argument to show that the closed disk is not homeomorphic to $\mathbb R^2$ is that the former is compact while the latter is not. This argument is not quite in line with the intuition you were trying to formalize. You intuition can be formalized to some degree by using some concepts from homotopy theory. The fundamental group of $\mathbb R^2$ with a point removed is $\mathbb Z$, while the fundamental group of the closed disk with a boundary point removed is trivial. Thus the two spaces are not even homotopy equivalent, let alone homeomorphic.
Thinking in terms of paths (basically the argument above but without assuming you know the fundamental group is), any path (that is a continuous function from $[0,1]$) in the disk with a boundary point removed can be continuously deformed into a point (just shrink the path to the centre of the disk). But, not every path in the plane with a point removed can be so contracted. If the path is a circle around the missing point then it can't be deformed to a point, the missing point keeps you from doing that.
The property of deformability of paths is an invariant of homeomorphisms (and of homotopy equivalences) and thus the two spaces are not homeomorphic. |
H: Determine number of function given two sets and properties
Let A={1,2,3,4,5,6,7} and B={v,w,x,y,z}. Determine the number of functions $f:A \rightarrow B$ where (i) f(A)={v, x}; (ii) |f(A)|=2
For (i) the answer key gives 2!S(7, 2) and (ii) $\binom{5}{2}[2!S(7,2)]$
I don't understand where these answers come from. So the sterling number of the second kind shows how many non empty subsets of size 2 can be made from the domain A but why is 2! used and where does the $\binom{5}{2}$ in (ii) come from?
AI: To answer your first question, the factor of $2!$ arises from the following. If you look at the subset of $A$ that maps to $v$, and similarly for $x$, you obtain a partition of $A$ into two parts, but those parts are unordered. Exchanging the roles of $v$ and $x$ gives a distinct function. How many ways are there to rearrange the two element set ${v, x}$?
$$2!$$
As for your second question, out of the five element set $B$, you must choose $2$ elements to be the range, and then the previous calculation applies. How many choices are there?
$$
\binom{5}{2}
$$ |
H: $f=g$ almost everywhere $\Rightarrow |f|=|g|$ almost everywhere?
Suppose $(X, \mathcal{M}, \mu)$ is a measure space. Assume $f: X\to\overline{\mathbb{R}}$ and $g=X\to\overline{\mathbb{R}}$ are measurable maps. Here $\overline{\mathbb{R}}$ denotes the set of extended real numbers. My question is:
If $f=g$ almost everywhere, does it follow that $|f|=|g|$ almost
everywhere?
I know the answer is "Yes" if $X$ is a complete measure space: If $f=g$ a.e. then $E=\{x\in X: f(x)\neq g(x)\}$ is a null set, i.e. $\mu(E)=0$. It is clear that
$$
F=\{x\in X : |f(x)|\neq |g(x)|\}\subseteq E
$$
Since $X$ is complete, all subsets of null sets are in $\mathcal{M}$, and so $\mu(F)=0$, and $|f|=|g|$ a.e.
What happens when $X$ is not complete?
Thanks for your time :)
AI: Since $f,g$ are measurable, so are $f-g$ and $f+g$. Let
$$
E=\{x\in X:\ f(x)=g(x)\}, \ \ E'=\{x\in X:\ f(x)=-g(x)\}.
$$
If $F$ is as in the question, the set where $|f|\neq|g|$, then we have $F^{c}=E\cup E'$. And then $F$ is measurable, because both $E$ and $E'$ are:
$$
E=(f-g)^{-1}(\{0\}),\ \ E'=(f+g)^{-1}(\{0\}).
$$ |
H: Proving a relation's inverse's properties by knowing the original's.
I'm getting fairly confused with two exercises related to proving a relation's inverse's properties by knowing the original's. I couldn't do either. Any hint is appreciated.
If $R$ is a symmetric relation over $A$ with $A \not = \emptyset$,
prove that $R^{-1}$ is also symmetric.
$R$ is symmetric, so we know that $(aRb \implies bRa)$.
We want to prove that $R^{-1}$ is symmetric. If we use the same $a,b$ as before, we could write it like this: $(bRa \implies aRb)$
So we want to prove $(bRa \implies aRb)$.
The rightmost part of that is $aRb$, which, according to our first premise, it implies $bRa$. I think we can replace it:
$$bRa \implies (aRb \implies bRa)$$
Uh, maybe we can simplify this with implication/disjunction:
$$b\not R a \lor (a\not Rb \lor bRa)$$
$$a\not Rb \lor V_0$$
This isn't quite right... But I'm way too lost - what should I have done?
If $R$ is a transitive relation over $A$ with $A \not = \emptyset$,
prove that $R \cap R^{-1}$ is also transitive.
Frankly, I'm lacking even the slightest clue here.
The only think I could observe was that since $R$ is transitive, $R^{-1}$ is transitive as well (is this okay to state, or do I have to prove it?). Not sure how would that help me though.
AI: I'm not sure what definitions you are using, but the idea should be something like this:
$R^{-1}=\{(b,a):(a,b)\in{R}\}$. Let $(a,b)\in{R^{-1}}$. As a result ${(b,a)\in{R}}$. Since $R$ is symmetric, $(a,b)\in{R}$. So by definition $(b,a)\in{R^{-1}}$. Hence we have that $R^{-1}$ is symmetric.
As for the other question: Suppose $(a,b),(b,c)\in{R\cap{R^{-1}}}$. We would like to show that $(a,c)\in{R\cap{R^{-1}}}$. Now $(a,b),(b,c),(b,a),(c,b)\in{R}$. Since $R$ is transitive we have $(a,c),(c,a)\in{R}$ by using transitivity on $R$. So we have $(a,c)\in{R^{-1}}$ which gives us what we need. |
H: How to compute the integral $\int_{-\infty}^\infty e^{-x^2}\,dx$?
How to compute the integral $\int_{-\infty}^\infty e^{-x^2}\,dx$ using polar coordinates?
AI: Hint: Let $I=\int_{-\infty}^\infty e^{-x^2}\,dx.$ Then $$I^2=\left(\int_{-\infty}^\infty e^{-x^2}\,dx\right)\left(\int_{-\infty}^\infty e^{-y^2}\,dy\right)=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-x^2}e^{-y^2}\,dx\,dy=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}\,dx\,dy.$$ Now switch to polar coordinates. |
H: Modular congruence rules
May take exponent of both sides of a modular congruence?
For instance, may I write
$$n^2\equiv-1\mod p \quad \rm as \quad(n^2)^{2k+1}\equiv (-1)^{2k+1}\mod p ?$$
AI: If $p|(a-b)$ does $p|a^k-b^k$? |
H: Proving that a certain set is convex
I'm trying to prove that the set $W = \{x \in \mathbb{R}^2 : 2x_{1}^{2} + 3x_{2}^{2} \leq 4\}$ is convex. I've been trying to do this using the definition of W being convex when for all $x, y \in W $ and every $\lambda \in [0,1] $ we have $\lambda x + (1-\lambda)y \in W $.
What I've done (I use $a$ in place of $\lambda$) thus far is rewrite $$2(ax_{1}+(1-a)y_{1})^{2}+3(ax_{2}+(1-a)y_{2})^{2}$$ to $$a^{2}(2x_{1}^{2}+3x_{2}^{2})+(1-a)^{2}(2y_{1}^{2}+3y_{2}^{2})+2a(1-a)(2x_{1}y_{1}+3x_{2}y_{2})$$
The part which I'm having trouble with is showing that $$2a(1-a)(2x_{1}y_{1}+3x_{2}y_{2}) \leq 2a(1-a)4 $$ or simply showing that
$$2x_{1}y_{1}+3x_{2}y_{2} \leq 4$$
How should I show that this is the case? Or if this is not necessary - how is it obvious that this is the case?
AI: Let $$u=\begin{bmatrix}x_1\sqrt2\\x_2\sqrt3\end{bmatrix}\qquad\text{and}\qquad v=\begin{bmatrix}y_1\sqrt2\\y_2\sqrt3\end{bmatrix}.$$
Cauchy-Schwarz inequality tells us that $$|u\cdot v|\le\sqrt{(u\cdot u)(v\cdot v)}\\|2x_1y_1+3x_2y_2|\le\sqrt{(2x_1^2+3x_2^2)(2y_1^2+3y_2^2)}.$$ Can you take it from there?
If you aren't familiar with Cauchy-Schwarz, then you can proceed in this way:
If $u=0,$ then the claim is easily shown, so suppose not. Note then that $$u\cdot\left(v-\frac{u\cdot v}{u\cdot u}u\right)=u\cdot v-\frac{u\cdot v}{u\cdot u}(u\cdot u)=u\cdot v-u\cdot v=0.$$ Hence, we have by dot-product properties that $$\begin{align}v\cdot v &= \left(v-\frac{u\cdot v}{u\cdot u}u+\frac{u\cdot v}{u\cdot u}u\right)\cdot\left(v-\frac{u\cdot v}{u\cdot u}u+\frac{u\cdot v}{u\cdot u}u\right)\\ &= \left(v-\frac{u\cdot v}{u\cdot u}u\right)\cdot\left(v-\frac{u\cdot v}{u\cdot u}u\right)-2\frac{u\cdot v}{u\cdot u}u\cdot\left(v-\frac{u\cdot v}{u\cdot u}u\right)+\left(\frac{u\cdot v}{u\cdot u}u\right)\cdot\left(\frac{u\cdot v}{u\cdot u}u\right)\\ &= \left(v-\frac{u\cdot v}{u\cdot u}u\right)\cdot\left(v-\frac{u\cdot v}{u\cdot u}u\right)+\frac{(u\cdot v)^2}{(u\cdot u)^2}(u\cdot u)\\ &= \left(v-\frac{u\cdot v}{u\cdot u}u\right)\cdot\left(v-\frac{u\cdot v}{u\cdot u}u\right)+\frac{(u\cdot v)^2}{u\cdot u}\\ &\geq \frac{(u\cdot v)^2}{u\cdot u},\end{align}$$ so $$(u\cdot u)(v\cdot v)\geq(u\cdot v)^2,$$ and so $$|u\cdot v|=\sqrt{(u\cdot v)^2}\le\sqrt{(u\cdot u)(v\cdot v)}.$$ |
H: Proving that $\mathbb{A}=\{\alpha \in \mathbb{C}: \alpha \text{ is algebraic over } \mathbb{Q} \}$ is not a finite extension
It is true that all finite field extensions are algebraic. It is not true however that all algebraic extensions are finite.
In lectures we were given the example of the field extension
$$\mathbb{A}=\{\alpha \in \mathbb{C}: \alpha \text{ is algebraic over } \mathbb{Q} \}$$
I want to prove to myself that this is indeed a non-finite algebraic extension.
I have shown that $\mathbb{A}$ is a subfield of $\mathbb{C}$. Now I consider roots of the equation $x^n-2$ for $n \geq 1$. Clearly such roots are in $\mathbb{A}$ by definition. I think that if I show that for a given $k$, the root of $x^k-2$ (say $\psi_k$) does not lie in $\mathbb{Q}(\psi_1,...,\psi_{k-1})$, then I have proven the extension is not finitie (since this will hold for all $n$).
Can someone please confirm this reasoning and suggest how the proof could be finished.
AI: Your idea is correct, and can be formalized quickly as follows. The polynomial $X^n-2$ is irreducible over $\mathbb Q$. Let $\alpha $ be a root. Then $[\mathbb Q(\alpha):\mathbb Q]=n$. But $\mathbb Q(\alpha)\subseteq \mathbb A$, and so $\mathbb A$ contains subfields of arbitrarily large degree over $\mathbb Q$, and so itself must be infinite dimensional over $\mathbb Q$, as needed. |
H: What is the name for this operator, and how can it be applied to multiple variables within the same equation?
My question is in two parts; the first is, what is the $|$ operator called? Here's an example of it in use:
$$(x + 5)|_{x=3} = 8$$
My second question is, how do I use this operator for more than one variable? For example, would I write $(x + y)|_{x=3,y=3}$? Or $(x + y)|_{x=3}|_{y=3}$? Or another variation?
Any help is appreciated.
AI: The operator $\mid$ usually reads as "such that" or "given that".
In your example, it works pretty clearly: Take $(x+5)$, given that $x=3$. Then $3+5 = 8$.
As another example, $A = \{x \mid x^2 = 4\}$ is the set of all $x$ such that $x^2 = 4$.
So to respond to your second question, you would write $(x+y)|_{x=3,y=3}$. (Given the informal definition I wrote above, the second variation wouldn't make a whole lot of sense.) |
H: What other definite integrals can be computed in a manner similar to $\int_{-\infty}^\infty e^{-x^2}dx$?
The technique for computing $\int_{-\infty}^\infty e^{-x^2} dx=\sqrt{\pi}$ by computing the integral squared using polar coordinates is well known. Are there any other integrals that can be computed in a similar way? My question is intentionally vague--any techniques however tangentially related are of interest!
(EDIT 1: To be more specific, what I am imagining is whether it is possible that some definite integrals appear algebraically out of a computation of a double or triple integral in some coordinate system. Or, if there are some heuristic arguments that this probably impossible. Or if someone has a better imagination as to a computation being similar to the example I gave is also good.)
(EDIT 2: It would also be interesting if someone could example how one might come up with the computation in the original example instead of it being seen as an after the factor sort of thing.)
AI: That is essentially the only integral that this trick is good for. See
http://www.unf.edu/~dbell/Poisson.pdf |
H: Why is $\mathbb R^n$ under the Zariski topology not a topological group?
Reasons that $(\mathbb R^n, +, \mathcal Z)$ is not a topological group: Given any two distinct points $\vec{p},\vec q \in \mathbb R ^n$ let $P$ be the unique hyperplane through $\vec p$ which is perpendicular to the vector $\vec p- \vec q$. $P$ is closed with respect to $\mathcal Z$ and hence its complement is an open set containing $\vec q$ but not $\vec p$. Therefore the space is $T_0$. If it were a topological group, this would imply it were Hausdorff, which is not the case.
Reasons that $(\mathbb R^n, +, \mathcal Z)$ should be a topological group. It is both a group an a topological space. Moreover all polynomials $(\mathbb R^n, \mathcal Z) \to (\mathbb R, \mathcal Z)$ are continuous. So any map of the form $\vec x \mapsto (p_1(\vec x), p_2(\vec x), \ldots, p_n(\vec x)) $ or $(\vec x, \vec y ) \mapsto (p_1(\vec x, \vec y), p_2(\vec x, \vec y), \ldots, p_n(\vec x, \vec y) $ where $p_i$ are polynomials, is continuous, and the operations of addition and negation are both of this form.
Could someone point out where my misapprehension?
AI: The problem is that the product topology is not the same as the Zariski topology. So, while maps $\mathbb{R}^{2n}\to\mathbb{R}$ are continuous, polynomial ones, this is in the Zariski topology, not the product topology.
In fact, every $T_0$ topological group is already Hausdorff. Every variety over $\mathbb{R}$ is $T_0$ but is not Hausdorff.
EDIT: Let me emphasize precisely what goes wrong. In your argument, you claim that the map $\mathbb{R}^n\times\mathbb{R}^n\to \mathbb{R}^n$ given by $(x,y)\mapsto x-y$ is continuous, since it has polynomial coefficients, where $\mathbb{R}^n\times\mathbb{R}^n$ is given the product topology. If this were true, then the preimage of $0$ would be closed, since $0$ is closed in $\mathbb{R}^n$ with the Zariski topology. But, the preimage is precisely the diagonal $\Delta_{\mathbb{R}^n}\subseteq\mathbb{R}^n\times\mathbb{R}^n$. But, the diagional being closed is equivalent to being Hausdorff, and since $\mathbb{R}^n$ is not Hausdorff with the Zariski topology, this is a contradiction. Thus, the map $(x,y)\mapsto x-y$ is not continuous. |
H: Find the series expansion of 2 multiplied functions
The first three terms in the series expansion of $(1+x)^m$ are $1 + mx + \dfrac{m(m-1)x^2}{2}$.
Find the first 3 terms in the series expansion of $(1+x)^{m+1}(1-2x)^m$.
I don't really know how to do this, because I have always done these binomial expansion using binomial coefficients and Pascal's triangle.
AI: The best way is to change your variables around so you don't get confused between $x$'s, $m$'s, and other variables. I recommend rewriting your original expansion as follows:
$${(1+x)}^m \implies {(1+u)}^t = 1 + tu + \frac{t(t-1)u^2}{2}$$
Then think of the second expression as using the arguments with particular values of $t$ and $u$:
$${(1+x)}^{m+1}{(1-2x)}^m \implies {(1+(x))}^{m+1}{(1+(-2x))}^{m} \\ = {(1+u_1)}^{t_1}{(1+u_2)}^{t_2}$$ where $u_1 = x; \quad t_1 = m+1; \quad u_2 = -2x; \quad t_2 = m$.
Then you can expand the expression above, using the first expansion, for the subscripted $u$'s and $t$'s to get:
$${(1+u_1)}^{t_1} = 1 + t_1u_1 + \frac{t_1[t_1-1]{u_1}^2}{2} \\ = 1 + (m+1)(x) + \frac{(m+1)[(m+1)-1](x)^2}{2} \\ = 1 + mx + x + \frac{(m+1)(m)x^2}{2}$$
And similarly for the second expression:
$${(1+u_2)}^{t_2} = 1 + t_2u_2 + \frac{t_2[t_2-1]{u_2}^2}{2} \\ = 1+ (m)(-2x) + \frac{(m)[(m)-1]{(-2x)}^2}{2} \\ = 1-2mx+\frac{m(m-1)4x^2}{2} \\ = 1-2mx+2m(m-1)x^2$$
Now all that needs to be done is multiply the two halves of the original product ${(1+u_1)}^{t_1}{(1+u_2)}^{t_2}$:
$${(1+u_1)}^{t_1}{(1+u_2)}^{t_2} = \left( 1 + mx + x + \frac{(m+1)(m)x^2}{2} \right) \left( 1-2mx+2m(m-1)x^2 \right)$$
At this point, the product is just tedium to find the first 3 terms of the expansion you're looking for. |
H: How many telephone numbers have no $0$ in the prefix (first three numbers)
I got that the total number of telephone numbers: $10^{10}$, but should I do the number of one $0$ in prefix, number of two $0$ in prefix and number of three $0$ in prefix, and subtract them. But I don't know how to count the number of $0$ in the prefix.
AI: If all numbers have $10$ digits, $3$ in the prefix and $7$ in the remainder, and your only limitation is that you can not have a $0$ in the prefix, you should have:
$$9^3\cdot 10^7$$
Possibilities. If say, you could not have either of two digits in the prefix (i.e. a choice of $8$ digits, you would simply replace the $9$ by an $8$. |
H: Integration by Substitution problem
I was given an integration problems sheet...with answers too but how a certain answer is to be obtained is obviuosly not stated.
Using integration by substitution integrate the following:
$$ \int \dfrac{5x +3}{\sqrt{3-x^2}} \, dx$$
And the answer at the back is:
$$ -5 \sqrt{3-x^2} +\arcsin \left( \frac{x}{\sqrt{3}}\right) +C$$
Any idea how i go about the substitution?
AI: First note that
$$ \int \frac{5x +3 }{\sqrt{3-x^2}}dx= \int \frac{5x }{\sqrt{3-x^2}}dx
+\int \frac{ 3 }{\sqrt{3-x^2}}dx$$
For the first integral you can use the substitution $u=3-x^2$ and for the second, express the denominator as $ \sqrt{1-(\frac{x}{\sqrt {3}})^2}
$ |
H: Non-empty interior
I would like to verify a certain condition for a theorem. It requires that $\{(\theta,\rho\theta):\rho>0,\theta>0\}$ has a non-empty interior. I have trouble visualizing such sets can you guys help me see if this has a non-empty interior (and if it does, why?)
AI: The set is the first quadrant with both axis removed which it is open, so in particular has a nonempty interior. |
H: An integrable Functions is almost everywhere finite
An integrable Functions is almost everywhere finite
Attempt: Let $X = \{ x : f(x) \, \text{ is infinite}\} $. We must show $m(X) = 0$. Suppose $m(X) > 0 $. then on $X$, we have
$$ \int\limits_X f \, dm > \infty$$
which implies $f$ cannot be integrable: contradiction.
Is this a correct solution? thanks
AI: Look at $[|f| > n]$; you have $|f| \ge |f|\chi_{[|f| > n]}$ and apply dominated convergence. |
H: Check to see if my isomorphism is correct
Is multiplication modulo $10$ isomorphic to addition modulo $4$?
$U(10) = \{1,3,7,9\}$,
the identity is $1$,
it is a cyclic group of order $4$, with generator $3$.
$\Bbb Z_4 = \{0,1,2,3\}$,
the identity is $0$,
it is a cyclic group of order $4$
\begin{gather*}
1 \mapsto 0 \\
3 \mapsto 1 \\
9 \mapsto 2 \\
7 \mapsto 3
\end{gather*}
(Is this map sufficient to show multiplication mod $10$ is isomorphic to addition mod $4$?
AI: As IBWiglin says in the comments, you need to check that this map $f$ is a group homomorphism. That means for every $a,b \in U(10)$ you need to check $f(ab) = f(a) + f(b)$. You can easily check that it is true here.
More generally if $a,b$ are coprime numbers, then $U(ab) \cong U(a) \times U(b)$ and $U(p)$ is isomorphic to $\mathbb{Z}_{p-1}$ for every prime number $p$. So in your case:
$$\begin{align*}
U(10) & \cong U(2) \times U(5) \\
& \cong \mathbb{Z}_1 \times \mathbb{Z}_4 \\
& \cong \mathbb{Z}_4
\end{align*}$$ |
H: Why isn't $2\log(-1)$ real?
In high school we learn that a $a\log[(x)] = \log (x^a)$
From this I would assume $2\log(-1) = \log [(-1)^2]$
However, the first is not real and the second is, according to my calculator and textbook. Why is this?
AI: The formula $a\log x=\log x^a$ requires that $\log x$ exists. Here you want $\log(-1)$; this would a number such that $e^{\log(-1)}=-1$. But it turns out that the exponential of real numbers is always positive, so to write $\log(-1)$ you somehow need to extend the log function to the complex plane.
Notice that if you keep going from your desired equality, you get
$$
2\log(-1)=\log(-1)^2=\log 1 = 0,
$$
which would imply that $\log(-1)=0$. What this shows is that the property $a\log x=\log x^a$ does not hold for arbitrary complex numbers when you extend the log function to the complex plane.
Regarding your first question, $\log(-1)$ cannot be real because the exponential of a real number is always positive. |
H: Use implicit function theorem to show $O(n)$ is a manifold
In class today our teacher mentioned that one can use the implicit function theorem to show that $O(n) \subseteq \mathbb{R}^{n^2}$ is a submanifold...that is, map $A \mapsto A^* A$, and set it equal to the identity matrix. There should be $n(n+1)/2$ equations in $n^2$ variables ($A^*A$ is always symmetric, so we only need $n(n+1)/2$ equations). When I asked how to show that the jacobian matrix of this transformation had full row rank (the condition of the implicit function theorem), my teacher said "take the derivative near the identity, and you see that the derivative is just a matrix plus its transpose ."
I wasn't sure what he meant. Can someone clarify?
AI: Let $f:\mathbb R^{n^2}\to Sym$ denote the map $A\mapsto AA^*$. Pick $A\in f^{-1}(I)$, $$Df_A(B)=\lim_{t\to 0} \frac{f(A+tB)-f(A)}{t}=\lim_{t\to 0} \frac{(A+tB)(A^*+tB^*)-I}{t}=\lim_{t\to 0} AB^*+BA^*+tBB^*$$
$$ \lim_{t\to 0} AB^*+BA^*+tBB^* = AB^*+BA^*$$
Thus, $Df_A(B)=AB^*+BA^*$. |
H: Logical representation of a prime number
Is it correct to represent a prime number like this?
$$\exists k \in \mathbb N,\, \exists n\in \mathbb N\, \Big((n\mid k) \land (n=k \lor n=1)\Big)$$
AI: What you have claimed is that there are two natural numbers, $n$ and $k$, such that $n$ divides $k$ and $n=k$ or $n=1$. This is true. For instance, let $k=5$ and $n=1$. What you want is that $k$ is prime if whenever $n$ divides $k$, $n$ must be $1$ or $k$. That is, if $k$ is prime then
$$
\forall n\in \mathbb{N}, \left((n\mid k)\Rightarrow (n=1)\vee (n=k)\right).
$$ |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.