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H: Continuity - Function of 2 variables
Verify if the following function is continuous:
$$ f(x,y) = \left\{\begin{matrix}
\sqrt{1-x^2-y^2},\; if \;\;x^2 + y^2 \leq 1\\
0,if \;\; x^2 + y^2 > 1
\end{matrix}\right.$$
I think the only possible "problematic point" is where $x^2 + y^2 = 1$. So what should I do? Calculate the limit when $x^2 + y^2 \to 1$ or we must consider more specific cases? I don't know how to write it formally.
Thanks!
AI: In the interior and exterior of the unit disk, the functions describing those cases are continuous. The only potential problem is on the unit circle. But at each point of the unit circle. you have a limit of zero and a function value of zero there, too. So the function is continuous on the entire plane. |
H: $\dim U + \dim U^\perp = \dim V$. Example for being wrong when not nondegenerate
$\beta\colon V \times V \to K$ is a nondegenerate symmetric bilinear form and $U$ is a linear subspace of $V$. Then this equation is proven: $\dim U + \dim U^\perp = \dim V$.
I need to find an example of
(1) it being wrong, when $\beta$ is not nondegenerate and
(2) why not in every case $V = U \oplus U^\perp$
I already made the proof of the equation being right, but I do not know any examples in which this is wrong.
AI: Consider the case when $V = \mathbb{R}^2$. Let $\beta(v,w)=\langle v, A w\rangle$ (the standard dot product) where $A$ is the $2 \times 2$ matrix given by
$$
A = \left(
\begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array}
\right).
$$
Now, take $U = V$. Then
$$
U^{\perp} = \operatorname{span}\left\{ \left( \begin{array}{c}0 \\1 \end{array} \right) \right\}
$$
and you get $\operatorname{dim}(U)+\operatorname{dim}(U^{\perp}) = 3 > \operatorname{dim}(V)$. Moreover, since in this case $U \cap U^{\perp} \neq \varnothing$, it is no longer true that $V = U \oplus U^{\perp}$. |
H: Let $\,g$ be the function defined on the set of real numbers by.....
I was thinking about the following problem which is as follows:
My Attempt: Case 1: Let $c$ be the arbitrary rational number.Let $\{x_n\}$ be the sequence of irrational numbers that converge to $c$. By density property that is $\exists$ an irrational number between any real numbers,assures us that such a sequence does exist.
Now,since $g(x_n)=0, \forall n \in \Bbb N$,we have $\lim_{n \to \infty} g(x_n)=e^{x_n}=e^c\,\,$ while $g(c)=1.$ Therefore,$g$ is not continuous at $c$. In the same way,we can show by taking $c$ be the arbitrary irrational number, $g$ is not continuous at $c$.
So, I think option 1 is the correct choice. Am I right?
My next question: Can someone explain how to find the number of ideals in the ring $R[x]\over {x^2-1}$ ?
AI: Your idea is correct - almost! First, your notation is a bit confusing. If $(x_n)$ is a sequence of irrational numbers wich converges to a rational number $c$, then you have
$$
\lim_{n\to\infty}g(x_n)=\lim_{n\to\infty}e^{x_n}=e^c.
$$
But $e^c\neq 1=g(c)$, UNLESS...? So your function is not continuous at $c$ UNLESS...? For $c$ being irrational you have indeed discontinuity. |
H: Why do we subtract 1 when calculating permutations in a ring?
$10$ persons are to be arranged in a ring shape. Number of ways to do that is $9!.$
I wonder why we subtarct $1$ in all such cases.
I can imagine that if A,B,C,D are sitting in a row then B,C,D,A would give me a different combination but had they been sitting on a circular table then both the above combinations would imply the same thing. Same for CDAB or DABC.
But how does it all lead to $(n-1)!$ formula? Individual cases, I am able to imagine but overall how are we generalizing it?
AI: Imagine that the table is arranged so that one of the seats is due north of the centre of the table. Seat the $n$ people around the table. Let $p_1$ be the person sitting in that north seat, and let the other $n-1$ be $p_2,p_3,\ldots,p_n$ clockwise around the table. Now rotate the table and the seats one place counterclockwise: $p_2$ is now in the north seat, and the others going clockwise around the table are $p_3,p_4\ldots,p_n,p_1$. Repeat: $p_3$ is now sitting to the north, and clockwise around the table we have $p_4,p_5,\ldots,p_n,p_1,p_2$. If we keep doing this, we bring each person in succession to the northern spot, and after $n$ of these rotations everything and everyone is back in its original position. From the standpoint of absolute compass directions we had $n$ different arrangements, but from the standpoint of the order of people around the table these $n$ arrangements were all the same. Thus, each cyclic permutation of the diners corresponds to $n$ different absolute permutations, i.e., permutations in which we care about the absolute seating position of each diner and not just who sits next to whom. Since there are $n!$ absolute permutations, there must be $\frac{n!}n=(n-1)!$ cyclic permutations.
Alternatively, you can think of it this way: given a seating of the diners around the table, you can designate any seat as the head of the table and list the diners clockwise around the table starting at the head. That gives $n$ different lists, but they all correspond to the same cyclic permutation of the diners. Since $n!$ different lists are possible, there must again be $\frac{n!}n=(n-1)!$ different cyclic permutations, each corresponding to $n$ of the lists. |
H: Differentiable manifolds, uniqueness of maximal atlases and definition of smooth manifolds maps.
I have proved that given an atlas for a topological space $M$ that a maximal atlas containing $M$ is unique. But my proof would fail to generalise to the statement that a maximal atlas conatining a chart is unique. Is this true?
Also given a map $f : M \mapsto N$. Then I have seen it written that the smoothness of $f$ is independant of the choice of chart. (Definition give here http://www.cis.upenn.edu/~cis610/cis61005sl7.pdf on page326, definition 6.15)
I can't see how this is independent from the way we 'chart' the topological spaces $M$ and $N$.
I would appreciate if someone could clear this up for me! Thanks!
AI: You are correct that there is not a unique maximal atlas containing a given chart. It is possible to take a system of charts (atlas) on a manifold and adjust some of them to get an incompatible system of charts (i.e. they extend to different maximal atlases), but both systems will have many charts in common. If you'd like, I can go into more detail about how to do this.
Regarding your second question, the smoothness of $f$ is independent of the choice of chart in a fixed atlas. In the definition, they pick a pair of charts $(U,\phi)$ and $(V,\psi)$ around $p$ and $q$, and test for $C^{k}$ differentiability there. However, if it works for one pair of charts, it works for any pair, so you can be sure that if you take $(U',\phi')$ and $(V',\psi')$ around $p$ and $q$ that you will get a $C^k$ map as well. You can prove this by using the chain rule and noting that the transition functions between charts are $C^k$ maps. |
H: Value of $ f(2012)$
$f(x) $ is an injective function . The definition of $f(x)$ is like following:
$$
f:[0, \infty[\to \Bbb R-\{0\},
f\left(x + \frac{1}{f(y)}\right) = \frac{f(x)f(y)}{f(x) + f(y)} $$
If $f(0) = 1$ then what is the value of $ f(2012)$?
Can you help me to solve this problem ?
AI: $f\left(x+ \frac{1}{f(y)} \right) = \frac{f(x)f(y)}{f(x)+f(y)}$ and $f(0)=1$
i.e., for $x=-\frac{1}{f(y)}$ we have $\frac{f(x).\frac{-1}{x}}{f(x)-\frac{1}{x}}=1$
$\frac{-f(x)}{xf(x)-1}=1$ i.e.,$-f(x)=xf(x)-1$ i.e., $f(x)(x+1)=1$ i.e., $f(x)=\frac{1}{x+1}$
This would give $f(x)$ for any $x$ in the domain.
in particular $f(2012)=\frac{1}{2013}$ |
H: Geometry Question - What is the length of the missing height?
Gary, $G$, can just see the top of a radio mast, $R$, over a wall $W$. Gary is $15\,\mathrm{m}$ from the wall. The wall is $45\,\mathrm{m}$ from the radio mast. The wall is $2.7\,\mathrm{m}$ high. Calculate the height of the radio mast, marked $h$ on the diagram.
How would you found the missing length $h$? I was thinking of doing Tan, Cos and Sin but I'm not that sure and I also had the idea of doing $15$ $+$ $45$ $=$ $60$ and then $2.7$ something. I know the answer is $10.79$ but I have no idea how that was worked out. Thanks.
AI: The smallest triangle is similar to the larger triangle, so we can use ratios to solve for the height $h$ of the larger triangle. $$\dfrac {\text{(height)}_\text{large}}{\text{(base)}_{\text{large}}} = \dfrac {\text{(height)}_{\text{small}}}{\text{(base)}_{\text{small}}}$$
$$\dfrac{h}{15+45} = \dfrac{2.7}{15}$$ |
H: Find $p$ and $q$ so that the integral converges
Find all values of $p$ and $q$ so that the below integral converges:
$$
I=\int_{0}^{1} x^p \left(\log\frac{1}{x}\right)^q\;\mathrm{d}x
$$
I tried and got the solution as:
$q\geq0$ and $p>q-1$
$-1<q<0$ and $p>-1$
Is it correct?
Solution:
After Substituting $x=e^{-y}$ the integral becomes:
$$
I=\int_{0}^{\infty} e^{-(p+1)y} y^q\;\mathrm{d}x
$$
AI: Hint:
1) $\ln (\frac{1}{x}) =-\ln(x) $
2) make the change of variables $\ln x = y$. |
H: How is the norm of a partition related to the norm of a vector?
Just finished a course in linear algebra, where the norm of a vector essentially was described as the length of the vector. In calculus, we just started talking about the definite integral of a function, where the norm of a partition came up, being defined as the max size of a subinterval given a set of subintervals.
Are these two uses of norm related, and if so how?
AI: The max size of the absolute value of a finite list of numbers is the infinity-norm of that list. This is an example of a p-norm. For $p=2$ you get the Euclidean norm that you're familiar with. |
H: Right Triangles and Altitudes
I am once again stuck on a question about geometry, this problem is about altitudes that crate right triangles:
Let there be a triangle that has side lengths of 13, 20, and 21. Given this, find the length of the altitude drawn to the side of length 21.
I have drawn the following picture to make my understanding clearer:
However, I am still not sure how to get the length of this altitude. My definition of an altitude is a segment drawn from one vertex to a point on the line opposite the point such that this segment is perpendicular to the line opposite the vertex.
I am very sure that to find the length, I will have to use the pythagorean theorem at some point, but I am not sure how to start this.
If anybody could give a starting point or hint that would be great :)
Note: I cannot use the cosine rule.
Thank you in advance.
AI: Let $a$ be altitude, and x be the base of the right-most right triangle.
$$20^2 = a^2 + x^2$$ $$13^2 = a^2 + (21 -x)^2$$
$$a^2 = \color{blue}{\bf 20^2 - x^2 = 13^2 - (21-x)^2 }$$
$$400 - x^2 = 169 - (441 - 42x + x^2) = -272 + 42x - x^2$$
$$672 = 42x \iff x = 16.$$
$$x = 16 \implies a = \sqrt{400 - x^2} = \sqrt{400 - 256} = 12$$ |
H: Are these conditions enough to specify a unique number? (counting sylow p-subgroups)
Sylow's third theorem gives the two facts that the number of sylow p-subgroups $n_p$ of a group $G$, whose order we can write as $|G| = p^rm$ such that $p\not |\ m$ will satisfy both $n_p | m$ and $n_p \equiv 1\ (\mod\ p)$. I'm just wondering if this is enough to specify $n_p$ uniquely, or if there may still be ambiguity.
If you're wondering why I'm asking what might be such a basic question; I know next to nothing about number theory!
AI: No, it's not enough. Take $|G| = 12 = 2^2 \cdot 3$, for example. Then $n_2 \equiv 1 \pmod{2}$ and $n_2 | 3$ gives that $n_2 \in \{1, 3\}$.
(In fact, there's no reason to take $2^2 \cdot 3$; $|G| = 6$ works just as well). |
H: Calculating a derivative using the chain rule vs. differentiating the composite
I have two functions $f:\mathbb{R} \rightarrow \mathbb{R}^2$ , $t \mapsto (t^3,t^2)$
and $g : \mathbb{R}^2 \rightarrow \mathbb{R}$ $(x,y) \mapsto (x^2+y^2)^{\alpha}$
Then we are asked to calculate $ (g \circ f)'(t)$ by using two different methods: First, the chain rule and second: by directly inserting f in g and then just differentiating with respect to t.
After that, we are supposed to say, what the difference is, but unfortunately I get in both cases $(g \circ f)'(t) = \alpha (t^6 + t ^4 )^{\alpha - 1} (6 t^5+ 4 t^3)$
Maybe it is something about the existence of the derivatives...
Unfortunately nobody here left a comment about the choice of $\alpha$. Can anybody comment on this, whether this could be a problem?
AI: Daniel Fischer noted in a comment on my now-deleted answer that if $\frac14<\alpha<1$, then $(t^6+t^4)^\alpha$ is differentiable, but $g$ is not, so your concerns were justified. |
H: Intuitive explanation of sum^2
The following equation can be easily proved with induction (using $(a+b)^2=a^2+2ab+b^2$):
$$\left(\sum\limits_{i=1}^n a_i\right)^2 = \sum\limits_{i=1}^n a_i^2 + 2\sum\limits_{1\leq i<j\leq n} a_ia_j$$
Do you have some intuitive explanation why is this equation true?
AI: Added an illustration of Arthur's answer:
"[..] draw up a square with sides ∑ai long and subdivided for each i. Then you can just see what the big square is built up from. Your identity is two different ways of calculating the area of the square (LHS is just side^2, RHS is the sum of all the small squares and rectangles)" |
H: Integral constraints for positive function
Let $C={\cal C}([0,1],(0,\infty))$ denote the set of all continuous
maps $[0,1]\to (0,\infty)$. Let $g_1,g_2 \in C$ ; one can then define
$$
\begin{array}{rcl}
\Phi &: C& \to (0,\infty)^2 \\
f &\mapsto& \bigg(\int_{[0,1]} fg_1,\int_{[0,1]} fg_2\bigg) \\
\end{array}
$$
Obviously, when $g_1$ and $g_2$ are not linearly independent, say $g_1=ag_2$ for some $a>0$, the image of $f$ is the diagonal
$\lbrace (i,ai) | i > 0\rbrace$. I believe that when $g_1$ and $g_2$ are linearly independent,
$\Phi$ is always surjective. Can anyone help me to show this ?
AI: Your conjecture is false.
Let $m$, $M$ be positive numbers such that $m \le g_1$ and $g_2 \le M$. Then, for all $f \in C$,
$$\frac{\int f g_1}{\int f g_2} \ge \frac{m \int f}{M \int f} = \frac{m}{M}.$$
In particular,
$$\frac{\int f g_1}{\int f g_2} \ge \frac{\min g_1}{\max g_2},$$
and, by the same kind of manipulations,
$$\frac{\int f g_1}{\int f g_2} \le \frac{\max g_1}{\min g_2}.$$
Hence, the couple $(\int fg_1, \int fg_2)$ always live in a cone. What you can show or study is the following:
you can show that the image of $\Phi$ is always a cone: it will be convex, and it has a decomposition in half-lines.
when is the cone is open, closed, semi-open? This will depend on, among other things, whether the maximum/minimum above are reached on an open set or not.
find when the estimates above are optimal, i.e. give the boundary of the cone.
The answer are rather easy when the sets where $g_1$ and $g_2$ reach their maximum / minimum are disjoint, but seems to be thorny otherwise (for instance if $g_1$ and $g_2$ reach their maximum at the same point). |
H: Subharmonic functions in the punctured disk
I want to prove the following (exercise from Ahlfors' text):
If $\Omega$ is the punctured disk $0<|z|<1$ and if $f$ is given by $f(\zeta)=0$ for $|\zeta|=1$, $f(0)=1$, show that all functions $v \in \mathfrak B(f)$ are $\leq0$ in $\Omega$.
Here $\mathfrak B(f)$ is the class of all functions $v$ which satisfy:
(a) $v$ is subharmonic in $\Omega$.
(b) $\limsup_{z \to \zeta} v(z) \leq f(\zeta)$ for all $\zeta \in \partial \Omega$.
After quite some time with no success, a short google search gave me this , which has some hints, but I need to get the full book in order to understand the references.
Any help on solving this problem will be highly appreciated!
Thanks!
AI: Given an arbitrary $v \in \mathfrak{B}(f)$, consider the functions
$$u_\varepsilon(z) = v(z) + \varepsilon\cdot \log \lvert z\rvert$$
for $\varepsilon > 0$. Since $\log \lvert z\rvert$ is harmonic in the punctured disk, $u_\varepsilon$ is subharmonic. Since $\limsup\limits_{z\to 0} v(z) < \infty$, we have $\lim\limits_{z\to 0} u_\varepsilon(z) = -\infty$. Also, $\log \lvert z\rvert \equiv 0$ on the unit circle.
By the maximum principle(1), $u_\varepsilon \leqslant 0$ in the punctured disk. Let $\varepsilon \to 0$.
(1) If we had $u_\varepsilon(z_0) = c > 0$ for some $z_0 \in \Omega$, then $u_\varepsilon$ would attain a global maximum in $\Omega$, hence be constant ($\equiv -\infty$):
Since $\limsup_{z\to \zeta} \leqslant 0$ for all $\zeta\in \partial \Omega$, each $\zeta\in\partial\Omega$ has an open neighbourhood $U_\zeta$ with $u_\varepsilon \leqslant c/2$ on $U_\zeta$.
$$K = \Omega \setminus \bigcup_{\zeta\in\partial\Omega} U_\zeta$$
is then a compact subset of $\Omega$ containing $z_0$, in particular non-empty. The upper semicontinuous (or even continuous) function $u_\varepsilon\lvert_K$ attains its maximum in $K$, say at $z_1$. Then $u_\varepsilon$ attains its global maximum in $\Omega$ in the point $z_1$. |
H: Peculiarities about Finding Absolute Values
It is known that the formula $\sqrt{x^2}$ is equal to the value of $|x|$. In my spare time last night, I wondered about $\sqrt[3]{x^3}$. After some thought and some graphing, I came up with this:
If $x<0$, $\Im(\sqrt[3]{x^3})$
If $x=0$, $0$
If $x>0$, $\Re(\sqrt[3]{x^3})$
This system is equal to $|x|$.
Why is this?
AI: But you're wrong: For $x\in \mathbb R$, $$\sqrt[3]{x^3} = x$$ plain and simple.
Take a simple example: $$\sqrt[3]{(-1)^3} = -1 \neq |-1| = 1$$ |
H: Proof of a Lemma guaranteeing the existence of the Borel-measurable functional calculus
In my lecture I had the following Lemma, which guarantees the existence of the Borel-measurable functional calculus:
Le $(H,<,>)$ be a complex Hilbert space and let $q:H\rightarrow \mathbb{C}$ be a function s.t.
1) $\vert q(x)\vert\leq C\cdot\Vert x \Vert^2$, for some $C>0$ and all $x\in H$
2) $q(x+y)+q(x-y)=2q(x)+2q(y)$, for all $x,y\in H$
3) $q(\lambda x)=\vert \lambda \vert^2 q(x)$, for all $x\in H$ and $\lambda\in\mathbb{C}$.
Then there exists a unique bounded operator $T:H\rightarrow H$, s.t. $q(x)=<Tx,x>, \forall x\in H$.
Unfortunately I do not understand the proof given in the lecture. First of all we defined $g(x,y):=\frac{1}{4}\left( q(x+y)-q(x-y)+i\cdot q(x+iy)-i\cdot q(x-iy) \right)$. Then we defined $Tx:=\sum_{i\in I}g(x,e_i)e_i$, whereas $(e_i)_{i\in I}$ is an ONB. Now my professor said, that the following equation is satisfied: $g(x,y)=<Tx,y>, \forall x,y\in H$. But why is this the case? How can I calculate this?
I hope, you can help me.
AI: Suppose $y \in H$, then
$$
y = \sum_j \langle y,e_j\rangle e_j \qquad\qquad\text{(1)}
$$
and so
$$
\langle Tx,y\rangle = \sum_i g(x,e_i)\langle e_i, y\rangle
$$
$$
= \sum_i \sum_j g(x,e_i) \overline{\langle y,e_j\rangle} \langle e_i, e_j\rangle
$$
$$
= \sum_i g(x,e_i)\overline{\langle y,e_i\rangle}
$$
Now if you use the continuity of $g(x,\cdot)$, and (1), then you will see that
$$
g(x,y) = \sum_i g(x,e_i)\overline{\langle y,e_i\rangle}
$$
(You need to check first that $g(x,\alpha y) = \overline{\alpha}g(x,y)$ for any $\alpha \in \mathbb{C}$)
Remark : Perhaps it is better to phrase this lemma in 2 parts :
Given a function $q:H\to \mathbb{C}$ such that the conditions in your question hold, then, $g(x,y)$ given by that equation defines a continuous conjugate linear form on $H$
Given a continuous conjugate linear form $g$ on $H$, there is a bounded linear operator $T$ on $H$ such that
$$
g(x,y) = \langle Tx,y\rangle
$$
The second lemma is then proved using the Riesz Representation theorem, and the entire lemma is perhaps a little more transparent (there is no need to refer to the series expansion as used above). |
H: Projection matrix that sum to identity are orthogonal
Can anyone help me to show if $Z_1$, $Z_2$, and $Z_3$ are projection matrices (i.e. idempotent and symmetric) and if $Z_1 + Z_2 + Z_3 = I_n$, then we can conclude:
$$\forall i \ne j \text{ ,} Z_i Z_j = 0$$
It seems to be an easy problem but I was not able to solve it.
Thanks,
AI: In fact if $P+Q$ is a projection, then $PQ = 0$, because $(P+Q)^2 = (P+Q)$, and so
$$
PQ(x) + QP(x) = 0 \Rightarrow PQ = -QP
$$
and hence for any $x \in ran(PQ)$,
$$
PQx = x \Rightarrow Px = x
$$
and similarly,
$$
-QPx = x \Rightarrow Qx = x
$$
and hence $x = -QPx = -Qx = -x$, and so $x = 0$, which means that $PQ = 0$
Now apply this to $P = Z_i, Q=Z_j$ for $i\neq j$ |
H: Complex value of a divergent series
Given the series:
$$S=\sum_{k=1}^{\infty}\frac{2^k}{k^2}$$
the sum obviously doesn't converge.
'Maple' gives for the value of the series:
$$S(a)=\sum_{k=1}^{\infty}\frac{a^k}{k^a}$$
$S(a)=Li_a(a)$ with $Li$ polylogarithmic function.
For $a=2$, for example, $Li_2(2)\approx 2.46-2.17i$.
The question is: if the series is not convergent and is a sum of real number, how is it possible to get a complex value for $S(a)$?
Thanks.
AI: Lets take a look at $S_i(x) = \sum\limits_k \frac 1 {k^i} x^k$. This is a power series with radius of convergence $\lim\limits_{k \to \infty}\frac { (\frac 1 {k+1})^i} {(\frac 1 {k})^i} = 1$. So the representation $S_i(x) = \operatorname{Li}_i(x)$ is only valid for $|x| \leq 1$, and not for $x=2$.
However, through analytic continuation, we can extend this function to $|x| > 1$, but are not guaranteed a real valued output for real valued input. |
H: Proving basic lemmas about categories with finite products and terminal/initial objects.
I would expect that in any category $\mathcal{C}$ with finite products and a terminal object $1$, the isomorphism $X \times 1 \cong X$ should hold, but I have a rather hard time finding the proof of this.
In my attempt, I use the fact that there are arrows $1_X : X \rightarrow X$ and $\top_X : X \rightarrow 1$, from which the universal property of $X \times 1$ gives me a unique $u : X \rightarrow X \times 1$ such that $\pi_1 \circ u = 1_X$ (and $\pi_2 \circ u = \top_X$). This gets me halfway, but I still need to prove that $u \circ \pi_1 = 1_{X \times 1}$ to prove that $u$ is indeed an iso.
I can easily obtain that $\pi_1 \circ u \circ \pi_1 = \pi_1 = \pi_1 \circ 1_{X \times 1}$, so as far as I can see, the problem can be reduced to proving that $\pi_1$ is a mono.
However, how to get there? Is this even provable, or do I need extra assumptions about $\mathcal{C}$? If it is provable, would $X \times 0 \cong 0$ be provable as well (assuming initial objects)?
Thanks!
AI: To show that $u \circ \pi_1 = id$ consider that $u \circ \pi_1$ is a morphism from $X\times 1 \to X\times 1$ such that $\pi_1 \circ u \circ \pi_1 = \pi_1$ and $\top_{X\times 1} = \top_{X\times 1} \circ u \circ \pi_1$. Since these equations also hold if you replace $u \circ \pi_1$ by $1_{X\times 1}$, you get by uniqueness that $u \circ \pi_1 = 1_{X\times 1}$.
The property $X \times 0 = 0$ does not hold in general. Consider the category of groups, where $1 = 0$, so $X \times 1 = X$ and since there are nontrivial groups the property clearly doesn't hold.
The property does however hold if, for instance, the category is cartesian closed. |
H: Solution Verification - Combinatorial Card-Picking
I have a problem as such:
How many ways are there to choose nine cards out of a standard deck of 52 cards in such a way that every suit is represented in the selection at least twice?
Here's my solution:
Partition the set into the four suits - then pick two cards from each suit: there are ${13 \choose 2}^4$ ways to do this. Then there are $44$ cards left over, of which one has to be picked. ${44 \choose 1} = 1$. So the total number of ways the goal can be achieved is $44 \cdot {13 \choose 2}^4$.
The trouble is that this gives me something like $1.6$ billion ways, which seems unrealistically large. Have I made a mistake?
AI: If you need that "every suit is represented in the selection at least twice" in the selection of 9 cards, so you have 3 cards of one suit and 2 cards of each other suit.
you can pick 2 cards out of 13 in ${13 \choose 2}$ different ways and you can pick 3 cards out of 13 in ${13 \choose 3}$ different ways. And if you have 4 different suits, so you need to multiply the answer by 4. There are $4 \cdot {13 \choose 2}^3 \cdot {13 \choose 3}$ different ways :)
542887488 :) |
H: Finding the image of $f(x)=\frac{1}{1+x^{2}}$
$f(x)=\frac{1}{1+x^{2}}$ and $x\geq0$
To find the image:
$y=f(x)$
$y=\frac{1}{1+x^{2}}$
$x=y^{-1}$
$x=\sqrt{y^{-1}-1}$
$y\geq1$
Then the image of $f(x)=\frac{1}{1+x^{2}}$ is $y\geq1$
Is this correct? If not, what information is missing?
AI: For $\sqrt{y^{-1}-1}$ to be defined, $y^{-1}\ge 1$, i.e. $\frac{1}{y}\ge 1$ or $y\le 1$. The range is indeed bounded above by 1; it is bounded below by zero since $1+x^2\ge 1>0$. $0$ is not in the range, since the fraction is never zero; however we can get arbitrarily close.
Hence the range is: $$(0,1]$$ |
H: You can write ${\left( {\frac{1}{2}} \right)^x}$ as ${2^{ - x}}$ , can the same be done with ${\left( {\frac{2}{3}} \right)^x}$?
You can write ${\left( {\frac{1}{2}} \right)^x}$ as ${2^{ - x}}$ as:
${\left( {\frac{1}{2}} \right)^x} = {({2^{ - 1}})^x} = {2^{ - x}}$
But what about ${\left( {\frac{2}{3}} \right)^x}$?
Can it be written so that it has a negative $x$ exponent? I am asking this because for base 'b' of an exponential function where $0<b<1$ is makes sense for me to write it this way as it is intuitive, i.e. ${\left( {\frac{1}{2}} \right)^x} = {2^{ - x}}$, it is intuitive because it allows me to think of the graph of this function as the reflection of the graph of the function $f(x) = {2^x}$ in the y-axis. Is it possible for me to to think of functions such as: $${\left( {\frac{2}{3}} \right)^x},{\left( {\frac{4}{9}} \right)^x},{\left( {\frac{2}{5}} \right)^x},{\left( {\frac{6}{7}} \right)^x},etc$$
In this way?
AI: It comes down to $a^{-1}=\frac{1}{a}$. This gives you
$$ 2^{-x} = (2^x)^{-1} = \frac{1}{2^x} = \left(\frac 1 2\right)^x.$$
From this you also get
$$ \left(\frac{2}{3}\right)^x = \left(\frac{3}{2}\right)^{-x}. $$ |
H: Entire function that decays faster than exponential on reciprocals of integers is $0$
If $f(z)$ is entire and $|f(1/n)|\le e^{-n}$ for all $n\in\mathbb{N}$ then $f=0$.
My idea is to express $f$ as a power series centered at $0$ that converges on the entire complex plane, then look at
$$\left|\sum a_k/ n^k \sum n^k/k!\right|\le 1$$
which requires $\lim\limits_{m\to\infty}\left|\sum_{k = 0}^m \frac{a_k}{n^k}\frac{n^{m-k}}{(m-k)!}\right| = 0$, but then I get stuck. Could someone suggest a good way to prove this?
AI: Hint: If $f \not\equiv 0$, then there are $c_k \ne 0$ and $\delta > 0$ such that $|f(z)| = \left|z^k(c_k + c_{k+1}z +\ldots)\right|\ge \dfrac{|c_k|}{2}|z|^k$ for $|z| < \delta$. |
H: Intermediate Value Property
I am trying show that the function $f:[0,1]\to \mathbb{R}$ defined by $f(x)=\sin \dfrac{1}{x}$ if $x\neq 0$ and $f(0)=0$ possesses IVP. Though it looks easy, but I am not getting any clue how to start with. Any help would be appreciated.
AI: Show that there exists a subset $A$ of $(0,1]$ such that $f(A)=f([0,1])$ and such that $f|_A$ ($f$ restricted to the domain $A$) is a continuous function. You may then apply the intermediate value theorem to $f|_A$.
Note that the above proves that $f$ has the property that you mentioned in the comments, but this is not what one would usually call the IVP. The usual intermediate value property is that for any two values $a$ and $b$ in the domain of $f$, and any $y$ between $f(a)$ and $f(b)$, there is some $c$ between $a$ and $b$ with $f(c) = y$. We call functions which satisfy the IVP Darboux functions. This question highlights the fact that the set of Darboux functions $[0,1]\rightarrow\mathbb{R}$ is a proper superset of the set of continuous functions $[0,1]\rightarrow \mathbb{R}$.
Assuming $a<b$, the above proves the IVP for $a=0$, $b=1$. For $b<1$, you will need to show that an $A$ exists as above, but such that $A\subset (0,b)$. It also remains to show that the IVP holds for $a\neq 0$ but this case is handled rather easily by simply restricting $f$ to the interval $[a,b]$ and noting that $f|_{[a,b]}$ is continuous. |
H: Banach Algebra: $\sigma(xy)\cup\{0\} = \sigma(yx)\cup\{0\}$
It is Rudin excercise 10.4 where we aim to prove $\sigma(xy)\cup\{0\} = \sigma(yx)\cup \{0\}$ for elements $x,y\in A$ a Banach-algebra.( $\sigma$ being the spectrum)
In (a) we prove that $e-yx$ invertible $\Leftrightarrow e-xy$ invertible.
Following the hint: Put $z= (e-xy)^{-1}$, write $z$ as geometric series (assume $\left\|x\right\| < 1, \left\|y\right\|< 1$), and use the identity $(xy)^n = x(yx)^{n-1}y$ to obtain a finite formula $(e-yx)^{-1}$ in terms of $x,y,z $. Then show that this formula works without any restrictions on $\left\|x\right\|$ or $\left\|y\right\|$.
Ok, so lets put
\begin{align*}
z = (e-xy)^{-1} &= \sum_{k=0}^{\infty} (xy)^k = e+\sum_{k=1}^{\infty} (xy)^k\\ & =e + \sum_{k=1}^{\infty}x(yx)^{k-1}y = e+ x(e-yx)^{-1}y
\end{align*}
Is this right? But I dont see how i can single out out $(e-yx)^{-1}$. And how to make this work without restrictions on $\left\|x\right\|$ or $\left\|y\right\|$?
And second (b) I want to show that if $\lambda\neq 0$ and $\lambda \in \sigma(xy)$ then $\lambda \in \sigma(yx)$. Thus $\sigma(xy)\cup\{0\} = \sigma(yx)\cup \{0\}$. Then also, how do we see that $\sigma(xy)$ doesn't have to be equal to $\sigma(yx)$.
Thanks for tips and suggestions.
AI: Suppose you know $e-yx$ is invertible with inverse $w$. You want to show that $(e-xy)$ has an inverse that equals $e+xwy$. You can check this directly. For example,
$(e-xy)(e+xwy)=e-xy+xwy-xyxwy=e-x(e-w+yxw)y=e-x(e-(e-yx)w)y=e.$
We do not need to place any restrictions on the norms of $x$ and $y$.
Suppose $\lambda\in \sigma (xy)$ with $\lambda\neq 0$. Then $xy-\lambda e$ is not invertible, so $(1/\lambda)xy-1$ is not invertible, which means $(1/\lambda)yx-1$ is not invertible by the above, so $\lambda\in \sigma (yx)$. Repeating this argument for the other direction, we find that $\sigma(xy)\cup \{0\}=\sigma(yx) \cup \{0\}$.
To show that $\sigma(xy)$ and $\sigma(yx)$ can differ, you should construct an example of where this is the case. The above work tells you that this can only happen for $\lambda =0$. Consider the space $l^2$, the unilateral shift $S$, and its adjoint $S^*$. |
H: Is axiom of completeness an axiom?
The following statement is the axiom of completeness:
Every non empty subset of $\mathbb R$ that is bounded above has a least upper bound.
So I was wondering: is it an axiom or can it be proven? One can construct $\mathbb R$ as the set of equivalence classes of Cauchy sequence in $\mathbb Q$. Two sequences are equivalent iff the have the same limit. Then, I tried to prove that a non empty set bounded above has a least upper bound but couldn't find a proof. But I think it is possible because $\mathbb R$ is defined by construction and then there can not be axioms in it?
AI: It depends on your choice of axioms. Many introductory texts define the real numbers axiomatically, as a set of objects that obey certain axioms. One of the required axioms is the least upper bound axiom you mention. However, using more fundamental (and abstract) axioms, one can construct the real number system from even more primitive notions. In that case, it would be a theorem. In that scenario it is often called the least upper bound property of the reals.
I see you have edited your question. Yes, you can prove the least upper bound property from the construction you mention. You can find a thorough treatment of that construction in many analysis textbooks. There is also a brief treatment on Wikipedia. |
H: If $R$ is a UFD then $R[X,X^{-1}]$ is a UFD
Prove that the ring $R[X,X^{-1}]$ of Laurent polynomials over a UFD $R$ is a UFD.
I'd like someone to give a full proof.
AI: If $R$ is UFD, then $R[X]$ is UFD (see any textbook).
If $R$ is UFD and $f \in R \setminus \{0\}$, then $R[\frac{1}{f}]$ is UFD. The prime elements are those of $R$ which don't divide $f$. Proof: They are prime because of the classification of prime ideals of localizations. If $0 \neq a \in R[\frac{1}{f}]$, say $a=x/f^k$, then $x$ is a product of prime elements. Those which divide $f$ are units in $R[\frac{1}{f}]$. Thus $a$ is associated to a product of the distinguished prime elements. |
H: LinAlg Vector Graphing
I'm looking for a graphing calculator to graph vectors - NOT vector fields. A Google search turned up many great calculators with vector-valued function capabilities. However, I'm looking for something to help me visualize the concepts I'm learning in Linear Algebra, e.g. that combinations of certain vectors lie in a plane.
An example of what I'm looking for: the parameter $[3, 6, 5]$ would be represented as a vector from the origin to the
point $(3, 6, 5)$
Thanks for your help!
AI: I usually use Maple to do this job. Here, are the codes, you can count on them:
[> with(plots):
arrow([3,6,5], axes = boxed, difference, color = blue, shape = cylindrical_arrow, fringe = 'red');
For a relevant well-design post see this one. |
H: If $ a_1 a_2a_3 ...a_{20} = 2^x * y! $ Then what is the value of (x+y)?
Let us consider a series with $ a_1 = 2012 $ and $ a_n = \frac{n}{a_{n-1}} $ . If $ a_1 a_2a_3 ...a_{20} = 2^x * y! $ Then what is the value of (x+y) ?
AI: HINT:
As $a_na_{n-1}=n, a_{2r+1}a_{2r+2}=2r+2=2(r+1) $
$$\prod_{1\le r\le 20}a_r=\prod_{0\le r\le9}a_{2r+1}a_{2r+2}=2^{10}\prod_{0\le r\le9}(r+1)=2^{10}(10)!$$
Now, we can use this, to find the highest power of $2$ in $10!$ and also the highest power of other primes $<10$ which constitute $y$ |
H: Expectation using a dirac measure as the probability measure
If $\varepsilon_a(dx)$ is the point mass at point $a\in\mathbb{R}$, I want to calculate $E_a[X]$ where $X$ is a real valued random variable and $E_a$ is expectation with respect to $\varepsilon_a(dx)$.
I'm not sure if my understanding is correct:
$$E_a[X]=\int_{\Omega} X\cdot\varepsilon_a(dx)=\int X1_{\{a\}}(x)dx=a$$
i.e. only when X takes the value of $a$ does the integrand be anything other than 0. Since $a$ is a point on real line, there is only one such point?
AI: Let $\Omega=\mathbb R$ (otherwise the question makes no sense). For every $a$ in $\Omega$ and every random variable $X:\Omega\to\mathbb R^n$, by definition of the Dirac measure $\delta_a$,
$$
\int_\Omega X\,\mathrm d\delta_a=\int_\Omega X(\omega)\,\mathrm d\delta_a(\omega)=X(a).
$$ |
H: $\sin ^6x+\cos ^6x=\frac{1}{8}\left(3\cos 4x+5\right)$, Any quick methods?
How to prove the following equation by a quick method?
\begin{eqnarray}
\\\sin ^6x+\cos ^6x=\frac{1}{8}\left(3\cos 4x+5\right)\\
\end{eqnarray}
If I use so much time to expand it and take extra care of the calculation process, I can find the answer. Since it is so easy to get mistakes, I have to try it 3 time to obtain the correct answer. Therefore, I wonder there are other easy ways to deal with this question. Is it right?
Thank you for your attention
AI: Using $a^3+b^3=(a+b)^3-3ab(a+b),$
$$(\sin^2x)^3+(\cos^2x)^3=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)$$
$$=1-3\frac{(2\sin x\cos x)^2}4$$
$$=1-\frac34(\sin^22x)\text{ using }\sin2x=2\sin x\cos x$$
$$=1-\frac34\frac{(1-\cos4x)}2 \text{ using }\cos2y=1-2\sin^2y$$ |
H: Strong explanation of Strong Form of Mathematical Induction
I don't quite understand induction well, and was wondering if you could explain to me what induction is and what the strong form of induction is.
AI: In regular induction, you assume some property holds true at some arbitrary index value, say $k$. You might not know whether it does or doesn't, but you say, "let's imagine it does, and see what happens."
Then, given your assumption at that value, you need to show that it must hold true at the next value, $k+1$. In other words, a consequence of your assumption is that the property holds true for the next value in your set.
But, you must always prove the base case as well, so you have to have a sort of anchor to say "it definitely holds at this base value, therefore, I have a foundation to build on." Often times, the base case is $k=1$, but not always. Sometimes it makes more sense to start at $k=2$, or $k=50$, or whatever. It all depends on your proof.
Strong induction is the same thing, except instead of assuming the property holds at some single arbitrary index value $k$, you assume that it holds for a range of them $k, k+1, k+2, \ldots, k+n$. |
H: Steady state and DE
Let $x_t = f (x_{t-1})$ a difference equation of order $1$, with $f(x) = ux(1-x)$; $u \in (0; 4)$.
Show that $x^* = 0$ is always a steady state for all $u$ .
Compute its positive steady state (meaning x? > 0) depending on .
Give two different values of $u$ such that in one case the steady state is stable and another for which it is unstable.
AI: First question: $x^* = 0$ is a steady state
You can check this just by inserting $x=0$ in $f(x)$, it is always zero regardless of $u$.
Second question: positive steady states
We look for $x > 0$ such that $x = f(x) = ux(1-x)$... it turns out to be $x=1 - \frac{1}{u}$, which is positive only for $u > 1$.
Third question: stability
In order to have stability, we need the derivative of $f(x)$ in $x^*$ to be lower than one. Given the formula for steady state in point 2, $f'(x^*)$ is equal to $2 - u$. So, any example with $u < 1$ will have an unstable steady state (and the other way around, $u > 1$ implies stable steady state). |
H: How to simplify this equations?
A equation is given,
$$(7x-6)^3- (5x-6)^3-6x(7x-6)(5x-6)$$.
Will i use the formula $a^3 - b^3$ to simplify the above?
Any tips or solution will be appreciated.
AI: Observe that $7x-6-(5x-6)=2x$
so $6x(7x-6)(5x-6)$ can be written as $3(7x-6)(5x-6)\{7x-6-(5x-6)\}$
Use $(a-b)^3=a^3-b^3-3ab(a-b)$ formula
Here $a=7x-6,b=5x-6$ |
H: Question on singular homology
please where i can found the prove of this:
If $X$ is a topological space and $(X_{\alpha})_{\alpha\in I}$ is the family of it's path connected components.
Prove that for each $n\in \mathbb{N}$, $$H_n(X;\mathbb{A})=\bigoplus_{\alpha\in I} H_n(X_{\alpha};\mathbb{A})$$
Please help me
thank you.
AI: Let $n\in\mathbb N$ and $s$ be a singular $n$-simplex of $X$, i.e. $s:\Delta^n\to X$, where $\Delta^n$ is the standard $n$-simplex. Since $\Delta^n$ is path-connected, so is its image $s(\Delta^n)$, hence there is an $\alpha\in I$ such that $s(\Delta^n)\subseteq X_\alpha$.
If $c$ is a singular $n$-chain then we can write $\displaystyle c=\sum_{\alpha\in I} c_\alpha$, where each $c_\alpha$ is a chain with support in $X_\alpha$.
Let $\displaystyle z=\sum_{\alpha\in I}z_\alpha$ be a $n$-cycle. Since the $X_\alpha$'s are disjoint sets, so are the supports of each $z_\alpha$ and the boundaries $\partial z_\alpha$ as well, hence
$$0=\partial z=\partial\left(\sum_{\alpha\in I}z_\alpha\right)=\sum_{\alpha\in I}\partial z_\alpha\Rightarrow \forall\alpha\in I,\ \partial z_\alpha=0$$
i.e. each $z_\alpha$ is a $n$-cycle (with support) in $X_\alpha$.
If you identify each $n$-simplex $s:\Delta^n\to X_\alpha$ with the map $s:\Delta^n\to X$, then you can write $z_\alpha\in Z_n(X_\alpha,\mathbb A)$ and $[z_\alpha]\in H_n(X_\alpha,\mathbb A)$.
Using similar arguments with the singular $(n+1)$-boundaries, you can prove that the application below is well defined and is an isomorphism.
$$\Phi_n:\left\{\begin{array}{rl}H_n(X,\mathbb A)&\longrightarrow \bigoplus_{\alpha\in I} H_n(X_\alpha,\mathbb A) \\ [z]&\longmapsto \oplus_\alpha[z_\alpha]\end{array}\right.$$
Also, I invite you to watch this video (and other videos of the same author as well). |
H: Proving a function is onto and one to one
I'm reading up on how to prove if a function (represented by a formula) is one-to-one or onto, and I'm having some trouble understanding.
To prove if a function is one-to-one, it says that I have to show that for elements $a$ and $b$ in set $A$, if $f(a) = f(b)$, then $a = b$. I understand this to mean that if two elements in a domain map to the the same element in a codomain, then for the function to be one-to-one, they must be the same element because by definition, a one-to-one function has at most one element in the domain mapped to a particular element in the co-domain. Did I understand this correctly?
Then to prove that the function is onto, I'm reading an example that says "let's prove that $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = 5x+2$ is onto, where $\mathbb{R}$ denotes the real numbers. We let $y$ be a typical element of the codomain and set up the equation $y =f(x)$. then, $y = 5x+2$ and solving for $x$ we get $x ={y-2\over 5}$. Since $y$ is a real number, then ${y-2\over 5}$ is a real number and $f({y-2\over 5})=5({y-2\over 5})+2=y.$
I'm not really seeing how that proves anything, so can anybody explain this to me?
AI: Yes, your understanding of a one-to-one function is correct.
A function is onto if and only if for every $y$ in the codomain, there is an $x$ in the domain such that $f(x) = y$.
So in the example you give, $f:\mathbb R \to \mathbb R,\quad f(x) = 5x+2$, the domain and codomain are the same set: $\mathbb R.\;$ Since, for every real number $y\in \mathbb R,\,$ there is an $\,x\in \mathbb R\,$ such that $f(x) = y$, the function is onto. The example you include shows an explicit way to determine which $x$ maps to a particular $y$, by solving for $x$ in terms of $y.$ That way, we can pick any $y$, solve for $f'(y) = x$, and know the value of $x$ which the original function maps to that $y$.
Side note:
Note that $f'(y) = f^{-1}(x)$ when we swap variables. We are guaranteed that every function $f$ that is onto and one-to-one has an inverse $f^{-1}$, a function such that $f(f^{-1}(x)) = f^{-1}(f(x)) = x$. |
H: Solving for coefficients of a polynomial in terms of roots of another
This was a homework assignment a while back (already turned in) that I'm not totally sure how to approach. I think it might be simple but I can't wrap my mind around it.
If someone could give me a tip that would be great.
I'm given a polynomial $f=x^3+ax^2+bx+c$ with roots $\alpha_1,\alpha_2,\alpha_3$.
How do I find the coefficients of the following two polynomials, in terms of a,b, and c:
the monic cubic with roots $\alpha_1^2,\alpha_2^2,\alpha_3^2$
the monic cubic with roots $\alpha_1+\alpha_2,\alpha_1+\alpha_3,\alpha_2+\alpha_3$
What I know:
I understand that by viete's formula, $a = \alpha_1+\alpha_2+\alpha_3$, $b=\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3$, and $c=\alpha_1\alpha_2\alpha_3$.
But is there an elegant way to solve for the relationship between the coefficients for (1) and (2) using the coefficients of $f$? It seems resistant to a clean solution, since the relationship isn't necessarily linear.
Cheers.
AI: There is a general method to express symmetric poynomials in terms of the elementary symmetric polynomials.
For example for the cubic with roots $\alpha_1^2, \alpha_2^2,\alpha_3^2$ on needs to express $\alpha_1^2+\alpha_2^2+\alpha_3^2$, $\alpha_1^2\alpha_2^2+\alpha_2^2\alpha_3^2+\alpha_1^2\alpha_3^2$, and $\alpha_1^2\alpha_2^2\alpha_3^2$ in terms of $\alpha_1+\alpha_2+\alpha_3$, $\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3$, and $\alpha_1\alpha_2\alpha_3$. One can do so by eliminating from highest to lowest power of the polynomial with the most distinct factors. For example,
$$\begin{align}\alpha_1^2+\alpha_2^2+\alpha_3^2&=(\alpha_1+\alpha_2+\alpha_3)^2-2\alpha_1\alpha_2-2\alpha_1\alpha_3-2\alpha_2\alpha_3\\
&=a^2-2b\end{align} $$
$$\begin{align}\alpha_1^2\alpha_2^2+\alpha_2^2\alpha_3^2+\alpha_1^2\alpha_3^2&=(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3)^2-2\alpha_1^2\alpha_2\alpha_3-2\alpha_1\alpha_2^2\alpha_3-2\alpha_1\alpha_2\alpha_3^2\\
&=b^2-2\alpha_1\alpha_2\alpha_3(\alpha_1+\alpha_2+\alpha_3)\\&=b^2-2ca\end{align} $$ |
H: I have ten professors, and need to pick four of them for a committee.
I have ten professors, and need to pick four of them for a committee. It is a bad idea for Hatfield and McCoy to serve together. It is a bad idea for El and Luthor to serve together. How many possible committee assignments are there?
I was thinking 10 choose 4 for all the possible committee combinations and now subtract 2*10 choose 2 for the committees Hatfield and McCoy serve together as well as the committees that El and Luthor serve together. Then add one back on the end for the committee that all four would serve on together. Not sure if that will work to count the committees. Let me know your thoughts.
AI: There are three major ways to approach this:
First is purely additive, finding and summing combinations that we know will be valid. There are 6 professors who can serve in any combination of four, and there are 15 such possible combinations of those 6 profs taken 4 at a time (the order in which the profs are chosen is irrelevant, so these are combinations, not permutations). On top of that, there are 20 combinations of these 6 profs taken 3 at a time, to which you can add any one of the four "restricted" profs, so there are 80 additional combinations of four professors which includes one restricted prof. Finally, there are 15 possible unique pairs of the 6 unrestricted profs, to which you could add either El and Hatfield, El and McCoy, Luthor and Hatfield or Luthor and McCoy, so there are 60 combinations with a valid pair of the restricted profs. 15 + 80 + 60 = 155.
You could also approach it from the opposite direction, starting with all possible combinations and removing those we know will be invalid. This was your chosen strategy if I read your OP right. There are 210 total possible combinations of the 10 profs taken 4 at a time. Of these, some would be invalid because they have one of the two invalid pairs: El and Luthor or Hatfield and McCoy. There are 15 possible pairs for the other two profs, and so 30 combinations that have an invalid pair. Other combinations would be invalid because they have an invalid pair plus another of the restricted profs; there are four of these invalid combinations (basically 4 nCr 3; E-L-H, E-L-M, H-M-E, H-M-L) and six possibilities for the last person, or 24 combinations. Finally, there is one combination with both invalid pairs comprising all four profs on the committee. 210 - 30 - 24 - 1 = 155.
One more way to build it up: if you take one valid pair of the restricted profs (Hatfield-El, Hatfield-Luthor, McCoy-El, McCoy-Luthor) and exclude the other pair, now you have 8 profs for which any combination of four is valid, or 70 possibilities. There are, as shown, four combinations of includable pairs (or excludable ones depending on your POV) and so you'd multiply by 4 (280); however, this total now includes certain overlapping sets. There are 15 combinations (6 nCr 4) that don't have any restricted profs, and these combinations will be present in all four groups, so we should remove three of those groups' duplicates from our total. Similarly, for each single restricted prof, there are two groups with 20 combinations each (6 nCr 3) that have that professor, and so for each of those four profs we subtract 20 duplicate possibilities. So, 280 - 3(15) - 4(20) = 155.
Obviously, all three strategies will get you to the right answer (and there are others); the "best" strategy will depend on how easy it is for you to ensure you have included all possibilities when including or excluding groups. |
H: How come this represents the height of a trapezoid?
Why does this formula
$$h= \frac{\sqrt{(-a+b+c+d)(a-b+c+d)(a-b+c-d)(a-b-c+d)}}{2|b-a|}$$
represent the height of a trapezoid?
Source: http://en.wikipedia.org/wiki/Trapezoid
Thanks!
AI: Draw a trapezoid with bases $a$ and $c$, so that the other two sides are $b$ and $d$. Divide your trapezoid into two right triangles (each of which has a side being $h$) and a rectangle by dropping perpendiculars from $a$ to $c$ in appropriate places. Now place the triangles side by side so they share the side $h$. The base of this triangle is $a-c$ and its other two sides are $b$ and $d$. Now write down the area of this triangle in two different ways. First $A=\frac{1}{2}(a-c)h$, and second using Heron's formula. This gives the desired relationship between $a,b,c,d$ and $h$. |
H: why minimum of these functions happen at a special place?
why minimum of these functions happen at a special place? how to use derivative to find the minimum of these functions?
$$|x-1| + |x-2| + \dots + |x-9|$$ minimum is for $x = 5$
$$|x-1| + |x-2| + \dots + |x-99|$$ minimum is for $x = 50$
$$|x-1| + |x-2| + \dots + |x-999|$$ minimum is for $x = 500$ $$ \dots$$ why it happens?
AI: Suppose that $a_1\lt a_2\lt a_3\lt \cdots\lt a_9$. Consider the function
$$f(x)=|x-a_1|+|x-a_2|+\cdots +|x-a_9|.$$
Then $f(x)$ is the sum of the distances from $x$ to the $a_i$.
Imagine the points $a_1,a_2,\dots,a_9$ are on the $x$-axis. A tiny bug is on the $x$-axis, well to the left of $a_1$, and starts walking in the rightward direction.
For every tiny step $s$ that she takes, the sum of her distances from the $a_i$ decreases by $9s$. This continues until she hits $a_1$. Now for every tiny step $s$ that she takes, her distance from each of $a_2,a_3,\dots,a_9$ decreases by $s$, and her distance from $a_1$ increases by $s$, for a total decrease of $7s$.
This continues until she hits $a_2$. Then for every step $s$ that she takes, her distance from each of $a_3$ to $a_9$ decreases by $s$, and her distance from each of $a_1$ and $a_2$ increases by $s$, for a total decrease of $5s$.
When she hits $a_3$, for each a tiny step to the right, there is a total decrease of $3s$. When she hits $a_4$, there is a total decrease of $s$ for each tiny step to the right. When she hits $a_5$ and takes a tiny step, there is an increase of $s$ for each tiny step to the right. And things get worse as she continues to the right. So the minimum value of $f(x)$ was reached at $x=a_5$.
Exactly the same idea shows that if we have $a_1\lt a_2\lt\cdots \lt a_{2m-1}$, and $f(x)=|x-a_1|+\cdots+|x-a_{2m-1}|$, then $f(x)$ reaches a minimum at $x=a_m$.
Remark: The situation is only slightly different is we have an even number of points $a_1\lt a_2\lt \cdots\lt a_{2m}$. Then the minimum value of $f(x)$ is reached at all points $x$ such that $a_m\le x\le a_{m+1}$.
To put it in other words, if we have an odd number of terms, then $f(x)$ is minimized at the (unique) median of the $a_i$. The same is true with an even number of terms, except that there are infinitely many medians.
Note that by contrast $g(x)=(x-a_1)^2+(x-a_2)^2+\cdots +(x-a_n)^2$ is minimized at the mean $\frac{a_1+a_2+\cdots+a_n}{n}$ of the $a_i$. |
H: Finding a combinatorial proof of this identity: $n!=\sum_{i=0}^n \binom{n}{n-i}D_i$
Can someone prove this. Let $D_n$ be the number of derangements of $n$ objects. Find a combinatorial proof of the following identity:
$$n!=\sum_{i=0}^n \binom{n}{n-i}D_i$$
AI: You can divide all $n!$ permutations of $n$ objects onto $n+1$ disjoint classes of permutations with exactly $i$ dearrangements where $i=0,\ldots,n$. Amount of permutations with exactly $i$ dearrangements equals to the product of ways to choose objects that will be dearranged i.e. ${n \choose i}$ and amount of ways to dearrange them i.e. $D_i$. Thus
$$
n!
=\sum\limits_{i=0}^n {n \choose i} D_i
=\sum\limits_{i=0}^n {n \choose n-i} D_i
$$ |
H: Compute C1, C2, C3, C4, C5
I have a question
C1 = 0, C$n$ = C$\lfloor n /2\rfloor$ + $n^2$ for all $n > 1$
Compute C1, C2, C3, C4
So what I did is:
C2 = C$\lfloor 2/2\rfloor + 2^2$ = 1 + 4 = 5 -- but that is wrong because the answer is supposed to be 4...
C4 = C$\lfloor 4/2\rfloor + 4 ^2$ = 2 + 16 = 18 -- but that is wrong as well, because the answer is supposed to be 20.
Am i not doing something right? Because I do know that Floor of 1 is 1...
AI: $C2 = C\lfloor2/2\rfloor+2^2 = 1 + 4 = 5$ -- but that is wrong because the answer is supposed to be 4...
On this line, for instance, you say that $C\lfloor 2/2 \rfloor = 1$, but $C\lfloor 2/2 \rfloor = C1 = 0$. |
H: I have reached a rut in my understanding of control systems. How do I cross this?
A little background here. I'm an undergrad in the final year. I have decided academia as my career path. My grades are not high but my research caliber is good and I have ongoing projects that are promising. My field of interest is control systems.
My question:
I had a course in basic control systems. I did well, I can understand the basics. For my literature survey for my ongoing project, I went through the current trends in control systems and it's completely greek and latin to me. It's disheartening but I really want to try.
I cannot understand how one goes from MIMO, SISO systems to sliding mode control, fuzzy logic control and all the other advance stuff that seems to be the talk of the town.
How do I cross this? Are there any standard books, literature bridging this gap or at least elaborating on the prerequisites to understand the current state of art in control system math.
AI: First, get a decent grasp of linear algebra. Most control systems techniques use linear algebra extensively.
Second, don't worry much at first about not understanding much of the mathematics. The truth is that most controls researchers also don't understand the underlying mathematics. At least not the real, hard math. They might understand what L2 minimization is, but it's rare to see any controls folks actually get into the functional analysis of their problem. (As a mathematician working in controls, it's often frustrating.)
Ultimately, I don't think there's a single book that bridges these concepts. Your best bet is to work on numerical applications of controls systems, and try to read progressively more advanced books, and also to work out examples.
If you look hard enough, you can find some nice canned examples of some canonical controls problems -- a 5-state aircraft model, for example -- implemented in MATLAB/Simulink.
The best thing you can do to educate yourself is to try to implement the techniques against a model you can understand. In controls, learning is often accomplished by doing. Truthfully, the field is not as mathematically rigorous as it should be. But that's OK, because it's very useful (and sometimes very easy) to develop good, working solutions to real problems without requiring the mathematical rigor. Once you can conceptually handle some of the more advanced methods, you can then work on understanding the mathematical rigor. |
H: Notation for integer between two values
This may be a silly question, but it has been a long time since I have used set notation to any real extent. How would I write that $i$ is an integer ranging from $1$ to $N$?
My (possibly faulty) recollection is that this is expressed as $i \in \{ \mathbb{Z}: [1,N]\}$. Is this correct?
AI: Notationally, to write $i$ is an integer within a given interval, you could write several different things:
$$i\in\Bbb Z:i\in[1,N]\tag 1$$
$$i\in\Bbb Z:1\le i\le N\tag 2$$
$$i\in\Bbb Z\cap[1,N]\tag 3$$
where each is read as follows:
$(1)$ "$i$ is an integer such that $i$ is within the interval $1,N$"
$(2)$ "$i$ is an integer such that $1$ is less than or equal to $i$ is less than or equal to $N$"
$(3)$ "$i$ is an element of the set intersection of the integers with the interval $1,N$"
Each would be considered a valid representation, and each may be considered more appropriate for given circumstances or writing styles than the others.
As mentioned elsewhere, it is also common to write $i=1,2,\dots,N$ especially when $i$ is an indexing element. |
H: Probability of a result from 3d6, lowest to highest
I have a fascination with tabletop sports games, and through that, I've developed an interest in probability. That said, it's not my strong suit, so I wanted to pose this question because I think involves a few different probability principles to solve it.
Here's the premise...the game uses a roll of 3d6 to determine the result of a play. The dice are read from lowest to highest, as in:
a roll of 3, 6, 2 would be 2-3-6
a roll of 1, 4, 1 would be 1-1-4
etc.
That roll is then looked up on a chart to determine the result.
I was curious about the probability of different results coming up so that if I wanted to make a house rule, I would know which result would be the best place to modify.
So, what I do know is that for probability, you multiply chances together, correct? So without the low to high rule I discussed above, any number would have a 1/6 * 1/6 * 1/6 chance of occurring, correct?
How would I apply this principle to any result? My guess is something like:
1-1-4 = 1/6 * 1/6 * 5/6
only one chance for the first 2 die, and the last can be anything but 1, so 5 remaining numbers
2-3-6 = 1/6 * 4/6 * 4/6
first number can be 2 only = 1/6
second number can be any number but 2 or 6 = 4/6
second number can be any number but 2 or 3 = 4/6
Am I understanding this correctly?
AI: Your answer's not correct. For 1-1-4, you have multiplied by $\frac56$, but it's not clear to me why. The correct calculation goes like this: there is a $\frac16$ probability of getting a 1 on the first die, a $\frac16$ probability of getting a 1 on the second die, and a $\frac16$ (not $\frac56$) probability of getting a 4 on the third die. This multiplies out to $\frac1{216}$. But there are three different orders in which the rolls could occur to give a result of 1-1-4, since 1-4-1 or 4-1-1 give the same result. So you must multiply the $\frac1{216}$ by 3, for a final result of $\frac1{72}$.
Similarly, the result for 2-3-6 is again $\frac1{216}$, but this time multiplied by 6 because there are 6 different orders in which the 2, 6, and 3 can appear; the final result is $6\cdot\frac1{216}=\frac1{36}$.
In general, the answer is:
If the pattern you're looking up is XXX, with all three dice the same, the probability is $\frac1{216}$.
If the pattern is XXY or XYY, the probability is $\frac1{72}$.
If the pattern is XYZ, the probability is $\frac1{36}$. |
H: Show that $(\phi_{n}^{(n)})^{-1}= -(\sum_{i=0}^{n-1}(\phi_{i}^{(n)})^{-1})$
So I have $n+1$ points $x_{0},x_{1},...,x_{n} \in \mathbb{R}$ and a following quasi-function:
$\phi_{j}^{(n)}=\prod_{i=0,i \neq j}^{n}(x_{j}-x_{i})$
Show that $(\phi_{n}^{(n)})^{-1}= -(\sum_{i=0}^{n-1}(\phi_{i}^{(n)})^{-1})$
AI: Hint: look at what happens when $n=2$, and see if you can generalize:
$$-\sum_{i=0}^{2-1} \left(\phi_i^{(n)}\right)^{-1} = -\left(\frac{1}{(x_0-x_1)(x_0-x_2)} + \frac{1}{(x_1-x_0)(x_1-x_2)}\right)$$
$$ = -\frac{(x_1-x_2)-(x_0-x_2)}{(x_0-x_1)(x_0-x_2)(x_1-x_2)}$$
$$= \frac{1}{(x_2-x_0)(x_2-x_1)}$$
$$=\left(\phi_2^{(2)}\right)^{-1}$$ |
H: What situations/models require calculating the area under a curve?
Besides inferring distance traveled from a velocity chart,
can anyone name some graphs where you need to know the area under the curve?
For example, I know in Statistics (bell curve), the "probability density function" is used to determine what percentage of the graph is b/w 1, 2, and 3 standard deviations, for example.
http://www.wolframalpha.com/input/?i=y%3D%281%2Fsqrt%282pi%29%29%28e%5E%28%28-x%5E2%29%2F2%29%29
http://en.wikipedia.org/wiki/Cumulative_distribution_function
http://en.wikipedia.org/wiki/Probability_density_function
Any other big topics (or specific examples) where the area under a graph comes into play?
AI: You mention distance and velocity. Yes, in this case finding the area under a curve is used. However, this situation can be generalized. The relationship between velocity of distance is that velocity is the first derivative of distance (displacement) with respect to time. Therefore, integrating (finding the area under the curve) of velocity with respect to time gives you change in displacement.
In general, when you have an quantity changing with respect to time - a rate - (or with respect to anything, technically), you would integrate that (area under curve) to find how much the quantity changed.
For example, I could say that that the amount of water in a tank increases at a certain rate (a number of gallons per minute). I could then integrate (find the area under) that curve to find how much the water in the tank increased over the given time.
I could take it further and say that the rate by which the amount of water in the tank is changing, is changing. This is the second derivative (akin to acceleration when talking about displacement) of the amount of water in the tank. Find the area under this curve would tell you how much the rate of water into the tank changed over the given time.
And so on. This is calculus. |
H: Inequality with the supremum
I am trying to prove the following statement for $A\subset \mathbb{R},~\epsilon>0$ with $A$ bounded above:
$\sup(A)-\epsilon<a\leq\sup(A)$, for some $a \in A$
I have tried dividing it into two cases.
Case 1: $\sup(A)\in A$. Then take $a=\sup(A)$ and we are done.
Case 2: $\sup(A)\notin A$. I tried to formalise the idea that if we started writing down elements of $A$ in increasing order, then the maximum element of these lists must tend to $\sup(A)$. This is because if it didn't, then there would be some element at which we would have to stop before we got as close as we wanted to $\sup(A)$. This element would be the least upper bound, but $\sup(A)\notin A $ which is a contradiction. Hence there exists $\epsilon$ such that $|a_n-\sup(A)|<\epsilon$ for some $n$ and the result follows.
Is this proof valid? Are there any better ones?
Thanks.
AI: If there is no such $a$, then $\sup(A) - \epsilon$ is a new l.u.b., contradiction. |
H: Computing the condition to solve a set membership problem
I am stuck with the following problem for the last few weeks
Alex , Bob and Charlie stand for local election.
Given fractions
a - fraction of voters prefer Alex to Bob
b - fraction of voters prefer Bob to Charlie
c - fraction of voters prefer Charlie to Alex
Which of the following cannot be values of (a,b,c)
a. (0.51 ,0.51, 0.51)
b. (0.68 ,0.68, 0.68)
c. (0.49 ,0.49, 0.49)
I reasoned (initially) that using formal notions of set theory -
$a = A\cap\bar B$
$b = B\cap\bar C$
$c = C\cap\bar A$ where
A = fraction of voters that will vote Alex
B = ..............do...................Bob
C = ................do..................Charlie
Then undertook some tedious/clumsy usage of venn diagrams to formulate some condition that may strike the odd one out.I am clueless after many tries.
Any idea/hints will be helpful.
For your information, I think my intial assumptions given above wont work.
And, the answer to this question was given as option (b)
AI: $v=v_{ABC}+v_{ACB}+v_{BAC}+v_{BCA}+v_{CAB}+v_{CBA}$ where for instance
$v_{ABC}$ stands for the number of voters that have Alex on top,
Bob on second and Charlie on third position.
Then we have fractions $f_{AB}=\frac{v_{ABC}+v_{ACB}+v_{CAB}}{v}$,
$f_{BC}=\frac{v_{ABC}+v_{BAC}+v_{BCA}}{v}$ and $f_{CA}=\frac{v_{BCA}+v_{CAB}+v_{CBA}}{v}$
This leads to $f_{AB}+f_{BC}+f_{CA}=\frac{v_{ABC}+v_{BCA}+v_{CAB}+v}{v}\leq\frac{2v}{v}=2$
so option b (that is: $f_{AB}=f_{BC}=f_{CA}=0.68$) cannot be true. |
H: Is *njwildberger* wrong about area and circumference of a circle?
In this video, njwildberger says that the area and circumference of a circle are proof-less theorems. But I heard that we can derive both the area and circumference of a circle using calculus? So are the area and circumference of a circle proof-less theorems?
AI: There is no such thing as a proof-less theorem. I just watched the video and he's fast and loose with his terminology. What he means to say is that these theorems are presented to high school students without proof. He doesn't mean there is no proof. Then again, lots of things are presented to students without proof early in their education, so that's not really saying much. |
H: Why do I get 251 square dm the correct answer is 252 square dm? (Error by a fraction)
I am sorry if I am bothering you folks, I've recently started to play with Trigonometry, it's really cool, but trying to understand what mistakes I am making, anyhow, I am guessing that I shall use the Pythagorean Theorem.
Here's what I have done:
$(a)^2 + (10.9)^2 = (25.6)^2$
$a^2 + 118.81 = 655.36$
$a^2 + 118.81 -118.81 = 655.36 - 118.81$
$a^2 = 536.55 \implies \sqrt{a} = \pm \sqrt{536.55} = 23$
$a = 23$
$\text{Area} = 10.9 \times 23 \approx 251 (250.7)$
What's wrong, a fraction is missing or maybe I shall use Trigonometry instead.
AI: You rounded $\sqrt{536.66}$ to 23, while you should have left it as $23.16$ (or $23.2$). |
H: Splitting field of $x^{4 }-3$ over $\mathbb{Q}$
I'm having trouble in understanding "the form" of a splitting field.
The problem is: Construct the splitting field over $\mathbb{Q}$ of the following polynomials.
One of the polynomials is $x^{4}-3$
So, the roots of this polynomial are $\pm \sqrt[4]{3}$ and $\pm \sqrt[4]{3}i$. Therefore, the splitting field over $\mathbb{Q}$ of this polynomial is $\mathbb{Q}[\sqrt[4]{3},i]$, right?
But what is the form of these elements?
Like, we have that $\mathbb{Q}[\sqrt{2}]=\{a_1+a_2\sqrt{2};a_1,a_2 \in \mathbb{Q}\}$. So, $\mathbb{Q}[\sqrt[4]{3},i]$ = ?
AI: Remember that $\;Q(\sqrt[4]3\,,\,i)\;$ is a vector space over $\;Q\;$ of dimension $\;4\cdot 2=8\;$ , and it has a very nice basis (putting $\;w:=\sqrt[4]3\;$ for simplicity, we get):
$\;\{1\,,\,w\,,\,w^2\,,\,w^3\,,\,i\,,\,wi\,,\,w^2i\,,\,w^3i\}\;$ , so you can conveniently write any element in the above field in the form
$$a+bw+cw^2+dw^3+ewi+fw^2i+gw^3i\;,\;\;a,b,c,d,e,f,g\in\Bbb Q$$ |
H: Prove that isomorphic rings have the same characteristic
If $\phi: A \to B$ is a ring isomorphism, I have to proof that $\operatorname{char} A= \operatorname{char} B$.
I know that an isomorphism $\phi$ is bijective and:
$$\phi(x+y)=\phi(x)+ \phi(y)$$
$$\phi(x*y)=\phi(x)*\phi(y)$$
$$\phi(1)=1$$
I have supposed that $\operatorname{char} A=n$, so: $n1=\underset{n\text{ times}}{\underbrace{1+1+\ldots+1}}=0$ in $A$.
As $\phi$ is surjective, $n1\in A$ exists, where $x=\phi(n1)$.
Now, $\phi(n1)=\phi(0)=\phi(1+1+\ldots+1)=\phi(1)+\ldots+\phi(1)=n*\phi(1)=n$.
I don't know if this is correct and if this is how to finish the proof. Could you help me please?
AI: Part 1: Take $1_B$ and show that if you add it up $n$ times you get $0_B$.
Part 2: Show that if you add up $1_B$ a total of $m$ times for $m < n$ then you don't get $0_B$.
For the first part, the proof could be something like:
Observe $1_B = \phi(1_A)$. Then
$$1_B + \cdots + 1_B = \phi(1_A) + \cdots + \phi(1_A) = \phi(1_A + \cdots + 1_A) = \phi(0_A) = 0_B$$
where each sum has $n$ terms. Can you justify each of the equalities above? |
H: A new restaurant has opened. They have 100 items to chose from! Of those items…
I need some help with this problem.
A new restaurant has opened. They have 100 items to chose from! Of those items…
45 are fattening
45 are gross
44 are ice-cold
10 are both fattening and gross
18 are both fattening and ice-cold
13 are both gross and ice-cold
3 are gross, ice-cold, and fattening
How many items do they have that are not fattening, not gross, and not ice-cold?
I think to start with 100 - 3 - 13 - 18 - 10 then add on the overlaps maybe. Not sure if this is right or really how to attack this problem.
AI: Let's look at it this way. We know the total is 100 and there are three basic groups described.
We first need to count up how many items are in each group i.e. the 45 fattening, the 45 gross, and the 44 ice-cold.
So:
45 + 45 + 44 = 134
We over counted so now we have to subtract how many items are shared by two of the groups i.e. 10 are both fattening and gross, 18 are both fattening and ice-cold, and 13 are both gross and ice-cold
So:
134 - 10 - 18 - 13 = 93
We now have under counted and have to add the items that fit into each group i.e. 3 are gross, ice-cold, and fattening
So:
93 + 3 = 96
Now we know how many items there are (100) and we know how many can be described by the 3 groups. So the difference is the number of items not in the group.
100 - 96 = 4
So 4 are not fattening, not gross, and not ice-cold. |
H: True or False? Continuous Functions
If the function $f+g:\mathbb{R}\rightarrow \mathbb{R}$ is continuous, then the functions $f:\mathbb{R}\rightarrow \mathbb{R}$ and $g:\mathbb{R}\rightarrow \mathbb{R}$ are also continuous.
False;
Let $f(x)=\begin{cases} -1 \text{ if } x<0 \\ 1 \text{ if } x\ge 0 \end{cases}$ $\hspace{10pt}$ and $\hspace{10pt}$ $g(x)=\begin{cases} 1 \text{ if } x<0 \\ -1 \text{ if } x\ge 0 \end{cases}$
Then $(f+g)(x)=0 \hspace{10pt}\forall x$.
Here, $f+g:\mathbb{R}\rightarrow \mathbb{R}$ is continuous, but the functions $f:\mathbb{R}\rightarrow \mathbb{R}$ and $g:\mathbb{R}\rightarrow \mathbb{R}$ are not continuous.
Is this a good example?
AI: (CW response so this doesn't appear as unanswered.)
The example is correct, but (depending on the class) you may wish to provide a proof that $f$ and $g$ are not continuous. In particular, you could prove that $f$ is not continuous at $0$, and then observe $g$ is not continuous either, since $g = -f$.
Proving the $0$ function is continuous shouldn't be too much trouble. |
H: Is this function twice differentiable at $0$?
I have a function $f(x)$:
$$f(x)=\frac{\exp(-|x|)}{1-0.5|\tanh(2x)|}$$
If I try differentiating it in Mathematica (taking $|x|=(2\theta(x)-1)x$ where $\theta(x)$ is Heaviside step function), I get answer in terms of Heaviside functions. But looking at the derivative I can't really see any jump there, which is why I start thinking that it has a second derivative too. But differentiating once more, I get answer with Dirac functions, which don't disappear when evaluating the result at $x=0$.
If I instead represent $|x|=\sqrt{x^2}$, then for the first derivative I get an expression which has $0$ as its limit as $x\to0$, and for second derivative the limit appears to be 1. So from this I could make the conclusion that $f(x)$ is twice differentiable at $x=0$.
So the question: is this function differentiable at $x=0$? Is it twice differentiable?
AI: Since we are dealing with real $x$ (we are, aren't we?), we can write
$$\lvert \tanh (2x)\rvert = \tanh (2\lvert x\rvert).$$
We have $\frac12\tanh (2\lvert x\rvert) = \lvert x\rvert + O(\lvert x\rvert^3)$, so we can calculate
$$\begin{align}
\frac{e^{-\lvert x\rvert}}{1 - \frac12\tanh(2\lvert x\rvert)} &= \left(1 - \lvert x\rvert + \frac{\lvert x\rvert^2}{2} + O(\lvert x\rvert^3)\right)\left(1 + \lvert x\rvert + \lvert x\rvert^2 + O(\lvert x\rvert^3)\right)\\
&= 1 + \frac{\lvert x\rvert^2}{2} + O(\lvert x\rvert^3).
\end{align}$$
Since the two halves of the function are analytic, it follows that the function is twice continuously differentiable.
Generally, if we have a function of the form
$$f(x) = \begin{cases}g(x) &, x \geqslant 0\\ h(x) &, x \leqslant 0 \end{cases}$$
where $g$ and $h$ are continuously differentiable functions with $g(0) = h(0)$ - so that the definition of $f$ has no conflict - then $f$ is differentiable in $0$ if and only if $g'(0) = h'(0)$, and then $f$ is continuously differentiable with $f'(0) = g'(0) = h'(0)$.
If $g$ and $h$ are analytic, i.e. we have power series representations
$$g(x) = \sum_{n=0}^\infty a_n x^n,\qquad h(x) = \sum_{n=0}^\infty b_n x^n,$$
then $f$ is $k$ times (continuously) differentiable if and only if $a_n = b_n$ for $0 \leqslant n \leqslant k$. The computation above shows that the two functions
$$g(x) = \frac{e^{-x}}{1-\frac12\tanh(2x)}\quad\text{and}\quad h(x) = \frac{e^x}{1+\frac12\tanh(2x)}$$
both have power series expansions starting with $1 + \frac12 x^2$, so $f$ is twice continuously differentiable. |
H: True or False? Continuous Functions #2
If functions $f+g:\mathbb{R}$ and $g:\mathbb{R} \rightarrow \mathbb{R}$ are continuous, then so is the function $f:\mathbb{R} \rightarrow \mathbb{R}$.
I feel like this is true, but I'm not sure how to justify my answer.
AI: You have only to remember that the (algebraic) sum of continuous functions is also a continuous function. |
H: What's wrong with this computation of Laurent Series?
I was solving this problem: find the two Laurent Series representations of $f: \mathbb{C}\setminus\{0,i,-i\}\to\mathbb{C}$ given by $f(z) = 1/z(z^2+1)$ in the correct domains.
My approach was: I've used partial fractions to rewrite $f$ as
$$f(z)=\dfrac{1}{z}-\dfrac{1}{2}\dfrac{1}{z-i}-\dfrac{1}{2}\dfrac{1}{z+i},$$
then I've done as follows, if $0<|z|<1$, then both $z \mapsto 1/(z-i)$ and $z\mapsto 1/(z+i)$ are analytic and can be expressed in their Taylor Series. For that matter, I rewrite
$$-\dfrac{1}{2}\dfrac{1}{z-i}=\dfrac{1}{2i}\dfrac{1}{1-z/i}=\dfrac{1}{2i}\sum_{n=0}^\infty \dfrac{z^n}{i^n}=\dfrac{1}{2}\sum_{n=0}^{\infty}\dfrac{z^n} {i^{n+1}},$$
and also
$$-\dfrac{1}{2}\dfrac{1}{z+i}=\dfrac{1}{2}\dfrac{1}{-z-i}=\dfrac{1}{2i}\dfrac{1}{-1-z/i}=\dfrac{i}{2}\dfrac{1}{1-(iz)}=\dfrac{i}{2}\sum_{n=0}^\infty i^n z^n=\dfrac{1}{2}\sum_{n=0}^\infty i^{n+1}z^n,$$
thus we should have
$$f(z) = \dfrac{1}{z} + \dfrac{1}{2}\sum_{n=0}^{\infty}\dfrac{z^n} {i^{n+1}}+\dfrac{1}{2}\sum_{n=0}^\infty i^{n+1}z^n,$$
however the answer is $1/z+ \sum_{n=0}^{\infty}(-1)^{n+1}z^{2n+1}$. Now, what I've done wrong there? Is there something wrong I did, or there's just some tricky to simplify this last expression I didn't see?
Thanks very much in advance!
AI: Why won't you expand directly?! If you have two series you will have to check very carefully whether some summands of one of them add/cancel with some summands of the other one (and this is what happens here...).
$$|z|<1\implies\frac1z\frac1{1+z^2}=\frac1z\sum_{n=0}^\infty(-1)^nz^{2n}$$
based on the powers expansion
$$\frac1{1+z}=\sum_{n=0}^\infty (-1)^nz^n\;,\;\;|z|<1$$ |
H: Can every function which can be described by words, be formulated as well?
Almost one year ago i was amused when i saw this page. It was the generation of the prime numbers using the floor function, mostly. I became more interested about the things we can do with the floor function. For instance to calculate IsPrime$(n)$, we can use Wilson's Theorem like this (Wilson's theorem states that n is prime if and only if it divides $(n - 1)! + 1$): IsPrime$(n)$=IsInteger$(\frac{(n-1)!+1}{n})$=$[\frac{(n-1)!+1}{n}]+[\frac{(n-1)!+1}{-n}]$+1 which i know calculating it will take more time than to see if it is prime or not with the prime numbers definition, but that's not my point now. To find another example, I came up with a idea to formulate $d(x)$ which is the number of divisors of $x$. I will give the necessary information of how i found it:
By a definition, $d(x)=(a_1+1)(a_2+1)...(a_k+1)$ where $a_1,a_2,\cdots,a_k$ are all of the powers of the distinct prime factors in the prime factorization of $n$. What i should do now is to find a formula for the highest power of each prime factor of $n$. I first calculate which powers of every prime number are in the factorization of $n$, so that counting them up will give us the highest power of $p$ in $n$. This is a formulated version of what i just said: PowerofP$(p,n)=\sum_{i=1}^{[\log_px]}{\left([\frac{x}{a^i}]+[\frac{-x}{a^i}]+1)\right)}$ And now using the definition of $d(x)$ we can formulate the overall function like this: $\prod_{\substack{
p=1 \\
p \in \mathbb P}}^\infty{(\text{PowerofP}(p,n)+1)}$
I even managed to formulate functions like: "Number of ways to write $2n$ as sum of $2$ primes" or "IsSquareFree$(n)$". Now my question is: Can every function which can be described by words, be formulated as well?
Restrictions:1) In our formulas, everything i used in my examples are allowed. Limits are not allowed. 2)Mapping and plotting are not counted as the functions i stated in my question, as one can find easy counterexamples for them.
AI: I don't think the question is well-formed, really, but since you mentioned Wilson's theorem and are asking about notions vs. notation, perhaps you would enjoy this quotation:
In his one-page proof of the long-unproven Wilson’s prime number theorem, first published by Edward Waring, Gauss noted that “neither of them was able to prove the theorem, and Waring confessed that the demonstration seemed more difficult because no notation can be devised to express a prime number. But in our opinion truths of this kind should be drawn from notions rather than from notations.” |
H: Compactness of closure
Let $A$ be a subset of a metric space $X$.
Assume that each sequence in $A$ has a convergent subsequence with limit in the closure $\overline{A}$.
Does this imply that $\overline{A}$ is compact? I read it on wikipedia but I couldn't find a proof of it anywhere.
AI: Yes, that does imply that $\overline{A}$ is compact. For metric spaces, compactness is equivalent to sequential compactness, so let's consider an arbitrary sequence $(x_n)$ in $\overline{A}$. For each $n$, choose an $y_n \in A$ with $d(y_n,x_n) < 2^{-n}$. By the premise, there is an $x \in \overline{A}$ and a subsequence $(y_{n_k})$ with $\lim\limits_{k\to\infty} y_{n_k} = x$. Then we also have $\lim\limits_{k\to\infty} x_{n_k} = x$ (since $d(x,x_{n_k}) \leqslant d(x,y_{n_k}) + d(y_{n_k},x_{n_k}) < d(x,y_{n_k}) + 2^{-n_k}$), so $\overline{A}$ is sequentially compact. |
H: How can I Prove $\frac{2xy}{x+y}\leq \sqrt{xy}\leq \frac{x+y}{2}$
for $x,y>0$ prove that
$\frac{2xy}{x+y}\leq \sqrt{xy}\leq \frac{x+y}{2}$
I have tried to develop $(x+y)^2=$ and to get to an expression that must be bigger than those above
Thanks!
AI: For example
$$\frac{2xy}{x+y}\le\sqrt{xy}\iff 4x^2y^2\le xy(x^2+2xy+y^2)\iff$$
$$ \iff xy(x^2+2xy+y^2-4xy)\ge 0\iff xy(x-y)^2\ge 0$$
And since the last rightmost inequality is obvious we're done. |
H: prove that $T^2=T$ diagonalizeable without using Jordan
Question:
Given that $T:V \to V$ and $T^2=T$ prove that $T$ is diagonalizable.
What I know:
$T^2-T=0=T(T-I)$.
$\operatorname{Im}(T-I) \subseteq\operatorname{Ker}(T)$ therefore
$\dim V=\dim\operatorname{Ker}(T-I)+\dim\operatorname{Im}(T-I)\leqslant \dim\operatorname{Ker}(T-I)+\dim\operatorname{Ker}T$
Can I just say now that since the the sum of the dimensions of these two kernels is $\dim V\leqslant$ so it's equal to $\dim V$ and that the map is diagonalizable? I feel like something is missing with this explanation.
Thanks for any enlightening comments.
AI: Hint: show instead that every element of V can be uniquely written as $u+v$ where $u$ is in the kernel and $v$ is in the image. How does $T$ act on such a decomposition? |
H: Symmetric matrix with zero patern
I'm wondering if I can find some general formula for the inverse of such symmetric matrix :
$$\begin{bmatrix}1 & -k & 0 &-k\\ -k & 1 & -k & 0\\0 & -k & 1 & -k\\-k & 0 & -k & 1 \end{bmatrix}$$
which would still be valid for NxN generalization. The first line would be $$\begin{bmatrix}1 & -k & 0 & \dots & 0 & -k\end{bmatrix}$$
Or some base transformation that'd simplify the thing.
AI: The inverse is simple:
$$\frac{1}{1-4k^2} \begin{bmatrix}
1-2 k^2 & k & 2 k^2 & k \\
k & 1-2 k^2 & k & 2 k^2 \\
2 k^2 & k & 1-2 k^2 & k \\
k & 2 k^2 & k & 1-2 k^2
\end{bmatrix}.$$
Your generalization, however, should be precisely defined. It is not obvious from the above what would an $n \times n$ matrix look like.
Edit: Your general form is a circulant matrix. For the inverse, you might want to look at the paper by Lin Fuyong, "The inverse of circulant matrix", or Durmuş Bozkurt's "On the Determinants and Inverses of Circulant Matrices with a General Number Sequence". |
H: Proof by Induction Question with regard to the Knight's Tour
I have to prove that the formula $4n^2-12n+8$ gives the number of edges on a knight graph, where n is the number of vertices horizontally and vertically and n^2 is the number of vertices.
I've proved it for $n=4$ (the smallest possible value of $n$ with which the knight graph works) and I've assumed that the result is true for n=k but where do I go from here? How do I prove it works for $n=k+1$?
AI: The idea of induction is that given your answer for n=k, you can show that the same holds (in this case that the number of edges equals 4n^2-12n+8) for n=k+1.
To show this we must look at the number of edges that are added when you take a chessboard of k+1 x k+1 instead of k x k. So lets add a new row and column below and left of the old kxk board. I call this the new board.
From the corners of this board you have 2 knight moves to the old board (and ofcourse 2 from the old board to these corners, but as we are counting edges and not moves we will only look at moves starting on the new board). The spaces next to the edges each have 3 options. All other new spaces have 4 options.
So the extra number of edges looks to be 3*2 (for the corners) + 4*3 (for the spaces next to the corners) + 2*((k+1)-4)*4 (for all other new spaces) = 6+12+8(k-3)=8k-6.
Note however that from the spaces next to the lower right corner you can also make a move to spaces on the new board, i.e. we counted them double. Therefore we have to substract 2 from the above answer.
Adding this all together; The new board containts all the moves the old one had + 8k-8 extra moves. I.e. 4k^2-12k+8 + 8k-8 = 4(k+1)^2-12(k+1)+8 |
H: Finding the limit of the recursive sequence $r_{n+1} = \sqrt{2 + r_n}$
In the example I am given, I am told that $r_n$ is defined as:
$$
r_n = \begin{cases}
r_0 = \sqrt{2} \\
r_{n + 1} = \sqrt{2 + r_n} \\
\end{cases}
$$
I was told to calculate $r_3$ and I found that to be:
$$
r_3 = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}}
$$
Then I am told to find
$$
\lim\limits_{n \rightarrow \infty} r_n
$$
So if I let that equal L I get
$$
L = \sqrt{2 + L}
$$
This is where I get stuck. I know that the limit is 2 as I have added a 10 terms together but that will not help me in my exam. How should I go about proving this?
AI: First, you need to solve $L=\sqrt {2+L}$ Square it and you have a quadratic. Once you get the solutions, plug them into the original equation to see which one is not spurious. You also have to show that the limit exists. In this case, you can show that (if $r_n \lt 2$, then $r_{n+1} \lt 2$) and $r_{n+1} \gt r_n$ so you have a monotonic sequence bounded above. That gives you convergence. |
H: Accumulation points of a countable set in [0,1]
Let A be a countable subset of [0,1]. Denote the set of accumulation points of A by A'. Can the set of accumulation points of A' be nonempty (i.e. can we have that A'' is nonempty)?
Thanks for your help
AI: HINT: Take $A=\Bbb Q\cap[0,1]$. |
H: Simplify $e^{\frac {(-\ln 2)}{2}}$
Simplify $\displaystyle e^{\large \frac {(-\ln2)}{2}}$
I know that $(-\ln 2)$ is $\ln\left(\frac{1}{2}\right)$ and the rule $e^{\ln x}=x$. How do I simplify with the fraction at the bottom?
AI: $$ \exp(\frac{- \ln 2}{2} ) = \exp( \ln2^{-{1/2}}) = 2^{-1/2} = \frac{1}{\sqrt{2}}$$ |
H: Derivation of Wallis's Formula
Use Euler's product formula $\Gamma (z)={1\over z}\prod_{n=1}^\infty({1+{1\over n}})^z({1+{z\over n}})^{-1}$ the fact that $\Gamma ({1\over 2})=\sqrt\pi$ to prove Wallis's Formula ${\pi\over 4}={2\over 3}\cdot {4\over 3}\cdot{4\over 5}\cdot{6\over 5}\cdots{2n\over 2n+1}\cdot{2n+2\over 2n+1}\cdots$.
AI: Consider Euler's product formula $\Gamma (z)={1\over z}\prod_{n=1}^\infty(1+{1\over n})^z (1+{z\over n})^{-1}$ and $\Gamma ({1\over 2})=\sqrt \pi$. Thus $$ \sqrt \pi = 2\prod_{n=1}^\infty (1+{1\over n})^{1\over 2} ({1+{1\over 2n}})^{-1}. $$ Multiplying both sides of the equation by ${1\over 2}$ and rewriting the right-hand side we obtain $${\sqrt \pi\over 2}=\prod_{n=1}^\infty ({n+1\over n})^{1\over 2}({2n\over 2n+1}).$$ Squaring both sides of the equation gives us $${\pi\over 4}=\prod_{n=1}^\infty({n+1\over n})({4n^2\over (2n+1)^2}).$$ Simplifying the right-hand side we see that $${\pi\over 4}=\prod_{n=1}^\infty({4n^2+4n\over (2n+1)^2}).$$ Thus factoring out $2n$ on the numerator and rewriting the right-hand side we have Wallis's Formula $${\pi\over 4}=\prod_{n=1}^\infty({2n\over 2n+1})({2n+2\over 2n+1}).$$ |
H: How to find the probability there are 11 or more cars?
During rush hour the number of cars passing through a particular intersection has a Poisson distribution with an average of 540 per hour.
Find the probability there are 11 or more cars?
The answer is 0.006669.
I don not know how to deal with this kind of question.
Please help.Thank you very much.
AI: Let $X$ be a Poisson random variable that describes the number of cars in one hour with $\lambda = 540$. Then:
$P(X\ge11)=1-P(X<11)=1-P(X=10)-P(X=9)-...-P(X=0)$
where $P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}$ |
H: Scheme-Theoretic Nakayama's Lemma
Let $X$ be a noetherian scheme and $\mathscr{F}$ a coherent $\mathscr{O}_{X}$-module. For a point $x \in X$, let $k(x)=\mathscr{O}_{X,x}/\mathfrak{m}_{x}$ be the residue field at $x$.
(a) Suppose $x \in X$ is a point such that $\mathscr{F}_{x} \otimes_{\mathscr{O}_{X,x}}k(x)=0$. Show that there exists an open neighborhood $U$ of $x$ such that $\mathscr{F} \mid_{U}=0$.
(b) Suppose $U$ is an open neighborhood of $x$ and $a_{1},...,a_{n} \in \mathscr{F}(U)$ are such that the images $\bar{a}_{1},...,\bar{a}_{n}$ generate $\mathscr{F}_{x} \otimes_{\mathscr{O}_{X,x}}k(x)$. Show that there exists an open neighborhood $U_{0} \subset U$ of $x$ such that $a_{1},...,a_{n}$ generate $\mathscr{F}\mid_{U_{0}}$.
(c) For $x \in X$ define $e(x)=dim_{k(x)}(\mathscr{F}_{x} \otimes_{\mathscr{O}_{X,x}}k(x))$. Show that $e$ is upper semi-continuous, i.e. $\left\{x:e(x)\leq r\right\}$ is open for all $r$.
Initial thoughts: (c) $e(x)$ is simply the minimal number of generators required to generate $\mathscr{F}_{x}$ over $\mathscr{O}_{X,x}$. Suppose that $e(x)=n$ and let $s_{1},...,s_{n}$ be the sections of $\mathscr{F}$ over some neighborhood $U$ of $x$ such that their images in $\mathscr{F}_{x}$ generate $\mathscr{F}_{x}$ over $\mathscr{O}_{X,x}$. Then I thought to define a morphism from $\mathscr{O}_{U}^n \rightarrow \mathscr{F}$ which is given by these sections and then show that the cokernel of this morphism is zero in some neighborhood of $x$. I think this should work.
I came across these various formulations of Nakayama's Lemma in my reading. How might I prove these? Any help would be appreciated!
AI: Let $X$ be an arbitrary locally ringed space and $\mathcal{F}$ be an arbitrary $\mathcal{O}_X$-module of finite type.
(a) Choose some open neighborhood of $x$ on which the restriction of $\mathcal{F}$ is generated by sections $s_1,\dotsc,s_n$. Then $\mathcal{F}_x$ is generated by their stalks $(s_1)_x,\dotsc,(s_n)_x$. Since $\mathcal{F}_x \otimes k(x)=0$, by Nakayama we get $\mathcal{F}_x=0$. In particular, $(s_1)_x=\dotsc=(s_n)_x=0$. It follows that $s_1,\dotsc,s_n$ vanish on some small neighborhood of $x$. Then also $\mathcal{F}$ vanishes there.
(b) Apply (a) to $F|_U / \langle a_1,\dotsc,a_n \rangle$.
(c) Follows immediately from (b). |
H: Help on understanding Schwartz space
Can someone give an example of Schwartz space function that doesn't decay exponentially?
AI: Consider your favorite bump function construction and replace the exponential function by $$e(x)=\sum_{k=0}^\infty \frac{x^k}{\sqrt{k!}}$$ |
H: Set theory: what is other way to represent A\B?
I try to represent basic set operations using other operations with some limitations.
For example $A \setminus B$ using only $\cup$ and $\oplus$ (Symmetric difference):
$A \setminus B = A \oplus (A \oplus B) \oplus (A \cup B)$
But I stuck with the same set operation but using onlu $\oplus$ and $\cap$.
I came to something like: $A \setminus B = (A \oplus B) \oplus (A \cap B) \cap B$
Is it correct or did I miss something?
AI: $A\setminus B=B\oplus(A\cup B)=A\oplus(A\cap B)$ |
H: Prove that $b^2=a^2$
Let $G$ be a group of order $8$. Assume that there exists $a \in G$ such that $\lvert a\rvert =4$ and that no elements of $G$ has order $8$. Assume $\langle a \rangle \lhd G$, $b \notin \langle a\rangle$ and $b^2 \in \langle a\rangle$. Suppose that $\lvert b\rvert=4$, then Prove that $b^2=a^2$.
Not sure how to do it. Help appreciated.
AI: If $b\in \langle a\rangle$ then $b^{2}=a^{k}$ for some $0\leq k\leq 3$. But we know $|b|=4$ then $b^{2}$ has order 2. You can check that the only element of order 2 in $\langle a\rangle$ is $a^{2}$. Hence $b^{2}=a^{2}$. Ι do not see where the hypotheses that $\langle a\rangle \triangleleft G$ and $b\notin \langle a\rangle$ are required though. |
H: Maximal ideals of polynomial rings in infinitely many variables
Let $k$ be an algebraically closed field. Nullstellensatz states that the maximal ideals of the polynomial ring $R=k[X_1,\dots,X_n]$ are precisely those of the form $\langle X_1-a_1,\dots,X_n-a_n\rangle$, with $(a_1,\dots,a_n)\in k^n$.
What if we are working with infinitely many variables, say $\{X_i\}_{i\in I}$, being $I$ an infinite set? Again, the ideals of the form $\bigl\langle\{X_i-a_i: i\in I\}\bigr\rangle$, with $(a_i)_{i\in I}\subseteq k$ are maximal (even if $k$ is not algebraically closed), but what about the converse?
AI: Let $k$ be a field and let $K$ be an extension of $k$ of degree larger than $1$. Let $R=k[X_i, i\in K]$ be a polynomial ring with one variable per element of $K$. There is a $k$-linear ring homomorphism $\phi:R\to K$ which maps $X_i$ to $i$ for all $i\in K$, and it is surjective. It follows that the kernel of $\phi$ is maximal. Clearly, it is not of the form you mentioned: indeed, the quotient of $R$ by all ideals of that form is $1$-dimensional as a $k$-algebra. |
H: Cofactors and conjugates of $SU(3)$.
I was playing around with some equations and noticed the following: Let $A$ be an element of $SU(3)$ with components $A_{ij}$. If $C_{ij}$ is the $(i,j)$ cofactor of $A$ then $C_{ij} = \overline{A_{ij}}$.
Now this is probably not surprising. Since $A \in SU(3)$ we know that its rows and columns form an orthonormal basis for $\mathbb C^3$, so that
$$ |A_{i1}|^2 + |A_{i2}|^2 + |A_{i3}|^2 = |A_{1j}|^2 + |A_{2j}|^2 + |A_{3j}|^2 = 1, \qquad \text{ for all } i,j \in \{1,2,3\}.$$
Furthermore, since $\det A = 1$, the cofactor expansion of the determinant tells us that for any $i,j$
\begin{align}
1 &= A_{i1} C_{i1} + A_{i2} C_{i2} + A_{i3}C_{i3} \\
&= A_{1j}C_{1j} + A_{2j}C_{2j} + A_{3j}C_{3j}, \tag{1}
\end{align}
and this corresponds to the fact that the determinant is invariant under choice of expansion. If $C_{ij} = \overline{A_{ij}}$ then everything works out nicely, but it seems reasonable to expect that there could be other solutions to $(1)$. Perhaps the fact that so many equations need to be simultaneously satisfied enforces a rigidity condition?
In any case, I came across this fact via a rather complicated computation (there are bundles and tori involved), but it seems to me to be sufficiently elementary and fundamental that a much simpler proof should exist, and should generalize to $SU(n)$. Maybe I'm just being foolish, but I cannot seem to see it.
AI: We have $A^\ast = A^{-1} = \frac{\operatorname{adj}(A)}{\det(A)} = \operatorname{adj}(A)$. Taking transpose on both sides, we obtain $\bar{A} = \operatorname{adj}(A)^\top$. Hence $\bar{A}_{ij}=C_{ij}$. |
H: A norm for Lipschitz-continuous functions
I am a first year undergrad math student, and I am struggling with a proof. Let the set $C_{\text{Lip}}:= \left \{ f:\mathbb{R}\rightarrow \mathbb{R}: f \text{ is Lipschitz continuous} \right \}$ be a Vectorspace over $\mathbb{R}$. For $f \in C_{\text{Lip}}$ we define
$$
c_f:=\sup\left \{ \frac{\left |f(x)-f(y) \right |}{\left |x-y \right |} : x,y \in \mathbb{R}, x\neq y \right \}.
$$
Show that:
$$
\left \| f \right \|:=\left | f(0) \right |+c_f
$$
is a norm on $C_{\text{Lip}}$.
What is the best way to go about this? Any help is appreciated.
Cheers
@TBrendle:
Tanzania, in my home country brother!
AI: When one has to show that something is a certain mathematical object, there is only one way to go: to verify the definition. So, if you want to show the assigned function is a norm, you have to verify the axioms for a norm.
$\lVert f \rVert \geq 0$. Obvious, since $\lVert f \rVert$ is sum of two nonnegative contributions.
$\lVert f \rVert = 0 \iff f = 0$. Suppose $f = 0$. Then
\begin{equation}
\lVert f \rVert = \lvert f(0) \rvert + c_f = 0,
\end{equation}
since $f(x) = 0$ for all $x$ in the domain. Note that $c_f$ is well defined. Conversely, suppose $\lVert f \rVert = 0$. Since it is sum of two nonnegative contributions, this is possible if and only if $f(x) = 0$ for all $x$ in $\mathbb R$.
$\lVert a f \rVert = \lvert a \rvert \lVert f \rVert$ ($a \in \mathbb R$). Obvious.
Triangle inequality. Let $f,g \in C_{Lip}$. Then
\begin{equation}
\lVert f + g \rVert = \lvert (f + g)(0) \rvert + c_{f+g} \leq \lvert f(0) \rvert + \lvert g(0) \rvert + c_f + c_g = \lVert f \rVert + \lVert g \rVert.
\end{equation}
The only non completely trivial passage is $c_{f+g} \leq c_f + c_g$. We have
\begin{equation}
\begin{split}
c_{f+g} &= \sup\left\{ \frac{\lvert (f + g)(x)-(f + g)(y) \rvert}{\lvert x - y \rvert} : x,y \in \mathbb R , x \neq y \right\} \\
& =\sup \left\{ \frac{\lvert ( f(x) - f(y) ) + (g(x)-g(y)) \rvert}{\lvert x - y \rvert} : x,y \in \mathbb R , x \neq y \right\} \\
& \leq \sup\left\{ \frac{\lvert (f(x)- f(y) \rvert}{\lvert x - y \rvert} : x,y \in \mathbb R , x \neq y \right\} + \sup\left\{ \frac{\lvert g(x)- g(y) \rvert}{\lvert x - y \rvert} : x,y \in \mathbb R , x \neq y \right\} \\
& = c_f + c_g.
\end{split}
\end{equation}
Hence the result. |
H: proof- can NOT be a linear combination
How can I prove that $X^2-Y,X-Y^2$ CAN NOT be written as a combination of $<X^3-Y^3,X^2Y-X>$ ?
AI: Suppose $X^2-Y=p(X,Y)(X^3-Y^3)+q(X,Y)(X^2Y-X)$. Evaluating at $X=0$ gives
$Y=p(0,Y)Y^3$, which is impossible. The same technique works for $X-Y^2$. |
H: Evaluation of Standard Normal Integral
I have always wondered how we calculate the percentiles of the Standard Normal Distribution given that the CDF cannot be obtained in closed form:
$$F(x)=\int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{t^2}{2}} $$
Do we approximate the function somehow? I have seen some texts that employ the Taylor Series of $e^x$, i.e. $e^x = \sum_0^{\infty} \frac{x^n}{n!} $ but this is not very practical. Is there another way to approximate the integral at a satisfactory level? Thank you.
AI: You may want to take a look at some of the approximations of the error function in Abramowitz and Stegun. Some of these may be derived by guessing the approximate form and fitting coefficients of the form to various numerically derived data points. This is evident in formulae 7.1.25-7.1.28. Unfortunately, much of applied mathematics can be a little on the ugly side like this; the results, however, are incredibly practical. |
H: Taking the cross product of a cross product? Proving an identity that involves gradients and vectors?
Problem 20:
Solution:
I am having difficulty understanding how the boxed is not equal to 0. The derivative of 1 is equal to 0.
AI: Let's back up. What you're really having trouble with is understanding
$$\nabla \times (a \times r) = (\nabla \cdot r) a - (\nabla \cdot a) r$$
The notation could be considered unclear. The product rule tells us that both $a$ and $r$ must be differentiated here; there are actually four terms. One way to denote this is to use overdots; I'll consider only the second term:
$$(\nabla \cdot a) r = (\dot \nabla \cdot \dot a) r + (\dot \nabla \cdot a) \dot r$$
The overdot means "if $\nabla$ has a dot over it, differentiate only that which also has a dot over it".
Let's look at the first term; it's just a divergence.
$$(\dot \nabla \cdot \dot a) r =r ( \mathrm{div} \, a) = 0$$
Because $a$ is a constant vector.
The second term is more clearly written as a directional derivative.
$$(\dot \nabla \cdot a) \dot r = (a \cdot \dot \nabla) \dot r = a$$
This is the boxed term. It's a known identity that the directional derivative of the position vector is just the direction the derivative is taken in.
Unfortunately, without any overdots or other decorations to denote what is to be differentiated, it becomes very unclear whether to apply the product rule or not. I would always regard a statement like $(\nabla \cdot a) r$ to differentiate only the vector $a$, for instance, but the authors of this problem did not choose to do that. |
H: Does the limit of this sequence exist?
While practicing for an exam, I encountered this question:
(e) For a sequence $\lbrace b_n \rbrace_{n=1}^\infty \subset \mathbb{R}$ is given: $\forall n \in \mathbb{N}$, $\: b_n < b_{n+1} < 2$.
Does $\lim_{n\to\infty} b_n$ exist? If $\lim_{n\to\infty}$ exists, give $\lim_{n\to\infty}$and explain the answer. If $\lim_{n\to\infty}$ does not exist, explain the answer.
My thought process: The limit exists $\Longleftrightarrow $ if $\exists n$ such that $b_n=b_{n+1}$. And because this contradicts with the fact that $b_n < b_{n+1}$. Hence, the limit does not exist.
A classmate on mine had an other answer: "Since $b_{n+1}>b_n$ but $b_{n+1} < 2$, we know that $b_{n+1}$ is closer to 2 than $b_n$. He deduces from this that the limit is 2.
Who of us is right?
AI: Your sequence is increasing and bounded. Thus it is convergent.
The limit $l$ satisfies $l \leq 2$, but as it was pointed it is not necessarily $2$. For each $ l\leq 2$
$$b_n =l -\frac{1}{n}$$
is an example of a sequence satisfying the conditions and converging to $l$. |
H: If a closed, smooth $m-1$ form, $\omega$ is nonzero at a point, there are local coordinates $x^i$ with $\omega = dx^2 \wedge\cdots \wedge dx^m.$
This is a problem on an old qualifying exam.
Let $\omega$ be a smooth, closed $m-1$ form on a smooth $m$-dimensional manifold $M$. If $\omega \neq 0$ at a point $p\in M$ then there is a coordinate system $(x^1,...,x^m)$ on a neighborhood of $p$ in which
$$\omega = dx^2 \wedge dx^3 \cdots \wedge dx^m.$$
I am thinking they want us to use the Frobenius theorem but I am having trouble making any progress.
AI: Well, Frobenius, offhand, applies to a differential system generated by a family of $1$-forms. But let's be clever. In standard coordinates $(y^1,\dots,y^m)$, we can write $$\omega = \iota_X dy^1\wedge\dots\wedge dy^m$$
for some vector field $X$ with $X(p)\ne 0$. By the flowbox theorem, we can choose local coordinates $(z^1,z^2,\dots,z^m)$ so that $X = \partial/\partial z^1$, so $\omega = f\,dz^2\wedge\dots\wedge dz^m$ for some nonzero function $f$. Since $d\omega = 0$, we infer that $\dfrac{\partial f}{\partial z^1} = 0$, and so $f=f(z^2,\dots,z^m)$. Now set
$$x^2=\int f(z^2,\dots,z^m)\,dz^2\,,$$ so that
$$dx^2 = f\,dz^2 + \sum_{j=3}^m \left(\int \dfrac{\partial f}{\partial z^j}\,dz^2\right)\,dz^j\,.$$
It follows that if we set $x^1=z^1, x^3=z^3,\dots, x^m=z^m$, then $\omega = dx^2\wedge\dots\wedge dx^m$. |
H: Transitive closure of a union
I have a quick question regarding transitive closures. In the text I'm currently reading, (Kunen - Set Theory, 2011), the transitive closure of a set $x$ is defined as trcl$(x) = \{ a : a \in^* x \}$, where $a \in^* x$ means there is an $\in$-path from $a$ to $x$ (I think the text implies that this is a finite path).
I'm curious as to whether or not the following is true for any two sets $x$ and $y$:
$$ \mbox{trcl}(x \cup y) = \mbox{trcl}(x) \cup \mbox{trcl}(y),$$
or if there is some sort of counter-example that shows this is false? Would this hold for and infinite union as well? Thanks in advance!
AI: HINT: Note that if $a\in\operatorname{trcl}(x)$ then either $a\in x$ or $a\in\operatorname{trcl}(x')$ for some $x'\in x$. Conclude equality holds (for the infinite case as well).
[You can also note that $\operatorname{trcl}(x)$ is the smallest transitive $y$ set such that $x\subseteq y$. In that case it becomes easier to prove these equalities.] |
H: How can I prove that 4k^2 mod 3 is always = 1
I have a statement $n \in N, \;n^2 \mod 3 = \{0, 1\}$, which basically says that any natural number $n$ when squared will have a remainder after dividing by $3$ of either $0$ or $1$.
From here I expended my proof into two cases
$n = 2k, n^2 = 4k^2$
and
$n = 2k + 1, n^2 = 4k^2 +4k + 1$
From here I have a sense that $4 \mod 3 = 1$
and
$4 \mod 3 + 4 \mod 3 + 1 \mod 3 = 3 \mod 3 = 0$
But how do I prove this?
AI: You're starting with the wrong idea. Write $n=3k$ or $n=3k+1$ or $n=3k+2$.
Working with residue classes modulo $3$ is of course the best strategy: if $[n]$ denotes the residue class of $n$, then $[n]=[0]$ or $[n]=[1]$ or $[n]=[2]$; since
\begin{align}
[0]^2&=[0],\\
[1]^2&=[1],\\
[2]^2&=[4]=[1],
\end{align}
the claim is proved.
This is not at all mysterious: Gauss's idea of using the residue classes is just formally forgetting the multiples of $3$ (or the modulo one uses) as computations go along. |
H: Closure of the range of a compact operator
Let $X$ be an infinite-dimensional Banach space, and let $Y$ be a banach. Let $T$ be a compact operator from $X$ to $Y$, ie. if $(x_n)$ is a sequence in $X$ then there is a subsequence s.t. $T(x_{n(k)})$ converges. We wish to show that 0 is in the closure of $\{Tx,||x||=1\}$.
I've been working on this problem for a while, and have been entirely unable to solve it without resorting to the spectral theorem, but I've been told there is an elementary solution involving only the fact that we can find a sequence of unit vectors $X$ s.t. that the distance between the elements is greater than or zero (ie. Riesz Lemma).
AI: Suppose you have a continuous linear operator $A\colon X \to Y$ with $X$ and $Y$ as in the question, such that $0$ is not in the (norm) closure of $\{ Ax : \lVert x\rVert_X = 1\}$. Then
$$\delta := \inf \{\lVert Ax\rVert_Y : \lVert x\rVert_X = 1\} > 0.$$
Hence $A$ is injective, and $A$ is an open mapping to $\mathcal{R}(A)$; we have $\delta\lVert x\rVert_X \leqslant \lVert Ax\rVert_Y \leqslant \lVert A\rVert\cdot\lVert x\rVert_X$ for all $x\in X$, so $A^{-1} \colon \mathcal{R}(A) \to X$ is continuous, with $\lVert A^{-1}\rVert = \delta^{-1}$. Therefore $\mathcal{R}(A)$ is a Banach space (if $(Ax_n)$ is a Cauchy sequence in $\mathcal{R}(A)$, then $(x_n)$ is a Cauchy sequence in $X$, hence convergent, $x_n \to x_\ast$, and then $Ax_n \to Ax_\ast$ by the continuity of $A$). Since $A$ is injective, $\mathcal{R}(A)$ is infinite-dimensional. Hence $\overline{A(B_X)}$, which is a neighbourhood of $0$ in $\mathcal{R}(A)$ is not compact (since only finite-dimensional Hausdorff topological vector spaces are locally compact [Riesz]). Therefore $A$ is not compact. |
H: What is the probability there are two or more claims?
In a group of policy holders for house insurance,the average number of claims per $100$ policies per year is $\lambda=8.0$. The number of claims for an individual policy holder is assumed to follow a Poisson distribution.
In a group of $20$ policy holders,what is the probability there are two or more claims?
AI: $\lambda=np$ so if n=100 and $\lambda$=8, what is p?
Now take that p to find your new $\lambda$ when n=20. Then let $X$ be a Poisson random variable with that new $\lambda$. Then you can compute:
$P(X\ge2)=1-P(X<2)=1-P(X=1)-P(X=0)$ where $P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}$ |
H: Proving a Property of a Set of Positive Integers
I have a question as such:
A set $\{a_1, \ldots , a_n \}$ of positive integers is nice iff
there are no non-trivial (i.e. those in which at least one component
is different from $0$) solutions to the equation $$a_1x_1 + \ldots +
> a_nx_n = 0$$ with $x_1 \ldots x_n \in \{ -1, 0, 1 \}$. Prove that any
nice set with $n$ elements necessarily contains at least one element
that is $\geq \frac{2^n}{n}$.
Here's my work so far:
Let $A = \{a_1, \ldots, a_n \}$. Let $P = a_1x_1 + \ldots + a_nx_n$. $A$ is nice iff there are only trivial solutions to $P$.
There is a nontrivial solution to $P$ iff $\exists B,C \subset A$ s.t. $sum(B)=sum(C)$. Note that $B\cap C = \emptyset$, but $B\cup C$ does not necessarily equal $A$, as once there are such subsets,
the other elements of $A$ can be set to multiply with $x_i = 0$ s.t. Moreover, either $B$ or $C$ has to be multiplied by $x_i = -1$ s.t. $B+C = 0$.
The contrapositive of the initial statement is that if $\forall a \in A, a < \frac{2^n}{n}$, then $A$ is not nice.
We note that a non-nice set can contain $x \geq \frac{2^n}{n}$. It is only that if a set is nice that it necessarily contains $x \geq \frac{2^n}{n}$. Just the fact that a set contains such an element doesn't actually tell us whether the set is nice or not.
The question that I arrived at is this: why is that if $\forall a\in A, a < \frac{2^n}{n}$, then $A$ has a non-trivial solution? If I can show why this is true, it would constitute proof of the contrapositive, and I would be done.
By the way, if anyone can think of a better title; please do suggest it.
AI: Hint: If the sums of two distinct subsets are equal, then the set is not nice.
There are $2^n$ subsets. If the numbers are all $\lt \frac{2^n}{n}$, then the sum of all the numbers is less than $2^n$. Now use the Pigeonhole Principle. |
H: Converting second order equation to first order equation
How would I convert the following second order equation to a first order?
$$ u'' + 3u' - 4.5u = -2.5\sin(3t) $$
I have let $v=u'(t)$, but not sure what $v'(t)$ would look like.
$$ u' = v $$
$$ v' = ??? $$
My attempt:
$$ u'' = v' = -2.5sin3t - 3v + 4.5u$$
Surely the $4.5u$ piece can be substituted in terms of v?
AI: Hint: since $v=u^\prime$, we have $v^\prime=u^{\prime\prime}$. Thus we must solve for $u^{\prime\prime}$ in the original equation:
$$
v^\prime=u^{\prime\prime}=-2.5\sin(3t)-3u^\prime+4.5u
$$
This should get you started. |
H: Is it possible for a relation to be symmetric, antisymmetric, but NOT reflexive?
If $A$ is a set $\{2,4,6,8\}$, and we are asked to give a relation on $A$ that is:
symmetric, antisymmetric, but not reflexive, is this possible?
If we were to say $\{(2,2),(4,4)\}$, it would indeed be symmetric and antisymmetric, but it would also be reflexive.
If we were to say $\{(2,2),(4,4),(6,8)\}$, this would not be symmetric.
I'm thinking that this isn't possible, but would like to know others' thoughts.
AI: Recall that being symmetric and anti-symmetric are internal properties of the relation. They can be defined regardless to any external set (such as $A$). These properties merely say "If the ordered pair ... is in the relation, then there is another ordered pair ..." or so.
On the other hand, reflexivity is an external property. It must refer to an external set. We say that $R$ is a reflexive relation on $A$ if every $a\in A$ satisfies $(a,a)\in R$. So for example $\varnothing$ is a reflexive relation on $\varnothing$, but not on $\{a\}$. |
H: Why is the empty family linearly independent?
Why is the empty family linearly independent?
How can you prove this? Is it right to say that it can not be written as a linear combination of the others vectors?
AI: Any subset of a linearly independent set is linearly independent. This is almost obvious for non empty subsets; why shouldn't it be for the empty subset?
In order to prove that a set is linearly dependent you have to exhibit vectors in the set and coefficients (not all $0$) so the linear combination gives $0$. Can you for the empty set?
If you want to see it differently, can a vector in the empty set be written as linear combination of the other vectors in the empty set? Of course not. For the same reason, a single non zero vector forms a linearly independent set, because it can't be written as a linear combination of the other vectors (there's none remaining).
But it's easier if one takes a step further and defines that the only linear combination of the empty set is the zero vector. So a singleton $\{v\}$ with $v\ne0$ is linearly independent, because the only linear combination of the vectors in the set different from $v$ is $0\ne v$; conversely $\{0\}$ is linearly dependent, because $0$ is a linear combination of the “other” vectors, namely of the empty set.
If you don't trust this (but you should), take it as a definition by convention, so also the zero vector space has a basis (the empty set) and every set containing the zero vector is linearly dependent. |
H: Decreasing sequence of measurable functions
Suppose $f_1(x),f_2(x),\ldots:[0,1]\rightarrow\mathbb{R}$ are measurable functions such that $f_1(x)\geq f_2(x)\geq\ldots$. (infinite sequence) and $\lim_{n\rightarrow\infty}f_n(x)=0$. Is it true that $\lim_{n\rightarrow \infty}\int_0^1 f_n(x)dx=0$?
I wanted to apply the dominated convergence theorem but cannot, because $f_i(x)$ is not necessarily integrable. So is there a counterexample for this?
AI: $$
f_n(x) = \begin{cases}
\dfrac{\chi_{[0, 1/n]}}{x} &: x \in (0, 1] \\
0 &: x = 0\end{cases}
$$ |
H: Greatest lower bound and meet-semilattice of set
I am wondering if I am understanding this concept correctly.
A partially ordered set $S$ is called a meet-semilattice if, for all $x,y \in S$, a unique greatest lower bound of $x$ and $y$ exists.
So let's say we have the set $S = \{1,2,3,12,36\}$, and we want to ask if this set, when partially ordered by divisibility, is a meet-semilattice.
I am inclined to say that NO it is not. If we take $x = 12$ and $y=36$, then there are three lower bounds ($1,2,3$). But, none of them is the greatest since $2$ and $3$ do not divide each other.
Is my logic/approach correct?
AI: The set $\{12,36\}$ actually has $4$ lower bounds; the one you missed is $12$, which also happens to be the greatest lower bound. In general, if $x$ and $y$ are comparable, they automatically have a greatest lower bound (and a least upper bound). You only need to check incomparable pairs, in this case, $\{2,3\}$. |
H: Integrals of probability density functions and their inter-relationships
I am a little bit confused about the relationship between marginal probability density functions (pdfs), joint pdfs (jpdfs), and conditional pdfs (cpdfs), and their integrals. Let me define the following pdfs: $f_X\left(x\right), f_Y\left(y\right)$. Their jpdf is denoted $f_{X,Y}\left(x,y\right)$. And their cpdf is denoted $f_{X|Y}\left(x|y\right)$. I know the following relationship holds:
$$f_{X,Y}\left(x,y\right)=f_X\left(x\right)f_{Y|X}\left(y|x\right)=f_Y\left(y\right)f_{X|Y}\left(x|y\right)$$
What I'm unsure about is what the integrals of these functions are. Intuitively I would expect:
$$\int f_{X|Y}\left(x|y\right)dy=f_X\left(x\right)$$
But, what about:
$$\int f_{X,Y}\left(x,y\right)dy=?$$
or
$$\int f_{X|Y}\left(x|y\right)dx=?$$
AI: The first integral you have is not correct. If I plug in the (correct) definition you give for the conditional PDF, you can see the integral is not so simple:
$$
\int f_{X|Y}(x|y) dy = \int \frac{f_{X,Y}(x,y)}{f_{Y}(y)} dy.
$$
The right hand side does not reduce to the marginal PDF for $X$. Your second integral, however, is precisely the marginal PDF for $X$:
$$
\int f_{X,Y}(x,y) dy = f_X(x).
$$
The third integral evaluates to one,
$$
\int f_{X|Y}(x|y) dx = 1,
$$
because a conditional PDF is still a probability density function and must have this normalization.
By normalization, I mean the result of integrating over the entire support of $X$. |
H: The closure of the complement of $A \subseteq \mathbb{R}^d$ with Lebesgue measure zero is $\mathbb{R}^d$?
I have been working on an excercise in measure theory for a few hours now, and although I have learned a lot, the answer to this problem avoids me. It concerns proving the following assertion:
Let $A \subseteq \mathbb{R}^d$ have Lebesgue measure zero. Then $\overline{\mathbb{R}^d \setminus A} = \mathbb{R}^d$,
where $\overline{A}$ is the closure of $A$ with respect to the standard topology on $\mathbb{R}^d$.
I have come to the conclusion that if I can show that $m(A) = 0$ implies that $A$ is nowhere dense in $\mathbb{R}^d$, the result follows, but I can't seem to prove this.
AI: Since $A$ has Lebesgue measure zero, it cannot contain any open ball. Therefore $A$ cannot be a neighborhood of any of its points. It interior is empty, so the closure of its complement is all of $\mathbb{R}^d$. |
H: Integral Of $\int \frac{2\cdot \cos^2(x)}{x^2}dx$
I`m trying to integrate the following:
$$\int\frac{2\cdot \cos^2(x)}{x^2}dx$$
what I did first is:
$$\int \frac{2\cdot (\frac{1}{2}+\frac{cos2x}{2})}{x^2}dx=\int \frac{1+cos2x}{x^2}dx$$
now what? any suggestions? thanks!
AI: Using a simple integration by parts, we arrive at the following expression, which, as has already mentioned above, is irreducible to elementary functions: $$F(x)=-2\left[\frac{\cos^2x}x+\text{Si}(2x)\right]$$ where $\text{Si}(t)=\int\frac{\sin x}x dx$. |
H: Question about basic exponential/logarithm properties
Solve for $k$: $$e^{k/2}=a$$
Solution: $$e^{2k}=a$$
$$ k/2 = \mathbf{ln}a$$
$$ k=2\mathbf{ln}a$$
$$= \mathbf{ln}a^2$$
My question is: why does $2\mathbf{ln}a = \mathbf{ln}a^2$? Why can you transfer the $2$ to be an exponent of $a$?
AI: Because logarithms map multiplication to addition:
$\ln ab= \ln a+\ln b$
So in particular:
$\ln a^2= \ln a+\ln a=2\ln a$ |
H: Intersection of a plane with an infinite right circular cylinder by means of coordinates
So, I started studying analytic geometry and I must say I'm finding it much harder than "classic" geometry, because of the equations without help from diagrams... Still, I wanted to see how to use it to get alternative proofs of geometric theorems.
The theorem that "the intersection of a plane not parallel nor perpendicular to the axis of an infinite right circular cylinder is an ellipse" looked appropriate to be proved via analytic geometry because imagining in three dimensions is harder than in the plane and with equations one does not have to imagine, just compute (at least this is what I'm thinking right now).
Some texts take that as the definition of an ellipse, but then one needs to show that it defines the same shape as the definition "the set of points in the plane such that the sum of the distances from the point to each of two fixed points is constant".
There is that famous synthetic proof in many texts with the auxiliary spheres, as can be found, for example, in Hilbert, D. and Cohn-Vossen, Geometry and the Imagination.
But I wanted to handle the problem by means of equations without having to imagine too much, but also without too long calculations :-)
So, since the equation of an infinite right circular cylinder is x² + y² - R² = 0 and the equation of a plane is Ax + By + Cz + D = 0, the points that satisfy both equations should form an ellipse, right? I was not sure it was correct reasoning, but I tried solving the system composed by these two equations by substitution of x for sqrt(R²-y²) on the second equation, but how to show it's the same as the equation of an ellipse (y/T)² + (z/U)² - 1 = 0 I do not know...
I'm confused, sorry if my question does not make sense.
AI: The slick analytic way to go about this is to (prove and then) appeal to the general theorem classifying second-degree curves in the plane:
Let $P(x,y)$ be a polynomial of degree 2. The locus of points $(x,y)$ with $P(x,y)=0$ is similar to the solution set of exactly one of
$$ \tag{ellipse or circle} x^2 + (y/a)^2 = 1 $$
$$ \tag{hyperbola} x^2 - (y/b)^2 = 1 $$
$$ \tag{parabola} x^2 = y $$
$$ \tag{intersecting lines} x^2 - (y/a)^2 = 0 $$
$$ \tag{parallel lines} x^2 = 1 $$
$$ \tag{one line} x^2 = 0 $$
$$ \tag{one point} x^2 + y^2=0 $$
$$ \tag{nothing} x^2 = -1 $$
for some $a\in (0,1]$ or $b\in(0,\infty)$.
In the case of a plane cutting a circular cylinder, if you write the plane as a parametric equation you get a coordinate system of the plane for free, and a bit of algebra easily shows that the equation of the intersection is a quadratic one.
It is also clear on geometric grounds that the solution set is bounded, and contains at least two points, which means that the only possible among the above cases is $x^2+(y/a)^2 = 1$.
We can then prove once and for all that this equation defines an ellipse in the plane. |
H: Functions are big only over exponentially small sets
Given $f:[0,1]\rightarrow\mathbb{R}$. Suppose $\mu(\{x\in[0,1]\mid |f_n(x)|>1/n\})\leq 1/2^n$ for all integers $n\geq 1$. Is it true that $\lim_{n\rightarrow\infty}f_n(x)=0$ for almost every $x$?
I tried to apply the Chebyshev's inequality, but it doesn't work out. I can't find any counterexample either.
AI: Here is what you want. Suppose $\mu$ is a finite measure on $\Omega,\Sigma)$ and that you have meeasurable sets $E_n$ so that
$$\sum_n \mu(E_n) < \infty.$$
Then $\mu(\limsup_{n\to\infty} E_n) = 0$. Recall that
$$\limsup_{n\to\infty} E_n = \{x\in\Omega\,|\, x\in E_n\,\,\hbox{infinitely often}\}.$$ |
H: Why is $2^x>2x$ when $x>2$?
I came across this inequality while trying to prove that a function was increasing for $x>2$. I checked it graphically and it's true but I don't have a particularly good proof of it.
So, can you tell me why $2^x>2x$ when $x>2$?
AI: You can just take the derivative: $\frac d{dx}(2^x-2x)=2^x \log 2 -2$. For $x \gt 2$ we have $2^x \log 2 -2\gt 4 \log 2 -2 \gt 0.772$ |
H: How to factor these monomials?
This is the original problem:
$x^3+x^2y+xy^2+y^3$
Answer:
$(x+y)(x^2+y^2)$
I understand that the answer is correct, but I can't figure out how to get to it.
AI: $$\underbrace{x^3+x^2y}_{\text{Group these two}}+\overbrace{xy^2+y^3}^{\text{Group these two}} = x^2(x+y) + (x+y)y^2 = (x+y)(x^2+y^2)$$ |
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