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Evaluate $\int \theta\sec\theta \tan\theta \ d\theta$ | Let $c = \cos \frac {\theta}2; s= \sin \frac {\theta}2$
Note that $$\sec \theta +\tan \theta = \frac {1+\sin \theta}{\cos \theta}=\frac {c^2+s^2+2cs}{c^2-s^2}=\frac {(c+s)^2}{(c+s)(c-s)}=\frac {c+s}{c-s}$$ and that should help you to reconcile the two answers. |
Defining region $D*$ for double interal in polar coordinates. | For the curve $y = x^2$, we simply substitute $$(x,y) = (r \cos \theta, r \sin \theta)$$ to obtain $$r \sin \theta = r^2 \cos^2 \theta,$$ hence $$r(r \cos^2 \theta - \sin \theta) = 0.$$ Since $r = 0$ corresponds to the origin, the desired relationship is $$r = \frac{\sin \theta}{\cos^2 \theta}, \quad 0 \le \theta \le \frac{\pi}{4}.$$ When $\theta = \pi/4$, we get $r = \sqrt{2}$ which corresponds to $(x,y) = (1,1)$ as expected. Thus the region of integration in polar coordinates satisfies the inequalities $$0 \le r \le \frac{\sin \theta}{\cos^2 \theta}, \quad 0 \le \theta \le \frac{\pi}{4}.$$ |
In $\triangle ABC$, if $AB=x+2$, $BC=x$, $AC=x-2$, and $C = 120^\circ$, then $x=5$ | It was a silly mistake. Already fixed it.
$(x + 2)^2 = x^2 + (x-2)^2 -2x(x-2)\cos 120°$ |
Probability of choosing a of b numbers in exactly c guesses | Note that this is can be modeled by a hypergeometric distribution for $a-1$ successes out of the first $c-1$ guesses, along with a $\frac{1}{b-c+1}$ chance of getting the last number on the next guess. Plugging in to the formula should give us $\frac{\binom{a}{a-1}\binom{b-a}{c-a}}{\binom{b}{c-1}} \times \frac{1}{b-c+1}$.
Alternatively, this can be seen as the probability of winning in at most $c$ turns but not winning in at most $c-1$ turns. There are $\binom{a}{a}\binom{b-a}{c-a}$ ways to win in $c$ turns, out of a total of $\binom{b}{c}$ possible turns, so this gives us
$$\frac{\binom{a}{a}\binom{b-a}{c-a}}{\binom{b}{c}} - \frac{\binom{a}{a}\binom{b-a}{c-a-1}}{\binom{b}{c-1}}$$
We check to make sure that this matches the above answer:
\begin{align*}
&= \frac{\frac{(b-a)!}{(c-a)!(b-c)!}}{\frac{b!}{c!(b-c)!}} - \frac{\frac{(b-a)!}{(c-a-1)!(b-c+1)!}}{\frac{b!}{(c-1)!(b-c+1)!}}\\
&=\frac{(b-a)!c!}{(c-a)!b!} - \frac{(b-a)!(c-1)!}{(c-a-1)!b!}\\
&=\frac{(b-a)!c!- (c-a)(b-a)!(c-1)!}{(c-a)!b!}\\
&=\frac{(b-a)!(c - (c-a))(c-1)!}{(c-a)!b!}\\
&=\frac{a(b-a)!(c-1)!}{(c-a)!b!}\times\frac{\frac{1}{(b-c)!}}{\frac{1}{(b-c)!}}\\
&=\frac{a\frac{(b-a)!}{(c-a)!(b-c)!}}{\frac{b!}{(c-1)!(b-c)!}}\\
&=\frac{a\binom{b-a}{c-a}}{\frac{b!}{(c-1)!(b-c+1)!}}\times\frac{1}{b-c+1}\\
&=\frac{a\binom{b-a}{c-a}}{\binom{b}{c-1}}\times\frac{1}{b-c+1}\\
\end{align*}
as desired. |
The homology group of the projective space of dimension $2$ | All that is going on is a consolidation of adjacent triangles to give cells. As @Mariano and @Jim Conant pointed out in the comments, picture 2 is a triangulation, while picture 3 is an equivalent CW-decomposition.
Consider first that every point interior to the rectangle represents a distinct point of $\mathbb{R} P^2$ (this follows directly for the characterization given above in terms of the hemisphere). Now consider the points on the left side of the rectangle. In order from bottom to top, we have $P_0, P_1, P_2, P_3$. Taken together, we may label the segment $\overline{P_0P_3} =: a_1$. Indeed, the direction is important, so $a$ is labeled (in the third picture) together with an arrow to indicate the specified orientation of $a_1$. Notice that the right side of the rectangle in the second picture has the same vertices $P_0, P_1, P_2, P_3$, taken from top to bottom now. Again, $\overline{P_0P_3} =: a$, but the orientation is opposite. We must think of the left and right sides of the rectangle as containing the same points, identified is such a way as to make the arrows go the same direction. A similar analysis identifies the top and bottom by $\overline{P_3P_4P_5P_0} =: a_2$, but with opposite orientations. Finally, $a_3 := \overline{P_0P_8P_{10}P_7P_0}$ is included in the third picture to make sure the result is a "Delta complex." (very much like a triangulation, except that one may identify faces in more flexible ways). Note, it is a standard result (see Hatcher, e.g.) that the Delta-complex defines the same homology as the original triangulation of a space. |
Evaluating the integral $\int_A y e^{-(x^2+y^2)/2}d(x, y)$ where $A$ is a certain set | $$\int_A y e^{-(x^2+y^2)/2}d(x, y) = \int_0^\infty e^{-x^2/2}\int_0^xye^{-y^2/2} dy\;dx$$
From there a simple substitution and evaluation of the inner integral reduces to a sum of known integrals. |
Other Useful Series Tests | Here are some references. The bibliographies in these will lead you to many more references. Based on your comments, Bonar/Khoury's book is probably the best fit for you right now.
Bonar/Khoury, Real Infinite Series (2006)
Bromwich, An Introduction to the Theory of Infinite Series (1908) [a 1925 2nd edition also exists]
Hirschman, Infinite Series (1962/2014)
Knopp, Theory and Application of Infinite Series (1954)
Knopp, Infinite Sequences and Series (1956)
Osgood, Introduction to Infinite Series (1897)
Rainville, Infinite Series (1967) |
Cartesian product of two odd and even numbers | $A\times B = \{(2n, 2m+1)\mid n\in \mathbb{N}, m\in \mathbb{N}\}$ is one way to do it. |
What are the possible values of the last digit of $4^m$, where m is a natural number? | You could use mathematical induction on $m$. That is, you show that if $m = 1$, then the last digit of $4^m$ is either $4$ or $6$. Then show that if $m$ is an integer and the last digit of $4^m$ is either $4$ or $6$, then the last digit of $4^{m+1}$ is either $4$ or $6$. |
continuity and differentiability of function of two variables | $$
f(x,y) = x - y^3\sin(.) = x + o(x,y)
$$
because $$\lim_{(x,y)\to (0,0)}\frac {y^3\sin(.)}{\sqrt{x^2 + y^2}} =0$$
For continuity:
$$\lim_{(0,0)} f(x,y) = \lim_{(0,0)}x + \lim_{(0,0)}o(x,y) =0
$$
then $f$ is continunous in $(0,0)$.
For differentiability:
Now identify with the definition of the derivative:
$$
f(x,y) = f(0,0) + Df(0,0)\cdot(x,y) + o(x,y)
$$
gives you the result:
$f$ is diffrentiable in in $(0,0)$ and $Df(0,0)\cdot(x,y) = x$. |
$N$ identical Balls to be placed in $P$ non-identical boxes with maximum $K$ in each box | Let's call the the number of ways $f(N,P,K)$. Then, thinking about the final box, you have $$f(N,P,K) = \sum_{m=1}^K f(N-m,P-1,K)$$ starting with $f(0,0,K)=1$ and $f(N,0,K)=0$ when $N \not = 0$. This is related to a generalised form of Pascal's triangle (slightly skewed)
So for example, if $K=3$, you would get a table which starts like this
P 0 1 2 3
N
0 1 0 0 0
1 0 1 0 0
2 0 1 1 0
3 0 1 2 1
4 0 0 3 3
5 0 0 2 6
6 0 0 1 7
7 0 0 0 6
8 0 0 0 3
9 0 0 0 1
and these are known as trinomial coefficients, with each value the sum of the three values immediately above in the previous column. With $K=2$ there would be binomial coefficients, and with $K=4$ there would be quadrinomial coefficients, and so on for larger $K$ |
Example 24.4 Tu's An Introduction to Manifolds | I assume by $H^k(S^1) = 0$ is actually meant $H^k(S^1) = \{0\}$, right?
Yes. It is a common abuse of notation. The first $0$ denotes the zero vector space and the second zero denotes the zero vector inside the zero vector space $\{0\}$.
Why does $\varphi(\omega)\neq 0$ implies such map is onto?
This is linear algebra. If a linear functional $f\colon V \to \Bbb K$ is non-zero, it is surjective. Take $\lambda \in \Bbb K$. Since there is $v \in V$ with $f(v) \neq 0$, we have that $f(\lambda v/f(v)) = \lambda$, so the range of $f$ is the whole $\Bbb K$.
How exactly is Stokes' theorem used to prove that exact $1$-forms are in $\ker \varphi$?
If $M^n$ is compact and without boundary, and $\beta \in \Omega^{n-1}(M)$, then $$\int_M {\rm d}\beta = \int_{\partial M} \beta = \int_\varnothing \beta = 0.$$
Where does $H^1(S^1) = \Bbb R$ comes from?
Integration is a surjective map $\int_{S^1}\colon Z^1(S^1) \to \Bbb R$ with kernel $B^1(S^1)$. This is what he proved. So the first isomorphism theorem says that $H^1(S^1) = Z^1(S^1)/B^1(S^1) \cong \Bbb R$. |
What do $\Bbb N^*$ and $\Bbb Z(p^n)$ mean in this paper? | primary component means the subgroup of all elements whose order is a power of $p$ and yes $\mathbb Z(p^n)$ is $\mathbb Z$/$p^n$$\mathbb Z$
$N^*$ should be non negative integer. |
Proof that no non-trivial integer solution exists? | One has $$(ab+bc+ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a+b+c) = a^2b^2 + b^2c^2 + c^2a^2 + 2(2ab+2bc+2ca)(\frac{1}{2}(ab+bc+ca))$$
Then $$a^2b^2 + b^2c^2 + c^2a^2 = -(ab+bc+ca)^2$$
So, one gets $ab=bc=ca=ab+bc+ca=0$,or $a=b=c=0$.
EDIT: i make it clear at the last step as @peter's comment:
One gets from $ab=bc=ca=0$ that $a=b=0$, but note that $c=a+b+c=\frac{1}{2}(ab+bc+ca)=0$. |
Couples sitting at a round table | EDIT: Oops, I misread this as circular permutations, never mind.
I think it would be easier to define your properties, or sets, however you like, to be the ones you don't want to count and then remove them. Thus, define your properties as:
$p_1=$ couple 1 sits together
$p_2=$ couple 2 sits together
$p_3=$ couple 3 sits together
Where, of course, your looking at the set of all possible circular permutations of 6 people, i.e. $5!$ What you want to find is the number of circular permutations with exactly none of those properties.
If one property holds, you pair a couple together, so you really are only ordering 5 people, so there are $4!$ circular permutations possible. However, there are also 2 possible configurations on how the couple sits, namely who is on the left, so you must also include a factor of 2.
Similarly, for some set X of properties, there are $(6-|X|-1)!*2^{|X|}$ possible ways to order the couples with at least those properties.
By PIE, there are $\sum_{i=0}^{3} \binom{3}{i} * (-1)^i * (6-i-1)! * 2^i $ ways to to seat these couples with none of the stated properties. Presumably, they'll have a slice of pie. |
How to calculate this sum: $\sum_{k=0}^{+\infty}\frac{\lambda^k}{k!}$? | To elaborate on the comment, The Taylor expansion for the function $e^x$ is given by $$e^x=\sum_{k=0}^{\infty} \frac {x^k}{k!}$$
Some authors use this as the definition of $e^x$. Alternatively, you can remark that the derivative of the power series is equal to the original series (taking care to verify absolute convergence for all $x$).
We see, of course, that your expression is then $\boxed {e^{\lambda}}$. |
20 identical balls, 4 distinct boxes | Since the balls are indistinguishable, and you need to put an even number of them into each box, you might as well glue them together in pairs. Now you just want to know the number of ways to distribute these glued pairs amongst the four boxes. If you let $x,y,z$, and $w$ be the numbers of glued pairs in the four boxes, then clearly you must have $x+y+z+w=10$, and each of $x,y,z$, and $w$ must be a non-negative integer. Conversely, if you have four non-negative integers $x,y,z$, and $w$ whose sum is $10$, they give you a possible distribution of the glued pairs: $x$ in the first box, $y$ in the second box, and so on. Thus, the number of ways to distribute the glued pairs is the same as the number of solutions of $x+y+z+w=10$ in non-negative integers. This is a standard stars-and-bars problem, with (as you found) the solution
$$\binom{10+4-1}{4-1}=\binom{13}3=286\;.$$
The reasoning behind the this result is described quite well in the linked article, but I’ll repeat here the special case that you need.
Think of laying out the $10$ glued pairs in a row on the table:
$$\begin{array}{c}\infty&\infty&\infty&\infty&\infty&\infty&\infty&\infty&\infty&\infty\end{array}$$
They’re completely indistinguishable, so it doesn’t matter which one is which. Now insert three dividers to mark the breaks between the pairs in the first and second boxes, those in the second and third boxes, and those in the third and fourth boxes. For instance, if there are $3$ pairs in the first box, none in the second, $5$ in the third, and $2$ in the fourth, you get this arrangement:
$$\begin{array}{c}\infty&\infty&\infty&|&|&\infty&\infty&\infty&\infty&\infty&|&\infty&\infty\end{array}$$
In any such arrangement of pairs and dividers you have a string of $10+3=13$ objects. There are $\binom{13}3$ ways to pick the $3$ positions for the dividers, and each choice of positions for the dividers corresponds to exactly one of the possible distributions of the glued pairs. Conversely, each distribution can be represented by one of these arrangements of $10$ pairs and $3$ dividers. Thus, there are $\binom{13}3$ distributions of the glued pairs. And as we saw at the beginning, each of them corresponds to exactly one of the possible distributions of balls having an even number of balls in each box. |
Eigenvalues of $A=\mathbf{u}_1\mathbf{u}_1^T + \mathbf{u}_2\mathbf{u}_2^T$ | What is $A\textbf{u}_1$? What is $A\textbf{u}_2$? If you have a vector $\textbf{v}$ that is orthogonal to both $\textbf{u}_1$ and $\textbf{u}_2$, what is $A\textbf{v}$? |
Mandelbrot fractal: How is it possible? | Let $m(z,c) = z^2 + c$, consider the sequence of polynomials: $m(z,z),\,$ $m(m(z,z),z)$, ... , which are $$z^2+z,\quad z^4 + 2z^3 + z^2 + z,\quad z^8 + 4z^7 + 6z^6 + 6z^5 + 5z^4 + 2z^3 + z^2 + z,\quad...\quad.$$
Note that in terms of complex numbers the transformation $z \mapsto m(z,c)$ can be seen a way of twisting and squashing the sphere over itself.. if you keep kneading something you are bound to get tearing and crumpled filaments and such like.
Here are graphs of the first seven (produced by the software here): |
Using the rounding function to compute $x=0.21045$ should the output be $0.210$ or $0.211$ within a certain set of floating points? | The definition says you just look at the first digit past what you want to round to. If it is $5$ or greater you increase the last digit you want to keep by $1$. If it is $0$ to $4$ you leave the last digit the same. In your example of $21037$ being rounded to three places, the first digit to remove is $3$, so you round down. You do not even look at the $7$. Similarly, in $21045$ the first digit you want to remove is $4$, which is less than $5$. You don't even look at the $5$. You also round this to $210$. This makes sense because you are always picking the closer value unless the part to remove is precisely $0.50000\ldots$. In that case there is no closer value. Either way you round is the same distance away.
When we round fixed point numbers we worry about an upward bias if we round all the exactly $0.5$ values up to $1$. Often we round exactly $0.5$ to the even number. For real numbers getting exactly $0.5$ is not supposed to happen or not enough to matter. |
Problem with integration limits using spherical substitution | Transforming to spherical coordinates, we find that
$$\begin{align}
\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{-\sqrt{1-x^{2}-y^{2}}}^{\sqrt{1-x^{2}-y^{2}}}\left(x^{2}+y^{2}+z^{2}\right)^{3/2}dzdydx
&=\int_0^{2\pi}\int_0^\pi\int_0^1 r^5\,dr\sin(\theta)\,d\theta\,d\phi\\\\&=(2\pi)(2)\left(\frac16\right)\\\\
&=\frac{2\pi}{3}\end{align}$$
as expected! |
How do you solve $\lim\limits_{n \to \infty} n( \sqrt[n] {a_1}+\sqrt[n] {a_2}+...\sqrt[n] {a_p}-p) $ | By L-Hospital's rule $$L=\lim_{x \rightarrow 0}\frac{a^x-1}{x}=\ln a$$
So firther from your step, you get the limit as $$\ln a_1 + \ln a_2+...+\ln a_p =\ln (a_1 a_2 a_3...a_p)$$ |
Prove that $f$ has finite number of roots | Combining the comments:
Assume by contradiction that $f$ has infinitely many zeroes in $[0,1]$. Then, we can find a sequence $x_n$ so that $f(x_n)=0$.
As $[0,1]$ is compact, $x_n$ has a converging subsequence. Lets call this subsequence $y_n$ and let $y=\lim_n y_n$.
Then by continuity $f(y)=0$ and $f'(y) =\lim_{x \to y} \frac{f(x)-f(y)}{x-y}=\lim_{n \to \infty} \frac{f(y_n)-f(y)}{y_n-y}=0$ contradiction. |
Two properties on $f^{-1}(A)$ | On $X=\{1,2\}$ let $f(1)=f(2)=0$ viewed as a function into $Y=\mathbb R$. Then $f^{-1}(\{0,1\})=f^{-1}(\{0,2\})$ but $\{0,1\} \neq \{0,2\}$ |
The sum of consecutive strings of positive integers of length $2*n$ | I would start by trying to find a formula for the last digits of the $n$th series. The last digits are $1, 1+2*1, 1+2*1+2*2, ...$ which you can see is equal to
$$1+\sum_{k=1}^n 2k= 1+2\sum_{k=1}^n k=1+n(n+1)$$
(look up triangular numbers if you don’t understand the last step).
$a(n)$ is equal to the sum of the integers less than or equal to the $n$th last digit, minus the sum of the integers less than or equal to the $n-1$st last digit. In other words, the $1+n(n+1)$st triangular number minus the $1+n(n-1)$st triangular number, which is equal to
$$(1+n(n+1))(2+n(n+1))/2-(1+n(n-1))(2+n(n-1))/2$$
and simplifies to the formula you found. |
Taylor Series of an arctangent function | Notice that after simplifications,
$$\left( \arctan \frac {2x} {1-x^2} \right) ' = \frac 1 {1 + \frac {4x^2} {(1-x^2)^2}} \frac {2(1-x^2) - 2x(-2x)} {(1-x^2)^2} = \frac 2 {1+x^2} = (2 \arctan x)' ,$$
whence it follows that
$$\arctan \frac {2x} {1-x^2} - 2 \arctan x = C$$
with $C \in \Bbb R$ a constant. To find out $C$ just evaluate the equality at $x=0$, to get
$$\arctan 0 - 2 \arctan 0 = C ,$$
whence $C=0$ and thus
$$\arctan \frac {2x} {1-x^2} = 2 \arctan x$$
and the required series is obvious now if you now the series of $\arctan$.
Alternatively, use the formula
$$\tan (u+v) = \frac {\tan u + \tan v} {1 - \tan u \tan v}$$
with $u = v$ to get
$$\tan (2u) = \frac {2 \tan u} {1 - \tan^2 u} ,$$
whence
$$2u = \arctan \frac {2 \tan u} {1 - \tan^2 u}$$
whence, denoting $x = \tan u$, we get
$$2 \arctan x = \arctan \frac {2x} {1-x^2} .$$
These trigonometric computations are a bit formal, since I haven't payed attention to the domain of definition of the above functions, but they can be made perfectly rigorous with just a bit of supplementary work. |
Nonstandard independent solutions of the Airy equation | $C_2$ is the easier of the two, and is where the Fundamental Theorem of Calculus comes in. Suppose $t = Re^{5i\pi/6}$. What's $\lim_{R\rightarrow\infty}e^{i(t^3/3+zt)}$? What about when $t = Re^{3i\pi/2}$?
For $C_1$, you should look at the wedge-shaped contours $\{se^{5i\pi/6}|0\le s\le R\} \cup \{Re^{i\theta}|5\pi/6\le\theta\le\pi\}\cup [-R, 0]$ and $[0, R]\cup \{Re^{i\theta}|0\le\theta\le\pi/6\} \cup \{se^{i\pi/6}|R\ge s\ge 0\}$. The radial contours along $5\pi/6$ and $\pi/6$ can still be handled by FTC. For the circle contours, show that the integrand decays quickly with $R$, and thus the integral over the whole arc vanishes as $R\rightarrow \infty$. |
order of an integral on the real line | Integration by parts after substituting $x\mapsto x^{1/2}$
$$
\begin{align}
\int_{n^{1/4}}^\infty x^{d-1}e^{-x^2}\,\mathrm{d}x
&=\frac12\int_{n^{1/2}}^\infty x^{d/2-1}e^{-x}\,\mathrm{d}x\\
&=\frac12\left(n^{d/4-1/2}e^{-n^{1/2}}-\left(\frac d2-1\right)\int_{n^{1/2}}^\infty x^{d/2-2}e^{-x}\,\mathrm{d}x\right)\\
&=\bbox[5px,border:2px solid #C0A000]{\frac12n^{d/4-1/2}e^{-n^{1/2}}\left(1+O\left(n^{-1/2}\right)\right)}
\end{align}
$$
since
$$
\int_{n^{1/2}}^\infty x^{d/2-2}e^{-x}\,\mathrm{d}x
\le n^{-1/2}\int_{n^{1/2}}^\infty x^{d/2-1}e^{-x}\,\mathrm{d}x
$$
Furthermore
$$
\frac12n^{d/4-1/2}e^{-n^{1/2}}=O\left(n^{d/4-1/4}e^{-n^{1/2}}\right)
$$ |
Classification of $u_{xx} + 2u_{xy} = 0$. | What you did is correct but the final result is missing.
$$u(x,y)=f(y)+g(y-2x)$$
with arbitrary functions $f$ and $g$.
Another method to find the above result :
$$u_{xx}+2u_{xy}=\frac{\partial}{\partial x}(u_x+2u_y)=0$$
Integrate :
$$u_x+2u_y=\phi(y)\quad\text{any function}\quad \phi$$
First order linear PDE through method of characteristics :
$$u(x,y)=\int \frac12 \phi(y)dy+g(y-2x)$$
Let $\frac12\int \phi(y)dy=f(y)$
$$u(x,y)=f(y)+g(y-2x)$$ |
5 fair coin flips probability | For two heads you missed $HTTH$. With the starting tail, that is the $\frac 1{32}$ that is missing. |
A problem with an infinite multitude of numbers that follow some rules | So $n=a^2+b^2+c^2$ with $a\mid n$, $b\mid n$, $c\mid n$.
If any of $a,b,c$ is a multiple of $3$, then so clearly is $n$.
In all other cases, $a^2\equiv b^2\equiv c^2\equiv 1\pmod 3$ so that $n\equiv 1+1+1\equiv 0\pmod 3$.
Start with any three distinct naturals $a_0,b_0,c_0$ and let $n_0=a_0^2+b_0^2+c_0^2$.
In general, $a_0\nmid n_0$. Let $\frac{n_0}{a_0}=\frac uv$ in shortest terms and assume that the denominator $v$ is $>1$. Let $a_1=va_0$, $b_1=vb_0$, $c_1=vc_0$, $n_1=v^2n_0$. Then $a_1,b_1,c_1$ are three distinct naturals and $n_1=a_1^2+b_1^2+c_1^2$, but now $a_1\mid b_1$. Also note that the denominator of $\frac{n_1}{b_1}$ is not bigger than that of $\frac{n_0}{b_0}$ etc. Thus we can repeat the above for $b$ and for $c$, arriving ultimately at a special number.
More concretely, $126=3^2+6^2+9^2$ is special and so is $126k^2=(3k)^2+(6k)^2+(9k)^2$ for all $k$. |
To control first derivative with the function itself: $f'(x)^2\leq Cf(x)$ near where $f(x_0)=f'(x_0)=f''(x_0)=0$. | For such $f$ there holds an estimate $f'(x)^2\le 2Mf(x)$ for all $x\in \mathbb R$, where $M=\max|f''|$. See lemma 5.6 in Yu. V. Egorov, Linear Differential Equations of Principal Type.
The idea is to consider $f$ on an interval $(x_1,x_2)$ where $x_1$ and $x_2$ are consecutive zeros of $f$.
If the maximum of $f'^2/f$ on $[x_1,x_2]$ is attained at some $c\in(x_1,x_2)$, then
$$2f''(c)f(c)-f'^2(c)=0$$ and $f'^2(x)\le 2f(x)|f''(c)|\,$ on $[x_1,x_2]$.
For endpoints L'Hopital's rule is used:
$$
\lim_{x\to x_1}\frac{f'(x)^2}{f(x)}=\lim_{x\to x_1}\frac{2f''(x)f'(x)}{f'(x)}=2f''(x_1).
$$ |
simplify expectation definition Hidden Markov Model | For a hint write
\begin{eqnarray*}
\sum_{d = 1}^{\infty} a^d & = & \frac{1}{1 - a}
\end{eqnarray*}
Take derivatives on both sides with respect to $a$ (I will not say rigorously why it is allowed
here)
\begin{eqnarray*}
\sum_{d = 1}^{\infty} da^{d - 1} & = & \frac{1}{\left( 1 - a \right)^2}
\end{eqnarray*}
Multiply by $\left( 1 - a \right)$ on both sides to conclude $$E[d]=\frac{1}{1-a}$$ |
Basic limit/upper bound Analysis question. | Let $\lim_{x \to a} f(x) = c$. We wish to show that $c \leq L$.
Without loss of generality assume that $c, L > 0$. By definition, $\forall \epsilon > 0, \exists \delta > 0\mid |x - a| < \delta \implies |f(x) - c| < \epsilon$.
Then,
$$ c \leq |c - f(x)| + |f(x)| < \epsilon + L
$$
Since that holds for all $\epsilon > 0$, we have $c \leq L$. |
Prove $m^3 \leq 2^m$ for $m \geq 10$ with induction | Your second equivalence is wrong. It has to be:
$$k^3 + 3k^2 + 3k + 1 \leq 2^k + 2^k \impliedby k^3 \leq 2^k \land 3k^2 + 3k + 1 \leq 2^k$$
Now $k^3 \leq 2^k$ by the induction hypothesis.
For the last inequality, it is much easier to prove $3k^2+3k+1\leq k^3$, because $k^3 \leq 2^k$ by the induction hypothesis.
Now you just have to prove $-k^3+3k^2+3k+1\leq 0$ (for $k \geq 10$) which can be done by simple function analysis. |
Given 2 analytic functions $ f(x) $ and $ g(x) $ | True because of the identity theorem (obviously, the set $\Bbb R$ has a cluster point in $\Bbb C$). |
On the Fourier transform of $f(x)=\ln(x^2+a^2)$ | If $\left| x \right|^{-1}$ is the distribution defined as
$$\left( \left| x \right|^{-1}, \phi \right) =
\int_{\left| x \right| < 1} \frac {\phi(x) - \phi(0)} {\left| x \right|} dx +
\int_{\left| x \right| > 1} \frac {\phi(x)} {\left| x \right|} dx,$$
then
$${\mathcal F}\!\left[ \ln\left( x^2 +a^2 \right) \right] =
\frac 1 {\sqrt{2 \pi}}
\int_{-\infty}^{\infty} \ln\left( x^2 +a^2 \right) e^{-i k x} dx =
-\sqrt{2 \pi} \left( \frac { e^{-a \left| k \right|}} {\left| k \right|} +
2 \gamma \delta(k) \right),$$
where $\gamma$ is Euler's constant. Probably the easiest way to prove it is to compute the inverse transform directly from the definition of $\left| x \right|^{-1}$. |
Probability of winning after x coin tossses | This is a binomial distribution. The probability of getting 7 heads and five tails is as follows: $\binom{12}{7} (p_{head})^{7} * (1 - p_{head})^{5}$. So you choose your heads, then multiply by the probability of exactly $7$ heads out of $12$. If this is a fair coin, $p_{head} = 1 - p_{head} = \frac{1}{2}$. |
Find the locus represented by | As you find out we have $$|PA| + |PB| = |AB|$$
so by triangle (in)equality we see that $P$ is on the segment $AB$. So $y=0$ and $|x|\leq 2$. |
How do I write minimal negations for the following statements? | Your reasoning is not quite correct. As it happens all of your answers are unsatisfactory in some way. Let's go through each:
A) There is somebody who was kung-fu fighting for not 10 hours
This is almost correct. If “$\neg$(Everybody is doing X)” then there has to be somebody who is not doing X. However, by saying “not 10 hours” you are mistaken: the correct negation is “less than 10 hours”.
B) There is not a river with at least two tributaries
Whilst this is a technically correct negation, it has not been simplified at all. You're likely expected to produce a sentence of the form “All rivers (...)”. I'll leave you to work this one out yourself; you can probably do so, maybe after reading the tips later on in this answer.
C) All baboons wear bowler hats
This is not a correct negation. To see why, try to negate the false statement “No cats are ginger”. By your logic, the negation, a true statement, should be “All cats are ginger”—a clearly false statement! The correct negation is “There is a baboon that wears a bowler hat”.
The general rules for solving this sort of problem can be written as so:
$\neg(\text{All $X$s have property $Y$}) \Leftrightarrow (\text{There is an $X$ without property $Y$})$.
$\neg(\text{There is an $X$ with property $Y$}) \Leftrightarrow (\text{All $X$s do not have property $Y$})$
Clearly, a) falls under rule 1, whereas b) falls under rule 2. It may be hard to see, but c) actually also falls under rule 1—can you see why? This should help you understand this kind of statement better.
If you'd like more justification, I recommend reading up on logical quantifiers. If these problems are part of a book or course on logic, you will likely encounter these very soon. Logical quantifiers allow rule 1, for instance, to be written as $\neg(\forall x, P(x)) \Leftrightarrow \exists x, \neg P(x)$. |
How to do $arcsin(1/2)$ by hand? | Consider an equilateral triangle if side length $2$. An altitude cuts one of the $\pi/3$ angles into two copies of $\pi/6$, in hypotenuse-$2$ right-angled triangles with opposite $1$. Thus $\pi/6=\arcsin 1/2$. |
How is the sample space of a random variable defined? | The formal definition of a (real) random variable is a function $X:\Omega\to\mathbb R$, where $\Omega$ is the sample space. The sample space is, in informal terms, where the "randomness" of probability comes from.
For example, say you flip a fair coin $5$ times. Then the sample space $\Omega$ is the set of your possible outcomes, e.g. $HHHHH$, $THTTH$ etc.
A random variable $X$ in this space might e.g. be the number of heads you toss. So for example, $X(HHHHH)=5$.
It makes no sense to add two random variables with different sample spaces in the same way it makes no sense to add two functions with different domains. This is because just as with functions, we define $X+Y$ by $$(X+Y)(\omega)=X(\omega)+Y(\omega),$$ for $\omega\in\Omega$. Evidently we require $X$ and $Y$ to have the same sample space.
As an example, consider $X$ and $\Omega$ as defined by the coin tosses, and consider a new random variable $X':\Omega'\to\mathbb R$. The sample space $\Omega'$ is the set of points on Earth, and $X'$ is the current temperature at a point on the Earth. What is $X+X'$?
The situtation can be salvaged though. Suppose $X$, $Y$ are real random variables with sample spaces $\Omega_X$, $\Omega_Y$. Define a random variable $Z:\Omega_X\times\Omega_Y\to\mathbb R$ by $$Z(\omega_X,\omega_Y)=X(\omega_X)+Y(\omega_Y).$$
Then the projections $\bar{X}:(\omega_X,\omega_Y)\mapsto X(\omega_X)$ and $\bar{Y}:(\omega_X,\omega_Y)\mapsto Y(\omega_Y)$ have the same distributions as $X$ and $Y$ respectively, and clearly $\bar{X}+\bar{Y}=Z$ is defined. In fact, $\bar{X}$ and $\bar{Y}$ are even independent. |
Why choose a prime number as the number of slots for hashing function that uses divison method? | If you know nothing about the distribution of the key $k$, then there is absolutely no argument one could formulate for (or against) taking $m$ prime. There is much of a "feel-good" element in using primes: many authors will promote the practice (often without giving the "technical details" that would explain it) and Hey! even Knuth mentions prime numbers in the context of hashing (forget that he does so in a very particular setting that does not match yours), nobody ever gives an argument against using prime numbers, and there are plenty of them around, so why not choose a prime number? I'll repair one of the points I mentioned: choosing $m=2$ is almost always a very poor choice, even though $2$ is a perfectly bona fide prime number; it just is too small.
If your hashing by just reducing modulo $m$ (which is applicable only if your keys are integers, which severely limits the application contexts), then a hash modulus $m$ is relatively bad (with respect to its size) if somehow your keys are very unevenly distributed modulo $m$. Without knowing anything about the origin of the keys there is no way to know whether this is the case; if the keys are uniformly distributed in some interval much larger than $m$ it will never be the case. However if one surmises that keys are have a non-uniform distribution modulo certain unknown "cursed" numbers, then choosing some $m$ amounts to a gamble that this $m$ happens to not be one of those. It is always possible to lose out on this gamble. One can argue that if $m$ is cursed then so will any multiple of $m$, so choosing $m$ prime in some sense limits the risk to one gamble, while choosing a composite number exposes to the risk of a curse on any of its divisors as well. But you can't make a good argument out of this if one does not substantiate the fear that there are any cursed numbers in the first place. And if a set of (prime) numbers are known to be non-cursed, then taking their product is a safer bet then taking an unknown prime number of about the same magnitude.
But all this is mostly beside the point, because if you don't know the origin of your keys, then hashing by modular reduction is just not a good idea. In practice one should first "scramble" the data in the keys (this is what "hashing" means, actually), and then perform a reduction to map to the size of the table. I've written a general-purpose hashing class whose self-adjusting table size is always a power of $2$, which is an advantage in modular reduction and resizing (no need to hunt for an appropriate new size, just double). Used with a bit of attention to writing data-specific hash functions (if you know some lower order bits are likely to be zero, just don't use them) that ensure a uniform distribution, this works quite well in very large practical computations. |
Does the norm of the product give information about the norm of the matrices? | Suppose that $A$ is a singular matrix and $B=null(A)$. Then, $AB=0$, even if $\|A\|>0$ and $\|B\|>0$.
Basically, nothing can be said about the product of norms, because it completely depends on the volume of the subspace shared by $A$ and $B$. |
p value : Confusion in basics | Now for a more serious answer.
You are correct that you take $2p$ where $p$ is the value from a table of normal distributions. This value represents the area under the curve of one tail of the normal distribution. With a double sided test, we need the area under the curve of both tails and so multiply by 2. |
Treewidth of complete bipartite graph using chordal graph characterisation? | Let $1\le m\le n$ and the bipartition of the graph $G=K_{m,n}$ consists of $m$ red vertices and $n$ blue vertices.
Let $H$ be the graph with added edges between all pairs of red vertices. The graph $H$ is chordal because it has the following perfect elimination scheme (see, for instance, p.99 in the slides of a lecture “Colorings, Cliques and Independent Sets” by Joachim Spoerhase and Alexander Wolff). First we remove all blue vertices in any order and next all red vertices in any order. Since any maximal clique of $H$ consists of all red vertices and one blue vertex, $\omega(H)=m+1$.
On the other hand, let $H’$ be a chordal graph containing $G$. I claim that either a subgraph $K’_m$ of $G$ induced on all red vertices of $H’$ is complete, or a subgraph $K’_n$ of $G$ induced on all blue vertices of $H’$ is complete. Indeed, if this doesn’t hold then there exist two non-adjacent red vertices $v_1$ and $v_2$ and two non-adjacent blue vertices $u_1$ and $u_2$. But then $v_1-u_1-v_2-u_2-v_1$ is a chordless cycle of $H’$ of length $4$, a contradiction. If the graph $K’_m$ is complete then $H$ contains a graph $H$ from the previous paragraph. So $\omega(H)\ge \omega(H’)=m+1$. If the graph $K’_n$ then we similarly show that $\omega(H)\ge n+1$. Anyway, $\omega(H)\ge m+1$. |
Simplify a rational identity | Multiply with $\frac{ab}{ab}$ as first step:
$$\frac{\frac ab-\frac ba}{1+\frac ba}=\frac{a^2-b^2}{ab+b^2}=\frac{(a+b)(a-b)}{(a+b)b}=\frac{a-b}b=\frac ab-1 $$
(provided $a+b\ne0$, but in that case the original fraction would not be defined) |
Prove $\sum_{k=0}^{n}\binom{n}{k} = 2^{n}$ combinatorially | A combinatorial argument is one that establishes an equality by counting the same things two different ways. You have done one side already-as you say, $2^n$ is the number of subsets of a set of size $n$. Hint: think about categorizing the subsets by their size-does that help you with the other side of the equality? |
If then, sufficient for, necessary condition | They are the same because their meaning in English is the same. Each says, in a slightly different way, that $q$ must be true when $p$ is. In a math paper or text an author can use any of them, choosing which sounds better in a particular context. |
Properties of Markov chains | A) You did not consider the case say (12) and (34) are in two seperate communicating class such that
$p_{12}=p_{21}=p_{34}=p_{43}=1/2$, then there is no unique stationary distribution. If there is an only 1 closed communicating class, $A$, and all other states are transient, what you said is true.
If we want to talk about one positive recurrent class instead of one irreducible chain, does this method of transferring the results ( by restricting the result to the subset A) always work?
Yes.
B) I do not really understand you are trying to pull here? Take an example of a chain with two communicating classes. A possible solution to $\pi = \pi^T P$ is setting 0 for all states in one class and find the unique distribution for the other class.
Then the set of equilibrium measures are linear combinations of these vectors. You cannot use $\pi = \pi^T P$ unless you know that there is only $1$ closed communicating class (by same reason as question 1)
A finite state Markov Chain has a unique stationary distribution iff it contains only one closed communicating class. (so what you said is true, if you assume there is just 1 closed communicating class, everything else is open)
C) Not as far as I know. To see a state is periodic, find two path of different length on which it returns itself such that the highest greatest common factor. This is normally quite easy for a simple Markov Chain.
D) consider something like
1 -> 2 <-> 3
then 2 and 3 are closed. 1 is open. You know that open and closed are class properties, so are transient and recurrence. In a finite chain, recurrence and closed are the same thing.
The only difference occurs in infinite state Chains.
E) For a uncountable states, you can just integrate $k$ times
For example
$$P(X_2\in A|X_0=x) = \int_{y\in\Omega}\int_{z\in A} q(x,y)q(y,z)dydz$$
where $\Omega$ is the entire state space. For $X_3$, you just need to integrate 3 times etc. Notice that summation (used for countable state) is just integral with respect to counting measure.
F) you stated the sufficient and necessary condition for recurrence. There is a standard textbook proof, and yes verifying this for 1 pair of $i, j$ is enough because recurrence and transience are class properties. In fact often it is verified for $i=j$. |
Derivative of magnitude of position vector | You can make the following substitution
$$\vec r\cdot\vec r=u,$$
and perform differentiation to get the required result.
$$\frac{dr}{dt}= \frac{d}{dt} \sqrt u $$
$$=\frac{1}{2\sqrt u}\frac{du}{dt}$$
$$=\frac{1}{2r}\frac{d}{dt}(\vec r\cdot\vec r)$$
$$=\frac{1}{2r}(2\vec r\cdot\frac{d\vec r}{dt})$$
$$=\frac{\vec r}{r}\cdot\frac{d\vec r}{dt}$$ |
Show $\mathbb{Q}( \sqrt{5},\sqrt{7} ) = \mathbb{Q}( \sqrt{5} + \sqrt{7} )$ | $$\sqrt7-\sqrt{5}=\frac{2}{\sqrt7+\sqrt5}\in\mathbb Q(\sqrt7+\sqrt5).$$
Thus, $$\sqrt7=\frac{\sqrt7+\sqrt5+\sqrt7-\sqrt5}{2}\in\mathbb Q(\sqrt7+\sqrt5)$$ and $$\sqrt5\in\mathbb Q(\sqrt7+\sqrt5).$$
Also, it's obvious that $$\mathbb Q(\sqrt7+\sqrt5)\subset\mathbb Q(\sqrt7,\sqrt5)$$ |
primitive root mod25 | Your understanding is correct, though there are faster ways to prove that.
$2$ is a primitive root $\bmod 25$ iff $20$ is the smaller positive exponent $n$ such that $2^n \equiv 1 \bmod 25$.
Since $a^{20} \equiv 1 \bmod 25$ for all $a$ coprime with $25$, you only need to prove that $2^{10} \equiv -1 \bmod 25$, which is immediate, since $2^{10}=1024$. |
Combinatorial Proof for Composite/Nested Binomial Coefficient | Both sides count the number of pairs of edges in the complete graph $K_n$. The LHS is clear. For the RHS, consider whether the edges share a common vertex (the second term) or not (the first term). |
Let $R$ be a ring and $X$ a non-empty set. Considering $A=R^X$, describe the ring $(A,+,\cdot)$ | 1) The zero divisors of $R^X$ are those functions which either vanishes somewhere or admits a zero divisor as value.
2) $D$ is not stable under addition. Take for example
$$f(x)=\begin{cases} 1 \; \text{if } x\in A \\
0 \; \text{otherwise} \end{cases} \qquad \text{and} \qquad g(x)=\begin{cases} 1 \; \text{if } x\in \bar{A} \\
0 \; \text{otherwise} \end{cases},$$
for some $\emptyset \subsetneq A \subsetneq X.$
Then $f,g\in D$ while $f+g\notin D.$
3) A function in $f\in R^X$ is uniquely determined by its values on $\{0,1\}$. So if $f(0)=x$ and $f(1)=y$ you can define $\phi : R^X \longrightarrow R \times R, \, \phi(f)=(x,y) .$ This is a ring homomorphism from the definition of $R^X$, it has inverse
$\psi : R \times R \longrightarrow R^X, (x,y)\mapsto (f: f(0)=x, f(1)=y)$. |
Linear Algebra 2 practice Exams | You might find some things on a website of mine for a past course, which also used Axler: http://math.berkeley.edu/~scanez/courses/math110/ |
Expected number of lines in use in call centre (markov process: queuing theory) | Let $X_t$ be the number of busy servers in a $M/M/\infty$ queuing system. You want to compute $E(X_t|X_0=n)$. This can be done by considering the sum
$$X_t=Y_t+Z_t,$$
where $\{Y_t|X_0=n\}\sim Binomial(n,e^{-\mu t})$ is the number of servers that did not complete service in the interval $(0,t]$ (out of the $n$ that were busy), and $Z_t\sim Poisson\left(\frac{\lambda}{\mu}(1-e^{-\mu t})\right)$ is the number of servers that became busy during the interval $(0,t]$ and have not completed service.
The distribution of $Y_t|X_0=n$ should be straightforward but what I claimed for $Z_t$ requires proof. Suppose they are both true then the expected value you seek is:
$$E(X_t|X_0=n)=E(Y_t|X_0=n)+EZ_t=ne^{-\mu t}+\frac{\lambda}{\mu}(1-e^{-\mu t}).$$
We are left with proving the distribution of $Z_t$. You should be familiar with the basic properties of the Poisson process (uniform property of the arrival times), and with the probability generating function of Bernoulli, Binomial and Poisson random variables. If you are not, I suggest looking those up.
Let $N_t\sim Poisson(\lambda t)$ be the number of arrivals to the system during the interval. For $i=1,\ldots,N_t$, let $S_i$ denote a Bernoulli random variable specifying whether arrival $i$ is still in the system at time $t$. Note that $Z_t=\sum_{i=0}^{N_t}S_i$ where $S_0:=0$. We use the fact that given the total number of arrivals during $(0,t]$, the actual arrival times are independent and uniformly distributed on the interval:
$$P(S_i=1|N_t=k)=\int_0^t \frac{1}{t}e^{-\mu(t-u)}du=\frac{1}{\mu t}(1-e^{-\mu t}) .$$
Using the law of iterated expectation we compute the probability generating function of $Z_t$ for some $0<z<1$:
$$
\begin{eqnarray}
E(z^{Z_t}) &=& EE(z^{Z_t}|N_t)=EE(z^{\sum_{i=0}^{N_t}S_i}|N_t) \\
&=& E(1-P(S_1=1)+P(S_1=1)z)^{N_t} \\
&=& E(1+\frac{1}{\mu t}(1-e^{-\mu t})(z-1))^{N_t} \\
&=& e^{\frac{\lambda t}{\mu t}(1-e^{-\mu t})(z-1)}.
\end{eqnarray} $$
We can conclude that $Z_t\sim Poisson\left(\frac{\lambda}{\mu}(1-e^{-\mu t})\right)$, as required. |
Calculate covariant of X and Y whose density function is given | As far as $E[X]$:
$$E[X]=\int_0^{\infty}xf_X(x)\ dx=\int_0^{\infty}x\int_0^{\infty}\frac1y e^{-y-\frac yx}\ dy \ dx=$$
$$=\int_0^{\infty}\frac1ye^{-y}\int_0^{\infty}xe^{-\frac xy}\ dx\ dy.$$
Integrating by part in the case of the internal integral we get
$$\int_0^{\infty}xe^{-\frac xy} \ dx=y^2.$$
So, integrating by part again
$$E[X]=\int_0^{\infty}ye^{-y}\ dy=1.$$ |
What does an inverse matrix abstracts? | A $2\times 2$ matrix corresponds to a map of the plane to itself. Not surprisingly, the inverse matrix of that matrix corresponds to the inverse map.
For instance, the matrix
$$
\begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \phantom-\cos \theta \\
\end{bmatrix}
$$
corresponds to a rotation of angle $\theta$ around the origin. The inverse matrix is
$$
\begin{bmatrix}
\phantom-\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta \\
\end{bmatrix}
$$
and corresponds to a rotation of angle $-\theta$ around the origin. |
Differentiation and multiplication of polynomials can't be both continuous | $MD(x^n)=nx^n$ for all $n$, therefore $$\sup_{p\in \Bbb R[x]\setminus \{0\}}\frac{\lVert MD(p)\rVert}{\lVert p\rVert}\ge \sup_{n\in\Bbb N} n=\infty$$
Since $M\circ D$ is not continuous, at least one of the two functions has not to be as well. |
Problems with this reasoning (Gambling) | This is certainly not formal, but maybe it will help you understand the system flaws a bit better.
Consider $X$ to be the waiting-time random variable representing the time until you finally win a bet and profit.
If (as mentioned in the comments) you have an infinite supply of money to bet with, then so long as $X$ is a proper random variable, i.e.
$$P(X < \infty) = 1$$ you will eventually profit with a net gain of 1 unit. With an infinite supply of money, no casino limits, an unlimited amount of time to continue gambling, etc. one could theoretically win as much money as they desired with this strategy given any game where $X$ is a proper random variable, simply by adjusting their first bet (representing their unit).
Realistically however, it doesn't work out so nicely. In order for an individual to assure themselves a profit, they would need $X$ such that
$$P(X<k) = 1$$
where $k$ represents the limiting number of gambles before they either
run out of money to gamble with
run out of time in which to gamble
...any other constraint on how long they can gamble that may apply
You can begin to now see the flaws with this system. Finite supplies of money and time as well as other constraints render $P(X<k)<1$, and expose the gambler to the risk of losing many more units of money than the 1 that they are attempting to win with this system.
Even if an individual has enough money and time such that they can assure themselves a profit, their first bet (i.e. unit they are attempting to gain) is more than likely so small that by the expected time at which they win their returns will be minimal, either as a result of how small they were forced to make their first bet, or how long it took them to finally win.
So in short, I presume unless your friend is incredibly rich, or attempting to win some small, small amount of money, or some combination of the two, the odds are still against him. |
$\pi$ and $e$ as irrational linear combinations | As finite sums, no. If you could, you'll get that $\pi$ and $e$ are algebraic numbers, which is known to be false.
As infinite sums:
$$e = \sum_{n=0}^{\infty} \frac1{n!}$$
$$\pi = 4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$$ |
How many different combinations animals can be crossbred? | Put all the heads on a table and all the bottoms on another table. Now choose the first head. You can pair with any of the bottoms from the second table and that gives $10$ combinations. You now have made all the possible different combinations with that head, so you may as well remove it from the table.
Repeat the process with the second head. You can pair it with any of the $10$ bottoms, and that gives again $10$ different combinations. Again, there are no other combinations that can be formed with this head, so you can remove it from the table.
You can see that the argument is the same for all the heads, and that gives
$$10 \times 10 = 100$$
different combinations. Note that $10$ of these combinations are the original animals. You may or may not want to exclude them. |
Why is the first $p$-adic congruence subgroup a pro-$p$ group? | For any whole number $n$ there is a surjection $\mathrm{GL}_2(\mathbb{Z}/p^n\mathbb{Z})\to\mathrm{GL}_2(\mathbb{Z}/p\mathbb{Z})$, Let $K_n$ be the kernel of this map. Then we have an infinite commutative diagram
$$\require{AMScd}
\begin{CD}
& & \vdots \\
& K_3 @>>> \mathrm{GL}_2(\mathbb{Z}/p^3\mathbb{Z}) \\
@VVV @VVV \\
& K_2 @>>> \mathrm{GL}_2(\mathbb{Z}/p^2\mathbb{Z}) \\
@VVV @VVV \\
& K_1 @>>> \mathrm{GL}_2(\mathbb{Z}/p\mathbb{Z})
\end{CD}$$
The kernel of $\mathrm{GL}_2(\mathbb{Z}_p)\to\mathrm{GL}_2(\mathbb{Z}/p\mathbb{Z})$ will be the inverse limit $\varprojlim K_n$.
Now compute $|K_n|$ to find it is a $p$-group for all $n$. Note all matrices in $K_n$ are expressible in the form $I+pA$ where $I$ is the identity matrix and $A$ is an arbitrary $2\times 2$ matrix over $\mathbb{Z}/p^n\mathbb{Z}$. |
Showing a limit f is differentiable | You might need to assume $f_n'$ to be continuous or replace the statement with $f' = \lim_n f_n'$ almost everywhere.
Hints:
Under the given assumption the fundamental theorem of calculus applies to $f_n$. That is
$$ f_n(x) = f_n(x_0) + \int_{x_0}^x f_n'(t) dt . $$
Use a limit theorem to show that the
$$ \lim_n f_n(x) = \lim_n f_n(x_0) + \int_{x_0}^x \lim_n f_n'(t) dt =: f(x) . $$
The fundamental theorem of calculus than yields
$$ f' = \lim_n f_n' $$
(almost everywhere). |
Let $f$ be continuous and $U \subset \mathbb{R}^n$ open, if $f: U \rightarrow \mathbb{R}^m $ is injective then $n \leq m$? | Yes, this is "invariance of domain" from basic algebraic topology. It is not easy to prove in any elementary way for merely-continuous $f$. For $f$ at least once-differentiable, it is elementary, because then dimension of tangent spaces (a genuinely vector-space notion of "dimension") is sufficient. |
Solving particle in vertical motion with air resistance using conservation of energy | Air resistance (also any sort of friction) renders the use of conservation of energy useless for the problem, because these constitute the "loss" of energy to some ignored sink. (For example, air resistance heats up the air, but you never carry that into your calculations, it would be impossible.)
This is why they call these "nonconservative" forces.
Of course, energy is conserved in the real world, but in just the problem you are considering, energy is magically "lost" because of the nonconservative force of friction or resistance. |
Given that $u$ is harmonic. Prove $\Delta v \geq 0$ where $v = |\nabla u |^2$ | Useful general formula: If $u,v\in C^2,$ then
$$\Delta (uv) = u\Delta v + v\Delta u + 2 \langle \nabla u, \nabla v\rangle.$$
Thus if $u$ is harmonic, then $\Delta (u^2) = 2 |\nabla u|^2.$ In your problem, apply this to each $(D_k u)^2,$ and recall that if $u$ is harmonic, then so is $D_ku.$ |
What do we call a lattice that does not have a sublattice the shape of the diamond $M_3$? | Let $\mathbf K$ be the class of all lattices not containing $M_3$ (the $5$-element nondistributive modular lattice) as a sublattice; in other words, lattices in which every modular sublattice is distributive. It is easy to see that $\mathbf K$ can be characterized as the class of all lattices satisfying the following sentence $\varphi$:$$\forall u\forall v\forall a\forall b\forall c[(ab=ac=bc=u)\wedge(a+b=a+c=b+c=v)\rightarrow(u=v)]$$or equivalently$$\forall a\forall b\forall c[(ab=ac=bc)\wedge(a+b=a+c=b+c)\rightarrow(ab=a+b)].$$Inasmuch as $\varphi$ is a universal Horn sentence, it follows that $\mathbf K$ is closed under taking sublattices and direct products.
On the other hand, $\mathbf K$ can not be characterized by identities, and is not closed under taking homomorphic images. Bjarni Jónsson [Sublattices of a free lattice, Canad. J. Math. 13 (1961), 256-264, Lemma 2.6(i)] observed that, as a corollary of P. M. Whitman's work, elements $u,a,b,c$ of a free lattice satisfy the condition: if $u=ab=ac$, then $u=a(b+c)$. It follows that $M_3$ is not a sublattice of a free lattice, i.e, the class $\mathbf K$ contains all free lattices. Thus every lattice is a homomorphic image of a member of $\mathbf K$, and every identity which holds in all $M_3$-free lattices is a consequence of the lattice axioms. In view of this, there is not likely to be any "nicer" algebraic characterization of the $M_3$-free lattices than the one in the previous paragraph. |
Intersection of Open Connected Sets in $\mathbb{R}^2$ | What about the intersection of a fat circle with a fat line? Is it connected in general? |
Solving exponential equation $e^{x^2+4x-7}(6x^2+12x+3)=0$ | $e^{x^2+4x-7}(6x^2+12x+3)=0 \Rightarrow e^{x^2+4x-7}=0 \text{ or } \ 6x^2+12x+3=0$
$$\text{It is known that } e^{x^2+4x-7} \text{ is non-zero }$$
therefore,you have to solve :
$$6x^2+12x+3=0$$
The solutions are:
$$x=-1-\frac{1}{\sqrt{2}} \\ x=-1+\frac{1}{\sqrt{2}}$$ |
Postfix notation and the equals operator | Equality is a relation, not an operator. However, one can expresss relations as well as operators in postfix notation; in your example one gets the string $$\text{12+3=}\;.$$ |
Derivative of an integral function using FTC | By the fundamental theorem if we denote $g(x)=\int_a^xf(t)\text{d}t$ then
$$g'(x)=\frac{\text{d}}{\text{d}x} \int_a^xf(t)\text{d}t=f(x)$$
Now it becomes a problem of using the chain rule
$$\frac{\text{d}}{\text{d}x}g(\tan x) = g'(\tan x) \tan'(x)=f(\tan x)\sec^2x$$ |
Maximal Distinct Roots in $F_q$ | Hints: Show that for every $a,b \in F_q$,
$r(x^i - ax) \le r(x^i - x)$.
$r(x^i - ax - b) \le r(x^i - ax)$.
For 1), factor out an $x$, then use the fact that the roots of $x^{i-1}-1$ form a multiplicative subgroup. For 2), use the fact that the roots of $x^i - ax$ form an additive subgroup. |
How to group combinations minimizing the number of elements in groups | A 2-design (BIBD) should meet your requirements. Let the points of the design be the set $S$ elements you wish to process, and let the blocks of the design be the computers. A point is in a block if and only if the associated object is transferred to the computer. Each computer then processes all of the pairs of objects it receives, and there is no risk of duplication due to the structure of the two design.
As an example, using the Fano plane with points $0 \ldots 7$ and blocks:
3,4,6
2,3,5
1,5,6
0,2,6
0,1,3
0,4,5
1,2,4
In this example, each of 7 computers gets three objects and processes three pairs.
There exist designs for a variety of parameters. For relatively small numbers, I recommend the Handbook of Combinatorial Designs (Colbourn and Dinitz, ed.), which lists explicit examples of designs.
UPDATE: (responding to comment)
It depends on the parameters. In design theory terms, you will know the values of $v$, the number of points (test cases); $b$, the number of blocks (computers); and $\lambda$, the replication number (1, since you only need to do each calculation once). Given $v$, you must pick $k$, the number of test cases given to each computer (which you will want to minimize) and $r$, the number of computers that get each test case to satisfy the relationships
$v-1 = r(k-1)$, and
$bk=vr$
The field of design theory is very rich and has a variety of constructions, but I don't think there is a general "pick v and b, and here's your design" construction. The Handbook of Combinatorial Designs I referenced earlier has the best list of known designs of relatively small order that I know of. If you provide specific parameters, I might be able to help more. |
Linear algebra, polynomial problem | Alright, for part a), you have to look at the definition of a subspace. That is, it must contain the additive identity (the zero-polynomial), which is trivial. It must be closed under multiplication by scalar, which it is, since its degree will not change, regardless of what real number $C$ you multiply a polynomial $P$ with. Lastly, we need it to be closed under addition, which it is, by similar argument as for multiplication by scalar.
For part b), you simply need to check. $q(1) = 1 - 2 + 1 = 0$, $q'(x) = 1 - 4x + 3x^2, q'(1) = 1 - 4 + 3 = 0$. So yes, it is in $S$. |
Question regarding matrices with same image | Suppose that $A$ and $B$ are $m \times n$ integer matrices with the same image. So, for any integer vector $x$, there exists an integer vector $y$ such that
$$
Ax = By
$$
Now, take $x \in \mathcal B = \{e_1,\dots,e_n\}$ (the standard basis), and let $y_j$ denote the vector $y$ corresponding to $e_j$ as above. Let $P$ be the matrix whose columns are $y_1,\dots,y_j$.
We note that for every $x \in \mathcal B$, we have
$$
Ax = BPx
$$
Since $\mathcal B$ is a generating set, we may conclude that $Ax = BPx$ for all $x \in \Bbb Z^n$. That is, $A = BP$ as desired.
Note, however, that $P$ as chosen here need not be invertible. Perhaps you can improve this construction (i.e. choose $y$ more carefully) in order to guarantee $P$ is invertible as a matrix over $\Bbb Q$. Then (since $B = AP'$ for a matrix $P'$), you can show that $P$ must also be invertible as a matrix over $\Bbb Z$.
For the case of $\Bbb Q$ matrices: it is sufficient to note that if $A$ and $B$ have the same image, they can be "column reduced" (as opposed to row-reduced) to the same form. |
Do there exist $2k+1$ irrational numbers such that their product and sum are both rational? | You could take $\{j\:\sqrt[2k+1]{2}|1\le j\le 2k\}\cup\{-k(2k+1)\:\sqrt[2k+1]{2}\}$. The sum is $0$; the product is $-(2k+1)!2k$. |
A wrong proof that the kernel and image are always complementary | The direct sum of two subspaces of $E$ is an abstract vector space that has a canonical map to $E$, but that map can fail to be injective, and it will precisely when the two subspaces have non-zero intersection. So your mistake is in assuming that the direct sum is a subspace of $E$.
More precisely, if $W_1,W_2$ are subspaces of $E$, then there is $W_1+W_2$, the sum of the two subspaces, which is by definition the subset of $E$ consisting of all sums $w_1+w_2$ with $w_1\in W_1$ and $w_2\in W_2$. This is a subspace of $E$. Then there is the direct sum, $W_1\oplus W_2$, which is, by definition, the set $W_1\times W_2$ with scalar multiplication done component-wise. There is a linear map $W_1\oplus W_2\rightarrow E$ given by $(w_1,w_2)\mapsto w_1+w_2$, so the image is $W_1+W_2\subseteq E$, but there could be a kernel. Namely, if $w$ is a non-zero element of $W_1\cap W_2$, then $(w,-w)\in W_1\oplus W_2$ is non-zero and maps to $w-w=0$. Conversely, if $(w_1,w_2)\mapsto 0$, then $w_1=-w_2\in W_1\cap W_2$. So the induced surjective linear map $W_1\oplus W_2\rightarrow W_1+W_2$ will be an isomorphism if and only if $W_1\cap W_2=\{0\}$. |
A question about a ring in the form of a direct product. | Yes
and 3.: Addition and multiplication is defined component-wise. So $a_1 \oplus a_2 = (a_1, a_2, 0, ..., 0)$. |
Determinant of matrix $A\in\mathbb{R}^{n,n}$ | Subtract every column other than the last by the last column. Then move the last column to the first (this gives you a factor of $(-1)^{n-1}$ in the determinant). The result is a lower triangular matrix whose main diagonal is $(x,-x,-x,\ldots,-x)$. Thus the determinant of the original matrix is $(-1)^{n-1}x(-x)^{n-1}=x^n$. |
Flaw with proving that $S$ is a basis of $V$ iff $\|v\|^2=\sum_{i=1}^{n}{c_i}^2$ | Looking at the second direction (i.e., assuming that $\|v\|^2 = \sum_{i=1}^n \langle v,v_i \rangle^2$ for all $v \in V$), all you've proved is that $\{v_1,\dotsc,v_n\}$ is linearly independent, but this is a general fact about orthogonal sets proved by precisely your argument. The essential point is proving that $\{v_1,\dotsc,v_n\}$ spans $V$. So, let $v \in V$, define
$$
v_0 := \sum_{i=1}^n \frac{\langle v_i,v\rangle}{\langle v_i,v_i\rangle} v_i = \sum_{i=1}^n \langle v_i,v\rangle v_i, \quad v_1 := v - v_0,
$$
and observe that $v = v_0 + v_1$ with $\langle v_0,v_1 \rangle = 0$; what you need to show is that $v_1 = 0$. But then, $$\|v\|^2 = \langle v,v \rangle = \langle v_0+v_1,v_0+v_1\rangle,$$ so what do you get when you expand out $\langle v_0+v_1,v_0+v_1 \rangle$? |
Showing the Sorgenfrey Line is Paracompact | HINT: Prove that the Sorgenfrey line is Lindelöf, and use the fact that a regular Lindelöf space is paracompact. To prove that it’s Lindelöf, start with a basic open cover $\mathscr{U}$ (i.e., a cover by sets of the form $[a,b)$). Show that $\{(a,b):[a,b)\in\mathscr{U}\}$ covers all but a countable subset of $\Bbb R$, and use the fact that $\Bbb R$ with the usual topology, being second countable, is hereditarily Lindelöf. |
Why do we use 2-valued logic (True / False) instead of for example 4-valued logic (Boolean Algebra on two elements powerset)? | You can interpret classical propositional logic (CPL) into any Boolean algebra. We also have soundness and completeness with respect to this interpretation. That is, a theorem of CPL is provable (with respect to some typical proof system, e.g. the propositional fragment of the sequent calculus LK) if and only if it evaluates to the unit/top element of any Boolean algebra. You can actually prove completeness from soundness and the Lindenbaum-Tarski algebra with this approach to semantics.
That said, it is much harder to (directly) show that that some formula interprets to the top element for every Boolean algebra than it is to show it for some particular Boolean algebra. In particular, the Boolean algebra on a two element set is extremely simple. It also turns out that CPL with usual proof systems is complete with respect to it too. In other words, instead of checking all Boolean algebras, it suffices to check this one extremely simple one.
And that's pretty much it. It is definitely useful to know that you can interpret CPL into any Boolean algebra, but, for the purposes of logic, no (closed) CPL formula can distinguish amongst them. If all you care about is establishing provability/validity, there's no reason to consider anything other than the two element Boolean algebra.
(As a matter of terminology, we don't use $2$-valued logic in proofs1, we use proof systems that don't care what the semantics are at all. You could say we use classical logic, but as you've just illustrated, that doesn't depend on being $2$-valued.)
1 Also, as a fairly subtle point, you can have a $2$-valued logic that isn't classical. |
Explaining A Mathematical Statement In Limitation Notation To A Layman | I like to think about limits to infinity this way: No matter what $\varepsilon$ you choose (as long as it's not the limit $k$ of the function), you can always find a large enough $x$ (denoted $x > M$), so that $f(x)$ will always lie between $k$ and $k\pm \varepsilon$ for all $x>M$. Thus, as you choose $\varepsilon$ to be arbitrarily small, $f(x)$ gets arbitrarily close to $k$.
(Obviously this isn't a rigorous statement, just an intuitive look at them).
In the case of $f(x)$ having a horizontal asymptote at $y=k$, $f(x)-k$ will get closer and closer to $0$ by taking a large enough value of $x$. |
Learning Roadmap for Borel - Weil - Bott Theorem | The two, broadly defined things you need to know are Lie theory and (complex) differential geometry. The specific things from each topic are
Lie groups/algebras
Highest weight theory of compact Lie groups/complex semisimple Lie groups. One needs to build up to the theorem that irreducible complex representations of a Lie algebra are parametrized by dominant positive weights.
Most books (including Brain Hall's) build up to this. I personally like Compact Lie Groups by Sepanski if one already understand the basics of manifolds, as its very concise, has good exercises, and concludes with a proof of the Borel-Weil theorem.
Differential geometry
One needs to know what a holomorphic vector bundle over a complex manifold is and their Dolbeault cohomology. Thus the following general differential geometry background is needed, all of which the last can be found in Lee's Smooth Manifolds
Vector bundles over a manifold (I think Lee mostly talks about real vector bundles but complex ones are just replacing $\mathbb R$ with $\mathbb C$).
Differential forms.
de Rham theory.
Connections on vector bundles (maybe not necessary but I think it's useful).
After having a solid differential geometry background, in my opinion the best place to learn the necessary complex geometry is part 1 of the freely available notes by Moroianu, titled Lectures on Kahler Geometry.
For the Borel-Weil-Bott theorem proper, I would see Sepanski's text and also the paper Representations in Dolbeault Cohomology by Zierau. |
A problem about conditional probability. | The distributions of $(\xi,\eta)$ and $(\xi',\eta')=(\eta,\xi)$ coincide hence, for every measurable bounded $u$,
$$
\mathbb E(\xi u(\xi+\eta))=\mathbb E(\xi' u(\xi'+\eta'))=\mathbb E(\eta u(\xi+\eta)).
$$
In particular, if $A$ is in $\sigma(\xi+\eta)$, there exists a Borel subset $B$ such that $A=[\xi+\eta\in B]$ hence, for $u=\mathbf 1_B$, the identity above yields $\mathbb E(\xi\mathbf 1_A)=\mathbb E(\eta\mathbf 1_A)$. |
values of sums of consecutive integer squares | A characterization is available for $n$ being a square:
When is a sum of consecutive squares equal to a square?
Also, OEIS A001422 gives the full list of
natural numbers not expressible as the sum of distinct squares at all:
$$
\begin{array}{c}
2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33,
\\
43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128.
\end{array}
$$
So all numbers $n\ge 129$ are a sum of distinct squares. A similar result might be true for consecutive squares; related literature is given in the article On Integers which are are representable as sums of large consecutive squares and the references therein. |
Finding value of an algebraic expression. | I do not think there is any special trick in solving the problem.
$$x=\frac{a}{2}-\frac{1}{2a}$$
$$y=\frac{b}{2}-\frac{1}{2b}$$
$$xy=\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab}$$
$$x^2+1 = \frac{a^2}{4}-\frac{1}{2}+\frac{1}{4a^2}+1=\frac{a^2}{4}+\frac{1}{2}+\frac{1}{4a^2}=(\frac{a}{2}+\frac{1}{2a})^2$$
$$y^2+1 = \frac{b^2}{4}-\frac{1}{2}+\frac{1}{4b^2}+1=\frac{b^2}{4}+\frac{1}{2}+\frac{1}{4b^2}=(\frac{b}{2}+\frac{1}{2b})^2$$
Putting in the above values
$$xy+\sqrt{(x^2+1)(y^2+1)}=(\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab})+\sqrt{(\frac{a}{2}+\frac{1}{2a})^2*(\frac{b}{2}+\frac{1}{2b})^2}$$
Assuming that $=\frac{a}{2}+\frac{1}{2a}>0$ and $\frac{b}{2}+\frac{1}{2b}>0$
$$=(\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab})+(\frac{a}{2}+\frac{1}{2a})*(\frac{b}{2}+\frac{1}{2b})$$
$$=(\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab})+(\frac{ab}{4}+\frac{a}{4b}+\frac{b}{4a}+\frac{1}{4ab})$$
$$=\frac{ab}{4}+\frac{1}{4ab}+\frac{ab}{4}+\frac{1}{4ab}$$
So,
$$xy+\sqrt{(x^2+1)(y^2+1)}=(\frac{ab}{2}+\frac{1}{2ab})$$ |
Integer Y with N Repeating Digits of X? | If the goal is to get a simple mathematical representation for the number in question:
The key observation is that $$10^n-1=\overbrace{9\dots 9}^{n\,times} $$
It follows that
$$\frac {10^n-1}9=\overbrace{1\dots 1}^{n\,times} $$
And hence that
$$x \times \frac {10^n-1}9=\overbrace{x\dots x}^{n\,times} $$
For $x\in\{1,2,3,4,5,6,7,8,9\}$
As remarked in the comments, this is impractical from a coding point of view (as exponentiation to large powers is a very costly operation). |
Discontinuity of the function at a point by the Cauchy definition | There is no limit $L$ that works. Suppose we take some possible limit $L < 2.$ No matter how small $\delta > 0$ is, there are points with $-\delta < x < 0$ such that $f(x) > \frac{5}{2},$ so violates the limit condition for any $0 < \varepsilon < \frac{1}{2}$
Similar for $L \geq 2,$ switching to $0 < x < \delta$ |
Prove complex logarithm problem involving Euler Mascheroni constant | Here's one way the Laplace transform can be derived:
$$
\begin{align}
f(t) &= cos(t)/t \\
t f(t) &= cos(t) \\
\mathcal{L} \Bigg\{t f(t)\Bigg\} &= \mathcal{L} \Bigg\{ cos(t) \Bigg\} \\
-\frac{d}{ds} \mathcal{L} \Bigg\{ f(t)\Bigg\} &= \frac {s}{s^2 + 1} \\
\mathcal{L} \Bigg\{ f(t)\Bigg\} &= - \int \frac {s}{s^2 + 1} ds \\
u &= s^2 + 1 \\
\mathcal{L} \Bigg\{ f(t)\Bigg\} &= - \frac{1}{2} \int \frac {1}{u} du \\
&= - \frac{1}{2} ln(u) + C \\
&= - \frac{1}{2} ln(s^2 + 1) + C \\
&= ln(\frac{1}{\sqrt{s^2 + 1}}) + C
\end{align}
$$
IVT/FVT to find the value of the integration constant.
You'll notice that both of your forms are very close to each other:
$$
-\frac{1}{2} ln (\frac{1}{s^2}+1) - ln(s) \\
-\frac{1}{2} ln (\frac{s^2 + 1}{s^2}) - ln(s) \\
\frac{1}{2} ln (\frac{s^2}{s^2+1}) - ln (s) \\
ln (\frac{s}{\sqrt{s^2+1}}) - ln(s) \\
ln (\frac{1}{\sqrt{s^2+1}})
$$
So the question comes down to the integration constant $-\gamma$ |
Definite Integral of the following term with $e^e$ | Hint. By making the change of variable
$$
u=e^{-\lambda t},\, (\lambda>0)\qquad du= -\lambda e^{-\lambda t}dt, \qquad -\lambda t=\ln u,
$$ you obtain
$$
\int^\infty_0 t(n\lambda e^{-ne^{-\lambda t} - \lambda t})dt=-\frac{n}{\lambda}\int_0^1\ln u \:e^{-nu}du
$$ then by the change of variable $v=nu$ you get an answer in terms of the Euler constant and in terms of the incomplete gamma function. |
If $x\sqrt {1-y^2} + y\sqrt {1-x^2}=a$, show that | HINT:
WLOG $x=\sin A,y=\sin B$ where $\dfrac\pi2\le A,B\le\dfrac\pi2$
$$\implies\sin(A+B)=a$$
$$ \arcsin a=\arcsin x+\arcsin y$$ |
Analogue to Fixed Point Theorem for Compact metric spaces | the proof I learned from Henno Brandsma's answer: Let $X$ be compact Hausdorff (no metric is needed), and define $A_0 = X$, $A_{n+1} = f[A_n]$; then all $A_n$ are compact non-empty, and the $A_n$ are decreasing. Try to show that $A = \cap_n A_n$, which is also compact and non-empty, satisfies $f[A] = A$.
Another non-constructive way to show this is to consider the poset $\mathcal{P} = \{ A \subset X \mid A, \mbox{closed, non-empty and } f[A] \subset A \}$, ordered under reverse inclusion. Then an upper bound for a chain from $\mathcal{P}$ is the (non-empty) intersection, and a maximal element (by Zorn one exists) is a set $A$ with $f[A] = A$. |
Equivalence of computing trace norm of matrix | I don't know if it is OK in stackexchange but I am trying to answer my own question to explain more about the derivation of the equivalence of trace norm. The reference is mostly from section 4.2.3.2 of MOSEK Modeling Cookbook.
We first prove that $||A||_{2} \le 1$ is equivalent to $A^{T} A\preceq I$.
\begin{align*}
& ||A||_{2} \le 1\\
\Leftrightarrow \quad & \max_{||x||_{2} =1} ||Ax||_{2} \le 1\\
\Leftrightarrow \quad & \max_{||x||_{2} =1} x^{T} A^{T} Ax\le 1\\
\Leftrightarrow \quad & \max_{||x||_{2} =1} x^{T}\left( A^{T} A-I\right) x\le 0\\
\Leftrightarrow \quad & A^{T} A\preceq I
\end{align*}
Then our problem
\begin{array}{ll} \mathop{\text{maximize}}\limits_{Y\in\mathbb{R}^{m\times n}} & \text{tr}(X^TY)\\ \text{subject to} & Y^TY \preceq I_n\end{array}
becomes $$\sup _{||Y||_{2} \leq 1}\mathrm{tr} (X^{T} Y)$$
Then we do SVD decomposition to $X$, $X=U\Sigma V^T$,
\begin{equation*}
\begin{aligned}
\sup _{||Z||_{2} \leq 1}\mathrm{tr} (X^{T} Z) & =\sup _{||Z\| _{2} \leq 1}\mathrm{tr} (V\Sigma ^{T} U^{T} Z)\\
& =\sup _{||Z\| _{2} \leq 1}\mathrm{tr} (\Sigma ^{T} U^{T} ZV)\\
& =\sup _{\| U^{T} ZV\| _{2} \leq 1}\mathrm{tr} (\Sigma ^{T} U^{T} ZV)\\
& =\sup _{\| Y\| _{2} \leq 1}\mathrm{tr} (\Sigma ^{T} Y)
\end{aligned}
\end{equation*}
$\mathrm{tr} (V\Sigma ^{T} U^{T} Z) = \mathrm{tr} (\Sigma ^{T} U^{T} ZV)$ follows from the invariant under cyclic permutations property of trace. And the equivalence of constraint $||Z||_2 \le 1 \Leftrightarrow ||U^T Z V ||_2 \le 1$ follows from the definition of $\ell_2$ norm of matrix.
And using the unitary invariance of the norm $||\cdot||_2$ again. We can consider $Y=\mathbf{diag}(y_1, \dots, y_p)$.
$$\sup_{\|Z\|_2\leq 1} \mathrm{tr}(X^T Z) = \sup_{|y_i| \leq 1} \sum_{i=1}^p \sigma_i y_i = \sum_{i=1}^p \sigma_i$$ |
What can be said about the leaves of a regular foliation? | The leaves can change topologically, you can have compact and non compact leaves as in the suspension. Let $M$ be a compact manifold and $\hat M$ its universal cover Let $h:\pi_1(M)\rightarrow Diff(F)$ consider the manifold $N$ which is the quotient of of $\hat M\times F$ by the diagonal action of $\pi_1(M)$. The foliation $\hat M\times y$ of $\hat M\times F$ is preserved by $h$ and defines on $N$ a foliation ${\cal F}$. Suppose that $F$ is not compact and $h$ has a unique fixed point $x_0$ and the stabilizer of every element different of $x_0$ is $\{1\}$. Let $p:\hat M\times F\rightarrow N$ be the projection, the unique compact leaf of ${\cal F}$ is the leaf through $p(x_0)$. An example of this is a flat vector bundle $p:V\rightarrow M$, only the zero section is a compact leaf of the horizontal foliation. You can eventually construct examples for which the orbit of every element by $h$ is finite. In this case, all the leaves are compact and are not necessarily diffeomorphic if you have a fixed point by the $h$ and if there are points which are not fixed by $h$. This can be achieved if $\pi_1(M)$ is finite.
This situation is quite the general situation. Haefliger (I believe in his thesis) has described the neighborhood of the leaf (compact) of every foliation by using the holonomy representation. Take a compact leaf ${\cal F}_x$ of the foliation ${\cal F}$ through $x$ on a compact manifold. Let $T$ be a local transveral of the foliation, you can describe a neighborhood $U$ of ${\cal F}_x$ which saturated as the suspention of $ \hat {\cal F}_x\times T$ by the holonomy representation.
The fact that the leaves are isomorphic is related to the fact that the holonomy of every leaf is trivial. I believed that Epstein has conjectured if all the leaves of a foliation on a manifold are compact and without holonomy, then they are diffeomorphic. |
Rank Theorem question | if $n<m$ then the rank cannot be $m$ (linear algebra), if $n\geqslant m$ then your local representation
$$
(x_1,\dots,x_n) \mapsto (x_1,\dots,x_m)
$$
proves the assertion |
Fermat's Little Theorem and Euler's Theorem | For the first problem you will need a little bit of Chinese Remainder Theorem. You want to find the remainder of the stacked exponential modulo $10^5 = 2^5 \times 5^5$. Consider the two prime divisors separately.
As $\phi(2^5) = 16$ we have that if $r_1$ is the remainder of $5^{5^{5^{5}}}$ modulo $\phi(2^5) = 16$ then $5^{5^{5^{5^{5}}}} \equiv 5^{r_1} \pmod {2^5}$. Now we have to find $r_1$, which is a solution of $5^{5^{5^{5}}} \equiv r_1 \pmod {2^4}$. Repeat this algorithm few times and you will get rid of the exponents and you will find a value such that: $5^{5^{5^{5^{5}}}} \equiv r \pmod {2^5}$. Now use that $5^{5^{5^{5^{5}}}} \equiv 0 \pmod {5^5}$ and glue the two solutions with Chinese Remainder Theorem.
The second one can be solved using similar method, but this time you won't need the Chinese Remainder Theorem, as $(7,100) = 1$. Actually this is easier as $7^4 \equiv 1 \pmod {100}$. |
A and B have negative correlation, so -A and -B have positive? | If $A$ and $B$ are negatively correlated, increasing $A$ decreases $B$.
But since increasing $A$ is equivalent to decreasing $-A$ and decreasing $B$ is equivalent to increasing $-B$, the previous statement is equivalent to “Decreasing $-A$ increases $-B$”, which means $-A$ and $-B$ are also negatively correlated. |
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