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Suppose $f: \mathbb{R} \to \mathbb{R}$ is continuous such that for any real $x$,
$|f(x) - f(f(x))| \leq \frac{1}{2} |f(x) -x|$.
Must $f$ have a fixed point?
The question seems to invite an eventual application of the standard contraction mapping theorem. But this approach has not led me to showing there is a fixed point.
Is it true that there is a fixed point, and if so why? |
If I understood your question correctly, you're essentially asking how one can find the equation for the directrix if one only has the equation for an ellipse with a given eccentricity.
You start with the equation below$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\qquad(1)$$where $a$ and $b$ are positive real-valued constants.
If you then
define two points on the $x$ axis, $F$ and $F'$, by their coordinates,$$F \equiv (\epsilon a, 0)\qquad(2)$$$$F' \equiv (-\epsilon a, 0)\qquad(3)$$for some as-yet undetermined real value $\epsilon \ge 0$, and then require that the sum of the distances from $F$ to $P$ and from $P$ to $F'$ be equal to $2a$ - for any arbitrary point $P$ on the ellipse - you'll find out that that is only possible if you choose $\epsilon$ to satisfy$$\epsilon = \sqrt{1 - \frac{b^2}{a^2}}\qquad(4)$$
So far, the above has nothing to do with the directrix. It's just a way to construct the foci and find the eccentricity, starting from some equation and then requiring that the curve described by that equation must have the fundamental property of an ellipse (namely, that the sum of the distances from any point on the ellipse to the two foci is a constant).
Now, let's see how we can prove the directrix property. Imagine that there is a vertical line at $x=d$ for some as-yet-undetermined real value $d \ge 0$ and let's see if we can find a value of $d$ such that the directrix property is satisfied. The focus for positive $x$ is $F$ (see its coordinates above), and an arbitrary point $D$ in the would-be directrix has coordinates $D = (d, y)$. The directrix property mandates that$$\epsilon = \frac{\overline{PF}}{\overline{PD}}\qquad(5)$$
Now, let's look at the square of the distances involved, since that gets rid of the square roots:$$\overline{PF}^{\,2} = (x - \epsilon a)^2 + (y - 0)^2 = (x - \epsilon a)^2 + y^2$$
But $P$ lies on the ellipse so $(x,y)$ satisfies $(1)$. Therefore, eliminating $x^2$ in favour of $y^2$, but leaving the $x$ term alone, we have$$\overline{PF}^{\,2} =a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax$$
How about $\overline{PD}^{\,2}$? Note that $D$ and $P$ have the same $y$ coordinate so$$\overline{PD}^{\,2} = (x - d)^2 + (0)^2 = (x - d\,)^2$$
Here's the crux now.
Can we find a value of $d$ such that $(5)$ is true?Imposing$$\frac{\overline{PF}^{\,2}}{\overline{PD}^{\,2}} =\frac{a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax}{(x - d\,)^2} = \epsilon^2$$we get$$a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax = \epsilon^2\,(x - d\,)^2$$
Note that the complicated expression above is something like this:$$(\mbox{terms independent of $x$ and $y$}) + (\mbox{terms involving $y^2$}) - 2\epsilon ax + 2\epsilon^2 xd = 0$$because the $x^2$ terms can be removed using $(1)$. Since the above has to be true for all $x$, the only hope for $(5)$ to be possible is if we choose $d$ such that$$2\epsilon ax - 2\epsilon^2 xd = 0\qquad\Rightarrow\qquadd = \frac{a}{\epsilon}$$
With this choice, it's then not hard to show that the terms not written above also vanish so, indeed,$$\epsilon = \frac{\overline{PF}}{\overline{PD}}$$is true, provided we make that choice for $d$.
So, what's the conclusion? The conclusion is that an ellipse has
two equivalent properties, namely, (A) the sum of the distances from the foci to any arbitrary point on the ellipse is a constant, and (B) the ratio of (the distance from an arbitrary point on the ellipse to one of the foci) to (the distance from that point to a fixed vertical line parallel to the semi-minor axis) is equal to the eccentricity of the ellipse.
And we also proved that $\epsilon = a/d$.
(Technically, I only proved that (A) above implies (B). We'd also have to prove that (B) implies (A) for the equivalence I referred to be proven. That's an exercise I leave for the reader...) |
what is the method for solving a differential equation when a summation is involved from the start ?
ex. what method is required to find a particular solution to
$$(1-x^2)y''-2xy'=\sum_{n=1}^{\infty}\tfrac{P_n(x)}{2^n}$$
I should point out also that
$\tfrac{2}{(5-4x)^{1/2}}=\sum_{n=0}^{\infty}\tfrac{P_n(x)}{2^n}$.
Edit: $|x| \leq 1$
My attempt :
$\tfrac{2}{(5-4x)^{1/2}}=\sum_{n=0}^{\infty}\tfrac{P_n(x)}{2^n} \Rightarrow \sum_{n=0}^{\infty}\tfrac{2^{n+1}}{(5-4x)^{1/2}}=\sum_{n=0}^{\infty}P_n(x)$. ( I don't know if you can do this though.....)
then we assume that $y=\sum_{n=0}^{\infty}a_nP_n(x)=\sum \tfrac{a_n2^{n+1}}{(5-4x)^{1/2}}$
$y'=\sum \tfrac{a_n2^{n+2}}{(5-4x)^{3/2}}$
$y''=\sum \tfrac{3a_n2^{n+3}}{(5-4x)^{5/2}}$
Is this anyway right ? |
How can one prove that $$\int_0^1 \tan^{-1}\left[\frac{\tanh^{-1}x-\tan^{-1}x}{\pi+\tanh^{-1}x-\tan^{-1}x}\right]\frac{dx}{x}$$ $$=\frac{\pi}{8}\ln\frac{\pi^2}{8}?$$
I wouldn't characterize my answer as a "solution to the integral", at least in the expected sense. What I will do, (also because it is a pity for this question to not have even one answer), is to use various transformations related to mathematical statistics and random variables, to transform the problem into one of proving existence and uniqueness of an expected (specific) value. At least in the end, no integral will be in sight.
Following notation established in the comments, let $S(x)=\tanh^{-1}(x)-\tan^{-1}(x)$. Our integral can be written (to prepare also for integration by parts)
$$I=\int_0^1 \tan^{-1}\left[\frac{S(x)}{\pi+S(x)}\right]\left(\frac{d\ln x}{dx}\right)dx$$ Integration by parts gives
$$I = \tan^{-1}\left[\frac{S(x)}{\pi+S(x)}\right]\ln x\Big|_0^1 - \int_0^1 \frac {d\tan^{-1}\left[\frac{S(x)}{\pi+S(x)}\right]}{dx}\ln xdx$$
$$= 0-\int_{0}^{1}\frac{\pi}{(\pi+S(x))^{2}+S(x)^{2}}\frac {dS(x)}{dx}\ln xdx$$
where for later use $\frac {dS(x)}{dx}=\frac {2x^2}{1-x^4}$.
Now consider the variable $Z=S(X)$. We first show that the term $\frac{1}{(\pi+z)^{2}+z^{2}} $ is the density of a Cauchy random variable. In general, this density is
$$f_Z(z) = \frac 1{\pi}\frac {\gamma}{(z-m)^2+\gamma^2} $$ where $m$ is the median/mode and $\gamma >0$ is a real scale parameter. If we set $m=-\pi/2,\;\; \gamma = \pi/2$ we obtain
$$f_Z(z;m=-\pi/2,\gamma=\pi/2) = \frac 1{\pi}\frac {\pi/2}{(z+\pi/2)^2+\pi^2/4}=\frac{1}{(\pi+z)^{2}+z^{2}}$$
So indeed, the variable $Z=S(X)$ can be seen as a Cauchy$(m=-\pi/2,\gamma=\pi/2)$ random variable with the above density. Now reverse the direction of thought: if $Z$ is a random variable, so is $X$, defined by $X=S^{-1}(Z)$. When we define a random variable as a function of another, if the function is strictly monotone, we have available the change-of-variable formula to derive the density of the former. The function $S(x)$ is strictly increasing and therefore so is its inverse. Inverting the relation we have $Z = (S^{-1})^{-1}(X) = S(X)$. The change-of-variable formula gives
$$f_X(x) = \left|\frac {dS(x)}{dx}\right|\cdot f_Z(S(x)) = \frac {dS(x)}{dx}\cdot f_Z(S(x))=\frac {dS(x)}{dx}\frac{1}{(\pi+S(x))^{2}+S(x)^{2}}$$
But this last expression exists already in our integral. Substituting we obtain
$$I = -\pi\int_{0}^{1}f_X(x)\ln xdx$$
This last expression
could be the expected value of $\ln x$, (by the so called "Law of Unconscious Statistician"), if only the density $f_X(x)$ integrates to unity over $[0,1]$.
Note that $$\int_{0}^{1}f_X(x)dx = \int_{0}^{\infty}f_Z(z)dz = \int_{0}^{\infty}\frac{1}{(\pi+z)^{2}+z^{2}}dz $$ Adopting to our case a formula from Gradshteyn and Ryzhik 7th ed. (3.252(1), p.325) we find that
$$\int_{0}^{\infty}f_Z(z)dz = \frac 14 = \int_{0}^{1}f_X(x)dx \Rightarrow \int_{0}^{1}4f_X(x)dx =1$$
So it is the random variable with density $\tilde f_X(x) = 4f_X(x)$ that has the (truncated) support $[0,1]$ that validates the treatment of our intergal as an expected value: $$I = -\frac {\pi}{4}\int_{0}^{1}4f_X(x)\ln xdx = -\frac {\pi}{4}E[\ln X]$$
What have we accomplished (if anything)? We have transformed the problem: from
"prove that $I =\frac{\pi}{8}\ln\frac{\pi^2}{8}$"
we now must somehow prove that
"There exists a random variable $X$ with support $[0,1]$ whose logarithm has expected value equal to $-\frac{1}{2}\ln\frac{\pi^2}{8}$".
If this sentence can be proven in general (for existence and uniqueness), then we have "solved the integral".
Another way to try to prove this is to try to match the $\tilde f_X(x)$ density with some known distribution. As an example, we observe that the integral is evaluated in $[0,1]$ which is the support of a Beta distribution. Also, if we can match our density with a Beta distribution, then we know the expression for the expected value of its logarithm : $E[\ln X]=\psi(a) - \psi(a+b)$, where $\psi()$ is the digamma function. Therefore, if we postulate a Beta density for $X$,$$4f_X(x) = \frac {x^{a-1}(1-x)^{b-1}}{\operatorname{B}(a,b)},\; a,b>0$$ the integral should satisfy
$$I = -\frac {\pi}{4}\Big(\psi(a) - \psi(a+b)\Big)$$
Here, the task has become to determine $a^*>0,b^*>0$ such that
$$ -\frac {\pi}{4}\Big(\psi(a^*) - \psi(a^*+b^*)\Big) = \frac{\pi}{8}\ln\frac{\pi^2}{8}\qquad [A]$$ and that they also satisfy
$$\frac {x^{a^*-1}(1-x)^{b^*-1}}{\operatorname{B}(a^*,b^*)} = \frac{4}{(\pi+S(x))^{2}+S(x)^{2}}\frac {2x^2}{1-x^4},\;\; \forall x\in [0,1]\qquad [B]$$
Note that the second equation must hold for the whole interval $x\in [0,1]$. If existence and uniqueness of the solution to this system of non-linear equations can be proven, we have essentially "solved the integral". If non-existence is proved, then the postulate of a beta-distributed $X$ is rejected. A comment has already shown that this postulate should be rejected, but I leave it as an example of the approach.
But the general approach, and the transformation of the integral into an expected value of $\ln X$ remains valid. |
I need to find the volume of ellipsoid: $$5x^2 + {y^2\over25} + {3z^2\over4} = 1$$ if the ellipsoid is bounded by $z={-1\over2}$ and $z=1$ planes.
I was able to find the total volume of the ellipsoid using the formula $V={4\pi\over3}*abc={40\pi\over3\sqrt15}$. But I don't think i can use this in any way to find the volume of the ellipsoid that is bounded by 2 planes.
To find the actual volume, I'm pretty sure I need to solve this: $V=\int_{-1\over2}^1S(x)dx$, where $S(x)$ is the area of the cross section of the ellipsoid, which is an ellipse. Now I think that the right move here would be to get the cross sections that are parallel to the $z$-axis. However my question is, how can I find these areas of the ellipses to plug into the above formula?
Any suggestions, would be greatly appreciated! |
I am reading the paper Slightly Superexponential Parameterized Problems at the moment and have two questions about it:
First question: The paper gives a proof of the following statement
Theorem 2.1: Assuming ETH, there is no $ 2^{o(klogk)}$ time algorithm for $ k \times k$ -Clique.
They prove this statement by sophisticated reduction from $ 3$ -coloring. They then state that this construction runs in time polynomial in $ k$ . They state:
The graph $ G$ has $ k^2$ vertices and the time required to construct G is polynomial in $ k$ . […] Therefore, the total running time is $ 2^{o(k \log(k)} \cdot k^{O(1)}$
Why is this not $ 2^{o(k \log(k)}) + k^{O(1)}$ instead? As far as I can see, we have only have to construct the graph once, and then we can run the presumed $ 2^{o(k \log(k)}$ algorithm.
Second question: From this theorem, it follows that $ k \times k$ -Clique can not have a $ k^{o(k)}$ algorithm under the Exponential Time Hypothesis. This follows from the abstract, which states that $ k^{O(k)} = 2^{O(k \log(k))}$ . What is a good way to proof this statement?
In this question, a
program means a parameterized multithreaded program with the interleaving semantics, a finite number of per-thread states (which may depend on the number of threads), and finite number of shared states (which may depend on the number of threads).
A
shared state is a valuation of the shared memory, and a per-thread (in other terminology, local) state is the valuation of thread-local memory (we assume no stack). Interleaving semantics means that the actions of the threads are interleaved on a single processor and a thread has rw-access to the shared memory and its own local memory and no access to the local memories of the other threads. Parameterized means that we conside a family of programs generated from a finite-desciption template such that the $ n$ th member of the family has $ n$ threads (which typically coincide up to the thread identifier).
To the best of my knowledge, for such a program, the size of the shared state-space is anywhere between constant (e.g., for a single boolean lock variable) and exponential (e.g., Peterson mutual exclusion protocol) in the number of the threads $ n$ .
Is there any well-known academic program in which the size of the shared state-space grows superexponentially in $ n$ ? |
I would like to consider the trace of the time evolution operator $e^{-\frac{i}{\hbar}\hat{H}t}$
Apparently in single-particle quantum mechanics is can be represented as
$$ tr \ e^{-\frac{i}{\hbar}\hat{H}t}= \int d^nr \left< \textbf{r}| e^{-\frac{i}{\hbar}\hat{H}t} | \textbf{r} \right>= \int d^nr K(\textbf{r},\textbf{r},t) $$
The second equality follows from the definition of a propagator but I cannot see how first equality holds. |
Here is a rectangle made out of 2x1 dominoes:
It can be divided along a line into two smaller rectangles:
What is the smallest (in area) rectangle of (edit) multiple 2x1 dominoes that
cannot be split along a line into two smaller rectangles?
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The best you can do is one with an area of 30 (5 x 6):
2 x 2 and 2 x 3 2 x anything
In any rectangle with a side length $l$ of 2, you can't have any domino aligned so that its long side is aligned with $l$ because that would provide an immediate cut point. The only other option is to have the long side of all dominoes 90° compared to $l$, which provides cut points after each stacked pair. (Or if you try to offset the pairs, leads to an unfillable gap of 1x1 at the ends.)
3 x 3
Is an odd number of tiles (9) and thus can't be tiled by 2 x 1 dominoes.
3 x 4
A 3 x 4 rectangle has one end (let's make it the top) of length three, and thus must begin like this:
But whichever way we try to fill the second position in the second row (horizontal or vertical) we end up with a cut site. 3 x anything
A side length of 3 will never work. Based on the second picture above (3x4), you must always start the top the same way, and then must always alternate magenta dominoes down the side. If it's an even number (3x6, 3x8, ...) you'll end up with a cut site like the one shown in the 3x4 diagram, on the second vertical. If it's an odd number, you'll end up with an unfillable 1x1 square at the bottom right.
4 x 4
Let's put a domino into the top left to get started (this covers all possible orientations as well). Once we've done that, we can't fill box 3 with a horizontal domino, because that would create a cut site on the first horiztonal. So we put a vertical domino in box 3, which means there must be a vertical one in box 4. Now box 12 can't be filled with a vertical domino, because that would create a cut site along the third vertical, so we must go horiztonal to fill box 12. And since box 16 is only height 1, we have to put another horizontal one there, which leaves a cut site on the second vertical. So there's no way to do this one.
4 x 5
In this case, there are two "top" configurations to consider. For the first (hor-vert-vert), we can see then that box 6 must have a vertical domino, otherwise we'd be able to cut on the second horiztonal. That means block 5 must also have a vertical since it's width one. By the same reasoning, blocks 11 and 12 must have vertical dominoes, and again 14 and 13. But that allows us to cut right down the middle, so no dice here.
The other top configuration (excluding the obviously wrong hor-hor and vert-vert-vert-vert, as well as reflection of the above one) is vert-hor-vert. In this case, we then see that box 6 (and 7) must have vertical dominoes to eliminate the cut at the second horizontal. That means squares 9 and 12 must also have vertical dominoes since there's no other way to fill them. But then whichever way we try to put a domino to cover box 14, we allow a cut site. If it's horizontal, you can cut at the 4th horizontal line. If it's vertical, you can cut at the first vertical line. 4 x 6
In the first top configuration (same as 4 x 5 above), you'll have the same problem, just one step further down as you continue (either way you fill box 19, you'll introduce a cut site).
In the second top configuration (same as 4 x 6 above), you must place the domino on box 14 vertically to avoid the 4th horizontal cut site, but that means below the domino in box 9 must be another vertical one, which allows a cut on the first vertical. Again, failure. 5 x 5
Again, odd number of squares (25) and so can't be tiled.
So that's definitely not an elegant mathematical proof, but I think it'll hold up. And since the image at the top of this answer shows that 5x6
can be done, that is the smallest possible answer.
Here is a proof that 5x6 is the smallest possible rectangle.
A rectangle of size $x$ by $y$ has $\frac{xy}{2}$ dominoes and $x+y-2$ potential lines. All of these lines must be blocked by at least one domino which has one square on each side. However, if the line divides the rectangle into two even areas, then one domino blocking it would leave an odd area on each side, so there must be a second domino also blocking that line. A domino can't block more than one line at once, so there must be at least one domino for every line making odd areas and two for every line making even areas.
Either both dimensions of the rectangle are even or one is even and one is odd. Suppose both are even, so the rectangle is $2m$ by $2n$ units for positive integers $m$ and $n$. Then every line is even, so at least $2(2m+2n-2)$ dominoes are required. There are $2mn$ dominoes, so $2mn\ge2(2m+2n-2)$. Rearranging, $(m-2)(n-2)\ge2$. The smallest possible $m$ and $n$ are $4$ and $3$, making an $8$-by-$6$ rectangle with an area of $48$.
Otherwise, one dimension is odd and the rectangle is $2m+1$ by $2n$ units for positive integers $m$ and $n$. Now there are $n$ odd lines (every other line running parallel to the odd edge) and $2m+n-1$ even ones, so $(2m+1)n\ge2(2m+n-1)+n$, which rearranges to $(m-1)(n-2)\ge1$. Now the smallest possible values are $m=2$, $n=3$, making a $5$-by-$6$ rectangle with an area of $30$.
I'll one-up you guys and prove this stronger statement: If $m\geq n$ are the dimensions of a rectangle that admits a nonsplittable domino tiling using more than one domino, then $m \geq 6$ and $n \geq 5$.
It's stronger because it also proves the nonexistence of very long yet thin nonsplittable domino tilings, such as $60\times3$ or $999 \times 4$.
Proof:
First we'll show that if a rectangle admits a nonsplittable tiling using more than one domino, neither dimension may be $4$ or less.
Let $m \geq n$ be the dimensions of a rectangle, and assume that the rectangle is aligned in portrait. Note that $m \cdot n$ must be even, otherwise you can't cover the rectangle with an integer number of dominoes (each of area $2$). We'll show case-by-case that $n$ can be neither $1, 2, 3$ nor $4$.
Case $n = 1$: For any tiling of a $m\times1$ rectangle, there is a split along the bottom edge of the topmost domino. Case $n = 2$: For any tiling of a $m\times2$ rectangle, there needs to be at least one horizontal domino, otherwise there would be a vertical split along the middle. Now either the top edge or the bottom edge of this domino induces a horizontal split. Case $n = 3$: By necessity, there are at least two horizontal dominoes, one to prevent each vertical split. If such a horizontal domino were not touching either the top or bottom edge of the rectangle, either the top or bottom edge of this domino induces a split (since the two dominoes required to block both splits would overlap). So we need exactly two horizontal dominoes, in opposite corners of the rectangle. Every other domino must be vertical.
Now consider the left column of the rectangle. One square of this column (either the top or the bottom square) is covered by one of the two horizontal dominoes, the other $m-1$ squares are covered by vertical dominos, so $m-1$ is even. But since $m\cdot n$ is even and $n = 3$, $m$ is even too, which gives a contradiction.
Case $n = 4$:
Suppose a horizontal split (at height $k$) were prevented by a single domino. This split divides the rectangle into two pieces, the bottom piece (of even area $4k$) would then be covered by a half domino and some whole dominoes, so it would have odd area. Contradiction.
So we know that each horizontal split is prevented by at least two (vertical) dominoes. Since there are $m-1$ such splits, there must be at least $2m-2$ vertical dominoes, and since there are $2m$ dominoes in total, there can be at most $2$ horizontal dominoes, which are not enough to prevent the $3$ possible vertical splits.
Okay, so now we've done all four cases, we have $m \geq n \geq 5$ for nonsplittable tilings. Since $25$ is odd, $m = n = 5$ isn't coverable in an integer amount of dominoes, so we get $m\geq 6$ and $n \geq 5$ as a new lower bound. $~~~~\square$
Concerning the 5x4 rectangle:
Here is a proof that the 5x4 rectangle cannot be tiled in a "split-free" way. Consider a grid that is four columns wide and five rows tall: If the filling is to be "split-free", then there must be at least one domino spanning Column 2 and Column 3. This domino cannot be placed in Rows 2, 3, or 4, because of the effects of the other two dominoes that also overlap this row. They would either both "point" up from this row or both point down from this row, creating a split; or one would "point" up and the other down. This latter case would divide the grid into two disconnected regions with an odd number of tiles remaining, neither of which could be tiled by dominoes. Thus, if such a tiling exists, there must be a domino in the center of the top row or the bottom row. Without loss of generality, assume that it is in the top row.
Now note that
there must also be at least one domino spanning Column 1 and Column 2, and another domino spanning Column 3 and Column 4. Neither of these can be in the second row, since we have already placed a domino in the center of the first row, and placing either of our new dominoes in an adjacent row would leave a single corner tile isolated and uncovered. Similarly, our two new dominoes cannot be in adjacent rows or in the same row, since this would necessarily create a split between two of the rows. Thus, these two new dominoes can only be in Row 3 and Row 5, respectively.
Once you have done this,
It is not hard to see that the remaining grid cannot be filled without creating a split between the second and third rows. |
Let $$f(x)=\sum_{n=1}^{\infty}\dfrac{\cos{nx}}{\sqrt{n^3+n}}$$ and $F(x)=\int_{0}^{x}f(t)\,\mathrm dt,F(0)=0$.
Show that: $$\dfrac{\sqrt{2}}{2}-\dfrac{1}{15}<F\left(\dfrac{\pi}{2}\right)<\dfrac{\sqrt{2}}{2}$$
Find the value $F\left(\dfrac{\pi}{2}\right)$.
Since the series $f$ converges normally on $\mathbb{R}$, we can integrate term-by-term. This gives $$ F\left(\frac{\pi}{2}\right)=\sum_{n\geq 1}\frac{1}{\sqrt{n^3+n}}\int_0^{\pi/2}\cos (nx)dx=\sum_{k\geq 0}\frac{(-1)^k}{\sqrt{2}(2k+1)^\frac{3}{2}(2k^2+2k+1)^\frac{1}{2}}. $$ The series converges absolutely to $S$. But this is an alternating series which satisfies Leibniz criterion (the relevant function is indeed decreasing on $(0,+\infty)$). So the sequence of partial sums $S_n$ alternates about $S$ and satisfies $$ S_{2n+1}< S< S_{2n} \qquad \forall n\geq 0. $$ In particular, we get $$ \frac{\sqrt{2}}{2}-\frac{1}{15}<\frac{\sqrt{2}}{2}-\frac{1}{3\sqrt{5}\sqrt{6}}=S_1<S<S_0=\frac{\sqrt{2}}{2}. $$ This answers 1.
I don't know if there is a closed form for $S$ and I am not alone.
I was delayed in posting this, so it is little more than a comment to julien's answer.
Since the series $\sum\limits_{n=1}^\infty\frac1{\sqrt{n^3+n}}$ is convergent, we can integrate term by term $$ F(x)=\sum_{n=1}^\infty\frac{\sin(nx)}{n\sqrt{n^3+n}} $$ $\sin((2k)\pi/2)=0$ and $\sin((2k+1)\pi/2)=(-1)^k$; therefore, $$\begin{align} F(\pi/2) &=\frac1{\sqrt2}\sum_{k=0}^\infty\frac{(-1)^k}{\sqrt{(2k+1)^3(2k^2+2k+1)}}\\ &=\frac1{\sqrt2}-\frac1{\sqrt{270}}+\dots \end{align} $$ By the alternating series test, the final sum is between $\frac1{\sqrt2}$ and $\frac1{\sqrt2}-\frac1{\sqrt{270}}$. That is, $$ \frac1{\sqrt2}-\frac1{15}\lt\frac1{\sqrt2}-\frac1{\sqrt{270}}\lt F(\pi/2)\lt\frac1{\sqrt2} $$
This evaluates to $0.6587329279592957$, but the ISC does not find a closed form. |
A table does not sublimate, and nor does a spoon. Ice does, however. What is the fundamental difference?
This relates to the difference between how ice, iron spoons and wooden tables stick together.
In wood, most or all of the bonds between the individual atoms and fibre units are covalent; making them very strong. Similarly, a spoon is an alloy of iron, nickel, chrome and carbon, also a very coherent structure.
In ice, on the other hand, the units of the crystal is held together with comparatively weak hydrogen bonds, meaning that not a lot of energy is required for a surface molecule of water to escape, in a process called sublimation.
Usually, once a water molecule escapes the ice, it is, given a relatively constant temperature, very unlikely to rejoin the body of ice; meaning that especially in open air and windy conditions, ice will sublimate at sub-zero temperatures, although as the temperature of the ice drops, this effect becomes steadily less noticeable.
Edit: It is worth noting that given enough time, a table or a spoon can indeed sublimate as well, at any temperature, but the odds against it are utterly staggering.
One term I'm missing so far in the discussion is the vapor pressure.
You may say that the water vapor pressure above ice is far higher than the iron vapor pressure above iron (at "normal" temperatures), or the vapor pressure of the various substances forming the table over that table.
According to this table in Wikipedia, you need to heat iron well above the melting point to reach such iron vapor pressures. They also cite a formula for the iron vapor pressure above solid iron:
$\log (P/\mathrm{Pa}) = 12.106 - 21723 / (T/\mathrm{K}) + 0.4536~\log (T/\mathrm{K}) - 0.5846 (T/\mathrm{K})^{-3}$ with the specification that this could be used from 298 K upwards.
Plugging in 298 K, I get $p_{Fe} \approx 10^{-60}$ Pa.
This is so low that lots of other processes will lead to more substantial loss of iron from the spoon via the gas phase (e.g. according to this $\ce{FeCl3}$ has a vapor pressure of 1 mmHg $\approx$ 133 hPa at 194 °C).
<<
This answer was posted before the question was completely rephrased and originally asked "why does a table not evaporate" >>
Who said a spoon doesn't evaporate?
Ice is a solid below its melting point ($0~^\circ\mathrm{C}$) but above its melting point it goes through a phase transition and melts to become a liquid (so it's liquid at room temp). Then when it's heated above room temperature to its boiling point ($100~^\circ\mathrm{C}$), it goes through another phase transition and it evaporates to become gas.
Iron (assuming an iron spoon here to keep it simple) is a solid (at room temperature) then when it's heated above room temperature to its melting point ($1538~^\circ\mathrm{C}$) it becomes a liquid. Then when it's heated above its melting point to its boiling point ($\approx2861~^\circ\mathrm{C}$), it evaporates. |
I am learning for the first time the Pumping Lemma for CFL, and I thought I understood how it works until I came across this example:
"Show that $L = \{a^m b^m c^n \mid m \leq n\}$ is not a CFL."
My problem is that I find a scenario that the above language can be CFL. I am showing my proof below:
Opponent picks $p$ We pick $z = a^p b^p c^{2p}$ where $|z| \geq p$ Opponent divides the string in $z = uvwxy$ where $|vwx| \leq p$ and $vx$ different than $\epsilon$ We have 5 scenarios to consider: -1- $vwx$ is all $a$'s -2- $vwx$ is all $b$'s -3- $vwx$ is all $c$'s -4- $vwx$ is a combination of $a$'s and $b$'s -5- $vwx$ is a combination of $b$'s and $c$'s
For scenario -3-, note that if we pump $v$ and $x$, we still get a word that is part of our given language (the number of $c$'s increases, and that does not affect $a$ or $b$). Am I doing something wrong, or is there a flaw in my logic? |
There is an interesting discussion to be had about the possible efficiency of wind turbines which presents opportunities for the instructor to look at several different kinds of computations. The discussion ends up with an optimization problem which could be approached by the classical methods of calculus – if the students have that available to them – or could be approximated simply by drawing a graph of one critical function.
We start by asking: how much energy is available to a wind turbine? Two key parameters govern this: the area \( S\) swept out by the turbine blades, and the speed \(v\) of the incoming wind – the turbine should rotate so that the plane of the blades is perpendicular to the wind. If \(\rho\approx 1.2 \text{kg}/\text{m}^3\) denotes the density of air, then in one second a cylinder of air of cross-section \(S\) and area \(v\) approaches the machine. The mass of this amount of air is \(\rho S v\) and the kinetic energy it contains is half the mass times the square of the speed, that is
\[ {\textstyle \frac12} \rho S v^3. \]
This is the energy per unit second, i.e. the
power (in watts), theoretically available in the wind incident on the turbine.
That is the first “theory” part of the lesson and one ought now do to some specific calculations. For example, a modern wind turbine might have a blade length of 30 meters; what is the theoretically available power at a wind speed of 10 meters per second? Using the formula \(S = \pi r^2 \) for the area of a circle, \(r \) being 30 meters in this example, the formula above gives
\[ 0.5 \times 1.2 \times \pi \times 30^2 \times 10^3 \approx 1.8 \times 10^6 W, \]
or nearly two megawatts. One should work things out for a few other examples.
Now comes the second theoretical kicker, though: it is impossible to extract
all that kinetic energy. Why? Remember that kinetic energy is the energy of air (or something) in motion; if you extract all its kinetic energy, it will stop. So to extract all the kinetic energy from the incoming wind we’d have to imagine the air just stopping when it got to the turbine and piling up without going anywhere. This is clearly ridiculous.
To get a better theoretical picture, we have to imagine what happens to a column of air as it moves through the turbine. As the air moves through the turbine it must slow down some (that is what extracting kinetic energy means, even if – as we have seen – we can’t slow it down to zero!) As it slows down it must spread out, so that there is the same total volume of moving air. (Remember, the volume of air moving per second is the area times the speed – if the speed decreases, the area must increase.) Therefore we get a picture like this: the wind arrives at a high speed \(v_1\) and departs at a slower speed \(v_2\). It passes through the turbine at a speed \(v\) which is between \(v_1\) and \(v_2\); in fact, we will assume that
\[ v = {\textstyle\frac12} (v_1+v_2). \]
(Though we treat this as an assumption, it is in fact a consequence of the physical laws of conservation of energy and momentum; students with sufficient background in physics would be able to follow the proof, which can be found in many places.) If \(V=Sv\) is the volume flow rate then the kinetic energies of the incoming and outgoing air are \(\frac12 Vv_1^2 \) and \(\frac12 Vv_2^2\) respectively; the difference, which is the energy available at the turbine, is (remembering the definition of \(V\))
\[ {\textstyle\frac12} S\left(\frac{v_1+v_2}{2}\right)(v_1^2-v_2^2) = {\textstyle\frac14}Sv_1^3 (1+\lambda-\lambda^2-\lambda^3), \]
where \(\lambda=v_2/v_1 \in (0,1)\). In this equation, \(S\) and \(v_1\) are given quantities, but the ratio \(\lambda\) can be changed by adjusting the blade pitches and so on. It therefore is reasonable to ask what choice of \(\lambda\) will maximize the energy output. Using either elementary calculus or graph-plotting estimates one discovers that the maximum value of \(1+\lambda-\lambda^2-\lambda^3\) occurs when \(\lambda=\frac13\); at this point
\[1+\lambda-\lambda^2-\lambda^3 = \frac{32}{27}\]
and the maximum power available at the turbine is \( {\displaystyle\frac{16}{27}} \approx 59\% \) of the naive estimate \({\frac12}S v_1^3\). This is called the
Betz limit on turbine efficiency. Thus in the example we used for calculation above, only about 1 megawatt of the theoretical 1.8 megawatts can actually be extracted by the turbine and turned into useful energy. |
Which kinds of fields of mathematics do I have to know about in order to understand the Riemann hypothesis millenium prize problem?
Just to understand the$^\dagger$
statement of the problem, you would have to be familiar with complex analysis and analytic number theory. The $\zeta$ function itself is an analytic object from number theory and to understand its significance (just on the surface!) you would have to study it in these realms. Of course it is also a function on $\Bbb C$ after analytic continuation - attained using a functional equation - with a simple pole at $1$, and understanding what this means and how to manipulate the function deftly will mean studying complex analysis.
$^\dagger$I refer to the statement that $\zeta(s)$ has all nontrivial zeros on the critical line. There are actually a lot of equivalent statements that require very little knowledge of complex analysis (you'll still need to pick up a few definitions of arithmetic functions from analytic NT for many of them, these aren't too hard). You can find a lot of equivalences listed here for example.
Beyond that, to understand the modern
approaches to RH and related or generalized conjectures and all of the theory there is surrounding this creature, you must go much further in algebraic number theory at the very least, and travel to many other worlds like modular forms, differential geometry, quantum theory and random matrices, etc. - basically at least a basic knowledge of most advanced subjects in analysis, algebra and geometry, and then especially deeply in pertinent areas.
I used to try to explain the complex analysis part. I have given up on that. See http://en.wikipedia.org/wiki/Prime-counting_function for details...
The short version is that the prime counting function $\pi(x)$ is approximated pretty well by $$ \frac{x}{\log x}. $$ It is known that a better approximation is given by $$ \mbox{li} \, (x). $$
The Riemann hypothesis is equivalent to a specific form of the statement that $ \mbox{li} \, (x) $ is a
see http://en.wikipedia.org/wiki/Prime-counting_function#The_Riemann_hypothesis and TABLE really good approximation,
Definitely, the number theory. You can find more connections to other fields in "The Millenium Problems" by Keith Devlin.
The Riemann Hypothesis is just a conjecture about the zeros of a function. Understanding this is simple.
However, understanding why this is important and what it means to number theory is far from simple. |
Let $X$ be an uncountable set. We define the cocountable topology $\tau$ as the set of all subsets $U\subseteq X$ such that either $U=\emptyset$ or $X\setminus U$ is countable.
I am interested in the conditions under which a set $E\subseteq X$ will be dense in $X$, and I honestly don't know where to start.
I have already shown that $(X,\tau)$ defines a topological space. I know that by definition, a set $E$ is dense in $X$ with respect to $\tau$ if for every $x\in X$ and every open set $U \subseteq X$ with $x\in U$, there exists an element $y\in E$ such that $y\in U$, or equivalently that $E$ is dense in $X$ with respect to $\tau$ if for every nonempty open set $U\subseteq X$, $E\cap U \neq \emptyset$.
I would like to see a complete proof if possible.
EDIT: I first want to show that: If $E\subseteq X$ is dense, then $X\setminus E$ is countable.
I attempted to prove this by the contrapositive (i.e. show that if $X\setminus E$ is uncountable, then $E$ is not dense in $X$). Suppose that $X\setminus E$ is uncountable. Then, $X \setminus E$ is open with respect to $\tau$, and so by definition $X\setminus (X\setminus E)=E$ is closed in $X$ with respect to $\tau$. Then, $E=\overline {E}$, where $\overline {E}$ denotes the closure of $E$. Since $E$ is countable and $X$ is uncountable, $E\neq X$. I am struggling to jump from here to the fact that $E$ is not dense in $X$. |
I want to find out if the following integrals converge and if possible, find their values.
$$(a) \int_{1}^{2} \frac{\mathrm dx}{\log x}$$
$$(b) \int_{0}^{+\infty} \left| \frac{\sin x}{x} \right| \mathrm dx$$
For $(a)$ I have $$\int_{1}^{2} \frac{1}{\log x}\mathrm dx = \operatorname{li}\left(x\right) + C$$ which is the logarithmic integral function.
An online function calculator gave me the antiderivative $-\operatorname{\Gamma}\left(0,-\ln\left(x\right)\right)$ and apparently it diverges. I don't understand what is meant with this antiderivative here: How does one get that, and why can the value of this integral not be calculated?
Regarding $(b)$ I used an online calculator as well, which gave me
$$\int_{0}^{+\infty} \left| \frac{\sin x}{x} \right| \mathrm dx= -\dfrac{\mathrm{i}\operatorname{\Gamma}\left(0,\mathrm{i}x\right)-\mathrm{i}\operatorname{\Gamma}\left(0,-\mathrm{i}x\right)}{2}$$ and the value ${\pi}/{2}$, but again I don't know if that is correct and how to get that antiderivative.
Can somebody maybe explain that? |
Let ${(I_t)}_{t\geq 0}$ be a stochastic integral defined by $$ I_t=\int_{0}^{t}\theta_sdW_t, $$ where $W$ is a standard Brownian motion defined on $(\Omega,\mathcal{F},{(\mathcal{F}_t)}_{t\geq 0},\mathbb{P})$ and $\theta$ a stochastic process adapted to $\mathcal{F}_t$ satisfying the follows condition of integrability $$ E\left(\int_{0}^{t}\theta_s^2 ds\right)<\infty\;\;\ \forall t> 0. $$
We define the first passage time at $a$ for Brownian motion $W$ by the following random variable $$ \tau_a = \inf\{t\geq 0,W_t\geq a\}, $$ where $a>0$.
It is possible to show that $\tau_a$ is a stopping time. Moreover, By virtue of the reflection principle, we know that the following process
\begin{equation*} Z_t = \begin{cases} W_t \qquad & if \qquad 0 \leq t \leq \tau_a \\ 2a-W_t \qquad & if \qquad t > \tau_a \end{cases} \end{equation*}
also follows a standard Brownian motion under $\mathbb{P}$.
My question is as follows : Is it possible to rewrite the process $I$ in relation to the process $Z$?
I would like your opinion on this issue, thank you in advance. |
I am facing hard time understanding min-entropy.
Fix $v \in \{0,1\}^{n/2}$, let $F\colon \{0,1\}^n \to \{0,1\}^{n/2}$ be chosen randomly, and let $X_v$ be a string chosen uniformly at random among $F^{(-1)}(v)$ (or a special string $\epsilon$ if $F^{(-1)}(v) = \emptyset$). We want to compute the probability that $H_\infty(X_v) \leq n/4$ (over the choice of $F$), where $H_\infty(X_v)=-\log(\max_{x}\{\Pr[X_v=x]\})$ is the
min-entropy of $X_v$.
It's not obvious what approach is needed to solve such things! It would be helpful if one could explain where to start... |
I'm trying to prove the in-completeness of $\mathcal{C}[0,1]$ with the $L_{1}$ norm. To do so, I've considered the function $f_{n}(x) = |\sin{(2\pi/x)}|$ for $x \in [1/(n+1), 1/n]$ and $0$ elsewhere, as well as the theorem that completeness is equivalent to every absolutely summable series in a space being summable.
This seems intuitively to me like it should work as a counter-example, as simply plotting the sine waves and considering the intervals I don't see how the series of functions could sum to some continuous function on $[0,1]$ (
it looks like it would be some step function to me), however it is clear that the series of functions is absolutely summable with $L_{1}$ simply by bounding $|\sin{(2\pi/x)}|$ by $1$ in the integral. My problem is, I don't know how to rigorously show that the series of functions is not summable to a continuous function to complete my proof. I've taken the definition of summability as well as the definition of continuity, and tried to formulate some contradiction by supposing that $\sum_{n=1}^{\infty} f_{n} = f$ for some $f \in \mathcal{C}[0,1]$, but I'm not sure where the actual argument involving the function should come into place. I'm starting to think that this approach with a contradiction may not be working altogether.
How can I formalize this part of my argument to complete the proof? |
This question already has an answer here:
I'm studying algebraic topology and got stuck.
a. $X_n\in \mathbb{R}^3$ is the union of $n$ distinct lines through the origin. Find $\pi_1(\mathbb{R}^3-X_n)$ for each $n$.
b. Let $X$ be the sum of two tori $S_1\times S_1$ by identifying a circle $S_1\times x_0$ in a torus with $S_1\times x_0$ of the other torus. Find $\pi_1(X)$.
For (a), $\pi_1(\mathbb{R}^3-X_1)=\mathbb{Z}.$ But I cannot compute $n\ge 2$ cases.. for (b), I guess the group is $<a,b,c|ab=ba,bc=cb>$ from Van Kampen. Is it correct? |
I tried to solve it by Wilson theorem and I got 41 but answer is 1927.
closed as off-topic by Shaun, Claude Leibovici, JMP, user26857, user91500 Aug 28 '17 at 9:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
" This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Claude Leibovici, JMP, user26857, user91500
You've probably found $$\frac{50!}{47}\equiv 41\pmod{47}$$ but you were looking for $$50!\equiv 41\cdot 47\equiv 1927\pmod{47^2}$$ |
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If cos3AcosA\dfrac {\cos {3A}}{\cos {A}}cosAcos3A =12= \dfrac {1}{2}=21, then the value of sin3AsinA\dfrac {\sin {3A}}{\sin {A}}sinAsin3A can be expressed as ab\dfrac {a}{b}ba, where aaa and bbb are coprime positive integers, find a+ba+ba+b.
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Here's a method that gets close. Do a continued fraction expansion of $\frac{m}{n}$:
$$\frac{m}{n} = \frac{1}{a + \frac{1}{b + \epsilon}} \approx \frac{1}{a + \frac{1}{b}} = \frac{1}{a} - \frac{1}{a(1 + ba)}$$
If $\epsilon$ is small, then this is probably very close to optimal if not optimal in most cases.
The method fails when the truncated continued fraction approximation isn't very good. Take $\frac{15403}{26685}$ (a rational approximation to the Euler-Mascheroni constant) as an example. The continued fraction approximation is:
$$\frac{15403}{26685} = \frac{1}{1 + \frac{1}{1 + \epsilon}}$$
Which suggests:
$$\frac{15403}{26685} \approx \frac{1}{1} - \frac{1}{2}$$
But clearly $\frac{1}{2} + \frac{1}{14}$ is closer.
I would wager that the worst case for any method is the conjugate golden ratio $\phi' = \frac{\sqrt{5} - 1}{2}$. This isn't rational (in fact it's the "most" irrational number in a technical sense; find its continued fraction expansion if you're curious), but you can get arbitrarily close by choosing $m$ and $n$ to be consecutive Fibonacci numbers, e.g. $\frac{4181}{6765}$. The best approximation I could find is $\frac{1}{2} + \frac{1}{9}$. |
Let $f:X\rightarrow X$ a bijective map between topological spaces (the same space X). A priori not known to be continuous. If we know $f\circ f=id$ does it mean that $f$ has to be continuous map and hence a homeomorphism ?
No. For instance, consider the topological space $X = \{a, b\}$ with the topology $$ \tau = \{\varnothing, \{a\}, X\} $$ and the bijection $f:X\to X$ with $f(a) = b, f(b) = a$.
Then $f^{-1}(\{a\}) = \{b\}$ is not open, so $f$ is not continuous.
For a more "interesting" example (i.e. closer to something you may reasonably encounter more often), consider $X = \Bbb R$ with the standard topology, and $$ f(x) = \cases{\frac1x & if $x<0$\\x&otherwise} $$ which is a discontinuous, self-inverse bijection.
If $X$ is the Sorgenfrey line $\Bbb S$, i.e. $\Bbb R$ in the topology generated by the sets $[a,b)$ (aka as the lower limit topology) and $f(x)=-x$ then $f$ is self inverse but continuous at no point, while $X$ is hereditarily normal etc., so quite nice. |
I've only found a recursive algorithm of the extended Euclidean algorithm. I'd like to know how to use it by hand. Any idea?
Perhaps the easiest way to do it by hand is in analogy to Gaussian elimination or triangularization, except that, since the coefficient ring is not a field, one has to use the division / Euclidean algorithm to iteratively descrease the coefficients till zero. In order to compute both $\rm\,gcd(a,b)\,$
and its Bezout linear representation $\rm\,j\,a+k\,b,\,$ we keep track of such linear representations for each remainder in the Euclidean algorithm, starting with the trivial representation of the gcd arguments, e.g. $\rm\: a = 1\cdot a + 0\cdot b.\:$ In matrix terms, this is achieved by augmenting (appending) an identity matrix that accumulates the effect of the elementary row operations. Below is an example that computes the Bezout representation for $\rm\:gcd(80,62) = 2,\ $ i.e. $\ 7\cdot 80\: -\: 9\cdot 62\ =\ 2\:.\:$ See this answer for a proof and for conceptual motivation of the ideas behind the algorithm (see the Remark below if you are not familiar with row operations from linear algebra).
For example, to solve m x + n y = gcd(m,n) one begins withtwo rows [m 1 0], [n 0 1], representing the twoequations m = 1m + 0n, n = 0m + 1n. Then one executesthe Euclidean algorithm on the numbers in the first column,doing the same operations in parallel on the other columns,Here is an example: d = x(80) + y(62) proceeds as: in equation form | in row form ---------------------+------------ 80 = 1(80) + 0(62) | 80 1 0 62 = 0(80) + 1(62) | 62 0 1 row1 - row2 -> 18 = 1(80) - 1(62) | 18 1 -1 row2 - 3 row3 -> 8 = -3(80) + 4(62) | 8 -3 4 row3 - 2 row4 -> 2 = 7(80) - 9(62) | 2 7 -9 row4 - 4 row5 -> 0 = -31(80) +40(62) | 0 -31 40The row operations above are those resulting from applyingthe Euclidean algorithm to the numbers in the first column, row1 row2 row3 row4 row5namely: 80, 62, 18, 8, 2 = Euclidean remainder sequence | |for example 62-3(18) = 8, the 2nd step in Euclidean algorithmbecomes: row2 -3 row3 = row4 when extended to all columns.
In effect we have row-reduced the first two rows to the last two.The matrix effecting the reduction is in the bottom right corner.It starts as 1, and is multiplied by each elementary row operation, hence it accumulates the product of all the row operations, namely:
$$ \left[ \begin{array}{ccc} 7 & -9\\ -31 & 40\end{array}\right ] \left[ \begin{array}{ccc} 80 & 1 & 0\\ 62 & 0 & 1\end{array}\right ] \ =\ \left[ \begin{array}{ccc} 2\ & \ \ \ 7\ & -9\\ 0\ & -31\ & 40\end{array}\right ] \qquad\qquad\qquad\qquad\qquad $$
Notice row 1 is the particular solution 2 = 7(80) - 9(62)Notice row 2 is the homogeneous solution 0 = -31(80) + 40(62),so the general solution is any linear combination of the two: n row1 + m row2 -> 2n = (7n-31m) 80 + (40m-9n) 62The same row/column reduction techniques tackle arbitrarysystems of linear Diophantine equations. Such techniquesgeneralize easily to similar coefficient rings possessing aEuclidean algorithm, e.g. polynomial rings F[x] over a field, Gaussian integers Z[i]. There are many analogous interestingmethods, e.g. search on keywords: Hermite / Smith normal form, invariant factors, lattice basis reduction, continued fractions,Farey fractions / mediants, Stern-Brocot tree / diatomic sequence.
Remark $ $ As an optimization, we can omit one of the Bezout coefficient columns (being derivable from the others). Then the calculations have a natural interpretation as modular fractions (though the "fractions" are multi-valued), e.g. follow the prior link.
Below I elaborate on the row operations to help readers unfamiliar with linear algebra.
Let $\,r_i\,$ be the Euclidean remainder sequence. Above $\, r_1,r_2,r_3\ldots = 80,62,18\ldots$ Given linear combinations $\,r_j = a_j m + b_j n\,$ for $\,r_{i-1}\,$ and $\,r_i\,$ we can calculate a linear combination for $\,r_{i+1} := r_{i-1}\bmod r_i = r_{i-1} - q_i r_i\,$ by substituting said combinations for $\,r_{i-1}\,$ and $\,r_i,\,$ i.e.
$$\begin{align} r_{i+1}\, &=\, \overbrace{a_{i-1} m + a_{i-1}n}^{\Large r_{i-1}}\, -\, q_i \overbrace{(a_i m + b_i n)}^{\Large r_i}\\[.3em] {\rm i.e.}\quad \underbrace{r_{i-1} - q_i r_i}_{\Large r_{i+1}}\, &=\, (\underbrace{a_{i-1}-q_i a_i}_{\Large a_{i+1}})\, m\, +\, (\underbrace{b_{i-1} - q_i b_i}_{\Large b_{i+1}})\, n \end{align}$$
Thus the $\,a_i,b_i\,$ satisfy the same recurrence as the remainders $\,r_i,\,$ viz. $\,f_{i+1} = f_{i-1}-q_i f_i.\,$ This implies that we can carry out the recurrence in parallel on row vectors $\,[r_i,a_i,b_i]$ representing the equation $\, r_i = a_i m + b_i n\,$ as follows
$$\begin{align} [r_{i+1},a_{i+1},b_{i+1}]\, &=\, [r_{i-1},a_{i-1},b_{i-1}] - q_i [r_i,a_i,b_i]\\ &=\, [r_{i-1},a_{i-1},b_{i-1}] - [q_i r_i,q_i a_i, q_i b_i]\\ &=\, [r_{i-1}-q_i r_i,\ a_{i-1}-q_i a_i,\ b_{i-1}-b_i r_i] \end{align}$$
which written in the tabular format employed far above becomes
$$\begin{array}{ccc} &r_{i-1}& a_{i-1} & b_{i-1}\\ &r_i& a_i &b_i\\ \rightarrow\ & \underbrace{r_{i-1}\!-q_i r_i}_{\Large r_{i+1}} &\underbrace{a_{i-1}\!-q_i a_i}_{\Large a_{i+1}}& \underbrace{b_{i-1}-q_i b_i}_{\Large b_{i+1}} \end{array}$$
Thus the
extended Euclidean step is: compute the quotient $\,q_i = \lfloor r_{i-1}/r_i\rfloor$ then multiply row $i$ by $q_i$ and subtract it from row $i\!-\!1.$ Said componentwise: in each column $\,r,a,b,\,$ multiply the $i$'th entry by $q_i$ then subtract it from the $i\!-\!1$'th entry, yielding the $i\!+\!1$'th entry. If we ignore the 2nd and 3rd columns $\,a_i,b_i$ then this is the usual Euclidean algorithm. The above extends this algorithm to simultaneously compute the representation of each remainder as a linear combination of $\,m,n,\,$ starting from the obvious initial representations $\,m = 1(m)+0(n),\,$ and $\,n = 0(m)+1(n).\,$
This is more a comment on the method explained by Bill Dubuque then a proper answer in itself, but I think there is a remark so obvious that I don’t understand that it is hardly ever made in texts discussing the extended Euclidean algorithm. This is the observation that you can save yourself half of the work by computing only
one of the Bezout coefficients. In other words, instead of recording for every new remainder $r_i$ a pair of coefficients $k_i,l_i$ so that $r_i=k_ia+l_ib$, you need to record only $k_i$ such that $r_i\equiv k_ia\pmod b$. Once you will have found $d=\gcd(a,b)$ and $k$ such that $d\equiv ka\pmod b$, you can then simply put $l=(d-ka)/b$ to get the other Bezout coefficient. This simplification is possible because the relation that gives the next pair of intermediate coefficients is perfectly independent for the two coefficients: say you have$$\begin{aligned} r_i&=k_ia+l_ib\\ r_{i+1}&=k_{i+1}a+l_{i+1}b\end{aligned}$$and Euclidean division gives $r_i=qr_{i+1}+r_{i+2}$, then in order to get$$ r_{i+2}=k_{i+2}a+l_{i+2}b$$one can take $k_{i+2}=k_i-qk_{i+1}$ and $l_{i+2}=l_i-ql_{i+1}$, where the equation for $k_{i+2}$ does not need $l_i$ or $l_{i+1}$, so you can just forget about the $l$'s. In matrix form, the passage is from$$ \begin{pmatrix} r_i&k_i&l_i\\ r_{i+1}&k_{i+1}&l_{i+1}\end{pmatrix} \quad\text{to}\quad \begin{pmatrix} r_{i+2}&k_{i+2}&l_{i+2}\\ r_{i+1}&k_{i+1}&l_{i+1}\end{pmatrix}$$by subtracting the second row $q$ times from the first, and it is clear that the last two columns are independent, and one might as well just keep the $r$'s and the $k$'s, passing from$$ \begin{pmatrix} r_i&k_i\\ r_{i+1}&k_{i+1}\end{pmatrix} \quad\text{to}\quad \begin{pmatrix} r_{i+2}&k_{i+2}\\ r_{i+1}&k_{i+1}\end{pmatrix}$$instead.
A very minor drawback is that the relation $r_i=k_ia+l_ib$ that should hold for every row is maybe a wee bit easier to check by inspection than $r_i\equiv k_ia\pmod b$, so that computational errors could slip in a bit more easily. But really, I think that with some practice this method is just as safe and faster than computing both coefficients. Certainly when programming this on a computer there is no reason at all to keep track of both coefficients.
A final bonus it that in many cases where you apply the extended Euclidean algorithm you are only interested in one of the Bezout coefficients in the first place, which saves you the final step of computing the other one. One example is computing inverse modulo a prime number $p$: if you take $b=p$, and $a$ is not divisible by it, then you know beforehand that you will find $d=1$, and the coefficient $k$ such that $d\equiv ka\pmod p$ is just the inverse of $a$ modulo $p$ that you were after.
The way to do this is due to Blankinship "A New Version of the Euclidean Algorithm", AMM 70:7 (Sep 1963), 742-745. Say we want $a x + b y = \gcd(a, b)$, for simplicity with positive $a$, $b$ with $a > b$. Set up auxiliary vectors $(x_1, x_2, x_3)$, $(y_1, y_2, y_3)$ and $(t_1, t_2, t_3)$ and keep them such that we always have $x_1 a + x_2 b = x_3$, $y_1 a + y_2 b = y_3$, $t_1 a + t_2 b = t_3$ throughout. The algorithm itself is:
(x1, x2, x3) := (1, 0, a)(y1, y2, y3) := (0, 1, b)while y3 <> 0 do q := floor(x3 / y3) (t1, t2, t3) := (x1, x2, x3) - q * (y1, y2, y3) (x1, x2, x3) := (y1, y2, y3) (y1, y2, y3) := (t1, t2, t3)
At the end, $x_1 a + x_2 b = x3 = \gcd(a, b)$. It is seen that $x_3$, $y_3$ do as the classic Euclidean algorithm, and easily checked that the invariant mentioned is kept all the time.
One can do away with $x_2$, $y_2$, $t_2$ and recover $x_2$ at the end as $(x_3 - x_1 a) / b$.
Just to complement the other answers, there's an alternative form of the extended Euclidean algorithm that requires no backtracking, and which you might find easier to understand and apply. Here's how to solve your problem* using it: $\newcommand{\x}{\phantom} \newcommand{\r}{\color{red}} \newcommand{\g}{\color{green}} \newcommand{\b}{\color{blue}}$
*) …from a duplicate question that I originally wrote this answer for.
$$\begin{aligned} \g{ 0} \cdot 19 + \r{\x+1} \cdot 29 &= \b{ 29} && (1) \\ \g{ 1} \cdot 19 + \r{\x+0} \cdot 29 &= \b{ 19} && (2) \\ \g{-1} \cdot 19 + \r{\x+1} \cdot 29 &= \b{ 10} && (3) = (1) - (2) \\ \g{ 2} \cdot 19 + \r{ -1} \cdot 29 &= \b{\x09} && (4) = (2) - (3) \\ \g{-3} \cdot 19 + \r{\x+2} \cdot 29 &= \b{\x01} && (5) = (3) - (4) \end{aligned}$$
…and now you have your solution: $\g{-3} \cdot 19 \equiv \b{1} \pmod{29}$, so the inverse of $19$ modulo $29$ is $-3$ (or $29 - 3 = 27$, if you prefer a non-negative solution).
In effect, what we're doing is trying to find a solution to the linear equation $\g x \cdot 19 + \r k \cdot 29 = \b r$ with the smallest possible $\b r > 0$. We do this by starting with the two trivial solutions $(1)$ and $(2)$ above, and then generate new solutions with a smaller and smaller $\b r$ by always subtracting both sides of the last solution from the one before it
as many times as needed to get a smaller $\b r$ than we have so far. (In your example, that's only once each time, but I'll show another example below where we that's not the case.)
Eventually, we'll either find a solution with $\b r = 1$, in which case the corresponding $\g x$ coefficient is the inverse we want,
or we'll end up with $\b r = 0$, in which case the previous solution's $\b r > 1$ is the greatest common divisor of the number we're trying to invert and the modulus, and thus no inverse exists. (Of course, we could also just keep going until $\b r = 0$ in any case, and then check the previous line to see if $\b r$ there equals $1$ or not, but that would be extra work we can easily avoid.)
Also, it's worth noting that we're not actually using the $\r k$ coefficients for anything, so if all you're interested in is finding the modular inverse (and/or the GCD), you don't actually have to calculate those. But showing them makes it clearer
why the algorithm works. (Also, if you do calculate $\r k$, it's easy to verify that you didn't make any mistakes just by checking that the last equation with $\b r = 1$ really holds.)
Anyway, here's a couple more worked examples to illustrate some cases that your example doesn't. To start with, let's try to find the inverse of $13$ modulo $29$:
$$\begin{aligned} \g{ 0} \cdot 13 + \r{\x+1} \cdot 29 &= \b{ 29} && (1) \\ \g{ 1} \cdot 13 + \r{\x+0} \cdot 29 &= \b{ 13} && (2) \\ \g{-2} \cdot 13 + \r{\x+1} \cdot 29 &= \b{\x03} && (3) = (1) - 2 \cdot (2) \\ \g{ 9} \cdot 13 + \r{ -4} \cdot 29 &= \b{\x01} && (4) = (2) - 4 \cdot (3) \\ \end{aligned}$$
This time, we could (and needed to) subtract solution $(2)$ from $(1)$ twice, since $\lfloor 29 \mathbin/ 13 \rfloor = 2$. And, similarly, we could subtract $(3)$ from $(2)$ four times, since $\lfloor 13 \mathbin/ 3 \rfloor = 4$. And we can verify that $\g 9$ is indeed the inverse of $13$ modulo $29$ just by checking that $9 \cdot 13 - 4 \cdot 29$ indeed equals $1$.
Now let's try an example where the inverse does
not exist, like trying to find the inverse of $15$ modulo $27$:
$$\begin{aligned} \g{ 0} \cdot 15 + \r{\x+1} \cdot 27 &= \b{ 27} && (1) \\ \g{ 1} \cdot 15 + \r{\x+0} \cdot 27 &= \b{ 15} && (2) \\ \g{-1} \cdot 15 + \r{\x+1} \cdot 27 &= \b{ 12} && (3) = (1) - (2) \\ \g{ 2} \cdot 15 + \r{ -1} \cdot 27 &= \b{\x03} && (4) = (2) - (3) \\ \g{-9} \cdot 15 + \r{\x+5} \cdot 27 &= \b{\x00} && (5) = (3) - 4 \cdot (4) \\ \end{aligned}$$
Oops. The last solution, with $\b r = 0$, is quite useless to us (except for checking that we did the arithmetic right). From the previous solution $(4)$, however, we can read that $\gcd(15, 27) = 3$, and even that $\g x = 2$ is a solution to the "generalized inverse" equation $\g x \cdot 15 \equiv \gcd(15, 27) \pmod{27}$.
Ps. See also this answer I wrote on crypto.SE last year, explaining the same algorithm from a slightly different viewpoint. |
Recently the question If $\frac{d}{dx}$ is an operator, on what does it operate? was asked on mathoverflow. It seems that some users there objected to the question, apparently interpreting it as an elementary inquiry about what kind of thing is a differential operator, and on this interpretation, I would agree that the question would not be right for mathoverflow. And so the question was closed down (and then reopened, and then closed again….
sigh). (Update 12/6/12: it was opened again,and so I’ve now posted my answer over there.)
Meanwhile, I find the question to be more interesting than that, and I believe that the OP intends the question in the way I am interpreting it, namely, as a logic question, a question about the nature of mathematical reference, about the connection between our mathematical symbols and the abstract mathematical objects to which we take them to refer. And specifically, about the curious form of variable binding that expressions involving $dx$ seem to involve. So let me write here the answer that I had intended to post on mathoverflow:
————————-
To my way of thinking, this is a serious question, and I am not really satisfied by the other answers and comments, which seem to answer a different question than the one that I find interesting here.
The problem is this. We want to regard $\frac{d}{dx}$ as an operator in the abstract senses mentioned by several of the other comments and answers. In the most elementary situation, it operates on a functions of a single real variable, returning another such function, the derivative. And the same for $\frac{d}{dt}$.
The problem is that, described this way, the operators $\frac{d}{dx}$ and $\frac{d}{dt}$ seem to be the
same operator, namely, the operator that takes a function to its derivative, but nevertheless we cannot seem freely to substitute these symbols for one another in formal expressions. For example, if an instructor were to write $\frac{d}{dt}x^3=3x^2$, a student might object, “don’t you mean $\frac{d}{dx}$?” and the instructor would likely reply, “Oh, yes, excuse me, I meant $\frac{d}{dx}x^3=3x^2$. The other expression would have a different meaning.”
But if they are the same operator, why don’t the two expressions have the same meaning? Why can’t we freely substitute different names for this operator and get the same result? What is going on with the logic of reference here?
The situation is that the operator $\frac{d}{dx}$ seems to make sense only when applied to functions whose independent variable is described by the symbol “x”. But this collides with the idea that what the function is at bottom has nothing to do with the way we represent it, with the particular symbols that we might use to express which function is meant. That is, the function is the abstract object (whether interpreted in set theory or category theory or whatever foundational theory), and is not connected in any intimate way with the symbol “$x$”. Surely the functions $x\mapsto x^3$ and $t\mapsto t^3$, with the same domain and codomain, are simply different ways of describing exactly the same function. So why can’t we seem to substitute them for one another in the formal expressions?
The answer is that the syntactic use of $\frac{d}{dx}$ in a formal expression involves a kind of binding of the variable $x$.
Consider the issue of
collision of bound variables in first order logic: if $\varphi(x)$ is the assertion that $x$ is not maximal with respect to $\lt$, expressed by $\exists y\ x\lt y$, then $\varphi(y)$, the assertion that $y$ is not maximal, is not correctly described as the assertion $\exists y\ y\lt y$, which is what would be obtained by simply replacing the occurrence of $x$ in $\varphi(x)$ with the symbol $y$. For the intended meaning, we cannot simply syntactically replace the occurrence of $x$ with the symbol $y$, if that occurrence of $x$ falls under the scope of a quantifier.
Similarly, although the functions $x\mapsto x^3$ and $t\mapsto t^3$ are equal as functions of a real variable, we cannot simply syntactically substitute the expression $x^3$ for $t^3$ in $\frac{d}{dt}t^3$ to get $\frac{d}{dt}x^3$. One might even take the latter as a kind of ill-formed expression, without further explanation of how $x^3$ is to be taken as a function of $t$.
So the expression $\frac{d}{dx}$ causes a binding of the variable $x$, much like a quantifier might, and this prevents free substitution in just the way that collision does. But the case here is not quite the same as the way $x$ is a bound variable in $\int_0^1 x^3\ dx$, since $x$ remains free in $\frac{d}{dx}x^3$, but we would say that $\int_0^1 x^3\ dx$ has the same meaning as $\int_0^1 y^3\ dy$.
Of course, the issue evaporates if one uses a notation, such as the $\lambda$-calculus, which insists that one be completely explicit about which syntactic variables are to be regarded as the independent variables of a functional term, as in $\lambda x.x^3$, which means the function of the variable $x$ with value $x^3$. And this is how I take several of the other answers to the question, namely, that the use of the operator $\frac{d}{dx}$ indicates that one has previously indicated which of the arguments of the given function is to be regarded as $x$, and it is with respect to this argument that one is differentiating. In practice, this is almost always clear without much remark. For example, our use of $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ seems to manage very well in complex situations, sometimes with dozens of variables running around, without adopting the onerous formalism of the $\lambda$-calculus, even if that formalism is what these solutions are essentially really about.
Meanwhile, it is easy to make examples where one must be very specific about which variables are the independent variable and which are not, as Todd mentions in his comment to David’s answer. For example, cases like
$$\frac{d}{dx}\int_0^x(t^2+x^3)dt\qquad
\frac{d}{dt}\int_t^x(t^2+x^3)dt$$
are surely clarified for students by a discussion of the usage of variables in formal expressions and more specifically the issue of bound and free variables. |
I am trying to find an elementary proof that if $\phi$ is a linear map from $\mathbb{Z}^n\rightarrow \mathbb{Z}^m$ represented by an $m \times n$ matrix $A$, then the map is surjective iff the gcd of the determinants of all the $m\times m$ minors of $A$ is $1$.
I know that for there to be surjectivity between $\mathbb{Z}^n$ and $\mathbb{Z}^m$ $n$ must be greater than or equal to $ m$ and for there to even be $m \times m$ minors $n$ again must be greater than or equal to $ m$, so I just assume this throughout.
I sort of have one direction $\Leftarrow$
i) Greatest Common Divisor =1 implies surjectivity: First observe that the if $ | \mathbb{Z}^m/ Im(M)| < \infty$ then $|\det M| = | \mathbb{Z}^m/ Im(M) |$ otherwise $\det(M) = 0$ where $M$ is an $m\times m$ matrix. We can consider the $n$ columns of $A$ as column vectors $v_1, v_2, \ldots, v_n$. These $n$ column vectors live in $\mathbb{Z}^m$. Let $S'' = \{ v_i\}$ and then let $S'$ be subsets of $S''$ of cardinality $m$ and lastly let $S$ be the elements of $S'$ such that when the $m$ $v_i$ vectors are considered as $m\times m$ matrices, the determinant is not zero, thus $S$ consists of all $m\times m$ minors of $A$ with non-zero determinant (we ignore zeroes since they do not affect gcd). For each $s\in S$ define a map $i_s: \mathbb{Z}^m \rightarrow \mathbb{Z}^n$ that maps the standard basis of $\mathbb{Z}^m$ to the basis elements $e_k \mathbb{Z}^n$ such that $v_k \in s$. That is, $\phi \circ i_s$ gives the matrix created by the column vectors of $s$. Let $\Lambda$ be the lattice Im$\phi \supset \sum_{s\in S}$ Im $\phi\circ i_s =\sum_{s \in S} \Lambda_s$. Thus $\forall s \in S$, $\Lambda_s \subset \Lambda \subset \mathbb{Z}^m$. Thinking in terms of group theory, we have that $\Lambda$ is a subgroup of $\mathbb{Z}^m$ and all the $\Lambda_s$ are subgroups of $\Lambda$. Thus by Lagrange's Theorem, we have $|\mathbb{Z}/\Lambda| \Big\vert |\mathbb{Z}^m/\Lambda_s|$ Since $|\mathbb{Z}^m/\Lambda_s|$ are the determininants of the $m\times m$ minors and the definition of the common divisor of several integers is the greatest positive integer dividing all of them. Thus by hypothesis $|\mathbb{Z}/\Lambda| \leq 1$ and so $|\mathbb{Z}/\Lambda| =1$ and we have that Im$A=\Lambda = \mathbb{Z}^m$ so the map is surjective.
I was hoping to get a more elementary proof that doesn't rely on the observation that the if $ | \mathbb{Z}^m/ Im(M)| < \infty$ then $|\det M| = | \mathbb{Z}^m/ Im(M) |$ otherwise $\det(M) = 0$ where $M$ is an $m\times m$ matrix or normal forms.
Thanks! |
I tried using a very specific counterexample here where I select a surjective function for which the compositions are equal but the functions within are not.
This is probably off-base, but it's what I've got so far.
Assume $f \circ g = f \circ h$.
Consider the surjective function $f:\mathbb{R} \rightarrow \mathbb{R}$ given by $f(x) = x*sin(x)$.
Should I prove this is surjective before proceeding?
Suppose for the sake of contradiction that $g \neq h$ given by
$g(x) = 0$ and $h(x) = 2\pi$.
Can I choose these constant functions? Do I need to define domains and codomains?
$(f \circ g)(x) = f(g(x)) = f(0) = 0 * sin(0) = 0$
$(f \circ h)(x) = f(h(x)) = f(2\pi) = 2\pi * sin(2\pi) = 0$
Observe that $f \circ g = f \circ h$ $\land$ $x_1 \neq x_2$.
Thus we have given a counterexample to disprove the statement. Thus surjectivity of $f$ is not a sufficient condition for the statement to be true.
I understand the proof completely now and understand I have it correct, thank you for your responses. |
What is the largest (by area) rectangle that can be inscribed in a circular sector wiith radius $r$ and central angle $\alpha$?
I think I have an answer to this question, but would like it confirmed.
It is well known that the largest rectangle in a semi-circle (where the central angle $\alpha = \pi$) has the dimensions $x= \sqrt{2}r$ and $y=\frac{\sqrt{2}}{2}r$. For sector angles larger than $\pi$, i.e for $\pi \lt \alpha \lt 2\pi$, this remains the largest rectangle, as no rectangle can bend around or encompass the circle's center (see figure below).
For sectors smaller than a semi-circle, i.e. for $0 \lt \alpha \lt \pi$ we have the following situation:
After setting up equations for x and y (dependent on $\theta$ and $\alpha$), setting area $A = xy$ and differentiating, I find that the maximum area rectangle occurs when $$\theta = \frac{\alpha}{4}$$ Can anyone confirm this is correct?
As a test I set $\alpha = \pi$ (i.e the semi-circle) and this gives $\theta = \frac{\pi}{4}$ which is correct.
An interesting sidenote (if the formula is correct) is that when $\alpha = \frac{\pi}{3}$ the maximum area rectangle is a square, for the first time since $\alpha = 2 \pi$.
A follow-on question: Is the largest inscribed rectangle also the largest possible rectangle? I.e. could there be larger rectangles where every corner does not touch the sector? |
Doesn't have to be "exactly in line with the first". The impossible (probability zero) thing would be for initially-linearly-polarized light to subsequently pass through a (plane) polarizing filter that's
orthogonal to the initial plane. Otherwise, if the second polaroid's rotated through an angle $\theta$, photons have the usual $\cos^2\theta$ probability of passing through (being "accepted").
That probability's just $\left|\left\lt\psi|\phi\right\gt\right|^2$, generally speaking. And that'll be zero only when the two (linearly polarized) states are orthogonal.
>>Edit<< attempting to address @AndrewMarzban 's comment below "haven't taken quantum mechanics". However, he apparaently has a mechanical engineering degree.
Okay, so a state, e.g., https://en.wikipedia.org/wiki/Quantum_state ,typically labelled $\left|a\right\gt$ or $\left|b\right>$ or $\left|\psi\right>$, etc, is simply a
"complete description" of the system/object/whatever under consideration, in this case a "photon". (For our purposes here, "photon" is okay, but that's ultimately an oversimplification of the quantum nature of the E&M field.)
So what does
"complete description" mean??? That's still somewhat of a mystery, but an intuitive operational definition might be a reproducible laboratory procedure for the preparation of the system/object/whatever under consideration. In this case that procedure could be starting with any old source of randomly polarized light, and then passing it through your initial polaroid filter.
In addition to "preparations", there are "tests". A prepared system (in this case initially-polarized photon) is subjected to a test apparatus/procedure, which either "accepts" or "rejects" it. In this case, photons from your first polaroid impinge on the second polaroid, and either pass through it or don't.
But all this doesn't get us much of anywhere as far as mathematical calculations, predicting probabilities, etc, is concerned. For that kind of purpose, the theory lets us associate a mathematical_function/"state" with that preparation procedure, which is typically a complex function (of space and time coordinates, and frequently of other stuff), our so-called $\left|\psi\right>$. And then $\left<\psi\right|$ is the notation for its complex conjugate (though this is again somewhat of an oversimplification -- slightly more accurately, but without discussion, $\left|\psi\right>$ is a function in the "state space of the system", and $\left<\psi\right|$ is its corresponding functional). And note that states are normalized so that $\left|\left<\psi|\psi\right>\right|^2=1$.
And then, suppose you have two different preparations/states, $\left|\psi\right>$ and $\left|\phi\right>$, and you subject the $\left|\psi\right>$-prepared photon to a $\left|\phi\right>$-accepting-test. Then the whole theoretical/mathematical machinery has been constructed so that the probability that the $\left|\phi\right>$-test will accept the $\left|\psi\right>$-prepared photon is (drum roll...) $\left|\left<\phi|\psi\right>\right|^2$. And as far as your photons are concerned, if the test-polaroid is rotated by $\theta$ relative to the preparation-polaroid, that'll be $\cos^2\theta$.
So now let's re-visit your whole question, which is, from above,...
"So how is it that light that has been passed though a polarizing filter and now composed of waves parallel to a single plane able to pass through another filter that is not exactly in line with the first?"
But let's just slightly rephrase that, using our preceding terminology, as follows: How is it that photons prepared by a $\left|\psi\right>$-preparation procedure can subsequently be accepted by a
different $\left|\phi\right>$-test procedure? Yeah, well, one way or another, that's just how it is, and the probability for that to happen is just $\left|\left<\phi|\psi\right>\right|^2$ as previously stated. And for linear photon polarization, that'll be zero if and only if the two polarizing filters are rotated by $90^o$ relative to each other.
That leaves one remaining question: how do you develop the appropriate mathematical expressions representing our polarized photon states $\left|\phi\right>$ and $\left|\psi\right>$? A really, really, really (did I say
really?) good discussion of this is the 39-page Chapter 1 (aptly titled "Photon Polarization") of Gordon Baym's almost-classic "Lectures on Quantum Mechanics", https://books.google.com/books?id=1125sVZ2_GcC&pg=PA1 And I mention "39-page" to point out that such a discussion is beyond the scope of a stackexchange discussion. If interested, try reading it, and followup with any specific questions. It's basically a first-year graduate text, but the Chapter 1 discussion is pretty much ab initio, assuming only prerequisite knowledge of some basic undergraduate E&M, which I think you have. |
Morley's Miracle A. Robson's Proof
Theorem
The three points of intersection of the adjacent trisectors of the angles of any triangle form an equilateral triangle.
Proof
With a reference to the diagram, let $BRL\;$ cut $AQ\;$ in $U;\;$ $AQ\;$ produced cuts $BP\;$ in $N\;$ and $CP\;$ in $V;\;$ $CP\;$ cuts $AR\;$ in $M;\;$ $QM\;$ cuts $RN\;$ in $O.\;$
$N(BRLU)=Q(CVPM)=Q(PMCV),$
As $R\;$ is the incenter of $\Delta ANB,\;$ $\angle ARN=90^{\circ}+\frac{1}{3}B.\;$
As $Q\;$ is the incenter of $\Delta AMC,\;$ $\angle RMQ=90^{\circ}-\frac{1}{3}A-\frac{1}{3}C,\;$
so that the difference, viz., $\angle ROM=60^{\circ}.\;$ Similarly, the other angles at $O\;$ are $60^{\circ};\;$ since they have a common base and equal angles at each of its extremities, the triangles $ORL\;$ and $OQL\;$ are congruent, and so are the triangles $PRL,\;PQL,$ making $\Delta PQR\;$ equilateral.
Eurika's Editor's notes on Robson's proof
The proof uses the idea of isogonal rays. If two rays through the vertex of an angle make equal angles with its sides, they are said to be isogonal. They are then mirror images in the bisector of the angle. The theorem used by Robson on isogonals is:
If three lines from the vertices of a triangle are concurrent, their isogonals are also concurrent.
He then uses the idea of projective pencils, pencils with equal cross-ratios, and the theorem that if two projective pencils with distinct vertices have a self-corresponding ray, the three intersections of corresponding rays are collinear. He also uses the theorem that in a cross-ratio the interchange of a pair of elements together with the interchange of the other pair does not affect the value of the cross-ratio. Robson's proof is as short as anyone could desire, and it avoids elaborate initial constructions.
Acknowledgment
I am deeply indebted to Roger Smyth for bringing the 1978 Eurika's issue cited below to my attention.
References A. Robson, The Mathematical Gazette, 1922-1923, pp 310-311 A. Robson, Eurika, v 3, n 10, 1978, p 280 On Morley and his theorem Doodling and Miracles Morley's Pursuit of Incidence Lines, Circles and Beyond On Motivation and Understanding Of Looking and Seeing Backward proofs J.Conway's proof D. J. Newman's proof B. Bollobás' proof G. Zsolt Kiss' proof Backward Proof by B. Stonebridge Morley's Equilaterals, Spiridon A. Kuruklis' proof J. Arioni's Proof of Morley's Theorem Trigonometric proofs Bankoff's proof B. Bollobás' trigonometric proof Proof by R. J. Webster A Vector-based Proof of Morley's Trisector Theorem L. Giugiuc's Proof of Morley's Theorem Dijkstra's Proof of Morley's Theorem Synthetic proofs Another proof Nikos Dergiades' proof M. T. Naraniengar's proof An Unexpected Variant Proof by B. Stonebridge and B. Millar Proof by B. Stonebridge Proof by Roger Smyth Proof by H. D. Grossman Proof by H. Shutrick Original Taylor and Marr's Proof of Morley's Theorem Taylor and Marr's Proof - R. A. Johnson's Version Morley's Theorem: Second Proof by Roger Smyth Proof by A. Robson Algebraic proofs Invalid proofs
Copyright © 1966-2016 Alexander Bogomolny
65618267 |
Revista Matemática Iberoamericana
Full-Text PDF (2137 KB) | Metadata | Table of Contents | RMI summary
Volume 10, Issue 3, 1994, pp. 467–505 DOI: 10.4171/RMI/159
Published online: 1994-12-31
On singular integrals of Calderón-type in $\mathbb R^n$, and BMODorina Mitrea
[1](1) University of Missouri, Columbia, United States
We prove $L^p$ (and weighted $L^p$) bounds for singular integrals of the form $$\rm p.v. \int_{\mathbb R^n} E \lgroup \frac{A(x)–A(y)}{|x–y} \rgroup \frac{\Omega(x–y)}{|x–y|^n} f(y)dy,$$ where $E(t) =$ cos $t$ if $\Omega$ is odd, and $E(t) =$ sin $t$ if $\Omega$ is even, and where $\bigtriangledown A \in$ BMO. Even in the case that $\Omega$ is smooth, the theory of singular integrals with "rough" kernels plays a key role in the proof. By standard techniques, the trigonometric function $E$ can then be replaced by a large class of smooth functions $F$. Some related operators are also considered. As a further application, we prove a compactness result for certain layer potentials.
No keywords available for this article.
Mitrea Dorina: On singular integrals of Calderón-type in $\mathbb R^n$, and BMO.
Rev. Mat. Iberoam. 10 (1994), 467-505. doi: 10.4171/RMI/159 |
Let $X$ be a smooth projective variety and $\mathcal{E}$ a locally free sheaf of rank $r$ on it. Now we consider the relative Grassmannian $\mathrm{Grass}_X(l,\mathcal{E})$. What is the canonical bundle of the relative Grassmannian?
If you know the calculation of the canonical bundle of the usual Grassmannian (nice writeup here), you can adapt it directly to the relative situation. The result is that $K_G=\mathcal O_{\mathbf P}(-r)_{|G}$, where $\mathbf P=\mathbf P_X \left( \bigwedge^l \mathcal E \right)$ is the projective bundle into which the relative Grassmannian is embedded by the Plücker embedding.
Let $\pi:Grass(\mathcal{E})\rightarrow X$ be the projection, and $\mathbf{q}:\pi^*\mathcal{E}\twoheadrightarrow \mathcal{Q}$ the universal quotient rank $l$ quotient. One has an exact sequence
$0\rightarrow T_{G/X}\rightarrow T_{G}\rightarrow \pi^* T_{X}\rightarrow 0$.
Since the tangent space to the ordinary Grassmannian at a point $q:\mathbb{C}^n\twoheadrightarrow Q$ is $\text{Hom}(\ker q, Q)$, one finds that the relative tangent bundle $T_{G/X}$ is isomorphic to $\underline{Hom}(\ker\mathbf{q},\mathcal{Q})$. Now the exact sequence above allows us to express $\det T_G=K_G^{-1}$ as the tensor product of the determinants of the left and right guys in the sequence. Since $\det T_{X/G}=(\det\ker\mathbf{q})^{-l}\otimes(\det\mathcal{Q})^{r-l}$, we get an explicit formula for the canonical bundle on $G$ in terms of $K_X$ and the tautological bundles on the Grassmannian.
Note that in the case where $\det\mathcal{E}=\mathcal{O}$ one has an $\det\ker\mathbf{q}^{-1}=\det\mathcal{Q}$, so that one obtains $\det T_{G/X}=(\det\mathcal{Q})^r$. In other words, in terms of the Plucker embedding, $\det T_{G/X}=\mathcal{O}_G(r)$. In this special case, the
relative canonical bundle is what was given in the previous answer. |
For fixed number of vertices $n$, I want to iterate through all vertex-transitive simple graphs to check for some properties. A nice way to find vertex-transitive graphs is to iterate binary vectors $\vec v \in \{0;1\}^n$ and for each vector $\vec v$ obtain a "nice" adjacency matrix by right-shifting the indices in row $i$ by $i-1$ like so:
$$M = \begin{bmatrix} v_{1} & v_{2} & \dots & \cdots & v_{n} \\ v_{n} & v_1 & v_2 & \dots & v_{n-1} \\ \dots & \dots & \dots & \dots & \dots \\ v_2 & v_3 & \dots & v_{n} & v_1\\ \end{bmatrix}$$
This will yield a simple undirected graph if $\vec v$ is of the form $(0,a_1,a_2,..,a_k,a_{k-1},..,a_1)$. Of course a lot of the graphs are isomorphic, but I am not too worried about that (unless anyone has an idea on how to improve on that).
However, it seems that not all vertex-transitive graphs have such a nice representation. For instance on 8 vertices, with degree 3, the only good vectors (by the above restriction) are $(0,1,0,0,1,0,0,1), (0,0,1,0,1,0,1,0)$ and $(0,0,0,1,1,1,0,0)$. None of these correspond to the following graph:
How can I find the remaining graphs, and is there an understandable property that divides vertex-transitive graphs into these two groups? Does my iteration cover most or almost none of the vertex-transitive graphs? |
Feynman diagrams provide a very compact and intuitive way of representing interactions between particles. These diagrams can be included into LaTeX documents thanks to a few packages. One of the older packages is
feynmf which uses MetaPost in order to generate the diagrams. More recently, a new package called Ti
kZ-Feynman has been published which uses Ti kZ in order to generate Feynman diagrams.
Contents
Ti
kZ-Feynman is a LaTeX package allowing Feynman diagrams to be easily generated within LaTeX with minimal user instructions and without the need of external programs. It builds upon the Ti kZ package and its graph drawing algorithms in order to automate the placement of many vertices. Ti kZ-Feynman still allows fine-tuned placement of vertices so that even complex diagrams can be generated with ease.
Currently, Ti
kZ-Feynman is too new to have made it into ShareLaTeX's installation, but we are working to get it included soon. In the meantime, it is possible to include the package files manually in a ShareLaTeX project as shown in this template.
After installing the package, the Ti
kZ-Feynman package can be loaded with
\usepackage{tikz-feynman} in the preamble. It is recommend that you also specify the version of Ti
kZ-Feynman to use with the
compat package option:
\usepackage[compat=1.0.0]{tikz-feynman}. This ensures that any new versions of Ti
kZ-Feynman do not produce any undesirable changes without warning.
Feynman diagrams can be declared with the
\feynmandiagram command. It is analogous to the
\tikz command from Ti
kZ and requires a final semi-colon (
;) to finish the environment. For example, a simple
s-channel diagram is:
\feynmandiagram [horizontal=a to b] { i1 -- [fermion] a -- [fermion] i2, a -- [photon] b, f1 -- [fermion] b -- [fermion] f2, };
Let's go through this example line by line:
\feynmandiagram introduces the Feynman diagram and allows for optional arguments to be given in the brackets
[<options>]. In this instance,
horizontal=a to b orients the algorithm outputs such that the line through vertices
a and
b is horizontal.
i1,
a and
i2) and connecting them with edges
--. Just like the
\feynmandiagram command above, each edge also take optional arguments specified in brackets
[<options>]. In this instance, we want these edges to have arrows to indicate that they are fermion lines, so we add the
fermion style to them. As you will see later on, optional arguments can also be given to the vertices in exactly the same way.
a and
b with an edge styled as a photon. Since there is already a vertex labelled
a, the algorithm will connect it to a new vertex labeled
b.
f1 and
f2. It re-uses the previously labelled
b vertex.
;) is important.
The name given to each vertex in the graph does not matter. So in this example,
i1,
i2 denote the initial particles;
f1,
f2 denotes the final particles; and
a,
b are the end points of the propagator. The only important aspect is that what we called
a in line 2 is also
a in line 3 so that the underlying algorithm treats them as the same vertex.
The order in which vertices are declared does not matter as the default algorithm re-arranges everything. For example, one might prefer to draw the fermion lines all at once, as with the following example (note also that the way we named vertices is completely different):
\feynmandiagram [horizontal=f2 to f3] { f1 -- [fermion] f2 -- [fermion] f3 -- [fermion] f4, f2 -- [photon] p1, f3 -- [photon] p2, };
As a final remark, the calculation of where vertices should be placed is usually done through an algorithm written in Lua. As a result, LuaTeX is required in order to make use of these algorithms. If LuaTeX is not used, Ti
kZ-Feynman will default to a more rudimentary algorithm and will warn the user instead.
So far, the examples have only used the
photon and
fermion styles. The Ti
kZ-Feynman package comes with quite a few extra styles for edges and vertices which are all documented over in the package documentation. For example, it is possible to add momentum arrows with
momentum=<text>, and in the case of end vertices, the particle can be labelled with
particle=<text>. To demonstrate how they are used, we take the generic
s-channel diagram from earlier and make it a electron-positron pairs annihilating into muons:
\feynmandiagram [horizontal=a to b] { i1 [particle=\(e^{-}\)] -- [fermion] a -- [fermion] i2 [particle=\(e^{+}\)], a -- [photon, edge label=\(\gamma\), momentum'=\(k\)] b, f1 [particle=\(\mu^{+}\)] -- [fermion] b -- [fermion] f2 [particle=\(\mu^{-}\)], };
In addition to the style keys documented below, style keys from Ti
kZ can be used as well:
\feynmandiagram [horizontal=a to b] { i1 [particle=\(e^{-}\)] -- [fermion, very thick] a -- [fermion, opacity=0.2] i2 [particle=\(e^{+}\)], a -- [red, photon, edge label=\(\gamma\), momentum'={[arrow style=red]\(k\)}] b, f1 [particle=\(\mu^{+}\)] -- [fermion, opacity=0.2] b -- [fermion, very thick] f2 [particle=\(\mu^{-}\)], };
For a list of all the various styles that Ti
kZ provides, have a look at the Ti kZ manual; it is extremely thorough and provides many usage examples.
By default, the
\feynmandiagram and
\diagram commands use the
spring layout algorithm to place all the edges. The
spring layout algorithm attempts to `spread out' the diagram as much as possible which—for most simpler diagrams—gives a satisfactory result; however in some cases, this does not produce the best diagram and this section will look at alternatives. There are three main alternatives:
draw=none. The algorithm will treat these extra edges in the same way, but they are simply not drawn at the end;
The underlying algorithm treats all edges in exactly the same way when calculating where to place all the vertices, and the actual drawing of the diagram (after the placements have been calculated) is done separately. Consequently, it is possible to add edges to the algorithm, but prevent them from being drawn by adding
draw=none to the edge style.
This is particularly useful if you want to ensure that the initial or final states remain closer together than they would have otherwise as illustrated in the following example (note that
opacity=0.2 is used instead of
draw=none to illustrate where exactly the edge is located).
% No invisible to keep the two photons together \feynmandiagram [small, horizontal=a to t1] { a [particle=\(\pi^{0}\)] -- [scalar] t1 -- t2 -- t3 -- t1, t2 -- [photon] p1 [particle=\(\gamma\)], t3 -- [photon] p2 [particle=\(\gamma\)], };
% Invisible edge ensures photons are parallel \feynmandiagram [small, horizontal=a to t1] { a [particle=\(\pi^{0}\)] -- [scalar] t1 -- t2 -- t3 -- t1, t2 -- [photon] p1 [particle=\(\gamma\)], t3 -- [photon] p2 [particle=\(\gamma\)], p1 -- [opacity=0.2] p2, };
The graph drawing library from Ti
kZ has several different algorithms to position the vertices. By default,
\diagram and
\feynmandiagram use the
spring layout algorithm to place the vertices. The
spring layout attempts to spread everything out as much as possible which, in most cases, gives a nice diagram; however, there are certain cases where this does not work. A good example where the
spring layout doesn't work are decays where we have the decaying particle on the left and all the daughter particles on the right.
% Using the default spring layout \feynmandiagram [horizontal=a to b] { a [particle=\(\mu^{-}\)] -- [fermion] b -- [fermion] f1 [particle=\(\nu_{\mu}\)], b -- [boson, edge label=\(W^{-}\)] c, f2 [particle=\(\overline \nu_{e}\)] -- [fermion] c -- [fermion] f3 [particle=\(e^{-}\)], };
% Using the layered layout \feynmandiagram [layered layout, horizontal=a to b] { a [particle=\(\mu^{-}\)] -- [fermion] b -- [fermion] f1 [particle=\(\nu_{\mu}\)], b -- [boson, edge label'=\(W^{-}\)] c, c -- [anti fermion] f2 [particle=\(\overline \nu_{e}\)], c -- [fermion] f3 [particle=\(e^{-}\)], };
You may notice that in addition to adding the
layered layout style to
\feynmandiagram, we also changed the order in which we specify the vertices. This is because the
layered layout algorithm does pay attention to the order in which vertices are declared (unlike the default
spring layout); as a result,
c--f2, c--f3 has a different meaning to
f2--c--f3. In the former case,
f2 and
f3 are both on the layer below
c as desired; whilst the latter case places
f2 on the layer above
c (that, the same layer as where the W-boson originates).
In more complicated diagrams, it is quite likely that none of the algorithms work, no matter how many invisible edges are added. In such cases, the vertices have to be placed manually. Ti
kZ-Feynman allows for vertices to be manually placed by using the
\vertex command.
The
\vertex command is available only within the
feynman environment (which itself is only available inside a
tikzpicture). The
feynman environment loads all the relevant styles from Ti
kZ-Feynman and declares additional Ti kZ-Feynman-specific commands such as
\vertex and
\diagram. This is inspired from PGFPlots and its use of the
axis environment.
The
\vertex command is very much analogous to the
\node command from Ti
kZ, with the notable exception that the vertex contents are optional; that is, you need not have
{<text>} at the end. In the case where
{} is specified, the vertex automatically is given the
particle style, and otherwise it is a usual (zero-sized) vertex.
To specify where the vertices go, it is possible to give explicit coordinates though it is probably easiest to use the
positioning library from Ti
kZ which allows vertices to be placed relative to existing vertices. By using relative placements, it is possible to easily tweak one part of the graph and everything will adjust accordingly—the alternative being to manually adjust the coordinates of every affected vertex.
Finally, once all the vertices have been specified, the
\diagram* command is used to specify all the edges. This works in much the same way as
\diagram (and also
\feynmandiagram), except that it uses an very basic algorithm to place new nodes and allows existing (named) nodes to be included. In order to refer to an existing node, the node must be given in parentheses.
This whole process of specifying the nodes and then drawing the edges between them is shown below for the muon decay:
\begin{tikzpicture} \begin{feynman} \vertex (a) {\(\mu^{-}\)}; \vertex [right=of a] (b); \vertex [above right=of b] (f1) {\(\nu_{\mu}\)}; \vertex [below right=of b] (c); \vertex [above right=of c] (f2) {\(\overline \nu_{e}\)}; \vertex [below right=of c] (f3) {\(e^{-}\)}; \diagram* { (a) -- [fermion] (b) -- [fermion] (f1), (b) -- [boson, edge label'=\(W^{-}\)] (c), (c) -- [anti fermion] (f2), (c) -- [fermion] (f3), }; \end{feynman} \end{tikzpicture}
The
feynmf package lets you easily draw Feynman diagrams in your LaTeX documents. All you need to do is specify the vertices, the particles and the labels, and it will automatically layout and draw your diagram for you.
Let's start with a quick example:
\begin{fmffile*}{diagram} \begin{fmfgraph}(40,25) \fmfleft{i1,i2} \fmfright{o1,o2} \fmf{fermion}{i1,v1,o1} \fmf{fermion}{i2,v2,o2} \fmf{photon}{v1,v2} \end{fmfgraph} \end{fmffile*}
The
fmffile* environment must be put around all of your Feynman diagrams. You can use fmffile environment for multiple diagrams, so you can put one around your whole document and forget about it. The second argument to the fmffile environment tells LaTeX where to write the files that it uses to store the diagram. You can name this whatever you want, but you need to run metafont on your diagram between LaTeX runs in order for your diagram to show up (ShareLaTeX does this automatically):
The 'fmfgraph' environment starts a Feynman diagram, and the figures in brackets afterwards specify the width and height of the diagram.
The first thing you need to do is specify your external vertices, and where they should be positioned. You can name your vertices anything you like, and say where they should be positioned with the commands
\fmfleft, \fmfright, \fmftop, \fmfbottom. For example
% Creates two vertices on the left called i1 and i2 \fmfleft{i1,i2} % Creates two vertices on the right called o1 and o2 \fmfright{o1,o2}
You can connect up vertices with the
\fmf, which will create new vertices if you pass in names that haven't been created yet. For example
% Will create a fermion line between i1 and % the newly created v1, and between v1 and o1. \fmf{fermion}{i1,v1,o1} % Will create a photon line between v1 and the newly created v2 \fmf{photon}{v2,v2}
A vertex can be labelled using the
\fmflabel command, which takes two arguments: the label to apply to the vertex, and the name of the vertex to apply it to. For example, in the above diagram, if we add in the following labels, we get the updated diagram below:
Note that math mode can used inside the vertex labels, as we have done above.
We've seen the 'photon' and 'fermion' line styles above, but the
feynmf package support many more.
Appearance Name(s) gluon, curly dbl_curly dashes scalar, dashes_arrow dbl_dashes dbl_dashes_arrow dots ghost, dots_arrow dbl_dots dbl_dots_arrow phantom phantom_arrow vanilla, plain fermion, electron, quark, plain_arrow double, dbl_plain double_arrow, heavy, dbl_plain_arrow boson, photon, wiggly dbl_wiggly zigzag dbl_zigzag
For more information see: |
Is there a "simple" mathematical proof that is fully understandable by a 1st year university student that impressed you because it is beautiful?
closed as primarily opinion-based by Daniel W. Farlow, Najib Idrissi, user91500, LutzL, Jonas Meyer Apr 7 '15 at 3:40
Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.
Here's a cute and lovely theorem.
There exist two irrational numbers $x,y$ such that $x^y$ is rational.
Proof. If $x=y=\sqrt2$ is an example, then we are done; otherwise $\sqrt2^{\sqrt2}$ is irrational, in which case taking $x=\sqrt2^{\sqrt2}$ and $y=\sqrt2$ gives us: $$\left(\sqrt2^{\sqrt2}\right)^{\sqrt2}=\sqrt2^{\sqrt2\sqrt2}=\sqrt2^2=2.\qquad\square$$
(Nowadays, using the Gelfond–Schneider theorem we know that $\sqrt2^{\sqrt2}$ is irrational, and in fact transcendental. But the above proof, of course, doesn't care for that.)
How about the proof that
$$1^3+2^3+\cdots+n^3=\left(1+2+\cdots+n\right)^2$$
I remember being impressed by this identity and the proof can be given in a picture:
Edit: Substituted $\frac{n(n+1)}{2}=1+2+\cdots+n$ in response to comments.
Cantor's Diagonalization Argument, proof that there are infinite sets that can't be put one to one with the set of natural numbers, is frequently cited as a beautifully simple but powerful proof. Essentially, with a list of infinite sequences, a sequence formed from taking the diagonal numbers will not be in the list.
I would personally argue that the proof that $\sqrt 2$ is irrational is simple enough for a university student (probably simple enough for a high school student) and very pretty in its use of proof by contradiction!
Prove that if $n$ and $m$ can each be written as a sum of two perfect squares, so can their product $nm$.
Proof: Let $n = a^2+b^2$ and $m=c^2+d^2$ ($a, b, c, d \in\mathbb Z$). Then, there exists some $x,y\in\mathbb Z$ such that
$$x+iy = (a+ib)(c+id)$$
Taking the magnitudes of both sides are squaring gives
$$x^2+y^2 = (a^2+b^2)(c^2+d^2) = nm$$
I would go for the proof by contradiction of an infinite number of primes, which is fairly simple:
Assume that there is a finite number of primes. Let $G$ be the set of allprimes $P_1,P_2,...,P_n$. Compute $K = P_1 \times P_2 \times ... \times P_n + 1$. If $K$ is prime, then it is obviously notin $G$. Otherwise, noneof its prime factors are in $G$. Conclusion: $G$ is notthe set of allprimes.
I think I learned that both in high-school and at 1st year, so it might be a little too simple...
By the concavity of the $\sin$ function on the interval $\left[0,\frac{\pi}2\right]$ we deduce these inequalities: $$\frac{2}\pi x\le \sin x\le x,\quad \forall x\in\left[0,\frac\pi2\right].$$
The first player in Hex has a winning strategy.
There are no draws in hex, so one player must have a winning strategy. If player two has a winning strategy, player one can steal that strategy by placing the first stone in the center (additional pieces on the board never hurt your position) then using player two's strategy.
You cannot have two dice (with numbers $1$ to $6$) biased so that when you throw both, the sum is uniformly distributed in $\{2,3,\dots,12\}$.
For easier notation, we use the equivalent formulation "You cannot have two dice (with numbers $0$ to $5$) biased such that when you throw both, the sum is uniformly distributed in $\{0,1,\dots,10\}$."
Proof:Assume that such dice exist. Let $p_i$ be the probability that the first die gives an $i$ and $q_i$ be the probability that the second die gives an $i$. Let $p(x)=\sum_{i=0}^5 p_i x^i$ and $q(x)=\sum_{i=0}^5 q_i x^i$.
Let $r(x)=p(x)q(x) = \sum_{i=0}^{10} r_i x^i$. We find that $r_i = \sum_{j+k=i}p_jq_k$. But hey, this is also the probability that the sum of the two dice is $i$. Therefore, $$ r(x)=\frac{1}{11}(1+x+\dots+x^{10}). $$ Now $r(1)=1\neq0$, and for $x\neq1$, $$ r(x)=\frac{(x^{11}-1)}{11(x-1)}, $$ which clearly is nonzero when $x\neq 1$. Therefore $r$ does not have any real zeros.
But because $p$ and $q$ are $5$th degree polynomials, they must have zeros. Therefore, $r(x)=p(x)q(x)$ has a zero. A contradiction.
Given a square consisting of $2n \times 2n$ tiles, it is possible to cover this square with pieces that each cover $2$ adjacent tiles (like domino bricks). Now imagine, you remove two tiles, from two opposite corners of the original square. Prove that is is now no longer possible to cover the remaining area with domino bricks.
Proof:
Imagine that the square is a checkerboard. Each domino brick will cover two tiles of different colors. When you remove tiles from two opposite corners, you will remove two tiles with the
samecolor. Thus, it can no longer be possible to cover the remaining area.
(Well, it may be
too "simple." But you did not state that it had to be a university student of mathematics. This one might even work for liberal arts majors...)
One little-known gem at the intersection of geometry and number theory is Aubry's reflective generation of primitive Pythagorean triples, i.e. coprime naturals $\,(x,y,z)\,$with $\,x^2 + y^2 = z^2.\,$ Dividing by $\,z^2$ yields $\,(x/z)^2+(y/z)^2 = 1,\,$ so each triple corresponds to a rational point $(x/z,\,y/z)$ on the unit circle. Aubry showed that we can generate all such triples by a very simple geometrical process. Start with the trivial point $(0,-1)$. Draw a line to the point $\,P = (1,1).\,$ It intersects the circle in the
rational point $\,A = (4/5,3/5)\,$ yielding the triple $\,(3,4,5).\,$ Next reflect the point $\,A\,$ into the other quadrants by taking all possible signs of each component, i.e. $\,(\pm4/5,\pm3/5),\,$ yielding the inscribed rectangle below. As before, the line through $\,A_B = (-4/5,-3/5)\,$ and $P$ intersects the circle in $\,B = (12/13, 5/13),\,$ yielding the triple $\,(12,5,13).\,$ Similarly the points $\,A_C,\, A_D\,$ yield the triples $\,(20,21,29)\,$ and $\,(8,15,17),\,$ We can iterate this process with the new points $\,B,C,D\,$ doing the same we did for $\,A,\,$ obtaining further triples. Iterating this process generates the primitive triples as a ternary tree
$\qquad\qquad$
Descent in the tree is given by the formula
$$\begin{eqnarray} (x,y,z)\,\mapsto &&(x,y,z)-2(x\!+\!y\!-\!z)\,(1,1,1)\\ = &&(-x-2y+2z,\,-2x-y+2z,\,-2x-2y+3z)\end{eqnarray}$$
e.g. $\ (12,5,13)\mapsto (12,5,13)-8(1,1,1) = (-3,4,5),\ $ yielding $\,(4/5,3/5)\,$ when reflected into the first quadrant.
Ascent in the tree by inverting this map, combined with trivial sign-changing reflections:
$\quad\quad (-3,+4,5) \mapsto (-3,+4,5) - 2 \; (-3+4-5) \; (1,1,1) = ( 5,12,13)$
$\quad\quad (-3,-4,5) \mapsto (-3,-4,5) - 2 \; (-3-4-5) \; (1,1,1) = (21,20,29)$
$\quad\quad (+3,-4,5) \mapsto (+3,-4,5) - 2 \; (+3-4-5) \; (1,1,1) = (15,8,17)$
See my MathOverflow post for further discussion, including generalizations and references.
I like the proof that there are infinitely many Pythagorean triples.
Theorem:There are infinitely many integers $ x, y, z$ such that $$ x^2+y^2=z^2 $$ Proof:$$ (2ab)^2 + ( a^2-b^2)^2= ( a^2+b^2)^2 $$
One cannot cover a disk of diameter 100 with 99 strips of length 100 and width 1.
Proof: project the disk and the strips on a semi-sphere on top of the disk. The projection of each strip would have area at most 1/100th of the area of the semi-sphere.
If you have any set of 51 integers between $1$ and $100$, the set must contain some pair of integers where one number in the pair is a multiple of the other.
Proof: Suppose you have a set of $51$ integers between $1$ and $100$. If an integer is between $1$ and $100$, its largest odd divisor is one of the odd numbers between $1$ and $99$. There are only $50$ odd numbers between $1$ and $99$, so your $51$ integers can’t all have different largest odd divisors — there are only $50$ possibilities. So two of your integers (possibly more) have the same largest odd divisor. Call that odd number $d$. You can factor those two integers into prime factors, and each will factor as (some $2$’s)$\cdot d$. This is because if $d$ is the largest divisor of a number, the rest of its factorization can’t include any more odd numbers. Of your two numbers with largest odd factor $d$, the one with more $2$’s in its factorization is a multiple of the other one. (In fact, the multiple is a power of $2$.)
In general, let $S$ be the set of integers from $1$ up to some even number $2n$. If a subset of $S$ contains more than half the elements in $S$, the set must contain a pair of numbers where one is a multiple of the other. The proof is the same, but it’s easier to follow if you see it for a specific $n$ first.
The proof that an isosceles triangle ABC (with AC and AB having equal length) has equal angles ABC and BCA is quite nice:
Triangles ABC and ACB are (mirrored) congruent (since AB = AC, BC = CB, and CA = BA), so the corresponding angles ABC and (mirrored) ACB are equal.
This congruency argument is nicer than that of cutting the triangle up into two right-angled triangles.
Parity of sine and cosine functions using Euler's forumla:
$e^{-i\theta} = cos\ (-\theta) + i\ sin\ (-\theta)$
$e^{-i\theta} = \frac 1 {e^{i\theta}} = \frac 1 {cos\ \theta \ + \ i\ sin\ \theta} = \frac {cos\ \theta\ -\ i\ sin\ \theta} {cos^2\ \theta\ +\ sin^2\ \theta} = cos\ \theta\ -\ i\ sin\ \theta$
$cos\ (-\theta) +\ i\ sin\ (-\theta) = cos\ \theta\ +i\ (-sin\ \theta)$
Thus
$cos\ (-\theta) = cos\ \theta$
$sin\ (-\theta) = -\ sin\ \theta$
$\blacksquare$
The proof is actually just the first two lines.
I believe Gauss was tasked with finding the sum of all the integers from $1$ to $100$ in his very early schooling years. He tackled it quicker than his peers or his teacher could, $$\sum_{n=1}^{100}n=1+2+3+4 +\dots+100$$ $$=100+99+98+\dots+1$$ $$\therefore 2 \sum_{n=1}^{100}n=(100+1)+(99+2)+\dots+(1+100)$$ $$=\underbrace{101+101+101+\dots+101}_{100 \space times}$$ $$=101\cdot 100$$ $$\therefore \sum_{n=1}^{100}n=\frac{101\cdot 100}{2}$$ $$=5050.$$ Hence he showed that $$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}.$$
If $H$ is a subgroup of $(\mathbb{R},+)$ and $H\bigcap [-1,1]$ is finite and contains a positive element. Then, $H$ is cyclic.
Fermat's little theorem from noting that modulo a prime p we have for $a\neq 0$:
$$1\times2\times3\times\cdots\times (p-1) = (1\times a)\times(2\times a)\times(3\times a)\times\cdots\times \left((p-1)\times a\right)$$
Proposition (No universal set): There does not exists a set which contain all the sets (even itself)
Proof: Suppose to the contrary that exists such set exists. Let $X$ be the universal set, then one can construct by the axiom schema of specification the set
$$C=\{A\in X: A \notin A\}$$
of all sets in the universe which did not contain themselves. As $X$ is universal, clearly $C\in X$. But then $C\in C \iff C\notin C$, a contradiction.
Edit: Assuming that one is working in ZF (as almost everywhere :P)
(In particular this proof really impressed me too much the first time and also is very simple)
Most proofs concerning the Cantor Set are simple but amazing.
The total number of intervals in the set is zero.
It is uncountable.
Every number in the set can be represented in ternary using just
0 and
2. No number with a
1 in it (in ternary) appears in the set.
The Cantor set contains as many points as the interval from which it is taken, yet itself contains no interval of nonzero length. The irrational numbers have the same property, but the Cantor set has the additional property of being closed, so it is not even dense in any interval, unlike the irrational numbers which are dense in every interval.
The Menger sponge which is a 3d extension of the Cantor set
simultaneously exhibits an infinite surface area and encloses zero volume.
The derivation of first principle of differentiation is so amazing, easy, useful and simply outstanding in all aspects. I put it here:
Suppose we have a quantity $y$ whose value depends upon a single variable $x$, and is expressed by an equation defining $y$ as some specific function of $x$. This is represented as:
$y=f(x)$
This relationship can be visualized by drawing a graph of function $y = f (x)$ regarding $y$ and $x$ as Cartesian coordinates, as shown in Figure(a).
Consider the point $P$ on the curve $y = f (x)$ whose coordinates are $(x, y)$ and another point $Q$ where coordinates are $(x + Δx, y + Δy)$.
The slope of the line joining $P$ and $Q$ is given by:
$tanθ = \frac{Δy}{Δx} = \frac{(y + Δy ) − y}{Δx}$
Suppose now that the point $Q$ moves along the curve towards $P$.
In this process, $Δy$ and $Δx$ decrease and approach zero; though their ratio $\frac{Δy}{Δx}$ will not necessarily vanish.
What happens to the line $PQ$ as $Δy→0$, $Δx→0$? You can see that this line becomes a tangent to the curve at point $P$ as shown in Figure(b). This means that $tan θ$ approaches the slope of the tangent at $P$, denoted by $m$:
$m=lim_{Δx→0} \frac{Δy}{Δx} = lim_{Δx→0} \frac{(y+Δy)-y}{Δx}$
The limit of the ratio $Δy/Δx$ as $Δx$ approaches zero is called the derivative of $y$ with respect to $x$ and is written as $dy/dx$.
It represents the slope of the tangent line to the curve $y=f(x)$ at the point $(x, y)$.
Since $y = f (x)$ and $y + Δy = f (x + Δx)$, we can write the definition of the derivative as:
$\frac{dy}{dx}=\frac{d{f(x)}}{dx} = lim_{x→0} [\frac{f(x+Δx)-f(x)}{Δx}]$,
which is the required formula.
This proof that $n^{1/n} \to 1$ as integral $n \to \infty$:
By Bernoulli's inequality (which is $(1+x)^n \ge 1+nx$), $(1+n^{-1/2})^n \ge 1+n^{1/2} > n^{1/2} $. Raising both sides to the $2/n$ power, $n^{1/n} <(1+n^{-1/2})^2 = 1+2n^{-1/2}+n^{-1} < 1+3n^{-1/2} $.
Can a Chess Knight starting at any corner then move to touch every space on the board exactly once, ending in the opposite corner?
The solution turns out to be childishly simple. Every time the Knight moves (up two, over one), it will hop from a black space to a white space, or vice versa. Assuming the Knight starts on a black corner of the board, it will need to touch 63 other squares, 32 white and 31 black. To touch all of the squares, it would need to end on a white square, but the opposite corner is also black, making it impossible.
The Eigenvalues of a skew-Hermitian matrix are purely imaginary.
The Eigenvalue equation is $A\vec x = \lambda\vec x$, and forming the vector norm gives $$\lambda \|\vec x\| = \lambda\left<\vec x, \vec x\right> = \left<\lambda \vec x,\vec x\right> = \left<A\vec x,\vec x\right> = \left<\vec x, A^{T*}\vec x\right> = \left<\vec x, -A\vec x\right> = -\lambda^* \|\vec x\|$$ and since $\|\vec x\| > 0$, we can divide it from left and right side. The second to last step uses the definition of skew-Hermitian. Using the definition for Hermitian or Unitarian matrices instead yields corresponding statements about the Eigenvalues of those matrices.
I like the proof that not every real number can be written in the form $a e + b \pi$ for some integers $a$ and $b$. I know it's almost trivial in one way; but in another way it is kind of deep. |
Limit ordinal Properties
All limit ordinals are equal to their union.
All limit ordinals contain an ordinal $\alpha$ if and only if they contain $\alpha + 1$.
$\omega$ is the smallest nonzero limit ordinal, and the smallest ordinal of infinite cardinal number.
$(\omega + \omega)$, also written $( \omega \cdot 2 )$, is the next limit ordinal. $( \omega \cdot \alpha )$ is a limit ordinal for any ordinal $\alpha$.
Types of Limits
A limit ordinal $\alpha$ is called
additively indecomposable (or a $\gamma$ number) if it cannot be the sum of $\beta<\alpha$ ordinals less than $\alpha$. These numbers are any ordinal of the form $\omega^\beta$ for $\beta$ an ordinal. The smallest is written $\gamma_0$, and the smallest larger than that is $\gamma_1$, etc.
A limit ordinal $\alpha$ is called
multiplicatively indecomposable (or a $\delta$ number) if it cannot be the product of $\beta<\alpha$ ordinals less than $\alpha$. These numbers are any ordinal of the form $\omega^{\omega^{\beta}}$. The smallest is written $\delta_0$, and the smallest larger than that is $\delta_1$, etc.
Interestingly, this pattern does not continue with
exponentially indecomposable (or $\varepsilon$ numbers) ordinals being $\omega^{\omega^{\omega^\beta}}$, but rather $\varepsilon_0=sup_{n<\omega}f^n(0)$ with $f(\alpha)=\omega^\alpha$ and $f^n(\alpha)=f(f(...f(\alpha)...))$ with $n$ iterations of $f$. It is the smallest fixed point of $f$. The next $\varepsilon$ number (i.e. the next fixed point of $f$) is then $\varepsilon_1=sup_{n<\omega}f^n(\varepsilon_0+1)$, and more generally the $(\alpha+1)$th fixed point of $f$ is $\varepsilon_{\alpha+1}=sup_{n<\omega}f^n(\varepsilon_\alpha+1)$, also $\varepsilon_\lambda=\cup_{\alpha<\lambda}\varepsilon_\alpha$ for limit $\lambda$.
The
tetrationally indecomposable ordinals (or $\zeta$ numbers) are then the ordinals $\zeta$ such that $\varepsilon_\zeta=\zeta$. These are obtained similarly as $\varepsilon$ numbers by taking $f(\alpha)=\varepsilon_\alpha$. Pentationally indecomposable ordinals (or $\eta$ ordinals) are then obtained by taking $f(\alpha)=\zeta_\alpha$, and so on.
This pattern continues on with the Veblen Hiearchy, continuing up to the Feferman-Schütte ordinal $\Gamma_0$, the smallest ordinal such that this process does not generate any larger kind of ordinals. |
In continuum mechanics,
wave action refers to a conservable measure of the wave part of a motion. [2] For small-amplitude and slowly varying waves, the wave action density is: [3] \mathcal{A} = \frac{E}{\omega_i},
where E is the intrinsic wave energy and \omega_i is the intrinsic frequency of the slowly modulated waves – intrinsic here implying: as observed in a frame of reference moving with the mean velocity of the motion.
[4]
The action of a wave was introduced by Sturrock (1962) in the study of the (pseudo) energy and momentum of waves in plasmas. Whitham (1965) derived the conservation of wave action – identified as an adiabatic invariant – from an averaged Lagrangian description of slowly varying nonlinear wave trains in inhomogeneous media:
\frac{\partial}{\partial t}\mathcal{A} + \boldsymbol{\nabla} \cdot \boldsymbol{\mathcal{B}} = 0,
where \boldsymbol{\mathcal{B}} is the wave-action density flux and \boldsymbol{\nabla}\cdot\boldsymbol{\mathcal{B}} is the divergence of \boldsymbol{\mathcal{B}}. The description of waves in inhomogeneous moving media was further elaborated by Bretherton & Garrett (1968) for the case of small-amplitude waves; they also called the quantity
wave action (by which name it has been referred to subsequently). For small-amplitude waves the conservation of wave action becomes: [3] [4] \frac{\partial}{\partial t}\left( \frac{E}{\omega_i} \right) + \boldsymbol{\nabla} \cdot \left[ \left( \boldsymbol{U} + \boldsymbol{c}_g \right)\, \frac{E}{\omega_i} \right] = 0, using \mathcal{A} = \frac{E}{\omega_i} and \boldsymbol{\mathcal{B}} = \left( \boldsymbol{U} + \boldsymbol{c}_g \right) \mathcal{A},
where \boldsymbol{c}_g is the group velocity and \boldsymbol{U} the mean velocity of the inhomogeneous moving medium. While the
total energy (the sum of the energies of the mean motion and of the wave motion) is conserved for a non-dissipative system, the energy of the wave motion is not conserved, since in general there can be an exchange of energy with the mean motion. However, wave action is a quantity which is conserved for the wave-part of the motion.
The equation for the conservation of wave action is for instance used extensively in wind wave models to forecast sea states as needed by mariners, the offshore industry and for coastal defense. Also in plasma physics and acoustics the concept of wave action is used.
The derivation of an exact wave-action equation for more general wave motion – not limited to slowly modulated waves, small-amplitude waves or (non-dissipative) conservative systems – was provided and analysed by Andrews & McIntyre (1978) using the framework of the generalised Lagrangian mean for the separation of wave and mean motion.
[4] Notes References
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I wrote an entire blog post explaining the answers to 2.3 but Blogger decided to eat it. I don't want to redo those answers so here is 3.1:
For now on I will title my posts with the section number as well to help Google.
Question 3.1-1:Let $f(n)$ and $g(n)$be asymptotically non-negative functions. Using the basic definition of $\theta$-notation, prove that $\max(f(n) , g(n)) \in \theta(f(n) + g(n))$ .
CLRS defines $\theta$ as $\theta(g(n))= \{ f(n) :$ there exists some positive constants $c_1, c_2$, and $n_0,$ such that $0 \leq c_1g(n) \leq f(n) \leq c_2g(n)$ for all $n \geq n_0\}$ Essentially we must prove that there exists some $c_1$ and $c_2$ such that $c_1 \times (f(n) + g(n)) \leq \max(f(n), g(n)) \leq c_2 \times (f(n) + g(n))$ There are a variety of ways to do this but I will choose the easiest way I could think of. Based on the above equation we know that $\max(f(n), g(n)) \leq f(n) + g(n)$ (as f(n) and g(n) must both me non-negative) and we further know that $\max(f(n), g(n))$ can't be more than twice f(n)+g(n). What we have then are the following inequalities: $$\max(f(n), g(n)) \leq c_1 \times (f(n) + g(n))$$ and $$c_2 \times (f(n) + g(n)) \leq 2 \times \max(f(n), g(n))$$ Solving for $c_1$ we get 1 and for $c_2$ we get $\frac {1} {2}$
Question 3.1-2:Show for any real constants $a$ and $b$ where $b \gt 0$ that $(n+a)^b \in \theta(n^b)$ Because $a$ is a constant and the definition of $\theta$ is true after some $n_0$ adding $a$ to $n$ does not affect the definition and we simplify to $n^b \in \theta(n^b)$ which is trivially true Question 3.1-3:Explain why the statement "The running time of $A$ is at least $O(n^2)$," is meaningless.
I'm a little uncertain of this answer but I think this is what CLRS is getting at when we say a function $f(n)$ has a running time of $O(g(n))$ what we really mean is that $f(n)$ has an asymptotic upper bound of $g(n)$. This means that $f(n) \leq g(n)$ after some $n_0$. To say a function has a running time of at least g(n) seems to be saying that $f(n) \leq g(n) \And f(n) \geq g(n)$ which is a contradiction.
Question 3.1-4:Is $2^{n+1} = O(2^n)$? Is $2^{2n} = O(2^n)$?
$2^{n+1} = 2 \times 2^n$. which means that $2^{n+1} \leq c_1 \times 2^n$ after $n_0$ so we have our answer that $2^{n+1} \in o(2^n)$ Alternatively we could say that the two functions only differ by a constant coefficient and therefore the answer is yes.
There is no constant such that $2^{2n} = c \times 2^n$ and thefore $2^{2n} \notin O(2^n)$
Question 3.1-5:Prove that for any two functions $f(n)$ and $g(n)$, we have $f(n) \in \theta(g(n)) \iff f(n) \in O(g(n)) \And f(n) \in \Omega(g(n))$
This is an "if an only if" problem so we must prove this in two parts:
Firstly, if $f(n) \in O(g(n))$ then there exists some $c_1$ and $n_0$ such that $f(n) \leq c_1 \times g(n)$ after some $n_0$. Further if $f(n) \in Omega(g(n))$ then there exists some $c_2$ and $n_0$ such that $f(n) \geq c_2 \times g(n)$ after some $n_0$.
If we combine the above two statements (which come from the definitions of $\Omega$ and O) than we know that there exists some $c_1, c_2, and n_0,$ such that $c_1g(n) \leq f(n) \leq c_2g(n)$ for all $n \geq n_0\}$
We could do the same thing backward for the other direction: If $f(n) \in \theta(g(n))$ then we could split the above inequality and show that each of the individual statements are true.
Question 3.1-6:Prove that the running time of an algorithm is $\theta(g(n)) \iff$ its worst-case running time is $O(g(n))$ and its best case running time $\Omega(g(n))$. I'm going to try for an intuitive proof here instead of a mathematical one. If the worst case is asymptotically bound above in the worst case by a certain function and is asymptotically bound from below in the best case which means that the function is tightly bound by both those functions. f(n) never goes below some constant times g(n) and never goes above some constant times g(n). This is what we get from the above definition of $\theta(g(n)))$ A mathematical follows from question 3.1-5. Question 3.1-7:Prove that $o(g(n)) \cap \omega(g(n)) = \varnothing$
little o and little omega are defined as follows: \[o(g(n)) = \{ f(n) : \forall c > 0 \exists n_0 \text{such that } 0 \leq f(n) \leq c \times g(n) \forall n \gt n_0\] and \[\omega(g(n)) = \{ f(n) : \forall c > 0 \exists n_0 \text{such that } 0 \leq c \times g(n) \leq f(n) \forall n \gt n_0\]
In other words
$$f(n) \in o(g(n)) \iff \lim_{n \to \infty} \frac {f(n)} {g(n)} = 0$$ and $$f(n) \in \omega(g(n)) \iff \lim_{n \to \infty} \frac {f(n)} {g(n)} = \infty$$It is obvious that these can not be true at the same time. This would require that $0 = \infty$ |
This is the exercise 5 in page 109 of
Analysis III of Amann and Escher.
Let $f_j,f:X\to\overline{\Bbb R}^+$ measurable functions such that $(f_j)\to f$ in measure. Show that
$$\int_Xf\,d\mu\le\varliminf\int_X f_j\,d\mu$$
Note: $X$ is a $\sigma$-finite space.
From a previous result I know that if $(f_j)\to f$ in measure then there is a subsequence $(f_{j_k})\to f$ almost everywhere.
My work so far: I set $g_j:=\inf_{k\ge j} f_j$. Then $(g_j)$ is increasing and by the existence of $(f_{j_k})$ we knows that $g_j\le f$ almost everywhere, thus WLOG we can consider that $g_j\le f$ in $X$.
Then by Fatou's lemma we have that $$ \int_X\varliminf f_j\,d\mu=\int_X \lim g_j\,d\mu\le\varliminf\int_X f_j\,d\mu\tag1 $$ Then if we shows that $(g_j)\to f$ almost everywhere we are done. However I was unable to prove it, I set $$ L:=\{x\in X: \lim |f(x)-g_j(x)|=0\}\\ A_{j,n}:=\{x\in X:|f(x)-f_j(x)|\ge 1/n\}\\ B_n:=\{x\in X:|f(x)-g_j(x)|\ge 1/n,\;\forall j\in\Bbb N\} $$ Then I tried to show that $L^\complement$ is a null set from the convergence in measure of $(f_j)$, trying to see if the $B_n$ are null, but I didnt find a way, and Im thinking that it is not necessarily true.
Then I started a different approach: if $\int_X f\le\varliminf\int_X f_j$ then eventually $\int_X f\le\int_X f_j$. This means that if $\int_X f=\infty$ then eventually $\int_X f_j=\infty$, and if $\int_X f=K<\infty$ then eventually $\int_X f_j\ge K$.
But from here I doesn't find something useful. Some help will be appreciated, thank you. |
Square root of exponential \(\!\!\!\!\!\!\!\!\!\!\!\!\!(1) ~ ~ ~ \varphi(\varphi(z))=z\)
Such a \(\varphi\) is assumed to be holomorphic function for some domain of values of \(z\).
Function \(\sqrt{\exp}\) should not be confused with \(z\mapsto \exp(z)^{1/2}\); in the range of holomorphism, the last can be reduced to \(\exp(z/2)\).
History \(\!\!\!\!\!\!\!\!\!\!\!\!\!(2) ~ ~ ~ \displaystyle \varphi(z)=\sqrt{\exp}(z) = \exp^{1/2}(z)=\mathrm{tet}\Big(\frac{1}{2}+\mathrm{ate}(z)\Big)\)
and namely this function is considered as default
square foot of exponential. Expression (2) is just special case of the general expression of the \(c\)th iteration of some transfer function \(T\) through its superfunction \(F\) and the corresponding Abel function \(G=F^{-1}\): \(\!\!\!\!\!\!\!\!\!\!\!\!\!(3) ~ ~ ~ \displaystyle T^c(z)=F(c+G(z))\)
In the equation (3), the number \(c\) of iteration can no need to be integer; it can be fractional, irrational or even complex. However, \(T^c(z)\) should not be confused with \(T(z)^c\). Equation (2) comes from (3) at \(T\!=\!\exp\), \(F\!=\!\mathrm{tet}\) and \(G\!=\!\mathrm{ate}\).
The square root of exponential seems to be the first non-trivial function for which the non-trivial non-integer iterations were reported, a "Holy Graal cup"
[3] that opened the research of various superfunctions. In the similar way, the half-iteration of the logistic sequence [4]is constructed in terms of the superfunction and the Abel function, and similarly, the square root of factorial [5] can be expressed in therms of the SuperFactorial and AbelFactorial (or "ArcSuperFactorial") functions. Uniqueness
The additional conditions on the behavior of the ArcTetration in vicinity of the fixed points of the logrithm provide the uniqueness of the ArcTetration
[6], and, therefore, the uniqueness of the suggested real-holomorphic square root of the exponential. Generalization
Equation (2) defines the square root of exponential as the \(\frac{1}{2}\)th iteration of the exponentis. It is special case of equation (3).
The similar expression for the evaluation of the half-iteration can be used also for the exponentials to various values of base, \(\exp_b\) for \(b\!>\!1\), as soon as the corresponding tetration and arctetration are implemented
[7] [8] [9].
Square root of various functions can be interpreted as a halfiteration. However, the writing \(T^c(z)\) should not be confused with \(T(z)^c\), nor with \(T(z^c)\); these are pretty different expressions.
Keywords References http://www.digizeitschriften.de/dms/img/?PPN=GDZPPN002175851 H.Kneser. Reelle analytische Lösungen der Gleichung \(\varphi(\varphi(x))=e^x\). Equationes Mathematicae, Journal fur die reine und angewandte Mathematik 18756–67 (1950) http://www.ils.uec.ac.jp/~dima/PAPERS/2009analuxpRepri.pdf D.Kouznetsov. Analytic solution of F(z+1)=exp(F(z)) in complex z-plane. Mathematics of Computation, v.78 (2009), 1647-1670. http://en.wikipedia.org/wiki/Holy_Grail http://www.springerlink.com/content/u712vtp4122544x4/ D.Kouznetsov. Holomorphic extension of the logistic sequence. Moscow University Physics Bulletin, 2010, No.2, p.91-98. http://www.ils.uec.ac.jp/~dima/PAPERS/2009supefae.pdf D.Kouznetsov, H.Trappmann. Superfunctions and square root of factorial. Moscow University Physics Bulletin, 2010, v.65, No.1, p.6-12. http://www.springerlink.com/content/u7327836m2850246/ H.Trappmann, D.Kouznetsov. Uniqueness of Analytic Abel Functions in Absence of a Real Fixed Point. Aequationes Mathematicae, 81, p.65-76 (2011) http://www.ils.uec.ac.jp/~dima/PAPERS/2010vladie.pdf D.Kouznetsov. Superexponential as special function. Vladikavkaz Mathematical Journal, 2010, v.12, issue 2, p.31-45. http://www.ams.org/journals/mcom/2010-79-271/S0025-5718-10-02342-2/home.html D.Kouznetsov, H.Trappmann. Portrait of the four regular super-exponentials to base sqrt(2). Mathematics of Computation, 2010, v.79, p.1727-1756. http://www.ils.uec.ac.jp/~dima/PAPERS/2011e1e.pdf H.Trappmann, D.Kouznetsov. Computation of the Two Regular Super-Exponentials to base exp(1/e). Mathematics of computation, in press, 2011. |
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Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE
(Elsevier, 2017-11)
Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ...
Investigations of anisotropic collectivity using multi-particle correlations in pp, p-Pb and Pb-Pb collisions
(Elsevier, 2017-11)
Two- and multi-particle azimuthal correlations have proven to be an excellent tool to probe the properties of the strongly interacting matter created in heavy-ion collisions. Recently, the results obtained for multi-particle ... |
1,766 69 1. Homework Statement
Let ##V## be a Banach space. Let ##f:V\to V## and ##g:V\to V## be two ##q##-contracting maps, ##q\in(0,1)##. Assume they are uniformly close to each other. Show the distance between fixed points of ##f,g## is at most ##\epsilon/(1-q)##.
2. Homework Equations
Definitions:
Uniformly close implies ##\forall \, \epsilon>0, v\in V## we have ##\| f(v) - g(v) \| < \epsilon##.
A map ##f## is ##q##-contracting if ##\exists \, q \in (0,1):\forall x,y\in\mathbb R^n, \| f(x) - f(y) \| \leq q\| x-y\|##.
3. The Attempt at a Solution
My idea is to iterate one map starting at the fixed point of the other map and show after sufficient amount of iterations, you will be within ##\epsilon## of the other fixed point. Since we are proving the distance between fixed points, the Banach Fixed Point Theorem seems relevant: $$\| x-x_n \| \leq \frac{q^n}{1-q}\| x_1-x_0 \|$$ where ##x## is the fixed point and ##x_n## is the ##n##th iteration.
So trying to put all this together, if ##x_f## is the fixed point of ##f##, then consider
$$\| g(x_f) - f(x_f) \| = \| g(x_f) - x_f \| \implies\\
\| g(x_f) - x_f \| < \epsilon
$$
But this can't be right. Can someone help me smooth it out? |
This page presents the
count model library proposed by Lixoft from the 2019 version on. It includes a short introduction on count data, the different ways to model this kind of data, and typical models. What is count data Formatting of count data in the MonolixSuite Important concepts: the probability mass function Encoding of count models in the MonolixSuite Library of parametric models for count data Case studies What is count data
Count data come from counting something, e.g., the number of trials required for completing a given task. The task can for instance be repeated several times (longitudinal count data) and the individuals performance followed.
Count data can also represent the number of events happening in regularly spaced intervals, e.g the number of seizures every week. If the time intervals are not regular, the data may be considered as repeated time-to-event interval censored, or the interval length can be given as regressor to be used to define the probability distribution in the model. Formatting of count data in the MonolixSuite Count data can take only non-negative integer values. The counts for each individual are recorded in the OBSERVATION column-type. If the data is longitudinal (over time), the times at which the counts have been recorded are indicated in the TIME column-type. If the data is not longitudinal, the TIME column is still necessary and a time of 0 can be used for instance. ID TIME Y 1 0 8 1 4 9 1 8 11 2 0 6 2 4 6 2 8 7
In the example above, the values in the time column can either indicate the time at which the counts have been counted, or the start or end time of the period over which the number of events are recorded.
In the Monolix GUI, the user must indicate that the data is of type Count/Categorical:
Important concept: the probability mass function
Count data are described by their probability mass function, which gives the probability that the discrete random variable is exactly equal to some value: \( \mathbb{P}(y_{ij}=k) \) with \(y_{ij}\) the observed count number for individual \(i\) at time \(j\) and \(k>0\).
For instance the probability mass function for a Poisson distribution is:
\(\mathbb{P}(y_{ij}=k)=\frac{\lambda^k e^{-\lambda}}{k!} \)
If \(\lambda=3\), then the probability of having a count of 0 is 4.9%, a count of 1 is 14.9%, a count of 2 is 22.4%, a count of 3 is 22.4%, etc.
Encoding of count models with the MonolixSuite
In Monolix, a model for count data is defined via the probability mass function, which in a population approach typically depends on individual parameters: \( \mathbb{P}(y_{ij}=k) = f(t_j,\psi_i)\).
The typical syntax to define the random variable representing the count number is the following: DEFINITION: CountNumber = {type=count, P(CountNumber=k) = ...}
CountNumber is the name of the random variable and can be replaced by another name. On the opposite, k is a mandatory name for the values. The random variable
CountNumber is then listed in the outputs, to be matched to the observed data.
The probability mass functions often involve ratios of factorials. While the ratio is often a reasonable value, the numerator and denominator can be huge values that are difficult to handle numerically.
It is therefore better to work with the log of the factorials. The function for log factorial in Mlxtran language is
factln() (for integers). To extend the factorials to the continuous domain, the gamma function can be useful: \( \Gamma(n)=(n-1)!\). The log of the gamma function corresponds to the
gammaln() function in Mlxtran. It is possible to define directly the log of the probability mass function with:
DEFINITION: CountNumber = {type=count, log(P(CountNumber=k)) = ...}
For instance to define a binomial distribution, we can write:
DEFINITION: Y = {type=count, log(P(Y=k))= gammaln(n+1) - factln(k) - gammaln(n-k+1) + k*log(p) + (n-k)*log(1-p)}
where
gammaln() is used for terms that can be non-integers and
factln() for terms that are integers.
It is possible to use “if” statements related to \(k\) in the definition of the probability mass function by using the syntax below. This can be useful to define zero-inflated models for instance. The “if” statements relating to time, regressors, or parameters values can be put in a
EQUATION: section before the
DEFINITION: block.
DEFINITION: CountNumber = {type=count, if k==0 Pk = ... else Pk = ... end P(CountNumber=k) = Pk}
The parameters involved in the definition of the probability mass function (for instance \(\lambda\) for a Poisson distribution) can be identical for all individuals, or vary from individual to individual. This is defined in the section “Individual model” of the graphical interface as usual. These parameters can also evolve over time or be a function of other variables such as drug concentration or tumor burden for instance. See the demo “PKcount_project.mlxtran” for instance in the Monolix demo folder, section 4.2.
Library of models for count data
To describe the various shapes that a count distribution can take, we have developed a library of models. The library includes the most typical distributions, their zero-inflated counterpart, and several options for the evolution of time of the main parameter:
The probability mass functions and a description of the parameters is given below:
Case studies
[coming soon] |
12th SSC CGL Tier II level Question Set, topic Trigonometry 3
This is the 12th question set for the 10 practice problem exercise for SSC CGL exam and 3rd on topic Trigonometry. The answers to the questions and link to the detailed solutions are given at the end.
We repeat the method of taking the test. It is important to follow result bearing methods even in practice test environment.
Method of taking the test for getting the best results from the test: Before start,you may refer to our tutorial or any short but good material to refresh your concepts if you so require. Basic and rich Trigonometric concepts and applications Answer the questionsin an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes. When the time limit of 12 minutes is over,mark up to which you have answered, but go on to complete the set. At the end,refer to the answers given at the end to mark your score at 12 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time. Identify and analyzethe problems that you couldn't doto learn how to solve those problems. Identify and analyzethe problems that you solved incorrectly. Identify the reasons behind the errors. If it is because of your shortcoming in topic knowledgeimprove it by referring to only that part of conceptfrom the best source you can get hold of. You might google it. If it is because of your method of answering,analyze and improve those aspects specifically. Identify and analyzethe problems that posed difficulties for you and delayed you. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques. Give a gapbefore you take a 10 problem practice test again.
Important:both and practice tests must be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance. mock tests Resources that should be useful for you You may refer to: or 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests to access all the valuable student resources that we have created specifically for SSC CGL, but section on SSC CGL generally for any hard MCQ test.
If you like,you may to get latest content from this place. subscribe 12th question set- 10 problems for SSC CGL Tier II exam: 3rd on Trigonometry - testing time 12 mins Problem 1.
If $2abcos \theta + (a^2-b^2)sin \theta=a^2+b^2$ then the value of $tan \theta$ is,
$\displaystyle\frac{1}{2ab}(a^2+b^2)$ $\displaystyle\frac{1}{2}(a^2-b^2)$ $\displaystyle\frac{1}{2}(a^2+b^2)$ $\displaystyle\frac{1}{2ab}(a^2-b^2)$ Problem 2.
$\displaystyle\frac{\sin^2 \theta}{\cos^2 \theta}+\displaystyle\frac{\cos^2 \theta}{\sin^2 \theta}$ is equal to,
$\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta}$ $\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta} -2$ $\displaystyle\frac{1}{\tan^2 \theta - \cot^2 \theta}$ $\displaystyle\frac{\sin^2 \theta}{\cot \theta - \sec \theta}$ Problem 3.
If $cos \theta + sec \theta = 2$, then the value of $cos^5 \theta + sec^5 \theta$ is,
$-2$ $2$ $1$ $-1$ Problem 4.
$sin(\alpha + \beta -\gamma)=cos(\beta + \gamma -\alpha)=\displaystyle\frac{1}{2}$ and $tan(\gamma + \alpha -\beta)=1$. If $\alpha$, $\beta$ and $\gamma$ are positive acute angles, value of $2\alpha + \beta$ is,
$105^0$ $110^0$ $115^0$ $120^0$ Problem 5.
If $sin \theta + sin^2 \theta=1$, then the value of $cos^{12} \theta + 3cos^{10} \theta + 3cos^{8} \theta + cos^6 \theta - 1$ is,
$1$ $0$ $2$ $3$ Problem 6.
The value of $sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right) - 2tan^2 \theta$ is,
4 0 2 1 Problem 7.
If $4cos^2 \theta - 4\sqrt{3}cos \theta + 3=0$ and $0^0 \leq \theta \leq 90^0$, then the value of $\theta$ is,
$60^0$ $90^0$ $30^0$ $45^0$ Problem 8.
If $sin \theta + cos \theta=\sqrt{2}sin(90^0 - \theta)$ then the value of $cot \theta$ is,
$\sqrt{2}-1$ $\sqrt{2}+1$ $-\sqrt{2}+1$ $-\sqrt{2}-1$ Problem 9.
If $x=asin \theta-bcos \theta$ and $y=acos \theta + bsin \theta$, then which of the following is true?
$x^2+y^2=a^2+b^2$ $\displaystyle\frac{x^2}{a^2}+ \displaystyle\frac{y^2}{b^2}=1$ $x^2+y^2=a^2-b^2$ $\displaystyle\frac{x^2}{y^2}+ \displaystyle\frac{a^2}{b^2}=1$ Problem 10.
If $tan \theta=\displaystyle\frac{a}{b}$, then the value of $\displaystyle\frac{asin^3 \theta - bcos^3 \theta}{asin^3 \theta + bcos^3 \theta}$ is,
$\displaystyle\frac{a^4-b^4}{a^4+b^4}$ $\displaystyle\frac{a^3+b^3}{a^3-b^3}$ $\displaystyle\frac{a^3-b^3}{a^3+b^3}$ $\displaystyle\frac{a^4+b^4}{a^4-b^4}$
For detailed solutions refer to the companion
SSC CGL Tier II Solution set 12 Trigonometry 3, questions with solutions.
You may also watch the video solutions at the two-part video below.
Part 1: Q1 to Q5 Part 2: Q6 to Q10
The answers to the questions are given below.
Answers to the questions. Problem 1. Answer: Option d: $\displaystyle\frac{1}{2ab}(a^2-b^2)$. Problem 2. Answer: Option b: $\displaystyle\frac{1}{\sin^2 \theta.cos^2 \theta} -2$. Problem 3. Answer: Option b: $2$. Problem 4. Answer: Option d: $120^0$. Problem 5. Answer: Option b: $0$. Problem 6. Answer: Option c: 2. Problem 7. Answer: Option c: $30^0$. Problem 8. Answer: Option b: $\sqrt{2}+1$. Problem 9. Answer: Option a: $x^2 + y^2=a^2 + b^2$. Problem 10. Answer: Option a: $\displaystyle\frac{a^4-b^4}{a^4+b^4}$. Resources on Trigonometry and related topics
You may refer to our useful resources on Trigonometry and other related topics especially algebra.
Tutorials on Trigonometry General guidelines for success in SSC CGL Efficient problem solving in Trigonometry A note on usability: The Efficient math problem solving sessions on School maths are equally usable for SSC CGL aspirants, as firstly, the "Prove the identity" problems can easily be converted to a MCQ type question, and secondly, the same set of problem solving reasoning and techniques have been used for any efficient Trigonometry problem solving. SSC CGL Tier II level question and solution sets on Trigonometry SSC CGL Tier II level Question set 12 Trigonometry 3, questions with answers |
The linked discussion thread comes from
comp.compilers, which at one time was the definitive discussion forum for questions of this form (disclaimer: I was a constant reader and occasional contributor to that forum, so my opinion might be biased.)
The rather interesting discussion thread in question dates from October 2002, and was an attempt to discover if there were unambiguous languages (
not grammars) which could not be parsed with any $LR$ algorithm. In the course of the discussion, Sönke Kannapinn noted that the $LR(k)$, $LALR(k)$ and $SLR(k)$ sets of languages are identical, for all $k\ge 0$. (That's the message cited in the question, but in a more readable presentation.)
Kannapinn cites the source of his assertion: the 1990 text
Parsing Theory Vol II by S. Sippu & E. Soisalon-Soininen, which definitely should be considered a "standard" textbook (as Kannapinn notes in an earlier message in the same thread). The question is addressed in Chapter 6.6 ( SLR(k) Parsing, from pages 70-84), which terminates with Theorem 6.71:
For any fixed $k>=0$, the families of $LR(k)$ languages, LALR(k) languages and SLR(k) languages are all equal.
The proof is too long to reproduce here, but the textbook should be available in any good university library. (If it isn't, demand that they acquire it.)
Coincidentally, there was another similar discussion thread the next year, in which I offered a simpler proof outline. I reproduce it here (without modification aside from an attempt to fight with MathJax and Markdown to make it a bit more readable), in case it's useful to someone, although of course it might be wrong since discussion posts in comp.compiler are not subject to the thorough review accorded to Springer Monographs. So if you find an error, let me know, and if you find the formatting ugly, feel free to fix it.
Take an $LR(1)$ grammar $$G = <N, T, P, S>$$ and create $$G' = <N', T, P', <S,\$> >$$ where $N'$ is a subset of $N \times T$ (that is, the non-terminals in G' are the cartesian product of the original non-terminals and the original terminals.)
$N'$ and $P'$ are constructed basically with the LR construction:
Start with $N' = \{<S, \$>\}$ and repeat the following until no new non-terminals are added:
It is clear that this must terminate, because all the sets are finite. Also, the function $FIRST$ used above never yields epsilon.
Let $<A, a>$ be an element of $N'$ for which there are no productions in $P'$. Now, for each production in $P$ of the form:
$$A \to w_1 w_2 ... w_n$$
add all possible productions
$$<A, a> \to w'_1 w'_2 ... w'_n$$
where
$$w'_i = w_i \text{ if } w_i \in T$$
$$w'_i = <w_i, b> \text{ for each } b \in FIRST(w_{i+1}...w_n ++ a)
\text{ if } w_ \in N$$
The second definition is multiple valued, so you might have to add a lot of rules :)
Furthermore, if any new rule mentions a non-terminal $<W, w>$ which is not yet in $N'$, add it to $N'$
Now, I assert that $G'$ recognises exactly the same language as $G$, and
furthermore that if $G$ is $LR(1)$ then $G'$ is $SLR(1)$, because it is impossible to have a reduce/reduce conflict between $<W, a>$ and $<W, b>$.
Of course, all this has done is move the lookahead sets into the definition of the non-terminals, so the resulting $SLR$ table is going to be just as big
as the original $LR$ table (bigger, actually, since no merging of lookaheads
has been done), but I think it does answer the original question. |
You can look at:
A. Grzegorczyk.
Some classes of recursive functions. Rozprawy Matematyczne, (IV), 1953.
R. W. Ritchie.
Classes of predictably computable functions. Trans. A.M.S., 106:139–173,1963.
(I don't know if you can find them for free in Internet).
Or download the Robert Daley's Lecture Notes "
Introduction to Theory of Computation" that contains a detailed explanation of the equivalence $PR \equiv LOOP \setminus WHILE$ (LOOP is a simple programming language).
Proof sketch
If $f \in PR$ then you can build a TM that computes $f$ and is time-bounded by a primitive recursive function. To prove it, you can use the derivation of $f$: start from
constant/successor/projection functions and then use induction on the operators composition/primitive recursion.
Conversely, if you have a $TM_{PR}$ that computes a function in $O(g)\ with\ g\ \in PR$ then you can build a LOOP program that emulates TM on each input $x$: if $m$ is the string representation of $TM_{PR}$ then the LOOP program will:
Prog:
1) decode the transition table from m
2) build a representation of the tape / head position / current state
3) FOR g(|x|) DO
3.1) simulate a single step of the TM (or do nothing if in a final state)
4) output the current representation of the tape
each step of Prog is primitive recursive (i.e. doesn't contain while / until); note that g used in the FOR statement is by hypothesis primitive recursive.
Prog is guarantee to end and correctly output $TM_{PR}(x)$ because $g(|x|)$ is a time bound for $TM_{PR}$.
Since $LOOP \setminus WHILE \equiv PR$ then the function computed by $TM_{PR}$ is in $PR$ |
To define the Schouten bracket I need to be able to sum over a cyclic permutation of the indices: $$ [\Phi,\Xi]_S=\mathfrak S_{i,j,k} \left(\Phi^{is}\partial_s\Xi^{jk}+ \Xi^{is}\partial_s\Phi^{jk}\right)\,,$$
where $\mathfrak S_{i,j,k} $ denotes the cyclic sum.
I am aware of the related posts however trying it myself I am not sure the implementation is correct:
ps = Permute[{i, j, k}, CyclicGroup[3]];Sch[A_, B_] := Sum[Sum[A[[ps[[r]][[1]],s]]D[B[[ps[[r]][[2]],ps[[r]][[3]]]],X[[s]]]+B[[ps[[r]][[1]],s]] D[A[[ps[[r]][[2]],ps[[r]][[3]]]],X[[s]]],{s,4},{r,3}],{i,4},{j,4},{k,4}]
The indices $i,j,k$ range over 1 to 4, and $X$ are my local coordinates.
EDIT: Indeed I was to brief, I won't to check that for example the following matrix $\Pi$:
{{0, 0, (1 + z1 zb1), z1 zb2}, {0, 0, z2 zb1, (1 - z1 zb1)}, {-(1 + z1 zb1), -z2 zb1, 0, 0}, {-z1 zb2, -(1 - z1 zb1), 0, 0}}
defines a Poisson bivector for the 2-dimensional projective space i.e. $[\Pi,\Pi]_S=0$. The local coordinates are $z_1,\bar z_1,z_2,\bar z_2$.
The reference that I am using is M. V. Karasev and V. P. Maslov - Nonlinear Poisson Brackets. Geometry and Quantization, this is the pertinent excerpt |
12th SSC CGL Tier II level Question Set, topic Trigonometry 3
This is the 12th question set for the 10 practice problem exercise for SSC CGL exam and 3rd on topic Trigonometry. The answers to the questions and link to the detailed solutions are given at the end.
We repeat the method of taking the test. It is important to follow result bearing methods even in practice test environment.
Method of taking the test for getting the best results from the test: Before start,you may refer to our tutorial or any short but good material to refresh your concepts if you so require. Basic and rich Trigonometric concepts and applications Answer the questionsin an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes. When the time limit of 12 minutes is over,mark up to which you have answered, but go on to complete the set. At the end,refer to the answers given at the end to mark your score at 12 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time. Identify and analyzethe problems that you couldn't doto learn how to solve those problems. Identify and analyzethe problems that you solved incorrectly. Identify the reasons behind the errors. If it is because of your shortcoming in topic knowledgeimprove it by referring to only that part of conceptfrom the best source you can get hold of. You might google it. If it is because of your method of answering,analyze and improve those aspects specifically. Identify and analyzethe problems that posed difficulties for you and delayed you. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques. Give a gapbefore you take a 10 problem practice test again.
Important:both and practice tests must be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance. mock tests Resources that should be useful for you You may refer to: or 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests to access all the valuable student resources that we have created specifically for SSC CGL, but section on SSC CGL generally for any hard MCQ test.
If you like,you may to get latest content from this place. subscribe 12th question set- 10 problems for SSC CGL Tier II exam: 3rd on Trigonometry - testing time 12 mins Problem 1.
If $2abcos \theta + (a^2-b^2)sin \theta=a^2+b^2$ then the value of $tan \theta$ is,
$\displaystyle\frac{1}{2ab}(a^2+b^2)$ $\displaystyle\frac{1}{2}(a^2-b^2)$ $\displaystyle\frac{1}{2}(a^2+b^2)$ $\displaystyle\frac{1}{2ab}(a^2-b^2)$ Problem 2.
$\displaystyle\frac{\sin^2 \theta}{\cos^2 \theta}+\displaystyle\frac{\cos^2 \theta}{\sin^2 \theta}$ is equal to,
$\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta}$ $\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta} -2$ $\displaystyle\frac{1}{\tan^2 \theta - \cot^2 \theta}$ $\displaystyle\frac{\sin^2 \theta}{\cot \theta - \sec \theta}$ Problem 3.
If $cos \theta + sec \theta = 2$, then the value of $cos^5 \theta + sec^5 \theta$ is,
$-2$ $2$ $1$ $-1$ Problem 4.
$sin(\alpha + \beta -\gamma)=cos(\beta + \gamma -\alpha)=\displaystyle\frac{1}{2}$ and $tan(\gamma + \alpha -\beta)=1$. If $\alpha$, $\beta$ and $\gamma$ are positive acute angles, value of $2\alpha + \beta$ is,
$105^0$ $110^0$ $115^0$ $120^0$ Problem 5.
If $sin \theta + sin^2 \theta=1$, then the value of $cos^{12} \theta + 3cos^{10} \theta + 3cos^{8} \theta + cos^6 \theta - 1$ is,
$1$ $0$ $2$ $3$ Problem 6.
The value of $sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right) - 2tan^2 \theta$ is,
4 0 2 1 Problem 7.
If $4cos^2 \theta - 4\sqrt{3}cos \theta + 3=0$ and $0^0 \leq \theta \leq 90^0$, then the value of $\theta$ is,
$60^0$ $90^0$ $30^0$ $45^0$ Problem 8.
If $sin \theta + cos \theta=\sqrt{2}sin(90^0 - \theta)$ then the value of $cot \theta$ is,
$\sqrt{2}-1$ $\sqrt{2}+1$ $-\sqrt{2}+1$ $-\sqrt{2}-1$ Problem 9.
If $x=asin \theta-bcos \theta$ and $y=acos \theta + bsin \theta$, then which of the following is true?
$x^2+y^2=a^2+b^2$ $\displaystyle\frac{x^2}{a^2}+ \displaystyle\frac{y^2}{b^2}=1$ $x^2+y^2=a^2-b^2$ $\displaystyle\frac{x^2}{y^2}+ \displaystyle\frac{a^2}{b^2}=1$ Problem 10.
If $tan \theta=\displaystyle\frac{a}{b}$, then the value of $\displaystyle\frac{asin^3 \theta - bcos^3 \theta}{asin^3 \theta + bcos^3 \theta}$ is,
$\displaystyle\frac{a^4-b^4}{a^4+b^4}$ $\displaystyle\frac{a^3+b^3}{a^3-b^3}$ $\displaystyle\frac{a^3-b^3}{a^3+b^3}$ $\displaystyle\frac{a^4+b^4}{a^4-b^4}$
For detailed solutions refer to the companion
SSC CGL Tier II Solution set 12 Trigonometry 3, questions with solutions.
You may also watch the video solutions at the two-part video below.
Part 1: Q1 to Q5 Part 2: Q6 to Q10
The answers to the questions are given below.
Answers to the questions. Problem 1. Answer: Option d: $\displaystyle\frac{1}{2ab}(a^2-b^2)$. Problem 2. Answer: Option b: $\displaystyle\frac{1}{\sin^2 \theta.cos^2 \theta} -2$. Problem 3. Answer: Option b: $2$. Problem 4. Answer: Option d: $120^0$. Problem 5. Answer: Option b: $0$. Problem 6. Answer: Option c: 2. Problem 7. Answer: Option c: $30^0$. Problem 8. Answer: Option b: $\sqrt{2}+1$. Problem 9. Answer: Option a: $x^2 + y^2=a^2 + b^2$. Problem 10. Answer: Option a: $\displaystyle\frac{a^4-b^4}{a^4+b^4}$. Resources on Trigonometry and related topics
You may refer to our useful resources on Trigonometry and other related topics especially algebra.
Tutorials on Trigonometry General guidelines for success in SSC CGL Efficient problem solving in Trigonometry A note on usability: The Efficient math problem solving sessions on School maths are equally usable for SSC CGL aspirants, as firstly, the "Prove the identity" problems can easily be converted to a MCQ type question, and secondly, the same set of problem solving reasoning and techniques have been used for any efficient Trigonometry problem solving. SSC CGL Tier II level question and solution sets on Trigonometry SSC CGL Tier II level Question set 12 Trigonometry 3, questions with answers |
Difference between revisions of "Timeline of prime gap bounds"
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2,114,964#? [m=3] ([http://terrytao.wordpress.com/2013/12/20/polymath8b-iv-enlarging-the-sieve-support-more-efficient-numerics-and-explicit-asymptotics/#comment-258451 Sutherland])
2,114,964#? [m=3] ([http://terrytao.wordpress.com/2013/12/20/polymath8b-iv-enlarging-the-sieve-support-more-efficient-numerics-and-explicit-asymptotics/#comment-258451 Sutherland])
+ +
| [http://math.mit.edu/~drew/admissible_35146_395154.txt 395,154]? [m=2] ([http://terrytao.wordpress.com/2013/12/20/polymath8b-iv-enlarging-the-sieve-support-more-efficient-numerics-and-explicit-asymptotics/#comment-258305 Sutherland])
| [http://math.mit.edu/~drew/admissible_35146_395154.txt 395,154]? [m=2] ([http://terrytao.wordpress.com/2013/12/20/polymath8b-iv-enlarging-the-sieve-support-more-efficient-numerics-and-explicit-asymptotics/#comment-258305 Sutherland])
Line 758: Line 760:
[http://math.mit.edu/~drew/admissible_1628944_24462790.txt 24,462,790]? [m=3] ([http://terrytao.wordpress.com/2013/12/20/polymath8b-iv-enlarging-the-sieve-support-more-efficient-numerics-and-explicit-asymptotics/#comment-258452 Sutherland])
[http://math.mit.edu/~drew/admissible_1628944_24462790.txt 24,462,790]? [m=3] ([http://terrytao.wordpress.com/2013/12/20/polymath8b-iv-enlarging-the-sieve-support-more-efficient-numerics-and-explicit-asymptotics/#comment-258452 Sutherland])
+ +
| A numerical precision issue was discovered in the earlier m=4 calculations
| A numerical precision issue was discovered in the earlier m=4 calculations
|}
|}
Revision as of 20:27, 22 December 2013 [math]H = H_1[/math] is a quantity such that there are infinitely many pairs of consecutive primes of distance at most [math]H[/math] apart. Would like to be as small as possible (this is a primary goal of the Polymath8 project). [math]k_0[/math] is a quantity such that every admissible [math]k_0[/math]-tuple has infinitely many translates which each contain at least two primes. Would like to be as small as possible. Improvements in [math]k_0[/math] lead to improvements in [math]H[/math]. (The relationship is roughly of the form [math]H \sim k_0 \log k_0[/math]; see the page on finding narrow admissible tuples.) More recent improvements on [math]k_0[/math] have come from solving a Selberg sieve variational problem. [math]\varpi[/math] is a technical parameter related to a specialized form of the Elliott-Halberstam conjecture. Would like to be as large as possible. Improvements in [math]\varpi[/math] lead to improvements in [math]k_0[/math], as described in the page on Dickson-Hardy-Littlewood theorems. In more recent work, the single parameter [math]\varpi[/math] is replaced by a pair [math](\varpi,\delta)[/math] (in previous work we had [math]\delta=\varpi[/math]). These estimates are obtained in turn from Type I, Type II, and Type III estimates, as described at the page on distribution of primes in smooth moduli.
In this table, infinitesimal losses in [math]\delta,\varpi[/math] are ignored.
Date [math]\varpi[/math] or [math](\varpi,\delta)[/math] [math]k_0[/math] [math]H[/math] Comments 10 Aug 2005 6 [EH] 16 [EH] ([Goldston-Pintz-Yildirim]) First bounded prime gap result (conditional on Elliott-Halberstam) 14 May 2013 1/1,168 (Zhang) 3,500,000 (Zhang) 70,000,000 (Zhang) All subsequent work (until the work of Maynard) is based on Zhang's breakthrough paper. 21 May 63,374,611 (Lewko) Optimises Zhang's condition [math]\pi(H)-\pi(k_0) \gt k_0[/math]; can be reduced by 1 by parity considerations 28 May 59,874,594 (Trudgian) Uses [math](p_{m+1},\ldots,p_{m+k_0})[/math] with [math]p_{m+1} \gt k_0[/math] 30 May 59,470,640 (Morrison)
58,885,998? (Tao)
59,093,364 (Morrison)
57,554,086 (Morrison)
Uses [math](p_{m+1},\ldots,p_{m+k_0})[/math] and then [math](\pm 1, \pm p_{m+1}, \ldots, \pm p_{m+k_0/2-1})[/math] following [HR1973], [HR1973b], [R1974] and optimises in m 31 May 2,947,442 (Morrison)
2,618,607 (Morrison)
48,112,378 (Morrison)
42,543,038 (Morrison)
42,342,946 (Morrison)
Optimizes Zhang's condition [math]\omega\gt0[/math], and then uses an improved bound on [math]\delta_2[/math] 1 Jun 42,342,924 (Tao) Tiny improvement using the parity of [math]k_0[/math] 2 Jun 866,605 (Morrison) 13,008,612 (Morrison) Uses a further improvement on the quantity [math]\Sigma_2[/math] in Zhang's analysis (replacing the previous bounds on [math]\delta_2[/math]) 3 Jun 1/1,040? (v08ltu) 341,640 (Morrison) 4,982,086 (Morrison)
4,802,222 (Morrison)
Uses a different method to establish [math]DHL[k_0,2][/math] that removes most of the inefficiency from Zhang's method. 4 Jun 1/224?? (v08ltu)
1/240?? (v08ltu)
4,801,744 (Sutherland)
4,788,240 (Sutherland)
Uses asymmetric version of the Hensley-Richards tuples 5 Jun 34,429? (Paldi/v08ltu) 4,725,021 (Elsholtz)
4,717,560 (Sutherland)
397,110? (Sutherland)
4,656,298 (Sutherland)
389,922 (Sutherland)
388,310 (Sutherland)
388,284 (Castryck)
388,248 (Sutherland)
387,982 (Castryck)
387,974 (Castryck)
[math]k_0[/math] bound uses the optimal Bessel function cutoff. Originally only provisional due to neglect of the kappa error, but then it was confirmed that the kappa error was within the allowed tolerance.
[math]H[/math] bound obtained by a hybrid Schinzel/greedy (or "greedy-greedy") sieve
6 Jun 387,960 (Angelveit)
387,904 (Angeltveit)
Improved [math]H[/math]-bounds based on experimentation with different residue classes and different intervals, and randomized tie-breaking in the greedy sieve. 7 Jun
26,024? (vo8ltu)
387,534 (pedant-Sutherland)
Many of the results ended up being retracted due to a number of issues found in the most recent preprint of Pintz. Jun 8 286,224 (Sutherland)
285,752 (pedant-Sutherland)
values of [math]\varpi,\delta,k_0[/math] now confirmed; most tuples available on dropbox. New bounds on [math]H[/math] obtained via iterated merging using a randomized greedy sieve. Jun 9 181,000*? (Pintz) 2,530,338*? (Pintz) New bounds on [math]H[/math] obtained by interleaving iterated merging with local optimizations. Jun 10 23,283? (Harcos/v08ltu) 285,210 (Sutherland) More efficient control of the [math]\kappa[/math] error using the fact that numbers with no small prime factor are usually coprime Jun 11 252,804 (Sutherland) More refined local "adjustment" optimizations, as detailed here.
An issue with the [math]k_0[/math] computation has been discovered, but is in the process of being repaired.
Jun 12 22,951 (Tao/v08ltu)
22,949 (Harcos)
249,180 (Castryck) Improved bound on [math]k_0[/math] avoids the technical issue in previous computations. Jun 13 Jun 14 248,898 (Sutherland) Jun 15 [math]348\varpi+68\delta \lt 1[/math]? (Tao) 6,330? (v08ltu)
6,329? (Harcos)
6,329 (v08ltu)
60,830? (Sutherland) Taking more advantage of the [math]\alpha[/math] convolution in the Type III sums Jun 16 [math]348\varpi+68\delta \lt 1[/math] (v08ltu) 60,760* (Sutherland) Attempting to make the Weyl differencing more efficient; unfortunately, it did not work Jun 18 5,937? (Pintz/Tao/v08ltu)
5,672? (v08ltu)
5,459? (v08ltu)
5,454? (v08ltu)
5,453? (v08ltu)
60,740 (xfxie)
58,866? (Sun)
53,898? (Sun)
53,842? (Sun)
A new truncated sieve of Pintz virtually eliminates the influence of [math]\delta[/math] Jun 19 5,455? (v08ltu)
5,453? (v08ltu)
5,452? (v08ltu)
53,774? (Sun)
53,672*? (Sun)
Some typos in [math]\kappa_3[/math] estimation had placed the 5,454 and 5,453 values of [math]k_0[/math] into doubt; however other refinements have counteracted this Jun 20 [math]178\varpi + 52\delta \lt 1[/math]? (Tao)
[math]148\varpi + 33\delta \lt 1[/math]? (Tao)
Replaced "completion of sums + Weil bounds" in estimation of incomplete Kloosterman-type sums by "Fourier transform + Weyl differencing + Weil bounds", taking advantage of factorability of moduli Jun 21 [math]148\varpi + 33\delta \lt 1[/math] (v08ltu) 1,470 (v08ltu)
1,467 (v08ltu)
12,042 (Engelsma) Systematic tables of tuples of small length have been set up here and here (update: As of June 27 these tables have been merged and uploaded to an online database of current bounds on [math]H(k)[/math] for [math]k[/math] up to 5000). Jun 22 Slight improvement in the [math]\tilde \theta[/math] parameter in the Pintz sieve; unfortunately, it does not seem to currently give an actual improvement to the optimal value of [math]k_0[/math] Jun 23 1,466 (Paldi/Harcos) 12,006 (Engelsma) An improved monotonicity formula for [math]G_{k_0-1,\tilde \theta}[/math] reduces [math]\kappa_3[/math] somewhat Jun 24 [math](134 + \tfrac{2}{3}) \varpi + 28\delta \le 1[/math]? (v08ltu)
[math]140\varpi + 32 \delta \lt 1[/math]? (Tao)
1,268? (v08ltu) 10,206? (Engelsma) A theoretical gain from rebalancing the exponents in the Type I exponential sum estimates Jun 25 [math]116\varpi+30\delta\lt1[/math]? (Fouvry-Kowalski-Michel-Nelson/Tao) 1,346? (Hannes)
1,007? (Hannes)
10,876? (Engelsma) Optimistic projections arise from combining the Graham-Ringrose numerology with the announced Fouvry-Kowalski-Michel-Nelson results on d_3 distribution Jun 26 [math]116\varpi + 25.5 \delta \lt 1[/math]? (Nielsen)
[math](112 + \tfrac{4}{7}) \varpi + (27 + \tfrac{6}{7}) \delta \lt 1[/math]? (Tao)
962? (Hannes) 7,470? (Engelsma) Beginning to flesh out various "levels" of Type I, Type II, and Type III estimates, see this page, in particular optimising van der Corput in the Type I sums. Integrated tuples page now online. Jun 27 [math]108\varpi + 30 \delta \lt 1[/math]? (Tao) 902? (Hannes) 6,966? (Engelsma) Improved the Type III estimates by averaging in [math]\alpha[/math]; also some slight improvements to the Type II sums. Tuples page is now accepting submissions. Jul 1 [math](93 + \frac{1}{3}) \varpi + (26 + \frac{2}{3}) \delta \lt 1[/math]? (Tao)
873? (Hannes)
Refactored the final Cauchy-Schwarz in the Type I sums to rebalance the off-diagonal and diagonal contributions Jul 5 [math] (93 + \frac{1}{3}) \varpi + (26 + \frac{2}{3}) \delta \lt 1[/math] (Tao)
Weakened the assumption of [math]x^\delta[/math]-smoothness of the original moduli to that of double [math]x^\delta[/math]-dense divisibility
Jul 10 7/600? (Tao) An in principle refinement of the van der Corput estimate based on exploiting additional averaging Jul 19 [math](85 + \frac{5}{7})\varpi + (25 + \frac{5}{7}) \delta \lt 1[/math]? (Tao) A more detailed computation of the Jul 10 refinement Jul 20 Jul 5 computations now confirmed Jul 27 633 (Tao)
632 (Harcos)
4,686 (Engelsma) Jul 30 [math]168\varpi + 48\delta \lt 1[/math]# (Tao) 1,788# (Tao) 14,994# (Sutherland) Bound obtained without using Deligne's theorems. Aug 17 1,783# (xfxie) 14,950# (Sutherland) Oct 3 13/1080?? (Nelson/Michel/Tao) 604?? (Tao) 4,428?? (Engelsma) Found an additional variable to apply van der Corput to Oct 11 [math]83\frac{1}{13}\varpi + 25\frac{5}{13} \delta \lt 1[/math]? (Tao) 603? (xfxie) 4,422?(Engelsma)
12 [EH] (Maynard)
Worked out the dependence on [math]\delta[/math] in the Oct 3 calculation Oct 21 All sections of the paper relating to the bounds obtained on Jul 27 and Aug 17 have been proofread at least twice Oct 23 700#? (Maynard) Announced at a talk in Oberwolfach Oct 24 110#? (Maynard) 628#? (Clark-Jarvis) With this value of [math]k_0[/math], the value of [math]H[/math] given is best possible (and similarly for smaller values of [math]k_0[/math]) Nov 19 105# (Maynard)
5 [EH] (Maynard)
600# (Maynard/Clark-Jarvis) One also gets three primes in intervals of length 600 if one assumes Elliott-Halberstam Nov 20 Optimizing the numerology in Maynard's large k analysis; unfortunately there was an error in the variance calculation Nov 21 68?? (Maynard)
582#*? (Nielsen])
59,451 [m=2]#? (Nielsen])
42,392 [m=2]? (Nielsen)
356?? (Clark-Jarvis) Optimistically inserting the Polymath8a distribution estimate into Maynard's low k calculations, ignoring the role of delta Nov 22 388*? (xfxie)
448#*? (Nielsen)
43,134 [m=2]#? (Nielsen)
698,288 [m=2]#? (Sutherland) Uses the m=2 values of k_0 from Nov 21 Nov 23 493,528 [m=2]#? Sutherland Nov 24 484,234 [m=2]? (Sutherland) Nov 25 385#*? (xfxie) 484,176 [m=2]? (Sutherland) Using the exponential moment method to control errors Nov 26 102# (Nielsen) 493,426 [m=2]#? (Sutherland) Optimising the original Maynard variational problem Nov 27 484,162 [m=2]? (Sutherland) Nov 28 484,136 [m=2]? (Sutherland Dec 4 64#? (Nielsen) 330#? (Clark-Jarvis) Searching over a wider range of polynomials than in Maynard's paper Dec 6 493,408 [m=2]#? (Sutherland) Dec 19 59#? (Nielsen)
10,000,000? [m=3] (Tao)
1,700,000? [m=3] (Tao)
38,000? [m=2] (Tao)
300#? (Clark-Jarvis)
182,087,080? [m=3] (Sutherland)
179,933,380? [m=3] (Sutherland)
More efficient memory management allows for an increase in the degree of the polynomials used; the m=2,3 results use an explicit version of the [math]M_k \geq \frac{k}{k-1} \log k - O(1)[/math] lower bound. Dec 20
55#? (Nielsen)
36,000? [m=2] (xfxie)
175,225,874? [m=3] (Sutherland)
27,398,976? [m=3] (Sutherland)
Dec 21 1,640,042? [m=3] (Sutherland) 429,798? [m=2] (Sutherland) Optimising the explicit lower bound [math]M_k \geq \log k-O(1)[/math] Dec 22 1,628,944? [m=3] (Castryck)
75,000,000? [m=4] (Castryck)
3,400,000,000? [m=5] (Castryck)
5,511? [EH] [m=3] (Sutherland)
2,114,964#? [m=3] (Sutherland)
309,954? [EH] [m=5] (Sutherland)
395,154? [m=2] (Sutherland)
1,523,781,850? [m=4] (Sutherland)
82,575,303,678? [m=5] (Sutherland)
A numerical precision issue was discovered in the earlier m=4 calculations Legend: ? - unconfirmed or conditional ?? - theoretical limit of an analysis, rather than a claimed record * - is majorized by an earlier but independent result # - bound does not rely on Deligne's theorems [EH] - bound is conditional the Elliott-Halberstam conjecture [m=N] - bound on intervals containing N+1 consecutive primes, rather than two strikethrough - values relied on a computation that has now been retracted
See also the article on
Finding narrow admissible tuples for benchmark values of [math]H[/math] for various key values of [math]k_0[/math]. |
A notebook is a collection of runnable cells (commands). When you use a notebook, you are primarily developing and running cells.
All notebook tasks are supported by UI actions, but you can also perform many tasks using keyboard shortcuts. Toggle the shortcut display by clicking the icon or selecting
? > .
This section describes how to develop notebook cells and navigate around a notebook.
In this section:
A notebook has a toolbar that lets you manage the notebook and perform actions within the notebook:
and one or more cells (or commands) that you can run:
At the far right of a cell, the cell actions , contains three menus:
Run, Dashboard, and Edit:
— —
and two actions: Hide and Delete .
To add a cell, mouse over a cell at the top or bottom and click the icon, or access the notebook cell menu at the far right, click , and select
Add Cell Above or Add Cell Below.
Go to the cell actions menu at the far right and click (Delete).
The primary language for each cell is shown in
( ) next to the notebook name:
You can override the primary language by specifying the language magic command
%<language> at the beginning of a cell. The supported magic commands are:
%python,
%r,
%scala, and
%sql.
Note
When you invoke a language magic command, the command is dispatched to the REPL in the execution context for the notebook. Variables defined in one language (and hence in the REPL for that language) are not available in the REPL of another language. REPLs can share state only through external resources such as files in DBFS or objects in object storage.
Additionally:
%sh
Allows you to execute shell code in your notebook. Add the
-eoption in order to fail the cell (and subsequently a job or a run all command) if the shell command has a non-zero exit status.
%fs
Allows you to use
dbutilsfilesystem commands. For more information, see dbutils.
%md
Allows you to include various types of documentation, including text, images, and mathematical formulas and equations.
In this section:
To include documentation in a notebook you can use the
%md magic command to identify
Markdown markup. The included Markdown markup is rendered into HTML. For example, this Markdown snippet contains markup for a level-one heading:
%md # Hello This is a Title
It is rendered as a HTML title:
Cells that appear after cells containing Markdown headings can be collapsed into the heading cell. The following image shows a level-one heading called
Heading 1 with the following two cells collapsed into it.
To expand and collapse headings, click the
+ and -.
Also see Hide and show cell content.
To display images stored in the FileStore, use the syntax:
%md
For example, let’s say you have the Databricks logo image file in FileStore:
dbfs ls dbfs:/FileStore/databricks-logo-mobile.png
When you include the following code in a Markdown cell:
the image is rendered in the cell:
Notebooks support KaTeX for displaying mathematical formulas and equations. For example,
%md\\(c = \\pm\\sqrt{a^2 + b^2} \\)\\(A{_i}{_j}=B{_i}{_j}\\)$$c = \\pm\\sqrt{a^2 + b^2}$$\\[A{_i}{_j}=B{_i}{_j}\\]
renders as:
and
%md\\( f(\beta)= -Y_t^T X_t \beta + \sum log( 1+{e}^{X_t\bullet\beta}) + \frac{1}{2}\delta^t S_t^{-1}\delta\\)where \\(\delta=(\beta - \mu_{t-1})\\)
renders as:
You can include HTML in a notebook by using the function
displayHTML. See HTML, D3, and SVG in Notebooks for an example of how to do this.
Note
The
displayHTML iframe is served from the domain
databricksusercontent.com and the iframe sandbox includes the
allow-same-origin attribute.
databricksusercontent.com must be accessible from your browser. If it is currently blocked by your corporate network, it will need to be whitelisted by IT.
You can have discussions with collaborators using command comments.
To toggle the Comments sidebar, click the
Comments button at the top right of a notebook.
To add a comment to a command:
Highlight the command text and click the comment bubble:
Add your comment and click
Comment.
To edit, delete, or reply to a comment, click the comment and choose an action.
To show line numbers or command numbers, click
View > Show line numbers or View > show command numbers. Once they’re displayed, you can hide them again from the same menu. You can also enable line numbers with the keyboard shortcut Control+L.
If you enable line or command numbers, Databricks saves your preference and shows them in all of your other notebooks for that browser.
Command numbers above cells link to that specific command. If you click on the command number for a cell, it updates your URL to be anchored to that command. If you want to link to a specific command in your notebook, right-click the command number and choose
copy link address.
To find and replace text within a notebook, select
File > Find and Replace.
The current match is highlighted in orange and all other matches are highlighted in yellow.
You can replace matches on an individual basis by clicking
Replace.
You can switch between matches by clicking the
Prev and Next buttons or pressing shift+enter and enter to go to the previous and next matches, respectively.
Close the find and replace tool by clicking the
x button or by pressing esc.
You can use Azure Databricks autocomplete features to automatically complete code segments as you enter them in cells. This reduces what you have to remember and minimizes the amount of typing you have to do. Azure Databricks supports two types of autocomplete in your notebook: local and server.
Local autocomplete completes words that exist in the notebook. Server autocomplete is more powerful because it accesses the cluster for defined types, classes, and objects, as well as SQL database and table names. To activate server autocomplete, you must attach your attach a notebook to a cluster and run all cells that define completable objects.
Important
Server autocomplete in R notebooks is blocked during command execution.
You trigger autocomplete by pressing
Tab after entering a completable object. For example, after you define and run the cells containing the definitions of
MyClass and
instance, the methods of
instance are completable, and a list of valid completions displays when you press
Tab.
Type completion and SQL database and table name completion work in the same way.
— —
This section describes how to run one or more notebook cells.
In this section:
The notebook must be attached to a cluster. If the cluster is not running, the cluster is started when you run one or more cells.
In the cell actions menu at the far right, click and select
Run Cell, or press shift+Enter.
Important
The maximum size for a notebook cell, both contents and output, is 16MB.
Note
By default, when you run a cell, the notebook automatically attaches to a running cluster without prompting. To change this setting, select
> User Settings > Notebook Settings.
For example, try executing these Python code snippets that reference the predefined variables.
spark
sqlContext
sc
Now that you’ve seen the pre-defined variables, run some real code:
1+1 # => 2
To run all cells above or below a cell, go to the cell actions menu at the far right, click , and select
Run All Above or Run All Below. Run All Below includes the cell you are in. Run All Above does not.
To run all the cells in a notebook, select
Run All in the notebook toolbar.
Important
Do not do a
Run All if steps for mount and unmount are in the same notebook. It could lead to a race condition and possibly corrupt the mount points.
Python and Scala notebooks support error highlighting. That is, the line of code that is throwing the error will be highlighted in the cell. Additionally, if the error output is a stacktrace, the cell in which the error is thrown is displayed in the stacktrace as a link to the cell. You can click this link to jump to the offending code.
Notifications alert you to certain events, such as which command is currently running during Run all cells and which commands are in error state. When your notebook is showing multiple error notifications, the first one will have a link that allows you to clear all notifications.
Notebook notifications are enabled by default. You can disable them under
> User Settings > Notebook Settings.
Databricks Advisor automatically analyzes commands every time they are run and displays appropriate advice in the notebooks. The advice notices provide information that can assist you in improving the performance of workloads, reducing costs, and avoiding common mistakes.
A blue box with a lightbulb icon signals that advice is available for a command. The box displays the number of distinct pieces of advice.
Click the lightbulb to expand the box and view the advice. One or more pieces of advice will become visible.
Click the
Learn more link to view documentation providing more information related to the advice.
Click the
Don’t show me this again link to hide the piece of advice. The advice of this type will no longer be displayed. This action can be reversed in Notebook Settings.
Click the lightbulb again to collapse the advice box.
Access the Notebook Settings page by selecting
> User Settings > Notebook Settings or by clicking the gear icon in the expanded advice box.
Toggle the
Turn on Databricks Advisor option to enable or disable advice.
The
Reset hidden advice link is displayed if one or more types of advice is currently hidden. Click the link to make that advice type visible again.
You can run a notebook from another notebook by using the
%run <notebook> magic command. This is roughly equivalent to a
:load command in a Scala REPL on your local machine or an
import statement in Python. All variables defined in
<notebook> become available in your current notebook.
%run must be in a cell
by itself, because it runs the entire notebook inline.
Note
You
cannot use
%run to run a Python file and
import the entities defined in that file into a notebook. To import from a Python file you must package the file into a Python library, create an Azure Databricks library from that Python library, and install the library into the cluster you use to run your notebook.
Example
Suppose you have
notebookA and
notebookB.
notebookA contains a cell that has the following Python code:
x = 5
Even though you did not define
x in
notebookB, you can access
x in
notebookB after you run
%run notebookA.
%run /Users/path/to/notebookAprint(x) # => 5
To specify a relative path, preface it with
./ or
../. For example, if
notebookA and
notebookB are in the same directory, you can alternatively run them from a relative path.
%run ./notebookAprint(x) # => 5
%run ../someDirectory/notebookA # up a directory and into anotherprint(x) # => 5
For more complex interactions between notebooks, see Notebook Workflows.
In this section:
To clear the notebook state and results, click
Clear in the notebook toolbar and select the action:
You can download a cell result that contains tabular output to your local machine. Click the button at the bottom of a cell.
A CSV file named
export.csv is downloaded to your default download directory.
By default Azure Databricks returns 1000 rows of a DataFrame. When there are more than 1000 rows, a down arrow is added to the button. To download all the results of a query:
Click the down arrow next to and select
Download full results.
Select
Re-execute and download.
After you download full results, a CSV file named
export.csvis downloaded to your local machine and the
/databricks-resultsfolder has a generated folder containing full the query results.
Cell content consists of cell code and the result of running the cell. You can hide and show the cell code and result using the cell actions menu at the top right of the cell.
To hide cell code:
Click and select Hide Code
To hide and show the cell result, do any of the following:
Click and select Hide Result Select Type Esc > Shift + o
To show hidden cell code or results, click the
Show links:
See also Collapsible headings.
Notebook isolation refers to the visibility of variables and classes between notebooks. Azure Databricks supports two types of isolation:
Variable and class isolation Spark session isolation
Note
Since all notebooks attached to the same cluster execute on the same cluster VMs, even with Spark session isolation enabled there is no guaranteed user isolation within a cluster.
Variables and classes are available only in the current notebook. For example, two notebooks attached to the same cluster can define variables and classes with the same name, but these objects are distinct.
To define a class that is visible to
all notebooks attached to the same cluster, define the class in a package cell. Then you can access the class by using its fully qualified name, which is the same as accessing a class in an attached Scala or Java library.
Every notebook attached to a cluster running Apache Spark 2.0.0 and above has a pre-defined variable called
spark that represents a
SparkSession.
SparkSession is the entry point for using Spark APIs as well as setting runtime configurations.
Spark session isolation is enabled by default. You can also use
global temporary views to share temporary views across notebooks. See Create View. To disable Spark session isolation, set
spark.databricks.session.share to
true in the Spark configuration.
Important
Setting
spark.databricks.session.share true breaks the monitoring used by both streaming notebook cells and streaming jobs. Specifically:
The graphs in streaming cells are not displayed. Jobs do not block as long as a stream is running (they just finish “successfully”, stopping the stream). Streams in jobs are not monitored for termination. Instead you must manually call
awaitTermination().
Calling the display function on streaming DataFrames doesn’t work.
Cells that trigger commands in other languages (that is, cells using
%scala,
%python,
%r, and
%sql) and cells that include other notebooks (that is, cells using
%run) are part of the current notebook. Thus, these cells are in the same session as other notebook cells. By contrast, a notebook workflow runs a notebook with an isolated
SparkSession, which means temporary views defined in such a notebook are
not visible in other notebooks.
Azure Databricks has basic version control for notebooks. You can perform the following actions on revisions:
To access notebook revisions, click
Revision History at the top right of the notebook toolbar.
Azure Databricks also integrates with these version control tools:
To add a comment to the latest revision:
Click the revision.
Click the
Save nowlink.
In the Save Notebook Revision dialog, enter a comment.
Click
Save. The notebook revision is saved with the entered comment.
To restore a revision:
Click the revision.
Click
Restore this revision.
Click
Confirm. The selected revision becomes the latest revision of the notebook.
To delete a notebook’s revision entry:
Click the revision.
Click the trash icon .
Click
Yes, erase. The selected revision is deleted from the notebook’s revision history.
To clear a notebook’s revision history:
Select
File > Clear Revision History.
Click
Yes, clear. The notebook revision history is cleared.
Warning
Once cleared, the revision history is not recoverable. |
I'm trying to wrap my head around ways to minimize total computational error (defined as a sum of the bounds on the truncation and rounding errors) by taking a differentiable function $f : \mathbb{R} \rightarrow \mathbb{R}$ and a finite difference approximation of its second derivative
$$ f''(x) = \frac{f(x + h) - 2f(x) + f(x-h)}{h^2} $$
I know that, by Taylor's Theorem
$$ f(x + h) = f(x) + f'(x)h + f''(x)\frac{h^2}{2} + f'''(\theta)\frac{h^3}{6} $$
for some $\theta \in [x, x + h]$.
How would you determine the value of $h$ for which a bound of the total computational error is minimized? |
Having $$f(n) = \sum_{k=0}^n g_n(k), \; g_n(x) = \min(2^x, 2^{2^{n-x}})$$ I want to know whether $\mathcal O(f(n)) \subsetneq \mathcal O(2^n)$. Since $g_n(x) \le 2^x$ it is at least $f(n) \in \mathcal O(2^{n+1}-1) = \mathcal O(2^n)$.
Let $x_n$ be the maximum point of $g_n(x)$, which is where $2^x = 2^{2^{n-x}}$. We get $$f(n) = \sum_{k=0}^{\lfloor x_n \rfloor} 2^k + \sum_{k=0}^{n - \lfloor x_n \rfloor} 2^{2^k}$$
Since $x_n$ solves $x + \log_2(x) = n$ we get for the second sum $$\sum_{k=0}^{n - \lfloor x_n \rfloor} 2^{2^k} = \sum_{k=0}^{ \lfloor \log_2(x_n) \rfloor} 2^{2^k} \le \sum_{k=0}^{ \lfloor x_n \rfloor} 2^k$$
Therefore $f(n) \in \mathcal O \big( \sum_{k=0}^{\lfloor x_n \rfloor} 2^k\big) = \mathcal O \big(2^{\lfloor x_n \rfloor} \big)$.
I'm quite unsure about the following lines, especially since this is the first time I've come across the Lambert-W-function, which is needed to solve $x + \log_2(x) = n$ (according to WolframAlpha): $$x_n = {W(2^n \log(2)) \over\log(2)}$$
From this paper, section 2.8, I've got $$ W(x) \in \mathcal O \Bigg(\log(x)-\log(\log(x))+{\log(\log(x)) \over \log(x)}\Bigg)$$
And plugging everything together (omitting constants):
$$ f(n) \in \mathcal O\big(2^{ W(2^n)}\big) = \mathcal O\Bigg(2^{ \log(2^n)-\log(\log(2^n))+{\log(\log(2^n)) \over \log(2^n)}}\Bigg) = \mathcal O\Big(2^n \cdot 2^{-\log(n)}\cdot \sqrt[n]{2^{\log(n)}}\Big) = \mathcal O\Big(2^n \cdot {1 \over n} \cdot 1\Big) = \mathcal O\Big({2^n \over n}\Big)$$
Is this correct? Is there some easy way to do it without Lambert-W? |
This is a really tough inequality (at least for me).
Can anyone help me show: $$\frac{1}{c}(1-(1-x)^c)^{c^{n}} + \frac{c-1}{c}(1-(1-x)^c)^c + (1-x)^{c-1}(1-x^{c^{n}}) \leq 1$$ within the range $0<x<1$ and $c \geq 4$ and $n \geq 2$ (and $c$ and $n$ are both integers).
I have plotted it over $x = 0 \text{ to } 1$ and it looks like this is completely true for all values I enter of $c$ and $n$, so long as $c$ is $\geq$ 4.
It is related to this question in that I believe proving this inequality here is sufficient for proving a small variation on the linked question over a subset of the desired range, and I can manually calculate the rest. I am posting it as a separate question, however, since it's not really the same thing. |
I'm trying to solve the limit of this inequality. The question goes as follows: If $$4x - 9 \leq f(x) \leq x^2 - 4x + 7$$ for $x \geq 0$, find $\lim_{x\to 4} f(x)$.
I'm not really sure how to go about this problem. I tried solving it using one-sided limits and got my answer as 7 but I'm not sure how to elegantly present my answer. I want to know how to present my answer in an appropriate mathematical way.
But I'm not sure if there is a way to solve it using Sandwich theorem. I would like to know if it's possible, as well.
Thanks. |
12th SSC CGL Tier II level Question Set, topic Trigonometry 3
This is the 12th question set for the 10 practice problem exercise for SSC CGL exam and 3rd on topic Trigonometry. The answers to the questions and link to the detailed solutions are given at the end.
We repeat the method of taking the test. It is important to follow result bearing methods even in practice test environment.
Method of taking the test for getting the best results from the test: Before start,you may refer to our tutorial or any short but good material to refresh your concepts if you so require. Basic and rich Trigonometric concepts and applications Answer the questionsin an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes. When the time limit of 12 minutes is over,mark up to which you have answered, but go on to complete the set. At the end,refer to the answers given at the end to mark your score at 12 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time. Identify and analyzethe problems that you couldn't doto learn how to solve those problems. Identify and analyzethe problems that you solved incorrectly. Identify the reasons behind the errors. If it is because of your shortcoming in topic knowledgeimprove it by referring to only that part of conceptfrom the best source you can get hold of. You might google it. If it is because of your method of answering,analyze and improve those aspects specifically. Identify and analyzethe problems that posed difficulties for you and delayed you. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques. Give a gapbefore you take a 10 problem practice test again.
Important:both and practice tests must be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance. mock tests Resources that should be useful for you You may refer to: or 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests to access all the valuable student resources that we have created specifically for SSC CGL, but section on SSC CGL generally for any hard MCQ test.
If you like,you may to get latest content from this place. subscribe 12th question set- 10 problems for SSC CGL Tier II exam: 3rd on Trigonometry - testing time 12 mins Problem 1.
If $2abcos \theta + (a^2-b^2)sin \theta=a^2+b^2$ then the value of $tan \theta$ is,
$\displaystyle\frac{1}{2ab}(a^2+b^2)$ $\displaystyle\frac{1}{2}(a^2-b^2)$ $\displaystyle\frac{1}{2}(a^2+b^2)$ $\displaystyle\frac{1}{2ab}(a^2-b^2)$ Problem 2.
$\displaystyle\frac{\sin^2 \theta}{\cos^2 \theta}+\displaystyle\frac{\cos^2 \theta}{\sin^2 \theta}$ is equal to,
$\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta}$ $\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta} -2$ $\displaystyle\frac{1}{\tan^2 \theta - \cot^2 \theta}$ $\displaystyle\frac{\sin^2 \theta}{\cot \theta - \sec \theta}$ Problem 3.
If $cos \theta + sec \theta = 2$, then the value of $cos^5 \theta + sec^5 \theta$ is,
$-2$ $2$ $1$ $-1$ Problem 4.
$sin(\alpha + \beta -\gamma)=cos(\beta + \gamma -\alpha)=\displaystyle\frac{1}{2}$ and $tan(\gamma + \alpha -\beta)=1$. If $\alpha$, $\beta$ and $\gamma$ are positive acute angles, value of $2\alpha + \beta$ is,
$105^0$ $110^0$ $115^0$ $120^0$ Problem 5.
If $sin \theta + sin^2 \theta=1$, then the value of $cos^{12} \theta + 3cos^{10} \theta + 3cos^{8} \theta + cos^6 \theta - 1$ is,
$1$ $0$ $2$ $3$ Problem 6.
The value of $sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right) - 2tan^2 \theta$ is,
4 0 2 1 Problem 7.
If $4cos^2 \theta - 4\sqrt{3}cos \theta + 3=0$ and $0^0 \leq \theta \leq 90^0$, then the value of $\theta$ is,
$60^0$ $90^0$ $30^0$ $45^0$ Problem 8.
If $sin \theta + cos \theta=\sqrt{2}sin(90^0 - \theta)$ then the value of $cot \theta$ is,
$\sqrt{2}-1$ $\sqrt{2}+1$ $-\sqrt{2}+1$ $-\sqrt{2}-1$ Problem 9.
If $x=asin \theta-bcos \theta$ and $y=acos \theta + bsin \theta$, then which of the following is true?
$x^2+y^2=a^2+b^2$ $\displaystyle\frac{x^2}{a^2}+ \displaystyle\frac{y^2}{b^2}=1$ $x^2+y^2=a^2-b^2$ $\displaystyle\frac{x^2}{y^2}+ \displaystyle\frac{a^2}{b^2}=1$ Problem 10.
If $tan \theta=\displaystyle\frac{a}{b}$, then the value of $\displaystyle\frac{asin^3 \theta - bcos^3 \theta}{asin^3 \theta + bcos^3 \theta}$ is,
$\displaystyle\frac{a^4-b^4}{a^4+b^4}$ $\displaystyle\frac{a^3+b^3}{a^3-b^3}$ $\displaystyle\frac{a^3-b^3}{a^3+b^3}$ $\displaystyle\frac{a^4+b^4}{a^4-b^4}$
For detailed solutions refer to the companion
SSC CGL Tier II Solution set 12 Trigonometry 3, questions with solutions.
You may also watch the video solutions at the two-part video below.
Part 1: Q1 to Q5 Part 2: Q6 to Q10
The answers to the questions are given below.
Answers to the questions. Problem 1. Answer: Option d: $\displaystyle\frac{1}{2ab}(a^2-b^2)$. Problem 2. Answer: Option b: $\displaystyle\frac{1}{\sin^2 \theta.cos^2 \theta} -2$. Problem 3. Answer: Option b: $2$. Problem 4. Answer: Option d: $120^0$. Problem 5. Answer: Option b: $0$. Problem 6. Answer: Option c: 2. Problem 7. Answer: Option c: $30^0$. Problem 8. Answer: Option b: $\sqrt{2}+1$. Problem 9. Answer: Option a: $x^2 + y^2=a^2 + b^2$. Problem 10. Answer: Option a: $\displaystyle\frac{a^4-b^4}{a^4+b^4}$. Resources on Trigonometry and related topics
You may refer to our useful resources on Trigonometry and other related topics especially algebra.
Tutorials on Trigonometry General guidelines for success in SSC CGL Efficient problem solving in Trigonometry A note on usability: The Efficient math problem solving sessions on School maths are equally usable for SSC CGL aspirants, as firstly, the "Prove the identity" problems can easily be converted to a MCQ type question, and secondly, the same set of problem solving reasoning and techniques have been used for any efficient Trigonometry problem solving. SSC CGL Tier II level question and solution sets on Trigonometry SSC CGL Tier II level Question set 12 Trigonometry 3, questions with answers |
I am quite new to differential equations and I have to solve the following $a(t)+b(t)C(t)+s a(t)C'(t)=0$, where $s$ is some constant. I read about differential equations and at this moment my main difficulty is that $C'(t)$ is multiplied by $s a(t)$ (the examples of differential equations I have seen so far have $s a(t)=1$). Does anyone have a hint about this equation (maybe it turns out to be a "famous" equation and I am not aware about it) and how to solve it?
What you did in your comment is correct up to (except a negative sign)
$$\exp{K(t)}\frac{b(t)C(t)}{sa(t)}+\exp{K(t)}C'(t)=-\exp{K(t)}s^{-1}$$
After that $$\left(\exp{K(t)}C(t)\right)'=-\exp{K(t)}s^{-1}$$
Integrate both sides $$\exp{K(t)}C(t)=-s^{-1}\int\exp{K(t)}\,dt+D\\ C(t)=-s^{-1}\exp{(-K(t))}\int\exp{K(t)}\,dt+D\exp{(-K(t))}$$ where $D$ is a constant. |
In physics books on classical field theory, the authors usually define the action as $$ S = \int\mathcal{L(\phi,\partial_\mu\phi)d^4x} $$ where $\mathcal{L}$ is the lagrangian density. Then, they say that the principle of least action states that systems evolve along a path in configuration space for which $S$ is an extremum. So far, so good.
Then, when deriving Euler-Lagrange Equations, they write $$ \delta S=0 \\ \iff \int\{\mathcal{\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi+ \frac{\partial L}{\partial(\partial_\mu\phi)}}\delta(\partial_\mu \phi) \} d^4x =0 $$
But they do not justify why this is true. Also, I do not have a clear, rigorous notion of what this "variation" $\delta$ is. Also: I have read somewhere that the principle of least action "is valid only in compact supported variations of the fields". Is this true? Although I know what compactly supported means, I do not know how that concept applies to this case.
Note: This argument can be found, for example, in Peskin and Schroeder's book on QFT. |
Suppose $M$ is a smooth manifold equipped with a Riemannian metric $g$. Given a curve $c$, let $P_c$ denote parallel transport along $c$. Now suppose you consider a new metric $g'=fg$ where $f$ is a smooth positive function. Let $P_c'$ denote parallel transport along $c$ with respect to $g'$. How are $P_c$ and $P_c$ related?
A similar question is: let $K:TTM \rightarrow TM$ denote the
connection map associated to $g$ and $K'$ the one associated to $g'$. How are $K$ and $K'$ related?
In case it's helpful, recall the definition of $K$: given $V\in T_{(x,v)}TM$, let $z(t)=(c(t),v(t))$ be a curve in $TM$ such that $z(0)=(x,v)$ and $\dot{z}(0)=V$. Then set $$K(V):=\nabla_{t}v(0).$$ |
12th SSC CGL Tier II level Question Set, topic Trigonometry 3
This is the 12th question set for the 10 practice problem exercise for SSC CGL exam and 3rd on topic Trigonometry. The answers to the questions and link to the detailed solutions are given at the end.
We repeat the method of taking the test. It is important to follow result bearing methods even in practice test environment.
Method of taking the test for getting the best results from the test: Before start,you may refer to our tutorial or any short but good material to refresh your concepts if you so require. Basic and rich Trigonometric concepts and applications Answer the questionsin an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes. When the time limit of 12 minutes is over,mark up to which you have answered, but go on to complete the set. At the end,refer to the answers given at the end to mark your score at 12 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time. Identify and analyzethe problems that you couldn't doto learn how to solve those problems. Identify and analyzethe problems that you solved incorrectly. Identify the reasons behind the errors. If it is because of your shortcoming in topic knowledgeimprove it by referring to only that part of conceptfrom the best source you can get hold of. You might google it. If it is because of your method of answering,analyze and improve those aspects specifically. Identify and analyzethe problems that posed difficulties for you and delayed you. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques. Give a gapbefore you take a 10 problem practice test again.
Important:both and practice tests must be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance. mock tests Resources that should be useful for you You may refer to: or 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests to access all the valuable student resources that we have created specifically for SSC CGL, but section on SSC CGL generally for any hard MCQ test.
If you like,you may to get latest content from this place. subscribe 12th question set- 10 problems for SSC CGL Tier II exam: 3rd on Trigonometry - testing time 12 mins Problem 1.
If $2abcos \theta + (a^2-b^2)sin \theta=a^2+b^2$ then the value of $tan \theta$ is,
$\displaystyle\frac{1}{2ab}(a^2+b^2)$ $\displaystyle\frac{1}{2}(a^2-b^2)$ $\displaystyle\frac{1}{2}(a^2+b^2)$ $\displaystyle\frac{1}{2ab}(a^2-b^2)$ Problem 2.
$\displaystyle\frac{\sin^2 \theta}{\cos^2 \theta}+\displaystyle\frac{\cos^2 \theta}{\sin^2 \theta}$ is equal to,
$\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta}$ $\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta} -2$ $\displaystyle\frac{1}{\tan^2 \theta - \cot^2 \theta}$ $\displaystyle\frac{\sin^2 \theta}{\cot \theta - \sec \theta}$ Problem 3.
If $cos \theta + sec \theta = 2$, then the value of $cos^5 \theta + sec^5 \theta$ is,
$-2$ $2$ $1$ $-1$ Problem 4.
$sin(\alpha + \beta -\gamma)=cos(\beta + \gamma -\alpha)=\displaystyle\frac{1}{2}$ and $tan(\gamma + \alpha -\beta)=1$. If $\alpha$, $\beta$ and $\gamma$ are positive acute angles, value of $2\alpha + \beta$ is,
$105^0$ $110^0$ $115^0$ $120^0$ Problem 5.
If $sin \theta + sin^2 \theta=1$, then the value of $cos^{12} \theta + 3cos^{10} \theta + 3cos^{8} \theta + cos^6 \theta - 1$ is,
$1$ $0$ $2$ $3$ Problem 6.
The value of $sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right) - 2tan^2 \theta$ is,
4 0 2 1 Problem 7.
If $4cos^2 \theta - 4\sqrt{3}cos \theta + 3=0$ and $0^0 \leq \theta \leq 90^0$, then the value of $\theta$ is,
$60^0$ $90^0$ $30^0$ $45^0$ Problem 8.
If $sin \theta + cos \theta=\sqrt{2}sin(90^0 - \theta)$ then the value of $cot \theta$ is,
$\sqrt{2}-1$ $\sqrt{2}+1$ $-\sqrt{2}+1$ $-\sqrt{2}-1$ Problem 9.
If $x=asin \theta-bcos \theta$ and $y=acos \theta + bsin \theta$, then which of the following is true?
$x^2+y^2=a^2+b^2$ $\displaystyle\frac{x^2}{a^2}+ \displaystyle\frac{y^2}{b^2}=1$ $x^2+y^2=a^2-b^2$ $\displaystyle\frac{x^2}{y^2}+ \displaystyle\frac{a^2}{b^2}=1$ Problem 10.
If $tan \theta=\displaystyle\frac{a}{b}$, then the value of $\displaystyle\frac{asin^3 \theta - bcos^3 \theta}{asin^3 \theta + bcos^3 \theta}$ is,
$\displaystyle\frac{a^4-b^4}{a^4+b^4}$ $\displaystyle\frac{a^3+b^3}{a^3-b^3}$ $\displaystyle\frac{a^3-b^3}{a^3+b^3}$ $\displaystyle\frac{a^4+b^4}{a^4-b^4}$
For detailed solutions refer to the companion
SSC CGL Tier II Solution set 12 Trigonometry 3, questions with solutions.
You may also watch the video solutions at the two-part video below.
Part 1: Q1 to Q5 Part 2: Q6 to Q10
The answers to the questions are given below.
Answers to the questions. Problem 1. Answer: Option d: $\displaystyle\frac{1}{2ab}(a^2-b^2)$. Problem 2. Answer: Option b: $\displaystyle\frac{1}{\sin^2 \theta.cos^2 \theta} -2$. Problem 3. Answer: Option b: $2$. Problem 4. Answer: Option d: $120^0$. Problem 5. Answer: Option b: $0$. Problem 6. Answer: Option c: 2. Problem 7. Answer: Option c: $30^0$. Problem 8. Answer: Option b: $\sqrt{2}+1$. Problem 9. Answer: Option a: $x^2 + y^2=a^2 + b^2$. Problem 10. Answer: Option a: $\displaystyle\frac{a^4-b^4}{a^4+b^4}$. Resources on Trigonometry and related topics
You may refer to our useful resources on Trigonometry and other related topics especially algebra.
Tutorials on Trigonometry General guidelines for success in SSC CGL Efficient problem solving in Trigonometry A note on usability: The Efficient math problem solving sessions on School maths are equally usable for SSC CGL aspirants, as firstly, the "Prove the identity" problems can easily be converted to a MCQ type question, and secondly, the same set of problem solving reasoning and techniques have been used for any efficient Trigonometry problem solving. SSC CGL Tier II level question and solution sets on Trigonometry SSC CGL Tier II level Question set 12 Trigonometry 3, questions with answers |
12th SSC CGL Tier II level Question Set, topic Trigonometry 3
This is the 12th question set for the 10 practice problem exercise for SSC CGL exam and 3rd on topic Trigonometry. The answers to the questions and link to the detailed solutions are given at the end.
We repeat the method of taking the test. It is important to follow result bearing methods even in practice test environment.
Method of taking the test for getting the best results from the test: Before start,you may refer to our tutorial or any short but good material to refresh your concepts if you so require. Basic and rich Trigonometric concepts and applications Answer the questionsin an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes. When the time limit of 12 minutes is over,mark up to which you have answered, but go on to complete the set. At the end,refer to the answers given at the end to mark your score at 12 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time. Identify and analyzethe problems that you couldn't doto learn how to solve those problems. Identify and analyzethe problems that you solved incorrectly. Identify the reasons behind the errors. If it is because of your shortcoming in topic knowledgeimprove it by referring to only that part of conceptfrom the best source you can get hold of. You might google it. If it is because of your method of answering,analyze and improve those aspects specifically. Identify and analyzethe problems that posed difficulties for you and delayed you. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques. Give a gapbefore you take a 10 problem practice test again.
Important:both and practice tests must be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance. mock tests Resources that should be useful for you You may refer to: or 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests to access all the valuable student resources that we have created specifically for SSC CGL, but section on SSC CGL generally for any hard MCQ test.
If you like,you may to get latest content from this place. subscribe 12th question set- 10 problems for SSC CGL Tier II exam: 3rd on Trigonometry - testing time 12 mins Problem 1.
If $2abcos \theta + (a^2-b^2)sin \theta=a^2+b^2$ then the value of $tan \theta$ is,
$\displaystyle\frac{1}{2ab}(a^2+b^2)$ $\displaystyle\frac{1}{2}(a^2-b^2)$ $\displaystyle\frac{1}{2}(a^2+b^2)$ $\displaystyle\frac{1}{2ab}(a^2-b^2)$ Problem 2.
$\displaystyle\frac{\sin^2 \theta}{\cos^2 \theta}+\displaystyle\frac{\cos^2 \theta}{\sin^2 \theta}$ is equal to,
$\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta}$ $\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta} -2$ $\displaystyle\frac{1}{\tan^2 \theta - \cot^2 \theta}$ $\displaystyle\frac{\sin^2 \theta}{\cot \theta - \sec \theta}$ Problem 3.
If $cos \theta + sec \theta = 2$, then the value of $cos^5 \theta + sec^5 \theta$ is,
$-2$ $2$ $1$ $-1$ Problem 4.
$sin(\alpha + \beta -\gamma)=cos(\beta + \gamma -\alpha)=\displaystyle\frac{1}{2}$ and $tan(\gamma + \alpha -\beta)=1$. If $\alpha$, $\beta$ and $\gamma$ are positive acute angles, value of $2\alpha + \beta$ is,
$105^0$ $110^0$ $115^0$ $120^0$ Problem 5.
If $sin \theta + sin^2 \theta=1$, then the value of $cos^{12} \theta + 3cos^{10} \theta + 3cos^{8} \theta + cos^6 \theta - 1$ is,
$1$ $0$ $2$ $3$ Problem 6.
The value of $sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right) - 2tan^2 \theta$ is,
4 0 2 1 Problem 7.
If $4cos^2 \theta - 4\sqrt{3}cos \theta + 3=0$ and $0^0 \leq \theta \leq 90^0$, then the value of $\theta$ is,
$60^0$ $90^0$ $30^0$ $45^0$ Problem 8.
If $sin \theta + cos \theta=\sqrt{2}sin(90^0 - \theta)$ then the value of $cot \theta$ is,
$\sqrt{2}-1$ $\sqrt{2}+1$ $-\sqrt{2}+1$ $-\sqrt{2}-1$ Problem 9.
If $x=asin \theta-bcos \theta$ and $y=acos \theta + bsin \theta$, then which of the following is true?
$x^2+y^2=a^2+b^2$ $\displaystyle\frac{x^2}{a^2}+ \displaystyle\frac{y^2}{b^2}=1$ $x^2+y^2=a^2-b^2$ $\displaystyle\frac{x^2}{y^2}+ \displaystyle\frac{a^2}{b^2}=1$ Problem 10.
If $tan \theta=\displaystyle\frac{a}{b}$, then the value of $\displaystyle\frac{asin^3 \theta - bcos^3 \theta}{asin^3 \theta + bcos^3 \theta}$ is,
$\displaystyle\frac{a^4-b^4}{a^4+b^4}$ $\displaystyle\frac{a^3+b^3}{a^3-b^3}$ $\displaystyle\frac{a^3-b^3}{a^3+b^3}$ $\displaystyle\frac{a^4+b^4}{a^4-b^4}$
For detailed solutions refer to the companion
SSC CGL Tier II Solution set 12 Trigonometry 3, questions with solutions.
You may also watch the video solutions at the two-part video below.
Part 1: Q1 to Q5 Part 2: Q6 to Q10
The answers to the questions are given below.
Answers to the questions. Problem 1. Answer: Option d: $\displaystyle\frac{1}{2ab}(a^2-b^2)$. Problem 2. Answer: Option b: $\displaystyle\frac{1}{\sin^2 \theta.cos^2 \theta} -2$. Problem 3. Answer: Option b: $2$. Problem 4. Answer: Option d: $120^0$. Problem 5. Answer: Option b: $0$. Problem 6. Answer: Option c: 2. Problem 7. Answer: Option c: $30^0$. Problem 8. Answer: Option b: $\sqrt{2}+1$. Problem 9. Answer: Option a: $x^2 + y^2=a^2 + b^2$. Problem 10. Answer: Option a: $\displaystyle\frac{a^4-b^4}{a^4+b^4}$. Resources on Trigonometry and related topics
You may refer to our useful resources on Trigonometry and other related topics especially algebra.
Tutorials on Trigonometry General guidelines for success in SSC CGL Efficient problem solving in Trigonometry A note on usability: The Efficient math problem solving sessions on School maths are equally usable for SSC CGL aspirants, as firstly, the "Prove the identity" problems can easily be converted to a MCQ type question, and secondly, the same set of problem solving reasoning and techniques have been used for any efficient Trigonometry problem solving. SSC CGL Tier II level question and solution sets on Trigonometry SSC CGL Tier II level Question set 12 Trigonometry 3, questions with answers |
(HP-65) The destructive bond pull test
01-09-2019, 02:30 PM
Post: #1
(HP-65) The destructive bond pull test
Extracts from the ABSTRACT and the Introduction of The Destructive Bond Test, US Commerce Dept., NBS Pub. 400-18, FEB 1976.
ABSTRACT
This report summarizes the work done at NBS on the destructive bond pull test as applied to small-diameter (approximately 1 mil or 25 μm) ultrasonically bonded aluminum wire … report begins with a,brief summary of the calculation of the resolution-of-forces operative in the bond system during the application of the pulling force …
INTRODUCTION
This report is an edited summary of the NBS effort on the destructive bond pull test.
The rationale for this work lies in the widespread use of this test method in the electro nics industry to evaluate the mechanical strength of wire bonds in semiconductor devices and in the large gap between the use of the pull test and the acquisition of reproducible calculable quantities which could be used to quantify the test results … to carry out this program, It was necessary to correlate the
measured pull strength as determined in the pull test with the stress in the wire, which depends on the geometry of the bond system. This was done through the resolution-of-forces calculation. An extensive series of experiments was undertaken in order to relate observed pull strengths to the results of the resolution-of-forces calculation … calculation of the resolution of the forces operative in the bond system during the application of the pulling force. A more complete derivation is presented in Appendix A and programs (in several different languages) which may be used for numerical calculation of the resolution-of-forces equations are given in Appendix B …
B1 . HP-65 Program for Resolution-of-Forces Calculation … 34 …
BEST!
SlideRule
01-12-2019, 12:45 PM
Post: #2
RE: (HP-65) The destructive bond pull test
For those who want to run the program in an emulator:
Code:
001: 23 : LBL
Quote:The angular variables \(\theta_t\) and \(\theta_d\), are often difficult to measure. However, the angles may be related to the variables \(h\), \(H\), \(d\), and \(\epsilon\) which are more easily accessible to measurement.
However it's much easier to calculate these angular variables using rectangle-to-polar coordinate transformation:
\((\epsilon d,h)\rightarrow(r_t, \theta_t)\)
\(((1-\epsilon)d,h+H)\rightarrow(r_d, \theta_d)\)
And then use the original formulas to calculate:
\(
F_{wt}=F\frac{\cos(\theta_d-\phi)}{\sin(\theta_t+\theta_d)}
\)
\(
F_{wd}=F\frac{\cos(\theta_t+\phi)}{\sin(\theta_t+\theta_d)}
\)
Here's my attempt:
Code:
001: 34 01 : RCL 1 h
Example:
Registers
01: 11.6
02: 3.5
03: 38.5
04: 0.375
05: 100
06: 20
R/S
103.48
x<>y
54.85
In addition to the that the angles \(\theta_t\) and \(\theta_d\) can be found in registers:
07: 38.78059606
08: 32.10960583
The exact values are:
54.85453818
103.4771281
Compare these to the result of the original program:
54.85453822
103.4771281
Cheers
Thomas
User(s) browsing this thread: 1 Guest(s) |
Solar Industry Growth and Affordability
Solar is playing an increasingly important role in the transition to a world powered by renewable energy. Over the past decade, the number of solar installations has grown at an accelerating rate and with increasing affordability. In the first quarter of 2016, over 29 GW of solar was installed in the United States.
Figure 1 illustrates the correlation of the increasing number of installations with the decreasing costs. It shows that the price of a solar installation is now less than a third of what it was in 2009, while annual installations have grown more than tenfold during the same period of time.
SEIA states that the solar industry is a powerful engine for economic growth. The US solar industry currently employs over 200,000 people, twice as many as in 2010 and now employs more people than the coal, or the oil and gas industries. As installed capacity continues to increase, SEIA predicts that the solar workforce will expand to 420,000 by 2020.
Solar Energy, Power, and Irradiance
Solar panels convert the energy of photons, or light particles, from the sun into electricity. Photovoltaic devices, such as solar panels, permit the incoming photons to transfer their energy to electrons. These energized electrons begin to flow, creating an electric current. We use the terms irradiance or insolation to refer to the power density of sunlight on a surface.
We typically measure energy in kilowatt-hours (kWh), and power (the rate at which energy is produced) in kilowatts (kW).
$$\text{Energy} = \text{Power }\cdot \text{Time} = 1\mathrm{kW} \cdot 1 \mathrm{hour} = 1\mathrm{kWh}$$
In solar, we usually define the size of a solar installation in terms of its power (in kW).
Irradiance is typically reported in units of kilowatt-hours per meter squared per day (kWh/m
2-d). The amount of irradiance hitting the surface of the earth is often quoted in terms of the number of hours of “full-sun” of solar energy. A "full-sun" is defined as 1 kW/m 2.
Quantity Units Definition Power kW Rate of energy production/output Energy kWh Capacity to do work Irradiance kWh/m 2-d Hours of full-sun for a square meter each day Solar Resource of a Rooftop
We can estimate the solar potential of a rooftop using its area and the local irradiance. NREL, the National Renewable Energy Laboratory, publishes irradiance data in its report
Solar Radiation Data Manual for Flat-Plate and Concentrating Collectors.
It is fairly straightforward to calculate rooftop solar potential of a rooftop using this data. For example, a south-facing roof plane of a home in Palo Alto, CA (Figure 5) receives an average irradiance of approximately 1,900 kWh/m
2/year. Dividing the annual irradiance value by the number of days in a year yields the average daily irradiance.
$$\text{Average Daily Irradiance} = \frac{\text{Annual Irradiance}}{\text{days/year}} = \frac{1900 \mathrm{kWh/m^2year}}{365 \mathrm{days/year}} = \mathrm{5.2 \mathrm{ kWh/m^2day}} $$
To calculate the amount of solar energy available on a roof face, multiply its area by the average irradiance value.
$$\text{Rooftop Energy}\mathrm{[\frac{kWh}{day}]} = \text{Irradiance}\mathrm{[\frac{kWh}{m^2 \cdot day}]} \times \text{Area}\mathrm{[m^2]}$$
If the rooftop has an area of approximately 150m
2, the solar energy available on the rooftop is as follows.
$$\text{Rooftop Energy} = 5.2 \frac{\mathrm{kWh}}{\mathrm{m^2}\cdot\mathrm{day}} \times 150\mathrm{m^2} = 780 \mathrm{\frac{kWh}{day}}$$
Besides the solar irradiance, Figure 5 also displays information on three additional quantities related to the solar resource: Solar Access, TOF, and TSRF:
Solar Access: This is the ratio of the actual solar energy available — taking into account shading cast by objects in the environment — to the solar energy that would be available in the absence of shading. You can learn more about the effects of shading on PV systems here.
$$\text{Solar Access} = \frac{\text{Energy with Shade}}{\text{Energy without shade}}$$
TOF (Tilt and Orientation Factor): This is the ratio of the amount of solar energy a location receives to the amount it would receive if the orientation of the roof were optimal. $$\text{TOF} = \frac{\text{Energy with actual tilt and orientation}}{\text{Energy with optimal tilt and orientation}}$$ TSRF (Total Solar Resource Factor): This is the percentage of the available solar resource that a location receives as compared to what it would receive with optimal orientation and without shading. TSRF is equivalent to the Solar Access multiplied by the Tilt and Orientation Factor. $$\text{TSRF} = \text{Solar Access}\times \text{TOF}$$ About Solar PV Education 101
Article 1: The Beginner's Guide to Solar Energy
Article 2: How a Photovoltaic System Produces Electricity Article 3: Reading Your Electricity Bill: A Beginner’s Guide Article 4: How to Size a PV System from an Electricity Bill Article 5: Shade Losses for PV Systems, and Techniques to Mitigate Them Article 6: The Basic Principles that Guide PV System Costs |
Answer
$\sqrt{17}(\cos(166^{\circ})+i\sin(166^{\circ}))$ (answer given to three significant figures)
Work Step by Step
To find the trigonometric form of a complex number from its Cartesian form, two things must be done. Firstly, the modulus of the complex number must be found. Secondly, the argument of the complex number must be found. To find the modulus of the complex number: Apply Pythagoras theorem to the coefficients of the complex number. i.e. $r=\sqrt{(-4)^{2}+(1)^{2}}\\=\sqrt{17}$ To find the argument of the complex number: First, find the basic argument of the complex number this is done by finding $\arctan$ of the fraction of the absolute value of the imaginary number coefficient over the coefficient of the real number. Therefore, the basic argument is $\arctan({\frac{1}{4}})\\=14.03624...^{\circ}$ Second, identify the quadrant that the line graph of the complex number lie in. Below are the conditions for the four quadrants that the complex number can lie in. First quadrant: both coefficient of real and imaginary parts are positive [argument=basic argument] Second quadrant: coefficient of real part is negative while coefficient of imaginary part is positive [argument=$180^{\circ}$-basic argument] Third quadrant: coefficient of real and imaginary part is negative [argument=$180^{\circ}$+basic argument] Fourth quadrant: coefficient of real part is positive while coefficient of imaginary part is negative [argument=$360^{\circ}$-basic argument] Since both the coefficient of the real part is negative and imaginary part is positive, the Cartesian equation lies in the second quadrant and the argument for the complex number trigonometric form is thus $180^{\circ}-14.03624...^{\circ}=165.9637...^{\circ}$ Thus, the trigonometric form of the equation of the Cartesian complex number $-4+i$ is $\sqrt{17}(\cos(166^{\circ})+i\sin(166^{\circ}))$ (answer given to three significant figures) |
Suppose $(\mathbb{R},\Sigma(m),m)$ is our measure space, where $m$ is Lebesgue measure. Also, suppose $f : \mathbb{R} \to [-\infty, \infty]$ is a Lebesgue measurable function.
The problem:
Prove that $f$ is equal almost everywhere to a Borel measurable function.
My attempt: We only have to prove this assertion for a non-negative Lebesgue measurable function $f$ because the result will follow for all Lebesgue measurable functions.
Furthermore, since any Lebesgue measurable function can be approximated by a sequence of non-negative, monotonically increasing, Lebesgue measurable simple functions $s_{n}$, we only need to prove the claim for an arbitrary Lebesgue measurable simple function.
So, let $s: \mathbb{R} \to [-\infty, \infty] $ be a Lebesgue measurable simple function. We can write $s$ canonically as $$ s(x) = \sum \limits_{i = 1}^{n} \alpha_{i} \chi_{A_{i}}(x)$$ where $\alpha_{i} \in [-\infty, \infty]$, $\bigcup \limits_{i = 1}^{n} A_{i} = \mathbb{R}$, and $A_{i} \cap A_{j} = \emptyset$ if $i \neq j$.
For each $i$, since $A_{i}$ is Lebesgue measurable, we can find a Borel set $B_{i}$ such that $A_{i} \subseteq B_{i}$ and $m(B_{i} \setminus A_{i}) = 0$. Clearly, this implies that $\bigcup \limits_{i = 1}^{n} B_{i} = \mathbb{R}$. Now we just need the $B_{i}$'s to be pairwise disjoint, with each $B_{i}$ still retaining $A_{i}$.
To make them pairwise disjoint, I constructed the following sets:
$\tilde{B_{i}} = [B_{i} \setminus (\bigcup \limits_{j \neq i} B_{j})] \cup A_{i}$. This construction gives us that the $\tilde{B_{i}}$'s are pairwise disjoint (I think....) and $A_{i} \subseteq \tilde{B_{i}}$. But I don't know that $\tilde{B_{i}}$ is necessarily still a Borel set. :( :( Am I approaching this problem all wrong? |
Difference between revisions of "Moser-lower.tex"
Line 33: Line 33:
least two.
least two.
(Recall Section \ref{notation-sec} for definitions),
(Recall Section \ref{notation-sec} for definitions),
− +
leads to the lower bound
\begin{equation}\label{cn3-low}
\begin{equation}\label{cn3-low}
c'_{n,3} \geq \binom{n}{i-1}
c'_{n,3} \geq \binom{n}{i-1}
Line 168: Line 168:
the bound for $n=6$ is inferior to the lower bound $c'_{6,3}\geq 344$ given above.
the bound for $n=6$ is inferior to the lower bound $c'_{6,3}\geq 344$ given above.
−
A modification of the construction in \eqref{cn3-low} leads to a slightly better lower bound. Observe that if $B \subset \Delta_n$, then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a Moser set as long as $B$ does not contain any ``isosceles triangles'' $(a+r,b,c+s), (a+s,b,c+r), (a,b+r+s,c)$ for any $r,s \geq 0$ not both zero; in particular, $B$ cannot contain any ``vertical line segments'' $(a+r,b,c+r), (a,b+2r,c)$. An example of such a set is provided by selecting $0 \leq i \leq n-3$ and letting $B$ consist of the triples $(a, n-i, i-a)$ when $a \neq 3 \mod 3$, $(a,n-i-1,i+1-a)$ when $a \neq 1 \mod 3$, $(a,n-i-2,i+2-a)$ when $a=0 \mod 3$, and $(a,n-i-3,i+3-a)$ when $a=2 \mod 3$. Asymptotically, this set occues about two thirds of the spheres $S_{n,i}$, $S_{n,i+1}$ and one third of the spheres $S_{n,i+2}, S_{n,i+3}$ and (setting $i$ close to $n/3$) gives a lower bound \eqref{cpn3} with $C = 2 \times \
+
A modification of the construction in \eqref{cn3-low} leads to a slightly better lower bound. Observe that if $B \subset \Delta_n$, then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a Moser set as long as $B$ does not contain any ``isosceles triangles'' $(a+r,b,c+s), (a+s,b,c+r), (a,b+r+s,c)$ for any $r,s \geq 0$ not both zero; in particular, $B$ cannot contain any ``vertical line segments'' $(a+r,b,c+r), (a,b+2r,c)$. An example of such a set is provided by selecting $0 \leq i \leq n-3$ and letting $B$ consist of the triples $(a, n-i, i-a)$ when $a \neq 3 \mod 3$, $(a,n-i-1,i+1-a)$ when $a \neq 1 \mod 3$, $(a,n-i-2,i+2-a)$ when $a=0 \mod 3$, and $(a,n-i-3,i+3-a)$ when $a=2 \mod 3$. Asymptotically, this set occues about two thirds of the spheres $S_{n,i}$, $S_{n,i+1}$ and one third of the spheres $S_{n,i+2}, S_{n,i+3}$ and (setting $i$ close to $n/3$) gives a lower bound \eqref{cpn3} with $C = 2 \times \{\{9}{4\pi}$, which is thus superior to the previous constructions.
An integer program was run to obtain the optimal lower bounds achievable by the $A_B$ construction (using \eqref{cn3}, of course). The results for
An integer program was run to obtain the optimal lower bounds achievable by the $A_B$ construction (using \eqref{cn3}, of course). The results for
Revision as of 11:16, 19 June 2009
\section{Lower bounds for the Moser problem}\label{moser-lower-sec}
In this section we discuss lower bounds for $c'_{n,3}$. Clearly we have $c'_{0,3}=1$ and $c'_{1,3}=2$, so we focus on the case $n \ge 2$. The first lower bounds may be due to Koml\'{o}s \cite{komlos}, who observed that the sphere $S_{i,n}$ of elements with exactly $n-i$ 2 entries (see Section \ref{notation-sec} for definition), is a Moser set, so that \begin{equation}\label{cin} c'_{n,3}\geq \vert S_{i,n}\vert \end{equation}
holds for all $i$. Choosing $i=\lfloor \frac{2n}{3}\rfloor$ and
applying Stirling's formula, we see that this lower bound takes the form \begin{equation}\label{cpn3} c'_{n,3} \geq (C-o(1)) 3^n / \sqrt{n} \end{equation} for some absolute constant $C>0$; in fact \eqref{cin} gives \eqref{cpn3} with $C := \sqrt{\frac{9}{4\pi}}$. In particular $c'_{3,3} \geq 12, c'_{4,3}\geq 24, c'_{5,3}\geq 80, c'_{6,3}\geq 240$. Asymptotically, the best lower bounds we know of are still of this type, but the values can be improved by studying combinations of several spheres or
semispheres or applying elementary results from coding theory.
Observe that if $\{w(1),w(2),w(3)\}$ is a geometric line in $[3]^n$, then $w(1), w(3)$ both lie in the same sphere $S_{i,n}$, and that $w(2)$ lies in a lower sphere $S_{i-r,n}$ for some $1 \leq r \leq i \leq n$. Furthermore, $w(1)$ and $w(3)$ are separated by Hamming distance $r$.
As a consequence, we see that $S_{i-1,n} \cup S_{i,n}^e$ (or $S_{i-1,n} \cup S_{i,n}^o$) is a Moser set for any $1 \leq i \leq n$, since any two distinct elements $S_{i,n}^e$ are separated by a Hamming distance of at least two. (Recall Section \ref{notation-sec} for definitions), This leads to the lower bound \begin{equation}\label{cn3-low}
c'_{n,3} \geq \binom{n}{i-1}
2^{i-1} + \binom{n}{i} 2^{i-1} = \binom{n+1}{i} 2^{i-1}. \end{equation} It is not hard to see that $\binom{n+1}{i+1} 2^{i} > \binom{n+1}{i} 2^{i-1}$ if and only if $3i < 2n+1$, and so this lower bound is maximised when $i = \lfloor \frac{2n+1}{3} \rfloor$ for $n \geq 2$, giving the formula \eqref{binom}. This leads to the lower bounds $$ c'_{2,3} \geq 6; c'_{3,3} \geq 16; c'_{4,3} \geq 40; c'_{5,3} \geq 120; c'_{6,3} \geq 336$$ which gives the right lower bounds for $n=2,3$, but is slightly off for $n=4,5$. Asymptotically, Stirling's formula and \eqref{cn3-low} then give the lower bound \eqref{cpn3} with $C = \frac{3}{2} \times \sqrt{\frac{9}{4\pi}}$, which is asymptotically $50\%$ better than the bound \eqref{cin}.
The work of Chv\'{a}tal \cite{chvatal1} already contained a refinement of this idea which we here translate into the usual notation of coding theory: Let $A(n,d)$ denote the size of the largest binary code of length $n$
and minimal distance $d$.
Then \begin{equation}\label{cnchvatal} c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right). \end{equation}
With the following values for $A(n,d)$: {\tiny{ \[ \begin{array}{llllllll} A(1,1)=2&&&&&&&\\ A(2,1)=4& A(2,2)=2&&&&&&\\ A(3,1)=8&A(3,2)=4&A(3,3)=2&&&&&\\ A(4,1)=16&A(4,2)=8& A(4,3)=2& A(4,4)=2&&&&\\ A(5,1)=32&A(5,2)=16& A(5,3)=4& A(5,4)=2&A(5,5)=2&&&\\ A(6,1)=64&A(6,2)=32& A(6,3)=8& A(6,4)=4&A(6,5)=2&A(6,6)=2&&\\ A(7,1)=128&A(7,2)=64& A(7,3)=16& A(7,4)=8&A(7,5)=2&A(7,6)=2&A(7,7)=2&\\ A(8,1)=256&A(8,2)=128& A(8,3)=20& A(8,4)=16&A(8,5)=4&A(8,6)=2 &A(8,7)=2&A(8,8)=2\\ A(9,1)=512&A(9,2)=256& A(9,3)=40& A(9,4)=20&A(9,5)=6&A(9,6)=4 &A(9,7)=2&A(9,8)=2\\ A(10,1)=1024&A(10,2)=512& A(10,3)=72& A(10,4)=40&A(10,5)=12&A(10,6)=6 &A(10,7)=2&A(10,8)=2\\ A(11,1)=2048&A(11,2)=1024& A(11,3)=144& A(11,4)=72&A(11,5)=24&A(11,6)=12 &A(11,7)=2&A(11,8)=2\\ A(12,1)=4096&A(12,2)=2048& A(12,3)=256& A(12,4)=144&A(12,5)=32&A(12,6)=24 &A(12,7)=4&A(12,8)=2\\ A(13,1)=8192&A(13,2)=4096& A(13,3)=512& A(13,4)=256&A(13,5)=64&A(12,6)=32 &A(13,7)=8&A(13,8)=4\\ \end{array} \] }}
Generally, $A(n,1)=2^n, A(n,2)=2^{n-1}, A(n-1,2e-1)=A(n,2e), A(n,d)=2$, if $d>\frac{2n}{3}$. The values were taken or derived from Andries Brower's table at\\ http://www.win.tue.nl/$\sim$aeb/codes/binary-1.html \textbf{include to references? or other book with explicit values of $A(n,d)$ }
For $c'_{n,3}$ we obtain the following lower bounds: with $k=2$ \[ \begin{array}{llll} c'_{4,3}&\geq &\binom{4}{0}A(4,3)+\binom{4}{1}A(3,2)+\binom{4}{2}A(2,1) =1\cdot 2+4 \cdot 4+6\cdot 4&=42.\\ c'_{5,3}&\geq &\binom{5}{0}A(5,3)+\binom{5}{1}A(4,2)+\binom{5}{2}A(3,1) =1\cdot 4+5 \cdot 8+10\cdot 8&=124.\\ c'_{6,3}&\geq &\binom{6}{0}A(6,3)+\binom{6}{1}A(5,2)+\binom{6}{2}A(4,1) =1\cdot 8+6 \cdot 16+15\cdot 16&=344. \end{array} \] With k=3 \[ \begin{array}{llll} c'_{7,3}&\geq& \binom{7}{0}A(7,4)+\binom{7}{1}A(6,3)+\binom{7}{2}A(5,2) + \binom{7}{3}A(4,1)&=960.\\ c'_{8,3}&\geq &\binom{8}{0}A(8,4)+\binom{8}{1}A(7,3)+\binom{8}{2}A(6,2) + \binom{8}{3}A(5,1)&=2832.\\ c'_{9,3}&\geq & \binom{9}{0}A(9,4)+\binom{9}{1}A(8,3)+\binom{9}{2}A(7,2) + \binom{9}{3}A(6,1)&=7880. \end{array}\] With k=4 \[ \begin{array}{llll} c'_{10,3}&\geq &\binom{10}{0}A(10,5)+\binom{10}{1}A(9,4)+\binom{10}{2}A(8,3) + \binom{10}{3}A(7,2)+\binom{10}{4}A(6,1)&=22232.\\ c'_{11,3}&\geq &\binom{11}{0}A(11,5)+\binom{11}{1}A(10,4)+\binom{11}{2}A(9,3) + \binom{11}{3}A(8,2)+\binom{11}{4}A(7,1)&=66024.\\ c'_{12,3}&\geq &\binom{12}{0}A(12,5)+\binom{12}{1}A(11,4)+\binom{12}{2}A(10,3) + \binom{12}{3}A(9,2)+\binom{12}{4}A(8,1)&=188688.\\ \end{array}\] With $k=5$ \[ c'_{13,3}\geq 539168.\]
It should be pointed out that these bounds are even numbers, so that $c'_{4,3}=43$ shows that one cannot generally expect this lower bound gives the optimum.
The maximum value appears to occur for $k=\lfloor\frac{n+2}{3}\rfloor$, so that using Stirling's formula and explicit bounds on $A(n,d)$ the best possible value known to date of the constant $C$ in equation \eqref{cpn3} can be worked out, but we refrain from doing this here. Using the Singleton bound $A(n,d)\leq 2^{n-d+1}$ Chv\'{a}tal \cite{chvatal1} proved that the expression on the right hand side of \eqref{cnchvatal} is also $O\left( \frac{3^n}{\sqrt{n}}\right)$, so that the refinement described above gains a constant factor over the initial construction only.
For $n=4$ the above does not yet give the exact value. The value $c'_{4,3}=43$ was first proven by Chandra \cite{chandra}. A uniform way of describing examples for the optimum values of $c'_{4,3}=43$ and $c'_{5,3}=124$ is the following:
Let us consider the sets $$ A := S_{i-1,n} \cup S_{i,n}^e \cup A'$$ where $A' \subset S_{i+1,n}$ has the property that any two elements in $A'$ are separated by a Hamming distance of at least three, or have a Hamming distance of exactly one but their midpoint lies in $S_{i,n}^o$. By the previous discussion we see that this is a Moser set, and we have the lower bound \begin{equation}\label{cnn} c'_{n,3} \geq \binom{n+1}{i} 2^{i-1} + |A'|. \end{equation} This gives some improved lower bounds for $c'_{n,3}$:
\begin{itemize} \item By taking $n=4$, $i=3$, and $A' = \{ 1111, 3331, 3333\}$, we obtain $c'_{4,3} \geq 43$; \item By taking $n=5$, $i=4$, and $A' = \{ 11111, 11333, 33311, 33331 \}$, we obtain $c'_{5,3} \geq 124$. \item By taking $n=6$, $i=5$, and $A' = \{ 111111, 111113, 111331, 111333, 331111, 331113\}$, we obtain $c'_{6,3} \geq 342$. \end{itemize}
This gives the lower bounds in Theorem \ref{moser} up to $n=5$, but the bound for $n=6$ is inferior to the lower bound $c'_{6,3}\geq 344$ given above.
A modification of the construction in \eqref{cn3-low} leads to a slightly better lower bound. Observe that if $B \subset \Delta_n$, then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a Moser set as long as $B$ does not contain any ``isosceles triangles
$(a+r,b,c+s), (a+s,b,c+r), (a,b+r+s,c)$ for any $r,s \geq 0$ not both zero; in particular, $B$ cannot contain any ``vertical line segments $(a+r,b,c+r), (a,b+2r,c)$. An example of such a set is provided by selecting $0 \leq i \leq n-3$ and letting $B$ consist of the triples $(a, n-i, i-a)$ when $a \neq 3 \mod 3$, $(a,n-i-1,i+1-a)$ when $a \neq 1 \mod 3$, $(a,n-i-2,i+2-a)$ when $a=0 \mod 3$, and $(a,n-i-3,i+3-a)$ when $a=2 \mod 3$. Asymptotically, this set occues about two thirds of the spheres $S_{n,i}$, $S_{n,i+1}$ and one third of the spheres $S_{n,i+2}, S_{n,i+3}$ and (setting $i$ close to $n/3$) gives a lower bound \eqref{cpn3} with $C = 2 \times \sqrt{\frac{9}{4\pi}}$, which is thus superior to the previous constructions.
An integer program was run to obtain the optimal lower bounds achievable by the $A_B$ construction (using \eqref{cn3}, of course). The results for $1 \leq n \leq 20$ are displayed in Figure \ref{nlow-moser}:
\begin{figure}[tb] \centerline{ \begin{tabular}{|ll|ll|} \hline n & lower bound & n & lower bound \\ \hline 1 & 2 &11& 71766\\ 2 & 6 & 12& 212423\\ 3 & 16 & 13& 614875\\ 4 & 43 & 14& 1794212\\ 5 & 122& 15& 5321796\\ 6 & 353& 16& 15455256\\ 7 & 1017& 17& 45345052\\ 8 & 2902&18& 134438520\\ 9 & 8622&19& 391796798\\ 10& 24786& 20& 1153402148\\ \hline \end{tabular}} \caption{Lower bounds for $c'_n$ obtained by the $A_B$ construction.} \label{nlow-moser} \end{figure}
More complete data, including the list of optimisers, can be found at {\tt http://abel.math.umu.se/~klasm/Data/HJ/}.
This indicates that greedily filling in spheres, semispheres or codes is no longer the optimal strategy in dimensions six and higher. The lower bound $c'_{6,3} \geq 353$ was first located by a genetic algorithm: see Appendix \ref{genetic-alg}.
\begin{figure}[tb] \centerline{\includegraphics{moser353new.png}} \caption{One of the examples of $353$-point sets in $[3]^6$ (elements of the set being indicated by white squares).} \label{moser353-fig} \end{figure}
Actually it is possible to improve upon these bounds by a slight amount. Observe that if $B$ is a maximiser for the right-hand side of \eqref{cn3} (subject to $B$ not containing isosceles triangles), then any triple $(a,b,c)$ not in $B$ must be the vertex of a (possibly degenerate) isosceles triangle with the other vertices in $B$. If this triangle is non-degenerate, or if $(a,b,c)$ is the upper vertex of a degenerate isosceles triangle, then no point from $\Gamma_{a,b,c}$ can be added to $A_B$ without creating a geometric line. However, if $(a,b,c) = (a'+r,b',c'+r)$ is only the lower vertex of a degenerate isosceles triangle $(a'+r,b',c'+r), (a',b'+2r,c')$, then one can add any subset of $\Gamma_{a,b,c}$ to $A_B$ and still have a Moser set as long as no pair of elements in that subset is separated by Hamming distance $2r$. For instance, in the $n=10$ case, the set $$B = \{(0 0 10),(0 2 8 ), (0 3 7 ),(0 4 6 ),(1 4 5 ),(2 1 7 ),(2 3 5 ),(3 2 5 ),(3 3 4 ),(3 4 3 ),(4 4 2 ),(5 1 4 ),(5 3 2 ),(6 2 2 ),(6 3 1 ),(6 4 0 ),(8 1 1 ),(9 0 1 ),(9 1 0 ) \}$$
generates the lower bound $c'_{10,3} \geq 24786$ given above (and, up to reflection $a \leftrightarrow c$, is the only such set that does so); but by adding twelve elements from $\Gamma_{5,0,5}$ one can increase the lower bound slightly to $24798$.
However, we have been unable to locate a lower bound which is asymptotically better than \eqref{cpn3}. Indeed, any method based purely on the $A_B$ construction cannot do asymptotically better than the previous constructions:
\begin{proposition} Let $B \subset \Delta_n$ be such that $A_B$ is a Moser set. Then $|A_B| \leq (2 \sqrt{\frac{9}{4\pi}} + o(1)) \frac{3^n}{\sqrt{n}}$. \end{proposition}
\begin{proof} By the previous discussion, $B$ cannot contain any pair of the form $(a,b+2r,c), (a+r,b,c+r)$ with $r>0$. In other words, for any $-n \leq h \leq n$, $B$ can contain at most one triple $(a,b,c)$ with $c-a=h$. From this and \eqref{cn3}, we see that $$ |A_B| \leq \sum_{h=-n}^n \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!}.$$ From the Chernoff inequality (or the Stirling formula computation below) we see that $\frac{n!}{a! b! c!} \leq \frac{1}{n^{10}} 3^n$ unless $a,b,c = n/3 + O( n^{1/2} \log^{1/2} n )$, so we may restrict to this regime, which also forces $h = O( n^{1/2}/\log^{1/2} n)$. If we write $a = n/3 + \alpha$, $b = n/3 + \beta$, $c = n/3+\gamma$ and apply Stirling's formula $n! = (1+o(1)) \sqrt{2\pi n} n^n e^{-n}$, we obtain $$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - (\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) - (\frac{n}{3}+\beta) \log (1 + \frac{3\beta}{n} ) - (\frac{n}{3}+\gamma) \log (1 + \frac{3\gamma}{n} ) ).$$ From Taylor expansion one has $$ (\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) = -\alpha - \frac{3}{2} \frac{\alpha^2}{n} + o(1)$$ and similarly for $\beta,\gamma$; since $\alpha+\beta+\gamma=0$, we conclude that $$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{2n} (\alpha^2+\beta^2+\gamma^2) ).$$ If $c-a=h$, then $\alpha^2+\beta^2+\gamma^2 = \frac{3\beta^2}{2} + \frac{h^2}{2}$. Thus we see that $$ \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!} \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{4n} h^2 ).$$ Using the integral test, we thus have $$ |A_B| \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \int_\R \exp( - \frac{3}{4n} x^2 )\ dx.$$ Since $\int_\R \exp( - \frac{3}{4n} x^2 )\ dx = \sqrt{\frac{4\pi n}{3}}$, we obtain the claim. \end{proof} |
Let $K$ be a complete field w.r.t discrete absolute value $|\cdot|_K$,
$\mathcal O_K=\{x\in K:|x|_K\leq 1\}$.
$L$ is an extension field with $[L:K]< \infty$ and let $\mathcal O_L$ be the integral closure of $\mathcal O_K$ in $L$.
Also let $\pi$ be the generator of the unique maximal ideal of the DVR $\mathcal O_K$.
If $\pi \mathcal O_L=\mathcal P^e$, with $\mathcal P$ the unique non zero prime in $\mathcal O_L$, then is it true that the absolute value $|\cdot|_{\mathcal P}$ on $L$ extends $|\cdot|_K$ on $K$?
Many thanks for your help. |
This is a naive answer.
The intersection over an empty collection of sets does not make sense (to me). The reason is this:
Suppose $S = \emptyset$. Now let $x$ be any comprehensible object (set, element whatever). Now I ask you why is $x$ not in $\bigcap S$. The only reason for $x$ to not be in $ \bigcap S $ is if $ (\forall A \in S, x \in A) $ is false. For this formula to be false there must exist a set $A'\in S$ such that $ x \not \in A' $ i.e. the negation of the "set-builder" formula is $ (\exists A' \in S \in , x \not \in A')$. But there can be no such $A'$ in $S$ since $S$ is empty.
Hence, anything belongs to $\bigcap S$. Whether this is a problem is for you to decide for yourself. It depends on the level of set-theoretic formlisation you wish to incorporate into your mathematics. If you believe in a "Universal Set" then no problem! But most would have seen big fat issues like Russell's Paradox resulting due to the thought of inclusive sets. |
So I am currently writing a computer program which among other things computes huge binary prime numbers. I am testing it on 16 digit numbers. So here is my question. So I generate 100 random odd numbers (aka a 16 digit string of binary numbers that begin and end with one). Then using the fact that $$ \pi(x) \approx \frac{x}{\log x} $$ Then obviously a rough approximation of the number of primes I will generate find in this range with 100 random guess is $$ 2 * 100\frac{\left(\frac{2^{16}}{\log{2^{16}}}-\frac{2^{15}-1}{\log(2^{15}-1)} \right)}{2^{16}-2^{15}+1} $$ since it would be twice the number of primes in the range $[2^{15}-1,2^{16}]$ since I am only considering odd numbers. However this gives me about ~38 primes and my code is generating consistently 16-25. So is my math wrong or is this approximation not good for this (relatively) small values of $\pi(x)$.
In case the exact number helps, Mathematica can compute
PrimePi[2^16] - PrimePi[2^15 - 1] to be $3030$. Choosing one-hundred odd integers uniformly at random from $[2^{15},2^{16}]$, the expected number of primes among them is $20200/10923\approx 18.4931$.
Calculating the approximation with the prime number theorem, I get approximately $16.8315$. The logarithms are supposed to be base-$e$: if I redo the calculation with base-$10$ logarithms, I get approximately $38.7559$. |
The Annals of Probability Ann. Probab. Volume 15, Number 2 (1987), 610-627. Large Deviations for Processes with Independent Increments Abstract
Let $\mathscr{X}$ be a topological space and $\mathscr{F}$ denote the Borel $\sigma$-field in $\mathscr{X}$. A family of probability measures $\{P_\lambda\}$ is said to obey the large deviation principle (LDP) with rate function $I(\cdot)$ if $P_\lambda(A)$ can be suitably approximated by $\exp\{-\lambda \inf_{x\in A}I(x)\}$ for appropriate sets $A$ in $\mathscr{F}$. Here the LDP is studied for probability measures induced by stochastic processes with stationary and independent increments which have no Gaussian component. It is assumed that the moment generating function of the increments exists and thus the sample paths of such stochastic processes lie in the space of functions of bounded variation. The LDP for such processes is obtained under the weak$^\ast$-topology. This covers a case which was ruled out in the earlier work of Varadhan (1966). As applications, the large deviation principle for the Poisson, Gamma and Dirichlet processes are obtained.
Article information Source Ann. Probab., Volume 15, Number 2 (1987), 610-627. Dates First available in Project Euclid: 19 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aop/1176992161 Digital Object Identifier doi:10.1214/aop/1176992161 Mathematical Reviews number (MathSciNet) MR885133 Zentralblatt MATH identifier 0624.60045 JSTOR links.jstor.org Subjects Primary: 60F10: Large deviations Secondary: 60J30 60E07: Infinitely divisible distributions; stable distributions Citation
Lynch, James; Sethuraman, Jayaram. Large Deviations for Processes with Independent Increments. Ann. Probab. 15 (1987), no. 2, 610--627. doi:10.1214/aop/1176992161. https://projecteuclid.org/euclid.aop/1176992161 |
String Searching by KMP algorithm (Knuth Morris Pratt algorithm) Motivation Problem: Given $$2$$ strings $$P$$(pattern) and $$T$$(text), find the number of occurrences of $$P$$ in $$T$$. Basic / Brute Force solution:
One obvious and easy to code solution that comes to mind is this: For each index of $$T$$, take it as a starting point and find if $$T_{i,i+1,...,i+|P|-1}$$ is equal to $$P$$.
for i = 0 to length(T)-length(P): Found = true for j = i to i + length(P) - 1: if P[j-i] not equal to T[j]: Found = False if Found = True answer = answer + 1
This brute force takes $$O(|P| \cdot |T|)$$ time in the worst case, which is obviously too slow for large strings.
Knuth Morris Pratt Algorithm:
Suppose for each index $$i$$ of some string $$Z$$, the longest suffix in $$Z_{0,1,...,i}$$ that is also a prefix of $$Z_{0,1,...,i}$$, be known. Formally, a length $$F_i$$ is known such that $$Z_{0,1,...,F_i-1}$$ = $$Z_{i-F_i+1,...,i}$$. Let these lengths be stored in array $$F$$. The suffix needs to be proper(whole string is not a proper suffix).
Then the solution to the motivation problem can be found as follows:
Define a string $$V = P + '#' + T$$, where $$'#'$$ is a delimiter that is not present in either of $$P$$ or $$T$$. Now, if the above information is known, all occurrences of $$P$$ in $$T$$ can be found as follows: If at some index $$i$$, $$F_i = |P|$$, then there is an occurrence of Pattern $$P$$ at position $$i-|P|+1$$. All such indices from $$|P|+1$$ [0 based indexing, the index just after '#'], need to be checked.
The main part of KMP algorithm calculates the array $$F$$, which is also called the prefix function. If calculation of $$F$$ or the prefix function can be done efficiently, then we have an efficient solution to the motivation problem. KMP algorithm finds the prefix function in $$O(length of String)$$ time.
To find the prefix function, best possible use of previous values of array $$F$$ is made, so that calculations aren't done again and again. Suppose all $$F_i$$ have been calculated, and now $$F_{i+1}$$ is to be calculated. It is to be noted that, value of $$F_{i+1}$$ can be at most 1 greater than $$F_i$$. Here is a proof by contradiction:
Suppose $$F_{i+1} > F_i + 1$$. Now, if the $$(i+1)^{th}$$ character is removed, we obtain a suffix ending at index $$i$$ that is of length $$F_{i+1} - 1$$, which is greater than $$F_i$$. This is a contradiction, hence proved.
Observe that if $$Z_{i+1} = Z_{F_i}$$, then the value of $$F_{i+1} = F_i + 1$$. If not, a smaller suffix ending at index $$i$$ is to be found, that is also a prefix of $$Z_{0,1...,i}$$. Let the length of such a suffix be $$j$$, then if $$Z_{i+1} = Z_{j}$$ then $$F_{i+1} = j + 1$$. If again the equality doesn't hold true, smaller and smaller suffixes that end at index $$i$$, which are also prefixes of $$Z_{0,1...,i}$$ need to be found.
The only thing remaining is, how to find the length of next smaller suffix ending at index $$i$$, that is also a prefix? This is also pretty simple. Observe that due to the property of $$F$$, the segment $$Z_{0,1,...,F_i-1}$$ is equal to the segment $$Z_{i-F_i+1,...,i}$$. So to find the next smaller suffix ending at index $$i$$, the longest suffix ending at $$F_i - 1$$ can be found which is $$F_{F_i-1}$$, and this suffix will be the next smaller suffix ending at index $$i$$. If this suffix also doesn't satisfy our criteria, then smaller suffixes can be found with the same process, here it will be $$F_{F_{F_i-1} - 1}$$. Note that, if at some point the length becomes $$0$$, the process is stopped.
This completes $$KMP$$ algorithm. Below is the code:
vector<int> prefix_function (string Z) { int n = (int) Z.length(); vector<int> F (n); F[0]=0; for (int i=1; i<n; ++i) { int j = F[i-1]; while (j > 0 && Z[i] != Z[j]) j = F[j-1]; if (Z[i] == Z[j]) ++j; F[i] = j; } return F;}
Finding the $$F$$ array for
"ABCABC"
Initially, $$F_0 = 0$$.
Index $$1 \rightarrow F_{0} = 0$$, $$j$$ does not go into while loop and $$Z_j \neq Z_i$$, therefore value of $$F_i = 0$$.
Index $$2 \rightarrow F_{1} = 0$$, $$j$$ does not go into while loop and $$Z_j \neq Z_i$$, therefore value of $$F_i = 0$$.
Index $$3 \rightarrow F_{2} = 0$$, $$j$$ does not go into while loop and $$Z_j = Z_i$$, therefore value of $$F_i = 1$$.
Index $$4 \rightarrow F_{3} = 1$$, $$j$$ satisfies while loop condition but $$Z_j = Z_i$$, hence does not go into while loop, therefore value of $$F_i = 2$$.
Index $$5 \rightarrow F_{4} = 2$$, $$j$$ satisfies while loop condition but $$Z_j = Z_i$$, therefore value of $$F_i = 3$$. |
Recently the question If $\frac{d}{dx}$ is an operator, on what does it operate? was asked on mathoverflow. It seems that some users there objected to the question, apparently interpreting it as an elementary inquiry about what kind of thing is a differential operator, and on this interpretation, I would agree that the question would not be right for mathoverflow. And so the question was closed down (and then reopened, and then closed again….
sigh). (Update 12/6/12: it was opened again,and so I’ve now posted my answer over there.)
Meanwhile, I find the question to be more interesting than that, and I believe that the OP intends the question in the way I am interpreting it, namely, as a logic question, a question about the nature of mathematical reference, about the connection between our mathematical symbols and the abstract mathematical objects to which we take them to refer. And specifically, about the curious form of variable binding that expressions involving $dx$ seem to involve. So let me write here the answer that I had intended to post on mathoverflow:
————————-
To my way of thinking, this is a serious question, and I am not really satisfied by the other answers and comments, which seem to answer a different question than the one that I find interesting here.
The problem is this. We want to regard $\frac{d}{dx}$ as an operator in the abstract senses mentioned by several of the other comments and answers. In the most elementary situation, it operates on a functions of a single real variable, returning another such function, the derivative. And the same for $\frac{d}{dt}$.
The problem is that, described this way, the operators $\frac{d}{dx}$ and $\frac{d}{dt}$ seem to be the
same operator, namely, the operator that takes a function to its derivative, but nevertheless we cannot seem freely to substitute these symbols for one another in formal expressions. For example, if an instructor were to write $\frac{d}{dt}x^3=3x^2$, a student might object, “don’t you mean $\frac{d}{dx}$?” and the instructor would likely reply, “Oh, yes, excuse me, I meant $\frac{d}{dx}x^3=3x^2$. The other expression would have a different meaning.”
But if they are the same operator, why don’t the two expressions have the same meaning? Why can’t we freely substitute different names for this operator and get the same result? What is going on with the logic of reference here?
The situation is that the operator $\frac{d}{dx}$ seems to make sense only when applied to functions whose independent variable is described by the symbol “x”. But this collides with the idea that what the function is at bottom has nothing to do with the way we represent it, with the particular symbols that we might use to express which function is meant. That is, the function is the abstract object (whether interpreted in set theory or category theory or whatever foundational theory), and is not connected in any intimate way with the symbol “$x$”. Surely the functions $x\mapsto x^3$ and $t\mapsto t^3$, with the same domain and codomain, are simply different ways of describing exactly the same function. So why can’t we seem to substitute them for one another in the formal expressions?
The answer is that the syntactic use of $\frac{d}{dx}$ in a formal expression involves a kind of binding of the variable $x$.
Consider the issue of
collision of bound variables in first order logic: if $\varphi(x)$ is the assertion that $x$ is not maximal with respect to $\lt$, expressed by $\exists y\ x\lt y$, then $\varphi(y)$, the assertion that $y$ is not maximal, is not correctly described as the assertion $\exists y\ y\lt y$, which is what would be obtained by simply replacing the occurrence of $x$ in $\varphi(x)$ with the symbol $y$. For the intended meaning, we cannot simply syntactically replace the occurrence of $x$ with the symbol $y$, if that occurrence of $x$ falls under the scope of a quantifier.
Similarly, although the functions $x\mapsto x^3$ and $t\mapsto t^3$ are equal as functions of a real variable, we cannot simply syntactically substitute the expression $x^3$ for $t^3$ in $\frac{d}{dt}t^3$ to get $\frac{d}{dt}x^3$. One might even take the latter as a kind of ill-formed expression, without further explanation of how $x^3$ is to be taken as a function of $t$.
So the expression $\frac{d}{dx}$ causes a binding of the variable $x$, much like a quantifier might, and this prevents free substitution in just the way that collision does. But the case here is not quite the same as the way $x$ is a bound variable in $\int_0^1 x^3\ dx$, since $x$ remains free in $\frac{d}{dx}x^3$, but we would say that $\int_0^1 x^3\ dx$ has the same meaning as $\int_0^1 y^3\ dy$.
Of course, the issue evaporates if one uses a notation, such as the $\lambda$-calculus, which insists that one be completely explicit about which syntactic variables are to be regarded as the independent variables of a functional term, as in $\lambda x.x^3$, which means the function of the variable $x$ with value $x^3$. And this is how I take several of the other answers to the question, namely, that the use of the operator $\frac{d}{dx}$ indicates that one has previously indicated which of the arguments of the given function is to be regarded as $x$, and it is with respect to this argument that one is differentiating. In practice, this is almost always clear without much remark. For example, our use of $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ seems to manage very well in complex situations, sometimes with dozens of variables running around, without adopting the onerous formalism of the $\lambda$-calculus, even if that formalism is what these solutions are essentially really about.
Meanwhile, it is easy to make examples where one must be very specific about which variables are the independent variable and which are not, as Todd mentions in his comment to David’s answer. For example, cases like
$$\frac{d}{dx}\int_0^x(t^2+x^3)dt\qquad
\frac{d}{dt}\int_t^x(t^2+x^3)dt$$
are surely clarified for students by a discussion of the usage of variables in formal expressions and more specifically the issue of bound and free variables. |
How can we evaluate the following?
$$\int e^{\frac{y}{x}}\ \mathrm dy$$
An explanation of the answer would be helpful.
The answer I got is $ x e^{y/x}$.
But not sure about the steps used for obtaining the answer...
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How can we evaluate the following?
$$\int e^{\frac{y}{x}}\ \mathrm dy$$
An explanation of the answer would be helpful.
The answer I got is $ x e^{y/x}$.
But not sure about the steps used for obtaining the answer...
$$\begin{align} \int_{- \infty} e^{\frac{y}{x}}dy &= \int_{- \infty}^y e^{\frac{t}{x}}dt \\ &= \int_{- \infty}^y e^{\frac{xu}{x}}d(xu) \\ &= x \int_{- \infty}^{y \over x} e^u d u \\ &= x \vert_{u = {- \infty}}^{y \over x} e^u \\ &= x \ e^{y \over x} \end{align}$$
Since this a task on finding a primitive, you can always check you candidate answer by taking (partial in your case) derivative.
First note that $$\int a^t\ \mathrm dt=\frac{a^t}{\ln a}+C$$ Therefore $$\int e^{\frac{y}{x}}\ \mathrm dy=\int \left(e^{\frac{1}{x}}\right)^y\ \mathrm dy$$ $$=\frac{\left(e^{\frac{1}{x}}\right)^y}{\ln e^{\frac{1}{x}}}+C=xe^{\frac{y}{x}}+C$$ |
For massless spinors case we can decompose momentum into Weyl sub-parts as
$$p = \lambda_{a}\tilde \lambda_{\dot a}.$$
But for the case of massive fermions can I do something like this? Decompose them into Weyl subparts with some additional terms? If so, how?
Why do I need it? I am performing a twistor transform for the equation of the process $q\bar q \to gg$ so I have to write the amplitude and the 4d delta function of the momentum and then Fourier transform the $\lambda$ and $\tilde \lambda$ separately. |
Answer
$6$
Work Step by Step
Let $x$ be the number of copy machines for 23 secretaries. $\frac{4}{15} = \frac{x}{23}$ $x=\frac{4}{15}\times23$ $x=\frac{92}{15}$ $x\approx6$
You can help us out by revising, improving and updating this answer.Update this answer
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First of all, I going to give you a quick explanation how Iterative Learning Control(ILC) is. But first a quick history.
Iterative Learning Control is a very modern control technique who is most used for system who repeat its dynamical system. For example: Robotics.
ILC was invented at the 70's in Japan, but it took about 30 years for the idea to be used in the industry. Yes, ILC is an industry controller, not only for experiment.
The idea behind ILC is to remember the "past disturbances" in form of past tracking error and past input signal. Example: If you are holding a pencil and writing a round circle on a paper, when you have write the half circle, somebody push you and your circle didn't become 100% round. Then to try again to write the round circle and you remember that when you are at the half circle position, you are going to be peppered for that push again, so you can finish your circle and have it 100% round as you planned to have it in the beginning.
That how ILC works. In other words - ILC is feed forwarding the past disturbances into the system to handle the coming disturbance.
So let's assume that we have a feedback system:
$$ G_f(s) = \frac{u_{k+1}G(s)}{1+u_{k+1}G(s)}$$
Where $u_{k+1}$ is our controller output signal. In this case, I'm using transfer functions. But you can use state space models if you want. ILC is a SISO-contoller, but you can have multiple ILC controller for a MIMO system.
Anyway!
Our $u_{k+1}$ is:
$$u_{k+1} = Q(s)(u_k(t) + L(s)e_k(t)) + C(s)e_{k+1}(t)$$
So what does this mean? Well: $Q(s)$ is a low pass filter, for example butterworth. $u_k(s)$ is our past input signal and $e_k(t)$ is also our past tracking error. That means that we need to record our current input signal $u_{k+1}$ and current tracking error $e_{k+1}(t)$ for the system.
$k$ is our iteration state. We begin always with $k = 0$.
The transfer function $L(s)$ is called our learning algorithm. I don't know why.
$$L(s) = \delta s^\kappa$$
Where
$$0 < \delta \leq 1$$
So how should you interpret $$L(s) = \delta s^\kappa$$
??
The factor $\delta$ is a scale factor only for learning gain. The factor $\kappa$ is telling the system that "OK! Now we are going to use acceleration if $\kappa = 2$". In other words $\kappa$ is a time shift gain.
$L(s)$ works as an amplifier for the past error $e_{k}(t)$
So assume that we have $\kappa = 1$ and $\delta = 1$. Then we would have $L(s) = s$ as a transfer function for derivative for frequencies. And in this case, frequencies is our disturbances. You can see the difference for $L(s)$ if we are using $\kappa =1, 2$
And the last thing! $Cs(s)$ is our real controller. In this case, $C(s)$ can be a P, PI, PD, PID controller. Example:
$$C(s) = K_P + K_I\frac{1}{s} + K_Ds$$
So from this:
$$u_{k+1} = Q(s)(u_k(t) + L(s)e_k(t)) + C(s)e_{k+1}(t)$$
The feedback controller is: $$C(s)e_{k+1}(t)$$
And the feed forward compensator for disturbance is:
$$Q(s)(u_k(t) + L(s)e_k(t))$$
Where the current error $e_k(t)$ is
$$e_k(t) = r(t) - y(t)$$
You may wonder: Why use $Q(s)$? Its for cutting of high frequency signals back to the controller. In some cases, you can set $Q(s) = 1$.
If this was not clear how ILC works. Have a look at this picture. This picture explains that for every iteration, the controller can stand against disturbance more and more.
Or look at this video: https://www.youtube.com/watch?v=sWilGsWQ1jo
Now for the question:
How do I find robustness in this loop transfer function?
$$0 = 1 + \begin{bmatrix} Q(s)G(s) & Q(s)L(s)G(s) & C(s)G(s) \end{bmatrix} \begin{bmatrix} u_k(t)\\ e_k(t)\\ e_{k+1}(t) \end{bmatrix}$$
It's from this feedback model:
$$ G_f(s) = \frac{u_{k+1}G(s)}{1+u_{k+1}G(s)}$$
Where if $u_{k+1}G(s)$ goes to $-1$, them our controller is not robust and not stable at all. |
Q. In the cube of side 'a' shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be:
Solution:
$\vec{r}_{g} = \frac{a}{2} \hat{i} + \frac{a}{2} \hat{k} $ $ \vec{r}_{H} =\frac{a}{2} \hat{j} + \frac{a}{2} \hat{k} $ $\vec{ r}_{H} - \vec{r}_{g} = \frac{a}{2} \left(\hat{j} - \hat{i}\right) $ Questions from JEE Main 2019 4. One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop $(B_L)$ to that at the centre of the coil $(B_C)$ , i.e., $R \frac{B_L}{B_C}$ will be 9. The magnetic field associated with a light wave is given, at the origin, by $B = B_0 [\sin(3.14 \times 10^7)ct + \sin(6.28 \times 10^7)ct]$.
If this light falls on a silver plate having a work
function of 4.7 eV, what will be the maximum
kinetic energy of the photo electrons ?
$(c = 3 \times 10^{8} ms^{-1}, h = 6.6 \times 10^{-34} J-s)$ Physics Most Viewed Questions 1. An AC ammeter is used to measure current in a circuit. When a given direct current passes through the circuit, the AC ammeter reads 3 A. When another alternating current passes through the circuit, the AC ammeter reads 4 A. Then the reading of this ammeter, if DC and AC flow through the circuit simultaneously, is 9. A cannon of mass I 000 kg, located at the base of an inclined plane fires a shelI of mass 100 kg in a horizontal direction with a velocity 180 $kmh^{-1}$. The angle of inclination of the inclined plane with the horizontal is 45$^{\circ}$. The coefficient of friction between the cannon and the inclined plane is 0.5. The height, in metre, to which the cannon ascends the inclined plane as a result of the recoiI is (g = 10 $ms^{-1})$ Latest Updates Top 10 Medical Entrance Exams in India Top 10 Engineering Entrance Exams in India JIPMER Results Announced NEET UG Counselling Started NEET Result Announced KCET Result Announced KCET College Predictor JEE Main Result Announced JEE Advanced Score Cards Available AP EAMCET Result Announced KEAM Result Announced UPSEE – Online Applications invited from NRI &Kashmiri Migrants MHT CET Result Announced NEET Rank Predictor Questions Tardigrade |
ISSN:
1930-8337
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All Issues
Inverse Problems & Imaging
February 2010 , Volume 4 , Issue 1
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Abstract:
We extend the classical spectral estimation problem to the infinite-dimensional case and propose a new approach to this problem using the Boundary Control (BC) method. Several applications to inverse problems for partial differential equations are provided.
Abstract:
Let $z=Au+\gamma$ be an ill-posed, linear operator equation. Such a model arises, for example, in both astronomical and medical imaging, in which case $\gamma$ corresponds to background, $u$ the unknown true image, $A$ the forward operator, and $z$ the data. Regularized solutions of this equation can be obtained by solving
$R_\alpha(A,z)= arg\min_{u\geq 0} \{T_0(Au;z)+\alpha J(u)\},$
where $T_0(Au;z)$ is the negative-log of the Poisson likelihood functional, and $\alpha>0$ and $J$ are the regularization parameter and functional, respectively. Our goal in this paper is to determine general conditions which guarantee that $R_\alpha$ defines a
regularization scheme for $z=Au+\gamma$. Determining the appropriate definition for regularization scheme in this context is important: not only will it serve to unify previous theoretical arguments in this direction, it will provide a framework for future theoretical analyses. To illustrate the latter, we end the paper with an application of the general framework to a case in which an analysis has not been done. Abstract:
In the context of scattering problems in the harmonic regime, we consider the problem of identification of some Generalized Impedance Boundary Conditions (GIBC) at the boundary of an object (which is supposed to be known) from far field measurements associated with a single incident plane wave at a fixed frequency. The GIBCs can be seen as approximate models for thin coatings, corrugated surfaces or highly absorbing media. After pointing out that uniqueness does not hold in the general case, we propose some additional assumptions for which uniqueness can be restored. We also consider the question of stability when uniqueness holds. We prove in particular Lipschitz stability when the impedance parameters belong to a compact subset of a finite dimensional space.
Abstract:
We consider the interior transmission eigenvalue problem corresponding to the inverse scattering problem for an isotropic inhomogeneous medium. We first prove that transmission eigenvalues exist for media with index of refraction greater or less than one without assuming that the contrast is sufficiently large. Then we show that for an arbitrary Lipshitz domain with constant index of refraction there exists an infinite discrete set of transmission eigenvalues that accumulate at infinity. Finally, for the general case of non constant index of refraction we provide a lower and an upper bound for the first transmission eigenvalue in terms of the first transmission eigenvalue for appropriate balls with constant index of refraction.
Abstract:
It is proved that, in two dimensions, the Calderón inverse conductivity problem in Lipschitz domains is stable in the $L^p$ sense when the conductivities are uniformly bounded in any fractional Sobolev space $W^{\alpha,p}$ $\alpha>0, 1 < p < \infty$.
Abstract:
For finite dimensional CMV matrices the classical inverse spectral problems are considered. We solve the inverse problem of reconstructing a CMV matrix by its Weyl's function, the problem of reconstructing the matrix by two spectra of CMV operators with different "boundary condition'', and the problem of reconstructing a CMV matrix by its spectrum and the spectrum of the CMV matrix obtained from it by unitary truncation.
Abstract:
We consider the tomography problem of recovering a covector field on a simple Riemannian manifold based on its weighted Doppler transformation over a family of curves $\Gamma$. This is a generalization of the attenuated Doppler transform. Uniqueness is proven for a generic set of weights and families of curves under a condition on the weight function. This condition is satisfied in particular if the weight function is never zero, and its derivatives along the curves in $\Gamma$ are never zero.
Abstract:
The inverse problem for time-harmonic acoustic wave scattering to recover a sound-soft obstacle from a given incident field and the far field pattern of the scattered field is considered. We split this problem into two subproblems; first to reconstruct the shape from the modulus of the data and this is followed by employing the full far field pattern in a few measurement points to find the location of the obstacle. We extend a nonlinear integral equation approach for shape reconstruction from the modulus of the far field data [6] to the three-dimensional case. It is known, see [13], that the location of the obstacle cannot be reconstructed from only the modulus of the far field pattern since it is invariant under translations. However, employing the underlying invariance relation and using only few far field measurements in the backscattering direction we propose a novel approach for the localization of the obstacle. The efficient implementation of the method is described and the feasibility of the approach is illustrated by numerical examples.
Abstract:
Three dimensional anisotropic attenuating and scattering media sharing the same albedo operator have been shown to be related via a gauge transformation. Such transformations define an equivalence relation. We show that the gauge equivalence is also valid in media with non-constant index of refraction, modeled by a Riemannian metric. The two dimensional model is also investigated.
Abstract:
We present a comparison of three methods for the solution of the magnetoencephalography inverse problem. The methods are: a linearly constrained minimum variance beamformer, an algorithm implementing multiple signal classification with recursively applied projection and a particle filter for Bayesian tracking. Synthetic data with neurophysiological significance are analyzed by the three methods to recover position, orientation and amplitude of the active sources. Finally, a real data set evoked by a simple auditory stimulus is considered.
Abstract:
Wavelet inpainting problem consists of filling in missed data in the wavelet domain. In [17], Chan, Shen, and Zhou proposed an efficient method to recover piecewise constant or smooth images by combining total variation regularization and wavelet representations. In this paper, we extend it to nonlocal total variation regularization in order to recover textures and local geometry structures simultaneously. Moreover, we apply an efficient algorithm framework for both local and nonlocal regularizers. Extensive experimental results on a variety of loss scenarios and natural images validate the performance of this approach.
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Question 11.1-1:
Suppose that a dynamic set $S$ is represented by a direct address table $T$ of length $m$. Describe a procedure that finds the maximum element of $S$. What is the worst case performance of your procedure?
Assuming the addresses are sorted by key: Start at the end of the direct address table and scan downward until a non-empty slot is found. This is the maximum and if not:
Initialize $max$ to $-\infty$ Start at the first address in the table and scan downward until a used slot is found. If you reach the end goto #5 Compare key to $max$. If it is greater assign it to $max$ Goto #2 Return $max$
The performance of this algorithm is $\Theta(m)$. A slightly smaller bound can be found in the first case of $\Theta(m - max)$
Question 11.1-2:
Describe how to use a bit vector to represent a dynamic set of distinct elements with no satellite data. Dictionary operations should run in $O(1)$ time.
Initialize a bit vector of length $|U|$ to all $0$s. When storing key $k$ set the $k$th bit and when deleting the $k$th bit set it to zero. This is $O(1)$ even in a non-transdichotomous model though it may be slower. Question 11.1-3:
Suggest how to implement a direct address table in which the keys of stored elements do not need to be distinct and the elements can have satellite data. All three dictionary operations must take $O(1)$ time.
Each element in the table should be a pointer to the head of a linked list containing the satellite data. $nul$ can be used for non-existent items. Question 11.1-4:
We wish to implement a dictionary by using direct addressing on a large array. At the start the array entries may contain garbage, and initializing the entire array is impractical because of its size. Describe a scheme for implementing a direct address dictionary on the array. Dictionary operations should take $O(1)$ time. Using an additional stack with size proportional to the number of stored keys is permitted.
On insert the array address is inserted into a stack. The array element is then initialized to the value of the location in the stack.
On search the array element value is to see if it is pointing into the stack. If it is the value of the stack is checked to see if it is pointing back to the array.
[1]
On delete, the array element can be set to a value not pointing the stack but this isn't required. If the element points to the value of the stack, it is simply popped off. If it is pointing to the middle of the stack, the top element and the key element are swapped and then the pop is performed. In addition the value which the top element was pointing to must be modified to point to the new location
Question 11.2-1:
Suppose we have use a hash function $h$ to hash $n$ distinct keys into an array $T$ of length $m$. Assuming simple uniform hashing what is the expected number of collisions?
Since each new value is equally likely to hash to any slot we would expect $n/m$ collisions. Question 11.2-2:
Demonstrate the insertion of the keys: $5, 28, 19, 15, 20, 33, 12, 17, 10$ into a hash table with 9 slots and $h(k) = k \mod{9}$
[2] hash values 1 28 -> 19 -> 1 2 20 3 12 5 5 6 15 -> 33 17 8 Question 11.2-3:
If the keys were stored in sorted order how is the running time for successful searches, unsuccessful searches, insertions, and deletions affected under the assumption of simple uniform hashing?
Successful and unsuccessful searches are largely unaffected although small gains can be achieved if if the search bails out early once the search finds a key later in the sort order than the one being searched for.
Insertions are the most affected operation. The time is changed from $\Theta(1)$ to $O(n/m)$
Deletions are unaffected. If the list was doubly linked the time remains $O(1)$. If it was singly linked the time remains $O(1 + \alpha)$
Question 11.2-4:
Suggest how storage for elements can be allocated and deallocated within the ash table by linking all unused slots into a free list. Assume one slot can store a flag and either one element or two pointers. All dictionary operations should run in $O(1)$ expected time.
Initialize all the values to a singly linked free list (flag set to false) with a head and tail pointer. On insert, use the memory pointed to by the head pointer and set the flag to true for the new element and increment the head pointer by one. On delete, set the flag to false and insert the newly freed memory at the tail of the linked list. Question 11.2-5:
Show that if $|U| > nm$ with $m$ the number of slots, there is a subset of $U$ of size $n$ consisting of keys that all hash to the same slot, so that the worst case searching time for hashing with chaining is $\Theta(n)$
Assuming the worst case of $|U|$ keys in the hash tabe assuming the optimial case of simple uniform hashing all m slots will have $|U|/m = n$ items. Removing the assumption of uniform hashing will allow some chains to become shorter at the expense of other chains becoming longer. There are more items then the number of slots so at least one slot must have at least $n$ items by the pigeon hole principle. Question 11.3-1:
Suppose we wish to search a linked list of length $n$, where every element contains a key $k$ along with a hash value $h(k)$. Each key is a long character string. How might we take advantage of the hash values when searching the list for an element of a given key?
You can use $h(k)$ to create a bloom filter of strings in the linked list. This is an $\Theta(1)$ check to determine if it is possible that a string appears in the linked list.
Additionally, you can create a hash table of pointers to elements in the linked list with that hash value. this way you only check a subset of the linked list. Alternatively, one can keep the hash of the value stored in the linked list as well and compare the hash of the search value to the hash of each item and only do the long comparison if the hash matches.
Question 11.3-2:
Suppose that a string of length $r$ is hashed into $m$ slots by treating it as a radix-128 number and then using the division method. The number $m$ is easily represented as a 32 bit word but the string of $r$ character treated as a radix-128 number takes many words. How can we apply the division method to compute the hash of the character string without using more than a constant number of words outside of the string itself?
Instead of treating the word as a radix-128 number some form of combination could be used. For example you may add the values of each character together modulus 128. Question 11.3-4:
Consider a hash table of size $m = 1000$ and a corresponding hash function $h(k) = \lfloor m (k A \mod{1})\rfloor$ for $ A = \frac{\sqrt{5} - 1}{2}$ Compute the locations to which the keys 61, 62, 63, 64, 65 are mapped.
key hash 61 700 62 318 63 936 64 554 65 172 Monday, May 7, 2012
I've taken a brief break from blogging about my Cormen readings but I decided to write up the answers to chapter 11. Note that the chapters and question numbers may not match up because I'm using an older edition of the book. |
Given $A_n\rightarrow \infty$ almost surely (a.s). Show that $\forall\ N > 0$, $P\left\{A_n<N\ \text{infinitely often}\right\} = 0$.
My thought: By the sake of contradiction, assume there exists an $\infty> N >0$ such that $P\left\{A_n<N\ \text{infinitely often}\right\} > 0$. This is equivalent to: $\lim_{n\rightarrow \infty} P\left\{\cup_{k\geq n} (A_k-N) < 0\right\} > 0$. But this means there exists a $k_1\geq n$ such that $A_{k_1} < N$ (1)
On the other hand, since $A_n\rightarrow \infty$ a.s, for every $\epsilon > 0$, $\lim_{n\rightarrow \infty} P\left\{\cup_{k\geq n} |A_k- \infty| \geq \epsilon\right\} = 0$. This means $|A_k-\infty|<\epsilon$ for every $k\geq n$. Since $k_1\geq n$, $\infty - A_{k_1} < |A_k- \infty| < \epsilon$ for any $\epsilon > 0$. Let $\epsilon = N$, we get $\infty < N+A_{k_1} < 2N$ (due to
(1)). This is a contradiction, since $\infty > N$. My question: I don't find my mathematical notation above legit, because I am doing subtraction with $\infty$. But I would like to know if my solution above is correct, so could someone please help verify? Also, I think there should be a shorter/slicker way to solve it, as my solution above is quite complicated and use more of real analysis. Anyone wants to give it a try? Any thoughts about my solution or this problem would be really appreciated anyway. |
Given the linear system of equations: $$ \begin{cases} x_1 + x_2 + x_3 = n\\ x_1 + x_2 + x_3 + x_4 + x_5 = 3n\\ 2x_1 +2x_2 +2x_3 +x_4 +x_5 +3x_6 +3x_7 +3x_8 =10n \end{cases} $$ how many solutions are in $\mathbb N\cup\{0\}$?
The solution must not be using sum notation like $\sum y$.
I know how to find the number of solutions to the regular equations like $x_1+x_2+x_3+\dots=n$ but I'm not sure how to do this for a system of equations. I thought of substituting some $x$'s: $$ x_1 + x_2 + x_3 =3n - (x_4 + x_5)\\ x_4 +x_5=10n-2(x_1 +x_2 +x_3) -3(x_6 +x_7 +x_8)\\ \implies x_1 + x_2 + x_3=3n-(10n-2(x_1 +x_2 +x_3) -3(x_6 +x_7 +x_8))\\ \implies x_1+x_2+x_3+3(x_6+x_7+x_8)=7n \quad * $$ As far as I understand finding the number of solutions for the system is equivalent to finding the number of solutions to the equation *.
The only next step from here I can think of is using generating functions: $$ (1+x+x^2+\dots)^3(1+x^3+x^6+x^9+\dots)^3 $$ and we need to find the coefficient of $x^{7n}$.
From the closed form identities we have: $$ \sum_{k=0}^{\infty} {3-1+k\choose k}x^k\cdot \sum_{i=0}^{\infty} {3-1+i\choose i}x^{3i} $$ But I have no idea now how to find the coefficient of $7n$ from here and certainly not without using some kind of sum notation. |
Later note: My first guess, below, as to what was desired, appears to be incorrect. See the further "later note" below.
If you mean "as $a\to\infty$" (the only guess I've got), then observe that$$
0 < \int_0^1 \frac{t^{a-1}}{e^t}\;dt < \int_0^1 t^{a-1}\;dt
$$and let $a\to\infty$.
If you didn't mean as $a\to\infty$, then you'd better re-write your question to make it clear what you meant.
Later note: One of your other question suggests something about what you meant. Your phrase "for $0 < a,t<1$" is confusing. $a$ is a parameter that remains fixed as $t$ goes from $0$ to $1$. The variable $t$ is already "bound" by the expression $\int_0^1\cdots\cdots dt$, and anything that binds the variable $a$, such as "for $0
Let's try again:$$
0<\int_0^1\frac{t^{a-1}}{e^t}\;dt < \int_0^1 t^{a-1}\;dt
$$and this is finite if $-1<a-1<0$. |
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In fluid dynamics, Luke's variational principle is a Lagrangian variational description of the motion of surface waves on a fluid with a free surface, under the action of gravity. This principle is named after J.C. Luke, who published it in 1967.[1] This variational principle is for incompressible and inviscid potential flows, and is used to derive approximate wave models like the so-called mild-slope equation,[2] or using the average-Lagrangian approach for wave propagation in inhomogeneous media.[3]
Luke's Lagrangian formulation can also be recast into a Hamiltonian formulation in terms of the surface elevation and velocity potential at the free surface.[4][5][6] This is often used when modelling the spectral density evolution of the free-surface in a sea state, sometimes called wave turbulence.
Both the Lagrangian and Hamiltonian formulations can be extended to include surface tension effects.
Luke's Lagrangian formulation is for non-linear surface gravity waves on an—incompressible, irrotational and inviscid—potential flow.
The relevant ingredients, needed in order to describe this flow, are:
The Lagrangian \mathcal{L}, as given by Luke, is:
From Bernoulli's principle, this Lagrangian can be seen to be the integral of the fluid pressure over the whole time-dependent fluid domain V(t). This is in agreement with the variational principles for inviscid flow without a free surface, found by Harry Bateman.[7]
Variation with respect to the velocity potential Φ(x,z,t) and free-moving surfaces like z=η(x,t) results in the Laplace equation for the potential in the fluid interior and all required boundary conditions: kinematic boundary conditions on all fluid boundaries and dynamic boundary conditions on free surfaces.[8] This may also include moving wavemaker walls and ship motion.
For the case of a horizontally unbounded domain with the free fluid surface at z=η(x,t) and a fixed bed at z=−h(x), Luke's variational principle results in the Lagrangian:
The bed-level term proportional to h2 in the potential energy has been neglected, since it is a constant and does not contribute in the variations. Below, Luke's variational principle is used to arrive at the flow equations for non-linear surface gravity waves on a potential flow.
The variation \delta\mathcal{L}=0 in the Lagrangian with respect to variations in the velocity potential Φ(x,z,t), as well as with respect to the surface elevation η(x,t), have to be zero. We consider both variations subsequently.
Consider a small variation δΦ in the velocity potential Φ.[8] Then the resulting variation in the Lagrangian is:
Using Leibniz integral rule, this becomes, in case of constant density ρ:[8]
The first integral on the right-hand side integrates out to the boundaries, in x and t, of the integration domain and is zero since the variations δΦ are taken to be zero at these boundaries. For variations δΦ which are zero at the free surface and the bed, the second integral remains, which is only zero for arbitrary δΦ in the fluid interior if there the Laplace equation holds:
with Δ=∇·∇ + ∂2/∂z2 the Laplace operator.
If variations δΦ are considered which are only non-zero at the free surface, only the third integral remains, giving rise to the kinematic free-surface boundary condition:
Similarly, variations δΦ only non-zero at the bottom z = -h result in the kinematic bed condition:
Considering the variation of the Lagrangian with respect to small changes δη gives:
This has to be zero for arbitrary δη, giving rise to the dynamic boundary condition at the free surface:
This is the Bernoulli equation for unsteady potential flow, applied at the free surface, and with the pressure above the free surface being a constant — which constant pressure is taken equal to zero for simplicity.
The Hamiltonian structure of surface gravity waves on a potential flow was discovered by Vladimir E. Zakharov in 1968, and rediscovered independently by Bert Broer and John Miles:[4][5][6]
where the surface elevation η and surface potential φ — which is the potential Φ at the free surface z=η(x,t) — are the canonical variables. The Hamiltonian \mathcal{H}(\varphi,\eta) is the sum of the kinetic and potential energy of the fluid:
The additional constraint is that the flow in the fluid domain has to satisfy Laplace's equation with appropriate boundary condition at the bottom z=-h(x) and that the potential at the free surface z=η is equal to φ: \delta\mathcal{H}/\delta\Phi\,=\,0.
The Hamiltonian formulation can be derived from Luke's Lagrangian description by using Leibniz integral rule on the integral of ∂Φ/∂t:[6]
with \varphi(\boldsymbol{x},t)=\Phi(\boldsymbol{x},\eta(\boldsymbol{x},t),t) the value of the velocity potential at the free surface, and H(\varphi,\eta;\boldsymbol{x},t) the Hamiltonian density — sum of the kinetic and potential energy density — and related to the Hamiltonian as:
The Hamiltonian density is written in terms of the surface potential using Green's third identity on the kinetic energy:[9]
where D(η) φ is equal to the normal derivative of ∂Φ/∂n at the free surface. Because of the linearity of the Laplace equation — valid in the fluid interior and depending on the boundary condition at the bed z=-h and free surface z=η — the normal derivative ∂Φ/∂n is a linear function of the surface potential φ, but depends non-linear on the surface elevation η. This is expressed by the Dirichlet-to-Neumann operator D(η), acting linearly on φ.
The Hamiltonian density can also be written as:[6]
with w(x,t) = ∂Φ/∂z the vertical velocity at the free surface z = η. Also w is a linear function of the surface potential φ through the Laplace equation, but w depends non-linear on the surface elevation η:[9]
with W operating linear on φ, but being non-linear in η. As a result, the Hamiltonian is a quadratic functional of the surface potential φ. Also the potential energy part of the Hamiltonian is quadratic. The source of non-linearity in surface gravity waves is through the kinetic energy depending non-linear on the free surface shape η.[9]
Further ∇φ is not to be mistaken for the horizontal velocity ∇Φ at the free surface:
Taking the variations of the Lagrangian \mathcal{L}_H with respect to the canonical variables \varphi(\boldsymbol{x},t) and \eta(\boldsymbol{x},t) gives:
provided in the fluid interior Φ satisfies the Laplace equation, ΔΦ=0, as well as the bottom boundary condition at z=-h and Φ=φ at the free surface.
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I am trying to wrap my head around these asymptotic notations. Given $f(n)$ and $g(n)$, one can write $f(n) = \Omega(g(n))$ as shorthand for $f(n) \geq c*g(n), n\geq n_0$. But what happens when $n<n_0$. What can we then say about $f(n)$ and $g(n)$? Do we just assume that $f(n)$ is a constant and the relation still works? Or does the question not even make sense to ask?
Nothing. If all you know about two functions is their $O$/$\Omega$ relations, you know
nothing about how they relate to each other on any finite prefix.
Yes, this makes asymptotics of, say, runtime functions absolutely useless in terms of questions like, "which algorithm should I use in scenario X?" On first sight, that is.
Luckily, we usually
do analyse with more detail but throw them away when formulating the theorem with coarse notation like Landau symbols. Many results about "practical" algorithms implicitly say, "and this behaviour can be observed for reasonable $n$".
Sometimes you get better theorems which use Landau notation for "error terms" only; authors like Knuth do not say "algorithm A runs in time $O(n \log n)$", but rather "algorithm A takes $2n\log n - 2n + O(1)$ comparisons". This does not completely do away with your issue, but mitigates it to some extent. If you know that that $O(1)$ stands for $\leq 1377$, you can make informed decisions based on that result.
Some of our reference material discusses facets of this issue. |
Let $(X,\mathcal O_X)$ (here $\mathcal O_X$ is the sheaf of holomorphic functions) be a compact Riemann surface and let $D=\sum_{x\in X} a_x[x]$ be a divisor on $X$. The invertible sheaf associated to $D$ is $$\mathcal O_X(D)(U)=\{f\in\mathscr M_X(U)^\ast: \operatorname{ord}_x(f)+a_x\ge 0 \quad\text{for any $x\in U$}\}\cup\{0\}$$ where $\mathscr M_X$ is the sheaf of meromorphic functions. Now let $\mathcal L$ be an invertible sheaf on $X$ an let $s\in\mathcal (\mathcal L\otimes_{\mathcal O_X}\mathcal M_X)(X)$ be a non-zero meromorphic section of $\mathcal L$. Then we can associate a divisor to $s$ that we denote with $\operatorname{div}(s)$.
Informally $s$ is given as a colletion of "compatible" meromorphic functions on a trivialization covering for $\mathcal L$ and the order of $\operatorname{div}(s)$ at a point $x\in X$ is defined as the order of the right merormophic function at that point (I can be more precise on the definition of $\operatorname{div}(s)$ if you want).
Problem: I don't understand why we have that $\mathcal O_X(\operatorname{div}(s))\cong \mathcal L$.
Can you help me in proving this? |
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Abstract: We employ gravitational-wave radiometry to map the gravitational waves
stochastic background expected from a variety of contributing mechanisms and
test the assumption of isotropy using data from Advanced LIGO's first observing
run. We also search for persistent gravitational waves from point sources with
only minimal assumptions over the 20 - 1726 Hz frequency band. Finding no
evidence of gravitational waves from either point sources or a stochastic
background, we set limits at 90% confidence. For broadband point sources, we
report upper limits on the gravitational wave energy flux per unit frequency in
the range $F(f, \Theta) < (0.1 - 56) \times 10^{-8}$ erg cm$^{-2}$ s$^{-1}$
Hz$^{-1}$ (f/25 Hz)$^{\alpha-1}$ depending on the sky location $\Theta$ and the
spectral power index $\alpha$. For extended sources, we report upper limits on
the fractional gravitational wave energy density required to close the Universe
of $\Omega(f,\Theta) < (0.39-7.6) \times 10^{-8}$ sr$^{-1}$ (f/25 Hz)$^\alpha$
depending on $\Theta$ and $\alpha$. Directed searches for narrowband
gravitational waves from astrophysically interesting objects (Scorpius X-1,
Supernova 1987 A, and the Galactic Center) yield median frequency-dependent
limits on strain amplitude of $h_0 <$ (6.7, 5.5, and 7.0) $\times 10^{-25}$
respectively, at the most sensitive detector frequencies between 130 - 175 Hz.
This represents a mean improvement of a factor of 2 across the band compared to
previous searches of this kind for these sky locations, considering the
different quantities of strain constrained in each case. |
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The orbital period is the time taken for a given object to make one complete orbit around another object.
When mentioned without further qualification in astronomy this refers to the sidereal period of an astronomical object, which is calculated with respect to the stars.
There are several kinds of orbital periods for objects around the Sun (or other celestial objects):
Table of synodic periods in the Solar System, relative to Earth:
In the case of a planet's moon, the synodic period usually means the Sun-synodic period, namely, the time it takes the moon to complete its illumination phases, completing the solar phases for an astronomer on the planet's surface. The Earth's motion does not determine this value for other planets because an Earth observer is not orbited by the moons in question. For example, Deimos's synodic period is 1.2648 days, 0.18% longer than Deimos's sidereal period of 1.2624 d.
According to Kepler's Third Law, the orbital period T\, (in seconds) of two bodies orbiting each other in a circular or elliptic orbit is:
where:
For all ellipses with a given semi-major axis the orbital period is the same, regardless of eccentricity.
When a very small body is in a circular orbit barely above the surface of a sphere of any radius and mean density ρ (in kg/m3), the above equation simplifies to (since M = \rho V = \rho {\frac {4}{3}} \pi a^3):
So, for the Earth as the central body (or any other spherically symmetric body with the same mean density, about 5515 kg/m3)[2] we get:
and for a body made of water (ρ≈1000 kg/m3)[3]
Thus, as an alternative for using a very small number like G, the strength of universal gravity can be described using some reference material, like water: the orbital period for an orbit just above the surface of a spherical body of water is 3 hours and 18 minutes. Conversely, this can be used as a kind of "universal" unit of time if we have a unit of mass, a unit of length and a unit of density.
In celestial mechanics, when both orbiting bodies' masses have to be taken into account, the orbital period T\, can be calculated as follows:[4]
Note that the orbital period is independent of size: for a scale model it would be the same, when densities are the same (see also Orbit#Scaling in gravity).
In a parabolic or hyperbolic trajectory, the motion is not periodic, and the duration of the full trajectory is infinite.
When two bodies orbit a third body in different orbits, and thus different orbital periods, their respective, synodic period can be found. If the orbital periods of the two bodies around the third are called P_1 and P_2, so that P_1 < P_2, their synodic period is given by
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I'm reading about Convolutional Neural Networks (CNNs) in
Deep Learning by Ian Goodfellow.
CNNs are different from traditional neural networks in that they use convolution in place of general matrix multiplication in at least one of their layers. The convolution is introduced as follows:
Suppose that we are tracking the location of a spaceshift with a laser sensor. Our laser provides a single output $x(t)$, the position of the spaceship at time $t$. Both $x$ and $t$ are real-valued, that is, we can get a different reading from the laser sensor at any instant in time. Now suppose our laser is somewhat noisy. To obtain a less noisy estimate of the spaceship's position, we would like to average several measurements. Of course, more recent measurements are more relevant, so we will want this to be a weighted average that gives more weight to recent measurements. We can do this with a weighting function $w(a)$, where $a$ is the age of a measurement. If we apply such a weighted average operation at every moment, we obtain a new function $s$ providing a smoothed estimate of the position of the spaceship: $$\displaystyle s(t) = \int x(a)w(t - a)da$$ This operation is called
convolution
$\ldots$
In convolutional network terminology, the first argument (in this example, the function $x$) to the convolution is often referred to as the
input, and the second argument (in this example, the function $w$) as the kernel. The output is sometimes referred to as the feature map.
$\ldots$
In machine learning applications, the input is usually a multidimensional array of data, and the kernel is usually a multidimensional array of parameters that are adapted by the learning algorithm. We will refer to those multidimensional arrays as tensors. Because each element of the input and kernel must be explicitly stored separately, we usually assume that these functions are zero everywhere but in the finite set of points for which we store the values. This means that in practice, we can implement the infinite summation as a summation over a finite number of array elements. Finally, we often use convolutions over more than one axis at the time. For example, if we use a two-dimensional image $I$ as our input, we probably also want to use a two-dimensional kernel $K$: $$S(i,j) = (I*K)(i,j) = \sum_m\sum_nI(m,n)K(i-m,j-n)$$
(I assume that $S(i,j)$ means the feature map at point $(i,j)$).
The author then gives an example of $2$-D convolution with the following image:
I don't understand how this image illustrates what the author explains earlier. If we consider the input an image, then $a$ would resemble $I(0,0)$ right? Using the given definition of the feature map I find that $S(0,0) = \sum_m\sum_nI(m,n)K(0-m,0-n) = I(0,0)K(0,0) = aw$. Since e would resemble $I(1,0)$ I find that $$S(1,0) = \sum_m\sum_nI(m,n)K(1-m,0-n) = I(0,0)K(1,0) + I(1,0)K(0,0) = ay + ew$$ However, according to the image the output would be $aw + bx + ey + fz$.
Question: Why does the output equal $aw + bx + ey + fz$? Edit: If the displayed outputs (feature maps) are only the outputs corresponding to the internal points on the grid, then I think I understand the figure. I would mean that the highlighted output corresponds to $S(1,1)$ and that the outputs $S(0,0), S(1,0), \ldots$ are simply not shown here right?
Thanks in advance! |
A question in study of subconvexity topic puzzles me for a long time, which mabe a stupid question for many experts. I really wish someone to help me out, and any advice will be highly appreciated.
Let g a Hecke-Maass form for SL3(Z) which do not come from a symmetric square lift, and f be a Hecke-Maass cusp form for SL2(Z) of level $q$. The Rankin-Selberg L-function is defined by $$L( s,g\times f)=\sum_{m,n\ge 1}\frac{A(m,n)a(n)}{(m^2n)^s}.$$
My question is
how to prove subconvexity bound on level aspect
$$L(\frac{1}{2},g\times f)\ll q^{3/4-\epsilon},\quad \text{some constant }\epsilon>0 \hskip2em ?$$ And who has studied it? please show me their works.
Remark: Suppose $f_1,f_2$ be Hecke-Maass cusp forms on SL2(Z). In many papers, the upperbound for $L(\frac{1}{2},\text{sym}^2(f_1)\times f_2)$ has greatly improved. However there was few literature involving in L-function with general GL(3) Hecke-Maass form twisted by a GL(2) cusp form, i.e. $L(\frac{1}{2},g\times f)$.
So far I know that Rizwanur Khan ( link his paper here) prove a conditional result, he proved that suppose $f$ be holomorphic, and $\sum_{n<L}a(n)^2\gg L^{1-\epsilon} \text{ for} L>q^{\frac{1}{4}+\frac{1}{2001}}$, then $L(\frac{1}{2},g\times f)\ll q^{3/4-1/2001}.$ If there are other literature studying $L(\frac{1}{2},g\times f)$, please guide me their names or papers.
Another stupid little question is let $g$ be a a Hecke-Maass form for SL3(Z) which is self-dual, I'm not sure whether $g$ must come from a symmetric square lift or not? |
This is several years late, but it may be helpful nonetheless.
As alluded to by Buschi, Olivia has given an explicit answer to this in Theorem 2.1.29 of her monograph
Theories, Sites and Toposes:
Let $\mathbb{T}$ be a geometric theory, $\mathcal{E}$ a Grothendieck
topos and $M$ a model of $\mathbb{T}$ in $\mathcal{E}$. Then
$\mathcal{E}$ is a classifying topos of $\mathbb{T}$ and $M$ is a
universal model (i.e. generic model) of $\mathbb{T}$ iff the following
conditions are satisfied:
The family $F$ of objects which can be built from the interpretations in $M$ of the sorts, function symbols and relation
symbols over the signature of $\mathbb{T}$ by using geometric logic
constructions (i.e. the objects given by the domains of the
interpretations in $M$ of geometric formulae over the signature of
$\mathbb{T}$) is separating for $\mathcal{E}$.
The model $M$ is conservative for $\mathbb{T}$, that is for any geometric sequent $\sigma$ over the signature of $\mathbb{T}$, $\sigma$ is valid in $M$ if and only if it is provable in
$\mathbb{T}$.
Any arrow $k$ in $\mathcal{E}$ between objects $A$ and $B$ in the family $F$ of condition (1) is definable; that is, if $A$ (resp. $B$)
is equal to the interpretation of a geometric formula $\phi(\vec{x})$
(resp. $\psi(\vec{y})$) over the signature of $\mathbb{T}$, there
exists a $\mathbb{T}$-provably functional formula $\theta$ from
$\phi(\vec{x})$ to $\psi(\vec{y})$ such that the interpretation of
$\theta$ in $M$ is equal to the graph of $k$.
I'm not sure if you were looking necessary and sufficient conditions on $M$ and $\mathcal{E}$, or just merely sufficient conditions, but since this Theorem gives an 'iff' result, one might try and prove the sufficiency of certain (perhaps more intuitive) critiera on $M$ and $\mathcal{E}$ by checking against the conditions listed in this theorem, i.e. by proving results of the flavour: 'If $M$ and $\mathcal{E}$ satisfy condition $X$, then they satisfy the 3 conditions of this theorem.'
Extending this thought, I am curious to see how these conditions relate to the special case mentioned in Dylan's comment. In particular, how does weak contractibility relate to the conditions spelt out by Olivia? This is not obvious to me, but I haven't taken the time to properly work through the details. |
Answer
The signal can be received in locations that are east or west of the origin up to a distance of 200 miles away. The signal can be received in regions which are within $45^{\circ}$ of the horizontal axis. The signal can not be received in any locations north or south of the origin, or regions located within $45^{\circ}$ of the vertical axis.
Work Step by Step
$r^2 = 40,000~cos~2\theta$ We can see the graph for the values of $\theta$ such that $0 \leq \theta \leq 360^{\circ}$ Note that the graph only includes points where $cos~2\theta \geq 0$ That is: $0 \leq \theta \leq 45^{\circ}$ $135 \leq \theta \leq 225^{\circ}$ $315 \leq \theta \leq 360^{\circ}$ When $\theta = 0^{\circ}$, then $r = \sqrt{40,000~cos~0^{\circ}} = 200$ When $\theta = 15^{\circ}$, then $r = \sqrt{40,000~cos~30^{\circ}} = 186$ When $\theta = 30^{\circ}$, then $r = \sqrt{40,000~cos~60^{\circ}} = 141$ When $\theta = 45^{\circ}$, then $r = \sqrt{40,000~cos~90^{\circ}} = 0$ When $\theta = 135^{\circ}$, then $r = \sqrt{40,000~cos~270^{\circ}} = 0$ When $\theta = 150^{\circ}$, then $r = \sqrt{40,000~cos~300^{\circ}} = 141$ When $\theta = 180^{\circ}$, then $r = \sqrt{40,000~cos~360^{\circ}} = 200$ When $\theta = 225^{\circ}$, then $r = \sqrt{40,000~cos~450^{\circ}} = 0$ When $\theta = 315^{\circ}$, then $r = \sqrt{40,000~cos~630^{\circ}} = 0$ When $\theta = 330^{\circ}$, then $r = \sqrt{40,000~cos~660^{\circ}} = 141$ When $\theta = 345^{\circ}$, then $r = \sqrt{40,000~cos~690^{\circ}} = 186$ When $\theta = 360^{\circ}$, then $r = \sqrt{40,000~cos~720^{\circ}} = 200$ We can see this graph below. Since locations where the signal can be received correspond to the interior of the curve, the signal can be received in locations that are east or west of the origin up to a distance of 200 miles away. Note that the signal can be received in regions which are within $45^{\circ}$ of the horizontal axis. The signal can not be received in regions outside of the curve, so the signal can not be received in any locations north or south of the origin, or regions located within $45^{\circ}$ of the vertical axis. |
There are three elements in a $C_3$ group and there are three classes in this group cause it is an Abelian group.
\begin{array}{c|ccc|cc} \hline C_3 & E & C_3 & C_3^2 & & \varepsilon = \exp(2\pi\mathrm{i}/3) \\ \hline \mathrm{A} & 1 & 1 & 1 & z, R_z & x^2+y^2, z^2 \\ \mathrm{E} & \left\{ \begin{aligned}1 \\ 1\end{aligned} \right. & \begin{aligned}\varepsilon \\ \varepsilon^* \end{aligned} & \left. \begin{aligned}\varepsilon^* \\ \varepsilon \end{aligned}\right\} & (x, y), (R_x, R_y) & (x^2-y^2,xy), (xz, yz) \\ \hline \end{array}
According to group representation theory, the number of classes of a group should be the same as the number of irreducible representations of the group, which means there should be 3 irreducible representations for $C_3$ group.
In a character table, this means there should be 3 rows.
Further more, according to $$\sum_{i}l_{j}^2= h$$ in which $l_{j}$ stands for the dimension of the $j-th$ irreducible representation, and $h$ stands for the order of the group.
All of the three irreducible representation of $C_3$ group should be of 1 dimension.
But in the link above, for $C_3$ group, there are only two irreducible representations, and one of the irreducible representation is of 2 dimension, so it seems it violates the basic results of group representation theory.
So, what I have misunderstood? |
Suppose $\Psi$ is an eigenstate of observable $\text H$ with eigenvalue $E_1$. Then uncertainty in the value of $\text H$,
$(\Delta E)^2=\langle E^2\rangle-\langle E\rangle^2$ which gives, $(\Delta E)^2=E_1^2\bigg(\langle\Psi_1|\Psi_1\rangle-\big|\langle\Psi_1|\Psi_1\rangle\big|^2\bigg)$ If the eigenstate is not normalized then the right hand side is not zero. But the measurement of $\text H$ on $\Psi$ must yield $E_1$ with no uncertainty.
Suppose $\Psi$ is an eigenstate of observable $\text H$ with eigenvalue $E_1$. Then uncertainty in the value of $\text H$,
It is part of the postulates of quantum mechanics that the expectation value of the observable corresponding to the hermitian operator $A$ in the normalized state $|\psi\rangle$ is given by $\langle A\rangle_\psi =\langle\psi|A|\psi\rangle$. Alternatively, you can postulate that the expectation value is given by $\langle A \rangle_\psi = \frac{\langle\psi|A|\psi\rangle}{\langle\psi|\psi\rangle}$. See, for example, the Dirac--von Neumann axioms
Either way the normalization is necessary, because otherwise you have to give physical meaning to the normalization.
Edit: So the reason your calculation is giving a strange result is you are calculating expectation values wrong. |
In Delta of binary option, I do not see how to prove that the limit of $\partial C_t/\partial S_t$ is equal to $+\infty$ as $t \rightarrow T$. Can someone help ?
The value of a bond binary call in the Black-Scholes model is given by
\begin{equation} B_t = e^{-r (T - t)} \mathcal{N} \left( d_- \right), \end{equation}
where
\begin{equation} d_- = \frac{1}{\sigma \sqrt{T - t}} \left( \ln \left( \frac{S_t}{K} \right) + \left( r - \frac{1}{2} \sigma^2 \right) (T - t) \right). \nonumber \end{equation}
The delta is
\begin{equation} \frac{\partial B_t}{\partial S_t} = e^{-r (T - t)} \mathcal{N}' \left( d_- \right) \frac{1}{S_t \sigma \sqrt{T - t}}. \end{equation}
We now want to take the limit as $t \rightarrow T$. First note that
\begin{equation} \lim_{t \rightarrow T} d_- = \begin{cases} -\infty & \text{if } S_t < K\\ 0 & \text{if } S_t = K\\ +\infty & \text{if } S_t > K \end{cases}. \end{equation}
Thus
\begin{equation} \lim_{t \rightarrow T} \mathcal{N}' \left( d_- \right) = \begin{cases} 0 & \text{if } S_t \neq K\\ 1 / \sqrt{2 \pi} & \text{if } S_t = K \end{cases} \end{equation}
and
\begin{equation} \lim_{t \rightarrow T} \frac{\partial B_t}{\partial S_t} = \begin{cases} 0 & \text{if } S_t \neq K\\ +\infty & \text{if } S_t = K \end{cases}. \end{equation}
In the last step we used that the exponential in $\mathcal{N}' \left( d_- \right)$ approaches zero faster than the $1 / \sqrt{T - t}$ approaches plus infinity in the limit when $S_t \neq K$.
Alternatively to LocalVolatility's already very nice answer, here's an approach to see that this result does not only hold under the Black-Scholes dynamics.
The $t$-value of a binary call expiring at $T$ can be written as $$ C_t = \Bbb{E}_t^\Bbb{Q} \left[ e^{-r(T-t)} {\bf{1}}\{S_T \geq K \} \right] $$
Its "delta" is defined as $$\Delta = \frac{\partial C_t}{\partial S_t}$$Under some light conditions (discussed in e.g.
Monte Carlo Methods in Financial Engineering, Glasserman, 2004), you can permute the expectation and differential operators to write:\begin{align}\Delta_t &= \frac{\partial}{\partial S_t} \Bbb{E}_t^\Bbb{Q} \left[ e^{-r(T-t)} {\bf{1}}\{S_T \geq K \} \right] \\&= \Bbb{E}_t^\Bbb{Q} \left[ e^{-r(T-t)} \frac{\partial}{\partial S_t}{\bf{1}}\{S_T \geq K \} \right] \\&= \Bbb{E}_t^\Bbb{Q} \left[ e^{-r(T-t)} \delta(S_T-K) \frac{\partial S_T}{\partial S_t} \right] \tag{1}\end{align}where we've used the chain rule ($S_T$ functionally depends on $S_t$) and the fact that the derivative of the Heaviside function ${\bf{1}}(x \geq a)$ is a Dirac impulse at $a$, i.e. $\delta(x-a)$.
It should then clear that: $$ \lim_{t \to T} \Delta_t = \delta(S_t-K) $$ hence the result.
You're short a digital call struck at 100. Your payoff : -\$1 above 100, \$0 below.
1 second before expiry, spot is 99.9999. If its stays there you owe nothing, if it goes a touch higher you owe $1 to the option's buyer.
You need to replicate this payoff via delta hedging : how much of the underlying do you need to hold to generate a \$1 gain, and offset your \$1 loss, when the spot moves from 99.9999 to 100.0?
Answer : a lot of it. |
Q. A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V bv the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90 %, the output current would be :
Solution:
$\eta = \frac{P_{out }}{P_{in}} = \frac{V_{S}I_{S}}{V_{P}I_{P}} $ $ \Rightarrow 0.9 = \frac{23 \times I_{S}}{230 \times5} $ $ \Rightarrow I_{S} = 45A $ Questions from JEE Main 2019 3. The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.
The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is : 10. A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is : Physics Most Viewed Questions 1. An AC ammeter is used to measure current in a circuit. When a given direct current passes through the circuit, the AC ammeter reads 3 A. When another alternating current passes through the circuit, the AC ammeter reads 4 A. Then the reading of this ammeter, if DC and AC flow through the circuit simultaneously, is Latest Updates Top 10 Medical Entrance Exams in India Top 10 Engineering Entrance Exams in India JIPMER Results Announced NEET UG Counselling Started NEET Result Announced KCET Result Announced KCET College Predictor JEE Main Result Announced JEE Advanced Score Cards Available AP EAMCET Result Announced KEAM Result Announced UPSEE – Online Applications invited from NRI &Kashmiri Migrants MHT CET Result Announced NEET Rank Predictor Questions Tardigrade |
2018-09-11 04:29
Proprieties of FBK UFSDs after neutron and proton irradiation up to $6*10^{15}$ neq/cm$^2$ / Mazza, S.M. (UC, Santa Cruz, Inst. Part. Phys.) ; Estrada, E. (UC, Santa Cruz, Inst. Part. Phys.) ; Galloway, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; Gee, C. (UC, Santa Cruz, Inst. Part. Phys.) ; Goto, A. (UC, Santa Cruz, Inst. Part. Phys.) ; Luce, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; McKinney-Martinez, F. (UC, Santa Cruz, Inst. Part. Phys.) ; Rodriguez, R. (UC, Santa Cruz, Inst. Part. Phys.) ; Sadrozinski, H.F.-W. (UC, Santa Cruz, Inst. Part. Phys.) ; Seiden, A. (UC, Santa Cruz, Inst. Part. Phys.) et al. The properties of 60-{\mu}m thick Ultra-Fast Silicon Detectors (UFSD) detectors manufactured by Fondazione Bruno Kessler (FBK), Trento (Italy) were tested before and after irradiation with minimum ionizing particles (MIPs) from a 90Sr \b{eta}-source . [...] arXiv:1804.05449. - 13 p. Preprint - Full text Notice détaillée - Notices similaires 2018-08-25 06:58
Charge-collection efficiency of heavily irradiated silicon diodes operated with an increased free-carrier concentration and under forward bias / Mandić, I (Ljubljana U. ; Stefan Inst., Ljubljana) ; Cindro, V (Ljubljana U. ; Stefan Inst., Ljubljana) ; Kramberger, G (Ljubljana U. ; Stefan Inst., Ljubljana) ; Mikuž, M (Ljubljana U. ; Stefan Inst., Ljubljana) ; Zavrtanik, M (Ljubljana U. ; Stefan Inst., Ljubljana) The charge-collection efficiency of Si pad diodes irradiated with neutrons up to $8 \times 10^{15} \ \rm{n} \ cm^{-2}$ was measured using a $^{90}$Sr source at temperatures from -180 to -30°C. The measurements were made with diodes under forward and reverse bias. [...] 2004 - 12 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 533 (2004) 442-453 Notice détaillée - Notices similaires 2018-08-23 11:31 Notice détaillée - Notices similaires 2018-08-23 11:31
Effect of electron injection on defect reactions in irradiated silicon containing boron, carbon, and oxygen / Makarenko, L F (Belarus State U.) ; Lastovskii, S B (Minsk, Inst. Phys.) ; Yakushevich, H S (Minsk, Inst. Phys.) ; Moll, M (CERN) ; Pintilie, I (Bucharest, Nat. Inst. Mat. Sci.) Comparative studies employing Deep Level Transient Spectroscopy and C-V measurements have been performed on recombination-enhanced reactions between defects of interstitial type in boron doped silicon diodes irradiated with alpha-particles. It has been shown that self-interstitial related defects which are immobile even at room temperatures can be activated by very low forward currents at liquid nitrogen temperatures. [...] 2018 - 7 p. - Published in : J. Appl. Phys. 123 (2018) 161576 Notice détaillée - Notices similaires 2018-08-23 11:31 Notice détaillée - Notices similaires 2018-08-23 11:31
Characterization of magnetic Czochralski silicon radiation detectors / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) Silicon wafers grown by the Magnetic Czochralski (MCZ) method have been processed in form of pad diodes at Instituto de Microelectrònica de Barcelona (IMB-CNM) facilities. The n-type MCZ wafers were manufactured by Okmetic OYJ and they have a nominal resistivity of $1 \rm{k} \Omega cm$. [...] 2005 - 9 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 548 (2005) 355-363 Notice détaillée - Notices similaires 2018-08-23 11:31
Silicon detectors: From radiation hard devices operating beyond LHC conditions to characterization of primary fourfold coordinated vacancy defects / Lazanu, I (Bucharest U.) ; Lazanu, S (Bucharest, Nat. Inst. Mat. Sci.) The physics potential at future hadron colliders as LHC and its upgrades in energy and luminosity Super-LHC and Very-LHC respectively, as well as the requirements for detectors in the conditions of possible scenarios for radiation environments are discussed in this contribution.Silicon detectors will be used extensively in experiments at these new facilities where they will be exposed to high fluences of fast hadrons. The principal obstacle to long-time operation arises from bulk displacement damage in silicon, which acts as an irreversible process in the in the material and conduces to the increase of the leakage current of the detector, decreases the satisfactory Signal/Noise ratio, and increases the effective carrier concentration. [...] 2005 - 9 p. - Published in : Rom. Rep. Phys.: 57 (2005) , no. 3, pp. 342-348 External link: RORPE Notice détaillée - Notices similaires 2018-08-22 06:27
Numerical simulation of radiation damage effects in p-type and n-type FZ silicon detectors / Petasecca, M (Perugia U. ; INFN, Perugia) ; Moscatelli, F (Perugia U. ; INFN, Perugia ; IMM, Bologna) ; Passeri, D (Perugia U. ; INFN, Perugia) ; Pignatel, G U (Perugia U. ; INFN, Perugia) In the framework of the CERN-RD50 Collaboration, the adoption of p-type substrates has been proposed as a suitable mean to improve the radiation hardness of silicon detectors up to fluencies of $1 \times 10^{16} \rm{n}/cm^2$. In this work two numerical simulation models will be presented for p-type and n-type silicon detectors, respectively. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 2971-2976 Notice détaillée - Notices similaires 2018-08-22 06:27
Technology development of p-type microstrip detectors with radiation hard p-spray isolation / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) ; Díez, S (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) A technology for the fabrication of p-type microstrip silicon radiation detectors using p-spray implant isolation has been developed at CNM-IMB. The p-spray isolation has been optimized in order to withstand a gamma irradiation dose up to 50 Mrad (Si), which represents the ionization radiation dose expected in the middle region of the SCT-Atlas detector of the future Super-LHC during 10 years of operation. [...] 2006 - 6 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 566 (2006) 360-365 Notice détaillée - Notices similaires 2018-08-22 06:27
Defect characterization in silicon particle detectors irradiated with Li ions / Scaringella, M (INFN, Florence ; U. Florence (main)) ; Menichelli, D (INFN, Florence ; U. Florence (main)) ; Candelori, A (INFN, Padua ; Padua U.) ; Rando, R (INFN, Padua ; Padua U.) ; Bruzzi, M (INFN, Florence ; U. Florence (main)) High Energy Physics experiments at future very high luminosity colliders will require ultra radiation-hard silicon detectors that can withstand fast hadron fluences up to $10^{16}$ cm$^{-2}$. In order to test the detectors radiation hardness in this fluence range, long irradiation times are required at the currently available proton irradiation facilities. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 589-594 Notice détaillée - Notices similaires |
I've been reading Jeffrey Lee's,
Manifolds and Differential Geometry and John Lee's, Introduction to smooth manifolds.
In the first book (here, in page 31), after introducing partition of unity, there's an exercise that says:
Exercise 1.74.Show that if a function is smooth on an arbitrary set $S\subset M$ as defined earlier, then it has a smooth extension to an open set that contains $S$.
Where he says "
as defined earlier", I assumed he meant that $M$ was paracompact, but maybe I missed something else.
On the sencond book (check here, in page 45) there's the Extension Lemma for Smooth Functions, and it says:
Lemma 2.26 (Extension Lemma for Smooth functions).Suppose $M$ is a smooth manifold with or without boundary, $A \subset M$ is a closed subset, and $f:A\to\Bbb R^k$ is a smooth function. For any open subset $U$ containing $A$, there exists a smooth function $\hat f:M\to\Bbb R^k$ such that $\hat f|_A=f$ and $supp(f) \subset U$.
And after proving the lemma there is this exercise that made me very confused.
Exercise 2.27.Give a counterexample to show that the conclusion of the extension lemma can be false if A is not closed.
Doesn't exercise 2.27 imply that exercise 1.74 is incorrect? Are those trick questions? Like when you are asked to prove something right but it turns out is not? Or, more likely, did I missed something in Jeffrey Lee's book? |
Revista Matemática Iberoamericana
Full-Text PDF (200 KB) | Metadata | Table of Contents | RMI summary
Volume 27, Issue 1, 2011, pp. 253–271 DOI: 10.4171/RMI/635
Published online: 2011-04-30
Cluster solutions for the Schrödinger-Poisson-Slater problem around a local minimum of the potentialDavid Ruiz
[1]and Giusi Vaira [2](1) Universidad de Granada, Spain
(2) SISSA, Trieste, Italy
In this paper we consider the system in $\mathbb{R}^3$ \begin{equation} \left\{ \begin{array}{l} -\varepsilon^2 \Delta u + V(x)u + \phi(x)u = u^p, \\ -\Delta \phi = u^2, \end{array} \right. \end{equation} for $p\in (1,5)$. We prove the existence of multi-bump solutions whose bumps concentrate around a local minimum of the potential $V(x)$. We point out that such solutions do not exist in the framework of the usual Nonlinear Schrödinger Equation.
Keywords: Nonlinear analysis, Schrödinger-Poisson-Slater problem, variational methods, singular perturbation method, multi-bump solutions.
Ruiz David, Vaira Giusi: Cluster solutions for the Schrödinger-Poisson-Slater problem around a local minimum of the potential.
Rev. Mat. Iberoam. 27 (2011), 253-271. doi: 10.4171/RMI/635 |
Journal of Spectral Theory
Full-Text PDF (372 KB) | Metadata | Table of Contents | JST summary
Volume 6, Issue 2, 2016, pp. 373–413 DOI: 10.4171/JST/127
Published online: 2016-04-06
Approximate zero modes for the Pauli operator on a regionDaniel M. Elton
[1](1) Lancaster University, UK
Let $\mathcal{P}_{\Omega,tB}$ denoted the the Pauli operator on a bounded open region $\Omega\subset\mathbb{R}^2$ with Dirichlet boundary conditions and magnetic potential$ $A scaled by some $t > 0$. Assume that the corresponding magnetic field $B = \mathrm {curl} A$ satisfies $B \in L \mathrm {log} L (\Omega) \cap C^\alpha (\Omega_0)$ where $\alpha>0$ and $\Omega_0$ is an open subset of $\Omega$ of full measure (note that, the Orlicz space $L\log L(\Omega)$ contains $L^p(\Omega)$ for any $p>1$). Let $\mathsf{N}_{\Omega,tB}(\lambda)$ denote the corresponding eigenvalue counting function. We establish the strong field asymptotic formula $$\mathsf{N}_{\Omega,tA}(\lambda(t))=\frac{t}{2\pi}\int_{\Omega} \lvert B(x) \rvert\, dx\;+o(t)$$ as $t \to +\infty$, whenever $\lambda (t) = Ce^{-ct^\sigma}$ for some $\sigma \in (0,1)$ and $c,C > 0$. The corresponding eigenfunctions can be viewed as a localised version of the Aharonov–Casher zero modes for the Pauli operator on $\mathbb{R}^2$.
Keywords: Pauli operator, eigenvalue asymptotics, approximate zero modes
Elton Daniel: Approximate zero modes for the Pauli operator on a region.
J. Spectr. Theory 6 (2016), 373-413. doi: 10.4171/JST/127 |
2018-08-25 06:58
Recent developments of the CERN RD50 collaboration / Menichelli, David (U. Florence (main) ; INFN, Florence)/CERN RD50 The objective of the RD50 collaboration is to develop radiation hard semiconductor detectors for very high luminosity colliders, particularly to face the requirements of the possible upgrade of the large hadron collider (LHC) at CERN. Some of the RD50 most recent results about silicon detectors are reported in this paper, with special reference to: (i) the progresses in the characterization of lattice defects responsible for carrier trapping; (ii) charge collection efficiency of n-in-p microstrip detectors, irradiated with neutrons, as measured with different readout electronics; (iii) charge collection efficiency of single-type column 3D detectors, after proton and neutron irradiations, including position-sensitive measurement; (iv) simulations of irradiated double-sided and full-3D detectors, as well as the state of their production process.. 2008 - 5 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 596 (2008) 48-52 In : 8th International Conference on Large Scale Applications and Radiation Hardness of Semiconductor Detectors, Florence, Italy, 27 - 29 Jun 2007, pp.48-52 Rekord szczegółowy - Podobne rekordy 2018-08-25 06:58 Rekord szczegółowy - Podobne rekordy 2018-08-25 06:58
Performance of irradiated bulk SiC detectors / Cunningham, W (Glasgow U.) ; Melone, J (Glasgow U.) ; Horn, M (Glasgow U.) ; Kazukauskas, V (Vilnius U.) ; Roy, P (Glasgow U.) ; Doherty, F (Glasgow U.) ; Glaser, M (CERN) ; Vaitkus, J (Vilnius U.) ; Rahman, M (Glasgow U.)/CERN RD50 Silicon carbide (SiC) is a wide bandgap material with many excellent properties for future use as a detector medium. We present here the performance of irradiated planar detector diodes made from 100-$\mu \rm{m}$-thick semi-insulating SiC from Cree. [...] 2003 - 5 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 509 (2003) 127-131 In : 4th International Workshop on Radiation Imaging Detectors, Amsterdam, The Netherlands, 8 - 12 Sep 2002, pp.127-131 Rekord szczegółowy - Podobne rekordy 2018-08-24 06:19
Measurements and simulations of charge collection efficiency of p$^+$/n junction SiC detectors / Moscatelli, F (IMM, Bologna ; U. Perugia (main) ; INFN, Perugia) ; Scorzoni, A (U. Perugia (main) ; INFN, Perugia ; IMM, Bologna) ; Poggi, A (Perugia U.) ; Bruzzi, M (Florence U.) ; Lagomarsino, S (Florence U.) ; Mersi, S (Florence U.) ; Sciortino, Silvio (Florence U.) ; Nipoti, R (IMM, Bologna) Due to its excellent electrical and physical properties, silicon carbide can represent a good alternative to Si in applications like the inner tracking detectors of particle physics experiments (RD50, LHCC 2002–2003, 15 February 2002, CERN, Ginevra). In this work p$^+$/n SiC diodes realised on a medium-doped ($1 \times 10^{15} \rm{cm}^{−3}$), 40 $\mu \rm{m}$ thick epitaxial layer are exploited as detectors and measurements of their charge collection properties under $\beta$ particle radiation from a $^{90}$Sr source are presented. [...] 2005 - 4 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 546 (2005) 218-221 In : 6th International Workshop on Radiation Imaging Detectors, Glasgow, UK, 25-29 Jul 2004, pp.218-221 Rekord szczegółowy - Podobne rekordy 2018-08-24 06:19
Measurement of trapping time constants in proton-irradiated silicon pad detectors / Krasel, O (Dortmund U.) ; Gossling, C (Dortmund U.) ; Klingenberg, R (Dortmund U.) ; Rajek, S (Dortmund U.) ; Wunstorf, R (Dortmund U.) Silicon pad-detectors fabricated from oxygenated silicon were irradiated with 24-GeV/c protons with fluences between $2 \cdot 10^{13} \ n_{\rm{eq}}/\rm{cm}^2$ and $9 \cdot 10^{14} \ n_{\rm{eq}}/\rm{cm}^2$. The transient current technique was used to measure the trapping probability for holes and electrons. [...] 2004 - 8 p. - Published in : IEEE Trans. Nucl. Sci. 51 (2004) 3055-3062 In : 50th IEEE 2003 Nuclear Science Symposium, Medical Imaging Conference, 13th International Workshop on Room Temperature Semiconductor Detectors and Symposium on Nuclear Power Systems, Portland, OR, USA, 19 - 25 Oct 2003, pp.3055-3062 Rekord szczegółowy - Podobne rekordy 2018-08-24 06:19
Lithium ion irradiation effects on epitaxial silicon detectors / Candelori, A (INFN, Padua ; Padua U.) ; Bisello, D (INFN, Padua ; Padua U.) ; Rando, R (INFN, Padua ; Padua U.) ; Schramm, A (Hamburg U., Inst. Exp. Phys. II) ; Contarato, D (Hamburg U., Inst. Exp. Phys. II) ; Fretwurst, E (Hamburg U., Inst. Exp. Phys. II) ; Lindstrom, G (Hamburg U., Inst. Exp. Phys. II) ; Wyss, J (Cassino U. ; INFN, Pisa) Diodes manufactured on a thin and highly doped epitaxial silicon layer grown on a Czochralski silicon substrate have been irradiated by high energy lithium ions in order to investigate the effects of high bulk damage levels. This information is useful for possible developments of pixel detectors in future very high luminosity colliders because these new devices present superior radiation hardness than nowadays silicon detectors. [...] 2004 - 7 p. - Published in : IEEE Trans. Nucl. Sci. 51 (2004) 1766-1772 In : 13th IEEE-NPSS Real Time Conference 2003, Montreal, Canada, 18 - 23 May 2003, pp.1766-1772 Rekord szczegółowy - Podobne rekordy 2018-08-24 06:19
Radiation hardness of different silicon materials after high-energy electron irradiation / Dittongo, S (Trieste U. ; INFN, Trieste) ; Bosisio, L (Trieste U. ; INFN, Trieste) ; Ciacchi, M (Trieste U.) ; Contarato, D (Hamburg U., Inst. Exp. Phys. II) ; D'Auria, G (Sincrotrone Trieste) ; Fretwurst, E (Hamburg U., Inst. Exp. Phys. II) ; Lindstrom, G (Hamburg U., Inst. Exp. Phys. II) The radiation hardness of diodes fabricated on standard and diffusion-oxygenated float-zone, Czochralski and epitaxial silicon substrates has been compared after irradiation with 900 MeV electrons up to a fluence of $2.1 \times 10^{15} \ \rm{e} / cm^2$. The variation of the effective dopant concentration, the current related damage constant $\alpha$ and their annealing behavior, as well as the charge collection efficiency of the irradiated devices have been investigated.. 2004 - 7 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 530 (2004) 110-116 In : 6th International Conference on Large Scale Applications and Radiation Hardness of Semiconductor Detectors, Florence, Italy, 29 Sep - 1 Oct 2003, pp.110-116 Rekord szczegółowy - Podobne rekordy 2018-08-24 06:19
Recovery of charge collection in heavily irradiated silicon diodes with continuous hole injection / Cindro, V (Stefan Inst., Ljubljana) ; Mandić, I (Stefan Inst., Ljubljana) ; Kramberger, G (Stefan Inst., Ljubljana) ; Mikuž, M (Stefan Inst., Ljubljana ; Ljubljana U.) ; Zavrtanik, M (Ljubljana U.) Holes were continuously injected into irradiated diodes by light illumination of the n$^+$-side. The charge of holes trapped in the radiation-induced levels modified the effective space charge. [...] 2004 - 3 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 518 (2004) 343-345 In : 9th Pisa Meeting on Advanced Detectors, La Biodola, Italy, 25 - 31 May 2003, pp.343-345 Rekord szczegółowy - Podobne rekordy 2018-08-24 06:19
First results on charge collection efficiency of heavily irradiated microstrip sensors fabricated on oxygenated p-type silicon / Casse, G (Liverpool U.) ; Allport, P P (Liverpool U.) ; Martí i Garcia, S (CSIC, Catalunya) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Turner, P R (Liverpool U.) Heavy hadron irradiation leads to type inversion of n-type silicon detectors. After type inversion, the charge collected at low bias voltages by silicon microstrip detectors is higher when read out from the n-side compared to p-side read out. [...] 2004 - 3 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 518 (2004) 340-342 In : 9th Pisa Meeting on Advanced Detectors, La Biodola, Italy, 25 - 31 May 2003, pp.340-342 Rekord szczegółowy - Podobne rekordy 2018-08-23 11:31
Formation and annealing of boron-oxygen defects in irradiated silicon and silicon-germanium n$^+$–p structures / Makarenko, L F (Belarus State U.) ; Lastovskii, S B (Minsk, Inst. Phys.) ; Korshunov, F P (Minsk, Inst. Phys.) ; Moll, M (CERN) ; Pintilie, I (Bucharest, Nat. Inst. Mat. Sci.) ; Abrosimov, N V (Unlisted, DE) New findings on the formation and annealing of interstitial boron-interstitial oxygen complex ($\rm{B_iO_i}$) in p-type silicon are presented. Different types of n+−p structures irradiated with electrons and alpha-particles have been used for DLTS and MCTS studies. [...] 2015 - 4 p. - Published in : AIP Conf. Proc. 1583 (2015) 123-126 Rekord szczegółowy - Podobne rekordy |
I am writing a program which will compute $\exp(z)$.
Originally I used the Taylor series, which worked fine. However, continued fractions can converge more quickly than power series, so I decided to go that route.
I found this continued fraction in multiple places. It's from
The Application of Continued Fractions and Their Generalizations to Problems in Approximation Theory by A. N. Khovanskii (1963), pg 114.
$${e^{z}=1+{\cfrac {2z}{2-z+{\cfrac {z^{2}}{6+{\cfrac {z^{2}}{10+{\cfrac {z^{2}}{14+\ddots }}}}}}}}}$$
It can be represented as
$${e^{z}=1+\cfrac{2z}{2-z +}\cfrac{z^2/6}{1 +}\sum_{m=3}^{\infty}\left({\cfrac{{a_m}^{z^2}}{1}}\right)}$$
(Sorry, first time using Mathjax.)
Annie A.M. Cuyt, Vigdis Petersen, Brigitte Verdonk, Haakon Waadeland, and William B. Jones. 2008.
Handbook of Continued Fractions for Special Functions (1 ed.). Springer Publishing Company, Incorporated, pg 194.
I'm computing the continued fraction using the modified Lentz algorithm from
Numerical Recipes in C: The Art of Scientific Computing (ISBN 0-521-43105-5), pg 171. The issue: it only works for a small set of numbers. (From what I can tell, [-30, 30].)
So, my question: is this expected? I'm relatively new to continued fractions, so while I
think I grasp them, I'm not entirely sure.
Given a "generator", C++'s boost library can compute continued fractions. Essentially each call to the "generator" returns the next term in the CF.
Here's what I used (where $z$ is the input and $m$ is the current term index):
Term 0: A: $0$ B: $1$ Term 1: A: $2z$ B: $2-z$ Term 2: A: $z^2$ B: $1$ Term 3..N: A: $1 / (4 * (2m - 3) * (2m - 1))) * z^2$ B: $1$
Given $z = 38.5$, the Lentz algorithm provides the following (each line is $f_j$):
-1.10958904109589041.3657233326736593-1.86362005926024172.81700268519061-4.7118082829461678.70960008496205-17.76540820053092439.92118587023142-98.65273729066934267.59129323043885-795.12705695171952582.9903072474267-9154.5172645927935324.19248620738-148091.74385797494673154.6572372171-3.310843427214524e+061.758455218773504e+07-1.0065603099002995e+086.197709770714103e+08-4.097272719854685e+092.9029669729927063e+10-2.2004739456295242e+111.7812408854416082e+12-1.5389163725064727e+131.40311430366499e+14-1.483409360122725e+158.317237533312957e+152.273093139324771e+161.964661705307002e+161.9877542316211204e+161.985862955594415e+161.9860080303623144e+161.985997463174032e+161.9859981941770148e+161.9859981460940616e+161.9859981491045844e+161.9859981489249772e+161.9859981489351984e+161.9859981489346428e+161.9859981489346716e+161.98599814893467e+16
So, it converges at $1.98599814893467e+16$ when the actual answer is supposed to be: $\approx 52521552285925160$ |
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