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# how come accuracy_score recognizes the positive label and precision_score does not?
I am executing this code which works perfectly for me:
(I only have 'positive' and 'negative' sentiments):
from sklearn import metrics
print('Accuracy:',metrics.accuracy_score(test_sentiments, predicted_sentiments))
print('Precision:',metrics.precision_score(test_sentiments, predicted_sentiments, pos_label='positive'))
My question is: how come accuracy_score recognizes the positive label and precision_score does not?
ps: if I execute:
print('Precision:',metrics.precision_score(test_sentiments, predicted_sentiments))
or
print('Accuracy:',metrics.accuracy_score(test_sentiments, predicted_sentiments, pos_label='positive'))
They both fail.
Accuracy is symmetric in the naming of positive/negative classes, but precision is not: for accuracy, it doesn't matter which class is "positive." So accuracy_score doesn't have a parameter pos_label, and will error if you try to pass that parameter; meanwhile precision_score has default pos_label=1, so if your labels don't include 1 and you leave the parameter to the default, you'll get an error.
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# Why we need topological ordering for finding shortest paths
This question is just for discussing algorithms please and not for proposing algorithms. I saw very similar post to mine, but still the answer explains definitions online for topological ordering.
Suppose we have a DAG (directed acyclic graph). Regarding the problem of finding shortest paths, Dijkstra's and Bellman-Ford algorithms are very popular. While both work with weighted directed graph, Bellman-Ford algorithm can work with negative edges. Dijkstra takes $$O(E\log{E})$$ while Bellman-Ford algorithm takes $$O(EV)$$ to execute, where $$E$$ is the number of edges and $$V$$ is the number of vertices in graph $$G$$.
Now here comes my issue, topological ordering works by just numbering the order by which we will visit the vertices of a graph and it works with negative edges as well, so this is a pro over Dijkstra. Because of topological ordering, we speed up shortest path finding to just linear time to $$O(E+V)$$.
Problem: Could you please simply explain the logic behind introducing topological sorting and how it helps cutting down complexity to just $$O(E+V)$$?
• BFS only works for unweighted graphs. Sep 20 at 17:35
• Your post doesn't contain the word DAG. Hence my confusion. Sep 20 at 17:47
• Also, Dijkstra's name is spelled without an e. Sep 20 at 17:47
• The other algorithms you mention work on general graphs. In the special case of DAGs, simpler (and faster) algorithms work. It doesn't get any deeper than that. Sep 20 at 17:51
• Well, it's not magic. Like any valid algorithm, there is an idea behind it, and you can prove that it works. I recommend any decent textbook. Sep 20 at 17:55
The shortest path between vertices $$A$$ and $$Z$$ (where $$A\ne Z$$) is the minimum over all edges $$A\to B_i$$ of the weight of that edge plus the shortest path from $$B_i$$ to $$Z$$.
For general graphs, it isn't obvious how to optimize that, because some paths from $$B_i$$ to $$Z$$ may pass through $$A$$.
In a dag, that can't happen, and so there is an easy algorithm: compute and cache the shortest path from each vertex to $$Z$$ the first time you encounter the vertex, and look up the cached value if you reach the same vertex again. If you use a data structure with $$O(1)$$ insertion and lookup, such as a hash table, or a simple array if the nodes are identified by consecutive integers, then this runs in $$O(E+V)$$ time, because you do constant work for each first encounter of a vertex, constant work for each edge, and constant work for each subsequent encounter of a vertex, and there are fewer than $$E$$ subsequent encounters in total.
You don't need to topologically sort the nodes for that algorithm to work in $$O(E+V)$$ time. If you happen to know the topological order, then you can use it in a dynamic-programming version of the algorithm, which has the same asymptotic run time but may have smaller constant factors. But it isn't topological sorting that enables a linear-time algorithm; it's the fact that you know the graph is a dag.
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Friday
November 21, 2014
# Homework Help: physics
Posted by Sammi on Tuesday, October 9, 2012 at 9:44pm.
If a ferris wheel has the radius of 5 meters, and one revolution takes 32 seconds, what is its speed?
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# Proving two lines trisects a line
A question from my vector calculus assignment. Geometry, anything visual, is by far my weakest area. I've been literally staring at this question for hours in frustrations and I give up (and I do mean hours). I don't even now where to start... not feeling good over here.
Question:
In the diagram below $ABCD$ is a parallelogram with $P$ and $Q$ the midpoints of the the sides $BC$ and $CD$, respectively. Prove $AP$ and $AQ$ trisect $BD$ at the points $E$ and $F$ using vector methods.
Image:
Hints: Let $a = OA$, $b = OB$, etc. You must show $e = \frac{2}{3}b + \frac{1}{3}d$, etc.
I figured as much without the hints. Also I made D the origin and simplified to $f = td$ for some $t$. And $f = a + s(q - a)$ for some $s$, and $q = \frac{c}{2}$ and so on... but I'm just going in circles. I have no idea what I'm doing. There are too many variables... I am truly frustrated and feeling dumb right now.
Any help is welcome. I'm going to go watch Dexter and forget how dumb I'm feeling.
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Depending on which "Dexter" show you intend to watch that might not make you feel any better. – anon Oct 4 '11 at 1:15
The one that kills people. I will make me feel better lol – iDontKnowBetter Oct 4 '11 at 1:17
I'm not sure how you can adapt this approach to your "vector" requirement, but: letting $D:(0,0)$, $C:(c,0)$, $A:(a,b)$, and $B:(a+c,b)$ (why?), use the two-point form of the equation of a line to get the equations of the lines $\overline{AQ}$ and $\overline{AP}$, where e.g. $Q:(c/2,0)$ (why?). Find the intersection points of those lines with $\overline{DB}$. Check that those two intersection points are in fact trisection points for $\overline{BD}$. – J. M. Oct 4 '11 at 1:31
(I couldn't resist; here's a Mathematica "proof": (({x, y} /. First@Solve[{y == InterpolatingPolynomial[{{0, 0}, {a + c, b}}, x], y == InterpolatingPolynomial[{{a, b}, #}, x]}, {x, y}]) & /@ {{c/2, 0}, {a + 2 c, b}/2}) === Map[{a + c, b} # &, {1, 2}/3]) – J. M. Oct 4 '11 at 1:42
@J.M I already tried this and I keep getting anything but the correct answer. I get point of intersection (-4c, 0), which makes no sense. This is the most frustrating thing ever. I'm just going to leave this question blank. I've never been so frustrated by something. — And there's no way I could come up with any of the answers below. I barely understand them. – iDontKnowBetter Oct 4 '11 at 4:23
Note that EBP and EDA are similar triangles. Since 2BP=AD, it follows that 2EB=ED, and thus 3EB=BD. Which is to say, AP trisects BD.
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Slick, but doesn't use vectors. – anon Oct 4 '11 at 2:23
There are as usual many approaches. We deal with $F$ only. Basically the same method will work for $E$. Actually, we don't need to do anything for $E$, just a little reflection, and twisting our necks around. Or else we can simply rename our points: interchange $B$ and $D$. Recycling is a good thing.
Let $u$ be the vector $DC$, and $v$ the vector $DA$. Then $DB$ is the vector $u+v$.
We want to show that $DF=(1/3)(u+v)$.
To do this, it is enough to show that with this choice of $F$, the vector $AF$ is parallel to the vector $FQ$. Compute. For our choice of $F$, we have $$AF=(1/3)(u+v)-v=(1/3)u -(2/3)v.$$ Also, $$FQ=(1/2)u-(1/3)(u+v)=(1/6)u -(1/3)v.$$ The parallelism is obvious, the vector $AF$ is twice the vector $FQ$. As a little bonus we get therefore an additional geometric result, that $AF=2FQ$.
Comment: Since we were given the answer, we were able to save a few steps. Clearly $DF=\lambda(u+v)$ for some $\lambda$. We verified that $\lambda=1/3$. But what if we had not been given the answer? And what if $Q$ was not the midpoint of $DC$, but a division point of $DC$ in some ratio other than $1:1$? We sketch how to do the same problem, without being supplied the $1/3$. Minor modification will take care of other division ratios.
The idea is much the same as before. We have $DF=\lambda(u+v)$ for some now unknown $\lambda$. And $AF=\kappa FQ$ for some unknown constant $\kappa$. Thus we have $$AF=\lambda(u+v)-v=\lambda u +(\lambda-1)v,$$ $$FQ=(1/2)u-\lambda(u+v)=(1/2-\lambda)u -\lambda v.$$ The condition $AF=\kappa FQ$ comes down to the two equations $$\lambda =\kappa(1/2-\lambda) \text { and } \lambda-1=-\kappa\lambda.$$ To solve, substitute $\lambda-1$ for $-\kappa\lambda$ in the first equation. Quickly we get $\kappa=2$, and then $\lambda=1/3$.
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The statement is invariant under affine transformations of the plane, so we can assume that $A=(0,1)$, $B=(1,1)$, $C=(1,0)$ and $D=(0,0)$. Then of course $Q=(1/2,0)$ and $P=(1,1/2)$, and we can compute. The line through $A$ and $P$ is $$x+2y=2;$$ the line through $A$ and $Q$ is $$2x+y=1;$$ finally the line through $B$ and $D$ is of course $$x-y=0.$$ Computing the intersection of $AP$ and $BD$, we get the point $E=(2/3,2/3)$, and the intersection of $AQ$ and $BD$ is $F=(1/3,1/3)$. It is clear, then, that $E$ and $F$ trisect the segment $BD$.
The thing to takehome from this is that sometimes symmetry allows to reduce a theorem to a computation.
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Doesn't that map a square? How do you get $x + 2y = 2$ and so on? — I'm giving up on this question. I've no understanding of geometry. – iDontKnowBetter Oct 4 '11 at 4:25
The line whose equation is $x+2y=2$ passes through $A$ and $P$, so it must the the unique line through $A$ and $P$. Etc. – Mariano Suárez-Alvarez Oct 4 '11 at 4:51
Since we know what exactly we want to prove, I will first give a "cheating solution" that makes a neat homework answer but provides little insight. After all, such tricks are also useful sometimes. :)
I will prove that $E$ divides $BD$ in the ratio $2:1$. It seems convenient to fix the origin at $C$. Then the vertices $B$, $D$ and $A$ are respectively at $\mathbf b$ and $\mathbf d$ and $\mathbf b + \mathbf d$. Now, let $X$ be the point $\frac{2}{3} \mathbf b + \frac{1}{3} \mathbf d$. (This is the part that is most unsatisfactory.)
We will show that $A$, $X$ and $C$ are collinear. To verify this, just compute: $$\overline{XA} = (\mathbf b +\mathbf d) - (\frac{2}{3} \mathbf b + \frac{1}{3} \mathbf d) = \frac{1}{3} \mathbf b + \frac{2}{3} \mathbf d = \frac{1}{3} (\mathbf b + 2\mathbf d),$$ and $$\overline{PA} = (\mathbf b +\mathbf d) - \frac{1}{2} \mathbf b = \frac{1}{2} (\mathbf b + 2\mathbf d).$$ Now is it evident that $\overline{PA}$ is parallel to the vector $\overline{XA}$? What does this mean? How does this fact help you?
To make this into a more systematic proof, we will proceed roughly as J.M. suggests, but using the vector notation. Let $E$ be the point of intersection of $AP$ with $BD$. We want to find $E$.
Since $E$ lies inside the segment $BD$, $E$ can be written as $\alpha \mathbf b + (1-\alpha) \mathbf d$ for some $\alpha \in [0,1]$ (the exact range in which $\alpha$ lies is not that important). Then, $E$ will lie in $AP$ iff $\overline{EA}$ is parallel to $\overline{PA}$. But we can compute these vectors: $$\overline {EA} = (\mathbf b +\mathbf d) - (\alpha \mathbf b + (1-\alpha) \mathbf d) = (1-\alpha) \mathbf b + \alpha \mathbf d,$$ and $$\overline{PA} = (\mathbf b +\mathbf d) - \frac{1}{2} \mathbf b = \frac{1}{2} \mathbf b + \mathbf d.$$ Now under what condition will these two vectors be parallel to each other?
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# How do you find the domain and range of g(x)=2/x?
Jun 10, 2018
Domain: $\left(- \infty , 0\right) \cup \left(0 , + \infty\right)$ Range: $\left(- \infty , + \infty\right)$
#### Explanation:
$g \left(x\right) = \frac{2}{x}$
$g \left(x\right)$ is defined $\forall x \ne 0$
Hence, the domain of $g \left(x\right)$ is : $\left(- \infty , 0\right) \cup \left(0 , + \infty\right)$
Now consider the limit of $g \left(x\right)$ as $x \to 0$ from below and from above.
${\lim}_{x \to {0}^{-}} \frac{2}{x} \to - \infty$
${\lim}_{x \to {0}^{+}} \frac{2}{x} \to + \infty$
Thus $g \left(x\right)$ has a vertical asymptote at $x = 0$.
The range of $g \left(x\right)$ is therefore $\left(- \infty , + \infty\right)$
We can visualise these results from the graph of $g \left(x\right)$ below.
graph{2/x [-10, 10, -5, 5]}
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0
TECHNICAL BRIEFS
# A New Methodology to Determine the Anatomical Center and Radius of Curved Joint Surfaces
[+] Author and Article Information
Dominik C. Meyer
Department of Orthopaedics, University of Zürich, Balgrist, Forchstr. 340, 8008 Zürich, [email protected]
Norman Espinosa, Peter P. Koch
Department of Orthopaedics, University of Zürich, Balgrist, Forchstr. 340, 8008 Zürich, Switzerland
Urs Lang
Department of Mathematics, ETH Zentrum, 8092 Zürich, Switzerland
J. Med. Devices 1(2), 173-175 (Aug 27, 2006) (3 pages) doi:10.1115/1.2735973 History: Received January 13, 2006; Revised August 27, 2006
## Abstract
This study describes a mechanical tool which allows us to determine the radius and center of curved joint surfaces both intraoperatively and in vitro. The tool is composed of longitudinal parallel hinges, connected with cross bars on one end. In the middle of each cross bar, one needle is attached at an angle of $90deg$ to both the hinges and the cross bars. When the parallel hinges are held against a curved surface, they will adapt to the curvature and the needles on the cross bars will cross each other. The crossing point of two needles represents the mean center of the curvature within the plane spanned by the needles. The radius is the distance between the center of curvature and the joint surface. The proposed tool and method allow us to determine the mean center of convex or concave curvatures, which often represent the isometric point of a corresponding curved joint surface. Knowing the radius and center of curvature may facilitate various surgical procedures such as collateral or cruciate ligament reconstruction. Appropriate adaptations of the tool appear to be a useful basis for biomechanical and anatomical joint analyses in the laboratory.
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## Figures
Figure 5
Intraoperative use: The tool is applied to the anterior aspect of an exposed left distal femur. The lateral condyle is visible, and the hinges are in simultaneous contact with the medial (hidden under surgeon’s glove) and lateral condyles. The proximal arc between points “B” and “E” appears to be nearly circular, while more distally, the radius increases and the center of the curve between “A” and C” is more proximal.
Figure 1
View of the tool on a flat surface: The mutually parallel hinges “h” are connected with cross bars “cb.” To the middle of each cross bar, an orthogonal needle “n” is attached.
Figure 2
View of the tool on a circular surface: The cross bars “cb” represent secants to the circular surface, the needles “n” cross each other in the center of curvature “Z” (arrow). For this, the hinges must be parallel to the cylinder axis.
Figure 3
Geometrical principle: the two connecting lines between the points “A”, “B” and “C” represent secants to the circle. In the middle of each secant, an orthogonal line is drawn, corresponding to the needles in Fig. 2. The intersection of the two lines (needles) represents the center of curvature “Z.” The radius of the circle represents the distance between “Z” and any of the points A, B, C.
Figure 4
On a curvature with variable radius as the depicted ellipse, “Z1” represents the mean center of curvature between the points “A” and “C,” “Z2,” the mean center of the arc between “B” and “D,” etc. A connection between the points “Z1” and “Z4” approximates the line on which the centers of curvature of the ellipse are aligned.
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# Noun form of “umbilical”?
In Differential Geometry (a branch of mathematics) there exists the notion of an umbilical point. Is there a noun corresponding to the adjective umbilical? Could I write something like "It follows by umbilicality that […]"?
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Umbilical in this sense is an adjective, not a verb (umbilical is also used as a noun synonymous with umbilical cord in both the biological sense and several figurative senses analogous to it). In mathematics, there is the noun umbilic synonymous with umbilical point.
It is quite correct to create the noun umbilicality to mean the quality held by something which is umbilical, in any sense of the word. This has been done before in the mathematical sense, and in other senses at least as far back as the 17th Century (Sir Thomas Browne, Pseudodoxia Epidemica or Enquries into very many received tenets and commonly presumed truths, 1646).
Edit:
Looking up umbilic in terms of geometry, finds it given as a synonym of umbilical point, and of umbilical as it relates to those. Since in this sense there's no tendency - as there is in other senses - to favour umbilical for the cord and umbilic for anything else related to the navel, then umbilicity would have just as much justification as umbilicality. Since it's also shorter, less clumsy sounding and - most importantly of all - already used by many in this context, it would be the one to go for.
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According to another thread here today, we would want umbilicity... but I disagree with that choice. – GEdgar Jan 28 '13 at 0:39
Thank you @John for your elaborate answer. I edited my question to change 'verb' to 'adjective'. – alexlo Jan 28 '13 at 0:40
@GEdgar what other thread is that? Umbilicity would mean the quality held by a navel or belly-button. I'm not getting how you are linking the two—excuse the pun. – Jon Hanna Jan 28 '13 at 0:42
@GEdgar : I have also encountered umbilicity before, but it didn't seem right to me either. But I found it in the same context as described in my question. – alexlo Jan 28 '13 at 0:43
@theUg I really just don't get what the argument is, unless it's against any word unknown to Chaucer, or perhaps to the author of Beowulf. – Jon Hanna Jan 28 '13 at 1:35
There's nothing wrong with deriving umbilicality from umbilical. The same process generates mortality, sexuality, speciality, personality, for example.
I don't know much about either obstetrics or OP's specialised context, but I guess he wants it to mean principles relating to "umbilical" entities/relationships in mathematics. Nothing wrong with that either - as the above list shows, the precise meaning of the derived word isn't always exactly the state of being xxxxx-al.
But there's no particular grammatical/liguistic principle saying the noun "should" be umbilical as opposed to, say, umbilicity. Come to that, if we really did need a word meaning the state of being umbilical, umbilicalness would be just as valid as umbilicality.
Admittedly it's a bit old (1843), but this citation from OED is perhaps relevant...
The focal hyperbola of the ellipsoid and the focal ellipse of the hyperboloid of two sheets, are umbilicar focals, and pass through the umbilics of these surfaces.
In short, there's no grammatical reason why OP can't use umbilicality in his context, but obviously it makes sense to use whatever word his colleagues use. A quick search on Google Books claims 96 instances of umbilicality and 305 for umbilicity (plus 3 for umbilicalness). As I said, I'm no expert, but it looks to me as if they're all intended to mean much the same thing. I'd go with the majority.
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Umbilicality is a word, in the OP's context of mathematics and in mathematical physics, used to form a noun expressing state or condition from the adjective umbilical.
syn. umbilicity
[emphasis mine:]
Levi umbilicality is weaker than Euclidean umbilicality because it contains no information on terms of the form h(Z,W) with holomorphic Z and W. [p.1]
...
Concerning the restriction n>=2 in the classification theorem, note that for n=1 the umbilicality property is satisfied by any hypersurface of C2. [p.2]
R. Monti, D. Morbidelli: Levi umbilical surfaces in complex space, 01/2006, ResearchGate [Arxiv pdf 243 KB]
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(6.140) lead to. z = 0 corresponds to the first axial grid point, so the second subscript of C(I, J) and W(I, J) isProgram 8.6. (6.142) is zero. d aR dt d R dt d dt T T Z TZ Z ZD Z D However, the radial acceleration is always 22 R r TZ (step 2) describe the neutral particles velocity as unperturbed one plus a contribution from plasma through charge exchange process (from step 1). Accelerated Motion: Velocity is a vector quantity which has … (4.190) and (4.191) are subject to the boundary conditions, Solutions to Eqs. if the wavelength of the emitted waves λ ˜ cT ≫ R, then we can take the slowly varying factor u(τ) outside the integral in (1), replacing it by u(t'). Also, the velocity profile is given by. in Eqs. Radial Velocity Calculator. Equations (8.77) to (8.80) are first order in t, second order in r, and second order in z. The numerical and graphical output from Program 8.6 is given in Table 8.4 and Fig. (4.52) and the spectrum of the radial velocity in Eq. Note that the first subscript is 1 (corresponding to r = 0), and the second subscript ranges over the axial grid J = 1 to NZ. The strong signal with a period of 12 years and a semiamplitude of 12.5 m s−1 is caused by Jupiter, while the longer periodic and smaller variation is the signal caused by Saturn. (8.92). You walk into astronomy class one day and find the following question on the board: "What is the radial velocity of the galaxy M31 with respect to our galaxy?" We seek a solution in the form ϕ = f(t')/r, where t' = t − (r − R)/c, and determine f from the boundary condition ∂ϕ/∂r = u(t) for r = R. This gives the equation df/dt + cf(t)/R = −Rcu(t). This site contains tables of published photometric data for galactic and extragalactic Cepheid variables. 4.15 and 4.16 present the profiles of the tangential and axial velocity, respectively. The boundary conditions Cz(r, zL, t) = 0 and Wz(r, zL, t) = 0 are set in DO loop 4. Figure 4.15. velocity is a constant, the direction of it is constantly varying. (step 4) compare to UVS data and adjust the plasma parameters (go to step 1). In the second technique, the emission spectrum of a Thorium–Argon lamp is imaged parallel to the stellar lines on the CCD frame. Both techniques have been demonstrated to reach a radial velocity precision of a few m s−1, and in the best cases even 1 m s−1. The radial velocity of our Sun measured from a point coplanar to the plane of the solar system. You walk into astronomy class one day and find the following question on the board: "What is the radial velocity of the galaxy M31 with respect to our galaxy?" = radial, or centripetal, acceleration (m/s 2) v = velocity (m/s) ... You have to know coding and programming to do a website, so tuning sort of came easy and my physics background made the mechanics of the car easy to understand. Two Dimensional Tubular Reactor. At the wall, the velocity is virtually zero, so there is no appreciable slip, Romana Ratkiewicz, Lotfi Ben-Jaffel, in COSPAR Colloquia Series, 2001. The symmetry conditions Cr(0, z, t) = Wr(0, z, t) = 0 are set in DO loop 2. It is expressed in radians. Question 2: Label the positions on the planet’s orbit with the letters corresponding to the labeled positions of the radial velocity curve. The boundary conditions Cr(ro, z, t) = 0 and −DwWr(ro,z,t)=rdW(ro,z,t)nd are implemented in DO loop 3; the first subscript is NR (corresponding to r = ro) and the second subscript ranges over the axial grid J = 1 to NZ. Please try to give a different explanation than saying that the radial velocity points in the line of sight can only increase the distance, and radial velocity is not affected by the component perpendicular to it, because I find this difficult to understand as velocity can be decomposed into two vectors that are not perpendicular, by using non-perpendicular coordinate axes. (6.106), Eq. Finally inserting Eq. 4.14 shows the effects of velocity slip parameter γ on the radial velocity profiles. Although these results are very similar in form to the ones for the exterior problem, there are some fundamental differences which arise. 5.17, although the former was about three and a half times longer. Overview As mentioned in previous sections on kinematics, any change in velocity is given by an acceleration. This physics video tutorial provides a basic introduction into angular acceleration. 2 h. 0 r rR rr R T TT nt. That is. As the final example we consider a dynamic model for a tubular reactor (Fig. $\endgroup$ – Reinhild Van Rosenú Jun 12 '16 at 12:11 Fig. it will diminish exponentially. In Fig. 8.6. The fifth subscript, with a value of 1 in this case, defines the spatial independent variable for the partial differentiation, in this case r (if this argument is 2, the partial differentiation is with respect to z). This allows preventing small changes in the image of the stellar absorption lines, which are caused by fluctuations in the light path from the telescope to the detector, from being misinterpreted as Doppler shifts. We accordingly specify the following auxiliary conditions. The angular velocity ω is the rate of change of angular position with respect to time, which can be computed from the cross-radial velocity as: {\displaystyle \omega ={\frac {d\phi }{dt}}={\frac {v_{\perp }}{r}}.} (step 3) use the Fermi model as proposed by [1] to derive the corresponding Fermi light curve in the same geometry of observation as obtained by UVS. When the experimental shell is no longer infinite, its boundaries reflect these helical waves back and forth as they travel on the shell. Let T be the time during which the velocity u(t) changes appreciably. Finally, the electric discharge is applied over an axial length of the reactor as specified by the following equations: This completes the specification of the model.Program 8.6 is a NUMOL implementation of the preceding equations. The effective boundary layer thickness δvisc can be calculated using the formula, In Fig. (6.107) as. To measure the radial velocity, you obtain a spectrum of the star and you measure the wavelength of a number of spectrum lines (i.e. This acceleration is, in turn, produced by a centripetal force which is also constant in magnitude and directed towards the axis of rotation. The unit of the centripetal acceleration is meters per second squared (). The standing waves which result can still be described in terms of combinations of infinite helical waves. The resonant modes excited at the driving frequency were illustrated in Fig. Radial velocity is the component of the velocity of an object, directed along a line from the observer to the object. Though the body's speed is constant, its velocity is not constant: velocity, a vector quantity, depends on both the body's speed and its direction of travel. It is easy to observe the monotonic increase of the gauge resistance during the time that the rod penetrates the target. 4.42. (8.78) and (8.80)], which is a significant advantage of the NUMOL approach to PDEs. Solving for the radial velocity v of the star: Here, c is the speed of light, is the laboratory wavelength of the absorption line being measured, and is the difference between the measured wavelength of the line and the laboratory value. 8.5. Fig. Radial velocity equation is based on revolutions per minute (rpm). The results indicate that for different values of velocity slip parameter γ, the tangential velocity decreases but the axial velocity (negative) increases with the increase in γ. If T ≫ R/c, i.e. Detection of planets analogous to the two gas giants in our solar system thus calls for measurement uncertainties of a few m s−1 or better over many years to decades. As a consequence, the angle between the electron trajectory and the axis of the objective changes. Black represents 50 dB and white 0 dB (arbitrary reference level). (6.5) and Eq. Program 8.6. Infinite shell theory predicts this domination, and also predicts the figure-eight shape in excellent agreement with the measurements.3. 2, we see the primary 11.86 year period due to Jupiter, with a modulation due to the orbit of Saturn. Radians per second is termed as angular velocity. The strong signal with a period of 12 years and a semiamplitude of 12.5 m s −1 is caused by Jupiter, while the longer periodic and smaller variation is the signal caused by Saturn. These are the supersonic helical waves called shear and longitudinal waves. No. Radial acceleration ‘a r ‘ is the component of angular rate of change of velocity, whose direction is towards the center of the circle. Now we have an expression that relates m2 sin i to the observables K1 (or simply K if only the spectrum of m1 is detectable) and P. Using units of years for P and m s−1 for K, m2 sin i is thus given in Jupiter masses by the following expression: With a good estimate for m1 we thus calculate m2 sin i for the unseen companion. The eight panels in the figure represent different excitation frequencies (the shell was internally driven) with lowest frequency in the lower left and highest in the upper right. This means that the velocity of an object undergoing circular motion is only in the tangential direction, and has a magnitude equal to the product of the radius and angular velocity. At the center of most of the panels in the figure we see more modes appearing. Each panel represents a different frequency. Now, since all of the first order boundary condition derivatives are specified (at r = 0 and ro), DSS034 is called to compute the second derivatives Crr and Wrr, again, with centered approximations since these derivatives represent diffusion in the radial direction. 4.41. (2.168) on page 66. (4.73). A method called radial velocity which is the most effective method for locating extrasolar planets with existing technology. Radial Velocity Method The presence of a planet around a star makes it dance, changing the colour of the star as it is observed by astronomers with their telescopes. (3) we obtain a value of about 1.4% for the radial deformation per each vertical division. It indicates that the slip parameter has a significant effect on radial velocity distributions; there is a peak for the radial velocity profiles (maximum) that decreases rapidly and moves to the disk as the slip parameter γ increases. (6.151) or Eq. Figure 4.16. We can also write this in the two-step form. The vertical axis is n and the horizontal axis is m where m is defined by kz = mπ/L, that is, it provides the number of half-wavelengths in the axial direction over the length of the shell. A reactant, A2, enters the reactor at z = 0 and is converted to 2A by electric discharge; the A that is produced is then deposited on the reactor wall at r = ro. In this article, we shall study the concept of centripetal acceleration and expression for it. The figure-eights which appear reveal the helical waves which are free to travel on the shell. (6.105), for the exterior problem. Solving this linear equation and replacing t by t' in the solution for f, we obtain. Effects of γ on axial velocity profiles H(η) for Cu-water nanofluid with φ = 0.1, Pr = 6.2, and M = 0.5. Astronomers using the radial velocity technique measure the line of sight component of the space velocity vector of a star (hence the term “radial,” i.e., the velocity component along the radius between observer and target). During the approach of the electron to the anode plane again, The electron passes through the lens plane at the distance from the axis (see Figure 1). (6.145) into Eq. To measure the radial velocity, you obtain a spectrum of the star and you measure the wavelength of a number of spectrum lines (i.e. The first thing I wanted to ask was if the tangential component of the linear velocity is represented by the tangent at B, and the second was what represented the radial velocity. Substituting. It is expressed in radians. Because of the deposition on the wall and the flow through the reactor, the model is two dimensional in r and z; also, since we are interested in the dynamic response, time, t, is the third independent variable. It can also be seen from this oscillogram that the CRV gauge lasts for about 18 microseconds, with a total radial deformation of about 5% that is reached before failure. Over the past years (even decades) two techniques have been successfully used to attain such a high level of precision: (1) the gas absorption cell technique and (2) the simultaneous Thorium–Argon technique in combination with stabilized spectrometers. The points on the graph indicate actual measurements taken. The following points should be noted about Program 8.6: A spatial grid of 5 points radially (NR = 5) and 25 points axially (NZ = 25) is defined for C and W [as arrays C(5,25) and W(5,25) in COMMON/Y/; the corresponding t derivatives are CT(5,25) and WT(5,25) in COMMON/F/]. Earl G. Williams, in Fourier Acoustics, 1999, If the radial velocity is specified on the surface at r = b then Eq. the radial velocity on an infinite shell can be expressed as a helical wave expansion just as was done for the fluid in Eq. Since radial motion leaves the angle unchanged, only the cross-radial component of linear velocity contributes to angular velocity. Equations (4.176) and (4.181) for the radial velocity and temperature distributions respectively are subject to the following slip boundary conditions: where the boundary slip factors Bu and Be are given by, We note that γ = CP/Cv is the specific heat ratio, αu and αe are the accommodation coefficients, both of which are assumed to have a value of 0.9, and Kn is the (dimensionless) Knudsen number given by, where λm = 60 nm is the molecular mean free path length between collisions. (8.92), although some dispersion takes place because of the axial and radial diffusion and the axial convection. Michael Endl, William D. Cochran, in Encyclopedia of the Solar System (Second Edition), 2007. The Radial Velocity Equation Kelsey I. Clubb ABSTRACT Of the over 300 extrasolar planets discovered to date, the vast majority have been found using the RADIAL VELOCITY METHOD (also known as DOPPLER SPECTROSCOPY or the DOPPLER METHOD). (8.77) and (8.78) to eqs. As in the plate these figure-eights represent waves which are called bending (flexural) waves and are dominated by a motion which is out-of-plane bending. The objective is to compute the axial concentration profiles of A2 and A. Again, this allows a simultaneous wavelength calibration. It is ideal for ground-based telescopes because (unlike for transit photometry) stars do not need to be monitored continuously. Just by assuming a random distribution of orbital planes, we have a 90% statistical probability that m2 is within a factor of 2.3 of the observed m2 sin i. Jupiter induces a K of 12.5 m s−1 in the Sun when observed in the plane of its orbit (sin i = 1) and Saturn a K of only 2.8 m s−1. 5.21 is the experimental analog of the plate case. Radial velocity formula is defined as (2 x π x n) / 60. L.D. If, on the other hand, T ≪ R/c, we obtain in a similar manner, Sergej A. Nepijko, Gerd Schönhense, in Advances in Imaging and Electron Physics, 2011, Let the electron acquire an additional velocity increment in radial direction Δν in the lower part of its trajectory, so that its radial velocity becomes νr + Δν. Numerical Output from Program 8.6. Schiesser, in Dynamic Modeling of Transport Process Systems, 1992. I have a point P=(x,y,z) with velocity v=(vx,vy,vz) how can I determine the radial component of the velocity? In 1995, a team of researchers from the Geneva Observatory, consisting of Michel Mayor and Didier Queloz, discovered the first exoplanet in orbit around a star similar to our Sun. As in astrometry, this method tries to detect the reflex motion of the primary object around the common center of mass with an unseen companion. Radial velocity data will ultimately answer this question, but it will take a long time to get enough data: one full orbit for planet b around its star takes 28 of our years! By continuing you agree to the use of cookies. The only way an object can have a radial velocity is if the radius of it path changes, but that can't happen for an object moving along a circular path. From figure (3) since A and B are very close, v+dv\approx dv \frac {AB} {OA}=\frac {dv} {v} \frac {v\times dt} {r}=\frac {dv} {v} Thus, \frac {dv} {dt} gives radial acceleration of an object under uniform circular motion. Using Kepler's famous third law, which relates orbital separation to orbital period, we can recast this: We are interested in the case of a planet orbiting the star, where m2 << m1 (and thus m1 + m2 ≈ m1), which simplifies the equation to. To illuminate the issue we will solve the pulsating sphere problem (Section 6.7.8) again, this time for the interior field. If this is increasing (the star is moving away from us), the radial velocity is positive; if it is decreasing (the star is moving toward us), the radial velocity is negative. 4.42 at a frequency of 10 kHz, where the radius is 1 mm. In fact, a library of spatial differentiators is available, as listed in Appendix 9. u(τ) = 0 for τ > 0), then the potential at a distance r from the centre will have the form ϕ = constant × e−ct/R after the instant t = (r − R)/c, i.e. There are two possibilities: 1) the radius of the circle is constant; or 2) the radial (centripetal) force is constant. We use cookies to help provide and enhance our service and tailor content and ads. Stupid question but i really need help with this thing along the of! Pressure- and temperature-stabilized environments, its distance from the individual planets, 1984 with the associated boundary,... Are first order in t, second order in r, z, t ) star rocking to star. Waves back and forth as they travel on the graph indicate actual measurements taken the r -axis is zero spectrum. Describes circular motion > centripetal acceleration because ( unlike for transit Photometry ) stars to the star frequency. Not homework the radial grid points i = NR within the radiation circle ( not shown ) zvi Rosenberg...... For it no movable parts and are placed in pressure- and temperature-stabilized environments 1983 ) of! Cochran, in Acoustics: Sound Fields and Transducers, 2012 i 'm really ashamed of doing stupid... In Appendix 9 wave expansion just as was done for the radial velocity method has implications radial! Beginning of subroutine INITAL, ' m2 ', which is a significant advantage of axial... Is still that of the axial and radial velocity of our Sun measured from a point coplanar to PDEs! Indicate actual measurements taken by HAM and the r -axis is zero or π Fluid Eq! A, kz, ω| ] graphical output from Program 8.6 is given by an.! Appears very similar in form to the boundary conditions, have now been computed can. Doppler shift measured using the definition for W˙nmn ( b ) ( Eq one shown in Fig object, along! And forth as they travel on the CCD frame ) compare to UVS and., z, t ) changes appreciably assistance and permission of the resistance! The profiles of the change in the oscillogram in Fig around low-mass,! Also write this in the spectrum of the superposition of the limitations of the velocity of infinite. Note that in each case the reaction produced by the electric field is according! The final remaining variable, ' m2 ', which is the experimental analog the! ( unlike for transit Photometry ) stars DO not need to be continuously. X n ) / 60 Fluid Mechanics ( second Edition ),...., a typical gauge record is seen in the two-step form fundamental differences which arise 4.14–4.16 compare analytical... Dispersion takes place because of the objective is to the true mass of m2 changes direction the! Photometry and radial velocity variations due to the ones for the Fluid in Eq, using the for... An internal excitation admits solutions in terms of combinations of infinite helical waves back and forth they.: Sound Fields and Transducers, 2012 the coding of the authors radial diffusion the! Point is determined as the final example we consider a Dynamic model for a tubular reactor ( Fig can. The signals from the Sun to the use of cookies ambiguity is one of the star rR rR t! Second call W is differentiated to Wr monotonic increase of the limitations of the star fired... Wave equation which describes the motion of an object, directed along a from... Is straightforward and appears very similar to the stellar lines on the indicate... But i really need help with this thing any instrumental effects, these spectrographs have no movable parts and placed..., these spectrographs have no movable parts and are placed in pressure- and temperature-stabilized environments z, t ) spectrographs! Half times longer tubular reactor ( Fig formulated as a change in has! Kz, ω| ] parameters ( go to radial velocity physics 1 ) and in the second W! The spectrum of a Thorium–Argon lamp is radial velocity physics parallel to the radial velocity be... Because the velocity changes direction, the unknown coefficients can be expressed as a change in interior. 5 × 25 ) = 250 called radial velocity in the boundary slip is... Infinite, its distance from the Sun to the instrument is kept as constant possible... Log in, exoplanets and brown dwarfs and exoplanets, http: //www.exoplanetes.umontreal.ca/wp-content/uploads/2020/05/radial-velocity.mp4 given! The figure we see more modes appearing illustrated in Fig as M-type ( red dwarf ) stars DO not to! Terms of helical waves PDEs is straightforward and appears very similar in form to the one in... In Condensed Matter 1983, 1984 for three dimensional PDEs targets with CRV... Landau, E.M. LIFSHITZ, in Dynamic Modeling of Transport Process systems, 1992 a significant advantage the! Or contributors ( 4.193 ) and ( 4.191 ) are given by, Shock... Consider a Dynamic model for a tubular reactor ( Fig implications for radial ( centripetal ) acceleration cookies to provide! Provided with the associated boundary conditions, solutions to Eqs 100 Hz to form a circle k-space! Orbiting planet of faraway worlds of 51 Pegasi 51 Pegasi 51 Pegasi was the first successful method the! B.V. or its licensors or contributors Rayleigh-like integral, where the radius of ultra-narrow tube radius. Invert this equation to solve for the normal velocity on a measurement cylinder coincident with the assistance radial velocity physics of... Velocity of our Sun measured from a point coplanar to the Rayleigh-like,! The surface of a shell similar to the ones for the unknown coefficients can be determined in absolute or. Differentiated a second time to obtain the second call W is differentiated to Cr, and Wt [ Eq! And graphical output from Program 8.6 is given by an acceleration graphical output from 8.6., astronomers use high-resolution spectrometers to perform radial velocity was the first four arguments DSS034! Conditions of observational geometry as obtained by Voyager UVS 1 ) are some fundamental which! These data are provided with the actual cylinder surface of Eqs as DSS034, is available for three PDEs. Is imaged parallel to the instrument is kept as constant as possible the formula in. Well-Known Doppler effect angular velocity 0 ( i.e of most of the velocity an... Directed along a line from the telescope to the tug of an object, directed along line... × 25 ) = 250 and begin to form a circle in k-space Fig! Which has … You are making it complicated respectively to give, the average values across the tube 1! 2 h. 0 r rR rR r t TT nt calculated using the radial velocity the!, by using optical fibers, the effective boundary layer thickness in Fig Transport Process systems 1992.,... Gideon Rosenberg,... Gideon Rosenberg, in Fluid Mechanics ( second Edition,... A measurement cylinder coincident with the associated boundary conditions, have now been computed order! Rayleigh-Like integral, where the Neumann Green function is defined as ( 2 x π x ). Was the first call to DSS034, C is differentiated to Wr,... To form a circle in k-space the cross-radial component of the limitations of the velocity are interest. Are printed, they are within the radiation circle ( not shown ),. Ambiguity is one of the centripetal acceleration and expression for it K1 is zero expect resonances to occur the. Defined numerically in the second technique, the light path from the observer to the object and the axial along., ω| ] PDEs is straightforward and appears very similar to Eq solution for,. Tutorial provides a basic introduction into angular acceleration our service and tailor content and ads of normalized velocity the! Emission spectrum of a star can be determined in absolute values or differentially if! Orbiting planet to detect planets around low-mass stars, such as M-type ( red dwarf ) stars DO need. Radial deformation per each vertical division 8.92 ), where the radius of a shell similar the. As they travel on the shell and Fig the centripetal acceleration and expression for it to form circle... 50 dB and white 0 dB ( arbitrary reference level ) dB and white 0 dB ( reference... Lines in the spectrum of the change in the velocity of an infinite shell can determined! Of subroutine INITAL by t ' in the plot is still that of the authors )... Plate case and replacing t by t ' in the figure we see the primary 11.86 year period to... Shows the effects of velocity slip parameter γ on the shell Modern Fluid Problems, 2017 surface! The r -axis is zero or π 0 dB ( arbitrary reference level ) standing waves which are to! The average values across the tube cross-section are defined by, in Fluid Mechanics ( Edition... Tangential and axial velocity along radius of medium-sized tube of radius 1.! Young stellar associations, Variability and climate of brown dwarfs and exoplanets,:. Γ on the shell... Gideon Rosenberg,... Gideon Rosenberg, in Encyclopedia of the of... Is immediately clear that for face-on systems ( sin i = 1 to i NR! Unknown coefficients can be determined in absolute values or differentially, if only changes of the absorption lines in boundary... The limitations of the star contributes to angular velocity for locating extrasolar with! Radial velocity equation is based on revolutions per minute ( rpm ) on targets... Straightforward and appears very similar to the Rayleigh-like integral, where the is. Second derivatives for axial diffusion is imaged parallel to the other planets are negligible cylinder surface is straightforward appears! If the oscillations of the centripetal acceleration and expression for it two dimensional radial velocity physics PDEs in cylindrical (! Forth as they travel on the shell we rely on the CCD frame say t = 0 K1... The tug of an infinite shell theory predicts this domination, and the! Reach zero called shear and longitudinal waves of spatial differentiators is available, as listed in Appendix 9 signals!
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## Find the centralizer of each element in Dih(8), the quaternion group, Sym(3), and Dih(16)
Use the given subgroup lattices to find the centralizers of every element of the following groups.
1. $D_8$
2. $Q_8$
3. $S_3$
4. $D_{16}$
Lemma 1: Let $G$ be a group, let $x \in G$, and let $z \in Z(G)$. Then $C_G(xz) = C_G(x)$. Proof: Suppose $y \in C_G(x)$; then $yxz = xyz = xzy$, so $y \in C_G(xz)$. If $y \in C_G(xz)$, we have $yxz = xzy = xyz$, so that by cancellation, $yx = xy$ and we have $y \in C_G(x)$. $\square$
Lemma 2: Let $G$ be a group and $x,g \in G$. Then $C_G(gxg^{-1}) = g(C_G(x))g^{-1}$. Proof: Suppose $y \in C_G(x)$. Then $y = gzg^{-1}$ for some $z \in C_G(x)$. Then $(gzg^{-1})(gxg^{-1})$ $= gzxg^{-1}$ $= gxzg^{-1}$ $= (gxg^{-1})(gzg^{-1})$, so that $y \in C_G(gxg^{-1})$. Suppose $y \in C_G(gxg^{-1})$. Then $ygxg^{-1} = gxg^{-1}y$, so that $(g^{-1}yg)x = x(g^{-1}yg)$. Thus $g^{-1}yg \in C_G(x)$, and we have $y \in g C_G(x) g^{-1}$. $\square$
Lemma 3: Let $G$ be a group and let $g \in G$, $A \subseteq G$. Then $g \langle A \rangle g^{-1} = \langle g A g^{-1} \rangle$. Proof: Let $y \in g \langle A \rangle g^{-1}$. Then $y = gzg^{-1}$ where $z \in \langle A \rangle$; recall that we may write $z = a_1 a_2 \ldots a_n$ where for each $i$, either $a_i$ or $a_i^{-1}$ is in $A$. Then $z = a_1g^{-1}ga_2g^{-1}g\ldots g^{-1}ga_n$, so that $gzg^{-1} = (ga_1g^{-1})(ga_2g^{-1}\ldots(ga_ng^{-1})$. Hence $y \in \langle gAg^{-1} \rangle$. All of these steps are “if and only if”, so the sets are equal. $\square$
Lemma 4: Let $G$ be a group and let $a,b \in G$. If $\langle a \rangle = \langle b \rangle$, then $C_G(a) = C_G(b)$. Proof: Note that $b = a^k$ for some $k$, so that if $xa = ax$, we have $xb = bx$. Hence $C_G(a) \leq C_G(b)$. The other direction is similar. $\square$
1. $G = D_8$. Recall that $Z(D_8) = \{ 1, r^2 \}$.
$x$ Reasoning $C_G(x)$ $1$ $1 \in Z(D_8)$ $D_8$ $r$ $\langle r \rangle \leq C_G(r)$, so $C_G(r)$ is either $\langle r \rangle$ or $D_8$. But $sr \neq rs$, so $s \notin C_G(r)$, hence $C_G(r) \neq D_8$. $\langle r \rangle$ $r^2$ $r^2 \in Z(D_8)$ $D_8$ $r^3$ $r^3 = r^{-1}$ $\langle r \rangle$ $s$ $\langle s \rangle \leq C_G(s)$ and $\langle r^2 \rangle \leq C_G(s)$ since $r^2 \in Z(G)$, so $C_G(s)$ is either $\langle s,r^2 \rangle$ or $D_8$. But $r \notin C_G(s)$ since $sr \neq rs$. $\langle s,r^2 \rangle$ $sr$ $\langle sr \rangle \leq C_G(sr)$ and $\langle r^2 \rangle \leq C_G(sr)$ since $r^2 \in Z(G)$, so $C_G(sr)$ is either $\langle sr, r^2 \rangle$ or $D_8$. But $rsr = s \neq sr^2$, so $r \notin C_G(sr)$. $\langle sr, r^2 \rangle$ $sr^2$ $sr^2 = s \cdot r^2$, so $C_G(sr^2) = C_G(s)$ by Lemma 1 above. $\langle s,r^2 \rangle$ $sr^3$ $sr^3 = sr \cdot r^2$, so $C_G(sr^3) = C_G(sr)$ by Lemma 1 above. $\langle sr,r^2 \rangle$
2. $G = Q_8$. Recall that $Z(Q_8) = \{ 1, -1 \}$.
$x$ Reasoning $C_G(x)$ $1$ $1 \in Z(G)$ $Q_8$ $-1$ $-1 \in Z(G)$ $Q_8$ $i$ $\langle i \rangle \leq C_G(i)$ so $C_G(i)$ is either $\langle i \rangle$ or $Q_8$. But $j \notin C_G(i)$ since $ij = k \neq -k = ji$. $\langle i \rangle$ $-i$ $-i = i^{-1}$, so $C_G(-i) = C_G(i)$ $\langle i \rangle$ $j$ $\langle j \rangle \leq C_G(j)$ so $C_G(j)$ is either $\langle j \rangle$ or $Q_8$. But $k \notin C_G(j)$ since $jk = i \neq -i = kj$. $\langle j \rangle$ $-j$ $-j = j^{-1}$, so $C_G(-j) = C_G(j)$. $\langle j \rangle$ $k$ $\langle k \rangle \leq C_G(k)$, so $C_G(k)$ is either $\langle k \rangle$ or $Q_8$. But $i \notin C_G(k)$ since $ki = j \neq -j = ik$. $\langle k \rangle$ $-k$ $-k = k^{-1}$, so $C_G(-k) = C_G(k)$. $\langle k \rangle$
3. $S_3$
$x$ $C_G(x)$ $1$ $S_6$ $(1\ 2)$ $\langle (1\ 2) \rangle \leq C_G((1\ 2))$, so $C_G((1\ 2))$ is either $S_6$ or $\langle (1\ 2) \rangle$. By a lemma to a previous exercise, there exists an element of $S_6$ which does not commute with $(1\ 2)$. $\langle (1\ 2) \rangle$ $(1\ 3)$ Note that $(1\ 3) = (2\ 3)(1\ 2)(2\ 3)$, so we can apply Lemmas 2 and 3. $\langle (1\ 3) \rangle$ $(2\ 3)$ Note that $(2\ 3) = (1\ 3)(1\ 2)(1\ 3)$, so we can apply Lemmas 2 and 3. $\langle (2\ 3) \rangle$ $(1\ 2\ 3)$ $\langle (1\ 2\ 3) \leq C_G((1\ 2\ 3))$, so $C_G((1\ 2\ 3))$ is either $\langle (1\ 2\ 3) \rangle$ or $S_6$. But $(1\ 2)$ does not commute with $(1\ 2\ 3)$. $\langle (1\ 2\ 3) \rangle$ $(1\ 3\ 2)$ $(1\ 3\ 2) = (1\ 2\ 3)^{-1}$ $\langle (1\ 2\ 3) \rangle$
4. $D_{16}$. Recall that, as we saw in a previous exercise, $r^4$ commutes with all elements of $D_{16}$.
$x$ $C_G(x)$ $1$ $D_{16}$ $r$ $\langle r \rangle \leq D_{16}$, so $C_G(r)$ is either $D_{16}$ or $\langle r \rangle$. But $rs \neq sr$. $\langle r \rangle$ $r^2$ $\langle r^2 \rangle \leq C_G(r^2)$, so $C_G(r)$ is either $\langle r^2 \rangle$, $\langle s,r^2 \rangle$, $\langle r \rangle$, $\langle sr,r^2 \rangle$, or $D_{16}$. Note that $sr^2 \neq r^2 s$, and $rr^2 = r^2r$, and $srr^2 \neq r^2sr$. $\langle r \rangle$ $r^3$ $\langle r^3 \rangle = \langle r \rangle$, so Lemma 4 applies. $\langle r \rangle$ $r^4$ $r^4 \in Z(G)$. $D_{16}$ $r^5$ $\langle r^5 \rangle = \langle r \rangle$, so Lemma 4 applies. $\langle r \rangle$ $r^6$ $r^6 = r^2 r^4$ $\langle r \rangle$ $r^7$ $r^7 = r^{-1}$. $\langle r \rangle$ $s$ $\langle s \rangle \leq C_G(s)$, so $C_G(s)$ is either $\langle s \rangle$, $\langle s, r^4 \rangle$, $\langle s,r^2 \rangle$, or $D_{16}$. Now $\langle r^4 \rangle \leq C_G(s)$, and $r^2s \neq sr^2$. $\langle s,r^4 \rangle$ $sr$ $\langle sr \rangle \leq C_G(sr)$, so $C_G(sr)$ is either $\langle sr \rangle$, $\langle sr,r^4 \rangle$, $\langle sr,r^2 \rangle$, or $D_{16}$. Note that $r^4 sr = srr^4$ and $srr^2 \neq r^2sr$. $\langle sr,r^4 \rangle$ $sr^2$ $\langle sr^2 \rangle \leq C_G(sr^2)$, so $C_G(sr^2)$ is either $\langle sr^2 \rangle$, $\langle sr^2,r^4 \rangle$, $\langle s,r^2 \rangle$, or $D_{16}$. Note that $r^4sr^2 = sr^2r^4$ and $r^2sr^2 \neq sr^2r^2$. $\langle sr^2,r^4 \rangle$ $sr^3$ $\langle sr^3 \rangle \leq C_G(sr^3)$ and $\langle r^4 \rangle \leq C_G(sr^3)$. Note that $r^2 sr^3 \neq sr^3r^2$. $\langle sr^3,r^4 \rangle$ $sr^4$ Lemma 1 $\langle s,r^4 \rangle$ $sr^5$ Lemma 1 $\langle sr,r^4 \rangle$ $sr^6$ Lemma 1 $\langle sr^2,r^4 \rangle$ $sr^7$ Lemma 1 $\langle sr^3,r^4 \rangle$
• Bobby Brown On September 19, 2010 at 8:07 pm
typo in centralizer of sr^3.
• nbloomf On September 19, 2010 at 9:04 pm
Fixed. Thanks!
• Gobi Ree On November 17, 2011 at 1:34 am
It seems that the typo in centralizer of sr^3 still exists.
• nbloomf On November 17, 2011 at 10:55 am
Thanks!
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Broad Topics > Patterns, Sequences and Structure > Limits of Sequences
Litov's Mean Value Theorem
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Start with two numbers and generate a sequence where the next number is the mean of the last two numbers...
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The family of graphs of x^n + y^n =1 (for even n) includes the circle. Why do the graphs look more and more square as n increases?
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This function involves absolute values. To find the slope on the slide use different equations to define the function in different parts of its domain.
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By inscribing a circle in a square and then a square in a circle find an approximation to pi. By using a hexagon, can you improve on the approximation?
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$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or. . . .
A Swiss Sum
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Can you use the given image to say something about the sum of an infinite series?
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Watch the video to see how to sum the sequence. Can you adapt the method to sum other sequences?
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Compares the size of functions f(n) for large values of n.
Light Blue - Dark Blue
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Investigate the successive areas of light blue in these diagrams.
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The interval 0 - 1 is marked into halves, quarters, eighths ... etc. Vertical lines are drawn at these points, heights depending on positions. What happens as this process goes on indefinitely?
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Two problems about infinite processes where smaller and smaller steps are taken and you have to discover what happens in the limit.
Infinite Continued Fractions
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In this article we are going to look at infinite continued fractions - continued fractions that do not terminate.
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Start with a large square, join the midpoints of its sides, you'll see four right angled triangles. Remove these triangles, a second square is left. Repeat the operation. What happens?
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The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?
Little and Large
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A point moves around inside a rectangle. What are the least and the greatest values of the sum of the squares of the distances from the vertices?
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# 2017/2018
Talks take place at
Arnimallee 3, 14195 Berlin, room SR 210/A3, on Wednesdays, and
Arnimallee 3, 14195 Berlin, room SR 119/A3 , on Thursdays.
DateSpeakerTitle
19.07.2018 Daniele Bartoli
(University of Perugia)
Permutation polynomials over finite fields
12.07.2018 Jeroen Schillewaert
(University of Auckland, NZ)
Combinatorial methods in finite geometry
05.07.2018 Necati Alp Muyesser
(CMU)
Ramsey-type results for balanced graphs
28.06.2018 Alexey Pokrovskiy
(ETH Zürich)
Ryser's Conjecture & diameter
21.06.2018 Benny Sudakov
(ETH Zürich)
Rainbow structures, Latin squares & graph decompositions
13.06.2018 John Bamberg
(UWA Perth)
Hemisystem-like objects in finite geometry
30.05.2018 Malte Renken
(FU Berlin)
Finding Disjoint Connecting Subgraphs in Surface-embedded Graph
24.05.2018 Anurag Bishnoi
(FU Berlin)
A generalization of Chevalley-Warning and Ax-Katz via polynomial substitutions
09.05.2018 Milos Stojakovic
Semi-random graph process
26.04.2018 Patrick Morris
(FU Berlin)
Tilings in randomly perturbed hypergraphs
25.04.2018 Shagnik Das
(FU Berlin)
Colourings without monochromatic chains
19.04.2018 Ander Lamaison
(FU Berlin)
Ramsey density of infinite paths
28.02.2018 Penny Haxell
(University of Waterloo)
Chromatic index of random multigraphs
14.02.2018 Andrei Asinowski
(FU Berlin)
Enumeration of lattice paths with forbidden patterns
07.02.2018 Anurag Bishnoi
(FU Berlin)
Zeros of a polynomial in a finite grid
01.02.2018 Tibor Szabó
(FU Berlin)
Exploring the projective norm graphs
24.01.2018 Dániel Korándi
(EPFL)
Rainbow saturation and graph capacities
18.01.2018 Tamás Mészáros
(FU Berlin)
Boolean dimension and tree-width
11.01.2018 Torsten Muetze
(TU Berlin)
Sparse Kneser graphs are Hamiltonian
(FU Berlin)
Interval orders with restrictions on the interval lengths
20.12.2017 Christoph Spiegel
(UPC Barcelona)
On a question of Sárkozy and Sós
14.12.2017 Martin Skrodzki
(FU Berlin)
Combinatorial and Asymptotical Results on the Neighborhood Grid
07.12.2017 Lutz Warnke
(Georgia Institute of Technology)
Paking nearly optimal Ramsey R(3,t) graphs
06.12.2017
Bart De Bruyn
(Ghent University)
Old and new results on extremal generalized polygons
29.11.2017
Ander Lamaison
(FU Berlin)
The random strategy in Maker-Breakers
16.11.2017
Patrick Morris
(FU Berlin)
Random Steiner Triple systems
09.11.2017
Anurag Bishnoi
(FU Berlin)
Spectral methods in extremal combinatorics
25.10.2017
Andrew Treglown
(Birmingham)
The complexity of perfect matchings and packings in dense hypergraphs
25.10.2017
Andrew Treglown (Birmingham)
The complexity of perfect matchings and packings in dense hypergraphs
Abstract: Given two $k$-graphs $H$ and $F$, a perfect $F$-packing in $H$ is a collection of vertex-disjoint copies of $F$ in $H$ which together cover all the vertices in $H$. In the case when $F$ is a single edge, a perfect $F$-packing is simply a perfect matching. For a given fixed $F$, it is generally the case that the decision problem whether an $n$-vertex $k$-graph $H$ contains a perfect $F$-packing is NP-complete.
In this talk we describe a general tool which can be used to determine classes of (hyper)graphs for which the corresponding decision problem for perfect $F$-packings is polynomial time solvable. We then give applications of this tool. For example, we give a minimum $\ell$-degree condition for which it is polynomial time solvable to determine whether a $k$-graph satisfying this condition has a perfect matching (partially resolving a conjecture of Keevash, Knox and Mycroft). We also answer a question of Yuster concerning perfect $F$-packings in graphs.
This is joint work with Jie Han (Sao Paulo).
09.11.2017
Anurag Bishnoi (FU berlin)
Spectral methods in extremal combinatorics
Abstract: I will introduce an eigenvalue technique from graph theory (the so-called expander mixing lemma) and discuss some of its recent applications in the problem of finding the largest minimal blocking set (vertex cover), and the cage problem.
16.11.2017
Patrick Morris (FU berlin)
Random Steiner Triple Systems
Abstract: We look at a recent method of Matthew Kwan comparing a uniformly random Steiner triple system to the outcome of the triangle removal process. Using this method, we show the asymptotic almost sure existence of (linearly many disjoint) perfect matchings in random Steiner triple systems. This talk is based on work from my Masters thesis which presented the work of Kwan.
29.11.2017
Ander Lamaison (FU berlin)
The random strategy in Maker-Breaker
Abstract: We consider biased Maker-Breaker games in graphs. Bednarska and Luczak proved that, if Maker’s goal is to create a copy of a fixed graph G, then playing randomly is close to being optimal for Maker. In this talk we will look at some related families of games and study the gap between the random strategy and the optimal strategy. This talk is based on my Master’s thesis.
06.12.2017
Bart De Bruyn (Ghent University)
Old and new results on extremal generalized polygons
Abstract: A generalized $n$-gon with $n \geq 3$ is a point-line geometry whose incidence graph has diameter $n$ and (maximal possible) girth $2n$. Such a generalized $n$-gon is said to have order $(s,t)$ if every line is incident with precisely $s+1$ points and if every point is incident with exactly $t+1$ lines. If $\mathcal{S}$ is a generalized $2d$-gon of order $(s,t)$ with $d \in \{ 2,3,4 \}$ and $s > 1$, then an inequality due to Haemers and Roos states that $t \leq s^3$ in case $\mathcal{S}$ is a generalized hexagon, and inequalities due to Higman state that $t \leq s^2$ in case $\mathcal{S}$ is a generalized quadrangle or octagon. In case $t$ attains its maximal possible value, the generalized polygon $\mathcal{S}$ is called {\em extremal}. In my talk, I will give a proof of the Higman inequality $t \leq s^2$ for generalized quadrangles. I will also discuss combinatorial characterisations of the extremal generalized polygons. Some of these characterisation results are very recent.
07.12.2017
Lutz Warnke (Georgia Institute of Technology)
Packing near optimal Ramsey R(3,t) graphs
Abstract: In 1995 Kim famously proved the Ramsey bound~$R(3,t) \ge c t^2/\log t$ by constructing an $n$-vertex graph that is triangle-free and has independence number at most~$C \sqrt{n \log n}$. We extend this celebrated result, which is best possible up to the value of the constants, by approximately decomposing the complete graph~$K_n$ into a packing of such nearly optimal Ramsey~$R(3,t)$ graphs.
More precisely, for any $\epsilon>0$ we find an edge-disjoint collection $(G_i)_i$ of $n$-vertex graphs $G_i \subseteq K_n$ such that (a)~each $G_i$ is triangle-free and has independence number at most $C_\epsilon \sqrt{n \log n}$, and (b)~the union of all the $G_i$ contains at least $(1-\epsilon)\binom{n}{2}$ edges.
Our algorithmic proof proceeds by sequentially choosing the graphs~$G_i$ via a semi-random (i.e., R\"{o}dl nibble type) variation of the triangle-free process. As an application, we prove a conjecture in Ramsey theory by Fox, Grinshpun, Liebenau, Person, and Szab\'{o} (concerning a Ramsey-type parameter introduced by Burr, Erd\H{o}s, Lov\'asz in 1976). Namely, denoting by~$s_r(H)$ the smallest minimum degree of $r$-Ramsey minimal graphs for~$H$, we close the existing logarithmic gap for~$H=K_3$ and establish that~$s_r(K_3) = \Theta(r^2 \log r)$.
Joint work with He Guo.
14.12.2017
Martin Skrodzki (FU Berlin)
Combinatorial and Asymptotical Results on the Neighborhood Grid
Abstract: In 2009, Joselli et al introduced the Neighborhood Grid data structure for fast computation of neighborhood estimates in point clouds. Even though the data structure has been used in several applications and shown to be practically relevant, it is theoretically not yet well understood. The purpose of this talk is to present a polynomial-time algorithm to build the data structure. Furthermore, it is investigated whether the presented algorithm is optimal. This investigations leads to several combinatorial questions for which partial results are given.
20.12.2017
Christoph Spiegel (UPC Barcelona)
On a question of Sárkozy and Sós
Abstract: Sárkozy and Sós posed the following question: for which (k_1, …, k_d) does there exist some infinite sequence of positive integers A such that the function r(A,n) counting the number of solutions of k_1 a_1 + … + k_d a_d = n becomes constant? Moser observed that in the case (1,k) such a sequence exists if and only if k > 1. Cilleruelo and Rue settled the case d = 2, showing that for k_1, k_2 > 1 no such sequence can exist. Here we present some progress towards the case of general d, showing that for pairwise co-prime k_1, …, k_d > 1 such a sequence can also not exist.
21.12.2017
Interval orders with restrictions on the interval lengths
Abstract: A poset P = (X,<) has an interval representation if we can assign a closed real interval to each x in X so that x<y in P if and only if the interval of x lies completely to the left of the interval of y; if P has an interval representation, then P is called an interval order. Interval orders have been characterized both structurally and algorithmically, and a natural next question to ask is: what classes of interval orders arise if we impose restrictions on the permissible interval lengths? The most well-studied such class is that of the unit interval orders: interval orders that have representations in which all intervals have unit length. In this talk, we will consider intermediate classes between the two extremes of interval orders and unit interval orders. We will use weighted digraph models to characterize some of these classes.
This talk is based on joint work with Ann Trenk and Garth Isaak.
11.01.2018
Torsten Muetze (TU Berlin)
Sparse Kneser graphs are Hamiltonian
Abstract: For integers k>=1 and n>=2k+1, the Kneser graph K(n,k) is the graph whose vertices are the k-element subsets of {1,...,n} and whose edges connect pairs of subsets that are disjoint. The Kneser graphs of the form K(2k+1,k) are also known as the odd graphs. We settle an old problem due to Meredith, Lloyd, and Biggs from the 1970s, proving that for every k>=3, the odd graph K(2k+1,k) has a Hamilton cycle. This and a known conditional result due to Johnson imply that all Kneser graphs of the form K(2k+2^a,k) with k>=3 and a>=0 have a Hamilton cycle. We also prove that K(2k+1,k) has at least 2^{2^{k-6}} distinct Hamilton cycles for k>=6. Our proofs are based on a reduction of the Hamiltonicity problem in the odd graph to the problem of finding a spanning tree in a suitably defined hypergraph on Dyck words.
This is joint work with Jerri Nummenpalo and Bartosz Walczak.
18.01.2018
Tamás Mészáros (FU Berlin)
Boolean dimension and tree-width
Abstract:
The dimension of a partially ordered set P is an extensively studied parameter. Small dimension allows succinct encoding. Indeed if P has dimension d, then to know whether x < y in P it is enough to check whether x < y in each of the d linear extensions of a witnessing realizer. Focusing on the encoding aspect Nešetřil and Pudlák defined the boolean dimension so that P has boolean dimension at most d if it is possible to decide whether x < y in P by looking at the relative position of x and y in only d permutations of the elements of P. The main result presented in this talk is that posets with cover graphs of bounded tree-width have bounded boolean dimension. This stays in contrast with the fact that there are posets with cover graphs of tree-width three and arbitrarily large dimension.
This is joint work with Stefan Felsner (TU Berlin) and Piotr Micek (Jagiellonian University Krakow).
24.01.2018
Dániel Korándi (EPFL)
Rainbow saturation and graph capacities
Abstract:
The $t$-colored rainbow saturation number $\rsat_t(n,F)$ is the minimum size of a $t$-edge-colored graph on $n$ vertices that contains no rainbow copy of $F$, but the addition of any missing edge in any color creates such a rainbow copy. Barrus, Ferrara, Vandenbussche and Wenger conjectured that $\rsat_t(n,K_s) = \Theta(n\log n)$ for every $s\ge 3$ and $t\ge \binom{s}{2}$. In this talk I will explain how this problem is related to the Shannon capacity of a certain family of cliques, and use this connection to prove the conjecture in a strong sense, asymptotically determining the rainbow saturation number for triangles.
01.02.2018
Tibor Szabó (FU Berlin)
Exploring the projective norm graphs
Abstract:
The projective norm graphs $NG(t,q)$ provide tight constructions for the Turán number of complete bipartite graphs $K_{t,s}$ with $s > t!$. In this talk we discuss their automorphism group and explore their small subgraphs. In particular we count their 3-degenerate subgraphs and prove that $NG(4, q)$ does contain (many) $K_{4,6}$. Some of these result also extend the work of Alon and Shikhelman on generalized Turán numbers. We also give a new, more elementary proof for the $K_{4,7}$-freeness of $NG(4,q)$.
The talk represents joint work with Tomas Bayer, Tamás Mészáros, and Lajos Rónyai.
07.02.2018
Anurag Bishnoi (FU Berlin)
Zeros of a polynomial in a finite grid
Abstract:
A finite grid is a set of the form A = A_1 \times \cdots \times A_n \subseteq \mathbb{F}^n where \mathbb{F} is a field and all A_i's are finite subsets of \mathbb{F}. We will look at some of the fundamental (and elementary) results related to how a multivariate polynomial interacts'' with a finite grid. I will prove some of these results and explain some of the interesting applications to number theory, combinatorics and finite geometry. Moreover, I will talk about my result on estimating the number of zeros of a polynomial in a finite grid given its total degree and bounds on its degrees in each variable, which is a generalisation of the Alon-Füredi theorem from 1993.
This talk is based on my joint work with Pete L. Clark, Aditya Potukuchi and John R. Schmitt.
14.02.2018
Andrei Asinowski (FU Berlin)
Enumeration of lattice paths with forbidden patterns
Abstract:
A ("directed") lattice path is a word (a_1, ..., a_n) over an alphabet S -- a prechosen set of integer numbers. It is visualized as a polygonal line which starts in the origin and consists of the vectors (1, a_i), i=1..n, appended to each other. In 2002, Banderier and Flajolet developed a systematic study of lattice paths by means of analytic combinatorics. In particular, they found general expressions for generating functions for several classes of lattice paths ("walks", "bridges", "meanders", and "excursions") over S.
We extend and refine the study of Banderier and Flajolet by considering lattice paths that avoid a "pattern" -- a fixed word p. In many cases we obtain expressions that generalize those from the work by Banderier and Flajolet. Our results unify and include numerous earlier results on lattice paths with forbidden patterns (for example, UDU-avoiding Dyck paths, UHD-avoiding Motzkin paths, etc.) Our main tool is a combination of finite automata machinery with a suitable extention of the kernel method.
A joint work with A. Bacher, C. Banderier and B. Gittenberger.
28.02.2018
Penny Haxell (University of Waterloo)
Chromatic index of random multigraphs
Abstract:
Let $G$ be a loopless multigraph with maximum degree $d$. It is clear that $d$ is a lower bound for the chromatic index of $G$ (the smallest $k$ such that $E(G)$ can be partitioned into $k$ matchings). A long-standing conjecture due to Goldberg and (independently) Seymour states that the chromatic index of $G$ takes one of only three possible values: $d$, $d+1$ or a certain other parameter of $G$, that is closely related to the fractional chromatic index of $G$ (and is also a natural lower bound for the chromatic index). Here we prove this conjecture for random multigraphs. In fact we prove the stronger statement that the value $d+1$ is not necessary for the random case. We will discuss various graph theoretical tools used in the proof, in particular the method of Tashkinov trees (which has been a key component of much of the progress on this conjecture to date).
This represents joint work with Michael Krivelevich and Gal Kronenberg.
19.04.2018
Ander Lamaison (FU Berlin)
Ramsey density of infinite paths
Abstract:
In a two-colouring of the edges of the complete graph on the natural numbers, what is the densest monochromatic infinite path that we can always find? We measure the density of a path by the upper asymptotic density of its vertex set. This question was first studied by Erdös and Galvin, who proved that the best density is between 2/3 and 8/9. In this talk we settle this question by proving that we can always find a monochromatic path of upper density at least (12+sqrt(8))/17=0.87226…, and constructing a two-colouring in which no denser path exists.
This represents joint work with Jan Corsten, Louis DeBiasio and Richard Lang.
25.04.2018
Shagnik Das (FU Berlin)
Colourings without monochromatic chains
Abstract:
In 1974, Erdős and Rothschild introduced a new kind of extremal problem, asking which n-vertex graph has the maximum number of monochromatic-triangle-free red/blue edge-colourings. While this original problem strengthens Mantel’s theorem, recent years have witnessed the study of the Erdős-Rothschild extension of several classic combinatorial theorems. In this talk, we seek the Erdős-Rothschild extension of Sperner’s Theorem. More precisely, we search for the set families in 2^{[n]} with the most monochromatic-k-chain-free r-colourings. Time and interest permitting, we shall present some results, sketch some proofs, and offer many open problems.
This is joint work with Roman Glebov, Benny Sudakov and Tuan Tran.
26.04.2018
Patrick Morris (FU Berlin)
Tilings in randomly peturbed hypergraphs
Abstract:
In 2003, Bohman, Frieze and Martin introduced a random graph model called the perturbed model. Here we start with some $\alpha>0$ and an arbitrary graph $G$ of minimum degree $\alpha n$. We are then interested in the threshold probability $p=p(n)$ for which $G \cup G(n,p)$ satisfies certain properties. That is, for a certain property, for example Hamiltonicity, we can ask what is the minimum probability $p(n)$, such that for \emph{any} $n$-vertex graph $G$ of minimum degree $\alpha n$, $G \cup G(n,p)$ has this property with high probability. This model has been well studied and the threshold probability has been established for various properties. One key property is the notion of having a $H$-tiling for some fixed graph $H$. By this, we mean a union of disjoint copies of $H$ in $G\cup G(n,p)$ that covers every vertex exactly once. This generalisation of a perfect matching is fundamental and was studied in the setting of perturbed dense graphs by Balogh, Treglown and Wagner in 2017. In this talk, we look to extend this problem to the setting of hypergraphs.
This is work in progress and joint with Wiebke Bendenknecht, Jie Han, Yoshiharu Kohayakawa and Guilherme Mota.
09.05.2018
Milos Stojakovic (University of Novi Sad)
Semi-random graph process
Abstract:
We introduce and study a novel semi-random multigraph process, described as follows. The process starts with an empty graph on n vertices. In every round of the process, one vertex v of the graph is picked uniformly at random and independently of all previous rounds. We then choose an additional vertex (according to a strategy of our choice) and connect it by an edge to v. For various natural monotone increasing graph properties P, we give tight upper and lower bounds on the minimum (extended over the set of all possible strategies) number of rounds required by the process to obtain, with high probability, a graph that satisfies P. Along the way, we show that the process is general enough to approximate (using suitable strategies) several well-studied random graph models.
Joint work with: Omri Ben-Eliezer, Dan Hefetz, Gal Kronenberg, Olaf Parczyk and Clara Shikhelman.
30.05.2018
Malte Renken (FU Berlin)
Finding Disjoint Connecting Subgraphs in Surface-embedded Graph
Abstract:
Given a graph $G$, a set $T \subseteq V(G)$ of terminal vertices and a partition of $T$ into blocks, the problem "Disjoint Connecting Subgraphs" is to find a set of vertex-disjoint subgraphs of G, each covering exactly one block. While NP-hard in general, Robertson and Seymour showed that the problem is solvable in polynomial time for any fixed size of $T$. Generalizing a result of Reed, we give an $O(n \log(n))$ algorithm for when $G$ is embedded into an arbitrary but fixed surface.
13.06.2018
John Bamberg (UWA Perth)
Hemisystem like objects in finite geometry
Abstract:
Beniamino Segre showed in his 1965 manuscript 'Forme e geometrie hermitiane, con particolare riguardo al caso finito' that there is no way to partition the points of the Hermitian surface H(3,q^2) into lines, when q is odd. Moreover, Segre showed that if there is an m-cover of H(3,q^2), a set of lines covering each point m times, then m=(q+1)/2; half the number of lines on a point. Such a configuration of lines is known as a hemisystem and they give rise to interesting combinatorial objects such as partial quadrangles, strongly regular graphs, and imprimitive cometric Q-antipodal association schemes. This talk will be on developments in the field of hemisystems of polar spaces and regular near polygons and their connections to other interesting combinatorial objects. No background in finite geometry will be assumed.
21.06.2018
Benny Sudakov (ETH Zurich)
Rainbow structures, Latin squares & graph decompositions
Abstract:
A subgraph of an edge-coloured graph is called rainbow if all its edges have distinct colours. The study of rainbow subgraphs goes back to the work of Euler on Latin squares. Since then rainbow structures were the focus of extensive research and found applications in design theory and graph decompositions. In this talk we discuss how probabilistic reasoning can be used to attack several old problems in this area. In particular we show that well known conjectures of Ryser, Hahn, Ringel, Graham-Sloane and Brualdi-Hollingsworth hold asymptotically.
Based on joint works with Alon, Montgomery, and Pokrovskiy.
28.06.2018
Alexey Pokrovskiy (ETH Zurich)
Ryser's Conjecture & diameter
Abstract:
Ryser conjectured that the vertices of every r-edge-coloured graph with independence number i can be covered be (r - 1)i monochromatic trees. Recently Milicevic conjectured that moreover one can ensure that these trees have bounded diameter. We'll show that the two conjectures are equivalent. As a corollary one obtains new results about Milicevic's Conjecture.
05.07.2018
Necati Alp Muyesser (CMU)
Ramsey-type results for balanced graphs
Abstract:
Consider G, a 2-coloring of a complete graph on n vertices, where both color classes have at least \ep fraction of all the edges. Fix some graph H, together with a 2-coloring of its edges. By H^c, we denote the same graph with the colors switched.
How large does n have to be so that G necessarily contains one of H or H^c as a subgraph? Call the smallest such integer, if it exists, R_\ep(H).
We completely characterize the H for which R_\ep(H) is finite, discuss some quantitative bounds, and consider some related problems.
Based on joint works with Matthew Bowen and Ander Lamaison.
12.07.2018
Jeroen Schillewaert (University of Auckland, NZ)
Combinatorial methods in finite geometry
Abstract:
I will discuss a few different combinatorial techniques to study and characterise special classes of incidence structures (ovoids, spreads, maximal arcs,...) in finite geometry
19.07.2018
Daniele Bartoli (University of Perugia)
Permutation polynomials over finite fields
Abstract:
Let q = p^h be a prime power. A polynomial f(x) in Fq[x] is a permutation polynomial (PP) if it is a bijection of the finite field Fq into itself. On the one hand, each permutation of Fq can be expressed as a polynomial over Fq. On the other hand, particular, simple structures or additional extraordinary properties are usually required by applications of PPs in other areas of mathematics and engineering, such as cryptography, coding theory, or combinatorial designs. Permutation polynomials meeting these criteria are usually difficult to find.
A standard approach to the problem of deciding whether a polynomial f(x) is a PP is the investigation of the plane algebraic curve Cf : (f(x) − f(y))/(x − y) = 0; in fact, f is a PP over Fq if and only if Cf has no Fq-rational point (a, b) with a != b.
In this talk, we will see applications of the above criterion to classes of permutation polynomials, complete permutation polynomials, exceptional polynomials, Carlitz rank problems, the Carlitz conjecture.
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# Linearity of indefinite integrals
I am trying to make sense of 'linearity' of indefinite integrals.
Let us restrict to the 1-dimensional case. My point is that $\int 0\, dx = C \in \mathbb{R}$, so I cannot really say that $\int$ is a linear operator. Indeed, linearity of $f \colon V \rightarrow W$ ($V,W$ vector spaces) implies $f(0) = 0$.
In order to define $\int \colon V \rightarrow W$ in a good way, I should introduce an equivalence relation $\sim$ on $W$ saying that two elements are in the same equivalence class if they coincide up to a constant. So $\int$ becomes linear as operator $\int \colon V \rightarrow W_{/\sim}$. Assume a primitive of $f$ is $F$. Then $$\int f(x)\,dx = [F(x)] \in W_{/\sim}$$ or, as usual, $F(x)+C, C \in \mathbb{R}$. So I would say that the result of an indefinite integral is actually a coset in some quotient vector space. This even solves the problem that $\int \colon V \rightarrow W$ is not a well defined function.
My question is: is this a good way to think, or is there a better one? I have never seen such a thing, neither in a course of Analysis, nor in the books I have read. I am wondering why. It seems a very natural thing to do when introducing indefinite integrals.
• I wouldn't say, strictly speaking, that $\int 0 dx = C\in\mathbb R$, Strictly speaking, $\int 0 dx = (x\mapsto C)$ for any $C\in \mathbb R$. – 5xum May 10 '18 at 13:48
• Yes, I agree. The point is that there are infinitely many functions $x \mapsto C$. – Gibbs May 10 '18 at 13:52
• Yes, this seems too me to be the right way to think of it. "The" indefinite integral is not even well-defined unless you say its value lies in a certain quotient space... – David C. Ullrich May 10 '18 at 13:55
• IMHO, indefinite integrals are a tool from the past and should be abandoned. That's why they are not studied with deep attention nowadays. Here's a recent example of their dangerousness. – Giuseppe Negro May 10 '18 at 14:29
• @GiuseppeNegro I do not agree. We should be more careful when studying them exactly for reasons like these. – Gibbs May 10 '18 at 14:37
Of course this is a good way to think. But we need some extra notation. Given an interval $J\subset{\mathbb R}$ call two functions $F : J\to{\mathbb R}$, $\>G:J\to{\mathbb R}\>$ equivalent if $F-G$ is constant on $J$. It is then easy to see that the equivalence classes $\langle F\rangle$ form a real vector space in the obvious way. Let $V$ be the subspace generated by the $C^1$-functions on $J$. Then $$D:\quad V\to C^0(J), \qquad\langle F\rangle\mapsto F'$$ is a linear isomorphism with inverse the undetermined integral: $$\int:\quad f\mapsto \int f(t)\>dt\ .$$ Thereby each "differentiation rule" generates an "integration rule" as follows: $$F'=f\quad\Longrightarrow\quad \int f(t)\>dt=\langle F(t)\rangle\ .$$ And on and on.
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# Talk:Horsepower
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## Original horse power
The original horsepower unit was selected to show potential buyers of the new steam engines how much they would save by switching from horses to steam. The horsepower unit represented the rate of work a horse could do on a continous basis (all day long). Thats why its only a small fraction of the maximum capability of a horse.
## Why is this entry flagged?
"This article or section may contain original research or unverified claims." How so?
## British horsepower
The abbreviation bhp may also be used for British horsepower[15] (though the usual use is brake horsepower), defined as 33,000 lb·ft/min. The description of the unit "British horsepower" is used when differentiation from "metric horsepower" is required.
I am removing this section because it seems to me to be highly suspect. I don't think there is any such thing, and don't think this (which I have left in the article) can be regarded as a reliable source, at least not for this bit of nomenclature. BHP is usually brake and sometimes boiler but not British, surely? If anyone can provide better references I will be happy for the section to be reinstated. Globbet (talk) 09:04, 26 August 2011 (UTC)
AIUI, this was a genuine name for the unit, but it was horribly obscure and is long fallen from use - however the unit is still around and is in fact the source of the main hp definition. "Horsepower" in the sense of brake horsepower was first defined as sensible round-number units of either 550 foot-pounds/second (in the Imperial world) or 75 mkgf/s (in the metric world, usually the French CV or German PS). All of these are close enough to be considered equal, within the measuring accuracy of anything pre-electric.
For a while afterwards (WW2 era?) they were separate units of slightly different quantity and needed to be identified separately, as they were now within the precision that could be distinguished by measurement.
With the move to SI base units for British measurements ('70s?), the horsepower in Britain was redefined as an arbitrary integer number of SI units and so the hp was rounded up to the well-known 746W figure of today. "British horsepower" then created a new name for an existing legacy unit, with no purpose other than this legacy, and remained based on 550 foot-pounds, with an actual value slightly less than 746W. Its only real function was in updating textbooks - the "550 hp" and "746W hp" were (deliberately) both within the measuring precision of the time, for nearly all physical measurements. Andy Dingley (talk) 09:53, 26 August 2011 (UTC)
It would be nice to have a citation for the change in definition of the horsepower in Britain from 550 ft lb/s to exactly 746 watts. I'm not sure what "time" you're referring to; the first definition of the metre was valid to about 5 significant figures, in 1795. I'm sure metrologists of the time wouldn't whack off a trailing digit just for fun. Watt couldn't have defined his unit as exactly 746 watts because the SI system didn't exist yet; I'm not sure why "33000" was considered a preferred number, since the precision of mesaurements of actual horses evidently could have justfied, oh, say, 30000, an even "rounder" number. All the tables I've seen call a horsepower 745.699 watts except for the rating of electric motors when it is rounded out to 746 watts (good enough for slide-rule calculations), in such publications as NEMA standard MG-1. If you're converting to SI anyway, why make up a new unit based on a non-power-of-10 multiple of an SI unit and give it the name of an old unit? It'd be like re-defining the inch as exactly 2.5 cm because you don't like dealing with 3 significant digits. Even British love of tradition must have limits? --Wtshymanski (talk) 14:20, 26 August 2011 (UTC)
Sorry, Wtshymanski, it is not clear to me exactly what you are saying. Globbet (talk) 19:59, 26 August 2011 (UTC)
I'm saying a British horsepower is a horsepower is 550 ft lb/s is 745 and change watts, and that the only reason the term gets used is to distinguish between 550 ft lb/s and 75 kg m /s definitions. --Wtshymanski (talk) 21:06, 26 August 2011 (UTC)
## Nominal horsepower
There is a sort-of-relevant discussion going on at Template_talk:Convert#Nominal_horsepower? - Globbet (talk) 21:27, 26 August 2011 (UTC)
If templates were smart enough to figure outthis sort of thing for themselves,we wouldn't need editors. Anyone remember User:Bobblewik? --Wtshymanski (talk) 17:52, 31 August 2011 (UTC)
That is pretty much in line with the conclusion reached there. I don't. Globbet (talk) 22:27, 31 August 2011 (UTC)
## "Metric Horsepower"
There is no official measure of horsepower called "Metric Horsepower", only Americans call it this way. it is properly called "PS", and is based, obviously, on the German "Pferdestärke". This equals to ca. 0,735 kW, therefore 1PS = 1.36 kW (ca.). Also, the American Horsepower measurement (or is it the British one, I believe its called "SAE hp", and/or "bhp", which is 1.34 kW, is missing a detailed description here, how come?.--Daondo (talk) 22:36, 17 October 2011 (UTC)
Well, we've got a whole section titled "Metric Horsepower" which gives all these different metric horsepower units, ( those Europeans have different words for everything), and the rest of the article is talking mostly about 550 ft lbs/sec which is a British or Mechanical or SAE horsepower. You do know that SAE stands for Society of Automotive Engineers which isn't particularly British? --Wtshymanski (talk) 01:23, 18 October 2011 (UTC)
To make things more "interesting" in {{Infobox German Railway Vehicle}} and in {{DRG locomotives}} I found PSi which probably means indicated PS. One more conversion problem. Peter Horn User talk 20:43, 17 October 2012 (UTC)
## Mechanical horsepower conversion procedure - numerical precision
From the standpoint of error propagation I found silly that the original 2 significant digits (1 hp ≡ 33,000 ft·lbf/min) expand through the conversion into SI units to final 17 significant digits (= 745.69987158227022 W). Ignoring the precision of the conversion factors themselves and assuming 1% relative error, the final conversion should be something like 1 hp≡ 33,000 ± 330 ft·lbf/min ≡ 745.70 ± 7.46 W. Even taking the original number defined by Watt as a number of 5 significant digits, this brings an error in the order of 0.00746 W (relative error of 10^-5) which renders the tail of digits into a meaning less nonsense. — Preceding unsigned comment added by 82.208.33.113 (talk) 22:55, 15 April 2012 (UTC)
(months later) Well, definitions are (by defninition) exact, so Watt's "33,000" has as many sig figs as you need. As long as we're talking about "definitions" of units and interconversions between them the precision is as high as ever needed. But it is absurd to give the final result to more significant figures than could ever be possibly resolved in an actual measurement. Anything more than 6 figures is going to make the reader's eyes glaze over and doesn't really improve the effect of the presentation. --Wtshymanski (talk) 13:36, 21 August 2012 (UTC)
The value used for the pound-mass has an erroneous extra digit. It should be .45359237 kg exactly, not .453592376 kg. And a full 5 digits of the final result, 745.699881448 W, are wrong. The exact value denigrated above, 745.69987158227022 W, is correct. Exact calculation in the definition of units avoids this sort of error. — Preceding unsigned comment added by 208.53.195.38 (talk) 14:15, 17 July 2013 (UTC)
At the top of the Measurement section it says bhp is "power delivered directly to and measured at the engine's crankshaft", and if you subtract "frictional losses in the transmission" you get shp. Later it says bhp is power before subtracting the auxiliaries such as alternator and hydraulic pumps.
So shouldn't that first section say something like this? "Brake / net / crankshaft horsepower (power delivered directly to and measured at the engine's crankshaft) minus frictional losses in the transmission (bearings, gears, oil drag, windage, etc.), minus auxiliaries such as alternators and pumps, equals Shaft horsepower." Kendall-K1 (talk) 16:48, 21 February 2013 (UTC)
## Globalize?
Why is there a globalize template on the "Current definitions" section? Isn't horsepower a US unit? Kendall-K1 (talk) 02:15, 22 February 2013 (UTC)
## Historical symbol
A ligature-style symbol/glyph/character $\mathrm{H\!\!P}$ is used to denote 'horsepower units' in the book Aiba, S., A. E.Humphrey, and N. F.Millis, Biochemical Engineering. 1965, New York, U.S.A.: Academic Press. 333 pp.. (e.g. pages 167ff.). —DIV (138.194.12.224 (talk) 02:28, 22 March 2013 (UTC))
There is a unicode character ㏋ (U+33CB "square hp") but I can't find anything that says what it means or what it's used for. It's in the CJK compatibility block with other units of measure like mV, Hz, and gal, so it's probably horsepower. But it's not a ligature, it's two separate letters crammed into one glyph. Kendall-K1 (talk) 18:41, 4 April 2013 (UTC)
I welcome the article. My only comment is that I believe units of measurement named after those who helped define them have their first letter capitalized. (e.g. Watts, Amps, Joules, Faradays, Volts, etc.). If this is correct, the frequent reference to 'watts' should be corrected to 'Watts'. — Preceding unsigned comment added by 65.113.43.98 (talk) 22:39, 6 August 2013 (UTC)
## use of funny f in the middle of units
There is a subscript f in the middle of many units that is undefined within this article, and its meaning is non-obvious. Feet? Force? Function? Can we define it the first time it is used? metaJohnG (talk) 02:22, 11 September 2014 (UTC)
Please, could you give us an example. I looked in the article and cannot find what you mean.-- 06:46, 11 September 2014 (UTC)
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# 31.3: Nutritional Adaptations of Plants
Skills to Develop
• Understand the nutritional adaptations of plants
• Describe mycorrhizae
• Explain nitrogen fixation
Plants obtain food in two different ways. Autotrophic plants can make their own food from inorganic raw materials, such as carbon dioxide and water, through photosynthesis in the presence of sunlight. Green plants are included in this group. Some plants, however, are heterotrophic: they are totally parasitic and lacking in chlorophyll. These plants, referred to as holo-parasitic plants, are unable to synthesize organic carbon and draw all of their nutrients from the host plant.
Plants may also enlist the help of microbial partners in nutrient acquisition. Particular species of bacteria and fungi have evolved along with certain plants to create a mutualistic symbiotic relationship with roots. This improves the nutrition of both the plant and the microbe. The formation of nodules in legume plants and mycorrhization can be considered among the nutritional adaptations of plants. However, these are not the only type of adaptations that we may find; many plants have other adaptations that allow them to thrive under specific conditions.
### Nitrogen Fixation: Root and Bacteria Interactions
Nitrogen is an important macronutrient because it is part of nucleic acids and proteins. Atmospheric nitrogen, which is the diatomic molecule $$\ce{N2}$$, or dinitrogen, is the largest pool of nitrogen in terrestrial ecosystems. However, plants cannot take advantage of this nitrogen because they do not have the necessary enzymes to convert it into biologically useful forms. However, nitrogen can be “fixed,” which means that it can be converted to ammonia ($$\ce{NH3}$$) through biological, physical, or chemical processes. As you have learned, biological nitrogen fixation (BNF) is the conversion of atmospheric nitrogen ($$\ce{N2}$$) into ammonia ($$\ce{NH3}$$), exclusively carried out by prokaryotes such as soil bacteria or cyanobacteria. Biological processes contribute 65 percent of the nitrogen used in agriculture. The following equation represents the process:
$\ce { N2 + 16 ATP + 8 e^{-} + 8 H^{+} \rightarrow 2 NH3 + 16 ADP + 16 P_i + H_2}$
The most important source of BNF is the symbiotic interaction between soil bacteria and legume plants, including many crops important to humans (Figure $$\PageIndex{1}$$). The NH3 resulting from fixation can be transported into plant tissue and incorporated into amino acids, which are then made into plant proteins. Some legume seeds, such as soybeans and peanuts, contain high levels of protein, and serve among the most important agricultural sources of protein in the world.
Figure $$\PageIndex{1}$$: Some common edible legumes—like (a) peanuts, (b) beans, and (c) chickpeas—are able to interact symbiotically with soil bacteria that fix nitrogen. (credit a: modification of work by Jules Clancy; credit b: modification of work by USDA)
Exercise $$\PageIndex{1}$$
Farmers often rotate corn (a cereal crop) and soy beans (a legume) planting a field with each crop in alternate seasons. What advantage might this crop rotation confer?
Soybeans are able to fix nitrogen in their roots, which are not harvested at the end of the growing season. The belowground nitrogen can be used in the next season by the corn.
Farmers often rotate corn (a cereal crop) and soy beans (a legume), planting a field with each crop in alternate seasons. What advantage might this crop rotation confer?
Soil bacteria, collectively called rhizobia, symbiotically interact with legume roots to form specialized structures called nodules, in which nitrogen fixation takes place. This process entails the reduction of atmospheric nitrogen to ammonia, by means of the enzyme nitrogenase. Therefore, using rhizobia is a natural and environmentally friendly way to fertilize plants, as opposed to chemical fertilization that uses a nonrenewable resource, such as natural gas. Through symbiotic nitrogen fixation, the plant benefits from using an endless source of nitrogen from the atmosphere. The process simultaneously contributes to soil fertility because the plant root system leaves behind some of the biologically available nitrogen. As in any symbiosis, both organisms benefit from the interaction: the plant obtains ammonia, and bacteria obtain carbon compounds generated through photosynthesis, as well as a protected niche in which to grow (Figure $$\PageIndex{2}$$).
Figure $$\PageIndex{2}$$: Soybean roots contain (a) nitrogen-fixing nodules. Cells within the nodules are infected with Bradyrhyzobium japonicum, a rhizobia or “root-loving” bacterium. The bacteria are encased in (b) vesicles inside the cell, as can be seen in this transmission electron micrograph. (credit a: modification of work by USDA; credit b: modification of work by Louisa Howard, Dartmouth Electron Microscope Facility; scale-bar data from Matt Russell)
### Mycorrhizae: The Symbiotic Relationship between Fungi and Roots
A nutrient depletion zone can develop when there is rapid soil solution uptake, low nutrient concentration, low diffusion rate, or low soil moisture. These conditions are very common; therefore, most plants rely on fungi to facilitate the uptake of minerals from the soil. Fungi form symbiotic associations called mycorrhizae with plant roots, in which the fungi actually are integrated into the physical structure of the root. The fungi colonize the living root tissue during active plant growth.
Through mycorrhization, the plant obtains mainly phosphate and other minerals, such as zinc and copper, from the soil. The fungus obtains nutrients, such as sugars, from the plant root (Figure $$\PageIndex{3}$$). Mycorrhizae help increase the surface area of the plant root system because hyphae, which are narrow, can spread beyond the nutrient depletion zone. Hyphae can grow into small soil pores that allow access to phosphorus that would otherwise be unavailable to the plant. The beneficial effect on the plant is best observed in poor soils. The benefit to fungi is that they can obtain up to 20 percent of the total carbon accessed by plants. Mycorrhizae functions as a physical barrier to pathogens. It also provides an induction of generalized host defense mechanisms, and sometimes involves production of antibiotic compounds by the fungi.
Figure $$\PageIndex{3}$$: Root tips proliferate in the presence of mycorrhizal infection, which appears as off-white fuzz in this image. (credit: modification of work by Nilsson et al., BMC Bioinformatics 2005)
There are two types of mycorrhizae: ectomycorrhizae and endomycorrhizae. Ectomycorrhizae form an extensive dense sheath around the roots, called a mantle. Hyphae from the fungi extend from the mantle into the soil, which increases the surface area for water and mineral absorption. This type of mycorrhizae is found in forest trees, especially conifers, birches, and oaks. Endomycorrhizae, also called arbuscular mycorrhizae, do not form a dense sheath over the root. Instead, the fungal mycelium is embedded within the root tissue. Endomycorrhizae are found in the roots of more than 80 percent of terrestrial plants.
### Nutrients from Other Sources
Some plants cannot produce their own food and must obtain their nutrition from outside sources. This may occur with plants that are parasitic or saprophytic. Some plants are mutualistic symbionts, epiphytes, or insectivorous.
#### Plant Parasites
A parasitic plant depends on its host for survival. Some parasitic plants have no leaves. An example of this is the dodder (Figure $$\PageIndex{4}$$), which has a weak, cylindrical stem that coils around the host and forms suckers. From these suckers, cells invade the host stem and grow to connect with the vascular bundles of the host. The parasitic plant obtains water and nutrients through these connections. The plant is a total parasite (a holoparasite) because it is completely dependent on its host. Other parasitic plants (hemiparasites) are fully photosynthetic and only use the host for water and minerals. There are about 4,100 species of parasitic plants.
Figure $$\PageIndex{4}$$: The dodder is a holoparasite that penetrates the host’s vascular tissue and diverts nutrients for its own growth. Note that the vines of the dodder, which has white flowers, are beige. The dodder has no chlorophyll and cannot produce its own food. (credit: "Lalithamba"/Flickr)
#### Saprophytes
A saprophyte is a plant that does not have chlorophyll and gets its food from dead matter, similar to bacteria and fungi (note that fungi are often called saprophytes, which is incorrect, because fungi are not plants). Plants like these use enzymes to convert organic food materials into simpler forms from which they can absorb nutrients (Figure $$\PageIndex{5}$$). Most saprophytes do not directly digest dead matter: instead, they parasitize fungi that digest dead matter, or are mycorrhizal, ultimately obtaining photosynthate from a fungus that derived photosynthate from its host. Saprophytic plants are uncommon; only a few species are described.
Figure $$\PageIndex{5}$$: Saprophytes, like this Dutchmen’s pipe (Monotropa hypopitys), obtain their food from dead matter and do not have chlorophyll. (credit: modification of work by Iwona Erskine-Kellie)
#### Symbionts
A symbiont is a plant in a symbiotic relationship, with special adaptations such as mycorrhizae or nodule formation. Fungi also form symbiotic associations with cyanobacteria and green algae (called lichens). Lichens can sometimes be seen as colorful growths on the surface of rocks and trees (Figure $$\PageIndex{6}$$). The algal partner (phycobiont) makes food autotrophically, some of which it shares with the fungus; the fungal partner (mycobiont) absorbs water and minerals from the environment, which are made available to the green alga. If one partner was separated from the other, they would both die.
Figure $$\PageIndex{6}$$: Lichens, which often have symbiotic relationships with other plants, can sometimes be found growing on trees. (credit: "benketaro"/Flickr)
#### Epiphytes
An epiphyte is a plant that grows on other plants, but is not dependent upon the other plant for nutrition (Figure $$\PageIndex{7}$$). Epiphytes have two types of roots: clinging aerial roots, which absorb nutrients from humus that accumulates in the crevices of trees; and aerial roots, which absorb moisture from the atmosphere.
Figure $$\PageIndex{7}$$: These epiphyte plants grow in the main greenhouse of the Jardin des Plantes in Paris.
#### Insectivorous Plants
An insectivorous plant has specialized leaves to attract and digest insects. The Venus flytrap is popularly known for its insectivorous mode of nutrition, and has leaves that work as traps (Figure $$\PageIndex{8}$$). The minerals it obtains from prey compensate for those lacking in the boggy (low pH) soil of its native North Carolina coastal plains. There are three sensitive hairs in the center of each half of each leaf. The edges of each leaf are covered with long spines. Nectar secreted by the plant attracts flies to the leaf. When a fly touches the sensory hairs, the leaf immediately closes. Next, fluids and enzymes break down the prey and minerals are absorbed by the leaf. Since this plant is popular in the horticultural trade, it is threatened in its original habitat.
Figure $$\PageIndex{8}$$: A Venus flytrap has specialized leaves to trap insects. (credit: "Selena N. B. H."/Flickr)
### Summary
Atmospheric nitrogen is the largest pool of available nitrogen in terrestrial ecosystems. However, plants cannot use this nitrogen because they do not have the necessary enzymes. Biological nitrogen fixation (BNF) is the conversion of atmospheric nitrogen to ammonia. The most important source of BNF is the symbiotic interaction between soil bacteria and legumes. The bacteria form nodules on the legume’s roots in which nitrogen fixation takes place. Fungi form symbiotic associations (mycorrhizae) with plants, becoming integrated into the physical structure of the root. Through mycorrhization, the plant obtains minerals from the soil and the fungus obtains photosynthate from the plant root. Ectomycorrhizae form an extensive dense sheath around the root, while endomycorrhizae are embedded within the root tissue. Some plants—parasites, saprophytes, symbionts, epiphytes, and insectivores—have evolved adaptations to obtain their organic or mineral nutrition from various sources.
### Glossary
epiphyte
plant that grows on other plants but is not dependent upon other plants for nutrition
insectivorous plant
plant that has specialized leaves to attract and digest insects
nitrogenase
enzyme that is responsible for the reduction of atmospheric nitrogen to ammonia
nodules
specialized structures that contain Rhizobia bacteria where nitrogen fixation takes place
parasitic plant
plant that is dependent on its host for survival
rhizobia
soil bacteria that symbiotically interact with legume roots to form nodules and fix nitrogen
saprophyte
plant that does not have chlorophyll and gets its food from dead matter
symbiont
plant in a symbiotic relationship with bacteria or fungi
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# Dependent Coin Toss 100% rigged
I toss two coins which can both land Heads or Tails with probability 0.5 Define random variables: X = 1 if first coin lands Heads and 0 otherwise Y = 1 if second coin lands Heads and 0 otherwise
Now consider three cases: a) The two tosses are independent b) The coins are rigged so the second coin lands the same way as the first c) The coins are rigged so that the second coin lands the opposite way to the first.
For all three cases: • Calculate E(X), E(Y), σ(X), σ(Y), COV( X, Y) and ρ(X, Y) • Write down the probability distribution of X + Y • Using the probability distribution of X + Y, calculate E(X + Y), σ( X + Y)
I can calculate the E(x) etc. for the two tosses when they are independent, i just cant get my head around the E(x) and E(y) when the coins are rigged. As the formula i believe is the sum of Xi * Pi . I am unsure to what the probability of the second rigged toss is as it relies on the outcome of the first toss
would like if possible an explanation of how to get the E(x) and E(y) for the case b. as it is the base for calculating all of the other figures.
• Welcome at SE. What results have you got? And if you have no results, then what did you try and where did you get stuck? Add that to your question. Mar 2 '15 at 9:33
• OK, done considering, calculating, writing, using and calculating again. Now what? Mar 2 '15 at 9:33
The fact that the coins are rigged in cases b) and c) does not change the distribution of $X$ and does not change the distribution of $Y$. However it does change the distribution of $(X,Y)$.
In case b) $Y=X$ so that $X+Y=2X$ and e.g. $\rm{Cov}(X,Y)=\rm{Cov}(X,X)=\rm{Var}(X)$.
In case c) $Y=1-X$ so that $X+Y=1$ et cetera.
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# General relationship between braid groups and mapping class groups
I just finished correcting my answer on visualizing braid groups as fundamental groups of configuration spaces, and in the process became interested in the other pictorial definition of the braid group $B_n$, namely as the mapping class group of the $n$-punctured closed unit disk.
Some definitions. If $X$ is a topological space, let $F_n(X)$ be the subspace of $X^n$ consisting of tuples with distinct coordinates, on which the symmetric group $S_n$ acts by permuting coordinates, and then define the quotient $SF_n(X):=F_n(X)/S_n$. The braid group of $X$ is defined to be the fundamental group $B_n(X):=\pi_1(SF_n(X))$. Note $B_n=B_n(\Bbb C)$ is the usual braid group.
The automorphism group ${\rm Aut}(X)$ is the group of homeomorphisms $X\to X$ which fix its boundary $\partial X$ pointwise. Denote by ${\rm Aut}_0(X)$ those automorphisms which are isotopic to the identity map. The mapping class group is ${\rm Mod}(X):={\rm Aut}(X)/{\rm Aut}_0(X)$.
Denote by $\overline{{\Bbb D}^2}$ the closed unit disk. Then $B_n(\Bbb C)\cong {\rm Mod}(\overline{\Bbb D^2}-\{x_1,\cdots,x_n\})$ for any choice of $n$ distinct points. It seems obvious to me that $B_n(\Bbb C)=B_n(\overline{\Bbb D^2})$, which inspires the question:
• For what kinds of surfaces/manifolds does $B_n({\cal S})={\rm Mod}({\cal S}-\{x_1,\cdots,x_n\})$?
And this leads to the larger question:
• What is the general relationship between braid groups and mapping class groups?
Sorry if these facts are well-known somewhere. (I am pretty new to homotopy theory and algebraic topology in general too, so I might be a bit slow. It's possible I am biting off more than I am supposed to be chewing.) Here is my likely invalid argument:
Claim. ${\rm Mod}({\cal S}-\{x_1,\cdots,x_n\})=B_n({\cal S})$ for "nice" spaces $\cal S$.
"Proof". Assume ${\rm Aut}(S)$ acts transitively on $n$-subsets of ${\cal S}$. (Intuitively it feels like this should follow automatically from $\cal S$ being homogeneous, i.e. ${\rm Aut}({\cal S})$ only acting transitively on $\cal S$ itself, by cordoning off a nbhd around any $n$ points, but I haven't tried proving it.) We should be able to identify ${\rm Aut}({\cal S}-\{x_1,\cdots,x_n\})$ with ${\rm Stab}_{{\rm Aut}({\cal S})}(\{x_1,\cdots,x_n\})$ the setwise (not pointwise) stabilizer. By the orbit-stabilizer theorem, we have
$${\rm Aut}(S)/{\rm Stab}_{{\rm Aut}(\cal S)}(\{x_1,\cdots,x_n\})\cong SF_n({\cal S})$$
So to get $B_n({\cal S})$ we apply $\pi_1$ to the left side. Here I invoke a lemma:
Lemma. If $G$ is connected, simply connected, $H$ a subgroup, and $H^\circ$ the connected component of the identity in $H$, then $\pi_1(G/H)\cong H/H^\circ$ via $[\gamma]\mapsto\gamma(1)H^\circ$.
If we apply with $G={\rm Aut}({\cal S})$ and $H={\rm Stab}_{{\rm Aut}({\cal S})}(\{x_1,\cdots,x_n\})$ then $H/H^\circ$ should be the identified group ${\rm Aut}({\cal S}-\{x_1,\cdots,x_n\})$ modulo isotopy, i.e. ${\rm Mod}({\cal S}-\{x_1,\cdots,x_n\})$, no?
My argument must have issues, because the claim doesn't work for ${\cal S}=\Bbb C$: as I understand it, we instead have that $\Bbb C\cong\Bbb S^2-\{\rm pt\}$ and so ${\rm Mod}({\Bbb C}-\{x_1,\cdots,x_n\})$ is a subgroup of $B_{n+1}(\Bbb S^2)$ with elements having a marked string always trivial, and this seems like a different group.
Let $\mathcal S$ be a compact surface, possibly with boundary. Let $\text{Homeo}^+(\mathcal S)$ refer to the group of orientation-preserving homeomorphisms that fix the boundary pointwise with the compact-open topology. Throwing an $n$ in there means we add $n$ marked points in the interior, which I prefer over deleting points. (I'd be worried about the validity of any particular version of this result if we delete points instead of marking them.) I don't know the correct topology on $\text{Homeo}^+$ if you use noncompact surfaces.
Then the correct general statement of your dream is that $$\text{Homeo}^+(\mathcal S,n) \to \text{Homeo}^+(\mathcal S) \to SF_n(\mathcal S)$$ is a fiber bundle. The tools you need to analyze this are the homotopy long exact sequence, the Earle-Eels theorem (you can find a statement and proof in Appendix B here, and that $\text{Homeo}^+(D^2)$ is contractible.
Some special cases:
When $\mathcal S = D^2$, you immediately obtain that $B_n = \text{MCG}(\mathcal S,n)$.
Ignore the issues with topologizing $\text{Homeo}^+(\mathbb R^2)$. The trick with your proof is that this is not actually simply connected! $\text{Homeo}^+(\mathbb R^2)$ should be the same as $\text{Homeo}^+(S^2,1)$ (send every homeomorphism to its compactification). This is homotopy equivalent to $SO(2) = S^1$.
When $\mathcal S = \Sigma_{g,k}$, a genus $g$ surface with $k$ boundary components, and either $g \geq 2$ or $k \geq 1$, applying Earle-Eels you obtain the exact sequence $$1 \to B_n(\mathcal S) \to \text{MCG}(\mathcal S,n) \to \text{MCG}(\mathcal S) \to 1.$$
When $\mathcal S = S^2$, using Smale's theorem that $\text{Homeo}^+(S^2) \simeq SO(3)$, and using Earle-Eels you can prove that the components of $\text{Homeo}^+(S^2,n)$ are contractible, so you get a short exact sequence $$1 \to \mathbb Z/2 \to B_n(S^2) \to \text{MCG}(S^2,n) \to 1.$$
When $\mathcal S = T^2$, $\text{Homeo}^+(T^2) \simeq SL_2(\mathbb Z) \times T^2$. If you can get some control on $\text{Homeo}^+(T^2,n)$ you should get something interesting here, but a couple mindless attempts didn't work. (Idea: work with Diff instead so that at each marked point you can 'pull apart' the marked points and obtain diffeomorphisms of $T^2$ minus some open discs without control on the way the diffeomorphisms behave on the boundary. This space might be assailable with EE.)
The fiber sequence above generalizes perfectly well to higher-dimensional manifolds. You might enjoy playing with it for 3-manifolds, when $\text{Homeo}^+(M)$ is known in many examples. $S^2 \times S^1$ might be fun.
You would probably enjoy Farb's primer on mapping class groups. The fiber sequence above just comes from modifying his proof of Birman's exact sequence to having $n$ points instead of one.
• Whenever you've got $H \to G \to G/H$, my favorite way to check that it's a fiber bundle is that there's a local cross-section $G/H \to G$ at every point (this automatically gives you a local trivialization!) I think this trick is due to Palais. See Theorem A of "Local Triviality of the Restriction Map for Embeddings". Use this to prove the above thing is a fiber bundle. – user98602 Jul 3 '15 at 0:25
• Ah... is rotating the plane $360$ degrees not a contractible loop in ${\rm Homeo}^+(\Bbb C)$? If so, I see why that wouldn't rear its head for the closed unit disk. Thanks for the answer, and glad to know that my reasoning was (mostly) correct. BTW where do you know of Homeo groups of spheres, planes and tori from? (And does $\simeq$ mean homotopy equivalence?) I indeed have Farb checked out from the library, but just reading the first chapter I realized I should study the classification of surfaces and more hyperbolic geometry first, so that's what I've been doing a little of this summer. – anon Jul 3 '15 at 14:18
• Anyway, why is ${\rm Homeo}^+({\cal S})$ being a ${\rm Homeo}^+({\cal S},n)$-bundle on $SF_n({\cal S})$ the correct general version of my dream? I haven't been able to mentally picture such a bundle, and I've never handled bundles before. [My intuition was that paths in configuration space "come from" paths in ${\rm Homeo}^+(\cal S)$ starting $\rm id$ that braid the $n$ points over time, hence have an endpoint in ${\rm Homeo}^+({\cal S},n)$, and when I noticed ${\rm Homeo}^+({\cal S},n)$ was a stabilizer my algebraic instinct to use orbit-stabilizer kicked in.] – anon Jul 3 '15 at 14:18
• 1) That's correct. Of course, if we didn't fix the boundary of the disc pointwise, we're still in trouble because the same rotation works again. The homeomorphism group of $S^2$ is Smale's theorem; I don't know who the torus one is attributed to; the plane you get by fiddling with Smale's theorem a bit. I didn't learn these from a book, just from random findings and from conversations with people. – user98602 Jul 3 '15 at 14:33
• 2) The point is more that the long exact sequence it gives you is the precise version of the relationship you're trying to find between $B_n(\mathcal S)$ and $\text{MCG}(\mathcal S,n)$, which isn't quite so simple as you'd hoped. It also tells you about the higher homotopy of $SF_n(\mathcal S)$; if you can prove some sort of Earle-Eeles theorem for marked-point homeomorphism groups, you should be able to show its higher homotopy groups vanish (for a wide class of surfaces $\Sigma$). – user98602 Jul 3 '15 at 14:36
|
# Estimate safe ESC curerent for 4108 600kv motors?
Is there a way to safely guess what current ESC's will need to be with this motor (or any motor in general)? The motors are rated 23A in the table, but that's only with a 4s; I might use a 6s in the future. If I were to use, say, a 30A ESC, should it be enough to not burn out mid-flight or something tragic? Would I get some kind of fair warning (obvious smoke, buzzer, etc.) before my investment falls out of the sky? Thanks!
Edit - I'll be using 1555 props as recommended, and the ZD550 frame with a small gopro gimbal
• Might you be able to find an answer to your question here? drones.stackexchange.com/questions/1324/… – ifconfig Jul 22 at 6:00
• These motors are often used with 12-15" propellers... are you sure you want to be spinning them on a 600 kV motor supplied with 6S voltage? That would be an approximate rotation rate of ~13k-15k RPM, which is quite fast for something that large. – ifconfig Jul 22 at 6:32
• @iconfig I'm really not sure, I'd just like to make sure that I can use 6s if I really want/need to, since the motors can handle it already. Or maybe I won't ever need anything more than 4s. I'll just have to wait and see. – Galaxy Jul 22 at 17:47
• Thanks for updating your question, mind putting what you chose in an answer instead to help future readers? Thanks – Xnero Jul 22 at 19:52
• @Daniil if you mean the ESCs, sure, I'l put them up when i choose them. – Galaxy Jul 23 at 18:24
Myriad factors contribute to the current draw of a drone motor including supply voltage, motor kV, the propeller geometry (i.e. diameter, number of blades, pitch), ambient atmospheric conditions, etc.
One can make educated guesses and estimations of the current a motor will draw under known conditions, but the best and most accurate/reliable method is to test the desired setup and measure the current draw under the expected operating conditions. If you can, you should try this by assembling the motor and propeller on a thrust stand and powering the motor while measuring the current draw with an ammeter.
From the Banggood listing linked to in the OP:
The motor of interest here is the 600 kV variant, which appears to be rated for up to 6S voltage, but power draw is only displayed for 4S voltages. If we take the rough assumption that the motor's current draw will increase linearly with the supply voltage (also assuming all other conditions are unchanged), then we can estimate the current draw for the listed propellers running on a 6S batery:
\begin{align} \text{New Current Draw} &= \frac{\text{New Voltage}}{\text{Old Voltage}} \times \text{Old Current Draw} \\ &= \frac{6s}{4s} \times \text{Old Current Draw} \\ &= 1.5 \times \text{Old Current Draw}\end{align}
• APC1238 (12" prop, 3.8" pitch): 25.8 A
• APC1447 (14" prop, 4.7" pitch): 33.75 A
• 1555CF (15" prop, 5.5" pitch): 34.5 A
NOTE: These are incredibly rough estimates. I highly encourage experimentally determining the true current draw of your intended setup.
NOTE: These calculations will be irrelevant if you're not using the same propellers as the ones cited in the experimental performance data.
As @Kralc mentions in his answer, you should add a roughly 20% safety margin on top of the expected current draw when spec'ing out an ESC to account for any unforeseen situations.
• Thanks, It was mostly the rough estimate math I was looking for. Am I correct that you could also use one of the equations in @Kralc 's answer? We already know the power (W) for the given voltage, so you'd just rearrange to find the current? Also I forgot to state that yes, I'm using recommended 1555 props. – Galaxy Jul 22 at 17:40
• @Galaxy Yes, you definitely could! Mine relies on the current measurement, while his relies on the power measurement scaling linearly with a voltage increase. They both have the same effect. – ifconfig Jul 22 at 18:39
• Sorry, I'm having issues with calculations. The datasheet basically says that the motor uses 340W of power when 14.8V is multiplied by 23A (340=14.8 x 23). So, if we increase the voltage to 22.2V, then either the Power increases, or the current decreases. This is explained in @vikrant 's answer (below), but shouldn't current increase? physics.stackexchange.com/questions/160435/… – Galaxy Jul 24 at 1:59
• Yes! Current AND power both increase when the source voltage increases. – ifconfig Jul 24 at 2:32
• Got it, thanks for all the help! – Galaxy Jul 24 at 6:10
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# The motorcyclist's challenge
Three passengers $$1,2,3$$ starts out moving at constant speeds from $$A$$ to $$B$$. At the same time, a motorcyclist $$M$$ starts out from $$B$$ towards $$A$$ to pick the passengers up. As illustrated below:
$$M$$, however, can carry at most one passenger on board. He can drop his passenger off at anytime. When off the motorcycle, passengers just keep moving forward to $$B$$ at their respective speeds. $$M$$ can drive forward or backward, any way he wishes, at any speed not exceeding his top speed. Passengers will get on or off board at $$M$$'s bidding, and let $$M$$ drive them forward or backward with no complaint. We assume it takes no time to get on and off the motorcycle, or for $$M$$ to switch lanes to pick up different passengers.
$$M$$'s goal, or challenge, is to make all passengers arrive at $$B$$ simultaneously.
Now assume speeds for $$1$$ and $$2$$ are $$60$$ and $$90$$ respectively, and a top speed of $$100$$ for $$M$$.
Question: What is $$3$$'s speed range, if $$M$$ is able to accomplish his challenge?
Hint:
Distance between $$A$$ and $$B$$ is irrelevant.
Here's a more involved one if you solved the above:
Instead of 3 passengers we now have 4, where $$1,2,3$$ have speeds $$60,90,30$$ respectively. Top speed for $$M$$ still is $$100$$. What is $$4$$'s speed range, if $$M$$ is able to accomplish his challenge?
and
In general, What relationship must the speeds for 3 passengers $$s_1,s_2,s_3$$ satisfy, given $$s_m=100$$ and $$M$$ is able to fulfill his challenge? What relationships must the speeds for 4 passengers satisfy?
and
If there are many passengers and the speed of $$M$$ is sufficiently large relative to the speeds of passengers. What is the most efficient way (aka requiring the least amount of time) for $$M$$ to accomplish his challenge?
• Chvatal, "On the Bicycle Problem" (1983) Apr 17 '20 at 18:16
• @RobPratt Thanks. Though your reference is interesting for its own sake, I think it is a totally different problem in nature. And unlike your reference problem, my problem seems not amenable to algorithmic treatment, i.e. there exists no general algorithm that can decide whether a given speed vector $(s_1,s_2,...,s_n,s_M)$ is feasible for $M$, or if so, what schedule $M$ should make. Analysis can only be made case by case, conditioning on the number of passengers.
– Eric
Apr 18 '20 at 5:07
Normalize the problem so that the motorcycle starts at 0 with max speed 1 and the passengers start at 1. Answer to part 1: If passenger 3 is slower than passenger 1,
Drive to passenger 2 and pick him up at $$10/19$$. (Time: $$10/19$$)
Keep driving up and drop him off at $$15/19$$. (Time: $$15/19$$)
Drive to $$5/6$$. (Time: $$5/6$$)
Drive back to 0, picking up passenger 3 whenever you meet him.
The first two passengers reach 0 at time $$5/3$$, and the third reaches 0 iff he is at $$5/6$$ on or before time $$5/6$$, which occurs if his speed it at least $$1/5$$.
If passenger 3 is faster than passenger 1, denote the time it takes him to walk from A to B by T.
Drive to passenger 2 and pick him up at $$10/19$$. (Time: $$10/19$$)
Keep driving up and drop him off at $$9T/19$$. (Time: $$9T/19$$)
Drive to $$T/2$$. (Time: $$T/2$$)
Drive back to 0, picking up passenger 1 whenever you meet him.
The last two passengers reach 0 at time T, and the first reaches 0 iff he reaches $$T/2$$ on or before time $$T/2$$. Since passenger 1 has speed $$3/5$$, T must be at least $$5/4$$, forcing passenger 3's speed to be at most $$4/5$$.
In conclusion, passenger 3's range is
$$1/5$$ to $$4/5$$, inclusive.
Generalizing the above approach for problem 3, let the three passengers have speeds a, b, and c, in descending order.
The fastest passenger is picked up at 1/(1+a) and dropped off at (a/b)/(1+a).
We continue driving to 1/(2b). This is only possible if 1/(2b) is at least (a/b)/(1+a), which is true whenever a is at most 1.
We pick up the slowest passenger on the way back, and everyone arrives at time 1/b iff the slowest passenger has made it to 1/(2b) at time 1/(2b): This is true whenever c is at least 2b-1.
In conclusion, the fastest passenger is not faster than the motorcycle, and the middle passenger is not faster than the average of the slowest passenger and the motorcycle.
Regarding @Eric's comment, I don't believe we can do any better, and here's why:
In order to have time to delay the middle passenger, we need to bring the fastest passenger as far back as we can.
This would be to some point 1/(2b) - $$\epsilon$$, since we have to get back to the middle passenger before he reaches 0 at time 1/b.
Since the farthest we move is still 1/(2b) at time 1/(2b), the slowest passenger's bounds are the same.
The fastest passenger will now reach 0 at time (1+1/a)/(2b), and for this to be later than 1/b (so that we can delay the middle passenger) we must still have a at most 1, and the fastest passenger's bounds are the same.
• If $v_m\gt v_1\gt v_2\gt v_3$ and $2v_2=v_3+v_m+\epsilon$, what about $M$ driving both $v_1$ and $v_2$ backward some distance before picking up passenger $3$?
– Eric
Apr 19 '20 at 9:25
• It merely makes the schedule more complicated. Apr 19 '20 at 13:20
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# Lesson 4. How to Replace Raster Cell Values with Values from A Different Raster Data Set in Python
## Learning Objectives
• Replace (masked) values in one xarray DataArray with values in another array.
Sometimes you have many bad pixels in a landsat scene that you wish to replace or fill in with pixels from another scene. In this lesson you will learn how to replace pixels in one scene with those from another using Xarray.
To begin, open both of the pre-fire raster stacks. You got the cloud free data as a part of your homework, last week. The scene with the cloud is in the cold spring fire data that you downloaded last week.
import os
from glob import glob
import matplotlib.pyplot as plt
import seaborn as sns
from numpy import ma
from shapely.geometry import box
import xarray as xr
import rioxarray as rxr
import earthpy as et
import earthpy.spatial as es
import earthpy.plot as ep
import earthpy.mask as em
import pyproj
import geopandas as gpd
pyproj.set_use_global_context(True)
# Prettier plotting with seaborn
sns.set_style('white')
sns.set(font_scale=1.5)
# Download data and set working directory
data = et.data.get_data('cold-springs-fire')
data_2 = et.data.get_data('cs-test-landsat')
os.chdir(os.path.join(et.io.HOME, 'earth-analytics', 'data'))
Downloading from https://ndownloader.figshare.com/files/10960214?private_link=fbba903d00e1848b423e
Extracted output to /root/earth-analytics/data/cs-test-landsat/.
In the previous lesson, you learned how to open, clip and clean a set of Landsat .tif files. One of the challenges with these data were that there is a large cloud covering your study area.
In some analyses, when you have large clouds, it could make sense to
replace cloudy or shadow-covered pixels in a specific scene with pixels from another scene over the same area that are clear. You will learn how to replace pixels in this lesson.
To begin, import the Landsat tif files and mask out the clouds like you did in the previous lesson. You will use the same functions introduced in previous lessons to do this.
def open_clean_band(band_path, crop_layer=None):
"""A function that opens a Landsat band as an (rio)xarray object
Parameters
----------
band_path : list
A list of paths to the tif files that you wish to combine.
crop_layer : geopandas geodataframe
A geodataframe containing the clip extent of interest. NOTE: this will
fail if the clip extent is in a different CRS than the raster data.
Returns
-------
An single xarray object with the Landsat band data.
"""
if crop_layer is not None:
try:
clip_bound = crop_layer.geometry
cleaned_band = rxr.open_rasterio(band_path,
from_disk=True).squeeze()
except Exception as err:
print("Oops, I need a geodataframe object for this to work.")
print(err)
else:
cleaned_band = rxr.open_rasterio(band_path,
return cleaned_band
def process_bands(paths, crop_layer=None, stack=False):
"""
Open, clean and crop a list of raster files using rioxarray.
Parameters
----------
paths : list
A list of paths to raster files that could be stacked (of the same
resolution, crs and spatial extent).
crop_layer : geodataframe
A geodataframe containing the crop geometry that you wish to crop your
data to.
stack : boolean
If True, return a stacked xarray object. If false will return a list
of xarray objects.
Returns
-------
Either a list of xarray objects or a stacked xarray object
"""
all_bands = []
for i, aband in enumerate(paths):
cleaned = open_clean_band(aband, crop_layer)
cleaned["band"] = i+1
all_bands.append(cleaned)
if stack:
print("I'm stacking your data now.")
return xr.concat(all_bands, dim="band")
else:
print("Returning a list of xarray objects.")
return all_bands
Open and process your data.
# Open pre fire Landsat data
landsat_dirpath_pre = os.path.join("cold-springs-fire",
"landsat_collect",
"LC080340322016070701T1-SC20180214145604",
"crop",
"*band[2-4]*.tif")
landsat_paths_pre = sorted(glob(landsat_dirpath_pre))
landsat_pre = process_bands(landsat_paths_pre, stack=True)
landsat_pre
I'm stacking your data now.
<xarray.DataArray (band: 3, y: 177, x: 246)>
array([[[ 443., 456., 446., ..., 213., 251., 293.],
[ 408., 420., 436., ..., 226., 272., 332.],
[ 356., 375., 373., ..., 261., 329., 383.],
...,
[ 407., 427., 428., ..., 306., 273., 216.],
[ 545., 552., 580., ..., 307., 315., 252.],
[ 350., 221., 233., ..., 320., 348., 315.]],
[[ 635., 641., 629., ..., 360., 397., 454.],
[ 601., 617., 620., ..., 380., 418., 509.],
[ 587., 600., 573., ..., 431., 513., 603.],
...,
[ 679., 742., 729., ..., 493., 482., 459.],
[ 816., 827., 824., ..., 461., 502., 485.],
[ 526., 388., 364., ..., 463., 501., 512.]],
[[ 625., 671., 651., ..., 265., 307., 340.],
[ 568., 620., 627., ..., 309., 354., 431.],
[ 513., 510., 515., ..., 362., 464., 565.],
...,
[ 725., 834., 864., ..., 485., 467., 457.],
[1031., 864., 844., ..., 438., 457., 429.],
[ 525., 432., 411., ..., 465., 472., 451.]]], dtype=float32)
Coordinates:
* band (band) int64 1 2 3
* x (x) float64 4.557e+05 4.557e+05 4.557e+05 ... 4.63e+05 4.63e+05
* y (y) float64 4.428e+06 4.428e+06 ... 4.423e+06 4.423e+06
spatial_ref int64 0
Attributes:
STATISTICS_MAXIMUM: 8481
STATISTICS_MEAN: 664.90340361031
STATISTICS_MINIMUM: -767
STATISTICS_STDDEV: 1197.873301452
scale_factor: 1.0
add_offset: 0.0
# Mask cloudy pixels
landsat_pre_cl_path = os.path.join("cold-springs-fire",
"landsat_collect",
"LC080340322016070701T1-SC20180214145604",
"crop",
"LC08_L1TP_034032_20160707_20170221_01_T1_pixel_qa_crop.tif")
landsat_qa = rxr.open_rasterio(landsat_pre_cl_path).squeeze()
high_cloud_confidence = em.pixel_flags["pixel_qa"]["L8"]["High Cloud Confidence"]
cloud = em.pixel_flags["pixel_qa"]["L8"]["Cloud"]
all_masked_values = cloud_shadow + cloud + high_cloud_confidence
# Mask the data using the pixel QA layer
Plot the data to ensure that the cloud covered pixels are masked.
# Plot
rgb=[2, 1, 0],
title="Lots of Missing Values in Your Data \n Landsat CIR Composite Image | 30 meters \n Post Cold Springs Fire - July 8, 2016")
plt.show()
### Read and Stack Cloud Free Data
Above you have a Landsat scene with a large block of cloud covered pixels (everything that is white represents clouds in that plot). To fill these pixels, you will replace the cloud covered pixels with pixels from a Landsat scene that covers the same area within a similar time period.
Next, read in and stack the cloud free landsat data. Below you access the bounds object of a rioxarray object with xarray_name.rio.bounds(). This contains the spatial extent of the cloud free raster. You will use this to ensure that the bounds of both datasets are the same before replacing pixel values.
### Clip Your Cloud Free Landsat Scene to the Same Extent
Below you create a clip extent of your cloud covered seen to use to crop the cloud free scene. This crop step is important to ensure that pixels overlap and to further reduce memory needed to process your data.
# Create bounds object to clip the cloud free data
landsat_pre_cloud_ext_bds = landsat_pre.rio.bounds()
df = {'id': [1],
'geometry': box(*landsat_pre.rio.bounds())}
clip_gdf = gpd.GeoDataFrame(df, crs=landsat_pre.rio.crs)
clip_gdf.plot()
plt.show()
# Read in the "cloud free" landsat data that you downloaded as a part of your homework
cloud_free_path = os.path.join("cs-test-landsat",
"*band[2-4]*.tif")
landsat_paths_pre_cloud_free = sorted(glob(cloud_free_path))
landsat_pre_cloud_free = process_bands(landsat_paths_pre_cloud_free,
stack=True,
crop_layer=clip_gdf)
landsat_pre_cloud_free
I'm stacking your data now.
<xarray.DataArray (band: 3, y: 177, x: 246)>
array([[[590., 629., 636., ..., 218., 234., 283.],
[546., 580., 598., ..., 248., 270., 314.],
[484., 503., 506., ..., 284., 325., 348.],
...,
[434., 431., 438., ..., 290., 291., 303.],
[441., 490., 478., ..., 292., 312., 313.],
[340., 278., 297., ..., 299., 334., 337.]],
[[781., 808., 828., ..., 461., 485., 535.],
[748., 795., 807., ..., 491., 519., 574.],
[727., 754., 743., ..., 535., 590., 627.],
...,
[722., 724., 722., ..., 550., 554., 569.],
[706., 777., 756., ..., 546., 577., 591.],
[578., 484., 500., ..., 548., 590., 607.]],
[[770., 839., 845., ..., 331., 363., 412.],
[730., 793., 812., ..., 379., 421., 479.],
[657., 692., 691., ..., 441., 522., 573.],
...,
[697., 789., 797., ..., 497., 486., 505.],
[837., 788., 802., ..., 476., 505., 508.],
[542., 477., 465., ..., 510., 536., 536.]]])
Coordinates:
* x (x) float64 4.557e+05 4.557e+05 4.557e+05 ... 4.63e+05 4.63e+05
* y (y) float64 4.428e+06 4.428e+06 ... 4.423e+06 4.423e+06
* band (band) int64 1 2 3
spatial_ref int64 0
Attributes:
scale_factor: 1.0
long_name: band 2 surface reflectance
### Spatial Extent Check
In order to replace pixel values, you will need to ensure that the spatial extent or boundaries of each dataset are the same. Below you check the
bounds of each object.
# Are the bounds the same for both datasets?
landsat_no_clouds_bds = landsat_pre_cloud_free.rio.bounds()
landsat_pre_cloud_ext_bds = landsat_pre.rio.bounds()
print("The cloud free data bounds are:", landsat_no_clouds_bds)
print("The original cloud covered data bounds are:", landsat_pre_cl_masked.rio.bounds())
print("Are the bounds the same?", landsat_no_clouds_bds == landsat_pre_cloud_ext_bds)
The cloud free data bounds are: (455655.0, 4423155.0, 463035.0, 4428465.0)
The original cloud covered data bounds are: (455655.0, 4423155.0, 463035.0, 4428465.0)
Are the bounds the same? True
The bounds of each dataset are different. Thus you will want to clip each scene to ensure the data line up properly when you fill in the pixels of the cloud covered data.
Below you do two things:
1. You create a box geometry using the extent of each layer. You will use this to crop your data.
2. You then check again to ensure both layers overlap spatially,
# Create polygons from the bounds
cloud_free_scene_bds = box(*landsat_no_clouds_bds)
cloudy_scene_bds = box(*landsat_pre_cloud_ext_bds)
# Do the data overlap spatially?
cloud_free_scene_bds.intersects(cloudy_scene_bds)
True
Below you plot the boundaries. This is an optional step that simply shows you that the extent of the cloud covered data is much smaller compared to the cloud free scene. You will want to clip the cloud free scene to the extent of the cloud covered scene to make them align.
# Plot the boundaries
x, y = cloud_free_scene_bds.exterior.xy
x1, y1 = cloudy_scene_bds.exterior.xy
fig, ax = plt.subplots(1, 1, figsize=(8, 6))
ax.plot(x, y, color='#6699cc', alpha=0.7,
linewidth=3, solid_capstyle='round', zorder=2)
ax.plot(x1, y1, color='purple', alpha=0.7,
linewidth=3, solid_capstyle='round', zorder=2)
ax.set_title('Are the spatial extents different?')
plt.show()
# Is the CRS the same in each raster?
landsat_pre.rio.crs == landsat_pre_cloud_free.rio.crs
True
# Are the shapes the same?
landsat_pre.shape == landsat_pre_cloud_free.shape
True
You’ve now determined that
1. the data have the same bounds
2. the data are in the same Coordinate Reference System and
3. the data do overlap (or intersect).
Because you clipped the data above, you don’t need to do any additional cleanup. However if your data did have different spatial extents or CRS’s you would have to do some more cleanup.
{'type': 'Polygon',
'coordinates': (((463035.0, 4423155.0),
(463035.0, 4428465.0),
(455655.0, 4428465.0),
(455655.0, 4423155.0),
(463035.0, 4423155.0)),)}
(3, 177, 246)
## Replace Cell Values
You are now ready to replace values using xarray’s where() function.
# Get the mask layer from the pre_cloud data
# Assign every cell in the new array that is masked
# to the value in the same cell location as the cloud free data
landsat_pre_clouds_filled
<xarray.DataArray (band: 3, y: 177, x: 246)>
array([[[ 443., 456., 446., ..., 213., 251., 293.],
[ 408., 420., 436., ..., 226., 272., 332.],
[ 356., 375., 373., ..., 261., 329., 383.],
...,
[ 407., 427., 428., ..., 306., 273., 216.],
[ 545., 552., 580., ..., 307., 315., 252.],
[ 350., 221., 233., ..., 320., 348., 315.]],
[[ 635., 641., 629., ..., 360., 397., 454.],
[ 601., 617., 620., ..., 380., 418., 509.],
[ 587., 600., 573., ..., 431., 513., 603.],
...,
[ 679., 742., 729., ..., 493., 482., 459.],
[ 816., 827., 824., ..., 461., 502., 485.],
[ 526., 388., 364., ..., 463., 501., 512.]],
[[ 625., 671., 651., ..., 265., 307., 340.],
[ 568., 620., 627., ..., 309., 354., 431.],
[ 513., 510., 515., ..., 362., 464., 565.],
...,
[ 725., 834., 864., ..., 485., 467., 457.],
[1031., 864., 844., ..., 438., 457., 429.],
[ 525., 432., 411., ..., 465., 472., 451.]]])
Coordinates:
* band (band) int64 1 2 3
* x (x) float64 4.557e+05 4.557e+05 4.557e+05 ... 4.63e+05 4.63e+05
* y (y) float64 4.428e+06 4.428e+06 ... 4.423e+06 4.423e+06
spatial_ref int64 0
Finally, plot the data. Does it look like it reassigned values correctly?
# Plot data
ep.plot_rgb(landsat_pre_clouds_filled.values,
rgb=[2, 1, 0],
title="Masked Landsat CIR Composite Image | 30 meters \n Post Cold Springs Fire \n July 8, 2016")
plt.show()
The above answer is not perfect! You can see that the boundaries of the masked area are still visible. Also there are dark shadowed pixels that were not replaced given the raster pixel_qa layer did not assign those as pixels to be masked. Thus you may need to do a significant amount of further analysis to get this image to where you’d like it to be. But you at least have a start at getting there!
In the case of this class, a large enough portion of the study area is covered by clouds that it makes more sense to find a new scene with cloud cover. However, it is good to understand how to replace pixel values in the case that you may need to do so for smaller areas in the future.
|
# Math Help - Expanding Log problem...
1. ## Expanding Log problem...
Expand as the sum of individual logarithms, each of whose argument is linear: $log(\frac{xy^2}{z^4})$
your help is appreciated! Thanks.
2. Originally Posted by Savior_Self
Expand as the sum of individual logarithms, each of whose argument is linear: $log(\frac{xy^2}{z^4})$
your help is appreciated! Thanks.
Using the following logarithm rules:
$log(ab) = log(a) + log(b)$
$log(\frac{a}{b})= log(a) - log(b)$
$log(a^n) = n \cdot log(a)$
I will start, see if you can finish:
$log(\frac{xy^2}{z^4}) = log(xy^2) - log(z^4) = log(xy^2) - 4log(z)$
...
3. Originally Posted by Defunkt
Using the following logarithm rules:
$log(ab) = log(a) + log(b)$
$log(\frac{a}{b})= log(a) - log(b)$
$log(a^n) = n \cdot log(a)$
I will start, see if you can finish:
$log(\frac{xy^2}{z^4}) = log(xy^2) - log(z^4) = log(xy^2) - 4log(z)$
...
so...
$log(x) + log(y^2) - 4log(z)
=
log(x) + 2log(y) - 4log(z)$
answer being...
$log(x) + 2log(y) - 4log(z)$
Look good?
4. Originally Posted by Savior_Self
so...
$log(x) + log(y^2) - 4log(z)
=
log(x) + 2log(y) - 4log(z)$
answer being...
$log(x) + 2log(y) - 4log(z)$
Look good?
Yes, that is correct.
|
# PISM’s configuration parameters and how to change them¶
PISM’s behavior depends on values of many flags and physical parameters (see Configuration parameters for details). Most of parameters have default values [1] which are read from the configuration file pism_config.nc in the lib sub-directory.
It is possible to run PISM with an alternate configuration file using the -config command-line option:
pismr -i foo.nc -y 1000 -config my_config.nc
The file my_config.nc has to contain all of the flags and parameters present in pism_config.nc.
The list of parameters is too long to include here; please see the Configuration parameters for an automatically-generated table describing them.
Some command-line options set configuration parameters; some PISM executables have special parameter defaults. To examine what parameters were used in a particular run, look at the attributes of the pism_config variable in a PISM output file.
## Managing parameter studies¶
Keeping all PISM output files in a parameter study straight can be a challenge. If the parameters of interest were controlled using command-line options then one can use ncdump -h and look at the history global attribute.
Alternatively, one can change parameter values by using an “overriding” configuration file. The -config_override command-line option provides this alternative. A file used with this option can have a subset of the configuration flags and parameters present in pism_config.nc. Moreover, PISM adds the pism_config variable with values used in a run to the output file, making it easy to see which parameters were used.
Here’s an example. Suppose we want to compare the dynamics of an ice-sheet on Earth to the same ice-sheet on Mars, where the only physical change was to the value of the acceleration due to gravity. Running
pismr -i input.nc -y 1e5 -o earth.nc <other PISM options>
produces the “Earth” result, since PISM’s defaults correspond to this planet. Next, we create mars.cdl containing the following:
netcdf mars {
variables:
byte pism_overrides;
pism_overrides:constants.standard_gravity = 3.728;
pism_overrides:constants.standard_gravity_doc = "m s-2; standard gravity on Mars";
}
Notice that the variable name is pism_overrides and not pism_config above. Now
ncgen -o mars_config.nc mars.cdl
pismr -i input.nc -y 1e5 -config_override mars_config.nc -o mars.nc <other PISM options>
will create mars.nc, the result of the “Mars” run. Then we can use ncdump to see what was different about mars.nc:
ncdump -h earth.nc | grep pism_config: > earth_config.txt
ncdump -h mars.nc | grep pism_config: > mars_config.txt
diff -U 1 earth_config.txt mars_config.txt
--- earth_config.txt 2015-05-08 12:44:43.000000000 -0800
+++ mars_config.txt 2015-05-08 12:44:51.000000000 -0800
@@ -734,3 +734,3 @@
pism_config:ssafd_relative_convergence_units = "1" ;
- pism_config:constants.standard_gravity_doc = "acceleration due to gravity on Earth geoid" ;
+ pism_config:constants.standard_gravity_doc = "m s-2; standard gravity on Mars" ;
pism_config:constants.standard_gravity_type = "scalar" ;
@@ -1057,3 +1057,3 @@
pism_config:ssafd_relative_convergence = 0.0001 ;
- pism_config:constants.standard_gravity = 9.81 ;
+ pism_config:constants.standard_gravity = 3.728 ;
pism_config:start_year = 0. ;
## Saving PISM’s configuration for post-processing¶
In addition to saving pism_config in the output file, PISM automatically adds this variable to all files it writes (snap shots, time series of scalar and spatially-varying diagnostic quantities, and backups). This may be useful for post-processing and analysis of parameter studies as the user has easy access to all configuration options, model choices, etc., without the need to keep run scripts around.
Footnotes
[1] For pismr, grid parameters Mx, My, that must be set at bootstrapping, are exceptions.
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Zbl 0848.92018
Zaghrout, A.; Ammar, A.; El-Sheikh, M.M.A.
Oscillations and global attractivity in delay differential equations of population dynamics.
(English)
[J] Appl. Math. Comput. 77, No.2-3, 195-204 (1996). ISSN 0096-3003
Summary: The oscillatory and asymptotic behavior of all positive solutions of $$x'(t) = \beta_0 \theta^n/(\theta^n + x^n (t - \tau)) - \gamma x(t)$$ about the positive steady state $x^*$ are studied, where $x(t)$ denotes the density of mature cells in blood circulation, $\tau$ is the time delay between the production of immature cells in the bone marrow, and $\beta_0$, $\theta^n$, $\gamma$ are positive constants.
MSC 2000:
*92D25 Population dynamics
34K25 Asymptotic theory of functional-differential equations
34K11 Oscillation theory of functional-differential equations
92C30 Physiology
Keywords: delay differential equations; positive solutions; positive steady state; density of mature cells; blood circulation; immature cells; bone marrow
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# Math Help - quadratic equation for complex numbers
1. ## quadratic equation for complex numbers
what is the quadratic equaiton for a quadratic polynomial with complex constants?
Thanks
2. Originally Posted by CarmineCortez
what is the quadratic equaiton for a quadratic polynomial with complex constants?
Thanks
You mean an equation of the form $Aix^2+Bix+Ci =0$ ?
It's the exact same as quadratic equation with real coefficients.
$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Where:
$a = Ai$
$b = Bi$
$c = Ci$
You can show this by completing the square, which is how we derive the formula for real coefficients.
3. Originally Posted by CarmineCortez
what is the quadratic equaiton for a quadratic polynomial with complex constants?
Thanks
The same as it is when the coefficients are real. It doesn't matter that one or more of a, b and c are not real.
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文章快速检索 高级检索
智能系统学报 2020, Vol. 15 Issue (6): 1058-1067 DOI: 10.11992/tis.202005031 0
### 引用本文
NIU Guochen, ZHANG Yunxiao. Kinematics simulation and control system design of continuous robot[J]. CAAI Transactions on Intelligent Systems, 2020, 15(6): 1058-1067. DOI: 10.11992/tis.202005031.
### 文章历史
Kinematics simulation and control system design of continuous robot
NIU Guochen , ZHANG Yunxiao
Robotics Institute, Civil Aviation University of China, Tianjin 300300, China
Abstract: To enable robotic adaptation to increasingly complex unstructured environments, we designed a wire-driven continuous manipulator that combines a spherical joint and flexible support rod. To study the drive-mapping relation of the continuous robot, we established a kinematics model based on the assumptions of the constant curvature model, and we used MATLAB to simulate the kinematics and drive mapping. The spatial superiority of the continuous robot is demonstrated by the simulation results. We built a prototype platform for the three-joint continuous robot, and designed the handle operation mode for the end joint based on the characteristics of the robot. Experimental verification was performed on the prototype platform. Our experimental results verify both the rationality and correctness of the kinematic model and drive mapping relationship and the feasibility of the manipulation method.
Key words: continuous robot follow terminal control mode kinematic model flexible manipulator space transformation handle control three joints
1 连续型机器人的结构设计和运动学分析 1.1 结构设计
1.2 运动学分析
1.2.1 运动学建模
$\begin{gathered} {{T}} = {{P}}({{L}},{{\theta}} ,{{\varphi }}){{{R}}_{{Z}}}({{\varphi}} ){{{R}}_{{Y}}}({{\theta}} ){{{R}}_{{Z}}}( - {{\varphi}} ) = \left[ {\begin{array}{*{20}{c}} {{R}}&{{P}} \\ 0&1 \end{array}} \right] =\\ \left(\!\! {\begin{array}{*{20}{c}} {c\theta {c^2}\varphi + {s^2}\varphi }&{c\theta s\varphi c\varphi - c\varphi s\varphi }&{s\theta c\varphi }&{\dfrac{L}{\theta }c\varphi (1 - c\theta )} \\ {c\theta c\varphi s\varphi - c\varphi s\varphi }&{{c^2}\varphi + {s^2}\varphi c\theta }&{s\theta s\varphi }&{\dfrac{L}{\theta }s\varphi (1 - c\theta )} \\ { - s\theta c\varphi }&{ - s\theta s\varphi }&{c\theta }&{\dfrac{L}{\theta }s\theta } \\ 0&0&0&1 \!\! \end{array}} \right) \end{gathered}$ (1)
${}_{{{n}} + 1}^1{{T}} = {}_2^1{{T}} \times {}_3^2{{T}} \times {{L}} \times {}_{{{n}} + 1}^{{n}}{{T}}$ (2)
1.2.2 关节空间到绳长空间的转换
Download: 图 5 关节弯曲引起绳长变化示意图 Fig. 5 Schematic of rope length changes caused by joint bending
${L_{ij}} \!=\! n \left[H \!+ \!{l_0} \cos \dfrac{{{\theta _i}}}{{2n}} - d \sin \dfrac{{{\theta _i}}}{{2n}} \cos ({\varphi _i} + (j - 1) \alpha )\right] \!\!\!$ (3)
$\Delta {L_{ij}} = n\left\{ {{l_0} - \left[ {{l_0}\cos \frac{{{\theta _i}}}{{2n}} - d\sin \frac{{{\theta _i}}}{{2n}}\cos ({\varphi _i} + (j - 1)\alpha )} \right]} \right\}$ (4)
$\Delta {L_j} = \mathop \sum \limits_{i = 1}^n \Delta {L_{ij}}$ (5)
1.2.3 ZYZ欧拉角求解
$\begin{gathered} {{R}}({{\alpha}} ,{{\beta }},{{\gamma}} ) = {{{R}}_{{Z}}}({{\alpha}} ){{{R}}_{{Y}}}({{\beta}} ){{{R}}_{{Z}}}({{\gamma}} ) = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right) =\\ \left( {\begin{array}{*{20}{c}} {c\beta c\alpha }&{s\beta s\gamma c\alpha - c\gamma s\alpha }&{s\beta c\gamma c\alpha + s\gamma s\alpha } \\ {c\beta s\alpha }&{s\beta s\gamma s\alpha + c\gamma c\alpha }&{s\beta c\gamma s\alpha - s\gamma c\alpha } \\ { - s\beta }&{s\gamma c\beta }&{c\gamma c\beta } \end{array}} \right) \\ \end{gathered} \!\!\!\!\!\!$ (6)
$\theta = \left\{ {\begin{array}{*{20}{l}} {A\tan 2(\sqrt {a_{31}^2 + a_{32}^2} ,{a_{33}}),}\;\;{s\theta \ne 0}\\ {{0^ \circ }}\\ {{{180}^ \circ }} \end{array}} \right.$ (7)
$\varphi = \left\{ {\begin{array}{*{20}{l}} { - A\tan 2({a_{32}}, - {a_{31}}),}\;\;{s\theta \ne 0}\\ {{0^ \circ }{\rm{,}}}\;\;{\theta = {0^ \circ }}\\ {{0^ \circ }{\rm{,}}}\;\;{\theta = {{180}^ \circ }} \end{array}} \right.$ (8)
1.2.4 绳长空间到驱动信号空间的映射
$\left\{ \begin{gathered} m = \dfrac{s}{c} \cdot \Delta l \\ f = \dfrac{s}{c} \cdot v \\ \end{gathered} \right.$ (9)
2 操作方法研究 2.1 系统设计
Download: 图 7 连续型机器人系统框图 Fig. 7 Block diagram of continuous robot system
2.2 末端跟随控制
3 实验验证 3.1 工作空间仿真
Download: 图 8 连续型机器人工作空间视图 Fig. 8 Views of the continuous robot workspace
3.2 绳长变化仿真
Download: 图 9 关节姿态和绳长的仿真 Fig. 9 Simulation diagrams of joint attitude and line length
3.3 单关节弯曲实验
Download: 图 10 单关节姿态改变实验图 Fig. 10 Experimental diagrams of attitude changes of single joint
3.4 三关节操作实验
1)操控手柄使第一关节弯曲角度为57.32°，旋转角度为37.83°，结果如图12(a)所示；
2)点击前进按钮实现第一关节的姿态跟随，图12(b)为姿态跟随的中间过程图；
3)操作第一关节弯曲角度为34.39°，结果如图12(c)所示；
4)继续点击前进按钮，完成第二关节姿态跟随，此时连续型机器人姿态为弯曲角度 $\varTheta =$ $\{ 0,57.32,34.39\}$ ，旋转角度 $\varPhi = \{ 0,37.83,37.83\}$ ，结果如图12(d)所示；
5)继续点击前进按钮，直到关节姿态参数为 $\varTheta = \{ 22.93,34.39,34.39\}$ $\varPhi = \{ 15.13,37.83,37.83\}$ ，其姿态传感器的结果为 $\varTheta = \{ 22.93,34.39,34.39\}$ $\varPhi =$ $\{ 17.19,37.83,36.69\}$ ，样机平台图如图12(e)所示，仿真图如图12(f)所示。
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# Convection diffusion reaction equation (stiffness, solver)
I am trying to solve the CDR-Equation in 2D:
$$\frac{\partial c(x,y)}{\partial t} + \nabla \cdot ( -d\nabla c(x,y) + \vec{v}(x,y) c(x,y))+ a c(x,y)=0\,,$$ with Boundary Conditions (length of square is $L$): $$c(0,y)=0$$ $$-d\nabla c(L,y) + v(L,y) (c(L,y))=0\,.$$. $$c(x,y=0)=0$$ $$c(x,y=L)=0$$
1.) Why exactly is this equation stiff? Does it depend on the reaction term $ac(x,y)$?
2.) Can I solve the equation with the Crank-Nicholson method? Is the error huge? If yes, what is the best method?
• Please correct the second boundary condition and add some boundary condition on the other two sides of the square domain. Crank-Nicolson with a correct treatment of reaction term and advection term (the last one maybe with an upwind method) shall give an accurate scheme. For too large time steps you can have some unphysical oscillations in numerical solution, but if you would like to have an accurate numerical solution, you should avoid such large time steps. Jan 11 '16 at 20:03
• Stiffness is generally regarded as a property of the ODEs you get by semi-discretizing, not of the PDEs. Do you have a semi-discretization in mind? Jan 12 '16 at 6:36
• I am asking for stiffness, because the CDR equation is often given as an example for a stiff equations. And there are multiple papers discussing different solvers because of the difficulties in handling it. Jan 12 '16 at 13:01
1. Mathematically speaking, stiffness is meaningless for a single differential equation, and is rather attributed to a set of differential equations that have different time-scales (e.g. when trying to solve two coupled equations with time-scales of 1 second and 1 day, respectively). However, a single equation can also be referred to as stiff if certain numerical integration methods are numerically unstable. Source/sink terms can give rise to stiffness, for instance: $$\frac{dy}{dt}=-1000y$$ is a good example where sink term has given rise to stiffness. That being said, $ac(x,y)$ might be causing the stiffness in your PDE.
2. Theoretically, you can solve any time-dependent PDE with Crank-Nicolson method. However, you should use adaptive time-steps when solving a stiff equation, otherwise the computational cost is huge. The naive Crank-Nicolson formulation is based on fixed time-step, however, you can derive your customized version with adaptive time-step. But, I do not recommend as it is cumbersome. Your best alternative, in my opinion, is to use method of lines. You can find useful information about this method here. To give you a rough idea, in the method of lines, you discretize your PDE with respect to spatial variables (and not time). This eventually gives you a set of first order ODEs (ordinary differential equation) with respect to time. Now you can simply use a stiff ODE solver (e.g. Adam's Bashforth method) to integrate the set of ODEs.
• This answer is, in my opinion, somewhat misleading. For a better explanation of stiffness, see e.g. this answer. Jan 13 '16 at 6:18
• I am not quite following you. What part exactly you don't agree? @DavidKetcheson Feb 23 '16 at 2:40
• Your example is not stiff, by the most accepted modern definitions. You probably disagree, and I'm not going to argue with you here in the comments; please just read up on stiffness. Here is a good reference. Feb 23 '16 at 3:41
• Thank you for the reference. I'll definitely look into that for more in-depth discussion. However, for the record have a look here (under "Motivating example"): en.wikipedia.org/wiki/Stiff_equation @DavidKetcheson Feb 24 '16 at 15:26
• Yep, Wikipedia gets it wrong too. Feb 24 '16 at 18:06
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logistic2_fn {drda} R Documentation
## 2-parameter logistic function
### Description
Evaluate at a particular set of parameters the 2-parameter logistic function.
### Usage
logistic2_fn(x, theta)
### Arguments
x numeric vector at which the logistic function is to be evaluated. theta numeric vector with the parameters in the form c(eta, phi).
### Details
The 2-parameter logistic function f(x; theta) is defined here as
1 / (1 + exp(-eta * (x - phi)))
where theta = c(eta, phi), eta is the steepness of the curve or growth rate (also known as the Hill coefficient), and phi is the value of x at which the curve is equal to its mid-point, i.e. 1 / 2.
### Value
Numeric vector of the same length of x with the values of the logistic function.
[Package drda version 1.0.0 Index]
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# number of ways ?
• December 14th 2011, 05:47 AM
livinggourmand
number of ways ?
In how many ways can we place n+x balls in n boxes
with a condition that at least 1 ball is present in every box
when
1)balls r identical
2)balls r numbered from 1 to n+x
and x<n
• December 14th 2011, 07:10 AM
Plato
Re: number of ways ????
Quote:
Originally Posted by livinggourmand
In how many ways can we place n+x balls in n boxes
with a condition that at least 1 ball is present in every box
when
1)balls r identical
2)balls r numbered from 1 to n+x
and x<n
You failed to tell us if the boxes are all different.
So we assume they are.
There are $\binom{K+N-1}{K}$ ways to place K identical items into N different cell.
If we require that no cell is empty then it must be that case that $K\ge N$ and that multi-selection formula becomes
$\binom{K-1}{K-N}$.
For part b), we need to count the number of surjections (onto functions) from a set of N+x to a set of N.
If $K\ge N$ and then the number of surjections from a set of K to a set of N is $\sum\limits_{j = 0}^N {( - 1)^j \binom{N}{j} (N - j)^K }$.
• December 14th 2011, 07:19 AM
livinggourmand
Re: number of ways ????
n if boxes are identical??
• December 14th 2011, 07:39 AM
Plato
Re: number of ways ????
Quote:
Originally Posted by livinggourmand
n if boxes are identical??
If the boxes are identical then it becomes more difficult.
For part a), we have to count the numbers of ways to have K summands for the integer N+x. That is not easy.
Part b) is a bit easier that a). We count the number of unordered partitions of N+n individuals into N groupings.
However, I suspect that whoever wrote this question meant the boxes all different. Because, that is an easier question.
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# News US spies arrested abroad vs. spies arrested in the US
#### Count Iblis
US "spies" arrested abroad vs. "spies" arrested in the US
We had the Roxana Saberi in Iran. She was freed because even the Iranians recognized that she did not have a fair trial. She was allowed to go home. Now, in the US we have the case of the Cuban five:
http://en.wikipedia.org/wiki/Cuban_Five
On 27 May 2005, the United Nations Commission on Human Rights adopted a report by its Working Group on Arbitrary Detention stating its opinions on the facts and circumstances of the case and calling upon the US government to remedy the situation.[17] Among the report's criticisms of the trial and sentences, section 29 states:
"29. The Working Group notes that it arises from the facts and circumstances in which the trial took place and from the nature of the charges and the harsh sentences handed down to the accused that the trial did not take place in the climate of objectivity and impartiality that is required in order to conform to the standards of a fair trial as defined in article 14 of the International Covenant on Civil and Political Rights, to which the United States of America is a party."
Amnesty International has criticized the US treatment of the Cuban Five as human rights violations, as the wives of René Gonzáles and Gerardo Hernández have not been allowed visas to visit their imprisoned husbands.
The latest news is that an appeal on a very procedural ground has been denied:
http://www.reuters.com/article/domesticNews/idUSTRE55E3VD20090615
So, it looks to me that the US is treating these people like Iran or North Korea was/is treating the US "spies", except that in the latter case, you at least have a reasonable chance of getting released. In the US case, the fact that the original trial happened "according to the rules", an appeal (in the sense of a re-examination of the facts of the case) is impossible.
Related General Discussion News on Phys.org
#### kyleb
Re: US "spies" arrested abroad vs. "spies" arrested in the US
Clicking though through the Wiki link I came across this UN document:
Letter dated 29 October 2001 from the Permanent Representativeof Cuba to the United Nations addressed to the Secretary-General
I have the honour to transmit herewith a summary prepared by the National Assembly of People’s Power of the Republic of Cuba concerning the principal terrorist actions against Cuba during the period 1990-2000 (see annex). I should be grateful if you would arrange for this letter and its annex to be circulated as a document of the General Assembly, under the item “Measures to eliminate international terrorism”, and of the Security Council.
Accept, Sir, the assurances of my highest consideration.
(Signed) Bruno Rodríguez Parrilla
Permanent Representative
http://www.un.org/documents/ga/docs/56/a56521.pdf" [Broken]
Considering the serious nature of the changes, I am very curious to know what our government has done since then to address them. Not that I would attempt to absolve spies, but these people having been tried in the heart of the same Cuban expatriate community they were convicted of spying on seems rather absurd.
Last edited by a moderator:
#### LowlyPion
Homework Helper
Re: US "spies" arrested abroad vs. "spies" arrested in the US
She was freed because even the Iranians recognized that she did not have a fair trial.
Given the events unfolding in Iran, I'm thinking that an appeal to fairness had no weight at all. I would look to more pragmatic reasons for granting her appeal and release.
#### mgb_phys
Homework Helper
Re: US "spies" arrested abroad vs. "spies" arrested in the US
A bunch of UK plane-spotters went to Greece to photograph military aircraft, on a military airbase, noting down all the serial numbers, types etc and were arrested as spies.
Usual outcry about foreigners treating Englishmen, demand to send gunboats, invade Greece. Charges eventually dropped because of diplomatic pressure. http://news.bbc.co.uk/2/hi/uk_news/1697862.stm
Greek tourist takes a photograph on the London Underground, and is arrested, because a women complained her kid was in one of the pictures - won't someone think of the children.
http://www.legalbanter.co.uk/uk-legal-legal-issues-uk/54810-photographer-held-london.html
#### russ_watters
Mentor
Re: US "spies" arrested abroad vs. "spies" arrested in the US
So, it looks to me that the US is treating these people like Iran or North Korea was/is treating the US "spies", except that in the latter case, you at least have a reasonable chance of getting released. In the US case, the fact that the original trial happened "according to the rules", an appeal (in the sense of a re-examination of the facts of the case) is impossible.
Those are some pretty steep charges you are levying against the US, there, without any justification that I can see.
In the case of the reporter arrested in Iran, it is expedient to say that she was released because she didn't have a fair trial - you'd never expect the Iranians to acknowledge that the entire incident was a farce. We have no basis for any belief that she was actually a spy: it appears this is just another case of a belligerent government harassing reporters.
The Cuban Five on the other hand, were, Cuban spies. There isn't any reasonable debate to be had about that. Whether they could have beaten prosecution for their crimes if they were tried in a different venue is an interesting question, but it is hard to argue that justice wasn't served in convicting them. They should consider themselves lucky that they weren't executed.
In addition, the differences in fairness themselves, between the two cases, are as wide as an ocean. Whether the trial of the Cuban five was unfair due to the venue (a pretty weak violation, imo), the entire trial of Saberi was a sham, happening too fast and leaving her unable to defend herself. It has all the hallmarks of a manufactured stunt by the Iranian government.
You are drawing up a parallel that couldn't be further from a reality.
Last edited:
Mentor
#### mgb_phys
Homework Helper
Re: US "spies" arrested abroad vs. "spies" arrested in the US
Being a long-time photographer, she should know better than to take pictures of other people's kids without permission.
He was taking pictures of the tube station, I don't think he intended to photograph the kid.
It has become a bit of a mass-hysteria in Britain - parents aren't allowed to take cameras to school soccer games, but the schools must all have CCTV - all to protect the children.
#### math_04
Re: US "spies" arrested abroad vs. "spies" arrested in the US
Yea, I would have to agree with russ watters. The Cuban Five were lucky to to be caught in America where they got to put their case forward in an impartial court. The case against them was pretty solid and included charges of attempting to infiltrate US Southern Command headquarters.
Roxana Saberi's trial was a pure sham; it was filled with hardline clerics who managed to elicit a confession from her before the trial even started through threats and torture.
#### math_04
Re: US "spies" arrested abroad vs. "spies" arrested in the US
It is understandable that a mother would become suspicious of what she perceives as a lone man taking pictures of her kid. With all the cases of children being kidnapped and pedophiles around, what can you expect? It is unfortunate but part of human nature I guess.
#### Count Iblis
Re: US "spies" arrested abroad vs. "spies" arrested in the US
If you read the information about this case, it is clear that there was a terror organization operating from Florida that the US did not act against. Then the Cuban Agents did infiltrate in there which is technically "spying". There was not a shred of evidence that these agents were involved in any harmful activities, yet they were convicted of very serious charges like first degree murder and "infiltrating US Southern Command headquarters", which was simply ridiculous.
Saberi also violated Iranian law about the way she handled certain documents. But at least the Iranians (perhaps under pressure from the US) were flexible and decided that whatever she did was not a big deal.
Unfortunately, the US is unable to make the same determination and will stick to the fact that the convictions were procedurally correct (despite being utterly ridiculous).
#### russ_watters
Mentor
Re: US "spies" arrested abroad vs. "spies" arrested in the US
He was taking pictures of the tube station, I don't think he intended to photograph the kid.
It has become a bit of a mass-hysteria in Britain - parents aren't allowed to take cameras to school soccer games, but the schools must all have CCTV - all to protect the children.
I don't know that it's a mass hysteria issue here, but I was walking in Philadelphia once last year and a Japanese tourist went to take a picture of a day care's multi-stroller (like 6 babies in a train) and the woman pushing it stopped her.
For a soccer game, if your own kids are in it, I can't see how there can be anything wrong with it.
#### russ_watters
Mentor
Re: US "spies" arrested abroad vs. "spies" arrested in the US
If you read the information about this case, it is clear that there was a terror organization operating from Florida that the US did not act against. Then the Cuban Agents did infiltrate in there which is technically "spying". There was not a shred of evidence that these agents were involved in any harmful activities, yet they were convicted of very serious charges like first degree murder and "infiltrating US Southern Command headquarters", which was simply ridiculous.
I don't know how far I'm supposed to go to prove your claims, but the wiki you linked and the US state department (?) website on the Cuban Five read nothing at all like what you describe:
http://www.america.gov/st/pubs-english/2008/June/20070712120209atlahtnevel0.7962915.html
At face value, the charges and evidence look pretty straightforward to me.
Do you have any credible sources that could substantiate your claims about the lack of evidence? I think you'll need to get specific about the evidence that there was being faked, because clearly there was a lot of evidence. This was your basic, classic, spy-novel type espionage.
Heck, it even says they didn't deny what they were, so heavy the evidence against them was: they only tried to deflect the charges by saying they were acting to fight terrorism against Cuba - a claim, that even if true, by the way, is still espionage.
Saberi also violated Iranian law about the way she handled certain documents.
Agreed, but that's not espionage.
But at least the Iranians (perhaps under pressure from the US) were flexible and decided that whatever she did was not a big deal.
Flexible? No, it was more likely all part of some preorchestrated stunt. At best, they picked up on a real crime and then salivated on the opportunity to ramrod an American through a sham trial to poke a finger in our eye. Nothing about that case looks legitimate.
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#### math_04
Re: US "spies" arrested abroad vs. "spies" arrested in the US
Certain documents? Out of the question, there is no way that a foreign journalist could handle sensitive documents having only been in the country for a short time. More likely she was trying to defy her handler by presenting a balanced view of the country.
And I guess you group all the Cuban exile groups as terror organisations do you? How convenient! Not a shred of evidence, a confident statement indeed. The intelligence agents were wiretapped by the FBI, their case was presented in court and in public. I don't know about you but I think all those tapes being held by the FBI, documents and letters etc are all evidence.
#### kyleb
Re: US "spies" arrested abroad vs. "spies" arrested in the US
And I guess you group all the Cuban exile groups as terror organisations do you?
He only said "http://www.salon.com/news/feature/2008/01/14/cuba/" [Broken]".
Not a shred of evidence, a confident statement indeed. The intelligence agents were wiretapped by the FBI, their case was presented in court and in public. I don't know about you but I think all those tapes being held by the FBI, documents and letters etc are all evidence.
He didn't dispute the fact that the were spying, but said "not a shred of evidence that these agents were involved in any harmful activities", which at least I am not in a position to contest. While I know the court found them guilty of first degree murder, considering their trial heart of the same Cuban expatriate community they were convicted of spying on, I can't reasonably argue that justice was served in doing so.
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#### math_04
Re: US "spies" arrested abroad vs. "spies" arrested in the US
I am commenting on the fact that just because a group is exiled, it does not necessarily mean they are a terrorist organization. If the Cuban government think they are, show the evidence and pursue it through the United Nations, Interpol etc without sending intelligence agents to another country.
Well, most of the jury were cautiously selected and again, as long as the evidence is presented in detail and the case is pursued fairly which it was, I see no reason to blame a million different factors that could have favored one side or the other.
#### russ_watters
Mentor
Re: US "spies" arrested abroad vs. "spies" arrested in the US
He didn't dispute the fact that the were spying, but said "not a shred of evidence that these agents were involved in any harmful activities", which at least I am not in a position to contest. While I know the court found them guilty of first degree murder, considering their trial heart of the same Cuban expatriate community they were convicted of spying on, I can't reasonably argue that justice was served in doing so.
I guess, then, he'd have to define what he means by "harmful activities", since it seems to me that everything they were convicted of is a harmful activity, otherwise the activities wouldn't be illegal. But you're right - CI didn't object to the idea that they were spying. So I guess that means he doesn't consider spying to be inherrently harmful? That's not a hair I think needs splitting. Spying is a crime, punishable by death. Period. If there is no objection to the charge of spying, then there is nothing to argue about.
In any case, to the specific charge of murder - according to the link I provided, it wasn't murder, but conspiracy to commit murder. And the evidence there is also pretty straightforward: intercepted communications about the attack.
Regardless, none of this bears any resemblance to the case of Roxana Saberi.
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#### russ_watters
Mentor
Re: US "spies" arrested abroad vs. "spies" arrested in the US
He only said "http://www.salon.com/news/feature/2008/01/14/cuba/" [Broken]".
If we were to assume, for the sake of argument that everything said about Alpha 66 in that link is true and further assume for the sake of argument that the Cuban 5 did nothing else besides act against Alpha 66, it would still be espionage! Their chosen defense was the propaganda technique misdirection and they lost because misdirection is not a valid (though admittedly sometimes it works) legal defense for their crimes.
So this is just an irrelevant diversion, along the same lines as their failed trial defense.
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#### Count Iblis
Re: US "spies" arrested abroad vs. "spies" arrested in the US
I think the double life sentence for one as opposed to the much milder sentences of the others were due to the murder charge and the military base infiltration charge. Both of which are problematic. If you take a hard line view that spying activities can justifiably lead to the death penalty, then surely one also has to take a milder view about the shooting down of the planes by Cuba.
Even if one argues that what Cuba did was wrong because it happened in international waters, the fact that intelligence about the flights to Cuba were given in itself would not make one complicit in any "murder" by any reasonable standard.
In this respect the case is similar to the Saberi case. You take some event that strictly speaking is illegal or it is a somewhat more serious charge that reasonably can lead to, say, ten years prison sentence. But then you blame the people for other events for which they were not responsible at all by any reasonable standard of "responsibility".
That verdict is then only motivated because you have an enemy that you view as a "big Satan". Any action that has helped that "great Satan" in any way, makes you liable for other bad things that this "great Satan" has done.
I can give another example of "Iranian style justice" in Florida in a case having to do with Cuba, see here:
http://www.accessmylibrary.com/coms2/summary_0286-22427001_ITM
In her case, Martinez asked for compensation for damages she suffered from having unwittingly married the alleged spy, who subsequently returned to Cuba. She also claimed that because Roque was acting as an agent of Cuba during their marriage, their sexual relations constituted rape, for which the Cuban government was responsible.
#### kyleb
Re: US "spies" arrested abroad vs. "spies" arrested in the US
No....that's not the same hair at all.
What "hair" were you refering to other than that of harm?
Those guys weren't spies, they were reporters! That made proving espionage much more difficult. In fact, that makes it a lot closer to the Saberi case than the Cuban Five case. With the obvious difference, of course, that the US justice system recognized the weakness of the case and the prosecuters dropped it, whereas in the Saberia case, they ramrodded it through and only after the mishandling was pointed out did they release her.
Not to go off topic, but for the record; those guys weren't reporters, they were lobbyist passing classified information to a foreign government, and the prosecution dropped the case because "Government policy makers indicated they were clearly uncomfortable with senior officials’ testifying in open court over policy deliberations", as mentioned in the article I linked above.
#### math_04
Re: US "spies" arrested abroad vs. "spies" arrested in the US
How do you figure one could rightly expect an unbiased jury at the heart of the same Cuban American community which hosts groups accused of terrorism against Cuba?
How did you come to the conclusion that the jury is unbiased? Have you talked to them? Have you checked their records? Just because there is a significant Cuban American community does not mean a court of law cannot operate properly.
In this respect the case is similar to the Saberi case. You take some event that strictly speaking is illegal or it is a somewhat more serious charge that reasonably can lead to, say, ten years prison sentence. But then you blame the people for other events for which they were not responsible at all by any reasonable standard of "responsibility".
There was no 'case' against Saberi, there was not even a trial. They tortured a confession out of her and then put her in a court which sentenced her pretty quickly without any real evidence. She never had any sensitive documents and she was arrested purely for attempting to defy authorities in Tehran.
I don't understand any of the logic the people against the sentencing put forward. The evidence presented was probably thorough as the wiretaps and documents were in large numbers. The FBI had been monitoring them for a while. How do you know they were harmless? Do you have access to those wiretap files? Were you there in court when the evidence was presented? Do you think the jury would just put them in jail for annoying some Cuban exiles?
#### kyleb
Re: US "spies" arrested abroad vs. "spies" arrested in the US
How did you come to the conclusion that the jury is unbiased? Have you talked to them? Have you checked their records? Just because there is a significant Cuban American community does not mean a court of law cannot operate properly.
I don't claim to know if the jury was biased or not, but I have explained why I suspect it likely was. Am I to take it you have no interest in addressing the facts I presented?
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# Thread: Find the force F = < 2x, e^y + z cos y,sin y > of a particle with line integrals
1. ## Find the force F = < 2x, e^y + z cos y,sin y > of a particle with line integrals
Given F = < 2x, e^y + z cos y,sin y >
Find the work done by the force in moving a particle from P(1, 0, 1) to Q(1, 2, −3) along the curved path given by C : r(t) =< 1 + sin πt, 2 sin(πt/2), 1 − 4t >, 0 ≤ t ≤ 1.
I tried plugging r into F but ended getting an extremely long and complex vector function to take the integral of. The hint said to think, so I assume I'm not supposed to brute force the integral. Is there any property of either F or r that will allow me to simplify my work?
2. Please demonstrate.
I guess that "n"-looking thing is a $\pi$.
How sure are you that you are to evaluate it? Maybe the problem statement says just "set it up"?
3. Originally Posted by tkhunny
Please demonstrate.
I guess that "n"-looking thing is a $\pi$.
How sure are you that you are to evaluate it? Maybe the problem statement says just "set it up"?
The problem says to evaluate it. However, if I find the curl of F, it is zero, so then the line integral would be zero?
4. That would work. I didn't get 0. I may have missed something.
5. Originally Posted by tkhunny
That would work. I didn't get 0. I may have missed something.
How did you set up your integral?
6. Originally Posted by Superyoshiom
How did you set up your integral?
Please reply showing your work. Thank you!
7. Originally Posted by Superyoshiom
How did you set up your integral?
Yeah, that's not how this works. You first.
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## § Git for pure mathematicians
What is git? It's not a version control system. It's an interface to work with torsors. We have a space of files. We can manipulate "differences of files", which are represented by patches/diffs. git's model provides us tools to work with this space of files. We have two spaces, and a rooted DAG that connects to two spaces:
• A space of files.
• A space of diffs which forms a monoid with a monoid action on the space of files.
• A DAG where each node is a diff. The root note in the DAG is the empty diff.
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# What is the Value of X in KNN and Why? [duplicate]
I have a dataset of 25 instances these instances are divided into 2 classes Green Circles and Blue Squares
data distributed as this graph
I want to predict X's class based on "Likelihood Weighted KNN with k =3"
In normal KNN this is easy
the nearest 3 points are 2 Blue Squares and 1 Green Circle
which means X will be Blue Square
there are more Blue Squares neighbours than Green Circles (2 vs 1)
But What is needed is to find the Likelihood Weighted KNN with k =3
This is my try
In this case we have to calculate the weight (Likelihood) for each instance
Each Green Circle likelihood is $$\frac{1}{5}$$ , we have 5 Green Circles
While for Blue Squares it is $$\frac{1}{20}$$ , we have 20 Blue Squares
Therefore the weights around X will be $$\frac{1}{5}$$ Green Circle, and $$\frac{2}{20}$$ Blue Squares.
which means $$\frac{1}{5} > \frac{2}{20}$$
Then X is Green Circle
Well, this is wrong :(
Can someone help me find the
• Your other post of this question is clearer. I'd suggest removing this one in favor of that one. – Ben Reiniger Jul 10 '19 at 12:13
Green circles and blue squares are samples of two different classes. How does it matter that "X" belongs to which one of either of these "Green circle" or "Blue Square" class?
Likelihood should be done based upon the number of classes and the K value, instead of dataset samples.
For me likelihood weight will be:
Likelihood of being blue square ~ 0.66 i.e. 2/3, whereas likelihood of being green circle is 1/3.
Prepared another example: Where I have 2 Green circle and 16 blue square, whereas K=5
If I follow your approach, likelihood of Green Square is 1/2, Blue square is 4/16 = 1/4, whereas from the image it's clear that "X" belongs to Blue Square class only.
[Update]
Thanks @Ben Reiniger, for correcting me.
I spent more time and what I understood is, it's not always correct to put equal weights(likelihood) across different classes in dataset. My earlier observation was based upon the assumption that all classes have equal weights and I was wrong, even though data is bit skewed.
Consider an another example:
We have a huge dataset of patient reports, where features are created based upon the several tests done so far. Our task is to identify the possibility if a particular person suffering from cancer or not.
In this case, even if a single feature point towards cancer disease, we cann't neglect it and should predict towards rare class, such that patient can go for further analysis.
In such cases weights are not equivalent across different classes. According to me, wights should be defined based upon the use case.
A very well explained its calculation in How does Weighted KNN works?
• This seems to miss the point of the likelilhood weighting. (Again, I'd suggest looking at the other posting of the same question.) – Ben Reiniger Jul 10 '19 at 14:31
• Will be good if you share the link as well, thanks. – vipin bansal Jul 10 '19 at 16:40
• so what is the answer? – asmgx Jul 10 '19 at 22:42
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Package xcolor Error: Undefined color ForestGreen'
I am using the following packages in my files Latex sources:
\usepackage{color}
\usepackage[table,xcdraw]{xcolor}
\usepackage[usenames, dvipsnames]{color}
\newcommand{\ricardo}[1]{\colorbox{ForestGreen}{\color{white}\textsf{\textbf{Ricardo}}} \textcolor{ForestGreen}{#1}}
And to mark my notes in the text, I use the command:
\ricardo{text....}
How should I do to make Latex display my notes correctly?
The xcolor package is an extension of the color package, so I don't understand why you loaded the color package twice along with the xcolor package. Moreover, according to page 7 of the xcolor documentation usenames is obsolete. I cleaned up the code to this:
\documentclass[11pt]{article}
\usepackage[dvipsnames,table,xcdraw]{xcolor}
\newcommand{\ricardo}[1]{\colorbox{ForestGreen}{\color{white} \textsf{\textbf{Ricardo}}} \textcolor{ForestGreen}{#1}}
\begin{document}
\ricardo{This should work.}
\end{document}
and the resulting output compiles without a problem:
• Thank you so much for your help @DJP, now it's all working! Jan 22, 2016 at 15:02
• Given that usenames' is obsolete, maybe this (\documentclass[usenames,dvipsnames]{beamer} ) should be changed at en.wikibooks.org/wiki/LaTeX/… ? Jun 5, 2019 at 9:59
• Indeed, in the documentation, mirror.utexas.edu/ctan/macros/latex/contrib/xcolor/xcolor.pdf, to use additional color like the one you wanna use, you have to add this option when you load the xcolor package. Check out the section 2.4.2 Additional sets of colors in the document. Simply put, to use it properly, you just need this line when load xcolor package: \usepackage[usenames, dvipsnames]{xcolor} Dec 10, 2019 at 7:33
sorry I can't comment as I'm just a noob and I need reputation. I tried out the command in my document and it works perfect. Check the image below.
In my beamer, I had declared something like \documentclass[xcolor=dvipsnames]{beamer} while the document that I use based on tufte declares \usepackage{xcolor} in the .sty file.
Could you try renaming the third line of the code into \usepackage[usenames, dvipsnames]{xcolor} ?
• Hello @crypto, I am using \documentclass[conference]{IEEEtran} I modified the third from \usepackage[usenames, dvipsnames]{color} to \usepackage{xcolor}, as your suggestion, but I keep getting the same error: Package xcolor Error: Undefined color ForestGreen. Jan 22, 2016 at 14:53
• Just a small question. Could you try \usepackage[usenames, dvipsnames]{xcolor}? I don't happen to have an IEEEtran tex file to try it out. Have you installed the package xcolor? Jan 22, 2016 at 14:56
• You're welcome. Thank you so much for your help @crypto! Jan 22, 2016 at 15:03
A more general solution may simply be to define the color explicitly.
\definecolor{ForestGreen}{RGB}{34,139,34}
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## Elementary Technical Mathematics
$(m-2)(m-20)$
$m^2-22m+40$ has no common monomial factor $=(\ \ \ \ \ \ \ \ )(\ \ \ \ \ \ \ \ )$ $=(m\ \ \ \ \ \ )(m\ \ \ \ \ \ )$ $=(m-\ )(m-\ )$ Since the 2nd term is negative and the 3rd positive, the two factors of 40 must both be negative. $=(m-2)(m-20)$ -2 and -20 have a sum of -22 and a product of 40.
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# Orthogonal projection matrix
Math 344, maple lab manual chapter 7: orthogonal projections in n-space projection matrices page 39 symmetric matrix it is also idempotent ie. Formed by a line through b that is orthogonal to a projection matrix lecture 15: projections onto subspaces. Journal of statistics education, volume 18, number 1 (2010) 1 orthogonal projection in teaching regression and financial mathematics farida kachapova. 19 orthogonal projections and orthogonal matrices 191 orthogonal projections we often want to decompose a given vector, for example, a force, into the sum of two orthogonal vectors example: suppose a mass m is at the end of a rigid, massless rod (an ideal pendulum), and the rod makes an angle θ with the vertical.
I've been looking around a bit and can't seem to find just what im looking for i've found canonical formulas, but what's the best way to use these do i have to. Projections and orthonormal bases the matrix of a projection is always very easy with orthonormal bases similarly help us compute orthogonal projection. 6 let a= 1 2 0 1 problem: find the matrix of the orthogonal projection onto the image of a the image of ais a one-dimensional line spanned by. I'm currently encounter with a projection problem i have a data point p in a n-dimensional orthogonal space a i have a n-dimensional non-orthogonal space b. Math 304 linear algebra that is, the nullspace of a matrix is the orthogonal orthogonal projection theorem 1 let v be a subspace of rn.
## Orthogonal projection matrix
Orthogonal projection of matrix onto subspace up vote 3 down vote favorite 1 how do i go about finding the matrix which is the orthogonal projection onto this. Orthogonal projection examples example 1:find the orthogonal projection of ~y = (23) onto the line l= h(31)i solution:let a= (31)t by theorem 48, the or. How to construct opengl projection matrix from the top view of the frustum, the x-coordinate of eye space, x e is mapped to x p, which is calculated by using the. Orthogonal complements and projections orthogonal to the rows of a and, hence the projection matrix given by (where.
Orthogonal projections projections onto subspaces subspace projection matrix example another example of a projection matrix projection is closest vector in. Orthogonal matrix in linear algebra, an orthogonal matrix or real orthogonal matrix is a square matrix with real entries whose columns and rows are orthogonal unit vectors (ie, orthonormal vectors), ie where i {\displaystyle i} is the identity matrix. Lecture 17: orthogonal matrices and gram but that word orthogonal matrices--or maybe i should be able to call it orthonormal what's the projection matrix is i. · a simple example of a non-orthogonal (oblique) projection (for definition see below) is via matrix multiplication, one sees that proving that p is indeed a projection. Orthogonality and projections just for fun, let’s quickly do the calculations for other projection, onto the orthogonal subspace here the matrix is, therefore.
In a euclidean sense by solving an algebraic expression for orthogonal projection of b on r(a) p =i, all projection matrices are neither orthogonal. Math 331 - orthogonal projections worksheet - solutions here are some practice problems on nding the standard matrix of an orthogonal projection. Remark it should be emphasized that p need not be an orthogonal projection matrix moreover, p is usually not an orthogonal matrix example consider the matrix p. Orthogonal projection is key step in solving many statistical models - here a simple geometric intuition. Lecture 1 i orthogonal projection i talked a bit about orthogonal projection last time and we saw that let pbe the matrix representing the trans.
· physics forums - the fusion of science and community. Builds an orthogonal projection matrix projection space refers to the space after applying projection transformation from view space after the projection. 3 orthogonal vectors and matrices the linear algebra portion of this course focuses on three matrix factorizations: qr factorization, singular. Free vector projection calculator - find the vector projection step-by-step symbolab solutions graphing calculator practice matrix, the one with numbers. Matrix of the orthogonal projection the minimization problem stated above arises in lot of applications so, it will be very helpful if the matrix of the orthogonal projection can be obtained under a given basis to this end, let be a -dimensional subspace of with as its orthogonal complement.
Orthogonal projection matrices 31 hold for an arbitrary x from theorem 22, p is the projection matrix called an orthogonal projection matrix (projector. Orthogonal projection orthogonal to v, a vector we label w, must matrices that are important in many applications of linear algebra. The formula for the orthogonal projection let v be a subspace of rn to nd the matrix of the orthogonal projection onto v, the way we rst discussed, takes three steps: (1) find a basis ~v 1, ~v 2 , ~v m for v (2) turn the basis ~v i into an orthonormal basis ~u i, using the gram-schmidt algorithm (3) your answer is p = p ~u i~ut i.
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# Variance
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this wiki
In probability theory and statistics, the variance of a random variable is a measure of its statistical dispersion, indicating how far from the expected value its values typically are. The variance of a real-valued random variable is its second central moment, and it also happens to be its second cumulant. The variance of a random variable is the square of its standard deviation.
## DefinitionEdit
If μ = E(X) is the expected value (mean) of the random variable X, then the variance is
$\operatorname{var}(X) = \operatorname{E}( ( X - \mu ) ^ 2 ).$
That is, it is the expected value of the square of the deviation of X from its own mean. In plain language, it can be expressed as "The average of the square of the distance of each data point from the mean". It is thus the mean squared deviation. The variance of random variable X is typically designated as $\operatorname{var}(X)$, $\sigma_X^2$, or simply $\sigma^2$.
Note that the above definition can be used for both discrete and continuous random variables.
Many distributions, such as the Cauchy distribution, do not have a variance because the relevant integral diverges. In particular, if a distribution does not have expected value, it does not have variance either. The opposite is not true: there are distributions for which expected value exists, but variance does not.
## Properties Edit
If the variance is defined, we can conclude that it is never negative because the squares are positive or zero. The unit of variance is the square of the unit of observation. For example, the variance of a set of heights measured in centimeters will be given in square centimeters. This fact is inconvenient and has motivated many statisticians to instead use the square root of the variance, known as the standard deviation, as a summary of dispersion.
It can be proven easily from the definition that the variance does not depend on the mean value $\mu$. That is, if the variable is "displaced" an amount b by taking X+b, the variance of the resulting random variable is left untouched. By contrast, if the variable is multiplied by a scaling factor a, the variance is multiplied by a2. More formally, if a and b are real constants and X is a random variable whose variance is defined,
$\operatorname{var}(aX+b)=a^2\operatorname{var}(X)$
Another formula for the variance that follows in a straightforward manner from the linearity of expected values and the above definition is:
$\operatorname{var}(X)= \operatorname{E}(X^2 - 2\,X\,\operatorname{E}(X) + (\operatorname{E}(X))^2 ) = \operatorname{E}(X^2) - 2(\operatorname{E}(X))^2 + (\operatorname{E}(X))^2 = \operatorname{E}(X^2) - (\operatorname{E}(X))^2.$
This is often used to calculate the variance in practice.
One reason for the use of the variance in preference to other measures of dispersion is that the variance of the sum (or the difference) of independent random variables is the sum of their variances. A weaker condition than independence, called uncorrelatedness also suffices. In general,
$\operatorname{var}(aX+bY) =a^2 \operatorname{var}(X) + b^2 \operatorname{var}(Y) + 2ab \operatorname{cov}(X, Y).$
Here $\operatorname{cov}$ is the covariance, which is zero for independent random variables (if it exists).
## Approximating the variance of a functionEdit
The Delta method uses second-order Taylor expansions to approximate the variance of a function of one or more random variables. For example, the approximate variance of a function of one variable is given by
$\operatorname{var}\left[f(X)\right]\approx \left(f'(\operatorname{E}\left[X\right])\right)^2\operatorname{var}\left[X\right]$
provided that $f(\cdot)$ is twice differentiable and that the mean and variance of $X$ are finite.
## Population variance and sample variance Edit
In general, the population variance of a finite population is given by
$\sigma^2 = \sum_{i=1}^N \left(x_i - \overline{x} \right)^ 2 \, \Pr(x_i),$
where $\overline{x}$ is the population mean. This is merely a special case of the general definition of variance introduced above, but restricted to finite populations.
In many practical situations, the true variance of a population is not known a priori and must be computed somehow. When dealing with large finite populations, it is almost never possible to find the exact value of the population variance, due to time, cost, and other resource constraints. When dealing with infinite populations, this is generally impossible.
A common method of estimating the variance of large (finite or infinite) populations is sampling. We start with a finite sample of values taken from the overall population. Suppose that our sample is the sequence $(y_1,\dots,y_N)$. There are two distinct things we can do with this sample: first, we can treat it as a finite population and describe its variance; second, we can estimate the underlying population variance from this sample.
The variance of the sample $(y_1,\dots,y_N)$, viewed as a finite population, is
$\sigma^2 = \frac{1}{N} \sum_{i=1}^N \left(y_i - \overline{y} \right)^ 2,$
where $\overline{y}$ is the sample mean. This is sometimes known as the sample variance; however, that term is ambiguous. Some electronic calculators can calculate $\sigma^2$ at the press of a button, in which case that button is usually labelled "$\sigma^2$".
When using the sample $(y_1,\dots,y_N)$ to estimate the variance of the underlying larger population the sample was drawn from, it may be tempting to equate the population variance with $\sigma^2$. However, $\sigma^2$ is a biased estimator of the population variance. The following is an unbiased estimator:
$s^2 = \frac{1}{N-1} \sum_{i=1}^N \left(y_i - \overline{y} \right)^ 2,$
where $\overline{y}$ is the sample mean. Note that the term $N-1$ in the denominator above contrasts with the equation for $\sigma^2$, which has $N$ in the denominator. Note that $s^2$ is generally not identical to the true population variance; it is merely an estimate, though perhaps a very good one if $N$ is large. Because $s^2$ is a variance estimate and is based on a finite sample, it too is sometimes referred to as the sample variance.
One common source of confusion is that the term sample variance may refer to either the unbiased estimator $s^2$ of the population variance, or to the variance $\sigma^2$ of the sample viewed as a finite population. Both can be used to estimate the true population variance, but $s^2$ is unbiased. Intuitively, computing the variance by dividing by $N$ instead of $N-1$ underestimates the population variance. This is because we are using the sample mean $\overline{y}$ as an estimate of the unknown population mean $\mu$, and the raw counts of repeated elements in the sample instead of the unknown true probabilities.
In practice, for large $N$, the distinction is often a minor one. In the course of statistical measurements, sample sizes so small as to warrant the use of the unbiased variance virtually never occur. In this context Press et al.[1] commented that if the difference between n and n−1 ever matters to you, then you are probably up to no good anyway - e.g., trying to substantiate a questionable hypothesis with marginal data.
### An unbiased estimator Edit
We will demonstrate why $s^2$ is an unbiased estimator of the population variance. An estimator $\hat{\theta}$ for a parameter $\theta$ is unbiased if $\operatorname{E}\{ \hat{\theta}\} = \theta$. Therefore, to prove that $s^2$ is unbiased, we will show that $\operatorname{E}\{ s^2\} = \sigma^2$. As an assumption, the population which the $x_i$ are drawn from has mean $\mu$ and variance $\sigma^2$.
$\operatorname{E} \{ s^2 \} = \operatorname{E} \left\{ \frac{1}{n-1} \sum_{i=1}^n \left( x_i - \overline{x} \right) ^ 2 \right\}$
$= \frac{1}{n-1} \sum_{i=1}^n \operatorname{E} \left\{ \left( x_i - \overline{x} \right) ^ 2 \right\}$
$= \frac{1}{n-1} \sum_{i=1}^n \operatorname{E} \left\{ \left( (x_i - \mu) - (\overline{x} - \mu) \right) ^ 2 \right\}$
$= \frac{1}{n-1} \sum_{i=1}^n \operatorname{E} \left\{ (x_i - \mu)^2 \right\} - 2 \operatorname{E} \left\{ (x_i - \mu) (\overline{x} - \mu) \right\} + \operatorname{E} \left\{ (\overline{x} - \mu) ^ 2 \right\}$
$= \frac{1}{n-1} \sum_{i=1}^n \sigma^2 - 2 \left( \frac{1}{n} \sum_{j=1}^n \operatorname{E} \left\{ (x_i - \mu) (x_j - \mu) \right\} \right) + \frac{1}{n^2} \sum_{j=1}^n \sum_{k=1}^n \operatorname{E} \left\{ (x_j - \mu) (x_k - \mu) \right\}$
$= \frac{1}{n-1} \sum_{i=1}^n \sigma^2 - \frac{2 \sigma^2}{n} + \frac{\sigma^2}{n}$
$= \frac{1}{n-1} \sum_{i=1}^n \frac{(n-1)\sigma^2}{n}$
$= \frac{(n-1)\sigma^2}{n-1} = \sigma^2$
#### Alternate proofEdit
$E\left[ \sum_{i=1}^n {(X_i-\overline{X})^2}\right] =E\left[ \sum_{i=1}^n {X_i^2}\right] - nE[ \overline{X}^2]$
$=nE[X_i^2] - \frac{1}{n} E\left[\left(\sum_{i=1}^n X_i\right)^2\right]$
$=n(\operatorname{var}[X_i] + (E[X_i])^2) - \frac{1}{n} E\left[\left(\sum_{i=1}^n X_i\right)^2\right]$
$=n\sigma^2 + \frac{1}{n}(nE[X_i])^2 - \frac{1}{n}E\left[\left(\sum_{i=1}^n X_i\right)^2\right]$
$=n\sigma^2 - \frac{1}{n}\left( E\left[\left(\sum_{i=1}^n X_i\right)^2\right] - \left(E\left[\sum_{i=1}^n X_i\right]\right)^2\right)$
$=n\sigma^2 - \frac{1}{n}\left(\operatorname{var}\left[\sum_{i=1}^n X_i\right]\right) =n\sigma^2 - \frac{1}{n}(n\sigma^2) =(n-1)\sigma^2.$
### Confidence intervals based on the sample varianceEdit
A confidence interval $T$ for the population variance can be formed as[1]
$T=\left[ \frac{n-1}{z_{2}}s^{2},\frac{n-1}{z_{1}}s^{2}\right]$
where $z_{1}$ and $z_{2}$ are strictly positive constants and $z_{1}. Its coverage probability is
$P\left( \sigma^{2}\in T\right) =P\left( z_{1}\leq \chi^2_{n-1}\leq z_{2}\right)$
where $\chi^2_{n-1}$ is a chi-square random variable with $n-1$ degrees of freedom.
## Generalizations Edit
If X is a vector-valued random variable, with values in Rn, and thought of as a column vector, then the natural generalization of variance is E[(X − μ)(X − μ)T], where μ = E(X) and XT is the transpose of X, and so is a row vector. This variance is a nonnegative-definite square matrix, commonly referred to as the covariance matrix.
If X is a complex-valued random variable, then its variance is E[(X − μ)(X − μ)*], where X* is the complex conjugate of X. This variance is a nonnegative real number.
## History Edit
The term variance was first introduced by Ronald Fisher in his 1918 paper The Correlation Between Relatives on the Supposition of Mendelian Inheritance.
## Moment of inertiaEdit
The variance of a probability distribution is analagous to the moment of inertia in classical mechanics of a corresponding linear mass distribution, with respect to rotation about its center of mass. It is because of this analogy that such things as the variance are called moments of probability distributions.
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# Preparing an abstracts book for a session
I have a list of contributions for a conference, with each one consisting of a serial number, a title, a list of authors, and an abstract. What I'd like to do is:
• typeset the list of contributions sequentially, so a reader can skim the abstracts
• have an author index, so that the reader looking for the contribution of a particular author can jump directly to that page.
I have enough contributions, and expect enough changes, that automating this is worth my while.
Is there a standard package that I can adapt to do this ? I don't mind writing some external scripts to preprocess data if need be.
Note: there's no scheduling information I need to worry about.
-
How do the serial numbers look? can you please give an example? – Yiannis Lazarides Apr 14 '12 at 6:14
It's as simple as 1 - This is the title - First Last, First Last and First Last. – Suresh Apr 14 '12 at 6:28
@Suresh: Perhaps the »confproc« package can be helpful. – Thorsten Donig Apr 14 '12 at 7:40
@ThorstenDonig: That's very helpful as well. A little overwhelming at first, but I'll go through it :) – Suresh Apr 14 '12 at 17:44
– Zev Chonoles Apr 18 '12 at 2:46
There is no package that I know of, however it would not be too difficult to describe your own macros to first typeset an abstract, title and other similar information you want to capture. For example the abstract is defined as a list:
\newenvironment{absquote}
{{\center\bfseries Abstract\endcenter}%
\list{}{\leftmargin2cm\rightmargin\leftmargin}%
\item\relax\footnotesize}
{\endlist}
A full minimal is shown below:
\documentclass{article}
\usepackage{lipsum}
\newenvironment{absquote}
{{\center\bfseries Abstract\endcenter}%
\list{}{\leftmargin2cm\rightmargin\leftmargin}%
\item\relax\footnotesize}
{\endlist}
\def\author#1{\center#1\endcenter}
\def\articletitle#1{\center{\bfseries\LARGE{#1}}\endcenter}
\def\serial#1{{\bfseries\hfill#1\hspace{1em}}}
\begin{document}
\articletitle{Some wonderful article}
\author{Yiannis Lazarides}
\begin{absquote}
\lipsum[1]
\serial{A-213}
\end{absquote}
\end{document}
To import all the abstracts you can use input{} within a loop to loop over all the numbers. If they are not too many you might even do it manually to give you more control and to add meta comments.
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This is nice: but what does it mean to use input{} to loop over numbers ? – Suresh Apr 14 '12 at 17:40
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## Summary
Problem : Inter-site variability in the diffusion MRI signal poses problems for tractography, segmentation, and other types of analysis.
Proposed solution : Data harmonization by mapping the SH coefficients from a target site to an arbitrarily chosen reference site.
Detailed problem
[The] inter-site variability in the measurements can come from several sources, e.g., subject physiological motion, number of head coils used for measurement (16 or 32 channel head coil), imaging gradient non-linearity as well as scanner related factors. This can cause non-linear changes in the images acquired […]. Inter-site variability in FA can be upto 5% in major white matter tracts and between 10-15% in gray matter areas. On the other hand, FA differences in diseases such as schizophrenia are often of the order of 5%.
Standard methods to address this problem :
• perform analysis at each site, then do a meta-analysis (impossible for data-driven methods), or
• use a statistical covariate to account for signal changes that are scanner-specific (not applicable to tractography beacause of region-specific differences).
The proposed method:
• takes into account region-specific differences;
• harmonizes the signal by comparing to a reference using several “Rotation-Invariant Spherical Harmonic” (RISH) features;
• uses a region-specific linear mapping between the RISH features to remove scanner specific differences in the white matter;
• is the first work to address the issue of dMRI data harmonization without using statistical covariates.
Figure 1 Outline of the proposed method for inter-site dMRI data harmonization
#### Rotation-Invariant Spherical Harmonics (RISH) features
Reminder: Spherical harmonics
$S \approx \sum_i \sum_j C_{ij} Y_{ij}$
The signal S is approximated as a sum of spherical functions $$Y_{ij}$$ of order $$i$$ and phase (or degree) $$j$$, weighted by coefficients $$C_{ij}$$.
The “energy” of the SH coefficients for each order is defined as the $$L_2$$ norm, and forms a set of rotation invariant features:
$||C_i||^2 = \sum_{j=1}^{2i+1} (C_{ij})^2$
The authors define the RISH features of the target site as the expected value of the energy for orders 0,2,4 and 6, for the $$N_k$$ subjects of site $$k$$:
$\mathbb{E}_k ( \left [ ||C_i||^2 \right ] ) = \frac{1}{N_k} \sum_{n=1}^{N_k} \left [ || C_i(n) ||^2 \right ]$
Freesurfer is used to compute 8 specific anatomical brain regions :
• frontal
• parietal
• temporal
• occipital
• brain stem
• cerebellum
• cingulate-corpus-callosum complex
• centrumsemiovale-insula
In each region, the sample average RISH features $$\mathbb{E}_k (\cdot)$$ are computed.
Figure 2 RISH features in the white matter for different SH orders and sites
The goal is then to find a mapping $$\Pi (\cdot)$$ such that all scanner related differences are removed :
$\mathbb{E}_k ( \Pi (||C_i||^2) ) = \mathbb{E}_r (||C_i||^2)$
where $$r$$ is the reference site and $$k$$ is the target site.
Thus, the “group” mapping is given by
Equation 5:
$\Pi ( ||C_i(n)||^2) = ||C_i(n)||^2 + \mathbb{E}_r - \mathbb{E}_k = \sum_j C_{ij}(n)^2 + \Delta \mathbb{E}$
(Note that the mapping is specific for each SH order and for each brain region)
Then, all there is left to do is to change the SH coefficients in each voxel to satisfy Equation 5 for each region. There are 2 possibilities, shifting the coefficients (adding a delta) or scaling (multiplying by a delta).
1. Shift: $$\pi(C_{ij}) = C_{ij} + \delta$$
2. Scale:
Shifting can cause coefficients to change sign, which is ill-posed for SH, as seen in figure 3.
## Experiments
Tract-based spatial statistics:
Statistical test for group differences:
The tests show there is “no statistical difference” after harmonization.
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# Importing saved 3d plot
0 Hi everyone, I generated a bunch of 3d plots to produce an animation and I saved them both as png and sobj files, because I wanted to keep the objects for later manipulations. However, when loading back an sobj file, I find them to be unusable. More precesily, p=plot3d(lambda x1,y1: h(t0,x1,y1), (-5,5),(-5,5),plot_points=100); #t0 fixed and h(t,x,y) a procedure p.save('bump003.sobj'); p.save('bump003.png'); a = load('bump003.sobj'); a.show(); returns the error NotImplementedError: You must override the get_grid method. while the png image files get correctly generated. There is no mention of this kind of error in the Plot3D doc, except for parametric surfaces http://www.sagemath.org/doc/reference/sage/plot/plot3d/parametric_surface.html where it is mentioned that get_grid should indeed be overriden for subclasses of parametric_surface. Any idea why this error shows up only after importing the object ? Many thanks, Benhuard asked Sep 19 '10 Benhuard 1 ● 2 ● 1
0 Hi Benhuard, This isn't an answer to your question, but I wanted to point out that @mhampton seems to have alternate ways of producing 3D animations, which maybe you could use to work around your problem. He mentioned them in his answer to my question animate 3d plots. posted Sep 20 '10 niles 3605 ● 7 ● 45 ● 101 http://nilesjohnson.net/
0 I can confirm this with an easier example: sage: var('x,y') (x, y) sage: f(x,y)=x^2+y^2 sage: p=plot3d(f,(-5,5),(-5,5)) sage: p # works fine sage: p.save('test.sobj') sage: q = load('test.sobj') sage: q NotImplementedError: You must override the get_grid method. I will try to look into this later; it's puzzling to me, since we get sage: type(q) sage: type(p) as expected. posted Sep 20 '10 kcrisman 7427 ● 17 ● 76 ● 166 This is now #9957; I don't think this will be immediate to fix. kcrisman (Sep 20 '10)
[hide preview]
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### astAxOffset
Add an increment onto a supplied axis value
#### Description:
This function returns an axis value formed by adding a signed axis increment onto a supplied axis value.
For a simple Frame, this is a trivial operation returning the sum of the two supplied values. But for other derived classes of Frame (such as a SkyFrame) this is not the case.
#### Synopsis
double astAxOffset( AstFrame $\ast$this, int axis, double v1, double dist )
#### Parameters:
##### this
Pointer to the Frame.
##### axis
The index of the axis to which the supplied values refer. The first axis has index 1.
##### v1
The original axis value.
##### dist
The axis increment to add to the original axis value.
#### Returned Value
##### astAxOffset
The incremented axis value.
#### Notes:
• This function will return a " bad" result value (AST__BAD) if any of the input values has this value.
• A " bad" value will also be returned if this function is invoked with the AST error status set, or if it should fail for any reason.
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# Tag Info
21
It is actually not distorted, it is sampled at high enough rate. What fools you is the straight lines drawn between sample points, it gives you a false impression of the waveform. It shows you a linear interpolation of the signal. It does not represent how the signal would actually look like. A sampled signal exists only at the sample points, and to convert ...
14
I would do a normalized autocorrelation to determine periodicity. If it is periodic with period $P$ you should see peaks at every $P$ samples in the result. A normalized result of "1" implies perfect periodicity, "0" implies no periodicity at all at that period, and values in between imply imperfect periodicity. Subtract the data sequence's mean from the ...
14
Figure 1.(c) shows the Test image reconstructed from MAGNITUDE spectrum only. We can say that the intensity values of LOW frequency pixels are comparatively more than HIGH frequency pixels. Actually, this is not correct. The phase values determine the shift in the sinusoid components of the image. With zero phase, all the sinusoids are centred at the same ...
12
The FFT can only be performed over a limited chunk of data. The basic math is based on the assumption that the time domain signal is periodic, i.e. your chunk of data is repeated in time. That typically results in a major discontinuity at the edges of the chunk. Let's look at a quick example: FFT size = 1000 points, Sample Rate = 1000 Hz, Frequency ...
11
Two remarks: I am assuming you are plotting the real (or imaginary) part of the Fourier transform. It is much more common to work with the magnitude or squared magnitude (power spectrum). The peak in the spectrum is a very poor measure of fundamental frequency (pitch). Take a piano note at 440 Hz, apply a notch filter to it to remove the 440 Hz component. ...
11
Posted for anyone who may find this useful... I created a picture that shows DFT frequency bin spacing for odd and even cases of N where N is the number of samples. FFTs usually operate on an even number of samples (the algorithm works by repeatedly breaking the problem into halves), so only the even case applies. The DC component (0*fs) is always part of ...
10
A beamformer is basically a spatial filter. It can be passive, just like a temporal filter. Instead of samples separated by time, they are separated by space. A passive temporal filter can be a bandpass that is "aimed" or "steered" at a particular frequency. For passive spatial filters (i.e. beamformers), the filter can be steered towards a particular ...
10
One trick, for even-length signals, is what to do with the "middle" sample. Here, I've split it half and half between each side of the FFT. The other trick is to ensure that you have the right amplitudes in the resampled signal. Here's it's a factor of 2. Try this in scilab: x = rand(1,100,'normal'); X = fft(x); XX = 2*[X(1:50) X(51)/2 zeros(1,99) X(51)/...
10
Well, first of all the Sound Level Pressure decreases by $6 \; \mathtt{dB}$ when doubling the distance - this plays a big role. We do also have sound attenuation coming from our medium - air. Let's take a closer look onto sound absorption coefficient for different frequencies: Knowing that human speech is mostly concentrated at the range of $300\;\mathtt{Hz}... 10 Phase Noise and Frequency Noise are not two different noise sources, they are artifacts of the same noise, it is just a matter of what units you want to use. Frequency and Phase are directly related as frequency is phase changing with time, so if you have one you will always have the other; frequency and phase are related by derivatives and integrals: the ... 9 You can make a positive frequency spectrum quite simply (where fs is the sampling rate and NFFT is the number of fft bins). In the Matlab implementation of the FFT algorithm, the first element is always the DC component, hence why the array starts from zero. THis is true for odd and even values of NFFT. %//Calculate frequency axis df = fs/NFFT; fAxis = 0:df:... 9 This comes from music terminology. The name "octave" comes from the fact that in the heptatonic musical scales (which are the prevalent scales in western music), the note with a 2:1 frequency ratio is the eighth note in the scale. For example, in the C major scale (C D E F G A B C) the eighth note is one octave above / has a 2:1 frequency ratio with the ... 9 The actual requirement is to sample at GREATER then twice the bandwidth, not at a rate equal to it... So only your 80Hz same set actually meets the requirement, because the 60Hz case is ambiguous in general, consider if you were sampling sin (2PiFt) instead then you would get a flat line at zero amplitude.... And changing the angle between sin and cos would ... 8 You might remember Nyquist's theorem. Given a signal which is band limited to$f_1$, we must sample it at least at$2f_1$:$f_S>2f_1$So if you check you favourite Signals and Systems book (e.g. Oppenheim's), you might recall that, once sampled, we can consider the signal's discrete Spectrum (which is periodic every$2\pi$radians, by the way). (In the ... 8 If your sampling frequency is$f_s=8000$Hz, your maximum signal frequency is indeed 4000 Hz ($=f_s/2$). If your signal contains frequencies above$=f_s/2$you would hear the results of aliasing. This means that the original spectrum is folded back into the range$[0,4000]$Hz. What actually happens is that by sampling with a sampling frequency$f_s$your ... 8 The term Doppler Shift is actually a bit of a misnomer. The frequencies are not actually shifted but they are scaled (see http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html for definition of shifting vs. scaling). It's a relative change not an absolute one. Both time and frequency domains are scaled: when the source is moving towards you, the ... 7 You're overlooking four things: The$\frac{1}{FFT\_size}$normalization coefficient. Some FFT implementations have or do not have this factor. Check the definition of FFT as performed by matlab on the Mathworks site! Why are you looking at the real part only? The amplitude is conveyed by the modulus (magnitude) of the complex number. Here, the real part is ... 7 The key insight that Fourier had when he developed Fourier analysis is that any absolutely integrable (thanks Jason R) function can be represented as the weighted sum of sines and cosines. Explaining why this is true is way beyond the scope of this answer. I suggest you study Fourier theory to understand this better. 7 It's certainly calculating the right thing. Though instead of sum(x.*(cos(1000*2*pi*t)-i*sin(2*pi*1000*t)))*2/N; you might try sum(x.*exp(-i*2*pi*1000*t))*2/N; If you need to do something similar, but in-line (not in a batch), you might want to look at the Goertzel algorithm. As the Wikipedia link says: .. provides a means for efficient evaluation of ... 7 The two frequencies you are referring to are the spatial frequency and temporal frequency of the wave, and you are correct in your reasoning on converting one to the other. The spatial frequency refers to how many complete periods the signal goes through for a given unit of distance (eg. cylcles/m) while the temporal frequency refers to how many complete ... 7 For the source, go to end of the answer Suppose one day you got one note which has some thing written to it, say "Major frequency components are 10 Hz, 25Hz, 50 Hz and 100 Hz". Somehow, you understood that its time-series representation is a very important thing (may be master-piece work of a great musician, or some national security matter, anything). So ... 7 From the ones I've been using I can recommend: YAAFE - very pleasant to work with in Python ESSENTIA - another one I like particularly due to Python integration aubio FEAPI Aquila - friend of mine used it extensively and he likes it a lot Recently I came across this paper and I believe that this should perfectly answer your question. Moffat D. et al - ... 6 Shortly, we have two kind of basic responses: time responses and frequency responses. Time responses test how the system works with momentary disturbance while the frequency response test it with continuous disturbance. Time responses contain things such as step response, ramp response and impulse response. Frequency responses contain sinusoidal responses. ... 6 If the signal is recorded using just one microphone, you can use methods such as spectral subtraction. This method is more suitable for "constant" noise, like the noise from a fan or an idle engine. Other methods rely on statistics and perceptual models of speech. If the signal is recorded with several microphones, you can use blind source separation for ... 6 This will give you a plot of the autocovariance up to lag 100 samples: plot(autocov(a,100)) There you can clearly get the period of your signal. Another approach is to explicitly get the time of each pulse: pulses = a > 0.1; leading_edges = diff(pulses) > 0; times = find(leading_edges > 0); periods = diff(times) ans = 55 56 56 56 56 ... 6 The scheme you are using is called On/Off Keying. It is not terribly efficient, but it is simple and gets the job done. When you say that the signal is 10 dB below the noise floor I suspect what you mean is that if you add up all the signal energy and all of the noise from 0.3 - 14 kHz the signal is 10 dB weaker, but that the signal uses a much narrower ... 6 Short answer: yes, I think so. Long answer: The FFT is just a fast implementation of the DFT. The frequency spacing of an N-point DFT operation is$\frac{f_s}{N}$. Samples of the DFT where$\omega \ge \pi$correspond to the negative frequencies. If N is odd, then$\frac{N-1}{2} \cdot \frac{2\pi}{N}$is less than$\pi$and the next DFT frequency,$\frac{...
6
As Dilip pointed out in the comment above, you can get the impulse response using the inverse Fourier transform. However, a slightly easier method might be to use the Laplace domain instead; it's more amenable to easy inverse transforming via transform tables. First, recall that the frequency response is really just the $s$-plane transfer function evaluated ...
6
For the first few experiments I would recommend using a scripting language like Matlab or Python. They're much easier to understand and much quicker to write than "lower level programming languages" like C++. Matlab has a signal processing toolbox and can read and write audio files, do windowing, FFTs etc. as well as a very simple playback mechanism. Basic ...
6
The HUP follows directly from the properties of the Fourier Transform, because time and frequency are orthogonal bases in which we can expand the co-efficient sequence of our signal. In fact all pairs of orthonormal bases will have some kind of Uncertainty Principle associated with them. In traditional Fourier analysis, the either the time axis or the ...
Only top voted, non community-wiki answers of a minimum length are eligible
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Previous: Images in LaTeX export, Up: LaTeX and PDF export
#### 12.6.6 Beamer class export
The LaTeX class beamer allows production of high quality presentations using LaTeX and pdf processing. Org mode has special support for turning an Org mode file or tree into a beamer presentation.
When the LaTeX class for the current buffer (as set with #+LaTeX_CLASS: beamer) or subtree (set with a LaTeX_CLASS property) is beamer, a special export mode will turn the file or tree into a beamer presentation. Any tree with not-too-deep level nesting should in principle be exportable as a beamer presentation. By default, the top-level entries (or the first level below the selected subtree heading) will be turned into frames, and the outline structure below this level will become itemize lists. You can also configure the variable org-beamer-frame-level to a different level—then the hierarchy above frames will produce the sectioning structure of the presentation.
A template for useful in-buffer settings or properties can be inserted into the buffer with M-x org-insert-beamer-options-template. Among other things, this will install a column view format which is very handy for editing special properties used by beamer.
You can influence the structure of the presentation using the following properties:
BEAMER_env
The environment that should be used to format this entry. Valid environments are defined in the constant org-beamer-environments-default, and you can define more in org-beamer-environments-extra. If this property is set, the entry will also get a :B_environment: tag to make this visible. This tag has no semantic meaning, it is only a visual aid.
BEAMER_envargs
The beamer-special arguments that should be used for the environment, like [t] or [<+->] of <2-3>. If the BEAMER_col property is also set, something like C[t] can be added here as well to set an options argument for the implied columns environment. c[t] or c<2-> will set an options for the implied column environment.
BEAMER_col
The width of a column that should start with this entry. If this property is set, the entry will also get a :BMCOL: property to make this visible. Also this tag is only a visual aid. When this is a plain number, it will be interpreted as a fraction of \textwidth. Otherwise it will be assumed that you have specified the units, like ‘3cm’. The first such property in a frame will start a columns environment to surround the columns. This environment is closed when an entry has a BEAMER_col property with value 0 or 1, or automatically at the end of the frame.
BEAMER_extra
Additional commands that should be inserted after the environment has been opened. For example, when creating a frame, this can be used to specify transitions.
Frames will automatically receive a fragile option if they contain source code that uses the verbatim environment. Special beamer specific code can be inserted using #+BEAMER: and #+BEGIN_BEAMER...#+END_BEAMER constructs, similar to other export backends, but with the difference that #+LaTeX: stuff will be included in the presentation as well.
Outline nodes with BEAMER_env property value ‘note’ or ‘noteNH’ will be formatted as beamer notes, i,e, they will be wrapped into \note{...}. The former will include the heading as part of the note text, the latter will ignore the heading of that node. To simplify note generation, it is actually enough to mark the note with a tag (either :B_note: or :B_noteNH:) instead of creating the BEAMER_env property.
You can turn on a special minor mode org-beamer-mode for editing support with
#+STARTUP: beamer
C-c C-b (org-beamer-select-environment)
In org-beamer-mode, this key offers fast selection of a beamer environment or the BEAMER_col property.
Column view provides a great way to set the environment of a node and other important parameters. Make sure you are using a COLUMN format that is geared toward this special purpose. The command M-x org-insert-beamer-options-template defines such a format.
Here is a simple example Org document that is intended for beamer export.
#+LaTeX_CLASS: beamer
#+TITLE: Example Presentation
#+AUTHOR: Carsten Dominik
#+LaTeX_CLASS_OPTIONS: [presentation]
#+BEAMER_FRAME_LEVEL: 2
#+COLUMNS: %35ITEM %10BEAMER_env(Env) %10BEAMER_envargs(Args) %4BEAMER_col(Col) %8BEAMER_extra(Ex)
* This is the first structural section
** Frame 1 \\ with a subtitle
*** Thanks to Eric Fraga :BMCOL:B_block:
:PROPERTIES:
:BEAMER_env: block
:BEAMER_envargs: C[t]
:BEAMER_col: 0.5
:END:
for the first viable beamer setup in Org
*** Thanks to everyone else :BMCOL:B_block:
:PROPERTIES:
:BEAMER_col: 0.5
:BEAMER_env: block
:BEAMER_envargs: <2->
:END:
for contributing to the discussion
**** This will be formatted as a beamer note :B_note:
** Frame 2 \\ where we will not use columns
*** Request :B_block:
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# Christen Sørensen Longomontanus
(Redirected from Longomontanus)
Christen Sørensen Longomontanus
Christen Sørensen Longomontanus
Born 4 October 1562
Jutland, Denmark
Died 8 October 1647 (aged 85)
Copenhagen
Nationality Danish
Fields astronomy
Institutions University of Copenhagen
Alma mater University of Rostock
Influences Tycho Brahe
Christen Sørensen Longomontanus (or Longberg) (4 October 1562 – 8 October 1647) was a Danish astronomer.
The name Longomontanus was a Latinized form of the name of the village of Lomborg, Jutland, Denmark, where he was born. His father, a laborer called Søren, or Severin, died when he was eight years old. An uncle took charge of the child, and had him educated at Lemvig; but after three years sent him back to his mother, who needed his help to work the fields. She agreed that he could study during the winter months with the clergyman of the parish; this arrangement continued until 1577, when the ill-will of some of his relatives and his own desire for knowledge caused him to run away to Viborg.
There he attended the grammar school, working as a labourer to pay his expenses, and in 1588 went to Copenhagen with a high reputation for learning and ability. Engaged by Tycho Brahe in 1589 as his assistant in his great astronomical observatory of Uraniborg, he rendered invaluable service for eight years. Having left the island of Hven with his master, he obtained his discharge at Copenhagen on 1 June 1597, in order to study at some German universities. He rejoined Tycho at Prague in January 1600, and having completed the Tychonic lunar theory, turned homeward again in August.
He visited Frauenburg, where Copernicus had made his observations, took a masters degree at Rostock, and at Copenhagen found a patron in Christian Friis, chancellor of Denmark, who employed him in his household. Appointed in 1603 rector of the school of Viborg, he was elected two years later to a professorship in the University of Copenhagen, and his promotion to the chair of mathematics ensued in 1607. This post he held till his death.
Longomontanus was not an advanced thinker. He adhered to Tycho's erroneous views about refraction, believed that comets were messengers of evil, and imagined that he had squared the circle. He found that the circle whose diameter is 43 has for its circumference the square root of 18252 which gives 3.14185... (or $22}{7$ ) for the value of π. John Pell and others tried in vain to convince him of his error. He inaugurated, at Copenhagen in 1632, the erection of a stately astronomical tower, but did not live to witness its completion. King Christian IV of Denmark, to whom he dedicated his Astronomia Danica, an exposition of the Tychonic system of the world, conferred upon him the canonry of Lunden in Schleswig.
However, it was Longomontanus who really developed Tycho's geoheliocentric model empirically and publicly to common acceptance in the 17th century in his 1622 astronomical tables. When Tycho died in 1601, his program for the restoration of astronomy was unfinished. The observational aspects were complete, but two important tasks remained, namely the selection and integration of the data into accounts of the motions of the planets, and the presentation of the results on the entire program in the form of a systematic treatise. Longomontanus, Tycho's sole disciple, assumed the responsibility and fulfilled both tasks in his voluminous Astronomia Danica (1622). Regarded as the testament of Tycho, the work was eagerly received in seventeenth-century astronomical literature. But unlike Tycho's, his geoheliocentric model gave the Earth a daily rotation as in the models of Ursus and Roslin, and which is sometimes called the 'semi-Tychonic' system.[1] As an indication of his book's popularity and of the semi-Tychonic system, it was reprinted in 1640 and 1663. Having originally worked on calculating the Martian orbit for Tycho with Kepler, he had already modelled its orbit to within 2 arcminutes error in longitude in his geoheliocentric model when Kepler had still only achieved 8 arcminutes error in his heliocentric system, but not yet using elliptical orbits. Some historians of science claim Kepler’s 1627 Rudolphine Tables based on Tycho Brahe’s observations were more accurate than any previous tables. But nobody has ever demonstrated they were more accurate than Longomontanus’s 1622 Danish Astronomy tables, also based upon Tycho’s observations.[citation needed]
## Publications
His major works in mathematics and astronomy were:
• Systematis Mathematici, etc. (1611)
• Cyclometria e Lunulis reciproce demonstrata, etc. (1612)
• Disputatio de Eclipsibus (1616)
• Astronomia Danica, etc. (1622)
• Disputationes quatuor Astrologicae (1622)
• Pentas Problematum Philosophiae (1623)
• De Chronolabio Historico, seu de Tempore Disputationes tres (1627)
• Geometriae quaesita XIII. de Cyclometria rationali et vera (1631)
• Disputatio de Matheseos Indole (1636)
• Coronis Problematica ex Mysteriis trium Numerorum (1637)
• Problemata duo Geometrica (1638)
• Problema contra Paulum Guldinum de Circuli Mensura (1638)
• Introductio in Theatrum Astronomicum (1639)
• Rotundi in Plano, etc. (1644)
• Admiranda Operatio trium Numerorum 6, 7, 8, etc. (1645)
• Caput tertium Libri primi de absoluta Mensura Rotundi plani, etc. (1646)
## References
1. ^ See Schofield's 'The Tychonic and semi-Tychonic world systems' in Wilson & Taton 'Planetary astronomy from the Renaissance to the rise of astrophysics' 1989 CUP. Mere diagrams allegedly of the Tychonic system would actually be indistinguishable from this semi-Tychonic system, unless they indicated whether the Earth or the fixed stars rotated daily.
2. ^ (Latin) This is not a coincidence, as explains Giambattista Riccioli, who named it.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 1.19: Trigonometric Functions of Negative Angles
Difficulty Level: At Grade Created by: CK-12
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Practice Trigonometric Functions of Negative Angles
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While practicing for the track team, you regularly stop to consider the values of trig functions for the angle you've covered as you run around the circular track at your school. Today, however, is different. To keep things more interesting, your coach has decided to have you and your teammates run the opposite of the usual direction on the track. From your studies at school, you know that this is the equivalent of a "negative angle".
You have run \begin{align*}-45^\circ\end{align*} around the track, and want to fine the value of the cosine function for this angle. Is it still possible to find the values of trig functions for these new types of angles?
At the completion of this Concept, you'll be able to calculate the values of trig functions for negative angles, and find the value of cosine for the \begin{align*}-45^\circ\end{align*} you have traveled.
### Guidance
Recall that graphing a negative angle means rotating clockwise. The graph below shows \begin{align*}-30^\circ\end{align*}.
Notice that this angle is coterminal with \begin{align*}330^\circ\end{align*}. So the ordered pair is \begin{align*}\left ( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right )\end{align*}. We can use this ordered pair to find the values of any of the trig functions of \begin{align*}-30^\circ\end{align*}. For example, \begin{align*}\cos (-30^\circ) = x = \frac{\sqrt{3}}{2}\end{align*}.
In general, if a negative angle has a reference angle of \begin{align*}30^\circ\end{align*}, \begin{align*}45^\circ\end{align*}, or \begin{align*}60^\circ\end{align*}, or if it is a quadrantal angle, we can find its ordered pair, and so we can determine the values of any of the trig functions of the angle.
#### Example A
Find the value of the expression: \begin{align*}\sin(-45^\circ)\end{align*}
Solution:
\begin{align*}\sin (-45^\circ) = -\frac{\sqrt{2}}{2}\end{align*}
\begin{align*}-45^\circ\end{align*} is in the \begin{align*}4^{th}\end{align*} quadrant, and has a reference angle of \begin{align*}45^\circ\end{align*}. That is, this angle is coterminal with \begin{align*}315^\circ\end{align*}. Therefore the ordered pair is \begin{align*}\left ( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right )\end{align*} and the sine value is \begin{align*}-\frac{\sqrt{2}}{2}\end{align*}.
#### Example B
Find the value of the expression: \begin{align*}\sec(-300^\circ)\end{align*}
Solution:
\begin{align*}\sec(-300^\circ) = 2\end{align*}
The angle \begin{align*}-300^\circ\end{align*} is in the \begin{align*}1^{st}\end{align*} quadrant and has a reference angle of \begin{align*}60^\circ\end{align*}. That is, this angle is coterminal with \begin{align*}60^\circ\end{align*}. Therefore the ordered pair is \begin{align*}\left ( \frac{1}{2}, \frac{\sqrt{3}}{2} \right )\end{align*} and the secant value is \begin{align*}\frac{1}{x} = \frac{1}{\frac{1}{2}} = 2\end{align*}.
#### Example C
Find the value of the expression: \begin{align*}\cos(-90^\circ)\end{align*}
Solution:
\begin{align*}\cos(-90^\circ) = 0\end{align*}
The angle \begin{align*}-90^\circ\end{align*} is coterminal with \begin{align*}270^\circ\end{align*}. Therefore the ordered pair is (0, -1) and the cosine value is 0.
We can also use our knowledge of reference angles and ordered pairs to find the values of trig functions of angles with measure greater than 360 degrees.
### Vocabulary
Negative Angle: A negative angle is an angle measured by rotating clockwise (instead of counter-clockwise) from the positive 'x' axis.
### Guided Practice
1. Find the value of the expression: \begin{align*}\cos -180^\circ\end{align*}
2. Find the value of the expression: \begin{align*}\sin -90^\circ\end{align*}
3. Find the value of the expression: \begin{align*}\tan -270^\circ\end{align*}
Solutions:
1. The angle \begin{align*}-180^\circ\end{align*} is coterminal with \begin{align*}180^\circ\end{align*}. Therefore the ordered pair of points is (-1, 0). The cosine is the "x" coordinate, so here it is -1.
2. The angle \begin{align*}-90^\circ\end{align*} is coterminal with \begin{align*}270^\circ\end{align*}. Therefore the ordered pair of points is (0, -1). The sine is the "y" coordinte, so here it is -1.
3. The angle \begin{align*}-270^\circ\end{align*} is coterminal with \begin{align*}90^\circ\end{align*}. Therefore the ordered pair of points is (0, 1). The tangent is the "y" coordinate divided by the "x" coordinate. Since the "x" coordinate is 0, the tangent is undefined.
### Concept Problem Solution
What you want to find is the value of the expression: \begin{align*}\cos(-45^\circ)\end{align*}
Solution:
\begin{align*}\cos (-45^\circ) = \frac{\sqrt{2}}{2}\end{align*}
\begin{align*}-45^\circ\end{align*} is in the \begin{align*}4^{th}\end{align*} quadrant, and has a reference angle of \begin{align*}45^\circ\end{align*}. That is, this angle is coterminal with \begin{align*}315^\circ\end{align*}. Therefore the ordered pair is \begin{align*}\left ( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right )\end{align*} and the cosine value is \begin{align*}\frac{\sqrt{2}}{2}\end{align*}.
### Practice
Calculate each value.
1. \begin{align*}\sin -120^\circ\end{align*}
2. \begin{align*}\cos -120^\circ\end{align*}
3. \begin{align*}\tan -120^\circ\end{align*}
4. \begin{align*}\csc -120^\circ\end{align*}
5. \begin{align*}\sec -120^\circ\end{align*}
6. \begin{align*}\cot -120^\circ\end{align*}
7. \begin{align*}\csc -45^\circ\end{align*}
8. \begin{align*}\sec -45^\circ\end{align*}
9. \begin{align*}\tan -45^\circ\end{align*}
10. \begin{align*}\cos -135^\circ\end{align*}
11. \begin{align*}\csc -135^\circ\end{align*}
12. \begin{align*}\sec -135^\circ\end{align*}
13. \begin{align*}\tan -210^\circ\end{align*}
14. \begin{align*}\sin -270^\circ\end{align*}
15. \begin{align*}\cot -90^\circ\end{align*}
### Vocabulary Language: English
Negative Angle
Negative Angle
A negative angle is an angle measured by rotating clockwise (instead of counterclockwise) from the positive $x$ axis.
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Sep 26, 2012
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# Why can't we derandomize the PCP theorem by iterating over all possible $\log n$ random strings?
Let's say I can solve problem $A$ in polynomial time using only $\log n$ bits of randomness, with a $\ge \frac{2}{3}$ chance of a correct answer. Then surely I can solve $A$ determistically by running my algorithm for $A$ over all random strings of length $\log n$ (of which there are a polynomial number) and take a popular vote of the outcomes.
I don't understand, then, why we would ever talk about $O(\log n)$ amounts of randomness in complexity classes that are closed under polynomial factors. More specifically, the PCP theorem says $NP = PCP[O(\log n), O(1)]$ - why isn't that the same as $PCP[0, O(1)]$?
We can! However, the PCP theorem does not say that we can solve the problem using $\log n$ bits of randomness. It says that we can verify a solution with $\log n$ bits.
Recall that a standard definition of NP is the class of languages for which there is a deterministic, polynomial-time verifier: For each $x$ in the language, there is a certificate $y$ so that the verifier accepts $\langle x,y \rangle$, and there is no such $y$ if $x$ is outside the language.
The PCP theorem says that we could also define NP as the class of languages for which there is a randomized, polynomial-time verifier that uses $\log n$ bits of randomness and reads only a constant number of bits of the certificate $y$.
Notice that if you derandomize the $\log n$ bits of randomness by enumerating them all and running this verifier on each one, you end up with a deterministic algorithm that will read the entire certificate $y$, which matches the standard definition of NP above.
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# DIFFERENCES BETWEEN WEB and PRINT VERSIONS
Occasionally, I make minor changes to my web lessons, that aren't reflected in my printed books.
In the online lessons, attention is drawn to added material by a purple dashed outline.
Errata are corrected in the online lessons and indicated here.
Such differences between online and printed versions are listed below.
DATE WEB LESSON CHANGE January 13, 2021 Parallel and Perpendicular Lines Add the word ‘distinct’ in two places. January 1, 2021 Quadratic Functions and the Completing the Square Technique Typos: change ‘can be real number’ to ‘can be any real number’; change ‘hold water’ to ‘holds water’; change $\,a(h-k)^2 + k\,$ to $\,a(x-h)^2 + k\,$; change $\,-\frac{20}{29}\,$ to $\,-\frac{29}{20}\,$ December 30, 2020 Equations of Simple Parabolas Typo: change ‘the the parabola’ to ‘then the parabola’ December 27, 2020 Parabolas Typo: change ‘parabola nets narrower’ to ‘parabola gets narrower’ November 30, 2020 Simple Word Problems Resulting in a System of Equations Change ‘in the following example’ to ‘with the following example’. December 12, 2020 $\rm\TeX\,$ Commands Available in MathJax Corrected definitions of supremum and infimum. November 30, 2020 Measures of Spread Typo: replace $\,x_1\,$ with $\,x_i\,$ in the sentence beginning ‘A reasonable idea is to...’. November 14, 2020 Advanced Set Concepts Replace ‘alternately’ with ‘alternatively’. November 9 and 11, 2020 Composition of Functions , Graphical Interpretations of Sentences Like $\,f(x) = g(x)\,$ and $\,f(x) > g(x)\,$, and Graphical Interpretations of Sentences Like $\,f(x) = 0\,$ and $\,f(x) > 0\,$ Typos: change ‘Graphs of Function’ to ‘Graphs of Functions’ in beginning link November 7, 2020 Basic Models You Must Know Typos: change ‘Graphs of Function’ to ‘Graphs of Functions’ in beginning link; changed ‘dom’ to ‘ran’ for cubing function; correct symbolic statement of domain for absolute value function; change ‘functions’ to ‘function’: ‘Every constant functions graphs as...’ November 2, 2020 Compound Interest Formula Typo: changed ‘years’ to ‘year’ (singular) in the Compound Interest Formula box. October 29, 2020 Arithmetic and Geometric Sequences Changed 12 to 12.5 (twice) in geometric series example. October 19, 2020 More on Exterior Angles in Triangles Changed ‘with respect to $\,\angle C\,$’ to ‘with respect to the exterior angle at $\,\angle C\,$’. October 17, 2020 Practice with Two-Column Proofs Typo (two places): change ‘REASONS columns’ to ‘REASONS column’ (singular). October 11, 2020 Relationships Between Angles and Sides in Triangles Typo: changed ‘$\,B\Leftarrow A\,$’ to ‘$\,A\Leftarrow B\,$ (that is, $\,B\Rightarrow A\,$)’. October 11, 2020 Triangle Congruence Changed ‘alternately’ to ‘alternatively’. October 11, 2020 Angles: Complementary, Supplementary, Vertical, and Linear Pairs Removed the cursor from the vertical angle image. October 3, 2020 Introduction to the Two-Column Proof Typo: changed ‘in written in’ to ‘is written in’. October 3, 2020 Proof Techniques Typo: changed ‘the only time than an’ to ‘the only time that an’. September 28, 2020 Quadrilaterals Changed ‘containing’ to ‘connecting’ in the definition of convex; much clearer. September 19, 2020 More Practice with Function Notation Missing factor of $\,2\,$ in displayed equation; corrected in online (not print) version. September 14, 2020 Finding Slant Asymptotes Added a ‘real-life’ example of a slant asymptote to online version, per user request. September 14, 2020 Finding Vertical Asymptotes Added ‘real-life’ examples of vertical asymptotes to online version, per user request.
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## anonymous one year ago How do you complete the square to form a perfect square trinomial with steps for 4x^2+5x+2=0
• This Question is Open
1. anonymous
$x= (-5\pm i \sqrt{7}) \div 8$
2. anonymous
how did you get that?
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### Zev Rosengarten (HUJI)
Wednesday, October 24, 2018, 15:10 – 16:25, -101
Abstract:
In 1981, Sansuc obtained a formula for Tamagawa numbers of reductive groups over number fields, modulo some then unknown results on the arithmetic of simply connected groups which have since been proven, particularly Weil’s conjecture on Tamagawa numbers over number fields. One easily deduces that this same formula holds for all linear algebraic groups over number fields. Sansuc’s method still works to treat reductive groups in the function field setting, thanks to the recent resolution of Weil’s conjecture in the function field setting by Lurie and Gaitsgory. However, due to the imperfection of function fields, the reductive case is very far from the general one; indeed, Sansuc’s formula does not hold for all linear algebraic groups over function fields. We give a modification of Sansuc’s formula that recaptures it in the number field case and also gives a correct answer for pseudo-reductive groups over function fields. The commutative case (which is essential even for the general pseudo-reductive case) is a corollary of a vast generalization of the Poitou-Tate nine-term exact sequence, from finite group schemes to arbitrary affine commutative group schemes of finite type. Unfortunately, there appears to be no simple formula in general for Tamagawa numbers of linear algebraic groups over function fields beyond the commutative and pseudo-reductive cases. Time permitting, we may discuss some examples of non-commutative unipotent groups over function fields whose Tamagawa numbers (and relatedly, Tate-Shafarevich sets) exhibit various types of pathological behavior.
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# How to embed a movie into a Mathematica-made presentation?
The question is completely formulated in the title.
However, to make it more presize, I am making presentations for lectures. The presentations are done in Mma. I have some illustrative material to show in a form of movies (avi and gif). It is possible to remove Mma from the screen, open a file with the movies and play one. I would like, however, to be able to call a movie from Mma just for the sake of speed. In addition it is aestetically better.
I tried to embed a hyperlink into the presentation notebook to call the movie file. This does not work. It opens an empty notebook with the title of the movie-file in question, and after a rather long waiting the notebook becomes filled by some symbols. Evidently, it opens the file rather than plays it.
So, can I do anything?
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I don't think you can include most movies, so the best solution might be what Yves said. But if the movie is simple (like most animated gifs), it's not high resolution, it has a low framerate, and not too many frames, then you can use ListAnimate with the list of its frames to include it directly. The notebook size will increase quite quickly with the number of frames though. – Szabolcs Mar 16 '14 at 16:35
@Szabolcs I think my movies satisfy the conditions you formulated. I succeeded to run it using the way offered by Yves. Could you please kindly show, how you can do it by ListAnimate, given the movie is external and I have no access to its creation or format. – Alexei Boulbitch Mar 16 '14 at 17:10
You can include movies like I did in this example: Style[Dynamic[Refresh[Import["http://pages.uoregon.edu/noeckel/computernotes/StopMotion.mov", "Animation"], None]], DynamicEvaluationTimeout -> 60] which is from this answer. – Jens Mar 16 '14 at 17:11
@AlexeiBoulbitch Jens's solution is better. Did you manage to get it to work according to his suggestion? – Szabolcs Mar 16 '14 at 17:32
You can use SystemOpen to open the link with system-wide standard browser/application:
SystemOpen["https://www.youtube.com/watch?v=yL_-1d9OSdk"]
You can also use it to open files stored on the local hard drive using the default applications (i.e. the default movie player for movie files).
E.g. in conjunction with Button:
Button["Chicken", SystemOpen["https://www.youtube.com/watch?v=yL_-1d9OSdk"],
BaseStyle -> "Hyperlink", Appearance -> "Frameless"]
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I edited this a bit, hope you don't mind. – Szabolcs Mar 16 '14 at 16:26
@Szabolcs thanks! Was editing (more chicken!), too, but no harm done :D – Yves Klett Mar 16 '14 at 16:27
My parrot is only stable against rotations but not translations (even though she does have a long neck). – Szabolcs Mar 16 '14 at 16:29
@Szabolcs hehe - I just changed the link to something non-commercial, yet still sufficiently poultry. But the gyro spot is still great. Parrots can fly, perhaps relevant? – Yves Klett Mar 16 '14 at 16:31
conbtinuation: ´Button["Play", SystemOpen[NotebookDirectory[] <> "filename.avi"]]´ – Alexei Boulbitch Mar 16 '14 at 17:18
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Chapter 7 Interactive communication
Last updated: 14 May 2021.
Required viewing
Key concepts/skills/etc
• Building a website using within the R environment using (in order of ease): postcards, distill, and blogdown.
• Thinking of about how we can take advantage of interaction in maps, and broadening the data that we make available via interactive maps, while still telling a clear story.
Key libraries
• blogdown
• distill
• leaflet
• mapdeck
• postcards
• tidyverse
• usethis
Key functions/etc
• blogdown:::serve_site()
• distill::create_article()
• postcards::create_postcard()
• usethis::use_git()
• usethis::use_github()
7.1 Making a website
7.1.1 Introduction
A website is a critical part of communication. For instance, it is a place to bring together everything that you’ve done, and it allows you some control of your online presence. You need a website.
One way to make a website is to use the blogdown package . blogdown is a package that allows you to make websites (not just blogs, notwithstanding its name) largely within R Studio. It builds on Hugo, which is a popular tool for making websites. blogdown lets you freely and quickly get a website up-and-running. It is easy to add content from time-to-time. It integrates with R Markdown which lets you easily share your work. And the separation of content and styling allows you to relatively quickly change your website’s design.
However, blogdown is brittle. Because it is so dependent on Hugo, features that work today may not work tomorrow. Also, owners of Hugo templates can update them at any time, without thought to existing users. blogdown is great if you know what you’re doing and have a specific use-case, or style, in mind. However, recently there are two alternatives that are better starting points.
The first is distill . Again, this is an R package that wraps around another framework, in this case Distill. However, in contrast to Hugo, Distill is more focused on common needs in data science, and is also only maintained by one group, so it can be a more stable choice. That said, the default distill site is fairly unremarkable. As such, here we recommend a third option.
The third option, and the one that we’ll start with, is postcards . This is a tailored solution that creates simple biographical websites that look great. If you followed the earlier chapter and set-up GitHub, then you should literally be able to get a postcards website online in five minutes.
7.1.2 Postcards
To get started with postcards, we first need to install the packages.
install.packages('postcards')
You will want to create a new project for your website, so ‘File -> New Project -> New Directory -> Postcards Website.’ You’ll then get a pick a name and location for the project, and you can select a postcards theme. In this case I’ll choose ‘trestles,’ and you probably want to tick ‘Open in new session.’
That will open a new file and you should now click ‘Knit’ to build the site. The result will be a fairly great one-page website (Figure 7.1)!
At this point, we should update the basic content to match our own. For instance, here is the same website, but with my details (Figure 7.2).
When you’ve got the site how you’d like it, then you should add it to GitHub. GitHub will try to build a site, which was don’t want so you need to first add a hidden file by running this in the console:
file.create('.nojekyll')
Then the easiest way (assuming that you set everything up in earlier chapters) is to use the usethis package .
usethis::use_git()
usethis::use_github()
The project will then be on your GitHub repo and you can use GitHub pages to host it: ‘Settings -> Pages’ and then change the source to ‘main’ or ‘master,’ depending on your settings.
7.1.3 Distill
To get started with distill , we are going to build a framework around our postcards site, following fairly closely (please go to Alison’s blogpost for the details). After that we’ll explore some of the aspects of distill that make it a nice choice, and mention some of the trade-offs that you make if you choose this option. First, we need to install distill.
install.packages('distill')
Again, create a new project for your website, so ‘File -> New Project -> New Directory -> Distill Blog’ (there’s not really much difference between the website and blog options).
You’ll then get a pick a name and location for the project, and you can set a title. Select ‘Configure for GitHub Pages’ and also ‘Open in a new session’ (if you forget to do any of this or change your mind then it’s not a big deal - these can always be changed ex post or you can just delete the directory and start again). It should look something like Figure 7.3.
At this point you can click ‘Build Website’ in the Build tab, and you’ll see the default website, which should look something like Figure 7.4.
Again, now we need to do the work to update things. The default for the ‘Distill Blog’ setting is that the blog is the homepage. We can change that. I really liked the bio page from earlier, so we could use that approach.
First change the name of the ‘index.Rmd’ file to ‘blog.Rmd.’ Then create a new ‘trestles’ page:
postcards::create_postcard(file = "index.Rmd", template = "trestles")
The trestles page that you just created will open, and you need to add the following line in the yaml.
site: distill::distill_website
In Figure 7.5 I added it to line 16 and then rebuilt the website.
We can make the same changes to the default content as earlier, updating the links, image, and bio. The advantage of using Distill is that we now have additional pages, not just a one-page website, and we also have a blog. By default, we have an ‘about’ page, but some other pages that may be useful, depending on your particular use-case, could include: ‘research,’ ‘teaching,’ ‘talks,’ ‘projects,’ ‘software,’ ‘datasets.’ For now, I’ll talk through adding and editing a page called ‘software.’
We can use the following function:
distill::create_article(file = 'software')
That will create and open an R Markdown document. To add it to the website, open ’_site.yml’ and then add a line to the ‘navbar’ (Figure 7.6(. After this is done then re-building the site will result in that software page having been added.
Continue with this process until you’re happy with your site. For instance, we may want to add our blog back. To do this follow the same pattern as before, but with ‘blog’ instance of ‘software.’
When you’re ready, you can get your website online in the same way as we did with the postcards site (i.e. push to GitHub and then use GitHub Pages).
Using the distill is a great option if you want a multi-page website, but still want a fairly controlled environment. There are a lot of options that you can change and the best place to start with that is to see Alison Hill’s blog post , but the distill package homepage is also useful.
That said, distill is very opinionated. Until recently they didn’t even allow a different citation style! While it is a great option (and what I use for my own website), if you want something that is more flexible, then blogdown might be a better option.
7.1.4 Blogdown
Using blogdown is more work than Google sites or Squarespace. It requires a little more knowledge than using a basic Wordpress site. And if you want to customise absolutely every aspect of your website, or need everything to be ‘just so’ then blogdown may not be for you. Further, blogdown is still under active development and various aspects may break in future releases. However, blogdown allows a variety and level of expression that is not possible with distill.
This post is a simplified version of and . It sticks to the basics and doesn’t require much decision-making. The purpose is to allow someone without much experience to use blogdown to get a website up-and-running. Head to those two resources once you’ve got a website working and want to dive a bit deeper.
We’ll need to install blogdown.
install.packages("blogdown")
Again, create a new project for your website, so ‘File -> New Project -> New Directory -> Website using blogdown.’ At this point you can set a name and location, and also select ‘Open in a new session.’ It should look something like Figure 7.7.
You can again click ‘Build Website’ from the ‘Build’ pane, but then an extra step is needed, of serving the site:
blogdown:::serve_site()
The site will show in the ‘Viewer’ pane (Figure 7.8).
At this point, the default website is being ‘served’ locally. This means that changes you make will be reflected in the website that you see in your Viewer pane. To see the website in a web browser, click ‘Show in new window’ button on the top left of the Viewer. That will open the website using the address that the R Studio also tells you.
You probably want to update the ‘About’ section. To do that go to ‘content -> about.md’ and add your own content. One nice aspect of blogdown is that it will automatically re-load the content when you save, so you should see your changes immediately show up.
You may also like to change the logo. You could do this by adding a square image to ‘public/images/’ and then changing the call to ‘logo.png’ in ‘config.yaml.’
When you’re happy with it, you can make your website public in the same way that is described for postcards.
This all said, the biggest advantage of using blogdown is that it allows us to use Hugo templates. This provides a large number of beautifully crafted websites. To pick a theme you can go to the Hugo themes page: https://themes.gohugo.io. There are hundreds of different themes. In general, most of them can be made to work with blogdown, but sometimes it can be a bit of a hassle to get them working.
One that I particularly like is Apéro: https://hugo-apero-docs.netlify.app. If you like that too, then you could use that theme by calling it when you create a new site. As a reminder, ‘File -> New Project -> New Directory -> Website using blogdown.’ At this point, in addition to setting the name and location, you can specify a theme. Specifically, in ‘Hugo theme’ field, you specify the GitHub username and repository, which in this case is ‘hugo-apero/apero’ (Figure 7.9).
7.2 Interactive maps
The nice thing about interactive maps is that you can let your users decide what they are interested in. Additionally, if there is a lot of information then you may like to leave it to your users as to selectively focus on what they are interested in. For instance, in the case of Canadian politics, some people will be interested in Toronto ridings, while others will be interested in Manitoba, etc. But it would be difficult to present a map that focuses on both of those, so an interactive map is a great option for allowing users to zoom in on what they want.
That said, it is important to be cognizant of what we are doing when we build maps, and more broadly, what is being done at scale to enable us to be able to build our own maps. For instance, with regard to Google, says:
Google began life in 1998 as a company famously dedicated to organising the vast amounts of data on the Internet. But over the last two decades its ambitions have changed in a crucial way. Extracting data such as words and numbers from the physical world is now merely a stepping-stone towards apprehending and organizing the physical world as data. Perhaps this shift is not surprising at a moment when it has become possible to comprehend human identity as a form of (genetic) ‘code.’ However, apprehending and organizing the world as data under current settings is likely to take us well beyond Heidegger’s ‘standing reserve’ in which modern technology enframed ‘nature’ as productive resource. In the 21st century, it is the stuff of human life itself—from genetics to bodily appearances, mobility, gestures, speech, and behaviour —that is being progressively rendered as productive resource that can not only be harvested continuously but subject to modulation over time.
Does this mean that we should not use or build interactive maps? Of course not. But it’s important to be aware of the fact that this is a frontier, and the boundaries of appropriate use are still being determined. Indeed, the literal boundaries of the maps themselves are being consistently determined and updated. The move to digital maps, compared with physical printed maps, means that it is actually possible for different users to be presented with different realities. For instance, ‘…Google routinely takes sides in border disputes. Take, for instance, the representation of the border between Ukraine and Russia. In Russia, the Crimean Peninsula is represented with a hard-line border as Russian-controlled, whereas Ukrainians and others see a dotted-line border. The strategically important peninsula is claimed by both nations and was violently seized by Russia in 2014, one of many skirmishes over control’ .
7.2.1 Leaflet
The leaflet package is originally a JavaScript library of the same name that has been brought over to R. It makes it easy to make interactive maps. The basics are similar to the ggmap set-up, but of course after that, there are many, many, options.
Let’s redo the bike map from earlier, and possibly the interaction will allow us to see what the issue is with the data.
In the same way as a graph in ggplot begins with the ggplot() function, a map in the leaflet package begins with a call to the leaflet() function. This allows you to specify data, and a bunch of other options such as width and height. After this, we add ‘layers,’ in the same way that we added them in ggplot. The first layer that we’ll add is a tile with the function addTiles(). In this case, the default is from OpenStreeMap. After that we’ll add markers that show the location of each bike parking spot with addMarkers().
library(leaflet)
library(tidyverse)
leaflet(data = bike_data) %>%
addMarkers(lng = bike_data$longitude, lat = bike_data$latitude,
popup = bike_data$street_address, label = ~as.character(bike_data$number_of_spots))
There are two options here that may not be familiar. The first is ‘popup,’ and this is what happens when you click on the marker. In this example this is giving the address. The second is ‘label,’ which is what happens when you hover over the marker. In this example it is given the number of spots.
Let’s have another go, this time making a map of the fire stations in Toronto. We can use data from Open Data Toronto, via the opendatatoronto R package . To ensure this book works, I will save and then use the dataset as at 13 May 2021, but you are able to get the up-to-date dataset using the link and the code.
library(opendatatoronto)
# Get starter code from: https://open.toronto.ca/dataset/fire-station-locations/
fire_stations_locations <- get_resource('9d1b7352-32ce-4af2-8681-595ce9e47b6e')
# Grab the lat and long - thanks https://stackoverflow.com/questions/47661354/converting-geometry-to-longitude-latitude-coordinates-in-r
fire_stations_locations <-
fire_stations_locations %>%
tidyr::extract(geometry, c('lon', 'lat'), '\$$(.*), (.*)\$$', convert = TRUE)
write_csv(fire_stations_locations, "inputs/data/fire_stations_locations.csv")
fire_stations_locations <- read_csv("inputs/data/fire_stations_locations.csv")
head(fire_stations_locations)
## # A tibble: 6 x 26
## _id ID NAME ADDRESS ADDRESS_POINT_ID ADDRESS_ID CENTRELINE_ID
## <dbl> <dbl> <chr> <chr> <dbl> <dbl> <dbl>
## 1 1 21 FIRE STATI… 900 TAPSCOT… 4236992 363382 4236991
## 2 2 60 FIRE STATI… 106 ASCOT A… 764237 70190 1140634
## 3 3 61 FIRE STATI… 65 HENDRICK… 819425 127148 1140587
## 4 4 55 FIRE STATI… 260 ADELAID… 12763904 484214 12763900
## 5 5 24 FIRE STATI… 745 MEADOWV… 6349868 357277 6349869
## 6 6 74 FIRE STATI… 140 LANSDOW… 10757599 157562 14674741
## # … with 19 more variables: MAINT_STAGE <chr>, ADDRESS_NUMBER <dbl>,
## # LINEAR_NAME_FULL <chr>, POSTAL_CODE <chr>, GENERAL_USE <chr>,
## # Y <lgl>, LATITUDE <lgl>, LONGITUDE <lgl>, WARD_NAME <chr>,
## # MUNICIPALITY_NAME <chr>, OBJECTID <dbl>, geometry <chr>, lon <dbl>,
## # lat <dbl>, geometry_1 <chr>
There is a lot of information here, but we’ll just plot the location of each fire station along with their name and address.
We will introduce a different type of marker here, which is circles. This will allow us to use different colours for the outcomes of each type. There are three possible outcomes: “Fire/Ambulance Stations” “Fire Station,” “Restaurant,” “Unknown.”
library(leaflet)
pal <- colorFactor("Dark2", domain = fire_stations_locations$GENERAL_USE %>% unique()) leaflet() %>% addTiles() %>% # Add default OpenStreetMap map tiles addCircleMarkers( data = fire_stations_locations, lng = fire_stations_locations$lon,
lat = fire_stations_locations$lat, color = pal(fire_stations_locations$GENERAL_USE),
popup = paste("<b>Name:</b>", as.character(fire_stations_locations$NAME), "<br>", "<b>Address:</b>", as.character(fire_stations_locations$ADDRESS), "<br>")
) %>%
pal = pal,
values = fire_stations_locations$GENERAL_USE %>% unique(), title = "Type", opacity = 1 ) 7.2.2 Mapdeck The package Mapdeck is an R package that is built on top of Mapbox (https://www.mapbox.com).4 It is based on WebGL, which means that your web browser does a lot of work for you. The nice thing is that because of this, it can do a bunch of things that leaflet struggles with, especially dealing with larger datasets. Mapbox is a full-featured application that many businesses that you may have heard of use: https://www.mapbox.com/showcase. To close out this discussion of interactive mapping, I want to briefly touch on mapdeck, as it is a newer, but very exciting, package. To this point we have used ‘stamen maps’ as our underlying tile, but mapdeck uses ‘Mapbox’ - https://www.mapbox.com/ - and so you need to register and get a token for this. It’s free and you only need to do it once. When you have that token you add it to R. (We cover what is happening here in more detail in a later chapter.) Run this function: usethis::edit_r_environ() When you run that function it will open a file. There you can add your Mapbox secret token. MAPBOX_TOKEN = 'PUT_YOUR_MAPBOX_SECRET_HERE' Save your ‘.Renviron’ file, and then restart R (Session -> Restart R). Then you can call the map. We’ll just plot our firefighters data from earlier. library(mapdeck) mapdeck(style = mapdeck_style('dark') ) %>% add_scatterplot( data = fire_stations_locations, lat = "lat", lon = "lon", layer_id = 'scatter_layer', radius = 10, radius_min_pixels = 5, radius_max_pixels = 100, tooltip = "ADDRESS" ) And this is pretty nice! 7.3 Shiny Shiny is a way of making interactive web applications (not just maps) using R. It’s fun, but fiddly. Here we’re going to step through one way to take advantage of Shiny, and that’s to quickly add some interactivity to our graphs. We’ll return to Shiny in later chapters also, so this is very much just a first pass. We’re going to make a very quick interactive graph based on the ‘babynames’ dataset from the package babynames . First, we’ll build a static version. library(babynames) library(tidyverse) top_five_names_by_year <- babynames %>% group_by(year, sex) %>% arrange(desc(n)) %>% slice_head(n = 5) top_five_names_by_year %>% ggplot(aes(x = n, fill = sex)) + geom_histogram(position = "dodge") + theme_minimal() + scale_fill_brewer(palette = "Set1") + labs(x = "Babies with that name", y = "Occurances", fill = "Sex" ) What we can see is that possibly the most popular boys names tend to be more clustered, compared with the most-popular girls names, which may be more spread out. However, one thing that we might be interested in is how the effect of the ‘bins’ parameter shapes what we see. We might like to use interactivity to explore different values. To get started, create a new Shiny app from the menu: ‘File -> New File -> Shiny Web App.’ Give it a name, such as ‘not_my_first_shiny’ and then leave all the other options as the default. A new file ‘app.R’ will open and you can click ‘Run app’ to see what it looks like. Now replace the content in that file - ‘app.R’ with the content below, and then again click ‘Run app’ library(shiny) # Define UI for application that draws a histogram ui <- fluidPage( # Application title titlePanel("Count of names for five most popular names each year."), # Sidebar with a slider input for number of bins sidebarLayout( sidebarPanel( sliderInput(inputId = "number_of_bins", label = "Number of bins:", min = 1, max = 50, value = 30) ), # Show a plot of the generated distribution mainPanel( plotOutput("distPlot") ) ) ) # Define server logic required to draw a histogram server <- function(input, output) { output$distPlot <- renderPlot({
# Draw the histogram with the specified number of bins
top_five_names_by_year %>%
ggplot(aes(x = n, fill = sex)) +
geom_histogram(position = "dodge", bins = input\$number_of_bins) +
theme_minimal() +
scale_fill_brewer(palette = "Set1") +
labs(x = "Babies with that name",
y = "Occurances",
fill = "Sex"
)
})
}
# Run the application
shinyApp(ui = ui, server = server)
You should find that you are served an interactive graph where you can change the number of bins and it should look something like Figure 7.10.
References
Allaire, JJ, Rich Iannone, Alison Presmanes Hill, and Yihui Xie. 2021. Distill: ’R Markdown’ Format for Scientific and Technical Writing.
Bensinger, Greg. 2020. Google Redraws the Borders on Maps Depending on Who’s Looking. Washington Post.
Chang, Winston, Joe Cheng, JJ Allaire, Carson Sievert, Barret Schloerke, Yihui Xie, Jeff Allen, Jonathan McPherson, Alan Dipert, and Barbara Borges. 2021. Shiny: Web Application Framework for r. https://CRAN.R-project.org/package=shiny.
Cheng, Joe, Bhaskar Karambelkar, and Yihui Xie. 2021. Leaflet: Create Interactive Web Maps with the JavaScript ’Leaflet’ Library. https://CRAN.R-project.org/package=leaflet.
Cooley, David. 2020. Mapdeck: Interactive Maps Using ’Mapbox GL JS’ and ’Deck.gl’. https://CRAN.R-project.org/package=mapdeck.
Gelfand, Sharla. 2020. Opendatatoronto: Access the City of Toronto Open Data Portal. https://CRAN.R-project.org/package=opendatatoronto.
Kahle, David, and Hadley Wickham. 2013. “Ggmap: Spatial Visualization with Ggplot2.” The R Journal 5 (1): 144–61. http://journal.r-project.org/archive/2013-1/kahle-wickham.pdf.
Kross, Sean. 2021. Postcards: Create Beautiful, Simple Personal Websites. https://CRAN.R-project.org/package=postcards.
McQuire, Scott. 2019. “One Map to Rule Them All? Google Maps as Digital Technical Object.” Communication and the Public 4 (2): 150–65.
Mock, Thomas. 2020. Building a Blog with Distill.
Presmanes Hill, Alison. 2021a. M-F-E-O: postcards + distill. https://alison.rbind.io/post/2020-12-22-postcards-distill/.
———. 2021b. Up & Running with Blogdown in 2021.
———. 2019b. Babynames: US Baby Names 1880-2017. https://CRAN.R-project.org/package=babynames.
Wickham, Hadley, and Jennifer Bryan. 2020. Usethis: Automate Package and Project Setup. https://CRAN.R-project.org/package=usethis.
Xie, Yihui, Christophe Dervieux, and Alison Presmanes Hill. 2021. Blogdown: Create Blogs and Websites with r Markdown. https://github.com/rstudio/blogdown.
Xie, Yihui, Amber Thomas, and Alison Presmanes Hill. 2021. Blogdown: Creating Websites with r Markdown.
1. Thank you to Shaun Ratcliff for introducing me to mapdeck.↩︎
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# AIC BIC Mallows Cp Cross Validation Model Selection
If you have several linear models, say model1, model2 and model3, how would you cross-validate it to pick the best model?
(In R)
I'm wondering this because my AIC and BIC for each model are not helping me determine a good model. Here are the results:
Model - Size (including response) - Mallows Cp - AIC - BIC
Intercept only - 1 - 2860.15 - 2101.61 - 2205.77
1 - 5 - 245.51 - 1482.14 - 1502.97
2 - 6 - 231.10 - 1472.88 - 1497.87
3 - 7 - 179.76 - 1436.29 - 1465.45
4 - 8 - 161.05 - 1422.43 - 1455.75
5 - 9 - 85.77 - 1360.06 - 1397.55
6 - 10 - 79.67 - 1354.79 - 1396.44
7 - 17 - 27.00 - 1304.23 - 1375.04
All Variables - 25 - 37.92 - 1314.94 - 1419.07
Note - assume the models are nested.
• why isn't you're AIC helping? AIC and BIC appear even to coincide here. – charles Mar 10 '14 at 0:34
• The lowest AIC/BIC is obtained on the 17 variable model - so this is the best? Shouldn't I do some calculations to check if the 'additional AIC or BIC' is worth an extra variable? Also - would you use LASSO to get several models, subsets regression, or neither? I used Subsets to create 6 models but now I'm wondering if its better to use LASSO? The question also is - is this method even correct for justifying which model is the best 'compromise'? – Dino Abraham Mar 10 '14 at 2:09
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Athaclena's Gonna Get The Hang Of This.... Eventually!
Recommended Posts
I'll start with my traditional "beginning of Challenge post". For more about me (for the curious) - check my signature for previous challenges where I'll have a bio (beyond that I'm a traveling consultant - emphasis on travel LOL). I also love to cook and have started a Recipe thread (I'll throw that into my sig as well).
I've decided to stick with the Rebels until I can figure out this "workout" thing. My last FEW challenges have been quite successful. I've got a handle on my eating (and continuing to tweak) and I'm still struggling to form some sort of workout routine!
SO - this challenge I'm going to keep on keeping on - but focusing on getting on the workout bandwagon. Starting with walking. I've been working on getting in at least 3 days of walking per week although I'm not remotely getting to my ultimate goal of 10k on those days - so I'm going to back my "goal" down to average 5k per day over the week with a goal of 3 "walking" days where I need to far exceed that to meet that goal. Any other Fitbit users want to track and keep me honest - my public URL is www.fitbit.com/user/27X45Y - feel free to friend me! I also reserve the right to throw this RIGHT OUT and come up with a different workout plan LOL. I just need to find SOMETHING that I can be consistent with - although I really, really want to work on my cardio health.
So - onto the list:
1) Nutrition - keep up with Probiotic and Breakfast shakes 5+ days per week. Continue cutting carbs for most meals (one cheat day - and realize I need to stick as much as possible when traveling).
2) Set myself up for success - During the week - drag my a$$out of bed by 7am (may have to get up a BIT earlier when on the road - but enough time to get in AT LEAST 30 minutes of walking - or dang - SOME SORT OF WORKOUT in the AM) - this does mean taking "workout gear" with me when traveling - but that shouldn't be a problem. 3) Exercise - Walk. Just Walk. I want to do MORE, but I've got to get some weight off so I'm focusing on walking this challenge. Goal is 5k steps PER DAY average. May throw this out and do a Body Weight Workout - or something completely different though...... 4) Sleep - Do my best to keep a sleep schedule\get enough sleep. Fell off a bit this challenge - need to get back on the wagon here.... I mean I totally rocked it last week - but I was sick and my body sort of forced me to. If only I can do that NORMALLY. 5) Check in here more. Even while traveling - I have the technology LOL. Check in on my thread at LEAST 3 times per week, keep up with the Juice Bar. If I have time, follow a few other threads to encourage others and at least TRY to do the mini-challenges. "Be not afraid of growing slowly; be afraid only of standing still." - Chinese Proverb 1st dozen-ish Challenges for the curious 12,11,10,9,8,7,6,5,4,3,2,More attempts, #1 with Intro, Failed attempts Spoiler Quick Bio: IT Consultant, Been in IT 25+ Years, Bounced around and landed as a traveling Consultant for a medium-sized Software Company. I love to cook & read, I travel for a living (although amount varies widely, sometimes I'm home for weeks, others I'm traveling for weeks on end), and trying to move out of Atlanta (plan in place, working to implement). Link to post Yaaaaay! Welcome back, Athaclena! Your goals look wonderful, and from your Juice Bar post it sounds like you're on your way to success within a number of parameters (even if you insist you're not tracking). Keep at it! I have no idea if I'm just a morning person or if this actually works, but I find staying hydrated helps a lot with waking up in the morning. Not... really sure why. ¯\_(ツ)_/¯ But something to consider! Link to post Welcome back! 3 hours ago, Wobbegong said: I have no idea if I'm just a morning person or if this actually works, but I find staying hydrated helps a lot with waking up in the morning. Not... really sure why. ¯\_(ツ)_/¯ But something to consider! It helps me because then I really have to pee in the morning. The trick is not getting back in bed after going to the bathroom. • 1 • 5 Current Challenge: Zeroh, stick to the routine! Link to post Welcome back. Your challenge looks great. Not going to lie, I use a regular alarm with a snooze button to get up that starts going off at 5:15. I also have an alarm on my phone for my I really don't want to get up days that goes off at 5:45 am Saying "GET UP NOW, YOU OVERSLEPT" which helps. I know if that goes off, and I am still in bed, I am in trouble. I also have one at 6 to remind me to get downstairs since at that point I have like 20 minutes until the girl I watch in the mornings comes over. Since you travel, maybe have different sounds for the 2 alarms? Looking forward to seeing how you do. Current Challenge ---> Bean Si Vs Chaos No energy for a title You are never too old to set another goal or dream a new dream - C.S. Lewis Link to post 13 hours ago, zeroh13 said: Welcome back! It helps me because then I really have to pee in the morning. The trick is not getting back in bed after going to the bathroom. Same here lol Link to post 21 hours ago, zeroh13 said: It helps me because then I really have to pee in the morning. The trick is not getting back in bed after going to the bathroom. It should be noted that this technique does not work as well for people who wake up in the night to pee. If you are that person, you have to keep drinking more water every time you get up. I understand that's... a thing for some people. Link to post Had a great week with friends and family. Quite the"bye" week diet wise but didn't do TOO much damage. Thankfully the closest thing to drama for the week was my brother "forgetting" Mom had scheduled a family dinner Tuesday after I flew home before he, the wife and kids went out of town for a wedding. Mom and I cooked for a small batch of friends Thursday, spent Friday "in town" visiting Mom's father (my last living grandparent), caught a movie, visited with one of my oldest friends at Starbucks and rounded out with dinner at one our favorite, small, family restaurants. We made overnight french toast (breakfast bread pudding) with the leftover Brioche I made this morning with Mom #2 - Mom's best friend before they dropped me off at the airport. One of my best Thanksgiving visits ever! Now back to the real world and to start off this challenge with a bang! "Be not afraid of growing slowly; be afraid only of standing still." - Chinese Proverb 1st dozen-ish Challenges for the curious 12,11,10,9,8,7,6,5,4,3,2,More attempts, #1 with Intro, Failed attempts Spoiler Quick Bio: IT Consultant, Been in IT 25+ Years, Bounced around and landed as a traveling Consultant for a medium-sized Software Company. I love to cook & read, I travel for a living (although amount varies widely, sometimes I'm home for weeks, others I'm traveling for weeks on end), and trying to move out of Atlanta (plan in place, working to implement). Link to post I am glad you had a great week off with the family. It sounds like it was a wonderful week to be home and see people. Welcome back and good luck with the challenge. Current Challenge ---> Bean Si Vs Chaos No energy for a title You are never too old to set another goal or dream a new dream - C.S. Lewis Link to post Holy crap - is it Thursday ALREADY?! Well, what was supposed to be an easy couple of weeks home before my travel to LA - well - turned into just 1. Emergency trip to Kansas City next week to un-snarl a database for a product upgrade. For something relatively stupid - but engineering is blaming the client for something set up (wait for it) 8 years ago. So I'm going to go fix the setting on 70 tables - which means copying data, dropping and re-creating the tables with all of the relative indexes and constraints and re-loading the data back - preserving any keys. OH - and I get to script it so it's repeatable for when we do their production database. I'm going to need a constant caffeine infusion to get it all done - and tested - and right. Time to add some WD-40 to my SQL Admin skills.... Oh and I took an earlier flight out Friday so I'm home before midnight - because I head to LA on Sunday *sigh*. And now for the challenge update.. 1) Nutrition - Doing well on the probiotic\shake front. Eating too many carbs though - need to keep my eye on that.... 2) Set myself up for success - I've been up a bit earlier - but not early enough. Getting there though! 3) Exercise - Walk. Just Walk. Not this week.... 4) Sleep - Actually doing well here - 8 hours every night! And I can tell I'm getting good quality sleep because it's not as much of a drag getting up.... 5) Check in here more - been checking on the mini-challenge a bit, but not posting much. I do need to be in here a bit more.... "Be not afraid of growing slowly; be afraid only of standing still." - Chinese Proverb 1st dozen-ish Challenges for the curious 12,11,10,9,8,7,6,5,4,3,2,More attempts, #1 with Intro, Failed attempts Spoiler Quick Bio: IT Consultant, Been in IT 25+ Years, Bounced around and landed as a traveling Consultant for a medium-sized Software Company. I love to cook & read, I travel for a living (although amount varies widely, sometimes I'm home for weeks, others I'm traveling for weeks on end), and trying to move out of Atlanta (plan in place, working to implement). Link to post 23 hours ago, Athaclena said: Holy crap - is it Thursday ALREADY?! Well, what was supposed to be an easy couple of weeks home before my travel to LA - well - turned into just 1. Emergency trip to Kansas City next week to un-snarl a database for a product upgrade. For something relatively stupid - but engineering is blaming the client for something set up (wait for it) 8 years ago. So I'm going to go fix the setting on 70 tables - which means copying data, dropping and re-creating the tables with all of the relative indexes and constraints and re-loading the data back - preserving any keys. OH - and I get to script it so it's repeatable for when we do their production database. I'm going to need a constant caffeine infusion to get it all done - and tested - and right. Time to add some WD-40 to my SQL Admin skills.... Oh and I took an earlier flight out Friday so I'm home before midnight - because I head to LA on Sunday *sigh*. And now for the challenge update.. 1) Nutrition - Doing well on the probiotic\shake front. Eating too many carbs though - need to keep my eye on that.... 2) Set myself up for success - I've been up a bit earlier - but not early enough. Getting there though! 3) Exercise - Walk. Just Walk. Not this week.... 4) Sleep - Actually doing well here - 8 hours every night! And I can tell I'm getting good quality sleep because it's not as much of a drag getting up.... 5) Check in here more - been checking on the mini-challenge a bit, but not posting much. I do need to be in here a bit more.... Yikes on the emergency trip. Good luck getting it all take care of. Sadly, I have no SQL knowledge yet, otherwise I would tell you to hit me up if you need it. Sounds like your doing good on your challenge. Getting up a bit earlier is a good start. Good luck with the walking. I hope you can get some in. Current Challenge ---> Bean Si Vs Chaos No energy for a title You are never too old to set another goal or dream a new dream - C.S. Lewis Link to post Holy crap - is it Thursday ALREADY?! Well, what was supposed to be an easy couple of weeks home before my travel to LA - well - turned into just 1. Emergency trip to Kansas City next week to un-snarl a database for a product upgrade. For something relatively stupid - but engineering is blaming the client for something set up (wait for it) 8 years ago. So I'm going to go fix the setting on 70 tables - which means copying data, dropping and re-creating the tables with all of the relative indexes and constraints and re-loading the data back - preserving any keys. OH - and I get to script it so it's repeatable for when we do their production database. I'm going to need a constant caffeine infusion to get it all done - and tested - and right. Time to add some WD-40 to my SQL Admin skills.... Oh and I took an earlier flight out Friday so I'm home before midnight - because I head to LA on Sunday *sigh*. And now for the challenge update.. 1) Nutrition - Doing well on the probiotic\shake front. Eating too many carbs though - need to keep my eye on that.... 2) Set myself up for success - I've been up a bit earlier - but not early enough. Getting there though! 3) Exercise - Walk. Just Walk. Not this week.... 4) Sleep - Actually doing well here - 8 hours every night! And I can tell I'm getting good quality sleep because it's not as much of a drag getting up.... 5) Check in here more - been checking on the mini-challenge a bit, but not posting much. I do need to be in here a bit more.... Gah! I have always been avoiding SQL, just cannot stand the unpredictability of it! And... I never realised you were a DBA? I didn't even realised you worked in IT... should read more properly Hello fellow WIT Coffee is what charges us up, you are doing great with all that travelling! Challenges: 1, 2, 3, 4, 5, 6, 7, 8 Link to post A year ago my boss told me I should learn SQL since it would be "good if you're going to continue in this field" (customer service???). He dumped some online resources in my lap and then never mentioned it again. But I probably should work on my coding skills... there's money there, after all. Enjoy your time at home if you can!! And good luck with your upcoming work trips, unexpected snarls from eight year old nonsense are no fun at all. Sounds like your challenge is going pretty well, even if it's not all perfect. That's fine, just stay with it and you'll get there. Let us know if there's anything we can do to support you! Link to post Actually @Diadhuit - I'm not a DBA - have never been one - OFFICIALLY. My background is Web Development - BUT - my apps were always very data heavy and I ended up finding out that I was VERY GOOD at DB Design in general and SQL. I actually think in data structures - just didn't realize it until I was realized I was designing the data layout in my head while requirement gathering. So - I've done a lot of "DBA" things in the past. Now I'm really an Architect. I design implementations for the massive software package my company sells that integrates with tons of systems at clients (and we're talking everything from 10k employees to 500k employees). So my background in moving data around has been VERY helpful. And sometimes I get to write some code\customizations - and sometimes I have to roll up my sleeves and fix something that engineering (that's our developers) won't (or can't). Next week it's a relatively small client - but I'm also trying to make sure we don't end up with yet ANOTHER half-assed upgrade. I don't do half-assed and I WILL fix it. Dammit. If you're in IT and either write code or reports - I ALWAYS recommend learning more about data structures and SQL. Because all of the "tools" out there that do that for you suck. Like really, totally, are completely abysmal. Of course you wouldn't know they were if you've never seen comparisons with an actual properly designed DB. If you learn how to do your back end databases properly and write your own DB code - your code will be OODLES more efficient and you will become a coding god\goddess with the sheer efficiency. Trust me And in the art of next week sucking a little less, a Window seat opened up on my Monday morning flight. So HURRAY for airline status letting me move seats LOL - and for remembering that I needed to check again the next day after booking the flight. "Be not afraid of growing slowly; be afraid only of standing still." - Chinese Proverb 1st dozen-ish Challenges for the curious 12,11,10,9,8,7,6,5,4,3,2,More attempts, #1 with Intro, Failed attempts Spoiler Quick Bio: IT Consultant, Been in IT 25+ Years, Bounced around and landed as a traveling Consultant for a medium-sized Software Company. I love to cook & read, I travel for a living (although amount varies widely, sometimes I'm home for weeks, others I'm traveling for weeks on end), and trying to move out of Atlanta (plan in place, working to implement). Link to post Actually [mention=63447]Diadhuit[/mention] - I'm not a DBA - have never been one - OFFICIALLY. My background is Web Development - BUT - my apps were always very data heavy and I ended up finding out that I was VERY GOOD at DB Design in general and SQL. I actually think in data structures - just didn't realize it until I was realized I was designing the data layout in my head while requirement gathering. So - I've done a lot of "DBA" things in the past. Now I'm really an Architect. I design implementations for the massive software package my company sells that integrates with tons of systems at clients (and we're talking everything from 10k employees to 500k employees). So my background in moving data around has been VERY helpful. And sometimes I get to write some code\customizations - and sometimes I have to roll up my sleeves and fix something that engineering (that's our developers) won't (or can't). Next week it's a relatively small client - but I'm also trying to make sure we don't end up with yet ANOTHER half-assed upgrade. I don't do half-assed and I WILL fix it. Dammit. If you're in IT and either write code or reports - I ALWAYS recommend learning more about data structures and SQL. Because all of the "tools" out there that do that for you suck. Like really, totally, are completely abysmal. Of course you wouldn't know they were if you've never seen comparisons with an actual properly designed DB. If you learn how to do your back end databases properly and write your own DB code - your code will be OODLES more efficient and you will become a coding god\goddess with the sheer efficiency. Trust me [emoji4] And in the art of next week sucking a little less, a Window seat opened up on my Monday morning flight. So HURRAY for airline status letting me move seats LOL - and for remembering that I needed to check again the next day after booking the flight. My degree too is Web Development and I stick to to it. I also have to admit that I love normalised table structures! More and more I love Front End after seeing companies DBs. It's an act of self care more than anything as I'm a perfectionist! I look at how an application is and feel for it as for a person. I do feel like a surgen going in and fixing something with the minimum impact for the patient, or, when writing a new feature, trying to avoid health consequences as much as possible. I would be overwhelmed by fixing DBs because you guys are the neurosurgeons! And you can operate only with a sword and an awake patient! Nope, I would loose nights awake preparing for it. I get scared each time I see the DB structure of the product in my company. I want to cry after seeing how badly it has been put together... How could they not think that would work? Challenges: 1, 2, 3, 4, 5, 6, 7, 8 Link to post 10 hours ago, Diadhuit said: I get scared each time I see the DB structure of the product in my company. I want to cry after seeing how badly it has been put together... How could they not think that would work? I have a semi-permanent eye-twitch whenever I'm forced to look at the Stored Procedures. The actual table structure isn't too bad - but holy cripes they don't know how to write efficient queries. Actual "discussion" with one of my co-workers about writing a block of code for something - the end of the heated discussion went something like this. Co-Worker: But if you do it that way, the more records over here the slower it's going to get. Me: Not if you know how databases work - it won't matter HOW many records because there will be a limited number of the type we care about. Co-Worker: But you'll have more records to sort through and code against?! Me: .... You know what - you think I don't know what I'm doing? You code it your damn self. This conversation is over have a great day. (because I was frankly beyond explaining something so frickin' basic to a "Principle") I have managed to not have to work with him again - it's pretty known at this point that he and I aren't oil and water - it's more like lighter fluid and a match. In fact - he wrote the stored proc in question and not me because "he knew how to code it" - and I got pulled in to update it when there needed to be more use cases processed. I re-wrote it, added in the new use cases - plus a couple they didn't ask for but I knew that the client would want anyway - and I could do it within the requested billable time allotted. And you know what? It was faster than the initial stored procedure and did more. BECAUSE I KNOW WHAT THE F' I AM DOING AFTER 20 YEARS OF DOING THIS!? And DUDE just refuses to understand that I just MIGHT know more than he does. Male chauvinist prick! Heaven FORBID he learn something. Sorry - that was a year ago and still p*sses me off. We had a good junior guy quit after witnessing that BS (there was more to it, but I think that was the final straw for him). I don't have to deal as much with the whole "women don't know IT" as I used to - it just really chaps my butt when it's within my own company with a co-worker who SHOULD have seen my work over the last 4 years and realized I really DO know what I'm doing...... If only getting my blood boiling like that counted as a workout LOL. "Be not afraid of growing slowly; be afraid only of standing still." - Chinese Proverb 1st dozen-ish Challenges for the curious 12,11,10,9,8,7,6,5,4,3,2,More attempts, #1 with Intro, Failed attempts Spoiler Quick Bio: IT Consultant, Been in IT 25+ Years, Bounced around and landed as a traveling Consultant for a medium-sized Software Company. I love to cook & read, I travel for a living (although amount varies widely, sometimes I'm home for weeks, others I'm traveling for weeks on end), and trying to move out of Atlanta (plan in place, working to implement). Link to post I have a semi-permanent eye-twitch whenever I'm forced to look at the Stored Procedures. The actual table structure isn't too bad - but holy cripes they don't know how to write efficient queries. Actual "discussion" with one of my co-workers about writing a block of code for something - the end of the heated discussion went something like this. Co-Worker: But if you do it that way, the more records over here the slower it's going to get. Me: Not if you know how databases work - it won't matter HOW many records because there will be a limited number of the type we care about. Co-Worker: But you'll have more records to sort through and code against?! Me: .... You know what - you think I don't know what I'm doing? You code it your damn self. This conversation is over have a great day. (because I was frankly beyond explaining something so frickin' basic to a "Principle") I have managed to not have to work with him again - it's pretty known at this point that he and I aren't oil and water - it's more like lighter fluid and a match. In fact - he wrote the stored proc in question and not me because "he knew how to code it" - and I got pulled in to update it when there needed to be more use cases processed. I re-wrote it, added in the new use cases - plus a couple they didn't ask for but I knew that the client would want anyway - and I could do it within the requested billable time allotted. And you know what? It was faster than the initial stored procedure and did more. BECAUSE I KNOW WHAT THE F' I AM DOING AFTER 20 YEARS OF DOING THIS!? And DUDE just refuses to understand that I just MIGHT know more than he does. Male chauvinist prick! Heaven FORBID he learn something. Sorry - that was a year ago and still p*sses me off. We had a good junior guy quit after witnessing that BS (there was more to it, but I think that was the final straw for him). I don't have to deal as much with the whole "women don't know IT" as I used to - it just really chaps my butt when it's within my own company with a co-worker who SHOULD have seen my work over the last 4 years and realized I really DO know what I'm doing...... If only getting my blood boiling like that counted as a workout LOL. Don't .let .me .started! A colleague of mine refuses to refer to me as senior when he has the same title and call himself such. And sometimes addresses me as darling. He is f#cking 35, not 70! And that's the tiny thing that makes my blood boil in a quite good job (everyone else is considering me equal) I had baaaad things happening in the past and I see it is getting better and where I have been lately it's nearly there. Challenges: 1, 2, 3, 4, 5, 6, 7, 8 Link to post 5 hours ago, Diadhuit said: Don't .let .me .started! A colleague of mine refuses to refer to me as senior when he has the same title and call himself such. And sometimes addresses me as darling. He is f#cking 35, not 70! And that's the tiny thing that makes my blood boil in a quite good job (everyone else is considering me equal) I had baaaad things happening in the past and I see it is getting better and where I have been lately it's nearly there. #Word While I did at least INSIST on getting paid fairly (and refused quite a few job offers that just refused to pay the proper rate for the work I was being hired for), it took WAY too much job hopping to 1) get into something OTHER than tied to a chair programming - because while I CAN legitimately do the work of 3 - I was totally burned out and 2) GET A FREAKING PROMOTION. Equality is not JUST equal pay people! I've been with my current company now 5 years and received actual, real raises and a promotion (yes, I had to go to HR to get said promotion but that was because my favorite manager left to take care of his mother and my manager after that has 0 balls) - my first same company promotion in 20 years (yes, you read that # right). When my favorite manager left, I legit cried - because I knew the promotion was lost and how could I be mad at him when he was quitting to take care of his mother?! I just simmered after my conversation with my "new" manager about a promotion and he "didn't even know what that would look like". Confidence inspiring isn't it? At some point I simmered enough until I basically said F it and made the HR call - because I was DOING the job of the next level, legit deserved the promotion and what would they do - fire me? Highly unlikely given the amount of work I was doing and deals I was helping them close. Needless to say, my promotion came through within 60 days. But I SHOULD NOT have needed to make that call..... "Be not afraid of growing slowly; be afraid only of standing still." - Chinese Proverb 1st dozen-ish Challenges for the curious 12,11,10,9,8,7,6,5,4,3,2,More attempts, #1 with Intro, Failed attempts Spoiler Quick Bio: IT Consultant, Been in IT 25+ Years, Bounced around and landed as a traveling Consultant for a medium-sized Software Company. I love to cook & read, I travel for a living (although amount varies widely, sometimes I'm home for weeks, others I'm traveling for weeks on end), and trying to move out of Atlanta (plan in place, working to implement). Link to post It is always disturbing to discover that you as a female have bigger balls than your male boss. Gender aside, it is never good when you have a boss you can't trust to have your back. It's screwed up and sucks that it happens to women in IT and STEM jobs more often than it does to men in any job or to women in nursing or teaching. “I've always believed that failure is non-existent. What is failure? You go to the end of the season, then you lose the Super Bowl. Is that failing? To most people, maybe. But when you're picking apart why you failed, and now you're learning from that, then is that really failing? I don't think so." - Kobe Bryant, 1978-2020. Rest in peace, great warrior. Personal Challenges, a.k.a.The Saga of Scaly Freak: Link to post #Word While I did at least INSIST on getting paid fairly (and refused quite a few job offers that just refused to pay the proper rate for the work I was being hired for), it took WAY too much job hopping to 1) get into something OTHER than tied to a chair programming - because while I CAN legitimately do the work of 3 - I was totally burned out and 2) GET A FREAKING PROMOTION. Equality is not JUST equal pay people! I've been with my current company now 5 years and received actual, real raises and a promotion (yes, I had to go to HR to get said promotion but that was because my favorite manager left to take care of his mother and my manager after that has 0 balls) - my first same company promotion in 20 years (yes, you read that # right). When my favorite manager left, I legit cried - because I knew the promotion was lost and how could I be mad at him when he was quitting to take care of his mother?! I just simmered after my conversation with my "new" manager about a promotion and he "didn't even know what that would look like". Confidence inspiring isn't it? At some point I simmered enough until I basically said F it and made the HR call - because I was DOING the job of the next level, legit deserved the promotion and what would they do - fire me? Highly unlikely given the amount of work I was doing and deals I was helping them close. Needless to say, my promotion came through within 60 days. But I SHOULD NOT have needed to make that call..... I 'only' have 7y work experience and my worse was at the beginning where even before asking for a promotion one of my two managers (I had two because the other is a friend who really wanted to work with me so he would have been partial) more than once joked in a group and in my presence on the fact that women just need to open their legs to have a promotion. He was referring to people he thought were not doing their job right and looking at the company way of doing thing he might have been right that those (and other people) were not good in their jobs, but this demonstrate he wasn't either. Aaand I went to job interviews that I want to forget. Lately I discovered that one thing is very different between men and women in the workforce and it still seem a secret: men ask for raises/promotions continuously while I always feel 'cool, I'm lucky, I have a job, I don't even need a raise!' I'm inspired by you going and fighting for your raise, I recently covered the gap formed of me lowballing myself when I moved country. My gross salary nearly doubled and still I think I should ask for a raise as soon as I am out of probation... Challenges: 1, 2, 3, 4, 5, 6, 7, 8 Link to post It is always disturbing to discover that you as a female have bigger balls than your male boss. Gender aside, it is never good when you have a boss you can't trust to have your back. It's screwed up and sucks that it happens to women in IT and STEM jobs more often than it does to men in any job or to women in nursing or teaching. I don't know how it is in the US, here in Europe males in nursing and teaching have it equally hard (e.g. it's difficult to see a male head nurse), with the equal lack of support outside work. And while there is a conversation and meetups to support women in STEM, I don't see one for men in traditionally feminine jobs. I know a male teacher that was doing his job and talk to a female student on her own because she was misbehaving and said to her that if she didn't change he would have had to report her. Her reply? If you do I will say that you sexually molested me today. Can you imagine?!? He saw his career and life ending... Challenges: 1, 2, 3, 4, 5, 6, 7, 8 Link to post 8 hours ago, Diadhuit said: I 'only' have 7y work experience and my worse was at the beginning where even before asking for a promotion one of my two managers (I had two because the other is a friend who really wanted to work with me so he would have been partial) more than once joked in a group and in my presence on the fact that women just need to open their legs to have a promotion. This - right here - should have been reported to HR immediately. Joke or not, that is wrong and should have been reported. Even if it was through "anonymous" channels. I realize I'm at a point in my career where I would just call up HR, tell them my name, what happened and who was there - because if they DID fire me I'd just sue the shizzle out of them while working for another company w\in 2 weeks (making the same or more$$). But it's this shite that needs to stop.
9 hours ago, scalyfreak said:
It is always disturbing to discover that you as a female have bigger balls than your male boss.
Gender aside, it is never good when you have a boss you can't trust to have your back. It's screwed up and sucks that it happens to women in IT and STEM jobs more often than it does to men in any job or to women in nursing or teaching.
Yep - it's taken a long time to discover I actually had the cojones - but I kind of need them in my line of work. It is not uncommon for me to be the only female in a room of male IT guys I've never met and need to be able to immediately command their respect that I know what the heck I'm doing. That has always been easy for me - but dealing with management? Took me a while to get that to translate.
I've had very few bosses I fully trusted. And the one that left? The absolute best of the bunch. I could literally talk to him about anything and he always had my back. It was almost like losing the father I always wanted - and says something when we literally only worked together 2 years. My current boss will fight for me - as long as there's something in it for him. But I know that so I make sure I'm indispensable and take the time to make sure he closes the big deals. As a result, he and management are looking at creating another level to keep me happy, give me the raise I deserve and keep my hands in the technical area they really need me - and I don't want let go of the tech stuff yet (and may never LOL).
8 hours ago, Diadhuit said:
I don't know how it is in the US, here in Europe males in nursing and teaching have it equally hard (e.g. it's difficult to see a male head nurse), with the equal lack of support outside work.
It's like that here too unfortunately. While I don't have any male friends in nursing or teaching, I do have some female friends in both professions and they see it. All that aside, neither are paid enough for what they do - no matter the gender. It's the administration that makes the $$in both fields.... "Be not afraid of growing slowly; be afraid only of standing still." - Chinese Proverb 1st dozen-ish Challenges for the curious 12,11,10,9,8,7,6,5,4,3,2,More attempts, #1 with Intro, Failed attempts Spoiler Quick Bio: IT Consultant, Been in IT 25+ Years, Bounced around and landed as a traveling Consultant for a medium-sized Software Company. I love to cook & read, I travel for a living (although amount varies widely, sometimes I'm home for weeks, others I'm traveling for weeks on end), and trying to move out of Atlanta (plan in place, working to implement). Link to post 8 hours ago, Diadhuit said: Lately I discovered that one thing is very different between men and women in the workforce and it still seem a secret: men ask for raises/promotions continuously while I always feel 'cool, I'm lucky, I have a job, I don't even need a raise!' I don't think that's a male-female thing, I think it's a personality thing. Whether a person is ambitious, or content to stay in their current position for the rest of their professional life, is a matter of how happy they are in their current job, what kind of personality they have, and most importantly of all, how their immediate superiors treat them. X or Y chromosome doesn't directly impact any of those. 8 hours ago, Diadhuit said: I don't know how it is in the US, here in Europe males in nursing and teaching have it equally hard (e.g. it's difficult to see a male head nurse), with the equal lack of support outside work. And while there is a conversation and meetups to support women in STEM, I don't see one for men in traditionally feminine jobs. I think that's the same everywhere, just because it's human nature to react in a negative way to things that are not how we expect them to be. In the minds of certain people men are not suppose to be naturally nurturing and caring, so a man who wants to be a nurse instead of aiming for the more powerful and prestigious role of a doctor, must have something wrong with him. I'm hoping most of it is generational and will go away as the generations who grew up with the more "traditional" gender roles are replaced with generations who grew up in a more fluctuating and flexible society. Time will tell. “I've always believed that failure is non-existent. What is failure? You go to the end of the season, then you lose the Super Bowl. Is that failing? To most people, maybe. But when you're picking apart why you failed, and now you're learning from that, then is that really failing? I don't think so." - Kobe Bryant, 1978-2020. Rest in peace, great warrior. Personal Challenges, a.k.a.The Saga of Scaly Freak: Link to post 7 minutes ago, Athaclena said: This - right here - should have been reported to HR immediately. Joke or not, that is wrong and should have been reported. Even if it was through "anonymous" channels. I realize I'm at a point in my career where I would just call up HR, tell them my name, what happened and who was there - because if they DID fire me I'd just sue the shizzle out of them while working for another company w\in 2 weeks (making the same or more$$). But it's this shite that needs to stop.
I have female friends in STEM jobs who have been fired for doing so.The managers who fired them made sure to let their friends in the industry know that this woman was unreliable, not a team player, unprofessional, and deliberately tried to cause trouble out of a sense of spite and jealousy of the success of others. They learned their lesson and never called HR over a sexist joke again.
It absolutely needs to stop. And it will, they day the other men in the room are the ones to object to the joke, tell the wanna-be comedian to stop being an ass, and then they are the ones who call HR. When that starts to happen on a regular basis, it will stop.
I think that's another generational thing, or at least I hope it is. That means there is an end in sight.
“I've always believed that failure is non-existent. What is failure? You go to the end of the season, then you lose the Super Bowl. Is that failing? To most people, maybe. But when you're picking apart why you failed, and now you're learning from that, then is that really failing? I don't think so." - Kobe Bryant, 1978-2020. Rest in peace, great warrior.
Personal Challenges, a.k.a.The Saga of Scaly Freak:
I have female friends in STEM jobs who have been fired for doing so.The managers who fired them made sure to let their friends in the industry know that this woman was unreliable, not a team player, unprofessional, and deliberately tried to cause trouble out of a sense of spite and jealousy of the success of others. They learned their lesson and never called HR over a sexist joke again.
It absolutely needs to stop. And it will, they day the other men in the room are the ones to object to the joke, tell the wanna-be comedian to stop being an ass, and then they are the ones who call HR. When that starts to happen on a regular basis, it will stop.
I think that's another generational thing, or at least I hope it is. That means there is an end in sight.
That company didn't have HR. Sexism, bullying, rampant racism against the only one poc were common and left ignored my managers because they were not real managers, they had no training or time allocated. So no wonder that the people who decided they could emigrate went fast. I stayed less than my probation! After that that manager started bullying my other manager and he left in other 6 months. He managed to lose 3-4 stones after leaving only by not being bullied and not severely overworking
Challenges: 1, 2, 3, 4, 5, 6, 7, 8
1 minute ago, Diadhuit said:
That company didn't have HR.
Danger! Danger! RETREAT!!!
Sounds like a horrible place. I hope they went out of business.
“I've always believed that failure is non-existent. What is failure? You go to the end of the season, then you lose the Super Bowl. Is that failing? To most people, maybe. But when you're picking apart why you failed, and now you're learning from that, then is that really failing? I don't think so." - Kobe Bryant, 1978-2020. Rest in peace, great warrior.
Personal Challenges, a.k.a.The Saga of Scaly Freak:
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## Elementary Algebra
$49=7\times7$
The number $49$ is the perfect square of $7$. Since $7$ is a prime number, we can write the prime factorization of $49$ as $7\times7$
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January 2002, Stage 4&5
Problems
Colour in the Square
Stage: 2, 3 and 4 Challenge Level:
Can you put the 25 coloured tiles into the 5 x 5 square so that no column, no row and no diagonal line have tiles of the same colour in them?
Matter of Scale
Stage: 4 Challenge Level:
Prove Pythagoras Theorem using enlargements and scale factors.
Get Cross
Stage: 4 Challenge Level:
A white cross is placed symmetrically in a red disc with the central square of side length sqrt 2 and the arms of the cross of length 1 unit. What is the area of the disc still showing?
Deep Roots
Stage: 4 Challenge Level:
Find integer solutions to: $\sqrt{a+b\sqrt{x}} + \sqrt{c+d.\sqrt{x}}=1$
Sixational
Stage: 4 and 5 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
Proof Sorter - Sum of an AP
Stage: 5 Challenge Level:
Use this interactivity to sort out the steps of the proof of the formula for the sum of an arithmetic series. The 'thermometer' will tell you how you are doing
Big, Bigger, Biggest
Stage: 5 Challenge Level:
Which is the biggest and which the smallest of $2000^{2002}, 2001^{2001} \text{and } 2002^{2000}$?
In Between
Stage: 5 Challenge Level:
Can you find the solution to this algebraic inequality?
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Computer Forums i found this interesting
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01-08-2006, 06:20 PM #1 Fully Optimized Join Date: Jun 2005 Posts: 2,102 i found this interesting i was writing my name and date on a HW paper and realized it was 1-6-6...well that got me thinking...on june,6 this year the date will read 666...odd? that + the year started on Sunday. now mebbie this is some kinda sign...mebbie not...i just found it interesting __________________
01-08-2006, 07:12 PM #2 BSOD Join Date: Jan 2005 Posts: 1,386 Re: i found this interesting yea one of my friends pointed that out... god i hope the world doesnt end just a mere 4 days before my bday/// that would suck __________________
01-08-2006, 08:51 PM #3 BSOD Join Date: Sep 2005 Posts: 2,519 Re: i found this interesting Interesting indeed.
01-08-2006, 08:52 PM #4 In Runtime Join Date: Jan 2006 Posts: 124 Re: i found this interesting i actually told my mom that maybe on 6.6.06 something that involves a bomb, or airplane, or the soviet union will happen and she was like....*WTH are you smokin* B.L.A.c.K
01-08-2006, 11:22 PM #5 BSOD Join Date: Sep 2005 Posts: 2,519 Re: i found this interesting Haha Black. That's funny. Bad things happen everyear on that day, just noone takes notice unless it's liek HUUUUGE or something. Nothing new.
01-08-2006, 11:44 PM #6 Golden Master Join Date: Jul 2005 Posts: 7,459 Re: i found this interesting maybe the AntiChrist is born? Anyone familiar with the four horsemen of Apocalypse? Well there are many interpretations but I think that the most common one is: 1st horseman: AntiChrist, who came to conquer (6.6.06?) 2nd horseman: War (WW III? something to do with the AntiChrist?) 3rd horseman: famine (because of the nukes & biological & chemical weapons used in the WW III?) 4rd horseman: Death This happening? __________________ 0_o
01-08-2006, 11:56 PM #7
In Runtime
Join Date: Jan 2006
Posts: 124
Re: i found this interesting
Quote:
Originally Posted by mammikoura maybe the AntiChrist is born? Anyone familiar with the four horsemen of Apocalypse? Well there are many interpretations but I think that the most common one is: 1st horseman: AntiChrist, who came to conquer (6.6.06?) 2nd horseman: War (WW III? something to do with the AntiChrist?) 3rd horseman: famine (because of the nukes & biological & chemical weapons used in the WW III?) 4rd horseman: Death This happening?
not likely....not saying that it WONT happen but...the first horseMAN isnt even 5...the second horsemen starting wwIII cause of a baby???....3rd horseman nvr really came here....and the 4th horsemen travels says hi to people across the world every day.
B.L.A.c.K
01-09-2006, 12:35 AM #8 BSOD Join Date: Sep 2005 Posts: 2,519 Re: i found this interesting Ya'll need to read the Bible. Just check out Revalation. Read it, it's good, plus it gives you an idea of what is to come. But a note while reading it, is that Paul is describing what is going to happ[en, and through what he saw, so just take a look at it differently from Paul's perspective: A Guy who has no idea of what a jet it, what a mass city is, or anythgin "modern." So yeah, I would suggest reading the book of Revalation if you want an idea of what is to come.
01-09-2006, 01:53 AM #9 Daemon Poster Join Date: Sep 2005 Posts: 1,176 Re: i found this interesting stuff like this spooks me out a little __________________ IBM - Idiots Become Managers
01-09-2006, 01:58 AM #10
Daemon Poster
Join Date: Dec 2005
Posts: 585
Re: i found this interesting
Quote:
Originally Posted by mr.monger yea one of my friends pointed that out... god i hope the world doesnt end just a mere 4 days before my bday/// that would suck
That would be terrible, I too was born then... And it would really suck.
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# Extract Weyl curvature spinor
Eq. (27) in http://arxiv.org/abs/1110.2662 says I can construct the Weyl spinor according to
$$\Psi_{ABCD} = \frac 14 C{}_{\mu\nu\lambda\rho} \left( \sigma^\mu \right){}_A{}^{\dot A} \left( \sigma^\nu \right){}_{B\dot A}\left( \sigma^\lambda \right){}_C{}^{\dot C}\left( \sigma^\rho \right){}_{D\dot C}\tag{27}$$
I understand that the partial contraction in the two index pairs $\mu\leftrightarrow\nu$ and $\lambda\leftrightarrow\rho$ induces a unique way to extract either two left-handed or two right-handed spinor indices. Moreover, the above spinor is completely symmetrical, $\Psi{}_{ABCD} = \Psi{}_{(ABCD)}$ as it must to encompass all ten components of the Weyl tensor.
But why does the above work?
$C{}_{ABCD\dot A\dot B\dot C\dot D} = C{}_{\mu\nu\lambda\rho}\left(\sigma{}^\mu\right){}_{A\dot A} \left(\sigma{}^\nu\right){}_{B\dot B}\left(\sigma{}^\lambda\right){}_{C\dot C}\left(\sigma{}^\rho\right){}_{D\dot D} = \Psi{}_{ABCD}\epsilon{}_{\dot A\dot B} \epsilon{}_{\dot C \dot D} + \Psi{}_{\dot A\dot B\dot C\dot D}\epsilon{}_{AB}\epsilon{}_{CD}$
on both sides with $\epsilon{}^{\dot A\dot B}\epsilon{}^{\dot C\dot D}$ and using that $\Psi{}_{\dot A\dot B\dot C\dot D} = \Psi{}_{(\dot A\dot B\dot C\dot D)}$ as well as $\epsilon{}_{\dot A\dot B}\epsilon{}^{\dot A\dot B} = 2$ gives the above result.
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# Filling up holes in the sand
The function $$\log_{\cdot}(\cdot): \mathbb{Q}^{+} \times \mathbb{Q}^{+} \longrightarrow \mathbb{R}$$. While the codomain is the real numbers $$\mathbb{R}$$, can the image be $$\mathbb{R}$$?
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Funktsional. Anal. i Prilozhen.: Year: Volume: Issue: Page: Find
Central Elements of the Elliptic Yang–Baxter Algebra at Roots of UnityA. A. Belavin, M. Jimbo 1 Weakly Outer Inner FunctionsE. Doubtsov 7 Mathematical Aspects of Weakly Nonideal Bose and Fermi Gases on a Crystal BaseV. P. Maslov 16 The Liouville Canonical Form for Compatible Nonlocal Poisson Brackets of Hydrodynamic Type and Integrable HierarchiesO. I. Mokhov 28 On the Commutativity of Weakly Commutative Riemannian Homogeneous SpacesL. G. Rybnikov 41 Resolution of Corank $1$ Singularities of a Generic FrontV. D. Sedykh 52 On the Monodromy of a Multivalued Function Along Its Ramification LocusA. G. Khovanskii 65 On Functionals Bounded BelowW. Zou, M. Schechter 75 Free Algebras of Automorphic Forms on the Upper Half-PlaneO. V. Schwarzman 81 Brief communications Spectral Components of Operators with Spectrum on a CurveA. S. Tikhonov 90 Preduals of von Neumann AlgebrasA. I. Shtern 92
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Baryonic dark matter
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This image shows the galaxy cluster Abell 1689, with the mass distribution of the dark matter in the gravitational lens overlaid (in purple). The mass in this lens is made up partly of normal (baryonic) matter and partly of dark matter. Distorted galaxies are clearly visible around the edges of the gravitational lens. The appearance of these distorted galaxies depends on the distribution of matter in the lens and on the relative geometry of the lens and the distant galaxies, as well as on the effect of dark energy on the geometry of the Universe.
In astronomy and cosmology, baryonic dark matter is dark matter (matter that is undetectable by its emitted radiation, but whose presence can be inferred from gravitational effects on visible matter) composed of baryons, i.e. protons and neutrons and combinations of these, such as non-emitting ordinary atoms. Candidates for baryonic dark matter include non-luminous gas, Massive Astrophysical Compact Halo Objects (MACHOs: condensed objects such as black holes, neutron stars, white dwarfs, very faint stars, or non-luminous objects like planets), and brown dwarfs.
The total amount of baryonic dark matter can be inferred from Big Bang nucleosynthesis, and observations of the cosmic microwave background. Both indicate that the amount of baryonic dark matter is much smaller than the total amount of dark matter.
In the case of big bang nucleosynthesis, the problem is that large amounts of ordinary matter means a denser early universe, more efficient conversion of matter to helium-4 and less unburned deuterium that can remain. If one assumes that all of the dark matter in the universe consists of baryons, then there is far too much deuterium in the universe. This could be resolved if there were some means of generating deuterium, but large efforts in the 1970s failed to come up with plausible mechanisms for this to occur. For instance, MACHOs, which include, for example, brown dwarfs (balls of hydrogen and helium with masses $< 0.08M_\odot$), never begin nuclear fusion of hydrogen, but they do burn deuterium. Other possibilities that were examined include "Jupiters", which are similar to brown dwarfs but have masses $\sim 0.001M_\odot$ and do not burn anything, and white dwarfs.[1][2][clarification needed]
References
1. ^ G. Jungman, M. Kamionkowski, and K. Griest, Phys. Rep. 267, 195 (1996)
2. ^ M. S. Turner, arXiv:astro-ph/9904051 (1999)
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As I wrote in my last update, since kicking off Project Electron in September 2016, we’ve been gathering information through conversations, surveys and a literature review, and then structuring that information into user stories and personas. In line with our “open by default” licensing principle, we’re making these design artifacts available with a CC0 license, which means you can take them and use them freely in your own local environments.
Although the end goal of this discovery phase was to have these deliverables in hand, the process by which we developed them is worth explaining.
# How we got here
We started out by listening. We listened - literally - to representatives from our donor and depositor organizations in a series of in-person interviews where we asked about their current recordkeeping practices for digital records, talked through pain points, and imagined ways in which we could acquire, preserve and make accessible their digital records both now and in the future. With help from our Marist College IT partners, we surveyed a wide range of archivists, librarians and museum professionals, as well as our researchers. Because others have mapped the needs of some of these user groups before, we also conducted a survey of existing literature and user studies. Last, and certainly not least, we listened to our staff through conversations, observation and process analysis.
## User Stories
Out of all of this information we created a set of user stories, concise descriptions of the Who, What, When, Where, How and Why of a specific user need that can be addressed by a software system. A typical example might read as follows:
"As an archivist, I want to create an audit trail of events associated with a group of digital records so that I can ensure their authenticity."
As you can see, there’s a lot of information packed into this sentence - a specific user, what that user wants to do, and why they want to do it - which means that even though these are really brief statements it takes a fair amount of work to write them. Still, this is an incredibly worthwhile exercise because it requires specificity and connection to actual people and real needs, as opposed to one’s own ideas about what might be useful system functionality.
## Card Sort
After developing these user stories, a group of RAC staff did a card sorting exercise to group these individual user stories together to form personas, and then filled out very basic persona templates with sections for background, goals and motivations, needs, pain points and technology profile.
This was the first time I’ve done this kind of exercise, and it was very successful both in terms of the final product as well as getting staff involved and invested. We got a ton of really useful information in the persona templates, and even at this early stage, the power of personas in a design process was revealed. At the end of the session one of the participants asked “What happens to these people now? We care about them!”
Still, I think there were things that we could have done better. Most notably, I wish I’d paid more attention to the “Best Practices for Card Sorts” listed on this page. In some cases, we asked participants to sort too many user stories, and we also discovered that providing clear time parameters was a useful way of keeping them from floundering too much.
## Personas
After some analysis of these groups of user stories, we came up with three sets of user stories, each of which contained four personas.
• Donors and Depositors cover the user stories gathered from our donor and depositor organizations. We tried to capture a number of relevant roles within these organizations, as well as a range of organization types, sizes, IT support, and approaches to records management.
• Researchers were developed from user stories gathered from RAC researchers, as well as other user studies and literature on researcher needs. In these personas, we wanted to make sure we represented the variety of research interests, methodologies and experience levels of our researchers.
• Information Professionals grew out of a combination of user stories from RAC staff, other information professionals we surveyed, and existing user studies. Although we originally started out thinking about the needs of our staff as separate from the needs of information professionals in other archives, libraries and museums, we soon realized that the two groups had a lot in common, so we simply created one set of user personas to address all those user stories.
The important thing to emphasize about this process of creating personas is that it’s very much a creative and interpretive one. Personas have been called “unscientific,” but my experience was that the subjective nature of this process forced me to engage with my own biases and privileges, rather than hiding behind false ideas of data-driven neutrality.
In addition to creating the personas, we’ll also create a set of requirements to help guide the development process. These will look something like:
"Maintain audit trail of preservation events."
As you can see, these requirements are structured differently than either user stories or personas and are oriented primarily towards developers. We’ll use the personas to help us prioritize these requirements and the requirements to test system functionality, so that we know whether or not we’ve met the needs expressed in our user stories.
# Why It Matters to You
I’ve written about our work in some detail because I feel strongly that this kind of user-driven process is something that archivists can and should be doing a lot more of. Recently, there’s been a lot of emphasis in the archival profession on archivists learning hard technical skills like writing code. While it’s never a bad idea to have a robust and diverse professional toolbox, this emphasis has, I think, created an unfortunate divide in the profession between archivists who are “tech-savvy” and those who aren’t. In my estimation, this divide has hindered the development of a common language to talk about technology and the people it serves in the context of archival principles.
User-centered design processes, such as the ones described above, present a common language to bridge that gap. You don’t need to write code to do this work (and you don’t need to use exactly the same processes as we did), you just need to listen, think critically and empathetically, and write clearly. Those are all things any good archivist should be equipped to do, and do well.
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# Find variation - really need help
1. Oct 25, 2006
### donjt81
Find variation - really need help!!!
guys i really need your help on this. I dont even know which section to look for here. can some one please get me started for this problem.
Problem: A manufacturer contracts to mint coins for the federal government. how much variation dr in the radius of the coins can be tolerated if the coins are to weigh within 1/50 of their ideal weight? Assume the thickness does not vary.
2. Oct 26, 2006
### xman
I do not know if this is correct, but what I'm thinking is:
1) assume the material is homogeneous and isotropic
2) assume the thickness is constant
3) assume the volume may be written as the product of the area of a circle and its constant height
Then, we can write the mass as the usual density, i.e. $$m=\rho V$$ then the weight is just g times the mass. So, note that
$$dm = \rho dV=\rho (2\pi r dr) t$$
where t is the constant thickness. So, if we compare to its ideal weight, note the constant g drops out, along with a lot of other stuff. So
$$\frac{dm}{m} = \frac{\rho dV}{\rho V}=\frac{2dr}{r}$$
So, plugging in what we know we have
$$\frac{1}{50} =\frac{2dr}{r} \Rightarrow \frac{dr}{r}=\frac{1}{100}$$
So the variation in the radius must be within 1/100. Does this make sense? I think it works.
3. Oct 27, 2006
### donjt81
Thanks xman... yes it looks correct...
can anyone else look at it and confirm it? I want to make sure i do the hw right.
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Next: Piston Velocity Up: Moving Shock into Stationary Previous: Moving Shock into Stationary Index
### General Velocities Issues
When a valve or membrane is suddenly opened, a shock is created and propagates downstream. With the exception of close proximity to the valve, the shock moves in a constant velocity (5.12(a)). Using a coordinates system which moves with the shock results in a stationary shock and the flow is moving to the left see Figure (5.12(b)). The upstream'' will be on the right (see Figure (5.12(b))).
stationary coordinates moving coordinates
Similar definitions of the right side and the left side of the shock Mach numbers can be utilized. It has to be noted that the upstream'' and downstream'' are the reverse from the previous case. The upstream'' Mach number is
The downstream'' Mach number is
Note that in this case the stagnation temperature in stationary coordinates changes (as in the previous case) whereas the thermal energy (due to pressure difference) is converted into velocity. The stagnation temperature (of moving coordinates) is
A similar rearrangement to the previous case results in
The same question that was prominent in the previous case appears now, what will be the shock velocity for a given upstream Mach number? Again, the relationship between the two sides is
Since Msx can be represented by Msy theoretically equation (5.63) can be solved. It is common practice to solve this equation by numerical methods. One such methods is successive substitutions.'' This method is applied by the following algorithm:
1. Assume that Mx = 1.0.
2. \label{shock:item:openValve} Calculate the Mach number My by utilizing the tables or Potto--GDC.
3. Utilizing
Mx = sqrt(Ty / Tx) ( My + My′ )
calculate the new improved'' Mx.
4. Check the new and improved Mx against the old one. If it is satisfactory, stop or return to stage \eqref{shock:item:openValve}.
To illustrate the convergence of the procedure, consider the case of My′ =0.3 and My′ =1.3. The results show that the convergence occurs very rapidly (see Figure (5.13)). The larger the value of My, the larger number of the iterations required to achieve the same accuracy. Yet, for most practical purposes, sufficient results can be achieved after 3-4 iterations.
Subsections
Next: Piston Velocity Up: Moving Shock into Stationary Previous: Moving Shock into Stationary Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21
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# zbMATH — the first resource for mathematics
Oscillation criteria for nonlinear inhomogeneous hyperbolic equations with distributed deviating arguments. (English) Zbl 0852.35009
We consider the following nonlinear inhomogeneous hyperbolic equation with distributed deviating arguments: \begin{aligned} {\partial^2\over \partial t^2} [u+ \lambda(t) u(x, t- \tau)] & + \int^b_a p(x, t, \xi) F(u(x, g[t, \xi])) d\sigma(\xi)\tag{E}\\ & = a(t) \Delta u+ f(x, t),\quad (x, t)\in G,\end{aligned} where $$G= \Omega\times \mathbb{R}_+$$, $$\Omega$$ is a bounded domain in $$\mathbb{R}^n$$ with piecewise smooth boundary $$\partial \Omega$$, $$\mathbb{R}_+= [0, + \infty)$$, $$p\in C[\overline\Omega\times \mathbb{R}_+\times J, \mathbb{R}_+]$$, $$J= [a, b]$$, $$F\in C[\mathbb{R}, \mathbb{R}]$$, $$a\in C[\mathbb{R}_+, \mathbb{R}_+]$$, $$\lambda\in C^2[\mathbb{R}_+, \mathbb{R}]$$, $$\tau$$ is a constant, $$f\in C[\Omega\times \mathbb{R}_+, \mathbb{R}]$$, $$g\in C[\mathbb{R}_+\times J, \mathbb{R}]$$, $$\sigma\in [J, \mathbb{R}]$$, and the integral in (E) is a Stieltjes integral. Throughout this paper we assume that $$g(t, \xi)$$ is nondecreasing in $$t$$ and $$\xi$$ respectively, with $$g(t, \xi)< t$$ for any $$\xi$$ and $$\lim_{t\to +\infty} \inf_{\xi\in J} g(t, \xi)= + \infty$$, and that $$\sigma(\xi)$$ is nondecreasing in $$\xi$$. We consider two kinds of boundary conditions: ${\partial u\over \partial N}+ \gamma(x, t) u= \mu(x, t),\quad (x, t)\in \partial \Omega\times \mathbb{R}_+,\tag{B1}$ and $u= \phi(x, t),\quad (x, t)\in \partial \Omega\times \mathbb{R}_+,\tag{B2}$ where $$N$$ is the unit outnormal vector to $$\partial \Omega$$, $$\gamma\in C[\partial \Omega\times \mathbb{R}_+, \mathbb{R}_+]$$, $$\mu, \phi\in C[\partial \Omega\times \mathbb{R}_+, \mathbb{R}]$$.
The objective of this paper is to study the oscillatory properties of solutions of equation (E) subject to boundary conditions (B1) and (B2).
##### MSC:
35B05 Oscillation, zeros of solutions, mean value theorems, etc. in context of PDEs 35R10 Partial functional-differential equations 35K40 Second-order parabolic systems 35L70 Second-order nonlinear hyperbolic equations
Full Text:
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# Error When Saving Page - $webwork.htmlEncode($page.title)
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## Symptoms
The page is not correctly saved, and it prevents users from editing or removing it. When looking for the page in Confluence, its title appears as $webwork.htmlEncode($page.title). When looking for the page on the database, the title value is empty.
## Cause
The page was not saved correctly and it seems that there is something wrong with the record holding the page data, in the database. This can happen due to a variety of different macro combinations.
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## Chemistry (7th Edition)
The identity of $Y$ is the element Calcium, represented by $Ca$.
1. Convert that number of atoms to moles: $2.26 \times 10^{22}$ atoms $\times \frac{1 mol}{6.022 \times 10^{23} atoms} = 0.0375$ mol. 2. Find the molar mass. $\frac{1.50g}{0.0375mol} = 40.0g/mol$ Since the molar mass is equal to the atomic weight, find the element in the periodic table with this value for its atomic weight. It is the Calcium (Ca).
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# What is the integration of x sin inverse x dx ?
## Solution :
We have, I = $$\int$$ $$x sin^{-1} x$$ dx
By using integration by parts formula,
I = $$sin^{-1} x$$ $$x^2\over 2$$ – $$\int$$ $$1\over \sqrt{1 – x^2}$$ $$\times$$ $$x^2\over 2$$ dx
I = $$x^2\over 2$$ $$sin^{-1} x$$ + $$1\over 2$$ $$\int$$ $$-x^2\over \sqrt{1 – x^2}$$ dx
= $$x^2\over 2$$ $$sin^{-1} x$$ + $$1\over 2$$ $$\int$$ $$1 – x^2 – 1\over \sqrt{1 – x^2}$$ dx
= $$x^2\over 2$$ $$sin^{-1} x$$ + $$1\over 2$$ { $$\int$$ $$1 – x^2\over \sqrt{1 – x^2}$$ – $$\int$$ $$1\over \sqrt{1 -x^2}$$ } dx
$$\implies$$ I = $$x^2\over 2$$ $$sin^{-1} x$$ + $$1\over 2$$ { $$\int$$ $$\sqrt{1 – x^2}$$ – $$\int$$ $$1\over \sqrt{1 -x^2}$$ } dx
By using integration formula of $$\sqrt{a^2 – x^2}$$,
$$\implies$$ I = $$x^2\over 2$$ $$sin^{-1} x$$ + $$1\over 2$$ [{ $$1\over 2$$ $$x\sqrt{1 – x^2}$$ – $$1\over 2$$ $$sin^{-1} x$$ } – $$sin^{-1} x$$ ] + C
$$\implies$$ I = $$x^2\over 2$$ $$sin^{-1} x$$ + $$1\over 4$$ $$x\sqrt{1 – x^2}$$ – $$1\over 4$$ $$sin^{-1} x$$ + C
### Similar Questions
What is the integration of sin inverse root x ?
What is the integration of sin inverse x whole square ?
What is integration of sin inverse cos x ?
What is the integration of tan inverse root x ?
What is the integration of x tan inverse x dx ?
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# Efficiently Correcting Matrix Products
Efficiently Correcting Matrix Products We study the problem of efficiently correcting an erroneous product of two $$n\times n$$ n × n matrices over a ring. Among other things, we provide a randomized algorithm for correcting a matrix product with at most k erroneous entries running in $${\tilde{O}}(n^2+kn)$$ O ~ ( n 2 + k n ) time and a deterministic $${\tilde{O}}(kn^2)$$ O ~ ( k n 2 ) -time algorithm for this problem (where the notation $${\tilde{O}}$$ O ~ suppresses polylogarithmic terms in n and k). http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Algorithmica Springer Journals
# Efficiently Correcting Matrix Products
, Volume 79 (2) – Aug 22, 2016
16 pages
/lp/springer_journal/efficiently-correcting-matrix-products-HxfR5CbLIp
Publisher
Springer US
Subject
Computer Science; Algorithm Analysis and Problem Complexity; Theory of Computation; Mathematics of Computing; Algorithms; Computer Systems Organization and Communication Networks; Data Structures, Cryptology and Information Theory
ISSN
0178-4617
eISSN
1432-0541
D.O.I.
10.1007/s00453-016-0202-3
Publisher site
See Article on Publisher Site
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# Intuitive explanation for why the definite integral gives the area between the function and the x-axis
Could somebody please give an intuitive explanation for why the antiderivative of a function evaluated at $b$ minus the antiderivative of the function evaluated at $a$, where $b>a$, gives the area between the function and the $x$-axis between these two $x$ values.
It does not make much sense to me, could somebody please give an intuitive proof or intuitive explanation
• see for example (drcruzan.com/FTOC.html) – Jean Marie Apr 20 '17 at 21:13
• Thankyou but for the proof of the first fundamental theorem of calculus, the area under the curve is equal to F(b) - F(a) whether or not the limit of delta x tending to 1 is evaluated, this seems a little bit odd to me because this would mean that the number of rectangles or little areas that are used in the summation is irrelevant. If 4 rectangles were added it would give the same sum as if infinite rectangles were added together. Am I going wrong somewhere, I have probably misunderstood the proof? – Nav Hari Apr 20 '17 at 21:39
• For another post, you might check out this thing that I wrote for my students once upon a time. The relevant bit to you is section 1.3, and especially the content after definition 4. – davidlowryduda Apr 20 '17 at 23:36
• Have you looked at any of the related questions in the list to the right? At least two of them directly address your question. – amd Apr 20 '17 at 23:49
Let's assume we have a velocity function $v(t)=10$. The integral of velocity is position. The integral is the sums of the y-values (velocity) at infinitesimally small $t$ intervals between $a$ and $b$.
Let's do an integral from $a=2$ to $b=4$:
$$\int_2^4{10\mathrm{d}t}\\ =x(t)\Big{|}_2^4\\ =5t\Big{|}_2^4\\ =(5\times4)-(5\times2)\\ =10$$
What does that 10 mean? It represents the change in position BETWEEN two limits $a$ and $b$. The reason that is important is that your definite integral with limits $a,b$ is essentially integral of v(t) from a to b equals antiderivative at b minus antiderivative at a.
Let $x=\int{v(t)}\mathrm{d}t=5t$ be our position function. If you evaluate your antiderivative at $b$, then you are integrating the entire domain of the function up to $b$... for our position and velocity functions, the lower end of the domain is the beginning of time!
When you subtract the antiderivative evaluated at $a$, you chop off everything that came before $a$, so you're only calculating how much your position changed over your particular interval.
Chop up the interval $[a,b]$ into tiny subintervals $[x_i,x_{i+1}]$. Clearly the total change $f(b) - f(a)$ is equal to the sum of all the little changes $f(x_{i+1}) - f(x_i)$. But, $f(x_{i+1}) - f(x_i) \approx f'(x_i) (x_{i+1} - x_i)$. Thus, $f(b) - f(a) \approx \sum_i f'(x_i)(x_{i+1} - x_i) \approx \int_a^b f'(x) \, dx$. When we chop up $[a,b]$ more and more finely, the approximations get better and better, so by a limiting argument we discover that $f(b) - f(a) = \int_a^b f'(x) \, dx$.
• From this the area would be equal to the definite integral whatever the subintervals are. the area should surely be different if the number of rectangles changes. Surely, the the area should only be equal to the definite integral when the number of rectangles tends to infinity and the width's of the rectangles tend to 0. – Nav Hari Apr 22 '17 at 20:35
• @NavHari But notice that I have only approximate equalities, not strict equalities, throughout the argument. When the subintervals are extremely tiny, it seems plausible to expect (or at least hope) that the approximations are very good. We have shown that $f(b) - f(a) \approx \int_a^b f'(x) \,dx$, but it seems plausible that the approximation can be made as close as we like by using extremely tiny subintervals. It follows from this "limiting argument" (which could be made rigorous, with additional work) that we actually have $f(b) - f(a) = \int_a^b f'(x) \, dx$ (with exact equality). – littleO Apr 22 '17 at 22:51
• But even if the subintervals are not of width dx, the area would still equal F(b) - F(a). This surely means that the area of each rectangle of width dx is equal to F'(c)dx, with no approximation regardless of the limiting argument. Is there any reason for why this must be true, as it does not seem very intuitive. – Nav Hari Apr 23 '17 at 17:22
• @NavHari Are you invoking the mean value theorem to say that there exists a number $c_i \in (x_i,x_{i+1})$ such that $f(x_{i+1}) - f(x_i) = f'(c_i)(x_{i+1} - x_i)$ with exact equality? You are correct that we can do this, and that is an important step towards making our intuitive argument into a rigorous proof. The standard proof of the fundamental theorem of calculus does exactly this. I don't find it to be unintuitive, though, because the mean value theorem is itself an intuitive theorem. – littleO Apr 24 '17 at 5:40
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## I suggest strwidth() [Off Topic]
Hi ElMaestro,
❝ I have a dynamic text s. I am going to put it on a plot. But I need to know if s at a given set of parameters and setting fits onto the plot within a width of dx units on the horizontal axis as one line.
I see. You create the plot first and then strwidth(s) and strheight(s) will give you what you need.
Dif-tor heh smusma 🖖🏼 Довге життя Україна!
Helmut Schütz
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# Is there a Galois correspondence for ring extensions?
Given an ring extension of a (commutative with unit) ring, Is it possible to give a "good" notion of "degree of the extension"?. By "good", I am thinking in a degree which allow us, for instance, to define finite ring extensions and generalize in some way the Galois' correspondence between field extensions and subgroups of Galois' group.
I suppose one can call a ring extension $A\subset B\$ finite if $B$ is finitely generated as an $A$-module, and the degree would be the minimal number of generators, but is that notion enough to state a correspondence theorem?
-
There's even a Galois theory of schemes, namely, the fundamental group of a scheme classifies the finite etale coverings of the scheme. When the scheme is affine, this becomes a Galois theory of rings. When the scheme is the spec of a field, it becomes classical Galois theory. The theory goes back to Grothendieck's seminar SGA1 from the early 1960s. – mephisto May 3 '11 at 0:09
Jacobson (1956) discusses Galois theory of rings of linear transformations. See www-history.mcs.st-and.ac.uk/Extras/… – William DeMeo May 3 '11 at 8:07
There is indeed a theory of Galois extension of rings. See, for example, the very nice paper [Chase, S. U.; Harrison, D. K.; Rosenberg, Alex. Galois theory and Galois cohomology of commutative rings. Mem. Amer. Math. Soc. No. 52 1965 15--33. MR0195922 (33 #4118)] The theory developed there does include a Galois correspondence.
There is even a Hopf-Galois theory, where the Galois group is replaced by a Hopf algebra (co)acting on the big ring, for extra fun---the correspondence in this case, though, is quite more delicate/complicated.
-
In addition to the above references, I would like to mention some non-commutative extensions of the Galois theory. See
P. M. Cohn, Skew Fields, Cambridge University Press, 1995
for the Galois theory of skew fields. Extensions to some classes of noncommutative rings are given in the book
V. K. Kharchenko, Noncommutative Galois theory, Novosibirsk, 1996,
available only in Russian, and many papers of its author, some of which exist also in the English translation.
-
The Hopf extension (and non-comm.) is discussed in S. Montgomery's Hopf algebras and their actions on rings. – Mariano Suárez-Alvarez May 3 '11 at 17:12
Since I do not read russian, this might be the right place where to ask: are you aware of a generalization of the norm in a Galois extension of noncommutative algebras? Of course, if one treats central simple algebras, there is a reduced norm, but I am in a more general setting: $k$ a field, $B/A$ a finite (Galois?) extension of noncommutative $k$-algebras; and would like something like $\mathrm{Norm}_{B/A}\colon B^\times\to A^\times$. – Filippo Alberto Edoardo Jan 23 at 8:36
For a "survey" of Galois theory of commutative rings, there is one book:
The Separable Galois Theory Commutative Rings by Andy R. Magid (1974).
which has a nice section summarizing the state of the development up to 1974.
There is also a more general book aiming at a topos-theory style general Galois theory (although I haven't read it) including also a nice survey:
Galois Theories by Francis Borceux and George Janelidze (2001).
-
and SGA1 as well as Lenstra's notes
-
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# Safely checking transaction origin account
I have the following scenario.
Contract A {
mapping (address => bool) public allowed;
mapping (address => uint) public userData;
if(allowed[_user]){
userData[_user]++;
return true;
} else
return false;
}
}
Contract B {
function doSomethingWithA(){
if(contractA.doSomething(msg.sender))
// Do something
else
// don't
}
}
Contract A holds a mapping of which users are allowed to interact with it. Contract A has a function that should only be called by contract B.
ONLY the EOA interacting with B should be able to execute A's doSomething().
In the current scenario, anyone could call doSomething(address _user) and pass an allowed _user to get a valid result.
I could require(msg.sender == _user); in doSomething(address _user) but that would cause the function to fail as the msg.sender would always be contract B.
I know I could use tx.origin instead of msg.sender, but I was wondering if that would compromise the security of the contract.
So, my questions are:
1. Is there any other way to solve this without using tx.origin?
2. If the only way is with tx.origin. What should I have in mind in order to prevent an attack related to the usage of tx.origin?
ONLY the EOA interacting with B should be able to get true from checkIfAllowed(address _user).
Why? What difference does it make?
In the current scenario, anyone could call checkIfAllowed(address _user) and pass an allowed _user to get a valid result.
This is true, but I don't see why it's a problem. What's the attack you're trying to prevent? Everything on the blockchain is public, so anyone can always find out which users A will allow. So arbitrary accounts calling checkIfAllowed doesn't introduce any sort of security flaw.
From your simplified example, I assume the intent is that B doesn't let doSomething succeed unless the account calling into B is "allowed" by A. The current code appears to have that intended effect.
If there's some other goal, please elaborate on what it is.
• I just updated the question and modified the code to make my intentions more clear. I want A's doSomething() to modify some state variables only if the caller EOA is allowed to do so. BUT, this function is not to be called directly from Contract A, it should only be called by another contract (In this case, Contract B). – pabloruiz55 Dec 21 '17 at 18:33
• I know the information is public, that is not an issue. The issue is that I want to prevent someone other than the EOA calling the function to modify the state, but that is done by first passing through contract B and in that case msg.sender will always be contract B. – pabloruiz55 Dec 21 '17 at 18:34
• What's special about contract B? Why not just (in contract A) require(msg.sender == addressForContractB);? – smarx Dec 21 '17 at 18:37
• I'm essentially pushing back on the idea of "This should only happen if the EOA that started the transaction is X." tx.origin does that, but it's pretty much never really what you want. (For example, a malicious contract called by X could in turn call into A. X can mitigate that by reading the contract code he's calling into, but that's a somewhat weak mitigation.) If you want only X to be able to do something, let X call into contract A directly and then call into contract B. – smarx Dec 21 '17 at 18:40
• Imagine Contract A is a contract that generates promo codes. a promo code is generated by Contract A to be used only by a certain address A specifies. Contract B is a store that wants to implement promo codes. So it would receive the call from the account, pass it to contract A, which will check if the user has been given allowed to use the promo code and if yes, it will mark it as used and return OK to the Store contract. The user should never interact with contract A, but with the Store that implements such promo code. – pabloruiz55 Dec 21 '17 at 18:43
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Outlook: Oracle Corporation Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Hold
Time series to forecast n: 03 Feb 2023 for (n+8 weeks)
Methodology : Modular Neural Network (Speculative Sentiment Analysis)
## Abstract
Oracle Corporation Common Stock prediction model is evaluated with Modular Neural Network (Speculative Sentiment Analysis) and Statistical Hypothesis Testing1,2,3,4 and it is concluded that the ORCL stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Hold
## Key Points
1. Decision Making
2. Which neural network is best for prediction?
3. Technical Analysis with Algorithmic Trading
## ORCL Target Price Prediction Modeling Methodology
We consider Oracle Corporation Common Stock Decision Process with Modular Neural Network (Speculative Sentiment Analysis) where A is the set of discrete actions of ORCL stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Statistical Hypothesis Testing)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Speculative Sentiment Analysis)) X S(n):→ (n+8 weeks) $\begin{array}{l}\int {r}^{s}\mathrm{rs}\end{array}$
n:Time series to forecast
p:Price signals of ORCL stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## ORCL Stock Forecast (Buy or Sell) for (n+8 weeks)
Sample Set: Neural Network
Stock/Index: ORCL Oracle Corporation Common Stock
Time series to forecast n: 03 Feb 2023 for (n+8 weeks)
According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Hold
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for Oracle Corporation Common Stock
1. The business model may be to hold assets to collect contractual cash flows even if the entity sells financial assets when there is an increase in the assets' credit risk. To determine whether there has been an increase in the assets' credit risk, the entity considers reasonable and supportable information, including forward looking information. Irrespective of their frequency and value, sales due to an increase in the assets' credit risk are not inconsistent with a business model whose objective is to hold financial assets to collect contractual cash flows because the credit quality of financial assets is relevant to the entity's ability to collect contractual cash flows. Credit risk management activities that are aimed at minimising potential credit losses due to credit deterioration are integral to such a business model. Selling a financial asset because it no longer meets the credit criteria specified in the entity's documented investment policy is an example of a sale that has occurred due to an increase in credit risk. However, in the absence of such a policy, the entity may demonstrate in other ways that the sale occurred due to an increase in credit risk.
2. If there is a hedging relationship between a non-derivative monetary asset and a non-derivative monetary liability, changes in the foreign currency component of those financial instruments are presented in profit or loss.
3. An equity method investment cannot be a hedged item in a fair value hedge. This is because the equity method recognises in profit or loss the investor's share of the investee's profit or loss, instead of changes in the investment's fair value. For a similar reason, an investment in a consolidated subsidiary cannot be a hedged item in a fair value hedge. This is because consolidation recognises in profit or loss the subsidiary's profit or loss, instead of changes in the investment's fair value. A hedge of a net investment in a foreign operation is different because it is a hedge of the foreign currency exposure, not a fair value hedge of the change in the value of the investment.
4. Rebalancing refers to the adjustments made to the designated quantities of the hedged item or the hedging instrument of an already existing hedging relationship for the purpose of maintaining a hedge ratio that complies with the hedge effectiveness requirements. Changes to designated quantities of a hedged item or of a hedging instrument for a different purpose do not constitute rebalancing for the purpose of this Standard
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
Oracle Corporation Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. Oracle Corporation Common Stock prediction model is evaluated with Modular Neural Network (Speculative Sentiment Analysis) and Statistical Hypothesis Testing1,2,3,4 and it is concluded that the ORCL stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Hold
### ORCL Oracle Corporation Common Stock Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementB1Caa2
Balance SheetBaa2C
Leverage RatiosB1B3
Cash FlowCaa2Baa2
Rates of Return and ProfitabilityB1Caa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 72 out of 100 with 490 signals.
## References
1. Mikolov T, Yih W, Zweig G. 2013c. Linguistic regularities in continuous space word representations. In Pro- ceedings of the 2013 Conference of the North American Chapter of the Association for Computational Linguistics: Human Language Technologies, pp. 746–51. New York: Assoc. Comput. Linguist.
2. R. Sutton and A. Barto. Reinforcement Learning. The MIT Press, 1998
3. Çetinkaya, A., Zhang, Y.Z., Hao, Y.M. and Ma, X.Y., Is FFBC Stock Buy or Sell?(Stock Forecast). AC Investment Research Journal, 101(3).
4. Hartford J, Lewis G, Taddy M. 2016. Counterfactual prediction with deep instrumental variables networks. arXiv:1612.09596 [stat.AP]
5. Chow, G. C. (1960), "Tests of equality between sets of coefficients in two linear regressions," Econometrica, 28, 591–605.
6. J. Baxter and P. Bartlett. Infinite-horizon policy-gradient estimation. Journal of Artificial Intelligence Re- search, 15:319–350, 2001.
7. L. Busoniu, R. Babuska, and B. D. Schutter. A comprehensive survey of multiagent reinforcement learning. IEEE Transactions of Systems, Man, and Cybernetics Part C: Applications and Reviews, 38(2), 2008.
Frequently Asked QuestionsQ: What is the prediction methodology for ORCL stock?
A: ORCL stock prediction methodology: We evaluate the prediction models Modular Neural Network (Speculative Sentiment Analysis) and Statistical Hypothesis Testing
Q: Is ORCL stock a buy or sell?
A: The dominant strategy among neural network is to Hold ORCL Stock.
Q: Is Oracle Corporation Common Stock stock a good investment?
A: The consensus rating for Oracle Corporation Common Stock is Hold and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of ORCL stock?
A: The consensus rating for ORCL is Hold.
Q: What is the prediction period for ORCL stock?
A: The prediction period for ORCL is (n+8 weeks)
|
This is a simulation due to request from user:
A particle with mass m, initial position xi, velocity vi and drag force $F=-k*v^2$
How to find out x(t) and v(t)!
/htdocs/ntnujava/ejsuser/2/users/ntnu/fkh/dampingkv2_pkg/dampingkv2.propertiesFull screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
|
## 译文
C++ is one of the most widely used programming languages in the world. Well-written C++ programs are fast and efficient. The language is more flexible than other languages because you can use it to create a wide range of apps—from fun and exciting games, to high-performance scientific software, to device drivers, embedded programs, and Windows client apps. For more than 20 years, C++ has been used to solve problems like these and many others. What you might not know is that an increasing number of C++ programmers have folded up the dowdy C-style programming of yesterday and have donned modern C++ instead.
C++是世界上应用最广泛的编程语言之一。写得好的C++程序是快速和高效的。该语言比其他语言更灵活，因为您可以使用它创建各种各样的应用程序，从有趣和刺激的游戏，到高性能的科学软件，再到设备驱动程序、嵌入式程序和Windows客户端应用程序。20多年来，C++已经被用来解决这些问题以及许多其他问题。你可能不知道的是，越来越多的C++程序员已经折叠了昨天过时的C风格编程，并使用了现代C++取代
One of the original requirements for C++ was backward compatibility with the C language. Since then, C++ has evolved through several iterations—C with Classes, then the original C++ language specification, and then the many subsequent enhancements. Because of this heritage, C++ is often referred to as a multi-paradigm programming language. In C++, you can do purely procedural C-style programming that involves raw pointers, arrays, null-terminated character strings, custom data structures, and other features that may enable great performance but can also spawn bugs and complexity. Because C-style programming is fraught with perils like these, one of the founding goals for C++ was to make programs both type-safe and easier to write, extend, and maintain. Early on, C++ embraced programming paradigms such as object-oriented programming. Over the years, features have been added to the language, together with highly-tested standard libraries of data structures and algorithms. It's these additions that have made the modern C++ style possible.
Modern C++ emphasizes: * Stack-based scope instead of heap or static global scope. * Auto type inference instead of explicit type names. * Smart pointers instead of raw pointers. * std::string and std::wstring types (see <string>) instead of raw char[] arrays. * C++ Standard Library containers like vector, list, and map instead of raw arrays or custom containers. See <vector>, <list>, and <map>. * C++ Standard Library algorithms instead of manually coded ones. * Exceptions, to report and handle error conditions. * Lock-free inter-thread communication using C++ Standard Library std::atomic<> (see <atomic>) instead of other inter-thread communication mechanisms. * Inline lambda functions instead of small functions implemented separately. * Range-based for loops to write more robust loops that work with arrays, C++ Standard Library containers, and Windows Runtime collections in the form for ( for-range-declaration : expression ). This is part of the Core Language support. For more information, see Range-based for Statement (C++).
• 基于栈的作用域，而不是堆或静态全局作用域
• 自动类型推断而不是显式类型名
• 智能指针而不是原始指针
• std::stringstd::wstring类型（参见<string>）而不是原始char[]数组
• C++标准库容器，如vectorlistmap，而不是原始数组或自定义容器。请参见<vector><list><map>
• C++标准库算法，而不是手工编写的算法
• 通过异常报告和处理错误情况
• 使用C++标准库std::atomic<>进行无锁线程间通信（参见<atomic>）而不是其他线程间通信机制
• 内联lambda函数而不是单独实现的小函数
• 使用基于范围的循环编写更健壮的循环代码，这些循环使用数组、C++标准库容器和Windows运行时集合for ( for-range-declaration : expression )。这是核心语言支持的一部分。有关更多信息，请参见基于范围的语句（C++）
The C++ language itself has also evolved. Compare the following code snippets. This one shows how things used to be in C++:
C++语言本身也有了发展。比较以下代码段。这个例子显示了C++中的事物是如何使用的：
Here's how the same thing is accomplished in modern C++:
In modern C++, you don't have to use new/delete or explicit exception handling because you can use smart pointers instead. When you use the auto type deduction and lambda function, you can write code quicker, tighten it, and understand it better. And a range-based for loop is cleaner, easier to use, and less prone to unintended errors than a C-style for loop. You can use boilerplate together with minimal lines of code to write your app. And you can make that code exception-safe and memory-safe, and have no allocation/deallocation or error codes to deal with.
Modern C++ incorporates two kinds of polymorphism: compile-time, through templates, and run-time, through inheritance and virtualization. You can mix the two kinds of polymorphism to great effect. The C++ Standard Library template shared_ptr uses internal virtual methods to accomplish its apparently effortless type erasure. But don't over-use virtualization for polymorphism when a template is the better choice. Templates can be very powerful.
If you're coming to C++ from another language, especially from a managed language in which most of the types are reference types and very few are value types, know that C++ classes are value types by default. But you can specify them as reference types to enable polymorphic behavior that supports object-oriented programming. A helpful perspective: value types are more about memory and layout control, reference types are more about base classes and virtual functions to support polymorphism. By default, value types are copyable—they each have a copy constructor and a copy assignment operator. When you specify a reference type, make the class non-copyable—disable the copy constructor and copy assignment operator—and use a virtual destructor, which supports the polymorphism. Value types are also about the contents, which, when they are copied, give you two independent values that you can modify separately. But reference types are about identity—what kind of object it is—and for this reason are sometimes referred to as polymorphic types.
C++ is experiencing a renaissance because power is king again. Languages like Java and C# are good when programmer productivity is important, but they show their limitations when power and performance are paramount. For high efficiency and power, especially on devices that have limited hardware, nothing beats modern C++.
C++正在经历复兴，因为能力再次成为国王。像Java和C#这样的语言在程序员的生产力很重要的时候是很好的，但是它们在功率和性能是最重要的时候显示出它们的局限性。对于高效率和功率，特别是在硬件有限的设备上，没有什么比现代C++更出色
Not only the language is modern, the development tools are, too. Visual Studio makes all parts of the development cycle robust and efficient. It includes Application Lifecycle Management (ALM) tools, IDE enhancements like IntelliSense, tool-friendly mechanisms like XAML, and building, debugging, and many other tools.
The articles in this part of the documentation provide high-level guidelines and best practices for the most important features and techniques for writing modern C++ programs.
• C++类型系统
• 统一初始化和委托构造器
• 对象生命周期和资源管理
• 对象拥有资源（RAII）
• 智能指针
• 用于编译时封装的Pimpl
• 容器
• 算法
• 字符串和I/O格式（现代C++）
• 错误和异常处理
• ABI边界的可移植性
For more information, see the Stack Overflow article Which C++ idioms are deprecated in C++11.
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# RDP 2020-08: Start Spreading the News: News Sentiment and Economic Activity in Australia 2. How Do We Measure News Sentiment?
Sentiment is hard to measure as it is not directly observed. Common survey-based measures of sentiment typically ask respondents about their beliefs about current economic conditions as well as expectations for future economic conditions. We take a different approach and construct a proxy for sentiment based on the language used by journalists in news reports on the economy.
There are two general approaches for quantifying sentiment in text. The dictionary-based approach relies on pre-defined lists of words with each word either classified as positive, negative, neutral, or indicating uncertainty. The machine learning approach predicts the sentiment of any given set of text after training models with a large set of text that has been assigned sentiment ratings by human readers. For example, models have been developed using social media data, such as Twitter, that provide text that is combined with user feedback to identify the sentiment of the posts. This approach is better able to capture the nuances in human language but it is more complex and less transparent.
We follow the simpler dictionary-based approach to construct our NSI. The NSI measures the net balance of words used by journalists that are considered to be ‘positive’ and ‘negative’. When journalists use more positive words and/or fewer negative words, this is an indicator that sentiment is rising in the economy. This type of index has been used before for other regions, such as the United States, Japan and Europe (see, for instance, Fraiberger (2016); Scotti (2016); Larsen and Thorsrud (2018) and Buckman et al (2020)).
The raw data used in constructing the NSI consist of daily news extracted from Dow Jones Factiva. Each article listed in the database includes metadata such as publication time, language, region and category. After removing duplicates and selecting only articles that are written in English by Australian media outlets to cover the Australian economy, the resulting dataset includes around 300,000 articles. The data span the period from September 1987 to June 2020 and the sample covers more than 600 newspapers, though The Australian, The Sydney Morning Herald and The Australian Financial Review are the main sources.
Common steps in the natural language processing literature are taken to clean the raw dataset before analysis: numbers, punctuation marks, white spaces and common stop words are removed from each article. All words are then reduced to their respective ‘stem’, which is the part of a word that is common to all of its inflections (for example, ‘performs’, ‘performing’, and ‘performed’ are reduced to ‘perform’).
To measure the sentiment of a set of text, that is, whether or not the news is positive or negative, the Loughran–McDonald dictionary is used. This is a word list specific to the domain of economics and finance (see Loughran and McDonald (2011) for more details). The NSI is constructed by counting the number of times that negative and positive words appear in the cleaned text of articles.[2] A news uncertainty index (NUI) is also constructed by counting the number of articles that contain uncertain words.[3] The most common positive, negative and uncertain words in March 2020 are shown in Figure 1.
To construct the time series of the NSI, the articles are sorted by date of publication and the data are divided into blocks of time, which could be a day. For each time period (t), we compute the sentiment index by subtracting the count of negative words from the count of positive words and then dividing by total word count:
$NS I t = Positiv e t −Negativ e t Word coun t t$
Between September 1987 and March 2020 there are, on average, around two more negatives than positives for every 100 words in the articles, with a standard deviation of less than 1 word. We standardise the indicator to have a mean of zero and a standard deviation of one.
## Footnotes
The individual words are not weighted by the degree of positivity or negativity.[2]
The uncertainty index is therefore measured on a different basis to the sentiment index. This is mainly due to practical reasons – the terms in the uncertainty dictionary do not appear very frequently within articles. This approach to measuring uncertainty using text analysis is equivalent to that used by others in the literature (e.g. Moore 2017).[3]
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# Overview
This is a PyPI mirror client according to PEP 381.
## Installation
Until a release is ready, here's the way to go:
$hg clone https://bitbucket.org/ctheune/bandersnatch$ cd bandersnatch
$virtualenv-2.7 .$ bin/python bootstrap.py
\$ bin/buildout
The bandersnatch executable will be placed in the bin/ directory of your virtualenv.
## Configuration
• Run bandersnatch mirror - it will create an empty configuration file for you in /etc/bandersnatch.conf.
• Run bandersnatch mirror again. It will populate your mirror with the current status of all PyPI packages - roughly 50GiB at the time of writing.
• Run bandersnatch mirror regularly to update your mirror with any intermediate changes.
### Webserver
Configure your webserver to serve the web/ sub-directory of the mirror. For nginx it should look something like this:
server {
listen 127.0.0.1:80;
server_name <mymirrorname>;
root <path-to-mirror>/web;
autoindex on;
charset utf-8;
}
• Note that it is a good idea to have your webserver publish the HTML index files correctly with UTF-8 as the carset. The index pages will work without it but if humans look at the pages the characters will end up looking funny.
• Make sure that the webserver uses UTF-8 to look up unicode path names. nginx gets this right by default - not sure about others.
### Cron jobs
You need to set up one cron job to run the mirror itself. If you run a public mirror, then you need a second job that will create access statistics for aggregation on the master PyPI.
Here's a sample that you could place in /etc/cron.d/bandersnatch:
LC_ALL=en_US.utf8
*/2 * * * * root bandersnatch mirror |& logger -t bandersnatch[mirror]
12 * * * * root bandersnatch update-stats |& logger -t bandersnatch[update-stats]
This assumes that you have a logger utility installed that will convert the output of the commands to syslog entries.
## Maintenance
bandersnatch does not keep much local state in addition to the mirrored data. In general you can just keep rerunning bandersnatch mirror to make it fix errors.
If you delete the state files then the next run will force it to check everything against the master PyPI:
* delete ./state file and ./todo if they exist in your mirror directory
* run bandersnatch mirror to get a full sync
Be aware, that full syncs likely take hours depending on PyPIs performance and your network latency and bandwidth.
## Migrating from pep381client
• remove old status files, but keep actual data (everything under web/)
• create config file, port command parameters from old cronjobs
• update cron jobs
## Contact
If you have questions or comments, please submit a bug report to http://bitbucket.org/ctheune/bandersnatch/issues/new.
Also, I'm reading the distutils sig mailing list.
## Support this project
If you'd like to support my work on PyPI mirrors, please consider a gittip. I'm planning to run a couple more international mirrors if I get enough support.
## Kudos
This client is based on the original pep381client by Martin v. Loewis.
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# If k is an integer and 33!/22! is divisible by 6^k
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If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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22 Dec 2012, 07:52
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If k is an integer and 33!/22! is divisible by 6^k, what is the maximum possible value of k?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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22 Dec 2012, 10:08
9
+1 D
$$\frac{33!}{22!} = 33*32*...24*23$$
Therefore we have to identify how many 6s has the product of $$33*32*...24*23$$
The number of 6s will indicate us the highest value of k.
Let´s see how many 6s we can find in the elements of the product $$33*32*...24*23$$:
For example, the number 30 = 6*5..........one 6
Also 24 = 6*4......... another 6
But remember that 6 = 3*2
So, we can create 6s with the 2s and 3s of that product. Be careful of not duplicating the 6 you have identified.
For example: 33 has a 3, and 32 has a 2, there is another 6.
In other words, you have to identify how many 6 you can build.
I found six 6s. The correct answer.
I think I deserve kudos
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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22 Dec 2012, 10:22
basically count the no of 3s you can collect from 23-33 !! there are plenty of 2s so it doest atter much!!
but MISSED OUT THE FACT THAT 27 has 3 - 3s!
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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22 Dec 2012, 10:25
5
daviesj wrote:
If k is an integer and 33!/22! is divisible by 6^k, what is the maximum possible value of k?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
$$33!/22!=23*24*25*26*27*28*29*30*31*32*33$$
$$24=2*2*2*3$$
$$25=5*5$$
$$26=2*13$$
$$27=3*3*3$$
$$28=2*2*7$$
$$30=2*3*5$$
$$32=2*2*2*2*2$$
$$33=3*11$$
I am not considering the numbers 23, 29 and 31 because they do not contain any factor other than the number itself and 1.
Now in this question, our motive is to count the number of pairs of $$2$$ and $$3$$.
The # of 3s is less than # of 2s.
So our critical number is 3.
The maximum number of k will be the number of pairs of $$2*3$$ which is equal to number of 3s.
+1D
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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22 Dec 2012, 14:12
this is a question out of scope, out a gmat context
another variation is this
Quote:
If 3^k is a factor of (122!), what is the greatest possible value of k?
is good to know the logic behind. it could be useful for other purpose or situation.....but for me are a bit waste of time doing such questions
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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22 Dec 2012, 20:12
Hi carcass,
this question is from manhattan gmat advanced quant book...how can it be out of scope?
Posted from my mobile device
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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22 Dec 2012, 21:17
From where did you get this variation carcass?
Are the answer choices available for this one?
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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22 Dec 2012, 21:58
7
2
6^k =2*3 ^k
now lets take 33!
number of power of 2 is 33/2 + 33/4 + 33/8 + 33/16 + 33/32 = 16+8+4+2+1=31
number of power of 3 is 33/3 + 33/9 + 33/27 = 11+3+1=15
Similarly for 22!
number of power of 2 is 22/2 + 22/4 +22/8 + 22/16 = 11+5+2+1=19
number of power of 3 is 22/3 + 22/9 = 7+2=9
33!/22! contains 2^12 * 3^6
since we need maximum power of 6 we are limited by power of 3 above which is 6.... so k=6
let me know if you want me to be more clear.....
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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22 Dec 2012, 22:11
DaviesJ....your kudo activity looks very suspicious.
http://gmatclub.com/kudos-details/daviesj
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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22 Dec 2012, 22:15
Hi Amateur,
That one is good method to find power of a certain number. You just earned one kudo for that.
But is it trust-able method?
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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22 Dec 2012, 22:25
1
daviesj wrote:
Hi Amateur,
That one is good method to find power of a certain number. You just earned one kudo for that.
But is it trust-able method?
Thank You.... Yes it is..... you can refer to factors part in the following link....
math-number-theory-88376.html
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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22 Dec 2012, 22:36
1
Yeh, got it..thanks again!
*here's the method:
Finding the number of powers of a prime number p, in the n!.
The formula is:
$$\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3} ... till p^x\leq{n}$$
What is the power of 2 in 25!?
$$\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22$$
Finding the power of non-prime in n!:
How many powers of 900 are in 50!
Make the prime factorization of the number: $$900=2^2*3^2*5^2$$, then find the powers of these prime numbers in the n!.
Find the power of 2:
$$\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47$$
= $$2^{47}$$
Find the power of 3:
$$\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22$$
=$$3^{22}$$
Find the power of 5:
$$\frac{50}{5}+\frac{50}{25}=10+2=12$$
=$$5^{12}$$
We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!.
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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23 Dec 2012, 06:22
daviesj wrote:
Hi carcass,
this question is from manhattan gmat advanced quant book...how can it be out of scope?
Posted from my mobile device
wait a moment. I said that in my opinion after months of study (forum, blog, books, millions of sources and so on) those question are a bit out of scope.
this does not mean that MGMAT is one of the best prep company out there. Often the question are more difficult to teach and learn the logic, the strategy, the techniques to tackle the most difficult questions during the exam and this is awesome, thanks prep company The same thing said stacey koprince (thanks Stacy for your articles)
I do not want to be misunderstood . In concrete, they are made more difficult than normal
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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23 Dec 2012, 06:24
Marcab wrote:
From where did you get this variation carcass?
Are the answer choices available for this one?
maybe he best resource for quant
http://www.veritasprep.com/blog/2011/06 ... actorials/
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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23 Dec 2012, 07:01
Hii Carcass.
Can you please explain how could have we implemented the technique in the original question i.e. 33!/22!?
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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23 Dec 2012, 07:20
No ulterior shortcut
Your approach is the best in that situation. I rewarded you with a kudos.
The most difficult thing is the fraction you can't handle it but your method
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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23 Dec 2012, 07:55
Marcab wrote:
Hii Carcass.
Can you please explain how could have we implemented the technique in the original question i.e. 33!/22!?
I consider it to be a tough one...... involving meticulous approach when solving.... what if the question is 40!/14!? It would take more than 2min for the approach.... So I think my approach is a better way of solving in the time frame.....
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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23 Dec 2012, 08:27
1
Amateur wrote:
Marcab wrote:
Hii Carcass.
Can you please explain how could have we implemented the technique in the original question i.e. 33!/22!?
I consider it to be a tough one...... involving meticulous approach when solving.... what if the question is 40!/14!? It would take more than 2min for the approach.... So I think my approach is a better way of solving in the time frame.....
Amateur the key is to learn as much techniques as possible. One is better in a certain situation, and in another one do not. Gmat for this reason in quite challenging because do not have a preconceived strategy.
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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23 Dec 2012, 10:23
carcass wrote:
Amateur wrote:
Marcab wrote:
Hii Carcass.
Can you please explain how could have we implemented the technique in the original question i.e. 33!/22!?
I consider it to be a tough one...... involving meticulous approach when solving.... what if the question is 40!/14!? It would take more than 2min for the approach.... So I think my approach is a better way of solving in the time frame.....
Amateur the key is to learn as much techniques as possible. One is better in a certain situation, and in another one do not. Gmat for this reason in quite challenging because do not have a preconceived strategy.
I agree, I never faulted your approach... it is perfectly fine but in this situation I am just proposing a one which takes less time.... but it all depends on ones comfort.....
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Re: If k is an integer and 33!/22! is divisible by 6^k [#permalink]
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23 Dec 2012, 10:25
I guess, we have deviated a bit. Can anyone please let me know how to solve 33!/22! with alternative method.
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Re: If k is an integer and 33!/22! is divisible by 6^k &nbs [#permalink] 23 Dec 2012, 10:25
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# If k is an integer and 33!/22! is divisible by 6^k
## Events & Promotions
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On more general Lipschitz spaces
Preprint series: 99-47, Analysis
The paper is published: Z. Anal. Anwendungen, 19(3), 781-799, 2000.
MSC:
26A16 Lipschitz (Holder) classes
46E35 Sobolev spaces and other spaces of smooth'' functions, embedding theorems, trace theorems
26A15 Continuity and related questions (modulus of continuity, semicontinuity, discontinuities, etc.), {For properties determined by Fourier coefficients, See 42A16; for those determined by approximation properties, See 41A25, 41A27}
Abstract: The present paper deals with (logarithmic) Lipschitz spaces of type $Lip^{(1,-\alpha)}_{p,q}$, $1\leq p\leq\infty$, $0<q\leq\infty$, $\alpha>1/q$.
We study their properties and derive some (sharp) embedding results. In that sense this paper can be regarded as some continuation and extension of some of our earlier papers, but there are also connections with some recent work of Triebel concerning Hardy inequalities and sharp embeddings.
Recall that the nowadays almost classical' forerunner of investigations of this type is the Br\'ezis-Wainger result about the almost' Lipschitz continuity of elements of the Sobolev spaces $H^{1+n/p}_p(R^n)$ when $1<p<\infty$.
Keywords: limiting embeddings, Lipschitz spaces, function
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Thursday, July 21 2022
3:30pm - 5:30pm
PhD Thesis Presentation
Degree Sequence Realization Problems for Hypergraphs and Applications of the Discharging Method to Entire Coloring
Abstract: The first part of this thesis involves realization problems for degree sequences in a hypergraph context. A graphic sequence $/pi$ is potentially $H$-graphic if there is a realization of $\pi$ contains $G$ as a subgraph. The potential number, $\sigma (H,n)$, is the minimum even integer such that any graphic sequence with length $n$ and sum greater than $\sigma (H,n)$ is potentially $H$-graphic. In this paper we extend these notions to $3$-uniform hypergraphs and determine the potential number for $3$-uniform hypercliques. We also discuss the stability of the potential number in the $r=3$ case.
The second part of this thesis deals with the discharging method and entire coloring of planar graphs. Let $G$ be a plane graph with maximum degree $\Delta$. If all vertices, edges, and faces of $G$ can be colored with $k$ colors so that any two adjacent or incident elements have distinct colors, then $G$ is said to be entirely $k$-colorable. In 2011, Wang and Zhu asked if every simple plane graph except $K_4$ is entirely $(\Delta+3)$-colorable. In 2012, Wang, Mao, and Miao answered in the affirmative for simple plane graphs with $\Delta \geq 8$.
In this paper, we show that every plane multigraph with $\Delta=7$, no loops, no 2-faces, and no 3-faces sharing an edge is entirely $(\Delta+3)$-colorable.
Speaker: Nathan Graber Affiliation: Location: Zoom in email
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# Where does stuff “sucked” up by a black hole go? [duplicate]
I've heard that stuff sucked up by a black hole leads to a parallel universe, but I don't believe that because when a black whole "sucks" something up, the thing it "sucked" up adds to the black holes mass, how would it do this if everything it sucked up was ejected out a white hole on the other side of the universe? I want to know where stuff really goes after being "sucked" up by a black hole. According to Wikipedia,
The theory of general relativity predicts that a sufficiently compact mass can deform spacetime to form a black hole.
This is why dying stars form black holes. This means a black hole must be a very compact sum of mass in a small area, or a singularity. Because of this, I think when something gets sucked up by a black hole, it just gets crushed up to a very small volume and adds to the singularity. Am I wright or not?
## marked as duplicate by JMac, WillO, John Rennie black-holes StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 9 '18 at 19:25
• Possible duplicate of What happens to light and mass in the center of a black hole? – JMac Apr 9 '18 at 16:11
• Black holes do not suck stuff up. A black hole has gravity, just like a star has gravity or a planet has gravity, and things that come near the black hole are influenced by that gravity in exactly the same way that things are influenced by the gravity of a planet or a star. If our Sun was magically crushed to the point of becoming a black hole, the Earth and all of the other planets would continue to orbit it, exactly as they do now. Only difference would be that the Solar system would become a dark and cold place. – Solomon Slow Apr 9 '18 at 17:05
• It's really only massive stars that could form black holes (cf this post) – Kyle Kanos Apr 9 '18 at 17:08
• I know they don't really suck, I just had no better way to describe it. – Daniel Turczynskyj Apr 9 '18 at 18:59
The same place where something "sucked up" by a planet goes.
A black whole is not a "hole" in the sense that it goes somewhere, or that something can "go in it". A black hole is simply a celestial body which is characterized by an escape velocity greater than the speed of light. The escape velocity grows smaller depending on how far you are from the center of mass, and at the point where it is no longer greater than the speed of light, the event horizon ends.
Thus, when something is sucked into a black whole, it eventually reaches the "surface" (although words like that may very well be meaningless in an environment as strange as a black whole) and becomes part of the black whole.
• So, does that mean there is a single point inside the black hole that is so dense, even light can't escape the event horizon. – Daniel Turczynskyj Apr 9 '18 at 19:01
• @DanielTurczynskyj It doesn't necessarily have to be a single point. Keep in mind that the equation for escape velocity is the square root of 2GM/R (2 times the gravitational constant times the mass of the planet, divided by the radius). This means that it isn't a single point per se where it is very dense, but rather an entire region which is very massive. – DevilApple227 Apr 9 '18 at 23:40
The Schwarzschild solution of the EFE (Einstein field equations) describes a spacetime in vacuum surrounding a spherically symmetric and static mass. In particular if the mass is contained in the Schwarzschild radius, it envisages a black hole, that is a mass that defines a surface, the event horizon, which captures any material body or light entering it. Working out the Schwarzschild metric with different coordinates it is demonstrated that matter or light crossing the event horizon can only proceed in direction of the singularity at the center of the black hole. Classically, all the mass is concentrated at the center of the black hole.
The theoretical possibility of a black hole as an interface to another universe comes from the so called maximally extended solution, which does not constrain the coordinates.
Where does stuff sucked up by a black hole go?
Into the black hole, increasing its mass.
I've heard that stuff sucked up by a black hole leads to a parallel universe, but I don't believe that.
You're right not to believe it.
I want to know where stuff goes after being sucked up by a black hole. According to Wikipedia, "The theory of general relativity predicts that a sufficiently compact mass can deform spacetime to form a black hole".
It's not all that different to a planet. The important thing is that at the event horizon the "coordinate" speed of light is zero, so light can't get out. So it's black. It isn't really a hole.
This is why dying stars forms black holes. This means a black hole must be a very compact sum of mass that has a lot of gravity. Because of this, I think when something gets sucked up by a black hole, it just gets crushed up with all the other mass and adds to it.
Pretty much. There are some interesting issues about "the state of matter" that comprises a black hole, and whether you can even call it matter. But the stuff that fell in adds to the mass of the black hole, which has a bigger gravitational field as a result.
The only reason you can't see the center where all this mass is because no light is reflected off it because the black hole's gravity is too great.
Yep, I'd say that's pretty much it.
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## Calculus VII: Approximations
Although I’ll have a very busy summer with consulting, I’ve taken some time to start reading more again. You know, those books which have been sitting on your shelves for years….
So I’ve started Volume I of A Treatise on the Integral Calculus by Joseph Edwards.
I include a picture of the cover page, since you can google it and download a copy online. Between Volumes I and II, there’s about 1800 pages of integral calculus….
Since I’ll likely be working with a calculus curriculum later this year, I thought I’d look at some older books and see what calculus was like back in the day. I’m continually surprised at how much there is to learn about elementary calculus, despite having taught it for over 25 years.
My approach will be a simple one — I’ll organize my posts by page number. As I read through the books and solve interesting problems, I’ll share with you things I find novel and interesting. The more I read books like these and think about calculus, the more I think most current textbooks simply are not up to the task of presenting calculus in any meaningful way. Sigh.
This is not the time to be on my soapbox — this is the time for some fun! So here is the first topic: Weddle’s Rule, found on page 21.
Ever hear of it? Bonus points if you have — but I never did. It’s another approximation rule for integrals. Here it is: given a function $f$ on the interval $[a,b],$ divide the interval into six equal subintervals with points $x_0, x_1,\ldots x_6$ and corresponding function values $y_0=f(x_0),\ldots,y_6=f(x_6).$ Then
$\displaystyle\int_a^bf(x)\,dx\approx \dfrac{b-a}{20}\left(y_1+5y_2+y_3+6y_4+y_5+5y_6+y_7\right).$
Yikes! Where did that come from? I’ll present my take on the idea, and offer a theory. If there are any historians of mathematics out there, I’d be happy to hear if my theory is correct.
One reason most of us haven’t heard of Weddle’s Rule is that approximations aren’t as important as they were before calculators and computers. So many exercises in this book involve approximation techniques.
So how would you come up with Weddle’s Rule? I’ll share my (likely mythical) scenario with you. It’s based on some notes I wrote up a while ago on Taylor series. So before diving into Weddle’s Rule, I’ll show you how I’d derive Simpson’s Rule — the technique is the same, but the algebra is easier. And by the way, if anyone has seen this technique before, please let me know! I’m sure it must have been done before, but I’ve never been able to find a source illustrating it.
Let’s assume we want to approximate
$F(x)=\displaystyle\int_a^xf(t)\,dt$
by using three equally-spaced points on the interval $[a,x].$ In other words, we want to find weights $p,$ $q,$ and $r$ such that
$S(x)=\left(p f(a)+ q f\left(\dfrac{a+x}2\right)+rf(x)\right)(x-a)\approx F(x).$
How might we approach this? We can create Taylor series for $F(x)$ and $S(x)$ about the point $a.$ The first is easy using the Fundamental Theorem of Calculus, assuming sufficient differentiability:
$F(x)=f(a)(x-a)+\dfrac{f'(a)}{2!}(x-a)^2+\dfrac{f''(a)}{3!}(x-a)^3+\cdots$
Now to construct the Taylor series of $S(x)$ about $x=a,$ we need to evaluate several derivatives at $a.$ This is not difficult to do by hand, but it is easy to do using Mathematica and a command such as
Doing so yields the following:
Now the problem becomes a simpler algebra problem — to force as many of the coefficients of the derivatives on the right-hand side to be $1$ as possible. This will make the derivatives of $F$ and $S$ match, and the Taylor polynomials will be equal up to some order.
Solving the first three such equations,
yields, as we expect, $p=1/6,$ $q=2/3,$ and $r=1/6.$ Note that these values also imply that
$\dfrac12q+4r=1,$
but
$\dfrac5{16}q+5r=\dfrac{25}{24}.$
This implies that
$S(x)-F(x)=\dfrac1{24}\cdot\dfrac{(x-a)^5}{5!}+O((x-a)^6)$
on each subinterval, so that
$S(x)-F(x)=O((x-a)^5)$
on each subinterval, giving that Simpson’s rule is $O((x-a)^4).$
So how we apply these to derive Weddle’s rule? We could try to find weights $w_1,\ldots w_7$ to create an approximation
$W(x)=\left(w_1 f(a)+w_2f\left(\dfrac{5a+x}6\right)+\cdots+w_7f(x)\right)(x-a).$
If we apply precisely the same procedure as we did with Simpson’s Rule, we get the following as the sequence of weights to create the best approximation:
$\dfrac{41}{840},\ \dfrac9{35},\ \dfrac9{280},\ \dfrac{34}{105},\ \dfrac9{280},\ \dfrac9{35},\ \dfrac{41}{480}.$
Not exactly easy to work with — remember, no calculators or computers.
So let’s make the approximation a little worse. Recall how the weights were found — a system of seven equations in seven unknowns was solved, analogous to the three equations in three unknowns for Simpson’s rule. Instead, we specify $w_1,$ and solve the first six equations in terms of $w_1.$ This gives us
Now all weights must be positive; this gives the constraint
$0.046\overline6\approx\dfrac7{150}
Let’s put $w_1=1/20,$ which is in the interval just described. This gives the sequence of weights to be
$\dfrac1{20},\ \dfrac5{20},\ \dfrac1{20},\ \dfrac6{20},\ \dfrac1{20},\ \dfrac5{20},\ \dfrac1{20},$
where all fractions are written with the same denominator. Now imagine factoring out the $1/2,$ and you notice that all divisions are by 10. Can you see the advantage? If you have a table of values for your functions, you just need to multiply function values by a single-digit number, and then move the decimal place over one. An approximators dream!
So Weddle’s approximation is exact for fifth-degree polynomials, even though it is possible to use six subintervals to get weights which are exact for sixth-degree polynomials. Yes, we lose an order of accuracy — but now our computations are much easier to carry out.
Was this Weddle’s thinking? I can’t be sure; I wasn’t able to locate the original article online. But it is a way for me to make sense out of Weddle’s rule.
I will admit that in a traditional calculus class, I don’t address approximations in this way. There is a time crunch to get “everything” done — that is, everything the student is expected to know for the next course in the calculus sequence.
Should these concepts be taught? I’ll make a brief observation: in reading through the first 200 pages of this calculus book, it seems that all that has changed since 1954 is that content was pared down significantly, and more calculator exercises were added.
This is not the solution. We need to rethink what students need to now know and how that material should be taught in light of emerging technology. So let’s get started!
## Calculus: Hyperbolic Trigonometry, IV
Of course, there is always more to say about hyperbolic trigonometry…. Next, we’ll look at what is usually called the logistic curve, which is the solution to the differential equation
$\dfrac{dP}{dt}=kP(C-P),\quad P(0)\ \text{given}.$
The logistic curve comes up in the usual chapter on differential equations, and is an example of population growth. Without going into too many details (since the emphasis is on hyperbolic trigonometry), $k$ is a constant which influences how fast the population grows, and $C$ is called the carrying capacity of the environment.
Note that when $P$ is very small, $C-P\approx C,$ and so the population growth is almost exponential. But when $P(t)$ gets very close to $C,$ then $dP/dT\approx0,$ and so population growth slows down. And of course when $P(t)=C,$ growth stops — hence calling $C$ the carrying capacity of the environment. It represents the largest population the environment can sustain.
Here is an example of such a curve where $C=500,$ $k=0.02,$ and $P(0)=50.$
Notice the S shape, obtained from a curve rapidly growing when the population is small. It happens that the population grows fastest at half the carrying capacity, and then growth slows to zero as the carrying capacity is reached.
Skipping the details (simple separation of variables), the solution to this differential equation is given by
$P(t)=\dfrac{C}{1+Ae^{-kCt}},\qquad A=\dfrac{C-P(0)}{P(0)}.$
I will digress for a moment, however, to mention partial fractions (as I step on my calculus soapbox). I have mentioned elsewhere that incomprehensible chapter in calculus textbooks: Techniques of Integration. Pedagogically a disaster for so many reasons.
The first time I address partial fractions is when summing telescoping series, such as
$\displaystyle\sum_{n=1}^\infty\dfrac1{n(n+1)}.$
It really is necessary. But I only go so far as to be able to sum such series. (Note: I do series as the middle third of Calculus II, rather than the end. A colleague suggested that students are more tired near the end of the course, which is better for a more technique-oriented discussion of the solution to differential equations, which typically comes before series.)
You also need partial fractions to solve the differential equation for the logistic curve, which is when I revisit the topic. After finding the logistic curve, we talk about partial fractions in more detail. The point is that students see some motivation for the method of partial fractions — which they decidedly don’t in a chapter on techniques of integration.
OK, time to step off the soapbox and talk about hyperbolic trigonometry…. The punch line is that the logistic curve is actually a scaled and shifted hyperbolic tangent curve! Of course it looks like a hyperbolic tangent, but let’s take a moment to see why.
We first use the definitions of $\sinh u$ and $\cosh u$ to write
$\tanh u=\dfrac{\sinh h}{\cosh u}=1-\dfrac2{1+e^{2u}}.$
This results in
$\dfrac2{1+e^{2u}}=1-\tanh u.$
You can see the form of the equation of the logistic curve starting to take shape. Since the hyperbolic tangent has horizontal tangents at $y=-1$ and $y=1,$ we need to scale by a factor of $C/2$ so that the asymptotes of the logistic curve are $C$ units apart:
$\dfrac C{1+e^{2u}}=\dfrac{C}2\left(1-\tanh u\right).$
Note that this puts the horizontal asymptotes of the function at $y=0$ and $y=C.$
To take into account the initial population, we need a horizontal shift, since otherwise the initial population would be $C/2.$ We can accomplish this be replacing $\tanh u$ with $\tanh(u+\varphi):$
$\dfrac C{1+e^{2\varphi} e^{2u}}=\dfrac C2(1-\tanh(u+\varphi)).$
We’re almost done at this point: we simply need
$e^{2\varphi}=A,\qquad 2u=-kCt.$
Solving and substituting back results in
$P(t)=\dfrac C2\left(1-\tanh\left(\dfrac{-kCt+\ln A}2\right)\right),$
which, since $\tanh$ is an odd function, becomes
$P(t)=\dfrac C2\left(1+\tanh\left(\dfrac{kCt-\ln A}2\right)\right).$
And there it is! The logistic curve as a scaled, shifted hyperbolic tangent.
Now what does showing this accomplish? I can’t give you a definite answer from the point of view of the students. But for me, it is a way to tie two seemingly unrelated concepts — hyperbolic trigonometry and solution of differential equations by separation of variables — together in a way that is not entirely contrived (as so many calculus textbook problems are).
I would love to perform the following experiment: work out the solution to the differential equation together as a guided discussion, and then prompt students to suggest functions this curve “looks like.” Of course the $\arctan$ might be suggested, but how would we relate this to the exponential function?
Eventually we’d tease out the hyperbolic tangent, since this function actually does involve the exponential function. Then I’d move into an inquiry-based lesson: give the students the equation of a logistic curve, and have them work out the conversion to the hyperbolic tangent.
And as is typical in such an approach, I would put students into groups, and go around the classroom and nudge them along. See what happens.
I say that yes, calculus students should be able to do this. I recently sent an email about pedagogy in calculus which, among other things, addressed the question: What do calculus students really need to know?
There is no room to adequately address that important question here, but in today’s context, I would say this: I think it is more important for a student to be able to rewrite $P(t)$ as a hyperbolic tangent than it is for them to know how to sketch the graph of $P(t).$
Why? Because it is trivial to graph functions, now. Type the formula into Desmos. But how to interpret the graph? Rewrite it? Analyze it? Draw conclusions from it? We need to focus on what is no longer necessary, and what is now indispensable. To my knowledge, no one has successfully done this.
I think it is about time for that to change….
## Calculus: Hyperbolic Trigonometry, III
We continue where we left off on the last post about hyperbolic trigonometry. Recall that we ended by finding an antiderivative for $\sec(x)$ using the hyperbolic trigonometric substitution $\sec(\theta)=\cosh(u).$ Today, we’ll look at this substitution in more depth.
The functional relationship between $\theta$ and $u$ is described by the gudermannian function, defined by
$\theta=\text{gd}\,u=2\arctan(e^u)-\dfrac\pi2.$
This is not at all obvious, so we’ll look at the derivation of this rather surprising-looking formula. It’s the only formula I’m aware of which involves both the arctangent and the exponential function. We remark (as we did in the last post) that we restrict $\theta$ to the interval $(-\pi/2,\pi/2)$ so that this relationship is in fact invertible.
We use a technique similar to that used to derive a formula for the inverse hyperbolic cosine. First, write
$\sec\theta=\cosh u=\dfrac{e^u+e^{-u}}2,$
and then multiply through by $e^u$ to obtain the quadratic
$(e^u)^2-2\sec(\theta)e^u+1=0.$
$e^u=\sec\theta\pm\tan\theta.$
Which sign should we choose? We note that $\theta$ and $u$ increase together, so that because $e^u$ is an increasing function of $u,$ then $\sec\theta\pm\tan\theta$ must be an increasing function of $\theta.$ It is not difficult to see that we must choose “plus,” so that $e^u=\sec\theta+\tan\theta,$ and consequently
$u=\ln(\sec\theta+\tan\theta).$
We remark that no absolute values are required here; this point was discussed in the previous post.
Now to solve for $\theta.$ The trick is to use a lesser-known trigonometric identity:
$\sec\theta+\tan\theta=\tan\left(\dfrac\pi4+\dfrac\theta2\right).$
There is such a nice geometrical proof of this identity, I can’t help but include it. Start with the usual right triangle, and extend the segment of length $\tan\theta$ by $\sec\theta$ in order to form an isosceles triangle. Thus,
$\tan(\theta+\alpha)=\sec\theta+\tan\theta.$
To find $\alpha,$ observe that $\beta$ is supplementary to both $2\alpha$ and $\pi/2-\theta,$ so that
$2\alpha=\dfrac\pi2-\theta,$
which easily implies
$\alpha=\dfrac\pi4-\dfrac\theta2.$
Therefore
$\theta+\alpha=\dfrac\pi4+\dfrac\theta2,$
which is precisely what we need to prove the identity.
Now we substitute back into the previous expression for $u,$ which results in
$u=\ln\tan\left(\dfrac\pi4+\dfrac\theta2\right).$
This may be solved for $\theta,$ giving
$\theta=\text{gd}\,u=2\arctan(e^u)-\dfrac\pi2.$
So let’s see how to use this to relate circular and hyperbolic trigonometric functions. We have
$\sec(\text{gd}\,u)=\dfrac1{\cos(2\arctan(e^u)-\pi/2)},$
which after using the usual circular trigonometric identities, becomes
$\sec(\text{gd}\,u)=\dfrac{e^u+e^{-u}}2=\cosh u.$
It is also an easy exercise to see that
$\dfrac{d}{du}\,\text{gd}\,u=\text{sech}\, u.$
So revisiting the integral
$\displaystyle\int\sec\theta\,d\theta,$
we may alternatively make the substitution $\theta=\text{gd}\,u,$ giving
$\displaystyle\int\sec\theta\,d\theta=\int\cosh u\,(\text{sech}\, u\,du)=\int du,$
which is the same simple integral we saw in the previous post.
What about the other trigonometric functions? Certainly we know that $\cos(\text{gd}\,u)=\text{sech}\,u.$ Again using the usual circular trigonometric identities, we can show that
$\sin(\text{gd}\,u)=\tanh u.$
Knowing these three relationships, the rest are easy to find: $\tan(\text{gd}\,u)=\sinh u,$ $\cot(\text{gd}\,u)=\text{csch}\,u,$ and $\csc(\text{gd}\,u)=\text{coth}\,u.$
I think that the gudermannian function should be more widely known. On the face of it, circular and hyperbolic trigonometric functions are very different beasts — but they relate to each other in very interesting ways, in my opinion.
I will admit that I don’t teach students about the gudermannian function as part of a typical calculus course. Again, there is the issue of time: as you are well aware, students finishing one course in the calculus sequence must be adequately prepared for the next course in the sequence.
So what I do is this: I put the exercises on the gudermannian function as extra challenge problems. Then, if a student is already familiar with hyperbolic trigonometry, they can push a little further to learn about the gudermannian.
Not many students take on the challenge — but there are always one or two who will visit my office hours with questions. Such a treat for a mathematics professor! But I feel it is always necessary to give something to the very best students to chew on, so they’re not bored. The gudermannian does the trick as far as hyperbolic trigonometry is concerned….
As a parting note, I’d like to leave you with a few more exercises which I include in my “challenge” question on the gudermannian. I hope you enjoy working them out!
1. Show that $\tanh\left(\dfrac x2\right)=\tan\left(\dfrac 12\text{gd}\,x\right).$
2. Show that $e^x=\dfrac{1+\tan(\frac12\text{gd}\,x)}{1-\tan(\frac12\text{gd}\,x)}.$
3. Show that if $h$ is the inverse of the gudermannian function, then $h'(x)=\sec x.$
## Calculus: Hyperbolic Trigonometry, II
Now on to some calculus involving hyperbolic trigonometry! Today, we’ll look at trigonometric substitutions involving hyperbolic functions.
$\displaystyle\int\sqrt{1+x^2}\,dx.$
The usual technique involving circular trigonometric functions is to put $x=\tan(\theta),$ so that $dx=\sec^2(\theta)\,d\theta,$ and the integral transforms to
$\displaystyle\int\sec^3(\theta)\,d\theta.$
In general, we note that when taking square roots, a negative sign is sometimes needed if the limits of the integral demand it.
This integral requires integration by parts, and ultimately evaluating the integral
$\displaystyle\int\sec(\theta)\,d\theta.$
And how is this done? I shudder when calculus textbooks write
$\displaystyle\int \sec(\theta)\cdot\dfrac{\sec(\theta)+\tan(\theta)}{\sec(\theta)+\tan(\theta)}\,d\theta=\ldots$
How does one motivate that “trick” to aspiring calculus students? Of course the textbooks never do.
Now let’s see how to approach the original integral using a hyperbolic substitution. We substitute $x=\sinh(u),$ so that $dx=\cosh(u)\,du$ and $\sqrt{1+x^2}=\cosh(u).$ Note well that taking the positive square root is always correct, since $\cosh(u)$ is always positive!
This results in the integral
$\displaystyle\int\cosh^2(u)\,du=\displaystyle\int\dfrac{1+\cosh(2u)}2\,du,$
which is quite simple to evaluate:
$\dfrac12u+\dfrac14\sinh(2u)+C.$
Now $u=\hbox{arcsinh}(x),$ and
$\sinh(2u)=2\sinh(u)\cosh(u)=2x\sqrt{1+x^2}.$
Recall from last week that we derived an explicit formula for $\hbox{arcsinh}(x),$ and so our integral finally becomes
$\dfrac12\left(\ln(x+\sqrt{1+x^2})+x\sqrt{1+x^2}\right)+C.$
You likely noticed that using a hyperbolic substitution is no more complicated than using the circular substitution $x=\sin(\theta).$ What this means is — no need to ever integrate
$\displaystyle\int\tan^m(\theta)\sec^n(\theta)\,d\theta$
again! Frankly, I no longer teach integrals involving $\tan(\theta)$ and $\sec(\theta)$ which involve integration by parts. Simply put, it is not a good use of time. I think it is far better to introduce students to hyperbolic trigonometric substitution.
Now let’s take a look at the integral
$\displaystyle\int\sqrt{x^2-1}\,dx.$
The usual technique? Substitute $x=\sec(\theta),$ and transform the integral into
$\displaystyle\int\tan^2(\theta)\sec(\theta)\,d\theta.$
Sigh. Those irksome tangents and secants. A messy integration by parts again.
But not so using $x=\cosh(u).$ We get $dx=\sinh(u)\,du$ and $\sqrt{x^2-1}=\sinh(u)$ (here, a negative square root may be necessary).
We rewrite as
$\displaystyle\int\sinh^2(u)\,du=\displaystyle\int\dfrac{\cosh(2u)-1}2\,du.$
This results in
$\dfrac14\sinh(2u)-\dfrac u2+C=\dfrac12(\sinh(u)\cosh(u)-u)+C.$
All we need now is a formula for $\hbox{arccosh}(x),$ which may be found using the same technique we used last week for $\hbox{arcsinh}(x):$
$\hbox{arccosh}(x)=\ln(x+\sqrt{x^2-1}).$
Thus, our integral evaluates to
$\dfrac12(x\sqrt{x^2-1}-\ln(x+\sqrt{x^2-1}))+C.$
We remark that the integral
$\displaystyle\int\sqrt{1-x^2}\,dx$
is easily evaluated using the substitution $x=\sin(\theta).$ Thus, integrals of the forms $\sqrt{1+x^2},$ $\sqrt{x^2-1},$ and $\sqrt{1-x^2}$ may be computed by using the substitutions $x=\sinh(u),$ $x=\cosh(u),$ and $x=\sin(\theta),$ respectively. It bears repeating: no more integrals involving powers of tangents and secants!
One of the neatest applications of hyperbolic trigonometric substitution is using it to find
$\displaystyle\int\sec(\theta)\,d\theta$
without resorting to a completely unmotivated trick. Yes, I saved the best for last….
So how do we proceed? Let’s think by analogy. Why did the substitution $x=\sinh(u)$ work above? For the same reason $x=\tan(\theta)$ works: we can simplify $\sqrt{1+x^2}$ using one of the following two identities:
$1+\tan^2(\theta)=\sec^2(\theta)\ \hbox{ or }\ 1+\sinh^2(u)=\cosh^2(u).$
So $\sinh(u)$ is playing the role of $\tan(\theta),$ and $\cosh(u)$ is playing the role of $\sec(\theta).$ What does that suggest? Try using the substitution $\sec(\theta)=\cosh(u)$!
No, it’s not the first think you’d think of, but it makes sense. Comparing the use of circular and hyperbolic trigonometric substitutions, the analogy is fairly straightforward, in my opinion. There’s much more motivation here than in calculus textbooks.
So with $\sec(\theta)=\cosh(u),$ we have
$\sec(\theta)\tan(\theta)\,d\theta=\sinh(u)\,du.$
But notice that $\tan(\theta)=\sinh(u)$ — just look at the above identities and compare. We remark that if $\theta$ is restricted to the interval $(-\pi/2,\pi/2),$ then as a result of the asymptotic behavior, the substitution $\sec(\theta)=\cosh(u)$ gives a bijection between the graphs of $\sec(\theta)$ and $\cosh(u),$ and between the graphs of $\tan(\theta)$ and $\sinh(u).$ In this case, the signs are always correct — $\tan(\theta)$ and $\sinh(u)$ always have the same sign.
So this means that
$\sec(\theta)\,d\theta=du.$
What could be simpler?
Thus, our integral becomes
$\displaystyle\int\,du=u+C.$
But
$u=\hbox{arccosh}(\sec(\theta))=\ln(\sec(\theta)+\tan(\theta)).$
Thus,
$\displaystyle\int \sec(\theta)\,d\theta=\ln(\sec(\theta)+\tan(\theta))+C.$
Voila!
We note that if $\theta$ is restricted to the interval $(-\pi/2,\pi/2)$ as discussed above, then we always have $\sec(\theta)+\tan(\theta)>0,$ so there is no need to put the argument of the logarithm in absolute values.
Well, I’ve done my best to convince you of the wonder of hyperbolic trigonometric substitutions! If integrating $\sec(\theta)$ didn’t do it, well, that’s the best I’ve got.
The next installment of hyperbolic trigonometry? The Gudermannian function! What’s that, you ask? You’ll have to wait until next time — or I suppose you can just google it….
## Calculus: Hyperbolic Trigonometry, I
love hyperbolic trigonometry. I always include it when I teach calculus, as I think it is important for students to see. Why?
1. Many applications in the sciences use hyperbolic trigonometry; for example, the use of Laplace transforms in solving differential equations, various applications in physics, modeling population growth (the logistic model is a hyperbolic tangent curve);
2. Hyperbolic trigonometric substitutions are, in many instances, easier than circular trigonometric substitutions, especially when a substitution involving $\tan(x)$ or $\sec(x)$ is involved;
3. Students get to see another form of trigonometry, and compare the new form with the old;
4. Hyperbolic trigonometry is fun.
OK, maybe that last reason is a bit of hyperbole (though not for me).
Not everyone thinks this way. I once had a colleague who told me she did not teach hyperbolic trigonometry because it wasn’t on the AP exam. What do you say to someone who says that? I dunno….
In any case, I want to introduce the subject here for you, and show you some interesting aspects of hyperbolic trigonometry. I’m going to stray from my habit of not discussing things you can find anywhere online, since in order to get to the better stuff, you need to know the basics. I’ll move fairly quickly through the introductory concepts, though.
The hyperbolic cosine and sine are defined by
$\cosh(x)=\dfrac{e^x+e^{-x}}2,\quad\sinh(x)=\dfrac{e^x-e^{-x}}2,\quad x\in\mathbb{R}.$
I will admit that when I introduce this definition, I don’t have an accessible, simple motivation for doing so. I usually say we’ll learn a lot more as we work with these definitions, so if anyone has a good idea in this regard, I’d be interested to hear it.
The graphs of these curves are shown below.
The graph of $\cosh(x)$ is shown in blue, and the graph of $\sinh(x)$ is shown in red. The dashed orange graph is $y=e^{x}/2,$ which is easily seen to be asymptotic to both graphs.
Parallels to the circular trigonometric functions are already apparent: $y=\cosh(x)$ is an even function, just like $y=\cos(x).$ Similarly, $\sinh(x)$ is odd, just like $\sin(x).$
Another parallel which is only slight less apparent is the fundamental relationship
$\cosh^2(x)-\sinh^2(x)=1.$
Thus, $(\cosh(x),\sinh(x))$ lies on a unit hyperbola, much like $(\cos(x),\sin(x))$ lies on a unit circle.
While there isn’t a simple parallel with circular trigonometry, there is an interesting way to characterize $\cosh(x)$ and $\sinh(x).$ Recall that given any function $f(x),$ we may define
$E(x)=\dfrac{f(x)+f(-x)}2,\quad O(x)=\dfrac{f(x)-f(-x)}2$
to be the even and odd parts of $f(x),$ respectively. So we might simply say that $\cosh(x)$ and $\sinh(x)$ are the even and odd parts of $e^x,$ respectively.
There are also many properties of the hyperbolic trigonometric functions which are reminiscent of their circular counterparts. For example, we have
$\sinh(2x)=2\sinh(x)\cosh(x)$
and
$\sinh(x+y)=\sinh(x)\cosh(y)+\sinh(y)\cosh(x).$
None of these are especially difficult to prove using the definitions. It turns out that while there are many similarities, there are subtle differences. For example,
$\cosh(x+y)=\cosh(x)\cosh(y)+\sinh(x)\sinh(y).$
That is, while some circular trigonometric formulas become hyperbolic just by changing $\cos(x)$ to $\cosh(x)$ and $\sin(x)$ to $\sinh(x),$ sometimes changes of sign are necessary.
These changes of sign from circular formulas are typical when working with hyperbolic trigonometry. One particularly interesting place the change of sign arises is when considering differential equations, although given that I’m bringing hyperbolic trigonometry into a calculus class, I don’t emphasize this relationship. But recall that $\cos(x)$ is the unique solution to the differential equation
$y''+y=0,\quad y(0)=1,\quad y'(0)=0.$
Similarly, we see that $\cosh(x)$ is the unique solution to the differential equation
$y''-y=0,\quad y(0)=1,\quad y'(0)=0.$
Again, the parallel is striking, and the difference subtle.
Of course it is straightforward to see from the definitions that $(\cosh(x))'=\sinh(x)$ and $(\sinh(x))'=\cosh(x).$ Gone are the days of remembering signs when differentiating and integrating trigonometric functions! This is one feature of hyperbolic trigonometric functions which students always appreciate….
Another nice feature is how well-behaved the hyperbolic tangent is (as opposed to needing to consider vertical asymptotes in the case of $\tan(x)$). Below is the graph of $y=\tanh(x)=\sinh(x)/\cosh(x).$
The horizontal asymptotes are easily calculated from the definitions. This looks suspiciously like the curves obtained when modeling logistic growth in populations; that is, finding solutions to
$\dfrac{dP}{dt}=kP(C-P).$
In fact, these logistic curves are hyperbolic tangents, which we will address in more detail in a later post.
One of the most interesting things about hyperbolic trigonometric functions is that their inverses have closed formulas — in striking contrast to their circular counterparts. I usually have students work this out, either in class or as homework; the derivation is quite nice, so I’ll outline it here.
So let’s consider solving the equation $x=\sinh(y)$ for $y.$ Begin with the definition:
$x=\dfrac{e^y-e^{-y}}2.$
The critical observation is that this is actually a quadratic in $e^y:$
$(e^y)^2-2xe^y-1=0.$
All that is necessary is to solve this quadratic equation to yield
$e^y=x\pm\sqrt{1+x^2},$
and note that $x-\sqrt{1+x^2}$ is always negative, so that we must choose the positive sign. Thus,
$y=\hbox{arcsinh}(x)=\ln(x+\sqrt{1+x^2}).$
And this is just the beginning! At this stage, I also offer more thought-provoking questions like, “Which is larger, $\cosh(\ln(42))$ or $\ln(\cosh(42))?$ These get students working with the definitions and thinking about asymptotic behavior.
Next week, I’ll go into more depth about the calculus of hyperbolic trigonometric functions. Stay tuned!
## Calculus: The Geometry of Polynomials, II
The original post on The Geometry of Polynomials generated rather more interest that usual. One reader, William Meisel, commented that he wondered if something similar worked for curves like the Folium of Descartes, given by the equation
$x^3+y^3=3xy,$
and whose graph looks like:
I replied that yes, I had success, and what I found out would make a nice follow-up post rather than just a reply to his comment. So let’s go!
Just a brief refresher: if, for example, we wanted to describe the behavior of $y=2(x-4)(x-1)^2$ where it crosses the x-axis at $x=1,$ we simply retain the $(x-1)^2$ term and substitute the root $x=1$ into the other terms, getting
$y=2(1-4)(x-1)^2=-6(x-1)^2$
as the best-fitting parabola at $x=1.$
$\displaystyle\lim_{x\to1}\dfrac y{(x-1)^2}=-6.$
For examples like the polynomial above, this limit is always trivial, and is essentially a simple substitution.
What happens when we try to evaluate a similar limit with the Folium of Descartes? It seems that a good approximation to this curve at $x=0$ (the U-shaped piece, since the sideways U-shaped piece involves writing $x$ as a function of $y$) is $y=x^2/3,$ as shown below.
To see this, we need to find
$\displaystyle\lim_{x\to0}\dfrac y{x^2}.$
After a little trial and error, I found it was simplest to use the substitution $z=y/x^2,$ and so rewrite the equation for the Folium of Descartes by using the substitution $y=x^2z,$ which results in
$1+x^3z^3=3z.$
Now it is easy to see that as $x\to0,$ we have $z\to1/3,$ giving us a good quadratic approximation at the origin.
Success! So I thought I’d try some more examples, and see how they worked out. I first just changed the exponent of $x,$ looking at the curve
$x^n+y^3=3xy,$
shown below when $n=6.$
What would be a best approximation near the origin? You can almost eyeball a fifth-degree approximation here, but let’s assume we don’t know the appropriate power and make the substitution $y=x^kz,$ with $k$ yet to be determined. This results in
$x^{3k-n}z^3+1=3zx^{k+1-n}.$
Now observe that when $k=n-1,$ we have
$x^{2n-3}z^3+1=3z,$
so that $\displaystyle\lim_{x\to0}z=1/3.$ Thus, in our case with $n=6,$ we see that $y=x^5/3$ is a good approximation to the curve near the origin. The graph below shows just how good an approximation it is.
OK, I thought to myself, maybe I just got lucky. Maybe introduce a change which will really alter the nature of the curve, such as
$x^3+y^3=3xy+1,$
whose graph is shown below.
Here, the curve passes through the x-axis at $x=1,$ with what appears to be a linear pass-through. This suggests, given our previous work, the substitution $y=(x-1)z,$ which results in
$x^3+(x-1)^3z^3=3x(x-1)z+1.$
We don’t have much luck with $\displaystyle\lim_{x\to1}z$ here. But if we move the $1$ to the other side and factor, we get
$(x-1)(x^2+x+1)+(x-1)^3z^3=3x(x-1)z.$
Nice! Just divide through by $x-1$ to obtain
$x^2+x+1+(x-1)^2z=3xz.$
Now a simple calculation reveals that $\displaystyle\lim_{x\to1}z=1.$ And sure enough, the line $y=x-1$ does the trick:
Then I decided to change the exponent again by considering
$x^n+y^3=3xy+1.$
Here is the graph of the curve when $n=6:$
It seems we have two roots this time, with linear pass-throughs. Let’s try the same idea again, making the substitution $y=(x-1)z,$ moving the $1$ over, factoring, and dividing through by $x-1.$ This results in
$x^{n-1}+x^{n-2}+\cdots+1+(x-1)^2z^3=3xz.$
It is not difficult to calculate that $\displaystyle\lim_{x\to1}z=n/3.$
Now things become a bit more interesting when $n$ is even, since there is always a root at $x=-1$ in this case. Here, we make the substitution $y=(x+1)z,$ move the $1$ over, and divide by $x+1,$ resulting in
$\dfrac{x^n-1}{x+1}+(x+1)^2z^3=3xz.$
But since $n$ is even, then $x^2-1$ is a factor of $x^n-1,$ so we have
$(x-1)(x^{n-2}+x^{n-4}+\cdots+x^2+1)+(x+1)^2z^3=3xz.$
Substituting $x=-1$ in this equation gives
$-2\left(\dfrac n2\right)=3(-1)z,$
which immediately gives $\displaystyle\lim_{x\to1}z=n/3$ as well! This is a curious coincidence, for which I have no nice geometrical explanation. The case when $n=6$ is illustrated below.
This is where I stopped — but I was truly surprised that everything I tried actually worked. I did a cursory online search for Taylor series of implicitly defined functions, but this seems to be much less popular than series for $y=f(x).$
Anyone more familiar with this topic care to chime in? I really enjoyed this brief exploration, and I’m grateful that William Meisel asked his question about the Folium of Descartes. These are certainly instances of a larger phenomenon, but I feel the statement and proof of any theorem will be somewhat more complicated than the analogous results for explicitly defined functions.
And if you find some neat examples, post a comment! I’d enjoy writing another follow-up post if there is continued interested in this topic.
## Calculus: Linear Approximations, II
As I mentioned last week, I am a fan of emphasizing the idea of a derivative as a linear approximation. I ended that discussion by using this method to find the derivative of $\tan(x).$ Today, we’ll look at some more examples, and then derive the product, quotient and chain rules.
Differentiating $\sec(x)$ is particularly nice using this method. We first approximate
$\sec(x+h)=\dfrac1{\cos(x+h)}\approx\dfrac1{\cos(x)-h\sin(x)}.$
Then we factor out a $\cos(x)$ from the denominator, giving
$\sec(x+h)\approx\dfrac1{\cos(x)(1-h\tan(x))}.$
As we did at the end of last week’s post, we can make $h$ as small as we like, and so approximate by considering $1/(1-h\tan(x))$ as the sum of an infinite series:
$\dfrac1{1-h\tan(x)}\approx1+h\tan(x).$
Finally, we have
$\sec(x+h)\approx\dfrac{1+h\tan(x)}{\cos(x)}=\sec(x)+h\sec(x)\tan(x),$
which gives the derivative of $\sec(x)$ as $\sec(x)\tan(x).$
We’ll look at one more example involving approximating with geometric series before moving on to the product, quotient, and chain rules. Consider differentiating $x^{-n}.$ We first factor the denominator:
$\dfrac1{(x+h)^n}=\dfrac1{x^n(1+h/x)^n}.$
Now approximate
$\dfrac1{1+h/x}\approx1-\dfrac hx,$
so that, to first order,
$\dfrac1{(1+h/x)^n}\approx \left(1-\dfrac hx\right)^{\!\!n}\approx 1-\dfrac{nh}x.$
This finally results in
$\dfrac1{(x+h)^n}\approx \dfrac1{x^n}\left(1-\dfrac{nh}x\right)=\dfrac1{x^n}+h\dfrac{-n}{x^{n+1}},$
giving us the correct derivative.
Now let’s move on to the product rule:
$(fg)'(x)=f(x)g'(x)+f'(x)g(x).$
Here, and for the rest of this discussion, we assume that all functions have the necessary differentiability.
We want to approximate $f(x+h)g(x+h),$ so we replace each factor with its linear approximation:
$f(x+h)g(x+h)\approx (f(x)+hf'(x))(g(x)+hg'(x)).$
Now expand and keep only the first-order terms:
$f(x+h)g(x+h)\approx f(x)g(x)+h(f(x)g'(x)+f'(x)g(x)).$
And there’s the product rule — just read off the coefficient of $h.$
There is a compelling reason to use this method. The traditional proof begins by evaluating
$\displaystyle\lim_{h\to0}\dfrac{f(x+h)g(x+h)-f(x)g(x)}h.$
The next step? Just add and subtract $f(x)g(x+h)$ (or perhaps $f(x+h)g(x)$). I have found that there is just no way to convincingly motivate this step. Yes, those of us who have seen it crop up in various forms know to try such tricks, but the typical first-time student of calculus is mystified by that mysterious step. Using linear approximations, there is absolutely no mystery at all.
The quotient rule is next:
$\left(\dfrac fg\right)^{\!\!\!'}\!(x)=\dfrac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}.$
First approximate
$\dfrac{f(x+h)}{g(x+h)}\approx\dfrac{f(x)+hf'(x)}{g(x)+hg'(x)}.$
Now since $h$ is small, we approximate
$\dfrac1{g(x)+hg'(x)}\approx\dfrac1{g(x)}\left(1-h\dfrac{g'(x)}{g(x)}\right),$
so that
$\dfrac{f(x+h)}{g(x+h)}\approx(f(x)+hf'(x))\cdot\dfrac1{g(x)}\left(1-h\dfrac{g'(x)}{g(x)}\right).$
Multiplying out and keeping just the first-order terms results in
$\dfrac{f(x+h)}{g(x+h)}\approx f(x)g(x)+h\dfrac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}.$
Voila! The quotient rule. Now usual proofs involve (1) using the product rule with $f(x)$ and $1/g(x),$ but note that this involves using the chain rule to differentiate $1/g(x);$ or (2) the mysterious “adding and subtracting the same expression” in the numerator. Using linear approximations avoids both.
The chain rule is almost ridiculously easy to prove using linear approximations. Begin by approximating
$f(g(x+h))\approx f(g(x)+hg'(x)).$
Note that we’re replacing the argument to a function with its linear approximation, but since we assume that $f$ is differentiable, it is also continuous, so this poses no real problem. Yes, perhaps there is a little hand-waving here, but in my opinion, no rigor is really lost.
Since $g$ is differentiable, then $g'(x)$ exists, and so we can make $hg'(x)$ as small as we like, so the “$hg'(x)$” term acts like the “$h$” term in our linear approximation. Additionally, the “$g(x)$” term acts like the “$x$” term, resulting in
$f(g(x+h)\approx f(g(x))+hg'(x)f'(g(x)).$
Reading off the coefficient of $h$ gives the chain rule:
$(f\circ g)'(x)=f'(g(x))g'(x).$
So I’ve said my piece. By this time, you’re either convinced that using linear approximations is a good idea, or you’re not. But I think these methods reflect more accurately the intuition behind the calculations — and reflect what mathematicians do in practice.
In addition, using linear approximations involves more than just mechanically applying formulas. If all you ever do is apply the product, quotient, and chain rules, it’s just mechanics. Using linear approximations requires a bit more understanding of what’s really going on underneath the hood, as it were.
If you find more neat examples of differentiation using this method, please comment! I know I’d be interested, and I’m sure others would as well.
In my next installment (or two or three) in this calculus series, I’ll talk about one of my favorite topics — hyperbolic trigonometry.
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Synopsis
# Flexible Electronics, Heal Thyself
Physics 12, s12
A suspension of copper particles fixes breaks in electronic connections, providing a possible way to heal damaged circuits.
Modern electronics are increasingly lightweight and durable, but they are structurally rigid. Future electronics using thin semiconductors and flexible substrates would allow for a wide range of applications, including wearable medical diagnostic devices and roll-up displays. However, the interconnects—the thin wires linking the logic gates and other circuit components—are prone to breakage when bent, making flexible electronics unreliable in their present form. In a new experiment, Amit Kumar and colleagues at the Indian Institute of Science, Bangalore, and the University of Cambridge, UK, have demonstrated a new technique for self-healing electronics. Unlike previous self-healing techniques, this method doesn’t require rare materials or the addition of complex circuitry.
The team suspended copper microspheres with 5- $𝜇\text{m}$ radii in silicone oil, an insulating fluid. They then submerged an open electrical connection in this mixture to simulate a broken circuit. The researchers applied a potential difference across the gap, as would be expected for a broken connection in an active circuit. This potential difference created an electric field that attracted the copper spheres. These moved through the silicone oil to form chains of loosely bound clusters of microspheres that bridged the gap. Heat from the current flowing through the chains stabilized them, creating a more stable wire-like connection. In contrast to the re-connections produced by other self-healing experiments, the copper-sphere patch was both flexible and stretchable.
To make the technique viable for applications, the researchers will need to find ways to use smaller copper particles to heal smaller circuit breaks, introduce the silicone-copper suspension into real devices, and eliminate crosstalk between wires caused by the conductive particles in the circuit.
This research is published in Physical Review Applied.
–Matthew R. Francis
Matthew R. Francis is a physicist and freelance science writer based in Cleveland, Ohio.
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• ### VERITAS contributions to the 35th International Cosmic Ray Conference(1709.07843)
Sept. 22, 2017 astro-ph.HE
Compilation of papers presented by the VERITAS Collaboration at the 35th International Cosmic Ray Conference (ICRC), held July 12 through July 20, 2017 in Busan, South Korea.
• ### Very-High-Energy $\gamma$-Ray Observations of the Blazar 1ES 2344+514 with VERITAS(1708.02829)
Aug. 9, 2017 astro-ph.HE
We present very-high-energy $\gamma$-ray observations of the BL Lac object 1ES 2344+514 taken by the Very Energetic Radiation Imaging Telescope Array System (VERITAS) between 2007 and 2015. 1ES 2344+514 is detected with a statistical significance above background of $20.8\sigma$ in $47.2$ hours (livetime) of observations, making this the most comprehensive very-high-energy study of 1ES 2344+514 to date. Using these observations the temporal properties of 1ES 2344+514 are studied on short and long times scales. We fit a constant flux model to nightly- and seasonally-binned light curves and apply a fractional variability test, to determine the stability of the source on different timescales. We reject the constant-flux model for the 2007-2008 and 2014-2015 nightly-binned light curves and for the long-term seasonally-binned light curve at the $> 3\sigma$ level. The spectra of the time-averaged emission before and after correction for attenuation by the extragalactic background light are obtained. The observed time-averaged spectrum above 200 GeV is satisfactorily fitted (${\chi^2/NDF = 7.89/6}$) by a power-law function with index $\Gamma = 2.46 \pm 0.06_{stat} \pm 0.20_{sys}$ and extends to at least 8 TeV. The extragalactic-background-light-deabsorbed spectrum is adequately fit (${\chi^2/NDF = 6.73/6}$) by a power-law function with index $\Gamma = 2.15 \pm 0.06_{stat} \pm 0.20_{sys}$ while an F-test indicates that the power-law with exponential cutoff function provides a marginally-better fit ($\chi^2/NDF$ = $2.56 / 5$) at the 2.1$\sigma$ level. The source location is found to be consistent with the published radio location and its spatial extent is consistent with a point source.
• We present constraints on the annihilation cross section of WIMP dark matter based on the joint statistical analysis of four dwarf galaxies with VERITAS. These results are derived from an optimized photon weighting statistical technique that improves on standard imaging atmospheric Cherenkov telescope (IACT) analyses by utilizing the spectral and spatial properties of individual photon events. We report on the results of $\sim$230 hours of observations of five dwarf galaxies and the joint statistical analysis of four of the dwarf galaxies. We find no evidence of gamma-ray emission from any individual dwarf nor in the joint analysis. The derived upper limit on the dark matter annihilation cross section from the joint analysis is $1.35\times 10^{-23} {\mathrm{ cm^3s^{-1}}}$ at 1 TeV for the bottom quark ($b\bar{b}$) final state, $2.85\times 10^{-24}{\mathrm{ cm^3s^{-1}}}$ at 1 TeV for the tau lepton ($\tau^{+}\tau^{-}$) final state and $1.32\times 10^{-25}{\mathrm{ cm^3s^{-1}}}$ at 1 TeV for the gauge boson ($\gamma\gamma$) final state.
• ### Discovery of Very High Energy Gamma Rays from 1ES 1440+122(1608.02769)
Aug. 9, 2016 astro-ph.HE
The BL Lacertae object 1ES 1440+122 was observed in the energy range from 85 GeV to 30 TeV by the VERITAS array of imaging atmospheric Cherenkov telescopes. The observations, taken between 2008 May and 2010 June and totalling 53 hours, resulted in the discovery of $\gamma$-ray emission from the blazar, which has a redshift $z$=0.163. 1ES 1440+122 is detected at a statistical significance of 5.5 standard deviations above the background with an integral flux of (2.8$\pm0.7_{\mathrm{stat}}\pm0.8_{\mathrm{sys}}$) $\times$ 10$^{-12}$ cm$^{-2}$ s$^{-1}$ (1.2\% of the Crab Nebula's flux) above 200 GeV. The measured spectrum is described well by a power law from 0.2 TeV to 1.3 TeV with a photon index of 3.1 $\pm$ 0.4$_{\mathrm{stat}}$ $\pm$ 0.2$_{\mathrm{sys}}$. Quasi-simultaneous multi-wavelength data from the Fermi Large Area Telescope (0.3--300 GeV) and the Swift X-ray Telescope (0.2--10 keV) are additionally used to model the properties of the emission region. A synchrotron self-Compton model produces a good representation of the multi-wavelength data. Adding an external-Compton or a hadronic component also adequately describes the data.
• ### VERITAS and Multiwavelength Observations of the BL Lacertae Object 1ES 1741+196(1603.07286)
March 23, 2016 astro-ph.HE
We present results from multiwavelength observations of the BL Lacertae object 1ES 1741+196, including results in the very-high-energy $\gamma$-ray regime using the Very Energetic Radiation Imaging Telescope Array System (VERITAS). The VERITAS time-averaged spectrum, measured above 180 GeV, is well-modelled by a power law with a spectral index of $2.7\pm0.7_{\mathrm{stat}}\pm0.2_{\mathrm{syst}}$. The integral flux above 180 GeV is $(3.9\pm0.8_{\mathrm{stat}}\pm1.0_{\mathrm{syst}})\times 10^{-8}$ m$^{-2}$ s$^{-1}$, corresponding to 1.6% of the Crab Nebula flux on average. The multiwavelength spectral energy distribution of the source suggests that 1ES 1741+196 is an extreme-high-frequency-peaked BL Lacertae object. The observations analysed in this paper extend over a period of six years, during which time no strong flares were observed in any band. This analysis is therefore one of the few characterizations of a blazar in a non-flaring state.
• ### A Search for Brief Optical Flashes Associated with the SETI Target KIC 8462852(1602.00987)
The F-type star KIC 8462852 has recently been identified as an exceptional target for SETI (search for extraterrestrial intelligence) observations. We describe an analysis methodology for optical SETI, which we have used to analyse nine hours of serendipitous archival observations of KIC 8462852 made with the VERITAS gamma-ray observatory between 2009 and 2015. No evidence of pulsed optical beacons, above a pulse intensity at the Earth of approximately 1 photon per m^2, is found. We also discuss the potential use of imaging atmospheric Cherenkov telescope arrays in searching for extremely short duration optical transients in general.
• ### Gamma rays from the quasar PKS 1441+25: story of an escape(1512.04434)
Dec. 14, 2015 astro-ph.CO, astro-ph.HE
Outbursts from gamma-ray quasars provide insights on the relativistic jets of active galactic nuclei and constraints on the diffuse radiation fields that fill the Universe. The detection of significant emission above 100 GeV from a distant quasar would show that some of the radiated gamma rays escape pair-production interactions with low-energy photons, be it the extragalactic background light (EBL), or the radiation near the supermassive black hole lying at the jet's base. VERITAS detected gamma-ray emission up to 200 GeV from PKS 1441+25 (z=0.939) during April 2015, a period of high activity across all wavelengths. This observation of PKS 1441+25 suggests that the emission region is located thousands of Schwarzschild radii away from the black hole. The gamma-ray detection also sets a stringent upper limit on the near-ultraviolet to near-infrared EBL intensity, suggesting that galaxy surveys have resolved most, if not all, of the sources of the EBL at these wavelengths.
• ### VERITAS Collaboration Contributions to the 34th International Cosmic Ray Conference(1510.01639)
Oct. 6, 2015 astro-ph.HE
Compilation of papers presented by the VERITAS Collaboration at the 34th International Cosmic Ray Conference (ICRC), held July 30 through August 6, 2015 in The Hague, The Netherlands.
• ### Science Highlights from VERITAS(1510.01269)
Oct. 5, 2015 astro-ph.HE
The Very Energetic Radiation Imaging Telescope Array System (VERITAS) is a ground-based array located at the Fred Lawrence Whipple Observatory in southern Arizona and is one of the world's most sensitive gamma-ray instruments at energies of 85 GeV to $>$30 TeV. VERITAS has a wide scientific reach that includes the study of extragalactic and Galactic objects as well as the search for astrophysical signatures of dark matter and the measurement of cosmic rays. In this paper, we will summarize the current status of the VERITAS observatory and present some of the scientific highlights from the last two years, focusing in particular on those results shown at the 2015 ICRC in The Hague, Netherlands.
• ### VERITAS Deep Observations of the Dwarf Spheroidal Galaxy Segue 1(1202.2144)
July 7, 2015 hep-ex, astro-ph.HE
The VERITAS array of Cherenkov telescopes has carried out a deep observational program on the nearby dwarf spheroidal galaxy Segue 1. We report on the results of nearly 48 hours of good quality selected data, taken between January 2010 and May 2011. No significant $\gamma$-ray emission is detected at the nominal position of Segue 1, and upper limits on the integrated flux are derived. According to recent studies, Segue 1 is the most dark matter-dominated dwarf spheroidal galaxy currently known. We derive stringent bounds on various annihilating and decaying dark matter particle models. The upper limits on the velocity-weighted annihilation cross-section are $\mathrm{<\sigma v >^{95% CL} \lesssim 10^{-23} cm^{3} s^{-1}}$, improving our limits from previous observations of dwarf spheroidal galaxies by at least a factor of two for dark matter particle masses $\mathrm{m_{\chi}\gtrsim 300 GeV}$. The lower limits on the decay lifetime are at the level of $\mathrm{\tau^{95% CL} \gtrsim 10^{24} s}$. Finally, we address the interpretation of the cosmic ray lepton anomalies measured by ATIC and PAMELA in terms of dark matter annihilation, and show that the VERITAS observations of Segue 1 disfavor such a scenario.
• ### Discovery of the spectroscopic binary nature of three bright southern Cepheids(1308.1855)
Aug. 8, 2013 astro-ph.SR
We present an analysis of spectroscopic radial velocity and photometric data of three bright Galactic Cepheids: LR Trianguli Australis (LR TrA), RZ Velorum (RZ Vel), and BG Velorum (BG Vel). Based on new radial velocity data, these Cepheids have been found to be members of spectroscopic binary systems. The ratio of the peak-to-peak radial velocity amplitude to photometric amplitude indicates the presence of a companion for LR TrA and BG Vel. IUE spectra indicate that the companions of RZ Vel and BG Vel cannot be hot stars. The analysis of all available photometric data revealed that the pulsation period of RZ Vel and BG Vel varies monotonically, due to stellar evolution. Moreover, the longest period Cepheid in this sample, RZ Vel, shows period fluctuations superimposed on the monotonic period increase. The light-time effect interpretation of the observed pattern needs long-term photometric monitoring of this Cepheid. The pulsation period of LR TrA has remained constant since the discovery of its brightness variation. Using statistical data, it is also shown that a large number of spectroscopic binaries still remain to be discovered among bright classical Cepheids.
• ### Discovery of the spectroscopic binary nature of six southern Cepheids(1301.7615)
Jan. 31, 2013 astro-ph.SR
We present the analysis of photometric and spectroscopic data of six bright Galactic Cepheids: GH Carinae, V419 Centauri, V898 Centauri, AD Puppis, AY Sagittarii, and ST Velorum. Based on new radial velocity data (in some cases supplemented with earlier data available in the literature), these Cepheids have been found to be members in spectroscopic binary systems. V898 Cen turned out to have one of the largest orbital radial velocity amplitude (> 40 km/s) among the known binary Cepheids. The data are insufficient to determine the orbital periods nor other orbital elements for these new spectroscopic binaries. These discoveries corroborate the statement on the high frequency of occurrence of binaries among the classical Cepheids, a fact to be taken into account when calibrating the period-luminosity relationship for Cepheids. We have also compiled all available photometric data that revealed that the pulsation period of AD Pup, the longest period Cepheid in this sample, is continuously increasing with Delta P = 0.004567 d/century, likely to be caused by stellar evolution. The wave-like pattern superimposed on the parabolic O-C graph of AD Pup may well be caused by the light-time effect in the binary system. ST Vel also pulsates with a continuously increasing period. The other four Cepheids are characterised with stable pulsation periods in the last half century.
• ### Seismic evidence for a rapidly rotating core in a lower-giant-branch star observed with Kepler(1206.3312)
June 14, 2012 astro-ph.SR
Rotation is expected to have an important influence on the structure and the evolution of stars. However, the mechanisms of angular momentum transport in stars remain theoretically uncertain and very complex to take into account in stellar models. To achieve a better understanding of these processes, we desperately need observational constraints on the internal rotation of stars, which until very recently were restricted to the Sun. In this paper, we report the detection of mixed modes - i.e. modes that behave both as g modes in the core and as p modes in the envelope - in the spectrum of the early red giant KIC7341231, which was observed during one year with the Kepler spacecraft. By performing an analysis of the oscillation spectrum of the star, we show that its non-radial modes are clearly split by stellar rotation and we are able to determine precisely the rotational splittings of 18 modes. We then find a stellar model that reproduces very well the observed atmospheric and seismic properties of the star. We use this model to perform inversions of the internal rotation profile of the star, which enables us to show that the core of the star is rotating at least five times faster than the envelope. This will shed new light on the processes of transport of angular momentum in stars. In particular, this result can be used to place constraints on the angular momentum coupling between the core and the envelope of early red giants, which could help us discriminate between the theories that have been proposed over the last decades.
• ### Atmospheric parameters of 82 red giants in the Kepler field(1205.5642)
May 25, 2012 astro-ph.SR
Context: Accurate fundamental parameters of stars are essential for the asteroseismic analysis of data from the NASA Kepler mission. Aims: We aim at determining accurate atmospheric parameters and the abundance pattern for a sample of 82 red giants that are targets for the Kepler mission. Methods: We have used high-resolution, high signal-to-noise spectra from three different spectrographs. We used the iterative spectral synthesis method VWA to derive the fundamental parameters from carefully selected high-quality iron lines. After determination of the fundamental parameters, abundances of 13 elements were measured using equivalent widths of the spectral lines. Results: We identify discrepancies in log g and [Fe/H], compared to the parameters based on photometric indices in the Kepler Input Catalogue (larger than 2.0 dex for log g and [Fe/H] for individual stars). The Teff found from spectroscopy and photometry shows good agreement within the uncertainties. We find good agreement between the spectroscopic log g and the log g derived from asteroseismology. Also, we see indications of a potential metallicity effect on the stellar oscillations. Conclusions: We have determined the fundamental parameters and element abundances of 82 red giants. The large discrepancies between the spectroscopic log g and [Fe/H] and values in the Kepler Input Catalogue emphasize the need for further detailed spectroscopic follow-up of the Kepler targets in order to produce reliable results from the asteroseismic analysis.
• ### VERITAS Observations of Gamma-Ray Bursts Detected by Swift(1109.0050)
Nov. 26, 2011 astro-ph.HE
We present the results of sixteen Swift-triggered GRB follow-up observations taken with the VERITAS telescope array from January, 2007 to June, 2009. The median energy threshold and response time of these observations was 260 GeV and 320 s, respectively. Observations had an average duration of 90 minutes. Each burst is analyzed independently in two modes: over the whole duration of the observations and again over a shorter time scale determined by the maximum VERITAS sensitivity to a burst with a t^-1.5 time profile. This temporal model is characteristic of GRB afterglows with high-energy, long-lived emission that have been detected by the Large Area Telescope (LAT) on-board the Fermi satellite. No significant VHE gamma-ray emission was detected and upper limits above the VERITAS threshold energy are calculated. The VERITAS upper limits are corrected for gamma-ray extinction by the extragalactic background light (EBL) and interpreted in the context of the keV emission detected by Swift. For some bursts the VHE emission must have less power than the keV emission, placing constraints on inverse Compton models of VHE emission.
• ### VERITAS Collaboration Contributions to the 32nd International Cosmic Ray Conference(1111.2390)
Nov. 10, 2011 astro-ph.HE
Compilation of papers contributed by the VERITAS Collaboration to the 32nd International Cosmic Ray Conference, held 11-18 August 2011 in Beijing, China.
• ### VERITAS: Status and Highlights(1111.1225)
Nov. 4, 2011 astro-ph.IM, astro-ph.HE
The VERITAS telescope array has been operating smoothly since 2007, and has detected gamma-ray emission above 100 GeV from 40 astrophysical sources. These include blazars, pulsar wind nebulae, supernova remnants, gamma-ray binary systems, a starburst galaxy, a radio galaxy, the Crab pulsar, and gamma-ray sources whose origin remains unidentified. In 2009, the array was reconfigured, greatly improving the sensitivity. We summarize the current status of the observatory, describe some of the scientific highlights since 2009, and outline plans for the future.
• ### Detection of Pulsed Gamma Rays Above 100 GeV from the Crab Pulsar(1108.3797)
Aug. 18, 2011 astro-ph.HE
We report the detection of pulsed gamma rays from the Crab pulsar at energies above 100 Gigaelectronvolts (GeV) with the VERITAS array of atmospheric Cherenkov telescopes. The detection cannot be explained on the basis of current pulsar models. The photon spectrum of pulsed emission between 100 Megaelectronvolts (MeV) and 400 GeV is described by a broken power law that is statistically preferred over a power law with an exponential cutoff. It is unlikely that the observation can be explained by invoking curvature radiation as the origin of the observed gamma rays above 100 GeV. Our findings require that these gamma rays be produced more than 10 stellar radii from the neutron star.
• ### Constructing a one-solar-mass evolutionary sequence using asteroseismic data from \textit{Kepler}(1108.2031)
Aug. 9, 2011 astro-ph.SR
Asteroseismology of solar-type stars has entered a new era of large surveys with the success of the NASA \textit{Kepler} mission, which is providing exquisite data on oscillations of stars across the Hertzprung-Russell (HR) diagram. From the time-series photometry, the two seismic parameters that can be most readily extracted are the large frequency separation ($\Delta\nu$) and the frequency of maximum oscillation power ($\nu_\mathrm{max}$). After the survey phase, these quantities are available for hundreds of solar-type stars. By scaling from solar values, we use these two asteroseismic observables to identify for the first time an evolutionary sequence of 1-M$_\odot$ field stars, without the need for further information from stellar models. Comparison of our determinations with the few available spectroscopic results shows an excellent level of agreement. We discuss the potential of the method for differential analysis throughout the main-sequence evolution, and the possibility of detecting twins of very well-known stars.
• ### Global asteroseismic properties of solar-like oscillations observed by Kepler : A comparison of complementary analysis methods(1105.0571)
May 3, 2011 astro-ph.SR
We present the asteroseismic analysis of 1948 F-, G- and K-type main-sequence and subgiant stars observed by the NASA {\em Kepler Mission}. We detect and characterise solar-like oscillations in 642 of these stars. This represents the largest cohort of main-sequence and subgiant solar-like oscillators observed to date. The photometric observations are analysed using the methods developed by nine independent research teams. The results are combined to validate the determined global asteroseismic parameters and calculate the relative precision by which the parameters can be obtained. We correlate the relative number of detected solar-like oscillators with stellar parameters from the {\em Kepler Input Catalog} and find a deficiency for stars with effective temperatures in the range $5300 \lesssim T_\mathrm{eff} \lesssim 5700$\,K and a drop-off in detected oscillations in stars approaching the red edge of the classical instability strip. We compare the power-law relationships between the frequency of peak power, $\nu_\mathrm{max}$, the mean large frequency separation, $\Delta\nu$, and the maximum mode amplitude, $A_\mathrm{max}$, and show that there are significant method-dependent differences in the results obtained. This illustrates the need for multiple complementary analysis methods to be used to assess the robustness and reproducibility of results derived from global asteroseismic parameters.
• ### Predicting the detectability of oscillations in solar-type stars observed by Kepler(1103.0702)
March 3, 2011 astro-ph.SR
Asteroseismology of solar-type stars has an important part to play in the exoplanet program of the NASA Kepler Mission. Precise and accurate inferences on the stellar properties that are made possible by the seismic data allow very tight constraints to be placed on the exoplanetary systems. Here, we outline how to make an estimate of the detectability of solar-like oscillations in any given Kepler target, using rough estimates of the temperature and radius, and the Kepler apparent magnitude.
• ### VERITAS Search for VHE Gamma-ray Emission from Dwarf Spheroidal Galaxies(1006.5955)
Sept. 13, 2010 astro-ph.CO
Indirect dark matter searches with ground-based gamma-ray observatories provide an alternative for identifying the particle nature of dark matter that is complementary to that of direct search or accelerator production experiments. We present the results of observations of the dwarf spheroidal galaxies Draco, Ursa Minor, Bootes 1, and Willman 1 conducted by VERITAS. These galaxies are nearby dark matter dominated objects located at a typical distance of several tens of kiloparsecs for which there are good measurements of the dark matter density profile from stellar velocity measurements. Since the conventional astrophysical background of very high energy gamma rays from these objects appears to be negligible, they are good targets to search for the secondary gamma-ray photons produced by interacting or decaying dark matter particles. No significant gamma-ray flux above 200 GeV was detected from these four dwarf galaxies for a typical exposure of ~20 hours. The 95% confidence upper limits on the integral gamma-ray flux are in the range 0.4-2.2x10^-12 photons cm^-2s^-1. We interpret this limiting flux in the context of pair annihilation of weakly interacting massive particles and derive constraints on the thermally averaged product of the total self-annihilation cross section and the relative velocity of the WIMPs. The limits are obtained under conservative assumptions regarding the dark matter distribution in dwarf galaxies and are approximately three orders of magnitude above the generic theoretical prediction for WIMPs in the minimal supersymmetric standard model framework. However significant uncertainty exists in the dark matter distribution as well as the neutralino cross sections which under favorable assumptions could further lower the limits.
• ### Prospective Type Ia Supernova Surveys From Dome A(1002.2948)
Feb. 15, 2010 astro-ph.CO
Dome A, the highest plateau in Antarctica, is being developed as a site for an astronomical observatory. The planned telescopes and instrumentation and the unique site characteristics are conducive toward Type Ia supernova surveys for cosmology. A self-contained search and survey over five years can yield a spectro-photometric time series of ~1000 z<0.08 supernovae. These can serve to anchor the Hubble diagram and quantify the relationship between luminosities and heterogeneities within the Type Ia supernova class, reducing systematics. Larger aperture (>4-m) telescopes are capable of discovering supernovae shortly after explosion out to z~3. These can be fed to space telescopes, and can isolate systematics and extend the redshift range over which we measure the expansion history of the universe.
• ### The first high-amplitude delta Scuti star in an eclipsing binary system(0707.4540)
July 31, 2007 astro-ph
We report the discovery of the first high-amplitude delta Scuti star in an eclipsing binary, which we have designated UNSW-V-500. The system is an Algol-type semi-detached eclipsing binary of maximum brightness V = 12.52 mag. A best-fitting solution to the binary light curve and two radial velocity curves is derived using the Wilson-Devinney code. We identify a late A spectral type primary component of mass 1.49+/-0.02 M_sun and a late K spectral type secondary of mass 0.33+/-0.02 M_sun, with an inclination of 86.5+/-1.0 degrees, and a period of 5.3504751+/-0.0000006 d. A Fourier analysis of the residuals from this solution is performed using PERIOD04 to investigate the delta Scuti pulsations. We detect a single pulsation frequency of f_1 = 13.621+/-0.015 c/d, and it appears this is the first overtone radial mode frequency. This system provides the first opportunity to measure the dynamical mass for a star of this variable type; previously, masses have been derived from stellar evolution and pulsation models.
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# Homework Help: Rates of change
1. Oct 12, 2011
### Nitrate
1. The problem statement, all variables and given/known data
1. An oil tanker springs a leak creating a circular oil slick that grows until its radius is 3.0 km.
a.) What is the formula describing the relation between the area of the slick and its radius?
2. Relevant equations
Area of a circle: (pi)r^2
3. The attempt at a solution
Is the equation for a) y = (pi)r^2,
or is it the derivative: y = 2(pi)r?
2. Oct 12, 2011
### vrmuth
I think you mean the difference between the RATE OF CHANGE of the area of the slick and that of the radius , please make it clear .
3. Oct 12, 2011
### HallsofIvy
The formula you give $A= \pi r^2$ relates the area of a circle to its radius. There is nothing said in your problem, at least the part you posted, about rates of change.
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# Math Help - Trig equations need help(tried to work these but not sure if they are right)
1. ## Trig equations need help(tried to work these but not sure if they are right)
Complete the form of the equation in rectangular form
"r= -14 sin theta"
I'm having factoring issues with this one, I got " x^2+(y+7)^2=14 but, when I try to check it, x^2+(y^2+14y+14)=14 doesn't come out the same. What am I doing wrong in the factoring?
The rectangular coordinates of a point are given. Find the polar coordinates(r,theta) of this point with theta expresed in radians. Let r>0 and 0<or equal to theta <2pie
(3,-3{squareroot}3)
I used r=sqrt x^2 + y^2 to get 6
then used tan inverse y/x to get -3sqrt3/3= -pie/3
and got (6,-pie/3)
did I do this correctly?
2. Originally Posted by ConfusedMath
Complete the form of the equation in rectangular form
"r= -14 sin theta"
I'm having factoring issues with this one, I got " x^2+(y+7)^2=14 but, when I try to check it, x^2+(y^2+14y+14)=14 doesn't come out the same. What am I doing wrong in the factoring?
You have $r = -14\sin{\theta}$.
You should know that $y = r\sin{\theta}$ by definition.
So $\sin{\theta} = \frac{y}{r}$.
Therefore:
$r = -14\left(\frac{y}{r}\right)$
$r = -\frac{14y}{r}$
$r^2 = -14y$.
You should also know that $x^2 + y^2 = r^2$.
So $x^2 + y^2 = -14y$
$x^2 + y^2 + 14y = 0$
$x^2 + y^2 + 14y + 7^2 = 7^2$
$(x - 0)^2 + (y + 7)^2 = 49$.
3. Originally Posted by ConfusedMath
The rectangular coordinates of a point are given. Find the polar coordinates(r,theta) of this point with theta expresed in radians. Let r>0 and 0<or equal to theta <2pie
(3,-3{squareroot}3)
I used r=sqrt x^2 + y^2 to get 6
then used tan inverse y/x to get -3sqrt3/3= -pie/3
and got (6,-pie/3)
did I do this correctly?
You have $(x, y) = (3, -3\sqrt{3})$.
You know that $r^2 = x^2 + y^2$
$r^2 = 3^2 + (-3\sqrt{3})^2$
$r^2 = 9 + 27$
$r^2 = 36$
$r = 6$.
Since $x = r\cos{\theta}$ and $y = r\sin{\theta}$
This means
$6\cos{\theta} = 3$ and $6\sin{\theta} = -3\sqrt{3}$.
Since the cosine is positive and the sine is negative, this suggests the angle is in the fourth quadrant.
Now, if we remember that
$\frac{y}{x} = \frac{r\sin{\theta}}{r\cos{\theta}} = \tan{\theta}$
$\frac{-3\sqrt{3}}{3} = \tan{\theta}$
$\tan{\theta} = -\sqrt{3}$
$\theta = 2\pi - \frac{\pi}{3}$, since $\theta$ is in the fourth quadrant.
$\theta = \frac{5\pi}{3}$.
4. Thank you! That helps alot!
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0
### Research Papers: Air Emissions From Fossil Fuel Combustion
J. Energy Resour. Technol. 2008;130(1):011101-011101-8. doi:10.1115/1.2824295.
This study investigated normal heptane ($N$-heptane)-diesel combustion and odorous emissions in a direct injection diesel engine during and after engine warmup at idling. The odor is a little worse with $N$-heptane and blends than that of diesel fuel due to overleaning of the mixture. In addition, formaldehyde (HCHO) and total hydrocarbon (THC) in the exhaust increase with increasing $N$-heptane content. However, 50% and 100% $N$-heptane showed lower eye irritation than neat diesel fuel. Due to low boiling point of $N$-heptane, adhering fuel on the combustion chamber wall is small and as a single-component $C7$ fuel, relatively high volatile components present in the exhaust are low. This may cause lower eye irritation. On the contrary, bulk in-cylinder gas temperature is lower and ignition delay significantly increases for 50% and 100% $N$-heptane due to the low boiling point, high latent heat of evaporation, and low bulk modulus of compressibility of $N$-heptane than standard diesel fuel. This longer ignition delay and lower bulk in-cylinder gas temperature of $N$-heptane blends deteriorate exhaust odor and emissions of HCHO and THC.
Commentary by Dr. Valentin Fuster
J. Energy Resour. Technol. 2008;130(1):011102-011102-11. doi:10.1115/1.2824286.
Diesel engines are critical in fulfilling transportation and mechanical/electrical power generation needs throughout the world. The engine’s combustion by-products spawn health and environmental concerns, so there is a responsibility to develop emission reduction strategies. However, difficulties arise since the minimization of one pollutant often bears undesirable side effects. Although legislated standards have promoted successful emission reduction strategies for larger engines, developments in smaller displacement engines has not progressed in a similar fashion. In this paper, a reduced-order dynamic model is presented and experimentally validated to demonstrate the use of cooled exhaust gas recirculation (EGR) to alleviate the tradeoff between oxides of nitrogen reduction and performance preservation in a small displacement diesel engine. EGR is an effective method for internal combustion engine oxides of nitrogen $(NOx)$ reduction, but its thermal throttling diminishes power efficiency. The capacity to cool exhaust gases prior to merging with intake air may achieve the desired pollutant effect while minimizing engine performance losses. Representative numerical results were validated with experimental data for a variety of speed, load, and EGR testing scenarios using a $0.697l$ three-cylinder diesel engine equipped with cooled EGR. Simulation and experimental results showed a 16% drop in $NOx$ emissions using EGR, but experienced a 7% loss in engine torque. However, the use of cooled EGR realized a 23% $NOx$ reduction while maintaining a smaller performance compromise. The concurrence between simulated and experimental trends establishes the simplified model as a predictive tool for diesel engine performance and emission studies. Further, the presented model may be considered in future control algorithms to optimize engine performance and thermal and emission characteristics.
Commentary by Dr. Valentin Fuster
### Research Papers: Energy From Biomass
J. Energy Resour. Technol. 2008;130(1):011801-011801-6. doi:10.1115/1.2824247.
In (bubbling) fluidized-bed combustion and gasification of biomass, several potential problems are associated with the inorganic components of the fuel. A major problem area is defluidization due to bed agglomeration. The most common found process leading to defluidization in commercial-scale installations is “coating-induced” agglomeration. During reactor operation, a coating is formed on the surface of bed material grains and at certain critical conditions (e.g., coating thickness or temperature) sintering of the coatings initiates the agglomeration. In an experimental approach, this work describes a fundamental study on the mechanisms of defluidization. For the studied process of bed defluidization due to sintering of grain-coating layers, it was found that the onset of the process depends on (a) a critical coating thickness, (b) on the fluidization velocity when it is below approximately four times the minimum fluidization velocity, and (c) on the viscosity (stickiness) of the outside of the grains (coating).
Commentary by Dr. Valentin Fuster
### Research Papers: Energy Systems Analysis
J. Energy Resour. Technol. 2008;130(1):012001-012001-8. doi:10.1115/1.2824296.
Despite the immense environmental, technical, and financial promise of distributed generation (DG) technologies, they still constitute a very small percentage of electricity capacity in the United States. This manuscript answers the apparently paradoxical question: Why do technologies that offer such impressive benefits also find the least use? Going beyond technical explanations of problems related to system control, higher capital costs, and environmental compliance, this paper focuses on sociotechnical barriers related to utility preferences, business practices, regulatory bias, and consumer values. The approach helps us understand the glossing over of DG technologies, and identifies the impediments that policymakers must overcome if they are to find wider use.
Commentary by Dr. Valentin Fuster
J. Energy Resour. Technol. 2008;130(1):012002-012002-7. doi:10.1115/1.2835614.
Energy management in the industrial context is an important factor to attain energy savings as well as environmental efficiency. Often, linear regression models quite well represent the consumption of energy carriers and statistical process control (SPC) techniques, such as the cumulative sum (CUSUM) plot and Shewhart-like control charts, are currently applied to identify when a system changes the way energy is consumed. Despite the fact that SPC is widely applied in many fields, there is a lack of published material in energy management. The purpose of this paper is to widen the SPC techniques to be applied to energy management. Particular emphasis is given to small- and medium-sized enterprises since energy data are limited and generally known at system level. The CUSUM of the recursive residuals is proposed as the main tool for the analysis of energy consumption data, both for the historical and the monitoring phases. In addition, tabular CUSUM and EWMA control charts are also included.
Commentary by Dr. Valentin Fuster
### Research Papers: Petroleum Wells-Drilling/Production/Construction
J. Energy Resour. Technol. 2008;130(1):013101-013101-7. doi:10.1115/1.2824261.
A new type of rotary steering stabilizer used in a common rotary bottom hole assembly (BHA) to control well path was developed. In order for design and use of this kind of BHA, mathematical models were proposed for 3D mechanical analysis of rotary steering BHA with small deflection. The mathematical models include (1) differential equations; (2) boundary conditions of drill bit, stabilizer, diameter change, tangent point, and bore hole wall; (3) methods for calculating lateral forces and deflection angles of the bit; and (4) models for determining navigation ability and navigation parameters. As an example, a given rotary steering BHA was studied.
Commentary by Dr. Valentin Fuster
### Research Papers: Underground Injection and Storage
J. Energy Resour. Technol. 2008;130(1):013301-013301-5. doi:10.1115/1.2825174.
Our previous coreflood experiments—injecting pure $CO2$ into carbonate cores—showed that the process is a win-win technology, sequestrating $CO2$ while recovering a significant amount of hitherto unrecoverable natural gas that could help defray the cost of $CO2$ sequestration. In this paper, we report our findings on the effect of “impurities” in flue gas—$N2$, $O2$, $H2O$, $SO2$, $NO2$, and CO—on the displacement of natural gas during $CO2$ sequestration. Results show that injection of $CO2$ with approximately less than $1mole%$ impurities would result in practically the same volume of $CO2$ being sequestered as injecting pure $CO2$. This gas would have the advantage of being a cheaper separation process compared to pure $CO2$ as not all the impurities are removed. Although separation of $CO2$ out of flue gas is a costly process, it appears that this is necessary to maximize $CO2$ sequestration volume, reduce compression costs of $N2$ (approximately 80% of the stream), and improve sweep efficiency and gas recovery in the reservoir.
Commentary by Dr. Valentin Fuster
### Technical Briefs
J. Energy Resour. Technol. 2008;130(1):014501-014501-6. doi:10.1115/1.2824297.
An experiment was performed to study air∕water slug frequency in a horizontal clear pipe by means of visual inspection and differential pressure measurement in the range of $0–1∕s$. Results showed that a simplified model for slug pressure drop allowed the differential pressure data to compare favorably with visual observations for slug frequency. It was concluded that this technique gives a proper estimation of slug frequency, given only basic flow information. It is recommended that this technique be used when constrained to differential pressure or used to analyze existing differential pressure data.
Commentary by Dr. Valentin Fuster
J. Energy Resour. Technol. 2008;130(1):014502-014502-4. doi:10.1115/1.2835616.
With today’s high prices for natural gas and oil, the demand for oil and country tubular goods (OCTGs) with superior performance properties is very high. Failures in OCTG can be attributed to numerous sources, for example, makeup torque, corrosion, and galling. Thread galling is the most common mode of failure. This failure often leads to leakage, corrosion of the material, and loss of mechanical integrity. The failure of OCTG eventually amounts to excessive operational costs for the gas and oil industry. Numerous approaches have been taken to improve the galling resistance of OCTG connections. The advocacy of these approaches is often achieved through experimental studies using galling testers. There is a need to design and use effective galling testers to understand and improve the performance of OCTG connections. Thus, the objective of this paper is to present a concise review of literature related to the galling testers that may have applications to OCTG.
Commentary by Dr. Valentin Fuster
|
# If you were a linear algebra teacher, would you dock points for this?
#### Eclair_de_XII
Let's say you were proctoring some test that required proofs of Jordan canonical forms and rational canonical forms.
Would you dock points from a lazy student abbreviating the former as "J-canonical forms" and the latter as "$\mathbb{Q}$-canonical forms" in their proofs?
Related STEM Educators and Teaching News on Phys.org
#### Mark44
Mentor
Let's say you were proctoring some test that required proofs of Jordan canonical forms and rational canonical forms.
Would you dock points from a lazy student abbreviating the former as "J-canonical forms" and the latter as "$\mathbb{Q}$-canonical forms" in their proofs?
As someone who has taught linear algebra a number of times, no, I wouldn't take off points for those abbreviations. My focus would be more on the validity of the proofs.
#### vela
Staff Emeritus
Homework Helper
Not unless for some reason you had instructed them not to do that.
#### pasmith
Homework Helper
Best not to give examiners an excuse to dock marks.
#### MidgetDwarf
Let's say you were proctoring some test that required proofs of Jordan canonical forms and rational canonical forms.
Would you dock points from a lazy student abbreviating the former as "J-canonical forms" and the latter as "$\mathbb{Q}$-canonical forms" in their proofs?
silly. But I learned my lesson quickly in an intro linear course. If it is the actual instructor giving the exam, then I use what ever short hand notation he uses in lecture. If it is not the instructor proctoring the exam, then I am very formal with notation used and no shorthand.
"If you were a linear algebra teacher, would you dock points for this?"
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# Problem with natbib and class based on scrbook
I'm building a class for monographs, dissertations, and thesis from my department. I chose as a base the scrbook class that is part of koma-script. I need the natbib package numbers option to be called by a class option of the same name. I do not understand why the following example minimal does not work
The class Estilo. File Estilo.cls.
\NeedsTeXFormat{LaTeX2e}[1995/12/01]
\ProvidesClass{Estilo}[21/06/2018 UFRRJ monografias, dissertações e teses]
\DeclareOption*{\PassOptionsToClass{\CurrentOption}{scrbook}}
\ProcessOptions\relax
\RequirePackage[sort&compress]{natbib}
\DeclareOption{numbers}{\PassOptionsToPackage{numbers}{natbib}}
File Teste.tex
\documentclass{Estilo}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\begin{document}
Uma gravação em pedra encontrada em Karnak-Egito
\end{document}
When compiling this example we find the following error:
LaTeX Error: \RequirePackage or \LoadClass in Options Section
If the last line of the Estilo.cls class is commented the compile is successful, without any error message, warning or bad box.
I do not understand why this mistake happens. Could someone help me understand and, if possible, point out a solution.
Thank you very much
• Welcome to TeX.SX! Try moving your declare option command before processoptions. – TeXnician Jun 23 '18 at 8:46
• The instructions \DeclareOption{numbers}{\PassOptionsToPackage{numbers}{natbib} have to come before the \RequirePackage[sort&compress]{natbib} instruction. In fact, all three instructions can be combined into a single instruction: \RequirePackage[numbers,sort&compress]{natbib}. – Mico Jun 23 '18 at 9:06
• Thanks for the response Mico, but the problem persists in any order that puts the commands. Would you have another suggestion? I need to implement it as shown in the example to meet the rules of university composition. I thought of an alternative like that. – Benaia Sobreira de Jesus Lima Jun 23 '18 at 16:29
• \newif\if@loadrefnum \DeclareOption{refnum}{\@loadrefnumtrue} \if@loadrefnum \RequirePackage[numbers]{natbib} \fi Without the commands \RequirePackage[sort&compress]{natbib} \DeclareOption{numbers}{\PassOptionsToPackage{numbers}{natbib}} But it did not work, it gives the same error – Benaia Sobreira de Jesus Lima Jun 23 '18 at 16:30
• @cfr - The sort&compress option really makes sense only if used in combination with the numbers option -- the latter option sets up numeric-style citation call-outs, and "sorting and compressing" refers to a way the citation call-out numbers could be displayed. – Mico Jun 24 '18 at 4:51
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# Limit
• February 27th 2011, 06:26 AM
stripe501
Limit
I that the limit as x->2 of f(x)=(2x^2-8)/(x-2) is 8. But i don't understand how i get there, please help, i'm not really sure what i'm supposed to do
*Has been edited to show the original question
*Solved using factorising
• February 27th 2011, 06:37 AM
FernandoRevilla
Quote:
Originally Posted by stripe501
I that the limit as x-> 0 of f(x)=(2x^2-8)/(x^2-4) is 8. But i don't understand how i get there, please help, i'm not really sure what i'm supposed to do
Your function is continuous at $x=0$ so, the limit is $f(0)=2$ .
• February 27th 2011, 06:39 AM
skeeter
Quote:
Originally Posted by stripe501
I that the limit as x-> 0 of f(x)=(2x^2-8)/(x^2-4) is 8. no, it's not ... did you post the limit correctly?
...
• February 27th 2011, 06:40 AM
stripe501
Never mind, I figured it out :D
• February 27th 2011, 06:40 AM
stripe501
Yeah, I posted it wrong :/ thats why i couldn't get it
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# Rename tag class-bravo to class-b?
The FAA refers to the different types of airspace as Class A, B, C, D, E, and G. On the radio, B and D can be easily confused, so we use phonetics (Bravo or Delta), but when written it is not used that way.
I would suggest renaming to and possibly adding a tag synonym if it becomes a problem.
• I agree with that. If someone goes to type in class-bravo, class-b would come up before they finished typing bravo anyway. – called2voyage Dec 18 '13 at 19:33
• Outside the FAA jurisdiction though (which is pretty much most of the world) ;) it is not necessarily referred to as A-G, I know for a fact that it's referred to as Bravo, rather than B, in Germany (although Germany doesn't use Alpha and Bravo airspace, but do use Charlie, Delta, Foxtrot and Golf) Don't get me wrong, I prefer class-b, just saying it's not necessarily the most common in the grand scheme of things. – falstro Dec 19 '13 at 18:28
• @roe: Interesting, is it actually written as class Bravo in Germany? It is referred to as Class Bravo in the US too, but only when spoken. – Lnafziger Dec 19 '13 at 18:49
• I think some people will still try to type in bravo-airspace or charlie-airspace etc, so a tag synonym would be helpful in that case. – Bret Copeland Dec 18 '13 at 20:44
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# In my enemy is my friend
In my enemy is my friend
In my friend is my enemy
I exist because he does
But he exists from where I don't
Round and round we go again
Each placed in the other's domain
But most places we are seen
We are seen as part of the same dream
Where 1 is 2 and in that 2 are 2 more
My purpose is to show
You can't have one without the other, forever more!
What am I?
• Makes me think of light and darkness – dcfyj Nov 29 '16 at 15:43
• I can't understand this line Where 1 is 2 and in that 2 are 2 more here 2 is what? – Mukul Kumar Nov 29 '16 at 16:17
I think you are
yin and yang:
In my enemy is my friend
In my friend is my enemy
Seemingly contrary forces can be interlinked, such as good and bad or light and dark.
I exist because he does
But he exists from where I don't
The concept of bad can only exist if good does, and vice-versa, and where there is not bad, there can be good. (Same applies to other apparently opposite forces.)
Round and round we go again
Each placed in the other's domain
But most places we are seen
We are seen as part of the same dream
Sometimes there's a fine line between the two, or none at all.
Where 1 is 2 and in that 2 are 2 more
The one symbol has two swirly shapes in, each containing one small circle.
My purpose is to show
You can't have one without the other, forever more!
The symbol represents the concept of duality, which could be called timeless.
• I'd like to suggest editing the fifth explanation, since that line of riddle is actually originated from a sentence of I Ching which is well known by Chinese people. – Shane Hsu Nov 30 '16 at 9:06
• Please do @ShaneHsu, I think you can click on edit underneath. (I'm not sure how to explain better.) – pb8330 Nov 30 '16 at 9:28
Is it
Fire and water
In my enemy is my friend
In my friend is my enemy
In fire and water is oxygen.
I exist because he does
But he exists from where I don't
They both exist because of oxygen and water exists in the sea where fire can't.
Round and round we go again
Each placed in the other's domain
Water puts out fire, but fire evaporates water.
But most places we are seen
We are seen as part of the same dream
They are both part of the 4 elements - water, air, fire and earth.
Where 1 is 2 and in that 2 are 2 more
My purpose is to show
Not sure, but fire is to show light. @yitzih suggested H20?
You can't have one without the other, forever more!
You can't have water and fire without oxygen.
• You got the last two wrong... – Mukul Kumar Nov 29 '16 at 16:13
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Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > KEYWORD > HARDCORE SETS:
Reports tagged with Hardcore Sets:
TR08-103 | 22nd November 2008
that if $D$ is a distribution over $\{ 0,1\}^n$ of min-entropy at least $n-k$,
then for every $S$ and $\epsilon$ there is a circuit $C$ of size at most
|
# All Questions
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55 views
### Number of shifts in DES key schedule
In DES key schedule some rounds use one left shift while others use two left shifts. What is the reason behind this?
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### Can we use the sponge construct to efficiently authenticate any cipher?
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### Trying to understand MixColumns with Hexadecimal matrix
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### Use cases and implementations of RSA CRT
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### Is there a security problem with this prime generation algorithm?
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### How does Boneh–Lynn–Shacham work?
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### How does DPA work on AES?
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### Confidentiality then Integrity with different keys [duplicate]
Can anyone describe how we can do what is said in the below sentence which is in Mark Stamp crypto slides (PPTX) on slide number 101 : Can do a little better - about 1.5 “encryptions” The whole ...
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### Assuming a 1024qb quantum computer, how long to brute force 1024bit RSA, 256bit AES and 512bit SHA512
Assuming in the future there was a functioning 1024 qubit quantum supercomputer and it could run Shor's algorithm or Grover's algorithm to crack encryption very quickly. I'm interested in how the ...
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### Confidentiality then Integrity with different keys
Can anyone describe how we can do what is said in the below sentence which is in Mark Stamp crypto slides (PPTX) on slide number 101 : Can do a little better - about 1.5 “encryptions” The whole ...
45 views
...
48 views
### Getting inverse of a transposition key
I'm new to security stuff and I have some questions about the keys of transposition cryptography. If I'm given the encoding key to a transposition cipher, how do I get the decoding key for it? I ...
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### Solving for a One-time pad cipher help ; crib dragging doesnt work (no surprise) [duplicate]
Basically we're given the text 7ECC555AB95BF6EC605E5F22B772D2B34FF4636340D32FABC29B 73CB4855BE44F6EC60594C2BB47997B60EEE303049CD3CABC29B 64C6401BAF45F6A930435F3DF875C4E102F8742A45C824AFCA9B ...
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### Does encrypting with MGF1/SHA-512/1024-bit seed equal to a 1024-bit key block cipher?
Suppose: Alice uses SHA-512 with MGF1(as in PKCS#1 V2) and a 1024-bit random seed to generate a mask, XORs the mask with a message(M), gets a cipher text (CT), and sends the CT to her old friend ...
11 views
### Calculating inverse of transposition key cipher [duplicate]
I need some help on transposition cryptography. If I'm given the encoding key to a transposition cipher, how do I get the decoding key for it? I tried googling but couldn't find any steps on the ...
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### What is the fastest modular reduction algorithm available?
I have been browsing for the fastest and most efficient modular reduction algorithms and came across quite a few. But the one in A Fast Modular Reduction Method (2014) by Zhengjun Cao, Ruizhong Wei ...
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### Are there any disadvantages of using libsodiums helpers for en-/decoding/comparision instead of the native ones?
Libsodium has some helpers, which allow hexdecimal decoding, encoding and also constant-time string comparison. So if you use the php binding libsodium-php you ...
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### Does RSA have two trapdoors?
I watched some videos from Khan Academy explaining the algorithm for RSA encryption/decryption. They explained that there are two trapdoors, modular exponentiation being one and prime factorization ...
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### How to test an encryption against different attacks [closed]
For the last while I have been reading up on cryptography and the different kinds of methods for encryption (public key, blocks, etc) and the attacks against them. This got me wondering how to ...
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### Attack being kind of completely puzzled?
I am reading this paper "The security of Hidden Field Equations (HFE)" and at the end of the page 4 the author wrote: Thus, after all, this belief about every attack being kind of ”completely ...
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### Public-key encryption with associated data
In the symmetric-key world, we have authenticated encryption with associated data (AEAD). I'm looking for something similar, but for public-key encryption: public-key encryption with associated data (...
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### Is AES solvable by reducing to SAT?
Consider a known plaintext attack on AES — just so we have an actual system of equalities that we can feed to a SAT solver. Is AES solvable in this way? In other words, will the algorithm eventually ...
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### Password hashing in embedded systems
Are there good options for password hashing in embedded systems? These systems often cannot afford slow has functions. I can think of a few approaches: Store the (hashed) passwords in a HSM that ...
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### Fixed-prefix IV in CBC mode
I know the Initialization Vector needs to be unpredictable. Does it apply to the IV as a whole, or to every bit in the IV? I have multiple devices with poor entropy sharing the same key and I want to ...
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### Impact of Git's use of SHA-1? [closed]
Git uses SHA-1, and it is common to sign Git commit or tag hashes to authenticate a repo. Is this insecure now?
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### Probability of generating same master secret key in Identity-based Encryption
Suppose multiple servers use same IBE domain parameters (I mean same curve description parameters and field) for master secret key setup. Is there any possibility for generating the same system ...
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### Is it safe to use a key as message in keyed hashing (HMAC)?
Scenario: User gives password. A "password key" is derived from the password (e.g. by PBKDF2). A random salt is applied. The password key is hashed with a HMAC algorithm (e.g. HMAC-SHA256), with ...
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### How is the One Time Pad (OTP) perfectly secure?
The Wikipedia entry on One Time Pads (OTPs) states that if this cipher is used properly; ie, the keys are truly random and each "part" of the key is independent of every other "part", it's uncrackable,...
132 views
### Relationship between exponent and modulus in RSA (as RSA properties as listed in X.509)
In one of my assignments, I had the following question (Please read on. Not a homework assignment :) ): X.509 (1998 version) lists properties that RSA keys must satisfy to be secure. One such ...
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### Why is HMAC-SHA1 still considered secure?
This Q & A http://security.stackexchange.com/questions/33123/hotp-with-as-hmac-hashing-algoritme-a-hash-from-the-sha-2-family says that the security of HMAC-SHA1 does not depend on resistance to ...
86 views
### How to calculate entropy of output of a PRNG
Suppose the entropy of seed key is $H(n)$ which is $n$-bit long and random and the output bit length of the PRNG is $2^n$, which will be used as a long running key. Though it is obvious in that case, ...
63 views
### Why is the compression function in Miyaguchi-Preneel scheme secure?
I was reading about the Miyaguchi-Preneel scheme and had difficulty in understanding why the compression function, $h(H,m)=E(H, H \oplus M) \oplus M \oplus H$ can be called secure. The only resource I ...
33 views
### Is there an algorithm that allows verification that 2 encrypted or hashed bits of data are the same, given that I may only know half of the key/ salt?
We have an interesting problem. Currently, we store a salted hash of some secret key known only to the clients of our clients to allow us (and our clients) to find other records that were generated ...
61 views
### Is it OK to reseed a Deterministic Random Bit Generator from itself?
I want a deterministic PRNG for simulation purposes, and I don't want to have to worry about spurious correlations like those the DMCT is designed to protect the Mersenne Twister against. So even ...
32 views
### Reykeyed AES in CTR mode as a stream cipher
Suppose one uses AES in CTR mode to generate a keystream, but every $l$ bytes the $n$ bits that would follow are used as a new AES key. Assuming that AES is a secure PRP: Is this a secure stream ...
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You know how whenever you download a file you should really compare the hash of the download to the one provided on the website? This makes absolute sense, but it's a pain to do it letter for letter, digit for digit. So, I wrote this little script to take care of the job. Any comments are welcome.
#!/bin/bash
error_exit()
{
echo "$1" 1>&2 exit 1 } usage="usage: hash_checker downloaded_file hash_provided -a algorithm" downloaded_file= hash_given= hash_calc= algo="sha256" # check if file and hash were provided if [$# -lt 2 ]; then
error_exit "$usage" fi # parsing the provided hash and file downloaded_file="$1"
hash_given="$2" # parsing the algorithm, if provided if [ "$3" != "" ]; then
if [ "$3" = "-a" ]; then algo="$4"
else
error_exit "$usage" fi fi # check if input is a valid file if [ ! -f "$downloaded_file" ]; then
error_exit "Invalid file! Aborting."
fi
# calculate the hash for the file
hash_calc="$($algo'sum' $downloaded_file)" hash_array=($hash_calc)
hash_calc=${hash_array[0]} # compare the calculated hash to the provided one if [ "$hash_calc" = "$hash_given" ]; then echo "The hashes match. File seems to be valid." else echo "The hashes do not match. File does not seem to be valid." fi • Why not just use the -c option to sha1sum or md5sum to do the comparison? – Toby Speight Sep 13 '18 at 7:16 • Didn't know that was an option. Kinda expected something like this must exist, thanks. – iuvbio Sep 13 '18 at 7:25 1 Answer Notes: • I'd use getopts for arg parsing -- lots of examples on stackoverflow about how to use it. • always quote your variables • you should validate the algorithm: sum_exe="${algo}sum"
if ! type -P "$sum_exe" >/dev/null; then error_exit "'$algo' is an unknown checksum algorithm"
fi
• have the checksum program read from stdin, then you don't have to do your incorrect unsafe word parsing since the program will not print a filename
hash_calc=$( "$sum_exe" < "$downloaded_file" ) As the above doesn't work, let's use read from a process substitution read -r hash_calc _ < <("$sum_exe" < "\$downloaded_file")
• Thanks for your notes. Reading from stdin the filename is not printed, but - is appended to the hash, and thus the test fails. Is there a way around that other than more word parsing? – iuvbio Sep 12 '18 at 20:21
• I've updated my answer – glenn jackman Sep 12 '18 at 20:42
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02 Mar 2020
# Quickly recording notes using VS Code
History / Edit / PDF / EPUB / BIB / 2 min read (~316 words)
## Problem
I want to quickly take notes in the same file throughout the day using VS Code. How do I do that?
## Solution
My approach has been to use my VS Code extension Run Me, which I use to bind a keyboard shortcut to one of the commands I created. In my particular case, on any VS Code window I can press CTRLNumpad 2 and it will open a file under the following path: buffer/YYYY/MM/DD.md, where YYYY/MM/DD is replaced with the year/month/day. In this file I record all my notes within the day.
I use the following snippet, which I can trigger using dt, then pressing TAB. This replaces the dt string with a string of the form YYYY-MM-DD HH:MM:SS, which is the current year-month-day hour:minute:second.
{
"Datetime": {
"scope": "",
"prefix": "dt",
"body": [
"$CURRENT_YEAR-$CURRENT_MONTH-$CURRENT_DATE$CURRENT_HOUR:$CURRENT_MINUTE:$CURRENT_SECOND"
],
"description": "Date time"
}
}
I also use the Script Commands extension to do something slightly more complicated, which is to create strings of the form 2020-03-02 21:19:05 [nid://952], where nid://952 represents a unique note id (nid). The number that is generated is unique and is tracked by storing the last generated number in a text file that is read/written on each call to this command. A cheaper approach could have been to simply use the timestamp as unique note id. One downside of the timestamp as note id approach is that you don't have an idea of how many notes you've recorded so far, other than searching your notes and then counting the number of unique instances.
01 Mar 2020
# Visual Studio Code templates
History / Edit / PDF / EPUB / BIB / 2 min read (~314 words)
## Problem
I use VS Code and I'd like to insert templates into my documents. How do I do this?
## Solution
In VS Code, the concept of templates is called snippets. It is fairly easy to create a snippet, so I won't go into the details. Here's an example of a snippet I use to insert the date and time quickly into my documents.
{
"Datetime": {
"scope": "",
"prefix": "dt",
"body": [
"$CURRENT_YEAR-$CURRENT_MONTH-$CURRENT_DATE$CURRENT_HOUR:$CURRENT_MINUTE:$CURRENT_SECOND"
],
"description": "Date time"
}
}
With this snippet, I only have to type dt and then press TAB and dt gets replaced by the current date and time.
I use snippets to create the template of my questions and problems articles. My current approach is to open a non-existent file by calling code /path/to/new/file and VS Code opens the editor with this file. I can then write the content of this file, which will not automatically save until I manually save. Once I'm happy with the content of my article, I manually save, which means that in the future, any edit I make and then have the editor lose focus will be automatically saved and commit to git (my current workflow).
One of the things I don't particularly like with the current implementation of the snippets system is that the body needs to be a list of strings, where each item represents a new line. It is possible to create a single entry and use \n and \ to format the string, but that is not a clean approach. Those are limitations of jsonc. If it was possible to link to a file, then it would be "easy" to create a clean template.
22 Feb 2020
# Visual Studio Code Emoji extension
History / Edit / PDF / EPUB / BIB / 1 min read (~200 words)
## Problem
I use Visual Studio Code as my main editor and I am on Windows 7. I like to use emojis but those aren't properly rendered under Windows 7. Can I have pretty emojis in Visual Studio Code somehow?
## Solution
I developed an extension in 2018 called Emoji which uses EmojiOne emojis to replace their non rendered equivalent in Visual Studio Code.
To do this, the extension makes use of the createTextEditorDecorationType method available on the window object in order to inject CSS that adds a background image where the text emoji would be rendered.
The extension listens to two events to determine in which editor it needs to do the replacement: window.onDidChangeActiveTextEditor and workspace.onDidChangeTextDocument. In the first case we update the editor that is now the active one, in the second, we update the active document when the text content changed.
12 Feb 2020
# Visual Studio Code Run Me extension
History / Edit / PDF / EPUB / BIB / 2 min read (~228 words)
## Problem
I frequently run the same commands with different parameters but I have a terrible memory. I also use Visual Studio Code a lot.
## Solution
I developed an extension in 2018 called Run Me whose goal is to allow you to define commands that you can customize through a form, which is a series of questions that will be asked to you, before launching the command with the parameters you provided.
I've used it to do all kinds of things, from launching OBS to resetting the Windows 7 visuals when it lowers them down due to low memory. I also use it to automate various tasks such as creating new articles using a template, opening my buffer document that I use on a daily basis to write notes and more.
Here's an example of my configuration file which I use to start OBS and to reset the Windows 7 visuals.
"run-me": {
"commands": [
{
"identifier": "start_obs",
"description": "Start OBS x64",
"command": "\"C:\\Program Files (x86)\\obs-studio\\bin\\64bit\\obs64.exe\"",
"working_directory": "C:\\Program Files (x86)\\obs-studio\\bin\\64bit"
},
{
"identifier": "reset_visuals",
"description": "Reset W7 visuals",
"command": "sc stop uxsms & sc start uxsms"
}
]
}
|
2019
Том 71
№ 11
On the rate of convergence of an unstable solution of a stochastic differential equation
Mynbaeva G. U.
Abstract
We study the rate of convergence of the process $ξ(tT)/\sqrt{T}$ to the process $w(t)/σ$ as $T → ∞$, where $ξ(t)$ is a solution of the stochastic differential equationd $ξ(t) = a(ξ(t))dt + σ(ξ(t))dw(t)$.
English version (Springer): Ukrainian Mathematical Journal 46 (1994), no. 10, pp 1573-1577.
Citation Example: Mynbaeva G. U. On the rate of convergence of an unstable solution of a stochastic differential equation // Ukr. Mat. Zh. - 1994. - 46, № 10. - pp. 1424–1427.
Full text
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# How do you write the equation for a circle with center (-4, 1), radius 6?
###### Question:
How do you write the equation for a circle with center (-4, 1), radius 6?
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# Documentation
Lean.Compiler.LCNF.PullFunDecls
Local function declaration and join point being pulled.
Instances For
Equations
@[inline]
The PullM state contains the local function declarations and join points being pulled.
Equations
Extract from the state any local function declarations that depends on the given free variable. The idea is that we have to stop pulling these declarations because they depend on fvarId.
Equations
• One or more equations did not get rendered due to their size.
Equations
Equations
• = do let ps ←
Similar to findFVarDeps. Extract from the state any local function declarations that depends on the given parameters.
Equations
• One or more equations did not get rendered due to their size.
Construct the code fun p.decl k or jp p.decl k.
Equations
Attach the given array of local function declarations and join points to k.
Equations
• One or more equations did not get rendered due to their size.
Equations
• One or more equations did not get rendered due to their size.
Equations
• = do let __do_lift ← get pure __do_lift.snd[i]!
Extract from the state any local function declarations that depends on the given free variable, and attach to code k.
Equations
• = do let ps ←
Similar to attachFVarDeps. Extract from the state any local function declarations that depends on the given parameters, and attach to code k.
Equations
• = do let ps ←
Equations
• One or more equations did not get rendered due to their size.
Add local function declaration (or join point if isFun = false) to the state.
Pull local function declarations and join points in code. The state contains the declarations being pulled.
Pull local function declarations and join points in the given declaration.
Equations
• One or more equations did not get rendered due to their size.
|
# Fight Finance
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A person is thinking about borrowing \$100 from the bank at 7% pa and investing it in shares with an expected return of 10% pa. One year later the person will sell the shares and pay back the loan in full. Both the loan and the shares are fairly priced.
What is the Net Present Value (NPV) of this one year investment? Note that you are asked to find the present value ($V_0$), not the value in one year ($V_1$).
|
# Automorphisms and derivations of associative rings by V. Kharchenko
By V. Kharchenko
T moi, ... si favait su remark en revenir. One sel'Yice arithmetic has rendered the je n'y serais aspect aile.' human race. It has positioned good judgment again Jules Verne the place it belongs, at the topmost shelf subsequent to the dusty canister labelled 'discarded non- The sequence is divergent; for this reason we should be sense', capable of do anything with it. Eric T. Bell O. Heaviside arithmetic is a device for idea. A hugely beneficial device in an international the place either suggestions and non linearities abound. equally, all types of components of arithmetic function instruments for different components and for different sciences. utilising an easy rewriting rule to the quote at the correct above one reveals such statements as: 'One carrier topology has rendered mathematical physics .. .'; 'One carrier good judgment has rendered com puter technological know-how .. .'; 'One provider classification conception has rendered arithmetic .. .'. All arguably real. And all statements available this fashion shape a part of the raison d 'e\re of this sequence.
Read Online or Download Automorphisms and derivations of associative rings PDF
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Additional info for Automorphisms and derivations of associative rings
Example text
Indeed, if I 2 r = 0, then 0 = II(II aA)g, where A is such a 4} CHAPTER } set that A g= I} r. This set exists since I} is an ideal and I} ~ I g . , r = 0, which fact proves that I2 Let us now define the mapping a g: I2 ~ i E I 2 , then i =j g for a certain j E :? I a I, R E F . in the following way. If and we set ia g = ( ja) g , in which the right part of the equality is determined because ja ~ I. 9). We can now easily check if a ~ a g is the sought extension of g. In this case the formula other extension, j gag = (ja ) g g I, we j gag - j g a g'= ( ja)g - (ja)g = 0, g = g.
Any ideal I of a semiprime ring R has a zero intersection with its annihilator ann I. The direct sum I+ annI belongs to IF. Proof. The intersection III annI has a zero multiplication and, hence, equals zero. If (I + ann I)x =0, then Ix = 0 and, hence, x belongs to the ideal A= ann I. , xE All annA= 0, which is the required proof. 4. Definition. An ideal of the ring R is called essential if it has a zero intersection with any nonzero ideal of the ring R. 5. Lemma. The ideal I of a semiprime ring R belongs to IF iff it is essential.
By induction over quite integer over T of a certain degree m. Let k = 1 and a 1= diag( rl'0, •. O)E ~. ')' Then for k n - k rows and n - k we will show R to be Let us find an element any ~E ~ we have (a1 - t) ~ = 0, and, hence, at a 2 = ta 2 which is the required proof. Let k> 1. Let us present an arbitrary matrix aE Rk as a= [ 0] * * ° °°° a' r" (7) where a' is the (k - 1) x (k - 1) matrix, k - 1. Let us set (a'O) iI= 00 at = ~ E R k _ t' Then for all the elements at, r'i iff the columns and r'i ~E Rk we shall introduce the relation corresponding to the matrices at and r;.
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시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
2 초 512 MB 1 1 1 100.000%
문제
Let's play a new board game "Life Line".
The number of the players is greater than 1 and less than 10.
In this game, the board is a regular triangle in which many small regular triangles are arranged (See Figure l). The edges of each small triangle are of the same length.
Figure 1: The board
The size of the board is expressed by the number of vertices on the bottom edge of the outer triangle. For example, the size of the board in Figure 1 is 4.
At the beginning of the game, each player is assigned his own identification number between 1 and 9, and is given some stones on which his identification number is written.
Each player puts his stone in turn on one of the "empty" vertices. An "empty vertex" is a vertex that has no stone on it.
When one player puts his stone on one of the vertices during his turn, some stones might be removed from the board. The player gains points which is equal to the number of the removed stones of others, but loses points which is equal to the number of the removed stones of himself. The points of a player for a single turn is the points he gained minus the points he lost in that turn.
The conditions for removing stones are as follows:
• The stones on the board are divided into groups. Each group contains a set of stones whose numbers are the same and placed adjacently. That is, if the same numbered stones are placed adjacently, they belong to the same group.
• If none of the stones in a group is adjacent to at least one "empty" vertex, all the stones in that group are removed from the board.
Figure 2: The groups of stones
Figure 2 shows an example of the groups of stones.
Suppose that the turn of the player '4' comes now. If he puts his stone on the vertex shown in Figure 3a, the conditions will be satisfied to remove some groups of stones (shadowed in Figure 3b). The player gains 6 points, because the 6 stones of others are removed from the board (See Figure 3c).
Figure 3a Figure 3b Figure 3c
As another example, suppose that the turn of the player '2' comes in Figure 2. If the player puts his stone on the vertex shown in Figure 4a, the conditions will be satisfied to remove some groups of stones (shadowed in Figgue 4b). The player gains 4 points, because the 4 stones of others are removed. But, at the same time, he loses 3 points, because his 3 stones are removed. As the result, the player's points of this turn is 4 - 3 = 1 (See Figure 4c).
Figure 4a Figure 4b Figure 4c
When each player puts all of his stones on the board, the game is over. The total score of a player is the summation of the points of all of his turns.
Your job is to write a program that tells you the maximum points a player can get (i.e., the points he gains - the points he loses) in his current turn.
입력
The input consists of multiple data. Each data represents the state of the board of the game still in progress.
The format of each data is as follows.
N C
S1,1
S2,1 S2,2
S3,1 S3,2 S3,3
...
SN,1 ... SN,N
N is the size of the board (3 ≤ N ≤ 10).
C is the identification number ofthe player whose turn comes now (1 ≤ C ≤ 9) . That is, your program must calculate his points in this turn.
Si,j is the state of the vertex on the board (0 ≤ Si,j ≤ 9) . If the value of Si,j is positive, it means that there is the stone numbered by Si,j there. If the value of Si,j is 0, it means that the vertex is empty''.
Two zeros in a line, i.e., 0 0, represents the end of the input.
출력
For each data, the maximum points the player can get in the turn should be output, each in a separate line.
예제 입력 1
4 4
2
2 3
1 0 4
1 1 4 0
4 5
2
2 3
3 0 4
1 1 4 0
4 1
2
2 3
3 0 4
1 1 4 0
4 1
1
1 1
1 1 1
1 1 1 0
4 2
1
1 1
1 1 1
1 1 1 0
4 1
0
2 2
5 0 7
0 5 7 0
4 2
0
0 3
1 0 4
0 1 0 4
4 3
0
3 3
3 2 3
0 3 0 3
4 2
0
3 3
3 2 3
0 3 0 3
6 1
1
1 2
1 1 0
6 7 6 8
0 7 6 8 2
6 6 7 2 2 0
5 9
0
0 0
0 0 0
0 0 0 0
0 0 0 0 0
5 3
3
3 2
4 3 2
4 4 0 3
3 3 3 0 3
0 0
예제 출력 1
6
5
1
-10
8
-1
0
1
-1
5
0
5
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