text:
Share

The Following is the Distribution of Height of Students of a Certain Class in a Certain City: - CBSE Class 10 - Mathematics

Question

The following is the distribution of height of students of a certain class in a certain city:

 Height (in cm): 160 - 162 163 - 165 166 - 168 169 - 171 172 - 174 No. of students: 15 118 142 127 18

Find the median height.

Solution

First we prepare the following cummulative table to compute the median.

 Height (in cm) Frequency (f1) Cumulative Frequency (c.f) 160 - 162 15 15 163 - 165 118 133 166 - 168 142 275 169 - 171 127 402 172 - 174 18 420 N = 420

Now, N = 420

thereforeN/2=420/2=210

Thus, the cumulative frequency just greater than 210 is 275 and the corresponding class is 166 - 168.

Therefore, 166 - 168 is the median class.

l = 166, f = 142, F = 133 and h = 2

We know that,

Median =l+{(N/2-F)/f}xxh

=166+{(210-133)/142}xx2

=166+(77xx2)/142

=166+154/142

= 166 + 1.08

= 167.08

Hence, the median height is approximately 167.1 cm.

Is there an error in this question or solution?

Video TutorialsVIEW ALL [4]

Solution The Following is the Distribution of Height of Students of a Certain Class in a Certain City: Concept: Median of Grouped Data.
S


text:
# Stochastic reserving with R: ChainLadder 0.1.5-1 released

Today we published version 0.1.5-1 of the ChainLadder package for R. It provides methods which are typically used in insurance claims reserving to forecast future claims payments.

 Claims development and chain-ladder forecast of the RAA data set using the Mack method
The package started out of presentations given at the Stochastic Reserving Seminar at the Institute of Actuaries in 2007, 2008 and 2010, followed by talks at CAS meetings in 2008 and 2010.

Initially the package came with implementations of the Mack-, Munich- and Bootstrap Chain-Ladder methods. Since version 0.1.3-3 it also provides general multivariate chain ladder models by Wayne Zhang. Version 0.1.4-0 introduced new functions on loss development factor fitting and Cape Cod by Daniel Murphy following a paper by David Clark. Version 0.1.5-0 has added loss reserving models within the generalized linear model framework following a paper by England P. and Verrall R. (1999) implemented by Wayne Zhang.

For more details see the project web site: http://code.google.com/p/chainladder/ and an early blog entry about R in the insurance industry.

Changes in version 0.1.5-1:
• Internal changes to plot.MackChainLadder to pass new checks introduced by R 2.14.0.
• Commented out unnecessary creation of 'io' matrix in ClarkCapeCod function. Allows for analysis of very large matrices for CapeCod without running out of RAM. 'io' matrix is an integral part of ClarkLDF, and so remains in that function.
• plot.clark method
• Removed "conclusion" stated in QQplot of clark methods.
• Restore 'par' settings upon exit
• Slight change to the title
• Reduced the minimum 'theta' boundary for weibull growth function
• Added warnings to as.triangle if origin or dev. period are not numeric

Here is a little example using the googleVis package to display the RAA claims development triangle:

library(ChainLadder)
data(RAA) # example data set of the ChainLadder package
class(RAA) <- "matrix" # change the class from triangle to matrix
df <- as.data.frame(t(RAA)) # coerce triangle into a data.frame
names(df) <- 1981 : 1990
df\$dev <-  1:10
plot(gvisLineChart(df, "dev", options=list(gvis.editor="Edit me!", hAxis.title="dev. period")))

format: plain


text:
# American Institute of Mathematical Sciences

November  2009, 24(4): 1209-1213. doi: 10.3934/dcds.2009.24.1209

## On a question of Katok in one-dimensional case

 1 Department of Mathematics, Queens College of CUNY, Flushing, NY 11367

Received  August 2008 Revised  November 2008 Published  May 2009

We prove that a $C^{1}$ orientation-preserving circle endomorphism which is Hölder conjugate to a $C^{1}$ circle expanding endomorphism is itself expanding.
Citation: Yunping Jiang. On a question of Katok in one-dimensional case. Discrete & Continuous Dynamical Systems - A, 2009, 24 (4) : 1209-1213. doi: 10.3934/dcds.2009.24.1209
 [1] Wei Ouyang, Li Li. Hölder strong metric subregularity and its applications to convergence analysis of inexact Newton methods. Journal of Industrial & Management Optimization, 2021, 17 (1) : 169-184. doi: 10.3934/jimo.2019105 [2] Yunping Jiang. Global graph of metric entropy on expanding Blaschke products. Discrete & Continuous Dynamical Systems - A, 2021, 41 (3) : 1469-1482. doi: 10.3934/dcds.2020325 [3] Joel Kübler, Tobias Weth. Spectral asymptotics of radial solutions and nonradial bifurcation for the Hénon equation. Discrete & Continuous Dynamical Systems - A, 2020, 40 (6) : 3629-3656. doi: 10.3934/dcds.2020032 [4] Paul A. Glendinning, David J. W. Simpson. A constructive approach to robust chaos using invariant manifolds and expanding cones. Discrete & Continuous Dynamical Systems - A, 2020  doi: 10.3934/dcds.2020409

2019 Impact Factor: 1.338


text:
# Rectangular hyperbola example

Martin McBride, 2020-09-12
Tags parametric equation hyperbola
Categories coordinate systems pure mathematics

A rectangular hyperbola has the parametric equations:

\begin{align} x = c t\newline y = \frac{c}{t} \end{align}

Where $c$ is a positive constant, and $t$ is the independent variable.

We can plot this curve by calculating the values of $x$ and $y$ for various values of $t$, and drawing a smooth curve through them.

## Curve for c = 1

Assuming $a = 1$, the parametric equations simplify to:

\begin{align} x = t\newline y = \frac{1}{t} \end{align}

The values are shown in the following table, for $t$ in the range -3 to +3:

t x y
-4 -3 -0.25
-2 -2 -0.5
-1 -2 -1
-0.5 -0.5 -2
0 0 undefined
0.5 0.5 2
1 1 1
2 2 0.5
4 4 0.25

Here are the points plotted on a graph:

This curve is actually a standard reciprocal curve, as shown here.

The curve can be drawn by plotting the points and drawing a smooth line through them. Notice that the curve value is undefined for $t = 0$:


text:
concrete.sty for 2e

concrete.sty for 2e

Hi, what is Latex2e package which could do similar things to latex209
concrete.sty?

concrete.sty for 2e

beton package in macros/latex/contrib/supported

--

I have installed Latex2e and I answered yes to everything, now
according to the LaTeX companion, I sould be able to use the packages
euler.sty and concrete.sty to use the Euler math fonts and the Concrete
roman fonts respectively. Can I heck, these two files appear to be
missing, I have gone through the installation again, extra carefully,
and these files are not generated (thought the corresponding .fd files
are). To add to the confusion pandora.sty is generated. I have looked all over
uk.ac.tex, and can only find concrete.sty, and this is a rather old Latex2.09
NFSS1 file. I have a feeling that these should be generated by the
install process. Anyway where do I get/how do I generate these files.

JAB.

-------------------------------------------------------------------------------
Jonathan A. Buzzard,
Physics Department,           Email:-

United Kingdom.


text:
# Nelson complexity index

Flows in a very complex oil refinery

The Nelson complexity index (NCI) is a measure to compare the secondary conversion capacity of a petroleum refinery with the primary distillation capacity.[1] The index provides an easy metric for quantifying and ranking the complexity of various refineries and units.[2] To calculate the index, it is necessary to use complexity factors, which compare the cost of upgrading units to the cost of crude distillation unit.[3]

## History

It was developed by Wilbur L. Nelson in a series of articles that appeared in the Oil & Gas Journal[4] from 1960 to 1961 (Mar. 14, p. 189; Sept. 26, p. 216; and June 19, p. 109). In 1976, he elaborated on the concept in another series of articles, again in the Oil & Gas Journal (Sept. 13, p. 81; Sept. 20, p. 202; and Sept. 27, p. 83).

## Formula

${\displaystyle {\text{NCI}}=\sum _{i=1}^{N}F_{i}*{\frac {C_{i}}{C_{CDU}}}}$[5]

Where:

• ${\displaystyle F_{i}}$ is a complexity factor
• ${\displaystyle C_{i}}$ is a unit capacity
• ${\displaystyle C_{CDU}}$ is a capacity of crude distillation unit
• ${\displaystyle N}$ is a number of all units

The NCI assigns a complexity factor to each major piece of refinery equipment based on its complexity and cost in comparison to crude distillation, which is assigned a complexity factor of 1.0. The complexity of each piece of refinery equipment is then calculated by multiplying its complexity factor by its throughput ratio as a percentage of crude distillation capacity. Adding up the complexity values assigned to each piece of equipment, including crude distillation, determines a refinery’s complexity on the NCI.

The NCI indicates not only the investment intensity or cost index of the refinery but also its potential value addition. Thus, the higher the index number, the greater the cost of the refinery and the higher the value of its products.

In the second edition of the book Petroleum Refinery Process Economics (2000), author Robert Maples notes that U.S. refineries rank highest in complexity index, averaging 9.5, compared with Europe's at 6.5. The Jamnagar refinery belonging to India-based Reliance Industries Limited is now one of the most complex refineries in the world with a Nelson complexity index of 14.[6] The new refinery began trial production on December 25, 2008. The former BP Texas City, Texas refinery, newly acquired by Marathon Petroleum as their Galveston Bay Refinery has a Nelson complexity index of 15.3 in 2013.

The Oil and Gas Journal annually calculates and publishes a list of refineries with their associated Nelson complexity index scores.

## Complexity factors

Some factors for various processing units:

Unit 1998 Reports Older Reports
Distillation capacity 1.0 1.0
Asphalt 1.5 1.5
Vacuum distillation 2.0 2.0
Thermal processes 2.75 5.0
Catalytic hydrorefining 3.0 3.0
Catalytic reforming 5.0 5.0
Catalytic cracking 6.0 6.0
Catalytic hydrocracking 6.0 6.0
Alkylation / Polymerization 10.0 10.0
Oxygenates 10.0 10.0
Aromatics / Isomerisation 15.0 15.0
Lubes 60.0 60.0

## Example

If an oil refinery has a crude distillation unit (100 kbd), vacuum distillation unit (60 kbd), and catalytic reforming unit (30 kbd), then the NCI will be 1*(100/100) + 2*(60/100) + 5*(30/100) = 1.0 + 1.2 + 1.5 = 3.7.

## References

1. ^ "Nelson Complexity Index" (PDF). pakpas.org. Retrieved 3 November 2016.
2. ^ "Nelson Index". investopedia.com. Investopedia. Retrieved 3 November 2016.
3. ^ David C. Johnston; Daniel Johnston (2006). Introduction to Oil Company Financial Analysis. PennWell Books. p. 199. ISBN 9781593700447. Retrieved 3 November 2016.
4. ^ Johnston, Daniel (March 18, 1996). "Refining Report Complexity index indicates refinery capability, value". ogj.com. Oil and Gas Journal. Retrieved 3 November 2016.
5. ^ "PMI-Oman 2014". pmioman14.wordpress.com. Retrieved 3 November 2016.
6. ^ Nelson's Complexity Factor (PDF), Reliance Industries Ltd, archived from the original (PDF) on 2009-02-25, retrieved 2009-02-28


text:
# Finding power series and ROC of complex function

I have the function $f(z) = \frac{3iz-6i}{z-3}$

I need to find a power series $\sum c_n (z-1)^n$ about $z_0 = 1$

I can rewrite $f$ as $\frac{2i-iz}{1-\frac{z}{3}}$, where I'm guessing the ROC would then be $3$.

However I have it in the form $\frac{a}{1-r}$ and I'm not sure with $a$ being dependent on $z$ that I can get a correct power series, also I don't see how we get his power series about $1$.

I've considered Taylor Series, but it seems far too complicated with complex numbers.

I've tried writing it as $3i \frac{z-2}{z-3}$ where now I just need to get the power series for $\frac{z-2}{z-3}$ but still having trouble.

Any ideas what a general approach might be. I'm more interested in the method over the answer.

EDIT:

I've also tried writing it as

$3i(\frac{1}{1-\frac{3}{z}})+2i (\frac{1}{1-\frac{z}{3}})$

which would get me ROC = $3$ again. But I'm not sure how this would be about $z_0 = 1$

• You want a power series with centre $1$. Writing $z = w+1$ may help finding it. May 10, 2016 at 19:29
• Can you explain a little further. Sorry if I'm a bit confused May 10, 2016 at 19:44
• $w$ is $z-1$. So you want an expansion in terms of $w$. If you had for example $\frac{1}{z-5}$ and wanted to expand it as a power series in $z-2$, you'd set $z = w+2$, and in terms of $w$ the function becomes $\frac{1}{(w+2)-5} = \frac{1}{w-3} = -\frac{1}{3}\cdot \frac{1}{1-w/3}$. You know how to expand that into a geometric series (with common ratio $w/3$), and then you just replace $w$ with $z-2$. May 10, 2016 at 19:52
• Thanks Daniel. From this I get $3i(\frac{1}{1-\frac{2}{w}})-\frac{3i}{2}(\frac{1}{1-\frac{w}{2}})+3i(\frac{1}{1-\frac{w}{2}})$. However I'm not sure how to get the ROC for $\frac{1}{1-\frac{2}{w}}$ May 11, 2016 at 9:11

The function $f(z)=\frac{3iz-6i}{z-3}$ has a simple pole at $z=3$. We want to develop $f$ into a power series at $z_0=1$. Since the ROC is the distance from the center of the series to the nearest singularity or branch point we obtain $ROC=2$.
The idea is to transform the function to use a geometric series representation with center $z=z_0$. \begin{align*} \frac{1}{z-a}&=-\frac{1}{(a-z_0)-(z-z_0)}\\ &=-\frac{1}{a-z_0}\cdot\frac{1}{1-\frac{z-z_0}{a-z_0}}\\ &=-\frac{1}{a-z_0}\sum_{n=0}^{\infty}\frac{1}{(a-z_0)^n}(z-z_0)^n\qquad\qquad|z-z_0|<|a-z_0| \end{align*}
We obtain for $|z-1|<2$ \begin{align*} f(z)&=\frac{3iz-6i}{z-3}\\ &=-\frac{3iz-6i}{2-(z-1)}\\ &=-\frac{3iz-6i}{2}\cdot\frac{1}{1-\frac{z-1}{2}}\\ &=3i\frac{1-(z-1)}{2}\sum_{n=0}^{\infty}\frac{1}{2^n}(z-1)^n\\ &=\cdots \end{align*}


text:
Question

# The structure of $XeF{}_{6}$is: (A) pentagonal bipyramidal(B) distorted octahedral(C) capped octahedral(D) square pyramidal

Verified
128.7k+ views
Hint: $XeF{}_{6}$ is a noble gas compound , which is stable at normal temperature. $XeF{}_{6}$ is having hybridisation that is $s{{p}^{3}}{{d}^{3}}$. It is found to have a lone pair of electrons.

- Firstly, we have to learn how to determine the type of hybridisation and understand the formation of a certain molecular geometry in xenon hexafluoride along with the bond angles.

- We can determine the hybridisation in two ways:
Firstly, by knowing the number of bond pairs and lone pairs. During the formation of $XeF{}_{6}$, xenon has 8 electrons in its valence shell and it forms six bonds with the fluorine atoms and will have remaining two electrons. Further, we can say that the molecules will have one lone pair and 6 bond pairs. Now, if we take the steric number then it will be 7. And this can be interpreted as $s{{p}^{3}}{{d}^{3}}$.
- In the structure what happens here is that the fluorine atoms are placed in the vertices of the octahedron while lone pairs move in the space to avoid or we can say to reduce the repulsion. In simple words we can say that the six bond pairs are formed with fluorine and initially form a structure like octahedron but the lone pair that is present, gets its place in between the two fluorine atoms to reduce the repulsion. And hence, distort the structure of octahedrons, and therefore this structure is called a distorted octahedron.
-Secondly, we can also determine the hybridisation of xenon hexafluoride by using the common formula which is: $Hybridisation=\dfrac{1}{2}\times \left[ v+m-c+a \right]$
Here,
v =number of valence electrons,
m =monovalent
c =positive charge
a =negative charge
Let’s put the values according to the formula,
Hybridisation=
$Hybridisation=\dfrac{1}{2}\times \left[ 8+6-0+0 \right]$
\begin{align} & =\dfrac{1}{2}\times \left[ 14 \right] \\ & =7 \\ \end{align}
So we can say that, the hybridisation number is equal to 7. Now we can say that the hybridisation is $s{{p}^{3}}{{d}^{3}}$.
-VSEPR theory suggests the Pentagonal bipyramid geometry, but by electron diffraction experiments it is said to have octahedral geometry with fluorine atoms on the corners.
- Hence we can say that the option (a) and (b) both are correct.

- It is found that the complete hydrolysis of $XeF{}_{6}$produces$Xe{{O}_{3}}$ which is xenon trioxide. This trioxide is highly explosive and acts as a powerful oxidising agent in solution.
-It is also seen that $XeF{}_{6}$ is the strongest fluorinating agent.
- It is to be noted that $XeF{}_{6}$ is having hybridisation that is $s{{p}^{3}}{{d}^{3}}$. During the formation of, $XeF{}_{6}$ xenon has 8 electrons in its valence shell and it forms six bonds with the fluorine atoms and will have remaining two electrons. This is having distorted octahedral geometry.


text:
# Lab 9: Sprites and Walls¶

Goal: Create a landscape of wall objects that the user must navigate around to collect coins. This will help practice using for loops to create and position multiple items.

• Step 1: Start with one of the move with walls examples. Use either the Move with Walls example or the Move with a Scrolling Screen example.
• Step 2: If you start with the examples, delete the current wall placement code. You want to create your own.
• Step 3: Create a more complex arrangement of walls. Make sure the walls don’t allow the user to go off-screen. 6 points, based on how complex the arrangement. See Individually Placing Walls, Placing Walls With A Loop, and Placing Walls With A List for ideas. Just DON’T do the same thing as examples. Make it your own. Also, if you have more than one type of wall block, that’s great. But you don’t need more than one wall list. The physics engine requires all walls be kept in the same list.
• Step 4: Update the graphics. Use multiple types of blocks for the walls. Maybe change the character. 4 points, one point for each graphic used that wasn’t part of the base example. Remember to put a quick citation in your program just before you load graphics or sounds.
• Step 5: Add coins (or something) for the user to collect. 4 points, based on the complexity of the coin layout. Remember, you can place coins like we placed wall blocks. If you randomly place coins, you might end up with coins on top of walls. See the “Important Part” around line 75 or so of the example sprite_no_coins_on_walls.py for how to avoid this.
• Step 6: Keep score of how many coins were collected, and display on-screen. 4 points.
• Step 7: Add a sound to play each time the user collects a coin. 2 points.

Warning

Don’t move the player twice!

The command self.physics_engine.update() moves the player while checking for walls. The command self.all_sprites_list.update() will move the player WITHOUT checking for walls. Don’t do both commands. You’ll end up “walking through walls.” If you have other sprites to update, update only those sprites. For example: self.coin_list.update().


text:
# Splitting off finite groupoids

Yesterday, David Rydh in this comment found a wonderful approach to proving Lemma Tag 03FM. In fact I had already more or less given up hope that the lemma could be fixed! I am so exited about David’s idea that I am going to try to explain it in this post. Of course any mistakes are mine.

First we are going to discuss a completely general construction. Namely, suppose that f : X —> Y is a morphism of schemes which is separated and locally of finite type. On the small etale site Y_{etale} of Y consider the following presheaf

(*) V/Y |—> {Z \subset X_V open, Z —> V is finite}

Note that as we assumed X —> Y is separated such a Z is also closed in X_V. Also note that the empty set is closed and finite over any scheme. Finite morphisms satisfies descent for the etale topology. This implies that the presheaf above is a sheaf on Y_{etale}. Now, it is a general fact that any sheaf on Y_{etale} is representable by an algebraic space etale over Y. Hence we obtain an algebraic space, which I am going to denote f_!X which comes equipped with an etale morphism f_!X —> Y and we get a universal open and closed subspace Z_{univ} \subset f_!X \times_Y X which is finite over f_!X.

Before we say more about this construction let us point out that f_!X is really an algebraic space and not a scheme. For example suppose that X = Spec(C) and Y = Spec(R[t]) and the morphism f : X —> Y maps the unique point of X to (t = 0) inside Y. In that case f_!X is a copy of Y with {0} replaced by {0_1, 0_2} with residue fields R, respectively C (but of course f_!X is still etale over Y). The complement of {0_1} in f_!X is an open subspace which is a copy of Example Tag 03FN. And this is the simplest example of an algebraic space etale over a scheme which is not a scheme.

Now a very important property of the construction above is that it commutes with arbitrary base change. This is the analogue of the fact that cohomology with proper supports commutes with any base change. Namely, suppose that Y’ —> Y is any morphism of schemes, and denote f’ : X’ —> Y’ the base change of f. By universality of f’_!X’ we obtain a morphism f_!X \times_Y Y’ —> f’_!X’ over Y’. To prove this map is an isomorphism it suffices to show that the stalks of the sheaf (*) can be computed from the geometric fibres of the morphism X —> Y. And this is almost exactly the content of Lemma Tag 02LN.

As in Keel-Mori we are going to use a slight modification of f_! in the case that the morphism f : X —> Y comes equipped with a section e : Y —> X. Namely, in that case we consider the sub presheaf of (*) consisting of those Z \subset X_V with e(V) contained in Z. In the same manner as above you show that this is representable by an algebraic space fe_!X etale over Y whose formation commutes with base change.

OK, so how do we apply this construction to a groupoid scheme (U, R, s, t, c, e, i) with s, t separated and locally of finite type? Well, consider U’ = se_!R and the universal closed sub space Z \subset R \times_{U, s} U’. The inverse map of the groupoid scheme shows that U’ is canonically isomorphic to te_!R which gives that i(Z) \subset U’ \times_{U, t} R is universal. Compatibility with base change and the fact that R \times_{s, U, t} R is isomorphic to R \times_{s, U, s} R over (the second) R shows that there is a canonical isomorphism U’ \times_{U, t} R = R \times_{s, U} U’ over R. Then you check by chasing some diagrams that this isomorphism maps the universal closed subscheme Z into i(Z). Hence if

R’ = U’ \times_{U, t} R \times_{s, U} U’

is the restriction of R to U’, then we conclude that Z is an open sub space of R’ which is finite under both s’ : R’ \to U’ and t’ : R’ \to U’. (Here we use that U’  \times_{U, t} R = R \times_{s, U} U’ can be identified with an open in R’ as these spaces are etale over R.) Because we have e’ inside of Z by construction some more diagram chasing shows that Z is a sub groupoid space! Awesome.

Of course it is possible that U’ is empty. But, as in the statement of Lemma Tag 03FM if we have a point u of U where the set of points {r_1, …, r_n} of R mapping to u under both s and t is finite, and if s, t are quasi-finite at each of those points, then we get a canonical point u’ of U’ whose residue field is the same as the residue field of u, and which corresponds to an open and closed sub scheme of the fibre of R —> U whose points are exactly the set {r_1, …, r_n}.

OK, so this utterly general construction allows us to find an open subgroupoid which is a finite groupoid, at the cost of replacing U by an algebraic space etale over U. And, as pointed out by David Rydh in his further comments we can then use further properties of the original groupoid (U, R, s, t, c, e, i) to prove that U’ has additional properties. For example, in case the morphisms s, t are also of finite presentation and flat, then (I think) U’ agrees with an open of the Hilbert scheme used by Keel and Mori and it is a quasi-projective scheme over U. Wonderful.

# Students and the stacks project

So I was asked today if all of the “omitted” proofs and explanations in the stacks project are supposed to be filled in. The answer is: Yes, they are!

Since my goal was to build enough theory to get to the more advanced material on algebraic spaces and algebraic stacks, I have not always filled in all the proofs in the earlier parts. On the other hand, I claim that what I omitted is consistently of a relatively low difficulty (provided you read the definitions, and understand the material generally speaking). Of course my judgment could have been off and if you find a more difficult omitted proof or something you don’t understand, then please email stacks.project@gmail.com.

Now if you are an undergraduate or graduate student, then this is a perfect opportunity to hone your algebraic geometry skills. The key word to search for in the documents or in the book is “omit” (case insensitive). Find a spot where the proof is missing and you feel comfortable filling it in. Write up the proof (with references only to other results in the stacks project) and email the tex file to the address mentioned above. Your writeup will be reviewed and perhaps edited, and then added to the stacks project.

A quick search through the tex files in the stacks project shows there are 740 occurrences of the stem “omit”. So there are many things you can do!

And before somebody asks: All errors are supposed to be fixed. So email any typos, misspellings, mistakes, arrows pointing the wrong way, etc to the email address above. Don’t wait, just send any you find right away.

# Good quotients

Just a short post to point out that I just defined the notions of “coarse quotient”, “good quotient”, and “geometric quotient” in the stacks project. To see the details see the chapter on Quotients of Groupoids. Please let me know if you think the definitions aren’t what you think they should be.

Really the idea for this chapter is to have silly things like: “U reduced => X reduced if X is a categorical quotient of s, t : R —> U” (from GIT, paragraph 2, Chapter 0) proved in excruciating detail and in ridiculous generality. It seems to me this is somewhat worthwhile since the first thing you always read about this material is that chapter of GIT and it can be confusing.

Also, the level of generality where R is not necessarily a groupoid can be useful, for example when forming the MRC quotient of a variety U you look at families T <--- C ---> U of rational curves C_t on U and the pre-relation you consider is R = C \times_T C (which has two maps to U). There is absolutely no reason that this should define a transitive relation and of course in general it really doesn’t.

# Spaces and points

Let X be a reduced algebraic space with 1 point. Then is it true that X is representable by the spectrum of a field?

The answer is no in general. The space [Spec(\bar{Q})/Gal(\bar{Q}/Q)] of Example Tag 02Z6 in the stacks project is a counter example.

Currently the best positive result (in the stacks project) is Lemma Tag 047Z which says that this holds when X is a decent algebraic space.

An algebraic space X is called decent if every point of X corresponds to a quasi-compact monomorphism Spec(k) —> X with k a field. This is not currently the definition in the stacks project, but it can be shown to be equivalent. It turns out that this is a convenient class of algebraic spaces to work with. It contains all schemes, all quasi-separated algebraic spaces, and likely all locally separated algebraic spaces (David Rydh, private communication). On the other hand a decent algebraic space is a bit like a scheme and has “enough points” in some sense.

Because there are non-representable 1-point spaces it turns out that the notions of “radicial” and “universally injective” are not the same for morphisms of algebraic spaces. Namely, the morphism [Spec(\bar{Q})/Gal(\bar{Q}/Q)] —> Spec(Q) is universally injective but not radicial (with any reasonable definition of radicial I can think of). Again for decent morphisms this does not happen, see Lemma Tag 0484.

# Schemes, Spaces, and Points

Suppose you have two morphisms a, b : X —> Y and you want to know whether a = b. If

1. X, Y are schemes,
2. X is reduced,
3. a(x) = b(x) for all x in X, and
4. the induced maps on residue fields are the same too,

then a = b. If

1. X, Y are algebraic spaces,
2. Y is locally separated,
3. X is reduced, and
4. a(x) = b(x) in Y(K) for every x in X(K) and any field K,

then a = b. But the last statement does not hold if we replace condition 2 by the condition that Y is quasi-separated. Recall that quasi-separated algebraic spaces are the “usual” algebraic spaces, i.e., the ones in Knutson’s book, not some bizarre ultra general class of algebraic spaces.

This comes up when you consider quotient maps for groupoids in algebraic spaces, and it is just the first small sign that things get a little more confusing when dealing with algebraic spaces. Namely, the above means that if we have a groupoid in algebraic spaces (U, R, s, t, c) and a morphism U —> X then even if all of U,R,X are reduced to see whether U —> X is R-invariant (i.e. a quotient map), it is not enough to check that this holds on field valued points.

# Mumford’s GIT and its unfortunate typesetting error

My student Yanhong Yang today noticed this typo in the definition of a geometric quotient in GIT (Definition 0.6).It states that the morphism \phi should be submersive and then has a blank line, after which it says

U’ \subset Y’ is open if and only if \phi’^{-1}(U’) is open in X’.

without any further explanation. This is present in all the editions that I have been able to find, including the latest one. Kollar, Keel and Mori, and Rydh have concluded from this that what is meant is that \phi should be universally submersive. (Just look at their papers. Maybe they even called up Mumford to ask him — another possibility is that they looked at the proof of Remark (4) of Paragraph 2 of Chapter 0 where Mumford seems to be using the fact that it is universally submersive.) There are also plenty of places in the literature where authors do not use universally submersive, only submersive.

Not only that, there is also a huge variance in the literature as to what a geometric quotient is. In the two papers of Kollar dealing with quotients (on by actions of group schemes, the other by finite equivalence relations) his definitions are adapted to the problem at hand, and in particular include the condition that the quotient map is locally of finite type, or finite — either of which seems like the wrong thing to require when considering the problem in general. In Rydh’s paper he includes the condition that the quotient map is universally submersive in the constructible topology (although this is in almost all cases implied by the other conditions). In a paper of Sheshadri he requires the quotient morphism to be affine. And so on. (I even found an article where they use Remark (4) of Paragraph 2 in Chapter 0 but do not require universally submersive…)

Let (U, R, s, t, c) be a groupoid in algebraic spaces (so R = G \times U if we are talking about an action of a group algebraic space). Let \phi : U —> X be an R-invariant map of algebraic spaces. It seems to me, but I may be wrong, that in each of these references, except for those where the author misread the typo, everybody always at least requires the following:

• X is an orbit space, i.e., the maps U —> X and R —> U \times_X U are surjective,
• \phi is universally submersive, and
• the structure sheaf O_X of X is the sheaf of R-invariant functions.

(The last condition is thrown in to attempt to make the quotient unique, but that only holds if you work in the category of schemes.) I think that in analogy with the introduction of algebraic spaces, we should use these three conditions to define the notion of a geometric quotient in the stacks project. Then we can have fun and add adjectives to describe additional properties of geometric quotients. For example, I particularly like the condition, introduced in the paper of Rydh, that a quotient is strong if it has the property that R —> U \times_X U is universally submersive.

By the way, I really really dislike the numbering scheme in GIT. Don’t you?

# Group schemes over fields

In the last couple of days I have added a few results on group schemes over fields to the stacks project. I mainly wanted to add the result that group schemes locally of finite type over a characteristic zero field are smooth which I hope to use later in an idea I have relating to finite groupoids in characteristic zero.

The sheaf of differentials of a group scheme over a field  is free (this holds in any characteristic). But actually I am not sure that a scheme over a field of characteristic zero whose sheaf of differentials is free is even necessarily reduced. In fact, in a paper entitled “Algebraic group schemes in characteristic zero are reduced” (1966) Frans Oort asks: Is every group scheme over a field of characteristic zero reduced? I googled and tried mathscinet but this question seems to be still open.

Another question I have is: Does any group scheme over a field have an open subgroup scheme which is quasi-compact? It seems that this could be true… but maybe I simply do not know any of the truly enormous group schemes that exist out there?

Leave a comment if you have an idea about either of these questions.

# Geometric quotients for finite groupoids

In this comment David Rydh formulates the conjecture that for a finite affine groupoid (U, R, s, t, c) the spectrum of the ring of invariants may be a geometric quotient for the stack [U/R]. In fact, the same question came up in a recent conversation with Jarod Alper here in the department.

I have an idea for generating invariant functions, which sounds so familiar to me that I am sure it is in the literature (let me know if you have a reference), or maybe I have already tried using it in the past. First, recall that if s,t are finite locally free flat so B is finite locally free over A then for any element x of A gives rise to an invariant element y by taking y = Nm_s(t(x)). In words y is the norm of t(x) with respect to the finite locally free ring map s : A —> B. Thus, in the general case where s, t are finite we try to find an element y in A which behaves like the norm of t(x) with respect to s. Maybe a falsifiable version of the conjecture above would be to conjecture the existence of a y in A such that for every prime p of A the value of y in k(p) is a power of the Nm of t(x) restricted to B \otimes_{s, A} k(p)?

My idea is to try to do the following. Take a finite free extension phi : A —> B’ and a surjection pi : B —> B’ such that pi o phi = s. (It may be convenient for later arguments to allow only certain types of ring maps A —> B, such as my personal favorite: finite flat relative complete intersections.) Now for any element x of A we can let y = Nm_phi(x’) where x’ in B’ is any element with pi(x’) = t(x). It is clear that y will NOT be R-invariant in general, simply because we have put too little restrictions on B’. But on the other hand, I am pretty confident that the ideal generated by all y of the form Nm_phi(x’) will be R-invariant. Namely, it should just cut out the set of points which are R-equivalent to a zero of the function x.

However, if A is an Artinian ring, then we can choose B’ so that B’ and B have the same maximal ideals. In this case if A has positive residue characteristics then it is quite easy to show that y^{p^n} for large n is independent of the choice of x’ and presumably is an invariant element of A (I haven’t checked this completely). This could then be the start of a kind of induction argument in the Noetherian case. But in characteristic zero I do not even know how to produce enough invariant functions in the Artinian case.

# Artin’s trick revisited

Well, my claims in this post were a little premature and may not be correct after all. It turns out that an argument in the lemma on etale localization of groupoids was wrong and has to be fixed. It doesn’t matter much for the overall picture since I can certainly use the arguments in the paper of Keel and Mori to fix it, but I wanted to prove it using only “etale localization” which may be hard.

The problem is the following: Suppose we have a groupoid scheme (U, R, s, t, c) with s, t flat, locally of finite presentation, and locally quasi-finite. Then we want to find many etale morphisms U’ –> U such that the restriction R’ of R to U’ is a disjoint union R’ = P \coprod Rest, with P a subgroupoid which is finite over U’ under both s’ and t’. The technique that I used (wrongly) in the current version of proof of the lemma mentioned above does prove that when we take U’ to be the spectrum of the henselization of the local ring of U at a point of U. But this doesn’t give you an etale morphism U’ —> U! What I am trying to see is if there is a kind of limit argument to descend P from the henselization to a smaller ring…

Let me just state here for the record that I think this means we can use this version of the lemma (with the henselization I mean) to define the (strictly) henselian local ring of the coarse moduli space without knowing that the coarse moduli space exists! Namely, since s, t : P —> Spec(O_{U,u}^h) are finite, flat and locally of finite presentation, we obtain that P is affine, and the “usual” arguments show that O_{U,u}^h is integral over the subring C of P-invariant elements of O_{U, u}^h. Presumably C (or its strict henselization) is what we are looking for. I haven’t thought this through completely, however.

I’ll post more here when I figure out how to repair the lemma.


text:
### Categories

It’s kinda weird to use command lines in Windows (maybe a lot of people do it, who knows) but the reason I’m using it is that in Windows based Matlab, the “system” function will just call cmd. So if I need to run something inside a Matlab loop, I should learn how to use cmd.

Here is the command line I used to open a data file and run a python script in Tecplot:

C:\ /WAIT tec360 -b -datafile C:\something.plt -p C:\some\python.py

• /WAIT is to hold the program until it’s done
• -b is batch mode, include this otherwise you will see the GUI
• -datafile is reading file
• -p is executing python script


text:
# Eigenvalues and eigenvectors.

I need to find eigenvectors and eigenvalues of $T : M_{n\times n} \to M_{n\times n}$, $T(A) = A^T$ for $A\in M_{n\times n}$.

My intuition tells me that eigenvalues are $1$ and $-1$. $1$ for symmetric matrices and $-1$ for skew-symmetric matrices.

I can show that eigenvalue can only be $1$ if the entries on the diagonal are nonzero, from this I can say that if eigenvalue is $1$ then the matrix is symmetric.

For skew-symmetric matrices I can prove that eigenvalue is $-1$.

My problem for matrices which are not skew-symmetric and have only $0$ diagonal, how do I show that there is no eigenvalue ? In other words, how do I show that only eigenvalues possible is $-1$ and $1$ ?

$\lambda$ is an Eigenvalue when $T(A)=\lambda A=A^T$.

Then $T(T(A))=(A^T)^T=A=\lambda^2A$.

The Eigenmatrices are such that either $A^T=A$ or $A^T=-A$, i.e. all symmetric or antisymmetric matrices.

Hint: $(A^T)^T=A$ that is:

$$T^2=Id$$

Hence any eigenvalue must be $1$ or $-1$

• Are you saying that idempotent transformations can have only $-1$ and $1$ as eigenvalues ? – user8277998 Dec 12 '17 at 16:11
• @123 no involution. I have corrected thanks – Guy Fsone Dec 12 '17 at 16:15


text:
Browse Questions

# Two particles A and B are in motion . if the wavelength associated with particle A is $5\times10^{-8}$m.Calculate the wavelength associated with particle B if its momentum is half of A.

$(a)\;1500A^{\large\circ}\qquad(b)\;1200A^{\large\circ}\qquad(c)\;1000A^{\large\circ}\qquad(d)\;1100A^{\large\circ}$

$(mV)_A=\large\frac{h}{\lambda_A}$
$(mV)_B=\large\frac{h}{\lambda_B}$
$(mV)_B=\large\frac{1}{2}(mV)_A$
$\large\frac{h}{\lambda_B}=\large\frac{h}{2\lambda_A}$
$=2\times5\times10^{-8}$m
$=10\times10^{-8}m$
$=1000A^{\large\circ}$

+1 vote


text:
# How do you find the value for sin x = 0.723?

Oct 18, 2015

Find sin x = 0.723

Ans: $x = {46}^{\circ} 30$ and $x = {133}^{\circ} 70$

#### Explanation:

Use calculator.
sin x = 0.723 --> arc $x = {46}^{\circ} 30$.
Trig unit circle gives another arc x that has the same sin value -->
$x = \left(180\right) - 46.30 = {133}^{\circ} 70$


text:
Learning

is

simple

and

fun!

Register now! Browse subjects

### Theory:

Types of ions:
Ions are classified into two types. They are:
• Cations
• Anions
Cations:
If an atom loses one or more electrons during a chemical reaction, it gains a positive charge. These are referred to as cations (or positive radicals).
Anions:
If an atom gains one or more electrons during a chemical reaction, it gains a negative charge. These are referred to as anions(or negative radicals).
Types of ion: Anion and Cation

1. Sodium atom loses one electron to achieve stable electronic configuration, and it becomes a cation.

2. The sodium ion is described as $$Na^+$$.

3. Chlorine atom gains one electron to achieve stable electronic configuration, and it becomes anions.

4. The chlorine ion is described as $$Cl^-$$.

Ion Classification


text:
Generating example files locally

You can generate the example files using the generate_examples function. Instructions for using the generate_examples are given below.

In the following code snippets, output_directory is the directory where you want to save the generated example files. output_directory should NOT be an existing directory. If output_directory already exists, you should delete it before running the generate_examples function.

Generating scripts (.jl files)

To generate the examples as Julia scripts (.jl files), use the following code.

PredictMD.generate_examples(output_directory; scripts = true)

Generating IJulia/Jupyter notebooks (.ipynb files)

To generate the examples as IJulia/Jupyter notebooks (.ipynb files), use the following code. output_directory is the directory where you want to save the generated example files. output_directory should NOT be an existing directory. If output_directory already exists, you should delete it before running the generate_examples function.

PredictMD.generate_examples(output_directory; notebooks = true)


text:
<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are viewing an older version of this Concept. Go to the latest version.

# Scale Factor of Similar Polygons

%
Progress
Practice Scale Factor of Similar Polygons
Progress
%
Scale Factor of Similar Polygons

Have you ever wondered about the relationship between side lengths of a figure?

Jessie looked at a drawing with two quadrilaterals in it. The side lengths of the quadrilaterals were listed as follows:

MN = 3 inches QT = 6 inches

NO = 2 inches QR = 4 inches

MP = 4 inches TS = 8 inches

OP = 2 inches RS = 4 inches

Given the relationship between these sides, can you figure out the scale factor?

This Concept will teach you how to do exactly that. We'll revisit this problem at the end of the Concept.

### Guidance

Similar figures are shapes that exist in proportion to each other. They have congruent angles, but their sides are different lengths. Squares, for example, are similar to each other because they always have four $90^{\circ}$ angles and four equal sides, even if the lengths of their sides differ. Other figures can be similar too, if their angles are equal. Let’s look at some pairs of similar figures.

Notice that in each pair the figures look the same, but one is smaller than the other. As you can see, similar figures have congruent angles but sides of different lengths.

Even though similar figures have sides of different lengths, corresponding sides still have a relationship with each other. Each pair of corresponding sides has the same relationship as every other pair of corresponding sides, so that, altogether, the pairs of sides exist in proportion to each other. For instance, if a side in one figure is twice as long as its corresponding side in a similar figure, all of the other sides will be twice as long too.

In this Concept, we are going to use these relationships to find the measures of unknown sides. This method is called indirect measurement.

Let’s take a step back to similar figures to understand how indirect measurement works.

First, let’s make sure we can recognize corresponding parts of similar figures. Similar figures have exactly the same angles. Therefore each angle in one figure corresponds to an angle in the other.

These triangles are similar because their angles have the same measures. Which corresponds to which? Angle $B$ is $100^{\circ}$ . Its corresponding angle will also measure $100^{\circ}$ : that makes angle $Q$ its corresponding angle. Angles $A$ and $P$ correspond, and angles $C$ and $R$ correspond.

Similar figures also have corresponding sides, even though the sides are not congruent. Corresponding sides are not always easy to spot. We can think of corresponding sides as those which are in the same place in relation to corresponding angles. For instance, side $AB$ , between angles $A$ and $B$ , must correspond to side $PQ$ , because $A$ corresponds to $P$ and $B$ corresponds to $Q$ .

Corresponding sides also have lengths that are related, even though they are not congruent. Specifically, the side lengths are proportional. In other words, each pair of corresponding sides has the same ratio as every other pair of corresponding sides. Take a look at the rectangles below.

These rectangles are similar because the sides of one are proportional to the other. We can see this if we set up proportions for each pair of corresponding sides. Let’s put the sides of the large rectangle on the top and the corresponding sides of the small rectangle on the bottom.

$\frac{LM}{WX} &= \frac{8}{4}\\\frac{MN}{XY} &= \frac{6}{3}\\\frac{ON}{ZY} &= \frac{8}{4}\\\frac{LO}{WZ} &= \frac{6}{3}$

Now you can clearly see each relationship. To figure out if the pairs do indeed form a proportion, we have to divide the numerator by the denominator. If the quotient is the same, then the ratios each form the same proportion and the figures are similar.

$\frac{LM}{WX} &= \frac{8}{4} = 2\\\frac{MN}{XY} &= \frac{6}{3} = 2\\\frac{ON}{ZY} &= \frac{8}{4} = 2\\\frac{LO}{WZ} &= \frac{6}{3} = 2$

Each quotient is the same so these ratios are proportional. The side lengths are proportional and the figures are similar.

We call this the scale factor. The scale factor is the ratio that determines the proportional relationship between the sides of similar figures. For the pairs of sides to be proportional to each other, they must have the same scale factor. In other words, similar figures have congruent angles and sides with the same scale factor. A scale factor of 2 means that each side of the larger figure is twice as long as its corresponding side is in the smaller figure.

Now it's time for you to try a few on your own. Determine each scale factor.

#### Example A

$\frac{18}{6}$ and $\frac{24}{8}$

Solution: $3$

#### Example B

$\frac{12}{6}$ and $\frac{8}{4}$

Solution: $2$

#### Example C

$\frac{25}{5}, \frac{45}{9}, \frac{15}{3}$

Solution: $5$

Here is the original problem once again.

Jessie looked at a drawing with two quadrilaterals in it. The side lengths of the quadrilaterals were listed as follows:

MN = 3 inches QT = 6 inches

NO = 2 inches QR = 4 inches

MP = 4 inches TS = 8 inches

OP = 2 inches RS = 4 inches

Given the relationship between these sides, can you figure out the scale factor?

To figure out the scale factor, we can write each corresponding side as a ratio comparing side lengths.

$\frac{3}{6}$

$\frac{2}{4}$

$\frac{4}{8}$

$\frac{2}{4}$

The scale factor of these figures is $\frac{1}{2}$ .

### Vocabulary

Similar Figures
figures that have the same angle measures but not the same side lengths.
Scale Factor
the proportional relationship between two side lengths.
Proportions
two equal ratios.

### Guided Practice

Here is one for you to try on your own.

What is the scale factor of the figures below?

We need to find the scale factor, so let’s set up the proportions of the sides. Let’s put all the sides from the large figure on top and the sides from the small figure on the bottom. It doesn’t matter which we put on top, as long as we keep all the sides from one figure in the same place.

$\frac{QR}{HI} &= \frac{15}{5}\\\frac{TS}{KJ} &= \frac{21}{7}\\\frac{RS}{IJ} &= \frac{6}{3}\\\frac{QT}{HK} &= \frac{15}{5}$

Now all we have to do is divide to find the scale factor.

$\frac{QR}{HI} &= \frac{15}{5} = 3\\\frac{TS}{KJ} &= \frac{21}{7} = 3\\\frac{RS}{IJ} &= \frac{6}{3} = 3\\\frac{QT}{HK} &= \frac{15}{5} = 3$

The scale factor for these similar figures is 3. This means that the first quadrilateral is exactly three times bigger than the second.

What happens if we had written the numbers the other way around?

If we had put the measurements of the smaller figure on top and the larger figure on the bottom, we would have found a scale factor of $\frac{1}{3}$ . You can write them the other way too as long as you understand how to read the scale factor. This is another way of saying the scale factor is three. When we say that the larger figure is three times bigger than the small one, it’s the same as saying that the small figure is one-third the size of the larger one.

### Practice

Directions: Find the scale factor of the pairs of similar figures below.

Directions: Use each ratio to determine scale factor.

5. $\frac{3}{1}$

6. $\frac{8}{2}$

7. $\frac{2}{8}$

8. $\frac{10}{5}$

9. $\frac{12}{4}$

10. $\frac{16}{2}$

11. $\frac{15}{3}$

12. $\frac{24}{4}$

13. $\frac{4}{2}$

14. $\frac{6}{2}$

15. $\frac{3}{9}$

### Vocabulary Language: English

Perimeter

Perimeter

Perimeter is the distance around a two-dimensional figure.
Proportion

Proportion

A proportion is an equation that shows two equivalent ratios.
Scale Factor

Scale Factor

A scale factor is a ratio of the scale to the original or actual dimension written in simplest form.
Similar

Similar

Two figures are similar if they have the same shape, but not necessarily the same size.


text:
# For the function given below, find a formula for the Riemann sum obtained by dividing the interval [a,b] into n equal subintervals and using the right-hand endpoint for each c Subscript kck. Then take a limit of this sum as n right arrow→ infinity∞ to calculate the area under the curve over [a,b].f(x)equals=28 x squared plus28x2+28 x cubed28x3 over the interval left bracket negative 1 comma 0 right bracket[−1,0]Find a formula for the Riemann sum.Upper S Subscript nSnequals

Question
16 views

For the function given below, find a formula for the Riemann sum obtained by dividing the interval [a,b] into n equal subintervals and using the right-hand endpoint for each
c Subscript k
ck. Then take a limit of this sum as n
right arrow

infinity
∞ to calculate the area under the curve over [a,b].

f(x)
equals
=
28 x squared plus
28x2+
28 x cubed
28x3 over the interval
left bracket negative 1 comma 0 right bracket
[−1,0]
Find a formula for the Riemann sum.

Upper S Subscript n
Sn
equals

check_circle

Step 1

Consider the given function as,

Step 2

Calculate the Riemann sum for right-hand endpoints as follows.

Step 3

Now calculate the area under the curve over [–1,0] by taking the limit of ...

### Want to see the full answer?

See Solution

#### Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.
Tagged in


text:
1
Explain Fan-in, Fan-out, Power Dissipation and Noise Immunity with reference to Digital ICs

Mumbai University > Electronics Engineering > Sem 3 > Digital Circuits and Design

Marks: 5M

Year: Dec 2015

0
• Fan-in:-

Fan-in is defined as the number of input gate has. For example, a two input gate will have a fan-in equal to 2. Fan-out:-

Fan-out is a term that defines the maximum number of digital inputs that the output of a single logic gate can feed. Most transistor-transistor logic (TTL) gates can drive up to 10 other digital gates or devices.

Power Dissipation:-

As a result of applied voltage and currents flowing through the logic IC’s, some power will be dissipated in it in the form of heat. The power drawn by an IC from the power supply is given by,

$$P= Vcc X Icc$$

Noise Immunity:-

Noise Immunity is the ability of a device or component to operate in the presence of noise disturbance.


text:
# Free Web-Based Orbital Solver to look into the Walsh orbitals of a perturbed cyclopropane

There exists a free online orbital calculator.

When I draw cyclopropane it plots three molecular orbitals, but unfortunately it doesn't use the Walsh orbitals.

Are there any free online tools which can solve such the problem?

(It would be very nice if it would also be an android app. Please let me know if you know one.)

The section below is for interested chemists:

The reason why I want to do that:

Cyclopropane can, to a good approximation, be described with the following orbitals:

The question was how the molecule orbitals behave if we add an electron withdrawing group (a $\pi$ acceptor because some people call it the "CXZ model")

The solution says:

In the substituted case all energy levels are lying more down in energy.

I don't agree because I only see overlap with the Walsh Möbius set of $sp$ orbitals in the following sketch:

The Hückel set (a simpler way to put it: the energy lowest orbital in the scheme of cyclorpropane; used above) gives no overlap with the $p$ orbitals of an electron withdrawing group, e.g. X=COOH.

As I see from the orbitals' interaction, I believe the solution is wrong, I don't agree that all orbitals are lowered.

EDIT:

1. Sorry for the mistake in the sketch, there is of course only one of the two degenerate Möbius orbitals lowered.
2. I guess COOH not only has a $\pi$ acceptor effect but also a $\sigma$ acceptor effect which could lower all orbitals.
• Don't know if this help but you can compute the canonical MOs (at the RHF/STO-3G//PM3 level of theory) using molcalc.org – Jan Jensen Jun 19 '15 at 10:41
• – Martin - マーチン Jun 19 '15 at 12:08
• @JanJensen I tried your page, however I like to get a graphical interface similar as in my SHMO solver, so I don't have to manually plot the energy levels. If there exist a freeware for Windows, which can easily do that (or an Android app) I would also be interested in. – laminin Jun 19 '15 at 14:31


text:
# Quick antiderivative question

## Homework Statement

Find the antiderivative of y=tan^2x+sec^2x

N/A

## The Attempt at a Solution

Seems to be a simple question, but the answer is eluding me no matter what I do. My first try was to replace the tan^2 with sec^2-1, and then factor out a 1/2 from the resulting 2sec^2x-1, but after that I have no idea how to continue. I know that the answer I should wind up with is Y=2tanx-x+C, but have no idea how to go about getting this answer :tongue: Any help would be greatly appreciated, thanks.

Mark44
Mentor
Hint: d/dx(tan(x)) = sec2(x)
Every derivative formula gives you an antiderivative formula for free, if look at it the right way.

I also believe the answer you should end up with is tan(2x) - x + C, which is slightly different from what you show.

Ohh I see it now, thanks Mark, and no the answer is correct, as it is what I arrived at as well just now.
It might have been my mistake writing the question wrong, as it is y=(tanx)^2+(secx)^2. I just wrote it as y=tan^2x+sec^2x assuming people would know that, for example, tan^2x=(tanx)^2. I apologize for that mistake.

Mark44
Mentor
Yes, I misread what you wrote.


text:
## Rent Stability

As usual, I wasn’t thrilled when my landlord told me about this year’s rent increase.  As someone familiar with how exponential growth works, I understand the power of the small, consistent percent increase.

After some mild protests, the landlord came back with some good news:  he was going to reduce my increase by $25 per month! I wasn’t exactly thrilled with that, either, but it’s a gesture. And I thought to myself “At least I won’t be paying future increases on that$25!”

Which got me thinking:  if my rent goes up 3% every year, how much of a difference will that $25 make? The answer: not much. In my lifetime, anyway. At 3% per year, the$25 difference in rent now will be a $30 difference in 5 years. In 15 years, the difference in the two rents will be around$40.  And if I stick around for 50 years, the difference will be about $90. If I could stay here for 200 years, I’d see a big difference: almost$10,000!  But then again, my rent would be around $1 million per month. I’ll have to be content knowing that, while in the short term I don’t see much benefit, I win out as t goes to infinity. Click here to see more in Application. www.MrHonner.com ### 4 Comments 1. Fawn Nguyen says: Maybe I see a math problem for students. What is your rent now based on given info? 2. MrHonner says: It’s a good question, one that I was actually wondering myself initially. It’s also interesting (and perhaps surprising) that some of the given information tells you nothing about the current rent. 3. flyingcoloursmaths says: I think you’re measuring the wrong thing! You’ll save$300 this year, $309 next year and so on – a geometric series. Over n years, I reckon you save 10,000(1.03^n – 1). That’s$3,500 over a decade, which isn’t bad at all

4. MrHonner says:

I appreciate your positive outlook. One person’s exponential growth is another’s exponential decay, I guess!


text:
If a,b,c,d,e,f  are A.M. \ 's between 2 and 12, then find the sum: | Filo

Class 11

Math

Algebra

Sequences and Series

572
150

If   are between and , then find the sum: .

572
150

Connecting you to a tutor in 60 seconds.


text:
# Geo Search

To model position(s) in documents, use field(s) of type position. Use this to limit hits to within an area, or use the distance from a position in rank functions. Sample schema:

schema local {
document local {
field title type string {
indexing: index
}
field location type position {
indexing: attribute
}
}
fieldset default {
fields: title
}
}

[
{
"fields": {
"title" :    "Random pizza place",
"location" : "N37.401;W121.996"
}
}
]

## Restrict

The preferred API for adding a geographical restriction is to use a geoLocation clause in the YQL statement, example:

$curl -H "Content-Type: application/json" \ --data '{"yql" : "select * from sources * where title contains \"pizza\" and geoLocation(location, 37.416383, -122.024683, \"20 miles\");"}' \ http://localhost:8080/search/  You can also build or modify the query programatically by adding a GeoLocationItem anywhere in the query tree. To allow the query location to match any position, specify a negative radius (e.g. "-1 m"). But to use a position for ranking only (without any requirement for a matching position) you should also specify it as a ranking-only term. Use the rank() operation in YQL, or a RankItem when building the query programatically. There is also a set of legacy parameters in the query API that may be used to restrict query results, defining the allowed area using a position + radius or a bounding box. • Restrict search results using pos.ll and default radius (50km): $ curl -H "Content-Type: application/json" \
--data '{"yql" : "select * from sources * where title contains \"pizza\";"}' \
http://localhost:8080/search/?pos.ll=N37.416383%3BW122.024683

$curl -H "Content-Type: application/json" \ --data '{"yql" : "select * from sources * where title contains \"pizza\";"}' \ http://localhost:8080/search/?pos.ll=N37.416383%3BW122.024683&pos.radius=5km  • Restrict search results using pos.bb bounding box is also currently possible, but this feature is scheduled for removal in Vespa 8. Using a bounding box gives no meaningful data for distance ranking or summary-features, trying to compute a distance or closeness to a bounding box gives undefined results. $ curl -H "Content-Type: application/json" \
--data '{"yql" : "select * from sources * where title contains \"pizza\";"}' \
http://localhost:8080/search/?pos.bb=n=37.44899,s=37.3323,e=-121.98241,w=-122.06566

## Rank

The corresponding rank features for this example are distance(latlong) or closeness(latlong) or closeness(latlong).logscale - the last is probably the most useful when combining distance ranking with textual relevance ranking.

## Summary fields

See the reference for rendering options.

If the request specifies a position, the distance to this position is calculated and rendered in fieldname.distance. For documents with multiple positions in the attribute, the distance to the nearest position is returned.

### X/Y

"x" and "y" are integers - these are in millionths of a degree - about 10cm. Also note which is which of "x" and "y":

It's just putting a normal coordinate system on top of the world map, so "x" is the longitude (east-west) and "y" the latitude (north-south).

The third summary field is the distance, also as an integer, and also in millionths of degrees. When converting to internal units (millionths of degrees), the Earth polar radius is used, so degrees = 180.0 * meters / (Math.PI * 6356752.0); is the basic conversion formula.

### summary-features

Instead of sending the position and the computed distance in a summary field, it may be more practical to add some of the associated rank features as summary-features. In particular, distance(fieldname).latitude and distance(fieldname).longitude gives the geographical coordinates directly, in degrees. This is especially useful to use these programatically from a searcher, accessing feature values in results for further processing.

## Using multiple position fields

For some applications, it can be useful to have several position attributes that may be searched. For example an address book application could use positions for home address and work address. This is possible to declare without any special considerations in the schema file, but needs some extra handling on the query side. A single query item can only search in one of the position attributes, and with the legacy API must specify which attribute with a pos.attribute query parameter. If you want to have a search that spans several fields, use YQL to combine several geoLocation items inside an or clause, or combine several fields into a combined array field:

schema address {
indexing: summary | index
}
field homelatlong type position {
indexing: attribute
}
indexing: summary | index
}
field worklatlong type position {
indexing: attribute
}
field bothlatlong type array<position> {
indexing: attribute
}
}
}
}

Here we assume that the home fields will contain the address and position of your house, the work fields the address and position of your workplace, while the bothlatlong field is assumed to be filled with the positions of both house and workplace (before feeding data into Vespa, or in a document processor during feeding). In a query it's then possible to say
search/?query=homeaddress:sunnyvale&pos.attribute=homelatlong&pos.ll=N37.416383%3BW122.024683&pos.radius=5km

which is unlikely to give very many hits, since it's mostly a business district around Yahoo! headquarters, while
search/?query=workaddress:sunnyvale&pos.attribute=worklatlong&pos.ll=N37.416383%3BW122.024683&pos.radius=5km

would show lots of people working in Sunnyvale; use pos.attribute=bothlatlong for cases where it's uncertain if home address or work address position was wanted.

## Distance to path

This example provides an overview of the DistanceToPath rank feature. This feature matches document locations to a path given in the query. Not only does this feature return the closest distance for each document to the path, it also includes the length traveled along the path before reaching the closest point, or intersection. This feature has been nick named the gas feature because of its obvious use case of finding gas stations along a planned trip.

In this example we have been traveling from the US to Bangalore, and we are now planning our trip back. We have decided to rent a car in Bangalore that we are to return upon arrival at the airport in Chennai. We are already quite hungry and wish to stop for a meal once we are outside of town. To avoid having to pay an additional fueling premium, we also wish to refuel just before reaching the airport. We need to figure out what roads to take, what restaurants are available outside of Bangalore, and what fuel stations are available once we get close to Chennai. In figure 1 we have plotted our trip from Bangalore to the airport:

If we search for restaurants along the path, we only see a small subset of all restaurants present in the window of our quite large map. In figure 2 you see how the most relevant results are actually all in Bangalore or Chennai:

To find the best results, move the map window to just about where we expect to be eating, and redo the search:

This has to be done similarly for finding a gas station near the airport. This illustrates searching for restaurants in a smaller window along the planned trip without DistanceToPath. Next, we outline how DistanceToPath can be used to quickly and easily improve this type of planning to be more convenient for the user.

The nature of this feature requires that the search corpus contains documents with position data. A searcher component needs to be written that is able to pass paths with the queries that lie in the same coordinate space as the searchable documents. Finally, a rank-profile needs to defined that scores documents according to how they match some target distance traveled and at the same time lies close "enough" to the path.

### Query Syntax

This document does not describe how to write a searcher plugin for the JDisc Container, refer to the container documentation. However, let us review the syntax expected by DistanceToPath. As noted in the the rank features reference, the path is supplied as a query parameter by name of the feature and the path keyword:

yql=(…)&rankproperty.distanceToPath(name).path=(x1,y1,x2,y2,…,xN,yN)

Here name has to match the name of the position attribute that holds the positions data.

The path itself is parsed as a list of N coordinate pairs that together form N-1 line segments:

$$(x_1,y_1) \rightarrow (x_2,y_2), (x_2,y_2) \rightarrow (x_3,y_3), (…), (x_{N-1},y_{N-1}) \rightarrow (x_N,y_N)$$

The path is not in a readable (longitude, latitude) format, but is a pair of integers in the internal format (degrees multiplied by 1 million). If a transform is required from geographic coordinates to this, the search plugin must do it; note that the first number in each pair (the "x") is longitude (degrees East or West) while the second ("y") is latitude (degrees North or South), corresponding to the usual orientation for maps - opposite to the usual order of latitude/longitude.

### Rank profile

If we were to disregard our scenario for a few moments, we could suggest the following rank profile:

rank-profile default {
first-phase {
expression: nativeRank
}
second-phase {
expression: firstPhase * if (distanceToPath(ll).distance < 10000, 1, 0)
}
}

This profile will first rank all documents according to Vespa's nativeRank feature, and then do a second pass over the top 100 results and order these based on their distance to our path. It is very simple; if a document lies within 100 metres of our path it retains its relevancy, otherwise its relevancy is set to 0. Such a rank profile would indeed solve the current problem, but Vespa's ranking model allows for us to take this a lot further.

The following is a very simple rank profile that ranks documents according to a query-specified target distance to path and distance traveled:

rank-profile default {
first-phase {
expression {
max(0,    $distance - distanceToPath(ll).distance) * (1 - fabs($traveled - distanceToPath(ll).traveled))
}
}
}

The expression is two-fold; a first component determines a rank based on the document's distance to the given path as compared to the query parameter $distance. If the allowed distance is exceeded, this component's contribution is 0. The distance contribution is then multiplied by the difference of the actual distance traveled as compared to the query parameter $traveled. In short, this profile will include all documents that lie close enough to the path, ranked according to their actual distance and traveled measure.

DistanceToPath is only compatible with 2D coordinates because pathing in 1 dimension makes no sense.

### Results

For the sake of this example, assume that we have implemented a custom path searcher that is able to pass the path found by the user's initial directions query to Vespa's query syntax. There are then two more parameters that must be supplied by the user; $distance and $traveled. Vespa expects these parameters to be supplied in a scale compatible with the feature's output, and should probably also be mapped by the container plugin. The feature's distance output is given in Vespa's internal resolution, which is approximately 10 units per meter. The traveled output is a normalized number between 0 and 1, where 0 represents the beginning of the path, and 1 is the end of the path.

This illustrates how these parameters can be used to return the most appropriate hits for our scenario. Note that the figures only show the top hit for each query:

1. Searching for restaurants with the DistanceToPath feature. $distance = 1000,$traveled = 0.1
2. Searching for gas stations with the DistanceToPath feature. $distance = 1000,$traveled = 0.9


text:
# How to correctly run explore_pr2 package to let PR2 navigate in the environment automatically? [closed]

I want to use the pr2_exploration package to let PR2 navigate in the environment automatically, and I executed the commands in two terminals

In the first terminal: roslaunch pr2_gazebo pr2_wg_world.launch

In the second terminal: export ROBOT=sim & roslaunch explore_pr2 explore_pr2.launch

However, I got some errors, the. You can find overall output messages here.

It seems like the key lies in these message:

[ERROR] [1366390413.797578744, 131.566000000]: Client [/explore] wants topic /base_scan_marking to have datatype/md5sum [sensor_msgs/PointCloud/d8e9c3f5afbdd8a130fd1d2763945fca], but our version has [sensor_msgs/PointCloud2/1158d486dd51d683ce2f1be655c3c181]. Dropping connection.

Because I got repeated message like:

[ WARN] [1366390421.398585539, 134.533000000]: The /base_scan_marking observation buffer has not been updated for 3.01 seconds, and it should be updated every 0.20 seconds.
[ WARN] [1366390423.807004380, 135.433000000]: Costmap2DROS transform timeout. Current time: 135.4330, global_pose stamp: 135.1820, tolerance: 0.2000
[ WARN] [1366390423.807063444, 135.433000000]: Could not get robot pose, cancelling reconfiguration
[ WARN] [1366390424.058372392, 135.533000000]: The /base_scan_marking observation buffer has not been updated for 4.01 seconds, and it should be updated every 0.20 seconds.
[ WARN] [1366390426.880916825, 136.533000000]: The /base_scan_marking observation buffer has not been updated for 5.01 seconds, and it should be updated every 0.20 seconds.

But I am not sure and don't know how to solve it, does anyone know how to correct this?