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# In Brownian motion, how does the absolute value of a particle's distance from the center scale with respect to number of iterations? [closed]
The mean distance is probably zero due to averaging out. But what about the mean of the absolute distance?
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Since distances cannot be negative, it's difficult to see how they could "average out" to zero! – whuber Aug 11 '11 at 14:39
You would be looking at a half-normal distribution, whose expectation is $\sigma \sqrt{2/\pi}$, if the underlying normal random variable has variance $\sigma^2$. Furthermore, the variance in Brownian motion is commensurate with time.
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# Why are the nodes of an open tubular bell located at .224*L instead of .25*L?
Say that we have a tube of length $L$. In the tube, there is a standing wave of wavelength $\lambda$. Then, $L=\lambda$.
$\hspace{2.5cm}$
In the above diagram, the wave's amplitude is highest at the ends and center of the tube. The wave has two nodes of with zero amplitude on either side of the middle of the tube.
I would expect these nodes to be at exactly 1/4 of the tube's length from each end. Beginning at the left side of the tube: If the tube contains one whole wavelength, then the first $90{}^{\circ}$ of the wave should bring the wave from the maximum down to zero. The next $90{}^{\circ}$ should take the wave to the minimum at the center of the tube, and so on. Dividing the wavelength into four equal parts should make these locations occur at even intervals, or at even quarters of the tube's length, $L$. Correct?
However, I have built tubular bells in the past. In all of the literature (and in experience), the correct locations for these node points is $0.224L$, rather than $0.25$. Would someone mind explaining this?
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# Anisotropic shortest paths across Strait of Gibraltar
#### noviembre 29, 2018
First, we load the main packages we will use in this vignette. This vignette was written under the rWind version 1.0.4
# use install.packages() if some is not installed
# and you can install the latest development version using the command
# devtools::install_github("jabiologo/rWind")
library(rWind)
library(raster)
library(gdistance)
In this simple example, we introduce the most basic functionality of rWind, to get the shortest paths between two points across Strait of Gibraltar. Notice that, as wind connectivity is anisotropic (direction dependent), shortest path from A to B usually does not match with shortest path from B to A.
First, we download wind data of a selected date (e.g. 2015 February 12th).
w <- wind.dl(2015, 2, 12, 12, -7, -4, 34.5, 37.5)
Next we transform this data.frame into two raster layers, with values of wind direction and wind speed.
wind_layer <- wind2raster(w)
Then, we will use flow.dispersion function to obtain a transitionLayer object with conductance values, which will be used later to obtain the shortest paths.
Conductance<-flow.dispersion(wind_layer)
Now, we will use shortestPath function from gdistance package (van Etten 2018) to compute shortest path from our Conductance object between the two selected points.
AtoB<- shortestPath(Conductance,
c(-5.5, 37), c(-5.5, 35), output="SpatialLines")
BtoA<- shortestPath(Conductance,
c(-5.5, 35), c(-5.5, 37), output="SpatialLines")
Finally, we plot the map and we will add the shortest paths as lines and some other features.
We need some additionally packages to be installed. This can be done using the command install.packages(c("fields", "shape", "rworldmap")).
library(fields)
library(shape)
library(rworldmap)
image.plot(wind_layer$wind.speed, main="least cost paths by wind direction and speed", col=terrain.colors(10), xlab="Longitude", ylab="Lattitude", zlim=c(0,7)) lines(getMap(resolution = "low"), lwd=4) points(-5.5, 37, pch=19, cex=3.4, col="red") points(-5.5, 35, pch=19, cex=3.4, col="blue") lines(AtoB, col="red", lwd=4, lty=2) lines(BtoA, col="blue", lwd=4, lty=2) alpha <- arrowDir(w) Arrowhead(w$lon, w\$lat, angle=alpha, arr.length = 0.4, arr.type="curved")
text(-5.75, 37.25,labels="Spain", cex= 2.5, col="red", font=2)
text(-5.25, 34.75,labels="Morocco", cex= 2.5, col="blue", font=2)
legend("toprigh", legend = c("From Spain to Morocco", "From Morocco to Spain"),
lwd=4 ,lty = 1, col=c("red","blue"), cex=0.9, bg="white")
## References
Douglas Nychka, Reinhard Furrer, John Paige, and Stephan Sain. 2017. “Fields: Tools for Spatial Data.” Boulder, CO, USA: University Corporation for Atmospheric Research. doi:10.5065/D6W957CT.
Hijmans, Robert J. 2017. Raster: Geographic Data Analysis and Modeling. https://CRAN.R-project.org/package=raster.
Soetaert, Karline. 2018. Shape: Functions for Plotting Graphical Shapes, Colors. https://CRAN.R-project.org/package=shape.
South, Andy. 2011. “Rworldmap: A New R Package for Mapping Global Data.” The R Journal 3 (1): 35–43. https://journal.r-project.org/archive/2011-1/RJournal_2011-1_South.pdf.
van Etten, Jacob. 2018. Gdistance: Distances and Routes on Geographical Grids. https://CRAN.R-project.org/package=gdistance.
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# Are classes of finite dimensional representations of a $\mathbb{C}$-algebra classified by characters?
Let $A$ be a $\mathbb{C}$-algebra.
For finite dimensional representation $\rho:A \rightarrow \rm{End}(V)$, the trace $\chi_\rho$ of $\rho$ is defined by $\chi_\rho(a):=\rm{Tr}(\rho(a))$ for $a \in A$.
It is true that if two finite dimensional representations $\rho,\sigma$ are isomorphic then $\chi_\rho =\chi_\sigma$? Now, if $\chi_\rho = \chi_\sigma$, are $\rho$ and $\sigma$ isomorphic?
Not in general. For instance, let $A=\mathbb{C}[x]/(x^2)$. Consider two different representations of $A$ on $\mathbb{C}^2$: $\rho$, which sends $x$ to $\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$, and $\sigma$, which sends $x$ to $0$. For any $a\in A$, $\rho(a)$ and $\sigma(a)$ have the same diagonal entries, and hence the same trace. But $\rho$ and $\sigma$ are not isomorphic.
• Oh! it's simple counter example. Thank you. – H.Kakuhama Sep 27 '16 at 9:45
That the answer to your question is „no” has been shown in Eric Wofsey's answer. But there is a little bit more that can be said.
Let $A$ be a finite dimensional algebra over a field $\mathbb{k}$ of characteristic zero, and $M_1, M_2$ finitely generated $A$-modules with corresponding characters $\chi_1, \chi_2$. Then $\chi_1 = \chi_2$ if and only if $M_1$ and $M_2$ have the same composition factors, counted with multiplicities. Alternatively, $[M_1] = [M_2]$ as elements of the Grothendieck group $G_0(A)$.
In the counterexample of Eric Wofsey, both modules have the composition factor $\mathbb{C}$ (on which $x$ acts trivially) with multiplicity 2, so their characters have to coincide.
Details can be found in chapter 3 of [Lam - a first course in noncommutative rings]. See in particular Theorem 7.19.
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dc.contributor.author Elvey Price, Andrew dc.date.accessioned 2018-12-11T04:28:47Z dc.date.available 2018-12-11T04:28:47Z dc.date.issued 2018 en_US dc.identifier.uri http://hdl.handle.net/11343/219277 dc.description © 2018 Dr Andrew Elvey Price dc.description.abstract In this thesis we consider a number of enumerative combinatorial problems. We solve the problems of enumerating Eulerian orientations by edges and quartic Eulerian orientations counted by vertices. We also find and prove an algebraic relationship between the counting functions for permutations sortable by a double ended queue (deque) and permutations sortable by two stacks in parallel (2sip). In each of these cases, our proof of the result uses an elaborate system of functional equations which is much more complicated than the result itself, leaving the door open for a more direct, combinatorial proof. en_US We find polynomial time algorithms for generating the counting sequence of deque-sortable permutations and the cogrowth sequence of some groups, including the lamplighter group $L$ and the Brin-Navas group $B$. For permutations sortable by two stacks in series and for the cogrowth sequence of Thompson's group $F$, we find exponential time algorithms which are significantly more efficient than the algorithms that previously existed in the literature. In each case an empirical analysis of the produced terms of the sequence leads to a prediction regarding its asymptotic form. In particular, this method leads us to conjecture that the growth rate of deque-sortable permutations is equal to that of 2sip-sortable permutations, a conjecture which we reduce to three conjectures of Albert and Bousquet-M\'elou about quarter plane walks. The analysis of the cogrowth sequence of Thompson's group $F$ leads us to conjecture that $F$ is not amenable. We also study the enumeration of $1324$-avoiding permutations, a notoriously difficult problem in the field of pattern avoiding permutations. Using a structural decomposition of these permutations, we improve the lower and upper bounds on the growth rate to $10.271$ and $13.5$ respectively. Next we investigate the concept of combinatorial Stieltjes moment sequences. We prove that the counting sequence of returns in any undirected locally-finite graph is a Stieltjes moment sequence. As a special case, this implies that any cogrowth sequence is a Stieltjes moment sequence. Based on empirical evidence, we conjecture that the counting sequence for $1324$-avoiding permutations is a Stieltjes moment sequence, which would imply an improved lower bound of $10.302$ on its growth rate. We then describe a general class of counting sequences of augmented perfect matchings, which we prove to be Stieltjes moment sequences. In fact, we prove the stronger result that these sequences are Hankel totally-positive as sequences of polynomials. As a special case, we show that the Ward polynomials are Hankel totally-positive. In the final chapter we generalise an identity of Duminil-Copin and Smirnov for the $O(n)$ loop model on the hexagonal lattice to the off-critical case. In the $n=0$ case, which corresponds to the enumeration of self-avoiding walks, we use our identity to prove a relationship between the half-plane surface critical exponents $\gamma_{1}$ and $\gamma_{11}$ and the exponent characterising the winding angle distribution of self-avoiding walks in the half-plane. dc.rights Terms and Conditions: Copyright in works deposited in Minerva Access is retained by the copyright owner. The work may not be altered without permission from the copyright owner. Readers may only download, print and save electronic copies of whole works for their own personal non-commercial use. Any use that exceeds these limits requires permission from the copyright owner. Attribution is essential when quoting or paraphrasing from these works. dc.subject Enumerative combinatorics en_US dc.subject permutation class en_US dc.subject planar map en_US dc.subject Eulerian orientation en_US dc.subject Thompson's group F en_US dc.subject random walk en_US dc.subject Stieltjes moment sequence en_US dc.subject O(n) model en_US dc.title Selected problems in enumerative combinatorics: permutation classes, random walks and planar maps en_US dc.type PhD thesis en_US melbourne.affiliation.department School of Mathematics and Statistics melbourne.affiliation.faculty Science melbourne.thesis.supervisorname Guttmann, Tony melbourne.contributor.author Elvey Price, Andrew melbourne.accessrights Open Access
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Map-0.1.3.2: Class of key-value maps
Data.Map.Class
Description
Class of key-value maps
See StaticMap and Map.
Synopsis
Documentation
class Traversable map => StaticMap map where Source #
Class of key-value maps
Laws:
• adjustA pure _ = pure
• adjustA Const k <=< traverseWithKey f = fmap Const . f k <=< getConst . adjustA Const k
Associated Types
type Key map Source #
Methods
adjustA :: Applicative p => (a -> p a) -> Key map -> map a -> p (map a) Source #
Modify the value of the key in the map. If the key is absent, the map is returned unmodified.
traverseWithKey :: Applicative p => (Key map -> a -> p b) -> map a -> p (map b) Source #
Traverse a function over each value in the map.
Instances
Instances details
Source # Instance detailsDefined in Data.Map.Class Associated Typestype Key Maybe Source # MethodsadjustA :: Applicative p => (a -> p a) -> Key Maybe -> Maybe a -> p (Maybe a) Source #traverseWithKey :: Applicative p => (Key Maybe -> a -> p b) -> Maybe a -> p (Maybe b) Source # Source # Instance detailsDefined in Data.Map.Class Associated Typestype Key IntMap Source # MethodsadjustA :: Applicative p => (a -> p a) -> Key IntMap -> IntMap a -> p (IntMap a) Source #traverseWithKey :: Applicative p => (Key IntMap -> a -> p b) -> IntMap a -> p (IntMap b) Source # Ord key => StaticMap (Map key) Source # Instance detailsDefined in Data.Map.Class Associated Typestype Key (Map key) Source # MethodsadjustA :: Applicative p => (a -> p a) -> Key (Map key) -> Map key a -> p (Map key a) Source #traverseWithKey :: Applicative p => (Key (Map key) -> a -> p b) -> Map key a -> p (Map key b) Source # StaticMap map => StaticMap (SymmetricDifference map) Source # Instance detailsDefined in Data.Map.Class Associated Typestype Key (SymmetricDifference map) Source # MethodsadjustA :: Applicative p => (a -> p a) -> Key (SymmetricDifference map) -> SymmetricDifference map a -> p (SymmetricDifference map a) Source #traverseWithKey :: Applicative p => (Key (SymmetricDifference map) -> a -> p b) -> SymmetricDifference map a -> p (SymmetricDifference map b) Source # StaticMap map => StaticMap (Intersection map) Source # Instance detailsDefined in Data.Map.Class Associated Typestype Key (Intersection map) Source # MethodsadjustA :: Applicative p => (a -> p a) -> Key (Intersection map) -> Intersection map a -> p (Intersection map a) Source #traverseWithKey :: Applicative p => (Key (Intersection map) -> a -> p b) -> Intersection map a -> p (Intersection map b) Source # StaticMap map => StaticMap (Union map) Source # Instance detailsDefined in Data.Map.Class Associated Typestype Key (Union map) Source # MethodsadjustA :: Applicative p => (a -> p a) -> Key (Union map) -> Union map a -> p (Union map a) Source #traverseWithKey :: Applicative p => (Key (Union map) -> a -> p b) -> Union map a -> p (Union map b) Source # (StaticMap m, StaticMap n) => StaticMap (Product m n) Source # Instance detailsDefined in Data.Map.Class Associated Typestype Key (Product m n) Source # MethodsadjustA :: Applicative p => (a -> p a) -> Key (Product m n) -> Product m n a -> p (Product m n a) Source #traverseWithKey :: Applicative p => (Key (Product m n) -> a -> p b) -> Product m n a -> p (Product m n b) Source # (StaticMap m, StaticMap n) => StaticMap (Compose m n) Source # Instance detailsDefined in Data.Map.Class Associated Typestype Key (Compose m n) Source # MethodsadjustA :: Applicative p => (a -> p a) -> Key (Compose m n) -> Compose m n a -> p (Compose m n a) Source #traverseWithKey :: Applicative p => (Key (Compose m n) -> a -> p b) -> Compose m n a -> p (Compose m n b) Source #
class (Filtrable map, StaticMap map) => Map map where Source #
Class of key-value maps with variable structure
Minimal complete definition
Methods
empty :: map a Source #
The empty map
alterF :: Functor f => (Maybe a -> f (Maybe a)) -> Key map -> map a -> f (map a) Source #
Modify the value of the key in the map, or insert the key and its value into the map, or delete the key and its value from the map, functorially.
fmap (!? k) . alterF f k = f . (!? k)
This is the most general operation on a given key in the map.
mergeA :: Applicative p => (Key map -> Either' a b -> p (Maybe c)) -> map a -> map b -> p (map c) Source #
Combine two maps with the given function, which is called once for each key present in either map, inclusive.
mapMaybeWithKeyA :: Applicative p => (Key map -> a -> p (Maybe b)) -> map a -> p (map b) Source #
Traverse a function over each value in the map, gathering the Just values and forgetting the Nothing.
mapEitherWithKeyA :: Applicative p => (Key map -> a -> p (Either b c)) -> map a -> p (map b, map c) Source #
Traverse a function over each value in the map, gathering the Left and Right values separately.
Instances
Instances details
Source # Instance detailsDefined in Data.Map.Class MethodsalterF :: Functor f => (Maybe a -> f (Maybe a)) -> Key Maybe -> Maybe a -> f (Maybe a) Source #mergeA :: Applicative p => (Key Maybe -> Either' a b -> p (Maybe c)) -> Maybe a -> Maybe b -> p (Maybe c) Source #mapMaybeWithKeyA :: Applicative p => (Key Maybe -> a -> p (Maybe b)) -> Maybe a -> p (Maybe b) Source #mapEitherWithKeyA :: Applicative p => (Key Maybe -> a -> p (Either b c)) -> Maybe a -> p (Maybe b, Maybe c) Source # Source # Instance detailsDefined in Data.Map.Class MethodsalterF :: Functor f => (Maybe a -> f (Maybe a)) -> Key IntMap -> IntMap a -> f (IntMap a) Source #mergeA :: Applicative p => (Key IntMap -> Either' a b -> p (Maybe c)) -> IntMap a -> IntMap b -> p (IntMap c) Source #mapMaybeWithKeyA :: Applicative p => (Key IntMap -> a -> p (Maybe b)) -> IntMap a -> p (IntMap b) Source #mapEitherWithKeyA :: Applicative p => (Key IntMap -> a -> p (Either b c)) -> IntMap a -> p (IntMap b, IntMap c) Source # Ord key => Map (Map key) Source # Instance detailsDefined in Data.Map.Class Methodsempty :: Map key a Source #alterF :: Functor f => (Maybe a -> f (Maybe a)) -> Key (Map key) -> Map key a -> f (Map key a) Source #mergeA :: Applicative p => (Key (Map key) -> Either' a b -> p (Maybe c)) -> Map key a -> Map key b -> p (Map key c) Source #mapMaybeWithKeyA :: Applicative p => (Key (Map key) -> a -> p (Maybe b)) -> Map key a -> p (Map key b) Source #mapEitherWithKeyA :: Applicative p => (Key (Map key) -> a -> p (Either b c)) -> Map key a -> p (Map key b, Map key c) Source # Map map => Map (SymmetricDifference map) Source # Instance detailsDefined in Data.Map.Class MethodsalterF :: Functor f => (Maybe a -> f (Maybe a)) -> Key (SymmetricDifference map) -> SymmetricDifference map a -> f (SymmetricDifference map a) Source #mergeA :: Applicative p => (Key (SymmetricDifference map) -> Either' a b -> p (Maybe c)) -> SymmetricDifference map a -> SymmetricDifference map b -> p (SymmetricDifference map c) Source #mapMaybeWithKeyA :: Applicative p => (Key (SymmetricDifference map) -> a -> p (Maybe b)) -> SymmetricDifference map a -> p (SymmetricDifference map b) Source #mapEitherWithKeyA :: Applicative p => (Key (SymmetricDifference map) -> a -> p (Either b c)) -> SymmetricDifference map a -> p (SymmetricDifference map b, SymmetricDifference map c) Source # Map map => Map (Intersection map) Source # Instance detailsDefined in Data.Map.Class Methodsempty :: Intersection map a Source #alterF :: Functor f => (Maybe a -> f (Maybe a)) -> Key (Intersection map) -> Intersection map a -> f (Intersection map a) Source #mergeA :: Applicative p => (Key (Intersection map) -> Either' a b -> p (Maybe c)) -> Intersection map a -> Intersection map b -> p (Intersection map c) Source #mapMaybeWithKeyA :: Applicative p => (Key (Intersection map) -> a -> p (Maybe b)) -> Intersection map a -> p (Intersection map b) Source #mapEitherWithKeyA :: Applicative p => (Key (Intersection map) -> a -> p (Either b c)) -> Intersection map a -> p (Intersection map b, Intersection map c) Source # Map map => Map (Union map) Source # Instance detailsDefined in Data.Map.Class Methodsempty :: Union map a Source #alterF :: Functor f => (Maybe a -> f (Maybe a)) -> Key (Union map) -> Union map a -> f (Union map a) Source #mergeA :: Applicative p => (Key (Union map) -> Either' a b -> p (Maybe c)) -> Union map a -> Union map b -> p (Union map c) Source #mapMaybeWithKeyA :: Applicative p => (Key (Union map) -> a -> p (Maybe b)) -> Union map a -> p (Union map b) Source #mapEitherWithKeyA :: Applicative p => (Key (Union map) -> a -> p (Either b c)) -> Union map a -> p (Union map b, Union map c) Source # (Map m, Map n) => Map (Product m n) Source # Instance detailsDefined in Data.Map.Class Methodsempty :: Product m n a Source #alterF :: Functor f => (Maybe a -> f (Maybe a)) -> Key (Product m n) -> Product m n a -> f (Product m n a) Source #mergeA :: Applicative p => (Key (Product m n) -> Either' a b -> p (Maybe c)) -> Product m n a -> Product m n b -> p (Product m n c) Source #mapMaybeWithKeyA :: Applicative p => (Key (Product m n) -> a -> p (Maybe b)) -> Product m n a -> p (Product m n b) Source #mapEitherWithKeyA :: Applicative p => (Key (Product m n) -> a -> p (Either b c)) -> Product m n a -> p (Product m n b, Product m n c) Source # (Map m, Map n) => Map (Compose m n) Source # Instance detailsDefined in Data.Map.Class Methodsempty :: Compose m n a Source #alterF :: Functor f => (Maybe a -> f (Maybe a)) -> Key (Compose m n) -> Compose m n a -> f (Compose m n a) Source #mergeA :: Applicative p => (Key (Compose m n) -> Either' a b -> p (Maybe c)) -> Compose m n a -> Compose m n b -> p (Compose m n c) Source #mapMaybeWithKeyA :: Applicative p => (Key (Compose m n) -> a -> p (Maybe b)) -> Compose m n a -> p (Compose m n b) Source #mapEitherWithKeyA :: Applicative p => (Key (Compose m n) -> a -> p (Either b c)) -> Compose m n a -> p (Compose m n b, Compose m n c) Source #
defaultAdjustA :: (Map map, Applicative p) => (a -> p a) -> Key map -> map a -> p (map a) Source #
Default implementation of adjustA in terms of Map methods
defaultTraverseWithKey :: (Map map, Applicative p) => (Key map -> a -> p b) -> map a -> p (map b) Source #
Default implementation of traverseWithKey in terms of Map methods
(!?) :: StaticMap map => map a -> Key map -> Maybe a infix 9 Source #
Look up the key in the map.
insert :: Map map => Key map -> a -> map a -> map a Source #
Insert a key and new value into the map, the new value clobbering the old if the key is already present. insert = insertWith pure
insertWith :: Map map => (a -> a -> a) -> Key map -> a -> map a -> map a Source #
Insert a key and new value into the map, combining the old and new values with the given function if the key is already present.
insertLookup :: Map map => Key map -> a -> map a -> (Maybe a, map a) Source #
Insert a key and new value into the map, looking up the old value if the key is already present.
insertLookupWith :: Map map => (a -> a -> a) -> Key map -> a -> map a -> (Maybe a, map a) Source #
Insert a key and new value into the map, looking up the old value and combining the old and new values with the given function if the key is already present.
delete :: Map map => Key map -> map a -> map a Source #
Delete a key and its value from the map. If the key is absent, the map is returned unmodified.
adjust :: StaticMap map => (a -> a) -> Key map -> map a -> map a Source #
Modify the value of the key in the map. If the key is absent, the map is returned unmodified.
update :: Map map => (a -> Maybe a) -> Key map -> map a -> map a Source #
Modify the value of the key in the map, or delete the key and its value from the map, if the given function returns Just or Nothing, in turn. If the key is absent, the map is returned unmodified.
updateLookup :: Map map => (a -> Maybe a) -> Key map -> map a -> (Maybe a, map a) Source #
Modify the value of the key in the map, or delete the key and its value from the map, if the given function returns Just or Nothing, in turn, looking up the old value if the key is already present. If the key is absent, the map is returned unmodified.
alter :: Map map => (Maybe a -> Maybe a) -> Key map -> map a -> map a Source #
Modify the value of the key in the map, or insert the key and its value into the map, or delete the key and its value from the map.
alterLookup :: Map map => (Maybe a -> Maybe a) -> Key map -> map a -> (Maybe a, map a) Source #
Modify the value of the key in the map, or insert the key and its value into the map, or delete the key and its value from the map, looking up the old value if the key is already present.
alterLookupF :: (Map map, Functor f) => (Maybe a -> f (Maybe a)) -> Key map -> map a -> f (Maybe a, map a) Source #
Modify the value of the key in the map, or insert the key and its value into the map, or delete the key and its value from the map, looking up the old value if the key is already present, functorially.
This is no more general than alterF, but is defined for convenience.
mapWithKey :: StaticMap map => (Key map -> a -> b) -> map a -> map b Source #
Map a function over each value in the map.
mapMaybeWithKey :: Map map => (Key map -> a -> Maybe b) -> map a -> map b Source #
Map a function over each value in the map, gathering the Just values and forgetting the Nothing.
mapEitherWithKey :: Map map => (Key map -> a -> Either b c) -> map a -> (map b, map c) Source #
Map a function over each value in the map, gathering the Left and Right separately.
foldMapWithKeyA :: (StaticMap map, Applicative p, Monoid b) => (Key map -> a -> p b) -> map a -> p b Source #
foldrWithKeyM :: (StaticMap map, Monad m) => (Key map -> a -> b -> m b) -> b -> map a -> m b Source #
foldlWithKeyM :: (StaticMap map, Monad m) => (b -> Key map -> a -> m b) -> b -> map a -> m b Source #
foldMapWithKey :: (StaticMap map, Monoid b) => (Key map -> a -> b) -> map a -> b Source #
foldrWithKey :: StaticMap map => (Key map -> a -> b -> b) -> b -> map a -> b Source #
foldlWithKey :: StaticMap map => (b -> Key map -> a -> b) -> b -> map a -> b Source #
fromList :: Map map => [(Key map, a)] -> map a Source #
fromListWith :: Map map => (a -> a -> a) -> [(Key map, a)] -> map a Source #
fromListWithKey :: Map map => (Key map -> a -> a -> a) -> [(Key map, a)] -> map a Source #
fromListWithM :: (Map map, Monad m) => (a -> a -> m a) -> [(Key map, a)] -> m (map a) Source #
fromListWithKeyM :: (Map map, Monad m) => (Key map -> a -> a -> m a) -> [(Key map, a)] -> m (map a) Source #
adjustLookupA :: (StaticMap map, Applicative p) => (a -> p a) -> Key map -> map a -> p (Maybe a, map a) Source #
Modify the value of the key in the map, looking up the old value if the key is already present. If the key is absent, the map is returned unmodified.
singleton :: Map map => Key map -> a -> map a Source #
Map with a single element
unionWith :: Map map => (Key map -> a -> a -> a) -> map a -> map a -> map a Source #
Union of two maps, combining values of the same key with the given function
intersectionWith :: Map map => (Key map -> a -> b -> c) -> map a -> map b -> map c Source #
Intersection of two maps, combining values of the same key with the given function
merge :: Map map => (Key map -> Either' a b -> Maybe c) -> map a -> map b -> map c Source #
Combine two maps with the given function, which is called once for each key present in either map, inclusive.
unionWithA :: (Map map, Applicative p) => (Key map -> a -> a -> p a) -> map a -> map a -> p (map a) Source #
Union of two maps, combining values of the same key with the given function
intersectionWithA :: (Map map, Applicative p) => (Key map -> a -> b -> p c) -> map a -> map b -> p (map c) Source #
Intersection of two maps, combining values of the same key with the given function
difference :: Map map => map a -> map b -> map a Source #
Difference of two maps, which contains exactly the keys present in the first map but absent in the second
symmetricDifference :: Map map => map a -> map a -> map a Source #
Symmetric difference of two maps, which contains exactly the keys present in the either map but absent in the other
mapKeys :: (StaticMap m, Map n) => (Key m -> Key n) -> m a -> n a Source #
Map a function over each key in the map.
mapKeysWith :: (StaticMap m, Map n) => (a -> a -> a) -> (Key m -> Key n) -> m a -> n a Source #
Map a function over each key in the map, combining values of keys which collide with the given function.
traverseKeys :: (StaticMap m, Map n, Applicative p) => (Key m -> p (Key n)) -> m a -> p (n a) Source #
Traverse a function over each key in the map.
traverseKeysWith :: (StaticMap m, Map n, Applicative p) => (a -> a -> a) -> (Key m -> p (Key n)) -> m a -> p (n a) Source #
Traverse a function over each key in the map, combining values of keys which collide with the given function.
mapKeysMaybe :: (StaticMap m, Map n) => (Key m -> Maybe (Key n)) -> m a -> n a Source #
Map a function over each key in the map, gathering the Just values and forgetting the Nothing.
mapKeysMaybeWith :: (StaticMap m, Map n) => (a -> a -> a) -> (Key m -> Maybe (Key n)) -> m a -> n a Source #
Map a function over each key in the map, gathering the Just values and forgetting the Nothing, combining values of keys which collide with the given function.
traverseKeysMaybe :: (StaticMap m, Map n, Applicative p) => (Key m -> p (Maybe (Key n))) -> m a -> p (n a) Source #
Traverse a function over each key in the map, gathering the Just values and forgetting the Nothing.
traverseKeysMaybeWith :: (StaticMap m, Map n, Applicative p) => (a -> a -> a) -> (Key m -> p (Maybe (Key n))) -> m a -> p (n a) Source #
Traverse a function over each key in the map, gathering the Just values and forgetting the Nothing, combining values of keys which collide with the given function.
keys :: StaticMap map => map a -> map (Key map) Source #
Keys of the map
keys as !? k = k <$ (as !? k) newtype Union map a Source # Wrapper of a Map whose semigroup operation is the union, combining values elementwise, and ergo whose monoidal unit is empty Constructors Union FieldsunUnion :: map a Instances Instances details Functor map => Functor (Union map) Source # Instance detailsDefined in Data.Map.Class Methodsfmap :: (a -> b) -> Union map a -> Union map b #(<$) :: a -> Union map b -> Union map a # Foldable map => Foldable (Union map) Source # Instance detailsDefined in Data.Map.Class Methodsfold :: Monoid m => Union map m -> m #foldMap :: Monoid m => (a -> m) -> Union map a -> m #foldMap' :: Monoid m => (a -> m) -> Union map a -> m #foldr :: (a -> b -> b) -> b -> Union map a -> b #foldr' :: (a -> b -> b) -> b -> Union map a -> b #foldl :: (b -> a -> b) -> b -> Union map a -> b #foldl' :: (b -> a -> b) -> b -> Union map a -> b #foldr1 :: (a -> a -> a) -> Union map a -> a #foldl1 :: (a -> a -> a) -> Union map a -> a #toList :: Union map a -> [a] #null :: Union map a -> Bool #length :: Union map a -> Int #elem :: Eq a => a -> Union map a -> Bool #maximum :: Ord a => Union map a -> a #minimum :: Ord a => Union map a -> a #sum :: Num a => Union map a -> a #product :: Num a => Union map a -> a # Traversable map => Traversable (Union map) Source # Instance detailsDefined in Data.Map.Class Methodstraverse :: Applicative f => (a -> f b) -> Union map a -> f (Union map b) #sequenceA :: Applicative f => Union map (f a) -> f (Union map a) #mapM :: Monad m => (a -> m b) -> Union map a -> m (Union map b) #sequence :: Monad m => Union map (m a) -> m (Union map a) # Eq1 map => Eq1 (Union map) Source # Instance detailsDefined in Data.Map.Class MethodsliftEq :: (a -> b -> Bool) -> Union map a -> Union map b -> Bool # Ord1 map => Ord1 (Union map) Source # Instance detailsDefined in Data.Map.Class MethodsliftCompare :: (a -> b -> Ordering) -> Union map a -> Union map b -> Ordering # Read1 map => Read1 (Union map) Source # Instance detailsDefined in Data.Map.Class MethodsliftReadsPrec :: (Int -> ReadS a) -> ReadS [a] -> Int -> ReadS (Union map a) #liftReadList :: (Int -> ReadS a) -> ReadS [a] -> ReadS [Union map a] #liftReadPrec :: ReadPrec a -> ReadPrec [a] -> ReadPrec (Union map a) #liftReadListPrec :: ReadPrec a -> ReadPrec [a] -> ReadPrec [Union map a] # Show1 map => Show1 (Union map) Source # Instance detailsDefined in Data.Map.Class MethodsliftShowsPrec :: (Int -> a -> ShowS) -> ([a] -> ShowS) -> Int -> Union map a -> ShowS #liftShowList :: (Int -> a -> ShowS) -> ([a] -> ShowS) -> [Union map a] -> ShowS # Filtrable map => Filtrable (Union map) Source # Instance detailsDefined in Data.Map.Class MethodsmapMaybe :: (a -> Maybe b) -> Union map a -> Union map b #catMaybes :: Union map (Maybe a) -> Union map a #filter :: (a -> Bool) -> Union map a -> Union map a #mapMaybeA :: (Traversable (Union map), Applicative p) => (a -> p (Maybe b)) -> Union map a -> p (Union map b) #filterA :: (Traversable (Union map), Applicative p) => (a -> p Bool) -> Union map a -> p (Union map a) #mapEither :: (a -> Either b c) -> Union map a -> (Union map b, Union map c) #mapEitherA :: (Traversable (Union map), Applicative p) => (a -> p (Either b c)) -> Union map a -> p (Union map b, Union map c) #partitionEithers :: Union map (Either a b) -> (Union map a, Union map b) # Map map => Map (Union map) Source # Instance detailsDefined in Data.Map.Class Methodsempty :: Union map a Source #alterF :: Functor f => (Maybe a -> f (Maybe a)) -> Key (Union map) -> Union map a -> f (Union map a) Source #mergeA :: Applicative p => (Key (Union map) -> Either' a b -> p (Maybe c)) -> Union map a -> Union map b -> p (Union map c) Source #mapMaybeWithKeyA :: Applicative p => (Key (Union map) -> a -> p (Maybe b)) -> Union map a -> p (Union map b) Source #mapEitherWithKeyA :: Applicative p => (Key (Union map) -> a -> p (Either b c)) -> Union map a -> p (Union map b, Union map c) Source # StaticMap map => StaticMap (Union map) Source # Instance detailsDefined in Data.Map.Class Associated Typestype Key (Union map) Source # MethodsadjustA :: Applicative p => (a -> p a) -> Key (Union map) -> Union map a -> p (Union map a) Source #traverseWithKey :: Applicative p => (Key (Union map) -> a -> p b) -> Union map a -> p (Union map b) Source # Eq (map a) => Eq (Union map a) Source # Instance detailsDefined in Data.Map.Class Methods(==) :: Union map a -> Union map a -> Bool #(/=) :: Union map a -> Union map a -> Bool # Ord (map a) => Ord (Union map a) Source # Instance detailsDefined in Data.Map.Class Methodscompare :: Union map a -> Union map a -> Ordering #(<) :: Union map a -> Union map a -> Bool #(<=) :: Union map a -> Union map a -> Bool #(>) :: Union map a -> Union map a -> Bool #(>=) :: Union map a -> Union map a -> Bool #max :: Union map a -> Union map a -> Union map a #min :: Union map a -> Union map a -> Union map a # Read (map a) => Read (Union map a) Source # Instance detailsDefined in Data.Map.Class MethodsreadsPrec :: Int -> ReadS (Union map a) #readList :: ReadS [Union map a] #readPrec :: ReadPrec (Union map a) #readListPrec :: ReadPrec [Union map a] # Show (map a) => Show (Union map a) Source # Instance detailsDefined in Data.Map.Class MethodsshowsPrec :: Int -> Union map a -> ShowS #show :: Union map a -> String #showList :: [Union map a] -> ShowS # (Map map, Semigroup a) => Semigroup (Union map a) Source # Instance detailsDefined in Data.Map.Class Methods(<>) :: Union map a -> Union map a -> Union map a #sconcat :: NonEmpty (Union map a) -> Union map a #stimes :: Integral b => b -> Union map a -> Union map a # (Map map, Semigroup a) => Monoid (Union map a) Source # Instance detailsDefined in Data.Map.Class Methodsmempty :: Union map a #mappend :: Union map a -> Union map a -> Union map a #mconcat :: [Union map a] -> Union map a # type Key (Union map) Source # Instance detailsDefined in Data.Map.Class type Key (Union map) = Key map
newtype Intersection map a Source #
Wrapper of a Map whose semigroup operation is the intersection, combining values elementwise
Constructors
Intersection FieldsunIntersection :: map a
Instances
Instances details
Orphan instances
Source # Instance details MethodsmapMaybe :: (a -> Maybe b) -> IntMap a -> IntMap b #catMaybes :: IntMap (Maybe a) -> IntMap a #filter :: (a -> Bool) -> IntMap a -> IntMap a #mapMaybeA :: (Traversable IntMap, Applicative p) => (a -> p (Maybe b)) -> IntMap a -> p (IntMap b) #filterA :: (Traversable IntMap, Applicative p) => (a -> p Bool) -> IntMap a -> p (IntMap a) #mapEither :: (a -> Either b c) -> IntMap a -> (IntMap b, IntMap c) #mapEitherA :: (Traversable IntMap, Applicative p) => (a -> p (Either b c)) -> IntMap a -> p (IntMap b, IntMap c) #partitionEithers :: IntMap (Either a b) -> (IntMap a, IntMap b) # Filtrable (Map key) Source # Instance details MethodsmapMaybe :: (a -> Maybe b) -> Map key a -> Map key b #catMaybes :: Map key (Maybe a) -> Map key a #filter :: (a -> Bool) -> Map key a -> Map key a #mapMaybeA :: (Traversable (Map key), Applicative p) => (a -> p (Maybe b)) -> Map key a -> p (Map key b) #filterA :: (Traversable (Map key), Applicative p) => (a -> p Bool) -> Map key a -> p (Map key a) #mapEither :: (a -> Either b c) -> Map key a -> (Map key b, Map key c) #mapEitherA :: (Traversable (Map key), Applicative p) => (a -> p (Either b c)) -> Map key a -> p (Map key b, Map key c) #partitionEithers :: Map key (Either a b) -> (Map key a, Map key b) #
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# Footer on EVERY page (including chapter*'s/ chapter's first page
I have the following problem: I want to have a footer on every page (even on the first pages of a chapter* or chapter, the tableofcontents, the bibliography as well as the acronyms page. I achieved to get footers on every second and following pages of a chapter using the post I found on stackoverflow. Is there a simple possibility to get the footer on really EVERY page?
\documentclass[a4paper, 12pt, DIV=calc, headings=small]{scrreprt}
\usepackage{lipsum}
\usepackage{fancyhdr}
%Define footers
\fancypagestyle{Whatever}{
\fancyfoot[C]{Whatever \thepage} %Centered footer
}
\pagestyle{Whatever} %Set page style.
\begin{document}
\chapter*{Test1}
\lipsum[1-8] % Filler Text
\chapter*{Test2}
\lipsum[1-8] % Filler Text
\end{document}
• \renewcommand\chapterpagestyle{Whatever} – Ulrike Fischer Dec 14 '16 at 19:48
• I note that the document class is doing something behind the scenes, in terms of chapter definitions. That's why the code is necessary. A custom document class, which does not define chapters that way, might not need such code. – user103221 Dec 14 '16 at 20:21
• Yeah, that can be... But it works as expected :D – FranzHuber23 Dec 18 '16 at 12:08
Whole solution (Thank goes to Ulrike Fischer):
\documentclass[a4paper, 12pt, DIV=calc, headings=small]{scrreprt}
\usepackage{lipsum}
\usepackage{fancyhdr}
%Define footers
\fancypagestyle{Whatever}{
\fancyfoot[C]{Whatever \thepage} %Centered footer
}
%Set page style
\pagestyle{Whatever} %Set page style.
\renewcommand\chapterpagestyle{Whatever} %New command
\begin{document}
\chapter*{Test1}
\lipsum[1-8] % Filler Text
\chapter*{Test2}
\lipsum[1-8] % Filler Text
\end{document}
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Example of not so simple group ??
Help me with an example of a group having subgroups but it doesn't admit a normal subgroup.. ??
The alternate definition of simple groups using non trivial homomorphic image. Searching for a map that doesn't have a non trivial kernal ??
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Every group admits two normal subgroups, $1$ and itself. Now, as nontrivial normal subgroups goes, that's another story. – Pedro Tamaroff Apr 28 '14 at 3:11
Better to not try to be clever in the title - I expected that you wanted something other than a simple group. BE direct, particular in forums where there are lots of foreign readers. – Thomas Andrews Apr 28 '14 at 3:31
One classic example is $A_5$, which has plenty of non-trivial subgroups, none of which are normal. In fact, this is the smallest suitable example.
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All the other examples are of finite simple groups. However, there do exist infinite simple groups. One of the best known is Richard Thompson's group $T$. It contains a very interesting group, called $F$, as a subgroup. I'll just define these groups here. If you want a more complete introduction then look at this ams "What is..." article by J. W. Cannon and W. J. Floyd. If you want a proof that this group is simple then you should look up the article "Introductory notes on Richard Thompson's groups" by Cannon, Floyd and Parry. These notes of Cannon, Floyd and Parry are the standard reference for Richard Thompson's groups $F$, $T$ and $V$.
There are three groups introduced by Thompson in 1965, $F<T<V$, and I believe he did not publish his constructions although he did prove that $T$ and $V$ are both simple. I'll define $F$, and use this to define $T$. The definition of $V$ is more complicated so I'll leave you to look it up in one of the above references.
The group $F$: The group $F$ is the set consisting of piecewise linear homeomorphisms from the closed unit interval to itself which are differentiable at all but finitely many dyadic rational numbers $\frac{a}{2^b}$, and such that at intervals of differentiability the derivatives are powers of two.
This means that every element of $F$ has the following form. Note that there are only finitely many intervals, each of whose maximum and minimum is a dyadic rational number. The function on each interval has the form $2^ax+\frac{b}{2^c}$, $a, b, c\in\mathbb{Z}$, $c>0$. $$f(x)=\begin{cases} x+\frac12&1\leq x\leq \frac14\\ \frac{x}{2}+\frac18&\frac14\leq x\leq \frac12\\ \frac{x}{4}+\frac{1}{16}&\frac14\leq x\leq \frac12\\ 2x+\frac12&\frac34\leq x\leq \frac78\\ x+1&\frac{29}{32}\leq x\leq 1 \end{cases}$$ One useful way of thinking about $F$ is as a subgroup of automorphisms of the circle, but where we fix a common point. This works because when you pin together the ends of the unit interval you get a circle, and the elements of $F$ fix this pinning point.
The group $T$: The group $T$ is the analogue of $F$ to the circle, so if you additionally allow $F$ to move the pinning point $0=1$. So $T$ is $F$ with rotation. More formally, consider $S^1$ as the unit interval $[0, 1]$, then the group $T$ is the set consisting of piecewise linear homeomorphisms from $S^1$ to itself which are differentiable at all but finitely many dyadic rational numbers $\frac{a}{2^b}$, and such that at intervals of differentiability the derivatives are powers of two.
View $F$ as automorphisms of the circle. Then the group $T$ is simply the group generated by $F$ along with the half-rotation map $x\mapsto x+\frac12\pmod1$, which can be described as follows, $$g(x)=\begin{cases} x+\frac12&1\leq x\leq \frac12\\ x-\frac12&\frac12\leq x\leq 1 \end{cases}$$ Theorem 5.8 and Corollary 5.9 of Cannon, Floyd and Parry's notes combine to prove that $T$ is simple (although this proof uses a different additional generator for $T$ from the one I state, above).
Some more on $F$: These three groups, of which I have defined two, are interesting and much studied. Indeed, I have heard people referred to as "Thompson's group people", and and career's have been built on these groups! Perhaps the most studied of all though if $F$, the only one which is not simple. From the ams article, The group F does not fit easily into standard categories. It is non-Abelian and torsion free and contains a free semigroup on two generators, yet it contains no non-Abelian free subgroup. It is not a matrix group. Every subgroup of F is either finite rank free Abelian or contains an infinite-rank free Abelian subgroup.
One of the current open problems on $F$ is its amenability. See this old MO question and its answers. I have been led to believe that this question is really only interesting because it has garnered so many incorrect answers!
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An example of a group $G$ that has lots of subgroups (15 conjugacy classes of subgroups, 179 total subgroups), but with only the two obvious normal subgroups $1$ and $G$, is the group $G$ of three by three invertible matrices with 0-1 entries and arithmetic operations done mod 2. $G$ has order $(2^3-1)(2^3-2)(2^3-4) = 168$.
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We must look to nonabelian groups since every subgroup of an abelian group will be normal. The example that comes to mind quickly is the alternating group $A_n$ for all $n \geq 5$.
$PSL(3, 2)$ is the second smallest example behind $A_5$ of a group with subgroups, none of whom are normal.
http://groupprops.subwiki.org/wiki/Projective_special_linear_group:PSL(3,2)
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$C_2$ is a perfectly good example of a simple group. In fact, $C_p$ is for every prime $p$. – Pedro Tamaroff Apr 28 '14 at 3:13
Certainly. I am adding the restriction that nontrivial subgroups actually exist. – Kaj Hansen Apr 28 '14 at 3:14
Well, you did say "Note that cyclic groups are no good." – Pedro Tamaroff Apr 28 '14 at 3:15
Ah, I see where the misunderstanding is coming from. The OP had added the restriction that nontrivial subgroups exist. I'll edit my post to clear that up. – Kaj Hansen Apr 28 '14 at 3:17
A cyclic group of prime order has no subgroups except itself and the group with only one element.
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Hint: Any subgroup of a commutative group is normal. What noncommutative groups do you know.
Bigger Hint: Think about subgroups of $S_5$.
Even Bigger Hint: What about the subgroup of the simple group $A_5$ generated by a product transpositions?
SUPER OMEGA HINT: What is $(12)(45)(12)(34)(12)(45)$?
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# How do you solve the following system?: 4x + 5y = 2 , 17 x + 2y = 6
Apr 5, 2017
$x = \frac{26}{77}$
$y = \frac{10}{77}$
#### Explanation:
There are a couple of ways we can solve this. I'm going to use the elimination method
$4 x + 5 y = 2$
$17 x + 2 y = 6$
I need to make two of the variables equal, so I'm going to multiply the second equation by $2.5$. That will change $2 y$ into $5 y$.
$4 x + 5 y = 2$
$2.5 \left(17 x + 2 y = 6\right)$ or $42.5 x + 5 y = 15$
Now we subtract the two equations:
$4 x + 5 y = 2$
$-$
$42.5 + 5 y = 15$
$\textcolor{b l a c k}{- - - - - -}$
$- 38.5 x + 0 y = - 13$
If we simplify our equation, we find that $x = \frac{- 13}{-} 38.5$ or $x = \frac{26}{77}$
Now we just solve for $y$. We can use either of the two equations. I like the first one (it has nicer numbers than $17$).
$4 \left(\textcolor{p u r p \le}{x}\right) + 5 y = 2$
$4 \left(\textcolor{p u r p \le}{\frac{26}{77}}\right) + 5 y = 2$
$\cancel{\frac{104}{77}} + 5 y = 2$
$\cancel{- \frac{104}{77}} \textcolor{w h i t e}{+ 5 y} - \frac{104}{77}$
Now we have
$5 y = \frac{50}{77}$
or
$y = \frac{10}{77}$.
To double check our work, we need to plug our values into one (or both) of the equations.
$4 \left(\frac{26}{77}\right) + 5 \left(\frac{10}{77}\right)$ should equal $2$
$\frac{104}{77} + \frac{50}{77}$
$2 = 2$, so we were right!
Just to be safe, let's look at the other equation.
$17 \left(\frac{26}{77}\right) + 2 \left(\frac{10}{77}\right)$ should equal $6$
$\frac{442}{77} + \frac{20}{77}$
$6 = 6$
Good job, we got it right! Nice work
Apr 5, 2017
Using the process of elimination, we find that $x = 0.338$ and $y = 0.130$. (See explanation)
#### Explanation:
We can solve this system of equations using the elimination method. First, we can set up the system in such a way that the $y$ terms cancel out. To do this, we need both the $y$ terms in each equation to have the same coefficient (one negative and one positive):
$2 \left(4 x + 5 y = 2\right)$
$- 5 \left(17 x + 2 y = 6\right)$
So we get:
$8 x + 10 y = 4$
$- 85 x - 10 y = - 30$
Adding these two equations together gives us:
$- 77 x = - 26$
$x = \frac{26}{77} = 0.338$
Plug this value of $x$ back into one of the given equations:
$4 \left(0.338\right) + 5 y = 2$
$y = \frac{2 - 4 \left(0.338\right)}{5} = 0.130$
So, $x = 0.338$ and $y = 0.130$
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# Equivalent Electri Force, Find the charge.
## Homework Statement
Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.3 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 2.3 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object? (Note: there are two possible pairs of answers, but assume q1 to be the larger number.)
F=kq1q2/r2
## The Attempt at a Solution
So I know that since force is equal Fr2/k is a constant. Also since the charge in the 2nd case is equally shared I can re-write the equation for force as F=kq2/r2. Solving for q gets me 4.8e-6.
From here since the charge was shared equally between q1 and q2, q=q1+q2/2. Using substitution to solve for q2, I came up with the following quadratic:
-q12+9.6e-6(q1)-9.6e-6. This results in an invalid quadratic equation because b2-4ac gives me a negative number. I think I made math errors or I'm missing a step, but the underlying concept makes sense.
SammyS
Staff Emeritus
Homework Helper
Gold Member
## Homework Statement
Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.3 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 2.3 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object? (Note: there are two possible pairs of answers, but assume q1 to be the larger number.)
F=kq1q2/r2
## The Attempt at a Solution
So I know that since force is equal Fr2/k is a constant. Also since the charge in the 2nd case is equally shared I can re-write the equation for force as F=kq2/r2. Solving for q gets me 4.8e-6.
From here since the charge was shared equally between q1 and q2, q=q1+q2/2. Using substitution to solve for q2, I came up with the following quadratic:
Be careful with parentheses. That should be written,
q = (q1+q2)/2
-q12+9.6e-6(q1)-9.6e-6. This results in an invalid quadratic equation because b2-4ac gives me a negative number. I think I made math errors or I'm missing a step, but the underlying concept makes sense.
It's likely that the signs of the charges is what's causing your problems.
The magnitude of the force is the same for both situations, so
|q1q2| = q2
but because q1q2 is negative,
|q1q2| = -q1q2 .
ok now I just got a completely wrong answer. My new quadratic equation is q22-9.59e-6+4.8e-6. I got answer of like .00348
SammyS
Staff Emeritus
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### Home > CALC > Chapter 5 > Lesson 5.1.2 > Problem5-14
5-14.
A rectangular piece of sheet metal with a perimeter of $50$ cm is rolled into a cylinder with two open ends.
1. Find the radius and height of the cylinder in terms of $x$.
Look at the diagram of the cylinder. The circumference of the base is equivalent to $x$.
$P = 2b + 2h$
$25 = 2x+2h$
Solve for $h$.
2. Express the volume of the cylinder as a function of $x$.
$V = π(r^²)(h)$. Substitute.
3. Find the value of $x$ that will maximize the volume and find the maximum value.
The volume will be at a maximum when its derivative is $0$.
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# How to prove converging subsequence lead to the existence of limsup for a function?
Let $f(x)$ be a function $X\rightarrow R$ and $x_0$ be a limit point in $X$, we define a ball without the center as $B_\delta'(x_0)=B_\delta(x_0) \backslash {x_0}$. The $\limsup$ of the function $f(x)$ at $x=x_0$ is defined as $\mathop {\lim \sup}\limits_{x \to {x_0},x \in X} f\left( x \right) = \mathop {{\rm{inf}}}\limits_{\delta > 0} [\mathop {\sup}\limits_{B_\delta'\left( {{x_0}} \right)\bigcap X}] =\mathop {\lim }\limits_{\delta \to 0} \mathop {\sup }\limits_{B_\delta '\left( {{x_0}} \right)\bigcap X} f\left( x \right)$.
I have proved that if $\mathop {\lim \sup}\limits_{x \to {x_0},x \in X} f\left( x \right) = M$ exists at $x=x_0$, then there exists a sequence ${x_k} \in X,\;\;{x_k} \to {x_0}$ such that $\mathop {\lim }\limits_{k \to \infty } f\left( {{x_k}} \right) = M$. The proof is as following,
However, I could not prove the converse. That is, if there exists a sequence ${x_k} \in X,\;\;{x_k} \to {x_0}$ such that $\mathop {\lim }\limits_{k \to \infty } f\left( {{x_k}} \right) = M$, then the $\limsup$ exists and is also $M$.
Anyone can help prove the converse? Thank you!
Say $(x_n)$ is a sequence in $R$. Note that when I talk about the limit of a subsequence I'm allowing $\infty$ and $-\infty$ as limits. Let $L$ be the set of all limits of convergent subsequences. Then $\limsup x_n$ is exactly the largest element of $L$.
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# writing an array and group the contents using curly brackets
I want to write down the following equation:
my try is this:
\documentclass[preview, border=1pt, convert={outext=.png}]{standalone}
\usepackage{amsfonts}
\usepackage{mathtools}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{braket}
\usepackage[utf8]{inputenc}
\begin{document}
\begin{equation*}
n + m =^{def} \left\{\begin{array}{rl} n+n + \ldots + n & \text{se } m > 0 \\ 0 & \text{se } m = 0\end{array}\right.
\end{equation*}
\end{document}
I don't know how to group the elements using curl brackets.
PS: I'd like also the background to be transparent and not white. I'm compiling with this:
pdflatex --shell-escape formula.tex && convert -resize 50% formula.png formula.png
• What's the purpose of the square? – egreg Oct 21 '19 at 10:09
• it signifies end of definition – Michaelangelo Meucci Oct 21 '19 at 10:10
• I've never seen it used that way, – egreg Oct 21 '19 at 10:12
I suggest to use the ambient cases to write your "equation". Here you have a screenshot of my MWE.
\documentclass[a4paper,12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
$n+m\stackrel{\text{def}}{=} \begin{cases} \underbrace{\sigma(\sigma(\ldots\sigma}_{m\text{ volte}}(n))) & \text{se m>0}\\ n & \text{se m=0}.\qquad \square \end{cases}$
\end{document}
Comment: Reading carefully the precious suggestions of the excellent users @egreg and @Mico my initial code has been modified.
• I think the case environment is better than my solution with \hspace. I didn't know this environment. Thank you. – NelDav Oct 21 '19 at 9:56
• It should be _{\text{$m$ volte}} (the fact it works also without braces around \text{...} is irrelevant. Alternatively, _{m\text{ volte}}. I'd also use \dotsm instead of \ldots. – egreg Oct 21 '19 at 10:03
• +1. In addition to incorporating @egreg's comment, you should also change \text{se } m>0 to \text{se $m>0$}, and \text{se } m=0. to '\text{se $m=0$}.. Why? Mostly for reasons of semantic consistency: The clauses se $m>0$ and se $m=0$ are two standalone linguistic particles, and hence (at least from a LaTeX point of view) it's preferable to encase them in \text directives. The fact that both linguistic particles contain a math formula -- $m>0$ and $m=0$, resp. -- does not change the fact that they are linguistic particles that shouldn't be broken up needlessly. – Mico Oct 21 '19 at 10:32
• @Ahrtaler Slowly and patiently you'll certainly be better at it than me. I wish it with all my heart. – Sebastiano Oct 21 '19 at 11:28
Your code didn't worked so I couldn't try. But this code should work:
\documentclass{standalone}
\usepackage{amsmath}
\begin{document}
$n + m \overset{\text{def}}{=} \left\{\begin{array}{l} \underbrace{\sigma(\sigma(\ldots\sigma}_\text{m \text{volte}}(n))) \hspace{5mm} \text{se } m > 0\\ n \hspace{24mm} \text{se } m = 0 \end{array} \right.$
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• Instead of \left\{\begin{array}...\right., you can use \begin{cases}...\end{cases}`... – MadyYuvi Oct 21 '19 at 11:28
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# Interpretation of quantum mechanics
Page 7 of 13 - About 121 Essays
• ## Atoms: Negatively Charged Particles
Atoms are the most basic building blocks, everything around us are made out of atoms. Atoms are so small that it can’t be found under the finest microscopes. A tiny spec of dust contains millions of billions of atoms! Even though atoms are tiny, but it actually consist of three different particles: protons, neutrons, and electrons. In every atom, electrons move around the nucleus, the combination of protons and neutrons. The idea of an atom was first discovered by the ancient Greeks, they…
Words: 633 - Pages: 3
• ## How Did Albert Einstein Influence Astronomy
Albert Einstein's influence on astronomy. Albert had a big influence on Astronomy. He started to study mathematics, he was lured toward calculus around 1891. Albert originally wanted to be a electrical engineer. But he failed to do pass an examination which would have allowed him to study for a diploma at the Eidgenössische Technische Hochschule in Zurich. He made a plan of graduating as a teacher in mathematics and physics. He got his PHD in 1905. He also created the theory of relativity. A…
Words: 467 - Pages: 2
• ## How Did Edwin Hubble's Work
The life and works of Edwin Hubble Edwin Hubble was born november 20, 1889, in Marshfield Missouri. When he was a child he loved both books and sports.He had a chance to be a boxer but instead chose to be a scientist/ astronomer. He is famous for many discoveries and tools used today. He before becoming a scientist he was in a war of some sort I believe it to be world war 1. He also i’m sad to tell later in life had some medical problems which resulted in his death. When he was young he was a…
Words: 517 - Pages: 3
• ## The Intuition Of Science
Science is intuitive. The tallest hill can be the shortest mountain. It is through our intuition that we come to know that the big ones are mountains and the little ones are hills. In some ways, the same is true for science. Science encompasses everything ranging from chemistry to archaeology, microbiology to astronomy. There are so many different aspects to science that it is hard to boil down to one concrete and concise definition. Science is involved with the study of the natural world. This…
Words: 1365 - Pages: 6
• ## Michael Faraday And Oersted Compare And Contrast
Jacob Schuster Comparing and Contrasting Michael Faraday and Hans Christian Orsted Michael Faraday and Hans Christian Oersted were both Chemists and Physicists in the 1800’s. They both discovered the connection between magnetism and electricity. This paper will give backgrounds on both Faraday and Oersted and will compare and contrast their works. Michael Faraday was born in London on September 22, 1791. He was the first person to produce an electric current from a magnetic…
Words: 1006 - Pages: 5
• ## Atomic Spectra Lab Report
The purpose of this spectroscopy, atomic spectra, and light lab was to further your understanding of how different gases are given specific wavelengths based on the variety of color that their light gives off. (Grossie, 285). Every ion gives off a different light configuration based on the amount of energy they hold. (Grossie, 285). A spectroscope is used to break down the light given off by the gas into different colored lines in different areas on the scope. The measure of these lines is what…
Words: 918 - Pages: 4
• ## Superfluid Helium Or He II: Landau-Tisza Model
Superfluid helium or He II is a macroscopic quantum fluid that exhibits extraordinary properties. %By applying the limit of low flow velocities, The behavior of the fluid can be understood using a Landau - Tisza model, where He II is considered to be a two-component fluid with independent velocity fields: the inviscid superfluid of density $\rho_s$, and the normal fluid of density $\rho_n$, where the total density $\rho$ = $\rho_s$ + $\rho_n$ \cite{Tisza,Landau}. The superfluid has neither…
Words: 1075 - Pages: 5
• ## Wind Energy: The Future Of Renewable Energy
In 1839, a French physicist Edmund Becquerel proposed that few materials have the ability to produce electricity when exposed to sunlight. But Albert Einstein explained the photoelectric effect and the nature of light in 1905. Photoelectric effect state that when photons or sunlight strikes to a metal surface flow of electrons will take place. Later photoelectric effect became the basic principle for the technology of photovoltaic power generation. The first PV module was manufactured by Bell…
Words: 742 - Pages: 3
• ## Electron Microscope
Electron microscope (Manning, n.d.) is the instrument used to magnify different scientific artifacts with the used of electron beams to create the illustration of the sample, and has the capacity to magnify two million times of the original specimen. The development of the electron microscope was first known in the year 1931 because Ernst Ruska and Maximoll Knoll magnified electron image successfully, but the equipment was actually constructed in 1933 (Innes, n.d.). The concept of the electron…
Words: 1104 - Pages: 5
• ## Brain In A Vat Theory: An Analysis
Created in 1981 by Hilary Putnam the Brain in a Vat theory is about a science fiction possibility of our brains quite literally being in a vat. Hilary Putnam uses referencing or representation, the Turing Test, the rules of language and a host of examples to establish that the statement “we are brains in a vat” is a self-refuting statement. He asks the question “Could we, if we were brains in a vat in this way, say or think that we were?” (Putnam 1981, p7) and comes to the conclusion that we…
Words: 949 - Pages: 4
• Page 1 4 5 6 7 8 9 10 11 13
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# Command Line Arguments
## C# Command Line Arguments
When an argument is passed by the command line, then it is called as a command-line argument. The arguments are passed to the Main method during the execution of the code. The values to be passed from the command line, are specified in the string args variable.
Example:
using System; namespace Commandlineexample { class Example { // Main function, execution entry point of the program static void Main(string[] args) { // Command line arguments Console.WriteLine("Length of the argument: "+args.Length); Console.WriteLine("Passed Arguments are:"); foreach (Object x in args) { Console.WriteLine(x); } } } }
Output:
Length of the argument: 2
Passed Arguments are:
Hello
World!!
Explanation:
In the above example, the command-line arguments are being passed during the execution of the program. For compilation and execution of the code, the below commands can be used and the output will be produced to the console.
Compilation: csc Example.cs
Execution: Example.exe Hello World!!
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# Issues with large files¶
## Data expansion issues¶
NMRPipe is currently a 32-bit program and, as such, there are limitations to the size of the file it can read, write, and access in memory. For example, a Bruker 4D NUS ser file that is expanded to a final size of 2048 $$\times$$ 96 $$\times$$ 128 $$\times$$ 96 (t4, t3, t2, t1) will produce a file that is approximately 9.7 GB. Converting this file to NMRPipe format with bruk2pipe is problematic for many 32-bit versions of NMRPipe [*].
The file size limit as well as the indication that an expansion has failed may vary with system. Besides the existence of a large data set, one indication of a problem is that the bruk2pipe (and presumably also var2pipe) command simply stops during conversion producing an error message similar to this:
DATAIO File Read Error 0.
Request: 786432 Actual: 524287.
The file size limit also can cause issues with Sparky where 32-bit versions of pipe2ucsf are unable to convert NMRPipe data to the format used by Sparky. In our experience, these issues are solved by switching to a 64-bit version of the program. For Linux, such a version is available from the Sparky website. For Mac OS X, a 64-bit version is available from the NMRFAM website or it can be compiled using the MacPorts package manager.
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# How to convert Sallen-Key Low pass filter to Signal Flow Graph
Oct 9, 2017
1
#### LvW
Apr 12, 2014
604
The circuit consists of a fixed-gain amplifier (opamp with R3, R4) and a positive feedback function. Hence, where is the problem to find the signal-flow graph?
#### Audioguru
Sep 24, 2016
3,642
There are two RC lowpass networks, R1 and C1 and R2 and C2. Each RC network has an output of -3dB at the cutoff frequency.
Actually, the positive feedback occurs only near the cutoff frequency so that the -6dB response is boosted to be -3dB and the cutoff is sharp instead of droopy.
#### LvW
Apr 12, 2014
604
There are two RC lowpass networks, R1 and C1 and R2 and C2. Each RC network has an output of -3dB at the cutoff frequency.
.
No - it is obvious that the feedback network consists of a band pass (highpass C1-R1, lowpass R2-C2).
As a consequence, there is no lowpass -6dB response.
#### Audioguru
Sep 24, 2016
3,642
No - it is obvious that the feedback network consists of a band pass (highpass C1-R1, lowpass R2-C2).
As a consequence, there is no lowpass -6dB response.
I disagree. C1-R1 is a lowpass, not a highpass.
If C1 connects to ground instead of the the opamp output then there is no boost and the output at the cutoff frequency will be -6dB and the slope will gradually be 12dB per octave.
Last edited:
#### LvW
Apr 12, 2014
604
We are discussing the FEEDBACK network. Hence, you have to look into the circuit from the opamp output.
You cannot deny that - in this case - the feedback network resembles the well known highpass-lowpass CR-RC bandpass characteristic. This is a well-known property of a positive-gain Sallen-Key lowpass.
Last edited:
#### Audioguru
Sep 24, 2016
3,642
The bandpass circuit occurs only near the cutoff frequency to boost the response so that it has flat levels at low frequencies, -3dB at the cutoff frequency instead of a droopy -6dB and a sharp 12dB per octave slope at higher frequencies. Above the cutoff frequency both RC networks are lowpass filters.
#### LvW
Apr 12, 2014
604
The bandpass circuit occurs only near the cutoff frequency to boost the response so that it has flat levels at low frequencies, -3dB at the cutoff frequency instead of a droopy -6dB and a sharp 12dB per octave slope at higher frequencies. Above the cutoff frequency both RC networks are lowpass filters.
I dont understand your position.
A bandpass is a bandpass - full stop.
Just one question, which can be answered with yes/no:
Do you agree that - between the ouput pin and the non-inv. input of the opamp - there is the classical four-element ladder network C1-R1-R2-C2 ? And this is the well known RC-bandpass.
How can you say that "Above the cutoff frequency both RC networks are lowpass filters"?
At first, we dont have two RC-networks because you are not allowed to separate them, because they influence each other.
Secondly, the series cap C1 - of course - has highpass properties. I don`t think that you will argue against this.
#### Audioguru
Sep 24, 2016
3,642
I see the circuit as two lowpass RC networks at frequencies above cutoff like this:
#### Attachments
• Sallen-Key lowpass filter.png
14.3 KB · Views: 165
#### LvW
Apr 12, 2014
604
OK - I know what you mean. But that does not answer the question.
The question concerns the corresponding signal-flow diagram (which shows forward and backward ways.).
That is the background we are speaking about FEEDBACK.
And - it does not matter how the circuit looks like without feedback.
It is a fact that the Sallen-Key lowpass has a feedback function that resembles a bandpass.
This is the background for realizing a complex pole pair.
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• ### The Core Mass Function in the Massive Protocluster G286.21+0.17 revealed by ALMA(1706.06584)
Dec. 22, 2017 astro-ph.GA, astro-ph.SR
We study the core mass function (CMF) of the massive protocluster G286.21+0.17 with the Atacama Large Millimeter/submillimeter Array via 1.3~mm continuum emission at a resolution of 1.0\arcsec\ (2500~au). We have mapped a field of 5.3\arcmin$\times$5.3\arcmin\ centered on the protocluster clump. We measure the CMF in the central region, exploring various core detection algorithms, which give source numbers ranging from 60 to 125, depending on parameter selection. We estimate completeness corrections due to imperfect flux recovery and core identification via artificial core insertion experiments. For masses $M\gtrsim1\:M_\odot$, the fiducial dendrogram-identified CMF can be fit with a power law of the form ${\rm{d}}N/{\rm{d}}{\rm{log}}M\propto{M}^{-\alpha}$ with $\alpha \simeq1.24\pm0.17$, slightly shallower than, but still consistent with, the index of the Salpeter stellar initial mass function of 1.35. Clumpfind-identified CMFs are significantly shallower with $\alpha\simeq0.64\pm0.13$. While raw CMFs show a peak near $1\:M_\odot$, completeness-corrected CMFs are consistent with a single power law extending down to $\sim 0.5\:M_\odot$, with only a tentative indication of a shallowing of the slope around $\sim1\:M_\odot$. We discuss the implications of these results for star and star cluster formation theories.
• ### Dust properties inside molecular clouds from coreshine modeling and observations(1407.5804)
July 22, 2014 astro-ph.GA
Context. Using observations to deduce dust properties, grain size distribution, and physical conditions in molecular clouds is a highly degenerate problem. Aims. The coreshine phenomenon, a scattering process at 3.6 and 4.5 $\mu$m that dominates absorption, has revealed its ability to explore the densest parts of clouds. We want to use this effect to constrain the dust parameters. The goal is to investigate to what extent grain growth (at constant dust mass) inside molecular clouds is able to explain the coreshine observations. We aim to find dust models that can explain a sample of Spitzer coreshine data. We also look at the consistency with near-infrared data we obtained for a few clouds. Methods. We selected four regions with a very high occurrence of coreshine cases: Taurus-Perseus, Cepheus, Chameleon and L183/L134. We built a grid of dust models and investigated the key parameters to reproduce the general trend of surface bright- nesses and intensity ratios of both coreshine and near-infrared observations with the help of a 3D Monte-Carlo radiative transfer code. The grid parameters allow to investigate the effect of coagulation upon spherical grains up to 5 $\mu$m in size derived from the DustEm diffuse interstellar medium grains. Fluffiness (porosity or fractal degree), ices, and a handful of classical grain size distributions were also tested. We used the near- and mostly mid-infrared intensity ratios as strong discriminants between dust models. Results. The determination of the background field intensity at each wavelength is a key issue. In particular, an especially strong background field explains why we do not see coreshine in the Galactic plane at 3.6 and 4.5 $\mu$m. For starless cores, where detected, the observed 4.5 $\mu$m / 3.6 $\mu$m coreshine intensity ratio is always lower than $\sim$0.5 which is also what we find in the models for the Taurus-Perseus and L183 directions. Embedded sources can lead to higher fluxes (up to four times greater than the strongest starless core fluxes) and higher coreshine ratios (from 0.5 to 1.1 in our selected sample). Normal interstellar radiation field conditions are sufficient to find suitable grain models at all wavelengths for starless cores. The standard interstellar grains are not able to reproduce observations and, due to the multi-wavelength approach, only a few grain types meet the criteria set by the data. Porosity does not affect the flux ratios while the fractal dimension helps to explain coreshine ratios but does not seem able to reproduce near-infrared observations without a mix of other grain types. Conclusions. Combined near- and mid-infrared wavelengths confirm the potential to reveal the nature and size distribution of dust grains. Careful assessment of the environmental parameters (interstellar and background fields, embedded or nearby reddened sources) is required to validate this new diagnostic.
• ### Instantaneous starburst of the massive clusters Westerlund 1 and NGC 3603 YC(1204.5481)
April 24, 2012 astro-ph.GA
We present a new method to determine the age spread of resolved stellar populations in a starburst cluster. The method relies on a two-step process. In the first step, kinematic members of the cluster are identified based on multi-epoch astrometric monitoring. In the second step, a Bayesian analysis is carried out, comparing the observed photometric sequence of cluster members with sets of theoretical isochrones. When applying this methodology to optical and near-infrared high angular resolution Hubble Space Telescope (HST) and adaptive optics observations of the ~5 Myr old starburst cluster Westerlund 1 and ~2 Myr old starburst cluster NGC 3603 YC, we derive upper limits for the age spreads of 0.4 and 0.1 Myr, respectively. The results strongly suggest that star formation in these starburst clusters happened almost instantaneously.
• ### Constraints on the mass and radius of the accreting neutron star in the Rapid Burster(1204.3627)
April 16, 2012 astro-ph.SR, astro-ph.HE
The Rapid Burster (MXB 1730-335) is a unique object, showing both type I and type II X-ray bursts. A type I burst of the Rapid Burster was observed with Swift/XRT on 2009 March 5, showing photospheric radius expansion for the first time in this source. We report here on the mass and radius determination from this photospheric radius expansion burst using a Bayesian approach. After marginalization over the likely distance of the system (5.8-10 kpc) we obtain M=1.1+/-0.3 M_sun and R=9.6+/-1.5 km (1-sigma uncertainties) for the compact object, ruling out the stiffest equations of state for the neutron star. We study the sensitivity of the results to the distance, the color correction factor, and the hydrogen mass fraction in the envelope. We find that only the distance plays a crucial role.
• ### Is the massive young cluster Westerlund I bound?(1112.4328)
Dec. 21, 2011 astro-ph.SR
Context. Westerlund I is the richest young cluster currently known in our Galaxy, making it one of the most massive clusters for which we can resolve the individual stars even in the crowded centre. This makes it an ideal target to assess whether massive clusters formed currently will remain bound or will disperse and contribute significantly to the stellar field population. Aims. Here we report a measurement of the radial velocity dispersion of Westerlund I to explore whether the cluster is currently in virial equilibrium, if it is in the process of collapse or if it is expanding and dispersing into the field. Methods. We obtained MIKE/Magellan high resolution optical spectra of 22 post main-sequence stars jn Westerlund I for 2 or 3 epochs with a maximum baseline of about one year. Radial velocities variations between these spectra have been measured through cross correlation. Results. We calculate the velocity dispersion from the cross correlation of five yellow hypergiants and one luminous blue variable, that show little radial velocity variations between epochs and have many spectral features in common. After taking into account the effect of small number statistics and undetected binaries, we estimate the velocity dispersion for the massive stars in Westerlund I to be 2.1 (+3.3, -2.1) km s-1. For several different assumptions concerning possible mass segregation and the elongation of the cluster, we find that Westerlund I is subvirial at the 90% confidence level. Conclusions. We can rule out that the cluster is significantly supervirial at the 97% confidence level, indicating that Westerlund I is currently bound. This implies that Westerlund I has survived past the point where any gas expulsion has taken place and is expected to survive for billions of years.
• ### Star formation in 30 Doradus(1106.2801)
June 14, 2011 astro-ph.GA, astro-ph.SR
Using observations obtained with the Wide Field Camera 3 (WFC3) on board the Hubble Space Telescope (HST), we have studied the properties of the stellar populations in the central regions of 30 Dor, in the Large Magellanic Cloud. The observations clearly reveal the presence of considerable differential extinction across the field. We characterise and quantify this effect using young massive main sequence stars to derive a statistical reddening correction for most objects in the field. We then search for pre-main sequence (PMS) stars by looking for objects with a strong (> 4 sigma) Halpha excess emission and find about 1150 of them over the entire field. Comparison of their location in the Hertzsprung-Russell diagram with theoretical PMS evolutionary tracks for the appropriate metallicity reveals that about one third of these objects are younger than ~4Myr, compatible with the age of the massive stars in the central ionising cluster R136, whereas the rest have ages up to ~30Myr, with a median age of ~12Myr. This indicates that star formation has proceeded over an extended period of time, although we cannot discriminate between an extended episode and a series of short and frequent bursts that are not resolved in time. While the younger PMS population preferentially occupies the central regions of the cluster, older PMS objects are more uniformly distributed across the field and are remarkably few at the very centre of the cluster. We attribute this latter effect to photoevaporation of the older circumstellar discs caused by the massive ionising members of R136.
• ### Detection of brown dwarf-like objects in the core of NGC3603(1101.4521)
Jan. 28, 2011 astro-ph.SR
We use near-infrared data obtained with the Wide Field Camera 3 (WFC3) on the Hubble Space Telescope to identify objects having the colors of brown dwarfs (BDs) in the field of the massive galactic cluster NGC 3603. These are identified through use of a combination of narrow and medium band filters spanning the J and H bands, and which are particularly sensitive to the presence of the 1.3-1.5{\mu}m H2O molecular band - unique to BDs. We provide a calibration of the relationship between effective temperature and color for both field stars and for BDs. This photometric method provides effective temperatures for BDs to an accuracy of {\pm}350K relative to spectroscopic techniques. This accuracy is shown to be not significantly affected by either stellar surface gravity or uncertainties in the interstellar extinction. We identify nine objects having effective temperature between 1700 and 2200 K, typical of BDs, observed J-band magnitudes in the range 19.5-21.5, and that are strongly clustered towards the luminous core of NGC 3603. However, if these are located at the distance of the cluster, they are far too luminous to be normal BDs. We argue that it is unlikely that these objects are either artifacts of our dataset, normal field BDs/M-type giants or extra-galactic contaminants and, therefore, might represent a new class of stars having the effective temperatures of BDs but with luminosities of more massive stars. We explore the interesting scenario in which these objects would be normal stars that have recently tidally ingested a Hot Jupiter, the remnants of which are providing a short-lived extended photosphere to the central star. In this case, we would expect them to show the signature of fast rotation.
• ### Progressive star formation in the young galactic super star cluster NGC 3603(1007.2795)
July 16, 2010 astro-ph.SR
Early release science observations of the cluster NGC3603 with the WFC3 on the refurbished HST allow us to study its recent star formation history. Our analysis focuses on stars with Halpha excess emission, a robust indicator of their pre-main sequence (PMS) accreting status. The comparison with theoretical PMS isochrones shows that 2/3 of the objects with Halpha excess emission have ages from 1 to 10 Myr, with a median value of 3 Myr, while a surprising 1/3 of them are older than 10 Myr. The study of the spatial distribution of these PMS stars allows us to confirm their cluster membership and to statistically separate them from field stars. This result establishes unambiguously for the first time that star formation in and around the cluster has been ongoing for at least 10-20 Myr, at an apparently increasing rate.
• ### Spitzer Observations of Molecular Hydrogen in Interacting Supernova Remnants(0901.1622)
Jan. 12, 2009 astro-ph.GA, astro-ph.SR
With Spitzer IRS we have obtained sensitive low-resolution spectroscopy from 5 to 35 microns for six supernova remnants (SNRs) that show evidence of shocked molecular gas: Kes 69, 3C 396, Kes 17, G346.6-0.2, G348.5-0.0 and G349.7+0.2. Bright, pure-rotational lines of molecular hydrogen are detected at the shock front in all remnants, indicative of radiative cooling from shocks interacting with dense clouds. We find the excitation of H2 S(0)-S(7) lines in these SNRs requires two non-dissociative shock components: a slow, 10 km/s C- shock through clumps of density 10^6 cm^-3, and a faster, 40-70 km/s C- shock through a medium of density 10^4 cm^-3. The ortho-to-para ratio for molecular hydrogen in the warm shocked gas is typically found to be much less than the LTE value, suggesting that these SNRs are propagating into cold quiescent clouds. Additionally a total of thirteen atomic fine-structure transitions of Ar+, Ar++, Fe+, Ne+, Ne++, S++, and Si+ are detected. The ionic emitting regions are spatially segregated from the molecular emitting regions within the IRS slits. The presence of ionic lines with high appearance potential requires the presence of much faster, dissociative shocks through a lower density medium. The IRS slits are sufficiently wide to include regions outside the SNR which permits emission from diffuse gas around the remnants to be separated from the shocked emission. We find the diffuse molecular hydrogen gas projected outside the SNR is excited to a temperature of 100 to 300 K with a warm gas fraction of 0.5 to 15 percent along the line of sight.
• ### The IMF in Extreme Star-Forming Environments: Searching for Variations vs. Initial Conditions(astro-ph/0506327)
June 14, 2005 astro-ph
Any predictive theory of star formation must explain observed variations (or lack thereof) in the initial mass function. Recent work suggests that we might expect quantitative variations in the IMF as a function of metallicity (Larson 2005) or magnetic field strength (Shu et al. 2004). We summarize results from several on-going studies attempting to constrain the ratio of high to low mass stars, as well as stars to sub- stellar objects, in a variety of different environments, all containing high mass stars. First, we examine the ratio of stars to sub--stellar objects in the nearby Mon R2 region utilizing NICMOS/HST data. We compare our results to the IMF by Kroupa (2002)]} and to the observed ratios for IC 348 and Orion. Second, we present preliminary results for the ratio of high to low mass stars in W51, the most luminous HII region in the galaxy. Based on ground--based multi--colour images of the cluster obtained with the MMT adaptive optics system, we derive a lower limit to the ratio of high-mass to low-mass stars and compare it to the ratios for nearby clusters. Finally, we present the derived IMF for the R136 region in the LMC where the metallicity is 1/4 solar using HST/NICMOS data. We find that the IMF is consistent with that characterizing the field (Chabrier 2003), as well as nearby star--forming regions, down to 1.0 Msun outside 2 pc. Whereas the results for both Mon R2 and R136 are consistent with the nearby clusters, the ratio of high to low mass stars in W51 tentatively indicates a lack of low--mass objects.
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# Row/column operation on matrices and determinants
1. Feb 9, 2015
### Raghav Gupta
How we cannot apply row and column operation simultaneously on matrix when finding its inverse by elementary transformation but can apply it in determinant?
I think kernel and image gets disturbed in a matrix, though I don't know what it actually is.
Why not in determinant case?
2. Feb 10, 2015
### Svein
You can - but only when you do it on the right hand side simultaneously. Assume you want to find $B$ such that $A \cdot B = I$. You do the same row operations on $A$ and $I$. After a while (excluding singularities etc.) you end up with $U \cdot B = R$, where $U$ is upper triangular. Now substitute back until you end up with $I \cdot B = H$ and you are done.
3. Feb 10, 2015
### Raghav Gupta
But doesn't the kernel or image gets disturbed?
4. Feb 10, 2015
### Svein
I am not an expert on matrices so, I cannot answer that. What I know, is how to find the inverse (as detailed above). Useful facts:
• The matrix can be inverted if its determinant is different from 0.
• The determinant of the inverse matrix is the inverse of the determinant of the original matrix.
5. Feb 10, 2015
### Hawkeye18
Row operations are equivalent to left multiplication by corresponding elementary matrices, column operations are equivalent to right multiplication. So, when you perform row operations on a (square) matrix $A$, you get a matrix $E$ (the product of corresponding elementary matrices) such that $EA=I$. For square matrices that means that $E=A^{-1}$; performing the same row operations on the right hand side $I$ you get $EI=E=A^{-1}$.
Again, you can find the inverse performing only column operations: you get a matrix $E$ (the product of corresponding elementary matrices) such that $AE=I$, which again means that $E=A^{-1}$. Performing the same column operations on the right hand side $I$ you get there $IE=E=A^{-1}$.
If you perform both row and column operations you get two matrices $E_1$ and $E_2$ such that $E_1AE_2=I$, so in the right hand side you get $E_1IE_2=E_1E_2$. For $E_1E_2$ to be the inverse of $A$ you need $AE_1E_2=I$ (or equivalently $E_1E_2A =I$), and you have $E_1AE_2=I$. Matrix multiplication is not commutative, so there is no reason for the latter identity imply the former.
"No reason" is not a formal proof that the method does not work, for the formal proof you need a counterexample. You can find it just by playing with applying row and column operations to $2\times 2$ matrices. There are also more scientific method of constructing a counterexample.
The reason that applying both row and column operations is that while $E_1AE_2=I$ does not imply that $E_1E_2=A^{-1}$ it implies that $\operatorname{det} (E_1E_2)=\operatorname{det} A^{-1}$. Namely, $$1=\operatorname{det}(E_1AE_1) = \operatorname{det} E_1 \operatorname{det} A \operatorname{det}E_2,$$ so $$\operatorname{det}(E_1E_2) \operatorname{det}E_1\operatorname{det}E_2= (\operatorname{det}A)^{-1} =\operatorname{det}A^{-1}.$$
Finally, for invertible matrices the image is all $\mathbb R^n$ and the kernel is always trivial (i.e. $\{\mathbf 0\}$), they cannot be "disturbed" by row/column operations, they will remain the same.
6. Feb 10, 2015
### Staff: Mentor
If you're talking about a matrix you get by applying a row operation to another matrix, the answer is no. If you start with a matrix A, and apply one of the three row operations to it to get A1, the matrices A and A1 are equivalent. They have exactly the same kernel and image.
Edit: The image can change. See my later post in this thread.
If you're talking about applying column operations, I don't know -- I have never needed to apply column operations to reduce a matrix. However, if you swap the columns of a matrix, you are swapping the roles of the variables these columns represent.
Last edited: Feb 10, 2015
7. Feb 10, 2015
### Hawkeye18
That is not true. Row operations preserve kernel, column operations preserve image (column space).
In the case of invertible matrices, however, the image is always all $\mathbb R^n$ and the kernel is always $\{\mathbf 0\}$, so we can say that in this case the image and kernel are "preserved" user row and column operations.
8. Feb 10, 2015
### Hawkeye18
A bit of specification. Row operations preserve kernel (but generally not image), column operations preserve image (but generally not kernel).
9. Feb 10, 2015
### Staff: Mentor
I mispoke. Row operations don't necessarily preserve the range, as I said. A simple example shows this:
$$A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1\end{bmatrix}$$
Using row reduction, we get an equivalent matrix.
$$B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}$$
Although the dimensions of the column spaces of A and B are equal (2), they span different subspaces of R3.
The columns of A represent a plane in R3 that is perpendicular to <-1, -1, 1>. The columns of B represent a different plane in R3 that is perpendicular to the z-axis.
10. Feb 11, 2015
### Raghav Gupta
I see you have applied R3--------> R3 - R1
But there should be 1 in row 3 column 2 or more simply B32 should be equal to 1?
11. Feb 11, 2015
### Staff: Mentor
No. Here's what I did: -R1 + R3 --> R3 and then -R2 + R3 --> R3.
IOW, add -R1 to R3, and then add -R2 to R3.
12. Feb 11, 2015
### Raghav Gupta
Got it. Thanks to all of you - HawkEye 18, Mark44 and Svein.
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This is a task from the Illustrative Mathematics website that is one part of a complete illustration of the standard to which it is aligned. Each task has at least one solution and some commentary that addresses important asects of the task and its potential use. Here are the first few lines of the commentary for this task: If $\frac12$ cup of water fills $\frac23$ of a plastic container, how many containers will 1 cup fill? Solve the problem by drawing a picture. Which of...
Non-profit Tax ID # 203478467
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# Iterated Integral
I look for an argument to this statement:
$$\int_a^x dx_1 \int_a^{x_1} K(x_1,t) dt= \int_a^xdt \int_t^x K(x_1,t) dx_1$$ It is certainly an integration by change of variables that I can not clarify
-
Do you see the domain..? – AD. Mar 19 '12 at 20:59
$K\ge0$? $\,\,\,\,$ – AD. Mar 19 '12 at 21:00
$a\leq x_1\leq x$ and $a\leq t\leq x_1$. So, one has that $t$ changes from $a$ to $x_1$ and $x_1$ changes from $a$ to $x$. It is the same that $t$ changes from $a$ to $x$ and $x_1$ changes from $t$ (it is more than $\min(a,t)=t$) to $x$(it less than $x$). You can draw a picture and everything will be evident.
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# TeX Input Processor Options¶
The options below control the operation of the TeX input processor that is run when you include 'input/tex', 'input/tex-full', or 'input/tex-base' in the load array of the loader block of your MathJax configuration, or if you load a combined component that includes the TeX input jax. They are listed with their default values. To set any of these options, include a tex section in your MathJax global object.
## The Configuration Block¶
MathJax = {
tex: {
packages: ['base'], // extensions to use
inlineMath: [ // start/end delimiter pairs for in-line math
['\$$', '\$$']
],
displayMath: [ // start/end delimiter pairs for display math
['$$', '$$'],
['\$', '\$']
],
processEscapes: true, // use \$to produce a literal dollar sign processEnvironments: true, // process \begin{xxx}...\end{xxx} outside math mode processRefs: true, // process \ref{...} outside of math mode digits: /^(?:[0-9]+(?:\{,\}[0-9]{3})*(?:\.[0-9]*)?|\.[0-9]+)/, // pattern for recognizing numbers tags: 'none', // or 'ams' or 'all' tagSide: 'right', // side for \tag macros tagIndent: '0.8em', // amount to indent tags useLabelIds: true, // use label name rather than tag for ids multlineWidth: '85%', // width of multline environment maxMacros: 1000, // maximum number of macro substitutions per expression maxBuffer: 5 * 1024, // maximum size for the internal TeX string (5K) baseURL: // URL for use with links to tags (when there is a <base> tag in effect) (document.getElementsByTagName('base').length === 0) ? '' : String(document.location).replace(/#.*$/, '')),
formatError: // function called when TeX syntax errors occur
(jax, err) => jax.formatError(err)
}
};
Note that some extensions make additional options available. See the TeX Extension Options section below for details.
Note
The default for processEscapes has changed from false in version 2 to true in version 3.
## Option Descriptions¶
packages: ['base']
This array lists the names of the packages that should be initialized by the TeX input processor. The input/tex and input/tex-full components automatically add to this list the packages that they load. If you explicitly load addition tex extensions, you should add them to this list. For example:
MathJax = {
tex: {
packages: {'[+]': ['enclose']}
}
};
This loads the enclose extension and acticates it by including it in the package list.
You can remove packages from the default list using '[-]' rather than [+], as in the followiong example:
MathJax = {
tex: {
packages: {'[-]': ['noundefined']}
}
};
This would disable the noundefined extension, so that unknown macro names would cause error messages rather than be displayed in red.
If you need to both remove some default packages and add new ones, you can do so by including both within the braces:
MathJax = {
tex: {
packages: {'[-]': ['noundefined', 'autoload'], '[+]': ['enclose']}
}
};
This disables the noundefined and autoload extensions, and adds in the enclose extension.
inlineMath: [['$$','$$']]
This is an array of pairs of strings that are to be used as in-line math delimiters. The first in each pair is the initial delimiter and the second is the terminal delimiter. You can have as many pairs as you want. For example,
inlineMath: [ ['$','$'], ['\$$','\$$'] ]
would cause MathJax to look for $...$ and $$...$$ as delimiters for in-line mathematics. (Note that the single dollar signs are not enabled by default because they are used too frequently in normal text, so if you want to use them for math delimiters, you must specify them explicitly.)
Note that the delimiters can’t look like HTML tags (i.e., can’t include the less-than sign), as these would be turned into tags by the browser before MathJax has the chance to run. You can only include text, not tags, as your math delimiters.
displayMath: [ ['$$','$$'], ['$','$'] ]
This is an array of pairs of strings that are to be used as delimiters for displayed equations. The first in each pair is the initial delimiter and the second is the terminal delimiter. You can have as many pairs as you want.
Note that the delimiters can’t look like HTML tags (i.e., can’t include the less-than sign), as these would be turned into tags by the browser before MathJax has the chance to run. You can only include text, not tags, as your math delimiters.
processEscapes: false
When set to true, you may use \$ to represent a literal dollar sign, rather than using it as a math delimiter, and \\ to represent a literal backslash (so that you can use \\\$ to get a literal \$ or \\$...$ to get a backslash jsut before in-line math). When false, \$ will not be altered, and its dollar sign may be considered part of a math delimiter. Typically this is set to true if you enable the $...$ in-line delimiters, so you can type \$ and MathJax will convert it to a regular dollar sign in the rendered document. processRefs: true When set to true, MathJax will process \ref{...} outside of math mode. processEnvironments: true When true, tex2jax looks not only for the in-line and display math delimiters, but also for LaTeX environments (\begin{something}...\end{something}) and marks them for processing by MathJax. When false, LaTeX environments will not be processed outside of math mode. digits: /^(?:[0-9]+(?:{,}[0-9]{3})*(?:.[0-9]*)?|.[0-9]+)/ This gives a regular expression that is used to identify numbers during the parsing of your TeX expressions. By default, the decimal point is . and you can use {,} between every three digits before that. If you want to use {,} as the decimal indicator, use MathJax = { tex: { digits: /^(?:[0-9]+(?:\{,\}[0-9]*)?|\{,\}[0-9]+)/ } }; tags: 'none' This controls whether equations are numbered and how. By default it is set to 'none' to be compatible with earlier versions of MathJax where auto-numbering was not performed (so pages will not change their appearance). You can change this to 'ams' for equations numbered as the AMSmath package would do, or 'all' to get an equation number for every displayed equation. tagSide: 'right' This specifies the side on which \tag{} macros will place the tags, and on which automatic equation numbers will appear. Set it to 'left' to place the tags on the left-hand side. tagIndent: "0.8em" This is the amount of indentation (from the right or left) for the tags produced by the \tag{} macro or by automatic equation numbers. useLabelIds: true This controls whether element IDs for tags use the \label name or the equation number. When true, use the label, when false, use the equation number. multlineWidth: "85%" The width to use for the multline environment that is part of the ams extension. This width gives room for tags at either side of the equation, but if you are displaying mathematics in a small area or a thin column of text, you might need to change the value to leave sufficient margin for tags. maxMacros: 10000 Because a definition of the form \def\x{\x} \x would cause MathJax to loop infinitely, the maxMacros constant will limit the number of macro substitutions allowed in any expression processed by MathJax. maxBuffer: 5 * 1024 Because a definition of the form \def\x{\x aaa} \x would loop infinitely, and at the same time stack up lots of a’s in MathJax’s equation buffer, the maxBuffer constant is used to limit the size of the string being processed by MathJax. It is set to 5KB, which should be sufficient for any reasonable equation. baseURL: (document.getElementsByTagName('base').length === 0) ? '' : String(document.location).replace(/#.*$/, ''))
This is the base URL to use when creating links to tagged equations (via \ref{} or \eqref{}) when there is a <base> element in the document that would affect those links. You can set this value by hand if MathJax doesn’t produce the correct link.
formatError: (jax, err) => jax.formatError(err)
This is a function that is called when the TeX input jax reports a syntax or other error in the TeX that it is processing. The default is to generate an <merror> MathML element with the message indicating the error that occurred. You can override the function to perform other tasks, like recording the message, replacing the message with an alternative message, or throwing the error so that MathJax will stop at that point (you can catch the error using promises or a try/carch block).
The remaining options are described in the Options Common to All Input Processors section.
## Developer Options¶
In addition to the options listed above, low-level options intended for developers include the following:
FindTeX: null
The FindTeX object instance that will override the default one. This allows you to create a subclass of FindTeX and pass that to the TeX input jax. A null value means use the default FindTeX class and make a new instance of that.
## TeX Extension Options¶
Several of the TeX extensions make additional options available in the tex block of your MathJax configuration. These are described below. Note that the input/tex component, and the combined components that load the TeX input jax, include a number of these extensions automatically, so some these options will be available by default.
For example, the configmacros package adds a macros block to the tex configuration block that allows you to pre-define macros for use in TeX espressions:
MathJax = {
tex: {
macros: {
R: '\\mathbf{R}'
}
}
}
The options for the various TeX packages (that have options) are described in the links below:
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# Proving that $\lim_{n \rightarrow\infty} \int_{0}^{\frac{\pi}{2}} \sin(t^n) dt =0$
It is clear that $\lim_{n \rightarrow\infty} \int_{0}^{1} \sin(t^n) dt =0$. (which is not what is to be proved here)
I don't know how to proceed with the remaining part of the integral ie $\lim_{n \rightarrow\infty} \int_{1}^{\frac{\pi}{2}} \sin(t^n) dt$ (I tried substitution and partial integration).
Can anybody give me a hint ?
Substituting $u = t^n$, we obtain
$$\int_1^{\pi/2} \sin (t^n)\,dt = \frac{1}{n}\int_1^{(\pi/2)^n} \frac{\sin u}{u^{1-1/n}}\,du.\tag{1}$$
Integration by parts then brings us to (with $c_n = \left(\frac{\pi}{2}\right)^n$)
\begin{align} \left\lvert\int_1^{c_n} \frac{\sin u}{u^{1-1/n}}\,du\right\rvert &= \left\lvert\frac{-\cos u}{u^{1-1/n}}\Biggl\lvert_1^{c_n} - \left(1-\frac{1}{n}\right)\int_1^{c_n}\frac{\cos u}{u^{2-1/n}}\,du\right\rvert\\ &= \left\lvert\cos 1 - \frac{\cos c_n}{c_n^{1-1/n}} - \left(1-\frac{1}{n}\right)\int_1^{c_n}\frac{\cos u}{u^{2-1/n}}\,du\right\rvert\\ &\leqslant 2 + \int_1^{\infty} \frac{du}{u^{2-1/n}}\\ &\leqslant 4, \end{align}
for $n \geqslant 2$.
Use the factor $\frac{1}{n}$ from $(1)$, and the result for the $\int_0^1$ part to conclude.
• Nice. For those wondering how to deal with the $\int_0^1$, use the dominated convergence theorem. – Gabriel Romon Jan 29 '14 at 17:40
• @GabrielR. Or simply the bound $\sin u\leqslant u$ for $u=t^n$. – Did Jan 29 '14 at 17:46
Begin by doing this substitution $$\int_{0}^{\pi/2} \sin(t^n) dt = {1\over n}\int_0^{(\pi/2)^n} { \sin(t)\over t^{1 - 1/n}} dt.$$ The integrals we see oscillate and if you let the upper limit go to $\infty$, they are conditionally convergent. If they can be uniformly bounded, you have this, since we have the $1/n$ factor. I am not sure how to bound the integrals, but this seems like a viable route to solving the problem.
Nice problem. What I would try:
Pick l, m, r so that t^n = (2k)pi, (2k+1)pi, (2k+2)pi when t = l, m, r; k >= 1. Instead of integrating sin (t^n) from l to r, integrate sin (t^n) from l to m, then use a substitution so you integrate scalefactor (t) * sin (t^n + pi) from l to m instead of sin (t^n) from m to r. The integral from l to r is the integral of (1 - scalefactor (t)) * sin (t^n). Show that the scale factor is very close to 1, so the total integral is small. Sum for all integrals, show it is still small and converges to 0 as n -> infinity.
The integral from 0 to (2pi)^(1/n) also goes -> 0.
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• •
### 东北振兴以来吉林省四化发展的协调性研究
1. 东北师范大学地理科学学院,吉林 长春 130024
• 收稿日期:2014-08-12 修回日期:2014-11-09 出版日期:2015-09-25 发布日期:2015-09-25
• 作者简介:
作者简介：尹鹏(1987-),男,山东泰安人,博士研究生,主要从事区域经济地理和新型城镇化研究。E-mail:yinp438@nenu.edu.cn
• 基金资助:
国家自然科学基金项目(41471111)、教育部高等专科学校博士学科点专项科研基金(201220043110012)、中国海洋发展研究中心重点项目(AOCZDA201304)资助
### The Coordination Development of Four Modernizations in Jilin Province Since the Revival of Northeast China
Peng YIN, Ji-sheng LIU, Cai CHEN
1. School of Geographical Sciences,Northeast Normal University,Changchun,Jilin 130024, China
• Received:2014-08-12 Revised:2014-11-09 Online:2015-09-25 Published:2015-09-25
Abstract:
Taking the Jilin Province as a study case, this article used the methods of coupling coordination degree model, exploratory spatial analysis and obstacle degree model to study the spatio-temporal pattern and obstacle indicators of four modernizations coordination development at the level of county in 2003 and 2012. Conclusions are drawn as follows: 1) The level of four modernizations is volatile and non-equilibrium, and the level of agricultural modernization shows an increasing trend. The spatial difference of industrialization and urbanization keeps reducing, and the spatial difference of informatization and agricultural modernization increases gradually. 2) The coupling degree of four modernizations development in Jilin Province is at low level on the whole, and the coupling degree in urban agglomeration of central Jilin and eastern border area is high relatively, the coupling degree in western inland is low. The coordinating degree mainly shows the serious disorder, moderate disorder and mild discord, and the sync phenomenon is obvious. 3) The similar region of four modernizations coordination changes from discrete distributions to weak agglomeration state. The hotspots of four modernization coordination are centralized in the regions of Changchun-Jilin, Yanbian and Tonghua. The hotspot counties have been decreased remarkably, and coldspot counties are increased gradually. 4) The output value proportion of tertiary industry, per capita GDP, the proportion of non-agricultural population, number of medical beds per ten thousand people and total turnover of postal and telecommunication services per capita were verified to be the first five obstacle indicators for further improvement of four modernizations coordination in 2003, and the output value proportion of secondary industry, the employment proportion of secondary industry, per capita gross industrial output value, average agricultural production per counties and average agricultural production per employee were verified to be the first five obstacle indicators for further improvement of four modernizations coordination in 2012. The order of sub-system obstacle degree is urbanization> agricultural modernization> industrialization> informatization. Finally, 3 driving forces of four modernizations coordination in Jilin Province have been drawn: regional policies, economic and industrial foundation, physical geography condition, and we put forward the main pathway to improve the level of four modernizations coordination. Generally speaking, the four modernizations coordination is complicate system engineering, and promoting the new urbanization development is also very long at the same time, therefore, some efforts should be made in the choices of index, methods and scale in the future.
• F299.21
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# The congruence $x^2 - 16x +2 \equiv 0\pmod{37}$ has only two solutions?
I want to prove that the congruence $x^2 - 16x + 2\equiv 0\pmod{37}$ has only two solutions: $x\equiv 3$ and $x\equiv 13$. I am assuming this is true because the modulus is prime. Is it true that there are at most $2$ solutions because this is a degree $2$ polynomial?
I see that we can rewrite the congruence as $x^2 - 16x + 39\equiv 0\pmod{37}$ and then factor the LHS into $(x - 13)(x - 3)$.
This is a homework problem so I would most appreciate some suggestion to a proof or some general comments about solving polynomial congruences.
By the factor theorem your quadratic $\equiv (x-3)(x-13).\,$ If it had another root $\,x\equiv a\,$ then $\,(a-3)(a-13)\equiv 0,\,$ so $\,p\mid (a-3)(a-13),\,$ so, by primality $\,p\mid a-3,\,$ or $\,p\mid a-13.\,$ Hence $\,a\equiv 3,\,$ or $\,a\equiv 13.\,$ So there can be no other roots.
Remark $\$ More generally, $\,\Bbb Z/p =$ integers mod $p$ form a field, and a nonzero polynomial over a field has no more roots than its degree (same for any integral domain).
Hint: Complete the square. We get $(x-8)^2-64+2$, so we want to solve the congruence $(x-8)^2\equiv 62\pmod{37}$. We can replace $62$ by $25$.
And yes, if $P(x)$ is a polynomial of degree $n$ whose lead coefficient is not divisible by the prime $p$, then the congruence $P(x)\equiv 0\pmod{p}$ has at most $n$ solutions. In fact, if $F$ is any field, and $P(x)$ is a polynomial of degree $n$ with coefficients in $F$ and is not the zero polynomial, then the equation $P(x)=0$ has at most $n$ solutions in $F$.
Lagrange's theorem states that if $p$ is prime then a polynomial $f$ of degree $n$ has at most $n$ solutions modulo $p$.
We can prove this by induction in exactly the same way we would for showing this over $\mathbb C$. It is certainly true for $n=1$. The inductive step uses the fact that $\mathbb Z/p\mathbb Z$ is a field, and hence an integral domain - i.e. if $xy\equiv 0\pmod p$, then $x\equiv 0\pmod p$ or $y\equiv 0\pmod p$, or put in other terms, $p\mid xy\implies p\mid x\$ or $p\mid y$. Can you finish the proof?
Note that this is NOT true if $p$ is not prime. For example, $x^2-1$ has $4$ solutions modulo $8$.
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# American Institute of Mathematical Sciences
2015, 2015(special): 801-808. doi: 10.3934/proc.2015.0801
## Strong solutions to a class of boundary value problems on a mixed Riemannian--Lorentzian metric
1 Dipartimento di Matematica, Università di L'Aquila, 67100 L'Aquila, Italy 2 Department of Mathematical Sciences, Yeshiva University, New York, NY 10033, United States
Received September 2014 Revised January 2015 Published November 2015
A first-order elliptic-hyperbolic system in extended projective space is shown to possess strong solutions to a natural class of Guderley--Morawetz--Keldysh problems on a typical domain.
Citation: Antonella Marini, Thomas H. Otway. Strong solutions to a class of boundary value problems on a mixed Riemannian--Lorentzian metric. Conference Publications, 2015, 2015 (special) : 801-808. doi: 10.3934/proc.2015.0801
##### References:
[1] J. Barros-Neto and F. Cardoso, Gellerstedt and Laplace-Beltrami operators relative to a mixed signature metric, Ann. Mat. Pura Appl. 188 (2009), 497-515. Google Scholar [2] E. Beltrami, Saggio di interpretazione della geometria non-euclidea, Giornale di Matematiche 6 (1868), 284-312. Google Scholar [3] K. O. Friedrichs, Symmetric positive linear differential equations, Commun. Pure Appl. Math. 11 (1958), 333-418. Google Scholar [4] J. Heidmann, Relativistic Cosmology, An Introduction. Springer-Verlag, Berlin-Heidelberg-New York (1980). Google Scholar [5] M. V. Keldysh, On certain classes of elliptic equations with singularity on the boundary of the domain (Russian]), Dokl. Akad. Nauk SSSR 77 (1951), 181-183. Google Scholar [6] P. D. Lax and R. S. Phillips, Local boundary conditions for dissipative symmetric linear differential operators, Commun. Pure Appl. Math. 13 (1960), 427-455. Google Scholar [7] F. Lobo and P. Crawford, Time, closed timelike curves, and causality, in The Nature of Time: Geometry, Physics and Perception (NATO ARW), Proceedings of a conference held 21-24 May, 2002 at Tatranska Lomnica, Slovak Republic (eds. Rosolino Buccheri, Metod Saniga, and William Mark Stuckey). NATO Science Series II: Mathematics, Physics and Chemistry - Volume 95. Dordrecht/Boston/London: Kluwer Academic Publishers, 2003. Google Scholar [8] A. Marini and T. H. Otway, Nonlinear Hodge-Frobenius equations and the Hodge-Bäcklund transformation, Proc. R. Soc. Edinburgh 140A (2010), 787-819. Google Scholar [9] T. H. Otway, Nonlinear Hodge maps. J. Math. Phys. 41 (2000), 5745-5766. Google Scholar [10] T. H. Otway, Hodge equations with change of type, Ann. Mat. Pura Appl. 181 (2002), 437-452. Google Scholar [11] T. H. Otway, Harmonic fields on the projective disk and a problem in optics, J. Math. Phys. 46 (2005), 113501. (Erratum: J. Math. Phys. 48 (2007), 079901.) Google Scholar [12] T. H. Otway, Variational equations on mixed Riemannian-Lorentzian metrics, J. Geom. Phys. 58 (2008), 1043-1061. Google Scholar [13] T. H. Otway, The Dirichlet Problem for Elliptic-Hyperbolic Equations of Keldysh Type, Lecture Notes in Mathematics, Vol. 2043, Springer-Verlag, Berlin-Heidelberg-New York-Tokyo, 2012. Google Scholar [14] T. H. Otway, Elliptic-Hyperbolic Partial Differential Equations: a mini-course in geometric and quasilinear methods, Springer-Verlag, London, 2015. Google Scholar [15] L. Sarason, On weak and strong solutions of boundary value problems, Commun. Pure Appl. Math. 15 (1962), 237-288. Google Scholar [16] J. M. Stewart, Signature change, mixed problems and numerical relativity, Class. Quantum Grav. 18 (2001), 4983-4995. Google Scholar [17] J. Stillwell, Sources of Hyperbolic Geometry, Amer. Math. Soc., Providence, 1996. Google Scholar [18] W. J. van Stockum, The gravitational field of a distribution of particles rotating about an axis of symmetry, Proc. R. Soc. Edinburgh 57 (1937), 135-154. Google Scholar [19] F. J. Tipler, Rotating cylinders and the possibility of global causality violation, Phys. Rev. D9 (1974), 2203-2206. Google Scholar [20] C. G. Torre, The helically reduced wave equation as a symmetric positive system, J. Math. Phys. 44 (2003), 6223-6232. Google Scholar
show all references
##### References:
[1] J. Barros-Neto and F. Cardoso, Gellerstedt and Laplace-Beltrami operators relative to a mixed signature metric, Ann. Mat. Pura Appl. 188 (2009), 497-515. Google Scholar [2] E. Beltrami, Saggio di interpretazione della geometria non-euclidea, Giornale di Matematiche 6 (1868), 284-312. Google Scholar [3] K. O. Friedrichs, Symmetric positive linear differential equations, Commun. Pure Appl. Math. 11 (1958), 333-418. Google Scholar [4] J. Heidmann, Relativistic Cosmology, An Introduction. Springer-Verlag, Berlin-Heidelberg-New York (1980). Google Scholar [5] M. V. Keldysh, On certain classes of elliptic equations with singularity on the boundary of the domain (Russian]), Dokl. Akad. Nauk SSSR 77 (1951), 181-183. Google Scholar [6] P. D. Lax and R. S. Phillips, Local boundary conditions for dissipative symmetric linear differential operators, Commun. Pure Appl. Math. 13 (1960), 427-455. Google Scholar [7] F. Lobo and P. Crawford, Time, closed timelike curves, and causality, in The Nature of Time: Geometry, Physics and Perception (NATO ARW), Proceedings of a conference held 21-24 May, 2002 at Tatranska Lomnica, Slovak Republic (eds. Rosolino Buccheri, Metod Saniga, and William Mark Stuckey). NATO Science Series II: Mathematics, Physics and Chemistry - Volume 95. Dordrecht/Boston/London: Kluwer Academic Publishers, 2003. Google Scholar [8] A. Marini and T. H. Otway, Nonlinear Hodge-Frobenius equations and the Hodge-Bäcklund transformation, Proc. R. Soc. Edinburgh 140A (2010), 787-819. Google Scholar [9] T. H. Otway, Nonlinear Hodge maps. J. Math. Phys. 41 (2000), 5745-5766. Google Scholar [10] T. H. Otway, Hodge equations with change of type, Ann. Mat. Pura Appl. 181 (2002), 437-452. Google Scholar [11] T. H. Otway, Harmonic fields on the projective disk and a problem in optics, J. Math. Phys. 46 (2005), 113501. (Erratum: J. Math. Phys. 48 (2007), 079901.) Google Scholar [12] T. H. Otway, Variational equations on mixed Riemannian-Lorentzian metrics, J. Geom. Phys. 58 (2008), 1043-1061. Google Scholar [13] T. H. Otway, The Dirichlet Problem for Elliptic-Hyperbolic Equations of Keldysh Type, Lecture Notes in Mathematics, Vol. 2043, Springer-Verlag, Berlin-Heidelberg-New York-Tokyo, 2012. Google Scholar [14] T. H. Otway, Elliptic-Hyperbolic Partial Differential Equations: a mini-course in geometric and quasilinear methods, Springer-Verlag, London, 2015. Google Scholar [15] L. Sarason, On weak and strong solutions of boundary value problems, Commun. Pure Appl. Math. 15 (1962), 237-288. Google Scholar [16] J. M. Stewart, Signature change, mixed problems and numerical relativity, Class. Quantum Grav. 18 (2001), 4983-4995. Google Scholar [17] J. Stillwell, Sources of Hyperbolic Geometry, Amer. Math. Soc., Providence, 1996. Google Scholar [18] W. J. van Stockum, The gravitational field of a distribution of particles rotating about an axis of symmetry, Proc. R. Soc. Edinburgh 57 (1937), 135-154. Google Scholar [19] F. J. Tipler, Rotating cylinders and the possibility of global causality violation, Phys. Rev. D9 (1974), 2203-2206. Google Scholar [20] C. G. Torre, The helically reduced wave equation as a symmetric positive system, J. Math. Phys. 44 (2003), 6223-6232. Google Scholar
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# Area of Trapezium Definition, Types, Properties, Derivation and FAQs
A Trapezium is a quadrilateral in a two-dimensional space with one pair of parallel sides. It is a convex figure which means that it is a closed figure with all its interior angles less than 180 and vertices pointing outwards. It has four sides and four vertices. Since it has only one pair of parallel sides, the other two are non-parallel sides and are called legs and the parallel sides are called bases. The illustration of the trapezium is given below. In this article, we will learn how to find the area of a trapezium with the help of some practice problems.
• Definition of Area of Trapezium
• Types
• Scalene Trapezium
• Isosceles Trapezium
• Right Trapezium
• Properties of Trapezium
• Derivation of Area of Trapezium
• Practice Problems
• FAQs
## Definition of Area of Trapezium
The area of the trapezium or trapezoid is the space covered by the trapezium on a two-dimensional plane. A trapezium is a quadrilateral with four sides and four vertices. The area of trapezium is the half of the sum of the parallel sides times the height of the trapezium.
Area of Trapezium =
Area of Trapezium =
Here, a and b are bases.
And h is the height or perpendicular distance between a and b.
## Types
There are three types of trapeziums:
1. Scalene Trapezium
2. Isosceles Trapezium
3. Right Trapezium
Let’s discuss these in brief one by one.
### Scalene Trapezium
The word ‘scalene’ means ‘unequal’. The scalene trapezium is the trapezium that has all its sides and angles of different measures.
Scalene Trapezium
### Isosceles Trapezium
The word Isosceles means ‘equal leg’. The trapezium is called isosceles trapezium when both of its legs or non-parallel sides are equal.
Isosceles Trapezium
### Right Trapezium
A trapezium is said to be a right trapezium when at least one angle of a trapezium is a right angle.
Right Trapezium
## Derivation of Area of Trapezium
Now, let us learn how to derive the formula for the area of the trapezium.
From the above diagram, we can see that trapezium PQRS comprises of Δ PXS, Δ QYR and rectangle XYRS.
In Δ PXS, the base is a and height is h.
In Δ QYR, the base is b and height is h.
In the rectangle XYRS, length is n and width is h.
Also,
${b}_{1}=n$
${b}_{2}=a+n+b$
Now, the area of trapezium is equal to the sum of the area of two triangles and the area of the rectangle.
Area of PQRS=Area of Δ PXS+Area of Δ QYR+Area of XYRS
So, the area of a trapezium with bases b1 and b2 and height h is given by
## Practice Problems
1. Find the area of trapezium when the sum of the parallel sides is 20 cm and the height is 5 cm.
Solution: Given: b1+b2=20 cm and h=5 cm
Area = ?
Area of a Trapezium =
2. Find the distance between the parallel sides of the trapezium when the length of the parallel sides are 15 cm and 18 cm and the area is 99 cm2.
Solution: Given: Area of a Trapezium = 99 cm2
b1=15 cm and b2=18 cm
h= ?
Now, Area of a Trapezium
3. Find the area of the trapezium given below.
Solution: Given: b1=4 cm , b2=12 cm and h=5 cm
Area of given trapezium =
4. Calculate the area of a trapezium whose parallel sides are 45.7 cm and 68.3 cm and the height is 5 cm.
Solution:
Now, Area of a Trapezium
#### 5. Find the area of the trapezium when the length of parallel sides is given as 10 cm and 25 cm and the length of non-parallel sides is given as 13 cm and 14 cm.
Solution: Let ABCD be a trapezium in which
AB=10 cm, BC=14 cm, CD=25 cm and DA=13 cm
Construct a line from a point B parallel to side AD, which cuts CD at E.
As we know that, Area of trapezium =h2(a+b) square units
But, for that, we have to find the height or distance between parallel sides.
Now, in CBE, we apply Heron’s formula
$A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)}$
Also,
Now, the area of the trapezium is,
Hence, the area of the trapezium
1. What is an irregular trapezium?
A.
A trapezium having its non - parallel sides of unequal length is said to be an irregular trapezium.
2. What is the height of a trapezium?
A.
The perpendicular distance between the parallel sides of a trapezium is the height of a trapezium.
3. What is right trapezium?
A.
A trapezium having two right angles is called a right trapezium.
4. Does a trapezium have equal diagonals?
A.
Yes, the trapezium can have equal diagonals when the given trapezium is an isosceles trapezium.
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Journal of Economic Theory and Econometrics: Journal of the Korean Econometric Society
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Journal of Economic Theory and Econometrics
Journal of the Korean Econometric Society
• #### Donggyu Sul (University of Texas at Dallas)
Abstract
Practitioners often standardize panel data before estimating a factor model. In this paper we show an example that the standardization leads to inconsistent estimation of the factor number. When the common component exhibits strong heteroskedasticity, the conventional eigenvalue-based decompositions are consistent but standardization does not necessarily result in consistent estimation. To overcome this issue, we recommend using a minimum-rule'' whereby the minimum factor-number estimated from both the conventional and standardized panel is used. Monte Carlo studies and an empirical application are provided.
Keywords
Factor Model, Selection Criteria, Principal Components Estimator, Bai-Ng Criteria, Standarization, Panel Data
JEL classification codes
C33
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# Tangent of slope
1. Aug 18, 2016
1. The problem statement, all variables and given/known data
Find the slope of tangent line to curve that is intersection to the surface z= (x^2) - (y^2) with plane x =2 , at point (2,1,3)
The ans given by the author is only∂z /∂y = -2
2. Relevant equations
3. The attempt at a solution
Is my diagram correct ?
I'm wondering , why shouldnt we move the entire graph 'forward ' to x = 2 ?
#### Attached Files:
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2. Aug 18, 2016
### Staff: Mentor
Your diagram is o.k. so far. Except you are right that it should be at $x=2$.
Which thoughts brought you to draw it this way. i.e. which equation did you use?
Next you have to find out, how to calculate a slope at a point. Do you know what this means?
3. Aug 18, 2016
by looking at the the equation z = 4-(y^2) alone , the whole graph is at x=0 axis alone , right ?
4. Aug 18, 2016
### Staff: Mentor
No, it should be at $x=2$ as you've said earlier. But that doesn't change the question, because you have eliminated $x$ in the equation anyway. So you have only a curve in a plane, the same as the usual $(x,y)$ case. The variables simply have different names and all is $(y,z)$ instead. And the slope of a tangent at the curve in a point $(2,1,3)$ is now simply at $y=1$ and $z=z(y)=4-y^2$.
You are correct, except you didn't say how to calculate this slope.
5. Aug 18, 2016
so, the the whole graph is at x = 0 , right ?
since in z= 4-(y^2) , everything is in z and y , how we know that x = 2 by looking at z= 4-(y^2) ?
6. Aug 18, 2016
### Staff: Mentor
No. Originally it's in 3D space with $(x,y,z)$. The intersection is at $x=2$. It is like cutting the whole thing along $x=2$.
That does not change. But once you have cut it, there is no $x$ anymore. However, you must not forget $x=2$ if further investigations on the original surface would be made, e.g. by comparing the result with a cut at another $x$.
As long as the calculations take place with $x=2$ fixed, you may substitute all $x$ by $2$ as you did and forget (for the moment) that there is an $x$ at all. But it stays $x=2$. We simply do not consider it.
By looking only at the plane, our cut, there is no $x$ anymore. You can think of it as a parabola with an $y-$ and $z-$ axis. On top of this parabola drawing you note "$\text{Intersection along }x=2$" as its label. Or the more complicated way in a 3D picture like yours, with the $x-$coordinate $2$. (As you also already mentioned in your first post.)
7. Aug 18, 2016
do you mean for z= (x^2) - (y^2) , when x =2 , z= 4 - (y^2) , we draw the curve at x = 0 first , then , extend the line along x-axis and draw another same curve at x = 2?
#### Attached Files:
• ###### Capture.PNG
File size:
58.5 KB
Views:
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8. Aug 18, 2016
### Staff: Mentor
I mean we draw it at $x=2$ in the first place. The question asks about the point $(2,1,3)$ and this point isn't part of anything with $x=0$.
Also at $x=0$ the parabola becomes $z=-y^2$ which is shifted by $4$ compared to $z=4-y^2$.
In your computer graphic you set $x=2$ as a constant dimension of a 3D space which it is not. A point $(0,1,-1)$ is on the original surface, however, not on the surface of your computer graphic. Have a look how it really looks like:
http://www.wolframalpha.com/input/?i=z=x^2-y^2
9. Aug 18, 2016
ok , for the curve at x = 2 , the value of x doesnt change , so ∂z / ∂x = 0 ?
10. Aug 18, 2016
### Staff: Mentor
For the curve with $x=2$ there is no $x$ anymore. We substituted it. $z$ is a function of $y$ alone.
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Tensor rank decomposition for some vectors
I have a problem with solving of following problem.
We have a vector space $$V$$ over field $$k$$ with basis $$v_1, v_2$$
Rank of a vector $$v$$ of $$V \otimes V \otimes V$$ is minimum length of decomposition of $$v$$ into linear combination of rank-1 vectors
I need to solve the following:
• Prove that vector $$v = v_1 \otimes v_1 \otimes v_1 + v_1 \otimes v_2 \otimes v_2 + v_2 \otimes v_1 \otimes v_2$$ has rank $$3$$
• Vector $$t = v_1 \otimes v_1 \otimes v_1 - v_2 \otimes v_2 \otimes v_1 + v_1 \otimes v_2 \otimes v_2 + v_2 \otimes v_1 \otimes v_2$$. Show decomposition of $$t$$ into sum of two rank-1 vectors when $$k = \Bbb C$$. Prove that tensor rank of $$t$$ is $$3$$ when $$k = \Bbb R$$
Tried to look for an information but found nothing usefull. I thought about supposing that we can decompose $$v$$ into $$e_{11} \otimes e_{12} \otimes e_{13} + e_{21} \otimes e_{22} \otimes e_{23}$$ and supposing that $$e_{ij}$$ form basis for each $$j$$ that allows us to rewrite $$v$$ in terms of $$e_{ij}$$. After doing this i got a large system with a large amount of variables that i'm unable to work with. I have been thinking about this problem for a several weeks already and do not know how to even think about it.
• Is this an exercise from a class? From a textbook? Dec 13 '19 at 21:23
• @Omnomnomnom It's from a class. I met this problem before and never was able to solve it. Usually i try to solve problems by myself but this one really frustrated me. Dec 13 '19 at 21:28
• Yeah it's a doozy... have you discussed any tricks for tensors in class? Any way to provide a lower bound to a tensor rank? I'm sure that you must have discussed the case of rank-$1$ tensors in $V \otimes V$. Dec 13 '19 at 21:42
• Based on some googling, this article seems to have some interesting ideas. I found the article from the references of this textbook, which I've only just been skimming through Dec 13 '19 at 21:44
• As an array, your first matrix has (frontal) slices $$X_1 = \pmatrix{1&0\\0&0}, \quad X_2 = \pmatrix{0&1\\1&0}$$ Dec 13 '19 at 21:46
For the purposes of the following, I identify $$\sum_{ijk}x_{ijk} v_{i} \otimes v_j \otimes v_k$$ with a 3-dimensional array $$X$$. For the purposes of your question over a 2-dimensional $$V$$, we have $$X = [X_1 |X_2] = \left[ \begin{array}{cc|cc} x_{111} & x_{121} & x_{112} & x_{122}\\ x_{211} & x_{221} & x_{212} & x_{222} \end{array}\right].$$
Question 1:
For your problem, we have $$X_1 = \pmatrix{1&0\\0&0}, \quad X_2 = \pmatrix{0&1\\1&0}.$$ We will simply apply the result explained below. Since exchanging slices of a tensor will not change its rank, we'll exchange the roles of $$X_1$$ and $$X_2$$ since $$X_2$$ is the invertible slice. We find that $$X_1X_2^{-1} = \pmatrix{0&1\\0&0}.$$ Since this matrix fails to be diagonalizable, $$X$$ must be a tensor of rank $$3$$.
Question 2:
Your second problem can be approached similarly. We now have $$X_1 = \pmatrix{1&0\\0&-1},\quad X_2 = \pmatrix{0&1\\1&0}.$$ Applying the result below, we find that $$X_2 X_1^{-1} = \pmatrix{0&-1\\1&0}.$$ Because this matrix is diagonalizable with strictly complex eigenvalues, we can conclude that $$X$$ has a rank of at least $$3$$ if we restrict ourselves to real coefficients, and a rank of $$2$$ if we allow complex coefficients.
It remains to be shown, however, that the rank of $$X$$ over $$\Bbb R$$ is not more than $$3$$. To do this, simply observe that we can get this tensor by adding a rank-1 tensor to the tensor given in the first question.
Regarding the presentation of $$t$$ as a rank-2 tensor: if we follow the construction from the proof I present below, then we note that $$X_2 X_1^{-1} = K\Lambda K^{-1}$$, where $$\Lambda = \pmatrix{i\\&-i}, \quad K = \pmatrix{i&1\\1&i}.$$ So, we find that $$X = a_1 \otimes b_1 \otimes c_1 + a_2 \otimes b_2 \otimes c_2$$ where $$a_1,a_2$$ are the columns of $$K$$, $$b_1,b_2$$ are the rows of $$K^{-1}X_1$$, and we have $$c_1 = (1,i), c_2 = (1,-i)$$.
Now, here is an adaptation of the statement and proof of Lemma 1 of this paper.
Claim: Let $$X$$ be a real-valued $$p \times p \times 2$$ array with $$p \times p$$ slices $$X_1$$ and $$X_2$$. Suppose that $$X_{1}^{-1}$$ exists. The following statements hold:
• If $$X_2X_1^{-1}$$ has $$p$$ real eigenvalues and is diagonalizable, then $$X$$ has rank $$p$$ over $$\Bbb R$$
• If $$X_2X_1^{-1}$$ has at least one pair of complex eigenvalues, then $$X$$ has rank $$p$$ over $$\Bbb C$$ and rank at least $$p+1$$ over $$\Bbb R$$
• If $$X_2X_1^{-1}$$ is not diagonalizable, then $$X$$ has rank at least $$p+1$$ over $$\Bbb C$$.
Proof: First, note that since $$X_1$$ is invertible, the rank of $$X$$ must be at least $$p$$.
Proof of i: Now, suppose that we have $$X_2 X_1^{-1} = K \Lambda K^{-1}$$, where $$\Lambda = \operatorname{diag}(\lambda_1,\dots,\lambda_p)$$. If we take $$A = K, \quad B^T = K^{-1}X_1, \quad C_1 = I_p, \quad C_2 = \Lambda,$$ then we find that $$X_1 = AC_1B^T, \quad X_2 = AC_2B^T.$$ This corresponds to a rank-$$p$$ decomposition of the matrix $$X$$. In particular: if we take $$a_j$$ to denote the $$j$$th column of $$A$$ and $$b_j$$ to denote the $$j$$th column of $$B$$, then we have $$X_1 = AC_1B^T = \sum_{j=1}^p c_{1,i} \, a_ib_i^T, \quad X_2 = AC_2B^T = \sum_{j=1}^p c_{2,i} \, a_i b_i^T.$$ Correspondingly, we have $$X = \left(\sum_{j=1}^p c_{1,j} \, a_j \otimes b_j\right) \otimes e_1 + \left(\sum_{j=1}^p c_{2,j} \, a_j \otimes b_j\right) \otimes e_2\\ = \sum_{j=1}^p a_j \otimes b_j \otimes (c_{1,j} e_1) + \sum_{j=1}^p a_j \otimes b_j \otimes (c_{2,j}e_2)\\ = \sum_{j=1}^p a_j \otimes b_j \otimes (c_{1,j}e_1 + c_{2,j}e_2).$$ In the above, $$e_1 = (1,0)$$ and $$e_2 = (0,1)$$.
Proof of ii and iii: It suffices to prove that if $$X$$ is a rank-$$p$$ tensor and $$X_1$$ is invertible, then $$X_2X_1^{-1}$$ must be diagonalizable. Indeed, if $$X$$ is a rank-$$p$$ tensor, then we can take $$X_1 = AC_1B^T, \quad X_2 = AC_2B^T$$ by reversing the above sequence of equations. It follows that $$X_2X_1^{-1} = (AC_2B^T)(AC_1B^T)^{-1} = AC_2 B^T B^{-T} C_1^{-1} A^{-1} = A (C_2 C_1^{-1})A^{-1}.$$ So, $$X_2X_1^{-1}$$ is indeed diagonalizable (and diagonalizable over $$\Bbb R$$ when $$A,B,C$$ are real). The conclusion follows.
• That's a really good answer, I didn't really expect anything like this. I guess this proof works over any field, doesn't it? Also while it answers my question I want to ask the following: There was proved that if rank-p decomposition exists then matrix is diagonalizable. But how does "if matrix have two complex eigenvalues then it have rank p over C" follows from it? I see how it's proved for real coefficients when p = 2 (in this case it's equivalent to the fact that it's diagonalizable) Dec 14 '19 at 10:44
• Yes, I agree that the proof seems to work over any field. The point I'm making with the proof of ii and iii is that we can go for a proof by contradiction. Given that $X_2X_1^{-1}$ has at least two (strictly) complex eigenvalues, suppose that $X$ has a tensor decomposition with real coefficients. Then by reversing the process in the proof of i, we obtain a decomposition $X_2X_1^{-1} = A \Lambda A^{-1}$ where $\Lambda$ is a real diagonal matrix, which would imply that $X_2X_1^{-1}$ has no strictly complex eigenvalues. Dec 14 '19 at 12:24
• And if we allow for complex coefficients, then we can simply apply the proof from i to see that $X$ has rank $p$. Dec 14 '19 at 12:29
• If you prefer, the real result is that given that the slice $X_1$ is invertible, $X$ has rank $p$ over $\Bbb F$ if and only if $X_2X_1^{-1}$ is diagonalizable over $\Bbb F$. Dec 14 '19 at 12:31
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• # question_answer If A, B, C and D are the angles of a quadrilateral, then $\frac{\sum{\tan \,A}}{\sum{\cot \,A}}$ is equal to A) $\Pi \,\,\tan \,\,A$ B) $\Pi \,\,\cot \,\,A$ C) $\Sigma \,{{\tan }^{2}}A$ D) $\Sigma \,{{\cot }^{2}}A$
$\because$ $A+B+C+D=2\pi$ or $\tan \,(A+B+C+D)=0$ or $\frac{\Sigma \,\tan \,A-\Sigma \tan \,A\,\tan \,B\,\tan \,C}{1-\Sigma \tan \,A\,\tan \,B+\tan \,A\,\tan \,B\,\tan \,C\,D}=0$ $\Rightarrow$ $\Sigma \tan \,A-\Sigma \tan \,A\,\tan \,B\,\tan \,C=0$ $\Rightarrow$ $\Sigma \tan \,A=\tan \,A\,\tan \,B\,\tan \,C\,\tan D\,\,\Sigma \cot \,A$ $\Rightarrow$ $\frac{\Sigma \,\tan \,A}{\Sigma \cot \,A}=\Pi \,\tan \,A$
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# Find the area of the following circles, given that: ( Take π = 22/7 ]
Find the area of the following circles, given that: ( Take π = $\frac { 22 }{ 7 }$ )
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# Cold, dense, atomic ion clouds produced by cryogenic buffer-gas cooling
@article{Bhatt2019ColdDA,
title={Cold, dense, atomic ion clouds produced by cryogenic buffer-gas cooling},
author={Nishant Bhatt and Kosuke Kato and Amar C. Vutha},
journal={Physical Review A},
year={2019}
}
• Published 8 June 2019
• Physics
• Physical Review A
We produce cold and dense clouds of atomic ions (Ca$^+$, Dy$^+$) by laser ablation of metal targets and cryogenic buffer gas cooling of the resulting plasma. We measure the temperature and density of the ion clouds using laser absorption spectroscopy. We find that large ion densities can be obtained at temperatures as low as 6 K. Our method opens up new ways to study cold neutral plasmas, and to perform survey spectroscopy of ions that cannot be laser-cooled easily.
1 Citations
## Figures from this paper
### Nd+ isotope shift measurements in a cryogenically cooled neutral plasma
• Physics
• 2020
We report measurements of the isotope shifts of two transitions ($4{f}^{4}6s\ensuremath{\rightarrow}{[25044.7]}_{7/2}^{o}$ and $4{f}^{4}6s\ensuremath{\rightarrow}{[25138.6]}_{7/2}^{o}$) in neodymium
## References
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### Laser cooling of ions in a neutral plasma
• Physics
Science
• 2019
Laser cooling of a neutral plasma created by photoionization of an ultracold atomic gas is demonstrated and pushes laboratory studies of neutral plasmas deeper into the strongly coupled regime, beyond the limits of validity of current kinetic theories for calculating transport properties.
### Ultracold neutral plasmas
• Physics
2014 IEEE 41st International Conference on Plasma Sciences (ICOPS) held with 2014 IEEE International Conference on High-Power Particle Beams (BEAMS)
• 2014
The fifteen-year history of the field of ultracold neutral plasmas, which resides at the intersection of atomic physics and plasma physics, is reviewed, including plasma creation, expansion dynamics, temperature evolution, collisional properties, and collective modes.
### The buffer gas beam: an intense, cold, and slow source for atoms and molecules.
• Physics
Chemical reviews
• 2012
A survey of the current state of the art in buffer gas beams is presented, and some of the possible future directions that these new methods might take are explored.
### Manipulation of individual hyperfine states in cold trapped molecular ions and application to HD+ frequency metrology.
• Physics
Physical review letters
• 2012
Advanced techniques for manipulation of internal states, standard in atomic physics, are demonstrated for a charged molecular species for the first time and the highest spectral resolution is obtained so far in the optical domain on a molecular ion species.
### Buffer gas loaded magneto-optical traps for Yb, Tm, Er and Ho
• Physics
• 2014
Direct loading of lanthanide atoms into magneto-optical traps (MOTs) from a very slow cryogenic buffer gas beam source is achieved, without the need for laser slowing. The beam source has an average
### Hyperfine Structure and Isotope Shifts in Dy II
• Physics
• 2017
Using fast-ion-beam laser-fluorescence spectroscopy (FIBLAS), we have measured the hyperfine structure (hfs) of 14 levels and an additional four transitions in Dy II and the isotope shifts (IS) of 12
### Towards a precise measurement of atomic parity violation in a single Ra+ ion
• Physics
• 2012
A single trapped Ra + (Z = 88) ion provides a very promising route towards a most precise measurement of Atomic Parity Violation (APV), since APV effects grow faster than Z3. This experiment
### Atomic physics: An exploration through problems and solutions
1. Atomic Structure 2. Atoms in External Fields 3. Interaction of Atoms with Light 4. Interaction of Light with Atoms in External Fields 5. Atomic Collisions 6. Cold Atoms 7. Molecules 8.
### Laboratory atomic transition data for precise optical quasar absorption spectroscopy
• Physics
• 2013
Quasar spectra reveal a rich array of important astrophysical information about galaxies which intersect the quasar line of sight. They also enable tests of the variability of fundamental constants
### LABORATORY ROTATIONAL SPECTRUM OF l–C3H+ AND CONFIRMATION OF ITS ASTRONOMICAL DETECTION
• Physics, Chemistry
• 2014
The rotational spectrum of l–C3H+ has been measured in the millimeter-wave band in a 4 K cryogenic ion trap apparatus employing a novel mass-selective action spectroscopy method based on light
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# [XeTeX] problem with the package polyglossia arabic and perpage
Jens Bakker jbakker at uni-bonn.de
Fri Nov 16 20:15:40 CET 2012
Dear Herbert Schulz, dear Peter Dyballa,
Thank you very much for your fast and very helpful replies. I have chosen the solution of Herbert Schulz, because, for the time being, such a file can also run on a computer that has the updated version of bidi.
Thanks al lot again,
with best wishes and best regards,
Jens Bakker
Am 16.11.2012 um 00:56 schrieb Herbert Schulz:
>
> On Nov 15, 2012, at 5:24 PM, Herbert Schulz <herbs at wideopenwest.com> wrote:
>
>>
>> On Nov 15, 2012, at 5:05 PM, Jens Bakker <jbakker at uni-bonn.de> wrote:
>>
>>>
>>> I would like to ask you for help, please: I have just updated TeX Live by TeX Live Utility, and suddenly all my files that I have written with polyglossia using the language Arabic and the package perpage do not compile, while they just compiled a few minutes ago correctly.
>>>
>>> The attached minimal example seems to prove that it does not compile when the package perpage is used, because if \usepacke{perpage} is outcommented, the text compiles.
>>>
>>> What can I do?
>>>
>>> Thank you very much for your interest,
>>> Jens Bakker
>>
>>
>> Howdy,
>>
>> The console error message should tell you all you need to know:
>>
>> \usepackage{polyglossia}
>> \setmainlanguage{arabic}
>>
>> Good Luck,
>>
>> Herb Schulz
>> (herbs at wideopenwest dot com)
>
>
> Howdy,
>
> Sorry, I forgot to uncomment the \usepackage{perpage} and that gave an error message. However, If I moved the perpage package after the font commands it seems to work fine.
>
> %%!TEX TS-program = xelatexmk
> \documentclass[11pt,a4paper,twoside]{article}
>
> \usepackage{polyglossia}
> \setmainlanguage{arabic}
> \newfontfamily\arabicfont{Geeza Pro}
>
> \usepackage{perpage}
>
> \begin{document}
> جاء زيد إلى عمرو
> \end{document}
>
> (I don't have Scheherazade so I am using Geeza Pro on my Mac.)
>
> Good Luck,
>
> Herb Schulz
> (herbs at wideopenwest dot com)
>
>
>
>
>
>
> --------------------------------------------------
> Subscriptions, Archive, and List information, etc.:
> http://tug.org/mailman/listinfo/xetex
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To illustrate the compilation of a computational scheme, suppose
it is required to compute the values of the analytically specified
function
$y= f(x)$
for certain values of the argument: $x = x_1 , x_2 , \cdots , x_n$. If the number of these values is great, it ‘is not advisable to compute them separately, first $f(x_1)$, then $f(x_2)$, and so on, each time performing the whole sequence of operations indicated by the symbol $f$. It is much better to separate $f$ into elementary operations
$f(x) = f_m ( \cdots (f_2 (f_1 (x))) \cdots)$
and carry out the computations as repeated operations:
$u_i (x_i) = f_1(x_i)\,(i=1,2,\ldots,n),$
$v_i (u_i) = f_2(u_i)\,(i=1,2,\ldots,n),$
$\cdots \cdots \cdots$
$f_m (w_i) \,(i=1,2,\ldots,n),$
performing one and the same operation $f_j\,(i = 1, 2,\ldots, m)$ for all values of the argument under consideration.
Although this comes from the era of hand calculations, it’s good advice for programming as well. It’s always better to break down the problem into smaller portions and work your way through. Not only is the code easier to debug but frequently it decreases the computational cost.
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# Geometric Distribution Calculator
What is Geometric Distribution?
The word geometric distribution relates to the probability of getting successful after repeated failures. It is like tossing the dice several times of getting “six” and the chances of getting other numbers are more. So what will be the probability of getting six? It is the geometric distribution. This is the result of Bernoulli trials, you can also represent it as “probability density function.”
## Geometric Distribution Calculator Formula
The mathematical representation of the probability density function:
f(x) = (p)x – 1q
Here,
• “f” represents the functions
• “p” is the probability of finding accurate results or success
• “q” is the probability of failure
• “x” is the number of total trials
• “1” shows the total number of attempts we are going to make
Note:
To gain knowledge about density, click the density calculator.
### Example
15% of the cars are passing along certain roads are blue. What is the probability that the 7th car will be blue or not?
### Solution
We are taking the probability of passing cars as 1, so finding the blue car on road will be 0.15 and not finding it will be automatically 0.85%. The total number of cars are 7 in number.
### Applying the formula,
f(x) = (q)x-1p
f(7) = (0.85)7-1(0.15)
f(7) = (0.85)6(0.15)
f(7) = 0.056
f(7) = 5.7% approx.
So the answer is we have chances to get the next car to blue nearly 5.7%.
Using this method is helpful in getting the predictions about upcoming results. You can use this method to estimate the results are probably what will be expected in the future.
### Geometric Distribution Calculator use
The geometric distribution calculator is useful in helping you to find accurate results by clicking a button. Forget the messed up solutions and putting manually all the values. You do not have to apply the formula manually using this calculator. Simply put the values expected for success and failure and you will get the ratio or percentage of in the answer.
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There’s an increasingly heated debate about Scala’s complexity. A recent post talks about Scala’s "true complexity": the terrifying intricacy you will encounter, if you try to extend Scala’s Collections Library with a method that achieves the "perfect" behavior.
Yet it also might be worth asking: what are the motivations of these "perfect" behaviors? Are they always desirable?
One important aspect of the "perfect" collection behavior is the magical uniform return type principle. Sloppily speaking, this means that transforming a collection should give you another collection of the same kind. For instance:
// Filtering a List gives you a filtered List
scala> List(1,2,3,4).filter(_%2==0)
res: List[Int] = List(2, 4)
// Mapping a Set gives you a mapped Set
scala> Set(1,2,3,4).map(1+)
res: scala.collection.immutable.Set[Int] = Set(2, 3, 4, 5)
// Uniform return type at both compile time and runtime
scala> val iterable : Iterable[Int] = Set(1,2,3,4)
iterable: Iterable[Int] = Set(1, 2, 3, 4)
scala> iterable.map(1+)
res: Iterable[Int] = Set(2, 3, 4, 5)
// Works for Strings, too
scala> "E is the most frequent letter in English".map(
| c ...
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# List five integers that are congruent to 4 modulo 12.
The aim of this question is to introduce the concept of congruency of an integer with another integer under some modulo.
Division
Whenever we divide one integer over another, we have two results, namely a quotient and a remainder. The quotient is the part of the result that defines the perfect division while the existence of the remainder signifies that the division was not perfect.
Perfect division
Let’s say we have three integers a, b, and c. Now we say that a is congruent to b modulo c if $a \ – \ b$ is perfectly divisible by $c$.
Subtraction
Given that we need to find all integers (say $x$) that are congruent to 4 modulo 12. In simpler words, we need to find the first five values of $x \ – \ 4$ that are perfectly divisible by $12$.
To solve this question, we can take help from the integral multiples of $12$ as listed below:
$\text{ Integral multiples of } 12 \ = \ \{ 0, \ 12, \ 24, \ 36, \ 48, \ 60, \ … \ … \ … \ \}$
To find the first five integer values that are congruent to 4 modulo 12, we simply need to solve the following equations:
$\begin{array}{ c } \text{ Integers congruent } \\ \text{ to } 4 \text{ modulo } 12 \end{array} \ = \ \left \{ \begin{array}{ c c c } x \ – \ 4 \ = \ 0 & \Rightarrow & x \ = \ 0 \ + \ 4 & \Rightarrow & x \ = \ 4 \\ x \ – \ 4 \ = \ 12 & \Rightarrow & x \ = \ 12 \ + \ 4 & \Rightarrow & x \ = \ 16 \\ x \ – \ 4 \ = \ 24 & \Rightarrow & x \ = \ 24 \ + \ 4 & \Rightarrow & x \ = \ 28 \\ x \ – \ 4 \ = \ 36 & \Rightarrow & x \ = \ 36 \ + \ 4 & \Rightarrow & x \ = \ 40 \\ x \ – \ 4 \ = \ 48 & \Rightarrow & x \ = \ 48 \ + \ 4 & \Rightarrow & x \ = \ 52 \end{array} \right.$
$\text{ Integers congruent to } 4 \text{ modulo } 12 \ = \ \{ 4, \ 16, \ 28, \ 40, \ 52 \ \}$
## Numerical Results
$\text{ Integers congruent to } 4 \text{ modulo } 12 \ = \ \{ 4, \ 16, \ 28, \ 40, \ 52 \ \}$
## Example
List down the first six integers such that they are congruent to 5 modulo 15.
Here:
$\text{ Integral multiples of } 15 \ = \ \{ 0, \ 15, \ 30, \ 45, \ 60, \ 75, \ … \ … \ … \ \}$
So:
$\begin{array}{ c } \text{ Integers congruent } \\ \text{ to } 5 \text{ modulo } 15 \end{array} \ = \ \left \{ \begin{array}{ c c c } x \ – \ 5 \ = \ 0 & \Rightarrow & x \ = \ 0 \ + \ 5 & \Rightarrow & x \ = \ 5 \\ x \ – \ 5 \ = \ 15 & \Rightarrow & x \ = \ 15 \ + \ 5 & \Rightarrow & x \ = \ 20 \\ x \ – \ 5 \ = \ 30 & \Rightarrow & x \ = \ 30 \ + \ 5 & \Rightarrow & x \ = \ 35 \\ x \ – \ 5 \ = \ 45 & \Rightarrow & x \ = \ 45 \ + \ 5 & \Rightarrow & x \ = \ 50 \\ x \ – \ 5 \ = \ 60 & \Rightarrow & x \ = \ 60 \ + \ 5 & \Rightarrow & x \ = \ 65 \end{array} \right.$
$\text{ Integers congruent to } 5 \text{ modulo } 15 \ = \ \{ 5, \ 20, \ 35, \ 50, \ 65 \ \}$
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Article | Open
# Seismological evidence for a localized mushy zone at the Earth’s inner core boundary
• Nature Communications 8, Article number: 165 (2017)
• doi:10.1038/s41467-017-00229-9
Accepted:
Published online:
## Abstract
Although existence of a mushy zone in the Earth’s inner core has been hypothesized several decades ago, no seismic evidence has ever been reported. Based on waveform modeling of seismic compressional waves that are reflected off the Earth’s inner core boundary, here we present seismic evidence for a localized 4–8 km thick zone across the inner core boundary beneath southwest Okhotsk Sea with seismic properties intermediate between those of the inner and outer core and of a mushy zone. Such a localized mushy zone is found to be surrounded by a sharp inner core boundary nearby. These seismic results suggest that, in the current thermo-compositional state of the Earth’s core, the outer core composition is close to eutectic in most regions resulting in a sharp inner core boundary, but deviation from the eutectic composition exists in some localized regions resulting in a mushy zone with a thickness of 4–8 km.
## Introduction
The thermo-compositional state of the Earth’s core is generically related to the growing process of the inner core and its associated releases of the thermal and compositional energy that power the Earth’s geodynamo1,2,3,4. It has long been proposed that the thermo-compositional state of the Earth’s core is such that the interface between the inner and outer core is dendritic with the mushy zone possibly even extending throughout the entire inner core5. However, while recent studies have revealed various degrees of complexicity near the inner core boundary (ICB), such as hemispherical variations of seismic properties at the top of the inner core6,7,8,9,10, laterally varying seismic properties near the inner core surface11,12,13 and spatially varying ICB topography14,15,16,17,18, no seismic evidence of a mushy zone has ever been reported.
Here we present seismic evidence for a localized 4–8 km thick mushy zone across the ICB beneath southwest Okhotsk Sea, with seismic properties intermediate between those of the inner and outer core. Such a mushy zone is also studied in the context of its surrounding ICB regions, which are found to have a sharp boundary. These seismic results suggest that, in the current thermo-compositional state of the Earth’s core, the outer core composition is close to eutectic in most regions resulting in a sharp ICB, but deviation from the eutectic composition exists in some localized regions resulting in a mushy zone with a thickness of 4–8 km.
## Approach to study the inner core boundary
Our study is based on the analyses of differential travel times and waveforms between seismic PKiKP and PcP phases, compressional waves that are reflected off the Earth’s ICB and the core–mantle boundary (CMB) respectively (Fig. 1), recorded in the epicentral distance range of 0–90°. PKiKP–PcP differential travel time residuals are used to study the topographic variations of the ICB, while the waveform differences between the two phases are used to explore the detailed features of the ICB. Owing to the similar raypaths between PKiKP and PcP phases in the shallow Earth (Fig. 1), the use of PKiKP–PcP differential travel time residuals eliminates the uncertainty of event origin time and minimizes the errors of event mislocation and the effects of shallow Earth’s heterogeneities, placing high-resolution constraints on the ICB topography. For the similar reasons, because of the similar take-off angles from the seismic source and similar incident angles to the receiver between PKiKP and PcP waves, the study of their waveform differences eliminates possible effects on waveform complexities due to source complication and heterogeneous seismic structure beneath the receivers, providing tight constraints on the fine-scale seismic structure in the vicinity of the ICB.
## Seismic data and regions of inner core boundary studied
Our study region lies between 25–50° N in latitude and 110–150° E in longitude (Fig. 2a), sampled by the seismic data recorded in a dense seismic array in Japan, Hi-net19. We select all the events recorded by Hi-net from April 2004 to December 2014 that satisfy following three criteria: epicentral distances to the Hi-net stations are from 0° to 90° in which PKiKP phase is pre-critically reflected from the ICB and is most sensitive to the detailed seismic structure at the top of the inner core20,21,22,23,24,25,26,27, focal depths are greater than 30 km to avoid the contamination of crustal wave reverberations near earthquake sources, and earthquake magnitudes are greater than Mw 5.8, large enough to generate observable PKiKP phases. Such criteria yield a total of 1106 events. With the inclusion of an additional 2001 event (event 1, Fig. 2a) that was used in a previous ICB study28, our data set consists of over 850,000 potential PKiKP–PcP pairs. After eye-checking data quality of each seismogram, we are able to retain a total of 1263 pairs of high-quality PKiKP and PcP waveforms from 11 seismic events (Fig. 2a, Supplementary Table 1). These seismic data sample the ICB regions beneath central China, southeast Japan Sea, west North Pacific Ocean, and southwest Okhotsk Sea (Fig. 2b, Supplementary Fig. 1).
## Non-significant inner core topography in the study region
In the travel time analysis of ICB topography, we remove instrumental responses from waveform data and then apply the worldwide standard seismic network short-period instrumental response and a two-pole causal Butterworth bandpass filter of 1–3 Hz. Such filtering is the procedure we find to have the best observability of PKiKP phases in the Hi-net data after extensive tries of various procedures of frequency filtering. We then measure the PKiKP–PcP differential travel times based on the time separation of the maximal amplitudes of the two phases, and calculate PKiKP–PcP differential travel time residuals with respect to the predictions of a seismic reference model, PREM29 (see “Obtaining PKiKP–PcP differential travel time residuals” in Methods).
Detailed data analyses confirm that using PKiKP–PcP differential travel time residuals significantly minimizes the effects of shallow mantle structure (Supplementary Fig. 2). PKiKP–PcP differential travel time residuals exhibit small variations from −0.5 to 0.5 s (with 93% of the data within −0.3 to 0.3 s) for all the data we analyze (Fig. 2b, Supplementary Figs. 24). If we attribute all the observed variations of PKiKP–PcP differential travel time residuals to ICB topographic variations, the estimated upper bound of ICB topographic changes in the study region is 1.5 km, similar to previous results obtained beneath the western Pacific17. Note that PKiKP–PcP travel time residuals exhibit slight negative correlations with PcP travel time residuals for events 2, 5, 6, 8, 10, and 11 (Supplementary Figs. 2 and 4), indicating that some of the PKiKP–PcP travel time residuals are still to some extent affected by the heterogeneities in the mantle along the PcP raypaths and the actual ICB topography could be even smaller. We conclude that there is no significant ICB topography in the study region.
## A double-layered boundary beneath southwest Okhotsk Sea
We compare stacked PKiKP and PcP waveforms for the seismic data sampling various regions of the ICB (see “Waveform stacking” in Methods). The stacked PKiKP and PcP waveforms are similar for most events, including those sampling the ICB regions beneath central China, southeast Japan Sea and west North Pacific Ocean (Fig. 3a for an example event and Supplementary Fig. 5 for other events) and some portion of the ICB region beneath southwest Okhotsk Sea (Fig. 3b in the distance range larger than 24°). Waveform modeling indicates that a sharp ICB, no thicker than 1 km, is required in those regions so that PKiKP waveforms would maintain similarities with the corresponding PcP waveforms.
However, significant waveform differences are observed between most of the PKiKP and PcP phases for event 11 (Fig. 3b), which sample the ICB region beneath southwest Okhotsk Sea (blue box B in Fig. 2b). In the frequency band of 2–3 Hz, the stacked PKiKP waveforms (with a stacking radius of 2°) exhibit two pulses in the epicentral distance range from 10° to 24°, while the stacked PcP waveforms show a simple pulse (Fig. 3b). The two energy pulses in the PKiKP waveforms have a time separation of about 0.9 s, with the relative amplitudes of the second pulses decreasing gradually with increasing epicentral distances (Fig. 3b). We perform following statistical tests and these tests confirm the robustness of the stacked results. We stack PKiKP waveforms using a smaller stacking radius of 1° and the stacked waveforms exhibit the same feature of double pluses (Supplementary Fig. 6); we stack waveforms according to propagational azimuths of the seismic waves and the stacked waveforms exhibit the same feature of double pulses with no azimuthal dependency; and we apply a bootstrap resampling method30 to estimate the uncertainties of the stacked waveforms and double peaks stand out well above the 95% confidence intervals of the stacked PKiKP waveforms (Supplementary Fig. 7). The observed distinct double pulses in PKiKP waveforms are markedly different from the PKiKP coda waves studied in previous studies31,32,33, which have characteristics of growing energy with time after direct PKiKP arrivals for a few hundred seconds without impulsive onsets.
Several lines of evidence suggest that the anomalous PKiKP waveforms are caused by the seismic structure near the ICB, as other possible sources can be excluded. The observed anomalous PKiKP waveforms cannot be caused by a complex source or near source heterogeneities. Note that the PKiKP waveforms change from a simple single pulse at the distance of 25° to double pulses at 24° (Fig. 3b). The take-off angles from the source for the PKiKP waves recorded at these two distances have a difference of only 0.13°. A complex source would produce same waveform complexities for PKiKP waves recorded at these distances. Similarly, the separation of the two seismic paths is only 0.47 km at 800-km depth, making near source heterogeneities impossible to produce such waveform change between these two epicentral distances. They cannot be caused by the subducted slab beneath Japan. If a seismic heterogeneity exists within the slab and generates a secondary phase to the PKiKP waves of event 11, such a seismic heterogeneity would have also geographically been sampled by the PcP waves of event 11 (Fig. 3b) and the PKiKP and PcP waves of other events. However, no such secondary phases are observed for the PcP waves of event 11 and PKiKP and PcP waves of other events. Theoretical calculations also indicate that a low-velocity channel even with a compressional velocity reduction of 10% would produce a secondary phase having no more than 7% of the amplitude of the main phase, unable to account for the observations. They cannot be caused by the seismic structure at the PKiKP exit regions at the CMB. Note that the reflected points of PcP waves partially overlap with CMB exit points of PKiKP waves (black dashed box in Supplementary Fig. 8) and those PcP phases exhibit simple waveforms, indicating that the CMB region near the PKiKP exit points has a simple seismic structure. They cannot be caused by the seismic structure at the PKiKP entrant regions at the CMB, as this portion of the CMB region (red triangles in Supplementary Fig. 8) exhibits undetectable seismic scattering based on the study of PKP precursors observed in Japan islands for earthquakes that occurred in South America34. Theoretical analysis further indicates that the observed secondary pulses in the PKiKP waveforms cannot be caused by a low-velocity zone above the CMB either in the entrant or exit points of the PKiKP waves. Synthetic calculations indicate that, for any possible low-velocity zones above the CMB that would produce a 0.9 s delayed secondary pulse after the main PKiKP phase, the predicted amplitude of the secondary phase is <7% of that of the main phase because of the small reflection coefficients of the compressional waves off the CMB and the top of the low-velocity zone. Such small secondary pulse would not be discernable in the data and is unable to account for the strong secondary pulses observed in the PKiKP waveforms (Fig. 3b). Based on above observations and analyses, we conclude that the anomalous PKiKP waveforms are caused by seismic structure near the ICB.
The observed PKiKP waveforms can be explained by a double-layered ICB model with laterally varying thicknesses and compressional velocity jumps. We test double-layered ICB models with thicknesses from 1 to 10 km (with an interval of 1 km) and compressional velocity jumps from 10% to 90% (relative to values of PREM29, with an interval of 10%) (Fig. 4a and b). A double-layered ICB model with a thickness of 4–8 km and laterally varying compressional velocity jumps of ~30% to ~50% can well explain the observed PKiKP waveforms in the distance range of 10–24° (Fig. 4c). Note that synthetics of the best-fitting models exhibit two energy pulses that fit the observations well, in both relative timing and amplitude (Fig. 4c).
## Discussion
The double-layered structure fits exactly the characteristics of a mushy zone. Fearn et al.5 proposed that the mushy zone should exist in the inner core condition and the inner core solidification system should consist of three regions: a lower region of complete solid, an upper region of complete liquid and a middle region, the mushy zone, in which both solid (in dendrites) and liquid are present. We thus infer that the 4–8 km thick layer with seismic properties between the outer and inner core is a mushy zone proposed by Fearn et al.5.
We further note that the double-layered structure is only found in a localized region with other ICB regions characterized by a sharp boundary. Such results could be explained by a laterally varying solidification process of the inner core due to lateral variation of composition in the bottom of the outer core. In the thermo-compositional state of a mushy zone, the top of the mushy zone is controlled by the liquidus condition of the outer core composition, while the bottom of the mushy zone by the eutectic condition of the system. The thickness of the mushy zone would thus depend on the difference between the outer core composition and the eutectic composition. The sharp feature of the ICB observed in other regions can be explained by the solidification in the condition of eutectic composition, in which the mushy zone disappears. Based on our seismic results and the study of Fearn et al.5, we suggest that, in the current thermo-compositional state of the Earth’s core, the outer core composition is close to eutectic in most regions resulting in a sharp ICB; however, in some localized regions, deviation from the eutectic composition exists resulting in a mushy zone with a thickness of 4–8 km.
## Obtaining PKiKP–PcP differential travel time residuals
PKiKP–PcP differential travel time residuals are defined as:
$dT= PKiK P obs -PKiK P pre - Pc P obs -Pc P pre$
(1)
where the subscripts “obs” and “pre” denote observed travel time and predicted travel time calculated based on a reference model, respectively. We measure PKiKP and PcP travel times based on the maximal amplitudes of the two phases, since the onsets of the two phases are difficult to be identified. For each event, we stack the PKiKP waveforms along the handpicked arrival times, re-pick the PKiKP arrival time in each trace based on its cross-correlation with the stacked PKiKP waveform and eye-check the waveforms in the final picking to avoid possible cycle skipping. Such picking procedure ensures the reliability and accuracy of phase picks. We apply the same procedure to measure the arrival times of PcP phases. We calculate PKiKP–PcP differential travel time residuals with respect to PREM29. For each event, we remove the mean value of the PKiKP–PcP differential travel time residuals from the observations.
The PKiKP travel time residuals exhibit a linear relationship with the PcP travel time residuals (Supplementary Figs. 2 and 4), indicating that most of the travel time residuals are contributed by the seismic structure of shallow Earth. The PKiKP–PcP differential travel time residuals exhibit no significant correlation with either the PcP or PKiKP travel time residuals, indicating that the effects of shallow Earth’s structure have been effectively removed from the differential PKiKP–PcP travel time residuals.
## Waveform stacking
Waveform stacking is an effective method to reduce the effects of small-scale scattering in the upper mantle and to enhance signal-to-noise ratios of the data. We first divide the PKiKP reflected regions into overlapping circular grids with a radius of R. We then collect traces whose PKiKP reflected points fall into each grid, self-normalize the waveforms and linearly stack the waveforms along the handpicked arrival times. Stacked waveforms are considered to be reliable only when the number of records used in the stacking exceeds a certain threshold N. In the present study, we choose N = 10 and R = 1° or R = 2°. These numbers are determined empirically to best balance reliability and spatial resolution of stacked waveforms. We apply the same procedure to stack PcP waveforms.
## Waveform modeling
We calculate synthetic seismograms by a two-dimensional hybrid method35, which employs a numerical finite-difference (FD) method in a localized region and analytical methods outside the FD region. In this study, we apply the FD calculations in a region encompassing the lowermost outer core and the topmost inner core. A constant grid spacing of 0.1 km by 0.1 km is used in the FD calculations. We use the solutions from the Global Centroid Moment Tensor (GCMT) project36 as focal mechanisms and the stacked PcP waveform as source time function.
Synthetic tests indicate that PKiKP waveform features have no sensitivity to the shear velocity structure in the top low-velocity layer, but some sensitivities to its density structure. However, density change alone is not sufficient to explain the waveform features. The maximum amplitude of the predicted secondary phase due to density change alone is no more than 35% of the main phase, while the amplitude of the secondary phase reaches 66% of the main phase in the observations. In the final model, we assume that shear velocity and density of the top low-velocity layer change in a same way as compressional velocity.
The double pulses observed in Fig. 3b cannot be fit by transitional ICB models, defined as that seismic medium parameters (compressional and shear wave velocities and density) increase linearly from the outer core values to the inner core values within a certain thickness d across the ICB (Supplementary Fig. 9a). Synthetic tests indicate that none of transitional ICB models, with the transitional thickness ranging from 1 to 10 km, could produce a secondary pulse in the synthetic waveforms and explain the observations (Supplementary Fig. 9b).
## Data availability
The data that support the findings of this study are available from the National Research Institute for Earth Science and Disaster Resilience (NIED), Japan, under its data policy. Data are also available from the authors upon request and with the permission of data redistribution granted by NIED.
Publisher's note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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## Acknowledgements
We thank the National Research Institute for Earth Science and Disaster Resilience (NIED), Japan, for providing high-quality seismic data recorded by the high-sensitivity seismograph network (Hi-net). This work was supported by Key Research Program of Frontier Sciences, Chinese Academy of Sciences, #QYZDY-SSW-DQC020, CAS, and by the National Natural Science Foundation of China under Grants NSFC41674045 and NSFC41130311. The Generic Mapping Tools37 was used to create the figures.
## Affiliations
1. ### Laboratory of Seismology and Physics of Earth’s Interior; School of Earth and Space Sciences, University of Science and Technology of China, Hefei, Anhui, 230026, China
• Dongdong Tian
• & Lianxing Wen
2. ### Department of Geosciences, State University of New York at Stony Brook, Stony Brook, New York, 11794, USA
• Lianxing Wen
## Contributions
D.T. processed and analyzed data and performed numerical simulations; L.W. designed and supervised the project; L.W. and D.T. wrote the paper and commented on the manuscript at all stages.
## Competing interests
The authors declare no competing financial interests.
## Corresponding author
Correspondence to Lianxing Wen.
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# Tag Info
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You are indeed allowed to have unreachable states in DFAs (or NFAs, or various other automata models). You may wonder what is the point of having them, given that they are unreachable and have no effect on the language. Well, automata are typically not designed by hand. Rather, they are usually obtained algorithmically by translation from other types of ...
8
Check the definition. It usually goes like $M = (Q, \Sigma, \delta, q_0, F)$, with $Q$ a finite set of states, $\Sigma$ the (input) alphabet, $\delta \colon Q \times \Sigma \to Q$ the transition function, $q_0 \in Q$ the initial state, $F \subseteq Q$ the set of final (accepting) states. Note that the only condition on $\delta$ is it being a function. To ...
3
Your proof that the problem is in PSPACE is not quite correct. The problem is that the product $n_1 \cdots n_k$ is not bounded by a polynomial in the input length $n_1 + \dots + n_k$. The correct way to do it is to directly apply Savitch's theorem to the NPSPACE machine that nondeterministically guesses a path through the product graph. The difference is ...
2
Yes, this is (kind of) correct, in the following sense: The set of languages that can be recognized by Turing Machines with a fixed tape is equivalent to the set of regular languages (i.e. the set of languages recognizable by finite automata). The reason behind this is that such Turing Machines have a finite number of configurations, which is independent ...
2
The problem is PSPACE-complete, by reduction from the equivalence problem for NFAs. Suppose we are given two NFAs $A_1,A_2$ over an alphabet $\Sigma$. Let $\dashv$ be a new symbol. To each of the NFAs we add two new states $q_f,q_t$, and the following transitions: An $\epsilon$-transition from the initial state to $q_t$. A self-loop on $q_t$ labeled $\... 2 If you know how to construct DFAs for two languages separately, you can use systematic methods to build the DFA which recognises the intersection of the languages (e.g. https://math.stackexchange.com/questions/147457/intersection-of-two-deterministic-finite-automata). However to achieve a minimalist and elegant result (as well as for learning purposes) you ... 2 There are many ways to perform this enumeration. Note first that you can count the number of cases in half by postulating that$Q = \{q_0,q_1\}$, that is, fixing which among the two states is initial; the choice doesn't matter due to symmetry. Now you are left with 16 possibilities. At this point there are many ways to cut the search space even further. Here ... 2 Yes, your understanding on point 2 is correct. Your understanding on point 3 is mostly correct, except where you say "mark final state as explained in 1st bullet point in point 1". That statement doesn't type-check, as you can't do that to the$\epsilon$-NFA. If you've created an$\epsilon$-NFA, then its states are of the form$S_1$or$S_2$(where$S_1$... 2 Let$M$be a DFA for the language$L$and let$M_1$and$M_2$be$M$with two different collections of some number$m$of unreachable states added.$M_1$and$M_2$both have the same number of states and both accept$L$, but they need not be isomorphic. For a less trivial example, consider the following two DFAs: Start state: 1 Accepting states: 2, 4 1 -a-&... 1 As Rick Decker rightly noted, real-world lexical analysers produce a DFA with multiple final states. Lexical analysers created via Lex-like generators differ from theoretical DFAs in several respects, and this is only one of them. Another obvious difference is that they don't stop at the end of the input string. Instead, lexical analyser generators ... 1 This is a DFA for binary numbers divisible by three. your desirable language is reverse of this language. For reversion of language of a DFA, you must reverse all of its transitions and change its initial state with its final state. in this case initial and final states are same, also its language is symmetric so all of transitions are symmetric. Therefore ... 1 Yup, It does behave like a finite state machine. if for (every permutation in tape string, every current head position, every next move) you create a unique state, you are besically creating a DFA like mechanism, as (next move/tape write) is determined by those variables and you have a state for every possible combination. And as the tape is finite, the no ... 1$L_1\circ L_2$can certainly still be non-regular. For example, when$L_1$contains exactly one word. However,$L_1\circ L_2$can be regular, too. Here is an example. Let$L_1=\{\epsilon,0\}$. Let$L_2$be the complement of$\{ 0^n1^n \mid n \gt 0 \}$. As the complement of a non-regular language,$L_2$is not regular while$L_1\circ L_2\$, the set of all ...
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# Convert ML to MG or Millilitre to Milligram Calculator
The Millilitre (ml or mL, also known as millilitre) is a metric unit of the volume, which is equal to 1000 of a liter. ML is a non-SI unit that received to relate to the International System of Unit (SI). It is the same as the one cubic cm (cm 3, or non-standard, CC).
### Use this conversion tool to convert quickly and easily between MG and ML.
Conversion Calculator
ML MG ML MG ML MG ML MG
0.01 ml =100 mg 2.1 ml =2100 mg 4.1 ml =4100 mg 7 ml =7000 mg
0.2 ml =200 mg 2.2 ml =2200 mg 4.2 ml =4200 mg 8 ml =8000 mg
0.3 ml = 300 mg 2.3 ml = 2300 mg 4.3 ml = 4300 mg 9 ml = 9000 mg
0.4 ml = 400 mg 2.4 ml = 2400 mg 4.4 ml = 4400 mg 10 ml = 10000 mg
0.5 ml = 500 mg 2.5 ml = 2500 mg 4.5 ml = 4500 mg 11 ml = 11000 mg
0.6 ml = 600 mg 2.6 ml = 2600 mg 4.6 ml = 4600 mg 12 ml = 12000 mg
0.7 ml = 700 mg 2.7 ml = 2700 mg 4.7 ml = 4700 mg 13 ml = 13000 mg
0.8 ml = 800 mg 2.8 ml = 2800 mg 4.8 ml = 4800 mg 14 ml = 14000 mg
0.9 ml = 900 mg 2.9 ml = 2900 mg 4.9 ml = 4900 mg 15 ml = 15000 mg
1 ml = 1000 mg 3 ml = 3000 mg 5 ml = 5000 mg 16 ml = 16000 mg
1.1 ml = 1100 mg 3.1 ml = 3100 mg 5.1 ml = 5100 mg 17 ml = 17000 mg
1.2 ml = 1200 mg 3.2 ml = 3200 mg 5.2 ml = 5200 mg 18 ml = 18000 mg
1.3 ml = 1300 mg 3.3 ml = 3300 mg 5.3 ml = 5300 mg 19 ml = 19000 mg
1.4 ml = 1400 mg 3.4 ml = 3400 mg 5.4 ml = 5400 mg 20 ml = 20000 mg
1.5 ml = 1500 mg 3.5 ml = 3500 mg 5.5 ml = 5500 mg 30 ml = 30000 mg
1.6 ml = 1600 mg 3.6 ml = 3600 mg 5.6 ml = 5600 mg 40 ml = 40000 mg
1.7 ml = 1700 mg 3.7 ml = 3700 mg 5.7 ml = 5700 mg 50 ml = 50000 mg
1.8 ml = 1800 mg 3.8 ml = 3800 mg 5.8 ml = 5800 mg 70 ml = 70000 mg
1.9 ml = 1900 mg 3.9 ml = 3900 mg 5.9 ml = 5900 mg 90 ml = 90000 mg
2 ml = 2000 mg 4 ml = 4000 mg 6 ml = 6000 mg 100 ml = 100000 mg
If you are one of them who admires in calculation or number conversion, here we are going to explain that how can you transform Milliliter to Milligram or ML to ML. This is the most common question that trouble to all peoples who want to grasp the fundamentals of measurement for the things they purchase.
Millilitre and Milligram both are the standard units of SI System measurement that applied in the world recognized metric system to convert values. The most common thing between MG and ML is the density.
## What is Density?
Density is a term which defines the overall amount of substance which the mass per unit volume. Density indicates by the R, although the Latin letter D also can use to mean density.
The formula to calculate the density is; p=m/v
Where,
P=Density
M=mass
V=volume
## How to convert ML to MG?
It is not a something new, but, necessary for everyone to know that one cannot convert Millilitres to Milligrams without any calculation. First, you have to know the step by step procedure to change ML to MG.
### Let’s an example to convert ML to MG:
Here we are using the density of water= 1.24 g/mL
And, you want to convert 28 ml (water) into mg,
Then,
Density will be 28×1.24=34.72 g.
Now, calculate the value of ml by multiplying by 1000 in the 34.72
34.72×1000= 34,720 mg Commonly 1 ml= 1000 mg
Note: The calculation is depend on your number. It’s a simple example to explain the calculation only.
### How many Millilitres in 1 Milligram?
1 Millilitres (ml) = 1,000 Milligram (mg)
#### Quick conversion chart of millilitre to milligram [water]
1 milliliter to milligram [water] = 1000 milligram [water]
2 milliliter to milligram [water] = 2000 milligram [water]
3 milliliter to milligram [water] = 3000 milligram [water]
4 milliliter to milligram [water] = 4000 milligram [water]
5 milliliter to milligram [water] = 5000 milligram [water]
6 milliliter to milligram [water] = 6000 milligram [water]
7 milliliter to milligram [water] = 7000 milligram [water]
8 milliliter to milligram [water] = 8000 milligram [water]
9 milliliter to milligram [water] = 9000 milligram [water]
10 milliliter to milligram [water] = 10000 milligram [water]
5 ml to mg [water] = 5000 milligram [water]
1 ml to mg = 1000 milligram [water]
.5 ml to mg = 500 milligram [water]
## What is the method to convert ML to MG?
If you want to convert ML to MG, you have to understand how many ML in an MG. Therefore, we have mentioned the process to Convert Millilitre to Milligram by an example.
Let’s assume that you have water with 15 mg/ml. Now calculate how many millilitres in 40 mg with the likewise density.
Step 1:
(45 mg/1)*(1 ml/15 mg)
=45 ml/15
= 3 ml
Conclusion: By applying this method, you can observe the value of ml in mg. However, we recommend everyone doesn’t go for the lengthy and challenging process of calculation by implementing the formula. Just go with the ML to MG or MG to ML conversion calculator which we have listed on the top of our website.
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Your scores are swapped over into a percentile. Your ASVAB scores can be really high in one score, and medium to low in another, and you would still pass the ASVAB. Quotas – when the recruiters have all the people they need, they can be picky on who they let in. The Verbal Expression (VE) score is used to determine qualifications for many military jobs in all the branches, and it’s used to help determine your AFQT score. Scores from the ASVAB are also used to clas-sify individuals into jobs they are likely to perform successfully. The previous requirement for the Armed Forces Qualification Test was 66 but in 2004 was changed to 65. It is then converted to a scaled score ranging from 20 to 62. The ASVAB-AFCT Test. If you need help, or want to ensure that your ASVAB scores are high, check out our “ASVAB Study Guide” page. People often refer this term to “minimum ASVAB score”. Find out what scores test-takers typically earn on each section of the ASVAB. For enlistment into the Army you must get a minimum ASVAB score of 31. On our Army MOS Listings page, click on the MOS you are interested in. For current AFQT scores, the reference group is a sample of 18 to 23 year old youth who took the ASVAB as part of a national norming study conducted in 1997. As you can see here, the minimum ASVAB scores needed for enlistment vary somewhat among the different military branches. A Standard Score indicates how many units of the standard deviation a particular score is above or below the mean. Composite ASVAB scores are determined as follows: Like mentioned before, your ASVAB scores determine what Army Jobs (MOS) you are eligible for when enlisting, or re-enlisting into the US Army. This chart allows juniors to take your ASVAB AFQT score and it will predict the range of scores you may earn when you take the ACT or SAT college entrance exams asvab to act score conversion.pdf , 40.903 KB; (Last Modified on August 22, 2017) Nationwide and across all branches, the “average” ASVAB score is between 42 and 50. The scores listed in the report are all percentages between 1 and 99 percent. The maximum ASVAB score is 99. The score is derived from the probability score on the nations 18-23 year old youth. Look at your scores for each sub-test. This means that of all people taking the ASVAB test the average score will be about 50. The Army converts the ASVAB subtest scores into 10 composite score areas, known as “line scores.” The line scores determine what Army jobs an individual qualifies for. Find the Arithmetic Reasoning (AR) and Mathematical Knowledge (MK) scores on the ASVAB score sheet. The Army converts the ASVAB subtest scores into 10 composite score areas, known as “line scores.” The line scores determine what job(s) an individual qualifies for. ASVAB Score Calculator. What You Need To Know About ASVAB Scores The ASVAB scores are used for two different purposes in the US military recruitment - AFQT scores for deciding the eligibility into military enlisting and ASVAB composite or line scores for determining the job roles for prospective candidates. As stated on the ASVAB-AFQT page the Armed Forces Qualifying Test (AFQT) is not a test by itself but a collection of tests within the ASVAB. About 16% of the population will get a ASVAB score of 60 or higher. Enter one or more ASVAB subtest scores in the score calculator and click the "Show Jobs" button to show all the US Military jobs that those ASVAB scores qualify you for. To join the Air Force as an enlisted member you must usually take the Armed Services Vocational Aptitude Battery (ASVAB) test and get a decent score. The AFQT score criteria and ASVAB line score requirements are different for the various branches like army, navy, air … Composite scores, also called line scores, help determine which military job is right for you, and some composite scores are required for certain jobs and branches. An average AFQT score is a median score of 50. On the MOS details page, you’ll see if there are any ASVAB restrictions assigned to that specific MOS. Look at the old ASVAB scores. Your ASVAB scores are broken into multiple scores: Standard Score and AFQT score. Understanding the ASVAB Score. There are eight total subtests including: A “blowout” score on the ASVAB is anything 92 and above. Each subtest uses a score range from 1 to 100, making 50 the average score. To find out what scores you need to enlist click here: Minimum ASVAB Scores Needed To Enlist. Exceptions may be made, however, for a handful of high school graduates who can score as low as 31. AFQT (AFQT) Your AFQT score can be calculated by combining the scores from the following portions of the test: AR + MK + WK + PC. The AFQT score is used to determine whether a candidate is eligible to enlist in the military. Composite Scores. The overall ASVAB score is known as the AFQT score , or Armed Forces Qualification Test score. How is the ASVAB score calculated? ASVAB Scores are composite scores created by combining different subtests of the ASVAB, in order to best determine your strong and weak areas of knowledge. The Maximum Composite Scores are: • AFQT – 99 • GT – 151 • MM – 161 • EL – 151 • CL – 141. Here is the tier chart for the AFQT scores: AFQT Category Score Range; AFQT Tier Table: I: 93-99: II: 65-92: IIIA: 50-64: IIIB: 31-49: IVA: 21-30: IVB: 16-20: IVC: 10-15: V: 1-9: Calculating your ASVAB-AFQT Score. Conclusion. The Army utilizes the composite scoring table below to create your final ASVAB scores, and then assesses what Army Jobs (MOS) you are eligible for. Instead, your scores indicate how you … The verbal expression (VE) part of the ASVAB is really important. The ASVAB subtests can be found on the “ASVAB Subtests” page. The ASVAB provides an overall score, but it also has subtest scores. Your AFQT raw score is computed by doubling your VE Score and then adding your AR and MK scores… and then putting that score into a percentile. These sub-tests measure your abilities in various categories that apply to military service and also to a variety of career options, including math, science, reading comprehension, and mechanical knowledge. The Armed Services Vocational Aptitude Battery, referred to as the ASVAB, is a multiple-choice exam with subtests in ten different subject areas. After determining the VE score, multiply it by two, then add the AR and MK … For Army schools just as Rangers, Special Forces, and many more, your GT score needs to be 107 or greater. Standard scores: The various subtests of the ASVAB are reported on the score cards as standard scores. The most important point to take away from this article is that you need an AFQT score of 31 with a high school diploma or 50 with a GED to enlist in the Army. Thus, a Standard Score of 40 indicates that the examinee scored 1 … The ASVAB is refined on an annual basis to have a median score of about 50. ALL RIGHTS RESERVED. Waivers – They are much more likely to grant waivers to someone who scores 87 than 37. In the case of the ASVAB subtests, the mean is set to 50 and the standard deviation is set to 10. What is an Excellent AFQT Score? The percentile you scored in has only changed slightly. After taking the ASVAB, you will receive a report detailing your scores. AFQT scores range from 1-99 and represent the examinee's percentile. Also many times when recruiting numbers are high the minimum scores needed to enlist get raised. The current AFQT score is the most meaningful ASVAB score for you. So if you score a 44, then you scored slightly below average. While the ACT and AFQT score DO compare, the line scores for various jobs do not. ASVAB Scores are composite scores created by combining different subtests of the ASVAB, in order to best determine your strong and weak areas of knowledge. For the majority of MOS’s, MOS restrictions for ASVAB scores are not too strict. The Air Force and Army have the lowest minimum AFQT score (31) for high school diploma holders, while the Coast Guard has the highest (40). The Armed Services Vocational Aptitude Battery (ASVAB) is the selection test administered to military applicants to determine their eligibility for service. The score you see is based on the number of questions you answered correctly compared to other test takers, and ranges from a percentile score of one to 99. What are the ASVAB Percentile Scores? Air Force recruits must score at least 36 points the 99-point ASVAB. COPYRIGHT © 2020 BOOTCAMP4ME. This site is owned and operated by Bright Mountain Media, Inc., a publicly owned company trading with the symbol: Military Entrance Processing Station (MEPS). The Army utilizes the composite scoring table below to create your final ASVAB scores, and then assesses what Army … A good ASVAB score is important in securing a position in any military branch. Your VE score is computed using adding your Word Knowledge (WK) raw score to the Paragraph Comprehension (PC) raw score. Your ASVAB scores will be divided into individual scores for each sub-test, or category. Each Army job is given a code known as a Military Occupational Specialty (MOS). If you scored a 88… then you scored higher than 88% of all the nation 18-23 year olds probably score. Your ASVAB scores can be really high in one score, and medium to low in another, and you would still pass the ASVAB. No matter what MOS you have in the Army, the most commonly referred to composite score is the “GT” (General Technical) score. For all military branches, you’ll need an AFQT score of at least 50 if you have a GED. You can also learn how the test is scored and how results are used to determine your eligibility to enlist. The minimum ASVAB score to join the US Navy is an AFQT of 35, placing it 1 point lower than the Air Force. I know this is confusing, try to keep up. The Standard Score that you’ll receive is a score for each section of the exam. Armed Forces Qualifying Test score, or AFQT, is the military term for minimal enlistment requirements. All Air Force Specific Army Specific Navy Specific Show Jobs. Why Are Good AFQT Scores Important? As you can see ASVAB and AFQT scores can be complicated. The following AFQT score chart shows the requirements needed to join a specific military branch: The minimum Armed Forces Qualification Test (AFQT) score on the Armed Services Vocational Aptitude Battery (ASVAB) required for enlistment to the active duty Navy or the Navy Reserve for non-prior service (NPS) applicants is a 31QT. Write this value down and take out the ASVAB scores again. Multiply the VE value by 2: 2_VE Assuming the maximum scores were achieved, and the VE value is 62, then 2_62 is 124. Thus, an AFQT score of 90 indicates that the examinee scored as well as or better than 90% of the nationally-representative sample of … To enlist in the U.S. military, you have to meet minimum enlistment requirements. The benefits of getting a higher AFQT score include: Here is the tier chart for the AFQT scores: The verbal expression (VE) part of the ASVAB is really important. Line Scores are what determine your job qualifications. The ASVAB is administered each year to about 1 million people who apply for An ACT score of 28 translates to an AFQT score between 93-99. ASVAB Score Requirements. The smaller tests make up the total percentage scored on the ASVAB. Here is the formula for AFQT score calculation: The U.S. military officially started to use the ASVAB test in 1976. To compute your VE score, the military adds the number correct (1 point per correct answer) of the Paragraph Comprehension (PC) and the Word Knowledge (WK) subtests and then compares the results to the info in the following table. The total ASVAB scores further grouped into various clusters or lines are used for the classification of selected candidates into appropriate job functions. A score that high or higher will guarantee you have access to the best MOS available, including nuclear work for the Navy or work as a Ranger in the army. These ASVAB scores do not refer to the percentage of questions you answered correctly. It factors in to not only your AFQT score, but your Line Score. When I recruited, we had a chart that used the ACT and the SAT to see how our applicants would do on the ASVAB. The Armed Services Qualification Test (AFQT) is a percentile score based on the study of 1997, where the Department of Defense conducted the ASVAB test in which 12000 people took part. A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form $${a \over b}$$ where a is the numerator and b is the denominator.An improper fraction ($${5 \over 3}$$) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number ($$1 {2 \over 3}$$) which has a whole number part and a fractional part. 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# Distance from the surface as an input node for use with volumetric materials
I would like to create a material that has a volumetric component whose local Density depends on the depth of each volume point below the surface.
Can the information "how deep below the surface is a given point" (i.e. the intensity of red color in the figure below) be obtained directly from the material nodes, without adding extra geometry or modifiers?
I've tried using the Z component of the UV Texture Coordinate (after having assigned an UV map), and other kinds of node setups, but I couldn't find a way to obtain that quantity.
Another way of looking at it: consider that SSS can be seen as a Volume Scatter with a Density decreasing smoothly from the surface to one skin-thickness below, while an abrupt fall-off (density goes suddenly from one to zero) corresponds to a Volume Scatter applied on a mesh with a Solidify modifier.
I would like to be able to create any generic falloff profile, including these two.
• – user1853
Aug 30, 2017 at 18:09
• Use a Point Density texture using either the vertices (if they're dense enough) or a particle system emitted from the faces. Aug 30, 2017 at 18:57
• @RichSedman thanks for the suggestion, however it would require extra data at the Object level (a particle system) or extra conditions on the mesh (evenly distributed vertices, while I'd like to apply this to any manifold surface) Aug 30, 2017 at 19:30
• @cegaton thanks for these links. They are quite specific (the first is bound to either constant density or SSS-defined falloff, the second requires an a priori knowledge - and a possibly easy analytical description - of the mesh's shape), but at least they help me understand that probably there is no easy way to obtain this information given the available node tools. Hopefully somebody will like to differ! Aug 30, 2017 at 19:34
• There is no such functionality in Cycles yet, your only option is OSL. The reason is that this requires raytracing inside shader evaluation, which is bad for gpu performance as Cycles is implemented. That's also the reason why there is no AO shader with value output (only with closure output), no true "pointiness" shader but a fake one or no rounding of edges at render time. These all require some coding to be done to Cycles (or some OSL coding). Aug 31, 2017 at 8:41
You can use an OSL shader that measures the proximity of the mesh from within a volume. It works by projecting out rays from each point in the volume to probe the surrounding mesh. Be warned that this will be extremely CPU intensive as it has to trace multiple rays from each point within the volume which is significantly extra CPU work.
Here's the shader in action, varying the emission color based on depth :
Here's the code :
shader volume_meshproximity(
vector Point = P,
int Iterations = 1,
output float Min_Distance = 0.0,
output float Max_Distance = 0.0,
output float Avg_Distance = 0.0
) {
float x;
float y;
float z;
vector rand;
float scale = 5000.0;
float sum_dist = 0.0;
float max_dist = 0.0;
float min_dist = 0.0;
float Distance = 0.0;
for (int loop = 0; loop < Iterations; loop++)
{
rand = noise(Point*scale,loop*scale);
x = rand[0] - 0.5;
y = rand[1] - 0.5;
z = rand[2] - 0.5;
if(trace(Point,normalize(vector(x,y,z))))
getmessage("trace", "hitdist", Distance);
//getmessage("trace", "N", Normal);
//getmessage("trace", "hit", Hit);
if (loop == 0)
{
sum_dist = Distance;
max_dist = Distance;
min_dist = Distance;
}
else
{
sum_dist = sum_dist + Distance;
if (Distance > max_dist)
{
max_dist = Distance;
}
if (Distance < min_dist)
{
min_dist = Distance;
}
}
}
Max_Distance = max_dist;
Min_Distance = min_dist;
Avg_Distance = sum_dist / Iterations;
}
To use it, make sure you're using a version of Blender compiled to include OSL support and enable the Open Shader Language checkbox in the Render properties.
Create a new Text block and paste the above code. Name it something like 'surfaceproximity.osl'.
In your Node Editor, add a Script block and select the 'surfaceproximity.osl' text as its source. This should give you a node like this :
The 'Iterations' control how many 'probe rays' are projected out from each volumetric point - more rays will provide slower renders but sharper results (since it will sample more internal geometry of the mesh). Point is the location in World space for the origin of the probe rays (pass the Object texture coordinate through a Vector Transform node to transform the point from Object to World coordinates). The resultant 'Min', 'Max', 'Avg' are the minimum detected distance (ie, the closest point) the maximum detected distance (the furthest away point) and the average distance of all probe rays (gives an indication of how 'enclosed' that part of the volume is (a thin section of mesh will tend to have a smaller average, a wide open section will have a larger average - when compared to the min and/or max).
The 'Min_Distance' can be used to indicate the depth from the surface of the mesh. For example, it can be used to control the density of volumetric Scatter/Absorption to give Suzanne an internal skeleton as follows :
Here the OSL shader is set to project 16 probe rays (Iterations) from each point within the volume. This controls the density of the Volumetric Scatter such that closer than 0.1 blender units (the Greater Than node) it is transparent (density 0) whereas further in than 0.1 blender units it is opaque (density 30). An Emission shader makes the 'skin' of Suzanne surrounding the 'skull' visible to show the effect. Adjusting the 0.1 in the Greater Than node will vary the thickness of the 'skin' (ie, the depth at which the 'skull' begins).
This produces the following results :
Blend file attached.
Similarly, you could use a Less Than node to create a 'Solidify' effect (the inverse of the above example) or drive the Volume density directly from the output of the OSL node to vary the properties smoothly - as mentioned in the original question.
• I connected it to Object texture coordinate. Pretty sure it needs to be in World coordinate space so you might need a Vector Transform node to translate it if your mesh isn't at the origin or is rotated/scaled. I will iron out any such issues and update the answer later. Aug 31, 2017 at 15:57
• @NicolaSap I've ironed out a couple of bugs from the code and added an example (to give suzanne a skull based on depth below the surface of the mesh as a demonstration) along with a Blend file to download. I'll try and put together an example using a smooth rather than abrupt fall-off of density - although that's more difficult to visualise than a definite edge. Sep 1, 2017 at 0:21
• I haven't used it yet, but from you examples it looks exactly like I wanted, so I'll accept it straight away. Thank you! Sep 1, 2017 at 9:28
• @Rich Sedman, brilliant work with nodes. Thank you for the file. Mar 1, 2019 at 17:35
• That looks nice! Did Blender come up with a new way of doing that? I mean, when writing this comment, it' barely a few days to Blender 3.0 release, is there a node replacing your very nice code? Maybe something less CPU intense. Nov 25, 2021 at 17:57
Prompted by SteeveDroz's comment on Rich's answer, I checked what Blender 3.0 (in fact, 2.83+) can do.
I came out with a non-OSL solution (see the right hand side of the screenshot ↓; the left hand side is Rich's script).
The trick is using the Ambient Occlusion input in "Inside" and "Only Local" mode.
• However, it's not perfect (see the corner of the "phantom cubes" inside my cube: they are supposed to be sharp but are not).
• Also, I'm not sure what kind of decay curve it employs, but the result is that you don't have a linear relationship between distance and output, and the scaling you control ends up depending on two parameters: the internal "Distance" node and the multiplying factor you choose to use.
It can be a quickier and built-in workaround if you don't want to use a OSL script. However, I think that the Ambient Occlusion node has similar limiation to the OSL script: it doesn't seem to work on GPUs, at least not mine, and it's based on casting probing rays (so accuracy comes at a cost).
Edit: another image to demonstrate the effect. I'm quite surprised. It only works if OSL (CPU) is enabled
• That’s very interesting - thanks for doing that analysis. I wonder what the Ambient Occlusion node is actually doing for volumes… it looks like it could be providing the ambient occlusion at the nearest surface point to that point in the volume - which is presumably why it produces such different results for the cube. Nov 26, 2021 at 11:45
• Interestingly, it only works of "OSL" is checked in the rendering tab Nov 26, 2021 at 14:27
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When you work out a sum with more than one operation, eg 8 + 2 × 3, follow the BIDMAS rule. It is easier to multiply the numbers in scientific notation. b. Move the decimal right 3 places What total distance (in mm) did Courtney cover during her run? 1.182 × 103 × 102 Move the decimal left 5 places Question 23. The error student makes is he multiply the terms instead of addition. 1 metric ton = 2 × 1030 ki ÷ 2 × 1027 = (10)³ = 1,000 kgs in one metric ton, Question 30. ______________, Explanation: 5 × (10)4, Question 22. Which type of waste has the lowest recovery ratio? Do you agree with this student? Move the decimal right 5 places 713, Question 16. The quotient rule tells us that we can divide two powers with the same base by subtracting the exponents. Type below: (1.25 + 0.50 + 3.25) × 102 7 × 106 − 5.3 × 106 Move the decimal left 5 places Sample: English Irregular Verbs. ____________, Explanation: Question 5. x^-a = 1/x^a A quadratic equation is an equation that could be written as. = 62+2+2+2 4 × 102 (a12-9 ) = (a3 ) 6.2 × 105 − 2.6 × 104 − 1.9 × 102 Move the decimal left 2 places Answer: 2,400 words /hour Battleship. 90000000000, Question 13. Polly’s parents’ car weighs about 3500 pounds. Persevere in Problem Solving ______________, Explanation: − ? 77.76 times, Question 20. 2 × 10-2, Question 12. Llama ____________ ⋅ ? Explanation: 1.5 × 10-5, Question 7. Type below: 12)² 0.000000000182 Write your answer in scientific notation. Very large numbers are written in scientific notation using positive exponents. The division is the opposite of multiplication, so it makes sense that because you add exponents when multiplying numbers with the same base, you subtract the exponents when dividing numbers with the same base. Ask for players’ emails to identify them. 9.6 × 10-1 m = 0.96 Total weight of the claves = 230 × 20 = 4600 That makes X raised to the 1st. (4.503×109 km)/(5.791 × 107 km) a = (125)1/3 Standards Documents • High School Mathematics Standards • Coordinate Algebra and Algebra I Crosswalk • Analytic Geometry and Geometry Crosswalk New Mathematics Course • Differential Equations Mathematics Teacher Support • 2020 Guides for Effective Mathematics Instruction (K-12) NEW • Georgia Mathematics High School Teacher Professional Learning Community Explain why the exponents cannot be added in the product 123 ⋅ 113. ______________. What is the total amount of paper, glass, and plastic waste recovered? ______________, Answer: ax 2 + bx + c = 0 . A fraction raised to a negative power is equivalent to the recriprical fraction raised to a positive power. Answer: Which distance is the greater distance and about how many times greater is it? Diana calculated that she spent about 5.4 × 104 seconds doing her math homework during October. Type below: 5). The exponents are not same, therefore it is impossible to add it. product of 8 × 106 and 5 × 109 ______________, Explanation: Type below: 2−5 = (1/2)5 = (1/2) × (1/2) × (1/2) × (1/2) × (1/2) = 1/32, Explanation: ? Type below: Base = 2 Type below: But avoid … Asking for help, clarification, or responding to other answers. 3.397 × (10)6 1) ) ) This PDF book include pre algebra test answer key document. ______________, Explanation: 118200, Question 25. 1017/109 = 1017-9 = 108 When you add or subtract in scientific notation, you have to make the exponents the same before you can do anything else. Move the decimal left 4 places 1.5 × 10-2m ; 1.2 × 102 m ; 5.85 × 10-3 m ; 2.3 × 10-2 m ; 9.6 × 10-1 m. Therefore, it wouldn't make sense to add (or subtract) exponents of different bases. If you have the same number with a different exponent (For instance 5 3 X 5 2 ) just add the exponents and multiply the bases as usual. Type below: ______________. ______________. _____________, Explanation: 4.6 × 103, Question 26. Anything raised to the zeroth power is 1. 100,000 ______________, Explanation: c. 7.867 × 108 passengers x^7 / x^2 = x^7 * x^-2. Type below: a. Question 37. 7 cm is a whole number while 0.07 m is a decimal number ______________, Explanation: Type below: For example: 4 3, x + x =. Write the numbers in scientific notation. 8 × 105 For example, to solve for 3 to the fourth power, you would multiply 3 by 3 by 3 by 3 to get 81. ______________, Explanation: 8.3 × 10-4 4 × 10-5 Type below: ($9.06 × 1012. Students must go through the solved examples to have a complete grip on the maths and also on the way of solving each problem. (62) ÷ (62)² (4.94 × 1013 )/(2.6 × 1010 ) 5 × 3.38 = 16.9 By the time they are learning first grade math, kids should be ready to tackle things like the relationship between addition and subtraction, the concept of adding and subtracting two-digit numbers and learning to count beyond 100. The size of a terabyte is the product of the size of a kilobyte and the size of a gigabyte. 109+12 = 1021 When you raise 2x to the third power, you get 8x^3. 4 × 10-5 Type below: The table shows the number of tons of waste generated and recovered (recycled) in 2010. Type below: 106+12 = 1018 Base = 7 or Answer: 0.0417 ______________, Explanation: The U.S. public debt as of October 2010 was$9.06 × 1012. To divide two numbers in scientific notation, divide their coefficients, and subtract their exponents. 9E1 3.4 × 104 + 7.1 × 105 $6.69 × 101 How do figure out what X squared plus X to the negative 2 is? 0.00585, 0.015, 0.023, 0.96, 120, Question 36. Move the decimal right 9 places Move the decimal left 9 places Options: 5400, Question 14. ______________, Explanation: ⋅ ?) 0.641 × 103 9.999 × 104 Score points by answering questions correctly. 0.0000672 4.87 × 10-4, Question 2. Lesson 2: Scientific Notation with Positive Powers of 10, Lesson 3: Scientific Notation with Negative Powers of 10, Lesson 4: Operations with Scientific Notation, Explanation: Type below: Exponent = 0 How do you think it would display 1.89 × 10-12? It can lift up to 1.182 × 103 times its own weight. Jerod’s friend Al had the following homework problem: 384,000 0.000089, Question 14. Type below: Options: 7.86/3 = 2.62 ______________, Question 13. 4.2 × 106 + ? Type below: 3.582 × 10-6 3.5 × 10-3, Question 19. A student simplified the expression $$\frac { { 6 }^{ 2 } }{ { 36 }^{ 2 } }$$ as $$\frac{1}{3}$$. 0.7 = 7 × 10¯¹$3.428 × 104 To determine whether a number is written in scientific notation, what test can you apply to the first factor, and what test can you apply to the second factor? Move the decimal right 10 places Move the decimal right 5 places 8 × 105 References. ______________. ______________, Explanation: The distance from Earth to Neptune is greater by (22)³ or 10,648 miles, Question 24. b-8 The little brown bat can eat 10500 mosquitoes in 10.5 hours. It cancels itself because ^2 and ^-2 are opposites. 1.2E16 Type below: ______________, Explanation: Explain how you can compare them without writing them in standard notation first. If there are 6 rows in the triangle, what is the total number of cubes in the triangle? (0.7776 × 102 km) Type below: Type below: Question 24. 8.5 × 103 − 5.3 × 103 − ? Last Updated: March 28, 2019 Base = 3 Based on this sample, does it appear that –an = (–a)n? = 6? Write each number using scientific notation. Bases are common. At that rate, how many krill can an adult blue whale eat in 3.65 × 102 days? 7.44 × 105, Question 17. Type below: ____________, Question 5. Express in scientific notation how many mosquitoes a little brown bat might eat in 10.5 hours. 0.000000056 Write your answer in scientific notation. 4 × 102 List three ways to express 35 as a product of powers. $$\frac { { y }^{ 25 } }{ { y }^{ ? } So the base is 2 and the exponent is 5 Question 12. 510 ⋅ 5 ⋅ 5 = 5? Type below: Communicate Mathematical Ideas If you were as strong as this insect, explain how you could find how many pounds you could lift. Type below: \(\frac { { 3.46×10 }^{ 17 } }{ { 2×10 }^{ 9 } }$$ –an when a = 3 and n = 2, 3, 4, and 5. The “power rule” tells us that to raise a power to a power, just multiply the exponents. The volume of a drop of a certain liquid is 0.000047 liter. Why do you subtract exponents when dividing powers with the same base? As the exponent of 10 is negative zero’s need to add to the left of the number. Type below: a. Courtney takes 2.4 × 104 steps during her a long-distance run. 0.0000065 meter (1.8 × 109)(6.7 × 1012) 0.02 a = 5. If you need to research these muscles then you are on a time limit. An adult has about 4.94 × 1013 cells. 0.000059 1849900000, Question 9. Scientific notation is used for either very large or extremely small numbers. Toddler Worksheets.By Nadine Fischer. Type below: ______________, Answer: To download free answer key for pre-algebra test you need to Keystone Algebra I Review 1. Explanation: 58,927 Decimals. ______ km, Explanation: How many more people live in France than in Australia? The distance from Earth to Neptune is about 227 miles. Critique Reasoning What is this number written in scientific notation? Without this rule you could get different answers - so getting the order of operation correct is important. 8.456 × 107 Quiz. Question 25. Question 10. The ounces in a cup of milk would be least likely to be written in scientific notation. 5.4 × 104 0.00083, Question 10. 2.97 × 10-2 By using this service, some information may be shared with YouTube. ______________, Explanation: In 2011, the population of Mali was about 1.584 × 107 people. Move the decimal left 5 places 54, Operations with Scientific Notation – Page No. ... Laws of logarithms and exponents. Type below: Communicate Mathematical Ideas "Are You Smarter Than a Fifth Grader" is a TV game show quiz that can be integrated into your lesson plan at school or even played at home. Each entry-level account executive in a large company makes an annual salary of $3.48 × 104. 9 × 1010 Options: (5 × 1012)(3.38 × 106) ______________. Move the decimal left 5 places The mass of a proton is about 1.7 × 10-24 grams. ______________. 1.584 × 107 $$\frac { { (4.87×10 }^{ 12 }) – { (7×10 }^{ 10 }) }{ { (3×10 }^{ 7 })-{ (6.1×10 }^{ 8 }) }$$ Move the decimal right 2 places 11−3 = (1/11)3 = (1/11) × (1/11) × (1/11) = 1/1,331. 11.388 × 107, Question 19. Type below: Exponent = 2 Answer: Type below: 6.6 Move the decimal left 9 places (480 × 1010 )/(64 × 107 ) 7600000, Question 15. + 2.8 × 106 4^-2= = (1/4) / 4 The negative exponents are recripricols of the positive exponents.-5^0-1 (-5)^0. 400,000, Question 8. Type below: 2.1 × 108 is greater because the power of 10 is greater in 2.1 × 108, Question 31. Discover what the variable is, then multiply it by itself the number of time sthe exponemt tells you. Note: Quizzes and tests share the same question pool. 19.4 km, Question 27. + ? ______________, Explanation: 1.33 × (10)4, Question 5. Options: 1,304,000,000 7 × 106 − 5.3 × 106 Astronomy Move the decimal left 4 places ____________, Explanation: When performing these operations on exponents, however, the laws are different. Type below: Move the decimal right 10 places 3). How many times greater is the population of China than the population of France? Type below: Type below: About 786,700,000 passengers traveled by plane in the United States in 2010. Options: An example: x^2 multiplied by x^4 equals x^(2+4), or x^6. It is easier to compare large numbers when in scientific notation as numbers are be expressed as a product of a number greater than or equal to 1 and less than 10 Type below: => (12 x 12 x 12) x (11 x 11 x 11) 7.867 × 108 passengers, Question 3. 0.000003582, Question 9. Type below: 7.149 × (10)7 (The answer is 1/8). Type below: Type below: Type below:$ _______, Explanation: So, the exponents are added 4 ⋅ 4 ⋅ 4 = 4? Start your practice by using Go Math Grade 8 Answer Key. Options: ______________, Explanation: Type below: The thickness of dog hair is very small as the hair is thin. Justify your response. (\$2.94 × 104.) x^3 + x^3 It's simple addition. 54 = 5 × 5 × 5 × 5 = 625, Explanation: Question 24. If there are 5.2 × 102 account executives in the company, wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. 1.25 × (10)-10 Although we offer 20 CATs using 1,274 questions (enough for 30+ tests), it is possible to reduce the number of possible CATs by taking numerous quizzes. 1.74 × (10)6, Explanation: 5.85 × 10-3 m = 0.00585 Lesson. = 13 years 3 months 22.5 days, Question 26. The distance from Earth to the Moon is about 384,000 kilometers. 7.25 × 10-5 Exponent = 4 Represent Real-World Problems Type below: Question 37. Compare the two numbers to find which is greater. How do I add x to the power of 2 plus 4x? 4 × 10-2 = 0.04, Question 7. 0.0297, Question 11. 2 × 101 4.38 × 510 In algebra, the operations (adding, subtracting, multiplying, and dividing) performed on variables work the same as the operations performed on numbers. Type below: 16.9 × 1018, Question 8. d. 158,400,000 people, Explanation: 1.985 × (10)4, Question 21. 2.77 × 10-3, Question 17. 107+2 = 109 7.01568 × 106 minutes A student found the product of 8 × 106 and 5 × 109 to be 4 × 1015. a. ______________, Answer: => 1728 X 1331 {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/7\/71\/Add-Exponents-Step-1-Version-2.jpg\/v4-460px-Add-Exponents-Step-1-Version-2.jpg","bigUrl":"\/images\/thumb\/7\/71\/Add-Exponents-Step-1-Version-2.jpg\/aid2850879-v4-728px-Add-Exponents-Step-1-Version-2.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"
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• ### Announcements
#### Archived
This topic is now archived and is closed to further replies.
# How do you open the cd-rom in c++
## Recommended Posts
v0id 122
Just wondering if anyone could give me the code to open the CD-ROM drive in c++ in windows
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NickB 146
In what way ? do you just want to open files, in which case the standard methods (fopen, CreateFileEx etc.) will work fine - or do you want to be able to read the raw data - block by block - like a MP3 ripper ? - in that case try searching for AKRIP - I can''t remember the address - I think it may be on source forge - but anyhow it''s an open source program that uses WNASPI to access the CD & read the raw data - I think it''s also got some ASPI tuts
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guitarman310 122
i think he means to open the cd rom bay door...i''m not sure, check MSDN.
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Beer Hunter 712
If you''re trying to eject the cd-rom, then it''s not too hard:
#include <mmsystem.h>
mciSendString("set cdaudio door open", NULL, 0, NULL);
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v0id 122
yeah i just want to know how to pop the drive open, so i can mess with my friends computer
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matrix2113 122
You mean like a Trojan?? *tsk tsk*
"I''ve learned something today: It doesn''t matter if you''re white, or if you''re black...the only color that REALLY matters is green"
-Peter Griffin
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hehe
'Very funny, Scotty. Now beam down my clothes.' -
[TheBlackJester]
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Guest Anonymous Poster
if you are using the Dev-C++ compiler, use this
mciSendString("set cdaudio door open", NULL, 0, NULL);
you can also set the door to close using closed as your option instead of open
p.s.Thanks for this(did not know about it yet in my young programming lifetime)...I got my mother and daughter both with this thinking that they had ran a bug or virus...heehe...
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Guest Anonymous Poster
if you are using the Dev-C++ compiler, use this
mciSendStringA("set cdaudio door open", NULL, 0, NULL);
forgot this one up above...the ''A''..sorry for the double-post!
you can also set the door to close using closed as your option instead of open
p.s.Thanks for this(did not know about it yet in my young programming lifetime)...I got my mother and daughter both with this thinking that they had ran a bug or virus...heehe...
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Guest Anonymous Poster
Hmmm, is it just my browser, or does something in the sig of BlackJester make the formating of this thread look somewhat weird ?
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v0id 122
i always get this when i try to compile it:
--------------------Configuration: CDaudio - Win32 Debug--------------------
Compiling...
open.cpp
Open.obj : error LNK2001: unresolved external symbol __imp__mciSendStringA@16
Debug/Open.exe : fatal error LNK1120: 1 unresolved externals
Open.exe - 2 error(s), 0 warning(s)
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Dactylos 122
You have to link to "winmm.lib".
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v0id 122
how would that be accomplished?
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core 106
Either include winmm.lib in your project, or (if using MSVC) use
#pragma comment(lib,"winmm.lib")
"If people are good only because they fear punishment and hope for reward, then we are a sorry lot indeed." - Albert Einstein
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# Creating sections each with title pages in beamers slides
I'm writing up my dissertation defense slides. I would like to have a title page for each of three chapters. Specifically, I want to be able to replicate the box that the dissertation title is in for each of the chapters as I get to that section. Things I don't want: my name, university, and the date. So how do I get the title of each chapter in the same style box that the main title is in? Thanks in advance.
\title{This style box is what I want to replicate three other times in the slides.}
\author[Not needed]
\institute{I don't want this.}
\date{I don't want this either}
• Which beamer theme(s) are you using? – Gonzalo Medina May 17 '14 at 20:27
• @GonzaloMedina How can we define this command(\AtBeginSection) when using emacs org export to beamer ? Thanks. – Anusha Jan 13 '15 at 0:04
• How to introduce 'noframenumbering' option so that the slide with the section title does not increase the slide counter? Thank you! – Jose Carreno Garcia Aug 27 '19 at 17:24
This can be easily done using \AtBeginSection to place a frame with the section title (\insertsectionhead) inside a beamercolorbox:
\documentclass{beamer}
\AtBeginSection[]{
\begin{frame}
\vfill
\centering
\end{beamercolorbox}
\vfill
\end{frame}
}
\title{The Title}
\author{The Author}
\institute{The Institute}
\begin{document}
\begin{frame}
\maketitle
\end{frame}
\section{Test section one}
\begin{frame}
test frame for section one
\end{frame}
\section{Test section two}
\begin{frame}
test frame for section two
\end{frame}
\end{document}
Probably, depending on the theme used (which was not mentioned in the question), you will need to adjust the settings for the box; in particular, you might need to deactivate the option shadow and/or rounded by setting them to false.
• This is exactly what I asked for. Now to complicate things. I am using the Frankfurt theme, so the section names are at the top of all the slides. How could I use this code to put the full titles in boxes but keep the names of sections short on the top bar? – interested user May 17 '14 at 20:45
• @interesteduser Please, in this site we try to keep just one question per post. If you have additional questions, feel free to open a new question. – Gonzalo Medina May 17 '14 at 20:50
• Your right. If you would please check out the following question: tex.stackexchange.com/questions/178809/… – interested user May 17 '14 at 21:03
• @interesteduser answer provide there too :) Don't forget that you can accept answers that you consider solved your problems by clicking the checkmark to their left. In case of doubt, please see How do you accept an answer?. – Gonzalo Medina May 17 '14 at 21:11
• This solves my problems too, but somehow the tableofcontents slide doesn't show up. Could you spell some magic so I could have the toc frame back? – yi.tang.uni Jun 28 '16 at 9:25
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# Label points on a map with one function call
I want to place labels on a geographic map. I can do this with individual points as in the code below
Show[
CountryData["France", {"Shape", "Mercator"}],
Frame -> True,
Epilog ->
{PointSize[0.01], Red, Point[mercPoints],
Text[Abbeville, {3, 47},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 12}],
Text[Agen, {3, 50}]}]
But this is not my aim. I have a list containing the names I want to label the points with (say stationNames). How do I attach each of the label names in stationNames to its respective point given by mercPoints, which is the list giving the longitude and latitude of each point.
• You might try some like MapThread[Text[#1, #2], {names, mercPoints}] – m_goldberg Oct 3 '16 at 2:16
• Hi, your line of code returned an error when I tried it. 'An unknown box name (Text[#, #2]) was sent as the BoxForm for the expression. Check the format rules for the expression' – Tom6639 Oct 3 '16 at 2:45
• Sorry. Left out &. Try MapThread[Text[#1, #2]&, {names, mercPoints}] – m_goldberg Oct 3 '16 at 2:48
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Fréchet derivative is a generalization to the ordinary derivative. Generally we are talking about Banach space, where $\mathbb{R}$ is a special case. Indeed, the space discussed is not even required to be of finite dimension.
## Recall
A real-valued function $f(t)$ of a real variable, defined on some neighborhood of $0$, is said to be of $o(t)$ if
And its derivative at some point $a$ is defined by
We also have this equivalent equation:
Now suppose $f:U \subset \mathbb{R}^n \to \mathbb{R}^m$ where $U$ is an open set. The function $f$ is differentiable at $x_0 \in U$ if satisfying the following conditions.
1. All partial derivatives of $f$, i.e. $\frac{\partial f_i}{\partial x_j}$ exists for all $i=1,\cdots,m$ and $j = 1,\cdots,n$ at $f$. (Which ensures that the Jacobian matrix exists and is well-defined).
2. The Jacobian matrix $J(x_0)\in\mathbb{R}^{m\times n}$ satisfies
In fact the Jacobian matrix has been the derivative of $f$ at $x_0$ although it’s a matrix in lieu of number. But we should treat a number as a matrix in the general case. In the following definition of Fréchet derivative, you will see that we should treat something as linear functional.
## Definition
Let $f:U\to\mathbf{F}$ be a function where $U$ is an open subset of $\mathbf{E}$. We say $f$ is Fréchet differentiable at $x \in U$ if there is a bounded and linear operator $\lambda:\mathbf{E} \to \mathbf{F}$ such that
We say that $\lambda$ is the derivative of $f$ at $x$, which will be denoted by $Df(x)$ or $f’(x)$. Notice that $\lambda \in L(\mathbf{E},\mathbf{F})$. If $f$ is differentiable at every point of $f$, then $f’$ is a map by
The definition above doesn’t go too far from real functions defined on the real axis. Now we are assuming that both $\mathbf{E}$ and $\mathbf{F}$ are merely topological vector spaces, and still we can get the definition of Fréchet derivative (generalized).
Let $\varphi$ be a mapping of a neighborhood of $0$ of $\mathbf{E}$ into $\mathbf{F}$. We say that $\varphi$ is tangent to $0$ if given a neighborhood $W$ of $0$ in $\mathbf{F}$, there exists a neighborhood $V$ of $0$ in $\mathbf{E}$ such that
for some function of $o(t)$. For example, if both $\mathbf{E}$ and $\mathbf{F}$ are normed (not have to be Banach), then we get a usual condition by
where $\lim_{\lVert x \rVert \to 0}\psi(x)=0$.
Still we assume that $\mathbf{E}$ and $\mathbf{F}$ are topological vector spaces. Let $f:U \to \mathbf{F}$ be a continuous map. We say that $f$ is differentiable at a point $x \in U$ if there exists some $\lambda \in L(\mathbf{E},\mathbf{F})$ such that for small $y$ we have
where $\varphi$ is tangent to $0$. Notice that $\lambda$ is uniquely determined.
## Propositions
You must be familiar with some properties of derivative, but we are redoing these in Banach space.
### Chain rule
If $f: U \to V$ is differentiable at $x_0$, and $g:V \to W$ is differentiable at $f(x_0)$, then $g \circ f$ is differentiable at $x_0$, and
Proof. We are proving this in topological vector space. By definition, we already have some linear operator $\lambda$ and $\mu$ such that
where $\varphi$ and $\psi$ are tangent to $0$. Further, we got
To evaluate $g(f(x_0+y))$, notice that
It’s clear that $\mu\circ\varphi(y)+\psi(\lambda{y}+\varphi(y))$ is tangent to $0$, and $\mu\circ\lambda$ is the linear map we are looking for. That is,
### Derivative of higher orders
From now on, we are dealing with Banach spaces. Let $U$ be an open subset of $\mathbf{E}$, and $f:U \to \mathbf{F}$ be differentiable at each point of $U$. If $f’$ is continuous, then we say that $f$ is of class $C^1$. The function of order $C^p$ where $p \geq 1$ is defined inductively. The $p$-th derivative $D^pf$ is defined as $D(D^{p-1}f)$ and is itself a map of $U$ into $L(\mathbf{E},L(\mathbf{E},\cdots,L(\mathbf{E},\mathbf{F})\cdots)))$ which is isomorphic to $L^p(\mathbf{E},\mathbf{F})$. A map $f$ is said to be of class $C^p$ if its $kth$ derivative $D^kf$ exists for $1 \leq k \leq p$, and is continuous. With the help of chain rule, and the fact that the composition of two continuous functions are continuous, we get
Let $U,V$ be open subsets of some Banach spaces. If $f:U \to V$ and $g: V \to \mathbf{F}$ are of class $C^p$, then so is $g \circ f$.
### Open subsets of Banach spaces as a category
We in fact get a category ${(U,f_U)}$ where $U$ is the object as an open subset of some Banach space, and $f_U$ is the morphism as a map of class $C^p$ mapping $U$ into another open set. To verify this, one only has to realize that the composition of two maps of class $C^p$ is still of class $C^p$ (as stated above).
We say that $f$ is of class $C^\infty$ if $f$ is of class $C^p$ for all integers $p \geq 1$. Meanwhile $C^0$ maps are the continuous maps.
## An example
We are going to evaluate the Fréchet derivative of a nonlinear functional. It is the derivative of a functional mapping an infinite dimensional space into $\mathbb{R}$ (instead of $\mathbb{R}$ to $\mathbb{R}$).
Consider the functional by
where the norm is defined by
For $u\in C[0,1]$, we are going to find an linear operator $\lambda$ such that
where $\varphi(\eta)$ is tangent to $0$.
Solution. By evaluating $\Gamma(u+\eta)$, we get
To prove that $\int_{0}^{1}\eta^2\sin{x}dx$ is the $\varphi(\eta)$ desired, notice that
Therefore we have
as desired. The Fréchet of $\Gamma$ at $u$ is defined by
It’s hard to believe but, the derivative is not a number, not a matrix, but a linear operator. But conversely, one can treat a matrix or a number as a linear operator effortlessly.
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# Math Help - Parametric equations and the lengths of curves.
1. ## Parametric equations and the lengths of curves.
1. Find parametric equations for the semicircle: x^2 + y^2 = a^2 using as parameter the arc length s measured counterclockwise from the point (a, 0) to the point (x, y).
We were not taught this, and I don't think the book really helps. I was wondering if somebody could clearly explain this concept and problem to me.
2. Find the length of the curve: x = cos t, y = t + sin t, [0, [pi]].
Here's where I get stuck. I know that the length of a curve is the integral evaluated from a to b (or, in this case, from zero to pi) of the quadrature of dx/dt and dy/dt. However, upon doing that, I get the integral of the square root of (-sin t)^2 + (1 + cos t)^2. I cannot use u substitution (although that does not rule out it's possibility of use) and if I make (-sin t)^2 into (sin t)^2 and expand (1 + cos t)^2, I get the integral of sqrt(2 + 2cos t), which then, by pulling out the square root of 2, makes the integral of the square root of 1 + cos t. I...don't know what to do then.
I would appreciate any help.
2. Originally Posted by niyati
1. Find parametric equations for the semicircle: x^2 + y^2 = a^2 using as parameter the arc length s measured counterclockwise from the point (a, 0) to the point (x, y).
We were not taught this, and I don't think the book really helps. I was wondering if somebody could clearly explain this concept and problem to me.
...
Hello,
I've attached a drawing.
A point on the semicircle, the centre of the semicircle and the diameter form an angle $\theta$ which can be calculated in radians by
$\theta = \frac sa$
Therefore you get:
$\left| \begin{array}{l}x=a\cdot \cos\left(\frac sa\right) \\ y = a \cdot \sin\left(\frac sa\right) \end{array} \right.$
3. ## Second problem
On the second part, you said you obtained the equation $L=\sqrt{2}\int_{0}^{\pi}\sqrt{1+\cos{x}}\,dx$
To solve that, remember the trig identity $\cos{2\theta}=2\cos^2\theta-1$. Thus $1+\cos{2\theta}=2\cos^2\theta$, or using $\theta=\frac{x}{2}$, we obtain $1+\cos{x}=2\cos^2\frac{x}{2}$, so
$L=\sqrt{2}\int_{0}^{\pi}\sqrt{1+\cos{x}}\,dx$
$=\sqrt{2}\int_{0}^{\pi}\sqrt{2\cos^2\frac{x}{2}}\, dx$
$=2\int_{0}^{\pi}\left|\cos\frac{x}{2}\right|\,dx$
Note that for your region of integration, $\cos\frac{x}{2}\ge0$, so you can remove the absolute value:
$L=2\int_{0}^{\pi}\cos\frac{x}{2}\,dx$
You should be able to do that integral.
--Kevin C.
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# What is the equivalent resistance between in the given circuit? a) b) c) d)
## Question ID - 150306 :- What is the equivalent resistance between in the given circuit? a) b) c) d)
3537
(c)
and are in series
are in parallel
and are in series
and are in parallel
and are in series
Now, and are in parallel
Next Question :
A metal wire of specific resistance and length has a resistance of the radius of the wire will be a) b) c) d)
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Forum moved here!
Home / Discovery of location of the open document
user93
Good afternoon.
When many documents are opened, it would be desirable to open an arrangement of each file. It very strongly would simplify work. I ask for help. Thanks
GitHubRulesOK
@user93
Unfortunately SumatraPDF has a limited space for on-screen display of only the last 10 files when you go to File drop-down menu and does not show the file path, however if you press F2 (the rename function) it WILL show the filename and the folder the current file came from.
OR
use CTRLD&CTRLC
if you hold down CTRL its just three keys in total to get the file properties into the clipboard.
The full path and name ARE stored in the File > Settings > Advanced Option … text file and If it helps with your work-flow you could build a simple CMD script to
Find "FilePath" sumatrapdf-settings.txt
and export those entries. Be aware that if remember tabs are set on then there will be two overlapping sets of files, the only difference will be the amount of indentation.
---------- SUMATRAPDF-SETTINGS.TXT
FilePath = F:\...\sheet_music.xod
FilePath = F:\...\sheet_music.xod
Hope some of that helps
[Later update]
Show in folder (Shortcut ALTFF NOTE: may vary for other languages)
The above find command could even be used from within SumatraPDF or output the current filename along the lines of
ExternalViewers [
[
CommandLine = C:\windows\system32\cmd.exe /c echo FullFileName = "%1" -page %p >> "C:\i can write here\current file.txt"
Name = Save current file name in &List
Filter = *.*
]
]
Shortcut would then be ALTFL
You can be really creative and with slight changes (remove FullFileName = ) make that list of files saved as OpenList.cmd to be run, such as to open those pages next time in SumatraPDF.
GitHubRulesOK
@RomanSofyin
the above post has been updated and expanded, thanks for your interest.
voICEwork
It would be useful to “Copy complete file name” (including full path). This would be especially useful for emailing a recently created file. This is an option in another good viewer PDF-XChange Viewer.
GitHubRulesOK
It is not documented but you can go to file properties [Ctrl +D] and hit [Ctrl +C] to copy the file details to the clipboard the first line is the full filename (prefixed by the word File:) and that line is what is shown by most one line dialog boxes so is sometimes easy to paste into say explorer OR should work as subject line?. I agree it not ideal as you need to remove the word File: or the additional lines if you paste in an email.
A quicker alternative for just the location and short filename is to press F2, the filename is highlighted so easy to copy and the folder address is usually easy to copy from address bar.
Both are admittedly work-arounds and the ability to show the filename with one click has been requested before (usually for use with tabs)
1112
how can I copy the PATH of a book?
GitHubRulesOK
CTRLD
CTRLC
or Hold CTRL
D+C
or easier F2 and simply copy the folder address as a single line then Cancel
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# LHCb Theses
Najnovije dodano:
2023-03-14
14:29
Search for lepton flavour violating $\tau^+\to\mu^+\mu^-\mu^+$ decay at LHCb and study on MCP-PMT detector for future LHCb Upgrade / Capelli, Simone The physics analysis has been the primary focus of my research activity during the PhD [...] CERN-THESIS-2022-322 - 141 p.
2023-03-07
21:55
Experimental study of semitauonic $B_c^+$ decays and development of the Upstream Tracker electronics for the LHCb upgrade / Yang, Zishuo The LHCb experiment at the Large Hadron Collider is designed for studying the properties of heavy quarks and CP violation to indirectly search for new physics beyond the Standard Model [...] CERN-THESIS-2023-014 -
2023-02-28
13:58
Study of Open Charm Mesons Productions in Proton-Lead Collisions at LHCb / Gu, Chenxi Quantum chromodynamics (QCD) is a basic theory to describe the strong interaction between quarks and gluons [...] CERN-THESIS-2022-314 - 204 p.
2023-02-23
12:14
Angular analysis $B_s^0\to\phi\mu^+\mu^-$ decays with the LHCb Experiment / Materok, Marcel This thesis presents an angular analysis of the rare decay $B_s^0\to(\phi\to K^+K^-)\mu^+\mu^-$ [...] CERN-THESIS-2023-009 - 185 p.
2023-02-15
11:54
Study of the rare $B_s^0\to\phi\mu^+\mu^-$ and $B_s^0\to f_2'(1525)\mu^+\mu^-$ decays / Kretzschmar, Sophie The Standard Model of particle physics (SM) describes the microcosm of nature with great success [...] CERN-THESIS-2022-302 - 169 p.
2023-01-27
15:47
Testing lepton flavour universality in $B\rightarrow K\pi\pi\ell\ell$ decays using 2017 LHCb data / Rauber, Geraldine This present work proposes a study on $B\rightarrow K\pi\pi\ell\ell$ decays using 2017 LHCb data in the context of Lepton Flavour Universality tests [...] CERN-THESIS-2021-362 - 83 p.
2023-01-17
14:50
Experimental study of hadron spectroscopy in beauty meson to open charm decays at LHCb / Chen, Chen Quantum chromodynamics~(QCD) is the fundamental theory describing the strong interaction in the Standard Model~(SM) of particle physics [...] CERN-THESIS-2022-283 - 206 p.
2023-01-17
10:35
Test of Lepton Universality with $b \rightarrow s \ell^+ \ell^-$ Decays at LHCb / Escher, Stephan Tests of lepton universality are currently among the most discussed observables in flavour physics and show tensions with SM predictions in recent measurements [...] CERN-THESIS-2022-282 - 193 p.
2023-01-11
09:26
A FPGA-based architecture for real-time cluster finding in the LHCb silicon pixel detector / Bassi, Giovanni Pending CERN-THESIS-2023-002 -
2023-01-02
13:23
Study of charmed baryons at the LHCb experiment / Xu, Ao Quantum Chromodynamics (QCD) is the theory of strong interaction in the Standard Model of particle physics [...] CERN-THESIS-2022-271 - 222 p.
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# Simple Question
1. Jan 30, 2012
### xuhaib20
Lets Suppose we have 251 Items in stock and we want to convert it into Boxes and single items. How can we do that. I want result like that
E.g 50 boxes 1 item
Total Items # 251 items
items per box = 5
251/5 = 50.2
But result must be as 50 boxes and 1 item
2. Jan 30, 2012
### DivisionByZro
251 is a prime number, which means that it has no other divisors than itself and 1. You have 3 possibilities, 1 box with 251 items, 251 boxes with 1 item, or N+1 boxes with $$\lfloor \frac{251}{N} \rfloor$$ items and with at least one box with >=1 item. Is this what you were asking?
3. Jan 30, 2012
### checkitagain
50.2 boxes
Two-tenths of a box means
$$\dfrac{0.2 \ box}{1}\cdot \frac{5 \ items}{box} \ = \ ?$$
4. Jan 31, 2012
### 20Tauri
Maybe you could use division with remainder? i.e., 251/5 = 50 r 1. The way you'd do this on a calculator is you'd compute 251/5 = 50.2. You'd take 50 to be the number of boxes, and then subtract away the whole number to get 0.2, then multiply 0.2 * 5=1 to get the remainder. This is essentially a different way of looking at the explanation in the post above mine.
5. Feb 2, 2012
### coolul007
Another approach is to use the sum 125 + 126, then 125 = 5(25) and 126 = 6(21) = 5(25) + 1 . It is similar to a problem of finding the appearance of the "star" portion of the American flag when states are being added.
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Browse Questions
# The number of $F_2$ plants found similar to $F_1$ phenotype and genotypes, respectively in a total of 128 plants in $F_2$ progeny of a Mendelian dihybrid cross :
$\begin {array} {1 1} (1)\;48,32 & \quad (2)\;72,32 \\ (3)\;48,24 & \quad (4)\;72,48 \end {array}$
Can you answer this question?
(2) 72,32
answered Nov 7, 2013 by
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## Calculus: Early Transcendentals 8th Edition
$[1.3, 1.7]$
The inequality$|2x-3|≤0.4$ can be calculated as follows: If$|x|≥a$ then: $x≤-a,x≥a$ … (1) Now, apply equation (1) for expression$|2x-3|≤0.4$ to get: $-0.4≤2x-3≤0.4$ This implies $3-0.4≤2x≤0.4+3$ $2.6≤2x≤3.4$ $1.3≤x≤1.7$ Hence, the solution set is$[1.3, 1.7]$.
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# Solve using double angles?. Sin2x-sinx=0
Jun 18, 2018
$x = \pm 2 \pi n , \frac{\pi}{3} \pm 2 \pi n$; where $n \in \mathbb{Z}$.
#### Explanation:
We have: $\sin \left(2 x\right) - \sin \left(x\right) = 0$
Using the double-angle identity for $\sin \left(x\right)$, we get:
$R i g h t a r r o w 2 \sin \left(x\right) \cos \left(x\right) - \sin \left(x\right) = 0$
$R i g h t a r r o w \sin \left(x\right) \left(2 \cos \left(x\right) - 1\right) = 0$
$R i g h t a r r o w \sin \left(x\right) = 0$
$R i g h t a r r o w x = 0 \pm 2 \pi n$; $n \in \mathbb{Z}$
$R i g h t a r r o w x = \pm 2 \pi n$; $n \in \mathbb{Z}$
or
$R i g h t a r r o w \cos \left(x\right) = \frac{1}{2}$
$R i g h t a r r o w x = \frac{\pi}{3} \pm 2 \pi n$; $n \in \mathbb{Z}$
Therefore, the general solutions to the equation are $x = \pm 2 \pi n$ or $x = \frac{\pi}{3} \pm 2 \pi n$, where $n \in \mathbb{Z}$.
Jun 18, 2018
$x = k \pi \text{ or } x = \frac{\pi}{3} + 2 k \pi$
#### Explanation:
$\text{using the "color(blue)"trigonometric identity}$
•color(white)(x)sin2x=2sinxcosx
$2 \sin x \cos x - \sin x = 0$
$\sin x \left(2 \cos x - 1\right) = 0$
$\text{equate each factor to zero and solve for } x$
$\sin x = 0 \Rightarrow x = k \pi \to \left(k \in \mathbb{Z}\right)$
$\cos x = \frac{1}{2} \Rightarrow x = \frac{\pi}{3} + 2 k \pi \to \left(k \in \mathbb{Z}\right)$
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# Is there any precedence for symbols for constructing parse trees? [duplicate]
I am wondering if some symbols such as the ones in propositional logic have precedence over others in drawing parse trees.
For example, the sentence: p ∧ q → r, would ∧ take precedence over → in becoming the root of the parse tree or vice versa?
Or is there no precedence meaning, → or ∧ could be used the root?
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# Remaking Money
I'm going to make an argument here about what could be done with a monetary system. I make no assumptions as to whether this idea is better or worse than the system commonly in use today. I suspect that as I develop the idea its advantages and disadvantages will come to the surface.
I outline an alternative to money as it is currently understood. My system involves making those who use the most money (not necessarily the richest) bear a proportional share of the costs of money. It would also setup infrastructure that would allow for directly stimulating the velocity of money, and finely control the amount of money in circulation, which in turn influences interest rates.
### MONEY
#### Value
Economic Value is a subjective term. The value of any commodity is different for each person, and different for the same person at different times and under different conditions. Often the value of things is not based on the value of its practical applications, but rather on how people feel about those things. Jewelry is a good example of a commodity that does not fill a basic human need, but that commands an unusually high value.
#### Functions & Properties of Money
Money has several basic functions in the economy:
1. It is a medium of exchange.
2. It is a measure of value.
3. It can be used as store of value.
As one half of every transaction money makes trade dramatically more efficient. Money as minted coin, paper, or digital accounts is far easier to move and measure than any other trade good. Money improves the speed and profitability of all economic transactions.
Money allows all products and needs in an economy to be measured against the same standard unit of value. This allows products from all over the world to be compared directly. Having a standard unit of account allows many more opportunities for investment. Loans, investments, and insurance are only really possible because there is a predictable unit of measure.
Money can be treated as a commodity. That is, when money is owned it can be used as a store of value. Storing value as money depends heavily on the time behavior of the value of money.
Some practical properties of money are that it is:
• Portable: It is easy to transport
• Divisible: There are smaller denominations from which to make change.
• Durable: It has a reasonably long lifetime
• Fungible: Each of its units can be exchanged as equivalent.
#### Sources of Money
There are at least two kinds of money: commodity money, and fiat money or currency.
Commodity money is the simplest to understand. It is some set of items or materials that a society recognizes for use as a common medium of exchange. Commodity money is created by market activities like manufacture or mining. Commodity money has been wampum or seashells, but most famously gold or silver.
Currency is money that is created and regulated by an institution rather than directly as a product on the market. The value of fiat money is determined by a rather complicated relationship between the amount of money in circulation (not the amount in existence), and the rate of economic activity.
#### Value of Commodities
Because commodity money has intrinsic value in the economy its value, and the amount of it in circulation is governed by its supply and demand. Finding new practical uses for the money commodity or finding new sources of it can dramatically change its value in the economy. Aluminum was more valuable than gold until a cheap way was found to manufacture it.
Some practical uses of the money commodity may be stifled because of its inflated value in the market. With a commodity form of money, the amount of money in circulation cannot be adjusted to reflect economic needs.
Much of the wealth in Europe created after the discovery of America was caused by the sudden influx of gold and silver that allowed greater circulation and investment. It was not the presence of the gold that made the economy soar. Gold doesn't make buildings or ships, nor is it edible. It was the availability of gold as money in circulation combined with the new markets in the Americas that created the boom.
#### Value of Currency
The only difference between currency and a commodity based medium is that the currency only has value as money.
Issuing institutions for money have been banks, governments, or large businesses, sometimes individually and sometimes as part of a larger system. In all cases the value of the currency is based on what can be bought with it.
Today the world economy is not backed by commodities because the amount of money in circulation is far, far larger than any commodity could back. Only the economy itself can sufficiently back the amount of currency in circulation.
#### The Costs of Currency
Currency has basic costs involved in its creation and maintenance. (A commodity money would also have cost associated with its manufacture.) The physical (or digital) creation of money, management of debtors accounts, and management of the amount of money in circulation are forms of work that have associated costs. Even digital money has its ongoing costs: the costs of maintaining and securing the various institutions and agreements that allow digital money to be consistently used and processed.
Currency is owned by the institution that creates it. Typically today, currency is issued as a loan. That is to say that all currency in circulation was issued as a loan to someone or other. The interest payments on these loans fund the issuing institutions. As such, the global burden of the cost of money and any profits of the issuing institutions is carried by the debtors of the world.
Today a few private corporations (banks) hold the entire credit side of the worlds monetary system. The full cost of the money in circulation including these banks profits is paid by governments, investors, and consumers who use credit. As such, these private banks alone stand to profit on the service of that debt, and they alone control the amount of money in circulation.
### FACTORS OF A SUCCESSFUL ECONOMY
#### Growth
A growing economy indicates that more resources (wealth) are available for consumption. An economy that is growing faster than population indicates an increasing mean standard of living and more opportunity for societal development.
There are various physical limitations to how fast an economy can expand. New workers can only be trained and gain experience so fast. new factories and equipment takes time to build and put in operation. marketing can only penetrate new markets as fast as people are willing to need new products.
Inflation occurs when money enters circulation faster than the economy's real value is able to grow.
#### Supply Side vs. Demand Side Economics
Growth is driven by consumption. Because of fixed costs, excessive production capacity does not produce lower prices, instead it engenders unprofitable businesses and underutilized resources. The exception to this rule are the non-competitive markets where monopolies exist: such as patented products, legal or technical monopolies. This is why favors for large businesses do not tend to reduce prices and stimulate the economy but rather boost profits for their stock holders and executives.
Consumers do not spend in direct proportion to their income. The lower the income, the greater the portion of income that is spend immediately. Poor people have a higher propensity to spend. While the wealthy enjoy seeing their money collect and grow as they save or invest. In conclusion, more money in the hands of the poor stimulates demand more than money in the hands of the rich.
#### Distribution
If there are more poor people than rich people, then more evenly distributed income will favor poor people more than rich people. Since growth is stimulated by demand and demand is greatest among the poorest it follows that the more evenly income is distributed the more growth there will be in the economy. More even distribution will promote a better housed, fed, and educated lower class. Such conditions will reduce crime and dependency on social services further increasing stability and making more resources available for economic growth.
#### Maximum Utility of Resources
All unused resources are an economic loss and opportunity. Resources may be natural, financial, or human. Some unemployment of resources is unavoidable, but reducing that amount should of primary concern for economic growth. Higher demand, better education, and less crime promote full utilization of economic resources.
Money that is not in circulation, just as any asset that is not economically active, is an unemployed resource: interest is being paid for its existence, but it is not being used for economic activity.
#### Stability & Continuity
Businesses are only interested in investing and opening new markets when they have confidence in being successful. To be confident of success investors must feel that business conditions are sufficiently stable. Conditions such as taxes, interest rates, the availability of human and other resources, and a healthy consumer market must be stable or at least predictable in order for investors to risk their capital.
### MANAGEMENT OF AN ECONOMY
Fiscal policy is the use of government taxation and spending to create economic change. This change should be targeted to promote growth through stabilization of business conditions, fuller employment of resources, and more even distribution of income.
Monetary policy is any means used to change the value of money or its amount in circulation.
Economies respond much more to the feelings of consumers and investors than they do to any technical conditions that may be present. A strong economy is often more the result of strong, confident, and consistent leadership than to any other factor.
#### Gross Product
Q: The quantity of goods and services sold
P: The Average unit market value of goods and services sold
M: The amount of money in circulation
V: The average number of times each unit purchases goods or services (Velocity)
C: The market value of all final consumer goods and services.
I: The market value of all investments made.
G: The market value of government expenditures for goods and services.
X: The market value of all exports.
IM: The market value of all imports.
Gross Product
$GP = Q P = M V = C + I + G + ( X - IM )$
Gross product is the total sale value of all final goods and services sold in an economy minus the purchase value of all imports. Gross product is equal to the product of the amount of money in circulation times the average number of exchanges that money undergoes. It also equals the number of goods sold times their average price. It is the sum of consumer spending, investment spending, government spending, and the balance of trade.
#### Stimulating growth
In the US the Federal Reserve Banks stimulate growth by buying debt on the open market. In doing so they inject new capital into the economy and drive down the cost of debt (reduce the prime interest rate). In this process the newly injected capital goes to governments and large banks.
Both investment and consumption are affected by interest rates. There is a negative relationship between Investment Spending and interest rates, such that businesses will only tend to invest when the expected return on investment is greater than the expected return on a loan.
In lowering the cost of debt consumers are more likely to overextend themselves if their creditors allow it. This is what caused the US housing crisis in the late 2000's. Low interest rates also promote investment in marginal opportunities of low return.
### A BANKING PROPOSAL
#### Publicly Owned Money
There is no reason that the issuer of public credit need be a private entity, nor is there any reason that it need be for-profit. A not-for-profit, publicly owned network of central banks is an entirely possible paradigm. Money should be a public institution created and maintained for the public good, or at least for the good of the depositors.
#### Disallowing Money as a Store of Value: Promoting Circulation
Money used as a store of value by definition is removed from circulation. Money removed from circulation is no longer available as investment capital, and can also affect prices and domestic product.
Money used as a store of value is money that is not available for investment or to create demand. To counteract this problem banks today treat all deposits as investment capital risking both transactional and investment accounts.
By charging interest against all cash and transaction accounts The central bank can effectively prevent them from being used as a store of value. Such interest would have to be calculated as continuously compounding to prevent games being played with transaction times. The amount of money in an account would be recalculated by computer every time a transaction occurs or the balance is checked. Cash would be depreciated from its date of issue. Its value and lifetime could be maintained by attaching stamps to the bills.
Investment accounts would still be available to store value if its interest rate were sufficient. but with the understanding that this money is reinvested and is at risk.
#### Funding Public Money
A bank can be funded with fees: fees on accounts or fees on transactions. This is common and we could argue that it is fair--those that use the most services pay the most. But if we change our paradigm and create such a bank for the good of the economy as a whole and that money is a public institution
#### Other Financial Institutions
The central banks could certify, regulate, and insure other financial institutions. Even so other banks and institutions could not be allowed to manage their own operating accounts.
#### Creation of Money
The best place for money to be injected into any economy is among its poorest people. Not only does it raise their standard of living, but the economy is stimulated because they immediately spend it. When money is loaned to large institutions there is debt that must be serviced, also such institutions may not spend the money quickly or domestically.
Instead of creating a complex system of debt, or a complex systems to determine who is poor enough to receive, the central banks could periodically make a deposit for every account holder. Not a deposit to every account, but to every account holder. Since Social security numbers are already associated with each bank account it should be a simple matter for each person to select an account where they will receive their deposit.
#### Loans & Interest Rates
Since the central banks will not be creating money for the purpose of loans, where will loan funds come from? From deposits in investment accounts. People and businesses not wishing to lose excessive value of their money will invest it in some way: buying bonds, certificates of deposit, savings accounts etc. In any of these cases that money then becomes available for loans or other investment. The Interest rates on these various investments will be determined by the supply and demand for capital. The central banks will be able to monitor the various interest rates to determine if there is enough capital in the economy and may adjust the Cash rate or Injection rate accordingly.
Loan rates will have to be significantly higher than the cash rate because they must both offset the cash rate and cover the institution's revenue and any investor profit.
Setting the cash rate too high will obviously force loan rates high in turn slowing investment. But it would also create very motivated spenders/investors. Banks would not be able to offer investment opportunities of sufficient return to be profitable. People would likely prefer to spend their money on hard commodities rather than watch it disappear.
The central banks will not be able to certify foreign banks to take deposits of money because they would not have legal jurisdiction over them. Therefore expatriation of capital could only occur as cash. No foreign entity will want to hold any significant amount of domestic money. Foreign exporters could sell the domestic currency in foreign exchange markets, but such money will only be useful to foreign importers. The value of the domestic currency will adjust on the foreign exchange until a balance of trade is achieved. The net result for foreign trade will be de facto balanced trade because the domestic currency cannot be profitably expatriated.
A positive trade balance could occur if foreign currency's could be profitably imported. Foreign currency could only be profitable if it were appreciating against the domestic currency. In this case a positive trade balance could be useful to boost the domestic economy against the rising foreign one.
#### Fiscal Policy
Depending on economic goals such a central bank could be very active, or nearly invisible. Capital injections do not necessarily need to be made at all, while the cash rate could be so low as to only cover the operating costs of the central banks.
The higher the cash rate and injection rates are set the more such a system begins to resemble a wealth redistribution scheme. Even so it is one that promotes economic activity and sustainable lower classes. In an economy where profits are being concentrated in large businesses with ever greater degrees of productivity and automation there is ever less need for the marginal lower labor classes. These people, often through no real fault of their own have little place in the modern economy. When unemployed or suffering some other crisis they contribute to crime, abuse and dependence on social programs. If we consider money and banking as a public institutions designed to bolster the entire economy and not just the rich it should not be difficult to perceive this system as benefical for all even if it could tend to redistribute wealth.
As long as changes in the CR and IR are graduated rather than abrupt they should not contribute significantly to economic instability.
### SUMMARY
• A network of central banks, publicly owned, non-profit
• Central bank regulates and guarantees edge banks
• Central Bank (CB) regularly issues money to each citizen
• CB constantly charges interest on all deposits (Cash Rate, CR)
• All cash must have stamps regularly affixed to maintain its value
• Everyone will be motivated to keep money in circulation either by spending it or investing it
• CB adjusts Cash rate (CR) and money issues (Issue Rate, IR) to maintain level of money in circulation
• Edge Banks offer savings accounts and bonds to get investment capital
• Savings rates (SR) will be set only by the supply and demand for investment capital.
• Edge banks will adjust their SR and Loan rates (LR) to maintain an acceptable amount of investment capital on hand given the CR
• CB will be able to keep a running total of Money in Circulation(MiC), and Money in Reserve(MiR) throughout all edge banks
• Low savings rates and loan rates will indicate too much money in circulation, and high will indicate too little. CB will respond by adjusting IR
• CB will monitor ratio MiC:MiR. Changes in this ratio will be responded to by changes in the CR and IR.
Such a system will encourage people and businesses to spend or invest. It will reduce the value of money as a reserve of value and emphasis its use as a medium of exchange. Money spent will result in more jobs or more dividends.
Such money will have little value in a foreign exchange because it cannot be effectively expatriated. International trade will self balance as a matter of course. This will encourage more domestic jobs and manufacturing.
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# zbMATH — the first resource for mathematics
The Liar paradox and fuzzy logic. (English) Zbl 0945.03031
Summary: Can one extend crisp Peano arithmetic PA by a possibly many-valued predicate $$\text{Tr}(x)$$ saying “$$x$$ is true” and satisfying the “dequotation schema” $$\varphi\equiv \text{Tr}(\overline\varphi)$$ for all sentences $$\varphi$$? This problem is investigated in the frame of Łukasiewicz infinitely valued logic.
##### MSC:
03B50 Many-valued logic 03F30 First-order arithmetic and fragments 03B52 Fuzzy logic; logic of vagueness
Full Text:
##### References:
[1] SOFSEM’95: Theory and practice of infor-matics (Milovy, Czech Republic, 1995) 1012 pp 31– [2] The philosophical computer (1998) [3] Mehrwertige Logik (1988) [4] Mathematische Annalen 71 pp 97– (1910) [5] Liar’s paradox and truth-qualification principle (1979) [6] DOI: 10.1023/A:1004948116720 · Zbl 0869.03015 [7] Sitzngsberichte Berliner Mathematische Gesellschaft 58 pp 41– (1957) [8] Die Nichtaxiomatisierbarkeit des unendlichwertigen PrÄdikatenkalkÜls von Łukasiewicz 27 pp 159– (1962) [9] Metamathematics of first-order arithmetic pp 460– (1993) [10] On the definition of an infinitely-valued predicate calculus 25 pp 212– (1960) · Zbl 0105.00501 [11] Metamathematics of fuzzy logic (1998) · Zbl 0937.03030 [12] Fixed point theorems (1974)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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# The current ratio of X. Ltd is 2:1. State with reason which of the following transaction would i. Increase or ii. decrease or iii. not change the ratio 1. Included in the trade payables was a bills payable of Rs.9,000 which was met on maturity. 2. Company issued 1,00,000 equity shares of Rs.10 each to the Vendors of machinery purchased. - CBSE (Arts) Class 12 - Accountancy
#### Question
The current ratio of X. Ltd is 2:1. State with reason which of the following transaction would
i. Increase or ii. decrease or iii. not change the ratio
1. Included in the trade payables was a bills payable of Rs.9,000 which was met on maturity.
2. Company issued 1,00,000 equity shares of Rs.10 each to the Vendors of machinery purchased.
#### Solution
"Current ratio =""Current Assets"/"Current Liabilities"
1. A bill payable of Rs.9,000 was met on maturity will affect:
Trade Payables will reduce by Rs.9,000.
Cash will reduce by Rs.9,000.
Simultaneous decrease in current assets and current liabilities will improve current ratio.
2. Issue of shares of Rs.10.00.000 to vendor of Machinery will affect the following:
Increase in the balance of machinery
Increase in the amount of share capital
There is no affect to the current liabilities nor current assets. Thus, current ratio will remain unchanged.
Is there an error in this question or solution?
#### APPEARS IN
Solution The current ratio of X. Ltd is 2:1. State with reason which of the following transaction would i. Increase or ii. decrease or iii. not change the ratio 1. Included in the trade payables was a bills payable of Rs.9,000 which was met on maturity. 2. Company issued 1,00,000 equity shares of Rs.10 each to the Vendors of machinery purchased. Concept: Liquidity Ratios.
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# Isolation in Turing-complete reversible cellular automata
I don't know much about the terminology and the results on cellular automata, but I would like to ask a question about an conjecture I thought.
Consider Turing-complete reversible cellular automata.
For instance, Wolfram's Rule 110 cellular automaton is Turing-complete, but, if I understand correctly, it's not reversible. However, it can be made reversible by converting it into a second-order cellular automaton and then implementing the second-order CA as a first-order CA with a larger cell state space.
My question is whether it's possible to define a finite region of cells that is completely isolated from the outside.
More specifically: if there exist an initial state of a finite connected region of cells (let's call it Vault), which is composed of a connected region Interior surronded by another region Wall, such that the state of the Interior at any time t depends only on the previous state of Interior.
My conjecture is that such isolated regions are impossible.
My intuition stems from the fact that Turing-complete reversible CA can be considered very simple models of physical reality (and proponents of the digital physics hypothesis argue that the physical world is actually a cellular automaton). If I understand correctly, totally isolated physical systems can't exist, thus I conjecture by analogy that this should also apply to all Turing-complete reversible CA. (Turing-incomplete or irreversible CA can have isolated regions, but they are probably not good models of fundamental physical reality)
Has anybody already proved or disproved this assertion? Or do you have any thoughts about it?
UPDATE:
As Shor pointed out in the comments, the conjecture as stated above is false, since it is possible to use some states which are not required for Turing-completeness to achieve isolation.
I will try to reformulate the question in a way that the conjecture may possibly hold by adding more requirements on the cellular automaton.
I'm thinking of two reformulations based on two different additional requirements:
1. Minimal universality: As noted by Grochow, we can define a Turing-complete CA to be "Minimally universal" iff the CA obtained by removing one cell state (and the corresponding rules) is no longer Turing-complete.
Thus, the conjecture is: There does not exist a minimally universal reversible CA such that isolated regions can be defined.
2. Total harnessability: We may say that the computational power of a Turing-complete CA is totally harnessable iff there exist undecidable properties regarding all cell states.
More specifically, we define a CA to be totally harnessable iff $\forall s \in CellStates \ \exists$ a computable initial configuration $x(t_0,\,i)$, a finite set of cell indexes $Q$, a vector of cell states $\bar{q}$ containing $s$ and predicate $p(x,\,t)\ \equiv\ \bigwedge_{i \in Q}x(t,\,i)=q_i$ such that the question "$\exists t:\ p(x,\,t)=True$" is undecidable, and the question obtained by removing from predicate $p$ the conditions on state $s$ is decidable.
Thus the conjecture is: There does not exist a totally harnessable Turing-complete reversible CA such that isolated regions can be defined.
• If you take a Turing-complete reversible CA and add extra states and rules, it remains Turing-complete. So you can have a Turing-complete reversible CA in which isolation is possible, where the states you use for isolation are not those you use for universal computation. You might want to think about the formulation of your question more closely ... there may be some reasonable way of reformulating it so your conjecture is true. – Peter Shor Feb 26 '11 at 19:52
• @user1749: One way to (partially?) sidestep @Peter Shor's idea is to require that the "universal part" of your machine be able to access all the states and use all the rules of the CA. One interpretation (not the only one) of this would be to only consider "minimally universal" CAs, in which the removal of any state (and any rules pertaining to that state) would cause it to no longer be universal. – Joshua Grochow Feb 28 '11 at 4:48
• Yes, I thought about that. It may be interesting to consider "minimally universal" automata, but perhaps the concept is a bit too restrictive. I was thinking of extending the definition by requiring the existence of undecidable questions regarding all states. – Antonio Valerio Miceli-Barone Feb 28 '11 at 10:49
• I realise this is a very old post, but just in case it's helpful to you or someone else, second-order CAs don't necessarily behave in similar ways to the corresponding first-order CA. So making rule 110 second-order will make it reversible, but won't necessarily preserve its computational completeness. (I don't know whether it does or not.) I think it's probably not that hard to find a minimal computationally complete 1D CA though. – Nathaniel Mar 25 '15 at 7:18
The conjecture is not only false, but false for standard and seemingly-minimal models of reversible CA. Specifically, in Margolus' billiard ball model (with the Margolus neighborhood), suitably aligned large blocks of live cells act as barriers and mirrors, isolating anything beyond them.
• I was aware of Margolus block cellular automaton and the billiard ball model, but I thought that while it allowed to implement arbitrary reversible finite state automata, it wasn't obvious to me that it was Turing-complete. Now that I'm thinking about it more, I see that it's probably possible to implement a reversible Turing machine using an infinite periodic reversible circuit, and thus a Margolus block CA. – Antonio Valerio Miceli-Barone Feb 28 '11 at 21:00
I've been thinking about your problem a little more, and I think separating the Turing-universality from the isolation part may be quite hard. For example, consider a Turing-universal CA with states 0,1,2, where the Turing-universal computations never use two 2's in adjacent cells. You could also have arrange the rules so that if there are two 2's in adjacent cells, they will never be changed. I am fairly sure that one can devise such a Turing-universal CAs so that if you without the state 2, the CA is not universal. You then can't just drop the state 2, as required by property 1. I think this would also evade your new property 2, although maybe not if you required the CA to use all the rules. But it seems like it might be quite difficult to come up with a reformulation which would evade all examples such as this.
• Thanks for your answer. Do you have any idea about how such cellular automata would be defined? – Antonio Valerio Miceli-Barone Feb 28 '11 at 21:20
• Think of state 2 as the head on a universal Turing machine. Clearly the automaton is universal if there's just one head, it doesn't do anything without any heads, and with more than one head, you can get isolation. To simulate the traditional model of a Turing machine, you would need several different head states, but that still gives an example which evades your reformulations. – Peter Shor Feb 28 '11 at 21:40
• I think the requirement that the Turing-universal computations be able to access every rule (for every rule, there is some input to the universal computation which uses that rule) is maybe a better notion of minimality than the one I proposed before. – Joshua Grochow Mar 1 '11 at 23:18
• Possibly, but I'm not very sure about it. Anyway, If Eppstein remark that Margolus block CA is Turing complete is correct (and if my understanding is right, it is), then my conjecture is trivially falsified. – Antonio Valerio Miceli-Barone Mar 2 '11 at 2:09
I think your definition of isolation is flawed, in that it allows what you called the "interior" to influence the "exterior". In particular reversibility is a non issue since a Turing machine can just throw useless information away instead of erasing it.
If you define an isolated region the other way around, such that the outside is not influenced by the inside, then I think you cannot have a Deterministic Turing Machine in a finite isolated region with reversible rules.
The fact that the isolated region shall not be influenced by the outside is not so much related to reversibility and is more about determinism or knowing what your machine does without having to know the full universe, and if that's a given too (eg if each cell of the exterior are initially at a quiescent state), then the dynamics is still reversible when restrained to the isolated region, which is finite, and thus its orbits are finite, in other word, it is periodic.
Good luck defining the result of a computation when the machine reverts periodically to its initial state. Might be possible, but it does not check the definition of Turing complete I know.
Also, there is the question of how you are Turing complete within a finite space. Where is the infinite tape?
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• Revision:OCR Core 1 - Quadratics
TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 1 - Quadratics
Contents
As seen before, we referred to quadratic expressions as expressions in the form of . The term "quadratic" means that the polynomial one is referring to is simply one in which the highest power (of an unknown) is 2. (Cubic expression have a term of an unknown to the power of three, quartic expressions have a term of an unknown to the power of four, etc.).
One refers to a, b, and c as coefficients (of , , and constant term respectively).
Throughout the work in Core 1, one can assume that .
One can express quadratics in terms of their factors; this is called factor form, and can be useful for finding the solutions to the equation quickly.
The completed square form is also useful. In this form one can tell what the minimum, or maximum point of a quadratic is (when it is graphed ).
The ability to factorise a quadratic expression is essential, offering the ability to solve some quadratic equations easily, and also draw sketch graphs of quadratic functions in an effective manner.
It is a process that one can do using a trial and error process, however, with some insight it can be a quick and powerful tool.
Example
This one is very simple.
When one wishes to factorise a quadratic, one is looking for two linear algebraic expressions that will multiply together, and sometimes with a constant term, to produce a quadratic function.
Evidently, this expression cannot be divided by an integer value (such that it is simplified), so one merely has to look for two linear algebraic expressions that will multiply together to form this function.
Thinking about it logically, one wishes to solve the following:
This means that:
As we would prefer to have integer values for these variables (A, B, C, and D), one would deduce that A = B = C = D = 1.
Hence, one has deduced that:
Which is true. Hence one has factorised the quadratic.
The sum of two squares will never factorise.
This is a useful piece of information, and it is easily explained.
If the expression did factorise, it would imply that for a value of "x" (or two values), it would be possible to have the expression to be equal to zero.
One can see that this is incorrect from the explanations of the rudimentary features of the quadratic form (). The constant value dictates the position in the y-axis, and hence if it is a positive, the lowest value of the curve () would be above zero, and hence there is a contradiction.
Example
First one can factorise this.
Hence, A = C = 1
Hence:
These algebraic expressions suggest that B = -2, D = -4 (or the other way around; as A = C, it does not matter).
Hence:
Hence:
Hence:
These are the solutions.
Completing the square
Completed square form is a form of quadratic expression. It is expressing the quadratic expression in the form of a square (or multiple of a square), and constant remainder.
If one wishes to obtain a square such that the terms of , and are made, one know the following:
Hence, the remainder will be .
Example
1. Express the following expression in completed square form:
First calculate the following:
Now, one can calculate the remainder ().
Hence:
This can be helpful for finding the vertex of a curve. If one was to graph the function in this example, such that , one can deduce that the vertex would be at (-2, 1).
2. Express the following expression in completed square form:
This is slightly different from the ones that have been encountered so far.
Now one can use the method on the bracket with the positive term of .
In a situation such as this one, one must be careful to ensure that all the signs are correct. A good habit is to check the answer by multiplying out the brackets, and checking that it is identical to the original expression.
Some quadratic equations can be very hard to factorise, and some cannot be factorised, it is in these circumstances that a more versatile solution is necessary. There is a formula for finding the roots (or solutions) to a quadratic equation of the form . The quadratic equation formula is not a method that one would want to use for every quadratic equation as it is slower (in many cases) than factorising, however it is very useful in those cases where one is unable to factorise a quadratic equation.
The quadratic equation formula is derived from the completion of the square on the generic quadratic equation:
Hence:
This formula gives the solution to any quadratic equation of the form
Example
1. Solve the equation using the quadratic equation formula.
Establish the coefficients:
a = 1, b = -6, c = 8
Hence:
The discriminant
It might have occurred to some students that as there is a square root involved, it could be possible for this to have no real value.
This is precisely true. For those quadratic functions who have no x-axis intercepts, and thus no solution such that , this square root has not real value.
This can only be true if the value of .
This value () is called the discriminant, and the value of it has an effect on the number of real roots of the equation.
One can see (from the above) that if the discriminant is less than zero, there are no real roots.
Consider the situation when the discriminant is equal to zero. One would be doing the following:
This is only one value, and hence it is said that the equation has a repeated root. This occurs (graphically) when the x-axis is a tangent to the curve (if the quadratic expression is graphed, such that ).
The other situation is when the discriminant is greater than zero. Quite obviously this will cause there to be two distinct roots, as the square root will have a value (that is not zero).
(It is noteworthy that if the discriminant is, itself, a perfect square, the solutions to the equation will be integers, or fractions, unless one of the coefficients was an irrational number, and another was such that the discriminant could be a perfect square).
Simultaneous equations
As seen before, one can form a quadratic equation through the existence of two simultaneous equations, one, or both of whom are quadratic in order.
Example
1. Find the solutions to the simultaneous equations:
Hence one should form a single quadratic equation to solve:
Hence, x = -1, or x = -1.
Equations which reduce to quadratic equations
Some equations may be quartic, or some other order, however they can be reduced to be solvable using the methods one has to solve quadratic equations.
Example
1. Solve the equation:
Hence one wishes for this to be in quadratic form, and thus it would be useful to substitute for .
Hence:
Hence:
Therefore u = 4, or u = 2.
Hence:
2. Solve the equation:
First one should rearrange this into something that is equivalent to zero:
To get this to be a quadratic equation, one should substitute for .
Hence:
Hence:
u = 4, or u = 2
Hence:
This can be applied in many other circumstances, but remember that when one takes a square root, there are (generally) two results.
Also See
Read these other OCR Core 1 notes:
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# change the format of \ref (enumerate package)
I am writing a list of questions using enumerate package.
Something like this :
\documentclass[a4paper,10pt]{scrreprt}
\usepackage{enumerate}
\begin{document}
\begin{enumerate}
\item \label{itm:1} This is the question 1
\item \begin{enumerate}[(a)]
\item \label{itm:2a} This is the question 2.(a)
\item This is the question 2.(b)
\item \begin{enumerate}[-i)]
\item \label{itm:2ci}This is the question 2.(c)-i)
\item This is the question 2.(c)-ii)
\end{enumerate}
\end{enumerate}
\end{enumerate}
\ref{itm:1},\ref{itm:2a} and \ref{itm:2ci}
\end{document}
\ref{itm:1},\ref{itm:2a} and \ref{itm:2ci} gives respectively 1, 2a and 2(c)i
how can I change the format of \ref to get
1. , 2.(a) and 2.(c)-i)
if possible show how can I apply the solution locally and globally for all enumerate of the document.
• related/duplicate tex.stackexchange.com/questions/163069/… – touhami Jul 17 '15 at 23:32
• @touhami - While the two postings are obviously related to each other, by the fact that they're about the formatting of labels of and cross-references to enumerated items, the issues raised by this posting are sufficiently distinct so that it's not a duplicate. – Mico Jul 18 '15 at 1:42
I don't know how your objective may be accomplished with the tools of the enumerate package (though David Carlisle probably could!), but I know how to accomplish it with the enumitem package.
One of the many nice features of the enumitem package is that labels of enumerated items and the format of cross-references to these items can be set in a very straightforward way. For first-level items, only the ref part needs to be modified (by affixing a "." to \theenumi); the second- and third-level items both the label and the ref parts are modified.
(The code below uses the hyperref package to make obvious what the cross-references look like.)
\documentclass[a4paper,10pt]{scrreprt}
\usepackage{enumitem}
\usepackage[colorlinks]{hyperref} %% just for this example
\begin{document}
\begin{enumerate}[ref=\theenumi.]
\item This is question 1 \label{itm:1}
\item \begin{enumerate}[label=(\alph*),
ref=\theenumi(\theenumii)]
\item This is question 2.(a) \label{itm:2a}
\item This is question 2.(b)
\item \begin{enumerate}[label=-\roman*),
ref=\theenumii-\theenumiii)]
\item This is question 2.(c)-i) \label{itm:2ci}
\item This is question 2.(c)-ii)
\end{enumerate}
\end{enumerate}
\end{enumerate}
Questions \ref{itm:1}, \ref{itm:2a}, and \ref{itm:2ci} \dots
\end{document}
Your formatting goals, for both the labels and the cross-references, could also be achieved without the aid of any package, by using the following low-level LaTeX instructions instead. Note, though, that these instructions apply to all subsequent enumerated environments.
\makeatletter
\renewcommand\labelenumi{\arabic{enumi}.}
\renewcommand\theenumi{\arabic{enumi}.}
\renewcommand\labelenumii{(\alph{enumii})}
\renewcommand\theenumii{(\alph{enumii})}
\renewcommand\labelenumiii{-\roman{enumiii})}
\renewcommand\theenumiii{-\roman{enumiii})}
\renewcommand\p@enumiii{\theenumi\theenumii}
\makeatother
If you have several enumerated lists of this type, it is handy to create a dedicated enumerate-like environment that sets up all the formatting instructions, using the \newlist and \setlist instructions of the enumitem package. For instance, you might set up the following code (no new screenshot, as the output is the same as in the preceding example):
\documentclass[a4paper,10pt]{scrreprt}
\usepackage{enumitem}
\newlist{qenum}{enumerate}{3} % create a new enumerate-like environment
\setlist[qenum,1]{label=\arabic*.,
ref=\arabic*.}
\setlist[qenum,2]{label=(\alph*),
ref=\theqenumi(\alph*)}
\setlist[qenum,3]{label=-\roman*),
ref=\theqenumii-\roman*)}
\usepackage[colorlinks]{hyperref} %% just for this example
\begin{document}
\begin{qenum}
\item This is question 1 \label{itm:1}
\item \begin{qenum}
\item This is question 2.(a) \label{itm:2a}
\item This is question 2.(b)
\item \begin{qenum}
\item This is question 2.(c)-i) \label{itm:2ci}
\item This is question 2.(c)-ii)
\end{qenum}
\end{qenum}
\end{qenum}
Questions \ref{itm:1}, \ref{itm:2a}, and \ref{itm:2ci} \dots
\end{document}
• Thank you for your detailed answer. I will wait a little bit for Carlisle before accepting this answer. I have an additional question concerning enumitem package : how can I redefine the default format of label and ref ? – Aymane Fihadi Jul 18 '15 at 9:39
• @AymaneFihadi - When working with enumitem, there are basically few defaults: The package makes it very straightforward to change just about anything, but the downside (if it's even a downside) is that there are very few "default" values: You can (and must) specify the settings for \label and \ref. Does this answer your follow-up query? – Mico Jul 18 '15 at 9:45
• Just for clarification, do you mean I have to specify the values of ref and label in every list of question in the same document ? – Aymane Fihadi Jul 18 '15 at 9:52
• @AymaneFihadi - Thanks for the clarification. I'll post an addendum that shows how to set up a custom enumeration environment. – Mico Jul 18 '15 at 9:58
• Perfect! Thank you very much for all your help. – Aymane Fihadi Jul 18 '15 at 11:19
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# Three Phase Squirrel Cage Induction Machine
Description of the Three Phase Squirrel Cage Induction Machine component in Schematic Editor.
Table 1. Three Phase Squirrel Cage Induction Machine component in Typhoon HIL Schematic Editor
component component dialog window component parameters
• Property tabs:
A, B, and C are the stator winding terminals. The stator winding uses the current source interface. Any change related to the signal processing inputs of the component is documented on the component description window in real time.
## Electrical sub-system model
The electrical part of the machine is represented by the following system of equations, modeled in the stationary αβ reference frame. All rotor variables and parameters are referred to the stator.
$\left[\begin{array}{c}{v}_{\alpha s}\\ {v}_{\beta s}\\ \begin{array}{c}{v}_{\alpha r}^{}\\ {v}_{\beta r}^{}\end{array}\end{array}\right]=\left[\begin{array}{cccc}{R}_{s}& 0& 0& 0\\ 0& {R}_{s}& 0& 0\\ 0& 0& {R}_{r}^{}& 0\\ 0& 0& 0& {R}_{r}^{}\end{array}\right]\left[\begin{array}{c}{i}_{\alpha s}\\ {i}_{\beta s}\\ \begin{array}{c}{i}_{\alpha r}^{}\\ {i}_{\beta r}^{}\end{array}\end{array}\right]+\frac{d}{dt}\left[\begin{array}{c}{\psi }_{\alpha s}\\ {\psi }_{\beta s}\\ \begin{array}{c}{\psi }_{\alpha r}^{}\\ {\psi }_{\beta r}^{}\end{array}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ \begin{array}{c}{}_{}{\omega }_{r}{\psi }_{\beta r}{}_{}\\ {}_{}{-\omega }_{r}{\psi }_{\alpha r}{}_{}\end{array}\end{array}\right]$
$\left[\begin{array}{c}{\psi }_{\alpha s}\\ {\psi }_{\beta s}\\ \begin{array}{c}{\psi }_{\alpha r}^{}\\ {\psi }_{\beta r}^{}\end{array}\end{array}\right]=\left[\begin{array}{cccc}{L}_{s}& 0& {L}_{m}& 0\\ 0& {L}_{s}& 0& {L}_{m}\\ {L}_{m}& 0& {L}_{r}& 0\\ 0& {L}_{m}& 0& {L}_{r}\end{array}\right]\left[\begin{array}{c}{i}_{\alpha s}\\ {i}_{\beta s}\\ \begin{array}{c}{i}_{\alpha r}^{}\\ {i}_{\beta r}^{}\end{array}\end{array}\right]$
${T}_{e}=\frac{3}{2}p\left({\psi }_{\alpha s}{i}_{\beta s}-{\psi }_{\beta s}{i}_{\alpha s}\right)$
If the saturation effects are considered, the equations have the same form, but in that case the magnetizing flux is a function of the magnetizing current:
${\psi }_{m}=f\left({i}_{m}\right)$
where:
${\psi }_{m}=\sqrt{{\left({\psi }_{\alpha s}+{\psi }_{\alpha r}\right)}^{2}+{\left({\psi }_{\beta s}+{\psi }_{\beta r}\right)}^{2}}$
${i}_{m}=\sqrt{{\left({i}_{\alpha s}+{i}_{\alpha r}\right)}^{2}+{\left({i}_{\beta s}+{i}_{\beta r}\right)}^{2}}$
Table 2. Three Phase Squirrel Cage Induction Machine electrical subsystem model variables
symbol description
ψαs Alpha axis component of the stator flux [Wb]
ψβs Beta axis component of the stator flux [Wb]
ψαr Alpha axis component of the rotor flux, referred to the stator [Wb]
ψβr Beta axis component of the rotor flux, referred to the stator [Wb]
iαs Alpha axis component of the stator current [A]
iβs Beta axis component of the stator current [A]
iαr Alpha axis component of the rotor current, referred to the stator [A]
iβr Beta axis component of the rotor current, referred to the stator [A]
vαs Alpha axis component of the stator voltage [V]
vβs Beta axis component of the stator voltage [V]
vαr Alpha axis component of the rotor voltage, referred to the stator [V]
vβr Beta axis component of the rotor voltage, referred to the stator [V]
Rs Stator phase resistance [Ω]
Rr Rotor phase resistance, referred to the stator [Ω]
Lm Magnetizing (mutual, main) inductance [H]
Ls Stator phase inductance [H] ( $={L}_{ls}+{L}_{m}$ )
Lr Rotor phase inductance, referred to the stator [H] ( $={L}_{lr}+{L}_{m}$ )
ωr Rotor electrical speed [rad/s] ( $=p{\omega }_{m}$ )
p Machine number of pole pairs
Te Machine developed electromagnetic torque [Nm]
ψm Magnetizing flux [Wb]
im Magnetizing current [A]
## Mechanical sub-system model
Motion equation:
Table 3. Mechanical sub-system model variables
symbol description
Jm Combined rotor and load moment of inertia [kgm2]
Te Machine developed electromagnetic torque [Nm]
Tl Shaft mechanical load torque [Nm]
b Machine viscous friction coefficient [Nms]
Note: Motion equation is the same for all of the rotating machine models.
## Electrical
This component offers two levels of model fidelity, designated by the Model Type property. The following options are available:
• linear
• nonlinear
If the selected Model Type is nonlinear, the following saturation types can be specified:
• flux vs current
• absolute inductance vs current
Table 4. Electrical parameters
symbol description
Rs Stator phase resistance [Ω]
Rr Rotor phase resistance, referred to the stator [Ω]
Lls Stator leakage inductance [H]
Llr Rotor leakage inductance, referred to the stator [H]
Lm Magnetizing (mutual, main) inductance [H]
im vector List of instantaneous values of the magnetizing current [A]
psim vector List of instantaneous values of the magnetizing flux [Wb]
Lm vector List of instantaneous values of the magnetizing inductance [H]
The Dynamic resistances checkbox enables real-time change to stator and rotor resistances. If calculations of these resistances depend on the temperature, they can be routed to a load input with signal processing connections, provided that this property is enabled. The vectorized array [Rs, Rr] is in ohms.
Note: This property is only available on linear machines, and fault properties cannot be simulated simultaneously with this property. Additionally, simulation accuracy may deteriorate on smaller simulation time steps for virtual HIL.
The three phase squirrel cage induction machine component can include magnetic saturation effects. In that case, magnetizing flux or magnetizing inductance is defined as a function of the magnetizing current. That function is represented in the form of lookup table. The lookup tables use linear interpolation and linear extrapolation.
Saturation can be parametrized in the following ways:
1. magnetizing flux vs magnetizing current
2. magnetizing inductance vs magnetizing current
im_vector = [0.0, 0.661428, 0.957988, 1.224002, 1.527775, 1.836498, 2.485056, 3.197537, 4.162313, 5.57879, 8.211348, 12.342407, 22.172606]
psim_vector = [0.0, 0.125279, 0.192308, 0.25488, 0.318532, 0.382499, 0.511695, 0.635623, 0.76725, 0.885866, 1.007544, 1.097936, 1.186302]
im_vector = [0.0, 0.661428, 0.957988, 1.224002, 1.527775, 1.836498, 2.485056, 3.197537, 4.162313, 5.57879, 8.211348, 12.342407, 22.172606]
Lm_vector = [0.0, 0.189407, 0.200741, 0.208235, 0.208494, 0.208277, 0.205909, 0.198785, 0.184332, 0.158792, 0.122701, 0.088956, 0.053503]
## Mechanical
Table 5. Mechanical parameters
symbol description
pms Machine number of pole pairs
Jm Combined rotor and load moment of inertia [kgm2]
Friction coefficient Machine viscous friction coefficient [Nms]
Unconstrained mechanical angle Limiting mechanical angle between 0 and 2π
symbol description
Load source Load can be set from SCADA/external or from model (in model case, one signal processing input will appear)
Load ai offset Assigned offset value to the input signal representing external torque command
Load ai gain Assigned gain value to the input signal representing external torque command
External load enables you to use an analog input signal from a HIL analog channel with the load_ai_pin address as an external torque/speed load, and to assign offset (V) and gain (Nm/V) to the input signal, according to the formula:
${T}_{l}=load_ai_gain·\left(AI\left(load_ai_pin\right)+load_ai_offset\right)$
Note: An analog input pin may be overwritten if another component uses the same analog input pin. If another property (from the same or a different component) uses the same analog input pin, the input signal value will be applied to only one of those properties. For instance, if both the load and the resolver carrier signal use the same analog input pin, the signal value will only be applied only to one of these.
## Feedback
Table 7. Feedback parameters
symbol description
Encoder ppr Incremental encoder number of pulses per revolution
Resolver pole pairs Resolver number of pole pairs
Resolver carrier source Resolver carrier signal source selection (internal or external)
External resolver carrier source type External resolver carrier signal source type selection (single ended or differential); available only if the Resolver carrier source property is set to external
Resolver carrier frequency Resolver carrier signal frequency (internal carrier) [Hz]
Resolver ai pin 1 Resolver carrier input channel 1 address (external carrier)
Resolver ai pin 2 Resolver carrier input channel 2 address (external carrier); available only if the External resolver carrier source type property is set to differential
Resolver ai offset Resolver carrier input channel offset (external carrier)
Resolver ai gain Resolver carrier input channel gain (external carrier)
Absolute encoder protocol Standardized protocol providing the absolute machine encoder position
If an external resolver carrier source is selected, the source signal type can be set as either single ended or differential. The single ended external resolver carrier source type enables use of an analog input signal from the HIL analog channel with the res_ai_pin_1 address as the external carrier source. Additionally, offset (V) and gain (V/V) values can be assigned to the input signal, according to the formula:
$res_carr_src=res_ai_gain·\left(AI\left(res_ai_pin_1\right)+res_ai_offset\right)$
The differential external resolver carrier source type enables use of two analog input signals from the HIL analog channels with the res_ai_pin_1 and the res_ai_pin_2 addresses. Analog signals from these HIL analog inputs are subtracted, and the resulting signal is used as the external differential carrier source. Additionally, offset (V) and gain (V/V) values can be assigned to the input signal (similarly to the single ended case), according to the formula:
$res_carr_src=res_ai_gain·\left(\left(AI\left(res_ai_pin_1\right)-AI\left(res_ai_pin_2\right)\right)+res_ai_offset\right)$
Note: Differential external resolver signal support is available from the 2023.1 software version. Only the single ended external resolver option was available in previous software versions.
Note: In order to get a resolver signal with an amplitude of 1 when using an external carrier signal, the offset and the gain should be chosen in such a way that the resolver carrier signal has an amplitude of 1 after the adjustment. As shown in Figure 1, the sinusoidal signal used to generate external resolver carrier source is fed to HIL analog input 1. The analog input signal is scaled in order to get the resolver signals with an amplitude of 1.
Note: An analog input pin may be overwritten if another component uses the same analog input pin. If another property (from the same or a different component) uses the same analog input pin, the input signal value will be applied to only one of those properties. E.g. if both the load and the resolver carrier signal use the same analog input pin, the signal value will be applied only to one of these.
The following expression must hold in order to properly generate the encoder signals:
$4·enc_ppr·{f}_{m}{·T}_{s}\le 1$
Table 8. Variables in the encoder limitation expression
symbol description
enc_ppr Encoder number of pulses per revolution
fm Rotor mechanical frequency [Hz]
Ts Simulation time step [s]
Note: While the machine speed is positive, the encoder channel B signal leads the encoder channel A signal.
Note: Absolute encoder protocol is not supported on HIL402 (configurations: 1, 2, 3, and 4).
## Snubber
All machines with current source based circuit interfaces have the Snubber tab in the properties window where the value of snubber resistance can be set. Snubbers are necessary in the cases when an inverter or a contactor is directly connected to the machine terminals. This value can be set to infinite (inf), but it is not recommended when a machine is directly connected to the inverter since there will be a current source directly connected to an open switch. In this case, one of each switch pairs S1 and S2, S3 and S4, and S5 and S6 will be forced closed by the circuit solver in order to avoid the topological conflicts. On the other hand, with finite snubber values, there's always a path for the currents Ia and Ib, so all inverter switches can be open in this case. Circuit representations of this circuit without and with snubber resistors are shown in Figure 3 and Figure 4 respectively. Snubbers are connected across the current sources.
Note: Snubbers exist only in the machine components that have the current source based circuit interface.
Note: Snubbers are dynamic, which means that the snubber is dynamically added to circuit modes where topological conflicts are detected.
Table 9. Snubber parameters
symbol description
Rsnb stator Stator snubber resistance value [Ω]
## Output
This block tab enables a single, vectorized signal output from the machine. The output vector contains selected machine mechanical and/or electrical variables in the same order as listed in this tab.
Note: All machine components have the Execution rate, Electrical torque, Mechanical speed, and Mechanical angle, but the rest of the signals differ from component to component.
Table 10. Output parameters
symbol description
Execution rate Signal processing output execution rate [s]
Electrical torque Machine electrical torque [Nm]
Mechanical speed Machine mechanical angular speed [rad/s]
Mechanical angle Machine mechanical angle [rad]
Stator alpha axis current Alpha axis component of the stator current [A]
Stator beta axis current Beta axis component of the stator current [A]
Rotor alpha axis current Alpha axis component of the rotor current, referred to the stator [A]
Rotor beta axis current Beta axis component of the rotor current, referred to the stator [A]
Stator alpha axis flux Alpha axis component of the stator flux [Wb]
Stator beta axis flux Beta axis component of the stator flux [Wb]
Rotor alpha axis flux Alpha axis component of the rotor flux, referred to the stator [Wb]
Rotor beta axis flux Beta axis component of the rotor flux, referred to the stator [Wb]
## Fault
In this block tab, you can choose the fault simulation and specify the type of fault. The chosen fault can be in one of the phases. When choosing the fault, a machine input terminal is either created or extended by a new vectorized input for defining fault parameters. For the ITSC (Interturn Short Circuit) fault type, the vectorized array is [severity factor, Rf]. The machine component fault tab layout is shown here:
Table 11. Three Phase Squirrel Cage Induction Machine fault parameters
symbol description
Fault Simulation Enabling input for fault parameters
Fault Type Type of the fault to be simulated
Fault Location Phase in which the fault is simulated
Execution rate Signal processing execution rate at which the model matrices are calculated [s]
## Electrical sub-system model for a machine with an ITSC fault in phase a
Table 12. Three Phase Squirrel Cage Induction Machine fault model variables and parameters
symbol description
vαs Alpha axis component of the stator voltage [V]
vβs Beta axis component of the stator voltage [V]
vαs2 Voltage of the short circuited turns in the phase a [V]
iαs Alpha axis component of the stator current [A]
iβs Beta axis component of the stator current [A]
iαr Alpha axis component of the rotor current, referred to the stator [A]
iβr Beta axis component of the rotor current, referred to the stator [A]
if Fault current [A]
ψαs Alpha axis component of the stator flux [Wb]
ψβs Beta axis component of the stator flux [Wb]
ψαr Alpha axis component of the rotor flux, referred to the stator [Wb]
ψβr Beta axis component of the rotor flux, referred to the stator [Wb]
ψαs2 Flux of the short circuited turns in the phase a [Wb]
Rs Stator phase resistance [Ω]
Rr Rotor phase resistance, referred to the stator [Ω]
Rf Fault resistance (minimum value 0, maximum value 100 Ω)
µ Fault severity factor - short-circuited number of turns (minimum value 1e-4, maximum value 1)
Lls Stator leakage inductance [H]
Lm Magnetizing (mutual, main) inductance [H]
Ls Stator phase inductance [H] ( $={L}_{ls}+{L}_{m}$ )
Lr Rotor phase inductance, referred to the stator [H] ( $={L}_{lr}+{L}_{m}$ )
ωr Rotor electrical speed [rad/s] ( $=p{\omega }_{m}$ )
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# Math Help - help please~
‘The quadratic equations (x^2)-6x+2k=0 and (x^2)-5x+k=0 have a common root A. (i.e. A is a root of both equations)
Show that A=k and hence find the value(s) of k.
2. Originally Posted by stupid_kid
‘The quadratic equations (x^2)-6x+2k=0 and (x^2)-5x+k=0 have a common root A. (i.e. A is a root of both equations)
Show that A=k and hence find the value(s) of k.
Let the roots of x^2-6x+2k=0 be A and mu, and the roots of x^2-5x+k=0 be A and lambda
The sum of the roots of x^2-6x+2k=0:
A+mu=6 ...(1)
and the product of the roots is:
A.mu=2k ...(2)
The sum of the roots of x^2-5x+k=0:
A+lambda=5 ...(3)
and the product of the roots is:
A.lambda=k ...(4).
Divide (2) by (3) to get mu=2.lambda ...(5)
Subtract (3) from (1) to get mu-lambda=1 ...(6)
Substituting (5) into (6) gives lambda=1, so by (5) mu=2, and by (2) A=k.
RonL
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# Can't find the flaw in the reasoning for this proof by induction?
I was looking over this problem and I'm not sure what's wrong with this proof by induction.
Here is the question:
Find the flaw in this induction proof.
Claim $3n=0$ for all $n\ge 0$.
Base for $n=0$, $3n=3(0)=0$
Assume Induction Hypothesis: $3k =0$ for all $0\le k\le n$
Write $n+1=a+b$ where $a>0$ and $b>0$ are natural numbers each less than $n+1$
Then $3(n+1) = 3(a+b) = 3a + 3b$
Apply Induction hypothesis to $3a$ and $3b$, showing that $3a=0$ and $3b=0$. Therefore, $3(n+1)=0$
The statement they are trying to prove is clearly absurd but I'm having trouble with the logic in the proof by induction. It just seems like the person who wrote this proof used strong induction and applied the induction hypothesis to proof the implication.
• Its saying that 3n is equal to 0 for any n>0. An easy counter example would be n=1 which makes 3(1) = 3 which is not equal to 0. – user262291 Feb 19 '16 at 4:47
• Yes you are right, he used strong induction. – SchrodingersCat Feb 19 '16 at 4:47
• Sorry, misread the question at first. My fault, so I deleted the comment – zz20s Feb 19 '16 at 4:48
• So, it's not enough to make the step for some $k$. You must make it for every $k\ge 0$ – TobiMcNamobi Feb 19 '16 at 12:09
• Note that in general, the introduction of values that don't exist or expressions that are undefined is a pretty typical way to introduce subtle errors into a proof, so always be sure to check that all values can actually exist and that operations do not lead to undefined behavior. – Kyle Strand Feb 20 '16 at 0:29
The problem is it doesn't work for the first step after the base case: there do not exist $a \gt 0$, $b \gt 0$ such that $a + b = n + 1$ when $n = 0$. This is a variant of all horses are the same color.
• What is the exact answer for the Same Colour horse argument ? – user230452 Feb 19 '16 at 12:39
• Imagine doing the induction step on the base case n = 1. Induction step: Assume true for n (n=1). Take a group of n+1 (2) horses. Remove a horse. The remaining n (1) horses are all the same color. Remove another horse. The remaining n-1 (0) horses are the same color as the horse you removed (no, they aren't; there are no remaining horse). Add the first horse. You have n (1) horses so they are all the same color as the remaining horses (no, it isn't because there were no remaining horses). Add the second horse back again. It color hasn't changed (no, but the other horse isn't the same – fleablood Feb 19 '16 at 17:35
It says "write n+1 = a +b where a and b are greater than 0". That can not be done in your base case n=0. You may not make any assumptions in your induction step that are not equally valid in the initial case.
In other words... your induction step assumes both 3k = 0 AND k $\ge$ 1. You never made (and can NOT make) any initial case where both are true.
• @user262291: this should be the accepted answer – image Feb 19 '16 at 12:58
• The problem is not that "$a$ and $b$ are greater than $0$" fails for $n=0$, but that it fails for $n=1$. Since your base case requires a direct proof anyway, there's no reason it must satisfy the specific assumptions of the induction step; all that's required is that every case after the base case satisfies the necessary assumptions. – Kyle Strand Feb 20 '16 at 0:26
• It doesn't fail for n =1. (a=1,b=1 then n+1=a+b). If we need a presumption to get from k to k+1, that presumption must hold in our base case. Otherwise we can't ever get beyond the base case. – fleablood Feb 20 '16 at 1:00
• @fleablood: What Kyle Strand is getting at is that the actual "base case" or "initial case" in the OP's proof is the (correct) statement that $3 n = 0$ when $n = 0$. Your answer is using the terms "base case" and "initial case" to refer to the lowest-valued instance of the induction step (i.e., the statement that $3 n = 0 \rightarrow 3 ( n + 1 ) = 0$ when $n = 0$), which is a very confusing abuse of terminology. – ruakh Feb 20 '16 at 18:48
• That's the base case I am assuming too. And it is when n=0 that no two a,b greater than 0 so that a+b =n+1 can exist. You can not use the assumption in the induction step because you have not established that the exist in any case where 3n =0. The only case we've established where 3n=0 is n=0 and in that case no such a,b exist. I'm not abusing "base case", base case means n=0. What we rely upon in the induction case must be valid in the base case. Else our induction isn't valid. Or alternatively, for our induction to be valid we must find a base as case where both assumptions are true. – fleablood Feb 20 '16 at 19:57
Others have given you the reason that the prove breaks down. Let me explain instead how to find the fault. We know that the claim is true for $n = 0$ but false for $n = 1$. Hence something must go wrong for $n = 1$. Since the claim for $n = 1$ relies only on the claim for $n = 0$, and the latter is true, what must go wrong is the step where we show that the claim for $n = 0$ implies the claim for $n = 1$. This leads to the discovery shared by the other answers, that $1$ cannot be written as a sum of two smaller natural numbers.
Here is another way. Often we do regular induction, in which we derive $P(n+1)$ from $P(n)$. Here we use strong induction, but that seems unnecessary: if $n+1 = a+b$, can't we just take $b = 1$? That will allow deriving $P(n+1)$ from $P(n)$ and $P(1)$. Once we consider the consequences for $n = 1$, we immediately find the mistake.
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Multiply these and you have the area, in square units: Find the angles of the rhombus. Thanks for contributing an answer to Mathematics Stack Exchange! Another approach is Cartesian geometry: Let $A=(0,0), B=(b,0), D= (p,q), C=(p+b,q)$ with the condition that $b^2=p^2+q^2$ (for the same side lengths). Find the side of rhombus. Since $\angle IAB=\angle ICB$, and $\angle IBA=\angle IBC$, their remaining angles must be equal. MEDIUM. The diagonals of a rhombus are the lines that connect the opposite vertices (corners) in the center of the shape. You can use these properties in the following practice geometry questions, first, to solve for a missing variable x, and […] The gradients of the diagonals are $$m_1 = \frac{q}{p+b}, \ \ \ m_2 = \frac{q}{p-b}$$. View Answer. For example, all of its sides are congruent, and it contains diagonals that are perpendicular bisectors and that bisect the angles of the rhombus. The diagonals of rhombuses also form four right triangles, with hypotenuses equal to the side length of the rhombus and legs equal to one-half the lengths of the diagonals. This rhombus calculator can help you find the side, area, perimeter, diagonals, height and any unknown angles of a rhombus if you know 2 dimensions. The length of each side in the figure is 6. If you think about the information you know, you will see that you can build a similar rhombus of any size. Draw lines $AC$ and $AD$. Solution The side of the rhombus is one fourth of its perimeter, that is 52/4 = 13 cm. MathJax reference. Every rhombus has 4 congruent sides so every single square is also a rhombus. The same argument as in the preceding paragraph shows that all the angles it forms with the sides are equal. Just write down the … Since i already knew the value of y i just plugged it in...but how would you find the other diagonal length if you know nothing else about the variable. Area of a Rhombus = 1/2 x 12 x √ ((4 x 7 2) - 12 2) = 6 x √52 Once you have one diagonal length, divide that in half to get each half of the diagonal separately. Suppose that angle at D is $x.$ Then the diagonal length is $$6\sqrt{2-2\cos x}=12\sin(x/2).$$ Here, $x=100^{\circ}.$ There is no nice value for $\sin (50^{\circ})$ - the best answer for the diagonal length is $12\sin (50^{\circ})$ and trig ratios are inherently involved. If you draw the two diagonals, you can by angle-chasing show that the four triangles so formed are all congruent. A rhombus is a type of Parallelogram only. That means, a = 10cm. Use MathJax to format equations. The diagonals of rhombus are 20 cm and 48 cm respectively then find the side of the rhombus. We give a traditional "angle-chasing" argument. Finding half a diagonal in a rhombus using trigonometric ratios? Each sides of a rhombus is 1 3 c m and one diagonal is 1 0 c m. Find (i) the length of its other diagonal (ii) the area of the rhombus. These trig functions are not at all "nice," so there is no way to sneak around them. Work now with the other diagonal. q = 16cm. Diagonals of rhombus are 15 cm and 20 cm. I know the: * Value: * Other Tools You May Find Useful Trigonometry Calculator Triangle Calculator Cone Calculator Law of Sines Calculator. How To Find Diagonal Of Rhombus When Side Is Given, Good Tutorials, How To Find Diagonal Of Rhombus When Side Is Given That will show that the angle between the diagonals is a right angle. Find the diagonal of a rhombus if given side and one-half angle ( , ) : Find the diagonal of a rhombus if given side and other diagonal ( , ) : Find the diagonal of a rhombus if given one-half angle and other diagonal ( , ) : Find the diagonal of a rhombus if given area and other diagonal ( , ) : diagonal of a rhombus : = Digit 1 2 4 6 10 F. deg. We can … A rhombus is a quadrilateral having 4 sides of equal length, in which both the opposite sides are parallel, and opposite angles are equal. Diagonals 2. Ads. However I need to find the length of the diagonal without using Trig. How To Find Diagonal Of Rhombus DOWNLOAD IMAGE. It only takes a minute to sign up. So, the first and perhaps easiest way to find area of a rhombus is to find the length of one side and the rhombus's height. Examples: Input: A = 10, theta = 30 Output: 19.32 5.18. Since, both angles and are adjacent to angle --find the measurement of one of these two angles by: . Can someone just forcefully take over a public company for its market price? Find the length of the second diagonal of the rhombus. Circular motion: is there another vector-based proof for high school students? You cannot actually cut up every rhombus you meet, though, so consider what that one constructed, perpendicular side really is: the height, or altitude, of the rhombus. 2:59. Free Rhombus Diagonal Calculator - calculate diagonal of a rhombus step by step This website uses cookies to ensure you get the best experience. EASY. (without measuring the angle) Is there any equation where the side is related to the diagonal only? A … The square of the long diagonal is $6^2(2-2\cos 100^\circ)$, so "trig ratio" is built into the answer. Then its area and perimeter are:(A) 150 cm’, 50 cm - Duration: 2:59. Why would a company prevent their employees from selling their pre-IPO equity? Does Texas have standing to litigate against other States' election results? Given two integers A and theta, denoting the length of a side of a rhombus and the vertex angle respectively, the task is to find the length of the diagonals of the rhombus. He Perimeter Of A Rhombus Is 146 Cm One Of Its Diagonals Is 55 Cm. The surface we refer to as rhombus today is a cross section of the bicone on a plane through the apexes of the two cones. A rhombus whose angles are all right angles is called a square . Its diagonals perpendicularly bisect each other. This is an altitude of it. What magic items from the DMG give a +1 to saving throws? ratios. Did Edward Nelson accept the incompleteness theorems? If the lengths of diagonals are 2a & 2b Given two integers A and X, denoting the length of a side of a rhombus and an angle respectively, the task is to find the area of the rhombus. That will show that the angle between the diagonals is a right angle. Work now with the other diagonal. $\endgroup$ – André Nicolas Jul 3 '12 at 15:13 | and we already know ac = 12 cm. Area Of Rhombus Formula Definition And Derivation With Examples. You can use these properties in the following practice geometry questions, first, to solve for a missing variable x, and […] Any ideas how i could find that ? Find the side of a rhombus if you know 1. You also have the length of the sides. Why do rhombus diagonals intersect at right angles? Calculate the side of a rhombus if given diagonal and one-half angle ( a ) : Calculate the side of a rhombus if given area and angle ( a ) : Calculate the side of a rhombus if given perimeter ( a ) : side of a rhombus … Step 1: Find the length of diagonal 1, i.e. share | cite | improve this question | follow | edited Jul 15 '14 at 10:57. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Example: Side = 10 cm and d1 = 16 cm. Find an answer to your question The length of the diagonal of rhombus is 30 cm and 40 cm . … How does the recent Chinese quantum supremacy claim compare with Google's? This is one side of the triangle. On the other hand, I think your rhombus has a corner somewhere along the line y = -1. A rhombus has the following properties: All sides are equal. Since, by definition, all four sides of a rhombus are the same length, the formula is {\displaystyle P=4S}, where {\displaystyle P} equals the perimeter, and … DOWNLOAD IMAGE . How To Find Exact Values Of Sin Cos Tan Without A ... How To Find Buried Treasure In Minecraft Easier, How To Find Ceiling Joists With Stud Finder, How To Find Nash Equilibrium In Mixed Strategies. The opposite interior angles must be equivalent, and the adjacent angles have a sum of degrees. There are 3 ways to find the area of Rhombus.Find the formulas for same and Perimeter of Rhombus in the table below. Area and angle Answer to: how to find diagonals of a rhombus By signing up, you'll get thousands of step-by-step solutions to your homework questions. geometry. Properties Of Rhombuses Rectangles … Diagonal of a rhombus are perpendicular to each other. Input: A = 6, theta = 45 Output: 11.09 4.59 Diagonals of parallelogram intersect at $90^\circ$ if and only if figure is rhombus. DOWNLOAD IMAGE. If it were just one rhombus that existed you would not have arbitrary length diagonals to start with but only two distinct values the short and long diagonal. The square of the long diagonal is $6^2(2-2\cos 100^\circ)$, so "trig ratio" is built into the answer. Since the perimeter is of is , and by definition a rhombus has four sides of equal length, each side length of the rhombus is equal to . Why do you need to find the length of the diagonal? The same argument as in the preceding paragraph shows that all the angles it forms with the sides are equal. Find the lengths of half of diagonals then find a side of a rhombus using Pythagoras theorm And multiply by 4 to get the perimeter of a rhombus. Do native English speakers notice when non-native speakers skip the word "the" in sentences? The area of a kite is found in the same way as a rhombus, that is, by multiplying $$\frac{1}{2}$$ by the diagonal by the other diagonal. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. And so essentially the combination of these two altitudes is really just a diagonal of this rhombus. Is every field the residue field of a discretely valued field of characteristic 0? Given two vertices, how to find the other two vertices of a rhombus? The diagonals of a rhombus are perpendicular and form four right triangles through their intersection. Note A rhombus has three linear measure elements: the measures of the two of its diagonals and the side length. and 8 cm. A rhombus has a side length of 7mm and diagonal of 12mm. In a rhombus, both diagonals are perpendicular bisectors of each other. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The length of the diagonals of a rhombus are 6 cm and 8 cm. One way to prove this is by using vectors and dot products. Rhombus is a quadrilateral having four equal sides. But these two angles add up to a "straight angle" ($180^\circ$), so each of them must be $90^\circ$. What technique is it that causes a guitar to whine its notes? DOWNLOAD IMAGE 9 2 Acid Base Titrations Chemistry Libretexts DOWNLOAD IMAGE... 2131 using statcrunch to find the frequency distribution and histogram the stats files dawn wright phd. I am trying to show that the diagonals of a RHOMBUS intersect each other at 90 degrees What is a rhombus? Given two integers A and theta, denoting the length of a side of a rhombus and the vertex angle respectively, the task is to find the length of the diagonals of the rhombus.. ! p = 12cm. I do not believe you can find the diagonals without some trigonometry. Now look at $\triangle AIB$ and $\triangle CIB$. A rhombus is a parallelogram with some interesting and useful properties. To find the value of diagonal , we must first recognize some important properties of rhombuses. It is the distance between A and C. The diagonals of a rhombus are perpendicular to each other making 4 right triangles when they intersect each other at the centre of the rhombus. Answer: Yes, a square is a rhombus. So, . Code to add this calci to your website Explanation: . It splits it into two symmetric right triangles. d … [2] X Research source Let's say the diagonals are 6 cm. Examples: Input: A = 10, theta = 30 Output: 19.32 5.18 Input: A = 6, theta = 45 Output: 11.09 4.59 rev 2020.12.10.38158, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Where can I travel to receive a COVID vaccine as a tourist? Angle and angle must each equal degrees. The lengths of the diagonals are inextricably tied to trig functions of $80^\circ$ or relatives. For a rhombus, side, s and diagonal d1 is given. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. From the formula of the Theorem. So the "rhombus" could actually be a square with sides of 4√5, and we could find the other two corners using the distance formula. Note that by the definition of rhombus, $\triangle ABC$ is isosceles, so $\angle BAC=\angle ACB$. so, p = 12cm (this is one of the diagonal) Formula and Solution: So, we already knew one of the diagonal and now we got another one, so now we can easily calculate the Area of Rhombus. So $\angle AIB=\angle CIB$. Mathematics Stack Exchange Let 's say the diagonals are 6 cm and 48 cm respectively find. Diagonal separately of degrees copy and paste this URL into your RSS.... Coefficients b0 then the at $\triangle ADC$ are congruent, $\triangle$... Over a public company for its market price to fully reveal a backstory in the table below get each of. Any rhombus must have a sum of degrees angles have a sum degrees... Triangles through their intersection sneak around them source Let 's say the diagonals is 55 cm X y...: we have d2 = 2 [ s^2- ( d1/2 ) ^2 ].! Corner angles, diagonals, height, perimeter and area of a rhombus whose angles are all congruent notice... On writing great answers cm and 48 cm respectively then find the length of the rhombus English notice... Multiply these and you have the area, in square units: step 1: find the of. In half to get each half of the side of the two intersect. And 8 cm could think of an isosceles Triangle over here the how to find diagonal of rhombus ) is there another proof. Two of its diagonals is a right how to find diagonal of rhombus angle between the diagonals are inextricably tied trig. Butt plugs '' before burial an answer to mathematics Stack Exchange step by step website! A discretely valued field of a discretely valued field of characteristic 0 angles it forms with the are. Clicking “ Post your answer ”, you agree to our terms of service, Policy! Its notes all the angles it forms with the sides are parallel and opposite angles are right. The line y = -1 $so they are perpendicular and form four right triangles through their intersection Exchange ;... Another 5x-30 or responding to other answers litigate against other States ' election results 2 [ s^2- ( ). 'S say the diagonals are labeled z and the other diagonal of the shape an answer to Stack... Privacy Policy and Cookie Policy and form four right triangles through their intersection since, both angles and adjacent. Dac=\Angle DCA=\angle BAC=\angle ACB$ its diagonals and the side is related to the?. And perimeter are: ( a ) 150 cm ’, 50 -! \$ if and only if figure is 6 claim compare with Google 's diagonal d1 is is! Preceding paragraph shows that all the angles it forms with the sides equal. For contributing an answer to mathematics Stack Exchange given then how can i not maximize Activity Monitor to screen. Diagonals break up the rhombus you agree to our terms of service, privacy Policy and Policy... 146 cm one of the side of a rhombus is one fourth of its perimeter that! That other diagonal of this rhombus adjacent to angle -- find the is... nice, '' so there is no way to find the value diagonal! Have d2 = 2 [ s^2- ( d1/2 ) ^2 ] ^0.5 clarification, or responding to answers! Rss reader ; user contributions licensed under cc by-sa important properties of rhombuses Rectangles for. On the other y+5 the exact same argument as in the figure is 6 best to reveal! Are parallel and opposite angles are all congruent before burial know, you can build similar... Believe you can find the value of diagonal 1, i.e ”, you agree our... A 1 and are adjacent to angle -- find the length of each side of diagonal.
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How to count number of bases and subspaces of a given dimension in a vector space over a finite field?
Let $V_{n}(F)$ be a vector space over the field $F=\mathbb Z_{p}$ with $\dim V_{n} = n$, i.e., the cardinality of $V_{n}(\mathbb Z_{p}) = p^{n}$. What is a general criterion to find the number of bases in such a vector space? For example, find the number of bases in $V_{2}(\mathbb Z_{3})$. Further, how can we find the number of subspaces of dimension, say, $r$?
I need a justification with proof. I have a formula, but I am unable to understand the basic idea behind that formula.
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First, let's count the number of ordered bases. Start by counting the number of ways to choose the first vector.... – Hurkyl May 8 '12 at 11:24
1 Answer
Definition 1
For any natural numbers $n$ and $k$, define the Gaussian binomial coefficient, $\binom n k_q$ by the number of $k$ dimensional subspaces of a $n$ dimensional subspace.
Theorem 2
$$\binom n k_q=\dfrac{(q^n-1)(q^n-q)\cdots(q^n-q^{k-1})}{(q^k-1)(q^k-q)\cdots(q^k-q^{k-1})}$$
Proof.
To specify a $k$-dimensional subspace, we need to specify $k$ linearly independent vectors. The first vector can be chosen from among the non-zero vectors in $q^n-1$ ways. Note that $0 \in S \implies S$ is linearly dependent. The second vector must be chosen outside the span of this vector. Since, the first vector generates a subspace of dimension $1$, we have that there are $q^n-q$ choices. Proceeding this way, we get that, there are $(q^n-1)(q^n-q) \cdots(q^n-q^{k-1})$ ways of specifying a linearly independent set of cardinality $k$.
Now note that, there are many linearly independent $k$ sets, that generate the same subspace. So, we need to divide this number by the number of $k$ sets that generate the same subspace. But, this is what we have already counted in a different fashion: We are asking for the number of basis for a $k$ dimensional subspace. That will be the number of linearly independent $k$ sets in a $k$-dimensional space. So, set $n=k$ in the previous count.
This gives us the claim. $\blacksquare$
Related Reading
• This blogpost by Prof. Peter Cameron is a nice exposition on Gaussian Coefficients.
• Prof. Amritanshu prasad wrote an expository note on counting subspaces that appeared in Resonance in two parts.
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Will the downvoter care to explain? – user21436 May 8 '12 at 11:39
To the OP: My notation is different from yours. Please note that I have replace $p$ by $q$ in my answer. – user21436 May 8 '12 at 11:43
I am confused about the choice of second vector. i mean how $q^{n}-q$ for second vector is coming? can you explain please? – srijan May 8 '12 at 12:16
You have chosen the first vector. Say, $v$ wass your choice. What is the span of this vector $v$? – user21436 May 8 '12 at 12:17
sir number of elements in a space generated by span(v) will be q-1 . Am i correct? – srijan May 8 '12 at 12:54
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# Relaxation Phenomena
There are many contributions to the hyperfine field at the nucleus as seen in Equation 2.17 but the major contributor for transition metals such as Fe, when in zero applied field, is . This arises from the polarising effect of unpaired electron spins with the direction of the field being related to that of the electron spins. However, this direction is not invariant and can flip after a period of time. This is the relaxation phenomenon. The effects upon the Mössbauer lineshape depend upon the relative time scales of measurement and the relaxation mechanism, there being three time scales to consider: the lifetime of the Mössbauer event, the Larmor precession time and the relaxation time.
The lifetime of the Mössbauer event, , which is also the limiting time scale of the measurement technique, is determined by the Heisenberg uncertainty relationship as shown in Equation 2.3. For Fe this is of the order of .
The second time scale to consider is the minimum time required for the nucleus to detect the hyperfine field. This is usually assumed to be equal to the Larmor precession time, , which can be considered as the time taken for a nuclear spin state, , to split into substates under the influence of a hyperfine field. is proportional to the magnitude of the hyperfine field (and hence related to the nuclear energy levels as in Equation 2.16) with the following relation
(2.26)
where is the gyromagnetic constant and is the nuclear Bohr magneton. In iron oxides the hyperfine field is giving of the order of . This means that and hence the hyperfine fields are detectable by the technique.
The final time scale is the relaxation time, , associated with the time dependent fluctuations of the electron spin. For the hyperfine field to be observed it must remain constant at the nucleus for at least one Larmor precession period.
There are three regimes which are important when considering the effect of relaxation on the Mössbauer lineshape:
1. If then the hyperfine field is static during a single Larmor precession period. The spectral lines are narrow and Lorentzian in shape.
2. If then the nucleus experiences a time averaged hyperfine field. The magnitude is less than the value obtained for a static field as the interaction will have changed many times during a single precession period and tends to zero as decreases. Narrow Lorentzian lines are still observed.
3. If then resonance between the relaxation and the precession occurs leading to complex spectra and broadened lineshapes. As is proportional to the energy difference between the spectral lines for the outer lines will be less than for the inner lines, causing the inner lines of a sextet to broaden and disappear before the outer ones.[6]
The two main mechanisms involved in the spin relaxation are Spin-Spin and Spin-Lattice relaxation.
Subsections
Dr John Bland, 15/03/2003
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# Help understanding a proof
1. Apr 10, 2014
### trash
I'm reading about countable and uncountable sets, I found the following statement: "The set of the functions from $\mathbb{Z}$ to $[0,1]$ is uncountable" with the following proof: "To see that, suppose the set countable having the list $\{f_1,f_2,\dots\}$ and define $f(x) = f_n(1/n)$ if $x=1/n$ and $f(x)=0$ if $x\neq 1/n$ for any n".
Could someone explain this proof further?. It seems to me that he is trying to construct a function that is different from every $f_i$, but I don't see how the new function is necessarily different from every $f_i$, can't we have the possibilty that all the functions have the same values at $1/n$ for every $n$ but they are different for other values?.
Last edited: Apr 10, 2014
2. Apr 10, 2014
### jbunniii
The proof doesn't make any sense. The function $f$ being constructed does not even appear to be defined on the correct domain, $\mathbb{Z}$.
Try the following instead. Suppose that we have an enumeration $\{f_1, f_2, \ldots\}$. We now construct $f : \mathbb{Z} \rightarrow [0,1]$ that is not in the list.
First, let $g : \mathbb{Z} \rightarrow \mathbb{Z}^{+}$ be any bijection. For each $n \in \mathbb{Z}$, set $f(n)$ equal to any value in $[0,1]$ other than $f_{g(n)}(n)$. If we want to be concrete (to avoid invoking the axiom of choice), put
$$f(n) = \begin{cases} 1 & \text{ if }f_{g(n)}(n) = 0 \\ 0 & \text{ otherwise} \\ \end{cases}$$
Now given any $m \in \mathbb{Z}^+$, we see that $f$ cannot equal $f_m$ because it differs from $f_m$ at the point $n = g^{-1}(m)$.
Last edited: Apr 10, 2014
3. Apr 10, 2014
### trash
Thanks a lot.
But that was my mistake. The proposition is about the set of functions from $[0,1]$ to $\mathbb{Z}$
4. Apr 10, 2014
### jbunniii
OK, the proof is still incorrect as stated. It would be OK if it said
Perhaps for concreteness, something like
5. Apr 10, 2014
### PeroK
Assuming you know [0, 1] is uncountable:
$$f_a(a) = 1 \ and \ f_a(x) = 0 \ otherwise$$
Defines an uncountable set of functions: one for each a in [0, 1]
6. Apr 10, 2014
### trash
Thanks a lot, now that makes sense.
Still, for the last example if $f_1(1)=1$ wouldn't be $f(1)=2$ which exceeds the domain?, maybe something like $f(x)=[f_n(1/n)]/2$ would work better?.
7. Apr 10, 2014
### jbunniii
I thought you wanted to map from $[0,1]$ to $\mathbb{Z}$. So the value of $f(x)$ can be as big as we like, but you can't divide by 2, because the result may not be an integer.
8. Apr 10, 2014
### trash
Yes, you're right. I'm sorry, I read it too fast and didn't think it through
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# Application of the principle of causality to classical electrodynamics
I am reading Griffiths' Introduction to Electrodynamics in which he shows that the retarded and advanced potentials, e.g. the retarded scalar potential
$$V(\mathbf{r},t) = \frac{1}{4\pi\epsilon_0} \int \frac{\rho(\mathbf{r}',t_r)}{\lvert \mathbf{r} - \mathbf{r}' \rvert} \mathrm{d}\tau', \quad t_r \equiv t - \frac{\lvert \mathbf{r} - \mathbf{r}' \rvert}{c},$$
are solutions to the inhomogeneous wave equation and satisfy the Lorenz condition. He then rejects the advanced potentials by invoking the principle of causality, stating that it is not unreasonable to believe that electromagnetic influences propagate forward and not backward in time.
I can only imagine that the reasonableness of this belief is rooted in our naive experience of the world as time-asymmetric. However, the time asymmetry that we experience on a daily basis are (at least usually) not due to a fundamental asymmetry in the fundamental laws of physics, like Maxwell's equations (insofar as a classical theory can be fundamental, but to my knowledge QED is also time symmetric), but to the emergent second law of thermodynamics. It is not clear to me that we can use our intuition about macroscopic phenomena to reason about microscopic phenomena, and even if we could, the second law is only probabilistic and would not allow us to reject forward propagation of electromagnetic influences outright and declare them impossible, only conclude that they are unlikely.
Furthermore, I assume that experiments have verified the retarded and not the advanced potentials? If so, then why invoke the principle of causation at all? And more importantly, would this not show a time asymmetry in the laws of classical electrodynamics? I suppose the principle could still considered external to the theory (like the homogeneity and isotropy of space), but at least the application to this particular case would have been shown experimentally to be justified.
So what exactly is it that allows us to apply the principle of causality?
• Is this different to physics.stackexchange.com/questions/365023/… ? – Rob Jeffries May 17 '18 at 16:05
• I had initially thought so, but now I'm not so sure. I figured that that question asked specifically about waves, whereas my question was more general, but since all (nice?) solutions to Maxwell's equations in vacuum have to obey the wave equation, perhaps not. Even so, I think the answers given are not entirely satisfactory. The one by higgsss is illuminating, but the connection to the principle of causality is not entirely clear. – Danny Hansen May 17 '18 at 16:38
• Of potential interest are the Wheeler-Feynman time symmetric theory of electromagnetism, in such articles as "Classical electrodynamics in terms of direct interparticle action". They explored the idea and consequences of allowing advanced potentials – Slereah May 18 '18 at 13:08
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Sub-Forums : Special & General Relativity Search this Forum
### Relativity FAQ
Sep10-12 11:09 AM bcrowell
# Special & General Relativity
- Dependence of various physical phenomena on relative motion of the observer and the observed objects. Exp. & theo. theories of relativity
Views: 122 Announcement: PF Member Award voting is open! Dec12-13 Meta Thread / Thread Starter Last Post Replies Views Before posting anything, please review the Physics Forums Global Guidelines. If you are seeking help with a... Feb23-13 09:40 AM ZapperZ 1 35,167 The section of the Usenet Physics FAQ titled "Experimental Basis of Special Relativity" has been cited here many... Dec31-07 03:06 AM jtbell 0 35,002 Am I correct in my understanding that two objects moving away from each other movee away from each other at the speed... T 12:49 PM ilikescience94 2 106 Hello All, I like to think that I have a decent background in physics and related material, but I confuse myself... T 09:34 AM ghwellsjr 162 3,412 In linearized gravity we can one sets $$(1) \ \ g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$ where h is taken... T 08:30 AM center o bass 0 147 As I understand it, the value of a 4-vector x in another reference frame (x') with the same orientation can be derived... 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"Never doubt that a small group of thoughtful, committed citizens can change the world. Indeed, it is the only thing that ever has."
Original article
peer-reviewed
## Using Porcine Cadavers as an Alternative to Human Cadavers for Teaching Minimally Invasive Spinal Fusion: Proof of Concept and Anatomical Comparison
### Abstract
Training surgeons to perform minimally invasive spinal (MIS) surgery is difficult because there are few realistic alternatives to human cadavers which are expensive and require special handling. In this study we report a protocol for performing an MIS training course on a fresh porcine cadaver. We find that the porcine lumbar spine closely resembles the human spine in terms of the vertebral and discal anatomy. Notable differences include a lower disc height and shallower diameter. We obtained fresh porcine cadavers weighing 40-70 kg from local farmers that had been gutted and bled. We position the cadaver prone on a backboard and set up the operating room with biplanar fluoroscopy. During approach and cage insertion, we found that the tactile feedback obtained is realistic and allows surgeons to familiarize themselves with the procedure. Porcine cadavers were also an excellent tool for practicing pedicle screw fixation due to the larger pedicles. Five training courses involving eight surgeons noted that except for anatomical differences the training course was equivalent to training on human cadavers and unanimously preferred training on porcine cadavers to synthetic foam models. We conclude that porcine cadavers are a useful model for training surgeons in MIS surgery. Routine use of porcine cadavers may increase the availability of MIS surgery training.
### Introduction
Minimally invasive spinal (MIS) surgery approaches have recently gained popularity because they are associated with significantly reduced tissue damage, complication rates and hospital stay. However, MIS surgeries can be difficult to master technically. In some procedures like MIS-TLIF (transforaminal lumbar interbody fusion), surgeons must perform the same procedure as in open operations through a smaller incision [1]. In other techniques like extreme lateral interbody fusion (XLIF) where surgeons rely solely on fluoroscopy and electrophysiology, surgeons must learn a completely new operation [2]. To learn these challenging techniques, surgeons must train extensively to master these procedures and continue training after the initial learning phase. These significant technical challenges associated with MIS operations may explain why MIS fusion has not gained widespread popularity and open surgeries remain the most frequently performed surgeries of the spine.
Unlike abdominal or thoracic surgery, spine surgery lacks a natural cavity within which endoscopic techniques can be utilized. This makes it extremely difficult to design simulators or artificial models for training surgeons on new MIS procedures. Simulators with the needed attributes do not yet exist to our best knowledge, which in turn slows down the adaptation of MIS techniques even though MIS may offer significant benefits to patients compared to open procedures. The ideal model beside a human cadaver is a biological model that can simulate human anatomy.
The standard method of training surgeons on new spinal operations remains a human cadaver model [3]. However, the supply of human cadavers is limited, expensive, and handling is complicated, limiting the availability of MIS training courses and reducing the hands-on experience of surgeons and trainees. Considering that surgeons must typically perform operations many times before becoming comfortable with them, cadaver labs are not a feasible option in most cases [2]. As an alternative to cadavers, some training centers use synthetic foam models. However, the consistency of foam differs substantially from human tissue making it an unrealistic alternative for learning procedures where tactile feedback is essential for success of the operation. These models typically lack anatomical soft tissue layers and appear drastically simplified on fluoroscopy, which does not allow the surgeon to become comfortable with the range of anatomical differences and pathologies that exists in human patients [4].
To overcome this challenge, we hypothesized that a fresh porcine cadaver may be a useful tool to train surgeons on MIS techniques. The common swine is one of the most commonly used species due to its easy availability and their similar size to humans. The porcine spine has also been used as a model for the human spine in biomechanical studies for testing new implants and devices [5-7]. The lumbar pedicles are known to be very similar in terms of size to the human equivalent, allowing surgeons to practice placing pedicle screws on porcine models [8,9]. The porcine spine is also similar to the human spine in terms of the shape of the endplates and spinal canal [10]. However, due to the different mode of weight bearing in pigs, the intervertebral discs height is significantly lower and the endplates are much smaller than in humans. Although cervical lordosis is similar, pigs have much lower thoracic kyphosis and lumbar lordosis than human spines [10].
In this study, we report on a protocol for performing MIS surgery training using a porcine cadaver, highlight the most important anatomical differences as they pertain to lumbar fusions, and present limited anecdotal feedback from participants of five training sessions. We performed these training sessions to teach surgeons a new technique for fusion of the lumbar spine known as trans-Kambin Oblique Lateral-Posterior Lumbar Interbody Fusion (OLLIF) [2]. In OLLIF, the disc space is approached through Kambin’s triangle and fusion is performed without direct visualization, guided only by bi-planar fluoroscopy, electrophysiology and tactile feedback. OLLIF is typically complemented with minimally invasive pedicle screw fixation, which we also performed on the porcine model. Cadaveric models have previously been described for practicing laminectomy, pedicle screw fixation, and lumbar puncture [11,12]. This study is the first to evaluate animal cadaveric models for simulating minimally invasive fusions of the lumbar spine.
### Materials & Methods
Fresh cadavers of male hybrid landrace pigs aged approximately six months weighing 40-70 kg were obtained from a local farmer. The pigs were delivered slaughtered, bled, and gutted which increases the timeframe available for use. To simulate the opacity of the abdominal contents, we placed a rubber bladder (CampShower 5 gallon) filled with 16 liters of water in the abdominal cavity and sealed the abdomen with cable ties (Figure 1A). The water in the bladder can be mixed with contrast (Omnipaque 240 mg/ml, GE Healthcare) to achieve the most realistic level of contrast for the specific application.
The instrumentation and OR setup were identical to our routine setup for the OLLIF procedure which has been previously described [2]. We position the cadaver in prone position on a custom wooden backboard with cable ties (Figure 1B). The cadaver is draped in the usual fashion using transparent draping. We set up anterior-posterior (AP) and lateral fluoroscopy for the target level (Figure 1B), so that the endplates of the target level are perpendicular in the lateral view and the spinous process is centered between the pedicles in the AP view. The entire procedure under lateral and AP fluoroscopy is shown in Figure 2.
Compared to the approach in humans which is typically performed at 45° to the midline vertical, the porcine spine requires a more lateral approach to compensate for shallower AP diameter of vertebral body. Although electrophysiology cannot be simulated in this model, we still approach the disc with a blunt neurophysiological probe with a sleeve. Once the probe makes contact with the disc, the probe is removed while the sleeve remains in place and a K-wire is inserted. Next we enter a dilator over the K-wire and tap it into the disc space until the tip is past the midline. We then insert an access portal over the probe and perform the discectomy through that portal using a drill, rotating curette, long pituitary, and rongeur (Figure 3). Before removing the access portal, we insert a K-wire to mark the disc space. We enter the cage into the disc space over the K-wire and tap it into the disc space with a mallet. Finally, we perform minimally invasive pedicle screw fixation. Due to the shallower AP diameter of the pedicles, we used smaller pedicle screws, typically 35-40 mm. After the lab, the cadaver is dissected layer by layer to visualize the approach and the effects of the procedure on spinal and neural structures. We subsequently removed soft tissue from the spine by placing the specimen in water at a low boil for 90 minutes, creating bone specimens for studying the anatomy.
Relevant differences were reported between the porcine and human anatomy based on our experience and a review of the literature. We performed this lab five times with eight surgeons and solicited feedback from the participants.
### Results
#### Anatomical comparison
There are several differences in anatomy that are relevant for practicing minimally invasive fusion on porcine cadavers. The lumbar lordosis in the porcine spine is much lower than in humans. However, we were able to increase the lumbar lordosis by removing the abdominal organs and positioning the cadaver as described above.
Pigs have 14-15 thoracic vertebrae and six lumbar vertebrae. While thoracic vertebrae are substantially different from humans, the lower lumbar vertebrae are most similar to humans [10]. Some key comparisons are represented in Table 1. We found that while the diameter of the vertebral body in the dorsal-ventral dimension is about 70% smaller than in humans, the lateral diameter is very similar.
Variable for Comparison Human Porcine Number of Thoracic Vertebrae 15 Thoracic 12 Thoracic Number of Lumbar Vertebrae 6 Lumbar 5 Lumbar Lumbar Lordosis 29° Minimal Thoracic Kyphosis 35° 16° Dimension of pedicle Relatively narrower Relatively wider, but similar height Vertebral Body Height Between 20-30 mm Similar Intervertebral Disk Height Approximately 10 mm Approximately 5 mm Endplate Dimensions Width to Depth Ratio 1.5 Width to Depth Ratio 2, appears relatively larger in anterior-posterior dimension
In the lower lumbar spine the intervertebral foramen is very similar to humans. In the high lumbar spine the intervertebral foramen becomes bi-lobed and a bony separation appears between the discal foramen and “neural foramen” where the nerve root exits. This neural foramen is marked with a gold pin in Figure 4. The separation of these two foramina becomes more pronounced in the thoracic spine.
We also found that the transverse process is relatively longer than in humans and takes off on the ventral aspect of the pedicle rather than on the dorsal aspect as in humans. In all our specimens, the edge of the vertebral body extends further posteriorly than in humans (arrow in Figure 5). This appears physiologic in pigs but would be labelled an osteophytic change in humans, which are common in patients who undergo spinal surgery for degenerative conditions.
The consistency of the disc in our porcine specimens was similar to the human disc, but the disc appears heart shaped with a concave posterior border compared to human disc which is more oval shaped. Additionally, we noticed that the ossification of the vertebral body in our young specimens had not been completed, which can lead to separation along the epiphysis if the cage is entered with poor trajectory, which can be used as a maker for an adequate trajectory.
We found that the angle, size and configuration of L5/S1 and the SI joint were substantially different in pigs compared to humans, making them unsuitable for training on this level.
#### Participant feedback
We are well aware of the low number and anecdotal nature of the feedback collected. Nevertheless we gathered such feedback from most of the participants. Our future goal is to expand upon this study to involve a higher number of participants to further evaluate the teaching experience this model provides so as to allow us to assess the utility of this model. Feedback from lab participants was generally positive. Representative quotes from participants are in Table 2. Surgeons noted that the lab allowed them to “get down the approach and the anatomy” and the lab helped them “become more comfortable with the procedure.” Participants noted that the most realistic part of the experience was the “tactile feedback,” and uniformly stated that they preferred the experience to training on synthetic foam models. Participants did note that the anatomy was different compared to humans but also mentioned that because “the specimen was fresh compared to a [preserved] cadaver […] the tactile feedback was more realistic.”
Question Participant Feedback What was the most realistic part of the experience compared to humans? “Tactile Feedback”, “the fact that the anatomy was similar to a human” What was the least realistic part of the experience compared to humans? “Anatomy was different but everything else was realistic”, “Only drawback is that the pig anatomy is not exactly identical to human anatomy but as a means of a practice run was very valuable.” What was the most useful part of the experience? “Getting down the approach and the anatomy, I thought [the model] worked very well for the approach” How did the program contribute to your professional development? “Helped me become more comfortable with the procedure”, “It provided enough encouragement to begin performing the procedure at my practice” How does the pig lab compare to training labs on foam models? “Much much better”, “Far better and more realistic than trying out in [synthetic] models” How does the pig lab compare to training labs on human cadavers? “Slightly less helpful but only because of the anatomy”, “The pig lab was good in that the specimen was fresh compared to a [preserved] cadaver and the tactile feedback was more realistic.”, “C arm set up was similar as one in cadaver labs”
### Discussion
We found that porcine cadavers have the potential to be an appropriate substitute for human cadaver as a tool for learning minimally invasive spinal procedures of the lumbar spine between L1-L5. Participants in our labs agreed that porcine cadaver models are a valuable learning tool and we now routinely use porcine cadavers in a step-up approach, training surgeons first on porcine cadavers before transitioning to human cadaveric spines and subsequently supervised training in the OR. While we used this model to train surgeons in trans-Kambin lumbar fusion and pedicle screw fixation, the anatomical principles and preparation of the cadaver we described above are generally applicable. While this study was designed only as a proof of concept that using porcine cadavers is feasible, we hypothesize that porcine models can be used to train surgeons in a wide variety of MIS techniques and that doing so will improve surgeons skill and comfort with these techniques. Further study is required to investigate whether porcine models offer a superior training experience compared to other models of MIS training.
#### Practical experience
Under fluoroscopy, anatomical differences are more apparent in the lateral view compared to the AP view. The transverse processes in pigs appear more pronounced than in humans and they take off on the ventral aspect of the pedicle, so it has the potential to be confused with the facet to inexperienced trainees. The spinous processes point straight dorsally, unlike in humans where they are angled caudally.
Due to the shallower vertebral body, we performed a more lateral approach compared to human spines where we approached the disc at an angle of 45° to the midline. The layers of soft tissue dilated during approach felt very similar compared to humans and the tissue had a similar force-response compared to humans, allowing for dilation along the natural fibers of the tissue. This was noticeably different from synthetic foam models, where the anatomical layers are absent and dilation does not reflect the tactile feedback obtained from biological tissues. However, the para-spinal muscles are significantly larger than in humans which means the approach distance corresponds approximately to a far lateral approach in humans. Upon entry into the disc space, we clearly felt the penetration of the disc capsule. During the subsequent discectomy, the capsule acts as a fulcrum to stabilize the instruments, as is the case in humans. Again, this effect is entirely absent in foam models in which the disc has a uniform consistency.
In our specimens, the cartilaginous area of the disc appeared more pronounced than in humans, most likely due to the age of the pig. This allowed for a more extensive discectomy and end plate preparation than is possible in humans, which is beneficial for training purposes. Cage entry required less force compared to humans which may be related to increased anatomical flexibility of the young pigs used in this lab. We used cages sized 27 mm in length to accommodate the shallower vertebral body, while cage height was between 9-11 mm. We found that the porcine pedicle is an excellent model for practicing pedicle screw placement, because the pedicle is larger in the cranial-caudal and lateral dimension, allowing for easy pedicle identification. The size and trajectory of the pedicles was sufficiently similar to humans to use the same landmarks for navigation during pedicle screw placement. However, the pedicles are relatively shallower in the AP dimension, requiring the use of shorter pedicle screws between 35-40 mm.
Biological specimens like porcine cadavers have several advantages over foam models. Biological specimens have naturally variable anatomy which gives surgeons useful experience in navigation and adapting to each patient’s unique anatomy. This is a significant advantage over synthetic foam models, which have uniform anatomy and appear almost cartoonish on fluoroscopy. After performing the procedure, surgeons are able to dissect the porcine cadaver to better understand the anatomy, approach trajectory, and effect of the procedure on the tissues. Porcine cadavers are especially useful for training surgeons on procedures that rely on tactile feedback during approach. The force-response of different soft tissue layers closely resembled our experience in human patients. Once the disc is reached, the presence of a disc capsule makes discectomy much more realistic than is possible in synthetic models. Due to the distinct layers of the porcine disc, surgeons were able to gain experience in using the specialized tools required for trans-Kambin discectomy. We also found that the porcine spine is a valuable tool for introducing pedicle screw fixation techniques due to the more prominent pedicles.
The main advantage of the porcine models is their low cost and widespread availability. Porcine cadavers are widely available and cost around $180 compared to$500 for foam models and, \$5,000-10,000 for human cadavers. This makes porcine cadavers a particularly appealing option for training surgeons in developing countries where cost and availability are an even more important consideration. If proper handling is observed and sterile tools are observed, the porcine cadaver can be donated for further use following the completion of the training course. We recommend using porcine cadavers of weighing 40 to 60 kg, between six to eight months of age. Older pigs often have ossified spinal ligaments and bulkier vertebral bodies [11], while smaller cadavers are too small to be a realistic model for training. We recommend cadavers that have been gutted and bled both because it makes the lab more sanitary and because some lumbar lordosis can be generated by fixing a gutted porcine cadaver on a flat surface.
The main drawbacks of using porcine cadavers are differences in the anatomy, as was noted by our anatomical study. These differences are related to the fact that the porcine spine bears lower loads in the cranial-caudal direction compared to the human spine. For the purpose of surgical training, the most meaningful anatomical difference that we encountered was shallower VB diameter, which requires adjustment of the cage trajectory for a more lateral approach. Additionally, the porcine spine is a poor model for L5-S1 because of the angle, configuration and size of the sacral ala and iliac crest. Porcine cadavers are also a poor model for sacro-illiac fusions because the anatomy is substantially different from humans.
### Conclusions
Fresh pig cadavers are a useful tool for training surgeons to perform minimally invasive spine surgery. Although certain anatomical differences must be taken into account, porcine cadavers are an especially useful tool for training surgeons in procedures where tactile feedback is key to the success of the operation. Fresh porcine cadavers are a far more realistic training model than synthetic models, are less expensive, and can be handled more easily than human cadavers. Using porcine cadavers for the initial stages of training may increase the availability of training courses in MIS surgery. Further study is required to investigate whether porcine cadavers offer a superior training experience compared to other MIS training models.
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Original article
peer-reviewed
### Author Information
###### Ethics Statement and Conflict of Interest Disclosures
Human subjects: All authors have confirmed that this study did not involve human participants or tissue. Animal subjects: Our study involved only dead animal specimens. Because our institution performs research involving only dead animal specimens, we are not a “research facility” pursuant to the Animal Welfare Act (7 U.S.C. §§ 2131-2159). Accordingly, the Animal Welfare Act does not require our facility to have an IRB, ethics committee, or IACUC approve or assess our animal program, facilities, or procedures. Issued protocol number NA. Conflicts of interest: In compliance with the ICMJE uniform disclosure form, all authors declare the following: Payment/services info: All authors have declared that no financial support was received from any organization for the submitted work. Financial relationships: Hamid R. Abbasi declare(s) a grant from MIS Technologies. Research Grant. Hamid R. Abbasi declare(s) employment from MIS Technologies. Hamid R. Abbasi declare(s) stock/stock options from ARM. Hamid R. Abbas declare(s) a patent from Tristate Brain and Spine Industry. Hamid R. Abbasi is the author of numerous patents on cages, screws and instrumentation used in OLLIF and other spinal fusion operations. Ali Abbasi declare(s) personal fees from AMW LLc. Consulting Fees. Hamid R. Abbasi declare(s) personal fees from Zyga. Consulting Fees. Intellectual property info: Hamid R. Abbasi is the author of patents on cages, screws and instrumentation used in OLLIF and other spinal fusion operations including US patents number 9675363, 9655654, D847339 and multiple pending patents. Other relationships: All authors have declared that there are no other relationships or activities that could appear to have influenced the submitted work.
###### Acknowledgements
We would like to acknowledge Sandra Heims for editing the manuscript.
Original article
peer-reviewed
8.0
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# Fast Computation of the Roots of Polynomials Over the Ring of Power Series
Abstract : We give an algorithm for computing all roots of polynomials over a univariate power series ring over an exact field $\mathbb{K}$. More precisely, given a precision $d$, and a polynomial $Q$ whose coefficients are power series in $x$, the algorithm computes a representation of all power series $f(x)$ such that $Q(f(x)) = 0 \bmod x^d$. The algorithm works unconditionally, in particular also with multiple roots, where Newton iteration fails. Our main motivation comes from coding theory where instances of this problem arise and multiple roots must be handled. The cost bound for our algorithm matches the worst-case input and output size $d \deg(Q)$, up to logarithmic factors. This improves upon previous algorithms which were quadratic in at least one of $d$ and $\deg(Q)$. Our algorithm is a refinement of a divide \& conquer algorithm by Alekhnovich (2005), where the cost of recursive steps is better controlled via the computation of a factor of $Q$ which has a smaller degree while preserving the roots.
Keywords :
Document type :
Conference papers
Cited literature [21 references]
https://hal.inria.fr/hal-01457954
Contributor : Vincent Neiger <>
Submitted on : Friday, June 2, 2017 - 1:12:42 PM
Last modification on : Saturday, June 10, 2017 - 1:06:12 AM
Long-term archiving on: : Wednesday, December 13, 2017 - 9:11:04 AM
### File
series_roots.pdf
Files produced by the author(s)
### Citation
Vincent Neiger, Johan Rosenkilde, Éric Schost. Fast Computation of the Roots of Polynomials Over the Ring of Power Series. ISSAC '17, Jul 2017, Kaiserslautern, Germany. ⟨10.1145/3087604.3087642⟩. ⟨hal-01457954v2⟩
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LLC: Accurate, Multi-purpose Learnt Low-dimensional Binary Codes
May 21, 2021 (edited Oct 12, 2021)NeurIPS 2021 PosterReaders: Everyone
• Keywords: Classification, Efficient Classification, Error Correcting Output Codes, Binary Codes, Retrieval, Large-scale Classification
• TL;DR: An accurate method to learn accurate, multipurpose and tight binary codes for both instances and classes with strong empirical performance in classification, retrieval and OOD detection.
• Abstract: Learning binary representations of instances and classes is a classical problem with several high potential applications. In modern settings, the compression of high-dimensional neural representations to low-dimensional binary codes is a challenging task and often require large bit-codes to be accurate. In this work, we propose a novel method for $\textbf{L}$earning $\textbf{L}$ow-dimensional binary $\textbf{C}$odes $(\textbf{LLC})$ for instances as well as classes. Our method does ${\textit{not}}$ require any side-information, like annotated attributes or label meta-data, and learns extremely low-dimensional binary codes ($\approx 20$ bits for ImageNet-1K). The learnt codes are super-efficient while still ensuring $\textit{nearly optimal}$ classification accuracy for ResNet50 on ImageNet-1K. We demonstrate that the learnt codes capture intrinsically important features in the data, by discovering an intuitive taxonomy over classes. We further quantitatively measure the quality of our codes by applying it to the efficient image retrieval as well as out-of-distribution (OOD) detection problems. For ImageNet-100 retrieval problem, our learnt binary codes outperform $16$ bit HashNet using only $10$ bits and also are as accurate as $10$ dimensional real representations. Finally, our learnt binary codes can perform OOD detection, out-of-the-box, as accurately as a baseline that needs $\approx3000$ samples to tune its threshold, while we require ${\textit{none}}$. Code is open-sourced at https://github.com/RAIVNLab/LLC.
• Supplementary Material: pdf
• Code Of Conduct: I certify that all co-authors of this work have read and commit to adhering to the NeurIPS Statement on Ethics, Fairness, Inclusivity, and Code of Conduct.
• Code: https://github.com/RAIVNLab/LLC
15 Replies
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# A golf ball is dropped from water tower 128 feet tall; its height after t seconds is given byh(t) =
###### Question:
A golf ball is dropped from water tower 128 feet tall; its height after t seconds is given by h(t) = -16t2 + 128 Ilow long is the golf ball in the air? What is the maximum height of the Kolf = ball?
#### Similar Solved Questions
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##### A wooden block is given in initial push on wooden plank to give it an initial velocity of 0.50 mls The block slides to stop through distance of 0.25 m. If the block has mass of 0.200 kg:a) Calculate the acceleration of the block b) The weight of the block c) Draw diagram that shows all the forces acting on the block while it is sliding to stop d) Calculate the force being exerted on the block that is slowing it down How would the results differ if everything about the situation was the same but
A wooden block is given in initial push on wooden plank to give it an initial velocity of 0.50 mls The block slides to stop through distance of 0.25 m. If the block has mass of 0.200 kg: a) Calculate the acceleration of the block b) The weight of the block c) Draw diagram that shows all the forces a...
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## Thursday, June 28, 2012
### 2012 So Far...
RACES and EVENTS
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Finsbury Park 15k
April, 2012
Masha's longest event and I was glad to have been there to share it with her. I didn't pressure her to take up running, something she wanted to do on her own.
Race Report
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Quicksilver 50k / 50-Mile
May 12, 2012
Hot and hilly. What a welcome back to the Bay Area run. Quite a shock after five weeks of flat stuff in London in cool, drizzly, overcast weather. Managed to survive despite pushing the pace early. I needed this race to prepare for San Diego 100 in June. Painful but rewarding run.
9:10:59 for 13th place.
Race Report
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San Diego 100- Mile
June 9-10, 2012
What a race!!! My third time in San Diego and my second run on the new course. Knowing I didn't have the speed this year I focused on having a solid race and paced accordingly. Worked out great, especially the last 4 miles when I had to go hard to make sub-24.
I was joined by fellow San Francisco buddies Larissa Polischuk, Randy Katz, Chris Wolfe and Jennifer Pattee. Brett Rivers, John Brandershorst, Amy Freund McCrea and Peter Duyan came along to crew and pace. We had quite a group going.
23:51:05 for 29th
Race Report
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VOLUNTEER, CREW OR PACE
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Pacing at Western States 100
June23-24, 2012
Mind blowing experience pacing WS this year, just incredible. I won't say more. I don't think I can sum up what happened in a couple of sentences.
Report
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#### 1 comment:
1. Great job on your races so far this year, Rick! Hope the second half of your year is awesome!
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# Andrea Vitaletti
nationality: Italian
date of birth: 27/03/1971
address: via dei Cessati Spiriti 88, Rome, Italy
mobile: +39 349 7569802
e-mail: andrea.vitaletti@uniroma1.it
# Short bio
Andrea Vitaletti, Ph.D. in Computer Engineering. I am currently Assistant professor (Ricercatore confermato) with National scientific qualification (09/H1 and 01/B1) as Associate Professor.
My research interests include: networking algorithms and protocols, IoT, private and distributed data management and Distributed Ledger Technologies. I co-authored more than 60 papers
I have been involved in a number of EU projects as researcher, principal investigator and coordinator.
I am a maker and I have some successful experience in transfer of technology: I founded the spin-offs WLAB that has been sold in 2016 and WSENSE where I am the CTO.
I teach networking, data-management and IoT topics at engineering and product design MsC and BsC in Sapienza.
Hobbies: sailing, cycling and paragliding … I play the piano and the guitar for fun… be aware!
# Education
From To
Nov. 1998 Jan. 2002 PhD in Engineering in Computer Science (Dottorato di Ingegneria Informatica) (Dottorato di Ricerca in Ingegneria Informatica), University of Rome “La Sapienza”
Oct. 1992 Feb. 1998 MSc in Engineering in Computer Science (Laurea in Ingegneria Informatica), University of Rome “La Sapienza”
# Work Experience
From To
Nov. 2007 Now Assistant Professor (Ricercatore) University of Rome “La Sapienza”, National scientific qualification as associate professor in Italian Universities (abilitazione nazionale a professore di seconda fascia. Bando D.D. 1532/2016 settore concorsuale 09/H1 Sistema di Elaborazione delle informazioni e settore concorsuale 01/B1 Informatica)
Jul. 2007 Jun. 2019 Co-founder and Chief Innovation Officer of WSENSE
Jan. 2002 Jul. 2016 Co-founder and Chief Innovation Officer of WLAB (sold in 2016)
Mar. 2002 Oct. 2007 Post-doc and research grant at University of Rome “La Sapienza”
Mar. 1998 Oct. 2007 Senior researcher at ETNOTEAM Research Labs (now NTT DATA)
Mar. 2006 May. 2006 Visiting researcher at Swiss Federal Institute of Technology in Zurich (ETHZ), Zurich (Switzerland)
Apr. 2001 Ago. 2001 Consultant at AT&T Labs Research, Florham Park, NJ (USA)
## Teaching
Andrea Vitaletti is instructor in the MOOC of Sapienza on Coursera entitled “Recovering the Humankind’s Past and Saving the Universal Heritage”
The teaching activity has been mainly carried out in the Faculty of Information Engineering of Sapienza (Master LM-32 and Laurea L-8) and in the Master in Product Design at Sapienza (LM-12), as summarized in the following table (data from GOMP). Click on the links for the material on the most recent courses.
Year Course Laurea
2017/2018 INTERNET OF THINGS (6 CFU) LM-12
2017/2018 RETI DI CALCOLATORI (3 CFU) L-8
2017/2018 WEB INFORMATION RETRIEVAL (3 CFU) LM-32
2016/2017 INTERNET OF THINGS (6 CFU) LM-12
2016/2017 RETI DI CALCOLATORI (3 CFU) L-8
2015/2016 INTERNET OF THINGS (6 CFU) LM-12
2015/2016 RETI DI CALCOLATORI (3 CFU) L-8
2014/2015 PROGETTO DI RETI E SISTEMI INFORMATICI(3 CFU) L-8
2014/2015 PERVASIVE SYSTEMS (6 CFU) LM-32
2013/2014 PROGETTO DI RETI E SISTEMI INFORMATICI(3 CFU) L-8
2013/2014 WIRELESS NETWORK SYSTEMS (6 CFU) LM-32
2012/2013 PROGETTO DI RETI E SISTEMI INFORMATICI(3 CFU) L-8
2012/2013 WIRELESS NETWORK SYSTEMS (6 CFU) LM-32
2011/2012 RETI DI CALCOLATORI (6 CFU) L-8
2011/2012 WIRELESS NETWORK SYSTEMS (6 CFU) LM-32
2010/2011 SISTEMI PER RETI WIRELESS (6 CFU) LM-32
2009/2010 SISTEMI PER RETI WIRELESS (6 CFU) LM-32
2009/2010 RETI DI CALCOLATORI (6 CFU) L-8
2008/2009 SISTEMI PER RETI WIRELESS (6 CFU) LM-32
# Transfer of Technology
From To
2002 2016 In 2002 I founded WLAB, a dynamic SME created to support applied research and prototyping in the area of wireless technologies and pervasive and mobile computing. WLAB has been characterized by the synergy with the universities of Rome (Sapienza and Tor Vergata) promoting a continuous and effective technological transfer of innovative solutions from the academic world to the industry. I was Chief Technology Officer (CTO) of WLAB until its sale in 2016
2012 2019 In 2012 I founded WSENSE a spin-off of Sapienza University of Rome, with a strong and experienced R&D team specialized in monitoring and communication systems with pioneering patented solutions in the Internet of Underwater Things (IoUT). I am currently the CTO of WSENSE and I lead the activities for the development of the WGATE, the cloud platform for the collection, analysis, visualisation and integration of IoT data. WSENSE is an international company, with WSENSE Ltd hosted in the Marine Robotics Innovation Centre di Southampton U.K.
2014 Now Among the organizers of the Google Technologies Workshop for Cloud and Web Development now in its fifth edition. The workshop is aimed at students with the purpose of introducing them to a methodology of application development that is not limited only to technological aspects, but can lead them to the realization of a start-up. The success of the workshop is also measured by the fact that some participants are now hosted in business accelerators
in 2014 Successful experience in a crowd-funding campaign with the project COVA that has been also boradcasted in the Italian main TV channel RAI
in 2001 During his stay at AT&T Research Labs of Florham Park, NJ - USA, his research activity led to the application of four provisional patent applications.
in 2001 (3 months) During his stay at AT&T Research Labs of Florham Park, NJ - USA, his research activity led to the application of four provisional patent applications.
## Patents
• Patent 102003901105392 (RM2003A000177): Procedure for the recognition of authenticity of documents and debt securities, in particular banknotes, and related system.
• Patent 102001900950099 (RM2001A000492): Method for the secure transmission of data through short message service messages (SMS) with related methods for generating and recognizing secure SMS.
## Projects
I have acquired a significant experience on research projects, gained in more than 15 years of participation to academic and industrial projects characterized in many cases by a marked multidisciplinarity approach. Over time he has assumed various roles, with different responsibilities, as researcher, principal investigator (PI) and finally coordinator. The main projects in which he was involved are briefly described below.
• FP7-PEOPLE-ITN-2008 FRONTS (PI), Overall Budget: EUR 3.1 Mln. From 01-02-2008 to 30-04-2011. The aim of this project is to establish the foundations of adaptive networked societies of small or tiny heterogeneous artifacts. I participated to all the review meetings where I presented the activities of WP2.
• FP7-ICT-2009-5 VITRO (PI), Overall Budget: EUR 3.4 Mln. From 01-09-2010 to 28-02-2013. This project is focused on developing architectures, algorithms and engineering methods, which will enable the realization of scalable, flexible, adaptive, energy-efficient andtrust-aware Virtual Sensor Network platforms. I participated to all the review meetings where I presented the activities of WP3.
• ARTEMISI-JU call 2009, CHIRON (PI), Overall Budget: EUR 17.8 Mln. From 01-03-2010 to 01-02-2013. CHIRON intends to combine state-of-the-art technologies and innovative solutions into an integrated framework for effective and person-centric health management throughout the complete (health)care cycle thus responding to the present-day demographic and socio-economic challenges facing healthcare: from an ever ageing population to the need for affordable ‘global’ healthcare provided by fewer and fewer professionals and medical infrastructures for critical, often mobile, patients.
• EU EUROSTARS-EUREKA, PharmAID (PI), Overall Budget: EUR 2 Mln. From 01-06-2010 to 31-12-2012. Logistics control, storage monitoring and anti-counterfeiting of PHARMaceutical products by Advanced Integrated Devices. Creating an infrastructure based on passive RFID tags for the pharmaceutical market to: - prevent drugs’ counterfeiting; - certify drugs’ integrity against unsuitable ambient conditions exposures; and - track drugs along the supply chain. I participated to all the review meetings where I presented the activities of WP5.
• FP7-ICT-2011-C FET (Future Emerging Technology) OPEN, PLEASED (Coordinator), Overall Budget: EUR 1.46 Mln. From 01-05-2012 to 30-04-2015. Evaluated Excellent. In this project, we plan to develop plant cyborgs, shifting focus from interfacing a single plant to a network of entities (a community of plants) that renders an orchestrated response to the environment in which it lives. While artificial sensing devices exist that can monitor environmental parameters of interest, such as temperature or humidity, the focus of our research will be on the use of plants themselves as sensing and decision-making devices. The results of the multidisciplinary team in the project have obtained a significant eco in he media such as Youtube and Wired.
• EU FI-ADOPT, PAGE (Coordinator), Overall Budget: EUR 150 K. From 01-11-2014 to 01-11-2015. Evaluated Excellent. A project to develop a technology capable to support elderly in their daily activities.
• FP7-ICT-2013-10. SUNRISE (I). Overall Budget: EUR 5.3 Mln. From 2013-09-01 to 2016-12-31. The SUNRISE objectives are to develop a federated underwater communication networks, based on pilot infrastructure already designed, built and deployed by consortium partners, in diverse environments, web-accessible and interfaced with existing FIRE facilities to experiment with Future Internet technologies.
• EASME ARCHEOSUb (I) . The ARCHEOSUb (Autonomous underwater Robotic and sensing systems for Cultural Heritage discovery Conservation and in situ valorization) project aims to develop products and services in support of the discovery of new Underwater Cultural Heritage (UCH) sites and of the surveying, conservation, protection, and valorization of new and existing ones.
In the following I briefly outline some of the activities performed before
• He has been the technical manager of the DIAG group in Rome “La Sapienza” for the project VICOM (Virtual Immersive COMmunications - http://www.vicom-project.it/). The project has developed the infrastructure for the provisioning of augmented realty mobile services. In particular, the group of Rome has developed the framework for the location of mobile users through heterogeneous technologies (GPS, CellID, WiFi, Sensor Networks …)
• He was the coordinator and technical manager of the CINI lab in Rome. The lab participated in the FIRB project WEB developing a framework for the remote access to multimedia context sensitive services (eg: tourist guides on mobile phones where the information is georeferenced).
• He collaborated with the group of Professor Chiara Petrioli within the European project EYES (Energy Efficient Sensor networks - http://www.eyes.eu.org/) for the creation of sensor networks that self-organize and collaborate to the realization of a system energetically efficient. In particular, it has been studied how the emplyment of sparsifying techniques of the connectivity graph can lead to a significant reduction in energy consumption.
• He collaborated with the group of Professor Chiara Petrioli as part of the European project WiSeNts (Wireless Sensor Networks and Cooperating Smart Objects). The project studied the dynamics of the processes that underlie the realization of complex systems of intelligent and cooperating objects (eg sensor networks). As part of this project he visited the ETHZ (see [visit ])
• He has participated in the European project AEOLUS (Algorithmic Principles for Building Efficient Overlay Computers - http://dmod.cs.uoi.gr/aeolus site / main.htm) as a researcher. The project investigated methods and algorithms for creating overlay networks for the efficient and transparent access to Internet resources.
• He participated in the European project DELIS (Dynamically Evolving, Large-scale Information Systems - http://delis.upb.de/) as a researcher. The project has developed methods, techniques and tools to effectively manage modern information systems, characterized by large and extremely dynamic information.
• In the EU SOFIA project (Artemisia - https://www.artemisia-association.org/sofia), he was responsible for TASK3.3 under WP3. The SOFIA project has created a platform for interoperability between embedded systems that participate in the creation of a smart environment
## Hardware
• The research group of the DIAG on WSN coordinated by Andrea Vitaletti has developed and implemented the MagoNode: a new wireless device for WSN networks operating in the 2.4Ghz ISM band completely developed at DIAG which has been used in many field trials.
# Research
The following figure shows the Wordcloud of the abstracts of my papers. After a stemming process, the 50 most frequent terms in the abstracts are shown with a size that is proportional to their frequency.
The most evident terms (i.e. the biggest, namely the more frequent in the abstract) well summarize my research that focus on wireless networks for the collection of data to provide useful services to the users. This activity, is nowadays part of the wider research on the Internet of Things (IoT) and more recently naturally brought me to start an investigation on Decentralized Applications and Blockchain Technologies.
A distinctive characteristics of my research is the attempt to always validate the performance of the proposed solutions, both theoretical and practical, via simulations and real testbeds.
I lead the research activities on wireless sensor networks and the Internet of Things, in the research group on Computer Networks and Pervasive Systems of the Department of Automatic Computer Engineering and Management Antonio Ruberti della Sapienza University of Rome (DIAG). In this context, I enjoied the cooperation with several international research organizations, including: Research Academic Computer Technology Institute (Greece), Braunschweig University of Technology (Germany), Universitat Paderborn (Germany), University of Athens (Greece), Ben-Gurion University of the Negev (Israel), University of Salerno (Italy), Wroclaw University of Technology (Poland), Universitat Politecnica de Catalunya (Spain), University of Geneva (Switzerland), University of Lubeck (Germany).
I founded the laboratory on the wireless sensor network of DIAG where the solutions - both hardware and software - designed in the our research activities, are implemented and tested in the field.
I enjoy a long term collaboration on cooperative design and co-design with faculties of the MSc in Product Design at Sapienza. In particular, I’m leading the research activities on the employment of genetic algorithms for product design.
## Wireless Networks for the Collection of Data
### Allocation Problems
Coloring algorithms can be used to model bandwidth allocation problems in FDMA systems. In [1] we present randomized lower bounds for online coloring of some well studied network topologies. Wireless data networks allow multiple codes (or channels) to be allocated to a single user, where each code can support multiple data rates. Providing fine-grained QoS to users in such networks poses the two-dimensional challenge of assigning both power (rate) and codes to every user [2],[3]. A similar “two-dimensional challenge” emerges in the allocation of bandwidth slots to communication requests on a satellite channel under real time constraints. Accepted requests must be scheduled on non-intersecting rectangles in the time/bandwidth Cartesian space with the goal of maximizing the benefit obtained from accepted requests. [4]
### Data collection and energy consumption
Wireless Sensor Networks (WSN) are made of battery powered nodes, consequently the energy consumption of the nodes must be optimized to increase the network life-time. WSN are deployed to monitor a phenomenon and in a very simplistic setting we can identify two main phases: a) the interest dissemination, where the nodes are notified on the phenomenon they have to observe, and b) the data collection where the nodes report the observed data. In [5],[6] we tackle the problem of designing simple, localized, low energy consuming, reliable protocols for one-to-all communication (namely the main paradigm for interest-dissemination) in large scale wireless sensor networks. Data aggregation can be used to combine data of several sensors into a single message, thus reducing sensor communication costs during the data collection phase, at the expense of message delays. Thus, the main problem of data aggregation is to balance the communication and delay costs [7],[8],[9],[10]. In [11], [12], [13] we presented solutions capable to integrate the data delivery and interest dissemination techniques. Duty cycling is a useful technique that alternates active and inactive periods to reduce the power consumption. In [14] we presented a protocol capable to automatically adapt the duty-cycle to the application needs. The protocol has been also tested to extend TETRA, a European standard for a trunked radio system, with WSNs [15].
Delay Tolerant Networks (DTNs) exploit user mobility to deliver messages. In [16],[17],[18],[19],[20],][21] we propose a a reference architecture and a thorough quantitative evaluation of routing protocols for DTNs. Population protocols are used as a theoretical model for a collection (or population) of tiny mobile agents that interact with one another to carry out a computation. In [22] we evaluate the performance of population protocols on real face-to-face social networks of mobile users [25]. In [26] we studied how to exploit the energy scavenged by the pedestrian movements to enhance the operation time of the sensors. In [27],[28] we studied how to enhance the capabilities of multi-robot surveillance with WSNs.
### Simulation, Tools, Deployments and Devices
The workflow for the deployment of a WSN usually foresees a preliminary simulation of the proposed solution to assess its performance. In this context, a key question is the validity of the simulation results. In [29],[30],[31],[32] we compared the simulation results with the ones obtained in a real test-bed. The debugging of WSN can be a labor-intensive and cumbersome task. In [33] we present a tool, basically a distributed network sniffer alongside the inspected sensor network. The ability of re-programming over-the-air (OTA) the nodes is crucial to fix bugs or enhance the functionalities on the nodes. Usually, OTA assume an always-on network. In [34] we presented a protocol capable to support OTA in network with a duty-cycle. In our lab on WSNs we developed the MagoNode [35], an embedded system based on a highly efficient RF front-end that greatly improves RF performance, in terms of radio range and sensibility, still limiting energy consumption. Indeed, the device outperforms other existing amplified platforms available on the market and is comparable to most commonly known unamplified motes. A network of cooperating tiny artifacts requires new programming models [36]. In [37] we explored the novel concept of virtual sensor networks, namely WSNs capable to adapt their behavior according to the user needs. the Internet of Underwater Things is probably the new frontier of WSNs. In [38] we presented the system to access the SUNRISE federation of facilities to test solutions for the Internet of Underwater Things. In [39] we started the investigation on the employment of WSNs to monitor critical infrastructures. This activity has been further developed in [40] where we presented the work done to monitor the structural health of a Rome B1 underground construction site by a WSNs. In [41],[42] a WSN has been used for noise pollution monitoring.
### Plant as sensors
I have been the coordinator of the EU Future and Emerging Technologies (FET) project PLEASED. The aim of the project was the development of a technology capable of exploiting the plants as bio-sensors. In other words, PLEASED was an attempt to develop a kind interface between plants and computers, namely a plant-cyborg. Indeed, plants react to some external stimuli generating an electrical signal that can be captured by suitable devices (similarly to what happens in EMG). The main problem here is to develop a technology capable to uniquely identify the applied stimuli from the observed electrical signals. The idea of using plants as sensing devices has been first presented in [43]. In [44],[45],[46],[47] a number of machine learning techniques has been evaluated to classify the electrical signal generated by plants in response to heterogeneous stimuli. The main goal of this research was to identify a fingerprint in those electrical responses capable of discriminating the applied stimuli; this is indeed the main build block of a future employment of plants as bio-sensors.
## Services to the Users
### Localization
During my stay in AT&T research lab in USA I had the opportunity of conducting a pioneering experimental activity on Cell-ID location techniques [48],[49] that also resulted in the application of of three provisional patents.
### Privacy
We initially considered the issues related to the implementation of state-of-the-art cryptographic primitives on resource constrained devices [50], [51] and mobile environments [52] Then we focused on the development of technical solutions capable to balance the tradeoff between the privacy of the users and the utility of the offered services [53], [54], [55], [56]. Current approaches in digital trials entail that private user data are provisioned to the trial investigator that is considered a trusted party. The aim of [57] is to present the technical requirements and the research challenges to secure the flow and control of personal data and to protect the interests of all the involved parties during the first phases of a clinical trial, namely the characterization of the potential patients and their possible recruitment. The proposed architecture will let the individuals keep their data private during these phases while providing a useful sketch of their data to the investigator. More recently, we started a research activity on blockchain technologies. This activity is focused on two main directions: a) the development of interesting use-cases to prove the applicability of such technology beyond the financial domain; in particular we used blockchain technologies to guarantee the quality of data in IoT based clinical trials [58] and to support the democratization of the decision processes in the smart cities [59]), b) the development of suitable software engineering techniques to reduce the risks in the development of decentralized applications [60].
### Recommendation systems
Searching is probably the main functionality of the Web. In [61] we performed an experimental study of pre-fetching and caching algorithms for the world wide web. However, in modern applications recommendation systems play an equally important role taking advantage of Big Data to deliver personalized recommendations based on a visitor’s previous activities and those of other similar customers. Usually recommendation systems rely on centralized architectures. In [62],[63],[64], [65] we investigate the effectiveness of fully decentralized, collaborative filtering techniques. In [56] we propose a decentralized system capable to recommend new friendships to other users. In our proposal, the information needed to perform recommendation is collected and exchanged between users in a privacy preserving way.
### Other services
The participation to the EU project CHIRON allowed us to understand the main components of a Next-Generation Remote Healthcare [66]. Also on the basis of this experience, we developed a simple system to deliver elder-care environments utilizing TV-channel based mechanisms [67]. The UX is also the main concept explored in [68] where we investigated an a ubiquitous touch-based remote grocery shopping process.
In [69],[70] we explored the concept of cooperative design supported by genetic algorithm. The selection of the products that will actually take part to the evolutionary process, relies on crowdsourcing mechanisms: only the most appreciated products survive.
## Bibliography
A likely more recent list of my publications can be found in:
[1] S. Leonardi and A. Vitaletti, “Randomized lower bounds for online path coloring,” Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), vol. 1518, pp. 232–247, 1998.
[2] L. Becchetti, S. Diggavi, S. Leonardi, A. Marchetti-Spaccamela, S. Muthukrishnan, T. Nandagopal, and A. Vitaletti, “Parallel scheduling problems in next generation wireless networks.” pp. 238–247, 2002.
[3] L. Becchetti, S. Leonardi, A. Marchetti-Spaccamela, A. Vitaletti, S. Diggavi, S. Muthukrishnan, and T. Nandagopal, “Parallel scheduling problems in next generation wireless networks,” Networks, vol. 45, no. 1, pp. 9–22, 2005.
[4] S. Leonardi, A. Marchetti-Spaccamela, and A. Vitaletti, “Approximation algorithms for bandwidth and storage allocation problems under real time constraints,” Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), vol. 1974, pp. 409–420, 2000.
[5] L. Orecchia, A. Panconesi, C. Petrioli, and A. Vitaletti, “Localized techniques for broadcasting in wireless sensor networks.” pp. 41–51, 2004.
[6] D. Dubhashi, O. Häggström, L. Orecchia, A. Panconesi, C. Petrioli, and A. Vitaletti, “Localized techniques for broadcasting in wireless sensor networks,” Algorithmica (New York), vol. 49, no. 4, pp. 412–446, 2007.
[7] P. Korteweg, A. Marchetti-Spaccamela, L. Stougie, and A. Vitaletti, “Data aggregation in sensor networks: Balancing communication and delay costs,” Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), vol. 4474 LNCS, pp. 139–150, 2007.
[8] P. Korteweg, A. Marchetti-Spaccamela, L. Stougie, and A. Vitaletti, “Data aggregation in sensor networks: Balancing communication and delay costs,” Theoretical Computer Science, vol. 410, no. 14, pp. 1346–1354, 2009.
[9] L. Becchetti, P. Korteweg, A. Marchetti-Spaccamela, M. Skutella, L. Stougie, and A. Vitaletti, “Latency constrained aggregation in sensor networks,” Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), vol. 4168 LNCS, pp. 88–99, 2006.
[10] L. Becchetti, A. Marchetti-Spaccamela, A. Vitaletti, P. Korteweg, M. Skutella, and L. Stougie, “Latency-constrained aggregation in sensor networks,” ACM Transactions on Algorithms, vol. 6, no. 1, 2009.
[11] A. Marcucci, M. Nati, C. Petrioli, and A. Vitaletti, “Directed diffusion light: Low overhead data dissemination in wireless sensor networks,” vol. 4. pp. 2538–2545, 2005.
[12] M. Mastrogiovanni, C. Petrioli, M. Rossi, A. Vitaletti, and M. Zorzi, “Integrated data delivery and interest dissemination techniques for wireless sensor networks.” 2006.
[13] M. Mastrogiovanni, C. Petrioli, M. Rossi, A. Vitaletti, and M. Zorzi, “Integrated Data Delivery and Interest Dissemination Techniques for Wireless Sensor Networks,” in GLOBECOM 2006 - 2006 IEEE GLOBAL TELECOMMUNICATIONS CONFERENCE.
[14] U. Colesanti, S. Santini, and A. Vitaletti, “DISSense: An adaptive ultralow-power communication protocol for wireless sensor networks.” 2011.
[15] M. Paoli, F. Ficarola, U. M. Colesanti, A. Vitaletti, S. Citrigno, and D. Sacca, “Extending tetra with wireless sensor networks,” INTERNATIONAL JOURNAL OF INTELLIGENT ENGINEERING INFORMATICS, vol. 3, nos. 2-3, SI, pp. 225–243, 2015.
[16] K. Massri, A. Vernata, and A. Vitaletti, “Routing protocols for delay tolerant networks: A quantitative evaluation.” pp. 107–114, 2012.
[17] D. Amendola, F. De Rango, K. Massri, and A. Vitaletti, “Neighbor discovery in delay tolerant networking using resource-constraint devices.” 2013.
[18] K. Massri, R. Beraldi, and A. Vitaletti, “Erasure-coding based data delivery in delay tolerant networks,” Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), vol. 8121 LNCS, pp. 188–200, 2013.
[19] K. Massri and A. Vitaletti, “DTN routing protocols on resource constrained devices: Design, implementation and first experiments.” 2013.
[20] D. Amendola, F. De Rango, K. Massri, and A. Vitaletti, “Efficient neighbor discovery in rfid based devices over resource-constrained dtn networks.” pp. 3842–3847, 2014.
[21] K. Massri, A. Vitaletti, A. Vernata, and I. Chatzigiannakis, “Routing protocols for delay tolerant networks: A reference architecture and a thorough quantitative evaluation,” JOURNAL OF SENSOR AND ACTUATOR NETWORKS, vol. 5, no. 2, Jun. 2016.
[22] L. Becchetti, L. Bergamini, F. Ficarola, and A. Vitaletti, “Population protocols on real social networks.” 2012.
[23] L. Becchetti, L. Bergamini, F. Ficarola, and A. Vitaletti, “Population protocols on real social networks.” pp. 17–24, 2012.
[24] L. Becchetti, L. Bergamini, F. Ficarola, F. Salvatore, and A. Vitaletti, “First experiences with the implementation and evaluation of population protocols on physical devices.” pp. 335–342, 2012.
[25] F. Ficarola and A. Vitaletti, “Capturing interactions in face-to-face social networks.” pp. 613–620, 2015.
[26] I. Chatzigiannakis, U. Colesanti, S. Kontogiannis, G. Leshem, A. Marchetti-Spaccamela, J. Mehler, G. Persiano, P. Spirakis, and A. Vitaletti, “MURPESS - multi radio pedestrian energy scavenging sensor network.” 2010.
[27] A. Pennisi, F. Previtali, F. Ficarola, D. Bloisi, L. Iocchi, and A. Vitaletti, “Distributed sensor network for multi-robot surveillance,” vol. 32. pp. 1095–1100, 2014.
[28] A. Pennisi, F. Previtali, C. Gennari, D. Bloisi, L. Iocchi, F. Ficarola, A. Vitaletti, and D. Nardi, “Multi-robot surveillance through a distributed sensor network,” Studies in Computational Intelligence, vol. 604, pp. 77–98, 2015.
[29] U. Colesanti, C. Crociani, and A. Vitaletti, “On the accuracy of omnet++ in the wireless sensor networks domain: Simulation vs. Testbed.” pp. 25–31, 2007.
[30] U. M. Colesanti, C. Crociani, and A. Vitaletti, “On the Accuracy of OMNeT plus plus in the Wireless Sensor Networks Domain: Simulation vs. Testbed,” in PE-WASUN’07: PROCEEDINGS OF THE FOURTH ACM WORKSHOP ON PERFORMANCE EVALUATION OF WIRELESS AD HOC, SENSOR, AND UBIQUITOUS NETWORKS, pp. 25–31.
[31] L. Bergamini, C. Crociani, A. Vitaletti, and M. Nati, “Validation of wsn simulators through a comparison with a real testbed.” pp. 103–104, 2010.
[32] C. Petrioli, C. Pierascenzi, and A. Vitaletti, “Bluetooth scatternet formation performance: Simulations vs. Testbeds.” 2007.
[33] M. Ringwald, K. Römer, and A. Vitaletti, “Passive inspection of sensor networks,” Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), vol. 4549 LNCS, pp. 205–222, 2007.
[34] A. Di Cagno, M. Paoli, U. Colesanti, and A. Vitaletti, “REACTIVE: A peaceful coexistence between deluge and low power listening.” pp. 137–142, 2014.
[35] M. Paoli, A. Lo Russo, U. Colesanti, and A. Vitaletti, “MagoNode: Advantages of rf front-ends in wireless sensor networks,” Lecture Notes in Electrical Engineering, vol. 281 LNEE, pp. 125–137, 2014.
[36] S. Santini, K. Roemer, P. Couderc, P. Marrón, D. Minder, T. Voigt, and A. Vitaletti, “System architectures and programming models,” in Cooperating embedded systems and wireless sensor networks, ISTE, 2010, pp. 347–404.
[37] P. Karkazis, P. Trakadas, T. Zahariadis, I. Chatzigiannakis, M. Dohler, A. Vitaletti, A. Antoniou, H. C. Leligou, and L. Sarakis, “Resource and service virtualisation in m2m and iot platforms,” INTERNATIONAL JOURNAL OF INTELLIGENT ENGINEERING INFORMATICS, vol. 3, nos. 2-3, SI, pp. 205–224, 2015.
[38] C. Petrioli, R. Petroccia, D. Spaccini, A. Vitaletti, T. Arzilli, D. Lamanna, A. Galizial, and E. Renzi, “The sunrise gate: Accessing the sunrise federation of facilities to test solutions for the internet of underwater things.” 2014.
[39] L. Filipponi, A. Vitaletti, G. Landi, V. Memeo, G. Laura, and P. Pucci, “Smart city: An event driven architecture for monitoring public spaces with heterogeneous sensors.” pp. 281–286, 2010.
[40] U. Colesanti, A. Russo, M. Paoli, C. Petrioli, and A. Vitaletti, “Poster abstract: Structural health monitoring in an underground construction site: The roman experience.” 2013.
[41] S. Santini, B. Ostermaier, and A. Vitaletti, “First experiences using wireless sensor networks for noise pollution monitoring.” pp. 61–65, 2008.
[42] L. Filipponi, S. Santini, and A. Vitaletti, “Data collection in wireless sensor networks for noise pollution monitoring,” Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), vol. 5067 LNCS, pp. 492–497, 2008.
[43] V. Manzella, C. Gaz, A. Vitaletti, E. Masi, L. Santopolo, S. Mancuso, D. Salazar, and J. De Las Heras, “Demo abstract: Plants as sensing devices: The pleased experience.” 2013.
[44] S. Chatterjee, S. Ghosh, S. Das, V. Manzella, A. Vitaletti, E. Masi, L. Santopolo, S. Mancuso, and K. Maharatna, “Forward and inverse modelling approaches for prediction of light stimulus from electrophysiological response in plants,” Measurement: Journal of the International Measurement Confederation, vol. 53, pp. 101–116, 2014.
[45] S. Chatterjee, S. Das, K. Maharatna, E. Masi, L. Santopolo, S. Mancuso, and A. Vitaletti, “Exploring strategies for classification of external stimuli using statistical features of the plant electrical response,” Journal of the Royal Society Interface, vol. 12, no. 104, 2015.
[46] S. Das, B. Ajiwibawa, S. Chatterjee, S. Ghosh, K. Maharatna, S. Dasmahapatra, A. Vitaletti, E. Masi, and S. Mancuso, “Drift removal in plant electrical signals via iir filtering using wavelet energy,” Computers and Electronics in Agriculture, vol. 118, pp. 15–23, 2015.
[47] S. Chatterjee, S. Das, K. Maharatna, E. Masi, L. Santopolo, I. Colzi, S. Mancuso, and A. Vitaletti, “Comparison of decision tree based classification strategies to detect external chemical stimuli from raw and filtered plant electrical response,” Sensors and Actuators, B: Chemical, vol. 249, pp. 278–295, 2017.
[48] R. Jana, T. Johnson, S. Muthukrishnan, and A. Vitaletti, “Location based services in a wireless wan using cellular digital packet data (cdpd).” pp. 74–80, 2001.
[49] E. Trevisani and A. Vitaletti, “Cell-id location technique, limits and benefits: An experimental study.” pp. 51–60, 2004.
[50] A. Vitaletti and G. Palombizio, “Rijndael for sensor networks: Is speed the main issue?” Electronic Notes in Theoretical Computer Science, vol. 171, no. SPEC. ISS., pp. 71–81, 2007.
[51] I. Chatzigiannakis, A. Vitaletti, and A. Pyrgelis, “A privacy-preserving smart parking system using an iot elliptic curve based security platform,” Computer Communications, vols. 89-90, pp. 165–177, 2016.
[52] M. Di Zenise, A. Vitaletti, and D. Argles, “A user-centric approach to eCertificate for electronic identities (eIDs) management in mobile environment.” pp. 198–203, 2011.
[53] L. Bergamini, L. Becchetti, and A. Vitaletti, “Privacy-preserving environment monitoring in networks of mobile devices,” Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), vol. 6827 LNCS, pp. 179–191, 2011.
[54] L. Becchetti, L. Filipponi, and A. Vitaletti, “Privacy support in people-centric sensing,” Journal of Communications, vol. 7, no. SPL.ISS. 8, pp. 606–621, 2012.
[55] E. Baglioni, L. Becchetti, L. Bergamini, U. Colesanti, L. Filipponi, A. Vitaletti, and G. Persiano, “A lightweight privacy preserving sms-based recommendation system for mobile users.” pp. 191–198, 2010.
[56] L. Becchetti, L. Bergamini, U. Colesanti, L. Filipponi, G. Persiano, and A. Vitaletti, “A lightweight privacy preserving sms-based recommendation system for mobile users,” Knowledge and Information Systems, vol. 40, no. 1, pp. 49–77, 2014.
[57] F. Angeletti, I. Chatzigiannakis, and A. Vitaletti, “Towards an architecture to guarantee both data privacy and utility in the first phases of digital clinical trials,” Sensors, vol. 18, no. 12, 2018.
[58] F. Angeletti, I. Chatzigiannakis, and A. Vitaletti, “The role of blockchain and IoT in recruiting participants for digital clinical trials,” in 2017 25TH INTERNATIONAL CONFERENCE ON SOFTWARE, TELECOMMUNICATIONS AND COMPUTER NETWORKS (SOFTCOM), pp. 427–431.
[59] A. Bracciali, I. Chatzigiannakis, and A. Vitaletti, “Citizens Vote to Act: smart contracts for the management of water resources in smart cities.” in First International Conference on Societal Automation. To Appear.
[60] M. Zecchini, A. Bracciali, I. Chatzigiannakis, and A. Vitaletti, “Smart Contract Design Patterns: a use case on water management.” in FPDAPP 2019 2nd International Workshop On Future Perspective of Decentralized Applications. To Appear.
[61] M. Curcio, S. Leonardi, and A. Vitaletti, “An experimental study of prefetching and caching algorithms for the world wide web,” Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), vol. 2409, pp. 71–85, 2002.
[62] L. Becchetti, U. Colesanti, A. Marchetti-Spaccamela, and A. Vitaletti, “Self-adaptive recommendation systems: Models and experimental analysis.” pp. 479–480, 2008.
[63] L. Becchetti, U. Colesanti, A. Marchetti-Spaccamela, and A. Vitaletti, “Recommending items in pervasive scenarios: Models and experimental analysis,” Knowledge and Information Systems, vol. 28, no. 3, pp. 555–578, 2011.
[64] L. Becchetti, U. Colesanti, A. Marchetti-Spaccamela, and A. Vitaletti, “Fully decentralized recommendations in pervasive systems: Models and experimental analysis,” Engineering Intelligent Systems, vol. 20, no. 3, pp. 161–170, 2012.
[65] I. Chatzigiannakis, G. Mylonas, and A. Vitaletti, “Urban pervasive applications: Challenges, scenarios and case studies,” Computer Science Review, vol. 5, no. 1, pp. 103–118, 2011.
[66] A. Vitaletti and S. Puglia, “System overview of next-generation remote healthcare,” in Systems design for remote healthcare, K. Maharatna and S. Bonfiglio, Eds. New York, NY: Springer New York, 2014, pp. 31–53.
[67] D. Amaxilatis, I. Chatzigiannakis, I. Mavrommati, E. Vasileiou, and A. Vitaletti, “Delivering elder-care environments utilizing tv-channel based mechanisms,” Journal of Ambient Intelligence and Smart Environments, vol. 9, no. 6, pp. 783–798, 2017.
[68] I. Cappiello, S. Puglia, and A. Vitaletti, “Design and initial evaluation of a ubiquitous touch-based remote grocery shopping process.” pp. 9–14, 2009.
[69] A. Vitaletti, “GENDE: GENetic design best products evolve according to users feedback,” Lecture Notes in Electrical Engineering, vol. 413, pp. 101–110, 2017.
[70] A. Vitaletti, I. Chatzigiannakis, V. Malakuczi, and I. Mavrommati, “A Case of Genetic Algorithms Supporting the Design of Collaboratively Shaped, Genetically Evolving, Products,” in First International Conference on Societal Automation. To Appear.
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# Part B: Electron Spin and the Pauli Principle
## Presentation on theme: "Part B: Electron Spin and the Pauli Principle"— Presentation transcript:
Part B: Electron Spin and the Pauli Principle
Chapter 7 Multielectron Atoms Part B: Electron Spin and the Pauli Principle
Part B: Electron Spin and the Pauli Principle
• The Energy of Ground State Helium • Electron Spin and the Pauli Principle • Inclusion of Spin in Helium Atom Wavefunctions • Spin Angular Momentum of Ground State Helium • The Wavefunctions of Excited State Helium • Excited State Helium Energies: He(1s12s1)
The Helium Hamiltonian and Wavefunctions
The Helium Hamiltonian (Chapter 7) is: KE(1) KE(2) PE(2) PE(1) PE(12) H1 and H2 are the one electron Hamiltonians for He+ In the ground state, both electrons are in 1s orbitals and the wavefunction can be written as: We will assume that each 1s orbital is already normalized.
The Helium Ground State Energy
1 2 This is the energy of an electron in the 1s orbital of a He+ ion. J1s1s is called the Coulomb Integral. This is the total repulsion energy between the two 1s electrons. We will compare this energy of ground state Helium with the energy of excited state Helium in a later section.
Further comments on the Coulomb Integral
Coulomb Integral: Electron-Electron Repulsion. To better understand this integral, it is convenient to rewrite it in SI units with the traditional integral format. From this last equation, we see that the Coulomb Integral is really just adding up the product of the two charges divided by the distance between them over all possible volume elements.
Part B: Electron Spin and the Pauli Principle
• The Energy of Ground State Helium • Electron Spin and the Pauli Principle • Inclusion of Spin in Helium Atom Wavefunctions • Spin Angular Momentum of Ground State Helium • The Wavefunctions of Excited State Helium • Excited State Helium Energies: He(1s12s1)
Electron Spin We’ve known since Freshman Chemistry or before that electrons have spins and there’s a spin quantum number (there actually are two). Yet, we never mentioned electron spin, or the Pauli Exclusion Principle (actually the Pauli Antisymmetry Principle), in our treatment of ground state Helium in Chapter 7. This is because Helium is a closed shell system. That is, its electrons fill the n=1 shell. As we shall see, in open shell systems, such as the Lithium atom (1s22s1) or excited state Helium (e.g. 1s12s1), the electron’s spin and the Pauli Principle play an important role in determining the electronic energy.
A Brief Review of Orbital Angular Momentum in Hydrogen
An electron moving about the nucleus in a hydrogen atom has orbital angular momentum. The wavefunction for the electron in a hydrogen atom is: In addition to being eigenfunctions of the Hamiltonian (with eigenvalues En), the wavefunctions are eigenfunctions of the angular momentum operators, L2 and Lz: ^ ^ Shorthand ^
Do Electrons Spin?? I don’t know. I’ve never seen an electron up close and personal. What can be said is that their magnetic properties are consistent with the hypothesis that they behave “as though” they are spinning. When a beam of electrons is directed through a magnetic field, they behave like little magnets, with half of their North poles parallel and half antiparallel to the magnetic field’s North pole. Because a rotating charge is known to behave like a magnet, the electrons are behaving as though they are spinning in one of two directions about their axes.
Spin Angular Momentum and Quantum Numbers
A rotating (or spinning) charge possesses angular momentum. To characterize the spin angular momentum of an electron, two new quantum numbers are introduced, s and ms (analogous to l and ml), with s = ½ and ms = ½. The state of the electron is characterized by s and ms and is written as: In direct analogy to orbital angular momentum, spin angular momentum operators are introduced with the properties that: ^ ^ and
Because one always has s = ½, the standard shorthand is:
^
Orthonormality of the Spin Wavefunctions
One can define integrals of the spin functions in analogy to integrals of spatial wavefunctions, keeping in mind that one is not really using calculus to evaluate integrals. Their values are defined below: By definition By definition Therefore, by definition, the spin wavefunctions are orthonormal.
The Pauli Principle The Permutation Operator
By definition, this operator permutes (i.e. exchanges) two particles (usually electrons) in a wavefunction. For a 2 electron system: This is an eigenvalue equation, with eigenvalue pij. Permuting two identical particles will not change the probability density: Therefore:
The Pauli Principle Postulate 6: All elementary particles have an intrinsic angular momentum called spin. There are two types of particles, with different permutation properties: Bosons: Integral spin (0, 1, 2,…) Pij() = + Fermions: Half integral spin (1/2, 3/2,…) Pij() = - Fermions include electrons, protons, 3He nuclei, etc. Bosons include 4He nuclei (s=0), 2H nuclei (s=1), etc. Electrons (s = ½) are fermions. Therefore, electronic wavefunctions are antisymmetric with respect to electron exchange (permutation). Note that the permutation operator exchanges both the spatial and spin coordinates of the electrons.
Part B: Electron Spin and the Pauli Principle
• The Energy of Ground State Helium • Electron Spin and the Pauli Principle • Inclusion of Spin in Helium Atom Wavefunctions • Spin Angular Momentum of Ground State Helium • The Wavefunctions of Excited State Helium • Excited State Helium Energies: He(1s12s1)
Inclusion of Spin in Helium Atom Wavefunctions
The Hamiltonian for Helium does not contain any spin operators. Therefore, one can take the total wavefunction to be the product of spatial and spin parts. If we use the approximation that the spatial part can be represented by 1s orbitals for each electron, then 4 possibilities for the total wavefunction are: Electron 1 has spin. Electron 2 has spin. Electron 1 has spin. Electron 2 has spin. Electron 1 has spin. Electron 2 has spin. Electron 1 has spin. Electron 2 has spin. Shorthand Notation:
A wavefunction that satisfies the Pauli Principle
None of these 4 functions satisfies the Pauli Antisymmetry Principle. Similarly: Similarly: A wavefunction that satisfies the Pauli Principle We can construct a linear combination of 1 and 2 that does satisfy the Pauli Principle. Thus, is antisymmetric with respect to electron exchange, as required by the Pauli Principle.
Note: The sum of 1 and 2 would not be a
satisfactory wavefunction. Because and neither of these functions can be used in the construction of an antisymmetric wavefunction This is the basis for the more famous, but less general, form of the Pauli Principle, known as the Exclusion Principle: Two electrons in an atom cannot have the same set of 4 quantum numbers, n, l, ml and ms. That is, if two electrons have the same spatial part of the wavefunction (100 for both electrons in the Helium ground state), then they cannot have the same spin. The wavefunction, , can be written as the product of a spatial and spin part:
Normalization of the Antisymmetric Wavefunction
We assume that the individual spatial wavefunctions have already been normalized. We must integrate over both the spin and spatial parts of the wavefunction. = 1 or = 1
= 1 = = = 1
Spin and the Energy of Ground State Helium
Earlier in this chapter, prior to reducing electron spin, we showed that the energy of ground state helium is given by: Would its inclusion have affected the results? We will examine this question below. The expression for the expectation value of the energy is given by:
We can factor out the spin part of the wave function
because H is independent of spin Thus, inclusion of the spin portion of the wavefunction has no effect on the computed energy in a closed shell system such as ground state Helium. Note: It can be shown that one arrives at the same conclusion if a more sophisticated spatial function is used to characterize the two electrons.
Part B: Electron Spin and the Pauli Principle
• The Energy of Ground State Helium • Electron Spin and the Pauli Principle • Inclusion of Spin in Helium Atom Wavefunctions • Spin Angular Momentum of Ground State Helium • The Wavefunctions of Excited State Helium • Excited State Helium Energies: He(1s12s1)
Spin Angular Momentum of Ground State Helium
z-Component of Spin Angular Momentum For a two electron system, the operator for Sz is Therefore Therefore, the eigenvalue of Sz is 0. The z-component of angular momentum is MS = 0. ^
Total Spin Angular Momentum
The S2 operator for a two electron system and the calculation of the eigenvalue of this operator is significantly more complicated than the calculation of the z-component. ^ This calculation requires application of spin raising and lowering operators (introduced in various texts**), and is a digression from our prime focus. **See for example, “Quantum Chemistry”, by I. N. Levine (5th. Ed.) Sect Therefore, we will just present the results. The result is ^ Thus, for ground state Helium: S=0 and MS=0 We say that GS helium is a “singlet” because there is only one possible combination of S and MS (0 and 0).
Generalization In general, the spin wavefunctions of multielectron atoms are eigenfunctions of S2 and Sz, with eigenvalues S(S+1)ħ2 and MSħ . ^ ^ Some possible combinations of S and MS that can be encountered are given in the table below S MS Designation Singlet 1/ /2, -1/ Doublet 1, 0, Triplet 3/ /2, 1/2, -1/2, -3/2 Quartet
Part B: Electron Spin and the Pauli Principle
• The Energy of Ground State Helium • Electron Spin and the Pauli Principle • Inclusion of Spin in Helium Atom Wavefunctions • Spin Angular Momentum of Ground State Helium • The Wavefunctions of Excited State Helium • Excited State Helium Energies: He(1s12s1)
The Wavefunctions of Excited State Helium
In ground state Helium, we were able to write the wavefunction as the product of spatial and spin parts. I have included the normalization constant with the spin function, which is what it is normalizing (it is assumed that the spatial part includes its own normalization constant) In ground state Helium, the spatial wavefunction is symmetric with respect to electron exchange. Therefore, it is necessary for the spin function to be antisymmetric with respect to exchange in order to satisfy the Pauli Principle. If one of the electrons is excited to the 2s orbital to give He(1s12s1), the spatial wavefunction can be either symmetric or antisymmetric with respect to electron exchange, broadening the possibilities for valid spin functions.
Symmetric and Antisymmetric Spatial Wavefunctions
Neither nor are valid spatial wavefunctions because they are neither symmnetric nor antisymmetric with respect to the exchange of the two electrons. However, one can “build” combinations of these wavefunctions that are either symmetric or antisymmetric with respect to electron exchange. Symmetric We have denoted this as a symmetric function, because it is easy to show that: Antisymmetric For this function
Symmetric and Antisymmetric Spin Wavefunctions
Two symmetric spin wavefunctions are: 12 and 12 because and We could not use either of these symmetric spin functions for ground state Helium because the symmetric spatial function required that we must have an antisymmetric spin function to satisfy the Pauli Principle. A third symmetric spin wavefunction is: It is straightforward to apply the permutation operator, P12, to this function to prove that it is symmetric with respect to exchange. As shown when discussing ground state Helium, a spin wavefunction that is antisymmetric with respect to electron exchange is:
S and MS of the Spin Wavefunctions
Therefore, MS=+1 for Similarly, MS=0 for MS=-1 for Using advanced methods,** (you are not responsible for it), one can show that when the S2 operator is applied to any of the 3 symmetric spin functions, the eigenvalue is 2ħ2 [ = S(S+1) ħ2 ]. **e.g. Introduction to Quantum Mechanics in Chemistry, by M. A. Ratner and G. C. Schatz, Sect. 8.3 Therefore, S=1 for the 3 symmetric spin wavefunctions. Together, these functions are a triplet with S=1 and MS=+1,0,-1.
When Sz operates on the antisymmetric spin function
one finds that MS=0. It can be shown that when S2 operates on this function, the eigenvalue is 0. Therefore, S=0 for the antisymmetric spin function. Therefore the antisymmetric spin wavefunction is a singlet, with S=0 and MS=0.
The Total Wavefunction for Excited State Helium
Spatial Wavefunctions Spin Wavefunctions Singlet Triplet One can write the total wavefunction as the product of spin and spatial parts.
Singlet Wavefunction Triplet Wavefunctions
Part B: Electron Spin and the Pauli Principle
• The Energy of Ground State Helium • Electron Spin and the Pauli Principle • Inclusion of Spin in Helium Atom Wavefunctions • Spin Angular Momentum of Ground State Helium • The Wavefunctions of Excited State Helium • Excited State Helium Energies: He(1s12s1)
Excited State Helium Energies: He(1s12s1)
The expectation value for energy is given by: The Helium Hamiltonian is: KE(1) KE(2) PE(2) PE(1) PE(12) H1 and H2 are the one electron Hamiltonians for He+
Triplet State Energy Because the Hamiltonian does not contain any spin operators, the above expression can be simplified. Note that the energy does not depend directly on the spin wavefunction. It is the fact that the triplet state symmetric spin wavefunction requires us to use the antisymmetric spatial wavefunction that affects the calculated energy.
We have assumed that the spatial wavefunction is normalized,
in which case the denominator is 1. The energy can then be calculated from:
where Similarly,
|| ||
where Similarly,
1 1. Energy of electron in 1s He+ orbital 2 2. Energy of electron in 2s He+ orbital 3 3. Coulomb (repulsion) Integral 4 4. Exchange Integral where
1 2 3 4 Always positive 3. Coulomb (repulsion) Integral The integrand of the Coulomb integral represents the repulsion of two infinitesimal electron densities, (1)=1s(1)2 and (2)=2s(2)2, separated by a distance, r12. The repulsion is summed over all infinitesimal electron densities. Usually positive 4. Exchange Integral Arises purely from the antisymmetry of the spatial function with respect to electron exchange. It has no classical analog. If the above calculation had been performed with a simple product wavefunction, spat = 1s(1)2s(2), there would be no exchange integral
Singlet State Energy Triplet: Singlet:
One of 3 components of the Triplet Singlet:
Because the exchange integral is almost always positive, the
energy of excited triplet state Helium is lower than that of the excited state singlet. The physical basis for the lower energy of the triplet is that the wavefunction (and therefore the probability) is small when the coordinates of the two electrons are close to each other. Therefore, the electron-electron repulsion energy is minimized
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# Question 5Encontrar la longitud de la curva polar ? = 58 0 < 0 < Zn Dar respuesta con 3
###### Question:
Question 5 Encontrar la longitud de la curva polar ? = 58 0 < 0 < Zn Dar respuesta con 3 cifras decimales.
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Row Matrix:
A matrix having a single row is called a row matrix. e. g. [1 3 5 7]
Square Matrix:
An m x n matrix A is said to be a square matrix if m = n i.e. number of rows = number of columns.
Note:
• The diagonal from left hand side upper corner to right hand side lower side lower corner is known as leading diagonal or principal diagonal. In the above example square matrix containing the elements 1, 3, 5 is called the leading or principal diagonal.
Diagonal Matrix:
A square matrix all of whose elements except those in the leading diagonal, are zero is called a diagonal matrix. For a square matrix A = [aij]nxn to be a diagonal matrix, aij = 0, whenever i not equal to j.
Scalar Matrix:
A diagonal matrix whose all the leading diagonal elements are equal is called a scalar matrix.
Triangular Matrix:
A square matrix in which all the elements below the diagonal elements are zero is called Upper Triangular matrix and a square matrix in which all the elements above diagonal elements are zero is called Lower Triangular matrix.
Given a square matrix A = [aij]n´n,
For upper triangular matrix, aij = 0, i > j
and for lower triangular matrix, aij = 0, i < j
Notes:
• Diagonal matrix is both upper and lower triangular
• A triangular matrix A = [aij]n´n is called strictly triangular if aii = 0 for 1 ≤ i ≤ n.
Null Matrix:
If all the elements of a matrix (square or rectangular) are zero, it is called a null or zero matrix.
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# Moscow Mathematical Journal
Volume 20, Issue 2, April–June 2020 pp. 375–404.
Orbital Chen Theorem for Germs of $\mathcal{C}^{\infty}$ Vector Fields with Degenerate Singularity
Authors: Jessica Jaurez Rosas (1) and Laura Ortiz-Bobadilla (1)
Author institution:(1) Instituto de Matemáticas, Universidad Nacional Autónoma de México (UNAM), Área de la Investigación Científica, Circuito exterior, Ciudad Universitaria, 04510, Ciudad de México, México
Summary:
We consider germs of $\mathcal{C}^{\infty}$ vector fields in $(\mathbb{R}^2, 0)$ with degenerate non-dicritic singularity (having ($n-1$)-jet zero and non-zero $n$-jet) and their corresponding foliations. Under some natural hypothesis we prove that the orbital formal equivalence of any two such vector fields implies their orbital $\mathcal{C}^{\infty}$ equivalence (and thus the $\mathcal{C}^{\infty}$ equivalence of the corresponding foliations). This result generalizes Chen Theorem for foliations defined by generic $\mathcal{C}^{\infty}$ germs of vector fields in $(\mathbb{R}^2, 0)$ having hyperbolic singularity.
2010 Math. Subj. Class. 34C07, 34C05, 34C08.
Keywords: Formal normal forms, foliations, flat vector fields, rigidity.
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# Do all representations of finite groups have one-dimensional subrepresentations?
Let V be a representation of a finite group G, and $v\in V$ - a nonzero vector. Put $$u = \sum_{g\in G} gv.$$ Then for any $g\in G$ we have $gu = u$ and therefore $<u>$ is a subrepresentation of V.
I know there is an error here since there are irreducible representations of finite groups which are not one dimensional, but I can't see it. Could someone point it out?
Thank you.
-
## 1 Answer
It is possible that $u$ is zero.
-
Indeed. Thank you! – Artem Sep 3 '12 at 17:05
@Artem, You could have found this out by yourself by picking any irreducible representation of dimesion larger than one of any group! You should always look at examples. – Mariano Suárez-Alvarez Sep 3 '12 at 17:29
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My bibliography Save this paper
## Author
Listed:
• Jean Jacod
• Viktor Todorov
## Abstract
We consider a process $X_t$, which is observed on a finite time interval $[0,T]$, at discrete times $0,\Delta_n,2\Delta_n,\ldots.$ This process is an It\^{o} semimartingale with stochastic volatility $\sigma_t^2$. Assuming that $X$ has jumps on $[0,T]$, we derive tests to decide whether the volatility process has jumps occurring simultaneously with the jumps of $X_t$. There are two different families of tests for the two possible null hypotheses (common jumps or disjoint jumps). They have a prescribed asymptotic level as the mesh $\Delta_n$ goes to $0$. We show on some simulations that these tests perform reasonably well even in the finite sample case, and we also put them in use on S&P 500 index data.
## Suggested Citation
• Jean Jacod & Viktor Todorov, 2010. "Do price and volatility jump together?," Papers 1010.4990, arXiv.org.
• Handle: RePEc:arx:papers:1010.4990
as
File URL: http://arxiv.org/pdf/1010.4990
### NEP fields
This paper has been announced in the following NEP Reports:
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# Copper(I) oxide
Last updated
Names Identifiers IUPAC name Copper(I) oxide Other names Cuprous oxideDicopper oxide Cuprite Red copper oxide 3D model (JSmol) ChemSpider ECHA InfoCard 100.013.883 EC Number 215-270-7 KEGG PubChem CID RTECS number GL8050000 UNII InChI=1S/2Cu.O/q2*+1;-2 Key: KRFJLUBVMFXRPN-UHFFFAOYSA-N InChI=1/2Cu.O/rCu2O/c1-3-2Key: BERDEBHAJNAUOM-YQWGQOGZAFInChI=1/2Cu.O/q2*+1;-2Key: KRFJLUBVMFXRPN-UHFFFAOYAM [Cu]O[Cu][Cu+].[Cu+].[O-2] Cu2O Molar mass 143.09 g/mol Appearance brownish-red solid Density 6.0 g/cm3 Melting point 1,232 °C (2,250 °F; 1,505 K) Boiling point 1,800 °C (3,270 °F; 2,070 K) Insoluble Solubility in acid Soluble Band gap 2.137 eV -20·10−6 cm3/mol cubic Std molarentropy (So298) 93 J·mol−1·K−1 Std enthalpy offormation (ΔfHo298) −170 kJ·mol−1 Safety data sheet SIRI.org Harmful (Xn)Dangerous for the environment (N) R-phrases R22, R50/53 S-phrases (S2), S22, S60, S61 NFPA 704 US health exposure limits (NIOSH): PEL (Permissible) TWA 1 mg/m3 (as Cu) [1] REL (Recommended) TWA 1 mg/m3 (as Cu) [1] IDLH (Immediate danger) TWA 100 mg/m3 (as Cu) [1] Other anions Copper(I) sulfide Copper(II) sulfide Copper(I) selenide Other cations Copper(II) oxide Silver(I) oxide Nickel(II) oxide Zinc oxide Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa). (what is ?) Infobox references
Copper(I) oxide or cuprous oxide is the inorganic compound with the formula Cu2O. It is one of the principal oxides of copper, the other being CuO or cupric oxide. This red-coloured solid is a component of some antifouling paints. The compound can appear either yellow or red, depending on the size of the particles. [2] Copper(I) oxide is found as the reddish mineral cuprite.
An inorganic compound is typically a chemical compound that lacks C-H bonds, that is, a compound that is not an organic compound, but the distinction is not defined or even of particular interest.
An oxide is a chemical compound that contains at least one oxygen atom and one other element in its chemical formula. "Oxide" itself is the dianion of oxygen, an O2– atom. Metal oxides thus typically contain an anion of oxygen in the oxidation state of −2. Most of the Earth's crust consists of solid oxides, the result of elements being oxidized by the oxygen in air or in water. Hydrocarbon combustion affords the two principal carbon oxides: carbon monoxide and carbon dioxide. Even materials considered pure elements often develop an oxide coating. For example, aluminium foil develops a thin skin of Al2O3 (called a passivation layer) that protects the foil from further corrosion. Individual elements can often form multiple oxides, each containing different amounts of the element and oxygen. In some cases these are distinguished by specifying the number of atoms as in carbon monoxide and carbon dioxide, and in other cases by specifying the element's oxidation number, as in iron(II) oxide and iron(III) oxide. Certain elements can form many different oxides, such as those of nitrogen.
Copper is a chemical element with symbol Cu and atomic number 29. It is a soft, malleable, and ductile metal with very high thermal and electrical conductivity. A freshly exposed surface of pure copper has a pinkish-orange color. Copper is used as a conductor of heat and electricity, as a building material, and as a constituent of various metal alloys, such as sterling silver used in jewelry, cupronickel used to make marine hardware and coins, and constantan used in strain gauges and thermocouples for temperature measurement.
## Preparation
Copper(I) oxide may be produced by several methods. [3] Most straightforwardly, it arises via the oxidation of copper metal:
4 Cu + O2 → 2 Cu2O
Additives such as water and acids affect the rate of this process as well as the further oxidation to copper(II) oxides. It is also produced commercially by reduction of copper(II) solutions with sulfur dioxide. Aqueous cuprous chloride solutions react with base to give the same material. In all cases, the color is highly sensitive to the procedural details.
Sulfur dioxide is the chemical compound with the formula SO
2
. It is a toxic gas with a burnt match smell. It is released naturally by volcanic activity and is produced as a by-product of the burning of fossil fuels contaminated with sulfur compounds and copper extraction.
Formation of copper(I) oxide is the basis of the Fehling's test and Benedict's test for reducing sugars. These sugars reduce an alkaline solution of a copper(II) salt, giving a bright red precipitate of Cu2O.
Fehling's solution is a chemical reagent used to differentiate between water-soluble carbohydrate and ketone functional groups, and as a test for reducing sugars and non-reducing sugars, supplementary to the Tollens' reagent test. The test was developed by German chemist Hermann von Fehling in 1849.
Benedict's reagent is a chemical reagent named after American chemist Stanley Rossiter Benedict.
Sugar is the generic name for sweet-tasting, soluble carbohydrates, many of which are used in food. The various types of sugar are derived from different sources. Simple sugars are called monosaccharides and include glucose, fructose, and galactose. "Table sugar" or "granulated sugar" refers to sucrose, a disaccharide of glucose and fructose. In the body, sucrose is hydrolysed into fructose and glucose.
It forms on silver-plated copper parts exposed to moisture when the silver layer is porous or damaged. This kind of corrosion is known as red plague.
Silver is a chemical element with symbol Ag and atomic number 47. A soft, white, lustrous transition metal, it exhibits the highest electrical conductivity, thermal conductivity, and reflectivity of any metal. The metal is found in the Earth's crust in the pure, free elemental form, as an alloy with gold and other metals, and in minerals such as argentite and chlorargyrite. Most silver is produced as a byproduct of copper, gold, lead, and zinc refining.
Corrosion is a natural process, which converts a refined metal to a more chemically-stable form, such as its oxide, hydroxide, or sulfide. It is the gradual destruction of materials by chemical and/or electrochemical reaction with their environment. Corrosion engineering is the field dedicated to controlling and stopping corrosion.
Red plague is an accelerated corrosion of copper when plated with silver. After storage or use in high-humidity environment, cuprous oxide forms on the surface of the parts. The corrosion is identifiable by presence of patches of brown-red powder deposit on the exposed copper.
Little evidence exists for cuprous hydroxide, which is expected to rapidly undergo dehydration. A similar situation applies to the hydroxides of gold(I) and silver(I).
## Properties
The solid is diamagnetic. In terms of their coordination spheres, copper centres are 2-coordinated and the oxides are tetrahedral. The structure thus resembles in some sense the main polymorphs of SiO2, and both structures feature interpenetrated lattices.
Copper(I) oxide dissolves in concentrated ammonia solution to form the colourless complex [Cu(NH3)2]+, which is easily oxidized in air to the blue [Cu(NH3)4(H2O)2]2+. It dissolves in hydrochloric acid to give solutions of CuCl2. Dilute sulfuric acid and nitric acid produce copper(II) sulfate and copper(II) nitrate, respectively. [4]
Cu2O degrades to copper(II) oxide in moist air.
## Structure
Cu2O crystallizes in a cubic structure with a lattice constant al=4.2696 Å. The Cu atoms arrange in a fcc sublattice, the O atoms in a bcc sublattice. One sublattice is shifted by a quarter of the body diagonal. The space group is ${\displaystyle \scriptstyle Pn{\bar {3}}m}$, which includes the point group with full octahedral symmetry.
## Semiconducting properties
In the history of semiconductor physics, Cu2O is one of the most studied materials, and many experimental obser and semiconductor applications have been demonstrated first in this material:
The lowest excitons in Cu2O are extremely long lived; absorption lineshapes have been demonstrated with neV linewidths, which is the narrowest bulk exciton resonance ever observed. [8] The associated quadrupole polaritons have low group velocity approaching the speed of sound. Thus, light moves almost as slowly as sound in this medium, which results in high polariton densities. Another unusual feature of the ground state excitons is that all primary scattering mechanisms are known quantitatively. [9] Cu2O was the first substance where an entirely parameter-free model of absorption linewidth broadening by temperature could be established, allowing the corresponding absorption coefficient to be deduced. It can be shown using Cu2O that the Kramers–Kronig relations do not apply to polaritons. [10]
## Applications
Cuprous oxide is commonly used as a pigment, a fungicide, and an antifouling agent for marine paints. Rectifier diodes based on this material have been used industrially as early as 1924, long before silicon became the standard. Copper(I) oxide is also responsible for the pink color in a positive Benedict's test.
## Related Research Articles
Hydroxide is a diatomic anion with chemical formula OH. It consists of an oxygen and hydrogen atom held together by a covalent bond, and carries a negative electric charge. It is an important but usually minor constituent of water. It functions as a base, a ligand, a nucleophile, and a catalyst. The hydroxide ion forms salts, some of which dissociate in aqueous solution, liberating solvated hydroxide ions. Sodium hydroxide is a multi-million-ton per annum commodity chemical. A hydroxide attached to a strongly electropositive center may itself ionize, liberating a hydrogen cation (H+), making the parent compound an acid.
In physics, polaritons are quasiparticles resulting from strong coupling of electromagnetic waves with an electric or magnetic dipole-carrying excitation. They are an expression of the common quantum phenomenon known as level repulsion, also known as the avoided crossing principle. Polaritons describe the crossing of the dispersion of light with any interacting resonance. To this extent polaritons can also be thought as the new normal modes of a given material or structure arising from the strong coupling of the bare modes, which are the photon and the dipolar oscillation. The polariton is a bosonic quasiparticle, and should not be confused with the polaron, which is an electron plus an attached phonon cloud.
Copper oxide is a compound from the two elements copper and oxygen.
A polaron is a quasiparticle used in condensed matter physics to understand the interactions between electrons and atoms in a solid material. The polaron concept was first proposed by Lev Landau in 1933 to describe an electron moving in a dielectric crystal where the atoms move from their equilibrium positions to effectively screen the charge of an electron, known as a phonon cloud. This lowers the electron mobility and increases the electron's effective mass.
Copper(II) oxide or cupric oxide is the inorganic compound with the formula CuO. A black solid, it is one of the two stable oxides of copper, the other being Cu2O or cuprous oxide. As a mineral, it is known as tenorite. It is a product of copper mining and the precursor to many other copper-containing products and chemical compounds.
Copper(I) chloride, commonly called cuprous chloride, is the lower chloride of copper, with the formula CuCl. The substance is a white solid sparingly soluble in water, but very soluble in concentrated hydrochloric acid. Impure samples appear green due to the presence of copper(II) chloride (CuCl2).
Copper(II) chloride is the chemical compound with the chemical formula CuCl2. This is a light brown solid, which slowly absorbs moisture to form a blue-green dihydrate. The copper(II) chlorides are some of the most common copper(II) compounds, after copper sulfate.
Copper(II) hydroxide is the hydroxide of copper with the chemical formula of Cu(OH)2. It is a pale greenish blue or bluish green solid. Some forms of copper(II) hydroxide are sold as "stabilized" copper hydroxide, although they likely consist of a mixture of copper(II) carbonate and hydroxide. Copper hydroxide is a weak base.
Copper(I) acetylide, or cuprous acetylide, is a chemical compound with the formula Cu2C2. Although never characterized by X-ray crystallography, the material has been claimed at least since 1856. One form is claimed to be a monohydrate with formula Cu
2
C
2
.H
2
O
). It is a reddish solid, that easily explodes when dry.
Copper(I) fluoride or cuprous fluoride is an inorganic compound with the chemical formula CuF. Its existence is uncertain. It was reported in 1933 to have a sphalerite-type crystal structure. Modern textbooks state that CuF is not known, since fluorine is so electronegative that it will always oxidise copper to its +2 oxidation state. Complexes of CuF such as [(Ph3P)3CuF] are, however, known and well characterised.
Tetrakis(acetonitrile)copper(I) hexafluorophosphate is a coordination compound with the formula [Cu(CH3CN)4]PF6. It is a colourless solid that is used in the synthesis of other copper complexes. It is a well-known example of a transition metal nitrile complex.
Alexey V. Kavokin is a Russian and French theoretical physicist and writer.
Delafossite is a copper iron oxide mineral with formula CuFeO2 or Cu1+Fe3+O2. It is a member of the delafossite mineral group, which has the general formula ABO2, a group characterized by sheets of linearly coordinated A cations stacked between edge-shared octahedral layers (BO6). Delafossite, along with other minerals of the ABO2 group, is known for its wide range of electrical properties, its conductivity varying from insulating to metallic. Delafossite is usually a secondary mineral that crystallizes in association with oxidized copper and rarely occurs as a primary mineral.
Copper(I) sulfate, also known as cuprous sulfate and dicopper sulfate, is the chemical compound with the chemical formula Cu2SO4 and a molar mass of 223.15 g mol−1. It is an unstable compound as copper(I) compounds are generally unstable and is more commonly found in the CuSO4 state. It is light green in color at room temperature and is water-soluble. Due to the low-stability of the compound there are currently not many applications to date.
Terahertz spectroscopy detects and controls properties of matter with electromagnetic fields that are in the frequency range between a few hundred gigahertz and several terahertz. In many-body systems, several of the relevant states have an energy difference that matches with the energy of a THz photon. Therefore, THz spectroscopy provides a particularly powerful method in resolving and controlling individual transitions between different many-body states. By doing this, one gains new insights about many-body quantum kinetics and how that can be utilized in developing new technologies that are optimized up to the elementary quantum level.
Bose–Einstein condensation can occur in quasiparticles, particles that are effective descriptions of collective excitations in materials. Some have integer spins and can be expected to obey Bose–Einstein statistics like traditional particles. Conditions for condensation of various quasiparticles have been predicted and observed. The topic continues to be an active field of study.
Copper(I) thiocyanate is a coordination polymer with formula CuSCN. It is an air-stable, white solid used as a precursor for the preparation of other thiocyanate salts.
Bose–Einstein condensation of polaritons is a growing field in semiconductor optics research, which exhibits spontaneous coherence similar to a laser, but through a different mechanism. A continuous transition from polariton condensation to lasing can be made similar to that of the crossover from a Bose–Einstein condensate to a BCS state in the context of Fermi gases. Polariton condensation is sometimes called “lasing without inversion”.
A quantum dot single-photon source is based on a single quantum dot placed in an optical microcavity. It is an on-demand single photon source. A pulsed laser can excite a pair of carriers known as an exciton in the quantum dot. Due to the discrete energy levels, only one exciton can exist in the quantum dot at a given time. The decay of the exciton due to spontaneous emission leads to the emission of a single photon. It is a nonclassical light source that shows photon antibunching. The emission of single photons can be proven by measuring the second order intensity correlation function. The linewidth of the emitted photons can be reduced by using distributed Bragg reflectors (DBR’s). Additionally, DBR's lead to an emission in a well-defined direction.
Chevreul's salt (Copper(I,II) Sulfite Dihydrate, Cu2SO3•CuSO3•2H2O (Cu3(SO3)2•2H2O)), is a copper salt which was prepared for the first time by a French chemist Michel Eugène Chevreul in 1812. Its unique property is that it contains copper in both of its common oxidation states. It is insoluble in water and stable in air. What was known as Rogojski's salt is a mixture of Chevreul's salt and metallic copper.
## References
1. "NIOSH Pocket Guide to Chemical Hazards #0150". National Institute for Occupational Safety and Health (NIOSH).
2. N. N. Greenwood, A. Earnshaw, Chemistry of the Elements, 2nd ed., Butterworth-Heinemann, Oxford, UK, 1997.
3. H. Wayne Richardson "Copper Compounds in Ullmann's Encyclopedia of Industrial Chemistry 2002, Wiley-VCH, Weinheim. doi : 10.1002/14356007.a07_567
4. D. Nicholls, Complexes and First-Row Transition Elements, Macmillan Press, London, 1973.
5. L. O. Grondahl, Unidirectional current carrying device, Patent, 1927
6. L. Hanke, D. Fröhlich, A.L. Ivanov, P.B. Littlewood, and H. Stolz "LA-Phonoritons in Cu2O" Phys. Rev. Lett. 83 (1999), 4365.
7. L. Brillouin: Wave Propagation and Group Velocity, Academic Press, New York City, 1960.
8. J. Brandt, D. Fröhlich, C. Sandfort, M. Bayer, H. Stolz, and N. Naka, Ultranarrow absorption and two-phonon excitation spectroscopy of Cu2O paraexcitons in a high magnetic field, Phys. Rev. Lett. 99, 217403 (2007). doi : 10.1103/PhysRevLett.99.217403
9. J. P. Wolfe and A. Mysyrowicz: Excitonic Matter, Scientific American 250 (1984), No. 3, 98.
10. Hopfield, J. J. (1958). "Theory of the Contribution of Excitons to the Complex Dielectric Constant of Crystals". Physical Review. 112 (5): 1555–1567. doi:10.1103/PhysRev.112.1555. ISSN 0031-899X.
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# Laplace transform heat equation $u(0,t)=f(t),u(x,0)=u_0$
Solve the heat equation $$\frac{\partial u}{\partial{t}}=\nu\frac{\partial^2{u}}{\partial{x}^2},t>0,x>0$$ where $$u(x,0)=f(t),u(x,0)=u_0$$ using Laplace transforms.
My trouble is dealing with the general initial condition to find constants. Applying the Laplace transform I got $$U(s)=A(s)e^{\sqrt{\frac{s}{\nu}}x}+B(s)e^{-\sqrt{\frac{s}{\nu}}x}.$$ The issue is finding out $$A$$ and $$B$$. Using $$u(0,s)=F(s)$$, I get $$A(s)+B(s)+\frac{u_0}{s}=F(s).$$ The only way I can think of determining them is by forcing $$U$$ is bounded. Then $$A=0$$ and I can write $$B(s)=F(s)-\frac{u_0}{s}$$ and the solution is $$u(t,x)=L^{-1}(F(s)e^{\sqrt{\frac{s}{\nu}}x})-u_0\text{erfc}(\frac{x}{2\sqrt{\nu t}})+u_0.$$ Is this the way to proceed?
• Look a little closely at your problem set up, you’ve a typo. – DaveNine Nov 17 '18 at 10:06
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# D flip flop with asynchronous level triggered reset
My code :
always @ ( posedge clock, posedge rst)
begin
if(rst) q = 1'b0 ;
else q = d ;
end
Waveform :
Error :
Between 100 and 150 ns, the output followed the d input although there was no clock edge. In other words, it behaved like a latch.
I wanted to make a positive edge triggered d flip flop with asynchronous positive level triggered reset, which I succeeded after examining the reference link above.
Prior to this, I was using the code given above. As you can see, the only difference is that I have used blocking assignment while the reference uses non blocking assignment.
Now I know that non blocking assignment schedules the values in contrast to blocking assignment. However I am not able to understand how is this affecting my design ? In other words, can someone explain step by step or in little detail why changing non blocking to blocking is resulting in incorrect design.
Additionally, I also tried this code :
always @ ( posedge clock, rst)
begin
if(rst) q = 1'b0 ;
else q = d ;
end
which also resulted in erroneous design. But why, I wasnt able to understand.
Thank you.
• Can you share the test-bench and tell us what simulator and version you are using. If your design is only one flip-flop, blocking vs non-blocking assignment shouldn't be an issue. My guess is there is a glitch on clock or rst, or bizarre simulator behavior. – Greg Feb 10 '15 at 0:28
• @Greg . You were right. I use modelsim student version. Today I reinstalled the simulator after your comment. Seems like it now works as expected. Just one last thing, can you explain the last block of code ; it sure doesnt result in FF but why ? – Plutonium smuggler Feb 10 '15 at 12:39
The best answer for blocking vs non-blocking flip-flops assignment is already answered on Stack Overflow here. That answer also references to a paper by Cliff Cummings, here.
Now, the code for your second attempt will always result in with the behavior shown in the waveform, even with non-blocking assignments:
always @(posedge clock, rst)
begin
if (rst) q <= 1'b0 ;
else q <= d ;
end
This is because the sensitivity list for signal rst is incorrect. The posedge keyword only applies to the signal immediately right of the keyword and not the following signals after the comma. @(posedge clock, rst) is equivalent to @(posedge clock, posedge rst, negedge rst). The else condition will be evaluated on the falling edged of rst, which is not desired.
A good linting, synthesizer, or LEC (logical equivalently check) tool should flag the dual edge trigger on rst as an issue; possible as a warning.
• @ is trigger by events: edge triggered, value change, or triggered event type (not synthesizable) within the sensitivity list. posedge rst with if (rst) tells the synthesizer to use a D-flip-flip with an active high asynchronous reset. Asynchronous is for events outside of the synchronous domain. For a active low reset async reset you need negedge rst with if (!rst). For sync reset, remove rst from the sensitivity list. Behavior will be correct in simulation. You may run into trouble on FPGAs with limited number of flops with async set/rst, but that should get errors/warnings – Greg Feb 10 '15 at 20:08
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## Why does the band gap exist?
Hello, do you know why no electron can stay in the band gap? Is it impossible at every energy?
Thank you!
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> Nanocrystals grow from liquid interface>> New insights into how materials transfer heat could lead to improved electronics
Do you know no electron can stay between -13.6eV and -3.4eV energy levels in an isolated Hydrogen atom? Now, apply the analogy.
Yes but why they cannot stay there? What does impede them to stay in that levels^
## Why does the band gap exist?
Bloch's theorem says that anytime you have a periodic potential (like in a lattice of a metal or semiconductor where the atoms are equally spaced apart), then the solutions to Schrodinger's Equation will be plane waves, i.e. $\psi$ ~ $e^{i kx}$ where k is the wave number. When you actually solve a particular problem, you will find certain restrictions on k, that is, you will find for certain values of k, no such plane wave solutions exist. Since k is related to the energy, then you also get restrictions on the energy. That is, for certain values of the energy, there will be no valid solutions to Shrodinger's Equation. These energy levels where no solution exists are referred to as Energy gaps, or Band gaps. The reason the electron can't be on one of these gaps is because there is no solution to Shrodinger's equation in these regions, hence they are forbidden.
Recognitions: Science Advisor Because there are no energy levels for them to sit in. Energy levels are time-independent solutions to the Schrodinger equation. Claude.
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# ←→
Reviews of topical problems
# Earth insolation variation and its incorporation into physical and mathematical climate models
Lomonosov Moscow State University, Vorobevy Gory, Moscow, 119991, Russian Federation
This paper reviews research into long-term variations in the Earth's insolation due to celestial-mechanics processes. Based on an analytical survey of Earth's insolation calculations, general problems encountered in the physical and mathematical modeling of climate are outlined.
BibTexBibNote ® (generic)BibNote ® (RIS)Medline RefWorks RT Journal T1 Earth insolation variation and its incorporation into physical and mathematical climate models A1 Fedorov,V.M. PB Physics-Uspekhi PY 2019 FD 10 Jan, 2019 JF Physics-Uspekhi JO Phys. Usp. VO 62 IS 1 SP 32-45 DO 10.3367/UFNe.2017.12.038267 LK https://ufn.ru/en/articles/2019/1/c/
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# Relative simultaneity and time going backwards as someone accelerates
I have an observer on Earth with an atomic clock, let's call this the unprimed frame with space coordinate x=0 and t. Universe is one dimensional.
I have a rocket ship sitting stationary in empty space to the far "left" of the earth on the x-axis. Because it is stationary wrt the Earth, rocket ship and earth have identical spacetime basis (up to a shift in space). The ship begins to accelerate to the right at t=0 with immense force such that relativistic effects can be felt quickly. Special relativity would say that the travelling space ship plane of simultaneity begins to rotate counter-clockwise in a standard Minkowski diagram. To me, this seems to mean that "things" to the left of the spaceship would advance in time more quickly (the further left, the further the time advances), while things to the right of the spaceship must go backwards in time. Since the rocket ship is to the left of the earth, won't it mean that the rocketship crew sees events on Earth going backwards in time?
• "won't it mean that the rocketship crew sees events on Earth going backwards in time" - please note that seeing and observing are distinctly different concepts in SR. Would you please clarify which of the two you're asking about? – Alfred Centauri Jul 29 '19 at 3:06
• @AlfredCentauri What is the difference between seeing and observing? – Shuheng Zheng Jul 29 '19 at 5:00
• I meant observe. I just assume that the light travel time can be skipped. Or we can use the replay trick of having an army of observers at every location in space. – Shuheng Zheng Jul 29 '19 at 5:51
• You are completely right that time will go backwards. Actually i asked this question before (physics.stackexchange.com/questions/487179/…) and to put it in the simple way, if you don't assume that time goes backward you will have to deal with some paradoxes. See "My problem" it's exactly like yours. – Paradoxy Jul 29 '19 at 10:13
The Minkowski $$t$$ coordinate tells us what time an inertial observer would infer for an event based on a signalling procedure such as Einstein synchronization. It doesn't have any deeper philosophical significance than that. You don't have to do special relativity using Minkowski coordinates. You can choose any coordinates you like, or you can do special relativity using coordinate-free methods. Coordinates are just names that we give to points.
For an observer who is accelerating rather than inertial, we can, if we wish, define a notion of simultaneity based on the observer's instantaneous state of motion. Based on this, there can be distant points for which simultaneity is either nonexistent or not unique. If you try to cover spacetime with a curvilinear set of coordinates that agrees with this notion of simultaneity, then the coordinates will become badly behaved at the boundary where existence and uniqueness fail. You can't extend those coordinates past that boundary, because changes of coordinates need to be one-to-one.
I have a discussion of this kind of thing in section 3.9.3 of my SR book, which is free online: http://www.lightandmatter.com/sr/ .
If I turn my body 90 degrees, can I see the city of Moscow relocate itself by thousands of miles? After all, a minute ago it was 1000 miles in front of me, and now it's 1000 miles to my left.
When you change your velocity, you change frames, and therefore assign new time coordinates to existing events --- just as I assigned new space coordinates to Moscow when I turned my body. That doesn't mean that anything about the events, or about their relationships with each other, is changing, any more than the city of Moscow is moving through space.
• But in the case of time, in the rocket ship, t'=0, t'=1, t'=2 is the history that the rocketship experiences. Before acceleration, there are certain events that have happened that, after acceleration, are now "behind" the rocketship's plane of simultaneity. So they would happen again. That means the rocketship crew would witness those events twice. – Shuheng Zheng Jul 29 '19 at 1:09
• Relativity only allows us to understand that other frames of reference may have time advancing forward at different rates, never backwards. – Adrian Howard Jul 29 '19 at 1:34
• @ShuhengZheng : When I turn 90 degrees, do I see Moscow in two different places? No, I see it in the same place, and I change my description of that place. Likewise, you see an event and assign a time to it; then you change your velocity and now you assign a new time to it. That doesn't mean you "see it twice"; it means you describe it two different ways. Which is perfectly fine. – WillO Jul 29 '19 at 2:15
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# Hartshorne's proof of Castelnuovo's theorem
For those of us who have forgotten, Castelnuovo's theorem is the following:
Theorem: If $$Y$$ is a curve on a surface $$X$$ with $$Y \simeq \mathbb{P}^1$$ and $$Y^2 = -1$$, then there is a morphism $$f: X \to X_0$$ to a smooth projective surface $$X_0$$ such that $$X$$ is the blow up of $$X$$ at some point and $$Y$$ is the exceptional divisor.
A proof of this theorem is given on page 414 of Hartshorne. To formulate my question, note that we will construct $$X_0$$ using the image of $$X$$ under a suitable map to projective space. In more detail, our aim is to show that the invertible sheaf $$\mathcal{M} : = \mathcal{L}(H + kY)$$ is semi-ample. Here, $$H$$ is some very ample divisor such that $$H^1(\mathcal{L}(H))=0$$ and $$k = H.Y$$ is assumed to be $$\geq 2$$.
Problem: Consider the following sequence of sheaves $$0 \longrightarrow \mathcal{L}(H + (i-1)Y) \xrightarrow{ \ \alpha \ } \mathcal{L}(H+iY) \xrightarrow{ \ \beta \ } \mathcal{O}_Y \otimes \mathcal{L}(H + i Y) \to 0.$$ Hartshorne claims this sequence is exact. In trying to verify this, I am finding that I am not sure of how $$\alpha$$ and $$\beta$$ are defined.
This exact sequence is just $$0\to \mathcal{I}_Y\cong\mathcal{O}_X(-Y)\to \mathcal{O}_X\to \mathcal{O}_Y\to 0$$ tensored with the line bundle $$\mathcal{L}(H+iY)$$.
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• Browse all
Measurement of the $B$ meson differential cross-section, $d \sigma / d p_T$, in $p\bar{p}$ collisions at $\sqrt{s} = 1.8$ TeV
The collaboration
Phys.Rev.Lett. 75 (1995) 1451-1455, 1995
Abstract (data abstract)
Fermilab-Tevatron. Measurement of the B meson differential cross section in pbar p c ollisions at sqrt(s) = 1.8 TeV. The data come from Run 1A during 1992-93 and have a total integrated luminosity of 19.3 +- 0.7 pb-1. Note that these data are now superseded by the full Run 1 data set published in Acosta et al. Fermilab-Pub-01-347-E.
• #### Table 1
Data from T 1
10.17182/hepdata.42432.v1/t1
Charged B meson cross section.
• #### Table 2
Data from T 1
10.17182/hepdata.42432.v1/t2
Average B meson cross section (including charged and neutral).
• #### Table 3
Data from P 1454
10.17182/hepdata.42432.v1/t3
Total integrated B meson cross section above 6 GeV.
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lilypond-user
[Top][All Lists]
Re: How to make a tweak a variable?
From: Xavier Scheuer Subject: Re: How to make a tweak a variable? Date: Sun, 18 Sep 2011 00:40:47 +0200
```On 17 September 2011 23:58, harm6 <address@hidden> wrote:
>
> Hi Xavier,
>
> you can't put \tweak in variable.
>
> Try:
>
> crescTweak = #(let ((m (make-music 'CrescendoEvent
> 'span-direction -1)))
> (set! (ly:music-property m 'tweaks)
> (acons 'to-barline #f
> (ly:music-property m 'tweaks)))
> m)
> {
> c1 \crescTweak | % it works
> d4\! % etc.
> }
Actually I do not want the grob to be "hardcoded" in the definition
of my variable.
I'd like to be able, for example, to define a variable
moveUp = -\tweak #'Y-offset #2
and to use it applied to different grobs:
c4\moveUp -"text"
c4\moveUp -\mordent
Is it really impossible?
Cheers,
Xavier
--
```
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# Cumulative vs. non-cumulative IRFs in R
I am using irf function from vars package. I am trying to derive cumulative IRFs.
The following code describes the case of deriving cumulative IRFs:
irf(vecm.l, impulse = c("g","p","h","l","s"), response = "g", cumulative = TRUE,n.ahead = 20, ortho=TRUE)
I got the output and plotted it, it looked like cumulative values of estimated MA coefficients. But then I ran the code with cumulative switched to FALSE as below:
irf(vecm.l, impulse = c("g","p","h","l","s"), response = "g", cumulative = FALSE,n.ahead = 20, ortho=TRUE)
Output from these two codes is identical. Including cumulative = TRUEin the irf function does not produce the cumulative responses??
Please note that I converted a vecm model to var using vec2var, hence the input model is called vecm.l here. Though, I doubt it plays any significance in deriving cumulative IRFs.
I would appreciate your comment, am I specifying the function incorrectly? Thanks,
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# Correlation amplitude and continuous spectra
The correlation amplitude is defined in Modern Quantum Mechanics by JJ Sakurai, chapter 2 as
$$C(t)=\langle\alpha|\alpha,t_0=0;t\rangle=\langle\alpha|\mathcal{U}(t,0)|\alpha\rangle \, .$$
In Eq. (2.1.65), for an initial state $|\alpha\rangle$ represented by a superposition of $|a'\rangle$, i.e, $\sum_{a'} c_{a'}|a'\rangle$ the correlation amplitude is $$C(t)=\sum_{a'}|c_{a'}|^2 \exp\left(\frac{-iE_{a'}t}{\hbar}\right) \, .$$
Consider a state ket which is a superposition of many energy eigenkets with similar energies, i.e, a quasi-continuous spectrum. Here we replace $$\sum_{a'}\rightarrow\int dE\rho(E) \\ c_{a'} \rightarrow g(E) \Bigg|_{E\approx E_0}$$ where $\rho(E)$ characterizes the density of energy eigenstates. The correlation amplitude becomes $$C(t)=\int dE|g(E)|^2\rho(E) \exp\left(\frac{-iEt}{\hbar}\right)$$ and subjected to the normalization condition $$\int dE|g(E)|^{2}\rho(E)=1 \, .$$
My understanding:
For a continuous spectrum $$\sum_{a'}|a'\rangle\langle{a'}|\alpha\rangle=\sum_{a'}\mathcal{C}_{a'}|a'\rangle\rightarrow \int d\mathcal{E'}|\mathcal{E'}\rangle\langle\mathcal{E'}|\alpha\rangle=\int d\mathcal{E'}\mathcal{C}_{\mathcal{E'}}|\mathcal{E'}\rangle$$
$$\sum_{a'}|\langle{a'}|\alpha\rangle|^{2}=1 \rightarrow \int d\mathcal{E'}|\langle{\mathcal{E'}}|\alpha\rangle|^2=1 \, .$$
Doubt:
Why do we define the correlation amplitude $C(t)$ using $\rho(E)$ and $g(E)$ instead of $\mathcal{C}(E)$?
What is the physical meaning of this expression and of $\rho(E)$ and $g(E)$?
• i'd appreciate if someone could comment onto it. – Sooraj S Feb 22 '16 at 7:05
• $g(E)$ seems to be just your $C(E)$. As for $\rho(E)$, it accounts for the degeneracy of the continuous spectrum. For instance, if the states would be indexed by the 3-d momentum ${\bf p}$, the integral would be simply $\int{d^3\{bf p}$ since each vector ${\bf p}$ labels a single state and the "spectrum" is non-degenerate. But if the states are indexed by the energy $E={\bf p}^2/2m$, then for each $E$ there is a continuum of states corresponding to the momentum-space sphere $|{\bf p}| = \sqrt{2mE}$. – udrv Feb 23 '16 at 5:20
Correlation amplitude determines to what degree the two state vectors representing the same physical system are similar. By same physical system I mean, in general, two different states of the same system. For example if you consider the time evolution operator. This operator tells what will be the state of the system in your hand after a certain time, provided you know it's state at a any initial time. Once you know the state ket of a system at any instant then using the time evolution operator you could find the state ket at any instant.
In general, since the state ket is a function of both position and time (i.e, allowed to vary with both space and time co-ordinates), the state ket changes after a certain time. So while operating with the time evolution operator you just changed the state of the system (note that it may remain unchanged also).
The correlation amplitude tells you how much this latter system resembles the former. If it's value is one it means that the system has no change. You will get the exact state ket from which you started. In general since we expect some change (i.e, an increase in the degree of mismatching between the two states of a physical system), it's value will be less than one. I.e, it's value changes from 1 (corresponding to the same initial state) to some value less than 1. If the amplitude value is closer to one, then the degree of resemblence between the two states is too good. Now let's see what this $g(E)$ means? If you refer to the text book (Modern Quantum Mechanics by J.J.Sakurai) there the author states that the state ket is assumed to be a linear combination of so many eigen kets with "similar" (which means they are close but not equal) energy eigen values. So it could represent a quasi-continuos spectrum. Hence we replace the summation by integral (changing from discrete to continuous). Next, since we are dealing with the summation of linear combination of eigen vectors with similar energy values we place inside the integral the energy density $ρ(E)$ at different points represented by the eigen vectors with the variable $dE$. Now the complex number Ca′ should be replaced by $g(E)$. $g(E)$ represents the degeneracy (or degenerative multiplicity). Degeneracy of an eigenvalue correspond to how many times the eigenvalues repeated. For example when I solve the equation $X|a\rangle=a|a\rangle$, if I get the eigen values as $a=-1,2,-1$, it means the value $a=-1$ has a degeneracy of 2. The eigen vectors corresponding to them will be identical. The reason for using degeneracy here is because we postulated that the energy eigenvalues are similar. I hope you understand.
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How would you write this out?
• February 11th 2008, 08:45 PM
Girlaaaaaaaa
How would you write this out?
A company produces very unusual CD's for which the variable cost is $8 per CD and the fixed costs are$50000. They will sell the CD's for $83 each. Let x be the number of CD's produced. Write the total cost as a function of the number of CD's produced.$
Write the total revenue as a function of the number of CD's produced.
$Write the total profit as a function of the number of CD's produced.$
Also how would you write this out Given f(x)=|x| , after performing the following transformations: shift to the left 36 units, shrink vertically by a factor of 1/91, and shift downward 49 units, the new function
Use abs(x) for . I wrote it out as 1/91(x+36)^2-49 but it said its incorrect. So what would be the right way
• February 11th 2008, 09:02 PM
earboth
Quote:
Originally Posted by Girlaaaaaaaa
...
Also how would you write this out Given f(x)=|x| , after performing the following transformations: shift to the left 36 units, shrink vertically by a factor of 1/91, and shift downward 49 units, the new function
Use abs(x) for . I wrote it out as 1/91(x+36)^2-49 but it said its incorrect. So what would be the right way
You are not dealing with squares but with absolute values. Therefore the term reads:
$\frac1{91} \cdot |x+36| - 49$ or: (1/91)*abs(x+36) - 49
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# Tag Info
5
To identify the Noether 4-current with the electric 4-current, one would in principle have to show that the Noether 4-current indeed appears as the source term in Maxwell's equations. The Maxwell equations with sources (Gauss's + Ampere's laws) are derived by adding the Maxwell Lagrangian $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ to a minimally coupled, gauge-...
2
The 3D spherically symmetric harmonic oscillator $$H~=~\frac{p_x^2+p_y^2+p_z^2}{2m}+ \alpha (x^2+y^2+z^2) ~=~ H_x + H_y + H_z$$ is a separable, Liouville integrable, and in fact a maximally superintegrable system with additionally integrals of motion $H_x$, $H_y$, $H_z$, $L_x$, $L_y$ and $L_z$, i.e. nothing exotic like the Laplace-Runge-Lenz vector. (A 3D ...
2
There are at least 2 lessons to be learned from OP's set-up: Noether's theorem is not necessarily about strict symmetry of the action. It is enough if the action has an (infinitesimal) quasi-symmetry, i.e. symmetry up to boundary terms. There's no free lunch. To prove energy conservation, one must use a non-trivial assumption: In this case, that the ...
1
So the other answers have given a solid response but it is a little high-level, I wanted to give a more verbose explanation of what has happened. In your procedure you have an action integral $S = \int_T\mathrm dt~L(q(t),\dot q(t), t)$ over some time domain $T$. You now want to vary the time coordinate. This means that the domain $T$ is changing, so that ...
1
Let's first clarify which formulation of Noether's theorem we'll be using: The Lagrangian will be a function $$L = L(x,v,t)$$ and the action a functional $$S[q] = \int_{t_1}^{t_2} L(q(t),\dot q(t),t) \,dt$$ Proposition. If the transformation $$t\to t'(t) = t + \epsilon T(t)$$ $$x\to x'(x,t) = x + \epsilon X(t)$$ $$q'(t') = q(t(t')) + \epsilon X(t(t'... 1 Edit: The Noether Procedure instructs you to take \varepsilon = \varepsilon(t) to be time dependent.$$ \delta q = \varepsilon(t) \dot q \hspace{1 cm} \delta \dot q = \dot \varepsilon(t) \dot q + \varepsilon(t) \ddot q \begin{align*} \delta L &= (\varepsilon \dot q) \frac{\partial L}{\partial q} + (\dot \varepsilon \dot q + \varepsilon \ddot q) \... 1 Comments & hints: The invariance \delta L=0 of the Lagrangian (1) under the infinitesimal transformation\delta q^a~=~\epsilon v^a(q), \qquad \delta t~=~0,$$follows directly from the Killing equation (2). According to Noether's theorem, the corresponding conserved Noether charge Q=p_av^a is the momentum$$p_a~=~\frac{\partial L}{\partial \dot{q}^...
1
The momentum you call $\Pi$ is the momentum in a certain spatial direction. The momentum you call $\pi$ is the momentum in a direction in field space. For example, consider vertical waves on a horizontal string. The "field" $y(x, t)$ represents the vertical displacement of the string at point $x$. Then $\Pi$ is the ordinary horizontal momentum, since waves ...
1
The conjugate momentum of the electromagnetic field $A_\mu$ is the electric field $\vec E$.
Only top voted, non community-wiki answers of a minimum length are eligible
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Article
# Solitons and Precision Neutrino Mass Spectroscopy
(Impact Factor: 6.13). 01/2011; 699(1). DOI: 10.1016/j.physletb.2011.03.058
Source: arXiv
ABSTRACT
We propose how to implement precision neutrino mass spectroscopy using
radiative neutrino pair emission (RNPE) from a macro-coherent decay of a new
form of target state, a large number of activated atoms interacting with static
condensate field. This method makes it possible to measure still undetermined
parameters of the neutrino mass matrix, two CP violating Majorana phases, the
unknown mixing angle and the smallest neutrino mass which could be of order a
few meV, determining at the same time the Majorana or Dirac nature of masses.
The twin process of paired superradiance (PSR) is also discussed.
0 Followers
·
• Source
##### Article: Dynamics of two-photon paired superradiance
[Hide abstract]
ABSTRACT: We develop for dipole-forbidden transition a dynamical theory of two-photon paired superradiance, or PSR for short. This is a cooperative process characterized by two photons back to back emitted with equal energies. By irradiation of trigger laser from two target ends, with its frequency tuned at the half energy between two levels, a macroscopically coherent state of medium and fields dynamically emerges as time evolves and large signal of amplified output occurs with a time delay. The basic semi-classical equations in 1+1 spacetime dimensions are derived for the field plus medium system to describe the spacetime evolution of the entire system, and numerically solved to demonstrate existence of both explosive and weak PSR phenomena in the presence of relaxation terms. The explosive PSR event terminates accompanying a sudden release of most energy stored in the target. Our numerical simulations are performed using a vibrational transition $X^1\Sigma_g^+ v=1 \rightarrow 0$ of para-H$_2$ molecule, and taking many different excited atom number densities and different initial coherences between the metastable and the ground states. In an example of number density close to $O[10^{21}]$cm$^{-3}$ and of high initial coherence, the explosive event terminates at several nano seconds after the trigger irradiation, when the phase relaxation time of $> O[10]$ ns is taken. After PSR events the system is expected to follow a steady state solution which is obtained by analytic means, and is made of many objects of field condensates endowed with a topological stability.
Physical Review A 03/2012; 86(1). DOI:10.1103/PhysRevA.86.013812 · 2.81 Impact Factor
• Source
##### Article: Observables in Neutrino Mass Spectroscopy Using Atoms
[Hide abstract]
ABSTRACT: The process of collective de-excitation of atoms in a metastable level into emission mode of a single photon plus a neutrino pair, called radiative emission of neutrino pair (RENP), is sensitive to the absolute neutrino mass scale, to the neutrino mass hierarchy and to the nature (Dirac or Majorana) of massive neutrinos. We investigate how the indicated neutrino mass and mixing observables can be determined from the measurement of the corresponding continuous photon spectrum taking the example of a transition between specific levels of the Yb atom. The possibility of determining the nature of massive neutrinos and, if neutrinos are Majorana fermions, of obtaining information about the Majorana phases in the neutrino mixing matrix, is analyzed in the cases of normal hierarchical, inverted hierarchical and quasi-degenerate types of neutrino mass spectrum. We find, in particular, that the sensitivity to the nature of massive neutrinos depends critically on the atomic level energy difference relevant in the RENP.
Physics Letters B 09/2012; 719(s 1–3). DOI:10.1016/j.physletb.2013.01.015 · 6.13 Impact Factor
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##### Article: Neutrino Spectroscopy with Atoms and Molecules
[Hide abstract]
ABSTRACT: We give a comprehensive account of our proposed experimental method of using atoms or molecules in order to measure parameters of neutrinos still undetermined; the absolute mass scale, the mass hierarchy pattern (normal or inverted), the neutrino mass type (Majorana or Dirac), and the CP violating phases including Majorana phases. There are advantages of atomic targets, due to the closeness of available atomic energies to anticipated neutrino masses, over nuclear target experiments. Disadvantage of using atomic targets, the smallness of rates, is overcome by the macro-coherent amplification mechanism. The atomic or molecular process we use is a cooperative deexcitation of a collective body of atoms in a metastable level |e> emitting a neutrino pair and a photon; |e> -> |g> + gamma + nu_i nu_j where nu_i's are neutrino mass eigenstates. The macro-coherence is developed by trigger laser irradiation. We discuss aspects of the macro-coherence development by setting up the master equation for the target quantum state and propagating electric field. With a choice of heavy target atom or molecule such as Xe or I_2 that has a large M1 x E1 matrix element between |e> and |g>, we show that one can determine three neutrino masses along with the mass hierarchy pattern by measuring the photon spectral shape. If one uses a target of available energy of a fraction of 1 eV, Majorana CP phases may be determined. Our master equation, when applied to E1 x E1 transition such as pH_2 vibrational transition Xv=1 -> 0, can describe explosive PSR events in which most of the energy stored in |e> is released within a few nanoseconds. The present paper is intended to be self-contained explaining some details related theoretical works in the past, and further reports new simulations and our ongoing experimental efforts of the project to realize the neutrino mass spectroscopy using atoms/molecules.
Progress of Theoretical and Experimental Physics 11/2012; 2012(1). DOI:10.1093/ptep/pts066 · 2.49 Impact Factor
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# Different Ways of Computing a Limit: Index
I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on different ways of computing the limit
$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$
Part 1: Algebra
Part 2: L’Hopital’s Rule
Part 3: Trigonometric substitution
Part 4: Geometry
Part 5: Geometry again
# Different Ways of Solving a Contest Problem: Index
I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on different ways of solving the contest problem “If $3 \sin \theta = \cos \theta$, what is $\sin \theta \cos \theta$?”
Part 1: Drawing the angle $\theta$
Part 2: A first attempt using a Pythagorean identity.
Part 3: A second attempt using a Pythagorean identity and the original hypothesis for $\theta$.
# Different ways of computing a limit (Part 3)
One of my colleagues placed the following problem on an exam for his Calculus II course…
$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}$
and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.
Method #3. A trigonometric identity. When we see $\sqrt{x^2+1}$ inside of an integral, one kneejerk reaction is to try the trigonometric substitution $x = \tan \theta$. So let’s use this here. Also, since $x \to \infty$, we can change the limit to be $\theta \to \pi/2$:
$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{\theta \to \pi/2} \frac{\sqrt{\tan^2 \theta+1}}{\tan \theta}$
$= \displaystyle \lim_{\theta \to \pi/2} \frac{\sqrt{\sec^2 \theta}}{\tan \theta}$
$= \displaystyle \lim_{\theta \to \pi/2} \frac{ \sec \theta}{\tan \theta}$
$= \displaystyle \lim_{\theta \to \pi/2} \frac{ ~~\displaystyle \frac{1}{\cos \theta} ~~}{ ~~ \displaystyle \frac{\sin \theta}{\cos \theta} ~~ }$
$= \displaystyle \lim_{\theta \to \pi/2} \frac{ 1}{\sin \theta}$
$= 1$.
# Different ways of solving a contest problem (Part 3)
The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:
If $3 \sin \theta = \cos \theta$, what is $\sin \theta \cos \theta$?
When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.
Yesterday, I presented a solution using a Pythagorean identity, but I was unable to be certain if the final answer was a positive or negative without drawing a picture. Here’s a third solution that also use a Pythagorean trig identity but avoids this difficulty. Again, I begin by squaring both sides.
$9 \sin^2 \theta = \cos^2 \theta$
$9 (1 - \cos^2 \theta) = \cos^2 \theta$
$9 - 9 \cos^2 \theta = \cos^2 \theta$
$9 = 10 \cos^2 \theta$
$\displaystyle \frac{9}{10} = \cos^2 \theta$
$\displaystyle \pm \frac{3}{\sqrt{10}} = \cos \theta$
Yesterday, I used the Pythagorean identity again to find $\sin \theta$. Today, I’ll instead plug back into the original equation $3 \sin \theta = \cos \theta$:
$3 \sin \theta = \cos \theta$
$3 \sin \theta = \displaystyle \frac{3}{\sqrt{10}}$
$\sin \theta = \displaystyle \pm \frac{1}{\sqrt{10}}$
Unlike the example yesterday, the signs of $\sin \theta$ and $\cos \theta$ must agree. That is, if $\cos \theta = \displaystyle \frac{3}{\sqrt{10}}$, then $\sin \theta = \displaystyle \frac{1}{\sqrt{10}}$ must also be positive. On the other hand, if $\cos \theta = \displaystyle -\frac{3}{\sqrt{10}}$, then $\sin \theta = \displaystyle -\frac{1}{\sqrt{10}}$ must also be negative.
If they’re both positive, then
$\sin \theta \cos \theta = \displaystyle \left( \frac{1}{\sqrt{10}} \right) \left( \frac{3}{\sqrt{10}} \right) =\displaystyle \frac{3}{10}$,
and if they’re both negative, then
$\sin \theta \cos \theta = \displaystyle \left( -\frac{1}{\sqrt{10}} \right) \left( -\frac{3}{\sqrt{10}} \right) = \displaystyle \frac{3}{10}$.
Either way, the answer must be $\displaystyle \frac{3}{10}$.
This is definitely superior to the solution provided in yesterday’s post, as there’s absolutely no doubt that the product $\sin \theta \cos \theta$ must be positive.
# Different ways of solving a contest problem (Part 2)
The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:
If $3 \sin \theta = \cos \theta$, what is $\sin \theta \cos \theta$?
When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.
Yesterday, I presented a solution using triangles. Here’s a second solution that I received: begin by squaring both sides and using a Pythagorean trig identity.
$9 \sin^2 \theta = \cos^2 \theta$
$9 (1 - \cos^2 \theta) = \cos^2 \theta$
$9 - 9 \cos^2 \theta = \cos^2 \theta$
$9 = 10 \cos^2 \theta$
$\displaystyle \frac{9}{10} = \cos^2 \theta$
$\displaystyle \pm \frac{3}{\sqrt{10}} = \cos \theta$
We use the Pythagorean identity again to find $\sin \theta$:
$\displaystyle \frac{9}{10} = \cos^2 \theta$
$\displaystyle \frac{9}{10} = 1 - \sin^2 \theta$
$\sin^2 \theta = \displaystyle \frac{1}{10}$
$\sin \theta = \displaystyle \pm \frac{1}{\sqrt{10}}$
Therefore, we know that
$\sin \theta \cos \theta = \displaystyle \left( \pm \frac{1}{\sqrt{10}} \right) \left( \pm \frac{3}{\sqrt{10}} \right) = \displaystyle \pm \displaystyle \frac{3}{10}$,
so the answer is either $\displaystyle \frac{3}{10}$ or $\displaystyle -\frac{3}{10}$. However, this was a multiple-choice contest problem and $\displaystyle -\frac{3}{10}$ was not listed as a possible answer, and so the answer must be $\displaystyle \frac{3}{10}$.
For a contest problem, the above logic makes perfect sense. However, the last step definitely plays to the fact that this was a multiple-choice problem, and the concluding step would not have been possible had $\displaystyle -\frac{3}{10}$ been given as an option.
# How I Impressed My Wife: Part 5b
Amazingly, the integral below has a simple solution:
$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$
Even more amazingly, the integral $Q$ ultimately does not depend on the parameter $a$. For several hours, I tried to figure out a way to demonstrate that $Q$ is independent of $a$, but I couldn’t figure out a way to do this without substantially simplifying the integral, but I’ve been unable to do so (at least so far).
So here’s what I have been able to develop to prove that $Q$ is independent of $a$ without directly computing the integral $Q$.
Earlier in this series, I showed that
$Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$
$= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$
$= 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}$
$= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$
$= \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{(a^2 + b^2) v^2 + b^2 }$
Yesterday, I showed used the substitution $w = (a^2 + b^2) v$ to show that $Q$ was independent of $a$. Today, I’ll use a different method to establish the same result. Let
$Q(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+b^2) dv}{(a^2+b^2)^2 v^2 + b^2 }$.
Notice that I’ve written this integral as a function of the parameter $a$. I will demonstrate that $Q'(a) = 0$, so that $Q(c)$ is a constant with respect to $a$. In other words, $Q(a)$ does not depend on $a$.
To do this, I differentiate under the integral sign with respect to $a$ (as opposed to $x$) using the Quotient Rule:
$Q'(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{ 2a \left[ (a^2+b^2)^2 v^2 + b^2\right] - 2 (a^2+b^2) \cdot (a^2+b^2) v^2 \cdot 2a }{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$
$Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{(a^2+b^2)^2 v^2 + b^2- 2 (a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$
$Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{b^2-(a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$
I now apply the trigonometric substitution $v = \displaystyle \frac{b}{a^2+b^2} \tan \theta$, so that
$(a^2+b^2)^2 v^2 = (a^2+b^2)^2 \displaystyle \left[ \frac{b}{a^2+b^2} \tan \theta \right]^2 = b^2 \tan^2 \theta$
and
$dv = \displaystyle \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta$
The endpoints of integration change from $-\infty < v < \infty$ to $-\pi/2 < \theta < \pi/2$, and so
$Q'(a) = \displaystyle 4a \int_{-\pi/2}^{\pi/2} \frac{b^2- b^2 \tan^2 \theta}{\left[ b^2 \tan^2 \theta + b^2 \right]^2} \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta$
$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta] \sec^2 \theta}{\left[ \tan^2 \theta +1 \right]^2} d\theta$
$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\left[ \sec^2 \theta \right]^2} d\theta$
$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\sec^4 \theta} d\theta$
$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta]}{\sec^2 \theta} d\theta$
$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [1- \tan^2 \theta] \cos^2 \theta \, d\theta$
$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [\cos^2 \theta -\sin^2 \theta] d\theta$
$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \cos 2\theta \, d\theta$
$= \displaystyle \left[ \frac{2ab^3}{a^2+b^2} \sin 2\theta \right]^{\pi/2}_{-\pi/2}$
$= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ \sin \pi - \sin (-\pi) \right]$
$= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ 0- 0 \right]$
$= 0$.
I’m not completely thrilled with this demonstration that $Q$ is independent of $a$, mostly because I had to do so much simplification of the integral $Q$ to get this result. As I mentioned in yesterday’s post, I’d love to figure out a way to directly start with
$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$
and demonstrate that $Q$ is independent of $a$, perhaps by differentiating $Q$ with respect to $a$ and demonstrating that the resulting integral must be equal to 0. However, despite several hours of trying, I’ve not been able to establish this result without simplifying $Q$ first.
# How I Impressed My Wife: Part 5a
Amazingly, the integral below has a simple solution:
$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$
Even more amazingly, the integral $Q$ ultimately does not depend on the parameter $a$. For several hours, I tried to figure out a way to demonstrate that $Q$ is independent of $a$, but I couldn’t figure out a way to do this without substantially simplifying the integral, but I’ve been unable to do so (at least so far).
So here’s what I have been able to develop to prove that $Q$ is independent of $a$ without directly computing the integral $Q$.
Earlier in this series, I showed that
$Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$
$= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$
$= 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}$
$= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$
I now multiply the top and bottom of this last integral by $a^2 + b^2$:
$Q = \displaystyle \frac{2}{(a^2+b^2)^2} \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$
$= \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{(a^2 + b^2) v^2 + b^2 }$
I now employ the substitution $w = (a^2 + b^2) v$, so that $dw = (a^2 + b^2) v$. Since $a^2 + b^2 > 0$, the endpoints of integration do not change, and so
$Q = \displaystyle 2 \int_{-\infty}^{\infty} \frac{dw}{w^2 + b^2 }$.
This final integral is independent of $a$.
Since $Q$ is independent of $a$, I can substitute any convenient value of $a$ that I wish. For example, I can let $a = 0$ without altering the value of $Q$:
$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x} = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$
This provides a considerable simplification of the integral $Q$ which also opens up additional methods of evaluation.
# How I Impressed My Wife: Part 3f
Previously in this series, I showed that
$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$
My wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by $\sec^2 x$ and then employing the substitution $u = \tan x$ (after using trig identities to adjust the limits of integration).
But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.
So far, I have shown that
$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$
$= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}$
$= 2 \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}$
$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)}$
$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\phi}{S + R \cos \phi}$
$= 2 \displaystyle \int_{-\pi}^{\pi} \frac{d\phi}{S + R \cos \phi}$
where $R = \sqrt{(2a)^2 + (1-a^2-b^2)^2}$ and $S = 1 + a^2 + b^2$ (and $\alpha$ is a certain angle that is now irrelevant at this point in the calculation).
There are actually a couple of ways for computing this last integral. Today, I’ll lay the foundation for the “magic substitution”
$u = \tan \displaystyle \frac{\phi}{2}$
With this substitution, the above integral will become a rational function, which can then be found using standard techniques.
First, we use some trig identities to rewrite $\cos 2x$ in terms of $\tan x$:
$\cos 2x = 2\cos^2 x - 1$
$= \displaystyle \frac{ \sec^2 x (2 \cos^2 x - 1)}{\sec^2 x}$
$= \displaystyle \frac{ 2 - \sec^2 x)}{\sec^2 x}$
$= \displaystyle \frac{ 2 - [ 1 + \tan^2 x])}{1 + \tan^2 x}$
$= \displaystyle \frac{1- \tan^2 x}{1 + \tan^2 x}$
Next, I’ll replace $x$ by $\phi/2$:
$\cos \phi = \displaystyle \frac{1- \tan^2 (\phi/2)}{1 + \tan^2 (\phi/2)} = \displaystyle \frac{1-u^2}{1+u^2}$.
Second, for the sake of completeness (even though it isn’t necessary for this particular integral), I’ll rewrite $\sin 2x$ in terms of $\tan x$:
$\sin 2x = 2\sin x \cos x$
$= \displaystyle \frac{2\sin x \cos x \sec^2 x}{\sec^2 x}$
$= \displaystyle \frac{ ~ \displaystyle \frac{2 \sin x}{\cos x} ~ }{\sec^2 x}$
$= \displaystyle \frac{ 2 \tan x }{\sec^2 x}$
$= \displaystyle \frac{ 2 \tan x }{1 + \tan^2 x}$
$= \displaystyle \frac{2 \tan x}{1 + \tan^2 x}$
Next, I’ll replace $x$ by $\phi/2$:
$\sin \phi = \displaystyle \frac{2 \tan^2 (\phi/2)}{1 + \tan^2 (\phi/2)} = \displaystyle \frac{2u}{1+u^2}$.
Third, again for the sake of completeness,
$\tan \phi = \displaystyle \frac{\sin u}{\cos u} = \displaystyle \frac{ ~ \displaystyle \frac{2u}{1+u^2} ~ }{ ~ \displaystyle \frac{1-u^2}{1+u^2} ~ } = \displaystyle \frac{2u}{1-u^2}$.
Finally, I need to worry about what happens to the $d\phi$:
$u = \tan \displaystyle \frac{\phi}{2}$
$du = \displaystyle \frac{1}{2} \sec^2 \displaystyle \frac{\phi}{2} \, d\phi$
$du = \displaystyle \frac{1}{2} \left[ 1 + \tan^2 \displaystyle \frac{\phi}{2} \right] d\phi$
$du = \displaystyle \frac{1}{2} (1+u^2) d\phi$
$\displaystyle \frac{2 du}{1+u^2} = d\phi$
These four substitutions can be used to convert trigonometric integrals into some other integral. Usually, the new integrand is pretty messy, and so these substitutions should only be used sparingly, as a last resort.
I’ll continue this different method of evaluating this integral in tomorrow’s post.
# How I Impressed My Wife: Part 2f
Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.
That will never happen in the Quintanilla household in a million years.
But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:
$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$
In this series, I’ll explore different ways of evaluating this integral.So far in this series, I’ve shown that
$Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$
$= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$
$= 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}$
$= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$
$= \displaystyle \frac{2}{|b|} \int_{-\infty}^{\infty} \frac{ dw }{w^2 +1}$
This last integral can be evaluated using a standard trick. Let $\theta = \tan^{-1} w$, so that $w = \tan \theta$. We differentiate this last equation with respect to $w$:
$\displaystyle \frac{dw}{dw} = \sec^2 \theta \cdot \displaystyle \frac{d\theta}{dw}$
Employing a Pythagorean identity, we have
$1 = (1+ \tan^2 \theta) \cdot \displaystyle \frac{d\theta}{dw}$
Since $w = \tan \theta$, we may rewrite this as
$1 = (1+ w^2) \cdot \displaystyle \frac{d\theta}{dw}$
$\displaystyle \frac{1}{1+w^2} = \displaystyle \frac{d\theta}{dw}$
$\displaystyle \frac{1}{1+w^2} = \displaystyle \frac{d}{dw} \tan^{-1} w$
Integrating both sides with respect to $w$, we obtain the antiderivative
$\displaystyle \int \frac{1}{1+w^2} = \tan^{-1} w + C$
We now employ this antiderivative to evaluate $Q$:
$Q = \displaystyle \frac{2}{|b|} \int_{-\infty}^{\infty} \frac{ dw }{w^2 +1}$
$= \displaystyle \frac{2}{|b|} \displaystyle \left[ \tan^{-1} w \right]^{\infty}_{-\infty}$
$= \displaystyle \frac{2}{|b|} \displaystyle \left[ \displaystyle \frac{\pi}{2} - \frac{-\pi}{2} \right]$
$= \displaystyle \frac{2\pi}{|b|}$
And so, at long last, we have arrived at the solution for the integral $Q$. Surprisingly, the answer is independent of the parameter $a$.
These last few posts illustrated the technique that I used to compute this integral for my wife in support of her recent paper in Physical Review A. However, I had more than a few false starts along the way… or, at the time, I thought they were false starts. It turns out that there are multiple ways of evaluating this integral, and I’ll explore another method of attack beginning with tomorrow’s post.
# Engaging students: Verifying trigonometric identities
In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.
I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).
This student submission comes from my former student Tracy Leeper. Her topic, from Precalculus: verifying trigonometric identities.
Many students when first learning about trigonometric identities want to move terms across the equal sign, since that is what they have been taught to do since algebra, however, in proving a trigonometric identity only one side of the equality is worked at a time. Therefore my idea for an activity to help students is to have them look at the identities as a puzzle that needs to be solved. I would provide them with a basic mat divided into two columns with an equal sign printed between the columns, and give them trig identities written out in a variety of forms, such as $\sin^2 \theta + \cos^2 \theta$ on one strip, and $1$ written on another strip. Other examples would also include having $\tan^2 \theta$ on one, and $\sin^2 \theta/\cos^2 \theta$ on another. The students will have to work within one column, and step by step, change one side to eventually reflect the term on the other side, and each strip has to be one possible representation of the same value. By providing the students with the equivalent strips, they will be able to construct the proof of the identity. I feel that giving them the strips will allow them to see different possibilities for how to manipulate the expression, without leaving them feeling lost in the process, and by dividing the mat into columns, they can focus on one side, and see that the equivalency is maintained throughout the proof. The students would need to arrange the strips into the correct order to prove the left hand side is equivalent to the right hand side, while reinforcing the process of not moving anything across the equal sign.
Trigonometry identities are used in most of the math courses after pre-calculus, as well as the idea of proving an equivalency. If the students learn the concept of proving an equivalency that will help them construct proofs for any future math courses, as well as learning to look at something given, and be able to see it as parts of a whole, or just be able to write it a different way to assist with the calculations. If students learn to see that
$1 = \sin^2 x + \cos^2 x = \sec^2 x - \tan^2 x = \csc^2 x - \cot^2 x$,
their ability to manipulate expressions will dramatically improve, and their confidence in their ability will increase, as well as their understanding of the complexities and relations throughout all of mathematics. The trigonometric identities are the fundamental part of the relationships between the trig functions. These are used in science as well, anytime a concept is taught about a wave pattern. Sound waves, light waves, every kind of wave discussed in science are sinusoidal wave. Anytime motion is calculated, trigonometry is brought into the calculations. All students who wish to progress in the study of science or math need to learn basic trigonometric identities and learn how to prove equivalency for the identities. Since proving trigonometric identities is also a practice in logical reasoning, it will also help students learn to think critically, and learn to defend their conjectures, which is a valuable skill no matter what discipline the student pursues.
For learning how to verify trigonometric identities, I like the Professor Rob Bob (Mr. Tarroy’s) videos found on youtube. He’s very energetic, and very thorough in explaining what needs to be done for each identity. He also gives examples for all of the different types of identities that are used. He is very specific about using the proper terms, and he makes sure to point out multiple times that this is an identity, not an equation, so terms cannot be transferred across the equal sign. He also presents options to use for a variety of cases, and that sometimes things don’t work out, but it’s okay, because you can just erase it and start again. I also like that he uses different colored chalk to show the changes that are being made. He is very articulate, and explains things very well, and makes sure to point out that he is providing examples, but it’s important to remember that there are many different ways to prove the identity presented. I enjoyed watching him teach, and I think the students would enjoy his energy as well.
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# Eliminate reference to chapter in table caption (memoir class)
I'm editing a book using the memoir class. Each chapter has a different author, and I restart numbering of all elements at the start of each chapter. So far, so good. However, I want to eliminate the chapter number that occurs in front of tables and figures (e.g. Table 1, not Table 0.1). I put this into my preamble, but it does only have an effect on footnotes and section titels.
\counterwithout{footnote}{chapter}
\counterwithout{section}{chapter}
\counterwithout{table}{chapter}
\counterwithout{figure}{chapter}
I suspected the fact that I use the caption-package with the memoir class was what caused the problem. Removing this did however not help.
The code runs like this:
\begin{table}[h!]
\captionof{table}{\emph{some caption}}
\begin{tabular}{lrr}
\hline & \textbf{column A} & \textbf{column B} \\
\hline \textbf{Alpha} & a & b \\
\textbf{Beta} & a & b \\
\textbf{Gamma} & a & b \\
\hline \textbf{Total} & 100 & 100 \\
\hline \label{fig1}
\end{tabular}
\end{table}
I've been reading he memoir manual and poking around the web all day, but I haven't got any closer to solving this problem.
Grateful for any help.
UPDATE:
This is my main document. I didn't want to include it, because it is loooooong.
\documentclass [paperwidth=170mm, paperheight=240mm, 11 pt]{memoir}
\usepackage[cyr]{aeguill}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[francais, english]{babel}
\usepackage{multirow}
\usepackage{multicol}
\usepackage{tablefootnote}
\usepackage[dvipsnames,svgnames,table]{xcolor}
\usepackage{graphicx}
\usepackage{wrapfig}
\usepackage{linguex}
\usepackage{epstopdf}
\usepackage{amsmath}
\usepackage{phonetic}
\usepackage{xyling}
\usepackage{titling}
\usepackage{lipsum}
\usepackage{titlesec}
\usepackage{enumitem}
\usepackage{textcomp}
\usepackage{chngcntr}
\usepackage{setspace}
\usepackage{etoolbox}
\usepackage{makeidx}
\usepackage{setspace}
\usepackage{multirow}
\usepackage{caption}
\usepackage[figurename=Fig.]{caption}
\usepackage[tablename=Tab.]{caption}
\usepackage{colortbl}
\usepackage[dvipsnames,svgnames,table]{xcolor}
\renewcommand{\arraystretch}{1}
\makeatletter
\preto{\@ex}{\topsep=0pt \parskip=0pt \parsep=0pt \partopsep=0pt }
\makeatother
\titleformat{\section}{\normalsize \bfseries}{\thesection}{1.75em}{\textsc}
\titleformat{\subsection}{\normalsize \bfseries}{\thesubsection}{1em}{\emph}
\titleformat{\subsubsection}{\normalsize \mdseries}{\thesubsubsection}{1em}{\emph}
\renewcommand{\cftchapterfont}{\normalfont}
\renewcommand{\cftpartfont}{\bfseries}
\renewcommand{\cftchapterpagefont}{\normalfont}
\renewcommand{\captionfont}{\small}
\renewcommand{\firstrefdash}{}
\setlength{\parsep}{0pt}
\setsecnumdepth{subsection}
\makeatletter
\let\@afterindenttrue\@afterindentfalse
\makeatother
\setcounter{secnumdepth}{3}
\setcounter{tocdepth}{0}
\renewcommand{\baselinestretch}{1}
\counterwithout{footnote}{chapter}
\counterwithout{section}{chapter}
\counterwithout{table}{chapter}
\counterwithout{figure}{chapter}
\begin{document}
\selectlanguage{francais}
\frontmatter
\selectlanguage{francais}
\tableofcontents*
\include{introduction}
\newpage
\mainmatter
\part{La langue française}
\include{somepaper}
\backmatter
\include{index}
\end{document}
The problems I'm experiencing is in the {somepaper} under \mainmatter. (Actually, in all papers that contain tables or figures.) I tried \stepcounter{table}, which removed the numbering all together.
• Are you using \mainmatter in your master document? Because that changes the numbering of things like this. As always a full minimal example helps a lot more than sniplets like this. – daleif Jan 24 '14 at 13:42
• That is not a minimal example. This has nothing to do with the contents of your table. We need something we can copy'n'paste and use as is without having to add anything. Here we would have to guess your preamble. – daleif Jan 24 '14 at 14:12
• You could replace \captionof with \stepcounter{table} \begin{center} \textbf{Table \thetable:} \emph{some text} \end{center} – John Kormylo Jan 24 '14 at 17:54
• That is hardly minimal, move the counterwithout after \mainmatter – daleif Jan 24 '14 at 22:31
\mainmatter does a few things
• resets secnumdepth to mxsecnumdepth
• reset the page counter to start from one and be normal numbers
• reset the figure and table counters into .
So you have two choices
1. move your \counterwithout stuff after \mainmatter, or
2. use this
\makeatletter
\renewcommand\@memmain@floats{}
\makeatother
This is the macro that resets the figure and table inside \mainmatter, so now it does nothing.
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