text:
# Math Help - root integral question..

1. ## root integral question..

$
\int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{dx}{x(\sqrt{x}+1+\sqrt{x}+\sqrt[3]{x})}=
\int \frac{dx}{x(\sqrt{x}+\frac{(1-x)}{1-\sqrt[3]{x}})}
$

i got read of one root but instead i got another one
??

2. Originally Posted by transgalactic
$
\int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{dx}{x(\sqrt{x}+1+\sqrt{x}+\sqrt[3]{x})}=
\int \frac{dx}{x(\sqrt{x}+\frac{(1-x)}{1-\sqrt[3]{x}})}
$

i got read of one root but instead i got another one
??
If you let $x = u^6$ then

$\int \frac{6\,du}{u(u+1)(2u^2-u+1)}$

3. when i substituted i didnt get your expression
i got
$
\int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{6u^5du}{u^6(1+2u^3+u^2)}
$

4. Originally Posted by transgalactic
when i substituted i didnt get your expression
i got
$
\int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{6u^5du}{u^6(1+2u^3+u^2)}
$
I'm sure you see the $u^5$ cancels and that if $f(u) = 2u^3+u^2+ 1$ then $f(-1) = 0$ so $u+1$ is a factor and

$2u^3+u^2+ 1 = (u+1)(2u^2-u+1)$

5. i cant solve this last part:
$
\frac{1}{3}\int\frac{-27u+2}{2u^2-u+1}
$

even when i split it:
and take t=2u^2 -u +1

??

6. Originally Posted by transgalactic
i cant solve this last part:
$
\frac{1}{3}\int\frac{-27u+2}{2u^2-u+1}
$

even when i split it:
and take t=2u^2 -u +1

??
You should get $
\int\frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1}
$

You have to split into 2 parts

$\int\frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1} = -\frac{3}{8}\int\frac{4u-1}{2u^2-u+1} - \frac{1}{8}\int\frac{1}{2u^2-u+1}$

The first part is integrable in log
The second part is integrable in Arctan after setting t=u-1/4 then $v = \frac{4t}{\sqrt{7}}$

7. i cant get your last part
how you got this coefficients??

2A+2B+c=0
A-B+c+d=0
B=-D

A=6

i got these equation
and i cant get your coefficients
??

8. $\frac{1}{u(1+2u^3+u^2)} = \frac{1}{u} - \frac{\frac{1}{4}}{u+1} + \frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1}$

I forgot the "6" as numerator ... sorry
Multiply everything above by 6

9. its still does give the last part as i got it

where did my coefficientcient equations differ yours??

10. ## did i get the coefficient equations ..

$
\frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}\\
$

$
A(u+1)(2u^2-u+1)-Bu(2u^2-u+1)+(Cu+D)(u^2+u)=1\\
$

$
u^3:\\
$

$
2A-2B+C=0\\
$

$
u^2:\\
$

$
A+B+C+D=0\\
u:\\
$

$
-B+D=0\\
$

1:
$
A=1
$

11. Originally Posted by transgalactic
$
\frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}\\
$

$
A(u+1)(2u^2-u+1)-Bu(2u^2-u+1)+(Cu+D)(u^2+u)=1\\
$

$
u^3:\\
$

$
2A-2B+C=0\\
$

$
u^2:\\
$

$
A+B+C+D=0\\
u:\\
$

$
-B+D=0\\
$

1:
$
A=1
$
The denominator factorises to $u(1+2u^3+u^2) = u(u+1)(2u^2 - u +1)$.

Thus, The fraction becomes $\frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} + \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}$.

You have a sign error. $B$ is $+$ve, and not $-$ve. Your formula produced are correct, just a sign error has made them slightly differ.

You can continue to use gaussian elimination after you've made the correction.

There is a much easier way :

$\frac{1}{u(1+2u^3+u^2)} = \frac{1}{u(u+1)(2u^2-u+1)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}$

Multiply both sides by u

$\frac{1}{(u+1)(2u^2-u+1)} = A - \frac{Bu}{u+1} + \frac{u(Cu+D)}{2u^2-u+1}$

And set u to 0
$1 = A$

Multiply both sides by u+1
$\frac{1}{u(2u^2-u+1)} = \frac{A(u+1)}{u} - B + \frac{(u+1)(Cu+D)}{2u^2-u+1}$

And set u to -1
$-\frac{1}{4} = - B$

13. i tried again

here is my full way till i got the coefficients
where did i go wrong
??

14. C=-9 is OK
then -3+2D=0 gives D=3/2 and not 2/3

15. $
\int \frac{-9u+\frac{3}{2}}{2u^2-u+1}
$

but then the derivative of the denominator doesnt give us the numenator
??


text:
# CHEFGAMES - Editorial

Author: Tejas Pandey
Testers: Utkarsh Gupta, Jatin Garg
Editorialist: Nishank Suresh

365

None

# PROBLEM:

Given the decisions of four referees as to whether a ball was in or out, decide whether the ball was indeed in or not. It is considered to be in if and only if all 4 referees say that it is in.

# EXPLANATION

Simply check if all 4 numbers given in the input are 0 or not. If they are, print “In”, otherwise print “Out”.

This check can be done in several ways, for example:

• Apply an if condition to each input integer
• Or, find the sum of all 4 numbers and check if this is zero

# TIME COMPLEXITY

\mathcal{O}(1) per test case.

# CODE:

Editorialist's code (Python)
for _ in range(int(input())):
print('In' if sum(map(int, input().split())) == 0 else 'Out')

format: plain


text:
Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp01h989r641w
 Title: Essays on Accountability and Representation Authors: Pomirchy, Michael Advisors: Canes-Wrone, Brandice Contributors: Politics Department Keywords: accountabilityCongresselectionsresponsiveness Subjects: Political science Issue Date: 2022 Publisher: Princeton, NJ : Princeton University Abstract: In this dissertation, I examine accountability mechanisms in elections and how they affect legislative behavior. First, I construct a theory of intra-party factions in which legislators in a continuum of districts choose to join a faction and cast a roll-call vote, before going up for reelection. This theory shows that factions can have positive welfare effects for voters, in that they signal incumbents' ideological alignment and empower voters to throw dissonant types out of office. In some cases, factions can also encourage more congruent legislative behavior too. Second, I measure the electoral consequences of faction membership by analyzing original survey data from the 2020 CCES. In a private module of 1,000 respondents, I gauge constituent perceptions of their representative's faction membership and relate these perceptions to electoral outcomes like vote choice and incumbent approval. In the context of the 2020 general election, I find that incumbents garner higher reelection rates and approval if they are affiliated with a faction that is ideologically congruent with their constituents. Moreover, these electoral rewards are larger among Republican incumbents and for incumbents affiliated with factions situated closer to the median of the party. Finally, I consider the impact of electoral incentives on legislative behavior in state legislatures. Using a natural experiment in four state legislatures -- Arkansas, Illinois, Florida, and Texas -- where term length was randomly assigned, I compare the responsiveness to constituency opinion of those randomly assigned a two-year term to those assigned a four-year term on different issue areas, like the economy, environment, and crime. I find no evidence for an electoral proximity effect. In addition, in the Illinois State Senate, I see similar results when examining the effect of term length on several individual roll-call votes, including the legalization of medical marijuana and gay marriage. URI: http://arks.princeton.edu/ark:/88435/dsp01h989r641w Alternate format: The Mudd Manuscript Library retains one bound copy of each dissertation. Search for these copies in the library's main catalog: catalog.princeton.edu Type of Material: Academic dissertations (Ph.D.) Language: en Appears in Collections: Politics

Files in This Item:


text:
Last edited by Kigakazahn
Monday, August 3, 2020 | History

2 edition of Spaces of functions and sets found in the catalog.

Spaces of functions and sets

R. B. Sher

# Spaces of functions and sets

## by R. B. Sher

Written in English

Subjects:
• Topology.,
• Set theory.

• Edition Notes

Classifications The Physical Object Statement by R.B. Sher. Series Monographs in undergraduate mathematics -- v.1 Contributions Guilford College., Conference on Undergraduate Mathematics (1975 : Guilford College) LC Classifications QA611 .S54 Pagination 40 p. : Number of Pages 40 Open Library OL14810565M

Chapter 6 Functions Injective, Surjective, Bijective Function. Composite and Inverse of Functions. Finite-State Machines. Automata and Their Semigroups. Chapter 7 Groups of Permutations Symmetric Groups. Dihedral Groups. An Application of Groups to Anthropology. Chapter 8 Permutations of a Finite Set Decomposition of Permutations into Cycles. Stack Exchange network consists of Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share .

Publisher Summary. This chapter describes the concept of the power of a set in mapping. The mapping f: X → Y is said to be one-to-one if,(x 1 ≠ x 2) ⇒ [f(x 1) ≠ f(x 2)] or, equivalently, if, [f(x 1) = f(x 2)] ⇒ (x 1 = x 2).The chapter presents a theorem that states that the inverse of a one-to-one mapping is one-to-one for f = rically, the transition to the inverse. Basic Point-Set Topology 3 means that f(x) is not in the other hand, x0 was in f −1(O) so f(x 0) is in O was assumed to be open, there is an interval (c,d) about f(x0) that is contained in points f(x) that are not in O are therefore not in (c,d) so they remain at least a ﬁxed positive distance from f(x0).To summarize: there are points.

This book makes a significant inroad into the unexpectedly difficult question of existence of Fréchet derivatives of Lipschitz maps of Banach spaces into higher dimensional spaces. Because the question turns out to be closely related to porous sets in Banach spaces, it provides a bridge between descriptive set theory and the classical topic of. a set Ain Rn is the minimal convex set containing A, denoted conv(A). Equivalently, conv(A) is the set of all convex combinations of points from A. Recall that a convex set Kin Rnis called symmetric if K= K(i.e. x2Kimplies x2K). A bounded convex set with nonempty interior is called a convex body. Let Xbe an n-dimensional Banach space.

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### Spaces of functions and sets by R. B. Sher Download PDF EPUB FB2

There are many diﬁerent kinds of function spaces, and there are usually several diﬁerent topologies that can be placed on a given set of functions. These notes describe three topologies that can be placed on the set of all functions from a setXto a spaceY: the product topology, the box topology, and the uniform Size: KB.

Set topology, the subject of the present volume, studies sets in topological spaces and topological vector spaces; whenever these sets are collections of n-tuples or classes of functions, the book recovers well-known results of classical by: A metric space is a set Xtogether with a metric don it, and we will use the notation (X;d) for a metric space.

Often, if the metric dis clear from context, we will simply denote the metric space (X;d) by Xitself. Example 1. The set of real numbers R with the function d(x;y) = jx yjis a metric space. More. Equicontinuity between metric spaces. Let X and Y be two metric spaces, and F a family of functions from X to shall denote by d the respective metrics of these spaces.

The family F is equicontinuous at a point x 0 ∈ X if for every ε > 0, there exists a δ > 0 such that d(ƒ(x 0), ƒ(x)). A metric space is a non-empty set equi pped with structure determined by a well-defin ed notion of distan ce.

The term ‘m etric’ i s d erived from the word metor (measur e). The essential uniqueness of density functions can fail if the positive measure space $$(S, \ms S, \mu)$$ is not $$\sigma$$-finite. A simple example is given below. Our next result answers the question of when a measure has a density function with respect to $$\mu$$, and is the fundamental theorem of this section.

7 Functions of bounded variation and absolutely continuous functions Measure Spaces Algebras and σ–algebras of sets Notation and preliminaries We shall denote by X a nonempty set, by P(X) the set of all parts (i.e., subsets) of X, and by ∅ the empty set.

Let Aand Bbe sets. A function ffrom Ato Bis to be thought of as a rule which assigns to every element aof the set Aan element f(a) of the set B.

The set Ais called the domain of f and Bthe codomain (or target) of f. We use the notation ‘f:A→ B’ (read ‘f, from Ato B’) to mean that f is a function with domain Aand codomain B. We can now discuss various vector spaces of functions.

First, we know from our previous work with measure spaces, that the set $$\mathscr{V}$$ of all measurable functions $$f: S \to \R$$ is a vector space under our standard (pointwise) definitions of sum and scalar multiple. The spaces we are studying in this section are subspaces.

Introduction To Mathematical Analysis John E. Hutchinson Revised by Richard J. Loy /6/7 Department of Mathematics School of Mathematical Sciences. abstract spaces such as vector spaces, Hilbert spaces, etc. It finds diverse applications in modern physics, especially in quantum mechanics." The S.

Banach treatise Theorie des Operationes Lineares, printed half a century ago, inaugurated functional analysis as.

METRIC AND TOPOLOGICAL SPACES 3 1. Introduction When we consider properties of a “reasonable” function, probably the ﬁrst thing that comes to mind is that it exhibits continuity: the behavior of the function at a certain point is similar to the behavior of the function in a small neighborhood of the point.

Lingadapted from UMass LingPartee lecture notes March 1, p. 3 Set Theory Predicate notation. Example: {x x is a natural number and x set of all x such that x is a natural number and is less than 8” So the second part of this notation is a prope rty the members of the set share (a condition.

Topological spaces form the broadest regime in which the notion of a continuous function makes sense. We can then formulate classical and basic theorems about continuous functions in a much broader framework.

For example, an important theorem in optimization is that any continuous function f: [a;b]!R achieves its minimum at least one point x2. If fis a function from a set Ainto a set B;this means that to every x2 A there corresponds a point f(x) 2 Band we write f: A.

(see Dudley™s book [D]). In measure theory, inevitably one encounters 1:For example the real is a measurable space and the function: M. [0;1] is a positive measure, (X;M;) is called a positive measure space.

We identify functions that di er on a set of measure zero. For p= 1, the space L1() is the space of essentially bounded Lebesgue measurable functions on with the essential supremum as the norm. The spaces Lp() are Banach spaces for 1 p 1. Example The Sobolev spaces, Wk;p, consist of functions whose derivatives satisfy an integrability.

space of square integrable functions. In order of logical simplicity, the space. comes ﬁrst since it occurs already in the description of functions integrable in the Lebesgue sense. Connected to it via duality is the.

L ∞ space of bounded functions, whose supremum norm carries over from the more familiar space of continuous functions. In mathematics, a metric space is a set together with a metric on the metric is a function that defines a concept of distance between any two members of the set, which are usually called metric satisfies a few simple properties.

Informally: the distance from a point to itself is zero, the distance between two distinct points is positive. The set of complex functions on an interval x ∈ [0,L], form a vector space overC.

To better understand a vector space one can try to ﬁgure out its possible subspaces. A subspace of a vector space V is a subset of V that is also a vector space. To verify that a subset U of V is a. spaces for the beginning of the second semester. The book normally used for the class at UIUC is Bartle and Sherbert, Introduction to Real Analysis third edition [BS].

The structure of the beginning of the book somewhat follows the standard syllabus of UIUC Math. Vector spaces and signal space In the previous chapter, we showed that any L 2 function u(t) can be expanded in various orthog onal expansions, using such sets of orthogonal functions as the T-spaced truncated sinusoids or the sinc-weighted sinusoids.

Thus u(t) may be speciﬁed (up to L 2 equivalence) by a countably inﬁnite sequence such as {u. Let be a set equipped with the initial topology of certain functions: →, where () ∈ is a family of topological spaces. Let Z {\displaystyle Z} be another topological space.

A function g: Z → X {\displaystyle g:Z\to X} is continuous if and only if all the functions f α ∘ g {\displaystyle f_{\alpha }\circ g} (α ∈ A {\displaystyle.Linear algebra is the study of vectors and linear functions.

In broad terms, vectors are things you can add and linear functions are functions of vectors that respect vector addition.

The goal of this text is to teach you to organize information about vector spaces in a way that makes problems involving linear functions of many variables easy.


text:
## BER simulation in GNU Radio

David Rowe always insists that you should simulate the bit error rate for any modem you build. I’ve been intending to do some simulations of the decoders in gr-satellites since a while ago, and I’ve finally had some time to do so. I have simulated the performance of the LilacSat-1 decoder, both for uncoded BPSK and for the Viterbi decoder. This is just the beginning of the story, as the code can be adapted to simulate other modems. Here I describe some generalities about BER simulation in GNU Radio, the simulations I have done for LilacSat-1, and the results.

I am doing some BER simulations with GNU Radio (stay tuned for the next post), and during my experiments I have stumbled upon a bug in the “Decode CCSDS 27″ block. This block is a Viterbi decoder for the CCSDS convolutional code with $$r=1/2$$, $$k=7$$ (note that the convention used by this block is first POLYA then POLYB so it doesn’t match the NASA-DSN convention nor the CCSDS/NASA-GSFC conventions, as I have mentioned in another post).

The bug consists in the block entering a “degraded” state after it has processed many symbols (on the order of several millions). In this degraded state, it doesn’t decode properly, producing lots of bit errors even if no input symbols are in error. Fortunately, there is another block in GNU Radio which can decode the CCSDS convolutional code, the “CC Decoder” included in FECAPI. This block doesn’t seem to suffer this issue. Here I describe how to replicate the bug, how to replace “Decode CCSDS 27” by “CC Decoder” and some other miscellaneous things related to this bug.

## A first look at DSLWP SSDV downlink

The Chang’e 4 is a Chinese lunar mission that will land a rover on the far side of the Moon by the end of 2018. To support this mission, the Chang’e 4 relay satellite will be launched six months before and put into a halo orbit around the Earth-Moon Lagrange L2 point. The relay will provide four 256Kbps links with the rover and lander on X-band and a 2Mbps link with Earth on S-band using a 4.2m dish. Two CE-4 microsatellites will be launched together with the relay satellite. They will be put in a 200km x 9000km lunar elliptical orbit. The main mission of the CE-4 microsatellites is to perform HF interferometry of celestial bodies, using the Moon as a shield from the radiation of the Sun and Earth. The satellites also carry an Amateur radio system called DSLWP, which will provide telecommand, telemetry and image downlink.

A team at Harbin Institute of Technology is currently designing the Amateur radio payload. As it is the case with previous HIT satellites such as BY70-1 and LilacSat-1, the payload will have a camera which can be telecommanded by radio Amateurs, which can use it to take and download pictures. Yesterday, Wei BG2BHC has released some work in progress of the image downlink. Many important parts of the downlink will still change, but releasing the work in progress at this early stage is a very good idea. Probably it is not too late in the development process so that the Amateur community can contribute with ideas and improvements.

The release consists of an IQ recording of the signal containing a full image and a decoder in gr-lilacsat. The IQ recording is at 2ksamp/s, since the signal is FSK at 250baud. Note that the recording is almost 32 minutes long. It takes a while to transmit an image at such a low rate. However, a low baudrate and a good amount of FEC are needed for an effective downlink from the Moon, given the huge path loss of around 197dB in the 70cm band.

The good news about this work in progress is that SSDV is now used to transmit the image. SSDV is a packetised protocol based on JPEG, but which is tolerant to packet loss. In contrast, BY70-1 and LilacSat-1 send JPEG images in 64byte chunks, and a single lost chunk can destroy the image completely. SSDV was originally developed to transmit images from Amateur high altitude ballons, so it is a good idea to use it also for DSLWP.

The bad news is that the way that SSDV has been included into the downlink protocol is not very optimal. In the rest of this post I do an in-depth look at the protocol, point out the main problems and suggest some solutions. Hopefully the protocol can still be modified and improved.


text:
## Estimating hiking velocity with rHike

##### April 13, 2020
r sf gdistance rHike

updated: May 13, 2020 (replaced Laplace with Lorentz distribution at Campbell’s suggestion)

I wanted to learn how to write an R package and I had this recurrent use-case (estimating cost-distance on a grid), so I made rHike. It’s a super-lightweight package that builds on Jacob van Etten’s gdistance by implementing both Tobler’s (1993) and Campbell’s (2019) hiking functions, though others could be added. Mainly, it just simplifies the user interface. Development mostly followed the workflow suggested by Hadley Wickham and Jenny Bryan in their R Packages book. Mostly…

Where are we going? Intuitively, walking speed should vary as a function of slope. Campbell’s Hiking Function reflects this using an empirically derived version of the Lorentz distribution:

$f(x) = \frac{c}{\pi\, b\left(1+\left[\frac{x-a}{b}\right]^{2}\right)} + d + e\, x$

where $$x$$ is the slope at a location measured in degrees. The idea here is first to get an estimate of slope, then use a hiking function like Campbell’s to calculate velocity, and finally to use velocity to estimate travel time (among other cost-distances).

slope <- seq(-50, 50, by = 1)

velocity <- hf_campbell(slope, decile = 10)

#### R Preamble

library(data.table)
library(gdistance)
library(raster)
library(rHike)
library(sf)
library(tidyverse)
library(viridis)

Coordinate Reference System (CRS):

utm12 <- st_crs(26912)

It should really be emphasized that

the unit of projection should match the unit of the stored values.

In this case, elevation is in meters, so the projection should be in meters too.

#### Data Wrangling

First, we read in a DEM of the foothills in Salt Lake City, Utah:

dem <- raster(system.file("extdata/slc.tif", package = "rHike"))

dem <- projectRaster(dem, crs = utm12\$proj4string)

Next we generate an arbitrary point pattern.

start_points <- st_sf(id = 1:3,
typ = "start",
geometry = st_sfc(st_point(c(424350, 4514200)),
st_point(c(426000, 4515000)),
st_point(c(429000, 4516500)),
crs = utm12))

end_points <- st_sf(id = 1:2,
typ = "end",
geometry = st_sfc(st_point(c(426200, 4516400)),
st_point(c(427600, 4515600)),
crs = utm12))

Now we use the elevation data to generate a slope raster and calculate a velocity surface for the average hiker using Campbell’s Hiking Function.

slope <- hf_slope(dem)

velocity <- hf_velocity(slope, hf = "campbell", decile = 10)

#### Calculating Accumulated Cost

Accumulated cost-distance is the total time it takes to travel from one location to another; in this case, to all other locations in the landscape of interest.

cost <- hf_accCost(velocity,
fun = function(x) 1/mean(x),
points = start_points)

# convert seconds to minutes
cost <- cost/60

That looks about right, twenty’ish minutes for the average hiker to walk a half-mile or so around the SLC foothills.

#### Calculating Least-Cost Path

short_paths <- hf_shortPath(velocity, start_points, end_points)

#### Future Work

In this post, conversion between different units was done on the fly. It would be nice to introduce methods to do that in a more rigorous way, so I might look into integrating the units package at some point.

#### References

Tobler, Waldo R. (1993). “Three Presentations on Geographical Analysis and Modeling: Non-Isotropic Geographic Modeling, Speculations on the Geometry of Geography, and Global Spatial Analysis.” 93-1.

Campbell, Michael J., Philip E. Dennison, Bret W. Butler, and Wesley G. Page (2019). "Using crowdsourced fitness tracker data to model the relationship between slope and travel rates. 106, 93-107.

#### Plotting fieldwork with mapview

##### April 7, 2020
r sf mapview fieldwork


text:
# testing derivative approximations

I'm writing a library that involves some approximations of variational calculus problems. Whenever I implement routines to evaluate the derivative or Hessian of an action functional $A$, I write a test to check that these are working correctly. To do this, I evaluate the error in using the first and second derivatives of the action to compute a local quadratic approximation to the objective functional:

$E = A(u + \delta v) - \left(A(u) + \delta\cdot dA(u)\cdot v + \frac{1}{2}\delta^2(d^2A(u)v)\cdot v\right)$

and check that $E/\delta^3$ doesn't grow for $\delta = 2^{-1},\ldots,2^{-N}$ for some big enough $N$.

This works well for moderately sized $N$ (up to 16 or so), but for $N$ too large, truncation error begins to dominate and the error in the approximation gets worse.

What is a good way to test derivative approximations in the face of truncation error?

The way I see it, I have only bad options:

1. Write an initial version of the test that prints the approximation errors, find the $N$ for which truncation error takes over, then use this $N$ as the cutoff point in the final version of the unit test.
2. Have some way of automatically diagnosing where truncation error takes over and discard all the results thereafter.
3. Don't make my unit tests return failure unless it's really obvious. Graph the results and let the user decide.
4. Determine analytically where truncation error is likely to occur.

I don't like option 1 because it feels like cherry-picking my tests so that they will pass. I don't like option 2 because a genuinely failing implementation's inherent mathematical errors might be erroneously assessed as truncation error, so that automated regression testing would miss the bug. At that rate, I'd have to check the results manually, in which case I might as well go with option 3. But I don't like option 3 because I want automated testing. I think option 4 is practical for functions of a single variable -- evaluating $\sqrt{\epsilon|f(x)/f''(x)|}$ where $\epsilon$ is the machine epsilon gives you a pretty good upper bound -- but for big PDE problems where the unknown is a field with thousands of variables, not so much.

• Are you running any simulations with $N=16$? If not why test up to this limit? Also, what precision are you using? Single? Double? Quad? – Charles Mar 24 '17 at 4:58

Well, I can tell you what I do, but it's somewhat unsatisfactory.

1. Check $dA(u)$ using $A(u)$. I prefer a fourth-order finite difference scheme, but I check $dA(u)\cdot \delta u$ against the approximation

$$d A(u)\cdot \delta u \approx \frac{A(u - 2\epsilon \delta u)-8A(u - \epsilon \delta u)+8A(u +\epsilon \delta u)-A(u + 2\epsilon \delta u)}{12}$$

1. Once we know that $dA(u)$ is accurate, check $d^2A(u)\delta u$ versus the approximation $$d^2 A(u)\delta u \approx \frac{dA(u - 2\epsilon \delta u)-8dA(u - \epsilon \delta u)+8dA(u +\epsilon \delta u)-dA(u + 2\epsilon \delta u)}{12}$$ This is not that different from above, except we now have a vector instead of a scalar, so the error is calculated with a norm and not an absolute value.

As you mentioned above, it's hard to know when there's a switch between truncation and round-off error. As such, I run this for several values of $\epsilon$ and look for two things

1. The minimum error in the approximation over all attempted $\epsilon$ reaches an acceptable level

2. Typically, there's a curve where the error goes high-low-high as the value of $\epsilon$ is reduced

In a unit test, I typically only use 1. When validating a function by hand, I look for 2. Technically, 1 could give an erroneous pass in case we get lucky and everything cancels. I've never seen that happen, so I don't worry about it.


text:
# Let F(x)=x2 and D(x)=xâˆ’1. Then using long division find R(x)such that F(x)=D(x)â‹…Q(x)+R(x).Please help, I am so lost.

###### Question:

Let F(x)=x2 and D(x)=xâˆ’1. Then using long division find R(x) such that F(x)=D(x)â‹…Q(x)+R(x). Please help, I am so lost.

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# Homework Help: Sequence of functions, continuity, uniform convergence

1. Aug 26, 2009

### fmam3

1. The problem statement, all variables and given/known data
Let $$(f_n)$$ be a sequence of continuous functions on $$[a,b]$$ that converges uniformly to $$f$$ on $$[a,b]$$. Show that if $$(x_n)$$ is a sequence in $$[a,b]$$ and if $$x_n \to x$$, then $$\lim_{n \to \infty} f_n (x_n) = f(x)$$

2. Relevant equations
None

3. The attempt at a solution
I just want to double check whether my proof works! Any criticisms welcomed! :)

Since $$(f_n)$$ is continuous on a closed interval $$[a,b]$$, $$(f_n)$$ is uniformly continuous on $$[a,b]$$. And since $$\lim x_n = x$$, $$(x_n)$$ is a Cauchy sequence and since $$(f_n)$$ is uniformly continuous on $$[a,b]$$, it follows that $$(f_n (x_n) )$$ is a Cauchy sequence on $$[a,b]$$.

Let $$\varepsilon > 0$$. Since $$(f_n(x_n))$$ is Cauchy, it converges to $$f(x_n)$$, thus for some $$N$$, $$n > N$$ implies $$|f_n (x_n) - f(x_n)| < \varepsilon / 2$$. And since the uniform limit of continuous functions is continuous, that is since $$f_n \to f$$ uniformly and $$(f_n)$$ is continuous on $$[a,b]$$, it implies that $$f$$ is continuous on $$[a,b]$$. And since $$\lim x_n = x$$ and by the continuity of $$f$$, there exists some $$\delta > 0$$ such that $$|x_n - x| < \delta$$ implies $$|f(x_n) - f(x)| < \varepsilon / 2$$.

Thus, by the inequality $$|f_n (x_n) - f(x)| = |f_n(x_n) - f(x_n) + f(x_n) - f(x)| \leq |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)|$$, it follows that when $$n > N$$ and $$|x_n - x| < \delta$$, we have $$|f_n (x_n) - f(x)| < \varepsilon /2 + \varepsilon /2 = \varepsilon$$. This completes our proof.

2. Aug 28, 2009

### Billy Bob

It's not clear how you would conclude this. It's practically tantamount to what you are trying to prove.

Instead, I would try obtaining more directly $$|f_n (y) - f(y)| < \varepsilon / 2$$ for all y. Then take y=x_n.

The rest isn't too bad, although I would use more of an N_2 approach instead of a delta.

3. Aug 28, 2009

### fmam3

Thanks for the reply and feedback!

Actually I looked at this question again today and realized the exact same point you noted --- by writing $$(f_n(x_n))$$ is Cauchy, it's practically just stating what I want to show and hence doesn't work.

I reworked another approach and here it is:

For contradiction, suppose $$\lim_{n \to \infty} f_n(x_n) \ne f(x)$$. That is, $$\exists \varepsilon > 0$$, $$\forall N$$, $$\exists n > N$$ such that $$|f_n(x_n) - f(x)| \geq \varepsilon$$.

Since $$(x_n)$$ is a sequence in a closed and bounded set $$[a,b]$$, by the Bolzano-Weierstrass Theorem, there exists a convergent subsequence $$(x_{n_k})_{k \in \mathbb{N}}$$. And since $$\lim x_n = x$$ and if the sequence converges, every subsequence converges to the same limit, it implies we have $$\lim_{k \to \infty} x_{n_k} = x$$; that is, $$\forall \delta > 0$$, $$\exists N_0$$ such that $$\forall k > N_0$$, we have $$|x_{n_k} - x | < \delta$$, and furthermore, by the continuity of each $$f_n$$ on $$[a,b]$$, it implies $$|f_n(x_{n_k}) - f_n(x)| < \varepsilon$$ for each $$n \in \mathbb{N}$$. Note that since $$\{n_k : k > N_0\} \subseteq \{n: n \in \mathbb{N}\}$$, $$k > N_0$$ implies $$|f_{n_k}(x_{n_k}) - f_{n_k}(x)| < \varepsilon / 2$$

Since$$f_n \to f$$ uniformly on $$[a,b]$$, $$\exists N_1$$ such that $$|f_n(x) - f(x)| < \varepsilon / 2$$ for $$\forall n > N_1, \forall x \in [a,b]$$.

Thus, take $$k > \max\{N_0, N_1\}$$, which implies $$n_k \geq k > \max\{N_0, N_1\}$$ and by the inequality: $$|f_{n_k} (x_{n_k}) - f(x)| = |f_{n_k} (x_{n_k}) - f_{n_k}(x) + f_{n_k} (x) - f(x)| \leq |f_{n_k} (x_{n_k}) - f_{n_k}(x)| + |f_{n_k}(x) - f(x)| < \varepsilon / 2 + \varepsilon / 2 = \varepsilon$$ ---- contradiction, since we have $$|f_n(x_n) - f(x)| \geq \varepsilon$$ for $$\forall n$$.

Perhaps I'm missing something here but can you explain a little bit more in detail on this. Since the problem already has $$f_n \to f$$ uniformly, the statement $$|f_n (y) - f(y)| < \varepsilon / 2, \forall y$$ is already true without doing any work. The most difficult part, at least to me, is that we are trying to show that $$f_n(x_n)$$ converges to something (i.e. $$f(x)$$), where $$f_n(x_n)$$ is indexed by $$n \in \mathbb{N}$$ on BOTH the sequence of functions and the sequence of numbers. This is in contrast to showing $$f_n(x) \to f(x)$$, where only the sequence of function is indexed but the input value $$x$$ is fixed.

Or it could be I'm just talking total rubbish above. I would appreciate any of your thoughts :)

4. Aug 28, 2009

### fmam3

Oh sorry, on reading what you wrote again, I think I'd misinterpreted your comment. My apologies. I thought the result was clear but anyways, the statement is the following (even though using this approach in my original question would not work):
(1)If $$(x_n)$$ is a Cauchy sequence in $$S$$ and $$f$$ is uniformly continuous on $$S$$, then $$(f(x_n))$$ is a Cauchy sequence in $$S$$.

The proof is quite straight forward. Let $$\forall \varepsilon > 0$$. Since $$f$$ is uniformly continuous on $$S$$, $$\exists \delta > 0$$ such that $$\forall x,y \in S$$ and $$|x - y| < \delta$$, we have $$|f(x) - f(y)| < \varepsilon$$.

Since $$(x_n)$$ is Cauchy, $$\exists N$$ such that $$\forall n,m > N$$, we have $$|x_n - x_m| < \delta$$. Thus, by the above on $$f$$ is uniformly continuous, $$n,m > N$$ implies $$|f(x_n) - f(x_m)| < \varepsilon$$ and thus $$(f(x_n))$$ is a Cauchy sequence.

The second statement that I'd used is:
(2)If $$f$$ is a continuous functions on a closed interval $$[a,b]$$, then $$f$$ is uniformly continuous on $$[a,b]$$. I'll skip the proof of this as it's essentially just an application of the Bolzano-Weierstrauss theorem.

And combining (1) and (2) together, I concluded that $$(f_n(x_n))$$ is Cauchy. But this would not work in my original question --- I think ---- is because while we know that $$(f_n(x_n))$$ is Cauchy, we do not know what the limiting function is and thus, it is insufficient to prove the result.

Last edited: Aug 28, 2009
5. Aug 28, 2009

### snipez90

Instead of resorting to Cauchy sequences, there is a much simpler way to do this. Let e > 0. Probably the first result in any treatment of uniform convergence tells us that the hypotheses in this problem imply that f is continuous on [a,b]. In particular, f is continuous at x and since x_n approaches x, f(x_n) approaches f(x), so there exists N such that n > N implies |f(x_n) - f(x)| < e/2 (*). By uniform convergence, there exists M such that n > M implies |f_n(y) - f(y)| < e/2 for all y in [a,b]. In particular, |f_n(x_n) - f(x_n)| < e/2 (**). If n > max{N,M}, then by (*) and (**) and the triangle inequality, we have |f_n(x_n) - f(x)| < e, as desired.

*EDIT* Essentially you had the last step of the proof figured out. It remained to estimate $$|f_n(x_n) - f(x_n)|$$ via uniform convergence and $$|f(x_n) - f(x)|$$ via continuity of f.

Last edited: Aug 28, 2009
6. Aug 29, 2009

### fmam3

That's what I'd thought too before I wrote my proof version 1 (the first post) and proof version 2 (the post after). However, I was quite hesitant on this for one single reason --- closed interval $$[a,b]$$ (more of a student mentality --- they don't usually include non-essential information in statements so everything must be somehow used). If this result holds for any interval (open or closed), then there's no need to be explicit about writing $$[a,b]$$ as part of the hypothesis of the problem. I'm actually thinking that the result actually does not hold for an open interval $$(a,b)$$ since if $$\lim x_n = x$$ and $$x \notin (a,b)$$, then all the results of continuity goes out the window. Thus, I in my proof version 2, I chose to use a subsequential argument to explicitly make use of the fact that we're dealing with a closed interval and thus we can use the Bolzano-Weierstrauss.

What do you think?

7. Aug 29, 2009

### snipez90

Hmm, well when I encountered this problem before, the problem didn't state explicitly that (x_n) was a sequence in [a,b]. It was a fair assumption I think, since without it you can't really do much with the bound from uniform convergence. Since you're told explicitly that (x_n) is a sequence in the closed interval [a,b], then the limit is in [a,b], so I think the argument holds.

8. Aug 29, 2009

### fmam3

I just thought about it and here's a simple example to illustrate my point about the importance of closed interval $$[a,b]$$.

Suppose we have a sequence $$(x_n)$$ defined as $$x_n = 1 / (n + 1)$$ for $$\forall n$$ and clearly, $$(x_n)_{n \in \mathbb{N}} \subseteq (0,1)$$ and that $$x_n \to 0$$ but note that $$0 \notin (0,1)$$. And suppose we define the sequence of functions $$f_n(x) = \frac{x + n}{n}$$ and naturally $$(f_n)$$ is continuous on (0,1) for $$\forall n$$. It is easy to show that $$f_n \to 1$$ uniformly on $$(0,1)$$. But observe that $$f_n(x_n) = \frac{1}{n} \cdot \large( \frac{1}{n + 1} + n\large) = \frac{n^2 + n + 1}{n}$$ and we clearly have that $$f_n(x_n) \to +\infty$$. Yet, we have that the limiting function is $$f(x) = 1$$. Thus, we have that $$\lim_{n \to \infty} f_n(x_n) \ne f(x)$$ because simply that $$(0,1)$$ is an open interval.

Hence, I think that we are forced to use the closed interval property and we can do so by the Bolzano-Weierstrauss. What do you think? And thanks for your reply above!

9. Aug 29, 2009

### snipez90

Hmmm, yes I agree that we need a closed interval, but I'm not sure if we necessarily need the Bolzano-Weierstrass theorem. It should be easy to prove directly from simple epsilon-delta arguments that if we have a sequence in [a,b], which by definition means that each term is in [a,b], then the limit must also be in [a,b].

10. Aug 29, 2009

### fmam3

You know... actually rereading my own proof version 2, I'm actually just making the exact same proof as yours (in fact, nearly word for word) but substituted everything from a plain vanilla sequence to a subsequence. Hmm... I guess I just thought too much about closed intervals and whenever I see them, I start thinking about the beloved Bolzano-Weierstrass. Another reason that I'd approached this problem with subsequences is motivated by a similar proof method of Dini's Theorem, and in that proof, the subsequential argument was essential.

But thanks for the input :)

11. Aug 29, 2009

### snipez90

Hmmm that's weird, is this problem from Spivak (probably stolen from Rudin)? I think the most difficult problem set I had during the spring quarter this year asked us to prove Dini's theorem, the problem you posted, and the converse of the problem you stated. Dini's theorem took awhile, since I had never actually seen how to apply Bolzano-Weierstrass.

12. Aug 29, 2009

### fmam3

Haha... that's interesting. Dini's Theorem was actually just an exercise the instructor given out in a problem set sheet and not from a book.

But just for fun, let's prove Dini's Theorem. Before we prove Dini's Theorem, let's show this lemma and the original Dini's Theorem comes out immediately afterwards:

Suppose $$(f_n)$$ is a sequence of continuous functions on a closed interval $$[a,b]$$ and $$(f_n(x))$$ is nonincreasing. And suppose $$f_n \to 0$$ pointwise. Then $$f_n \to 0$$ uniformly.

------------------------------------------------------------------------------------------------

Here's the proof that I'd worked out for my problem set. Any comments again appreciated!

First, we'll show that $$f_n(x) \geq 0$$ for $$\forall x \in [a,b], \forall n \in \mathbb{N}$$. Suppose not, then $$\exists N_0, \exists x \in [a,b]$$ such that $$f_{N_0}(x) < 0$$. Since $$(f_n)$$ is nonincreasing $$n > N_0$$ implies $$f_n(x) \leq f_{N_0}(x) < 0$$. But since $$f_n \to 0$$ pointwise, there exists $$N_1$$ such that $$n > \max\{N_0, N_1\}$$ implies $$0 \leq f_{N_0}(x) \leq 0$$ or that $$f_{N_0}(x) = 0$$ --- which is not possible since we assumed $$f_N(x) < 0$$.

For contradiction, suppose $$f_n \to 0$$ uniformly on $$[a,b]$$ does not hold. That is, $$\exists \varepsilon > 0, \forall N$$ we have $$|f_n(x) - 0| = f_n(x) \geq \varepsilon$$ for $$\exists n > N, \exists x \in [a,b]$$. Now, we construct a sequence of numbers from $$[a,b]$$ as follows. We claim that for each $$n \in \mathbb{N}, \exists x_n \in [a,b]$$ such that $$f_n(x_n) \geq \varepsilon$$. Suppose this does not hold. Then it implies $$\exists n_0$$ such that $$\forall x \in [a,b]$$, we have $$f_{n_0}(x) < \varepsilon$$. But since $$(f_n)$$ is nonincreasing, $$n \geq n_0$$ implies $$f_n(x) \leq f_{n_0}(x) < \varepsilon$$ --- this is not possible as it contradicts $$f_n(x) \geq \varepsilon$$ as shown above.

13. Aug 29, 2009

### fmam3

(continue from above post)

Now, since $$(x_n)_{n \in \mathbb{N}} \subseteq [a,b]$$, by the Bolzano-Weierstrass Theorem, there exists a convergent subsequence, say, $$(x_{n_k})$$ converging to, say, $$x_0 \in [a,b]$$. Thus, $$\lim_{k \to \infty} x_{n_k} = x_0$$ implies that $$\forall \delta > 0, \exists K$$ such that $$|x_{n_k} - x_0| < \delta$$ when $$k > K$$. Now, since $$f_n \to 0$$ pointwise, $$\forall x \in [a,b], \exists m \in \mathbb{N}$$ such that $$|f_m(x) - 0| = f_m(x) < \varepsilon$$. But by the continuity of $$(f_n)$$, $$\lim_{k \to \infty}x_{n_k} = x$$ implies $$\lim_{k \to \infty}f_m(x_{n_k}) = f_m(x_0) < \varepsilon$$.

Thus, for $$k > \max\{K, m\}$$, implying $$n_k \geq k > \max\{K, m\}$$ and again by the fact that $$(f_n)$$ is nonincreasing, we have that $$f_{n_k}(x_{n_k}) \leq f_m(x_{n_k}) < \varepsilon$$ ---- contradiction, since we had assumed $$|f_n(x) - 0| = f_n(x) \geq \varepsilon$$ would hold. This completes the proof.

---------

Dini's theorem is simply applying the above statement to $$g_n = f_n - f$$ where $$f$$ is the limiting function.

Would you mind sharing how your class went about proving Dini's Theorem? :)

14. Aug 29, 2009

### snipez90

That looks like a good proof. There was a review session where my teacher spent over an hour trying to work out the proof (the instructor took the course as an undergrad at the same institution, but I guess he was a little rusty). Personally, I'm terrible at making the bookkeeping work out for the last part where you have to use the nonincreasing sequence hypothesis. But for the most part, this is more or less the proof some of us learned.

Here are a few comments. Once I negate a statement and it seems clear enough to me that I can generate a sequence, I usually just apply the Bolzano-Weierstrass theorem without further explanation, but this is a minor difference. I think you chose clearer notation then I did before, and your explicit use of continuity is probably a lot better than the argument I had before. I was more hand-wavy in mine I think, using the continuity of f_n to argue that f_n(y) < epsilon for all y close enough to x_0 (as you had labeled it). This is correct, but not entirely rigorous. I think you made everything clear and explicit and it looks correct, even though I'm pretty tired.

15. Aug 29, 2009

### fmam3

Thanks for your reply. When we went over this particular problem, it was actually quite tricky on how they had to negate several statements to obtain certain desired properties (i.e. on how $$f_n(x) \geq 0$$ and how the sequence $$(x_n)$$ was obtained). But I agree with you --- once the sequence $$(x_n)$$ is obtained, which in my opinion is actually the trickiest part, the rest just follows from the Bolzano-Weierstrauss, continuity and the works.

Thanks for the input and this great dialogue!

16. Aug 29, 2009

### Billy Bob

For your "Dini lemma," I prefer a Heine-Borel approach instead of B-W. All the contradictions in the B-W make my head spin, plus it is hard to work out on the spot. Your "Dini lemma" is a great oral exam question.

First, by hypothesis, for each x, $$f_1(x)\ge f_2(x)\ge f_3(x)\ge\dots\ge0$$.

Let $$\epsilon>0$$.

Fix x.

Since $$f_n(x)$$ decreases to 0 as n increases, there exists $$N_x$$ such that $$f_{N_x}(x)$$ is small.

But $$f_{N_x}$$ is continuous. Thus there exists an open neighborhood $$B_x$$ of x, on which $$f_{N_x}$$ is small.

Now you have an open cover by $$\{B_x\}$$. Apply Heine-Borel.

Can you fill in the details?


text:
# Compiler Flags

As programmers, we tend to be smarter than compilers. More often than not, we can’t. Nowadays, compilers comes with hundreds of flags to choose from. To make lives easier, compiler developers provide some neat flags for most common optimizations, namely -Ox series. I am not a fan of compiler flags, nor do I pretend that I can be smarter by compiling the code beyond the default compiler settings. Recently I have been testing Vlasiator on multiple platforms. I am shocked by the differences caused simply by GCC optimization flags. Here are a few things I have learned so far.

1. Standard optimization flags are safer

On my Ubuntu laptop with GCC 7.5 (which is pretty old for sure), the results are consistent with standard -O0, -O1, -O2 and even -O3 flags. As promised, the default flags are guaranteed to work in most scenarios.

1. Be cautious about -ffast-math

IEEE standard is there for a reason. -ffast-math is a higher level flag which consists of multiple low level flags to speeds floating point operations up by loosening the IEEE restrictions. On one specific test, adding this flag gives me 10% speed boost at the cost of inconsistent results: 0.2% relative difference in plasma bulk speed values for one spatial cell running for 3600s simulation time.

1. Be cautious about avx instructions

In my understanding, the avx series of instructions simply means vectorization of more operations in the cache. In the current Vlasiator Makefile, there is a -mavx flag together with specific vectorization class from an external C++ library.

In other recommended flag settings, many people mentioned the -march flag. For example, -march=native tried to detect the CPU architecture you are compiling on and add hardware-specific instructions, mostly avx-related flags, to your program.

Surprisingly, this also affects the results quite significantly. Although the speed improved by 10% beyond pure -O3, the plasma bulk speed difference is also about 0.2%.

1. Be careful about -Ofast

On top of the previous two observations, I need to be really careful about the most aggressive -Ofast flag, which includes -ffast-math.

## Vlasiator

However, even if I turn off all the optimization flags, Vlasiator still shows different result on different platforms. This indicates that there are even more hidden facts underneath.

Tags:

Categories:

Updated:


text:
Question: What is lfcSE and how is it calculated?
0
28 days ago by
lizzid0
lizzid0 wrote:

After callings res() in DESeq2, I see a column for lfcSE. I've read in multiple places that this is the standard error the of the log2 fold change and read in other places that it is not. I'm wondering what this column is for and how it is calculated exactly.

deseq2 standarderror • 73 views
modified 28 days ago • written 28 days ago by lizzid0
Answer: What is lfcShrink and how is it calculated?
0
28 days ago by
Michael Love26k
United States
Michael Love26k wrote:

Can you provide your exact code? There isn't a res function in DESeq2, and there isn't a column lfcShrink in the results table. I need to know specifically what you are referring to. Note that there is a lot of details about this in the vignette, so that would be a good place for you to start.

My apologies. I meant that I called the results() function and see a column for lfcSE which is seen in the edit of my post. The vignette says that lfcSE is the standard error of log2foldchange but I'm wondering what samples and variables that are being taken into account when calculating lfcSE.


text:
Getting the cells from the clusters Seurat has found
1
0
Entering edit mode
2.7 years ago
suvratha ▴ 60

Hello, I've been using Seurat v3.1 for single-cell analysis of mouse data. I've clustered the cells and Seurat has found 12 different clusters for my data. Now, I need to name the clusters so that I know which particular cell type has clustered where on the UMAP plot. Is there any other way to do this other than using the FindMarkers() function and then naming them based on the most differentially expressed gene in the cluster? Also, is there a way to retrieve the data where we know which cell/data has gone into the UMAP predicted cluster so that I can at least I know which cell type has gone where in the plot based on a metadata file I have for my data?

Thank you.

RNA-Seq Seurat single cell R • 2.2k views
0
Entering edit mode

you can plot the known marker gene to check the cluster.

1
Entering edit mode
2.6 years ago
Evan ▴ 170

Hello, I got approximately the same question recently and I succeed to find the answer. Here is the command in Seurat to get a file with two columns, one with cell sequence and a second column with associated cluster. Each line of this file will be one cell.

write.table(yourdata@active.ident, file='Cell_Cluster.tsv', quote=FALSE, sep='\t', col.names = FALSE)


text:
# 11.1 Systems of linear equations: two variables  (Page 8/20)

 Page 8 / 20

$\begin{array}{c}5x-y=4\text{\hspace{0.17em}}\\ x+6y=2\end{array}$ and $\text{\hspace{0.17em}}\left(4,0\right)$

and $\left(-6,1\right)$

Yes

$\begin{array}{c}3x+7y=1\text{\hspace{0.17em}}\\ 2x+4y=0\end{array}$ and $\text{\hspace{0.17em}}\left(2,3\right)$

and $\left(-1,1\right)$

Yes

$\begin{array}{c}x+8y=43\text{\hspace{0.17em}}\\ 3x-2y=-1\end{array}$ and $\text{\hspace{0.17em}}\left(3,5\right)$

For the following exercises, solve each system by substitution.

$\left(-1,2\right)$

$\begin{array}{l}4x+2y=-10\\ 3x+9y=0\end{array}$

$\left(-3,1\right)$

$\begin{array}{l}2x+4y=-3.8\\ 9x-5y=1.3\end{array}$

$\begin{array}{l}\hfill \\ \begin{array}{l}\\ \begin{array}{l}-2x+3y=1.2\hfill \\ -3x-6y=1.8\hfill \end{array}\end{array}\hfill \end{array}$

$\left(-\frac{3}{5},0\right)$

No solutions exist.

$\begin{array}{l}\frac{1}{2}x+\frac{1}{3}y=16\\ \frac{1}{6}x+\frac{1}{4}y=9\end{array}$

$\left(\frac{72}{5},\frac{132}{5}\right)$

$\begin{array}{l}\\ \begin{array}{l}-\frac{1}{4}x+\frac{3}{2}y=11\hfill \\ -\frac{1}{8}x+\frac{1}{3}y=3\hfill \end{array}\end{array}$

For the following exercises, solve each system by addition.

$\left(6,-6\right)$

$\begin{array}{l}6x-5y=-34\\ 2x+6y=4\end{array}$

$\left(-\frac{1}{2},\frac{1}{10}\right)$

$\begin{array}{l}7x-2y=3\\ 4x+5y=3.25\end{array}$

No solutions exist.

$\begin{array}{l}\frac{5}{6}x+\frac{1}{4}y=0\\ \frac{1}{8}x-\frac{1}{2}y=-\frac{43}{120}\end{array}$

$\left(-\frac{1}{5},\frac{2}{3}\right)$

$\left(x,\frac{x+3}{2}\right)$

For the following exercises, solve each system by any method.

$\left(-4,4\right)$

$\begin{array}{l}6x-8y=-0.6\\ 3x+2y=0.9\end{array}$

$\begin{array}{l}5x-2y=2.25\\ 7x-4y=3\end{array}$

$\left(\frac{1}{2},\frac{1}{8}\right)$

$\begin{array}{l}\\ \begin{array}{l}7x-4y=\frac{7}{6}\hfill \\ 2x+4y=\frac{1}{3}\hfill \end{array}\end{array}$

$\left(\frac{1}{6},0\right)$

$\begin{array}{l}3x+6y=11\\ 2x+4y=9\end{array}$

$\left(x,2\left(7x-6\right)\right)$

$\begin{array}{l}\frac{1}{2}x+\frac{1}{3}y=\frac{1}{3}\\ \frac{3}{2}x+\frac{1}{4}y=-\frac{1}{8}\end{array}$

$\begin{array}{l}2.2x+1.3y=-0.1\\ 4.2x+4.2y=2.1\end{array}$

$\left(-\frac{5}{6},\frac{4}{3}\right)$

## Graphical

For the following exercises, graph the system of equations and state whether the system is consistent, inconsistent, or dependent and whether the system has one solution, no solution, or infinite solutions.

$\begin{array}{l}3x-y=0.6\\ x-2y=1.3\end{array}$

Consistent with one solution

Consistent with one solution

Dependent with infinitely many solutions

## Technology

For the following exercises, use the intersect function on a graphing device to solve each system. Round all answers to the nearest hundredth.

$\begin{array}{l}\hfill \\ \begin{array}{l}-0.01x+0.12y=0.62\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.15x+0.20y=0.52\hfill \end{array}\hfill \end{array}$

$\left(-3.08,4.91\right)$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.5x+0.3y=4\hfill \\ 0.25x-0.9y=0.46\hfill \end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.15x+0.27y=0.39\hfill \\ -0.34x+0.56y=1.8\hfill \end{array}$

$\left(-1.52,2.29\right)$

$\begin{array}{l}\begin{array}{l}\\ -0.71x+0.92y=0.13\end{array}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.83x+0.05y=2.1\hfill \end{array}$

## Extensions

For the following exercises, solve each system in terms of $\text{\hspace{0.17em}}A,B,C,D,E,\text{}$ and $\text{\hspace{0.17em}}F\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}A–F\text{\hspace{0.17em}}$ are nonzero numbers. Note that $\text{\hspace{0.17em}}A\ne B\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}AE\ne BD.$

$\begin{array}{l}x+y=A\\ x-y=B\end{array}$

$\left(\frac{A+B}{2},\frac{A-B}{2}\right)$

$\begin{array}{l}x+Ay=1\\ x+By=1\end{array}$

$\begin{array}{l}Ax+y=0\\ Bx+y=1\end{array}$

$\left(\frac{-1}{A-B},\frac{A}{A-B}\right)$

$\begin{array}{l}Ax+By=C\\ x+y=1\end{array}$

$\begin{array}{l}Ax+By=C\\ Dx+Ey=F\end{array}$

$\left(\frac{CE-BF}{BD-AE},\frac{AF-CD}{BD-AE}\right)$

## Real-world applications

For the following exercises, solve for the desired quantity.

A stuffed animal business has a total cost of production $\text{\hspace{0.17em}}C=12x+30\text{\hspace{0.17em}}$ and a revenue function $\text{\hspace{0.17em}}R=20x.\text{\hspace{0.17em}}$ Find the break-even point.

A fast-food restaurant has a cost of production $\text{\hspace{0.17em}}C\left(x\right)=11x+120\text{\hspace{0.17em}}$ and a revenue function $\text{\hspace{0.17em}}R\left(x\right)=5x.\text{\hspace{0.17em}}$ When does the company start to turn a profit?

They never turn a profit.

A cell phone factory has a cost of production $\text{\hspace{0.17em}}C\left(x\right)=150x+10,000\text{\hspace{0.17em}}$ and a revenue function $\text{\hspace{0.17em}}R\left(x\right)=200x.\text{\hspace{0.17em}}$ What is the break-even point?

A musician charges $\text{\hspace{0.17em}}C\left(x\right)=64x+20,000,\text{}$ where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is the total number of attendees at the concert. The venue charges \$80 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point?

$\left(1,250,100,000\right)$

A guitar factory has a cost of production $\text{\hspace{0.17em}}C\left(x\right)=75x+50,000.\text{\hspace{0.17em}}$ If the company needs to break even after 150 units sold, at what price should they sell each guitar? Round up to the nearest dollar, and write the revenue function.

find the sum of 28th term of the AP 3+10+17+---------
I think you should say "28 terms" instead of "28th term"
Vedant
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
Give me the reciprocal of even number
Aliyu
The reciprocal of an even number is a proper fraction
Jamilu
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve


text:
## Quantum frameness for Charge-Parity-Time inversion symmetry    [PDF]

Michael Skotiniotis, Borzu Toloui, Ian T. Durham, Barry C. Sanders
We develop a theory of charge-parity-time (CPT) frameness resources to circumvent CPT-superselection. We construct and quantify such resources for spin~0, $\frac{1}{2}$, 1, and Majorana particles and show that quantum information processing is possible even with CPT superselection. Our method employs a unitary representation of CPT inversion by considering the aggregate action of CPT rather than the composition of separate C, P and T operations, as some of these operations involve problematic anti-unitary representations.
View original: http://arxiv.org/abs/1306.6114


text:
# Photon 2015

15-19 June 2015
Asia/Novosibirsk timezone
Home > Timetable > Session details > Contribution details

# Mesurement of the π+π- cross section at BESIII

## Speakers

• Mr. Benedikt KLOSS

## Content

Precise hadronic cross section measurements are an important input for the standard model prediction of $(g-2)_mu$.

Especially, the most important hadronic cross section as input for $(g-2)mu$, $sigma_{pipi} = sigma(e^+e^-rightarrow pi^+pi^-)$, has been measured over decades with ever increasing accuracy at accelerators in Novosibirsk, Orsay, and Frascati. More recently, the two most accurate measurements have been obtained by the KLOE collaboration in Frascati, and the BABAR collaboration at SLAC. Both experiments claim an accuracy of better than 1% in the energy range below 1 GeV, in which the $rho(770)$ resonance is dominating the hadronic cross section. However, a discrepancy of approximately 3% on the peak of the $rho(770)$ resonance is observed. The discrepancy is even increasing towards higher energies.

Unfortunately, this discrepancy is limiting our current knowledge of $a_mu equiv (g-2)mu/2$, which is a famous precision observable of the Standard Model (SM). The accuracy of the SM prediction of $(g-2)mu$, is entirely limited by the knowledge of the hadronic vacuum polarization contribution, which is obtained in a dispersive framework by using experimental R-ratio data on $sigma(e^+e^-rightarrow rm hadrons)$. The cross section $sigma_{pipi}$ contributes more than 70% to this dispersion relation and, hence, is the by far most important exclusive hadronic channel of the total hadronic cross section. Currently, a discrepancy of 3.6 standard deviations is found between the direct measurement of $a_mu$ and its SM prediction. However, the discrepancy reduces to 2.4$sigma$, when only BABAR data is used as input to the dispersion relation, which shows the need for a new measurement. In this talk, we are presenting a new measurement of the cross section $sigma_{pipi}$ at the BESIII experiment in Beijing.


text:
Looking for Algebra worksheets?
Check out our pre-made Algebra worksheets!
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##### Browse Questions

You can create printable tests and worksheets from these Grade 3 Function and Algebra Concepts questions! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page.

1 2 3
Using the commutative property, what is $2xx3xx8$ equal to?
1. $2xx6$
2. $2-3-8$
3. $2+3+8$
4. $3xx2xx8$
Grade 3 Number Properties CCSS: 3.OA.B.5
Using the distributive property, what is 7(8 + 2) equal to?
1. 70
2. (7 x 8) + (7 x 2)
3. (8 x 7) + (8 x 2)
Grade 3 Number Properties CCSS: 3.OA.B.5
Grade 3 Number Properties CCSS: 3.OA.B.5
Using the distributive property, what is 2(9 + 7) equal to?
1. (2 x 9) + (2 x 7)
2. (9 x 2) + (9 x 7)
3. 32
Grade 3 Number Properties CCSS: 3.OA.B.5
Grade 3 Number Properties CCSS: 3.OA.B.5
Which statement explains the equation $(6 xx 3) xx 2 = 6 xx (3 xx 2)$?
1. Changing the grouping of addends changes the sum.
2. Changing the grouping of factors changes the product.
3. Changing the grouping of the addends does not change the sum.
4. Changing the grouping of factors does not change the product.
Which correctly compares the pictures?

1. $4 < 3$
2. $4 > 3$
3. $3 = 3$
4. $4 = 4$
Grade 3 Number Properties CCSS: 3.OA.B.5
Which multiplication property is shown?

$6 xx 7 = 7 xx 6$
1. Identity
2. Distributive
3. Associative
4. Commutative
Grade 3 Number Properties CCSS: 3.OA.B.5
Grade 3 Number Properties CCSS: 3.OA.B.5
Grade 3 Number Properties CCSS: 3.OA.B.5
states that you can add two or more numbers in any order and get the same sum.
1. The Commutative Property of Addition
2. The Associative Property of Addition
3. Compatible numbers
4. The Identity Property of Addition
states that when you add zero to any number, the sum is that number.
1. The Commutative Property of Addition
2. The Associative Property of Addition
3. Compatible numbers
4. The Identity Property of Addition
states that you can group addends in different ways and still get the same sum. It is also called the Grouping Property.
1. The Commutative Property of Addition
2. The Associative Property of Addition
3. Compatible numbers
4. The Identity Property of Addition
Which property shows
(7 + 4) + 6 = 7 + (4 + 6)?
1. distributive property
2. associative property
3. identity property
4. commutative property
4 + 5 = 5 + 4
1. distributive property
2. associative property
3. identity property
4. commutative property


text:
PISA 2012: Calculating variance explained by ESCS

jenjen91

New Member
Hi there! I was wondering if anyone could help me figure out why I am getting the wrong results!

I am using the PISA dataset of 2012 to calculate the variance explained by ESCS (socio-economic status).

I followed the steps in the PISA DATA ANALYSES Manuel 2nd edition for SAS:

1. SAS® syntax for preparing a data file for the multilevel analysis (Box 16.3 at p.238-239).

2. SAS® syntax for running multilevel models with the PROC_MIXED_PV macro (Box 16.6 at p.242).

3. Calculating the results:
I used following equation:

1-(T2/T1)

T2 = is the intercept/between-school variance in model 2 and;
T1 = is the intercept/between-school variance in the empty model

________________________________________________________

First I tried using SPSS, than SAS, but either way I am not getting the same results as in the examples given by PISA.
For example: According to PISA, in Norway, 7% of variance in math performance is explained by ESCS. According to my calculations that is wrong, as much as 20% is explained by ESCS, and 25% by ESCS and MU_ESCS.

Interestingly when I run a one-way ANOVA, than the R-square is (almost) the same percentages as in the examples by PISA.

I only have a few more days left before submitting my BA Paper, and I really need to calculate the variance for countries that are not included in the examples on PISA's website.

jenjen91

New Member
Thanks for trying to help out!!!
From the PROC_MIXED_PV.sas file proved by PISA it lookes like all the 80 different weights are being used.

Lazar

Phineas Packard
Ok so I have the data to hand and I get variance explained of 7.41%

Note that I don't know sas at all but here is my R script. Note that I have all pisa waves in an sqlite3 database:
Code:
library(RSQLite)
library(mitools)
library(survey)
#set working directory
setwd("/Users/phparker/Dropbox/Databases/LSAY_DATSETS")
# Extract 2003 data
setwd("/Users/phparker/Dropbox/Databases/LSAY_DATSETS/")
pisa <- dbConnect(SQLite(), dbname="LSAY.sqlite")
dbListTables(pisa)

dbListFields(pisa, "PISA2012")

norway <- dbGetQuery(pisa, "SELECT * FROM PISA2012 WHERE cnt = 'NOR'")
brr <- norway[,grep("W_FST.*", names(norway))]

norway2 <- list()
for (i in 1:5){
tmp <- paste0("PV",i,"MATH")
tmpData<-	norway[,c("W_FSTUWT", "ESCS", "StIDStd", tmp)]
names(tmpData)[4] <- "math"
norway2[[i]] <- tmpData
}

norway2 <- imputationList(norway2)

dclust <- svrepdesign(ids = ~StIDStd, weights = ~W_FSTUWT,
data = norway2, repweights = brr[,-1],
type = "Fay", rho = 0.5)

M1 <- with(dclust, svyglm(scale(math) ~ scale(ESCS) - 1) )
summary(MIcombine(M1))[1,1]^2
Which gives:
Code:
Multiple imputation results:
with(dclust, svyglm(scale(math) ~ scale(ESCS) - 1))
MIcombine.default(M1)
results         se    (lower    upper) missInfo
scale(ESCS) 0.2723315 0.02066994 0.2317947 0.3128684      5 %
[1] 0.07416447

jenjen91

New Member
Wow, thats is very interesting! I tried to replicate your code in R, but I hard a hard time figure out how R works. Thanks for the help, I will try to compare the codes and see if I can find the difference.

Lazar

Phineas Packard
Ok so there are a few issues here:

1. You have to at least use the W_FSTUWT weight.
2. The PISA results that you allude to (the same as what I got) are fixed effects model not multilevel models. They deal with the complex nature of the data using replicate weights. In addition they are also running each analysis 5 times. Once for each plausible value and then they integrate the results afterwards using Rubin's multiple imputation rules. SAS and SPSS should handle all of this.
3. What it looks like you want to do is you want to do a contextual model? That is look at the degree to which escs and school average escs predicts math achievement? Is this right?