text:
# getData -- get the underlying samples

## Synopsis

• Usage:
Data = getData(Sample)
• Inputs:
• S, an instance of the type Sample, to extract data from
• Outputs:
• Data, a list, of all samples in object

## Description

An object of type Sample contains several pieces of information about the random sample, namely: the name of the model used to generate the sample, values of the model's parameters used to generate the sample, the size of the sample (that is, the number of data points in it), and the data itself. The function getData is used to extract the sample data.

In the example below, we obtain a sample of size $5$ from the ER model with parameter values $(3,4,0.1)$

 i1 : s = sample(ER(3,4,0.1),5) o1 = Sample{...4...} o1 : Sample i2 : getData s 2 2 2 3 2 3 2 2 3 2 3 o2 = {{x , x , x x , x x }, {x , x , x x , x x x , x x , x x x }, {x x , 3 2 1 3 1 2 3 2 2 3 1 2 3 1 2 1 2 3 1 2 ------------------------------------------------------------------------ 3 2 2 x x , x x }, {x , x x }, {x x , x x x }} 1 3 2 3 2 2 3 1 2 1 2 3 o2 : List

## Ways to use getData :

• "getData(Sample)"

## For the programmer

The object getData is .

format: plain


text:
## Toll Index October 2017 – Jamaican growth

German GDP growth in the third quarter of 2017 came in at +.8% (season and calendar adjusted, +2.8% only with calendar adjustment) on a quarter on quarter basis. The Toll Index for October shows that growth continues as the outbound truck activity rose +4.3% calendar adjusted in comparison to September 2017 and +6.1% compared to September 2016.

The data is available at the Data Repository of the IDSC – Research Data Center of IZA. The paper we first studied the properties of these data is published at the Journal of Forecasting.

The Toll Index has been widely covered in national and international press (selection):

## Toll-Index-September-2017: exceptional both in level and in trend

The Toll Index (based on fresh data just released by the German BAG)  shows that German border crossing heavy truck activity is coming out of a seasonal August deep in a very impressive fashion recording the best September ever. On a per working day basis the index rose 17.68% month on month and 9.5% year on year. In September, on average the index rose by 14% month on month in the past so this is an exceptionally steep rise.

The data is available at the Data Repository of the IDSC – Research Data Center of IZA. The paper we first studied the properties of these data is published at the Journal of Forecasting.

The Toll Index has been widely covered in national and international press (selection):

Posted in data, nowcasting | | Comments Off on Toll-Index-September-2017: exceptional both in level and in trend

## German Export in August and the Toll Index

German Exports in the month of August were announced today by Destatis. The data is still preliminary but it does suggest an increase of 7.2% with respect to August 2016. According to the German Press (for example finanzen.net) these results “surprised German Economists”.

On September 21 I published the Toll index for the month of August. How much less surprised would you have been if you used the Toll Index to forecast exports?

If $E$ are monthly exports (in billions of Euros, $T$ is the Toll Index (border crossing trucks), $\Delta$ is the monthly difference operator and $m_i: i= 1,\ldots,12$ monthly dummies  and you estimated a model like this:

$\Delta(E) = \alpha \Delta(T)+\displaystyle{\sum_{i=1}^{12}} \beta_{i}m_i,$

you would have come up with +5.7% which would imply you would be much less surprised. There is still some difference to 7.2% but first this is a preliminary estimate, second there is some noise in the world and third the model is build and run in under 5 minutes. Of course you would be almost just as informed if you just saw this graph:

Posted in Big Data, data, economics, nowcasting, technology | Tagged , , | Comments Off on German Export in August and the Toll Index

## Galton’s Crowd forecasts #BTW2017 outcome better than Pollsters

Some time on Saturday September 23 2017 wahlrecht.de put up an elections prediction market which closed on Sunday September 24 2017 at 14:00 hrs, four hours before the voting stopped. A total of 4,144 people provided their guesses. How did the crowd do? Much better than the pollsters.

Before we go on to the results let me remind you of the Diversity Prediction Theorem,  a very easy to verify fact. If $N$ individuals are asked to guess a true measurement $T$ and $s_i: i=1,\ldots,N$ are their single guesses and $c$ is the mean of their guesses (also called the crowd guess) then the theorem states:

$(c-T)^2=\frac{1}{n} \displaystyle{\sum_{i=1}^{N}} (s_i-T)^2 - \frac{1}{n} \displaystyle{\sum_{i=1}^{N}} (s_i-c)^2$

which simply reads that the square error of the crowd’s guess equals the mean squared individual error minus the variance of the individual errors. Which means that the crowd does better than any of its members as variance increases.

This works well when the crowd members are uncorrelated and fails miserably when not as manias, information cascades and bubbles show.

On a party by party basis we can thus compute the mean individual squared error vs the crowd squared error and we can do so for the crowd of $4,144$ laymen as well as for the crowd of the 13 pollsters whose last polls can be found here together with the guesses of the $4,144$ lay people.

In each cell the left hand side number is the mean individual (polling firm or layperson) squared error and and on the right you see the crowd’s (of firms or person) squared error. Clearly with the exception of Die Grünen and Die Linke the crowd guessed much better than the pollsters.

One can legitimately ask whether the pollsters should ask people to predict the elections instead of (or in addition to) asking them to reveal their preference. The former is immune to strategic response whereas the latter is not. Greetings from Galton.

Caveats: The laymen were still guessing on Sunday perhaops giving them an advantage over pollsters. The crowd of the pollsters is 13 strong whereas the laymen are 4,144 also an advantage for the laypeople. On the other hand the pollsters have the entire statistical toolkit at their disposal and a larger sample than each of the laymen.

For the convenience of Stata users here are the datasets:

Posted in Big Data, data, markets, nowcasting, science | | Comments Off on Galton’s Crowd forecasts #BTW2017 outcome better than Pollsters


text:
# Source code for picamera.frames

# vim: set et sw=4 sts=4 fileencoding=utf-8:
#
# Python camera library for the Rasperry-Pi camera module
# Copyright (c) 2013-2015 Dave Jones <dave@waveform.org.uk>
#
# Redistribution and use in source and binary forms, with or without
# modification, are permitted provided that the following conditions are met:
#
#     * Redistributions of source code must retain the above copyright
#       notice, this list of conditions and the following disclaimer.
#     * Redistributions in binary form must reproduce the above copyright
#       notice, this list of conditions and the following disclaimer in the
#       documentation and/or other materials provided with the distribution.
#     * Neither the name of the copyright holder nor the
#       names of its contributors may be used to endorse or promote products
#       derived from this software without specific prior written permission.
#
# THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
# AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
# IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
# ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE
# LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
# CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
# SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
# INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
# CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
# ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE
# POSSIBILITY OF SUCH DAMAGE.

from __future__ import (
unicode_literals,
print_function,
division,
absolute_import,
)

# Make Py2's str and range equivalent to Py3's
str = type('')

import warnings
from collections import namedtuple

from picamera.exc import (
mmal_check,
PiCameraError,
PiCameraMMALError,
PiCameraValueError,
PiCameraRuntimeError,
)

[docs]class PiVideoFrameType(object):
"""
This class simply defines constants used to represent the type of a frame
in :attr:PiVideoFrame.frame_type. Effectively it is a namespace for an
enum.

.. attribute:: frame

Indicates a predicted frame (P-frame). This is the most common frame
type.

.. attribute:: key_frame

Indicates an intra-frame (I-frame) also known as a key frame.

Indicates an inline SPS/PPS header (rather than picture data) which is
typically used as a split point.

.. attribute:: motion_data

Indicates the frame is inline motion vector data, rather than picture
data.

"""
frame = 0
key_frame = 1
motion_data = 3

[docs]class PiVideoFrame(namedtuple('PiVideoFrame', (
'index',         # the frame number, where the first frame is 0
'frame_type',    # a constant indicating the frame type (see PiVideoFrameType)
'frame_size',    # the size (in bytes) of the frame's data
'video_size',    # the size (in bytes) of the video so far
'split_size',    # the size (in bytes) of the video since the last split
'timestamp',     # the presentation timestamp (PTS) of the frame
'complete',      # whether the frame is complete or not
))):
"""
This class is a :func:~collections.namedtuple derivative used to store
information about a video frame. It is recommended that you access the
information stored by this class by attribute name rather than position
(for example: frame.index rather than frame[0]).

.. attribute:: index

Returns the zero-based number of the frame. This is a monotonic counter
that is simply incremented every time the camera starts outputting a
new frame. As a consequence, this attribute cannot be used to detect
dropped frames. Nor does it necessarily represent actual frames; it
will be incremented for SPS headers and motion data buffers too.

.. attribute:: frame_type

Returns a constant indicating the kind of data that the frame contains
(see :class:PiVideoFrameType). Please note that certain frame types
contain no image data at all.

.. attribute:: frame_size

Returns the size in bytes of the current frame. If a frame is written
in multiple chunks, this value will increment while :attr:index
remains static. Query :attr:complete to determine whether the frame
has been completely output yet.

.. attribute:: video_size

Returns the size in bytes of the entire video up to this frame.  Note
that this is unlikely to match the size of the actual file/stream
written so far. This is because a stream may utilize buffering which
will cause the actual amount written (e.g. to disk) to lag behind the
value reported by this attribute.

.. attribute:: split_size

Returns the size in bytes of the video recorded since the last call to
either :meth:~PiCamera.start_recording or
:meth:~PiCamera.split_recording. For the reasons explained above,
this may differ from the size of the actual file/stream written so far.

.. attribute:: timestamp

Returns the presentation timestamp (PTS) of the frame. This represents
the point in time that the Pi received the first line of the frame from
the camera.

The timestamp is measured in microseconds (millionths of a second).
When the camera's clock mode is 'reset' (the default), the
timestamp is relative to the start of the video recording.  When the
camera's :attr:~PiCamera.clock_mode is 'raw', it is relative to
the last system reboot. See :attr:~PiCamera.timestamp for more
information.

.. warning::

Currently, the camera occasionally returns "time unknown" values in
this field which picamera represents as None. If you are
querying this property you will need to check the value is not
None before using it. This happens for SPS header "frames",
for example.

.. attribute:: complete

Returns a bool indicating whether the current frame is complete or not.
If the frame is complete then :attr:frame_size will not increment
any further, and will reset for the next frame.

.. versionchanged:: 1.5
Deprecated :attr:header and :attr:keyframe attributes and added the
new :attr:frame_type attribute instead.

.. versionchanged:: 1.9
Added the :attr:complete attribute.
"""

__slots__ = () # workaround python issue #24931

@property
def position(self):
"""
Returns the zero-based position of the frame in the stream containing
it.
"""
return self.split_size - self.frame_size

@property
def keyframe(self):
"""
Returns a bool indicating whether the current frame is a keyframe (an
intra-frame, or I-frame in MPEG parlance).

.. deprecated:: 1.5
Please compare :attr:frame_type to
:attr:PiVideoFrameType.key_frame instead.
"""
warnings.warn(
'PiVideoFrame.keyframe is deprecated; please check '
'PiVideoFrame.frame_type for equality with '
return self.frame_type == PiVideoFrameType.key_frame

@property
"""
Contains a bool indicating whether the current frame is actually an
SPS/PPS header. Typically it is best to split an H.264 stream so that
it starts with an SPS/PPS header.

.. deprecated:: 1.5
Please compare :attr:frame_type to
:attr:PiVideoFrameType.sps_header instead.
"""
warnings.warn(


text:
Environmental

The key to a comprehensive environmental assessment is the subsurface investigation. Ground penetrating radar (GPR) plays an integral part by providing a non-intrusive means of examining the subsurface for environmental hazards such as soil contamination, underground storage tanks and drums. GPR can delineate landfills and pathways for contaminant flow, as well as conduct hydrogeologic investigations such as water table mapping.
Environmental Assessment
• Site Assessment
• Underground Storage Tanks (USTs) and Drums
• Water Table Mapping
Environmental Utility Locating
• Utility Mapping
• Drilling Clearance
• Landfill Delineation

Environmental Assessment

Environmental Assessment: Site Assessments

Environmental scientists and land developers use GPR and EM to assist in their redevelopment efforts. These proven geophysical methods allow professionals to conduct a non-invasive investigation of the surface at a relatively low cost.

This data shows a clearly defined boundary of potential toxic chemicals. The red line is a pipe that runs along the left edge of the data set. This data was collected with a 400 MHz antenna and post-processed in RADAN 7.

Environmental Assessment: Underground Storage Tanks (USTs) and Drums

Civil engineers, environmental consultants and environmental remediation specialists use GPR and EM to locate the position and impact of underground storage tanks.

2D GPR profile shows 4 USTs from a Mobil gas station in Colchester, Vermont. Traditional USTs were constructed in steel, recently they have been modernized to fiberglass. This data image is an example of fiberglass USTs, the GPR can denote the top of the UST as well as the ‘product’ levels it contains.

Environmental Assessment: Water Table Mapping

Hydrogeologists use GPR to determine the depth to water table and to identify potential pathways for subsurface flow.

This data illustrates a well-defined water table. Elevation data has been corrected using topography data in RADAN 7. Data was collected with the SIR 4000 and 200 MHz antenna.

Environmental Utility Locating

Environmental Utility Locating: Utility Mapping

Utility locators and engineers can locate the depth and position of metallic and non-metallic pipes in real time using ground penetrating radar technology. GPR can enhance one’s overall understanding of subsurface targets and obstructions.

Data illustrates a 3-dimensional view of a survey of existing and abandoned utilities presented in RADAN 3D. Note the broken linear feature that denotes an abandoned utility. Data collected with a 400 MHz antenna.

Environmental Utility Locating: Drilling Clearance

Ground penetrating radar can detect what lies beneath the surface before drilling and trenching efforts. GPR technology allows users to safely identify subsurface features and utilities, and avoid costly or dangerous hits.

Data set shows three utilities at varying depths. Utilities are located just above a clearly defined bedrock horizon. This data was collected with a 400 MHz antenna.


text:
# Lesson 2

Multiplying Powers of Ten

### Lesson Narrative

Students make use of repeated reasoning to discover the exponent rule $$10^n \boldcdot 10^m = 10^{n+m}$$ (MP8). At this time, students develop rules for positive exponents. In subsequent lessons, students will extend the exponent rules to cases where the exponents are zero or negative. Students reason abstractly and quantitatively when applying exponent rules, pausing to consider the meaning of quantities, not just how to compute them (MP2).

### Learning Goals

Teacher Facing

• Generalize a process for multiplying exponential expressions with the same base, and justify (orally and in writing) that $10^n \boldcdot 10^m = 10^{n+m}$.

### Student Facing

Let’s explore patterns with exponents when we multiply powers of 10.

### Required Preparation

Create visual displays of the exponent rule $$10^n \boldcdot 10^m = 10^{n+m}$$ to be displayed for all to see throughout the unit. Here is a sample visual display:

### Student Facing

• I can explain and use a rule for multiplying powers of 10.

Building On

Building Towards


text:
# How can I plot a circular region and lines?

I am having difficulty plotting something along these lines in Mathematica.

I will explain what is going on here. Basically, I am solving an interval equation $ax=b$, in the first example $a = [1, 20]$ and $b =[3, 4]$ and therefore $b/a = [3/20, 4]$. As you can see from the graph I want a circle representing the projective line. A straight line will come off the two end points of the interval which represents the solutions and those lines will go through the center of the circle and the region bounded by the lines should be shaded in.

Similarly the other example below is an unbounded one where $a = [-1,2]$ and $b = [5,100]$ and hence $b/a = (-\infty, -5] \cup [5/2, \infty]$. Same type of plot though, the lines come off $-5$ and $5/2$ on the x axis and go through the center of the circle and the bounded regions are shaded.

-
How are the circles constructed? Are they unit circles centered at {0, -1}. –  seismatica Aug 18 '14 at 18:54
Yes they are :) –  Kadir Aug 18 '14 at 19:17
Oh also letting you know that my solution to the interval equation will be given as a system of inequalities so for the first example above I will be given the following: x1 >= 0 && 1*x1 <= 4 && 20*x1 >= 3 || x1 <= 0 && 20*x1 <= 4 && 1*x1 >= 3 –  Kadir Aug 18 '14 at 19:20
It would help to know what you have tried, to avoid unnecessarily spending time explaining things you already did. –  Jens Aug 18 '14 at 19:30
I am completely new to mathematica I have no clue to be honest and need this desperately as soon as possible. Could someone please help? My idea is to figure out the points from the inequalities somehow then make a function that computes a straight line that goes through that point on the x axis and (0,-1) then fill in the region :/ –  Kadir Aug 18 '14 at 19:39

This is not a complete solution but it may help.

Manipulate[
Graphics[{{Circle[{0, -1}, 1]}, {Blue,
Disk[{0, -1}, 1, {ArcTan[1/b], ArcTan[1/a]}]}, {Blue,
Disk[{0, -1},
1, \[Pi] + {ArcTan[1/b],
If[b < 0, \[Pi], 0] + ArcTan[1/a]}]}, {Green,
Line[{{{a, 0}, {0, -1}}, {{b, 0}, {0, -1}}}]}}, Frame -> True,
PlotRange -> {{-5, 5}, {-3, 1}}, Axes -> True], {a, .1, 3}, {b, -5,
5}]

-
This is incredible thanks! How do I modify it so I can see the Circle itself as well rather than just the shaded in areas? –  Kadir Aug 18 '14 at 20:06
If i substitute in -5, and 5/2 for a and b respectively, it doesn't exactly plot what I am looking for. Doesn't shade the bottom part? Looks really close though. Any idea how to fix that? Thanks] –  Kadir Aug 18 '14 at 20:13
@Kadir (1) add Circle[{0, -1}] to the list of directives. –  Mr.Wizard Aug 18 '14 at 20:26
@Kadir I have already add the circle. for the range -5 and 5/2, you may change the plot range to include the far distances of a or b. I have already did that. –  Algohi Aug 18 '14 at 20:30
@Kadir Don't forget to accept the answer if you like it. –  Jens Aug 18 '14 at 20:43


text:
# What is the derivative of  (cot^(2)x)?

Dec 18, 2015

$- 2 {\csc}^{2} x \cot x$

#### Explanation:

Treat ${\cot}^{2} x$ like ${u}^{2}$, where $u = \cot x$.

According to the chain rule,

$\frac{d}{\mathrm{dx}} \left[{u}^{2}\right] = 2 u \cdot u '$

Therefore,

$\frac{d}{\mathrm{dx}} \left[{\cot}^{2} x\right] = 2 \cot x \cdot \frac{d}{\mathrm{dx}} \left[\cot x\right]$

Know that $\frac{d}{\mathrm{dx}} \left[\cot x\right] = - {\csc}^{2} x$

Thus,

$\frac{d}{\mathrm{dx}} \left[{\cot}^{2} x\right] = - 2 {\csc}^{2} x \cot x$


text:
Tag Info

Hot answers tagged matplotlib

13

I'll just be cheeky and post a slightly modified version of my SO answer here. So first things first, you want to get rid of the loops. They are slow to execute. The first loop: for x in range(rows): for y in range(cols): if Z[x][y] == 1: if (N[x][y] < 2) or (N[x][y] > 3): Z[x][y] = 0 ...

11

matplotlib adding stuff to the current figure is because you are not using the OO-interface. It is slightly clunkier, but allows way more freedom In my view, you are mixing several things up. I would seperate the generation of the timings, the aggregation of these results and the plotting. In this way, if you want to change to another plotting library (...

10

I've never used numpy or matplotlib, so I can only speak to issues of style. You're allowing for far too much nesting here. Your code consists of a giant, dense, deeply nested chunk. As a result, the eyes have very few good landmarks to rest on; making your code hard to read. Put your existing function definitions at the very top level outside of any ...

8

You're not really using numpy at the moment. If we use the simple 'translation table' below, your code would work if you just replaced the NumPy function with the Python equivalent: np.arange -> range, assuming your domain is just integers, np.fabs -> abs, np.cos -> math.cos, and, np.pi -> math.pi. Instead you want to take advantage of NumPy. Take ...

8

First, a style comment. On the internet, especially in programming, and in particular on this website, English is the lingua franca. So you should avoid mixing other languages and English. This way your code is the most transferable, re-usable and readable. Second, a comment on the algorithm itself. Your algorithm (and this includes any changes I make to it ...

8

I'm going to put this code into an editor, "proofread" it from top to bottom, give notes as I go, and then paste the final result to show the effect of the edits. Code editors don't usually wrap lines since linebreaks mean things in code. Consequently, most coding style guidelines suggest limiting your column width so the reader doesn't need to scroll ...

7

The biggest issue I see is code organization. There are two large tasks: Numerical Simulation Data Visualization They are mixed together in ways that create arbitrary dependencies and make extending and maintaining the code difficult. Names Assigning meaningful names to numbers will improve readability and reduce comments: right = 1 left = 2 up = 3 ...

7

This is my first Python script using argparse instead of sys for argument handling. I'm especially interested in whether the way I implemented it is up to par. You nailed it :-) Under the argument parsing I have a couple of BOLD_SNAKE_CASE variables which are pseudo constants. There's probably a better way to do this. That's the common practice in ...

7

Well, unfortunately you can't get any better than this (as far as I have read). And why would you use an iterator in this case ? I'd however change a bit of the structure of your code, which allows one to easily change the code if there's any need. I didn't changed too many things, just separated the logic into three different functions and added a for loop ...

7

You need not declare # -*- coding: utf-8 -*- in Python 3: UTF-8 is the default. Writing I = np.logspace(…) inside your curve_equation function is bad practice, because it hard-codes certain x-values for your plot. In fact, your plot is wrong, because the calculations are for $I_s$ values ranging from 102 A to 105 A (due to the inner I = np.logspace(np....

7

Nice. Here are some observations: consider using an IntEnum: This helps remove "magic" numbers from the source code. (A year from now, will you remember that if self.selected_type == 3 is a check to see if the mode is 'File'?) from enum import auto, IntEnum class SoundType(IntEnum): NOTE = auto() DESIGN = auto() FILE = auto() Then later ...

6

Style Your docstring at the top of the seq function should be moved underneath the function signature, like this: def seq(start, end, step): """Function to iterate with decimal step size""" ... The variable map, should be written like this, to improve clarity and readability: map = Basemap( projection="merc", resolution = 'h', ...

6

DISCLAIMER: I don't know cython and have never used it, so if any of my advice doesn't apply because of cython limitations, feel free to disregard it. Counting neighbors in a game board is very easy to do via convolution with the proper kernel. Below, I used SciPy's convolve function, not the much faster fftconvolve, because the latter only works on floats ...

6

This code is really nice, and the output is very good looking! I do have a few nitpicky things that I want to cover. Why are you storing the usage in a variable named usage? This should be in the file's main docstring. You don't need to print usage under if __name__ == "__main__":. Is # encoding: utf=8 really needed? It doesn't seem like you're using any ...

6

First off, this: Q = 1.0 ; T = 1.0 Should be expanded to this: Q = 1.0 T = 1.0 And, if Q and T aren't constants, they should be renamed to q and t. Secondly, you're missing whitespace in lots of places. For example, this: def v(self,i,j,k): Should be, again, expanded to this: def v(self, i, j, k): You should also have whitespace between mathematical ...

6

I want to start this review by explicitly saying that I don't have a deep knowledge when it comes to matplotlib but I'll give it a try anyway Have in mind that I just reviewed the code as is, keeping the same logic / functionality it has at the moment. Unused imports Don't import modules you're not using (decimal, pandas, numpy). Imports grouping From ...

6

coding style Your code is almost pep-8 compliant. There are a few spaces missing after comma's, but all in all this is not too bad. I myself use black to take care of this formatting for me. some of the variables names can be clearer. What does nc1 mean for example magic numbers The number 3, 2 and 6 are the number of rows and columns on the grid. Better ...

5

Using % for string formatting is deprecated, you should be using str.format instead. The str.format methods also allows you to do some cool things as well. Here's an example of it in action: # No positional or named arguments print "{} {}".format("Hello", "world") # Positional arguments (*args) print "{1} {0}".format("world", "Hello") # Named arguments (**...

5

Let's start with the obvious remarks about whitespace: Whitespace is important in Python. You got trailing whitespaces all over the place and you use an indentation of 2 spaces where 4 is prescribed by the official PEP8 Style Guide. When talking about sticking to best practices in Python, starting with PEP8 is a good idea. There's a lot more violations ...

5

Superficial issues """Function to iterate…""" is a docstring, which belongs inside the function definition, not before it. """Divide global area into grid cells""" should be written as a comment (# Divide global area into grid cells). The way you formatted map = Basemap(…) makes it hard to see that it's one statement over five lines. The four ...

5

Let us start with some code review, before improving the performance, and finish off with a comparision against the original code. Code review As a gesture to us reviewing your code, it would have been nice to know which modules where external to a default installation, as I had to plunder a little to install the astropy and jplephem module. However I got ...

5

First of all some style comments: Please use spaces after commas and around operators – It is hard to read your sequence of values for the X and y lists when they are all compressed. And the same applies for the ([2,7]) when setting limits. It reads a lot better with [3, 3.2, 3.5, ..., 5.8, 5.9, 5] and ([2, 7]) Choose better names for variables – Why have ...

5

Compute legendre(n) once per loop iteration instead of twice. Use np.hypot instead of np.sqrt(x**2 + z**2). Don't use phivec, just pass the arrays X and Y directly to phi, which is already almost vectorized. The only change that's needed is to use np.minimum and np.maximum instead of min and max respectively. Revised code: def phi(x, z, nlimit=35): r = ...

5

Here's my reworking of the calculation functions. Graphs look similar (didn't check details) when using the arange version of equidistant, but I think the linspace version is more balanced. The main thing when trying to use numpy is keeping track of dimensions. In this case points is an array of 8-20 values, x an array of 2001. def equidistant_points(...

5

Usually, when using for loops and numpy together, you’re probably doing it wrong. Some things you could replace in your code: Use np.arange(256) instead of np.zeros + for i in range(256); Use slicing instead of manually extracting values out of an array; Use comparisons and .sum() instead of manually counting the number of pixels of a certain color ...

5

Your function proc can be greatly reduced by using the fact that numpy functions are vectorized and most of them take the axis to act upon as an argument. This is certainly true for numpy.median and numpy.percentile. import numpy as np def proc(data): return (np.median(data, axis=0), np.percentile(data, 25, axis=0), np....

5

Your directions dictionary is only used by get_direction; why not put it inside the function rather than have it as a global? I'd maybe even put the dict inside get_chance_of_positions() and drop the get_direction() function entirely; it's short, only used once and doesn't require a docstring (its obvious how it works). Put dirs straight into the line as ...

5

Shebang #!/opt/anaconda3/bin/python This is suspicious. Usually you should just #!/usr/bin/env python and make sure that your environment uses the correct Python. Among other things, it'll make it easier for you to switch to system Python if you need it. Enums RABBIT = 0 FOX = 1 It's good that you're assigning symbols for these values, but you should ...

4

Why did you write your own functions for mean and weighted_median? It looks like you are doing fairly standard things with them, so I would use implementations from numpy or scipy instead. (And calling the median of a vector a times w a "weighted median" of vector a makes some sense but it is really just the regular vanilla median of a*w.) Your Game class ...

4

Try using matplotlib's LineCollection class. Here's an example. In your case, you might do: from matplotlib import pyplot as plt from matplotlib.collections import LineCollection ax = plt.gca() pts = myPath.reshape((-1,2)) # make a matrix of (x,y) pairs edges = LineCollection(pts) ax.add_collection(edges) plt.show()

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# A Global Characterization of the Fucik Spectrum for a System of Ordinary Diﬀerential Equations

(1)

## a System of Ordinary Differential Equations

Eugenio Massa a,∗,† and Bernhard Ruf b

a Departamento de Matem´atica, Instituto de Ciˆencias Matem´aticas e de Computa¸ao,

Universidade de S˜ao Paulo - Campus de S˜ao Carlos, Caixa Postal 668, 13560-970, S˜ao Carlos SP, Brazil.

e-mail: eugenio@icmc.usp.br

bDipartimento di Matematica, Universit`a degli Studi di Milano, Via Saldini 50, 20133 Milano, Italy.

e-mail: ruf@mat.unimi.it

Abstract

The Fuˇc´ık spectrum for systems of second order ordinary differential equations with Dirichlet or Neumann boundary values is considered: it is proved that the Fuˇc´ık spectrum consists of global C1 surfaces, and that through each eigenvalue of the linear system pass exactly two of these surfaces. Further qualitative, asymptotic and symmetry properties of these spectral surfaces are given. Finally, related problems with nonlinearities which cross asymptotically some eigenvalues, as well as linear-superlinear systems are studied.

Keywords and phrases: systems of ordinary differential equations, Fuˇc´ık spectrum for sys- tems, symmetry properties

### 1 Introduction

The notion of Fuˇc´ık spectrum was introduced for the scalar Laplace problem in [1] and [2]. It is defined as the set Σscal⊆ R2 of the points (λ+, λ) for which there exists a non trivial solution of the problem

−u00= λ+u+− λu in (0, 1)

Bu = 0 in {0; 1}

; (1.1)

here u±(x) = max{0, ±u(x)} and Bu = 0 represents Dirichlet or Neumann boundary conditions.

The Fuˇc´ık spectrum Σscalfor equation (1.1) is explicitly known and consists of unbounded curves in R2: more precisely, from every diagonal point (λk, λk) ∈ R2 (where λk, k ∈ N, denote the eigenvalues of −u00 with boundary conditions B) emanate two curves Ck±, with the property that the corresponding solutions have exactly k − 1 nodal points and are positive resp. negative

Corresponding author

The author was partially supported by Fapesp / Brazil.

1

(2)

near zero. The two curves may coincide, and in fact they always do for the Neumann problem, and for k even in the Dirichlet problem.

The corresponding partial differential equation

−∆u = λ+u+− λu in Ω

Bu = 0 in ∂Ω

, (1.2)

where Ω ⊂ Rn denotes a bounded domain, is more difficult to handle and the corresponding Fuˇc´ık spectrum is only partially known, namely:

- the so-called trivial part of the spectrum, corresponding to positive or negative solutions;

- if λk is a simple eigenvalue, then the Fuˇc´ık spectrum in (λk−1, λk+1)2 ⊂ R consists of two curves (maybe coincident) which pass through the point (λk, λk), see [3], [4];

- the first nontrivial curve, passing through the point (λ2, λ2), see [5].

In [6] the following generalization of the Fuˇc´ık spectrum to nonlinear elliptic systems was introduced:









−∆u = λ+v+− λv in Ω

−∆v = µ+u+− µu in Ω

Bu = Bv = 0 on ∂Ω

. (1.3)

It was shown there that, due to intrinsic symmetries of the system, one may assume that two of the four values λ+, λ, µ+, µcoincide. Thus, setting λ+ = µ+, the Fuˇc´ık spectrum of (1.3) can be defined as: bΣ = {(λ+, λ, µ) ∈ R3 ; (1.3) has a nontrivial solution }. It was also proved that near a simple eigenvalue of the system the Fuˇc´ık spectrum consists of exactly two (maybe coincident) 2-dimensional surfaces, which can be parametrized as λ+1,2, µ). Furthermore, by a variational characterization, the existence of an unbounded continuum of points in bΣ containing the point (λ2, λ2, λ2) was given.

Finally, it was shown that many of the properties of the scalar Fuˇc´ık spectrum continue to hold for its system counterpart (1.5), in particular, the properties related to the solvability of problems with “jumping nonlinearities” (that is nonlinearities with asymptotically linear, but different, growths at +∞ and −∞). In short, in dependence of the position of these asymptotic slopes with respect to the Fuˇc´ık spectrum bΣ one has: if these slopes lie in the regions in R3\ bΣ which contain points λ+= λ = µ of the diagonal, then one has solvability for every forcing term; on the other hand, if these slopes lie between two spectral surfaces emanating from the same eigenvalue (λk, λkλk) (in the case they are distinct), one obtains existence or nonexistence of solutions in dependence of the forcing terms.

These properties show the importance of a good knowledge of the shape of the Fuˇc´ık spec- trum.

In this paper we consider the Fuˇc´ık spectrum for a system of second order ordinary differential equations: we consider the problem









−u00= λ+v+− λv in (0, 1)

−v00= λ+u+− µu in (0, 1) Bu = Bv = 0 in {0; 1}

, (1.4)

(3)

and we call Fuˇc´ık spectrum the set

Σ =b (λ+, λ, µ) ∈ R3 such that λ±, µ≥ 0 and (1.4) has nontrivial solutions . (1.5) Throughout the paper we will always denote by 0 ≤ λ1 < λ2≤ λ3 ≤ ... ≤ λk≤ ... the eigenvalues of −dxd22 in H01(0, 1), when dealing with the Dirichlet problem, and in H1(0, 1) for the Neumann problem; moreover {φk}k=1,2,.. will be the corresponding eigenfunctions, taken orthogonal and normalized with kφkkL2 = 1 and φ1 > 0.

We recall here that the linear spectrum of a system coupled like (1.4), namely









−u00= λv in (0, 1)

−v00= λu in (0, 1) Bu = Bv = 0 in {0; 1}

, (1.6)

is composed of the eigenvalues λk and −λk (k = 1, 2, ..), with corresponding eigenfunctions the pairs (φk, φk) and (φk, −φk), respectively. As a consequence, the points (λk, λk, λk) belong to bΣ for k ≥ 1.

When restricting to dimension one, the techniques for ordinary differential equations become available, and this will allow to obtain a much more detailed description of bΣ than was possible for (1.3). In particular, the techniques which we use here are inspired from [7], where the interest was the Fuˇc´ık spectrum for a fourth order problem in dimension one with so-called Navier boundary conditions. Indeed, we remark that this equation is a special case of system (1.4) which is obtained by setting µ+= µ .

The results presented here will be in different directions. First (in section 3) we will derive some qualitative properties of the nontrivial solutions of (1.4):

Theorem 1.1. If (λ+, λ, µ) ∈ bΣ and (u, v) is a corresponding nontrivial solution of problem (1.4), then u and v have only simple zeros, of the same number, and of the same sign both in a neighborhood of 0 and of 1.

The main result of the paper will be to obtain a good description of the Fuˇc´ık spectrum Σ: first, through the implicit function theorem, we will obtain a local description of bb Σ in a neighborhood of one of its points (see lemma 4.1), and then, as a consequence, also a global description (see proposition 4.4).

Finally, by joining the global description with the results in theorem 1.1 and with the know- ledge of the linear spectrum (1.6) and of the scalar Fuˇc´ık spectrum Σscal, we will be able to establish

Theorem 1.2. Through any point (λk, λk, λk) ∈ R3, with k ≥ 2, pass two C1 surfaces in bΣ:

• the two surfaces corresponding to k ≥ 2 are always coincident with Neumann boundary conditions, and they are coincident if and only if k is even with Dirichlet boundary con- ditions;

• if the two surfaces corresponding to k ≥ 2 do not coincide, then their intersection contains an unbounded curve which passes through (λk, λk, λk);

(4)

• the surfaces corresponding to different values of k are disjoint;

• each of the surfaces is characterized by the number of (simple) interior zeros (in fact, k −1) of the corresponding nontrivial solutions and by their sign in a neighborhood of zero; they will therefore be denoted by bΣ+k and bΣk;

• these surfaces may be represented expressing the variable λ+ in terms of the other two, and they are monotone decreasing in the two variables and unbounded in the three directions;

• all points in bΣ belong to one of these surfaces (except for those corresponding to solutions which do not change sign, see relation (2.8)).

• in the Neumann case, if (λ+, λ, µ) ∈ bΣ±k with k ≥ 2, then pλµ> λk/4 and λ+ > λk/4 ;

in the Dirichlet case: if (λ+, λ, µ) ∈ bΣ+k (resp. bΣk) with k ≥ 2, then pλµ> λ[k/2] and λ+> λ[(k+1)/2]

resp.

µ> λ[(k+1)/2] and λ+> λ[k/2]

(here [ · ] is the integer part of its argument).

In section 4.3, we also present a result about the symmetries of the non trivial solutions:

Proposition 1.3. The nontrivial solutions corresponding to points in bΣ±k with k odd are sym- metric, in the sense that (u, v)(x) = (u, v)(1 − x).

Moreover, for Neumann boundary conditions, one has that if j ≥ 0 and k ≡ 1 (MOD 2j+1), then (u, v) are (1/2)j periodic in [0, 1].

Finally, in section 5, we will analyze some properties of the set (R+)3\ bΣ, motivated by the fact that this set is related to the solvability of nonlinear problems with jumping nonlinearities (see there for more details). Indeed, proposition 5.3 presents regions where we may guarantee solvability, as given in theorem 5.4.

We remark that also problems with nonlinearities which are asymptotically linear at −∞

and superlinear at +∞ may be treated by exploiting the properties of the Fuˇc´ık spectrum, as done for the scalar equation in [8] and in [9]: what is important for these applications is the knowledge of the asymptotic behavior of the curves in Σscal; in particular, the key point in order to have the possibility to obtain existence results for such a nonlinearity, is the presence of a

“gap at infinity” between the surfaces in bΣ corresponding to consecutive eigenvalues; the results in proposition 5.5 will guarantee the existence of such gaps, at least for Neumann boundary conditions.

(5)

A result for the first of these gaps was already obtained in [10], that is, existence of solution when the slopes at −∞ are positive and satisfy pλµ < λ2/4. We are confident that the better insight on bΣ obtained here will prove useful to improve this result.

To conclude this introduction, we remark that, although the results obtained here are much more detailed than for the multidimensional case [6], we cannot obtain an explicit description of bΣ as is possible for the scalar problem in dimension one. This is due to the fact that the interaction of the two functions u and v gives rise to more complicated phenomena: just to give an idea, observe that the nontrivial solutions of the scalar Fuˇc´ık problem in dimension one can be built by suitably gluing bumps of the form sin(√

λ+x) and − sin(√

λx); on the other hand, for the system, there appear intervals where u and v have opposite sign, and there the solutions have a component of the form of the hyperbolic functions sinh and cosh.

### 2 Some useful results about the Fuˇ c´ık spectrum

For the scalar problem in dimension one, it is known that the Fuˇc´ık spectrum Σscal may be explicitly calculated: for Dirichlet boundary conditions it is composed by the curves

Σ+2i≡ Σ2i: iπ

λ+ + iπ

λ = 1 , i ≥ 1 (2.1)

Σ+2i−1: iπ

λ+ +(i − 1)π

λ = 1 , Σ2i−1: (i − 1)π

λ+ + iπ

λ = 1, i ≥ 1 ; (2.2) for Neumann boundary conditions it is given by the curves

Σ+1 : λ+ = 0 , Σ1 : λ= 0 , (2.3) Σ+j ≡ Σj : (j − 1)π

2√

λ+ +(j − 1)π 2√

λ = 1 j ≥ 2, (2.4)

where the curve Σ+k (resp. Σk) corresponds to nontrivial solutions having k − 1 internal zeros and starting positive (resp. negative); observe that Σ±k passes through the point (λk, λk) (recall that λk= (kπ)2 in the Dirichlet case and λk = ((k − 1)π)2 in the Neumann case). Observe also that the asymptotes of these curves are located, for the Dirichlet case, at the values

λ = λi f or Σ2i−1, Σ2i, Σ+2i+1, (2.5) λ+ = λi f or Σ+2i−1, Σ2i, Σ2i+1, (2.6) and for the Neumann case at

λ= λj

4 and λ+= λj

4 for Σj. (2.7)

These simple computations may be found in [1], for Dirichlet boundary conditions, and in [8], for the periodic case, which is very similar to the Neumann case.

We also refer to [2], where the most important properties of the (scalar) Fuˇc´ık spectrum are presented for what concerns its application to the solvability of related nonlinear equations.

(6)

A survey of all these results was given in [11].

Since we are going to use several results from [6], we devote the last part of this section to briefly summarize them.

First, we remark that in (1.4) (and hence in ˆΣ) we are considering only nonnegative coeffi- cients, and identical coefficients for u+ and v+; the reasons for this choice are widely discussed in [6] and rely on the observation that, even if we use four parameters, the “interesting” points of the spectrum have coefficients which are all of the same sign and which, through a suitable rescaling, may be transformed into an equivalent point of this form (i.e. lying in ˆΣ).

Furthermore, for points in ˆΣ, the following lemma was proved:

Lemma 2.1. Let (λ+, λ, µ) ∈ bΣ and (u, v) be a corresponding nontrivial solution, then:

• both u and v change sign or none of the two;

• if none of the two changes sign then λ+= λ1 or pλµ= λ1;

• if both change sign, then λ+> λ1 and pλµ> λ1.

This lemma implies that bΣ is composed of a trivial part

Σbt= n

+, λ, µ) ∈ R3 such that: λ+= λ1 orp

λµ= λ1

o

, (2.8)

corresponding to solutions which do not change sign, and a nontrivial part ˆΣnt (to which we may concentrate our attention) which we define as

Σbnt=(λ+, λ, µ) ∈ R3 such that:

λ±, µ> 0 and (1.4) has nontrivial solutions which (both) change sign . (2.9) Finally, we remark that bΣnt has several symmetries, in particular we have

Lemma 2.2. If (λ+, λ, µ) ∈ bΣnt with corresponding nontrivial solutions (u, v), then 1. (λ+, µ, λ) ∈ bΣnt with corresponding nontrivial solutions (v, u);

2.

µ, λ+ qµ

λ, λ+ qλ

µ

∈ bΣnt, with corresponding nontrivial solutions (−u, − qλ

µv).

For what concerns the relationship between Σscal and bΣ we recall

Lemma 2.3. If (λ+, λ) ∈ Σscalwith corresponding nontrivial solution u, then (λ+, λ, λ) ∈ bΣ with corresponding nontrivial solutions (u, u).

(7)

### 3 Some properties of the nontrivial solutions

In this section we will obtain some preliminary lemmas about the non trivial solutions of a system like (1.4); by joining these results, we obtain theorem 1.1.

Lemma 3.1. Weak solutions of problem (1.4) are in fact classical solutions.

Proof. Weak solutions of problem (1.4) are by definition in H1(Ω), then the right hand side in the equations is in H1 too; so one gets, by a boot strap argument, that the solutions are in H3(Ω); this, in dimension one, gives solutions in C2( ¯Ω), and so they are classical solutions.

Lemma 3.2. Let c, d ∈ L(0, 1) with c, d > 0 a.e. and (u, v) be a nontrivial solution of the boundary value problem (BVP for short)









−u00= c(x)v in (0, 1)

−v00= d(x)u in (0, 1) Bu = Bv = 0 in {0; 1}

. (3.1)

Then for no point x ∈ [0, 1] may hold:

u(x) ≥ 0, u0(x) ≥ 0, v(x) ≤ 0, v0(x) ≤ 0 , (3.2) nor u(x) ≥ 0, u0(x) ≤ 0, v(x) ≤ 0, v0(x) ≥ 0 , (3.3) nor u(x) ≤ 0, u0(x) ≤ 0, v(x) ≥ 0, v0(x) ≥ 0 , (3.4) nor u(x) ≤ 0, u0(x) ≥ 0, v(x) ≥ 0, v0(x) ≤ 0 . (3.5) Moreover:

• for ¯x ∈ [0, 1],

(Ai) if u(¯x) = 0 (or v(¯x) = 0) then u0(¯x)v0(¯x) > 0, (Aii) if u0(¯x) = 0 (or v0(¯x) = 0) then u(¯x)v(¯x) > 0;

• for x1, x2 ∈ [0, 1], x1 < x2,

(Bi) if u0(x1) = u0(x2) = 0 and u0(x) 6= 0 for x ∈ (x1, x2) then u(x1)u(x2) < 0, (Bii) if u(x1) = u(x2) = 0 and u(x) 6= 0 for x ∈ (x1, x2) then u0(x1)u0(x2) < 0,

and the same holds for v.

Proof. For the system (3.1), uniqueness of solution for the corresponding initial value problem (IVP for short) holds and then we immediately see that a nontrivial solution may have u(x) = v(x) = u0(x) = v0(x) = 0 in no point x ∈ [0, 1].

(8)

By integrating the equations in (3.1) we get, for x, x0 ∈ [0, 1], u(x) = u(x0) + u0(x0)(x − x0) −

Z x x0

1

Z ξ1

x0

c(ξ2)v(ξ2)dξ2, (3.6) v(x) = v(x0) + v0(x0)(x − x0) −

Z x x0

1

Z ξ1

x0

d(ξ2)u(ξ2)dξ2, (3.7) u0(x) = u0(x0) −

Z x x0

c(ξ1)v(ξ1)dξ1, (3.8)

v0(x) = v0(x0) − Z x

x0

d(ξ1)u(ξ1)dξ1. (3.9)

We first prove (3.2) with strict inequalities: suppose the inequalities hold strictly in x < 1 and let x1 > x be the first point (if any) where one of the two functions has zero derivative;

observe that u and v maintain their sign in (x, x1] and then by (3.8) u0(x1) > 0 and by (3.9) v0(x1) < 0: this implies that u and v may never satisfy the boundary condition at 1.

By the same argument one sees that if (3.3) holds with strict inequalities in x > 0 then u and v may never satisfy the boundary condition at 0.

Then observe that if u(x) = 0, then in a right neighborhood of x it assumes the sign of u0 in the same neighborhood; if u0(x) = 0, then in a right neighborhood of xit assumes, by (3.8), the opposite sign of v in the same neighborhood, and so on; then one may relax all inequalities in (3.2) since even if only one of the four is strict then in a right neighborhood all become strict;

the same considerations for left neighborhoods of x allow to do the same with (3.3).

Finally, equations (3.4) and (3.5) follow by linearity (or by exchanging the role of u and v).

The proof of (3.2) and (3.4) in 1 and of (3.3) and (3.5) in 0 is superfluous by virtue of the boundary conditions: as an example, observe that in the Dirichlet case (3.2) is equivalent to (3.5) in 0 and 1.

Now, (Ai) and (Aii) follow by combining the four relations; we just give an example of this (all the functions in the following will be evaluated in ¯x): let u = 0; if v ≤ 0 use (3.2) and (3.3) and obtain that u0 = 0 gives contradiction for any v0, while u0 > 0 gives v0 > 0 by (3.2) and u0 < 0 gives v0< 0 by (3.3); for v ≥ 0 the same is obtained by using (3.4) and (3.5).

To prove (Bi) observe that, by (Aii), v(x1) 6= 0 6= v(x2) and, by (3.8), 0 = u0(x2) = −

Z x2

x1

c(ξ1)v(ξ1)dξ1, (3.10)

implying that v changes sign in (x1, x2); if it changes sign once then v(x1)v(x2) < 0 and then, by (Aii), u(x1)u(x2) < 0 as claimed. Now observe that u may have no more than one zero in (x1, x2) since otherwise there would exist also a point ¯x ∈ (x1, x2) with u0(¯x) = 0; then we just need to show that it may not happen that v changes sign more then once in (x1, x2) while u does not change sign: actually in this case, since u v > 0 in x1 and x2 by (Aii), there would exist an interval [x3, x4] ⊆ (x1, x2) such that u v < 0 in (x3, x4) and v(x3) = v(x4) = 0; but this contradicts (Aii) since then there would exist also a point ¯x ∈ (x3, x4) with v0(¯x) = 0 and u(¯x)v(¯x) < 0.

Finally, (Bii) is trivial since by (Ai) the zeros of u are simple and in the given hypotheses u has the same sign in a right neighborhood of x1 and in a left neighborhood of x2.

(9)

Proposition 3.3. If (λ+, λ, µ) ∈ bΣ and (u, v) is a corresponding nontrivial solution of prob- lem (1.4), then the same conclusions of the above lemma hold for (u, v); in particular u and v have only simple zeros and the same sign both in a neighborhood of 0 and of 1.

Proof. The non trivial solution satisfies problem (3.1) with c(x) = λ+χ{v≥0}(x) + λχ{v<0}(x) and d(x) = λ+χ{u≥0}(x) + µχ{u<0}(x). (Here and in the following χA(x) is the characteristic function of the set A, and then satisfies the hypotheses of Lemma 3.2).

In particular, (Ai) in lemma 3.2 implies that the zeros are always simple, while (Ai) and (Aii) imply that u v > 0 both in a neighborhood of 0 and of 1.

The results in the previous lemmas allow to obtain the following important result:

Proposition 3.4. If (λ+, λ, µ) ∈ bΣnt and (u, v) is a corresponding nontrivial solution of problem (1.4), then u and v have the same number of (simple) zeros.

Proof. Let x1 be the first stationary point of u: for Neumann boundary conditions this is 0, while for Dirichlet boundary conditions we may see that there is no zero of u or v in (0, x1]:

actually u and u0 maintain their sign in (0, x1), by (Ai) in lemma 3.2 u0(0)v0(0) > 0, and then if v had a zero x ∈ (0, x1] we would have u0(x)v0(x) ≤ 0 contradicting (Ai) in lemma 3.2.

Now let x2 be the second stationary point of u: by (Bi) in lemma 3.2 u(x1)u(x2) < 0 and then u has exactly one zero xu ∈ (x1, x2): we claim that v has a unique zero in [x1, x2] too:

actually it may not have an even number of zeros since by (Aii) in lemma 3.2 v(x1)v(x2) < 0, and if it had three or more, then it would also have two stationary points and for at least one of them vu ≤ 0, contradicting (Aii) in lemma 3.2. The same argument may be applied to the following intervals between stationary points and, finally, we have no zero between the last stationary point of u and 1 by the same argument used above.

Then, we conclude that the zeros of u and v are exactly one for each interval between two stationary points of u, and so we proved the claim.

Remark 3.5. We observe that the results in this section may be described in terms of the phase plane: actually, the equations from (3.2) to (3.5) simply say that u and v may never lie in opposite (closed) quadrants of the phase plane.

With this interpretation, it is simple to deduce intuitively the properties in lemma 3.2 and also proposition 3.4.

### 4 Construction of b Σ

nt

In this section, we will build the non trivial part of the Fuˇc´ık spectrum bΣnt; in order to do this, we will first construct a related set in R4. We consider the initial value problem

(IV P )









−u00 = λ+v+− λv

−v00= λ+u+− µu

(u, v, u0, v0)(0) = (u0, v0, u00, v00)

, with λ+, λ, µ ≥ 0 .

(10)

Then we define, for Dirichlet boundary conditions, the sets

Σe± =

+, λ, µ, s) ∈ (R+)3× R : the solution (u, v) of IVP with (u0, v0, u00, v00) = (0, 0, ±1, s) satisfies u(1) = v(1) = 0

(4.1)

and for Neumann boundary conditions

Σe±=

+, λ, µ, s) ∈ (R+)3× R : the solution (u, v) of IVP with (u0, v0, u00, v00) = (±1, s, 0, 0) satisfies u0(1) = v0(1) = 0

. (4.2)

Then we will denote by

Σb±= {(λ+, λ, µ) ∈ (R+)3 : ∃ s ∈ R such that (λ+, λ, µ, s) ∈ eΣ± } , (4.3) and so bΣ = bΣ+∪ bΣ.

In particular, we only need to study one of the two components (say eΣ+), since the other one may be found by exploiting the symmetry 2 in lemma 2.2.

4.1 Local study of eΣ+

In the following lemma we will use the implicit function theorem to describe eΣ+ in a neighbor- hood of one of its points.

Lemma 4.1. Given (¯λ+, ¯λ, ¯µ, ¯s) ∈ eΣ+ such that the corresponding nontrivial solutions (¯u, ¯v) change sign (both), then eΣ+ is locally of the form (λ+, µ), λ, µ, s(λ, µ)), where (for a suitable ε > 0)

λ+, s

: (¯λ− ε, ¯λ+ ε) × (¯µ− ε, ¯µ+ ε) → R2 (4.4) is a C1 function of λ and µ; moreover

∂λ+

∂λ(¯λ, ¯µ) = −R1 0(¯v)2 R1

0(¯u+)2+ (¯v+)2 < 0 , (4.5)

∂λ+

∂µ(¯λ, ¯µ) = −R1 0(¯u)2 R1

0(¯u+)2+ (¯v+)2 < 0 . (4.6) Finally, the related nontrivial solutions have all the same number of (simple) zeros (hence both change sign) and the both have same sign in a neighborhood of 0 and of 1.

Proof. We will give the proof for Dirichlet boundary conditions.

We will denote by (u, v) [λ+, λ, µ, s] (x) the solution of the IVP









−u00 = λ+v+− λv in (0, 1)

−v00= λ+u+− µu in (0, 1) (u, v, u0, v0)(0) = (0, 0, 1, s)

(4.7)

(11)

and we will apply the implicit function theorem to the system

(u, v)λ+, λ, µ, s (1) = (0, 0) , (4.8) that is, we want to solve locally the set of its zeros with respect to the variables λ, µ.

We remark that (u, v) [λ+, λ, µ, s] (x) is a C1function of the five variables (λ+, λ, µ, s) ∈ N and x ∈ [0, 1], where N is a suitable neighborhood of the point (¯λ+, ¯λ, ¯µ, ¯s); actually the derivatives may be calculated through the differential equation, where the nonlinearities λ+v+− λv and λ+u+− µu are C1 functions of the variables λ±, µ.

Denote by (¯u, ¯v) = (u, v)¯λ+, ¯λ, ¯µ, ¯s; since the zeros of ¯u and ¯v are simple by lemma 3.2, we may restrict the neighborhood N such that this property is maintained for all the functions (u, v) [λ+, λ, µ, s] with (λ+, λ, µ, s) ∈ N and such that the internal zeros do not disappear and the sign in a neighborhood of 0 and of the last zero does not change. We remark also that since ¯u and ¯v change sign then this property too is maintained in N .

Now let c(x) = ¯λ+χv>0}+ ¯λχv<0} and d(x) = ¯λ+χu>0} + ¯µχu<0}: then (¯u, ¯v) also satisfies the IVP









−¯u00= c(x)¯v in (0, 1)

−¯v00= d(x)¯u in (0, 1) (¯u, ¯v, ¯u0, ¯v0)(0) = (0, 0, 1, ¯s)

. (4.9)

Then let (us, vs)(x) = ∂s (u, v)λ¯+, ¯λ, ¯µ, ¯s (x): differentiating (4.9) with respect to s we get (the dependence on s is just in the initial condition):









−u00s = c(x)vs in (0, 1)

−v00s = d(x)us in (0, 1) (us, vs, u0s, v0s)(0) = (0, 0, 0, 1)

; (4.10)

note here that by using equations (3.6) and (3.7) one obtains that vs(x) > 0 and us(x) < 0 for x ∈ (0, 1].

Now, multiplying the first equation in (4.10) by ¯v, the second by ¯u and summing we have R −u00sv − v¯ 00su =¯ R c(x)vs¯v + d(x)usu; integrating by parts two times we obtain¯

Z 1 0

−us00− vs00+ [−u0sv − v¯ s0u + u¯ s0+ vs0]10= Z 1

0

c(x)vs¯v + d(x)usu .¯ (4.11)

Since (¯u, ¯v) is solution of the BVP too, it satisfies

¯

u(0) = ¯v(0) = ¯u(1) = ¯v(1) = 0 (4.12) and so in equation (4.11) only the following term survives:

(us¯v0+ vs0)(1) = 0 . (4.13)

(12)

In the same way, let (uλ+, vλ+)(x), (uλ, vλ)(x) and (uµ, vµ)(x) be defined like (us, us)(x) above: differentiating (4.9) with respect to λ+, λ and µ respectively, we get (the dependence on λ+, λ and µ is in the coefficients c(x) and d(x)):









−u00λ+ = c(x)vλ++ ¯v+ in (0, 1)

−vλ00+ = d(x)uλ++ ¯u+ in (0, 1) (uλ+, vλ+, u0λ+, vλ0+)(0) = (0, 0, 0, 0)

, (4.14)









−u00λ= c(x)vλ− ¯v in (0, 1)

−vλ00 = d(x)uλ+ in (0, 1) (uλ, vλ, u0λ, v0λ)(0) = (0, 0, 0, 0)

, (4.15)









−u00µ= c(x)vµ in (0, 1)

−v00µ = d(x)uµ− ¯u in (0, 1) (uµ, vµ, u0µ, vµ0)(0) = (0, 0, 0, 0)

; (4.16)

then, proceeding as for (4.13), we get

(uλ+0+ vλ+0)(1) = Z 1

0

(¯u+)2+ Z 1

0

(¯v+)2 > 0 , (4.17)

(uλ¯v0+ vλ0)(1) = Z 1

0

(¯v)2> 0 , (4.18)

(uµ0+ vµ0)(1) = Z 1

0

(¯u)2 > 0 . (4.19) We deduce by the above computations that the vector (us(1), vs(1)) is not zero, and by (4.13) it is orthogonal to (¯v0(1), ¯u0(1)), while (uλ+(1), vλ+(1)) is not orthogonal to it; then

det

uλ+(1) vλ+(1) us(1) vs(1)

6= 0 , (4.20)

which is indeed the condition we need to apply the implicit function theorem to system (4.8), that is to obtain the claimed function (λ+, s) (λ, µ).

Now we may also obtain the derivatives ∂λ∂λ+(¯λ, ¯µ) and ∂λ∂µ+(¯λ, ¯µ): by differentiating (4.8) we get

 ∂(u, v)

∂(s, λ+)

  ∂(s, λ+)

∂(λ, µ)

 +

 ∂(u, v)

∂(λ, µ)

= 0 , (4.21)

that is

us(1) uλ+(1) vs(1) vλ+(1)

∂s

∂λ(¯λ, ¯µ) ∂µ∂s(¯λ, ¯µ)

∂λ+

∂λ(¯λ, ¯µ) ∂µ∂λ+(¯λ, ¯µ)

= −

uλ(1) uµ(1) vλ(1) vµ(1)

; (4.22)

(13)

by multiplying on the left by the vector [¯v0(1), ¯u0(1)] we get

h

(usv + v¯ su(1)¯ (uλ+¯v + vλ+u(1)¯ i

∂s

∂λ(¯λ, ¯µ) ∂µ∂s(¯λ, ¯µ)

∂λ+

∂λ(¯λ, ¯µ) ∂µ∂λ+(¯λ, ¯µ)

= − h

(uλ¯v0+ vλ0)(1) (uµv + v¯ µ0)(1) i

, (4.23) that is (by equations (4.13), (4.17), (4.18) and (4.19))

h

0 R1

0(¯u+)2+ (¯v+)2 i

∂s

∂λ(¯λ, ¯µ) ∂µ∂s(¯λ, ¯µ)

∂λ+

∂λ(¯λ, ¯µ) ∂µ∂λ+(¯λ, ¯µ)

= − h

R1

0(¯v)2 R1 0(¯u)2

i (4.24) and then we get equations (4.5) and (4.6).

The same results may be obtained for Neumann boundary conditions by applying the implicit function theorem to the system

(u0, v0)λ+, λ, µ, s (1) = (0, 0) , (4.25) where now (u, v) is the solution of the IVP with (u, v, u0, v0)(0) = (1, s, 0, 0)

We remark that the hypothesis that the nontrivial solutions change sign was used to guar- antee the condition (4.20): in fact, the implicit function theorem (at least in this form) may not be applied to bΣt.

4.2 Global study of eΣ±

Now, we want to exploit the local information obtained in lemma 4.1, in order to obtain a qualitative description of the set eΣ+; to do this we will need the following lemma and its corollary 4.3:

Lemma 4.2. Given {(λ+n, λn, µn, sn)} ⊆ eΣ+ with (λ+n, λn, µn) → (λ+0, λ0, µ0) ∈ R3, there exists a subsequence sn→ s0∈ R such that (λ+0, λ0, µ0, s0) ∈ eΣ+.

Moreover, if the sequence (un, vn) of the corresponding nontrivial solutions is composed by functions all with the same number of (simple) zeros, then the functions (u0, v0) corresponding to the point (λ+0, λ0, µ0, s0) have this number of (simple) zeros too.

Proof. Consider the Dirichlet boundary conditions. As seen in lemma 3.1, the functions un, vn∈ H2(0, 1) ⊆ C1,1/2([0, 1]).

Let (Un, Vn) =

un

kunkH2,kvvn

nkH2

: then, up to a subsequence, (Un, Vn) → (U0, V0) weakly in [H2(0, 1)]2 and strongly inC1([0, 1])2

. The variational equation for (Un, Vn) is R1

0 Un0φ0+ Vn0ψ0 =R1

0+nVn+− λnVn)φ + (λ+nUn+− µnUn)ψ for all φ, ψ ∈ H01; taking the limit gives

Z 1 0

U00φ0+ V00ψ0= Z 1

0

+0V0+− λ0V0)φ + (λ0+U0+− µ0U0)ψ f or all φ, ψ ∈ H01, (4.26)

(14)

that is (U0, V0) is a solution of (1.4) with coefficients (λ+0, λ0, µ0).

Since the solutions are strong too by lemma 3.1, we have Z 1

0

(−Un00− U000)2= Z 1

0

(λ+nVn+− λnVn) − (λ+0V0+− λ0V0)2

, (4.27)

where the right hand side goes to zero and so Un → U0 and (by the same argument) Vn → V0 strongly in H2; this implies that kU0kH2 = kV0kH2 = 1 and so (U0, V0) is a non trivial solution, that is (λ+0, λ0, µ0,VU000(0)

0(0)) ∈ eΣ+ (observe that U00(0), V00(0) > 0 by lemma 3.2 and the C1 convergence).

Moreover, since sn= vn0(0) = UVn00(0)

n(0), its limit exists (by the C1 convergence) and it is indeed

V00(0)

U00(0), so we proved the first part of the lemma.

To conclude, observe that (Un(x), Vn(x)) = (un(x), vn(x)) Un0(0) and Un0(0) is (up to a sub- sequence) bounded away from zero; this implies that if the sequence (un, vn) is composed by functions all with the same number of simple zeros then the same is true for (Un, Vn) and, by the C1 convergence, also for (U0, V0).

For Neumann boundary conditions, the proof goes in the same way, by proving that (up to a subsequence) sn= vn(0) = UVn(0)

n(0)VU0(0)

0(0) and that (λ+0, λ0, µ0,UV0(0)

0(0)) ∈ eΣ+. Corollary 4.3. The set bΣ+ is closed.

Proof. Actually, the previous lemma implies that if a sequence {(λ+n, λn, µn)} ⊆ bΣ+is such that (λ+n, λn, µn) → (λ+0, λ0, µ0) ∈ R3, then (λ+0, λ0, µ0) ∈ bΣ+.

Now, a first qualitative description of the set eΣ+ may be given:

Proposition 4.4. Let C be a connected component of eΣ+ such that the corresponding nontrivial solutions change sign; then there exists a connected open set Λ ⊆ {(λ, µ) : λ, µ> 0} such that C is of the form C = {(λ+, µ), λ, µ, s(λ, µ)) : (λ, µ) ∈ Λ} and the following assertions hold:

1. (λ+, s) : Λ → R2 is a C1 function of λ, µ and for any (λ, µ) ∈ Λ

∂λ+

∂λ, µ) < 0 , (4.28)

∂λ+

∂µ, µ) < 0 . (4.29)

2. There exist a real ρ ≥ 0 and a non increasing function D : (ρ, +∞) → R+ such that for any eλ> 0,

Λ ∩ n

(eλ, µ) : µ∈ R+o

=

∅ for eλ≤ ρ

n

(eλ, µ) : µ> D(eλ)o

for eλ> ρ

; (4.30)

in the case eλ> ρ we have

lim

µ→D(eλ)

λ+(eλ, µ) = +∞ , (4.31)

(15)

and

lim

µ→+∞λ+(eλ, µ) = E(eλ) (4.32) defines a non increasing function E : (ρ, +∞) → R+.

Finally, the same holds if we exchange the role of the variables λ and µ.

3. The corresponding nontrivial solutions of (1.4) have all the same number of (simple) zeros.

4. ∃ ! (λ, µ) ∈ Λ such that λ+, µ) = λ = µ, and then λ = µ = λk for some k ≥ 2.

Proof. ¿From lemma 4.1 we deduce that C may be expressed as a function of λ, µdefined on a open connected set and we obtain the point 3 and equations (4.28) and (4.29); moreover, by point 3 and lemma 2.1 we deduce that λ, µ and λ+, µ) are all strictly positive.

Now remark that, by lemma 4.2, if lim)→(λ00)λ+, µ) = λ+0 ∈ R then for some s0 ∈ R the point (λ+0, λ0, µ0, s0) ∈ C and then (λ0, µ0) would be an interior point of Λ by lemma 4.1.

Once fixed fλ > 0, the variable µ has to take values in an open subset of (0, +∞), but the function λ+( fλ, µ) has to be positive and decreasing in µ; this and the above remark imply that this open set has to be a halfline, that the limit of λ+( ˜λ0, · ) at +∞ exists and is nonnegative while the limit at the initial point D( fλ) of the halfline must be +∞, proving the first part of point 2.

The fact that one may exchange the role of the variables λand µfollows by the symmetries of the Fuˇc´ık spectrum and this implies that the function D is non increasing (otherwise the sections in direction λ would not be halflines). Finally E is non increasing by (4.28).

To prove point 4, consider the halfline {(λ, µ) : λ= µ> 0} (observe that by point 2 it may not have empty intersection with Λ): since the function λ+ is C1(Λ), by equations (4.28) and (4.29) we have that it is decreasing along this halfline and so we may deduce as above that it has a nonnegative limit for λ = µ → +∞ and an asymptote where the halfline enters Λ;

this implies that there exists a (unique) point ξ where λ+(ξ, ξ) = ξ, proving point 4 (in fact in this case problem (1.4) reduces to (1.6) and then ξ ∈ σ(−∆)\ {λ1}).

Proposition 4.4 allows us to characterize all the connected components of eΣ±: actually, the complete knowledge of the linear spectrum (1.6) and property 4 in Proposition 4.4 imply the following:

Proposition 4.5. There exists a one to one relation between the eigenvalues {λk}k≥2 of the Laplacian and those connected components of eΣ+ (respectively of eΣ) which correspond to non- trivial sign changing solutions.

Proof. For any eigenvalue λk there exists a unique value s (namely s = 1) such that the point (λk, λk, λk, s) ∈ eΣ+ and, by lemma 4.1, through this point passes one connected component of Σe+; conversely, by point 4 in proposition 4.4, we have that any connected component in eΣ+ passes through a point of the form (λk, λk, λk, s) for some k ≥ 2, s ∈ R.

These facts give the one to one relation.

Updating...

## References

Related subjects :


text:
# What would the second degree equation be?

Find a second degree differential equation describing family of circles: $$(x-a)^2 + (y-b)^2 = b^2.$$ I differentiated once and obtained: $$x'(t) (x(t) - a) + y'(t) (y(t) - b) = 0,$$ then differentiated for the second time and got: $$(x'(t))^2 + x(t)x''(t) - ax''(t) + (y'(t))^2 + y(t)y'(t) - by''(t) =0.$$ And...what? It doesn't make my work any easier, it doesn't simplify my first equation really... How to solve it? I would really appreciate any help.

-
Are you supposed to find a differential equation of the form $F(x'(t),y'(t),t)=0$? Or of the form $F(y'(x),x)=0$? –  Avi Steiner Mar 3 '13 at 19:01
You want to get rid of the constants $a$ and $b$ in your resulting second order ODE by combining the derivatives. –  vonbrand Mar 3 '13 at 19:01

Why don't you treat $y$ as a function of $x$ and differentiate three times with respect to $x$? Differentiating once gives $(x - a) + y'(y - b) = 0$. Differentiating again gives $$1 + y''(y - b) + (y')^2 = 0 \tag{1} \\$$ $$y - b = -\frac{1 + (y')^2}{y''}. \tag{2}$$ Differentiating $(1)$ gives the final equation we need $$y'''(y - b) + y''y' + 2y'y'' = 0$$ $$y - b = -\frac{3y'y''}{y'''}. \tag{3}$$ We can equate $(2)$ and $(3)$ to get $$\frac{1 + (y')^2}{y''} = \frac{3y'y''}{y'''} \\ y'''(1 + (y')^2) = 3y'(y'')^2 \\ y'''(1 + (y')^2) - 3y'(y'')^2 = 0$$ whichever form you prefer.
@Did Of course not. But if you want the differential equation for the family of all circles (which I did) then there are three constant: $a, b, r$, where $r$ is the radius (I am assuming that the equation written in the question has a typo for the radius squared). Thus, you can't have a second order equation. If the question did not have a typo, then it represents a smaller family of circles and to find a differential equation which are only satisfied by those circles would require a different approach. –  Pratyush Sarkar Mar 6 '13 at 1:00
$$x''(t)=-y'(t)\qquad y''(t)=x'(t)$$ $$\text{or, equivalently,}$$ $$(x''(t)+y'(t))^2+(y''(t)-x'(t))^2=0$$


text:
# Bounded Height in Families of Dynamical Systems

01 Nov 2017

Let a, b ∈ $\bar{\mathbb{Q}}$ be such that exactly one of a and b is an algebraic integer, and let f$_{t}$(z) := z$^{2}$ + t be a family of polynomials parameterized by t ∈ $\bar{\mathbb{Q}}$. We prove that the set of all t ∈ $\bar{\mathbb{Q}}$ for which there exist m, n ≥ 0 such that f$^{m}_{t}$(a) = f$^{n}_{t}$(b) has bounded height. This is a special case of a more general result supporting a new bounded height conjecture in arithmetic dynamics.


text:
### Hallway Borders

What are the possible dimensions of a rectangular hallway if the number of tiles around the perimeter is exactly half the total number of tiles?

### Not a Polite Question

When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square...

Four rods, two of length a and two of length b, are linked to form a kite. The linkage is moveable so that the angles change. What is the maximum area of the kite?

# Grow up Fast

##### Age 11 to 14 ShortChallenge Level

Using the number of years that have passed
 Julie Megan Zoey today 32 4 1 in $n$ years 32 + $n$ 4 + $n$ 1 + $n$

32 + $n$ = 4 + $n$ + 1 + $n$
= 5 + 2$n$
27 = $n$

In 27 years Julie will be 32 + 27 = 59

Using algebra and Julie's age
Julie's age is $J$

$32-4=28$ so Megan's age is $J-28$
$32-1=31$ so Zoey's age is $J-31$
\begin{align}(J-28)+(J-31)&=J\\ \Rightarrow 2J-59&=J\\ \Rightarrow J&=59\end{align} So Julie will be $59$.

Using trial
 Julie Megan Zoey M + Z 32 4 1 42 14 11 25 too small 52 24 21 45 too small 62 34 31 65 too big 61 33 30 63 too big 59 31 28 59 perfect

Julie is 59

You can find more short problems, arranged by curriculum topic, in our short problems collection.


text:
## Celmusa 2 years ago A disc of moment of inertia I=125KG.m squared is rotating with angular velocity of 90rad/s and is brought in contact with a stationary dic of moment of inertia I2= 80kg.m squared such that their axis of rotation coincide. find the final angular velocity reached by the two discs?

• This Question is Open
1. sumanth4phy

consevarion of angular momentum$I_{1} \omega_{1} = I_{1} \omega_{2} + I_{2} \omega_{2}$


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Ellipses

## Collection of points whose sum of distances from two foci is constant.

Estimated8 minsto complete
%
Progress
Practice Ellipses

MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated8 minsto complete
%
Degenerate Conics

Feel free to modify and personalize this study guide by clicking “Customize.”

### Vocabulary

##### Complete the chart.
 Word Definition ____________ the distance from the center of the ellipse to the furthest point on the ellipse; represented by the letter  \begin{align*}a\end{align*} ____________ the longest distance from end to end of an ellipse Semi-minor axis ______________________________________________________________ Ellipse ______________________________________________________________ ____________ the two points that the ellipse curves around Eccentricity ______________________________________________________________ Hyperbola ______________________________________________________________ Degenerate conic ______________________________________________________________

### Ellipses

How many foci do ellipses have? ______________

For every point on the ellipse, the sum of the distances to each foci is _______________.

On the following image, label the major, semi-major, and minor axes:

What is the general equation of an ellipse? _______________________

The coefficient  \begin{align*}a\end{align*} always comes from the length of the ____________ axis and the coefficient  \begin{align*}b\end{align*} always comes from the length of the _____________ axis.

How do you find the locations of the two foci? What is the equation? _______________________

.

In your own words, describe eccentricity. ________________________________

What two numbers is the eccentricity between? ____________

What happens when the number gets closer to 1? ________________________________

What is the equation for the distance from the center of the ellipse to each directrix? _______________________

.

#### Hyperbolas

What is the equation for a hyperbola that opens side to side? _______________________

Up and down? _______________________

Note:  \begin{align*}a\end{align*} goes with the positive term and  \begin{align*}b\end{align*} goes with the negative term.  It does not matter which constant is larger.

The conjugate axis travels ______________ to the transverse axis through the ____________ and has length _______ .

.

How do you measure eccentricity in hyperbolas? ____________________________

What numbers can the eccentricity be for hyperbolas? _________________

.

#### Degenerate Conics

The general equation of a conic is \begin{align*}Ax^2+Bxy+Cy^2+Dx+Ey+F=0\end{align*} . What does this form encompass? ____________________________________________

The three types of degenerate conics are:

1. singular point , which is of the form: _____________________ . You can think of a singular point as a circle or an ellipse with an infinitely small radius.
2. line , which has coefficients  \begin{align*}A=B=C=0\end{align*} in the general equation of a conic.  The remaining portion of the equation is ________________ , which is a line.
3. degenerate hyperbola , which is of the form:  __________________ .  The result is two intersecting lines that make an “X” shape.  What is the equation for the slopes of the intersecting lines forming the X? ___________


text:
OpenKattis
Reykjavik University

Cuckoo Hashing

One of the most fundamental data structure problems is the dictionary problem: given a set ${D}$ of words you want to be able to quickly determine if any given query string $q$ is present in the dictionary ${D}$ or not. Hashing is a well-known solution for the problem. The idea is to create a function $h:\Sigma ^*\rightarrow [0..n-1]$ from all strings to the integer range $0,1,..,n-1$, i.e. you describe a fast deterministic program which takes a string as input and outputs an integer between $0$ and $n-1$. Next you allocate an empty hash table $T$ of size $n$ and for each word $w$ in ${D}$, you set $T[h(w)]=w$. Thus, given a query string $q$, you only need to calculate $h(q)$ and see if $T[h(q)]$ equals $q$, to determine if $q$ is in the dictionary. Seems simple enough, but aren’t we forgetting something? Of course, what if two words in ${D}$ map to the same location in the table? This phenomenon, called collision, happens fairly often (remember the Birthday paradox: in a class of 24 pupils there is more than $50$% chance that two of them share birthday). On average you will only be able to put roughly $\sqrt {n}$-sized dictionaries into the table without getting collisions, quite poor space usage!

A stronger variant is Cuckoo Hashing1. The idea is to use two hash functions $h_1$ and $h_2$. Thus each string maps to two positions in the table. A query string $q$ is now handled as follows: you compute both $h_1(q)$ and $h_2(q)$, and if $T[h_1(q)]=q$, or $T[h_2(q)]=q$, you conclude that $q$ is in ${D}$. The name “Cuckoo Hashing” stems from the process of creating the table. Initially you have an empty table. You iterate over the words $d$ in ${D}$, and insert them one by one. If $T[h_1(d)]$ is free, you set $T[h_1(d)]=d$. Otherwise if $T[h_2(d)]$ is free, you set $T[h_2(d)]=d$. If both are occupied however, just like the cuckoo with other birds’ eggs, you evict the word $r$ in $T[h_1(d)]$ and set $T[h_1(d)]=d$. Next you put $r$ back into the table in its alternative place (and if that entry was already occupied you evict that word and move it to its alternative place, and so on). Of course, we may end up in an infinite loop here, in which case we need to rebuild the table with other choices of hash functions. The good news is that this will not happen with great probability even if ${D}$ contains up to $n/2$ words!

Input

On the first line of input is a single positive integer $1\leq t \leq 50$ specifying the number of test cases to follow. Each test case begins with two positive integers $1 \leq m \leq n \leq 10000$ on a line of itself, $m$ telling the number of words in the dictionary and $n$ the size of the hash table in the test case. Next follow $m$ lines of which the $i$:th describes the $i$:th word $d_ i$ in the dictionary ${D}$ by two non-negative integers $h_1(d_ i)$ and $h_2(d_ i)$ less than $n$ giving the two hash function values of the word $d_ i$. The two values may be identical.

Output

For each test case there should be exactly one line of output either containing the string “successful hashing” if it is possible to insert all words in the given order into the table, or the string “rehash necessary” if it is impossible.

Sample Input 1 Sample Output 1
2
3 3
0 1
1 2
2 0
5 6
2 3
3 1
1 2
5 1
2 5
successful hashing
rehash necessary

Footnotes

1. Cuckoo Hashing was suggested by the danes R. Pagh and F. F. Rödler in 2001


text:
R-CNN 是使用深度学习解决目标检测问题的里程碑，让我们感受一些这篇开山之作，回顾历史的开端。

核心思想

• 区域选择不再是滑窗提取特征，而是采用启发式的候选区域生成算法（Selective Search）
• 提取特征也从手工，变成了卷积神经网络自动提取特征，增强了特征的鲁棒性

流程步骤

1、使用 Selective Search 算法生成候选区域
2、使用 CNN 提取候选区域图像的特征
3、将特征输入到 SVM 分类器，判别输入类别
4、以回归的方式精修候选框

核心部分

• 使用一种过分割手段，将图像分割成小区域
• 查看现有小区域，合并可能性最高的两个区域，重复直到整张图像合并成一个区域位置
• 输出所有处理完的区域，即候选区域

合并规则

• 颜色（颜色直方图）相近的
• 纹理（梯度直方图）相近的
• 合并后总面积小的，并保证合并操作的尺度较为均匀
• 合并后，总面积在其 Bbox 中所占比例大的，并保证形状比较规则

Selective Search 为例的候选区域生成算法，都是基于 Low Level 的图像特征，比如纹理、颜色等，比较低等不具有抽象性，效果比较局限。

Feature extraction

Features are computed by forward propagating a mean-subtracted 227 × 227 RGB image through five convolutional layers and two fully connected layers

Neural Networks

Aside from replacing the CNN’s ImageNet-specific 1000-way classification layer with a randomly initialized (N + 1)-way classification layer (where N is the number of object classes, plus 1 for background), the CNN architecture is unchanged.

Once features are extracted and training labels are applied, we optimize one linear SVM per class. Since the training data is too large to fit in memory, we adopt the standard hard negative mining method [17, 37]. Hard negative mining converges quickly and in practice mAP stops increasing after only a single pass over all images.

The overlap threshold, 0.3, was selected by a grid search over {0,0.1,…,0.5} on a validation set.

Classification

This performance drop likely arises from a combination of several factors including that the definition of positive examples used in fine-tuning does not emphasize precise localization and the softmax classifier was trained on randomly sampled negative examples rather than on the subset of “hard negatives” used for SVM training.

Bounding Box Refine

1、先做平移 $(\delta{x},\delta{y})$，$\delta{x}=P_{w}d_{x}(P)$，$\delta{y}=P_{h}d_{y}(P)$

$$\hat{G_{x}} = P_{w}d_{x}(P)+P_{x}$$

$$\hat{G_{y}} = P_{h}d_{y}(P)+P_{y}$$

2、再做尺度缩放 $(S_{w},S_{h})$，$S_{w}=P_{w}exp(d_{w}(P))$，$S_{h}=P_{h}exp(d_{h}(P))$

$$\hat{G_{w}} = P_{w}exp(d_{w}(P))$$

$$\hat{G_{h}} = P_{h}exp(d_{h}(P))$$

$$w_{*} = \arg\min_{\hat{w_{*}}} \sum_{i}^{n} (t_{*}^{i} − w_{*}^{T} * \phi_{5}(P^{i}))^2 + \lambda ||\hat{w_{*}}||^2$$

$t_{x}=(G_{x} - P_{x})/P_{w}$
$t_{y}=(G_{y} - P_{y})/P_{h}$
$t_{w}=log(G_{w}/P_{w})$
$t_{h}=log(G_{w}/P_{h})$


text:
• entries
338
2045
• views
1168603

# Oh yes...

313 views

I got my new comp all setup [grin]

All for under $1,900 w/taxes and all. I tried to buy this computer @ Dell, Gateway, Alienware, Falcon northwest, etc. every time it's$3,200+ for various reasons, and Dell doesn't even offer the 8800, as far as I could tell.

I got it from a local Intrex computer store, my only bet for pricing this thing better...would have been to get each part off newegg.com and assemble it myself. That takes time, and I wouldn't have a 3 year warranty. This comp. was assembled, and in my possession in under 30 hours notice. Which is good service [grin]

Specs -
Quote:
 nVidia GeForce 8800GTS DDR 640MBIntel Core2Duo 2.4GHz duel core, 4MB cache, 64-bit2GB DDR2 RAM PC2-5300 (667Mhz)350GB HD 7200RPM500W power supply.And a case w/window [so I can see the 8800], 3 fans, thermal gauges, and lights inside.

I'm busy installing many SDKs, software, games, tools, etc. etc. this is gonna take forever...

Here are some pics...man I'm so impressed with the GeForce 8800 the thing is huge. It takes up two slots, one is used for the fan...here is a picture...

Some more pics of the comp...

I decided to get a new mouse as well...the Logitech G5 was the obvious choice to compliment my G15 keyboard...this thing is crazy. It has a 'weight cartridge' where you can insert up to 8 different weights into the thing to tweak the weight, it has a dynamic sensitivity adjustment w/LED display, and is the best 'feeling' mouse I've ever come across.

I also got all 6 Game Programming Gems books for X-Mas...these are gonna be some good reading. No more hit-and-miss getting articles off the internet. I now have the full library :-d

Alright I'm gonna put the 8800 through my little tests :-D To gauge how much faster this is than my last comp. I'll report my findings lol.

- Dan

I am interested to hear your results - I am considering an upgrade myself. The 8800 is very appealing right now, so is the DX10 work that could go with it...

I find the two random bodybuilding books hidden in with the game programming books to be hilarious. :)

Newegg: ~$1,200 + Shipping, no sales tax unless you live in CA, TN or NJ. That 8800 is a beast sitting in that case! I'm anxious to hear how it works out. #### Share this comment ##### Link to comment Quote:  Original post by Jesse Chounard I find the two random bodybuilding books hidden in with the game programming books to be hilarious. :) "Bodybuilding Encyclopedia", now that is....hahahahahahahaa. #### Share this comment ##### Link to comment lol, yea it's not exactly the most diverse library, just programming books, a lot of WW2 books, and those bodybuilding books [grin]. Dave - was it really that cheap, is that w/motherboard, CD drive, and other little things, or just the 6 things listed? That is pretty damn cheap. I did look up the 8800, the GTX was about$600, but I did see some GTSs for like \$400.

I was getting like 300 FPS in counter-strike with full options and, Battlefield 2142 is a little too easy now that it doesn't lag like a mofo. I was also getting like 10,000 FPS in some of the basic DirectX SDK apps. Oblivion also ran like a dream. The card itself came with a copy of Dark Messiah of Might & Magic (which uses the source engine, so that ran about as good as CS).

It was around that figure. Including a MOBO, CD/DVD Drive.

I am usually not one to gloat but I spot a book on your shelves that contains one of my published articles. [wink]

## Create an account

Register a new account


text:
# The angle between two 3D lines

1. Jul 5, 2006

### somy

Can anyone tell me how can I calculate the angle between two lines in 3D. I need in my codings.
I know it has a simple solution. but I really have mixed up!!!
Thanks.

2. Jul 5, 2006

3. Jul 6, 2006

### HallsofIvy

If the two lines actually intersect, then the angle is given by
$$cos(\theta)= \frac{u\cdot v}{|u||v|}$$ where u and v are any vectors pointing in the direction of each line. You can use the point of intersection and one point on each line to define those vectors.

If the line do not intersect (the "typical" case in 3d) then there is no angle between them.

Last edited by a moderator: Jul 6, 2006
4. Jul 6, 2006

### somy

Thank you. I'll try it!!!


text:
For water
Question:

For water $\Delta_{\text {vap }} \mathrm{H}=41 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $373 \mathrm{~K}$ and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is $\mathrm{kJ} \mathrm{mol}^{-1}$

$\left[\right.$ Use $\left.: \mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right]$

Solution:

Given equation is

$\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}): \Delta \mathrm{H}=41 \frac{\mathrm{kJ}}{\mathrm{mol}}$

$\Rightarrow$ From the relation $: \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$

$\Rightarrow 41 \frac{\mathrm{kJ}}{\mathrm{mol}}=\Delta \mathrm{U}+(1) \times \frac{8.3}{1000} \times 373$

$\Rightarrow \mathrm{DU}=41-3.0959$

$=38 \mathrm{~kJ} / \mathrm{mol}$


text:
# Bayesian Change Point Dectection under Complex Time Series in Python Machine Learning Client for SAP HANA

A complex time series in real life usually has many change points inside it. When dealing with such data, simply applying traditional seasonality test to it may not render a convincing decomposition result. In this blog post, we will show how to use Bayesian Change Point Detection in the Python machine learning client for SAP HANA(hana-ml) to detect those change points and decompose the target time series.

In this blog post, you will learn:

• Decomposition for complex time series
• Change point detection with hana-ml

# Introduction

Time series may not ideally contain monotonic trend and seasonal waves after decomposition. On the contrary, it may include a great many inner change points in those parts.

Fig1

As illustrated above, we can see an obvious changing trend and seasonal wave from the time series. Currently, most algorithms are not able to extract them correctly due to the lack of change point analysis. In SAP HANA PAL and hana-ml, we provide BCPD to tackle that.

In this blog post, we will focus on the task of detecting the change points within the varying trend and seasonal components of complex time series.

# Solutions

Bayesian Change Point Detection(BCPD), to some extent, can been seen as an enhanced version of seasonality test in additive mode. Similarly, it decomposes a time series into three components: trend, seasonal and random, but with a remarkable difference that it is capable of detecting change points within both trend and season parts, using a quasi RJ-MCMC method.

Like the additive decomposition in seasonality test, we treat the time series Y(t) as an addition of  trend part T and seasonal part S  along with random noise:

$Y(t)=T(t,\theta_T)+S(t,\theta_S) +\epsilon(t)$

where $\theta_T$ are the parameters in the trend part, composed of the positions of trend change points and the coefficients $\gamma$ for each separated segment. Specifically, any trend segment can been written as

$trend\text{\_}segment(t) = \sum_{i=0}^{trend\text{\_}order}\gamma_i*t^i$

Likewise, $\theta_S$ are the parameters in the seasonal part, composed of the positions of seasonal change points and the coefficients for each seasonal segment, i.e. $(\alpha\text{,} \beta)$ and we may have

$season\text{\_}segment(t) = \sum_{l=1}^{harmonic\text{\_}order}[\alpha_l*sin(\frac{2\pi lt}{period})+\beta_l*cos(\frac{2\pi lt}{period})]$

One notable thing is that periods in different seasonal segments can vary from one to another, which expands our algorithm to much wider scenarios.

All source code in examples of the following context will use Python machine learning client for SAP HANA Predictive Analsysi Library(PAL).

## Connect to SAP HANA

import hana_ml
from hana_ml import dataframe

### Use Case I : Detecting Changing Trend

In this use case, we will focus on detecting the change points in the trend part only,

The mocking data is stored in database in a table with name ‘PAL_MOCKING_BCPD_DATA_1_TBL’, we can use the table() function of ConnectionContext to create a corresponding hana_ml.DataFrame object for it.

mocking_df = cc.table('PAL_MOCKING_BCPD_DATA_1_TBL')

The collect() function of hana_ml.DataFrame can help to fetch data from database to the python client and the data is illustrated as follows:

plt.plot(mocking_df.collect()["SERIES"])

Fig2

The data is of length 40, then we import the BCPD algorithm from hana-ml and apply it to the mocking dataset:

from hana_ml.algorithms.pal.tsa.changepoint import BCPD
bcpd = BCPD(max_tcp=5, max_scp=0, random_seed=1)
#tcp: location of trend change points
#scp: location of seasonal change points
#period: period of each seasonal segment
#components: decomposition values of the time series
tcp, scp, period, components = bcpd.fit_predict(data=mocking_df)

Again we can use collect() to get the final results from the database. Since we are only interested in the trend part, we can visualize that using the following code:

print(tcp.collect())
plt.plot(mocking_df.collect()["SERIES"], label='data')
plt.plot(components.collect()["TREND"], label='trend')
for cp in list(tcp.collect()["TREND_CP"]):
plt.axvline(x=cp, color="red", linestyle='dashed')
plt.legend(['original series', 'trend component'])
plt.title("Trend component")
plt.show()

Fig3

### Use Case II : Detecting Changing Trend and Season

In this use case, we are to apply BCPD to the data shown in Fig1, in which the trend and season are changing.

Similarly, the data is stored in database in a table with name ‘PAL_MOCKING_BCPD_DATA_2_TBL’ and we need to adjust our parameters for this use case

# detailed introduction of parameters can be found on our user manual page
bcpd = BCPD(max_tcp=5, max_scp=5, max_harmonic_order=1, max_period=10, max_iter=10000, interval_ratio=0.2, random_seed=1)
mocking_df = cc.table('PAL_MOCKING_BCPD_DATA_2_TBL') # data shown in Fig1
tcp, scp, period, components = bcpd.fit_predict(data=mocking_df)

The trend visualization code is the same as Use case I and the trend plot rendered is:

Fig4

Further, we use the following code to visualize the seasonal part:

print(scp.collect())
print(period.collect())
plt.plot(components.collect()["SEASONAL"], label='trend')
for cp in scp.collect()["SEASON_CP"]:
plt.axvline(x=cp, color="orange", linestyle='dashed')
plt.title("Seasonal component")

Fig5

The above plots reveal that BCPD is able to give decent decomposition results on both trend and seasonal parts from the time series.

### Use Case III : Sensor Data Abrupt Change Detection and Denoising

In this use case, we are going to apply BCPD to real life sensor data to detect potential abrupt change points and to cancel the random term after decomposition for denoising .

Assume the data is stored in a dataframe named sensor_df , we firstly visualize the data using the following code:

figure(num=None, figsize=(10, 3))
plt.plot(sensor_df.collect()["SERIES"])
print(sensor_df.collect())

Fig6

In order to obtain a better fit, we are going to use a second-order trend in BCPD for this use case :

# detailed introduction of parameters can be found on our user manual page
bcpd = BCPD(trend_order=2, max_tcp=10, max_scp=10, max_harmonic_order=10, mmin_period=50, max_period=50, max_iter=15000, interval_ratio=0.01, random_seed=1)
tcp, scp, period, components = bcpd.fit_predict(data=sensor_df)

After the algorithm finishes, we use the following code to show potential abrupt change points:

figure(num=None, figsize=(16, 4), dpi=80, facecolor='w', edgecolor='k')
plt.plot(sensor_df.collect()["SERIES"])
for cp in list(tcp.collect()["TREND_CP"]):
plt.axvline(x=cp, color="orange", linestyle='dashed')
for cp in scp.collect()["SEASON_CP"]:
plt.axvline(x=cp, color="orange", linestyle='dashed')
plt.legend(['sensor data', 'potential abrupt change'])
plt.show()

Fig7

Denoised time series can be restored by simply adding the trend and season parts after decomposition:

Fig8

# Discussion and Summary

In the blog post, we introduced a new SAP HANA ML algorithm for detecting change points in the time series with several use cases under Python machine learning client for SAP HANA(hana-ml).

BCPD can be applied to different scenarios: trend test, seasonality test, change points detection, signal noise cancellation, etc.

Weibull Analysis using Python machine learning client for SAP HANA

Outlier Detection using Statistical Tests in Python Machine Learning Client for SAP HANA

Outlier Detection by Clustering using Python Machine Learning Client for SAP HANA

Anomaly Detection in Time-Series using Seasonal Decomposition in Python Machine Learning Client for SAP HANA

Outlier Detection with One-class Classification using Python Machine Learning Client for SAP HANA

Learning from Labeled Anomalies for Efficient Anomaly Detection using Python Machine Learning Client for SAP HANA

Additive Model Time-series Analysis using Python Machine Learning Client for SAP HANA

Time-Series Modeling and Analysis using SAP HANA Predictive Analysis Library(PAL) through Python Machine Learning Client for SAP HANA

Import multiple excel files into a single SAP HANA table

COPD study, explanation and interpretability with Python machine learning client for SAP HANA


text:
RUS  ENG JOURNALS   PEOPLE   ORGANISATIONS   CONFERENCES   SEMINARS   VIDEO LIBRARY   PACKAGE AMSBIB
 General information Latest issue Archive Impact factor Subscription Guidelines for authors License agreement Submit a manuscript Search papers Search references RSS Latest issue Current issues Archive issues What is RSS

 TMF: Year: Volume: Issue: Page: Find

 TMF, 1974, Volume 20, Number 3, Pages 338–352 (Mi tmf3844)

Partial-wave analysis of dual amplitude with mandelstam analyticity

L. L. Enkovskii

Abstract: The partial-wave projection of the dual amplitude with Mandelstam analyticity is studied. It is shown that the resonances in the model can be completely Reggeized. From the point of view of the structure of the $J$ plane, the preferable class of trajectories consists of those that satisfy the asymptotic bound $O(\ln^{1+\varepsilon}s)\leqslant\vert\alpha(s)\vert\leqslant O(s^{1/2})$, for which the singularities in the $J$ plane are exhausted by poles on the parent and on the daughter trajectory and also at nonsense points with wrong signature. From the point of view of two-particle unitarity, trajectories that take a half-integral value at the threshold are preferable.

Full text: PDF file (1645 kB)
References: PDF file   HTML file

English version:
Theoretical and Mathematical Physics, 1974, 20:3, 862–872

Bibliographic databases:

Citation: L. L. Enkovskii, “Partial-wave analysis of dual amplitude with mandelstam analyticity”, TMF, 20:3 (1974), 338–352; Theoret. and Math. Phys., 20:3 (1974), 862–872

Citation in format AMSBIB
\Bibitem{Enk74} \by L.~L.~Enkovskii \paper Partial-wave analysis of dual amplitude with mandelstam analyticity \jour TMF \yr 1974 \vol 20 \issue 3 \pages 338--352 \mathnet{http://mi.mathnet.ru/tmf3844} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=468773} \transl \jour Theoret. and Math. Phys. \yr 1974 \vol 20 \issue 3 \pages 862--872 \crossref{https://doi.org/10.1007/BF01040167}


text:
Conference paper Open Access

# Creation of an OpenFOAM fuel performance class based on FRED and integration into the GeN-Foam multi-physics code

Fiorina Carlo; Pautz Andreas; Mikityuk Konstantin

### Dublin Core Export

<?xml version='1.0' encoding='utf-8'?>
<oai_dc:dc xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:oai_dc="http://www.openarchives.org/OAI/2.0/oai_dc/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/oai_dc/ http://www.openarchives.org/OAI/2.0/oai_dc.xsd">
<dc:creator>Fiorina Carlo</dc:creator>
<dc:creator>Pautz Andreas</dc:creator>
<dc:creator>Mikityuk Konstantin</dc:creator>
<dc:date>2018-09-26</dc:date>
<dc:description>The FRED code is an in-house tool developed at the Paul Scherrer Institut for the so-called 1.5-D nuclear fuel performance analysis. In order to extend its field of application, this code has been re-implemented as a class of the OpenFOAM numerical library. A first objective of this re-implementation is to provide this tool with the parallel scalability necessary for full-core analyses. In addition, the use of OpenFOAM as base library allows for a straightforward interface with the standard Open-FOAM CFD solvers, as well as with the several OpenFOAM-based applications developed by the nuclear engineering community. In this paper, the newly developed FRED-based Open-FOAM class has been integrated in the GeN-Foam multi-physics code mainly developed at the E´cole polytechnique fe´de´rale de Lausanne and at the Paul Scherrer Institut. The paper presents the details of both the re-implementation of the FRED code and of its integration in GeN-Foam. The performances and parallel scalability of the tool are preliminary investigated and an example of application is provided by performing a full-core multiphysics analysis of the European Sodium Fast Reactor.</dc:description>
<dc:identifier>https://zenodo.org/record/1435696</dc:identifier>
<dc:identifier>10.5281/zenodo.1435696</dc:identifier>
<dc:identifier>oai:zenodo.org:1435696</dc:identifier>
<dc:relation>info:eu-repo/grantAgreement/EC/H2020/754501/</dc:relation>
<dc:relation>doi:10.5281/zenodo.1435695</dc:relation>
<dc:rights>info:eu-repo/semantics/openAccess</dc:rights>
<dc:title>Creation of an OpenFOAM fuel performance class based on FRED and integration into the GeN-Foam multi-physics code</dc:title>
<dc:type>info:eu-repo/semantics/conferencePaper</dc:type>
<dc:type>publication-conferencepaper</dc:type>
</oai_dc:dc>

135
122
views


text:
Show that the points A (1, 2, 3), B (–1, –2, –1), C (2, 3, 2) and D (4, 7, 6) are the vertices of a parallelog

### Question Asked by a Student from EXXAMM.com Team

Q 3019345219.     Show that the points A (1, 2, 3), B (–1, –2, –1), C (2, 3, 2) and D (4, 7, 6) are the vertices of a parallelogram ABCD, but it is not a rectangle.

#### HINT

(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

#### Access free resources including

• 100% free video lectures with detailed notes and examples
• Previous Year Papers
• Mock Tests
• Practices question categorized in topics and 4 levels with detailed solutions
• Syllabus & Pattern Analysis