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# Math Help - gauss vs gauss jordan

1. ## gauss vs gauss jordan

Hi Guys,

I have a simple and basic question....

I know what Gauss elimination is, and what we use it for, and also knows what Gauss Jordan is...what I wanted to know, is when do I use each one ? I mean, in order to solve a system of equations, Gauss is enough, so why do I need Gauss Jordan for ? Why making the extra move of transferring the matrix to a canonical form ?

2. ## Re: gauss vs gauss jordan

Originally Posted by WeeG
Hi Guys,

I have a simple and basic question....

I know what Gauss elimination is, and what we use it for, and also knows what Gauss Jordan is...what I wanted to know, is when do I use each one ? I mean, in order to solve a system of equations, Gauss is enough, so why do I need Gauss Jordan for ? Why making the extra move of transferring the matrix to a canonical form ?
It's just to make the actual calculations easier...

format: plain


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One day, I decided to calculate the density of air, so I searched for the individual densities of the 3 main elements that forms breathable air. I found the following data, at $\pu{0^\circ C}$ and $\pu{100 kPa}$:

\begin{array}{lrS} \text{Component} &\% &\rho,\,\pu{g cm-3}\\ \hline \text{Nitrogen} &78 &0.001251 \\ \text{Oxygen} &21 &0.00142897 \\ \text{Argon} &1 &0.001784 \end{array}

I'm calculating the density of $\pu{1 cm3}$ of air. Individual masses in $\pu{1 cm3}$ of air:

\begin{align} &\text{Oxygen} &\pu{0.0003000837 g}\\ &\text{Nitrogen} &\pu{0.00097578 g} \\ &\text{Argon} &\pu{0.000001784 g} \\ \end{align}

The final density is equal to $\pu{0.0012937037 g cm-3}$

Then I realized that the density of ozone should be lighter due to the ozone layer to be able to exist, otherwise, the ozone gas, as it is fairly more dense than air, would pass directly through the atmosphere. The problem is that the density of ozone gas at the same conditions (STP), have a density of $\pu{0.00214 g cm-3}$, which is $65.41654785\%$ heavier than air.

With the context explained, my question is: if is really a ozone layer up there in the atmosphere, how it is there, being heavier than the base gas mix that is below it?


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# Coordinate geometry of the circle question

1. Feb 21, 2009

### Ed Aboud

1. The problem statement, all variables and given/known data
Fairly straight forward question but I just can't see whats wrong.

A circle passes through the point (2,-4) and touches both the x-axis and the y-axis. Find the equations of the two circles which satisfy these conditions.

2. Relevant equations

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

with a centre point $c (-g,-f)$

$$r = \sqrt{g^2 + f^2 - c}$$

3. The attempt at a solution

After drawing a diagram I concluded that r=g and r=f therefore g=f.

$$g = \sqrt{g^2 + f^2 - c}$$

$$g^2 = g^2 + f^2 - c$$

$$f^2 = c$$

Since the point (2,-4) is on the circle it will satisfy :

$x^2 + y^2 + 2gx + 2fy + c = 0$

$$(2)^2 + (-4)^2 + 2g(2) + 2f(-4) + c = 0$$

$$4 + 16 + 4g -8f + c = 0$$

$$20 + 4g -8f + c = 0$$

But $f^2 = c$ and $f = g$

So

$$20 + 4f - 8f + f^2 = 0$$
$$20 -4f + f^2 = 0$$
$$f^2 -4f +20 = 0$$

This quadratic only has complex roots.

Thanks for any help!

2. Feb 21, 2009

### HallsofIvy

If the center is $(x_0,y_0)$, that is, the equation of the circle is $(x-x_0)^2+ (y-y_0)^2= R^2$ and it "touches" the axes- by which I assume you mean it is tangent to them, then $(x_0,0)$ and $(0,y_0)$ are the points at which the circle is tangent to the axes. Putting $x= x_0$, y= 0 into the equation, $(-y_0)^2= R^2$ and putting $y= y_0$, x= 0, $(-x_0)^2= R^2$ so we must have $x_0= y_0= R$ so we can write the equation as $(x- R)^2+ (y- R)^2= R^2$. Adding the condition that (2,-4) lies on that circle gives $(2-R)^2+ (-4-R)^2= R^2$ which you can solve for R.

3. Feb 23, 2009

### Ed Aboud

Yeah so you get $$R^2 + 4R + 20 = 0$$

this has complex roots as well....

4. Feb 23, 2009

### Dick

The center of the circle isn't at (R,R). That's in the wrong quadrant. You have to pick y0 to be the negative root of y0^2=R^2. Put it at (R,-R).

5. Feb 23, 2009

### Ed Aboud

Ok cool thanks so it is

$$R^2 -12R +20 = 0$$

$$R=2$$ and $$R=10$$

Therefore the equations of the circles are

$$(x- R)^2+ (y+ R)^2= R^2$$

$$(x- 2)^2+ (y+ 2)^2= R^2$$

and

$$(x- 10)^2+ (y+ 10)^2= R^2$$

Just out of curiosity what was wrong with my original attempt?

6. Feb 23, 2009

### Dick

Exactly the same thing. You said f=g=r. But that's not true. |f|=|g|=r is what is true. You have to figure out the signs based on the location of (2,-4).


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Magyar Information Contest Journal Articles
 Current issue Previous issues Order Form Special issues Archives

# English Issue, December 2002

Previous pageContentsNext pageORDER FORM

## Solutions of problems for physics P

M. 222. The angle of refraction of a puck bouncing back from a board is usually not identical to its angle of incident. (The motion of a puck can be modelled using the lid of a jar.) How does the angle of refraction depend on the speed of the puck (lid) and on the angle of incidence?

Solution. For an adequate measurement one needs a puck launching device' which can repeat puck launches at the same initial speed (within measurement error). This technical problem was resolved by most competitors, using a spring or rubber operated catapult'. Biró István (10th form student of Bolyai Farkas Secondary School, Marosvásárhely) constructed a clever device, which, in addition to the repeatable launches, solved the problem of recording the path of the puck (Fig. 1). He bore a hole into the middle of the puck, into which he put a toothpick wrapped in sewing thread. He had black ink permeated into the thread, and so the end of the sewing thread, just touching a glass plate, draw the path of the centre of mass of the puck.

Figure 1

Figure 2

Two competitors, Nagy Ádám (12th form student of Szent István Secondary Grammar School, Budapest) and Biró István, noticed that the reflected path of the puck bends', but they did not use this fact in the determination of the reflection angle. All of the competitors calculated the $\displaystyle \beta$2 reflection angle from the position of the puck after it had stopped, but this differs slightly from angle $\displaystyle \beta$1 which characterizes the collision better though it is more difficult to measure (Fig. 2).

Figure 3

A crucial point in the completion of the measurement is the selection of the sliding surface. Most competitors chose a smooth surface. In the velocity measurement they had to take friction into account. Jurányi Zsófia (12th form student of Leöwey Klára Secondary Grammar School, Pécs) carried out her measurements on a panel board oiled with co-oking oil. She stated that on a board like that friction is negligibly small. Geresdi Attila (11th form student of Árpád Fejedelem Secondary Grammar School, Pécs) made his experiments on a 1 m2 air-cushioned table. The competitors usually used rulers and bevels for measuring the angles, but there were other methods as well: Sükösd Zsuzsanna (7th form student of Orsolya Rendi Szent Angéla Elementary and Secondary Grammar School, Budapest,) used chequered' floor-plates as a co-ordinate system to define the position and the motion of the puck. Almost all the competitors got a result where the reflection angle was greater than the angle of incidence, but the difference was small and decreased while approaching the perpendicular direction. This was demonstrated by the graph drawn by Szilágyi Péter (9th form student of Kossuth L. Secondary Grammar School, Debrecen) containing the data for 3 different velocities (Figure 3). Nagy Ádám plotted the ratio of the reflection and incidence angles as a function of the incidence angle (Figure 4).

Figure 4

Figure 5

The experimental examination of velocity dependence is more difficult than measuring the angles. Using slopes the attainable velocity differences of the pucks started from different heights are not big enough - with considerable measurement errors - to find a definite connection between the angles and velocities. (Using catapult devices the velocity of the impact can be greater, but the measurement of velocity becomes more inaccurate.) The data measured by Jurányi Zsófia is shown in Figure 5. She plotted the $\displaystyle \beta$ reflection angle (the accuracy of the measurement is also indicated) against different v impact velocities at fixed $\displaystyle \alpha$=45o incidence angle. It can be seen that in the investigated interval $\displaystyle \beta$ is independent of v.

 Támogatóink: Morgan Stanley


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# check_units function¶

(Shortest import: from brian2 import check_units)

brian2.units.fundamentalunits.check_units(**au)[source]

Decorator to check units of arguments passed to a function

Raises

DimensionMismatchError
In case the input arguments or the return value do not have the expected dimensions.

Notes

This decorator will destroy the signature of the original function, and replace it with the signature (*args, **kwds). Other decorators will do the same thing, and this decorator critically needs to know the signature of the function it is acting on, so it is important that it is the first decorator to act on a function. It cannot be used in combination with another decorator that also needs to know the signature of the function.

Examples

>>> from brian2.units import *
>>> @check_units(I=amp, R=ohm, wibble=metre, result=volt)
... def getvoltage(I, R, **k):
...     return I*R

You don’t have to check the units of every variable in the function, and you can define what the units should be for variables that aren’t explicitly named in the definition of the function. For example, the code above checks that the variable wibble should be a length, so writing

>>> getvoltage(1*amp, 1*ohm, wibble=1)
Traceback (most recent call last):
...
DimensionMismatchError: Function "getvoltage" variable "wibble" has wrong dimensions, dimensions were (1) (m)

fails, but

>>> getvoltage(1*amp, 1*ohm, wibble=1*metre)
1. * volt

passes. String arguments or None are not checked

>>> getvoltage(1*amp, 1*ohm, wibble='hello')
1. * volt

By using the special name result, you can check the return value of the function.


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×

# A really awesome situation

Solving differential equations to solve physical problems is always fun, but it is never as fun as the situation i am going to present in this note,

Think about the situation shown in the figure, a tiny block is moving upon and along a rod that is hinged at the middle and can rotate in the vertical plane only,

Firstly you will observe that the rod rotates because the blocks weight has a net torque about it (when it is not at centre)

Let the distance along the rod be measured as 'x' and the angle the rod makes with the horizontal is 'a'

Qualitatively here are the Three situations you will expect would happen

1) The block moves too fast (at t=0) and so it slides off the rod very soon

2) The block moves too slow and comes to rest at the middle (ofcourse this can never practically happen, imagine the limiting case)

3) The block exactly reaches the centre when the rod becomes horizontal and both also reach the extremum at the same time, which means the system will be in oscillatory motion forever

Ok now let us see what The Math of the situation says,,

First let me make some assumptions so the differential equation we have to solve comes out to be simple

namely

1) The angle the rod rotates is very small, and thus this also tells that the systems initial energy is very small and hence we can neglect coriolis and centripetal forces in rotating frame (or their accelerations in ground frame)

2) All motions very slow

While the assumptions greatly limit us, i am forced to do so but the results are still awesome and Diverse

So, Let us analyse -

The torque on the rod is

$$Nx\quad =\quad I\ddot { a }$$

(worry not the directions and signs, they all depend on initial conditions and we can use negative numbers)

(also 'a' is the angle)

Now, for small angles, the normal reaction is roughly the weight of the block itself,

hence

$$Nx\quad =\quad I\ddot { a } =mgx$$

Now the force on the block along the rod assuming no friction is

$$mgsin(a)-m{ w }^{ 2 }x=m\ddot { x } \simeq \quad mga$$

Now to solve the two equations, i differentiate second one twice and substitute to get

$$\\ m{ g }^{ 2 }x\quad =\quad I\ddddot { x } \\ \\ m{ g }^{ 2 }a\quad =\quad I\ddddot { a }$$

Solving them yields me

$$x\quad =\quad A{ e }^{ wt }+B{ e }^{ wt }+Csin(wt+D)\\ \\ a\quad =\quad A{ e }^{ wt }+B{ e }^{ wt }+Csin(wt+D)$$

where $$w=\left( \frac { m{ g }^{ 2 } }{ I } \right) ^{ \frac { 1 }{ 4 } }$$

Which tell me all the information i need,

Now observe, if A and B are 0 as per initial conditions,

, Then the rod executes simple harmonic motion along with the block ,

If C and D and B are 0, then Quickly the block acquires enormous speed and distance and the block flies off or slides off (technically this is wrong as under my approximation this is impossible)

and motion dies off if A=C=D=0 as it exponentially decays

and there are several other cases, by setting some initial conditions, you can easily check it.

Note by Mvs Saketh
2 years ago

Sort by:

This Interesting Situation , And I want to analyse it ! And Then Join the Discussion ! · 2 years ago

yes please analyse it, and do try to solve it without approximation, which seems too complex for me,, · 2 years ago

I thought exactly the same situation some months ago and wanted to make a question on it but can't because of the so complex differential equation, but you did good job with the approximations, I didn't even thought about making approximations and then analysing the situation. @Mvs Saketh · 2 years ago

making approximations seems reasonable to me as far as the question will be about SHM, cause we dont expect SHM beyond the "slightly displaced" zone anyway · 2 years ago

I gave up, I make all Possible efforts But Fails every time! It is Too too complex for me too ! Your Approximations are great! Now I feel that how Little I know about "how to solve an diff. equation "! · 2 years ago

Actually i saw a similar question in the US physics team webpage, where the same question was posted except we were to assume that they are performing SHM and find the condition, and even there approximations were allowed however as we could directly assume SHM so one doesnt have to solve any differential equation,, but then when i analysed the differential equation , all these extra terms showing different kinds of motion than just SHM turned up But yes as you said actual differential equation is way too complex. · 2 years ago


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3. Creating configurations (gen3sis input)

26.10.2021

The configuration object defines the core processes of the system modeled. By defining some run settings the configuration object is essential to define the model to be simulated and facilitates formalization of eco-evolutionary processes.

• Core functions: include a custom initialization, observer, speciation, dispersal, evolution and ecology functions. Altogether, these six functions are applied as defined in the simulation engine. The possibility to customize these functions confers the high flexibility and generality of gen3sis in terms of including a wide range of theoretical knowledge.
• Settings: include the ecological traits considered in the simulation; whether a random seed is used thus enabling simulation reproducibility; start and end times of the simulation; and rules about aborting the simulation, including the maximum global or local species number permitted.

This vignette will guide you trough the config input object. It can be either an R object or a file, that can be easily shared and modified.

Here we will:

1. Create an empty config using the write_config_skeleton function.

2. Explain an example config using the config from introduction vignette.

3. Modify a config from within R.

library(gen3sis)

1. Create an empty config

In order to create an empty config, we use the function write_config_skeleton and define the output location. For that we have to define where the empty config will be stored.

# get data path
datapath <- system.file(file.path("extdata", "EmptyConfig"), package="gen3sis")
# set config_empty.R file path
config_file_path <- file.path(tempdir(), "config_empty.R")

And then create an empty config.

#writes out a config skeleton
write_config_skeleton(config_file_path)

This is how the config_empty.R looks like. We present them by main areas and each part is described briefly. For more information and a hand’s on example, see next session.

Content of config_empty.R

Metadata

# Version: 1.0
# Author:
# Date:
# Landscape:
# Publications:
# Description:

Settings

# set the random seed for the simulation.
random_seed = NA
# set the starting time step or leave NA to use the earliest/highest time-step.
start_time = NA
# set the end time step or leave as NA to use the latest/lowest time-step (0).
end_time = NA
# maximum total number of species in the simulation before it is aborted.
max_number_of_species = 25000
# maximum number of species within one cell before the simulation is aborted.
max_number_of_coexisting_species = 2500
# a list of traits to include with each species.
# a "dispersal" trait is implicitly added in any case.
trait_names = c("dispersal")
# ranges to scale the input environments with:
# not listed variable:         no scaling takes place
# listed, set to NA:           env. variable will be scaled from [min, max] to [0, 1]
# listed with a given range r: env. variable will be scaled from [r1, r2] to [0, 1]
environmental_ranges = list( )

Observer

end_of_timestep_observer = function(data, vars, config){
# the list of all species can be found in data$all_species. # the current landscape can be found in data$landscape.
}

Initialization

# the initial abundance of a newly colonized cell, both during setup and later when
# colonizing a cell during the dispersal.
initial_abundance = 1
# place species in the landscape:
create_ancestor_species <- function(landscape, config) {
stop("create the initial species here")
}

Core Processes

### Dispersal
# the maximum range to consider when calculating the distances from local distance inputs.
max_dispersal <- Inf
# returns n dispersal values.
get_dispersal_values <- function(n, species, landscape, config) {
stop("calculate dispersal values here")
}
### Speciation
# threshold for genetic distance after which a speciation event takes place.
divergence_threshold = NULL
# factor by which the divergence is increased between geographically isolated population.
# can also be a matrix between the different population clusters.
get_divergence_factor <- function(species, cluster_indices, landscape, config) {
stop("calculate divergence factor here")
}
### Evolution
# mutate the traits of a species and return the new traits matrix.
apply_evolution <- function(species, cluster_indices, landscape, config) {
stop("mutate species traits here")
}
### Ecology
# called for every cell with all occurring species, this function calculates abundances and/or who survives for each sites
# returns a vector of abundances.
# set the abundance to 0 for every species supposed to die.
apply_ecology <- function(abundance, traits, environment, config) {
stop("calculate species abundances and deaths here")
}

Note that the main core functions have stop warnings by default, this forces the configuration to be edited according to the rules and assumptions (model) that is to be created. You can do this from scratch or by modifying an existing config.

2. Explain example config

Here we will go bit by bit of the config used at the introduction vignette vignette. We will follow the structure presented above.

Metadata

The first section of the config file contains metadata. It informs the user when and by whom the file has been created, which landscape it was made for and informs about associated publications and implemented processes or exploration intentions of the specific configuration.

# Version: 1.0
# Author: Oskar Hagen
# Date: 26.10.2020
# Landscape: SouthAmerica
# Publications: R-package gen3sis
# Description: Example config used at the introduction vignette and similar to case study global configs in Hagen et al. 2020.
# O. Hagen, B. Flück, F. Fopp, J.S. Cabral, F. Hartig, M. Pontarp, T.F. Rangel, L. Pellissier. gen3sis: the general engine for eco-evolutionary simulations on the origins of biodiversity.

Settings

We start our simulation 40 Ma, which corresponds to time step 40 since 1 time-step = 1 myr. We end it a the latest available time-step at the landscape, thus start_time is set to 40 and end_time is set to NA. We limit the maximum total number of species alive in a simulation to 50000 with max_number_of_species and the maximum number of species within one site to 10000 with max_number_of_coexisting_species. Like so, simulations that could possible go off hand and generate too many species are stopped and properly flagged at the sgen3sis summary object. We define which traits we will consider in our simulation with traits_names, in our example, species only have a dispersal ability and a optimum temperature traits. For our model to work, we normalize the environmental values, according to the range informed at environmental_ranges.

random_seed = 6
start_time = 40
end_time = NA
max_number_of_species = 50000
max_number_of_coexisting_species = 10000
trait_names = c("temp",  "dispersal")
environmental_ranges = list("temp" = c(-45, 55), "area"=c(2361.5, 12923.4),
"arid"=c(1,0.5))

Observer

With the observer function, changes over time in any abiotic or biotic information of the virtual world can be recorded by defining the outputs that are saved at specified time steps, and results can be saved and plotted in real-time as the model runs. In our observer function we save the species and plot the richness. Like so, plots are printed as the simulation progresses and also saved inside the plot folder inside output.

end_of_timestep_observer = function(data, vars, config){
save_species()
plot_richness(data$all_species, data$landscape)
}

Combined plots can be generated inside the observer function as for example:

end_of_timestep_observer = function(data, vars, config){
par(mfrow=c(2,3))
plot_raster_single(data$landscape$environment[,"temp"], data$landscape, "temp", NA) plot_raster_single(data$landscape$environment[,"arid"], data$landscape, "arid", NA)
plot_raster_single(data$landscape$environment[,"area"], data$landscape, "area", NA) plot_richness(data$all_species, data$landscape) plot_species_presence(data$all_species[[1]], data$landscape) plot(0,type='n',axes=FALSE,ann=FALSE) mtext("STATUS",1) } Initialization The initialization function creates the ancestor species at the start of the simulation. Users can define the number of ancestor from a single to multiple species, their distribution within the paleolandscape and their trait values. In our example, the initial abundance of the 10 ancestor species, as well as the abundance of species on newly colonized sites is 1. We restrict the region of colonization to South America by limiting the range to a spatial extent of c(-95, -24, -68, 13). We then create ten different species, each of which is randomly inhabiting one of the sites of the continent. Thereby all of the sites have an equal probability of being inhabited by a species in the beginning of the simulation. However since the sites vary in their size, the habitation probability per area is not the same for all regions. If you want to correct for that you can consider using an equal area transformation of your coordinates or tweeking sampling functions to account for area correction. We set the optimum temperature trait temp of each population of the initial species to the environmental temperature. This means, that each population is adapted to the local conditions, as we will see later again at the ecology function. Dispersal is set to one, since it is not allowed to evolve, we use it’s maximum value. The function create_ancestor_species is expected to return a list of new species, i.e. the ancestor(s) one(s). initial_abundance = 1 create_ancestor_species <- function(landscape, config) { range <- c(-95, -24, -68, 13) co <- landscape$coordinates
selection <- co[, "x"] >= range[1] &
co[, "x"] <= range[2] &
co[, "y"] >= range[3] &
co[, "y"] <= range[4]
new_species <- list()
for(i in 1:10){
initial_cells <- rownames(co)[selection]
initial_cells <- sample(initial_cells, 1)
new_species[[i]] <- create_species(initial_cells, config)
#set local adaptation to max optimal temp equals local temp
new_species[[i]]$traits[ , "temp"] <- landscape$environment[initial_cells,"temp"]
weight_abd <- species$abundance[cells_cluster]/mean_abd traits[cells_cluster, "temp"] <- mean(traits[cells_cluster, "temp"]*weight_abd) } #mutations mutation_deltas <-rnorm(length(traits[, "temp"]), mean=0, sd=trait_evolutionary_power) traits[, "temp"] <- traits[, "temp"] + mutation_deltas return(traits) } Ecology The ecology function determines the abundance or presence of the populations in occupied sites of each species. The function iterates over all occupied sites and updates the species population abundances or presences on the basis of local environmental values, updated co-occurrence patterns and species traits. The function takes as input the species abundance, species trait, species divergence and clusters, and the landscape values. In our example, we calculate abundances in a site based on how close the species is to the site temperature. We scale the values to avoid small numbers and apply a carrying capacity based on aridity and temperature corrected by the area of a site. If abundances are below 1, species are considered extinct, if total abundance in a site is above the carrying capacity, small abundances are removed progressively and randomly distributed across the present species until total abundance is smaller or equal to the carrying capacity. apply_ecology <- function(abundance, traits, landscape, config) { abundance_scale = 10 abundance_threshold = 1 #abundance treashold survive <- abundance>=abundance_threshold abundance[!survive] <- 0 abundance <- (( 1-abs( traits[, "temp"] - landscape[, "temp"]))*abundance_scale)*as.numeric(survive) #abundance thhreashold abundance[abundance<abundance_threshold] <- 0 k <- ((landscape[,"area"]*(landscape[,"arid"]+0.1)*(landscape[,"temp"]+0.1))*abundance_scale^2) total_ab <- sum(abundance) subtract <- total_ab-k if (subtract > 0) { # print(paste("should:", k, "is:", total_ab, "DIFF:", round(subtract,0) )) while (total_ab>k){ alive <- abundance>0 loose <- sample(1:length(abundance[alive]),1) abundance[alive][loose] <- abundance[alive][loose]-1 total_ab <- sum(abundance) } #set negative abundances to zero abundance[!alive] <- 0 } return(abundance) } 3. Modify a config A configuration object can be modified by editing a file or the config object in R. Here we will load the config_southamerica.R as an R object and modify the initialization of species. Instead of having multiple species that occupy one site each, we will now initialize one species that occurs in all habitable sites and starts with abundance 10. # get data path datapath <- system.file(file.path("extdata", "SouthAmerica"), package="gen3sis") # creates config object from config file config_object <- create_input_config(file.path(datapath, "config/config_southamerica.R")) # modify the initialization function config_object$gen3sis$initialization$create_ancestor_species <- function(landscape, config) {
range <- c(-95, -24, -68, 13)
co <- landscape$coordinates selection <- co[, "x"] >= range[1] & co[, "x"] <= range[2] & co[, "y"] >= range[3] & co[, "y"] <= range[4] initial_cells <- rownames(co)[selection] new_species <- create_species(initial_cells, config) #set local adaptation to max optimal temp equals local temp new_species$traits[ , "temp"] <- landscape$environment[initial_cells,"temp"] new_species$traits[ , "dispersal"] <- 1
new_species$abundance <- new_species$abundance*10
return(list(new_species))
}

Additionally, we will now implement a simpler ecology function in our config_object. Abundances in a site are defined by how close the species population optimum temperature is to the site temperature. The abundance_threshold parameter defines the niche width cutoff. The higher the value, the narrower is the temperature range of the niche.

# modify the ecology function
config_object$gen3sis$ecology$apply_ecology <- function(abundance, traits, landscape, config, abundance_scale = 10, abundance_threshold = 8) { #abundance threshold abundance <- as.numeric(!abundance<abundance_threshold) abundance <- (( 1-abs( traits[, "temp"] - landscape[, "temp"]))*abundance_scale)*abundance #abundance threshold abundance[abundance<abundance_threshold] <- 0 return(abundance) } Like so, functions and parameters can be conveniently changed and simulated in gen3sis. Remember that run_simulation works either with a path to a config file or config object, so we can run the old and the new modified config. To compare the initial modified config, we will load the old config and set the old and the new config to run only for 3 time-steps and plot species ranges plot_ranges config_object_old <- create_input_config(file.path(datapath, "config/config_southamerica.R")) # define observer config_object_old$gen3sis$general$end_of_timestep_observer <- function(data, vars, config){
plot_ranges(data$all_species, data$landscape, disturb=0, max_sps=10)
}

config_object$gen3sis$general$end_of_timestep_observer <- config_object_old$gen3sis$general$end_of_timestep_observer

# define only 3 time-steps
config_object_old$gen3sis$general$end_time <- 38 config_object$gen3sis$general$end_time <- 38

Before using the modified configs, test if it is valid. If the function verify_config returns TRUE, you are good to go.

verify_config(config_object_old)
#> [1] TRUE
verify_config(config_object)
#> [1] TRUE

Run the modified old config

sim_old <- run_simulation(config = config_object_old,
landscape = file.path(datapath, "landscape"),
output_directory=tempdir(),
call_observer = 4,
verbose=0) # no progress printed

And run the modified new config

sim_new <- run_simulation(config = config_object,
landscape = file.path(datapath, "landscape"),
verbose=0, # no progress printed
output_directory=tempdir())


text:
# Another Analysis question: continuity and compactness

1. ### TaylorWatts

16
Let I = [0,1] be the closed unit interval. Suppose f is a continuous mapping from I to I. Prove that for one x an element of I, f(x) = x.

Proof:

Since [0,1] is compact and f is continuous, f is uniformly continuous.

This is where I'm stuck. I'm wondering if I can use the fact that since max {|x-y| = 1} if |x-y| = 1 then f(x) - f(y) < max {episolin}. This of course only occurs when WLOG x=1 y=0.

Stuck as far as the rest of it goes.

2. ### Moo Of Doom

367
Consider the function $g : I \to \mathbb{R}$ given by $g \left( x \right) = f \left( x \right) - x$. What happens if $g \left( x \right) = 0$ for some $x$? How can you show that this has to happen in $I$? I'm sure you must know a nice theorem that you can use here.


text:
+0

# what is the greatest common factor of 1134218 and 1135620

0
454
3

what is the greatest common factor of 1134218 and 1135620 77 pnxpw

Guest Nov 23, 2014

#3
+90023
+5

Heureka's method is from Euclid.....

It says that, if a - b = c  and c / b, then c / a   and c is the greatest common factor of a and b

So

1135620

- 1134218

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

1420

And 1420 / 1134218  →    so,  1420 / 1135620     and is the greatest common factor of both numbers......

CPhill  Nov 24, 2014
#1
+17745
+5

Since both 1 134 218 and 1 135 620 are both even, 2 divides into both

1 134 218 ÷ 2 = 567 109

1 134 620 ÷ 2 = 567 810

After going online and finding a list of prime numbers, trying them one by one, you can get to 701:

567 109 ÷ 701 = 809

567 810 ÷ 701 = 810

Since 809 and 810 are relatively prime (no whole number divides into both of them), we can stop.

The greatest common factor is 2 x 701 = 1402.

Actually, I didn't do that. I have a calculator that, if you divide 1134218 by 1135620, you get .9987654321...

Then, by using a button that changes decimal fractions to common fractions, it gives the answer 809/810.

Dividing 1134218 by 809, the result is 1402, which is 701 x 2. Then, I went to an online list of prime numbers and finding 701, I knew that I was finished.

geno3141  Nov 23, 2014
#2
+20025
+5

what is the greatest common factor of 1134218 and 1135620

Use the euklid algorithmn:

$$a= 1135620 \ and \ b= 1134218 \ and \ a> b\\ \\ \begin{array}{c|c|c|c|c|c} \hline a & b & q & r \\ \hline 1135620 & 1134218 & \frac{a}{b}={1} & {1402} & 1135620 = {1} * 1134218 + {1402} \\ 1134218 (old\ b) & 1402(old\ r) & \frac{a}{b}={809} & {0} &1134218 = {809} * 1134218 + {0}& r= 0 \ {Stop }\\ \hline \end{array}$$

if  r = 0 then in the line before r = 1402 is the greatest common factor

heureka  Nov 24, 2014
#3
+90023
+5

Heureka's method is from Euclid.....

It says that, if a - b = c  and c / b, then c / a   and c is the greatest common factor of a and b

So

1135620

- 1134218

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

1420

And 1420 / 1134218  →    so,  1420 / 1135620     and is the greatest common factor of both numbers......

CPhill  Nov 24, 2014


text:
# PR2 head control segmentation fault

Hello !

I'm running PR2 in simulation with visual SLAM algorithm (in ROS electric), and a strange behaviour appears when I spawn the robot as his head falls :

When I did the simple_head tutorial, duplicate the point head node to create a raise head node that uses the lookAt method to set the head in the right direction. Everything compiled and worked well till yesterday. I changed a value in the call to lookAt to ajust the direction and compiled again, but now the node is segfaulting at launch. Investigating a bit more showed that the segfault is in instantiating the PointHeadClient in constructor of RobotHead class. I checked actionlib, controller messages, but they're built so they should be effective.

I think the problem may be correlated as I moved from boost 1.40 to 1.49 by copying files in boost_1_49_0/boost in /usr/include/boost as property tree are required by the visual SLAM I use.

Thanks, Erwan

EDIT : here is the code :

#include <ros/ros.h>

#include <actionlib/client/simple_action_client.h>

// Our Action interface type, provided as a typedef for convenience

{
private:

public:
//! Action client initialization
{
//Initialize the client for the Action interface to the head controller
//wait for head controller action server to come up
{
ROS_INFO("Waiting for the point_head_action server to come up");
}
}

{
}

//! Points the high-def camera frame at a point in a given frame
void lookAt(std::string frame_id, double x, double y, double z)
{
//the goal message we will be sending

//the target point, expressed in the requested frame
geometry_msgs::PointStamped point;
point.point.x = x; point.point.y = y; point.point.z = z;
goal.target = point;

//we are pointing the high-def camera frame
//(pointing_axis defaults to X-axis)
goal.pointing_frame = "high_def_frame";

//take at least 0.5 seconds to get there
goal.min_duration = ros::Duration(0.5);

//and go no faster than 1 rad/s
goal.max_velocity = 1.0;

//send the goal

//wait for it to get there (abort after 2 secs to prevent getting stuck)
}

//! Shake the head from left to right n times
{
int count = 0;
while (ros::ok() && ++count <= n )
{
//Looks at a point forward (x=5m), slightly left (y=1m), and 1.2m up

//Looks at a point forward (x=5m), slightly right (y=-1m), and 1.2m up
}
}
};

int main(int argc, char** argv)
{
//init the ROS node
ros::init(argc, argv, "robot_driver");

// The only difference in raisehead is that the previous line is replaced by :
}

edit retag close merge delete

Sort by » oldest newest most voted

Problem solved using boost 1.41.0 but I don't think it's a good solution. If someone can update this tutorial with recent informations, it may help keeping ROS usable.

BtW, version 1.50 of boost looks absolutely not compatible with electric.

EDIT : as having some header-only libraries (and others to be compiled), Boost header-only files must be placed in a folder that is parsed when the compiler look for libraries and headers. The path :

/usr/include/

is an acceptable place for that (I don't know Linux well enough, but I think boost comes with the recent distributions - at least of Ubuntu - so there'll be a folder called boost in this directory). You just have to copy the boost_1_xx_y/boost subdirectory content into /usr/include/boost folder. In my case, the upper code doesn't work with version 1.46 and more. As far as I know, it's a problem in the class instanciation. I need property_tree from boost to use the OpenRatSLAM implementation, that are available from boost 1.41, so I downgraded to that version. I think it's not a good solution as there are fixes in newer implementations of boost that are probably useful. Maybe tutorials would be revised to update the code with more up-to-date ROS classes and features. In my case, I'm only using a low importance program (raising head), but it's required to run RatSLAM (as it is visual SLAM).

more

So what did you do exactly? Even if it is a workaround, other people might benefit from it if you described it some more.

( 2012-07-05 20:40:32 -0500 )edit

The fact that Boost 1.50 is incompatible with electric shouldn't come as a surprise, see reps/rep-0003 for more information. 1.50 is more than 10 minor releases newer than what Electric was developed for / with. Incompatibilities expected.

( 2012-07-05 20:43:39 -0500 )edit

( 2012-07-05 23:15:35 -0500 )edit

Boost is not a pure header-only library. There are a bunch of .so files, including threads, regexp and filesystem. Since many ros libraries and nodes link against at least boost thread, it's always a bad idea to use different headers in an own project without recompiling whole ROS.

( 2012-07-05 23:36:21 -0500 )edit

Ok so you mean that maybe the problem comes from the fact that I updated the header files and not the compiled librairies (I had some troubles understanding how to install boost the first time I had a problem of property_tree) ? - updated my answer to be more precise.

( 2012-07-05 23:44:01 -0500 )edit

@Lorenz: certain parts of boost are header only, but you are right, some are not. And as you point out (and I did earlier), mixing and matching versions is not a good idea, no matter how small the version difference (major / minor versions).

( 2012-07-05 23:44:38 -0500 )edit

@Erwan R.: no, Lorenz is telling you that you cannot (in principle) mix different versions of Boost for use with ROS. The ROS packages are linked against a specific version of Boost, your nodes are then linked against ROS and against your newer version of Boost.

( 2012-07-05 23:46:30 -0500 )edit

@Erwan R.: it might be possible with header only libraries and taking some care with linking, but this is not trivial, and not a supported use case (I assume).

( 2012-07-05 23:49:47 -0500 )edit

I think the problem may be correlated as I moved from boost 1.40 to 1.49 by copying files in boost_1_49_0/boost in /usr/include/boost as property tree are required by the visual SLAM I use.

So you copied a set of headers from a Boost v1.49 distribution to /usr/include/boost on a system that originally came with v1.40?

Depending on what exactly is using those new headers, and how isolated they are (ie: do they depend on other bits of Boost which you didn't copy), mixing & matching two Boost versions might not be such a good idea.

Everything compiled and worked well till yesterday. I changed a value in the call to lookAt to ajust the direction and compiled again, but now the node is segfaulting at launch.

If you added the Boost v1.49 headers before recompiling your node, perhaps see if things start working again if you remove them and recompile? Segfaults are one possibility of having version(header) != version(library) for instance.

EDIT after the added answer and more info by @Erwan R.: See 'Using concurrently 2 versions of boost' on stackoverflow for some additional info on the problem. As I mentioned: it can be done, but probably requires quite some work.

more

I went back to v1.40 and the segfault disappeared, but I strongly need a boost version that supports property_tree for another algorithm. I'm gonna investigate v1.50 or degrade to a version that is compatible with this node and has property_tree. (sry for 4-times comment)

( 2012-07-05 03:18:22 -0500 )edit


text:
Get A+ with YaClass!
Register now to understand any school subject with ease and get great results on your exams!

### Theory:

Calculation of dust particles in the air:

The dust particles present in the air at a particular place can be easily calculated with a graph sheet.
Dust particles on graph sheet
1. In a graph sheet, using a marker pen, a square of $$\text{5} \times \text{5}\ cm$$ is drawn.
2. A thin film of grease is applied to the graph sheet to serve as a dust collector.
3. Likewise, four or five graph sheets are made.
4. Each dust collector is placed in different places, and the time to collect the dust particles is scheduled.
5. After the scheduled time, the paper is removed, and the number of collected dust particles in the marked area is counted using a magnifying glass.
The mean number of dust particles in the marked area can be calculated using the formula,

$$\text{Mean} = \frac{\text{Total number of dust particles on collector}}{\text{Number of squares on collector}}$$

The calculation of dust range is done with the help of the formula,

$$Range$$ $$=$$ $$Maximum\ value\ –$$ $$minimum\ value$$

From the collected details, the recordings are tabulated in the given table format.

 Location of dust collector Mean number of dust particles per small square Range

If the collector sheets are kept on the roadside or any windy places, then the range of dust will be more. If they are kept inside closed places like buildings or houses, the range of dust will be lesser, comparatively.
Water vapour:
Have you ever noticed some water droplets on the outside of glass containing cold water in it? If not, try to notice from now.

A glass containing ice cubes or cold water will form tiny water droplets over their outer surface. This is because air contains water vapour in it.

Water droplets on the glass tumbler

When the air comes in contact with a cold surface, the water vapour present in it condenses on the glass and then drops of water appear on the outer surface.

Water droplets on the glass

Water vapour in the air plays a vital role in the process of water cycle. Water vapour rises in the air, gets cooled and condenses to form clouds that bring us rain.

Water cycle
Reference:
https://pxhere.com/en/photo/1382808


text:
# Trial solution, undetermined coefficients, 2nd order non-homogeneous equation

Tags:
1. Aug 5, 2015

### kostoglotov

1. The problem statement, all variables and given/known data

Write a trial solution for the method of undetermined coefficients. Do not determine the coefficients.

For: $y'' + 2y' + 10y = x^2e^{-x}\cos{3x}$

There's a modification performed and I'm not 100% confident as to why.

2. Relevant equations

3. The attempt at a solution

The complementary equation gives a solution of

$$y_c = e^{-x}\left(c_1\cos{3x} + c_2\sin{3x}\right)$$

Now, before considering a modification to the particular solution $y_p$, let's consider the standard trial solution by the method of undetermined coefficients. I would say it's

$$y_p = e^{-x}\left[(Ax^2+Bx+C)\cos{3x} + (Dx^2+Ex+F)\sin{3x}\right]$$

Now I know that the modification is performed so that "no term in the particular solution can be a solution for the complementary equation"...I do know basically what this means...and I would have thought that for the particular trial solution above, that neither term would suitably solve the complementary equation. Because the particular solution has a polynomial of degree 2 and the complementary solution has a polynomial of degree 0.

So I've thought, maybe we could solve the complementary equation with the trial solution $y_p$ if we set A,B and/or D,E to = 0,0, so that the polynomials in the trial solutions were reduced from 2nd degree to 0 degree.

Is this why we need to modify $y_p$ by multiply it by x? Because we can set A,B,D,E to zero in order to solve for the complementary equation? Thus, it won't really matter what degree a polynomial is in the solutions as far as deciding to modify is concerned?

2. Aug 6, 2015

### ehild

As e-xcos(3x) is solution of the homogeneous (complementary) equation, you have to modify the particular solution by multiplying it with x.

Last edited: Aug 6, 2015
3. Aug 6, 2015

### kostoglotov

Yes I understand that, but in my particular solution I've already multiplied it by a polynomial of 2nd degree...why does it need to be brought up to a polynomial of 3rd degree?

4. Aug 6, 2015

### ehild

You had to multiply with second order polynomials as there is the factor x2 in the inhomogeneous term. One more x to multiply with is because e-xcos(3x) is solution of the homogeneous equation. Try to find a solution with your method; you won't be able.

Last edited: Aug 6, 2015
5. Aug 7, 2015

### vela

Staff Emeritus
Denote the derivative operator by D. The operator D+r acting on a term of the form $x^p e^{-rx}$ yields
$$(D+r) (x^p e^{-rx}) = (p x^{p-1} e^{-rx} - r x^p e^{-rx}) + r x^p e^{-rx} = p x^{p-1} e^{-rx}.$$ The effect of the operator is basically to reduce the power of $x$ by 1 if $p\ne 0$. If p=0, we have $(D+r)e^{-rx} = 0$. In other words, $e^{-rx}$ is a solution of the homogeneous differential equation y' + ry = 0.

Now consider the differential equation $(D+r)y = x^n e^{-rx}$. It has a homogeneous solution $y_h = e^{-rx}$. From the result above, you should be able to see that in order to get a term like $x^n e^{-rx}$, $y$ must contain a term of the form $x^{n+1} e^{-rx}$. It doesn't matter that $x^n e^{-rx}$, …, and $x e^{-rx}$ are independent from $y_h$. When D+r acts on those terms, you end up with $x^{n-1}e^{-rx}$, …, and $e^{-rx}$, and there's no linear combination of them that is equal to $x^n e^{-rx}$.

Last edited: Aug 7, 2015


text:
Equations and Inequalities

# Linear Inequalities and Absolute Value Inequalities

### Learning Objectives

In this section you will:

• Use interval notation.
• Use properties of inequalities.
• Solve inequalities in one variable algebraically.
• Solve absolute value inequalities.

It is not easy to make the honor role at most top universities. Suppose students were required to carry a course load of at least 12 credit hours and maintain a grade point average of 3.5 or above. How could these honor roll requirements be expressed mathematically? In this section, we will explore various ways to express different sets of numbers, inequalities, and absolute value inequalities.

### Using Interval Notation

Indicating the solution to an inequality such ascan be achieved in several ways.

We can use a number line as shown in (Figure). The blue ray begins atand, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4.

We can use set-builder notation:which translates to “all real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set.

The third method is interval notation, in which solution sets are indicated with parentheses or brackets. The solutions toare represented asThis is perhaps the most useful method, as it applies to concepts studied later in this course and to other higher-level math courses.

The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.” A few examples of an interval, or a set of numbers in which a solution falls, areor all numbers betweenandincludingbut not including$\left(-1,0\right),$all real numbers between, but not includingandandall real numbers less than and including(Figure) outlines the possibilities.

Set Indicated Set-Builder Notation Interval Notation
All real numbers between a and b, but not including a or b
All real numbers greater than a, but not including a
All real numbers less than b, but not including b
All real numbers greater than a, including a
All real numbers less than b, including b
All real numbers between a and b, including a
All real numbers between a and b, including b
All real numbers between a and b, including a and b
All real numbers less than a or greater than b
All real numbers

### Using Interval Notation to Express All Real Numbers Greater Than or Equal to a

Use interval notation to indicate all real numbers greater than or equal to

Use a bracket on the left ofand parentheses after infinity:The bracket indicates thatis included in the set with all real numbers greater thanto infinity.

### Try It

Use interval notation to indicate all real numbers between and includingand

### Using Interval Notation to Express All Real Numbers Less Than or Equal to a or Greater Than or Equal to b

Write the interval expressing all real numbers less than or equal toor greater than or equal to

We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins atand ends atwhich is written as

The second interval must show all real numbers greater than or equal towhich is written asHowever, we want to combine these two sets. We accomplish this by inserting the union symbol,between the two intervals.

### Try It

Express all real numbers less thanor greater than or equal to 3 in interval notation.

### Using the Properties of Inequalities

When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol.

### Properties of Inequalities

These properties also apply to$a>b,$and

Illustrate the addition property for inequalities by solving each of the following:

• (a)
• (b)
• (c)

The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.

Solve:

### Demonstrating the Multiplication Property

Illustrate the multiplication property for inequalities by solving each of the following:

Solve:

### Solving Inequalities in One Variable Algebraically

As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.

### Solving an Inequality Algebraically

Solve the inequality:

Solving this inequality is similar to solving an equation up until the last step.

The solution set is given by the intervalor all real numbers less than and including 1.[/hidden-answer]

### Try It

Solve the inequality and write the answer using interval notation:

### Solving an Inequality with Fractions

Solve the following inequality and write the answer in interval notation:

We begin solving in the same way we do when solving an equation.

The solution set is the interval[/hidden-answer]

### Try It

Solve the inequality and write the answer in interval notation:

### Understanding Compound Inequalities

A compound inequality includes two inequalities in one statement. A statement such asmeansandThere are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods.

### Solving a Compound Inequality

Solve the compound inequality:

The first method is to write two separate inequalities:andWe solve them independently.

Then, we can rewrite the solution as a compound inequality, the same way the problem began.

In interval notation, the solution is written as

The second method is to leave the compound inequality intact, and perform solving procedures on the three parts at the same time.

### Try It

Solve the compound inequality:

### Solving a Compound Inequality with the Variable in All Three Parts

Solve the compound inequality with variables in all three parts:

Let’s try the first method. Write two inequalities:

The solution set isor in interval notationNotice that when we write the solution in interval notation, the smaller number comes first. We read intervals from left to right, as they appear on a number line. See (Figure).

### Try It

Solve the compound inequality:

### Solving Absolute Value Inequalities

As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located athas an absolute value ofas it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero.

An absolute value inequality is an equation of the form

Where A, and sometimes B, represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of all values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.

There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.

Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of$600. We can solve algebraically for the set of x-values such that the distance betweenand 600 is less than 200. We represent the distance betweenand 600 asand therefore,or

This means our returns would be between $400 and$800.

To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.

### Absolute Value Inequalities

For an algebraic expression X, andan absolute value inequality is an inequality of the form

These statements also apply toand

### Determining a Number within a Prescribed Distance

Describe all valueswithin a distance of 4 from the number 5.

We want the distance betweenand 5 to be less than or equal to 4. We can draw a number line, such as in (Figure), to represent the condition to be satisfied.

The distance fromto 5 can be represented using an absolute value symbol,Write the values ofthat satisfy the condition as an absolute value inequality.

We need to write two inequalities as there are always two solutions to an absolute value equation.

If the solution set isandthen the solution set is an interval including all real numbers between and including 1 and 9.

### Try It

Describe all x-values within a distance of 3 from the number 2.

Solve .

### Using a Graphical Approach to Solve Absolute Value Inequalities

Given the equation determine the x-values for which the y-values are negative.

We are trying to determine wherewhich is whenWe begin by isolating the absolute value.

Next, we solve for the equality

Now, we can examine the graph to observe where the y-values are negative. We observe where the branches are below the x-axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis atandand that the graph opens downward. See (Figure).

### Try It

Solve

orin interval notation, this would be

Access these online resources for additional instruction and practice with linear inequalities and absolute value inequalities.

### Key Concepts

• Interval notation is a method to indicate the solution set to an inequality. Highly applicable in calculus, it is a system of parentheses and brackets that indicate what numbers are included in a set and whether the endpoints are included as well. See (Figure) and (Figure).
• Solving inequalities is similar to solving equations. The same algebraic rules apply, except for one: multiplying or dividing by a negative number reverses the inequality. See (Figure), (Figure), (Figure), and (Figure).
• Compound inequalities often have three parts and can be rewritten as two independent inequalities. Solutions are given by boundary values, which are indicated as a beginning boundary or an ending boundary in the solutions to the two inequalities. See (Figure) and (Figure).
• Absolute value inequalities will produce two solution sets due to the nature of absolute value. We solve by writing two equations: one equal to a positive value and one equal to a negative value. See (Figure) and (Figure).
• Absolute value inequalities can also be solved by graphing. At least we can check the algebraic solutions by graphing, as we cannot depend on a visual for a precise solution. See (Figure).

### Section Exercises

#### Verbal

When solving an inequality, explain what happened from Step 1 to Step 2:

When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.

When solving an inequality, we arrive at:

Explain what our solution set is.

When writing our solution in interval notation, how do we represent all the real numbers?

When solving an inequality, we arrive at:

Explain what our solution set is.

Describe how to graph

We start by finding the x-intercept, or where the function = 0. Once we have that point, which iswe graph to the right the straight line graphand then when we draw it to the left we plot positive y values, taking the absolute value of them.

#### Algebraic

For the following exercises, solve the inequality. Write your final answer in interval notation.

For the following exercises, solve the inequality involving absolute value. Write your final answer in interval notation.

All real numbers

No solution

For the following exercises, describe all the x-values within or including a distance of the given values.

Distance of 5 units from the number 7

Distance of 3 units from the number 9

Distance of10 units from the number 4

Distance of 11 units from the number 1

#### Graphical

For the following exercises, graph the function. Observe the points of intersection and shade the x-axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation.

It is never less than zero. No solution.

For the following exercises, graph both straight lines (left-hand side being y1 and right-hand side being y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the y-values of the lines.

Where the blue line is above the orange line; point of intersection is

Where the blue line is above the orange line; always. All real numbers.

#### Numeric

For the following exercises, write the set in interval notation.

For the following exercises, write the interval in set-builder notation.

For the following exercises, write the set of numbers represented on the number line in interval notation.

#### Technology

For the following exercises, input the left-hand side of the inequality as a Y1 graph in your graphing utility. Enter y2 = the right-hand side. Entering the absolute value of an expression is found in the MATH menu, Num, 1:abs(. Find the points of intersection, recall (2nd CALC 5:intersection, 1st curve, enter, 2nd curve, enter, guess, enter). Copy a sketch of the graph and shade the x-axis for your solution set to the inequality. Write final answers in interval notation.

Where the blue is below the orange; always. All real numbers.

Where the blue is below the orange;

#### Extensions

Solve

Solve

$x\ne -7$

is a profit formula for a small business. Find the set of x-values that will keep this profit positive.

#### Real-World Applications

In chemistry the volume for a certain gas is given bywhere V is measured in cc and T is temperature in ºC. If the temperature varies between 80ºC and 120ºC, find the set of volume values.

84 mi

#### Complex Numbers

For the following exercises, use the quadratic equation to solve.

For the following exercises, name the horizontal component and the vertical component.

horizontal componentvertical component

For the following exercises, perform the operations indicated.

For the following exercises, solve the quadratic equation by factoring.

For the following exercises, solve the quadratic equation by using the square-root property.

For the following exercises, solve the quadratic equation by completing the square.

For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No real solution.

For the following exercises, solve the quadratic equation by the method of your choice.

#### Other Types of Equations

For the following exercises, solve the equations.

#### Linear Inequalities and Absolute Value Inequalities

For the following exercises, solve the inequality. Write your final answer in interval notation.

No solution

For the following exercises, solve the compound inequality. Write your answer in interval notation.

For the following exercises, graph as described.

Graph the absolute value function and graph the constant function. Observe the points of intersection and shade the x-axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation.

Graph both straight lines (left-hand side being y1 and right-hand side being y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the y-values of the lines. See the interval where the inequality is true.

Where the blue is below the orange line; point of intersection is

### Chapter Practice Test

Graph the following:

x y
0 2
2 5
4 8

Find the x- and y-intercepts for the following:

Find the x- and y-intercepts of this equation, and sketch the graph of the line using just the intercepts plotted.

$\left(4,0\right)$

Find the exact distance betweenandFind the coordinates of the midpoint of the line segment joining the two points.

Write the interval notation for the set of numbers represented by

Solve for x:

Solve for x:

Solve for x:

Solve for x:

$x=\frac{-5}{2},1$

The perimeter of a triangle is 30 in. The longest side is 2 less than 3 times the shortest side and the other side is 2 more than twice the shortest side. Find the length of each side.

Solve:

Solve:

Solve:

For the following exercises, find the equation of the line with the given information.

Passes through the pointsand

Has an undefined slope and passes through the point

Passes through the pointand is perpendicular to

Simplify:

Multiply:

Divide:

Solve this quadratic equation and write the two complex roots inform:

Solve:

Solve:

Solve:

Solve:

Solve:

Solve:

For the following exercises, find the real solutions of each equation by factoring.

### Glossary

compound inequality
a problem or a statement that includes two inequalities
interval
an interval describes a set of numbers within which a solution falls
interval notation
a mathematical statement that describes a solution set and uses parentheses or brackets to indicate where an interval begins and ends
linear inequality
similar to a linear equation except that the solutions will include sets of numbers


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• ### The THESEUS space mission concept: science case, design and expected performances(1710.04638)

March 27, 2018 astro-ph.IM, astro-ph.HE
THESEUS is a space mission concept aimed at exploiting Gamma-Ray Bursts for investigating the early Universe and at providing a substantial advancement of multi-messenger and time-domain astrophysics. These goals will be achieved through a unique combination of instruments allowing GRB and X-ray transient detection over a broad field of view (more than 1sr) with 0.5-1 arcmin localization, an energy band extending from several MeV down to 0.3 keV and high sensitivity to transient sources in the soft X-ray domain, as well as on-board prompt (few minutes) follow-up with a 0.7 m class IR telescope with both imaging and spectroscopic capabilities. THESEUS will be perfectly suited for addressing the main open issues in cosmology such as, e.g., star formation rate and metallicity evolution of the inter-stellar and intra-galactic medium up to redshift $\sim$10, signatures of Pop III stars, sources and physics of re-ionization, and the faint end of the galaxy luminosity function. In addition, it will provide unprecedented capability to monitor the X-ray variable sky, thus detecting, localizing, and identifying the electromagnetic counterparts to sources of gravitational radiation, which may be routinely detected in the late '20s / early '30s by next generation facilities like aLIGO/ aVirgo, eLISA, KAGRA, and Einstein Telescope. THESEUS will also provide powerful synergies with the next generation of multi-wavelength observatories (e.g., LSST, ELT, SKA, CTA, ATHENA).
• ### Observations of Gamma-ray Bursts with ASTRO-H and Fermi(1503.01182)

March 4, 2015 astro-ph.HE
ASTRO-H, the sixth Japanese X-ray observatory, which is scheduled to be launched by the end of Japanese fiscal year 2015 has a capability to observe the prompt emission from Gamma-ray Bursts (GRBs) utilizing BGO active shields for the soft gamma-ray detector (SGD). The effective area of the SGD shield detectors is very large and its data acquisition system is optimized for short transients such as short GRBs. Thus, we expect to perform more detailed time-resolved spectral analysis with a combination of ASTRO-H and Fermi LAT/GBM to investigate the gamma-ray emission mechanism of short GRBs. In addition, the environment of the GRB progenitor should be a remarkable objective from the point of view of the chemical evolution of high-z universe. If we can maneuver the spacecraft to the GRBs, we can perform a high-resolution spectroscopy of the X-ray afterglow of GRBs utilizing the onboard micro calorimeter and X-ray CCD camera.
• We evaluate the exposure during nadir observations with JEM-EUSO, the Extreme Universe Space Observatory, on-board the Japanese Experiment Module of the International Space Station. Designed as a mission to explore the extreme energy Universe from space, JEM-EUSO will monitor the Earth's nighttime atmosphere to record the ultraviolet light from tracks generated by extensive air showers initiated by ultra-high energy cosmic rays. In the present work, we discuss the particularities of space-based observation and we compute the annual exposure in nadir observation. The results are based on studies of the expected trigger aperture and observational duty cycle, as well as, on the investigations of the effects of clouds and different types of background light. We show that the annual exposure is about one order of magnitude higher than those of the presently operating ground-based observatories.
• Contributions of the JEM-EUSO Collaboration to the 32nd International Cosmic Ray Conference, Beijing, August, 2011.
• ### Gamma-Ray Burst Polarimeter - GAP - aboard the Small Solar Power Sail Demonstrator IKAROS(1010.5305)

Oct. 26, 2010 astro-ph.IM
The small solar power sail demonstrator "IKAROS" is a Japanese engineering verification spacecraft launched by H-IIA rocket on May 21, 2010 at JAXA Tanegashima Space Center. IKAROS has a huge sail with 20 m in diameter which is made of thin polyimide membrane. This sail converts the solar radiation-pressure into the propulsion force of IKAROS and accelerates the spacecraft. The Gamma-Ray Burst Polarimeter (GAP) aboard IKAROS is the first polarimeter to observe the gamma-ray polarization of Gamma-Ray Bursts (GRBs) during the IKAROS cruising phase. GAP is a tinny detector of 3.8 kg in weight and 17 cm in size with an energy range between 50-300 keV. The GAP detector also plays a role of the interplanetary network (IPN) to determine the GRB direction. The detection principle of gamma-ray polarization is the anisotropy of the Compton scattering. GAP works as the GRB polarimeter with the full coincidence mode between the central plastic and the surrounding CsI detectors. GAP is the first instrument, devoted for the observation of gamma-ray polarization in the astronomical history. In this paper, we present the GAP detector and its ground and onboard calibrations.
• ### X-Ray Spectral Variability of the Seyfert Galaxy NGC 4051 Observed with Suzaku(0811.0692)

Nov. 5, 2008 astro-ph
We report results from a Suzaku observation of the narrow-line Seyfert 1 NGC 4051. During our observation, large amplitude rapid variability is seen and the averaged 2--10 keV flux is 8.1x10^-12 erg s^-1 cm^-2, which is several times lower than the historical average. The X-ray spectrum hardens when the source flux becomes lower, confirming the trend of spectral variability known for many Seyfert 1 galaxies. The broad-band averaged spectrum and spectra in high and low flux intervals are analyzed. The spectra are first fitted with a model consisting of a power-law component, a reflection continuum originating in cold matter, a blackbody component, two zones of ionized absorber, and several Gaussian emission lines. The amount of reflection is rather large (R ~ 7, where R=1 corresponds to reflection by an infinite slab), while the equivalent width of the Fe-K line at 6.4 keV is modest (140 eV) for the averaged spectrum. We then model the overall spectra by introducing partial covering for the power-law component and reflection continuum independently. The column density for the former is 1x10^23 cm^-2, while it is fixed at 1x10^24 cm-2 for the latter. By comparing the spectra in different flux states, we identify the causes of spectral variability. (abridged)
• ### Hard X-ray Detector (HXD) on Board Suzaku(astro-ph/0611232)

Nov. 7, 2006 astro-ph
The Hard X-ray Detector (HXD) on board Suzaku covers a wide energy range from 10 keV to 600 keV by combination of silicon PIN diodes and GSO scintillators. The HXD is designed to achieve an extremely low in-orbit back ground based on a combination of new techniques, including the concept of well-type active shield counter. With an effective area of 142 cm^2 at 20 keV and 273 cm2 at 150 keV, the background level at the sea level reached ~1x10^{-5} cts s^{-1} cm^{-2} keV^{-1} at 30 keV for the PI N diodes, and ~2x10^{-5} cts s^{-1} cm^{-2} keV^{-1} at 100 keV, and ~7x10^{-6} cts s^{-1} cm^{-2} keV^{-1} at 200 keV for the phoswich counter. Tight active shielding of the HXD results in a large array of guard counters surrounding the main detector parts. These anti-coincidence counters, made of ~4 cm thick BGO crystals, have a large effective area for sub-MeV to MeV gamma-rays. They work as an excellent gamma-ray burst monitor with limited angular resolution (~5 degree). The on-board signal-processing system and the data transmitted to the ground are also described.
• ### In-Orbit Performance of the Hard X-ray Detector on board Suzaku(astro-ph/0611233)

Nov. 7, 2006 astro-ph
The in-orbit performance and calibration of the Hard X-ray Detector (HXD) on board the X-ray astronomy satellite Suzaku are described. Its basic performances, including a wide energy bandpass of 10-600 keV, energy resolutions of ~4 keV (FWHM) at 40 keV and ~11% at 511 keV, and a high background rejection efficiency, have been confirmed by extensive in-orbit calibrations. The long-term gains of PIN-Si diodes have been stable within 1% for half a year, and those of scintillators have decreased by 5-20%. The residual non-X-ray background of the HXD is the lowest among past non-imaging hard X-ray instruments in energy ranges of 15-70 and 150-500 keV. We provide accurate calibrations of energy responses, angular responses, timing accuracy of the HXD, and relative normalizations to the X-ray CCD cameras using multiple observations of the Crab Nebula.
• ### Wide band X-ray Imager (WXI) and Soft Gamma-ray Detector (SGD) for the NeXT Mission(astro-ph/0410402)

Oct. 27, 2004 astro-ph
The NeXT mission has been proposed to study high-energy non-thermal phenomena in the universe. The high-energy response of the super mirror will enable us to perform the first sensitive imaging observations up to 80 keV. The focal plane detector, which combines a fully depleted X-ray CCD and a pixellated CdTe detector, will provide spectra and images in the wide energy range from 0.5 keV to 80 keV. In the soft gamma-ray band up to ~1 MeV, a narrow field-of-view Compton gamma-ray telescope utilizing several tens of layers of thin Si or CdTe detector will provide precise spectra with much higher sensitivity than present instruments. The continuum sensitivity will reach several times 10^(-8) photons/s/keV/cm^(2) in the hard X-ray region and a few times10^(-7) photons/s/keV/cm^(2) in the soft gamma-ray region.
• ### GRB Formation Rates inferred from the Spectral Peak Energy - Luminosity Relation(astro-ph/0309217)

Sept. 8, 2003 astro-ph
• ### Non-Equilibrium Ionization States of GRB Environments(astro-ph/0107330)

July 18, 2001 astro-ph
Iron spectral features are thought to be the best tracer of a progenitor of gamma-ray bursts (GRBs). The detections of spectral features such as an iron line and/or a Radiative Recombination edge and Continuum (RRC) were reported in four X-ray afterglows of GRBs. However their properties were different each other burst by burst. For example, Chandra observation of GRB 991216 reported both the strong H-like iron line together with its RRC. On the contrary, Yoshida et al. (2001) report only a detection of the strong RRC in GRB 970828 with ASCA. Since it is difficult to produce the strong RRC, we have to consider special condition for the line and/or the RRC forming region. In this paper, we point out a possibility of a non-equilibrium ionization state'' for the line and the RRC forming region.


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### Introduction

From an applications viewpoint, eigenvalue problems are probably the most important problems that arise in connection with matrix analysis. In this Section we discuss the basic concepts. We shall see that eigenvalues and eigenvectors are associated with square matrices of order $n×n$ . If $n$ is small (2 or 3), determining eigenvalues is a fairly straightforward process (requiring the solutiuon of a low order polynomial equation). Obtaining eigenvectors is a little strange initially and it will help if you read this preliminary Section first.

#### Prerequisites

• have a knowledge of determinants and matrices
• have a knowledge of linear first order differential equations

#### Learning Outcomes

• obtain eigenvalues and eigenvectors of $2×2$ and $3×3$ matrices
• state basic properties of eigenvalues and eigenvectors

1.1 Determinants

2.1 Case 1


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In the previous post we discussed the number of primes less than a given number and derived some very poor estimates for this quantity. In this post, using no extra technical machinery whatsoever, we derive a slightly better estimate.

Previously, we used Euclid’s proof of the infinitude of the primes as inspiration for a way of arriving at an upper bound for the number of primes less than a given number. Here, we will come up with a slightly cleverer, more quantitative proof of Euclid’s Theorem and, as before, our estimate will grow out of the proof of Euclid’s Theorem, the improvement in the bound being testament to the fact that the proof is more illuminating.

In Euclid’s proof, one constructs a number which is not divisible by any of the first n primes, but in fact it is clear we are done if we prove the existence of a number that has a prime factor not among the first n primes. To do this we need to show that it cannot be the case that every number has all of its prime factors among the first n primes. And, luckily, it does indeed seem very unlikely that every number has all of its prime factors among the first n primes. So the quantitative question we ask is: “How many numbers less than or equal to x have all of their prime factors among the first n primes?”. Call this number $N(x).$

### Reductio

Let’s estimate $N(x)$: Let k be some number less than or equal to x, all of whose prime factors are among the first n primes. We can write

$k = k_1^2m$,

where m is square-free, that is it is not divisible by the square of any prime, and where $k_1 \leq \sqrt{k} \leq \sqrt{x}$. The question is: “How many such numbers are there?”. We can get away with a crude estimate obtained by counting the two factors entirely independently: Firstly, there are at most $\sqrt{x}$ choices for $k_1$ simply because it is less than $\sqrt{x}$. What about m? Well, we know that m must be a product of primes (this follows from the fact that every number is divisible by a prime – see the previous post) and that all its prime factors are among the first n primes so

$m = p_1^{\alpha_1}\cdot\dots \cdot p_n^{\alpha_n}$,

where, since m is not divisible by the square of any prime, each exponent $\alpha_j$ is either equal to one or equal to zero. Counting all possible choices of these exponents tells us there are at most $2^n$ possible values of m. This counting work tells us that $N(x) \leq 2^n\sqrt{x}$.

If we assume for the sake of contraction that there are only n primes, then $N(x) = x-1$ for every x, because every number (except one) has a prime divisor and therefore every number (except one) has a prime divisor among the first n primes. This would mean that $x-1 \leq 2^n\sqrt{x}$ for every $x$, which is clearly nonsense. Therefore, there are infinitely many primes.

### An Estimate and a Sum

As promised, we use the simple idea of this proof to get an estimate on the number of primes less than a given number. Fix $x$ and write $n = \pi(x)$. This means we are calling the biggest prime less than or equal to $x$ the $n^{th}$ prime. So the first $n$ primes are actually all the primes less than or equal to x. Thus $N(x)$ is counting the number of numbers less than or equal to $x$ all of whose prime factors are less than or equal to x. On the one hand this is obviously equal to $x -1$, but on the other hand we have the bound obtained in the previous section, i.e.$N(x) \leq 2^{\pi(x)}\sqrt{x}$, and so $x-1 \leq 2^{\pi(x)}\sqrt{x}$. We can deduce from this that

$\pi(x) \geq C\log x$.

While this is a bit of an improvement on $\log \log x$, it is still very bad. For example, it says that the number of primes less than one billion is greater than 14.9 when in actual fact it is greater than fifty million.

Another thing that comes out of the work in this post is a very quick and neat proof that the sum of the reciprocals of the prime nunbers diverges, i.e. this sum:

$\sum_{p\ \text{prime}}\frac{1}{p}$.

Our observation is the following one: For any prime number $p$, only every $p^{th}$ number is divisible by $p$. Therefore the number of numbers less than or equal to $x$ which are divisible by a given prime $p$ is at most $x/p$. We use this observation in the context of the quantity $x - N(x)$. This is the number of numbers less than or equal to $x$ which have a prime divisor greater than $p_n$, i.e. which are divisible by one or more of $p_{n+1},\ p_{n+2},\ ...$. This is not more than

$\frac{x}{p_{n+1}} + \frac{x}{p_{n+2}} + \frac{x}{p_{n+3}} + ...$

If the sum $\sum_{p\ \text{prime}}\frac{1}{p}$ converged, then the tail of the sum could be made arbitrarily small and so by choice of n, we would have that

$\frac{x}{p_{n+1}} + \frac{x}{p_{n+2}} + \frac{x}{p_{n+3}} + ... < \frac{x}{2}.$

Thus $x - N(x) \leq x/2$, or $x/2 \leq N(x)$ which is clearly nonsense again. Therefore the sum diverges.


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# SAT Math : How to use the inverse variation formula

## Example Questions

### Example Question #361 : Algebra

The square of  varies inversely with the cube of . If  when , then what is the value of  when

Explanation:

When two quantities vary inversely, their products are always equal to a constant, which we can call k. If the square of x and the cube of y vary inversely, this means that the product of the square of x and the cube of y will equal k. We can represent the square of x as x2 and the cube of y as y3. Now, we can write the equation for inverse variation.

x2y3 = k

We are told that when x = 8, y = 8. We can substitute these values into our equation for inverse variation and then solve for k.

82(83) = k

k = 82(83)

Because this will probably be a large number, it might help just to keep it in exponent form. Let's apply the property of exponents which says that abac = ab+c.

k = 82(83) = 82+3 = 85.

Next, we must find the value of y when x = 1. Let's use our equation for inverse variation equation, substituting 85 in for k.

x2y3 = 85

(1)2y3 = 85

y3 = 85

In order to solve this, we will have to take a cube root. Thus, it will help to rewrite 8 as the cube of 2, or 23.

y3 = (23)5

We can now apply the property of exponents that states that (ab)c = abc.

y3 = 23•5 = 215

In order to get y by itself, we will have the raise each side of the equation to the 1/3 power.

(y3)(1/3) = (215)(1/3)

Once again, let's apply the property (ab)c = abc.

y(3 • 1/3) = 2(15 • 1/3)

y = 25 = 32

### Example Question #5 : Direct And Inverse Variation

varies directly as and inversely as .

and .

Which of the following is true about ?

varies inversely as the square of and directly as the fourth power of .

varies directly as both the square root of and the fourth root of .

varies inversely as the square root of and directly as the fourth root of .

varies directly as the square of and inversely as the fourth power of .

varies directly as the square root of and inversely as the fourth root of .

varies inversely as the square of and directly as the fourth power of .

Explanation:

varies directly as and inversely as , so for some constant of variation ,

.

We can square both sides to obtain:

.

, so .

By substitution,

.

Using as the constant of variation, we see that varies inversely as the square of and  directly as the fourth power of .

### Example Question #3 : Direct And Inverse Variation

The radius of the base of a cylinder is ; the height of the same cylinder is ; the cylinder has volume 1,000.

Which of the following is a true statement?

Assume all quantities are positive.

varies dorectly as the square root of .

varies inversely as .

varies inversely as the square of .

varies inversely as the square root of .

varies directly as the square of .

varies inversely as .

Explanation:

The volume of a cylinder can be calculated from its height and the radius of its base using the formula:

, so ;

, so .

The volume is 1,000, and by substitution, using the other equations:

If we take as the constant of variation, we get

,

meaning that varies inversely as .

### Example Question #1 : How To Use The Inverse Variation Formula

varies directly as the square of and the cube root of , and inversely as the fourth root of . Which of the following is a true statement?

varies directly as the square root of , and inversely as the eighth root of and the sixth root of .

varies directly as the square root of , and inversely as the eighth power of  and the sixth power of .

varies directly as the square root of and the eighth power of , and inversely as the sixth root of .

varies directly as the square root of and the eighth root of , and inversely as the sixth root of .

varies directly as the square root of and the eighth root of , and inversely as the sixth power of .

varies directly as the square root of and the eighth root of , and inversely as the sixth root of .

Explanation:

varies directly as the square of and the cube root of , and inversely as the fourth root of , so, for some constant of variation ,

We take the reciprocal of both sides, then extract the square root:

Taking as the constant of variation, we see that varies directly as the square root of and the eighth root of , and inversely as the sixth root of .

### Example Question #1 : How To Use The Inverse Variation Formula

If  varies inversely as , and  when , find  when .


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Question

# The given solids are made up of small cubes of side 1 cm. Which of these solids are formed of more number of small cubes?

Open in App
Solution

## We know volume of cube = side × side × side Volume of the bigger cube = 8 × 8 × 8 = 512 cu. cm Volume of the small cube = 1 × 1 × 1 = 1 cu. cm Number of small cubes used to make the bigger cube = 512 ÷ 1 = 512 We know, Volume of cuboid = length × breadth × height Volume of the cuboid = 16 × 8 × 4 = 512 cu. cm Number of small cubes used to make the cuboid = 512 ÷ 1 = 512 So, we see that same number of small cubes are used to form both the cube and the cuboid.

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# Find the Differential Equation Relating Input Voltage and Output Voltage

1. Nov 10, 2012

### tesla93

Hey all,

I'm stuck on a dynamic systems question, it's attached as a jpeg

I started off by writing nodal equations for each node:

Node 1: 1/R1(ei-eA)=C1D(eA-eo)+1/R2(eA-eB)

Node 2:1/R2(eA-eB)=C2D(eB)

I know that I have to isolate for ei and eo but I'm really confused with the D^2 that's in the answer. Any help is greatly appreciated! :)

#### Attached Files:

• ###### dynamic systems question.jpg
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2. Nov 10, 2012

### Staff: Mentor

Write the node equations in the state space domain (Laplace); The impedance of a capacitor is given by 1/(s*C). The state variable 's' translates to the differential operator D → $\frac{d}{dt}$ in the time domain.

3. Nov 10, 2012

### aralbrec

It's the second derivative DDy -> D2y

For example, if you took the derivative of your second node equation:

1/R2(eA-eB) = C2D(eB)
D(1/R2(eA-eB)) = D(C2D(eB))
1/R2 D(eA-eB) = C2 D2(eB)

4. Nov 11, 2012

### tesla93

Ohhhh okay that makes sense. Took a while to wrap my head around it but I got the right answer in the end. Thanks! :)

5. Nov 11, 2012

### tesla93

While I'm at it, I'm having trouble with another question like the one above, but this time it's a mechanical system. It's the jpeg below.

I started off with the writing equations for each mass

Mass 1: K1x1 + DB(x) + k2(x1-x2) = 0

Mass 2: m2D^2x = DB(x1-x2) + k2(x1-x2)

First off, can someone confirm that those equations are right?

After that, I'm supposed to sub in the given values and get everything in terms of fa(t) and x1, but I feel like my equations are wrong because it's not working out.

Thanks for looking!

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6. Nov 11, 2012

### Staff: Mentor

You've used variables x, x1, and x2 in your equations. I can see x1 and x2 indicated on your diagram, but what is x?

The method that I usually prefer for these sorts of problems is to convert the mechanical system into its equivalent electrical analog. Then I have a available all the analytical apparatus for electrical circuits (including circuit simulators). Your problem "translates" into:

Note that potentials at V1 and V2 will represent the velocities of the masses. In the Laplace domain, position x1(s) can be found by integrating velocity V1(s). Integration is trivially accomplished by dividing by the Laplace state variable: x1(s) = V1(s)/s.

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# Question #92256

Aug 4, 2015

See explanation

#### Explanation:

Break this into two parts, firstly the inner part:

${e}^{x}$

This is positive and increasing for all real numbers and goes from 0 to $\infty$ as $x$ goes from $- \infty$ to $\infty$

The we have:

$\arctan \left(u\right)$

The has a right horizontal asymptote at $y = \frac{\pi}{2}$. Going from $u = 0 \rightarrow \infty$, at $u = 0$ this function is positive and increasing over this domain, takes a value of 0 at $u = 0$, a value of $\frac{\pi}{4}$ at $u = 1$ and a value of $\frac{\pi}{2}$ at $u = \infty$.

These points therefore get pulled to $x = - \infty , 0 , \infty$ respectively and we end up with a graph looking like this as a result:

graph{arctan(e^x) [-10, 10, -1.5, 3]}

Which is the positive part of the $\arctan$ function stretch over the entire real line with the left value being stretch into a horizontal asymptote at $y = 0$.

Aug 4, 2015

See explanation

#### Explanation:

Domain is $\mathbb{R}$

Symmetry
Neither with respect to the $x$ axis nor w.r.t the origin.

$\arctan \left({e}^{- x}\right)$ does not simplify to $\arctan \left({e}^{x}\right)$
nor to $- \arctan \left({e}^{x}\right)$

Intercepts
$x$ intercepts: none

We cannot get $y = 0$ because that would require ${e}^{x} = 0$
But ${e}^{x}$ is never $0$, it only approaches $0$ as $x \rightarrow - \infty$.
So, $y \rightarrow 0$ as $x \rightarrow - \infty$ and the $x$ axis os a horizontal
asymptote on the left.

$y$ intercept: $\frac{\pi}{4}$

When $x = 0$, we get $y = \arctan \left(1\right) = \frac{\pi}{4}$

Asymptotes:
Vertical : none

$\arctan$ is between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$ by definition, so never goes to $\infty$

Horizontal:
Left: $y = 0$ as discussed above

Right: $y = \frac{\pi}{2}$

We know that, as $\theta \rightarrow \frac{\pi}{2}$ with $\theta < \frac{\pi}{2}$, we get $\tan \theta \rightarrow \infty$
so, as $x \rightarrow \infty$, we get ${e}^{x} \rightarrow \infty$, so $y = \arctan \left({e}^{x}\right) \rightarrow \frac{\pi}{2}$

First derivative

$y ' = {e}^{x} / \left(1 + {e}^{2 x}\right)$ is never $0$ and never undefined, so there are no critical numbers.

For every $x$ we have $y ' > 0$ so the function is increasing on $\left(- \infty , \infty\right)$

There are no local extrema.

Second derivative

$y ' ' = \frac{{e}^{x} \left(1 + {e}^{2 x}\right) - {e}^{x} \left(2 {e}^{2 x}\right)}{1 + {e}^{2 x}} ^ 2$

$= \frac{{e}^{x} + {e}^{3 x} - 2 {e}^{3 x}}{1 + {e}^{2 x}} ^ 2$

$= \frac{{e}^{x} \left(1 - {e}^{2 x}\right)}{1 + {e}^{2 x}} ^ 2$

$y ' '$ is never undefined, and it is $0$ at $x = 0$

Sign of $y ' '$:
On $\left(- \infty , 0\right)$, we get ${e}^{2 x} < 1$ so $y ' ' > 0$ and the graph is concave up

On $\left(0 , \infty\right)$, we get ${e}^{2 x} > 1$ so $y ' ' < 0$ and the graph is concave down

The concavity changes at $x = 0$, so the inflection point is:
$\left(0 , \frac{\pi}{4}\right)$

Now sketch the graph


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# Recent content by kirbylee

1. ### For Sale Minimax project

I still got everything except the prop. But I have another brand new one I may sell. I also think there are several other small parts that would save alot of time. If you would like to call, I could probably tell you what else I have. I should have listed everything. I'll try and check my...

Yes I do.
5. ### For Sale Minimax project

I have a complete set of wings less paint. Ailerons Landing gear. Wind shield. Seat back. Push pull cables. Several wooden, and aluminum brackets. Hours of work. Elevator. Rudder with hinges. Horz stab. Verticle fin. Control stick. Wheels, & Tires. 3 blade Ultra Prop. Wood to...
6. ### Wanted JC Propeller Design Software

Did you ever find a copy of the software?
7. ### Minimax wings on an Affordaplane

You're probably right, I was going to build the Affordaplane wings, but just ran into this Minimax project along the way. Mainly bought it for the engine. You're also right about the confusion being an enemy.
8. ### Minimax wings on an Affordaplane

The grass strip in Jonesboro? If your talking about the one out on highway 18 with all the hangers that Marty Bunch built, yes. I used to go out there all the time. Planning on heading that way again, maybe tomorrow. Might even pitch one of those wings in the truck and take it with me. Sometimes...
9. ### Minimax wings on an Affordaplane

Actually I used to hang around a crop duster's strip a long time ago. Learned a lot of things about airplanes from those guys. Watched him cover part of a Thrush, just watched. Got to be good friends with him and several other guys. His hair was starting to turn grey then, now both of us are...
10. ### Minimax wings on an Affordaplane

I wouldn't be able to fly anything as questionable as these wings, and have enjoy it. I bought the wings, landing gear, engine, and several other parts, from a friend of mine that I thought was I could trust. I'm not sure if the wings have flown. I asked him about the guy he got it from. He said...
11. ### Minimax wings on an Affordaplane

I understand. It was nice thinking I wasn't going to have to build the wings. Thanks for helping.
12. ### Minimax wings on an Affordaplane

The plywood gusset that the aileron fitting is bolted to is raised up off of the rib. I don't know if it was sloppy work, and was installed that way, or a weak bond and the aileron forced it up. There is no sign of glue that I can see. I have a few more photos, and would like to know what you...
13. ### Minimax wings on an Affordaplane

Seems I keep running into problems. At first glance all the dust makes the wood look awful, but a feather duster takes care of this easily. In most all the places where I picked and scraped on both wings have a good healthy crisp sound with no soft spots at all. Several dark spots, mainly...
14. ### Thou shall not stall for the ground may come up and smite thee

Thou shall not stall for the ground may come up and smite thee
15. ### Minimax wings on an Affordaplane

Someone else mentioned that. Sure would make things easier. I'm not sure how to get in there to get to it without cutting a hole in the fabric. I have almost zero experience covering airplanes, but its something I'm going to have to learn anyway. I already have one tear I have to fix, and may...


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###### Isolating on the Right

Note that in solving the equation in Exercise 2.3.16 we found that $$-5=a\text{,}$$ which is equivalent to $$a=-5\text{.}$$ We did not write $$a=-5$$ as an extra step though, as $$-5=a$$ identified the solution.

in-context


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# Why is the argument of $i$ equal to $\pi/2$?

So it's obvious geometrically that the argument of $z=i$ is $\pi/2$.

However the method of getting the argument is $\arctan(y/x)$. And when in the case of $z=i$, $y/x = 1/0$ which is undefined...

So when you want to find the argument of a complex number is this the correct process -

1. $\operatorname{Argument}(z) = \arctan(y/x)$ if $x\neq0$.
2. If $x = 0$, then $\operatorname{Argument}(z) = \pi/2\text{ or }-\pi/2$

Is that the way I should be approaching it?

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You will find this wiki article very relevant. – Sasha Jan 24 '12 at 1:12
Are you sure the arctangent rule is the definition of "argument" in the presentation you're following? That would be unusual; commonly one defines the argument in some different way, and then the arctangent rule is just a computational technique that works if (and only if) the real part is positive. – Henning Makholm Jan 24 '12 at 1:38

Although 1/0 is undefined, $\lim_{x \to 0^+}\operatorname{tan}(y/x)=\pi/2$ for any y. This corresponds with the geometric justification. One considers the limit as x goes to 0 from above because this is what happens when x rotates counterclockwise from 1 to i. To get the argument of -i, one still considers the counterclockwise rotation, which leads to the limit from below, which gives instead $-\pi/2$. The arguments of multiples of i and -i work out the same way.


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# $A^A$ in category of graphs

(reference is Lawvere/Schanuel, Session 31, Ex. 1)

I'm trying to calculate the exponential object $A^A$ and its evalution map $e \colon A \times A^A \to A$ in the category of graphs, where $A$ is the "arrow graph" (ie. one arrow and two dots).

In the following, $D$ is the graph with one dot and no arrows, $1$ is the terminal object in this category (graph with one dot and one arrow, the loop).

So far I have:

• The points of $\mathbf{1}\to A^A$ correspond to the maps $A\to A$ (via two standard isomorphisms), and since $\mathbf{1}$ is the loop, and there is one map of graphs $A \to A$, there is one loop in $A^A$.
• The dots $D\to A^A$ correspond to the maps $A \times D \to A$, of which there are four, hence four dots in $A^A$.
• The arrows $A \to A^A$ correspond to the maps $A \times A \to A$, of which there are four, hence four arrows in $A^A$.

But I'm stuck on how to put these together to constitute $A^A$ and its evaluation map.

You're looking good so far. The only information you're missing is which arrows go with which dots. So let's introduce some notation. Write $A=0\to 1$, so that $A\times D=\{0,1\}$ and $A\times A$ has vertices $\{(0,0),(0,1),(1,0),(1,1)\}$ and sole edge $\{((0,0),(1,1))\}$. Denote the four maps $D\times A\to A$ by $00,01,10,11$, where for instance $10(0)=1$ and $10(1)=0$. Write $p_x,p_y,m_1,m_0$ for the four maps $A\times A\to A$, sending respectively $(0,1)\mapsto 0$ and $(1,0)\mapsto 1$; $(0,1)\mapsto 1$ and $(1,0)\mapsto 0$; $(0,1),(1,0)\mapsto 1$; $(0,1),(1,0)\mapsto 0$. To determine their associated vertices we should precompose $p_x,p_y,m_1,m_0$ with the two inclusions $0,1:D\to A$. Under transposition, these become the two maps $a,b:D\times A\to A\times A$ with $a(0)=(0,0),a(1)=(0,1),b(0)=(1,0),b(1)=(1,1)$. (If you have unusual conventions for your Cartesian product, you could end up with the two "horizontal" maps, rather than the vertical ones, here.)
So we want to compose each of $p_x,...,m_0$ with each of $a,b$. This is now a straightforward computation: $p_xa=00,p_xb=11 , p_ya=01,p_yb=01,m_0a=00,m_0b=01,m_1a=01,m_1b=11$. Thus $A^A=(\{00,01,10,11\},\{(01,01),(00,11),(00,01),(01,11)\})$.