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# Resistors and the water analogy

The the most prevalent analogy for understanding electricity is the water hose one. I personally like this because it's very understandable but I have a question about how this works with resistors.

I understand that resistors will limit current which because of ohms law will affect the voltage but in a hose, decreasing the hose diameter (current) would increase the water pressure (voltage). Am I wrong on my understanding of this principle?

• current is not the hose diameter, it's how much water flows per second. – 1p2r3k4t Oct 15 '13 at 14:42
• Let us assume inlet of hose is connected with the over head tank (voltage source) and the outlet of hose is connected with the ground tank.we know water will flow from higher pressure to lower pressure point;so water will flow from OHT to ground tank(if the pressure at the outlet of hose is also same as that in the inlet of hose there wont be any water flow) – yogece Oct 15 '13 at 14:43
• Water as an analogy for resistive electricity, like most analogies, an imperfect comparison for the sake of teaching. – Passerby Oct 15 '13 at 15:30

I'm guessing this is the phenomena you are imagining:

You hold the end of a garden hose with nothing attached, and turn on the spigot. Some water gushes out, with high volume, but low velocity. Then, you restrict the end of the hose with your thumb. The volume of the water is less, but the pressure in the hose rises, and it shoots much farther.

Hydraulic systems can make good analogies for electric systems. The problem is that not all hydraulic systems have an electric analog, and some people suck at making hydraulic analogies.

Here's one problem with mapping this hydraulic system to an electric system: charge is conserved, both in the universe as a whole, and practically speaking, most circuits. With your garden hose, an unlimited supply of water is magically created by the city water system, and when it exits the hose and hits the ground, it's no longer in the circuit.

To get the electric analog of this, you need a circuit capable of shooting charged particles out into space. You also need something capable of supplying charged particles. Things like this do exist (for example, a CRT flings electrons through a near vacuum at the phosphor coating on the screen) but they typically require high voltages, and you aren't going to build anything like this with just a resistor and a battery. If you connect a resistor across a battery, the electric charge is pumped around the circuit. No charge enters, and no charge leaves.

Another problem: very commonly, electric circuits are designed to maintain a constant voltage. The water supply system is also designed to maintain a (roughly) constant pressure. However, since nothing (electric or hydraulic) can supply unlimited current, all of these voltage/pressure regulation systems have limits. In the case of your garden hose, the water supply system can't supply enough water to keep the pressure at the end of the hose at the target, say 30 psi. With nothing attached to the hose, there isn't enough resistance for the supply to work against to build the pressure. It is analogous to an electric short circuit.

If you were to block the hose entirely, you would find the pressure inside the hose (and indeed, everywhere in your hydraulic system, if the hight differences are irrelevant) will be 30 psi. If you were to open the end just a little bit, you'd find that the pressure is still pretty much 30 psi. Only until quite a lot of water is flowing would the pressure drop from 30 psi; this is because at high currents, the friction of the water flowing in the hose becomes significant, and increasingly more of the pressure is lost over the hose's resistance as the current increases.

If we wanted to model the hose-thumb system electrically, we'd need to take into account that the hose has some friction. Maybe something like this:

simulate this circuit – Schematic created using CircuitLab

This is called a voltage divider. When your finger is not on the hose, ($0\Omega$), the $5\Omega$ from the hose is very significant, since it's the biggest resistance in the system. When your finger is blocking most of the hose (let's say $1000\Omega$), then the additional $5\Omega$ from the hose makes very little difference.

but in a hose, decreasing the hose diameter (current) would increase the water pressure (voltage).

So with that explained, we can circle back to your confusion. It depends on where you measure the pressure. You probably know that if you totally block the hose, the water pressure does not rise without bound (your pipes would burst!). Using thinner or longer pipes actually decreases the water pressure available at the spigot (or appliance, or whatever is connected), because more pressure is lost to friction between the supply and the spigot. There's an electric analog: higher currents require fatter wire.

In the water analogy voltage is presented by water level, not pressure (though there's a relationship between the two). A 12 V across the resistor is equivalent to a (for instance) 12 m water level difference between the ends of the hose. If there's no water level difference ( no voltage difference) then there's no current, no matter what the pressure is.

• So using this analogy, the current could be slowed down... (the flow) but the voltage could remain the same if it was a constant supply (the water level)? But there is also voltage consumed by the resistor correct? – G Thompson Oct 15 '13 at 14:41
• There is more than one water analogy. – Phil Frost Oct 15 '13 at 14:43
• @GThompson: no, voltage is not consumed, but current is. The voltage is across the resistor, but doesn't necessarily change, while there's a current flowing. The water flow represents that consumption, and if the power source is a capacitor the flow will decrease the voltage difference, just like water flowing out of a tank will decrease the water level there. – Johan.A Oct 15 '13 at 14:43
• @Johan.A I appreciate it. So say I have a 12volt power supply and I have a 5volt DC motor. I can use resitors to bring the voltage down enough to keep it safe correct? Would I need to know the resistance of the Motor itself? – G Thompson Oct 15 '13 at 14:54
• Not really, no; the bad idea of using resistor dividers for this purpose keeps coming up and is still a bad idea. – pjc50 Oct 15 '13 at 15:03

Every analogy can be dangerous because is difficult to consider all things. It helps you understand, but you cannot consider it as a perfect model. Resistor will limit current considering you have a fixed source voltage. The ideal source voltage will not variate.

In the water analogy, a resistor is a constriction in a pipe. If you externally always force a certain flow rate (current) thru the pipe, then yes, the pressure will increase. Real resistors work that way too. If you always force a certain current thru a resistor, the voltage (pressure) will go up if the resistor goes up.

In the case of a fixed pressure (voltage) accross the ends of the pipe, the flow rate (current) will go down as the constriction gets more narrow (resistance goes up), just like it works with real resistors too.


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Question: SCAN.UPC : lost too many samples from GSE (BrainArray + ConvThreshold)
1
7 months ago by
s.goncalvesclaro10 wrote:

Hello everyone !

This is my first post here, and i thank you all for the time that you will give to solve my problem.

I'm a trainee and i'm working on a subject based on personnalized-medicine. I have to use gene expression data from the GEO database. To preprocess these raw data, i'm using the package SCAN.UPC, maintained by Stephen Picollo, which is a great tool and really effective for the next part of my work.

But i have a problem concerning the number of samples at the end of the preprocessing part. For example for the GSE26639 (REMAGUS02-trial), i get 64 patients from the 226 in total. I have followed the workflow proposed in the pdf available in Bioconductor, and mapped the probe to genes using the BrainArray package available for the specific platform. I'm using the latest version of BrainArray CDF, the 23.0.0.

I'm also tuning the convThreshold parameter, which allows me to gain some samples or to lost them. For example :

• convThreshold = 0.50 => 64 samples
• convThreshold = 0.90 => 45 samples
• convThreshold = 0.01 => 24 samples

It would be good if i can get between 60 and 75 % of the samples. Is there a way to achieve this goal ?

This a part of my code where i get the raw CEL files from GEO, get the corresponding ENTREZ BrainArray probe mapping, and process it all with SCAN.

# Get Raw Cel files and pkg name CDF BrainArray
raw_data <- get_CEL(gseName,PATH_RAW)
colnames(raw_data)[1]
celPath <- paste(PATH_RAW,gseName,"/data", sep = "")
pkgName <- InstallBrainArrayPackage(paste(celPath,colnames(raw_data)[1], sep = "/"),"23.0.0","hs", "entrezg")
# celFilesPath <- file.path(celPath, "*.CEL*")

# SCAN UPC part
gc()
scan_norm <- SCAN(gseName,probeSummaryPackage = pkgName,convThreshold = 0.40)
dim(t(exprs(scan_norm)))

Thank you and i hope it was clear.

modified 7 months ago • written 7 months ago by s.goncalvesclaro10

Thanks for your question. I may be misunderstanding, but GSE26636 only has 13 samples (see https://www.ncbi.nlm.nih.gov/geo/query/acc.cgi?acc=GSE26636). Is this the one you meant?

Thanks for the reply, sorry this is my mistake it is GSE26639 for REMAGUS02. There are 226 patients in this dataset. https://www.ncbi.nlm.nih.gov/geo/query/acc.cgi?acc=GSE26639

I have edited it in my main post.

Answer: SCAN.UPC : lost too many samples from GSE (BrainArray + ConvThreshold)
0
7 months ago by
s.goncalvesclaro10 wrote:

I solved my problem after further investigation. It was the registerDoParallel without any arguments like cores. After several redo of my workflow, i found out that the problem was not due to SCAN but to R with its memory management.

As R had some troubles before (abort computation, freeze), i let the registerDoParallel() like this. I hoped that R would automatically manage the parallel jobs considering my CPUs. But it is important to precise at least the cores argument like this : registerDoParallel(cores = 2), as described in the SCAN.UPC vignette.

Now the convThreshold argument has an effect on the number of sample that i get at the end of my workflow. I hope my mistake will help to prevent this error in other project using SCAN.UPC.

Thank you Stephen for your support. I'm going to use SCAN.UPC for my workflow, it is really trustworthy.

1

Glad you got it working. Thanks for the detailed description of how you solved it!

format: plain


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# Why is there no FPTAS for the maximum independent set problem?

I want to prove that the NP-hardness of Maximum Independent Set implies that there is no FPTAS for the Maximum Independent Set problem unless $P=NP$.

I found the following approach after some research online:

Assuming that such an FPTAS exists which computes a $(1 - \epsilon)$-approximation for Maximum Independent Set on a graph G. I want to use this assumed algorithm as a subroutine in a new algorithm to solve the maximum independent set problem exactly. Now proving that this new algorithm runs in polynomial time would be a contradiction (assuming $P \neq NP)$ since this would mean that I solved the maximum independent set problem in polynomial time. Therefore there exists no FPTAS.

I can't figure out a polynomial time algorithm which uses the above assumed subroutine. How does this algorithm look like, how to use the assumed subroutine (for example what $\epsilon$ to choose?).

• Hint: If $\varepsilon$ is sufficiently small compared to the optimum, a $(1-\varepsilon)$-approximation will necessarily give you the optimum (as solutions to the Independent Set problem are always integers). – user53923 Sep 15 '17 at 17:55
• @user53923 Let $opt$ denote the size of the maximal independent set. My goal is to set $\epsilon$ such that I get a "1-approximation" which would be an exact solution which would give me the contradiction I'm looking for. I would then choose $\epsilon = 1/opt$ which would yield $1-\epsilon = 1-(1/opt) = (opt-1)/opt = opt$ where the last equality comes from the fact that opt is an integer. Does this make sense? I'm not sure about my answer, the last step where I use the fact that opt is an integer bothers me, what if $opt=1$ for example? – Anna Vopureta Sep 15 '17 at 22:40
• You want that $(1-\varepsilon)opt > opt-1$. Your idea for $\varepsilon$ is almost there. Also note that you do not know $opt$, so instead, use an appopriate bound on $opt$. – user53923 Sep 16 '17 at 22:13


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## Truth

esr speaks it:

All data, including primary un-”corrected” datasets, must be available for auditing by third parties. All modeling code must be published. The assumptions made in data reduction and smoothing must be an explicitly documented part of the work product.

These requirements would kill off AGW alarmism as surely as a bullet through the head.

Transparency would kill off AGW denialism, too.

## What is a catastrophe?

In the previous post, I mentioned two types of catastrophe: cliff-diving and gradual. Suppose the first is a sudden major decrease in output. How much would returns to capital have to decline to get 1% average yearly returns over a century (given 6% “usual” returns)? This is the solution to this problem:
$\left \[(1.06)^{99} (1 + x) \right ]^{1/100} = 1.01$

If I got my sums right, this implies about a 99% reduction in returns to capital in the year of the catastrophe.

Now, suppose the catastrophe plays out over a decade. How negative would growth in capital productivity have to be to get an average 1% returns for a century? Doing a similar calculation, returns to capital would have to be about -35% every year for a decade to get average returns that low over the century.

These examples make the catastrophes of the 20th century pale in comparison. In the US in the 1930s, GDP growth was, on average, positive. Austria was the biggest loser during WWII, in terms of GDP, and it only decreased by about 10% a year from 1938 to 1945.

In Weitzman’s calibration, there’s a 1 in 200 chance of having 1% yearly returns over the next century. These examples make this calibration look rather unrealistic.

## But innovation is endogenous, too

Risk aversion usually makes investments look less attractive. In the standard story, then, as risk aversion goes up, we would spend less money today to avert future catastrophes. Martin Weitzman argues, however, this relationship reverses when we’re uncertain about future productivity levels. For if it turns out that productivity is low in the future (e.g. because of a climate catastrophe) then consumption will be low. Low consumption means high marginal utility and so low discount rates.

My purpose here is to focus sharply on clarifying the long-run discounting issue by using a
super-simple super-crisp formulation… There is no good substitute for seeing clearly before ones eyes the basic structure of a model laid bare.

Suppose [there’s uncertainty about discount rates in the future and] that discount rate ri > 0 will occur with probability-like weight wi > 0, where sum wi = 1… The [expected] discount rates … decline over time starting from their average value and going down over time to approach their lowest possible value. Over time, the impact of the higher discount rates … diminishes because the higher rates effectively discount themselves exponentially out of existence, leaving the field to the lower discount rates (and, eventually, to the lowest).

[A]ny given value of [future productivity level] subsequently determines the endogenous future growth rate … and endogenous consumption level … as the solution to a Ramsey optimal growth problem (given that value of [productivity]). The paper shows that when future productivity … is uncertain, then higher values of [risk aversion] are associated with lower future discount rates, thereby reversing the conventional wisdom.

[An example using standard parameter values] indicate[s] that the risk-aversion effects of uncertain future productivity on lowering distant-future discount rates might be quite powerful. The driving force is a ”fear factor” associated with the possibility of low-probability but catastrophically-high permanent damages to future productivity.

I personally would be inclined toward a much lower climate-change discount rate than 6% per annum, but the ultimate goal of this paper … will be to show that under uncertainty, even with expected discount rates as high as 6%, the effective discount rate, which ought to be used, can be much lower than 6%.

The results in the example he gives depend very much on the distribution of future productivity, especially on its variance. Here’s his assumed distribution:

In his model, this distribution should be interpreted as the distribution of the average productivity over a very long time (i.e. the next century or so). To me, his assumed uncertainty about productivity is too high (i.e. the variance of the distribution is too wide). Is it even remotely possible that the capital-output ratio will average 50 or 100 for the next century? Even if there’s a sudden, cliff diving catastrophe, I can’t imagine people won’t invent their way around it thus increasing the productivity of capital in the long run. And if catastrophe happens in slow motion, certainly people will be adjust fast enough to keep returns on capital high on average.

The most striking stylized fact about growth is its constancy. For whatever reason, people route around the particular circumstances of their time and space and invent their way to 2% growth. Why would climate change pose fundamentally different obstacles to innovation than what has been seen over the last 200 years? Do the open systems studied in Diamond’s Collapse tell us anything about the closed global system?

## Cool down, cool down

Even if The Emails discredit all the “science” on tree-rings, there’s still ice cores and bore hole data. And even if all paleoclimatology is BS, thermometer readings of temperatures show a sharp spike in the last couple of decades.

One piece of evidence for AGW has, perhaps!, been discredited. Not all evidence has been discredited. We still know C02 is a green house gas, humans have been pumping tons of it into the air and there’s the direct evidence of warming. There is no reason at all to think the AGW theory is a “scam” or a “fraud”.

## Other people’s emails

I agree with McArdle who agrees with Cowen and Hanson. There appears to be consensus in the literature…

Lesson one: scientists are people, people are jerks and that Mann guys seems to be an especially big person.

Lesson two: its way too fun reading other people’s email.

Lesson three (the real lesson): data and methods should be freely available. Reading all the nonsense that goes on trying to replicate policy-important studies (like the Mann et al 1999 paper) is infuriating. Luckily, in my experience, we don’t have this problem in economics.

Contrary to McArdle and Cowen, my prior on the joint claim “the last century was the hottest of the millennium” or “tree ring data is a good proxy for temperature” has decreased reading some of the replication studies. While thermometer readings show a pronounced increase in temperatures in recent decades, the tree ring data probably doesn’t. There’s even a good chance the tree ring data, if a good temperature proxy, show several periods in the past 1000 years that were warmer than the last century.

Here’s a paper in Science that uses a different data set from Mann paper and shows the so-called “Medieval Warm Period” probably had temperatures at least as high as the 20th century.

I haven’t seen a good explanation for why thermometer readings show different trends than tree rings (or other temperature proxies). Of course, I trust thermometers over proxies, but still.

## How much is a “big change” in the climate?

Here’s the density of temperature changes over centuries. I used these data and calculate the average change in temperature per century.

Last century’s temperature increase of 0.8 degrees C was an outlier (but not an extreme outlier). About 95% of temperature changes were slower than last century’s temperature change. If the climate models are correct and the world sees a 2.5 degree increase, this would be an extreme outlier. Only about 1 in a three or four hundred centuries sees that dramatic of temperature changes.

This, of course, doesn’t guarantee catastrophe, but it suggests we should at least insure ourselves against the possibility of catastrophe.

## Mass outbreak of illiteracy?

I just read The Chapter Five (in this book) and I’ll go as far as to say the critics don’t even have a point.

The chapter motivates geo-engineering without saying anything outlandish about the science (as far as this semi-informed lay-person can tell). There’s not even as much as an implication in the whole chapter that global warming isn’t a big deal. In fact, such a level-headed discussion of solutions to the problem actually bumped up my subjective probability of global warming being an issue to worry about. There’s no use fretting over something we can’t, practically, do anything about. It appears more likely to me, now, that we can do something about global warming.

The authors haven’t given up on marginal analysis, they seem to get that people respond to incentives and they appear to be aware of Pigovian-type taxes. They explicitly address the problems with carbon pricing and move the discussion forward with their description of one alternative policy (i.e. subsidized research in bio-engineering).

The reaction to this chapter was just plain strange.

## Does this argument really work?

Tim Lambert, responding to this opinion piece ((why in the deuce is science being argued in the opinion pages!)), says:

If the hot spot really is missing is does not prove that CO2 is not causing warming, but it would indicate something wrong with the models. (Which might mean that things are worse than what the models predict.)

That parenthetical kills me. Somebody should try it in a macro seminar. “No my model doesn’t show the characteristic hump shaped inflation response, but this means my model is wrong. The correct model might make my conclusion even stronger!”

But what really pisses me off about the piece is its title. Yeah, yeah, I know its a blog post, but its a blog post at “Scienceblogs” which I would expect to be more scientific-y or something. Anyway, how is a facts-driven, plausible sounding critical review a “war on science”?

In the backwoods where I’m from, facts-driven plausible sounding stories are science.


text:
By Linda Chuss, MathWorks

If you have ever lost your Command History for an entire session due to a system failure, you can avoid this by changing your preferences to save the history throughout the session instead of only at shutdown. By default, MATLAB saves the Command History upon shutdown. If an abnormal termination like a power failure occurs, however, the history is lost. This is especially troublesome if you do not shut down MATLAB daily.

To avoid that problem, select File -> Preferences -> Command History. Set the Saving option to Save after n commands and click OK. For example, set n to 1. MATLAB then saves the history after every line you execute.

If you want a record of system output as well as input, use the diary function. To automatically use the diary for every session, put this function in your startup.m file. Also, consider the logfile startup option for automatically writing output to the specified file. For more information about the Command History and related features, see the online documentation in the Help browser, or follow these links:

• Command History Preferences , including not saving consecutive duplicate entries to the history
• The diary function and startup.m file
• The logfile startup option for Windows and UNIX platforms
• More about the Command History, including searching the history
Figure 1: Change preferences to save your Command History after every statement. In the event of a system failure, you will only lose one statement instead of hours, days, or even weeks of history.

Published 2003


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# Kramers-Kronig relations for limited data point

1. Jul 18, 2011

### physengineer

Hello,

I need to measure the complex-optical conductivity of some materials. The problem is that I can only measure the imaginary part of the complex conductivity only for limited wavelengths between 1030 nm and 2300 nm.

From Kramers-Kronig relations, we know that the real and imaginary parts of the optical conductivity are related but the required integrations are from $-\infty$ to
$+\infty$.

Is there any way to still be able to use Kramers-Kronig relations when I only know the imaginary part for just an interval rather than the whole spectrum.

I would appreciate any help in this regard!

2. Jul 22, 2011

### chrisbaird

If you have enough data, you may be able to reasonably extrapolate to find the other values. Fit a curve to your data and use the curve for values you don't have.

Also, in practice, we can never acquire an infinite number of points in order to satisfy certain relations such as Kramers-Kronig or Fourier Transforms, and thus edge effects are inevitable. We can minimize edge effects by applying http://en.wikipedia.org/wiki/Window_function" [Broken].

Last edited by a moderator: May 5, 2017


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Meeting Details

Title: Grouping and rearranging terms in infinite series PMASS Colloquium Joe Roberts, Penn State When studying infinite series of real numbers, it is tempting to expect the familiar properties of addition to hold, and in some cases they do. Series that converge absolutely are as well behaved as finite series --- they are associative and commutative. However, series that are not absolutely convergent have very different properties. The Riemann Rearrangement Theorem states that any conditionally convergent series can be reordered to sum to any real number or to diverge (illustrating a failure of commutativity), and there are divergent series for which one can choose a subsequence of partial sums that converge to any arbitrary real number (a failure of associativity). I will give examples of these kinds of unexpected outcomes, eventually building to a construction due to Sierpinski of a single power series whose terms can be grouped to converge uniformly to any continuous function on [0,1] vanishing at 0 --- a "universal Taylor series".

Room Reservation Information

Room Number: MB113 04 / 03 / 2014 02:30pm - 03:20pm


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## Cryptology ePrint Archive: Report 2016/574

Structure vs Hardness through the Obfuscation Lens

Nir Bitansky and Akshay Degwekar and Vinod Vaikuntanathan

Abstract: Much of modern cryptography, starting from public-key encryption and going beyond, is based on the hardness of structured (mostly algebraic) problems like factoring, discrete log, or finding short lattice vectors. While structure is perhaps what enables advanced applications, it also puts the hardness of these problems in question. In particular, this structure often puts them in low (and so called structured) complexity classes such as NP$\cap$coNP or statistical zero-knowledge (SZK).

Is this structure really necessary? For some cryptographic primitives, such as one-way permutations and homomorphic encryption, we know that the answer is yes — they imply hard problems in NP$\cap$coNP and SZK, respectively. In contrast, one-way functions do not imply such hard problems, at least not by black-box reductions. Yet, for many basic primitives such as public-key encryption, oblivious transfer, and functional encryption, we do not have any answer.

We show that the above primitives, and many others, do not imply hard problems in NP$\cap$coNP or SZK via black-box reductions. In fact, we first show that even the very powerful notion of Indistinguishability Obfuscation (IO) does not imply such hard problems, and then deduce the same for a large class of primitives that can be constructed from IO.

Category / Keywords: Indistinguishability Obfuscation, Statistical Zero-knowledge, Structured Hardness, Collision-Resistant Hashing

Date: received 3 Jun 2016, last revised 30 Nov 2016

Contact author: akshayd at mit edu

Available format(s): PDF | BibTeX Citation

Note: Fixed typo in Def 4.2.

Short URL: ia.cr/2016/574

[ Cryptology ePrint archive ]


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How do you factor the expression 91-90t-16t^2?

Dec 2, 2015

Answer:

Factor y = -16t^2 - 90t + 91

Ans:$- \left(8 t - 7\right) \left(2 t + 13\right)$

Explanation:

I use the new AC Method to factor trinomials (Socratic Search).
$y = - \left(16 {t}^{2} + 90 t - 91\right) =$ -16(x + p)(x + q)
Converted trinomial $y ' = - \left({t}^{2} + 90 t - 1456\right) =$ -(t + p')(t + q').
p' and q' have opposite signs.
Factor pairs of (-1456) --> ...(-8, 182)(-14, 104). This sum is (- 14 + 104) = 90 = b. Then, p' = -14 and q' = 104.
Therefor, $p = \frac{p '}{a} = - \frac{14}{16} = - \frac{7}{8}$ and $q = \frac{q '}{a} = \frac{104}{16} = \frac{13}{2}$.
Factored form: $y = - 16 \left(t - \frac{7}{8}\right) \left(t + \frac{13}{2}\right) = - \left(8 t - 7\right) \left(2 t + 13\right) .$

NOTE. This method is systematic. There are no guessing and no lengthy factoring by grouping.


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# Issues with manual culling, using BMesh

I have a camera and a monkey (suzanne) in the scene. I'm attempting to manually cull the back faces from the monkey.

The problem is that certain faces get culled, despite its normal seen from the camera's view port.

1. The monkey in object mode.
2. The monkey after culling (using the function below). I transform the monkey's vertices into world space, then transform the vertices to face the camera (camera's inverse matrix). This assumes the camera is at (0,0,0).

I'm using the internal renderer. Blender 2.78.

def cull_manual(camera_obj, object_obj):
matrix = camera_obj.matrix_local.inverted() * object_obj.matrix_local

delete_faces = []

mesh = object_obj.data
new_bmesh = bmesh.from_edit_mesh(mesh)

for face in new_bmesh.faces:
verts = [(matrix * bmvert.co) for bmvert in face.verts]
a = (verts[1] - verts[0])
b = (verts[2] - verts[0]) if (len(verts) == 3) else (verts[3] - verts[0])
normal = a.cross(b).normalized()

v = verts[0].normalized()

visibility = (-v).dot(normal)
if visibility <= 0.0:
delete_faces.append(face)

bmesh.ops.delete(new_bmesh, geom = delete_faces, context = 5)
bmesh.update_edit_mesh(mesh)

camera_obj = find_obj(scene, "Camera.05")
object_obj = find_obj(scene, "Suzanne")
cull_manual(camera_obj, object_obj)

I devised a way to do it the opposite way, where I move the object to the origin, and move the camera by applying the object's inverse transform:

def cull_from_camera(camera_obj, object_obj):
matrix = object_obj.matrix_world.inverted()
camera_matrix = matrix * camera_obj.matrix_world
location, _, _ = camera_matrix.decompose()
from_camera = location.normalized()
delete_faces = []
mesh = object_obj.data
new_bmesh = bmesh.new()
new_bmesh.from_mesh(mesh)
for face in new_bmesh.faces:
visibility = from_camera.dot(face.normal)
E = 1.0e-05
R = -0.5 # HERE
if visibility <= (R + E):
delete_faces.append(face)
bmesh.ops.delete(new_bmesh, geom = delete_faces, context = 5)
new_bmesh.to_mesh(mesh)
mesh.update()

It works, but there are still some aggressive culling happening, unless I change the visibility check from 0.0 to -0.5.

I can't think of a way to truly cull only the faces that are not visible to the camera.

I have tried two implementations, and I wasn't too satisfied. I should say that the second implementation is definitely the better of the two since I need to simply cull faces that block the camera (think camera outside facing a wall of a room).

The most accurate seems to be in NDC space. For every face, project each vertex, transform to NDC space, then construct a normal from the 2D triangle for that ace. If the z component of the normal is negative, cull face.

The important bit here is that we do not clip in either clip or NDC space. Clipping is necessary when vertices are behind the camera (in homogeneous coordinates, w ≤ 0). The result is the vertex being "mirrored" in front of the camera, resulting in normals flipping! So we take a shortcut, forget clipping altogether, and simply move the vertices as close to w as possible (1.0e-05).

projection_matrix = camera_obj.calc_matrix_camera(scene.render.resolution_x,
scene.render.resolution_y,
scene.render.pixel_aspect_x,
scene.render.pixel_aspect_y)
project_matrix = projection_matrix * (camera_obj.matrix_world.inverted() * (object_obj.matrix_world))
delete_faces = []
mesh = object_obj.data
new_bmesh = bmesh.new()
new_bmesh.from_mesh(mesh)
for face in new_bmesh.faces:
face_ndc = []
for vert in face.verts:
# Project point
pvert_4d = project_matrix * vert.co.to_4d()
# Perspective divide; transform to viewport space (NDC).

# We need to consider the case where where w < 0.
# For each 4D vertex, if w < 0, the vertex will be
# flipped when performing the perspective divide,
# resulting in the normal being flipped.

# We could do go through and implement clipping using
# the Liang-Barsky line clipping algorithm, but we only
# care about backface culling. Clamping w to 0.0 (close
# to) does result in distorting the triangle, but the
# normal will remain the same as the normal's magnitude
# is proportional to the area of the triangle, but
# independent of the direction of the normal.
w = 1.0e-05 if (pvert_4d.w < 0.0) else pvert_4d.w
ndc_vert = pvert_4d / w
face_ndc.append(ndc_vert)
# Calculate the normal
z = calc_normal(face_ndc)
if z < 0.0:
delete_faces.append(face)
bmesh.ops.delete(new_bmesh, geom = delete_faces, context = 5)
new_bmesh.to_mesh(mesh)
mesh.update()


text:
# When A Goats Vulva Start To Expand During Pregnancy

Learning about when a goats vulva start to expand during pregnancy is very important. Especially, if you have pregnant doe in your herd which is going to give birth of kids soon. If your doe is pregnant, then you have to make sure you don’t miss any signs that her kid/kids are ok. Usually goat’s gestation period takes not more than five months (155 days). So you have a lots of time to get everything ready after the mating of your does.

# When A Goats Vulva Start To Expand During Pregnancy

Usually a healthy goat’s vulva should appear relatively tight until birth is impending. During a few days prior to kidding, the vulva starts opening and becomes slacker. This is a clear sign that a or a few new kids are going to see the world soon.

## Bagging Up

Bagging up means your doe’s udder fills with milk and become ready to feed the newborn kid/kids. Bagging up is the first sign of kidding. You can determine the delivery date is coming closer when your doe starts bagging up. Bagging up is an obvious sign, rather than something subtle or quick that you might miss. Some does don’t bag up until the day before delivery. And some goats might bag up a week or more before kidding. If you have numerous goats in your herd, then separate the pregnant doe once you see her bagging up. Keep her in a completely separate area where other goat’s won’t disturb her or in such an area where you intend to have her kidding.

## Days Before Birth

The doe becomes restless, when the kid/kids are about to born within a day or two. She starts pawing and getting up and down frequently and appears nervous. During this period, you might see a mucus discharge emanating from her vulva. Usually the discharge from her vulva should be thin and relatively clear. You need to consult with your vet, if you see greenish discharge. Because it indicates infection. Discharge of mucus is also an indicator for your does that her vulva is beginning to expand. When she starts bleating and doesn’t stop, that’s the key that labor is beginning and your doe is going to give birth of kids soon. Sometimes some doe might go on making noise for up to 12 hours before delivery.

## Kidding

When kidding is very closer, your doe will start walking around nervously and start getting up and down constantly. And once the kid/kids move into the birth canal, the doe’s flanks appear sunken in. The kids are not far behind, when you will notice a thick, stringy vaginal discharge. Your doe will begin strong constractions. And you will see a reddish fluid bag emanating from her vulva. And then she has more contractions. After a few contractions, you should see the front feet and nose emerging. After a few more contractions, the first kid is born. If your doe has more than one kid, then any others will be born within intervals of a few minutes.

## Afterwards

After kidding you can wipe off some of the amniotic fluid from the kid’s nose (if you attend the birth). Open the sac if it didn’t break. Usually your doe will lick off and clean her babies naturally. The placenta should come out within half an hour after delivery. You can dispose of it, or your doe can eat it. If the placenta doesn’t emerge within two hours of the birth, then it’s an emergency. And you have to call the vet as soon as possible. If this happen, then there is a possibility a dead kid is still inside your doe. So you are in need of help of a vet urgently.


text:
# How do you find the limit of x(a^(1/x)-1) as x->oo?

Jun 1, 2017

${\log}_{e} a$

#### Explanation:

$x \left({a}^{\frac{1}{x}} - 1\right) = \frac{{a}^{0 + \frac{1}{x}} - {a}^{0}}{\frac{1}{x}}$ now calling $\frac{1}{x} = h$ we have

${\lim}_{x \to \infty} x \left({a}^{\frac{1}{x}} - 1\right) = {\lim}_{h \to 0} \frac{{a}^{0 + h} - {a}^{0}}{h} = {a}^{0} {\log}_{e} a = {\log}_{e} a$

Jun 1, 2017

$\ln \left(a\right)$

#### Explanation:

Rewrite the limit as

${\lim}_{x \to \infty} \frac{{a}^{\frac{1}{x}} - 1}{\frac{1}{x}}$

So that it produces the indeterminate form $\frac{0}{0}$.

Now use L'hopital's rule (and chain rule as a result):

${\lim}_{x \to \infty} \frac{{a}^{\frac{1}{x}} - 1}{\frac{1}{x}} = {\lim}_{x \to \infty} \frac{{a}^{\frac{1}{x}} \cdot \ln \left(a\right) \cdot \frac{- 1}{x} ^ 2}{\frac{- 1}{x} ^ 2}$

$= {\lim}_{x \to \infty} {a}^{\frac{1}{x}} \ln \left(a\right) = {a}^{0} \ln \left(a\right) = \ln \left(a\right)$


text:
## College Physics (4th Edition)

We can rank the flywheels in order of radial acceleration of a point at the rim, from largest to smallest: $c \gt d \gt a \gt b \gt e$
In general: $\omega = \frac{2\pi}{T}$ $v = \omega~r$ $a_c = \frac{v^2}{r}$ We can find the radial acceleration of a point at the rim of each flywheel: (a) $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.0040~s} = 1571~rad/s$ $v = (1571~rad/s)(0.080~m) = 125.7~m/s$ $a_c = \frac{(125.7~m/s)^2}{0.080~m} = 197,000~m^2/s$ (b) $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.0040~s} = 1571~rad/s$ $v = (1571~rad/s)(0.020~m) = 31.4~m/s$ $a_c = \frac{(31.4~m/s)^2}{0.020~m} = 49,000~m^2/s$ (c) $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.0010~s} = 6283~rad/s$ $v = (6283~rad/s)(0.080~m) = 502.6~m/s$ $a_c = \frac{(502.6~m/s)^2}{0.080~m} = 3,158,000~m^2/s$ (d) $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.0010~s} = 6283~rad/s$ $v = (6283~rad/s)(0.020~m) = 125.7~m/s$ $a_c = \frac{(125.7~m/s)^2}{0.020~m} = 790,000~m^2/s$ (e) $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.0040~s} = 1571~rad/s$ $v = (1571~rad/s)(0.010~m) = 15.7~m/s$ $a_c = \frac{(15.7~m/s)^2}{0.010~m} = 25,000~m^2/s$ We can rank the flywheels in order of radial acceleration of a point at the rim, from largest to smallest: $c \gt d \gt a \gt b \gt e$


text:
+0

# Periodic Compound Interest

0
244
4
+233

How much should you deposit at the end of each month into an investment account that pays 8.5 %compounded monthly to have$3 million when you retire in 40 years? How much of the$ 3 million comes from interest?

Oct 14, 2019

#1
+1

Since your are given the Financial Formulas to calculate all your posted questions, you should be able to use them and arrive at the right answer. I, however, will use an online financial calculator to give you the answers. You can use the calculator here: https://arachnoid.com/finance/

The monthly deposit will be = $742.82$3,000,0000 - [$742.82 x 480] =$2,643,446.40 - This is the interest portion of the investment.

Oct 14, 2019
edited by Guest  Oct 14, 2019
#3
+233
0

thank you

skye25  Oct 14, 2019
#2
+1910
+3

You should use the formula. This will help to give you an “intuitive” understanding of how the TVM works. Save the online financial calculators for checking your work. If you just use financial calculators or software to solve these, it will alter your DNA, and you’ll become a Blarney Banker.

$$\text {Here is the formula ...}\\ \text { }\\ \dfrac{FV*\left(\dfrac{r}{n}\right)} {\left(1+\dfrac{r}{n}\right)^{nt} -1} =D\\ \text{ }\\ \small \dfrac{3000000*\left(\dfrac{0.085}{12}\right)} {\left(1+\dfrac{0.085}{12}\right)^{(12*40)} -1} = 742.82\\ \text{ }\\ \small \text {A monthly deposit of 742.82 dollars for 40 years and the accrued compounded interest will have a future value of 3,000,000 dollars. }\\$$

Paste this into the forum’s calculator:

f=3000000;r=0.085;n=12;t=40; f*(r/n)/ ((1+r/n)^(n*t) -1)

GA

Oct 14, 2019
#4
+233
+1

thank you

skye25  Oct 14, 2019


text:
# Print invisible text

Given a string as input, output a number of whitespace characters (0x0A and 0x20) equal to the length of the string.

For example, given the string Hello, World! your code would need to output exactly 13 whitespace characters and nothing else. These can be any mix of spaces and newlines.

## Testcases:

     Input      -> Amount of whitespace to output
"Hello, World!" -> 13
"Hi"            -> 2
"   Don't
Existing
Whitespace!   " -> 45
""              -> 0
"             " -> 13
"
"               -> 1

## Scoring:

This is so fewest bytes wins!

• I don't get what you mean with that “0x0A”. Where should that be output? Should that be kept, so “a␠b␊c” becomes “␠␠␠␊␠”? – manatwork May 25 '17 at 12:56
• @manatwork 0x0A and 0x20 are the hexadecimal values for the Newline and Space characters respectively – Skidsdev May 25 '17 at 12:58
• “output a number of whitespace characters (0x0A and 0x20)” – Where in the output should those newline characters be? – manatwork May 25 '17 at 13:00
• These can be any mix of spaces and newlines Your output can be any mix of spaces and newlines, you can just output spaces if you want, like everyone else, or you can just output newlines. It's up to you – Skidsdev May 25 '17 at 13:05
• Got it. Thanks. – manatwork May 25 '17 at 13:06

# Common Lisp, 32 bytes

(format t"~va"(length(read))#\ )

Try it online!

# Carrot, 12 bytes

#^//()/gS" "

Try it online! (append a ^@^v@ after the code to see the spaces bounded between @s)

### Explanation

#^            Set the stack-string to be equal to the input
/             Get matches of this regex
/()/g         any position (not character) in the string (shorter than /\b|\B/ by 3 bytes)
If the length of the string is 3, this returns a 4-element array
consisting of empty strings
S" "          Join on spaces, so the example 4-element array will result in 3 spaces

The space can even be replaced with a literal newline or tab for the same bytecount.

# Befunge-98 (PyFunge), 5 bytes

#@~a,

Try it online!

For every character in the input (#@~), this prints a new line (a,).

# dc, 18 bytes

?Zd[9P1-d0<r]sr0<r

Try it online!

# RProgN 2, 3 bytes

L•*

Uncreative solution is uncreative.

Just gets the length of the input, and multiplies •, which is predefined to be a space, by it.

Try it online!

# Ly, 9 bytes

iy[' o,];

Try it online!

Explanation:

iy[' o,];

iy        # push length of input
[    ]  # while loop
' o    # output a space
,   # decrement input length
; # terminate (avoids implicit output)

def L:length;$x|if L<1then empty elif L<2then""else L-1|.*" "end  This is longer then usual to compensate for newline jq normally emits and the odd behavior of * with N<2 Sample runs: $ jq -Mrn --arg x 'Hello, World!' 'def L:length;$x|if L<1then empty elif L<2then""else L-1|.*" "end'$ jq -Mrn --arg x 'Hello, World!' 'def L:length;$x|if L<1then empty elif L<2then""else L-1|.*" "end' | wc -c 13$ echo -n 'def L:length;$x|if L<1then empty elif L<2then""else L-1|.*" "end' | wc -c 64  Shout out to Jonathan Frech for finding 3 superfluous spaces! • I do not know jq, but as then""else is syntactically valid, it seems like you could also omit the space in " " end. – Jonathan Frech Sep 19 '17 at 7:07 • As variables probably cannot begin with a digit, the spaces in <1 then and <2 then may also be superfluous. – Jonathan Frech Sep 19 '17 at 7:40 # Vim, 4 bytes VGr  (a trailing space) # Julia 0.6, 15 bytes s->" "^endof(s)  Try it online! ^ applied to strings is the repetition operator, and endof gives the last index of the string, which is equal to the length of the string (since Julia indexing is 1-based). # Yabasic, 37 bytes Takes input, and outputs as many new lines as the length of the input Line Input""s$
For i=1TO Len(s\$)?Next

Try it online!

# x86-16 ASM, IBM PC DOS, 11 bytes

Binary:

00000000: 8a0e 8000 49b8 200a cd10 c3              ....I. ....

Unassembled:

D1 EE       SHR  SI, 1          ; SI to 80H (SI intialized at 100H)
AC          LODSB               ; load string length into AL
91          XCHG AX, CX         ; put input string length into CX
49          DEC  CX             ; remove leading whitespace from length
AC          LODSB               ; load whitespace delimiter into AL
B4 0A       MOV  AH, 0AH        ; BIOS "write character CX number of times" function
CD 10       INT  10H            ; call BIOS, display to console

Explanation:

Input is via command line, though all that's important is the length. Command line input length is always stored at memory address DS:0080H in DOS, so put that into CX. DOS includes the space between the executable name and the command line args string in this number.

For example: in FOO.COM Hello, length is 6 and command line string is " Hello", or calling as FOO.COM/Hello, command line string is "/Hello" (Note: those are the the only valid characters for the character immediately after the executable name). This first character (will be a space when called normally) is what is displayed as the "invisible text" for output. This builds in a handy little "debug mode" where you can use a slash instead of a space to actually be able to test your output is the right length.

Then, use the IBM PC BIOS's INT 10H "Write character only at cursor position" (0AH) function that writes the same character CX number of times.

Example Output:

Admittedly, displaying 13 chars of whitespace is not very interesting in a screenshot. However, by using a slash instead of a space ("debug mode") you can actually see that you are displaying the right number of chars.

# C# .NET, 84 bytes

class P{static void Main(string[]a){System.Console.Write("".PadLeft(a[0].Length));}}


text:
# remove backwards motion completely

Hi all,

We are trying integrate the TEB Local Planner ros package in our system and customize this package for the physics of our robot.

Our robot has a restriction that it doesn't move backwards. Are there any way to remove backwards motion completely from planner?. We already have weight_kinematics_forward_drive on max, and still appear backwards motion some time.

The robot can move forward and rotate in a radius of 0.5 m.

Modifying weight, we can't get the desired behavior, so we thought modify the source code to customize the planner. Is it possible? Any suggestions about where make the modifications in the code?

Thanks!

edit retag close merge delete

Sort by » oldest newest most voted

First of all, in the current version, it is not possible to guarantee that the underlying optimizer chooses forward solutions only.

The configuration is much easier for differential-drive robots than for carlike robots. Some important notes for selecting a penalty for backward motions (parameter weight_kinematics_forward_drive:

• local minima (sensitive w.r.t. initialization): The initial trajectory (especially its orientation part) is very important since our optimizer seeks for the next local minimum. In the case of a differential drive robot, the robot can always turn in place which can be easily handled during optimization. For a car-like robot, the trajectory usually consists of backward and forwards arc segments which introduce further local minima.

• feasibility: according to your radius of 0.5 m, I assume that you have a carlike robot. The feasible set of possible configurations is significantly reduced if you prohibit backward motions (especially in cluttered environments). You might check Reeds and Chepp curves to get an idea.

However, this does not solve your problem at all. Sometimes it is possible to bypass this behavior, e.g. we could rotate the robot towards the goal before starting the actual trajectory planning. But I prefer to keep the number of additional rules on top of the optimization low if possible. Or we can implement different initialization strategies.

If you have possible solutions or suggestions in mind, I would be happy to discuss them with in our issue tracker.

Some side notes: (hard) constraints are currently treated as penalties and giving their corresponding terms appropriate weights might enforce them (extremely large weights can cause slower convergence due to a bad matrix condition). However, since the optimization problem size is medium, penalties should be sufficient (maybe we can change the solver backend to some fast sqp or interior point solver someday to handle hard constraints).

more

Thank you very much Christoph!

( 2017-03-24 12:39:21 -0500 )edit


text:
# How to calculate the Back-EMF energy on an Inductor?

The Back-EMF generated by an inductor in Volts is: $V = -L\dfrac{di}{dt}$

Where $L$ is the self-inductance and $\dfrac{di}{dt}$ the rate of current change.

Now, let's say I want to store those Volts on a capacitor $(F, V_0)$ where $F$ is the capacitor capacity in Farads and $V_0$ is the initial voltage on the capacitor.

How can I find out the final voltage on the capacitor? How much energy stored on the Back-EMF field?

• Inductor is current source and capacitor is voltage source. So an energy transformation is necessary. This transformation done by the equivalent resistor of both. – GR Tech Dec 18 '14 at 15:38

The instantaneous energy stored in an inductor is

$$E = \frac{1}{2}L I^2$$

The energy stored in a capacitor is

$$E = \frac{1}{2}C V^2$$

You can see that there's a tradeoff between the capacitance value and the voltage required to store a particular amount of energy.

Assume there is a perfect diode between L and C, and that C is large.

Switch off the current source, and $$\frac{dI}{dT}$$ will be negative, so the EMF V will be positive.

At V=V0, the diode turns on, so $$\frac{dI}{dT} = \frac{-V_0}{L}$$.
(approximately constant because we assumed C is large)

At I=0 there is no more energy stored in the inductor so the diode turns off, occurring at $t = I_0 \cdot L/V_0$.

Charge transferred Q is the integral of the current transferred over time,

$$Q = \frac{I_0 \cdot t}{2}$$

$$Q = \frac{I_0^2 \cdot L}{ (2 \cdot V_0)}$$

which will add dV = Q/C to the capacitor voltage.
(if dV is large, our assumption about C was incorrect)

The energy transferred in this cycle is simply $$E = Q \cdot V_0$$ or as Dave Tweed says, $$E = \frac{I_0^2 \cdot L}{2}$$


text:
# Independence of Filtration

Given a sequence of random variables $\{X_{t}\}_{t=1}^{T}$ , and a filtration $\mathcal{F}_{T}$, what does it mean that $X_{t+1}$ is independent of $\mathcal{F}_{t}$? Is it simply that $X_{t+1}$ is independent of any $X_{j}$, when j $\leq$ t, or is it more sufisticated?

• If $\mathcal{F}_t=\sigma(X_j\mid j\leq t)$, then yes. Otherwise, no. – Stefan Hansen Jun 12 '14 at 11:28
• @StefanHansen Sorry but $X_2$ can be independent of $X_0$ and independent of $X_1$ without being independent of $F_1=\sigma(X_0,X_1)$. – Did Jun 12 '14 at 11:41
• @Did Sure. I (mis)read it as $X_{t+1}$ should be simultaneously independent of $X_j$ for $j\leq t$. I guess this is not what the OP meant. – Stefan Hansen Jun 12 '14 at 11:43
• I don't know if this is implicit in the question but it needs to be stated that $\{X_t\}_{t=1}^T$ is an adapted process, right? – Calculon Aug 21 '14 at 9:50

It just means that $X_{t+1}$ is independent of every ${\cal F}_t$-measurable random variable.
In the case ${\cal F}_t = \sigma(X_0, \dots, X_t)$, it means that $X_{t+1}$ is independent of $(X_0, \dots, X_t)$.
In practice, you often have ${\cal F}_t \supseteq \sigma(X_0, \dots, X_t)$ but not always equality.
• Regarding the second sentence of your post, note that $X_2$ can be independent of $X_0$ and independent of $X_1$ without being independent of $F_1=\sigma(X_0,X_1)$. – Did Jun 12 '14 at 11:42


text:
##### Notes

This printable supports Common Core Mathematics Standard HSS-CP.A.6

##### Print Instructions

NOTE: Only your test content will print.
To preview this test, click on the File menu and select Print Preview.

See our guide on How To Change Browser Print Settings to customize headers and footers before printing.

# Conditional Probability as Fraction of Outcomes (Grade 10)

Print Test (Only the test content will print)

## Conditional Probability as Fraction of Outcomes

1.
Emily orders two cheese pizzas from Toni's Pizza for herself and some friends. However, Toni's Pizza accidentally put some sausage on a few of the slices of each pizza (shown in the pictures as the shaded slices). The sausage is not visible when just looking at the slice of pizza. Given that one of Emily's friends grabs the pizza containing the three slices of sausage, leaving only the other pizza for Emily to choose from, what is the probability that Emily chooses a slice of sausage pizza?

1. 1/4
2. 3/8
3. 5/16
4. 6/16
2.
Peter is dividing a regular deck of 52 cards in half and choosing a card at random. What is the probability of choosing a Jack, if he happens to divide the deck perfectly into black cards and red cards and then chooses from the red cards?
1. 1/13
2. 2/13
3. 1/2
4. 1/4
3.
An integer is randomly chosen from the following number line. Given that the number lies on $bar{AD}$, what is the probability that it also lies on $bar{BC}$?
1. 1/3
2. 2/9
3. 1/9
4. 1/4
4.
An integer is randomly chosen from the following number line. Given that the number lies on $bar{AD}$, what is the probability that it also lies on $bar{CF} ?$
1. 8/8
2. 1/2
3. 8/11
4. 5/8
5.
An integer is randomly chosen from the following number line. Given that the number lies on $bar{BE}$, what is the probability that it also lies on $bar{AF} ?$
1. 7/11
2. 2/11
3. 1
4. 4/11
6.
Jane has a collection of marbles, consisting of 3 red, 4 green, 4 blue, and 3 orange marbles. She has offered two of her friends, David and Malcolm, that they may take one marble each. What is the probability that Malcolm chooses a red marble, given that David has already chosen a red marble?
1. 1/7
2. 2/13
3. 3/14
4. 1/4
7.
Given the following set of numbers, ${1,2,3,4,5,6,7,8,9,10}$, what is the probability of choosing an even number, given that a multiple of 3 is chosen?
1. 1/2
2. 1/3
3. 0
4. 3/4
8.
Given the following set of numbers, ${1,1,1,1,2,3,4,5,5,5,5,6,7,8,8,8,8,8,9,10}$, what is the probability of choosing a prime number, given that a factor of 36 is chosen?
1. 7/20
2. 2/5
3. 1/3
4. 2/9
9.
Janet, Cam, and Mike arrive at a school party and there are only four cupcakes left. Everyone else has eaten enough cupcakes and will eat no more. Below is table of all possible ways these four cupcakes could be eaten by these three people. How likely it is that Janet will eat more cupcakes than Cam and Mike, given that before Janet gets to the cupcake table, Cam and Mike have already each eaten a cupcake?

 $"Janet"$ $"Cam"$ $"Mike"$ $4$ $0$ $0$ $3$ $1$ $0$ $3$ $0$ $1$ $2$ $2$ $0$ $2$ $0$ $2$ $2$ $1$ $1$ $1$ $3$ $0$ $1$ $0$ $3$ $1$ $2$ $1$ $1$ $1$ $2$ $0$ $4$ $0$ $0$ $0$ $4$ $0$ $3$ $1$ $0$ $1$ $3$ $0$ $2$ $2$
1. 4/15, because there are only four possible scenarios in which Janet eats more cupcakes out of the total 15 possibilities.
2. 3/14, since for row 1, neither Cam nor Mike has had a cupcake.
3. 1/12, because rows 1-3 don't count, as they have Mike and Cam eating only one cupcake between them (and two need to have been already eaten).
4. 1/6, since rows 1-5, 7, 8, 11, 12 aren't included as they don't have Cam and Mike as having eaten at least one cupcake each.
10.
Mia goes to East York Collegiate Institute. She is given the following information:

There are 150 grade 12 students at East York Collegiate Institute.
70 of the grade 12 students drive a car at the school.
Half of all students with a grade average of 90 or higher have a car.

She is asked to find the probability of randomly choosing a grade 12 student with a car, given that the student has an average grade of 90 or higher. She reasons initially that, since there are 70 grade 12 students out of 150 who have a car, it would be 7/15. However, since it is a conditional probability, she needs to multiply by 1/2, since the probability that a student with a grade average of 90 or higher is 1/2. This results in a probability of 7/30. Is she correct or not, and why?
1. Yes, and her reasoning is correct.
2. No, she didn't need to multiply by 1/2 and the probability is 7/15. The probability that a student has a car can be assumed to be independent of their grade average, and so no multiplication is needed.
3. No, there is insufficient information to find this probability.
4. No, she needs to state that it can be assumed that half of grade 12 students have an average of 90 or higher. Therefore, half of 70 is 35, half of 150 is 75, and the probability is 7/15.
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text:
RFPTA Recent Articles
Volume 2019

# On new hyperstability results for the generalized $p$-radical functional equation in quasi-Banach spaces with the illustrative example-by Laddawan Aiemsomboon, Wutiphol Sintunavarat

Abstract: The goal of this paper is to investigate the general solution and the hyperstability results of the generalized $p$-radical functional equation
\begin{equation*}
f\left(\sqrt[p]{\sum_{i=1}^kx_i^p}\right)=\sum_{i=1}^kf(x_i)
\end{equation*}
for all $x_1,\dots,x_k \in \mathbb{R}$,
where $f: \mathbb{R} \to Y$ is a mapping, $p$ is a natural number, $k$ is a natural number such that $k \geq 2$, and $Y$ is a quasi-Banach space. Furthermore, the hyperstablity result for the inhomogeneous generalized $p$-radical functional equations is derived from the main results. Finally, we give the illustrative example showing the validity of the obtained results while the main results of Almahalebi et al. is not applicable.

Keywords: Fixed point; hyperstability; quasi-Banach space; $p$-radical functional equation.

DOI: 10.30697/rfpta-2019-012

______
How to cite this article: Laddawan Aiemsomboon, Wutiphol Sintunavarat, On new hyperstability results for the generalized $p$-radical functional equation in quasi-Banach spaces with the illustrative example, Results in Fixed Point Theory and Applications, Vol. 2019, Article ID 2019012, 14 pages, 2019.  DOI: https://doi.org/10.30697/rfpta-2019-012

______
Copyright © 2018 SUMA Publishing Group., This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


text:
• Redundant internal coordinates, compliance constants and non-bonded interactions - some new insights

• # Fulltext

https://www.ias.ac.in/article/fulltext/jcsc/125/01/0009-0015

• # Keywords

Redundant internal coordinates; compliance constants; non-bonded interactions; normal modes; complete basis.

• # Abstract

A long standing problem in normal mode analysis is identifying the right internal coordinates given only the cartesian coordinates, the masses of the atoms and the cartesian force constants without using any other additional chemical information. A possible solution is suggested here as drawing the normal modes obtained from the mass weighted cartesian force constant matrix and identifying the correct bonds and angles from the normal mode pictures. If chosen properly, the internal coordinates will have minimum mixing in the normal mode representation. This can in principle lead to an automation algorithm. A complete basis of internal coordinates is defined as the minimum number of valence internal coordinates that describe all the normal modes as completely as possible. It was shown in the literature that the relaxed force constants could be used as a measure of bond order in all atom-atom distance coordinates. Some of the bonded and non-bonded atom pairs can have similar values of the relaxed force constants and hence to use the relaxed force constant as a measure of bond order we need to separate the bonded pairs from the non-bonded ones. This needs extra chemical information of which pairs are bonded. The new definition of complete basis of non-redundant valence internal coordinates helps to identify the bonded pairs effectively without extra information. The hydrogen bonded water clusters (H2O)$_n$, n = 2-6, methane dimer and methane-water complex are used as examples to verify that the relaxed force constants of bonded pairs are indeed a measure of bond order.

• # Author Affiliations

1. Department of Chemistry, Indian Institute of Technology Kanpur, Kanpur 208 016, India

• # Journal of Chemical Sciences

Volume 135, 2023
All articles
Continuous Article Publishing mode

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019

Click here for Editorial Note on CAP Mode

© 2022-2023 Indian Academy of Sciences, Bengaluru.


text:
## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 5

#### Answer

$$2.24$$

#### Work Step by Step

\eqalign{ & - 3{x^3} + 5{x^2} + 3x + 2 = 0;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{in the interval }}\left[ {2,3} \right] \cr & {\text{let }}f\left( x \right) = - 3{x^3} + 5{x^2} + 3x + 2,{\text{ so that}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( { - 3{x^3} + 5{x^2} + 3x + 2} \right) \cr & f'\left( x \right) = - 9{x^2} + 10x + 3 \cr & {\text{interval }}\left[ {2,3} \right]{\text{ then }}a = 2{\text{ and }}b = 3 \cr & {\text{check }}f\left( a \right){\text{ and }}f\left( b \right) \cr & f\left( 2 \right) = - 3{\left( 2 \right)^3} + 5{\left( 2 \right)^2} + 3\left( 2 \right) + 2 = 4 > 0 \cr & f'\left( 3 \right) = - 3{\left( 3 \right)^3} + 5{\left( 3 \right)^2} + 3\left( 3 \right) + 2 = - 25 < 0 \cr & f\left( 2 \right){\text{ and }}f\left( 3 \right){\text{ have opposite signs}}{\text{, then}} \cr & {\text{there is a solution for the equaion in the interval }}\left( {2,3} \right) \cr & {\text{As an initial guess }}{c_1} = 2 \cr & {\text{A better guess}}{\text{, }}{c_2},{\text{ can be found as follows }} \cr & {c_{n + 1}} = {c_n} - \frac{{f\left( {{c_n}} \right)}}{{f'\left( {{c_n}} \right)}} \cr & {\text{then }} \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 2 - \frac{{ - 3{{\left( 2 \right)}^3} + 5{{\left( 2 \right)}^2} + 3\left( 2 \right) + 2}}{{ - 15{{\left( 2 \right)}^2} + 10\left( 2 \right) + 3}} = 2.3076 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = 2.3076 - \frac{{ - 3{{\left( {2.3076} \right)}^3} + 5{{\left( {2.3076} \right)}^2} + 3\left( {2.3076} \right) + 2}}{{ - 15{{\left( {2.3076} \right)}^2} + 10\left( {2.3076} \right) + 3}} = 2.2473 \cr & {\text{In the same way}} \cr & {c_4} = 2.2473 - \frac{{ - 3{{\left( {2.2473} \right)}^3} + 5{{\left( {2.2473} \right)}^2} + 3\left( {2.2473} \right) + 2}}{{ - 15{{\left( {2.2473} \right)}^2} + 10\left( {2.2473} \right) + 3}} = 2.2445 \cr & {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr & {\text{Then}} \cr & {\text{A solution for the given interval is }}x \approx 2.24 \cr}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# Important formulas of Optics Class 12 Physics JEE & NEET

Here you will get all the Optics  formulas for class 12, JEE and NEET examination. You can quickly revise entire chapter here in a quick overview.


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Address 207 W Main St, Independence, VA 24348 (276) 773-9922 http://acecomputingva.com

# calculating standard error of linear regression Elk Creek, Virginia

standard-error inferential-statistics share|improve this question edited Mar 6 '15 at 14:38 Christoph Hanck 9,13332149 asked Feb 9 '14 at 9:11 loganecolss 5531926 stats.stackexchange.com/questions/44838/… –ocram Feb 9 '14 at 9:14 more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed Take-aways 1. An unbiased estimate of the standard deviation of the true errors is given by the standard error of the regression, denoted by s.

Formulas for standard errors and confidence limits for means and forecasts The standard error of the mean of Y for a given value of X is the estimated standard deviation The reason N-2 is used rather than N-1 is that two parameters (the slope and the intercept) were estimated in order to estimate the sum of squares. For the BMI example, about 95% of the observations should fall within plus/minus 7% of the fitted line, which is a close match for the prediction interval. How to Find an Interquartile Range 2.

Figure 1. Kind regards, Nicholas Name: Himanshu • Saturday, July 5, 2014 Hi Jim! Confidence intervals The formulas given in the previous section allow one to calculate the point estimates of α and β — that is, the coefficients of the regression line for the Thank you once again.

Under such interpretation, the least-squares estimators α ^ {\displaystyle {\hat {\alpha }}} and β ^ {\displaystyle {\hat {\beta }}} will themselves be random variables, and they will unbiasedly estimate the "true A model does not always improve when more variables are added: adjusted R-squared can go down (even go negative) if irrelevant variables are added. 8. An Error Occurred Unable to complete the action because of changes made to the page. The coefficients and error measures for a regression model are entirely determined by the following summary statistics: means, standard deviations and correlations among the variables, and the sample size. 2.

In the mean model, the standard error of the mean is a constant, while in a regression model it depends on the value of the independent variable at which the forecast Since the conversion factor is one inch to 2.54cm, this is not a correct conversion. In light of that, can you provide a proof that it should be $\hat{\mathbf{\beta}} = (\mathbf{X}^{\prime} \mathbf{X})^{-1} \mathbf{X}^{\prime} \mathbf{y} - (\mathbf{X}^{\prime} \mathbf{X})^{-1} \mathbf{X}^{\prime} \mathbf{\epsilon}$ instead? –gung Apr 6 at 3:40 1 Step 4: Select the sign from your alternate hypothesis.

more hot questions question feed default about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation S provides important information that R-squared does not. Formulas for R-squared and standard error of the regression The fraction of the variance of Y that is "explained" by the simple regression model, i.e., the percentage by which the As the sample size gets larger, the standard error of the regression merely becomes a more accurate estimate of the standard deviation of the noise.

It is well known that an estimate of $\mathbf{\beta}$ is given by (refer, e.g., to the wikipedia article) $$\hat{\mathbf{\beta}} = (\mathbf{X}^{\prime} \mathbf{X})^{-1} \mathbf{X}^{\prime} \mathbf{y}.$$ Hence  \textrm{Var}(\hat{\mathbf{\beta}}) = (\mathbf{X}^{\prime} \mathbf{X})^{-1} \mathbf{X}^{\prime} That works. up vote 6 down vote favorite 2 I have the following model and want to make a table with the interpretation of the interaction effects as suggested by Bambor and Clark There are various formulas for it, but the one that is most intuitive is expressed in terms of the standardized values of the variables.

how to find them, how to use them - Duration: 9:07. How much should I adjust the CR of encounters to compensate for PCs having very little GP? Show more Language: English Content location: United States Restricted Mode: Off History Help Loading... price, part 1: descriptive analysis · Beer sales vs.

I actually haven't read a textbook for awhile. Todd Grande 1,477 views 13:04 Standard Error - Duration: 7:05. Note: The TI83 doesn't find the SE of the regression slope directly; the "s" reported on the output is the SE of the residuals, not the SE of the regression slope. Similarly, the confidence interval for the intercept coefficient α is given by α ∈ [ α ^ − s α ^ t n − 2 ∗ ,   α ^ +

Bionic Turtle 159,719 views 9:57 How to Calculate t test Using Excel for Unrelated Groups (Independent groups) - Duration: 13:49. For all but the smallest sample sizes, a 95% confidence interval is approximately equal to the point forecast plus-or-minus two standard errors, although there is nothing particularly magical about the 95% Hence, it is equivalent to say that your goal is to minimize the standard error of the regression or to maximize adjusted R-squared through your choice of X, other things being S. (1962) "Linear Regression and Correlation." Ch. 15 in Mathematics of Statistics, Pt. 1, 3rd ed.

Does using OpenDNS or Google DNS affect anything about security or gaming speed? Bozeman Science 171,662 views 7:05 What does r squared tell us? The simple regression model reduces to the mean model in the special case where the estimated slope is exactly zero. For example, let's sat your t value was -2.51 and your b value was -.067.


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# Find the next term: 1/5, 2/8, 2/12, etc

Find the next term in the following sequence:

$1/5$, $~~2/8$, $~~2/12$, $~~4/18$, $~~2/24$, $~~?$

It's my first question so it might come off as a bit too easy.

• Probably would have made it harder by reducing the fractions as much as you could. Oct 3 '15 at 18:03
• @Spencerkatty Reducing in simplest form would have given multiple choices for each term .... it will also make it much tougher to go in the right direction ... Oct 3 '15 at 23:12

The given number sequence is derived from

the sequence of prime numbers:

$~~~~$ $2$, $~3$, $~5$, $~7$, $~11$, $~13$, $~17$, $~19$, $~23$, $~\ldots$

The two entries of every pair are

the difference and the sum of consecutive prime numbers

This yields

3-2/3+2, $~$5-3/5+3, $~$7-5/7+5, $~$11-7/11+7, $~$13-11/13+11, $~$17-13/17+13,

so that the next term will be $~$4/30.

• You cracked it!! Oct 3 '15 at 17:20
• What was the inspiration behind the answer? Oct 3 '15 at 17:21
• He may have searched the series of numerators on OEIS. Oct 4 '15 at 10:33

Just a guess..

$\frac{2}{28}$

because

Sum of numerator and denominator of a term equals difference between numerator and denominator of the next term.

This probably implies that

The next term is of the type $\frac{a}{a+26}$

but I don't know why

I'm guessing $a=2$.

A more complete answer will be required.

• No. That is not the answer. Going by this method can give numerous answer. Oct 3 '15 at 16:34
• The actual logic gives one clear answer. Oct 3 '15 at 16:35
• He was right on the 26 bit, his reasoning is connected with how the numbers are calced. He just lacked the top number. By this I mean his answer is 'incomplete' rather then 'based on a wrong method'. Oct 3 '15 at 19:22


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# Please give me more hints

## Homework Statement

Prove that every uncountable subset of the real lline has a limit point. There is a hint: The union of a countable family of finite sets is countable.

## The Attempt at a Solution

Please give me more hints. Thanks a lot!

Dick
Homework Helper

## Homework Statement

Prove that every uncountable subset of the real lline has a limit point. There is a hint: The union of a countable family of finite sets is countable.

## The Attempt at a Solution

Please give me more hints. Thanks a lot!

If a closed bounded interval contains an infinite number of points, then it contains a limit point. Do you know this or can you prove it?

If a closed bounded interval contains an infinite number of points, then it contains a limit point. Do you know this or can you prove it?

Yes. I can. It is pretty easy.

Dick
Homework Helper
Yes. I can. It is pretty easy.

Well, then, if the set doesn't have a limit point, then every interval must contain a finite number of points, right?

Well, then, if the set doesn't have a limit point, then every interval must contain a finite number of points, right?

I think it is right for the bounded interval. But now, the interval is actually a set of some finite discrete points, isn't it?

So I should first suppose that the opposite statement is true. Then show that every uncountable subset of real line can be decomposed into countably many subsets. Some of them are intervals. Then I should prove that any interval has a limit point. So the uncountable subset consists of countable subsets which are finite. Is that right?

Dick
Homework Helper
I think it is right for the bounded interval. But now, the interval is actually a set of some finite discrete points, isn't it?

So I should first suppose that the opposite statement is true. Then show that every uncountable subset of real line can be decomposed into countably many subsets. Some of them are intervals. Then I should prove that any interval has a limit point. So the uncountable subset consists of countable subsets which are finite. Is that right?

The real line is a countable union of bounded intervals. Intersect your set with each interval.

Last edited:
The real line is a countable union of bounded intervals. Intersect your set with each interval.

Oh, I see. Thank you very much!


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# zbMATH — the first resource for mathematics

The structure of spaces of Toeplitz matrices. (English) Zbl 0894.57016
A finite Toeplitz matrix is a matrix $$t_n = (a_{i,j})$$ with the coefficients of the form $$a_{i,j} = a_{i-j}$$. Consider the variety of projective equivalence classes of nonzero $$n\times n$$-Toeplitz matrices $$(t_n\equiv \alpha t_n$$ for all nonzero $$\alpha\in\mathbb{C}$$), which are parameterized by the complex projective space $$\mathbb{C} P^{2n-2}$$. The paper is devoted to a study of subvarieties $$T_{n,k}$$ consisting of those $$t_n$$ with $$\dim(\text{kernel}(t_n))\geq k$$, and the open varieties $${\mathcal T}_{n,k} = T_{n,k}-T_{n,k+1}$$. The results are summarized in the following three theorems.
Theorem A. The space $${\mathcal T}_{n,0}$$ of nonsingular $$n\times n$$-Toeplitz matrices is homeomorphic to the orbit space under the action of $$GL_2(\mathbb{C})$$ on the space of pairs $$(p_1(z),p_2(z))$$ of coprime polynomials with $$\max(\text{deg}(p_1(z)),\text{deg}(p_2(z)))=n$$, where the action is given as $(p_1\;p_2) {{a\;c}\choose {b\;d}}= (ap_1+ bp_2\;cp_1+ dp_2).$ Theorem B. Let $$t_n\in {\mathcal T}_{n,k}$$. Then there is a vector $$w\in\text{Ker}(t_n)$$ so that $\text{Ker}(t_n)=\langle w,s(w),\dots, s^{k -1}(w)\rangle,$ where $$s$$ is a shift operator. Moreover, (1) the first $$k-1$$ coordinates of $$w$$ are zero; (2) $$w$$ is unique up to multiplication by a nonzero scalar.
Theorem C. \begin{aligned} H^*({\mathcal T}_{n,k};\mathbb{Q})& \cong H^*(S^2;\mathbb{Q}), k\geq 1\\ H^*({\mathcal T}_{n,0};\mathbb{Q})& \cong H^*(pt;\mathbb{Q}).\end{aligned} The rational cohomology of the unprojectivized versions of $${\mathcal T}_{n,k}$$ are also computed. These results lead to a complete determination of the rational cohomology of the strata in a stratification of the moduli spaces $${\mathcal M}_k$$ of gauge equivalence classes of SU(2)-Yang-Mills instantons on $$S^4$$. This stratification was described in [C. P. Boyer, J. C. Hurtubise, B. M. Mann and R. J. Milgram, Ann. Math., II. Ser. 137, No. 3, 561-609 (1993; Zbl 0816.55002)].

##### MSC:
 57N65 Algebraic topology of manifolds 32G13 Complex-analytic moduli problems 55P35 Loop spaces 58D27 Moduli problems for differential geometric structures
Full Text:


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Harmonic oscillator

1. May 5, 2009

Ofey

The frequency of a harmonic oscillator is (as you know)

$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

I am wondering if this equation only applies for massless harmonic oscillators (or oscillators oscillating sideways)?

The proof for the equation above is

$$\sum {F=ma}$$

$$-kx=ma$$

$$-kx=-m\omega^2x$$

$$k=m\omega^2$$

Since

$$\omega=2\pi f$$

We get the familiar equation above

$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

But as you can see we have neglected the force of gravity (if we assume that such a force exists). I solved an assignment today using this formula above. It gave the wrong answer when I just used the formula which neglected gravity. But when I went back and proved the formula "again", this time assuming the existense of gravity I ended up in the right answer. This would be a bummer, since I have always been taught that the frequency for a harmonic oscillator is:

$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

No matter what.

2. May 5, 2009

deferro

No, the m in all the equations is for the mass term; this is determinate (use a weighing balance to measure it).
The m takes care of gravity for a discrete body with sufficient mass, so you have it in the equations.
But quantum oscillators are more complicated.

3. May 5, 2009

KLoux

There is no such thing as a "massless harmonic oscillator"...

Yes, this relationship applies even in the presence of gravity. The difference in the analysis of a mass hanging on a spring in a gravitational field and a horizontal mass-spring system on a frictionless surface is the "neutral" length of the spring.

Maybe you can show us this question that you got wrong and we can try to figure out where the problem is.

-Kerry

4. May 6, 2009

Ofey

A computer used in space shuttles has to work during extreme circumstances. The computer has to work during an acceleration equivalent to 10G without breaking. The computer is tested by placing it on an oscillating horizontal plane. The amplitude of the oscillation is 0.10m. Determine the frequency of the oscillation.

The correct solution

$$x=Asin(\omega t)$$

$$v=\frac{dx}{dt}=\frac{d}{dt}(Asin(\omega t))$$

$$a=\frac{dv}{dt}=\frac{d}{dt}(A\omega cos(\omega t))=-A\omega^2sin(\omega t)$$

Since

$$sin(\omega t) \in [-1,1]$$

The maximal acceleration is given by

$$A\omega^2=A(2\pi f)^2=10g$$

Solving for frequency yields

$$f=\frac{1}{2\pi}\sqrt{\frac{10g}{A}}$$

My first attempt

This is the situation when the computer is at the lowest position in the oscillation. This is when the acceleration is at its largest. Chosing upwards to be the positive direction

$$-kx-mg=ma$$

$$-kx=m(g+a)$$

Since

$$a=10g$$

$$kx=-11mg$$

$$\frac{k}{m}=\frac{-11g}{x}$$

Placing this in the well-known equation gives

$$f=\frac{1}{2\pi}\sqrt{\frac{-11g}{x}}$$

Where $$x=-A$$

Then I tried to prove the frequency for a harmonic oscillator again, this time assuming gravitational effects.

$$\Sigma F=ma$$

(I mean it says the sum of all forces, gravity does act on the body so surely it has to be added?)

$$-kx-mg=ma$$

$$-kx=m(g-\omega^2x)$$

$$\frac{k}{m}=\frac{\omega^2x-g}{x}$$

Again, since

$$\omega=2\pi f$$

$$f=\frac{1}{2\pi}\sqrt{\frac{kx+mg}{mx}}$$

Since

$$kx=-11mg$$

$$f=\frac{1}{2\pi}\sqrt{\frac{-11mg+mg}{mx}}$$

$$f=\frac{1}{2\pi}\sqrt{\frac{-10g}{x}}$$

Since

$$x=-A$$

We get

$$f=\frac{1}{2\pi}\sqrt{\frac{10g}{A}}$$

Which is correct? What am I missing

Last edited: May 6, 2009
5. May 6, 2009

Born2bwire

If it's oscillating in a horizontal plane, why would you have a force from gravity acting along the direction of oscillation?

6. May 6, 2009

Ofey

I apologize for my english.

The computer is placed on a horizontal plane which oscillates in the vertical direction.

7. May 6, 2009

KLoux

Ofey,

It sounds like you're introducing a spring into a problem that doesn't necessarily have a spring (unless harmonic oscillator implies that it is a mass-spring system instead of something driven by a motor).

Anyway - you're adding 10g acceleration to something that is already experiencing 1g due to Earth's gravity. This will give you 11g acceleration. You should only add 9g to get the 10g that you want.

If the platform isn't moving, the object is at 1g. If the platform is accelerated upwards relative to the ground at 1g, the object is at 2g. If it accelerates downwards at 1g, the object is at 0g.

It's difficult to give a better explanation than this without a diagram showing your system, but it seems like there's probably something going on with reaction forces between the object and the platform? Anyway, as you showed with the correct solution, it is not necessary to have any additional information to solve the problem.

I hope this helps.

-Kerry

8. May 8, 2009

Ofey

One question. In wikipedia it says:

"If F* is the only force acting on the system, the system is called a simple harmonic oscillator"

But F isn't the only force if gravity also acts on the system?

*F=-kx

In all proofs of harmonic motion gravity is neglected. I don't understand why we do that. :(

9. May 8, 2009

davieddy

It is taken into account by choosing x = 0 to be the point
where the spring force and mg are equal and opposite.

Then you find the net force is -kx

10. May 8, 2009

Ofey

Now I understand. Thank you :)


text:
## Dynamic stochastic general equilibrium models for policy analysis

What are DSGE models?

Dynamic stochastic general equilibrium (DSGE) models are used by macroeconomists to model multiple time series. A DSGE model is based on economic theory. A theory will have equations for how individuals or sectors in the economy behave and how the sectors interact. What emerges is a system of equations whose parameters can be linked back to the decisions of economic actors. In many economic theories, individuals take actions based partly on the values they expect variables to take in the future, not just on the values those variables take in the current period. The strength of DSGE models is that they incorporate these expectations explicitly, unlike other models of multiple time series.

DSGE models are often used in the analysis of shocks or counterfactuals. A researcher might subject the model economy to an unexpected change in policy or the environment and see how variables respond. For example, what is the effect of an unexpected rise in interest rates on output? Or a researcher might compare the responses of economic variables with different policy regimes. For example, a model might be used to compare outcomes under a high-tax versus a low-tax regime. A researcher would explore the behavior of the model under different settings for tax rate parameters, holding other parameters constant.

In this post, I show you how to estimate the parameters of a DSGE model, how to create and interpret an impulse response, and how to compare the impulse response estimated from the data with an impulse response generated by a counterfactual policy regime. Read more…

Categories: Statistics Tags:

## Ermistatas and Stata’s new ERMs commands

27 March 2018

Ermistatas is our most popular t-shirt these days. See it and you will understand why.

We call the character Ermistatas and he is thinking—Ermistatas cogitatu. Notice the electricity bolts being emitted and received by his three antennae.

The shirt is popular even among those who do not use Stata and it’s leading them to ask questions. “Who or what is Ermistatas and why is he, she, or it deserving of a t-shirt?”. Then they add, “And why three and not the usual two antennae?”

Ermistatas is the creation of our arts-and-graphics department to represent Stata 15’s new commands for fitting Extended Regression Models—a term we coined. We call it ERMs for short. The new commands are Read more…

## How to create animated graphics to illustrate spatial spillover effects

This post shows how to create animated graphics that illustrate the spatial spillover effects generated by a spatial autoregressive (SAR) model. After reading this post, you could create an animated graph like the following.

This post is organized as follows. First, I estimate the parameters of a SAR model. Second, I show why a SAR model can produce spatial spillover effects. Finally, I show how to create an animated graph that illustrates the spatial spillover effects. Read more…

Categories: Tags:

## The book that Stata programmers have been waiting for

27 February 2018

“The book that Stata programmers have been waiting for” is how the Stata Press describes my new book on Mata, the full title of which is

The Stata Press took its cue from me in claiming that it this the book you have been waiting for, although I was less presumptuous in the introduction:

This book is for you if you have tried to learn Mata by reading the Mata Reference Manual and failed. You are not alone. Though the manual describes the parts of Mata, it never gets around to telling you what Mata is, what is special about Mata, what you might do with Mata, or even how Mata’s parts fit together. This book does that.

I’m excited about the book, but for a while I despaired of ever completing it. I started and stopped four times. I stopped because the drafts were boring. Read more…

Categories: Tags:

## Programming an estimation command in Stata: Writing a Java plugin

This post is the fourth in a series that illustrates how to plug code written in another language (like C, C++, or Java) into Stata. This technique is known as writing a plugin or as writing a dynamic-link library (DLL) for Stata.

In this post, I write a plugin in Java that implements the calculations performed by mymean_work() in mymean11.ado, discussed in Programming an estimation command in Stata: Preparing to write a plugin, and I assume that you are familiar with that material.

This post is analogous to Programming an estimation command in Stata: Writing a C plugin and to Programming an estimation command in Stata: Writing a C++ plugin. The differences are due to the plugin code being in Java instead of C or C++. I do not assume that you are familiar with the material in those posts, and much of that material is repeated here.

This is the 32nd post in the series Programming an estimation command in Stata. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series. Read more…

Categories: Programming Tags:

## Programming an estimation command in Stata: Writing a C++ plugin

This post is the third in a series that illustrates how to plug code written in another language (like C, C++, or Java) into Stata. This technique is known as writing a plugin or as writing a dynamic-link library (DLL) for Stata.

In this post, I write a plugin in C++ that implements the calculations performed by mymean_work() in mymean11.ado, discussed in Programming an estimation command in Stata: Preparing to write a plugin. I assume that you are familiar with the material in that post.

This post is analogous to Programming an estimation command in Stata: Writing a C plugin. The differences are due to the plugin code being in C++ instead of C. I do not assume that you are familiar with the material in that post, and you will find much of it repeated here.

This is the 31st post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series. Read more…

Categories: Programming Tags:

## Programming an estimation command in Stata: Writing a C plugin

This post is the second in a series that illustrates how to plug code written in another language (like C, C++, or Java) into Stata. This technique is known as writing a plugin or as writing a dynamic-link library (DLL) for Stata.

In this post, I write a plugin in C that implements the calculations performed by mymean_work() in mymean11.ado, discussed in Programming an estimation command in Stata: Preparing to write a plugin. I assume that you are familiar with the material in that post.

This is the 30th post in the series Programming an estimation command in Stata. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series. Read more…

Categories: Programming Tags:

## Programming an estimation command in Stata: Preparing to write a plugin

This post is the first in a series that illustrates how to plug code written in another language (like C, C++, or Java) into Stata. This technique is known as writing a plugin or as writing a dynamic-link library (DLL) for Stata.

Plugins can be written for any task, including data management, graphical analysis, or statistical estimation. Per the theme of this series, I discuss plugins for estimation commands.

In this post, I discuss the tradeoffs of writing a plugin, and I discuss a simple program whose calculations I will replace with plugins in subsequent posts.

This is the 29th post in the series Programming an estimation command in Stata. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series. Read more…

Categories: Programming Tags:

## Importing Facebook data into Stata

In a previous post, we released a new command to import Twitter data into Stata. We have now added another new command, facebook2stata, that imports Facebook data. To install facebook2stata, type

net install https://www.stata.com/users/kcrow/facebook2stata, replace

Categories: Data Management Tags:


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# tangent line equation from a gradient vector

• March 16th 2009, 07:24 AM
Frostking
tangent line equation from a gradient vector
Throughout our vector calculus class, we have used
tangent line equation = f(a,b) + f sub x (a,b)(x-a) + f sub y(a,b) (y-b)

On an old final the question is asked to find the equation for the tangent line to a function who's output at (1,3) is = 7 and who's gradient vector (1,3) = 2 i - 5 j

Needless to say since I do not have the function I can not use my nice canned equation to solve this one!

The question is then asked to find the tangent plane to the surface of z = f(x,y) at the point (1,3,7). NO other information is provided?????

I believe in the first question that my gradient vector would be providing how x and y are changing so I should be able to get a slope for my line out of it. I am really not sure how to proceed. Any nudges or pushes toward the solution would be appreciated very much! Frostking
• March 16th 2009, 11:38 AM
running-gag
Hi

The gradient vector at (1,3) is 2i - 5j
Therefore $\frac{\partial f}{\partial x}(1,3) = 2$ and $\frac{\partial f}{\partial y}(1,3) = -5$