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# Trigonometry worked solutions

Systems of trigonometric equations how to calculate solutions of systems of sine and cosine functions translate sine and cosine graphs : how to perform vertical translations on sine and cosine graphs. Section 71 solving trigonometric equations and identities 411 example 2 solve 02 t t 3sec ( ) 5sec( ) 2 for all solutions t 0 2 since the left side of this equation is quadratic in secant, we can try to factor it, and. Guided textbook solutions created by chegg experts learn from step-by-step solutions for over 22,000 isbns in math, science, engineering, business and more. Proving trigonometric identities (page 1 of 3) proving an identity is very different in concept from solving an equation though you'll use many of the same techniques, they are not the same, and the differences are what can cause you problems.

Derivatives of trigonometric functions this free handout has eight basic to graph it is an introduction into trig chart equations center hw 33 worksheet six using the. 1 introduction this unit looks at the solution of trigonometric equations in order to solve these equations we shall make extensive use of the graphs of the functions sine, cosine and tangent. Solution: in this triangle, we need to find the lengths of two sideswe can find the length of one side using a trig ratio then we can find the length of the third side either using a trig ratio, or the pythagorean theorem.

Maths tutorial: trigonometry law of sines / sine rule find this pin and more on math by dirk wagner maths tutorials, geometry and trigonometry the law of sines (also known as the sine rule) is a method for working out the angle or side length in a non righ. Trigonometric equations and identities part 1: pythagorean identities recall that, in the section on the unit circle, we established that given any angle $\theta$, $\left(\cos\left(\theta\right),\sin\left(\theta\right)\right)$ are the coordinates of a point on the unit circle. Transform the given trig equation into a trig equation having only one unique trig function as variable there are a few tips on how to select the appropriate variable the common variables to select are: sin x = t cos x = t cos 2x = t, tan x = t and tan (x/2) = t. Here is a collection of past paper questions with detailed worked solutions on trigonometry for the cie a-level mathematics course ideal for testing or independent practise for your pupils. Perhaps we have mr britt to thank for providing the final pieces of the algebra and trigonometry jigsaw puzzle in this solution manual for students stuck with the sullivan text if so, my hat is off to the poor guy for having to find a way to make the rambling sullivan make sense at last for too many students.

The trigonometry worksheets are randomly created and will never repeat so you have an endless supply of quality trigonometry worksheets to use in the classroom or at home we have trigonometric ratios, inverse trigonometric ratios, solving right triangles, and multi-step problems worksheets for your use. Solution 15 : find an equation of the line tangent to the graph of at x=-1 if x = -1 then so that the tangent line passes through the point (-1, 0 ) the slope of the tangent line follows from the derivative. Here is a set of practice problems to accompany the integrals involving trig functions section of the applications of integrals chapter of the notes for paul dawkins calculus ii course at lamar university. Learn trigonometry for free—right triangles, the unit circle, graphs, identities, and more full curriculum of exercises and videos.

Trigonometry practice problems try solving these as much as you can on your own, and if you need help, look at the hidden solutions you may use a calculator. After completing the problems you can check your work by viewing the video playlist of problems linked in the table when you click on the link, the problems will appear in the center column for each problem you have the option of watching different people present the solution. Hi danielle thanks for the feedback i couldn't agree more that it needed to be changed so all trig functions and pythag were required in the solution. A guide to trigonometric equations one for each trig ratio from these specific solutions within a given range can be questions are worked through these are. Trigonometric identities worked solutions trigonometric identities she loves math, here are some examples of simple identity proofs with reciprocal and quotient identities typically, to do these proofs, you must.

## Trigonometry worked solutions

This book contains fully worked solutions for every question in each chapter of our textbook mathematics for australia 10a in this book, where applicable, each worked solution is modelled on the worked example in the textbook. Learn how to solve trigonometric equations and how to use trigonometric identities to solve various problems learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Worked-out solutions calc chat offers free worked-out solutions to all odd numbered exercises for ron larson textbooks you will also have access to free math tutors for additional help.

• When we work with inverse trig functions it is especially important to draw a triangle since the output of the inverse trig function is an angle of a right triangle.
• Review sheets trigonometry math 112 a summary of concepts needed to be successful in mathematics the following sheets list the key concepts which are taught in the specified math course.
• Example 1 use the basic identities to show that tan csc u5u sec u for all values of foru which each side of the equation is defined solution each of the functions in the given equation can be written in terms of sin u.

How to solve word problems using trigonometry: sine, cosine, tangent, angle of elevation, with examples and step by step solutions, calculate the height of a building, balloon, length of ramp, altitude, angle of elevation, questions and answers. 1 trigonometry of right triangles 11 degree and radian measure 12 trigonometric functions in a right triangle 13 extending the domains of trigonometric functions to any angles 14 the law of cosines 15 the law of sines 16 areas of triangles. ©5 n2k001 q2j rkzuzt ta y dsjo sfdt 2w3awr1ef ylzlecu7 u 5a vlklw crbi gxhktxsb grnezsfedrwvde5dmz v gmmaydte 2 owuiat0h q vi5naf 2i4ncift2e m pahlqgbeub hroaz 62 ee worksheet by kuta software llc.

Trigonometry worked solutions
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# 5.6: Conclusion

In this chapter we have reviewed the basic motivation theories that have been developed to explain motivated behavior. Several theories view motivated behavior as attempts to satisfy needs. Based on this approach, managers would benefit from understanding what people need so that the actions of employees can be understood and managed. Other theories explain motivated behavior using the cognitive processes of employees. Employees respond to unfairness in their environment, they learn from the consequences of their actions and repeat the behaviors that lead to positive results, and they are motivated to exert effort if they see their actions will lead to outcomes that would get them desired rewards. None of these theories are complete on their own, but each theory provides us with a framework we can use to analyze, interpret, and manage employee behaviors in the workplace.

5.6: Conclusion is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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# Define an atlas on the set of all pointed lines in $\mathbb{R}^2$.

Define $M$ as the set of all pairs $(p, r)$ where $r$ is an affine line $r \subset \mathbb{R}^2$ and $p \in r$.

I am given a map $P \colon \mathbb{R}^3 \to M$ where $(u_1, u_2, u) \mapsto (p,r)$ where $p = (u_1, u_2)$ and $r = \{(x,y) \in \mathbb{R}^2 \mid y-u_2 = u(x-u_1)\}$ and I am asked the following

Show that $P$ is injective, find its image $\mathcal{U}\subset M$, define a chart $(\mathcal{U}, \varphi = P^{-1})$ of $M$ (and similarly define another chart $(\mathcal{V}, \psi)$ so that $\{(\mathcal{U}, \varphi), (\mathcal{V}, \psi)\}$ is an atlas. Prove that there is a unique riemannian metric $g$ on $M$ such that its local expression in the chart $(\mathcal{U}, \varphi)$ si $$g^{\varphi} = du_1^2 + du_2^2 + \frac{1}{(1+u^2)^2}du^2$$

Showing that $P$ is injective has been no problem, but I am unable to find it's image. Also I can't find the inverse map $P^{-1}$. I believe that if I had at least $P^{-1}$ I could prove the last question about the local representation of the metric $g$, but I am stuck at the beginning. Thank you.

• Given an arpitrary $(p,r)\in M$, can you find $(u_1,u_2,u)\in \Bbb R^3$ that masp to $(p,r)$? What is a possible obstacle? – Hagen von Eitzen Jun 26 '18 at 18:29
• This raises the interesting question of endowing the set of affine lines with a « natural topology ». Some points on that question here. – mathcounterexamples.net Jun 26 '18 at 20:20
• @HagenvonEitzen Well, the first two coordinates $(u_1, u_2)$ have to be precisely the coordinates of the point $p$. Since $r$ is a line that goes through $p$, the las coordinate should be the slope of $r$, maybe? – user313212 Jun 27 '18 at 7:22


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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

I just discovered a handy function in R to produce a scatterplot matrix of selected variables in a dataset. The base graphics function is pairs(). Producing these plots can be helpful in exploring your data, especially using the second method below.

Try it out on the built in iris dataset. (data set gives the measurements in cm of the variables sepal length and width and petal length and width, respectively, for 50 flowers from each of 3 species of iris. The species are Iris setosa, versicolor, and virginica).

# Load the iris dataset.
data(iris)

# Plot #1: Basic scatterplot matrix of the four measurements
pairs(~Sepal.Length+Sepal.Width+Petal.Length+Petal.Width, data=iris)
Looking at the pairs help page I found that there’s another built-in function, panel.smooth(), that can be used to plot a loess curve for each plot in a scatterplot matrix. Pass this function to the lower.panel argument of the pairs function. The panel.cor() function below can compute the absolute correlation between pairs of variables, and display these in the upper panels, with the font size proportional to the absolute value of the correlation.

# panel.smooth function is built in.
# panel.cor puts correlation in upper panels, size proportional to correlation
panel.cor <- function(x, y, digits=2, prefix="", cex.cor, ...)
{
usr <- par("usr"); on.exit(par(usr))
par(usr = c(0, 1, 0, 1))
r <- abs(cor(x, y))
txt <- format(c(r, 0.123456789), digits=digits)[1]
txt <- paste(prefix, txt, sep="")
if(missing(cex.cor)) cex.cor <- 0.8/strwidth(txt)
text(0.5, 0.5, txt, cex = cex.cor * r)
}

# Plot #2: same as above, but add loess smoother in lower and correlation in upper
pairs(~Sepal.Length+Sepal.Width+Petal.Length+Petal.Width, data=iris,
lower.panel=panel.smooth, upper.panel=panel.cor,
pch=20, main="Iris Scatterplot Matrix")

Finally, you can produce a similar plot using ggplot2, with the diagonal showing the kernel density.

# Plot #3: similar plot using ggplot2
# install.packages("ggplot2") ## uncomment to install ggplot2
library(ggplot2)
plotmatrix(with(iris, data.frame(Sepal.Length, Sepal.Width, Petal.Length, Petal.Width)))

See more on the pairs function here.

...

Update:  A tip of the hat to Hadley Wickham (@hadleywickham) for pointing out two packages useful for scatterplot matrices. The gpairs package has some useful functionality for showing the relationship between both continuous and categorical variables in a dataset, and the GGally package extends ggplot2 for plot matrices.


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## Lua encoding

General discussion about LÖVE, Lua, game development, puns, and unicorns.
Coder567
Prole
Posts: 42
Joined: Sun Apr 07, 2019 12:32 pm

### Re: Lua encoding

zorg wrote:
Wed Oct 09, 2019 11:51 am
Coder567 wrote:
Wed Oct 09, 2019 11:41 am
My projects are always so big I would like to have some protection. Then after some time after the release I could release the code so people can mod it but at start I would like to have the source closed. Or partly open source, for modding.

I tried luajit to compile lua file but love couldn't read the compiled file
I think i mentioned it previously, but you need to use the same version of LuaJIT that the Löve version you have uses.
That must be the issue, I used luaJIT 2.1.0-beta3
I guess love is still using 2.0

Coder567
Prole
Posts: 42
Joined: Sun Apr 07, 2019 12:32 pm

### Re: Lua encoding

Got the code encoding working now with love. I wonder how well can people actually decode the files because I haven't found any good decoders from google. Found just this one (luajit-decomp) which doesn't do too good job

original source:

Code: Select all

local test = "Test string "

print (test .. 123)

function love.draw()
love.graphics.print("Hello World!", 400, 300)
end

Decoded:

Code: Select all

function someFunc0()
love.graphics.print("Hello World!",  400 ,  300 )
return
end

function someFunc1()
local var_1_2 = "Test string "
var_1_2 = var_1_2 ..  123
print(var_1_2)
local randomFunction0 = function() end -- starts at  test.lua:0
love.draw = randomFunction0
return
end

Nelvin
Party member
Posts: 115
Joined: Mon Sep 12, 2016 7:52 am
Location: Germany

### Re: Lua encoding

Coder567 wrote:
Wed Oct 09, 2019 11:41 am
My projects are always so big I would like to have some protection. Then after some time after the release I could release the code so people can mod it but at start I would like to have the source closed. Or partly open source, for modding.
May I ask to see any of your previous projects just to get a feeling for what you mean with your projects are always so big?

Coder567
Prole
Posts: 42
Joined: Sun Apr 07, 2019 12:32 pm

### Re: Lua encoding

Nelvin wrote:
Thu Oct 10, 2019 5:41 pm
Coder567 wrote:
Wed Oct 09, 2019 11:41 am
My projects are always so big I would like to have some protection. Then after some time after the release I could release the code so people can mod it but at start I would like to have the source closed. Or partly open source, for modding.
May I ask to see any of your previous projects just to get a feeling for what you mean with your projects are always so big?
My previous projects are mostly (the big ones) mods for games. I decided to make game of my own this time
And it's pretty big again

### Who is online

Users browsing this forum: icekiller8002 and 35 guests


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Lesson 6, Topic 1
In Progress

# Grouped Frequency Distribution

Lesson Progress
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Grouped Discrete Data

Recall that in a grouped discrete data, the data are usually whole numbers. Also, in grouped discrete data, the intervals do not overlap. So it is inappropriate to use such intervals. However, when we group discrete data, we are actually treating it as though it was continuous.

Grouped Continuous Data

When using data that are continuous, the variable is measured on a continuous scale. Note that for continuous data, we use the class boundaries and not the class intervals.

Class Limits: Given a class interval of 40 – 44, the number 40 is the lower class limit, while the number 44 is the upper class limit.

Class Boundaries: Given a class interval of 45 – 50, the class boundaries become 44.5 – 50.5

That is to obtain the lower class boundary subtract 0.5 from the lower class limit.

Mid-Class Value (Class Mark)

The Mid-value of a class is known as the class mark. For instance, the mid-class value is exactly half way between the lower limit and the upper limit. For example the mid-class value for 39.5 – 45.5 is given as

$$\frac{39.5 + 44.5}{2} \scriptsize = 42.5$$

error:


text: A conventional drained triaxial compression test was conducted on a normally consolidated clay sample under an effective confining pressure of $200$ kPa. The deviator stress at failure was found to be $400$ kPa. An identical specimen of the same clay sample is isotropically consolidated to a confining pressure of $200$ kPa and subjected to standard undrained triaxial compression test. If the deviator stress at failure is $150$ kPa, the pore pressure developed (in kPa, up to one decimal place) is ________


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# Affine Transformations#

## API#

Perform 3D affine transformation of image given a 4x4 transformation matrix

optional dzyx parameter specifies the voxel size of the image (dz, dy, dx). If it is provided, it will be used to transform the image from intrinsic coordinates to world coordinates prior to transformation, e.g.:

$$x = 0.5 + (x - 0.5) * dx$$

and then back to intrinsic coords afterwards… e.g.:

$$tu = 0.5 + (tu - 0.5) / dx$$

Parameters
• im (np.ndarray) – 3D input volume

• tmat (np.ndarray) – Affine transformation matrix

• dzyx (3-tuple of int) – Voxel size of input volume (dz, dy, dx). If provided, the transformation matrix is assumed to be in units of sample space. otherwise the transformation is performed in image coordinates. by default None

Returns

volume – Transformed volume

Return type

np.ndarray

Raises

ValueError – If the dimensions of the transformation matrix do not the input volume. For instance, a 3D input volume requires a 4 x 4 tranforamtion matrix.

Examples

Perform simple translation

>>> nx, ny, nz = (10, 20, 3)
>>> T = np.array([[1, 0, 0, nx],
...               [0, 1, 0, ny],
...               [0, 0, 1, nz],
...               [0, 0, 0, 1]])
>>> rotated = affineGPU(im, T)

Perform a rotation about the Y axis… (this is the underlying code for rotateGPU())

>>> theta = angle * np.pi / 180
>>> nz, ny, nx = im.shape
>>> xzRatio = dxdata / (np.deg2rad(angle) * dzdata)
>>> # first translate the middle of the image to the origin
>>> T1 = np.array([[1, 0, 0, nx / 2],
>>>                [0, 1, 0, ny / 2],
>>>                [0, 0, 1, nz / 2],
>>>                [0, 0, 0, 1]])
>>> # then scale (resample) the Z axis the dz/dx ratio
>>> S = np.array([[1, 0, 0, 0],
>>>               [0, 1, 0, 0],
>>>               [0, 0, xzRatio, 0],
>>>               [0, 0, 0, 1]])
>>> # then rotate theta degrees about the Y axis
>>> R = np.array([[np.cos(theta), 0, -np.sin(theta), 0],
>>>               [0, 1, 0, 0],
>>>               [np.sin(theta), 0, np.cos(theta), 0],
>>>               [0, 0, 0, 1]])
>>> # then translate back to the original origin
>>> T2 = np.array([[1, 0, 0, -nx / 2],
>>>                [0, 1, 0, -ny / 2],
>>>                [0, 0, 1, -nz / 2],
>>>                [0, 0, 0, 1]])
>>> T = np.eye(4)
>>> T = np.dot(np.dot(np.dot(np.dot(T, T1), S), R), T2)
>>> rotated = affineGPU(im, T)

Deskew data acquired in stage-scanning mode on GPU

Simple affine transform variant to perform a shear operation to correct for volume shearing.

Parameters
• im (np.ndarray) – Image volume to deskew

• dxdata (float, optional) – XY Pixel size of image volume, by default 0.1

• dzdata (float, optional) – Z-step size in image volume. In a typical light sheet stage-scanning aquisition, this corresponds to the step size that the stage takes between planes, NOT the final Z-step size between planeds after deskewing along the optical axis of the detection objective, by default 0.5

• angle (float, optional) – Deskew angle (usually, angle between sheet and axis of stage motion), by default 31.5

• width (int, optional) – If not 0, crop output image to specified width, by default 0

• shift (int, optional) – If not 0, shift image center by this value, by default 0

• pad_val (int, optional) – Value to pad image with when deskewing. If None the median value of the last Z plane will be used, by default None

Returns

Deskewed volume

Return type

np.ndarray

Rotate image around Y axis by some angle

This is a convenience function that will apply the appropriate affine transformation for rotating a volume around the Y axis by some angle. This is typically done with images acquired on inverted light sheet microscopes where the image plane is not parallel to the coverslip (such as lattice light sheet, or diSPIM microscopes), in order to change the coordinate system of the image volume such that the Z axis is orthogonal to the coverslip

Parameters
• im (np.ndarray) – 3D volume to be rotated

• dzdata (float) – Z-step size in microns of the image volume

• dxdata (float) – XY pixel size of the volume, by default 0.1

• angle (float) – Angle to rotate around Y axis, by default 31.5

• reverse (bool) – Rotate in the opposite direction, by default False

Returns

Rotated 3D volume

Return type

np.ndarray


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## Thursday, July 05, 2012

### RBI vision document on payments: An evaluation

by Madhavi Pundit and Suyash Rai.

On June 27, RBI published its Payment System Vision Document (2012-15). The document shows RBI's vision and mission for the payment system, and specifies the objectives, approaches and courses of action emanating from the same. It is a laudable step taken by RBI to discuss its plans for payments in India. All regulatory activities, such as banking and capital account liberalisation should have vision documents to similarly show the road map. They bring clarity to market participants, and helps everyone plan better.

A vision document is an opportunity to think from first principles, and to dream about the payments landscape. The document does not do this sufficiently.

Going by the ideas of the vision document, it is difficult to hold RBI accountable to it or to evaluate its role as a regulator, because it is not clear what we should expect the payment system to achieve, say, five years from now. Though the vision itself may be a general, aspirational statement, it should be accompanied by quantifiable goals that can be achieved by Indian payments system (regulator + industry). For example, the regulator can set objectives that by 2015, cash will be x% of transactions, or that cheques will be phased out by 2020. RBI can then, as a regulator, take steps to facilitate the achievement of such goals, with the expectation that other participants will play their part. Such sharp statements are absent in the document.

Once a suitably ambitious, but quantifiable, vision for payments has been stated, achieving it requires looking beyond incremental modifications. This brings us to the kind of steps RBI has proposed in the document; the things that it will do to achieve the vision over the next three years. For a vision document, the proposed steps are rather tactical and operational. From this document, it seems RBI is running all payment systems, with incidental cooperation from the private sector. It is difficult to imagine how industry participants should work towards the vision. For an example of how this can be done, contrast the RBI document with the Strategic Review of Innovation in the Payments System, recently released by the Reserve Bank of Australia. Unlike RBI's document, the focus of this document is purely strategic, and on removing barriers that prevent market participants from innovating.

The document should clearly state what RBI sees as its role in payment systems - which is above all, that of a regulator - and what it sees as the role of market participants. While the document focuses on the development of certain types of electronic payment systems, certain standards, authorisation methods etc., there are a host of other ways the market could innovate. The document's approach precludes other means by which the same goal of higher electronic payments can be achieved.

All regulation must be rooted in market failures that damage the interests of consumers or threaten systemic crises. A specific regulation must address a specific market failure and thus tangibly further consumer protection or systemic stability:
1. In payment systems, consumer protection would entail measures that ensure transparency and disclosure by payment providers, so that consumers receive what they are promised.
2. In addition, providers must be subjected to micro-prudential regulations, such as capital requirements, risk management and investment restrictions that ensure their safety and soundness, based on the risks they take. The risk based approach means that small value systems with real-time payments need less regulation than large value systems that hold clients' funds for a certain amount of time.
3. For systemic stability, enhanced regulation and supervision of systemically important payment systems, especially back-end infrastructure, such as RTGS, is required.

RBI ought to focus on these regulatory objectives, where it would deliver public goods, rather than take on `private goods' functions that can be handled ably by the market. Systems such as NEFT and ECS, which essentially require capabilities that go beyond a regulator's core competence, can be run well by the private sector, under RBI's regulations. In such systems, competition is of essence.

From this perspective, the vision document starts looking less impressive. It is tied to the existing ways of doing things, and intent on incrementally improving them, rather than questioning the existing paradigm. This is unsatisfactory, for a paradigm shift is what India most requires.

Perhaps that is why there seems to be a lack of clarity of purpose. For example, RBI talks about investing in cheque systems and electronic systems at the same time. Developing a grid system to replace clearing houses as suggested is expensive. If the objective is to phase out cheques and promote electronic payments, the revamp of the cheque clearing system has no place in the vision document for electronic payments.

The emphasis on electronic payments is welcome. It is time for India to become a less-cash society, and ultimately a cash-less society. Myriad inexpensive, safe and useful electronic technologies are available, and more are being developed as we write. Hence the extensive use of cash and other paper-based instruments is not acceptable. They are expensive and inconvenient, and cost the most to those with the least - who pay for using these instruments and also face value erosion due to inflation. More needs to be done to move to electronic payments, and soon.

Competition and innovation are both important for this goal. The document talks about the dilemma the regulator faces with regard to pricing. To us, there is no such dilemma. To a large extent, the regulator should not intervene in business decisions such as pricing. In terms of market structure, there are two types of charges in payments - by retail payment providers and by infrastructure providers in the system. At the front end, innovative and cost effective payments products and gateways can develop if there is competition and there is no case for regulatory intervention here. It is a serious issue, and as experience from credit markets would suggest, a cap on pricing usually leads to more exclusion than inclusion.

Anti-competitive actions by players can be taken up to the Competition Commission. At the same time, it is important to note that in industries that are network based, there may be a need for monopolies or duopolies in infrastructure provision which require modification of the standard approaches of competition law (example).

Under these circumstances, if there is evidence of supernormal profits, there may be role for regulating prices. But even here, price determination should be done transparently, based on a full analysis of costs and reasonable returns, and in consultation with industry participants. For example, in the recent announcement of a cap on merchant discount rate on debit cards, there is no explanation from the regulator for how the amount 0.75 per cent was decided, and what are its costs and benefits to the system.

The role for non-banks is conspicuous by its absence in the vision statement. Currently, regulations tie the hands of non-bank payment providers. Take the example of Airtel money, which is a semi-closed mobile wallet. This means money can be transferred to other Airtel customers and transactions can take place with certain merchants, but there is no possibility for cashing out. Vodafone has partnered with a bank, and hence allows cash out from retail points; but these registered points have to be within 30 km of the bank partner.

A key insight that should guide the way forward is that payments is a separate business from banking, and should have its own regulation. Decoupling them could help achieve the twin goals of innovation and inclusion. An electronic payments revolution can take place when small value transactions are done electronically, i.e., customers in every nook and corner of the country can access secure, efficient and low cost retail payments services that can be considered cash substitutes. E-money in many countries has exploded on the backs of non-bank led payments systems such as telecom companies and retail chains, and their reach has been impressive. Easing restrictions on non-bank payments systems in India is required to really take advantage of their vast networks that have already penetrated unbanked areas. There are risks, but nothing that a forward thinking regulator who recognises the immense potential cannot creatively address. (See How to achieve safety in payments for an example.)

Finally, for large value electronic payments systems, RBI's vision should be to bring them up to world standards and integrated with global systems. Cross-border payments are an important facet of international trade and integration, and this can lead to settlement/ Herstatt risks. RBI should address operational and regulatory issues to minimise these risks. For example, RTGS should be brought as close as possible to a 24 by 7 settlement system to ensure overlaps with corresponding systems in other countries and time zones.

Additionally, in light of recent data that shows that the INR is the third most traded emerging market currency, these and other steps should be taken so that the INR becomes an eligible currency for settlement in the Continuous Linked Settlement (CLS) system, alongside the other international currencies already on CLS. In conclusion, it is commendable that RBI has released a payments vision document. Such a document gives an opportunity for us to understand the mind of a government agency, and discuss and debate its priorities and actions. But writing a vision statement is a chance to step away from the familiarity of set ways and ask the big questions. RBI should not squander this opportunity.

Please note: LaTeX mathematics works. This means that if you want to say $10 you have to say \$10.


text:
# Homework Help: Conditional Probability & Independence

1. Jan 30, 2010

### exitwound

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

One satellite is scheduled to be launched form Cape Canaveral in Florida, and another launching is scheduled for Vandenberg AFB in California. Let A denote the event that the Vandenberg launch goes off on schedule, and let B represent the event that the Cape Canaveral launch goes off on schedule. If A and B are independent events with P(A)>P(B) and P(A u B) = .626, P(A ∩ B) = .144, determine the values of P(A) and P(B).

3. The attempt at a solution

This is a practice problem. I have the answers of .45 & .32. I have no idea how to get them.

I know that if P(A|B) = P(A), then the two probabilities are independent.

I also know that P(A|B)P(B) = P(A ∩ B) .

I don't know how to do this problem.

2. Jan 30, 2010

### rock.freak667

Well then you know that P(A|B)P(B) = P(A ∩ B) simplifies to P(A∩B)=P(A)P(B).

P(A u B) should be the same as P(A)+P(B)

3. Jan 30, 2010

### vela

Staff Emeritus
That's not correct. You'd be double-counting when A and B occur simultaneously. It should be

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$.

4. Jan 30, 2010

### exitwound

I actually got those two equations:

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
and
$P(A\cap B)=P(A)P(B)$

I end up with two equations with two unknowns, but I can't come up with any answers that work.

5. Jan 30, 2010

### vela

Staff Emeritus
Solve for P(A) in the second equation and plug it into the first, then multiply through by P(B). You'll end up with a quadratic equation.

6. Jan 30, 2010

### exitwound

Yup. And it comes out with no solutions. I've tried like 10 times.

7. Jan 30, 2010

### vela

Staff Emeritus
You must be making an algebra error somewhere. I did the same thing and got the expected answer. Why don't you post your work so we can see where you might have gone astray?

8. Jan 30, 2010

### exitwound

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
and
$P(A\cap B)=P(A)P(B)$

$$P(A)=\frac{P(A\cap B)}{P(B)}$$

$$P(A \cup B)=\frac{P(A\cap B)}{P(B)}+P(B)-P(A \cap B)$$

$$P(A \cup B)=P(A\cap B)+P(B)^2-P(A \cap B)P(B)$$

$.626=.144+P(B)^2-.144P(B)$

$P(B)^2-.144P(B)-.770=0$

9. Jan 30, 2010

### vela

Staff Emeritus
You forgot to multiply the lefthand side by P(B).

10. Jan 30, 2010

### exitwound

$$P(B)P(A \cup B)=P(A\cap B)+P(B)^2-P(A \cap B)P(B)$$

$$.626P(B)=.144+P(B)^2-.144P(B)$$
$$0 = .144+P(B)^2-.770P(B)$$

Still no solutions.

11. Jan 31, 2010

### vela

Staff Emeritus
Well, your equation is correct, and it has the aforementioned solutions. All I can suggest is you check your work again carefully.

12. Jan 31, 2010

### exitwound

Oh. I'm a doofus. The coefficient of P(B)^2 is 1, not 0. Thanks for the help.


text:
# 3.2: Solve Percent Applications

##### Learning Objectives

By the end of this section, you will be able to:

• Translate and solve basic percent equations
• Solve percent applications
• Find percent increase and percent decrease
• Solve simple interest applications
• Solve applications with discount or mark-up
##### Be prepared

Before you get started, take this readiness quiz.

1. Convert 4.5% to a decimal.
If you missed this problem, review [link].
2. Convert 0.6 to a percent.
If you missed this problem, review [link].
3. Round 0.875 to the nearest hundredth.
If you missed this problem, review [link].
4. Multiply (4.5)(2.38).
If you missed this problem, review [link].
5. Solve 3.5=0.7n.
If you missed this problem, review [link].
6. Subtract 50−37.45.
If you missed this problem, review [link].

## Translate and Solve Basic Percent Equations

We will solve percent equations using the methods we used to solve equations with fractions or decimals. Without the tools of algebra, the best method available to solve percent problems was by setting them up as proportions. Now as an algebra student, you can just translate English sentences into algebraic equations and then solve the equations.

We can use any letter you like as a variable, but it is a good idea to choose a letter that will remind us of what you are looking for. We must be sure to change the given percent to a decimal when we put it in the equation.

##### Exercise $$\PageIndex{1}$$

Translate and solve: What number is 35% of 90?

 Translate into algebra. Let n= the number. Remember "of" means multiply, "is" means equals. Multiply. 31.5 is 35% of 90
##### Exercise $$\PageIndex{2}$$

Translate and solve:

What number is 45% of 80?

36

##### Exercise $$\PageIndex{3}$$

Translate and solve:

What number is 55% of 60?

33

We must be very careful when we translate the words in the next example. The unknown quantity will not be isolated at first, like it was in Example. We will again use direct translation to write the equation.

##### Exercise $$\PageIndex{23}$$

Find the percent decrease. (Round to the nearest tenth of a percent.)

The population of North Dakota was about 672,000 in 2010. The population is projected to be about 630,000 in 2020.

6.3%

##### Exercise $$\PageIndex{24}$$

Find the percent decrease.

Last year, Sheila’s salary was $42,000. Because of furlough days, this year, her salary was$37,800.

10%

## Solve Simple Interest Applications

Do you know that banks pay you to keep your money? The money a customer puts in the bank is called the principal, P, and the money the bank pays the customer is called the interest. The interest is computed as a certain percent of the principal; called the rate of interest, r. We usually express rate of interest as a percent per year, and we calculate it by using the decimal equivalent of the percent. The variable t, (for time) represents the number of years the money is in the account.

To find the interest we use the simple interest formula, I=Prt.

##### SIMPLE INTEREST

If an amount of money, P, called the principal, is invested for a period of t years at an annual interest rate r, the amount of interest, I, earned is

$\begin{array}{lllll} {} &{} &{I} &{=} &{\text { interest }}\\ {I = Prt} &{\text{where}} &{P} &{=} &{\text { principle }}\\ {} &{} &{r} &{=} &{\text { rate }}\\ {} &{} &{t} &{=} &{\text { time }} \end{array}$

Interest earned according to this formula is called simple interest.

Interest may also be calculated another way, called compound interest. This type of interest will be covered in later math classes.

The formula we use to calculate simple interest is I=Prt. To use the formula, we substitute in the values the problem gives us for the variables, and then solve for the unknown variable. It may be helpful to organize the information in a chart.

$7020 There may be times when we know the amount of interest earned on a given principal over a certain length of time, but we don’t know the rate. To find the rate, we use the simple interest formula, substitute in the given values for the principal and time, and then solve for the rate. ##### Exercise $$\PageIndex{28}$$ Loren loaned his brother$3,000 to help him buy a car. In 4 years his brother paid him back the $3,000 plus$660 in interest. What was the rate of interest?

$\begin{array}{lll} {I} &{=} &{\ 660} \\ {P} &{=} &{\ 3000} \\ {r} &{=} &{?} \\ {t} &{=} &{4 \text { years } }\end{array}$

$$\begin{array} {ll} {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the rate of interest}} \\\\ {\textbf{Step 3. Name} \text{ what we are looking for. Choose}} &{\text{Let r = the rate of interest.}} \\ {\text{ a variable to represent that quantity}} &{\text{}} \\\\ {\textbf{Step 4.} \text{ Translate into an equation.}} &{} \\ {\qquad\text{Write the formula.}} &{I = Prt} \\ {\qquad\text{Substitute in the given information.}} &{660 = (3000)r(4)} \\ \\ {\textbf{Step 5. Solve} \text{ the equation.}} &{} \\ {} &{660 = (12000)r} \\ {\text{Divide.}} &{0.055 = r} \\ {\text{Change to percent form.}} &{5.5\% = r} \\\\ {\textbf{Step 6. Check} \text{: Does this make sense?}} &{} \\\\ {I = Prt} &{} \\ {660 \stackrel{?}{=} (3000)(0.055)(4)} &{} \\ {660 = 660\checkmark} &{} \\ {\textbf{Step 7. Answer} \text{ the question with a}} &{\text{The rate of interest was }5.5%} \\ {\text{complete sentence.}} &{} \end{array}$$

Notice that in this example, Loren’s brother paid Loren interest, just like a bank would have paid interest if Loren invested his money there.

4%

$9,600 ## Solve Applications with Discount or Mark-up Applications of discount are very common in retail settings. When you buy an item on sale, the original price has been discounted by some dollar amount. The discount rate, usually given as a percent, is used to determine the amount of the discount. To determine the amount of discount, we multiply the discount rate by the original price. We summarize the discount model in the box below. ##### DISCOUNT $\begin{array}{l}{\text { amount of discount }=\text { discount rate } \times \text { original price }} \\ {\text { sale price }=\text { original price - amount of discount }}\end{array}$ Keep in mind that the sale price should always be less than the original price. ##### Exercise $$\PageIndex{34}$$ Elise bought a dress that was discounted 35% off of the original price of$140. What was ⓐ the amount of discount and ⓑ the sale price of the dress?

1. $$\begin{array} {lll} {\text{Original price}} &{=} &{140} \\ {\text{Discount rate}} &{=} &{35\%} \\ {\text{Discount?}} &{=} &{?} \end{array}$$
$$\begin{array} {ll} \\ {\textbf{Step 1. Read} \text{ the problem.}} &{} \\ \\ {\textbf{Step 2. Identify} \text{ what we are looking for.}} &{\text{the amount of discount}} \\\\ {\textbf{Step 3. Name} \text{ what we are looking for. }} &{\text{}} \\ {\text{Choose a variable to represent that quantity.}} &{\text{Let d = the amount of discount.}} \\\\ {\textbf{Step 4. Translate} \text{ into an equation. Write a}} &{} \\ {\text{sentence that gives the information to find it.}} &{} \\ {\text{Translate into an equation}} &{d = 0.35(140)} \\ \\ {\textbf{Step 5. Solve} \text{ the equation.}} &{d = 49} \\ \\ {\textbf{Step 6. Check} \text{: Does this make sense?}} &{} \\ \\ {\text{Is a }49\text{ discount reasonable for a}} &{} \\ {140\text{ dress? Yes.}} &{} \\\\ {\textbf{Step 7. Write} \text{ a complete sentence to answer}} &{\text{The amount of discount was }49} \\ {\text{the question.}} &{} \ \end{array}$$

2.

 Step 1. Identify what we are looking for. the sale price of the dress Step 2. Name what we are looking for. Choose a variable to represent that quantity. Let s= the sale price. Step 3. Translate into an equation. Write a sentence that gives the information to find it. Translate into an equation. Step 4. Solve the equation. Step 5. Check. Does this make sense? Is the sale price less than the original price? Yes, $91 is less than$140. Step 6. Answer the question with a complete sentence. The sale price of the dress was $91. ##### Exercise $$\PageIndex{35}$$ Find ⓐ the amount of discount and ⓑ the sale price: Sergio bought a belt that was discounted 40% from an original price of$29.

ⓐ $11.60 ⓑ$17.40

##### Exercise $$\PageIndex{36}$$

Find ⓐ the amount of discount and ⓑ the sale price:

2. 33%
##### Exercise $$\PageIndex{39}$$

Find

1. the amount of discount and
2. the discount rate.

Nick bought a multi-room air conditioner at a sale price of $340. The original price of the air conditioner was$400.

1. $60 2. 15% Applications of mark-up are very common in retail settings. The price a retailer pays for an item is called the original cost. The retailer then adds a mark-up to the original cost to get the list price, the price he sells the item for. The mark-up is usually calculated as a percent of the original cost. To determine the amount of mark-up, multiply the mark-up rate by the original cost. We summarize the mark-up model in the box below. ##### MARK-UP $\begin{array}{l}{\text { amount of mark-up }=\text { mark-up rate } \times \text { original cost }} \\ {\text { list price }=\text { original cost }+\text { amount of mark up }}\end{array}$ Keep in mind that the list price should always be more than the original cost. ##### Exercise $$\PageIndex{40}$$ Adam’s art gallery bought a photograph at original cost$250. Adam marked the price up 40%. Find the

1. amount of mark-up and
2. the list price of the photograph.

1.

 Step 1. Read the problem. Step 2. Identify what we are looking for. the amount of mark-up Step 3. Name what we are looking for. Choose a variable to represent it. Let m= the amount of markup. Step 4. Translate into an equation. Write a sentence that gives the information to find it. Translate into an equation. Step 5. Solve the equation. Step 6. Check. Does this make sense? Yes, 40% is less than one-half and 100 is less than half of 250. Step 7. Answer the question with a complete sentence. The mark-up on the photograph was $100. 2.  Step 1. Read the problem again. Step 2. Identify what we are looking for. the list price Step 3. Name what we are looking for. Choose a variable to represent it. Let p= the list price. Step 4. Translate into an equation. Write a sentence that gives the information to find it. Translate into an equation. Step 5. Solve the equation. Step 6. Check. Does this make sense? Is the list price more than the net price? Is$350 more than $250? Yes. Step 7. Answer the question with a complete sentence. The list price of the photograph was$350.
##### Exercise $$\PageIndex{41}$$

Find

1. the amount of mark-up and
2. the list price.

Jim’s music store bought a guitar at original cost $1,200. Jim marked the price up 50%. Answer 1.$600
2. $1,800 ##### Exercise $$\PageIndex{42}$$ Find 1. the amount of mark-up and 2. the list price. The Auto Resale Store bought Pablo’s Toyota for$8,500. They marked the price up 35%.

1. $2,975 2.$11,475

## Key Concepts

• Percent Increase To find the percent increase:
1. Find the amount of increase. increase=new amount−originalamountincrease=new amount−originalamount
2. Find the percent increase. Increase is what percent of the original amount?
• Percent Decrease To find the percent decrease:
1. Find the amount of decrease. decrease=original amount−newamountdecrease=original amount−newamount
2. Find the percent decrease. Decrease is what percent of the original amount?
• Simple Interest If an amount of money, P, called the principal, is invested for a period of t years at an annual interest rate r, the amount of interest, I, earned is

\begin{aligned} I &=P r t \\ \text { where } I &=\text { interest } \\ P &=\text { principal } \\ r &=\text { rate } \\ t &=\text { time } \end{aligned}

• Discount
• amount of discount is discount rate ·· original price
• sale price is original price – discount
• Mark-up
• amount of mark-up is mark-up rate ·· original cost
• list price is original cost + mark up

## Glossary

amount of discount
The amount of discount is the amount resulting when a discount rate is multiplied by the original price of an item.
discount rate
The discount rate is the percent used to determine the amount of a discount, common in retail settings.
interest
Interest is the money that a bank pays its customers for keeping their money in the bank.
list price
The list price is the price a retailer sells an item for.
mark-up
A mark-up is a percentage of the original cost used to increase the price of an item.
original cost
The original cost in a retail setting, is the price that a retailer pays for an item.
principal
The principal is the original amount of money invested or borrowed for a period of time at a specific interest rate.
rate of interest
The rate of interest is a percent of the principal, usually expressed as a percent per year.
simple interest
Simple interest is the interest earned according to the formula I=Prt.

This page titled 3.2: Solve Percent Applications is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax.


text:
# Interpretation of EN 61000-4-5 Surge requirement on a DC Power Port

I am trying to understand how to treat the DC power port on my industrial device (read: will be used in CNC machine shops and industrial processing plants and the like) should be tested according to EN61000-6-2 for the surge test specified in EN 61000-4-5.

Table 3 in the standard calls out the tests required for DC power ports. Line 3.2 calls out 500V surge testing applied between line to ground and also between line to earth. So far there is nothing confusing.

However, note (c) is called out under "remarks" for surge testing, and it states that surge is not applicable under certain conditions. Most of the conditions don't apply to my situation, but the final line of the note states "DC Ports, which are not intended to be connected to a DC Distribution network, are treated as signal ports".

So now the question is - what is a DC Distribution network? My device is intended to run off of either 12 or 24 V DC power, but it's unclear to me if a voltage regulator on the same CNC machine or process equipment as my device is considered a "network", or if it must be some centralized regulator that is run throughout a plant, or some other definition. Can anyone here provide a working definition or clarity on what this means?

If I can treat the DC Power port as a signal port, it would greatly simplify my transient protection scheme.

• It appears to be Class 0 – Sunnyskyguy EE75 Dec 11 '18 at 4:24

The surge test according to IEC EN 61000-4-5 is performed on equipment whose power supply connections are geometrically far away from the device itself, in order to test its resistance to the three following stroke related events (accordig to this nice survey):

1. Direct lightning strokes to outdoor circuits injecting high currents and producing overvoltages.
2. Indirect lightning strikes, i.e. strikes between or within clouds or to nearby objects which produce electromagnetic fields that induce voltages/currents on the conductors outside and/or inside a building.
3. Lightning ground current flows, resulting from nearby direct-to-earth discharges coupling into the common ground paths of the grounding system of the installation.

Note that all three events above involve long cable lengths and distances: specifically the third one, which is closely related to the necessity of making a 500V surge test, requires the presence of a long ground path in order to induce a dangerous EMF on nearby cables by high value pulsed currents flowing through it (by the Faraday-Neumann-Lentz law).

Said that, if you equipment is powered from a DC power supply placed within a few meters (precisely within $$\3\$$ meters of cable, but I can possibly remember a wrong value) from it, the three events above are considered as very low probability events to happen on the DC power input: the DC port is thus considered as a signal port. Therefore you can avoid the IEC EN 61000-4-5 test on the DC input of your equipment, but

1. you must do it on the input of your DC power supply (or choose a a power supply tested accordingly) and ship it to be used with your equipment or
2. state a maximum cable length for the DC supply power cables (customarily less than the maximum one allowed).
3. Edit: something that I remembered later is that if your device is designed to be embedded inside a larger equipment, you can require your customer to provide you a nice power supply rail, fully protected from surges. Such devices are sometimes called OEM (Original Equipment Manufacturing) devices.

If you choose to not proceed according to the above guidelines, it is implicitly assumed that your equipment can be powered by a DC rail, involving a single power supply placed several (cable) meters away from the DC input of your equipment and which possibly powers other plant equipment as well: then you have to be sure that it survives a surge event, so you have to do the test.


text:
# Tetration Forum

Full Version: Taylor series of i[x]
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Pages: 1 2
I have been interested in that what the taylor series of i[x] is for long years.
There are infinity base units from i[0]=1 through i[pi] to i[10000...]. Their multiplication with each other usually is -1, but we can be sure only when x is an integer bigger than 0.
Here is a multiplication table for base units from i[0] to i[15]: https://en.wikipedia.org/wiki/Sedenion
(So we xor their indexes: n^m where ^ is the xor binary operator giving somehow a sign ... unfortunately I do not know what the form for this is.)
But the biggest question is that how we can get its taylor series. Any idea?
(01/12/2017, 04:50 PM)Xorter Wrote: [ -> ]I have been interested in that what the taylor series of i[x] is for long years.
There are infinity base units from i[0]=1 through i[pi] to i[10000...]. Their multiplication with each other usually is -1, but we can be sure only when x is an integer bigger than 0.
Here is a multiplication table for base units from i[0] to i[15]: https://en.wikipedia.org/wiki/Sedenion
(So we xor their indexes: n^m where ^ is the xor binary operator giving somehow a sign ... unfortunately I do not know what the form for this is.)
But the biggest question is that how we can get its taylor series. Any idea?

I'm not familiar with Sedenion, but it is an abstract algebra concept, not a complex analytic function, right?  So by definition, unless there is some mapping to a complex function, then it would not have a Taylor series...
(01/12/2017, 08:50 PM)sheldonison Wrote: [ -> ]I'm not familiar with Sedenion, but it is an abstract algebra concept, not a complex analytic function, right?  So by definition, unless there is some mapping to a complex function, then it would not have a Taylor series...

Well, I believe it is differentialable and if it is, it will have an exact Taylor series.
We just need to get its derivative. I could start to determine it, but I could not finish it, just look at it:
$di_x / dx = lim (i_{x+h}-i_x)/h = i_x lim (1-i_x / i_{x+h})/h$, where h approaches to 0.
So I know that its derivative will have multiplicative part as i[x] and another thing which is questionable. What is lim (1 - i[x]/i[x+h])/h.
I think it is absolutely solvable ... but how?
(01/12/2017, 08:50 PM)sheldonison Wrote: [ -> ]I'm not familiar with Sedenion, but it is an abstract algebra concept, not a complex analytic function, right?  So by definition, unless there is some mapping to a complex function, then it would not have a Taylor series...

Ah, yes, and I could recognise something interesting.
I tried to take i[x] into the Carleman matrix and to square it with itself, it seems successfully, maybe. Well, I got that the new Carleman matrix 2nd coloumn have I[1,0], I[1,1], I[1,2], ... where
$I_{1,N} = 1/N! \sum_{k=1}^{\infty} i^{(N)k}_x i^{(k)}_x / k! |_{x=0}$
So
$i_x ^{o2} = i_{i_x} = \sum_{k=0}^{\infty} I_{1,k} x^k =
= \sum_{k=0}^{\infty} i^{(k)}_x / k! + x \sum_{k=0}^{\infty} i'^{k}_x i^{(k)}_x / k! + x^2 \sum_{k=0}^{\infty} i''^{k}_x i^{(k)}_x / k! + ... |_{x=0}$

Now I feel we are closer than before.
Here is another interesting fact:
The Taylor series of i[x] is the following:
$i_x = \sum_{k=0}^{\infty} i^{(k)}_x x^k / k! = i_x + x i'_x + x^2 1/2 i''_x + x^3 1/6 i'''_x + ...$
and
$i_x^{o2} = i_{i_x} = \sum_{k=0}^{\infty} i^{(k)}_x / k! + x \sum_{k=0}^{\infty} i'^k_x i^{(k)}_x / k! + x^2 \sum_{k=0}^{\infty} i''^k_x i^{(k)}_x / k! + ... |_{x=0}$
So i[i[x]] has a part as 1 looking like this:
$i_{i_x} = 1 + x \sum_{k=0}^{\infty} i'^k_x i^{(k)}_x / k! + x^2 \sum_{k=0}^{\infty} i''^k_x i^{(k)}_x / k! + ... |_{x=0}$
Therefor di[x]/dx cannot be 0, then what can it be?
(01/12/2017, 04:50 PM)Xorter Wrote: [ -> ]I have been interested in that what the taylor series of i[x] is for long years.
There are infinity base units from i[0]=1 through i[pi] to i[10000...]. Their multiplication with each other usually is -1, but we can be sure only when x is an integer bigger than 0.
Here is a multiplication table for base units from i[0] to i[15]: https://en.wikipedia.org/wiki/Sedenion
(So we xor their indexes: n^m where ^ is the xor binary operator giving somehow a sign ... unfortunately I do not know what the form for this is.)
But the biggest question is that how we can get its taylor series. Any idea?

Well you haven't really defined terms but I take it by "base units" you mean the imaginary units that appear in the Cayley-Dickson construction and want to see if it would make sense to define those for non-integer indices somehow.

Well, the XOR bit actually provides what would probably be the most natural way: if $e_x$ is a base unit with some non-negative real number $x$ which is not a dyadic fraction (i.e. we declare an uncountable set of base units $B = \{ e_x : x \in \mathbb{R}\ \mathrm{and}\ \ x \ge 0 \}$), then that $x$ has a unique binary expansion and if we have two of those units we could define $e_x e_y = \mathrm{sign}(x, y) e_{x\ \mathrm{xor}\ y}$ where the XOR operation is just XORing the infinite binaries of $x$ and $y$ and $\mathrm{sign}$ is some unknown generalization of the signing for when $x$ and $y$ are integer. Dyadic fractions are a problem though, because they have two different binary expansions, one ending in endless 0s and another in endless 1s. But we could say that the integers can all be considered as such, and the XOR we use there corresponds to doing it with representations ending in all 0s, thus the most natural rule would then be to just resolve them by taking their binaries to end in all 0s.

Thus we'd get that, say, $e_{1/3} e_{1/5}$ should be $e_{2/5}$ times some unknown sign, and $e_{1/2}e_{1/3}$ is $e_{5/6}$ times some unknown sign. Not sure of the sign bit yet but I'm sure there is some sort of method which can be used to generate it, and that would paint the way toward its extension.

These functions will, of course, be non-continuous when viewed as functions of the indices, so will not be analytic and have no Taylor series. But this is a rather different kind of interpolation/generalization animal than tetration, so you can't expect the same rules to apply. I also suspect that were one to craft an analytic interpolation it would be rather cumbersome and highly arbitrary, and not have any neat algebraic properties, like this might. Unlike tetration, we actually have algebra rules that point toward a generalization, like in exponentiation (e.g. $e^{x + y} = e^x e^y$ -- that's real-number "e" not unital "e"). So in some sense, this is actually a much easier problem.

What I'm saying is that Taylor series are probably the wrong way to go about this, and the answer is really a lot simpler.

ADD!: I was googling "sedenions xor" and got this

https://arxiv.org/pdf/math/0011260.pdf

where they seem to touch on this a bit and suggests the indexing is arbitrary. This would make sense -- you can throw an isomorphism on top. Which means we could relabel the elements to make the indices follow some other rule. But what that really goes to show is the real "meat" here is in the signs. IOW, still not close. But I really don't think Taylor series are of any use here.
(01/23/2017, 07:38 AM)mike3 Wrote: [ -> ]Well you haven't really defined terms but I take it by "base units" you mean the imaginary units that appear in the Cayley-Dickson construction and want to see if it would make sense to define those for non-integer indices somehow.

Well, the XOR bit actually provides what would probably be the most natural way: if $e_x$ is a base unit with some non-negative real number $x$ which is not a dyadic fraction (i.e. we declare an uncountable set of base units $B = \{ e_x : x \in \mathbb{R}\ \mathrm{and}\ \ x \ge 0 \}$), then that $x$ has a unique binary expansion and if we have two of those units we could define $e_x e_y = \mathrm{sign}(x, y) e_{x\ \mathrm{xor}\ y}$ where the XOR operation is just XORing the infinite binaries of $x$ and $y$ and $\mathrm{sign}$ is some unknown generalization of the signing for when $x$ and $y$ are integer. Dyadic fractions are a problem though, because they have two different binary expansions, one ending in endless 0s and another in endless 1s. But we could say that the integers can all be considered as such, and the XOR we use there corresponds to doing it with representations ending in all 0s, thus the most natural rule would then be to just resolve them by taking their binaries to end in all 0s.

Thus we'd get that, say, $e_{1/3} e_{1/5}$ should be $e_{2/5}$ times some unknown sign, and $e_{1/2}e_{1/3}$ is $e_{5/6}$ times some unknown sign. Not sure of the sign bit yet but I'm sure there is some sort of method which can be used to generate it, and that would paint the way toward its extension.

These functions will, of course, be non-continuous when viewed as functions of the indices, so will not be analytic and have no Taylor series. But this is a rather different kind of interpolation/generalization animal than tetration, so you can't expect the same rules to apply. I also suspect that were one to craft an analytic interpolation it would be rather cumbersome and highly arbitrary, and not have any neat algebraic properties, like this might. Unlike tetration, we actually have algebra rules that point toward a generalization, like in exponentiation (e.g. $e^{x + y} = e^x e^y$ -- that's real-number "e" not unital "e"). So in some sense, this is actually a much easier problem.

What I'm saying is that Taylor series are probably the wrong way to go about this, and the answer is really a lot simpler.

ADD!: I was googling "sedenions xor" and got this

https://arxiv.org/pdf/math/0011260.pdf

where they seem to touch on this a bit and suggests the indexing is arbitrary. This would make sense -- you can throw an isomorphism on top. Which means we could relabel the elements to make the indices follow some other rule. But what that really goes to show is the real "meat" here is in the signs. IOW, still not close. But I really don't think Taylor series are of any use here.

Yes, we have to xor the indeces to get a new one of the multiplication, but the method to determine the sign is unkown, I could not find any useful formula neither in that Arxiv paper nor anywhere.
I have written a program in PARI/gp which can xor two rational numbers, and yes, your speculation is correct, so i[1/3] i[1/5] = sgn() i[2/5] and i[1/2] i[1/3] = sgn() i[5/6].
For instance, here is the result of my programme code:
https://www.dropbox.com/s/5rcwawcfyisvcf...).bmp?dl=0
y = 1 xor x
And as you mentioned at the whole numbers the function is problematic, but it just has a cut-off point like sgn(x) at 0 or d|x|/dx at 0. It is on my graph you can see above.
And it is really simple thing to extrapolate onto complexes and interpolate onto reals. But this is not my question.
My question is that how we can determine the sign of multiplication of base units. I feel, intuitively, that these "signs" will be have a form like exp(ia). So these will be complex signs, if I can say it.
Now we are closer than ever before. I believe it is expansionable to Taylor series, but this is just my opinion.
Okey, here is a website: http://jwbales.us/ where you can calculate the sign by a programme. I checked the source of it, and I take the important part of it out. You can read it at the bottom.
I have made an own java html code, and it works. So we have a recursive formula for the signs. The next question is how to make just one formula and how to interpolate (to reals) and extrapolate (to complexes).
We are really close to have its Taylor series!

Code:
function bitValue(number,bit){ var result = (number & (1<<bit)) >>> bit; return result; } function twist(p,q){ var state="A0"; var pbit=0; var qbit=0; var result=1; var p0=p; var q0=q; var pplus=1+p0; var qplus=1+q0; var stateA0 = []; var stateA  = []; var stateB  = []; var stateNB = []; var stateC  = []; var stateNC = []; var loopIndex=Math.ceil(Math.LOG2E*Math.log((Math.max(pplus,qplus)))); var pXORq =eval(p0) ^ eval(q0); stateA0[0] = "A0" stateA0[1] = "A" stateA0[2] = "B" stateA0[3] = "NB" stateA[0]  = "A"; stateA[1]  = "A"; stateA[2]  = "C"; stateA[3]  = "NC"; stateB[0]  = "B"; stateB[1]  = "NC"; stateB[2]  = "B"; stateB[3]  = "C"; stateNB[0] = "NB"; stateNB[1] = "C"; stateNB[2] = "NC"; stateNB[3] = "NB"; stateC[0]  = "C"; stateC[1]  = "NC"; stateC[2]  = "NC"; stateC[3]  = "NC"; stateNC[0] = "NC"; stateNC[1] = "C"; stateNC[2] = "C"; stateNC[3] = "C"; if (p < 0 || q < 0){ p=p0;   q=q0;   str = "Values of p and q cannot be negative!";   alert(str); } for (var i = 0; i < loopIndex; i++){    var j = loopIndex - i - 1;    var k = 2*bitValue(p,j)+bitValue(q,j);    if (state == 'C'){        state = stateC[k];        continue;    }    if (state == 'NC'){        state = stateNC[k];        continue;    }    if (state == 'B'){        state = stateB[k];        continue;    }    if (state == 'NB'){        state = stateNB[k];        continue;    }    if (state == 'A'){        state = stateA[k];        continue;    }    if (state == 'A0'){        state = stateA0[k];        continue;    } }    if (state=="NB" || state=="NC") {        result="-1";    }    return result;     }
Okey, I programmed it and I got the next graph for y = Y(1;x)*(1 xor x):
https://www.dropbox.com/s/81rxif99w7j50u...D.bmp?dl=0

And this is my PARI/gp code:
Code:
twist(p,q)={ L=ceil(1.4426950408889634*log((max(q+1,p+1)))); res=1;sta=1; if(round(p)!=p||round(q)!=q,p*=2^60;q*=2^60;L+=60); p=round(p);q=round(q); T=[1,2,3,4;2,2,5,6;3,6,3,5;4,5,6,4;5,6,6,6;6,5,5,5]; for(i=0,L,j=L-i-1;k=2*bittest(p,j)+bittest(q,j)+1;sta=T[sta,k]); if(sta==4||sta==6,return(-1),return(1)); } xor(x,y)={return(bitxor(round(x*2^60),round(y*2^60))/2^60)}

I checked it in a really little interval (from 0 to 10^-8 ) and it does not look like something which is differentialable. So should I say it has not a derivative and Taylor series?
Anyway, can anyone determine the integral of it from 0 to x? It would be really interesting, I think so.
Okey, I have some news!
I got a little bit closer than before.
Let us get the length of the curve of points of the base units from i[exp(0)] to i[exp(2ipi)]:
In polar-coordinate system:
L = .5 [ sqrt(i[x]'^2 - 1) ] from 0 to 2pi

In cartesian coordinate system:
L = [ sqrt( i[exp(ix)]'^2 + 1 ) ] from 0 to 2pi

So
.5 [ sqrt(i[x]'^2 - 1) ] from 0 to 2pi = [ sqrt( i[exp(ix)]'^2 + 1 ) ] from 0 to 2pi
It looks a promising integral and differential equation.
What do you think, what should be the next step?
Pages: 1 2


text:
# How to improve maintenance of starred/non starred versions of commands or environments

All documentations I've read about starred/non starred versions of commands recommend more or less the following way of doing (following for instance the UK List of TeX FAQ):

\newcommand{\mycommand}{\@ifstar\mycommandStar\mycommandNoStar}
\newcommand{\mycommandStar}{%
<few lines of code only for starred mycommand>
<many lines of code common to starred/non starred mycommand>
}
\newcommand{\mycommandNoStar}{%
<few lines of code only for non starred mycommand>
<many lines of code common to starred/non starred mycommand>
}

For environments, the scheme is similar (following for instance this answer):

\newenvironment{myenvironment}{%
<few lines of code only for non starred myenvironment>
<many lines of code common to starred/non starred myenvironment>
}{%
<few lines of code only for non starred myenvironment>
<many lines of code common to starred/non starred myenvironment>
}
\newenvironment{myenvironment*}{%
<few lines of code only for starred myenvironment>
<many lines of code common to starred/non starred myenvironment>
}{%
<few lines of code only for starred myenvironment>
<many lines of code common to starred/non starred myenvironment>
}

But, most of the case, there are only slight differences between starred and non starred versions and this way of doing requires to copy all changes in the common code between the two versions, which is a pain of maintenance (particularly if the common code is long).

Is there a more efficient way of doing?

It mostly depends on what the commands should do. If \mycommand* differs from \mycommand just because some different code has to be executed at the start, the following approach should work:

\newcommand{\mycommand}{\@ifstar{\@tempswatrue\@mycommand}{\@tempswafalse\@mycommand}}
\newcommand{\@mycommand}{%
\if@tempswa
<few lines of code only for starred mycommand>
\else
<few lines of code only for non starred mycommand>
\fi
<many lines of code common to starred/non starred mycommand>
}

With xparse this becomes easier:

\usepackage{xparse}
\NewDocumentCommand{\mycommand}{s}
{\IfBooleanTF{#1}
{<few lines of code only for starred mycommand>}
{<few lines of code only for non starred mycommand>}%
<many lines of code common to starred/non starred mycommand>%
}

For the environment form there are additional complications, because there's no \@ifstar available.

\newenvironment{myenvironment}{%
<few lines of code only for non starred myenvironment>
\@myenvironmentstart
}{%
<few lines of code only for non starred myenvironment>
\@myenvironmentfinish
}
\newenvironment{myenvironment*}{%
<few lines of code only for starred myenvironment>
\@myenvironmentstart
}{%
<few lines of code only for starred myenvironment>
\@myenvironmentfinish
}

\newcommand{\@myenvironmentstart}{%
<many lines of code common to starred/non starred myenvironment>
}
\newcommand{\@myenvironmentfinish}{%
<many lines of code common to starred/non starred myenvironment>
}

No real simplification is possible with xparse.

• It's not the first time I'm advised to make use of xparse package. The LaTeX class I'm currently working on makes use of standard TeX/LaTeX2e tools (including etoolbox package): is there a drawback to mix xparse package with non LaTeX3 tools? – Denis Bitouzé Jul 13 '13 at 14:31
• @DenisBitouzé The only drawback is that it requires a fairly up-to-date TeX distribution. – egreg Jul 13 '13 at 14:40

The xparse package, part of the LaTeX3 bundle, is very handy in this regard. For example:

\documentclass{article}
\usepackage{xparse}
\NewDocumentCommand{\foo}{s}{This is the \IfBooleanTF{#1}{starred }{}foo command.}
\begin{document}
\foo{}

\foo*{}
\end{document}


text:
Taxonomy information in nr database
1
0
Entering edit mode
3.0 years ago
navela78 ▴ 60

How is taxonomy information injected into BLAST databases?

My application logic is requiring me to rebuild nr from the fasta file (ftp://ftp.ncbi.nlm.nih.gov/blast/db/FASTA/) because I need to make some custom changes to the sequence headers:

In that file the headers does not seem to have taxonomy information other than the name of the taxonomy rank in brackets like this [Bacillus]. That doesn't seem to be enough to perform extractions using blasdbcmd like this

$blastdbcmd -db nr -entry all -outfmt "%g %T" | \ awk ' { if ($2 == 9606) { print $1 } } ' | \ blastdbcmd -db nr -entry_batch - -out human_sequences.txt  There is an option called taxid_map in makeblastdb but where do I get the mapping file? I guess a simpler way to ask my question is what command does NCBI use to make their nr database from the nr fasta file? nr blast • 2.5k views ADD COMMENT 1 Entering edit mode 3.0 years ago You have to have to download the taxonomy database as well. You can use the update script for that: update_blastdb.pl taxdb --decompress  Then ensure that blast can automatically access that information as well: export BLASTDB=$BLASTDB:~/location/of/taxdb

0
Entering edit mode

Thanks for your suggestion. Your suggestion works for preformatted nr database downloaded from NCBI. However I need to make the nr database from scratch from the fasta file (because I need to add some information to sequence headers in the nr fasta file).

My question is similar to this one how to makeblastdb with taxon id's

Any thoughts on this?

0
Entering edit mode

I believe that as long as the accession numbers are the same you'd get the same behavior, hence you would not need to do anything in particular. My expectation is that the blast TaxDB is indexed by accession numbers.

If you also wanted to build your own custom taxonomy - then you'd have to build it with makeblastdb as you suspect. The mapping files are at:

ftp://ftp.ncbi.nlm.nih.gov/pub/taxonomy/

0
Entering edit mode

Also, there's a lot things in here. What should i download to get the [-taxid TaxID] [-taxid_map TaxIDMapFile] needed for makeblastdb?

0
Entering edit mode

Hello Istvan,

I would also like to do the blast with taxon information as well using the swissprot database however i haven't found detailed instruction of doing it. What I did first is to download the taxonomy database as per instruction but i got error.

Can you help me with what should I do first because it seems like you know how to do this. Thank you!


text:
Jet Set Go! All about Aeroplanes Jet Set Go! All about Aeroplanes

# Salt Hydrolysis Questions

The neutralisation reaction between an acid and a base produces salt. In general, they ionise in water, producing cations and anions. The cations and anions formed during salt ionisation either exist in aqueous solutions as hydrated ions or interact with water to regenerate the acids and bases. The interaction of salt anions or cations with water is referred to as salt hydrolysis.

 Definition: It is defined as the reaction of a salt with water to produce acid and base.

## Salt Hydrolysis Chemistry Questions with Solutions

Q-1: Which of the following statements about the degree of hydrolysis for strong base and weak acid is incorrect?

1. It decreases with dilution
2. It is dependent of dilution
3. It increases with increase in temperature
4. It increases with decrease in Ka of the bases

Answer: a) It decreases with dilution

Explanation: We know that for strong base and weak acid, degree of hydrolysis is given by

$$\begin{array}{l}h=\sqrt{\frac{K_{W}}{c\times K_{a}}}\end{array}$$

As a result of dilution, concentration decreases. The relationship clearly shows that the degree of hydrolysis is inversely proportional to concentration. As a result, as concentration decreases, the degree of hydrolysis increases. This means it gets stronger with dilution.

Q-2: Which of the salts undergoes cationic hydrolysis?

1. NaH2PO4
2. CH3COONa
3. GaCl3
4. Li2CO3

Explanation: Cationic hydrolysis occurs in salts of strong acids and weak bases. Because GaCl3 is a salt of both a weak base (Ga(OH)3) and a strong acid (HCl), it will undergo cationic hydrolysis. Rest salts are salts of a strong base and a weak acid, so they will not undergo this type of hydrolysis.

Q-3: In which case both pH and hydrolysis constant are independent of the concentration of ions involved in hydrolysis?

1. WA+SB
2. SA+WB
3. WA+WB
4. SA+SB

Explanation: For a salt of weak acid(WA) and weak base(WB), the expression for pH is

pH = ½(pKW+pKa-pKb)

And for hydrolysis constant, is

Kh = KW/(Ka×Kb)

We can clearly see that neither the hydrolysis constant nor the pH expressions contain concentration terms. This means that it is unaffected by the concentration of an ion involved in hydrolysis.

Q-4: The degree of hydrolysis of a salt of a weak acid and a weak base in 0.1M solution is found to be 30%. If the molarity of the solution is 0.2 M, the salt’s percent hydrolysis should be

1. 90%
2. 30%
3. 65%
4. 10%

Explanation: The degree of hydrolysis of a weak acid and weak salt is independent of the concentration of the solution. Hence, when the molarity of the solution is 0.2 M, the percentage degree of hydrolysis will remain the same,that is, 30%.

Q-5: At 25 degrees Celsius, the sodium salt of a weak monobasic organic acid is hydrolyzed to a 3 percent extent in its 0.1M solution. Given that the ionic product of water at this temperature is 10-14, what is the acid’s dissociation constant?

Answer: Given % hydrolysis,h = 3% = 0.03

For a sodium salt of weak acid, formula for acid’s dissociation constant(Ka) is

Ka = KW/Kh

Where KW is the ionic product of water and Kh is the hydrolysis constant.

Kh= h2c = 0.032 ×0.1 = 9×10-5

Substitute Kw = 10-14 and Kh=9×10-5.in the formula for Ka

Ka = KW/Kh

Ka = 10-14 / 9×10-5

= 10-10( approximately)

Q-6: Calculate the % hydrolysis in a 10-2 M solution of NaCN (Ka = 6.2×10-10).

Answer: NaCN is a salt of weak acid(HCN) and strong base(NaOH). The degree of hydrolysis(h) for such as salt can be calculated by using the following formulae:

$$\begin{array}{l}h= \sqrt{\frac{K_{W}}{K_{ a}\times C}}\end{array}$$
$$\begin{array}{l}h= \sqrt{\frac{10^{-14}}{6.2\times 10^{-10}\times 10^{-2}}} = 0.0401\end{array}$$

Thus, % hydrolysis = %h = 0.0401×100 = 4.0%

Q-7: NH4Cl has a hydrolysis constant (Kh) of 5.6×10-10.Calculate the pH at equilibrium for 0.1 M NH4Cl solution.

Answer: NH4Cl is a salt of weak base (NH4OH) and strong acid (HCl). The hydrolysis constant for such a salt has the following formulae:

$$\begin{array}{l}K_{h}=\frac{K_{w}}{K_{b}}\end{array}$$

Substitute Kw = 10-14 and Kh= 5.6×10-10 in the above formula and calculate for Kb.

$$\begin{array}{l}5.6\times 10^{-10}=\frac{10^{-14}}{K_{b}}\end{array}$$

On solving, Kb= 1.78 × 10-5

pKb= -log(Kb)= 4.74

The pH for this salt can be calculated using the below equation:

pH = ½[pKw-pKb-log c] =½ [14-4.74-log(0.1)] = 5.13

Q-8: Salt hydrolysis is a chemical reaction that occurs when salt reacts with _______

1. Acid
2. Base
3. Water
4. Salt

Q-9: AlBr3 is formed by the reaction of a strong acid and a weak base. What is the nature of the solution,when this salt dissociates?

1. Basic
2. Neutral
3. Basic and acidic
4. Acidic

Q-10: Give the hydrolysis reaction for magnesium nitride.

Answer: The hydrolysis reaction for magnesium nitride(Mg3N2) is shown below:

$$\begin{array}{l}Mg_{3}N_{2}+ 6H_{2}O\to 3Mg(OH)_{2}+2NH_{3}\end{array}$$

Q-11: Which nitrogen oxides are formed on the hydrolysis of N2O3?

1. N2O + NO
2. NO2 + N2O
3. N2O + NO2
4. None of the above

Q-12: The compound with highest pH among the following is

1. CH3COONH4
2. Na2CO3
3. NH4Cl
4. NaNO3

Explanation: The basic salt has the highest pH. Na2CO3 is the salt of NaOH (a strong base) and H2CO3 (weak acid). Overall, the salt will be basic, resulting in the highest pH.

Q-13: Kb for NH4OH is 1.8 ×10-5 and Ka for CH3COOH is 1.8×10-5. Ammonium acetate’s pH will be

1. 7.08
2. 6.75
3. 7.0
4. Between 6 and 7

Explanation: Ammonium acetate is a salt of weak acid and weak base. For a salt of weak acid(WA) and weak base(WB), pH can be calculated using the below formula:

pH = ½(pKW+pKa-pKb)

We can see Kb and Ka values are the same, it means their pKa and pKb values are also the same.

pH= ½(pKW) = 14/2 = 7

Q-14: The pH of a 0.02 M solution of pyridinium hydrochloride is 3.44. Calculate the pyridine’s ionisation constant.

Answer: Pyridinium hydrochloride is a weak base and strong acid(HCl) salt. We need to calculate ionisation constant(Kb).

The pH formula for this salt is shown below:

pH = ½[pKw-pKb-log c]

3.44= ½[14-pKb-log(0.02)]

pKb.= 8.82

pKb = -logKb

Kb = 1.51 × 10-9

Q-15: The hydrolysis of weak acid and strong base is called

1. Cationic hydrolysis
2. Anionic hydrolysis
3. Non ionic hydrolysis
4. Amphoteric hydrolysis

## Practise Questions on Salt Hydrolysis

Q-1: HX is a weak acid( Ka= 10-5).When it reacts with caustic soda, it forms the salt NaX(0.1 M). The degree of hydrolysis of NaX is

1. 0.1%
2. 0.000001%
3. 0.01%
4. 0.5%

Q-2: Determine whether the following salts’ aqueous solutions are acidic, basic, or neutral.

1. NH4F
2. Na2HPO4
3. NH4Cl

Q-3: A salt hydrolysis reaction occurs when

1. Salts dissociate in any liquid solvent to produce proton and hydroxide ions.
2. Salt dissociates in water to produce acidic and basic solutions.
3. The salt releases anions and cations, resulting in acidic solutions.
4. The salt releases anions and cations, resulting in basic solutions.

Q-4: Which of the following salt’s degree of hydrolysis is independent of solution concentration?

1. NH4CN
2. CH3COONa
3. (NH4)2SO4
4. None of these

Q-5: The pH of 0.4M aqueous NaCN solution will be (Given pKb of CN = 4.70)

1. 2.548
2. 3.513
3. 10.213
4. 11.450

Click the PDF to check the answers for Practice Questions.


text:
$$\require{cancel}$$

# 16.7: Damped Harmonic Motion

A guitar string stops oscillating a few seconds after being plucked. To keep a child happy on a swing, you must keep pushing. Although we can often make friction and other non-conservative forces negligibly small, completely undamped motion is rare. In fact, we may even want to damp oscillations, such as with car shock absorbers.

For a system that has a small amount of damping, the period and frequency are nearly the same as for simple harmonic motion, but the amplitude gradually decreases as shown in Figure $$\PageIndex{2}$$. This occurs because the non-conservative damping force removes energy from the system, usually in the form of thermal energy. In general, energy removal by non-conservative forces is described as $W_{nc} = \Delta (KE + PE),$ where $$W_{nc}$$ is work done by a non-conservative force (here the damping force). For a damped harmonic oscillator, $$W_{nc}$$ is negative because it removes mechanical energy (KE + PE) from the system.

If you gradually increase the amount of damping in a system, the period and frequency begin to be affected, because damping opposes and hence slows the back and forth motion. (The net force is smaller in both directions.) If there is very large damping, the system does not even oscillate—it slowly moves toward equilibrium. Figure $$\PageIndex{3}$$shows the displacement of a harmonic oscillator for different amounts of damping. When we want to damp out oscillations, such as in the suspension of a car, we may want the system to return to equilibrium as quickly as possible Critical damping is defined as the condition in which the damping of an oscillator results in it returning as quickly as possible to its equilibrium position The critically damped system may overshoot the equilibrium position, but if it does, it will do so only once. Critical damping is represented by Curve A in Figure $$\PageIndex{3}$$. With less-than critical damping, the system will return to equilibrium faster but will overshoot and cross over one or more times. Such a system is underdamped; its displacement is represented by the curve in Figure $$\PageIndex{2}$$. Curve B in Figure $$\PageIndex{3}$$ represents an overdamped system. As with critical damping, it too may overshoot the equilibrium position, but will reach equilibrium over a longer period of time.

Critical damping is often desired, because such a system returns to equilibrium rapidly and remains at equilibrium as well. In addition, a constant force applied to a critically damped system moves the system to a new equilibrium position in the shortest time possible without overshooting or oscillating about the new position. For example, when you stand on bathroom scales that have a needle gauge, the needle moves to its equilibrium position without oscillating. It would be quite inconvenient if the needle oscillated about the new equilibrium position for a long time before settling. Damping forces can vary greatly in character. Friction, for example, is sometimes independent of velocity (as assumed in most places in this text). But many damping forces depend on velocity—sometimes in complex ways, sometimes simply being proportional to velocity.

Example $$\PageIndex{1}$$: Damping an Oscillatory Motion: Friction on an Object Connected to a Spring

Damping oscillatory motion is important in many systems, and the ability to control the damping is even more so. This is generally attained using non-conservative forces such as the friction between surfaces, and viscosity for objects moving through fluids. The following example considers friction. Suppose a 0.200-kg object is connected to a spring as shown in Figure $$\PageIndex{4}$$, but there is simple friction between the object and the surface, and the coefficient of friction $$\mu_k$$ is equal to 0.0800. (a) What is the frictional force between the surfaces? (b) What total distance does the object travel if it is released 0.100 m from equilibrium, starting at $$v = 0$$? The force constant of the spring is $$k = 50.0 \, N/m$$.

Strategy

This problem requires you to integrate your knowledge of various concepts regarding waves, oscillations, and damping. To solve an integrated concept problem, you must first identify the physical principles involved. Part (a) is about the frictional force. This is a topic involving the application of Newton’s Laws. Part (b) requires an understanding of work and conservation of energy, as well as some understanding of horizontal oscillatory systems.

Now that we have identified the principles we must apply in order to solve the problems, we need to identify the knowns and unknowns for each part of the question, as well as the quantity that is constant in Part (a) and Part (b) of the question.

Solution a

1. Choose the proper equation: Friction is $$f = \mu_kmg$$.
2. Identify the known values.
3. Enter the known values into the equation: $f = (0.0800)(0.200 \, kg)(9.80 \, m/s^2).$
4. Calculate and convert units: $$f = 0.157 \, N$$.

Discussion a

The force here is small because the system and the coefficients are small.

Solution b

Identify the known:

• The system involves elastic potential energy as the spring compresses and expands, friction that is related to the work done, and the kinetic energy as the body speeds up and slows down.
• Energy is not conserved as the mass oscillates because friction is a non-conservative force.
• The motion is horizontal, so gravitational potential energy does not need to be considered.
• Because the motion starts from rest, the energy in the system is initially $$PE_{el,i} = (1/2)kX^2$$. This energy is removed by work done by friction $$W_{nc} = -fd$$, where $$d$$ is the total distance traveled and $$f = \mu_kmg$$ is the force of friction. When the system stops moving, the friction force will balance the force exerted by the spring, so $$PE_{el,f} = (1/2)kx^2$$, where $$x$$ is the final position and is given by $F_{el} = f$$kx = \mu_kmg.$$x = \dfrac{\mu_kmg}{k}$
1. By equating the work done to the energy removed, solve for the distance $$d$$.
2. The work done by the non-conservative forces equals the initial, stored elastic potential energy. Identify the correct equation to use: $W_{nc} = \Delta (KE + PE) = PE_{el,f} - PE_{el,i} = \dfrac{1}{2}k \left(\left(\dfrac{\mu_kmg}{k}\right)^2 - X^2\right).$
3. Recall that $$W_{nc} = -fd$$.
4. Enter the friction as $$f = \mu_kmg$$ into $$W_{nc} = - fd$$, thus $W_{nc} = \mu_kmgd.$
5. Combine these two equations to find $\dfrac{1}{2}k \left( \left(\dfrac{\mu_kmg}{k}\right)^2 - X^2\right) = - \mu_kmgd.$
6. Solve the equation for $$d$$:$d = \dfrac{k}{2\mu_kmg} \left(X^2 - \left(\dfrac{\mu_kmg}{k}\right)^2\right).$
7. Enter the known values into the resulting equation: $d = \dfrac{50.0 \, N/m}{2(0.0800)(0.200 \, kg)(9.80 \, m/s^2)}\left((0.100 \, m)^2 - \left(\dfrac{(0.0800)(0.200 \, kg)(9.80 \, m/s^2)}{50.0 \, N/m} \right)^2 \right).$
8. Calculate $$d$$ and convert units: $d = 1.59 \, m$

Discussion b

This is the total distance traveled back and forth across $$x = 0$$, which is the undamped equilibrium position. The number of oscillations about the equilibrium position will be more than $$d/X = (1.59 \, m)(0.100 \, m) = 15.9$$ because the amplitude of the oscillations is decreasing with time. At the end of the motion, this system will not return to $$x = 0$$ for this type of damping force, because static friction will exceed the restoring force. This system is underdamped. In contrast, an overdamped system with a simple constant damping force would not cross the equilibrium position $$x = 0$$ a single time. For example, if this system had a damping force 20 times greater, it would only move 0.0484 m toward the equilibrium position from its original 0.100-m position.

This worked example illustrates how to apply problem-solving strategies to situations that integrate the different concepts you have learned. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknowns using familiar problem-solving strategies. These are found throughout the text, and many worked examples show how to use them for single topics. In this integrated concepts example, you can see how to apply them across several topics. You will find these techniques useful in applications of physics outside a physics course, such as in your profession, in other science disciplines, and in everyday life.

Exercise $$\PageIndex{1}$$:Check Your Understanding

Why are completely undamped harmonic oscillators so rare?

Friction often comes into play whenever an object is moving. Friction causes damping in a harmonic oscillator.

Exercise $$\PageIndex{2}$$:Check Your Understanding

Describe the difference between overdamping, underdamping, and critical damping.

An overdamped system moves slowly toward equilibrium. An underdamped system moves quickly to equilibrium, but will oscillate about the equilibrium point as it does so. A critically damped system moves as quickly as possible toward equilibrium without oscillating about the equilibrium.

## Summary

• Damped harmonic oscillators have non-conservative forces that dissipate their energy.
• Critical damping returns the system to equilibrium as fast as possible without overshooting.
• An underdamped system will oscillate through the equilibrium position.
• An overdamped system moves more slowly toward equilibrium than one that is critically damped.

## Glossary

critical damping
the condition in which the damping of an oscillator causes it to return as quickly as possible to its equilibrium position without oscillating back and forth about this position
over damping
the condition in which damping of an oscillator causes it to return to equilibrium without oscillating; oscillator moves more slowly toward equilibrium than in the critically damped system
under damping
the condition in which damping of an oscillator causes it to return to equilibrium with the amplitude gradually decreasing to zero; system returns to equilibrium faster but overshoots and crosses the equilibrium position one or more times


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# All Questions

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## Irreversible and Reversible Processes

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Joanne Guan 1B
Posts: 30
Joined: Sat Jul 22, 2017 3:01 am

### Irreversible and Reversible Processes

I understand that a reversible process is one that can be reversed by an infinitely small change and an irreversible process is expansion against an external pressure that differs by a finite amount from the system's pressure. Can someone give examples for these two processes?

Alexia Joseph 2B
Posts: 56
Joined: Thu Jul 27, 2017 3:01 am

### Re: Irreversible and Reversible Processes

In class, the example Dr. Lavelle used was with two systems compressed with pistons.

Irreversible expansion: will have sudden expansion (results in a large change in volume)
Pressure of external atmosphere- 1 atm
Pressure of system- 2 atm

Reversible expansion: system will change slowly (results a very small change in V)
Pressure of external atmosphere- 1 atm
Pressure of system- 1 atm

804899546
Posts: 51
Joined: Sat Jul 22, 2017 3:00 am

### Re: Irreversible and Reversible Processes

So if the difference is that reversible reactions occur at equilibrium, what makes a reaction that occurs at non-equilibrium irreversible?

Kevin Tabibian 1A
Posts: 42
Joined: Fri Sep 29, 2017 7:06 am

### Re: Irreversible and Reversible Processes

An irreversible process would be as if I took a single 10g weight off of a balloon, allowing it to quickly expand. While a reversible process would be as if I individually removed from a balloon infinitely small weights that all collectively added up to 10g. The work in the second case would be much larger.

504754253
Posts: 45
Joined: Fri Jun 23, 2017 11:39 am

### Re: Irreversible and Reversible Processes

can someone explain what makes a reaction that occurs at non-equilibrium irreversible? and give an example


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# Work and Kinetic Energy Question

1. Jun 4, 2008

### MaZnFLiP

[SOLVED] Work and Kinetic Energy Question

The Problem/Question

The amount of work to pull an archery bow back to "Full Draw" is 12 joules. The arrow has a mass of 0.0185kg.
- How fast will the arrow leave the bow?
- How high will the arrow go if it's shot straight up?
- How far would it go if it was shot at a 35 degree angle above the horizon?
- If the arrow sticks into a 2.40kg box that is suspended from the ceiling with negligible friction, how fast will the box & arrow be moving just after the collision?

Relevant Equations

W = F$$\Delta$$x = $$\Delta$$KE
$$\Delta$$KE = KE$$_{}F$$ - KE$$_{}I$$

Known Information

Work = 12 Joules
Mass = 0.0185kg
Velocity = ?

The Attempt

W = F$$\Delta$$x = $$\Delta$$KE

(12J) = KEf - KEi

KE = 0.5mv^2

0.5(0.0185kg)v^2 - 0.5(0.0185kg)v^2

(12J) = 0.5(0.0185kg)v^2 - 0.5(0.0185kg)v^2

(12J) = v^2 - v^2?

This is where I'm getting confused because if I use the equation W = F$$\Delta$$x, it doesn't give me a displacement so I cant use that. I can also try to use $$\Delta$$KE. But that doesn't work for me either because I end up getting 12J = 0 which is not what I'm even looking for. I'm trying to find velocity but the way I've been doing it doesn't work out. Could someone please help me out on this?

2. Jun 4, 2008

### rock.freak667

well all the energy stored in the bow (12J) is converted into kinetic energy. Initially it is at rest (u=0),

$$\Delta E_k=\frac{1}{2}mv^2-\frac{1}{2}mu^2$$

3. Jun 4, 2008

### MaZnFLiP

Oh! I completely forgot about that. Okay so that now means that everything after the minus sign is canceled right? so that leaves us with $$\Delta E_k=\frac{1}{2}mv^2$$

So: $$W=\Delta KE=\frac{1}{2}mv^2$$

So that would mean that
$$12J = \frac{1}{2}(0.0185kg)v^2$$

After Solving that I now got:

$$36\frac{m}{s}^2$$

After looking at the answer sheet that is the right answer. Thanks so much for the help!

Last edited: Jun 4, 2008