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Home > Relative Error > Absolute And Relative Error Formulas

# Absolute And Relative Error Formulas

## Contents

Chemistry Expert Share Pin Tweet Submit Stumble Post Share By Anne Marie Helmenstine, Ph.D. In plain English: The absolute error is the difference between the measured value and the actual value. (The absolute error will have the same unit label as the measured quantity.) Relative You can compare your own results to get Absolute Error, which measures how far off you were from the expected results. So you know that your measurement is accurate to within + or - 1 mm; your absolute error is 1 mm. this contact form

when measuring we don't know the actual value! Then the relative error is defined by where is the absolute error. Back to Top The relative error formula is given byRelative error =$\frac{Absolute\ error}{Value\ of\ thing\ to\ be\ measured}$ = $\frac{\Delta\ x}{x}$.In terms of percentage it is expressed asRelative error = $\frac{\Delta\ The essential idea is this: Is the measurement good to about 10% or to about 5% or 1%, or even 0.1%? http://www.regentsprep.org/regents/math/algebra/am3/LError.htm ## How To Find Relative Error Formula Wolfram|Alpha» Explore anything with the first computational knowledge engine. Video Tips Make sure that your experimental value and real value are all expressed in the same unit of measurement. The precision of a measuring instrument is determined by the smallest unit to which it can measure. It is the difference between the result of the measurement and the true value of what you were measuring. This fact gives us a key for understanding what to do about random errors. Relative ErrorProblems Related Concepts Relative Error Formula Absolute Error Formula Formula for Sampling Error formula for relative density Relations Margin Error Type I Error and Type II Error Physical Chemistry and By continuing to use our site, you agree to our cookie policy. How Do You Find Absolute Error What if some of the experimental values are negative? Find the absolute error, relative error and percent of error of the approximation 3.14 to the value , using the TI-83+/84+ entry of pi as the actual value. Solution: Given: The measured value of metal ball xo = 3.97 cm The true value of ball x = 4 cm Absolute error$\Delta$x = True value - Measured Back to Top To calculate the relative error use the following way:Observe the true value (x) and approximate measured value (xo). navigate here Scalar Component of Vector Conceptual Physics Answers Relative Error Formula What is Relative Error? and Stegun, I.A. (Eds.). How To Compute Absolute Error Create an account EXPLORE Community DashboardRandom ArticleAbout UsCategoriesRecent Changes HELP US Write an ArticleRequest a New ArticleAnswer a RequestMore Ideas... this is about accuracy. Percent Error Definition Quick Review of Experimental Error More from the Web Powered By ZergNet Sign Up for Our Free Newsletters Thanks, You're in! ## Relative Error Vs Absolute Error Incorrect measuring technique: For example, one might make an incorrect scale reading because of parallax error. How to Calculate the Relative Error? How To Find Relative Error Formula Show more unanswered questions Ask a Question Submit Already answered Not a question Bad question Other If this question (or a similar one) is answered twice in this section, please click How To Find Absolute Error And Relative Error Although random errors can be handled more or less routinely, there is no prescribed way to find systematic errors. Moreover, you should be able to convert one way of writing into another. weblink For example, let's say you managed to measure the length of your dog L to be 85 cm with a precision 3 cm. You already know the convention for reporting Then find the absolute deviation using formulaAbsolute deviation$\Delta$x = True value - measured value = x - xoThen substitute the absolute deviation value$\Delta\$ x in relative error formula If you are measuring a 200 foot boat, and miss the measurement by 2 feet, your percentage error will be much lower than missing the 20 foot tree measurement by 2 How To Calculate Relative Error Chemistry

Your absolute error is 20 - 18 = 2 feet (60.96 centimeters).[3] 2 Alternatively, when measuring something, assume the absolute error to be the smallest unit of measurement at your disposal. Paper Boat Creative, Getty Images By Anne Marie Helmenstine, Ph.D. The absolute error of the measurement shows how large the error actually is, while the relative error of the measurement shows how large the error is in relation to the correct navigate here Answer this question Flag as...

It takes 10ml drops of water to cause a reaction, but his dropper claims it is "+/- .5ml." The Absolute Error in his measurements must be: +/- .5ml 5 Understand what Relative Error Chemistry Relative ErrorProblems Back to Top Below are given some relative error examples you can go through it: Solved Examples Question1: John measures the size of metal ball as 3.97 cm but Examples: 1.

## But, if you are measuring a small machine part (< 3cm), an absolute error of 1 cm is very significant.

How to Calculate the Relative Error? Calculate the absolute and relative errors? Measure under controlled conditions. Absolute And Relative Error In Numerical Methods Learn how.

Note, however that this doesn't make sense when giving percentages, as your error is not 10% of 2 feet. Please select a newsletter. Relative values are ratios, and have no units. his comment is here If you measure a voltage with a meter that later turns out to have a 0.2 V offset, you can correct the originally determined voltages by this amount and eliminate the

Skeeter, the dog, weighs exactly 36.5 pounds. The quantity 0.428 m is said to have three significant figures, that is, three digits that make sense in terms of the measurement. When weighed on a defective scale, he weighed 38 pounds. (a) What is the percent of error in measurement of the defective scale to the nearest tenth? (b) If Millie, the The ratios are commonly expressed as fractions (e.g. 0.562), as percent (fraction x 100, e.g. 56.2%), as parts per thousand (fraction x 1000, e.g. 562 ppt), or as parts per million

Note that absolute error is reported in the same units as the measurement.Alternatively, you may have a known or calculated value and you want to use absolute error to express how Computerbasedmath.org» Join the initiative for modernizing math education. Get the best of About Education in your inbox. A low relative error is, of course, desirable.


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# Tradeoff between error and rates of quantum communication

Suppose Alice and Bob share $$n$$ copies of a noiseless quantum channel $$I_{A\rightarrow B}$$ which can be used to send quantum states and $$H_A\cong H_B$$ i.e. the input and output Hilbert spaces are the same dimension, $$d$$. For simplicity, we can assume that $$d=2$$ so we are dealing with transmitting qubits.

Suppose there exists a protocol that uses $$I^{\otimes n}_{A\rightarrow B}$$ (and possibly other no-signaling resources such as shared entanglement) to simulate $$I_{A\rightarrow B}^{\otimes m}$$, where $$m>n$$ but this simulation occurs with some error $$\varepsilon\in (0,1)$$.

What theorem in quantum information, if any, bounds achievable $$\varepsilon$$? It seems natural that the case of $$\varepsilon = 0$$ is definitely out of the question since otherwise, we could simply repeat the process and communicate arbitrary amounts of information starting with just a few perfect channels. I am not sure about the case where we allow $$\varepsilon\in (0,1)$$.

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SMTP Mail Sender is an application that allows you to send messages through an SMTP server. It lets you send messages if you do not have a mail application installed, or need to send from an account that is only used for sending messages without worrying about responses. It also runs from a USB thumb drive, so you can send emails from anywhere you have access to a computer with Internet access.
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Discussion
Wolf
August 27, 2021
Hi, Joey,
I forgot how to hide the Reply-To label and field
in the hide.txt file.
Joey Hazlett
August 27, 2021
Hi Wolf,

You just need to edit the hide.txt file and remove the two # characters in front of the ReplyTo field objects:
# "Reply-To" label and input box
#tbReplyTo
becomes
# "Reply-To" label and input box
tbReplyTo
Motschz77
July 17, 2020
Hi Joey,
I am trying to send an email with the following command line.But your madness creates an error. What did I do wrong? Is the% date% format not supported? :
"C:\iFix Project\SSI01\APP\SMTP\APP_2020\SMTPMailSender.exe" -subject "Daily Report" -from "somewhere@somewehere.com" -to "somewhere@somewehere.com" -attachment "D:\Proficy\Dailyreport\SILOS\%date% Daily Report Silo 1" -body "Automation Report" -send

regards, Sascha
Joey Hazlett
July 17, 2020
Hi Sascha,

When I do the following, it reports a date in an invalid file format:
C:\Users\Joey>echo %date%
Fri 07/17/2020

Since those forward slashes are in the variable, I think that is what would cause this problem. If you have a date format that doesn't use the slashes, then it may be literally reading %date% as a string, and looking for a directory of the same name. While that is a valid file format, I highly doubt you have a directory named %date% containing your files. :)


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# How can I save text to a file in R?

I have a R function which can generate the LaTeX code (the output is the LaTex code) by using cat(), while now I want to save these LaTeX code, but I don't know which function can save these LaTeX code...

-
Are you sure Sweave would be of no help? –  Roman Luštrik Jun 3 '11 at 7:09

I like to use the sink() function:

latex.code <- function(){
cat("\\begin{align}\n")
cat("[X'X]^{-1}X'y\n")
cat("\\end{align}\n")
}
sink(file='ols.txt')
latex.code()
sink()

Edit: Obviously, you can choose the file path where the file will be saved by changing the sink argument such as: sink(file='c:/Users/Eva/Desktop/ols.txt'), or sink(file='~/ols.txt')

-

If it happens that you have your output in a character vector (i.e. you are using something like cat(<something>) to have it written to the console), you can use writeLines function, like this:

writeLines(<something>,"filename.txt")

However the best way to make a LaTeX file in R is to use either Sweave or make a brew template.

-
@Eva All files are saved, provided absolute path is not set, to setwd(). –  Roman Luštrik Jun 3 '11 at 7:09


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# Chapter 7 Exponential and Logarithmic Functions - 7.4 Evaluate Logarithms and Graph Logarithmic Functions - 7.4 Exercises - Skill Practice - Page 503: 19

2

#### Work Step by Step

For a logarithmic expression, we know the following: $$log_{b}\ y=x \Rightarrow b^x=y$$ Using this, we find that the expression becomes: $$11^x=121 \\ x=2$$

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


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# Laplace Operator

1. Dec 13, 2011

### tsamocki

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

The Laplace operator Δ is defined by: Δ=

Show in polar coordinates r and Θ, that the Laplace operator takes the following form:

2. Relevant equations

x=rcos(Θ), y=rsin(Θ), r ≥ 0, Θ ∈ [0,2∏]

3. The attempt at a solution

It seems simple; convert the x and y in terms of polar coordinates, and differentiate twice. However, when i attempt to differentiate twice, in respect to r and theta, my r disappears and the trig function simply reverts.

2. Dec 13, 2011

### I like Serena

Can you show your work where you differentiate the first time?

3. Dec 14, 2011

### tsamocki

Sorry for the delayed response, been studying for other finals.

Here is where i am stuck.

For the x partial derivative:

$\partial$ / $\partial$x= -rsinθ+cosθ

For the y partial derivative:

$\partial$ / $\partial$y= sinθ+rcosθ

I know this is incorrect because i was informed by my professor that there is a nested chain ruled inside a product rule in the second differentiation (or something similar). d/dr d/dx +d/dθ d/dx is the format for the x partial derivative, and just replacing x with y gives you the format for the y partial derivative.

Where am i going wrong?

4. Dec 15, 2011

### I like Serena

What did you differentiate? That seems to be missing.
And how did you get this result?

Yes, the chain rule for multivariable differentiation is:
$${\partial \over \partial x}f(u(x,y), v(x,y))={\partial f \over \partial u}{\partial u\over \partial x}+{\partial f \over \partial v}{\partial v \over \partial x}$$

You should do something similar to f(x(r,θ), y(r,θ))...


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19th Ave New York, NY 95822, USA

# 数学代写|偏微分方程作业代写Partial Differential Equations代考|Some Final Remarks

my-assignmentexpert™ 偏微分方程Partial Differential Equations作业代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。my-assignmentexpert™， 最高质量的偏微分方程Partial Differential Equations作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于统计Statistics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此偏微分方程Partial Differential Equations作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

my-assignmentexpert™ 为您的留学生涯保驾护航 在数学Mathematics作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的偏微分方程Partial Differential Equations代写服务。我们的专家在数学Mathematics代写方面经验极为丰富，各种偏微分方程Partial Differential Equations相关的作业也就用不着 说。

## 数学代写|偏微分方程作业代写Partial Differential Equations代考|Nonsmooth Solutions

3.12.1. Nonsmooth Solutions. When solving the IVP (3.5) to arrive at D’Alembert’s formula, we implicitly made some assumptions on the initial data $\phi$ and $\psi$. Alternatively, in order to show that D’Alembert’s formula does indeed solve the wave equation, we would need $\phi$ to be $C^{2}$ and $\psi$ to be $C^{1}$. On the other hand, we do not need these assumptions to just plug $\phi$ and $\psi$ into D’Alembert’s formula: Indeed, in our examples of the plucked string and the hammer blow, we simply applied D’Alembert’s formula to, respectively, a $\phi$ which was not $C^{1}$ and a $\psi$ which was not even continuous. While we discussed the resulting solutions and plotted profiles, in what sense do we have a solution? Do the discontinuities count or can they be ignored? The answer to these questions will be revealed once we introduce the notion of a solution in the sense of distributions (cf. Chapter 9). This more integral-focused notion of a solution will indeed allow us to conclude that our placement of these nonsmooth functions in D’Alembert’s formula does indeed yield a solution, albeit a distributional one.

## 数学代写|偏微分方程作业代写Partial Differential Equations代考|Heterogeneous Media and Scattering

Wave propagation through a heterogeneous string leads to a wave equation in which the speed parameter $c$ is spatially dependent:
$$u_{t t}=c^{2}(x) u_{x x}$$
This spatial dependence of $c$ is determined by the physical nature of the string. With certain general assumptions on the coefficient function $c(x)$, the resulting IVPs are well-posed. However, closed form solution formulas are not usually available. An interesting feature here is the scattering of waves, whereby aspects of the inhomogeneous medium can cause incoming waves to have a change in their outgoing behavior. For example, consider the $1 \mathrm{D}$ wave equation with
$$c(x)= \begin{cases}c_{1}, & x<0 \\ c_{2}, & x \geq 0\end{cases}$$ where $c_{1}>0, c_{2}>0$ with $c_{1} \neq c_{2}$. You can think of this as a model for a vibrating string made of two materials of differing mass densities, one for $x<0$ and the other for $x>0$. A signal concentrated at say $x_{0}<0$ will propagate in the positive $x$ direction until it interacts with the change in medium at $x=0$. This interaction will cause the wave to be scattered and give rise to both a change in its amplitude and speed. In Exercise 3.22, We give more details and outline the steps in order to find an exact solution.

## 数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|Finite Propagation Speed, Other “Wave” Equations, and Dispersion

The solution to the wave equation exhibits finite propagation speed. This was illustrated in the domain of influence where we saw that information can propagate through space $(x)$ with different speeds, but no speed greater than $c$. The name “the wave equation” is mainly for historical reasons and one often loosely calls any timedependent PDE in which information (data) propagates at finite speeds “a wave equation”. For example, the simple first-order transport equations of Chapter 2 can be thought of as wave equations. On the other hand, there are also far more complicated linear and nonlinear PDEs for wave propagation, for example,

• the inviscid Burgers equation which we saw in Section 2.3,
• the Schrödinger, Klein-Gordon, Telegraph, and KdV equations.
The latter two PDEs exhibit a fundamental process associated with the propagation of waves known as dispersion. Dispersion for our vanilla wave equation proves rather trivial. It is true that light propagation (as an electromagnetic wave) satisfies the 3D wave equation with $c$ being the speed of light; however, it does so in a vacuum. This is rather uninteresting as there is literally nothing that “can be seen”. On the other hand, when lights hits a glass prism, the waves of different wavenumbers have different velocities and the resulting dispersion produces a spectrum of colors. In general, dispersion describes the way in which an initial disturbance of the wave medium distorts over time, explicitly specifying how composite plane wave components of an initial signal evolve temporally for different wavenumbers.

## 数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|NONSMOOTH SOLUTIONS

3.12.1。不光滑的解决方案。解决IVP时3.5为了得出 D’Alembert 公式，我们隐含地对初始数据做了一些假设φ和ψ. 或者，为了证明 D’Alembert 公式确实解决了波动方程，我们需要φ成为C2和ψ成为C1. 另一方面，我们不需要这些假设来插入φ和ψ进入 D’Alembert 公式：确实，在我们的拨弦和锤击示例中，我们简单地将 D’Alembert 公式分别应用于φ这不是C1和一个ψ这甚至不是连续的。虽然我们讨论了最终的解决方案和绘制的配置文件，但在什么意义上我们有解决方案？不连续性是否重要，还是可以忽略不计？一旦我们在分布的意义上引入解决方案的概念，这些问题的答案就会揭晓CF.CH一种p吨和r9. 这种更注重积分的解决方案概念确实使我们能够得出结论，我们将这些非光滑函数置于 D’Alembert 公式中确实会产生一个解决方案，尽管是一个分布解决方案。

## 数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|HETEROGENEOUS MEDIA AND SCATTERING

C(X)={C1,X<0C2,X≥0在哪里C1>0,C2>0和C1≠C2. 您可以将其视为由两种不同质量密度的材料制成的振动弦的模型，一种用于X<0另一个为X>0. 一个信号集中在sayX0<0将向正面传播X方向直到它与介质的变化相互作用X=0. 这种相互作用将导致波散射并引起其幅度和速度的变化。在练习 3.22 中，我们给出了更多细节并概述了步骤，以便找到一个精确的解决方案。

## 数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|FINITE PROPAGATION SPEED, OTHER “WAVE” EQUATIONS, AND DISPERSION

• 我们在第 2.3 节中看到的无粘性 Burgers 方程，
• 薛定谔、克莱因-戈登、电报和 KdV 方程。
后两个偏微分方程表现出与称为色散的波传播相关的基本过程。我们的普通波动方程的色散证明是微不足道的。光的传播是真的一种s一种n和l和C吨r这米一种Gn和吨一世C在一种在和满足 3D 波动方程C是光速；但是，它是在真空中进行的。这是相当无趣的，因为实际上没有“可以看到”的东西。另一方面，当光线照射到玻璃棱镜上时，不同波数的波具有不同的速度，由此产生的色散会产生光谱。一般来说，色散描述了波介质的初始扰动随时间扭曲的方式，明确指定了初始信号的复合平面波分量如何针对不同的波数随时间演变。


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# 7.3: Wave Shaping

#### Active Clampers

Clampers are used to add a specific amount of DC to a signal. Generally, the amount of DC will be equal to the peak value of the signal. Clampers are intelligent in that they can adjust the amount of DC if the peak value of the input signal changes. Ideally, clampers will not change the shape of the input signal; rather, the output signal is simply a “vertically shifted” version of the input, as shown in Figure 7.19. One common application of the clamper is in television receivers. Here, a clamper is referred to as a DC restorer. Certain portions of the video signal, such as sync pulses, are required to be at specific levels. After the video signal has been amplified by AC coupled gain stages, the DC restorer returns the video to its normal orientation. Without the clamping action, the various parts of the signal cannot be decoded properly. Clampers are commonly made with passive components, just as rectifiers are. Like rectifiers, simple clampers produce errors caused by the diode's forward voltage drop. Active clampers remove this error.

Conceptually, a clamper needs to sense the amplitude of the input signal and create a DC signal of equal value. This DC signal is then added to the input, as shown in Figure 7.20. If the peak value is known, and does not change, it is possible to create this function with a summing amplifier. If the signal level is dynamic, a different course needs to be taken. The trick is in getting the DC source to properly track input level changes. Obviously, a simple DC supply is not appropriate; however, a charged capacitor will fit the bill nicely - as long as its discharge time constant is much longer than the period of the input wave. All that needs to be done is to have the capacitor charge to the peak level of the input. When the capacitor voltage is added to the input signal, the appropriate DC shift will result.

Figure 7.21 shows an active positive clamper. On the first negative going peak of the input, the source will attempt to pull current through the capacitor in the direction shown. This forces the inverting input of the op amp to go slightly negative, thus creating a positive op amp output. The result of this action will be to forward-bias the diode and supply charging current to the capacitor. The capacitor will charge to the negative peak value of the input. The output resistance of the op amp is very low, so charging is relatively fast, only being limited by the maximum output current of the op amp. When the source changes direction, the op amp will produce a negative output, thus turning off the diode and effectively removing the op amp from the circuit. Because the discharge time for $$C$$ is much longer than the input period, its potential stays at roughly $$V_{p-}$$. It now acts like a voltage source. These two sources add, so we can see that the output must be:

$$V_{out} = V_{in} + |V_{p-}| \label{7.1}\tag{7.1}$$

The op amp will remain in saturation until the next negative peak, at which point the capacitor will be recharged. During the charging period, the feedback loop is closed, and thus, the diode's forward drop is compensated for by the op amp. In other words, the op amp's output will be approximately 0.6 to 0.7 V above the inverting input's potential. The discharge time of the circuit is set by the load resistor, $$R_l$$. If particularly long time constants are required, a buffer stage may be used, along with an FET input clamping op amp. The resistor $$R$$ is used to prevent possible damage to the op amp from capacitor discharge. This value is normally in the low kilohm region. In order to make a negative clamper, just reverse the polarity of the diode.

Sometimes, it is desirable to clamp a signal and add a fixed offset as well. An example output waveform of this type is shown in Figure 7.22.

This function is relatively easy to add to the basic clamper. In order to include the offset, all that needs to be done is to change the op amp's reference point. In the basic clamper, the noninverting input is tied to ground. Consequently, this establishes the point at which the charge/discharge cycle starts. If this reference is altered, the charge/discharge point is altered too. To create a positive offset, a DC signal equal to the offset is applied to the noninverting input, as shown in Figure 7.23. A similar arrangement may be used for negative clampers.

#### Example 7.2

An active clamper is shown in Figure 7.24. An LF412 op amp is used (a dual LF411). Determine the capacitor's voltage and verify that the time constant is appropriate for the input waveform. Also, sketch the output waveform and determine the maximum differential input voltage for the first op amp.

Because this is a positive clamper, the capacitor's voltage is the sum of the offset potential and the negative peak potential of the input waveform. $$V_{offset}$$ is set by the 10 kΩ/1.5 kΩ divider.

$$V_{offset} = V_{CC} \frac{R_2}{R_1+R_2} \notag$$
$$V_{offset} = 15 V \frac{1.5 k}{10 k+1.5 k} \notag$$

$$V_{offset} = 1.96 V \notag$$

$$V_c = V_{offset} + |V_{p-}| \notag$$

$$V_c = 1.96 V+|−1 V| \notag$$

$$V_c = 2.96 V \notag$$

This is the amount of DC added to the input signal. The output waveform will look just like the input, except that it will be shifted up by 2.96 V. This is shown in Figure 7.25.

The discharge time constant should be much larger than the period of the input wave. The input wave has a frequency of 1 kHz, and thus, a period of 1 ms. The capacitor discharges through the load resistor. As FET input op amps are used, their effect on the discharge rate is minimal.

$$T = R_l C \notag$$

$$T = 10 M×10nF \notag$$
$$T = 0.1s \notag$$

The discharge rate is 100 times longer than the input period, so the capacitor droop will be satisfactory.

The maximum differential input signal is the worst-case difference between the inverting and noninverting input potentials. The noninverting input is tied directly to 1.96 V. The inverting input sees the output waveform. The maximum value of the output is 5.96 V. So the difference is

$$V_{in-diff} = V_{in +} − V_{in -} \notag$$
$$V_{in-diff} = 1.96 V−5.96 V \notag$$

$$V_{in-diff} = −4 V \notag$$

The LF412 will have no problem with a differential input of this size.

#### Computer Simulation

A simulation of the clamper circuit of Example 7.2 is shown in Figure 7.26. In order to simplify the simulation somewhat, the input signal has been altered modestly. Instead of the shifted 4 volt peak-to-peak full-wave rectified signal originally used, a shifted 4 volt peak-to-peak sine wave is used.

Both the input and output waveforms are plotted in the Transient Analysis. The circuit requires about 1 cycle of the waveform before the output stabilizes. After that point the clamping action and shift of nearly 2 volts is quite evident in the output waveform. There is no noticeable droop, scaling error, or obvious distortion in the output either, although some minor aberrations can be seen at the negative peaks on occasion.

#### Active Limiters

A limiter is a circuit that places a maximum restriction on its output level. The output of a limiter can never be above a specific preset level. In one sense, all active circuits are limiters, in that they will all eventually clip the signal by going into saturation. A true limiter, though, prohibits the signal at levels considerably lower than saturation. The exact level is fairly easy to set. Limiters can be used to protect following stages from excessive input levels. They can also be seen as a type of wave shaper. One form of limiter was shown in Chapter Six, in the Pocket Rockit's schematic. In that circuit, a limiter was used to purposely clip the music signal for artistic effect. The limitation of that form is that the limit potential is locked at ±.7 V by the parallel signal diodes. For a more general form, an arbitrary limit potential is desired. Instead of using parallel signal diodes, a series combination of Zener diodes will prove useful.

An example limiter is shown in Figure 7.27. It is based upon the inverting voltage amplifier form. As long as the output signal is below the Zener potential, the output equals the input times the voltage gain. If the output voltage tries to rise above the Zener potential, one of the diodes will go into Zener conduction, and the other diode will go into forward bias. Once this happens, the low dynamic resistance of the diodes will disallow any further increase in output potential. The output will not be allowed to move outside of the Zener potential (plus the 0.7 V turn-on for the second diode):

$$|V_{out}| ≤ V_{zener} + 0.7V \label{7.2}\tag{7.2}$$

Or we might reword this as

$$V_{limit} = ±(V_{zener} + 0.7 V) \label{7.3}\tag{7.3}$$

This effect is shown graphically in Figure 7.28, the limiter's transfer characteristic.

#### Example 7.3

If the input to the circuit of Figure 7.29 is a 4 V peak triangle wave, sketch $$V_{out}$$.

Without the Zener limit diodes, this amplifier would simply multiply the input by a gain of two and invert it.

$$V_{out} = A_v V_{in} \notag$$

$$V_{out} = −2×4 V (peak) \notag$$

$$V_{out} = −8V (peak) \notag$$

So, an inverted 8 V peak triangle would be the output. With the inclusion of the diodes, the maximum output is

$$V_{limit} = ±(V_{zener} + 0.7 V) \notag$$

$$V_{limit} = ±(5.1 V + 0.7 V) \notag$$

$$V_{limit} = ±5.8 V \notag$$

Consequently, the output wave is clipped at 5.8 V, as shown in Figure 7.30.

By using different Zeners, it is also possible to produce asymmetrical limiting. For example, positive signals might be limited to 10 V, whereas negative signals could be limited to -5 V.


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Multi stage centrifugal pumps.

When a centrifugal pump consists of two or more impellers, the pump is called as Multi stage centrifugal pump.

Important functions of a Multi stage centrifugal pump:

1] To produce high heads.

2] To discharge a large quantity of liquid.

Multi stage centrifugal pumps for high heads:

The fig show number of impellers mounted in series.

In this, the water from suction pipe enters the $1^{st}$ impeller at inlet and is discharged at outlet with increased pressure. The water with increased pressure from the outlet of $1^{st}$ impeller is taken to the inlet of $2^{nd}$ impeller with the help of a connecting pipe as shown in figure. At the outlet of the $2^{nd}$ impeller, the pressure of water will be more than the pressure of water at the outlet of $1^{st}$ impeller. Thus if more impellers are mounted on the same shaft, the pressure at the outlet will be increased further.

Total Head Developed = n x $H_m$.

Where, n = No. of identical impellers.

$H_m$ = Head developed by each impeller.

Multi stage centrifugal pumps for high discharge:

For obtaining high discharge, the pumps should be connected in parallel. each pump lifts water from a common pump and discharges water to a common pipe to which the delivery pipes of each pump is connected, each pump is working against the same head.

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# Integral Solution

Find the general solution for integral values of $$x,y,z$$ in the equation $$6x+10y+15z=-1$$.

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# How do solve 1/x^2-1<=0 graphically?

Jan 9, 2018

$x \le - 1 \mathmr{and} x \ge 1$

#### Explanation:

$\frac{1}{x} ^ 2 - 1 \le 0$

First, add 1 to both sides so that you end up with $\frac{1}{x} ^ 2$ on the left, to make graphing a bit easier

$\frac{1}{x} ^ 2 \le 1$

Then, plot the graph of $y = \frac{1}{x} ^ 2$

graph{1/x^2 [-10, 10, -5, 5]}

To find the range of x for which $\frac{1}{x} ^ 2 \le 1$ is true, just find the range of x for which $y \le 1$ in the graph above

graph{(y-1/x^2)(y-1)=0 [-10, 10, -5, 5]}

As you can see from this graph, $\frac{1}{x} ^ 2$ drops below $y = 1$ when $x \le - 1$ and $x \ge 1$

Thus the solution to $\frac{1}{x} ^ 2 \le 1$, and thus $\frac{1}{x} ^ 2 - 1 = 0$, is $x \le - 1 \mathmr{and} x \ge 1$


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QUESTION

# HIM 435 Week 6 DQs

This file of HIM 435 Week 6 DQs contains answers on:

We identified in the lecture the various clinical databases required by the federal government. What is the purpose of requiring these databases? What makes the information being collected important?

Data have been collected upon completion of the surgical informed consent form. More than 200 records have been reviewed for the presence of the following data items on the consent form: date and time the consent was completed and explained to the patient, the type of surgery performed, risks or possible complications of surgery, the patient's signature, the date and time of signing, the physician or surgeon's signature, the date and time of signing, the witness' or nurse's signature, and the date and time of signing. In order to set up a database with the information obtained, what column heads do you create, and how do you document the presence or absence of a data item on the consent form?

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## Introduction

Sea-level budget assessments quantify the different physical processes contributing to sea-level change1. Since 1993, sea-level budget assessments generally combine satellite altimetry estimates of total global-mean sea-level (GMSL) rise and, more recently, Argo float-derived estimates of global-mean thermosteric sea-level rise with satellite gravimetric measurements of barystatic contributions (i.e., from land-ice and land water) and process model-derived information2. For example, the GMSL budget (1993–2018) indicates the thermosteric contribution (1.3 mm yr−1) was dominant with additional contributions from glaciers and ice sheets (~0.7 mm yr−1 each)3. The 20th century GMSL budget has been estimated with the additional aid of tide-gauge data and direct measurements of changing mass balance of glaciers and ice sheets with a total rate of rise of 1.2 ± 0.2 mm yr−1 from 1901–19904. Studies on preinstrumental timescales are limited to single time slices, such as quantifying the ice sheet contributions to the GMSL lowstand during the Last Glacial Maximum5. A further limitation of sea-level budgets is the paucity of regional and local relative sea-level change assessments6. Relative sea level (RSL) differs from GMSL because of driving processes such as glacial isostatic adjustment (GIA); ocean dynamic sea-level change; gravitational, rotational, and deformational (GRD) responses to barystatic changes; tectonics; and sediment compaction7,8. The driving processes are spatially variable and cause RSL change to vary in rate and magnitude among regions9.

Sea-level budgets for the Common Era (0–2000 CE) are unknown, but proxy RSL reconstructions have extended the instrumental record back before the 19th century10 and have improved understanding of magnitudes, rates, and driving processes of regional sea-level change at centennial to multidecadal timescales11. For example, along the U.S. Atlantic coast, GIA has been a significant driving factor in RSL rise through the Common Era, creating a spatially variable signal due to the region’s proximity to the former Laurentide Ice Sheet12,13. Larger uncertainty is associated with the remaining RSL processes occurring on different spatial scales, including common global signals driven by thermosteric and barystatic changes; regional signals such as ocean dynamic sea-level change and GRD changes; and local site-specific signals such as tidal range change and sediment compaction.

Here, we estimate site-specific Common Era sea-level budgets along the U.S. Atlantic coast. We complete a new high-resolution (decimeter vertical scale, decadal temporal scale) RSL record in northern New Jersey (Fig. 1), filling in a spatial data gap between RSL records in southern New Jersey14 and New York City15. Integrating this new record into an updated global database of instrumental and proxy sea-level records11,16 (Supplementary Data), we use a spatiotemporal empirical hierarchical model11,17 to examine magnitudes and rates of Common Era RSL in the northern New Jersey record and five other published records along the U.S. Atlantic coast (Fig. 1a). The high concentration of these high-resolution RSL reconstructions over a 700 km stretch of coastline allows the records to be decomposed into process-related signals on different spatial scales, assisting in interpretation of processes that drive both spatial and temporal patterns of sea-level changes.

## Results

### Common Era sea-level trends

To examine regional RSL trends over the Common Era for the U.S. Atlantic coast, we compare 100-year average rates from Cheesequake State Park in northern New Jersey to published proxy-based sea-level reconstructions (Fig. 1a).

The last millennium reconstruction of RSL from northern New Jersey (Fig. 1d, Supplementary Figs. 19) reveals that RSL rose continuously from 1000–2000 CE at an average rate of 1.5 ± 0.2 mm yr−1 (2σ). Over the preindustrial Common Era from 0–1700 CE, RSL in northern New Jersey rose at a rate of 1.3 ± 0.2 mm yr−1. This rate is consistent with the spatial gradient from the other U.S. Atlantic coast sites, with the fastest rates of rise occurring in southern New Jersey with 1.6 ± 0.1 mm yr−1 at Leeds Point (LP) and 1.5 ± 0.1 mm yr−1 at Cape May Courthouse (CMC) (Fig. 2). In New York City, there were slightly slower rates of rise of 1.2 ± 0.1 mm yr−1, and the slowest rates of rise occurred in Connecticut and North Carolina with 1.0 ± 0.1 and 1.1 ± 0.1 mm yr−1, respectively. The spatial gradient of Common Era RSL rates is a result of each location’s position relative to the time-evolving position of the former Laurentide Ice Sheet13.

The reconstructions over the last 300 years show elevated RSL rates at all six sites, ranging from 1.7 ± 0.3 mm yr−1 in Connecticut to 2.5 ± 0.3 mm yr−1 in southern New Jersey (CMC). It is extremely likely (probability P ≥ 0.98) that the average rate of rise from 1700–2000 CE at all six sites was faster than during any preceding 300-year period during the preindustrial Common Era (0–1700 CE). Furthermore, it is virtually certain (P > 0.99) that the 20th century rate of rise at all six sites was faster than during any preceding century in the Common Era.

The updated global database of instrumental and proxy records of the Common Era (Fig. 3a) illustrates a global-scale component, or globally uniform signal, of RSL trends similar to previous results11,16. Global sea level gradually rose from 0 to 500 CE at a rate of 0.1 ± 0.1 mm yr−1 (1σ), but then gradually fell from 500 to 1300 CE at a rate of −0.1 ± 0.1 mm yr−1 (Fig. 3b). It then rose from 1300 to 1600 CE at a rate of 0.1 ± 0.2 mm yr−1 and fell again from 1600 to 1800 CE at a rate of −0.1 ± 0.2 mm yr−1. Global sea-level rise beginning in the 19th century reached an average rate of 1.3 ± 0.1 mm yr−1 in the 20th century. It is virtually certain (P > 0.999) that the global 20th century rate of rise was faster than during any preceding century during the Common Era.

### Sea-level budgets: global component

The global component of RSL, identified as a signal common to all of the records in the Common Era database, is relatively small (i.e., centimeter-scale) and fairly stable without any consistent periodicity at multidecadal to multicentennial timescales until the onset of modern rates of rise (Fig. 3). Rates of the global component fluctuated between −0.2 and 0.2 mm yr−1 from 0 to 1800 CE, with an increasing rate of rise since 1800 CE, reaching 1.3 ± 0.1 mm yr−1 in the 20th century (Fig. 4, Table 1).

Our global sea-level estimate using the globally uniform signal is consistent with GMSL budgets for the 20th century, which found total rates of rise of 1.2 ± 0.2 mm yr−1 (1901–1990) (90% confidence interval)18 and 1.56 ± 0.33 mm yr−1 (1900–2018) (90% confidence interval)19. Prior to the 20th century, there is no evidence for global sea-level rate changes associated with the time period of the Medieval Climate Anomaly; however, there is a negative global contribution during the latter part of the Little Ice Age (Fig. 5a), which coincides with a period of decreased global air and sea surface temperatures and the most extensive glacial advances in the Common Era (Fig. 5b, c)20,21. The enhanced rates of rise and greatest contribution of the global component that followed in the 20th century are caused primarily by increased ocean mass and volume from glacier and ice sheet melt and thermal expansion on a global scale in response to greenhouse forcing of warming sea surface and surface air temperatures (Fig. 5b, c)22. The increasing influence of the global component is the most significant change in the sea-level budgets at all six sites through the Common Era (Fig. 4). By 1950, global sea-level change was responsible for 36–50% of RSL change at each site.

### Sea-level budgets: regional linear component

The dominant processes driving RSL change at U.S. Atlantic coast sites are regional-scale and temporally linear, contributing 2.1–3.2 m of rise over the Common Era (Fig. 4). In northern New Jersey, the linear rate contribution is 1.3 ± 0.08 mm yr−1 (Table 1). The linear component exhibits a spatially variable signal, with the greatest contributions in southern New Jersey of 1.6 ± 0.02 mm yr−1 (LP) and 1.5 ± 0.05 mm yr−1 (CMC) and smaller contributions to the north and south, with the smallest in Connecticut at 1.0 ± 0.03 mm yr−1. The spatial variability of the dominant linear contribution is consistent with the effects of GIA. As the Laurentide ice sheet retreated, the peripheral forebulge began to collapse, causing land subsidence and maximum rates of RSL rise in the mid-Atlantic region (e.g., New Jersey)12. A similar spatial pattern and magnitude of land subsidence, likewise attributed to GIA, has been described on the U.S. Atlantic coast from the late Holocene through the 20th century23. Coastal plain locations (e.g., New Jersey) have also experienced higher rates of RSL rise than locations underlain by bedrock (e.g., New York City) due to the natural compaction of unconsolidated Late Quaternary coastal plain sediments, with an average 20th century compaction rate of 0.16 mm yr−1 (90% confidence interval, 0.06–0.32 mm yr−1)24. Therefore, coastal plain subsidence may also contribute to the greater linear contribution in New Jersey. At all sites, the temporally linear processes provide the dominant contribution to sea-level rise throughout the Common Era, while the nonlinear signals from global, regional, and local-scale processes yield comparatively smaller contributions (Supplementary Figs. 10, 11). Although the combined nonlinear signal varies on multidecadal to multicentennial timescales, we do not identify any consistent periodicity, nor any rapid, large magnitude changes in rate as have been described recently25. Instead the linear contribution clearly dominates RSL rise through the Common Era until it is matched or surpassed by the global contribution in the 20th century.

### Sea-level budgets: regional nonlinear component

Regional-scale nonlinear contributions are similar across U.S. Atlantic coast sites and have an amplitude of <15 cm through the Common Era (Fig. 4, Supplementary Fig. 12). For example, in northern New Jersey, regional nonlinear processes contributed a negative influence from 0–500 CE, followed by a period of stability from 500–1000 CE. A second interval of negative influence occurred from ~1000 CE until 1600 CE, after which the regional nonlinear contribution became positive and increased until present. The contribution of the regional nonlinear component to northern New Jersey RSL fluctuated between −0.2 and 0.2 mm yr−1 during the preindustrial Common Era before increasing to a rate of 0.4 ± 0.2 mm yr−1 in the 20th century (Table 1). Across the U.S. Atlantic coast sites, the regional nonlinear contribution is nearly identical from Connecticut to southern New Jersey and differs only slightly in North Carolina, where rates fluctuated between −0.3 and 0.2 mm yr−1 during the preindustrial Common Era before increasing to a rate of 0.4 ± 0.2 mm yr−1 in the 20th century (Fig. 4).

The regional nonlinear contribution is likely explained by a combination of changing physical processes through the Common Era. The negative regional nonlinear influence from 0–500 CE could be driven by regional-scale steric effects26 associated with long-term cooling broadly in the Northern Hemisphere and within the North Atlantic following early to mid-Holocene maxima16,27. The stabilization of the regional nonlinear contribution from 500–1000 CE could indicate a reduction in long-term cooling or, alternatively, additional processes driving an opposing influence on regional sea-level trends to offset the cooling effects. Changes in atmosphere and ocean circulation that have altered prevailing winds and ocean currents can drive regional sea-level changes16. For example, changes in the strength of the Atlantic Meridional Overturning Circulation (AMOC) and the Gulf Stream can cause sea-level changes on centimeter to decimeter scales on the U.S. Atlantic coast28. Additionally, proxy reconstructions of the North Atlantic Oscillation (NAO) provide evidence for changing atmospheric circulation over the Common Era, which could manifest in centimeter-scale changes in regional sea level29. The shift from a negative to positive regional nonlinear contribution around 1600 CE (Fig. 5a) may be explained by broad climate transitions in the North Atlantic. Beginning around 1400 CE, the NAO changed from a sustained positive phase to a negative phase and AMOC experienced a weakening (Fig. 5d)30. If a decrease of 1 Sv in overturning transport can cause an increase in sea level of ~1.5 cm on the northeast U.S. Atlantic coast31, then the positive contributions from the regional nonlinear component could at least be partially explained by an AMOC weakening over the last 500 years. There is additional evidence for a weakening AMOC over the industrial period (Fig. 5d)32, which may also correspond to the positive regional nonlinear contribution specifically in the last century.

Additionally, the evolving mass of ice sheets could contribute to regional-scale nonlinear RSL trends in the mid-Atlantic over the Common Era. However, the behaviour of the Antarctic Ice Sheet is poorly constrained over this time period33 and while the Greenland Ice Sheet is better constrained, its potential influence on the regional nonlinear trends is uncertain34. The Greenland Ice Sheet may have advanced and reached a peak in mass by the end of the Little Ice Age and subsequently began melting and losing mass to the global ocean35. However, other studies show minimal variability in ice-mass loss over the Common Era (Fig. 5c)36, which is supported by the fact that nonlinear changes in mass of the Greenland Ice Sheet should produce a spatially variable fingerprint along the U.S. Atlantic coast, increasing from north to south, and such a fingerprint is not discernible in the six sites analysed here. Additionally, the effects of the ice sheet may be too small or overprinted by other processes that exhibit a greater nonlinear signal (e.g., steric effects, ocean circulation changes) to be detected in the regional-scale nonlinear trends16. As multiple processes may act simultaneously and with opposing influences on regional sea-level trends, the methods used here cannot fully distinguish and quantify the relative magnitude of each individual process comprising the regional nonlinear contribution.

### Sea-level budgets: local component

The local-scale contribution comprises those trends unique to each individual site that are not observed over larger spatial scales. The local component is spatially and temporally variable; however, its interpretation is limited by the extent and resolution of the available Common Era RSL data at each site (Fig. 4, Supplementary Fig. 13). In northern New Jersey, the amplitude of local-scale sea-level change is <2 cm and it has the smallest contribution to RSL, with rates ranging from −0.1 to 0.1 mm yr−1 throughout the Common Era (Table 1). At all sites, the amplitude of local-scale sea-level change is <16 cm and its contribution to RSL change ranges from −0.3 to 0.3 mm yr−1, with persistent positive or negative contributions for several centuries at a time, and the greatest contributions in New York City and North Carolina.

Local-scale sediment compaction can contribute to RSL change, but the sequences from these sites have been decompacted (e.g., Supplementary Fig. 6), and furthermore, sediment compaction has been shown to have a minimal effect on RSL reconstructions from continuous sequences of high salt-marsh peat with small overburden16,37. Anthropogenic groundwater withdrawal can cause local-scale RSL differences across geographically proximal locations24. Coastal New Jersey has been shown to experience up to ~0.7 mm yr−1 of subsidence due to groundwater withdrawal in the 20th century24, which could explain some of the local contribution observed in northern and southern New Jersey in the last century. The effects of groundwater withdrawal-induced subsidence are likely highly localized, however, considering the NOAA-operated tide gauge at Sandy Hook, New Jersey (station number 8531680), located ~20 km from Cheesequake State Park (Fig. 1b, d), shows a rate of rise of 4.1 ± 0.1 mm yr−1 from 1940–2000 CE compared to the proxy-based reconstruction at Cheesequake of 3.2 ± 0.8 mm yr−1.

Tidal range changes through time could be a contributing factor to local differences in RSL because the foraminifera indicators used as a proxy to reconstruct RSL are linked to modern tidal levels38,39. Therefore, if the local tidal range differed in the past, the RSL reconstructions using these tide-level indicators will not match the true RSL curve40. Changing bathymetric depths or coastline shapes from the effects of sedimentation can affect tidal ranges over centuries to millennia41,42. More recent tidal range changes, such as those observed at tide-gauge sites worldwide in the last century, due to natural or anthropogenic changes such as a loss of wetlands, dredging, or changes in sedimentation due to deforestation could also contribute to changes in the local component over the 20th century43. For example, in New York City, the local component underwent an abrupt shift during the last century. The six U.S. Atlantic coast sites analysed here were not corrected for tidal range changes, but they were reconstructed from depositional environments with small tidal ranges that have been shown to produce the most precise RSL reconstructions44; therefore, even a large percentage change in tidal range would have a minimal influence on the absolute elevation of foraminiferal indicators. Further, paleotidal modelling efforts on the U.S. Atlantic coast have demonstrated that past tidal range changes have minimal influence on rates of sea-level change, at least on basin scales; nevertheless, site-specific factors, such as the location of a site within a bay system, must be considered41,42. For example, in North Carolina, Kemp et al.45 explores the influence of tidal range changes on RSL reconstructions from changes in paleogeography through the opening and closing of inlets46 in the Outer Banks barrier islands. In the case of a doubling of tidal range through the opening of a barrier, the RSL reconstruction is altered on centimeter scales45. In New York City, Kemp et al.15 used hydrodynamic modelling to examine the influence of tidal range changes as RSL rise likely increased the tidal range in Long Island Sound over the past 1500 years. A reconstruction adjusted for past tidal range change results in an average difference from the original reconstruction of 0.05 m (up to 0.11 m); however, the overall RSL trends remain the same as these changes are within the boundary of uncertainties15. Therefore, there is potential for tidal range change to influence reconstructed RSL rates, but the magnitude of RSL change is relatively small and the extent of its influence is site specific. In particular, the precise local-scale geomorphologic evolution at each individual site is needed to understand the full effects of potential tidal range changes driven by geomorphologic changes that could contribute to the local component of sea-level change.

In this study, we use the distinct spatial scales of processes driving sea-level change to distinguish their varying contributions through time. Along the U.S. Atlantic coast, Common Era RSL change is dominated by regional-scale linear processes due to the effects of GIA, with the fastest rates of rise of 1.6 ± 0.02 mm yr−1 in southern New Jersey. Regional nonlinear changes from atmosphere/ocean dynamics and gravitational, rotational, and deformational effects of land-ice change are similar among the sites, contributing between −0.3 and 0.4 mm yr−1 at each site through the Common Era. The consistency of this signal north of the North Carolina site argues that atmosphere/ocean dynamics are the dominant driver of this term. The local signal, likely primarily due to tidal range changes and anthropogenic groundwater withdrawal, vary spatially and temporally among sites between −0.3 and 0.3 mm yr−1. The most significant feature of each sea-level budget is the recent redistribution of budget components due to the increasing contribution of the common global component, associated with the global-mean effects of thermosteric sea-level change and land-ice-mass loss, which matches or surpasses the influence of GIA at each site in the 20th century, when the global signal reaches a rate of 1.3 ± 0.1 mm yr−1. Based on these results on the U.S. Atlantic coast, this method could be applied to more sites globally to produce site-specific Common Era sea-level budgets, which could be used to resolve the spatially distinct processes in even greater detail.

## Methods

### Relative sea-level reconstructions

The RSL reconstructions along the U.S. Atlantic coast (Fig. 1a) use salt-marsh foraminifera as a proxy because their modern distributions exhibit vertical zonation in relation to tidal levels38,39. Foraminiferal-based transfer functions utilize a modern foraminifera training set to quantify species assemblages’ relationship with elevation, which is then applied to sediment core fossil assemblages to produce continuous records of sea level at decadal and decimeter scale resolution39.

We constructed a new Common Era RSL record from a salt-marsh site at Cheesequake State Park in northern New Jersey off of Raritan Bay, ~4 miles from the southern tip of Staten Island and ~23 miles from The Battery (Fig. 1, Supplementary Fig. 1). The foraminiferal-based transfer function for northern New Jersey uses a Bayesian approach that employs foraminifera, as well as bulk sediment δ13C measurements as an additional constraint to reduce vertical uncertainty15,47. Stable carbon isotope geochemistry (δ13C) in bulk sediment represents the dominant vegetation type and can be used as a proxy for sea level because the transition between C3- and C4-dominated salt-marsh plant communities has been shown to occur at the mean higher high water (MHHW) tidal datum on the U.S. mid-Atlantic coast48. The Bayesian transfer function (BTF) was developed using a New Jersey modern training set of salt-marsh foraminifera and δ13C in southern New Jersey from Kemp et al. (2013), in addition to 32 modern samples collected at Cheesequake State Park (Supplementary Figs. 2, 3). We measured bulk sediment δ13C on modern and down-core samples at Bryn Mawr College using cavity ring-down laser spectroscopy using techniques following Balslev-Clausen et al.49. We calibrated the BTF using the combined New Jersey modern training set and evaluated its performance using cross-validation47 (Supplementary Fig. 4). We formally accounted for temporal and spatial variability of modern foraminifera distributions in the BTF by including informative foraminifera variability priors for individual species using data from a monitoring study of modern foraminifera in southern New Jersey50. The BTF was applied to sediment core foraminifera and δ13C data to provide paleomarsh elevation (PME) estimates with 95% credible intervals for each core sample (Supplementary Fig. 5). We calculated RSL by subtracting the PME estimates from the sample altitude. In addition, we used a geotechnical model51,52 to correct the RSL record for post-depositional lowering through sediment compaction (Supplementary Fig. 6) that has been used previously to correct salt-marsh RSL reconstructions for compaction16.

We reconstructed RSL using the transfer function estimates of PME in combination with a sediment core chronology. A sediment core chronology was constructed using Accelerator Mass Spectrometry radiocarbon (14C) dating on identifiable plant macrofossils (stems and rhizomes) in the sediment core (Supplementary Table 1). A plateau in the radiocarbon calibration curve often results in radiocarbon dated material from the past ~300 years having multimodal age estimates and large uncertainties53. Therefore, to provide a chronology for the last several hundred years, we used changes in Ambrosia pollen abundances, regional-scale pollution markers (recognized in changes in down-core concentrations of lead, copper, cadmium, and nickel), the ratio of lead isotopes (206Pb:207Pb), and 137Cs activity (Supplementary Fig. 7). Radiocarbon dates, pollen abundances, pollution markers, and 137Cs activity were compiled using the Bchron package in R54,55, which uses a Bayesian framework to produce an age-depth model and estimates ages with associated uncertainties for every 1-cm-thick interval in the core (Supplementary Fig. 8). The age estimates and uncertainties from Bchron were applied to all core samples with a reconstructed PME (Supplementary Fig. 9). The data for the northern New Jersey RSL record can be found in the Supplementary Materials.

### Spatiotemporal statistical model

To estimate past RSL and rates of RSL change and their associated uncertainties, we used a spatiotemporal empirical hierarchical model11,17 with a sea-level database comprising proxy sea-level records with high-resolution chronologies from 36 regions around the world, including the new northern New Jersey RSL record (Fig. 3a). The 2274 individual data points in the database use proxies such as foraminifera, diatoms, testate amoebae, coral microatolls, archaeological evidence, and sediment geochemistry. We have updated the database (Supplementary Data) from Kemp et al. (2018) to include 390 new RSL data points from northern New Jersey, USA (this study); Croatia56; French Polynesia57; Israel58; Quebec59; Connecticut25; Maine25; and Nova Scotia25. We compared the new northern New Jersey record with other published records along the U.S. Atlantic coast from southern New Jersey (Leeds Point and Cape May Courthouse14), New York City15, Connecticut60, and North Carolina45 to examine regional variability in magnitudes and rates of past RSL change, as well as variability in regional Common Era sea-level budgets.

As in Kopp et al.11 and Kemp et al.16, decadal-average values from instrumental tide-gauge records in the Permanent Service for Mean Sea Level (PSMSL61) were included in the analysis, provided they were either (1) longer than 150 years, (2) within 5 degrees distance of a proxy site and longer than 70 years, or (3) the nearest tide gauge to a proxy site that is longer than 20 years11,16. We also include multicentury records from Amsterdam (1700–1925 CE)62, Kronstadt (1773–1993 CE)63, and Stockholm (1774–2000 CE)64, as compiled by PSMSL. The input data also include the global-mean sea-level reconstruction of Hay et al.18 from tide-gauge records for 1880–2010.

The model has (1) a process level that characterizes RSL over space and time and (2) a data level that links RSL observations (reconstructions) to the RSL process. Hyperparameters characterize prior expectations of dominant spatial and temporal scales of RSL variability, set through maximum-likelihood optimization.

At the process level, the RSL field $$f\left( {{\boldsymbol{x}},t} \right)$$ is modelled as the sum of seven components16:

$$f\left( {{\boldsymbol{x}},t} \right) = g_f\left( t \right) + g_s\left( t \right) + m\left( {\boldsymbol{x}} \right)\left( {t - t_0} \right) + r_s\left( {{\boldsymbol{x}},t} \right) + r_f\left( {{\boldsymbol{x}},t} \right) + l_s\left( {{\boldsymbol{x}},t} \right) + l_f({\boldsymbol{x}},t)$$
(1)

where $${\boldsymbol{x}}$$ represent geographic location, $$t$$ represents time, and $$t_0$$ is a reference time point (2000 CE). The seven components include fast and slow common global (or globally uniform) terms ($$g_f\left( t \right)$$ and $$g_s\left( t \right)$$), a regional linear term ($$m\left( {\boldsymbol{x}} \right)\left( {t - t_0} \right)$$), fast and slow regional nonlinear terms ($$r_f\left( {{\boldsymbol{x}},t} \right)$$ and $$r_s\left( {{\boldsymbol{x}},t} \right)$$), and fast and slow local terms ($$l_f({\boldsymbol{x}},t)$$ and $$l_s\left( {{\boldsymbol{x}},t} \right)$$). The regional linear term uses predictions from the ICE5G–VM2– 90 Earth-ice model65 as prior means. We also ran the model using predictions from the ICE6G–VM5a model66, but the differences in regional linear rates were negligible (Supplementary Table 3).

The data level includes the RSL reconstructions with observations, $$y_i$$, where:

$$y_i = f\left( {{\boldsymbol{x}}_i,t_i} \right) + y_0\left( {{\boldsymbol{x}}_i} \right) + \varepsilon _i + w\left( {{\boldsymbol{x}}_{\boldsymbol{i}},t_i} \right)$$
(2)
$$t_i = \widehat {t_i} + \delta _i$$
(3)

where $$f\left( {{\boldsymbol{x}}_i,t_i} \right)$$ is the true RSL value at location $${\boldsymbol{x}}$$ and time $$t$$, $$y_0({\boldsymbol{x}}_i)$$ is a site-specific vertical datum correction to ensure that the RSL reconstructions are directly comparable to one another, $$\varepsilon _i$$ is the vertical uncertainty, which is treated as independent and normally distributed (with a standard deviation for each data point from the original publication), and $$w\left( {{\boldsymbol{x}}_{\boldsymbol{i}},t_i} \right)$$ is supplemental white noise. The true age of a RSL observation ($$t_i$$) is the mean estimate ($$\widehat {t_i}$$) and its error ($$\delta _i$$).

The hyperparameters characterize prior expectations of amplitudes and spatial and temporal scales of RSL variability (Supplementary Table 2). These amplitudes and scales are estimated using a maximum-likelihood optimization. The nonlinear terms were characterized by three spatial scales (global, regional, and local) and two temporal scales (fast and slow). These different spatial and temporal scales allow RSL to be decomposed into common global, regional temporally linear, regional nonlinear, and local components. As in Kopp et al.11 and Kemp et al.16, we apply a constraint on the model that mean global sea level over −100–100 CE is equal to mean global sea level over 1600–1800 CE because a constant global rate could also be interpreted as a regional linear trend (Supplementary Fig. 14). The decomposed components are used to produce Common Era sea-level budgets for northern New Jersey and five other sites along the U.S. Atlantic coast using 100-year average rates in 20-year timesteps to examine the evolving contribution of each component through time.

We also conduct several sensitivity tests on the model. We ran a sensitivity check to test the prior specifications by fixing the prior amplitudes for all nonlinear terms to be the same and then optimizing all of the hyperparameters; the objective is to maximize the log likelihood and in this case with new hyperparameter values, the posterior likelihood decreased (log likelihood changed from −10352 to −10438), and RSL trends remained largely consistent. Additionally, we assess the effect of initial values (within the upper to lower bounds) of each hyperparameter on the optimized maximum-likelihood values. We find that different initial values have very little influence on the optimized hyperparameters: differences in timescale estimates were <2 years, differences in length scale estimates were <0.03 degrees, and differences in prior SD were <0.3 cm. We also performed a “leave-one-site-out” cross-validation of the six study sites using the original optimized hyperparameters, individually removing the data from each site and predicting the RSL change at each site given the rest of the data. Overall the model is reasonably well calibrated with a coverage probability based on a “leave-one-site-out” validation of 88%. The mean error is −0.105 m, suggesting that the model tends to over predict on average and the mean absolute error is 0.165 m (Supplementary Table 4). Further, the rate predictions at the other five sites remained consistent and the uncertainties increased by <0.01 mm/yr (Supplementary Fig. 15). Therefore, the model results are generally robust to the removal of site data. Additionally, Supplementary Fig. 16 illustrates a comparison of RSL predictions for northern New Jersey using different model variations, including removing the northern New Jersey data and predicting RSL, using only the northern New Jersey data, and reoptimizing hyperparameters using only the northern New Jersey data.


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# Factoring 3rd degree polynomials

• Dec 5th 2010, 02:52 PM
l flipboi l
Factoring 3rd degree polynomials
Hello,

Can you please show me how to factor this out x^3 - 6x^2 + 11x - 6?

I would like to see how its done, step-by-step if possible.

Thanks!
• Dec 5th 2010, 02:59 PM
Jerry99
Id start out by finding a zero on my cacluator. For example, 2 is a zero in that function. And, then Id do synthetic division to make it quadratic and then factor it. Are you familier with synthetic division?
• Dec 5th 2010, 03:03 PM
rtblue
Well, from intuition we can gather that x=1 is a root. Now we have:

$(x-1)(x^2-5x+6)=0$

$(x-1)(x-3)(x-2)=0$

$x=1,x=3,x=2$
• Dec 5th 2010, 03:10 PM
l flipboi l
Thanks! i just figured it out too by using synthetic division.

I guess my question now is, would plugging in numbers to find it equal to zero be the best approach when dealing with 3rd degree polynomials or higher?
• Dec 5th 2010, 06:37 PM
rtblue
There are many approaches to solving polynomials of a high degree. Synthetic division is probably the best one. Lets call the leading coefficient of an n degree polynomial Q and lets call the last term of the same polynomial P. Now, calculate all the values of P/Q and use synthetic division, with those values (only those values could possibly work). Have you heard of Descarte's Rule of Signs? It gives you some direction as to where the roots might be.

Descartes' Rule of Signs
• Dec 6th 2010, 05:05 AM
HallsofIvy
Quote:

Originally Posted by l flipboi l
Thanks! i just figured it out too by using synthetic division.

I guess my question now is, would plugging in numbers to find it equal to zero be the best approach when dealing with 3rd degree polynomials or higher?

That depends upon what you mean by "plugging in numbers". You have to choose those numbers carefully. When rtblue said "from intuition we can gather that x=1 is a root" I suspect he meant that it is easy to see that 1- 6+ 11- 6= 12- 12= 0. You can also use the "rational root theorem"- if the leading coefficient is a and the constant term of a polynomial is b, then any rational root if of the form m/n where n divides a and m divides b. Here, the leading coefficient is 1 and the constant term is 6 so that the only possible rational roots are the integers $\pm 1$, $\pm 2$, $\pm 3$, and $\pm 6$. If NONE of those turned out to be a root, there would be no way to factor this with integer coefficients.
• Dec 6th 2010, 01:24 PM
l flipboi l
Thanks! i'm gonna read the rules of signs.


text:
# Reduce spacing between items [duplicate]

Possible Duplicate:
Vertical space in lists

Please, how can I reduce spacing between items ?

\begin{itemize}

\item[\textbullet]  first
\item[\textbullet]  second
\item[\textbullet] third

\end{itemize}

You could use the enumitem package and set the itemsep parameter to some suitable value. E.g., to have no extra spacing between items (whether bulleted, enumerated, or otherwise styled), you'd type
\setlist{itemsep=0pt}

after issuing the \usepackage{enumitem} statement.
• @researcher - I'm not sure what you mean by "the item on the left": by default, the items are organized in a vertical list. Please consult the manual of the enumitem package (available by typing "texdoc enumitem" at a command line) to learn more about the extensive customization options that this package offers. Alternatively, you could post a new question, with a full minimum working example and not just a code snippet, to inquire about the issue you're trying to solve. – Mico Sep 9 '12 at 10:24


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## why segmentation fault is arising , because of these three lines?

The code is giving segmentation fault. But it is working fine, :

1. if i remove the first weird line “q++;”. Or
2. if i call solve(s,1) in main function instead of solve(s,0) at third weird line.
3. the code is working fine if i use “solve(s,q)” at second weird line.
 ''' #include using namespace std;  >bool solve(string s,int q) {     q++; //first weird line     if(q==10)         return true;     solve(s,q+1);//second weird line     return false; }  int main() {     string s;     cin>>s;     solve(s,0);//third weird line      return 0; }'''

## Channel coding and Error probability. Where are these probabilities from?

From where are the following probabilities?

We consider BSCε with ε = 0,1 and block code C = {c1, c2} with the code words c1 = 010 and c2 = 101. On the received word y we use the decoder D = {D1,D2} which decodes the word to the code word which has the lowest hamming distance to y. Determine D1 and D2 and the global error probability ERROR(D) if the code words have the same probability. Hint: To an output y there exists only one x which gets to a failing decoding. (Y = 100 will only gets wrong decoded if we sent message is x = c1 = 010.) So the term (1− p(D(y)|y)) is equal to Pr[X = x|Y = y] for a suitable x.

Nun

\begin{aligned} &\text { Hamming-Distance: }\ &\begin{array}{c|cc} \text { Code } & 010 & 101 \ \hline \hline 000 & 1 & 2 \ \hline 001 & 2 & 1 \ \hline 010 & 0 & 3 \ \hline 011 & 1 & 2 \ \hline 100 & 2 & 1 \ \hline 101 & 3 & 0 \ \hline 110 & 1 & 2 \ \hline 111 & 2 & 1 \ \hline \end{array} \end{aligned} $$\left.D_{1}=\{000,010,011,110\} \text { (Decides for } 010\right) \left.D_{2}=\{001,100,101,111\} \text { (Decides for } 101\right)$$ \begin{aligned} E R R O R(D) &=\sum_{y \in \Sigma_{A}^{3}} p(y)(1-p(D(y) | y)) \ &=\overbrace{2 \cdot p(y)(1-p(D(y) | y))}+\quad \overbrace{6 \cdot p(y)(1-p(D(y) | y))}^{ } \ &=2 \cdot\left(\frac{729}{2000}+\frac{1}{2000}\right)\left(\frac{7}{250}\right)+6 \cdot\left(\frac{81}{2000}+\frac{9}{2000}\right)\left(\frac{757}{1000}\right) \end{aligned} How do I get to the probabilities $$\frac{7}{250}$$ and $$\frac{757}{1000}$$??

I don’t get this calculation. It should be right. But I don’t get how to get to these probabilities.

Could someone explain me this?

## What is the subspecies name for the standard race choices in the D&D 3.5 Players Handbook? What is appearance do these subspecies, and a few others?

I am playing D&D on a Neverwinter Nights Enhanced Edition module using the D&D 3.5 ruleset. Please note that this is not a Neverwinter Nights game question. This is a D&D 3.5 lore question.

I am having difficultly finding the subspecies name for the standard race choices offered in the Players Handbook. I am also having difficulty finding descriptions and images of their appearance online. I am avoiding 4e or 5e images and information because some of the lore has changed.

Here is my list of questions. Hair and skin is all I need for description. If you can have a picture link that would be very helpful. I will also gladly look at any online resource that answers my questions and saves people time from writing out their answers.

What is the standard elf race subspecies name in D&D 3.5? What is their suggested appearance?

What is the standard gnome race subspecies name in D&D 3.5? What is their suggested appearance?

What is the standard dwarf race subspecies name in D&D 3.5? What is their suggested appearance?

What is the standard halfling subspecies name race in D&D 3.5? What is their suggested appearance?

What does a deep dwarf look like?

What does a wild elf look like?

What does a wood elf look like?

What does a gray elF look like?

What does a forest gnome look like?

What does a lightfoot halfling look like?

What does a tallfellow halfling look like?

What is a tribal orc? What does it look like?

What is a deep orc? What does it look like?

## These pages showing url too long and poorly formatted in uber suggest in terms of seo.How to solve this problem

I have a word press site and doing SEO with yoast premium plugin all other SEO related issue are solved but these pagination problem is a headache I search the internet but didn’t find solution for that

## Which of these images is better?

Which of these images is better?

Attached are 2 files "arrow3" and "arrow4"
Each has 3 sets of arrow shapes in it.

The effect I am going for is:
Left arrow: rising from ground up
Middle arrow: falling from sky
Right arrow: simple shadow, like photoshop

In first image, arrow3,
and in second image, arrow4,

My question:
Which of these arrows is better suited for a web logo etc

Thanks

## What is the difference in time-complexity for sorting these 2-d arrays?

Let $$A$$ have $$n/10$$ rows, $$10$$ columns and $$n$$ overall elements

Let $$B$$ have 10 rows, $$n/10$$ columns and $$n$$ overall elements.

It is given that each row in sorted in ascending order, Can you sort each of these in $$O(n\log(n))$$ or better using comparison sort?

I’m leaning towards k-way merge implementing a min-heap following this implementation merging sorted arrays, but I can’t seem to figure out what the difference between this cases is.

$$B$$ for example will have $$10$$ elements constantly in the min-heap, so the time complexity will be $$10n \log(n) \in O(n)$$? Is this even possible in comparison sorts?

While $$A$$ would have $$n/10$$ elements in the min-heap, but are the run times equivalent?

## Which of these 2 images is better?

Which of these 2 images is better?
(for a music website)

http://tvyotv.com/images/dpRocks3D.png

http://tvyotv.com/images/dpRocks3D2.png

The bottom gray reverse gradients are shadows, like on a 3d object.
Thanks

## Are these two homebrew shields balanced?

I remember being impressed with this question, although it was closed as too broad for asking about too many things in one go. I left a comment at the time about asking less, and coming back to this later, I’ve decided to have a go at asking this. Note that I’ve also taken a look at this question about a homebrew tower shield.

This is partly based on NWN2, a video game that uses 3.5e rules, so I assume shields were broken up similarly in 3.5e, but I only know that game. In that game, there were “light shields” (i.e. “bucklers”), “heavy shields” and “tower shields”.

In short, I wanted to introduce a “light shield” and “tower shield”, keeping the RAW 5e shield as a “heavy shield”. See below for the table comparing AC and pricing, similar to the official table and the above linked question about a tower shield.

$$\begin{array}{|c|c|c|c|c|} \hline \textbf{Armor Name} & \textbf{Cost} & \textbf{Armor Class} & \textbf{Strength} & \textbf{Stealth} & \textbf{Weight} \ \hline \text{Light shield} & 5\,\text{gp} & +1 & \text{N/A} & \text{N/A} & 2\,\text{lb.} \ \hline \text{Heavy shield} & 10\,\text{gp} & +2 & \text{N/A} & \text{N/A} & 6\,\text{lb.} \ \hline \text{Tower shield} & 50\,\text{gp} & +3 & 15 & \text{Disadvantage} & 45\,\text{lb.} \ \hline \end{array}$$

In addition to the above table:

• I wanted it to be possible for the light shield to be donned and doffed as a bonus action (since otherwise why would anyone want it over a heavy shield?)
• I wanted the tower shield to have the restriction that you needed to be proficient with heavy armor and shields, otherwise you cannot benefit from it and trying to use it imposes the same penalties as with not having proficiency with armor (from Armor Proficiency):

Armor Proficiency. Anyone can put on a suit of armor or strap a shield to an arm. Only those proficient in the armor’s use know how to wear it effectively, however. Your class gives you proficiency with certain types of armor. If you wear armor that you lack proficiency with, you have disadvantage on any ability check, saving throw, or attack roll that involves Strength or Dexterity, and you can’t cast spells.

So, my question is, are my new light shield and tower shield balanced? Meaning, are there any problems from having a light shield that you can don with a bonus action for the cost of -1 AC compared to a RAW shield (what I’m now calling the heavy shield)? Are my restrictions on the tower shield enough to justify basically being able to buy a +1 shield (albeit a nonmagical one), or is it still too strong even so?

(I also considered adding a Strength requirement to the “heavy shield”, like 13 or even just 10, but that kinda screws over Dexterity based medium armor wearing shield users, so I decided against it, but I could be easily swayed into reconsidering it; possibly a separate question to this one, though.)

## Is the usage for asymptotic notation for these algorithms correct? [duplicate]

So after reading a lot of information around asymptotic analysis of algorithms and the use of Big O / Big Ω and Θ, I’m trying to grasp how to utilise this in the best way when representing algorithms and operations on data structures.

For example there is a recommended website where I got this screenshot from describing Quicksort and I’ve noticed a few issues that stand out to me based on what I’ve learnt.

1. Is it possible for all notations to represent “Best” “Average” and “Worst” cases? and if so how is this possible? For example for a “Worst” case, How can Big Ω represent the Upper bound. The upper bound is tied to Big O.
2. I thought in order to find Theta Θ, Big O and Big Ω had to be the same values? In the screenshot “Best” case is n log(n) and Worst case is n^2 so how can Θ(n log(n))?
3. Take for instance a Hash Table data structure, if you were to perform an analysis on the time complexity for insertion of an element. Would I be correct is saying you could interchangeably say Ω(1) and O(N) or conversely “Average Case is O(1)” and “Worst Case is O(N)”?


text:
# ankursinha.in/blog

neuroscience/fedora/musings

Thu 23 August 2012

# Compiling the cp210x module for use with the NAO

Posted by ankur in Tech (356 words, approximately a 2 minute read)

I've been recently working on trying to get a NavChipto work with the NAO. (The NAO is very cool btw!) To start with, I was delighted to see that the NAO uses a Linux OS on it's Geode processor. (The newer NAOs are using an Atom processor though) The issue was that this kernel that Aldebaran (the manufacturer) uses is customized. Specifically, it lacks the cp210x driver that the NavChip requires. I've been trying to build/compile this driver for the past week, and the one before. I'm really not a kernel developer. In fact, this is the first time I've come close to touching the kernel sources and well, running make!

Anyway, as a note to self (yes, another!), and as some documentation for folks who'd try to use the NAO with more hardware, here's what I've done so far. I haven't tested the final module yet, but it looks okay. I'll only know tomorrow when I try it out on the robot if I were successful.

Get the kernel configuration from the Nao. It should be /boot/config on the file system. Copy this to your downloaded kernel source tree as .config (DOT config).

Run:

make oldconfig ARCH=i386 #My system is an x86_64, so the ARCH argument is needed

The Nao's kernel is 2.6, so you need to use the Makefile26 file in the cp210x directory. Run configure. It'll create a defaults.mk file. Modify this file to correct the KVER etcetera variables (You can also modify the configure script instead. Whatever you prefer):

# Makefile.config
# Automatically generated
KVER=2.6.29.6-rt24
KVER1=2
KVER2=6
KVER3=29.6-rt24
EXT=.ko
KOFILE=cp210x.ko
KDIR=/home/ankur/Documents/work/code/NAO/OS/linux-aldebaran
MODFILE=cp2101.ko
MODDIR=/home/ankur/Documents/work/code/NAO/OS/linux-aldebaran
PWD=/home/ankur/Documents/work/code/NAO/OS/linux-aldebaran/cp210x-3.1.0/cp210x

In the Makefile, under "default:", I added ARCH=i386 (again, since my host is an x86_64). Now, just run

make

That should be it. Check if your module is okay using "file" and "modinfo". It looks okay. I've got to try it on the Nao tomorrow though to confirm.


text:
| English | 简体中文 |

# 496. Next Greater Element I

## Description

The next greater element of some element x in an array is the first greater element that is to the right of x in the same array.

You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.

For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1.

Return an array ans of length nums1.length such that ans[i] is the next greater element as described above.

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.

Constraints:

• 1 <= nums1.length <= nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 104
• All integers in nums1 and nums2 are unique.
• All the integers of nums1 also appear in nums2.

Follow up: Could you find an O(nums1.length + nums2.length) solution?


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# 5.4 Equation of a Straight Line

5.4 Equation of a Straight Line: y = mx + c

1. Given the value of the gradient, m, and the y-intercept, c, an equation of a straight line
y = mx + can be formed.

2. If the equation of a straight line is written in the form y = mx + c, the gradient, m, and the
y-intercept, c, can be determined directly from the
equation.

Example:
Given that the equation of a straight line is y = 3 – 4x. Find the gradient and y-intercept of the line?

Solution:
y= 3 – 4x
y= – 4x + 3 ← (y = mx + c)
Therefore, gradient, m = – 4
y-intercept, c = 3

3. If the equation of a straight line is written in the form ax + by + c = 0, change it to the form
y = mx + c before finding the gradient and the
y-intercept.

Example:
Given that the equation of a straight line is 4x + 6y– 3 = 0. What is the gradient and y-intercept of the line?

Solution:
4x + 6y – 3 = 0
6y = –4x + 3

$\begin{array}{l}y=-\frac{2}{3}x+\frac{1}{2}←\overline{)y=mx+c}\\ \therefore \text{Gradient}m=-\frac{2}{3}\\ \text{}y-\text{intercept},\text{}c=\frac{1}{2}\end{array}$


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# Why is the Big Bang Theory considered a "theory" and not a fact?

May 1, 2018

We do not and cannot know that what is described in Big Bang theory actually happened.

#### Explanation:

In the natural sciences we make observations and construct models.

If those models are consistent with our observations then we can make predictions from those models and test them against more observations.

If some observations contradict our models, then we can tell that our models are wrong or in need of modification.

For example, Newton's laws of physics provide pretty good models sufficient in accuracy to allow us to calculate how to land a man on the moon. Are they therefore true? Not exactly. Newton's laws fail to work well at velocities approaching the speed of light. So we can tell that they need some adjustment.

Einstein's special and general theory of relativity provide more accurate models. So are they true? They predict all sorts of strange behaviour which we have actually observed. They seem to provide useful models, but they do not explain everything. For example, the extreme conditions in a black hole or those we think existed at the start of the universe provide some challenges to General Relativity.

From a practical perspective it is useful to operate within theoretical frameworks that we treat as facts. We need to make assumptions in order to reach substantial conclusions. So we often act as if something like Big Bang theory is a fact, but we cannot know it to be so.

Essentially in any natural science we cannot prove anything to be true. We can demonstrate that theories are false. This is the honesty of science.

Consider an alternative theory: The universe and everything you "know" actually started $10$ minutes ago.

You may protest that you remember what you did yesterday. In my theory I can explain that these are simply implanted memories, not real. What about radiometric dating? What about the Andromedra galaxy being $2.5$ million light years away, so the light we see from it started travelling in our direction $2.5$ million years ago? In my theory I can claim that these are all an elaborate implanted fiction.

There is no way to prove that this theory is true and probably no way to prove that it is false.


text:
# (d) to 2.80 6. Suppose that you are testing the following hypothe ses where the variance...

###### Question:

(d) to 2.80 6. Suppose that you are testing the following hypothe ses where the variance is unknown: Ho: 100 H <100 The sample size is n 25. Find bounds on t P-value for the following values of the test statisti (a) =-2.80 (b) to=-1.75 (c) to- -2.54 (d) to =-2.05

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