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## Ranking Investment Proposals

Read this section about capital budgeting and ranking investment proposals. These are important when businesses are comparing similar real estates investments with the intent of picking the investment that yields the highest return. This section presents several methods that are commonly used to rank investment proposals, including net present value (NPV), internal rate of return (IRR), profitability index (PI), and accounting rate of return (ARR). It also gives formulas and examples that demonstrate how to apply these formulas in the real world.

Several methods are commonly used to rank investment proposals, including NPV, IRR, PI, payback period, and ARR.

#### LEARNING OBJECTIVE

• Analyze investment proposals by ranking them using different methods

#### KEY POINTS

• The higher the NPV, the more attractive the investment proposal.
• The higher a project's IRR, the more desirable it is to undertake the project.
• As the value of the profitability index increases, so does the financial attractiveness of the proposed project.
• Shorter payback periods are preferable to longer payback periods.
• The higher the ARR, the more attractive the investment.

#### TERMS

• time value of money

The time value of money is the value of money, figuring in a given amount of interest earned over a given amount of time.

• discounted cash flow

In finance, discounted cash flow (DCF) analysis is a method of valuing a project, company, or asset using the concepts of the time value of money.

The most valuable aim of capital budgeting is to rank investment proposals. To choose the most valuable investment option, several methods are commonly used:

#### Net Present Value (NPV):

NPV can be described as the "difference amount" between the sums of discounted:cash inflows and cash outflows. In the case when all future cash flows are incoming, and the only outflow of cash is the purchase price, the NPV is simply the PV of future cash flows minus the purchase price (which is its own PV). The higher the NPV, the more attractive the investment proposal. NPV is a central tool in discounted cash flow(DCF) analysis and is a standard method for using the time value of money to appraise long-term projects. Used for capital budgeting and widely used throughout economics,finance, and accounting, it measures the excess or shortfall of cash flows, in present value terms, once financing charges are met.

$\operatorname{NPV}(i, N)=\sum_{t=1}^{N} \frac{R_{t}}{(1+i)^{t}}$

NPV formula: Each cash inflow/outflow is discounted back to its present value (PV). Then they are summed. Therefore, NPV is the sum of all terms.

In financial theory, if there is a choice between two mutually exclusive alternatives, the one yielding the higher NPV should be selected. The rules of decision making are:

• When NPV > 0, the investment would add value to the firm so the project may be accepted
• When NPV < 0, the investment would subtract value from the firm so the project should be rejected
• When NPV = 0, the investment would neither gain nor lose value for the firm. We should be indifferent in the decision whether to accept or reject the project. This project adds no monetary value. Decision should be based on other criteria (e.g., strategic positioning or other factors not explicitly included in the calculation).

An NPV calculated using variable discount rates (if they are known for the duration of the investment) better reflects the situation than one calculated from a constant discount rate for the entire investment duration.

#### Internal Rate of Return (IRR)

The internal rate of return on an investment or project is the "annualized effective compounded return rate" or "rate of return" that makes the net present value (NPV as NET*1/(1+IRR)^year) of all cash flows (both positive and negative) from a particular investment equal to zero.

IRR calculations are commonly used to evaluate the desirability of investments or projects. The higher a project's IRR, the more desirable it is to undertake the project. Assuming all projects require the same amount of up-front investment, the project with the highest IRR would be considered the best and undertaken first.

#### Profitability Index (PI)

It is a useful tool for ranking projects, because it allows you to quantify the amount of value created per unit of investment. The ratio is calculated as follows:

Profitability index = PV of future cash flows / Initial investment

As the value of the profitability index increases, so does the financial attractiveness of the proposed project. Rules for selection or rejection of a project:

• If PI > 1 then accept the project
• If PI < 1 then reject the project

#### Payback Period

Payback period intuitively measures how long something takes to "pay for itself. " All else being equal, shorter payback periods are preferable to longer payback periods. Payback period is widely used because of its ease of use despite the recognized limitations: The time value of money is not taken into account.

#### Accounting Rate of Return (ARR)

The ratio does not take into account the concept of time value of money. ARR calculates the return, generated from net income of the proposed capital investment. The ARR is a percentage return. Say, if ARR = 7%, then it means that the project is expected to earn seven cents out of each dollar invested. If the ARR is equal to or greater than the required rate of return, the project is acceptable. If it is less than the desired rate, it should be rejected. When comparing investments, the higher the ARR, the more attractive the investment. Basic formulae:

ARR = Average profit / Average investment

Where: Average investment = (Book value at beginning of year 1 + Book value at end of user life) / 2

Source: Boundless


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# Blog: flow and load duration curves

### by Beth Allen and François Birgand

started 2011-09-11 and updated 2019-01-14

### Keywords

• hydrograph
• chemograph
• duration curves

## In search of additional water quality indicators

Lately, we have been researching and developing techniques of analyzing continuous water quality and hydrology data in order to explain hydrological and biogeochemical processes controlling stream chemistry in watersheds. One of the techniques in which we have applied to an entire year of water quality and hydrology data is a method that assesses how relatively reactive nutrient/sediment loading is to flow and the relative flashiness of a watershed. This method, referred to as load and flow duration curves, determines the percentages of the total load or Mass (Mk%) and volume of Water (Wk%) that occur in a percentage of the total sampling time, termed probability of occurrence.

Mk% and Wk% duration curves can be plotted as a function of the probability of occurrence, which provides an interesting way of demonstrating how loading relates to flow and varies among water quality constituents. This method can also be applied to individual storm events to assess if there is a first flush response.

## Plotting hydrographs and chemographs in R

There are lots of great new (in 2017) plotting tools in R, such as ggplot, qqplot, or plotly. These tools have developped to manipulate data as function of factors, parameter, etc. But as far as we can tell, they are not well designed for handling time series of different units. This is too bad because they are otherwise rather attractive. As a result, for continuous flow and concentration data, we are still forced to use the more basic R graphics. But maybe me are missing something. In the meantime, it is important to be able to have some basic knowledge on how to plot continuous hydrographs and chemographs.

In the code below, we have added quite a bit of comments to explain what code line does what.

data<-read.csv(file="https://raw.githubusercontent.com/francoisbirgand/francoisbirgand.github.io/master/data/sample_1hr_QC_data.csv",header = TRUE) #Reads file into table format

WSarea<-24.2 #Area of watershed in km2
WS<-"Maudouve at Saint-Donan, France"
names(data)=c("datetime","Q","C")   # renames the columns in simpler names
data<-as.data.frame(data)
data$datetime<-as.POSIXct(strptime(data$datetime, "%Y-%m-%d %H:%M:%S")) # transforms characters into date values understood by R
D<-data$datetime Q<-data$Q   #Defines Q as the flow value (m3/s)
C<-data\$C   #Defines C as the Concentration value (mg NO3-N/L)
L<-Q*C
N=nrow(data)   #Sets N to the value equal to the number of total rows in the table

# definition of the x and y axes limits
xlim = as.POSIXct(c(D[1],D[N]))  # this renders the first and last date understandable for plotting purposes
ylimQ = c(0,max(Q))           # ylim for flow
ylimC = c(0,max(C))           # ylim for concentrations

ScaleF = 1.2                  # scaling factor for size of fonts and other things

y1lab<-expression("Flow rate (" * m^3 * "/s)")  # defines the label for flow
y2lab<-substitute(paste("Nitrate concentration (mg ",NO[x]^{y},"-N)",sep=""),list(x=3,y="-")) # defines the label for concentrations

par(mar=c(4.5,4.5,4,4.5))     # defines the sizes, in number of lines, for the margins (bottom, left, top, right)

ltyp=c(1,2)

plot(D,Q,col="blue",type="l",cex=0.1,yaxt="n",
lty=ltyp[1],xaxt="n",xlab="",ylab="",xlim=xlim,ylim=ylimQ)
# we are taking all the default addition of axis tick marks and numbers out by using xaxt and yaxt = "n"
# and setting the axis labels at nothing using xlab = "" and ylab = ""
abline(h=0)
axis.POSIXct(1, at=seq(D[1], D[N], by="month"), format="%m/%d",cex.axis=ScaleF)
# this tells R that we want the X axis ticks and values to be displayed as dates, be added on a monthly basis,
# using the month/day format
axis(2,cex.axis=ScaleF)
# this tells R that the first Y axis ticks can be displayed  (that function was repressed earlier by 'yaxt="n" ')
par(new=TRUE)
# this tells R that a new plot has already been opened, in other words you are telling R to keep adding things
# on the existing plot

ColElmt="deeppink1"
plot(D,C,col=ColElmt,type="l",cex=0.1,yaxt="n",
lty=ltyp[1],xaxt="n",xlab="",ylab="",xlim=xlim,ylim=ylimC)
# plots the concentration data
axis(4,cex.axis=ScaleF)
# this tells R that the second Y axis ticks can be displayed (that function was repressed earlier by 'yaxt="n" ')
par(new=TRUE)

mtext("Dates in 1997-1998",side=1,line=3,cex=ScaleF) # add in the margin the defined labels and title
mtext(y1lab,side=2,line=3,cex=ScaleF)
mtext(y2lab,side=4,line=3,cex=ScaleF)
mtext(WS,side=3,line=1.5,cex=ScaleF)

legend("topleft",c("Flow","Conc"),lty = c(1,1), col = c("blue",ColElmt))

## Sorting flow and load values

Flow duration curves represent the percentage of the total flow that occurred in x% of the time corresponding to the highest flows. The same applies for loads. This might sound a bit merky, but hopefully it will not with the further explanations below. To get there, one first needs to order flow and loads in descending order.

QSort=sort(Q,decreasing = TRUE)   #Sorts instantaneous flow rates in descending order
LSort=sort(L,decreasing = TRUE)   #Sorts instantaneous flux values in descending order

The blue hydrograph from above now becomes:

par(mar=c(4.5,4.5,1,1))
plot(QSort,col="blue",type="l",cex=0.1,yaxt="n",
lty=ltyp[1],xaxt="n",xlab="",ylab="",ylim=ylimQ)
abline(h=0)
axis(2,cex.axis=ScaleF)
mtext(y1lab,side=2,line=3,cex=ScaleF)
mtext("Cumulative number of flow values",side=1,line=3,cex=ScaleF)
legend("topright",c("Sorted Flow"),lty = c(1), col = c("blue"))

The next thing to do is to integrate under the sorted curves to obtain the cumulative flow as a function for each represented flow value. To do this, there is a really nice way to calculate this using the cumcum() function as below. All values are put in percentage values

cumQSort<-c(0,(cumsum(QSort[-1])+cumsum(head(QSort,-1)))/2)
cumQSort<-cumQSort/tail(cumQSort,1)*100
cumLSort<-cumLSort/tail(cumLSort,1)*100

So now the cumulative flow volume curve corresponding to the highest flow rates as a function of the cumulative number of flow values looks like this:

par(mar=c(4.5,5.5,1,1))
plot(cumQSort,col="blue",type="l",cex=0.1,yaxt="n",
lty=ltyp[1],xaxt="n",xlab="",ylab="", ylim=c(0,100))
abline(h=0)
axis(2,cex.axis=ScaleF)
mtext("Cumulative number of flow values",side=1,line=1,cex=ScaleF)
mtext("%age of cumulative flow corresponding to\n the highest sorted flow rates ",side=2,line=3,cex=ScaleF)

 # the \n in the text allows for line break in the title

Notice that there is still no unit added for x axis because I decided that the cumulative number of flow value does not really add a lot to the analysis. However, it becomes very interesting to transform these values in probability of occurence. Each value has 1/N the probability to occur. We can also calculate the cumulative probability of occurence of flow values. Flow and load duration curves are thus derived this way.

In more details, the cumulative discharge calculated at each instantaneous flow rate can be calculated as a percentage of the total discharge yielding Wk% values corresponding to the kth cumulative probability and the time elapsed at each point can be calculated as a percentage of the total time. This works because even though flow rates are rearranged, the same amount of data points exist within the dataset with the same time increment occurring between each value. Wk% values can then be plotted as a function of the percentage of the total time. This is what is referred to as Flow Duration Curves. This provides a way of demonstrating of how relatively flashy the watershed may be, either relatively to other watersheds or to previous years.

The flashiness of a watershed refers to how rapidly flow is altered as a result of storm events/varying conditions. More frequent spikes in flow in response to precipitation events, in which flow increases and decreases more greatly and rapidly, are typically indicative of watersheds with predominant portions of streamflow being influenced by surface runoff, a quicker responding contributor of water to streamflow.

Wk=quantile(cumQSort,probs=seq(0.01,1,0.01))/tail(cumQSort,1) #Calculates and assigns Wk% values to a probability occurring in 1-100% of the total time
Mk=quantile(cumLSort,probs=seq(0.01,1,0.01))/tail(cumLSort,1)

## Flow and Load duration curves

This method allows us to see the percentage of the total discharge that occurs in a fraction of the total time with the lowest probabilities of occurrence corresponding with the highest flow rates associated with event flow. Therefore, if one watershed produces a majority of the total discharge in 50% of the time versus a watershed that produces a majority of the total discharge in 80% of the time, that watershed may be considered relatively flashier because a greater portion of the total discharge occurs in association with higher flow rates. In other words, streamflow would be considered more reactive to event water because the event hydrograph rises and recedes more quickly than the other watershed. This quick rise and recession allows for most flow to occur in a smaller percentage of the time versus the watershed that has a much wider event hydrograph spanning across a greater range of instantaneous flow values over a greater period of time. Visually, this method can provide a relative comparison of the flashiness of multiple watersheds. In the example above, the shape of the curve in the first watershed would have a greater slope towards the lower percentages/probabilities of occurrence and the curve for the second watershed would be somewhat flatter.

    xlim=c(0,100);ylim=c(0,100);
plot(1:100,Wk*100,xlab="Prob of Occurrence %",ylab="Mk% & Wk%",xlim=xlim,ylim=ylim,pch=21,col="black",bg="blue")
par(new=TRUE)
plot(1:100,Mk*100,xlab="Prob of Occurrence %",ylab="Mk% & Wk%",xlim=xlim,ylim=ylim,pch=22,col="black",bg=ColElmt)
par(new=TRUE)
abline(1,1,xlab="Prob of Occurrence %",ylab="Mk% & Wk",col="black",lty="dashed",xlim=xlim,ylim=ylim)
par(new=TRUE)
pch=c(19,19),
col=c("blue",ColElmt),
bg="white")
title(main="Nitrate Load and flow duration curves") # the \n in the text allows for line break in the title

    #Code for plotting the double cumulative plot using the normal distribution probabilities to zoom in on the lower tails
v<-c(1,2,3,5,10,25,50,75,90,95,97,98,99,100) #v is a set of user defined values of percentages of the total time to be used for calculating and plotting specific corresponding Wk% and Mk% values
x<-v/100  #Sets x as values of v in percentages
y<-Mk[v]  #Sets y as the set of Mk% values corresponding with percentages of the total time defined by values in x
xx<-qnorm(x) #qnorm takes a given probability (x values in this case) and returns the corresponding cumulative distribution value (Z-score) based on a normal distribution curve
yy<-qnorm(y) #Here, qnorm is taking the percentages of the load that occur associated with percentages of the total time defined by values in x and calculating associated Z-scores according to a normal distribution curve
yV<-Wk[v] #Sets yV as the set of Wk% values corresponding with percentages of the total time defined by values in x
yyV<-qnorm(yV) #qnorm takes percentages of the flow occurring associated with percentages of the total time defined by x values and assigns Z-score values to each probability based on the normal distribution curve
yy<-cbind(yy,yyV) #Binds both sets of Z-score values corresponding with Mk% (yy) and Wk% (yV)
xx<-cbind(xx,xx) #Binds both sets of Z-scores correspdonding with defined percentages of the total time (They are the same, this is for plotting purposes)
color<-c(ColElmt,"blue") #Defines the plot colors for yy and yV

for (i in 1:2){
plot(xx[,i],yy[,i],xlab="Probability of Occurrence (%)",ylab="Mk% & Wk%",type="o",xaxt="n",yaxt="n",xlim=c(-2.5,2.5),ylim=c(-2.5,2.5),pch=i,col=color[i])#Plots yy and yV values as a function of xx values
par(new=TRUE)
}

axis(1,at=xx[,1],labels=x*100)
axis(2,at=xx[,1],labels=x*100)
title(main="Double cumulative probability plot for nitrate load \nand for flow volume") # the \n in the text allows for line break in the title
abline(h=xx,lty=3,col="grey") #Plots gridlines
abline(v=xx,lty=3,col="grey")
bg="white")
It is helpful to use the qnorm function in R to zoom in on the very low and high probabilities of occurrence, or on the tails of the normal distribution curve.

format: plain


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## energy storage efficiency, $$\eta$$

https://doi.org/10.1351/goldbook.E02114
The rate of the Gibbs energy storage in an endothermic photochemical reaction divided by the incident irradiance.


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# How is the uniqueness theorem is used in this example?

I was going through example 2.10 in Griffiths(Introduction to Electrodynamics 4th edition), everything seemed okay, but at last he makes an assertion, the proof for which I am not able to understand.

Nope: As we'll see in the uniqueness theorems of Chapter 3, electrostatics is very stingy with its options; there is always precisely one way -- no more -- of distributing the charge on a conductor so as to make the field inside zero. Having found a possible way, we are guaranteed that no alternative exists, even in principle.

I searched for uniqueness theorems in Chapter 3, and I got two of them, but none of them seem to directly imply that there is always precisely one way- no more--of distributing the charge on a conductor so as to make the field inside zero.

First uniqueness theorem states that:

The solution to Laplace's equation in some volume $$V$$ is uniquely determined if $$V$$ is specified on the boundary surface S.

and the second one states that:

In a volume V surrounded by conductors and containing a specified charge density p, the electric field is uniquely determined if the total charge on each conductor is given (Fig. 3.6). (The region as a whole can be bounded by another conductor, or else unbounded.)

My question: How is the assertion made in example 2.10 justified by the uniqueness theorems?

A paraphrase of Aaron Stevens' answer.

• The second uniqueness theorem states:

In a volume $$V$$ surrounded by conductors and containing a specified charge density $$\rho$$, the electric field is uniquely determined if the total charge on each conductor is given. (The region as a whole can be bounded by another conductor, or else unbounded.)

• Apply this theorem for the cavity region.
The cavity as a whole is bounded by the conductor in question and there are no other conductors inside the cavity. The charge density inside the cavity is specified: $$\rho(\mathbf{r})=q\delta^3(\mathbf{r}-\mathbf{r}_0)$$ where $$\mathbf{r}_0$$ is the point where the charge $$q$$ is located.
The theorem states that there is a unique electric field $$\mathbf{E}(\mathbf{r})$$ inside the cavity.
• This electric field $$\mathbf{E}(\mathbf{r})$$ also uniquely determines the surface charge distribution on the boundary surface (wall of the cavity) using the following electrostatic boundary condition (ref. eq. (2.33) in Griffiths):
$$\mathbf{E}_{\text{above}}-\mathbf{E}_{\text{below}}=\frac{\sigma}{\epsilon_0}\hat{n}$$ "Above" and "Below" are with reference to which way the $$\hat{n}$$ points. Since the electric field inside the conductor is zero, $$\mathbf{E}(\mathbf{r'})=\frac{\sigma(\mathbf{r'})}{\epsilon_0} \hat{n}$$ where, $$\mathbf{r'}$$ is a point on the boundary surface and $$\hat{n}$$ is the inward pointing unit normal vector.
• One can see that the electric field inside the cavity and the surface charge distribution on the wall of the cavity are uniquely determined based on the shape of the cavity and the position of the charge inside the cavity alone. Everything else, such as the shape of the conductor that houses the cavity, is irrelevant.
• I think my answer is a paraphrase of yours. You have way more detail :) – BioPhysicist Feb 11 at 14:23

It relates to the second uniqueness theorem.

In a volume $$V$$ surrounded by conductors and containing a specified charge density $$\rho$$, the electric field is uniquely determined if the total charge on each conductor is given (Fig. 3.6). (The region as a whole can be bounded by another conductor, or else unbounded.)

Therefore, if you specify the charge on the conductor, you are guaranteed that there is only one unique field configuration (and hence potential function (up to a constant)). By Gauss's law, this then means that there is only one unique way to distribute the charges on the conductor.

Then jump back to chapter 2. Griffiths has found one solution that works. Since we are guaranteed there is only one unique solution by the uniqueness theorem, this is the only solution.

This is the whole idea behind the third chapter. As long as you find some way to come up with a solution, you know that it is the solution. Even if it seems like a convoluted method such as the method of images, getting an infinite sum of separable solutions, etc. you know you got it, and you don't need to keep looking.

• @SriramGoutamP I am not sure what you are confused about, because you seem to be answering your own question in your comment. We have "If the total charge is specified, then the field is uniquely determined". Therefore, when Griffiths has a scenario where the total charge is specified, then we know that the field is uniquely determined. – BioPhysicist Feb 11 at 3:24
• sorry I misread your answer. My confusion is about "Since we are guaranteed there is only one unique solution by the uniqueness theorem, this is the only solution". The second uniqueness theorem says about the uniqueness of electric field, whereas I think you are saying about uniqueness of charge distribution for that electric field. – Sriram Goutam P Feb 11 at 6:02
• @SriramGoutamP $\nabla\cdot E=\rho/\epsilon_0$ – BioPhysicist Feb 11 at 12:18

In a static situation, there can be no field inside a conductor. If there were, charges would move until there was no field. This means that every point within a conductor (including points on the surface surrounding a cavity) is at the same potential. With a fixed total charge, any rearrangement of the charge would change the potential at every point, and that would require a non-zero field. If there was a charge inside a cavity in the conductor, Gauss's law would require an equal and opposite charge on the inner surface of the conductor that surrounds the cavity.

• This doesn't address the question of the OP in addressing this in terms of uniqueness theorems – BioPhysicist Feb 10 at 16:56


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# Estimation, bias and mean squared error

1. Jan 13, 2012

### stukbv

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

(x1,x2,...xn) is modelled as observed values of independent random variables X1,X2,....Xn each with the distribution 1/θ for x in [0,θ] and 0 otherwise.
A proposed estimate of θ is m = max(x1,....xn) Calculate the distribution of the random variable M=max(X1,X2,....Xn) and considering M as an estimator for θ, its bias and mean squared error.

2. The attempt at a solution
P(max(X1,...Xn)≤m) = P(X1≤m)P(X2≤m)....P(Xn≤m)
via independent of the Xi's.

Then since they have the same distribution this is just
(m/θ)n

So to get the distribution do I just differentiate with respect to θ
Which would give me

n(m/θ)n-1 * (-m/(θ2))

Is this the right way to think about it ?

Thank you

Last edited: Jan 13, 2012
2. Jan 13, 2012

### Ray Vickson

You have the (cumulative) distribution function F(m) = Pr{M <= m}. How do you get the density function of M from that? There is a standard formula; you just need to use it.

RGV

3. Jan 14, 2012

### stukbv

I know that you differentiate to get the density function but I cant work out whether its with respect to theta (which is what I did above) or with respect to m?

4. Jan 14, 2012

### I like Serena

$f(m) = {d \over dm} F(m)$.

5. Jan 14, 2012

### Ray Vickson

The standard formula would tell you exactly what to do---no confusion!

RGV

6. Jan 14, 2012

### stukbv

I see thamk you!


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# No maverick elements in a free space

This keeps cropping up, and every time it takes me some effort to rediscover the proof. It happened first with free groups, then with free vector spaces and again with free Lie algebras.

I’ll formulate the issue with groups, but it is easy to carry it over to other cases.

A free group on set is defined by its universal property; furthermore, a particular specimen can be constructed. This construction is essential for proving the existence of a free group, but that should be its only role.

The question is: Is the subgroup of generated by necessarily itself? Can’t there be any “maverick” elements in , that are not in that subgroup? It seems that there shouldn’t be any, because they would be “too free”; when applying the universal property, the value given to them by the supposedly unique group morphism from to the “test group” might not be uniquely determined. But how do we prove this?

If we examine the constructed specimen ‑ in the case of free groups, the set of “strings” of elements of considered as “letters” plus their “primed versions”, with a certain composition rule ‑ the answer is that indeed, the whole of is generated by . And since all free groups on a given set are isomorphic, this answers our question.

But having to make use of a particular specimen doesn’t seem right to me. I want a proof with the universal property alone! Here it is, for my own repeated reference.

Let be a set, and any free group on . Let be the subgroup of generated by .

Now since, by definition of a generated subgroup, , is also a mapping from to ; or rather, we will call it as the codomain has changed. Thus is the mapping .

We thus have a group and a mapping ; the universal property of free groups satisfied by implies the existence of a group morphism such that .

Now can also be seen as a mapping from to ; or rather, since again we have changed the codomain, we define the new mapping . Since is a group morphism, so is .

Our equality  can be rewritten with the larger codomain , yielding . So is a group morphism from to such that . We know that there is only one such group morphism. Since we also have , necessarily .

For any , we have . Hence .


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# weibull survival function excel

Warning: this is a very technical, hands-on post.It turns out Weibull distribution is quite common among statistical distributions of lead times in software development and IT projects. The most accepted method for evaluation of Weibull Parameter is Maximum Likelihood Method. Time Variable(s) One (or two in the case of interval data) variable is … Although it has been replaced, the Weibull function is still available in Excel 2010 (stored in the list of compatibility functions), to allow compatibility with earlier versions of Excel. This insight belongs to… Weibull analysis works well, even with small samples (less than 20). The above chart on the right shows the Weibull Probability Density Function with the shape parameter, alpha set to 3 and the scale parameter, beta set to 1. It must be greater than or equal to zero. To get this, the value of cumulative should be true. The Weibull characteristic life parameter ($$\eta$$) estimate is 606.5280 and the shape parameter ($$\beta$$) estimate is 1.7208. The survivor function simply indicates the probability that the event of in-terest has not yet occurred by time t; thus, if T denotes time until death, S(t) denotes probability of surviving beyond time t. Note that, for an arbitrary T, F() and S() as de ned above are right con-tinuous in t. For continuous survival time T, both functions are continuous For more information about the new function, see the Weibull_Dist method. I want to fit to Weibull function S(t)=EXP(-alpha*Time^gamma). If θ 1 and θ 2 are the scale and shape parameters, respectively, then one may write α 0(t,θ) = θ 1θ 2tθ 2−1 or θθ 2 1 θ 2t θ 2−1 or θ 1t θ 2−1 or probably several other things. A description for the deletion of some functions is added. I just noted that in this definition of the Weibull function the role of alpha and beta is reversed compared to the Excel Weibull.dist() function. 4. If you get an error from the Excel Weibull function this is likely to be one of the following: (adsbygoogle = window.adsbygoogle || []).push({}); The value at which the function is to be calculated (must be ≥ 0). 1.3 Weibull Tis Weibull with parameters and p, denoted T˘W( ;p), if Tp˘E( ). Example 2 – Weibull Cumulative Distribution Function. short 6 minute step by step tutorial for using excel to determine weibull modulus for test data. Both are the parameters to the function. The above value calculates Weibull Cumulative Distribution. Weibull models are used to describe various types of observed failures of components and phenomena. This article describes the formula syntax and usage of the WEIBULL.DIST function in Microsoft Excel. =WEIBULL.DIST(x,alpha,beta,cumulative) The WEIBULL.DIST function uses the following arguments: 1. Although the function is new in Excel 2010, it is simply an updated version of the Weibull function, which isavailable in earlier versions of Excel. The Weibull shape parameter, called, indicates whether the failure rate is increasing, constant or decreasing. But I got a different alpha and gamma. The Excel WEIBULL function calculates the Weibull Probability Density Function or the Weibull Cumulative Distribution Function for a supplied set of parameters. It must be greater than 0. And Weibull distribution function in excel of two types: The only difference between the two types of Weibull distribution function is the cumulative logical argument. Hazard and survival functions for a hypothetical machine using the Weibull model. Not all data is normally distributed (i.e., bell-shaped). The Shape parameter to the distribution (must be > 0). Let's understand these arguments with the function syntax as explained below. Use this distribution in reliability analysis, such as calculating a device's mean time to failure. 3. Function Description The Excel Weibull.Dist function calculates the Weibull Probability Density Function or the Weibull Cumulative Distribution Function for a supplied set of parameters. The equation for the Weibull cumulative distribution function is: The equation for the Weibull probability density function is: When alpha = 1, WEIBULL returns the exponential distribution with: Example . The Weibull distribution is a probability distribution that is frequently used in engineering. A logical argument which denotes the type of distribution to be used; Occurs if any of the supplied arguments is non-numeric. The value above calculated is Weibull Probability Density. Since F(t) can be computed in Excel using the WEIBULL.DIST function, we can calculate MRL(t) in Excel using the following formula =α*EXP(GAMMALN(1+1/β))*(1-GAMMA.DIST((t/α)^β,1+1/β,1,TRUE))/ (1–WEIBULL.DIST(t,TRUE))-t. For those of you with a calculus background, Gamma Function Advanced explains all of this in more detail. Syntax. The chart on the right shows the the Weibull Cumulative Distribution Function with the shape parameter, alpha set to 5 and the scale parameter, beta set to 1.5. If we insert False at the cumulative, it gives us Weibull Probability Density Value. The Weibull shape parameter, called, indicates whether the failure rate is increasing, constant or decreasing. So excel provides the inbuilt statistical function to get weibull distribution function results. If I use a method tranforming probability to Ln(Ln(1/S(t)) which will be a linear function of Ln (t). UPDATE: The contents of this post are still valid, but there is a new, complementary post: How to Match to Weibull Distribution without Excel. We will use an example dataset from Stata’s data library. Parametric survival models or Weibull models A parametric survival model is a well-recognized statistical technique for exploring the relationship between the survival of a patient, a parametric distribution and several explanatory variables. It will complete the function name and enter the very first parenthesis. To use Weibull distribution, we need to have three values, which are X, Alpha, And Beta. Syntax: = WEIBULL.DIST (x,alpha,beta,cumulative) We have already learned that Weibull distribution is a continuous probability distribution. We have seen that Inserting True in the value of cumulative gives us Weibull Cumulative Distributive Value. Inverse Survival Function The formula for the inverse survival function of the Weibull distribution is Survival Function The formula for the survival function of the Weibull distribution is $$S(x) = \exp{-(x^{\gamma})} \hspace{.3in} x \ge 0; \gamma > 0$$ The following is the plot of the Weibull survival function with the same values of γ as the pdf plots above. Weibull Cumulative Distribution Function takes True as a cumulative argument, whereas Weibull Probability Density Function takes False as a cumulative argument. Let us go with the first example. Weibull Analysis Template in Excel Source : www.qimacros.com. This scenario is typical of "infant mortality" and indicates that the product is failing during its "burn-in" period. #NUM! They are widely used in reliability and survival analysis. However, you can use method of moments because it is in closed form, easy to use and reliable. Excel is among the most popular spreadsheet software on the marketplace today. X (required argument) – This is the value at which the function is to be calculated. : This error comes when the value of x is less than zero. Weibull survival function A key assumption of the exponential survival function is that the hazard rate is constant. The assumption of constant hazard may not be appropriate. The Weibull distribution is a special case of the generalized extreme value distribution.It was in this connection that the distribution was first identified by Maurice Fréchet in 1927. Weibull Analysis in Excel with QI Macros Use Weibull Analysis with Failure Rate Data. Estimate the survival function based on the coordinates from the previous step using a pre-built template. Over 1 Million Random Numbers, also available for 64bit version. Alpha And Beta, which are the parameter to the function the also need to be equals to or greater than zero. However, if backward compatibility is not required, you should consider using the new functions from now on, because they more accurately describe their functionality. Alpha & Beta are both the parameter of the function. A <1.0 indicates that the product has a decreasing failure rate. The Weibull probability density function is: where x is the independent variable, α is the shape parameter, and β is the scale parameter. WEIBULL.DIST Function in Excel. Excel Weibull distribution is widely used in statistics to obtain a model for several data sets, the original formula to calculate weibull distribution is very complex but we have an inbuilt function in excel known as Weibull.Dist function which calculates Weibull distribution. The syntax of the function is: WEIBULL (x, alpha, beta, cumulative) WEIBULL.DIST(x,alpha,beta,cumulative) 2013 by Statpoint Technologies, Inc. Weibull Analysis - 14 Survival Function The Survival Function plots the estimated probability that an item will survive until time t: Weibull Distribution 1000 10000 100000 Distance 0 0.2 0.4 0.6 0.8 1 y It decreases from 1.0 at to 0.0 at large values of X. In an example given above, the proportion of men dying each year was constant at 10%, meaning that the hazard rate was constant. Here we discuss how to use WEIBULL.DIST Function, along with practical examples and downloadable Excel templates. It allows us to estimate the parameters of the distribution. This is the probability that an individual survives beyond time t. This is usually the first quantity that is studied. This has been a guide to Weibull distribution in Excel. Weibull analysis is especially suited to failure rates (e.g., how long does a TV, PC, ball bearing or whatever operate before failing). #Value! Let us again use this function in excel. Beta (required argument) – This is the scale parameter to the Excel Weibull distribution and it must be greater than 0. The cumulative hazard is ( t) = ( t)p, the survivor function is S(t) = expf ( t)pg, and the hazard is (t) = pptp 1: The log of the Weibull hazard is a linear function of log time with constant plog + … STAT 525 Notes on the Weibull hazard and survreg in R There are quite a few ways to parameterize a Weibull hazard function. I have a survival dataset (see below). We have seen that X is valued at which we evaluate the function. Generate a Weibull curve that closely resembles the survival function and whose parameters can be easily incorporated into a simple three-state Markov model MOTIVATING EXAMPLE. By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy, Download Weibull Distribution Excel Template, New Year Offer - All in One Excel VBA Bundle (35 Courses with Projects) View More, You can download this Weibull Distribution Excel Template here –, All in One Excel VBA Bundle (35 Courses with Projects), 35+ Courses | 120+ Hours | Full Lifetime Access | Certificate of Completion. At first I tripped over this and couldnt understand why the calculated Fit is “so bad”, but after exchanging the values in Weibull.dist it now makes sense. Powerful NtRand3.2. It is the shape parameter to the distribution. Login details for this Free course will be emailed to you, This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy. Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. Standing on the shoulders of Giants : NtRand3.3 rises. A =1.0 indicates a constant failure rate. The regression allows for one potentially interval-censored and an arbitrary number of non-censored covariates. Let's understand how to use the WEIBULL.DIST function to work as an weibull distribution calculator. WEIBULL.DIST is a statistical function which returns the weibull distribution at a particular value. Achieving the precision of Excel 2010. NtRand 3.1 Ultimate Random Number Generator for Excel-Addin Just Released! This scenario is typical of "infant mortality" and indicates that the product is failing during its "burn-in" period. Excel Weibull distribution is widely used in statistics to obtain a model for several data sets, the original formula to calculate weibull distribution is very complex but we have an inbuilt function in excel known as Weibull.Dist function which calculates Weibull distribution. Alpha (required argument) – This is a parameter to the distribution. A <1.0 indicates that the product has a decreasing failure rate. Function Description The Excel WEIBULL function calculates the Weibull Probability Density Function or the Weibull Cumulative Distribution Function for a supplied set of parameters. X, which is a value to the function, is a non-negative number and cannot be zero, so it should be greater than zero. Computes estimators for the shape and scale parameter of the Weibull distribution, as well as for the vector of regression parameters in a parametric survival model with potentially right-censored time-to-event endpoint distributed according to a Weibull distribution. And Weibull distribution function in excel of two types: Weibull Cumulative Distribution Function; Weibull Probability Density Function; The only difference between the two types of Weibull distribution function is the cumulative logical argument, Weibull Cumulative Distribution Function takes True as a cumulative argument, whereas Weibull Probability Density Function takes False as a cumulative argument. It takes the value and the two parameters named alpha and beta along with the type of distribution required (cdf or pdf). 2. Next, we perform a regression analysis for a survival model assuming that failure times have a Weibull distribution. First, I’ll set up a function to generate simulated data from a Weibull distribution and censor any observations greater than 100. Cumulative (required argu… A=1.0 indicates a constant failure rate. Let us use this function in excel. WEIBULL.DIST Function in Excel WEIBULL.DIST is a statistical function which returns the weibull distribution at a particular value. As time goes on, it becomes more and more likely that the machine will fail … You may learn more about excel from the following articles –, Copyright © 2021. The Weibull cumulative distribution function is: In Excel 2010, the WEIBULL function has been renamed the WEIBULL.DIST function. Returns the Weibull distribution. The Scale parameter to the distribution (must be > 0). This function is still available for compatibility with earlier versions of Excel. CFA Institute Does Not Endorse, Promote, Or Warrant The Accuracy Or Quality Of WallStreetMojo. : This error comes if any arguments given are non-numeric. AS we know, X is valued at which we evaluate the function, Alpha & Beta. If I use Excel Solver, alpha=0.00367 and gamma=2.32. If you want to use Excel to calculate the value of this function at x = 2, this can be done with the Weibull function, as follows: For further information and examples of the Excel Weibull function, see the Microsoft Office website. I set the function up in anticipation of using the survreg() function from the survival package in R. The syntax is a little funky so … If you want to calculate the value of this function at x = 1, this can be done with the Excel Weibull function, as follows: The above chart on the right shows the Weibull Cumulative Distribution Function with the shape parameter, alpha set to 5 and the scale parameter, beta set to 1.5. Survival function, S(t) or Reliability function, R(t). Survival data are somewhat more difficult to enter because of the presence of various types of censoring. CFA® And Chartered Financial Analyst® Are Registered Trademarks Owned By CFA Institute.Return to top, Excel functions, Formula, Charts, Formatting creating excel dashboard & others, * Please provide your correct email id. Home » Excel-Built-In-Functions » Excel-Statistical-Functions » Excel-Weibull-Function. Of Excel we insert False at the cumulative, it gives us Weibull cumulative distribution is. At which we evaluate the function the also need to be calculated are both the parameter to distribution... May learn more about Excel from the previous step using a pre-built template required argument ) – this the... Required ( cdf or pdf ) and Beta, which are x, alpha, Beta... Regression analysis for a supplied set of parameters Giants: NtRand3.3 rises Probability that an individual survives time... Hazard may not be appropriate previous step using a pre-built template is.! Number of non-censored covariates, and paste it in cell A1 of a new worksheet! 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text:
How do you write 54,000,000 in scientific notation?

Mar 25, 2018

$5.4 \cdot {10}^{7}$

Explanation:

We have $54 000 000$
It can be translated as $5.4 \cdot 10000000$
Which is $5.4 \cdot 10000 \cdot 1000$
A power of $10$ is $10$ at the power of how many $0$s we have, if that makes sense.

So $1000$ is ${10}^{3}$ since we have 3 $0$s.
$10000$ is ${10}^{4}$ because there are 4 $0$s.
So we have $5.4 \cdot {10}^{4} \cdot {10}^{3} = 5.4 \cdot {10}^{4 + 3} = 5.4 \cdot {10}^{7}$


text:
# [Tex/LaTex] Smaller font size of the algorithm description

algorithmsfontsize

\small or other tag seems can not change the font size of the line number.

The package I use is \usepackage{algorithmic}.

Use \algsetup{linenosize=\small}. See p. 12 of the documentation of the algorithms bundle for details.

\documentclass{article}

\usepackage{algorithmic}
\algsetup{linenosize=\small}

\begin{document}

\begin{algorithmic}[1]
\IF{some condition is true}
\STATE do some processing
\ELSE
\STATE do the default actions
\ENDIF
\end{algorithmic}

\end{document}


text:
# What is the opposite of -49?

May 24, 2018

49

#### Explanation:

The opposite of a negative number would be its positive number, which in this case, is 49

May 24, 2018

The opposite of $- 49$ is its additive inverse $49$.

#### Explanation:

The sum of a number and its additive inverse is $0$.

The additive inverse property is represented as:

$\textcolor{b l u e}{a} + \textcolor{g r e e n}{- a} = \textcolor{p u r p \le}{0} = \textcolor{g r e e n}{- a} + \textcolor{b l u e}{a}$


text:
Find the sum

Algebra Level 3

$\large 7, 77, 777, 7777, 77777,\ldots$ What is the sum of the first 13 terms of the above sequence?

×


text:
User david farmer - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T06:02:51Z http://mathoverflow.net/feeds/user/19964 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/127550/hejhals-algorithm-and-computational-methods-for-non-classical-maass-wave-forms/127996#127996 Answer by David Farmer for Hejhal's algorithm and computational methods for non-classical Maass wave forms David Farmer 2013-04-18T17:46:50Z 2013-04-18T17:46:50Z
I don't think it is practical to directly find higher rank Maass forms along the lines of what Hejhal did, because they are functions of several variables and their Fourier expansions involve multiple sums. Not to mention the need to implement the appropriate special functions that appear. Even if you have the Fourier expansion on the computer (which has been done for SL(3,Z) (by Boris Mezhericher) but not Sp(4,Z)), using the relations to make equations in the coefficients and/or the analog of integrating along horocycles is probably computationally prohibitive.
 http://mathoverflow.net/questions/127851/central-critical-special-values-of-l-functions-terminology central/critical/special values of L-functions terminology David Farmer 2013-04-17T14:06:02Z 2013-04-17T16:44:54Z
I have a question about the terminology for special values of L-functions. Is the following a correct description of standard usage:

Suppose L(s) is an L-function which satisfies a functional equation relating $s$ to $w+1-s$, where $w$ is the (motivic) weight. ADDED LATER: I am assuming the L-function is motivic, otherwise (please correct me if I am wrong) there is nothing special about the value at any integer.

1) If $m$ is an integer then $L(m)$ is a special value of the L-function.

2) If $m$ is an integer and neither $m$ nor $w+1-m$ is a pole of a $\Gamma$-factor of the L-function, then $m$ is a critical point and $L(m)$ is a critical value of the L-function.

3) $L(\frac{w+1}{2})$ is the central value of the L-function.

4) If $\frac{w+1}{2}$ is not an integer, then the central value is not a special value.

I am pretty sure 2) is correct, unless Deligne's notion of critical point is not the only one. I am also pretty sure 3) is correct, since the central point of the functional equation is pretty unambiguous. It is 1) and 4) that I am hoping the experts can clarify.
 http://mathoverflow.net/questions/124572/is-there-an-explicit-expression-for-the-imaginary-part-of-some-non-trivial-zero-o/124885#124885 Answer by David Farmer for Is there an explicit expression for the imaginary part of some non-trivial zero of zeta, David Farmer 2013-03-18T18:16:27Z 2013-03-20T15:25:14Z
Write $\rho = \frac12 + i \gamma$ for a nontrivial zero of a primitive L-function. ("Primitive" means that it can't be written as the product of other L-functions.)

It is generally believed that:

a) If $\gamma\not=0$ then $\gamma$ is transcendental.

b) If $\gamma\not=0$ then $\gamma$ is algebraically independent of every well-known constant and every other zero of every primitive L-function (except when the L-function has real coefficients, in which case $\frac12 - i \gamma$ is also a zero).

As far as I know, nobody has any clue how to prove these conjectures.

Clarification added later: what definition of L-function are we using?

Greg Martin's comment (below) refers to $L(s+ i y)$ where $L(s)$ is an L-function and $y$ is real. While it is true that for some definitions of "L-function" the set of L-functions is closed under that operation, that is not what I intended.

For the L-functions in my answer above, the Euler product axiom can be written as:

 There is a Dirichlet character $\chi$, the "central character" of the L-function, such that $$L(s)= \prod_{p \, {\rm prime}} F_p(p^{-s})^{-1},$$ where $F_p$ is a polynomial of the form $$F_p(z)=1-a_p z + \cdots + (-1)^d\chi(p) z^d .$$

Here $d$ is the degree of the L-function. Note that I have normalized the L-function so that the functional equation relates $s$ to $1-s$.

All known L-functions satisfy that axiom, and this formulation tells you how to select the distinguished member of the family $L(s+i y)$.
 http://mathoverflow.net/questions/122891/small-index-subgroups-of-sl3-z Small index subgroups of SL(3,Z) David Farmer 2013-02-25T14:58:19Z 2013-02-28T19:43:36Z
I would like to know the smallest index subgroups of SL(3,Z).

The smallest I could find has even entries $a_{3,1}$ and $a_{3,2}$, along the bottom row. I could not figure out whether there are subgroups of index 2 or 3.

A search found lots of information about SL(2,Z) but not SL(3,Z).
 http://mathoverflow.net/questions/90577/are-there-known-non-real-zeros-of-derivatives-of-riemann-zeta-with-0-res-1/91824#91824 Answer by David Farmer for Are there known non-real zeros of derivatives of Riemann zeta with 0 < Re(s) < 1/2? David Farmer 2012-03-21T13:37:17Z 2012-03-21T13:37:17Z
Mathematica claims that the 5th derivative of the Riemann zeta function has a zero at approximately

0.2876 + 4.6944 i.

I don't think it should be too hard to resolve the case of the next several derivatives. The techniques in the paper referenced by you or by Micah Milinovich should let you find an explicit upper bound on the imaginary part of the nonreal zeros of $\zeta^{(k)}(s)$ in the left half-plane. Then it is just a numerical calculation to find all the zeros below that bound.
 http://mathoverflow.net/questions/75960/distinct-simple-zeros-of-dirichlet-l-functions/90061#90061 Answer by David Farmer for Distinct simple zeros of Dirichlet L-functions David Farmer 2012-03-02T17:53:17Z 2012-03-02T17:53:17Z
You probably guessed this from the lack of responses, but the answer is that there is no hope of making progress on this question using current methods.

One measure of the complexity of an L-function is its degree, where the Riemann zeta function and Dirichlet L-functions have degree 1, the L-function of a holomorphic cusp form has degree 2, the standard L-function of a GL(n) automorphic form has degree n, etc. The precise definition of degree is the number of $\Gamma$-factors in the functional equation, where a $\Gamma(s+A)$ counts as two $\Gamma$-factors.

These days we have very few tools for dealing with degree 3 and higher. Your product of Dirichlet L-functions is like one degree $r$ L-function, and so you are stuck once $r$ is bigger than 2. In particular, the fact it is a product doesn't seem to help much.

It also doesn't help that you want only one simple zero, unless you have a particular explicit case in mind, in which case you can show it by direct calculation.
 http://mathoverflow.net/questions/127550/hejhals-algorithm-and-computational-methods-for-non-classical-maass-wave-forms/127996#127996 Comment by David Farmer David Farmer 2013-04-19T10:05:19Z 2013-04-19T10:05:19Z I think the paper on his home page is the same as the one he posted to the ArXiv: sites.google.com/site/mboris http://mathoverflow.net/questions/127851/central-critical-special-values-of-l-functions-terminology/127860#127860 Comment by David Farmer David Farmer 2013-04-17T16:41:15Z 2013-04-17T16:41:15Z I wrote the question in the "arithmetic" normalization, with $s$ going to $w+1-s$ in the functional equation. I know that "most" L-functions are not motivic, but my understanding is that the concept of special/critical values only makes sense for motivic L-functions. You wouldn't say, for example, that every integer is a critical point for the L-function of a Maass form. There are times when you want to normalize the L-function so that $s$ goes to $1-s$, but this is not one of those times. http://mathoverflow.net/questions/124572/is-there-an-explicit-expression-for-the-imaginary-part-of-some-non-trivial-zero-o/124885#124885 Comment by David Farmer David Farmer 2013-03-21T19:26:49Z 2013-03-21T19:26:49Z Greg, Not even close: we can't even prove that for degree 2. It is not even known that the members of the Selberg class, of degree 2 or more, have Euler products whose local factors are the reciprocal of a polynomial. (I saw something along those lines -- I forget the details -- from Kaczorowski and Perelli, but what I wrote is definitely true for "degree more than 2.") http://mathoverflow.net/questions/124572/is-there-an-explicit-expression-for-the-imaginary-part-of-some-non-trivial-zero-o/124885#124885 Comment by David Farmer David Farmer 2013-03-18T20:28:12Z 2013-03-18T20:28:12Z Paul, It is conjectured that the factorization of L-functions into primitive L-functions is unique. This follows from the Selberg Orthonormality Conjecture. Unique factorization implies that you can't "split up" the cancellation of zeros of one L-function by part of the zeros of several different L-functions. So (assuming appropriate conjectures) the Artin L-function can only be entire if there is a cancellation of actual L-functions in the numerator and denominator. http://mathoverflow.net/questions/124572/is-there-an-explicit-expression-for-the-imaginary-part-of-some-non-trivial-zero-o/124885#124885 Comment by David Farmer David Farmer 2013-03-18T19:06:59Z 2013-03-18T19:06:59Z If an Artin L-function was genuinely a ratio, and not a product, of other L-functions (in other words, if it had infinitely many poles) then it does not deserve to be called an L-function! For this discussion I think it makes sense to assume all reasonable conjectures. http://mathoverflow.net/questions/122891/small-index-subgroups-of-sl3-z/123108#123108 Comment by David Farmer David Farmer 2013-02-28T18:59:32Z 2013-02-28T18:59:32Z Great answer. I wish I could "accept" two answers. http://mathoverflow.net/questions/122891/small-index-subgroups-of-sl3-z/122907#122907 Comment by David Farmer David Farmer 2013-02-25T20:55:42Z 2013-02-25T20:55:42Z A step I am missing is why the subgroup of $SL(3,Z)$ has to be the pullback of a subgroup of $SL(3,n)$. Is maximality being used? If $G$ is the subgroup, I get that there exists $H<G$ so that $H$ is the matrices congruent to the identity mod $n$, for some integer $n$. Probably I am missing something easy for the next step.


text:
# A tough integral : $\int_{0}^{\infty }\frac{\sin x \text{ or} \cos x}{\sqrt{x^{2}+z^{2}}}\ln\left ( x^{2}+z^{2} \right )\mathrm{d}x$

Recently, I found these two interesting integrals in Handbook of special functions page 141. $$\mathcal{I}=\int_{0}^{\infty }\frac{\sin(ax)}{\sqrt{x^{2}+z^{2}}}\ln\left ( x^{2}+z^{2} \right )\mathrm{d}x$$ $$\mathcal{J}=\int_{0}^{\infty }\frac{\cos(ax)}{\sqrt{x^{2}+z^{2}}}\ln\left ( x^{2}+z^{2} \right )\mathrm{d}x$$ In this book, it gives the answer below $$\mathcal{I}=\frac{\pi }{2}\left (\ln\frac{z}{2a}-\gamma \right )\left [ I_0\left ( az \right )- \mathbf{L}_0\left ( az \right )\right ]+\frac{1}{4\pi }G_{24}^{32}\left ( \frac{a^{2}z^{2}}{4}\middle|\begin{matrix} \dfrac{1}{2},\dfrac{1}{2} \\ 0,0,\dfrac{1}{2},\dfrac{1}{2} \end{matrix} \right )~~~,~~~\left (a,\Re z>0 \right )$$ $$\mathcal{J}=\left ( \ln\frac{z}{2a}-\gamma \right )K_0\left ( az \right )~~~,~~~\left ( a,\Re z>0 \right )$$ where $I_0(\cdot)$ is modified bessel function of the first kind, $\mathbf{L}_0(\cdot)$ is modified struve function, $G_{pq}^{mn}(\cdot)$ is meijer-G function and $K_0(\cdot)$ is bessel function of rhe second kind.

So, I tried to figure out how to get the answer.

My attempt: Let $x=z\tan t$, we have \begin{align*} \mathcal{I}&=2\int_{0}^{\frac{\pi }{2}}\sin\left ( az\tan t \right )\ln\left ( z\sec t \right )\sec t\, \mathrm{d}t\\ &=2\int_{0}^{\frac{\pi }{2}}\sin\left ( az\tan t \right )\ln\left ( \sec t \right )\sec t\, \mathrm{d}t+2\ln z\int_{0}^{\frac{\pi }{2}}\sin\left ( az\tan t \right )\sec t\, \mathrm{d}t \end{align*} Hence, define $$\mathcal{I}\left ( m \right )=\int_{0}^{\frac{\pi }{2}}\sin\left ( az\tan t \right )\sec^mt\, \mathrm{d}t$$ then using the taylor series of $\sin x$ we get \begin{align*} \mathcal{I}\left ( m \right )&=\sum_{k=0}^{\infty }\left ( -1 \right )^{k}\frac{\left ( az \right )^{2k+1}}{\left ( 2k+1 \right )!}\int_{0}^{\frac{\pi }{2}}\tan^{2k+1}t\sec^mt\, \mathrm{d}t \\ &=\sum_{k=0}^{\infty }\left ( -1 \right )^{k}\frac{\left ( az \right )^{2k+1}}{\left ( 2k+1 \right )!}\int_{0}^{\frac{\pi }{2}}\sin^{2k+1}t\cos^{-2k-m-1}t\, \mathrm{d}t \end{align*} By using the same way we get \begin{align*} \mathcal{J}\left ( m \right )&=\sum_{k=0}^{\infty }\left ( -1 \right )^{k}\frac{\left ( az \right )^{2k}}{\left ( 2k \right )!}\int_{0}^{\frac{\pi }{2}}\tan^{2k}t\sec^mt\, \mathrm{d}t \\ &=\sum_{k=0}^{\infty }\left ( -1 \right )^{k}\frac{\left ( az \right )^{2k}}{\left ( 2k \right )!}\int_{0}^{\frac{\pi }{2}}\sin^{2k}t\cos^{-2k-m}t\, \mathrm{d}t \end{align*} But how to evaluate the last integral, it seems can't be expressed by Beta function.

If I'm doing the wrong way, is there another way to solve the problem.

Any help will be appreciated!

• perhaps $\sin^2(t) \to \rho$ and $\cos^2(t) \to (1-\rho)$ will help – Francesco Alem. Jan 17 '17 at 8:53
• the first integral with the Meijer-G isn't really appealing to me – tired Jan 17 '17 at 8:55
• For the second integral, have you tried something along this lines? math.stackexchange.com/questions/1954866/… – tired Jan 17 '17 at 9:45
• Furthermore the prefactor in front of $K_0(z)$ looks suspicious like its small $z$ expansion – tired Jan 17 '17 at 9:58
• And I was thinking $K_0$ is the elliptic integral of the first kind. – Zaid Alyafeai Jan 17 '17 at 10:11

For the second integral

Note that

$$K_\nu(az)=\frac{\Gamma(\nu+1/2)(2z/a)^\nu}{\sqrt{\pi}}\int_0^\infty\frac{\cos at }{(t^2+z^2)^{\nu+1/2}} dt$$

By differentiation with respect to $\nu$

\begin{align} \frac{\partial K_\nu(az)}{ \partial \nu} &=(\Gamma'(\nu+1/2)+\log(2z/a) )K_\nu(az)\\&-\frac{\Gamma(\nu+1/2)(2z/a)^\nu}{\sqrt{\pi}}\int_0^\infty\frac{\cos a t }{(t^2+z^2)^{\nu+1/2}} \log(x^2+z^2)dt \end{align}

Note that in

$$\left|\frac{\partial K_\nu(z)}{ \partial \nu} \right|_{\nu=0} = 0$$

This implies

$$\int_0^\infty\frac{\cos a t }{\sqrt{t^2+z^2)}} \log(x^2+z^2)\,dt = (\Gamma'(1/2)+\log(2z/a) )K_0(az)$$

Note that

$$\Gamma'(1/2)+\log(2z/a) = -\gamma -2\log(2)+\log(2)+\log(z/a) =\log(z/2a) -\gamma$$

Hence

$$\mathcal{J}=\left ( \log (z/2a)-\gamma \right )K_0\left ( az \right )$$

$$K_\nu (z) = \int^\infty_0 e^{-z\cosh t} \cosh(\nu t)\,dt$$

This implies

$$\frac{\partial K_\nu(z)}{ \partial \nu} = \int^\infty_0t e^{-z\cosh t} \sinh(\nu t)\,dt$$

Hence we have

$$\left|\frac{\partial K_\nu(z)}{ \partial \nu} \right|_{\nu=0} = 0$$

• uhhh, nice i didn't know this particular integral rep.... an additional proof of your last statement would be awesome! – tired Jan 17 '17 at 10:43
• Thx! But for the first one, it seems more difficult! Does it have a similar method like this? – Renascence_5. Jan 17 '17 at 10:59
• @tired, see the addendum. – Zaid Alyafeai Jan 17 '17 at 11:28
• @Renascence_5 , I think we can apply it to the first integral but this will be tedious. – Zaid Alyafeai Jan 17 '17 at 11:29


text:
## Tokyo Journal of Mathematics

### $S^{1}$-equivariant CMC-hypersurfaces in the Hyperbolic 3-space and the Corresponding Lagrangians

Keiichi KIKUCHI

#### Abstract

A family of $S^{1}$-equivariant hypersurfaces of constant mean curvature can be obtained by using the Lagrangians with suitable potentials in the hyperbolic 3-space. The conservation law is effectively applied to the construction of $S^{1}$-equivariant hypersurfaces of constant mean curvature in the hyperbolic 3-space.

#### Article information

Source
Tokyo J. Math., Volume 36, Number 1 (2013), 207-213.

Dates
First available in Project Euclid: 22 July 2013

https://projecteuclid.org/euclid.tjm/1374497520

Digital Object Identifier
doi:10.3836/tjm/1374497520

Mathematical Reviews number (MathSciNet)
MR2976548

Zentralblatt MATH identifier
1277.53054

#### Citation

KIKUCHI, Keiichi. $S^{1}$-equivariant CMC-hypersurfaces in the Hyperbolic 3-space and the Corresponding Lagrangians. Tokyo J. Math. 36 (2013), no. 1, 207--213. doi:10.3836/tjm/1374497520. https://projecteuclid.org/euclid.tjm/1374497520

#### References

• J. Eells, The surfaces of Delaunay, Math. Intell. 9 (1987), 53–57.
• J. Eells and A. Ratto, Harmonic maps and minimal immersions with symmetries, Ann. Math. Studies 130, 1993.
• E. V. Ferapontov and A. P. Fordy, Separable Hamiltonians and integrable systems of hydrodynamic type, J. Geometry and Physics 21 (1997), 169–182.
• D. Ferus and F. Pedit, $S^{1}$-equivariant Minimal Tori in $S^{4}$ and $S^{1}$-equivariant Willmore Tori in $S^{3}$, Math. Z. 204 (1990), 269–282.
• W-Y. Hsiang, On generalization of theorems of A. D. Alexandrov and C. Delaunay on hypersurfaces of constant mean curvature, Duke Math. J. 49 (1982), 485–496.
• W-Y. Hsiang and H. B. Lawson, Minimal submanifolds of low cohomogeneity, J. Diff. Geom. 5 (1971), 1–38.
• K. Kenmotsu, Kyokumenron Kougi (Japanese), Baifukan, 2000.
• K. Kikuchi, The construction of rotation surfaces of constant mean curvature and the corresponding Lagrangians, Tsukuba J. Math. Vol. 36, No. 1 (2012), 43–52.
• T. Otsuki, Minimal hypersurfaces in a Riemannian manifold of constant curvature, Amer. J. Math. 92 (1970), 145–173.
• C. Von Westenholz, Differential forms in Mathematical Physics, Chap. 12 Hamiltonian mechanics and geometry, Series Studies in Mathematics and its Applications, Vol. 3, North-Holland, 1981.


text:
# custom slug for custom post type

I want to insert the post author into my custom post type slug.

Example: http://example.com/charts/%author%/

Any ideas how to accomplish this?

Here is my custom post type:

register_post_type('charts', array(
'label' => 'Charts',
'description' => '',
'public' => true,
'show_ui' => true,
'capability_type' => 'post',
'hierarchical' => false,
'rewrite' => array('slug' => '/charts/author'),
'query_var' => true,
'supports' => array(
'title',
'editor',
'custom-fields',
'author',
),
'labels' => array (
'name' => 'Charts',
'singular_name' => 'Charts',
),
));

bless jnz

-

I figured out a solution and decided I'd share because its nice to be nice. This works for me and is based on a solution by Jonathan Brinley. If anyone has any suggestions or corrections please feel free to let me know.

First, create your custom post type and set it up like this (this is just an example, remember to make it fit your own needs. The slug setting is important!)

register_post_type('charts', array(
'label' => 'Whatever',
'description' => '',
'public' => true,
'show_ui' => true,
'capability_type' => 'post',
'hierarchical' => true,
'rewrite' => array('slug' => '/whatever/%author%'),
'query_var' => true,
'supports' => array(
'title',
'editor',
'custom-fields',
'author'
)
));

Next, setup a function for your filter (in functions.php):

function my_post_type_link_filter_function($post_link,$id = 0, $leavename = FALSE) { if (strpos('%author%',$post_link) === FALSE) {
$post = &get_post($id);
$author = get_userdata($post->post_author);
return str_replace('%author%', $author->user_nicename,$post_link);
}
}

Then activate the filter (also in functions.php):

add_filter('post_type_link', 'my_post_type_link_filter_function', 1, 3);

Like I said, I'm not sure this is the best way to do this but it works for me :)

-
That’s an interesting solution. Please mark your question as answered. –  toscho Feb 25 '12 at 15:18


text:
# Ordinal corresponding to well-quasi-order on graphs

Let $K$ be an infinite cardinal. Then, by the Robertson–Seymour theorem, the set of graphs with fewer than $K$ vertices and edges form a well-quasi-order.

In terms of $K$, what is the maximal order type of this well-quasi-order?

(The maximal order type of a well-quasi-order $(X,\le_X$) is the supremum of the ordinals that embed into $\le_X$.)

• This question rests on a false premise: the second sentence in the OP is false: by [Robin Thomas, A counter-example to 'Wagner' conjecture' for infinite graphs, Math. Proc. Camb. Phil. Soc. (1988), 103, 55-57], if $K=2^{\aleph_0}$, then your set of graphs $G$ with at most $K$ vertices is not a well-quasi-order. The question will have to be corrected. I don't know how though. An obvious idea would be to impose an upper bound on $K$, but making it $K<2^{\aleph_0}$ would be rather uninformative, for reasons related to forcing and the unnknown size of the cardinality of the continuum. – Peter Heinig Feb 22 '18 at 11:27
• Also worth pointing out: to say "with fewer than $K$ vertices and edges" is redundant, because by basic cardinal arithmetic, a graph having at most $K$ vertices, necessarily has at most $K$ edges. – Peter Heinig Feb 22 '18 at 13:28


text:
## Hydrogen

Henry Cavendish discover hydrogen in 1766 by the action of dilute H2O4 on iron and named ‘inflammable air’, Lavoisier gave the name hydrogen (Creek: Hydra = water, gennas = producer]. These occurs in free state as well as in the combined state.

Lavoisier name hydrogen. Hydrogen is the 9th most abundant element in the earth’s crust.

Hydrogen exists in diatomic state but in the triatomic state it is Hyzone. Systematic name of water is oxidane.

### Position of Hydrogen in the Periodic Table:

Hydrogen resembles with alkali metals (group I) as well as halogens (group 17), At the same time, it differs from both in certain characteristics. That is why hydrogen if; called “rogue element”.

However. it has been placed in group 1 on the basis of its configuration 1s1, which is the basis of modern classification of elements.

• It forms monovalent H+ ion like Li+, Na+
•  valency is also 1.
• oxide (H2O) is stable as Li2O, Na2O.
• good reducing agent (In atomic as well as molecular state) like Na, Li…
• Hydrogen also resembles halogens (Group VII A)
• diatomic (H2)like F2, Cl2

#### Isotopes of Hydrogen

 Isotopes Relative atomic  mass Density Nature Protium $${H^1_1}$$ 1.0078 0.09 Non Radioactive Deuterium $${H^2_1}$$ 2.0141 0.18 Non Radioactive Tritium $${H^3_1}$$ 3.016 0.27 Radioactive

## Dihydrogen [H2]

Methods of Preparation:

### a) Lab methods

i) Zn + H2SO4 (aq) → ZnSO4 (aq) + H2 (g)

Metals which have reduction potential lesser than H, can liberate H2 from acids.

• Pure zinc is not used because it reacts slowly. The presence of some impurities increases the rate of reaction due to the formation of electrochemical couples.
• Conc sulphuric acid is also not used because it oxidises, H2 formed into H2O.

Zn + 2H2SO4 (conc.) → ZnSO4 + SO2 + 2H2O

ii) The reaction of zinc with aqueous alkali.

Zn + NaOH $$\overset{\Delta}{\rightarrow}$$ Na2ZnO2 + H2

#### b) Commercial production of dihydrogen

i) By the electrolysis of acidified water

 Electrode H2O ⇌ H+  +  OH– Ionisation At Cathode, H+  +  e– → H• Reduction H•  +  H• → H2 At Anode, 4OH– → 4OH  +  4e– Oxidation 4OH → 2H2O +  O2

ii) From water gas (Bosch process)

CO + H2 + H2O  $$\overset{Fe_2O_3 + Cr_2O_3}{\longrightarrow}$$ CO2  + O2

Carbon monoxide + hydrogen gas makes (Water-gas) +(Steam).Above reaction occurs in 773K

Carbon dioxide comes out by dissolving it in water under pressure (20-25 atm) and hydrogen left behind is obtain.

iii)  From steam (Lane’s process) Superheated steam is passed over iron filings heated to about 1023-1073 K when hydrogen is formed.

3Fe + 4H2O (Steam) $$\overset{1023 – 1073 K}{\rightarrow}$$ Fe3O4  + H2O

iv) Highly pure (> 99.95%)dihydrogen is obtained by electrolysing warm aqueous barium hydroxide solution between nickel electrodes.

v) From hydrocarbons by partial oxidation

CH4  + H2O $$\overset{Ni- Cr Catalyst}{\longrightarrow}$$ CO + 3H2

CH4  + H2O ( Natural gas + Steam). Above reaction occurs in (1270K)

vi) It is also obtained as a by-product in the manufacture of NaOH and chlorine by the electrolysis of brine solution.

During electrolysis, the reactions that take place are:

At Anode,  2Cl(aq) → Cl2(g) + 2e

At Cathode, 2H2O(l) +2e– →  H2(g) + 2OH(aq)

The overall reactions by separating Na+ ions,

2Na+(aq) + 2Cl(aq) + 2H2O(l) → Cl2(g) + H2(g)+ 2Na+ + 2OH (aq)

## Physical Properties of Dihydrogen

Dihydrogen is a colourless, odourless, tasteless, combustible gas. It is lighter than air and insoluble in water. It is neutral to litmus.

### Chemical Properties of Dihydrogen

i) Reactivity: The relative inertness of dihydrogen at room temperature is because of Its high enthalpy of H-H bond i.e.. high bond dissociation energy. So its reactions take place under specific conditions only (at high temperature).

ii) Action with non-metals:

2H2 + O2 $$\overset{970K}{\longrightarrow}$$ 2H2O(l); $$\Delta$$Ho = -285.9 kJ mol-1

N2 + 3H+ $$\overset{673K/200atm}{\longrightarrow}$$ 2NH3(g);$$\Delta$$Ho = -92.6kJ mol-1

H + X2 $$\overset{Dark}{\longrightarrow}$$ 2HX (X = halogens)

Order of reactivity of halogens:

F2 > Cl2 > Br2 > I2

iii) Reaction with metals Here H2 acts as oxidising agent.

2Na + H$$\overset{\Delta}{\rightarrow}$$ 2NaH

Ca + H $$\overset{\Delta}{\rightarrow}$$ CaH(Hydrolith)

iv) Reducing action of dihydrogen:

CuO + H$$\overset{\Delta}{\rightarrow}$$ Cu + H2O

v) Reactions with metal ions and metal oxides:

H2(g) + Pd+2(aq) → Pd(s) + 2H+(aq)

yH2(g) + MxOy(s) → xM(s) + yH2O(l)

vi) Reaction with organic compounds:

a)Veg.oil + H2  $$\overset{Ni/400K}{\longrightarrow}$$ Veg.ghee

b) R—CH = CH2 + H+CO  $$\overset{[Co(CO_4)_2]}{\longrightarrow}$$

RCH2CH2CHO + H $$\overset{Ni}{\longrightarrow}$$ RCH2CH2CH2OH

#### Uses of Dihydrogen

• In the manufacture of CH3OH.
• CO(g) + 2H(g)  $$\overset{Co}{\rightarrow}$$ CH3OH(l)
• It produces a temperature of 2850°C and oxy-atomic hydrogen flame produces a temperature of 4000°C, so it is used in oxy-hydrogen flame.
• The largest single use of H2 is in the synthesis of NH3 which is used in the manufacture ofHNO3 and fertilizers.
• We use liquid hydrogen (LH2) in fuel
• H2 is used as a reducing agent in extraction of metals.
• H2 is used in fuel cell for generating electrical energy.
• We is used in the manufacture of synthetic petrol.

(By heating H2 with coal and heavy oils under very high pressure in the presence of catalyst.)

## Different Forms of Hydrogen

### Atomic Hydrogen

H$$\overset{Ni/400K}{\rightarrow}$$ 2H $$\Delta$$Ho = 105.4 kcal mol-1

It is very reactive and its half-life period is 0.33 s.

#### Nascent Hydrogen

Freshly prepared hydrogen is known as nascent hydrogen and is more reactive than ordinary hydrogen. It causes the reduction of certain compounds which IS not possible with ordinary hydrogen. It can never be isolated.

Zn + H2SO4  → ZNSO4  +  2[H]

Adsorption of hydrogen at the metal surface is called occlusion. This hydrogen brings out many chemical changes such as reduction and hydrogenation. Occlusion decreases with rise in temperature.

###### Ortho and Para Hydrogen

When in hydrogen molecule, the nuclear spins are in the same direction, it is known as ortho hydrogen. On the other hand when the nuclear spins arc in the opposite direction. it is known as para hydrogen. At room temperature hydrogen consists of 75% ortho and 25% para hydrogen.

## Hydrogen Economy

The hydrogen economy is the use of liquid hydrogen as an alternative source of energy. The technology involves the production, transportation and storage of energy in the form of liquid hydrogen. Large scale production of hydrogen can be done by electrolysis of water or by thermochemical reaction cycle. Storage of hydrogen in liquid form can be done in vacuum insulated cryogenic tanks or in a metal or in all alloy like iron-titanium alloy as an interstitial hydride. Hydrogen fuel has many advantage over conventional fuels in that it is non-polluting and it liberates a large amount of energy on combustion.

Photohydrogen is used to obtain renewable energy from sunlight by using microscopic organism such as bacteria or algae.

## Hydrides

The compounds of hydrogen with metals and non-metals are called hydrides.

### Ionic Hydrides

These are formed by elements of group I, II, (except Be and Mg) by heating them in hydrogen.

These are white colourless solids (crystalline) having high m.p. and b.p. easily decomposed by water, CO2 or SO2.

CaH2  +  2H2O → Ca  + (OH)2 + 2H2

CaH2 + 2CO2 → (HCOO)2Ca

They are strong reducing agents. Alkali metal hydrides are used for making LiAlH4, NaBH4, etc and for removing last traces of water from organic compounds.

#### Molecular or Covalent Hydrides

These are formed by elements of p-block having higher electronegativity than hydrogen.

Electron deficient hydrides: are the hydrides which do not have a sufficient number of electrons needed to form normal covalent bonds, e.g., BH3, AlH3, etc.
Precise: Which have exact number of electrons needed to form normal covalent bonds. e.g. hydrides of group 14 (CH4, SiH4, etc.)
Rich: They have greater number of electrons than require to form normal covalent bonds. e.g., hydrides of group 15, 16, 17, (NH3, PH3 ,H2S, HF, HCl, etc). The excess electrons in these hydrides are present as lone pairs of electrons.

##### Metallic or Interstitial Hydrides

The transition metals and rare earth metals combine with hydrogen to form interstitial hydrides. They exhibit metallic properties and are powerful reducing agents. They are non-stoichiometric hydrides and their composition varies with temperature and pressure for e.g., LaH2.76 TiH1.73

Metals of group 7, 8 and 9 do not form hydrides and this region of the Periodic Table is called hydride gap.

###### Polymeric Hydrides and Complex Hydrides

Polymeric hydrides are formed by’ elements having electronegativity in the range 1.4 to 2.0, e.g., (BeH2)n, (AlH3)n, etc. In complex hydrides, H– acts as ligand and is attached to the central metal atom, e.g., LiAlH4, LiBH4, etc

## Water

Water is the most abundant and widely distributed on the earth. It occurs in all three physical states. H2O is a covalent molecule in which oxygen is sp3 hybridised. It has a bent structure.

### Physical Properties of Water

• Water is a colourless, odourless, tasteless liquid. It has abnormally high b.p., f.p., the heat of vaporisation due to hydrogen bonding.
• Pure water is not a good conductor so it is made conductor by adding small amount of acid or alkali.
• The density of ice (which is mass per unit volume) is lesser than that of water and it floats over water.
• Water has a maximum density at 4°0.
• Water is a highly polar solvent with high dielectric constant 78.39. It interacts with polar or ionic substances effectively with the release of a considerable amount of energy due to ion-dipole interaction.
• The dissolution of covalent compounds like urea, glucose and C2H5OH, etc is due to the tendency of these molecules to form hydrogen bond with water.

#### Chemical Properties of Water

1) Water is amphoteric in nature.

H2O(l) + HCl(aq) ⇌  H3O+(aq) + Cl(aq)

H2O(l) + NH3(aq) ⇌ NH4+(aq) + OH(aq)

2) In redox reactions,water reacts with metals and non- metals both.

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

3) In hydrated salts, water may remain in five types such as coordinated water, hydrogen bonded water, lattice water, clathrate water and zeolite water.

4) A number of compounds such as calcium hydride, calcium phosphide. etc ., undergo hydrolysis with water.

## Purification of Water

It involves two processes

a) Removal of suspended impurities
b) Destroying the bacteria.
Suspended particles are removed by coagulation with alum followed by filtration.

Exposure to sunlight, boiling, chlorination (treatment with liquid Cl2 or bleaching powder), ozonisation and addition of CuSO5 are some processes which are employed to destroy bacteria.

## Heavy Water [D2O]

It was discovered by Urey in 1932 and can be prepared by exhaustive electrolysis of ordinary water using nickel electrodes. They are colourless, odourless, tasteless liquid.

### Chemical Reactions of Heavy Water

 D2O reacts → Na SO3 CaC2 Al4C3 Ca3P2 Heavy Water NaOD +$$\frac{1}{2}$$ D2 D2SO4 Ca(OD)2 +C2D2 4Al(OD)3 + CD4 3Ca(OD)2 + 2PD3

#### Uses of Heavy Water

1. In nuclear reactors to slow down the speed of neutrons and moderator.
2. As a tracer compound to study the mechanisms of many reactions.

## Soft and Hard Water

The water which produces a large amount of lather with soap is known as soft water and which forms scum with soap is known as hard water.

### Types of Hardness of Water in hydrogen

Temporary hardness: It is due to the presence of bicarbonates of calcium and magnesium.
Permanent hardness: It is due to the presence of chlorides and sulphates of calcium and magnesium.

#### Removal of Temporary Hardness

a) By boiling The soluble bicarbonates are converted into insoluble carbonates.

M(HCO3)2 → MCO3 ↓ + H2O + CO2 (M = Ca , Mg)

b) By Clark’s process By adding lime water or milk of lime.

M(HCO3)2 + Ca(OH)2 → MCO3 ↓ + 2H2O + CaCO3

##### Removal of Permanent Hardness

a) By adding washing soda The calcium or magnesium salts are precipitate as carbonates.

MgCl2 + Na2CO3 → MgCO3 ↓ + 2NaCl

b) By adding caustic soda The temporary and permanent hardness can be removed by adding caustic soda.

CaSO4  + 2NaOH →  Ca(OH)2 ↓ + Na2SO4

MgCl2 + 2NaOH →  Mg(OH)2 ↓ + 2NaCl

c) By adding sodium phosphate (Na3PO4) The phosphates of calcium and magnesium are precipitate.

3CaCl2 + 2Na3PO4 →  Ca3(PO4)2 ↓ + 6NaCl

3CaSO4 + 2Na3PO4 → Ca3(PO4)2 ↓ + 3Na2SO4

Similarly, magnesium also precipitate out in the form of magnesium phosphate, Mg3(PO4)2.

d) Calgon’s process Calgon is sodium hexa metaphosphate (Na6P6O18). This calgon when added to hard water form soluble complex.

2CaSO4 + Na2[Na4(PO3)6] → 2 Na2SO4 + Na2[Ca2(PO3)6]

Similarly. Mg2+ can also precipitate as Na2[Mg2(PO3)6] and water becomes free from Ca2+ and Mg2+ Ions.

e) Permutit process Permutit is hydrated sodium aluminium silicate Na2Al2Si2O8.xH2O. It exchanges its sodium ions for divalent ions Ca2+ and Mg2+

Na2Al2Si2O8 + MCl2 → MAl2Si2O8 + 2NaCl  (M = Ca , Mg)

Permutit when fully exhausted can be regenerated by treating with 10% solution of sodium chloride. It is most efficient method to gel water with zero degree of hardness.

###### By synthetic resins

These are of two types:

•  Cation exchange resins are big molecules containing sulphonic acid group (-SO3H). It is first changed into sodium salt with the general formula $$\overset{-}{R}$$ $$\overset{-}{Na}$$. The hard water is passed through it so Ca2+ and M2+ are exchanged and removed.

2RNa+ + Ca2+ → R2Ca + 2Na+

2RNa+ + Mg2+ → R2Mg + 2Na+

The resins like permutit can be regenerated with a solution of NaCl.

• Anion exchange resins are also big molecules and can exchange anions. They contain an amino group.

RNH2 + H2O → R$$\overset{+}{N}$$H3$$\overset{-}{O}$$H

The water is first passed through cation resins and then through anion resin and pure distilled water is form.

## Measurement of Degree of Hardness

Degree of hardness is the number of parts of calcium carbonate or equivalent to various calcium and magnesium salts present in one million parts of water by mass. It is expressed in ppm.

Degree of hardness (in ppm) = $$\frac{(wt. of CaCO_3(g))}{(wt. of hard water (g))}\times{10^6}$$

The molecular wt. of Ca(HCO3)2, Mg(HCO3)2, CaCl2, MgCl2, CaSO4 and MgSO4 is 162, 146, 111, 95, 136 and 120 respectively. The mol. wt. of CaCO3 is 100.

Thus, 162 g Ca(HCO3)2, 146 g Mg(HCO3)2, 111g CaCl2, 95g MgCl2 136g CaSO4 and 120 g MgSO4 are equivalent to 100 g CaCO3.

## Hydrogen Peroxide [H2O2]

J.L. Thenard discover H2O2 in 1818. Hydrogen Peroxide used in pollution control treatment of domestic and industrial effluents.

### Methods of Preparation

i) BaO2 . 8H2O(s) + H2SO4(aq) → BaSO4(s) + H2O2(aq) + 8H2O(l)

ii) 2HSO4(aq) $$\overset{Electrolysis}{\longrightarrow}$$ HO3SOOSO3H(aq) $$\overset{H_2O}{\longrightarrow}$$ 2HSO4(aq) + 2H+(aq) + H2O2(aq)

iii) 2 −ethylanthraquinol $$\overset{O_2(air)}{\longrightarrow}$$ H2O2 + Oxidized product

#### Strength of Hydrogen Peroxide

The most common method to express the strength of H2O2 is in terms of the volume (in mL) of oxygen liberate at NTP by decomposition or 1 mL of that sample of H2O2.A solution of H2O2 take as ’10 volume’ actually means “1 mL of such a solution of H2O2 on decomposition by heat produces 10 mL of oxygen at NTP”.

##### Storage of Hydrogen Peroxide (H2O2)

It is store in the presence of traces of alcohol, acetanilide or sodium pyrophosphate which slow down the rate of decomposition of hydrogen peroxide.

###### Chemical Properties of H2O2

Acidic nature: It is weakly acidic in nature and pure hydrogen peroxide turns blue litmus red.

Oxidising agent: It acts as a strong oxidising agent in acidic as well as in basic medium.

Bleaching properties:Its bleaching action is due to oxidation by atomic oxygen and permanent.

H2O2 rarr; H2O + [O]

dye + [O] → dye is oxidised and bleached.

## Important Question & Answers in Hydrogen:

1.) Which pair does not show hydrogen isotopes

a) Ortho hydrogen and para hydrogen

b) Protium and deuterium

c) Deuterium and tritium

d) Tritium and protium

ANS:

Ortho and Para hydrogen show different spin in a hydrogen molecule it does not show hydrogen isotopes.

2.) On reaction with Mg, very dilute nitric acid produces:

a) NH3

b) Nitrous oxide

c) Nitric oxide

d) Hydrogen

ANS:

Mg + 2HNO3 → Mg(NO3)2 + H2

3.) Hydrogen resembles in many of its properties

a) Halogen

b) Alkali metals

c) Both a and b

d) None of these

4.) Hydrogen does not combine with:

a) Antimony

b) Sodium

c) Bismuth

d) Helium

ANS:

Helium is a noble gas and does not combine with hydrogen.

5.) Which of the following statements is most applicable to hydrogen

a) It can act as a reducing agent

b) It can act as an oxidising agent

c) It can act both as oxidising and reducing agent

d) It can neither act as oxidising nor as a reducing agent

ANS:

It acts both as a reducing agent and oxidising agent.


text:
# Math learning via a virtual world

By Murray Bourne, 29 Jul 2006

This is an interesting article from Purdue University Virtual Reality World Teaches Deaf Children Math Skills. They are using virtual reality to teach math to deaf chidren.

This is so true:

"Environments are more stimulating when students are able to interact with the subject and travel through the scenes," said Adamo-Villani, assistant professor of computer graphics. "Hands-on experiences equate to a better understanding of mathematical concepts in real- world situations."

But this is not only the case for deaf students. I think virtual reality has a lot of potential for learning - and it has been largely unrealised so far.

The following aims are once again relevant for all learners:

The virtual reality program is designed to provide early elementary school age students with disabilities with a number of active, individualized learning conditions:
- The ability to control their environment.
- The ability to engage in learning activities at their own pace.
- The ability to repeat activities as needed.
- The ability to see or feel items or processes in concrete terms.
- The ability to practice daily living tasks in a safe and barrier- free environment.
- Motivation to succeed.

It's a great concept for the deaf kids. But let's expand it for a much wider audience of learners.

Be the first to comment below.

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text:
## Maximum Area of Rectangle Between Parabola and x Axis

What is the maximum area of a rectangle enclosed between the parabola
$y=16-x^2$
and the
$x$
axis?
From the diagram, the base of the rectangle is
$3x$
and the height is
$16-x^2$
os the area is
$A=x(16-x^2)=16x-x^3$
.
$\frac{dA}{dx}=16-3x^2$
.
Set
$\frac{dA}{dx}=0$
and solve to give
$x= \frac{4 \sqrt{3}}{3}$
.
The area is then
$A=16 \frac{4 \sqrt{3}}{3} - (\frac{4 \sqrt{3}}{3})^3= \frac{128 \sqrt{3}}{9}$
.


text:
This content will become publicly available on February 1, 2023

Neutron transfer reactions on the ground state and isomeric state of a ${}^{130}\mathrm{Sn}$ beam
Authors:
; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; more »
Award ID(s):
Publication Date:
NSF-PAR ID:
10317789
Journal Name:
Physical Review C
Volume:
105
Issue:
2
ISSN:
2469-9985