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# Homework Help: Invariance of a Lagrangian under Transformation

1. Jan 8, 2013

### Fragezeichen

1. The problem statement, all variables and given/known data
Show that the Lagrangian
$\mathcal{L}=\frac{m}{2}\vec{\dot{r}}^2 \, \frac{1}{(1+g \vec{r}^2)^2}$
is invariant under the Transformation

$\vec{r} \rightarrow \tilde{r}=\vec{r}+\vec{a}(1-g\vec{r}^2)+2g\vec{r}(\vec{r} \cdot \vec{a})$

where b is a constant and $\vec{a}$ are infinitesimal parameters.

2. The attempt at a solution
$(1+g\vec{\tilde{r}}^2)^2=(1+g\vec{r}^2)^2 (1+4g(\vec{r} \cdot \vec{a}))$
$\frac{d \vec{\tilde{r}}}{dt}=\vec{\dot{r}}-2g\vec{a}(\vec{r}\cdot \vec{\dot{r}})+2g(\underbrace{\vec{\dot{r}} (\vec{a}\cdot \vec{r})+\vec{r}(\vec{a}\cdot\vec{\dot{r}}}_{?}))$

Can you tell me, wheter this is OK so far?

2. Jan 9, 2013

### clamtrox

Looks OK to me... What happens when you plug those things into the Lagrangian?

3. Jan 9, 2013

### Fragezeichen

$(\frac{d \vec{\tilde{r}}}{dt})^2$ should be at least something like $\vec{\dot{r}}^2(1+4g(\vec{a}\cdot \vec{r}))$

But i can´t get it into this form, especially the underbraced part of it makes me think there´s sth. wrong...

4. Jan 9, 2013

### clamtrox

Yeah that's what you get.. It's not terribly difficult to see either... Two of the cross terms vanish because they are proportional to a2. Then two of the terms proportional to a are equal with opposite signs so they cancel, and you're left with exactly what you were looking for

5. Jan 9, 2013

### Fragezeichen

Seems like i do the same mistake every time i try it...

$(\frac{d \vec{\tilde{r}}}{dt})^2=(\vec{\dot{r}}^2-2 \vec{a} g(\vec{r}\cdot \vec{\dot{r}})+2g \vec{\dot{r}}(-\vec{a}\cdot \vec{r})+2g\vec{r}(\vec{a}\cdot \vec{\dot{r}}))^2$
$=(\vec{\dot{r}}-2g \underbrace{(\vec{a}(\vec{r}\cdot \vec{\dot{r}})-\vec{\dot{r}}(\vec{a}\cdot \vec{r})-\vec{r}(\vec{a}\cdot \vec{\dot{r}})}_{?} )^2$
How to deal with the underbraced term without expanding like this:

$((\vec{a}(\vec{r}\cdot \vec{\dot{r}})-\vec{\dot{r}}(\vec{a}\cdot \vec{r})-\vec{r}(\vec{a}\cdot \vec{\dot{r}}))^2=\underbrace{(\vec{a}(\vec{r} \cdot \vec{\dot{r}}))^2+(\vec{\dot{r}}(\vec{a}\cdot \vec{r}))^2+(\vec{r}(\vec{a}\cdot \vec{\dot{r}}))^2}_{vanishes}-2(((\vec{a}(\vec{r}\cdot \vec{\dot{r}}))\cdot (\vec{\dot{r}}(\vec{a}\cdot \vec{r})) )-2((\vec{a}(\vec{r}\cdot \vec{\dot{r}}))\cdot(\vec{r}(\vec{a}\cdot \vec{\dot{r}})) )+2( (\vec{\dot{r}}(\vec{a}\cdot \vec{r}))\cdot (\vec{r}(\vec{a}\cdot \vec{\dot{r}})) )$

But how to manage the remaing term?

Last edited: Jan 9, 2013
6. Jan 11, 2013

### clamtrox

$$\frac{d \vec{\tilde{r}}}{dt}=\vec{\dot{r}}-2g\vec{a}(\vec{r}\cdot \vec{\dot{r}})+2g(\vec{\dot{r}} (\vec{a}\cdot \vec{r})+\vec{r}(\vec{a}\cdot\vec{\dot{r}}))$$

When you calculate $\tilde{r}^2$, all the cross terms between things containing $\vec{a}$ vanish. Therefore

$$\left(\frac{d \vec{\tilde{r}}}{dt} \right)^2 = \vec{\dot{r}}^2 + 2\vec{\dot{r}}\cdot(-2g\vec{a}(\vec{r}\cdot \vec{\dot{r}})+2g(\vec{\dot{r}} (\vec{a}\cdot \vec{r})+\vec{r}(\vec{a}\cdot\vec{\dot{r}})) + \mathcal{O}(a^2)$$

7. Jan 15, 2013

### Fragezeichen

Thank´s for your answer, now i get it. :)
Now i´m trying to apply Noether-Theorem.

At first i´d try to get the transformation in such a form:
$\vec{r} \longmapsto \vec{r}+\vec{a}\cdot \vec{\psi}(\vec{r})$

Therefore:
$\vec{\psi}(\vec{r})=1-g\vec{r}^2+\underbrace{2g \vec{r}}$
But the underbraced term only produces a $2g \vec{a}\cdot \vec{r}$, but i need an additional $\vec{r}$ to get $2g\vec{r}(\vec{a}\cdot \vec{r})$.

8. Jan 15, 2013

### clamtrox

You, or at least your notation, are going badly wrong here. Note that your transformation has to be a vector, and what you are suggesting is a scalar product between two vectors, so a scalar. Likewise your suggestion for ψ contains vectors and scalars added together, and this is clearly incorrect.

You could think that maybe your transformation is of the form $$\vec{r}\rightarrow \vec{r} + \psi(\vec{r}) \vec{a}.$$ However, this is not possible as you need also a component along $\vec{r}$. Next possibility would be $$\vec{r}\rightarrow \vec{r}+ \vec{\psi}(\vec{r}) \times \vec{a}$$ but this doesn't work either, as the cross product is perpendicular to $\vec{a}$

9. Jan 15, 2013

### Fragezeichen

I´m sorry i meant your first option.
Do you have any clue how to get ψ ?

Or at least a little hint... :-)

Last edited: Jan 15, 2013


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WordPress LaTeX Issues

This is just a brief notice. There appears to be a new problem with WordPress.com displaying LaTeX maths equations. The issue appears to be sporadic, with some of the formulas on both old and new posts now showing “Formula does not parse”. It is affecting other blogs hosted on WordPress too. Thanks to Timothy Gowers on twitter for pointing this out.
The WordPress team are looking into it. Hopefully this will be fixed soon. Update: This issue has now been fixed.

For now, the following method can be used to display the equations using MathJax instead. Click here, then drag the displayed link to the address bar of your browser. Then click on it while viewing any page on this blog, and all formulas will be displayed with MathJax. Magic!

Some tests follow. At the time of writing, some of the following formulas show an error:
Test: ${X^n\xrightarrow{\rm ucp}X}$.

Another test: ${X^n\xrightarrow{\rm ucp}X}$

Yet another test: ${X^n\rightarrow X}$

And another: ${X^n X}$

And another: $X^n X$

And another: ${[S]_\infty}$

yet more: ${[S]_\infty}$

and more: $[S]_\infty$

and more: $[S]$

more: $S_\infty$

What’s in a Name?

That which we call a rose, by any other name would smell as sweet.
You may have noticed, if you pay attention to the address bar, that the domain of this site has changed, and it is pretty sweet! We are now almostsuremath.com, not almostsure.wordpress.com.
It would have been sweeter still to be almostsure.com, but the owner of that domain is not wanting to give it up. The new url contains the name of this site, and is descriptive, so is still pretty good.
It is not just a name change though. Ads have gone away! I dug into my pocket and found the £3 a month to get rid of ads and map my own domain here. Also, switching to the new domain opens up possibilities…such as a self hosted wordpress site, more customization, mathjax, etc.

Is this a Blog?

Well, according to the tagline, it is a “random mathematical blog”, and is hosted on the popular blogging platform, WordPress.com. According to Wikipedia, a blog is a

discussion or informational website published on the World Wide Web consisting of discrete, often informal diary-style text entries (posts).

Most of the posts on this site are not especially discrete, consist of some quite formal maths, and are not at all diary style. According to wix.com, blogs are

regularly updated websites that provide essential insights into a certain topic.

I like to think think that this site provides essential insights into stochastic calculus and probability theory, but how regular is regular? Daily? To qualify, it should probably be weekly, at least. Here, I can go a month or so without updating. It has sometimes been significantly longer between updates. I try and write posts containing proper rigorous mathematics, and to explain things as well as I can. The style that I aim for in many of the posts here are not unlike you might see in published maths papers or in textbooks. These take some time, and I cannot just rush out a post containing detailed proofs and explanations of mathematical theory. It is not my job, just something that I like to do. I like to work through advanced maths subjects, prove results, and to get a good understanding of these. It would be possible to post weekly, but the style would not be the same and the mathematical content would be much reduced and at a shallower level, which is not what I would enjoy doing.

So, no, this is not a blog!

I should probably change the tagline. This is a random mathematical website. I would however like to change the style of the site a bit. The idea is to update with detailed maths posts, as always, but not on a weekly basis. However, I am inclined to also do some weekly updates. Just short updates on mathematical subjects, or anything related to this website. I’ll see how it goes…

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## aRpsDCA examples¶

aRpsDCA isn't picky about data formats; all the functions work in terms of simple vectors of time and rate or volume data. We also provide some functions for unit conversion.

For example, let's generate some data following a hyperbolic decline, add some noise, and try to recover it using the best fit function.

First, we'll need to import the package.

In [32]:
require(aRpsDCA)

I also have to toggle graphics mode to work around a bug in the R kernel for IPython; you won't need to worry about this.

I'm also going to make the plots a little smaller than default.

In [33]:
options(jupyter.plot_mimetypes = 'image/png')
options(repr.plot.width=7, repr.plot.height=6)

Now, let's generate some data following a hyperbolic decline with $q_i$ = 1000 bbl/d, $D_i$ = 75% tangent effective per year (note that many DCA packages present units in terms of secant effective decline: these can be converted following the equations in SPEE monograph #3, available online), and $b$ = 1.5.

In [34]:
t <- seq(0, 365) # one year, in days
q <- hyperbolic.q(
qi=1000, # units must match for rate, time, decline
# we use helper functions from aRpsDCA to handle
# effective -> nominal conversion and change units
Di=as.nominal(0.75, from.period="year", to.period="day"),
b=1.5,
t)
plot(q ~ t, log="y",
main="Generated Data", xlab="Time (days)", ylab="Rate (bbl/d)",
type="l")

Ok, let's add some random (Gaussian) noise to make this a nice example data set to try to fit. We'll see how close we come to recovering the original parameters.

I'm setting a fixed seed so this example comes out the same every time.

In [35]:
set.seed(555)

q.noised <- q + rnorm(length(q), mean=0, sd=10)
plot(q ~ t, log="y",
main="Generated Data", xlab="Time (days)", ylab="Rate (bbl/d)",
type="l")
lines(q.noised ~ t, col="red", lty="dashed")

Now, let's see how well we can recover the curve parameters with the best hyperbolic fit function.

In [36]:
# to be honest, I am not sure where the vapply(...) error comes from below;
# I can't reproduce it outside jupyter and it doesn't alter the result, so
# I'm chalking it up to some issue in the IR kernel.

fit.curve <- best.hyperbolic(q.noised, t)
fit.curve

Error in vapply(seq_along(mapped), function(i) {: values must be length 1,
but FUN(X[[1]]) result is length 0

Out[36]:
$decline [1] "Arps hyperbolic decline: <qi = 1000.598, Di = 0.003811069, b = 1.512116>"$sse
[1] 37651.8

The object-oriented design of aRpsDCA also lets us apply this fit result directly against the time data. Note that the printed $q_i$ and $D_i$ values are in daily units (nominal daily for $D_i$).

In [37]:
fit.forecast <- arps.q(fit.curve$decline, t) plot(q ~ t, log="y", main="Fit Data", xlab="Time (days)", ylab="Rate (bbl/d)", type="l") lines(q.noised ~ t, col="red", lty="dashed") lines(fit.forecast ~ t, col="blue", lty="dotted", lwd=2) legend("top", lty=c("solid", "dashed", "dotted"), col=c("black", "red", "blue"), bty="n", legend=c("Original Data", "Noised Data", "Best Fit"))  Finally, we'll convert the best fit$D_i$value back to a yearly tangent effective rate to compare to the original. In [39]: as.effective(fit.curve$decline\$Di, from.period="day", to.period="year")

Out[39]:
0.751420575465941

Pretty close!


text:
## Discord 3 years ago 3m - 2n + 4p = 15 m - n + p = 3 m + 4n - 5p = 0 System of equations

1. micahwood50

Hello there, Discord. Rainbow Dash is here to help. Isolate any variable in the first equation then plug it to either second or third equation. Does this make sense?

2. Discord

3. Discord

And I could use some clarification

4. micahwood50

Saying, I want to isolate m in first equation.$3m - 2n + 4p = 15$$3m = 15 + 2n - 4p$$m = \frac{15 + 2n - 4p}{3} = 5 + \frac{2}{3}n - \frac{4}{3}p$ Ok, now I know m. Plug it into second or third equation. Does that make sense?

5. Discord

Wait what does m equal?

6. micahwood50

$m = 5 + \frac{2}{3}n - \frac{4}{3}p$

7. Discord

Oh okay.

8. micahwood50

You know what to do?

9. Discord

I think so. Lemme try and solve it.


text:
## Time And Work Questions MCQs

### Total Questions : 226

Page 1 of 12 pages
Question 1. A, B and C completed a work costing Rs. 1800. A worked for 6 days, B for 4 days and C for 9 days. If their daily wages are in the ratio of 5 : 6 : 4, how much amount will be received by A ?
1.    Rs. 600
2.    Rs. 750
3.    Rs. 800
4.    Rs. 900
Question 2. P and Q together can do a job in 6 days. Q and R can finish the same job in $$\frac{{60}}{7}$$ days. P started the work and worked for 3 days. Q and R continued for 6 days to finish the work. Then the difference of days in which R and P can complete the alone is P can complete the job alone is ?
1.    10 days
2.    8 days
3.    12 days
4.    15 days
Question 3. X can do a piece of work in 24 days. When he had worked for 4 days, Y joined him. If complete work was finished in 16 days, Y can alone finish that work in ?
1.    27 days
2.    36 days
3.    42 days
4.    18 days
Question 4. Working efficiencies of P and Q for completing a piece of working are in the ratio 3 : 4. The number of days to be taken by them to complete the work will be in the ratio ?
1.    3 : 2
2.    2 : 3
3.    3 : 4
4.    4 : 3
Question 5. If the expenditure of gas on burning 6 burners for 6 hours a day for 8 days is Rs. 450, then how many burners can be used for 10 days at 5 hours a day for Rs. 625 ?
1.    12
2.    16
3.    4
4.    8
Question 6. A and B can complete a piece of work in 12 and 18 days respectively. A begins to do the work and they work alternatively one at a time for one day each. The whole work will be completed in ?
1.    $${\text{14}}\frac{1}{3}{\text{ days}}$$
2.    $${\text{15}}\frac{2}{3}{\text{ days}}$$
3.    $${\text{16}}\frac{1}{3}{\text{ days}}$$
4.    $${\text{18}}\frac{2}{3}{\text{ days}}$$
Question 7. A can do a piece of work in 90 days, B in 40 days and C in 12 days. They work for a day each in turn i.e., first day A does it alone, B does it the second day and C the third day. After that A does it for another day, and so on. After finishing the work they get Rs. 240. If the wages are divided in proportion to the work done by them, find what each will get ?
1.    A Rs. 24, B Rs. 54, C Rs. 162
2.    A Rs. 22, B Rs.50, C Rs. 132
3.    A Rs. 26, B Rs. 52, C Rs. 142
4.    A Rs. 20, B Rs. 44, C Rs. 182
Question 8. P is thrice as good a workman as Q and therefore able to finish a job in 48 days less than Q. Working together, they can do it in ?
1.    18 days
2.    24 days
3.    30 days
4.    12 days
Question 9. A job can be done by 3 skilled workmen in 20 days or by 5 boys in 30 days. How many days will they take if they work together ?
1.    8 days
2.    10 days
3.    11 days
4.    12 days
Question 10. A does 20% less work than B. If A can complete a piece of work in $$7\frac{1}{2}$$ hours, then B can do it in ?
1.    $${\text{6}}\frac{1}{2}{\text{ hours}}$$
2.    6 hours
3.    $${\text{5}}\frac{1}{2}{\text{ hours}}$$
4.    5 hours
Question 11. A can do a piece of work in 6 days. B is 25% more efficient than A. How long would B alone take to finish this work ?
1.    $${\text{ 4}}\frac{4}{5}{\text{ days}}$$
2.    $${\text{ 3}}\frac{1}{3}{\text{ days}}$$
3.    $${\text{ 5}}\frac{1}{4}{\text{ days}}$$
4.    $${\text{2}}\frac{2}{3}{\text{ days}}$$
Question 12. A takes three times as long as B and C together to do a job. B takes four times as long as A and C together to do the work. If all the three, working together can complete the job in 24 days, then the number of days, A alone will take to finish the job is = ?
1.    100 days
2.    96 days
3.    95 days
4.    90 days
Question 13. Five men are working to complete a work in 15 days. After five days 10 women are accompanied by them to complete the work in next 5 days. If the work is to be done by women only, then in how many days could the work be over if 10 women have started it ?
1.    10 days
2.    12 days
3.    15 days
4.    18 days
Question 14. A contractor undertake to do a piece of work in 40 days. He engages 100 men at the beginning and 100 more after 35 days and completes the work in stipulated time. If he had not engaged the work in men, how many days behind the scheduled the work should have been finished ?
1.    5
2.    6
3.    7
4.    8
Question 15. A man, a woman and a boy can do a piece of work in 6, 9 and 18 days respectively. How many boys must assist one man and one woman to do the work in 1 day ?
1.    5
2.    6
3.    9
4.    13
Question 16. A man is twice as fast as a women and a woman is twice as fast as a boy in doing a work. If all of them, a man , a woman and a boy can finish the work in 7 days, a boy will do it alone ?
1.    49 days
2.    7 days
3.    6 days
4.    42 days
Question 17. A's 2 days work is equal to B's 3 days work. If A can complete the work in 8 days, then to complete the work B will take ?
1.    15 days
2.    12 days
3.    16 days
4.    14 days
Question 18. If A, B and C can complete a piece of work in 6 days. If A can work twice faster than B and thrice faster then C, than the number of days C alone can complete the work is ?
1.    44 days
2.    33 days
3.    22 days
4.    11 days


text:
## Intermediate Algebra (12th Edition)

$3p^{2}(2p-1-3p^{2})$
The greatest common factor of $6p^{3}-3p^{2}-9p^{4}$ is $3p^{2}$. We can use the distributive property to factor out the greatest common factor. 1. $6p^{3}-3p^{2}-9p^{4}$ 2. $3p^{2}\times2p+3p^{2}\times-1+3p^{2}\times-3p^{2}$ 3. $3p^{2}(2p-1-3p^{2})$


text:
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# Levi-Civita symbol

Article Id: WHEBN0000195982
Reproduction Date:

 Title: Levi-Civita symbol Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:

### Levi-Civita symbol

In mathematics, particularly in linear algebra, tensor analysis, and differential geometry, the Levi-Civita symbol represents a collection of numbers; defined from the sign of a permutation of the natural numbers 1, 2, …, n, for some positive integer n. It is named after the Italian mathematician and physicist Tullio Levi-Civita. Other names include the permutation symbol, antisymmetric symbol, or alternating symbol, which refer to its antisymmetric property and definition in terms of permutations.

The standard letters to denote the Levi-Civita symbol are the Greek lower case epsilon ε or ϵ, or less commonly the Latin lower case e. Index notation allows one to display permutations in a way compatible with tensor analysis:

\varepsilon_{i_1 i_2 \cdots i_n}

where each index i1, i2, …, in takes values 1, 2, …, n. There are nn indexed values of \varepsilon_{i_1i_2\cdots i_n}, which can be arranged into an n-dimensional array. The key definitive property of the symbol is total antisymmetry in all the indices. When any two indices are interchanged, equal or not, the symbol is negated:

\varepsilon_{\cdots i_p \cdots i_q \cdots }=-\varepsilon_{\cdots i_q \cdots i_p \cdots } .

If any two indices are equal, the symbol is zero. When all indices are unequal, we have:

\varepsilon_{i_1 i_2 \cdots i_n} = (-1)^p \varepsilon_{1 2 \cdots n} ,

where p (called the parity of the permutation) is the number of interchanges of indices necessary to unscramble i1, i2, …, in into the order 1, 2, …, n, and the factor (−1)p is called the sign or signature of the permutation. The value ε12…n must be defined, else the particular values of the symbol for all permutations are indeterminate. Most authors choose ε12…n = +1, which means the Levi-Civita symbol equals the sign of a permutation when the indices are all unequal. This choice is used throughout this article.

The term "n-dimensional Levi-Civita symbol" refers to the fact that the number of indices on the symbol n matches the dimensionality of the relevant vector space in question, which may be Euclidean or non-Euclidean, pure space or spacetime. The values of the Levi-Civita symbol are independent of any metric tensor and coordinate system. Also, the specific term "symbol" emphasizes that it is not a tensor because of how it transforms between coordinate systems, however it can be interpreted as a tensor density.

The Levi-Civita symbol allows the determinant of a square matrix, and the cross product of two vectors in 3d Euclidean space, to be expressed in index notation.

## Contents

• Definition 1
• Two dimensions 1.1
• Three dimensions 1.2
• Four dimensions 1.3
• Generalization to n dimensions 1.4
• Properties 2
• Two dimensions 2.1
• Three dimensions 2.2
• n dimensions 2.3
• Proofs 2.4
• Applications and examples 3
• Determinants 3.1
• Vector cross product 3.2
• Cross product (two vectors) 3.2.1
• Triple scalar product (three vectors) 3.2.2
• Curl (one vector field) 3.2.3
• Tensor density 4
• Levi-Civita tensors 5
• Example: Minkowski Space 5.1
• Notes 7
• References 8

## Definition

The common dimensionalities of the Levi-Civita symbol are in 3d and 4d, and to some extent 2d, so it is useful to see these definitions before the general one in any number of dimensions.

### Two dimensions

The two-dimensional Levi-Civita symbol is defined by:

\varepsilon_{ij} = \begin{cases} +1 & \text{if } (i,j) \text{ is } (1,2) \\ -1 & \text{if } (i,j) \text{ is } (2,1) \\ \;\;\,0 & \text{if }i=j \end{cases}

The values can be arranged into a 2 × 2 antisymmetric matrix:

\begin{pmatrix} \varepsilon_{11} & \varepsilon_{12} \\ \varepsilon_{21} & \varepsilon_{22} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}

Use of the 2d symbol is relatively uncommon, although in certain specialized topics like supersymmetry[1] and twistor theory[2] it appears in the context of 2-spinors. The 3d and higher-dimensional Levi-Civita symbols are used more commonly.

### Three dimensions

For the indices (i, j, k) in εijk, the values 1, 2, 3 occurring in the cyclic order (1,2,3) (yellow) correspond to ε = +1, while occurring in the reverse cyclic order (red) correspond to ε = −1, otherwise ε = 0.

In three dimensions, the Levi-Civita symbol is defined as follows:[3]

\varepsilon_{ijk} = \begin{cases} +1 & \text{if } (i,j,k) \text{ is } (1,2,3), (2,3,1) \text{ or } (3,1,2), \\ -1 & \text{if } (i,j,k) \text{ is } (3,2,1), (1,3,2) \text{ or } (2,1,3), \\ \;\;\,0 & \text{if }i=j \text{ or } j=k \text{ or } k=i \end{cases}

i.e. \varepsilon_{ijk} is 1 if (i, j, k) is an even permutation of (1,2,3), −1 if it is an odd permutation, and 0 if any index is repeated. In three dimensions only, the cyclic permutations of (1,2,3) are all even permutations, similarly the anticyclic permutations are all odd permutations. This means in 3d it is sufficient to take cyclic or anticyclic permutations of (1,2,3) and easily obtain all the even or odd permutations.

Analogous to 2d matrices, the values of the 3d Levi-Civita symbol can be arranged into a 3×3×3 array:

where i is the depth, j the row and k the column.

Some examples:

\varepsilon_{\color{BrickRed}{1}\color{Violet}{3}\color{Orange}{2}} = -\varepsilon_{\color{BrickRed}{1}\color{Orange}{2}\color{Violet}{3}} = - 1
\varepsilon_{\color{Violet}{3}\color{BrickRed}{1}\color{Orange}{2}} = -\varepsilon_{\color{Orange}{2}\color{BrickRed}{1}\color{Violet}{3}} = -(-\varepsilon_{\color{BrickRed}{1}\color{Orange}{2}\color{Violet}{3}}) = 1
\varepsilon_{\color{Orange}{2}\color{Violet}{3}\color{BrickRed}{1}} = -\varepsilon_{\color{BrickRed}{1}\color{Violet}{3}\color{Orange}{2}} = -(-\varepsilon_{\color{BrickRed}{1}\color{Orange}{2}\color{Violet}{3}}) = 1
\varepsilon_{\color{Orange}{2}\color{Violet}{3}\color{Orange}{2}} = -\varepsilon_{\color{Orange}{2}\color{Violet}{3}\color{Orange}{2}} = 0

### Four dimensions

In four dimensions, the Levi-Civita symbol is defined as:

\varepsilon_{ijkl } = \begin{cases} +1 & \text{if }(i,j,k,l) \text{ is an even permutation of } (1,2,3,4) \\ -1 & \text{if }(i,j,k,l) \text{ is an odd permutation of } (1,2,3,4) \\ 0 & \text{otherwise} \end{cases}

These values can be arranged into a 4×4×4×4 array, although in 4d and higher this is difficult to draw.

Some examples:

\varepsilon_{\color{BrickRed}{1}\color{RedViolet}{4}\color{Violet}{3}\color{Orange}{\color{Orange}{2}}} = -\varepsilon_{\color{BrickRed}{1}\color{Orange}{\color{Orange}{2}}\color{Violet}{3}\color{RedViolet}{4}} = - 1
\varepsilon_{\color{Orange}{\color{Orange}{2}}\color{BrickRed}{1}\color{Violet}{3}\color{RedViolet}{4}} = -\varepsilon_{\color{BrickRed}{1}\color{Orange}{\color{Orange}{2}}\color{Violet}{3}\color{RedViolet}{4}} = -1
\varepsilon_{\color{RedViolet}{4}\color{Violet}{3}\color{Orange}{\color{Orange}{2}}\color{BrickRed}{1}} = -\varepsilon_{\color{BrickRed}{1}\color{Violet}{3}\color{Orange}{\color{Orange}{2}}\color{RedViolet}{4}} = -(-\varepsilon_{\color{BrickRed}{1}\color{Orange}{\color{Orange}{2}}\color{Violet}{3}\color{RedViolet}{4}}) = 1
\varepsilon_{\color{Violet}{3}\color{Orange}{\color{Orange}{2}}\color{RedViolet}{4}\color{Violet}{3}} = -\varepsilon_{\color{Violet}{3}\color{Orange}{\color{Orange}{2}}\color{RedViolet}{4}\color{Violet}{3}} = 0

### Generalization to n dimensions

The Levi-Civita symbol can be generalized to n dimensions:[4]

\varepsilon_{a_1 a_2 a_3 \ldots a_n} = \begin{cases} +1 & \text{if }(a_1 , a_2 , a_3 , \ldots , a_n) \text{ is an even permutation of } (1,2,3,\dots,n) \\ -1 & \text{if }(a_1 , a_2 , a_3 , \ldots , a_n) \text{ is an odd permutation of } (1,2,3,\dots,n) \\ 0 & \text{otherwise} \end{cases}

Thus, it is the sign of the permutation in the case of a permutation, and zero otherwise.

Using the capital Pi notation \prod for ordinary multiplication of numbers, an explicit expression for the symbol is:

\begin{align} \varepsilon_{a_1 a_2 a_3 \ldots a_n} & = \prod_{1\leq i < j \leq n} \sgn ( a_j-a_i ) \\ & = \sgn(a_2 - a_1)\sgn(a_3 - a_1)\ldots\sgn(a_n - a_1)\sgn(a_3 - a_2)\sgn(a_4 - a_2)\ldots\sgn(a_n - a_2)\ldots\sgn(a_n - a_{n-1}) \end{align}

where the product is totally antisymmetric in all indices, and the sign function (denoted by "sgn") extracts the sign of each difference discarding the absolute value. The formula is true for all index values, and for any n (when n = 1 or 0, this is the empty product). However, it is seldom used in practice since interchanging indices is quicker.

## Properties

A tensor whose components in an orthonormal basis are given by the Levi-Civita symbol (a tensor of covariant rank n) is sometimes called a permutation tensor. It is actually a pseudotensor because under an orthogonal transformation of jacobian determinant −1 (i.e., a rotation composed with a reflection), it acquires a minus sign. As the Levi-Civita symbol is a pseudotensor, the result of taking a cross product is a pseudovector, not a vector.[5]

Under a general coordinate change, the components of the permutation tensor are multiplied by the jacobian of the transformation matrix. This implies that in coordinate frames different from the one in which the tensor was defined, its components can differ from those of the Levi-Civita symbol by an overall factor. If the frame is orthonormal, the factor will be ±1 depending on whether the orientation of the frame is the same or not.[5]

In index-free tensor notation, the Levi-Civita symbol is replaced by the concept of the Hodge dual.

In a context where tensor index notation is used to manipulate tensor components, the Levi-Civita symbol may be written with its indices as either subscripts or superscripts with no change in meaning, as might be convenient. Thus, one could write

\varepsilon^{ij\dots k} = \varepsilon_{ij\dots k} .

In these examples, superscripts should be considered equivalent with subscripts.

Summation symbols can be eliminated by using Einstein notation, where an index repeated between two or more terms indicates summation over that index. For example

\varepsilon_{ijk} \varepsilon^{imn} \equiv \sum_{i=1,2,3} \varepsilon_{ijk} \varepsilon^{imn}.

In the following examples, Einstein notation is used.

### Two dimensions

In two dimensions, when all i, j, m, n each take the values 1 and 2,[3]

\varepsilon_{ij} \varepsilon^{mn} = \delta_i {}^m \delta_j {}^n - \delta_i {}^n \delta_j {}^m

(1)

\varepsilon_{ij} \varepsilon^{in} = \delta_j{}^n

(2)

\varepsilon_{ij} \varepsilon^{ij} = 2.

(3)

### Three dimensions

Index and symbol values

In three dimensions, when all i, j, k, m, n each take values 1, 2, and 3:[3]

\varepsilon_{ijk} \varepsilon^{imn}=\delta_j{}^{m}\delta_k{}^n - \delta_j{}^n\delta_k{}^m

(4)

\varepsilon_{jmn} \varepsilon^{imn}=2\delta^i_j

(5)

\varepsilon_{ijk} \varepsilon^{ijk}=6.

(6)

Product

The Levi-Civita symbol is related to the Kronecker delta. In three dimensions, the relationship is given by the following equations (vertical lines denote the determinant):[4]

\begin{align} \varepsilon_{ijk}\varepsilon_{lmn} & = \begin{vmatrix} \delta_{il} & \delta_{im}& \delta_{in}\\ \delta_{jl} & \delta_{jm}& \delta_{jn}\\ \delta_{kl} & \delta_{km}& \delta_{kn}\\ \end{vmatrix}\\ & = \delta_{il}\left( \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}\right) - \delta_{im}\left( \delta_{jl}\delta_{kn} - \delta_{jn}\delta_{kl} \right) + \delta_{in} \left( \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl} \right). \end{align}

A special case of this result is (4):

\sum_{i=1}^3 \varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}

sometimes called the "contracted epsilon identity".

In Einstein notation, the duplication of the i index implies the sum on i. The previous is then denoted: \varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}\,.

\sum_{i=1}^3 \sum_{j=1}^3 \varepsilon_{ijk}\varepsilon_{ijn} = 2\delta_{kn}

### n dimensions

Index and symbol values

In n dimensions, when all i1,...,in, j1,...,jn take values 1, 2,..., n:

\varepsilon_{i_1 \dots i_n} \varepsilon^{j_1 \dots j_n} = n! \delta_{}^{j_n} = \delta^{j_1 \dots j_n}_{i_1 \dots i_n}

(7)

\varepsilon_{i_1 \dots i_k~i_{k+1}\dots i_n} \varepsilon^{i_1 \dots i_k~j_{k+1}\dots j_n}= k!(n-k)!~\delta_{}^{j_{k+1}} \dots \delta_{i_n ]}{}^{j_n} = k!~\delta^{j_{k+1} \dots j_n}_{i_{k+1} \dots i_n}

(8)

\varepsilon_{i_1 \dots i_n}\varepsilon^{i_1 \dots i_n} = n!

(9)

where the exclamation mark (!) denotes the factorial, and δα…
β…
is the generalized Kronecker delta. For any n, the property

\sum_{i,j,k,\dots=1}^n \varepsilon_{ijk\dots}\varepsilon_{ijk\dots} = n!

follows from the facts that

• every permutation is either even or odd,
• (+1)2 = (−1)2 = 1, and
• the number of permutations of any n-element set number is exactly n!.
Product

In general, for n dimensions, one can write the product of two Levi-Civita symbols as:

\varepsilon_{i_1 i_2 \dots i_n} \varepsilon_{j_1 j_2 \dots j_n} = \begin{vmatrix} \delta_{i_1 j_1} & \delta_{i_1 j_2} & \dots & \delta_{i_1 j_n} \\ \delta_{i_2 j_1} & \delta_{i_2 j_2} & \dots & \delta_{i_2 j_n} \\ \vdots & \vdots & \ddots & \vdots \\ \delta_{i_n j_1} & \delta_{i_n j_2} & \dots & \delta_{i_n j_n} \\ \end{vmatrix} .

### Proofs

For (1), both sides are antisymmetric with respect of ij and mn. We therefore only need to consider the case ij and mn. By substitution, we see that the equation holds for \varepsilon_{12} \varepsilon^{12}, i.e., for i = m = 1 and j = n = 2. (Both sides are then one). Since the equation is antisymmetric in ij and mn, any set of values for these can be reduced to the above case (which holds). The equation thus holds for all values of ij and mn.

Using (1), we have for (2)

\varepsilon_{ij} \varepsilon^{in} = \delta_i{}^i \delta_j{}^n - \delta_i{}^n \delta_j{}^i = 2 \delta_j{}^n - \delta_j{}^n = \delta_j{}^n \,.

Here we used the Einstein summation convention with i going from 1 to 2. Next, (3) follows similarly from (2).

To establish (5), notice that both sides vanish when ij. Indeed, if ij, then one can not choose m and n such that both permutation symbols on the left are nonzero. Then, with i = j fixed, there are only two ways to choose m and n from the remaining two indices. For any such indices, we have

\varepsilon_{jmn} \varepsilon^{imn} = (\varepsilon^{imn})^2 = 1

(no summation), and the result follows.

Then (6) follows since 3! = 6 and for any distinct indices i, j, k taking values 1, 2, 3, we have

\varepsilon_{ijk} \varepsilon^{ijk}=1 (no summation, distinct i, j, k ).

## Applications and examples

### Determinants

In linear algebra, the determinant of a 3 × 3 square matrix A = (aij) can be written[6]

\det(\mathbf{A}) = \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \varepsilon_{ijk} a_{1i} a_{2j} a_{3k}

Similarly the determinant of an n × n matrix A = (aij) can be written as[5]

\det(\mathbf{A}) = \varepsilon_{i_1\cdots i_n} a_{1i_1} \cdots a_{ni_n},

where each ir should be summed over 1,..., n, or equivalently:

\det(\mathbf{A}) = \frac{1}{n!} \varepsilon_{i_1\cdots i_n} \varepsilon_{j_1\cdots j_n} a_{i_1 j_1} \cdots a_{i_n j_n},

where now each ir and each jr should be summed over 1,.., n. More generally, we have the identity[5]

\sum_{i_1,i_2,\dots}\varepsilon_{i_1\cdots i_n} a_{i_1 \, j_1} \cdots a_{i_n \, j_n} = \det(\mathbf{A}) \varepsilon_{j_1\cdots j_n}

### Vector cross product

#### Cross product (two vectors)

If a = (a1, a2, a3) and b = (b1, b2, b3) are vectors in \mathbb{R}^3 (represented in some right-handed coordinate system using an orthonormal basis), their cross product can be written as a determinant:[5]

\mathbf{a \times b} = \begin{vmatrix} \mathbf{e_1} & \mathbf{e_2} & \mathbf{e_3} \\ a^1 & a^2 & a^3 \\ b^1 & b^2 & b^3 \\ \end{vmatrix} = \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \varepsilon_{ijk} \mathbf{e}_i a^j b^k

hence also using the Levi-Civita symbol, and more simply:

(\mathbf{a \times b})_i = \sum_{j=1}^3 \sum_{k=1}^3 \varepsilon_{ijk} a^j b^k.

In Einstein notation, the summation symbols may be omitted, and the ith component of their cross product equals[4]

(\mathbf{a\times b})_i = \varepsilon_{ijk} a^j b^k.

The first component is

(\mathbf{a\times b})_1 = a^2 b^3-a^3 b^2\,,

then by cyclic permutations of 1, 2, 3 the others can be derived immediately, without explicitly calculating them from the above formulae:

(\mathbf{a\times b})_2 = a^3 b^1-a^1 b^3\,,
(\mathbf{a\times b})_3 = a^1 b^2-a^2 b^1\,.

#### Triple scalar product (three vectors)

From the above expression for the cross product, we have:

\mathbf{a\times b} = -\mathbf{b\times a}.

If c = (c1, c2, c3) is another vector, then the triple scalar product equals

\mathbf{a}\cdot(\mathbf{b\times c}) = \varepsilon_{ijk} a^i b^j c^k.

From this expression, it can be seen that the triple scalar product is antisymmetric when exchanging any pair of arguments. For example,

\mathbf{a}\cdot(\mathbf{b\times c})= -\mathbf{b}\cdot(\mathbf{a\times c}).

#### Curl (one vector field)

If F = (F1, F2, F3) is a vector field defined on some open set of \mathbb{R}^3 as a function of position x = (x1, x2, x3) (using Cartesian coordinates). Then the ith component of the curl of F equals[4]

(\nabla \times \mathbf{F})^i(\mathbf{x}) = \varepsilon^{ijk}\frac{\partial}{\partial x^j} F^k(\mathbf{x}),

which follows from the cross product expression above, substituting components of the gradient vector operator (nabla).

## Tensor density

In any arbitrary curvilinear coordinate system and even in the absence of a metric on the manifold, the Levi-Civita symbol as defined above may be considered to be a tensor density field in two different ways. It may be regarded as a contravariant tensor density of weight +1 or as a covariant tensor density of weight −1. In n dimensions using the generalized Kronecker delta,[7][8]

\varepsilon^{\mu_1 \cdots \mu_n} = \delta^{\mu_1 \cdots \mu_n}_{\,1 \,\cdots \,n} \,
\varepsilon_{\nu_1 \cdots \nu_n} = \delta^{\,1 \,\cdots \,n}_{\nu_1 \cdots \nu_n} \,.

Notice that these are numerically identical. In particular, the sign is the same.

## Levi-Civita tensors

On a pseudo-Riemannian manifold, one may define coordinate-invariant covariant and contravariant tensor fields whose coordinate representations agree with the Levi-Civita symbol wherever the coordinate system is such that the basis of the tangent space is orthonormal with respect to the metric and matches a selected orientation. These tensors should not be confused with each other, nor should they be confused with the tensor density fields mentioned above. The covariant Levi-Civita tensor (also known as the Riemannian volume form) in a given coordinate system is

E_{a_1\cdots a_n} = \sqrt{|\det [g_{ab}]|} \varepsilon_{a_1\cdots a_n} \,,

where g_{ab} is the representation of the metric in that coordinate system. This tensor may be converted to the contravariant tensor by raising the indices with the metric as usual, but a minus sign is needed if the metric signature contains an odd number of negatives.

E^{a_1\cdots a_n} = (-1)^s E_{b_1\cdots b_n} \prod_{i=1}^n g^{a_i b_i} \,,

where s is the number of negatives in the signature. This results in the following:

E^{a_1\cdots a_n} = \frac 1 {\sqrt{|\det [g_{ab}]|}} \varepsilon^{a_1\cdots a_n} \,

### Example: Minkowski Space

In Minkowski space (the four-dimensional spacetime of special relativity), the covariant Levi-Civita tensor is

E_{\alpha \beta \gamma \delta} = \sqrt{|\det [g_{\mu \nu}]|} \varepsilon_{\alpha \beta \gamma \delta} \,,

but the contravariant Levi-Civita tensor is

E^{\alpha \beta \gamma \delta} = - g^{\alpha \zeta} g^{\beta \eta} g^{\gamma \theta} g^{\delta \iota} E_{\zeta \eta \theta \iota} \,.

Notice the minus sign. The following are identities.

E^{\alpha \beta \gamma \delta} E^{\rho \sigma \mu \nu} = - g^{\alpha \zeta} g^{\beta \eta} g^{\gamma \theta} g^{\delta \iota} \delta^{\rho \sigma \mu \nu}_{\zeta \eta \theta \iota} \,
E_{\alpha \beta \gamma \delta} E_{\rho \sigma \mu \nu} = - g_{\alpha \zeta} g_{\beta \eta} g_{\gamma \theta} g_{\delta \iota} \delta^{\zeta \eta \theta \iota}_{\rho \sigma \mu \nu} \,
E^{\alpha \beta \gamma \delta} E_{\rho \beta \gamma \delta} = 6 \delta^{\alpha}_{\rho} \,.
E^{\alpha \beta \gamma \delta} E_{\rho \sigma \gamma \delta} = 2 \delta^{\alpha \beta}_{\rho \sigma} \,.
E^{\alpha \beta \gamma \delta} E_{\rho \sigma \theta \delta} = \delta^{\alpha \beta \gamma}_{\rho \sigma \theta} \,.

## Notes

1. ^ P. Labelle (2010). Supersymmetry. Demystified. McGraw-Hill. pp. 57–58.
2. ^ F. Hadrovich. "Twistor Primer". Retrieved 03/09/2013.
3. ^ a b c J.R. Tyldesley (1973). An introduction to Tensor Analysis: For Engineers and Applied Scientists. Longman.
4. ^ a b c d D.C. Kay (1988). Tensor Calculus. Schaum’s Outlines, McGraw Hill (USA).
5. ^ a b c d e K.F. Riley, M.P. Hobson, S.J. Bence (2010). Mathematical methods for physics and engineering. Cambridge University Press.
6. ^ S. Lipcshutz, M. Lipson (2009). Linear Algebra (4th ed.). Schaum’s Outlines, McGraw Hill (USA).
7. ^ Murnaghan, F. D. (1925), "The generalized Kronecker symbol and its application to the theory of determinants", Amer. Math. Monthly 32: 233–241
8. ^ David Lovelock, Hanno Rund (1989). Tensors, Differential Forms, and Variational Principles. Courier Dover Publications. p. 113.

## References

• J.A. Wheeler, C. Misner, K.S. Thorne (1973).
• D. E. Neuenschwander (2015). Tensor Calculus for Physics. Johns Hopkins University Press. pp. 11, 29, 95.


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# VWAP Tutorial: Calculation, Uses, and Limitations

The Volume Weighted Average Price (VWAP) is simple to calculate and has a variety of uses. While a Hedge Fund or Mutual fund uses it to guide their decision while buying a substantial number of shares, a retail trader would use it to check if the price at which he traded was a good price or not. There are also intraday traders who will use it as an indicator and buy when the price is below the VWAP.

But exactly what is VWAP?

In this blog, we will understand the VWAP and also how to calculate it. Along the way, we will also compare it with another simple indicator, i.e. the moving average and understand its advantages and disadvantages, We will see if we can create an intraday trading strategy using VWAP. We will cover the following points in this blog

## What is VWAP?

Let’s face it, at its fundamental level, if we had to compare two seemingly good securities, more often than not, we would check its price trend and the trading volume. Price is obvious, but why the volume?

Volume is important as we don’t want to get stuck with a stock which has few takers, even if you think it is priced attractively. Thus, the VWAP was created to take into account both volume as well as Price so that the potential investor would make the trading decision or not.

In simple terms, the Volume Weighted Average price is the cumulative average price with respect to the volume.

The formula for calculating VWAP is as following:

VWAP = (Cumulative (Price * Volume)) / (Cumulative Volume)

While we can go through the formula easily, we thought we would understand VWAP by going through an example itself.

## Calculating the VWAP in Excel

To calculate VWAP, we take the daily minute-by-minute data of Tesla, which has the dubious distinction of being one of the most volatile stocks. You can get sample historical data from Alpha Vantage. We have used the daily data for the date of 18 October 2019. A sample of the data is as follows:

 Date Open High Low Close Volume 2019-10-10 9:31:00 245.2903 245.516 244.7652 244.8702 103033 2019-10-10 9:32:00 245.0807 245.0807 244.55 244.66 21168 2019-10-10 9:33:00 244.58 245.8 244.55 245.6 36544 2019-10-10 9:34:00 245.7097 246.09 245.57 245.92 30057 2019-10-10 9:35:00 245.62 245.62 245.62 245.62 26301 2019-10-10 9:36:00 245.7126 246.44 245.7126 246.188 31494 2019-10-10 9:37:00 246.46 246.46 246.45 246.45 24271 2019-10-10 9:38:00 246.755 246.755 246.25 246.25 37951 2019-10-10 9:39:00 246.2818 246.655 246.2818 246.655 15324 2019-10-10 9:40:00 246.78 246.78 246.56 246.762 23285 2019-10-10 9:41:00 246.75 246.75 246.38 246.5 23365 2019-10-10 9:42:00 246.17 246.17 246.17 246.17 16130 2019-10-10 9:43:00 246.135 246.135 245.82 245.82 27227 2019-10-10 9:44:00 245.9335 245.9335 245.91 245.91 14464 2019-10-10 9:45:00 246.41 246.41 246.41 246.41 17156 2019-10-10 9:46:00 246.44 246.46 246.1683 246.1683 23938 2019-10-10 9:47:00 246.2857 246.57 246.2857 246.57 70833 2019-10-10 9:48:00 246.6 247.47 246.6 247.47 59743 2019-10-10 9:49:00 247.49 247.65 247.49 247.65 71995 2019-10-10 9:50:00 247.685 247.801 247.65 247.69 46038 2019-10-10 9:51:00 247.95 248.74 247.95 248.74 103773 2019-10-10 9:52:00 248.56 248.56 247.95 247.95 73810 2019-10-10 9:53:00 247.93 247.93 247.6614 247.6614 29784 2019-10-10 9:54:00 247.74 247.76 247.65 247.76 37138 2019-10-10 9:55:00 247.93 248.03 247.93 248.03 53166 2019-10-10 9:56:00 247.91 248.44 247.91 248.44 40789 2019-10-10 9:57:00 248.52 248.52 248.3154 248.3154 51988 2019-10-10 9:58:00 248.4409 248.62 248.4409 248.62 53405 2019-10-10 9:59:00 248.9199 248.9199 248.9199 248.9199 85348 2019-10-10 10:00:00 248.91 249.08 248.42 248.72 58270

### Step 1: Find the average or Typical price

To get a reliable estimate of the price at which a security was traded for a given period, we take the average of the values, in this case, the average of the high, low, and close price.

Thus, for 9:31, the average price = (245.516 + 244.7652 + 244.8702)/3 = 245.0504667.

Similarly, for 9:32, the average price = (245.0807+ 244.55+ 244.66) / 3 = 244.7635667

### Step 2: Multiply the Typical price with the volume for that period and add the cumulative total of the previous period

Since we are looking for a period of 9:31, the volume traded was 103033.

Thus, (Price * Volume) = 245.0504667 * 103033 = 25248284.73

Since it was the first period of the day, it was a simple multiplication. From the second column onwards, we take the cumulative total, ie adding the previous period’s value to the current value.

Thus, for 9:32, with volume at 21168, the cumulative average price = ((Average price at 9:32) * volume at 9:32) + cumulative total at 9:31

= [244.7635667 * 21168] + 25248284.73 = 30429439.91

### Step 2.5: Find the cumulative total volume

Since we found the cumulative average price * volume, we have to keep a running total of the volume of the security traded.

Hence, for 9:31, it will just be 103033 as it is the first period of the day.

For 9:32, it will be (Volume at 9:32) + cumulative volume of the previous period, ie (21168 + 103033) = 124201.

### Step 3: Find VWAP

We simply divide the cumulative price * volume by the cumulative volume.

Thus, for 9:31, VWAP = 25248284.73 / 103033 = 245.0504667

For 9:32, VWAP = 30429439.91 / 124201 = 245.0015693

Thus, the excel sheet would look something like this:

 Date Open High Low Close Volume Average Price Average price*Volume Cumulative Volume VWAP 2019-10-10 9:31:00 245.2903 245.516 244.7652 244.8702 103033 245.0504667 25248284.73 103033 245.0504667 2019-10-10 9:32:00 245.0807 245.0807 244.55 244.66 21168 244.7635667 30429439.91 124201 245.0015693 2019-10-10 9:33:00 244.58 245.8 244.55 245.6 36544 245.3166667 39394292.18 160745 245.073204 2019-10-10 9:34:00 245.7097 246.09 245.57 245.92 30057 245.86 46784106.2 190802 245.1971478 2019-10-10 9:35:00 245.62 245.62 245.62 245.62 26301 245.62 53244157.82 217103 245.2483744 2019-10-10 9:36:00 245.7126 246.44 245.7126 246.188 31494 246.1135333 60995257.44 248597 245.3579787 2019-10-10 9:37:00 246.46 246.46 246.45 246.45 24271 246.4533333 66976926.29 272868 245.4554081 2019-10-10 9:38:00 246.755 246.755 246.25 246.25 37951 246.4183333 76328748.46 310819 245.5729812 2019-10-10 9:39:00 246.2818 246.655 246.2818 246.655 15324 246.5306 80106583.37 326143 245.6179755 2019-10-10 9:40:00 246.78 246.78 246.56 246.762 23285 246.7006667 85851008.4 349428 245.6901233 2019-10-10 9:41:00 246.75 246.75 246.38 246.5 23365 246.5433333 91611493.38 372793 245.7435987 2019-10-10 9:42:00 246.17 246.17 246.17 246.17 16130 246.17 95582215.48 388923 245.761283 2019-10-10 9:43:00 246.135 246.135 245.82 245.82 27227 245.925 102278015.5 416150 245.7719944 2019-10-10 9:44:00 245.9335 245.9335 245.91 245.91 14464 245.9178333 105834971 430614 245.776893 2019-10-10 9:45:00 246.41 246.41 246.41 246.41 17156 246.41 110062381 447770 245.80115 2019-10-10 9:46:00 246.44 246.46 246.1683 246.1683 23938 246.2655333 115957485.3 471708 245.8247163 2019-10-10 9:47:00 246.2857 246.57 246.2857 246.57 70833 246.4752333 133416065.5 542541 245.9096465 2019-10-10 9:48:00 246.6 247.47 246.6 247.47 59743 247.18 148183340.2 602284 246.035658 2019-10-10 9:49:00 247.49 247.65 247.49 247.65 71995 247.5966667 166009062.3 674279 246.202332 2019-10-10 9:50:00 247.685 247.801 247.65 247.69 46038 247.7136667 177413304 720317 246.2989268 2019-10-10 9:51:00 247.95 248.74 247.95 248.74 103773 248.4766667 203198473.2 824090 246.5731573 2019-10-10 9:52:00 248.56 248.56 247.95 247.95 73810 248.1533333 221514670.7 897900 246.7030523 2019-10-10 9:53:00 247.93 247.93 247.6614 247.6614 29784 247.7509333 228893684.5 927684 246.7366954 2019-10-10 9:54:00 247.74 247.76 247.65 247.76 37138 247.7233333 238093633.7 964822 246.7746731 2019-10-10 9:55:00 247.93 248.03 247.93 248.03 53166 247.9966667 251278624.4 1017988 246.8384936 2019-10-10 9:56:00 247.91 248.44 247.91 248.44 40789 248.2633333 261405037.5 1058777 246.893385 2019-10-10 9:57:00 248.52 248.52 248.3154 248.3154 51988 248.3836 274318004.1 1110765 246.9631327 2019-10-10 9:58:00 248.4409 248.62 248.4409 248.62 53405 248.5603 287592367 1164170 247.036401 2019-10-10 9:59:00 248.9199 248.9199 248.9199 248.9199 85348 248.9199 308837182.6 1249518 247.1650529 2019-10-10 10:00:00 248.91 249.08 248.42 248.72 58270 248.74 323331262.4 1307788 247.2352265

Great! We have just understood how to find the VWAP for a security. If we plot the VWAP with the closing price for the whole day, we will get the graph as seen below:

Now you must be wondering why we have used 1-minute data for calculating the VWAP. The truth is, we can calculate the VWAP on different time periods, be it a 5 minute, 10 minute time period etc.

However, a point to note is that VWAP is only calculated for the day and thus cannot be used for periods ranging to multiple days. We will understand why this is so in the next section of the VWAP Tutorial.

## How to use VWAP

### VWAP as trend confirmation

We have mentioned before how VWAP gives us information related to both volume as well as price. It also helps us confirm the presence of any trend which might be emerging in the day.

Let’s understand this with the VWAP of the Tesla share we calculated previously.

If you see the graph for VWAP, despite the frequent swings in the closing price, we can clearly see that the VWAP is rising. The rising VWAP indicates that there are more buyers than sellers.

Similarly, a rising VWAP can indicate a bullish phase whereas a decreasing VWAP will indicate a bearish phase.

### VWAP as a trade execution strategy

VWAP is also used by institutional buyers who need to buy or sell a large number of shares but do not want to cause a spike in the volume as it attracts attention and affects the price.

To explain this further, let’s say an institution is interested in buying 10000 shares of Tesla. If it puts an order of 10,000, the immediate action would be a spike in the price as the exchange fills the order. Now, if other traders know that there is a big demand for the share, they would try to buy the share at a higher price than the bid price of the institution and sell it back at a higher price, effectively increasing the ask price of the share.

To avoid this scenario, these institutions develop an automated trading strategy to divide the number of shares into smaller amounts and bid for the shares in such a way that their trades do not let the closing prices go far from the VWAP. Since VWAP acts as a guideline on which certain traders base their trading decisions on, it helps to keep the closing price as close to the VWAP as possible.

### VWAP as an Indicator

Among intraday traders, the VWAP indicator can be used in a trading strategy too. There are conflicting theories on how exactly you should use the VWAP as an indicator, and thus we will try to understand this aspect in greater detail. We usually consider scenarios when the closing price crosses the VWAP as a signal, and thus, a VWAP cross can be used to enter or exit the trade depending on your risk profile.

Before we look at the different scenarios, let’s step back and understand that VWAP can actually be self-fulfilling when it comes to traders. As seen previously, certain institutional traders would try to execute trades in such a way that the closing price doesn’t go farther than the VWAP. This can influence other traders who would look at the closing price and take a trading decision thinking that the closing price is bound to get close to the VWAP eventually.

Hence, when the closing price starts moving up and farther from the VWAP, there is pressure among the traders to sell, due to the logic that the other would sell at any time. This creates a situation where the general belief might be that the stock is overvalued. Similarly, when the closing price starts moving down and further from the VWAP, there is a belief that the stock is undervalued and there is a pressure among traders to buy the stock.

In this way, we can call VWAP as self-fulfilling. Of course, depending on the mindset of the community, there can be different scenarios and thus, one cannot depend on VWAP alone to make a trading decision.

Let us now look at a few other scenarios.

Some traders prefer the VWAP cross as an indicator and buy the stock when the closing price crosses the VWAP and climbs higher, indicating a bullish trend. One will then either wait for the closing price to reach the high of the day at which point they sell and exit the trade. Other traders will exit as soon as the closing price shows signs of reversing.

At the other end, some traders would short the stock when the closing price crosses the VWAP and keeps going down. Once the closing price reaches the low of the day, they would then close the trade.

Now, some traders would prefer a price below VWAP as a good price to buy and a price above VWAP could indicate that it is a good time to sell. Taking the previous VWAP chart for Tesla, you can see as the price goes above the VWAP there is a small period where the price keeps increasing and then the price decreases.

It is however seen that for the trading strategy, traders consider the crossover of the closing price with the VWAP as a signal. However, one should note that the VWAP lags behind the closing price and thus should not be the sole indicator in a trading strategy.

### VWAP as a check of Profitability

Once traders ave closed their trade, they look at the VWAP to check if their trade was profitable or not. For example, if a trader had bought a share of Tesla at $248 and the VWAP at the end of the day was$250, then the trader actually bought the stock at a good price and can make a profit on this trade.

We have so far seen some of the uses of VWAP. However, while going through the article, did you feel some sort of deja vu or realized you have read about something similar with a different name? Do you think VWAP is just another variation of a moving average?

## VWAP versus Moving Average

If you remember Moving averages, at its basic level, it is simply an average of 10 (or 20 depending upon your choice of the period) recent average prices. In fact, after 500 minutes, you can say that the VWAP is comparable to the 499 period moving average.

However, to say that the VWAP is similar to the moving average will not be right due to the simple fact that VWAP starts fresh at the open whereas the moving average contains past data as well.

Thus, while the moving average would be similar to VWAP at the end of the day, it will not be the same throughout the day.

Great! Not only do we know how to calculate the VWAP, but we also saw its uses and compared it with another popular indicator. But are there any limitations to VWAP? Let’s find out.

## Limitations of VWAP

As we have mentioned earlier. VWAP is a lagging indicator and thus, if you try to use it for more than a day, it will not be able to portray the correct trend.

Thus, it should be used only for intraday.

Furthermore, there are cases where certain stocks (or the market itself) are in a strong bullish phase and thus there will be no crossovers for the entire day, which in turn portrays very little information to the traders as well as institutions. In a way, the major drawback of VWAP is it cannot be used for more than a day, and thus, not able to provide much information from a historical point of view.

## Conclusion

We have understood that the VWAP is a combination of both price and volume, and thus provides valuable information, compared to the moving averages. We also learned how to calculate the VWAP in Excel and how to interpret it when used alongside the closing price. In the end, we also understood its limitation as a tool only for intraday traders and not for a long term investor.


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# Effects of increase on intermolecular forces

Intermolecular forces between intramolecular and intermolecular bonds in liquids and solids are therefore often called van der waals forces. Intermolecular forces: van der waals interaction (history of intermolecular forces from newton medium involve polarization effects 3 quantum mechanical forces. What happens to these physical properties as the strength of intermolecular forces increases increase or decrease a) melting point b) vapor pressure c) boiling point d) viscosity e) surface tension and how did u go about getting this answer thanks. Effect of surface: in hydrocarbons, the forces increase over surface area where the strength of dispersion forces depends on the contact of as many regions of the molecules as possible hydrocarbons with lower number of carbons and shorter chains. Start studying effects of intermolecular forces learn vocabulary, terms, and more with flashcards, games, and other study tools. This proved that geckos stick to surfaces because of dispersion forces—weak intermolecular dipole-dipole forces and their effects london forces increase. Surface tension is the energy required to increase the surface area of a liquid by a unit amount and varies greatly from liquid to liquid based on the nature of the intermolecular forces, eg, water with hydrogen bonds has a surface tension of 729 x 10-2 j/m 2 (at 20°c), while mercury with metallic (electrostatic) bonds has as surface tension that is 15-times higher: 486 x 10-1 j/m 2 (at 20°c.

Start studying chapter 11 questions and problems 1110 list the types of intermolecular forces mercury in a glass flask is a good example of the effects of. Describe the types of intermolecular forces possible between atoms or molecules in condensed phases (dispersion forces increase substantially the effect of. Effects of intermolecular forces tags: the ke would not increase if if forces increased similar threads for effects intermolecular forces. Surface tension surface tension is measured as the energy required to increase the surface area of a liquid by a unit of area the surface tension of a liquid results from an imbalance of intermolecular attractive forces, the cohesive forcesbetween molecules.

The effects of methyl and amino the whole process manifested the weakening of intermolecular forces intramolecular hydrogen bond can increase density. Surface tension is the energy required to increase the which depends on intermolecular forces surface tension is the the effects of the high surface tension. Get an answer for 'what are the effects of intermolecular forces and temperature on density ' and find homework help for other science questions at enotes.

The effects of intermolecular forces the following five optional experiments investigate the effect of various physical properties (evaporation, surface tension, solubility, boiling point and capillarity) of substances and determine how these properties relate to intermolecular forces. Why do molecules like alkanes with higher relative molecular mass ($m_\mathrm r$) have stronger intermolecular forces for example, methane ($\ce{ch4}$) has a weaker intermolecular force. Effect of dipole-dipole intermolecular forces in the vibration spectra of organic compounds i effect on frequency: theoretical discussion. Rule 1 - longer carbon chains = stronger intermolecular forces in this case the strength of intermolecualr forces can be seen through the increased boiling points the picture below shows the.

## Effects of increase on intermolecular forces

Alcohol’s intermolecular forces are van der wal forces and hydrogen bonds this experiment is designed to determine whether the increase in relative molecular mass (mr) in alcohols (methanol, ethanol and propanol) will increase the strength of the intermolecular forces, specifically the van der wall forces.

Deviation of gas from ideal behavior at low temperatures where the contribution of intermolecular forces or instantaneous dipole forces, tend to increase. Intermolecular forces are the forces that hold molecules together (like in liquids and solids) dipole-dipole forces polar molecules have a positively-charged portion and a negatively-charged portion. Intermolecular forces the increase in temperature is a comparison of the boiling points of the noble gases demonstrates this effect. How do intermolecular forces affect viscosity chemistry phases of matter how intermolecular forces affect phases of what are some causes and effects of sediment. Intermolecular forces the repulsive force chiefly has the effect of keeping two molecules from when a gas is compressed to increase its. If intermolecular forces are strong, the solubility is reduced in other words higher the intermolecular forces lower the solubility is for example compounds containing covalent bond (double or triple) are less soluble because they possess strong intermolecular forces. This site might help you re: how does the strength of intermolecular forces effect evaporation i need to know weather the stronger intermolecular forces cause the evaporation to be faster or slower and vice versa with the weaker forces.

The purpose of this lab was to explore intermolecular forces and solubility the first step in achieving this purpose was to complete a variety of chemical reactions between alcohols and solvents and determine which were miscible/immiscible. • surface tension may increase 3) strength of intermolecular forces effect of external pressure on boiling point lecture 3 intermolecular forces. At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Dispersion forces are found in all substances the strengths of these forces increase with increased molecular weight and also depend on molecular shapes dipole-dipole forces add to the effect of dispersion forces and are found in polar molecules hydrogen bonds, which are recognized by the presence of h atoms bonded to f, o, or n, also add to the effect of dispersion forces. (a) critical temperature as intermolecular forces increase, critical temperature increases as intermolecular forces increase, critical temperature decreases. Intermolecular forces , it turns out that it can form a special type intermolecular interaction called the as these intermolecular forces increase. These charged particles alter the intermolecular forces between water molecules why does adding salt increase the boiling point of water.

Effects of increase on intermolecular forces
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# Why of probability in QM

Is there an amateur level book that talks about why QM depends on probability (assuming that is the actual teaching)?


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# Complexity!

Algebra Level 3

If z = $$i^{1/2}$$, and z = x + iy (where x and y are real numbers), find xy:

×


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Home » » The maximum value of the function of y = 2 + 3x - 4x$$^2$$ is

# The maximum value of the function of y = 2 + 3x - 4x$$^2$$ is

### Question

The maximum value of the function of y = 2 + 3x - 4x$$^2$$ is

A) -23/16

B) 5/15

C) 41/16

D) 59/16

### Explanation:

$$y = -4x^2 + 3x + 2$$
$$a = -4, b = 3, c = 2$$
max $$= c - \left( \cfrac{b^2}{4a} \right)$$
$$= 2 - \left( \cfrac{3^2}{4(-4)} \right)$$
$$= 2 + \cfrac{9}{16}$$
$$= \cfrac{32 + 9}{16} = \frac{41}{16}$$

## Discussion (2)

• $$y = -4x^2 + 3x + 2$$
$$a = -4, b = 3, c = 2$$
max $$= c - \left( \cfrac{b^2}{4a} \right)$$
$$= 2 - \left( \cfrac{3^2}{4(-4)} \right)$$
$$= 2 + \cfrac{9}{16}$$
$$= \cfrac{32 + 9}{16} = \frac{41}{16}$$

• Why are there no explanationations for the mathematics questions?


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# nLab weak excluded middle

Weak excluded middle

# Weak excluded middle

## Definition

In logic, the principle of weak excluded middle says that for any proposition $P$ we have $\neg P \vee \neg\neg P$.

This follows from the full principle of excluded middle, which says that for any $Q$ we have $Q \vee \neg Q$ (take $Q = \neg P$). Thus, in classical mathematics weak excluded middle is just true. But in constructive mathematics (i.e. intuitionistic logic), it is a weaker assumption than full excluded middle.

## Equivalent forms

### De Morgan’s Law

In intuitionistic logic, de Morgan’s law often refers to the one of de Morgan's four laws that is not an intuitionistic tautology, namely $\neg (P \wedge Q) \to (\neg P \vee \neg Q)$ for any $P,Q$.

###### Theorem

De Morgan’s law is equivalent to weak excluded middle.

###### Proof

If de Morgan’s law holds, then since $\neg (P \wedge \neg P)$, we have $\neg P \vee \neg\neg P$, as desired. Conversely, if weak excluded middle holds and we have $\neg (P\wedge Q)$, then from weak excluded middle we get $\neg P \vee \neg\neg P$ and $\neg Q \vee \neg\neg Q$ which give four cases. In three of those cases $\neg P \vee \neg Q$ holds, while in the fourth we have $\neg\neg P$ and $\neg\neg Q$, which together imply $\neg\neg(P\wedge Q)$ (see the first lemma here and its proof), contradicting the assumption of $\neg (P\wedge Q)$; so the fourth case is impossible.

Last revised on May 30, 2019 at 17:48:46. See the history of this page for a list of all contributions to it.


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# Quotient map $f$ from the $n$-sphere to the $n$-disk [duplicate]

I'm trying to construct a quotient map $$f$$ from the $$n$$-sphere $$S^n$$ to the $$n$$-disk $$D^n$$, in which $$f(x_1,x_2,\ldots,x_n,x_{n+1})=f(x_1,x_2,\ldots,x_n,-x_{n+1})$$. The goal is to show that $$S^n/\sim_f$$ is homeomorphic to $$D^n$$. (Is this the right way to go about it?)
## marked as duplicate by Henno Brandsma general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 11 '18 at 15:57
• If you're willing to go the other way, you can get the homeomorphism directly: $x\mapsto (2\sqrt {1-\|x\|^2}x,2\|x\|^2-1)$ – Matematleta Dec 11 '18 at 16:14


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# Homework Help: Question about the properties of the rank of a matrix

1. Mar 13, 2012

### skyturnred

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

Matrix A is a 4 row by 5 column matrix. Matrix B is a column vector in R$^{4}$. We are supposed to decide whether the following are (a) no solution, (b) one solution, or (c) infinitely many solutions, or whether (d) the data do not give enough information to tell, or (e) the data are impossible.

For the following six cases:

rank(A) rank([A | b])
(i) 4 5
(ii) 4 4
(iii) 4 3 .
(iv)3 4
(v) 3 5
(vi)3 3

***numbers on the right are the rank for augmented matrix, numbers on left are for matrix A alone

2. Relevant equations

3. The attempt at a solution

I THINK I know the answer to all of them.. but I am not entirely certain. I have always found rank just a little confusing. Can someone please tell me where I went wrong in the following and why?

(i) The data is impossible: rank exceeds number of rows.
(ii) There are 5 unkowns (because matrix A has 5 columns) and the rank is 4. So that means there is ONE free variable, so there are infinitely many solutions.
(iii) Not possible: augmenting Matrix A by a column vector B cannot possible DECREASE the rank. Can only be equal to or ONE more than rank(A).
(iv) No solutions. Since augmenting by a column vector increased the rank by one, that means that the augmented matrix is inconsistent.
(v)Not possible: rank exceeds number of rows
(vi)Infinitely many solutions: 5 unknowns, rank is 3, so 2 free variables.

Thanks!


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# How can I find specific heat in thermochemistry of phase of changes?

$q = m \cdot H {\Delta}_{f}$
$q = m {C}_{s} \Delta T$