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# How to find first derivative of function $y=x \ln(x)$ by limit definition using this formula $y'=\lim_{h\to 0}\frac{f(x+h)-f(x)}h$?

How to find first derivative of function $$y=x \ln(x)$$ by limit definition, that is using this formula $$y'=\lim_{h\to 0}\frac{f(x+h)-f(x)}h$$

not product rule or L'Hopital rule are allowed.

• Without directly using the product rule, you can start by using the same "trick" often used in the derivative of the rule:$$\lim_{h\to0}\frac{(x+h)\ln(x+h)-x\ln x}h=\lim_{h\to0}\frac{(x+h)\ln(x+h)\color{red}{{}-x\ln(x+h)+x\ln(x+h)}-x\ln x}h$$ – user170231 Oct 14 '20 at 14:20
• @user170231 thank you so much for your response but how can i find from your last step that first derivative of y=x*ln(x) is y'=ln(x)+1 ? :) – John Oct 14 '20 at 14:30

We have that

$$\lim_{h \to 0}\frac{(x+h)\log (x+h)-x\log x}{h}=\lim_{h \to 0}\frac{x\left(\log (x+h)-\log x\right)+h\log (x+h)}{h}=$$

$$=\lim_{h \to 0} \left(x \cdot \frac{\log (x+h)-\log x}{h} +\log (x+h)\right)=x \cdot \frac1x +\log x$$

indeed

$$\frac{\log (x+h)-\log x}{h}=\frac1x\frac{\log \left(1+\frac hx\right)}{\frac hx} \to \frac1x$$

indeed by $$y= \frac x h \to \infty$$

$$\frac{\log \left(1+\frac hx\right)}{\frac hx}=\log \left(1+\frac1y\right)^y \to \log e=1$$

• thank you so much for your response but how can i find from your last step that first derivative of y=x*ln(x) is y'=ln(x)+1 ? :) , i mean how to go to the result y'=ln(x)+1 :) – John Oct 14 '20 at 14:31
• @John Are you given that $\lim_{h \to 0} \frac{\log (x+h)-\log x}{h}=\frac1x$? – user Oct 14 '20 at 14:33
• Thank you so much sir now i completely understand problem and solution.You were so fast and accurate to this problem :) – John Oct 14 '20 at 14:52
• @John You are very welcome! Bye – user Oct 14 '20 at 15:03

$$\begin{multline} \lim_{h\to\infty}\frac{(x+h)\ln(x+h)-x\ln x}{h}=\lim_{h\to\infty}\frac{x\ln\left(\frac{x+h}{x}\right)+h\ln(x+h)}{h}\\=x\ln\left[\lim_{h\to\infty}\left(1+\frac{x}{h}\right)^\frac1h\right]+\lim_{h\to\infty}\ln(x+h)=1+\ln x \end{multline}$$ where $$x\ln\left[\lim_{h\to\infty}\left(1+\frac{x}{h}\right)^\frac1h\right]=1$$ follows from the well known limit: $$\lim_{h\to\infty}\left(1+\frac{x}{h}\right)^\frac1h=e^{\frac1x}$$

• Thank you so much for your answer :) – John Oct 14 '20 at 15:21

Continuing with the suggestion in my comment, you might do

\begin{align} \lim_{h\to0}\frac{(x+h)\ln(x+h)-x\ln x}h&=\lim_{h\to0}\frac{(x+h)\ln(x+h)-x\ln(x+h)+x\ln(x+h)-x\ln x}h\\[1ex] &=\left(\lim_{h\to0}\ln(x+h)\right)\left(\lim_{h\to0}\frac{(x+h)-x}h\right)+\left(\lim_{h\to0}x\right)\left(\lim_{h\to0}\frac{\ln(x+h)-\ln x}h\right)\\[1ex] &=\ln x\lim_{h\to0}\frac{(x+h)-x}h+x\lim_{h\to0}\frac{\ln(x+h)-\ln x}h \end{align}

$$(x+h)-x=h$$, so the first limit is $$1$$. For the other limit, see the other answers here, or methods shown here.

• Thank you so much for your answer :) – John Oct 14 '20 at 14:55


text:
How do you simplify 2m^-2 and write it using only positive exponents?

Oct 16, 2016

$2 \cdot \frac{1}{m} ^ 2$

Explanation:

The thing to remember about exponents is that they essentially switch signs every time you move terms from one side of the fraction bar to the other, i.e. from the numerator to the denominator, or vice versa.

Take, for example, $2$. As you know, this can be written as

$2 = \frac{2}{1} = {2}^{1} / 1$

If you want to write $2$ with a negative exponent, you have to do two things

• move $2$ from the numerator of ${2}^{1} / 1$ to the denominator
• switch the sign of its exponent

You would have

${2}^{1} / 1 = \frac{1}{2} ^ \textcolor{red}{- 1}$

Now try this for ${m}^{- 2}$. To write it using a positive exponent, you must move $m$ from the numerator of ${m}^{- 2} / 1$ to the denominator and switch its sign.

${m}^{- 2} = \frac{1}{m} ^ \textcolor{red}{2}$

And there you have it. Put it all together to find

$2 \cdot {m}^{- 2} = 2 \cdot \frac{1}{m} ^ \left(2\right)$

Do one more for good practice. How would you write ${a}^{- 2} / 3$ using only positive exponents?

Move $a$ from the numerator to the denominator and switch its sign -- keep in mind that $3$ already has a positive exponent, ${3}^{1}$, so leave it as-is!

a^(-2)/3 = 1/3 * 1/a^color(red)(2) = 1/(3a^color(red)(2)

format: plain


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# RMO Problem Discussion

Upload the pdf of RMO 2013 and discuss problems under this thread ONLY AFTER COMPLETION of the exam ( i.e. 4pm). Best of luck to all of you,buddies.

Note by Priyatam Roy
6 years, 2 months ago

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CRMO-Paper 3

- 6 years, 2 months ago

I took this paper. I was sitting behind Sreejato, and I got only questions 1, 2, and 4 right. I hope I make the cutoff!

- 6 years, 2 months ago

Isn't RMO a contest held in India? That's the impression that I had at least

edit: looking at the problems, I have to say problems 3 and 5 sure are tough.

- 6 years, 2 months ago

Yes, RMO is specifically for Indians. And problem 5 is pretty easy if you can find the recursion.

- 6 years, 2 months ago

hey....Sreejato in which class do u study????

- 6 years, 2 months ago

well 3rd one is simple. Menelaus theorem you see.... For maharashtra region... :P

- 6 years, 2 months ago

We are talking about CRMO Paper 3.

- 6 years, 2 months ago

I thought you live in USA. :O

- 6 years, 2 months ago

But I saw Cody entering ISI :P

- 6 years, 2 months ago

Have you been dreaming when you gave RMO? xD

- 6 years, 2 months ago

Dreaming that I had enough time to write up the solutions :|

- 6 years, 2 months ago

Oh!How I wish that you were seated next to me ,so that I could have a good shot>>>>Counterattack :p

- 6 years, 2 months ago

Downvoted.

- 6 years, 2 months ago

Any idea about the cutoff of paper 3 ?

- 6 years, 2 months ago

That's what I got! The problems weren't very hard, but the time limit (3 hours) put pressure on me. Apparently I solved the first four problems, but I won't get full credit for the third one, as I couldn't complete it (lack of time! >:( ). I had the fifth question on scratch, but didn't get the time to present it carefully. :'(

- 6 years, 2 months ago

Solved the fifth one now, only 4 lines more and my solution would have been complete >.< time was really a factor and that I did not have practice in solution writing was really a major fault from my side

- 6 years, 2 months ago

The fifth one was completely solved in my rough work. Unfortunately I won't get marks for it. -.-

- 6 years, 2 months ago

:| You could have atleast shown the recursive relation and give outline of the proof

- 6 years, 2 months ago

I had only like 1 minute left when I solved #4. :(

- 6 years, 2 months ago

Sreejato.. keno pakamo marish... :P emniteo peye jabi.. chap nis na.

- 6 years, 2 months ago

guarantee nei :(

- 6 years, 2 months ago

It was quite easy though.

- 6 years, 2 months ago

* Mumbai RMO-Paper*

- 6 years, 2 months ago

- 4 years, 6 months ago

There has to be a 2013 problem or solution.Therefore note that 2013=3 X 11 X 61

- 6 years, 2 months ago

True dat.

- 6 years, 2 months ago

You were right!

- 6 years, 2 months ago

About what..........exactly?Mine didn't have any 2013 related problem

- 6 years, 2 months ago

You can find the question papers and solutions here. I'm creating subthreads for CRMO 1-4.

- 6 years, 2 months ago

CRMO-Paper 1

- 6 years, 2 months ago

CRMO-Paper 2

- 6 years, 2 months ago

I got 3 questions correct in CRMO paper 2. Anyone else get CRMO paper 2?

- 6 years, 2 months ago

I got this paper :) Felt it was easy after finishing the test and found the solutions(at home)

- 6 years, 2 months ago

CRMO-Paper 4

- 6 years, 2 months ago

Was there any question common in CRMO 1, 2, 3 or 4?

- 6 years, 2 months ago

I don't think so.

- 6 years, 2 months ago

nope....all were miles away.....

- 6 years, 2 months ago

Mine was CRMO Paper 2....And there is a 2013 related problem....

- 6 years, 2 months ago

RMO Maharashtra and Goa region question paper and answers are not available online. They never upload it!

- 6 years, 2 months ago

Well then we can type the questions and start the discussion. Well Maharashtra paper was easy rhan other papers, I hope so.

- 6 years, 2 months ago

Really it was easier than expected...... R U really going to type the whole paper?? I have already typed it as a new discussion

- 6 years, 2 months ago

- 6 years, 2 months ago

mine was crmo 1..

- 6 years, 2 months ago

mine too....how many were u able to solve...and from which region r u...i was able to solve only the 1, 2 and 5.....and yeah the 6th one partly

- 6 years, 2 months ago

i was able to solve 1,3 and 4 and rest was quite tough..time limitation ..:P

- 6 years, 2 months ago

- 6 years, 2 months ago

- 6 years, 2 months ago

well i was able to solve 4 questions completely but i was unable to solve the last question due to less time.so i want to know whats the cutoff for delhi region

- 6 years, 2 months ago

I am in class 9( goint to class 10 this year). I will be giving Pre RMO and hopefully RMO this year, can someone please give me some tips on how to prepare and what to prepare for both. Do suggest some books also please....

- 2 years ago

this rmo paper is tough, but much easier than the previous years!!

- 6 years, 2 months ago


text:
Protocol John Doe 2.0

Back in January 2019, Cubariqueño Cigar Company released its third limited edition cigar known as the Protocol John Doe. John Doe is a name that fits in with Cubariqueño’s theme of law enforcement. It is the name given to an unidentified male body until identification can be made by next of kin. In a way, this was a fitting name because the story behind Protocol John Doe is that Cubariqueño has said they came into these unidentified and unknown blends. At the time the cigars were announced, with only the size (Robusto) and packaging disclosed of this unknown blend, Cubariqueño sold through 300 ten-count bundles in less than a day. That was one of the most remarkable small batch runs to date. The John Doe was followed up late in 2019 by a Protocol Jane Doe cigar, and now in 2020, there is a Protocol John Doe 2.0. Today, we turn our attention to the Protocol John Doe 2.0 cigar.

Protocol John Doe 2.0 is the seventh limited edition cigar by Cubariqueño. It is unique because the other limited edition cigars were released as shop exclusives to stores in New Jersey or Pennsylvania. This time the release is going to Omerta Cigar Company, which is based in Oklahoma City. This reflects how Cubariqueño has grown as a company over the past five years. Last year, Cubariqueño released four limited editions, but because of the COVID-19 pandemic, things have slowed down all around and this is Cubariqueño’s first 2020 release.

Cubariqueño has truly kept its limited editions as “one and done.” Below is the list of Cubariqueño Cigar Company’s limited edition shop exclusives:

Without further ado, let’s break down the Protocol John Doe 2.0 and see what this cigar brings to the table:

# SPECIFICATIONS

## Blend and Origin

As mentioned, the blend details have not been disclosed for the Protocol John Doe 2.0. The cigar originates from Erik Espinosa’s La Zona factory in Estelí, Nicaragua.

Wrapper: Not Disclosed
Binder: Not Disclosed
Filler: Not Disclosed
Country of Origin: Nicaragua
Factory: La Zona

## Vitolas Offered

Protocol Jane Doe was released in one size – a 5 x 52 Robusto. It is packaged in ten-count bundles. The number of bundles offered to Omerta is 225.

Protocol John Doe 2.0 Bundle Packaging

## Appearance

The wrapper of the Protocol John Doe 2.0 had a medium brown color with a rosado tint to it. Upon close examination, there was some darker mottling on the surface of the wrapper. There wasn’t much in the way of oil on the wrapper. There were some thin visible veins and thin visible wrapper seams. Overall this was a wrapper that was on the smoother side.

The band on the Protocol John Doe 2.0 is a blank white paper band. I’ll infer this represents the “unidentified” person of Jane Doe.

# PERFORMANCE

## Pre-Light Draw

A straight cut was used to remove the cap of the Protocol John Doe 2.0. Once the cap was removed, it was on to the pre-light draw experience. The cold draw delivered notes of dusty earth, fruit sweetness, and black pepper. Overall this was a satisfactory pre-light draw. At this time I moved on to toast the foot of the Protocol John Doe 2.0 and see what the smoking experience would have in store.

## Tasting Notes

The Protocol John Doe 2.0 started off with more notes of dusty earth, fruit sweetness, and black pepper. There also were some notes of natural tobacco. Early on the fruit and pepper notes took center stage as both notes alternated in intensity. The earth and natural tobacco notes settled in the background. The pepper was powerful on the retro-hale, with a mix of black and red pepper.

The fruit and pepper notes remained in the forefront going into the second third. Both the fruit and pepper continued to alternate at various points during the early stages of the second third. During this stage, the earth notes increased and displaced the fruit notes by the midway point. The natural tobacco notes were more distant.

As the cigar moved into the final third, the pepper notes took control as the primary note. The earth notes were a close secondary note. There still were notes of fruit sweetness and natural tobacco present. This is the way the Protocol John Doe 2.0 came to a close. The resulting nub was soft to the touch and cool in temperature.

## Burn

The burn of the Protocol John Doe 2.0 started out relatively well keeping on a straight track. As the cigar progressed maintaining a straight burn line became more difficult. By the second half, there were more touch-ups necessary. These touch-ups were on the cosmetic side, but the cigar was never in danger of coming off the burn path. The resulting ash was firm early on and became slightly less firm in the second half. This was an ash that was light gray in color. Meanwhile, the burn rate and burn temperature were ideal.

Burn of the Protocol John Doe 2.0

## Draw

The draw of the Protocol John Doe 2.0 followed a similar pattern to the burn. The early part of the cigar saw the cigar have a touch of resistance – which is something that I like. In the second half, I found the resistance to get tighter past the sweet spot. The cigar never went out, but there were a few points where it became more work to derive the flavor.

## Strength and Body

Simply put, this cigar is a power-bomb. It’s the boldest cigar to come out from Cubariqueño since the company’s inception. This cigar is full-strength and full-bodied right out of the gate – and it doesn’t let up. It actually gets bolder along the way.

The strength and the body do a good job of balancing each other. Neither had a significant edge of the other.

# OVERALL ASSESSMENT

## Final Thoughts

As mentioned upfront, Protocol John Doe 2.0 is by far the boldest cigar to come out from Cubariqueño Cigar Company to date. When the original Protocol (Blue) cigar came out, it was billed as a very strong cigar. The Protocol John Doe 2.0 is significantly bolder than that cigar. Protocol John Doe 2.0 isn’t the most complex of blends, and it does have a straight-forward flavor profile. The fact the flavor is not sacrificed on this cigar is a positive.

When it has come to the regular production cigars of Cubariqueño versus the limiteds, I’ve seen a significant gap – the regular production cigars are better. While I’d put Protocol John Doe 2.0 as one of the better limiteds Cubariqueño has released, it still isn’t up there with the company’s regular production stuff. Still, if you like a full-throttle, full strength and body cigar with flavor, this is going to be the cigar for you. It’s a cigar that I’d recommend to the experienced cigar enthusiast. As for myself, this is a cigar I would smoke again – and it’s one I’d recommend one to buy.

## Summary

Key Flavors: Dusty Earth, Pepper, Fruit Sweetness, Natural Tobacco
Burn: Very Good
Draw: Very Good
Complexity: Medium Minus
Strength: Full
Body: Full
Finish: Good

Score: 89

## References

News: Protocol John Doe 2.0 Coming in July
Price: \$9.20
Source: Cubariqueño Cigar Company
Brand Reference: Cubariqueño Cigar Company

Photo Credits: Cigar Coop


text:
Palindromes in infinite ternary words
RAIRO - Theoretical Informatics and Applications - Informatique Théorique et Applications, Volume 43 (2009) no. 4, pp. 687-702.

We study infinite words $𝐮$ over an alphabet $𝒜$ satisfying the property $𝒫:\phantom{\rule{3.33333pt}{0ex}}𝒫\left(n\right)+𝒫\left(n+1\right)=1+#𝒜\phantom{\rule{4pt}{0ex}}\mathrm{for}\phantom{\rule{4pt}{0ex}}\mathrm{any}\phantom{\rule{4pt}{0ex}}n\in ℕ$, where $𝒫\left(n\right)$ denotes the number of palindromic factors of length $n$ occurring in the language of $𝐮$. We study also infinite words satisfying a stronger property $\mathrm{𝒫ℰ}:\phantom{\rule{3.33333pt}{0ex}}\mathrm{every}\phantom{\rule{4pt}{0ex}}\mathrm{palindrome}\phantom{\rule{4pt}{0ex}}\mathrm{of}\phantom{\rule{4pt}{0ex}}𝐮\phantom{\rule{4pt}{0ex}}\mathrm{has}\phantom{\rule{4pt}{0ex}}\mathrm{exactly}\phantom{\rule{4pt}{0ex}}\mathrm{one}\phantom{\rule{4pt}{0ex}}\mathrm{palindromic}\mathrm{extension}\phantom{\rule{4pt}{0ex}}\mathrm{in}\phantom{\rule{4pt}{0ex}}𝐮.$ For binary words, the properties $𝒫$ and $\mathrm{𝒫ℰ}$ coincide and these properties characterize sturmian words, i.e., words with the complexity $𝒞\left(n\right)=n+1$ for any $n\in ℕ$. In this paper, we focus on ternary infinite words with the language closed under reversal. For such words $𝐮$, we prove that if $𝒞\left(n\right)=2n+1$ for any $n\in ℕ$, then $𝐮$ satisfies the property $𝒫$ and moreover $𝐮$ is rich in palindromes. Also a sufficient condition for the property $\mathrm{𝒫ℰ}$ is given. We construct a word demonstrating that $𝒫$ on a ternary alphabet does not imply $\mathrm{𝒫ℰ}$.

DOI: 10.1051/ita/2009016
Classification: 68R15
Keywords: ternary infinite words, palindromes, generalized sturmian words, rich words
@article{ITA_2009__43_4_687_0,
author = {Balkov\'a, L'ubom{\'\i}ra and Pelantov\'a, Edita and Starosta, \v{S}t\v{e}p\'an},
title = {Palindromes in infinite ternary words},
journal = {RAIRO - Theoretical Informatics and Applications - Informatique Th\'eorique et Applications},
pages = {687--702},
publisher = {EDP-Sciences},
volume = {43},
number = {4},
year = {2009},
doi = {10.1051/ita/2009016},
zbl = {1191.68476},
mrnumber = {2589989},
language = {en},
url = {http://www.numdam.org/articles/10.1051/ita/2009016/}
}
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%A Starosta, Štěpán
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Balková, L'ubomíra; Pelantová, Edita; Starosta, Štěpán. Palindromes in infinite ternary words. RAIRO - Theoretical Informatics and Applications - Informatique Théorique et Applications, Volume 43 (2009) no. 4, pp. 687-702. doi : 10.1051/ita/2009016. http://www.numdam.org/articles/10.1051/ita/2009016/

[1] P. Arnoux and G. Rauzy, Représentation géométrique de suites de complexité $2n+1$. Bull. Soc. Math. France 119 (1991) 199-215. | EuDML | Numdam | MR | Zbl

[2] P. Baláži, Z. Masáková and E. Pelantová, Factor versus palindromic complexity of uniformly recurrent infinite words. Theoret. Comput. Sci. 380 (2007) 266-275. | MR | Zbl

[3] L'. Balková, E. Pelantová and W. Steiner, Sequences with constant number of return words. Monatsh. Math. 155 (2008) 251-263. | MR | Zbl

[4] M. Bucci, A. De Luca, A. Glen and L.Q. Zamboni, A connection between palindromic and factor complexity using return words. Adv. Appl. Math. 42 (2009) 60-74. | MR | Zbl

[5] J. Cassaigne, Complexity and special factors. Bull. Belg. Math. Soc. Simon Stevin 4 (1997) 67-88. | EuDML | MR | Zbl

[6] X. Droubay, G. Pirillo, Palindromes and Sturmian words. Theoret. Comput. Sci. 223 (1999) 73-85. | MR | Zbl

[7] J. Justin and G. Pirillo, Episturmian words and episturmian morphisms. Theoret. Comput. Sci. 276 (2002) 281-313. | MR | Zbl

[8] M. Morse and G.A. Hedlund, Symbolic dynamics II - Sturmian trajectories. Amer. J. Math. 62 (1940) 1-42. | JFM | MR

[9] G. Rote, Sequences with subword complexity $2n$. J. Number Theor. 46 (1993) 196-213. | MR | Zbl

[10] L. Vuillon, A characterization of Sturmian words by return words. Eur. J. Combin. 22 (2001) 263-275. | MR | Zbl

Cited by Sources:


text:
Semiconductor Devices

# Small-signal diode models

The idea of small-signal modelling is that the short-cut part of the current-voltage characteristic of the semiconductor diode is not precise enough in an exact approximation and it is logical to represent the diode as a resistor at the short-cut. Using the circuit analysis technique we can find the small-signal resistance of the diode. Kirchhof equations with a diode as a load will help to calculate the ${I}_{diode}$ and  ${V}_{diode}$. These values of voltage and current for the diode are called the operating point of the diode.

Piecewise linear diode model: In this model the diode is considered as an open circuit when it is off, and as a linear resistor with the offset voltage in series, when the diode is on. And close to the operating point, discussed above, the diode behaves as a small-signal resistance with: ${r}_{diode}=\frac{{{\partial }^{i}}_{diode}}{{{\partial }^{v}}_{diode}}$ calculated in the operating point.  This method helps to use the semiconductor diode more accurately in its states.


text:
[–] 23 points24 points

sorry, this has been archived and can no longer be voted on

Here is an earlier conference proceedings version (pdf) of the paper. (edit: just an extended abstract with theorem statements but no proofs)

And just to be clear, since almost everyone who has commented so far seems quite confused:

• Travelling salesman problem is NP-hard (its decision version is NP-complete). The special case of TSP in the euclidean plane is also NP-hard. The authors do not solve an NP-hard variant of TSP. P vs NP has not been resolved.

• This paper solves the special case of Euclidean TSP (called XY-TSP) in which all cities live on the x axis or y axis.

It is an interesting result. One thing I'm not sure about is how important this unsolved XY-TSP problem was considered in the algorithms community.

[–] 4 points5 points

sorry, this has been archived and can no longer be voted on

Horrible article. Does a poor job describing the problem after a terrible intro talking about P=NP... and then doesn't even show which paper it is, just says its behind a paywall. Lots of us (not me at the moment... but lots of us) are at institutions with access to those journals and might be able to check it out but instead, we get this bullshit.

[–] 0 points1 point

sorry, this has been archived and can no longer be voted on

[–] 3 points4 points

sorry, this has been archived and can no longer be voted on

What this means is that solving the decision version in polynomial time would provide a fast solution to all NP problems.

Uh, isn't this completely not true?

The only problem is that the paper describing how it all works is behind a paywall and costs $31.50 to read - which is an ongoing disgrace. So do we have any good reason to believe the solution is actually correct? [–] 17 points18 points sorry, this has been archived and can no longer be voted on The article isn't very clear, but it's correct. Ignore everything you just read and read this: The Traveling Salesman Problem (TSP) is an NP-complete problem. The problem is, given a set of points and the distances between them, does there exist a path visiting all of them of length <= k? Notice that this is a decision problem (yes/no) rather than a "find the shortest path" problem. Only decision problems can be NP-Complete. It is absolutely true that a polynomial-time algorithm for any NP-Complete problem such as TSP would provide a polytime algorithm for every problem in NP, which would be a miracle and would be highly disbelieved. Now, the XY Traveling Salesman Problem is the following problem. Given a set of points guaranteed to lie on the x and y axes of a graph, is there a path visiting all of them of length <= k? This problem is not NP-Complete and can be solved in polynomial time. So we may as well think of the XY TSP as a totally different problem. It does sound similar to describe, but other than that and the name, it is not the same as TSP. [–] 0 points1 point sorry, this has been archived and can no longer be voted on Uh, isn't this completely not true? No, that's completely true. That's exactly what it means for a problem to be NP-hard, which TSP is. So do we have any good reason to believe the solution is actually correct? Well, it passed peer review for one thing. Do you have any good reason to not believe the solution is actually correct? [–] 0 points1 point sorry, this has been archived and can no longer be voted on a fast solution As others have said, this is true, though I'd like to mention that it's only provided that you interpret "fast solution" as "polynomial time solution". Just because something runs in an amount of time which is bounded by an arbitrary polynomial of the input size doesn't imply that it is fast for any size in practical terms. [–] 0 points1 point sorry, this has been archived and can no longer be voted on Just because something runs in an amount of time which is bounded by an arbitrary polynomial of the input size doesn't imply that it is fast for any size in practical terms. They found a solution that is quadratic in the size of the input. Considering this is a variant of TSP, I think it's fair to call that fast. [–] 0 points1 point sorry, this has been archived and can no longer be voted on Yes, in that case, that's likely reasonably fast, but the bit of text I'm replying to is not talking about the restricted problem solved by the paper, but the general TSP and hence the P=NP problem. Even if P=NP is true, there is a reasonable chance that the polynomials involved are very high degree, or the constant factors involved are sufficient to make it completely impractical. [–] -5 points-4 points sorry, this has been archived and can no longer be voted on Yeah I really dislike it when people post stuff that reads: "BIG PROBLEM SOLVED!!!" ........ "but only a very specific, special case of the problem, when properly crafted and constructed, can almost be solved in blah blah blah and we ain't gonna give you any details, only a catchy title to mislead you into reading the article". [–] 7 points8 points sorry, this has been archived and can no longer be voted on I don't understand the problem with the article. The title is "XY Traveling Salesman Problem Solved". That's about as exactly correct as a title can be. They solved the XY Traveling Salesman Problem. Not a restricted special case of it. They solved it. The article described exactly what the problem was and how quickly their algorithm runs. What's the problem? [–] 0 points1 point sorry, this has been archived and can no longer be voted on The full text is available here (it seems a reader of the linked article bought it & rehosted it, it's in the comments) [PDF Warning] Link I have confirmed that it is the actual paper, and I doubt that the guy would've put a virus in it. But still, take at your own risk. [–] -2 points-1 points sorry, this has been archived and can no longer be voted on 31$ for an article??!!!!!!!

[–] -3 points-2 points

sorry, this has been archived and can no longer be voted on

The general traveling salesman is NP complete which means that any NP problem can be translated into a traveling salesman problem in polynomial time. In other words, if you can solve one NP complete problem in polynomial time you can solve all NP problems in polynomial time. As for the \$31.50 part, here is the paper. EDIT Upon reading the link I posted, I am not sure that is actually the paper as it has no justifications for its "theorems"

[–] 0 points1 point

sorry, this has been archived and can no longer be voted on

Upon reading the link I posted, I am not sure that is actually the paper as it has no justifications for its "theorems"

It's not; it seems to be some extremely abbreviated version of the paper. The first three pages of the actual paper are available here: http://www.opt.math.tu-graz.ac.at/~cela/papers/xyReni1.pdf


text:
anonymous one year ago Two boys want to use a seesaw, and they need to move the seesaw so that their weights will balance out. The formula is given by w1 • d1 = w2 • d2 where w1 = weight of the first boy, d1 = distance of the first boy from the fulcrum, w2 = weight of the second boy, and d2 = distance of the second boy from the fulcrum. Rewrite the formula to solve for d2. Show all steps in your work.

1. welshfella

w1 . d1 = w2 . d2 divide both sides by w2 and you have it.

2. anonymous

Oh ok that's not that hard at all so u understand when one thing multiplies on one side it divides on other

3. anonymous

So what they are saying make d2 the subject

4. anonymous

Which means d2 must be left alone in one side so you w2 is multiplying by d2 just divide it in the other side

5. anonymous

w1 x d1 = w2 x d2 1 = 4 ?? im soz im confused by this

6. anonymous

Where did u get 1=4

7. anonymous

well multiplying idek how ;-;

8. Goose.

Do you have any numbers to plug in?

9. anonymous

No you won't get 1=4 you will get d2= ( w1 times d1)/ w2

10. anonymous

There are no numbers in question which u need to plug in

11. welshfella

you are give no values yor answer will be a formula with d2 as the subject

12. imqwerty

$w _{1}\times d _{1}=w _{2}\times d _{2}$divide both sides by w2 u get-$\frac{ (w _{1} \times d _{1}) }{ w _{2} }=d _{2}$

13. welshfella

exactly - that's what i said to do in my first post

14. anonymous

so then its just w2 = d2 thats the final answer

15. anonymous

?*

16. welshfella

No the answer is the formula that imqwerty gave in his post

17. welshfella

|dw:1442499349177:dw|

18. welshfella

that's it

19. anonymous

oh ok thx soz i just suck at math su thanks XD

20. welshfella

yw

Find more explanations on OpenStudy


text:
Magma - Maple Programming Help

Home : Support : Online Help : Mathematics : Algebra : Magma : Magma/IsomorphicCopy

Magma

 IsomorphicCopy
 produce an isomorphic copy of a magma

 Calling Sequence IsomorphicCopy( src, p )

Parameters

 src - Array; source magma whose Cayley table is to be moved p - permlist; permutation of 1..n to effect the isomorphic copy

Description

 • The IsomorphicCopy( 'src', 'p' ) command uses the permutation p to produce an isomorphic copy of the source magma src, in such a way that the permutation p is then an isomorphism from src to the isomorphic copy. The isomorphic copy is returned.
 • The operation of this command is effected by calling TransportStructure.
 • If you want to generate a number of isomorphic copies of a magma (or several magmas), you can use the TransportStructure command instead, which allows you to re-use an Array which you pass to the command.
 • If the permutation p is not supplied, then a randomly generated permutation is used.

Examples

 > $\mathrm{with}\left(\mathrm{Magma}\right):$
 > $m≔⟨⟨⟨1|2|3⟩,⟨2|3|1⟩,⟨3|1|2⟩⟩⟩$
 ${m}{≔}\left[\begin{array}{ccc}{1}& {2}& {3}\\ {2}& {3}& {1}\\ {3}& {1}& {2}\end{array}\right]$ (1)
 > $p≔\left[2,1,3\right]$
 ${p}{≔}\left[{2}{,}{1}{,}{3}\right]$ (2)
 > $\mathrm{m2}≔\mathrm{IsomorphicCopy}\left(m,p\right)$
 ${\mathrm{m2}}{≔}\left[\begin{array}{ccc}{3}& {1}& {2}\\ {1}& {2}& {3}\\ {2}& {3}& {1}\end{array}\right]$ (3)
 > $\mathrm{AreIsomorphic}\left(m,\mathrm{m2}\right)$
 ${\mathrm{true}}$ (4)

Compatibility

 • The Magma[IsomorphicCopy] command was introduced in Maple 16.
 • For more information on Maple 16 changes, see Updates in Maple 16.


text:
Trigonometry - Fourier Transform

The Fourier transform decomposes a function of time (e.g., a periodic signal) into the frequencies that make it up, similarly to how a musical chord can be expressed as the amplitude (or loudness) of its constituent notes.

With Fourier tranforms, you are able to create any waveform out of other combined waveforms.

• frequency $\frac{1}{P}$

Documentation / Reference

Data Science
Data Analysis
Statistics
Data Science
Linear Algebra Mathematics
Trigonometry


text:
# biweight_scale¶

astropy.stats.biweight.biweight_scale(data, c=9.0, M=None, axis=None, modify_sample_size=False, *, ignore_nan=False)[source]

Compute the biweight scale.

The biweight scale is a robust statistic for determining the standard deviation of a distribution. It is the square root of the biweight midvariance. It is given by:

$\zeta_{biscl} = \sqrt{n} \ \frac{\sqrt{\sum_{|u_i| < 1} \ (x_i - M)^2 (1 - u_i^2)^4}} {|(\sum_{|u_i| < 1} \ (1 - u_i^2) (1 - 5u_i^2))|}$

where $$x$$ is the input data, $$M$$ is the sample median (or the input location) and $$u_i$$ is given by:

$u_{i} = \frac{(x_i - M)}{c * MAD}$

where $$c$$ is the tuning constant and $$MAD$$ is the median absolute deviation. The biweight midvariance tuning constant c is typically 9.0 (the default).

If $$MAD$$ is zero, then zero will be returned.

For the standard definition of biweight scale, $$n$$ is the total number of points in the array (or along the input axis, if specified). That definition is used if modify_sample_size is False, which is the default.

However, if modify_sample_size = True, then $$n$$ is the number of points for which $$|u_i| < 1$$ (i.e. the total number of non-rejected values), i.e.

$n = \sum_{|u_i| < 1} \ 1$

which results in a value closer to the true standard deviation for small sample sizes or for a large number of rejected values.

Parameters
dataarray_like

Input array or object that can be converted to an array. data can be a MaskedArray.

cfloat, optional

Tuning constant for the biweight estimator (default = 9.0).

Mfloat or array_like, optional

The location estimate. If M is a scalar value, then its value will be used for the entire array (or along each axis, if specified). If M is an array, then its must be an array containing the location estimate along each axis of the input array. If None (default), then the median of the input array will be used (or along each axis, if specified).

axisNone, int, or tuple of int, optional

The axis or axes along which the biweight scales are computed. If None (default), then the biweight scale of the flattened input array will be computed.

modify_sample_sizebool, optional

If False (default), then the sample size used is the total number of elements in the array (or along the input axis, if specified), which follows the standard definition of biweight scale. If True, then the sample size is reduced to correct for any rejected values (i.e. the sample size used includes only the non-rejected values), which results in a value closer to the true standard deviation for small sample sizes or for a large number of rejected values.

ignore_nanbool, optional

Whether to ignore NaN values in the input data.

Returns
biweight_scale

The biweight scale of the input data. If axis is None then a scalar will be returned, otherwise a ndarray will be returned.

References

1

Beers, Flynn, and Gebhardt (1990; AJ 100, 32) (https://ui.adsabs.harvard.edu/abs/1990AJ….100…32B)

2

https://www.itl.nist.gov/div898/software/dataplot/refman2/auxillar/biwscale.htm

Examples

Generate random variates from a Gaussian distribution and return the biweight scale of the distribution:

>>> import numpy as np
>>> from astropy.stats import biweight_scale
>>> rand = np.random.default_rng(12345)
>>> biscl = biweight_scale(rand.standard_normal(1000))
>>> print(biscl)
1.0239311812635818


text:
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# If c = 3.905 X 10^x and x is a negative integer, what is x?

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If $$c = 3.905*10^x$$ and x is a negative integer, what is x?

(1) $$0.00005 < c < 0.0005$$

(2) $$10^{-4} < c <10^{-3}$$

Originally posted by Mo2men on 18 Jul 2017, 09:14.
Last edited by Bunuel on 18 Jul 2017, 09:18, edited 1 time in total.
Edited the question.
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18 Jul 2017, 10:25
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If $$c = 3.905*10^x$$ and x is a negative integer, what is x?

(1) $$0.00005 < c < 0.0005$$
clearly C must be 0.0003905...
Or $$5*10^{-5}<c<5*10^{-4}$$. So x is -4
Sufficient

(2) $$10^{-4} < c <10^{-3}$$
this means x is between 0.0001 and 0.001...
Also 3.905*10^{-4} will be > 10^-4 and <10*10^-4...
Again x is -4
Sufficient

D
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text:
# Calculus/Vector calculus identities

 ← Vector calculus Calculus Inverting vector calculus operators → Vector calculus identities

In this chapter, numerous identities related to the gradient (${\displaystyle \nabla f}$), directional derivative (${\displaystyle (\mathbf {V} \cdot \nabla )f}$, ${\displaystyle (\mathbf {V} \cdot \nabla )\mathbf {F} }$), divergence (${\displaystyle \nabla \cdot \mathbf {F} }$), Laplacian (${\displaystyle \nabla ^{2}f}$, ${\displaystyle \nabla ^{2}\mathbf {F} }$), and curl (${\displaystyle \nabla \times \mathbf {F} }$) will be derived.

## Notation

To simplify the derivation of various vector identities, the following notation will be utilized:

• The coordinates ${\displaystyle x,y,z}$ will instead be denoted with ${\displaystyle x_{1},x_{2},x_{3}}$ respectively.
• Given an arbitrary vector ${\displaystyle \mathbf {F} }$, then ${\displaystyle F_{i}}$ will denote the ${\displaystyle i^{\text{th}}}$ entry of ${\displaystyle \mathbf {F} }$ where ${\displaystyle i=1,2,3}$. All vectors will be assumed to be denoted by Cartesian basis vectors (${\displaystyle \mathbf {i} ,\mathbf {j} ,\mathbf {k} }$) unless otherwise specified: ${\displaystyle \mathbf {F} =F_{1}\mathbf {i} +F_{2}\mathbf {j} +F_{3}\mathbf {k} }$.
• Given an arbitrary expression ${\displaystyle f:\{1,2,3\}\to \mathbb {R} }$ that assigns a real number to each index ${\displaystyle i=1,2,3}$, then ${\displaystyle (i,f(i))}$ will denote the vector whose entries are determined by ${\displaystyle f}$. For example, ${\displaystyle \mathbf {F} =(i,F_{i})}$.
• Given an arbitrary expression ${\displaystyle f:\{1,2,3\}\to \mathbb {R} }$ that assigns a real number to each index ${\displaystyle i=1,2,3}$, then ${\displaystyle \sum _{i}f(i)}$ will denote the sum ${\displaystyle f(1)+f(2)+f(3)}$. For example, ${\displaystyle \nabla \cdot \mathbf {F} =\sum _{i}{\frac {\partial F_{i}}{\partial x_{i}}}}$.
• Given an index variable ${\displaystyle i\in \{1,2,3\}}$, ${\displaystyle i+1}$ will rotate ${\displaystyle i}$ forwards by 1, and ${\displaystyle i+2}$ will rotate ${\displaystyle i}$ forwards by 2. In essence, ${\displaystyle i+1=\left\{{\begin{array}{cc}i+1&(i=1,2)\\1&(i=3)\end{array}}\right.}$ and ${\displaystyle i+2=\left\{{\begin{array}{cc}3&(i=1)\\i-1&(i=2,3)\end{array}}\right.}$. For example, ${\displaystyle \mathbf {F} \times \mathbf {G} =(i,F_{i+1}G_{i+2}-F_{i+2}G_{i+1})}$.

As an example of using the above notation, consider the problem of expanding the triple cross product ${\displaystyle \mathbf {F} \times (\mathbf {G} \times \mathbf {H} )}$.

${\displaystyle \mathbf {F} \times (\mathbf {G} \times \mathbf {H} )=\mathbf {F} \times (i,G_{i+1}H_{i+2}-G_{i+2}H_{i+1})}$

${\displaystyle =(i,F_{i+1}(G_{i}H_{i+1}-G_{i+1}H_{i})-F_{i+2}(G_{i+2}H_{i}-G_{i}H_{i+2}))}$

${\displaystyle =(i,G_{i}(F_{i+1}H_{i+1}+F_{i+2}H_{i+2})-(F_{i+1}G_{i+1}+F_{i+2}G_{i+2})H_{i})}$

${\displaystyle =(i,G_{i}(F_{i}H_{i}+F_{i+1}H_{i+1}+F_{i+2}H_{i+2})-(F_{i}G_{i}+F_{i+1}G_{i+1}+F_{i+2}G_{i+2})H_{i})}$

${\displaystyle =(i,G_{i}(\mathbf {F} \cdot \mathbf {H} )-(\mathbf {F} \cdot \mathbf {G} )H_{i})}$

${\displaystyle =(\mathbf {F} \cdot \mathbf {H} )\mathbf {G} -(\mathbf {F} \cdot \mathbf {G} )\mathbf {H} }$

Therefore: ${\displaystyle \mathbf {F} \times (\mathbf {G} \times \mathbf {H} )=(\mathbf {F} \cdot \mathbf {H} )\mathbf {G} -(\mathbf {F} \cdot \mathbf {G} )\mathbf {H} }$

As another example of using the above notation, consider the scalar triple product ${\displaystyle \mathbf {F} \cdot (\mathbf {G} \times \mathbf {H} )}$

${\displaystyle \mathbf {F} \cdot (\mathbf {G} \times \mathbf {H} )=\mathbf {F} \cdot (i,G_{i+1}H_{i+2}-G_{i+2}H_{i+1})}$

${\displaystyle =\sum _{i}F_{i}(G_{i+1}H_{i+2}-G_{i+2}H_{i+1})}$

${\displaystyle =(\sum _{i}F_{i}G_{i+1}H_{i+2})-(\sum _{i}F_{i}G_{i+2}H_{i+1})}$

The index ${\displaystyle i}$ in the above summations can be shifted by fixed amounts without changing the sum. For example, ${\displaystyle \sum _{i}F_{i}G_{i+1}H_{i+2}=\sum _{i}F_{i+1}G_{i+2}H_{i}=\sum _{i}F_{i+2}G_{i}H_{i+1}}$. This allows:

${\displaystyle (\sum _{i}F_{i}G_{i+1}H_{i+2})-(\sum _{i}F_{i}G_{i+2}H_{i+1})=(\sum _{i}F_{i+2}G_{i}H_{i+1})-(\sum _{i}F_{i+1}G_{i}H_{i+2})=(\sum _{i}F_{i+1}G_{i+2}H_{i})-(\sum _{i}F_{i+2}G_{i+1}H_{i})}$

${\displaystyle \implies \mathbf {F} \cdot (i,G_{i+1}H_{i+2}-G_{i+2}H_{i+1})=\mathbf {G} \cdot (i,H_{i+1}F_{i+2}-H_{i+2}F_{i+1})=\mathbf {H} \cdot (i,F_{i+1}G_{i+2}-F_{i+2}G_{i+1})}$

${\displaystyle \implies \mathbf {F} \cdot (\mathbf {G} \times \mathbf {H} )=\mathbf {G} \cdot (\mathbf {H} \times \mathbf {F} )=\mathbf {H} \cdot (\mathbf {F} \times \mathbf {G} )}$

which establishes the cyclical property of the scalar triple product.

Given scalar fields, ${\displaystyle f}$ and ${\displaystyle g}$, then ${\displaystyle \nabla (f+g)=(\nabla f)+(\nabla g)}$.

Derivation

${\displaystyle \nabla (f+g)=(i,{\frac {\partial }{\partial x_{i}}}(f+g))}$ ${\displaystyle =(i,{\frac {\partial f}{\partial x_{i}}}+{\frac {\partial g}{\partial x_{i}}})}$ ${\displaystyle =(i,{\frac {\partial f}{\partial x_{i}}})+(i,{\frac {\partial g}{\partial x_{i}}})}$ ${\displaystyle =(\nabla f)+(\nabla g)}$

Given scalar fields ${\displaystyle f}$ and ${\displaystyle g}$, then ${\displaystyle \nabla (fg)=(\nabla f)g+f(\nabla g)}$. If ${\displaystyle f}$ is a constant ${\displaystyle c}$, then ${\displaystyle \nabla (cg)=c(\nabla g)}$.

Derivation

${\displaystyle \nabla (fg)=(i,{\frac {\partial }{\partial x_{i}}}(fg))}$ ${\displaystyle =(i,{\frac {\partial f}{\partial x_{i}}}g+f{\frac {\partial g}{\partial x_{i}}})}$ ${\displaystyle =(i,{\frac {\partial f}{\partial x_{i}}})g+f(i,{\frac {\partial g}{\partial x_{i}}})}$ ${\displaystyle =(\nabla f)g+f(\nabla g)}$

Given vector fields ${\displaystyle \mathbf {F} }$ and ${\displaystyle \mathbf {G} }$, then ${\displaystyle \nabla (\mathbf {F} \cdot \mathbf {G} )=((\mathbf {F} \cdot \nabla )\mathbf {G} +\mathbf {F} \times (\nabla \times \mathbf {G} ))+((\mathbf {G} \cdot \nabla )\mathbf {F} +\mathbf {G} \times (\nabla \times \mathbf {F} ))}$

Derivation

${\displaystyle \nabla (\mathbf {F} \cdot \mathbf {G} )=(i,{\frac {\partial }{\partial x_{i}}}(\mathbf {F} \cdot \mathbf {G} ))}$ ${\displaystyle =(i,{\frac {\partial }{\partial x_{i}}}(\sum _{j}F_{j}G_{j}))}$ ${\displaystyle =(i,\sum _{j}({\frac {\partial F_{j}}{\partial x_{i}}}G_{j}+F_{j}{\frac {\partial G_{j}}{\partial x_{i}}}))}$ ${\displaystyle =(i,\sum _{j}F_{j}{\frac {\partial G_{j}}{\partial x_{i}}})+(i,\sum _{j}G_{j}{\frac {\partial F_{j}}{\partial x_{i}}})}$

${\displaystyle =(i,F_{i}{\frac {\partial G_{i}}{\partial x_{i}}}+F_{i+1}{\frac {\partial G_{i+1}}{\partial x_{i}}}+F_{i+2}{\frac {\partial G_{i+2}}{\partial x_{i}}})}$ ${\displaystyle +(i,G_{i}{\frac {\partial F_{i}}{\partial x_{i}}}+G_{i+1}{\frac {\partial F_{i+1}}{\partial x_{i}}}+G_{i+2}{\frac {\partial F_{i+2}}{\partial x_{i}}})}$

${\displaystyle =(i,(F_{i}{\frac {\partial G_{i}}{\partial x_{i}}}+F_{i+1}{\frac {\partial G_{i}}{\partial x_{i+1}}}+F_{i+2}{\frac {\partial G_{i}}{\partial x_{i+2}}})+((F_{i+1}{\frac {\partial G_{i+1}}{\partial x_{i}}}-F_{i+1}{\frac {\partial G_{i}}{\partial x_{i+1}}})+(F_{i+2}{\frac {\partial G_{i+2}}{\partial x_{i}}}-F_{i+2}{\frac {\partial G_{i}}{\partial x_{i+2}}})))}$ ${\displaystyle +(i,(G_{i}{\frac {\partial F_{i}}{\partial x_{i}}}+G_{i+1}{\frac {\partial F_{i}}{\partial x_{i+1}}}+G_{i+2}{\frac {\partial F_{i}}{\partial x_{i+2}}})+((G_{i+1}{\frac {\partial F_{i+1}}{\partial x_{i}}}-G_{i+1}{\frac {\partial F_{i}}{\partial x_{i+1}}})+(G_{i+2}{\frac {\partial F_{i+2}}{\partial x_{i}}}-G_{i+2}{\frac {\partial F_{i}}{\partial x_{i+2}}})))}$

${\displaystyle =(i,\sum _{j}F_{j}{\frac {\partial G_{i}}{\partial x_{j}}})+(i,F_{i+1}({\frac {\partial G_{i+1}}{\partial x_{i}}}-{\frac {\partial G_{i}}{\partial x_{i+1}}})-F_{i+2}({\frac {\partial G_{i}}{\partial x_{i+2}}}-{\frac {\partial G_{i+2}}{\partial x_{i}}}))}$ ${\displaystyle +(i,\sum _{j}G_{j}{\frac {\partial F_{i}}{\partial x_{j}}})+(i,G_{i+1}({\frac {\partial F_{i+1}}{\partial x_{i}}}-{\frac {\partial F_{i}}{\partial x_{i+1}}})-G_{i+2}({\frac {\partial F_{i}}{\partial x_{i+2}}}-{\frac {\partial F_{i+2}}{\partial x_{i}}}))}$

${\displaystyle =(i,(\mathbf {F} \cdot \nabla )G_{i})+\mathbf {F} \times (i,{\frac {\partial G_{i+2}}{\partial x_{i+1}}}-{\frac {\partial G_{i+1}}{\partial x_{i+2}}})+(i,(\mathbf {G} \cdot \nabla )F_{i})+\mathbf {G} \times (i,{\frac {\partial F_{i+2}}{\partial x_{i+1}}}-{\frac {\partial F_{i+1}}{\partial x_{i+2}}})}$

${\displaystyle =((\mathbf {F} \cdot \nabla )\mathbf {G} +\mathbf {F} \times (\nabla \times \mathbf {G} ))+((\mathbf {G} \cdot \nabla )\mathbf {F} +\mathbf {G} \times (\nabla \times \mathbf {F} ))}$

Given scalar fields ${\displaystyle f_{1},f_{2},\dots ,f_{n}}$ and an ${\displaystyle n}$ input function ${\displaystyle g(y_{1},y_{2},\dots ,y_{n})}$, then ${\displaystyle \nabla (g(f_{1},f_{2},\dots ,f_{n}))={\frac {\partial g}{\partial y_{1}}}{\bigg |}_{y_{1}=f_{1}}(\nabla f_{1})+{\frac {\partial g}{\partial y_{2}}}{\bigg |}_{y_{2}=f_{2}}(\nabla f_{2})+\dots +{\frac {\partial g}{\partial y_{n}}}{\bigg |}_{y_{n}=f_{n}}(\nabla f_{n})}$.

Derivation

${\displaystyle \nabla (g(f_{1},f_{2},\dots ,f_{n}))=(i,{\frac {\partial }{\partial x_{i}}}(g(f_{1},f_{2},\dots ,f_{n})))}$ ${\displaystyle =(i,{\frac {\partial g}{\partial y_{1}}}{\bigg |}_{y_{1}=f_{1}}{\frac {\partial f_{1}}{\partial x_{i}}}+{\frac {\partial g}{\partial y_{2}}}{\bigg |}_{y_{2}=f_{2}}{\frac {\partial f_{2}}{\partial x_{i}}}+\dots +{\frac {\partial g}{\partial y_{n}}}{\bigg |}_{y_{n}=f_{n}}{\frac {\partial f_{n}}{\partial x_{i}}})}$ ${\displaystyle ={\frac {\partial g}{\partial y_{1}}}{\bigg |}_{y_{1}=f_{1}}(i,{\frac {\partial f_{1}}{\partial x_{i}}})+{\frac {\partial g}{\partial y_{2}}}{\bigg |}_{y_{2}=f_{2}}(i,{\frac {\partial f_{2}}{\partial x_{i}}})+\dots +{\frac {\partial g}{\partial y_{n}}}{\bigg |}_{y_{n}=f_{n}}(i,{\frac {\partial f_{n}}{\partial x_{i}}})}$ ${\displaystyle ={\frac {\partial g}{\partial y_{1}}}{\bigg |}_{y_{1}=f_{1}}(\nabla f_{1})+{\frac {\partial g}{\partial y_{2}}}{\bigg |}_{y_{2}=f_{2}}(\nabla f_{2})+\dots +{\frac {\partial g}{\partial y_{n}}}{\bigg |}_{y_{n}=f_{n}}(\nabla f_{n})}$

## Directional Derivative Identities

Given vector fields ${\displaystyle \mathbf {V} }$ and ${\displaystyle \mathbf {W} }$, and scalar field ${\displaystyle f}$, then ${\displaystyle ((\mathbf {V} +\mathbf {W} )\cdot \nabla )f=(\mathbf {V} \cdot \nabla )f+(\mathbf {W} \cdot \nabla )f}$.

When ${\displaystyle \mathbf {F} }$ is a vector field, it is also the case that: ${\displaystyle ((\mathbf {V} +\mathbf {W} )\cdot \nabla )\mathbf {F} =(\mathbf {V} \cdot \nabla )\mathbf {F} +(\mathbf {W} \cdot \nabla )\mathbf {F} }$.

Derivation

For scalar fields:

${\displaystyle ((\mathbf {V} +\mathbf {W} )\cdot \nabla )f=\sum _{i}((V_{i}+W_{i}){\frac {\partial f}{\partial x_{i}}})}$ ${\displaystyle =\sum _{i}(V_{i}{\frac {\partial f}{\partial x_{i}}}+W_{i}{\frac {\partial f}{\partial x_{i}}})}$ ${\displaystyle =\sum _{i}(V_{i}{\frac {\partial f}{\partial x_{i}}})+\sum _{i}(W_{i}{\frac {\partial f}{\partial x_{i}}})}$ ${\displaystyle =(\mathbf {V} \cdot \nabla )f+(\mathbf {W} \cdot \nabla )f}$

For vector fields:

${\displaystyle ((\mathbf {V} +\mathbf {W} )\cdot \nabla )\mathbf {F} =(i,((\mathbf {V} +\mathbf {W} )\cdot \nabla )F_{i})}$ ${\displaystyle =(i,(\mathbf {V} \cdot \nabla )F_{i}+(\mathbf {W} \cdot \nabla )F_{i})}$ ${\displaystyle =(\mathbf {V} \cdot \nabla )\mathbf {F} +(\mathbf {W} \cdot \nabla )\mathbf {F} }$

Given vector field ${\displaystyle \mathbf {V} }$, and scalar fields ${\displaystyle v}$ and ${\displaystyle f}$, then ${\displaystyle ((v\mathbf {V} )\cdot \nabla )f=v((\mathbf {V} \cdot \nabla )f)}$.

When ${\displaystyle \mathbf {F} }$ is a vector field, it is also the case that: ${\displaystyle ((v\mathbf {V} )\cdot \nabla )\mathbf {F} =v((\mathbf {V} \cdot \nabla )\mathbf {F} )}$.

Derivation

For scalar fields:

${\displaystyle ((v\mathbf {V} )\cdot \nabla )f=\sum _{i}(vV_{i}{\frac {\partial f}{\partial x_{i}}})}$ ${\displaystyle =v\sum _{i}(V_{i}{\frac {\partial f}{\partial x_{i}}})}$ ${\displaystyle =v((\mathbf {V} \cdot \nabla )f)}$

For vector fields:

${\displaystyle ((v\mathbf {V} )\cdot \nabla )\mathbf {F} =(i,((v\mathbf {V} )\cdot \nabla )F_{i})}$ ${\displaystyle =(i,v((\mathbf {V} \cdot \nabla )F_{i}))}$ ${\displaystyle =v((\mathbf {V} \cdot \nabla )\mathbf {F} )}$

Given vector field ${\displaystyle \mathbf {V} }$, and scalar fields ${\displaystyle f}$ and ${\displaystyle g}$, then ${\displaystyle (\mathbf {V} \cdot \nabla )(f+g)=(\mathbf {V} \cdot \nabla )f+(\mathbf {V} \cdot \nabla )g}$.

When ${\displaystyle \mathbf {F} }$ and ${\displaystyle \mathbf {G} }$ are vector fields, it is also the case that: ${\displaystyle (\mathbf {V} \cdot \nabla )(\mathbf {F} +\mathbf {G} )=(\mathbf {V} \cdot \nabla )\mathbf {F} +(\mathbf {V} \cdot \nabla )\mathbf {G} }$.

Derivation

For scalar fields:

${\displaystyle (\mathbf {V} \cdot \nabla )(f+g)=\sum _{i}(V_{i}{\frac {\partial }{\partial x_{i}}}(f+g))}$ ${\displaystyle =\sum _{i}(V_{i}{\frac {\partial f}{\partial x_{i}}}+V_{i}{\frac {\partial g}{\partial x_{i}}})}$ ${\displaystyle =\sum _{i}(V_{i}{\frac {\partial f}{\partial x_{i}}})+\sum _{i}(V_{i}{\frac {\partial g}{\partial x_{i}}})}$ ${\displaystyle =(\mathbf {V} \cdot \nabla )f+(\mathbf {V} \cdot \nabla )g}$

For vector fields:

${\displaystyle (\mathbf {V} \cdot \nabla )(\mathbf {F} +\mathbf {G} )=(i,(\mathbf {V} \cdot \nabla )(F_{i}+G_{i}))}$ ${\displaystyle =(i,(\mathbf {V} \cdot \nabla )F_{i}+(\mathbf {V} \cdot \nabla )G_{i})}$ ${\displaystyle =(\mathbf {V} \cdot \nabla )\mathbf {F} +(\mathbf {V} \cdot \nabla )\mathbf {G} }$

Given vector field ${\displaystyle \mathbf {V} }$, and scalar fields ${\displaystyle f}$ and ${\displaystyle g}$, then ${\displaystyle (\mathbf {V} \cdot \nabla )(fg)=((\mathbf {V} \cdot \nabla )f)g+f((\mathbf {V} \cdot \nabla )g)}$

If ${\displaystyle \mathbf {G} }$ is a vector field, it is also the case that: ${\displaystyle (\mathbf {V} \cdot \nabla )(f\mathbf {G} )=((\mathbf {V} \cdot \nabla )f)\mathbf {G} +f((\mathbf {V} \cdot \nabla )\mathbf {G} )}$

Derivation

For scalar fields:

${\displaystyle (\mathbf {V} \cdot \nabla )(fg)=\sum _{i}V_{i}{\frac {\partial }{\partial x_{i}}}(fg)}$ ${\displaystyle =\sum _{i}V_{i}({\frac {\partial f}{\partial x_{i}}}g+f{\frac {\partial g}{\partial x_{i}}})}$ ${\displaystyle =(\sum _{i}V_{i}{\frac {\partial f}{\partial x_{i}}})g+f(\sum _{i}V_{i}{\frac {\partial g}{\partial x_{i}}})}$ ${\displaystyle =((\mathbf {V} \cdot \nabla )f)g+f((\mathbf {V} \cdot \nabla )g)}$

For vector fields:

${\displaystyle (\mathbf {V} \cdot \nabla )(f\mathbf {G} )=(i,(\mathbf {V} \cdot \nabla )(fG_{i}))}$ ${\displaystyle =(i,((\mathbf {V} \cdot \nabla )f)G_{i}+f((\mathbf {V} \cdot \nabla )G_{i}))}$ ${\displaystyle =((\mathbf {V} \cdot \nabla )f)\mathbf {G} +f((\mathbf {V} \cdot \nabla )\mathbf {G} )}$

Given vector fields ${\displaystyle \mathbf {V} }$, ${\displaystyle \mathbf {F} }$, and ${\displaystyle \mathbf {G} }$, then ${\displaystyle (\mathbf {V} \cdot \nabla )(\mathbf {F} \cdot \mathbf {G} )=((\mathbf {V} \cdot \nabla )\mathbf {F} )\cdot \mathbf {G} +\mathbf {F} \cdot ((\mathbf {V} \cdot \nabla )\mathbf {G} )}$

Derivation

${\displaystyle (\mathbf {V} \cdot \nabla )(\mathbf {F} \cdot \mathbf {G} )=\sum _{i}V_{i}{\frac {\partial }{\partial x_{i}}}(\mathbf {F} \cdot \mathbf {G} )}$ ${\displaystyle =\sum _{i}V_{i}{\frac {\partial }{\partial x_{i}}}\sum _{j}(F_{j}G_{j})}$ ${\displaystyle =\sum _{i}\sum _{j}V_{i}{\frac {\partial }{\partial x_{i}}}(F_{j}G_{j})}$ ${\displaystyle =\sum _{i}\sum _{j}V_{i}({\frac {\partial F_{j}}{\partial x_{i}}}G_{j}+F_{j}{\frac {\partial G_{j}}{\partial x_{i}}})}$ ${\displaystyle =\sum _{j}((\sum _{i}V_{i}{\frac {\partial F_{j}}{\partial x_{i}}})G_{j})+\sum _{j}(F_{j}(\sum _{i}V_{i}{\frac {\partial G_{j}}{\partial x_{i}}}))}$ ${\displaystyle =\sum _{j}(((\mathbf {V} \cdot \nabla )F_{j})G_{j})+\sum _{j}(F_{j}((\mathbf {V} \cdot \nabla )G_{j}))}$ ${\displaystyle =((\mathbf {V} \cdot \nabla )\mathbf {F} )\cdot \mathbf {G} +\mathbf {F} \cdot ((\mathbf {V} \cdot \nabla )\mathbf {G} )}$

Given vector fields ${\displaystyle \mathbf {V} }$, ${\displaystyle \mathbf {F} }$, and ${\displaystyle \mathbf {G} }$, then ${\displaystyle (\mathbf {V} \cdot \nabla )(\mathbf {F} \times \mathbf {G} )=((\mathbf {V} \cdot \nabla )\mathbf {F} )\times \mathbf {G} +\mathbf {F} \times ((\mathbf {V} \cdot \nabla )\mathbf {G} )}$

Derivation

${\displaystyle (\mathbf {V} \cdot \nabla )(\mathbf {F} \times \mathbf {G} )=(i,(\mathbf {V} \cdot \nabla )(F_{i+1}G_{i+2}-F_{i+2}G_{i+1}))}$ ${\displaystyle =(i,\sum _{j}V_{j}{\frac {\partial }{\partial x_{j}}}(F_{i+1}G_{i+2}-F_{i+2}G_{i+1}))}$ ${\displaystyle =(i,\sum _{j}V_{j}(({\frac {\partial F_{i+1}}{\partial x_{j}}}G_{i+2}+F_{i+1}{\frac {\partial G_{i+2}}{\partial x_{j}}})-({\frac {\partial F_{i+2}}{\partial x_{j}}}G_{i+1}+F_{i+2}{\frac {\partial G_{i+1}}{\partial x_{j}}})))}$ ${\displaystyle =(i,(\sum _{j}V_{j}{\frac {\partial F_{i+1}}{\partial x_{j}}})G_{i+2}-(\sum _{j}V_{j}{\frac {\partial F_{i+2}}{\partial x_{j}}})G_{i+1})+(i,F_{i+1}(\sum _{j}V_{j}{\frac {\partial G_{i+2}}{\partial x_{j}}})-F_{i+2}(\sum _{j}V_{j}{\frac {\partial G_{i+1}}{\partial x_{j}}}))}$ ${\displaystyle =(i,((\mathbf {V} \cdot \nabla )F_{i+1})G_{i+2}-((\mathbf {V} \cdot \nabla )F_{i+2})G_{i+1})+(i,F_{i+1}((\mathbf {V} \cdot \nabla )G_{i+2})-F_{i+2}((\mathbf {V} \cdot \nabla )G_{i+1}))}$ ${\displaystyle =((\mathbf {V} \cdot \nabla )\mathbf {F} )\times \mathbf {G} +\mathbf {F} \times ((\mathbf {V} \cdot \nabla )\mathbf {G} )}$

## Divergence Identities

Given vector fields ${\displaystyle \mathbf {F} }$ and ${\displaystyle \mathbf {G} }$, then ${\displaystyle \nabla \cdot (\mathbf {F} +\mathbf {G} )=(\nabla \cdot \mathbf {F} )+(\nabla \cdot \mathbf {G} )}$.

Derivation

${\displaystyle \nabla \cdot (\mathbf {F} +\mathbf {G} )=\sum _{i}({\frac {\partial }{\partial x_{i}}}(F_{i}+G_{i}))}$ ${\displaystyle =(\sum _{i}{\frac {\partial F_{i}}{\partial x_{i}}})+(\sum _{i}{\frac {\partial G_{i}}{\partial x_{i}}})}$ ${\displaystyle =(\nabla \cdot \mathbf {F} )+(\nabla \cdot \mathbf {G} )}$

Given a scalar field ${\displaystyle f}$ and a vector field ${\displaystyle \mathbf {G} }$, then ${\displaystyle \nabla \cdot (f\mathbf {G} )=(\nabla f)\cdot \mathbf {G} +f(\nabla \cdot \mathbf {G} )}$. If ${\displaystyle f}$ is a constant ${\displaystyle c}$, then ${\displaystyle \nabla \cdot (c\mathbf {G} )=c(\nabla \cdot \mathbf {G} )}$. If ${\displaystyle \mathbf {G} }$ is a constant ${\displaystyle \mathbf {C} }$, then ${\displaystyle \nabla \cdot (f\mathbf {C} )=(\nabla f)\cdot \mathbf {C} }$.

Derivation

${\displaystyle \nabla \cdot (f\mathbf {G} )=\sum _{i}{\frac {\partial }{\partial x_{i}}}(fG_{i})}$ ${\displaystyle =\sum _{i}({\frac {\partial f}{\partial x_{i}}}G_{i}+f{\frac {\partial G_{i}}{\partial x_{i}}})}$ ${\displaystyle =\sum _{i}({\frac {\partial f}{\partial x_{i}}}G_{i})+f\sum _{i}{\frac {\partial G_{i}}{\partial x_{i}}}}$ ${\displaystyle =(\nabla f)\cdot \mathbf {G} +f(\nabla \cdot \mathbf {G} )}$

Given vector fields ${\displaystyle \mathbf {F} }$ and ${\displaystyle \mathbf {G} }$, then ${\displaystyle \nabla \cdot (\mathbf {F} \times \mathbf {G} )=(\nabla \times \mathbf {F} )\cdot \mathbf {G} -\mathbf {F} \cdot (\nabla \times \mathbf {G} )}$.

Derivation

${\displaystyle \nabla \cdot (\mathbf {F} \times \mathbf {G} )=\sum _{i}{\frac {\partial }{\partial x_{i}}}(F_{i+1}G_{i+2}-F_{i+2}G_{i+1})}$ ${\displaystyle =\sum _{i}(({\frac {\partial F_{i+1}}{\partial x_{i}}}G_{i+2}+F_{i+1}{\frac {\partial G_{i+2}}{\partial x_{i}}})-({\frac {\partial F_{i+2}}{\partial x_{i}}}G_{i+1}+F_{i+2}{\frac {\partial G_{i+1}}{\partial x_{i}}}))}$ ${\displaystyle =\sum _{i}(({\frac {\partial F_{i+2}}{\partial x_{i+1}}}G_{i}+F_{i}{\frac {\partial G_{i+1}}{\partial x_{i+2}}})-({\frac {\partial F_{i+1}}{\partial x_{i+2}}}G_{i}+F_{i}{\frac {\partial G_{i+2}}{\partial x_{i+1}}}))}$ ${\displaystyle =\sum _{i}(({\frac {\partial F_{i+2}}{\partial x_{i+1}}}-{\frac {\partial F_{i+1}}{\partial x_{i+2}}})G_{i}-F_{i}({\frac {\partial G_{i+2}}{\partial x_{i+1}}}-{\frac {\partial G_{i+1}}{\partial x_{i+2}}}))}$ ${\displaystyle =\sum _{i}(\nabla \times \mathbf {F} )_{i}G_{i}-\sum _{i}F_{i}(\nabla \times \mathbf {G} )_{i}}$ ${\displaystyle =(\nabla \times \mathbf {F} )\cdot \mathbf {G} -\mathbf {F} \cdot (\nabla \times \mathbf {G} )}$

In the above derivation, the third equality is established by cycling the terms inside a sum. For example: ${\displaystyle \sum _{i}{\frac {\partial F_{i+1}}{\partial x_{i}}}G_{i+2}=\sum _{i}{\frac {\partial F_{i+2}}{\partial x_{i+1}}}G_{i}}$ by replacing ${\displaystyle i}$ with ${\displaystyle i+1}$. Different terms can be cycled independently: ${\displaystyle \sum _{i}({\frac {\partial F_{i+1}}{\partial x_{i}}}G_{i+2}+F_{i+1}{\frac {\partial G_{i+2}}{\partial x_{i}}})=\sum _{i}({\frac {\partial F_{i+2}}{\partial x_{i+1}}}G_{i}+F_{i}{\frac {\partial G_{i+1}}{\partial x_{i+2}}})}$

The following identity is a very important property regarding vector fields which are the curl of another vector field. A vector field which is the curl of another vector field is divergence free. Given vector field ${\displaystyle \mathbf {F} }$, then ${\displaystyle \nabla \cdot (\nabla \times \mathbf {F} )=0}$

Derivation

${\displaystyle \nabla \cdot (\nabla \times \mathbf {F} )=\nabla \cdot (i,{\frac {\partial F_{i+2}}{\partial x_{i+1}}}-{\frac {\partial F_{i+1}}{\partial x_{i+2}}})}$ ${\displaystyle =\sum _{i}{\frac {\partial }{\partial x_{i}}}({\frac {\partial F_{i+2}}{\partial x_{i+1}}}-{\frac {\partial F_{i+1}}{\partial x_{i+2}}})}$ ${\displaystyle =\sum _{i}({\frac {\partial ^{2}F_{i+2}}{\partial x_{i}\partial x_{i+1}}}-{\frac {\partial ^{2}F_{i+1}}{\partial x_{i}\partial x_{i+2}}})}$ ${\displaystyle =\sum _{i}{\frac {\partial ^{2}F_{i+2}}{\partial x_{i}\partial x_{i+1}}}-\sum _{i}{\frac {\partial ^{2}F_{i+1}}{\partial x_{i+2}\partial x_{i}}}}$ ${\displaystyle =\sum _{i}{\frac {\partial ^{2}F_{i+2}}{\partial x_{i}\partial x_{i+1}}}-\sum _{i}{\frac {\partial ^{2}F_{i+2}}{\partial x_{i}\partial x_{i+1}}}}$ ${\displaystyle =0}$

## Laplacian Identities

Given scalar fields ${\displaystyle f}$ and ${\displaystyle g}$, then ${\displaystyle \nabla ^{2}(f+g)=(\nabla ^{2}f)+(\nabla ^{2}g)}$

When ${\displaystyle \mathbf {F} }$ and ${\displaystyle \mathbf {G} }$ are vector fields, it is also the case that: ${\displaystyle \nabla ^{2}(\mathbf {F} +\mathbf {G} )=(\nabla ^{2}\mathbf {F} )+(\nabla ^{2}\mathbf {G} )}$

Derivation

For scalar fields:

${\displaystyle \nabla ^{2}(f+g)=\sum _{i}{\frac {\partial ^{2}}{\partial x_{i}^{2}}}(f+g)}$ ${\displaystyle =\sum _{i}({\frac {\partial ^{2}f}{\partial x_{i}^{2}}}+{\frac {\partial ^{2}g}{\partial x_{i}^{2}}})}$ ${\displaystyle =(\sum _{i}{\frac {\partial ^{2}f}{\partial x_{i}^{2}}})+(\sum _{i}{\frac {\partial ^{2}g}{\partial x_{i}^{2}}})}$ ${\displaystyle =(\nabla ^{2}f)+(\nabla ^{2}g)}$

For vector fields:

${\displaystyle \nabla ^{2}(\mathbf {F} +\mathbf {G} )=(i,\nabla ^{2}(F_{i}+G_{i}))}$ ${\displaystyle =(i,(\nabla ^{2}F_{i})+(\nabla ^{2}G_{i}))}$ ${\displaystyle =(\nabla ^{2}\mathbf {F} )+(\nabla ^{2}\mathbf {G} )}$

Given scalar fields ${\displaystyle f}$ and ${\displaystyle g}$, then ${\displaystyle \nabla ^{2}(fg)=(\nabla ^{2}f)g+2(\nabla f)\cdot (\nabla g)+f(\nabla ^{2}g)}$

When ${\displaystyle \mathbf {G} }$ is a vector field, it is also the case that ${\displaystyle \nabla ^{2}(f\mathbf {G} )=(\nabla ^{2}f)\mathbf {G} +2((\nabla f)\cdot \nabla )\mathbf {G} +f(\nabla ^{2}\mathbf {G} )}$

Derivation

For scalar fields:

${\displaystyle \nabla ^{2}(fg)=\sum _{i}{\frac {\partial ^{2}}{\partial x_{i}^{2}}}(fg)}$ ${\displaystyle =\sum _{i}{\frac {\partial }{\partial x_{i}}}({\frac {\partial f}{\partial x_{i}}}g+f{\frac {\partial g}{\partial x_{i}}})}$ ${\displaystyle =\sum _{i}({\frac {\partial ^{2}f}{\partial x_{i}^{2}}}g+2{\frac {\partial f}{\partial x_{i}}}{\frac {\partial g}{\partial x_{i}}}+f{\frac {\partial ^{2}g}{\partial x_{i}^{2}}})}$ ${\displaystyle =(\sum _{i}{\frac {\partial ^{2}f}{\partial x_{i}^{2}}})g+2\sum _{i}({\frac {\partial f}{\partial x_{i}}}{\frac {\partial g}{\partial x_{i}}})+f(\sum _{i}{\frac {\partial ^{2}g}{\partial x_{i}^{2}}})}$ ${\displaystyle =(\nabla ^{2}f)g+2(\nabla f)\cdot (\nabla g)+f(\nabla ^{2}g)}$

For vector fields:

${\displaystyle \nabla ^{2}(f\mathbf {G} )=(i,\nabla ^{2}(fG_{i}))}$ ${\displaystyle =(i,(\nabla ^{2}f)G_{i}+2(\nabla f)\cdot (\nabla G_{i})+f(\nabla ^{2}G_{i}))}$ ${\displaystyle =(i,(\nabla ^{2}f)G_{i})+2(i,((\nabla f)\cdot \nabla )G_{i})+(i,f(\nabla ^{2}G_{i}))}$ ${\displaystyle =(\nabla ^{2}f)\mathbf {G} +2((\nabla f)\cdot \nabla )\mathbf {G} +f(\nabla ^{2}\mathbf {G} )}$

## Curl Identities

Given vector fields ${\displaystyle \mathbf {F} }$ and ${\displaystyle \mathbf {G} }$, then ${\displaystyle \nabla \times (\mathbf {F} +\mathbf {G} )=(\nabla \times \mathbf {F} )+(\nabla \times \mathbf {G} )}$

Derivation

${\displaystyle \nabla \times (\mathbf {F} +\mathbf {G} )=(i,{\frac {\partial }{\partial x_{i+1}}}(F_{i+2}+G_{i+2})-{\frac {\partial }{\partial x_{i+2}}}(F_{i+1}+G_{i+1}))}$ ${\displaystyle =(i,({\frac {\partial F_{i+2}}{\partial x_{i+1}}}+{\frac {\partial G_{i+2}}{\partial x_{i+1}}})-({\frac {\partial F_{i+1}}{\partial x_{i+2}}}+{\frac {\partial G_{i+1}}{\partial x_{i+2}}}))}$ ${\displaystyle =(i,{\frac {\partial F_{i+2}}{\partial x_{i+1}}}-{\frac {\partial F_{i+1}}{\partial x_{i+2}}})+(i,{\frac {\partial G_{i+2}}{\partial x_{i+1}}}-{\frac {\partial G_{i+1}}{\partial x_{i+2}}})}$ ${\displaystyle =(\nabla \times \mathbf {F} )+(\nabla \times \mathbf {G} )}$

Given scalar field ${\displaystyle f}$ and vector field ${\displaystyle \mathbf {G} }$, then ${\displaystyle \nabla \times (f\mathbf {G} )=(\nabla f)\times \mathbf {G} +f(\nabla \times \mathbf {G} )}$. If ${\displaystyle f}$ is a constant ${\displaystyle c}$, then ${\displaystyle \nabla \times (c\mathbf {G} )=c(\nabla \times \mathbf {G} )}$. If ${\displaystyle \mathbf {G} }$ is a constant ${\displaystyle \mathbf {C} }$, then ${\displaystyle \nabla \times (f\mathbf {C} )=(\nabla f)\times \mathbf {C} }$.

Derivation

${\displaystyle \nabla \times (f\mathbf {G} )=(i,{\frac {\partial }{\partial x_{i+1}}}(fG_{i+2})-{\frac {\partial }{\partial x_{i+2}}}(fG_{i+1}))}$ ${\displaystyle =(i,({\frac {\partial f}{\partial x_{i+1}}}G_{i+2}+f{\frac {\partial G_{i+2}}{\partial x_{i+1}}})-({\frac {\partial f}{\partial x_{i+2}}}G_{i+1}+f{\frac {\partial G_{i+1}}{\partial x_{i+2}}}))}$ ${\displaystyle =(i,{\frac {\partial f}{\partial x_{i+1}}}G_{i+2}-{\frac {\partial f}{\partial x_{i+2}}}G_{i+1})+f(i,{\frac {\partial G_{i+2}}{\partial x_{i+1}}}-{\frac {\partial G_{i+1}}{\partial x_{i+2}}})}$ ${\displaystyle =(\nabla f)\times \mathbf {G} +f(\nabla \times \mathbf {G} )}$

Given vector fields ${\displaystyle \mathbf {F} }$ and ${\displaystyle \mathbf {G} }$, then ${\displaystyle \nabla \times (\mathbf {F} \times \mathbf {G} )=((\nabla \cdot \mathbf {G} )\mathbf {F} +(\mathbf {G} \cdot \nabla )\mathbf {F} )-((\nabla \cdot \mathbf {F} )\mathbf {G} +(\mathbf {F} \cdot \nabla )\mathbf {G} )}$

Derivation

${\displaystyle \nabla \times (\mathbf {F} \times \mathbf {G} )=\nabla \times (i,F_{i+1}G_{i+2}-F_{i+2}G_{i+1})}$ ${\displaystyle =(i,{\frac {\partial }{\partial x_{i+1}}}(F_{i}G_{i+1}-F_{i+1}G_{i})-{\frac {\partial }{\partial x_{i+2}}}(F_{i+2}G_{i}-F_{i}G_{i+2}))}$ ${\displaystyle =(i,(({\frac {\partial F_{i}}{\partial x_{i+1}}}G_{i+1}+F_{i}{\frac {\partial G_{i+1}}{\partial x_{i+1}}})-({\frac {\partial F_{i+1}}{\partial x_{i+1}}}G_{i}+F_{i+1}{\frac {\partial G_{i}}{\partial x_{i+1}}}))-(({\frac {\partial F_{i+2}}{\partial x_{i+2}}}G_{i}+F_{i+2}{\frac {\partial G_{i}}{\partial x_{i+2}}})-({\frac {\partial F_{i}}{\partial x_{i+2}}}G_{i+2}+F_{i}{\frac {\partial G_{i+2}}{\partial x_{i+2}}})))}$ ${\displaystyle =(i,F_{i}({\frac {\partial G_{i+1}}{\partial x_{i+1}}}+{\frac {\partial G_{i+2}}{\partial x_{i+2}}})-({\frac {\partial F_{i+1}}{\partial x_{i+1}}}+{\frac {\partial F_{i+2}}{\partial x_{i+2}}})G_{i}-(F_{i+1}{\frac {\partial G_{i}}{\partial x_{i+1}}}+F_{i+2}{\frac {\partial G_{i}}{\partial x_{i+2}}})+({\frac {\partial F_{i}}{\partial x_{i+1}}}G_{i+1}+{\frac {\partial F_{i}}{\partial x_{i+2}}}G_{i+2}))}$

${\displaystyle =(i,F_{i}({\frac {\partial G_{i}}{\partial x_{i}}}+{\frac {\partial G_{i+1}}{\partial x_{i+1}}}+{\frac {\partial G_{i+2}}{\partial x_{i+2}}})-({\frac {\partial F_{i}}{\partial x_{i}}}+{\frac {\partial F_{i+1}}{\partial x_{i+1}}}+{\frac {\partial F_{i+2}}{\partial x_{i+2}}})G_{i}}$ ${\displaystyle -(F_{i}{\frac {\partial G_{i}}{\partial x_{i}}}+F_{i+1}{\frac {\partial G_{i}}{\partial x_{i+1}}}+F_{i+2}{\frac {\partial G_{i}}{\partial x_{i+2}}})+({\frac {\partial F_{i}}{\partial x_{i}}}G_{i}+{\frac {\partial F_{i}}{\partial x_{i+1}}}G_{i+1}+{\frac {\partial F_{i}}{\partial x_{i+2}}}G_{i+2}))}$

${\displaystyle =(i,F_{i}(\nabla \cdot \mathbf {G} )-(\nabla \cdot \mathbf {F} )G_{i}-(\mathbf {F} \cdot \nabla )G_{i}+(\mathbf {G} \cdot \nabla )F_{i})}$ ${\displaystyle =(\nabla \cdot \mathbf {G} )\mathbf {F} -(\nabla \cdot \mathbf {F} )\mathbf {G} -(\mathbf {F} \cdot \nabla )\mathbf {G} +(\mathbf {G} \cdot \nabla )\mathbf {F} }$ ${\displaystyle =((\nabla \cdot \mathbf {G} )\mathbf {F} +(\mathbf {G} \cdot \nabla )\mathbf {F} )-((\nabla \cdot \mathbf {F} )\mathbf {G} +(\mathbf {F} \cdot \nabla )\mathbf {G} )}$

The following identity is a very important property of vector fields which are the gradient of a scalar field. A vector field which is the gradient of a scalar field is always irrotational. Given scalar field ${\displaystyle f}$, then ${\displaystyle \nabla \times (\nabla f)=\mathbf {0} }$

Derivation

${\displaystyle \nabla \times (\nabla f)=\nabla \times (i,{\frac {\partial f}{\partial x_{i}}})}$ ${\displaystyle =(i,{\frac {\partial }{\partial x_{i+1}}}({\frac {\partial f}{\partial x_{i+2}}})-{\frac {\partial }{\partial x_{i+2}}}({\frac {\partial f}{\partial x_{i+1}}}))}$ ${\displaystyle =(i,{\frac {\partial ^{2}f}{\partial x_{i+1}\partial x_{i+2}}}-{\frac {\partial ^{2}f}{\partial x_{i+2}\partial x_{i+1}}})}$ ${\displaystyle =(i,0)}$ ${\displaystyle =\mathbf {0} }$

The following identity is a complex, yet popular identity used for deriving the Helmholtz decomposition theorem. Given vector field ${\displaystyle \mathbf {F} }$, then ${\displaystyle \nabla \times (\nabla \times \mathbf {F} )=\nabla (\nabla \cdot \mathbf {F} )-\nabla ^{2}\mathbf {F} }$

Derivation

${\displaystyle \nabla \times (\nabla \times \mathbf {F} )=\nabla \times (i,{\frac {\partial F_{i+2}}{\partial x_{i+1}}}-{\frac {\partial F_{i+1}}{\partial x_{i+2}}})}$ ${\displaystyle =(i,{\frac {\partial }{\partial x_{i+1}}}({\frac {\partial F_{i+1}}{\partial x_{i}}}-{\frac {\partial F_{i}}{\partial x_{i+1}}})-{\frac {\partial }{\partial x_{i+2}}}({\frac {\partial F_{i}}{\partial x_{i+2}}}-{\frac {\partial F_{i+2}}{\partial x_{i}}}))}$ ${\displaystyle =(i,({\frac {\partial ^{2}F_{i+1}}{\partial x_{i}\partial x_{i+1}}}+{\frac {\partial ^{2}F_{i+2}}{\partial x_{i}\partial x_{i+2}}})-({\frac {\partial ^{2}F_{i}}{\partial x_{i+1}^{2}}}+{\frac {\partial ^{2}F_{i}}{\partial x_{i+2}^{2}}}))}$ ${\displaystyle =(i,({\frac {\partial ^{2}F_{i}}{\partial x_{i}\partial x_{i}}}+{\frac {\partial ^{2}F_{i+1}}{\partial x_{i}\partial x_{i+1}}}+{\frac {\partial ^{2}F_{i+2}}{\partial x_{i}\partial x_{i+2}}})-({\frac {\partial ^{2}F_{i}}{\partial x_{i}^{2}}}+{\frac {\partial ^{2}F_{i}}{\partial x_{i+1}^{2}}}+{\frac {\partial ^{2}F_{i}}{\partial x_{i+2}^{2}}}))}$ ${\displaystyle =(i,{\frac {\partial }{\partial x_{i}}}({\frac {\partial F_{i}}{\partial x_{i}}}+{\frac {\partial F_{i+1}}{\partial x_{i+1}}}+{\frac {\partial F_{i+2}}{\partial x_{i+2}}})-\nabla ^{2}F_{i})}$ ${\displaystyle =(i,{\frac {\partial }{\partial x_{i}}}(\nabla \cdot \mathbf {F} )-\nabla ^{2}F_{i})}$ ${\displaystyle =\nabla (\nabla \cdot \mathbf {F} )-\nabla ^{2}\mathbf {F} }$

## Basis Vector Identities

The Cartesian basis vectors ${\displaystyle \mathbf {i} }$, ${\displaystyle \mathbf {j} }$, and ${\displaystyle \mathbf {k} }$ are the same at all points in space. However, in other coordinate systems like cylindrical coordinates or spherical coordinates, the basis vectors can change with respect to position.

In cylindrical coordinates, the unit-length mutually perpendicular basis vectors are ${\displaystyle {\hat {\mathbf {\rho } }}=(\cos \phi )\mathbf {i} +(\sin \phi )\mathbf {j} }$, ${\displaystyle {\hat {\mathbf {\phi } }}=(-\sin \phi )\mathbf {i} +(\cos \phi )\mathbf {j} }$, and ${\displaystyle {\hat {\mathbf {z} }}=\mathbf {k} }$ at position ${\displaystyle (\rho ,\phi ,z)}$ which corresponds to Cartesian coordinates ${\displaystyle (\rho \cos \phi ,\rho \sin \phi ,z)}$.

In spherical coordinates, the unit-length mutually perpendicular basis vectors are ${\displaystyle {\hat {\mathbf {r} }}=(\sin \theta \cos \phi )\mathbf {i} +(\sin \theta \sin \phi )\mathbf {j} +(\cos \theta )\mathbf {k} }$, ${\displaystyle {\hat {\mathbf {\theta } }}=(\cos \theta \cos \phi )\mathbf {i} +(\cos \theta \sin \phi )\mathbf {j} +(-\sin \theta )\mathbf {k} }$, and ${\displaystyle {\hat {\mathbf {\phi } }}=(-\sin \phi )\mathbf {i} +(\cos \phi )\mathbf {j} }$ at position ${\displaystyle (r,\theta ,\phi )}$ which corresponds to Cartesian coordinates ${\displaystyle (r\sin \theta \cos \phi ,r\sin \theta \sin \phi ,r\cos \theta )}$.

It should be noted that ${\displaystyle {\hat {\mathbf {\phi } }}}$ is the same in both cylindrical and spherical coordinates.

This section will compute the directional derivative and Laplacian for the following vectors since these quantities do not immediately follow from the formulas established for the directional derivative and Laplacian for scalar fields in various coordinate systems.

${\displaystyle {\hat {\mathbf {\rho } }}}$ which is the unit length vector that points away from the z-axis and is perpendicular to the z-axis.
${\displaystyle {\hat {\mathbf {\phi } }}}$ which is the unit length vector that points around the z-axis in a counterclockwise direction and is both parallel to the xy-plane and perpendicular to the position vector projected onto the xy-plane.
${\displaystyle {\hat {\mathbf {r} }}}$ which is the unit length vector that points away from the origin.
${\displaystyle {\hat {\mathbf {\theta } }}}$ which is the unit length vector that is perpendicular to the position vector and points "south" on the surface of a sphere that is centered on the origin.

The following quantities are also important:

${\displaystyle \rho }$ which is the perpendicular distance from the z-axis.
${\displaystyle \phi }$ which is the azimuth: the counterclockwise angle of the position vector relative to the x-axis after being projected onto the xy-plane.
${\displaystyle r}$ which is the distance from the origin.
${\displaystyle \theta }$ which is the angle of the position vector to the z-axis.

### Vector ${\displaystyle {\hat {\mathbf {\rho } }}}$

${\displaystyle {\hat {\mathbf {\rho } }}}$ only changes with respect to ${\displaystyle \phi }$: ${\displaystyle {\frac {\partial {\hat {\mathbf {\rho } }}}{\partial \phi }}={\hat {\mathbf {\phi } }}}$.

Given vector field ${\displaystyle \mathbf {V} =\mathbf {V} _{\perp }+v_{\phi }{\hat {\mathbf {\phi } }}}$ where ${\displaystyle \mathbf {V} _{\perp }}$ is always orthogonal to ${\displaystyle {\hat {\mathbf {\phi } }}}$, then ${\displaystyle (\mathbf {V} \cdot \nabla ){\hat {\mathbf {\rho } }}={\frac {v_{\phi }}{\rho }}{\hat {\mathbf {\phi } }}}$

Derivation

Using cylindrical coordinates, let ${\displaystyle \mathbf {V} _{\perp }=v_{\rho }{\hat {\mathbf {\rho } }}+v_{z}{\hat {\mathbf {z} }}}$

The cylindrical coordinate version of the directional derivative gives:

${\displaystyle (\mathbf {V} \cdot \nabla ){\hat {\mathbf {\rho } }}=((v_{\rho }{\hat {\mathbf {\rho } }}+v_{\phi }{\hat {\mathbf {\phi } }}+v_{z}{\hat {\mathbf {z} }})\cdot \nabla ){\hat {\mathbf {\rho } }}}$ ${\displaystyle =v_{\rho }{\frac {\partial {\hat {\mathbf {\rho } }}}{\partial \rho }}+{\frac {v_{\phi }}{\rho }}{\frac {\partial {\hat {\mathbf {\rho } }}}{\partial \phi }}+v_{z}{\frac {\partial {\hat {\mathbf {\rho } }}}{\partial z}}}$ ${\displaystyle =v_{\rho }\mathbf {0} +{\frac {v_{\phi }}{\rho }}{\hat {\mathbf {\phi } }}+v_{z}\mathbf {0} }$ ${\displaystyle ={\frac {v_{\phi }}{\rho }}{\hat {\mathbf {\phi } }}}$

${\displaystyle \nabla ^{2}{\hat {\mathbf {\rho } }}=-{\frac {1}{\rho ^{2}}}{\hat {\mathbf {\rho } }}}$

Derivation

Using the cylindrical coordinate version of the Laplacian,

${\displaystyle \nabla ^{2}{\hat {\mathbf {\rho } }}={\frac {1}{\rho }}{\frac {\partial }{\partial \rho }}(\rho {\frac {\partial {\hat {\mathbf {\rho } }}}{\partial \rho }})+{\frac {1}{\rho ^{2}}}{\frac {\partial ^{2}{\hat {\mathbf {\rho } }}}{\partial \phi ^{2}}}+{\frac {\partial ^{2}{\hat {\mathbf {\rho } }}}{\partial z^{2}}}}$ ${\displaystyle ={\frac {1}{\rho }}{\frac {\partial }{\partial \rho }}(\rho \mathbf {0} )+{\frac {1}{\rho ^{2}}}{\frac {\partial {\hat {\mathbf {\phi } }}}{\partial \phi }}+{\frac {\partial \mathbf {0} }{\partial z}}}$ ${\displaystyle =-{\frac {1}{\rho ^{2}}}{\hat {\mathbf {\rho } }}}$

### Vector ${\displaystyle {\hat {\mathbf {\phi } }}}$

${\displaystyle {\hat {\mathbf {\phi } }}}$ only changes with respect to ${\displaystyle \phi }$: ${\displaystyle {\frac {\partial {\hat {\mathbf {\phi } }}}{\partial \phi }}=-{\hat {\mathbf {\rho } }}}$.

Given vector field ${\displaystyle \mathbf {V} =\mathbf {V} _{\perp }+v_{\phi }{\hat {\mathbf {\phi } }}}$ where ${\displaystyle \mathbf {V} _{\perp }}$ is always orthogonal to ${\displaystyle {\hat {\mathbf {\phi } }}}$, then ${\displaystyle (\mathbf {V} \cdot \nabla ){\hat {\mathbf {\phi } }}=-{\frac {v_{\phi }}{\rho }}{\hat {\mathbf {\rho } }}}$

Derivation

Using cylindrical coordinates, let ${\displaystyle \mathbf {V} _{\perp }=v_{\rho }{\hat {\mathbf {\rho } }}+v_{z}{\hat {\mathbf {z} }}}$

The cylindrical coordinate version of the directional derivative gives:

${\displaystyle (\mathbf {V} \cdot \nabla ){\hat {\mathbf {\phi } }}=((v_{\rho }{\hat {\mathbf {\rho } }}+v_{\phi }{\hat {\mathbf {\phi } }}+v_{z}{\hat {\mathbf {z} }})\cdot \nabla ){\hat {\mathbf {\phi } }}}$ ${\displaystyle =v_{\rho }{\frac {\partial {\hat {\mathbf {\rho } }}}{\partial \phi }}+{\frac {v_{\phi }}{\rho }}{\frac {\partial {\hat {\mathbf {\phi } }}}{\partial \phi }}+v_{z}{\frac {\partial {\hat {\mathbf {\phi } }}}{\partial z}}}$ ${\displaystyle =v_{\rho }\mathbf {0} +{\frac {v_{\phi }}{\rho }}(-{\hat {\mathbf {\rho } }})+v_{z}\mathbf {0} }$ ${\displaystyle =-{\frac {v_{\phi }}{\rho }}{\hat {\mathbf {\rho } }}}$

${\displaystyle \nabla ^{2}{\hat {\mathbf {\phi } }}=-{\frac {1}{\rho ^{2}}}{\hat {\mathbf {\phi } }}}$

Derivation

Using the cylindrical coordinate version of the Laplacian,

${\displaystyle \nabla ^{2}{\hat {\mathbf {\phi } }}={\frac {1}{\rho }}{\frac {\partial }{\partial \rho }}(\rho {\frac {\partial {\hat {\mathbf {\phi } }}}{\partial \rho }})+{\frac {1}{\rho ^{2}}}{\frac {\partial ^{2}{\hat {\mathbf {\phi } }}}{\partial \phi ^{2}}}+{\frac {\partial ^{2}{\hat {\mathbf {\phi } }}}{\partial z^{2}}}}$ ${\displaystyle ={\frac {1}{\rho }}{\frac {\partial }{\partial \rho }}(\rho \mathbf {0} )-{\frac {1}{\rho ^{2}}}{\frac {\partial {\hat {\mathbf {\rho } }}}{\partial \phi }}+{\frac {\partial \mathbf {0} }{\partial z}}}$ ${\displaystyle =-{\frac {1}{\rho ^{2}}}{\hat {\mathbf {\phi } }}}$

### Vector ${\displaystyle {\hat {\mathbf {r} }}}$

${\displaystyle {\hat {\mathbf {r} }}}$ changes with respect to ${\displaystyle \theta }$ and ${\displaystyle \phi }$: ${\displaystyle {\frac {\partial {\hat {\mathbf {r} }}}{\partial \theta }}={\hat {\mathbf {\theta } }}}$ and ${\displaystyle {\frac {\partial {\hat {\mathbf {r} }}}{\partial \phi }}=(\sin \theta ){\hat {\mathbf {\phi } }}}$

Given vector field ${\displaystyle \mathbf {V} =v_{r}{\hat {\mathbf {r} }}+v_{\theta }{\hat {\mathbf {\theta } }}+v_{\phi }{\hat {\mathbf {\phi } }}}$, then ${\displaystyle (\mathbf {V} \cdot \nabla ){\hat {r}}={\frac {1}{r}}(v_{\theta }{\hat {\mathbf {\theta } }}+v_{\phi }{\hat {\mathbf {\phi } }})}$

Derivation

The spherical coordinate version of the directional derivative gives:

${\displaystyle (\mathbf {V} \cdot \nabla ){\hat {\mathbf {r} }}=((v_{r}{\hat {\mathbf {r} }}+v_{\theta }{\hat {\mathbf {\theta } }}+v_{\phi }{\hat {\mathbf {\phi } }})\cdot \nabla ){\hat {\mathbf {r} }}}$ ${\displaystyle =v_{r}{\frac {\partial {\hat {\mathbf {r} }}}{\partial r}}+{\frac {v_{\theta }}{r}}{\frac {\partial {\hat {\mathbf {r} }}}{\partial \theta }}+{\frac {v_{\phi }}{r\sin \theta }}{\frac {\partial {\hat {\mathbf {r} }}}{\partial \phi }}}$ ${\displaystyle =v_{r}\mathbf {0} +{\frac {v_{\theta }}{r}}{\hat {\mathbf {\theta } }}+{\frac {v_{\phi }}{r\sin \theta }}(\sin \theta {\hat {\mathbf {\phi } }})}$ ${\displaystyle ={\frac {1}{r}}(v_{\theta }{\hat {\mathbf {\theta } }}+v_{\phi }{\hat {\mathbf {\phi } }})}$

${\displaystyle \nabla ^{2}{\hat {\mathbf {r} }}=-{\frac {2}{r^{2}}}{\hat {\mathbf {r} }}}$

Derivation

The spherical coordinate version of the Laplacian gives:

${\displaystyle \nabla ^{2}{\hat {\mathbf {r} }}={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}(r^{2}{\frac {\partial {\hat {\mathbf {r} }}}{\partial r}})+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}(\sin \theta {\frac {\partial {\hat {\mathbf {r} }}}{\partial \theta }})+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}{\hat {\mathbf {r} }}}{\partial \phi ^{2}}}}$ ${\displaystyle ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}(r^{2}\mathbf {0} )+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}(\sin \theta {\hat {\mathbf {\theta } }})+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial }{\partial \phi }}(\sin \theta {\hat {\mathbf {\phi } }})}$ ${\displaystyle ={\frac {1}{r^{2}\sin \theta }}(\cos \theta {\hat {\mathbf {\theta } }}+\sin \theta (-{\hat {\mathbf {r} }}))+{\frac {1}{r^{2}\sin \theta }}(-\sin \theta {\hat {\mathbf {r} }}-\cos \theta {\hat {\mathbf {\theta } }})}$ ${\displaystyle =-{\frac {2}{r^{2}}}{\hat {\mathbf {r} }}}$

### Vector ${\displaystyle {\hat {\mathbf {\theta } }}}$

${\displaystyle {\hat {\mathbf {\theta } }}}$ changes with respect to ${\displaystyle \theta }$ and ${\displaystyle \phi }$: ${\displaystyle {\frac {\partial {\hat {\mathbf {\theta } }}}{\partial \theta }}=-{\hat {\mathbf {r} }}}$ and ${\displaystyle {\frac {\partial {\hat {\mathbf {\theta } }}}{\partial \phi }}=(\cos \theta ){\hat {\mathbf {\phi } }}}$

Given vector field ${\displaystyle \mathbf {V} =v_{r}{\hat {\mathbf {r} }}+v_{\theta }{\hat {\mathbf {\theta } }}+v_{\phi }{\hat {\mathbf {\phi } }}}$, then ${\displaystyle (\mathbf {V} \cdot \nabla ){\hat {\theta }}={\frac {1}{r}}(-v_{\theta }{\hat {\mathbf {r} }}+\cot \theta v_{\phi }{\hat {\mathbf {\phi } }})}$

Derivation

The spherical coordinate version of the directional derivative gives:

${\displaystyle (\mathbf {V} \cdot \nabla ){\hat {\mathbf {\theta } }}=((v_{r}{\hat {\mathbf {r} }}+v_{\theta }{\hat {\mathbf {\theta } }}+v_{\phi }{\hat {\mathbf {\phi } }})\cdot \nabla ){\hat {\mathbf {\theta } }}}$ ${\displaystyle =v_{r}{\frac {\partial {\hat {\mathbf {\theta } }}}{\partial r}}+{\frac {v_{\theta }}{r}}{\frac {\partial {\hat {\mathbf {\theta } }}}{\partial \theta }}+{\frac {v_{\phi }}{r\sin \theta }}{\frac {\partial {\hat {\mathbf {\theta } }}}{\partial \phi }}}$ ${\displaystyle =v_{r}\mathbf {0} +{\frac {v_{\theta }}{r}}(-{\hat {\mathbf {r} }})+{\frac {v_{\phi }}{r\sin \theta }}(\cos \theta {\hat {\mathbf {\phi } }})}$ ${\displaystyle ={\frac {1}{r}}(-v_{\theta }{\hat {\mathbf {r} }}+\cot \theta v_{\phi }{\hat {\mathbf {\phi } }})}$

${\displaystyle \nabla ^{2}{\hat {\mathbf {\theta } }}=-{\frac {1}{r^{2}\sin \theta }}(2\cos \theta {\hat {\mathbf {r} }}+\csc \theta {\hat {\mathbf {\theta } }})=-{\frac {1}{r^{2}\sin ^{2}\theta }}(\sin(2\theta ){\hat {\mathbf {r} }}+{\hat {\mathbf {\theta } }})}$

Derivation

The spherical coordinate version of the Laplacian gives:

${\displaystyle \nabla ^{2}{\hat {\mathbf {\theta } }}={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}(r^{2}{\frac {\partial {\hat {\mathbf {\theta } }}}{\partial r}})+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}(\sin \theta {\frac {\partial {\hat {\mathbf {\theta } }}}{\partial \theta }})+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}{\hat {\mathbf {\theta } }}}{\partial \phi ^{2}}}}$ ${\displaystyle ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}(r^{2}\mathbf {0} )+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}(\sin \theta (-{\hat {\mathbf {r} }}))+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial }{\partial \phi }}(\cos \theta {\hat {\mathbf {\phi } }})}$ ${\displaystyle =-{\frac {1}{r^{2}\sin \theta }}(\cos \theta {\hat {\mathbf {r} }}+\sin \theta {\hat {\mathbf {\theta } }})+{\frac {\cos \theta }{r^{2}\sin ^{2}\theta }}(-\sin \theta {\hat {\mathbf {r} }}-\cos \theta {\hat {\mathbf {\theta } }})}$ ${\displaystyle =-{\frac {1}{r^{2}\sin \theta }}(2\cos \theta {\hat {\mathbf {r} }}+(\sin \theta +{\frac {\cos ^{2}\theta }{\sin \theta }}){\hat {\mathbf {\theta } }})}$ ${\displaystyle =-{\frac {1}{r^{2}\sin \theta }}(2\cos \theta {\hat {\mathbf {r} }}+\csc \theta {\hat {\mathbf {\theta } }})}$


text:
# How to set page margin of 9“ by 6”

I am writing solution manual of matric mathematics book. To design this book in latex I am using this geometry.

\documentclass[10pt,a4paper,twoside]{report}

\usepackage[twocolumn]{geometry}

\setlength{\columnseprule}{0.4pt}

\usepackage{tikz-cd}

%\usepackage{fullpage}

\usepackage{amssymb}

%\usepackage{amused}

\usepackage{mathrsfs}

%\usepackage{eureka}

\usepackage[centertags]{amsmath}
\usepackage{amsfonts}

\usepackage{amssymb}

\usepackage{amsthm}

\newtheorem{theorem}{Theorem}

%\usepackage[all]{xy}

\usepackage{epsfig}

\usepackage{graphicx}

\usepackage{amsthm}

\usepackage{pgf,tikz,pgfplots}

\pgfplotsset{compat=1.15}
\usepackage{mathrsfs}

\usetikzlibrary{arrows}

%\usepackage{tikzpicture}

%\usepackage{breqn}

%\usepackage{verbatim}

%\usepackage{apst-all}

%\usepackage{xy-pic}

\usepackage{amssymb,latexsym}

\usepackage{amsfonts,amsmath}

\usepackage[utf8]{inputenc}

\usepackage[T1]{fontenc}

\usepackage{textcomp}

%\usepackage{gensymb}

\usepackage{amsmath,amssymb}

%\usepackage{dsfont}\let\mathbb\mathds
\usepackage{latexsym}

\usepackage{amssymb}

\usepackage{amsfonts}

%\usepackage{fancyhdr}

\usepackage[arrow,frame,matrix]{xy}

\usepackage{amsmath}

%\usepackage{lipsum}

%\usepackage{eqlist}

%\usepackage{fixltx2e}

\usepackage{xcolor}

%\usepackage[demo]{graphicx}

%\usepackage{mwe}

%\usepackage{breqn}

\renewcommand{\baselinestretch}{1.5}

%\theoremstyle{plain}
%\newtheorem{Thm}{Theorem}[section]
%\newtheorem{lem}[Thm]{Lemma}
%\newtheorem{pro}[Thm]{Proposition}
%\newtheorem{de}[Thm]{Definition}
%\newtheorem{re}[Thm]{Remark}
%\newtheorem{ex}[Thm]{Example}
%\newtheorem{cor}[Thm]{Corollary}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcommand\aug{\fboxsep=-\fboxrule\!\!\!\fbox{\strut}\!\!\!}

\theoremstyle{definition}

\newtheorem{Thm}{Theorem}[section]

\newtheorem{lem}[Thm]{Lemma}

\newtheorem{pro}[Thm]{Proposition}

\newtheorem{de}[Thm]{Definition}

\newtheorem{re}[Thm]{Remark}

\newtheorem{ex}[Thm]{Example}

\newtheorem{cor}[Thm]{Corollary}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newcommand{\BibUrl}{\url}
%\def\be{$$%\def\ee$$}
%\def\bea{\begin{eqnarray}$$%\def\ee$$a{\end{eqnarray}

\numberwithin{equation}{section}

\definecolor{uuuuuu}{rgb}{0.26666666666666666,0.26666666666666666,0.26666666666666666}

\begin{document}

\tableofcontents
\end{document}

But the press wants me set the page margin of 9" by 6". The maximum empty space on the top and bottom should be 1".

I do not know how to set this setting. I just know how to write in latex but how to set the mention page size margin? please help!

• You are already using the geometry package, read its manual. It has a super easy user interface. – daleif Jun 23 at 7:26
• @daleif I try this $\usepackage[total={6in, 9in}, twocolumn, margin=1in, twoside]{geometry}$ but when i run and get the dpf, then zoom page level in pdf to 100 it has still page margin 11 by 8.5. – Noor Aslam Jun 23 at 11:09
• Show your work. And please make your example minimal 85% of what you list here are not related to your question. – daleif Jun 23 at 11:23
• Also your question does not make much sense. Is it the text block that needs to be 6" by 9", 9" margins make no sense. – daleif Jun 23 at 11:26
• When you say total={6in,9in},margin=1in, the margin=... is negating the total=..., See the example just before section 3 in the manual. margin sets all the margins and recalculates the text block from that. But you just set attempted to set the text block via total. So instead of margin use left and say top as in the example in the marnual. – daleif Jun 23 at 11:30

You are not using geometry correctly. total={6in,9in} sets the size of the text block to be 6in by 9in. But your subsequent margin=1in recalculates the text block to have the width of the paper minus 2in and the height of the paper minus 2in.

See the example in the geometry manual right before section 3.

Here is an MWE with some added tikz to show where things are. Try for your self to see what happens with \usepackage[total={6in, 9in}, twocolumn, margin=1in, twoside]{geometry}

\documentclass[10pt,a4paper,twoside]{report}

\usepackage[twocolumn,total={6in,9in},left=1in,top=1in]{geometry}
\usepackage{kantlipsum}

\usepackage{tikz}
\usetikzlibrary{calc}

\begin{document}

\kant[2-5]

\begin{tikzpicture}[remember picture,overlay]

\draw ($(current page.north west)+(1in,0)$)     -- ++(0,-30cm);
\draw ($(current page.north west)+(1in+6in,0)$) -- ++(0,-30cm);

\draw ($(current page.north west)+(0,-1in)$)     -- ++(30cm,0);
\draw ($(current page.north west)+(0,-1in-9in)$) -- ++(30cm,0);

\end{tikzpicture}

\end{document}

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}

\node[minimum width=1cm,draw,circle] (a) at (0,0) {$\alpha$};
\node[minimum width=1cm,draw,circle] (b) at ($(a)+(2,0)$) {$\beta$};
\node[minimum width=1cm,draw,circle] (c) at ($(b)+(2,0)$) {$\gamma$};

\draw[->] (a) -- (b);
\draw[->] (c) -- (b);

\end{tikzpicture}
\end{document}

• @dalief Please brother set it for me I tried a lot but i can not get the desire. – Noor Aslam Jun 23 at 17:05