text:
# Ex-wizard that pretends he still is a wizard

I am pondering creating an epic-level wizard character, who through excessive use of Mordenkainen's Disjunction on artifacts has permanently lost all casting ability as well as all his class levels and drawn the ire of at least one god.

So now he has to start over from scratch with a non-casting class, but he likes to continue pretending to be the big-shot wizard he was, and occasionally even forgets he has lost anything at all.

"You fools shall feel the power of my fireball!"
- nothing happens -
"Oh, right, forgot about that..." - take out wand of fireball; use magic device
"Now, feel the power of my fireball!"

To be clear, this character has probably no hope of permanently fooling anyone but the most hapless rubes. The idea is not that he is a con artist, just that he got a little unhinged after his loss of power, and has his own ways of coping (mostly denial and rejection of reality).

I was thinking I need a rogue with high ranks in use magic device and a lot of wands and scrolls, with a pimped-out pet rat as his ex-familiar. But I am a little fuzzy on the details.

For example, can he bluff someone into being charmed, and what other options does he have to fake high-level spellcasting powers? He also need some sort of half-believable excuse for whenever his magic fails to materialize at all.

Which feats and items do I need to make this concept viable so that he can keep up with the rest of the party? (I would like to limit this to mostly core, perhaps with a sprinkling of extras, but nothing too obscure)

Is this even feasible with the costs of scrolls and wands at low levels?

Are there better ways to accomplish this concept instead of a rogue with use-magic-device?

• You ask about "low levels" and "epic". Do you mean low spell levels? What do you mean by core, SRD? Is this character concept intended to be playable? How many levels of "after the wizard" do we have to work with? Jul 19 '14 at 9:06
• @BrianBallsun-Stanton Well ,the character used to be a high-level wizard, so he needs a way to reliably cast low level spells and occasionally fake high level spells. regarding core/srd, I am a little unsure on the correct terminology. Basically I want to avoid too obscure sources that a lot of players will not know or have. Jul 19 '14 at 9:13
• For scrolls, though. You'll need to be able to make a DC 20 + caster level, which at Level 10, would be a DC 30. You'll have a 23 in UMD, not counting CHA bonus, which I've been assuming to be 0. Jul 19 '14 at 9:28
• I think this might be funnier if he fully knows he's faking it and embraces this. Having him behave alzheimer-like doesn't smell very fun.
– o0'.
Jul 19 '14 at 11:15
• Well, playing a cunning and deceiving trickster may have its charms, but the way I am imagining this is more a character who got a little unhinged after the devastating loss of his life work, and his way of coping is pretending it is just a minor setback, or simply ignoring it. Jul 22 '14 at 21:57

You won't be an effective combatant on UMD alone. You just won't, not without substantial splat support.

To be effective:

## Just build a mundane badass.

You don't need to fight with Fireballs for people to think you are a big-shot magic-slinger. Display your badassitude by knowing things (accomplished with skill checks) and taking charge (accomplished with role-playing).

In combat, you can rely on fighting with a sword or staff (think Gandalf in like 90% of the fights he gets in). Build your character around being proficient in this area.

When asked, why doesn't the character regularly use magic in combat? He doesn't need to. Using the high art of magic for mere killing is vulgar. Casting spells attracts the eyes of Hidden Evil Guys. Excuses abound, and he may even believe some of them himself, if you want.

In any case, I suggest you focus on making the character functional without even a scrap of magic. This gets much easier the more splatbooks you add. For some things you could add to your wishlist:

• The Warblade class from the Tome of Battle. Supreme intelligent fighter.
• The Factotum class from Dungeonscape. Has a tiny bit of spells, but overall it's a class for smart characters.
• The Knowledge Devotion feat from Complete Champion. Use your knowledge to hit harder.

## But when you really need magic, go all in.

You can maximize your UMD check on more or less any character. This allows you to keep some really big spell bombs in reserve.

This is different from regularly fighting with UMD. This is buying a few high level scrolls as soon as you can afford and reliably activate them. This should happen at a level where casting a single high-level spell completely ends the encounter. Have you seen what a casting of Black Tentacles does to low CR enemies? They usually just die.

Overall: Talk big, fight smart, carry an underpriced nuke.

Being a caster-from-items isn't possible on any sort of realistic budget.

Assuming that this is a character backstory concept and not literally the case, I would either play this as the wizard choosing to be level-drained back down to 1. As an intelligent "not a spellcaster" I can only recommend the Factotum from Dungeonscape. While it's not a pure "caster-as-items" it's a skillful intelligence based class that can take Arcane Familiar (Complete Arcane) after level 3. (Arcane dilettante can be represented as you learning a few spells again, despite the "curse" of disjunction."

# A Somewhat Obscure but Semi-official Option

The prestige class charlatan appears on pages 62-6 of Dragon #335. The Brilliant Gameologists' forum has a very short charlatan handbook that describes (but doesn't reproduce) many of the class's features.

The gist of the class is that, instead of casting spells, the charlatan convinces folks that he cast spells, is casting spells, and will cast spells by, respectively, taking credit for everyday occurrences, using misdirection and alchemical items, and making threats. Although not combat focused, it remains a pretty spiffy prestige class.

It's only 5 levels long, and a now-spell-less wizard would certainly find it appealing. The class even has the skill Use Magic Device as a class skill.

## An evaluation of being a Rogue depending purely on UMD.

There's not really much to say except UMD check success rate.

In regards to wands
Level 1 wands will be your friend.
Level 2+ wands will make you cry.
Look into getting a staff.

Feats
Skill Focus (Use Magic Device)
Magical Aptitude

You must be able to make a DC 20 UMD check regularly, and this is only really possible by Level 10 (that's when you start to make the check even on a natural 1). As a level 1, you'll fail half the time.

I suggest you make magic items in the mean-time, if your DM will allow you to take Craft Wondrous Items as an Ex-Wizard. Making an item that gives a +10 bonus to UMD costs 10,000 gp or 5,000 gp if you make it yourself.

Bonus Needed | Level | Cost to Buy | Cost to Make

+10 UMD | Level 1 | 10,000gp | 5,000gp
+9 UMD | Level 2 |  8,100gp | 4,050gp
+8 UMD | Level 3 |  6,400gp | 3,200gp
+7 UMD | Level 4 |  4,900gp | 2,450gp
+6 UMD | Level 5 |  3,600gp | 1,800gp

So, looking at this, assuming you took both feats I listed, the soonest you could get a 100% chance at casting wands is Level 5. It is the soonest that you can afford such an item, while still being within the guidelines that no one item you own is over half your expected wealth for your level.

If you make the item yourself (half cost), you can get the item by level 4. However, you lack a caster level to qualify for Craft Wondrous Items. It's up to your GM or a party crafter to get you where you need to be by level 4.

This is all assuming a CHA bonus of 0.

• Isn't there a limit on skill checks bonuses over which the item becomes epic ? Anyway, it could be interesting to mix +UMD and +CHA since +1 CHA for example is quite cheap (only 1000 gp), which is great when your CHA score is odd. Jul 21 '14 at 12:38

A wizard who lost their spellcasting still has wizard spells on their spell list and would not need UMD checks to activate appropriate scrolls/wands.

• well, this is supposed to be my character's backstory. No matter how nifty it is, a backstory cannot give me skills; but that is easily explained away by saying that this wizard's actions angered the gods so much that not only his spellcasting but all his class abilities were taken away. Jul 22 '14 at 21:54
• Hi pareus, welcome to RPG.StackExchange.com. We're not a forum, so this sort of answer isn't helpful. Please take a look at our About page to understand how we work. Jul 22 '14 at 23:24
• @HugoRune, that should be in the question. As is, this answer actually makes a good point, even if it needs to expand on it (I think), if you are indeed a Wizard who's simply lost their casting. Jul 23 '14 at 7:49


text:
# Use Newton’s method to find the solution of $2x^3+5x-6=0$ by using the following steps

Use Newton’s method to find the solution of $$2x^3+5x-6=0$$ by using the following steps

a) Find, $f(0)$ where $$f(x)=2x^3+5x-6$$

b) Find, $f(1)$ where $$f(x)=2x^3+5x-6$$

c) Choose a guess point $$x_0$$

d) Perform the iterations, until the values stabilize to the 8-digits after the decimal sign.

my work

for a)

newton formula

$$x_{n+1} = x_n -\frac{f(x_n )}{f'(x_n )}$$ , for $n = 0,1,2,3,...$

$$f' (x)=6x^2+5$$

$$x_1=x_0-\frac{-6}{5}$$

$$F(0)= x_0- \frac{-6}{5}$$

for b)

$$x_2=x_1-\frac{-6}{5}$$

for c)

Let $$x_0=1$$

$$x_1=1-\frac{-6}{5}=\frac{11}{5}$$

for d)

$$x_1=x_0-\frac{-6}{5}$$

$$x_2=x_1-\frac{-6}{5}$$

$$x_3=x_2-\frac{-6}{5}$$

$$x_4=x_3-\frac{-6}{5}$$

$$x_5=x_4-\frac{-6}{5}$$

is that correct ???

• I would like to add that Newton's method does not, in general, give the roots of a polynomial. It approximates it to any desired degree of accuracy (doubling in decimal accuracy roughly each iteration) but it won't find the EXACT solution. – Trogdor Jun 15 '15 at 7:03
• The goal of (a) and (b) is to see that $f(0)<0$ and $f(1)>0$, so there is a root of the polynomial in the interval $[0,1]$. Once you know this, you can choose a point $x_0$ accordingly, (c). – hjhjhj57 Jun 15 '15 at 7:07

As you wrote, Newton method updates the initial guess $x_0$ according to $$x_{n+1} = x_n -\frac{f(x_n )}{f'(x_n )}$$ So, if $f(x)=2 x^3+5 x-6$, $f'(x)=6 x^2+5$, this gives, after simplification $$x_{n+1} =\frac{4 x_n^3+6}{6 x_n^2+5}$$ As AjmalW answered, choose $x_0=1$ and start repeating the iterations. You will find $x_1=\frac{10}{11}$, $x_2=\frac{11986}{13255}$ and so on until the convergence criteria is met.
a) You have to determine $f(0)$, that is $f(0) = 2(0)^3+5(0)-6=\cdots$.
b) You have to determine $f(1)$. See above.
c) You have to guess a number, say $x_{0}=1$, since it's close to the zero point.
d) You have to solve $f(x)=0$ using the Newton's method. Since $f'(x_{0})=6x^{2}+5$, you have $$x_{1} = x_0 -\frac{f(x_0 )}{f'(x_0 )} = 10/11$$ where $x_{0}=1$. Find also $x_{2}$, $x_{3}$ etc.

format: plain


text:
# How do you find the integral of int [arccos(x)]^n if is an integer?

Oct 20, 2015

${I}_{n} = x {\left(\arccos x\right)}^{n} - n {\left(\arccos x\right)}^{n - 1} \sqrt{1 - {x}^{2}} - n \left(n - 1\right) {I}_{n - 2}$

#### Explanation:

$u = {\left(\arccos x\right)}^{n} \implies \mathrm{du} = n {\left(\arccos x\right)}^{n - 1} \cdot \left(- \frac{1}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}$

$\mathrm{dv} = \mathrm{dx} \implies v = x$

${I}_{n} = x {\left(\arccos x\right)}^{n} - \int x \cdot n {\left(\arccos x\right)}^{n - 1} \cdot \left(- \frac{1}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}$

${I}_{n} = x {\left(\arccos x\right)}^{n} + n \int {\left(\arccos x\right)}^{n - 1} \cdot \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

$u = {\left(\arccos x\right)}^{n - 1} \implies \mathrm{du} = \left(n - 1\right) {\left(\arccos x\right)}^{n - 2} \cdot \left(- \frac{1}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}$

$\mathrm{dv} = \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

$v = \int \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

$1 - {x}^{2} = t \implies - 2 x \mathrm{dx} = \mathrm{dt} \implies x \mathrm{dx} = - \frac{\mathrm{dt}}{2}$

$v = - \frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{t}} = - \frac{1}{2} {t}^{\frac{1}{2}} / \left(\frac{1}{2}\right) = - \sqrt{t} = - \sqrt{1 - {x}^{2}}$

${I}_{n} = x {\left(\arccos x\right)}^{n} + n \left[- {\left(\arccos x\right)}^{n - 1} \sqrt{1 - {x}^{2}} - \int \left(- \sqrt{1 - {x}^{2}}\right) \left(n - 1\right) {\left(\arccos x\right)}^{n - 2} \cdot \left(- \frac{1}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}\right]$

${I}_{n} = x {\left(\arccos x\right)}^{n} + n \left[- {\left(\arccos x\right)}^{n - 1} \sqrt{1 - {x}^{2}} - \left(n - 1\right) \int {\left(\arccos x\right)}^{n - 2} \mathrm{dx}\right]$

${I}_{n} = x {\left(\arccos x\right)}^{n} - n {\left(\arccos x\right)}^{n - 1} \sqrt{1 - {x}^{2}} - n \left(n - 1\right) {I}_{n - 2}$

We have recurrent formula, we only have to find ${I}_{1}$ and ${I}_{2}$ and that's trivial using Integration By Parts.


text:
Programming project 7

QuickSelect

Consider the following function select(A: List[String], k: Int): String:

def select(A: List[String], k: Int): String = {
val B = A.sorted
B(k)
}

This function returns the $$k$$th smallest value in the list. For instance, if k = 0, it returns the smallest element, when k = A.length - 1, it returns the largest element.

Note that even if the input list A is not sorted, the function still selects the $$k$$th element correctly.

Your task is to write a function quickSelect(A: List[String], k: Int): String that returns the same result as select, but which does not sort the list. Your code should be faster than sorting on average.

Hint: Use the idea of Quicksort. Pick a pivot, and partition the list into pieces based on the pivot. Then recursively look for an element in one of the pieces.

Write test code that measures the running time of select and quickSelect for a large, unsorted list, and prints out timing information. Which function is faster?

Mergesort without recursion

Mergesort can be implemented without recursion by using a queue of queues. You start by creating an empty queue $$Q$$. For each element $$x$$ of the input list $$L$$, make a queue with one element and enqueue it in $$Q$$. Then, as long as $$Q$$ contains more than one element, dequeue the first two elements, merge them, and enqueue the result. Finally return the contents of the single element remaining in $$Q$$.

Submit a Scala file queuesort.scala containing a function with the following argument and result type:

  def mergeSort(data: List[Int]): List[Int] = {
// fill in
}

The function should implement the strategy above for sorting the input list. Use the Scala library Queue.

Inside the source file, explain (in a comment) why this algorithm correctly sorts the list, and analyse its running time. What is the loop invariant?

In this project, we implement merge sort for a doubly-linked list.

• median: Node[T] This method returns the node in the middle of the list:
scala> a
scala> a.median.el
res5: Int = 2
scala> a.append(4)
scala> a.median.el
res7: Int = 2
scala> a.append(5)
scala> a.median.el
res9: Int = 3

Note that when the length of the list is odd, the median is the one exactly in the middle. When the length of the list is even, it's the last element of the left half.

You must not compute the length of the list—that is less efficient. Start by walking from the front and the rear of the list at the same time until the two references meet in the middle.

• split(n: Node[T]): DoublyLinkedList[T] This method splits the list after the node n. The front part of the list until (and including) node n remains in this list object, the remainder is returned as a new DoublyLinkedList object.

The method must run in $$O(1)$$ time. It should not create any new nodes (except for the sentinels of the new list)—all the original list nodes should be used.

It's okay for n to be the front sentinel of this list, but of course not the rear sentinel.

Here are some examples:

scala> a
scala> val n = a.first.next.next
n: Node[String] = Node@4a9a6cc8
scala> n.el
res9: String = fun
scala> val b = a.split(n)
scala> a
scala> a.split(a.last)
scala> a
scala> val c = b.split(b.first)
scala> b
scala> val d = a.split(a.first.previous) // split on front sentinel
scala> a

• steal(other: DoublyLinkedList[T]) This method takes the first node of the other list, removes it from the other list, and appends it to this list.

This method must run in constant time, and does not create any new nodes (it "steals" the node from the other list and uses it for this list).

For example:

scala> a
scala> b
scala> a.steal(b)
scala> a
scala> b
scala> a.steal(b)
scala> a
scala> b
scala> b.steal(a)
scala> b
scala> a

• merge(other: DoublyLinkedList[T]) This method assumes that this list is sorted in non-decreasing order, and that the other list is also sorted in non-decreasing order. It merges the nodes of both lists into this list, "stealing" them from the other list (so at the end the other list is empty).

This method must run in $$O(n + m)$$ time, where $$n$$ is the size of this list and $$m$$ is the size of the other list. It must not create new nodes, and instead use the nodes of the other list.

For example:

scala> a
scala> b
scala> a.merge(b)
scala> a
scala> b

When you are done with these three methods, the sort method that is already implemented will work correctly:

  def sort() {
// is length <= 1 ?
if (isEmpty || front.next.next == rear)
return
val other = split(median)
sort()
other.sort()
merge(other)
}

Note how it determines in constant time if the length of the list is at most one—calling length instead would have taken linear time.

I have made a test suite checklistsort.scala to test the four methods. As usual, you run it as

$fsc doublylinkedlist.scala$ fsc checklistsort.scala
\$ scala org.scalatest.run CheckListSortSuite
Run starting. Expected test count is: 5
...


text:
# PNEC

> Glossar > PNEC
6. November 2020

Short for predicted no effect concentration, predicted concentration of a hazardous substance in the environment, which shows no impact on the environment (effect assessment). If this concentration is thus below the PNEC, it should not have any negative effects. PNECs find application especially with PECs in the risk quotient in environmental risk assessment.


text:
# Homework Help: Space Mechanics

1. Oct 22, 2012

### Biosyn

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

The dry mass fraction, F when multiplied by the propellant mass gives the dry mass of the vehicle (not counting its payload). This dry mass faction F, is a function of system design and lightness of the materials employed for its construction. So if F = 0.1, a ship carrying 90 tonnes of propellant would have a dry mass of 9 tonnes. If the mass ratio of the system had to be 10 to perform a certain ΔV, this would allow the craft to carry 1 tonne of payload. But if F = 0.11, the dry mass of the system would be 9.9 tonnes and the payload would fall by a factor of 10, to 0.1 tonne. If F = 0.12, the dry mass of the vehicle would be 10.8 tonnes, and the mission would be impossible even without any payload.

I need help understanding where 0.1 tonne comes from. I know how they got 1 tonne of payload.

2. Relevant equations

$\frac{M+P}{P}$= e$^{ΔV/M}$

M = Dry mass of vehicle without payload
P = Mass of propellant

3. The attempt at a solution

(9 + 90)/9 - X = 10

X = 1 tonne

(9.9 + 90)/9.9 - X = 10

X = 0.09090909~ tonne

Last edited: Oct 22, 2012
2. Oct 22, 2012

### Simon Bridge

You mean the 0.1T cargo for F=0.01?

mass ratio has to be 10 to make a delta-vee.
then a ship with 90T of fuel can still do the delta-vee if the combined dry-mass and cargo-mass is smaller than a certain amount. What is this amount?

What is the dry-mass for a ship with F=0.01 and 90T of fuel?
How much does this leave for cargo?

3. Oct 22, 2012

### Biosyn

Sorry, where did you get F=0.01 from? I'm confused.

The amount of payload for a ship with F=0.1 and 90T of fuel will be 1T.

I'm not sure where they came up with 0.1T of cargo for a ship with F=0.11 , 90T of fuel , and 9.9T of dry mass.

The mass ratio is 10 = (M+P)/M right?

Last edited: Oct 22, 2012
4. Oct 22, 2012

### Simon Bridge

F=0.11 is the example given but you seem to have found that confusing. The math is the same for any F. Here's the bit I guessed we were talking about:
mass ratio = R
mass of fuel = P
dry mass = D
cargo mass = C
then mass-ratio is R = (P+D+C)/(D+C)

so for P=90T, D+C=10T to get a mass-ratio: 100/10 = 10
as a consequence: C = 10-D.

If you like: since F=P/D, R=[(1+F)P+C]/[FP+C] solve for C
But the above example does it in 3 steps ... thus:

1. Given R and P you get D+C.
2. Given F and P you get D
3. combine 1 and 2 to get C.

5. Oct 22, 2012

### Biosyn

Does
$\frac{90T + 9.9T + C}{9.9T + C}$ = 10T ?

Then solving for C
....

C = 0.1T

6. Oct 22, 2012

### Simon Bridge

Now you know where that 0.1 comes from?

7. Oct 23, 2012

### Biosyn

Yup, but uh..I'm not entirely sure where 10T comes from.

Is D+C always going to be 10T when P=90T?

EDIT:
Or is it C = 0.1 because the units cancel out. Therefore that's why the Ratio is equal to [(1+F)P+C]/[FP+C].
Sorry, I think I'm just going around in circles. ._.

Last edited: Oct 23, 2012
8. Oct 25, 2012

### Simon Bridge

There's a bunch of definitions:
Dry mass, and mass ratio. These are separate.
The 10T comes from a mass-ratio of 10, when the propellant mass is 90T.
(You'd think a MR of 10 would mean "10:1" wouldn't you? But it doesn't.)

Using above notation:
if M = D+C+P is the total mass of the craft.
MR = M/(M-P)

So a 100T ship with MR=10 has P=90T and D+C=10T.

Rocket design is actually quite complicated and these wee numbers help engineers make design decisions.


text:
# Lagrange multipliers finding the maximum and minimum.

Consider the two-variable function $f(x, y) = x^ 2 + 2y^2$

(a) Find the maxima and minima of $f(x, y)$ on the unit circle $x^2 +y^2 = 1$

I have used lagrange multipies to get landa= to $1$ or $2$. Given the points$(1,0),(-1,0),(0,1),(0,-1)$. so the min would be at $(0,0)$ and the max at $(0,-1)$ and $(0,1)$

Am i right in thinking this?

Secondandly how can i sketch a contour plot of this question to show graphically mypoints of maximum and minimum. I am aware its an ellipse contour.

thanks

• $(0,0)$ can't be part of the answer because it doesn't lie on the unit circle. – zipirovich Mar 24 '17 at 12:28

For that case we have:

$$f(x,y)=x^2+y^2+y^2=1+y^2$$

because we are looking for minumim and maximum using the constrain $x^2+y^2=1$.

But $y^2=1-x^2$ so the maximum of $y^2$ is $1$ (when $x=0$) and the minimum is $0$ (when $x=\pm1$).

So

$$f_{max}=1+1=2\\ f_{min}=1+0=1$$

• thank you very much! does this mean my method of lagrange multipliers is correct its just the outcome i have written to be wrong> i taught the method to my self so not 100% confident – Shem Katz Mar 24 '17 at 20:56
• You are very welcome! – Arnaldo Mar 24 '17 at 21:09


text:
# Relativistic invariance of canonical/Hamiltonian field quantisation

When I read Mukanov's book in "Quantum effects in gravity", I found the following interesting remark at p. 57.

Remark: Lorentz invariance.To quantize a field theory, we use the Hamiltonian formalism which explicitly separates the time coordinate $t$ from the spatial coordinate $x$. However, if the classical theory is relativistic (Lorentz-invariant), the resulting quantum theory is also relativistic.

This is a statement. I would like to ask: is there any proof, or any simple way to see that it is true? Namely, why by a recipe that is not relativistically invariant (at least it looks so), one can be sure that all the consequences are relativistically variant.

I can't see why it should be true.

PS: This question hits my scar. When I asked exactly that question in my university, I could not get any answer, neither from the lecturer, nor from any book I found. I felt so bad that I even abandoned the quantum field theory course after that.

• Just look at the eventual results for that what is physically relevant: Expectation values, S-matrix elements. They all are properly covariant. – ACuriousMind Jul 29 '16 at 11:10
• @ACuriousMind but that is the question: why starting from some recipes that is not covariant, one gets the covariant expectation values, S-matrix elements... Or asking in another way, how to prove that there is no consequence from the recipes that violate relativity (expectation, S-matrix do not cover all consequences, I suppose) – ophelia Jul 29 '16 at 11:16
• Well, the "proof" is that everything you can conceive of that is physically meaningful is covariant, even when computed with the Hamiltonian recipes. Are you also bothered by the ADM formalism giving covariant results? If we are free to choose coordinates how we want, then choosing specific ones must be allowed. I'm not sure what exactly bothers you about that. – ACuriousMind Jul 29 '16 at 11:19
• @ACuriousMind But how do we know that we have conceived everything we can? I am not familiar with the ADM formalism, but I suppose it can be proved to be equivalent to other covariant forms. Likewise, I have no problem with going from Lagrangian to Hamiltonian in relativistic mechanics too. But quantisation is rather an extrapolation, it is not derived from the classical field equation. – ophelia Jul 29 '16 at 11:30
• Maybe have a look at physics.stackexchange.com/q/14481/50583 – ACuriousMind Jul 29 '16 at 11:45

A good/intuitive discussion of how to take a classical field theory and obtain a quantum field theory in the Heisenberg picture is discussed in Bjorken and Drell [BD] section 11.3. A discussion on how classical symmetries like Lorentz covariance are to be expressed on a quantum level is provided in BD 11.4, directly motivating the above-mentioned material in Srednicki. Discussions of Lorentz covariance of specific quantum field theories are provided in BD 12.1, 13.5, 14.3 and in 15.3. Covariance of e.g. classical Bosonic string theory only goes to the quantum level for D = 26, as one checks analogously to the BD references given, so it's not as straightforward as the quoted passage makes it seem.

The way this is done is to take $$U(a,b) \psi_r(x) U^{-1}(a,b) = S^{-1}_{rs}(a) \psi_s(ax+b)$$ (motivated in [11.4]) to first order for translations and Lorentz transformations, i.e. $$i[\hat{P}^{\mu},\psi_r(x)] = \partial^{\mu} \psi_r(x)$$ and $$i[\hat{M}^{\mu \nu},\psi_r(x)] = x^{\mu} \partial^{\nu} \psi_r(x) - ...$$ and to show that these transformation laws, which are expressions of the classical displacement/Lorentz invariance of the classical theory [11.4], do not upset the canonical commutation relations, or rather, the theory with the commutation relations imposed is still Lorentz invariant, which is shown by showing the commutation relations still hold for transformed fields, as is shown in e.g. [13.5] explicitly.

I would like to refer to the presentation of canonical quantization of scalar fields in chapter 3 of Quantum Field Theory, Srednicki (2007). To clarify this covers canonical quantization in standard Quantum Field Theory (not Quantum Gravity) which I understand is the question.

Srednicki is very careful to use a Lorentz invariant differential so you can see clearly that Lorentz invariance is maintained throughout the quantization procedure (moving from classical variables to quantum operators).

The "proof" then (for the free theory) is to show at the end that once you have created the Fock space of particle states then these all transform as expected under Lorentz transformation. For example for a one-particle state $|k\rangle$ you can prove that

$$U(\Lambda)|k\rangle = |\Lambda k\rangle$$

where $U(\Lambda)$ is the unitary operator associated with the classical Lorentz transformation $\Lambda$.

Problem 3.3 works you through this.

• Thanks, I will have a looked at the book. Sounds like a careful formulation. – ophelia Jul 29 '16 at 16:16


text:
# Quotient rule

## Definition

The quotient rule is a formal rule for differentiating problems where one function is divided by another.

## Solved Exercises

### Struggling with math?

Access detailed step by step solutions to millions of problems, growing every day!


text:
# Math Help - Finding the number of ways

1. ## Finding the number of ways

Hi, if we have an initial arrangement say 12345 where all elements are unique, then in how many ways can one re-arrange them so that in the final configurations none of the elements which were adjacent to each other in the given arrangement are adjacent now in the final configurations.?
Does this involve some series?
E.g for 5 element say 12345, the number of such arrangements is 14.
Thanks.

2. ## Re: Finding the number of ways

Originally Posted by pranay
Hi, if we have an initial arrangement say 12345 where all elements are unique, then in how many ways can one re-arrange them so that in the final configurations none of the elements which were adjacent to each other in the given arrangement are adjacent now in the final configurations.? Does this involve some series?
E.g for 5 element say 12345, the number of such arrangements is 14.
How did you get that 14?

Note that for $123$ there are no cases.

How many are there for $1234$?
Is this one of those, $3142~?$
Given that string, there are three places to put a 5: $53142,~35142,~31425$.

So if we know how may valid cases there are for $1234$ we can build valid cases for $12345$.

I will let you work on that.

3. ## Re: Finding the number of ways

if the initial configuration is ABCDE , then the 14 required configurations are:
ACEBD
BDACE
BDAEC


text:
# GPS time accuracy (SIM808 or similar) - process between NMEA sentences and AT command response

In an SIM808 chip or similar you can get the GPS information with this AT command:

AT+CGNSINF
+CGNSINF:1,1,20150327014838.000,31.221783,121.354528,114.600,0.28,0.0,1,,1.9,22,1.0,,8,4,,,42,,

the UTC time is in the third data field as: yyyyMMddhhmmss.sss

However in the manual the function of AT+CGNSSEQ is described as: Define the last NMEA sentence that parsed.

My question is in general:

• how accurate is this time when I receive it from the serial interface of the chip?

Sub questions:

• Should I assume that the time between the AT command answer and the last NMEA message that was received by the GPS unit is variable?
• Can I compensate for delays in the systems and how?

Delays I expect:

• Difference between NMEA sentence coming in and AT command?

• Time to process NMEA sentence

• Time to read AT command and prepare response

• Serial communication time

The message reporting the time issued by a GPS receiver generally comes out a few hundred milliseconds after the actual time stated inside the message itself. That reported time is actually the beginning of the current measurement cycle — and if the receiver has a 1PPS output, it would correspond to the leading edge of that pulse.

Unless you take special care, the polling you do with your AT commands runs asynchronously with respect to GPS time — and the times at which the messages are issued by the receiver. Therefore, you'll always have an uncertainty that's equal to your polling period added to whatever other communication delays are in the system.

Therefore, if you're polling once a second, the time you eventually see in the message is going to be "stale" by anywhere from about 0.1 second to 1.1 second. You can reduce the upper bound by polling more often and paying attention to when the time value changes from one result to the next.

The SIM808 has a 1 pps output on pad 37, so for higher accuracy you could capture this in an interrupt and use it to discipline your MCU's internal sense of time.

Based on Dave's suggestion that the serial time information is "stale" it seems like you might want to do something like to take that value, prune the fraction, add 1 second, and cache it so that the next interrupt will know which second is being marked.

Or if you can configure an MCU hardware timer to reset on external input from the PPS signal and count up otherwise, you may be able to do a query and replace the fractional seconds in the result with a conversion to fractional seconds of the timer's current count.

In either case you probably want to take care if your issue one of these queries near the end of a second, where it could be ambiguous if the reply you get is reporting the old whole seconds or the new. Saving the internal sense of time from when the query is issued might work - if the reply you calculate is "earlier" than the time you asked, you probably need to add a second to fix it.


text:
Q&A

# Posts tagged planets

This tag doesn't have any usage information yet.

This tag doesn't have a detailed wiki yet.

1.4k posts
80%
+6 −0

1 answer  ·  posted 5mo ago by celtschk  ·  last activity 5mo ago by dsr

66%
+2 −0

5 answers  ·  posted 7y ago by Liath  ·  last activity over 1 year ago by Canina

60%
+1 −0

1 answer  ·  posted over 6 years ago by HDE 226868  ·  last activity over 1 year ago by Canina

Question planets mines
66%
+2 −0

5 answers  ·  posted 6y ago by Monica Cellio  ·  edited over 1 year ago by Monica Cellio

50%
+0 −0

3 answers  ·  posted almost 6 years ago by Monica Cellio  ·  edited over 1 year ago by Monica Cellio

Question planets stars
60%
+1 −0

2 answers  ·  posted almost 2 years ago by Zavael  ·  last activity over 1 year ago by HDE 226868

66%
+2 −0

4 answers  ·  posted over 1 year ago by Mister Onion  ·  last activity over 1 year ago by HDE 226868

66%
+2 −0

3 answers  ·  posted over 2 years ago by TheDyingOfLight  ·  edited over 1 year ago by HDE 226868

50%
+0 −0

6 answers  ·  posted almost 7 years ago by Monica Cellio  ·  edited over 1 year ago by Canina

50%
+0 −0

0 answers  ·  posted over 1 year ago by A.E. Stephenson  ·  last activity over 1 year ago by A.E. Stephenson

66%
+2 −0

2 answers  ·  posted over 1 year ago by Mister Onion  ·  last activity over 1 year ago by Mister Onion

Question planets
50%
+0 −0

2 answers  ·  posted over 1 year ago by Praearcturus  ·  last activity over 1 year ago by Praearcturus

50%
+0 −0

4 answers  ·  posted over 1 year ago by user76027  ·  last activity over 1 year ago by user76027

50%
+0 −0

1 answer  ·  posted over 1 year ago by RMurphy-Keith  ·  last activity over 1 year ago by RMurphy-Keith

50%
+0 −0

1 answer  ·  posted over 1 year ago by Vigilant  ·  last activity over 1 year ago by Vigilant

50%
+0 −0

1 answer  ·  posted over 1 year ago by galactic_analyzer  ·  last activity over 1 year ago by galactic_analyzer

50%
+0 −0

6 answers  ·  posted over 1 year ago by galactic_analyzer  ·  last activity over 1 year ago by galactic_analyzer

50%
+0 −0

8 answers  ·  posted over 1 year ago by The Weasel Sagas  ·  last activity over 1 year ago by The Weasel Sagas

50%
+0 −0

8 answers  ·  posted over 1 year ago by Chickenpeep Chickenpeep  ·  last activity over 1 year ago by Chickenpeep Chickenpeep

50%
+0 −0

2 answers  ·  posted over 1 year ago by Khwarezm  ·  last activity over 1 year ago by Khwarezm

60%
+1 −0

1 answer  ·  posted over 1 year ago by Sam D. Jones  ·  last activity over 1 year ago by Sam D. Jones

Question planets chemistry
50%
+0 −0

4 answers  ·  posted over 1 year ago by Faz  ·  last activity over 1 year ago by Faz

50%
+0 −0

3 answers  ·  posted over 1 year ago by chaeltom  ·  last activity over 1 year ago by chaeltom

50%
+0 −0

2 answers  ·  posted over 1 year ago by Ryan Williamson  ·  last activity over 1 year ago by Ryan Williamson

50%
+0 −0

5 answers  ·  posted over 1 year ago by Space_Cadet  ·  last activity over 1 year ago by Space_Cadet

50%
+0 −0

6 answers  ·  posted over 1 year ago by Zjerzy  ·  last activity over 1 year ago by Zjerzy

50%
+0 −0

4 answers  ·  posted over 1 year ago by LonelyCryptid  ·  last activity over 1 year ago by LonelyCryptid

50%
+0 −0

2 answers  ·  posted over 1 year ago by Slarty  ·  last activity over 1 year ago by Slarty

50%
+0 −0

1 answer  ·  posted over 1 year ago by user72655  ·  last activity over 1 year ago by user72655

50%
+0 −0

2 answers  ·  posted over 1 year ago by Ichthys King  ·  last activity over 1 year ago by Ichthys King

Question planets ocean
60%
+1 −0

2 answers  ·  posted over 1 year ago by RandySavage  ·  last activity over 1 year ago by RandySavage

50%
+0 −0

2 answers  ·  posted over 1 year ago by James Grossmann  ·  last activity over 1 year ago by James Grossmann

50%
+0 −0

2 answers  ·  posted over 1 year ago by Carlos Zamora  ·  last activity over 1 year ago by Carlos Zamora

50%
+0 −0

3 answers  ·  posted over 1 year ago by QUESTER42  ·  last activity over 1 year ago by QUESTER42

50%
+0 −0

12 answers  ·  posted over 1 year ago by Faz  ·  last activity over 1 year ago by Faz

50%
+0 −0

4 answers  ·  posted over 1 year ago by Julianne  ·  last activity over 1 year ago by Julianne

50%
+0 −0

2 answers  ·  posted over 1 year ago by In the name of the story  ·  last activity over 1 year ago by In the name of the story

50%
+0 −0

3 answers  ·  posted over 1 year ago by RJ0001  ·  last activity over 1 year ago by RJ0001

50%
+0 −0

3 answers  ·  posted over 1 year ago by N. Bleac  ·  last activity over 1 year ago by N. Bleac

50%
+0 −0

5 answers  ·  posted over 1 year ago by The Weasel Sagas  ·  last activity over 1 year ago by The Weasel Sagas

50%
+0 −0

2 answers  ·  posted over 1 year ago by The Weasel Sagas  ·  last activity over 1 year ago by The Weasel Sagas

50%
+0 −0

3 answers  ·  posted over 1 year ago by Covision  ·  last activity over 1 year ago by Covision

60%
+1 −0

1 answer  ·  posted over 1 year ago by CAE Jones  ·  last activity over 1 year ago by CAE Jones

50%
+0 −0

9 answers  ·  posted over 1 year ago by The Weasel Sagas  ·  last activity over 1 year ago by The Weasel Sagas

50%
+0 −0

6 answers  ·  posted over 1 year ago by Wax  ·  last activity over 1 year ago by Wax

50%
+0 −0

1 answer  ·  posted over 1 year ago by Vigilant  ·  last activity over 1 year ago by Vigilant

50%
+0 −0

2 answers  ·  posted over 1 year ago by SentiCarter  ·  last activity over 1 year ago by SentiCarter

50%
+0 −0

11 answers  ·  posted over 1 year ago by Primordial  ·  last activity over 1 year ago by Primordial

50%
+0 −0

4 answers  ·  posted over 1 year ago by CatSharkSnake  ·  last activity over 1 year ago by CatSharkSnake

50%
+0 −0

4 answers  ·  posted over 1 year ago by user6760  ·  last activity over 1 year ago by user6760

This community is part of the Codidact network. We have other communities too — take a look!


text:
# American Institute of Mathematical Sciences

• Previous Article
On the initial boundary value problem for certain 2D MHD-$\alpha$ equations without velocity viscosity
• CPAA Home
• This Issue
• Next Article
Existence and uniqueness for $\mathbb{D}$-solutions of reflected BSDEs with two barriers without Mokobodzki's condition
July  2016, 15(4): 1157-1178. doi: 10.3934/cpaa.2016.15.1157

## On the $L^p-$ theory of Anisotropic singular perturbations of elliptic problems

 1 Académie de Grenoble, Lycée Saint-Marc, Nivolas-Vermelle, 38300, France

Received  April 2015 Revised  January 2016 Published  April 2016

In this article we give an extension of the $L^2-$theory of anisotropic singular perturbations for elliptic problems. We study a linear and some nonlinear problems involving $L^{p}$ data ($1 < p < 2$). Convergences in pseudo Sobolev spaces are proved for weak and entropy solutions, and rate of convergence is given in cylindrical domains.
Citation: Ogabi Chokri. On the $L^p-$ theory of Anisotropic singular perturbations of elliptic problems. Communications on Pure & Applied Analysis, 2016, 15 (4) : 1157-1178. doi: 10.3934/cpaa.2016.15.1157
##### References:
 [1] R.A. Adams and John J.F. Fournier, Sobolev Spaces,, Pure and Applied Mathematics, (2003).   Google Scholar [2] S. Agmon, A. Douglis and L. Nirenberg, Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions I,, \emph{Comm. Pure Appl. Math}, 12 (1959), 623.   Google Scholar [3] L. Boccardo, T. Gallouët and J.L. Vazquez, Nonlinear elliptic equations in $R^n$ without growth restrictions on the data,, \emph{Journal of Differential Equations}, 105 (1993), 334.  doi: 10.1006/jdeq.1993.1092.  Google Scholar [4] Ph. Bénilan, Philippe, L. Boccardo, Th. Gallouët, R. Gariepy, M. Pierre and J.L. Vazquez, An $L^{1}$-theory of existence and uniqueness of solutions of nonlinear elliptic equations,, \emph{Ann. Scuola. Norm. Sup. Pisa Cl. Sci}, 22 (1995), 241.   Google Scholar [5] M. Chipot, Elliptic Equations, An Introductory Cours,, Birkhauser, (2009), 978.  doi: 10.1007/978-3-7643-9982-5.  Google Scholar [6] M. Chipot, On some anisotropic singular perturbation problems,, \emph{Asymptotic Analysis}, 55 (2007), 125.   Google Scholar [7] M. Chipot and K. Yeressian, Exponential rates of convergence by an iteration technique,, \emph{C. R. Acad. Sci. Paris}, 346 (2008), 21.  doi: 10.1016/j.crma.2007.12.004.  Google Scholar [8] M. Chipot and S. Guesmia, On the asymptotic behaviour of elliptic, anisotropic singular perturbations problems,, \emph{Com. Pur. App. Ana}, 8 (2009), 179.  doi: 10.3934/cpaa.2009.8.179.  Google Scholar [9] M. Chipot, S. Guesmia and M. Sengouga, Singular perturbations of some nonlinear problems,, \emph{J. Math. Sci}, 176 (2011), 828.  doi: 10.1007/s10958-011-0439-y.  Google Scholar [10] M. Chipot and S. Guesmia, On a class of integro-differential problems,, \emph{Commun. Pure Appl. Anal.}, 9 (2010), 1249.  doi: 10.3934/cpaa.2010.9.1249.  Google Scholar [11] P. Enflo, A counterexample to the approximation problem in Banach spaces,, \emph{Acta Mathematica}, 130 (1973), 309.   Google Scholar [12] S. Fucik, O. John and J. Necas, On the existence of Schauder basis in Sobolev spaces,, \emph{Comment. Math. Univ. Carolin}, 13 (1972), 163.   Google Scholar [13] T. Gallouet and R. Herbin, Existence of a solution to a coupled elliptic system,, \emph{Appl.Math. Letters}, 7 (1994), 49.  doi: 10.1016/0893-9659(94)90030-2.  Google Scholar [14] C. Ogabi, On a class of nonlinear elliptic, anisotropic singular perturbations problems,, \textit{Preprint:} \url{https://hal.archives-ouvertes.fr/hal-01074262}., ().   Google Scholar [15] J. Serrin, Pathological solutions of elliptic differential equations,, \emph{Ann. Sc. Norm. Sup. Pisa}, 18 (1964), 385.   Google Scholar

show all references

##### References:
 [1] R.A. Adams and John J.F. Fournier, Sobolev Spaces,, Pure and Applied Mathematics, (2003).   Google Scholar [2] S. Agmon, A. Douglis and L. Nirenberg, Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions I,, \emph{Comm. Pure Appl. Math}, 12 (1959), 623.   Google Scholar [3] L. Boccardo, T. Gallouët and J.L. Vazquez, Nonlinear elliptic equations in $R^n$ without growth restrictions on the data,, \emph{Journal of Differential Equations}, 105 (1993), 334.  doi: 10.1006/jdeq.1993.1092.  Google Scholar [4] Ph. Bénilan, Philippe, L. Boccardo, Th. Gallouët, R. Gariepy, M. Pierre and J.L. Vazquez, An $L^{1}$-theory of existence and uniqueness of solutions of nonlinear elliptic equations,, \emph{Ann. Scuola. Norm. Sup. Pisa Cl. Sci}, 22 (1995), 241.   Google Scholar [5] M. Chipot, Elliptic Equations, An Introductory Cours,, Birkhauser, (2009), 978.  doi: 10.1007/978-3-7643-9982-5.  Google Scholar [6] M. Chipot, On some anisotropic singular perturbation problems,, \emph{Asymptotic Analysis}, 55 (2007), 125.   Google Scholar [7] M. Chipot and K. Yeressian, Exponential rates of convergence by an iteration technique,, \emph{C. R. Acad. Sci. Paris}, 346 (2008), 21.  doi: 10.1016/j.crma.2007.12.004.  Google Scholar [8] M. Chipot and S. Guesmia, On the asymptotic behaviour of elliptic, anisotropic singular perturbations problems,, \emph{Com. Pur. App. Ana}, 8 (2009), 179.  doi: 10.3934/cpaa.2009.8.179.  Google Scholar [9] M. Chipot, S. Guesmia and M. Sengouga, Singular perturbations of some nonlinear problems,, \emph{J. Math. Sci}, 176 (2011), 828.  doi: 10.1007/s10958-011-0439-y.  Google Scholar [10] M. Chipot and S. Guesmia, On a class of integro-differential problems,, \emph{Commun. Pure Appl. Anal.}, 9 (2010), 1249.  doi: 10.3934/cpaa.2010.9.1249.  Google Scholar [11] P. Enflo, A counterexample to the approximation problem in Banach spaces,, \emph{Acta Mathematica}, 130 (1973), 309.   Google Scholar [12] S. Fucik, O. John and J. Necas, On the existence of Schauder basis in Sobolev spaces,, \emph{Comment. Math. Univ. Carolin}, 13 (1972), 163.   Google Scholar [13] T. Gallouet and R. Herbin, Existence of a solution to a coupled elliptic system,, \emph{Appl.Math. Letters}, 7 (1994), 49.  doi: 10.1016/0893-9659(94)90030-2.  Google Scholar [14] C. Ogabi, On a class of nonlinear elliptic, anisotropic singular perturbations problems,, \textit{Preprint:} \url{https://hal.archives-ouvertes.fr/hal-01074262}., ().   Google Scholar [15] J. Serrin, Pathological solutions of elliptic differential equations,, \emph{Ann. Sc. Norm. Sup. Pisa}, 18 (1964), 385.   Google Scholar
 [1] Yichen Zhang, Meiqiang Feng. A coupled $p$-Laplacian elliptic system: Existence, uniqueness and asymptotic behavior. Electronic Research Archive, 2020, 28 (4) : 1419-1438. doi: 10.3934/era.2020075 [2] Mokhtar Bouloudene, Manar A. Alqudah, Fahd Jarad, Yassine Adjabi, Thabet Abdeljawad. Nonlinear singular $p$ -Laplacian boundary value problems in the frame of conformable derivative. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020442 [3] Hoang The Tuan. On the asymptotic behavior of solutions to time-fractional elliptic equations driven by a multiplicative white noise. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020318 [4] Yongxiu Shi, Haitao Wan. Refined asymptotic behavior and uniqueness of large solutions to a quasilinear elliptic equation in a borderline case. Electronic Research Archive, , () : -. doi: 10.3934/era.2020119 [5] Zuliang Lu, Fei Huang, Xiankui Wu, Lin Li, Shang Liu. Convergence and quasi-optimality of $L^2-$norms based an adaptive finite element method for nonlinear optimal control problems. Electronic Research Archive, 2020, 28 (4) : 1459-1486. doi: 10.3934/era.2020077 [6] Maoding Zhen, Binlin Zhang, Vicenţiu D. Rădulescu. Normalized solutions for nonlinear coupled fractional systems: Low and high perturbations in the attractive case. Discrete & Continuous Dynamical Systems - A, 2020  doi: 10.3934/dcds.2020379 [7] Hua Chen, Yawei Wei. Multiple solutions for nonlinear cone degenerate elliptic equations. Communications on Pure & Applied Analysis, , () : -. doi: 10.3934/cpaa.2020272 [8] Craig Cowan, Abdolrahman Razani. Singular solutions of a Lane-Emden system. Discrete & Continuous Dynamical Systems - A, 2021, 41 (2) : 621-656. doi: 10.3934/dcds.2020291 [9] Pierre-Etienne Druet. A theory of generalised solutions for ideal gas mixtures with Maxwell-Stefan diffusion. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020458 [10] Shiqiu Fu, Kanishka Perera. On a class of semipositone problems with singular Trudinger-Moser nonlinearities. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020452 [11] Dorothee Knees, Chiara Zanini. Existence of parameterized BV-solutions for rate-independent systems with discontinuous loads. Discrete & Continuous Dynamical Systems - S, 2021, 14 (1) : 121-149. doi: 10.3934/dcdss.2020332 [12] Claudianor O. Alves, Rodrigo C. M. Nemer, Sergio H. Monari Soares. The use of the Morse theory to estimate the number of nontrivial solutions of a nonlinear Schrödinger equation with a magnetic field. Communications on Pure & Applied Analysis, 2021, 20 (1) : 449-465. doi: 10.3934/cpaa.2020276 [13] Anna Canale, Francesco Pappalardo, Ciro Tarantino. Weighted multipolar Hardy inequalities and evolution problems with Kolmogorov operators perturbed by singular potentials. Communications on Pure & Applied Analysis, 2021, 20 (1) : 405-425. doi: 10.3934/cpaa.2020274 [14] Zedong Yang, Guotao Wang, Ravi P. Agarwal, Haiyong Xu. Existence and nonexistence of entire positive radial solutions for a class of Schrödinger elliptic systems involving a nonlinear operator. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020436 [15] Yi-Long Luo, Yangjun Ma. Low Mach number limit for the compressible inertial Qian-Sheng model of liquid crystals: Convergence for classical solutions. Discrete & Continuous Dynamical Systems - A, 2021, 41 (2) : 921-966. doi: 10.3934/dcds.2020304 [16] Gongbao Li, Tao Yang. Improved Sobolev inequalities involving weighted Morrey norms and the existence of nontrivial solutions to doubly critical elliptic systems involving fractional Laplacian and Hardy terms. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020469 [17] Luca Battaglia, Francesca Gladiali, Massimo Grossi. Asymptotic behavior of minimal solutions of $-\Delta u = \lambda f(u)$ as $\lambda\to-\infty$. Discrete & Continuous Dynamical Systems - A, 2021, 41 (2) : 681-700. doi: 10.3934/dcds.2020293 [18] Riccarda Rossi, Ulisse Stefanelli, Marita Thomas. Rate-independent evolution of sets. Discrete & Continuous Dynamical Systems - S, 2021, 14 (1) : 89-119. doi: 10.3934/dcdss.2020304 [19] Tahir Aliyev Azeroğlu, Bülent Nafi Örnek, Timur Düzenli. Some results on the behaviour of transfer functions at the right half plane. Evolution Equations & Control Theory, 2020  doi: 10.3934/eect.2020106 [20] George W. Patrick. The geometry of convergence in numerical analysis. Journal of Computational Dynamics, 2021, 8 (1) : 33-58. doi: 10.3934/jcd.2021003

2019 Impact Factor: 1.105


text:
usecase.tex 6.85 KB
 Dennis Müller committed Apr 25, 2018 1 2 The Viewfinder algorithm is implemented in the MMT system and exposed within the jEdit-IDE, allowing us to realize the use case stated in the introduction. A screenshot of Jane's theory of beautiful sets is given in Figure \ref{fig:use:source}; it  Dennis Müller committed Apr 27, 2018 3 is based on the (basic higher-order logic) foundation of the Math-in-the-Middle library (\cite{ODK:mitm:16} developed natively in MMT.  Dennis Müller committed Apr 25, 2018 4 5  \begin{figure}[ht]\centering  Dennis Müller committed Apr 27, 2018 6  \fbox{\includegraphics[width=\textwidth]{beautysource}}  Dennis Müller committed Apr 25, 2018 7 8 9 10  \fbox{\includegraphics[width=\textwidth]{results}} \caption{A Theory of Beautiful Sets'' in MMT Surface Syntax and Results of the Viewfinder}\label{fig:use:source} \end{figure}  Dennis Müller committed Apr 25, 2018 11 Right-clicking anywhere within the theory allows Jane to select \cn{MMT} $\to$ \cn{Find\ Views\ to...} $\to$ \cn{MitM/smglom} (the main Math-in-the-Middle library), telling her (within less than one second) that two views have been found, the most promising of which points to the theory  Dennis Müller committed Apr 25, 2018 12 13 14 15 16 17 18 \cn{matroid\_theory} (see Figure \ref{fig:use:target}) in the library. \begin{figure}[ht]\centering \fbox{\includegraphics[width=0.6\textwidth]{matroids}} \caption{The Theory of Matroids in the MitM Library}\label{fig:use:target} \end{figure}  Dennis Müller committed Apr 26, 2018 19 20 21 22 23 24 25 \begin{newpart}{DM: Moved} \section{Across-Library Viewfinding}\label{sec:across} We have so far assumed one fixed meta-theory for all theories involved; we will now discuss the situation when looking for views between theories in different libraries (and built on different foundations). Obviously, various differences in available foundational primitives and library-specific best practices and idiosyncracies can prevent the algorithm from finding desired matches. There are two approaches to increasing the number of results in these cases: \begin{itemize}  Dennis Müller committed Apr 27, 2018 26  \item In many instances, the translations between two foundations is too complex to be discovered purely syntactically. In these cases we can provide arbitrary translations between theories, which are applied before computing the abstract syntax trees (e.g. alignment-based translations as in \cite{MRLR:alignments:17})  Dennis Müller committed Apr 26, 2018 27 28 29 30 31 32 33 34 35 36 37 38  \item We can do additional transformations before preprocessing theories, auch as normalizing expressions, eliminating higher-order abstract syntax encodings or encoding-related redundant information (such as the type of a typed equality, which in the presence of subtyping can be different from the types of both sides of an equation), or elaborating abbreviations/definitions. \end{itemize} When elaborating definitions, it is important to consider that this may also reduce the number of results, if both theories use similar abbreviations for complex terms, or the same concept is declared axiomatically in one theory, but definitionally in the other. For that reason, we can allow \textbf{several abstract syntax trees for the same constant}, such as one with definitions expanded and one as is''. Similarly, certain idiosyncracies -- such as PVS's common usage of theory parameters -- call for not just matching symbol references, but also variables or possibly even complex expressions. To handle these situations, we additionally allow for \textbf{holes} in the constant lists of an abstract syntax tree, which may be unified with any other symbol or hole, but are not recursed into. The subterms that are to be considered holes can be marked as such during preprocessing. \subsection{Normalization}\label{sec:preproc} The common logical framework used for all the libraries at our disposal -- namely LF and extensions thereof -- makes it easy to systematically normalize theories built on various logical foundations. We currently use the following approaches to preprocessing theories: \begin{itemize} \item Free variables in a term, often occurences of theory parameters as e.g. used extensively in the PVS system, are replaced by holes. \item For foundations that use product types, we curry function types $(A_1 \times\ldots A_n)\to B$ to $A_1 \to \ldots \to A_n\to B$. We treat lambda-expressions and applications accordingly.  Dennis Müller committed Apr 27, 2018 39 40  For example: $f = \lambda (n,m) : \mathbb N \times \mathbb N .\; n + m\text{ becomes } f = \lambda n : \mathbb N.\; \lambda m : \mathbb N .\; n + m$ $f(\langle a, b\rangle)\text{ becomes }f(a,b).$  Dennis Müller committed Apr 26, 2018 41 42  \item Higher-order abstract syntax encodings are eliminated by raising atomic types, function types, applications and lambdas to the level of the logical framework. This eliminates (redundant) implicit arguments that only occur due to their formalization in the logical framework.  Dennis Müller committed Apr 27, 2018 43 44 45 46 47  For example, the apply-operator in the PVS meta-theory has the LF-type $(A : \cn{psvtype}) \to (B : \cn{pvstype}) \to \cn{pvsterm}_{A\Rightarrow B} \to \cn{pvsterm}_A \to \cn{pvsterm}_B$ Hence every application of a function in pvs -- as \mmt-term: $\oma{\cn{pvsapply}}{A,B,f,a}$ -- has two additional implicit arguments, which we eliminate by replacing this expression by $\oma{f}{a}$. This prevents false negatives due to mismatching type arguments in the presence of subtyping.  Dennis Müller committed Apr 27, 2018 48  \item Since judgments -- in our imports -- correspond to constants of type $\vdash P$ for some proposition $P$, we can use the curry-howard correspondence to equate the type $\vdash (P \Rightarrow Q)$ with the function type $(\vdash P) \to (\vdash Q)$, as well as the judgment $\vdash \forall x : A.\;P$ with the function type $(x:A) \to \vdash P$.  Dennis Müller committed Apr 27, 2018 49 50 51  Since both styles of formalization are more or less preferred in different systems, we replace each occurence of the former by the latter. \item For classical logics, we rewrite all remaining logical connectives using their usual definitions on the basis of negation and conjunction only. Double negations are eliminated.  Dennis Müller committed Apr 26, 2018 52  \item Typed Equalities are transformed to untyped ones; again getting rid of the redundant type argument of the equality.  Dennis Müller committed Apr 27, 2018 53 54 55  For example, the typed equality $\oma{\cn{equals}}{\mathbb N,2,1+1}$ becomes $\oma{\cn{equals}}{2,1+1}$, allowing to match the same equation on e.g. the real numbers. \item The arguments of conjunctions and equalities are ordered (currently only by their number of subterms).  Dennis Müller committed Apr 26, 2018 56 57 58 59 \end{itemize} \subsection{Implementation}\label{sec:pvs} \paragraph{} Using the above normalization methods, we can examplary write down a theory for a commutative binary operator using the Math-in-the-Middle foundation, while targeting e.g. the PVS Prelude library -- allowing us to find all commutative operators, as in Figure \ref{fig:use:pvs}.  Dennis Müller committed Apr 25, 2018 60   Dennis Müller committed Apr 25, 2018 61 62 \begin{figure}[ht]\centering \fbox{\includegraphics[width=\textwidth]{pvs}}  Dennis Müller committed Apr 25, 2018 63 \caption{Searching for Commutative Operators in PVS}\label{fig:use:pvs}  Dennis Müller committed Apr 26, 2018 64 \end{figure}  Dennis Müller committed Apr 26, 2018 65 66 67 68 69 \ednote{8 results for NASA, but NASA doesn't work in jEdit because of limited memory} This example also hints at a way to iteratively improve the results of the viewfinder: since we can find properties like commutativity and associativity, we can use the results to in turn inform a better normalization of the theory by exploiting these properties. This in turn would potentially allow for finding more views. \end{newpart}


text:
💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# For a fish swimming at a speed $v$ relative to the water, the energy expenditure per unit time is proportional to $v^3$. It is believed that migrating fish try to minimize the total energy required to swim a fixed distance. If the fish are swimming against a current $\mu(\mu < v)$, then the time required to swim a distance $L$ is $L/(v - \mu)$ and the total energy $E$ required to swim the distance is given by$$E(v) = av^3 \cdot \dfrac{L}{v - \mu}$$where $a$ is the proportionality constant.(a) Determine the value of $v$ that minimizes $E$.(b) Sketch the graph of $E$.Note: This result has been verified experimentally; migrating fish swim against a current at a speed of 50% greater than the current speed.

## a) $v=\frac{3}{2} u$b) $\left(\frac{3}{2} u, E\left(\frac{3}{2} u\right)\right)$

Derivatives

Differentiation

Volume

### Discussion

You must be signed in to discuss.
##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp

### Video Transcript

in this problem we have to use our knowledge of differentiation Thio optimize a specific function were given and the function that were given deals with various things like distance and, um speed. So we're given this function will call Eve equals a V cubed times l over v minus you. Well, we can simplify that a little bit to say e of e equals a V cubed times l Oliver v minus you. What we want is we want the derivative equal to zero. So we'll take the derivative of this function. We will set it equal Thio zero we'll have to a V cubed times l minus three a u L V squared equal to zero. We could simplify a little bit and cancel out some terms will have to v cubed minus three u V squared equal to zero and then we can factor out a common a common factor. Essentially, we have the square and will multiply that by the quantity to V minus three You equal to zero and then we can solve for V pretty easily. These three over to you And now for be, we're told. Well, let's find this information by using a graph. We need that coordinate point of where? Yeah, our values of X and Y satisfy the equation. So this is a little bit tricky, really. You can test different values and see what you get. These values work well. We have l equal 100. You equal five and a equal 50.1 And this is the graph that we get. So this coordinate point is essentially the answer to part B. This satisfies the specific speeds and the derivative, Um, that we need. So I hope that this problem helped you understand a little bit more about optimization and how we can go about optimizing a function using differentiation and its associated graph.

University of Denver

#### Topics

Derivatives

Differentiation

Volume

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp