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 Equations/213735: solve algebraically, using one variable. find three consecutive intergers such that three times the sum of the first two intergers is three less than seven times the third. 1 solutions Answer 161462 by jim_thompson5910(29613)   on 2009-09-15 01:01:17 (Show Source): You can put this solution on YOUR website!Recall that consecutive integers are of the form: x, x+1, x+2, x+3, etc... The statement: "three times the sum of the first two intergers is three less than seven times the third" translates to Start with the given equation. Remove unneeded parenthesis. Combine like terms. Distribute. Combine like terms on the right side. Subtract from both sides. Subtract from both sides. Combine like terms on the left side. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the solution is This means that the three numbers are -8, -7, and -6
 Expressions-with-variables/213734: (-3)^2-4^3-2(-3)(-1)^21 solutions Answer 161461 by jim_thompson5910(29613)   on 2009-09-15 00:56:33 (Show Source): You can put this solution on YOUR website! Start with the given expression. Cube 4 to get 64 Square -3 to get 9 Square -1 to get 1 Multiply Combine like terms. So
 Equations/213733: 1 solutions Answer 161460 by jim_thompson5910(29613)   on 2009-09-15 00:39:49 (Show Source): You can put this solution on YOUR website! Start with the given equation. Multiply both sides by 5. Multiply So the solution is
 Exponential-and-logarithmic-functions/213724: Recall that the formula for compounding interest is A = P(1+ r)t where A is the total accumulated amount in an account with principal P compounded at a annual interest rate r once per year for t years. a. How many years will it take for a principal of P =1000 to reach an amount of A = 5000 when compounded once a year at an annual interest rate of 6%? b. If the interest rate is doubled to 12%, what will the number of years be for the principal of P =1000 to reach an amount of A = 5000? 1 solutions Answer 161453 by jim_thompson5910(29613)   on 2009-09-14 23:28:07 (Show Source): You can put this solution on YOUR website!a) Start with the given equation. Plug in , , and Add Divide both sides by 1000. Divide Take the log of both sides. Pull down the exponent. Evaluate the left side Evaluate the log on the right side Divide both sides by 0.02531. Divide and rearrange the equation. So it will take about 27.61 years (a little over 27 and a half years) for $1,000 to become$5,000 at an interest rate of 6% ------------------------------------ b) Start with the given equation. Plug in , , and Add Divide both sides by 1000. Divide Take the log of both sides. Pull down the exponent. Evaluate the left side Evaluate the log on the right side Divide both sides by 0.04922. Divide and rearrange the equation. So it will take about 14.2 years (a little over 14 years) for $1,000 to become$5,000 at an interest rate of 12%
 Human-and-algebraic-language/213503: In isosceles trapezoid ABCD, the longer base, AB, is one and one half times as long as the short base, CD. The other two sides, AD and BC, are both 13 cm long. A. If the perimeter is 76 cm, find the lengths of AB and CD. this is what i did: AD = 13 BC = 13 CD = x AB = x* 1 1/2 the equation i got : 13+13+x+1 1/2*x=76 but when i follow through with the equation i can't get 76, help please, i have no idea what i'm doing wrong.1 solutions Answer 161310 by jim_thompson5910(29613)   on 2009-09-14 15:57:31 (Show Source): You can put this solution on YOUR website!Note: the mixed fraction is equivalent to the improper fraction . Ie. . So the equation becomes If you multiply EVERY term by the LCD 2, you'll clear out the fractions to get . Now you have a much simpler equation to solve. I'll let you do that.
 Permutations/213492: Can you give me a formula to work out the amount of squares within a square eg. a chess board 1 solutions Answer 161292 by jim_thompson5910(29613)   on 2009-09-14 14:28:00 (Show Source): You can put this solution on YOUR website!If you're asking just about the number of 1x1 squares, then the last post is correct. If you're asking about larger squares (2x2, 3x3, etc..), then there's more to this problem. Check out this page to see the answer.
 Functions/213488: I am still unsure when it comes to functions. I have a question that is as follows: Is f(x) = x^2 + 2x - 3 a one-to-one function? Explain your answer. When figuring out the problem if f(a) = f(b) then my answer would be no as a^2 + 2a - 3 = b^2 + 2b -3 does not factor down to a = b. Am I figuring this correctly? Thank you = Lori1 solutions Answer 161291 by jim_thompson5910(29613)   on 2009-09-14 14:15:01 (Show Source): You can put this solution on YOUR website!You're on the right track. A function is one-to-one if only means . In other words, if two outputs of a function are equal, then the inputs must be equal for the function to be one-to-one. So if you think that this function is NOT one-to-one, then simply pick a counter-example to prove this claim false. Let's say that . Plugging this into gets us . Now solve for 'x' to get or So if we plug in either 0 or -2 into f(x), we'll get -3 as an output. Ie. . Since , this means that the function is NOT one-to-one. So you are correct.
 Distributive-associative-commutative-properties/213060: This question is from textbook Prentice PRE-ALGEBRA Trying to simplify this equation and Justify each step. w+3w+4(5+w-3w) Here is what I have tried w+3w+4(5+w-3w) w+3w+4(2w) w+3w+8w w+11w1 solutions Answer 160950 by jim_thompson5910(29613)   on 2009-09-12 20:03:03 (Show Source): You can put this solution on YOUR website! Start with the given expression. Combine like terms. Distribute. Multiply Combine like terms. So
 Graphs/212993: If an object on Earth is projected upward with an initial velocity of 32 feet per second, then its height after (t) seconds is given by: h(t)=32t-16t^2. Find the maximum height attained by the object and the number of seconds it takes to hit the ground. I began like this: h(t)=-16t^2 + 32t -32/2(-16)=-32/32=1 -16(1)^2 + 32(1) -16+32=16 and now I am stuck!1 solutions Answer 160900 by jim_thompson5910(29613)   on 2009-09-12 14:52:08 (Show Source): You can put this solution on YOUR website!You're pretty much done. What you just did was find the vertex of . The vertex is (1,16) and at the vertex is the max value of h(t) (ie the height). So the max height is 16 ft (which occurs at t=1 second).
 Linear-equations/212930: How do I find the equation of the line which passes through the points (-3,2) and (3,-2)1 solutions Answer 160891 by jim_thompson5910(29613)   on 2009-09-12 13:54:24 (Show Source): You can put this solution on YOUR website!First let's find the slope of the line through the points and Note: is the first point . So this means that and . Also, is the second point . So this means that and . Start with the slope formula. Plug in , , , and Subtract from to get Subtract from to get Reduce So the slope of the line that goes through the points and is Now let's use the point slope formula: Start with the point slope formula Plug in , , and Rewrite as Distribute Multiply Add 2 to both sides. Combine like terms. Simplify So the equation that goes through the points and is Notice how the graph of goes through the points and . So this visually verifies our answer. Graph of through the points and
Polynomials-and-rational-expressions/212774: need help to find the slope of the line containing the points
p1(5,-3) and p2(3,-1)
1 solutions

Answer 160691 by jim_thompson5910(29613)   on 2009-09-11 12:17:55 (Show Source):
You can put this solution on YOUR website!
 Solved by pluggable solver: Finding the slope Slope of the line through the points (5, -3) and (3, -1) Answer: Slope is

Linear-equations/212770: need help with the slope of the line containing points
p1(-1,2) and p2(3,3)
1 solutions

Answer 160690 by jim_thompson5910(29613)   on 2009-09-11 12:17:00 (Show Source):
You can put this solution on YOUR website!
 Solved by pluggable solver: Finding the slope Slope of the line through the points (-1, 2) and (3, 3) Answer: Slope is

 Graphs/212735: Hi there. I need to see if I am on the right track. I need to graph the following y = -2 abs(x+1)-4. I need to construc a sequence of five graphs to get from one to another. I am setting up my 5 graphs as follows: (1) y = abs(x) (2) y = abs(x+1) (3) y = -2 abs (x+1) (4) y = 2 abs (x+1) and (5) y = -2 abs (x+1)-4. Am I on the right track with the above equations? Thanks - Lori1 solutions Answer 160671 by jim_thompson5910(29613)   on 2009-09-11 11:05:06 (Show Source): You can put this solution on YOUR website!To graph , simply follow the given steps below: Step 1) Graph Graph of Note: the graph of is simply comprised of the lines and (just don't graph the portion that lies below the x-axis). Step 2) Shift 1 unit to the left to get Graph of (red) and (green) Step 3) Vertically stretch by a factor of 2 to get Graph of (red) and (green) Step 4) Reflect over the x axis to get Graph of (red) and (green) Step 5) Shift the entire graph of down 4 units to get Graph of (red) and (green) So the graph of is Graph of
 absolute-value/212385: This question is from textbook precalculus explain why the graphs of y=x and y=(root x)^2 are not the same1 solutions Answer 160400 by jim_thompson5910(29613)   on 2009-09-10 00:13:06 (Show Source): You can put this solution on YOUR website!Let's say that 'x' can be negative. So let's plug in The first equation is simply (after plugging in ) which is a real number; however, the second equation does not produce a real number. So the two are not equal when . They are not equal when . If we said that , then would be true.
 Radicals/212370: I’m really confusing with this factoring, Can anyone please help me with step by step? Thank you very much! Simplify by factoring. 1.4*sqrt(405x^8y^10)=? 2.5*sqrt(64x^12y^15)=? 1 solutions Answer 160399 by jim_thompson5910(29613)   on 2009-09-10 00:10:04 (Show Source): You can put this solution on YOUR website!I'll do the first one to get you going in the right direction. # 1 Start with the given expression. Break up the square root using the identity Factor 405 into 81*5 Rewrite 81 as Rewrite as Rewrite as Break up the first square root (using that same identity) Take the square root of to get 9 (notice how the squares and the square roots 'undo' each other). Take the square root of to get . Take the square root of to get . Rearrange the terms. So where every variable is nonnegative.
 Square-cubic-other-roots/212356: I am so confused with this problem. Express as a sum, difference, and product of logarithims, without using exponents. 3 log b √ x^8/y^2z^5 I could not get it to show the way the problem does. It reads log (small b below) square root with 3 above the square root symbol. Inside the square root symbol it has x^8 over y^2z^5. 1 solutions Answer 160385 by jim_thompson5910(29613)   on 2009-09-09 22:27:11 (Show Source): You can put this solution on YOUR website! ... Start with the given expression. ... Convert to rational exponent notation. ... Pull down the exponent using the identity ... Break up the log using the identity ... Break up the second log using the identity Distribute ... Pull down the exponents using the identity ... Distribute So assuming that every variable is positive.
 Quadratic_Equations/212341: given f(x) = ax^2 + bx + 5. Find a and b such that f(x + 1) - f(x) = 8x + 31 solutions Answer 160377 by jim_thompson5910(29613)   on 2009-09-09 22:12:59 (Show Source): You can put this solution on YOUR website!Since , this means that . Or simply ---------------------------------------------- Start with the given equation. Plug in and Distribute Combine like terms. So we then get that which means or Also, we find that . Plug in to get and solve to get
 Angles/212281: I have to figure out the measure of the angles. The problem is "find the measure of angle T if the measure of angle T is 20 more than four times its supplement" I know the answer is 148 but I can't figure out how to get to that. I tried solving it by setting up the equation 4t+20=180 but I got 40. And I just really can't figure out how to so this. Please help!!1 solutions Answer 160322 by jim_thompson5910(29613)   on 2009-09-09 19:39:32 (Show Source): You can put this solution on YOUR website!Let S = measure of supplement So by definition (since supplemental angles add to 180 degrees). Also, since "the measure of angle T is 20 more than four times its supplement", this means Start with the first equation. Plug in Combine like terms on the left side. Subtract from both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the solution is This means that the supplemental angle is 32 degrees while angle T is degrees.
 Radicals/212277: 2 SQRT over 3 * 6 SQRT over 6 And then you get 12 outside the radical, cause 2 * 6 = 12 But what do you get inside.1 solutions Answer 160321 by jim_thompson5910(29613)   on 2009-09-09 19:37:51 (Show Source): You can put this solution on YOUR website! Start with the given expression. Group like terms. Multiply 2 and 6 to get 12 Combine the roots using the identity Multiply Factor 18 into 9*2 Break up the root using the identity Take the square root of 9 to get 3. Multiply. So
 Radicals/212275: 2 SQRT over6^2 + 6 SQRT over 2^2 = c^2 My friend said that this was the answer.. 12 + 12 = c^2 c=SQRT over 24 But I don't know where she got it or if its right. I'm really not good with Square roots.1 solutions Answer 160320 by jim_thompson5910(29613)   on 2009-09-09 19:34:27 (Show Source): You can put this solution on YOUR website!Is the problem ??? Start with the given equation. Evaluate the square root of to get 6. Evaluate the square root of to get 2. Multiply Combine like terms. Rearrange the equation. Take the square root of both sides. or Break up the 'plus/minus' or Factor 24 to get 4*6 or Break up the square root. or Take the square root of 4 to get 2. So the solutions are or
 sets-and-operations/212276: using only 3 nines, create an equation that equals 12. you can use addition, subtraction, multiplying, dividing, and square roots 1 solutions Answer 160319 by jim_thompson5910(29613)   on 2009-09-09 19:28:45 (Show Source): You can put this solution on YOUR website!Here's one way to do it:
 Angles/212249: find the measure of angle T if the measure of angle T is 20 more than four times its supplement.1 solutions Answer 160302 by jim_thompson5910(29613)   on 2009-09-09 18:21:30 (Show Source): You can put this solution on YOUR website!Let S = measure of supplement So by definition (since supplemental angles add to 180 degrees). Also, since "the measure of angle T is 20 more than four times its supplement", this means Start with the first equation. Plug in Combine like terms on the left side. Subtract from both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the solution is This means that the supplemental angle is 32 degrees while angle T is degrees.
 sets-and-operations/212248: Using only 3 nines and no other numbers, create an equation that equals 4. You may use addition, subtraction, multiplying, dividing, and square roots.1 solutions Answer 160298 by jim_thompson5910(29613)   on 2009-09-09 18:17:49 (Show Source): You can put this solution on YOUR website!Here's one equation:
 Linear-equations/211922: -5(2x-1)= 3(x+4)1 solutions Answer 160099 by jim_thompson5910(29613)   on 2009-09-08 17:17:15 (Show Source): You can put this solution on YOUR website! Start with the given equation. Distribute. Subtract from both sides. Subtract from both sides. Combine like terms on the left side. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the solution is which approximates to .
 absolute-value/211857: Please could you help find the value of x in the following equation: 2(x+1)=3x Many thanks1 solutions Answer 160056 by jim_thompson5910(29613)   on 2009-09-08 13:47:44 (Show Source): You can put this solution on YOUR website! Start with the given equation Distribute Subtract 2x from both sides. Combine like terms. Rearrange the equation So the solution is
Matrices-and-determiminant/211775: find the value of
|5...3..-2|
|1...0...4|
|4...1...2|
1 solutions

Answer 160048 by jim_thompson5910(29613)   on 2009-09-08 13:17:26 (Show Source):
You can put this solution on YOUR website!
 Solved by pluggable solver: Finding the Determinant of a 3x3 Matrix If you have the general 3x3 matrix:the determinant is: Which further breaks down to:Note: , and are determinants themselves. If you need help finding the determinant of 2x2 matrices (which is required to find the determinant of 3x3 matrices), check out this solver--------------------------------------------------------------From the matrix , we can see that , , , , , , , , and Start with the general 3x3 determinant. Plug in the given values (see above) Multiply Subtract Multiply Combine like terms.======================================================================Answer:So , which means that the determinant of the matrix is

 Trigonometry-basics/211815: If tan x = .88, what is cot x?1 solutions Answer 160047 by jim_thompson5910(29613)   on 2009-09-08 13:15:14 (Show Source): You can put this solution on YOUR website!Since , this means that (where the '36' repeats)
1 solutions

Answer 160046 by jim_thompson5910(29613)   on 2009-09-08 13:12:01 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .

Now multiply the first coefficient by the last term to get .

Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?

To find these two numbers, we need to list all of the factors of (the previous product).

Factors of :

1,2,4,7,8,14,16,28,56,112

-1,-2,-4,-7,-8,-14,-16,-28,-56,-112

Note: list the negative of each factor. This will allow us to find all possible combinations.

These factors pair up and multiply to .

1*112 = 112
2*56 = 112
4*28 = 112
7*16 = 112
8*14 = 112
(-1)*(-112) = 112
(-2)*(-56) = 112
(-4)*(-28) = 112
(-7)*(-16) = 112
(-8)*(-14) = 112

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :

First NumberSecond NumberSum
11121+112=113
2562+56=58
4284+28=32
7167+16=23
8148+14=22
-1-112-1+(-112)=-113
-2-56-2+(-56)=-58
-4-28-4+(-28)=-32
-7-16-7+(-16)=-23
-8-14-8+(-14)=-22

From the table, we can see that there are no pairs of numbers which add to . So cannot be factored.

===============================================================

So doesn't factor at all (over the rational numbers).

So is prime.

So make sure that you have the correct problem.


text:
# 7.3: Flow Rate

Flow is the path that a certain liquid (water) takes for a specific time. It is commonly referred to as Flow Rate. A garden hose, a water main, an aqueduct, a gutter, have flow rates. You must know the area of the vessel or container that is flowing the liquid and you must also know the velocity that the liquid is moving. Flow Rate is simply expressed as volume over time.

$\dfrac{\text { Volume }}{\text { Time }} \nonumber$

The volume can be gallons, million gallons, cubic feet or any other volume unit. The time can be any time such as, seconds, minutes, hours, etc. However, when solving Flow Rate ALWAYS use the UNTIS – CUBIC FEET PER SECOND, where Q represents Flow Rate.

Expressed as

$\mathrm{Q}=\dfrac{\text { Volume }}{\text { Time }}=\dfrac{\text { cubic feet }}{\text { second }} \nonumber$

In order to calculate a Flow Rate, you need to know the Area of the vessel that the liquid is traveling through and the Velocity in which the liquid is traveling. The formula for Flow Rate is

$\mathrm{Q}=\mathrm{A} \times \mathrm{V} \nonumber$

Where the Area is expressed as – $$0.785 \times D^2$$ (for circles) - $$L \times H$$ (for rectangles) and $$\dfrac{b 1+b 2}{2} \times H$$ (for trapezoids)

And

Where Velocity is expressed as – $$\dfrac{\text { Distance }}{\text { Time }}$$

The UNITS for AREA should always be in SQUARE FEET and the UNITS for VELOCITY should always be in FEET PER SECOND.

Therefore, the equation should look like this

$\dfrac{\text { cubic }}{\text { feet }}=\dfrac{\mathrm{ft}^{2}}{1} \times \dfrac{\mathrm{ft}}{\text { second }} \quad \rightarrow \quad \dfrac{\mathrm{ft} \times \mathrm{ft} \times \mathrm{ft}}{\text { second }}=\dfrac{\mathrm{ft} \times \mathrm{ft}}{1} \times \dfrac{\mathrm{ft}}{\text { second }} \nonumber$

If you like using the “Pie Wheel” the following can be applied for Flow Rate.

## Exercise 7.3

1. What is the pressure in pounds per square inch at the bottom of a 45 foot tall water storage tank?
2. An 80 foot tall standpipe is ¾ full. What is the pressure in psi at the bottom?
3. A 30 foot water tank has a pressure gauge installed 2 feet from the bottom of the tank. If the pressure gauge reads 11 psi, what is the level in the tank?
4. A 12” pipeline is flowing water at a velocity of 3.5 fps. What is the corresponding flow rate?
5. A 36” diameter pipe is flowing water at a velocity of 5 fps. How many gallons of water will flow through it in 2 hours?
6. An 18” pipe is flowing 1,500 gpm. What is the velocity?
7. What is the area of a box culvert that is flowing 4.5 MGD at a velocity of 2.45 fps?
8. An 8” diameter pipe is flowing water at a velocity of 7.65 fps. What is the flow rate in gallons per minute?
9. A 28 foot tall water tank is sitting on a 75 foot tall hill. What is the pressure in a home at the bottom of the hill? (Assume the pressure gauge is 3 feet from the bottom of the home and the tank is full)
10. What is the velocity through a 20” diameter pipe that flows 2.5 MGD?
11. A home that is receiving water from a water tank that is on top of a 115 foot hill has a pressure of 57 psi. What is the water level in the tank?
12. What is the area of a pipe that flows 2,550 gpm at a velocity of 5 fps?

7.3: Flow Rate is shared under a CC BY license and was authored, remixed, and/or curated by LibreTexts.

format: plain


text:
# model deterministic command

Syntax

model deterministic <default> b

Set the deterministic mode. Deterministic mode is reset to the default setting whenever a model new command is given. The optional default keyword is used to set the default setting to b. When default is not used then the current deterministic mode only is set. On installation (that is, prior to any user change) the program’s default determinism setting is on. The default setting can also be changed by using “Determinism by default” checkbox in the Options Dialog.

When a calculation is performed in parallel over multiple CPUs or cores, the order in which specific computations occur can and will change. This can cause roundoff errors to accumulate differently. For many models, the final result of these differences is nearly impossible to detect. But for a wide class of models these differences can magnify, resulting in a situation where a different result occurs every time a given model is run.

It is important to note that all of these results are equally valid, as they are all within the range of finite-precision floating point error. Deterministic mode sacrifices efficiency and memory to force the parallel computation to result in the same answer every time, regardless of the number of CPUs it is performed on or their relative speed.

The following are the effects of turning determinism off.

• The model will run faster. How much faster will be code- and model-dependent. FLAC3D is always deterministic; setting determinism off with this command is ignored. 3DEC will be approximately 10% faster. PFC will be about 15% faster.
• Results will not be repeatable. In some cases where small changes propagate, very different results will occur every time the same model is run.
• Data files (and project bundle) will no longer reliably reproduce results. For archival purposes, users should store the resulting save files along with input data.
• For a problem requiring support, Itasca may not be able to replicate the issue. In this case the model should be rerun with determinism on. If the problem no longer occurs, there might be very little that can be done to resolve the issue.


text:
# +1 AC or +20 HP--which is better for a nimble, frontline character?

I've planned a 1 Monk, 19 Draconic Sorcerer, and I have the option of setting Con to 14 and Wis to 12 (providing +1 HP at each level up), or Wis to 14 (+1 AC, given Monk Unarmored Defense) and Con to 12 (dipping in Barbarian instead of Monk is not an option).

Which would provide more survivability in the frontlines, going from 17 AC to 18 AC (14 Wis), or +20 HP (14 Con)?

I am not yet really familiar with the fifth edition, but as far as I know, it seems important to me to also consider that +2 WIS is also a boost to your save and some skills which can drastically affect your survavibility.

Instead of giving you my very own answer which may not be your depending your style of play, I will give you list of pros for each option.

Pros to +2 Wis

• +1 to perception checks means that you will less likely be surprised by a sneaky character, which makes quite a difference for a squishy character.
• +1 to save is always good to take even more at low levels... Yet it is a weak save for you, so at lvl 20 I doubt it will make a huge difference.
• +1 AC is pretty good, it ends up as a 5% miss chance which, on the long run, is better. Well, before lvl 20, enemies tend to hit significantly harder than 20 pesky damages. It won't do much better than "buffering", at best, a single attack.

Pro to +2 Con

• It gives you +20 HP so about 21 % more HP. It is a good bonus to your survivability yet it won't save you from a bruiser's beating.

CON 12: 8 (monk) + 3.5 * 19 (average sorcerer HP roll) +20 (Con) = 94

CON 14: 8 (monk) + 3.5 * 19 (average sorcerer HP roll) +40 (Con) = 114

• +1 to one of your strong saves, which is good and probably better on the long run, as the system seems to encourage you to be good in your area of expertise and count on friends to compensate your weaknesses.

Bonus:
As told previously, +1 CON won't net you the best result in term of sheer HP bonus, if you are really looking for a boost, for your HP, the feat 'Tough' is a better alternative.

Unarmored Defenses and Draconic Resilience are both AC calculations and thus do not stack. Only one of the two will apply, unless there was something else you wanted a dip into monk for that is suboptimal.

You might also consider the tough feat if your table is allowing feat use as it provides +2 hp per level of survivability over +1 to hp or AC from a stat boost.

The plus 1 AC gives more survivability. At level 20 where you would actually gain 20 hp, monsters are doing so much damage that 20 hp is not worth the extra 5% of missed hits.

Extra AC is always better due to flat bonuses on attacks and bounded accuracy, while damage is a scaling increase based on CR. Saves are based on abilities now so increasing Wis is always bettter than extra HP.

This is assuming that you use Array and put your highest att on Dex > Cha > Wis due to Sorc Multiclassing.


text:
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lewisje commented on a post in r/badmathematics
38
Number154 23 points

Obviously areas aren’t additive. If you cut a square into four smaller squares the sum of their areas is half the area of the large square. The large square gets extra area synergistically from the interactions of sets of points spread across the different smaller squares.

lewisje 8 points

because Banach–Tarski or sth

etherealeminence 3 points

I read something about Calabi-Yau manifolds and I think there were like 6 dimensions? This proves that math is a fraud.

lewisje 1 point

# TIME CUBE

lewisje commented on a post in r/learnmath
0
dukes_haven 0 points

The fractions cancel out and you're left with int dt which is equal to

Wait.. I don't get this part, what fractions?

lewisje 1 point

When you substitute x=sin(t) into the integral of dx/√(1-x2), the numerator becomes cos(t)dt and the denominator becomes √(1-sin(t)2); when you work it out, stuff cancels out of the numerator and denominator.

lewisje commented on a post in r/learnmath
3
lewisje 1 point
• u_(n+1)=int(sin(t)/t,t,(n+1)π,(n+2)π).
• The substitution T=t-π means t=T+π and dt=dT, and then the limits of integration become nπ to (n+1)π; also, sin(T+π)=-sin(T):
• u_(n+1)=-int(sin(T)/(T+π),T,nπ,(n+1)π).
• After re-labelling the dummy variable, notice that because t>0 and t+π>t, this means 1/(t+π)<1/t.

Then use the alternating-series test to finish part b.

As for part c, there are ways to calculate this integral directly, but another way to estimate it might be approximating the sin(t) factor for each term with a piecewise-linear factor, or with a quadratic factor; that does get messy, though.

For piecewise-linear, your "outer estimates" would be tangent to the sine curve at each multiple of π, and your "inner estimates" would still be zero at each multiple of π but would also hit the same peak as the sine curve; for quadratic, your "inner estimates" would be tangent to the sine curve at each multiple of π, and your "outer estimates" would still be zero at each multiple of π but would instead be tangent to the sine curve at its peak.

The "outer estimate" for the first term would be (x if 0<x<π/2, π-x if π/2<x<π) for piecewise-linear and 4(π-x)x/π2 for quadratic, and the "inner estimate" would be (2x/π if 0<x<π/2, 2-2x/π if π/2<x<π) for piecewise-linear and (π-x)x/π for quadratic; then shift them across the terms as needed.

I tried the quadratic approximations and they looked simple enough, but Wolfram|Alpha couldn't calculate them.

lewisje commented on a post in r/mathbooks
5
lewisje 1 point

FTR there are many many textbooks on Ordinary Differential Equations that have been re-printed by Dover (most Dover books are re-prints, but there are a few original publications); this particular one is quite good, and another one you might consider is An Introduction to Ordinary Differential Equations by Coddington, a more proof-oriented book.

lewisje commented on a post in r/learnmath
okewp 3 points

While insightful I believe your post doesn't quite solve my problem. You see, if I have a simple matrix, let's say (1, 1; 2, 2); its determinant is zero as the vectors in it are LD. This transformation over the vector (1,0) results in (1; 2) and the transformation over (0, 1) also gives (1; 2). The transformation over those two LI vectors that I picked resulted on two LD vectors... I had a square and now I have a line. The same thing you said basically.

My question was if I could affirm that some functions are LD if the determinant of the Jacobian is zero. The answer is no as /u/lewisje, said bellow. Searching further I found out that I can say that some functions (and not their gradients!) are LD if the Wronskian is zero. TIL.

Thanks!

lewisje 2 points

Yeah, I was suspecting that you were really thinking about the Wronskian, which is something that you find for scalar-valued functions of one variable; also, I think you meant "not just their gradients", because if functions are linearly dependent, so are their gradients, by the linearity of the gradient operation (also, for functions of one variable, the gradient is the same as the ordinary derivative).

okewp 2 points

But does it mean that the functions are linearly dependent if the Jacobian is zero the same way it does with vectors?

lewisje 2 points

If the Jacobian determinant is just zero at a point, no; if the Jacobian determinant is identically zero, that means that the gradients of the functions are linearly dependent, and that the vectors consisting of the partial derivatives of each function with respect to the same variable are linearly dependent, but the functions themselves can still be linearly independent, as a simple example will show:

• Let f(x,y)=x2+y2+1, g(x,y)=2x2+2y2-1, and r(x,y)=[;\begin{bmatrix}f(x,y)\\g(x,y)\end{bmatrix};]; then clearly f and g are linearly independent (neither being a multiple of the other), but
• ∂f/∂x=2x, ∂f/∂y=2y, ∂g/∂x=4x, and ∂g/∂y=4y
• so ∇f=[2x 2y] and ∇g=[4x 4y], so ∇g=2∇f, so the gradients are linearly dependent,
• and ∂r/∂x=[;\begin{bmatrix}2x\\4x\end{bmatrix};] and ∂r/∂y=[;\begin{bmatrix}2y\\4y\end{bmatrix};], so if x=y=0, then the partial derivatives are equal, and otherwise, y∂r/∂x-x∂r/∂y=0, a nontrivial linear combination, so in either case, the partial derivatives are linearly dependent, and
• det(Jr)=[;\begin{vmatrix}2x&2y\\4x&4y\end{vmatrix};]=8xy-8xy=0.

A curious question to consider is whether, if the gradients of n scalar functions of n variables are linearly dependent, this means that there exist n constants such that the set consisting of each function plus one of the constants is also linearly dependent; clearly, for the example above, two constants that work are -1 and 1, but can they be found in general?

lewisje commented on a post in r/math
1
csguy124 2 points

I have a meeting coming up this week with the prof of numerical analysis!

I also took diffeq and linear algebra and they were so easy that I like literally slept through class and learned everything the night before the exams...but naturally forgot everything right after

Perhaps now im missing some fundamentals..?

lewisje 5 points

You probably are, because both linear algebra and ODEs are key for understanding dynamical systems.

lewisje commented on a post in r/math
elseifian 10 points

It's really unfortunate that a lot of people seem to get the impression that complex numbers come from magically declaring i to be the square root of negative one.

The important step is writing down axioms we want our number system to satisfy and then showing that the complex numbers satisfy those axioms.

There's nothing stopping us from inventing a number system where "j" or whatever is the result of division by zero. The problem is that it wouldn't be a very good number system, because division by zero isn't compatible with the other properties we want a number system to have.

lewisje 3 points

I'll admit, I didn't understand at first why some textbooks started by defining complex addition and multiplication without the "treat i like some thing that squares to -1" mnemonic, which makes the multiplication in particular much easier to deal with.

lewisje commented on a post in r/learnmath
1
lewisje 1 point

Your explanation is adequate, but keep in mind that 1 is also a member of the equivalence class of 1, and that 7 is not a member of the equivalence class of 1, because 13≡1 (mod 7), but 73≡0 (mod 7).

It turns out that the integers mod 7 have no zero-divisors (that is, two integers that are not equivalent to 0 mod 7, with a product equivalent to 0 mod 7), and this is because 7 is prime; therefore, you can simplify the problem of finding the equivalence classes of 1 under R a bit:

• x3≡1 (mod 7) iff x3-1≡0 (mod 7).
• x3-1=(x-1)(x2+x+1), so x-1≡0 (mod 7) or x2+x+1≡0 (mod 7).
• From this, x≡1 (mod 7) or -(x+1)x≡1 (mod 7).
• Because -1≡6 (mod 7), the latter equation is (x+1)x≡6 (mod 7).
• The only positive solutions to the first of these equations are 1, 8, 15, ...; for the second one, clearly 0 doesn't work, and then
1. 2*1=3 ❌
2. 3*2=6 ✔
3. 4*3=12≡5 (mod 7) ❌
4. 5*4=20≡6 (mod 7) ✔
5. 6*5=30≡2 (mod 7) ❌
6. 7*6≡0 (mod 7) ❌
• Therefore, the smallest positive solutions to the second equation are 2, 4, 9, 11, ...
• From this, the four smallest positive members of the equivalence class of 1 under R are 1, 2, 4, and 8.
• You can check that they are in the class:
• 13=1
• 23=8≡1 (mod 7)
• 43=64≡1 (mod 7)
• 83=512=7*73+1≡1 (mod 7).

It is curious that the members of the equivalence class of 1 consist of integer multiples of 7 plus whole-number powers of 2.

EDIT: Also, because it turns out that the integers mod 7 form a field, and that the discriminant of x2+x+1 has a square root in this field, you can use the quadratic formula too:

• The solutions to x2+x+1≡0 (mod 7) are
• x≡(-1±√(1-4))/2≡(-1±√(-3))*4≡4(-1±√4)
• ≡4(-1±2)≡(4*(-3) or 4*1)≡(4*4 or 4)≡(16 or 4)≡(2 or 4) (mod 7).
• Keep in mind that -3≡4 (mod 7) and that 4*2=8≡1 (mod 7), so the multiplicative inverse of 2 mod 7 is 4.

The integers with square roots (mod 7) are quadratic residues mod 7, and they are the integers equivalent to 0, 1, 2, or 4 (mod 7).

lewisje commented on a post in r/speedrun
89
TimpZ91 2 points

that's unfortunately not good enough for TAS:es :(.

lewisje 5 points

Do we have 120Hz games or something?

lewisje commented on a post in r/badmathematics
222
Puddleduck24 38 points

Oh man. I teach maths I have so many. My least favourites are:

• Cancelling sigmas on quotients of sums.
• sin(2x)/x = sin(2)
lewisje 7 points

For real, I once thought something like if H⊴G then H×(G/H)≅G in group theory.

(As a general counterexample, consider Z4 and its normal subgroup isomorphic to Z2; you can't even get a semidirect product out of that.)

Alternatingwaffle 2 points

As someone who never even questioned that falsehood I'm suddenly deeply perturbed with the state of my algebra knowledge. Especially the first semester was for me, by far the least rigorous, proof based course I took in my math undergrad.

So... why doesn't this work? Exactly when DOES it work? In my head I always thought of quotienting a group as rewriting the group into elements of the form hg (well defined if H normal), then just delete all the 'h's and play around with whatever symbols are left with the logical binary operation. On the other hand, a direct product is kind of like for each g, making 5 new elements h1g h2g h3g h4g h5g, so didn't it just add back all of the information I just ripped out? Somehow this process rearranges some of the cycles in G, but... why?

lewisje 2 points

You can get a deeper understanding by looking at outer semidirect products; a key fact about Z4 is that it has only one subgroup isomorphic to Z2 (call it N), so there is no subgroup of Z4 that is isomorphic to Z4/N that has a trivial intersection with N.

Another example involves S3 vs. Z6; both are distinct semidirect products of the form Z3⋊Z2, but the former involves the non-trivial automorphism on Z3 sending each element to its inverse and the latter, involving the trivial automorphism, is the direct product.

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lewisje commented on a post in r/math
R-Module 3 points

But then simple propositional logic should be taught. I know well too many students here in Germany who know what an integral is but do not know what a contraposition/quantifier is. I think logic is far more important in life than calculus

lewisje 3 points

In some American high schools, something like this is taught in Geometry, although without the actual symbols and truth tables; at least terms like "converse" and "contrapositive" are introduced.

ShearedSheep 2 points

My precalculus course in high school, for some reason, spent the first chunk of the class on exactly this (truth tables, formal logic, etc.).

lewisje 1 point

If it wasn't in Geometry, it ought to be covered by then; my school didn't cover that at all, but it wasn't close to a top school.

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lewisje commented on a post in r/math
50
julesjacobs 10 points

Structure and Interpretation of Computer Programs. I was hooked after I learned how to calculate the n-th Fibonacci number in O(log n) arithmetic operations.

lewisje 5 points
lewisje commented on a post in r/learnmath
1
facebookhatingoldguy 2 points

Hard to know if the OP means Arctan(x) which is normally written tan-1(x), or if they mean (tan(x))-1 which would be cot(x), or if they mean tan(x-1) = tan(1/x). It's probably not the last one since that has no antiderivative in terms of elementary functions.

lewisje 1 point

There's no way OP meant tan(x-1). (nvm OP had edited it)

facebookhatingoldguy 3 points

Which is pretty much what I said. I've just seen a lot of crazy things on this sub over the years (so have you for that matter), so it seemed bad to assume anything. Just curious, were you the one that downvoted my comment? I have tremendous respect for you as a contributor here, so if you think what I said was inappropriate, I'd gladly take some constructive criticism. (I couldn't care less about the internet points).

lewisje 3 points

I downvoted it at first, but then someone else's reply made me realize that OP originally wrote it even more ambiguously.

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lewisje commented on a post in r/math
7
lewisje 3 points

This is where the theory of manifolds comes in (particularly Riemannian manifolds); my understanding is that if you can represent a particular subset of R3 as a subset of a Riemannian manifold of finite area, you can look at the inverse images of this subset in R2 (under the various charts in the atlas of the manifold) to check their Jordan measure, and then use the metric tensor to figure out how this area, if it exists, changes as each of these inverse images is mapped to the aforementioned manifold.

15master 2 points

Then how can you decide if a manifold has a finite area or not?

lewisje 1 point

with the metric tensor, reducing the problem to the one of finite integrals over R2

lewisje commented on a post in r/learnmath
5
ktrprpr 3 points

a circle from B to A which intersected at yellow point

This is the problem. If the A isn't choose properly, the right circle at radius BA will not intersect yellow point. You can choose A very far away and see what happens.

lewisje 1 point

I misread the line you quoted in my top-level reply and thought it was "a circle centered at B through the yellow point", which actually would work out.

lewisje 2 points

I'm not sure it's wrong, even though it's not the usual construction:

1. Draw the circle centered at the yellow point through A.
2. Mark its other point of intersection with the line as B.
3. Draw the circle centered at A through the yellow point.
4. Draw the circle centered at B through the yellow point.
5. Draw the line through the yellow point and the other intersection between the circles centered at A and B that you drew earlier.

The construction that I described marks four points in the shape of a rhombus, while yours (the one I misread you as making) marks four points in the shape of a kite, and both shapes have the property that their diagonals are perpendicular.

EDIT: I misread your construction as drawing a circle centered at B through the yellow point, which would have been fine; a circle centered at B through A wouldn't necessarily work.

lewisje commented on a post in r/learnmath
Yeazelicious 3 points

I just looked up Stokes' theorem. My only understanding of it is that it's Green's theorem but for 3-dimensional objects? Other than that, I can't even begin to understand what's going on. Something to do with surfaces and vector fields?

Either way, after nearly a half hour of trying to figure it out in even a shallow capacity, I understand what a tall order I put in. Yikes.

lewisje 3 points

A closer analogue to "Green's theorem but for 3-dimensional objects" is Gauss's theorem (a.k.a. the divergence theorem), which relates the integral of the divergence of a vector field over a closed and bounded region to the integral of the vector field itself over the boundary of that region.

In the language of differential forms (see my top-level reply), it's the integral of a 2-form over a closed surface in space compared with the integral of its exterior derivative (a 3-form, which in 3D space is like a scalar) over the region of space inside that surface.

Stokes's theorem is quite amazing though: If you can find a "primitive" for a vector field (specifically a vector field whose curl is the given vector field, similar to the concept of an antiderivative), then the line integral of that primitive over a curve in space equals the surface integral of the original vector field over any surface bounded by that curve, so long as the primitive is differentiable all along that surface.

(Because the curl is involved, that's why this theorem is about the swirly; Gauss's theorem, BTW, is about the flow.)

lewisje 5 points

It's a slight generalization of the Fundamental Theorem of Calculus: Instead of considering the values of a differentiable function at the endpoints of an interval and the integral of its derivative over that interval, you consider the line integral of a vector-valued function over the boundary of a region in the plane and the surface integral of its exterior derivative over that region.

The line integral is like calculating the total work done along a path, and a surface integral over a region in the plane is like calculating the volume under a surface over that region; "surface integral" is also often used for integrating over a surface in space, and that's like calculating total flux, or charge given a charge-density.

I'm getting a little into the language of differential forms here, but it's like you have a function f of x and y, and then its exterior derivative is the derivative with respect to x times this abstract thing called "dx" plus its y-derivative times "dy".

• That is, df = ∂f/∂x dx + ∂f/∂y dy
• The weird script d means a "partial derivative": To calculate ∂f/∂x, work as if y were just some constant, and vice-versa for ∂f/∂y.

Then the exterior derivative of something like f dx is

• d(f dx) = df∧dx = (∂f/∂x dx + ∂f/∂ y dy)∧dx
• = ∂f/∂x dx∧dx + ∂f/∂y dy∧dx.
• Then the special rules are dx∧dx = 0, dy∧dy = 0, and dy∧dx = -dx∧dy, so you end up with...
• d(f dx) = -∂f/∂y dx∧dy.

This sort of thing is why the statement of Green's theorem looks the way it does:

• You start with the differential form F dx + G dy, and then its exterior derivative, after working out all the computations, is (∂G/∂x-∂F/∂y) dx∧dy.
• Green's theorem says that if the first thing is differentiable over a region in the plane, then its line integral along the boundary of that region equals the integral of the second thing over the region itself.
• In more formal language, if F dx + G dy is differentiable over D, a compact region bounded by C, then
• [;\displaystyle\oint_\limits{C}F\operatorname{d}\! x+G\operatorname{d}\! y=\iint_\limits{D}\left(\frac{\partial G}{\partial x}-\frac{\partial F}{\partial y}\right)\operatorname{d}\! x\wedge\operatorname{d}\! y;]

Usually, when you encounter this in Calculus III, the language of differential forms will not be used, and the double integral on the right will just be written with "dx dy" at the end, with less motivation for where that minus sign came from.

This duality with regions and their boundaries vis-à-vis functions and their derivatives is why, in more advanced mathematics, ∂ denotes the boundary of a region.

In its greatest generality, you get the generalized Stokes's theorem, which says that the integral of a differential form over a closed and oriented manifold equals the integral of its exterior derivative over a manifold-with-boundary that is bounded by that closed and oriented manifold.

• It looks elegant enough using the language of differential forms:
• [;\displaystyle\int_\limits{\partial D}\omega=\int_\limits{D}\operatorname{d}\!\omega;]

The "oriented" bit is why the minus sign appears in the Fundamental Theorem of Calculus, BTW.

In vector calculus, you'll see this stuff written in terms of "curl" or ∇×, "div" or ∇·, and so on, which is more confusing than seeing it with differential forms first and translating to vector calculus later.

lewisje commented on a post in r/learnmath
3
lewisje 1 point

The search term is "functional differential equation"; for this particular one, you can differentiate both sides twice to get f[4](x)=4kf''(2x)=4kf(4x), and again to get f[6](x)=64kf''(4x)=64kf(8x), and the following pattern appears to emerge:

• f[2n](x)=(2n)n-1kf(2nx)
• but substituting n=0 doesn't help, because it implies f(x)=0-1kf(0); this might work as a "limiting case" if f(0)=0 though...

Generally, FDEs are difficult to solve analytically.

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# James Edward Lewis II

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Home > Cleartool Error > Cleartool Error Unable To Access File Name Too Long

# Cleartool Error Unable To Access File Name Too Long

Martina Don't Postpone Joy - Have Fun ReleaseTEAM Inc http://www.releaseteam.com IBM Rational Certified Consultants IBM Business Partner SystemAdmin 110000D4XK 47283 Posts Re: Unable to transfer a file: Filename too long. 2012-02-09T20:53:38Z For users with access privilege to secured folder update is fine Have a nice day. it's fine after running the cleartool setcs, but when it runs maven commands, it's downhill after that... Looks like you're trying to execute the process from a non existing folder. http://tubee.net/cleartool-error/cleartool-error-unable-to-access-file.html

Am I hitting a Windows pathname limit or a Perforce pathname limit? The absolute pathname for the file is 263 characters. Submit feedback to IBM Support 1-800-IBM-7378 (USA) Directory of worldwide contacts Contact Privacy Terms of use Accessibility I searched the archives and didn't see this so i hope it's OK to Thanks in advance.. http://www-01.ibm.com/support/docview.wss?uid=swg21580643

If you have received this message in error please notify the sender by return e-mail and delete it from your system. WARNING XXXXXX: Could not create temp dir in M:/xxxxxx_view7: couldn't create directory "M:\xxxxxx_view7\xxxxx_tmp_3840_0": The media is write protected (error code 19) Whenever I open a DOS command line window (via cmd.exe) Perforce client apps do not hold connections open. Watson Product Search Search None of the above, continue with my search Our apologies...

ex: 1) /vob/customer/project1 2) /vob/customer/project2 - config spec #1 - element * LABEL1 load /vob/customer/$Unknown macro: {PROJECT} - end config spec - This way we could create a job for n!x wrote: > Any idea, for the solution?? > For my other friends its going fine. United States English English IBM® Site map IBM IBM Support Check here to start a new keyword search. check-in & undo-checkout the parent to force a creation of the Branch Type on that folder. [by the way ... Changes to read-only components are not detected and loaded in to the view 2. Since the NUL terminator has to be included, the file path itself must be no longer than 259, and there are further restrictions. Command line="cleartool mkview -snapshot ", actual exit code=1 java.io.IOException: Clear tool did not return the expected exit code. https://www.ibm.com/developerworks/community/forums/thread.jspa?threadID=417575 Or if within the US, Spirent Communications, 26750 Agoura Road, Calabasas, CA, 91302, USA. I'm using CC6/Maven2, wit... The problem is that, the output of it should not be shown in the actual job's log, since it's not directly related with the build process.A command like "cleartool lsview$CLEARCASE_VIEWNAME" I also had trouble using localhost, and needed to use my machine name. If you are not the intended recipient, you are hereby notified that any disclosure, copying, distribution and / or the taking of any action based upon reliance on the contents of

If I tell it to use another view (tla2_vu), the line causing the error changes from [tla_vu] to [tla2_vu]. learn this here now All rights reserved. Please re-enable javascript to access full functionality. 1 The file name is too long Started by Mailman Sync, Mar 02 2010 03:26 PM Page 1 of 2 1 2 Next Please Or it might be impossible to re-implement: I've heard that there are some machine-dependent things that they refuse to put into the Java specification, though I don't know what they might

But I have got "Error: Unabled to change configuration specification: Permission denied" error, when jenkins set the config spec of view with "cleartool setcs ... " command. check over here cleartool: Error: Unable to access "/net/foo-blablablabla.xyz.hhh.ttt.com/vol/clearcase_area/vobstore_abc/vob-1.vbs": File name too long. Permalink Apr 16, 2009 Park Su Jin says: Hello I would like to get source in Clear Case with polling mode using Hudson. ... A good solution for this would be a checkbox "Use existing view" which doesnt need a config-spec.

Only users of a special group can access that subfolder. Hide Permalink tla612 added a comment - 2009/Apr/29 2:45 PM It was a configuration error on my part. Host-local path: goblin:/.2.vws Global path:     /.2.vws It has the following rights: User :   : rwx Group: : his comment is here It lets you associate a drive letter with a path on the current machine, as in subst m: c:\this\is\a\very\long\path\that\would\otherwise\get\me\in\trouble Then all references to m: point to that c:\stuff, though you

Set the config spec that should be used. This is the accepted answer. For example, in case C:\ClearCaseWorkSpace-CashViews5.1 is your viewroot, you could m net use K: C:\ClearCaseWorkSpace-CashViews5.1 And then update your view from the drive you have assigned the viewroot to (in this

## Multi-site poll buffer Optionally, if you wish to have polling check for changes since the last build plus an additional period of time, in case of changes made at another site

Back to top #12 Mailman Sync Mailman Sync Advanced Member Maillist Aggregator 2427 posts Posted 02 March 2010 - 10:47 PM Originally posted to the perforce-user mailing list by: Tim McDaniel Tel No. +44 (0) 1293 767676 Fax No. +44 (0) 1293 767677 Registered in England Number 470893 Registered at Northwood Park, Gatwick Road, Crawley, West Sussex, RH10 9XN, United Kingdom. Or if within the US, Spirent Communications, 26750 Agoura Road, Calabasas, CA, 91302, USA. Try searching and/or posting on the user mailiing list....?

Do you have a directory called: tla_vu or is that missing? In the Advanced section in the configuration it is possible to specify one or several paths in the VOB path(s) field that will be used when retrieving the history. If the config spec contains "vobs/gtx2" then the VOB path(s) field should be set to gtx2. (Mail thread, issue #1053) Todo list Add documentation on how to install a build trigger weblink cleartool: Error: Unable to access "\n%c\n'": ClearCase object not found.

Can anybody tell me how to copy the clearcase plugin source codes without copying it manually from Fisheye repository. Forgot I hadn't logged in. This is why we tell all our users to put the root of their workspace as close to the root of a drive as possible. (The default that P4V likes to If you require assistance, please contact our IT department at helpdesk at spirent.com.

Am I hitting a Windows pathname limit or a Perforce pathname limit? but we just about to expand to 4, and using "Hudson 1.339" and "Clearcase 1.1.1" and "ClearCase UCM Baseline 1.3" in our configuration. This will allow the protectvob command to run as expected. cleartool: Error: Unable to access "\gfx_Development\Test\ghal3d": Permission denied.

As others have indicated, connections to the server (or client processes) are established at the onset of a command and terminated when it's done. I don't know what Perforce is doing in the current case to know whether it's Perforce's own client code, or whether it's using something else, so I can only talk in Can shorten the workspace name and get under the limit?

E-mail confidentiality. -------------------------------- This e-mail contains confidential and / or privileged information belonging to Spirent Communications plc, its affiliates Advanced configuration Click the "Advanced" button to reveal the advanced options, including the dynamic view options.

Can you please check if the following fix with plug-in version 1.1 is really fixed? Thanks and Regards Permalink Sep 14, 2011 Nagaraju says: Hi, When i build my project i got success mail to my mail id .In m...


text:
Inertia example of moment tensor

Moment of Inertia Tensor Physics Forums

The Moment of Inertia Tensor SpringerLink. moment of inertia tensor. to see how our tensor notation works, consider example 9.1.1 on p.340. the products of inertia are evaluated as integrals,, moment of inertia 5 an example of this is the concrete t-beam composite is coincident, the moments of inertia are additive. examples for moment of inertia).

In this post, weвЂ™ll see where the inertia tensor, moments of inertia, and products of inertia come from, In the example above, the product of inertia \(I_ Lecture 36: Principal Axes of Inertia three non-zero inertia tensor elements the principal moments of inertia вЂ“ For example,

Lecture 11 2D Motion of Rigid Bodies: Finding Moments of Inertia, Rolling Cylinder with Hole Example Finding Moments of Inertia Figure 1: Rigid Body. In that example, one principal axis, the one corresponding to eigenvalue , was Thus the moment of inertia tensor can be diagonalized to look like

are called the moments of inertia about the x 1-,x 2-, and x 3-axes Example Calculate the inertia tensor for a homogeneous cube of density !, mass M, and side The Symmetric Inertia Tensor block creates an inertia tensor from moments and products of inertia.

Lecture 11 2D Motion of Rigid Bodies: Finding Moments of Inertia, Rolling Cylinder with Hole Example Finding Moments of Inertia Figure 1: Rigid Body. Moment of Inertia 5 An example of this is the concrete T-beam composite is coincident, the moments of inertia are additive. Examples for Moment of Inertia

zz are called moments of inertia with respect to the x, y and z axis, respectively, and are given by I xx = (y the tensor of inertia is diagonal, Rotation & Inertia Tensors For example, think of an oddly Once we know this we can approximate the moment of inertia of a go stone by breaking it up into a

Moment of inertia tensor. To see how our tensor notation works, consider Example 9.1.1 on p.340. The products of inertia are evaluated as integrals, Moments of inertia and inertia tensors are well-known characteristics of rigid bodies in physics and applied mathematics. How to calculate them in Mathematica? Is

The Moment of Inertia Tensor SpringerLink

Symmetric Inertia Tensor MathWorks. the tensor of inertia of a solid sphere is $i_{ii}=\frac{2}{5}mr^2$ about an axis passing through its cm. why would the tensor of inertia of each hemisphere about, physical applications of triple integrals. tensor of inertia. example 4. find the moment of inertia of a right circular homogeneous cone about its axis.); obviously, for this example, principal axes form of moment of inertia tensor. we already know that the transformed matrix is diagonal, so its form has to be., axis and izz mass moment inertia about the z i want to show you one more example, this inertia property matrix is called a second-order tensor and in.

Example of moment of inertia calculation. uh.edu

Lecture 34 Principal Axes of Inertia physics.arizona.edu. moments of inertia and inertia tensors are well-known characteristics of rigid bodies in physics and applied mathematics. how to calculate them in mathematica? is, moment of inertia tensor consider a rigid body rotating with fixed angular velocity about an axis which passes through the origin--see figure 28.).

Moment of Inertia Examples Engineering Society

Rotation & Inertia Tensors Gaffer On Games. what is moment of inertia? letвђ™s take a common example; one can determine moment of inertia as a tensor quantity., for more examples, see the list of moments of inertia. the moment of inertia tensor is a convenient way to summarize all moments of inertia of an object with one).

What is the moment of inertia tensor? How is it derived

Moment of Inertia Examples Engineering Feed. for more examples, see the list of moments of inertia. the moment of inertia tensor is a convenient way to summarize all moments of inertia of an object with one, are called the moments of inertia about the x 1-,x 2-, and x 3-axes example calculate the inertia tensor for a homogeneous cube of density !, mass m, and side).

Moment of Inertia of an Equilateral Triangular Plate

10.24 ** (a) If Icm Denotes The Moment Of Inertia. in this post, weвђ™ll see where the inertia tensor, moments of inertia, and products of inertia come from, in the example above, the product of inertia \(i_, answer to 10.24 ** (a) if icm denotes the moment of inertia tensor of a rigid body (mass m) about its cm, and i the corresponding).

physics Calculate inertia tensors - Mathematica Stack

Definition of Moment of Inertia m i в€‘mi UVic. calculating moments of inertia lana sheridan 1 de nitions the moment of inertia, 2 an example: moment of inertia of a right circular cone, in this post, weвђ™ll see where the inertia tensor, moments of inertia, and products of inertia come from, in the example above, the product of inertia \(i_).

16/04/2015В В· Example of moment of inertia Apr the moment of inertia is a second order tensor.) That being said, the moment of inertia is not what is important for whether your zz are called moments of inertia with respect to the x, y and z axis, respectively, and are given by I xx = (y the tensor of inertia is diagonal,

Rotation Kinematics, Momentof Inertia, andTorque Mathematically, give you a few examples of moments of inertia for bodies of particularly simple shapes: axis and IZZ mass moment inertia about the Z I want to show you one more example, This inertia property matrix is called a second-order tensor and in

Moments of inertia and inertia tensors are well-known characteristics of rigid bodies in physics and applied mathematics. How to calculate them in Mathematica? Is Moments of inertia and inertia tensors are well-known characteristics of rigid bodies in physics and applied mathematics. How to calculate them in Mathematica? Is

Lecture 34: Principal Axes of Inertia three non-zero inertia tensor elements the principal moments of inertia вЂ“ For example, P442 вЂ“ Analytical Mechanics - II The Tensor of Inertia c Alex R. Dzierba We will work out some speciп¬Ѓc examples of problems using moments of inertia. Example 1 x y A

LetвЂ™s talk about where the moment of inertia tensor came from. One of the confusing aspects, I think, is that it seems like a completely separate quantity. What 1 CHAPTER 2 MOMENT OF INERTIA 2.1 Definition of Moment of Inertia Consider a straight line (the "axis") and a set of point masses m1, m2, m3,K such that the

Rotation Kinematics, Momentof Inertia, andTorque Mathematically, give you a few examples of moments of inertia for bodies of particularly simple shapes: MomentOfInertia[reg, pt, v] computes the moment of inertia for the region reg rotating around an axis through the point pt in direction v. MomentOfInertia[reg

Moment of Inertia Examples Engineering Society


text:
# The equation for the transverse wave on a string is y = 4sin 2π (t/0.05 -x/50) with length expressed if centimetre and time In second.

14 views

The equation for the transverse wave on a string is y = 4sin 2π (t/0.05 -x/50) with length expressed if centimetre and time In second. Calculate the wave velocity and maximum particle velocity.

selected by

Given

Particle velocity


text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

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On the approximation of the boundary layers for the controllability problem of nonlinear singularly perturbed systems. (English) Zbl 1250.93034
Summary: A new systematic approach to the construction of approximate solutions to a class of nonlinear singularly perturbed feedback control systems using the boundary layer functions especially with regard to the possible occurrence of the boundary layers is proposed. For example, problems with feedback control, such as the steady-states of the thermostats, where the controllers add or remove heat, depending upon the temperature registered in another place of the heated bar, can be interpreted with a second-order ordinary differential equation subject to a nonlocal three-point boundary condition. The $O\left(ϵ\right)$ accurate approximation of behavior of these nonlinear systems in terms of the exponentially small boundary layer functions is given. At the end of this paper, we formulate the unsolved controllability problem for nonlinear systems.
##### MSC:
 93B05 Controllability 93C10 Nonlinear control systems 93C73 Perturbations in control systems 93B52 Feedback control
##### References:
 [1] Munch, A.; Zuazua, E.: Numerical approximation of null controls for the heat equation: ill-posedness and remedies, Inverse problems 26, No. 8 (2010) · Zbl 1203.35015 · doi:10.1088/0266-5611/26/8/085018 [2] Castro, C.; Zuazua, E.: Unique continuation and control for the heat equation from an oscillating lower dimensional manifold, SIAM journal on control and optimization 43, No. 4, 1400-1434 (2004) · Zbl 1101.93010 · doi:10.1137/S0363012903430317 [3] Micu, S.; Zuazua, E.: Regularity issues for the null-controllability of the linear 1-d heat equation, Systems control letters 60, No. 6, 406-413 (2011) [4] Alvarez, J. J.; Gadella, M.; Glasser, L. M.; Lara, L. P.; Nieto, L. M.: One dimensional systems with singular perturbations, Journal of physics: conference series 284, 012009 (2011) [5] Nguyen, T.; Gajic, Z.: Finite horizon optimal control of singularly perturbed linear systems: a differential Lyapunov equation approach, IEEE transactions on automatic control 55, 2148-2152 (2010) [6] Kokotovic, P.; Khali, H. K.; O’reilly, J.: Singular perturbation methods in control, analysis and design, (1986) · Zbl 0646.93001 [7] Artstein, Z.: Stability in the presence of singular perturbations, Nonlinear analysis TMA 34, No. 6, 817-827 (1998) · Zbl 0948.34029 · doi:10.1016/S0362-546X(97)00574-9 [8] Artstein, Z.: Singularly perturbed ordinary differential equations with nonautonomous fast dynamics, Journal of dynamics and differential equations 11, 297-318 (1999) · Zbl 0937.34044 · doi:10.1023/A:1021981430215 [9] Artstein, Z.; Gaitsgory, V.: The value function of singularly perturbed control system, Applied mathematics and optimization 41, 425-445 (2000) · Zbl 0958.49019 · doi:10.1007/s002459911022 [10] Gaitsgory, V.: On a representation of the limit occupational measures set of a control system with applications to singularly perturbed control systems, SIAM journal on control and optimization 43, No. 1, 325-340 (2004) · Zbl 1101.49023 · doi:10.1137/S0363012903424186 [11] Gaitsgory, V.; Nguyen, M. T.: Multiscale singularly perturbed control systems: limit occupational measures sets and averaging, SIAM journal on control and optimization 41, No. 3, 954-974 (2002) · Zbl 1027.34071 · doi:10.1137/S0363012901393055 [12] Bouzaouache, H.; Braiek, N. B.: On guaranteed global exponential stability of polynomial singularly perturbed control systems, International journal of computers, communications control 1, No. 4, 21-34 (2006) [13] Bouzaouache, H.; Ennaceur, B. H. B.; Benrejeb, M.: Reduced optimal control of nonlinear singularly perturbed systems, Systems analysis modelling simulation 43, No. 1, 75-87 (2003) · Zbl 1057.93034 · doi:10.1080/0232929031000116353 [14] Christofides, P. D.; Teel, A. R.: Singular perturbations and input-to-state stability, Institute of electrical and electronics engineers, transactions on automatic control 41, No. 11 (1996) · Zbl 0864.93086 · doi:10.1109/9.544001 [15] Mei, P.; Cai, Ch.; Zou, Y.: A generalized KYP lemma-based approach for H$\infty$ control of singularly perturbed systems, Circuits, systems, and signal processing (2009) [16] Meng, B.; Jing, Y-W.: Robust semiglobally practical stabilization for nonlinear singularly perturbed systems, Nonlinear analysis 70, 2691-2699 (2009) · Zbl 1161.93022 · doi:10.1016/j.na.2008.03.056 [17] I. Boglaev, Robust monotone iterates for nonlinear singularly perturbed boundary value problems, Boundary Value Problems Volume 2009, Article ID 320606, doi:10.1155/2009/320606. · Zbl 1177.65107 · doi:10.1155/2009/320606 [18] Jankowski, T.: Positive solutions of three-point boundary value problems for second order impulsive differential equations with advanced arguments, Applied mathematics and computation 197, 179-189 (2008) · Zbl 1145.34355 · doi:10.1016/j.amc.2007.07.081 [19] Khan, R. A.: Positive solutions of four-point singular boundary value problems, Applied mathematics and computation 201, 762-773 (2008) · Zbl 1152.34016 · doi:10.1016/j.amc.2008.01.014 [20] Khan, R. A.; Webb, J. R. L.: Existence of at least three solutions of a second-order three-point boundary value problem, Nonlinear analysis 64, 1356-1366 (2006) · Zbl 1101.34005 · doi:10.1016/j.na.2005.06.040 [21] Xu, X.: Positive solutions for singular m-point boundary value problems with positive parameter, Journal of mathematical analysis and applications 291, 352-367 (2004) · Zbl 1047.34016 · doi:10.1016/j.jmaa.2003.11.009 [22] Lin, X.; Liu, W.: Singular perturbation of a second-order three-point boundary value problem for nonlinear systems, Journal of the Shanghai university (English edition) 13, No. 1, 16-19 (2009) · Zbl 1212.34027 · doi:10.1007/s11741-009-0104-3 [23] Chang, K. W.; Howes, F. A.: Nonlinear singular perturbation phenomena: theory and applications, (1984) [24] De Coster, C.; Habets, P.: Two-point boundary value problems: lower and upper solutions, Mathematics in science and engineering 205 (2006) [25] Vrabel, R.: Nonlocal four-point boundary value problem for the singularly perturbed semilinear differential equations, Boundary value problems 2011, 9 (2011) · Zbl 1209.34018 · doi:10.1155/2011/570493 [26] Miller, J. J. H.; Oriordan, E.; Shishkin, G. I.: Fitted numerical methods for singular perturbation problems: error estimates in the maximum norm for linear problems in one and two dimensions, (1996) [27] Riordan, E. O’; Quinn, J.: Parameter-uniform numerical methods for some linear and nonlinear singularly perturbed convection diffusion boundary turning point problems, BIT numerical mathematics 51, 317-337 (2011) · Zbl 1229.65130 · doi:10.1007/s10543-010-0290-4 [28] Shishkin, G. I.: Grid approximation of singularly perturbed parabolic convection–diffusion equations subject to a piecewise smooth initial condition, Computational mathematics and mathematical physics 46, No. 1, 49-72 (2006) · Zbl 1210.35148 · doi:http://www.maik.rssi.ru./abstract/commat/6/commat1_6p49abs.htm


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## Chemistry (4th Edition)

(a) X = $^{23}_{11}Na$ (b) X = $^1_1H$ (c) X = $^1_0n$ (d) X = $^{56}_{26}Fe$ (e) X = $^{0}_{-1}\beta$
1. Find the atomic number of X: (a) 12 + 1 = 2 + x x = 11 X = Na (b) 27 + 1 = 27 + x x = 1 X = H (c) 92 + 0 = 36 + 56 + 3x 3x = 0 x = 0 X = n (d) 24 + 2 = 0 + x x = 26 X = Fe (e) 8 = 9 + x x = -1 X = $\beta$ 2. Find the atomic mass of X: (a) 26 + 1 = 4 + x x = 23 (b) 59 + 2 = 60 + x x = 1 (c) 235 + 1 = 94 + 139 + 3x 3x =3 x = 1 (d) 53 + 4 = 1 + x x = 56 (e) 20 = 20 + x x = 0


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# Is there any graphical methods by which we can directly notice whether the $\{f_k\}$ converges uniformly?

Thanks to books, many pdf files on google and this web-site, I understood somewhat about uniform and pointwise convergence. This question may be the last question about this part.

Now, I can check whether the sequence of functions $f_k$ does not converge, converges pointwise, or converges uniformly by using the definitions.

However, I cannot check it, only depending on graph.

Let me give you some examples.

$$f_n(x)=x^n, ~~~~~ \mbox{for all}~~x \in [0, 1)$$

$\lim_{n\to\infty}f_n(x)=0$ because $|x|\lt1$. Hence, $\{f_n\}$ converges pointwise, whose pointwise-limit function $f$ is the zero function on the domain $D=[0, 1)$.

Now, I am checking whether the sequence $\{f_n\}$ converges uniformly or not.

for $\displaystyle\varepsilon=\frac{1}{3}$, $~~\nexists N\in\mathbb{R}$ such that $\displaystyle\left|f_n(1-\frac{1}{n})-f(1-\frac1n)\right|=\left(1-\frac1n\right)^n\lt\varepsilon~~~$ for every $n\gt{N}$ and for every $x\in{D}$.

The reason why $N$ does not exist is that even if $n$ approaches to $\infty$, $\displaystyle\left(1-\frac1n\right)^n=\frac{1}{e}\gt\varepsilon=\frac{1}{3}$.

Therefore, the sequence of functions $\{f_n\}$ converges pointwise but doesn't uniformly.

I tried to draw the functions by using Matlab.

However, I failed to find intuitively that it converges only pointwise only by seeing graph. Is there any methods to know $\{f_n\}$ converges uniformly or pointwise with only graph?

• Just look at the picture-it doesn't matter how big you make $n$, for some $x$es, $f_n(x)$ is still far from $0$! – Kevin Arlin Mar 14 '16 at 0:58
• Try comparing the plots you have to the plots of $g_n(x):=\frac{\sin(x)}{n}$, and see if you can see a difference. – snulty Mar 14 '16 at 1:18
• Do the pictures here and here help? I searched in Google Images for uniformly convergent site:math.stackexchange.com and sup norm site:math.stackexchange.com. Maybe you will find some other relevant posts among the results. – Martin Sleziak May 6 '16 at 4:07

Examples of sequence of functions that converge uniformly

In this case:

1. Limit function must be continuous

2. sequence of functions eventually converges to the function (lies in an $\epsilon$-tube) as $N \to \infty$

Examples of sequence of functions that converge pointwise but not uniformly

Note (b) has an error it should be $f_n(x) = \exp(-nx)$

In this case:

1. The limit function do not have to be continuous

2. The sequence of functions do not have to lie fully in an $\epsilon$-tube of the limit function as $N \to \infty$

The latter is because in the definition of pointwise convergence:

$\forall \epsilon > 0, \forall x \in [a,b], \exists N\in \mathbb{N}$ s.t. $\forall n\in \mathbb{N}, n \geq N, \|f_n - f\| < \epsilon$

We can pick our $N$ large enough to make the latter condition $\|f_n - f\| < \epsilon$ true for a particular $x$, but not for all $x$ at once.

• Many examples are helpful. Thank you. – Danny_Kim Mar 14 '16 at 5:01

According to the definition, sequence $f_n \to 0$ uniformly on interval $I$ if for every $\epsilon > 0$ there is $N$ such that $|f_n(x)| < \epsilon$ for all $n > N$ and all $x \in I$.

Graphically, what this means is that if you take an arbitrarily narrow horizontal band around the $x$ axis, ($\{(x,y): |y| < \epsilon\}$), for large enough $n$ the graph of $f_n$ will lie in this band.

In your picture, the graph of each function $f_n$ will poke up out of the band for $x$ near $1$. Therefore the sequence does not converge uniformly to $0$.

• Or, in otherwords, do the $\epsilon$-neighborhoods of each point converge to an $\epsilon$-neighborhood of the corresponding limit point? Notice how that fails to happen for $x=1$ in the given example. – Justin Benfield Mar 14 '16 at 1:03


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# Name/definition of $\int \log F(x) \cdot g(x)dx$?

We know that:

$$-\int \log f(x) \cdot g(x)dx,$$ where $$f$$ and $$g$$ are density functions, is known as the cross entropy. Does $$-\int \log F(x) \cdot g(x)dx,$$ where $$F$$ is the cumulative distribution function of $$f$$, have a name?

• Could you explain in what problem or analysis this expression arises? – whuber May 8 at 14:47


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# The inclusion $E(K)/2E(K)\hookrightarrow K^*/(K^*)^2$ for Weak Mordell-Weil theorem

I'm reading a master's dissertation where the author proves the weak Mordell-Weil theorem for $\mathbb{Q}$ (i.e.: for any elliptic curve $E:y^2=f(x)=x^3+Ax+B$ over $\mathbb {Q}$, the group $E(\mathbb{Q})/2E(\mathbb{Q})$ is finite). He begins the proof by saying this:

Let $K:=\mathbb{Q}(\theta)=\mathbb{Q}[x]/f(x)$ and $H$ the subgroup of $K^*$ consisting of elements $a\in K^*$ such that $N_{K/\mathbb{Q}}(a)$ is a square in $\mathbb{Q}$. Define the map: \begin{align*} \mu:E(\mathbb{Q}) &\to \overline{H}<K^*/(K^*)^2\\ (x:y:1) &\mapsto \overline{x-\theta}\\ O &\mapsto \overline{1} \end{align*}

He immediatelly goes on to prove that $\mu$ is a group homomorphism among other things.t

But why is $N_{K/\mathbb{Q}}(x-\theta)$ necessarily is a square in $\mathbb{Q}$ in the first place, and why is $im(\mu)=\overline{H}$?

• For $x \in \mathbb{Q}, N(x-\theta) = \prod_{i=1}^3 \sigma_i(x-\theta))=\prod_{i=1}^3 (x-\sigma_i(\theta)) = f(x)$. – reuns Nov 19 '17 at 19:35
• I know $\theta$ is a linear combination of the roots of $f$, but why are $\sigma_i(\theta)$ exactly the roots of $f$? – rmdmc89 Nov 19 '17 at 23:26
• What is your definition of the field norm ? For $x \in \mathbb{Q}$, $N_{K/\mathbb{Q}}(x-a) = \det(xI-A)$ where $A$ is the matrix representing the multiplication by $a$ in the vector space $K$. And $\det(xI-A)$ doesn't depend on the basis we choose for $K$, in particular it stays the same when using the $n=[K:\mathbb{Q}]$ embeddings $\sigma_i : K \to \mathbb{C}$ to embed $K$ in $\mathbb{C}^n$, where $A$ becomes diagonal so that $\det(xI-A)= \prod_{i=1}^n (x-\sigma_i(a))$ – reuns Nov 19 '17 at 23:31
• If $f \in \mathbb{Q}[x]$ is irreducible with roots $\theta_1,\ldots,\theta_m$ then $\mathbb{Q}(\theta_1)\cong \mathbb{Q}(\theta_j) \cong \mathbb{Q}[x]/(f(x))$. I don't think you understand what I said about $N_{K/\mathbb{Q}}$. – reuns Nov 27 '17 at 14:45
• It is the splitting field when $disc(f) =-4A^3-27 B^2$ is a square – reuns Nov 27 '17 at 16:18

$$N_{K\Bbb Q}(x-\theta)=x^3+Ax+B=y^2$$ so is a square in $\Bbb Q$. I'm not sure that the image of $\mu$ is all of $\overline H$.
• $K^*/(K^*)^2$ is the set of quadratic extensions of $\mathbb{Q}(\theta)$, is $\overline{H}$ much smaller ? – reuns Nov 19 '17 at 19:43
• Maybe the image isn't exactly $\overline{H}$. But I've just realized that's ok, because for this proof we only need the fact that the image is a subgroup of $K^*/(K^*)^2$ – rmdmc89 Nov 27 '17 at 14:03


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Is most mass from KE or PE?

I know most invariant mass is from the interaction of quarks with the gluon field, but is that kinetic energy or potential energy, or does this question not make sense? Am a college freshman but I want to understand this. Thank you!

Is the question sensible?

Let's start by examining the question

"Does it even make sense to ask what fraction of the mass comes from potential and kinetic energies?"

in general before we try to work out the answer for the strong interaction in particular

Long-range forces

For gravity, electrostatics, and the nuclear force (AKA "strong nuclear force" or "residual strong force") the force and potential vanishes at large distance. The result is that systems held together by these forces (i.e. bound) have negative potential energy.

But of course the kinetic energy is positive.

So the mass defect of these systems is caused by the potential energy but partially relieved by kinetic energy.

The question has—at best—an unsatisfying answer for long-range forces.

The strong force

The real strong force (the one that holds baryons together) is different. It exhibits both confinement (which requires that the potential grow rapidly at large distances) and asymptotic freedom (which means the potential is constant at very short distances). Moreover it is experimentally determined to have a potential roughly of the form $$U_\text{strong}(r) \approx kr \;. \tag{1}$$ As a result in this system both the kinetic and potential energy are positive.

The question makes sense for the strong interaction.

A tool to find the answer

The Virial theorem is a relationship in classical mechanics between the mean kinetic energy of a system and a particular average over the inner product of the forces on particles and the position of the particles (which is called the "virial of system") as long as the system meets certain requirements (many bound systems qualify).

If the force laws at work between the particles are central and proportional to a power of distance between the particles (that is of the form $$U = kr^n$$), then the theorem can be extended to give a relationship between the average kinetic energy of the system and the average potential energy of the system.

Using the force law in (1) to get the relationship for the strong force we find1 $$\langle T \rangle = \frac{1}{2} \langle U \rangle \;,$$

and if we write the dynamic energy $$E_d$$ of the system2 we get \begin{align} E_d &= \langle T \rangle + \langle U \rangle \\ &= \frac{3}{2} \langle U \rangle \;. \end{align}

So, long story short one-third of the dynamic energy is kinetic and two-thirds is potential.

1 Writing $$T$$ for kinetic energy and $$U$$ for potential energy.

2 Which is to say the part of the mass that comes from the behavior of the partons rather than the mass of a valence quarks.

• 1) If potential energy can be negative, can something have negative mass? Is this possibly how virtual particles have negative energy? – Alexander Wu Jun 2 '18 at 2:06
• 2) In classical mechanics you can set an arbitrary zero value for potential energy. But that would mean mass is arbitrary. Does that no longer work in relativity? – Alexander Wu Jun 2 '18 at 2:08
• The mass of a system made from multiple parts can be less than the total mass of all the parts. In the context of nuclear physics the difference is called the "mass defect", but this mechanism neither allows negative absolute masses nor explains the virtual particles of quantum field theory (which should be treated as "real" only with care and within well defined contexts physics.stackexchange.com/q/162845/520 physics.stackexchange.com/q/185110/520 ). – dmckee Jun 2 '18 at 2:45
• And in relativity you can still take potential energies to have arbitrary zeros for many purposes, but there is a correct choice for computing masses. – dmckee Jun 2 '18 at 2:46