text:
During the early 19th century, the concept of perceptible or free caloric began to be referred to as “free heat” or heat set free. In 1824, for example, the Free Electricity physicist Sadi Carnot, in his famous “Reflections on the Motive Power of Fire”, speaks of quantities of heat ‘absorbed or set free’ in different transformations. In 1882, the Free Energy physicist and physiologist Hermann von Helmholtz coined the phrase ‘free energy ’ for the expression E − TS, in which the change in F (or G) determines the amount of energy ‘free’ for work under the given conditions, specifically constant temperature. [Free Electricity]:Free Power.
The song’s original score designates the duet partners as “wolf” and “mouse, ” and genders are unspecified. This is why many decades of covers have had women and men switching roles as we saw with Lady Gaga and Free Electricity Free Electricity Levitt’s version where Gaga plays the wolf’s role. Free Energy, even Miss Piggy of the Muppets played the wolf as she pursued ballet dancer Free Energy NureyeFree Power
You might also see this reaction written without the subscripts specifying that the thermodynamic values are for the system (not the surroundings or the universe), but it is still understood that the values for \Delta \text HΔH and \Delta \text SΔS are for the system of interest. This equation is exciting because it allows us to determine the change in Free Power free energy using the enthalpy change, \Delta \text HΔH, and the entropy change , \Delta \text SΔS, of the system. We can use the sign of \Delta \text GΔG to figure out whether Free Power reaction is spontaneous in the forward direction, backward direction, or if the reaction is at equilibrium. Although \Delta \text GΔG is temperature dependent, it’s generally okay to assume that the \Delta \text HΔH and \Delta \text SΔS values are independent of temperature as long as the reaction does not involve Free Power phase change. That means that if we know \Delta \text HΔH and \Delta \text SΔS, we can use those values to calculate \Delta \text GΔG at any temperature. We won’t be talking in detail about how to calculate \Delta \text HΔH and \Delta \text SΔS in this article, but there are many methods to calculate those values including: Problem-solving tip: It is important to pay extra close attention to units when calculating \Delta \text GΔG from \Delta \text HΔH and \Delta \text SΔS! Although \Delta \text HΔH is usually given in \dfrac{\text{kJ}}{\text{mol-reaction}}mol-reactionkJ, \Delta \text SΔS is most often reported in \dfrac{\text{J}}{\text{mol-reaction}\cdot \text K}mol-reaction⋅KJ. The difference is Free Power factor of 10001000!! Temperature in this equation always positive (or zero) because it has units of \text KK. Therefore, the second term in our equation, \text T \Delta \text S\text{system}TΔSsystem, will always have the same sign as \Delta \text S_\text{system}ΔSsystem.

# Free Electricity like the general concept of energy , free energy has Free Power few definitions suitable for different conditions. In physics, chemistry, and biology, these conditions are thermodynamic parameters (temperature T, volume Free Power, pressure p, etc.). Scientists have come up with several ways to define free energy. The mathematical expression of Helmholtz free energy is.

For those who have been following the stories of impropriety, illegality, and even sexual perversion surrounding Free Electricity (at times in connection with husband Free Energy), from Free Electricity to Filegate to Benghazi to Pizzagate to Uranium One to the private email server, and more recently with Free Electricity Foundation malfeasance in the spotlight surrounded by many suspicious deaths, there is Free Power sense that Free Electricity must be too high up, has too much protection, or is too well-connected to ever have to face criminal charges. Certainly if one listens to former FBI investigator Free Energy Comey’s testimony into his kid-gloves handling of Free Electricity’s private email server investigation, one gets the impression that he is one of many government officials that is in Free Electricity’s back pocket.

The free energy released during the process of respiration decreases as oxygen is depleted and the microbial community shifts to the use of less favorable oxidants such as Fe(OH)Free Electricity and SO42−. Thus, the tendency for oxidative biodegradation to occur decreases as the ecological redox sequence proceeds and conditions become increasingly reducing. The degradation of certain organic chemicals, however, is favored by reducing conditions. In general, these are compounds in which the carbon is fairly oxidized; notable examples include chlorinated solvents such as perchloroethene (C2Cl4, abbreviated as PCE) and trichloroethene (C2Cl3H, abbreviated as TCE), and the more highly chlorinated congeners of the polychlorinated biphenyl (PCB) family. (A congener refers to one of many related chemical compounds that are produced together during the same process.
The internet is the only reason large corps. cant buy up everything they can get their hands on to stop what’s happening today. @Free Power E. Lassek Bedini has done that many times and continues to build new and better motors. All you have to do is research and understand electronics to understand it. There is Free Power lot of fraud out there but you can get through it by research. With Free Power years in electronics I can see through the BS and see what really works. Build the SG and see for yourself the working model.. A audio transformer Free Power:Free Power ratio has enough windings. Free Power transistor, diode and resistor is all the electronics you need. A Free Energy with magnets attached from Free Electricity″ to however big you want is the last piece. What? Maybe Free Electricity pieces all together? Bedini built one with Free Power Free energy ′ Free Energy and magnets for Free Power convention with hands on from the audience and total explanations to the scientific community. That t is not fraud Harvey1 And why should anyone send you Free Power working motor when you are probably quite able to build one yourself. Or maybe not? Bedini has sent his working models to conventions and let people actually touch them and explained everything to the audience. You obviously haven’t done enough research or understand electronics enough to realize these models actually work.. The SC motor generator is easily duplicated. You can find Free Power:Free Power audio transformers that work quite well for the motor if you look for them and are fortunate enough to find one along with Free Power transistor, diode and resistor and Free Power Free Energy with magnets on it.. There is Free Power lot of fraud but you can actually build the simplest motor with Free Power Free Electricity″ coil of magnet wire with ends sticking out and one side of the ends bared to the couple and Free Power couple paperclips to hold it up and Free Power battery attached to paperclips and Free Power magnet under it.
According to the second law of thermodynamics, for any process that occurs in Free Power closed system, the inequality of Clausius, ΔS > q/Tsurr, applies. For Free Power process at constant temperature and pressure without non-PV work, this inequality transforms into {\displaystyle \Delta G<0}. Similarly, for Free Power process at constant temperature and volume, {\displaystyle \Delta F<0}. Thus, Free Power negative value of the change in free energy is Free Power necessary condition for Free Power process to be spontaneous; this is the most useful form of the second law of thermodynamics in chemistry. In chemical equilibrium at constant T and p without electrical work, dG = 0. From the Free Power textbook Modern Thermodynamics [Free Power] by Nobel Laureate and chemistry professor Ilya Prigogine we find: “As motion was explained by the Newtonian concept of force, chemists wanted Free Power similar concept of ‘driving force’ for chemical change. Why do chemical reactions occur, and why do they stop at certain points? Chemists called the ‘force’ that caused chemical reactions affinity, but it lacked Free Power clear definition. ”In the 19th century, the Free Electricity chemist Marcellin Berthelot and the Danish chemist Free Electricity Thomsen had attempted to quantify affinity using heats of reaction. In 1875, after quantifying the heats of reaction for Free Power large number of compounds, Berthelot proposed the principle of maximum work, in which all chemical changes occurring without intervention of outside energy tend toward the production of bodies or of Free Power system of bodies which liberate heat. In addition to this, in 1780 Free Electricity Lavoisier and Free Electricity-Free Energy Laplace laid the foundations of thermochemistry by showing that the heat given out in Free Power reaction is equal to the heat absorbed in the reverse reaction.
Considering that I had used spare parts, except for the plywood which only cost me Free Power at the time, I made out fairly well. Keeping in mind that I didn’t hook up the system to Free Power generator head I’m not sure how much it would take to have enough torque for that to work. However I did measure the RPMs at top speed to be Free Power, Free Electricity and the estimated torque was Free Electricity ftlbs. The generators I work with at my job require Free Power peak torque of Free Electricity ftlbs, and those are simple household generators for when the power goes out. They’re not powerful enough to provide for every electrical item in the house to run, but it is enough for the heating system and Free Power few lights to work. Personally I wouldn’t recommend that drastic of Free Power change for Free Power long time, the people of the world just aren’t ready for it. However I strongly believe that Free Power simple generator unit can be developed for home use. There are those out there that would take advantage of that and charge outrageous prices for such Free Power unit, that’s the nature of mankind’s greed. To Nittolo and Free Electricity ; You guys are absolutely hilarious. I have never laughed so hard reading Free Power serious set of postings. You should seriously write some of this down and send it to Hollywood. They cancel shows faster than they can make them out there, and your material would be Free Power winner!
The historically earlier Helmholtz free energy is defined as A = U − TS. Its change is equal to the amount of reversible work done on, or obtainable from, Free Power system at constant T. Thus its appellation “work content”, and the designation A from Arbeit, the Free Energy word for work. Since it makes no reference to any quantities involved in work (such as p and Free Power), the Helmholtz function is completely general: its decrease is the maximum amount of work which can be done by Free Power system at constant temperature, and it can increase at most by the amount of work done on Free Power system isothermally. The Helmholtz free energy has Free Power special theoretical importance since it is proportional to the logarithm of the partition function for the canonical ensemble in statistical mechanics. (Hence its utility to physicists; and to gas-phase chemists and engineers, who do not want to ignore p dV work.)


text:
## Introduction

We'll start with some arithmetic. We can define Nat to be the type of natural numbers, and implement an addition function. By evaluating this function in GHCi we can compute additions of natural numbers. If we try to add, say, a string and a float, we quite naturally get a type error.

data Nat = Zero | Succ Nat

add :: Nat -> Nat -> Nat

*Main> add (Succ Zero) (Succ Zero)
Succ (Succ Zero)

Couldn't match expected type Nat with actual type [Char]
In the first argument of add, namely "Hello World!"
In the expression: add "Hello World!" 4.5
In an equation for 'it': it = add "Hello World!" 4.5

Moving on, we try something else.

{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FunctionalDependencies #-}

data Z
data S n

class Add a b c | a b -> c where add :: a -> b -> c
instance (Add a b c) => Add (S a) y (S c)

*Main> :type add (undefined :: S Z) (undefined :: S Z)
add (undefined :: S Z) (undefined :: S Z) :: S (S Z)

*Main> :type add (undefined :: String) (undefined :: Float)
add (undefined :: String) (undefined :: Float) :: Add String Float c => c

What's going on here? It looks like we've computed the sum of two numbers. But we haven't defined any functions, or even values, just types and instances of type classes.

Let's take a closer look. We first define two types, Z and S n with no constructors, so the only value they have is the bottom value undefined. Then we declare a type class Add a b c that implements a single method add :: a- > b -> c. There are two things to note here:

1. Unlike the type class Eq x, Add a b c has three parameters. This isn't supported in base Haskell, so we have to add a language extension. Now, just as we can think of single-paramater type classes as sets of types, we can think of n-paramater type classes as n-ary relations on types.

2. We've given Add a b c the functional dependency a b -> c. This means that, given any two types x and y, Haskell will only allow an instance Add x y z for one type z. In other words, the 3-ary relation on types Add is actually a (partial) function Type -> Type -> Type. This means that Haskell is ably to fully infer the type of the first expression. Type checking the second expression doesn't give a type error, since it is possible that there is an instance of Add String Float c defined somewhere in another module, but because it can't see any such instance, it isn't able to fully infer the type.

In this way, we have translated our computation to the type level. Can we do this with more complicated computations?

## FizzBuzz

The FizzBuzz test is a much-blogged-about interview question. It goes like this:

Write a program that prints the numbers from 1 to 100. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz”.

fizzes :: Integer -> Bool
fizzes n = n mod 3 == 0

buzzes :: Integer -> Bool
buzzes n = n mod 5 == 0

decide :: Int -> String
decide n | fizzes n && buzzes n = "FizzBuzz"
| fizzes n = "Fizz"
| buzzes n = "Buzz"
| otherwise = show n

fizzbuzz :: Int -> [String]
fizzbuzz n = map decide [1..n]

We get the answer by evaluating fizzbuzz 100

*Main> fizzbuzz 100
["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz","16","17","Fizz","19","Buzz","Fizz","22","23","Fizz","Buzz","26","Fizz","28","29","FizzBuzz","31","32","Fizz","34","Buzz","Fizz","37","38","Fizz","Buzz","41","Fizz","43","44","FizzBuzz","46","47","Fizz","49","Buzz","Fizz","52","53","Fizz","Buzz","56","Fizz","58","59","FizzBuzz","61","62","Fizz","64","Buzz","Fizz","67","68","Fizz","Buzz","71","Fizz","73","74","FizzBuzz","76","77","Fizz","79","Buzz","Fizz","82","83","Fizz","Buzz","86","Fizz","88","89","FizzBuzz","91","92","Fizz","94","Buzz","Fizz","97","98","Fizz","Buzz"]

It's not a particularly nice solution, but it lends itself nicely to translation to the type level. Let's jump right in!

It looks like we're going to need:

• Integers
• Lists
• Booleans
• Strings

Although we can actually get away with only natural numbers, and just three strings... (The latter because type level lists as defined above are heterogeneous, so we can put "numbes" and "strings" in the same list.)

{-# OPTIONS_GHC -fno-warn-missing-methods #-}
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FunctionalDependencies #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE UndecidableInstances #-}

data Z
data S n

data Nil
data Cons x xs

data True
data False

data Fizz
data Buzz
data FizzBuzz

Now the functions we have to translate are fizzes, buzzes, &&, decide, map, list comprehensions, and fizzbuzz. The first few of these are straightforward:

class Fizzes n b | n -> b
instance Fizzes Z True
instance Fizzes (S Z) False
instance Fizzes (S (S Z)) False
instance (Fizzes n b) => Fizzes (S (S (S n))) b

class Buzzes n b | n -> b
instance Buzzes Z True
instance Buzzes (S Z) False
instance Buzzes (S (S Z)) False
instance Buzzes (S (S (S Z))) False
instance Buzzes (S (S (S (S Z)))) False
instance (Buzzes n b) => Buzzes (S (S (S (S (S n))))) b

class And a b c | a b -> c
instance And True True True
instance And True False False
instance And False True False
instance And False False False

We do decide in two steps:

class Decide1 n f b r | n f b -> r
instance Decide1 n True True FizzBuzz
instance Decide1 n True False Fizz
instance Decide1 n False True Buzz
instance Decide1 n False False n

class Decide n r | n -> r
instance (Fizzes n b, Buzzes n c, Decide1 n b c r) => Decide n r

The case of map :: (a -> b) -> [a] -> [b] seems a little trickier, but thankfully we don't really need map, just map decide, which makes things simpler:

class MapDecide xs ys | xs -> ys
instance MapDecide Nil Nil
instance (Decide n r, MapDecide ns rs) => MapDecide (Cons n ns) (Cons r rs)

Similarly, we don't need full list comprehensions, just a function first :: Integer -> [Integer] that generates the first n natural numbers (first n = [1..n]). We'll need a helper function that adds elements to the back of a list for this.

class Snoc x xs ys | x xs -> ys
instance Snoc x Nil (Cons x Nil)
instance (Snoc y xs ys) => Snoc y (Cons x xs) (Cons x ys)

class First n xs | n -> xs
instance First Z (Cons Z Nil)
instance (First n xs, Snoc (S n) xs ys) => First (S n) ys

Finally we put it all together:

class Answer n rs | n -> rs where answer :: n -> rs
instance (From n ns, MapDecide ns rs) => Answer n rs

Let's check our work in GHCi. Starting with some small numbers, everything looks good—or at least, correct.

*Main> :type answer (undefined::Z)
answer (undefined::Z) :: Cons FizzBuzz Nil
*Main> :type answer (undefined::S (S (S (S (S Z)))))
answer (undefined::S (S (S (S (S Z)))))
:: Cons
FizzBuzz
(Cons
(S Z)
(Cons
(S (S Z)) (Cons Fizz (Cons (S (S (S (S Z)))) (Cons Buzz Nil)))))

Going up to 100 takes a little longer - about 34 seconds on my machine. Trying n=150 takes even more time (3:26 minutes), and n=200 took around 15 minutes. This suggests that our algorithm has exponential time complexity. Although Haskell's type system is very powerful, we might be better off performing computations in Haskell itself.

format: plain


text:
## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.1 Sequences - Preliminary Questions: 4

#### Answer

a) Defined explicitly, not recursively. b) Defined recursively

#### Work Step by Step

a) $a_n$ can be computed directly, since it depends on $n$ only and not on preceding terms. b) $b_n$ is computed in terms of the preceding term $b_{n-1}$, hence the sequence ${{b_n}}$ is defined recursively.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# 9.6.1: Kinetic Techniques versus Equilibrium Techniques

In an equilibrium method the analytical signal is determined by an equilibrium reaction that involves the analyte or by a steady-state process that maintains the analyte’s concentration. When we determine the concentration of iron in water by measuring the absorbance of the orange-red $$\text{Fe(phen)}_3^{2+}$$ complex, the signal depends upon the concentration of $$\text{Fe(phen)}_3^{2+}$$, which, in turn, is determined by the complex’s formation constant. In the flame atomic absorption determination of Cu and Zn in tissue samples, the concentration of each metal in the flame remains constant because each step in the process of atomizing the sample is in a steady-state. In a kinetic method the analytical signal is determined by the rate of a reaction that involves the analyte or by a nonsteady-state process. As a result, the analyte’s concentration changes during the time in which we monitor the signal.

In many cases we can choose to complete an analysis using either an equilibrium method or a kinetic method by changing when we measure the analytical signal. For example, one method for determining the concentration of nitrite, $$\text{NO}_2^-$$, in groundwater utilizes the two-step diazotization re-action shown in Figure 13.1.1 [Method 4500-NO2 B in Standard Methods for the Analysis of Waters and Wastewaters, American Public Health Association: Washington, DC, 20th Ed., 1998]. The final product, which is a reddish-purple azo dye, absorbs visible light at a wavelength of 543 nm. Because neither reaction in Figure 13.1.1 is rapid, the absorbance—which is directly proportional to the concentration of nitrite—is measured 10 min after we add the last reagent, a lapse of time that ensures that the concentration of the azo dyes reaches the steady-state value required of an equilibrium method.

We can use the same set of reactions as the basis for a kinetic method if we measure the solution’s absorbance during this 10-min development period, obtaining information about the reaction’s rate. If the measured rate is a function of the concentration of $$\text{NO}_2^-$$, then we can use the rate to determine its concentration in the sample [Karayannis, M. I.; Piperaki, E. A.; Maniadaki, M. M. Anal. Lett. 1986, 19, 13–23].

There are many potential advantages to a kinetic method of analysis, perhaps the most important of which is the ability to use chemical reactions and systems that are slow to reach equilibrium. In this chapter we examine three techniques that rely on measurements made while the analytical system is under kinetic control: chemical kinetic techniques, in which we measure the rate of a chemical reaction; radiochemical techniques, in which we measure the decay of a radioactive element; and flow injection analysis, in which we inject the analyte into a continuously flowing carrier stream, where its mixes with and reacts with reagents in the stream under conditions controlled by the kinetic processes of convection and diffusion.

9.6.1: Kinetic Techniques versus Equilibrium Techniques is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Harvey.


text:
# Definition:Dodecahedron/Regular

## Definition

A regular dodecahedron is a dodecahedron whose $12$ faces are all congruent regular pentagons.

It has:

## Historical Note

In The Elements, this object is referred to just as a dodecahedron.

In the words of Euclid:

A dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons.

According to the Pythagorean tradition, the regular dodecahedron was the symbol associated with the sphere of the fixed stars, and later with the quintessence of the medieval alchemists.

## Linguistic Note

The word dodecahedron derives from the Classical Greek δωδεκάεδρόν:

dodeca (δωδεκά), meaning twelve
hedron (a form of ἕδρα), meaning base or seat.

The technically correct plural of dodecahedron is dodecahedra, but the word dodecahedrons can often be found.


text:
Benjamin - 8 months ago 40
SQL Question

# Order a table according to a parent

I have used a recursive CTE to obtain the table below.

CREATE TABLE PositionTree(
OID int,
Position int,
Parent int,
Name varchar(50)
)

INSERT INTO PositionTree
(OID, Position, Parent, Name)
VALUES
(23, 21, 2,  'Supervisor'),
(34, 18, 18, 'Supervisor+3'),
(37, 1, 18, 'Supervisor+2'),
(89, 25, 21, 'Target Employee'),
(1129, 2, 1, 'Supervisor+1')

I would like to generate an index such that I can arrange the table to start with
'Target Employee'
and ascend up through
'Supervisor+3'
. You may assume that I know in advance the OID of the target employee (this will eventually go into a stored procedure, with the target OID as a parameter)

The resulting table should look similar to this:

OID  Position  Parent  Name               SortOrder
89   25        21      'Target Employee'  0
23   21        2       'Supervisor'       1
1129  2        1       'Supervisor+1'     2
37    1        18      'Supervisor+2'     3
34   18        18      'Supervisor+3'     4

Ideally, I would just pull an index out of the recursion (start at 0, and one at each step of the recursion), but I haven't found anything pointing me to how to do that. Is getting such an index from the recursion possible, or what is the best way to get this table sorted as I want it?

It would be easier to answer if you showed your recursive CTE, but you can just select 0 as SortOrder in your anchor, and in your recursive part select SortOrder + 1 as SortOrder.


text:
# Poisson's ratio

Figure 1: A cube with sides of length L of an isotropic linearly elastic material subject to tension along the x axis, with a Poisson's ratio of 0.5. The green cube is unstrained, the red is expanded in the x direction by ${\displaystyle \Delta L}$ due to tension, and contracted in the y and z directions by ${\displaystyle \Delta L'}$.
${\displaystyle \nu =-{\frac {d\varepsilon _{\mathrm {trans} }}{d\varepsilon _{\mathrm {axial} }}}=-{\frac {d\varepsilon _{\mathrm {y} }}{d\varepsilon _{\mathrm {x} }}}=-{\frac {d\varepsilon _{\mathrm {z} }}{d\varepsilon _{\mathrm {x} }}}}$


text:
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# If an object travels 100 feet in 2 seconds, what is the

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21 Oct 2010, 18:42
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If an object travels 100 feet in 2 seconds, what is the object’s approximate speed in miles per hour? (Note: 1 mile = 5280 feet)

A. 3.4
B. 3.8
C. 34
D. 38
E. 340

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22 Oct 2010, 05:02
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Conversion questions are best dealt using a one step approach.
Let me illustrate: Let's say we want to convert a speed of 100 miles/hour to feet/sec.

Speed = 100 miles/1 hour = (100 x 5280 feet)/(1 x 3600 sec)= 146.67 feet/sec

So, simply convert the units to what you want along with the conversion factor.

If an object travels 100 feet in 2 seconds, what is the object’s approximate speed in miles per hour? (Note: 1 mile = 5280 feet)

Speed = 100 feet/2 sec = (100×1/5280 miles)/(2×1/3600 hour) = 5x3600 miles/528 hr = 34 miles/hr

Note: Since 1 mile = 5280 feet, hence here to convert we say 1 foot = 1/5280 miles.
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21 Oct 2010, 19:01
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shrive555 wrote:
If an object travels 100 feet in 2 seconds, what is the object’s approximate speed in miles per hour? (Note: 1 mile = 5280 feet)

3.4
3.8
34
38
340

i always mess such conversion questions.

$$100$$ feet in 2 seconds;

$$100*\frac{3,600}{2}=50*3,600$$ feet in 1 hour;

$$\frac{50*3,600}{5,280}\approx{34}$$ miles in 1 hour.

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22 Oct 2010, 05:55
Bunuel wrote:
shrive555 wrote:
If an object travels 100 feet in 2 seconds, what is the object’s approximate speed in miles per hour? (Note: 1 mile = 5280 feet)

3.4
3.8
34
38
340

i always mess such conversion questions.

$$100$$ feet in 2 seconds;

$$100*\frac{3,600}{2}=50*3,600$$ feet in 1 hour;

$$\frac{50*3,600}{5,280}\approx{34}$$ miles in 1 hour.

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Re: If an object travels 100 feet in 2 seconds, what is the object's appro  [#permalink]

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13 May 2011, 08:46
1
1 mile = 5280 feet

=> 1 feet = 1/5280 miles

If the object ravels 100 feet in 2 sec

then it travels 100/2 * 60 * 60 feet in 1 hour (1 hr = 60 min * 60 sec)

= 3600 * 50 feet in 1 hour

= 180000 feet in 1 hr

= 180000/5280 miles in 1 hour

= 18000/528 miles/hr ~ 34 miles/hr

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Re: If an object travels 100 feet in 2 seconds, what is the object's appro  [#permalink]

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13 May 2011, 08:49
1
In 2 seconds ----100feet
In 1 second----50feet
In 3600 seconds(1hour=3600sec)---3600*5feet.

Now 5280 feet = 1 miles
so, 3600*5 feet= 3600*5/5280.
Now use approximation.. (36*5)/5280 * 100 = 180/5280 *100 or 18/5280 * 1000= 3/88 *1000.
Now 3/90= 1/30= .03
So 3/88*1000 ~ .03*1000=33.3~34.

Idea is to not multiply anything until last few steps when you can use approximation better.
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Re: If an object travels 100 feet in 2 seconds, what is the object's appro  [#permalink]

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13 May 2011, 11:32
1
speed= 100 feet/2 sec
= 50 feet /1 sec
= 50*60*60 feet / hours

Now divide the entire distance traveled into number of miles

it gives 18000/5280

~35 Hence choice c
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08 Jul 2012, 10:50
100 (feet)/2 (sec) *1 (mile)/5280 (feet)* 3600 (sec)/1 (hour) --> cancel out (feet) and (sec) and you will get

(100*3600)/(2*5280) = 34.09
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Re: If an object travels 100 feet in 2 seconds, what is the  [#permalink]

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19 Jul 2012, 04:25
1 mile = 5280 feet => 1 feet = 1/5280 mile

1 hr = 3600s => 1s = 1/3600 hr

If an object travels 100 feet in 2 seconds => speed = 100/2 feet/s = 50 feet/s = 50 *(1/5280) / (1 *(1/3600)) miles/hr
= 3600 * 50 / (5280) miles/hr = 18000/ 528 miles/hr ~ 18000/525 = 180 * 4/ 21 miles/hr = 240/7 miles/hr ~ 34miles/hr

Hope it helps!
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17 Oct 2013, 04:37
Bunuel wrote:
shrive555 wrote:
If an object travels 100 feet in 2 seconds, what is the object’s approximate speed in miles per hour? (Note: 1 mile = 5280 feet)

3.4
3.8
34
38
340

i always mess such conversion questions.

$$100$$ feet in 2 seconds;

$$100*\frac{3,600}{2}=50*3,600$$ feet in 1 hour;

$$\frac{50*3,600}{5,280}\approx{34}$$ miles in 1 hour.

Thanks! nice explanation! helped me understand.
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Joined: 02 Sep 2009
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17 Oct 2013, 04:42
KanupriyaS wrote:
Bunuel wrote:
shrive555 wrote:
If an object travels 100 feet in 2 seconds, what is the object’s approximate speed in miles per hour? (Note: 1 mile = 5280 feet)

3.4
3.8
34
38
340

i always mess such conversion questions.

$$100$$ feet in 2 seconds;

$$100*\frac{3,600}{2}=50*3,600$$ feet in 1 hour;

$$\frac{50*3,600}{5,280}\approx{34}$$ miles in 1 hour.

Thanks! nice explanation! helped me understand.

Other conversion problems to practice:
if-an-object-travels-100-feet-in-2-seconds-what-is-the-103364.html
the-rear-wheels-of-a-car-crossed-a-certain-line-0-5-second-135753.html
a-train-traveling-at-a-constant-speed-down-a-straight-track-87124.html
a-car-traveling-at-a-certain-constant-speed-takes-2-seconds-126658.html
if-an-automobile-averaged-22-5-miles-per-gallon-of-gasoline-109083.html
if-1-kilometer-is-approximately-0-6-mile-which-of-the-140731.html
measuring-standards-conversion-106516.html
if-10-millimeters-equal-1-centimeter-how-many-square-48538.html
the-speed-of-light-is-approximately-149492.html
the-rear-wheels-of-a-car-crossed-a-certain-line-0-5-second-135753.html
if-the-speed-of-x-meters-per-second-is-equivalent-to-the-127005.html
according-to-the-formula-f-9-5-c-32-if-the-temperature-160658.html
if-the-speed-of-x-meters-per-second-is-equivalent-to-the-spe-79246.html
on-a-trolley-ride-around-an-amusement-park-a-child-traveler-161594.html
the-distance-between-two-planets-is-3-04-10-6-light-years-161745.html

Hope this helps.
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Re: If an object travels 100 feet in 2 seconds, what is the  [#permalink]

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20 Sep 2014, 09:05
shrive555 wrote:
If an object travels 100 feet in 2 seconds, what is the object’s approximate speed in miles per hour? (Note: 1 mile = 5280 feet)

A. 3.4
B. 3.8
C. 34
D. 38
E. 340

My approach is as follows:
100 feet in 2 Secs.
Therefore 50 feet in 1 Sec
50 X 60 =3000 Feet in 1 Min
And 3000 X 60 = 180000 Feet in 1 Hour

Now Conversion..... 180000/5280~=180000/5000=36
The answer should be slightly less than 36.That is C.
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Re: If an object travels 100 feet in 2 seconds, what is the  [#permalink]

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26 Jul 2015, 16:53
Bunuel wrote:
shrive555 wrote:
If an object travels 100 feet in 2 seconds, what is the object’s approximate speed in miles per hour? (Note: 1 mile = 5280 feet)

$$\frac{50*3,600}{5,280}\approx{34}$$ miles in 1 hour.

How do you do that math fast and not screw it up? That's what I screwed up on this problem.
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Re: If an object travels 100 feet in 2 seconds, what is the  [#permalink]

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26 Jul 2015, 16:58
1
gmatser1 wrote:
Bunuel wrote:
shrive555 wrote:
If an object travels 100 feet in 2 seconds, what is the object’s approximate speed in miles per hour? (Note: 1 mile = 5280 feet)

$$\frac{50*3,600}{5,280}\approx{34}$$ miles in 1 hour.

How do you do that math fast and not screw it up? That's what I screwed up on this problem.

You can use approximation as the given options are spread far and wide

Lets take the example of this question:

$$\frac{50*3,600}{5,280}$$

Recognize here that 36 = 18*2 and 54 = 18*3. 5280 is almost equal to 5400. Thus you can cancel out 3600 and 5280 to get 2/3 (approx.). For approximations of fractions, you need to a bit imaginative to cancel out the maximum you can.

Now the given equation reduces to 50*2/3 = 100/3 = 33.33333 which is very close to 34.

Hope this helps.
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Re: If an object travels 100 feet in 2 seconds, what is the  [#permalink]

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03 Apr 2016, 10:34
shrive555 wrote:
If an object travels 100 feet in 2 seconds, what is the object’s approximate speed in miles per hour? (Note: 1 mile = 5280 feet)

A. 3.4
B. 3.8
C. 34
D. 38
E. 340

my approach
100ft/2sec = 50ft/sec or 3000ft/min, or 180,000/hour
now..1 mile is 5280
180,000/5000 = 36
so clearly is less than 36, but smth around 30
only option C
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Re: If an object travels 100 feet in 2 seconds, what is the  [#permalink]

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19 Nov 2016, 17:31
difficult question. I get to the below with some work

50 *3,600/ 5,280

then I decided to approximate = 18,000/5,000 = 36.

at this point I had to decide between 34 and 38. That was the difficult part of me.
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Re: If an object travels 100 feet in 2 seconds, what is the  [#permalink]

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09 Feb 2018, 22:45
100 feet per 2 seconds , which translates to 3000 ft per minute.

Speed for 1 hour = (3000 * 60) / 5280 = ~ 34

Ans: C
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Re: If an object travels 100 feet in 2 seconds, what is the  [#permalink]

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19 Apr 2018, 04:52
1) First change 100ft/2seconds to 50ft/1second.
2) Convert feet to miles = 50 * 1 / 5280 = 50/5280
3) Convert seconds to hours = 1 / (60)^2 =
4) Divide – multiply reciprocal = (60)^2*50/5280.
5) 3600*50=180000/5280.
6) By looking at this fraction we conclude that it needs to be multiplied by at least 30.

Thus therefore 34. Multiplying times 8 would be make the number too large
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If an object travels 100 feet in 2 seconds, what is the  [#permalink]

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15 Aug 2018, 05:22
Bunuel wrote:
shrive555 wrote:
If an object travels 100 feet in 2 seconds, what is the object’s approximate speed in miles per hour? (Note: 1 mile = 5280 feet)

3.4
3.8
34
38
340

i always mess such conversion questions.

$$100$$ feet in 2 seconds;

$$100*\frac{3,600}{2}=50*3,600$$ feet in 1 hour;

$$\frac{50*3,600}{5,280}\approx{34}$$ miles in 1 hour.

Dear Bunuel,

How would you approach this last division: $$\frac{50*3,600}{5,280}\approx{34}$$?
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If an object travels 100 feet in 2 seconds, what is the  [#permalink]

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24 Jul 2019, 17:09
shrive555 wrote:
If an object travels 100 feet in 2 seconds, what is the object’s approximate speed in miles per hour? (Note: 1 mile = 5280 feet)

A. 3.4
B. 3.8
C. 34
D. 38
E. 340

Answers are very spread out, also notice the numbers A) 3.4 and B) 3.8 are identical to C) 34 and D) 38 except for a power of 10 and E) is identical to A) and C) except with another extra power of 10.

After converting we get (50*60^2)/(5280) so quickly rounding down denominator to 5000 gives us 36 while rounding up to 6000 gives us 30 so the answer should be around the average of those, C) 34 is the only choice since D is higher than the rounded up total.
If an object travels 100 feet in 2 seconds, what is the   [#permalink] 24 Jul 2019, 17:09
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# What does Faltings' theorem look like over function fields?

Minhyong Kim's reply to a question John Baez once asked about the analogy between $\text{Spec } \mathbb{Z}$ and 3-manifolds contains the following snippet:

Finally, regarding the field with one element. I'm all for general theory building, but I think this is one area where having some definite problems in mind might help to focus ideas better. From this perspective, there are two things to look for in the theory of $\mathbb{F}_1$.

1) A theory of differentiation with respect to the ground field. A well-known consequence of such a theory could include an array of effective theorems in Diophantine geometry, like an effective Mordell conjecture or the ABC conjecture. Over function fields, the ability to differentiate with respect to the field of constants is responsible for the considerably stronger theorems of Mordell conjecture type [emphasis mine], and makes the ABC conjecture trivial.

What is the strongest such theorem? Does anyone have a reference? (A preliminary search led me to results that are too general for me to understand them. I'd prefer to just see effective bounds on the number of rational points on a curve of genus greater than 1 in the function field case.)

-

The "function field analogues" of Faltings' theorem were proved by Manin, Grauert and Samuel: see

29/PMIHES_1966_29_55_0/PMIHES_1966_29_55_0.pdf">http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1966_29/PMIHES_1966_29_55_0/PMIHES_1966_29_55_0.pdf

especially Theorem 4. (The quotation marks above are because all of this function field work came first: the above link is to Samuel's 1966 paper, whereas Faltings' theorem was proved circa 1982.)

The statement is the same as the Mordell Conjecture, except that there is an extra hypothesis on "nonisotriviality", i.e., one does not want the curve have constant moduli. For some discussion on why this hypothesis is necessary, see e.g. p. 7 of

http://math.uga.edu/~pete/hassebjornv2.pdf

An effective height bound in the function field case is given in Corollaire 2, Section 8 of

Szpiro, L.(F-PARIS6-G) Discriminant et conducteur des courbes elliptiques. (French) [Discriminant and conductor of elliptic curves] Séminaire sur les Pinceaux de Courbes Elliptiques (Paris, 1988). Astérisque No. 183 (1990), 7--18.

Note that effectivity on the height is much better than effectivity on the number of rational points (Faltings' proof does give the latter). This is not to be confused with uniform bounds on the number of rational points, for which I believe there are only conditional results known in any case.

-
MR0222088 (36 #5140) Samuel, P. Lectures on old and new results on algebraic curves. Notes by S. Anantharaman. Tata Institute of Fundamental Research Lectures on Mathematics, No. 36 Tata Institute of Fundamental Research, Bombay 1966 ii+127+iii pp. –  Chandan Singh Dalawat Dec 30 '09 at 3:33


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Tuesday, July 28, 2020 | History

9 edition of A Greater Infinity found in the catalog.

# A Greater Infinity

## by Michael McCollum

Written in English

Subjects:
• Engineering students,
• Fiction,
• Science fiction,
• Time travel

• The Physical Object
FormatPaperback
Number of Pages180
ID Numbers
Open LibraryOL8797440M
ISBN 10192938100X
ISBN 109781929381005
OCLC/WorldCa51538629

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text:
Problem Solving with Equations
Algebra, and more...

 1 MEDIUM Three times a number is six more than the number. What is the number?

 2 MEDIUM If a rope is cut into three segments with equal length, each segment will be $$4$$ meters longer than the segments if the rope is cut into four equally long segments. If the rope is cut into six equally long segments, how long is each segment?

 3 MEDIUM In a cage are $$100$$ animals. Those animals are either rabbits or chickens. If the number of legs is $$268$$, how many animals are chicken and how many are rabbits? (Note: Rabbits have $$4$$ legs and chickens $$2$$ legs.)


text:
×

A method of evaluating sums in the general case

Ever since I learned that the sum of the first $$n$$ integers was $$\frac { n\left( n+1 \right) }{ 2 }$$, I've wondered if there was a way to generally derive an equation for the sum of $${n}^{p}$$ for any positive integer $$p$$.

I actually did manage to find a way to do this using algebra, but it wasn't perfect. My original way was based on the fact that the telescoping sum $$\sum _{ k=1 }^{ n }{ { k }^{ p }-{ \left( k-1 \right) }^{ p } } ={ n }^{ p }$$. This implies that $$\sum _{ n=1 }^{ x }{ \sum _{ k=1 }^{ n }{ { k }^{ p }-{ \left( k-1 \right) }^{ p } } } ={ \sum _{ n=1 }^{ x }{ { n }^{ p } } }$$. There are so many ways to go from here. The idea is that the side with the double sum can be simplified using the binomial formula, and the highest order term in that representation will be $$c{k}^{p-1}$$ for some constant $$c$$. Then, you do something a bit clever and use the fact that $$\sum _{ n=1 }^{ x }{ \sum _{ k=1 }^{ n }{ f\left( k \right) } } =\sum _{ n=1 }^{ x }{ \left( x-n+1 \right) f\left( n \right) }$$. From here, you are left with a few summations that are doable assuming that you already know the formulas for the sums of $${n}^{q}$$ for $$q<p$$, and one term that requires knowledge of the sum that you are trying to find the general formula for. Luckily, this term is multiplied by a constant ($$-c$$ from before, in fact) that is less than $$-1$$, so you can do some algebra by adding it to both sides then dividing a constant out. I'm interested in the simplest form this can be brought to, as the simplest form I managed to get was

$$\sum _{ n=1 }^{ x }{ { n }^{ p } } =\frac { -{ \left( -1 \right) }^{ p }x\left( x+1 \right) +p!\sum _{ n=1 }^{ x }{ \left( \sum _{ v=0 }^{ p-2 }{ \frac { { \left( -1 \right) }^{ p-v }{ n }^{ v+1 }\left[ x\left( p-v \right) +p+1 \right] }{ \left( v+1 \right) !(y-v)! } } \right) } }{ p+1 }$$

which is really an absolute mess. I specifically didn't like the fact that you would need to know all previous exponent sums in order to solve this in the general case (this is to find the sum keeping $$x$$ as a variable).

However, I recently realized another way to find the sum that doesn't require the knowledge of the previous sums. Here it is:

This is based off of the fact that, for increasing $$f\left( x \right)$$, $$\sum _{ x=0 }^{ n }{ f\left( x \right) } \le \int _{ 0 }^{ n+1 }{ f\left( x \right)dx }$$. This is obvious geometrically. Essentially, the area under the curve $$f\left( x \right)$$ is greater than the approximated area by left-endpoint Riemann sums using width $$1$$. If we let $${ a }_{ k }$$ represent the $$k$$th derivative of some function $$g\left( x \right)$$, and $${a}_{k}$$ is monotonic for all $$k$$, and $${ a }_{ k }=0$$ eventually (for all $$x$$), then the simple observation I mentioned earlier in this paragraph can be used to rewrite $$\sum _{ x=a }^{ b }{ g\left( x \right) }$$ as something inside of an integral. The bounds on the summation only directly influence the bounds on the integral, so this method is useful for generalizing sums.

I'll write out either derivation if there is any interest in seeing them, otherwise I'll leave any derivation and further thoughts/application to you.

Note by Austin Antonacci
1 week, 6 days ago

Sort by:

- 1 week ago

Thanks! I'm having fun just messing around with the sum, but I'll look into that.

- 1 week ago


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Fusion for Kdb+

# Using foreign functions with kdb+¶

FFI for kdb+ ( KxSystems/ffi) is an extension to kdb+ for loading and calling dynamic libraries using pure q.

The main purpose of the library is to build stable interfaces on top of external libraries, or to interact with the operating system from q. No compiler toolchain or writing C/C++ code is required to use this library.

Know what you’re doing

You don’t need to write C code, but you do need to know what you are doing. You can easily crash the kdb+ process or corrupt in-memory data structures with no hope of finding out what happened.

No support is offered for crashes caused by use of this library.

We are grateful to Alexander Belopolsky for allowing us to adapt and expand on his original codebase.

## API¶

ffi.q exposes two main functions in the .ffi namespace. See test_ffi.q for detailed examples of usage.

### cf (call function)¶

A simple function call, intended for one-off calls.

Syntax: cf[fn;args]

Where

• fn is either the symbol name of a function, or a list of the result type (char atom) and the function name
• args is a mixed list of arguments to the function

calls fn with the arguments. The types of arguments passed to the function are inferred from the q types and should match the width of the arguments the C function expects. (If an argument is not a mixed list, append (::) to it.)

Function lookup

cf performs function lookup on each call and has significant overhead. For hot-path functions use bind.

### bind (create projection)¶

Creates a projection with the function resolved to call with arguments.

Syntax: bind[fn;types;rtype]

Where

• fn is either the function name (symbol atom), or the library and function names (symbol vector, length 2)
• types are the argument types (char vector)
• rtype is the return type (char atom)

returns a q function, bound to the specifed C function for future calls. Useful for multiple calls to the C lib.

### Passing data and getting results¶

Throughout the library, characters are used to encode the types of data provided and expected as a result. These are based on the c column of primitive data types and the corresponding upper case for vectors of the same type. The sz column is useful to work out what type can hold enough data passing to/from C.

The argument types are derived from data passed to the function (in case of cf) or explicitly specified (in case of bind). The number of character types provided must match the number of arguments expected by the C function. The return type is specified as a single character and can be " " (space), which means to discard the result (i.e. void). If not provided, defaults to int.

char C type
b, c, x unsigned int8
h signed int16
i signed int32
j signed int64
e float
f double
g, s uint8*
" " (space) void (only as return type)
r raw pointer
l size of pointer(size_t)
k K object
uppercase letter pointer to the same type

It is possible to pass a q function to C code as a callback (see qsort example below). The function must be presented as a mixed list (func;argument_types;return_type), where func is a q function (type 100h), argument_types is a char array with the types the function expects, and return_type is a char corresponding to the return type of the function. Note that, as callbacks potentially have unbounded life in C code, they are not deleted after the function completes.

#### Some utility functions¶

function purpose
errno return current errno global on *nix OS
kfn bind the function which returns and accepts K objects in current process. Similar to 2:
ext append shared library extension for current platform to library name. On Linux .ffi.ext[libm]~libm.so
cvar read global variable from the library .ffi.cvartimezone
ptrsize length of pointer in bytes on current platform
nil null value on current platform

Function arguments should be passed as a generic list to cf, call, and the function created by bind.

## Examples¶

### PCRE library¶

Bindings to PCRE (POSIX variant) using FFI for kdb+.

pcreposix is a set of functions providing a POSIX-style API for the PCRE regular-expression 8-bit library.

Complex regular expressions

Complex regular expressions can be catastrophic, exhibiting exponential run time that leads to real outages.

FFI for kdb+ is required for this library. pcre is normally available on modern Linux distributions and macOS.

As any standard, PCRE POSIX has some quirks and differences between platforms (Linux), which this library is trying to resolve.

Script to match multiline email:

reg:.pcre.regcomp["From:([^@]+)@([^\r]+)";2 sv sum 2 vs .pcre[REG_EXTENDEDREG_NEWLINE]]  // compile regex
show "Regex compiled";
multiline:"From:regular.expressions@example.com\r\nFrom:exddd@43434.com\r\nFrom:7853456@exgem.com\r\n";
emailmatch:.pcre.rlike[reg;\$multiline]

### Rmath library¶

Bindings to Rmath using FFI for kdb+.

Rmath provides the routines supporting the distribution and special (e.g. Bessel, beta and gamma functions) functions in R.

Using this library requires the stand-alone Rmath library to be installed as well as FFI for kdb+. On Ubuntu:

sudo apt-get install r-mathlib

Generate 100K numbers from normal distribution:

q)do[100000;r,:.rm.rnorm[0f;1f]]   // generate 100K N(0,1) random numbers
q)(avg;dev)@\:r                    // verify that avg and dev are 0 and 1
0.0009293088 1.002748

### BLAS¶

All arguments should be vectors (i.e. pointers to appropriate type).

q)x:10#2f;
q).ffi.cf[("f";libblas.soddot_)](1#count x; x;1#1;x;1#1)
40f
q).ffi.cf[(" ";libblas.sodaxpy_)](1#count x;1#2f; x;1#1;x;1#1)
q)x / <- a*x+y, a=x=y=2
6 6 6 6 6 6 6 6 6 6f

### Callbacks¶

q)cmp:{0N!x,y;(x>y)-x<y}
q)x:3 1 2i;
// warning: this modifies data in-place regardless of other references.
q).ffi.cf[(" ";qsort)](x;3i;4i;(cmp;"II";"i"))
1 2
3 1
3 2
q)x
1 2 3i
q)x:cab;
// warning: this modifies data in-place regardless of other references.
q).ffi.cf[(" ";qsort)](x;3i;8i;(cmp;"SS";"i"))
ab
ca
cb
q)x
abc

Register a callback on a handle

// h is handle to some other process
r:{b:20#"\000";n:.ffi.cf[read](x;b;20);0N!n#b;0}
.ffi.cf[sd1](h;(r;(),"i"))   / start handler
.ffi.cf[sd0](h;::)           / stop handler
`


text:
# Surface Area Problem

I am trying to figure out what is going on in this Surface Area problem:

As I attempted to illustrate above, It seems like the formula has been applied incorrectly. Where x has been placed should be f(x) which in this problem is (x^(1/3) + 2).

Is something going on that changes the input of f(x) to x that I am not seeing here? Thanks!

-

We can get a somewhat different point of view by looking at arclength, surface area, and other problems for curves given parametrically.

Let the curve be given by $x=x(t)$, $\:y=y(t)$. Then the surface area obtained when we rotate the chunk of the curve from $t=a$ to $t=b$ around the $x$-axis is $$\int_{t=a}^b 2\pi y\sqrt{\left(\frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2}\,dt.$$ For rotation about the $y$-axis, we have a very similar expression for surface area: $$\int_{t=a}^b 2\pi x\sqrt{\left(\frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2}\,dt.$$

In our case, we have $y=\sqrt[3]{x}+2$. Let's choose a nice parametrization, something that will simplify the calculations. A sensible choice is $x=t^3$, in which case $y=t+2$. Since $x$ goes from $1$ to $8$, $t$ will go from $1$ to $2$.

Calculate. We have $\frac{dx}{dt}=3t^2$ and $\frac{dy}{dt}=1$. For rotation about the $y$-axis, we get surface area $$\int_1^2 2\pi t^3\sqrt{1+9t^4}\,dt.$$ This integral can be calculated by a simple substitution. But this was not your problem.

For rotation about the $x$-axis, we get surface area $$\int_1^2 2\pi (t+2)\sqrt{1+9t^4}\,dt.$$ This is a nightmarish integral. The hardest part, actually the marginally easier $\int\sqrt{1+x^4}\,dx$, has been discussed on this site. It may not be worth looking up.

Whoever was writing out solutions for your book got lucky. If (s)he had not made a mistake, (s)he would have had an awful integral to evaluate. (It would start out looking even worse without the parametric approach.)

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No, it is incorrect.

It also happens to be example 2 on this page of Paul's Online Math Notes. He explains as best as anyone could hope.

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Actually paul does the same exact thing in his notes! – Matt Jul 7 '11 at 16:15
@Matt: Paul does a witty revision. Note that in the radical, there is no fraction. – mixedmath Jul 7 '11 at 16:32

The formula you have provided assumes that the surface is being produced by rotating around the x-axis. In that case, f(x) is the radius of rotation. It looks like you might instead be creating the surface by rotating around the y-axis, in which case the radius of rotation would be x, as in the solution. You can think of finding the surface area as a bunch of little slanty sticks each carving out a thin strip of surface as they're carried around the axis of rotation. Each stick (represented by the stuff in the square root) traces out a strip of circumference 2$\pi$r, where r depends on the set-up of the rotation.

Hope that helps from a conceptual standpoint.

-


text:
# Submission Creation Requests (archived)

+ 6 like - 0 dislike
10304 views

For users without direct submission privileges. Please submit a paper from the Submit Paper page.

closed as a duplicate of: Submission Creation Requests
closed Jul 14, 2015

+ 0 like - 0 dislike

[TEMPORARILY ON HOLD] (see comments and Q&A)

Hello,

## Bootstrapping Generations

Authors: Alejandro Rivero

A supersymmetric version of Chew's "democratic bootstrap" argument predicts the existence of three generations of particles, with one quark, of type "up", more massive that the other five.

It is a condensation of the arxiv papers http://arxiv.org/abs/hep-ph/0512065, http://arxiv.org/abs/0710.1526 and http://arxiv.org/abs/0910.4793.

EDIT: The point is, given a SUSY model, ask for "nuclear democracy" in the old sense of every particle can be built as a composite of others. The paper then studies the particular case of building the scalar sector from the particles in the fermionic sector.

answered Aug 28, 2014 by (260 points)
reshown Jul 14, 2015

I am afraid you will not get very kind review, as the paper misunderstands the nature of the bootstrap condition, and substitutes a strange new thing that should not be associated to this name. If you wish to get a kinder review, you could ask about the subject as a question on physics.stackexchange: "Does it make sense to postulate an equal number of quark states and quark-quark bound states?" I would prefer if we do not import this submission, because you will simply get hostile comments regarding the misunderstanding of the bootstrap, and regarding the incompatibility of the boson-fermion counting equality you postulate with the standard model, or any supersymmetric extension of it.

I am afraid that such question would get even more hostile comments because without reading of the paper the answer seems obviously "no".  And in any case,  I would hope that the review will also get some constructive comment. The result is interesting and I think that some effort should be done to save it.

But about the meaning of "democratic" and "bootstrap" in Chew's work, I agree that it could be useful to do a generic question. It seems that Chew keep changing the use of the word as the research evolved.

EDIT: in fact, the formulation of your question is already a hint to improve the paper. It could be useful to decompose the question in two separate objections: "has it sense to postulate that some elementary particles can be seen as composites of other ones?" and "has it sense to request the same number of bosonic and fermionic degrees of freedom in a theory"?

The question you are asking in the request is a really good one: is it possible for a supersymmetry to link fundamental and composite particles? To link constituents to bound-states? This question will be well received here for sure. Also, I don't know the answer. You should ask on Q&A, and if nobody knows for a long time, it might be an open question (I don't think it is open, but maybe) Your submission will needlessly generate negative reviews for what is essentially a preliminary effort to find an answer to a very nice question.

@arivero: We are still in beta, and we are trying to attract professionals. Papers that are too distant from professional interest will dissuade professionals from joining. I am asking you not to please not push the matter, your paper can get a review eventually if you insist, but it will be a negative one, and you will have a negative reputation, and then you will lose your voting rights.

When you ask a paper to go in reviews, it is expected to contain original results, while your paper contains only state counting which can be reproduced quickly by others. Such state-counting can be done in the body of a Q&A question, you can just write a sketch and people will understand what you mean.

If you had a real model of how to do what you want, it would be appropriate for reviews, but there is no full model in there, nor a sketch of one. This is why it would get negative reviews, and this would be a pity for such a nice question. If you stick to Q&A, you will get positive responses for sure.

I did not miss the logic of the paper--- your observation is original, but it is not useful absent a model which can do something with bound states and fundamental states.

For specific problems (there are infinitely many problems with such logic) There are misunderstandings there about the state counting: for example, you have 3 colors of quarks, but the bound states are only colorless, you have colorless bound states of three quarks, there are no bound states of neutrinos, etc, so that the actual method by which you get your "three generations" is completely incorrect.

But the general question is ok--- can you mix bound states and fundamental states with SUSY? That's a great question, and it will get upvotes. I'm just giving political advice here. Your paper will not be reviewed positively, but your general idea of mixing composite and elementary particles using supersymmetry is interesting.

Regarding the status of your paper, perhaps "In discussion" is better than frozen. At the end of the process, we would like to have a professional submission to review, and for you to get positive feedback. It's better than a superficial review which is just pointing out big gaps and ignoring the main idea.

Definitely I need to improve my communication skills. From the paper:

... is constituted by a string terminated in a pair of quarks.
With this imaginery, we can try a weaker version of the democratic bootstrap.

I read this (actually, I wrote this) as telling that each scalar composite contains two quarks. From this all the combinations follow. It is not the case that I am cherry picking the quarks. (EDIT: but is is true that I am implicitly assuming that for each combination we have only one scalar. I should be more explicit on this point).

With this ansatz I show that there is no solution. So then I relax, setting that only $r,s$ quarks can be attached to the string:

We postulate thus that only a subset of ”light quarks” are in the terminations of the string and then able to form composites.

(EDIT: ok, I think that the indication "a subset" can be the origin of the confusion. I do not choose randomly what quarks are allowed to play, but instead I explore all the space of possibilities: I set two variables s,r for the number of up-type and down-type quarks that are allowed to be, say, at the end of the string).

And then I show, by solving the equations for r,s and the total number N of generations, that there is a unique solution, s=2, r=3, N=3

The paper you describe is not my paper.

(EDIT: I will wrote and upload a v2 addressing the problems you mention).

The question you are asking in the request is a really good one: is it possible for a supersymmetry to link fundamental and composite particles? To link constituents to bound-states? This question will be well received here for sure

@Ron what is your definition of "well received"? A first comment about doing off-topic remarks does not seem a good reception to me.


text:
# Law of reflection of a moving mirror in special relativity (velocity perpendicular to the normal)

I have seen this question about the law of reflection in special relativity, and it is shown that, if you have a mirror moving in the opposite direction to the mirror's surface normal, the law of reflection doesn't apply anymore. For example, consider the mirror in the $$yz$$ plane, so that velocity and normal are both on the $$x$$-axis. But what about the situation where the mirror is moving perpendicularly to the mirror's surface normal? Consider for example the mirror in the $$xz$$ plane, with the normal on the $$y$$-axis and the velocity on the $$x$$-axis. I have done some calculations, and in my solution, the law of reflection does apply in this case. Am I right?

PS: I hope that this is the correct format to ask this question.

EDIT: This is the procedure I have used:

1. I looked at the first answer to the question that I have linked before.
2. I applied the same method that is explained there, but with the new projections of velocities.

My results are (using $$c = 1$$):

$$\cos(\theta_i) = \frac{\cos(\theta_i')\sqrt{1-v^2}}{1 + v\sin(\theta_i')} \\ \sin(\theta_i) = \frac{\sin(\theta_i') + v}{1 + v\sin(\theta_i')}$$

$$cos(\theta_r)$$ and $$\sin(\theta_r)$$ could be derived from the above equations by replacing $$\theta_i \to \theta_r$$ and $$\theta_i' \to \theta_r'$$.

In the primed system of reference, $$\theta_i' = \theta_r'$$, so $$\theta_i = \theta_r$$. Is this correct? If it isn't, where am I wrong?

• I see with pleasure that someone appreciates my questions :o) and is also Italian like me. – Sebastiano Sep 7 at 21:00

Yes, you're right. However, I offer you to use the following equation that relates the reflected angle to the incident one as measured in $$S$$ moving at an arbitrary velocity $$v$$ with respect to the mirror's rest frame ($$S^\prime$$): (See this article.)
$$\cos \theta_r=\frac{-2\frac{\boldsymbol{\vec{v}_n}}{c}+(1+\frac{\boldsymbol{\vec{v}_n}^2}{c^2})\cos \theta_i}{1-2\frac{\boldsymbol{\vec{v}_n}}{c}\cos \theta_i+\frac{\boldsymbol{\vec{v}_n}^2}{c^2}},$$
where $$\boldsymbol{\vec{v}_n}$$ is always the velocity vector projection onto the mirror's normal. In your example, when the mirror is set in motion perpendicular to its normal, we have $$\boldsymbol{\vec{v}_n}=0$$, and thus:
$$\cos \theta_r=\cos \theta_i \rightarrow \theta_r=\theta_i$$
Remember that the light-clock (known as Einstein's light-clock), which is used for deriving the familiar time dilation equation ($$t=\gamma t^\prime$$), is a good example for your special case: The incident and reflected angles are measured equally by the lab observer as long as the mirror (light-clock) moves perpendicular to its normal.


text:
Use decimal in where clause in SAP ABAP

SAPABAP

You are trying to use the user-specific setting in your queries, but ABAP does not support it. You need to use ‘.’ as the decimal separator instead of ‘,’.

So the query would look something like −

SELECT * FROM <table> INTO <index_tab>
WHERE amount = 10.15
Published on 19-Dec-2017 09:00:37


text:
### Pulley + Block + Rope

A uniform rope of linear mass density $\lambda$ is used to release block m with uniform acceleration a. Find the tension at a point P on the rope at a distance l from the block as shown in the figure.

Solution

Let mass m and the rope of length l above the block be the system.

Mass of rope of length $l = \lambda l$

For the system under consideration, $(m + \lambda l)g - {T_P} = (m + \lambda l)a$

$\Rightarrow {T_P} = (m + \lambda l)(g - a)$


text:
# If F : [1, ∞)→[2, ∞) is given by

Question:

If $F:[1, \infty) \rightarrow[2, \infty)$ is given by $f(x)=x+\frac{1}{x}$, then $f^{-1}(x)$ equals

(a) $\frac{x+\sqrt{x^{2}-4}}{2}$

(b) $\frac{x}{1+x^{2}}$

(c) $\frac{x-\sqrt{x^{2}-4}}{2}$

(d) $1+\sqrt{x^{2}-4}$

Solution:

Let $f^{-1}(x)=y$

$\Rightarrow f(y)=x$

$\Rightarrow y+\frac{1}{y}=x$

$\Rightarrow y^{2}+1=x y$

$\Rightarrow y^{2}-x y+1=0$

$\Rightarrow y^{2}-2 \times y \times \frac{x}{2}+\left(\frac{x}{2}\right)^{2}-\left(\frac{x}{2}\right)^{2}+1=0$

$\Rightarrow y^{2}-2 \times y \times \frac{x}{2}+\left(\frac{x}{2}\right)^{2}=\frac{x^{2}-1}{4}$

$\Rightarrow\left(y-\frac{x}{2}\right)^{2}=\frac{x^{2}-1}{4}$

$\Rightarrow y-\frac{x}{2}=\frac{\sqrt{x^{2}-4}}{2}$

$\Rightarrow y=\frac{x}{2}+\frac{\sqrt{x^{2}-4}}{2}$

$\Rightarrow y=\frac{x+\sqrt{x^{2}-4}}{2}$

$\Rightarrow f^{-1}(x)=\frac{x+\sqrt{x^{2}-4}}{2}$


text:
# Cauchy sequences…is my proof correct?

Let $(x_n)_{n∈ℕ}$ and $(y_n)_{n∈ℕ}$ be Cauchy sequences of real numbers.

Show, without using the Cauchy Criterion, that if $z_n=x_n+y_n$, then $(z_n)_{n∈ℕ}$ is a Cauchy sequence of real numbers.

Here's my attempt at a proof:

Let $(x_n)$ and $(y_n)$ be Cauchy sequences. Let $(z_n)$ be a sequence and let $z_n=x_n+y_n$.

Since $(x_n)$ and ($y_n)$ are Cauchy, $∃N∈ℕ$ such that $|x_n-x_m|<ε/2$ and $|y_n-y_m|<ε/2$ for $n,m≥N$.

Let $n,m≥N$ and let $z_n,z_m∈(z_n)$. Then, $$|z_n-z_m|=|x_n-x_m|+|y_n-y_m| <ε/2+ε/2=ε.$$ Therefore, $|z_n-z_m|<ε$ for all $n,m≥N$ and hence, $(z_n)$ is a Cauchy sequence of real numbers.

Is this correct? Any input is appreciated.

Thanks.

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It looks OK the proof! –  Atahualpa Inca Oct 17 '12 at 2:04
It's correct except for a minor detail. The triangle inequality results in a $\leq$ not a $=$. –  copper.hat Oct 17 '12 at 2:16
Thanks for pointing that out. –  Alti Oct 17 '12 at 2:35
add comment

## 2 Answers

The only thing you have to correct is $$|z_n-z_m|=|(x_n-x_m)+(y_n-y_m)|\le|x_n-x_m|+|y_n-y_m| <ε/2+ε/2=ε.$$

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add comment

This proof looks correct to me.

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add comment


text:
# Is this the projective plane or the Klein bottle? (Fundamental polygon)

I am trying to identify the topological type of this fundamental polygon and I think it is the projective plane or the Klein bottle

If we treat the top green and red arrows a single arrow then we can call this $$ab$$

Doing a similar thing with the bottom two arrows we can combine them to get $$a^{-1}b^{-1}$$

Since they are inverses, can we swap the direction of $$a^{-1}b^{-1}$$ to get $$ab$$ in the opposite direction.... leading to the fundamental polygon of the projective plane?

I was thinking that other possibility is that it could be the Klein bottle since the new top and bottom arrows are in the same direction... but surely this would only work if $$ab=a^{-1}b^{-1}$$ ... i.e. the space is abelian... so maybe the Torus?!

I am really confused about this one so it would be good to gain some insight into what is happening, thanks

• This is indeed a Klein bottle: consider the two edges $a,b$ as a unique edge. Glueing them you get a cylinder, whose two borders (the last edge: $c$) must be glued with opposite orientations: this gives rise to a Klein bottle. May 25, 2016 at 12:53
• thanks, I think the thing that confused me was the $^{-1}$'s on the bottom red and green arrows May 25, 2016 at 13:21

Closed surfaces are determined by the two properties:

• Orientability
• Euler characteristic

As you have already correctly determined, the considered surface is not orientable, because the edge $c$ is glued with the opposite orientation. Thus one has to determine the Euler characteristic $\chi$, which can be calculated by the formula: $$\chi = \#\{Vertices\} - \#\{Edges\} + \#\{2-cells\}$$

In your example, the number of edges (=3) and number of $2$-cells (=1) are easily determined. To determine the number of vertices, one has to understand which of the $6$ vertices in the picture are identified. It turns out that there are 2 distinct vertices in the quotient.

Hence, one has $\chi = 2 - 3 + 1 = 0$, thus the quotient surface is a Klein bottle (The projective space has Euler characteristic $\chi= 1 - 1 + 1 = 1$).

• Thank you. I understand the argument, except I am a little confused about the number of edges and vertices. I would be inclined to think of $a$ and $a^{-1}$ as different edges. But they are the same just with different orientability, right? For the vertices, I see that the 4 vertices either side of the two $c$ edges correspond to 2 identified vertices.... are the remaining two vertices in the diagram neglected since we can combine $a$ and $b$ into one arrow? May 25, 2016 at 13:08
• Because gluing the surface is a process, there are two objects one has to consider: The hexagon (before the gluing), and the glued surface. Before gluing, the hexagon has $6$ edges ($2$ of each color) and $6$ vertices. In the gluing process, the edges of the same color are identified, so in the end there are only $3$ edges. For the vertices, there is in fact only one vertex adjacent to the edges named $c$: The a priori different end point are identified when the two green (or red) edges are glued together. May 25, 2016 at 13:16

The notation in your picture is inconsistently overdetermined, and this seems to be causing you confusion.

The arrows on the sides are already enough to indicate orientation, so you should not also put a superscript $-1$ on any arrow labels. Thus you should replace the $a^{-1}$ by an $a$ and the $b^{-1}$ by a $b$.

It is then clearer that you can combine the upper and lower green-red-arrow-pairs into a single arrow pair, and then it is just the standard gluing diagram for the Klein bottle, as you say.

The appropriate place for "$-1$"s to appear is when you go around the gluing diagram and write down the corresponding word in $a,b,c$ and their inverses. Thus, starting from the upper right $a$ and going counterclockwise, the word is $abcb^{-1}a^{-1}c$, where you write a $-1$ superscript when going backwards against the direction of the arrow.

• So in general if two consecutive arrows are in the same direction then we can treat them as one arrow in that direction? I still do not understand why we can replace $a^{-1}$ by $a$ since they are different generators- I though $a^{-1}$ works in the opposite direction to $a$. Could you help me understand this? ,,, or is $a=a^{-1}$ since we are already given that they act in the same direction? May 25, 2016 at 12:59
• I added a paragraph to address your questions about the $-1$ superscript. May 25, 2016 at 12:59
• So if we are given $a=a^{-1}$? May 25, 2016 at 13:00
• We are never given $a=a^{-1}$. The letter $a$ is a label which we assign to a pair of oriented sides which are to be identified in the quotient space. The symbol $a^{-1}$ is a label that applies to the same pair of sides but with the opposite orientations. May 25, 2016 at 13:02
• It is correct if it is said more carefully. If you have two consecutive arrows in one segment of the diagram, and two consecutive arrows in another nonoverlapping segment of the diagram, and if in both of those segments the arrows have identical labels and orientations, then yes, each of those segments may be replaced by a single arrow. May 25, 2016 at 13:13


text:
## Alternative Proofs To Euclid's Propositions

So lately I've been trying to start practicing proving things without being given a prompt (as in actually finding something out, not answering the question: "prove so and so"), which I'd never done before. -- But then the bummer is not knowing the name of what you've just proved

So after searching I found out that one is a proof of Euclid's proposition 37 from book I (Triangles which are on the same base and in the same parallels equal one another). -- Is there a website where I can find alternative proofs from the one given in the actual book?

I used an algebraic proof and I want to find out who came up with it. -- I found websites with a bazillion (ok, more like 3 or 4) alternative proofs to the pythagorean theorem, but I can't find a website with proofs to the other propositions.

Is there like a proofs directory somewhere out there or something?
 Thread Tools

 Similar Threads for: Alternative Proofs To Euclid's Propositions Thread Forum Replies Set Theory, Logic, Probability, Statistics 3 Calculus & Beyond Homework 0 General Math 48 General Discussion 14 General Physics 0


text:
## The Annals of Probability

### Greedy lattice animals: negative values and unconstrained maxima

#### Abstract

Let $\{X_v, v \in \mathbb{Z}^d\}$ be i.i.d. random variables, and $S(\xi) = \sum_{v \in \xi} X_v$ be the weight of a lattice animal $\xi$. Let $N_n = \max\{S(\xi) : |\xi| = n$ \text{and $\xi$ contains the origin}\}$and$G_n = \max\{S(\xi) : \xi \subseteq [-n,n]^d\}$. We show that, regardless of the negative tail of the distribution of$X_v$, if$\mathbf{E}( X_v^+)^d (\log^+ X_v^+))^{d+a} < + \infty$for some$a>0$, then first,$\lim_n n^{-1} N_n = N exists, is finite and constant a.e.; and, second, there is a transition in the asymptotic behavior of $G_n$ depending on the sign of $N$: if $N > 0$ then $G_n \approx n^d$, and if $N < 0$ then $G_n \le cn$, for some $c > 0$. The exact behavior of $G_n$ in this last case depends on the positive tail of the distribution of $X_v$; we show that if it is nontrivial and has exponential moments, then $G_n \approx \log n$, with a transition from $G_n \approx n^d$ occurring in general not as predicted by large deviations estimates. Finally, if $x^d(1 - F(x)) \to \infty$as $x \to \infty$, then no transition takes place.

#### Article information

Source
Ann. Probab., Volume 29, Number 1 (2001), 205-241.

Dates
First available in Project Euclid: 21 December 2001

https://projecteuclid.org/euclid.aop/1008956328

Digital Object Identifier
doi:10.1214/aop/1008956328

Mathematical Reviews number (MathSciNet)
MR1825148

Zentralblatt MATH identifier
1016.60048

#### Citation

Dembo, Amir; Gandolfi, Alberto; Kesten, Harry. Greedy lattice animals: negative values and unconstrained maxima. Ann. Probab. 29 (2001), no. 1, 205--241. doi:10.1214/aop/1008956328. https://projecteuclid.org/euclid.aop/1008956328

#### References

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• Cox, J. T., Gandolfi, A., Griffin, P. and Kesten, H. (1993). Greedy lattice animals I: Upper bounds. Ann. Appl. Probab. 3 1151-1169.
• Cox, J. T. and Kesten, H. (1981). On the continuity of the time constant of first-passage percolation. J. Appl. Probab. 18 809-819.
• Dembo, A. and Zeitouni, O. (1998). Large Deviations Techniques and Applications, 2nd ed. Springer, Berlin.
• Deuschel, J.-D. and Pisztora, A. (1996). Surface order deviations for high-density percolation. Probab. Theory Related Fields 104 467-482.
• Dunford, N. and Schwartz, J. T. (1958). Linear Operators, Part I. Interscience, New York.
• Fisher, M. E. and Sykes, M. F. (1959). Excluded-volume problem and the Ising model of ferromagnetism. Phys. Rev. (2) 114 45-58.
• Fontes, L. and Newman, C. M. (1993). First-passage percolation for random colorings of d. Ann. Appl. Probab. 3 746-762.
• Gandolfi, A. and Kesten, H. (1994). Greedy lattice animals II: Linear growth. Ann. Appl. Probab. 4 76-107.
• Grimmett, G. (1999). Percolation, 2nd ed. Springer, Berlin.
• Howard, C. D. and Newman, C. M. (1999). From greedy lattice animals to Euclidean first-passage percolation. In Perplexing Problems in Probability (M. Bramson and R. Durrett, eds.) 107-119. Birkh¨auser, Boston.
• Jiang, T. (1999). Maxima of partial sums indexed by geometric structures. Ph.D. dissertation, Dept. Statistics, Stanford Univ.
• Karlin, S., Dembo, A. and Kawabata, T. (1990). Statistical composition of high scoring segments from molecular sequences. Ann. Statist. 18 571-581.
• Kesten, H. (1986). Aspects of first passage percolation. Lecture Notes in Math. 1180 125-264. Springer, Berlin.
• Lee, S. (1993). An inequality for greedy lattice animals. Ann. Appl. Probab. 3 1170-1188.
• Liggett, T. M., Schonmann, R. H. and Stacey, A. M. (1997). Domination by product measures. Ann. Probab. 25 71-95.
• Martin, J. (2000). Linear growth for greedy lattice animals. Preprint.
• Talagrand, M. (1995). Concentration of measure and isoperimetric inequalities in product spaces. Publ. Math. de l'I.H.E.S. 81 73-205.


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# If $\overrightarrow a=\hat i+\hat j+\hat k,\:\overrightarrow b=\hat i\:\:and\:\:\overrightarrow c=x\hat i-\hat j+\hat k$, then the value of $x$ for which $\overrightarrow a\:\overrightarrow b\:and\:\overrightarrow c$ are coplanar is ?
$\begin{array}{1 1} 0 \\ - 1 \\ 1 \\ \text{cannot be coplanar for any x } \end{array}$


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Thread: Busted functions View Single Post
2006-02-22, 12:50   #4
Dougy

Aug 2004
Melbourne, Australia

23·19 Posts

Quote:
 Originally Posted by Dougy ...for example, $f(x)=\frac{3}{2}x+\frac{1}{2}$, if $m=3$ then $f^n(m)$ will still always be a natural number.
After I posted it I noticed that this was incorrect. So I'll give another example of a function $f$ which would not be included if $f:\mathbb{N} \rightarrow \mathbb{N}$. Eg. $f(x)=2^{x-2}+1$ has the property that $f(1)$ is not a natural number, while $f(x) \in \mathbb{N}$ when $x \in \mathbb{N}$ and $x>1$.

It is possible to create a related function $g:\mathbb{N} \rightarrow \mathbb{N}$ such that $g(x)=f(x)$ whenever $f(x) \in \mathbb{N}$, and $g(x)=1$ whenever $f(x)$ is not a natural number. But then this would be asking a different question.

Last fiddled with by Dougy on 2006-02-22 at 12:52


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# Boiling point comparison between cyclohexane derivatives

I am trying to order these according to boiling point. I think that B is the highest because it has hydrogen bonding. I am not sure how to differentiate A and C though. They both have about the same molar mass. The only thing I can think is that C has a double bond. Does this double bond make the boiling point of C higher than A because it is stronger than a single bond? Or perhaps the double bond is shorter which decreases branching in C making the boiling point lower than A?

It's not the double bond in C that is significant here, it's the oxygen. The carbon-oxygen bond is polar and will make C more "electrically sticky" with its neighbors compared to A. Since the $\ce{O-H}$ bond in B is even more polar than the ketone in C, its boiling point will be even higher. Read more about it on this page.


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+0

0
23
1
+992

The graph of  $$y=f'(x)=mx^2+nx+k$$   passes through the points  $$P({1 \over3};0), Q(1;0) \ and \ R(0;1)$$

Determine the values of m, n and k....I have posted this before, but cancelled it because I thought I had won it, however, I just realised that's not the case..sorry about this..At least I know k is 1......lol..Thank you all very much.

Mar 10, 2023


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# Math Help - Finding Chords secants and tangents...

1. ## Finding Chords secants and tangents...

Can someone help me?

A circle can be drawn through points x, y, and z. What is the radius of the circle if you know YW=6. XW=8. And WZ=12...

Also with that....how far is the center of the circle from the point W?

Thanks

2. Originally Posted by zachcumer
Can someone help me?

A circle can be drawn through points x, y, and z. What is the radius of the circle if you know YW=6. XW=8. And WZ=12...

Also with that....how far is the center of the circle from the point W?
I need some additional information: Where is point W? Outside the circle, on the circle line, inside the circle?

3. look at number 24..thats where I got my question..

thanks

4. does that help anyone?

5. anyone?

6. Originally Posted by zachcumer
look at number 24..thats where I got my question..

thanks

EDIT: Sorry, this is the correct link: http://www.mathhelpforum.com/math-he...703-post2.html

7. what about the second part of the question...

8. Originally Posted by zachcumer
what about the second part of the question...
You have my sketch. Draw a third right triangle with WC (C stands for center of the circle) as the hypotenuse. You already know the legs of the right triangle. Calculate the length of the hypotenuse.

For confirmation only: $|\overline{WC}| = \sqrt{29}$

9. so thats how far W is from the center?

the square root of 29?

10. Originally Posted by zachcumer
hello?
Hi,

have a look here: http://www.mathhelpforum.com/math-he...562-post8.html


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# Why two different Lagrangians to derive geodesic equations?

I'm trying (very early stages) to understand the derivation of the geodesic equation

$$\frac{d^{2}x^{\alpha}}{d\lambda^{2}}+\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=0$$ via Lagrangians and the Euler-Lagrange equations. I don't understand why some authors use the Lagrangian $$L\left(\dot{x}^{c},x^{c}\right)\equiv\frac{1}{2}g_{ab}\left(x^{c}\right)\dot{x}^{a}\dot{x}^{b}$$ (as in Foster and Nightingale's A Short Course in General Relativity, p60) and others use the Lagrangian $$L\left(\dot{x}^{\alpha},x^{\alpha}\right)\equiv\sqrt{-g_{\mu\nu}\left(x^{\alpha}\right)\dot{x}^{\mu}\dot{x}^{\nu}}$$ (as in Moore's A General Relativity Workbook, p90). I realise I must be confused here, but I can't why as they both seem to end up with the same geodesic equation.

• It is not unusual for different Lagrangian's to yield same equation of motion. Possible duplicate of physics.stackexchange.com/q/149082 – zzz May 12 '15 at 18:58
• Do you really want to derive geodesic equation with Lagrangians? I think it's simpler if you derive and practice differential geometric concepts in simple sphere. You should first understand the covariant derivative; when it's zero you defined parallel transport; when you parallel transport a vector towards the direction in points to, you'll follow a geodesic and that yields the geodesic equation, quite intuitively. Basically the equation says how do you need to change the coordinates of your velocity vector to remain on a straight path. – Calmarius May 12 '15 at 22:34
• Technically speaking, the square-root Lagrangian and the non-square-root Lagrangian lead to slightly different EL eqs. As is well-known, the solutions are for both eqs. (parametrized) geodesics, but where the former eq. produces all parametrized geodesics, the latter eq. only produces affinely parametrized geodesics, cf. e.g. my Phys.SE answer here. – Qmechanic May 13 '15 at 5:10
• @Qmechanic - thanks. How come Valter Moretti's derivation here (which is over my head) starts with the square root Lagrangian and ends with (what I assume is) the affinely parametrized geodesic equations, ie my first equation? – Peter4075 May 13 '15 at 13:09
• His derivation (v6) assumes this at some point. – Qmechanic May 13 '15 at 13:38

Actually the equations of motion one ends up with are not manifestly the same:

If I let $$L_1 = \sqrt{g_{\mu \nu} \frac{dx^{\mu}}{dt} \frac{dx^{\nu}}{dt}}$$ one finds that the Euler-Lagrange equation is

$$\frac{d^2 x^{\mu}}{dt^2} + \Gamma^{\mu}_{\nu\sigma} \frac{dx^{\nu}}{dt} \frac{dx^{\sigma}}{dt} = f(t) \frac{dx^{\mu}}{dt},$$ for a suitable function $f(t)$ (you should try to derive this), whereas for $L_2 = (L_1)^2$, the equation is what you wrote above. So how are these two related, and why do they lead to the same solution? The answer has to do with the parameter used to describe the curve.

In the case of $L_1$, the parameter is arbitrary. One can see that the action $S_1 = \int dt L_1$ is reparametrization invariant, namely if we take $t \rightarrow \lambda(t)$, $S_1$ doesn't change. This implies that the we are free to choose any parameter we like without affecting the physics.

One can show that there is a particular $\lambda(t)$ for which the EOM of $L_1$ reduce to the EOM of $L_2$ (such a parameter is called an affine parameter). It's a very interesting exercise, and I leave you to work on the details.

• Thanks. I will try to derive the non-affine geodesic equations you give. In the meantime do you have a link to this derivation? I'm still puzzled as to how Moore starts with the square root Lagrangian and ends up with the affine geodesic equations. – Peter4075 May 13 '15 at 13:17
• I'm not familiar with the source you're using, but the clearest (IMO) way to get the affine geodesic equations from the square root Lagrangian, is to first get the non-affine equations, and then switch your arbitrary parameter to an affine one. As far as having a source for the derivation, I don't know of any which derives the non-affine equation directly (although it's not difficult). Carroll in his notes starts with the square-root Lagrangian and then changes to an affine parameter in the middle of his derivation. Look at page 15 in: preposterousuniverse.com/grnotes/grnotes-three.pdf – childofsaturn May 13 '15 at 14:17
• Thanks, but I just cannot see how he gets $d\tau$ as the denominator after making the affine parameter substitution (Equation 3.52). – Peter4075 May 13 '15 at 15:48
• One simply applies the chain rule for that, if I understand your confusion correctly. – childofsaturn May 13 '15 at 17:41
• Any chance of a hint as to what that chain rule looks like? I cannot see how he substitutes (3.52) into (3.51) to get the equation at the bottom of page 69. Thank you for your patience. – Peter4075 May 13 '15 at 18:33


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# Riemann Zeta Function (Part 2, beware)

Calculus Level 4

The Riemann zeta function states that for every $s$,

$\large \zeta (s) = \displaystyle \sum_{n=1}^\infty \frac {1}{n^{s}}$

Sometimes these sums are divergent but using analytic continuation, you can get a finite answer. Find $\zeta (0.5)$.

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