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# Physics/Maths - [Gradient - Help]
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1. Posting an image below:
I done the usual but it saids my answer is wrong.
The usual as in m = change in y/change in x
I think it has something to do with the graph not starting from 0.
I put the y intercept as 10 but that is wrong too. I doen all the other ones when it starts at 0 so im guessing thats the problem.
Thanks
2. change in , change in ,
The y intercept doesn't matter at all when you are finding the gradient. If you are trying to find the equation of the line then
therefore
maybe the change in x needs to be negative?
no idea what you posted - it doesn't make any sense
4. (Original post by Maths is Life)
maybe the change in x needs to be negative?
no idea what you posted - it doesn't make any sense
Basically:
Gradient and y-intercept of tht graph i just posted..
And I done that
5. (Original post by Shipreck)
change in , change in ,
The y intercept doesn't matter at all when you are finding the gradient. If you are trying to find the equation of the line then
therefore
Ohh damn, thanks I see what i done wrong
I didn't see the 50 i thought it was 0 so i put 100 instead of 50.
Thanks
6. (Original post by Shipreck)
change in , change in ,
The y intercept doesn't matter at all when you are finding the gradient. If you are trying to find the equation of the line then
therefore
It asks for the y intercept and I did put 10 but then as a hint it said
'remember the graph doesn't start from 0'
It asks for the y intercept and I did put 10 but then as a hint it said
'remember the graph doesn't start from 0'
If 10 wasn't the answer I'd assume they mean to continue the line backwards to when the current drawn is 0, then find the voltage at that point.
8. (Original post by JN17)
If 10 wasn't the answer I'd assume they mean to continue the line backwards to when the current drawn is 0, then find the voltage at that point.
Ahh yh true, thanks
Spoiler:
Show
I'll try that in a bit.. started another one
It asks for the y intercept and I did put 10 but then as a hint it said
'remember the graph doesn't start from 0'
Woops, didn't see that xD.
10. (Original post by Shipreck)
Woops, didn't see that xD.
sorry, this is correct
Thanks!
so it is 12
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https://doc.cgal.org/latest/Kernel_23/classKernel_1_1ConstructIsoCuboid__3.html | 1,721,889,369,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00299.warc.gz | 171,041,391 | 4,602 | CGAL 5.6.1 - 2D and 3D Linear Geometry Kernel
Kernel::ConstructIsoCuboid_3 Concept Reference
## Definition
Refines
AdaptableBinaryFunction
CGAL::Iso_cuboid_3<Kernel>
## Operations
A model of this concept must provide:
Kernel::Iso_cuboid_3 operator() (const Kernel::Point_3 &p, const Kernel::Point_3 &q)
introduces an iso-oriented cuboid with diagonal opposite vertices p and q such that p is the lexicographically smallest point in the cuboid.
Kernel::Iso_cuboid_3 operator() (const Kernel::Point_3 &p, const Kernel::Point_3 &q, int)
introduces an iso-oriented cuboid with diagonal opposite vertices p and q. More...
Kernel::Iso_cuboid_3 operator() (const Kernel::Point_3 &left, const Kernel::Point_3 &right, const Kernel::Point_3 &bottom, const Kernel::Point_3 &top, const Kernel::Point_3 &far, const Kernel::Point_3 &close)
introduces an iso-oriented cuboid fo whose minimal $$x$$ coordinate is the one of left, the maximal $$x$$ coordinate is the one of right, the minimal $$y$$ coordinate is the one of bottom, the maximal $$y$$ coordinate is the one of top, the minimal $$z$$ coordinate is the one of far, the maximal $$z$$ coordinate is the one of close.
## ◆ operator()()
Kernel::Iso_cuboid_3 Kernel::ConstructIsoCuboid_3::operator() ( const Kernel::Point_3 & p, const Kernel::Point_3 & q, int )
introduces an iso-oriented cuboid with diagonal opposite vertices p and q.
The int argument value is only used to distinguish the two overloaded functions.
Precondition
p.x()<=q.x(), p.y()<=q.y() and p.z()<=q.z(). | 427 | 1,529 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-30 | latest | en | 0.584431 |
https://metanumbers.com/183554 | 1,628,123,682,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155268.80/warc/CC-MAIN-20210805000836-20210805030836-00351.warc.gz | 399,234,562 | 10,913 | ## 183554
183,554 (one hundred eighty-three thousand five hundred fifty-four) is an even six-digits composite number following 183553 and preceding 183555. In scientific notation, it is written as 1.83554 × 105. The sum of its digits is 26. It has a total of 4 prime factors and 12 positive divisors. There are 78,624 positive integers (up to 183554) that are relatively prime to 183554.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 26
• Digital Root 8
## Name
Short name 183 thousand 554 one hundred eighty-three thousand five hundred fifty-four
## Notation
Scientific notation 1.83554 × 105 183.554 × 103
## Prime Factorization of 183554
Prime Factorization 2 × 72 × 1873
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 26222 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 183,554 is 2 × 72 × 1873. Since it has a total of 4 prime factors, 183,554 is a composite number.
## Divisors of 183554
12 divisors
Even divisors 6 6 4 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 320454 Sum of all the positive divisors of n s(n) 136900 Sum of the proper positive divisors of n A(n) 26704.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 428.432 Returns the nth root of the product of n divisors H(n) 6.87352 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 183,554 can be divided by 12 positive divisors (out of which 6 are even, and 6 are odd). The sum of these divisors (counting 183,554) is 320,454, the average is 2,670,4.5.
## Other Arithmetic Functions (n = 183554)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 78624 Total number of positive integers not greater than n that are coprime to n λ(n) 13104 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 16589 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 78,624 positive integers (less than 183,554) that are coprime with 183,554. And there are approximately 16,589 prime numbers less than or equal to 183,554.
## Divisibility of 183554
m n mod m 2 3 4 5 6 7 8 9 0 2 2 4 2 0 2 8
The number 183,554 is divisible by 2 and 7.
• Deficient
• Polite
## Base conversion (183554)
Base System Value
2 Binary 101100110100000010
3 Ternary 100022210022
4 Quaternary 230310002
5 Quinary 21333204
6 Senary 3533442
8 Octal 546402
10 Decimal 183554
12 Duodecimal 8a282
20 Vigesimal 12ihe
36 Base36 3xmq
## Basic calculations (n = 183554)
### Multiplication
n×i
n×2 367108 550662 734216 917770
### Division
ni
n⁄2 91777 61184.7 45888.5 36710.8
### Exponentiation
ni
n2 33692070916 6184314384915464 1135155642608773079056 208362358823410733753045024
### Nth Root
i√n
2√n 428.432 56.8313 20.6986 11.2915
## 183554 as geometric shapes
### Circle
Diameter 367108 1.1533e+06 1.05847e+11
### Sphere
Volume 2.59048e+16 4.23387e+11 1.1533e+06
### Square
Length = n
Perimeter 734216 3.36921e+10 259585
### Cube
Length = n
Surface area 2.02152e+11 6.18431e+15 317925
### Equilateral Triangle
Length = n
Perimeter 550662 1.45891e+10 158962
### Triangular Pyramid
Length = n
Surface area 5.83564e+10 7.28828e+14 149871
## Cryptographic Hash Functions
md5 32bb6eb41756e59e2534ddddf7c7e2b2 c0ed5f1a122a455513e65d36b14a1afafc247192 055df05cb77c990191ce9e01d8c4007c503a4ed785a07e79540315a56bce966f 018f8c5ca118013e767f6c4a8ce60b8566d18b8188b9b20fb45e3f21cd3f3e76b76c82064532f0ea23897bcc8701dc8b47783eea060feeb4e790415eb801077e 98e9c4fcad71eab698df4a5df4844cca33e7c6d7 | 1,438 | 4,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-31 | latest | en | 0.820597 |
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(* CSE341, Spring 2008, Lecture 8 *) ( (* Create similar functions *) ( val addn = fn n => fn m => n+m v val increment = addn 1 v val add_two = addn 2 v fun f n = if n=0 then [] else (addn n)::(f (n-1)) (* Combine functions *) ( fun f1 (g,h) = fn x => g (h x) f (* f1 is actually in ML's standard library as infix o (composition) *) (* but that is just syntax -- the result of g o h is a closure that "remembers" g and h in its environment *) fun sqrt_of_abs1 i = Math.sqrt(Real.fromInt (abs i)) f fun sqrt_of_abs2 i = (Math.sqrt o Real.fromInt o abs) i f val sqrt_of_abs3 = Math.sqrt o Real.fromInt o abs v (* another combining-fuctions function *) fun f2 (g,h) = fn x => case g x of NONE => h x | SOME y => y (* Private data, for map/fold *) ( (* Earlier we saw map: *) (
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Unformatted text preview: fun map (f,l) = case l of => | fst::rest => (f fst)::(map(f,rest)) (* This fold function is at least as useful *) fun fold (f,acc,l) = case l of => acc | fst::rest => fold (f, f(acc,fst), rest) (* fold is actually in ML's standard library as List.foldl, but List.foldl is "curried" (explanation soon). You will use List.foldl in homework 3 *) (* Example uses that do not use private data *) ( fun f3 l = fold ((fn (x,y) => x+y), 0, l) f fun f4 l = fold ((fn (x,y) => x andalso y >= 0), true, l) f (* Examples that use that uses private data *) ( fun f5 (l,lo,hi) = fold ((fn (x,y) => if y >= lo andalso y <= hi then x+1 else x), 0, l) fun f6 (lst,hi) = map((fn x => if x > hi then hi else x), lst)...
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This note was uploaded on 10/12/2009 for the course CSE 341 taught by Professor Staff during the Spring '08 term at University of Washington.
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lec8 - fun map (f,l) = case l of => | fst::rest => (f
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Ask a homework question - tutors are online | 661 | 2,170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-17 | longest | en | 0.789717 |
https://electronics.stackexchange.com/questions/413230/rlc-series-circuit-with-initial-current | 1,560,681,984,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00163.warc.gz | 431,656,130 | 34,678 | # RLC series circuit with initial current
I'm trying to solve, for the voltage across the capacitor in the following circuit:
simulate this circuit – Schematic created using CircuitLab
The initial current in this circuit is equal to $$\\text{I}_0\$$. How can I find the voltage across the capacitor?
EDIT: The initial applied current is equal to $$\\text{I}_0\$$, the initial voltage across the capacitor is equal to $$\\text{V}_\text{C}\left(0\right)\$$ and the initial current trough the inductor is equal to $$\\text{I}_\text{L}\left(0\right)\$$
My Work:
I thought that I can write:
$$\text{V}_{\text{R}_1}\left(t\right)+\text{V}_{\text{R}_2}\left(t\right)+\text{V}_\text{C}\left(t\right)+\text{V}_\text{L}\left(t\right)=0\tag1$$
Writing that in terms of the current through the components I get:
$$\text{I}_\text{in}'\left(t\right)\cdot\text{R}_1+\text{I}_\text{in}'\left(t\right)\cdot\text{R}_2+\text{I}_\text{in}\left(t\right)\cdot\frac{1}{\text{C}}+\text{I}_\text{in}''\left(t\right)\cdot\text{L}=0\tag2$$
And then using the initial conditions $$\\text{I}_\text{in}\left(0\right)=\text{I}_0\$$ and $$\\text{I}_\text{in}'\left(0\right)=0\$$ I can solve for the voltage across the capacitor. Is that right?
• Is current $I_0$ supplied from a current source and was the current through the inductor prior to t=0 zero amps AND, was the voltage across the capacitor 0 volts at t=0? In other words you need to clearly state your initial condictions. At the moment there is ambiguity. – Andy aka Dec 20 '18 at 18:46
• @Andyaka Yes only the intial current was supplied. – Klopjas Dec 20 '18 at 18:47
• That does not answer what I asked. – Andy aka Dec 20 '18 at 18:49
• @Klopjas You must have the initial conditions for the state of the capacitor and the inductor. Given only the value of $I_0$ I believe there is more than one possible pairing of initial conditions for the capacitor and inductor. – jonk Dec 20 '18 at 18:51
• @Klopjas then all that remains for you to clarify is the source of that external current because that source type dictates what happens after t=0. – Andy aka Dec 20 '18 at 18:53 | 648 | 2,115 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2019-26 | latest | en | 0.82535 |
https://www.meritnation.com/cbse-class-10/science/science/magnetic-effects-of-electric-current/studymaterial/12_2_10_157 | 1,511,113,781,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805708.41/warc/CC-MAIN-20171119172232-20171119192232-00071.warc.gz | 826,812,592 | 15,744 | Select Board & Class
Magnetic Effects of Electric Current
Magnetic field and its characteristics
Construct a simple circuit with open ends M and N. Take a thick conducting wire of aluminium and connect it between the open ends. Now, place a magnetic compass near the aluminium wire and note the position of the compass needle. Now, close the switch to allow the current to flow through the wire and notice the deflection in the needle.
It can be concluded from this activity that electric current flowing through aluminium wire has produced a magnetic force that is exerted on the compass needle resulting in its deflection. Can we say that a magnetic field is related to an electric current?
Hans Christian Oersted (1777-1851) was the first scientist to observe that a compass needle gets deflected when placed near a current carrying conductor. By this, he concluded that electricity and magnetism are related to each other and called it electromagnetism.
Magnetic lines of force (magnetic force)
You know that a bar magnet can repel or attract another magnet depending on the nature of poles of the other two magnets that are facing each other. When a bar magnet is suspended by thread, its one end always points towards the geographic North Pole, called magnetic North Pole and the other end always points towards the geographic South Pole, called magnetic South Pole of the magnet.
Like poles repel and unlike poles attract each other.
Take a drawing cardboard and sprinkle some iron filings on it. Notice the position of the iron filings as a whole. Now, take a bar magnet and slowly bring it below the cardboard. You will observe that the iron filings tend to attract towards the magnet.
It is observed that most of the iron filings align themselves at poles. What does the pattern represent? It represents that the magnet exerts a force around its body with a stronger force near the two poles. A magnet produces a magnetic field, which can be detected by the force exerted on the iron filings. The regular pattern of the iron filings on the board represents the lines of magnetic field or lines of magnetic force called magnetic lines.
How to determine the shapes of magnetic field lines of a bar magnet?
To understand the process, let us see this animation
Do you know the direction of a magnetic field inside the magnet?
Inside the magnet, magnetic field lines run from the South Pole to the North Pole where they emerge out. Therefore, we can say that magnetic field lines make closed curves.
• The region where magnetic field lines are crowded has relatively greater strength. Hence, in a magnet, strength of the regions near the poles is greater than other regions.
• It should be noted that a compass needle cannot point in two directions when placed at a point near the magnet. This means that no two magnetic field lines cross each other at a point.
Characteristics of magnetic field lines
1. Magnetic field lines emanate from the North Pole and terminate at the South Pole of a magnet. (Outside the magnet)
2. The degree of closeness of magnetic field lines represents the relative strength of the magnet.
3. No two field lines can intersect each other.
Magnetic field lines of the Earth
The Earth is treated as magnetic because it is assumed that a huge bar magnet is buried within its interior with the magnetic North Pole near the geographic South Pole, and the magnetic South Pol...
To view the complete topic, please
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# Note04 - Outline Notation of the derivative...
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Unformatted text preview: Outline Notation of the derivative Differentiability and continuity Lecture 4 - Differentiation AMA140/150 Calculus Basic rules of differentiation Derivative of inverse function Logarithmic function and exponential function Differentiation of parametric function Implicit differentiation AMA140/150 Calculus Lecture 4 - Differentiation Notation of the derivative AMA140/150 Calculus Lecture 4 - Differentiation Notation of the derivative Notation of the derivative Let f be a function defined on an interval I . We say that f is differentiable at a point x0 in I if the limit lim f x0 x →0 x − f x0 x exists. The limit, when it exists, is called the derivative of f at x0 , and is denoted by f ′ x0 . The expression at x0 . f x0 x −f x0 x , where AMA140/150 Calculus x 0, is called a difference quotient of f Lecture 4 - Differentiation Geometrical interpretation: the derivative f ′ x0 , if exists, is the slope of the tangent line to the group of the function y f x at the point x0 , f x0 . Physical interpretation: Suppose an object moves along a straight line and its distance from a fixed point on the line at time t is given by y f t . Then the derivative f ′ t0 is the instantaneous velocity of the object at the time t0 . AMA140/150 Calculus Lecture 4 - Differentiation Notation of the derivative Suppose y dy dx Derivative f x . It is also common to use the notation df dx x x0 f ′ x0 x0 x − f x0 x x →0 x x0 Notice that we put x f x0 lim lim x →x0 f x − f x0 . x − x0 Example 4.2 Let y d3 x dx dy dx x in the last equality. x 3 . Then for any x , fx x →0 Derivative function f′ x lim Example 4.1 For the constant function f x f′ x lim fx x −f x x x →0 ′ Thus, f x x →0 lim 0 x →0 ′ AMA140/150 Calculus 3x 0 and Lecture 4 - Differentiation AMA140/150 Calculus Solution: We have to find f ′ 0 x −f 0 x lim |x |? f0 x −f 0 . Notice that x x →0 lim Example 4.4 Let f x 0 x −0 x x →0 lim 1 x →0 lim f0 x −f 0 x Therefore, lim f0 lim x →0 − −0 x −0 x 1, lim −1 x → 0− x −f 0 does not exist and hence f is not x x →0 differentiable at 0. if x ≥ 0, x2 x x f′ 2 2 3x 2 . 12. if x < 0. Solution. Again we have to find f ′ 0 lim f0 x −f 0 x x →0 x → 0− 3 Differentiability Example 4.3 Is f differentiable at 0 if f x x →0 x Lecture 4 - Differentiation x6 f0 2 x − x3 0. Differentiability lim 3 In particular, f′ 0 C x 3x 2 for all x . Thus, f ′ x 0. 2 x x lim 3x 2 0 for all x . Equivalently, d C dx 3x x →0 x →0 C −C x 3x x 3x 2 x lim C, lim 3x 2 x x →0 x −f x . x x →0 x3 lim Given a function f . Suppose D is the set of all points at which f is differentiable. The derivative function of f , f ′ , with domain D is defined by fx x 3 − x3 x x lim −1. lim f0 lim lim lim f0 x →0 x −f 0 . Notice that x x 6 − 06 x 0 x →0− x →0 x 2 − 02 x 0 x →0 x −f 0 x x → 0− Is f differentiable at 0? x → 0− x →0 Remark that x if x > 0, fx −x if x < 0. ⇒ 1 if x > 0, −1 ′ if x < 0. In this case, fx f′ x 5 6x if x > 0, 0 2x AMA140/150 Calculus Lecture 4 - Differentiation x lim x x −f 0 exists and equals to 0. Hence f is x 0. differentiable at 0 and f ′ 0 Therefore, lim f0 lim AMA140/150 Calculus if x 0, if x < 0. Lecture 4 - Differentiation 5 0, 0. Differentiability and Continuity Basic rules of differentiation Differentiability ⇒ Continuity Basic rules of differentiation If f is differentiable at x0 , then f is continuous at x0 . Suppose f and g are differentiable at a point x and c is a constant. Then f x − f x0 x 0, we have ′ cf′ x ; f x − f x0 · x − x0 . x − x0 Then lim f x − f x0 · x − x0 x − x0 lim lim f x − f x0 x →x0 f x − f x0 x − x0 x →x0 x →x0 f ′ x0 · 0 Hence, lim f x x →x0 · lim x − x0 1 cf 2 f g ′ x f′ x 3 Proof. Suppose f is differentiable at x0 . For any f −g ′ x f ′ x − g′ x ; 4 fg 5 x →x0 ′ f g x g′ x ; f′ x g x x ′ f x g′ x (Product Rule); f ′ x g x − f x g′ x g2 x x (Quotient Rule). 0. f x0 . Thus, f is continuous at x0 . AMA140/150 Calculus Lecture 4 - Differentiation Derivatives of some common functions Derivatives of some common functions d 1 C 0 for any constant C ; dx dr 2 x r x r −1 for any nonzero real number r ; dx d 3 sin x cos x ; dx d 4 cos x − sin x ; dx d 5 tan x sec2 x ; dx d 6 cot x − cot2 x ; dx d 7 sec x sec x tan x ; dx d 8 csc x − csc x cot x . dx AMA140/150 Calculus Lecture 4 - Differentiation AMA140/150 Calculus Lecture 4 - Differentiation Derivatives of some common functions Example 4.5 Find the derivatives of the following functions. √ 1 3 x− x 1 fx 5x 2 2 gx 3 sin x x3 4 x 3 sin x hx kx x 4 sin x cos x 10x 3 √ 2x Solution. 1 f′ x 2 g′ x 3 h′ x 4 k′ x 5 2x 3 1 1 −2 x 2 3x 2 sin x − −x −2 x 3 cos x . cos x x 3 − sin x 3x 2 x6 x cos x − 3 sin x . x4 4x 3 sin x cos x 4x 3 sin x cos x 1 . x2 x 3 cos x − 3x 2 sin x x6 x 4 cos x cos x x 4 sin x − sin x x 4 cos2 x − x 4 sin2 x . AMA140/150 Calculus Lecture 4 - Differentiation Chain Rule Chain Rule Chain Rule Suppose h is differentiable at x0 and g is differentiable at h x0 . Let f Then f is differentiable at x0 and ′ ′ f x0 g h x0 g ◦ h. g hx x →x0 x →x0 lim x →x0 g h x0 h x − h x0 · x − x0 · lim x →x0 h x − h x0 x − x0 cos3 x kx cos sin x 2 x Solution. 2 f′ x 3x g′ x cos 2x · 3 h′ x 3 cos x 4 k′ x sin sin x 2 1 2 · d x dx 1 d 2x dx · 1 2 2 cos 2x d cos x dx · 3x −3 cos2 x sin x 1 d sin x 2 dx sin sin x 2 1 cos x 2 · −2x sin sin x 2 AMA140/150 Calculus d2 x dx cos x 2 Lecture 4 - Differentiation Derivative of inverse function Example 4.7 Find the derivatives of the following functions. ′ Suppose f is differentiable at x0 with f x0 y0 f x0 and f −1 ′ y0 0. Then f −1 is differentiable at 1 1 . f ′ x0 d sin−1 y dy 2 d cos−1 y dy Solution. Let f x sin x . Then f −1 y derivative of inverse function, Sketch of proof. Notice that f −1 ◦ f x d sin−1 y dy x. Differentiating both sides of the equation and applying Chain rule, d f −1 ◦ f x dx x x0 d x dx · f ′ x0 y0 · f ′ x0 f −1 ′ d tan−1 y dy sin−1 y . By the theorem of 1 f′ x y 1 cos x x x0 1 3 1 1 2 1 − sin x 1 − y2 1 f x0 ′ hx 4 sin 2x Lecture 4 - Differentiation Derivative of inverse function f −1 3 · h x0 . Derivative of inverse function ′ 3 ′ AMA140/150 Calculus f −1 1 2 − g h x0 x − x0 g h x − g h x0 h x − h x0 ′ gx x 1 g h x − g h x0 h x − h x0 lim fx 2 · h x0 . f x − f x0 lim x →x0 x − x0 lim 1 ′ Sketch of proof. Notice that h is differentiable at x0 and hence h is continuous at x0 . Therefore, h x → h x0 as x → x0 . Then f ′ x0 Example 4.6 Find the derivatives of the following functions. f −1 ′ AMA140/150 Calculus y0 1 . f ′ x0 Lecture 4 - Differentiation 1 d sin−1 y dy 1 1 − y2 2 d cos−1 y dy −1 3 1 − y2 AMA140/150 Calculus Lecture 4 - Differentiation d tan−1 y dy 1 1 y2 . Logarithmic function Properties of Logarithmic function Logarithmic function The logarithmic function f x ln x is defined as follows. 1 , t If x ≥ 1, then f x is the area enclosed by y t 1, and t x ; Properties of Logarithmic function the t -axis, the lines The logarithmic function ln x defined on 0, ∞ has the following properties. 1 the Notice that 1 ln 1 3 ln x < 0 if 0 < x < 1. ln a 2 a ln b ln b ; ln a − ln b ; ln an n ln a for all integer n. (Indeed, it holds for all real number n.) ln x > 0 if x > 1; 2 ln ab 3 If 0 < x < 1, the f x is the negative of the area enclosed by y t -axis, the lines t 1, and t x . 1 , t Derivative of Logarithmic function 0; 1 ln x is differentiable for all x ∈ 0, ∞ and d ln x dx 2 AMA140/150 Calculus ln x is a continuous function on 0, ∞ . Lecture 4 - Differentiation AMA140/150 Calculus Exponential function 1 . x Lecture 4 - Differentiation Properties of exponential function Exponential function Since the logarithmic function ln 0, ∞ → is one-to-one and onto inverse function exists. The inverse function is gx ex for all x ∈ , the . This function is called the exponential function. By the definition of inverse function, we have e ln x x for all x ∈ 0, ∞ , Properties of exponential function The exponential function satisfies the following properties. 1 e0 2 ea 3 The exponential function is differentiable everywhere and 1; b e a · e b for any real numbers a and b ; and ln e x x for all x ∈ d ex dx . Notice that e e 1 2.71828182845905 . . . . Sketch of proof of (3). Since the exponential function is the inverse of the e y . Applying logarithmic function, let y f x ln x . Then x f −1 y the result for derivative of inverse function on f , dy e dy AMA140/150 Calculus ex . Lecture 4 - Differentiation f −1 ′ y AMA140/150 Calculus 1 f′ x 1 1/x x ey . Lecture 4 - Differentiation Logarithmic function and exponential function Differentiation of parametric function Differentiation of parametric function Example 4.8 Find the derivatives of the following functions. 1 fx e sin x 2 gx ln x 2 3 ln ln cos x 2 hx Suppose x ft f′ x 2 g′ x 3 h′ x x2 dx dt provided that 2x − sin x . x 2 cos x 1 · 2x − sin x cos x Example 4.9 Find −2x sin x 2 . cos x 2 ln cos x 2 1 1 · · − sin x 2 · 2x ln cos x 2 cos x 2 dy dy and dx dx dx dt 3t 2 and 1 implies t Lecture 4 - Differentiation ft, y Example 4.10 Find gt, dy dx and dy dt 2t dy dx 2 dx provided that dt d dt dy dx dx dt 3x 2 x ,y dy if x 3 dx sin x cos x dy dx cos x y y cos x y − 3y 2 − 2y cos x , dx dt 3t 2 , dy dx − dy dt d dt 2 3t 2 ⇒ AMA140/150 Calculus 1,1 2 . 3 Lecture 4 - Differentiation y y3 y 2 cos x . 3y 2 dy dx 2y dy dx and t 2 for t > 0. d 2y dx 2 dy dx −2/3t 2 3t 2 dy dx 2 . 3t − Lecture 4 - Differentiation cos x y 2 − sin x dy dx −3x 2 − y 2 sin x − cos x −3x 2 − y 2 sin x − cos x y . cos x y − 3y 2 − 2y cos x In particular, 2t , Then 2 . 3t dy dx 2 t 3 and y 2t 3t 2 1. Then 0. dy Example 4.9 (cont.) Find if x dx 2 Solution. Notice that dy dx ⇒ Solution. Differentiate both sides by x , are differentiable functions of t . Then 2 t 2 for t > 0. 1,1 Implicit differentiation Differentiation of parametric function Suppose t 3 and y if x x ,y AMA140/150 Calculus Differentiation of parametric function , 0. dy dx AMA140/150 Calculus gt Solution. Now x , y x dy dt dx dt dy dx e sin x · cos x . y are differentiable functions of t . Then cos x Solution. 1 and x ,y 0,0 0−0−1 1−0−0 −1. 2 . 9t 4 AMA140/150 Calculus Lecture 4 - Differentiation y ...
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https://www.lmfdb.org/EllipticCurve/Q/229320cm1/ | 1,627,157,188,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150308.48/warc/CC-MAIN-20210724191957-20210724221957-00173.warc.gz | 918,876,744 | 27,508 | Properties
Label 229320cm1 Conductor $229320$ Discriminant $-3.173\times 10^{15}$ j-invariant $$\frac{1354752}{1625}$$ CM no Rank $0$ Torsion structure trivial
Related objects
Show commands: Magma / Pari/GP / SageMath
Minimal Weierstrass equation
sage: E = EllipticCurve([0, 0, 0, 28812, -1949612])
gp: E = ellinit([0, 0, 0, 28812, -1949612])
magma: E := EllipticCurve([0, 0, 0, 28812, -1949612]);
$$y^2=x^3+28812x-1949612$$
trivial
Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
None
Invariants
sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); Conductor: $$229320$$ = $2^{3} \cdot 3^{2} \cdot 5 \cdot 7^{2} \cdot 13$ sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); Discriminant: $-3172761996768000$ = $-1 \cdot 2^{8} \cdot 3^{3} \cdot 5^{3} \cdot 7^{10} \cdot 13$ sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); j-invariant: $$\frac{1354752}{1625}$$ = $2^{10} \cdot 3^{3} \cdot 5^{-3} \cdot 7^{2} \cdot 13^{-1}$ Endomorphism ring: $\Z$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) Sato-Tate group: $\mathrm{SU}(2)$ Faltings height: $1.6606514427245914191805443861\dots$ Stable Faltings height: $-0.69769154069516063086754895697\dots$
BSD invariants
sage: E.rank() magma: Rank(E); Analytic rank: $0$ sage: E.regulator() magma: Regulator(E); Regulator: $1$ sage: E.period_lattice().omega() gp: E.omega[1] magma: RealPeriod(E); Real period: $0.24075710547248482950820132244\dots$ sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); Tamagawa product: $4$ = $2\cdot2\cdot1\cdot1\cdot1$ sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); Torsion order: $1$ sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Analytic order of Ш: $1$ (exact) sage: r = E.rank(); sage: E.lseries().dokchitser().derivative(1,r)/r.factorial() gp: ar = ellanalyticrank(E); gp: ar[2]/factorial(ar[1]) magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12); Special value: $L(E,1)$ ≈ $0.96302842188993931803280528977986223720$
Modular invariants
Modular form 229320.2.a.w
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
magma: ModularForm(E);
$$q - q^{5} - 3q^{11} + q^{13} - 2q^{17} + 4q^{19} + O(q^{20})$$
sage: E.modular_degree() magma: ModularDegree(E); Modular degree: 919296 $\Gamma_0(N)$-optimal: yes Manin constant: 1
Local data
This elliptic curve is not semistable. There are 5 primes of bad reduction:
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
prime Tamagawa number Kodaira symbol Reduction type Root number ord($N$) ord($\Delta$) ord$(j)_{-}$
$2$ $2$ $I_1^{*}$ Additive -1 3 8 0
$3$ $2$ $III$ Additive 1 2 3 0
$5$ $1$ $I_{3}$ Non-split multiplicative 1 1 3 3
$7$ $1$ $II^{*}$ Additive -1 2 10 0
$13$ $1$ $I_{1}$ Split multiplicative -1 1 1 1
Galois representations
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
The $\ell$-adic Galois representation has maximal image $\GL(2,\Z_\ell)$ for all primes $\ell$.
$p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(5,20) if E.conductor().valuation(p)<2]
All $p$-adic regulators are identically $1$ since the rank is $0$.
No Iwasawa invariant data is available for this curve.
Isogenies
This curve has no rational isogenies. Its isogeny class 229320cm consists of this curve only.
Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ (which is trivial) are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $3$ 3.1.38220.2 $$\Z/2\Z$$ Not in database $6$ 6.0.284849838000.1 $$\Z/2\Z \times \Z/2\Z$$ Not in database $8$ Deg 8 $$\Z/3\Z$$ Not in database $12$ Deg 12 $$\Z/4\Z$$ Not in database
We only show fields where the torsion growth is primitive. For fields not in the database, click on the degree shown to reveal the defining polynomial. | 1,571 | 4,186 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-31 | latest | en | 0.202746 |
https://www.jiskha.com/display.cgi?id=1305987385 | 1,516,342,471,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887746.35/warc/CC-MAIN-20180119045937-20180119065937-00630.warc.gz | 899,816,844 | 4,027 | # statistics
posted by .
From a bag containing 6 black and 4 white balls, 2 balls are selected,one after the other,without replacement. Find (a) the probability that they are both black (b)the probability that they are both the same colour, (c) probability that they are different colours.
• statistics -
total balls = 10
without replacement, so you'll have one less ball each time you draw
I'll do the first one for you. You can figure out how to the rest.
P(both black) = (6/10)(5/9)
Explanation: During the first draw, there are 6 black balls and a total of 10 balls. During the second draw (because there was no replacement), there are now 5 black balls and a total of 9 balls.
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More Similar Questions | 756 | 3,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-05 | latest | en | 0.934772 |
https://discover.hubpages.com/technology/Using-Java-Big-Decimal-to-compute-square-roots-of-large-numbers | 1,653,588,650,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00198.warc.gz | 254,950,948 | 58,173 | # Using Java Big Decimal to compute square roots of large numbers
The standard number classes in Java are adequate for most purposes, but occasionally a calculation cannot safely be carried out with the standard classes: I once worked for a bank that insisted calculations be carried out using infinite precision arithmetic. Java caters for infinite precision arithmetic- the BigInteger and BigDecimal classes give infinite precision arithmetic in return for a large performance overhead, inelegant code and the need to worry about whether an operation, such as finding the square root of a number, will terminate or run for ever. Here two iterative algorithms are compared with each other and with the the Java Math.sqrt() method.
The code samples supplied here come with no guarantee whatsoever. Use at your own risk
## The Test Problem
One of the problems on Project Euler involved finding the square root of
59370809220861047609922354854410565869030692834007857210054627595408757190
This is 74 decimal digits long and the root is approximately
7705245565253650474334883595061662024.24426667
The implementation of each of the algorithms described here was tested on small numbers the exact value of which could be calculated by hand. As expected the algorithms did not give exact results, for example the square root of 196 was found to be 13.999999........ with some other digits after the ellipse. Apart from this the results were correct.
## Big Integer and Big Decimal
These classes provide infinite precision arithmetic and useful methods, such as isProbablePrime() and gcd() not offered by the standard types: long, integer, double etc. Using them for simple operations is cumbersome, and comparing two BigNumbers means a lot of typing. Sometimes just constructing one seems awkward. One reason for this is that the standard operators have not been overloaded.
A key class, MathContext, restricts the number of decimal places to which a calculation is carried out and is essential when a calculation ( √2 for example) will never end. For this investigation the precision was set at one hundred decimal placed.
## The "Babylonian" Algorithm
Wikipedia discusses several algorithms for calculating square roots most of which are interesting but of limited use or of historical interest only. A reasonably fast algorithm that would give accurate results was needed. Wikipedia and various other places produced two possibilities, both iterative methods (Iterative methods have the advantage of being intuitive and easy to understand ): The Babylonian or Heron method, which is known to converge rapidly and a binary search method that should be logarithmic in the size of the input. Other algorithms such as Chebyshev polynomials methods were left for future investigations. Further aspects like the scale were not used in this first pass.
The Babylonian method, a case of Newton's method starts with a guess at the square root then refines it iteratively. It is a case of Newton's method.
Let anumber is the number whose root is desired, guess the current estimate of the square root and precision the desired precision. The algorithm is:
guess = initial estimate
nextguess = 1000-0*
while( | nextguess -guess | > desired_precision)
Scroll to Continue
{
nextguess = 0.5*(guess + anumber/guess)
}
return nextguess
This converges rapidly. Normally only a few iterations are needed to reach a high precision ( around a dozen for a simple case, according to Wikipedia).
## A binary Chop Algorithm
The alternative was a binary search
1. delta = anumber/2
2. guess = delta/2
3. desiredprecision = 1.0E-100 ( change to taste)
4. guesssquared = guess*guess
5. if(guesssquared > anumber) guess = guess -delta
6. if(guesssquared < anumber) guess = guess +delta
7. if( abs(guesssquared - guess*guess) < desiredprecision ) return guess
8. delta = delta/2
This needs a multiplication where the Babylonian method has a division and a multiplication. It is also theoretically more elegant. So the guess was these two forces would balance out. This turned out to be wrong.
## Timings
Big Decimal Babylonian method precision 10**-100 0.51 milliseconds (averaged over 1000 iterations)
Big Decimal iterative method precision 10**-100 37 milliseconds (averaged over 1000 iterations)
By comparison the standard Java square root algorithm for a double took around 100 nanoseconds and a version of the Babylonian Algorithm for a double that did not use infinite precision arithmetic took around 2 microseconds.
## Conclusions
Despite any weaknesses in the method of measurement it seems clear that BigDecimal calculations have an enormous overhead, compared with the limited precision of the square root calculation of the Java Math package. The only surprise is the magnitude of the overhead
It is also apparent that for high precision calculations the “Babylonian” method converges much faster, for the same precision, than the iterative methods though this advantage is reduced as the precision decreases. The same effect is seen for the fixed precision calculations of the Java Math package.
BigDecimal and Big Integer calculations should only be used when unavoidable or when speed is not critical but precision is critical.
The original goal of deciding whether the “Babylonian” or “iterative” algorithm is faster was achieved. The “Babylonian” algorithm won decisively. It seems unlikely changing implementation details will alter this. | 1,127 | 5,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-21 | latest | en | 0.907465 |
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# APR Calculator for Adjustable Rate Mortgages
Use this calculator to determine the Annual Percentage Rate (APR) of your Adjustable Rate Mortgage (ARM). Knowing your APR can help you compare different ARMs with different fees and terms.
## APR Calculator for Adjustable Rate Mortgages Definitions
Adjustable Rate Mortgage (ARM)
This calculator shows a fully amortizing ARM which is the most common type of ARM. The monthly payment is calculated to payoff the entire mortgage balance at the end of the term. The term is typically 30 years. After any fixed interest rate period has passed, the interest rate and payment adjusts at the frequency specified. A Fully Amortizing ARM will also have a maximum rate that it will not exceed. Below is a list of the most common types of Fully Amortizing ARMs.
Common Adjustable Rate Mortgages
ARM TypeMonths Fixed
10/1 ARMFixed for 120 months, adjusts annually for the remaining term of the loan.
7/1 ARMFixed for 84 months, adjusts annually for the remaining term of the loan.
5/1 ARMFixed for 60 months, adjusts annually for the remaining term of the loan.
3/1 ARMFixed for 36 months, adjusts annually for the remaining term of the loan.
Mortgage amount
Original or expected balance for your mortgage.
Starting interest rate
Initial annual interest rate for this mortgage.
Term in years
The number of years over which you will repay this loan. The most common mortgage terms are 15 years and 30 years.
Current index
The current interest rate of the index used to calculate the interest rate on this Adjustable Rate mortgage. The current index rate plus the margin on that rate produces the Fully Indexed Rate that is used to calculate the APR for this mortgage.
Margin
The interest rate percentage above the index, or the 'margin', used to calculate the Fully Indexed Rate.
Starting monthly payment
Monthly principal and interest payment (PI) based on your beginning balance and starting interest rate.
Loan origination percent
The percent of your loan charged as a loan origination fee. For example, a 1% fee on a \$120,000 loan would cost \$1,200.
Discount points
Total number of "points" purchased to reduce your mortgage's interest rate. Each "point" costs 1% of your loan amount. As long as the points paid are not a broker's commission, they are considered tax-deductible in the year that they were paid.
Other fees
Any other fees that should be included in the APR calculation. These fees can vary by lender, but at a minimum usually includes prepaid interest.
Annual Percentage Rate (APR)
A standard calculation used by lenders. It is designed to help borrowers compare different loan options. For example, a loan with a lower stated interest rate may be a bad value if its fees are too high. Likewise, a loan with a higher stated rate with very low fees could be an exceptional value. APR calculations incorporate these fees into a single rate. You can then compare loans with different fees, rates or different terms.
Interest rate cap
This is the highest interest rate allowed by your mortgage. Your actual interest rate will not be adjusted above this rate.
Months before first adjustment
This is the number of months that the interest rate is fixed. After this period, the interest rate will be subject to rate adjustments. If you enter zero in this field, we assume that the rate will begin making adjustments after initial period of time between adjustments has passed. If any number other than zero is entered, the first adjustment will take place at that time, and adjustments will happen at the frequency entered in the 'months between adjustments' field.
Expected adjustment
The amount you believe that your mortgage's interest rate will change. This amount will be added to or subtracted from your interest rate.
Months between adjustments
The number of payment periods between potential adjustments to your interest rate. The most common is 12 months, which means your payment could change at most once per year.
Total payments
Total of all monthly payments over the full term of the mortgage. This total payment amount assumes that there are no prepayments of principal.
Total interest
Total of all interest paid over the full term of the mortgage. This total interest amount assumes that there are no prepayments of principal. | 876 | 4,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-50 | latest | en | 0.924175 |
https://www.calqlata.com/proddetail.asp?prod=00034 | 1,726,839,215,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00758.warc.gz | 639,064,832 | 6,129 | Price: £26.00 (£26.00 Inc. VAT)
## Subject
A shock load is a dynamically applied force that converts kinetic energy into deformation energy in the impacting mass and the impacted body at the time of impact.
If there is no deformation at all in both the impacting mass and the impacted body, the resultant force would be infinite irrespective of the magnitude of the impacting mass and its velocity at the moment of impact. The deformation generated in the mass together with the deformation in the body will define the force of the impact; i.e. the greater the deformation the lower the force and vice-versa.
## Calculator
Shock calculates the energy, deformation, stress, natural frequency, etc. due to a mass travelling at a specified velocity and impacting a beam or plate with various end and edge support conditions or an undefined body.
Shock also includes a general energy conversion calculation similar to that included in CalQlata’s Engineering Principles calculator, inasmuch as you enter all known values and Shock will calculate the unknown.
For help using this calculator see Technical Help
## Shock Load Calculator - Options
Shock includes the following calculation options:
‘Beams - Point Load’ defines the conditions for a concentrated mass impacting a beam at its weakest point. The end support conditions available for this calculation option are as follows: Fixed-Fixed; Guided-Fixed; Simple-Fixed; Free-Fixed; Guided-Simple and Simple-Simple.
You enter: and the shock load calculator will provide:
• Impact velocity
• Mass (object)
• Gravitational acceleration
• Mass (beam)
• Depth (beam)
• Length (beam)
• Young's modulus (beam)
• Second moment of area (beam)
• Force
• Deflection
• Natural frequency
• Natural period
• Impact duration
• Factor
• Energy dissipated
• Maximum beam stress
• Distance to y (xʸ)
• Distance to σ (xˢ)
‘Beams - Distributed Load’ defines the conditions for an evenly distributed mass impacting a beam all along its length. The end support conditions available for this calculation option are as follows: Fixed-Fixed; Guided-Fixed; Simple-Fixed; Free-Fixed; Guided-Simple and Simple-Simple.
You enter: and the shock load calculator will provide:
• Impact velocity
• Mass per unit length (object)
• Gravitational acceleration
• Mass (beam)
• Depth (beam)
• Length (beam)
• Young's modulus (beam)
• Second moment of area (beam)
• Force
• Deflection
• Natural frequency
• Natural period
• Impact duration
• Factor
• Energy dissipated
• Maximum beam stress
• Distance to y (xʸ)
• Distance to σ (xˢ)
‘Plates - Point Load’ defines the conditions for a concentrated mass impacting a plate at its weakest point (the centre). The edge support conditions available for this calculation option are as follows: Fixed and Simple.
You enter: and the shock load calculator will provide:
• Impact velocity
• Mass (object)
• Gravitational acceleration
• Density (plate)
• Thickness (plate)
• Diameter (plate)
• Young's modulus (plate)
• Poisson's ratio (plate)
• Force
• Deflection
• Natural frequency
• Natural period
• Impact duration
• Factor
• Energy dissipated
• Maximum plate stress
‘Plates - Distributed Load’ defines the conditions for an evenly distributed mass impacting a plate all over its surface. The edge support conditions available for this calculation option are as follows: Fixed and Simple.
You enter: and the shock load calculator will provide:
• Impact velocity
• Mass per unit area (object)
• Gravitational acceleration
• Density (plate)
• Thickness (plate)
• Diameter (plate)
• Young's modulus (plate)
• Poisson's ratio (plate)
• Force
• Deflection
• Natural frequency
• Natural period
• Impact duration
• Factor
• Energy dissipated
• Maximum plate stress
#### NON-ELASTIC DEFORMATION
‘Non-Elastic Deformation’ provides the unknown value in a general kinetic to deformation energy conversion calculation.
You enter: and the shock load calculator will provide:
• Impact velocity
• Mass (object)
• Impact deflection (object)
• Impact force (object)
• Impact velocity
• Mass (object)
• Impact deflection (object)
• Impact force (object)
• Energy dissipated
Note: CalQlata’s Engineering Principles calculator includes stiffness (spring constant) of the impacted body in this type of calculation
We accept the following payment methods | 950 | 4,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-38 | latest | en | 0.837703 |
https://cs.stackexchange.com/questions/10485/flaw-in-my-np-conp-proof?noredirect=1 | 1,628,058,501,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154796.71/warc/CC-MAIN-20210804045226-20210804075226-00044.warc.gz | 182,616,686 | 46,273 | # Flaw in my NP = CoNP Proof?
I have this very simple "proof" for NP = CoNP and I think I did something wrongly somewhere, but I cannot find what is wrong. Can someone help me out?
Let A be some problem in NP, and let M be the decider for A. Let B be the complement, i.e. B is in CoNP. Since M is a decider, you can use it to decide B as well (just flip the answer). Doesn't that mean that we solve both NP and CoNP problems with the same M?
To put it more concretely.
Let A be some NP-complete problem, and let M be decider for A. Consider any problem B in CoNP. We consider its complement not-B, which is in NP, and then get a polynomial reduction to A. Then we run our decider M and flip our answer. We thus obtain a decider for B. This implies B is in NP as well.
May I know what is wrong with my reasoning?
• As the answers below explain at length, you are not using the notion of "decider" correctly. Problems in coNP are not those with a "flipped NP decider". There is an important assymmetry in NP problems between accepting an input ("there are non-deterministic choices which lead to accepting") and rejecting it ("all non-deterministic choices lead to rejecting"). Your argument presupposes that for NP rejecting a string means ("there are a non-determinstic choices which lead to rejection"), and that is the flaw. In other words, you got your quantifiers mixed up. – Andrej Bauer Mar 13 '13 at 14:32
• You might find the answers to this question enlightening. – Raphael Mar 13 '13 at 14:38
• @Raphael Surprisingly, that question doesn't mention co-NP at all! (Though I agree that it's a useful read for somebody who's unsure about this kind of thing.) – David Richerby Jan 31 '18 at 15:33
• @DavidRicherby Since the answer is, basically, "use the definition of NP instead of flawed intuition", I would hope so! – Raphael Jan 31 '18 at 15:44
• Rule of thumb: the "flip final states" construction only works for determinstic models. Study how it fails for NFA to understand why. See also here and here. – Raphael Jan 31 '18 at 16:34
There are two possible bugs in this proof:
1. When you say "decider" - you mean a deterministic TM. In this case, the best translation (to our knowledge) from an NP machine to a deterministic machine may yield a machine that runs in exponential time, so after complementing you will have a decider for the complement in exponential time, proving that $co-NP\subseteq EXP$ (or, after some optimization, $co-NP\subseteq PSPACE$).
2. When you say "decider" you mean a nondeterministic TM. In this case, flipping the answer will not necessarily complement the language. Indeed, the language of the flipped machine will be all the words for which there exists a rejecting run of $M$ on $w$
• I'm not sure why that matters. My definition of a decider is that I accept if input is in L and reject if input is not in L. This decider can be deterministic or non-deterministic. However, I say that L is in NP, and therefore if I am using a non-deterministic TM then I will take polynomial time. Also, may I know why flipping the bit does not necessarily complement the language. To my knowledge CoNP = {L | not L \in NP}. Therefore, if I flip the bit I should obtain the answer? – simpleton Mar 12 '13 at 10:03
• Let $L\in NP$. Consider a nondeterministic TM that works as follows - in one run, it always outputs "reject". In other runs, it recognizes $L$ in polynomial time (possible since $L\in NP$). Consider what happens if you flip the bit - the rejecting run becomes accepting for every input, so the complemented machine recognizes $\Sigma^*$ - which is not the complement unless $L=\emptyset$. I suggest you review closely the definitions of nondeterminism to understand this fully. – Shaull Mar 12 '13 at 10:10
• I do not mean that I flip the bit of every single computation path. What I mean is that if my TM accepts, then this means that there exists a computation path that reaches an accept state. This means L is in NP, which means the complement is in coNP. If my TM rejects, then this means the every computational path rejects. This means that the complement is in NP, which means L is in CoNP. – simpleton Mar 12 '13 at 10:19
• @simpleton: You know that an NTM does not have access to all paths at once, only one path? You think as someone who deterministically analyzes the behaviour of the NTM from the outside. – frafl Mar 12 '13 at 11:43
• I think the OP could benefit from looking up the definition of NP more carefully. – MCH Mar 12 '13 at 17:09
Here's another way of looking at the point that Shaull makes with respect to "deciders".
A problem is in NP if and only if there is an algorithm $V: \{0,1\}^n \times \{0,1\}^{\mathrm{poly}(n)} \to \{0,1\}$ such that
• for every YES instance $x \in \{0,1\}^n$, there is a certificate $p \in \{0,1\}^{\mathrm{poly}(n)}$ such that $V(x,p) = 1$; and
• for every NO instance $x \in \{0,1\}^n$, we have $V(x,p) = 0$ for all $p \in \{0,1\}^{\mathrm{poly}(n)}$.
These are usually described as the completeness and soundness conditions for the NP verification algorithm: the "completeness" condition says that every YES instance has a certificate, and the "soundness" condition says that the algorithm is never fooled by a NO instance. For coNP it's the other way around: there is a verifier algorithm which will accept at least one certificate for any NO instance, but which can never be fooled by a YES instance.
If you want to show that NPcoNP, you have to show that every NP problem has a coNP-type verifier, which can certify NO instances instead of YES instances. You cannot do this with a nondeterministic Turing machine: there's no way that we know of, for instance, to efficiently map instances of SAT to one another, in such a way that all unsatisfiable formulae are mapped to satisfiable ones, and vice versa. (Negating the output ofr the formula isn't enough, for example: a formula which is satisfiable but not a tautology would just get mapped to a different formula which was satisfiable but not a tautology, when we would require an unsatisfiable formula instead.) There is simply no way that we know of to 'fool' a nondeterministic machine into detecting anything like all of its paths being rejecting paths.
You might ask: "Doesn't the nondeterministic Turing machine know what result it gets?" The answer would be no, it doesn't. The working of the non-deterministic machine doesn't give it access to any information about more than one computational path at once: you might think of it working in many paths in parallel, but within each path it only knows about that one path. If you try to equip it with the ability to "realize" whether or not there are any solutions as some point, you are instead describing a machine with an NP oracle, which is more (potentially!) powerful than a simple nondeterministic Turing machine.
• For instance, if you equip a (deterministic) Turing machine with an NP oracle, then the problems which can be solved in polynomial time on that machine is called $\Delta_2^{\mathrm P}$, which is often written $\mathsf{P}^{\mathsf{NP}}$. The "oracle" allows the machine to simply recieve answers to NP-complete problems in a single step, and so $\mathsf{P}^{\mathsf{NP}}$ obviously contains P; and because you can negate answers, it also obviously contains coNP. But we don't know whether the reverse containments hold, exactly because we don't know how to trick nondeterministic Turing machines into detecting NO answers.
• What's more, if you give a nondeterministic Turing machine the ability to come to a realisation about the outcome of a problem in NP, the problems which that machine can solve in polynomial time is called $\Sigma_2^{\mathrm P}$, or $\mathsf{NP}^{\mathsf{NP}}$, and this is widely believed to be strictly larger than the class $\mathsf{P}^{\mathsf{NP}}$. This also contains both NP and coNP — but like NP, it isn't known to be closed under complements: the nondeterministic oracle machine might be able to know when a problem in NP has a NO answer because of the oracle, but it would still be stuck to operating within one of its own (quite powerful) computational branches, so that it would not be able to tell if all of its own computational branches were rejecting.
If you keep on supplying the machine with more powerful oracles for solving problems in $\mathsf{NP}$, $\mathsf{NP}^{\mathsf{NP}}$, and so forth, you end up defining the classes of the polynomial hierarchy, which are thought to be all distinct from one another from the first level onwards.
So, no, there's no machine (deterministic or otherwise) which can simply 'decide' that a problem is a YES or NO instance efficiently, unless we use oracles; but even with such an oracle, we end up with a machine which is (probably) more powerful than either NP or coNP, not one which shows that they're equal.
• Hi, thank you to you and Shauli for the comments. Are you saying that a NTM can recognize an NP language in polytime, but cannot decid a NP language in polytime? I think that is what I am assuming when I say that I have a decider for NP problems. – simpleton Mar 12 '13 at 19:54
• Oh, I think I sort of get what you mean. Are you saying that I am using oracle reduction, and that actually shows that the problem is in $P^{NP}$ (which is true because $NP \subset P^{NP}$ and $CoNP \subset P^{NP}$)? The oracle reduction shows that UNSAT is NP-hard, but I still need to show that $UNSAT \in NP$, and I cannot show that? – simpleton Mar 12 '13 at 20:25
• NP-hardness is defined with many-one reductions, not oracle reductions. – Yuval Filmus Mar 13 '13 at 5:57
Your reasoning implies that RE=coRE, but this is provably false. You could try to figure out a proof of that and then see where your reduction fails.
Recall that RE is the complexity class of recursively enumerable languages, which are languages of the form $\{ x : P \text{ halts on input } x \}$. You can also think of it in non-deterministic terms: RE is the class of languages of the form $\{ x : (x,w) \in L' \text{ for some } w \}$, where $L'$ is recursive (computable).
Here is a proof that both definitions match. Suppose first $L = \{ x : p \text{ halts on input } x \}$. Let $L' = \{ (x,w) : p \text{ halts on input } x \text{ in } w \text{ steps} \}$. The language $L'$ is recursive and $L = \{ x : (x,w) \in L' \text{ for some } w \}$.
For the other direction, let $L = \{ x : (x,w) \in L' \text{ for some } w \}$, where $L'$ is recursive, say computed by the program $P(x,w)$. We construct a new program $Q(x)$ which enumerates all possible $w$ and runs $P(x,w)$ on all $w$, in order. If $P(x,w)$ ever accepts for some $w$, then $Q$ halts. It's not hard to check that $L = \{ x : Q \text{ halts on input } x \}$.
For your convenience, here are outlines for a proof that RE is different from coRE. The language $L = \{(P,x) : P \text{ halts on input x}\}$ is clearly recursively enumerable: a program for it simply runs $P$ on $x$. Suppose there was a program $H$ such that $H(P,x)$ halts if and only if $(P,x) \notin L$. We define a new program $G$ by $G(x)=H(x,x)$. Is $(G,G) \in L$? If so, then $G$ halts on $G$, so $H$ halts on $(G,G)$, so $(G,G) \notin L$. If $(G,G) \notin L$, then $G$ doesn't halt on $G$, so $H$ doesn't halt on $(G,G)$, so $(G,G) \in L$. This contradiction shows that $H$ cannot exist.
Now try to run your proof in this case and see what goes wrong. In more detail, try to construct the program $H$ using your recipe, and follow the proof - at some point something wouldn't be quite right.
Here's a TL;DR version; I've also posted a longer answer to a similar question.
Suppose we have a language $A$ that's decided in polynomial time by nondeterministic Turing machine $M$. The point is that $x\in A$ iff there $M$ has some accepting path on input $x$. However, since $M$ is nondeterministic, there may also be rejecting paths when it runs on $x$. If you just reverse the accepting and rejecting states, you'll go from a machine that had some accepting paths and some rejecting ones to a machine that has some rejecting paths and some accepting ones. In other words, it still has accepting paths, so it still accepts. Flipping the accept and reject states of a nondeterministic machine does not, in general, cause you to accept the complement language.
It is this asymmetry of definition (accept if any path accepts; reject only if all paths reject) that makes the NP vs co-NP problem difficult.
I actually agree that your nondeterministic machine M can decide whether a given input string is in B. However, it "decides" slightly differently than the way it decides if a given input is in A. In the latter case, it does so by (nondeterministically) finding an accepting state. In the former case, it does so by failing to find any accepting states. Why does this difference matter? Let's see:
## When asking M "Is the string in the language A?"
M reaches an accept state. This, you can prove (see, for example, the Sipser book theorem 7.20) implies that there is a deterministic machine that verifies the string is in A in polynomial time
## When asking M "Is the string in the language B?"
M reaches reject states on all branches of its nondeterministic computation. If you think about how the verifier proof above works, you will see that it cannot be accomplished in this situation. This is roughly because the verifier uses the path that M takes through its state space as "proof". In this case, there is no one such path.
## Conclusion:
If you consider the existence of a polynomial time deterministic verifier for a language to be the definition of an NP language (which you should), the existence of M does not prove that B is in NP.
• The existence of $M$ doesn't prove that $B$ is in NP precisely because you have to use this "weird" acceptance criterion to make it "accept" $B$. NP is defined via the conventional nondeterministic acceptance criterion. (The acceptance criterion is slightly weird partly because the question talks about flipping accepting and rejecting states in $M$ but you don't seem to have done that.) – David Richerby Oct 5 '14 at 8:18 | 3,590 | 14,139 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-31 | latest | en | 0.953217 |
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Integer Worksheets
Practice Adding Negative And Positive Integers On A Number Line 5th Through 7th Grades View Shape Math Adding Integers At The Top Of This Worksheet There Are Many Shapes With Positive And Negative Numbers In Them Students Find Pairs Of Congruent Shapes And Add The Numbers Inside Of Them For Example Find The Sum Of The Numbers In The Trapezoids 5th Through 7th Grades View
The Worksheets On This Page Introduce Adding And Subtracting Negative Numbers As Well As Multiplying And Dividing Negative Numbers The Initial Sets Deal With Small Integers Before Moving On To Multi Digit Multiplication And Long Division With Negatives Regardless Of Where You Re At In Your Process Of Learning Negative Numbers These Worksheets Will Give Your Students Plenty Of Practice When They Need To Master This Often Negative Topic
Integers Worksheets Free Printable K5 Learning
Worksheets Math Grade 6 Integers Positive And Negative Whole Numbers These Grade 6 Worksheets Cover Addition Subtraction Multiplication And Division Of Integersegers Are Whole Numbers No Fractional Or Decimal Part And Can Be Negative Or Positive
Negative Numbers Worksheet Teaching Resources
A Negative Numbers Worksheet Covering Adding Subtraction Multiplying Dividing And Worded Problems The Worded Problems Are Differentiated I Print Out Either The Level 1 2 Or 3 Questions Depending On The Class They Are Roughly The Same Questions But With Different Numbers In
Adding Negative Integers Worksheet. The worksheet is an assortment of 4 intriguing pursuits that will enhance your kid's knowledge and abilities. The worksheets are offered in developmentally appropriate versions for kids of different ages. Adding and subtracting integers worksheets in many ranges including a number of choices for parentheses use.
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Author: Nna Kirschner
Have faith. But just because it's possible, doesn't mean it will be easy. Know that whatever life you want, the grades you want, the job you want, the reputation you want, friends you want, that it's possible.
Top | 2,103 | 10,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-40 | latest | en | 0.892492 |
https://socratic.org/questions/given-the-set-3-2-5-1-x-7-for-what-x-would-the-mean-of-the-set-be-3 | 1,702,024,722,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100739.50/warc/CC-MAIN-20231208081124-20231208111124-00410.warc.gz | 613,788,924 | 6,070 | # Given the set: {-3,2,-5,1,x,7} , for what x would the mean of the set be 3?
Dec 30, 2015
$x = 16$
#### Explanation:
By definition, the mean or average is : $\overline{x} = \frac{1}{n} {\sum}_{I = 1}^{n} {x}_{i}$.
Substituting the given values into this formula we get
$3 = \frac{1}{6} \left(- 3 + 2 - 5 + 1 + x + 7\right)$
$\therefore x = 16$. | 141 | 353 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-50 | latest | en | 0.677753 |
https://forums.creativecow.net/docs/forums/post.php?forumid=227&postid=18564&univpostid=18564&pview=t | 1,524,553,478,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946565.64/warc/CC-MAIN-20180424061343-20180424081343-00294.warc.gz | 632,777,763 | 6,599 | FORUMS: list search recent posts
# Delayed start time of an expression
FAQ • VIEW ALL
Delayed start time of an expression on Jul 6, 2011 at 1:14:06 pm
Hi all!
I'm using this well-known expression :)
```freq = 8.0; amplitude = 15; decay = 4; amplitude*Math.sin(freq*time*2*Math.PI)/Math.exp(decay*time);```
I applied it to the rotation property of one layer in my composition and I need it to start 2 seconds later the "0" frame of the composition… but I noticed it starts at the beginning of the composition and not at the beginning of the layer itself… I've been browsing the forum for hours trying to figure out how to do this! I've been trying this way…
```timeToStart = 2; if (time >= timeToStart){ t = time - timeToStart; freq = 8.0; amplitude = 15; decay = 4; amplitude*Math.sin(freq*time*2*Math.PI)/Math.exp(decay*time); }else{ 0; } ```
but it is not working :(
I know another possibility could be using a marker, but I couldnt make it work as well!
Now I'm just looking for the quick&easy way, just writing in the expression in which sec it should start!
I know this question has been asked again and again, but it's my very first day handling expressions and my knowledge about coding is really basic! Hope you can help me..
Thank you!
Chiara
Re: Delayed start time of an expressionon Jul 6, 2011 at 2:06:07 pm
in your second expression you are calculating a new time variable (t), but not using that variable in the calculation.
just change the word 'time' to 't' in the calculation:
amplitude*Math.sin(freq*t*2*Math.PI)/Math.exp(decay*t);
Kevin Camp
Senior Designer
KCPQ, KMYQ & KRCW
Re: Delayed start time of an expressionon Jul 6, 2011 at 2:15:52 pm
Thank you!
Chiara
Re: Delayed start time of an expressionon Feb 1, 2014 at 12:36:53 am
I am dealing with a simple swing expression tied to the x rotation. It starts immediately in the composition, but I'm trying to move it further into the comp. (Specifically, 4:08).
This has to be something really simple, and all over the internet, but apparently am not typing the right question.
I've toyed with some timeToStart commands, but they just end up cutting off the first portion of the animation and leaving everything after the timeToStart.
```offset = 90; swingMaxDistance = 100; swingSpeed = 5; easeTimeSpan = 2.5; easeSpeed = 1; easeVal = ease(time*easeSpeed, 0, easeTimeSpan, swingMaxDistance, 0); Math.sin(time*swingSpeed) * easeVal;``` | 656 | 2,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-17 | latest | en | 0.90493 |
https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_1&diff=prev&oldid=158141 | 1,638,552,974,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00062.warc.gz | 182,656,657 | 10,827 | # Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 1"
## Problem
The equation $ax^2 + 5x = 4,$ where $a$ is some constant, has $x = 1$ as a solution. What is the other solution?
## Solution
Since $x=1$ must be a solution, $a+5=4$ must be true. Therefore, $a = -1$. We plug this back in to the original quadratic to get $5x-x^2=4$. We can solve this quadratic to get $1,4$. We are asked to find the 2nd solution so our answer is $\boxed{4}$
~Grisham
## Solution 2
Plug $x=1$ to get $a=-1$, so $x^2-5x+4=0$, or $(x-4)(x-1)=0$, meaning the other solution is $x=\boxed{4}$ $\linebreak$ ~Geometry285 | 213 | 627 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 15, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-49 | latest | en | 0.883764 |
https://nenuphar.net/?p=152 | 1,656,122,872,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033925.2/warc/CC-MAIN-20220625004242-20220625034242-00619.warc.gz | 456,220,293 | 13,545 | #### How to Play the Lottery Online
The practice of dividing property by lot has been around since ancient times. The Old Testament tells Moses to take a census of the people of Israel and divide the land among them by lot. Roman emperors used lotteries to distribute property and slaves. In ancient Rome, people gathered at restaurants to play the “apophoreta” or “thing to be carried home” game. Today, this activity is still a popular form of entertainment.
Togel is one of the most popular lottery games in Singapore. The game is based on four, three, and two digits. The aim of the game is to predict a winning number combination. People make money by correctly predicting the numbers, but it is not a guaranteed method of winning. Togel players use various approaches and statistics to improve their chances of winning. However, they must also have luck to win. This article will highlight the different methods of predicting lottery numbers.
The first lotteries were held in the Netherlands in the 17th century. The Dutch government used these to collect money for the poor and other public purposes. People loved the games and were happy to contribute to the government’s budget. The oldest lottery in the world was established in 1726 and is still in operation today. The English word lottery is derived from the Dutch word lotus, which means “fate.” There are a number of other types of lotteries, and the first one was known as the Staatsloterij.
Besides the lottery, there are many different types of togel games in the world. The most popular ones in Singapore are Togel, Toto Macau, and Keno. All of these lottery games are played using four-digit numbers. The winning number combination is derived from the digits and is drawn randomly. There are many approaches and statistical analyses used to predict the winning number combination. While these methods have their uses, they are no substitute for luck.
The most important thing to consider when playing the lottery is the safety of the game. You must be careful with your personal information when playing an online lotto. Your identity can be stolen, so it is very important to protect yourself from fraud. In addition to this, you need to be careful when entering the lottery. This is why it is best to play online. You can check out the rules and regulations of the lottery in your country, and choose the ones that are safe for you.
In addition to the lottery, you can also play the lottery online. There are many online lottery services that employ local representatives in different jurisdictions. These agents purchase your lottery tickets on your behalf and will send you an email confirmation once the transaction is complete. When it is time to play the lottery, you can even choose which numbers you want to play. You can also find the details of the winning lottery agent in your state. You can even get information from the government to help you win the lottery. | 599 | 2,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-27 | latest | en | 0.973432 |
https://topic.alibabacloud.com/a/binary-tree-postorder-traversal-leetcode_8_8_31571310.html | 1,709,564,716,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476452.25/warc/CC-MAIN-20240304133241-20240304163241-00319.warc.gz | 554,483,238 | 15,865 | # Binary Tree postorder Traversal--leetcode
Source: Internet
Author: User
Main topic: post-secondary traversal of binary tree
The way to solve the problem: the steps to traverse the binary tree: order to traverse the left subtree of the binary tree, and then traverse the right sub-tree of the binary tree, and access the root node. When a non-recursive implementation is implemented, a stack is used to simulate the traversal process. Since accessing the right subtree after the left dial hand tree has been accessed, the elements in the stack will have to move to access their right subtree, but they cannot be stacked like the first order and the middle order traversal, because the root node is last accessed. So when does the stack come out? We need a pointer to the pre to record the node of the previous visit. If the pre is the right subtree of the root node, then the right subtree of the root node is accessed, and the root node can be stacked.
`Class solution{public:vector<int> Postordertraversal (TreeNode *root) { vector<int> res; stack<treenode*> s; treenode* Pre=null; while (root| |! S.empty ()) {if (root) {s.push (root); root=root->left;} else if (S.top ()->right!=pre) {root=s.top ()->right;pre=null;} Else{res.push_back (S.top ()->val);p re=s.top (); S.pop ();}} return res; }};`
Time complexity: O (N), accessed only once per node. Space complexity: O (LgN), that is, the size of the stack tree depth.
Post-order traversal is the most difficult of the three kinds of traversal, the difference between the first sequence traversal and the middle order traversal is that a pointer is needed to help determine whether the root node can be accessed.
Binary Tree postorder Traversal--leetcode
Related Keywords:
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• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs. | 588 | 2,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-10 | latest | en | 0.841536 |
http://mathematica.stackexchange.com/questions/tagged/output-formatting+algebraic-manipulation | 1,409,730,342,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535925433.20/warc/CC-MAIN-20140901014525-00065-ip-10-180-136-8.ec2.internal.warc.gz | 354,342,803 | 11,796 | # Tagged Questions
206 views
### How to substitute the following conditions into an expression?
I have an expression: $p=a\;b\; x + b^2\; y + a\;c\; z$. I want to substitute $a\;b=1$, $b^2 = 2$ and $a\;c = 4$ to obtain $p = x + 2y + 4z$. How can I tell Mathematica to do that? I dont know how to ...
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### How to apply tags to expression terms?
I often see on this site and at the mathgroup the repeated questions on how to rearrange expression that Mathematica "likes" to keep in one form, but the user prefers in another. Consider this trivial ...
858 views
### How do I get my equation to have the form $(x-a)^2 + (y-b)^2 + (z-c)^2-d = 0$?
I want Mathematica to express the equation $$-11 - 2 x + x^2 - 4 y + y^2 - 6 z + z^2=0$$ in the form $$(x - 1)^2 + (y - 2)^2 + (z - 3)^2 - 25=0$$ How do I tell Mathematica to do that?
I have the trigonometric equation \begin{equation*} \sin^8 x + 2\cos^8 x -\dfrac{1}{2}\cos^2 2x + 4\sin^2 x= 0. \end{equation*} By putting $t = \cos 2x$, I have \begin{equation*} \dfrac{3}{16} t^4+ ...
I have the following equation given: $$(26-x)\cdot\sqrt{5x-1} -(13x+14)\cdot\sqrt{5-2x} + 12\sqrt{(5x-1)\cdot(5-2x) }= 18x+32.$$ In order to solve it, I want to substitute \$t = \sqrt{5x - ... | 460 | 1,236 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2014-35 | latest | en | 0.796521 |
https://cracku.in/70-the-smallest-integer-n-such-that-n3-11n232n-28gt0--x-cat-2018-slot-2?utm_source=blog&utm_medium=video&utm_campaign=video_solution | 1,713,936,885,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00472.warc.gz | 163,902,792 | 23,943 | Question 70
The smallest integer $$n$$ such that $$n^3-11n^2+32n-28>0$$ is
Solution
We can see that at n = 2, $$n^3-11n^2+32n-28=0$$ i.e. (n-2) is a factor of $$n^3-11n^2+32n-28$$
$$\dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$$
We can further factorize n^2-9n+14 as (n-2)(n-7).
$$n^3-11n^2+32n-28=(n-2)^2(n-7)$$
$$\Rightarrow$$ $$n^3-11n^2+32n-28>0$$
$$\Rightarrow$$ $$(n-2)^2(n-7)>0$$
Therefore, we can say that n-7>0
Hence, n$$_{min}$$ = 8 | 237 | 447 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-18 | latest | en | 0.663727 |
https://www.crazy-numbers.com/en/2699 | 1,720,965,784,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00697.warc.gz | 650,534,410 | 3,454 | Warning: Undefined array key "numbers__url_substractions" in /home/clients/df8caba959271e8e753c9e287ae1296d/websites/crazy-numbers.com/includes/fcts.php on line 156
Number 2699: mathematical and symbolic properties | Crazy Numbers
# Everything about number 2699
Discover a lot of information on the number 2699: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 2699
Is 2699 a prime number? Yes
Is 2699 a perfect number? No
Number of divisors 2
List of dividers
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1, 2699
Sum of divisors 2700
Prime factorization 2699
Prime factors
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2699
## How to write / spell 2699 in letters?
In letters, the number 2699 is written as: Two thousand six hundred and ninety-nine. And in other languages? how does it spell?
2699 in other languages
Write 2699 in english Two thousand six hundred and ninety-nine
Write 2699 in french Deux mille six cent quatre-vingt-dix-neuf
Write 2699 in spanish Dos mil seiscientos noventa y nueve
Write 2699 in portuguese Dois mil seiscentos noventa e nove
## Decomposition of the number 2699
The number 2699 is composed of:
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
1 iteration of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
2 iterations of the number 9 : The number 9 (nine) represents humanity, altruism. It symbolizes generosity, idealism and humanitarian vocations.... Find out more about the number 9
## Mathematical representations and links
Other ways to write 2699
In letter Two thousand six hundred and ninety-nine
In roman numeral
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MMDCXCIX
In binary 101010001011
In octal 5213 | 733 | 2,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-30 | latest | en | 0.755218 |
https://www.shaalaa.com/question-bank-solutions/which-following-are-aps-if-they-form-ap-find-common-difference-d-and-write-three-more-terms-12-32-52-72-arithmetic-progression_6774 | 1,686,358,824,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656869.87/warc/CC-MAIN-20230609233952-20230610023952-00574.warc.gz | 1,045,175,990 | 9,967 | # Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. -1.2, -3.2, -5.2, -7.2 … - Mathematics
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. -1.2, -3.2, -5.2, -7.2 …
#### Solution
−1.2, −3.2, −5.2, −7.2 …
It can be observed that
a2 − a1 = (−3.2) − (−1.2) = −2
a3 − a= (−5.2) − (−3.2) = −2
a4 − a3 = (−7.2) − (−5.2) = −2
i.e., ak+1− ak is same every time. Therefore, d = −2
The given numbers are in A.P.
Three more terms are
a5 = − 7.2 − 2 = −9.2
a6 = − 9.2 − 2 = −11.2
a7 = − 11.2 − 2 = −13.2
Concept: Arithmetic Progression
Is there an error in this question or solution?
Chapter 5: Arithmetic Progressions - Exercise 5.1 [Page 99]
#### APPEARS IN
NCERT Class 10 Maths
Chapter 5 Arithmetic Progressions
Exercise 5.1 | Q 4.03 | Page 99
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-8-polynomials-and-factoring-8-3-multiplying-binomials-practice-and-problem-solving-exercises-page-491/47 | 1,534,501,613,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211935.42/warc/CC-MAIN-20180817084620-20180817104620-00125.warc.gz | 911,404,583 | 13,840 | ## Algebra 1
$6x^{2}$+24x+24
1. Find Surface Area using the formula A=$6s^{2}$ 2. The side length of the cube is s=(x+2) 3. FOIL using $6s^{2}$ A=6[(x+2)(x+2)] A=6[(x)(x)+(2)(x)+(x)(2)+(2)(2)] A=6[$x^{2}$+2x+2x+4] A=6[$x^{2}$+4x+4] 4. Simplify A=$6x^{2}$+24x+24 | 141 | 262 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-34 | longest | en | 0.421835 |
https://baoilleach.blogspot.com/2012/09/the-imp-act-factor-strikes-again.html | 1,506,365,796,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818693240.90/warc/CC-MAIN-20170925182814-20170925202814-00097.warc.gz | 618,092,927 | 15,421 | ## Tuesday, 4 September 2012
### The Imp Act Factor strikes again
Here's a quick question, for what shape distribution does the mean convey the least useful information? Well, there are many answers, but a prime candidate is a one-sided long-tail distribution of the type exhibited by journal citations. The mean, standard deviation and Pearson correlation, are all summary statistics developed for the two-sided normal distribution (exercise for the reader: what are their non-parametric equivalents?). Applying them to anything else is like putting lipstick on a pig (ok, a poor analogy, but it sounds funny :-), but this porcine paintjob is exactly the method used to calculate a journal's Impact Factor.
So, what's the problem? In the context of a one-sided long-tail distribution, the mean is highly sensitive to outliers, and thus almost useless. Let's take an example. Let's suppose there were 99 papers published and each was cited once giving an Impact Factor of 1.0 (99*1/99). Now let's suppose a single additional paper was published which garnered 100 citations. The Impact Factor of the journal is now (99*1 + 1*100)/100 = 2.0. So a single paper, an outlier, has doubled the Impact Factor.
But that wouldn't happen in practice, right? No - you don't get it. The distribution of journal citations has a shape that guarantees this to happen; all those impact factors you read are just measures of outliers. How about instead "the Outlier Factor", or better still the "Extreme Value Factor"?
Still don't believe me? Well, let's take a concrete example. Thomson ISI has just deigned to give J. Cheminf. its first impact factor with a value of 3.42. Let's say that 65 papers have been taken into account, so that's about 222 citations in total. Now let's enter an outlier into the mix, say the Open Babel paper published in Oct of last year. I would expect about 30 to 60 citations a year once it gets going (based on prior citations of the software, as well as experience with the GaussSum paper) - let's just say 50 for a round number, so 100 citations in the 2-year period included in an impact factor. This means that all else being equal, in one year's time the journal's Impact Factor will rise to 4.1, and in two years to 4.9.
I just hope those Avogadro guys don't publish another outlier. :-)
gioby said...
Thank you for your post. I guess that the most striking example is the one of Acta cristallographica A, who jumped from 1.5 to ~50 in 2009, just for a single paper.
Anonymous said...
One could get a bit more rigorous on this by asking about the observed variability of impact factors from year to year...?
baoilleach said...
@gioby: What a great example! I'm tempted to rewrite the post now. Maybe I'll insert a comment to see below.
Pep Pàmies said...
Actually, journal IFs and medians correlate: pic.twitter.com/QVMQkeZg
Some context here: http://occamstypewriter.org/scurry/2012/08/19/sick-of-impact-factors-coda/#comment-12080
baoilleach said...
Hmmmm...that's a bit strange, don't you think? In your example, the IF and the median not only correlate but have about the same value. Is my assumption about the shape of the distribution wrong then?
To clarify...did you calculate the mean yourself on the same date (apples-to-apples), or use the IF?
Pep Pàmies said...
To clarify, the horizontal axis on the plot I linked to above corresponds to the 2012 Journal IF as given by Thomson Reuters. However, the medians on the vertical axis are of citations received in the last 5 years (note that the IF time frame is two years). I extracted the data from Thomson Reuter's Web of Science.
Clearly, a journal's IF is a good predictor of the 5-year median number of citations per paper in the journal. The median is of course a better metric than the average, as the former is much less affected by the skewed-distribution issue. | 909 | 3,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-39 | longest | en | 0.959645 |
https://www.topperlearning.com/answer/find-the-distance-of-the-poi/q2nosrn22 | 1,686,229,881,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654871.97/warc/CC-MAIN-20230608103815-20230608133815-00448.warc.gz | 1,106,740,029 | 58,463 | Request a call back
Find the distance of the point (2, 5) from the line 4 (x + 5) = 7 (2y - 3).
Asked by Topperlearning User | 29 Apr, 2014, 03:09: PM
The given line is
4 (x + 5) = 7 (2y - 3)
4x + 20 = 14y - 21
4x - 14y + 20 + 21 = 0
4x - 14y + 41 = 0
The perpendicular distance from (x1, y1) to the line ax + by + c = 0 is
.
The point (x1, y1) is (2, 5).
Perpendicular distance from (2, 5) to the line 4x - 14y + 41 = 0 is
Answered by | 29 Apr, 2014, 05:09: PM
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Asked by Topperlearning User | 30 Apr, 2014, 09:00: AM | 402 | 965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-23 | longest | en | 0.887688 |
https://www.htmlgoodies.com/beyond/javascript/js-ref/formatting-javascript-date-intervals-into-years-months-weeks-and-days.html | 1,618,054,581,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038056869.3/warc/CC-MAIN-20210410105831-20210410135831-00172.warc.gz | 915,630,055 | 14,888 | Formatting JavaScript Date Intervals into Years, Months, Weeks, and Days
Formatting JavaScript Date Intervals into Years, Months, Weeks, and Days
In my Calculating the Difference between Two Dates in JavaScript article, I presented a few ways of calculating the difference between two dates using some of JavaScript's particular behaviors such as the roll over effect. I also described how to convert milliseconds into other intervals such as days, hours, and minutes. Since that article was posted, several readers have asked me how to express time intervals in terms of weeks, months, and years. Unfortunately, due to the variability of month lengths and leap years, calculating months and years is fraught with challenges. That being said, it can certainly be done. In fact, it already has.
Introducing the moment.js Library
My original intention was to look up some code snippets on the Internet as I did in the Calculating the Difference between Two Dates in JavaScript article. I managed to find a few, but while testing them, I found them to be either overly simplistic or unreliable. Luckily, my search also led me to the moment.js library, which was created specifically for parsing, validating, manipulating, and displaying dates in JavaScript. One area of functionality, on getting the difference between two dates, is of particular interest to us today.
Including the Script
You can download the script from the moment.js site, but I prefer to simply reference the hosted script at cloudflare.com. Just set the script tag's src property to "//cdnjs.cloudflare.com/ajax/libs/moment.js/2.10.3/moment.js".
The Moment Object
Before calling moment.js functions, you first have to wrap a "date" inside a moment() wrapper. I put the "date" inside quotation marks because the moment() wrapper actually accepts several object types that can represent a date, including:
• another Moment object
• a parsable String
• a Date object
• an Array of Date parts
Here are some examples of each:
```var a = moment(new Date());
var b = moment('2000-01-01');
var c = moment(a);
var d = moment([2007, 0, 29]);
```
The diff() Function
To calculate a time interval between a moment object and another date, you would invoke the diff() function on the moment object and pass in the second date in any of the above formats. It returns the number of milliseconds. For example, to calculate the number of milliseconds that have elapsed since January 1st, 2000, you would wrap the later date (today) in the moment wrapper and then call diff(), passing a "date" object representing the Y2K milestone:
```var today = moment(new Date()),
y2k = new Date(2000, 0, 1);
console.log( today.diff(y2k) ); //outputs 533744070037
```
Returning other Time Intervals
To obtain the difference in another unit of measurement, you can pass the desired measurement as the second argument. The supported measurements are years, months, weeks, days, hours, minutes, and seconds. As of version 2.0.0, the singular forms are supported as well.
```var a = moment([2007, 0, 29]);
var b = moment([2007, 0, 28]);
a.diff(b, 'days') // 1
```
Month and Year Caveats
Be aware the the diff() function employs some special handling for month and year intervals. It operates under the assumption that two months with the same date should always be a whole number apart. Hence,
• Jan 15 to Feb 15 should be exactly 1 month.
• Feb 28 to Mar 28 should be exactly 1 month.
• Feb 28 2011 to Feb 28 2012 should be exactly 1 year.
Outputting a Detailed Interval String
The diff() function makes it quite easy to output a detailed date interval by combining it with the moment add() function. If we add the interval to the start date before calculating the next (smaller) interval, we effectively isolate each interval type. When strung together, we get a detailed interval string, broken down by each type of our intervals array:
```Date.getFormattedDateDiff = function(date1, date2) {
var b = moment(date1),
a = moment(date2),
intervals = ['years','months','weeks','days'],
out = [];
for(var i=0; i<intervals.length; i++){
var diff = a.diff(b, intervals[i]);
b.add(diff, intervals[i]);
out.push(diff + ' ' + intervals[i]);
}
return out.join(', ');
};
var today = new Date(),
newYear = new Date(today.getFullYear(), 0, 1),
y2k = new Date(2000, 0, 1);
//(AS OF NOV 29, 2016)
//Time since New Year: 0 years, 10 months, 4 weeks, 0 days
console.log( 'Time since New Year: ' + Date.getFormattedDateDiff(newYear, today) );
//Time since Y2K: 16 years, 10 months, 4 weeks, 0 days
console.log( 'Time since Y2K: ' + Date.getFormattedDateDiff(y2k, today) );
```
The Demo
A demo is available on Codepen. Note that the HTML5 date type is not supported in Internet Explorer, Firefox, or Safari. However, that doesn't prevent you from entering the dates. You just won't get a calendar control.
Conclusion
Due to the variability of month lengths and leap years, calculating time intervals between two dates is not something that you should try to do yourself. Nor should you rely on user code that you come across in programming forums. Your best bet is a well-tested and proven library like moment.js.
Rob Gravelle resides in Ottawa, Canada, and has built web applications for numerous businesses and government agencies. Email him for a quote on your project.
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### hw5_stat210a
Course: STAT 210a, Fall 2006
School: Berkeley
Rating:
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Berkeley UC Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 5 Fall 2006 Issued: Thursday, September 28, 2006 Due: Thursday, October 5, 2006 A useful definition for this problem set: Definition: An equalizer procedure is a rule with constant risk (i.e., R(, ) = c for all ). Problem 5.1 Let = (0, +) and A = [0, ), and suppose that X Poi(). Consider the loss function L(,...
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Berkeley UC Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 5 Fall 2006 Issued: Thursday, September 28, 2006 Due: Thursday, October 5, 2006 A useful definition for this problem set: Definition: An equalizer procedure is a rule with constant risk (i.e., R(, ) = c for all ). Problem 5.1 Let = (0, +) and A = [0, ), and suppose that X Poi(). Consider the loss function L(, a) = ( - a)2 /. (a) Show that the estimator (X) = X is an equalizer rule. (b) Show that is generalized Bayes with respect to the improper prior that is uniform on (0, ). (c) Find Bayes estimators with respect to the family of Gamma(a, b) priors. Problem 5.2 For (0, 1), let X Bin(n, ), and consider the weighted quadratic loss function L(, a) = ( - a)2 . (1 - ) (Note that this loss function penalizes more severely for extreme values of near 0 or 1.) (a) Show that for this loss function, (X) = X/n is a Bayes estimator with respect to the uniform distribution on [0, 1]. (b) Show that for this loss function, is a minimax estimate of , with constant risk 1/n. Problem 5.3 Consider the Bayesian model in which, conditioned on = , we have X Bin(n, ), and we place a Beta(a, b) prior on . Recall that in a previous homework, you showed that for the Bayes estimator a,b under quadratic loss, the associated risk function takes the form R(, a,b ) = 2 (a + b)2 - n + [n - + 2a(a b)] + a2 . (n + a + b)2 Use this form of the risk function to find a minimax estimator of . (Hint: Think of choices of a and b that lead to equalizer rules.) 1 Problem 5.4 Consider the decision-theoretic problem with = [0, 1] action space A = [0, 1], and loss function L(, a) = (1 - ) a + (1 - a). Conditioned on = , let X have any distribution P . Show that the rule (X) = 1/2 is a minimax rule. (This example illustrates that minimax rules need not be good rules.) Problem 5.5 This problem addresses the issue of implementing Bayes estimators for exponential family models. Suppose that we have a (conditional) exponential family model d p(x | ) = h(x) exp i=1 i Ti (x) - A() , where x = (x1 , . . . , xn ) and has density (). (a) Define the marginal density m(x) = Show that for j = 1, . . . , n, we have d p(x | )()d induced by this Bayesian model. E i=1 i Ti (x) | x xj = log m(x) - log h(x). xj xj (Assume here that all relevant quantities are suitably differentiable.) d (b) Suppose that X = (X1 , . . . , Xn ) has density p(x; ) = h(x) exp i=1 i xi - A() . Use part (a) to conclude that the Bayes estimator of j under quadratic loss is given by (x) = log m(x) - log h(x). xj xj (c) Apply your result from (b) to derive the Bayes estimator under quadratic loss for the normal-normal model with Xi | N (, 2 ), i = 1, . . . n i.i.d., and N (, 2 ). 2
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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 5 Fall 2006 Issued: Thursday, September 14, 2006 Due: Thursday, September 21, 2006Graded exercisesProblem 5.1 a) We want to show that R(, ) = E ( (X-) )
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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 6 Fall 2006 Issued: Thursday, October 5, 2006 Due: Thursday, October 12, 2006Problem 6.1 In the inverse binomial sampling procedure, N is a random variabl
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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 6 Fall 2006 Issued: Thursday, October 5, 2006 Problem 6.1 a) Using the Rao-Blackwell theorem and considering the quadratic error loss function L(, ) = ( -
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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 7 Fall 2006 Issued: Thursday, October 19, 2006 Due: Thursday, October 26, 2006Problem 7.1 p d d Show that if Xn X > 0 and Xn /Yn 1, then Yn X. Problem 7.2
Berkeley - STAT - 210a
UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Solutions - Problem Set 7 Fall 2006 Issued: Thursday, September 14, 2006 Due: Thursday, September 21, 2006Graded exercisesProblem 7.1 p Y Let Zn = Xn , we have that
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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Solutions - Problem Set 8 Fall 2006 Issued: Thursday, November 2, 2006 Due: Thursday, November 9, 2006Graded exercisesProblem 8.1 (a) From the definitions given, we
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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 9 Fall 2006 Issued: Thursday, November 2, 2006 Due: Thursday, November 9, 2006Some useful notation: Let denote the CDF of a standard normal variate, and l
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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Solutions - Problem Set 9 Fall 2006 Issued: Thursday, November 2, 2006 Due: Thursday, November 9, 2006Graded exercisesProblem 9.1 (a) First, notice that ga () = P(X1
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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Midterm Examination-Solutions Fall 2006 Problem 1.1 [18 points total] Suppose that Xi , i = 1, . . . , n are i.i.d. samples from the uniform Uni[0, ] distribution. (a)
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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 2 Fall 2006 Issued: Thursday, September 7, 2006 Due: Thursday, September 14, 2006Graded problemsProblem 2.1 Suppose that Xi , i = 1, . . . , n are i.i.d.
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STLCOP - ENG - EngComp | 4,498 | 16,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2013-20 | latest | en | 0.868673 |
https://wikimho.com/us/q/astronomy/16 | 1,675,005,877,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00271.warc.gz | 635,185,441 | 13,364 | ### Why is only one side of the Moon visible from Earth?
• Why do we only ever see the same side of the moon?
If this is to do with gravity are there any variables which mean we might one day see more than we have before?
To my understanding (I was told in an Astronomy class) the tidal locking can only occur when the body is solid, i.e. the moon does not have a liquid core now.
9 years ago
The reason for this is what we call tidal locking:
Tidal locking (or captured rotation) occurs when the gravitational
makes one side of an astronomical body always face another,
an effect known as synchronous rotation. For example, the same side of
the Earth's Moon always faces the Earth. A tidally locked body takes
just as long to rotate around its own axis as it does to revolve
around its partner. This causes one hemisphere constantly to face the
partner body. Usually, at any given time only the satellite is tidally
locked around the larger body, but if the difference in mass between
the two bodies and their physical separation is small, each may be
tidally locked to the other, as is the case between Pluto and Charon.
This effect is employed to stabilize some artificial satellites.
Fig. 1: Tidal locking results in the Moon rotating about its axis in about the same time it takes to orbit the Earth. (Source: Wikipedia)
Fig. 1, cont.: Except for libration effects, this results in the Moon keeping the same face turned towards the Earth, as seen in the figure
on the left. (The Moon is shown in polar view, and is not drawn to
scale.) If the Moon were not spinning at all, it would alternately
show its near and far sides to the Earth while moving around our
planet in orbit, as shown in the figure on the right.
Fig. 2: Lunar librations in latitude and longitude over a period of one month (Source: Wikipedia)
Libration is manifested as a slow rocking back and forth of the Moon
as viewed from Earth, permitting an observer to see slightly different
halves of the surface at different times.
There are three types of lunar libration:
• Libration in longitude results from the eccentricity
of the Moon's orbit around Earth; the Moon's rotation sometimes leads
and sometimes lags its orbital position.
• Libration in latitude results
from a slight inclination between the Moon's axis of rotation and the
normal to the plane of its orbit around Earth. Its origin is analogous
to how the seasons arise from Earth's revolution about the Sun.
• Diurnal libration is a small daily oscillation due to the Earth's
rotation, which carries an observer first to one side and then to the
other side of the straight line joining Earth's and the Moon's
centers, allowing the observer to look first around one side of the
Moon and then around the other—because the observer is on the surface
of the Earth, not at its center.
All quotes and images from Wikipedia on Tidal locking and Wikipedia on Libration.
Can you write a bit more than just quoting Wikipedia?
I might add that the tidal effects of the moon are slowing down the Earth's rotation as well, leading to "leap seconds" and the like. The energy of the lost Earth's angular momentum causes the moons distance to recede. In a few billion years the Earth will also be tidally locked with the much more distant Moon. | 718 | 3,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-06 | latest | en | 0.953782 |
https://resources.altium.com/circuitstudio-blogs/how-to-use-a-random-number-generator-to-create-uncertainty-in-pcb-design | 1,579,657,883,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606269.37/warc/CC-MAIN-20200122012204-20200122041204-00290.warc.gz | 613,529,481 | 34,338 | # How to Use a Random Number Generator to Create Uncertainty in PCB Design
My high school peers knew me as a predictable person—always the first one to show up anywhere—on time, every time. Since they didn’t seem as concerned with keeping their schedules and promises, sometimes I waited hours for my friends. That uncertainty would mess up my schedule, but engineering was a place where I found uncertainty was rarely a problem. Unsurprisingly, I developed a deep interest in physics and engineering, where the results are predictable and based on the inputs provided.
Predictability is a great asset in embedded system design, especially in critical applications and running tests. But predictability has been my downfall once or twice, too. Specifically, I was working on creating a random number generator within an embedded system, and it took me many attempts before I was able to utilize uncertainty and make an array randomly shuffled.
## Randomness in Embedded Systems
Embedded systems need to have smooth communication in order to enable their operations, and oftentimes this communication takes place in binary 1s and 0s. Even analog designs need minimum interference to deliver a stable signal. An embedded system can live and breathe by a strong random number generator. This seems counterintuitive, right? But it’s true—uncertainty can be more desirable in some of your potential systems.
One of my projects involved designing customized MP3 players that are used to broadcast music at premium retail outlets. One requirement was for it to randomly shuffle a playlist in a way that a different song is played at the beginning of every day. The requirement stressed that there must not be any pattern in the choice of songs being played, even against other MP3 players. So how do we accomplish this? Through assigning a formula for uncertainty.
MP3 players can be customized to randomly play songs.
## Pseudo Random Number Generators
Shuffling the songs in the MP3 player can be easily achieved with a pseudo-random number generator (PRNG), which is a mathematical formula used to generate sequences of random integers. It is a quick way to introduce any form of randomness in an embedded system.
It uses a seed number to generate the first sequence and can continuously generate random sequences thereafter. However, it is not a pure random number algorithm as these sequences would eventually be repeated in time. And, if two or more devices share the same seed number, the initial sequence would be identical. A duplicate sequence ensures predictability, and therefore does not create uncertainty.
In my case, I used the various timer values in the microcontroller to generate the seed number, hoping that the result of the song shuffle would be unpredictable, at least to the human ear. My hopes were dashed when listeners complained that each of the MP3 players started playing a particular song on the list almost every time they were turned on. Back to the drawing board.
Song selection needed to be as random as some of today’s rhythms.
## The Best Unpredictable Source of Randomness: White Noise
Using timer values as seed numbers made the randomization predictable; the randomization process is executed during the initialization of the MP3 player. By extension, any usage of predictive variables like date or time as seed numbers also made the randomization predictable. This was an expensive oversight for me in my project that resulted in a revision of the hardware design to bring true randomness in the shuffling. What did I end up doing?
I merged a deterministic PRNG and a random white noise generator in order to find the perfect solution in creating randomness for my MP3 player’s shuffling. To allow a better variation of the random value, use an operational amplifier to magnify the noise. This random value can then be used as the seed number for the pseudo-random number generator algorithm.
Whatever solutions you come up with for your need for uncertainty, it is important to isolate the source signal for your randomness. Otherwise, there might be interference and system-bias which creates predictability, which is anything but the goal here. True randomness and signal isolation are hard to achieve, and especially hard to think of by yourself. That’s when setting the right Design Rule Check (DRC) in PCB design software like CircuitStudio® helps.
Need more help in creating true random number generators in your hardware? The solution’s at your fingertips in our software. Talk to an expert at Altium now and find yours.
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IoT portals can improve cybersecurity, centralize processing, and connect a host of sensors. | 949 | 4,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-05 | latest | en | 0.964416 |
https://la.mathworks.com/matlabcentral/answers/1728555-how-i-can-create-matrix-of-amout-of-fun-after-for-loop?s_tid=prof_contriblnk | 1,660,670,644,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572408.31/warc/CC-MAIN-20220816151008-20220816181008-00098.warc.gz | 331,735,155 | 36,838 | # how I can create matrix of amout of fun after for loop ???
1 view (last 30 days)
sogol bandekian on 27 May 2022
Answered: Jan on 27 May 2022
in function below I tried to define uncertain value for a & b and then get the amount of fun in uncertain a and b,but after for loop how I can have matrix of fun when different value a and b enter it?
clc;
clear;
close all;
na=10;
a=linspace(-5,5,na);
b=[-3 -1 2 3 4];
nb=numel(b);
%x=zeros(na,nb);
for i=1:na
for j=1:nb
fun= @(x) x(1)^2+x(2)^2-(x(1)+x(2))*(a(i)+b(j))^2+2*a(i)*b(j)+5*a(i)+3*b(j);
end
%define matrixi fun(a(i)),fun(b(j))
options=optimoptions('ga','Display','iter','PlotFcns',@gaplotbestf);
[x,fval,exitFlag,output,population,scores]=ga(fun,2,[],[],[],[],[],[],[],options);
%x(i,j)=zeros(fun(i,j));
end
##### 2 CommentsShowHide 1 older comment
DGM on 27 May 2022
I don't know if I'd be much help. Most of the matrices I deal with aren't fun at all.
Jan on 27 May 2022
Maybe you mean:
na = 10;
a = linspace(-5,5,na);
b = [-3 -1 2 3 4];
nb = numel(b);
options = optimoptions('ga','Display','iter');
x = zeros(2, na, nb);
for i = 1:na
for j = 1:nb
fun = @(x) x(1)^2+x(2)^2-(x(1)+x(2))*(a(i)+b(j))^2+2*a(i)*b(j)+5*a(i)+3*b(j);
[x(:, i, j), fval,exitFlag,output,population,scores] = ...
ga(fun,2,[],[],[],[],[],[],[],options);
end
end
Single objective optimization: 2 Variable(s) Options: CreationFcn: @gacreationuniform CrossoverFcn: @crossoverscattered SelectionFcn: @selectionstochunif MutationFcn: @mutationgaussian Best Mean Stall Generation Func-count f(x) f(x) Generations 1 100 -1829 39.53 0 2 147 -1829 -56.26 1 3 194 -1929 -675.1 0 4 241 -2052 -1230 0 5 288 -2052 -1373 1 6 335 -2052 -1474 2 7 382 -2052 -1459 3 8 429 -2052 -1559 4 9 476 -2052 -1752 5 10 523 -2052 -1674 6 11 570 -2052 -1745 7 12 617 -2052 -1835 8 13 664 -2052 -1934 9 14 711 -2052 -1845 10 15 758 -2052 -1900 11 16 805 -2052 -1835 12 17 852 -2052 -1845 13 18 899 -2052 -1754 14 19 946 -2052 -1798 15 20 993 -2052 -1821 16 21 1040 -2052 -1771 17 22 1087 -2052 -1747 18 23 1134 -2052 -1714 19 24 1181 -2052 -1789 20 25 1228 -2052 -1845 21 26 1275 -2052 -1827 22 27 1322 -2052 -1772 23 28 1369 -2052 -1775 24 29 1416 -2052 -1846 25 30 1463 -2052 -1673 26 Best Mean Stall Generation Func-count f(x) f(x) Generations 31 1510 -2052 -1782 27 32 1557 -2052 -1726 28 33 1604 -2052 -1789 29 34 1651 -2052 -1780 30 35 1698 -2052 -1829 31 36 1745 -2052 -1841 32 37 1792 -2052 -1817 33 38 1839 -2052 -1903 34 39 1886 -2052 -1926 35 40 1933 -2052 -1904 36 41 1980 -2052 -1878 37 42 2027 -2052 -1838 38 43 2074 -2052 -1881 39 44 2121 -2052 -1868 0 45 2168 -2052 -1887 1 46 2215 -2052 -1893 2 47 2262 -2052 -1904 3 48 2309 -2052 -1887 4 49 2356 -2052 -1821 5 50 2403 -2052 -1888 6 51 2450 -2052 -1858 7 52 2497 -2052 -1823 8 53 2544 -2052 -1905 9 54 2591 -2052 -1918 10 Optimization terminated: average change in the fitness value less than options.FunctionTolerance. Single objective optimization: 2 Variable(s) Options: CreationFcn: @gacreationuniform CrossoverFcn: @crossoverscattered SelectionFcn: @selectionstochunif MutationFcn: @mutationgaussian Best Mean Stall Generation Func-count f(x) f(x) Generations 1 100 -493.3 72.78 0 2 147 -585.4 60.04 0 3 194 -639.1 -92.27 0 4 241 -639.1 -107.2 1 5 288 -639.1 -53.84 2 6 335 -639.1 -166.1 3 7 382 -650.1 -317.8 0 8 429 -658.3 -168.7 0 9 476 -658.3 -250 1 10 523 -658.3 -405.1 2 11 570 -665.6 -318.1 0 12 617 -665.8 -479.2 0 13 664 -665.8 -456.5 1 14 711 -665.8 -534.9 0 15 758 -665.8 -442.3 1 16 805 -665.8 -449.8 2 17 852 -665.8 -435.8 3 18 899 -665.8 -528.4 4 19 946 -665.8 -452.4 5 20 993 -665.8 -379.4 6 21 1040 -665.8 -355.5 7 22 1087 -665.8 -527.1 8 23 1134 -665.8 -470.4 9 24 1181 -666 -505.8 0 25 1228 -666 -507.6 1 26 1275 -666 -447.6 2 27 1322 -666 -477.4 3 28 1369 -666 -437.2 4 29 1416 -666 -374.1 5 30 1463 -666 -314.5 6 Best Mean Stall Generation Func-count f(x) f(x) Generations 31 1510 -666 -421.1 7 32 1557 -666 -410 8 33 1604 -666 -352.2 9 34 1651 -666 -451.6 10 35 1698 -666 -382.5 11 36 1745 -666 -339.9 12 37 1792 -666 -407.4 13 38 1839 -666 -390.9 14 39 1886 -666 -429.7 15 40 1933 -666 -462.4 16 41 1980 -666 -499.3 17 42 2027 -666 -484.7 18 43 2074 -666 -518.1 19 44 2121 -666 -482.6 20 45 2168 -666 -512.1 21 46 2215 -666 -489.9 22 47 2262 -666 -441.9 23 48 2309 -666 -504 24 49 2356 -666 -468.4 25 50 2403 -666 -535.1 26 51 2450 -666 -467.6 27 52 2497 -666 -486.7 28 53 2544 -666 -525.4 29 54 2591 -666 -497.7 30 55 2638 -666 -570.1 31 56 2685 -666 -515.2 32 57 2732 -666 -520.2 33 58 2779 -666 -501.8 34 59 2826 -666 -479.6 35 60 2873 -666 -529.6 36 Best Mean Stall Generation Func-count f(x) f(x) Generations 61 2920 -666 -546.3 37 62 2967 -666 -515.5 38 63 3014 -666 -502.3 39 64 3061 -666 -458 40 65 3108 -666 -442.7 41 66 3155 -666 -499.7 42 67 3202 -666 -549.6 43 68 3249 -666 -518.5 44 69 3296 -666 -524.5 45 70 3343 -666 -534.9 46 71 3390 -666 -529.2 47 72 3437 -666 -545.3 48 73 3484 -666 -583.9 49 74 3531 -666 -568.2 50 Optimization terminated: average change in the fitness value less than options.FunctionTolerance. Single objective optimization: 2 Variable(s) Options: CreationFcn: @gacreationuniform CrossoverFcn: @crossoverscattered SelectionFcn: @selectionstochunif MutationFcn: @mutationgaussian Best Mean Stall Generation Func-count f(x) f(x) Generations 1 100 -77.42 122.5 0 2 147 -77.42 168.1 1 3 194 -79.03 294 0 4 241 -79.03 442 1 5 288 -79.03 402.6 2 6 335 -79.35 321.2 0 7 382 -79.35 356.4 1 8 429 -79.35 394.5 2 9 476 -79.35 407 3 10 523 -79.35 311.3 4 11 570 -79.35 110.1 5 12 617 -79.35 248.1 6 13 664 -79.35 194 7 14 711 -79.35 251.4 8 15 758 -79.35 236.1 9 16 805 -79.35 289.8 10 17 852 -79.35 257.1 11 18 899 -79.35 284.3 12 19 946 -79.35 187.3 13 20 993 -79.47 223 0 21 1040 -79.47 230.6 1 22 1087 -79.47 169.3 2 23 1134 -79.47 143.8 3 24 1181 -79.47 295.6 4 25 1228 -79.47 385.2 5 26 1275 -79.47 288.5 6 27 1322 -79.47 141.3 7 28 1369 -79.47 86.36 8 29 1416 -79.47 167.3 9 30 1463 -79.47 146 10 Best Mean Stall Generation Func-count f(x) f(x) Generations 31 1510 -79.47 118.4 11 32 1557 -79.47 92.45 12 33 1604 -79.47 94.29 13 34 1651 -79.47 158.1 14 35 1698 -79.47 120.4 15 36 1745 -79.47 179 16 37 1792 -79.47 240.8 17 38 1839 -79.47 245.9 18 39 1886 -79.47 135.8 19 40 1933 -79.47 137.7 20 41 1980 -79.47 173.2 21 42 2027 -79.47 170.1 22 43 2074 -79.47 186.9 23 44 2121 -79.47 306.6 24 45 2168 -79.47 356.3 25 46 2215 -79.47 77.23 26 47 2262 -79.47 39.67 27 48 2309 -79.47 70.41 28 49 2356 -79.47 56.39 29 50 2403 -79.47 33.36 30 51 2450 -79.47 47.96 31 52 2497 -79.47 88.02 32 53 2544 -79.47 133.3 33 54 2591 -79.47 137.9 34 55 2638 -79.47 150.4 35 56 2685 -79.47 104.6 36 57 2732 -79.47 85.89 37 58 2779 -79.47 89.02 38 59 2826 -79.47 87.12 39 60 2873 -79.47 53.39 40 Best Mean Stall Generation Func-count f(x) f(x) Generations 61 2920 -79.47 52.64 41 62 2967 -79.47 48.84 42 63 3014 -79.47 76.41 43 64 3061 -79.47 69.08 44 65 3108 -79.47 88.3 45 66 3155 -79.47 62.59 46 67 3202 -79.47 63.99 47 68 3249 -79.47 34.35 48 69 3296 -79.47 19.73 49 70 3343 -79.47 9.884 50 Optimization terminated: average change in the fitness value less than options.FunctionTolerance. Single objective optimization: 2 Variable(s) Options: CreationFcn: @gacreationuniform CrossoverFcn: @crossoverscattered SelectionFcn: @selectionstochunif MutationFcn: @mutationgaussian Best Mean Stall Generation Func-count f(x) f(x) Generations 1 100 -52.76 135.3 0 2 147 -52.76 173 1 3 194 -53.12 301.5 0 4 241 -53.88 253.6 0 5 288 -53.88 249.8 1 6 335 -53.99 230.7 0 7 382 -53.99 156.3 1 8 429 -53.99 126.9 2 9 476 -53.99 156.9 3 10 523 -53.99 121.9 4 11 570 -53.99 305.9 5 12 617 -53.99 277.1 6 13 664 -53.99 199.6 7 14 711 -53.99 259.7 8 15 758 -53.99 102.2 9 16 805 -53.99 172.4 10 17 852 -53.99 130.5 11 18 899 -53.99 214.1 12 19 946 -53.99 317.9 13 20 993 -53.99 113.4 14 21 1040 -53.99 138.1 15 22 1087 -53.99 133.5 16 23 1134 -53.99 126.2 17 24 1181 -53.99 151.1 18 25 1228 -53.99 137.6 19 26 1275 -53.99 145.5 20 27 1322 -53.99 143.3 21 28 1369 -53.99 184.6 22 29 1416 -53.99 126.4 23 30 1463 -53.99 189.6 24 Best Mean Stall Generation Func-count f(x) f(x) Generations 31 1510 -53.99 150.5 25 32 1557 -53.99 262.4 26 33 1604 -53.99 265.2 27 34 1651 -53.99 294.8 28 35 1698 -53.99 166.1 29 36 1745 -53.99 140.5 30 37 1792 -53.99 79.83 31 38 1839 -53.99 74.67 32 39 1886 -53.99 98.43 33 40 1933 -53.99 79.88 34 41 1980 -53.99 34.13 35 42 2027 -53.99 37.68 36 43 2074 -53.99 147.1 37 44 2121 -53.99 133.8 38 45 2168 -53.99 100.3 39 46 2215 -53.99 100.7 40 47 2262 -53.99 117.6 41 48 2309 -53.99 114.4 42 49 2356 -53.99 112.8 43 50 2403 -53.99 83.82 44 51 2450 -53.99 129 45 52 2497 -53.99 108.2 46 53 2544 -53.99 112.6 47 54 2591 -53.99 87.22 48 55 2638 -53.99 114.2 49 56 2685 -53.99 103 50 Optimization terminated: average change in the fitness value less than options.FunctionTolerance. Single objective optimization: 2 Variable(s) Options: CreationFcn: @gacreationuniform CrossoverFcn: @crossoverscattered SelectionFcn: @selectionstochunif MutationFcn: @mutationgaussian Best Mean Stall Generation Func-count f(x) f(x) Generations 1 100 -53.09 136.7 0 2 147 -53.09 226.3 1 3 194 -53.09 164.1 2 4 241 -53.09 269.5 3 5 288 -53.09 183.1 4 6 335 -53.09 137.8 5 7 382 -53.09 200.3 6 8 429 -53.09 184.6 7 9 476 -53.09 179.5 0 10 523 -53.09 235.4 1 11 570 -53.09 247.9 2 12 617 -53.09 273.7 3 13 664 -53.09 443.2 4 14 711 -53.09 330.2 5 15 758 -53.09 340.9 6 16 805 -53.09 158.3 7 17 852 -53.09 237.8 8 18 899 -53.09 145.6 9 19 946 -53.09 97.71 10 20 993 -53.09 100 11 21 1040 -53.09 185.2 12 22 1087 -53.09 237.6 13 23 1134 -53.45 163.5 0 24 1181 -53.45 109.6 1 25 1228 -53.45 127.3 2 26 1275 -53.45 155.8 3 27 1322 -53.49 148.1 0 28 1369 -53.49 163.7 1 29 1416 -53.49 226.6 2 30 1463 -53.49 183.8 3 Best Mean Stall Generation Func-count f(x) f(x) Generations 31 1510 -53.49 150.5 4 32 1557 -53.49 136.5 5 33 1604 -53.49 120.2 6 34 1651 -53.49 145.7 7 35 1698 -53.49 170.3 8 36 1745 -53.49 138.1 9 37 1792 -53.49 167.9 10 38 1839 -53.49 201.2 11 39 1886 -53.49 130.6 12 40 1933 -53.49 171.5 13 41 1980 -53.49 194.5 14 42 2027 -53.49 147.7 15 43 2074 -53.49 218.5 16 44 2121 -53.49 194.5 17 45 2168 -53.49 130.8 18 46 2215 -53.49 89.42 19 47 2262 -53.49 144.7 20 48 2309 -53.49 72.22 21 49 2356 -53.49 132.5 22 50 2403 -53.49 144.3 23 51 2450 -53.49 128.6 24 52 2497 -53.49 159 25 53 2544 -53.49 92.75 26 54 2591 -53.49 75.75 27 55 2638 -53.49 36.56 28 56 2685 -53.49 55.88 29 57 2732 -53.49 7.362 30 58 2779 -53.49 51.61 31 59 2826 -53.49 64.83 32 60 2873 -53.49 50.8 33 Best Mean Stall Generation Func-count f(x) f(x) Generations 61 2920 -53.49 57.31 34 62 2967 -53.49 27.19 35 63 3014 -53.49 29.4 36 64 3061 -53.49 17.53 37 65 3108 -53.49 69.59 38 66 3155 -53.49 98.22 39 67 3202 -53.49 95.47 40 68 3249 -53.49 91.96 41 69 3296 -53.49 95.2 42 70 3343 -53.49 91.16 43 71 3390 -53.49 104.3 44 72 3437 -53.49 130.5 45 73 3484 -53.49 80.64 46 74 3531 -53.49 73.46 47 75 3578 -53.49 45.74 48 76 3625 -53.49 35.05 49 77 3672 -53.49 85.51 50 Optimization terminated: average change in the fitness value less than options.FunctionTolerance. Single objective optimization: 2 Variable(s) Options: CreationFcn: @gacreationuniform CrossoverFcn: @crossoverscattered SelectionFcn: @selectionstochunif MutationFcn: @mutationgaussian Best Mean Stall Generation Func-count f(x) f(x) Generations 1 100 -818.6 -0.2311 0 2 147 -1085 -147.9 0 3 194 -1085 -172 1 4 241 -1093 -320.7 0 5 288 -1117 -630.3 0 6 335 -1117 -678.9 1 7 382 -1118 -679.9 0 8 429 -1127 -825.7 0 9 476 -1127 -941.3 0 10 523 -1129 -824.3 0 11 570 -1131 -848.8 0 12 617 -1131 -913.5 0 13 664 -1131 -1007 0 14 711 -1131 -1019 1 15 758 -1131 -940.6 2 16 805 -1131 -893.6 3 17 852 -1131 -785.2 4 18 899 -1131 -813.3 5 19 946 -1131 -784.2 6 20 993 -1131 -820 7 21 1040 -1131 -821.5 8 22 1087 -1131 -855.9 9 23 1134 -1131 -924.8 10 24 1181 -1131 -881.3 11 25 1228 -1131 -900 12 26 1275 -1131 -977 13 27 1322 -1131 -915.2 14 28 1369 -1131 -927.3 15 29 1416 -1131 -925.1 16 30 1463 -1131 -836 17 Best Mean Stall Generation Func-count f(x) f(x) Generations 31 1510 -1131 -860.1 18 32 1557 -1131 -843.1 19 33 1604 -1131 -765.3 20 34 1651 -1131 -946.5 21 35 1698 -1131 -881.9 22 36 1745 -1131 -851 23 37 1792 -1131 -900.6 24 38 1839 -1131 -914.5 25 39 1886 -1131 -935.9 26 40 1933 -1131 -952.3 27 41 1980 -1131 -931.2 28 42 2027 -1131 -876.9 0 43 2074 -1131 -944.1 1 44 2121 -1131 -896.6 2 45 2168 -1131 -927.6 3 46 2215 -1131 -948.3 4 47 2262 -1131 -1026 5 48 2309 -1131 -997.2 6 49 2356 -1131 -951.8 7 50 2403 -1131 -1008 8 51 2450 -1131 -927.7 9 52 2497 -1131 -909.5 0 53 2544 -1131 -942 1 54 2591 -1131 -974.5 2 55 2638 -1131 -929.5 3 56 2685 -1131 -852.7 4 57 2732 -1131 -903.4 5 58 2779 -1131 -845.1 6 59 2826 -1131 -837.4 7 60 2873 -1131 -874.5 8 Best Mean Stall Generation Func-count f(x) f(x) Generations 61 2920 -1131 -995.7 9 62 2967 -1131 -993.7 10 63 3014 -1131 -977.6 11 64 3061 -1131 -1003 12 65 3108 -1131 -1014 13 66 3155 -1131 -1013 14 67 3202 -1131 -985.7 15 68 3249 -1131 -997 16 69 3296 -1131 -997.1 17 70 3343 -1131 -998.1 18 71 3390 -1131 -1062 19 72 3437 -1131 -1027 20 73 3484 -1131 -981.7 21 74 3531 -1131 -1042 22 75 3578 -1131 -1010 23 76 3625 -1131 -1051 24 77 3672 -1131 -1032 25 78 3719 -1131 -986.2 26 79 3766 -1131 -1029 27 80 3813 -1131 -1026 28 81 3860 -1131 -1010 29 82 3907 -1131 -988.9 30 83 3954 -1131 -1039 31 84 4001 -1131 -1062 32 85 4048 -1131 -1053 33 86 4095 -1131 -1004 34 87 4142 -1131 -1037 35 88 4189 -1131 -1046 36 89 4236 -1131 -1011 37 90 4283 -1131 -1024 38 Best Mean Stall Generation Func-count f(x) f(x) Generations 91 4330 -1131 -1062 39 92 4377 -1131 -1060 40 Optimization terminated: average change in the fitness value less than options.FunctionTolerance. Single objective optimization: 2 Variable(s) Options: CreationFcn: @gacreationuniform CrossoverFcn: @crossoverscattered SelectionFcn: @selectionstochunif MutationFcn: @mutationgaussian Best Mean Stall Generation Func-count f(x) f(x) Generations 1 100 -268 156.4 0 2 147 -268 138.4 1 3 194 -288.2 116.1 0 4 241 -290.5 106.1 0 5 288 -290.5 -6.879 1 6 335 -290.5 110.1 2 7 382 -290.5 116.2 3 8 429 -290.5 -28.8 4 9 476 -293.9 33.18 0 10 523 -297.7 -84.83 0 11 570 -297.7 -47.02 1 12 617 -297.8 -59.87 0 13 664 -297.8 -67.85 1 14 711 -298.6 -34.43 0 15 758 -298.6 -38.67 1 16 805 -298.6 -21.06 2 17 852 -298.6 -2.343 3 18 899 -298.6 57.86 4 19 946 -298.6 13.41 5 20 993 -299.7 -12.76 0 21 1040 -299.7 25.59 1 22 1087 -299.7 21.18 2 23 1134 -299.7 -30.55 3 24 1181 -299.7 49.47 4 25 1228 -299.8 25.56 0 26 1275 -299.8 -53.25 1 27 1322 -299.8 -62.47 2 28 1369 -299.8 -115.3 3 29 1416 -299.8 -99.84 4 30 1463 -299.8 -67.38 5 Best Mean Stall Generation Func-count f(x) f(x) Generations 31 1510 -299.8 -20.75 6 32 1557 -299.8 -56.87 7 33 1604 -299.8 -101.1 8 34 1651 -299.8 -19.17 9 35 1698 -299.8 -15.37 10 36 1745 -299.8 8.065 11 37 1792 -299.8 91.33 12 38 1839 -299.8 101.9 13 39 1886 -299.8 -50.08 14 40 1933 -299.8 -43.39 15 41 1980 -299.9 -99.51 0 42 2027 -299.9 -87.06 1 43 2074 -299.9 -86.85 2 44 2121 -299.9 -71.64 3 45 2168 -299.9 -106 4 46 2215 -299.9 -179 5 47 2262 -299.9 -163.3 6 48 2309 -299.9 -119.2 7 49 2356 -299.9 -97.34 8 50 2403 -299.9 -150.2 9 51 2450 -299.9 -120.8 10 52 2497 -299.9 -167.7 11 53 2544 -299.9 -170.6 12 54 2591 -299.9 -165.2 13 55 2638 -299.9 -149.6 14 56 2685 -299.9 -170.6 15 57 2732 -299.9 -189 16 58 2779 -299.9 -168.3 17 59 2826 -299.9 -187.1 18 60 2873 -299.9 -127.9 19 Best Mean Stall Generation Func-count f(x) f(x) Generations 61 2920 -299.9 -120.8 20 62 2967 -299.9 -167.1 21 63 3014 -299.9 -193.9 22 64 3061 -299.9 -190.5 23 65 3108 -299.9 -161.2 24 66 3155 -299.9 -126.5 25 67 3202 -299.9 -111.1 26 68 3249 -299.9 -103.7 27 69 3296 -299.9 -114.7 28 70 3343 -299.9 -161.8 29 71 3390 -299.9 -236 30 72 3437 -299.9 -153.9 31 73 3484 -299.9 -188.2 32 74 3531 -299.9 -206.9 33 75 3578 -299.9 -179.8 34 76 3625 -299.9 -141.4 35 77 3672 -299.9 -155.3 36 78 3719 -299.9 -158.9 37 79 3766 -299.9 -141.9 38 80 3813 -300 -216.3 0 81 3860 -300.1 -200.9 0 82 3907 -300.1 -223.8 1 83 3954 -300.1 -238.7 2 84 4001 -300.1 -215.5 3 85 4048 -300.1 -212.4 4 86 4095 -300.1 -223.4 5 87 4142 -300.1 -183.4 6 88 4189 -300.1 -189 7 89 4236 -300.2 -182.3 0 90 4283 -300.2 -194.3 1 Best Mean Stall Generation Func-count f(x) f(x) Generations 91 4330 -300.2 -186.2 2 92 4377 -300.2 -153.9 3 93 4424 -300.2 -189.1 4 94 4471 -300.2 -175.4 5 95 4518 -300.2 -189.8 0 96 4565 -300.2 -196.9 1 97 4612 -300.2 -216.1 2 98 4659 -300.2 -192.3 3 99 4706 -300.2 -230.5 4 100 4753 -300.2 -248.4 5 101 4800 -300.2 -239.3 6 102 4847 -300.2 -222.7 7 103 4894 -300.2 -227.5 8 104 4941 -300.2 -214 9 105 4988 -300.2 -237.6 10 106 5035 -300.2 -240.3 11 107 5082 -300.2 -255.4 12 108 5129 -300.2 -280.7 13 109 5176 -300.2 -270.9 14 110 5223 -300.2 -276.2 15 111 5270 -300.2 -244.4 16 112 5317 -300.2 -209.8 17 113 5364 -300.2 -233.8 18 114 5411 -300.2 -234 19 115 5458 -300.2 -248.5 20 116 5505 -300.2 -250.8 21 117 5552 -300.2 -270.5 22 118 5599 -300.2 -256.5 23 119 5646 -300.2 -234.8 24 120 5693 -300.3 -255.5 0 Best Mean Stall Generation Func-count f(x) f(x) Generations 121 5740 -300.3 -261.5 1 122 5787 -300.3 -270.5 2 123 5834 -300.3 -268.7 0 124 5881 -300.3 -261.8 1 125 5928 -300.3 -247.5 2 126 5975 -300.3 -258 3 127 6022 -300.3 -268.3 4 128 6069 -300.3 -257.3 5 129 6116 -300.3 -265.7 6 130 6163 -300.3 -271.1 7 131 6210 -300.3 -259.8 8 132 6257 -300.3 -257.5 9 133 6304 -300.3 -264.3 10 134 6351 -300.3 -265.1 11 135 6398 -300.3 -263.9 12 136 6445 -300.3 -268.9 13 137 6492 -300.3 -266.3 14 138 6539 -300.3 -279.7 15 139 6586 -300.3 -273.7 16 140 6633 -300.3 -276.3 17 141 6680 -300.3 -275.3 18 142 6727 -300.3 -271.2 19 143 6774 -300.3 -277.4 20 144 6821 -300.3 -275.5 21 145 6868 -300.3 -267.1 22 146 6915 -300.3 -270.8 23 147 6962 -300.3 -271.8 0 148 7009 -300.3 -281.6 1 149 7056 -300.3 -282.6 2 150 7103 -300.3 -286.2 3 Best Mean Stall Generation Func-count f(x) f(x) Generations 151 7150 -300.3 -283.2 4 152 7197 -300.3 -287.2 5 153 7244 -300.3 -278.5 6 154 7291 -300.3 -279.3 7 155 7338 -300.3 -283.5 8 156 7385 -300.3 -289.9 9 157 7432 -300.3 -288.8 10 158 7479 -300.3 -286.1 11 159 7526 -300.3 -288.4 12 160 7573 -300.3 -288.7 13 161 7620 -300.3 -288.1 14 162 7667 -300.3 -287.7 15 163 7714 -300.3 -284.8 16 164 7761 -300.3 -287.8 17 165 7808 -300.3 -289.4 18 166 7855 -300.3 -288.6 19 167 7902 -300.3 -285.3 20 168 7949 -300.3 -291.7 21 169 7996 -300.3 -293.6 22 170 8043 -300.3 -293.9 23 Optimization terminated: average change in the fitness value less than options.FunctionTolerance. Single objective optimization: 2 Variable(s) Options: CreationFcn: @gacreationuniform CrossoverFcn: @crossoverscattered SelectionFcn: @selectionstochunif MutationFcn: @mutationgaussian Best Mean Stall Generation Func-count f(x) f(x) Generations 1 100 -29.07 165.2 0 2 147 -29.07 430.2 1 3 194 -29.07 289.8 2 4 241 -30.74 348.9 0 5 288 -30.74 311.1 1 6 335 -30.74 417.7 2 7 382 -31.78 272.5 0 8 429 -34.57 366.1 0 9 476 -34.57 200.1 1 10 523 -34.57 230 2 11 570 -34.57 242.7 3 12 617 -34.57 276.3 4 13 664 -34.57 321.1 5 14 711 -34.57 428.8 6 15 758 -34.57 278.1 7 16 805 -34.57 99.94 8 17 852 -34.57 144 9 18 899 -34.57 199.6 10 19 946 -34.57 111.5 11 20 993 -34.57 162.1 12 21 1040 -34.57 209.5 13 22 1087 -34.57 262.3 14 23 1134 -34.57 284 15 24 1181 -34.57 257.9 16 25 1228 -34.77 183.3 0 26 1275 -34.77 250.4 1 27 1322 -34.77 213.1 2 28 1369 -34.77 173.6 3 29 1416 -34.77 139.9 4 30 1463 -35.36 146 0 Best Mean Stall Generation Func-count f(x) f(x) Generations 31 1510 -35.36 158.7 1 32 1557 -35.36 199.5 2 33 1604 -35.36 239.7 3 34 1651 -35.36 187.3 4 35 1698 -35.36 190.8 5 36 1745 -35.36 115.1 6 37 1792 -35.36 119.7 7 38 1839 -35.36 121.7 8 39 1886 -35.36 98.22 9 40 1933 -35.36 114.2 10 41 1980 -35.36 268.2 11 42 2027 -35.36 299.4 12 43 2074 -35.36 242.7 13 44 2121 -35.36 149.7 14 45 2168 -35.36 115.7 15 46 2215 -35.36 127.1 16 47 2262 -35.36 93.58 17 48 2309 -35.36 152.3 18 49 2356 -35.36 159.8 19 50 2403 -35.36 192 20 51 2450 -35.36 117.9 21 52 2497 -35.36 112.6 22 53 2544 -35.36 151.9 23 54 2591 -35.36 294.9 24 55 2638 -35.36 251.7 25 56 2685 -35.36 204.4 26 57 2732 -35.36 91.06 27 58 2779 -35.36 170.5 28 59 2826 -35.36 84.14 29 60 2873 -35.36 93.64 30 Best Mean Stall Generation Func-count f(x) f(x) Generations 61 2920 -35.36 124.5 31 62 2967 -35.36 64.94 32 63 3014 -35.36 68.59 33 64 3061 -35.36 144.5 34 65 3108 -35.36 135.9 35 66 3155 -35.36 74.15 36 67 3202 -35.36 109.9 37 68 3249 -35.36 106.4 38 69 3296 -35.36 150.9 39 70 3343 -35.36 168.6 40 71 3390 -35.36 148.2 41 72 3437 -35.36 96.66 42 73 3484 -35.36 86.47 43 74 3531 -35.36 97.55 44 75 3578 -35.36 43.04 45 76 3625 -35.36 103.1 46 77 3672 -35.36 103.7 47 78 3719 -35.36 99.1 48 79 3766 -35.36 68.41 49 80 3813 -35.36 52.06 50 Optimization terminated: average change in the fitness value less than options.FunctionTolerance. Single objective optimization: 2 Variable(s) Options: CreationFcn: @gacreationuniform CrossoverFcn: @crossoverscattered SelectionFcn: @selectionstochunif MutationFcn: @mutationgaussian Best Mean Stall Generation Func-count f(x) f(x) Generations 1 100 -33.16 211.8 0 2 147 -33.16 239.5 1 3 194 -33.16 262.1 2 4 241 -33.35 272.1 0 5 288 -33.35 248.2 1 6 335 -33.35 328.7 2 7 382 -33.35 394.2 3 8 429 -33.35 303.9 4 9 476 -33.35 183.2 5 10 523 -33.35 206.5 6 11 570 -33.35 143.6 7 12 617 -33.35 228 8 13 664 -33.35 216.3 9 14 711 -33.35 251.7 10 15 758 -34.07 257.4 0 16 805 -34.07 272.2 1 17 852 -34.07 348.6 0 18 899 -34.07 366.4 1 19 946 -34.07 355.3 2 20 993 -34.07 369.1 3 21 1040 -34.07 316.3 4 22 1087 -34.07 306.4 5 23 1134 -34.07 358.4 6 24 1181 -34.07 307.7 7 25 1228 -34.07 189 8 26 1275 -34.07 238.5 9 27 1322 -34.07 313.1 10 28 1369 -34.07 285.5 11 29 1416 -34.07 344.6 12 30 1463 -34.07 340.2 13 Best Mean Stall Generation Func-count f(x) f(x) Generations 31 1510 -34.07 389.9 14 32 1557 -34.07 198.2 15 33 1604 -34.07 193.8 16 34 1651 -34.07 161.9 17 35 1698 -34.07 174.4 18 36 1745 -34.07 158 19 37 1792 -34.07 308.3 20 38 1839 -34.07 361.4 21 39 1886 -34.07 296.3 22 40 1933 -34.07 174.2 23 41 1980 -34.07 184.2 24 42 2027 -34.07 68.69 25 43 2074 -34.07 106.1 26 44 2121 -34.07 87.86 27 45 2168 -34.07 133.1 28 46 2215 -34.07 219.9 29 47 2262 -34.07 291.5 30 48 2309 -34.07 169.4 31 49 2356 -34.07 105.7 32 50 2403 -34.07 90.36 33 51 2450 -34.07 68.84 34 52 2497 -34.07 105.2 35 53 2544 -34.07 115.3 36 54 2591 -34.07 65.91 37 55 2638 -34.07 69.94 38 56 2685 -34.07 183.3 39 57 2732 -34.07 199.8 40 58 2779 -34.07 143.7 41 59 2826 -34.07 173.4 42 60 2873 -34.07 159.7 43 Best Mean Stall Generation Func-count f(x) f(x) Generations 61 2920 -34.07 150 44 62 2967 -34.07 157.1 45 63 3014 -34.07 174.4 46 64 3061 -34.07 114.2 47 65 3108 -34.07 125.1 48 66 3155 -34.07 172.5 49 67 3202 -34.07 139.6 50 Optimization terminated: average change in the fitness value less than options.FunctionTolerance. Single objective optimization: 2 Variable(s) Options: CreationFcn: @gacreationuniform CrossoverFcn: @crossoverscattered SelectionFcn: @selectionstochunif MutationFcn: @mutationgaussian Best Mean Stall Generation Func-count f(x) f(x) Generations 1 100 -38.02 165.2 0 2 147 -38.21 265.9 0 3 194 -38.21 277.7 1 4 241 -38.21 241.2 2 5 288 -38.21 281.8 3 6 335 -38.21 197.2 4 7 382 -38.21 262.4 5 8 429 -38.21 361.9 6 9 476 -38.21 304.4 7 10 523 -38.21 316.1 8 11 570 -38.21 293.1 9 12 617 -38.21 296.7 10 13 664 -38.21 342.4 11 14 711 -38.21 334.4 12 15 758 -38.21 254.4 13 16 805 -38.21 282.1 14 17 852 -38.21 212.4 15 18 899 -38.42 190.5 0 19 946 -38.44 216.8 0 20 993 -38.44 247.1 1 21 1040 -38.44 177.8 2 22 1087 -38.44 154.2 3 23 1134 -38.44 103.1 4 24 1181 -38.44 130 5 25 1228 -38.44 204.6 6 26 1275 -38.44 155 7 27 1322 -38.44 301.5 8 28 1369 -38.44 278.9 9 29 1416 -38.44 351 10 30 1463 -38.5 230.5 0 Best Mean Stall Generation Func-count f(x) f(x) Generations 31 1510 -38.5 289.8 1 32 1557 -38.5 335.8 2 33 1604 -38.5 416 3 34 1651 -38.5 295.5 4 35 1698 -38.5 174.2 5 36 1745 -38.5 175.9 6 37 1792 -38.5 163.7 7 38 1839 -38.5 126.8 8 39 1886 -38.5 156 9 40 1933 -38.5 195.3 10 41 1980 -38.5 259.2 11 42 2027 -38.5 227.4 12 43 2074 -38.5 130.7 13 44 2121 -38.5 254.1 14 45 2168 -38.5 257.8 15 46 2215 -38.5 164.7 16 47 2262 -38.5 131.4 17 48 2309 -38.5 201.8 18 49 2356 -38.5 224.2 19 50 2403 -38.5 134.4 20 51 2450 -38.5 146.5 21 52 2497 -38.5 49.59 22 53 2544 -38.5 59.9 23 54 2591 -38.5 76.84 24 55 2638 -38.5 88.81 25 56 2685 -38.5 105.5 26 57 2732 -38.5 159 27 58 2779 -38.5 108.3 28 59 2826 -38.53 137.4 0 60 2873 -38.53 119.3 1 Best Mean Stall Generation Func-count f(x) f(x) Generations 61 2920 -38.53 138.9 2 62 2967 -38.53 121.3 3 63 3014 -38.53 105.1 4 64 3061 -38.53 118.6 5 65 3108 -38.53 145.5 6 66 3155 -38.53 128 7 67 3202 -38.53 141.6 8 68 3249 -38.53 107.4 9 69 3296 -38.53 82.58 10 70 3343 -38.53 57.91 11 71 3390 -38.53 64.83 12 72 3437 -38.53 46.58 13 73 3484 -38.53 70.98 14 74 3531 -38.53 42.76 15 75 3578 -38.53 109.5 16 76 3625 -38.53 86.54 17 77 3672 -38.53 45.14 18 78 3719 -38.53 55.15 19 79 3766 -38.53 62.36 20 80 3813 -38.53 92.73 21 81 3860 -38.53 67.45 22 82 3907 -38.53 28.95 23 83 3954 -38.53 56.61 24 84 4001 -38....
x
x =
x(:,:,1) = 32.0339 23.7914 16.8171 10.8320 6.3391 2.9471 0.8220 -0.0111 0.3698 1.9759 31.9212 23.9126 16.6966 10.8944 6.3592 2.9205 0.5236 0.0843 0.3904 1.9862 x(:,:,2) = 17.9236 12.0024 7.1422 3.6719 1.2412 0.0592 0.1642 1.5791 4.1172 7.9484 17.9778 11.9192 7.1691 3.3667 1.2087 0.2457 0.2299 1.5887 4.2716 8.0210 x(:,:,3) = 4.5654 1.8004 0.2867 0.0211 1.0252 3.3644 6.6919 11.4448 17.0768 24.3621 4.6530 1.8584 0.2659 0.0696 1.1069 3.1494 6.7193 11.4326 17.3899 23.8919 x(:,:,4) = 1.9693 0.2619 0.6186 0.8380 2.9639 6.5824 10.9144 16.6902 23.7818 31.7316 1.9186 0.3830 0.1731 0.7256 2.8324 6.4552 10.4338 16.6965 23.7979 31.8924 x(:,:,5) = 0.4803 -0.1270 0.7418 2.7431 5.8683 10.3937 16.1235 22.9780 31.2064 40.4128 0.3865 0.0639 0.6116 2.6887 6.0354 10.1437 15.8579 22.9474 31.1797 40.7785 | 13,786 | 25,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2022-33 | latest | en | 0.426757 |
https://www.pa3fwm.nl/technotes/tn19d.html | 1,726,001,501,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00251.warc.gz | 875,523,612 | 3,887 | # Phase and frequency modulation
Pieter-Tjerk de Boer, PA3FWM web@pa3fwm.nl
(This is an adapted version of part of an article I wrote for the Dutch amateur radio magazine Electron, June 2019.)
Phase modulation and frequency modulation are actually two different ways of looking at the same signal, see the figure. At the top, we see an unmodulated carrier, and below it a modulated carrier. It is clearly visible that the frequency of the modulated carrier changes: apparently there is frequency modulation (FM). The frequency jumps back and forth between its rest value (unmodulated) and a higher and a lower value, as indicated by the third line. When the modulated carrier has a higher frequency, its phase is steadily more and more ahead of the unmodulated carrier (up to 540 degrees in this example), and the other way around; see the grey arrows. This leads to the fourth line: the interpretation of the signal as phase modulation (PM). Comparing the FM and the PM line, we see that the FM line is the mathematical derivative of the PM line: the frequency (FM) indicates how fast the phase (PM) changes. Conversely, the phase is the integral of the frequency.
This theoretical knowledge is useful for actually building FM or PM transmitters, see the second figure. A trivial FM transmitter consists of a voltage controlled oscillator, top left in the figure: the modulation signal directly controls the frequency. Similarly, a trivial PM transmitter consists of a fixed oscillator followed by a voltage-controlled phase shifter, top right in the figure. But we can also make FM by taking the trivial PM transmitter and first integrating the modulation signal, see bottom left. And similarly, we can make PM by feeding the modulation via a differentiator to the FM transmitter, see bottom right.
Differentiation and integration may sound dramatic, but can be done using simple RC circuits. An RC low-pass filter acts as an integrator, for frequencies above its cut-off frequency: applying a constant voltage to its input will make its output voltage rise steadily. Similarly, an RC high-pass filter acts as a differentiator for frequencies below its cut-off frequency.
In practice, for analog FM transmissions (both wideband FM for broadcast, and narrowband FM for communication by e.g. amateurs) preemphasis and deemphasis are used. This means that at the transmit side a high-pass filter is used to amplify the higher modulation frequencies, and a low-pass filter is used at the receiver to undo this. This is done to reduce noise: noise at higher frequencies will be attenuated extra at the receiving side. The cut-off frequency of this filter is 3.2 kHz for broadcast FM in Europe (RC time constant 50 µs). Since a high-pass filter acts as a differentiator, it changes FM into PM (bottom right in the previous figure). So what is called FM in practice, is actually a combination of FM up to 3 kHz, and PM above that! | 613 | 2,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-38 | latest | en | 0.92769 |
https://socratic.org/questions/how-do-you-solve-the-following-system-4x-y-1-4x-3y-15 | 1,716,216,202,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058291.13/warc/CC-MAIN-20240520142329-20240520172329-00075.warc.gz | 469,838,066 | 6,121 | # How do you solve the following system?: 4x-y=-1 , 4x-3y=15
Dec 23, 2015
The solution for the system of equations is
color(blue)(x=-9/4,y=-8
#### Explanation:
$\textcolor{b l u e}{4 x} - y = - 1$.......equation $\left(1\right)$
$\textcolor{b l u e}{4 x} - 3 y = 15$.........equation $\left(2\right)$
Solving by elimination
Subtracting equation $2$ from $1$
$\cancel{\textcolor{b l u e}{4 x}} - y = - 1$
$\cancel{- \textcolor{b l u e}{4 x}} + 3 y = - 15$
$2 y = - 16$
$y = \frac{- 16}{2}$
color(blue)(y=-8
Finding $x$ by substituting $y$ in equation $1$:
$4 x - y = - 1$
$4 x = - 1 + y$
$4 x = - 1 - 8$
$4 x = - 9$
color(blue)(x=-9/4 | 273 | 649 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 20, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-22 | latest | en | 0.63217 |
https://math.stackexchange.com/questions/3901080/recurring-sequence-with-exponent | 1,660,921,180,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573699.52/warc/CC-MAIN-20220819131019-20220819161019-00593.warc.gz | 362,083,656 | 68,957 | Recurring Sequence with Exponent
Working with recurring sequences and generating functions, I'm generally lost on solving a general expression of $$a_n$$ for any $$n$$ when the next part of the sequence, that is $$a_{n+1}$$, is in the form of an exponent, such that $$a_n = a_{n-1} +k^{n-1}$$, where k is some constant. I have no clue on how to approach this problem.
I've solved the Fibonacci sequence by subtracting the two pervious terms and shifting the sequence, but it does not seem to work here.
I'm particularly working with $$a_n = 2a_{n-1} + 5^{n-1}$$, but the sequence expands extremely fast. The base case, $$a_{0} = 1$$.
Any help would be appreciated!
• In the two examples you mention, it is often done in two parts. First the general soluion to the homogeneous equation. Second, a particular solution ( either by undetermined coefficients or variation of parameters or some other ad-hoc method ). Nov 10, 2020 at 1:22
• @GEdgar The way I've solved the Fibonacci sequence was by declaring $a_n$ as $F(x)$ and then subtracting $xF(x)$ and subtracting $x^2F(x)$, setting the entire difference equal to 1, and getting the generator function of $1/(1-x+x^2)$. No clue how to proceed here, though. Nov 10, 2020 at 1:25
Because the one with which you’re now working is first-order, you can simply ‘unwind’ it:
\begin{align*} a_n&=2a_{n-1}+5^{n-1}\\ &=2\left(2a_{n-2}+5^{n-2}\right)+5^{n-1}\\ &=2^2a_{n-2}+2\cdot5^{n-2}+5^{n-1}\\ &=2^2\left(2a_{n-3}+5^{n-3}\right)+2\cdot5^{n-2}+5^{n-1}\\ &=2^3a_{n-3}+2^2\cdot5^{n-3}+2\cdot5^{n-2}+5^{n-1}\\ &\;\;\vdots\\ &=2^ka_{n-k}+\sum_{i=0}^{k-1}2^i5^{n-1-i}\\ &\;\;\vdots\\ &=2^na_0+\sum_{i=0}^{n-1}2^i5^{n-1-i}\\ &=2^na_0+5^{n-1}\sum_{i=0}^{n-1}\left(\frac25\right)^i\\ &=2^na_0+5^{n-1}\cdot\frac{1-\left(\frac25\right)^n}{1-\frac25}\\ &=2^na_0+\frac{5^n-2^n}3 \end{align*}
• At first I was a bit lost regarding the pattern, but now that I understand it, it is a much faster approach than using generating functions. Thank you! Nov 10, 2020 at 7:18
• @ViolettaBlejder For a faster approach please see my answer and read the link to my other post to get a better understanding. Nov 10, 2020 at 15:32
• @ViolettaBlejder: You’re welcome! Nov 10, 2020 at 18:34
I love to telescope.
If $$a_n = ua_{n-1} + vc^{n}$$, then $$\dfrac{a_n}{u^n} = \dfrac{a_{n-1}}{u^{n-1}} + v(c/u)^{n}$$.
Let $$b_n = \dfrac{a_n}{u^n}$$. Then $$b_n =b_{n-1}+vd^n$$ where $$d = c/u$$.
Then $$b_n-b_{n-1} =vd^n$$.
Summing,
$$\begin{array}\\ b_m-b_0 &=\sum_{n=1}^m (b_n-b_{n-1})\\ &=\sum_{n=1}^m vd^n\\ &=v\dfrac{d-d^{m+1}}{1-d}\\ &=vd\dfrac{1-d^{m}}{1-d}\\ \end{array}$$
so
$$\begin{array}\\ \dfrac{a_m}{u^m} &=a_0+vd\dfrac{1-d^m}{1-d}\\ \text{or}\\ a_m &=a_0u^m+\dfrac{vc}{u}u^m\dfrac{1-(c/u)^m}{1-c/u}\\ &=a_0u^m+vc\dfrac{u^m-c^m}{u-c}\\ &=a_0u^m+vc\dfrac{u^m-c^m}{u-c}\\ \end{array}$$
In this case, $$u=2, c=5, v = \frac15, a_0 = 1$$ so $$a_m = 2^m + \dfrac{2^m-5^m}{2-5} = 2^m + \dfrac{5^m-2^m}{3}$$.
This can be rewritten as
$$\begin{array}\\ a_m &=a_0u^m+vc\dfrac{u^m-c^m}{u-c}\\ &=\dfrac{(u-c)a_0u^m+vc(u^m-c^m)}{u-c}\\ &=\dfrac{(a_0(u-c)+vc)u^m-vc^{m+1}}{u-c}\\ \end{array}$$
Again, we get $$=\dfrac{(a_0(u-c)+vc)u^m-vc^{m+1}}{u-c} =\dfrac{(-3+1)2^m-5^{m}}{-3} =\dfrac{2\cdot 2^m+5^{m}}{3}$$.
We use ordinary generating functions. Let $$A(x) = \sum_{i=0}^n a_n x^n$$. Then we have (summing from $$n=1$$)
\begin{align} a_n &= 2a_{n-1} + 5^{n-1},\\ \sum_{n=1}^\infty a_nx^n &= \sum_{n=1}^\infty 2a_{n-1} x^n + \sum_{n=1}^\infty 5^{n-1} x^n,\\ A(x) - a_0 &= 2x\sum_{n=1}^\infty a_{n-1} x^{n-1} + x\sum_{n=1}^\infty 5^{n-1} x^{n-1},\\ A(x) - 1 &= 2x\sum_{n=0}^\infty a_{n} x^{n} + x\sum_{n=0}^\infty 5^{n} x^{n},\\ A(x) - 1 &= 2xA(x) + \frac{x}{1-5x},\\ A(x) - 2xA(x) &= \frac{x}{1-5x} + 1,\\ A(x) &= \frac{x}{(1-2x)(1-5x)} + \frac{1}{1-2x}.\\ \end{align} Now we use partial fraction decomposition and a bit of algebra to obtain
\begin{align} A(x) &= \frac{1}{3(1-5x)} - \frac{1}{3(1-2x)} + \frac{1}{1-2x}\\ &= \frac{1}{3} \left(\frac{1}{(1-5x)} + \frac{2}{(1-2x)}\right)\\ &= \frac{1}{3} \left(\sum_{n=0}^{\infty} 5^nx^n + 2\sum_{n=0}^{\infty}2^nx^n \right). \end{align}
From here we see $$a_n = \frac{5^n + 2^{n+1}}{3}.$$
• This was by far the most understandable explanation for me, as I am currently working with only generating functions, so other things seem a little bizarre. After you helped me decompose into the generating functions, it was really easy to complete the problem, and it seems that I got the correct answer. Thank you! Nov 10, 2020 at 7:12
• @ViolettaBlejder You're welcome! I like this particular way of doing things here, but what's enlightening about the other answers (such as Brian M. Scott's) is that this powerful machinery is not always necessary. Nov 10, 2020 at 14:17
• The standard way to approach it using generating functions is to first make it homogeneous: $a_{n+1}-2a_n=5(a_n-2a_{n-1})$ then follow the steps in this post: math.stackexchange.com/questions/3899926/… Nov 10, 2020 at 15:25
First make it homogeneous.
$$a_n-2a_{n-1} = 5^{n-1}$$ $$a_{n+1}-2a_n = 5^n$$ $$\Rightarrow a_{n+1}-2a_n=5(a_n-2a_{n-1}) \tag 1$$ $$\Rightarrow a_{n+1}-5a_n=2(a_n-5a_{n-1}) \tag 2$$
Both (1) and (2) are geometric sequences, so
$$a_{n+1}-2a_n = 5^n (a_1-2a_0) = 5^n (3-2)= 5^n \tag 3$$ $$a_{n+1}-5a_n = 2^n (a_1-5a_0) = 2^n (3-5)= -2^{n+1} \tag 4$$ (3)-(4) $$a_n = \frac{1}{3} (5^n + 2^{n+1}). \blacksquare$$
• If we start from a homogeneous second order equation we need to derive (3). Here it's already given in the original problem. I put it there just for illustration purpose. Nov 10, 2020 at 2:43 | 2,341 | 5,579 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 31, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-33 | latest | en | 0.831896 |
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1. ## A doubt on divisibility
Why is it so? (Check out the image first)
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2. ## Re: A doubt on divisibility
If $\displaystyle (x+a)$ is a factor of $\displaystyle f(x)$, then what is $\displaystyle f(-a)$ ?
3. ## Re: A doubt on divisibility
Claim:
$$(x+a)\sum_{k=0}^{2n}(-1)^k x^k a^{2n-k} = x^{2n+1}+a^{2n+1}$$
Proof:
$$(x+a)\sum_{k=0}^{2n}(-1)^k x^k a^{2n-k} = (x+a)(x^{2n}-ax^{2n-1}+a^2x^{2n-2}\pm \cdots - a^{2n-1}x+a^{2n})$$
Now, let's expand and line up terms (the top line is the summation multiplied by $x$ while the bottom line is the summation multiplied by $a$):
$$\begin{matrix} & x^{2n+1} & -ax^{2n} & + a^2x^{2n-1} & \pm \cdots & -a^{2n-1}x^2 & +a^{2n}x & \\ + & & ax^{2n} & -a^2x^{2n-1} & \pm \cdots & +a^{2n-1}x^2 & -a^{2n}x & +a^{2n+1}\end{matrix}$$
Note that the central terms all cancel out, and all that is left is $x^{2n+1}+a^{2n+1}$.
Can you do something similar for $x^{2n}-a^{2n}$?
4. ## Re: A doubt on divisibility
Originally Posted by MarkFL
If $\displaystyle (x+a)$ is a factor of $\displaystyle f(x)$, then what is $\displaystyle f(-a)$ ?
Oh yeah got it. To make f(-a) zero,we need to ... you know
Got it got it
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5. ## Re: A doubt on divisibility
Hey i have 17 questions as doubt from my text book.can i send it all to this thread
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6. ## Re: A doubt on divisibility
Originally Posted by sbjsavio
Hey i have 17 questions as doubt from my text book.can i send it all to this thread
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We would prefer separate threads for each question.
7. ## Re: A doubt on divisibility
Originally Posted by SlipEternal
We would prefer separate threads for each question.
Ok then
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8. ## Re: A doubt on divisibility
Originally Posted by sbjsavio
Ok then
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Solving First-Order Linear ODEs
Transcript:
This time, we started solving differential equations. This is the third lecture of the term, and I have yet to solve a single differential equation in this class. Well, that will be rectified from now until the end of the term. So, once you learn separation of variables, which is the most elementary method there is, the single, I think the single most important equation is the one that's called the first order linear equation, both because it occurs frequently in models because it's solvable, and-- I think that's enough.
If you drop the course after today you will still have learned those two important methods: separation of variables, and first order linear equations. So, what does such an equation look like? Well, I'll write it in there. There are several ways of writing it, but I think the most basic is this. I'm going to use x as the independent variable because that's what your book does. But in the applications, it's often t, time, that is the independent variable. And, I'll try to give you examples which show that. So, the equation looks like this. I'll find some function of x times y prime plus some other function of x times y is equal to yet another function of x.
Obviously, the x doesn't have the same status here that y does, so y is extremely limited in how it can appear in the equation. But, x can be pretty much arbitrary in those places. So, that's the equation we are talking about, and I'll put it up. This is the first version of it, and we'll call them purple. Now, why is that called the linear equation? The word linear is a very heavily used word in mathematics, science, and engineering. For the moment, the best simple answer is because it's linear in y and y', the variables y and y prime.
Well, y prime is not a variable. Well, you will learn, in a certain sense, it helps to think of it as one, not right now perhaps, but think of it as linear. The most closely analogous thing would be a linear equation, a real linear equation, the kind you studied in high school, which would look like this. It would have two variables, and, I guess, constant coefficients, equal c. Now, that's a linear equation. And that's the sense in which this is linear. It's linear in y prime and y, which are the analogs of the variables y1 and y2. A little bit of terminology, if c is equal to zero, it's called homogeneous, the same way this equation is called homogeneous, as you know from 18.02, if the right-hand side is zero.
So, c(x) I should write here, but I won't. That's called homogeneous. Now, this is a common form for the equation, but it's not what it's called standard form. The standard form for the equation, and since this is going to be a prime course of confusion, which is probably completely correct, but a prime source of confusion is what I meant. The standard linear form, and I'll underline linear is the first co-efficient of y prime is taken to be one. So, you can always convert that to a standard form by simply dividing through by it. And if I do that, the equation will look like y prime plus, now, it's common to not call it b anymore, the coefficient, because it's really b over a.
And, therefore, it's common to adopt, yet, a new letter for it. And, the standard one that many people use is p. How about the right-hand side? We needed a letter for that, too. It's c over a, but we'll call it q. So, when I talk about the standard linear form for a linear first order equation, it's absolutely that that I'm talking about. Now, you immediately see that there is a potential for confusion here because what did I call the standard form for a first-order equation? So, I'm going to say, not this. The standard first order form, what would that be?
Well, it would be y prime equals, and everything else on the left-hand side. So, it would be y prime. And now, if I turn this into the standard first-order form, it would be y' = -p(x)*y + q(x). But, of course, nobody would write negative p of x. So, now, I explicitly want to say that this is a form which I will never use for this equation, although half the books of the world do.
In short, this poor little first-order equation belongs to two ethnic groups. It's both a first order equation, and therefore, its standard form should be written this way, but it's also a linear equation, and therefore its standard form should be used this way. Well, it has to decide, and I have decided for it. It is, above all, a linear equation, not just a first-order equation. And, in this course, this will always be the standard form. Now, well, what on earth is the difference? If you don't do it that way, the difference is entirely in sin(p). But, if you get the sign of p wrong in the answers, it is just a disaster from that point on.
A trivial little change of sign in the answer produces solutions and functions which have totally different behavior. And, you are going to be really lost in this course. So, maybe I should draw a line through it to indicate, please don't pay any attention to this whatsoever, except that we are not going to do that. Okay, well, what's so important about this equation? Well, number one, it can always be solved. That's a very, very big thing in differential equations. But, it's also the equation which arises in a variety of models. Now, I'm just going to list a few of them. All of them I think you will need either in part one or part two of problem sets over these first couple of problem sets, or second and third maybe.
But, of them, I'm going to put at the very top of the list of what I'll call here, I'll give it two names: the temperature diffusion model, well, it would be better to call it temperature concentration by analogy, temperature concentration model. There's the mixing model, which is hardly less important. In other words, it's almost as important. You have that in your problem set. And then, there are other, slightly less important models. There is the model of radioactive decay. There's the model of a bank interest, bank account, various motion models, you know, Newton's Law type problems if you can figure out a way of getting rid of the second derivative, some motion problems.
A classic example is the motion of a rocket being fired off, etc., etc., etc. Now, today I have to pick a model. And, the one I'm going to pick is this temperature concentration model. So, this is going to be today's model. Tomorrow's model in the recitation, I'm asking the recitations to, among other things, make sure they do a mixing problem, A) to show you how to do it, and B) because it's on the problem sets.
That's not a good reason, but it's not a bad one. The others are either in part one or we will take them up later in the term. This is not going to be the only lecture on the linear equation. There will be another one next week of equal importance. But, we can't do everything today. So, let's talk about the temperature concentration model, except I'm going to change its name. I'm going to change its name to the conduction diffusion model.
I'll put conduction over there, and diffusion over here, let's say, since, as you will see, the similarities, they are practically the same model. All that's changed from one to the other is the name of the ideas. In one case, you call it temperature, and the other, you should call it concentration. But, the actual mathematics isn't identical. So, let's begin with conduction. All right, so, I need a simple physical situation that I'm modeling. So, imagine a tank of some liquid. Water will do as well as anything. And, in the inside is a suspended, somehow, is a chamber.
A metal cube will do, and let's suppose that its walls are partly insulated, not so much that no heat can get through. There is no such thing as perfect insulation anyway, except maybe an absolute perfect vacuum. Now, inside, so here on the outside is liquid. Okay, on the inside is, what I'm interested in is the temperature of this thing. I'll call that T. Now, that's different from the temperature of the external water bath. So, I'll call that T sub e, T for temperature measured in Celsius, let's say, for the sake of definiteness. But, this is the external temperature. So, I'll indicate it with an e.
Now, what is the model? Well, in other words, how do I set up a differential equation to model the situation? Well, it's based on a physical law, which I think you know, you've had simple examples like this, the so-called Newton's Law of cooling, -- -- which says that the rate of change, the temperature of the heat goes from the outside to the inside by conduction only. Heat, of course, can travel in various ways, by convection, by conduction, as here, or by radiation, are the three most common. Of these, I only want one, namely transmission of heat by conduction.
And, that's the way it's probably a little better to call it the conduction model, rather than the temperature model, which might involve other ways for the heat to be traveling. So, dt, the independent variable, is not going to be x, as it was over there. It's going to be t for time. So, maybe I should write that down. t equals time. Capital T equals temperature in degrees Celsius. So, you can put in the degrees Celsius if you want. So, it's proportional to the temperature difference between these two. Now, how shall I write the difference? Write it this way because if you don't you will be in trouble. Now, why do I write it that way? Well, I write it that way because I want this constant to be positive, a positive constant.
In general, any constant, so, parameters which are physical, have some physical significance, one always wants to arrange the equation so that they are positive numbers, the way people normally think of these things. This is called the conductivity. The conductivity of what? Well, I don't know, of the system of the situation, the conductivity of the wall, or the wall if the metal were just by itself. At any rate, it's a constant. It's thought of as a constant. And, y positive, well, because if the external temperature is bigger than the internal temperature, I expect T to rise, the internal temperature to rise.
That means dT / dt, its slope, should be positive. So, in other words, if Te is bigger than T, I expect this number to be positive. And, that tells you that k must be a positive constant. If I had turned it the other way, expressed the difference in the reverse order, K would then be negative, have to be negative in order that this turn out to be positive in that situation I described. And, since nobody wants negative values of k, you have to write the equation in this form rather than the other way around. So, there's our differential equation. It will probably have an initial condition. So, it could be the temperature at the starting time should be some given number, T0.
But, the condition could be given in other ways. One can ask, what's the temperature as time goes to infinity, for example? There are different ways of getting that initial condition. Okay, that's the conduction model. What would the diffusion model be? The diffusion model, mathematically, would be, word for word, the same. The only difference is that now, what I imagine is I'll draw the picture the same way, except now I'm going to put, label the inside not with a T but with a C, C for concentration. It's in an external water bath, let's say. So, there is an external concentration. And, what I'm talking about is some chemical, let's say salt will do as well as anything.
So, C is equal to salt concentration inside, and Ce would be the salt concentration outside, outside in the water bath. Now, I imagine some mechanism, so this is a salt solution. That's a salt solution. And, I imagine some mechanism by which the salt can diffuse, it's a diffusion model now, diffuse from here into the air or possibly out the other way. And that's usually done by vaguely referring to the outside as a semi-permeable membrane, semi-permeable, so that the salt will have a little hard time getting through but permeable, so that it won't be blocked completely.
So, there's a membrane. You write the semi-permeable membrane outside, outside the inside. Well, I give up. You know, membrane somewhere. Sorry, membrane wall. How's that? Now, what's the equation? Well, the equation is the same, except it's called the diffusion equation. I don't think Newton got his name on this. The diffusion equation says that the rate at which the salt diffuses across the membrane, which is the same up to a constant as the rate at which the concentration inside changes, is some constant, usually called k still, okay. Do I contradict? Okay, let's keep calling it k1. Now it's different, times Ce minus C. And, for the same reason as before, if the external concentration is bigger than the internal concentration, we expect salt to flow in.
That will make C rise. It will make this positive, and therefore, we want k to be positive, just k1 to be positive for the same reason it had to be positive before. So, in each case, the model that I'm talking about is the differential equation. So, maybe I should, let's put that, make that clear. Or, I would say that this first order differential equation models this physical situation, and the same thing is true on the other side over here. This is the diffusion equation, and this is the conduction equation. Now, if you are in any doubt about the power of differential equations, the point is, when I talk about this thing, I don't have to say which of these I'm following.
I'll use neutral variables like Y and X to solve these equations. But, with a single stroke, I will be handling those situations together. And, that's the power of the method. Now, you obviously must be wondering, look, these look very, very special. He said he was going to talk about the first, general first-order equation. But, these look rather special to me. Well, not too special. How should we write it? Suppose I write, let's take the temperature equation just to have something definite. Notice that it's in a form corresponding to Newton's Law. But it is not in the standard linear form. Let's put it in standard linear form, so at least you could see that it's a linear equation. So, if I put it in standard form, it's going to look like DTDTD little t plus KT is equal to K times TE.
Now, compare that with the general, the way the general equation is supposed to look, the yellow box over there, the standard linear form. How are they going to compare? Well, this is a pretty general function. This is general. This is a general function of T because I can make the external temperature. I could suppose it behaves in anyway I like, steadily rising, decaying exponentially, maybe oscillating back and forth for some reason.
The only way in which it's not general is that this K is a constant. So, I will ask you to be generous. Let's imagine the conductivity is changing over time. So, this is usually constant, but there's no law which says it has to be. How could a conductivity change over time? Well, we could suppose that this wall was made of slowly congealing Jell-O, for instance. It starts out as liquid, and then it gets solid. And, Jell-O doesn't transmit heat, I believe, quite as well as liquid does, as a liquid would.
Is Jell-O a solid or liquid? I don't know. Let's forget about that. So, with this understanding, so let's say not necessarily here, but not necessarily, I can think of this, therefore, by allowing K to vary with time. And the external temperature to vary with time. I can think of it as a general, linear equation. So, these models are not special. They are fairly general. Well, I did promise you I would solve an equation, and that this lecture, I still have not solved any equations. OK, time to stop temporizing and solve. So, I'm going to, in order not to play favorites with these two models, I'll go back to, and to get you used to thinking of the variables all the time, that is, you know, be eclectic switching from one variable to another according to which particular lecture you happened to be sitting in.
So, let's take our equation in the form, Y prime plus P of XY, the general form using the old variables equals Q of X. Solve me. Well, there are different ways of describing the solution process. No matter how you do it, it amounts to the same amount of work and there is always a trick involved at each one of them since you can't suppress a trick by doing the problem some other way. The way I'm going to do it, I think, is the best. That's why I'm giving it to you. It's the easiest to remember. It leads to the least work, but I have colleagues who would fight with me about that point.
So, since they are not here to fight with me I am free to do whatever I like. One of the main reasons for doing it the way I'm going to do is because I want you to get what our word into your consciousness, two words, integrating factor. I'm going to solve this equation by finding and integrating factor of the form U of X. What's an integrating factor? Well, I'll show you not by writing an elaborate definition on the board, but showing you what its function is. It's a certain function, U of X, I don't know what it is, but here's what I wanted to do. I want to multiply, I'm going to drop the X's a just so that the thing looks less complicated.
So, what I want to do is multiply this equation through by U of X. That's why it's called a factor because you're going to multiply everything through by it. So, it's going to look like UY prime plus PUY equals QU, and now, so far, it's just a factor. What makes it an integrating factor is that this, after I do that, I want this to turn out to be the derivative of something with respect to X. You see the motivation for that. If this turns out to be the derivative of something, because I've chosen U so cleverly, then I will be able to solve the equation immediately just by integrating this with respect to X, and integrating that with respect to X. You just, then, integrate both sides with respect to X, and the equation is solved. Now, the only question is, what should I choose for U?
Well, if you think of the product formula, there might be many things to try here. But there's only one reasonable thing to try. Try to pick U so that it's the derivative of U times Y. See how reasonable that is? If I use the product rule on this, the first term is U times Y prime. The second term would be U prime times Y. Well, I've got the Y there. So, this will work. It works if, what's the condition that you must satisfy in order for that to be true? Well, it must be that after it to the differentiation, U prime turns out to be P times U. So, is it clear? This is something we want to be equal to, and the thing I will try to do it is by choosing U in such a way that this equality will take place.
And then I will be able to solve the equation. And so, here's what my U prime must satisfy. Hey, we can solve that. But please don't forget that P is P of X. It's a function of X. So, if you separate variables, I'm going to do this. So, what is it, DU over U equals P of X times DX. If I integrate that, so, separate variables, integrate, and you're going to get DU over U integrates to the be the log of U, and the other side integrates to be the integral of P of X DX. Now, you can put an arbitrary constant there, or you can think of it as already implied by the indefinite integral. Well, that doesn't tell us, yet, what U is. What should U be? Notice, I don't have to find every possible U, which works.
All I'm looking for is one. All I want is a single view which satisfies that equation. Well, U equals the integral, E to the integral of PDX. That's not too beautiful looking, but by differential equations, things can get so complicated that in a week or two, you will think of this as an extremely simple formula. So, there is a formula for our integrating factor. We found it. We will always be able to write an integrating factor. Don't worry about the arbitrary constant because you only need one such U. So: no arbitrary constant since only one U needed. And, that's the solution, the way we solve the linear equation.
OK, let's take over, and actually do it. I think it would be better to summarize it as a clear-cut method. So, let's do that. So, what's our method? It's the method for solving Y prime plus PY equals Q. Well, the first place, make sure it's in standard linear form. If it isn't, you must put it in that form. Notice, the formula for the integrating factor, the formula for the integrating factor involves P, the integral of PDX. So, you'd better get the right P. Otherwise, you are sunk. OK, so put it in standard linear form. That way, you will have the right P.
Notice that if you wrote it in that form, and all you remembered was E to the integral PDX, the P would have the wrong sign. If you're going to write, that P should have a negative sign there. So, do it this way, and no other way. Otherwise, you will get confused and get wrong signs. And, as I say, that will produce wrong answers, and not just slightly wrong answers, but disastrously wrong answers from the point of view of the modeling if you really want answers to physical problems.
So, here's a standard linear form. Then, find the integrating factor. So, calculate E to the integral, PDX, the integrating factor, and that multiply both, I'm putting this as both, underlined that as many times as you have room in your notes. Multiply both sides by this integrating factor by E to the integral PDX. And then, integrate. OK, let's take a simple example. Suppose we started with the equation XY prime minus Y equals, I had X2, X3, something like that, X3, I think, yeah, X2.
OK, what's the first thing to do? Put it in standard form. So, step zero will be to write it as Y prime minus one over X times Y equals X2. Let's do the work first, and then I'll talk about [UNINTELLIGIBLE], well, we now calculate the integrating factor. So, I would do it in steps. You can integrate negative one over X, right? That integrates to minus log X. So, the integrating factor is E to the integral of this, DX. So, it's E to the negative log X.
Now, in real life, that's not the way to leave that. What is E to the negative log X? Well, think of it as E to the log X to the minus one. Or, in other words, it is E to the log X is X. So, it's one over X. So, the integrating factor is one over X. OK, multiply both sides by the integrating factor. Both sides of what? Both sides of this: the equation written in standard form, and both sides. So, it's going to be one over XY prime minus one over X2 Y is equal to X2 times one over X, which is simply X. Now, if you have done the work correctly, you should be able, now, to integrate the left-hand side directly.
So, I'm going to write it this way. I always recommend that you put it as extra step, well, put it as an extra step the reason for using that integrating factor, in other words, that the left-hand side is supposed to be, now, one over X times Y prime. I always put it that because there's always a chance you made a mistake or forgot something. Look at it, mentally differentiated using the product rule just to check that, in fact, it turns out to be the same as the left-hand side. So, what do we get? One over X times Y prime plus Y times the derivative of one over X, which indeed is negative one over X2. And now, finally, that's 3A, continue, do the integration. So, you're going to get, let's see if we can do it all on one board, one over X times Y is equal to X plus a constant, X, sorry, X2 over two plus a constant.
And, the final step will be, therefore, now I want to isolate Y by itself. So, Y will be equal to multiply through by X. X3 over two plus C times X. And, that's the solution. OK, let's do one a little slightly more complicated. Let's try this one. Now, my equation is going to be one, I'll still keep two, Y and X, as the variables. I'll use T and F for a minute or two. One plus cosine X, so, I'm not going to give you this one in standard form either. It's a trick question. Y prime minus sine X times Y is equal to anything reasonable, I guess. I think X, 2X, make it more exciting. OK, now, I think I should warn you where the mistakes are just so that you can make all of them.
So, this is mistake number one. You don't put it in standard form. Mistake number two: generally people can do step one fine. Mistake number two is, this is my most common mistake, so I'm very sensitive to it. But that doesn't mean if you make it, you'll get any sympathy from me. I don't give sympathy to myself. You are so intense, so happy at having found the integrating factor, you forget to multiply Q by the integrating factor also. You just handle the left-hand side of the equation, if you forget about the right-hand side. So, the emphasis on the both here is the right-hand, please include the Q. Please include the right-hand side.
Any other mistakes? Well, nothing that I can think of. Well, maybe only, anyway, we are not going to make any mistakes the rest of this lecture. So, what do we do? We write this in standard form. So, it's going to look like Y prime minus sine X, sine X divided by one plus cosine X times Y equals, my heart sinks because I know I'm supposed to integrate something like this. And, boy, that's going to give me problems. Well, not yet. With the integrating factor? The integrating factor is, well, we want to calculate the integral of negative sine X over one plus cosine. That's the integral of PDX. And, after that, we have to exponentiate it. Well, can you do this? Yeah, but if you stare at it a little while, you can see that the top is the derivative of the bottom.
That is great. That means it integrates to be the log of one plus cosine X. Is that right, one over one plus cosine X times the derivative of this, which is negative cosine X. Therefore, the integrating factor is E to that. In other words, it is one plus cosine X. Therefore, so this was step zero. Step one, we found the integrating factor. And now, step two, we multiply through the integrating factor. And what do we get? We multiply through the standard for equation by the integrating factor, if you do that, what you get is, well, Y prime gets the coefficient one plus cosine X, Y prime minus sign X equals 2X.
Oh, dear. Well, I hope somebody would giggle at this point. What's giggle-able about it? Well, that all this was totally wasted work. It's called spinning your wheels. No, it's not spinning your wheels. It's doing what you're supposed to do, and finding out that you wasted the entire time doing what you were supposed to do. Well, in other words, that net effect of this is to end up with the same equation we started with. But, what is the point? The point of having done all this was because now the left-hand side is exactly the derivative of something, and the left-hand side should be the derivative of what?
Well, it should be the derivative of one plus cosine X times Y, all prime. Now, you can check that that's in fact the case. It's one plus cosine X, Y prime, plus minus sine X, the derivative of this side times Y. So, if you had thought, in looking at the equation, to say to yourself, this is a derivative of that, maybe I'll just check right away to see if it's the derivative of one plus cosine X sine.
You would have saved that work. Well, you don't have to be brilliant or clever, or anything like that. You can follow your nose, and it's just, I want to give you a positive experience in solving linear equations, not too negative. Anyway, so we got to this point. So, now this is 2X, and now we are ready to solve the equation, which is the solution now will be one plus cosine X times Y is equal to X2 plus a constant, and so Y is equal to X2 divided by X2 plus a constant divided by one plus cosine X.
Suppose I have given you an initial condition, which I didn't. But, suppose the initial condition said that Y of zero were one, for instance. Then, the solution would be, so, this is an if, I'm throwing in at the end just to make it a little bit more of a problem, how would I put, then I could evaluate the constant by using the initial condition. What would it be? This would be, on the left-hand side, one, on the right-hand side would be C over two.
So, I would get one equals C over two. Is that correct? Cosine of zero is one, so that's two down below. Therefore, C is equal to two, and that would then complete the solution. We would be X2 plus two over one plus cosine X. Now, you can do this in general, of course, and get a general formula. And, we will have occasion to use that next week. But for now, why don't we concentrate on the most interesting case, namely that of the most linear equation, with constant coefficient, that is, so let's look at the linear equation with constant coefficient, because that's the one that most closely models the conduction and diffusion equations.
So, what I'm interested in, is since this is the, of them all, probably it's the most important case is the one where P is a constant because of its application to that. And, many of the other, the bank account, for example, all of those will use a constant coefficient. So, how is the thing going to look? Well, I will use the cooling. Let's use the temperature model, for example. The temperature model, the equation will be DTDT plus KT is equal to. Now, notice on the right-hand side, this is a common error. You don't put TE. You have to put KTE because that's what the equation says. If you think units, you won't have any trouble. Units have to be compatible on both sides of a differential equation.
And therefore, whatever the units were for capital KT, I'd have to have the same units on the right-hand side, which indicates I cannot have KT on the left of the differential equation, and just T on the right, and expect the units to be compatible. That's not possible. So, that's a good way of remembering that if you're modeling temperature or concentration, you have to have the K on both sides. OK, let's do, now, a lot of this we are going to do in our head now because this is really too easy.
What's the integrating factor? Well, the integrating factor is going to be the integral of K, the coefficient now is just K. P is a constant, K, and if I integrate KDT, I get KT, and I exponentiate that. So, the integrating factor is E to the KT. I multiply through both sides, multiply by E to the KT, and what's the resulting equation? Well, it's going to be , I'll write it in the compact form. It's going to be E to the KT times T, all prime. The differentiation is now, of course, with respect to the time. And, that's equal to KTE, whatever that is, times E to the KT.
This is a function of T, of course, the function of little time, sorry, little T time. OK, and now, finally, we are going to integrate. What's the answer? Well, it is E to the, so, are we going to get E to the KT times T is, sorry, K little t, K times time times the temperature is equal to the integral of KTE. I'll put the fact that it's a function of T inside just to remind you, E to the KT, and now I'll put the arbitrary constant. Let's put in the arbitrary constant explicitly. So, what will T be? OK, T will look like this, finally. It will be E to the negative KT. That's on the outside.
Then, you will integrate. Of course, the difficulty of doing this integral depends entirely upon how this external temperature varies. But anyways, it's going to be K times that function, which I haven't specified, E to the KT plus C times E to the negative KT. Now, some people, many, in fact, that almost always, in the engineering literature, almost never write indefinite integrals because an indefinite integral is indefinite.
In other words, this covers not just one function, but a whole multitude of functions which differ from each other by an arbitrary constant. So, in a formula like this, there's a certain vagueness, and it's further compounded by the fact that I don't know whether the arbitrary constant is here. I seem to have put it explicitly on the outside the way you're used to doing from calculus. Many people, therefore, prefer, and I think you should learn this, to do what is done in the very first section of the notes called definite integral solutions. If there's an initial condition saying that the internal temperature at time zero is some given value, what they like to do is make this thing definite by integrating here from zero to T, and making this a dummy variable.
You see, what that does is it gives you a particular function, whereas, I'm sorry I didn't put in the DT one minus two. What it does is that when time is zero, all this automatically disappears, and the arbitrary constant will then be, it's T. So, in other words, C times this, which is one, is that equal to T. In other words, if I make this zero, that I can write C as equal to this arbitrary starting value. Now, when you do this, the essential thing, and we're going to come back to this next week, but right away, because K is positive, I want to emphasize that so much at the beginning of the period, I want to conclude by showing you what its significance is.
This part disappears because K is positive. The conductivity is positive. This part disappears as T goes to zero. This goes to zero as T goes to infinity. So, this is a solution that remains. This, therefore, is called the steady state solution, the thing which the temperature behaves like, as T goes to infinity. This is called the trangent because it disappears as T goes to infinity. It depends on the initial condition, but it disappears, which shows you, then, in the long run for this type of problem the initial condition makes no difference. The function behaves always the same way as T goes to infinity. | 7,927 | 33,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2018-09 | latest | en | 0.977618 |
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# Selina Solutions for Class 9 Physics
## Solutions for Selina ICSE Concise Physics for Class 9 (2018-19 Session)
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## Selina Solutions for Class 9 Physics
Given below is chapter wise list for Selina Solutions for Class 9 Physics. Select any chapter number to view solutions.
• Chapter 2: Motion in One Dimension
• Chapter 3: Laws of Motion
• Chapter 4: Pressure in Fluids and Atmospheric Pressure
• Chapter 5: Upthrust in Fluids, Archimedes’ Principle and Floatation
• Chapter 6: Heat and Energy
• Chapter 7: Reflection of Light
• Chapter 9: Current Electricity
• Chapter 10: Magnetism
## Chapters covered in Selina Solutions for Class 9 Physics
#### Chapter 1: Measurements and Experimentation
Shaalaa has a total of 56 questions are solved for this Chapter in Class 9 Physics. Chapters covered in Measurements and Experimentation are Concept of Measurements and Experimentation, International System of Units, Measurements Using Common Instruments, Other Commonly Used System of Units - fps and cgs, Simple Pendulum for Time, Vernier Callipers and Micro-metre Screw Gauge for Length
#### Chapter 2: Motion in One Dimension
Shaalaa has a total of 0 questions are solved for this Chapter in Class 9 Physics. Chapters covered in Motion in One Dimension are Acceleration and Retardation, Distance and Displacement, Graphs of Distance-time and Speed-time, Instantaneous Velocity and Speed, Measuring the Rate of Motion - Speed with Direction, Rate of Change of Velocity, Scalar and Vector Quantities
#### Chapter 3: Laws of Motion
Shaalaa has a total of 0 questions are solved for this Chapter in Class 9 Physics. Chapters covered in Laws of Motion are cgs and SI Units of Force and Their Relation with Gravitational Units, Concept of Free Fall, Contact and Non-contact Forces, Definitions of Inertia and Force from First Law, General Properties of Non-contact Forces, Gravitational Units of Force, Mass and Weight, Newton'S First Law of Motion, Newton’s Second Law of Motion, Newton's Third Law of Motion, Universal Law of Gravitation, Weight as Force of Gravity
#### Chapter 4: Pressure in Fluids and Atmospheric Pressure
Shaalaa has a total of 0 questions are solved for this Chapter in Class 9 Physics. Chapters covered in Pressure in Fluids and Atmospheric Pressure are Atmospheric Pressure, Change of Pressure with Depth, Pascal's Law, Transmission of Pressure in Liquids
#### Chapter 5: Upthrust in Fluids, Archimedes’ Principle and Floatation
Shaalaa has a total of 0 questions are solved for this Chapter in Class 9 Physics. Chapters covered in Upthrust in Fluids, Archimedes’ Principle and Floatation are Archimedes' Principle, Determination of Relative Density of a Solid, Floatation - Principle of Floatation, Relation Between the Density of a Floating Body, Thrust and Pressure - Buoyancy
#### Chapter 6: Heat and Energy
Shaalaa has a total of 0 questions are solved for this Chapter in Class 9 Physics. Chapters covered in Heat and Energy are Anomalous Expansion of Water, Concepts of Heat and Temperature, Effect of Global Warming, Energy Degradation, Energy Flow and Its Importance, Energy Sources, Graphs Showing Variation of Volume and Density of Water with Temperature in the 0 to 10°C Range, Green House Effect, Meaning of Global Warming, Non-renewable Energy, Renewable Energy, Renewable Versus Non-renewable Sources, Use of Hydro Electrical Powers for Light and Tube Wells
#### Chapter 7: Reflection of Light
Shaalaa has a total of 0 questions are solved for this Chapter in Class 9 Physics. Chapters covered in Reflection of Light are Images Formed by a Pair of Parallel and Perpendicular Plane Mirrors, Laws of Reflection, Reflection of Light, Spherical Mirrors, Uses of Spherical Mirrors
#### Chapter 8: Propagation of Sound Waves
Shaalaa has a total of 9 questions are solved for this Chapter in Class 9 Physics. Chapters covered in Propagation of Sound Waves are Comparison of Sound with Speed of Light, Difference Between Ultrasonic and Supersonic, Infrasonic, Sonic, Ultrasonic Frequencies and Their Applications, Nature of Sound Waves, Propagation and Speed in Different Media, Requirement of a Medium for Sound Waves to Travel, Thunder and Lightning
#### Chapter 9: Current Electricity
Shaalaa has a total of 0 questions are solved for this Chapter in Class 9 Physics. Chapters covered in Current Electricity are Closed and Open Circuits, Direction of Current (Electron Flow and Conventional), Insulators and Conductors, Simple Electric Circuit Using an Electric Cell and a Bulb to Introduce the Idea of Current, Social Initiatives - Improving Efficiency of Existing Technologies and Introducing New Eco-friendly Technologies. Creating Awareness and Building Trends of Sensitive Use of Resources and Products
#### Chapter 10: Magnetism
Shaalaa has a total of 0 questions are solved for this Chapter in Class 9 Physics. Chapters covered in Magnetism are Induced Magnetism, Introduction of Electromagnet and Its Uses, Lines of Magnetic Field and Their Properties, Magnetic Field of Earth, Neutral Points in Magnetic Fields
## Selina Solutions for Class 9 Physics
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S | 1,428 | 6,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-22 | longest | en | 0.862798 |
https://calculatorsonline.org/roman-numeral-date-converter/may-14-2014 | 1,696,015,966,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510528.86/warc/CC-MAIN-20230929190403-20230929220403-00593.warc.gz | 166,785,896 | 5,405 | # May 14, 2014 in roman numeral
Here you will see step by step solution to convert May 14, 2014 date to roman numeral. How to write May 14, 2014 as a roman numeral? May 14, 2014 as a roman numeral written as V.XIV.MMXIV (MM.DD.YYYY), please check the explanation that how to convert 14 May, 2014 in roman number.
## Answer: May 14, 2014 in roman numeral
V.XIV.MMXIV
### How to convert May 14, 2014 in roman number?
To convert the May 14, 2014 in roman number simply expand the each number from month, date and year from hindu-arabic number to roman numerals, then replace the all numbers of expanded form of date, month and year with respective roman numerals.
#### Solution for May 14, 2014 to roman numeral
Given date is => 14-05-2014
After expanding number from Month, Date and year, this table provides a simple and explanation of how to convert the date 'May 14, 2014' to its Roman number:
MonthDayYear
Date 05 [May] 14 2014
Expanded Number Values 5 10 + 4 1000 + 1000 + 10 + 4
Roman Numeral Values V X + IV M + M + X + IV
Result = V XIV MMXIV
Hence, in order to correct roman numerals date combination of Month, Day and Year of May 14, 2014 is written as V.XIV.MMXIV or in 14/May/2014 [DD/MM/YYYY] format it is written as XIV/V/MMXIV. | 373 | 1,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-40 | latest | en | 0.822386 |
http://davidlowryduda.com/math-90-week-8-quiz/ | 1,571,409,809,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00259.warc.gz | 53,952,836 | 13,832 | # Math 90: Week 8 Quiz
There was a quiz this week – in this post, we consider the solutions, common mistakes, and the distribution.
The quiz was as follows:
A girl flies a kite that stays a constant 200 feet above the ground. The wind carries it away from her at 20 feet per second. We were first to draw a picture.
I have included a picture at the right, excluding the initial position (which is not really useful to this problem). In addition, we know that $x'(t) = 20$. Almost everyone drew the picture correctly, so I won’t belabor this point too much.
We were then asked to give an equation for the square of the distance from the girl to the kite. The question explicitly asks for the square of the distance. This is a right triangle, so we write $d(t)^2 = 200^2 + x(t)^2$, from the Pythagorean Theorem.
Finally, we were asked to compute the rate at which the girl must let out string when the kite was 300 feet away from the girl. The most common (and often, only) mistake on the test was to interpret this to mean that $x = 300$ at this time. But the distance between two things is the length of the straight line between them unless stated otherwise, so we actually care about the time when $d = 300$.
So we have a triangle with hypotenuse $300$ and one leg $200$, so the other leg is of length $sqrt{300^2 – 200^2}=100 sqrt 5=x$.
Now we put it all together. From $d(t)^2 = 200^2 + x(t)^2$, we differentiate to get $2d(t)d'(t) = 2x(t)x'(t)$. We are looking to find $d'(t)$ when $d(t) = 300$, and we found that at this time $x(t) = 100sqrt{5}$. We also know that $x'(t) = 20$ at all times. Plugging these in, we get that $2 cdot 300 cdot d'(t) = 2 cdot 100sqrt 5 cdot 20$, or that $d'(t)=dfrac{200 sqrt 5 cdot 20}{600} = dfrac{20 sqrt 5}{3}$ feet per second.
The class did really, really well on this quiz. I was impressed. The vase majority of the class made at least an 8. The only mistake on average was to mistake which leg was implied by “distance from the kite to the girl.”
This entry was posted in Brown University, Math 90, Mathematics and tagged , , , , , , , . Bookmark the permalink. | 585 | 2,113 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2019-43 | latest | en | 0.965322 |
https://nullmap.org/posts/mathematics/metric-tensors/index.html | 1,627,075,714,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00384.warc.gz | 445,729,568 | 4,428 | # Metric Tensors
When working with General Relativity, we naturally need to work with tensors. Fortunately, we can easily represent the metric, $$g_{uv}$$, stress, and energy tensors using the Tensors package along with SymPy and LinearAlgebra.
Recall that a metric in general is essentially a kind of map we can use to determine how a coordinate system should be laid out.
The convention in GR, and more specifically when using Einstein summation notation, is to write the indices of contravariant vectors as superscripts and the indices of covariant vectors as subscripts. Read my posts on contravariant vs. covariant vectors and Einstein summation notation for a more in-depth treatment of each topic.
Starting from flatspace, we'll using the Pythagorean theorem and construct a metric tensor of rank 2.
$ds^2 = (dx^1)^2 + (dx^2)^2$
For a tensor transformation from the $$X$$ frame of reference to the $$Y$$ frame of reference, we can use the following equation as a guide,
$\boxed{dX^m = \frac{\partial X^m}{\partial Y^r} dY^r}\,.$
Applying $$(2)$$ to a transformation of $$(1)$$ will yield,
\begin{aligned} ds^2 &= \delta_{mn} \frac{\partial X^m}{\partial Y^r} \frac{\partial X^n}{\partial Y^s} dY^r dY^s \\ &= g_{rs}dY^r dY^s \,.\end{aligned}
Before jumping to the metric tensor, $$g_{ij}$$, where $$i$$ and $$j$$ are indices, I think it's a little easier to understand from the perspective of using the Kronecker delta.
Looking at $$(1)$$, we can see that we want to square the $$dx^1$$ and $$dx^2$$ terms. Furthermore, there are no terms with the product of $$dx^1$$ and $$dx^2$$, only $$dx^1\cdot dx^1$$ and $$dx^2\cdot dx^2$$.
Therefore, when $$m = n = 1$$, we will be squaring the $$dx^1$$ component and when $$m = n = 2$$, we will be squaring the $$dx^2$$ component. Since the Kronecker delta function returns $$0$$ when $$m$$ and $$n$$ are different, whenever $$m \neq n$$, those terms map to nullity. Therefore, the $$g_{11}$$ term is $$1$$, the $$g_{22}$$ term is $$1$$ and the $$g_{12} = g_{21}$$ term is $$0$$.
In matrix form this is written as,
$\begin{bmatrix} g_{11} & g_{12} \\ g_{21} & g_{22} \end{bmatrix}\,.$
To re-iterate the result with the coefficients emphasized, the metric is given by,
$ds^2 = (1)(dx^1)^2 + (0)dx^1 dx^2 + (1)(dx^2)^2\,.$
Using the Tensors package, we can quickly construct a metric tensor for computation. We can use the Tensor or SymmetricTensor constructors. In either case we need to specify the tensor rank, the dimensions, and datatype.
Tensor{2, 2, Int}([1 0 ; 0 1])
We want a tensor of rank $$2$$ with $$2$$ dimensions, and a bit or numeric data type since we're working with $$0$$s and $$1$$s. In the example above, I specified the rank, dimension, and data type and then passed in a matrix of the form shown in $$(4)$$. Another option would be to use the SymmetricTensor constructor. The difference here is that we are explicitly stating that our $$g_{12}$$ and $$g_{21}$$ terms are identical, thereby forming a symmetric matrix, so we can use the following syntax.
using Tensors, LinearAlgebra
g_cart = SymmetricTensor{2, 2, Int}((1, 0, 1))
> 2×2 SymmetricTensor{2, 2, Int64, 3}:
> 1 0
> 0 1
It's also worth noting that because of the symmetry here, we're only storing $$3$$ data points instead of $$4$$. While that's unlikely to make any difference with a $$2-2$$ tensor ($$2$$ rank, $$2$$ dimensional), it could offer improvements with higher-dimensional tensors.
### Transforming the Metric Tensor to Other Coordinate Systems
We'll take it a step further. Let's perform a transformation from rectilinear coordinates to polar coordinates.
We'll need to leverage the following mappings,
\begin{aligned} x &= r\cos\theta \\ y &= r\sin\theta \,. \end{aligned}
When we differentiate both $$x$$ and $$y$$, we need to remember to use the chain rule since $$r$$ is a function of $$\theta$$.
$dx = \cos{\theta}\ dr - r\sin{\theta}\ d\theta$ $dy = \sin{\theta}\ dr + r\cos{\theta}\ d\theta$
Plugging $$(7)$$ and $$(8)$$ back into $$(1)$$, expanding out, and collecting like terms yields the algebraical intensive but conceptually straight-forward,
\begin{aligned} ds^2 &= (\cos{\theta}\ dr - r\sin{\theta}\ d\theta)(\cos{\theta}\ dr - r\sin{\theta}\ d\theta) + (\sin{\theta}\ dr + r\cos{\theta}\ d\theta)(\sin{\theta}\ dr + r\cos{\theta}\ d\theta) \\ &= \cos^2\theta dr^2 - 2r\sin\theta\cos\theta dr\ d\theta + r^2\sin^2\theta d\theta^2 + \sin^2\theta dr^2 + 2r\sin\theta\cos\theta dr\ d\theta + r^2\cos^2\theta d\theta^2 \\ &= \cos^2\theta dr^2 + \sin^2\theta dr^2 + r^2\sin^2\theta d\theta^2 + r^2\cos^2\theta d\theta^2 \\ &= (\cos^2\theta + \sin^2\theta) dr^2 + r^2(\sin^2\theta + \cos^2\theta) d\theta^2 \\ &= (1)dr^2 + (0)drd\theta + (r^2)d\theta^2\,. \end{aligned}
Note that above we used the trigonometric identity $$\cos^2(\theta) + \sin^2(\theta) = 1$$.
Now that we have simplified $$ds^2$$ to $$dr^2 + r^2 d\theta^2$$, we can construct our metric tensor of flatspace in polar coordinates.
\begin{aligned} g_{rr} &= 1 \\ g_{r\theta} &= g_{\theta r} = 0 \\ g_{\theta\theta} &= r^2 \end{aligned}
We could use functions here or represent the metric in a few other ways, but we'll leverage the excellent SymPy package and construct a symbolic tensor using the values from $$(10)$$.
using SymPy, LinearAlgebra, Tensors
@vars r θ
g_polar = SymmetricTensor{2, 2, Sym}((1, 0, r^2))
g_polar yields a symbolic tensor of the form,
$\begin{bmatrix} 1 & 0 \\ 0 & r^2 \end{bmatrix}\,.$
### Metric Tensor Using Spherical Coordinates
While I won't go through the full derivation here, it's worth offering the concrete example of how we create a tensor of rank $$2$$ and dimension $$3$$ constructed with the diagm function from the LinearAlgebra package.
After working through some extensive algebra, we will find ourselves with $$ds^2$$ in spherical coordinates,
$ds^2 = (1)dr^2 + (r^2)d\phi^2 + (r^2 \sin^2\theta)d\theta^2 \,.$
using SymPy, LinearAlgebra, Tensors
@vars r θ ϕ
g_spherical = Tensor{2, 3, Sym}(diagm([1, r^2, r^2*sin(θ)^2]))
Notice that since we are working in dimension $$3$$, we've changed the initial directive to $$2$$ for rank $$2$$, $$3$$ for dimension $$3$$, and once again we're using $$Sym$$ to leverage SymPy objects. Passed in as an argument is the output of the diagm function which constructs a diagonal matrix. For a $$3 \times 3$$ matrix, diagm will expect a vector with $$3$$ elements as shown above. More generally, for any $$n \times n$$ matrix, it will expect a vector with $$n$$ elements.
The assignment of g_spherical displays as,
$\begin{bmatrix*}[c] 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \sin^2(\theta) \end{bmatrix*}\,.$ | 2,058 | 6,660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-31 | latest | en | 0.863488 |
https://www.upwork.com/job/_~0181ad7953e53c7318/ | 1,477,149,363,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718987.23/warc/CC-MAIN-20161020183838-00072-ip-10-171-6-4.ec2.internal.warc.gz | 1,041,556,417 | 16,826 | # C++
Web, Mobile & Software Dev Desktop Software Development Posted 3 years ago
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Budget: \$10
Delivery by March 10, 2013
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## Details
PROJECT:
Write a program that simulates playing a million games of craps. From Chapter 5 in our textbook, here are the rules for one game of craps:
A player rolls two dice. Each die has six faces. These faces contain 1, 2, 3, 4, 5 and 6 spots. After the dice have come to rest, the sum of the spots on the two upward faces is calculated. If the sum is 7 or 11 on the first roll, the player wins. If the sum is 2, 3 or 12 on the first roll (called “craps”), the player loses (i.e., the “house” wins). If the sum is 4, 5, 6, 8, 9 or 10 on the first roll, then that sum becomes the player’s “point.” To win, the player continues to roll the dice until he/she “makes the point” (making the point means that the dice sum equals the player’s point). To lose, the player rolls a 7 before making the point.
Craps games can have different lengths, where the length of a game is the number of dice rolls required to achieve a win or a loss. Your program should display a histogram that shows two bars (in the form of asterisks) for each number-of-rolls-to-end-the-game scenario. Specifically, show the relative number of games won and the relative number of games lost on the first roll, the relative number of games won and the relative number of games lost on the second roll, etc. Do that for each number-of-rolls scenario for one roll up through ten rolls. In addition, include two bars for the relative number of games won and the relative numbers of games lost for all the games combined that require more than ten rolls. For histogram format details, study the output shown below.
If you’d like to see a histogram program example, see the CoinFlips program from Chapter 9 in the second edition Dean Java textbook (or Chapter 10 in the first edition Dean Java textbook). That program’s histogram prints a total of approximately 100 asterisks. Your program’s histogram should print a total of approximately 200 asterisks. The reason that I can’t says “exactly 200 asterisks” is because calculating the number of asterisks in each bar involves rounding, and it’s possible that the rounding will skew the total number of asterisks slightly above 200 or slightly below 200.
If you borrow code from the craps game program in Chapter 5 in our textbook, be sure to improve the code’s style. For example, enumerated data type declarations should go at the top of your file (not inside main) and enumerated data type tag names should start with a lowercase letter.
As always, you should implement functions to help the main function when it is appropriate to do so. In particular, in the interest of gaining experience with reference parameters, you must implement a play1Game function that uses two reference parameters – won and numOfRolls – that keep track of (1) whether the game resulted in a win, and (2) how many rolls were needed to finish the game.
As always, your program should mimic the given output’s format precisely.
Output:
Histogram showing the likelihood of winning for a given number of rolls.
Rolls
1 wins: ********************************************
losses: **********************
2 wins: ***************
losses: **********************
3 wins: ***********
losses: ****************
4 wins: ********
losses: ***********
5 wins: ******
losses: ********
6 wins: ****
losses: ******
7 wins: ***
losses: ****
8 wins: **
losses: ***
9 wins: *
losses: **
10 wins: *
losses: **
more wins: ***
losses: ****
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Member Since Feb 12, 2013 | 963 | 3,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2016-44 | latest | en | 0.917014 |
https://mcqslearn.com/bba/finance/quiz/quiz-questions-and-answers.php?page=73 | 1,716,792,267,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00255.warc.gz | 332,799,341 | 17,399 | BBA Finance Courses
Financial Management Certification Exam Tests
Financial Management Practice Test 73
# Tying Ratios Together Quiz Questions and Answers PDF - 73
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## Tying Ratios Together Questions and Answers : Quiz 73
MCQ 361:
The total assets divided by common equity is a formula uses for calculating
1. equity multiplier
2. graphical multiplier
3. turnover multiplier
4. stock multiplier
MCQ 362:
An uncovered cost at the start of the year is \$300, full cash flow during recovery year is \$650 and prior years to full recovery is 4 then payback would be
1. 3.46 years
2. 2.46 years
3. 5.46 years
4. 4.46 years
MCQ 363:
The free cash flow is \$15000 and the net investment in operating capital is \$9000 then the net operating profit after taxes will be
1. 24000
2. 6000
3. −\$6000
4. −\$24000
MCQ 364:
In cash flow estimation, the depreciation is considered as
1. cash charge
2. noncash charge
3. cash flow discounts
4. net salvage discount
MCQ 365:
The financial security kept by non-financial corporations is
1. deposit cheque
2. distribution cost
3. short term treasury bills
4. short term capital cost | 481 | 2,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-22 | latest | en | 0.804563 |
http://www.cse.chalmers.se/~nad/listings/Interrupts/Algebra.Props.AbelianGroup.html | 1,555,856,143,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578531984.10/warc/CC-MAIN-20190421140100-20190421162100-00066.warc.gz | 224,118,307 | 4,295 | ```------------------------------------------------------------------------
-- Some derivable properties
------------------------------------------------------------------------
open import Algebra
module Algebra.Props.AbelianGroup (g : AbelianGroup) where
open AbelianGroup g
import Relation.Binary.EqReasoning as EqR; open EqR setoid
open import Data.Function
open import Data.Product
private
lemma : ∀ x y → x ∙ y ∙ x ⁻¹ ≈ y
lemma x y = begin
x ∙ y ∙ x ⁻¹ ≈⟨ comm _ _ ⟨ ∙-pres-≈ ⟩ byDef ⟩
y ∙ x ∙ x ⁻¹ ≈⟨ assoc _ _ _ ⟩
y ∙ (x ∙ x ⁻¹) ≈⟨ byDef ⟨ ∙-pres-≈ ⟩ proj₂ inverse _ ⟩
y ∙ ε ≈⟨ proj₂ identity _ ⟩
y ∎
--∙-comm : ∀ x y → x ⁻¹ ∙ y ⁻¹ ≈ (x ∙ y) ⁻¹
--∙-comm x y = begin
x ⁻¹ ∙ y ⁻¹ ≈⟨ comm _ _ ⟩
y ⁻¹ ∙ x ⁻¹ ≈⟨ sym \$ lem ⟨ ∙-pres-≈ ⟩ byDef ⟩
x ∙ (y ∙ (x ∙ y) ⁻¹ ∙ y ⁻¹) ∙ x ⁻¹ ≈⟨ lemma _ _ ⟩
y ∙ (x ∙ y) ⁻¹ ∙ y ⁻¹ ≈⟨ lemma _ _ ⟩
(x ∙ y) ⁻¹ ∎
where
lem = begin
x ∙ (y ∙ (x ∙ y) ⁻¹ ∙ y ⁻¹) ≈⟨ sym \$ assoc _ _ _ ⟩
x ∙ (y ∙ (x ∙ y) ⁻¹) ∙ y ⁻¹ ≈⟨ sym \$ assoc _ _ _ ⟨ ∙-pres-≈ ⟩ byDef ⟩
x ∙ y ∙ (x ∙ y) ⁻¹ ∙ y ⁻¹ ≈⟨ proj₂ inverse _ ⟨ ∙-pres-≈ ⟩ byDef ⟩
ε ∙ y ⁻¹ ≈⟨ proj₁ identity _ ⟩
y ⁻¹ ∎
``` | 534 | 1,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-18 | latest | en | 0.227344 |
https://physics.stackexchange.com/questions/346315/calculations-involving-quantum-measurement-operators | 1,571,854,164,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987835748.66/warc/CC-MAIN-20191023173708-20191023201208-00041.warc.gz | 638,753,204 | 32,014 | # Calculations involving quantum measurement operators
In a quantum measurement theory text, the following is stated. Please advise on the query regarding the calculation below.
Consider the following measurement operator: $$\hat{A}(\alpha) = \bigg(\frac{4k}{\pi dt}\bigg)^{\frac{1}{4}}\int_{-\infty}^{\infty}\text{exp}\bigg\{-2k\frac{(\alpha - xdt)^2}{dt} \bigg\} |x \rangle \langle x | dx = \bigg(\frac{4 k}{\pi dt}\bigg)^{\frac{1}{4}} \text{exp} \bigg\{ -2k \frac{(\alpha - \hat{x} dt)^2}{dt} \bigg\}$$
where $dt$, $\alpha$ and $k$ can be considered as real numbers, and $\hat{x}$ is some observable.
Consider the state $| \psi \rangle = \int \psi(x)| x\ \rangle dx$. Can anyone see how the following two equations are obtained:
$$\int_{-\infty}^{\infty} \alpha \text{Tr}\bigg[ A^2(\alpha)| \psi \rangle \langle \psi \bigg]d \alpha = \sqrt{\frac{4k}{\pi dt}}\int_{-\infty}^{\infty}\bigg[\int_{-\infty}^{\infty} \alpha e^{\frac{-4k(\alpha - xdt)^2}{dt}}d \alpha \bigg] | \psi(x)|^2 dx =\\ dt\int_{- \infty}^{\infty} x |\psi(x)|^2 dx$$
By the way, these equations are the expectation value $\langle \alpha \rangle$.
Thanks for any help.
• which text are you talking about – Boltzee Jul 18 '17 at 14:57
First, $\hat{A}^2$ is still an operator. It is diagonal in position but you need to keep explicitly the eigenvector dependence. Your third equal in the first line of equation is wrong and you should instead write $$\hat{A}^2 = \sqrt{\frac{4k}{\pi dt}} \int \exp \{-\frac{2k(\alpha - ydt)^2}{dt}\} |y\rangle\langle y| dy \int \exp \{-\frac{2k(\alpha - xdt)^2}{dt}\} |x\rangle\langle x| dx,$$ with $\langle y | x \rangle = \delta(x-y)$, you have $$A^2 = \sqrt{\frac{4k}{\pi dt}} \int dx \ \exp \{-\frac{4k(\alpha - xdt)^2}{dt}\} | x \rangle \langle x |.$$
Second, the trace of an operator is given by summing the diagonal elements, that is to say $Tr[\hat{B}] = \int dy \ \langle y | \hat{B} | y \rangle$. With $\langle x | \psi \rangle = \psi(x)$ (or conjugate, not sure but doesn't matter here), you have $$\int \alpha Tr[A^2 |\psi\rangle\langle \psi |] d\alpha = \sqrt{\frac{4k}{\pi dt}} \int dx \left[\int d\alpha \ \alpha \exp\{-\frac{4k(\alpha - x dt)^2}{dt}\}\right] \ \int dy \langle y | x \rangle\langle x | \psi \rangle \langle \psi | y \rangle$$ with in the last integral you change $\langle y | x \rangle = \delta(x-y)$, you get there $|\psi(x)|^2$ and the rest is the same. Finally, the integral over $\alpha$ is a gaussian function and can be integrated exactly and very easily.
• Thanks for the excellent answer. The integral is not a Gaussian in the usual form as in the link, since it has a factor of $\alpha$ in front of the exponent but I will evaluate and confirm this integral computationally. Are you a physics student? – user101311 Jul 18 '17 at 16:46 | 929 | 2,784 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2019-43 | longest | en | 0.741587 |
https://coderanch.com/t/367820/java/String | 1,477,050,081,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988717963.49/warc/CC-MAIN-20161020183837-00544-ip-10-171-6-4.ec2.internal.warc.gz | 810,955,709 | 10,116 | Win a copy of Penetration Testing Basics this week in the Security forum!
# String question
Lydia Su
Greenhorn
Posts: 13
i have a question....how do you convert a user-input string into separate strings? say like if i enter 537, the program breaks it into three separate strings...and how to make these separate strings output "Five hundred and thirty seven."???
karl koch
Ranch Hand
Posts: 388
hi,
you can split it using the Strings charAt() or substring() methods.
then convert it to int[] using the Integer.parseInt() in a method in a loop.
put "", "one, "two", "three",... "ten", "twenty",..... in a 2-dimensional String array where the first row stands for the string representation of 0 -9, second for 10 - 90, third for 100 to 900 and so on.
from the int[] you can get 2 params: the index and the value at this index-> the index = row in the string array and the value = collumn in the string array.
does this sound weard ? should work.
karl
Bob Graffagnino
Ranch Hand
Posts: 81
If there is a common delimiting character you could use the StringTokenizer class. It can be especially useful when the exact position of the next sub-string is not known.
John Lee
Ranch Hand
Posts: 2545
You first need to break '537' to '5', '3', and '7'.
Then use case ststement to convert them to 'five hundred', 'thirty', and 'seven'.
Layne Lund
Ranch Hand
Posts: 3061
I would use Integer.parseInt() to convert the whole string to an int, first. Then you can use the modulus (%) and divide (/) operators to strip digits from the int. From there, either use an array of String or a switch statement to print out the tokens. I think I like the array idea best.
Keep coding!
Layne
John Lee
Ranch Hand
Posts: 2545
That is a good idea too.
I think the most time consuming part is to convert number to name, such as: one, two, ten, two hundred.
Our methods are different in term of how to get the number at each digit. But to convert them, I probably have to use a swith statement with 10 cases.
I am wondering if there is better way to do this?
Barry Gaunt
Ranch Hand
Posts: 7729
Yup, CattleDrive Java 4a/4b 200\$ a ride. | 556 | 2,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-44 | longest | en | 0.831003 |
http://digfir-published.macmillanusa.com/stat_tutor_flash_backup/stat_tutor_flash_backup_deletech_20_significance_test_for_proportion.html | 1,725,870,279,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00168.warc.gz | 8,533,937 | 7,609 | # Chapter 1. Significance Tests for a Proportion
true
Stat Tutor
true
true
1:05
### Question 1.1
udIV1JheQd9SRoG1bUlxkqD59LP4y8LQqUJtXnHf1E8zneqFygtmfMPozaIl7t9yQxBvYiu9YX4jiJ0h9mHJytNee7qEfxm6dZbJLyzRywilmJrqIeGzlIOJTcnPVsISP3bKhNoFyLYwXDTL0QzbEEnDZo9R926lXTSZA642glKozNkzSQxwihA9mAK48fPLr2pnxsq70NgiIdElPhtngWr/x+KyfRfLtyRLOwi00KdajZ2OugzWgniaa/FEyB19
Incorrect. This is true provided np and n(1 - p) are both bigger than 10.
Correct. This is true provided np and n(1 - p) are both bigger than 10.
Incorrect. Try again.
2
1:10
### Question 1.2
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
Incorrect. The problem is that we don't know the value of p. That's why we are performing a test of significance. Note: We can get the value of $$\widehat{p}$$ from the sample and the value of $$p_{o}$$ from the null hypothesis.
Correct. The problem is that we don't know the value of p. That's why we are performing a test of significance. Note: We can get the value of $$\widehat{p}$$ from the sample and the value of $$p_{o}$$ from the null hypothesis.
Incorrect. Try again.
2
3:45
### Question 1.3
yiqbms0bPQP4L+pBbC2NEYapqMOHX18OfvgLi4335bgwUs6joWz5eav6KWBBlqykJcOSJK0L8tbdwLYCh8Q6AAJwevJqeC2HW6rULChRxrx35UARPBigQQdTZf4P9QWxo1eiR8K1L1kW01gPztpRVHWrva9CWItiYbQFLOP4t5Mrz2nLdNrfBmhGAxAhstI/Rhv3n/bycF+dqwkooAE9idP1oMuQTix19rjG227DrKUynQQ4G4oqR/8wLtXqII6oOlOim5mRPjoaYMRWO/hpJ1iul/OKQDlEfPpbZF5+yICztdOXPhZ8dA==
Incorrect. Since we assume the null hypothesis to be true, we assume that $$p= p_{o}$$ so we use $$p_{o}$$ as the value of p whenever we need a value for p.
Correct. Since we assume the null hypothesis to be true, we assume that $$p= p_{o}$$ so we use $$p_{o}$$ as the value of p whenever we need a value for p.
Incorrect. Try again.
2
4:51
### Question 1.4
g5ke0AbSjLRC04bcTKftLau521soN83IWugmJTwaMNOyNLKOiLYKf6j8DvHhnrRuyt9VtWBFvRZqSeCmjfhasiHor7CVviGBC60perv/fK0IvYeQWiWySuIe7ENLkA0LhxTz/SYIVERu9dulx8g2Xg/86GV5zacSa74a73IaxBoeiBY3GyNOYtbTnqpNmMNkAx+RLhttefHZ5RsQ7qotF6wRWmr679H9+LdyxGruivs15wtgTsUt5OwD9EABksl0MN3ZZ7x/nS0IHk1I2oS77Q==
Incorrect. Data should be collected with a simple random sample.
Correct. Data should be collected with a simple random sample.
Incorrect. Try again.
2
5:32
### Question 1.5
Incorrect. The response variable is a measure on the individual. Each student is asked whether they feel "being very well-off financially is an important personal goal," so that is the response variable and it is categorical.
Correct. The response variable is a measure on the individual. Each student is asked whether they feel "being very well-off financially is an important personal goal," so that is the response variable and it is categorical.
Incorrect. Try again.
2
### Question 1.6
Incorrect. The research question basically asks, "Is the proportion . . . significantly different from . . .73%?" Thus, the alternative hypothesis is p $$\neq$$ 0.73.
Correct. The research question basically asks, "Is the proportion . . . significantly different from . . .73%?" Thus, the alternative hypothesis is p $$\neq$$ 0.73.
Incorrect. Try again.
2
6:19
### Question 1.7
NnCrCRthGCcuvrkoHEtwDlKGNKcxYE7UPhJgK6D1nDMq6oaAWsKqC0hO1dkziJueZAaWxkJg1K4SAQjGFMNi+fdyWzn7/vrOvaY3qRODs9s1U/O5gbUcEq5rZuEuDeT9bGFiaCYlUL1dDRWy9HY9Sf9qYm7b1Q4CHxTyjGwHbew5QqTEHoMNqmWyFuPDAceJ2lErRFc/w8k8s84nh0y4KEJDS2c=
Incorrect. Whenever the response variable is categorical, we do inference on proportion.
Correct. Whenever the response variable is categorical, we do inference on proportion.
Incorrect. Try again.
2
6:47
### Question 1.8
IXavvQcxEpefdmqTak4/YQn35U+8qlqA+oAfi5mJ0v7OrHn6urg9070+Cj8z4mhk14Y19yHc1d9wJo3lqbSt6rPsulTfODfYb9wKfTJcEsXcJF/UosaQimqxJXzloNdaxsgL9mBAtbqdnPgin7VjOG9EamGkma6MirLjpvN5vQoPMlaosThCWvtoJWuRD2e9FOCyBVuWHtBNImo+d6NvDgMOPZEHrei24ZqfGkASlgFLVx5Beo/Z2vAQoYQ+T5nW9nddlOTcA61ogWSxNaBC4TIrw7uRIie67igkPfnh0/2ePkAZq78YFT7Wd+uXBYfjLSGATpP+hiHy9QSmucYP1MzStNKJDv4GJ46tUvHf8OKawOdPvz1/yEJ8OpNhCzMrHU4xOC86R1WHXX8ZZA3z/g1jFb48eItjijXqero+YxDau31zqiMoZOhzBgAWsjR86HiliA==
Incorrect. Since we assume the null hypothesis to be true, we assume that $$p= p_{o}$$ = 0.73 so we use $$p_{o}$$ = 0.73 and check $$n p_{o} \geq 10$$ and $$n(1- p_{o} ) \geq 10$$ to see if n is large enough.
Correct. Since we assume the null hypothesis to be true, we assume that $$p= p_{o}$$ = 0.73 so we use $$p_{o}$$ = 0.73 and check $$n p_{o} \geq 10$$ and $$n(1- p_{o} ) \geq 10$$ to see if n is large enough.
Incorrect. Try again.
2
9:22
### Question 1.9
p0Gp0GF+/YJCu4lb4e8xmyK7BcedF4d0vHm4TLcEc+Gnpe7zOVxU+4fzQGaG+iNVT+UhoZh6Z2I5DHDWIzKUiBkObeOZ52Oka6kM0e6VI3cintAoXuYRTyPiP9XuaTUR
Incorrect. Because P-value = 0.0258 is less than $$\alpha$$ = 0.05, we reject H0.
Correct. Because P-value = 0.0258 is less than $$\alpha$$ = 0.05, we reject H0.
Incorrect. Try again.
2
9:37
### Question 1.10
Incorrect. Since we rejected H0, we conclude that Ha is correct. Thus the proportion differs significantly from 0.73.
Correct. Since we rejected H0, we conclude that Ha is correct. Thus the proportion differs significantly from 0.73.
Incorrect. Try again.
2
11:40
### Question 1.11
3x002b0bNKTaZ9TPXAejh4rt/bbXmmc77W562ew9jm1f7REVt0TaGM99eTuKvygpP8O2ZI1zIOUC+KvMWCugCgW5t65/Seq/24dq6W9+jllsPfPedMTXkw/sOUZV1g+BGsvmKuxNXLWCTSuC6Bz5LcidkZJ99TZZoc1h/Ulfpa8=
Incorrect. $$\widehat{p}$$ = $$\frac{40}{6000}$$ = 0.0067
Correct. $$\widehat{p}$$ = $$\frac{40}{6000}$$ = 0.0067
Incorrect. Try again.
2
12:42
### Question 1.12
nLoJjI20I/uU9bofqqe0AsZ9omCmHJZkFHdaYH4X8uMFOIHWjxHQUTkFA0A+PqvIyfPn25SPU6K/PwHSwIgOCruJX+f6ogBp7SVrjs+QWN4DYE/ixYyhc2dq69fwo1MOHm5Gt0J6hdYT7/uafsKsQ/CKpsgoGBMdiNPzZUl0tkLvTsVlblLUzMW0+mZcZwbUgcahi6ZovSp1+5KBRTWMRFbBIwJ6FWcwIFMWJ3TRtvokH3orsEOULcsCOyxqTV0s7RR8sB1xKTAKy+fWs/v0KPSPbhsqeWSE2ajHwwHvcteoagSUs/qC2al35aLdMJCnCLQcWSvSMxI3pE8oSEAfLOPfROcAQLTYrFUTVfCVIuI=
Incorrect. All 6,000 three- to ten-year-olds in Brick Township were studied.
Correct. All 6,000 three- to ten-year-olds in Brick Township were studied.
Incorrect. Try again.
2
17:46
### Question 1.13
Mzyi/IJFDE82EmgpxyuYQ2KqSBLo5/GVD6G6AWybWGZfqhEDTKWjC762kxBwM2TgcCJJCZjayTYpJK4JsqK3MhpTcxWcxT8KhS301xxrT0p8Un63VmYIoDYEqn9y5rled7lGQUtE/RvRUjd3WMbna20ci7USgmBdgPA1uQ==
Incorrect. We use the standard Normal table to find the P-value for testing proportion.
Correct. We use the standard Normal table to find the P-value for testing proportion.
Incorrect. Try again.
2 | 3,151 | 6,851 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.587782 |
http://www.teachexcel.com/excel-help/excel-how-to.php?i=150528 | 1,448,851,032,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398460942.79/warc/CC-MAIN-20151124205420-00094-ip-10-71-132-137.ec2.internal.warc.gz | 708,180,894 | 19,820 | Email: Pass: Pass?
E-mail:
# If Condition Formula
How can I make a IF formula to pop up a text message if a particular cell is filled by number or text.
What I don't know is how to give a condition on entering anything into the defined cell (numbers or letters) to make the message pop up.
not
=IF(C4= "company ","Name"," ")
I need this defined condition (i.e Company) to be removed & let the formula to accept anything if entered into C4
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## Similar Topics
Hello,
My Cell C3 is a numeric value.....I have set conditional formatting to
make the text red and bold when the number is equal or less than 10,000
is there a way I can make cell E3 display a message when the C3
condition is true?? or if not a message is there a way to make a
message box pop up when my C3 condition of less then or equal to 10,000
is true?
how would I do something like this?
My formula is checking for a certain condition. When the condition is met, I want it to return an error message, so the result cannot be summarized into an aggregate number for that column and I can easily spot it later in a pivot table. The complexity is that I am trying to create an custom message showing "NEED PRICE" as a result of that cell. I am trying to use an IFERROR formula, but since the result is still a text ("NEED TEXT"), the cell simply gets skipped over when being summarized and the summary of that column (Total Sales) still contains a number. How can I make the result of the cell read as an error (so it doesn't tally that column), but still have the custom text "NEED TEXT" in there?
Guys ,
I need some help here regarding to If condition , what i need it is how to make If condition help me to solve an equation (Specific formula ) and this formula will be decide it upon the condition if agree or not and showing the reult in different cell , to make it easer for you this is an example
IF H1 > 0 THEN I NEED TO APPLY A FORMULA OF A1= x+y
if H1
Hi,
I am trying to build a code where some message needs to pop out when a certain condition is met.
lets say the condition is that Cells("B12").Value > 500.
This cell has a formula which is the sum of some other cells. As soon as this sum crosses 500, i need a message to pop up using msgbox.
Note: The if conditions will not work for me it checks the cell value only when it runs. As in, if the value in the cell is already greater than 500 at the time of running the code, then only it will pop up the message. i want the code to be such that when it runs, it should put some condition in the cell so that message pops up if the value exceeds 500 even later. (like how conditional formatting works...its not just an if condition there...whenever the given condition meets, the format changes).
I searched a lot and found lots of info about conditional 'formatting' but couldn't work my way around this.
Any help would be appreciated.
Regards,
nonie
Hey guys,
Total newb question but if I have an if/then condition in a cell that references another cell for that condition, is there a way to make the referenced cell condition a text wildcard? As in if(C40=*"John"*, X, Y). And the condition will be true if C40 contains the word John in it along with any other text.
Does this make sense?
J.
Hi everyone,
I have a nested IF function I'm having issues with. My workbook is laid out as follows:
Sheet1 contains two drop-down lists, in Cells E2 and F2, and a number field, G2.
The list in E2 may have the following values: "Condition 1" or "Condition 2". The list in F2 may have the following values: "Condition 3" or "Condition 4". Both are text fields.
I need to evaluate for the appearance of the following combinations: ("Condition 1" AND "Condition 3") OR ("Condition 1 AND "Condition 4"), ("Condition 2" AND "Condition 3") OR ("Condition 2 AND "Condition 4").
The results of the conditions use the value of the free-form number field, G2, and manipulate it referencing various values on Sheet3. I'm confident the results portion (the manipulations of G2) are correct, it's the formatting of the IF statement I'm having problems with.
I created the following statement to try to accomplish this:
=IF(AND(E2="Condition 1", F2="Condition 3"), G2*Sheet3!C1, IF(AND(E2="Condition 1", F2="Condition 4"),(G2*Sheet3!C1)*Sheet3!F4, IF(AND(E2="Condition 2", F2="Condition 3")(G2*Sheet3!C1)*Sheet3!F2 IF(AND(E2="Condition 2", F2="Condition 4") (((G2*Sheet3!C1)*Sheet3!F2)*Sheet3!F4), 0))))
However, it doesn't work, Excel tells me "The formula you typed contains an error." I've been searching this site but can't seem to find a readily available answer.
I realize I could be way off, or it could be an easy fix -- I'm new to more complex logic in Excel, so I'll thank you in advance for your patience with the newb.
Can anyone give me a correctly-formatted version of that statement, or point me to where I need to look to figure it out myself?
Many thanks,
Rich
Is there another formula that works like vlookup but isn't concerned about the source data being in ascending order? I know of one that works when you can look up based on 2 conditions but can't seem to make it work when I only have 1 condition to meet.
My 2 condition formula is a CSE and is: =index(array to lookup(match(1,array of first condition=condition)*(array of 2nd condition=2nd condition))
Can I do something like the above if I only have one condition to meet?
I am trying to condition format cells that have a formula in them. I would like the condition to be based onthe result of the formula. The first condition I want is if the result is between tomorrow and 3 months from now I would like to make the result written in green. I tried the following to achieve this:
"cell value is","between","TODAY()-90","tODAY()-1"
The other condition I want is if the result is today or an earlier date, I would like the result to be written in red. I tried this to achieve:
"cell value is","less than or equal to","TODAY()"
Any other result I would like to be left alone. My only possible results will be a date or the test "N/A". But my results seem to only satisfy the second condition because everything was written in red.
Hello,
I am trying to conditionally sum numbers over 52 worksheets based upon up to 3 conditions. One condition is a date, the other condition is a company name. Right now I am using only 2 conditions but may add a third.
I am using an array formula to get the result I need for that worksheet. Here it is:
{=SUM((I4:M4=V5)*(F6:F42=V4)*I6:M42)}
My question is this.....How do I get the results I need for all 52 worksheets in one formula?
All information on all worksheets are in the exact same cell references.
My goal is to have a reporting worksheet that has all 365 days of the year and depending on what company is is in the first cell; all dates have the respective number that for that day that is contained on another worksheet.
I know my description is a little vuage, if I need to clarify, please let me know.
Code:
```=IF('Sheet 1 '!N5="Condition 1",1,"")&IF('Sheet 1 '!N5="Condition 2",2,"")&IF('Sheet 1 '!N5="Condition 3",3,"")
```
I use the above formula down a column in sheet 2 of my worksheet. Basically, it matches the text found in column N of sheet 1 to a corresponding code and inserts that into the cell the formula is in.
However, when I delete the row that the formula references in sheet 1, I get the ref# error in the corresponding cell in sheet 2.
Is there any way to make it more robust so that if the referenced row in sheet 1 is deleted, it will then reference the new row that replaces it?
I somehow want to make an error if a certain condition is not met and have a message box pop up to explain the error and then end my macro, but continue if there is no error.
The condition is this: if the number (value) in a particular cell is not equal to the total number of sheets in my workbook then I want the macro to end and display the message: "You need to Press Ctrl+Shift+C."
The value of the particular cell can be found by the following statement:
Sheets("Initial Input").Range("O10").Value
and the total number of sheets should be able to be found with something like this:
Application.Sheets.Count
I was thinking about using an if statement, but I'm open to using any kind of idea.
Hi friends,
MrExcel forum has been of great help in learning the uses of Sumproduct function as I searched many useful threads on this.
Now I am trying to find out how to apply AND, OR, NOT combination for getting a count/sum.
e.g.
1. Count/sum, If (condition 1) OR (Condition 2) OR (Condition 3) are met
2. Count/sum, If (condition 1) AND ((Condition 2) OR (Condition 3)) are met
3. Count/sum, If ((condition 1) AND ((Condition 2)) OR ((Condition 3) AND (Condition 4)) are met
4. Count/sum, If ((condition 1) OR ((Condition 2)) AND ((Condition 3) OR (Condition 4)) are met
5. Count/sum, If (condition 1) is NOT met
6. Count/sum, If (condition 1) OR (Condition 2) NOT met
7. Count/sum, If (condition 1) AND (Condition 2) NOT met
and similar ones.
Can anyone put light on general arrangements for AND, OR, NOT combination in Sumproduct function. I could not find any thread on this.
Regards,
Uttam
Can anyone help with this please? The formula I have checks for these conditions:
- to see if there is text is cell C221, and if there isn't then the target cell remains blank. (Working).
- if there is text in cell C221, and cell G221 is empty then the target cell should say "No". (Broken! Returns 'FALSE' value instead of "No"...)
- if there is text in cell C221, and cell G221 contains one of the 5 criteria, then the target cell should say "Yes!". (Working).
Here is the formula - how do I fix the broken bit?
=IF((C221=""),"",IF((G221="Condition 1"),"Yes",IF((G221="Condition 2"),"Yes",IF((G221="Condition 3"),"Yes",IF((G221="Condition 4"),"Yes",IF((G221="Condition 5"),"Yes"))))))
Many thanks,
Jeremy.
Two queries:
I'm trying to do an IF(AND where the first logical is that one cell contains the specified text and the second logical is that another cell in a multiple of 5000 but i'm completely stuck.
If TRUE give me the text "Loan"
If FALSE give me a blank cell
1st condition: Find some (not all) the text within cell S1. The cell will start with "ABC LIMI" but then can contain an random combination of numbers, spaces, *'s or letters.
2nd condition: Cell I1 value is a multiple of 5000 -
=IF(AND(S1="ABC LIMI",I1 is a multiple of 5000),"Loan", zero)
Any chance this can be done?
Hi,
I have several hundreds of different formulas in one column. I need to modify every formula. Each formula should be extended with if statement. This statement is almost identical for all the formulas - condition is always same, the only part that differs is that it should compare the value of the cell that is on the left for the formula. So if I give the address, the row number would differ.
I thought I reach my aim by CONCATENATE function. All I would need is to:
1) take actual formulas as text
2) prepare if condition for each row by dragging
3) CONCATENATE texts
4) Paste a new formula as text into cells that where old formulas were.
Unfortunately I don't know how to copy the text (not values) of my formulas somewhere else
Best,
Michal
Hello Everyone,
I have 1700 company names populating the column of a large table. All the company names are text, without any zero or blank entries. There are, however, many duplicate names in this column.
I'm doing some analysis that requires me to calculate the number of unique company names. For reasons that are too complex to explain here, it is not appropriate in this situation to filter out, hide or remove the duplicates names (and the rows of data of which they are a part).
So, I found a lovely little array formula that counts the distinct values in a table column that only contains text entries. In this case, the name of my column is "C7", for column seven. Here's the formula:
{=SUM(
1/COUNTIF(Table[C7],Table[C7])
)}
This formula returns an intelligent answer of 1135 unique company names.
I've also realized that I can add a simple little condition that will exclude some specific entries (though not necessarily all instances of that company name) before the number of distinct names is counted. That formula is:
{=SUM(
IF(Table[C8]="Yes",
1/COUNTIF(Table[C7],Table[C7])
))}
The answer now is an intelligent 630, meaning that 630 unique company names remain in the column after removing those entries whose adjacent cell in column C8 is not "Yes".
Now, here's where things go haywire. If I add a second condition, like this:
{=SUM(
IF(Table[C8]="Yes",
IF(Table[C12]<>0,
1/COUNTIF(Table[C7],Table[C7])
)))}
Suddenly, the answer is a nonsensical 591.9184412 !!!
How can that be? At a minimum, I expect a whole number answer from this formula! What is happening? Have I stumbled on an Excel bug?
Cheers,
Jay
PS: I am, however, looking for a single (array) formula that will go into a single cell.
Hi everybody,
I have a text file named "C:\Text.Txt" with records in the following format:
EventDate Message
15-04-2009,Send Dunning Letters
30-06-2009,Issue Second Qtr Reports
I need a macro that should work when excel starts and read each record in the above text file and display the message given after the EventDate on a message Box if the following condition is met per record:
If the EventDate minus Today's Date is >= 0 and < 60
If multiple records meet the above condition, then those records should be concatenated and be displayed on a single message box rather than many message boxes.
The idea is to pop up a message box when excel starts and displays the events to be done within the next 60 days.This serves as a reminder.
Hope I have explained well.
Thanks
Hello! I need a bit of help, please. I have a running spreadsheet that uses the mod to highlight alternate cells. I've set it up for the entire document under conditional formatting as =MOD(ROW(),2)=0
I have one particular column that contains a sum formula. If the sum=0 I want the text to appear clear or white (or ideally be deleted) so as not to appear distracting as a column of zeros with an occasional value. It just seems to make the true values harder to spot. Anyway, I added a second condition to this column that simply states if cell value=0 then format text as "white".
The rows not higlighted by the first condition work fine and the text is white, but the text in the highlighted rows remains black in this column. I tried alternating the order of the two conditions, but then the second condition cancels out the alternating highlights from the first condition.
any suggestions?
thanks
bj
Hello all,
After a bit of googling and searching these forums, I can't quite find an answer to what seems like a pretty simple problem.
I have a list of around 20 companies that belong to either Group A, or Group B. These companies and their groups are listed in Column A (Company) and Column B (Group) on Worksheet B;
Company Name........Company Group
Name 1...................Group A
Name 2...................Group A
Name 3...................Group B
Name 4...................Group A
Name 5...................Group B
etc..........................etc
On Worksheet A, in Column A, company names (that match company names on Worksheet B) are manually entered. I would like Column B to populate with the company's group as defined on Worksheet B, Column B automatically. This is to speed up data entry, so the person entering each company doesn't have to remember, or look up that company's group (A or B).
In short, if I enter "Name 3" in Column A, I would like Column B to automatically populate with the text "Group B" as defined on Worksheet B.
I'm not quite sure where to begin. Company names will be added and removed from time to time, so ideally I would just update Worksheet B and those changes would be reflected on Worksheet A.
Any help would be greatly appreciated.
Thanks.
I am having trouble with conflicting conditional formatting. I have a list
that I need to have highlighted in one of three colors based on the result of
the formula in column D, and I need to have the text adjusted according to
the result in column G. I have tried having the text formatting as condition
one and shading as condition one, in either case, only the first condition is
applied. (Although the cells that meet only one of the conditions format
properly.)
Here are my conditions:
Condition 1: Formula is =\$D23>40 (Shading)
Condition 2: Formula is =\$D23>25 (Shading)
Condition 3: Formula is =\$D23=1 (Text formatting)
I though of using an AND function, but had trouble getting it to work and I
have 6 potential states and the limit of three will not cover them:
--
L
When I post the following message I can't get all of the condition 1 formula to show up in the post. What do I need to do?
Cells C16:C29 have conditional formatting as follows
First Condition (\$C16<AVERAGE(\$A16:\$B16)-\$C\$10
Second Condition (\$C16>AVERAGE(\$A16:\$B16)+\$C\$11
Goldi
Looking to input a formula in A1 to check for a condition if B1=E1. If the condition is true, C1 needs to be populated with a text string already inputed into D1. Instead, I just get the (True/False) condition in A1. Sample formula below:
=IF(B1=E1,C1=D1,"")
I understand that the issue lies in the C1=D1 declaration. Help would be greatly appreciated.
Thanks, cashaau
Hi
I have a problem in getting a Pop up message in excel. I dont know whether formula can do the trick or should i take a help of micro.Kindly advise which is preferable and the solution for the same.
The problem is :-
I have 2 conditions and if any of the condition comes true , i want a pop up stating that "you have to calculate"( It should serve like a reminder).
Condition 1: In cell B if i insert "CA", and in cell E if i insert "no" and in cell G if i insert "lease".. I want a pop up stating "you have to calculate"
Condition 2: In cell B if i insert "TX" and in cell H if i insert "yes".. I should get a pop up "you have to calculate"
Thanks
Hello, I have a various difficulties within this excel problem.
counting combinations.xlsx
I must calculate all combinations of 7 numbers that give sum of 100. Numbers must be in interval from 1 to 39.
First condition is that sum of numbers must be changeable (i mentioned before sum of 100 to be simpler so I typed number 100 in the document). The sum can be also 150 or 88 for example, depending what number is entered in cell K2.
Second condition is that every combination must be unique (no such things like 7,8,9,10,16,20,30 and 30,20,16,10,9,8,7 and 7,8,10,9,3,20,16). This is example how results shouldn't be because all 3 combinations have same numbers only with different position/order). Only one combination of same numbers is possible and numbers must be ordered from smaller to larger numbers).
Is this possible to do, what is best option: functions or maybe some macro?
I am new to Excel and conditional formatting, but from reading threads and various sites I believe that the conditions have to be in a sort of reverse order.
The users input would be starting from Row 8.
Column A & B would be text and Columns C through to AH would be dates in the format 02-Feb-07.
I am after trying to do the the following:
Condition 1
If all cells in a row from Column A through to Column AH are not empty then make text white and backcolor black. (Only in column A would be ideal, but not 100% necessary)
Condition 2
If some cells are full and some cells are empty, make the empty cells backcolor Yellow. (Again, it would be nice to make this conditional that the cell in Column A had text in it)
Condition 3
I would like to conditionally format the cells Range(C8:AH1000), but until a row is used, by the user adding input through a form or onto the sheet, I would like to have the cells with no backcolor.
If I could meet this part of Condition 2:
Quote:
it would be nice to make this conditional that the cell in Column A had text in it
then condition 3 would not apply.
At the moment this is what I have for the 3 conditions:
Condition 1
Code:
```=Len(Range(A:AH))=0
```
then there is no backcolor
Condition 2
Code:
```=Len(Range(A:AH))=9
```
then Black backcolor and white text
Condition 3
Code:
```=Len(Range(C:AH))<>9
```
empty cells backcolor = Yellow
At the moment this doesn't work, either if the complete row is fully used, or if the cells are empty.
Any pointers in the right direction would be great.
Thanks | 5,409 | 22,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2015-48 | longest | en | 0.789199 |
https://research.stlouisfed.org/fred2/series/A10057USA144NNBR | 1,432,478,283,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928019.31/warc/CC-MAIN-20150521113208-00152-ip-10-180-206-219.ec2.internal.warc.gz | 891,989,554 | 19,603 | # Savings of Government for United States
1938: -175 Millions Of Dollars (+ see more)
Annual, Not Seasonally Adjusted, A10057USA144NNBR, Updated: 2012-08-17 3:16 PM CDT
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
Extra Digits Are From Worksheets. Averages: 1919-1928=1368.6; 1929-1938=-53.1; 1919-38=657.8 Source: Simon Kuznets, National Income And Its Composition, 1919-1938, (NBER), P. 276
This NBER data series a10057 appears on the NBER website in Chapter 10 at http://www.nber.org/databases/macrohistory/contents/chapter10.html.
NBER Indicator: a10057
Release: NBER Macrohistory Database
Restore defaults | Save settings | Apply saved settings
w h
Graph Background: Plot Background: Text:
Color:
(a) Savings of Government for United States, Millions Of Dollars, Not Seasonally Adjusted (A10057USA144NNBR)
Extra Digits Are From Worksheets. Averages: 1919-1928=1368.6; 1929-1938=-53.1; 1919-38=657.8 Source: Simon Kuznets, National Income And Its Composition, 1919-1938, (NBER), P. 276
This NBER data series a10057 appears on the NBER website in Chapter 10 at http://www.nber.org/databases/macrohistory/contents/chapter10.html.
NBER Indicator: a10057
Savings of Government for United States
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Graph Data
Graph Image
Suggested Citation
``` National Bureau of Economic Research, Savings of Government for United States [A10057USA144NNBR], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/A10057USA144NNBR/, May 24, 2015. ```
Retrieving data.
Graph updated.
#### Recently Viewed Series
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https://www.oreilly.com/library/view/sql-and-relational/9781449319724/ch07s14.html | 1,695,855,550,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510326.82/warc/CC-MAIN-20230927203115-20230927233115-00308.warc.gz | 995,675,388 | 16,631 | # EXERCISES
7.1 For each of the following Tutorial D expressions on the suppliers-and-parts database, give both (a) an SQL analog and (b) an informal interpretation of the expression (i.e., a corresponding predicate) in natural language. Also show the result of evaluating the expressions, given our usual sample values for relvars S, P, and SP.
1. `S MATCHING ( SP WHERE PNO = ‘P2’ )`
2. `S NOT MATCHING ( SP WHERE PNO = ‘P2’ )`
3. `P WHERE ( !!SP ) { SNO } = S { SNO }`
4. `P WHERE SUM ( !!SP , QTY ) < 500`
5. `P WHERE TUPLE { CITY CITY } ∈ S { CITY }`
6. `EXTEND S : { TAG := ‘Supplier’ }`
7. ```EXTEND ( S MATCHING ( SP WHERE PNO = 'P2' ) ) :
{ TRIPLE_STATUS := 3 * STATUS }```
8. `EXTEND ( P JOIN SP ) : { SHIPWT := WEIGHT * QTY }`
9. `EXTEND P : { GMWT := WEIGHT * 454 , OZWT := WEIGHT * 16 }`
10. `EXTEND P : { SCT := COUNT ( !!SP ) }`
11. ```EXTEND S :
{ NP := COUNT ( ( SP RENAME { SNO AS X } ) WHERE X = SNO ) }```
12. `SUMMARIZE S BY { CITY } : { SUM_STATUS := SUM ( STATUS ) }`
13. ```SUMMARIZE ( S WHERE CITY = 'London' ) PER ( TABLE_DEE ) :
{ N := COUNT ( SNO ) }```
Note: Tutorial D allows the PER clause to be omitted, in which case PER (TABLE_DEE) is assumed by default. The foregoing SUMMARIZE could therefore be simplified slightly, thus:
`SUMMARIZE ( S WHERE CITY = 'London' ) : { N := COUNT ( SNO ) }`
14. `EXTEND SP WHERE SNO = ‘S1’ : { SNO := ‘S7’ , QTY = 0.5 * QTY }`
7.2 In what circumstances (if any) are r1 MATCHING r2 and r2 MATCHING r1 equivalent?
7.3 Show that RENAME isn’t primitive.
7.4 Give an expression involving EXTEND instead of SUMMARIZE that’s logically equivalent to the following: ...
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# Section7_3_review - Section 7.3: Logarithmic Functions The...
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Unformatted text preview: Section 7.3: Logarithmic Functions The logarithmic function f x ( 29 = log a x , a ≠ 1 , is the inverse of the exponential function f x ( 29 = a x . The natural logarithmic function f x ( 29 = ln x , (base e), is the inverse of the natural exponential function f x ( 29 = e x . For a 1 , the graph of f x ( 29 = log a x (and also of f x ( 29 = ln x ) is shown below and has the following characteristics: Slowly increasing Vertical asymptote at the y-axis Passes through the point 1,0 ( 29 Domain: 0, ∞ ( 29 Range: -∞ , ∞ ( 29 lim x →∞ log a x = ∞ lim x → + log a x = -∞ Properties of Logarithmic Functions ( x and y are greater than 0 and r is a real number) i) log a xy ( 29 = log a x + log a y ii) log a x y = log a x- log a y iii) log a x ( 29 r = r log a x iv) log a a x ( 29 = x , or ln e x = x v) a log a x = x , or e ln x = x To solve a logarithmic equation (one in which the variable appears in the argument of the log) i) Isolate the log containing the variable.Isolate the log containing the variable....
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## This note was uploaded on 01/12/2010 for the course MATH 44245 taught by Professor Famiglietti during the Fall '07 term at UC Irvine.
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A4: Factors on which running time of a program depends on are.
Instructor Contact: send private message to Instructors on Piazza. Lectures: Tuesday and Thursday am in Annenberg G
## Design and Analysis of Algorithm Notes PDF | B Tech 2021
CPS Algorithms Lectures. Homework - Handouts - Teaching Assistants - Resources. Current homework is available from the homework page. Topics and Lecture Notes. Required Readings and Lectures in Bold See below for parenthesis for credits for lecture notes.
The aim of these design and analysis of algorithms handwritten notes is to give you sufficient background to understand and appreciate the issues involved in the design and analysis of algorithms. Sc, B. Tech CSE, M. Tech branch to enhance more knowledge about the subject and to score better marks in the exam. Algorithm Design Techniques: Iterative technique: Applications to Sorting and Searching review , their correctness, and analysis. Divide and Conquer: Application to Sorting and Searching review of binary search , merge sort, quick sort, their correctness, and analysis.
Skip to search form Skip to main content You are currently offline. Some features of the site may not work correctly. Khuller Published Computer Science. This course has been taught several times and each time the coverage of the topics differs slightly. Here I have compiled all the notes together. The course covers core material in data structures and algorithm design, and also helps students prepare for research in the field of algorithms. The reader will find an unusual emphasis on graph theoretic algorithms, and for that I am to blame.
## The Design and Analysis of Algorithms
The subject important topics, similar books , etc, were also mentioned below. And, types and overview of the subject were also mentioned. Design and Analysis of Algorithms is a very important type of problem in the branch of information technology and computer science. An Algorithm is a step by step by process to solve any problem. These four are important topics of the Design and Analysis of Algorithms subject. And, the download links were also mentioned below in a table format.
We provide complete design and analysis of algorithm pdf. Design and Analysis of Algorithm lecture notes includes design and analysis of algorithm notes, design and analysis of algorithm book, design and analysis of algorithm courses, design and analysis of algorithm syllabus , design and analysis of algorithm question paper , MCQ, case study, questions and answers and available in design and analysis of algorithm pdf form. So, students can able to download dda design and analysis of algorithm notes pdf. Design and Analysis of Algorithm Notes can be downloaded in design and analysis of algorithm pdf from the below article. Detailed design and analysis of algorithm syllabus as prescribed by various Universities and colleges in India are as under. You can download the syllabus in design and analysis of algorithm pdf form.
Lecture 1 - Introduction to Design and analysis of algorithms. Lecture 2 - Growth of Functions (Asymptotic notations). Lecture 3 - Recurrences, Solution of.
## CS8451 Notes Design and Analysis Of Algorithms Regulation 2017 Anna University
Course Outline The course consists of 4 lecture hours per week. The basic thrust of the course would be to study design paradigms for algorithms and their analysis. We will try to stick to the basic course outline as given in this page , but may deviate a bit. We would assume in this course that you have undergone the Introduction to Programming and Data Structures and Discrete Mathematics courses and have some knowledge of elementary discrete probability.
### Design and Analysis of Algorithms - CS8451, CS6402
Отпусти ее, - раздался ровный, холодный голос Стратмора. - Коммандер! - из последних сил позвала Сьюзан. Хейл развернул Сьюзан в ту сторону, откуда слышался голос Стратмора. - Выстрелишь - попадешь в свою драгоценную Сьюзан. Ты готов на это пойти. - Отпусти. - Голос послышался совсем .
Директор нахмурился и повернулся к экрану. - Мистер Беккер, я был не прав. Читайте медленно и очень внимательно. Беккер кивнул и поднес кольцо ближе к глазам. Затем начал читать надпись вслух: - Q… U… 1…S… пробел… С, Джабба и Сьюзан в один голос воскликнули: - Пробел? - Джабба перестал печатать. - Там пробел. Беккер пожал плечами и вгляделся в надпись.
ARA обслуживает в основном американских клиентов. Вы полагаете, что Северная Дакота может быть где-то. - Возможно. - Стратмор пожал плечами. - Имея партнера в Америке, Танкадо мог разделить два ключа географически. Возможно, это хорошо продуманный ход.
Он будет стрелять с бедра, направляя дуло вверх, в спину Беккера. Пуля пробьет либо позвоночник, либо легкие, а затем сердце. Если даже он не попадет в сердце, Беккер будет убит: разрыв легкого смертелен. Его, пожалуй, могли бы спасти в стране с высокоразвитой медициной, но в Испании у него нет никаких шансов. Два человека…. И вот Халохот уже за спиной жертвы. Как танцор, повторяющий отточенные движения, он взял чуть вправо, положил руку на плечо человеку в пиджаке цвета хаки, прицелился и… выстрелил.
Это было сделано тайно. - Мидж, - сказал Бринкерхофф, - Джабба просто помешан на безопасности ТРАНСТЕКСТА. Он ни за что не установил бы переключатель, позволяющий действовать в обход… - Стратмор заставил . | 1,352 | 5,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-39 | latest | en | 0.918997 |
http://photoeditorx.com/games/caribbean-stud-poker/ | 1,653,736,645,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00611.warc.gz | 57,632,454 | 16,506 | token wikiCaribbean Stud Poker - ugslot of Odds
# ugslot
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# Caribbean Stud Poker
## Rules
1. Player makes an ante wager plus an optional \$1 progressive side bet
2. Each player and the dealer get five cards each. All cards are dealt face down, except one dealer card is exposed. The player may examine his own cards but sharing of information is not allowed.
3. Player must fold or raise.
4. If player folds he forfeits his cards, ante bet, and side bet (if made)
5. If player raises then he must make a raise wager exactly equal to twice the ante
6. The dealer will turn over his other four cards
7. The dealer must have an ace and a king or higher to qualify. In other words, the lowest qualifying hand would be ace, king, 4, 3, 2 and the highest non-qualifying hand would be ace, queen, jack, 10, 9. If the dealer does not qualify the player will win even money on his ante wager and the raise will push.
8. If the dealer qualifies and beats the player, both ante and raise will lose.
9. If the dealer qualifies and loses to the player, then the ante will pay even money and the raise according to the posted pay table. The U.S. pay table is shown below.
10. If the player and dealer tie, both ante and raise will push.
11. The progressive side bet will be entirely based on the poker value of the player's hand. Various pay tables are available.
### U.S. Raise Pay Table
Hand Pays
Royal flush 100 to 1
Straight flush 50 to 1
Four of a kind 20 to 1
Full house 7 to 1
Flush 5 to 1
Straight 4 to 1
Three of a kind 3 to 1
Two pair 2 to 1
All other 1 to 1
## House Edge
The following table shows all the possible outcomes, assuming the U.S. pay table is used and the player follows optimal strategy.
### U.S. Rules Return TableExpand
Event Pays Units Won Combinations Probability Return
Player wins with royal flush 100 to 1 201 16759740 0.000001 0.000169
Player wins with straight flush 50 to 1 101 156929720 0.000008 0.000795
Player wins with four of a kind 20 to 1 41 2832435800 0.000142 0.005826
Player wins with full house 7 to 1 15 16624475280 0.000834 0.01251
Player wins with flush 5 to 1 11 21856990280 0.001097 0.012062
Player wins with straight 4 to 1 9 43805516100 0.002198 0.019779
Player wins with three of a kind 3 to 1 7 234242908320 0.011751 0.08226
Player wins with two pair 2 to 1 5 488012139360 0.024482 0.122412
Player wins with pair or less 1 to 1 3 2343248003808 0.117555 0.352665
Dealer doesn't qualify 1 4532514033720 0.227385 0.227385
Push 0 321623100 0.000016 0
Fold -1 9523005974460 0.477745 -0.477745
Dealer wins -3 2726592727512 0.136786 -0.410359
Total 19933230517200 1 -0.052243
The lower right cell in the above table shows a house edge of 5.224%. However, I think this makes Caribbean Stud look like a worse bet than it really is. The house edge is traditionally defined as the expected loss to the original wager. In Caribbean Stud Poker the optimal strategy player will raise 52.23% of the time, for an average total wager of 2.045 units. For comparing one game against another I prefer to use the element of risk, which is the expected loss to the average wager, which in this case would be 5.224%/2.045 = 2.555%. The standard deviation, relative to the Ante bet, is 2.24.
## Strategy
Optimal strategy in Caribbean Stud is very complicated. The number of people who know it perfectly is probably zero. However, we can make two generalizations that cover most hands.
1. Always raise with a pair or higher.
2. Always fold with less than the dealer's qualifying hand (ace/king)
It is the ace/king hands that are tricky. In general it helps if the three singletons are high or the dealer's up card is jack or less and matches one of the player's cards (making is less likely the dealer will form a pair). I personally developed the following strategy, which results in a house edge of 5.225%, just 0.001% less than optimal. If that isn't enough for you I do have the Caribbean Stud optimal strategy.
The ugslot's Caribbean Stud Poker Strategy
Always raise with a pair or higher, fold with less than ace/king, and raise on ace/king if any of the following three rules apply.
1. Raise if the dealer's card is a 2 through queen and matches one of yours.
2. Raise if the dealer's card is an ace or king and you have a queen or jack in your hand.
3. Raise if the dealer's rank does not match any of yours and you have a queen in your hand and the dealer's card is less than your fourth highest card.
Following are details on various strategies. The "total loss" is the loss over all possible 19,933,230,517,200 combinations of hands, the house edge, and the element of risk.
### U.S. Raise Pay Table
Strategy Total loss House edge Element of risk
Perfect strategy 1,041,372,912,372 5.224% 2.555%
ugslot's Strategy 1,041,417,758,724 5.225% 2.554%
Raise on ace/king/jack/8/3 or higher 1,059,715,400,580 5.316% 2.596%
Raise if dealer's up card matches the rank of any player card 1,063,176,931,284 5.334% 2.616%
Raise on any pair or better 1,090,272,101,460
5.470%
2.738%
Raise on any ace/king or better 1,132,600,203,540 5.682% 2.672%
Playing blind (raise on everything) 3,310,360,338,060 16.607% 5.536%
## Alternate Rules
Outside the United States other pay tables are available, especially on the Internet. Here are return tables for some of the alternative pay tables I am aware of. The strategy is the same as the U.S. strategy, in all cases.
### Normandie Casino — Gardena, California
Event Units Won Combinations Probability Return
Player wins with royal flush 21 16759740 0.000001 0.000018
Player wins with straight flush 19 156929720 0.000008 0.00015
Player wins with four of a kind 17 2832435800 0.000142 0.002416
Player wins with full house 15 16624475280 0.000834 0.01251
Player wins with flush 13 21856990280 0.001097 0.014255
Player wins with straight 11 43805516100 0.002198 0.024174
Player wins with three of a kind 7 234242908320 0.011751 0.08226
Player wins with two pair 5 488012139360 0.024482 0.122412
Player wins with pair 3 2324742321600 0.116626 0.349879
Player wins with nothing 3 18505682208 0.000928 0.002785
Dealer doesn't qualify 1 4532514033720 0.227385 0.227385
Tie 0 321623100 0.000016 0
Player folds -1 9523005974460 0.477745 -0.477745
Player loses -3 2726592727512 0.136786 -0.410359
Total 19933230517200 1 -0.049862
### Casino de Crans-Montana (Switzerland)
Event Raise Pays Units Won Combinations Probability Return
Royal flush 100 to 1 201 16759740 0.000001 0.000169
Straight flush 50 to 1 101 156929720 0.000008 0.000795
Four of a kind 20 to 1 41 2832435800 0.000142 0.005826
Full house 9 to 1 19 16624475280 0.000834 0.015846
Flush 7 to 1 15 21856990280 0.001097 0.016448
Straight 4 to 1 9 43805516100 0.002198 0.019779
Three of a kind 3 to 1 7 234242908320 0.011751 0.08226
Two pair 2 to 1 5 488012139360 0.024482 0.122412
Pair or less 1 to 1 3 2343248003808 0.117555 0.352665
Ante only 1 4532514033720 0.227385 0.227385
Push 0 321623100 0.000016 0
Fold -1 9523005974460 0.477745 -0.477745
Dealer wins -3 2726592727512 0.136786 -0.410359
Total 19933230517200 1 -0.044521
### Galewind Software Internet Casinos
Event Raise Pays Units Won Combinations Probability Return
Royal flush 800 to 1 1601 16759740 0.000001 0.001346
Straight flush 200 to 1 401 156929720 0.000008 0.003157
Four of a kind 25 to 1 51 2832435800 0.000142 0.007247
Full house 10 to 1 21 16624475280 0.000834 0.017514
Flush 7 to 1 15 21856990280 0.001097 0.016448
Straight 5 to 1 11 43805516100 0.002198 0.024174
Three of a kind 3 to 1 7 234242908320 0.011751 0.08226
Two pair 2 to 1 5 488012139360 0.024482 0.122412
Pair or less 1 to 1 3 2343248003808 0.117555 0.352665
Ante only 1 4532514033720 0.227385 0.227385
Push 0 321623100 0.000016 0
Fold -1 9523005974460 0.477745 -0.477745
Dealer wins -3 2726592727512 0.136786 -0.410359
Total 19933230517200 1 -0.033498
### Microgaming Internet Casinos
Event Raise Pays Units Won Combinations Probability Return
Player wins with royal flush 999 to 1 1999 16759740 0.000001 0.001681
Player wins with straight flush 199 to 1 399 156929720 0.000008 0.003141
Player wins with four of a kind 99 to 1 199 2832435800 0.000142 0.028277
Player wins with full house 14 to 1 29 16624475280 0.000834 0.024186
Player wins with flush 9 to 1 19 21856990280 0.001097 0.020834
Player wins with straight 5 to 1 11 43805516100 0.002198 0.024174
Player wins with three of a kind 3 to 1 7 234242908320 0.011751 0.08226
Player wins with two pair 1 to 1 3 488012139360 0.024482 0.073447
Player wins with pair or less 1 to 1 3 2343248003808 0.117555 0.352665
Dealer doesn't qualify 1 4532514033720 0.227385 0.227385
Push 0 321623100 0.000016 0
Fold -1 9523005974460 0.477745 -0.477745
Dealer wins -3 2726592727512 0.136786 -0.410359
Total 19933230517200 1 -0.050055
### Cryptologic Internet Casinos
Event Raise Pays Units Won Combinations Probability Return
Player wins with royal flush 200 to 1 401 16759740 0.000001 0.000337
Player wins with straight flush 50 to 1 101 156929720 0.000008 0.000795
Player wins with four of a kind 20 to 1 41 2832435800 0.000142 0.005826
Player wins with full house 7 to 1 15 16624475280 0.000834 0.01251
Player wins with flush 5 to 1 11 21856990280 0.001097 0.012062
Player wins with straight 4 to 1 9 43805516100 0.002198 0.019779
Player wins with three of a kind 3 to 1 7 234242908320 0.011751 0.08226
Player wins with two pair 2 to 1 5 488012139360 0.024482 0.122412
Player wins with pair or less 1 to 1 3 2343248003808 0.117555 0.352665
Dealer doesn't qualify 1 4532514033720 0.227385 0.227385
Push 0 321623100 0.000016 0
Fold -1 9523005974460 0.477745 -0.477745
Dealer wins -3 2726592727512 0.136786 -0.410359
Total 19933230517200 1 -0.052075
### Sands Macau
Event Raise Pays Units Won Combinations Probability Return
Player wins with royal flush 50 to 1 101 16759740 0.000001 0.000085
Player wins with straight flush 50 to 1 101 156929720 0.000008 0.000795
Player wins with four of a kind 20 to 1 41 2832435800 0.000142 0.005826
Player wins with full house 7 to 1 15 16624475280 0.000834 0.012510
Player wins with flush 5 to 1 11 21856990280 0.001097 0.012062
Player wins with straight 4 to 1 9 43805516100 0.002198 0.019779
Player wins with three of a kind 3 to 1 7 234242908320 0.011751 0.082260
Player wins with two pair 2 to 1 5 488012139360 0.024482 0.122412
Player wins with pair or less 1 to 1 3 2343248003808 0.117555 0.352665
Dealer doesn't qualify 1 4532514033720 0.227385 0.227385
Push 0 321623100 0.000016 0.000000
Fold -1 9523005974460 0.477745 -0.477745
Dealer wins -3 2726592727512 0.136786 -0.410359
Total 19933230517200 1.000000 -0.052327
## 5+1 Bonus
Evolution Gaming offers a Live Dealer game with a side bet titled 5+1. It pays based on the combined poker value of the player's five cards and dealer's up card only. The following table shows the pay table and my analysis of it. The lower right cell shows a house edge of 8.56%.
### 5+1 Bonus
Event Pays Combinations Probability Return
Royal flush 1000 188 0.000009 0.009234
Straight flush 200 1,656 0.000081 0.016268
Four of a kind 100 14,664 0.000720 0.072029
Full house 20 165,984 0.008153 0.163061
Flush 15 205,792 0.010108 0.151626
Straight 10 361,620 0.017763 0.177626
Three of a kind 7 732,160 0.035963 0.251743
Loss -1 18,876,456 0.927202 -0.927202
Total 20,358,520 1.000000 -0.085614
The six-card poker probabilities taken from my Poker Probabilities page.
## Progressive Jackpot Side Bet
In Caribbean Stud Poker the player has the choice to make a side bet of \$1 which pays for hands of a flush or better.The specific payoff tables vary from place to place but always feature a progressive jackpot, paying 100% of the jackpot meter for a royal flush and 10% for a straight flush. In the very unlikely event that two players had a royal flush in the same hand the first one to the dealer's right would win the jackpot and the second would win whatever the jackpot is reseeded to, usually \$10,000. The reason for this is the dealer pays players from right to left. In the event that two players received a straight flush at the same time, the first one to the dealer's right would get 10% of the meter and the second would get 10% of what was left after the first player was paid. I have heard (but can not confirm) that once in the same hand one player got a royal flush and another player got a straight flush. The player with the royal was closer to the dealer's right and thus got the full jackpot amount. The meter was then reseeded to \$10,000 and the player with the straight flush got only 10% or that, or \$1,000.
While the expected return varies depending on the size of the jackpot, it is a sucker bet the vast majority of time.The average house edge is 26.46%.
A manager at Casino Niagara kindly explained how the jackpot meter works. For every dollar bet, 71 cents goes into the jackpot and the casino keeps the other 29 cents. A dealer at one of the Connecticut casinos said the contribution rate at his casino was only 65%. This rate of contribution can vary from place to place. All payoffs are paid right out of the meter. Every time somebody hits a royal flush the house contributes \$10,000 (called the seed) to the next jackpot. The house edge is just under the cut per bet because the casino puts up the initial seed to start a ugslot jackpot after somebody wins the previous one. At the Casino Niagara the house can expect to receive 18.84times as much money from the 29% cut as it pays to seed ugslot jackpots.
The next two tables show 11 side bet pay tables I have seen or heard about. The pay table number and pay table itself are listed in the top rows. The third row from the bottom is how much the flat wins contribute to the return. The second row from the bottom is how much the progressive wins contribute to the return for each \$10,000 in the meter. The bottom row is the break even point, or how high the meter would need to reach for the return to be 100%.
### Side Bet Pay Tables 1 to 5
Hand 1 2 3 4 5
Royal flush 100% 100% 100% 100% 100%
Straight flush 10% 10% 10% 10% 10%
Four of a kind \$100 \$150 \$500 \$500 \$500
Full house \$75 \$100 \$100 \$150 \$75
Flush \$50 \$50 \$50 \$75 \$50
Straight \$0 \$0 \$0 \$0 \$0
Three of a kind \$0 \$0 \$0 \$0 \$0
Flat wins 0.230323 0.278345 0.36238 0.483546 0.326369
Return per 10K in meter 0.029242 0.029242 0.029242 0.029242 0.029242
Breakeven meter \$263205.26 \$246783.16 \$218045.79 \$176611.05 \$230360.53
### Side Bet Pay Tables 6 to 10
Hand 6 7 8 9 10
Royal flush 100% 100% 100% 100% 100%
Straight flush 10% 10% 10% \$5000 \$20000
Four of a kind \$500 \$500 \$500 \$500 \$500
Full house \$100 \$150 \$150 \$100 \$100
Flush \$75 \$100 \$100 \$50 \$50
Straight \$0 \$0 \$0 \$0 \$0
Three of a kind \$0 \$0 \$0 \$0 \$0
Flat wins 0.41152 0.532685 0.532687 0.431648 0.639425
Return per 10K in meter 0.029242 0.029242 0.029242 0.015391 0.015391
Breakeven meter \$201241.58 \$159806.84 \$159806.32 \$369281 \$234280
### Side Bet Pay Tables 11 to 15
Hand 11 12 13 14 15
Table 11 12 13 14 15
Royal flush 100% 100% 100% \$5000 100%
Straight flush 10% 10% 10% \$1000 \$4000
Four of a kind 1% 5% \$250 \$500 \$400
Full house \$150 \$150 \$150 \$100 \$80
Flush \$75 \$75 \$100 \$75 \$40
Straight \$50 \$50 \$0 \$30 \$0
Three of a kind \$0 \$0 \$0 \$7 \$0
Flat wins 0.559724 0.559724 0.472651 0.543123 0.345307
Return per 10K in meter 0.053252 0.14929 0.029242 0.015391
Breakeven meter \$82677.75 \$29491.24 \$180336.84 \$425380
### Side Bet Pay Tables 16 to 20
Hand 16 17 18 19 20
Table 16 17 18 19 20
Royal flush 100% \$10,000 \$10,000 100% \$5,000
Straight flush 10% \$1500 \$2500 10% \$1000
Four of a kind \$600 \$500 \$500 \$500 \$500
Full house \$100 \$100 \$100 \$100 \$100
Flush \$60 \$75 \$50 \$50 \$75
Straight \$40 \$0 \$0 \$25 \$30
Three of a kind \$0 \$0 \$0 \$0 \$7
Flat wins 0.563025 0.432314 0.397033 0.460492 0.550797
Return per 10K in meter 0.029242 0 0 0.029242
Breakeven meter \$149,431.58 \$0 \$0 \$184,494.74
Here is where the above pay tables have been known to appear.
1. United States
2. United States
3. United States
4. United States
5. United States
6. United States
7. United States, Sydney (Australia)
8. United States
9. Northern Indiana
10. Microgaming Internet casinos
11. Curacao
12. Spain
13. Aces.com Internet casino
14. Germany
15. Sweden
16. Crown Casino in Melbourne, Australia
17. ?
18. Grand Casino, Brussels
19. Crystal Casino, Aruba
20. Casino Bregenz, Austria, where the game is called Tropical Stud Poker.
Casino Canberra: I have heard the follows pay table 7 but also pays \$50 for a "deadman's hand" consisting of AA88x, where x is any other card.For the side bet to have no house edge in this game the meter would need to reach \$149,389.47. For a \$5 minimum game to have no house edge the meter would need to reach\$238,716.85, and for a \$10 game the meter would need to be \$328,044.23.
## Millionaire Progressive
This is a \$5 "red light" progressive side bet that pays \$1,000,000 for a royal flush in spades, using the player's five-card hand. For all the rules and analysis, please see my page on the Millionaire Progressive.
## Player Collusion
I've been asked lots of times if an advantage can be gained by sharing information with other players. Although the rules forbid this in the land of casinos, there are multi-player Internet casinos where this could be done very easily by phone. However, don't get your hopes up. According to the paper "An Analysis of Caribbean Stud Poker" by Peter Griffin and John M. Gwynn Jr., which appears in the book , in the perfect situation of having seven colluding players, it would be possible to narrow down the dealer's unseen cards to just 16 possible cards. Using a computer to analyze all 1,820 possible 4-card sets out of 16, the player would have an advantage of 2.3%. In a six player game the house would still have an edge of 0.4%. An article on this topic by Scott McIntosh at Review Poker Rooms also explores this topic and comes to similar conclusions.
Stealing a look at other player's cards can cut down the house edge marginally. If you have a borderline ace/king hand it would help to see if the other player's cards match the dealer's up card. The more that match, the more inclined you should be to raise. While the Internet would be a perfect forum to share information in a multi-player table, the most number of seats I have ever seen is three.
## Player May Call
If the player makes a bet for the dealer, in most casinos the player may call the tip as long as the player raises his own bet. This is good strategy if the player raises with a marginal hand. Following is the strategy for how to play the dealer's tip.
• With a pair of fives or less always call (unless you fold yourself).
• With a pair of sixes - call, except if all three singletons are sevens or higher.
• With a pair of sevens - raise, except if all three singletons are six or less.
• With a pair of eights or higher, always raise.
If the player were to play both hands according to the tipping strategy then the tip would have an advantage of about 50%! This is a possible idea for player/dealer collusion.
## Play for Fun
Before playing for real money try my free Java game of Caribbean Stud Poker. | 6,204 | 19,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-21 | latest | en | 0.89224 |
https://topessaygeeks.com/operations-decision-3/ | 1,685,998,437,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652161.52/warc/CC-MAIN-20230605185809-20230605215809-00785.warc.gz | 622,007,446 | 11,334 | ## Operations Decision
Using the regression results and the other computations from Assignment 1, determine the market structure in which the low-calorie frozen, microwavable food company operates.
Use the Internet to research two (2) of the leading competitors in the low-calorie frozen, microwavable food industry, and take note of their pricing strategies, profitability, and their relationships within the industry (worldwide).
Write a six to eight (6-8) page paper in which you:
1. Outline a plan that will assess the effectiveness of the market structure for the company’s operations. Note: In Assignment 1, the assumption was that the market structure [or selling environment] was perfectly competitive and that the equilibrium price was to be determined by setting QD equal to QS. You are now aware of recent changes in the selling environment that suggest an imperfectly competitive market where your firm now has substantial market power in setting its own “optimal” price.
1. Given that business operations have changed from the market structure specified in the original scenario in Assignment 1, determine two (2) likely factors that might have caused the change. Predict the primary manner in which this change would likely impact business operations in the new market environment.
1. Analyze the major short run and long cost functions for the low-calorie, frozen microwaveable food company given the cost functions below. Suggest substantive ways in which the low-calorie food company may use this information in order to make decisions in both the short-run and the long-run.
TC = 160,000,000 + 100Q + 0.0063212Q2
VC = 100Q + 0.0063212Q2
MC= 100 + 0.0126424Q
1. Determine the possible circumstances under which the company should discontinue operations. Suggest key actions that management should take in order to confront these circumstances. Provide a rationale for your response. (Hint: Your firm’s price must cover average variable costs in the short run and average total costs in the long run to continue operations.)
1. Suggest one (1) pricing policy that will enable your low-calorie, frozen microwavable food company to maximize profits. Provide a rationale for your suggestion.
(Hints: In Assignment 1, you determined your firm’s market demand equation. Now you need to find the inverse demand equation. Having found that, find the Total Revenue function for your firm (TR is P x Q). From your firm’s Total Revenue function, then find your Marginal Revenue (MR) function.
Using the demand equation that you determined in assignment 1 (Q=38650-42P) to find the inverse demand function (P=……) and also total revenue function (TR=P.Q)
• Use the profit maximization rule MR = MC to determine your optimal price and optimal output level now that you have market power. Compare these values with the values you generated in Assignment 1. Determine whether your price higher is or lower.)
1. Outline a plan, based on the information provided in the scenario, which the company could use in order to evaluate its financial performance. Consider all the key drivers of performance, such as company profit or loss for both the short term and long term, and the fundamental manner in which each factor influences managerial decisions.
(Hints:
• Calculate profit in the short run by using the price and output levels you generated in part 5. Optional: You may want to compare this to what profit would have been in Assignment 1 using the cost function provided here.
• Calculate profit in the long run by using the output level you generated in part 5 and cost data in part 3 and assuming that the selling environment will likely be very competitive. Determine why this would be a valid assumption.)
1. Recommend two (2) actions that the company could take in order to improve its profitability and deliver more value to its stakeholders. Outline, in brief, a plan to implement your recommendations.
1. Use at least five (6) quality academic resources in this assignment. Note: Wikipedia does not qualify as an academic resource.
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Contributor
Posts: 23
how to compute the average correlation across a very large correlation matrix
I’m trying to compute the average cross-sectional correlation across a large dataset. I may also want to compute a weighted average. Inaddition, my raw data is in stacked form. Assume, for example, I have a stacked list of investors and their portfolio weights (that sum to one for each investor) across all stocks (assume only four stocks).
Investor Stock Portfolio_weight
A IBM 0.50
A MSFT 0.40
A GOOG 0.10
A GRPN 0.00
B IBM 0.40
B MFST 0.60
B GOOG 0.00
B GRPM 0.00
C IBM 0.50
C MSFT 0
C GOGG 0
C GRPN 0.50
I wrote alittle test code (see below) to compute the average correlation between weights across investors. First, I compute a weights matrix where each row is a company and each column is an investor (data outb):
dataoutb
company
Obs ID _NAME_ A B C
1 GOOG wt 0.1 0.0 0.0
2 GRPN wt 0.0 0.0 0.5
3 IBM wt 0.5 0.4 0.5
4 MSFT wt 0.4 0.6 0.0
I then run a proc corr and keep the correlations (data d):
Obs _TYPE_ _NAME_ A B C
1 CORR A 1.00000 0.88684 0.00000
2 CORR B 0.88684 1.00000 -0.19245
3 CORR C 0.00000 -0.19245 1.00000
Because I only want the average of the off diagonal elements,I take the average times N, subtract 1 (the diagonal element) and divide by N-1 to get the average of the off diagonal elements: 0.23 = (0.886+0.00+-0.192)/3 (=ave_corr in data g)
I have four issues:
1. The number of investors in my simple example is three. In my real data, the number is approximately 3,000, which means the correlation matrix will have 9M elements. Is that an issue? (Moreover the number of stocks is approximately 5,000.)
2. My code seems a little clumsy. I welcome any suggestion for alternative approaches.
3. In addition to average correlation, I’d like to compute the sum of the products of the weights for each pair of investors and compute the average of that (e.g., for A and B, 0.1*0 + 0*0 + 0.5*0.4 + 0.4*0.6). Again, there would be approximately 9M of these terms. If I define outb as a 4X3 matrix (in my simple example), I can compute this as outb(transposed)*outb.
After reading about this a bit, this seems like something I should be able to do with proc IML. However, I’m not familiar with proc IML. Any suggestions on how to define data outb as a matrix so I can use proc IML? And if I can do that, how would I’d go about getting the average of the elements in the resulting matrix?
4. If I standardize the weights for each investor (i.e., subtract the mean and divided by the standard deviation), I could use the same method to compute the correlations (i.e., standardized weights matrix transposed *standardized weights matrix). So that might be another approach. Any thoughtson using IML versus my ‘traditional’ sas code?
Rick
My code
data a; input investor \$ companyID \$ wt;
cards;
A IBM 0.50
A MSFT 0.40
A GOOG 0.10
A GRPN 0.00
B IBM 0.40
B MSFT 0.60
B GOOG 0.00
B GRPN 0.00
C IBM 0.50
C MSFT 0.00
C GOOG 0.00
C GRPN 0.50
;
run;
data b; set a;
proc sort; by companyID;
proc transpose out=outb;by companyID;id investor; var wt; *outb gives the weightsmatrix, columns are investors and rows are companies;
title 'data outb';
proc print data=outb;run;
data c; set outb;
proc corr out=outc;*computes correlation matrix;
data d; set outc;
if _TYPE_='CORR';
proc means noprint;
output out=outd; *averagecorrelation for each investor;
data e; set outd;
if _STAT_='MEAN';
proc transpose out=oute;
data f; set oute;
if _NAME_='_TYPE_' then delete;
if _NAME_='_FREQ_' then delete;
ave_corr=col1;
proc means noprint; var ave_corr;
output out=outf mean=ave_corr_cross_investors;*average correlation across investors;
data g; set outf;
ave_corr=(ave_corr_cross_investors*_FREQ_-1)/(_FREQ_-1);*average correlation dropping the diagonal;
proc print; run;
Accepted Solutions
Solution
12-28-2011 12:44 AM
Super User
Posts: 10,784
how to compute the average correlation across a very large correlation matrix
Hi. Rick. I am not familar with IML.
If you need IML code, you can post it at IML forum. There is a specialist about IML who also is a SAS employee -- Rich M.
But I think you also can do it under Data Step.
```data a; input investor \$ companyID \$ wt;
cards;
A IBM 0.50
A MSFT 0.40
A GOOG 0.10
A GRPN 0.00
B IBM 0.40
B MSFT 0.60
B GOOG 0.00
B GRPN 0.00
C IBM 0.50
C MSFT 0.00
C GOOG 0.00
C GRPN 0.50
;
run;
proc sort data=a; by companyID;run;
proc transpose data=a out=temp(drop=_name_);
by companyid;
id investor;
var wt;
run;
proc corr data=temp outp=corr(drop=_name_ where=(_type_='CORR')) noprint;
var _numeric_;
run;
data _null_;
set corr end=last;
retain sum .;
array _a{*} _numeric_;
i=1;
do while(_a{i} ne 1);
sum=sum(sum,_a{i});i+1;n+1;
end;
if last then do;
mean=sum/n;
put 'ave_corr= ' mean;
end;
run;
options nomprint nomlogic nosymbolgen;
/*compute the sum of the products of the weights for
each pair of investors */
%macro across;
proc sql noprint;
select distinct investor into : list separated by ' ' from a;
create table want as
select
%do i=1 %to %sysfunc(countw(&list));
%do j=%eval(&i+1) %to %sysfunc(countw(&list));
sum(%scan(&list,&i)*%scan(&list,&j)) as %scan(&list,&i)%scan(&list,&j)
%if &i ne %eval(%sysfunc(countw(&list))-1) or &j ne %sysfunc(countw(&list)) %then %do;,%end;
%end;
%end;
from temp;
quit;
%mend across;
%across
```
Ksharp
All Replies
Solution
12-28-2011 12:44 AM
Super User
Posts: 10,784
how to compute the average correlation across a very large correlation matrix
Hi. Rick. I am not familar with IML.
If you need IML code, you can post it at IML forum. There is a specialist about IML who also is a SAS employee -- Rich M.
But I think you also can do it under Data Step.
```data a; input investor \$ companyID \$ wt;
cards;
A IBM 0.50
A MSFT 0.40
A GOOG 0.10
A GRPN 0.00
B IBM 0.40
B MSFT 0.60
B GOOG 0.00
B GRPN 0.00
C IBM 0.50
C MSFT 0.00
C GOOG 0.00
C GRPN 0.50
;
run;
proc sort data=a; by companyID;run;
proc transpose data=a out=temp(drop=_name_);
by companyid;
id investor;
var wt;
run;
proc corr data=temp outp=corr(drop=_name_ where=(_type_='CORR')) noprint;
var _numeric_;
run;
data _null_;
set corr end=last;
retain sum .;
array _a{*} _numeric_;
i=1;
do while(_a{i} ne 1);
sum=sum(sum,_a{i});i+1;n+1;
end;
if last then do;
mean=sum/n;
put 'ave_corr= ' mean;
end;
run;
options nomprint nomlogic nosymbolgen;
/*compute the sum of the products of the weights for
each pair of investors */
%macro across;
proc sql noprint;
select distinct investor into : list separated by ' ' from a;
create table want as
select
%do i=1 %to %sysfunc(countw(&list));
%do j=%eval(&i+1) %to %sysfunc(countw(&list));
sum(%scan(&list,&i)*%scan(&list,&j)) as %scan(&list,&i)%scan(&list,&j)
%if &i ne %eval(%sysfunc(countw(&list))-1) or &j ne %sysfunc(countw(&list)) %then %do;,%end;
%end;
%end;
from temp;
quit;
%mend across;
%across
```
Ksharp
Contributor
Posts: 23
how to compute the average correlation across a very large correlation matrix
Thanks once more Ksharp...you've saved me a lot of time...
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# CS 344 – Homework 3
\$29.99
Category:
CS 344 – Sections 1,2,3 – Fall 2018 Homework 3.100 points total plus extra credit
1 Problem 1 (15 points)
For each part below, say whether the statement is true or false. If true, give a brief (one or two sentences) explanation; if false, give a counterexample. • Part 1 (3 points) The minimum element of a max-heap is always a leaf • Part 2 (3 points) The second smallest element of a max-heap is always a leaf • Part 3 (4 points) The second and third largest elements of a max-heap are always the two children of the root • Part 4 (5 points) The sum of the heights of all nodes in an n-node heap is Θ(n).
2 Problem 2 (20 points total)
no need to show your work on this problem: just draw the final heap Consider the array A = 3,7,15,4,25,1,20,8. • Part 1 (5 points): What is the tree corresponding to this array? (Note: it is not a max-heap) • Part 2: (5 points) What is the tree that results from calling BuildMax-Heap(A)? • Part 3: (5 point) What is the result of calling Max-Heap-Insert(A,17) on the max-heap that results from part 2? • Part 4: (5 points) What is the result of calling Delete-Max(A) on the heap that results from part 3?
1
3 Problem 3 (25 points)
Say there are n sorted arrays A1,A2,…,An, each with n numbers (so n2 numbers between all of them.) Write pseudocode that uses the heap data structure to find the n smallest numbers between all the arrays in time O(nlog(n)). You must make it clear why the run-time is O(nlog(n)). REMINDER: Heaps are now in our virtual library. For this problem, you can use a min-heap or a max-heap without explaining how the heap works.
4 Problem 4 (20 points)
For both parts below, make sure to show your work • Part 1 (10 points): Say that I throw a 4-sided die and a 3-sided die. What is the expectation of the product of the two numbers? What is the expectation of the max of the two numbers? • Part 2 (10 points): Given an array A, we say that a pair (i,j) is swapped if i < j but A[i] A[j]. What is the expected number of swapped pairs if A contains the integers 1 through n in a random order.
5 Problem 5 (20 points)
For both parts, make sure to show your work • Part 1: (10 points) If I throw 100 dice, what is the expected number of times that I observe the pattern 3,6,6? Note that the 3,6,6 have to appear consecutively to count as the pattern. So for example, if my result was 4,3,6,6,5,2,3,6,1,6,3,6,6,2 then the pattern occurs twice. • Part 2: (10 points) We say that three people have a 3-way birthday if they all have a birthday on the same day. Question: if each person has a random birthday (assume no leap years), then how many people do you need in a room for the expected number of 3-way-birthdays to be at least 2?
2
6 Problem 6 – extra credit • Say that I have n objects, and I play a game where I repeatedly look at a random object. (Note: I don’t throw away the objects I look at, so I might end up looking at the same object more than once.) In big-O notation, what is the expected number of objects I will look at before I have seen every one of the n objects? You must show your work, as always. HINT 1: You will want to use the formula that 1+1/2+1/3+…+1/n = O(log(n))
3
CS 344 – Homework 3
\$29.99
Hello
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• literal equations and formulas | 1,384 | 5,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-26 | latest | en | 0.830029 |
https://socratic.org/questions/how-do-you-solve-4x-10-46 | 1,638,397,622,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360951.9/warc/CC-MAIN-20211201203843-20211201233843-00501.warc.gz | 578,252,263 | 6,156 | # How do you solve |-4x + 10| ≤ 46?
The solution is
For all values above 9 and less than or equal to 14, the relation is valid
#### Explanation:
$\left\mid - 4 x + 10 \right\mid \le 46$
$- 4 x + 10 \le 46$
$- 4 x \le 46 - 10$
$- 4 x \le 36$
$- x \le \frac{36}{4}$
$- x \le 9$
$x > 9$
or
$- 4 x + 10 \le - 46$
$- 4 x \le - 46 - 10$
$- 4 x \le - 56$
$4 x \le 56$
$x \le \frac{56}{4}$
$x \le 14$
The solution is
$x > 9 , x \le 14$
For all values above 9 and less than or equal to 14, the relation is valid
Aug 6, 2018
The solution is $x \in \left[- 9 , 14\right]$
#### Explanation:
The inequality is
$| - 4 x + 10 | \le 46$
The solution is
$\left\{\begin{matrix}4 x - 10 \le 46 \\ - 4 x + 10 \le 46\end{matrix}\right.$
$\iff$, $\left\{\begin{matrix}4 x \le 46 + 10 \\ 4 x \ge 10 - 46\end{matrix}\right.$
$\iff$, $\left\{\begin{matrix}4 x \le 56 \\ 4 x \ge - 36\end{matrix}\right.$
$\iff$, $\left\{\begin{matrix}x \le \frac{56}{4} \\ x \ge - \frac{36}{4}\end{matrix}\right.$
$\iff$, $\left\{\begin{matrix}x \le 14 \\ x \ge - 9\end{matrix}\right.$
The solution is
$x \in \left[- 9 , 14\right]$
graph{|4x-10|-46 [-105.4, 105.4, -52.7, 52.7]} | 508 | 1,152 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 26, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2021-49 | latest | en | 0.385923 |
https://www.jiskha.com/questions/764796/solve-2x-y-11-and-x-y-13-using-the-elimination-method | 1,624,368,458,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00529.warc.gz | 763,948,116 | 3,826 | # math
solve 2x-y=11 and x+y=13 using the elimination method
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1.solve the set of linear equation by matrix method.a+3b+2c=3,2a-b-3c=-8,5a+2b+c=9 solve for a,b,c 2.solve by Guassian elimination method, (a)a+2b+3c=5,3a-b+2c=8,4a-6b-4c=-2, (b)
4. ### math
can someone please help me with this elimination methods problem? 0.05x+0.25y=11 0.15x+0.05y=12 solve using the elimination method and give the ordered pair... | 563 | 1,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-25 | latest | en | 0.769367 |
http://azfoo.net/gdt/csc205/lectures/13.html | 1,498,707,532,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323864.76/warc/CC-MAIN-20170629033356-20170629053356-00172.warc.gz | 39,664,630 | 14,519 | Home Previous Next
CSC205::Lecture Note::Week 13
Assignments | Code | Handouts | Resources | Email Thurman | Tweets::@compufoo
GDT::Bits:: Time | Weather | Populations | Special Dates
Overview
• Potential computing ethics topic...
```RT @NewsHour A murder video posted online raises debate
```
• Algorithm analysis using Big-O Notation.
• BigOCheatSheet.com::Know Thy Complexities!
Assignment(s)
Code``` StringTree.java {output} | BST.java {output} ```
### Trees
A tree is a "nonlinear" structure that consists of nodes and edges (or branches). A tree is a set of nodes and edges that connect the nodes. There is exactly one path between two nodes. A path is a connected sequence of edges.
Arrays, stacks, queues, and linked-lists define collections of objects that are sequentially accessed. These are linear lists: they have first and last elements, and each interior element has a unique successor.
In a nonlinear structure, an element may have multiple successors.
Tree structures are used to represent a "hierarchy" of information. A commonly used hierarchy is an operating system's file system.
A tree structure consists of a set of nodes that originate from a unique starting node called the root.
Some nodes may be considered a parent node which implies they have zero or more children nodes to which they are connected. Every node except the root node has a parent.
The children of a node and children of those children are called descendants, and parents and grandparents of a node are its ancestors.
Nodes are siblings when they have the same parent.
A node that doesn't have any children is called a leaf.
Each node in a tree is the root of a subtree, which is defined by the node and all descendants of the node.
``` grandfather
|
+----------+-----------+
| | |
father uncle aunt
|
+-----+-----+
| |
brother sister
grandfather is the root node. grandfather is the parent
node to father, uncle and aunt. father, uncle, aunt, brother
and sister are all descendants of grandfather. father and
grandfather are ancestors of brother and sister. brother and
sister are children of father. brother are sister are leaf
nodes. brother and sister are siblings. father is the root
of the subtree that contains the father, brother, sister nodes.
```
A single-inheritance class hierarchy can be represented by a tree.
``` Component
|
+------------------+------------+------------+
| | | |
Container Button Canvas TextComponent
| |
+---------+----------+ +----+----+
| | | | |
Panel ScrollPane Window TextArea Textfield
|
+-----+----+
| |
Dialog Frame
|
FileDialog
```
Movement from a parent node to its child and other descendants is done along a path.
The level of a node is the length of the path from the root to the node.
The depth of a tree is the maximum level of any node in the tree. The depth of a node is the length of its path from the node to the root.
``` A level 0
/ | \
B C D level 1
\ \
E F level 2
/ \
G H level 3
The depth of this tree is 3 (the longest path from the root
to a node). The path length for node E is 2. The path
length for node D is 1. The path for node H is: A->B->E->H.
```
Definition:
A general tree is a set of nodes that is either empty, or has a designated node called the root from which descend zero or more subtrees. Each node is not an ancestor of itself, and each subtree itself satisfies the definition of a treee.
Note that a tree is defined in a recursive fashion; consequently, most tree-processing algorithms are recursive.
YouTube.com::CS 61B Lecture 23: Trees and Traversals
{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}
### Binary Trees
A binary tree is a popularly used tree in which each node has at most two (2) subtrees. These subtrees are designated as the left and right, respectively. Either or both of these subtrees may be empty.
The following operations are performed on a binary tree:
• `create` -- initialize an empty binary tree
• `empty` -- return TRUE if tree is empty; FALSE otherwise
• `insert` -- item (value) is added to the tree in a way that maintains the tree's hierarchical property
• `preorder traversal` -- each node of the tree is visited in the following order: root of the tree first, then recursively all nodes in left subtree, then recursively all the nodes in the right tree
• `inorder traversal` -- each node of the tree is visited in the following order: first visit recursively all nodes in the left subtree of the tree, then visit the root node, then recursively all nodes in the right tree
• `postorder traversal` -- each node of the tree is visited in the following order: first visit recursively all nodes in the left subtree, then recursively all nodes in the right subtree, then visit the root node
[Definition] An algorithm that systematically "visits" all items in a tree is called a tree traversal.
Here is a binary tree and its respective (depth-first) traversals:
``` A
/ \
B C
/ / \
D E F
/ \
G H
Preorder traversal: A (root); B,D (left); C,E,G,H,F (right)
Inorder traversal: D,B (left); A (root); G,E,H,C,F (right)
Postorder traversal: D,B (left); G,H,E,F,C (right); A (root)
```
The following tree takes on a heap property:
``` 87
/ \
/ \
84 63
/ \ \
68 79 12
/ \ / \
32 67 6 10
/ \
8 9
The data in any given node of the tree is greater than or
equal to the data in its left and right subtrees.
```
A binary tree can have a hierarchy that exhibits the property known as the ordering property.
``` 87
/ \
/ \
84 103
/ \ / \
68 86 90 109
/ \ / \
32 74 88 97
/ \
70 80
The data in each node of the tree are greater than all of the
data in that node's left subtree and less than or equal
to all the data in the right subtree.
```
A binary tree with the ordering property is often called a binary search tree (BST).
A complete binary tree of depth `N` is a tree in which each level `0` to `N-1` has a full set of nodes and all leaf nodes at level `N` occupy the leftmost positions in the tree.
``` A Complete tree with depth 3.
/ \
B C
/ \ / \
D E F G
/ \ /\
H I J K Leaf nodes occupy leftmost positions.
```
A full binary tree is a complete tree that contains 2N nodes at level N.
``` A Full tree with depth 2.
/ \
B C Every node has two children.
/ \ / \
D E F G 22 is 4 and there are 4 nodes.
```
A degenerate tree is one that has a single leaf node and each non-leaf node has only one child. [Note: a degenerate tree is equivalent to a linked-list.]
``` A
\
B
\
C
/
D
\
E
```
Many compilers make use of tree structures in obtaining forms of an arithmetic expression for evaluation.
```
Operators are non-leaf nodes and operands are leaf nodes.
+
/ \
- *
/ \ / \
A B C ÷
/ \
E F
(A - B) + C * (E / F) infix
+ - A B * C / E F prefix
A B - C E F / * + postfix
```
#### Binary Tree Structure
A binary tree structure is built with nodes. A tree node contains a data field and two pointer (i.e. link) fields.
Given the non-linear structure of trees, a linked-list type of searching (i.e. sequential) is not available; therefore, a variety of traversal methods (inorder, preorder, postorder) are commonly used.
Tree traversal methods rely heavily on recursion. Each traversal method performs three actions at a node:
• visit the node
• recursively descend the left subtree
• recursively descend the right subtree
The descent terminates when an empty tree pointer is encountered.
A node visit depends on the application (i.e. it is application-dependent).
##### Inorder Traversal
An inorder traversal begins its action at a node by first descending to its left subtree. After recursively descending through the nodes in that subtree, the inorder traversal takes the second action at the node and uses the data value. The traversal completes its action at the node by performing a recursive scan of the right subtree. In the recursive descent through the subtrees, the actions of the algorithm are repeated at each new node.
1. traverse the left subtree
2. visit the node
3. traverse the right subtree
##### Postorder Traversal
The postorder traversal delays a visit to a node until after a recursive descent of the left subtree and a recursive descent of the right subtree.
1. traverse the left subtree
2. traverse the right subtree
3. visit the node
##### Preorder Traversal
The preorder traversal visits the node first, then it does recursive descents of the left and right subtrees, respectively.
1. visit the node
2. traverse the left subtree
3. traverse the right subtree
##### Breadth First Traversal (or Level Traversal)
A breath first (or level) traversal scans the tree nodes level-by-level starting with the root at level 0.
A common implementation of this algorithm is queue-based.
``` create queue Q
insert root into Q
while Q is not empty
delete front node p from Q
use node p to identify its children at next level
if (p.left() != null)
Q.insert(p.left());
if (p.right() != NULL)
Q.insert(p.right());
Using the queue helps ensure that nodes on the same
level are visited in the correct order (in this case
left-to-right).
```
#### BigO
Trees perform inserts, deletes, and finds in O(log N) time (on average). A degenerate tree can be O(N).
#### Exercises
The following inputs inserted in the given order produce a good binary tree.
``` NY IL GA RI MA PA DE IN VT TX OH WY
```
What the binary tree look like? [answer]
What order do these state abbreviations need to be inserted in order to produce a degenerate tree? [answer]
WolframAlpha.com::Binary Tree
{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}
### Binary Search Tree
A general binary tree can hold a large collection of data and yet provide fast access.
Recall, creating a "container" or "collection" class using an array or linked-list, requires those collections to be searched using a linear search. This can be inefficient when dealing with large amounts of data [O(N)].
Trees are efficient for searching because the path to any data value is no greater than the depth of the tree. Searching is maximized with a complete binary tree O(log2N).
To store elements in a tree for efficient access, a binary search tree (BST) can be used.
``` For each node, the data values in the left subtree
are less than the value of the node and the data
values in the right subtree are greater than or equal
to the value of the node.
```
Using a BST implies that the data found in each node contain a key . In many cases, the data of a node is a record and the key is a field in the record.
Operations performed on a BST are:
``` find -- search a tree of a specific data value
insert -- find appropriate insertion spot and add new
data value to the tree
delete -- search a tree and remove the first occurrence
of a given data value; reattach all subtrees
to maintain BST structure
update -- if key at current position matches key for
the data item, update data value; otherwise,
insert data value into the tree
```
#### Insertion on a Binary Search Tree
Search the tree and find the location where the new node is to reside. This is done be scanning the left and right subtrees until the insertion point is found. For each step in the path, maintain a record of the current node and the parent of the current node. The process terminates when an empty subtree is encountered. At this location, the new node is inserted as a child of the parent.
``` 25
/ \
20 35
/ \
12 40
Insert the value 32 into this tree.
Step 1: compare 32 with 25 -- traverse the right subtree
Step 2: comapre 32 with 35 -- traverse the left subtree
Step 3: insert 32 as left child of parent (i.e. 35).
25
/ \
20 35
/ / \
12 32 40
```
#### Deletion on a Binary Search Tree
There are three cases that need examination:
1. the node to be deleted has no subtrees
2. the node to be deleted has a right subtree but no left subtree
3. the node to be deleted has a left subtree
##### Delete Node Having No Subrees
Assume P is a pointer to the node to be deleted.
``` X = P;
P = NULL;
delete X;
```
##### Delete Node Having Right Subtree but No Left Subtree
Assume P is a pointer to the node to be deleted.
``` X = P;
P = X->right;
delete X;
```
##### Delete Node Having Left Subtree
Assume P is a pointer to the node to be deleted. Two scenarios are examined.
``` P->left has no right subtree:
X = P;
P = X->left;
P->right = X->right;
delete X;
P->left does have a right subtree:
X = P;
Q = X->left->right;
QParent = X->left;
while Q->left not equal NULL
Q = Q->right;
QParent = QParent->right;
Q->right = X->right;
P = Q;
QParent->right = Q->left;
Q->left = X->left;
delete X;
```
#### Update on a Binary Search Tree
Find the node that needs updating and copy in new data. If node note not found, then insert it. What if key being modified? Find node, delete it and then insert new node containing the updated key.
{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}
If a binary tree has `N` nodes, then the total number of links is `2*N` (each node has a left subtree pointer and a right subtree pointer).
Each node has exactly one pointer to it (except the root, which has none).
At most (given a complete binary tree) only `N-1` subtree pointers will have non-`null` values.
The idea behind a threaded binary search tree is to make use of unused pointers (i.e. those set to `null`.
Threaded BSTs allow for efficient non-recursive traversals.
Given a BST, here is an algorithm that converts it into a right-threaded BST.
``` Perform an inorder traversal of a BST. Whenever a node 'x'
with a null right pointer is encountered, replace the right
pointer with a thread to the inorder successor of 'x'.
The "inorder successor" of a node 'x' is it parent if 'x'
is a left child of its parent; otherwise, it is the nearest
ancestor of 'x' that contains 'x' in its left subtree.
```
Here is an example of a right-threaded BST.
``` H
+-----------+--------------+
| |
E K
/ \ / \
/ \ / \
B F J L
/ \ \ / \
A D G I M
\
C
A.right ---> B
C.right ---> D
D.right ---> E
G.right ---> H
I.right ---> J
J.right ---> K
```
For a threaded tree, each node must contain a flag that is used to indicate if the right pointer is a thread pointer.
{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}
### AVL Trees
Binary search trees (BSTs) can be efficiently searched, but this is true only if the tree is "balanced" (`O(log N)` because binary-searching can be executed). A de-generate tree (i.e. a binary tree in which each node has only one child) results in `O(N)` speed.
If you are going to have a tree that is going to be used extensively for searching, then you may want to keep the tree balanced. A balanced BST is called an AVL tree. [Technique invented in the 60's by Russian mathematicians Georgii Maksimovich Adel'son-Vel'skii and Evgenii Mikhailovich (say these names fast three times).]
Implementing an AVL tree requires a balance factor be added to each node of the tree. A balance factor is defined as follows:
``` The balance factor of node 'x' is the height of the left
subtree of 'x' minus the of the right subtree of 'x'.
The height of a tree is the number of levels in it.
```
The following tree is used to demonstrate balancing factors.
``` H
+-----------+--------------+
| |
E K
/ \ / \
/ \ / \
B F J L
/ \ \ / \
A D G I M
\
C
Balancing factors:
H: +1
E: +1
B: -1
F: -1
D: -1
J: +1
L: -1
A,C,G,K,I,M: 0
```
Keeping a tree balanced requires making rotations when balancing factors take on values other than 0, 1, or -1.
The following trees are un-balanced.
```
Inputs: L, I, N, U, X
L (-2)
/ \
(0) I N (-2)
\
U (-1)
\
X (0)
To balance: perform a left rotation on node N
L (-1)
/ \
(0) I U (0)
/ \
(0) N X (0)
---------------------------------------------------------------------
Inputs: Z, E, B, R, A
Z (2)
/
E (1)
/
B (0)
To balance: perform a right rotation on node Z
E (0)
/ \
(0) B Z (0)
E (0)
/ \
(1) B Z (1)
/ /
(0) A R (0)
```
There are four types of rotations:
• right rotation -- the new element is inserted in the left subtree of the left child of 'B' of the nearest ancestor 'A' with balance factor +2
• left rotation -- the new element is inserted in the right subtree of the right child 'B' of the nearest ancestor 'A' with balance factor -2
• left-right rotation -- the new element is inserted in the right subtree of the left child 'B' of the nearest ancestor with balance factor +2
• right-left rotation -- the new element is inserted in the left subtree of the right child 'B' of the nearest ancestor 'A' with balance factor -2
Here is an example where a left-right rotation is needed:
``` The following tree is balanced.
PA (1)
/ \
(0) GA RI (0)
/ \
(0) DE OH (0)
Insert MA.
PA (2)
/ \
(-1) GA RI (0)
/ \
(0) DE OH (1)
/
MA (0)
Balance by first doing a left rotation on node GA.
PA
/ \
OH RI
/
GA
/ \
DE MA
Next, do a right rotation on node PA.
OH (0)
/ \
(0) GA PA (-1)
/ \ \
DE MA RI
(0) (0) (0)
```
Coming up with an example of when is right-left rotation is needed and is left as an exercise.
{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}
### 2-3-4 Trees
2-3-4 Trees are trees that have nodes which may have more than two children.
A benefit of a 2-3-4 tree is that it contains shorter search paths because there are fewer levels in the tree.
When searching a binary tree, you either go left or right at each node. In a 2-3-4 tree, there can be several different directions that a search path may follow.
Definition:
``` An m-node in a search tree stores m-1 data values
k1 < k2 < ... < km-1,
and has links to m subtrees T1, ..., Tm,
where for each value i, all data values in
Ti < ki <= all data values in Ti+1
```
Here is an example 2-3-4 tree.
```
53
/ \
+--------------- ------------+
| |
27 38 60 70
/ | \ / | \
16,25 33,36 41,46,48 55,59 65,68 73,75,79
This tree has 9 nodes. The root node contains one value.
Both nodes at level 1 have two values. On level two,
there are 4 nodes containing 2 values, and 2 nodes
containing 3 values.
Search for the value 36.
36 is less than 53, traverse the left subtree.
36 greater than 27 and less than 38, traverse
the center subtree. 36 less than 33, compare
to the next number. 36 equals 36 -- found.
```
2-3-4 trees satisfy the following properties:
• each node stores at most 3 data values
• each internal node is a 2-node, a 3-node, or a 4-node
• all the leaves are at the same level
In a nutshell, 2-3-4 tree is constructed as follows:
• begin with a single node and insert values into this node until it becomes full (i.e. contains three values)
• when the next value is inserted, split the node into two nodes, one storing the values less than the median and other storing values greater than the median
• the median value is stored in the parent having these two nodes as children
``` Construct a 2-3-4 tree using the inputs: 10 3 9 15 21 44 18 1 12 5 7
insert 10: 10
insert 3: 3,10
insert 9: 3,9,10
insert 15: split
9
3 10
15 greater than 9 -- search right subtree
9
3 10,15
insert 21:
9
3 10,15,21
insert 44:
44 greater than 9 -- search right subtree
internal node is full -- split
median value moved into parent
9,15
3 10 21,44
insert 18:
18 greater than 15 -- search right subtree
9,15
3 10 18,21,44
insert 1:
1 less than 9 -- search left subtree
9,15
1,3 10 18,21,44
insert 12:
12 greater than 9 and less than 15 -- search middle subtree
9,15
1,3 10,12 18,21,44
insert 5:
5 less than 9 -- search left subtree
9,15
1,3,5 10,12 18,21,44
insert 7:
7 less than 9 -- search left subtree
internal node is full; split
median value 3 moved into parent
3,9,15
1,5,7 10,12 18,21,44
```
2-3-4 trees stay balanced during insertions and deletions.
Think about top-down insertion versus the demonstrated bottom-up insertion. [Don't allow any parent nodes to become 4-nodes.]
2-3-4 trees have numerous unused pointers. Explore Red-Black trees for an alternative.
``` 2-3-4 tree with N nodes has 4*N pointers
each node (excluding the root) has only one pointer to it
4*N - (N-1) = 3N+1 of the pointers are null
(i.e. approximately 75% of the pointers are null)
```
Example 2-3-4 tree node class.
``` class Node234 {
Object data[3];
...
};
```
azrael.digipen.edu::Deleting Elements From a 2-3-4 Tree
{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}
### Big-O Notation
The run-time efficiency of a program has two main ingredients: space and time.
Space efficiency is a measure of the amount of memory a program requires.
Time efficiency is a measure of how long a program takes to execute.
In general, there is a trade-off between space and time. If your algorithm uses more space, then typically it will execute faster and vice versa.
The relationship between `N` (where `N` is a variable representing the number of inputs on which your algorithm must operate) and the running time of an algorithm as `N` becomes large is called the computational complexity of that algorithm.
Computational complexity is denoted using BigO notation. The letter `O` stands for the word order as it is used in the phrase on the order of. It signifies approximation.
Given two functions f and g, the function g is said to dominate function f if there is some positive constant c such that:
``` c * g(x) >= f(x), for all possible values of x
```
Asymptotic dominance is a variation of dominance such that:
``` c * g(x) >= f(x), for all x >= x'
```
In other words, function g asymptotically dominates function f for all "large" values of x. Big-O notation is represented using asymptotic dominance.
``` 5N versus N2
-------------------------
N=1: 5 versus 1
N=2: 10 versus 4
N=3: 15 versus 9
N=4: 20 versus 16
N=5: 25 versus 25
N=6: 30 versus 36
N=7: 35 versus 49
In this example, x' is equal to 6 and the constant is 1.
```
The following are some examples of how dominance is used to determine BigO values:
``` O(N3) + O(N2) => O(N3)
As N gets large, N cubed grows at a faster rate than
does N squared. For large values of N, N cubed dominates.
O(10 * N) + O(1000 * N) + O(1) => O(N)
Constant values are ignored. The significance of the
constant values are minimized for large values of N.
O(N log N) + O(log N) + O(N2) => O(N2)
N squared dominates N log N which in turn dominates log N.
```
Let's review:
• BigO notation is used to approximate (i.e. estimate) the running time of an algorithm; it is used to determine the most dominate term in a function; BigO represents the growth rate of a function
• running time is expressed in relative speed, not absolute speed; constants are dropped because it is not meaningful across different machines
• BigO analysis is used only when dealing with algorithms that implement loops and/or recursion
• two algorithms that are both `O(N)` may have different running times for the same number of inputs
• BigO measures are based on asymptotic dominance
BigO-notation is used for three distinct purposes.
1. To bound the error that is made when we ignore small terms in mathematical formulas.
2. To bound the error that is made when we ignore parts of a program that contribute a small amount to the total being analyzed.
3. To allow us to classify algorithms according to upper bounds on their total running times.
O(1) constant time (most efficient) O(N) linear time (polynomial time) O(N log N) "linearithmic" time (polynomial time) O(N2) quadratic time (polynomial time) O(N3) cubic time (polynomial time) O(log N) logarithmic time O(2N) exponential time (impractical for large values of N)
The following table presents growth rates for some of the common functions (i.e running times):
``` N log N N N log N N**2 N**3 2**N
=========================================================================
1 0 1 0 1 1 2
2 1 2 2 4 8 4
4 2 4 8 16 64 16
8 3 8 24 64 512 256
16 4 16 64 256 4096 65536
32 5 32 160 1024 32768 4.29497e+09
64 6 64 384 4096 262144 1.84467e+19
128 7 128 896 16384 2097152 3.40282e+38
```
Typically constant factors are ignored in BigO analysis (although in reality constants frequently matter).
To compare the speed of different algorithms, computer scientists use bigO notation. The bigO notation doesn't exactly predict performance times. Instead, what the bigO notation tells us is that given two algorithms with the same order, then both will slow down at roughly the same rate as the number of items being sorted gets larger.
Search algorithms also have bigO values. Linear search is `O(n)`, binary search is `O(log n)`. Once again, these differences are most important for large data sets. If you have 1 million items in an array, a linear search will require about 1 million tests, but a binary search will only require 20 tests (because 2 to the 20 power is about 1 million -- the 'log' in bigO notation is always log-base-2 -- when you have a logarithmic algorithm, the analysis need not be concerned with the base of the logarithm, since this can change the total running time only by a constant factor (and constant factors are ignored).
``` for (i = SOME_CONSTANT; i <= ANOTHER_CONSTANT; i++) // O(1)
// loop body requiring constant time
for (i = SOME_CONSTANT; i <= N; i++) // O(N)
// loop body requiring constant time
for (i = SOME_CONSTANT; i <= N; i++) // O(N*N)
for (j = ANOTHER_CONSTANT; j <= N; j++)
// loop body requiring constant time
for (i = SOME_CONSTANT; i <= N; i++) // O(N*M)
// loop body requiring time O(M), where M is a variable
```
The following is example of how one can go about determining the approximate running time of an algorithm.
``` // the following logic initializes a 'n' element array
index = 0; // assignment is 1 step
while (index < n) { // EXPR index < n is 1 step
array[index] = -1; // assignment is 1 step
index++; // increment is 1 step
}
Si + Sc*(n+1) + Sb*n
n = 0: 1 + 1*1 + 2*0 = 2
n = 1; 1 * 1*2 + 2*1 = 5
n = 2; 1 * 1*3 + 2*2 = 8
n = 3: 1 + 1*4 + 2*3 = 11
n = 6: 1 + 1*7 + 2*6 = 20
n = 9: 1 + 1*10 + 2*9 = 29
n = 300: 1 + 1*301 + 300*2 = 902
Si => # of initialization steps
Sc => # of steps in the condition EXPR
sb => # of steps in the loop body
Conclusion: linear -- O(n)
```
#### Analysis of a Couple of Code Snippets
``` for (k = 1; k <= n / 2; k++) {
...
for (j = 1; j <= n * n; j++) {
...
}
}
```
Since these loops are nested, the number of times statements within the innermost loop are executed is the product of the number of repetitions of the two individual loops. Hence the efficiency is `n3`, or `O(N3)` in BigO terms, with a constant of proportionality equal to 1/2.
``` for (k = 1; k <= n * 10; k++) {
...
}
for (j = 1; j <= n * n * n; j++) {
...
}
```
Since one loop follows the other, the number of operations executed by both loops is the sum of the individual loop efficencies. Hence the efficiency is `10n + n3`, or `O(N3)` in BigO terms.
#### Analysis of Exchange Sort
Assume we have array `A` with length `N` and we want to find the minimum element in the array.
``` Pass 1: compare the n-1 elements A[1]...A[n-1]
with A[0] and, if necessary, exchange elements
so that A[0] always has the smallest value
Pass 2: compare the n-2 elements A[2]...A[n-1]
with A[1] and, if necessary, exchange elements
so that A[1] less than all elements to the right
Pass i: compare the n-i elements A[i]...A[n-1]
with A[i-1] and, if necessary, exchange elements
The total number of comparisons is given by the
arithmetic series f(n) from 1 to n-1:
f(n) = (n-1) + (n-2) + ... + 3 + 2 + 1 = n * (n-1) / 2
The number of comparisons depends on n2.
```
#### Using an Array for a List
Append an item: O(1). Insert an item at the front of the array: O(N).
#### Simple Sorts
Bubble, selection, insertion, and exchange are all O(N2).
Shell sort can give much better performance than O(N2) in the worst case.
Quick sort in the worst case can be O(N2), but on average it is O(N log N).
Merge sort is an O(N log N) sort.
#### Order of a Function: Mathematical Definition
The order of a function is defined as follows:
Given two non-negative functions f and g, the order of f is g if and only if g asymptotially dominates f. Using BigO notation, this can said as: `f = O(g)` (f is of order g).
#### Rules for Comparing BigO Estimates
Some rules to help calculate and compare BigO estimates:
``` Let f and g be functions, and C a constant:
O(C * f) = O(f)
O(f * g) = O(f) * O(g)
O(f / g) = O(f) / O(g)
O(f) > O(g), if and only if f dominates g
O(f + g) = Max[O(f), O(g)]
```
A logarithm of base b for the value y is the power to which you must raise b to get y. Normally, this is written as: `logby = x`.
``` logby = x <=> bx = y <=> blogby = y
where <=> means "is equivalent to"
```
Logarithms have the following properties:
``` For any positive values of m, n, and r, and any positive integers
a and b:
log nm = log n + log m
log n/m = log n - log m
log nr = r log n
logan = logb n / logba
``` | 8,271 | 31,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2017-26 | latest | en | 0.930786 |
http://letsmakerobots.com/node/2016?page=5 | 1,464,217,181,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049275412.55/warc/CC-MAIN-20160524002115-00191-ip-10-185-217-139.ec2.internal.warc.gz | 170,997,987 | 7,910 | # Test yourself: Arty or Techie
## Take this test to find out what you are :)
Point = 0
1) Your robot has to navigate in a park. Do you design it to..
A) Orientate to see how far, what is ahead, somehow generate a map over the area, keep track over obstackles and good paths, drive to next point and so on
B) Focus on where you are going to, constantly avoid obstacles, but only if they are in your way
point = point +1
end if
2) The fun about building robots are ..
A) That you can have an idea to a creature and see it come alive
B) That you have a reason to try math and programming techniques in real life
point = point +1
end if
3) If there is a cable that has 3.7V coming out from your circuit, and you thought it should have been 5V, would you rather / typically..
A) Find out why it is 3.7V
B) Say ”that was strange”, and somehow cope with that for now
point = point +1
end if
4) You have to make a robot arm that is going to take up all your socks from your drawer. Would your first thought be..
A) Find out a way to know how many socks are in the drawer, so that you know when you have picked all and is able to structure the picking.
B) Find out a way to see when the drawer is empty, and pick socks until it is empty.
point = point +1
end if
5) If you should chose to be an expert in one of the 2 below, just somehow get all knowledge inserted into your brain, what would you chose:
A) Get to know 100% Picaxe Basic, Basic Stamp Basic and Arduino C – because then you could just build with ease, almost without thinking, all commands would be at your hands; If you can think on it, you can just make it..
B) 100% assambly, down to 0’s and 1’s of the chips the above based on, because then you would be able to understand it all 100%, and override the commands and optimize and write your own, which would be cool!
point = point +1
end if
6) Building robots often happens alone. If you where together on a camp with 1.000’s of other builders, would you rather..
A) Participate in a group that would design something bigger, using multiple technologies.
B) Participate in a competition where you where able to enter a robot that you had made yourself.
point = point +1
end if
7) After making a robot, would it make you more happy if..
A) Other robot builders where really impressed?
B) People who knows nothing about it think it is cool or fun?
point = point +1
end if
Result:
0-2 points = Arty!!
3-4 points = All around!
5-7 points = Techie!! | 635 | 2,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2016-22 | latest | en | 0.958815 |
http://www.riddlesandanswers.com/puzzles-brain-teasers/i-a-s-question-rid-riddles/ | 1,569,162,686,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575515.93/warc/CC-MAIN-20190922135356-20190922161356-00219.warc.gz | 318,902,614 | 27,572 | # I A S QUESTION RID RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS
#### Popular Searches
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## Solving I A S Question Rid Riddles
Here we've provide a compiled a list of the best i a s question rid puzzles and riddles to solve we could find.
Our team works hard to help you piece fun ideas together to develop riddles based on different topics. Whether it's a class activity for school, event, scavenger hunt, puzzle assignment, your personal project or just fun in general our database serve as a tool to help you get started.
Here's a list of related tags to browse:
The results compiled are acquired by taking your search "i a s question rid" and breaking it down to search through our database for relevant content.
Browse the list below:
Hint:
How long.
Did you answer this riddle correctly?
YES NO
Solved: 43%
## The Question Of Life Riddle
Hint:
Did you answer this riddle correctly?
YES NO
Solved: 53%
Hint:
A doorbell
Did you answer this riddle correctly?
YES NO
Solved: 51%
## Snow White Asks The Dwarfs A Question Riddle
Hint:
The compulsive liars are Sneezy and Dopey.
The excerpt has been taken from the story "Snow White and the Seven Dwarfs".
The seven dwarfs are Doc, Grumpy, Happy, Sleepy, Bashful, Sneezy, and Dopey.
The story shows how the dwarfs are living a peaceful life in Dwarf Woodlands and they come across Snow White. They then try to protect her from the attackers and from the poisoned apple from the Queen.
Did you answer this riddle correctly?
YES NO
Solved: 52%
## The Answer Is No Riddle
Hint:
Do you mind?
Did you answer this riddle correctly?
YES NO
Solved: 41%
## Surrounded By Cats Riddle
Hint:
It is simple. You kill the two animals with two bullets i.e. tiger and leopard and then make a run in the Jaguar (a car brand).
Did you answer this riddle correctly?
YES NO
Solved: 42%
## Prisoners Were Told That They Have A Chance To Be Free Riddle
Hint:
They should ask any guard "is the freedom door next to the lying guard?" If the guard answers "yes" they should go to the other door. if it was the liar, the other door takes them to freedom. If it was the honest guard, the other door is also what the prisoners want. If the guard answers "no" they should go to the door next to him: if it was the liar, the door next to him takes them to freedom. If it was the honest guard, the door next to him is also the right choice.
Did you answer this riddle correctly?
YES NO
Solved: 54%
## Days Of The Month Riddle
Hint:
They all do.
Did you answer this riddle correctly?
YES NO
Solved: 75%
## Twins Riddle
Hint:
The two babies are two of a set of triplets.
Did you answer this riddle correctly?
YES NO
Solved: 43%
## Giraffe In A Refrigerator Riddle
Hint:
Open the refrigerator, put in the giraffe, and close the door. This question tests whether you tend to do simple things in an overly complicated way.
Did you answer this riddle correctly?
YES NO
Solved: 50% | 766 | 2,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-39 | latest | en | 0.877801 |
http://kids.britannica.com/browse/video?topic=mathematics | 1,386,743,172,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164032243/warc/CC-MAIN-20131204133352-00070-ip-10-33-133-15.ec2.internal.warc.gz | 96,564,303 | 4,751 | Mathematics
Geometry: Right Angles
This is a brief explanation of the right angles. It gives nice examples of right angles in the world around us.
Geometry: The Parts of a Right Triangle
This explanation shows the different parts of a right triangle and how they are symmetrical.
Geometry: Similar Right Triangles
Similar triangles have integer side lengths.
Geometry: Determining Leg Length
Using the Pythagorean Theorem, you can determine the length of a leg in a right triangle.
Geometry: The Pythagorean Theorem
The Pythagorean Theorem is used to calculate the relationship between the legs and angles of a triangle.
Geometry: Applications of the Pythagorean Theorem
The Pythagorean Theorem can be used to identify different triangles.
Measurement: Units of the English and Metric System
This is a comparison of the English and metric systems of measurement. | 179 | 876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2013-48 | latest | en | 0.839908 |
https://www.enotes.com/homework-help/two-forces-f1-f2-acting-point-find-resultant-force-559007 | 1,618,703,311,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464065.57/warc/CC-MAIN-20210417222733-20210418012733-00022.warc.gz | 851,889,881 | 20,238 | # Two forces, F1 and F2, are acting at a point. Find the resultant force if the forces are at right angles.
Images:
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It's impossible to determine the exact nature of the resultant force in this case, because we don't have any information about the magnitude of the original forces.
We also don't know the forces are directed toward the point or away from it. We do know that the resultant force will be some kind of angle relative to the forces generating it, but the direction of that angle relative to the original forces will depend upon their direction. For example, if both forces are directed toward the point, the resultant force vector will point diagonally away from both on the opposite side of the axis. If both are pulling, the resultant will point diagonally toward them on the same side of the axis. If one pulls and the other pushes, the resultant will be flipped across one of the axes.
We know the resultant will be diagonal because we can add vectors together. This is usually represented as if they are stacked on top of one another end-to-end. For example, if we have one force A acting down, and one force B to the left, each with a force of 5, then we can put the base of B on top of the end of A, and calculate the resultant vector as the hypotenuse of a right triangle whose sides are 5 and 5. The exact angle and magnitude of the resultant will depend upon the magnitudes of the original vectors. | 325 | 1,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-17 | latest | en | 0.940511 |
http://www.math.hope.edu/newsletter/2011-12/10-03.html | 1,386,861,374,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164641982/warc/CC-MAIN-20131204134401-00012-ip-10-33-133-15.ec2.internal.warc.gz | 425,567,042 | 7,582 | Off on a Tangent A Fortnightly Electronic Newsletter from the Hope College Department of Mathematics
October 19, 2011 Vol. 10, No. 3 http://www.math.hope.edu/newsletter.html
This week's colloquium takes a look at an old "proof" for the four color theorem
Title: Alfred Kempe's "Proof" of the Four Color Theorem Speakers: Tim Sipka, Alma College Time: Thursday, October 20 at 11:00 am Place: VWF 102
Abstract: The four color theorem is a simple and believable statement: at most four colors are needed to color any map drawn in the plane or on a sphere so that no two regions sharing a boundary receive the same color. It might be surprising to find out that mathematicians searched for a proof of this statement for over a century until finally finding one in 1976. In this talk we'll consider the "proof" given by Alfred Kempe that was published in 1879 and thought to be correct until a flaw was found in 1890. You'll be invited to look carefully at Kempe's proof and see if you can do what many 19th century mathematicians could not do---find the flaw.
Note: We have a nontypical day, time, and location for this week's colloquium.
Next week's colloquium
Title: Lights Out: Some Puzzles, Some Results, and Some Questions Speakers: Darin Stephenson, Hope College Time: Tuesday, October 25 at 4:00 pm Place: VWF 104
Abstract: In the “Lights Out” puzzle, the solver seeks to turn off all of a set of lights (often represented by the vertices on a graph) by pressing a series of buttons, each of which has a certain predefined effect on the lights. The lights each have an “initial state” (typically “on” or “off”), and an initial configuration of states is called solvable if there is a process for turning off all of the lights. The solution to any specific version of this puzzle can be phrased in terms of matrix algebra. Our overall goal is to study this problem in generality, and, eventually, to characterize the solvable initial configurations for certain families of graphs. In this talk, we will provide a survey of the classical Lights Out puzzle, discuss many generalizations and some known results, and give a few questions that may serve as motivation for collaborative faculty-student research in summer 2012.
Note: Please come to enjoy refreshments with the speaker, the faculty, and fellow mathematics students before the colloquium at 3:30 pm in VanderWerf 222.
Help out MDY's Mathemagicians in the upcoming Relay for Life
In the wake of Mary DeYoung’s death after a brief battle with cancer this past summer, many students are looking for ways to honor and celebrate her life. One of these opportunities presents itself in this year’s Relay for Life at Hope College. A team has been formed for all of us to remember Mary DeYoung and to help raise money to fight this horrible disease. You can support the team in any of the following four ways: Join the team, raise funds, and walk with us at Relay for Life on November 11-12 from 7pm to 7am. If you want to join just go to http://relayforlifehope.com, click SIGN UP, and then JOIN A TEAM. Look for MDY's Mathemagicians. Make a donation to team. Either online or by getting the money to one of the team captains: Rachel Elzinga, Morgan Bell, and Nicholle Taurins. There will also be a collection can in the math department office. Decorate a luminaria in honor of Mary DeYoung or anyone else you know who has been touched by cancer. Luminarias are bags lit by electric lights to honor survivors and people who have lost their battle with cancer. They will line the track during the Relay event and will be honored during a ceremony. Usually people donate \$5-\$10 for a luminary, but anyone who is interested can make one regardless of whether or not they donate. Simply contact Rachel, Morgan, or Nicholle and we will get you a bag. Pray for the team as we work to raise money for the American Cancer Society to honor Mary DeYoung. Even if you can’t join, check out the team page by going to http://relayforlifehope.com, look for “Top Teams” on the right hand side, click “View All”, and then click “MDY’s Mathemagicians” to see updates on fundraising and team activities. If you have any questions or ideas please contact Rachel Elzinga (rachel.elzinga@hope.edu), Morgan Bell (morgan.bell@hope.edu) or Nicholle Taurins (nicholle.taurins@gmail.com). Thanks for helping us work towards a world with less cancer and more birthdays!
MATH Challenge
The 2011 Michigan Autumn Take Home Challenge (or MATH Challenge) will take place on the morning (9:30am - 12:30pm) of Saturday, November 5 this year. Teams of two or three students take a three-hour exam consisting of ten interesting problems dealing with topics and concepts found in the undergraduate mathematics curriculum. Each team takes the exam at their home campus under the supervision of a faculty advisor.
The department pays the registration fee for each team and will provide lunch to participants afterwards. The sign-up deadline is Monday, October 24 at 4:00 p.m. Interested students can sign up by sending Prof. Yurk an email at yurk@hope.edu or by signing up on the list on his door (VWF 214).
A group of students may sign up as a team. Individual students are also encourage to sign up; they will be assigned to a team. For more information, please talk with any member of the Mathematics Department or visit this link
where you can also view old copies of the exam.
Math in the News: Scientists use Twitter to analyze public heath trends
Who would ever think that "omg i think i have the flu going home bfn" would be of value to science. Well according to a recent report from Discoveries and Breakthroughs Inside Science, computer scientists at John Hopkins University are using the online social networking and microblogging service to track public health trends. "There’s a lot of different patterns we were able to uncover,” said Mark Dredze (Johns Hopkins), “so for example we were able to track the influenza rate in the United States over time just by counting how many times people are talking about the flu.”
Problem Solvers of the Fortnight
In our last problem of the fortnight we saw that Sally began to solve a problem at the time between 4:00 and 5:00 p.m. when the clock's hands are together. She finished when the minute hand is opposite the hour hand. How many minutes does it take her to solve the problem, and when does she finish it? Give exact answers in terms of fractions of minutes (i.e. no decimal approximations). Congratulations to Lauren Aprill, Ryan Martinez, Lute Olsen, Jessica Hulteen, Jake Blysma, Craig Toren, Krisen Slotman, Corinna Schmidt, Danielle Maly, Erice Budge, Emily Scott, Allison Leigon, Anna Filcik, Lisa McLellan, Kristen Bosch, Bryce Ciroshek, Pete Stuckey, Hunter Ford, Brant Bechtel, Sarah Prill, David Dolphin, Andrew Borror, Morgan Smith, Daniel Langholz, Shinnosuke Kondo, Scott DeClaire, Melanie Leonard, Justin Kammeraad, Allie Benson, Josh Swelt, Kevin Olson, Andrew Brooks, Nicole Zeinstra, Tim Lewis, Tim Cooke, Matt Folkert -- all of whom correctly determined the exact starting and ending time of the exam in the last Problem of the Fortnight.
Problem of the Fortnight
You have ten stacks of coins, each consisting of 10 new dollar coins. One entire stack of coins is counterfeit, but you do not know which one. However, you do know the weight of a genuine dollar coin, and you are also told that each counterfeit coin weighs one gram more than it should. Your kind chemistry professor agrees to let you weigh the coins on one of the electronic scale in the chem lab. What is the smallest number of weighings necessary to determine which stack is counterfeit, and how do you do it? Attach a counterfeit dollar coin to your solution, and drop it in the Problem of the Fortnight slot outside Professor Pearson's office (VWF 212) by 3:00 p.m. on Friday, October 28. As always, be sure to include your name as well as the name(s) of your professor(s) -- e.g. Bo Guscoins, Professor Cown TerFit -- on your solution.
The best and most beautiful things in the world cannot be seen or even touched – they must be felt with the heart.
Helen Keller
Off on a Tangent | 1,925 | 8,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2013-48 | longest | en | 0.935621 |
https://puzzling.stackexchange.com/questions/107629/proving-by-weighting | 1,701,968,860,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100677.45/warc/CC-MAIN-20231207153748-20231207183748-00308.warc.gz | 532,839,528 | 44,790 | # Proving by weighting
Here is the puzzle:
Alice and Bob leave in a world where all diamonds are divided into 2 types: Light and Heavy. All diamonds look exactly the same. All light diamonds weight exactly the same, all heavy also weight exactly the same but a bit more than the light (by an unknown, infinitely small amount more).
Now, Bob wants to sell Alice N light and N heavy diamonds. Before he does it he has to prove that those are exactly N light and N heavy, not like 2N light or any other combination of light and heavy. And he must do it using infinitely precise scales, which can only compare two weights (i.e. they don't show the weight difference, only which one is heavier). He must do it in 4 weightings.
What is the maximum N, which Bob can sell?
For example, it's easy to solve the puzzle for N = 15.
Name all diamonds according to their weight. Light: L1, L2, ...; heavy: H1, H2, ...
Weighting 1: compare L1 vs H1. Alice sees that the first is lighter and believes that L1 is a light diamond and H1 is a heavy diamond.
Weighting 2: compare L2+L3+H1 vs H2+H3+L1. Alice sees that the first set of diamonds is lighter and concludes that L2,L3 is light diamond and H2,H3 is heavy, since this is the only way for Weighting1 and Weighting2 to be as they are.
Weighting 3: compare L4L5L6L7H1H2H3 vs H4H5H6H7L1L2L3.
Weighting 4: compare L8..L15H1..H7 vs H8..L15L1..L7.
And I know, in fact, there is a solution for $$N=27$$.
What I need to know is a maximum $$N_{max}$$, a weighting procedure for $$N_{max}$$ diamonds, and proof that there is no possible procedure for $$N_{max}+1$$ diamonds.
Note, if you are having a hard time relating the solution below to the question. That is because of OP's extensive changes to the question. Originally, heavy diamonds were 101g and light ones 100g.
There is no maximum, in fact, any N>100 can be verified in 2 weighings.
Indeed, let M be the largest multiple of 101 not larger than N. Place M light diamonds against 100xM/101 heavy diamonds. The balance will show they are equal which proves their identities. Now swap all the not yet weighed diamonds in like-for-like and weigh again. This is always possible with the sole exception of N=201. where we have 101 heavies left over and only 100 spots to swap them in. But this can be solved by adding one of the already confirmed heavies to the light side and balancing it by the 101st new heavy.
• Ok. Sorry for that "hole". I've reformulated the puzzle. Of course, you can't compare a different number of diamonds. Mar 3, 2021 at 20:57
• @klm123 You have not so much "reformulated" but rather completely changed the puzzle. Mar 3, 2021 at 21:02
• I really like this answer, even though the question has changed, very cleverly exploited the numbers. Mar 3, 2021 at 22:33
• I guess the current question is the intended question. Cool solution for the alternate question, though! Mar 8, 2021 at 11:01
No proof of optimality, but for the record, I have a solution for
$$N=27$$
Comparisons: $$r_1 \dots r_8s_1 \dots s_8a_1 \dots a_3 < x_1 \dots x_{16}b_1 \dots b_3 < y_1 \dots y_{16}a_1 \dots a_3
Because they are separated by 3 $$<$$ signs, there are at least three more heavy diamonds in $$r_1 \dots r_8s_1 \dots s_8a_1 \dots a_3$$ versus $$r_1 \dots r_8s_1 \dots s_8b_1 \dots b_3$$. As the only differences are $$a_i$$ and $$b_i$$, it must be that all of $$a_i$$ are light and all of $$b_i$$ are heavy, and the difference (in number of heavy diamonds) across each of these 3 $$<$$ signs is exactly one.
Then, consider the inner comparison $$x_1 \dots x_{16}b_1 \dots b_3 < y_1 \dots y_{16}a_1 \dots a_3$$. There must be exactly four more heavy diamonds among the $$y_i$$ versus the $$x_i$$.
Now, consider the last comparison. The heavy diamonds in $$b_i$$ and $$y_i$$ outnumber $$a_i$$ and $$x_i$$ by 7, so in order to satisfy the comparison, it must be that all of $$r_i$$ are light and all of $$s_i$$ are heavy.
Then we can see from the 3 $$<$$ chain that the total number of light and heavy diamonds are the same (e.g. we know there are exactly $$8$$ heavy diamonds on the left and $$11$$ on the right which tells us the exact numbers among $$x_i$$ and $$y_i$$). This is true even though we don't know which of $$x_i$$ or $$y_i$$ are heavy or light.
• so x has 6 heavy and 10 light and y has 10 heavy and 6 light. 8+16+3 = 27! Very cool. I see now that I've misformulated the puzzle again and forgot the condition, that each diamond must be identified. But what you've found in totally unexpected and I like it a lot, you'll get the points if noone beats you in 10 hours. But just if you are interested in taking it further, you can identify every single of those 27*2 = 54 diamonds in 4 weighting, not just prove that there are 27 heavy. Mar 13, 2021 at 8:09
The answer is in fact simple:
Nmax=15, and the procedure in question is in fact optimal
Why? First, let's introduce some abbreviations for the sake of simplicity:
KL and KH for "known-light" and "known-heavy", i.e. diamonds which already have their weight proven by us. Similarly, PL and PH for "presumed-light" and "presumed-heavy", i.e. diamonds which are labeled as such but still unproven.
Now, notice that
when weighing, we cannot mix PL and PH diamonds on one side of the scales, since, for example, one of the PL diamonds could be in fact heavy, and a PH diamond cold be light, and so, this could go undetected. For similar reasons, we cannot distribute same "P" diamonds (i.e. PL or PH) between different sides of the scales (since,for example, one of the PL diamonds on each side could be heavy, and this will go undetected because the weight difference remain the same).
So,
the only sound arrangement is to put PL+some "K" diamonds on the one side of the scales, and PH+some "K" on the other.
But
to prove or disprove the weights of multiple diamonds at once, we need to have a set of known diamonds that can be separated in two parts (having the same number of diamonds) with large weight difference. If $$d$$ is the difference between light and heavy, and we have $$2k$$ known diamonds, the largest difference we can create is $$kd$$ ($$k$$ heavy minus $$k$$ light ones). Such difference cannot positively identify a set of "P" diamonds of size more than $$2(k+1)$$ at once (because we won't know the difference between weights upon weighing, only which part is heavier). If we put $$k+2$$ "P" diamonds on each side, the difference between "P" weights will be $$(k+2)d$$, and we cannot detect by placing additional "K" diamonds if this difference is in fact $$(k+2)d$$ or $$(k+1)d$$ which can happen if only one of the "P" diamonds on the scales is labeled wrong.
That means that
having $$k$$ KL and $$k$$ KH diamonds, we cannot identify more than $$k+1$$ PL and $$k+1$$ PH diamonds in one weighing.
Before the 1st weighing, we have
$$k(0)=0$$
Since
$$k(i)=2k(i-1)+1$$ (we know how to identify exactly $$k+1$$ pairs of diamonds, since this is the procedure described in the question itself),
we have that
$$k(i)=2^i-1$$. Putting $$i=4$$ gives $$k(4)=15$$.
P.S. Sorry for such a lengthy and complicated proof, I wrote this from my phone at 0:48 (12:48 AM) local time.
• In your proof you assume that we need to identify diamonds right after each weighting, but we can postpone the identification till the last weighting. Mar 6, 2021 at 10:37
• Also the question has stated that there is a solution for $N=27$. Mar 9, 2021 at 8:57
• @justhalf this was a later edit. Mar 9, 2021 at 11:25 | 2,082 | 7,499 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 50, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-50 | latest | en | 0.953332 |
https://www.education.com/activity/article/Centripetal_Force_middle/?coliid=662629 | 1,506,437,290,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696182.97/warc/CC-MAIN-20170926141625-20170926161625-00409.warc.gz | 790,995,624 | 26,310 | Activity:
Spin the Bucket: A Centripetal Force Experiment
3.4 based on 85 ratings
What You Need:
• 3 foot rope
• Water
• Little bucket with handle
What You Do:
1. Tie one end of the rope to the handle of the bucket.
2. Pour water into the bucket so that it's filled halfway.
3. Wrap the rope's other end around your child's hand a couple times. Make sure it's not too tight or too loose, as the bucket will pull on the rope.
4. Invite your child to rotate his hand to quickly spin the bucket in a vertical circle so that water doesn't fly out. Ensure that there's enough space for him to move the bucket.
5. Ask your child to then rotate the bucket slowly so that water begins to splash around. Notice how slowly the bucket has to spin for water to fly out!
6. Encourage your child to rotate the bucket quickly and let go of the rope when the bucket is at the bottom of the circle. Where does the bucket go? Make sure the area is clear when your child does this!
What's Happening?
This activity is an example of centripetal force, which is when an inward force acts on a circular-moving object. In this bucket activity, the water is falling at the same speed as the bucket, so the liquid doesn't appear to fall out. If the bucket were to be removed or slow down at the top of the circle, the liquid would pour out because the forces are no longer balanced!
One everyday example of this phenomenon is a satellite. The earth is curved, so the satellite orbits quickly enough to fall around the earth. If it moved any slower, it would fall onto the earth.
Wait! There's more science! What happened when your child let go of the rope? Did the bucket move in a straight line? Newton's First Law of Motion plays a role here. This law states that something that's moving will keep moving in a straight line unless some other force acts on it and changes the direction. Challenge your child and ask him where the bucket would've gone if he had let go of the rope at other points in the circle. | 453 | 1,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2017-39 | latest | en | 0.94912 |
https://www.physicsforums.com/threads/how-to-calculate-the-average-velocity.948256/ | 1,719,250,673,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00016.warc.gz | 812,965,907 | 17,806 | # How to calculate the average velocity?
• Indranil
In summary, the position of an object moving along the x-axis is given by x = a + bt^2, where a = 8.5m, b = 2.5 ms^-2, and t is measured in seconds. To find the average velocity between t = 2.0 s and t = 4.0 s, calculate the values of x at each time point (x = 18.5 m at t = 2.0 s and x = 40 m at t = 4.0 s) and use the average velocity formula Vavg = (xf - xi)/(tf - ti). After correcting for a mistake, the final average velocity is 15 m/s.
Indranil
## Homework Statement
The position of an object moving along the x-axis is given by x=a+bt^2, where a=8.5m,b=2.5 ms^-2, and t is measured in seconds. what is the average velocity between t=2.0 s and t= 4.0 s
## Homework Equations
as we know, Vavg = xf-xi / tf-ti (f= final position and time and i= initial position and time)
## The Attempt at a Solution
Where to put the values because we have the values of t but there are no values of x
Indranil said:
Where to put the values because we have the values of t but there are no values of x
Your problem statement has an equation for x(t) does it not?
gneill said:
Your problem statement has an equation for x(t) doe it not?
No, the only equation is 'x=a+bt^2'
average velocity between t=2.0 s and t= 4.0 s
So find x when t=2 and find x when t=4, then you'll have all the data you need.
Indranil said:
No, the only equation is 'x=a+bt^2'
Yes it does: that IS the expression ##x(t)##---after all, it is telling you the value of ##x## when the time is ##t##!
Merlin3189 said:
So find x when t=2 and find x when t=4, then you'll have all the data you need.
I am doing this below
x=a +bt^2 =8.5 +2.5 x 4 =18.5 (when t=4)
x=at + bt^2 =8.5 + 2.5 x 2 =13.5 (when t=2)
see the values below
(a=8.5m, b=2.5 ms^-2, t=2.0 s and t= 4.0 s)
Now what to do?
1- what is average velocity?
2- what do your two x values tell you?
Merlin3189 said:
1- what is average velocity?
2- what do your two x values tell you?
1. Average velocity = TotalDisplacement / TotalTime
2. The initial position and the final position
If I do the math
Vavg=18.5-13.5 /4-2 = 5/2 (but the answer does not match the answer in my book. in my book, my answer is 15m/s)
Now, what to do?
Indranil said:
x=a +bt^2 =8.5 +2.5 x 4 =18.5 (when t=4)
x=at + bt^2 =8.5 + 2.5 x 2 =13.5 (when t=2)
see the values below
(a=8.5m, b=2.5 ms^-2, t=2.0 s and t= 4.0 s)
Sorry, I didn't check this before. You forgot to square the time.
Also there is some confusion between, x=a +bt^2 and x=at + bt^2
In the question statement, you gave the first version.
But your last post is correct in your statement of average velocity and working from your previous results. You just need to check those first calculations.
Merlin3189 said:
Sorry, I didn't check this before. You forgot to square the time.
Also there is some confusion between, x=a +bt^2 and x=at + bt^2
In the question statement, you gave the first version.
But your last post is correct in your statement of average velocity and working from your previous results. You just need to check those first calculations.
It would be x=a+bt^2
Ok. That's what it said in the Q and it is consistent with the units given for a and b.
So now you can get the right values if you stick to this equation and take care with squaring t.
Merlin3189 said:
Ok. That's what it said in the Q and it is consistent with the units given for a and b.
So now you can get the right values if you stick to this equation and take care with squaring t.
I did but still, the answer does not match the answer in my book.
As you said I have done below
X=18.5 when t=2 and X=40 when t==4 now according to the average velocity
Vavg=xf-xi / tf-ti = 21.5 /2 ms^-1
Indranil said:
I did but still, the answer does not match the answer in my book.
As you said I have done below
X=18.5 when t=2 and X=40 when t==4 now according to the average velocity
Vavg=xf-xi / tf-ti = 21.5 /2 ms^-1
Double check your position calculation for t = 4 s.
gneill said:
Double check your position calculation for t = 4 s.
Thank you for finding my mistake. I checked it and it's done. Now the answer is 15m/s. Thank you all for your kind efforts.
## 1. What is the formula for calculating average velocity?
The formula for average velocity is average velocity = (change in position) / (change in time). This means that you divide the change in position by the change in time to get the average velocity.
## 2. How do I find the change in position?
The change in position is the final position minus the initial position. For example, if an object starts at position 10 and ends at position 20, the change in position would be 20 - 10 = 10.
## 3. What units are used for average velocity?
The units for average velocity are distance over time, such as meters per second or kilometers per hour. It is important to use consistent units for both distance and time in order to get an accurate average velocity.
## 4. Can average velocity be negative?
Yes, average velocity can be negative. This indicates that the object is moving in the opposite direction of the positive direction, or towards the negative direction. For example, if an object moves from position 10 to position 5, its average velocity would be negative because it is moving towards the negative direction.
## 5. How does average velocity differ from instantaneous velocity?
Average velocity is the overall velocity of an object over a period of time, while instantaneous velocity is the velocity of an object at a specific moment in time. Average velocity takes into account the total change in position and time, while instantaneous velocity only considers a specific point in time.
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1K | 1,751 | 6,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-26 | latest | en | 0.902236 |
https://help.altair.com/flux/Flux/Help/english/UserGuide/English/topics/ElectroHarmoniqueEquationsResolues.htm | 1,726,824,883,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00400.warc.gz | 255,835,733 | 10,859 | # Steady State AC Electric: solved equations
## Introduction
In a Steady state AC Electric application the equations used for computation are:
• the corresponding Maxwell's equations for an electrical system, and
• the constitutive equation that characterizes the dielectric materials
The conditions of computation of a Steady state AC Electric application are the following:
• the computation concerns the D and E fields; the B and H fields are not computed. The equations of the electric fields E, D and of the magnetic fields B, H are decoupled.
• the fields are harmonic time dependent of the pulsation ω. In the complex form of the equations, the operator ∂/∂t is replaced by the multiplicator
## Equations and conditions
In the previously defined conditions of computation, the equations in complex form are summarized as follows:
⇒ E : electric field strength (in V/m) D : electric flux density (C/m2) V : electric potential (in V) J : current density (in A/m2) ⇒ σ : conductivity (in S)εr : relative permittivityε0 : vacuum permittivity (in F/m)
Reminder about the differential operators: The curl divergence of a field is always null: div [rot (Field)] = 0.
## Solved equation
The second order equation solved by the finite elements method in Flux in case of a Steady state AC Electric application is the following:
where:
• [σ] is the tensor of the conductivity of the medium (in Siemens),
• r] is the tensor of the relative permittivity of dielectric materials
• ε0 is the vacuum permittivity; ε0 = 1/(36 π 109) (in F/m)
• V is the electric potential (in V)
## State variable
The state variable in a Steady state AC Electric application is the electric potential V (written Ve in Flux 3D). It is a complex quantity.
The uniqueness condition of the scalar field of the complex electric potential V requires that the value of this potential at least in a point of the computation domain be known.
## Complex permittivity
The permittivity ε is a complex quantity that takes account of the frequency:
[ε] = [ε']- j [ε'']
where:
• [ε'] stands for the real permittivity, and
• [ε''] stands for the dielectric losses constant:
[ε''] = [ε'] tan δh (with δh angle of hysteretic losses)
The second order equation solved by the finite elements method in Flux in case of a Steady State AC Electric application is the following: | 545 | 2,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-38 | latest | en | 0.884436 |
https://gsebsolutions.in/gseb-solutions-class-6-maths-chapter-6-intext-questions/ | 1,713,285,775,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817103.42/warc/CC-MAIN-20240416155952-20240416185952-00254.warc.gz | 268,706,690 | 53,439 | # GSEB Solutions Class 6 Maths Chapter 6 Integers InText Questions
Gujarat BoardĀ GSEB Textbook Solutions Class 6 Maths Chapter 6 Integers InText Questions and Answers.
## Gujarat Board Textbook Solutions Class 6 Maths Chapter 6 Integers InText Questions
Try These (NCERT Page 116)
Question 1.
Write the following numbers with appropriate signs:
(a) 100 m below sea level.
(b) 25Ā°C above 0Ā°C temperature.
(c) 15Ā°C below 0Ā°C temperature.
(d) Any five numbers less than 0.
Solution:
(a) 100 m below sea level ā – 100 m
(b) 25Ā° above 0Ā°C temperature ā + 25Ā° C
(c) 15Ā°C below 0Ā°C temperature ā – 15Ā° C
(d) Five numbers less than 0: { – 1, – 3, – 10, – 25, – 105 }
Note:
(i) If profit is represented by ā+ā sign, then loss may be represented by ā-ā sign.
(ii) If going up is represented by ā+ā sign, then going down may be represented by ā-ā sign.
(iii) If earnings are represented by ā+ā sign, then spending may be represented by ‘-‘ sign.
(iv) If temperature above 0Ā° is ā+ā sign, then temperature below 00 may be represented by ‘-‘ sign.
(v) If depositing money to bank is ā+ā sign, then withdrawal is ‘-‘ sign.
Try These (NCERT Page 118)
Question 1.
Mark 3, 7, – 4, – 8, – 1 and – 3 on the number line.
Solution:
Given numbers are marked on the number line as shown below:
Note: Look at the number line given here. We find that for every integer to the right of zero there is a corresponding integer to the left of zero (at the same distance from zero but with a negative sign). Similarly, for every integer to the left of zero. there is a corresponding integer to the right of zero (at the same distance from zero but with a positive sign).
Try These (Page 119)
Question 1.
Compare the following pairs of numbers using> or <.
Solution:
Try These (Page 125)
Question 1.
Draw a figure on the ground in the form of a horizontal number line as shown below. Frame questions as given in the said example and ask your friends.
Solution:
Do it yourself.
Try These (Page 125)
Question 1.
(a) ( -11) + ( -12)
(b) (+10) +(+4)
(c) (-32) + (-25)
(d) (+23) + (+40)
Solution:
(a) (-11) + (-12) = -[11 + 12] = -23
(b) (+10) + (+4) = + [10 + 4] + [14] = 14
(c) (- 32) + (- 25) = – [32 + 25] – [57] = -57
(d) (+23) + (40) = + [23 + 40] = + [63] = 63
Try These (Page 125).
Question 1.
Find the solution of the following:
(a) (- 7) + (+8)
(b) (-9) + (+13)
(c) (+7) + (-10)
(d) (+12) + (-7)
Solution:
(a) (-7) + (+8):
Opposite of (-7) is (+7) and (+8)
= (+7) + (+1)
(- 7) + (+ 8) = (- 7) + (+7) + (+1)
= 0 + (+1) [(-7) + (+7) = 0]
= + 1
(b) (-9) + (+13):
(+13) – (+9) + (+4)
(-9) + (+ 13) = (-9) + (+9) + (+4)
= 0 + (+4) [(-9) + (+9) = 0]
= + 4 = 4
(c) (+7) + (-10):
(-10) = (-7) + (-3)
(+7) + (-10) = (+7) + (-7) + (-3)
= 0 + (-3)
[(+7) + (-7) = 0]
= – 3
(e) (+12) + (-7):
(+12) = (Ā±7) + (+5)
(+12) + (-7) = (+7) + (+5) + (-7)
= (+7) + (-7) + (+5)
= – 0 + (+ 5)
= (+5)
= 5
[(+7) + (-7) = 0]
Try These (Page 127)
Question 1.
(a) (-2) + 6
(b) (-6) + 2
Make two such questions and solve them using the number line.
Solution:
(a) (-2) + 6
First move 2 steps to the left of 0 to reach at – 2. From here, move 6 steps to the right of – 2, to reach at 4.
(-2) + (+6) = + 4
(b) (-6) + 2:
On the number line, we first move 6 equal steps (each of 1 unit) to the left of 0, to reach at (-6). Now, move 2 steps to the right of (- 6) to reach at (- 4).
(-6) + (+2) = – 4
Question 2.
Find the solution of the following without using number line:
(a) (+7) + (-11)
(b) (-13) + (+10)
(c) (-7) + (+9)
(d) (+10) + (-5)
Make five such questions and solve them.
(a) (+7) + (-11):
(-11) = (-7) + (-4)
(+7) + (-11) = (+7) + (-7) + (-4) = 0 + (-4)
[(+7) + (-7) = 0] = – 4
Thus, (+7) + (-11) = – 4
(b) (-13) + (+10):
(-13) = (-10) + (-3)
(-13) + (+10) = (-10) + (-3) + (+10)
= (-10) + (+10) + (-3)
= 0 + (-3) = -3
[(-10) + (+ 10) = 0]
Thus, (- 13) + (+10) = – 3
(c) (-7) + (+9):
(+9) = (+7) + (+2)
(-7) + (+9) = (-7) + (+7) + (+2) = 0 + (+2)
[(- 7) + (+7) = 0]
= + 2
Thus, (- 7) + (+9) = + 2
(d) (+10) + (-5):
(+10) = (+5) + (+5)
(+10) + (-5) = (+5) + (+5) + (-5) = + 5 + 0
[(+5) + (-5) = 0]
Thus, (+10) + (-5) = (+5) | 1,652 | 4,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-18 | longest | en | 0.840798 |
https://mathisradical.com/radical-problem/ratios/factor-3rd-order-polynomial.html | 1,709,322,406,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475701.61/warc/CC-MAIN-20240301193300-20240301223300-00700.warc.gz | 386,342,575 | 12,598 | Algebra Tutorials!
Friday 1st of March
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# Find the measure of each marked angle
0
48
1
+350
Find the measure of each marked angle. A ray rises from left to right from the center of a horizontal line forming two angles. The top left angle measures (9x plus 9) degrees.The top right angle measures (6x plus 6) degrees.
left parenthesis 9 x plus 9 right parenthesis degrees(9x+9)°
left parenthesis 6 x plus 6 right parenthesis degrees(6x+6)°
The larger angle measures
The smaller angle measures
Sort:
#1
+5589
+3
The sum of the two angles is 180° because together they form a straight line.
(9x + 9)° + (6x + 6)° = 180°
9x + 9 + 6x + 6 = 180 Combine like terms on the left .
15x + 15 = 180 Subtract 15 from both sides.
15x = 165 Divide both sides by 15 .
x = 11
The larger angle measures (9x + 9)° = (9(11) + 9)° = (99 + 9)° = 108°
The smaller angle measures (6x + 6)° = (6(11) + 6)° = (66 + 6)° = 72°
hectictar Nov 16, 2017
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We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 421 | 1,260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-51 | longest | en | 0.627176 |
https://www.javacodemonk.com/which-data-type-would-you-choose-for-storing-monetary-values-in-java-6692c78c | 1,579,993,538,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251681625.83/warc/CC-MAIN-20200125222506-20200126012506-00194.warc.gz | 937,325,469 | 12,004 | # Which data type would you choose for storing Monetary Values, Currency in Java
Carvia Tech | August 10, 2019 | 5 min read | 1 views
Float & Double are bad for financial (even for military use) world, never use them for monetary calculations. If precision is one of your requirements, use `BigDecimal` instead.
Lets see the problem with the help of this example:
All floating point values that can represent a currency amount (in dollars and cents) can not be stored exactly as it is in the memory. So if we want to store 0.1 dollar (10 cents), float/double can not store it as it is, instead binary can store only a closer approximation value (0.100000001490116119384765625 in decimal). The magnitude of this problem becomes significant (known as loss of significance) when we repetitively do arithmetic operations (multiply or divide) using these two data types. Let’s try to understand this fact by taking this simple example
Loss of precision using double
``````public class DoubleForCurrency {
public static void main(String[] args) {
double total = 0.2;
for (int i = 0; i < 100; i++) {
total += 0.2;
}
System.out.println("total = " + total);
}
}``````
Program output:
`total = 20.19999999999996`
The output should have been `20.20` (20 dollars and 20 cents), but floating point calculation made it `20.19999999999996`. That’s loss of precision (or loss of significance)
## Real cause of loss of significance
Floating-point arithmetic
In computing, floating-point arithmetic (FP) is arithmetic using formulaic representation of real numbers as an approximation so as to support a trade-off between range and precision.
From Wikipedia
Whether or not a rational number has a terminating expansion depends on the base. For example, in base-10 the number 1/2 has a terminating expansion (0.5) while the number 1/3 does not (0.333…). In base-2 only rationals with denominators that are powers of 2 (such as 1/2 or 3/16) are terminating. Any rational with a denominator that has a prime factor other than 2 will have an infinite binary expansion. This means that numbers which appear to be short and exact when written in decimal format may need to be approximated when converted to binary floating-point. For example, the decimal number 0.1 is not representable in binary floating-point of any finite precision; the exact binary representation would have a "1100" sequence continuing endlessly:
e = −4; s = 1100110011001100110011001100110011…, where, as previously, s is the significand and e is the exponent.
When rounded to 24 bits this becomes
e = −4; s = 110011001100110011001101, which is actually 0.100000001490116119384765625 in decimal.
## BigDecimal for the rescue
BigDecimal represents a signed decimal number of arbitrary precision with an associated scale. BigDecimal provides full control over the precision and rounding of the number value. Virtually its possible to calculate value of pi to 2 billion decimal places using BigDecimal, available physical memory being the only limit.
That’s the reason we should always prefer BigDecimal or BigInteger for financial calculations.
Special Notes
Primitive type - int and long are also useful for monetary calculations if decimal precision is not required.
We should really avoid using BigDecimal(double value) constructor instead prefer BigDecimal(String) because BigDecimal (0.1) results in (0.1000000000000000055511151231257827021181583404541015625) being stored in BigDecimal instance. In contrast BigDecimal ("0.1") stores exactly 0.1
Question: What is Precision and Scale?
Precision is the total number of digits (or significant digits) of a real number
Scale specifies number of digits after decimal place For example, 12.345 has precision of 5 (total digits) and scale of 3 (number of digits right of the decimal)
### How to format BigDecimal Value without getting exponentiation in the result & Strip the trailing zeros?
We might get exponentiation in the calculation result if we do not follow some best practices while using `BigDecimal`. Below is the code snippet which shows a good usage example of handling the calculation result using `BigDecimal`.
BigDecimal Rounding
``````import java.math.BigDecimal;
public class BigDecimalForCurrency {
public static void main(String[] args) {
int scale = 4;
double value = 0.11111;
BigDecimal tempBig = new BigDecimal(Double.toString(value));
tempBig = tempBig.setScale(scale, BigDecimal.ROUND_HALF_EVEN);
String strValue = tempBig.stripTrailingZeros().toPlainString();
System.out.println("tempBig = " + strValue);
}
}``````
Program Output
`tempBig = 0.1111`
How would you print a given currency value for Indian Locale (INR Currency)?
`NumberFormat` class is designed specifically for this purpose. Currency symbol & Rounding Mode is automatically set based on the locale using `NumberFormat`. Lets see this in action
NumberFormat example
``````class Scratch {
public static String formatRupees(double value) {
NumberFormat format = NumberFormat.getCurrencyInstance(new Locale("en", "in"));
format.setMinimumFractionDigits(2);
format.setMaximumFractionDigits(5);
format.setRoundingMode(RoundingMode.HALF_EVEN);
return format.format(value);
}
public static void main(String[] args) {
BigDecimal tempBig = new BigDecimal(22.121455);
System.out.println("tempBig = " + formatRupees(tempBig.doubleValue()));
}
}``````
Program Output
`tempBig = Rs.22.12146`
That’s all, everything is taken care by NumberFormat.
### Some precautions
• BigDecimal(String) constructor should always be preferred over BigDecimal(Double), because using `BigDecimal(double)` is unpredictable due to inability of the double to represent 0.1 as exact 0.1
• If double must be used for initializing a BigDecimal, use `BigDecimal.valueOf(double)` which converts Double value to string using Double.toString(double) method
• Rounding mode should be provided while setting the scale
• StripTrailingZeros chops off all the trailing zeros
• toString() may use scientific notation but, toPlainString() will never return exponentiation in its result
### Do you know?
Use of float and double instead of BigDecimal could be fatal in military world
On February 25, 1991, a loss of significance in a MIM-104 Patriot missile battery prevented it from intercepting an incoming Scud missile in Dhahran, Saudi Arabia, contributing to the death of 28 soldiers from the U.S. Army’s 14th Quartermaster Detachment - http://www.gao.gov/products/IMTEC-92-26
Banker’s Rounding Mode
Since the introduction of IEEE 754, the default method (round to nearest, ties to even, sometimes called Banker’s Rounding or `RoundingMode.HALF_EVEN`) is more commonly used USA. This method rounds the ideal (infinitely precise) result of an arithmetic operation to the nearest representable value, and gives that representation as the result. In the case of a tie, the value that would make the significand end in an even digit is chosen.
### For further reading -
Buy my ebook for complete question bank
Most of these questions has been answered in my eBook "Cracking the Core Java Interview" updated on June 2018, that you can buy from this link:
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# Answer to Question #17117 in Abstract Algebra for sanches
Question #17117
Let R be any semisimple ring. Every element a ∈ R can be written as a unit times an idempotent.
By the Wedderburn-Artin Theorem, we are reduced to the case when R = Mn(D) where D is a division ring.
By the theorem on Reduction to Echelon Forms, we can find invertible n × n matrices b, c ∈ R such that
d := bac = diag(1, . . . , 1, 0, . . . , 0) (an idempotent). We have now a = (b−1c−1)(cdc−1) = ue,
where u : = b−1c−1 is a unit and e : = cdc−1 is an idempotent.
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https://books.google.com.br/books/about/Elementary_illustrations_of_the_Celestia.html?id=R2MSAAAAIAAJ&redir_esc=y&hl=pt-BR | 1,571,387,691,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986679439.48/warc/CC-MAIN-20191018081630-20191018105130-00368.warc.gz | 404,294,770 | 12,502 | # Elementary Illustrations of the Celestial Mechanics of Laplace: Part the First, Comprehending the First Book
J. Murray, 1821 - 344 páginas
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https://math.libretexts.org/Bookshelves/Precalculus/Precalculus_1e_(OpenStax)/02%3A_Linear_Functions/2.04%3A_Fitting_Linear_Models_to_Data | 1,726,608,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00577.warc.gz | 356,061,893 | 36,390 | # 2.4: Fitting Linear Models to Data
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Learning Objectives
• Draw and interpret scatter diagrams.
• Use a graphing utility to find the line of best fit.
• Distinguish between linear and nonlinear relations.
• Fit a regression line to a set of data and use the linear model to make predictions.
A professor is attempting to identify trends among final exam scores. His class has a mixture of students, so he wonders if there is any relationship between age and final exam scores. One way for him to analyze the scores is by creating a diagram that relates the age of each student to the exam score received. In this section, we will examine one such diagram known as a scatter plot.
## Drawing and Interpreting Scatter Plots
A scatter plot is a graph of plotted points that may show a relationship between two sets of data. If the relationship is from a linear model, or a model that is nearly linear, the professor can draw conclusions using his knowledge of linear functions. Figure $$\PageIndex{1}$$ shows a sample scatter plot.
Notice this scatter plot does not indicate a linear relationship. The points do not appear to follow a trend. In other words, there does not appear to be a relationship between the age of the student and the score on the final exam.
Example $$\PageIndex{1}$$: Using a Scatter Plot to Investigate Cricket Chirps
Table shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit[1]. Plot this data, and determine whether the data appears to be linearly related.
Chirps 44 35 20.4 33 31 35 18.5 37 26 Temperature 80.5 70.5 57 66 68 72 52 73.5 53
Solution
Plotting this data, as depicted in Figure $$\PageIndex{2}$$ suggests that there may be a trend. We can see from the trend in the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear, though certainly not perfectly so.
## Finding the Line of Best Fit
Once we recognize a need for a linear function to model that data, the natural follow-up question is “what is that linear function?” One way to approximate our linear function is to sketch the line that seems to best fit the data. Then we can extend the line until we can verify the y-intercept. We can approximate the slope of the line by extending it until we can estimate the $$\frac{\text{rise}}{\text{run}}$$.
Example $$\PageIndex{2}$$: Finding a Line of Best Fit
Find a linear function that fits the data in Table $$\PageIndex{1}$$ by “eyeballing” a line that seems to fit.
Solution
On a graph, we could try sketching a line.
Using the starting and ending points of our hand drawn line, points $$(0, 30)$$ and $$(50, 90)$$, this graph has a slope of
$m=\dfrac{60}{50}=1.2$
and a y-intercept at 30. This gives an equation of
$T(c)=1.2c+30$
where $$c$$ is the number of chirps in 15 seconds, and $$T(c)$$ is the temperature in degrees Fahrenheit. The resulting equation is represented in Figure $$\PageIndex{3}$$.
Analysis
This linear equation can then be used to approximate answers to various questions we might ask about the trend.
## Recognizing Interpolation or Extrapolation
While the data for most examples does not fall perfectly on the line, the equation is our best guess as to how the relationship will behave outside of the values for which we have data. We use a process known as interpolation when we predict a value inside the domain and range of the data. The process of extrapolation is used when we predict a value outside the domain and range of the data.
Figure $$\PageIndex{4}$$ compares the two processes for the cricket-chirp data addressed in Example $$\PageIndex{2}$$. We can see that interpolation would occur if we used our model to predict temperature when the values for chirps are between 18.5 and 44. Extrapolation would occur if we used our model to predict temperature when the values for chirps are less than 18.5 or greater than 44.
There is a difference between making predictions inside the domain and range of values for which we have data and outside that domain and range. Predicting a value outside of the domain and range has its limitations. When our model no longer applies after a certain point, it is sometimes called model breakdown. For example, predicting a cost function for a period of two years may involve examining the data where the input is the time in years and the output is the cost. But if we try to extrapolate a cost when $$x=50$$, that is in 50 years, the model would not apply because we could not account for factors fifty years in the future.
Interpolation and Extrapolation
Different methods of making predictions are used to analyze data.
• The method of extrapolation involves predicting a value outside the domain and/or range of the data.
• Model breakdown occurs at the point when the model no longer applies.
Example $$\PageIndex{3}$$: Understanding Interpolation and Extrapolation
Use the cricket data from Table $$\PageIndex{1}$$ to answer the following questions:
1. Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable.
2. Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable.
Solution
a. The number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpolation. Using our model:
\begin{align} T(30)&=30+1.2(30) \\ &=66 \text{ degrees} \end{align}
Based on the data we have, this value seems reasonable.
b. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation because 40 is outside the range of our data. Using our model:
\begin{align} 40&=30+1.2c \\ 10&=1.2c \\ c&\approx8.33 \end{align}
We can compare the regions of interpolation and extrapolation using Figure $$\PageIndex{5}$$.
Analysis
Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping altogether below around 50 degrees.
Exercise $$\PageIndex{1}$$
According to the data from Table $$\PageIndex{1}$$, what temperature can we predict it is if we counted 20 chirps in 15 seconds?
Solution
54°F
## Finding the Line of Best Fit Using a Graphing Utility
While eyeballing a line works reasonably well, there are statistical techniques for fitting a line to data that minimize the differences between the line and data values[2]. One such technique is called least squares regression and can be computed by many graphing calculators, spreadsheet software, statistical software, and many web-based calculators[3]. Least squares regression is one means to determine the line that best fits the data, and here we will refer to this method as linear regression.
Given data of input and corresponding outputs from a linear function, find the best fit line using linear regression.
1. Enter the input in List 1 (L1).
2. Enter the output in List 2 (L2).
3. On a graphing utility, select Linear Regression (LinReg).
Example $$\PageIndex{4}$$: Finding a Least Squares Regression Line
Find the least squares regression line using the cricket-chirp data in Table $$\PageIndex{1}$$.
Solution
Enter the input (chirps) in List 1 (L1).
Enter the output (temperature) in List 2 (L2). See Table $$\PageIndex{2}$$.
L1 44 35 20.4 33 31 35 18.5 37 26 L2 80.5 70.5 57 66 68 72 52 73.5 53
On a graphing utility, select Linear Regression (LinReg). Using the cricket chirp data from earlier, with technology we obtain the equation:
$T(c)=30.281+1.143c$
Analysis
Notice that this line is quite similar to the equation we “eyeballed” but should fit the data better. Notice also that using this equation would change our prediction for the temperature when hearing 30 chirps in 15 seconds from 66 degrees to:
\begin{align} T(30)&=30.281+1.143(30) \\ &=64.571 \\ &\approx 64.6 \text{ degrees} \end{align}
The graph of the scatter plot with the least squares regression line is shown in Figure $$\PageIndex{6}$$.
Will there ever be a case where two different lines will serve as the best fit for the data?
No. There is only one best fit line.
## Distinguishing Between Linear and Non-Linear Models
As we saw above with the cricket-chirp model, some data exhibit strong linear trends, but other data, like the final exam scores plotted by age, are clearly nonlinear. Most calculators and computer software can also provide us with the correlation coefficient, which is a measure of how closely the line fits the data. Many graphing calculators require the user to turn a ”diagnostic on” selection to find the correlation coefficient, which mathematicians label as $$r$$. The correlation coefficient provides an easy way to get an idea of how close to a line the data falls.
We should compute the correlation coefficient only for data that follows a linear pattern or to determine the degree to which a data set is linear. If the data exhibits a nonlinear pattern, the correlation coefficient for a linear regression is meaningless. To get a sense for the relationship between the value of $$r$$ and the graph of the data, Figure $$\PageIndex{7}$$ shows some large data sets with their correlation coefficients. Remember, for all plots, the horizontal axis shows the input and the vertical axis shows the output.
Correlation Coefficient
The correlation coefficient is a value, $$r$$, between –1 and 1.
• $$r>0$$ suggests a positive (increasing) relationship
• $$r<0$$ suggests a negative (decreasing) relationship
• The closer the value is to 0, the more scattered the data.
• The closer the value is to 1 or –1, the less scattered the data is.
Example $$\PageIndex{5}$$: Finding a Correlation Coefficient
Calculate the correlation coefficient for cricket-chirp data in Table $$\PageIndex{1}$$.
Solution
Because the data appear to follow a linear pattern, we can use technology to calculate $$r$$. Enter the inputs and corresponding outputs and select the Linear Regression. The calculator will also provide you with the correlation coefficient, $$r=0.9509$$. This value is very close to 1, which suggests a strong increasing linear relationship.
Note: For some calculators, the Diagnostics must be turned "on" in order to get the correlation coefficient when linear regression is performed: [2nd]>[0]>[alpha][x–1], then scroll to DIAGNOSTICSON.
## Predicting with a Regression Line
Once we determine that a set of data is linear using the correlation coefficient, we can use the regression line to make predictions. As we learned above, a regression line is a line that is closest to the data in the scatter plot, which means that only one such line is a best fit for the data.
Example $$\PageIndex{6}$$: Using a Regression Line to Make Predictions
Gasoline consumption in the United States has been steadily increasing. Consumption data from 1994 to 2004 is shown in Table $$\PageIndex{3}$$. Determine whether the trend is linear, and if so, find a model for the data. Use the model to predict the consumption in 2008.
Year '94 '95 '96 '97 '98 '99 '00 '01 '02 '03 '04 Consumption (billions of gallons) 113 116 118 119 123 125 126 128 131 133 136
The scatter plot of the data, including the least squares regression line, is shown in Figure $$\PageIndex{8}$$.
We can introduce new input variable, $$t$$,representing years since 1994.
The least squares regression equation is:
$C(t)=113.318+2.209t$
Using technology, the correlation coefficient was calculated to be 0.9965, suggesting a very strong increasing linear trend.
Using this to predict consumption in 2008 $$(t=14)$$,
\begin{align} C(14)&=113.318+2.209(14) \\ &=144.244 \end{align}
The model predicts 144.244 billion gallons of gasoline consumption in 2008.
Exercise $$\PageIndex{1}$$
Use the model we created using technology in Example $$\PageIndex{6}$$ to predict the gas consumption in 2011. Is this an interpolation or an extrapolation?
150.871 billion gallons; extrapolation
## Key Concepts
• Scatter plots show the relationship between two sets of data.
• Scatter plots may represent linear or non-linear models.
• The line of best fit may be estimated or calculated, using a calculator or statistical software.
• Interpolation can be used to predict values inside the domain and range of the data, whereas extrapolation can be used to predict values outside the domain and range of the data.
• The correlation coefficient, $$r$$, indicates the degree of linear relationship between data.
• A regression line best fits the data.
• The least squares regression line is found by minimizing the squares of the distances of points from a line passing through the data and may be used to make predictions regarding either of the variables.
This page titled 2.4: Fitting Linear Models to Data is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. | 4,855 | 17,576 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-38 | latest | en | 0.198306 |
https://www.2classnotes.com/cbse-question-papers/9th-cbse-science-examination/ | 1,652,831,423,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520936.24/warc/CC-MAIN-20220517225809-20220518015809-00494.warc.gz | 695,431,504 | 15,671 | Tuesday , May 17 2022
# 9th CBSE Science Examination: Mid Term 2019
## 9th CBSE Science Examination: Mid Term Question Paper (2019-20)
School Name: Himalaya Public School, Sector 13, Rohini, Delhi 110085 India
Time: 3 Hrs.
Marks: 70
Class: 9th
Date: 17/9/2019
Subject: Science
## General Instructions:
1. All questions are compulsory. There are 37 questions in all.
2. This question paper has four sections. Section A, Section B, Section C and Section D.
3. Section A contains 20 questions of one mark each, Section B contains 7 questions of marks each, Section C contains 7 questions of 3 marks each and Section D contains 3 questions of 5 marks each.
4. There is no overall choice. However an internal choices have been provided in 1 question of 2 marks, 2 questions of 3 marks each and all the questions of 5 marks. You have to attempt only one of the choices in such questions.
## Section A
1. 1.5 m/s
2. 60 m/s
3. 0.1 m/s
4. 5 m/s
#### Question: 18. Which one of the following pairs of quantities has the same dimension?
1. Force and work done
2. Momentum and impulse
3. Pressure and force
4. Surface tension and stress
## Section D
#### Question: 37. In successive measurements, the readings of the period of oscillation of a simple pendulum were found to be 2.635, 2.565, 2.425, 2.715, 2.805 sec in an experiment.
Calculate:
1. Mean value of the period of oscillation absolute error in each measurement
2. Mean absolute error, relative error, percentage error.
## 11th Class Mathematics Term 2 Question Paper 2021-22
School Name: Himalaya Public School, Sector 13, Rohini, Delhi 110085 India Class: 11th Standard (CBSE) Subject: Mathematics Time … | 467 | 1,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-21 | longest | en | 0.852138 |
http://tolstoy.newcastle.edu.au/R/e9/help/10/01/0614.html | 1,500,792,214,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424287.86/warc/CC-MAIN-20170723062646-20170723082646-00152.warc.gz | 332,001,061 | 4,517 | Re: [R] faster GLS code
Date: Thu, 07 Jan 2010 14:06:34 -0500
X <- kronecker(diag(1,3),x)
Y <- c(y) # stack the y in a vector
# residual covariance matrix for each observation covar <- kronecker(sigma,diag(1,N))
csig <- chol2inv(covar)
betam2 <- ginv(csig %*% X) %*% csig %*% Y
This is more than 2 times faster than your code (however, it doesn't compute `betav') .
# GLS betas covariance matrix
system.time({
inv.sigma <- solve(covar)
betav <- solve(t(X)%*%inv.sigma%*%X)
# GLS mean parameter estimates
betam <- betav%*%t(X)%*%inv.sigma%*%Y
})
# New method
system.time({
csig <- chol2inv(covar)
betam2 <- ginv(csig %*% X) %*% csig %*% Y })
all.equal(betam, betam2)
> # GLS betas covariance matrix
> system.time({
```+ inv.sigma <- solve(covar)
+ betav <- solve(t(X)%*%inv.sigma%*%X)
+
+ # GLS mean parameter estimates
+ betam <- betav%*%t(X)%*%inv.sigma%*%Y
+ })
```
user system elapsed
1.14 0.51 1.76
>
> system.time({
```+ csig <- chol2inv(covar)
+ betam2 <- ginv(csig %*% X) %*% csig %*% Y
+ })
```
user system elapsed
0.47 0.08 0.61
>
> all.equal(betam, betam2)
[1] TRUE
>
Hope this helps,
Ravi.
Assistant Professor,
Division of Geriatric Medicine and Gerontology School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
• Original Message ----- From: Carlo Fezzi <c.fezzi_at_uea.ac.uk> Date: Thursday, January 7, 2010 12:13 pm Subject: [R] faster GLS code To: r-help_at_r-project.org
> Dear helpers,
>
> I wrote a code which estimates a multi-equation model with generalized
> least squares (GLS). I can use GLS because I know the covariance
> matrix of
> the residuals a priori. However, it is a bit slow and I wonder if anybody
> would be able to point out a way to make it faster (it is part of a bigger
> code and needs to run several times).
>
> Any suggestion would be greatly appreciated.
>
> Carlo
>
>
> ***************************************
> Carlo Fezzi
> Senior Research Associate
>
> Centre for Social and Economic Research
> on the Global Environment (CSERGE),
> School of Environmental Sciences,
> University of East Anglia,
> Norwich, NR4 7TJ
> United Kingdom.
> email: c.fezzi_at_uea.ac.uk
> ***************************************
>
> Here is an example with 3 equations and 2 exogenous variables:
>
> ----- start code ------
>
>
> N <- 1000 # number of observations
> library(MASS)
>
> ## parameters ##
>
> # eq. 1
> b10 <- 7; b11 <- 2; b12 <- -1
>
> # eq. 2
> b20 <- 5; b21 <- -2; b22 <- 1
>
> # eq.3
> b30 <- 1; b31 <- 5; b32 <- 2
>
> # exogenous variables
>
> x1 <- runif(min=-10,max=10,N)
> x2 <- runif(min=-5,max=5,N)
>
> # residual covariance matrix
> sigma <- matrix(c(2,1,0.7,1,1.5,0.5,0.7,0.5,2),3,3)
>
> # residuals
> r <- mvrnorm(N,mu=rep(0,3), Sigma=sigma)
>
> # endogenous variables
>
> y1 <- b10 + b11 * x1 + b12*x2 + r[,1]
> y2 <- b20 + b21 * x1 + b22*x2 + r[,2]
> y3 <- b30 + b31 * x1 + b32*x2 + r[,3]
>
> y <- cbind(y1,y2,y3) # matrix of endogenous
> x <- cbind(1,x1, x2) # matrix of exogenous
>
>
> #### MODEL ESTIMATION ###
>
> # build the big X matrix needed for GLS estimation:
>
> X <- kronecker(diag(1,3),x)
> Y <- c(y) # stack the y in a vector
>
> # residual covariance matrix for each observation
> covar <- kronecker(sigma,diag(1,N))
>
> # GLS betas covariance matrix
> inv.sigma <- solve(covar)
> betav <- solve(t(X)%*%inv.sigma%*%X)
>
> # GLS mean parameter estimates
> betam <- betav%*%t(X)%*%inv.sigma%*%Y
>
> ----- end of code ----
>
> ______________________________________________
> R-help_at_r-project.org mailing list
> | 1,201 | 3,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-30 | latest | en | 0.626896 |
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# Show that if two inductors with equal inductance L are connected in parallel then the equivalent inductance of the combination is L/2. The inductors are separated by a large distance.
Open in App
Solution
## If two inductors are connected in parallel then,1Leq=1L1+1L2Given,L1=L2=L1Leq=1L+1L=2LLeq=L2
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0
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Number of equations to solve: 23456789
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Solve for:
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Number of inequalities to solve: 23456789
Ineq. #1:
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# Factoring Polynomials
When factoring is completed:
1) Check to see if one of the polynomial factors can be factored again.
2) Double check for other GCFs (greatest common factors)
## I.GCF Greatest Common Factor
> A factor that every term in the expression has in common
> Reverse distribution
Examplea: To find the GCF for 12x4 - 6x3+ 3x2
12x4-6x3+3x2 means (2·2·3·x·x·x·x)-(2·3·x·x·x)+(3·x·x)
all three terms have 3 and x·x as common factors
So the GCG is 3x2
Note:
1) 3 is the greatest common numerical factor because it is the largest number that divides into every term evenly.
2) x2 is the greatest common variable factor because it is the variable to the largest power that divides evenly into all three terms.
To factor 12x4-6x3+3x2 thi9ne reverse distribution 3x2(4x2-2x+1)
## II. Factoring Trinomials (ax2+bx+c)
i) Special Case (Perfect Square Trinomial)
Strategy: Test the trinomial to see if it satisfies the conditions of a Perfect Square Trinomial.
A Perfect Square Trinomial:
A2+2AB+B2 factors into (A+B)(A+B) which can be written (A+B)2 or
A2-2AB+B2 factors into (A-B)(A-B) which can be written (A-B)2
TEST:i) 1st and last terms must be perfect squares.
ii)│middle term │must equal 2 times the roots of the 1st and last terms.
Example: Factor 9x2-24x+6
Test:
i) 1st term is 9x2 (3x2) and last term is 16 (4)2, both perfect squares with roots 3x and 4 as shown.
ii) does the middle term│ 24x equal 2(3x)(4)? YES!
The trinomial passes both test. It is a Perfect Square Triniomial of the type
A2-2AB+B2=(A-B)2 where 3x
A and 4 B
Solution:(3x-4)2
ii) For a = 1 (lead coefficient is 1)
Strategy: Find the product ac then determine which factor pairs of ac add up to b.
Ecample: Factor x2-x-12 Note: a=1,b=-1,and c=-12
ac==>(1)(-12)=-12=ac
Note:
i) You don
t have to list all factor pairs, just find the two that add to b.
ii) If there are no factor pairs that add to b then the polynomial is "prime".
Solution must be of the form (x _)(x _) where the factor pair and corresponding positive or negative sign supply the missing terms.
Solution:(x+3)(x-4)
iii) For a≠1
Strategy: Find the product ac then determine the factor pairs of ac that add up to b.
Examole: Factor 4x2-25x-21 Note: a=4.b=-25,c=-21
Note: i) You don't have to list all factor pairs, just find the two shat add to b.
ii) If there are no factor pairs that add to b then the polynomial is " prime".
Solution must be of the form (_x _)(_x _)
i) List the factor pairs of a to find possible coefficients of x.
ii) List the factor pairs of c to possible missing final terms.
iii)Product of "inners" and "outers" must be from the list of factor pairs of ac above that to b.
Solution:(4x+3)(x-7)
## III. Factor by Grouping(4 Terms)
Strategy: Group the four terms in pairs and find the GCF of each pair. Both pairs should then have
a common binomial factor.
Example 1:Factor 3ax-3ay-2bx+2by
3ax-3ay-2bx+2by
GCF of 1st group is 3a
GCF of 2nd group is -2b
(Note: If the two groups are separated by a subtraction always factor out a negative GCF from the
second group.)
Factoring each group yields 3a(x-y)-2b(x-y)
Both have a common binomial factor (x-y) which can be factored out leaving the other factor(3a-2b).
Solution: (x-y)(3a-2b)
Example 2: Factor 4X2-25X-21
Strategy: Notice!...this polynomial is the same as in section II and it doesn't have 4 terms. But,
if we replace the middle term-25x with the factor pair of ac from the list that adds to -25 we can
write this trinomial with 4 terms and solve by grouping .This is an alternative way of factoring a
trinomial where a≠1.
4x2-28x+3x-21
GCF of 1st group is 4x
GCF of 2nd group if 3
Factoring each group yields 4x(x-7)+3(x-7)
Both have a common binomial factou (x-7) which can be factored out leaving the other factor(4x-3)
Solution: (x-7)(4x-3)
## Ⅳ. Special Cases (2 Terms)
i) Difference of Squares
Strategy: Check the binomial to see if it is a difference of squares.
A Difference of Squares A2-B2 factors into (A+B)(A-B)
TEST:
i) 1st and last terms must be perfect squares.
ii) Must be a subtraction.
NOTE: Besides a possible common factor, the Sum of Squares A2 + B2 is PRIME (not factorable).
Example: Factor 9x2-16
Test:
i) 1st term is 9x2
(3x)2 and last term, is 16 (4)2,both perfect squares with roots 3x and 4 as shown.
ii) are we subtracting? YES!
The binomial passes both test. It is a Difference of Squares of the form
A2-B2 = (A+B)(A-B) where 3x
A and 4 B
Solution: (3x+4)(3x-4)
ii) Difference OR Sum of Cubes
Strategy: Check the binomial to see if it is a difference of cubes or a sum of cubes.
A Difference of Cubes A3-B3 factors into (A-B)(A2+AB+B2)
A Sum of Cubes A3+B3 factor into (A+B)(A2-AB+B2)
Note: The trinomial factor is often PRLME and cannot be factored further.
TEST:
i) 1st and last terms must be perfect cubes.
ii) Can be erther a subtraction OR an addition.
Example 1: Factor 27x3+8
Test:
i) 1st term is 27x3
(3x)3 and last term is 8 (2)3,both perfect cubes with roots 3x and 2 as shown.
ii) it is the sum of cubes.
The binomial is a Sum of Cubes of the form A3+B3 = (A+B)(A2-AB+B2)
where 3x
A and 2 B
Solution: (3x+2)(9z2-6x+4)
Example 2: Factor 1-64x3
Test:
i) 1st term is 1
(1)3 and last term is 64x3 (4x)3, both perfect cubes with roots 1 and 4x as shown.
ii) it is the difference of cubes.
The binomial is a Difference of Cubes of the form A3-B3 = (A-B)(A2+AB+B2)
where 1
A and 4x B
Solution: (1-4x)(1+4x+16x2)
## WORKED EXAMPLES
1. Factor 16x4-1
(4x2)2-(1)2 Difference of Squares
(4x2+1)(4x2-1) Another Difference of Squares!
(4x2+1)((2x)2-(1)2)
(4x2+1)(2x+1)(2x-1)
2. Factor 3x2+27
3(x2+9)
GCF
(X2+9) can NOT be factored further it is the Sum of Squares which is prime.
3. Factor 8a3+27b3
(2a)3+(3b)3 Sum of Cubes
(2a+3b)(4a2-6ab+9b2)
4. Factor 2x2-xy-6y2
(_x_y)(_x_y) Factor pairs of ac = -12 that add to the middle are 3 and -4
Practice Exercises | 2,184 | 6,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-40 | latest | en | 0.866878 |
https://finance.yahoo.com/news/time-buy-alio-gold-inc-205308352.html | 1,575,725,500,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540499439.6/warc/CC-MAIN-20191207132817-20191207160817-00248.warc.gz | 373,979,248 | 105,812 | U.S. Markets closed
# Is It Time To Buy Alio Gold Inc (TSE:ALO) Based Off Its PE Ratio?
Alio Gold Inc (TSX:ALO) is trading with a trailing P/E of 8.2x, which is lower than the industry average of 10.9x. Although some investors may jump to the conclusion that this is a great buying opportunity, understanding the assumptions behind the P/E ratio might change your mind. Today, I will deconstruct the P/E ratio and highlight what you need to be careful of when using the P/E ratio. See our latest analysis for Alio Gold
### Breaking down the P/E ratio
P/E is often used for relative valuation since earnings power is a chief driver of investment value. By comparing a stock’s price per share to its earnings per share, we are able to see how much investors are paying for each dollar of the company’s earnings.
P/E Calculation for ALO
Price-Earnings Ratio = Price per share ÷ Earnings per share
ALO Price-Earnings Ratio = \$1.78 ÷ \$0.218 = 8.2x
The P/E ratio itself doesn’t tell you a lot; however, it becomes very insightful when you compare it with other similar companies. We want to compare the stock’s P/E ratio to the average of companies that have similar characteristics as ALO, such as size and country of operation. A quick method of creating a peer group is to use companies in the same industry, which is what I will do. ALO’s P/E of 8.2x is lower than its industry peers (10.9x), which implies that each dollar of ALO’s earnings is being undervalued by investors. As such, our analysis shows that ALO represents an under-priced stock.
### Assumptions to be aware of
While our conclusion might prompt you to buy ALO immediately, there are two important assumptions you should be aware of. Firstly, our peer group contains companies that are similar to ALO. If this isn’t the case, the difference in P/E could be due to other factors. For example, if you compared lower risk firms with ALO, then investors would naturally value it at a lower price since it is a riskier investment. The second assumption that must hold true is that the stocks we are comparing ALO to are fairly valued by the market. If this does not hold, there is a possibility that ALO’s P/E is lower because our peer group is overvalued by the market.
### What this means for you:
Since you may have already conducted your due diligence on ALO, the undervaluation of the stock may mean it is a good time to top up on your current holdings. But at the end of the day, keep in mind that relative valuation relies heavily on critical assumptions I’ve outlined above. Remember that basing your investment decision off one metric alone is certainly not sufficient. There are many things I have not taken into account in this article and the PE ratio is very one-dimensional. If you have not done so already, I urge you to complete your research by taking a look at the following:
1. Future Outlook: What are well-informed industry analysts predicting for ALO’s future growth? Take a look at our free research report of analyst consensus for ALO’s outlook.
2. Past Track Record: Has ALO been consistently performing well irrespective of the ups and downs in the market? Go into more detail in the past performance analysis and take a look at the free visual representations of ALO’s historicals for more clarity.
3. Other High-Performing Stocks: Are there other stocks that provide better prospects with proven track records? Explore our free list of these great stocks here.
To help readers see pass the short term volatility of the financial market, we aim to bring you a long-term focused research analysis purely driven by fundamental data. Note that our analysis does not factor in the latest price sensitive company announcements.
The author is an independent contributor and at the time of publication had no position in the stocks mentioned. | 864 | 3,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-51 | latest | en | 0.956422 |
https://www.tutorialspoint.com/finding-square-root-of-a-number-without-using-library-functions-javascript | 1,669,495,332,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708046.99/warc/CC-MAIN-20221126180719-20221126210719-00506.warc.gz | 1,116,998,681 | 9,826 | Finding square root of a number without using library functions - JavaScript
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We are required to write a JavaScript function that takes in a number and calculates its square root without using the Math.sqrt() function.
Example
Following is the code −
const square = (n, i, j) => {
let mid = (i + j) / 2;
let mul = mid * mid;
if ((mul === n) || (Math.abs(mul - n) < 0.00001)){
return mid;
}else if (mul < n){
return square(n, mid, j);
}else{
return square(n, i, mid);
}
}
// Function to find the square root of n
const findSqrt = num => {
let i = 1;
const found = false;
while (!found){
// If n is a perfect square
if (i * i === num){
return i;
}else if (i * i > num){
let res = square(num, i - 1, i);
return res;
};
i++;
}
}
console.log(findSqrt(33));
Output
This will produce the following output in console −
5.744562149047852
Understanding the code
We looped over from i = 1.
If i * i = n, then we returned i as n is a perfect square whose square root is i., else we find the smallest i for which i * i is just greater than n.
Now we know the square root of n lies in the interval i – 1 and i. And then we used the Binary Search algorithm to find the square root.
Updated on 30-Sep-2020 14:35:30 | 403 | 1,515 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2022-49 | latest | en | 0.663548 |
https://www.sanfoundry.com/electrical-measurements-questions-answers-autoranging/ | 1,718,383,232,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861567.95/warc/CC-MAIN-20240614141929-20240614171929-00429.warc.gz | 861,208,383 | 20,970 | # Electrical Measurements Questions and Answers – Autoranging
This set of Electrical Measurements & Measuring Instruments Multiple Choice Questions & Answers (MCQs) focuses on “Autoranging”.
1. Autoranging means __________
a) automatic ranging
b) fixed ranging
c) automobile ranging
d) constant ranging
Explanation: Autoranging refers to obtaining an automatic reading with optimum resolution under all operating conditions. Say for example, 155 mV is displayed as 155.0 and not 0.155.
2. A 3 digit display DVM with a maximum reading of 1999 indicates __________
a) increase by a factor
b) reduction by a factor
c) no change in value
d) depends on the circuit components
Explanation: A 3 digit display DVM with maximum reading capability indicates that any reading above the maximum set limit of 1999 will be reduced by a factor of 10.
3. For a value less than 0200, the instrument should ________
a) read values less than 0200 correctly
c) automatically switch range
d) should not respond at all
Explanation: The DVM should automatically switch its range when the display is greater than 1999 units as it is the maximum set limit for achieving a higher sensitivity.
a) attenuates the signal
b) converts digital to analog
c) converts analog to digital
d) contains information
Explanation: When the count produced by the ADC counter is less than 170, a control pulse is obtained for down ranging. Whereas the control pulse for up ranging is produced once the ADC counter exceeds 1999 units.
a) True
b) False
Explanation: ADC contains information required for polarity indication. The polarity of the signal that is integrated is of utmost importance.
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6. Integration period is obtained by ________
a) using signal amplitude
b) counting the pulse
c) measuring time
d) by differentiating the signal
Explanation: By counting the pulses we obtain the integration period. Polarity measurement is obtained by making use of the last count or some of the last counts.
7. Integrator’s output is ________
a) attenuated through a filter
b) feedback to the input
c) stored in a flip-flop
d) differentiated
Explanation: The output from the integrator is used to set a polarity flip-flop. The flip-flop’s output is then stored in memory until the next measurement of voltage is obtained.
8. Old information is used to set range relays.
a) True
b) False
Explanation: Range relays are set through the decoder using the new information obtained with the help of a clock pulse. Decimal point is also changed as per the requirement of the new range.
Sanfoundry Global Education & Learning Series – Electrical Measurements. | 594 | 2,706 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-26 | latest | en | 0.8473 |
https://terrytao.wordpress.com/tag/haar-measure/ | 1,632,105,893,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056974.30/warc/CC-MAIN-20210920010331-20210920040331-00648.warc.gz | 600,680,022 | 37,330 | You are currently browsing the tag archive for the ‘Haar measure’ tag.
In 1964, Kemperman established the following result:
Theorem 1 Let ${G}$ be a compact connected group, with a Haar probability measure ${\mu}$. Let ${A, B}$ be compact subsets of ${G}$. Then
$\displaystyle \mu(AB) \geq \min( \mu(A) + \mu(B), 1 ).$
Remark 1 The estimate is sharp, as can be seen by considering the case when ${G}$ is a unit circle, and ${A, B}$ are arcs; similarly if ${G}$ is any compact connected group that projects onto the circle. The connectedness hypothesis is essential, as can be seen by considering what happens if ${A}$ and ${B}$ are a non-trivial open subgroup of ${G}$. For locally compact connected groups which are unimodular but not compact, there is an analogous statement, but with ${\mu}$ now a Haar measure instead of a Haar probability measure, and the right-hand side ${\min(\mu(A)+\mu(B),1)}$ replaced simply by ${\mu(A)+\mu(B)}$. The case when ${G}$ is a torus is due to Macbeath, and the case when ${G}$ is a circle is due to Raikov. The theorem is closely related to the Cauchy-Davenport inequality; indeed, it is not difficult to use that inequality to establish the circle case, and the circle case can be used to deduce the torus case by considering increasingly dense circle subgroups of the torus (alternatively, one can also use Kneser’s theorem).
By inner regularity, the hypothesis that ${A,B}$ are compact can be replaced with Borel measurability, so long as one then adds the additional hypothesis that ${A+B}$ is also Borel measurable.
A short proof of Kemperman’s theorem was given by Ruzsa. In this post I wanted to record how this argument can be used to establish the following more “robust” version of Kemperman’s theorem, which not only lower bounds ${AB}$, but gives many elements of ${AB}$ some multiplicity:
Theorem 2 Let ${G}$ be a compact connected group, with a Haar probability measure ${\mu}$. Let ${A, B}$ be compact subsets of ${G}$. Then for any ${0 \leq t \leq \min(\mu(A),\mu(B))}$, one has
$\displaystyle \int_G \min(1_A*1_B, t)\ d\mu \geq t \min(\mu(A)+\mu(B) - t,1). \ \ \ \ \ (1)$
Indeed, Theorem 1 can be deduced from Theorem 2 by dividing (1) by ${t}$ and then taking limits as ${t \rightarrow 0}$. The bound in (1) is sharp, as can again be seen by considering the case when ${A,B}$ are arcs in a circle. The analogous claim for cyclic groups for prime order was established by Pollard, and for general abelian groups by Green and Ruzsa.
Let us now prove Theorem 2. It uses a submodularity argument related to one discussed in this previous post. We fix ${B}$ and ${t}$ with ${0 \leq t \leq \mu(B)}$, and define the quantity
$\displaystyle c(A) := \int_G \min(1_A*1_B, t)\ d\mu - t (\mu(A)+\mu(B)-t).$
for any compact set ${A}$. Our task is to establish that ${c(A) \geq 0}$ whenever ${t \leq \mu(A) \leq 1-\mu(B)+t}$.
We first verify the extreme cases. If ${\mu(A) = t}$, then ${1_A*1_B \leq t}$, and so ${c(A)=0}$ in this case (since ${\int_G 1_A*1_B = \mu(A)\mu(B) = t \mu(B)}$). At the other extreme, if ${\mu(A) = 1-\mu(B)+t}$, then from the inclusion-exclusion principle we see that ${1_A * 1_B \geq t}$, and so again ${c(A)=0}$ in this case.
To handle the intermediate regime when ${\mu(A)}$ lies between ${t}$ and ${1-\mu(B)+t}$, we rely on the submodularity inequality
$\displaystyle c(A_1) + c(A_2) \geq c(A_1 \cap A_2) + c(A_1 \cup A_2) \ \ \ \ \ (2)$
for arbitrary compact ${A_1,A_2}$. This inequality comes from the obvious pointwise identity
$\displaystyle 1_{A_1} + 1_{A_2} = 1_{A_1 \cap A_2} + 1_{A_1 \cup A_2}$
whence
$\displaystyle 1_{A_1}*1_B + 1_{A_2}*1_B = 1_{A_1 \cap A_2}*1_B + 1_{A_1 \cup A_2}*1_B$
and thus (noting that the quantities on the left are closer to each other than the quantities on the right)
$\displaystyle \min(1_{A_1}*1_B,t) + \min(1_{A_2}*1_B,t)$
$\displaystyle \geq \min(1_{A_1 \cap A_2}*1_B,t) + \min(1_{A_1 \cup A_2}*1_B,t)$
at which point (2) follows by integrating over ${G}$ and then using the inclusion-exclusion principle.
Now introduce the function
$\displaystyle f(a) := \inf \{ c(A) : \mu(A) = a \}$
for ${t \leq a \leq 1-\mu(B)+t}$. From the preceding discussion ${f(a)}$ vanishes at the endpoints ${a = t, 1-\mu(B)+t}$; our task is to show that ${f(a)}$ is non-negative in the interior region ${t < a < 1-\mu(B)+t}$. Suppose for contradiction that this was not the case. It is easy to see that ${f}$ is continuous (indeed, it is even Lipschitz continuous), so there must be ${t < a < 1-\mu(B)+t}$ at which ${f}$ is a local minimum and not locally constant. In particular, ${0 . But for any ${A}$ with ${\mu(A) = a}$, we have the translation-invariance
$\displaystyle c(gA) = c(A) \ \ \ \ \ (3)$
for any ${g \in G}$, and hence by (2)
$\displaystyle c(A) \geq \frac{1}{2} c(A \cap gA) + \frac{1}{2} c(A \cup gA ).$
Note that ${\mu(A \cap gA)}$ depends continuously on ${g}$, equals ${a}$ when ${g}$ is the identity, and has an average value of ${a^2}$. As ${G}$ is connected, we thus see from the intermediate value theorem that for any ${0 < \epsilon < a-a^2}$, we can find ${g}$ such that ${\mu(A \cap gA) = a-\epsilon}$, and thus by inclusion-exclusion ${\mu(A \cup gA) = a+\epsilon}$. By definition of ${f}$, we thus have
$\displaystyle c(A) \geq \frac{1}{2} f(a-\epsilon) + \frac{1}{2} f(a+\epsilon).$
Taking infima in ${A}$ (and noting that the hypotheses on ${\epsilon}$ are independent of ${A}$) we conclude that
$\displaystyle f(a) \geq \frac{1}{2} f(a-\epsilon) + \frac{1}{2} f(a+\epsilon)$
for all ${0 < \epsilon < a-a^2}$. As ${f}$ is a local minimum and ${\epsilon}$ is arbitrarily small, this implies that ${f}$ is locally constant, a contradiction. This establishes Theorem 2.
We observe the following corollary:
Corollary 3 Let ${G}$ be a compact connected group, with a Haar probability measure ${\mu}$. Let ${A, B, C}$ be compact subsets of ${G}$, and let ${\delta := \min(\mu(A),\mu(B),\mu(C))}$. Then one has the pointwise estimate
$\displaystyle 1_A * 1_B * 1_C \geq \frac{1}{4} (\mu(A)+\mu(B)+\mu(C)-1)_+^2$
if ${\mu(A)+\mu(B)+\mu(C)-1 \leq 2 \delta}$, and
$\displaystyle 1_A * 1_B * 1_C \geq \delta (\mu(A)+\mu(B)+\mu(C)-1-\delta)$
if ${\mu(A)+\mu(B)+\mu(C)-1 \geq 2 \delta}$.
Once again, the bounds are completely sharp, as can be seen by computing ${1_A*1_B*1_C}$ when ${A,B,C}$ are arcs of a circle. For quasirandom ${G}$, one can do much better than these bounds, as discussed in this recent blog post; thus, the abelian case is morally the worst case here, although it seems difficult to convert this intuition into a rigorous reduction.
Proof: By cyclic permutation we may take ${\delta = \mu(C)}$. For any
$\displaystyle (\mu(A)+\mu(B)-1)_+ \leq t \leq \min(\mu(A),\mu(B)),$
we can bound
$\displaystyle 1_A*1_B*1_C \geq \min(1_A*1_B,t)*1_C$
$\displaystyle \geq \int_G \min(1_A*1_B,t)\ d\mu - t (1-\mu(C))$
$\displaystyle \geq t (\mu(A)+\mu(B)-t) - t (1-\mu(C))$
$\displaystyle = t \min( \mu(A)+\mu(B)+\mu(C)-1-t )$
where we used Theorem 2 to obtain the third line. Optimising in ${t}$, we obtain the claim. $\Box$
In the last few notes, we have been steadily reducing the amount of regularity needed on a topological group in order to be able to show that it is in fact a Lie group, in the spirit of Hilbert’s fifth problem. Now, we will work on Hilbert’s fifth problem from the other end, starting with the minimal assumption of local compactness on a topological group ${G}$, and seeing what kind of structures one can build using this assumption. (For simplicity we shall mostly confine our discussion to global groups rather than local groups for now.) In view of the preceding notes, we would like to see two types of structures emerge in particular:
• representations of ${G}$ into some more structured group, such as a matrix group ${GL_n({\bf C})}$; and
• metrics on ${G}$ that capture the escape and commutator structure of ${G}$ (i.e. Gleason metrics).
To build either of these structures, a fundamentally useful tool is that of (left-) Haar measure – a left-invariant Radon measure ${\mu}$ on ${G}$. (One can of course also consider right-Haar measures; in many cases (such as for compact or abelian groups), the two concepts are the same, but this is not always the case.) This concept generalises the concept of Lebesgue measure on Euclidean spaces ${{\bf R}^d}$, which is of course fundamental in analysis on those spaces.
Haar measures will help us build useful representations and useful metrics on locally compact groups ${G}$. For instance, a Haar measure ${\mu}$ gives rise to the regular representation ${\tau: G \rightarrow U(L^2(G,d\mu))}$ that maps each element ${g \in G}$ of ${G}$ to the unitary translation operator ${\rho(g): L^2(G,d\mu) \rightarrow L^2(G,d\mu)}$ on the Hilbert space ${L^2(G,d\mu)}$ of square-integrable measurable functions on ${G}$ with respect to this Haar measure by the formula
$\displaystyle \tau(g) f(x) := f(g^{-1} x).$
(The presence of the inverse ${g^{-1}}$ is convenient in order to obtain the homomorphism property ${\tau(gh) = \tau(g)\tau(h)}$ without a reversal in the group multiplication.) In general, this is an infinite-dimensional representation; but in many cases (and in particular, in the case when ${G}$ is compact) we can decompose this representation into a useful collection of finite-dimensional representations, leading to the Peter-Weyl theorem, which is a fundamental tool for understanding the structure of compact groups. This theorem is particularly simple in the compact abelian case, where it turns out that the representations can be decomposed into one-dimensional representations ${\chi: G \rightarrow U({\bf C}) \equiv S^1}$, better known as characters, leading to the theory of Fourier analysis on general compact abelian groups. With this and some additional (largely combinatorial) arguments, we will also be able to obtain satisfactory structural control on locally compact abelian groups as well.
The link between Haar measure and useful metrics on ${G}$ is a little more complicated. Firstly, once one has the regular representation ${\tau: G\rightarrow U(L^2(G,d\mu))}$, and given a suitable “test” function ${\psi: G \rightarrow {\bf C}}$, one can then embed ${G}$ into ${L^2(G,d\mu)}$ (or into other function spaces on ${G}$, such as ${C_c(G)}$ or ${L^\infty(G)}$) by mapping a group element ${g \in G}$ to the translate ${\tau(g) \psi}$ of ${\psi}$ in that function space. (This map might not actually be an embedding if ${\psi}$ enjoys a non-trivial translation symmetry ${\tau(g)\psi=\psi}$, but let us ignore this possibility for now.) One can then pull the metric structure on the function space back to a metric on ${G}$, for instance defining an ${L^2(G,d\mu)}$-based metric
$\displaystyle d(g,h) := \| \tau(g) \psi - \tau(h) \psi \|_{L^2(G,d\mu)}$
if ${\psi}$ is square-integrable, or perhaps a ${C_c(G)}$-based metric
$\displaystyle d(g,h) := \| \tau(g) \psi - \tau(h) \psi \|_{C_c(G)} \ \ \ \ \ (1)$
if ${\psi}$ is continuous and compactly supported (with ${\|f \|_{C_c(G)} := \sup_{x \in G} |f(x)|}$ denoting the supremum norm). These metrics tend to have several nice properties (for instance, they are automatically left-invariant), particularly if the test function is chosen to be sufficiently “smooth”. For instance, if we introduce the differentiation (or more precisely, finite difference) operators
$\displaystyle \partial_g := 1-\tau(g)$
(so that ${\partial_g f(x) = f(x) - f(g^{-1} x)}$) and use the metric (1), then a short computation (relying on the translation-invariance of the ${C_c(G)}$ norm) shows that
$\displaystyle d([g,h], \hbox{id}) = \| \partial_g \partial_h \psi - \partial_h \partial_g \psi \|_{C_c(G)}$
for all ${g,h \in G}$. This suggests that commutator estimates, such as those appearing in the definition of a Gleason metric in Notes 2, might be available if one can control “second derivatives” of ${\psi}$; informally, we would like our test functions ${\psi}$ to have a “${C^{1,1}}$” type regularity.
If ${G}$ was already a Lie group (or something similar, such as a ${C^{1,1}}$ local group) then it would not be too difficult to concoct such a function ${\psi}$ by using local coordinates. But of course the whole point of Hilbert’s fifth problem is to do without such regularity hypotheses, and so we need to build ${C^{1,1}}$ test functions ${\psi}$ by other means. And here is where the Haar measure comes in: it provides the fundamental tool of convolution
$\displaystyle \phi * \psi(x) := \int_G \phi(y) \psi(y^{-1}x) d\mu(y)$
between two suitable functions ${\phi, \psi: G \rightarrow {\bf C}}$, which can be used to build smoother functions out of rougher ones. For instance:
Exercise 1 Let ${\phi, \psi: {\bf R}^d \rightarrow {\bf C}}$ be continuous, compactly supported functions which are Lipschitz continuous. Show that the convolution ${\phi * \psi}$ using Lebesgue measure on ${{\bf R}^d}$ obeys the ${C^{1,1}}$-type commutator estimate
$\displaystyle \| \partial_g \partial_h (\phi * \psi) \|_{C_c({\bf R}^d)} \leq C \|g\| \|h\|$
for all ${g,h \in {\bf R}^d}$ and some finite quantity ${C}$ depending only on ${\phi, \psi}$.
This exercise suggests a strategy to build Gleason metrics by convolving together some “Lipschitz” test functions and then using the resulting convolution as a test function to define a metric. This strategy may seem somewhat circular because one needs a notion of metric in order to define Lipschitz continuity in the first place, but it turns out that the properties required on that metric are weaker than those that the Gleason metric will satisfy, and so one will be able to break the circularity by using a “bootstrap” or “induction” argument.
We will discuss this strategy – which is due to Gleason, and is fundamental to all currently known solutions to Hilbert’s fifth problem – in later posts. In this post, we will construct Haar measure on general locally compact groups, and then establish the Peter-Weyl theorem, which in turn can be used to obtain a reasonably satisfactory structural classification of both compact groups and locally compact abelian groups.
James on 254A, Notes 2: Complex-analyti… Pierre de la Harpe on Goursat and Furstenberg-Weiss… Jacob Wakem on Work hard Anonymous on The Hardy–Littlewood… Anonymous on The Hardy–Littlewood… Raphael on The Hardy–Littlewood… Quill on The Hardy–Littlewood… Dwight Walsh on Why global regularity for Navi… Viki Esther Chang on About Siegelvsgrh on The Hardy–Littlewood… Terence Tao on The Hardy–Littlewood… Terence Tao on The Hardy–Littlewood… Terence Tao on The Hardy–Littlewood… Yaver Gulusoy on 275A, Notes 2: Product measure… Anonymous on The Hardy–Littlewood… | 4,267 | 14,718 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 173, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-39 | latest | en | 0.938634 |
https://youvegotthismath.com/tracing-number-12/ | 1,709,128,724,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474715.58/warc/CC-MAIN-20240228112121-20240228142121-00117.warc.gz | 1,101,093,871 | 62,289 | # 30+ Tracing Number 12 Pages | Free Printable
These tracing number 12 worksheets will help you identify and understand tracing numbers. Grade 1 students will learn to trace and write the number 12, which improves their ability to identify math numbers, with our free printable worksheets.
## Tracing Number 12 Using Arrow Worksheet
Students will get an arrow to trace the number 12. They will follow the directions and trace them.
## Tracing Number 12 without Arrow Worksheet
This part is similar to the previous part, but no arrow direction is given to trace number 12.
## Tracing and Coloring Number 12 Worksheet
Students will trace the number 12 and then color the circles that have the number 12 in them.
## Ocean-Themed Tracing Number 12 Worksheet
Students will count the ocean objects, then the hand signs, and finally trace the number 12.
## Insect-Themed Tracing Number 12 Worksheet
Students will count the insects and trace the number 12.
## Count and Trace Number 12 Worksheet
Students will first count different objects and then trace the number 12 and the word name.
## Finding and Tracing Number 12 Worksheet
Firstly, students find the circle with the number 12, then trace the numbers in the following.
## Tracing Number 12 and Word Name Worksheet
Students will learn the number 12 and its word name, and then they will trace both the number and the word name.
## Count, Trace, and Color Number 12 Worksheet
Students will first count some objects, trace the number 12, and finally color the number.
## Fall-Themed Tracing Number 12 Worksheet
Students will trace the number 12 on various fall-themed objects.
## Christmas-Themed Tracing Number 12 Worksheet
Students will first count Christmas-themed objects, trace the number 12, and finally color the number.
This whole article represents the methods of tracing number 12 worksheets. The worksheets attached to this article will trace, color the number and objects, count objects, trace according to the direction, and learn the number word name. Your grade 1 students will learn to trace 12 by practicing our worksheets. | 448 | 2,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-10 | latest | en | 0.835116 |
http://ankhufurha.com/laird/example-of-conduction-in-liquids.php | 1,590,360,022,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347385193.5/warc/CC-MAIN-20200524210325-20200525000325-00546.warc.gz | 7,940,592 | 12,553 | ## Unit 12 Conduction in Liquids and Gases [PPT Powerpoint]
Conduction of Electricity in Liquids Electrolysis Of. What are some examples of solids liquids and gases for kids? What are some examples of solids, liquids, What are examples of conduction and convection?, How is heat transferred? Convection occurs when warmer areas of a liquid or gas rise to cooler Water boiling in a pan is a good example of these convection.
### Unit 12 Conduction in Liquids and Gases PBworks
Conduction Simple English Wikipedia the free encyclopedia. Conduction in physics is about forms of energy, namely heat or electricity. Heat conduction takes place between two objects in contact with each other., Conduction of liquids Class practical If students have already passed an electric current through solid materials, they can now test the conductivity of liquids and.
How is heat transferred? Convection occurs when warmer areas of a liquid or gas rise to cooler Water boiling in a pan is a good example of these convection The most efficient method of heat transfer is conduction. This mode of heat Heat transfer from a solid to a fluid (liquid or for example, no
some examples of conduction are: The difference lies in the fact that these ions are not dissolved in a solvent but exist as chsrged ions in the liquid state. . 20 Examples of Heat Conduction liquid or gaseous state. Conduction is a heat transfer from a body with a higher temperature to a body with a lower temperature.
Background information Year 3, unit 1: 1. conduction or diffusion • is the transfer of heat by motion of liquids. •curs when, for example, What Type of Heat Transfer Occurs in Liquids & Gases? Conduction in Gases and Liquids. The sun is a classic example of radiative heat transfer:
Thermal Energy Transfer: Conduction, Convection, Radiation. Conduction; Convection; Radiation (liquids and gases) Example with boiling water: 10/11/2009В В· I'm working on a project and right now my idea fluid is to use Gallium, a liquid metal. Only major issue is its currently about \$1000 USD for 1kg of this...
### Conduction of Electricity in Liquids Electrolysis Of
Conduction and Breakdown in Commercial Liquids. Tab. 9.4 lists a selection of thermal diffusivity values for commonly encountered liquids. Heat conduction to a point is influenced by the For example, if we, For example, if A = 1 m 2 = 1 m electrical conduction happens not by band electrons or holes, Conduction in ionic liquids is also controlled by the movement.
NTTI Lesson CONDUCTION CONVECTION RADIATION OH MY. Read about Conduction of Electricity in Liquids and Electrolysis of water. Learn more about the conduction of electricity in liquids by visiting BYJU'S, Mechanisms of Heat Loss or Transfer. Example of Conduction in Regard to Residential Heating. Unlike convection or conduction, where energy from gases, liquids.
### Conduction of Electricity in Liquids Electrolysis Of
Heat Transfer Mechanisms Walter Scott Jr. College of. For example, if A = 1 m 2 = 1 m electrical conduction happens not by band electrons or holes, Conduction in ionic liquids is also controlled by the movement Heat Transfer: Conduction, Convection, and Radiation example of conduction. molecules in liquids and gases are spread further apart,.
Liquids and gases have their molecules farther apart and are generally poor conductors of heat. For an example of the conduction process, Convection vs Conduction. heat transfer due to bulk movement of molecules within fluids such as gases and liquids. Example – Convection vs Conduction
Thermal Energy Transfer: Conduction, Convection, Radiation. Conduction; Convection; Radiation (liquids and gases) Example with boiling water: some examples of conduction are: The difference lies in the fact that these ions are not dissolved in a solvent but exist as chsrged ions in the liquid state. .
Heat Transfer: Conduction, Convection, and Radiation example of conduction. molecules in liquids and gases are spread further apart, Mechanisms of Heat Loss or Transfer. Example of Conduction in Regard to Residential Heating. Unlike convection or conduction, where energy from gases, liquids
What are examples of liquids and gases? What are the examples of conduction in liquids and gases? That is an example of liquid conducting. Conduction occurs more readily in solids and liquids, where the particles are closer to together, than in gases,
What are examples of liquids and gases? What are the examples of conduction in liquids and gases? That is an example of liquid conducting. Unit 12 Conduction in Liquids and Gases Unit 12 Conduction in Liquids and Gases Objectives: An arc welder is an example of a practical application.
## Conduction of Electricity in Liquids Electrolysis Of
Unit 12 Conduction in Liquids and Gases [PPT Powerpoint]. 20 Examples of Heat Conduction liquid or gaseous state. Conduction is a heat transfer from a body with a higher temperature to a body with a lower temperature., Conduction Phenomena in Dielectric Liquids 105 5 and more. For example, in a solution of(iso-CsH 11 )4N'SCN" in benzene it ism= 5.47 at a concentration 3.8В·10"3 mole.
### Unit 12 Conduction in Liquids and Gases PBworks
Conduction and Breakdown in Commercial Liquids. What Type of Heat Transfer Occurs in Liquids & Gases? Conduction in Gases and Liquids. The sun is a classic example of radiative heat transfer:, Example Vibrational Modes in an Atomic and a Molecular Solid. Conduction in a Liquid (with convection suppressed) click image to view animation.
Unit 12 Conduction in Liquids and Gases Conduction in Liquids •An electrolyte is a solution that will conduct electricity. •An electrolytic solution may be Conduction in physics is about forms of energy, namely heat or electricity. Heat conduction takes place between two objects in contact with each other.
Thermal Energy Transfer: Conduction, Convection, Radiation. Conduction; Convection; Radiation (liquids and gases) Example with boiling water: Read about Conduction of Electricity in Liquids and Electrolysis of water. Learn more about the conduction of electricity in liquids by visiting BYJU'S
Learn about conduction, The liquid or gas in hot areas is less dense. than the liquid or gas in cold areas, As well as these examples, Thermal conduction in liquids and gases. Conduction in liquids In liquids, conduction heat transfer Just take an example of comparison of thermal conductivity
Historical Examples. of conduction. Sense of "conducting of a liquid through a channel" is from 1610s; in physics, of heat, etc., from 1814. Show More. Conduction and Breakdown in Commercial Liquids are not the breakdown strength depends on the pressure and the molecular structure of the liquid. For example,
10/11/2009В В· I'm working on a project and right now my idea fluid is to use Gallium, a liquid metal. Only major issue is its currently about \$1000 USD for 1kg of this... For example, the heat transfer by conduction through the bodywork of a car. for example, the density of liquid water is 1 Kg per liter).
For example, if A = 1 m 2 = 1 m electrical conduction happens not by band electrons or holes, Conduction in ionic liquids is also controlled by the movement Conduction Phenomena in Dielectric Liquids 105 5 and more. For example, in a solution of(iso-CsH 11 )4N'SCN" in benzene it ism= 5.47 at a concentration 3.8В·10"3 mole
Mechanisms of Heat Loss or Transfer. Example of Conduction in Regard to Residential Heating. Unlike convection or conduction, where energy from gases, liquids Heat transfer: conduction more quickly in solids than in liquids or gases. Conduction occurs more quickly sinking of water in the hot-water system is an example
Heat (Thermal) Energy and Heat Transfer Conduction takes place in solids, liquids and gases, but works best in solids as their atoms/molecules are located closer convection - a method by which heat is transferred by currents in a liquid or gas. Give examples of Radiation, Convection, Conduction.
### What are examples of liquids and gases? Quora
Science Dictionary Convection Liquid webquest.hawaii.edu. 20 Examples of Heat Conduction liquid or gaseous state. Conduction is a heat transfer from a body with a higher temperature to a body with a lower temperature., What Type of Heat Transfer Occurs in Liquids & Gases? Conduction in Gases and Liquids. The sun is a classic example of radiative heat transfer:.
Heat Transfer Mechanisms Walter Scott Jr. College of. Heat (Thermal) Energy and Heat Transfer Conduction takes place in solids, liquids and gases, but works best in solids as their atoms/molecules are located closer, Conduction and Breakdown in Commercial Liquids are not the breakdown strength depends on the pressure and the molecular structure of the liquid. For example,.
### CONDUCTION MODELS IN DIELECTRIC LIQUIDS 6.1 Introduction
What are some convection examples in liquids and gases. some examples of conduction are: The difference lies in the fact that these ions are not dissolved in a solvent but exist as chsrged ions in the liquid state. . Heat typically does not flow through liquids and gases by means of conduction. Liquids and gases are fluids; The two examples of convection discussed here.
The most efficient method of heat transfer is conduction. This mode of heat Heat transfer from a solid to a fluid (liquid or for example, no Why Doesn't Convection Occur i to transfer heat which is only possible in a fluid state of matter such as liquid or can transfer heat through conduction.
Thermal conduction in liquids and gases. Conduction in liquids In liquids, conduction heat transfer Just take an example of comparison of thermal conductivity Mechanisms of Heat Loss or Transfer. Example of Conduction in Regard to Residential Heating. Unlike convection or conduction, where energy from gases, liquids
some examples of conduction are: The difference lies in the fact that these ions are not dissolved in a solvent but exist as chsrged ions in the liquid state. . Examples Of Conduction. conduction tubes and instant cooling machines for liquid and semi-liquid food. is an example of SVT with abberant conduction to the
Why Doesn't Convection Occur i to transfer heat which is only possible in a fluid state of matter such as liquid or can transfer heat through conduction. Convection vs Conduction. heat transfer due to bulk movement of molecules within fluids such as gases and liquids. Example – Convection vs Conduction
What Type of Heat Transfer Occurs in Liquids & Gases? Conduction in Gases and Liquids. The sun is a classic example of radiative heat transfer: 10/11/2009В В· I'm working on a project and right now my idea fluid is to use Gallium, a liquid metal. Only major issue is its currently about \$1000 USD for 1kg of this...
Can anyone clarify fluid Conduction? I came across liquid and gas conduction in heat transfer when I was reading about conduction. for example, its thermal What are some examples of solids liquids and gases for kids? What are some examples of solids, liquids, What are examples of conduction and convection?
What are some examples of solids liquids and gases for kids? What are some examples of solids, liquids, What are examples of conduction and convection? Examples Of Conduction. conduction tubes and instant cooling machines for liquid and semi-liquid food. is an example of SVT with abberant conduction to the
What's Hot and What's Not? Quick Look. for example. Water is in a liquid Furthermore, conduction occurs between liquids and solids, Conduction Phenomena in Dielectric Liquids 105 5 and more. For example, in a solution of(iso-CsH 11 )4N'SCN" in benzene it ism= 5.47 at a concentration 3.8В·10"3 mole
Common examples of convection include the boiling of water The three mechanisms for heat transfer include conduction, What Are Examples of Solids, Liquids and Convection currents in boiling liquid. When water starts bubbling, these bubbles of hot water start rising to the surface. The heat is transferred from the hot water
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275363 | 2,776 | 12,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-24 | latest | en | 0.918463 |
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• Ball Mill Critical Speed Mineral Processing & Metallurgy
A Ball Mill Critical Speed (actually ball, rod, AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shell’s inside surface and no balls will fall from its position onto the shell. The imagery below helps explain what goes on inside a mill as speed varies. Use our online formula The mill speed is typically defined as the percent of the TheoreticalОнлайн-запрос
• Ball Mills Mine Engineer.Com
This formula calculates the critical speed of any ball mill. Most ball mills operate most efficiently between 65% and 75% of their critical speed. Photo of a 10 Ft diameter by 32 Ft long ball mill in a Cement Plant. Photo of a series of ball mills in a Copper Plant, grinding the ore for flotation. Image of cut away ball mill, showing material flow through typical ball mill. Flash viedo of JarОнлайн-запрос
• Ball mill Wikipedia
Critical speed can be understood as that speed after which the steel balls that are responsible for the grinding of particles start rotating along the direction of the cylindrical device, thus causing no further grinding. Ball mills are used extensively in the mechanical alloying process in which they are used for grinding and for cold welding, producing alloys from powders. The ball mill is aОнлайн-запрос
• Ball Mill Parameter Selection – Power, Rotate Speed, Steel
30/08/2019· 2.2 Rotation Speed Calculation of Ball Mill \ Critical Speed_ When the ball mill cylinder is rotated, there is no relative slip between the grinding medium and the cylinder wall, and it just starts to run in a state of rotation with the cylinder of the mill. This instantaneous speed of the mill is as follows:Онлайн-запрос
• TECHNICAL NOTES 8 GRINDING R. P. King
The critical speed of the mill, & c, is defined as the speed at which a single ball will just remain against the wall for a full cycle. At the top of the cycle =0 and Fc Fg (8.5) mp & 2 cDm 2 mpg (8.6) & c 2g Dm 1/2 (8.7) The critical speed is usually expressed in terms of the number of revolutions per second Nc & c 2 1 2 2g Dm 1/2 (2×9.81)1/2Онлайн-запрос
• Ball Mills an overview | ScienceDirect Topics
The critical speed n (rpm) when the balls are attached to the wall due to centrifugation: Figure 2.7. Displacement of balls in mill. n = 42.3 D m. where D m is the mill diameter in meters. The optimum rotational speed is usually set at 65–80% of the critical speed. These data are approximate and may not be valid for metal particles that tend to agglomerate by welding. View chapter PurchaseОнлайн-запрос
• range of speed of ore mill
Critical Speed Of Priron Ore List Ball Mill; TECHNICAL NOTES 8 GRINDING R. P. King. Figures 8.5 for the popular mill types. 3 c is the mill speed measured as a fraction of the critical speed. More reliable models for the prediction of the power drawn by ball, semi-autogenous and fully autogenous mills have been developed by Morrell and by Austin. (Morrell, S. Power draw of wet tumbling millsОнлайн-запрос
• SAGMILLING.COM .:. Mill Critical Speed Determination
The "Critical Speed" for a grinding mill is defined as the rotational speed where centrifugal forces equal gravitational forces at the mill shell's inside surface. This is the rotational speed where balls will not fall away from the mill's shell. Mill Inside Diameter: Enter the mill diameter inside the shell (excluding liners). Most mill sizes are reported by vendors are outside theОнлайн-запрос
• Ball Mill Parameter Selection – Power, Rotate Speed, Steel
30/08/2019· 2.2 Rotation Speed Calculation of Ball Mill \ Critical Speed_ When the ball mill cylinder is rotated, there is no relative slip between the grinding medium and the cylinder wall, and it just starts to run in a state of rotation with the cylinder of the mill. This instantaneous speed of the mill is as follows:Онлайн-запрос
• Mill Speed Critical Speed Paul O. Abbe
Mill Speed Critical Speed. Mill Speed . No matter how large or small a mill, ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar rolling mill, its rotational speed is important to proper and efficient mill operation. Too low a speed and little energy is imparted on the product. Too fast and inefficient media movement (knownОнлайн-запрос
• Tubular Ball Mills ScienceDirect
01/01/2016· A ball mill is to produce a grind of 34 μm (P 80) product from a feed size of 200 μm at a rate of 1.5 t/h. The grinding media used was 90% Al 2 O 3 ceramic ball of S.G. 3.5. The balls occupied 28% of the mill volume. The mill was rotated at 65% of the critical speed. The work index of the ore was 11.3 kWh/t. Estimate the size of the millОнлайн-запрос
• formula critical speed of ball mill in gambia
formula critical speed of ball mill in gambia. Feed Hammer Mill for Fine Grinding 2 It not only can be used for ordinary grinding but also for fine grinding This machine is mostly used for fine grinding 3 When it is used for fine grinding the fineness of the finished products can reach 50 meshes 4 We adopt direct connection transmission Both hammer arrangement and hammer sieve intervals areОнлайн-запрос
• 2020 | Ascot Resources Ltd.
15/10/2020· The order comprises a 22-foot diameter by 8-foot effective grinding length(egl) SAG mill and 14.5-foot diameter by 19.5-foot effective grinding length ball mill. Both mills will be driven by 2000 KW, low speed synchronous motors at 78% critical speed. The mills will be supported on 90-inch diameter hydrodynamic trunnion bearings which will be interchangeable. The mill lube systems will beОнлайн-запрос
• Critical Speed Of Priron Ore List Ball Mill
Critical Speed Of Priron Ore List Ball Mill; TECHNICAL NOTES 8 GRINDING R. P. King. Figures 8.5 for the popular mill types. 3 c is the mill speed measured as a fraction of the critical speed. More reliable models for the prediction of the power drawn by ball, semi-autogenous and fully autogenous mills have been developed by Morrell and by Austin. (Morrell, S. Power draw of wet tumbling millsОнлайн-запрос
• critical speed ball mill copper ore ball mill trader
Critical Speed Of A Ball Mill To Grind Ores. Critical Speed Of Ball Mill Ore Crusher Ore Crushi. 2020-4-13Critical Speed Of Ball Mill Ore Crusher Ore Crushi. We have critical speed of ball mill ore crusher ore crushiSep 11 2016 This is my homemade gold ore ball mill to process lode claim material Uses steel balls or i have a large round wheel made of steel to also crush This could also useОнлайн-запрос
• attentions to ore ball mills critical speed
Critical Speed Of A Ball Mill To Grind Ores. Ball mills mine engineerm the point where the mill becomes a centrifuge is called the "critical speed" and ball mills usually operate at 65 to 75 of the critical speedall mills are generally used to grind material 14 inch and. Chat Now; critical speed of ball mill ore crusher ore crushi . Critical Speed Of Ball Mill Ore Crusher Ore Crushi. We haveОнлайн-запрос
• history of grind ball mill grind machine centrifuge
The point where the mill becomes a centrifuge is called the "Critical Speed", and ball mills usually operate at 65% to 75% of the critical speed. Ball Mills are generally used to grind material 1/4 inch and finer, down to the particle size of 20 to 75 microns.get price. Ball Mill hiimac. Ball mill is widely used for the dry type or wet type grinding of all kinds of ores and other grind-ableОнлайн-запрос
• Tubular Ball Mills ScienceDirect
01/01/2016· A ball mill is to produce a grind of 34 μm (P 80) product from a feed size of 200 μm at a rate of 1.5 t/h. The grinding media used was 90% Al 2 O 3 ceramic ball of S.G. 3.5. The balls occupied 28% of the mill volume. The mill was rotated at 65% of the critical speed. The work index of the ore was 11.3 kWh/t. Estimate the size of the millОнлайн-запрос
• formula critical speed of ball mill in gambia
formula critical speed of ball mill in gambia. Feed Hammer Mill for Fine Grinding 2 It not only can be used for ordinary grinding but also for fine grinding This machine is mostly used for fine grinding 3 When it is used for fine grinding the fineness of the finished products can reach 50 meshes 4 We adopt direct connection transmission Both hammer arrangement and hammer sieve intervals areОнлайн-запрос
• critical speed of cement ball mill
CRITICAL SPEED OF THE BALL MILL. Let a grinding ball of mass m is in motion in a mill of diameter D meters. It is at a position making an angle a at the center (called the angle of repose).. The force acting on the ball are described in the figure. Ball Mill Critical Speed Mineral Processing & Metallurgy. Get price. formula for critical speed of ball mill pochiraju . Axial transport in dryОнлайн-запрос
• Critical Speed Of A Ball Mill To Grind Ores
Gold Ore Milling Machine Buy Milling Machineore Mill. The point where the mill becomes a centrifuge is called the critical speed and ball mills usually operate at 65 to 75 of the critical speed application ball mill is a key equipment to grind all kinds of ores and other is widely used in powdermaking production line including cement silicate newtype building materialОнлайн-запрос
• Critical Speed Of Priron Ore List Ball Mill
Critical Speed Of Priron Ore List Ball Mill; TECHNICAL NOTES 8 GRINDING R. P. King. Figures 8.5 for the popular mill types. 3 c is the mill speed measured as a fraction of the critical speed. More reliable models for the prediction of the power drawn by ball, semi-autogenous and fully autogenous mills have been developed by Morrell and by Austin. (Morrell, S. Power draw of wet tumbling millsОнлайн-запрос
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If the peripheral speed of the mill is too great, it begins to act like a centrifuge and the balls do not fall back, but stay on the perimeter of the mill. The point where the mill becomes a centrifuge is called the "Critical Speed", and ball mills usually operate at 65% to 75% of the critical speed. Ball Mills are generally used to grindОнлайн-запрос
• attentions to ore ball mills critical speed
Critical Speed Of A Ball Mill To Grind Ores. Ball mills mine engineerm the point where the mill becomes a centrifuge is called the "critical speed" and ball mills usually operate at 65 to 75 of the critical speedall mills are generally used to grind material 14 inch and. Chat Now; critical speed of ball mill ore crusher ore crushi . Critical Speed Of Ball Mill Ore Crusher Ore Crushi. We haveОнлайн-запрос
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Ball Mill Critical Speed & Working Principle . Jun 20, 2015 · The effect of Ball Mill RPM speed going from subcritical to supercritical helps understand the Ball Mill Working Principles of ballonball VS ballonshell grinding. The Motion of the Ball Charge . Get price. Page 1 Ball Milling Theory freeshell . involve grinding). With Lloyd''s ball milling book having sold over 2000 copies, thereОнлайн-запрос
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The point where the mill becomes a centrifuge is called the "Critical Speed", and ball mills usually operate at 65% to 75% of the critical speed. Ball Mills are generally used to grind material 1/4 inch and finer, down to the particle size of 20 to 75 microns.get price. Ball Mill hiimac. Ball mill is widely used for the dry type or wet type grinding of all kinds of ores and other grind-ableОнлайн-запрос
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Ball Mill Critical Speed Working PrincipleYouTube. Jun 20 2015 · The effect of Ball Mill RPM speed going from sub-critical to super-critical helps understand the Ball Mill Working Principles of ball-on-ball VS ball-on-shell grinding. The Motion of the Ball Charge . Chat Now; Wet Grinding in Planetary Ball MillsRETSCHYouTube. Aug 09 2016 · The extremely high centrifugal forces of a planetary | 2,876 | 11,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-17 | latest | en | 0.876357 |
http://www.brightstorm.com/qna/question/6777/ | 1,369,389,966,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704517601/warc/CC-MAIN-20130516114157-00005-ip-10-60-113-184.ec2.internal.warc.gz | 368,087,067 | 10,284 | Quick Homework Help
# -2x+y=2 -x+3y=-4 ⚑ Flag
by lisa468 at October 25, 2010
solve each system of equations using substitution.
-2x + y = 2...Equation A -x + 3y = -4..Equation B I will solve Equation A for y. y = 2x + 2 I will now plug 2x + 2 for y in Equation B to find the value of x. -x + 3(2x + 2) = -4 -x + 6x + 6 = -4 5x = -6 - 4 5x = -10 x = -10/5 x = -2 To find the value of y, replace x with -2 in EITHER given equation. -2x + y = 2 -2(-2) + y = 2 4 + y = 2 y = 2 - 4 y = -2 The answer is x = -2 and y = -2 which can be written as a point (-2,-2). fdr84@hotmail.com
mathguy October 25, 2010
(-2,-2)
yankeekid October 25, 2010
(-2, -2) is the answer.
ChromeRedCat October 25, 2010
(-2,2)
What_ok October 25, 2010
sorry (-2,-2)
What_ok October 25, 2010
(-2,-2
SATmaster October 26, 2010
(-2,-2)
fabianscorpio October 26, 2010
the answer to your Question is (-2,-2).
robotfighter October 26, 2010
the answer is (-2, -2)
maddiep October 27, 2010 | 400 | 1,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2013-20 | latest | en | 0.906088 |
https://socratic.org/questions/how-do-you-evaluate-81-3-4 | 1,585,716,627,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505366.8/warc/CC-MAIN-20200401034127-20200401064127-00426.warc.gz | 729,378,201 | 5,876 | # How do you evaluate -81^(3/4)?
Jun 22, 2015
=color(blue)( - 27
#### Explanation:
81 = color(blue)(3^4
the expression can be re written as :
$= - {\textcolor{b l u e}{\left({3}^{\cancel{4}}\right)}}^{\frac{3}{\cancel{4}}}$
$= - {\left(3\right)}^{3}$
=color(blue)( - 27 | 113 | 278 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-16 | longest | en | 0.62268 |
https://www.shaalaa.com/question-bank-solutions/an-electric-kettle-rated-3-kw-250-v-give-reason-whether-this-kettle-can-be-used-circuit-which-contains-13-fuse-fuses_37100 | 1,620,877,952,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00270.warc.gz | 1,011,396,322 | 8,830 | # An Electric Kettle is Rated 3 Kw, 250 V. Give Reason Whether this Kettle Can Be Used in a Circuit Which Contains a 13 a Fuse. - Physics
An electric kettle is rated 3 kW, 250 V. give reason whether this kettle can be used in a circuit
which contains a 13 A fuse.
#### Solution
Yes, this kettle can be used in a circuit which contains a 13 A fuse because safe limit of current
for kettle is,
I = (3000W)/(250V) = 12 A
Concept: Fuses
Is there an error in this question or solution?
#### APPEARS IN
Selina Concise Physics Class 10 ICSE
Chapter 9 Electrical Power and Household Circuits
Exercise 9 (B) | Q 12 | Page 227 | 174 | 624 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-21 | latest | en | 0.834208 |
https://testbook.com/question-answer/which-of-the-following-is-a-data-visualization-met--5feb27272d40eead91152fc2 | 1,627,202,217,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00212.warc.gz | 565,848,051 | 25,822 | # Which of the following is a data visualization method?
Free Practice With Testbook Mock Tests
## Options:
1. Line
2. Circle and Triangle
3. Pie chart and Bar chart
4. Pentagon
### Correct Answer: Option 3 (Solution Below)
This question was previously asked in
Official Paper 2: Held on 24 Sep 2020 Shift 2
## Solution:
Pie charts and Bar charts are considered data visualization methods.
Data visualization method:
• It is a graphical method of presenting data
• For this purpose, we use graphical elements like graphs, charts, maps, etc.
• Visualizations tools can be selected based on the size and type of data
1. Pie charts:
• It is a circle and sector diagram
• The values are shown as part of a 3600 circle
• The values are converted into percentage values before plot them into the chart
2. Bar charts:
• It uses to show mainly the frequency distribution graphically
• Sometimes we plot the percentage values
• Line, circle, triangle, and pentagon are shapes
• Line graph, circle and triangle diagram, pentagon graph, etc, are used to represent different data of different format | 250 | 1,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-31 | latest | en | 0.84206 |
https://www.watermedia.org/how-big-of-a-water-heater-do-i-need | 1,718,288,543,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00033.warc.gz | 938,310,627 | 13,738 | # How Big of a Water Heater Do I Need? A Comprehensive Guide
Greetings, dear readers! Having a hot water supply at home is essential, especially during colder months. However, choosing the right size of water heater can be confusing and overwhelming. In this article, we will provide you with a comprehensive guide on how big of a water heater you need.
## The Importance of Choosing the Right Water Heater Size
Before discussing how big of a water heater you need, it is crucial to know why it matters. Choosing an appropriately sized water heater not only ensures an adequate amount of hot water but also saves you money on energy bills. An oversized water heater will consume more energy and increase your bills, while an undersized one will not provide you with enough hot water.
### What Factors Affect the Water Heater Size I Need?
Several factors affect the size of water heater you need. These include:
Factor Explanation
Number of people in your household The more people, the larger the water heater you need.
Peak hot water demand If you use multiple appliances simultaneously, you need a bigger water heater.
Water usage habits If you take long showers or use hot water frequently, you need a larger water heater.
Climate Colder climates require a larger water heater to compensate for the lower groundwater temperature.
## How Big of a Water Heater Do I Need?
Now that you know why the size of a water heater matters and what factors affect it, let us discuss how big of a water heater you need.
### Step 1: Calculate Your Peak Hourly Hot Water Demand
The first step in determining the size of your water heater is to calculate your peak hourly hot water demand. This is the amount of hot water your household uses during the busiest hour of the day. To calculate this, you need to:
1. List all the hot water appliances you use simultaneously
2. Determine their flow rate in gallons per minute (GPM)
3. Multiply the flow rate with 60 minutes to get the hourly demand for each appliance
4. Add up the hourly demand for all appliances to get your total peak hourly hot water demand
### Step 2: Determine the Required Gallon Capacity
Once you have calculated your peak hourly hot water demand, you need to determine the required gallon capacity. This is the amount of hot water your water heater needs to supply during the busiest hour of the day. To determine this, you need to multiply your peak hourly hot water demand with the First Hour Rating (FHR) of the water heater. FHR is the number of gallons of hot water the water heater can supply per hour with a full tank when the groundwater temperature is at 135°F.
### Step 3: Choose the Right Size
Finally, you need to choose the right size of water heater that meets your required gallon capacity. Below is a table that provides you with an estimate of the water heater size you need based on the number of people in your household and their hot water usage habits.
Number of People Water Heater Size (Gallons)
1-2 30-40
2-3 40-50
3-4 50-60
5 or more 60 or more
## Advantages and Disadvantages of Choosing the Right Water Heater Size
👍 Ensures an adequate hot water supply
👍 Saves money on energy bills
👍 Reduces the chances of running out of hot water
👎 Initial cost can be higher for larger water heaters
👎 Takes up more space
## Frequently Asked Questions (FAQs)
### 1. Is a larger water heater always better?
No, a larger water heater may consume more energy and increase your bills. It is essential to choose the right size based on your hot water demand.
### 2. What happens if I choose an undersized water heater?
You may run out of hot water frequently and have to wait for the water heater to heat more water. This can be inconvenient and frustrating.
### 3. Can I use a water heater of a different size than what is recommended?
It is not recommended as it may lead to insufficient hot water or higher energy bills. Always choose the water heater size that meets your hot water demand.
### 4. How long does a water heater typically last?
Average lifespan varies based on the type of water heater. However, most last between 8 to 12 years.
### 5. Can I install a water heater myself?
It is recommended to have a professional plumber install your water heater to ensure it is installed correctly and safely.
### 6. What is the most energy-efficient water heater size?
The most energy-efficient water heater size is the one that meets your hot water demand. An oversize or undersize water heater may consume more energy and increase your bills.
### 7. How can I improve the energy efficiency of my water heater?
You can improve the energy efficiency of your water heater by installing insulation on the pipes and lowering the temperature to 120°F.
## Conclusion
Choosing the right size of water heater is essential to ensure an adequate hot water supply and save money on energy bills. By following the steps mentioned in this article, you can determine how big of a water heater you need. Remember, it is always better to choose a water heater that meets your hot water demand rather than choosing an oversized one. We hope this article helped you in making an informed decision about your water heater size.
Take action now and ensure your comfort during colder months by choosing the right size of water heater!
## Closing/Disclaimer
The information provided in this article is for educational purposes only and should not be considered as professional advice. Always consult a professional plumber to determine the appropriate water heater size for your household and follow all safety precautions when installing a water heater. The authors and website owners are not liable for any damages or losses caused by the use of the information provided in this article. | 1,195 | 5,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.90938 |
https://www.mapleprimes.com/questions/226588-Fsolve-Doesnt-Provide-The-Answer-For | 1,722,865,077,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640447331.19/warc/CC-MAIN-20240805114033-20240805144033-00202.warc.gz | 685,545,775 | 25,798 | Question:Fsolve doesn't provide the answer for highly non-linear system of equations
Question:Fsolve doesn't provide the answer for highly non-linear system of equations
Maple 2017
Hello everyone
I try to solve the system of nonlinear equations, where three variables are inside of integrals, for different values of the real parameter q.
It seems that not for all values of q the system has a solution because I substitute different q but the function fsolve does not give any numerical answer or the special message even after the long evaluation.
Here the code:
```restart;
q := 70;
k0 := 100; n := 3; r := 1.1; t := 0.1e-2; Eg := n/((n+1)^(3/2)+1)^(2/3); kF2 := (1/((n+1)^(3/2)+1))^(1/3);
R1 := int(a^2*tanh(sqrt((a^2-mu)^2+d1^2)/(2*t))/sqrt((a^2-mu)^2+d1^2), a = 0 .. k0);
R2 := int(a^2*tanh(sqrt((a^2-mu+Eg)^2+d2^2)/(2*t))/sqrt((a^2-mu+Eg)^2+d2^2), a = 0 .. k0);
R3 := int(a^2*(1-(a^2-mu)*tanh(sqrt((a^2-mu)^2+d1^2)/(2*t))/sqrt((a^2-mu)^2+d1^2)), a = 0 .. k0);
R4 := int(a^2*(1-(a^2-mu+Eg)*tanh(sqrt((a^2-mu+Eg)^2+d2^2)/(2*t))/sqrt((a^2-mu+Eg)^2+d2^2)), a = 0 .. k0);
Eq1 := evalf(k0+k0*(r-1)-(1/2)*Pi*r*kF2*q-R1);
Eq2 := evalf(k0-(1/2)*Pi*kF2*q-R2);
Eq3 := evalf(2/3-R3-R4);
fsolve({Eq1, Eq2, Eq3}, {d1, d2, mu});```
I know that this system has the solution for some values of q but why I can't get the solution or the answer using fsolve? | 530 | 1,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-33 | latest | en | 0.696178 |
http://mathoverflow.net/questions/18336?sort=newest | 1,371,642,060,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708739983/warc/CC-MAIN-20130516125219-00010-ip-10-60-113-184.ec2.internal.warc.gz | 154,836,706 | 14,511 | MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
## Background
Model categories are an axiomization of the machinery underlying the study of topological spaces up to homotopy equivalence. They consist of a category $C$, together with three distinguished classes of morphism: Weak Equivalences, Fibrations, and Cofibrations. There are then a series of axioms this structure must satisfy, to guarantee that the classes behave analogously to the topological maps they are named after. The axioms can be found here.
(As far as I know...) The main practical advantage of this machinery is that it gives a rather concrete realization of the localization category $C/\sim$ where the Weak Equivalences have been inverted, which generalizes the homotopy category of topological spaces. The main conceptual advantage is that it is a first step towards formalizing the concept of "a category enriched over topological spaces".
A discussion of examples and intuition can be found at this question.
## The Question
The examples found in the answers to Ilya's question, as well as in the introductory papers I have read, all have a model category structure that could be expected. They are all examples along the lines of topological spaces, derived categories, or simplicial objects, which are all conceptually rooted in homotopy theory and so their model structures aren't really surprising.
I am hoping for an example or two which would elicit disbelief from someone who just learned the axioms for a model category. Along the lines of someone who just learned what a category being briefly skeptical that any poset defines a category, or that '$n$-cobordisms' defines a category.
-
Great question. I have also wondered this in the past. – Kevin Lin Mar 16 2010 at 22:29
Why would someone be skeptical that a poset defines a category? – Mike Shulman Mar 17 2010 at 15:22
"Why would someone be skeptical that a poset defines a category?"-- for the same reason that they would be skeptical that a group defines a category. Usually when people first encounter categories they are given examples like the category of sets or vector spaces. These are places that mathematics happens. This at first seem very different than things like groups and posets which are usually the objects of study, not the context. – Chris Schommer-Pries Mar 17 2010 at 19:12
Here is an example that surprised me at some time in the past. Bisson and Tsemo introduce a nontrivial model structure on the topos of directed graphs. Here a directed graph is simply a $4$-tuple $(V,E,s,t)$ where an arc $e \in E$ starts at $s(e) \in V$ and ends at $t(e) \in V$. Fibrations are maps that induce a surjection on the set of outgoing arcs of each vertex, cofibrations are embeddings obtained by attaching a bunch of trees, and weak equivalences are maps that induce a bijection on the sets of cycles.
In this model structure fibrant objects are graphs without sinks and cofibrant objects are graphs with exactly one incoming arc for every vertex. Cofibrant replacement replaces a graph by the disjoint union of its cycles with the obvious morphism into the original graph.
We have a chain of inclusions of categories $A\to B \to C\to D$, where $D$ is the topos of directed graphs, $C$ is the full subcategory of $D$ consisting of all graphs with exactly one incoming arc for each vertex, $B$ is the full subcategory of $C$ consisting of all graphs with exactly one outgoing arc for each vertex, and $A$ is the full subcategory of $B$ consisting of all graphs such that $s=t$.
Each functor is a part of a Quillen adjunction and total left and right derived functors compute nontrivial information about graphs under consideration.
Two finite graphs are homotopy equivalent iff they are isospectral iff their zeta-functions coincide.
-
Excellent, thats exactly the kind of example I was looking for. Do you happen to know if this model structure is something more familiar, if one thinks of the directed graphs in terms of the associated quiver? – Greg Muller Mar 16 2010 at 13:51
The name is Tsemo, not Tserno. – Omar Antolín-Camarena Mar 16 2010 at 15:44
@Greg: I have no idea whether this model structure can be related in any way to quivers. – Dmitri Pavlov Mar 16 2010 at 19:27
The fibrations in this model structure at least are not surprising: they are very similar to the fibrations in the canonical (or folk) model structure on Cat. – David Roberts Jan 10 2011 at 4:41
The category of sets admits precisely nine model category structures, no more no less.
I learned this fact from Tom Goodwillie's comments on a different MO question. It always shocks people when I mention it to them, so I guess it is surprising. I am not sure which is more surprising, that you can actually compute all the model structures or that there are exactly nine of them. Working out the details is such a fun exercise that I don't want to spoil the fun here.
-
This is just great! – Thomas Nikolaus Mar 12 2012 at 23:11
(more detail on the answer by mmm: Gavrilovich in http://arxiv.org/abs/1006.4647 and then further works of Gavrilovich and Hasson http://arxiv.org/abs/1102.5562 and then Gavrilovich, Hasson and Kaplan http://arxiv.org/abs/1111.3489 explore in depth connections with pcf theory (a part of set theory, one could say) - in particular, they recover Shelah's covering number $cov(\lambda, \aleph_1,\aleph_1, 2)$)... How surprising it is depends (I guess) on how much you are used to "detect" homotopic content in areas where it was seemingly not present.
-
Here is an example of a poset which is a model category. The construction is set-theoretic and mentions Continuum Hypothesis.
-
The link is broken. – Rasmus May 7 2012 at 17:12
There is a series of paper by Philippe Gaucher on the arxiv that deal with model categories in the context of theoretical computer science. E.g.:
• Abstract homotopical methods for theoretical computer science (0707.1449)
The purpose of this paper is to collect the homotopical methods used in the development of the theory of flows initialized by author's paper A model category for the homotopy theory of concurrency''.
• A model category for the homotopy theory of concurrency (math/0308054)
We construct a cofibrantly generated model structure on the category of flows such that any flow is fibrant and such that two cofibrant flows are homotopy equivalent for this model structure if and only if they are S-homotopy equivalent. This result provides an interpretation of the notion of S-homotopy equivalence in the framework of model categories.
I guess it is just because of my ignorance, but to me this was unexpected.
-
There is a preprint (arXiv) by Finnur Larusson explaining a model structure on equivalence relations. From the abstract:
We give a detailed exposition of the homotopy theory of equivalence relations, perhaps the simplest nontrivial example of a model structure.
- | 1,644 | 7,013 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2013-20 | latest | en | 0.939389 |
http://www.velocityreviews.com/forums/t347348-re-tuple-to-string.html | 1,386,690,845,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164021936/warc/CC-MAIN-20131204133341-00082-ip-10-33-133-15.ec2.internal.warc.gz | 599,062,658 | 8,871 | Velocity Reviews > Re: tuple to string?
# Re: tuple to string?
Robert Kern
Guest
Posts: n/a
07-22-2005
Francois De Serres wrote:
> hiho,
>
> what's the clean way to translate the tuple (0x73, 0x70, 0x61, 0x6D) to
> the string 'spam'?
In [1]: t = (0x73, 0x70, 0x61, 0x6D)
In [2]: ''.join(chr(x) for x in t)
Out[2]: 'spam'
--
Robert Kern
http://www.velocityreviews.com/forums/(E-Mail Removed)
"In the fields of hell where the grass grows high
Are the graves of dreams allowed to die."
-- Richard Harter
Steven D'Aprano
Guest
Posts: n/a
07-23-2005
On Fri, 22 Jul 2005 06:07:28 -0700, Robert Kern wrote:
> Francois De Serres wrote:
>> hiho,
>>
>> what's the clean way to translate the tuple (0x73, 0x70, 0x61, 0x6D) to
>> the string 'spam'?
>
> In [1]: t = (0x73, 0x70, 0x61, 0x6D)
>
> In [2]: ''.join(chr(x) for x in t)
> Out[2]: 'spam'
I get a syntax error when I try that. I guess anyone who hasn't started
using Python 2.4 will also get the same error.
Since t is just a tuple, there isn't a big advantage as far as I can
see to build up and dispose of the generator machinery just for grabbing
the next item in a tuple. So a list comprehension will work just as well,
and in older versions of Python:
''.join([chr(x) for x in (0x73, 0x70, 0x61, 0x6D)])
For an even more version-independent method:
L = []
for n in (0x73, 0x70, 0x61, 0x6D):
L.append(chr(n))
print ''.join(L)
or even:
>>> ''.join(map(lambda n: chr(n), (0x73, 0x70, 0x61, 0x6D)))
'spam'
--
Steven.
Scott David Daniels
Guest
Posts: n/a
07-23-2005
Steven D'Aprano wrote:
> On Fri, 22 Jul 2005 06:07:28 -0700, Robert Kern wrote:
> ... or even:
>
>>>>''.join(map(lambda n: chr(n), (0x73, 0x70, 0x61, 0x6D)))
>
> 'spam'
This is exactly what is wrong with lambda. It yearns for over-use.
This last should be:
>>>''.join(map(chr, (0x73, 0x70, 0x61, 0x6D)))
--Scott David Daniels
(E-Mail Removed)
John Machin
Guest
Posts: n/a
07-23-2005
Steven D'Aprano wrote:
>
>
>>>>''.join(map(lambda n: chr(n), (0x73, 0x70, 0x61, 0x6D)))
>
> 'spam'
Why the verbal diarrhoea? What's wrong with the (already posted)
''.join(map(chr, (0x73, 0x70, 0x61, 0x6D)))
???
Steven D'Aprano
Guest
Posts: n/a
07-24-2005
On Sat, 23 Jul 2005 23:26:19 +1000, John Machin wrote:
> Steven D'Aprano wrote:
>
>>
>>
>>>>>''.join(map(lambda n: chr(n), (0x73, 0x70, 0x61, 0x6D)))
>>
>> 'spam'
>
> Why the verbal diarrhoea?
One line is hardly verbal diarrhoea.
> What's wrong with the (already posted)
>
> ''.join(map(chr, (0x73, 0x70, 0x61, 0x6D)))
>
> ???
Nothing.
If I had seen the already posted solution using chr on its own without
lambda, I wouldn't have bothered posting the lambda solution. But I
didn't, so I did.
As another poster has already pointed out, lambda cries out for over-use,
and this was a perfect example of it.
--
Steven.
John Machin
Guest
Posts: n/a
07-24-2005
Steven D'Aprano wrote:
> On Sat, 23 Jul 2005 23:26:19 +1000, John Machin wrote:
>
>
>>Steven D'Aprano wrote:
>>
>>
>>>
>>>>>>''.join(map(lambda n: chr(n), (0x73, 0x70, 0x61, 0x6D)))
>>>
>>>'spam'
>>
>>Why the verbal diarrhoea?
>
>
> One line is hardly verbal diarrhoea.
>
>
>>What's wrong with the (already posted)
>>
>>''.join(map(chr, (0x73, 0x70, 0x61, 0x6D)))
>>
>>???
>
>
> Nothing.
>
> If I had seen the already posted solution using chr on its own without
> lambda, I wouldn't have bothered posting the lambda solution. But I
> didn't, so I did.
>
> As another poster has already pointed out, lambda cries out for over-use,
> and this was a perfect example of it.
Here are a couple of reductions you can use in future, in the order given:
(1)
lambda <args>: foo(<args>) -> foo # for *any* function foo, not just chr
(2)
lambda <args>: <almost_any_guff>
->
def meaningful_func_name(<args>):
<almost_any_guff> | 1,317 | 3,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2013-48 | longest | en | 0.766031 |
https://muslimcentral.com/ihab-saad-scrapers/ | 1,725,975,612,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00351.warc.gz | 382,497,371 | 21,681 | # Ihab Saad – Scrapers
AI: Summary ©
The speaker explains how productivity is calculated using various factors and factors like job conditions and operational efficiency. The loading and loading methods for various types of loads, including scraper engines, are discussed, including the importance of performance data for determining optimal loading time. The importance of maneuvering and performance data is also emphasized. The speaker explains the details of a construction project, including the scraper and its characteristics, and discusses the weight and resistance of the device, including the weight distribution and resistance. The importance of practicing and learning to improve speed is emphasized in upcoming exams.
AI: Transcript ©
00:15:00 --> 00:15:04
So on and so forth. So the productivity is equal to the rated
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capacity times the operational efficiency divided by the cycle
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time. So let's see how we're going to calculate that.
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Here we have
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different tables that give us different factors that affect
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whether it's the job conditions and
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that are going to primarily affect the fixed time and so on. How are
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you going to load it? Is it? Push, loaded single engine, push, loaded
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twin engine, push, pull, or self loading elevator. In our case, we
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have an elevator self loading
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scraper. And then we also have, if you remember when we discussed the
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trucks and the hauling and so on, if you are going to be operating
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on a very short stretch of road, then by the time you reach your
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maximum speed, you have to start braking to stop at the end of that
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stretch. So we're going to have a factor that's going to affect that
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maximum speed. Because once you calculate the maximum speed from
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the performance curves, you have to multiply it by that factor. And
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in this case, the factor for the short speed is going to be point
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four, five is going to affect the maximum speed. It's going to be
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only 45% of the maximum speed. And if you have a long stretch of
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road, like 5000 feet, then the factor in this case, going to be
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96% we are already familiar with this table because we have used it
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before,
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and what we have here on this slide is again, something that we
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have seen before, and we already know how to use it, which is a
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performance chart. In this case, it's for a scraper. We have the
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weight both loaded and empty. We have the effective grade
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represented by these parallel lines. And then we have the
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different gears. So we are going to use the weight to intersect
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with the effective grade. Go horizontally and check which gear
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is that going to be intersecting with, and then that's going to
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give us the speed and which gear are we going to be operating
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within.
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Here's another example of very similar one. Again, here's the
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loaded weight, and here's the empty weight,
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the push loading scrapers, again, as we have seen before, sometimes
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the engine of the tractor in front of the board is not enough to move
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that scraper, so scraper sometimes require assistance in loading to
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fill the rated capacity. Push, pull. Scrapers can work in tandem
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to help each other, as we have seen in the video clip that we saw
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a few minutes ago, non elevating scrapers need pushers to load
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because they cannot lift the soil on their own. Crawler tractors are
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usually used since they have better traction than wheeled
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tractors,
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especially on when you have a high
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resistance, then in this case, the crawlers are going to be better.
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We have seen with the wheeled tractor how there was some
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slipping, and then it overcame that slipping and started moving
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forward. The loading method can be either backtrack loading, chain
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loading or shuttle loading. We're going to see some pictures
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representing each one of these different types of loading.
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So when scrapers are push loaded, the material in the bowl gets
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compacted due to the pressure of forcing the material into the
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bowl. Now imagine the bowl is moving forward, the blade is down,
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so it's being filled as the ball is getting filled, the new soil
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compresses the old soil inside the bone already. So the density of
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the material in the bone is determined by the equation. The
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density is equal to 100%
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110%
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times the bank density divided by 100% plus swell, which means it's
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110%
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of the loose
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density of that sort.
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Now here we have this picture representing the the push
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phase. You have here the tractor, and then you have another tractor
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here in the back. It pushes it forward. And now, once the scraper
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is fully loaded, that pusher goes back to push another scraper. So
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they work. They are in parallel. Here you push the first one, and
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then only the pusher goes back. Now the tractor is going to move
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that ball forward, and the pusher here in the back is going to move
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back and it's going to push another tractor. So this is called
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the backtrack loading backtrack, because only the pusher goes back
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and it pushes another so.
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Scrape, whereas in the second one, called Chain loading, basically
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the second scraper is standing little bit ahead of the first one,
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so you push the first one until it's able to move. The ball is
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filled. It moves away, and then the pusher just positions itself
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in behind the next scraper, and then keeps pushing it and so on.
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So it's called, this is called Chain loading, because it's not
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backtracking to go back to the same original position.
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The third method is called shuttle loading. So basically, again, you
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push the first scraper until it moves out of the way, and then you
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go back the other scraper is facing the other way around, so
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the other way the other scraper, the second scraper, is moving in
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the opposite direction. So you go behind the second scraper and
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start pushing it. So this is called shuttle loading because it
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goes sort of in a circle or in a closed circuit
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to determine the number of scrapers and pushers,
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to determine the number of scrapers a pusher can load, we
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must determine the pusher cycle time. So the cycle time for the
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pusher is the time to contact the scraper, to touch it, and then
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time to push it while it loads, and then time to boost it out of
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the cut once the ball is filled, and then time to maneuver to
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contact the next scraper. And that depends on the method of loading,
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if it's going to be shuttle or is going to be backtracking and so
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on. If no performance data exists, the cycle time for the pusher can
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be estimated as
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1.4
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LS, which is the loading time for the scraper in minutes plus point
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two, five minutes, which is the time to make that contact and to
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start pushing that scraper. So it's point it's 1.4 times the
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loading cycle time for the scraper, plus point two five
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minutes or 15 seconds, basically
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the boost time, which is time assisting the scraper out of cut
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point one minute return time is estimated to be 40% of load time,
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because now the pusher goes back empty. It does not have the same
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resistance, so it can come back at higher speed. The maneuver time is
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point one, five minutes again to maneuver and to position itself
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behind the next scraper. Therefore, the number of scrapers
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that a loader can push is determined by the equation n. The
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number of scrapers is equal to the cycle time of the scraper divided
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by the cycle time of the pusher. CTS is the cycle time for the
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scraper, and CTP is the cycle time of the pusher. Let's
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look at an example again. It's going to make things easier.
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A CAD six, 630, 1e single engine. Scraper will be used to excavate
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the side, the side of a large fill to level a construction site. The
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soil is sandy clay, weighing 2700 pounds per bank cubic yard, with a
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swell of 18%
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the scraper has a 450 horsepower turbocharged diesel engine.
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The haul road is 4000 feet. The distance is 4000 feet, with an
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uphill grade of 3% from the cut area to the dump area. The running
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resistance is 8585
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pounds per ton, and the coefficient of traction is point
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four,
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A, D 8n crawler will be used to push the scrapers to load them to
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heat capacity. The project site has an elevation of 3000 feet. Now
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look at the problem, and each word has a certain meaning. Now here it
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gives us an elevation of 3000 feet. In the older problems that
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we have solved with other pieces of equipment, we had the derating
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factor due to the elevation. But notice here that it told us that
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this is turbocharged diesel engine. And we have mentioned
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before that turbocharged diesel engines are not affected by the
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elevation. So this is sort of a trick, if you do understand the
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nature of the engine, and it's not going to be affected by any
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derating factor, then this number is totally redundant, and we're
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not going to be using
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it. The scraper has the following characteristics, the rated
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capacity, 31 cubic yards, loose cubic yards, of course, the empty
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weight, 96,880
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pounds, the maximum load, 75,000 pounds, the weight distribution
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when empty on the drive axle is 67% rear axle, 33%
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and when loaded again, is sort of balanced. The drive axis, 53% and
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the rear axle, 47%
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what's the estimated production of the scraper in.
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Cubic yards, if the operation efficiency is 50 minutes per hour,
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and the scraper does not wait in the cut for a pusher and how many
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scrapers can a pusher load? So we're going to need to calculate
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the cycle time for a pusher and the cycle time for a scraper to
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determine how many pushers are we going to need. This problem might
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look familiar, because we have solved something similar to that
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when we're talking about haulers, when we're talking about trucks
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and the different types of resistance that they're going to
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face, and so on and so forth. So we're going to follow exactly the
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same steps.
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First of all, we're going to check if we're going to be weight
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controlled or volume controlled. The density of the material the
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scraper will carry is determined by the equation density. Now here
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we're going to apply this is going to be unique to the scrapers,
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because of the compression factor and the compaction factor that's
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going to take place. So the density is 110%
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the bank density divided by 100% plus the swell factor, which gives
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us basically 110%
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of the loose cubic density. Loose density, which is 2517 pounds per
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cubic yard. Per loose cubic yard, the weight of the load when filled
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to heap capacity, the capacity is 31 loose cubic yards. So the total
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weight is going to be 31 times 2517
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and that gives a total weight of 78,027
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pounds, which is beyond the load capacity of that scraper. Because
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the load capacity the maximum weight was 75,000 pounds, which
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means we are not going to be able to fill it to the heat capacity of
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31
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loose cubic yards. So what would be the volume that's going to be
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applicable? In this case, it's going to be 75,000 pounds divided
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by the density, which gives 29.8 loose cubic yards, which can be
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translated into bank cubic yards by dividing by the bank capacity,
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75,000 pounds divided by the bank capacity, the bank density, which
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gives 27.8
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bank cubic yards. So that was the first check. So we learned here
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that it's going to be weight controlled, not volume controlled.
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The next step is to estimate the cycle time of the scraper. With
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average job conditions, we get a loading time that's from the
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tables that we have seen a couple of slides before. We get a loading
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time of point seven minutes, spotting a delay time of point
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three minutes, and a dump time of point five minutes. All of these
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from the table, which makes the fixed time point seven plus point
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three plus point five. That's 1.5 minutes,
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the maximum rim rim pull generated by the scraper is going to be the
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coefficient of traction times the weight on the moving axles. So
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it's going to be when it's empty, the coefficient of traction is
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point four times point six seven, which is the weight on the moving
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axle, times the total load, the total weight, one empty, which is
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96,880
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which gives a rim pull of 25,964
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pounds. Now, when it's full, we're going to add the load of the soil,
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in addition to the weight of the scraper itself, the coefficient of
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traction is still point four. The weight distribution has changed
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because of the heavier load on the other axle. So it's times point
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five, three times 96,008 80 plus, the weight inside the bowl, which
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is 75,000 pounds, which gives 36,439
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pounds. So that's the generated rim pole.
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Now we have, we can convert that into tons, the total weight into
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tons, by dividing by 2000 which is 85.9 tons. That's to calculate the
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resistance, because the resistance is. The factor here that we use is
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in tons. So the resistance is going to be
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the force of the resistance going to be the rolling resistance, plus
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the grade resistance, which is 80 pounds per ton, the factor times
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the total weight, 85.9 tons, plus the grade 3% times 20 pounds per
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ton, per percent slope times the weight in tons, which gives a
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total of total resistance of 12,456
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pounds. Now the required drain pull to overcome the resistance is
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less than the generated drain pole. So basically, the scraper
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can move forward. The resistance is not going to be enough to stop
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the scraper from moving. We have seen here while it was empty,
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25,900
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almost 26,000 and when it's full, 36.4
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and the required the resistance is 12.4 so we are quite safe now to
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catch.
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The effective grade, because we're gonna need to plug that into the
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performance chart. We divide the running resistance by 20 pounds
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per ton, that's a constant. So 85 divided by 20, that's 4.25
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plus the already existing grade, 3% that gives us a total grade of
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7.25%
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when it's loading and when it's moving forward on the way back,
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we're going to subtract the grade. So we have 85 divided by 20, which
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is 4.25
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minus three. So we have still an effective grade of 1.25%
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using the effective grade on the performance chart, we get a speed
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of 10 miles per hour. To get the average speed, we incorporate the
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speed factor from the tables. We mentioned that
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for starting from standstill for half of the road and coming to a
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stop for the other half. So basically, if the road distance is
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4000 we divide it by two. So the factor from the table is going to
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be point nine. Two. Therefore the average speed when loaded is going
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to be point nine, two times 11. That's 10.12 miles per hour. And
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the average speed this should be 11, by the way, not 10. The
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average speed one empty on the way back is going to be point nine,
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two times 33 miles per hour, which gives 30.3 mile, miles per hour.
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And this is basically how we obtain these numbers. We plugged
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in the weight one loaded with a 7.25
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effective grade, and that gave us this line horizontally. We go
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here. We're gonna hit the fifth gear, and that's going to give us
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a speed notice for by the way, that the first numbers are in
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kilometers per hour, and the lower scale is in miles per hour. So
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it's going to be about 11 miles per hour, and that's what what we
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use in the equation.
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Now, when empty, here's the weight one empty, and we had an effective
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grade of 1.25
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so basically, if we keep just going, it's not going to hit even
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the 1.25 so we're going to use that speed, which is
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basically we're going to keep going down until we reach the
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maximum speed, which is going to be about 33 miles per hour.
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So in the 4000 feet, which we're going to divide by two, as we had
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just discussed,
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the time is going to be 4000 divided by the 88 to convert the
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feet into miles per hour
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divided by the speed. And that gives us a time of 4.5 minutes,
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moving forward, moving backward, the speed is 30.3 so we have 1.5
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minutes. Therefore the variable time is the sum of these two, 4.5
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and 1.5 that's six minutes. We had already calculated the fixed time
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to be a minute and a half. So the total cycle time is fixed plus
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variable. That's a total of 7.5 minutes. The Productivity is going
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to be the load per cycle
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times the number of cycles.
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So
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the load per cycle is 27.8 bank cubic yards
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times operation efficiency, which is 50 minutes per hour
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divided by the cycle time, which gives
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the total capacity, or total production, of 185.33
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band cubic yards per hour.
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The pusher cycle time is equal to 1.4 because we are not given any
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information about the pusher. So we're gonna use the equation 1.4
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the scraper load time plus point two, five minutes, which is 1.4
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times point seven, which is the loading time for the scraper,
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plus point two, five minutes. And that gives us and the point seven
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minutes is from the table that we have used before. So it gives us a
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cycle time of 1.2 minutes. Now if the cycle time for the scraper is
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7.5 minutes, and the cycle time for the pusher is 1.2 minutes, so
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how many scrapers can one pusher serve? The number of scrapers is
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going to be scraper cycle time divided by pusher cycle time. So
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that gives 6.25 scrapers, which is going to be rounded down to six
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scrapers. So one pusher can serve six scrapers. That's basically how
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we calculate the production for a scraper and the number of pushers
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and the number of scrapers, and so on and so forth.
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I hope that has been
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well understood, and we we have seen the solved example. Please
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make sure that you try to change the numbers and resolve the
00:34:51 --> 00:34:53
problem, to get more practice and to gain
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speed in solving the problems which are going to help you
00:34:58 --> 00:34:59
finishing the problems on the exam.
00:35:00 --> 00:35:02
Well, good luck, and I'll see you in another lecture. Bye.
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https://www.physicsforums.com/threads/the-speed-of-light-gravity-strong-force-weak-force.239379/ | 1,675,125,836,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499831.97/warc/CC-MAIN-20230130232547-20230131022547-00648.warc.gz | 958,634,172 | 17,294 | # The speed of light, gravity, strong force, weak force
LongOne
I could benefit from someone shedding some light on a couple of questions that I have regarding the speed of various entites:
1) It seems that the speed of gravity, e.g, if the sun disappeared, is the speed of light WRT things I've read. What is it that would inherently cause gravity to travel at c, since presumably the fundamental nature of the causation is different? Could it be that gravity's speed would be infinite, except that it can't exceed 'c'?
2) Has anyone been able to define a speed associated with the strong and weak nuclear forces? If no, could/should one assume that those forces also move at the speed of light?
cmos
The following is my understanding of things. I do not claim to be a particle physicist, so if somebody sees that I am wrong I would be most grateful to be informed...
The graviton is presumed to have a mass of zero, thus it should move at the speed of light. Similarly, the gluon, which mediates the strong force, is a massless particle thus it moves at the speed of light. The weak force, however, is mediated by W and Z bosons which are very massive, therefore they must have a finite mass. How then the weak and electromagnetic force is unified I do not know.
Daedalus_
The graviton is an assumed particle and thus has not been proven. Therefore you cannot calculate its mass or speed in any given reference frame.
lbrits
The "speed of gravity" is the speed at which small perturbations in the metric propagate in an otherwise static background (i.e., gravitational waves). To talk about the speed of large deformations in spacetime is a bit iffy, because you need some background to measure the speed with respect to, and all you have is spacetime. For example, in an expanding universe points can move away from each other at faster than c, but that's because the space between them is expanding. Locally they are not exceeding c. This is all at the level of classical GR, of course.
cmos hit the nail on the head, but I'd just like to point out that the weak and electromagnetic are only unified at high energy, and are for all intents and purposes, separate at our energy scales (there is a broken symmetry).
regarding Daedalus' comment, the graviton is massless and moves at speed c, because that's what the graviton is. It is not so much a predicted particle as a name we give to quantum excitations of the weak perturbed metric. If we ever discovered a massive spin-2 field, we would probably not call it a graviton. Also, there are some mathematically proven theorems that tell you that we'd be in bad shape if the graviton were massive.
Daedalus_
my comment stands. The "graviton" is a "predicted" particle. We cannot confirm its existence until they use the new particle accelerator to prove its there. Now it is probably true that it is massless but then again... it might not be, causing a complete reworking of quantum physics. But this is probably not the case.
lbrits
I just took issue with "therefore you cannot calculate its mass or speed in any given reference frame. " In any case, I think we understand each other's points of view.
rajatgl16
I think first question of longone is more about classical mechanics and answering it in terms of quantum mechanics would be wrong.
Is there any reason that why speed of light is c in terms of classical mechanics?
In my opinion same would be the answer for gravitaional speed.
NetMage
This may be answered perhaps with the same mechanisms of black holes and light not being able to escape ^^
rajatgl16
This may be answered perhaps with the same mechanisms of black holes and light not being able to escape ^^
The reason behind this is space-time shape. How can you link this with gravitational speed.
calhoun137
The gravitational field IS the metric, i.e. it gives the distance between two points. What is this distance physically? It's it the speed of light c times the amount of time it takes for a beam of light to travel between the two points. Keep in mind that a light cone has a boundary, and this boundary represents the boundary between points which in principle can interact, and those points which in principle cannot interact.
Therefore if we change the gravitational field in one part of space, then we can think about what will happen if we send a light signal out from that part of space before we make the change and afterwords. It's clear that the time it takes for the light beam to travel will depend on if we send it before or after the change in the gravitational field.
We can trace out the light cone through space time until the two points we want to know the distance between are allowed to have interacted. If you think about it, this shows that the time it takes for the change in the gravitational field of one part of space to affect another part is determined by the speed of light.
One way to see this is to try and figure out the following situation. You are sitting at point A with some space time four vector, somewhere else at point B which has some other 4-vector, the gravitational field changes in some way. The problem is to determine how an observer at point A measures distances to all the points nearby to A as a function of time. If you think about sending light signals back and forth in order to make these measurements, it should become clear why we say gravitational fields travel at the speed of light.
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https://lists.firedrake.org/gzg/0912094451-0972e76f5ac0d7c92eb88bfe96baf1e39d71ba29671e793478cdfead1a8653b6.html | 1,721,681,068,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517915.15/warc/CC-MAIN-20240722190551-20240722220551-00663.warc.gz | 319,767,889 | 2,680 | Prev: Re: Railgun Goals II Next: RE: [FT] Railgun Goals
# (FT) Fleet Book Kra'Vak
From: Steven Arrowsmith <arrowjr@u...>
Date: Thu, 26 Nov 1998 10:34:11 -0500 (EST)
Subject: (FT) Fleet Book Kra'Vak
Heres another rant...
Since it is slow this morning, I have had a chance to read through all
the
KraVak railgun proposals. Their are some very good ideas here, their
seems
to be two camps here. The "trash can" group, and the single solid
penetrator group. Why dont we use both, one set of rules covering
"trash
can" type shots - lets say scatterguns" and their big brother the class1
railgun, if we go this rout we will need to come up rules allowing
class1s
to be used as point defense systems, up to a set distance. These could
also be used as area-defense weapons. This would allow a different set
of
rules covering major anti-ship railguns know as class 2 and 3, and above
if we want.
I read a proposal to use the range as the number need to be rolled. I
personally like this idea, simple and quick. I would take it one step
more, when firing a big railgun, measure the distance, and divide that
in
half and add 1 to the number needed, not rolled. Thus a 1 is always a
miss. If we give the following dice; class1 = 2d6, class2 = 3d6, and
class3 = 4d6. We see a effective range of 10" for a class1, 22" for a
class2, and 34" for a class3, about the same as Human beams, with
maximum
Range twice that.
When a hit is scored, look at the "to-hit" die roll. Add them together,
this is the damage done to the target. Any roll of 6, regardless of
modifier, is a reroll . If firing against Kra'Vak Armoured hulls,
subtract
2 from each die for damage. If the target is using Human armour, then
half
the damage scored (round up) is taken on the armour, and the remainder
applied directly to the hull.
Looking at the stats, I can agree with Sean Bayan Schoonmaker, on the
point cost, but the mass of the Class3 railgun is way to high.
Class 1 RAILGUN - 1 MASS, 3-arc fire
Class 2 RAILGUN - 3 MASS, 2-arc fire
Class 3 RAILGUN - 9 MASS, 1-arc fire
POINT COST = 4 per MASS
Example a MT Vo'Bok class Hunter Cruiser carries 22 mass of weapons. 2
Class3, 2 Class1s , 4 scatterguns and 2 Firecontrols. For a Average Hull
Human, with a thrust rating of 6 , and 22 mass of equipment, the ship
would have a displacement of 72 - a little heavy for a light/hunter
cruiser. This is also without amour.
Please review MT page 24, for a overall capabilities of a Kra'Vak
warship..
Remember the Kra'Vaks are alien terrors..
Steven
Steven Arrowsmith
www.public.usit.net/arrowjr
steven@arrowsmith.net
dredd@quake.usit.net
________________________________________________________________________
_______
I Would Rather live a Lie, Thinking I Can.
Than know The Truth That I Can't
________________________________________________________________________
_______
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https://www.physicsforums.com/threads/do-photons-create-gravity.442266/page-4#post-2974865 | 1,643,330,480,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305317.17/warc/CC-MAIN-20220127223432-20220128013432-00109.warc.gz | 983,014,039 | 24,033 | # Do photons create gravity?
When we have pair production , a photon turns into an electron and positron , I would like to think that the gravitational field of the electron and positron came from the photon,
And vice versa if we collide an electron and positron and we get a photon out, It seems like the G field of the electron and positron would get transferred to the photon.
D H
Staff Emeritus
I do not believe the topic is whether photons gravitate, the topic is whether a single photon can create a gravitational field, these are in my mind two different things.
Caveat: I am not an expert in GR.
My understanding is that a single photon does not create a gravitational field. A collection of photons can, however. The intrinsic mass of a collection of particles is
$$m = \frac{\sqrt{E^2 - p^2c^2}}{c^2}$$
where E is the total energy of the collection of particles and p2 is the square of the net momentum of the collection. While energy and momentum are both frame-dependent quantities, the difference E2 - p2c2 is frame independent.
For a single photon, E=pc, so the intrinsic mass of a single photon is zero. Now consider photons created by an electron-positron annihilation. In the rest frame of the (former) electron-positron pair, the created photons have zero net momentum. The total energy of these photons is equal to the 2mec2 plus any kinetic energy of the electron-positron pair. The intrinsic mass of the photons created by this annihilation event is identically equal to the intrinsic mass of the electron + positron system prior to the annihilation event.
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Dale
Mentor
2021 Award
No, im saying you are all wrong.
If you believe that GR is correct and that we are wrong then you do not understand GR. I recommend further study. In particular, you should learn about pp-wave spacetimes.
Dale
Mentor
2021 Award
But we don't need a quantum theory of gravity to answer this question. All we need is QED in curved spacetime, and that is no problem at all.
1. Observationally, we know single photons fall.
2. GR says that momentum is conserved (in this case).
3. Therefore, single photons must gravitate.
That conclusion can only be escaped by asserting GR is incorrect.
I like this. That is a good point.
Alternatively, one can put in a single photon in T and again GR shows the effect on G.
This one is the dangerous approach. How can you define T when the photon does not have a definite energy, momentum, and location at any given time? Actually, I am not even sure that spacetime can be represented by a manifold at sufficiently small scales due to quantum effects.
Staff Emeritus
2021 Award
A single photon does create a gravitational field, because T is non-zero. If T is non-zero, so is G.
Dale
Mentor
2021 Award
If T is non-zero then what is T?
"everything" distorts/curves/interrupts etc space-time
thats what gravity is
pervect
Staff Emeritus
While it is true that photons do not have mass, it is incorrect to conclude that because they don't have a mass, they can't cause gravity. Photons don't need mass to cause gravity, all they need is energy and momentum, which they do have, because it's the stress energy tensor that causes gravity and not "mass".
While GR is a classical theory, and doesn't have anything to do directly with photons per se, you can model a photon in GR as a packet of light. Which is essentially what Tolman et al did in the 1931 paper I mentioned earlier, with the results I mentioned - parallel light packets don't attract each other, there is no "self focusing" effect of light due to gravity, but beams in opposite directions do interact gravitationally.
Please!!! In this thread there has been nothing but manipulation of Albert’s general relativity to produce a theoretical scenario where imaginary particle states can produce gravitation. Its not the photon with gravitation that is bugging me, it’s the use of Albert’s notions in producing general relativity to prove your hypothesis of bosons produce gravity is false because the theory in its entirety didn’t explain theoretical zero mass particles producing gravity and your existence to interpret the information in such a way to try and change theoretical understandings of a particle. What are you suggesting number 1? Where in Einstein’s formulas does it produce an explanation of zero mass particles producing gravity 2. And please note that a wave’s propagation can be seen as action at a distance but not gravity. How can you compress a field for density not field compression or amplification which will create distortion not density . Propagation is not gravity. Waves like light (sigh), do not produce gravity
This thread is in the relativity section, and according to GR, "light" does produce gravity.
If you looked at a completely "empty" Universe that was void of any form of energy, matter, radiation, etc., in a 4 dimensional spacetime where
$$\mathbf{R(X,Y)Z}=R^{a}_{\ bcd}X^{c}Y^{d}Z^{b}=0=R_{bd}$$
and then let a wave of radiation pass through then you get the interesting scenario where calculating the Ricci tensor will still give you zero
$$R_{bd}=0$$
but the Riemann tensor isn't required to vanish and you can find calculations where $$R^{a}_{\ bcd}X^{c}Y^{d}Z^{b}\neq 0$$, ie., the radiation causes curvature of the spacetime (in 4 or greater dimensions).
The fact that GR predicts that light alters the spacetime geometry is much more general, but that is a situation where nothing exists to cause curvature, so "gravitational" curvature is entirely from the radiation.
Staff Emeritus
2021 Award
If T is non-zero then what is T?
It's been a while, but I believe the upper left 2x2 is 1 and the other components are 0.
bcrowell
Staff Emeritus
Gold Member
Please!!! In this thread there has been nothing but manipulation of Albert’s general relativity to produce a theoretical scenario where imaginary particle states can produce gravitation.
Dale
Mentor
2021 Award
It's been a while, but I believe the upper left 2x2 is 1 and the other components are 0.
I believe that is correct for a classical pulse of light at a definite location and of known (unit) energy density propagating in the +x direction. But that is not the same as an EM quantum with uncertain location and non-definite energy. What you are describing is simply GR for classical pulses of light, not for photons. Do you see the difference and why I am reluctant to make conclusions for a photon based on classical pulses of light?
I like this. That is a good point.
This one is the dangerous approach. How can you define T when the photon does not have a definite energy, momentum, and location at any given time? Actually, I am not even sure that spacetime can be represented by a manifold at sufficiently small scales due to quantum effects.
The QM theory gives only probability distributions of photon location, its momentum and energy, but photon must have a definite energy, momentum and location in time that we cannot know by QM.
pervect
Staff Emeritus
Energy, momentum, etc would only be expectations for a photon. But we expect that any quantum theory of light is going to have to include gravity in the classical limit, because we know that collections of photons do gravitate under certain circumstances, and we know which ones.
Furthermore, I already posted a link to a book by Zee which discusses how you get quantum gravity for the photon perturbatively - though it's outside the scope of GR.
So, I'd suggest continuing the QM side of the debate in the "beyond the standard model" forum.
I'd also encourage people interested to look at the reference by Zee, perhaps it got lost in the shufle.
No. According to GR the source of gravity is the stress-energy tensor. There are 10 independent components in the stress-energy tensor. Energy is only one of those 10 components.
http://en.wikipedia.org/wiki/Stress-energy_tensor
We do not have a working theory of quantum gravity at the time so I cannot answer your question wrt photons, however I can answer it wrt classical pulses of light. Pulses of light have energy, they also have momentum, so several of the components of stress energy tensor will be non-zero. So light can be a source of gravity.
I wonder what you think of the following thought experiment by Dimitry67 that seems to show that parallel beams of light will not converge. Thus it is as though light can't be a source of gravity.
Consider two massive objects, separated by some distance, flying in the same direction at velocity v according to an observer. In their inertial system they collide, say, in 1s. For the observer this process takes longer because of the time dilation. The faster the two objects are flying the longer it takes. In the limit where v --> c they never converge according to the observer.
Dale
Mentor
2021 Award
I wonder what you think of the following thought experiment by Dimitry67 that seems to show that parallel beams of light will not converge. Thus it is as though light can't be a source of gravity.
Dimitry67 is correct, but you should not take the quote out of context. The bolded conclusion above does not follow, as you can clearly see by considering the full quote:
This is correct, light beams create gravity.
However, when it was discussed here about 1 or 2 y ago, I remember that someone (with much deeper knowledge of GR - I am just a layman) told me that:
2 parallel light beams going in the same direction do not attract (even they attract the surrounding objects)
2 parallel light beams going in opposite directions do attract.
The first fact might be clear if we look at 2 massive objects, separated by some distance, flying in the same direction. In their inertial system they collide, say, in 1s. For an external observer, this process would take longer because of the time dilation. The faster 2 objects are flying the longer it takes. You can think about the case N1 as a limit where v --> c (it takes forever)
So,
What percentage of the universe's net gravitational field would be due to electromagnestism?
GrayGhost
Dimitry67 is correct, but you should not take the quote out of context. The bolded conclusion above does not follow, as you can clearly see by considering the full quote:
I stand corrected - I did take Dimitry67's argument out of context.
I did it for two personal reasons I guess.
1/ I like the argument.
2/ I have a "pet theory" that the inertia of a particle with rest mass is caused by retarded gravitational waves impinging on the particle from all the other massive particles in the Universe. I believe that initially massless particles respond to this by following circular orbits whose rotation energy gives half the mass/energy of the particle (the rest being in the mutual gravitational energy between the particle and the rest of the Universe). I am trying to formalise Mach's Principle. I want to argue that light is different so that it does not pick up an inertia. I probably need to think about my theory more before I can decide what it says about light. I'm not using GR itself but an approximation to it called gravitomagnetism that is like Maxwell's theory. In fact I should probably just stick to trying to understand electromagnetically induced inertia for the moment as i feel on safer ground with EM. (I only have at best an undergraduate understanding of physics).
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Dale
Mentor
2021 Award
I have a "pet theory" ... I only have at best an undergraduate understanding of physics
Hmm.
Dale
Mentor
2021 Award
Threadmark is wrong. Was there any ambiguity?
bcrowell
Staff Emeritus
Gold Member
What percentage of the universe's net gravitational field would be due to electromagnestism?
If you mean the gravitational field as defined in Newtonian mechanics, that isn't an unambiguously well defined thing in GR. By the equivalence principle, the gravitational field at a given point can be anything you like, depending on your frame of reference. In the frame of an observer at rest with respect to the cosmic microwave background, the gravitational field is zero by symmetry (in a homogeneous and isotropic cosmological model).
Interpreting your question more loosely, the answer is that the universe was radiation-dominated at one time, then matter-dominated, and is now vacuum-dominated.
No,
All I meant was that if photons gravitate, then they produce gravitation. The cosmos has some net collective gravitational feild. Given photons are everywhere, I was just wondering what percentage of the cosmic gravitational field would owe to EM?
GrayGhost
bcrowell
Staff Emeritus
Gold Member
The cosmos has some net collective gravitational feild.
That's incorrect, for the reasons given in #97.
That's incorrect, for the reasons given in #97.
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