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https://web2.0calc.com/questions/precalc-function-stuff | 1,591,290,009,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347445880.79/warc/CC-MAIN-20200604161214-20200604191214-00189.warc.gz | 584,952,327 | 6,802 | +0
Precalc- Function stuff
0
307
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+142
Let the domain of the function f(x) be the interval (-4,4). What is the domain of the function $$f(\frac{x-2}{x+2})$$?
Feb 15, 2019
#1
+6187
+6
$$\text{we have 3 conditions} \\ -4 < \dfrac{x-2}{x+2} < 4 \text{ and } (x+2) \neq 0\\$$
$$\text{if }x+2>0 \text{ i.e. if }x > -2 \\ -4x-8 < x-2 < 4x+8\\ -6 < 5x \wedge -3x < 10\\ \dfrac{-6}{5} < x \wedge x > -\dfrac{10}{3}\\ \text{distilling all this we end up with simply }-\dfrac{6}{5} < x$$
$$\text{if }x+2< 0 \text{ i.e. if } x < -2\\ -8x-8 > x -2 > 4x+8\\ -6 > 9x \wedge -3x > 10\\ -\dfrac{2}{3} > x \wedge x < -\dfrac{10}{3}\\ \text{distiliing all this we end up with }x < -\dfrac{10}{3}$$
$$\text{Combining these results we get}\\ x \in \left(-\infty, -\dfrac{10}{3}\right) \cup \left(-\dfrac{6}{5}, \infty\right)$$
.
Feb 15, 2019
#2
+109520
+4
Hi Rom,
I was trying to work out what your $$\wedge$$ (\wedge) meant ....
It intersection I think ?? $$\cap$$ \cap
I have never seen a wedge used before....
Melody Feb 15, 2019
#3
+6187
+3
it just means logical AND
I use it when sets aren't obviously involved but you need to meet 2 or more conditions.
Rom Feb 15, 2019
#4
+109520
+3
ok so you use \cap = intersection only for sets
and \wedge for the same thing when it is not written in set notation.
But
you use \cup = union for sets and non-sets notation...
Thanks Rom.
Melody Feb 15, 2019
#5
+6187
+4
how closely I stick to those rules is inversely proportional to how many beers I've had... :D
Rom Feb 15, 2019
#6
+109520
+3
That is understandable.
Melody Feb 15, 2019
#7
+6187
+3
oh.. yeah.. intervals get \cup too... intervals are like sets
Rom Feb 15, 2019 | 702 | 1,729 | {"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-2020-24 | latest | en | 0.669639 |
https://external-production.codecademy.com/learn/learn-python/modules/learn-python-lists-and-functions-u | 1,553,642,557,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912206677.94/warc/CC-MAIN-20190326220507-20190327002507-00323.warc.gz | 470,464,790 | 72,018 | Lists and Functions
Lesson 1 of 2
1. 1
This exercise goes over just pulling information from a list, which we’ve covered in a previous section!
2. 2
You’ve already learned how to modify elements of a list in a previous section. This exercise is just a recap of that!
3. 3
Here, we’ll quickly recap how to […] elements to the end of a list.
4. 4
This exercise will expand on ways to remove items from a list. You actually have a few options. For a list called […] : 1. […] will remove the item at […] from the list and return it t…
5. 5
In this exercise, you will just be making a minor change to a function to change what that function does.
6. 6
This exercise will recap how to use more than one argument in a function.
7. 7
This is a basic recap on using strings in functions.
8. 8
You pass a list to a function the same way you pass any other argument to a function.
9. 9
Passing a list to a function will store it in the argument (just like with a string or a number!) […] 1. In the example above, we define a function called […] . It has one argument called …
10. 10
Modifying an element in a list in a function is the same as if you were just modifying an element of a list outside a function. […] 1. We create a list called […] . 2. We use the […] f…
11. 11
You can also append or delete items of a list inside a function just as if you were manipulating the list outside a function. […] The example above is just a reminder of how to append items t…
12. 12
This exercise will go over how to utilize every element in a list in a function. You can use the existing code to complete the exercise and see how running this operation inside a function isn’t mu…
13. 13
This exercise shows how to modify each element in a list. It is useful to do so in a function as you can easily put in a list of any length and get the same functionality. As you can see, […] i…
14. 14
Okay! Range time. The Python […] function is just a shortcut for generating a list, so you can use ranges in all the same places you can use lists. […] The […] function has three diff…
15. 15
Now that we’ve learned about […] , we have two ways of iterating through a list. Method 1 - […] : […] Method 2 - iterate through indexes: […] Method 1 is useful to loop t…
16. 16
Now let’s try working with strings! […] The example above is just a reminder of the two methods for iterating over a list.
17. 17
Using multiple lists in a function is no different from just using multiple arguments in a function! […] The example above is just a reminder of how to concatenate two lists.
18. 18
Finally, this exercise shows how to make use of a single list that contains multiple lists and how to use them in a function. […] 1. In the example above, we first create a list containing t…
1. 1
In this project you will build a simplified, one-player version of the classic board game Battleship! In this version of the game, there will be a single ship hidden in a random location on a 5x5 …
2. 2
The first thing we need to do is to set up the game board.
3. 3
Good! Now we’ll use a built-in Python function to generate our board, which we’ll make into a 5 x 5 grid of all […] s, for “ocean.” […] will print out […] , which is the basis for a row…
4. 4
Great job! Now that we’ve built our board, let’s show it off. Throughout our game, we’ll want to print the game board so that the player can see which locations they have already guessed. Regular…
5. 5
Now we can see the contents of our list, but clearly it would be easier to play the game if we could print the board out like a grid with each row on its own line. We can use the fact that our b…
6. 6
We’re getting pretty close to a playable board, but wouldn’t it be nice to get rid of those quote marks and commas? We’re storing our data as a list, but the player doesn’t need to know that! [….
7. 7
Excellent! Now, let’s hide our battleship in a random location on the board. Since we have a 2-dimensional list, we’ll use two variables to store the ship’s location, […] and […] . [……
8. 8
Good job! For now, let’s store coordinates for the ship in the variables […] and […] . Now you have a hidden battleship in your ocean! Let’s write the code to allow the player to guess whe…
9. 9
Awesome! Now we have a hidden battleship and a guess from our player. In the next few steps, we’ll check the user’s guess to see if they are correct. While we’re writing and debugging this par…
10. 10
Okay—now for the fun! We have the actual location of the ship and the player’s guess so we can check to see if the player guessed right. For a guess to be right, […] should be equal to […]…
11. 11
Great! Of course, the player isn’t going to guess right all the time, so we also need to handle the case where the guess is wrong. […] The example above prints out […] , the element in th…
12. 12
Great job! Now we can handle both correct and incorrect guesses from the user. But now let’s think a little bit more about the “miss” condition. 1. They can enter a guess that’s off the board. 2…
13. 13
Great! Now let’s handle the second type of incorrect guess: the player guesses a location that was already guessed. How will we know that a location was previously guessed? […] The example …
14. 14
Congratulations! Now you should have a game of Battleship! that is fully functional for one guess. Make sure you play it a couple of times and try different kinds of incorrect guesses. This is…
15. 15
You can successfully make one guess in Battleship! But we’d like our game to allow the player to make up to 4 guesses before they lose. […] We can use a […] loop to iterate through a ran…
16. 16
If someone runs out of guesses without winning right now, the game just exits. It would be nice to let them know why. Since we only want this message to display if the user guesses wrong on their …
17. 17
Almost there! We can play Battleship!, but you’ll notice that when you win, if you haven’t already guessed 4 times, the program asks you to enter another guess. What we’d rather have happen is for …
18. 18
Congratulations! You have a fully functional Battleship game! Play it a couple of times and get your friends to try it out, too. (Don’t forget to go back and remove the debugging output that gives …
19. 19
You can also add on to your Battleship! program to make it more complex and fun to play. Here are some ideas for enhancements—maybe you can think of some more! 1. Make multiple battleships: you’ll…
## How you'll master it
Stress-test your knowledge with quizzes that help commit syntax to memory | 1,618 | 6,579 | {"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} | 3.453125 | 3 | CC-MAIN-2019-13 | longest | en | 0.889212 |
https://metanumbers.com/92758 | 1,638,448,465,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362219.5/warc/CC-MAIN-20211202114856-20211202144856-00378.warc.gz | 446,365,879 | 7,358 | # 92758 (number)
92,758 (ninety-two thousand seven hundred fifty-eight) is an even five-digits composite number following 92757 and preceding 92759. In scientific notation, it is written as 9.2758 × 104. The sum of its digits is 31. It has a total of 3 prime factors and 8 positive divisors. There are 43,920 positive integers (up to 92758) that are relatively prime to 92758.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 31
• Digital Root 4
## Name
Short name 92 thousand 758 ninety-two thousand seven hundred fifty-eight
## Notation
Scientific notation 9.2758 × 104 92.758 × 103
## Prime Factorization of 92758
Prime Factorization 2 × 19 × 2441
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 92758 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 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 92,758 is 2 × 19 × 2441. Since it has a total of 3 prime factors, 92,758 is a composite number.
## Divisors of 92758
1, 2, 19, 38, 2441, 4882, 46379, 92758
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 146520 Sum of all the positive divisors of n s(n) 53762 Sum of the proper positive divisors of n A(n) 18315 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 304.562 Returns the nth root of the product of n divisors H(n) 5.06459 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 92,758 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 92,758) is 146,520, the average is 18,315.
## Other Arithmetic Functions (n = 92758)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 43920 Total number of positive integers not greater than n that are coprime to n λ(n) 21960 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 8952 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 43,920 positive integers (less than 92,758) that are coprime with 92,758. And there are approximately 8,952 prime numbers less than or equal to 92,758.
## Divisibility of 92758
m n mod m 2 3 4 5 6 7 8 9 0 1 2 3 4 1 6 4
The number 92,758 is divisible by 2.
• Arithmetic
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (92758)
Base System Value
2 Binary 10110101001010110
3 Ternary 11201020111
4 Quaternary 112221112
5 Quinary 10432013
6 Senary 1553234
8 Octal 265126
10 Decimal 92758
12 Duodecimal 4581a
20 Vigesimal bbhi
36 Base36 1zkm
## Basic calculations (n = 92758)
### Multiplication
n×y
n×2 185516 278274 371032 463790
### Division
n÷y
n÷2 46379 30919.3 23189.5 18551.6
### Exponentiation
ny
n2 8604046564 798094151183512 74029617275480206096 6866839239238992957052768
### Nth Root
y√n
2√n 304.562 45.2672 17.4517 9.85077
## 92758 as geometric shapes
### Circle
Diameter 185516 582816 2.70304e+10
### Sphere
Volume 3.34305e+15 1.08122e+11 582816
### Square
Length = n
Perimeter 371032 8.60405e+09 131180
### Cube
Length = n
Surface area 5.16243e+10 7.98094e+14 160662
### Equilateral Triangle
Length = n
Perimeter 278274 3.72566e+09 80330.8
### Triangular Pyramid
Length = n
Surface area 1.49026e+10 9.40563e+13 75736.6
## Cryptographic Hash Functions
md5 acbc56c14340d44104f368d3330eb135 44207ea4a36ff38bd47a486a109cfd060644532e 0903b03f15c622f9254ad10f067216e3f143f08e9f874dd2b71946acb064f8b8 e079b0d933919998fbd3d5816aa9279687ce6da41b17e0a1d0f582458375facc967e73613556f9996e881028d52beb06af9e57f6037abb478cba0ec2204befb1 d539ba6d4d6c3e51ebc09a3c170d1f2e352de763 | 1,447 | 4,126 | {"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} | 3.59375 | 4 | CC-MAIN-2021-49 | latest | en | 0.810245 |
https://docu.ngsolve.org/latest/i-tutorials/unit-7-optimization/03_Shape_Derivative_Laplace_SemiAuto.html | 1,685,626,104,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647810.28/warc/CC-MAIN-20230601110845-20230601140845-00508.warc.gz | 254,137,277 | 20,253 | # 7.3 PDE-Constrained Shape Optimization (semi-automated)¶
We want to solve the PDE-constrained shape optimization problem
$\underset{\Omega\subset \mathsf{D}}{\mbox{min}} \; J(u) := \int_\Omega |u-u_d|^q \; dx, \quad q\ge 2$
subject to that $$(\Omega,u)$$ satisfy
$\int_\Omega \nabla u \cdot \nabla v \; dx = \int_\Omega f v \; dx \; \quad \text{ for all } v \in H_0^1(\Omega),$
where $$\Omega \subset \mathbb R^2$$ for given $$u_d, \, f \in C^1(\mathbb R^2)$$.
Here, we want to compute the shape derivative by differentiation of a suitably defined perturbed Lagrangian using automated differentiation. For details we refer to
1. Gangl, K. Sturm, M. Neunteufel, J. Schöberl. Fully and Semi-Automated Shape Differentiation in NGSolve, Struct. Multidisc. Optim., 63, pp.1579-1607, 2021.
[1]:
from ngsolve import *
from ngsolve.webgui import Draw
from netgen.geom2d import SplineGeometry
[2]:
geo = SplineGeometry()
mesh = Mesh(geo.GenerateMesh(maxh = 0.08))
[3]:
#given data of our problem (chosen such that \Omega^* = [0,1]^2 is the optimal shape)
f = CoefficientFunction(2*y*(1-y)+2*x*(1-x))
ud = x*(1-x)*y*(1-y)
grad_f = CoefficientFunction( (f.Diff(x), f.Diff(y) ) )
grad_ud = CoefficientFunction( (ud.Diff(x), ud.Diff(y) ) )
## State equation¶
[4]:
fes = H1(mesh, order=2, dirichlet=".*")
u, v = fes.TnT()
gfu = GridFunction(fes)
scene = Draw(gfu, mesh, "state")
a = BilinearForm(fes, symmetric=True)
fstate = LinearForm(fes)
fstate += f*v*dx
[5]:
def SolveStateEquation():
rhs = gfu.vec.CreateVector()
rhs.data = fstate.vec - a.mat * gfu.vec
update = gfu.vec.CreateVector()
update.data = a.mat.Inverse(fes.FreeDofs()) * rhs
gfu.vec.data += update
[6]:
a.Assemble()
fstate.Assemble()
SolveStateEquation()
scene.Redraw()
We set up the adjoint equation
$\mbox{Find } p \in H_0^1(\Omega): \int_\Omega \nabla w \cdot \nabla p \, \mbox dx = - \partial_u J(u)(w) \quad \text{ for all } w \in H_0^1(\Omega)$
where $$u$$ is the solution to the state equation. For $$J(u) = \int_\Omega |u-u_d|^q \mbox dx$$, we get
$\partial_u J(u)(w) = q \int_\Omega (u-u_d)^{d-1}w \,\mbox dx.$
However, we can also use the Diff(…) command:
[7]:
q=4
def Cost(u):
return (u-ud)**q*dx
p, w = fes.TnT()
gfp = GridFunction(fes)
scene = Draw (gfp, mesh, "adjoint")
[8]:
def SolveAdjointEquation():
rhs = gfp.vec.CreateVector()
rhs.data = fadjoint.vec - a.mat.T * gfp.vec
update = gfp.vec.CreateVector()
update.data = a.mat.Inverse(fes.FreeDofs()).T * rhs
gfp.vec.data += update
[9]:
fadjoint.Assemble()
scene.Redraw()
Note that (for linear problems) the operator on the left hand side of the adjoint equation is just the transpose of the state operator.
## Automatic Shape Differentiation¶
The formula for the shape derivative was derived as the partial derivative of the perturbed Lagrangian (brought back to the original domain):
$d\mathcal J(\Omega; X) = \frac{\partial}{\partial t} \left. \left( \int_\Omega \xi(t)|u - u_d^t|^q \mbox{d} x + \int_{\Omega} (F_t^{-\top}\nabla u) \cdot (F_t^{-\top} \nabla p) \xi(t) \, dx - \int_{\Omega} \xi(t) f^t p \,dx \right) \right\rvert_{t=0}$
where
• $$T_t(x)=x+tX(x)=y$$
• $$F_t = DT_t = I+t DX$$
• $$\xi(t) = \mbox{det}(F_t)$$
• $$u_d^t = u_d \circ T_t$$
• $$f^t = f \circ T_t$$
The integrand depends on the parameter $$t$$ only via $$\xi(t)$$, $$F_t$$ and $$T_t$$. Denoting the integrand by $$G^{u,p}$$, the derivative is given by
$\begin{split}\begin{array}{rl} d\mathcal J(\Omega; X) =& \frac{d G^{u,p}}{d \xi} \frac{d \xi}{d t} + \frac{d G^{u,p}}{d F} \frac{d F}{d t} + \frac{d G^{u,p}}{dy} \cdot \frac{d T_t}{dt} \\ =& \frac{d G^{u,p}}{d \xi} \mbox{div}(X) + \frac{d G^{u,p}}{d F} DX + \frac{d G^{u,p}}{dy} \cdot X \end{array}\end{split}$
We define the perturbed PDE and cost function involving parameters $$F$$ and $$J$$ representing the Jacobian of a shape transformation and its determinant, respectively. While their values are set to unity and thus does not influence the solution of the state or adjoint equation, this procedure allows us to differentiate with respect to them.
[10]:
J = Parameter(1)
F = Id(2)
#Finv = 2*Id(2) - F # consistent linearization at F = I (avoids calling Inv() for speedup)
def Equation(u,v):
def CostAuto(u):
return J*(u-ud)**q*dx
[11]:
VEC = H1(mesh, order=2, dim=2)
PHI, X = VEC.TnT()
Lagrangian = CostAuto(gfu) + Equation(gfu,gfp)
dJOmegaAuto = LinearForm(VEC)
dJOmegaAuto += Lagrangian.Diff(J, div(X))
dJOmegaAuto += Lagrangian.Diff(x, X[0]) + Lagrangian.Diff(y, X[1])
[12]:
b = BilinearForm(VEC)
gfX = GridFunction(VEC)
[13]:
def SolveDeformationEquationAuto():
rhs = gfX.vec.CreateVector()
rhs.data = dJOmegaAuto.vec - b.mat * gfX.vec
update = gfX.vec.CreateVector()
update.data = b.mat.Inverse(VEC.FreeDofs()) * rhs
gfX.vec.data += update
[14]:
b.Assemble()
dJOmegaAuto.Assemble()
SolveDeformationEquationAuto()
Draw(-gfX, mesh, "gfX")
[14]:
BaseWebGuiScene
[15]:
# gfset denotes the deformation of the original domain and will be updated during the shape optimization
gfset = GridFunction(VEC)
gfset.Set((0,0))
mesh.SetDeformation(gfset)
sceneSet = Draw(gfset,mesh,"gfset")
SetVisualization (deformation=True)
[16]:
gfset.Set((0,0))
mesh.SetDeformation(gfset)
print('Cost at initial design', Integrate (CostAuto(gfu), mesh))
scale = 0.5 / Norm(gfX.vec)
gfset.vec.data -= scale * gfX.vec
mesh.SetDeformation(gfset)
sceneSet.Redraw()
Cost at initial design 5.152244130814951e-09
[17]:
a.Assemble()
fstate.Assemble()
SolveStateEquation()
print('Cost at new design', Integrate (CostAuto(gfu), mesh))
Cost at new design 9.793644493069302e-10
Equation can also be used to define the bilinear form. The following defines the same bilinear form as above:
[18]:
aAuto = BilinearForm(fes, symmetric=True)
aAuto += Equation(u,v)
Thus, the user has to enter the PDE (in its transformed form) only once.
Finally, let us again run the full algorithm:
[19]:
#reset to and solve for initial configuration
gfset.Set((0,0))
mesh.SetDeformation(gfset)
scene = Draw(gfset,mesh,"gfset")
SetVisualization (deformation=True)
a.Assemble()
fstate.Assemble()
SolveStateEquation()
LineSearch = False
iter_max = 600
Jold = Integrate(CostAuto(gfu), mesh)
converged = False
for k in range(iter_max):
scene.Redraw()
print('cost at iteration', k, ': ', Jold)
mesh.SetDeformation(gfset)
a.Assemble()
fstate.Assemble()
SolveStateEquation()
b.Assemble()
dJOmegaAuto.Assemble()
SolveDeformationEquationAuto()
scale = 0.01 / Norm(gfX.vec)
gfsetOld = gfset
gfset.vec.data -= scale * gfX.vec
Jnew = Integrate(CostAuto(gfu), mesh)
if LineSearch:
while Jnew > Jold and scale > 1e-12:
scale = scale / 2
if scale <= 1e-12:
converged = True
break
gfset.vec.data = gfsetOld.vec - scale * gfX.vec
mesh.SetDeformation(gfset)
a.Assemble()
fstate.Assemble()
SolveStateEquation()
Jnew = Integrate(CostAuto(gfu), mesh)
if converged==True:
break
Jold = Jnew
cost at iteration 0 : 5.152244130814932e-09
cost at iteration 1 : 4.942040315052465e-09
cost at iteration 2 : 4.5850808690470135e-09
cost at iteration 3 : 4.249439885598259e-09
cost at iteration 4 : 3.934229274496262e-09
cost at iteration 5 : 3.6385740544274783e-09
cost at iteration 6 : 3.3616132927088307e-09
cost at iteration 7 : 3.102501038477676e-09
cost at iteration 8 : 2.8604072470352507e-09
cost at iteration 9 : 2.634518692682722e-09
cost at iteration 10 : 2.424039866929757e-09
cost at iteration 11 : 2.2281938583607805e-09
cost at iteration 12 : 2.0462232096623734e-09
cost at iteration 13 : 1.877390746276767e-09
cost at iteration 14 : 1.7209803697550375e-09
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[ ]: | 14,048 | 35,310 | {"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} | 3.015625 | 3 | CC-MAIN-2023-23 | longest | en | 0.507851 |
https://mathoverflow.net/questions/312453/a-sequence-in-generalized-order-spaces | 1,550,681,257,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247495147.61/warc/CC-MAIN-20190220150139-20190220172139-00242.warc.gz | 614,850,485 | 29,404 | # A sequence in generalized order spaces
Let $$X$$ be a GO-space with the topology $$\tau$$ and $$\lambda$$ be the usual open interval topology on $$X$$. Put $$R= \{x\in X: [x, \rightarrow) \in \tau\setminus \lambda \} \text{ and } L= \{x\in X: (\leftarrow,x] \in \tau\setminus \lambda \}.$$ Define $$X^* \subset X\times \mathbb{Z}$$ as follows: $$X^*=(X\times \{0\}) \cup (R \times \{k \in \mathbb{Z}: k<0\})\cup (L \times \{k \in \mathbb{Z}: k>0\}).$$ Let $$X^*$$ have the open interval topology generated by the lexicographical order. Then $$f: X \rightarrow X^*$$ defined by $$f(x)=\langle x, 0\rangle$$ is an order-preserving homeomorphism from $$X$$ onto the closed subspace $$X\times \{0\}$$ of $$X^*$$. So the space $$X^*$$ is a closed linearly ordered extension of $$X$$.
My questions are as follows: 1. Let $$\{a_n=\langle x_n, k_n\rangle\}$$ be a sequence of $$X^*$$ and let the sequence $$\{x_n\} \subset X$$ be convergent to $$x \in X$$. Does $$\{a_n\}$$ converge to the point $$\langle x, 0\rangle$$ of $$X^*$$?
1. Suppose that $$X$$ is first countable. Is also $$X^*$$ first countable?
1. Oddly enough, no: take $$x\in R$$ and let $$a_n=\langle x,-n\rangle$$ for all $$n$$ (so $$x_n=x$$ and $$k_n=-n$$) then $$\{x_n\}$$ is constant and converges to $$x$$, but $$\{a_n\}$$ does not converge at all.
2. Yes; just consider cases, say if $$x\in R\setminus L$$ and $$\{[x,c_n):n\in\mathbb{N}\}$$ is a local base for $$X$$ at $$x$$ then $$\{(\langle x,-1\rangle,\langle c_n,0\rangle):n\in\mathbb{N}\}$$ is a local base for $$X^*$$ at $$\langle x,0\rangle$$. The extra points are isolated so they pose no problem. | 587 | 1,625 | {"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": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 39, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-09 | latest | en | 0.669318 |
https://www.knowledgemax.org/what/what-numbers-should-you-hit-on-in-blackjack/ | 1,679,434,963,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00672.warc.gz | 966,719,355 | 11,992 | # What numbers should you hit on in blackjack?
### What numbers should you hit on in blackjack?
It is common practice to hit on eight or less, but stand on anything 12 or higher. When the dealer has a three, you should hit on anything eight or below and 12, while standing on anything 13 or over. If the dealer has a two it is best to hit on nine or less and stand on anything 13 or over.
### What does D mean on blackjack chart?
The actions in a blackjack strategy chart are: H – Hit. S – Stand. Dh – Double if possible, otherwise, hit. Ds – Double if possible, otherwise stand.Dec 5, 2020
### How do you understand blackjack strategy?
– Stand when your hand is 12-16 when the dealer has 2-6.
– Hit when your hand is 12-16 when the dealer has 7-Ace.
– Always split Aces and 8s.
– Double 11 versus the dealer’s 2-10.
– Hit or double Aces-6.
### How long does it take to learn basic blackjack strategy?
The most important idea I have, though, is that you should be able to learn basic strategy in less than an hour. Yes, there are close to 200 individual situations in blackjack where you need to know the correct move. The thing is, though, that many hands in many situations should be played the same way.Jun 16, 2019
### Which blackjack strategy is best?
– STRATEGY #1: ALWAYS DOUBLE DOWN ON A HARD 11. …
– STRATEGY #2: ALWAYS SPLIT A PAIR OF 8s AND ACES. …
– STRATEGY #3: NEVER SPLIT A PAIR OF 5s OR TENS. …
– STRATEGY #4: ALWAYS HIT A HARD 12 AGAINST A DEALER’S 2 OR 3 UPCARD. …
– STRATEGY #6: ALWAYS DOUBLE DOWN ON 10 WHEN THE DEALER’S UPCARD IS 9 OR LESS.
READ What is the use of Dokan plugin?
### How do I increase my chances of winning blackjack?
– Play games with liberal playing rules.
– Learn the basic playing strategy.
– Use a strategy card.
– Avoid making the insurance wager.
– Avoid progressive strategies.
– Don’t believe you are due to win.
– Don’t play on tables that use a continuous shuffler.
### Is it better to hit or stay on 16 in blackjack?
Never hit your 16. And you’ll lose nearly 70% of the time when you hit your 16. Here’s the statistics. If you hit on your 16, you’ll win 25.23% of the time, push 5.46% of the time, and you will lose 69.31% of the time. That’s a net loss of 44.08% when you hit your 16.May 6, 2013
### Do you hit a 12 vs a 3?
Bottom line: Even though you’ll never get rich on 12 against a 3, no matter how you play it, hitting is the better play, because in the long run it will save you money compared to standing. Play #4. Not Splitting 8s Against a Dealer’s 9, 10, or Ace.
### Do you hit 13 against a 2?
14 stands against dealer 2 through 6, otherwise hit. 13 stands against dealer 2 through 6, otherwise hit.
### When should you hit on 12 Blackjack?
When holding nine or less or 12-16 it’s best to hit, but stand on a total of 17 or more. If the dealer’s card is a four, five or six it is vital you do not bust. It is common practice to hit on eight or less, but stand on anything 12 or higher. | 812 | 2,968 | {"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} | 2.703125 | 3 | CC-MAIN-2023-14 | latest | en | 0.899495 |
https://www.coursehero.com/file/212298/072108/ | 1,516,329,807,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887692.13/warc/CC-MAIN-20180119010338-20180119030338-00131.warc.gz | 876,509,304 | 35,396 | 072108
# 072108 - Solubility Rules Solubility Rules Solubility Rules...
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Solubility Rules
Solubility Equilibria
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Solubility Equilibria y For each of the following compounds, write the equation for the dissociation of the ions in water. { Ag 2 CO 3 { PbCrO 4 { Al(OH) 3 { Hg 2 Cl 2
Solubility Equilibria y A particular saturated solution of PbI 2 has[Pb 2+ ] = 5.0 x 10 -3 M and [I - ] = 1.3 x 10 -3 M. { What is the value of K sp for PbI 2 ? { What is [I - ] in a saturated solution that has [Pb 2+ ] = 2.5 x 10 -4 M? { What is [Pb 2+ ] in a saturated solution that has [I - ] = 2.5 x 10 -4 M?
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Solubility Equilibria y A 1.00 L solution saturated at 25 o C with calcium oxalate, CaC 2 O 4 , is evaporated to dryness giving a 0.0061 g residue of CaC 2 O 4 . Calculate the solubility and the solubility product constant for this salt at 25 o C.
y Which of the following salts is the most soluble in 1.0 L of water? { Mg(OH)
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Ask a homework question - tutors are online | 528 | 1,878 | {"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-2018-05 | latest | en | 0.830125 |
https://www.kidsmathplay.com/category/math-games/page/2/ | 1,624,349,698,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488512243.88/warc/CC-MAIN-20210622063335-20210622093335-00271.warc.gz | 738,648,070 | 9,380 | # Math Games
###### Penguin Jump Multiplication
game of multiplication for kids which has facts to 12. to respond to the problem given you have to click on the iceberg which has the answer that is correct.
###### Penguin Hop States and capitals
penguin hop consists of states and capitals. 4 players can play this game at a time. This is very easy for the kids to learn the states names and their capitals.
###### Owl Planes Typing Words
owl planes game is to type the words that are long. The standard which is required for the game is to have sufficient knowledge about the skills of keyboarding.
###### Otter Rush Algebraic Exponent Expressions
The otters speed depends on the speed in which you give the correct answers. 12 players can play this game at a time.
orbit integer game is to add the integers and 4 players can play this game at a time. This is a game which is for free to teach integer addition to kids.
###### Octopus Feed Homonyms
octopus feed is a game which can be played by 4 players at a time and the content of the game is homonyms. It is a game which can be played all through the world and multi users can be able to operate it from different systems.
###### Furious Frogs Antonyms
The content of Furious Frog is to learn antonyms that it is the words which are opposite to the original word and possess the opposite meaning.
###### Meteor Multiplication Facts
meteor multiplication helps the candidates to learn multiplication and practice them. The meteros along with the multiplication equations will have a shift onward to a star in the large station
You can compete when you are practising on spelling. The number of members who can play this game at a time is 8.
###### Integer Warp Multiplying
this is a game of racing which can be played with multi-players so the students can race with the other and can use it from any place of the world when they are practising integers multiplication.
Hungry Puppies Decimal Addition is providing you with an option to compete with the other players at the time when you practice for decimal addition
###### Grand Prix Multiplication
Grand Prix Multiplication is a game which can be played by 4 members at a time. The students can play this game from any place in the globe.
###### Giraffe Karts Subject Verb Agreement
Giraffe karts consist of the content as an agreement of a verb and a subject. The game can be played with four players.
###### Elephant Synonyms Feed
Elephant Feed is to learn synonyms that is the words which are similar to the original word and possess the same meaning.
###### Drag Division Race
Drag Division Race multiplayer racing race improve the division skill, and race against other player
###### Division Derby Race
Devision Derby is a game of racing which can be played by multiple players which allows the child to play from anywhere in the globe. In this game they can race with one of the other players
###### Dirt Bike Proportions
Dirt bike proportion is a game of racing which can be played by multiple players can race with one of the other players who are racing at the time of finishing the proportions of the equivalents.
###### Canoe Penguins Multiplication Race
Canoe Penguins is a game which can be played by multiple players. They have a chance to compete against each other while they are practicing to multiply the number of two digits. | 694 | 3,385 | {"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} | 2.71875 | 3 | CC-MAIN-2021-25 | latest | en | 0.963497 |
https://www.ukessays.com/essays/mathematics/definite-integral.php | 1,601,280,929,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401598891.71/warc/CC-MAIN-20200928073028-20200928103028-00426.warc.gz | 1,041,591,625 | 14,446 | # Definite integral
1460 words (6 pages) Essay
1st Jan 1970 Mathematics Reference this
Disclaimer: This work has been submitted by a university student. This is not an example of the work produced by our Essay Writing Service. You can view samples of our professional work here.
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### DEFINITE INTEGRAL
Integration is an important concept in mathematics which, together with differentiation, forms one of the main operations in calculus. Given a function ƒ of a real variable x and an interval [a, b] of the real line, the definite integral, is defined informally to be the net signed area of the region in the xy-plane bounded by the graph of ƒ, the x-axis, and the vertical lines x = a and x = b.
### APPLICATIONS OF DEFINITE INTEGRAL
Definite integrals aren’t just for area any more
1. Any definite integral may be interpreted as a signed area.
2. Area, volume, arc length, work, mass, fluid pressure, and accumulated financial value are
quantities that may be calculated with definite integrals.
3. The most important components of these problems are constructing the correct integral and
Interpreting the results.n
### TWO VIEWS OF DEFINITE INTECRAL
When using the definite integral to solve various problems, it is useful to consider two different interpretations:
1. A limit of approximating sums: The definite integral is formally defined as a limit of approximating sums using right sums.
2. Accumulated change in an antiderivative: The Fundamental Theorem of Calculus states
where F is any antiderivative of f on [a ; b].
The difference F(b) – F(a) represents the accumulated change (or net change) in F over the
interval [a; b].
To find the accumulated change in F over [a; b], integrate f, the rate function associated with
F, over the interval [a ; b].
### WHICH VIEW IS BETTER : SUM OR ANTIDERIVATIVE ?
1. Often we need to decide which view (or interpretation) of the definite integral is the correct one for a given application. It could be that an approximating sum is acceptable or that a precise symbolic antiderivative is more appropriate.
2. If an integral is presented in symbolic form, then antidifferentiation seems reasonable.
3. For data given graphically or in a table, approximating sums are the logical choice.
### TRAPEZOIDAL RULE
The trapezoidal rule (also known as the trapezoid rule, or the trapezium rule in British English) is a way to approximately calculate the definite integral
The trapezoidal rule works by approximating the region under the graph of the function f(x) as a trapezoid and calculating its area. It follows that
To calculate this integral more accurately, one first splits the interval of integration [a,b] into n smaller subintervals, and then applies the trapezoidal rule on each of them. One obtains the composite trapezoidal rule:
Illustration of the composite trapezoidal rule (with a non-uniform grid)
This can alternatively be written as:
Where
(one can also use a non-uniform grid).
The trapezoidal rule is one of a family of formulas for numerical integration called Newton–Cotes formulas. Simpson’s rule is another, often more accurate, member of the same family. Simpson’s rule and other like methods can be expected to improve on the trapezoidal rule for functions which are twice continuously differentiable; however for rougher functions the trapezoidal rule is likely to prove preferable. Moreover, the trapezoidal rule tends to become extremely accurate when periodic functions are integrated over their periods, a fact best understood in connection with the Euler–Maclaurin summation formula. For non-periodic functions, however, methods with unequally spaced points such as Gaussian quadrature and Clenshaw–Curtis quadrature are generally far more accurate; Clenshaw–Curtis quadrature can be viewed as a change of variables to express arbitrary integrals in terms of periodic integrals, at which point the trapezoidal rule can be applied accurately
### SIMPSON RULE
In numerical analysis, Simpson’s rule is a method for numerical integration, the numerical approximation of definite integrals. Specifically, it is the following approximation:
Simpson’s rule can be derived by approximating the integrand f(x) (in blue) by the quadratic interpolant P(x) (in red).
### TRAPEZOIDAL METHOD
We have n=1 , x0 =a , x1=b and h=x1-x0.
Rn= ——(1)
Using eq 1 ,the rule can be made exact for polynomial of degree upto one.For f(x)=1 and x, we get the system of equations .
f(x)= 1: x1- x0 = + or = +
f(x) = x: ½ ( – ) = +
( ) ( ) = +
h( 2 + h ) = + ( )
h( 2 + h ) = ( + ) + h = h + h
h= , or =
From the first equation , we get h – = h /2 . The method becomes
= [ f( ) + f (]
The error constant is given by
C = – [ – ] – [ – ]
= [ 2 ( + 3 h + 3 + ) -2 -3 h -3h( + 2h + ) ]
= –
### SIMPSON` S METHOD
We have n = 2 , = a , = + h , = + 2h = b , h=(b – a )/2 .We write
= f( ) + f() + f( )
The rule can be made exact for polynomials of degree upto two .
For f(x) = 1, x , , we get the following system of equations.
f(x) = 1: – = + + , or 2h = + + ———–(2)
f(x) = x: ( – ) = + + —————-(3)
f(x) = : ( – ) = + + —————–(4)
From (3) , we get
( ) ( ) = + + h) + + 2h)
(2h) (2+ 2h) = ( + + ) + ( + 2 ) h
= 2h + ( + 2 ) h
2h = + 2 ————–(5)
From (4) , we get
[( + 6 h + 12 + 8 ) – ] = + ( + 2 h + ) +
( + 4 + 2 h + ) h + )
Or h = + 4 —————(6)
Solving (5) , (6) and (2) , we obtain = , = ,
The Method is given by ………..,
= [ f() + 4 f() + f ()
The error constant is given by
C =
= –
### COMPARISON BETWEEN TRAPEZOIDAL RULE AND SIMPSON’S RULE
Two widely used rules for approximating areas are the trapezoidal rule and Simpson’s rule. To motivate the new methods, we recall that rectangular rules approximated the function by a horizontal line in each interval. It is reasonable to expect that if we approximate the function more accurately inside each interval then a more efficient numerical scheme will follow. This is the idea behind the trapezoidal and Simpson’s rules. Here the trapezoidal rule approximates the function by a suitably chosen (not necessarily horizontal) line segment. The function values at the two points in the interval are used in the approximation. While Simpson’s rule uses a suitably chosen parabolic shape (see Section 4.6 of the text) and uses the function at three points.
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The Maple student package has commands trapezoid and simpson that implement these methods. The command syntax is very similar to the rectangular approximations. See the examples below. Note that an even number of subintervals is required for the simpson command and that the default number of subintervals is n=4 for both trapezoid and simpson.
> with(student):
> trapezoid(x^2,x=0..4);
> evalf(trapezoid(x^2,x=0..4));
22
> evalf(trapezoid(x^2,x=0..4,10));
21.44000000
> simpson(x^2,x=0..4);
> evalf(simpson(x^2,x=0..4));
21.33333333
> evalf(simpson(x^2,x=0..4,10));
21.33333333
### DEFINITE INTEGRAL
Integration is an important concept in mathematics which, together with differentiation, forms one of the main operations in calculus. Given a function ƒ of a real variable x and an interval [a, b] of the real line, the definite integral, is defined informally to be the net signed area of the region in the xy-plane bounded by the graph of ƒ, the x-axis, and the vertical lines x = a and x = b.
### APPLICATIONS OF DEFINITE INTEGRAL
Definite integrals aren’t just for area any more
1. Any definite integral may be interpreted as a signed area.
2. Area, volume, arc length, work, mass, fluid pressure, and accumulated financial value are
quantities that may be calculated with definite integrals.
3. The most important components of these problems are constructing the correct integral and
Interpreting the results.n
### TWO VIEWS OF DEFINITE INTECRAL
When using the definite integral to solve various problems, it is useful to consider two different interpretations:
1. A limit of approximating sums: The definite integral is formally defined as a limit of approximating sums using right sums.
2. Accumulated change in an antiderivative: The Fundamental Theorem of Calculus states
where F is any antiderivative of f on [a ; b].
The difference F(b) – F(a) represents the accumulated change (or net change) in F over the
interval [a; b].
To find the accumulated change in F over [a; b], integrate f, the rate function associated with
F, over the interval [a ; b].
### WHICH VIEW IS BETTER : SUM OR ANTIDERIVATIVE ?
1. Often we need to decide which view (or interpretation) of the definite integral is the correct one for a given application. It could be that an approximating sum is acceptable or that a precise symbolic antiderivative is more appropriate.
2. If an integral is presented in symbolic form, then antidifferentiation seems reasonable.
3. For data given graphically or in a table, approximating sums are the logical choice.
### TRAPEZOIDAL RULE
The trapezoidal rule (also known as the trapezoid rule, or the trapezium rule in British English) is a way to approximately calculate the definite integral
The trapezoidal rule works by approximating the region under the graph of the function f(x) as a trapezoid and calculating its area. It follows that
To calculate this integral more accurately, one first splits the interval of integration [a,b] into n smaller subintervals, and then applies the trapezoidal rule on each of them. One obtains the composite trapezoidal rule:
Illustration of the composite trapezoidal rule (with a non-uniform grid)
This can alternatively be written as:
Where
(one can also use a non-uniform grid).
The trapezoidal rule is one of a family of formulas for numerical integration called Newton–Cotes formulas. Simpson’s rule is another, often more accurate, member of the same family. Simpson’s rule and other like methods can be expected to improve on the trapezoidal rule for functions which are twice continuously differentiable; however for rougher functions the trapezoidal rule is likely to prove preferable. Moreover, the trapezoidal rule tends to become extremely accurate when periodic functions are integrated over their periods, a fact best understood in connection with the Euler–Maclaurin summation formula. For non-periodic functions, however, methods with unequally spaced points such as Gaussian quadrature and Clenshaw–Curtis quadrature are generally far more accurate; Clenshaw–Curtis quadrature can be viewed as a change of variables to express arbitrary integrals in terms of periodic integrals, at which point the trapezoidal rule can be applied accurately
### SIMPSON RULE
In numerical analysis, Simpson’s rule is a method for numerical integration, the numerical approximation of definite integrals. Specifically, it is the following approximation:
Simpson’s rule can be derived by approximating the integrand f(x) (in blue) by the quadratic interpolant P(x) (in red).
### TRAPEZOIDAL METHOD
We have n=1 , x0 =a , x1=b and h=x1-x0.
Rn= ——(1)
Using eq 1 ,the rule can be made exact for polynomial of degree upto one.For f(x)=1 and x, we get the system of equations .
f(x)= 1: x1- x0 = + or = +
f(x) = x: ½ ( – ) = +
( ) ( ) = +
h( 2 + h ) = + ( )
h( 2 + h ) = ( + ) + h = h + h
h= , or =
From the first equation , we get h – = h /2 . The method becomes
= [ f( ) + f (]
The error constant is given by
C = – [ – ] – [ – ]
= [ 2 ( + 3 h + 3 + ) -2 -3 h -3h( + 2h + ) ]
= –
### SIMPSON` S METHOD
We have n = 2 , = a , = + h , = + 2h = b , h=(b – a )/2 .We write
= f( ) + f() + f( )
The rule can be made exact for polynomials of degree upto two .
For f(x) = 1, x , , we get the following system of equations.
f(x) = 1: – = + + , or 2h = + + ———–(2)
f(x) = x: ( – ) = + + —————-(3)
f(x) = : ( – ) = + + —————–(4)
From (3) , we get
( ) ( ) = + + h) + + 2h)
(2h) (2+ 2h) = ( + + ) + ( + 2 ) h
= 2h + ( + 2 ) h
2h = + 2 ————–(5)
From (4) , we get
[( + 6 h + 12 + 8 ) – ] = + ( + 2 h + ) +
( + 4 + 2 h + ) h + )
Or h = + 4 —————(6)
Solving (5) , (6) and (2) , we obtain = , = ,
The Method is given by ………..,
= [ f() + 4 f() + f ()
The error constant is given by
C =
= –
### COMPARISON BETWEEN TRAPEZOIDAL RULE AND SIMPSON’S RULE
Two widely used rules for approximating areas are the trapezoidal rule and Simpson’s rule. To motivate the new methods, we recall that rectangular rules approximated the function by a horizontal line in each interval. It is reasonable to expect that if we approximate the function more accurately inside each interval then a more efficient numerical scheme will follow. This is the idea behind the trapezoidal and Simpson’s rules. Here the trapezoidal rule approximates the function by a suitably chosen (not necessarily horizontal) line segment. The function values at the two points in the interval are used in the approximation. While Simpson’s rule uses a suitably chosen parabolic shape (see Section 4.6 of the text) and uses the function at three points.
The Maple student package has commands trapezoid and simpson that implement these methods. The command syntax is very similar to the rectangular approximations. See the examples below. Note that an even number of subintervals is required for the simpson command and that the default number of subintervals is n=4 for both trapezoid and simpson.
> with(student):
> trapezoid(x^2,x=0..4);
> evalf(trapezoid(x^2,x=0..4));
22
> evalf(trapezoid(x^2,x=0..4,10));
21.44000000
> simpson(x^2,x=0..4);
> evalf(simpson(x^2,x=0..4));
21.33333333
> evalf(simpson(x^2,x=0..4,10));
21.33333333
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Related Services | 3,742 | 14,124 | {"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} | 4.125 | 4 | CC-MAIN-2020-40 | latest | en | 0.87653 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/114/2/ | 1,713,158,467,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816942.33/warc/CC-MAIN-20240415045222-20240415075222-00353.warc.gz | 797,763,424 | 94,659 | # Properties
Label 114.2 Level 114 Weight 2 Dimension 91 Nonzero newspaces 6 Newform subspaces 21 Sturm bound 1440 Trace bound 1
## Defining parameters
Level: $$N$$ = $$114 = 2 \cdot 3 \cdot 19$$ Weight: $$k$$ = $$2$$ Nonzero newspaces: $$6$$ Newform subspaces: $$21$$ Sturm bound: $$1440$$ Trace bound: $$1$$
## Dimensions
The following table gives the dimensions of various subspaces of $$M_{2}(\Gamma_1(114))$$.
Total New Old
Modular forms 432 91 341
Cusp forms 289 91 198
Eisenstein series 143 0 143
## Trace form
$$91 q + q^{2} + q^{3} + q^{4} + 6 q^{5} + q^{6} + 8 q^{7} + q^{8} + q^{9} + O(q^{10})$$ $$91 q + q^{2} + q^{3} + q^{4} + 6 q^{5} + q^{6} + 8 q^{7} + q^{8} + q^{9} + 6 q^{10} + 12 q^{11} - 5 q^{12} - 34 q^{13} - 28 q^{14} - 30 q^{15} + q^{16} - 18 q^{17} + q^{18} - 65 q^{19} - 30 q^{20} - 34 q^{21} - 42 q^{22} - 12 q^{23} + q^{24} - 41 q^{25} - 22 q^{26} - 14 q^{27} - 4 q^{28} - 6 q^{29} + 6 q^{30} - 4 q^{31} + q^{32} - 42 q^{33} + 18 q^{34} - 24 q^{35} + q^{36} + 2 q^{37} + 19 q^{38} - 40 q^{39} + 6 q^{40} + 6 q^{41} + 8 q^{42} - 28 q^{43} + 12 q^{44} + 6 q^{45} + 24 q^{46} + 12 q^{47} + 10 q^{48} + 21 q^{49} + 31 q^{50} + 99 q^{51} + 14 q^{52} + 54 q^{53} + 55 q^{54} + 72 q^{55} + 8 q^{56} + 109 q^{57} + 30 q^{58} + 60 q^{59} + 42 q^{60} + 50 q^{61} + 32 q^{62} + 74 q^{63} + q^{64} + 48 q^{65} + 84 q^{66} - 4 q^{67} + 18 q^{68} + 78 q^{69} + 48 q^{70} + 36 q^{71} + 10 q^{72} + 20 q^{73} + 38 q^{74} - 11 q^{75} + 19 q^{76} + 24 q^{77} - 22 q^{78} - 76 q^{79} + 6 q^{80} - 71 q^{81} - 30 q^{82} - 60 q^{83} - 46 q^{84} - 36 q^{85} - 28 q^{86} - 150 q^{87} + 12 q^{88} - 54 q^{89} - 84 q^{90} - 80 q^{91} - 48 q^{92} - 118 q^{93} - 96 q^{94} + 6 q^{95} + q^{96} - 10 q^{97} - 87 q^{98} - 69 q^{99} + O(q^{100})$$
## Decomposition of $$S_{2}^{\mathrm{new}}(\Gamma_1(114))$$
We only show spaces with even parity, since no modular forms exist when this condition is not satisfied. Within each space $$S_k^{\mathrm{new}}(N, \chi)$$ we list the newforms together with their dimension.
Label $$\chi$$ Newforms Dimension $$\chi$$ degree
114.2.a $$\chi_{114}(1, \cdot)$$ 114.2.a.a 1 1
114.2.a.b 1
114.2.a.c 1
114.2.b $$\chi_{114}(113, \cdot)$$ 114.2.b.a 2 1
114.2.b.b 2
114.2.b.c 2
114.2.b.d 2
114.2.e $$\chi_{114}(7, \cdot)$$ 114.2.e.a 2 2
114.2.e.b 2
114.2.h $$\chi_{114}(65, \cdot)$$ 114.2.h.a 2 2
114.2.h.b 2
114.2.h.c 2
114.2.h.d 2
114.2.h.e 4
114.2.h.f 4
114.2.i $$\chi_{114}(25, \cdot)$$ 114.2.i.a 6 6
114.2.i.b 6
114.2.i.c 6
114.2.i.d 6
114.2.l $$\chi_{114}(29, \cdot)$$ 114.2.l.a 18 6
114.2.l.b 18
## Decomposition of $$S_{2}^{\mathrm{old}}(\Gamma_1(114))$$ into lower level spaces
$$S_{2}^{\mathrm{old}}(\Gamma_1(114)) \cong$$ $$S_{2}^{\mathrm{new}}(\Gamma_1(19))$$$$^{\oplus 4}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(38))$$$$^{\oplus 2}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(\Gamma_1(57))$$$$^{\oplus 2}$$ | 1,404 | 2,855 | {"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} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.272623 |
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# IF it is true that x > -2 and x < 7, which of the
Author Message
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Director
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IF it is true that x > -2 and x < 7, which of the [#permalink] 03 Sep 2005, 05:05
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IF it is true that x > -2 and x < 7, which of the following must be true?
A. x > 2
B. x > -7
C. x < 2
D. -7 < x < 2
E. None of the above
_________________
Cheers, Rahul.
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No it is B.
x>-7
The range is -2 < x < 7
A and C split that range so it can't be right.
D has numbers other way round.
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E.
the values of x will be: -2<x<7
(-6 is greater than -7 but x can't be -6, so D is wrong)
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Re: PS Inequalities [#permalink] 03 Sep 2005, 20:10
rahulraao wrote:
IF it is true that x > -2 and x < 7, which of the following must be true?
A. x > 2
B. x > -7
C. x < 2
D. -7 < x < 2
E. None of the above
hmmm.... i thought i have answered it. Anyway, B is correct.
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B says x>-7, even 10 is greater than -7, but 10 cannot be the value ox x
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E should be the right answer.
Since x> -2 and x<7, x can be any number between these two values but x cannot be more than 7 or less than -2.
a) x>2 includes values that are more than 7, so it is not correct
b) x> -7 incudes values that are less than -2, so it is not correct
c) x< 2 includes values that can be less -2, so it is not always correct
d)x > -7 is inconsistent with the statement that x> -2
e) using poe E is the correct answer
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B is absolutely right. Rests are wrong.
It says "IF it is true that x > -2 and x < 7". In other words, -2<x<7.
THEREFORE, x logically is GREATER than -7, for all x.
D is wrong with x=5, for instance.
Hope this help.
tibeo - vietnam
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acid_burn wrote:
E should be the right answer.
Since x> -2 and x<7, x can be any number between these two values but x cannot be more than 7 or less than -2.
a) x>2 includes values that are more than 7, so it is not correct
b) x> -7 incudes values that are less than -2, so it is not correct
c) x< 2 includes values that can be less -2, so it is not always correct
d)x > -7 is inconsistent with the statement that x> -2
e) using poe E is the correct answer
This is a classical question and classical mistake in answering this kind of questions.
When it says x<7 then it MUST be true that x is less than 8, or 9, etc. But x>-2, so it must be true that it is greater than -3, -4 etc. X may or may not be less than 6 if all we know is that it is less then 7. It may or may not be less than -1 if all we know is that it is greater than -2.
If the question is asked differently, then answer would be different. For example if the question is like this:
Which of the following statement will ensure x is less than 7 and greater than -2?
Then you would need to look at it from another perspective. If x is less than 7, 6 or 5 etc., then we would be able to ensure x is less than 7. If x is greater than -2, -1 or 0 etc., then we would be able to ensure x is less than -2.
Basically all we are deriving at is if one (A) is true, then the other (B) is always true. But we just need to know which one is A and which one is B.
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Guys,
Nice explanations! I really liked HongHu's logic! Thanks man!
OA is B
OA reasoning is almost in line with HongHu's logic, but rather unclear!
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# Question about proc survey select
I have a data set with 15 observation hours for each of my subjects. I'm trying to use survey select to generate new data sets with decreasing numbers of hours, that I'm then comparing to the total hours. (i.e. what is the correlation between 14 and 15 hours? 13 and 15? 12 and 15? etc.) I have 2 strata, basically age and subject. My problem is that I'd like to keep the same set of hours for each subject, and I can't figure out how to do that. For example, when I generate a set containing 3 hours out of the 15, if the hours for subject 1 are 8, 10, and 12 then I want the hours for subject 2 (and all others) to also be 8, 10 and 12. Is there any way to do this?
Posts: 3,852
## Re: Question about proc survey select
> I have a data set with 15 observation hours for each
> of my subjects.
Does the data set produced by the following code model your data?
[pre]
proc plan seed=618029071;
factors
subjid = 10 ordered
agegroup = 1 of 3 random
time = 15 ordered
y1 = 1 of 200
/ noprint;
treatments y2=1 of 50 y3=1 of 30;
output out=plan;
run;
quit;
proc print;
run;
[/pre]
> I'm trying to use survey select to
> generate new data sets with decreasing numbers of
> hours, that I'm then comparing to the total hours.
> (i.e. what is the correlation between 14 and 15
> hours? 13 and 15? 12 and 15? etc.) I have 2 strata,
> basically age and subject. My problem is that I'd
> like to keep the same set of hours for each subject,
> and I can't figure out how to do that. For example,
> when I generate a set containing 3 hours out of the
> 15, if the hours for subject 1 are 8, 10, and 12 then
> I want the hours for subject 2 (and all others) to
> also be 8, 10 and 12. Is there any way to do this?
I don't think SURVERSELECT is going to work well this. SURVEYSELECT selects observations from data sets. It sounds like you want to select levels of a variable (TIME). In your example.
[pre]where time in(8,10,12) [/pre]
There are a number of ways to select (k of n) values at random. There is CALL RANPERK
[pre]
CALL RANPERK Routine
Randomly permutes the values of the arguments, and returns a permutation of k out of n values [/pre]
Also PROC PLAN in the FACTORS statement. I used this above to create sample data.
[pre]
name=m < OF n > < selection-type >
where
name
is a valid SAS name. This gives the name of a factor in the design.
m
is a positive integer that gives the number of values to be selected. If n is specified, the value of m must be less than or equal to n.
n
is a positive integer that gives the number of values to be selected from.
[/pre]
There are others, these are the ones I'm most familiar with.
If I am correct the details of which method(s) might be most appropriate depend on the output you desire. You mentioned correlation. If you describe (with sample data) how the data should look to produce the analysis this will help refine the solution.
Also, do you want to do this for all (n of m) subsets and do you want replication? That is replications of subsets of size n.
Not applicable
Posts: 0
## Re: Question about proc survey select
This is what my data looks like (very similar to what you posted)
proc plan;
factors
age = 3 ordered
subjid = 5 ordered
hour = 15 ordered
y1 = 1 of 200
/ noprint;
output out=plan;
run;
quit;
proc print;
run;
I'm having some trouble getting sample data output to look like what I want as the end result, but this is close:
proc plan;
factors
age = 3 ordered
subjid = 5 ordered
hour = 14 ordered
y1 = 1 of 200
/ noprint;
output out=sample1;
run;
quit;
proc print;
run;
What I actually want is a random selection of 14 out of the original 15 hours, instead of hours 1 to 14. My snag is that I want the same hours for each age and subject, so that the output would look like the above if hour 15 was the hour that was randomly chosen to be thrown out.
Another way I've thought to do this is to copy the original data 10 times and add a replication column, so that my input data would look something like:
proc plan;
factors
age = 3 ordered
subjid = 5 ordered
rep = 10 ordered
hour = 15 ordered
/ noprint;
output out=sample2;
run;
quit;
proc print;
run;
I would then use a random number generator to select the hours that I want to keep and have SAS delete all the other hours with
data sample2;
set sample2;
if rep = 1 and hour = 10 then delete;
if rep = 2 and hour = 5 then delete;
if rep = 3 and hour = 14 then delete;
run;
(Obviously, with the real data I would continue so it included all 10 replicates.)
The problem with this is that it gets very time consuming, since I want to do this, not just for 14 hours, but with 13 hours, 12 hours, 11 hours, all the way down to 1 hour.
I hope that made things clearer and not more confusing. If you have advice for either of these methods I'd really appreciate it.
Posts: 3,852
## Re: Question about proc survey select
I think the following may give you what you want. I used RANPERK to get a list of hours to KEEP. I generated IF statements with a data step because it seemed easy enough to do it that way. Writing wallpaper with code.
This produced a sample for each K from 14 to 1 by -1 the variable K will act as a BY variable in your analysis.
Let me know if this does what you want.
[pre]
dm 'clear log; clear output;';
proc plan seed=767318578;
factors
age = 3 ordered
subjid = 5 ordered
hour = 15 ordered
y1 = 1 of 200
/ noprint;
output out=plan;
run;
quit;
*proc print;
run;
filename FT85F001 temp;
data hours;
file FT85F001;
seed=1046482356;
array _h[15] (1:15);
do k = dim(_h)-1 to 1 by -1;
put +3 k= ';';
put +3 'if hour in(' @;
call ranperk(seed,k,of _h
• );
do _n_ = 1 to k;
put _h[_n_] 3. @;
end;
put ') then output;';
end;
run;
data hoursV / view=hoursV;
set plan;
%inc FT85F001 / source2;
run;
proc sort data=hoursV out=hours;
by descending k;
run;
proc print data=_last_(obs=100);
run;
[/pre]
• Not applicable
Posts: 0 | 1,669 | 5,931 | {"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} | 2.9375 | 3 | CC-MAIN-2018-30 | latest | en | 0.93948 |
https://forums.bohemia.net/forums/topic/62956-quick-way-to-test-if-point-is-inside-marker/ | 1,611,068,448,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519395.23/warc/CC-MAIN-20210119135001-20210119165001-00188.warc.gz | 345,457,144 | 25,544 | # Quick way to test if point is inside marker
## Recommended Posts
Hi
Is there any way to quickly test if a point is located anywhere within a marker? Rectangle or ellipse at any sizes and angle.
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You can create a trigger covering the same area, and use the list command.
It should be reasonably easy to create a script function that checks the given position against the marker's position, shape, angle and area, but I don't know about you, but my maths is a bit rusty
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I have code to do this for rectangular and elliptical markers which I use for a minefield script. One annoying problem is that there is no way to detect whether a marker is rectangular or elliptical (ie there is no 'markerShape' function) so you need to know in advance which function you are going to call.
For a rectangle the basic principle is to get the relative position of the point of interest from the center of the rectangle, rotate the resulting point by -1 times the angle of the rectangle, then test the x and y coordinates against the a and b dimensions of the rectangle.
For ellipses, the quickest method is to precompute the position of the two focii. Do this by calculating the distance of the focii from the center point
d= sqrt (b^2 - a^2)
The focii (F1,F2) lie at points d away from the center, in the direction the ellipse is rotated.
From then on , a point is inside the ellipse if distance from P to F1 added to distance from P to F2 is less than 2a.
I'll post code here over the weekend if you are interested.
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Quick way for all - no. Quick way for some - yes.
Check the UPS script, it does all these checks for any marker shape and angle.
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Yes, I thought about using triggers. But then I'd have to create an object and check its existence (?) within the trigger. Seemed a way over the top
I have managed to create random points within rectangular offsetted area at an angle, but still not a way to check. I guess the maths is similar, but yeah, maths skills deterioate quickly I guess
Naturally I have the ups, I'll check that one. Thanks for pointing this one out.
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Since I didn't know how to check the ellipse, I found out, and did it. My maths is pretty bad these days and they have not been checked for either logic or scripting mistakes, but they seem to work in my testing
Two functions, one for rectangles, one for elipses (elipsii?). They return true if given position is inside the marker area, otherwise false. Feel free to do anything with these, I'm sure they could be improved upon. Example of use:
<table border="0" align="center" width="95%" cellpadding="0" cellspacing="0"><tr><td>Code Sample </td></tr><tr><td id="CODE">insideMarkerR = compile preprocessFile "insideMarkerR.sqf";
_inMarker = [getPos player, "testMarker"] call insideMarkerR;
insideMarkerR.sqf
<table border="0" align="center" width="95%" cellpadding="0" cellspacing="0"><tr><td>Code Sample </td></tr><tr><td id="CODE">
//returns true if given position is within the given rectangular marker
//all angles are math style, cause I'm stupid :)
//rectangle version
private ["_posi","_markerName","_markerPos","_markerSize","_markerDir","_relDist","_relAngle","_markAngle"];
//_position is a 3d position array
_posi = _this select 0;
//marker name must be a string
_markerName = _this select 1;
//3d position array
_markerPos = markerPos _markerName;
_markerSize = markerSize _markerName;
//change dir to maths dir
_markerDir = 90 -(markerDir _markerName);
//distance between marker center and position
_relDist = (_markerPos distance _posi);
//check if distance is greater than 1.42 (approx sqrt 2)
if
((_relDist > (1.42*(_markerSize select 0))) && (_relDist > (1.42*(_markerSize select 1))))
exitWith
{false};
//
if
((_relDist < _markerSize select 0) && (_relDist < _markerSize select 1))
exitWith
{true};
//angle between 0 and a line between marker and posi
_relAngle = (90 - ((_posi select 0) - (_markerPos select 0)) atan2 ((_posi select 1) - (_markerPos select 1)));
if
(_relAngle > 180)
then
{_relAngle = -(360 - _relAngle)};
//angle between marker's direction and a line between marker and posi
_markAngle = 90 - (_markerDir - _relAngle);
if
(_markAngle > 180)
then
{_markAngle = -(360 - _markAngle)};
/*//sidechat feedback used in testing
player sideChat format ["rel angle is %1",_relAngle];
player sideChat format ["mark angle is %1",_markAngle];
player sideChat format ["relX is %1, relY is %2",(_relDist*(cos _markAngle)), (_relDist*(sin _markAngle))];*/
if
((abs (_relDist*(cos _markAngle)) < _markerSize select 0) && (abs (_relDist*(sin _markAngle)) < _markerSize select 1) )
exitWith
{true};
//if not inside, must be outside!
false
insideMarkerR.sqf
<table border="0" align="center" width="95%" cellpadding="0" cellspacing="0"><tr><td>Code Sample </td></tr><tr><td id="CODE">
//returns true if given position is within the given eliptical marker
//all angles are math style (north = +90), cause I'm stupid :)
//elipse version
private ["_posi","_markerName","_markerPos","_markerSize","_markerDir","_relDist","_relAngle","_markAngle"];
//_position is a 3d position array
_posi = _this select 0;
//marker name must be a string
_markerName = _this select 1;
//3d position array
_markerPos = markerPos _markerName;
_markerSize = markerSize _markerName;
//change dir to maths dir
_markerDir = 90 -(markerDir _markerName);
//distance between marker center and position
_relDist = (_markerPos distance _posi);
//check if distance is greater than large axis
if
(_relDist > ((_markerSize select 0) max (_markerSize select 1)))
exitWith
{false};
//check if distance is less than smaller axis
if
(_relDist < ((_markerSize select 0) min (_markerSize select 1)))
exitWith
{true};
//angle between 0 and a line between marker and posi
_relAngle = (90 - ((_posi select 0) - (_markerPos select 0)) atan2 ((_posi select 1) - (_markerPos select 1)));
if
(_relAngle > 180)
then
{_relAngle = -(360 - _relAngle)};
//angle between marker's direction and a line between marker and posi
_markAngle = 90 - (_markerDir - _relAngle);
if
(_markAngle > 180)
then
{_markAngle = -(360 - _markAngle)};
/*sidechat feedback used in testing
player sideChat format ["rel angle is %1",_relAngle];
player sideChat format ["mark angle is %1",_markAngle];
player sideChat format ["relX is %1, relY is %2",(_relDist*(cos _markAngle)), (_relDist*(sin _markAngle))];
player sideChat format["checkSum is %1", (((_relDist*(cos _markAngle))*(_relDist*(cos _markAngle)))/((_markerSize select 0)*(_markerSize select 0))) + (((_relDist*(sin _markAngle))*(_relDist*(sin _markAngle)))/((_markerSize select 1)*(_markerSize select 1)))]; */
//check if posi is in elipse. equation was found on wikipedia
if
((((_relDist*(cos _markAngle))*(_relDist*(cos _markAngle)))/((_markerSize select 0)*(_markerSize select 0))) + (((_relDist*(sin _markAngle))*(_relDist*(sin _markAngle)))/((_markerSize select 1)*(_markerSize select 1))) < 1)
exitWith
{true};
//if not inside, must be outside!
false
'love those brackets
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Thanks for the reply. But I've spent most of my working day with excel today, and I came up with this. It does appear to work just from quick testing in init.sqf on the mission. Easy to rewrite for more global use though I think. Here is the result:
<table border="0" align="center" width="95%" cellpadding="0" cellspacing="0"><tr><td>Code Sample </td></tr><tr><td id="CODE">
sleep 0.1;
//Begin Script
_px = position player select 0;
_py = position player select 1;
_mpx = getMarkerPos "mTest" select 0;
_mpy = getMarkerPos "mTest" select 1;
_msx = getMarkerSize "mTest" select 0;
_msy = getMarkerSize "mTest" select 1;
_ma = -markerDir "mTest";
_xmin = _mpx - _msx;
_xmax = _mpx + _msx;
_ymin = _mpy - _msy;
_ymax = _mpy + _msy;
//Now, rotate point to investigate around markers center in order to check against a nonrotated marker
_rpx = Â ( (_px - _mpx) * cos(_ma) ) + ( (_py - _mpy) * sin(_ma) ) + _mpx;
_rpy = (-(_px - _mpx) * sin(_ma) ) + ( (_py - _mpy) * cos(_ma) ) + _mpy;
//Now for the actual test against markersize.
hint format ["px %1\npy %2\nmpx %3\nmpy %4\nmsx %5\nmsy %6\nma %7\nxmin %8\nxmax %9\nymin %10\nymax %11\nrpx %12\nrpy %13", _px, _py, _mpx, _mpy, _msx, _msy, _ma, _xmin, _xmax, _ymin, _ymax, _rpx, _rpy];
if (((_rpx > _xmin) && (_rpx < _xmax)) && ((_rpy > _ymin) && (_rpy < _ymax))) then
{
  player globalChat "player inside marker";
}
else
{
  player globalChat "player outside marker";
};
//End of Script
And here is the end of the script but for elliptical markers, after rotated points _rpx and _rpy has been calculated:
<table border="0" align="center" width="95%" cellpadding="0" cellspacing="0"><tr><td>Code Sample </td></tr><tr><td id="CODE">
_res = (((_rpx-_mpx)^2)/(_msx^2)) + (((_rpy-_mpy)^2)/(_msy^2));
if ( _res < 1 ) then
{
player globalChat "player inside marker";
}
else
{
player globalChat "player outside marker";
};
Not quite sure how much parantheses is really needed.
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I've written this into function form. Maybe someone could test if it works properly? Does *seem* to work for me though. Have included enough comments for newbies to understand what is going on.
The point of it all is to check if a point; object or an icon  marker, lies inside a rectangular or elliptical area marker of any size and angle. In order to reduce the complexity of the angle maths, I rotate the point to inventigate the oposite amount of degrees the marker is rotated, about the markers center point. I needed this because I need No-Fire-Zones for certain types of munitions for my artillery in my mission, and these zones are marked with, well, area markers
The function:
<table border="0" align="center" width="95%" cellpadding="0" cellspacing="0"><tr><td>Code Sample </td></tr><tr><td id="CODE">
// call with check = ["markerpoint", "markertocheckagainst", "type"] call markerCheck; for markers
// call with check = [object, "markertocheckagainst", "type"] call markerCheck; for objects
// object is an object, i.e. player or uTankNumber3. "markerpoint" can be used for automatically generated markers, such as "mTargetMarker"
// "markertocheckagainst" is the name of the marker as a string that is the area you want to check against.
// Type is a string indicating "R" or "E" to enable rectangular or elliptical checks respectively.
//variables needs to be set private for function to work
private ["_p","_m", "_t", "_px", "_py", "_mpx", "_mpy", "_msx", "_msy", "_rpx", "_rpy", "_xmin", "_xmax", "_ymin", "_ymax", "_ma", "_res", "_ret"];
//description of variable names
//_p   point to investigate, typically an object or marker with a position
//_m   marker to check against
//_t   marker type, string "R" or "E" for rectangular or elliptical markers respectively
//_px  x value of point
//_py  y value of point
//_mpx  markerpos x
//_mpy  markerpos y
//_msx  markersize x
//_msy  markersize y
//_rpx  rotated position point x
//_rpy  rotated position point y
//_xmin minimum point x in a rectangular marker
//_xmax maximum point x in a rectangular marker
//_ymin minimum point y in a rectangular marker
//_ymax maximum point y in a rectangular marker
//_ma  marker angle
//_res  result carrier in elliptic marker
//_ret  return value
_p = _this select 0; //object
_m = _this select 1; //always a marker
_t = _this select 2; //marker shape, "E" or "R"
if (typeName _p == "OBJECT") then
{
  _px = position _p select 0;
  _py = position _p select 1;
}
else
{
  _px = getMarkerPos _p select 0;
  _py = getMarkerPos _p select 1;
};
_mpx = getMarkerPos _m select 0;
_mpy = getMarkerPos _m select 1;
_msx = getMarkerSize _m select 0;
_msy = getMarkerSize _m select 1;
_ma = -markerDir _m;
_rpx = ( (_px - _mpx) * cos(_ma) ) + ( (_py - _mpy) * sin(_ma) ) + _mpx;
_rpy = (-(_px - _mpx) * sin(_ma) ) + ( (_py - _mpy) * cos(_ma) ) + _mpy;
if (_t == "R") then
{
  _xmin = _mpx - _msx;
  _xmax = _mpx + _msx;
  _ymin = _mpy - _msy;
  _ymax = _mpy + _msy;
  if (((_rpx > _xmin) && (_rpx < _xmax)) && ((_rpy > _ymin) && (_rpy < _ymax))) then
  {
    _ret=true;
  }
  else
  {
    _ret=false;
  };
}
else
{
  _res = (((_rpx-_mpx)^2)/(_msx^2)) + (((_rpy-_mpy)^2)/(_msy^2));
  if ( _res < 1 ) then
  {
    _ret=true;
  }
  else
  {
    _ret=false;
  };
};
_ret;
In the init.sqf:
<table border="0" align="center" width="95%" cellpadding="0" cellspacing="0"><tr><td>Code Sample </td></tr><tr><td id="CODE">
markerCheck = compile preprocessFile "checkinsidemarker.sqf";
Check with following:
<table border="0" align="center" width="95%" cellpadding="0" cellspacing="0"><tr><td>Code Sample </td></tr><tr><td id="CODE">
check = ["mTest", "mR001_AB", "E"] call markerCheck; //check if marker lies inside an elliptical marker called mR001_AB
check = [player, "mR001_AB", "R"] call markerCheck; //check if player lies inside a rectangular marker called mR001_AB
Function returns true or false. Yippi, my first ever function Â
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I know that this thread is very old but it's still relevant and Nr. 1 displayed in google :)
The above code snippets can't be read and are outdated. Here a new improved version of it:
``` // call with check = ["markerpoint", "markertocheckagainst"] call fnc_isInMarker; for markers
// call with check = [object, "markertocheckagainst"] call fnc_isInMarker; for objects
fnc_isInMarker = {
private ["_p","_m", "_px", "_py", "_mpx", "_mpy", "_msx", "_msy", "_rpx", "_rpy", "_xmin", "_xmax", "_ymin", "_ymax", "_ma", "_res", "_ret"];
_p = _this select 0; // object
_m = _this select 1; // marker
if (typeName _p == "OBJECT") then {
_px = position _p select 0;
_py = position _p select 1;
} else {
_px = getMarkerPos _p select 0;
_py = getMarkerPos _p select 1;
};
_mpx = getMarkerPos _m select 0;
_mpy = getMarkerPos _m select 1;
_msx = getMarkerSize _m select 0;
_msy = getMarkerSize _m select 1;
_ma = -markerDir _m;
_rpx = ( (_px - _mpx) * cos(_ma) ) + ( (_py - _mpy) * sin(_ma) ) + _mpx;
_rpy = (-(_px - _mpx) * sin(_ma) ) + ( (_py - _mpy) * cos(_ma) ) + _mpy;
if ((markerShape _m) == "RECTANGLE") then {
_xmin = _mpx - _msx;_xmax = _mpx + _msx;_ymin = _mpy - _msy;_ymax = _mpy + _msy;
if (((_rpx > _xmin) && (_rpx < _xmax)) && ((_rpy > _ymin) && (_rpy < _ymax))) then { _ret=true; } else { _ret=false; };
} else {
_res = (((_rpx-_mpx)^2)/(_msx^2)) + (((_rpy-_mpy)^2)/(_msy^2));
if ( _res < 1 ) then{ _ret=true; }else{ _ret=false; };
};
_ret;
};```
Cheers and happy X-Mas
Edited by Armitxes
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Here's a quick method using locations (call as func):
```private ["_pos","_mark","_inside","_shape","_loc"];
_pos = _this select 0; _mark = _this select 1;
_inside = false;
_shape = markerShape _mark;
if (! (_shape == "ICON")) then {
_loc = createLocation (["Name", markerPos _mark] + (markerSize _mark));
if (_shape == "RECTANGLE") then { _loc setRectangular true };
_loc setDirection (markerDir _mark);
if (_pos in _loc) then { _inside = true };
deleteLocation _loc;
};
_inside
```
×
• Clubs | 4,616 | 15,390 | {"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-2021-04 | latest | en | 0.91505 |
https://ictbyte.com/b-sc-csit/bsc-csit-3rd-semester-dsa-old-questions/ | 1,657,172,137,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683683.99/warc/CC-MAIN-20220707033101-20220707063101-00101.warc.gz | 337,392,767 | 56,539 | # BSc.CSIT 3rd Semester Data Structure and Algorithms Old Questions
0
1437
## Subject: Data Structure and Algorithms
### Year: 2065 B.Sc.CSIT 3rd Semester Data Structure and Algorithms Old Questions
Attempt any TWO questions: (10×2=20)
1. What do you mean by binary tree? Explain the binary search tree with example.
2. What do you mean by recursion? Explain the implementation of factorial and Fibonacci sequences
with example.
3. Explain the implementation of stack and queue with example.
Section B
Attempt any EIGHT questions: (8×5=40)
1. What are the difference between two dimension array and multidimension array?
2. What are the major characteristics of algorithms?
3. How can you convert from infix to post fix notation?
4. How can you use Queue as ADT?
5. What is Post-order traversal?
6. What is sorting? Describe the Insertion.
7. Explain the binary searching.
8. Differentiate between Pre-order and In order traversal.
9. Explain the tower of Hanoi algorithm.
10. Explain the Kruskal‟s algorithm.
### B.Sc.CSIT 3rd Semester Data Structure and Algorithms Old Questions of Year : 2066
Attempt any TWO questions. (10×2=20)
1. Write a menu program to demonstrate the simulation of stack operations in array implementation.
2. State relative merits and demerits of contiguous list and Linked list. Explain the steps involved in
inserting and deleting a mode in singly linked list.
3. A binary tree T has 12 nodes. The in-order and pre-order traversals of T yield the following
sequence of nodes:
In-order : VPNAQRSOKBTM
Pre-order : SPVQNARTOKBM
Construct the Binary tree T showing each step. Explain, how you can arrive at solution in brief?
Section B
Attempt any EIGHT questions. (8×5=40)
1. Consider the function:
Void transfer (int n, char from, char to, char temp)
{ if (n > 0)
{ transfer ( n – 1, from, temp, to_;
Print if ( “In Move Disk % d from % C to % C” N, from, to);
Transfer ( n – 1, temp, to, from);
}
Trace the output with the function Cell!
Transfer ( 3, „R‟, „L‟, „C‟);
1. “To write an efficient program, we should know about data structures.” Explain the above
statement.
2. Write C function to display all the items in a circular queue in array implementation. Write
assumptions, you need.
3. Explain Divide and Conquer algorithm taking reference to Merge Sort.
1. Trace Binary Search algorithm for the data:
21, 36, 56, 79, 101, 123, 142, 203
And Search for the values 123 and 153.
2. Differentiate between tree and graph. What are spanning forest and spanning tree. Explain MST
(Minimum cost Spanning Tree) problem.
3. A file contains 100 symbols in which following character with their probability of occurrence.
Build a Huff man tree according to Greedy Strategy.
a 48
b 11
c 9
d 14
e 7
f 11
4. Explain the use of Big O notation in analyzing algorithms. Compare sorting time efficiencies of
Quick-Sort and Merge-Sort.
5. Explain CLL, DLL, DCLL (Circular, Doubly, Doubly Circular Linked List).
6. Write Short notes on (any two):
a) Hash function
b) External Sorting
### B.Sc.CSIT 3rd Semester Data Structure and Algorithms Old Questions of Year : 2067
Attempt any two questions. (2×10=20)
1. Define stack as ADT. Describe its primitive operations on Array implementation and linked list
implementation.
2. Describe properties of Binary Search Tree. Write recursive algorithms for constructing BST and its traversals.
Illustrate them with an example.
3. What are external and internal sorting? Explain partition strategies of Merge sort and Quick sort.
Trace these sort algorithms for following data:
11 45 61 33 55 9 83 25
Section B
Attempt any eight questions. (8×5=40)
1. Write recursive algorithm to get Fibonacci term. Illustrate it drawing recursion tree.
2. Construct an expression tree from the following postfix:
AB + C*DC – -FG + \$
3. Differentiate between Singly linked list, DLL, CLL and DCLL.
4. Describe circular Queue operations in array implementation.
5. Compare and Contrast between Binary searching and Binary tree searching.
6. State collision resolution techniques in hashing. Explain double hashing and quadratic probing techniques.
7. State MST (Minimum Cost Spanning Tree) problem and shortest path (single source and all other
destination) problem. Name the algorithms for solving these problems.
8. Justify the statement: “To write an efficient program, we should know about data structures and
algorithms”.
9. Discuss the merits and demerits of contiguous list and linked list.
10. What is priority queue? How it is best implemented?
### Year : 2068 B.Sc.CSIT 3rd Semester Data Structure and Algorithms Old Questions
Attempt any two questions:
1. Define Queue as an ADT. Write a program for basic operations in Linear queue in array implementation.
2. Why recursion is required? Explain with Tower-of-Hanoi example. How recursive algorithm makes program
effective? Write the merits and demerits of recursion in Programming.
3. Explain In-fix to Postfix Conversion Algorithm. Illustrate it with an example. What changes should be made for
converting postfix to prefix.
Section B
Attempt any eight questions: (8×5=40)
1. Explain Kruskal’s algorithm with example.
2. Write a program in C for bubble sorting.
3. Differentiate between contiguous list and linked list with examples.
4. Explain binary search. Illustrate it with example.
5. Explain hashing with example.
6. Explain why linked list is called dynamic list? Write the algorithm for deleting a new node before a node.
7. Explain the characteristics of Huffman’s algorithm and its application.
8. Write merits and demerits of recursive function over non-recursive function.
9. Write the steps involved in deleting a node in a Binary selection tree.
10. Discuss merge sort. How you rate this sorting from selection sort?
### B.Sc.CSIT 3rd Semester Data Structure and Algorithms Old Questions of Year : 2069
Attempt any two questions:
1. Define Queue as ADT. Describe its primitive operation on array implementation and linked list
implementation.
2. Describe the significance of Huffman tree. Describe procedure for construction of a Huffman tree. Illustrate
it with example. Describe different types of applications of Binary trees.
3. Explain the algorithms for infix to postfix conversion and evaluation of postfix expression. Trace the
algorithms with suitable example.
Section (B)
Attempt any eight questions:
(8×5=40)
1. State TOH problem. Write recursion tree when no. of disks are four.
2. Write about applications of Binary trees.
3. Compare partition strategies of Merge sort and Quick sort.
4. Explain Bubble sort algorithm. Illustrate it with an example.
5. How do you insert a nodes at last in doubly linked list? Explain.
6. Describe recursive procedure of Binary searching technique? Discuss about efficiency of Binary searching.
7. What are Hashing and collision? Write about any three hashing algorithms.
8. What is Big ‘O’ notation? Analyze any one sorting algorithm.
9. Describe strong and weekly connected graphs with examples. What is weighted graph?
10. State relative merits & demerits of contiguous list and linked list.
### Year : 2070 B.Sc.CSIT Data Structure and Algorithms Old Questions
Attempt any two questions:
1. Trace out Infix to Postfix conversion algorithm with given Infix expression.
A + (((B-C) * (D-E) + F)/G) \$ (H-I)
Evaluate the postfix expression acquired from above for the given values:
A = 6, B = 2, C = 4, D = 3, E = 8, F = 2, G = 3, H = 5, I = 1.
2. Explain the structure of Doubly Linked List (DLL). Differentiate the difference between DLL and Doubly Circular
Linked List (DCLL). Explain the procedures to insert a node in DLL at the beginning and at the last.
3. Define Binary Search Type (BST). Write an algorithm to insert a node in non-empty BST. Construct BST from
the data:
10, 20, 30, 25, 27, 7, 4, 23, 26, 21.
Section B
Attempt any eight questions: (5×8=40)
1. Write C function to insert an item circular queue in array implementation. Write assumptions, you need.
2. What is an algorithm? What is to analyze in algorithm? Define Big C = Oh notation for time complexity
measurement of algorithm.
3. State TOH problem. Explain a recursive algorithm to solve the problem.
4. Trace selection – sort algorithm for the following data:
42, 23, 74, 11, 65, 58, 94, 86
5. What is Hashing? What collision means? State collision resolution techniques. Explain one of them in brief.
6. What is weighted graph? Explain Depth-first traversal of a graph.
7. Create a Huffman tree for the following set of data:
Characters a b c d e f
Probability 48 13 11 16 07 05
Encode 0 101 100 111 1101 1100
### Year : 2071 B.Sc.CSIT 3rd Semester Old Questions
Attempt any two questions:
1. What is stack? How is it different from queue? Write a program to implement all stack operations.
2. What is linked list? Explain the process of inserting and removing nodes from a linked list.
3. What is graph traversal? Discuss depth-first traversal technique with suitable example.
Section (B)
Attempt any eight questions:
(8×5=40)
1. Discuss array as an ADT.
2. Transform the postfix expression AB − C + DEF − + \$ to infix.
3. What is recursion? Write a recursive program to find factorial of a number.
4. Explain almost complete binary tree with example.
5. Write a program to sort an array using selection sort.
6. Discuss binary search technique along with its efficiency.
7. Why do we need Hashing? Discuss linear probing in detail.
8. How to find complexity of algorithms? Explain.
9. Hand test the insertion sort algorithm with following array of numbers.
16 7 31 2 9 41 -10
1. Write short notes on:
a. Tree traversal
b. Circular queue
### Year : 2072 Old Questions of B.Sc.CSIT 3rd Semester Data Structure and Algorithms
Attempt any TWO questions: (2×10=20)
1. What is binary search tree? Explain with an example. Write an algorithm to search, insert and delete node in
binary search tree.
2. What is Postfix expression? Write an algorithm to evaluate value of postfix expression. Trace the following
expression into postfix expression.
(A+B*C) D E/ F) + −
3. What is circular queue? Write an algorithm and C function to implement Circular queue.
Section B
Attempt any EIGHT questions: (8×5=40)
1. What is Recursion? Write a recursive algorithm to implement binary search.
2. Differentiate between array and pointer with example.
3. What is an algorithm? Write down the features of an algorithm.
4. How stack as ADT? Explain with example.
5. Write an algorithm and C function to delete node in singly link list.
6. Write an algorithm and C function for merge sort.
7. What do you mean by graph traversal? Explain primes algorithm with example.
8. Differentiate between selection sort and bubble sort.
9. Write an algorithm to implement tower of Hanoi.
10. Write short notes on
a) Hashing | 2,588 | 10,788 | {"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} | 3.671875 | 4 | CC-MAIN-2022-27 | longest | en | 0.804181 |
https://math.stackexchange.com/questions/2474366/p-is-irreducible-in-mathbb-z-n-if-and-only-if-p2-divides-n/2474427 | 1,717,063,383,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059632.19/warc/CC-MAIN-20240530083640-20240530113640-00498.warc.gz | 313,558,978 | 37,155 | # $p$ is irreducible in $\mathbb Z_n$ if and only if $p^2$ divides $n$.
I'm reading Gallian's Contemporary Abstract Algebra and I came across this question:
Let $$p$$ be a prime divisor of a positive integer $$n$$. Prove that $$p$$ is irreducible in $$\mathbb Z_n$$ if and only if $$p^2$$ divides $$n$$.
Here's my attempt at a proof:
$$(\leftarrow)$$: Suppose that $$p^2\mid n$$ but $$p$$ is reducible in $$\mathbb Z_n$$. That is, $$p \equiv ab \pmod{n}$$ for some non-units $$a,b\in\mathbb Z_n$$. So we can write $$ab = p + kn$$ for some integer $$k$$. This means $$p|a$$ or $$p|b$$. Without loss of generality, we may suppose $$p|a$$, in which case $$p(1) \equiv p(jb) \pmod{n}$$ for some integer $$j$$. I'd like to conclude from this that $$b$$ is a unit but the problem is that $$\mathbb Z_n$$ need not be an Integral Domain, so the cancellation law need not hold.
For the other direction I'm not even sure where to begin.
Any help would be greatly appreciated,
Thanks.
($$\Rightarrow$$): If $$p^2\nmid n$$, then $$(p,\frac{n}{p})=1$$, so there are $$a,b\in\mathbb{Z}$$ such that $$ap+b\frac{n}{p}=1,$$ i.e., $$p=ap^2+bn,$$ hence $$p\equiv ap^2\pmod{n}.$$ Note that both $$ap$$ and $$p$$ are not invertible in $$\mathbb{Z}_n$$, a contradiction.
($$\Leftarrow$$): Suppose $$p\equiv ab\pmod{n}$$ and $$p|a$$, then $$1-ub\equiv 0\pmod{\frac{n}{p}}$$ where $$u=\frac{a}{p}$$. Since $$p|\frac{n}{p}$$, $$1-ub\equiv 0\pmod{p},$$therefore $$(1-ub)^2\equiv 0\pmod{n},$$ i.e., $$1\equiv -b(u^2b-2u)\pmod{n}$$, hence $$b$$ is a unit.
Lemma: If $$a, b, c \in \mathbb{Z}^+$$, $$\gcd(a, b) = 1$$ and $$a|bc$$, then $$a|c$$.
Suppose first that $$p^2|n$$. Let $$p = bc$$ in $$\mathbb Z_n$$. Suppose that $$p|b$$, then $$p\not \mid c$$. Otherwise $$p^2|bc$$. Then $$\exists k ,s, r \in \mathbb{Z}$$ such that $$p^2s = bc = p + kn = p + kp^2r$$. In particular, $$p^2|p$$, which is impossible.
Let $$d := \gcd(c, n)$$, then $$\gcd(p, d) = 1$$ by the argument above.
Since $$d|bc$$, $$d|p + kp^2r = p(1 + kpr)$$. By our Lemma, $$d|1 + kpr$$. Similarly, $$d|n = p^2r$$ gives that $$d|r$$. Putting these two equalities together, one get $$d|1$$. Hence $$d = 1$$ and $$c$$ is a unit.
Conversely, let $$p$$ be an irreducible and suppose for contradiction that $$p^2\not \mid n$$. Then $$p \not \mid \frac{n}{p}$$. In particular, $$\gcd(p, \frac{n}{p}) = 1$$. Hence $$\exists a, b \in \mathbb{Z}$$ s.t $$ap + bn/p = 1$$. ie, $$ap^2 + bn = 1$$. Then we have $$p(ap + bk) = 1$$, where $$n = pk$$ for some $$k$$. Since $$p$$ is not invertible in $$\mathbb Z_n$$, this is a contradiction.
• In $\mathbb Z_n$ the elements are residue classes modulo $n$, not numbers. I suggest to write $p\equiv bc\bmod n$. Jun 5, 2021 at 7:01 | 1,013 | 2,716 | {"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": 76, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-22 | latest | en | 0.711807 |
https://matholympiad.org.bd/forum/viewtopic.php?f=15&t=6333&p=23053&sid=395f26b11400dfc849b692d9610bea91 | 1,627,195,504,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151638.93/warc/CC-MAIN-20210725045638-20210725075638-00201.warc.gz | 390,487,928 | 6,246 | ## APMO 2021 P4 - There exists a good subset consisting of at least $666$ cells
Discussion on Asian Pacific Mathematical Olympiad (APMO)
Anindya Biswas
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### APMO 2021 P4 - There exists a good subset consisting of at least $666$ cells
Given a $32\times32$ table, we put a mouse (facing up) at the bottom left cell and a piece of cheese at several other cells. The mouse then starts moving. It moves forward except that when it reaches a piece of cheese, it eats a part of it, turns right, and continues moving forward. We say that a subset of cells containing cheese is good if, during this process, the mouse tastes each piece of cheese exactly once and then falls off the table. Show that:
1. No good subset consists of $888$ cells.
2. There exists a good subset consisting of at least $666$ cells.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann | 255 | 1,019 | {"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} | 2.765625 | 3 | CC-MAIN-2021-31 | latest | en | 0.904902 |
http://www.mathworks.se/help/physmod/sps/powersys/ref/idealswitch.html?nocookie=true | 1,404,951,440,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776400808.24/warc/CC-MAIN-20140707234000-00089-ip-10-180-212-248.ec2.internal.warc.gz | 245,354,141 | 8,950 | Accelerating the pace of engineering and science
# Documentation Center
• Trial Software
# Ideal Switch
Implement ideal switch device
## Library
Power Electronics
## Description
The Ideal Switch block does not correspond to a particular physical device. When used with appropriate switching logic, it can be used to model simplified semiconductor devices such as a GTO or a MOSFET, or even a power circuit breaker with current chopping. The switch is simulated as a resistor Ron in series with a switch controlled by a logical gate signal g.
The Ideal Switch block is fully controlled by the gate signal (g > 0 or g = 0). It has the following characteristics:
• Blocks any forward or reverse applied voltage with 0 current flow when g = 0
• Conducts any bidirectional current with quasi-zero voltage drop when g > 0
• Switches instantaneously between on and off states when triggered
The Ideal Switch block turns on when a positive signal is present at the gate input (g > 0). It turns off when the gate signal equals 0 (g = 0).
The Ideal Switch block also contains a series Rs-Cs snubber circuit that can be connected in parallel with the ideal switch (between nodes 1 and 2).
## Dialog Box and Parameters
Internal resistance Ron
The internal resistance of the switch device, in ohms (Ω). The Internal resistance Ron parameter cannot be set to 0.
Initial state
The initial state of the Ideal Switch block. The initial status of the Ideal Switch block is taken into account in the steady-state calculation.
Snubber resistance Rs
The snubber resistance, in ohms (Ω). Set the Snubber resistance Rs parameter to inf to eliminate the snubber from the model.
Snubber capacitance Cs
The snubber capacitance in farads (F). Set the Snubber capacitance Cs parameter to 0 to eliminate the snubber, or to inf to get a resistive snubber.
Show measurement port
If selected, add a Simulink® output to the block returning the ideal switch current and voltage.
## Inputs and Outputs
g
Simulink signal to control the opening and closing of the switch.
m
The Simulink output of the block is a vector containing two signals. You can demultiplex these signals by using the Bus Selector block provided in the Simulink library.
Signal
Definition
Units
1
Ideal switch current
A
2
Ideal switch voltage
V
## Assumptions and Limitations
The Ideal Switch block is modeled as a current source. It cannot be connected in series with an inductor, a current source, or an open circuit, unless its snubber circuit is in use.
Use the Powergui block to specify either continuous simulation or discretization of your electrical circuit containing ideal switches. When using a continuous model, the ode23tb solver with a relative tolerance of 1e-4 is recommended for best accuracy and simulation speed.
## Example
The power_switchpower_switch example uses the Ideal Switch block to switch an RLC circuit on an AC source (60 Hz). The switch, which is initially closed, is first opened at t = 50 ms (3 cycles) and then reclosed at t = 138 ms (8.25 cycles). The Ideal Switch block has 0.01 ohms resistance and no snubber is used.
Run the simulation and observe the inductor current, the switch current, and the capacitor voltage. Notice the high-frequency overvoltage produced by inductive current chopping. Note also the high switch current spike when the switch is reclosed on the capacitor at maximum source voltage.
## References
[1] Mohan, N., T.M. Undeland, and W.P. Robbins, Power Electronics: Converters, Applications, and Design, John Wiley & Sons, Inc., New York, 1995. | 793 | 3,587 | {"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-2014-23 | latest | en | 0.842483 |
http://jwgranit.pl/13qwv06e/7a3aa0-mass-number-for-the-tantalum-isotope-with-109-neutrons | 1,623,562,151,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487600396.21/warc/CC-MAIN-20210613041713-20210613071713-00380.warc.gz | 30,268,647 | 18,946 | Yugioh Tag Force 6, California Civil Code 880, Application Of Physics In Biochemistry, Tomato Production Guide, Flashing Check Engine Light, Taste Of Home Phone Number, " />
# Niepowtarzalne wzory nagrobków oraz pomników blisko miasta Łomży z graniu oraz marmuru
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# mass number for the tantalum isotope with 109 neutrons
Posted by on Gru 30, 2020 in Bez kategorii | 0 comments
The three isotopes of carbon can be referred to as carbon-12 (C 6 12), carbon-13 (C 6 13), and carbon-14 (C 6 14). The name carbon-14 tells us that this isotope's mass number is #14#. So a Muslim bizarre total number of neutrons So from the information given so as we can extract the information that the mass number is 90 91 92 94 or 96 respectively. To the nearest whole number, isotopic mass = mass number for a specific isotope. In a typical sample of carbon-containing material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. 4 Related Records Expand this section. Define the mass number of an isotope. Element Atomic Number Most Common Isotopes Atomic Mass Mass Number (of most common isotope) Number of (in one atom of the most common isotope): Electrons Protons Neutrons Titanium 22 46Ti, 47Ti, 48Ti, 49Ti, 47.9 Calcium 20 40Ca, 42Ca, 44Ca 40.1 Tantalum 73 180Ta, 181Ta 180.9 4 Self-Check Activity b. the calcium isotope with 22 neutrons. Atoms with the same atomic number but different mass numbers are isotopes of the same element. For example, carbon-12 , carbon-13 , and carbon-14 are three isotopes of the element carbon with mass numbers 12, 13, and 14, respectively. ANSWER: Correct Exercise 4.96 Silicon has three naturally occurring isotopes: with mass 27.9769 and a natural abundance of 92.21%, with mass 28.9765 and a natural abundance of 4.69%, and with mass 29.9737 and a natural … The atomic mass of these isotopes is 34.969 u and 36.966 u respectively. The number of protons (atomic number) for two different isotopes is identical. a. the hydrogen isotope with 2 neutrons. 1 Structures Expand this section. Add the mass of protons and neutrons to compute the atomic mass of a single atom of an element. If an element only has one isotope, relative atomic mass = relative mass of this isotope. Tantalum carbide (TaC) Tantalum chloride (TaCl 5) Tantalum nitride (TaN) Tantalum pentoxide (Ta 2 O 5) Tantalum telluride (TaTe 2) Interesting facts: It is highly corrosion-resistant. What is the atomic number for the following isotope: the calcium isotope with 22 neutrons. Carbon … Express your answer as an integer. Isotopes can be divided into two groups due to their stability: stable and unstable (otherwise known as radioactive). For our boron example, 11 (atomic mass) – 5 (atomic number) = 6 neutrons; Advertisement . C) The calcium isotope with 22 neutrons. Thorium 234. Periodic table is arranged by atomic number, number of protons in the nucleus which is same as number of electrons. a) What is the atomic number and the mass number for the following isotope: the hydrogen isotope with 2 neutrons. The numerical difference between the actual measured mass of an isotope and A is called either the mass excess or the mass defect (symbol Δ; see table). The symbol for an isotope lists the mass number as a superscript at the upper left. For hydrogen, 1.008 is closer to 1 than 2, so let's call it 1. For instance, it can be determined experimentally that neon consists of three isotopes: neon-20 (with 10 protons and 10 neutrons in its nucleus) with a mass of 19.992 amu and an abundance of 90.48%, neon-21 (with 10 protons and 11 neutrons) with a mass of 20.994 amu and an abundance of 0.27%, and neon-22 (with 10 protons and 12 neutrons) with a mass of 21.991 amu and an abundance of 9.25%. Thorium, isotope of mass 234. All fluorine atoms have a mass of 19 (19 F), therefore its relative atomic mass is 19 and no calculation is needed. DTXSID90891767. Potassium is a chemical element with atomic number 19 which means there are 19 protons in its nucleus.Total number of protons in the nucleus is called the atomic number of the atom and is given the symbol Z.The total electrical charge of the nucleus is therefore +Ze, where e (elementary charge) equals to 1,602 x 10-19 coulombs. b) What is the atomic number and the mass number for the following isotope: the chromium isotope with 28 neutrons. Atomic mass of Tantalum is 180.9479 u. Determine the number of protons in Express your answer as an integer. The net charge of an ion will appear as a small superscript number following the element. Isotopes are atoms that have the same atomic number but different mass numbers due to a change in the number of neutrons. the atomic number) from the atomic mass will give you the calculated number of neutrons in the atom. Atomic number is the number of protons. An ion is an atom that has a positive or negative charge due to the addition or removal of electrons. Subtract the atomic number from the atomic mass. Identify the element. Determine the atomic number and mass number for each isotope. The atoms of two different isotopes have different numbers of neutrons in the atomic nucleus. Atomic Number – Protons, Electrons and Neutrons in Potassium. 5 Use and Manufacturing Expand this section. From the periodic table you can see that carbon has an atomic number of 6, which is its proton number. Therefore you can directly look for atomic number 73 to find Tantalum on periodic table. Isotopes. For this reason, it is still the same chemical element. For 12 C the atomic mass is exactly 12u, since the atomic mass unit is defined from it. 131 I. b. A) Atomic number 1, Mass number 3 B) Atomic number 24, Mass number 52 C) Atomic number 20, Mass number 42 D) Atomic number 73, Mass number 182. [1] Number of protons and neutrons/nucleons in an atom/isotope; b. For example, 63 Cu (29 protons and 34 neutrons) has a mass number of 63 and an isotopic mass … It is easily fabricated and a good conductor of heat and electricity. D) The tantalum isotope with 109 neutrons. a. Consider an example of chlorine, which has two naturally-occurring isotopes: 35 Cl and 37 Cl. atoms of the same element with the same number of protons but different numbers of neutrons (have different mass numbers) (Ex: 12C, 13C, 14C) Atomic mass. c. the tantalum isotope with 109 neutrons. ANSWER: Correct Part H Determine the number of neutrons in Express your answer as an integer. Contents. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation $_\text{Z}^\text{A}\text{X}$, where X is the symbol for the element, A is the mass number, and Z is the atomic number (for example, $_6^{14}\text{C}$). The number of nucleons (both protons and neutrons) in the nucleus is the atom's mass number, and each isotope of a given element has a different mass number. 60. And then this podcast follows on to another to podcasts that kind of surround the same sort of information. [1] Zinc/Zn; c. State what processes P and Q are on the diagram. g. What is the atomic number for the following isotope: the tantalum isotope with 109 neutrons. Naturally occurring samples of most elements are mixtures of isotopes. Now write the isotopic notation for carbon-14. The mass number of an isotope is the total number of protons and neutrons in an atomic nucleus. For other isotopes, the isotopic mass usually differs and is usually within 0.1 u of the mass number. #""_6^14"C"# We can determine the number of neutrons as #14-6=8# neutrons. c) What is the atomic number and the mass number for the following isotope: the calcium isotope with 22 neutrons. 1. 2. Astatine-211 has a significantly higher energy than the previous isotope, because it has a nucleus with 126 neutrons, and 126 is a magic number corresponding to a filled neutron shell. a. Atomic Number of Tantalum. Molecular Weight: 234.0436 g/mol. Despite having a similar half-life to the previous isotope (8.1 hours for astatine-210 and 7.2 hours for astatine-211), the alpha decay probability is much higher for the latter: 41.81% against only 0.18%. A) The hydrogen isotope with 2 neutrons. Determine the atomic number and mass number for each isotope. f. What is the mass number for the isotope above? An isotope of any element can be uniquely represented as $${}_Z^{A}X$$where X is the atomic symbol of the element. Identify the net charge. F. Be. An element has a mass number of 68 and 38 neutrons. 2 Names and Identifiers Expand this section. Tantalum is a chemical element with atomic number 73 which means there are 73 protons and 73 electrons in the atomic structure. B) The chromium isotope with 28 neutrons. 10 mass number. Discovery of 155 192Ta Thirty-eight tantalum isotopes from A = 155{192 have been discovered so far; these include 1 stable, 26 neutron-de cient and 11 neutron-rich isotopes. So, what you need to do is round the atomic weight to the nearest whole number to get a mass number for your calculations. Example: Find the atomic mass of a carbon isotope which has 7 neutrons. Because of this variance in the neutron number, an atom of the same element can have a different atomic mass. 59. State the number of protons, neutrons, and electrons in neutral atoms of each isotope. Carbon-14 tells us that this isotope 's mass number for a specific isotope ends at right. Of neutrons in Potassium on to another to podcasts that kind of surround the same can... By atomic number 118 ) but different mass numbers are isotopes of the mass number is.... Heat and electricity neutrons, and electrons in neutral atoms of two different isotopes is identical charge an... – 5 ( atomic number ) from the periodic table starts at top (. Nucleus has 6 protons and neutrons in Potassium is a chemical element with atomic number and mass number the... State the number of protons, neutrons, then its mass number for the following isotope: the chromium with. ) and ends at bottom right ( atomic number and the mass number of neutrons in nucleus..., we mass number for the tantalum isotope with 109 neutrons that carbon has an atomic nucleus, isotopic mass usually differs and is usually within 0.1 of! It is the same element book \Mass Spectra and isotopes '' [ 13 ] numbers of in! Podcasts that kind mass number for the tantalum isotope with 109 neutrons surround the same element we can determine the number of 68 and neutrons... S book \Mass Spectra and isotopes '' [ 13 ] # neutrons then its mass number for each.! The symbol for tantalum is Ta.. atomic mass of a carbon isotope which has 7.! 37 Cl periodic table starts at top left ( atomic number is 1 is an atom the. As number of protons and 73 electrons in neutral atoms of two different isotopes is 34.969 u and u... The nucleus which is same as number of protons in the neutron number, an atom that has mass... 118 ) that has a positive or negative charge due to the whole... 73 which means there are 73 protons and neutrons in an atom/isotope ;.!: the tantalum isotope with 22 neutrons we see that carbon has an atomic number ) = neutrons! [ 1 ] number of neutrons in the nucleus of an atom 6 which! 6 protons and neutrons in Potassium 6 protons and neutrons in Potassium is identical which has two naturally-occurring isotopes 35. Is defined from it Find the atomic mass of a single atom the... Starts at top left ( atomic number and mass number is 12 number Submit Request answer Part atomic. 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Different atomic mass ) – 5 ( atomic number but different mass numbers are of. Is exactly 12u, since the atomic mass ) – 5 ( atomic mass = mass for! Are on the diagram ; c. state What processes P and Q are on the diagram of isotope! = mass number for each isotope the masses of all naturally occurring samples of most are... Isotopes have different numbers of neutrons in the neutron number, number of neutrons in Express your as... Mass usually differs and is usually within 0.1 u of the number of protons 73... Us that this isotope 's mass number for the following isotope: the tantalum isotope with 109 Express! Neutrons/Nucleons in an atom/isotope ; b follows on to another to podcasts that kind of surround same! 11 ( atomic number ( number of 6, which has two naturally-occurring isotopes: 35 Cl and Cl... 35 Cl and 37 Cl be divided into two groups due to their stability stable. Of a carbon isotope which has 7 neutrons its proton number same can. And 7 neutrons the following isotope: the chromium isotope with 2 neutrons will... Add the mass of protons in mass number for the tantalum isotope with 109 neutrons atom s book \Mass Spectra and isotopes [! The second edition of Aston ’ s book \Mass Spectra and isotopes '' [ ]. Give you the calculated number of protons in Express your answer as an integer neutrons, and electrons in atoms! Directly look for atomic number and the mass of mass number for the tantalum isotope with 109 neutrons ) = 6 neutrons and! Mass numbers are isotopes of an ion is an atom that has mass. 73 electrons in neutral atoms of two different isotopes have different numbers of neutrons in the atomic =. And a good conductor of heat and electricity and Q are on the diagram # we can the! Ions Present a small superscript number following the element and Q are on diagram! # '' C '' # we can determine the number of protons in atomic... Of these isotopes is 34.969 u and 36.966 u respectively a single atom of ion... Is 12 whole number, an atom of the same for mass number for the tantalum isotope with 109 neutrons isotopes of an atom:... Numbers are isotopes of the masses of all naturally occurring isotopes in an element Express your as! Chemical element the electrons with Ions Present 73 to Find tantalum on table! All isotopes of the same chemical element with atomic number of neutrons #! Easily fabricated and a good conductor of heat and electricity of tantalum has one isotope, relative atomic mass give! Of surround the same element stable and unstable ( otherwise known as radioactive ) example: Find atomic. An atom that has a mass number of an element as an.... That the atomic mass of these isotopes is identical in an atom/isotope ; b this. The same for all isotopes of the same chemical element same for all isotopes of the number of in. 22 neutrons you can see that the atomic structure have different numbers of neutrons in Potassium an number! Isotope: the hydrogen isotope with 109 neutrons for this reason, it is the sum of same! Single atom of an element has a positive or negative charge due to the addition removal. [ 1 ] number of protons and mass number for the tantalum isotope with 109 neutrons to compute the atomic unit... Isotopes can be divided into two groups due to the addition or of... ] number of protons, neutrons, then its mass number is 1 What processes P and are... Average of the same element Request answer Part G atomic number 73 means. Is identical has two naturally-occurring isotopes: 35 Cl and 37 Cl ends bottom... To podcasts that kind of surround the same element can have a different atomic mass of this variance the... Is the atomic mass ) – 5 ( atomic number 118 ) Zinc/Zn ; c. state What processes P Q... Which has 7 neutrons, then its mass number as a superscript at the upper left all isotopes the! ) from the periodic table you can directly look for atomic number is mass number for the tantalum isotope with 109 neutrons... Most elements are mixtures of isotopes tells us that this isotope 's mass.! On to another to podcasts that kind of surround the same sort of information are mixtures of isotopes of variance! A superscript at the upper left is an atom that has a mass number the! Table starts at top left ( atomic number and mass number for the following:... 14 # the chemical symbol for an isotope is the atomic number ) for different... ] Zinc/Zn ; c. state What processes P and Q are on the diagram same... Chromium isotope with 2 neutrons element - so for hydrogen atomic number and mass... And then this podcast follows on to another to podcasts that kind of the... C '' # carbon-14 tells us that this mass number for the tantalum isotope with 109 neutrons 's mass number the... Part mass number for the tantalum isotope with 109 neutrons determine the atomic number and the mass number for each isotope 14-6=8 neutrons! Numbers of neutrons in the atomic number for the following isotope: the tantalum isotope with 109 neutrons Express answer... Left ( atomic number – protons, neutrons, then its mass number for the isotope above left ( number... | 4,559 | 18,477 | {"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} | 2.59375 | 3 | CC-MAIN-2021-25 | latest | en | 0.703385 |
http://math.stackexchange.com/questions/71564/simplifying-a-series | 1,466,863,694,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783393332.57/warc/CC-MAIN-20160624154953-00157-ip-10-164-35-72.ec2.internal.warc.gz | 199,618,358 | 17,039 | # Simplifying a series
I have a series like this:
$$1 + \frac 1 {x^2} + \frac 1 {x^4} + \frac 1 {x^6} ....$$
Is this a known series? Can I simplify this to something?
Thanks.
-
It is a geometric series. – Phira Oct 10 '11 at 22:25
Your series is just a geometric series, $\sum_{n = 0}^{\infty} x^{-2n} = \sum_{n = 0}^{\infty} (x^{-2})^n$. – Adrián Barquero Oct 10 '11 at 22:27
If $|x|\gt 1$ your series will converge. You can put $y=x^2$ and it is a standard geometric series
You mean $|x| > 1$. – Robert Israel Oct 10 '11 at 22:29 | 208 | 537 | {"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} | 3.1875 | 3 | CC-MAIN-2016-26 | longest | en | 0.813356 |
https://www.physicsforums.com/threads/complex-integration-branch-cuts.389027/ | 1,653,595,814,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662625600.87/warc/CC-MAIN-20220526193923-20220526223923-00342.warc.gz | 1,087,193,116 | 18,554 | # Complex integration, branch cuts
Hi guys,
I need to show that:
$$\int_{0}^{\infty } \frac{x^{a}}{(x+1)^2} \dx = \frac{\pi a}{\sin(\pi a)}$$
, where -1<a<1.
The problem is, that although a hint is given,the path of integrating it, I have difficulty what they really mean with "cut line, branch points, multivalued functions" etc.
In the hint , the path of integration is the following, which I don't understand why they have chosen it like this, could some explain their reasoning? For example, why do they exclude z=-1?
http://img143.imageshack.us/img143/151/acf1.jpg [Broken]
EDIT:
I have now calculated this integral using my own , simpler, contour, which is basically the same as this one, except it includes the point z=-1 ( Which I still don't understand why my book excludes..).
By the way; the whole idea of branch cuts etc basically means that if you wounded one time around the origin, you have to use z=r*exp(i(theta+2pi)); is this right? I calculated using this and I got the correct answer..
Last edited by a moderator:
z = -1 does not need to be excluded.
There are two different approaches to attack these sorts of problems. The first approach invloves carefully chosing the branch cut, in the second approach you can more or less ignore the precise details about the branch cuts.
First, you need to understand what branch points and branch cuts are. You should study that from your complex analysis books. A simple explanation is that branch points are points where the function is non-analytic, such that moving around that point leads the value of the function to jump.
This then means that you have to account for this when developing complex function theory rigorously. The function z^a clearly has a branch point at z = 0. What this means is that you cannot apply the residue theorem as long as z = 0 is inside the contour.
Another thing is that whenever a point in the comlex plane is specified, you need to know what the value of the function is using some unambiguous prescription. But if going around the origin changes the value of the function, specifying that e.g. f(z) = z^a is not enough.
You need to be able to distinguish between a point z and the "same point" when you have gone around the origin. That you can do by introducing polar coordinates. At this point you have two main choices to proceed.
The first choice is to work with single valued functions. Thgis means that you have to prevent the function from jumping, so when going around the origin at some polar angle, you set the polar angle back by letting it jump by minus 2 pi. This then defines the so-called branch cut, it can be chosen arbitrarily.
Now, when you do contour integration, you can only apply the residue theorem if the function is analytic. This then means that not only do you have to make sure that there are no branch points inside thecontour, the contour also cannot intersect any branch cuts (as the function jumps there, so it discontinuous and thus not analytic). In your case, this means that you have to choose the branch cut along the positive real axis.
Another approach is to choose the branch cut in some arbitrary way and then to analytically continue the function across the branch cut. You chop up the contour in two pieces, so that the integral along the first contour integrates the original function avoiding its branch cut, while the second contour integral is along the second contour of the analytic continuation which has its branch cut chosen such that it doesn't interect the second contour (it will then intersect the first contour).
Now this analytic continuation and chopping up of the contour doesn't have to be actually performed. The argument is that you can do it, therefore you don't have to bother with the minutia of choosing the branch cut along the positive real axis and then having to infinitessimally avoiding it.
A more instructive problem involving branch points and cuts is to evaluate:
$$\int_{0}^{1}\sqrt[3]{x^2 - x^3}dx$$
Sorry to prolong this thread longer than possibly necessary, but how much complex analysis would one need to know to understand the tools used here? I'm fairly comfortable with power series and metric topology, but the extent to which I know complex analysis is basically an equivalence statement regarding the Cauchy-Riemann equations (i.e., I don't know anything).
Sorry to prolong this thread longer than possibly necessary, but how much complex analysis would one need to know to understand the tools used here? I'm fairly comfortable with power series and metric topology, but the extent to which I know complex analysis is basically an equivalence statement regarding the Cauchy-Riemann equations (i.e., I don't know anything).
Then you are infinitesimally close to understanding the tools used here. If f(z) is a complex differentiable function, then:
$$\oint f(z)dz=0$$
where the contour integral is over any arbitrary contour (Cauchy's theorem). The proof is rather tedious. Once you have arrived at this point, the rest is pretty much trivial. It is easy to show that:
f(a) = $$\frac{1}{2 \pi i}\oint \frac{f(z)}{z-a}dz$$ (1)
where the point a is inside the contour, by applying Cauchy's theorem to the contour that after almost moving around the point z= a moves toward a, encircles it and moves back (see the figure posted by the OP where z = 0 is removed in a similar way). The z = a is no longer in the contour, so the contour integral is zero as the integrand is complex differentiable everywhere inside the contour. You can compute the contour integral with a small radius around the singularity at z = a in the limit that the radius shrinks to zero, which leads to the result.
Eq. (1) (a.k.a. Cauchy's integral theorem) implies that complex differentiable functions are analytic. The proof is quite simple. It involves studying the derivative w.r.t. a and then noting that what you get is also complex differentiable. Since Eq. (1) is valid for any complex differentiable function, this means that complex differentiable functions are infinitely often differentiable. That f(z) is analytic also follows as a by product of this proof.
The residue theorem pretty much follows in a similar was as Eq. (1). If you consider functions that are analytic except for poles where they behave like 1/(z-a)^n for some integer n, then you can deal with that by removing the pole as in the proof of Eq. (1). The integral over a small circle encircling z= a of 1/(z-a)^n is zero unless n = 1, in which case it is 2 pi i. he resdue theorem that says that the contour integral is 2 pi i times the sum of all residues at the poles, where residue is at pole z = a is the coefficient of 1/(z-a), then follows.
That is an awesome explanation.
hey guys, i need to do the following contour integration...
\int_{-\infty}^{\infty} dz f(z)/\sqrt(z*z+a*a)
where f(z) is a well behaved function in the complex plane.
The problem is , how to choose the contour for the evaluation of this integration....
You need to start a new thread. I think resurrecting an old thread like this is called "necroposting" and is not allowed.
i was not aware of the fact...so my apologies!!!
Bacle2 | 1,625 | 7,180 | {"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} | 3.671875 | 4 | CC-MAIN-2022-21 | latest | en | 0.955535 |
https://www.geeksforgeeks.org/print-all-nodes-at-distance-k-from-given-node-iterative-approach/amp/?ref=rp | 1,604,067,793,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910815.89/warc/CC-MAIN-20201030122851-20201030152851-00128.warc.gz | 735,950,595 | 22,949 | # Print all nodes at distance K from given node: Iterative Approach
Given a binary tree, a target node and an integer K, the task is to find all the nodes that are at distance K from the given target node.
Consider the above-given Tree, For the target node 12.
Input: K = 1
Output: 8 10 14
Input: K = 2
Output: 4 20
Input: K = 3
output: 22
Approach:
There are generally two cases for the nodes at a distance of K:
1. Node at a distance K is a child node of the target node.
2. Node at a distance K is the ancestor of the target node.
The idea is to store the parent node of every node in a hash-map with the help of the Level-order traversal on the tree. Then, Simply Traverse the nodes from the Target node using Breadth-First Search on the left-child, right-child, and the parent node. At any instant when the distance of a node the from the target node is equal to K then print all the nodes of the queue.
Below is the implementation of the above approach:
// C++ implementation to print all // the nodes from the given target // node iterative approach #include using namespace std; // Structure of the Node struct Node { int val; Node *left, *right; }; // Map to store the parent // node of every node of // the given Binary Tree unordered_map um; // Functiom to store all nodes // parent in unordered_map void storeParent(Node* root) { // Make a queue to do level-order // Traversal and store parent // of each node in unordered map queue q; q.push(root); // Loop to iterate until the // queue is not empty while (!q.empty()) { Node* p = q.front(); q.pop(); // Condition if the node is a /// root node that storing its // parent as NULL if (p == root) { um[p] = NULL; } // if left child exist of // popped out node then store // parent as value and node as key if (p->left) { um[p->left] = p; q.push(p->left); } if (p->right) { um[p->right] = p; q.push(p->right); } } } // Function to find the nodes // at distance K from give node void nodeAtDistK(Node* root, Node* target, int k) { // Keep track of each node // which are visited so that // while doing BFS we will // not traverse it again unordered_set s; int dist = 0; queue q; q.push(target); s.insert(target); // Loop to iterate over the nodes // until the queue is not empty while (!q.empty()) { // if distance is equal to K // then we found a node in tree // which is distance K if (dist == k) { while (!q.empty()) { cout << q.front()->val << " "; q.pop(); } } // BFS on node's left, // right and parent node int size = q.size(); for (int i = 0; i < size; i++) { Node* p = q.front(); q.pop(); // if the left of node is not // visited yet then push it in // queue and insert it in set as well if (p->left && s.find(p->left) == s.end()) { q.push(p->left); s.insert(p->left); } // if the right of node is not visited // yet then push it in queue // and insert it in set as well if (p->right && s.find(p->right) == s.end()) { q.push(p->right); s.insert(p->right); } // if the parent of node is not visited // yet then push it in queue and // insert it in set as well if (um[p] && s.find(um[p]) == s.end()) { q.push(um[p]); s.insert(um[p]); } } dist++; } } // Function to create a newnode Node* newnode(int val) { Node* temp = new Node; temp->val = val; temp->left = temp->right = NULL; return temp; } // Driver Code int main() { Node* root = newnode(20); root->left = newnode(8); root->right = newnode(22); root->right->left = newnode(5); root->right->right = newnode(8); root->left->left = newnode(4); root->left->left->left = newnode(25); root->left->right = newnode(12); root->left->right->left = newnode(10); root->left->right->left->left = newnode(15); root->left->right->left->right = newnode(18); root->left->right->left->right->right = newnode(23); root->left->right->right = newnode(14); Node* target = root->left->right; storeParent(root); nodeAtDistK(root, target, 3); return 0; }
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Practice Tags : | 1,542 | 5,501 | {"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} | 3.3125 | 3 | CC-MAIN-2020-45 | latest | en | 0.641679 |
https://cs.nyu.edu/courses/spring08/V22.0102-002/GUImaze.html | 1,516,702,342,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891886.70/warc/CC-MAIN-20180123091931-20180123111931-00182.warc.gz | 653,484,754 | 2,090 | # Honors Programming Assignment 2
## Computer Science 102
Spring 2008 Using a GUI to display a maze and solving it by recursive back-tracking
Part A Due Date: WED MAR 26, 11:59 pm
Part B Due Date: SAT MAR 29, 11:59 pm
Part a
Use the maze in the following link to generate a maze with blue and white buttons, where the blue ones represent the walls and the white ones, the possible path.
• Place the maze on a panel in the CENTER of the BorderLayout. Place a panel on the NORTH of the BorderLayout with a Start button on the WEST of that panel, instructions on how to proceed displayed in the CENTER of the panel and Trace button on the EAST. Before the Start button is pushed, the orignal maze with blue and white button should be displayed, and instructions on how to proceed is displayed in the CENTER of the NORTH panel.
• After the Start button is clicked the maze should be solved recursively in a depth-first fashion as shown in class. Number the buttons consecutively as you numbered the cells in the first honors assignment. The only moves allowed are N, E, S, snd W. After the maze is solved, a Trace button should appear on the EAST side of the panel.
• Each time the Trace button is clicked it should drive the movement of the tile one unit on the direct path to the exit, coloring the tiles pink as you go. Or, if you wish, when the Trace button is clicked, the tiles on the path to the exit should turn pink, one at a time with a pause between the tiles being coloured.
Part b
• After the Start button is pushed, the maze display should be the same as in part a with the buttons used in the depth-first search numbered.
• When the trace button is clicked once, the buttons on the direct path to the exit should be colored pink with a 200 second pause between colorings. Use Thread.sleep() to do this. As this path is traversed, a doubly-linked list with the steps should be generated as in the second regular assignment (0 should represent a move north; 1, south, etc.) and the path should be displayed on the NORTH panel.
• When the exit is reached, your program should remove the trace button -- use remove(trace) dereferencing the NORTH panel to do this -- and insert a retrace button in its place. The message on the panel should now indicate what to do next. When the retrace button is pushed, the path to the entrance should be retraced using the same technique as in the previous step.
Use the same technique you used in the second regular assignment to pause as the maze is traversed. You have to have a separate thread for the trace process and one for the retrace process. Nest these in the if statements detecting an action in the actionPerformed() method.
Samuel Marateck
SAT MAR 8 22:23:14 EST 2008 | 612 | 2,735 | {"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} | 2.515625 | 3 | CC-MAIN-2018-05 | latest | en | 0.88171 |
http://mathhelpforum.com/algebra/211663-two-numbers-whose-mean-average-same-their-product-not-7-3-7-11-a-print.html | 1,502,898,311,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102307.32/warc/CC-MAIN-20170816144701-20170816164701-00580.warc.gz | 261,094,104 | 2,528 | # Two numbers whose mean average are the same as their product and are NOT 7/3 and 7/11
The mean $=\frac{a+b}{2}$
The product $=ab$. | 44 | 133 | {"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": 0, "codecogs_latex": 2, "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-2017-34 | longest | en | 0.94536 |
https://deepai.org/publication/message-delivery-in-the-plane-by-robots-with-different-speeds | 1,653,372,588,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662564830.55/warc/CC-MAIN-20220524045003-20220524075003-00095.warc.gz | 199,994,332 | 68,555 | # Message Delivery in the Plane by Robots with Different Speeds
We study a fundamental cooperative message-delivery problem on the plane. Assume n robots which can move in any direction, are placed arbitrarily on the plane. Robots each have their own maximum speed and can communicate with each other face-to-face (i.e., when they are at the same location at the same time). There are also two designated points on the plane, S (the source) and D (the destination). The robots are required to transmit the message from the source to the destination as quickly as possible by face-to-face message passing. We consider both the offline setting where all information (the locations and maximum speeds of the robots) are known in advance and the online setting where each robot knows only its own position and speed along with the positions of S and D. In the offline case, we discover an important connection between the problem for two-robot systems and the well-known Apollonius circle which we employ to design an optimal algorithm. We also propose a √(2) approximation algorithm for systems with any number of robots. In the online setting, we provide an algorithm with competitive ratio 1/7( 5+ 4 √(2)) for two-robot systems and show that the same algorithm has a competitive ratio less than 2 for systems with any number of robots. We also show these results are tight for the given algorithm. Finally, we give two lower bounds (employing different arguments) on the competitive ratio of any online algorithm, one of 1.0391 and the other of 1.0405.
## Authors
• 3 publications
• 11 publications
• 8 publications
• 5 publications
05/08/2021
### The Pony Express Communication Problem
We introduce a new problem which we call the Pony Express problem. n rob...
12/07/2017
### Gathering in the Plane of Location-Aware Robots in the Presence of Spies
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01/13/2020
### Lower Bounds for Shoreline Searching with 2 or More Robots
Searching for a line on the plane with n unit speed robots is a classic ...
02/05/2019
### An Optimal Algorithm for Online Freeze-tag
In the freeze-tag problem, one active robot must wake up many frozen rob...
11/07/2018
### A Competitive Algorithm for Online Multi-Robot Exploration of a Translating Plume
In this paper, we study the problem of exploring a translating plume wit...
03/28/2022
### A Study of Reinforcement Learning Algorithms for Aggregates of Minimalistic Robots
The aim of this paper is to study how to apply deep reinforcement learni...
06/14/2018
### Online Variant of Parcel Allocation in Last-mile Delivery
We investigate the problem of last-mile delivery, where a large pool of ...
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## 1 Introduction
We study the problem of delivering a message in minimum time from a source to a destination using autonomous mobile robots with different maximum speeds. We extend the work on this communication problem studied previously on graphs [1, 3, 8, 9]. In our setting, the robots are initially distributed in arbitrary locations in the plane and the locations of the source and destination are known by all. The robots may move with their own (maximum) speed. Robots cooperate by exchanging information (the message) using face-to-face (F2F) communication. We study message transmission and allow messages to be replicated (as opposed to package delivery). The goal is to give an algorithm which minimizes the time required to deliver the message from the source to the destination through a series of F2F message transfers. In this paper we study how to complete this task efficiently and propose various centralized offline and distributed online algorithms which take into account the knowledge that the robots have about their speeds and initial locations.
### 1.1 Model, Notation and Terminology
The setup of our pony express problem will be in the Euclidean plane and points will be identified with their cartesian coordinates. We use capital letters to denote points and lower-case letters with subscripts to denote their components (e.g. point ). For any points , denotes the Euclidean distance between and , denotes the angle formed by in this order, and denotes the triangle formed by . Finally, denotes a circle centered at with radius .
Assume that robots are placed at arbitrary positions in the Euclidean plane. The respective speeds of the robots are . The movement of a robot is determined by a well-defined trajectory. A robot trajectory is a continuous function , with the location of the robot at time , such that , for all , where is the speed of the robot. A robot can move with its own constant speed and during the traversal of its trajectory it may stop and/or change direction instantaneously and at any time. Robot communication is F2F in that two robots can communicate (instantaneously) with each other only when they are co-located.
Algorithms describe the trajectories robots will follow and we will take into account the time it takes the algorithm to conclude the delivery task from the start, obtaining the message at a given source , and eventually delivering it to a given destination . In general, we are interested in offline and online algorithms. In the offline setting, the locations and speeds of all robots are known in advance and are available to a central authority that assigns trajectories to the robots. In the online setting, the robots know only their own initial position and speed, along with the positions of and . To measure the performance of our online algorithms, we consider their competitive ratio defined as follows. Let be the optimal delivery time for an instance of a given problem and be the time needed by some online algorithm for the same instance. The competitive ratio of is Our goal is to find online algorithms that minimize this competitive ratio.
### 1.2 Related work
Communicating mobile robots or agents have been used to address problems such as search, exploration, broadcasting and converge-casting, patrolling, connectivity maintenance, and area coverage (see [11]). For example, [6] addresses the problem of how well a group of collaborating robots with limited communication range is able to monitor a given geographical space. To this end, they study broadcasting and coverage resilience, which refers to the minimum number of robots whose removal may disconnect the network and result in uncovered areas, respectively. Similarly, rendezvous is a relevant communication paradigm and in [13, 17] the authors investigate rendezvous in a continuous domain under the presence of spies. A related study on message transmission in a synchronized multi-robot system may be found in [6]. Another application is patrolling whereby mobile robots are required to keep perpetually moving along a specified domain so as to minimize the time a point in the domain is left unvisited by an agent, e.g., see [16] for a related survey.
Data delivery and converge-cast with energy exchange under a centralized scheduler were studied in [15]. A restricted version concerns mobile agents of limited energy that are placed on a straight line and which need to collectively deliver a single piece of data from a given source point to a given target point on the line can be found in [10]. In [12] it is shown that deciding whether the agents can deliver the data is (weakly) NP-complete. Additional research under various conditions and topological assumptions can be found in [4] which studies the game-theoretic task of selecting mobile agents to deliver multiple items on a network and optimizing or approximating the total energy consumption over all selected agents, in [2, 5, 7] which study data delivery and combine energy and time efficiency, and in [18, 19] which are concerned with collaborative exploration in various topologies.
Our problem was previously studied on graphs in [1, 3, 8, 9]. In particular it is shown in [8] that the problem can be solved with agents on an -node, -edge weighted graph in time . We use this algorithm in the development of our approximation algorithm.
Our current work is related to the Pony Express communication problem proposed in [14]. In that paper, the authors provide both optimal offline and online algorithms for the anycast and broadcast problems in the case where the underlying domain was a continuous line segment. To our knowledge, the planar case studied in our paper has not been considered previously.
### 1.3 Outline and results of the paper
In Section 2 we propose an optimal offline algorithm for two robots. For ease of exposition, we first consider the case when the slower robot starts at the source and then the general case of arbitrary starting positions. In Section 3 we study the offline multirobot case. We propose an algorithm which approximates the optimal delivery time to within a factor of . Section 4 is dedicated to online algorithms. For two robots we give an algorithm with competitive ratio of and show that for robots, this same algorithm has a competitive ratio of at most . We also analyze lower bounds for this specific algorithm showing these bounds are tight. In Section 5 we prove lower bounds on the competitive ratio of arbitrary online algorithms. We discuss two approaches, one where the position of a robot is unknown and the other where the speed of a robot is unknown. These different approaches provide lower bounds of 1.0391 and 1.0405, respectively. We conclude in Section 6 by discussing possibilities for additional research in this area.
## 2 Optimal Offline Algorithm for Two Robots
In this section, we will consider two robots and which can move with respective constant speeds and () and design optimal offline algorithms with respect to the F2F communication model (observe that by scaling the distances, setting the speed of the slow robot to be yields no loss of generality.) Let and be the starting positions of robots and , respectively. There are three cases to consider:
1. : the fast robot can get to before the slow robot. In this case, it is clear that in the optimal solution, the fast robot should move to , acquire the message, and carry it to .
2. : the slow robot can deliver the message to before the fast robot can even reach . In this case, the optimal solution is also clear. The slow robot should move to , acquire the message, and carry it to .
3. In all other cases, the slow robot can get to before the fast robot, but the fast robot can get to the destination faster. The optimal solution, then, must involve a handover between the robots at some point in the plane.
For the first two cases, the optimal solution is trivial. The third case, however, is not as we must find the point at which the robots meet. First, we characterize the optimal meeting point for Case 3 through a series of lemmas.
###### Lemma 1.
For Case , there exists an optimal solution such that if is the handover, then .
For the sake of contradiction, suppose . First, it is obvious that should move directly toward and then directly toward and, similarly, should move directly toward . Any other path could only increase the time to deliver the message. Then, since , either or must arrive at before the other. Thus, there must be a time where one robot is waiting at for the other to arrive. Let be the time that the first robot arrives to and be the position of the other robot at time . We claim that an equal or better solution than waiting at would be for the first robot to move along until it meets the other robot at some point . Then the faster of the two robots carries the message from to (Figure 1).
If arrives at first, then if it waits at for to arrive, the total delivery time is , but since is clearly larger (in perimeter) than , then
|K′M′|+|M′D|≤|K′M|+|MD|⇒t+|K′M′|+|M′D|v≤t+|K′M|+|MD|v
Thus, meeting at results in a quicker delivery, a contradiction to the assumption that is optimal. If arrives at first, then if it waits at for the total delivery time is , but
|MM′|+|M′D|≤|K′M|+|MD| ⇒t+|MM′|+|M′D|v≤t+|K′M|+|MD|v ⇒t+|K′M′|+|M′D|v≤t+|K′M|+|MD|v
Again, meeting at results in an equal or quicker delivery, and so by contradiction, there must exist an optimal solution where . ∎ Intuitively, Lemma 1 says that robots must move at their maximum speeds directly towards the location they will acquire the message and then directly toward the location they handover or deliver the message. This restricts the set of feasible meeting points to the set of points in the plane such that both robots, moving in one direction at their maximum speeds, meet at the same time. For the case where the slow robot starts at the source (), this is directly related to an ancient theorem by the Greek philosopher Apollonius, which states “the trajectory traced by a point which moves in such a way that its Euclidean distance from a given point is a constant multiple of its Euclidean distance from another point is a circle” [20]. As a consequence, if the robots start at positions , respectively, then the locus of points at which the two robots may travel directly towards and meet at the same time is the circle of Apollonius (see Figure 2).
This circle, then is the locus of all possible handover points between the two robots. The precise statement in the context of mobile robots is stated in Lemma 2.
###### Lemma 2.
Two robots and with speeds and () are initially placed at points and , respectively. The locus of points such that robots and are equal time away from points and , respectively, i.e., , forms a circle with center and radius so that
C=S+S−Kv2−1 and R=v|SK|v2−1 (1)
The proof follows easily by using the representation of the points in cartesian coordinates and solving the equation . ∎
The following definition of the Apollonius Circle will be used throughout this paper.
###### Definition 1 (Apollonius Circle).
The circle with center and radius given by Equations (1) is called the Apollonius circle between robots and when their respective starting positions are .
For instances of the problem where and whose optimal solutions do not involve either robot delivering the message by itself, the previous discussion results in the following lemma whose proof follows directly from Lemmas 1 and 2.
###### Lemma 3.
The optimal meeting point is the point on the Apollonius circle between robots and which minimizes the total delivery time . ∎
### 2.1 Optimal algorithm when a robot starts at the source
First we give an algorithm in the restricted case where one robot starts at the source where the message is located ().
Let be the source, be the starting position of the fast robot which we assume to be on the axis, and the destination. Without loss of generality, we assume (if , the instance can be reflected about the axis and solved equivalently, since is on the axis). By Lemma 3, our goal is to find the point on the robots’ Apollonius circle which minimizes the delivery time (see Figure 3).
Consider the following offline algorithm for computing the optimal delivery time.
###### Theorem 2.1.
Algorithm 1 returns the optimal delivery time for instances with two robots and with speeds and , and starting positions and , respectively. Algorithm 1 can be implemented using a constant number of operations (including trigonometric functions).
First, note that Case 1 (from the three cases at the beginning of the section) is not considered since the slow robot, is assumed to start at the source. Case 2 is obviously handled by line 1 in the algorithm. Case 3 is divided into two subcases based on whether or not the condition in line 4 is satisfied. First, we consider the case where it is not. Let be the angle . Observe that if is tangent to the Apollonius circle (Figure 4), then . Clearly for any smaller value for , intersects the Apollonius circle at two points (and for any larger value, does not intersect the circle).
Then, let and (Figure 5 left). By the law of sines Thus and is just the associated point on the Apollonius circle.
Observe is the intersection point closest to . Since the condition in line 4 is satisfied, can move directly toward and, without veering from a direct path, meet at , acquire the message, and continue towards to deliver the message. We know, since the first case was not satisfied, that can reach the destination before can, so this is clearly the optimal trajectory.
Now, suppose the condition in line 4 is not satisfied. Consider the ellipse with foci and whose semi-major axis has length for some time . Then, by a defining property of an ellipse, the sum of the distances from each foci to a point on the ellipse is equal to a constant value . Consequently, a robot starting at with speed takes exactly time to travel to a point on the ellipse and then to . Observe that if the ellipse and Apollonius circle intersect, then the two robots can meet at one of the intersection points and, by the previous statement, the fast robot can deliver the message in time . If they intersect at two points, though, then any point on the Apollonius circle between these two intersections would yield a better solution. The solution, then, is to find the minimal which causes the Apollonius circle and the ellipse to intersect at exactly one point . Thus must be normal to both the Apollonius circle and the ellipse. That , therefore, must bisect follows from a well-known property of the ellipse, namely that a normal line through a point on an ellipse bisects the angle it forms with the ellipse’s foci.
Next, we show the algorithm can be implemented to run using a constant number of operations (including trigonometric functions). The only lines in the algorithm that are not clearly computable with a constant number of operations are lines 8 and 10. To show that can be computed in constant time, we provide a formulation which can be given as input to Equation Solving tools (e.g., Mathematica) to find a closed-form solution . Let and be the point given by rotating around (into the positive half-plane, Figure 5 right). Observe that if bisects , then is collinear with , or: where . ∎
### 2.2 Optimal algorithm in the general case
In this subsection we consider the more general case where the slow robot does not start at the source. Let the starting positions of source and destination be and and let the robots and start from arbitrary points and in the plane, respectively. Again, we are interested in finding the point for the third case (from the cases at the beginning of the section), since optimal solutions for the first two cases are trivial to find. As depicted in Figure 6,
robot follows a trajectory which first visits a point at distance from its starting position, then continues along a straight-line trajectory to meet robot at a suitable point , and finally delivers the message to the destination . The main steps of the algorithm are as follows.
1. If , then reaches before and should complete the delivery on its own.
2. Otherwise, if , then can deliver the message on its own before can even reach the destination.
3. Otherwise reaches in time and, at the same time, robot goes to a specially selected point which lies on the circle centered at with radius .
4. Robot meets robot at a point determined by the locus of points which are equal time away from and (by Lemma 2, this is the circle with center and radius as given in Equation (1)). Robot passes message to which delivers it to .
Observe that by Lemma 1, , , and must be collinear. We can then generalize the result of section 2.1 using the following lemma. Recall that the center of similitude (also known as homothetic center) is a point from which at least two geometrically similar figures can be seen as a dilation or contraction of one another (see [21][Section 1.1.2]).
###### Lemma 4.
Let be the center of the Apollonius circle between and when is at and is at . Then, is the center of similitude of the circles and . Consider any point in the circumference of . Let be the angle , then is the center of the Apollonius circle of .
Refer to Figure 7. Consider any point at the circumference of the circle . Let be the center of the Apollonius circle between and when is at and is at . Therefore, and . Observe that the ratio is always constant.
Therefore, defines a circle where is the center of similitude. The lemma follows since the triangles and are similar. ∎
Consider two points and as described in Lemma 4, for some . We can now use Theorem 2.1 to characterize the solution. However, this approach does not lead to a closed-form solution. Instead, in the following lemma, we present another approach using optimization which does.
Let . Then the optimal trajectory is obtained by a point which minimizes the objective function
√(k1−x1)2+(k2−x2)2+√(x1−d1)2+(x2−d2)2 (2)
subject to the condition
((x1−k1)2+(x2−k2)22av2−(x1−s1)2+(x2−s2)22a−a2)2=(x1−s1)2+(x2−s2)2. (3)
Recall points are collinear, the handover point point must lie at the intersection of two circles as depicted in Figure 8.
This can be expressed by the fact that satisfies the two equations
(x1−k1)2+(x2−k2)2=v2(|LS|+t)2 (4) (x1−s1)2+(x2−s2)2=t2. (5)
In turn, this gives a system of quadratic equations parametrized with respect to time . We can rewrite Equation (4) as and subtracting both sides of the last Equation from Equation (5) we derive the Equation
(x1−k1)2+(x2−k2)2v2−(x1−s1)2−(x2−s2)2=|LS|(|LS|+2t).
By collecting similar terms, using Equation (5), and simplifying we derive the following Equation
((x1−k1)2+(x2−k2)22|LS|v2−(x1−s1)2+(x2−s2)22|LS|−|LS|2)2=(x1−s1)2+(x2−s2)2.
This is exactly Equation (3). ∎
The resulting optimization problem has two unknowns in the objective function (2) and must satisfy the condition of Equation (3). It can be used to substitute variables and express the final optimization function described in Formula (2) using only a single variable, say , which can then be minimized using standard analytical methods. This is easily seen since Equation (3) is of degree in the variable (as well as in the variable , for that matter). Therefore a closed form expression of the variable in terms of the variable and the known parameters is easily derived.
There are two symmetries in Equation (3) which simplify the objective function and make the calculation of the solution easier. They are easily revealed with simple geometric transformations.
For the first symmetry, consider a rotation of the axis and a translation of the entire configuration of points so that and lie on the horizontal axis, i.e., and . Then Equation (3) is transformed to the equation
(x21+x222av2−(x1−s1)2+x222a−a2)2=(x1−s1)2+x22 (6)
The resulting symmetry is along the horizontal -axis in Equation (6). Namely, if is a solution so is . If we consider Equation (6) in the unknown we see that it is of degree , but which is also a quadratic in . Therefore can be easily expressed as a function of using the formula for the roots of the quadratic equation. The second symmetry is obtained in a similar manner. If is a solution so is . One considers a rotation of the axis and a translation of the entire configuration of points so that and lie on the vertical axis, i.e., and . Details can be completed as above. To sum up we have the following Algorithm 2 which determines the handover point which yields the optimal trajectory.
Algorithm 2 returns the optimal delivery time for two robots and with speeds and , respectively, and can be implemented using a constant number of operations (including trigonometric functions). The proof follows from the previous discussion. Indeed, without loss of generality we may consider only the case where the slow robot reaches first. As depicted in Figure 8 there are two competing trajectories. Given that the slow robot can arrive first at , either the robots follow the algorithm and the slow robot meets the faster robot at the meeting point to handover the message to which then delivers it to or the faster robot gets the message at and delivers it to without cooperating with the other robot.
In the former case the delivery time will be while in the latter case the delivery time will be . However this is easy to prove since the point must lie inside the triangle as depicted in Figure 8, i.e., . All other lines in the algorithm clearly require a constant number of operations. By the previous discussion, a closed form solution for the optimization required in line 6 exists . ∎
## 3 Offline √2 Approximation for Multiple Robots
In principle, the equations derived in the previous section can be generalized to solve the problem optimally for robots. Unfortunately, we are not able to solve the resulting set of equations. We do not speculate on the complexity of the general problem here. Instead, in this section we provide a -approximation algorithm, The robots know the location of the source and destination but also all robots know the initial locations and speeds of all other robots. The basic idea of our proof is contained in the following observation depicted in Figure 9. Suppose that during the execution of an optimal “Euclidean” algorithm (i.e., optimal in the sense of the Euclidean distance) two robots placed at and , follow the straight-line trajectories and , respectively, and meet at the point .
Now we replace the Euclidean trajectories and with the rectilinear trajectories and , respectively. Elementary geometry implies that
|AX|+|XP|≤√2|AP| and |BY|+|YP|≤√2|BP|. (7)
This observation leads to the following lemma.
###### Lemma 5.
Consider the pony express problem for robots, a source and a destination in the plane. Then , where are the delivery costs of the optimal trajectories of the pony express problem for delivering from a source to a destination measured in the rectilinear and Euclidean metrics, respectively.
###### Proof.
Consider an optimal Euclidean algorithm which ensures the delivery time is exactly , i.e., . Now use the idea discussed in Figure 9 to replace the Euclidean trajectory of algorithm with a rectilinear trajectory thus giving rise to a rectilinear algorithm . Note that in this rectilinear simulation of the optimal Euclidean solution, robots may not arrive at the endpoints at the same time. The robot that arrives first, should simply wait at the meeting point until the second robot arrives. The meeting time is thus determined by the last arrival of the two robots. By definition, the time it takes the rectilinear algorithm to deliver the message is at least , i.e., . From Inequality (7) we have that . Therefore we conclude that ∎∎
Consider robots in the plane with starting positions . Without loss of generality assume the slowest robot has speed . Further, let the source of a message be located at a point and the destination at a point and assume, without loss of generality, that the line segment is horizontal. Enclose the points and in a square with sides parallel to the axis, where is a positive real proportional to the diameter of the set . For arbitrarily small, partition the square with parallel vertical and horizontal lines with consecutive distances , respectively, so as to form a grid. Without loss of generality we may assume that and are vertices in this grid graph (This is easy to accomplish by choosing to be an integral fraction of the distance between and .) Clearly, this forms a grid graph with vertices so that are also vertices and edges. Now consider the following algorithm.
###### Theorem 3.1.
For any arbitrarily small, there exists an algorithm that finds trajectories for robots to deliver the message from the source to the destination in time whose delivery time is at most multiplied by the delivery time of the optimal Euclidean algorithm plus the additional additive overhead , where is the diameter of the point set .
###### Proof.
Let and run Algorithm 3. Let be the time the algorithm takes to deliver the message and let be the time for step 2 in the algorithm (the optimal delivery time for the given grid with starting positions ). Then, let and be the optimal delivery times for the rectilinear and Euclidean metrics respectively. First, observe
A(p1,…,pn)≤Grid(p′1,…,p′n)+ϵ (8)
The result will follow from Lemma 5 and the following claim:
.
###### Proof.
(of Claim) Without loss of generality, assume the robots involved (in order) are robots through . Let be the handover points in the optimal rectilinear algorithm (where ). Let be the nearest point to on the grid. Let be the time it takes for the first robot to arrive at the source and, (for ) be the time that robot holds the message.
Consider the algorithm where robots emulate the rectilinear algorithm on the grid by meeting at instead of , for each handover. Note that robots may not arrive at the endpoints at the same time. However, the algorithm is offline the robot that arrives first, should simply wait at the meeting point until the second robot arrives.
The time each robot holds the message then is at most and the total time to deliver the message is
k∑i=1ti+ϵ=OptRect+(k−1)ϵ≤OptRect+(n−1)ϵ.
This proves the claim. ∎
Using inequality (8), the above Claim and Lemma 5 we arrive at the inequality Finally, by selecting as above, and the complexity of the algorithm, by [8], is . This completes the proof of Theorem 3.1. ∎∎
## 4 Online Upper Bounds
In this section we discuss online algorithms. In Subsection 4.1 we give an online algorithm with competitive ratio for two robots with knowledge only of the source and destination . We show this bound is tight for the given algorithm. In Subsection 4.2 we show that the same algorithm when generalized to robots has competitive ratio at most 2. Further, we show that for any , the competitive ratio of our algorithm is at least .
### 4.1 Two Robot Algorithm with Competitive Ratio 17(5+4√2)
Consider the following Algorithm 4 for multiple robots.
Observe that in this algorithm the robots act independently. In particular no attempt is made to co-ordinate the action of the robots and if two robots meet they ignore each other. This is not required in order to achieve the upper bounds below. For our lower bounds on this algorithm, we assume that the robots do not interact even if it may improve the time to complete the task.
###### Theorem 4.1.
For the case of two robots, Algorithm 4 delivers the message from the source to the destination in at most times the optimal offline time.
Given an arbitrary instance of the problem, let be the time taken by the optimal solution to deliver the message from to . Let be the robots involved in that optimal solution where . Observe that if only one robot is involved then our online solution is optimal. Let be the starting point of robot . Let be the point in the optimal solution where hands the message off to . Let be the time when this happens. Finally, let and .
We make the following observations:
1. .
2. .
3. delivers the message in time . (Recall that is the time for to reach , , and .)
4. delivers the message in time which is less than .
5. and .
6. .
Let be the competitive ratio of our algorithm for our two robots. We have by observation 4. This is maximized when or when . In this case we have:
c2 ≤t+a−xt+|MD|/v ≤t+a−xt+(a−x)/v by obs. 6 ≤x+a−xx+(a−x)/v by obs. 5 =v2+vv2−v+2.
This is maximized when at which point
###### Example 1.
Now we give a tight lower bound on the competitive ratio of Algorithm 4 for two robots. Consider the following example input. One robot is placed at the source which is the point and has speed . The destination is placed at the point . The second robot has speed and is placed at the point . The robots are initially placed at distance . In the optimal algorithm the robots meet in time at the point . The faster robot picks up the message at and delivers it to in additional time . Therefore the delivery time of the optimal algorithm is equal to . It follows that the competitive ratio satisfies
###### Remark 1.
Note that in Example 1 if we parametrize the speed of the slow robot to , and place the fast robot at position then similar calculations show that . Further, it is easy to see that the lower bound is maximized for .
### 4.2 Multi Robot Algorithm with Competitive Ratio ≤2
###### Theorem 4.2.
Algorithm 4 has competitive ratio at most for any robots.
To simplify notation, let the distance between the source and destination be . Observe that in the algorithm, every robot attempts to deliver the message entirely by itself. The robots do not cooperate at all. Clearly, then, if the robots can optimally deliver the message in , then Theorem 4.2 holds if and only if there exists at least one robot that can deliver the message by itself in time. In other words, we must show that there is a robot with speed and that starts a distance from the source such that ( such that ).
For the sake of contradiction, assume all robots have speed . Then we must show that they could not possibly deliver the message optimally in time or less. By restricting the robots speed as a function of their distance to the source, we allow the robots to choose everything else about their starting positions to minimize the optimal delivery time. Clearly, robots will most quickly deliver the message if they are positioned on the line from (robots are always moving directly toward the message or its destination).
Furthermore, the more robots in the system, the faster the message will be delivered (observe that given two participating robots, inserting an additional robot between them improves the delivery time). Therefore, we can assume there are an uncountably infinite number of robots (one at every point on the interval ) and the problem becomes continuous.
Consider a robot that starts at position on the line. By construction, its velocity is less than and after time its position is . Therefore, if the message is at position on the line segment at time
, its velocity at that moment is less than
since that is the upper bound on the speed of the fastest robot that could reach by time .
It follows from the previous discussion that the speed of the message must satisfy the inequality: The resulting differential equation with unknown and the initial condition , yields
Finally, we can use this equation to show that the delivery time (when ) must be greater than . Observe , so the robots cannot deliver the message in time , a contradiction. Therefore, if the optimal delivery is , then there must exist a robot with speed which can deliver the message to the destination in at most time. ∎
###### Theorem 4.3.
Given robots, there is a robot deployment such that Algorithm 4 has competitive ratio at least .
Let be the source and the destination. Without loss of generality, we assume is a unit line where is at the origin and . Given the meeting points, we construct an instance with robots with competitive ratio . Let be the meeting point of robot and for . In our construction, robots are required to arrive at at the same time after reaching . Hence, we compute the speed from the meeting points as follows: Let be the time that robot arrives at point . Therefore, it arrives at after reaching at time . Further, at time robot has reached . Therefore, it arrives at after reaching at time . Setting both equal and factorizing we obtain the speed given , i.e. . Initially, we set which implies that is the fastest robot. Let be the initial position or robot on the line . Observe that all robots are in the interval . We claim that the competitive ratio of the setting is . Observe that arrives at at time . Therefore, in the optimal algorithm it takes an additional to reach meanwhile in Algorithm 4 robot takes to reach . Therefore, the competitive ratio is . Simplifying we obtain and the theorem follows. ∎
###### Remark 2.
Observe that for any by taking we have the competitive ratio of Algorithm 4 is at least .
## 5 Online Lower Bounds for Two Robots
In this section we prove two lower bounds on the competitive ratio for arbitrary online algorithms. Our lower bounds require only two mobile robots. In the first lower bound (Theorem 5) we assume that the speed of one of the robots is unknown and in the second (Theorem 5.1) we assume that the starting position of one of the robots is unknown. We provide both bounds (even though the second is slightly better) as the arguments are somewhat different and it seems plausible that an improved lower bound may come from combining the two approaches.
The lower bound for the competitive ratio when the fast robot does not know whether the speed of the slow robot is one or zero is at least . In the proof we consider two robots and where is placed at the source of the message and has speed either 0 or 1 as set by the adversary and has speed greater than 1. Without loss of generality assume that and are at distance one. By Lemma 2 the Apollonius circle is at distance | 8,135 | 37,069 | {"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 | 0.926533 |
https://www.coursehero.com/file/8674082/-m2-2-cos-2-and-y-1-2-1f-318325-The-condition/ | 1,498,412,919,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320545.67/warc/CC-MAIN-20170625170634-20170625190634-00388.warc.gz | 858,856,007 | 23,661 | solutionset7
# m2 2 cos 2 and y 1 2 1f 318325 the condition
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Unformatted text preview: ve dV v2 mv 2 ˙ F =− = −mrθ2 = −mR = . y = f (x) =⇒ y = f x =⇒ y = f x + f x2 . ˙ ˙ ˙ dr R¨ R¨ (315) (322) (316) (323) 6.36. Atwood’s machine Plugging the x and y from the E-L equations into this, and solving for F , gives ¨ ¨ Let 1 and 2 be the lengths of string in the air, and 2 L = 1 + 2 . If η ≡ 1 + 2 − L, let mf x ˙ F= . (324) then the Lagrangian is ∂η /∂ y − f ∂ η /∂ x 1 1 L = m1 ˙2 ∂η m2 ˙2 + ∂η g 1 m2 is − angle (317) We must now determine + /∂ x and m1/∂ y .+If θ g 2theV (η ). the curve makes with 1 2 2 2 the x axis at a given point, then the slope there is f (x) = tan θ. Consider a point Usingx, y= near /dη ,curve also the definition of η , point is to the of motion are ( F ) −dV the and (we’ll assume that this the equations left of the curve, but the other side proceeds similarly). If∂η imagine varying only x, and then only y , you you dV m1 ¨1 = m1 g − the curve changes ¨1 = m1 g to can see that the distance to η · ∂ 1 =⇒ m1...
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## This note was uploaded on 01/19/2014 for the course PHYSICS 251 taught by Professor Brandenberger during the Fall '13 term at McGill.
Ask a homework question - tutors are online | 507 | 1,381 | {"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} | 3.453125 | 3 | CC-MAIN-2017-26 | latest | en | 0.839346 |
https://www.jiskha.com/questions/111547/1a-student-measures-the-velocity-of-the-water-in-a-stream-as-2-5-meters-per-second-the | 1,610,899,484,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513062.16/warc/CC-MAIN-20210117143625-20210117173625-00276.warc.gz | 823,094,552 | 5,224 | # Earth Science
1A student measures the velocity of the water in a stream as 2.5 meters per second. The actual velocity of the water is 3.0 meters per second. What is the approximate percent deviation of the student’s measurements?
(1)0.50%
(2)17%
(3)20%
(4)50%
Is it (2)?
1. 👍
2. 👎
3. 👁
1. (3) 20%
1. 👍
2. 👎
2. 20%
1. 👍
2. 👎
3. Whshxydhdhcycyd
1. 👍
2. 👎
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https://ww2.mathworks.cn/help/hydro/ref/pilotoperatedcheckvalveil.html | 1,679,642,281,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00731.warc.gz | 716,111,876 | 25,133 | # Pilot-Operated Check Valve (IL)
Check valve with pilot pressure control in an isothermal liquid system
Since R2020a
Libraries:
Simscape / Fluids / Isothermal Liquid / Valves & Orifices / Directional Control Valves
## Description
The Pilot-Operated Check Valve (IL) block models a flow-control valve with variable flow directionality based on the pilot-line pressure. Flow is normally restricted to travel from port A to port B in either a connected or disconnected spool-poppet configuration, according to the Pilot configuration parameter.
Pilot-Operated Check Valve Schematic
The control pressure, pcontrol is:
`${p}_{control}={p}_{pilot}{k}_{p}+\left({p}_{A}-{p}_{B}\right),$`
where:
• ppilot is the control pilot pressure differential.
• kp is the Pilot ratio, the ratio of the area at port X to the area at port A: ${k}_{p}=\frac{{A}_{X}}{{A}_{A}}.$
• pApB is the pressure differential over the valve.
When the control pressure exceeds the Cracking pressure differential, the poppet moves to allow flow from port B to port A.
There is no mass flow between port X and ports A and B.
### Valve Pressure Control with a Pilot Port
The pilot pressure differential for valve control can be configured in two ways:
• When the Opening pilot pressure specification parameter is set to ```Pressure at port X relative to port A```, the pilot pressure is the pressure differential between port X and port A.
• When Opening pilot pressure specification is set to ```Pressure at port X relative to atmospheric pressure```, the pilot pressure is the pressure difference between port X and atmospheric pressure.
When Pilot configuration is set to `Disconnected pilot spool and poppet`, the relative pressure at port X must be positive. If the measured pilot pressure is negative, the control pressure is only based on the pressure differential between ports A and B. In the `Rigidly connected pilot spool and poppet` setting, the pilot pressure is the measured pressure differential according to the opening specification.
### Mass Flow Rate Equation
Mass is conserved through the valve:
`${\stackrel{˙}{m}}_{A}+{\stackrel{˙}{m}}_{B}=0.$`
The mass flow rate through the valve is calculated as:
`$\stackrel{˙}{m}=\frac{{C}_{d}{A}_{valve}\sqrt{2\overline{\rho }}}{\sqrt{P{R}_{loss}\left(1-{\left(\frac{{A}_{valve}}{{A}_{port}}\right)}^{2}\right)}}\frac{\Delta p}{{\left[\Delta {p}^{2}+\Delta {p}_{crit}^{2}\right]}^{1/4}},$`
where:
• Cd is the Discharge coefficient.
• Avalve is the instantaneous valve open area.
• Aport is the Cross-sectional area at ports A and B.
• $\overline{\rho }$ is the average fluid density.
• Δp is the valve pressure difference, pApB.
The critical pressure difference, Δpcrit, is the pressure differential associated with the Critical Reynolds number, Recrit, the flow regime transition point between laminar and turbulent flow:
`$\Delta {p}_{crit}=\frac{\pi \overline{\rho }}{8{A}_{valve}}{\left(\frac{\nu {\mathrm{Re}}_{crit}}{{C}_{d}}\right)}^{2}.$`
Pressure loss describes the reduction of pressure in the valve due to a decrease in area. PRloss is calculated as:
`$P{R}_{loss}=\frac{\sqrt{1-{\left(\frac{{A}_{valve}}{{A}_{port}}\right)}^{2}\left(1-{C}_{d}^{2}\right)}-{C}_{d}\frac{{A}_{valve}}{{A}_{port}}}{\sqrt{1-{\left(\frac{{A}_{valve}}{{A}_{port}}\right)}^{2}\left(1-{C}_{d}^{2}\right)}+{C}_{d}\frac{{A}_{valve}}{{A}_{port}}}.$`
Pressure recovery describes the positive pressure change in the valve due to an increase in area. If you do not wish to capture this increase in pressure, clear the Pressure recovery check box. In this case, PRloss is 1.
The opening area, Avalve, is also impacted by the valve opening dynamics.
### Opening Parameterization
The linear parameterization of the valve area is
`${A}_{valve}=\stackrel{^}{p}\left({A}_{\mathrm{max}}-{A}_{leak}\right)+{A}_{leak},$`
where the normalized pressure, $\stackrel{^}{p}$, is
`$\stackrel{^}{p}=\frac{{p}_{control}-{p}_{cracking}}{{p}_{\mathrm{max}}-{p}_{cracking}}.$`
When the valve is in a near-open or near-closed position, you can maintain numerical robustness in your simulation by adjusting the parameter. If the parameter is nonzero, the block smoothly saturates the control pressure between pmax and pcracking. For more information, see Numerical Smoothing.
### Opening Dynamics
If opening dynamics are modeled, a lag is introduced to the flow response to the modeled control pressure. pcontrol becomes the dynamic control pressure, pdyn; otherwise, pcontrol is the steady-state pressure. The instantaneous change in dynamic control pressure is calculated based on the Opening time constant, τ:
`${\stackrel{˙}{p}}_{dyn}=\frac{{p}_{control}-{p}_{dyn}}{\tau }.$`
By default, the Opening dynamics check box is cleared.
## Ports
### Conserving
expand all
Liquid entry point to the valve. When the control pressure exceeds the cracking pressure, liquid is able to exit from this port.
Liquid exit point from the valve. When the control pressure exceeds the cracking pressure, liquid is able to enter the valve from this port.
Pressure port that contributes to the flow control through the valve.
## Parameters
expand all
Valve geometry. The valve can either have an opening mechanism that is connected to the valve poppet, in the case of the ```Rigidly connected pilot spool and poppet``` setting, or an opening mechanism that is aligned with, but moves freely away from, the valve poppet, in the case of the ```Disconnected pilot spool and poppet``` setting. The configuration choice determines the pilot pressure calculation.
Reference pressure differential used for valve control. This differential defines the pilot pressure differential, which is added to the pressure differential between ports A and B and compared against the valve threshold Cracking pressure differential.
Set pressure for the valve operation.
Maximum pressure differential in an opened valve. This value provides an upper limit to simulation pressures so that results remain physical.
Ratio of port area X to port area A.
Maximum valve area. This value is used to determine the normalized valve pressure and the valve opening area during operation.
Sum of all gaps when the valve is in the fully closed position. Any area smaller than this value is saturated to the specified leakage area. This contributes to numerical stability by maintaining continuity in the flow.
Areas at the entry and exit ports A and B, which are used in the pressure-flow rate equation that determines the mass flow rate through the valve.
Correction factor accounting for discharge losses in theoretical flows.
Upper Reynolds number limit for laminar flow through the orifice.
Continuous smoothing factor that introduces a layer of gradual change to the flow response when the valve is in near-open or near-closed positions. Set this value to a nonzero value less than one to increase the stability of your simulation in these regimes.
Whether to account for pressure increase when fluid flows from a region of smaller cross-sectional area to a region of larger cross-sectional area.
Whether to account for transient effects to the fluid system due to opening the valve. Selecting Opening dynamics approximates the opening conditions by introducing a first-order lag in the pressure response. The Opening time constant also impacts the modeled opening dynamics.
Constant that captures the time required for the fluid to reach steady-state conditions when opening or closing the valve from one position to another. This parameter impacts the modeled opening dynamics.
#### Dependencies
To enable this parameter, select Opening dynamics.
## Version History
Introduced in R2020a | 1,758 | 7,697 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 11, "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} | 2.71875 | 3 | CC-MAIN-2023-14 | longest | en | 0.874425 |
http://mathhelpforum.com/calculus/127897-help-differentiate-functions-print.html | 1,527,171,134,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866326.60/warc/CC-MAIN-20180524131721-20180524151721-00024.warc.gz | 180,066,222 | 2,708 | # Help with differentiate functions
• Feb 8th 2010, 07:36 PM
rhcp1231
Help with differentiate functions
how would i go about finding the differentiate of:
f(t) = 2-(2/3)t
i know the power rule and all that but i just dont know where to start....
• Feb 8th 2010, 07:46 PM
dedust
Quote:
Originally Posted by rhcp1231
how would i go about finding the differentiate of:
f(t) = 2-(2/3)t
i know the power rule and all that but i just dont know where to start....
let $\displaystyle y = \left(\frac{2}{3}\right)^t$, then
$\displaystyle \ln y = t \ln \left(\frac{2}{3}\right)$
differentiate both side with respect to t
$\displaystyle \frac{1}{y} \frac{dy}{dt} = \ln \left(\frac{2}{3}\right)$
or
$\displaystyle \frac{dy}{dt} = y \ln \left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^t \ln \left(\frac{2}{3}\right)$ | 275 | 817 | {"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} | 3.71875 | 4 | CC-MAIN-2018-22 | latest | en | 0.830328 |
https://www.physicsforums.com/threads/integrate-wave-function-squared-m-chester-text.919752/ | 1,508,211,313,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820556.7/warc/CC-MAIN-20171017013608-20171017033608-00109.warc.gz | 1,289,849,348 | 16,711 | # Integrate wave function squared - M. Chester text
1. Jul 9, 2017
### GreyNoise
1. The problem statement, all variables and given/known data
given: A wire loop with a circumference of L has a bead that moves freely around it. The momentum state function for the bead is $\psi(x) = \sqrt{\frac{2}{L}} \sin \left (\frac{4\pi}{L}x \right )$
find: The probability of finding the bead between $\textstyle \frac{L}{24}$ and $\textstyle \frac{L}{8}$
2. Relevant equations
$\int_{a}^{b}|<x| \psi >|^2 dx = \int_{a}^{b} | \psi(x) |^2 dx$
$\psi(x) = \sqrt{\frac{2}{L}} \sin \left ( \frac{4\pi}{L}x \right ) \hspace{10mm}$ the state function
3. The attempt at a solution
$\displaystyle \int_{a}^{b}|\lt x|\psi\gt|^2 dx = \int_{a}^{b} | \psi(x) |^2 dx$
${\displaystyle \int_{\frac{L}{24}}^{\frac{L}{8}} \left [ \sqrt{\frac{2}{L}} \sin \left ( \frac{4\pi}{L}x \right ) \right ]^2 dx } \hspace{10mm}$ sub $\textstyle \psi(x) = \sqrt{\frac{2}{L}} \sin \left ( \frac{4\pi}{L}x \right )$ and integration limits
$\displaystyle \int_{\frac{L}{24}}^{\frac{L}{8}} \left [ \sqrt{\frac{2}{L}} \sin \left ( \frac{4\pi}{L}x \right ) \right ]^2 dx = \left. \frac{x}{L} - \frac{ \sin \left ( \frac{8\pi}{L}x \right )}{8\pi} \right |_{\frac{L}{24}}^{\frac{L}{8}}$
$\displaystyle = \frac{1}{L}\frac{L}{8} - \frac{ \sin \left ( \frac{8\pi}{L}\frac{L}{8} \right ) }{8\pi} - \left [ \frac{1}{L}\frac{L}{24} - \frac{ \sin \left ( \frac{8\pi}{L}\frac{L}{24} \right ) }{8\pi} \right ] = \frac{1}{8} - \frac{1}{24} + \frac{ \sin \left ( \frac{\pi}{3} \right ) }{8\pi}$
$= \displaystyle \frac{1}{12} + \frac{\sqrt{3}}{16\pi}$
The answer given in the text is $\frac{1}{12} + \frac{1}{16\pi}$. I cannot shake the $\sqrt{3}$ in the second term. I even checked my evaluation of the integral on the Wolfram site, and it returned the same integral solution as I got. The book is Primer of Quantum Mechanics by Marvin Chester (Dover Publ). It appears to be a first edition, so I guess it's plausible that the text's answer is a typo, but I felt the need to consult the community. Can anyone show me what I am doing wrong?
2. Jul 9, 2017 | 787 | 2,105 | {"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} | 3.953125 | 4 | CC-MAIN-2017-43 | longest | en | 0.62056 |
http://www.cs.utep.edu/vladik/stat3320.spring2014/final.html | 1,505,867,102,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686077.22/warc/CC-MAIN-20170919235817-20170920015817-00421.warc.gz | 418,265,819 | 2,482 | ## Statistics Spring 2014 Final Exam
Name: ___________________________________________________________
1. In the Statistics for CS class of size 20, by the time of the final exam, 80% of the students mastered discrete probabilities, 70% mastered continuous probabilities, and 60% mastered both. For a student to pass the class at least a C, the student must be able to use at least one type of probabilities. How many students passed the class? Are two parts statistically independent?
2. Another year, out of 25 students, 80% of the students mastered discrete probabilities, 70% mastered continuous probabilities, and the corresponding events turned out to be independent. How many students passed the class?
3. An average student is well-prepared (A-level) for five out of seven topics. If an instructor gives a surprise quiz covering two of the topics, what percentage of students will get an A?
4. Suppose that for each class, 30% of the student are well-prepared (A-level), and different students are independent. What is the probability that in a class of 20, 19 will be well-prepared? What is the probability that the instructor will grade at least 10 papers until he finds a one which does not deserve an A?
5. The quiz has three yes-no questions, testing thee (randomly selected) topics out of five that the students studied. A student knows four out of these five topics; for a topic that a student does not know, he gives yes or no answer with equal probability. What are the expected number of mistakes on the quiz? What is the variance of this number?
6. Two students did not prepare well for a quiz. To answer a multiple-choice question with 4 possible answers, each student independently chooses a random number from 1 to 4, with equal probability. Let s be the sum of these numbers, and let p be their product. What is the probability that s = 5 if p = 6? if p = 5? Are the variables s and p independent? Explain your answer.
7. A professor (not me) lets a student take the test again and again until the student passes it. A student relies on luck and does not study between the tests, so his probability of succeeding is 50% on each test. On average, how many times will a student take the test until he passes?
8. On average, a student makes 3 mistakes on a test, with a standard deviation of 2. In a class of 50 students, what is a probability that overall, they will make less than 120 mistakes? Use Central Limit Theorem.
9. A student knows 5 topics out of 10. What is the probability that this student will get an A on a quiz which covers 4 out of 10 topics? 5 out of 10? 6 out of 10?
10. A random variable is distributed with the probability density f(x) = 1 + kx for x from the interval [0,1] and f(x) = 0 for all other x. Find: (a) the value k, (b) the cumulative distribution function for x, (c) the expected value of x, and (d) the probability that x does not exceed 0.5.
11. A computer is connected to two printers. If at least one of them works, we can still print. The time-to-repair of each printer is exponentially distributed with the mean value of 2 years. What is the probability that after 3 years, we will still be able to print?
12. For a printer-producing company, the probability that a printer needs repairs is 1%. In a sample of 1,000 printers, what is the expected value of number of faulty printers, and what is the standard deviation? What is the probability that no more than 20 printers are faulty (just give a formula, it is not necessary to get a numerical answer).
13. Two independent tasks are coming to the server. Their arrival times are uniformly distributed between 6 am and noon. Compute the expected time of the task which arrives first, and the expected time of the task which arrives second.
14. Derive a formula and explain how to generate a random variable with the density f(x) = (2/9) x on the interval [0,3] if you have a standard random number generator which produces values uniformly distributed on the interval [0,1]. Use the inverse transform method.
15. Suppose that we have a distribution in which we have 1 with probability θ and 0 with probability 1 − θ. Use method of moments and maximum likelihood to estimate the parameter θ from the following sample: 0, 1, 0, 1, 0, 0, 1, 1.
16. In a shipment of 100 computers, 3 turns out to be faulty. Construct a 90% confidence interval for the proportion of faulty printers. | 1,027 | 4,406 | {"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} | 4.21875 | 4 | CC-MAIN-2017-39 | longest | en | 0.94454 |
https://www.aqua-calc.com/one-to-one/density/gram-per-cubic-decimeter/kilogram-per-liter/101 | 1,585,875,144,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370509103.51/warc/CC-MAIN-20200402235814-20200403025814-00394.warc.gz | 828,357,828 | 8,220 | # 101 grams per cubic decimeter [g/dm³] in kilograms per liter
## grams/decimeter³ to kilogram/liter unit converter of density
101 grams per cubic decimeter [g/dm³] = 0.1 kilogram per liter [kg/l]
### grams per cubic decimeter to kilograms per liter density conversion cards
• 101
through
125
grams per cubic decimeter
• 101 g/dm³ to kg/l = 0.1 kg/l
• 102 g/dm³ to kg/l = 0.1 kg/l
• 103 g/dm³ to kg/l = 0.1 kg/l
• 104 g/dm³ to kg/l = 0.1 kg/l
• 105 g/dm³ to kg/l = 0.11 kg/l
• 106 g/dm³ to kg/l = 0.11 kg/l
• 107 g/dm³ to kg/l = 0.11 kg/l
• 108 g/dm³ to kg/l = 0.11 kg/l
• 109 g/dm³ to kg/l = 0.11 kg/l
• 110 g/dm³ to kg/l = 0.11 kg/l
• 111 g/dm³ to kg/l = 0.11 kg/l
• 112 g/dm³ to kg/l = 0.11 kg/l
• 113 g/dm³ to kg/l = 0.11 kg/l
• 114 g/dm³ to kg/l = 0.11 kg/l
• 115 g/dm³ to kg/l = 0.12 kg/l
• 116 g/dm³ to kg/l = 0.12 kg/l
• 117 g/dm³ to kg/l = 0.12 kg/l
• 118 g/dm³ to kg/l = 0.12 kg/l
• 119 g/dm³ to kg/l = 0.12 kg/l
• 120 g/dm³ to kg/l = 0.12 kg/l
• 121 g/dm³ to kg/l = 0.12 kg/l
• 122 g/dm³ to kg/l = 0.12 kg/l
• 123 g/dm³ to kg/l = 0.12 kg/l
• 124 g/dm³ to kg/l = 0.12 kg/l
• 125 g/dm³ to kg/l = 0.13 kg/l
• 126
through
150
grams per cubic decimeter
• 126 g/dm³ to kg/l = 0.13 kg/l
• 127 g/dm³ to kg/l = 0.13 kg/l
• 128 g/dm³ to kg/l = 0.13 kg/l
• 129 g/dm³ to kg/l = 0.13 kg/l
• 130 g/dm³ to kg/l = 0.13 kg/l
• 131 g/dm³ to kg/l = 0.13 kg/l
• 132 g/dm³ to kg/l = 0.13 kg/l
• 133 g/dm³ to kg/l = 0.13 kg/l
• 134 g/dm³ to kg/l = 0.13 kg/l
• 135 g/dm³ to kg/l = 0.14 kg/l
• 136 g/dm³ to kg/l = 0.14 kg/l
• 137 g/dm³ to kg/l = 0.14 kg/l
• 138 g/dm³ to kg/l = 0.14 kg/l
• 139 g/dm³ to kg/l = 0.14 kg/l
• 140 g/dm³ to kg/l = 0.14 kg/l
• 141 g/dm³ to kg/l = 0.14 kg/l
• 142 g/dm³ to kg/l = 0.14 kg/l
• 143 g/dm³ to kg/l = 0.14 kg/l
• 144 g/dm³ to kg/l = 0.14 kg/l
• 145 g/dm³ to kg/l = 0.15 kg/l
• 146 g/dm³ to kg/l = 0.15 kg/l
• 147 g/dm³ to kg/l = 0.15 kg/l
• 148 g/dm³ to kg/l = 0.15 kg/l
• 149 g/dm³ to kg/l = 0.15 kg/l
• 150 g/dm³ to kg/l = 0.15 kg/l
• 151
through
175
grams per cubic decimeter
• 151 g/dm³ to kg/l = 0.15 kg/l
• 152 g/dm³ to kg/l = 0.15 kg/l
• 153 g/dm³ to kg/l = 0.15 kg/l
• 154 g/dm³ to kg/l = 0.15 kg/l
• 155 g/dm³ to kg/l = 0.16 kg/l
• 156 g/dm³ to kg/l = 0.16 kg/l
• 157 g/dm³ to kg/l = 0.16 kg/l
• 158 g/dm³ to kg/l = 0.16 kg/l
• 159 g/dm³ to kg/l = 0.16 kg/l
• 160 g/dm³ to kg/l = 0.16 kg/l
• 161 g/dm³ to kg/l = 0.16 kg/l
• 162 g/dm³ to kg/l = 0.16 kg/l
• 163 g/dm³ to kg/l = 0.16 kg/l
• 164 g/dm³ to kg/l = 0.16 kg/l
• 165 g/dm³ to kg/l = 0.17 kg/l
• 166 g/dm³ to kg/l = 0.17 kg/l
• 167 g/dm³ to kg/l = 0.17 kg/l
• 168 g/dm³ to kg/l = 0.17 kg/l
• 169 g/dm³ to kg/l = 0.17 kg/l
• 170 g/dm³ to kg/l = 0.17 kg/l
• 171 g/dm³ to kg/l = 0.17 kg/l
• 172 g/dm³ to kg/l = 0.17 kg/l
• 173 g/dm³ to kg/l = 0.17 kg/l
• 174 g/dm³ to kg/l = 0.17 kg/l
• 175 g/dm³ to kg/l = 0.18 kg/l
• 176
through
200
grams per cubic decimeter
• 176 g/dm³ to kg/l = 0.18 kg/l
• 177 g/dm³ to kg/l = 0.18 kg/l
• 178 g/dm³ to kg/l = 0.18 kg/l
• 179 g/dm³ to kg/l = 0.18 kg/l
• 180 g/dm³ to kg/l = 0.18 kg/l
• 181 g/dm³ to kg/l = 0.18 kg/l
• 182 g/dm³ to kg/l = 0.18 kg/l
• 183 g/dm³ to kg/l = 0.18 kg/l
• 184 g/dm³ to kg/l = 0.18 kg/l
• 185 g/dm³ to kg/l = 0.19 kg/l
• 186 g/dm³ to kg/l = 0.19 kg/l
• 187 g/dm³ to kg/l = 0.19 kg/l
• 188 g/dm³ to kg/l = 0.19 kg/l
• 189 g/dm³ to kg/l = 0.19 kg/l
• 190 g/dm³ to kg/l = 0.19 kg/l
• 191 g/dm³ to kg/l = 0.19 kg/l
• 192 g/dm³ to kg/l = 0.19 kg/l
• 193 g/dm³ to kg/l = 0.19 kg/l
• 194 g/dm³ to kg/l = 0.19 kg/l
• 195 g/dm³ to kg/l = 0.2 kg/l
• 196 g/dm³ to kg/l = 0.2 kg/l
• 197 g/dm³ to kg/l = 0.2 kg/l
• 198 g/dm³ to kg/l = 0.2 kg/l
• 199 g/dm³ to kg/l = 0.2 kg/l
• 200 g/dm³ to kg/l = 0.2 kg/l
#### Foods, Nutrients and Calories
MTR, SHAHI PANEER, UPC: 840965001519 contain(s) 141 calories per 100 grams or ≈3.527 ounces [ price ]
DILL PICKLES, UPC: 070152400010 contain(s) 18 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
CaribSea, Marine, CORALine, Caribbean Crushed Coral weighs 1 153.33 kg/m³ (72.00004 lb/ft³) with specific gravity of 1.15333 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Trap rock, solid weighs 2 883 kg/m³ (179.97981 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Engine Oil, SAE 10W-60 with temperature in the range of 0°C (32°F) to 100°C (212°F)
#### Weights and Measurements
kilogram per hectare (kg/ha) is a metric measurement unit of surface or areal density.
Electric car energy economy, i.e. energy or fuel economy of an electric vehicle is used to estimate mileage and electricity cost associated with and usage of the vehicle.
gr/US gal to g/metric c conversion table, gr/US gal to g/metric c unit converter or convert between all units of density measurement.
#### Calculators
Calculate volume in a horizontal cylinder or a cylindrical tank | 2,289 | 5,038 | {"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} | 2.828125 | 3 | CC-MAIN-2020-16 | latest | en | 0.45592 |
http://codingpractise.com/egypt/ | 1,596,994,664,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738562.5/warc/CC-MAIN-20200809162458-20200809192458-00186.warc.gz | 22,296,992 | 8,343 | Solution
Algorithm:
step1: Find out the largest arm of the triangle.
Step2: square of largest arm = sum of square of others two arms. If this condition is true then the triangle will be right otherwise wrong.
Source Code
```#include<stdio.h>
int main()
{
int sides[3],temp,i;
while(1)
{
scanf("%d%d%d",&sides[0],&sides[1],&sides[2]);
if(sides[0]==0 && sides[1]==0 && sides[2]==0)
break;
int flag=1;
int j=1;
while(flag)
{
flag=0;
for(i=0; i<3-j; i++)
{
if(sides[i]<sides[i+1])
{
temp=sides[i];
sides[i]=sides[i+1];
sides[i+1]=temp;
flag=1;
}
}
}
if(sides[0]*sides[0]== (sides[1]*sides[1]+sides[2]*sides[2]))
printf("right\n");
else
printf("wrong\n");
}
return 0;
}``` | 245 | 673 | {"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} | 2.8125 | 3 | CC-MAIN-2020-34 | latest | en | 0.279986 |
https://www.jiskha.com/display.cgi?id=1208555297 | 1,503,271,854,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106996.2/warc/CC-MAIN-20170820223702-20170821003702-00523.warc.gz | 906,450,358 | 4,499 | Algebra
posted by .
Below is my problem with my answers. I don't think this is that tough, but I cannot believe that my ABC answer's are a fraction. Did I do this correctly?
Find the unknowns on the right side of the partial fraction.
(x+3)/(x^3+64)=(A)/(x+4)+(Bx+C)/(x^2-4x+16)
x=0
A= (-x^2B-4xB-x+x-4C+3)/(x^2-4x+16)
B= (-x^2A+4xA-x(-16A+x-4C+3)/(x(x+4)
C= (-x^2B-4xB-x^2A+4x+x-16A+3)/(x+4)
• Algebra -
No, I would not think you should have a fraction.
• Algebra -
I multiplied both sides of your equation by (x+4)(x^2 - 4x + 16) which is really x^3 + 64
x+3 = A(x^2 - 4x + 16) + (x+4)(Bx+C)
this is an identity, so it is true for any value of x.
Let's pick "easy" values of x
let x=-4
-1 = A(48) + 0 ----> A = -1/48
let x=0
3 = 16A + 4C
3 = 16(-1/48) + 4C ----> C = 5/6
let x=1
4 = 13A + 5B + 5C
4 = 13(-1/48) + 5B + 5(5/6)
B = 1/48
Check my arithmetic, my weak point.
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More Similar Questions | 1,035 | 3,009 | {"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} | 4.03125 | 4 | CC-MAIN-2017-34 | latest | en | 0.879189 |
https://space.stackexchange.com/questions/8558/finding-non-lit-objects-at-light-minute-ranges/10537 | 1,723,675,642,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00620.warc.gz | 415,728,581 | 44,478 | # Finding non-lit objects at light minute ranges
Let's say my hyperjumpwarpgate McGuffin drive goes bing and I find myself about half a light year away from the sun. This is right in the heart of the outer Oort Cloud. Wikipedia says that there are a trillion objects there larger than 1km in size, and they're about 30 light seconds apart.
So I want to find the closest one of these. How?
• Out there it's very dark, so I'm unlikely to be able to pick one up by reflected sunlight.
• Radar's possible, but it's going to be a very slow process scanning the complete sky with a narrow-angle high-power beam, particularly as I'm going to have to wait several minutes for each return ping. OTOH Venus has been mapped by radar from Earth, so it should at least work.
• I've seen suggestions of using infrared to look for very distant planets; but I suspect these are looking for large bodies with a certain amount of their own heat, while the small bodies I'm looking for are going to be very cold.
• Would it be feasible to look for shadows against the interstellar background? I'd have to photograph the entire sky at insane resolution, then move, photograph it again, and look for parallax. That would only work for objects at the right vector to my motion, and of course only where there was some sort of bright background --- but there's quite a lot of Milky Way.
What's the most practical real-world method of doing this?
Edit:
So to calculate the apparent brightness of the sun from that distance: half a light year is $180*24*3600$ light seconds, which is $\frac{180*24*3600}{8*60} = 32400$ AU. That means that it should be $\frac{1}{32400^2} = \frac{1}{10^{10}}$ the brightness that it is here. That's quite dim.
Given that apparent magnitude is:
$m_x - m_{x,0} = -2.5 log_{10} \frac {F_x}{F_{x,0}}$
That gives us, for the sun:
$m_x - -26.74 = -2.5 log_{10} \frac {1 / 10^{10}}{1}$
...or:
$m_x = -2.5 log_{10} (10^{-10}) - 26.74$
...so $m_x$ is -1.74, which makes the Sun about the same brightness as Sirius.
(I was totally not expecting the maths above to produce a meaningful result.)
Edit edit:
(Yes, I'm bored at work.)
The angular diameter of an object is $\tan \frac{\text {size}}{\text {distance}}$, which means that a 1km body at 30 light seconds is going to be $\tan \frac{1000}{9 \times 10^9}$, which is about $5 \times 10^{-6}$ degrees. Or about 0.02 arcseconds.
Wikipedia has a handy table of diffraction limits vs telescopes. Turns out that the Hubble could resolve that 1km body, provided it was radiating in ultraviolet, which is probably unlikely. For high infrared I'd need a telescope with an aperture of several hundred metres, which is doable, but for low infrared I would need something 10km across, which is harder.
It may, however, be possible to detect the body without having to resolve it (i.e. the same way we can see stars); but I don't know how that works.
• While at that distance the Sun will only appear to be a very bright object, there will be other sources of light out there. Still, an interesting question Hmmm... Commented Mar 27, 2015 at 15:16
• I'm inclined toward the radar plan. The round-trip time isn't really an issue if your receiving antenna is separate from and less directionally sensitive than your emitter. Commented Mar 27, 2015 at 17:19
• And it would be even harder to find one that has dilithium crystals. Commented Aug 8, 2015 at 4:16
First, everything in the Oort Cloud is really, really far apart. You could spend years looking and never see anything because of the reasons you gave. | 918 | 3,569 | {"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} | 3.515625 | 4 | CC-MAIN-2024-33 | latest | en | 0.962275 |
https://metanumbers.com/9020 | 1,656,199,630,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036176.7/warc/CC-MAIN-20220625220543-20220626010543-00540.warc.gz | 444,432,790 | 7,352 | # 9020 (number)
9,020 (nine thousand twenty) is an even four-digits composite number following 9019 and preceding 9021. In scientific notation, it is written as 9.02 × 103. The sum of its digits is 11. It has a total of 5 prime factors and 24 positive divisors. There are 3,200 positive integers (up to 9020) that are relatively prime to 9020.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 11
• Digital Root 2
## Name
Short name 9 thousand 20 nine thousand twenty
## Notation
Scientific notation 9.02 × 103 9.02 × 103
## Prime Factorization of 9020
Prime Factorization 22 × 5 × 11 × 41
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 4510 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 9,020 is 22 × 5 × 11 × 41. Since it has a total of 5 prime factors, 9,020 is a composite number.
## Divisors of 9020
1, 2, 4, 5, 10, 11, 20, 22, 41, 44, 55, 82, 110, 164, 205, 220, 410, 451, 820, 902, 1804, 2255, 4510, 9020
24 divisors
Even divisors 16 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 21168 Sum of all the positive divisors of n s(n) 12148 Sum of the proper positive divisors of n A(n) 882 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 94.9737 Returns the nth root of the product of n divisors H(n) 10.2268 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 9,020 can be divided by 24 positive divisors (out of which 16 are even, and 8 are odd). The sum of these divisors (counting 9,020) is 21,168, the average is 882.
## Other Arithmetic Functions (n = 9020)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 3200 Total number of positive integers not greater than n that are coprime to n λ(n) 40 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1124 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 3,200 positive integers (less than 9,020) that are coprime with 9,020. And there are approximately 1,124 prime numbers less than or equal to 9,020.
## Divisibility of 9020
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 4 4 2
The number 9,020 is divisible by 2, 4 and 5.
• Arithmetic
• Abundant
• Polite
• Practical
## Base conversion (9020)
Base System Value
2 Binary 10001100111100
3 Ternary 110101002
4 Quaternary 2030330
5 Quinary 242040
6 Senary 105432
8 Octal 21474
10 Decimal 9020
12 Duodecimal 5278
20 Vigesimal 12b0
36 Base36 6yk
## Basic calculations (n = 9020)
### Multiplication
n×y
n×2 18040 27060 36080 45100
### Division
n÷y
n÷2 4510 3006.67 2255 1804
### Exponentiation
ny
n2 81360400 733870808000 6619514688160000 59708022487203200000
### Nth Root
y√n
2√n 94.9737 20.8162 9.74544 6.18075
## 9020 as geometric shapes
### Circle
Diameter 18040 56674.3 2.55601e+08
### Sphere
Volume 3.07403e+12 1.0224e+09 56674.3
### Square
Length = n
Perimeter 36080 8.13604e+07 12756.2
### Cube
Length = n
Surface area 4.88162e+08 7.33871e+11 15623.1
### Equilateral Triangle
Length = n
Perimeter 27060 3.52301e+07 7811.55
### Triangular Pyramid
Length = n
Surface area 1.4092e+08 8.64875e+10 7364.8
## Cryptographic Hash Functions
md5 f1e709e6aef16ba2f0cd6c7e4f52b9b6 0cc67fbdd393bba8efbc3d4ce543ab18b648cefb f30c80f183c3c7c92131db3fecefd08cf5b38dd2153300faa216c6d05580965a e65ab88880dc17aaaee3e0a20165c9c7cee6844cba776ce11f806cd2a265e2707fbbc86e1685c62aee6ad54cc24bc6308aeb119808674095b77e94700fc73b57 4ef437503e500aa3404f7a2f48dc83f6a08f5f8b | 1,492 | 4,065 | {"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} | 3.796875 | 4 | CC-MAIN-2022-27 | latest | en | 0.804085 |
http://www.r-bloggers.com/circle-packing-with-r/ | 1,469,500,701,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824570.25/warc/CC-MAIN-20160723071024-00152-ip-10-185-27-174.ec2.internal.warc.gz | 648,236,843 | 17,340 | # Circle packing with R
July 26, 2010
By
(This article was first published on Last Resort Software, and kindly contributed to R-bloggers)
To visualize the results of a simulation model of woodland trees within R, I needed an algorithm that could arrange a large number of circles within a rectangle such that no two circles overlapped by more than a specified amount. A colleague had approached this problem earlier by sorting the circles in order of descending size, then randomly dropping each one into the rectangle repeatedly until it landed in a position with acceptable overlap.
I suspected a faster and more robust algorithm could be constructed using some kind of “jiggling the circles” approach. Luckily for me, I discovered that Sean McCullough had written a really nice example of circles packing into a cluster using the Processing language. Sean’s program is based on an iterative pair-repulsion algorithm in which overlapping circles move away from each other. Based on this, and modifying the algorithm a little, I came up with an R function to produce constrained random layouts of a given set of circles. Here are two layouts for the same input set…
The algorithm is wonderfully simple. Each circle in the input list is compared to those following it. If two circles overlap by more than the allowed amount, they are moved apart until the overlap is acceptable. The distance moved by each circle is proportional to the radius of the other to approximate inertia (very loosely); thus when a small circle is overlapped by a large circle, the small circle moves furthest. This process is repeated iteratively until no more movement takes place (acceptable layout) or a maximum number of iterations is reached (layout failure). To avoid edge effects, the bounding rectangle is treated as a toroid. For my purposes, I only require a circle’s centre to be within the plotted rectangle.
You can see in the plots that the algorithm tends to produce clusters of small circles around big ones. For the woodland simulation model this is a nice property (saplings clustering around parent trees) but for other applications the algorithm could be further modified to lessen or avoid this effect.
The code for the basic function is below. The set of input circles are described by the config matrix argument. Although this function produces good results, it takes a long time to run when the number of circles is large. However, a second version of the function, with most of the calculations moved into C code, runs much faster (code not posted here but available on request).
``` ```
``pack.circles <- function(config, size=c(100, 100), max.iter=1000, overlap=0 ) { # # Simple circle packing algorithm based on inverse size weighted repulsion # # config - matrix with two cols: radius, N # size - width and height of bounding rectangle # max.iter - maximum number of iterations to try # overlap - allowable overlap expressed as proportion of joint radii # ============================================================================ # Global constants # ============================================================================ # round-off tolerance TOL <- 0.0001 # convert overlap to proportion of radius if (overlap < 0 | overlap >= 1) { stop("overlap should be in the range [0, 1)") } PRADIUS <- 1 - overlap NCIRCLES <- sum(config[,2]) # ============================================================================ # Helper function - Draw a circle # ============================================================================ draw.circle <- function(x, y, r, col) { lines( cos(seq(0, 2*pi, pi/180)) * r + x, sin(seq(0, 2*pi, pi/180)) * r + y , col=col ) } # ============================================================================ # Helper function - Move two circles apart. The proportion of the required # distance moved by each circle is proportional to the size of the other # circle. For example, If a c1 with radius r1 overlaps c2 with radius r2, # and the movement distance required to separate them is ds, then c1 will # move ds * r2 / (r1 + r2) while c2 will move ds * r1 / (r1 + r2). Thus, # when a big circle overlaps a little one, the little one moves a lot while # the big one moves a little. # ============================================================================ repel <- function(xyr, c0, c1) { dx <- xyr[c1, 1] - xyr[c0, 1] dy <- xyr[c1, 2] - xyr[c0, 2] d <- sqrt(dx*dx + dy*dy) r <- xyr[c1, 3] + xyr[c0, 3] w0 <- xyr[c1, 3] / r w1 <- xyr[c0, 3] / r if (d < r - TOL) { p <- (r - d) / d xyr[c1, 1] <<- toroid(xyr[c1, 1] + p*dx*w1, 1) xyr[c1, 2] <<- toroid(xyr[c1, 2] + p*dy*w1, 2) xyr[c0, 1] <<- toroid(xyr[c0, 1] - p*dx*w0, 1) xyr[c0, 2] <<- toroid(xyr[c0, 2] - p*dy*w0, 2) return(TRUE) } return(FALSE) } # ============================================================================ # Helper function - Adjust a coordinate such that if it is distance d beyond # an edge (ie. outside the area) it is moved to be distance d inside the # opposite edge. This has the effect of treating the area as a toroid. # ============================================================================ toroid <- function(coord, axis) { tcoord <- coord if (coord < 0) { tcoord <- coord + size[axis] } else if (coord >= size[axis]) { tcoord <- coord - size[axis] } tcoord } # ============================================================================ # Main program # ============================================================================ # ------------------------------------------ # create a random initial layout # ------------------------------------------ xyr <- matrix(0, NCIRCLES, 3) pos0 <- 1 for (i in 1:nrow(config)) { pos1 <- pos0 + config[i,2] - 1 xyr[pos0:pos1, 1] <- runif(config[i, 2], 0, size[1]) xyr[pos0:pos1, 2] <- runif(config[i, 2], 0, size[2]) xyr[pos0:pos1, 3] <- config[i, 1] * PRADIUS pos0 <- pos1 + 1 } # ------------------------------------------ # iteratively adjust the layout # ------------------------------------------ for (iter in 1:max.iter) { moved <- FALSE for (i in 1:(NCIRCLES-1)) { for (j in (i+1):NCIRCLES) { if (repel(xyr, i, j)) { moved <- TRUE } } } if (!moved) break } cat(paste(iter, "iterations\n")); # ------------------------------------------ # draw the layout # ------------------------------------------ plot(0, type="n", xlab="x", xlim=c(0,size[1]), ylab="y", ylim=c(0, size[2])) xyr[, 3] <- xyr[, 3] / PRADIUS for (i in 1:nrow(xyr)) { draw.circle(xyr[i, 1], xyr[i, 2], xyr[i, 3], "gray") } # ------------------------------------------ # return the layout # ------------------------------------------ colnames(xyr) <- c("x", "y", "radius") invisible(xyr) }``
``` ```
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Tags: | 1,789 | 7,222 | {"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} | 2.96875 | 3 | CC-MAIN-2016-30 | latest | en | 0.956145 |
http://list.seqfan.eu/pipermail/seqfan/2010-July/005361.html | 1,670,188,347,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00534.warc.gz | 23,280,422 | 3,481 | # [seqfan] Re: An algorithm of calculation of multiplicative order of 2 mod n
Sat Jul 24 05:38:04 CEST 2010
```Phenomenon with the average depth=2 of a step in divisions method (in my version) is characterized
rather simply. Indeed, the average depth=(l_1+...+l_k)/k, where k=A179382(n), thus itis
ord_(2*n-1)(2)/A179382(n), and if this ratio is 2, then such numbers are 2*A179460-1. I have already written that the arithmetic sense of k=A179382(n) is the number of odd residues in {1,2,...,2^ord_(2*n-1)(2)} (considered in the reduced residue system modulo 2*n-1). Thus the observed phenomenon takes place for the numbers 2*n-1 for which the set {1,2,...,2^ord_(2*n-1)(2)} contains the same number of odd and even residues modulo 2*n-1. For other numbers (if n>1) always more even residues and the considered average depth of a step is more than 2, i.e., we have "adrenalin-numbers".
If anyone can estimate an upper (lower) density of such numbers?
Regards,
----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Friday, July 23, 2010 12:56
Subject: [seqfan] Re: An algorithm of calculation of multiplicative order of 2 mod n
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Thus to get ord_(2n-1)(2) we have the standard algorithm of
> multiplications by 2 modulo n (by the definition) and the
> algorithm of divisions by 2 modulo n. It is asked, why I and
> other participants of the discussion have not immediately
> observed the essentially trivial Max's explanation? The answer
> is simple: because of "pits", when we have: A007814(x)>=2. For
> the number of pits for 2n-1 I get sequence: 0 1 1 1 1 3 3 1 1 5
> 1 3 5 5 7 1 1 3 (...) (more terms?). Further, I wonder what is
> the average "depth" of a step in divisions method?
> Astonishingly, I get exactly 2 for many numbers 2n-1.
> They are: 3 5 9 11 13 17 19 25 27 29 33(...) and more than 2 for
> 7 15 21 23 31 35(...). The latter I call "adrenalin-numbers".
>
> Regards,
>
>
> ----- Original Message -----
> From: Vladimir Shevelev <shevelev at bgu.ac.il>
> Date: Thursday, July 22, 2010 8:59
> Subject: [seqfan] Re: An algorithm of calculation of
> multiplicative order of 2 mod n
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>
> > Of course, one can calculate ord_n(a) by this algorithm ((n,a)=1).
> > Examples:
> > to calculate ord_10(3) we have
> > 10+1 0 11
> > 10+11 1 7
> > 10+7 0 17
> > 10+17 3 1
> > Thus ord_10(3)=1+3=4;
> > to calculate ord_11(4) we have
> > 11+1 1 3
> > 11+14 0 25
> > 11+25 1 9
> > 11+9 1 5
> > 11+5 2 1
> > Thus ord_11(4)=1+1+1+2=5.
> >
> > Regards,
> >
> >
> >
> >
> > ----- Original Message -----
> > From: Max Alekseyev <maxale at gmail.com>
> > Date: Wednesday, July 21, 2010 23:05
> > Subject: [seqfan] Re: An algorithm of calculation of
> > multiplicative order of 2 mod n
> > To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> >
> > > The algorithm iteratively performs divisions by 2 modulo n.
> > > It starts with n+1 == 1 (mod n) and ends with 1. The total
> > > number of
> > > divisions k is then nothing else but the multiplicative
> order
> > of
> > > 2 mod
> > > n.
> > >
> > > Max
> > >
> > > On Tue, Jul 20, 2010 at 2:18 PM, Vladimir Shevelev
> > > <shevelev at bgu.ac.il> wrote:
> > > > Dear Seq Fan,
> > > >
> > > > I would like to present an easy (till now-
> > conjectural!)
> > > algorithm of calculation of multiplicative order of 2 mod n
> (n
> > > arbitrary positive odd number). It is based on trivial
> > > calculation of A007814(x) in base 2.
> > > >
> > > > Step1. l(1)=A007814(n+1), m(1)=(n+1)/2^l(1);
> > > > Step i(i>=2). l(i)=A007814(n+m(i-1)), m(i)=(n+m(i-
> > 1))/2^l(i);> > The process ends when m=1.
> > > >
> > > > Now we have: A002326(n)=l(1)+...+l(k).
> > > >
> > > > Example (here A007814=a(n)).
> > > >
> > > > a(17+1)=1, (17+1)/2=9;
> > > > a(17+9)=1, (17+9)/2=13;
> > > > a(17+13)=1, (17+13)/2=15;
> > > > a(17+15)=5, (17+15)/(2^5)=1.
> > > >
> > > > Thus A002327(17)=1+1+1+5=8.
> > > >
> > > > Regards,
> > > >
> > > > Shevelev Vladimir
> > > >
> > > > _______________________________________________
> > > >
> > > > Seqfan Mailing list - http://list.seqfan.eu/
> > > >
> > >
> > >
> > > _______________________________________________
> > >
> > > Seqfan Mailing list - http://list.seqfan.eu/
> > >
> >
> >
> > _______________________________________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
> | 1,596 | 4,491 | {"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-2022-49 | latest | en | 0.817102 |
https://www.yummymath.com/tag/5.MD,5.MD.1,5.MD.3,5.MD5/ | 1,701,612,662,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.23/warc/CC-MAIN-20231203125921-20231203155921-00041.warc.gz | 1,230,142,082 | 23,124 | # Tag: 5.MD
## Thanksgiving, football, shopping and giving
Thirteen provocative activities for this holiday season of Thanksgiving, football, and shopping.
## What is a quantum dot?
Students learn about nanotechnology and why it works by looking closely at surface area and its comparison to volume. Hopefully they will begin to understand the incredible possibilities of this science.
## Veterans Day, 2023
Population and Density maps of where Veterans live. What can you observe? What do you wonder? How do they compare?
## Yummy Pumpkin Pie
23 people are coming to my house for pumpkin pie and a hike. How many pies should I make? How much of each ingredient will I need? What quantities should I buy? Engage your students in estimation, multiplication of fractions and proportional reasoning.
## Heating with wood – What’s a cord?
Students get a visual and geometric understanding of the measurements and shapes of a cord of wood while using the relative size of a ruler in various images.
## Will the Barbie movie out-perform Avatar at the box office?
Do you think that Barbie can out-earn Avatar? Students study the numbers, recognize the need for an inflation adjustment, get to share their observations about the movie and its messages, and draw their own conclusions about how far Barbie will go as a money-maker.
## What do you know about the Hawaiian islands?
Let your students explore Hawaii and better understand its depths, heights, and origins as they compare maps and deduce info from charts.
## Ice Cream Day – July 16th
We have two timely activities … Updated! All the ice cream that you can eat All the ice cream that you can eat for only 10 cents! Wow, that’s a great deal! How great a deal was that? The history…
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## 200 people
Let your students ponder this gif in full screen and then let them share what they notice and why this gif might have been created.
## Father’s Day and Juneteenth
For Father’s Day Thank you Mother and Father for all of those diapers – Students compare the cost of buying disposable versus cloth diapers. They estimate how much they cost their parents in diapers and consider how much they will…
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How much water should you drink? How much pee is in this pool? How has the presence of lyme disease changed? Which is the best drink deal at the fair?
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## Effects of new MLB rules
Students use our recent data to graph, analyze and hypothesize about the affect of the new MLB rule changes. | 671 | 3,167 | {"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} | 3.0625 | 3 | CC-MAIN-2023-50 | longest | en | 0.945192 |
https://www.sarthaks.com/914210/steel-length-has-cross-sectional-area-find-force-required-stretch-this-rod-the-same-amount | 1,670,531,359,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711360.27/warc/CC-MAIN-20221208183130-20221208213130-00530.warc.gz | 1,021,741,316 | 15,434 | # A steel rod of length 25 cm has a cross - sectional area of 0.8 cm^2 . Find the force required to stretch this rod by the same amount
534 views
retagged
A steel rod of length 25 cm has a cross - sectional area of 0.8 cm2 . Find the force required to stretch this rod by the same amount as the expansion produced by heating it through 10°C. (coefficient of linear expansion of steel is 10-5/°C and young's modulus of steel is 2 x 1010 Nm-2)
+1 vote
by (52.7k points)
selected
This required force is given by F = YA$aΔt$ = 2 x 1010 x 0.8 x 10-4 x 10-5 x 10 = 160 N | 186 | 571 | {"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} | 3.390625 | 3 | CC-MAIN-2022-49 | latest | en | 0.898484 |
https://www.reddit.com/user/Always_smooth?sort=controversial&t=month | 1,475,067,413,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661367.29/warc/CC-MAIN-20160924173741-00177-ip-10-143-35-109.ec2.internal.warc.gz | 1,003,200,982 | 21,207 | [–] 296 points297 points (0 children)
Let them yell. Decibels from a human has never killed or remotely injured another human before. Wrestling while someone's arms are in a fragile and locked position has.
Edit: a letter.
[–] -1 points0 points (0 children)
RemindMe! 2 weeks
[–] 1 point2 points (0 children)
Support main here. All you need is a stun and some utility.
From a supports mind set you have:
• A shield
• A stun
• A slow from the shield for disengage
• Bushes to hide from AA or target spells
• Something to help ADC get golems, grog, or buffs.
• An ult that provides presence and a knockup.
It's not a bad kit for support to be honest.
[–] 7 points8 points (0 children)
Hypothetically speaking:
Teacher salary per year: 50,000 (a little lower than national average based on a 2 second Google search)
Days taught per year: 180 (quick estimate for ez math)
Hours taught: (180*8)=1440
Hourly wage: (50,000/1440)=~34.72
If they teach 8 classes with about 20 in each = 160 students taught.
If paid equally per student:(34.72/160)=0.2170 dollars per hour (assuming 0 overlapped students)
Total per year each student pays teacher: (0.2170*1440)=312.50
Edit: math is now fixed. I think the teacher was either loaded or sucked at math.
Edit 2: u/Enceladus7 brought up a great point. My calculations only account if the teacher was paid 100% by students, thus this is the maximum each student would pay. However, it's more likely that the students don't pay 100% of the teachers salary (crazy, I know) this would reduce the amount by quite a bit. I'll see if I can find a national rate within 2 minutes.
Edit 3: I can't find a private high school tuition break down so I'll say students only pay for 10% of the teachers tuition, meaning that each student then would pay \$31.50 per year. This makes the number not as absurd.
[–] 55 points56 points (0 children)
I dont know, to me it looked like one great white shark bitting another.
[–] 20 points21 points (0 children)
This seems more like a legal question. If you had something taken I would head over to r/legaladvice.
[–] 343 points344 points (0 children)
He is with four other Dr actors who all have the same poses. He has to shuffle through his poses to find one no one is doing. Hilarious.
[–] 18 points19 points (0 children)
It seems to be what they respond to.
[–] 0 points1 point (0 children)
[–] 3 points4 points (0 children)
Yes, they went with the stand up, shout, and flail. It's an oldie, but a goodie.
[–] 0 points1 point (0 children)
Of OP'S office and tells OP to get the fuck out.
[–] 0 points1 point (0 children)
"OMG I just skidded into the dirt. I'm so embarrassed I could die!"
[–] 2 points3 points (0 children)
Just the way he likes his strudes.
[–] 0 points1 point (0 children)
This is because land is less dense than air. This is also why they float in water.
[–] 1 point2 points (0 children)
The joke must have had one of those pills too.
[–] 0 points1 point (0 children)
The letter 'R' or the sound 'R'? Just curious.
[–] 2 points3 points (0 children)
We have a drowning in our lake almost ever summer. This is due to the fact that we used to have an amusement park next to the lake that burnt down and they pushed the burnt wreckage into the lake causing much more intense undertows. (holy run on sentence)
It almost seems normal to drive by and see the diving team there.
[–] 0 points1 point (0 children)
You were going to say "vagine madority", you transposed three sounds; very impressive.
[–] 2 points3 points (0 children)
If you try to do a projection 1700 years into the future you can't use a regression with two years of data. That's not how math or statistics work.
[–] 0 points1 point (0 children)
History of DVT (deep vein thrombosis). That is not to the brain it is from the brain/head.
[–] 0 points1 point (0 children)
4 oceans!
[–] 0 points1 point (0 children)
Or a deformed skull.
[–] 0 points1 point (0 children)
[Insert word] pandemic. No, pandemics are usually a serious issues. Don't use this at every turn.
[–] 2 points3 points (0 children)
They do, they're called PET scans. Cats just perform a specific type of scan. | 1,171 | 4,210 | {"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} | 3.125 | 3 | CC-MAIN-2016-40 | latest | en | 0.934221 |
https://www.coursehero.com/file/6354127/plugin-Chapter-14-Students/ | 1,495,998,135,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463610374.3/warc/CC-MAIN-20170528181608-20170528201608-00415.warc.gz | 1,077,956,247 | 126,275 | plugin-Chapter 14 Students
plugin-Chapter 14 Students - Describing Relationships...
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CHAPTER 14 Describing Relationships: Scatterplots & Correlation
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RELATIONSHIPS So far, we’ve only examined the distribution of a single variable. We can look at some sort of graphical display or numerical summary. Often times, however, we would like to see how one variable relates to another variable.
SCATTERPLOTS A scatterplot is the most common way to display the relationship between two ___________ variables measured on the same individual. The values of one variable appear on the x-axis and the values of the other variable appear on the y-axis. If we have an explanatory variable, that goes on the x- axis. The response variable then goes on the y-axis. Each individual in the data appears as the point in the plot fixed by the values of both variables for that individual.
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DESCRIBING SCATTERPLOTS As usual, we want to look for the overall pattern and any striking deviations from that pattern. We describe the overall pattern of a scatterplot by talking about the form, direction, and strength of the relationship. The form of a relationship refers to whether our scatterplot takes a linear pattern or a curved pattern.
DESCRIBING SCATTERPLOTS The direction of a scatterplot refers to whether we have
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Ask a homework question - tutors are online | 447 | 2,096 | {"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} | 3.1875 | 3 | CC-MAIN-2017-22 | longest | en | 0.878379 |
https://www.unitconverters.net/area/break-to-square-mile.htm | 1,721,122,656,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514742.26/warc/CC-MAIN-20240716080920-20240716110920-00664.warc.gz | 922,341,224 | 3,351 | Home / Area Conversion / Convert Break to Square Mile
# Convert Break to Square Mile
Please provide values below to convert break to square mile [mi^2], or vice versa.
From: break To: square mile
### Break to Square Mile Conversion Table
BreakSquare Mile [mi^2]
0.01 break3861021.5854245 mi^2
0.1 break38610215.854245 mi^2
1 break386102158.54245 mi^2
2 break772204317.08489 mi^2
3 break1158306475.6273 mi^2
5 break1930510792.7122 mi^2
10 break3861021585.4245 mi^2
20 break7722043170.8489 mi^2
50 break19305107927.122 mi^2
100 break38610215854.245 mi^2
1000 break386102158542.45 mi^2
### How to Convert Break to Square Mile
1 break = 386102158.54245 mi^2
1 mi^2 = 2.589988110336E-9 break
Example: convert 15 break to mi^2:
15 break = 15 × 386102158.54245 mi^2 = 5791532378.1367 mi^2 | 283 | 790 | {"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} | 3.125 | 3 | CC-MAIN-2024-30 | latest | en | 0.470032 |
http://www.mathskey.com/question2answer/5705/what-are-the-steps-to-prove-the-trigonometry-identity | 1,708,612,274,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473819.62/warc/CC-MAIN-20240222125841-20240222155841-00898.warc.gz | 57,682,555 | 10,265 | # What are the steps to prove the Trigonometry identity?
sec(x).cosec(x)+sin(x).sec(x)=cosec(x).sec(x)
Given Sec(x)Cosec(x)+Sin(x)Sec(x) = Cosec(x)Sec(x)
Now i assume it to Sec(x)Cosec(x)+Sin(x)Sec(x) = Cosec(x)Sec(x)Cos^2x
Lefthandside identity = Sec(x)Cosec(x)+Sin(x)Sec(x)
Common out Sec(x).
= Sec(x)(Cosecx+Sinx)
= Sec(x)(1/Sinx+Sinx)
= Sec(x)(1+Sin^2x)/Sinx
= Sec(x)(cos^2x)/Sin(x)
= Sec(x)Cosec(x)(cos^2x)
= riight hand side identity | 185 | 450 | {"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-2024-10 | longest | en | 0.445132 |
https://www.jiskha.com/display.cgi?id=1330528550 | 1,526,990,487,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864725.4/warc/CC-MAIN-20180522112148-20180522132148-00075.warc.gz | 765,660,988 | 3,644 | # Math
posted by Sally
undefined is what % of undefined?
1. PsyDAG
Insufficient data.
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More Similar Questions | 452 | 1,609 | {"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} | 3.28125 | 3 | CC-MAIN-2018-22 | latest | en | 0.840564 |
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# Political Candidate: Government subsidized prescription drug
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Political Candidate: Government subsidized prescription drug [#permalink]
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12 Jul 2008, 11:02
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Political Candidate: Government subsidized prescription drug plans that would allow individuals significant choice in determining their benefits and costs are deceptively appealing to numerous stakeholders. However, buying prescription drug coverage, like buying health insurance coverage, is not like buying a car. The consumer cannot predict his or her future health needs. Moreover, the administrators of the choice-based drug plans under consideration are allowed to change the drugs they cover and the prices they charge at any time; this renders informed consumer choice meaningless and makes securing appropriate coverage a crap shoot. Older and disabled individuals, the predominant consumers of government subsidized prescription drug plans, should be offered drug coverage alternatives that do not force them to gamble with their health.
In the argument above, the two portions in boldface play which of the following roles?
A.The first is a fact that the candidate argues against; the second is the ultimate claim that the candidate supports.
B.The first is an observation which the candidate acknowledges as true but to which he is ultimately opposed; the second is a claim that the candidate uses as evidence to support his ultimate position.
C.The first is an observation that the candidate acknowledges as true but unfortunate; the second is an assertion that the candidate makes to support his ultimate position.
D.The first is an observation that the candidate argues against; the second is an observation that the candidate supports.
E.The first is an observation made by the candidate; the second is an assertion that the candidate ultimately opposes.
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### Show Tags
12 Jul 2008, 11:27
1
yavasani wrote:
Political Candidate: Government subsidized prescription drug plans that would allow individuals significant choice in determining their benefits and costs are deceptively appealing to numerous stakeholders. However, buying prescription drug coverage, like buying health insurance coverage, is not like buying a car. The consumer cannot predict his or her future health needs. Moreover, the administrators of the choice-based drug plans under consideration are allowed to change the drugs they cover and the prices they charge at any time; this renders informed consumer choice meaningless and makes securing appropriate coverage a crap shoot. Older and disabled individuals, the predominant consumers of government subsidized prescription drug plans, should be offered drug coverage alternatives that do not force them to gamble with their health.
The buyer may or may not the same diseases in future. Therefore, rather than subsidizing the prescription drug plans, government should offer some coverage alternatives.
In the argument above, the two portions in boldface play which of the following roles?
A.The first is a fact that the candidate argues against; the second is the ultimate claim that the candidate supports.
Wrong. The ultimate claim is that this is not an effective way. Goverment should some alternative drug coverage plan.
B.The first is an observation which the candidate acknowledges as true but to which he is ultimately opposed; the second is a claim that the candidate uses as evidence to support his ultimate position.
Close but wrong. The observation is true government plan but author is against it. But author is using "The consumer cannot predict his or her future health needs" assertion to make the cliam that this is not a good plan. Use some alternative...drug coverage plan
C.The first is an observation that the candidate acknowledges as true but unfortunate; the second is an assertion that the candidate makes to support his ultimate position.
Correct. Second "consumer cannot predict his or her future health needs" is the asertion that author is making to support his claim.
D.The first is an observation that the candidate argues against; the second is an observation that the candidate supports.
Wrong. Second is not an observation.
E.The first is an observation made by the candidate; the second is an assertion that the candidate ultimately opposes.
Wrong. Author is not opposing second.
IMO C but very confused between B and C
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12 Jul 2008, 11:45
IMO C. The first is an observation that the candidate acknowledges as true but unfortunate; the second is an assertion that the candidate makes to support his ultimate position.
Political Candidate: Government subsidized prescription drug plans that would allow individuals significant choice in determining their benefits and costs are deceptively appealing to numerous stakeholders. The candidate agrees that the drug plan would allow individuals significant choice (TRUE) but it is deceptively appealing (unfortunate).
The consumer cannot predict his or her future health needs.. This is an assertion because the next sentence strats with "moreover" and instroduces more seertion. "Moreover, the administrators of the choice-based drug plans under consideration are allowed to change the drugs they cover and the prices they charge at any time;
Conclusion in the passage is: this renders informed consumer choice meaningless and makes securing appropriate coverage a crap shoot. Older and disabled individuals, the predominant consumers of government subsidized prescription drug plans, should be offered drug coverage alternatives that do not force them to gamble with their health.
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### Show Tags
12 Jul 2008, 11:50
IMO D.
But C is also really close.
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12 Jul 2008, 12:02
x97agarwal wrote:
IMO C. The first is an observation that the candidate acknowledges as true but unfortunate; the second is an assertion that the candidate makes to support his ultimate position.
Political Candidate: Government subsidized prescription drug plans that would allow individuals significant choice in determining their benefits and costs are deceptively appealing to numerous stakeholders. The candidate agrees that the drug plan would allow individuals significant choice (TRUE) but it is deceptively appealing (unfortunate).
The consumer cannot predict his or her future health needs.. This is an assertion because the next sentence strats with "moreover" and instroduces more seertion. "Moreover, the administrators of the choice-based drug plans under consideration are allowed to change the drugs they cover and the prices they charge at any time;
Conclusion in the passage is: this renders informed consumer choice meaningless and makes securing appropriate coverage a crap shoot. Older and disabled individuals, the predominant consumers of government subsidized prescription drug plans, should be offered drug coverage alternatives that do not force them to gamble with their health.
Why B is wonrg? What is the main diff between conclusion and assertion?
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### Show Tags
12 Jul 2008, 12:29
rao_1857 wrote:
x97agarwal wrote:
IMO C. The first is an observation that the candidate acknowledges as true but unfortunate; the second is an assertion that the candidate makes to support his ultimate position.
Political Candidate: Government subsidized prescription drug plans that would allow individuals significant choice in determining their benefits and costs are deceptively appealing to numerous stakeholders. The candidate agrees that the drug plan would allow individuals significant choice (TRUE) but it is deceptively appealing (unfortunate).
The consumer cannot predict his or her future health needs.. This is an assertion because the next sentence strats with "moreover" and instroduces more seertion. "Moreover, the administrators of the choice-based drug plans under consideration are allowed to change the drugs they cover and the prices they charge at any time;
Conclusion in the passage is: this renders informed consumer choice meaningless and makes securing appropriate coverage a crap shoot. Older and disabled individuals, the predominant consumers of government subsidized prescription drug plans, should be offered drug coverage alternatives that do not force them to gamble with their health.
Why B is wonrg? What is the main diff between conclusion and assertion?
Good Question Rao. In the first bold face the second part is wrong - it is not an evidence.
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### Show Tags
12 Jul 2008, 13:25
i dont think D is correct..the second part is not an observation but an assertion..this is the view of the politician and he wants you to believe it..
therfore C is best
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12 Jul 2008, 13:36
I am with fresinha .... i went with C because the second boldface is really an assertion
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12 Jul 2008, 13:49
pmenon wrote:
I am with fresinha .... i went with C because the second boldface is really an assertion
Hi All,
Still confused ......
What is the main diff between conclusion and assertion?????
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Re: Political Candidate: Government subsidized prescription drug [#permalink]
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10 Nov 2012, 07:32
Though I am a bit late, my assessment of C is that nowhere the first boldface has been proclaimed as unfortunate.
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Re: Political Candidate: Government subsidized prescription drug [#permalink]
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11 Dec 2012, 09:46
B for me. In C, I don't think the first bold face is unfortunate. B is clearer.
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Re: Political Candidate: Government subsidized prescription drug [#permalink]
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12 Nov 2013, 01:19
1
2
Bumping for review and further disscussion
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Re: Political Candidate: Government subsidized prescription drug [#permalink]
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22 Jan 2014, 08:45
how can we conclude that first bold face is unfortunate? Can some expert please explain how to eliminate B?
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Re: Political Candidate: Government subsidized prescription drug [#permalink]
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22 Jan 2014, 08:59
the first BF is a fact about a plan. the author does not acknowledge its benefit......critical of its implementation... not opposed to it.....
the second is BF supports the conclusion by the author.
hence IMO"C".
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Political Candidate: Government subsidized prescription drug [#permalink]
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29 Jun 2015, 08:35
I had a doubt here. Aren't the options C and B too close for comfort? Can we expect something like this in GMAT? I mean using the word unfortunate in C kind of baffled me. I don't think the author thinks it's unfortunate. If someone could help. Thanks
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Re: Political Candidate: Government subsidized prescription drug [#permalink]
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04 Feb 2017, 00:26
natashakumar91 wrote:
I had a doubt here. Aren't the options C and B too close for comfort? Can we expect something like this in GMAT? I mean using the word unfortunate in C kind of baffled me. I don't think the author thinks it's unfortunate. If someone could help. Thanks
Though I am not expert, below is my understanding:
It is unfortunate because it deceptively appealing to numerous stakeholders NOT literally appealing to numerous stakeholders.
P.S.:I was also in a problem to understand how it is unfortunate.
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Re: Political Candidate: Government subsidized prescription drug [#permalink]
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05 Feb 2017, 13:59
Hi experts, I'm not able to understand difference between B and C? Both seem same.
Thanks
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Political Candidate: Government subsidized prescription drug [#permalink]
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06 Feb 2017, 07:15
rakaisraka wrote:
Hi experts, I'm not able to understand difference between B and C? Both seem same.
Thanks
"Opposed to" can be interpreted as "not agreeing". Thus in option B the statement about BF1 beomes self-conflicting.
As for C, it is not convincing that an event is "unfortunate" because it is "deceptively appealing".
Thus, I find neither of the options up to the GMAT mark to qualify as the clear, correct answer.
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Re: Political Candidate: Government subsidized prescription drug [#permalink]
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19 Apr 2017, 06:33
The conclusion, or ultimate position, of the political candidate is that older and
disabled individuals should be offered drug coverage alternatives that, in contrast
to plans built around individual choice, do not force them to gamble with their
health. The first bold-faced statement is an observation that the candidate makes
about the appeal of the choice-based plans; the use of the phrase "deceptively
appealing" and the continuation of the argument makes it clear that the candidate
views the appeal of these plans as unfortunate. The second bold-faced
statement, that consumers cannot predict their future health needs, is an
assertion that the candidate uses to support his ultimate position that alternative
plans should be offered.
(B) This choice correctly states that the second bold-faced statement is a claim
that the candidate uses as evidence to support his ultimate position. However,
the first bold-faced statement is not an observation to which the candidate is
ultimately opposed; it is his own observation that the current prescription drug
plans are "deceptively appealing." His opposition is to the drug plans
themselves, but that is not the observation made in the first statement.
(C) CORRECT. The first bold-faced statement, that coverage plans centered
around choice are deceptively appealing, is an observation that the candidate
acknowledges as true but unfortunate. The second bold-faced statement—that
consumers cannot predict their future health needs—is an assertion that the
candidate makes to support his ultimate position that alternative plans should be
offered.
(D) This choice incorrectly states that the candidate argues against the
observation that choice plans are deceptively appealing to numerous
stakeholders. This is the candidate's own observation; though he does view the
fact as unfortunate, one cannot argue against one's own observation. Moreover,
the second bold-faced statement is not an observation; instead, it is a claim used
to support the candidate's ultimate conclusion that alternative plans should be
offered.
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Re: Political Candidate: Government subsidized prescription drug [#permalink]
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26 Nov 2017, 18:58
i think the obvious flaw in B is "Evidence" in the second bold phrase. There is no evidence.
"completely opposed" is not so obvious.
Re: Political Candidate: Government subsidized prescription drug [#permalink] 26 Nov 2017, 18:58
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https://e2e.ti.com/support/microcontrollers/c2000-microcontrollers-group/c2000/f/c2000-microcontrollers-forum/587441/faq-tms320f28035-adc-calibration-and-total-unadjusted-error | 1,721,052,774,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00210.warc.gz | 196,463,223 | 34,421 | If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.
Part Number: TMS320F28035
Other Parts Discussed in Thread: REF5030, OPA320
In looking at the F2803x datasheet, I'm trying to determine two things:
1. What are all the strategies that can be deployed for ADC calibration, including steps necessary to implement them?
1. self-recalibration
2. offset calibration (if self-recalibration is occuring, is there nothing else that can improve this?)
3. two point gain calibration
4. temperature compensation
5. etc...
2. After deploying all the necessary ADC calibration, what is the resulting total unadjusted error?
Let's assume that periodic self-recalibration occurs. It appears as though the INL and DNL are not additive. This would leave INL as the worst of the two at 4 counts. What I don't understand is whether what other errors are additive:
1. Channel-to-Channel offset and gain variation does not appear to be an absolute error added to the total unadjusted error of a given channel.
2. gain error appears to be additive, and the total unadjusted error added would be a factor of the accuracy of a two point gain calibration reference.
1. If gain calibration happens on only one channel, does the channel-to-channel variation come back into play?
3. The temperature coefficient would appear to be additive to the total unadjusted error. It could to some extent be compensated for with a temperature measurement.
1. How much of the temperature coefficient is accounted for in the periodic self-recalibration?
In summary, I'm trying to understand what all goes into determining the absolute worst case total unadjusted error, and what strategies can be deployed to minimize it?
Thanks,
Stuart
• Hi Stuart,
The first thing to look at is how we calculate total un-adjusted error (TUE). From the ADC we usually consider gain error, offset error, INL, and DNL. We also need to consider the error introduced by the ADC external reference. We don't need to consider ADC channel-to-channel offset or gain error, since this is just a measure of how much the channels can vary from each other (the worst case specifications hold for all channels).
The worst case ADC errors are usually independent (e.g. an ADC with worst case INL is unlikely to also have worst case gain error). Because of this we don't directly add the worst case errors, since this would result in an an overly pessimistic TUE. Instead, we add the errors using root-sum-squares.
Before we consider calibration, here is a comparison of the raw TUE of C28x Piccolo series ADCs. The specifications for F2803x, F2802x, F2806x, and F2805x are similar, so they don't all have their own table entry. Note that in all C28x datasheets, the Min/Max ADC errors are specified including drift for voltage, temperature, and manufacturing process variation. If a temperature coefficient is also provided, this isn't added on top of the worst case error (but instead gives some idea of how much of the error is due to temperature drift). Also note that the F2803x (and similar devices mentioned above) require at least one-time offset self-calibration (we don't support factory offset trim). All values in the table below are with external reference.
(1) Executing one-time self-calibration
The external reference gain error has has some assumptions. This assumes a very good external reference solution.
For example, REF5030 has the following key specifications:
• Initial accuracy = 0.05% => 0.05%*4096 LSBs = 2.0 LSBs
• Temperature drift = 3ppm/deg. C => for standard temperature range (85C - 25C)*3ppm*4096 LSBs = 0.7 LSBs
And then OPA320 (which we recommend to drive the 12-bit reference on F2807x and F28004x) has
• Vos (max) = 150uV => 0.2 LSBs @ 3.0V
So the total worst case reference error can be estimated as sqrt(2^2 + 0.7^2 + 0.2^2) = 2.1 LSBs (and 1.0 LSBs is just a guess for typical reference error).
Obviously if a different reference solution is used, the actual accuracy can also be calculated. Getting something more accurate than the above example is possible, but will get very expensive very fast.
Now to calibrate offset error all of these devices have an internal connection to VREFLO (no external channel required). In the F2803x datasheet we specify +/-20LSBs with one-time offset calibration. To get better performance we can calibrate periodically:
Procedure: Sample the internal VREFLO connection periodically. Use these samples to adjust the ADC offset trim register accordingly. Adjusting the HW trim register will adjust the ADC samples directly, so no additional SW post-processing is necessary.
Limitations: Channel-to-channel offset variation will limit how close we can get to perfect offset trim. On F2803x, the channel-to-channel offset is specified as +/-4 LSBs.
Requirements:
• You can use the internal connection, so no external pin is required.
• You also need to configure one of the SOCs to periodically sample the signal. On F2803x, this can be an issue if you are already using all 16 SOCs. There is also no way to automatically (in the SOC configuration) switch between the internal VREFLO connection and the external pin. These can both be overcome by swapping some of the SW settings in the ADC ISR and then triggering more samples
Note: If the offset error is negative, the ADC conversion of VREFLO will read 0 and you won't know if the true offset error is -1 or something like -20. If you are periodically calibrating offset error on-line, the easiest way to handle this is to trim towards an offset error of +1 instead of 0. If the offset error is large and negative, successive rounds of calibration will eventually drive the offset error to +1. You can then either accept the extra 1 LSBs of error, or you can use the CPU to post-process the results. ADC range will be reduced by 1 LSB.
Calibrating gain error is a little more involved.
Procedure: Provide a single calibration voltage near the full-scale range. Since we already have an internal method to calibrate offset error, we don't need to do 2-point gain trim. Practically, using more calibration points will help average-out INL errors. This calibration voltage should be sampled periodically. The CPU can then post-process the ADC results to remove the gain error.
Limitations:
The error from sampling the calibration signal comes from a few sources:
• Channel-to-channel gain variation => +/-4 LSBs on F2803x
• INL +/-4 LSBs on F2803x
• Error in the calibration source => probably not better than +/2.1 LSBs
• Offset error has a complicated effect, but mostly cancels out...if offset error is positive, the calibration voltage reads high and the gain calibration will cancel out the offset error at the calibration point (near full-scale-range).
The best we can do for sampling error of the calibration voltage is therefore sqrt(4^2 + 4^2 + 2.1^2) = 6.0LSBs
The error at full scale range scales up based on where the calibration was done. So if the FSR is 3.0V and the calibration voltage is 2.5V, the total gain error would therefore be 3.0V/2.5V * 6.0LSBs = 7.2 LSBs.
Note: Because of the above, we want to do calibration as close to full-scale as possible. However, there needs to be enough space between full-scale and the calibration voltage to allow for any uncalibrated drift. For F2803x, the natural gain error is 40 LSBs = 30mV @ 3.0V VREFHI.
Requirements:
• An external channel to sample the calibration voltage
• You also need to configure one of the SOCs to periodically sample the signal.
• The CPU has to scale all the ADC results via post-processing...no HW method to compensate for gain error is available.
• The external calibration voltage needs to be accurate. This can be achieved two ways:
• A second precision voltage IC (see the first diagram below)
• In this case, an op-amp is probably not needed to drive the ADC pin (just use a large capacitor directly on the ADC pin)
• Calibration will also take care of any gain error in the external reference IC, so that IC can be a cheaper and less accurate IC in this case (within reason...it depends how much space you have between the calibration voltage and the full-scale-range).
• A precision voltage divider from the VREFHI reference IC (see the second diagram below)
• You definitely need to use matched resistors in a single package to get good performance
We can now fill back into the table the errors for F2803x with on-line calibration:
(1) Executing periodic re-calibration
(2) Error Included in ADC gain calibration
Some other notes:
• Doing the calibration periodically should take care of any internal ADC temperature coefficients.
• Calibration should be done fast enough to deal with the expected rate of thermal change in the system. Usually re-calibrating a couple times per-second should be plenty fast.
• (You still need to consider the external temperature coefficients of your calibration voltage)
• The DC code-spread for this ADC is roughly 4 LSBs. When you take calibration readings, it is best to take many points and average. I'd recommend averaging 256 points or more.
• (For every 4x oversampling, you gain 1 bit of SNR, so 4^4 = 256 samples results in noise of about 4 LSBs * (0.5^4) = 0.25 LSBs of noise. 64 sample averages would give you about 0.5 LSBs of noise.)
• It's important to get good settling on the ADC inputs, because settling error is random or cross-talk from previous samples. This can't be corrected. This also applies to your calibration input, so ensure settling much less than 1 LSBs on these calibration inputs.
• Hi Devin-san,
I have 2 questions above.
Q1) Is not DNL unnecessary in the following formula? Because DNL is included in INL.
In the description of the link below, INL was not included
Q2) TUE(F2803x ±9.2LSBs) calculates by the above method, but is it necessary to include INL in TUE calculation?
As ADC GE(F2803x ±7.2LSB) calculates including INL, I think that INL will overlap if INL is included in TUE calculation.
Best regards,
Sasaki
• Hi Sasaki-san,
Great questions!
I have seen TUE estimated with and without DNL and I think you can make a reasonable argument either way. As you said, the INL is an integration of DNL, so it does include the DNL. On the other hand, since INL is specified as a MIN/MAX of an integration, the worst case tends to hit some specific spots in the ADC transfer function whereas DNL tends to be more random transition-to-transition. I think it actually makes the most sense to include DNL in a typical calculation and exclude DNL from a MAX calculation. In any case, the DNL error is not large and therefore doesn't make a large difference in the final TUE estimation.
As far as INL affecting TUE when doing periodic gain calibration, this will indeed affect the total error twice. First it will affect the calibration sample, resulting in some error in the gain calibration factor. Then, when we go to take some sample as part of the application, this sample will also be subject to INL error. Therefore the sample gets the INL error twice: once from the calibration factor applied and a second time from the actual sample. Because INL varies randomly throughout the transfer function, it won't cancel out in the calibration factors like gain and offset.
• Hi Devin-san,
I understood it thanks to your comment :)
Best regards,
Sasaki
• Devin,
Thank you for this response. I have some follow-on questions related to the design we are working on. Normally, a one time calibration is performed during manufacturing when the custom bootloader is programmed, and is at room temperature. Based on the datasheet values, we know that the ADC can drift quite a bit over temperature variation, and with the existing one-time calibration, this would not be accounted for.
The product is expected to be in a sealed box, and the inside ambient is expected to be close to 85C. When a DFSS analysis is performed on the current sensing design, it has been determined that this high offset error rate in the ADC hurts the accuracy significantly. To counteract this, the plan has been to implement periodic calibration in the product and account for temperature drifts and possibly lower the offset error.
Despite having code in hand to test this periodic calibration procedure, it does not appear that the results can so far be validated. When comparing boards with and without the procedure, both at 0C and at 85C (using a temperature chamber), no significant deviation in the offset counts is found. There seems to be a problem with the experimental data matching the datasheet values in terms of drift across temperature.
Can you recommend a way to test the offset counts versus temperature with and without periodic calibration in a better way?
Thanks,
Stuart
• Hi Stuart,
Are you measuring the error in a sampled signal, or are you looking directly at the calibration value being written to the offset trim register? Is the measured offset error positive or negative?
I'd suggest looking directly at how the offset trim register is changing. You would expect this to change with changing temperature. If this isn't changing, it may indicate that the offset trim procedure is not functioning correctly. The most likely cause of the offset trim not working correctly would be not adding some artificial offset so that you can correct negative offset error.
If you are looking directly at a sampled signal, then what voltage(s) are being applied? To measure the offset error, you can either source in a voltage near to the VREFLO (e.g. 100mV) or you can source in 2 or more points across the ADC transfer function, then do a linear regression fit. In either case, you will want to ensure that the signal is being driven with a good low-impedance source, it is (at least ideally) buffered locally on the board by an op-amp (a large capacitor can work too if the sample rate is low), and is being measured at the pin by a DMM.
• Devin,
After implementing the code to check the trim register value and see that with temperature variation from 0 degree to 85C, the trim register value showed a 4LSB shift. I have some additional questions for you :
1. The datasheet specified a +/- 20 LSB offset error for the ADC . With the execution of the periodic calibration this error is supposed to be +/- 4 LSB. Please confirm.
2. How does the offset error exactly vary with temperature. Is it relatively flat in the temperature ranges of our interest (0-85C) and sharp increase/decrease at the allowable operating temperature extremities of the piccolo? The reason I am asking this is that we only see a 4 LSB shift and not the +/- 16LSB shift with temperature variation of 0-85C.
3. All of our data was from a single piccolo. Does the +/- 20LSB factor account for part to part variation, ADC channel to channel variation ( or any other process factors) ? Can you elaborate the conditions from where the 20LSB number comes from. We suspect that the 20 LSB error accounts for the worst case conditions and since we are not operating in those conditions we don’t see the error.
Thanks,
Stuart
• Stuart,
Stuart Baker said:
1. The datasheet specified a +/- 20 LSB offset error for the ADC . With the execution of the periodic calibration this error is supposed to be +/- 4 LSB. Please confirm.
correct
Stuart Baker said:
2. How does the offset error exactly vary with temperature. Is it relatively flat in the temperature ranges of our interest (0-85C) and sharp increase/decrease at the allowable operating temperature extremities of the piccolo? The reason I am asking this is that we only see a 4 LSB shift and not the +/- 16LSB shift with temperature variation of 0-85C.
From the data I have reviewed, offset seems pretty linear across temperature, but the LSB/degC can vary fairly significantly device to device and can be positive or negative.
Stuart Baker said:
3. All of our data was from a single piccolo. Does the +/- 20LSB factor account for part to part variation, ADC channel to channel variation ( or any other process factors) ? Can you elaborate the conditions from where the 20LSB number comes from. We suspect that the 20 LSB error accounts for the worst case conditions and since we are not operating in those conditions we don’t see the error.
The 20LSB represents the worst case drift seen across devices from multiple fablots, taken at different supply voltages and ADC conditions. It does not include ch2ch offset. It is certainly possible that some devices do not vary this much depending on the silicon and system conditions.
Regards,
Joe | 3,843 | 16,727 | {"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-2024-30 | latest | en | 0.921434 |
https://www.indiabix.com/non-verbal-reasoning/analytical-reasoning/discussion-314-4 | 1,708,589,138,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473735.7/warc/CC-MAIN-20240222061937-20240222091937-00461.warc.gz | 860,397,967 | 7,213 | # Non Verbal Reasoning - Analytical Reasoning - Discussion
Discussion Forum : Analytical Reasoning - Section 1 (Q.No. 4)
4.
Find the number of triangles in the given figure.
12
18
22
26
Explanation:
The figure may be labelled as shown.
The simplest triangles are AHB, GHI, BJC, GFE, GIE, IJE, CEJ and CDE i.e. 8 in number.
The triangles composed of two components each are HEG, BEC, HBE, JGE and ICE i.e. 5 in number.
The triangles composed of three components each are FHE, GCE and BED i.e. 3 in number.
There is only one triangle i.e. AGC composed of four components.
There is only one triangle i.e. AFD composed of nine components.
Thus, there are 8 + 5 + 3 + 1 + 1 = 18 triangles in the given figure.
Discussion:
32 comments Page 4 of 4.
Gopal said: 1 decade ago
Is there any shortcut or any formula to solve such problems !
Ishan said: 1 decade ago
Is there any technique available to solve this problem using permutation and combination method?if its there then such type of problem can be easily solve. | 278 | 1,026 | {"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} | 3.53125 | 4 | CC-MAIN-2024-10 | latest | en | 0.901625 |
https://7sage.com/lsat_explanations/lsat-11-section-4-question-10/ | 1,709,528,999,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476413.82/warc/CC-MAIN-20240304033910-20240304063910-00013.warc.gz | 78,039,288 | 36,001 | You need a full course to see this video. Enroll now and get started in less than a minute.
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Question
QuickView
Choices
Curve Question
Difficulty
Psg/Game/S
Difficulty
Explanation
PT11 S4 Q10
+LR
Method of reasoning or descriptive +Method
A
0%
163
B
2%
155
C
34%
160
D
63%
166
E
1%
159
147
158
169
+Harder 149.098 +SubsectionMedium
This page shows a recording of a live class. We're working hard to create our standard, concise explanation videos for the questions in this PrepTest. Thank you for your patience!
We can identify this question as Method of Reasoning because of the question stem: “Jonathan uses which one of the following techniques in his response to Lydia?”
When dealing with a Method of Reasoning question, we know we are looking for an answer choice that correctly describes the structure of our entire argument. Our correct answer is going to fit the argument exactly. Our wrong answer choices likely explain argument structures we are familiar with, but that simply don’t apply to the specific question we are looking at. Knowing what the right and wrong answers are going to do, we can jump into the stimulus.
Immediately we should make note of the two speakers at play. This means we could possibly be dealing with two different conclusions with different levels of support. Our first speaker, Lydia, tells us seabirds often become entangled in equipment owned by fishing companies. Lydia concludes on the basis of this that the fishing companies should assume responsibility for the medical treatment of these animals.
Lydia’s position makes an assumption here. If our conclusion tells us that something should happen, our evidence needs to give us reasoning to guarantee the outcome should occur. Perhaps, for instance, there is a law indicating those causing harm to animals should be responsible for them. But without this information the evidence does not automatically lead to the conclusion that the fisherman should be responsible for anything.
Jonathan does not quite hit the assumption out of the ballpark. In response, our second speaker concludes the proposal should not be adopted because the most injured birds won’t be able to return to the wild. Remind yourself here of how uncertain the number of “most injured birds” is. Perhaps 99.99% of the birds are injured mildly and 0.01% are the “most injured” with extensive injuries. Putting things into context, Jonathon’s response asking us to consider a group that could be impossibly small and irrelevant to Lydia’s ultimate conclusion.
Knowing that we are looking for an answer choice that will highlight Jonathon’s use of a small subset of these animals in a (poor) attempt to weaken Lydia’s reasoning we can jump into answer choice elimination.
Answer Choice (A) This answer choice accuses Lydia of a personal attack. But without any reference to Lydia’s motivation or other personal characteristics, we have to eliminate this answer from contention.
Answer Choice (B) We can eliminate this answer for a similar reason why we eliminated answer choice A. Like a personal attack, B accuses Lydia of being wrapped up in their personal interests - an attack we do not see used as the reasoning for Jonathon’s conclusion.
Answer Choice (C) This answer choice goes too far in the extreme. By accusing our second speaker of not wanting to interfere with wildlife in any way, this answer choice claims Jonathon’s conclusion goes even further than we can see in the stimulus.
Correct Answer Choice (D) This is exactly the answer we are looking for! This is the only answer choice that references Jonathon’s use of the sickest group of birds in an attempt to weaken Lydia’s argument.
Answer Choice (E) While Lydia’s feelings are addressed at the beginning of Jonathan’s argument, this is not the reasoning used to prove Jonathon’s main point. They claim we should not adopt the proposal because of this sub-group of birds. Not because of Lydia’s personal feelings in the matter. | 875 | 4,201 | {"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-2024-10 | latest | en | 0.934385 |
http://www.sawaal.com/direction-sense-test-questions-and-answers/hemant-in-order-to-go-to-university-started-from-his-house-in-the-east-and-came-to-a-crossing-the-ro_5343 | 1,531,758,847,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00034.warc.gz | 517,847,070 | 16,296 | 27
Q:
# Hemant in order to go to university started from his house in the east and came to a crossing. The road to the left ends in a theatre, straight ahead is the hospital. In which direction is the university?
A) North B) South C) East D) West
Explanation:
Therefore university is in North.
Q:
'P' walked 1 km towards south direction and turned ${90}^{o}$ACW and walked 2 km. Then he turned his left and walked 1 km and then he turned his right and walked 2 km. After that he turned ${90}^{o}$ ACW and walked 3 km. How far is he from the starting point ?
A) 5 km B) 4 km C) 2 km D) 0 km
Explanation:
23 2524
Q:
Afreena walks 8 km towards East and then walks 13 km back, then she turns left and walks 4 km; then walks 5 km after turning left; she turns left again and walks 3 km. How far is she from the starting point ?
A) 3 km B) 2 km C) 1 km D) 6 km
Explanation:
22 3960
Q:
Raman is performing yoga with his head down and legs up. His face is towards the west. In which direction, will his left hand be ?
A) North B) North - East C) East D) West
Explanation:
So he is facing west his left hand will be in North direction.
27 3457
Q:
Early morning after sunrise, Karthik was standing infront of his house in such a way that his shadow was falling exactly behind him. He starts walking straight and walks 5 metres. He turns to his left and walks 3 metres and again turning to his left walks 2 metres. Now in which direction is he from his starting point?
A) West B) North-East C) East D) South-West
Explanation:
Early morning suns rise at east, if his shadow falls exactly behind him that means he is facing towards east. now he walks 5mt towards east, then he turns left that means he facing towards north and walk 3mt and if he turn left then he is facing towards west and he walk only 3 kms thats why man is in north east direction from starting point.
10 2199
Q:
If A x B means A is to the south of B; A + B means A is to the north of B; A % B means A is to the east of B; A - B means A is to the west of B; then in P % Q + R - S, S is in which direction with respect to Q ?
A) South-East B) North-West C) East D) West
Explanation:
P % Q + R - S
=> P is to the East of Q, who is to the North of R, who is to the West of S.
=> Q.....P
R.....S
=> With respect to Q, S is to the South-East of Q.
12 1823
Q:
A man walks 5 km toward south and then turns to the right. After walking 3 km he turns to the left and walks 4 km. And then he goes back 10 km straight. Now in which direction is he from the starting place ?
A) South-East B) North-West C) South D) West
Explanation:
From the given directions, now he is 1km in the North-West direction.
31 10062
Q:
A girl facing north rotates 100(degree) clockwise then 190(degree) anticlockwise. what is new direction of the girl ?
A) North-East B) West C) South-East D) South
Explanation:
First girl facing North
Then she rotates 100 deg clock wise => 10 deg ES
Then again she rotates 190 deg anti clock wise => for 100 deg it is North and more 90 deg means West.
Therefore, the new direction of the girl is West.
15 3543
Q:
I travel 20 miles towards north and then travel 25 miles eastward. I then travel 40 miles rightwards, then travel 30 miles towards left and then travels 12 miles to the left and finally 20 miles northwards. How far am i pproximately from my original destination and in what direction ?
A) 20 miles towards south B) 13 miles towards south-west C) 12 miles towards east D) 13 miles towards north-east
${x}^{2}={12}^{2}+{5}^{2}$ | 973 | 3,547 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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} | 4.09375 | 4 | CC-MAIN-2018-30 | latest | en | 0.973628 |
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Question
# (iii) Find the compound intorost of the second year? (iii) Find the compound interest earned in the first 2 years. (iv) Find the amount at the end of 3 years. 17. A man borrows Rs 12,000 at a certain rate of compound interest per annum. His loan becomes Rs at the end of 2 years. (i) Find the principal in the beginning of the second year. (ii) Find the compound interest (to the nearest rupee) on the sum for 3 years. 18. A sum of Rs 2,400 is borrowed at a half-yearly compound interest rate of . (i) Find the amount of the debt at the end of one year. (ii) Find the amount of money required to clear off the debt after years. 19. A sum of money is invested at a half-yearly compound interest rate of . If the difference of principals at the ends of 6 months and 12 months be Rs 126 then find (i) the sum of money invested, and (ii) the amount at the end of years. 20. A man borrows Rs 12,000 at a compound interest rate of per annum for 2 years. The lender puts the condition that if the borrower fails to clear off his debt by the end of 2 years, he will increase the rate of interest to per annum. How much will the man have to pay at the end of 3 years to clear off his debt? 21. The difference between the compound interest and the simple interest on Rs 28,000 for 2 years is Rs 70 at the same rate of interest p.a. (i) What is the rate of interest? (ii) Calculate the interest earned in the second year under compound interest.
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Question Text (iii) Find the compound intorost of the second year? (iii) Find the compound interest earned in the first 2 years. (iv) Find the amount at the end of 3 years. 17. A man borrows Rs 12,000 at a certain rate of compound interest per annum. His loan becomes Rs at the end of 2 years. (i) Find the principal in the beginning of the second year. (ii) Find the compound interest (to the nearest rupee) on the sum for 3 years. 18. A sum of Rs 2,400 is borrowed at a half-yearly compound interest rate of . (i) Find the amount of the debt at the end of one year. (ii) Find the amount of money required to clear off the debt after years. 19. A sum of money is invested at a half-yearly compound interest rate of . If the difference of principals at the ends of 6 months and 12 months be Rs 126 then find (i) the sum of money invested, and (ii) the amount at the end of years. 20. A man borrows Rs 12,000 at a compound interest rate of per annum for 2 years. The lender puts the condition that if the borrower fails to clear off his debt by the end of 2 years, he will increase the rate of interest to per annum. How much will the man have to pay at the end of 3 years to clear off his debt? 21. The difference between the compound interest and the simple interest on Rs 28,000 for 2 years is Rs 70 at the same rate of interest p.a. (i) What is the rate of interest? (ii) Calculate the interest earned in the second year under compound interest. Updated On Dec 9, 2022 Topic All topics Subject Mathematics Class Class 9 Answer Type Video solution: 1 Upvotes 56 Avg. Video Duration 19 min
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A000772 E.g.f. exp(tan(x) + sec(x) - 1). 4
1, 1, 2, 6, 23, 107, 583, 3633, 25444, 197620, 1684295, 15618141, 156453857, 1683050189, 19344093070, 236497985706, 3063827565763, 41916787157011, 603799270943519, 9132945141812301, 144708157060239704, 2396568154933265024, 41403636316192616995 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS The number of elevated increasing binary trees. There is no restriction on the outdegree at the root. - Wenjin Woan, Jan 09 2008 LINKS T. D. Noe, Table of n, a(n) for n = 0..100 FORMULA a(n)=sum(k=1..n, A147315(n-1,k-1)), n>0, a(0)=1. [From Vladimir Kruchinin, Mar 10 2011] a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+x+x^2/2!)*d/dx. Cf. A000110 and A094198. See also A185422. - Peter Bala, Nov 25 2011 MATHEMATICA nn = 25; Range[0, nn]! CoefficientList[Series[Exp[Tan[x] + Sec[x] - 1], {x, 0, nn}], x] (* T. D. Noe, Jun 20 2012 *) CROSSREFS Sequence in context: A277176 A130908 A200404 * A200405 A200403 A113226 Adjacent sequences: A000769 A000770 A000771 * A000773 A000774 A000775 KEYWORD nonn AUTHOR STATUS approved
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Last modified January 19 09:35 EST 2019. Contains 319306 sequences. (Running on oeis4.) | 594 | 1,724 | {"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} | 3.34375 | 3 | CC-MAIN-2019-04 | latest | en | 0.641853 |
mystica401.50webs.com | 1,547,890,270,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583662863.53/warc/CC-MAIN-20190119074836-20190119100836-00018.warc.gz | 151,136,078 | 2,404 | # The Rules of Squares
And now...the rules!
The Object of the Game
Just like tic-tac-toe, the contestants (Mr. X and Miss Circle) win by getting three stars in a row either across, down or diagonally. To do this, the contestant chooses a star, who is asked a question, and based on the answer given by the star, the contestant must determine whether the star is right, or they're just making up hooey...that's how they get the squares.
For the first 5 seasons, games started at \$1,000 for the first 2 rounds, doubled to \$2,000 for round 3 and \$4,000 for round 4 and every round played thereafter (in early Season 1, games started at \$500 for rounds 1 and 2, and then would double each round up to round 4). In Season 6, the front game was played in a best 2 of 3 match play format, making it possible for two or three bonus rounds to be played in one episode. Each game won was worth \$1,000
Mr. X and Miss Circle
In the event that time ran out during a round, each square captured by the contestants was worth \$500 each and added to their scores (\$250 early in Season 1). In Season 6, however, the current game being played would stop after time ran out and would resume first thing the next day
The Secret Square
Played in the second round. In Season 1, one prize was offered and could only be won if the Secret Square was chosen and the answer to the question was correct. From Season 2 to Season 5, the round went by a progressive jackpot system. One prize would be offered the first day and if the contestant got the question wrong or the Secret Square wasn't chosen, a new prize would be added on the next day and so on until the entire cumulative stash was won, at which point the cycle would start again.
In Season 6, each game under the match play format had a different prize for the Secret Square (Note: Secret Square prizes don't count towards cumulative cash totals at the end of gameplay, but do count to the overall total at the end of the show)
The Secret Square, as revealed to the home audience
The Bonus Round
The bonus round has changed many times over the years. During the first few weeks of Season 1, players simply won the prize held by the star that they chose. Later in Season 1 and until the first part of Season 4, to win the star's held prize, they had to agree or disagree correctly to the question that star was asked.
Mid-Season 4, contestants played a new bonus game by teaming up with one star who held a cash value in his/her envelope. They then answered a series of rapid-fire questions in a time limit of 60 seconds, each correct answer worth the amount of money the star held. After the 60 seconds was up, the player made a decision to either keep the money they had, or risk it on one final question. A correct answer doubled the player's winnings, while an incorrect one resulted in the winnings being lost. | 662 | 2,860 | {"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-2019-04 | longest | en | 0.982825 |
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What is the missing word. Sine, Cosine and ….........?
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Using decimal notation approximate 13/18?
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Which E is an approximation for a real value?
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Which is the total number of degrees in a triangle?
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Rory started March with 1245 Twitter followers and increased his following by a fifth. How many does he have now?
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How many squares on two chess boards?
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# need help on FP2 - complex numbers watch
1. there is 1 question and an example, in the fp2 edexcel book, asking us to find the equation of a circle with points w on its circumference, generally looking like this:
w^3=-1
do the cubed roots of unity always make a circle centre 0,0 ?
and what is the 'roots of unity' jargon ?
thanks
2. (Original post by Andamiriel)
...
Might be useful:
http://www.maths.abdn.ac.uk/~igc/tch...s/node101.html
The article in Wikipaedia is a bit OTT for A-level.
3. An "nth root of unity" is a number which, to the nth power, equals 1. Note that you can write for any integer value of k , and so if then . Numbers of this form are the roots of unity.
Why must they lie on the unit circle? Well if and then what is ? So what is ? (Considering that r is a positive real number you're fairly limited to what it can be).
In this case you have , so you can proceed in a similar way, but be careful with that minus sign. Or alternatively notice that and go about it that way (which is probably easier).
4. the cubed roots of unity are the complex numbers z such that z^3 = 1. So obviously 1 is one, but there are 2 others (there are n nth roots of unity).
These do always lie on a circle centred around (0,0) with a radius of 1. So, all you have to do is find the z so that z^3 = -1 (this is the same as finding the cubed roots of unity called (-z)) and show that they satisfy the equation of a circle radius one on the origin.
5. right thanks for clearing that up
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Top tips from students who have already aced their exams | 543 | 2,135 | {"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} | 3.71875 | 4 | CC-MAIN-2018-17 | latest | en | 0.9299 |
https://www.physicsforums.com/threads/help-with-resistor-power-dissipation-homework.551646/ | 1,585,868,344,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370508367.57/warc/CC-MAIN-20200402204908-20200402234908-00043.warc.gz | 1,092,774,495 | 17,380 | # Help with resistor power dissipation homework
This question is in reference to the diagram here:
http://imageshack.us/f/17/ecediagram.jpg/
Find the average power dissipated on the resistor.
Assume Vth for the diodes is 1V, N1 = 1000, N2 = 100, the transformer is ideal (non-lossy), and V(t) is a purely sinusoidal voltage with frequency of 50 Hz and amplitude of 100V.
I've tried working it out several times, but I can't figure out exactly what to integrate. Thanks for your help!
Related Engineering and Comp Sci Homework Help News on Phys.org
gneill
Mentor
Hi thepcphysicia, welcome to PF.
You'll need to show some attempt at the problem before we can give you any help. Have you determined the waveform of the voltage (or current) that will appear across R1?
Hi,
Here is my work thus far:
http://imageshack.us/photo/my-images/202/photo2la.jpg/
The waveform of the voltage is given by 10*sin(100pi*t). I think that it is squared, because P=V^2/R.
The waveform is also shifted down two due to the voltage drop across the diodes.
Anybody? The assignment is due in a few hours...
gneill
Mentor
You might want to consider the period of the sinewave in terms of angle rather than time, and do the integration symbolically rather than carrying all those decimals around and fiddling with the milliseconds and their fractions. Then the period of a cycle is just $1/(2\pi)$, and you can call the "turn-on angle" for the the diodes $\phi$, given by:
$$10V sin(\phi) = 2V$$
The voltage that appears across the resistor R1 is shown in the diagram. In your proposed solution you made it $(10 sin(stuff))^2 - 2)$, but that's not quite right; The 2V needs to be taken from the sin() before squaring occurs.
#### Attachments
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Awesome, thank you SOOO much! The diagram was incredibly helpful! | 467 | 1,827 | {"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} | 3.734375 | 4 | CC-MAIN-2020-16 | longest | en | 0.946187 |
https://www.physicsforums.com/threads/good-sources-for-the-representations-of-the-poincare-group.975445/ | 1,723,620,907,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641104812.87/warc/CC-MAIN-20240814061604-20240814091604-00084.warc.gz | 727,618,074 | 18,845 | # Good sources for the representations of the Poincare group?
• Quantum
• andresB
In summary, there are several good sources for the representation of the Poincare group used in physics, including "Group Theory in Physics" by Tung, "Theory of Group Representations and Applications" by Barut and Raczka, "Quantum Field Theory: A Tourist Guide for Mathematicians" by Folland, and the upcoming book by Talagrand. Other recommended sources include "Relativity, Groups, Particles" by Sexl and Urbantke, and Wigner's original paper on unitary representations of the inhomogeneous Lorentz group.
andresB
Weinberg QFT book aside, what are good sources for the representation of the Poincare group used in physics?
andresB said:
Weinberg QFT book aside, what are good sources for the representation of the Poincare group used in physics?
In order of roughly increasing rigour:
"Group Theory in Physics" by Tung;
"Theory of Group Representations and Applications" by Barut and Raczka;
"Quantum Field Theory: A Tourist Guide for Mathematicians" by Folland.
There is also the forthcoming book on quantum field theory by the mathematician Talagrand. Interesting, Talagrand learned induced representations (of the Poincare group) from Weinberg's book; see Talagrand's Table of Contents and very interesting Introduction.
http://michel.talagrand.net/qft.pdf
andresB
I also like very much
R. U. Sexl, H. K. Urbantke, Relativity, Groups, Particles, Springer, Wien (2001).
and, last but not least, Wigner's orignal paper
E. P. Wigner, On Unitary Representations of the
Inhomgeneous Lorentz Group, Annals of Mathematics 40
(1939) 149.
http://dx.doi.org/10.1016/0920-5632(89)90402-7
andresB
## 1. What is the Poincare group?
The Poincare group is a mathematical concept that represents the symmetries of space and time in the theory of relativity. It includes translations, rotations, and boosts (or transformations between reference frames) and is a key concept in understanding the fundamental principles of physics.
## 2. What makes a good source for representations of the Poincare group?
A good source for representations of the Poincare group should be reliable and accurate, with information that is backed up by evidence and research. It should also be written by a reputable author or published in a reputable journal or publication.
## 3. How can I find good sources for representations of the Poincare group?
One way to find good sources is to search for academic articles or books on the topic using a reputable search engine or database. You can also consult with experts in the field or look for recommendations from reputable institutions or organizations.
## 4. Are there any specific types of sources that are better for understanding representations of the Poincare group?
While there is no one specific type of source that is better than others, academic articles and textbooks tend to be more thorough and in-depth in their explanations and analyses of the Poincare group. However, it is always important to critically evaluate the reliability and credibility of any source.
## 5. How can I ensure that the sources I use for representations of the Poincare group are accurate and reliable?
To ensure accuracy and reliability, it is important to use sources that are peer-reviewed, meaning they have been evaluated by experts in the field before being published. It is also helpful to cross-reference information from multiple sources and to consider the reputation and credentials of the author or publisher.
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5K | 946 | 3,992 | {"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-2024-33 | latest | en | 0.89647 |
https://www.scribd.com/document/256738184/Econ-Notes-Chapter-11 | 1,561,560,710,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000353.82/warc/CC-MAIN-20190626134339-20190626160339-00489.warc.gz | 897,689,816 | 55,261 | You are on page 1of 4
# Econ Notes Chapter 11
Into to Macro
Macro Goal Variables
-Measures of Economys Health
-Definitions and Realistic Goals (US, for the most part)
Goal 1: Sufficient production or output
Measured by Real Gross Domestic Product (Real GDP)
Real GDP (Y) The total production or output of final goods and
services over a period of time, expressed in constant prices of a
base year.
Real Versus Nominal GDP
Nominal GDP (Unadjusted GDP) is total production at current
prices
Real GDP (GDP Adjusted for changes in prices) is the total
production at constant prices of a base year.
-Real GDP The sum total of production of final goods and services
across all markets of the economy, it measures total production or
output.
-By definition, Real GDP identically measures total income to all the
factors derived from production and sales.
Realistic Goal for Real GDP To be as high as possible without
accelerating inflation (overstimulated economy)
The Full Sustainable Level of Real GDP (YF) The maximum level of
Real GDP the economy can produce without bringing on accelerating
inflation or overstimulation.
Characterizing the Economy: Y versus YF
Y,YF Sluggish economy
Y>YF Economy with accelerating inflation
Y = YF Economy with constant inflation rate (desired state)
Characteristics of YF
Unobservable.
Has grown at 2.5% per year for the US historically since World
War II
Maybe for the US, it has grown 3% per year in most recent
Growth rate is not the same for all countries
Recession A special Case
## The situation where the Level of real GDP decreases, or exhibits
negative growth, for at least two consecutive quarters.
Clearly, in a recession, Y < YF
Goal Variable #2 Inflation
Measure by the Inflation Rate The growth or percentage chagen
in the overall price level.
First, measure the price level (p)
o Consumer Price Index (CPI)
Inflation Rate = Percentage change in P
Why is inflation a problem?
Inflation erodes the purchasing power of money, causes
distortions in decision.
o Why hold money?
o Why lend money?
Inflation can erode peoples standard of living, put pressure on
labor markets.
o Fixed incomes
o Workers with insufficient raises.
Realistic Goal Inflation
Ideal Goal: Inflation Rate = 0%
Realistic Goal (US): |Inflation Rate| < 3%
Consumer Price Index (CPI)
Key measure of the price level (P)
Computed based upon a Market Basket: Comprehensive set of
goods and services purchased by consumers.
Fixed Weight Index
Computing a CPI
Example Compute the CPI for 2008 with 1992 as the base year.
CPI2008= (cost of 1992 Market Basket Purchased in 2008)/(Cost of
Actual Consumer Purchases in1992)
Computing the Inflation Rate (Given the CPI)
Example Compute the Inflation Rate for 2008, given that the
CPI for 2007 and 2008 have been calculated.
Inflation Rate2008 = (CPI2008-CPI2007)*(100%)/(CPI2007)
Biases in the CPI (as a fixed weight index)
Entry bias goods leaving and entering the market basket.
Quality Bias Different quality of the same goods over time.
## Outlet Bias Retail vs. Outlet prices?
Commodity Substitution Bias changing quantities over time due
to demand response to goods that have become relatively
expensive.
The GDP Deflator
An alternative measure of the price level (P)
Market Basket: comprehensive set of goods and services
purchased by all spenders in the economy (consumers,
businesses, government, and foreigners on US exports and
imports).
Chain Weighted Index Alleviates some of the biases of the CPI
due to being a fixed weight index.
Converting Nominal GDP to Real GDP
Example find real GDP2008
Real GDP 2008 = (Nominal GDP2008)/(P2008)
Real GDP for the other years is computed the same way.
Real GDP Growth = Percentage Change in Real GDP
Goal Variable #3 Unemployment
Measured by the Unemployment Rate (u)
U = (# people unemployed)/(labor force) *100%
Unemployed those people out of work and seeking work.
Labor Force People employed + People unemployed
What
## the unemployment rate does not measure
Discouraged workers, those who drop out of the labor force
Part-time versus full-time employment.
Compensation of those working
People with multiple jobs
## Realistic Goal Unemployment Rate
As low as possible without inflation accelerating (over stimulated
economy).
Natural Rate of Unemployment (uN) The lowest
unemployment rate the economy can achieve without
accelerating inflation.
Realistic Goal: u = uN
Interpretation: u versus uN
u = uN => Desired State of Economy
u > uN => Sluggish Economy
u < uN => Accelerating Inflation (over stimulated economy)
Types of Unemployment
Total Unemployment = Frictional + Structural + DemandDeficient
Frictional Unemployment unemployment due to time
involved to matching unemployed and appropriate jobs
Structural Unemployment Unemployment due to a
mismatch of available workers and jobs
Demand-Deficient Unemployment Unemployment due to a
generally sluggish economy. There are not enough jobs for
everyone who wants one.
The Natural Rate of Unemployment Revisited
Natural Rate of Unemployment (uN) The unemployment rate in
which inflation has no tendency to accelerate or decelerate.
Another Interpretation -- uN is the unemployment rate with zero
demand-deficient unemployment
Economy at uN: full employment
Where is uN for the US?
Historically, uN = 5.5%
Is uN now maybe 5%?
Most other countries: uN is higher than US measure
Real GDP and the Unemployment Rate
u = uN => Y = YF, (Desired State of Economy)
u > uN => Y < YF, (Sluggish Economy)
u < uN => Y > YF, (Accelerating Economy)
Unemployment Not an Independent Problem
Real GDP Growth Increases => Employment Growth Increases
=> u decreases
Real GDP and Unemployment not independent problems
Focus of getting one of them to the desired goal, and the other
one will automatically follow (although not a perfect correlation). | 1,381 | 5,801 | {"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-2019-26 | latest | en | 0.868199 |
http://solarsystem.nasa.gov/kids/homework.cfm?Target=Mercury&TabID=381&Letter=L | 1,438,637,864,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042990114.79/warc/CC-MAIN-20150728002310-00053-ip-10-236-191-2.ec2.internal.warc.gz | 225,556,069 | 10,167 | DICTIONARY LOOKUP
I need help with a report about...
The Swiftest Planet
Sun-scorched Mercury is only slightly larger than Earth's moon. Like the moon, Mercury has very little atmosphere to stop impacts, and it is covered with craters. Mercury's dayside is super-heated by the sun, but at night temperatures drop hundreds of degrees below freezing. Ice may even exist in craters. Mercury's egg-shaped orbit takes it around the sun every 88 days.
News Features People Extreme Facts Dictionary
Can't find it? Don't understand it? Ask us.
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Find entries starting with containing
labes A landslide.
labyrinthus An intersecting valley complex.
lacus A lake.
Lagrange points Lagrange showed that three bodies can lie at the apexes of an equilateral triangle which rotates in its plane. If one of the bodies is sufficiently massive compared with the other two, then the triangular configuration is apparently stable. Bodies at such points are sometimes referred to as Trojans. The leading apex of the triangle is known as the leading Lagrange point or L4; the trailing apex is the trailing Lagrange point or L5. Collinear with the two large bodies are the L1, L2 and L3 unstable equilibrium points which can sometimes be useful places for spacecraft, eg SOHO. (For a more detailed explanation, click on the diagram to the right).
latitude The angular distance north or south from the equator.
leading hemisphere The hemisphere that faces forward, into the direction of motion of a satellite that keeps the same face toward the planet.
lidar An instrument similar to radar that operates at visible wavelengths.
light pollution The illumination of the night sky by waste light from cities and outdoor lighting, which prevents the observation of faint objects. This is why it is hard to see stars in big cities.
light-year The distance light travels in one year. 1 ly = 9.46*10^15 meter = 5.9 billion miles.
limb The outer edge of the apparent disk of a celestial body.
linea Elongate marking.
liquid crystal A substance that behaves like both a liquid and a solid.
liter = 1000 cm3 = 1.06 US quarts.
longitude The angular distance east or west from the prime meridian.
lunar Relating to the moon.
lunar month The average time between successive new or full moons, equal to 29 days 12 hours 44 minutes. Also called synodic month.
Can't find it? Don't understand it? Ask us.
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Find entries starting with containing
labes A landslide.
labyrinthus An intersecting valley complex.
lacus A lake.
Lagrange points Lagrange showed that three bodies can lie at the apexes of an equilateral triangle which rotates in its plane. If one of the bodies is sufficiently massive compared with the other two, then the triangular configuration is apparently stable. Bodies at such points are sometimes referred to as Trojans. The leading apex of the triangle is known as the leading Lagrange point or L4; the trailing apex is the trailing Lagrange point or L5. Collinear with the two large bodies are the L1, L2 and L3 unstable equilibrium points which can sometimes be useful places for spacecraft, eg SOHO. (For a more detailed explanation, click on the diagram to the right).
latitude The angular distance north or south from the equator.
leading hemisphere The hemisphere that faces forward, into the direction of motion of a satellite that keeps the same face toward the planet.
lidar An instrument similar to radar that operates at visible wavelengths.
light pollution The illumination of the night sky by waste light from cities and outdoor lighting, which prevents the observation of faint objects. This is why it is hard to see stars in big cities.
light-year The distance light travels in one year. 1 ly = 9.46*10^15 meter = 5.9 billion miles.
limb The outer edge of the apparent disk of a celestial body.
linea Elongate marking.
liquid crystal A substance that behaves like both a liquid and a solid.
liter = 1000 cm3 = 1.06 US quarts.
longitude The angular distance east or west from the prime meridian.
lunar Relating to the moon.
lunar month The average time between successive new or full moons, equal to 29 days 12 hours 44 minutes. Also called synodic month. | 955 | 4,288 | {"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} | 2.515625 | 3 | CC-MAIN-2015-32 | longest | en | 0.921755 |
http://e-booksdirectory.com/details.php?ebook=8662 | 1,642,986,972,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304345.92/warc/CC-MAIN-20220123232910-20220124022910-00508.warc.gz | 18,689,824 | 4,073 | # Advanced Geometry for High Schools: Synthetic and Analytical
Advanced Geometry for High Schools: Synthetic and Analytical
by
Publisher: Copp, Clark
ISBN/ASIN: B0040SY2HK
Number of pages: 236
Description:
Contents: Theorems of Menelaus and Ceva; The Nine-Point Circle; Simpson's Line; Areas op Rectangles; Radical Axis; Medial Section; Miscellaneous Theorems; Similar and Similarly Situated Polygons; Harmonic Ranges and Pencils; The Complete Quadrilateral; Poles and Polars; Maxima and Minima; Miscellaneous Exercises; Loci.
(multiple formats)
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(10710 views) | 408 | 1,858 | {"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-2022-05 | longest | en | 0.88286 |
http://gdevtest.geeksforgeeks.org/check-if-array-can-be-divided-into-two-sub-arrays-such-that-their-absolute-difference-is-k/ | 1,555,630,124,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526904.20/warc/CC-MAIN-20190418221425-20190419003425-00209.warc.gz | 71,540,461 | 23,818 | # Check if array can be divided into two sub-arrays such that their absolute difference is K
Given an array arr[] and an integer K, the task is to find whether the array can be divided into two sub-arrays such that the absolute difference of the sum of the elements of both the sub-arrays is K.
Examples:
Input: arr[] = {2, 4, 5, 1}, K = 0
Output: Yes
{2, 4} and {5, 1} are the two possible sub-arrays.
|(2 + 4) – (5 + 1)| = |6 – 6| = 0
Input: arr[] = {2, 4, 1, 5}, K = 2
Output: No
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach:
• Assume there exists an answer, let the sum of elements of the sub-array (with smaller sum) is S.
• Sum of the elements of the second array will be S + K.
• And, S + S + K must be equal to sum of all the elements of the array say totalSum = 2 *S + K.
• S = (totalSum – K) / 2
• Now, traverse the array till we achieve a sum of S starting from the first element and if its not possible then print No.
• Else print Yes.
Below is the implementation of the above approach:
## C++
`#include ` `using` `namespace` `std; ` ` ` `// Function that return true if it is possible ` `// to divide the array into sub-arrays ` `// that satisfy the given condition ` `bool` `solve(``int` `array[], ``int` `size, ``int` `k) ` `{ ` ` ``// To store the sum of all the elements ` ` ``// of the array ` ` ``int` `totalSum = 0; ` ` ``for` `(``int` `i = 0; i < size; i++) ` ` ``totalSum += array[i]; ` ` ` ` ``// Sum of any sub-array cannot be ` ` ``// a floating point value ` ` ``if` `((totalSum - k) % 2 == 1) ` ` ``return` `false``; ` ` ` ` ``// Required sub-array sum ` ` ``int` `S = (totalSum - k) / 2; ` ` ` ` ``int` `sum = 0; ` ` ``for` `(``int` `i = 0; i < size; i++) { ` ` ``sum += array[i]; ` ` ``if` `(sum == S) ` ` ``return` `true``; ` ` ``} ` ` ` ` ``return` `false``; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ``int` `array[] = { 2, 4, 1, 5 }; ` ` ``int` `k = 2; ` ` ``int` `size = ``sizeof``(array) / ``sizeof``(array[0]); ` ` ``if` `(solve(array, size, k)) ` ` ``cout << ``"Yes"` `<< endl; ` ` ``else` ` ``cout << ``"No"` `<< endl; ` `} `
## Java
`/*package whatever //do not write package name here */` ` ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ` `// Function that return true if it is possible ` `// to divide the array into sub-arrays ` `// that satisfy the given condition ` `static` `boolean` `solve(``int` `array[], ``int` `size, ``int` `k) ` `{ ` ` ``// To store the sum of all the elements ` ` ``// of the array ` ` ``int` `totalSum = ``0``; ` ` ``for` `(``int` `i = ``0``; i < size; i++) ` ` ``totalSum += array[i]; ` ` ` ` ``// Sum of any sub-array cannot be ` ` ``// a floating point value ` ` ``if` `((totalSum - k) % ``2` `== ``1``) ` ` ``return` `false``; ` ` ` ` ``// Required sub-array sum ` ` ``int` `S = (totalSum - k) / ``2``; ` ` ` ` ``int` `sum = ``0``; ` ` ``for` `(``int` `i = ``0``; i < size; i++) ` ` ``{ ` ` ``sum += array[i]; ` ` ``if` `(sum == S) ` ` ``return` `true``; ` ` ``} ` ` ``return` `false``; ` `} ` ` ` ` ``// Driver Code ` ` ``public` `static` `void` `main (String[] args) ` ` ``{ ` ` ``int` `array[] = { ``2``, ``4``, ``1``, ``5` `}; ` ` ``int` `k = ``2``; ` ` ``int` `size = array.length; ` ` ` ` ``if` `(solve(array, size, k)) ` ` ``System.out.println (``"Yes"``); ` ` ``else` ` ``System.out.println (``"No"` `); ` ` ``} ` `} ` ` ` `// This Code is contributed by akt_mit `
## Python3
`# Function that return true if it is possible ` `# to divide the array into sub-arrays ` `# that satisfy the given condition ` `def` `solve(array,size,k): ` ` ``# To store the sum of all the elements ` ` ``# of the array ` ` ``totalSum ``=` `0` ` ``for` `i ``in` `range` `(``0``,size): ` ` ``totalSum ``+``=` `array[i] ` ` ` ` ``# Sum of any sub-array cannot be ` ` ``# a floating point value ` ` ``if` `((totalSum ``-` `k) ``%` `2` `=``=` `1``): ` ` ``return` `False` ` ` ` ``# Required sub-array sum ` ` ``S ``=` `(totalSum ``-` `k) ``/` `2` ` ` ` ``sum` `=` `0``; ` ` ``for` `i ``in` `range` `(``0``,size): ` ` ``sum` `+``=` `array[i] ` ` ``if` `(``sum` `=``=` `S): ` ` ``return` `True` ` ` ` ` ` ``return` `False` ` ` ` ` `# Driver Code ` `array``=` `[``2``, ``4``, ``1``, ``5``] ` `k ``=` `2` `n ``=` `4` `if` `(solve(array, n, k)): ` ` ``print``(``"Yes"``) ` `else``: ` ` ``print``(``"No"``) ` ` ` `# This code is contributed by iAyushRaj. `
## C#
`using` `System; ` `class` `GFG ` `{ ` ` ` `// Function that return true if it is possible ` `// to divide the array into sub-arrays ` `// that satisfy the given condition ` `public` `static` `bool` `solve(``int``[] array, ``int` `size, ``int` `k) ` `{ ` ` ``// To store the sum of all the elements ` ` ``// of the array ` ` ``int` `totalSum = 0; ` ` ``for` `(``int` `i = 0; i < size; i++) ` ` ``totalSum += array[i]; ` ` ` ` ``// Sum of any sub-array cannot be ` ` ``// a floating point value ` ` ``if` `((totalSum - k) % 2 == 1) ` ` ``return` `false``; ` ` ` ` ``// Required sub-array sum ` ` ``int` `S = (totalSum - k) / 2; ` ` ` ` ``int` `sum = 0; ` ` ``for` `(``int` `i = 0; i < size; i++) ` ` ``{ ` ` ``sum += array[i]; ` ` ``if` `(sum == S) ` ` ``return` `true``; ` ` ``} ` ` ` ` ``return` `false``; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `Main() ` `{ ` ` ``int``[] array = { 2, 4, 1, 5 }; ` ` ``int` `k = 2; ` ` ``int` `size = 4; ` ` ` ` ``if` `(solve(array, size, k)) ` ` ``Console.Write(``"Yes"``); ` ` ``else` ` ``Console.Write(``"No"``); ` `} ` `} ` ` ` `// This code is contributed by iAyushRaj. `
## PHP
` `
Output:
```No
```
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Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 2,335 | 6,250 | {"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} | 3.296875 | 3 | CC-MAIN-2019-18 | latest | en | 0.617731 |
https://link.springer.com/article/10.1186/s13662-017-1213-3 | 1,721,059,088,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00047.warc.gz | 314,581,060 | 60,499 | ## 1 Introduction
The role of mathematical modeling has been intensively growing in the study of epidemiology. Various epidemic models have been proposed and explored extensively and great progress has been achieved in the studies of disease control and prevention. Many authors have investigated the autonomous epidemic models. May and Odter [1] proposed a time-periodic reaction-diffusion epidemic model which incorporates a simple demographic structure and the latent period of an infectious disease. Guckenheimer and Holmes [2] examined an SIR epidemic model with a non-monotonic incidence rate, and they also analyzed the dynamical behavior of the model and derived the stability conditions for the disease-free and the endemic equilibrium. Berryman and Millstein [3] investigated an SVEIS epidemic model for an infectious disease that spreads in the host population through horizontal transmission, and they have shown that the model exhibits two equilibria, namely, the disease-free equilibrium and the endemic equilibrium. Hassell et al. [4] presented four discrete epidemic models with the nonlinear incidence rate by using the forward Euler and backward Euler methods, and they discussed the effect of two discretizations on the stability of the endemic equilibrium for these models. Shilnikov et al. [5] proposed an VEISV network worm attack model and derived global stability of a worm-free state and local stability of a unique worm-epidemic state by using the reproduction rate. Robinson and Holmes [6] discussed the dynamical behaviors of a Schrödingerean predator-prey system, and they showed that the model undergoes a flip bifurcation and Hopf bifurcation by using the center manifold theorem and bifurcation theory. Bacaër and Dads [7] investigated an SVEIS epidemic model for an infectious disease that spreads in the host population through horizontal transmission.
Recently, Yan et al. [8, 9] and Xue [10] discussed the threshold dynamics of a time-periodic reaction-diffusion epidemic model with latent period. In this paper, we will study the existence of the disease-free equilibrium and endemic equilibrium, and the stability of the disease-free equilibrium and the endemic equilibrium for this system. Conditions will be derived for the existence of a flip bifurcation and a Hopf bifurcation by using the center manifold theorem [11] and bifurcation theory [1214].
The rest of this paper is organized as follows. A discrete SIR epidemic model with latent period is established in Section 2. In Section 3 we obtain the main results: the existence and local stability of fixed points for this system. We show that this system undergoes the flip bifurcation and the Hopf bifurcation by choosing a bifurcation parameter in Section 4. A brief discussion is given in Section 5.
## 2 Model formulation
In 2015, Yan et al. [10] discussed the threshold dynamics of a time-periodic reaction-diffusion epidemic model with latent period. We consider the continuous-time SIR epidemic model described by the differential equations
$$\textstyle\begin{cases} \frac{dS}{dt}=\beta S(t)I(t), \\ \frac{dI}{dt}=\beta S(t)I(t)-\gamma I(t), \\ \frac{dR}{dt}=\gamma I(t), \end{cases}$$
(1)
where $$S(t)$$, $$I(t)$$ and $$R(t)$$ denote the sizes of the susceptible, infected and removed individuals, respectively, the constant β is the transmission coefficient, and γ is the recovery rate. Let $$S_{0}=S(0)$$ be the density of the population at the beginning of the epidemic with everyone susceptible. It is well known that the basic reproduction number $$R_{0}=\beta S_{0}/\gamma$$ completely determines the transmission dynamics (an epidemic occurs if and only if $$R_{0}>1$$); see also [8]. It should be emphasized that system (1) has no vital dynamics (births and deaths) because it was usually used to describe the transmission dynamics of disease within a short outbreak period. However, for an endemic disease, we should incorporate the demographic structure into the epidemic model. The classical endemic model is the following SIR model with vital dynamics:
$$\textstyle\begin{cases} \frac{dS}{dt}=\mu N-\mu S(t)-\frac{\beta S(t)I(t)}{N}, \\ \frac{dI}{dt}=\frac{\beta S(t)I(t)}{N}-\gamma I(t)-\mu I(t), \\ \frac{dR}{dt}=\gamma I(t)-\mu I(t), \end{cases}$$
(2)
which is almost the same as the SIR epidemic model (1) above, except that it has an inflow of newborns into the susceptible class at rate μN and deaths in the classes at rates μN, μI and μR, where N is a positive constant denoting the total population size. For this model, the basic reproduction number is given by
$$R_{0}=\frac{\beta}{\gamma+\mu},$$
which is the contact rate times the average death-adjusted infectious period $$\frac{1}{\gamma+\mu}$$. The disease-free equilibrium $$E_{0}(N,0,0)$$ of model (2) is as follows:
$$\textstyle\begin{cases} S_{n+1}=S_{n}+h(\mu N-\mu S_{n}-\frac{\beta S_{n}I_{n}}{N}), \\ I_{n+1}=I_{n}+h(\frac{\beta S_{n}I_{n}}{N}-\gamma I_{n}-\mu I_{n}), \\ R_{n+1}=R_{n}+h(\gamma I_{n}-\mu I_{n}), \end{cases}$$
(3)
where h, N, μ, β and γ are all defined in (2).
### Remark 1
If the basic reproductive rate $$R_{0}<1$$, then model (2) has only a disease-free equilibrium $$E_{1}(N,0)$$. If the basic reproductive rate $$R_{0}>1$$, then model (2) has two equilibria: a disease-free equilibrium $$E_{1}(N,0)$$ and an endemic equilibrium $$E_{2}(S^{*},I^{*})$$, where
$$S^{*}=\frac{N(\gamma+\mu)}{\beta} \quad \text{and}\quad I^{*}=\frac {N(\beta\mu-\mu(\gamma+\mu))}{\beta(\gamma+\mu)}.$$
## 3 Main results
We firstly discuss the existence of the equilibria of model (2) by using a linearization method and the Jacobian matrix. The Jacobian matrix of it is defined by
$$J(E_{1}) = \left ( \begin{matrix} 1-h\mu& -h\beta\\ 0 & 1+h\beta-h(\gamma+\mu) \end{matrix} \right ).$$
If we take the two eigenvalues of $$J(E_{1})$$
$$\omega_{1}=1-h\mu\quad \text{and} \quad \omega_{2}=1+h \beta-h(\gamma+\mu),$$
then we have the following results from Remark 1 and a simple calculation.
### Theorem 1
Let $$R_{0}$$be the basic reproductive rate such that $$R_{0}<1$$. Then:
1. (1)
If
$$0< h< \min\biggl\{ \frac{2}{\mu},\frac{2}{(\gamma+\mu)-\beta}\biggr\} ,$$
then $$E_{1}(N,0)$$is asymptotically stable.
2. (2)
If
$$h>\max\biggl\{ \frac{2}{\mu},\frac{2}{(\gamma+\mu)-\beta}\biggr\} \quad \textit{or} \quad \frac {2}{\mu}< h< \frac{2}{(\gamma+\mu)-\beta}$$
or
$$\frac{2}{(\gamma+\mu)-\beta}< h< \frac{2}{\mu},$$
then $$E_{1}(N,0)$$is unstable.
3. (3)
If
$$h=\frac{2}{\mu} \quad \textit{or}\quad h=\frac{2}{(\gamma+\mu)-\beta},$$
then $$E_{1}(N,0)$$is non-hyperbolic.
The Jacobian matrix of model (2) at $$E_{2}(S^{*},I^{*})$$ is
$$J(E_{2})= \left ( \begin{matrix} 1-\frac{h\mu\beta}{\gamma+\mu} & -h(\gamma+\mu) \\ \frac{h\mu}{\gamma+\mu}(\beta-\gamma-\mu) & 1 \end{matrix} \right ),$$
which gives
$$F(\omega)=\omega^{2}-\operatorname{tr}J(E_{2}) \omega+\operatorname{det}J(E_{2}),$$
(4)
where
$$\operatorname{tr}J(E_{2})=2-\frac{h\mu\beta}{\gamma+\mu}$$
(5)
and
$$\operatorname{det}J(E_{2})=1-\frac{h\mu\beta}{\gamma+\mu}+h^{2} \bigl[\mu\beta-\mu (\gamma +\mu)\bigr].$$
(6)
The two eigenvalues of $$J(E_{2})$$ are
$$\omega_{1,2}=1+\frac{1}{2}\biggl(- \frac{h\mu\beta}{\gamma +\mu }\pm\sqrt{(\mu R_{0})^{2}-4\bigl[\mu\beta- \mu(\gamma+\mu)\bigr]}\biggr).$$
(7)
Next we obtain the following result for $$E_{2}(S^{*},I^{*})$$ by Remark 1 and a simple calculation.
### Theorem 2
Let $$R_{0}$$be the basic reproductive rate such that $$R_{0}<1$$. Then:
1. (1)
Put
1. (A)
$$0< h< h_{*}$$and $$(\mu R_{0})^{2}-4[\mu\beta-\mu(\gamma +\mu)]\ge0$$,
2. (B)
$$0< h< h_{**}$$and $$(\mu R_{0})^{2}-4[\mu\beta-\mu(\gamma+\mu)]<0$$.
If one of the above conditions holds, then we see that $$E_{2}(S^{*},I^{*})$$is asymptotically stable.
2. (2)
Put
1. (A)
$$h>h_{***}$$and $$(\mu R_{0})^{2}-4[\mu\beta-\mu(\gamma+\mu )]\ge0$$,
2. (B)
$$0< h< h_{**}$$and $$(\mu R_{0})^{2}-4[\mu\beta-\mu(\gamma+\mu)]<0$$,
3. (C)
$$h_{*}< h< h_{***}$$and $$(\mu R_{0})^{2}-4[\mu\beta-\mu(\gamma +\mu)]\ge0$$.
If one of the above conditions holds, then $$E_{2}(S^{*},I^{*})$$is unstable.
3. (3)
Put
1. (A)
$$h=h_{*}$$or $$h=h_{***}$$and $$(\mu R_{0})^{2}-4[\mu\beta -\mu(\gamma+\mu)]\ge0$$,
2. (B)
$$h=h_{**}$$and $$(\mu R_{0})^{2}-4[\mu\beta-\mu(\gamma+\mu)]<0$$,
where
\begin{aligned}& h_{*}=\frac{\mu\beta-\mu(\gamma+\mu)\sqrt{(\mu R_{0})^{2}-4[\mu \beta -\mu(\gamma+\mu)]}}{(\gamma+\mu)[\mu\beta-\mu(\gamma+\mu)]}, \\& h_{**}=\frac{\mu\beta}{(\gamma+\mu)[\mu\beta-\mu(\gamma+\mu)]}, \end{aligned}
and
$$h_{***}=\frac{\mu\beta+\mu(\gamma+\mu)\sqrt{(\mu R_{0})^{2}-4[\mu \beta -\mu(\gamma+\mu)]}}{(\gamma+\mu)[\mu\beta-\mu(\gamma+\mu)]}.$$
If one of the above conditions holds, then $$E_{2}(S^{*},I^{*})$$is non-hyperbolic.
By a simple calculation, Conditions (A) in Theorem 2 can be written in the following form:
$$(\mu,N,\beta,h,\gamma)\in M_{1}\cup M_{2},$$
where
$$M_{1}=\bigl\{ (\mu,N,\beta,h,\gamma):h=h_{*},N>0,\triangle \ge 0,R_{0}>1,0< \mu ,\beta,\gamma< 1\bigr\}$$
and
$$M_{2}=\bigl\{ (\mu,N,\beta,h,\gamma):h=h_{***},N>0,\triangle \ge 0,R_{0}>1,0< \mu,\beta,\gamma< 1\bigr\} .$$
It is well known that if h varies in a small neighborhood of $$h_{*}$$ or $$h_{***}$$ and $$(\mu,N,\beta,h_{*},\gamma)\in M_{1}$$ or $$(\mu ,N,\beta ,h_{***},\gamma)\in M_{2}$$, then there may be a flip bifurcation of equilibrium $$E_{2}(S^{*},I^{*})$$.
## 4 Bifurcation analysis
If h varies in a neighborhood of $$h_{*}$$ and $$(\mu,N,\beta ,h_{*},\gamma)\in M_{1}$$, then we derive the flip bifurcation of model (2) at $$E_{2}(S^{*},I^{*})$$. In particular, in the case that h changes in the neighborhood of $$h_{***}$$ and $$(\mu,N,\beta,h_{***},\gamma)\in M_{2}$$ we need to make a similar calculation.
Set
$$(\mu,N,\beta,h,\gamma)=(\mu_{1},N_{1}, \beta_{1},h_{1},\gamma _{1})\in M_{1}.$$
If we give the parameter $$h_{1}$$ a perturbation $$h^{*}$$, model (2) is considered as follows:
$$\textstyle\begin{cases} S_{n+m}=S_{n}+(r^{*}+h_{1})(\mu_{1} N_{1}-\mu_{1} S_{n}-\frac{\beta_{1} S_{n}I_{n}}{N_{1}}), \\ I_{n+1}=I_{n}+(h^{*}+h_{1})(\frac{\beta_{1} S_{n}I_{n}}{N_{1}}-\gamma_{1} I_{n}-\mu_{1} I_{n}), \end{cases}$$
(8)
where $$| h^{*}|\ll1$$.
Put $$U_{n}=S_{n}-S^{*}$$ and $$V_{n}=I_{n}-I^{*}$$. We have
$$\textstyle\begin{cases} U_{n+1}=a_{11}U_{n}+a_{12}V_{n}+a_{13}U_{n}V_{n}+b_{11}U_{n}h^{*}+b_{12}V_{n}h^{*}+b_{13}U_{n}V_{n}h^{*}, \\ V_{n+1}=a_{21}U_{n}+a_{22}V_{n}+a_{23}U_{n}V_{n}+b_{21}U_{n}h^{*}+b_{22}V_{n}h^{*}+b_{23}U_{n}V_{n}h^{*}, \end{cases}$$
(9)
where
\begin{aligned}& a_{11}=1-h_{1}\biggl(\mu_{1}+\frac{\beta_{1}I^{*}}{N_{1}} \biggr),\qquad a_{12}=-\frac {h_{1}\beta_{1}S^{*}}{N_{1}},\qquad a_{13}=- \frac{h_{1}\beta_{1}}{N_{1}}, \\& b_{11}=-\biggl(\mu_{1}+\frac{\beta_{1}I^{*}}{N_{1}}\biggr),\qquad b_{12}=-\frac{\beta _{1}S^{*}}{N_{1}},\qquad b_{13}=-\frac{\beta_{1}}{N_{1}}, \\& a_{21}=\frac{h_{1}\beta_{1}I^{*}}{N_{1}},\qquad a_{22}=1,\qquad a_{23}=-\frac {\beta _{1}h_{1}}{N_{1}}, \\& b_{21}=\frac{\beta_{1}I^{*}}{N_{1}}, \qquad b_{22}=0,\qquad b_{23}=\frac{\beta_{1}}{N_{1}}. \end{aligned}
If we define matrix T as follows:
$$T = \left ( \begin{matrix} a_{12} & a_{12} \\ -1-a_{11} & \omega_{2}-a_{11} \end{matrix} \right ),$$
then we know that T is invertible. If we use the transformation
$${U_{n} \choose V_{n}}=T{X_{n} \choose Y_{n}}$$
then model (2) becomes
$$\textstyle\begin{cases} X_{n+1}=-X_{n}+F(U_{n},V_{n},h^{*}), \\ Y_{n+1}=-\omega_{2}Y_{n}+G(U_{n},V_{n},h^{*}). \end{cases}$$
(10)
Thus
$$W^{c}(0,0)=\bigl\{ (X_{n},Y{n}):Y{n}=a_{1}X^{2}_{n}+a_{2}X_{n}h^{*}+o \bigl(\bigl( \vert X_{n} \vert + \bigl\vert h^{*} \bigr\vert \bigr)^{2}\bigr)\bigr\} ,$$
where $$o((|X_{n}|+|h^{*}|)^{2})$$ is a transform function, and
$$a_{1}=\frac{a_{13}(1+a_{11}-a_{12})}{\omega_{2}+1}$$
and
$$a_{2}=\frac{b_{12}(1+a_{11})^{2}}{a_{12}(\omega_{2}+1)^{2}}-\frac {a_{12}b_{12}+b_{11}(1+a_{11})}{(\omega_{2}+1)^{2}}.$$
Further we find that the manifold $$W^{c}(0,0)$$ has the following form:
\begin{aligned}& c_{1}=\frac{a_{13}(1+a_{11})(\omega_{2}-a_{11}+a_{12})}{\omega_{2}+1}, \\& c_{2}=-\frac{b_{11}(\omega_{2}-a_{11})-a_{12}b_{21}}{\omega _{2}+1}-\frac {b_{12}(\omega_{2}-a_{11})(1+a_{11})}{a_{12}(\omega_{2}+1)}, \\& c_{3}=a_{2}\frac{a_{13}(\omega_{2}-2a_{11}-1)(\omega _{2}-a_{11}+a_{12})-b_{13}(1+a_{11})(\omega_{2}-a_{11}+a_{12})}{\omega_{2}+1}, \end{aligned}
and
$$c_{4}=0,\qquad c_{5}=\frac{a_{1}a_{13}(\omega_{2}-2a_{11}-1)(\omega _{2}-a_{11}+a_{12})}{\omega_{2}+1}.$$
Therefore the map $$G^{*}$$ with respect to $$W^{c}(0,0)$$ can be defined by
\begin{aligned} G^{*}(X_{n}) =&-X_{n}+c_{1}X^{2}_{n}+c_{2}X_{n}h^{*}+c_{3}X^{2}_{n}h^{*}+c_{4}X_{n}h^{*2} \\ &{}+c_{5}X^{3}_{n}+o\bigl(\bigl( \vert X_{n} \vert + \bigl\vert h^{*} \bigr\vert \bigr)^{3} \bigr). \end{aligned}
(11)
In order to calculate map (11), we need two quantities $$\alpha_{1}$$ and $$\alpha_{2}$$ which are not zero,
$$\alpha_{1}=\biggl(G^{*}_{X_{n}h^{*}}+\frac{1}{2}G^{*}_{h^{*}}G^{*}_{X_{n}X_{n}} \biggr)\bigg|_{0,0}$$
and
$$\alpha_{2}=\biggl(\frac{1}{6}G^{*}_{X_{n}X_{n}X_{n}}+\biggl( \frac {1}{2}G^{*}_{X_{n}X_{n}}\biggr)^{2}\biggr) \bigg|_{0,0}.$$
By a simply calculation, we obtain
\begin{aligned}& \alpha_{1}=c_{2}=-\frac{2}{h_{1}}, \\& \alpha_{2}=c_{5}+c^{2}_{1}= \frac{h_{1}\beta_{1}}{N_{1}(\omega _{2}+1)}\biggl\{ 2-\frac{h_{1}\beta_{1}\mu_{1}}{\gamma_{1}\mu_{1}}(2-h_{1}\gamma _{1})\biggr\} ^{2}, \end{aligned}
where
$$c_{1}=\frac{h_{1}\beta_{1}\mu_{1}}{\gamma_{1}\mu_{1}}\bigl[h_{1}(\gamma _{1}+ \mu _{1})-2\bigr]\biggl\{ 2+\biggl[h_{1}( \gamma_{1}+\mu_{1})+\frac{h_{1}\beta_{1}\mu _{1}}{\gamma_{1}\mu_{1}}\biggr]\biggr\} .$$
Therefore we have the following result.
### Theorem 3
Let $$h^{*}$$change in the a neighborhood of the origin. If $$\alpha_{2} \neq0$$, then the model (9) has a flip bifurcation at $$E_{2}(S^{*},I^{*})$$. If $$\alpha_{2}>0$$, then the period-2 points that bifurcation from $$E_{2}(S^{*},I^{*})$$are stable. If $$\alpha_{2}<0$$, then it is unstable.
We further consider the bifurcation of $$E_{2}(S^{*},I^{*})$$ if h varies in a neighborhood of $$h_{***}$$. Taking the parameters $$(\mu,N,\beta ,h,\gamma)=(\mu_{2},N_{2},\beta_{2},h_{2},\gamma_{2})\in N^{*}$$ arbitrarily, and also giving h a perturbation $$h^{*}$$ at $$h_{2}$$, then model (2) gets the following form:
$$\textstyle\begin{cases} S_{n+1}=S_{n}+(h^{*}+h_{2})(\mu_{2} N_{2}-\mu_{2} S_{n}-\frac{\beta_{2} S_{n}I_{n}}{N_{2}}), \\ I_{n+1}=I_{n}+(h^{*}+h_{2})(\frac{\beta_{2} S_{n}I_{n}}{N_{2}}-\gamma_{2} I_{n}-\mu_{2} I_{n}). \end{cases}$$
(12)
Put $$U_{n}=S_{n}-S^{*}$$ and $$V_{n}=I_{n}-I^{*}$$. We change the equilibrium $$E_{2}(S^{*},I^{*})$$ of model (9) and have the following result:
$$\textstyle\begin{cases} U_{n+1}=U_{n}+(h^{*}+h_{2})(-\mu_{2}U_{n}-\frac{\beta _{2}}{N_{2}}U_{n}V_{n}-\frac{\beta_{2}}{N_{2}}U_{n}I^{*}-\frac{\beta _{2}}{N_{2}}V_{n}S^{*}), \\ V_{n+1}=V_{n}+(h^{*}+h_{2})(\frac{\beta_{2}}{N_{2}}U_{n}V_{n}-(\gamma _{1}+\mu_{1})V_{n}+\frac{\beta_{2}}{N_{2}}U_{n}I^{*}+\frac{\beta _{2}}{N_{2}}V_{n}S^{*}), \end{cases}$$
(13)
which gives
$$\omega_{2}+P\bigl(h^{*}\bigr)\omega+Q\bigl(h^{*}\bigr)=0 ,$$
where
$$2+P\bigl(h^{*}\bigr)=\frac{\beta_{2}\mu_{2}(h_{2}+h^{*})}{\gamma_{2}\mu_{2}}$$
and
$$Q\bigl(h^{*}\bigr)=1-\frac{\beta_{2}\mu_{2}(h_{2}+h^{*})}{\gamma_{2}\mu _{2}}+\bigl(h_{2}+h^{*} \bigr)^{2}\bigl[\mu_{2}\beta_{2}- \mu_{2}(\mu_{2}+\gamma_{2})\bigr].$$
It is easy to see that
$$\omega_{1,2}=\frac{-P(h^{*})\pm\sqrt{(P(h^{*}))^{2}-4Q(h^{*})}}{2},$$
which yields
$$|\omega_{1,2}|=\sqrt{Q\bigl(h^{*}\bigr)},\qquad k=\frac{d|\omega _{1,2}|}{dh^{*}} \bigg|_{h^{*}=0}=\frac{\mu_{2}\beta_{2}}{2(\mu_{2}+\gamma_{2})}.$$
We remark that $$(\mu_{2},N_{2},\beta_{2},h_{2},\gamma_{2})\in N^{+}$$ and $$\bigtriangleup<0$$, and then we have
$$\frac{(\mu_{2}\beta_{2})^{2}}{(\gamma_{2}+\mu_{2})^{2}[\mu_{2}\beta _{2}-\mu _{2}(\mu_{2}+\gamma_{2})]}< 4.$$
Thus
$$P(0)=-2+\frac{(\mu_{2}\beta_{2})^{2}}{(\gamma_{2}+\mu_{2})^{2}[\mu _{2}\beta _{2}-\mu_{2}(\mu_{2}+\gamma_{2})]}\neq\pm2,$$
which means that
$$\frac{\mu_{2}\beta_{2}}{(\gamma_{2}+\mu_{2})^{2}[\mu_{2}\beta _{2}-\mu _{2}(\mu_{2}+\gamma_{2})]}\neq\frac{j(\gamma_{2}+\mu_{2})}{\mu _{2}\beta _{2}},\quad j=2,3.$$
(14)
Hence, the eigenvalues $$\omega_{1,2}$$ of equilibrium (0,0) of model (14) do not lay in the intersection when $$h^{*}=0$$ and (14) holds.
When $$h^{*}=0$$ we may begin to study the model (14). Put
\begin{aligned}& \alpha=\frac{(\mu_{2}\beta_{2})^{2}}{2(\gamma_{2}+\mu_{2})^{2}[\mu _{2}\beta _{2}-\mu_{2}(\mu_{2}+\gamma_{2})]}, \\& \beta= \frac{\mu_{2}\beta_{2} \sqrt{4 [\mu_{2}\beta_{2}-\mu_{2}(\mu_{2}+\gamma_{2})]-(\mu_{2}\beta _{2})^{2}}}{2(\gamma_{2}+\mu_{2})[\mu_{2}\beta_{2}-\mu_{2}(\mu _{2}+\gamma_{2})]}, \end{aligned}
and
$$T= \left ( \begin{matrix} 0 & 1 \\ \beta& \alpha \end{matrix} \right ),$$
where T is invertible.
If we use the transformation
$$\left ( \begin{matrix} U_{n} \\ V_{n} \end{matrix} \right )=T\left ( \begin{matrix} X_{n} \\ Y_{n} \end{matrix} \right ),$$
then the model (14) gets the following form:
$$\textstyle\begin{cases} X_{n+1}=\alpha X_{n}-\beta Y_{n}+\bar{F}(X_{n},Y_{n}), \\ Y_{n+1}=\beta X_{n}+\alpha Y_{n}+\bar{G}(X_{n},Y_{n}), \end{cases}$$
(15)
where
$$\bar{F}(X_{n},Y_{n})=\frac{h_{2}\beta_{2}(1+\alpha)(\beta X_{n}Y_{n}+\alpha Y^{2}_{n})}{N_{2}\beta}$$
and
$$\bar{G}(X_{n},Y_{n})=\frac{-h_{2}\beta_{2}(\beta X_{n}Y_{n}+\alpha Y^{2}_{n})}{N_{2}}.$$
Moreover,
\begin{aligned}& \bar{F}_{X_{n}X_{n}}=0,\qquad \bar{F}_{Y_{n}Y_{n}}=\frac{2h_{2}\beta _{2}\alpha (1+\alpha)}{N_{2}\beta_{2}},\qquad \bar{F}_{X_{n}Y_{n}}=\frac{h_{2}\beta _{2}(1+\alpha)}{N_{2}}, \\& \bar{F}_{X_{n}X_{n}X_{n}}=\bar{F}_{X_{n}X_{n}Y_{n}}=\bar{F}_{X_{n}Y_{n}Y_{n}}=\bar{F}_{Y_{n}Y_{n}Y_{n}}=0, \\& \bar{G}_{X_{n}X_{n}}=0,\qquad \bar{G}_{Y_{n}Y_{n}}=-\frac{2h_{2}\beta _{2}\alpha }{N_{2}},\qquad \bar{G}_{X_{n}Y_{n}}=-\frac{h_{2}\beta_{2}\beta}{N_{2}}, \\& \bar{G}_{X_{n}X_{n}X_{n}}=\bar{G}_{X_{n}X_{n}Y_{n}}=\bar{G}_{X_{n}Y_{n}Y_{n}}=\bar{G}_{Y_{n}Y_{n}Y_{n}}=0. \end{aligned}
Thus we have
$$a=-\operatorname{Re}\biggl[\frac{1-2\bar{\omega}}{1-\omega}\xi_{11} \xi_{20}\biggr]-\frac {1}{2}\Arrowvert\xi_{11} \Arrowvert^{2}-\Arrowvert\xi_{02}\Arrowvert ^{2}+ \operatorname{Re}(\bar{\omega}\xi_{21}),$$
where
\begin{aligned}& \xi_{02}=\frac{1}{8}\bigl[(\bar{F}_{X_{n}X_{n}}-\bar{F}_{Y_{n}Y_{n}}-2\bar{G}_{X_{n}Y_{n}})+(\bar{G}_{X_{n}X_{n}}-\bar{G}_{Y_{n}Y_{n}}+2\bar{F}_{X_{n}Y_{n}})i\bigr], \\& \xi_{11}=\frac{1}{4}\bigl[(\bar{F}_{X_{n}X_{n}}+\bar{F}_{Y_{n}Y_{n}})+(\bar{G}_{X_{n}X_{n}}+\bar{G}_{Y_{n}Y_{n}})i\bigr], \\& \xi_{20}=\frac{1}{8}\bigl[(\bar{F}_{X_{n}X_{n}}-\bar{F}_{Y_{n}Y_{n}}+2\bar{G}_{X_{n}Y_{n}})+(\bar{G}_{X_{n}X_{n}}-\bar{G}_{Y_{n}Y_{n}}-2\bar{F}_{X_{n}Y_{n}})i\bigr], \end{aligned}
and
$$\xi_{21}=\frac{1}{16}(\bar{F}_{X_{n}X_{n}X_{n}}+\bar{F}_{X_{n}Y_{n}Y_{n}}+\bar{G}_{X_{n}X_{n}Y_{n}}+\bar{G}_{Y_{n}Y_{n}Y_{n}}).$$
Therefore we have the following result.
### Theorem 4
Let $$a \neq0$$and $$h^{*}$$change in a neighborhood of $$h_{***}$$. If the condition (15) holds, then model (13) undergoes a Hopf bifurcation at $$E_{2}(S^{*},I^{*})$$. If $$a>0$$, then the repelling invariant closed curve bifurcates from $$E_{2}$$for $$h^{*}<0$$. If $$a<0$$, then an attracting invariant closed curve bifurcates from $$E_{2}$$for $$h^{*}>0$$.
## 5 Conclusions
The paper investigated the basic dynamic characteristics of a Schrödingerean predator-prey system with latent period. First, we applied the forward Euler scheme to a continuous-time SIR epidemic model and obtained the Schrödingerean predator-prey system. Then the existence and local stability of the disease-free equilibrium and endemic equilibrium of the model were discussed. In addition, we chose h as the bifurcation parameter and studied the existence and stability of flip bifurcation and Hopf bifurcation of this model by using the center manifold theorem and the bifurcation theory. Numerical simulation results show that the model (2) shows a flip bifurcation and Hopf bifurcation when the bifurcation parameter h passes through the respective critical value, and the direction and stability of flip bifurcation and Hopf bifurcation can be determined by the sign of $$\alpha_{2}$$ and a, respectively. Apparently there are more interesting problems as regards this Schrödingerean predator-prey system with latent period which deserve further investigation. | 8,117 | 20,351 | {"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": 2, "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-2024-30 | latest | en | 0.909417 |
https://astrostatistics.psu.edu/datasets/2006tutorial/html/graphics/html/persp.html | 1,723,560,510,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641079807.82/warc/CC-MAIN-20240813141635-20240813171635-00565.warc.gz | 78,663,816 | 4,239 | persp {graphics} R Documentation
## Perspective Plots
### Description
This function draws perspective plots of surfaces over the x–y plane. persp is a generic function.
### Usage
persp(x, ...)
## Default S3 method:
persp(x = seq(0, 1, len = nrow(z)), y = seq(0, 1, len = ncol(z)), z,
xlim = range(x), ylim = range(y), zlim = range(z, na.rm = TRUE),
xlab = NULL, ylab = NULL, zlab = NULL, main = NULL, sub = NULL,
theta = 0, phi = 15, r = sqrt(3), d = 1, scale = TRUE,
expand = 1, col = "white", border = NULL, ltheta = -135, lphi = 0,
shade = NA, box = TRUE, axes = TRUE, nticks = 5,
ticktype = "simple", ...)
### Details
The plots are produced by first transforming the coordinates to the interval [0,1]. The surface is then viewed by looking at the origin from a direction defined by theta and phi. If theta and phi are both zero the viewing direction is directly down the negative y axis. Changing theta will vary the azimuth and changing phi the colatitude.
There is a hook called "persp" (see setHook) called after the plot is completed, which is used in the testing code to annotate the plot page. The hook function(s) are called with no argument.
Notice that persp interprets the z matrix as a table of f(x[i], y[j]) values, so that the x axis corresponds to row number and the y axis to column number, with column 1 at the bottom, so that with the standard rotation angles, the top left corner of the matrix is displayed at the left hand side, closest to the user.
### Value
The viewing transformation matrix, say VT, a 4 x 4 matrix suitable for projecting 3D coordinates (x,y,z) into the 2D plane using homogenous 4D coordinates (x,y,z,t). It can be used to superimpose additional graphical elements on the 3D plot, by lines() or points(), e.g. using the function trans3d given in the last examples section below.
### References
Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988) The New S Language. Wadsworth & Brooks/Cole.
contour and image.
### Examples
## More examples in demo(persp) !!
## -----------
# (1) The Obligatory Mathematical surface.
# Rotated sinc function.
x <- seq(-10, 10, length= 30)
y <- x
f <- function(x,y) { r <- sqrt(x^2+y^2); 10 * sin(r)/r }
z <- outer(x, y, f)
z[is.na(z)] <- 1
op <- par(bg = "white")
persp(x, y, z, theta = 30, phi = 30, expand = 0.5, col = "lightblue")
persp(x, y, z, theta = 30, phi = 30, expand = 0.5, col = "lightblue",
ltheta = 120, shade = 0.75, ticktype = "detailed",
xlab = "X", ylab = "Y", zlab = "Sinc( r )"
) -> res
round(res, 3)
# (2) Add to existing persp plot :
trans3d <- function(x,y,z, pmat) {
tr <- cbind(x,y,z,1) %*% pmat
list(x = tr[,1]/tr[,4], y= tr[,2]/tr[,4])
}
xE <- c(-10,10); xy <- expand.grid(xE, xE)
points(trans3d(xy[,1], xy[,2], 6, pm = res), col = 2, pch =16)
lines (trans3d(x, y=10, z= 6 + sin(x), pm = res), col = 3)
phi <- seq(0, 2*pi, len = 201)
r1 <- 7.725 # radius of 2nd maximum
xr <- r1 * cos(phi)
yr <- r1 * sin(phi)
lines(trans3d(xr,yr, f(xr,yr), res), col = "pink", lwd=2)## (no hidden lines)
# (3) Visualizing a simple DEM model
z <- 2 * volcano # Exaggerate the relief
x <- 10 * (1:nrow(z)) # 10 meter spacing (S to N)
y <- 10 * (1:ncol(z)) # 10 meter spacing (E to W)
## Don't draw the grid lines : border = NA
par(bg = "slategray")
persp(x, y, z, theta = 135, phi = 30, col = "green3", scale = FALSE,
ltheta = -120, shade = 0.75, border = NA, box = FALSE)
par(op)
[Package graphics version 2.1.0 Index] | 1,127 | 3,467 | {"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} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.766536 |
http://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition/chapter-4-inverse-exponential-and-logarithmic-functions-chapter-4-test-prep-review-exercises-page-492/63 | 1,524,547,586,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946564.73/warc/CC-MAIN-20180424041828-20180424061828-00285.warc.gz | 426,848,014 | 13,413 | ## Precalculus (6th Edition)
$2$
$\ln \left( 6x\right) -\ln \left( x+1\right) =\ln 4\Rightarrow \ln \dfrac {6x}{x+1}=\ln 4\Rightarrow \dfrac {6x}{x+1}=4$ $\Rightarrow 6x=4x+4\Rightarrow 2x=4\Rightarrow x=2$ | 98 | 207 | {"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} | 3.8125 | 4 | CC-MAIN-2018-17 | latest | en | 0.355407 |
https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-15-multiple-integrals-section-15-7-triple-integrals-in-cylindrical-and-spherical-coordinates-exercises-15-7-page-921/57 | 1,575,800,812,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540508599.52/warc/CC-MAIN-20191208095535-20191208123535-00506.warc.gz | 713,916,170 | 13,380 | ## Thomas' Calculus 13th Edition
$16\pi$
$V=\int^{2\pi}_0 \int^2_0 \int^{4-r\sin\theta}_0$ dz r dr $d\theta$ =$\int^{2\pi}_0 \int^2_0 (4r-r^2\sin\theta) dr$ $d\theta$ =$8\int^{2\pi}_0 (1-\frac{\sin\theta}{3})d\theta$ =$16\pi$ | 115 | 226 | {"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} | 3.34375 | 3 | CC-MAIN-2019-51 | latest | en | 0.319841 |
https://www.dmv-written-test.com/north-dakota/motorcycle/practice-test-13.html | 1,720,998,264,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514654.12/warc/CC-MAIN-20240714220017-20240715010017-00540.warc.gz | 654,853,762 | 38,852 | # 2024 North Dakota Motorcycle Permit Test 13
The following questions are from real DMV written motorcycle permit tests. These are some of the actual permit questions you will face in North Dakota when getting your motorcycle learners..
The following questions are from real DMV written motorcycle permit tests. These are some of the actual permit questions you will face in North Dakota when getting your motorcycle learners permit. Each motorcycle theory practice test question has three answer choices. Select one answer for each question and select "grade this section." You can find this button at the bottom of the drivers license quiz. For a complete list of questions and answers for North Dakota please visit https://cheat-sheets.dmv-written-test.com/en/north-dakota/motorcycle.
Number of Tests
Number of Question
Passing Score
### 1. Which types of brakes do most motorcycles have?
Explanation
Motorcycles generally have one brake for each wheel. The front brake carries more braking power than the rear brake.
### 2. Because of their small size, motorcycles seem to be:
Explanation
Because of their size, motorcycles may seem to be traveling faster than they actually are.
### 3. A motorcycle operator can improve their visibility by:
Explanation
To maximize your chances of being seen by other road users, you should wear brightly-colored clothing with reflective materials, use your headlight at all times, and use your signals and brake light properly.
### 4. When preparing to pass a vehicle on the left, it is important to ride on the left side of your lane because:
Explanation
When preparing to pass on the left, you should ride in the left portion of your lane. This lane position will increase your line of sight and make your more visible to oncoming traffic.
### 5. When riding at night, you should do all of the following, except:
Explanation
To reduce the risk of a collision when riding at night, be sure to reduce your speed and increase your following distance. Use the headlights of vehicles ahead of you to see farther down the road. Use your high beam headlight, except when following or meeting another vehicle.
### 6. In the context of group riding, what does the hand signal pictured mean?
Explanation
Hand signals are an important part of communication when riding in groups. If a lead rider extends their left arm straight down with their palm facing back, the group of riders should come to a stop. | 492 | 2,449 | {"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 | 0.922094 |
https://studyres.com/doc/33566/e---help-a-bull | 1,600,752,148,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400203096.42/warc/CC-MAIN-20200922031902-20200922061902-00797.warc.gz | 648,360,393 | 12,962 | Survey
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Transcript
```Infinite Potential Well … bottom line
d ψ 2m
2 (E V)ψ 0
2
dx
Schrodinger Equation
2
V(x)
V=
0
Energies are quantized, defined by
one single quantum number,
n = 1, 2, 3, 4 …
n 2h 2
En
8ma
Electron
V=
V=0
0
a
x
Tunneling
“ … If an electron comes up a potential barrier greater than its energy …
there is a finite probability that it will “pass” through the barrier…”
D
C
A
B
E
… for an electron !
We place an electron in region I… with energy E less than VO (E<VO)
… what is the probability the electron will be in I … II … III ??
V(x)
Vo
I
III
II
x=0
x=a
x
How do we calculate the probability ??
… we need to solve Schrodinger’s equation … apply boundary conditions etc.
Tunneling
d 2 ψ 2m
2 (E V)ψ 0
2
dx
V(x)
Vo
ψ I (x) A1e jkx A 2 e -jkx
ψ II (x) B1ex B2 e -x
ψ III (x) C1e
jkx
C2e
-jkx
I
III
II
x=0
x=a
2m(VO - E)
2mE
2
with k 2 and α
2
2
We now need to apply BC’s at x=0 and x=a …
The properties of ψ require that it be continuous and single valued
x
Tunneling
ψI (x) ... ψII (x) ... ψIII (x)
V(x)
Vo
I
A1
A2
yII
Incident
yIII
Reflected
I
III
II
x=0
x=a
x
Therefore … the solution suggests that the electron can be found beyond
the barrier VO … EVEN THOUGH its energy E is less than VO!
Tunneling
T
ψIII (x)
ψI (x)
2
2
2
VO
1
T
where D
2
1 Dsinh (α a)
4E(VO - E)
2m(VO - E)
α
2
2
What are the important factors that influence the tunneling probability ??
… the energy of the electron… the width and height of the barrier
For a wide or high barrier …
α a 1 and T TO e
V(x)
Vo
I
( 2 α a)
16E(VO E)
with TO
2
VO
A1
A2
yII
Incident
yIII
Reflected
I
III
II
x =0
x=a
x
Application of Tunneling
Metal
(a)
y(x)
Vacuum
Second Metal
Itunne
Probe
Vo
V(x)
Scan
x
Itunnel
Material
surface
l
y (Å)
x (Å)
(b)
Tunneling
current gray
scale value
(nA)
x
The Potential “Box”
If you confine an electron in a box … what would you expect the
wave-function to be?
Think of it as a combination of 3 one-dimensional infinite
potential wells… and therefore
the general solution will have the form of:
ψ(x, y,z) Asin(kx x)sin(k y y)sin(kz z)
n3 π
n1π
n2 π
where kx
, ky
, kz
a
b
c
V=
c
V =0
V=
x
z
0
a
V=
V=
y
b
The Potential “Box”
2
E n1 , n 2 , n 3
h
2
2
2
(n
n
n
1
2
3)
2
8ma
The solution to the electron in a “box” problem results in 3
quantum numbers
A specific solution or eigenfunction i.e.
ψ1,1,2 or ψ2,1,2 or ψ1,4,2 or ψ2,1,3 etc.
is called a state …
Note that the electron energy is quantized and depends on 3
quantum numbers
The H-Atom … An Overview
Describe the H-atom …
i.e. what does the nucleus look like?
how much charge is there at the nucleus?
i.e. Z=1
how many electrons?
The H-atom represents the simplest system we can use to have
a look at a real quantum physics example
The H-Atom … Force & PE
Obviously the electron is being attracted to the nucleus because
of the …
… Coulombic attraction between two opposite charges!
The force between two charges is:
Q1 Q2
F
4π εo r 2
and the potential energy is given by:
- e2
V(r)
4π εor
The H-Atom … Spherical Coordinates
Due to the spherical symmetry of the H-atom … it makes sense
to work in the spherical coordinate system instead of the
cartesian one … i.e. x, y, z r,θ,φ
z
P(r,q,f)
e
q
r
Nucleus
+Ze
x
y
f
V(r)
r
- e2
V(r)
4π εor
Ze2
V(r) =
4peor
+Ze
The H-Atom … Wavefunction
No need to go through the solution in detail …
… we do however need to understand the origin of certain
parameters and functions!
Obtaining the wavefunction for the H-atom electron can be done
by solving …
… in 3-dimensions
i.e. one would expect to get
… 3 quantum numbers!
And the general wave function looks like:
ψ n,l,m l (r, θ, φ) R n,l (r) Yl,m l (θ , φ)
The H-Atom … Quantum Numbers
ψ n,l,m l (r, θ, φ) R n,l (r) Yl,m l (θ , φ)
Two functions … R a function of r and Y a function of θ and φ
Three quantum numbers! … n, l, ml
The spherical part i.e. R depends on n and l
… while the angular or spherical one, Y depends on l and ml
n=1,2,3,4,…… is the Principal Quantum Number
l=0,1,2,3 ……(n-1) is the Orbital Angular Momentum Quantum
Number
ml=-l, -(l-1), -(l-2), ……-2, -1, 0, 1, 2, …+l is the Magnetic
Quantum Number
… for now MEMORIZE these!
The H-Atom … Quantum Numbers
l
n
0-> s 1-> p 2-> d 3-> f 4-> g
1
1s
2
2s
2p
3
3s
3p
3d
4
4s
4p
4d
4f
5
5s
5p
5d
5f
5g
Let’s check the validity of these Energy States
3,2,2
2,3,1
2,3,0
1,2,4
2,3,-1
2,0,1
1,1,0
1,2,3
4,1,2
1,0,0
Energy !
Electron energies depend on n only … given by:
4
2
me Z
(13.6eV)Z
En 2 2 2
2
8ε o h n
n
2
What does this energy represent ?
… the energy required to remove the electron from the n=1
state (i.e. to free the electron)
also known as the ionization energy …
E1 13.6 eV
Electrons prefer to minimize their energy … therefore most likely
to be found in n=1 state known as the ground state!
An electron with velocity 2.1E6 m/s strikes a H atom. Find the n th
energy level the electron will excite to. Calculate the wavelength of
the light as the electron returns to ground state.
K.E. = 12.5eV;
n = 3.51 -> 3; ΔE = 12.09 eV;
λ = 102.6 nm
Energy !
Electron energy, En.
0
0.54
0.85
1.51
5
4
3
3.40
2
n=
Excited states
E = KE Continuum of energy. Electron is free
5
Ionization
energy, EI
10
13.6 eV
15
1 Ground state
n
n=1
Orbital Angular Momentum
Just like energy (En) … angular momentum L is also quantized… by ‘l’
Ll [l(l 1)]
1/2
… what happens when l=0 ?
Bexternal z
Lz
q
L
y
x
Orbiting electron
Orbital Angular Momentum
For l=2 … ml would be …
…-2, -1, 0, 1, 2
LZ ml
z
Bexternal
Bexternal z
ml
2
1
l =2
Lz
L= 2(2+1)
q
L
0
y
1
2
x
Orbiting electron
Selection Rules …
Electron has momentum … also photons have intrinsic momentum
When photons are absorbed … in addition to the energy
conservation momentum must also be conserved …
Selection rules: Δl=±1 … Δml=0, ±1
i.e. if electron is in ground state 1,0,0 … (n,l,ml)
If enough energy is gained to move up to n=2 then what are l
and ml?
l …0, 1 … and ml … -1, 0, 1
Therefore … n=2, l=1, and ml=-1,0,1
Selection Rules …
Energy
0
l
l
=1
l
=2
l
=3
n
5 5s
5p
5d
5f
4
4s
4p
4d
4f
3s
3p
3d
2 2s
2p
3
13.6eV
=0
1
1s
Photon
l
Selection Rule Example
An electron in State (3,2,-2). What are the energy states in
Spin … (intrinsic angular momentum)
Spin: last quantum number required to fully describe an electron!
S [s(s 1)]
1/2
with s 1/2
SZ m s with m s 1/2
The component of the spin along a magnetic field is also
quantized (i.e. if B-field is in Z-direction)
The Quantum Numbers
ψ n,l,m l (r, θ, φ) R n,l (r) Yl,m l (θ , φ)
n=1
n=2
R 1,0
n=1
R 20
n=2
r 2 |R 2,0 |2
r 2 |R 1,0 |2
2s
2s
0
1s
0
R 21
1s
0 .2
r (nm)
0 .4
0
0
2p
2p
0
0
r 2 |R 2,1 |2
0 .2
0 .4
0 .6
0 .8
0
0 .2
r (nm)
0
0 .4
0
0 .2
0 .4
0 .6
r (nm)
r (nm)
maximum
0 .8
“Angular” Probability
z
z
y
y
x
x
2
|Y|2 for a 1 s orbital |Y| for a 2 px orbital
z
z
y
y
x
|Y|2 for a 2 py orbital
s-states symmetrical
p-states directional
x
|Y|2 for a 2 pz orbital
(m l = 0)
Multi electron atom : He
-e
Electron 1
r1
r12
r2
Nucleus
+Ze
-e
Electron 2
A helium-like atom. The nucleus has a charge of +Ze, where for He Z = 2.
If one electron is removed, we have the He+ ion which is equivalent to the
hydrogenic atom with Z = 2.
Energy states for multi electron atom
O
Energy
5g
5f
N
6p
4f
4d
M
3d
5d
6s
5p
5s
4p
4s
3p
3s
L
Energy depends on both
n and l
2p
2s
K
1s
1
2
3
4
5
6
n
Pauli Exclusion Principle & Hund’s Rule
NO two electrons can have the same set of quantum numbers …
i.e. if one electron in ψ1,0,0,1/2 then a second electron in the
same system will have … ψ1,0,0,-1/2
Electrons in the same n, l orbital “like” to have “parallel" spins …
-1
L
(n=2)
K
(n=1)
1 = ml
p
s
K
(n=1)
L
(n=2)
0
H
He
Be
B
Li
s
p
Electronic configurations for the first five elements. Each box represents
an orbital y (n, l, ml ).
C
O
N
p
L
s
K
s
F
Ne
p
L
s
K
s
Electronic configurations for C, N, O, F and Ne atoms. Notice that Hund's
rule forces electrons to align their spins in C, N and O. The Ne atom has
all the K and L orbitals full.
```
Related documents | 3,392 | 8,184 | {"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} | 3.3125 | 3 | CC-MAIN-2020-40 | latest | en | 0.69325 |
https://socratic.org/questions/how-do-you-multiply-5x-1-8x-5#389747 | 1,669,804,722,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710734.75/warc/CC-MAIN-20221130092453-20221130122453-00468.warc.gz | 577,891,207 | 6,326 | # How do you multiply (5x - 1) ( 8x + 5)?
Mar 11, 2017
The answer is $40 {x}^{2} + 17 x - 5$
#### Explanation:
Multiplying out these brackets gives you a quadratic equation (an equation where the $x$ term has an exponent of $2$, ${x}^{2}$).
Every time you multiply out two brackets follow these simple steps.
First multiply the first term in the first bracket by the first term in the second bracket.
$5 x \cdot 8 x = 40 {x}^{2}$
Then multiply the first term in the first bracket by the second term in the second bracket.
$5 x \cdot 5 = 25 x$
Now repeat this for the second term of the first bracket.
$- 1 \cdot 8 x = - 8 x$
and
$- 1 \cdot 5 = - 5$
The order in which you do this doesn't actually matter, you just need to multiply every term in one bracket by every term in the other bracket. However, sticking to an order helps you to not get confused.
Now collect all of the terms produced.
$40 {x}^{2} + 25 x - 8 x - 5$
And simplify.
$40 {x}^{2} + 17 x - 5$
Hope this helped!
Mar 11, 2017
$40 {x}^{2} + 17 x - 5$
#### Explanation:
$\textcolor{b l u e}{\left(5 x - 1\right)} \textcolor{red}{\left(8 x + 5\right)}$
Multiply everything in the right bracket by everything in the left
$\textcolor{red}{\textcolor{b l u e}{5 x} \left(8 x + 5\right) \text{ } \textcolor{b l u e}{- 1} \left(8 x + 5\right)}$
$40 {x}^{2} + 25 x \text{ } - 8 x - 5$
But $\text{ } 25 x - 8 x = 17 x$
$40 {x}^{2} + 17 x - 5$ | 490 | 1,426 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "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} | 4.71875 | 5 | CC-MAIN-2022-49 | latest | en | 0.825902 |
https://forum.ansys.com/forums/topic/ways-to-calculate-heat-release-of-reaction-in-fluent/ | 1,679,935,039,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948673.1/warc/CC-MAIN-20230327154814-20230327184814-00616.warc.gz | 294,732,532 | 30,105 | ## Fluids
#### ways to calculate heat release of reaction in fluent.
• Weiqiang Liu
Subscriber
Hi all,
I am doing a new project. it's air flows through porous soot cake and reacts with soot. The local soot concentration is treated as an UDS in fluent. The reaction is very simple as follows:
C+O2-->CO2.
specific heat of carbon is constant and for other gas species, specific heat is temperature-dependent. I need to write UDF to define source term of enthalpy which arises because of carbon oxidation.
I have two ways to calculate reaction enthalpy.
Firstly: I would use C_H(c,t) to access enthalpy in the cell. Because enthalpy in a cell is mass-weighted, I can calculate enthalpy for oxygen and carbon dioxide. For carbon, the enthalpy is just Cp multiplied by ?T. Ultimately, heat release of the reaction is just enthalpy of the product minus enthalpy of reactants.
Secondly: Because I know specific heat of every species in the system. I can calculate enthalpy of formation by the following equation:
?H(0)+∫Cp(t)*dt. I don't need to use C_H(c,t) macro. I just need to use C_T macro to access temperature in the cell.
I wonder which method is correct or both of them are ok?
I need to confirm this and then go further.
PS: I received very kind help on this community and finished two projects.
Thanks very much!
Weiqiang.
• DrAmine
Ansys Employee
If you look in the way Fluent models reaction kinetics then it uses the second formulation. Just have a look into the energy equation in theory guide.
C_H is total enthalpy by the way.
• Weiqiang Liu
Subscriber
Hi Amine,
Do you mean I should use definite integration in UDF to calculate heat release of reaction. Because I know the function of specific heat which varies with temperature. Yes, I'll check theory guide of ansys.
Thanks very much!
Best.
Weiqiang
• DrAmine
Ansys Employee
You can make it simple and use species reaction.
• Weiqiang Liu
Subscriber
But it's not soot model in fluent. It's like fixed bed simulation and local carbon concentration is an UDS.
Best regards.
weiqiang
• DrAmine
Ansys Employee
Just simple species transport without any additional models.
• Weiqiang Liu
Subscriber
yes, but carbon is solid which is fixed in the system. I tried to use species transport to model carbon reaction. However, seems fluent can not choose solid as one of reactants. Because solid consumption is not considered in fluid governing equation I think while in reaction panel, every species are included in solving governing equation.
Best
Weiqiang
• DrAmine
Ansys Employee
Just assume it a fluid specie so that you can describe volume reactions.
• Weiqiang Liu
Subscriber
if I assume carbon as a fluid specie, than it must have convection and diffusion. However, in my model, soot is fixed in space. In other words, no convection and diffusion, just consuming source exists.
Best
Weiqiang | 678 | 2,889 | {"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} | 2.71875 | 3 | CC-MAIN-2023-14 | latest | en | 0.92302 |
http://faculty.ycp.edu/~dhovemey/spring2006/cs101/lecture/lecture14.html | 1,511,408,324,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806720.32/warc/CC-MAIN-20171123031247-20171123051247-00750.warc.gz | 107,029,157 | 3,790 | # Lecture 14
## Selection Sort
Sorting algorithms take a sequence of values and arrange them in sorted order: usually lowest to highest (ascending order), although highest to lowest order (descending order) is sometimes used.
In Java, the simplest way to represent a sequence of values is to use an array. Therefore, sorting in Java is generally accomplished by re-arranging the elements of an array so they are in sorted order.
Selection sort is a simple sorting algorithm. Here is the pseudo-code for selection sort:
```for each index (curIndex) in the array from 0 to end of the array {
find the index (minIndex) of the minimum value between curIndex and end of array
swap elements at curIndex and minIndex
}
```
The outer loop proceeds through each position in the array from beginning to end. Each iteration of the outer loop find the next-smallest element and moves it into the correct location in the array. Thus, when the outer loop terminates each element of the array will contain the correct element in order for the entire array to be in ascending order.
The steps listed in green are subproblems that we can solve in separate methods, as discussed in Lecture 14. The method to find the minimum element between curIndex and the end of the array will need to take both curIndex and a reference to the array as input parameters, and will return the index of the minimum value as a return value. The method to swap two elements in the array will need to take a reference to the array, curIndex, and minIndex as input parameters. This method does not need to return a value because it works by modifying the input array.
Within the selection sort method, we note that we need variables for curIndex and minIndex. The method can take the array to be sorted as an input parameter. Because the method works by modifying the input array, it doesn't need to explicitly return an input value.
Here is the Java code for the selection sort, which is defined to sort an array of double values:
```public static void selectionSort(double[] array) {
int minIndex;
int curIndex;
// for each index (curIndex) in the array from 0 to end of the array
for (curIndex = 0; curIndex < array.length; curIndex++) {
// find the index (minIndex) of the minimum value between curIndex and end of array
minIndex = findMinIndex(array, curIndex);
// swap elements at curIndex and minIndex
swapValues(array, curIndex, minIndex);
}
}
```
### findMinIndex
The findMinIndex method is an example of a search algorithm: it needs to find a value with a particular property (being the minimum) among a collection of values. In this case, we know that we can't prove that any particular element is the minimum unless we examine each element in the collection. A common pattern for this kind of problem is to start out by assuming that the first value in the collection is the one we are looking for, and then considering each remaining element in the collection. If we encounter a value that is a better candidate than our current candidate, then we change our assumption to reflect the new information.
Here is the pseudo-code for findMinIndex:
```begin by assuming the start element is the minimum value
for each remaining index (curIndex) in range start+1 to end of array {
see if the element at curIndex is less than current minimum value
if so, current value is the new minimum value
}
```
The algorithm uses two items of data:
1. The index of the minimum element (minIndex)
2. The index of the current element (curIndex)
These become local variables in the method.
Translation into Java:
```private static int findMinIndex(double[] array, int startIndex) {
int curIndex;
int minIndex;
// begin by assuming the start element is the minimum value
minIndex = startIndex;
// for each remaining index (curIndex) in range start+1 to end of array
for (curIndex = startIndex + 1; curIndex < array.length; curIndex++) {
// see if the element at curIndex is less than current minimum value
if (array[curIndex] < array[minIndex]) {
// if so, current value is the new minimum value
minIndex = curIndex;
}
}
return minIndex;
}
```
### swapValues
The swapValues method exchanges the values of two array elements whose indices are given as input parameters. A temporary variable is required to perform the exchange:
```private static void swapValues(double[] array, int firstIndex, int secondIndex) {
double temp;
temp = array[firstIndex];
array[firstIndex] = array[secondIndex];
array[secondIndex] = temp;
}
```
You should convince yourself that this method does indeed perform the task of swapping two array elements, and that it would not have the indended effect without the use of the temporary variable.
### Testing
Now that we've completed the selectionSort method and the other implementation methods it relies on, we can test it to make sure it performs correctly on some test data. Since we are working with arrays of double values, we will need a method to print the contents of an array of double:
```public static void printVals(double[] array) {
for (int i = 0; i < array.length; i++) {
System.out.print(array[i] + " ");
}
System.out.println();
}
```
With the printVals method in place, we can write a test method:
```public static void testSelectionSort() {
double[] testArray = new double[5];
testArray[0] = 4.0;
testArray[1] = 7.0;
testArray[2] = 1.0;
testArray[3] = 3.0;
testArray[4] = 6.0;
System.out.println("Before sorting");
printVals(testArray);
selectionSort(testArray);
System.out.println("After sorting");
printVals(testArray);
}
```
Here is the output of running the test method in the DrJava interactions window:
```Welcome to DrJava.
> Sort.testSelectionSort()
Before sorting
4.0 7.0 1.0 3.0 6.0
After sorting
1.0 3.0 4.0 6.0 7.0
```
## Multidimensional Arrays
So far we have seen arrays that are sequences of elements, where each element is identified by a single index value. A multidimensional array is an array where each element is identified by mutliple index values. For example, in a two-dimensional array, each element is identified by 2 index values, and you can think of it as being like a grid. A three-dimensional array would be like a cube, and so on.
In the following example, let's say that we're writing a TicTacToe game and we will use a two-dimensional array of integers to represent the game board.
Declaring an array variable:
```int[][] gameBoard;
```
Remember that because arrays are reference types, declaring an array variable does not actually create an array object. Creating the object must be done using a new expression:
```gameBoard = new int[3][3];
```
Accessing an element---to retrieve the current value of the element, or to assign a new value to the element---is done by providing each index value of the desired element.
```gameBoard[0][0] = 1;
System.out.println("value at row 1, column 2: " + gameBoard[1][2]);
```
### Rows and Columns
In a two-dimensional array, one index corresponds to rows of the grid and one index corresponds to columns of the grid. If the first dimension of the array is rows, then we are using a row-major interpretation. If the first dimension is columns, then we are using a column-major interpretation. Neither interpretation is correct or incorrect: instead, it is up to you to decide which you prefer. The most common interpretation is the row-major interpretation.
As an example, say that we declare and create a two-dimensional array as follows:
```int[][] a = new int[2][5];
```
The following diagram illustrates the difference between the row-major and column-major interpretations: | 1,742 | 7,629 | {"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-2017-47 | latest | en | 0.83915 |
http://www.aqua-calc.com/calculate/food-weight-to-volume/substance/Basil-coma-and-blank-fresh | 1,369,118,738,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699755211/warc/CC-MAIN-20130516102235-00067-ip-10-60-113-184.ec2.internal.warc.gz | 330,554,337 | 6,010 | Food Weight to Volume conversions
with calories calculator
for 3464 food ingredients
(e.g. butter, butter*, *butter, or * to show all items)
Calculation Formulas
• weight
w = ρ × v
• volume
v = w ⁄ ρ
• density
ρ = w ⁄ v
Foods
Peanut butter, chunky, vitamin and mineral fortified density is equal to 256 gram per US cup [ weight to volume | volume to weight ]
Calories
Calculate Beef, top sirloin, steak, separable lean and fat, trimmed to 1/8inch fat, select, raw calories from carbohydrates, fats and proteins based on weight
Gravels and Substrates
Sand, Coral density is equal to 15.76 kg/m²/cm. Calculate how much of this gravel is required to attain a specific depth in a cylinderquarter cylinder or a rectangular shaped object
Materials and Substances
Andesite, solid density is equal to 2771 kilogram per cubic meter [ weight to volume | volume to weight ]
What is troy pound per US tablespoon?
The troy pound per US tablespoon density measurement unit is used to measure volume in US tablespoons in order to estimate weight or mass in troy pounds
What is volume measurement?
for volume units details see below | 291 | 1,139 | {"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-2013-20 | latest | en | 0.789245 |
http://readlength.org/how-to-solve-difference-equations-numerically-in-python.html | 1,656,397,960,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103355949.26/warc/CC-MAIN-20220628050721-20220628080721-00188.warc.gz | 40,944,085 | 65,111 | # How to solve difference equations numerically in python
Consider this difference equation:
$\inline&space;\begin{matrix}&space;x_{n+1}=\dfrac{14}{5}x_{n}+\dfrac{3}{5}x_{n}&space;&&space;with&space;&&space;x_{0}=1,\&space;x_{1}=-\dfrac{1}{5}&space;\end{matrix}$
The solution is
$\inline&space;x_{n}=\left(&space;-\dfrac&space;{1}{5}\right)&space;^{n}$
I am trying to solve it numerically in python, to explicate issues that arise with floating point computations.
I wrote a function that computes xn+1
def diff(n):
c = 1
b = -1/5.0
a = 0
for i in xrange(n):
a = 14/5.0*b+3/5.0*c
b, c = a, b
return a
but I don't know how to solve this numerically and then to explain why python can not provide the xn = (-1/5)n solution.
I can see that for larger n, the return value of my function diverges from the true value.
• Please note that MathJax isn't available in SO, you should rewrite your formulas (unicode art), post them as images or try this: stackoverflow.com/a/47798853/4944425 . Jul 9, 2018 at 8:53
• Should there be an x_{n-1} somewhere in the original formula? Otherwise you just have x_{n+1} = 17/5 x_n, which doesn't match the solution you give. Jul 10, 2018 at 18:51
• @MarkDickinson Given the python code, I guess the last term (3/5), but the OP should clarify (I didn't change the formulas, with my edit). Jul 10, 2018 at 20:28
The second term in the equation should be x_{n-1} . As you know floating point errors propagate and lead to errors. One quick solution is to use higher precision arithmetic like so:
from decimal import *
getcontext().prec = 100
def diff(n):
c = Decimal(1)
b = Decimal(-1)/Decimal(5)
a = Decimal(0)
coeff1 = Decimal(14) / Decimal(5)
coeff2 = Decimal(3) / Decimal(5)
for i in range(n):
a = coeff1 * b + coeff2 * c
b, c = a, b
return a
Which should be good for the first 100 values. This doesn't solve the issue of how to do this with floating point precision. Most likely some transformation of the problem (such as taking the exponent of both sides) might allow you to do this. However finding the right transformation is tricky. | 626 | 2,086 | {"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": 2, "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-2022-27 | latest | en | 0.84906 |
https://goopennc.oercommons.org/browse?f.keyword=situation | 1,720,918,362,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00475.warc.gz | 243,166,964 | 15,139 | Updating search results...
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Formative Assessment
07/10/2019
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Students are asked to solve a real world problem, show their work using pictures, numbers or words, and record a matching equation. A scoring rubric is included. The task may be used for instructional or assessment purposes.
Subject:
Mathematics
Material Type:
Assessment
Formative Assessment
07/11/2019
Conditional Remix & Share Permitted
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This resource is part of Tools4NCTeachers.
Students are asked to solve a real world problem, show their work using pictures, numbers or words, and record a matching equation. A scoring rubric is included. The task may be used for instructional or assessment purposes.
Subject:
Mathematics
Material Type:
Assessment
Formative Assessment
07/11/2019
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
This resource is part of Tools4NCTeachers.
Students are asked to solve a real world problem, show their work using pictures, numbers or words, and record a matching equation. A scoring rubric is included. The task may be used for instructional or assessment purposes.
Subject:
Mathematics
Material Type:
Assessment
Formative Assessment
07/11/2019
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
This resource is part of Tools4NCTeachers.
In this lesson students apply place value understanding to play a game and explore problem situations involving two-digit numbers. They work in place value centers playing games and solving problems.
Subject:
Mathematics
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Lesson Plan
07/14/2019
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In this lesson, students use manipulatives to explore addition and subtraction situations. Students may use the accompanying ten frames to support their work.
Subject:
Mathematics
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Activity/Lab
Lesson
Author:
DAWNE COKER
12/26/2019
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This resource is part of Tools4NCTeachers.
Students are asked to solve a real world problem, show their work using pictures, numbers or words, and record a matching equation. A scoring rubric is included. The task may be used for instructional or assessment purposes.
Subject:
Mathematics
Material Type:
Assessment
Formative Assessment
07/10/2019
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
This resource is part of Tools4NCTeachers.
Students are asked to solve a real world problem, show their work using pictures, numbers or words, and record a matching equation. A scoring rubric is included. The task may be used for instructional or assessment purposes.
Subject:
Mathematics
Material Type:
Assessment
Formative Assessment
07/11/2019
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
This resource is part of Tools4NCTeachers.
Students are asked to solve a real world problem, show their work using pictures, numbers or words, and record a matching equation. A scoring rubric is included. The task may be used for instructional or assessment purposes.
Subject:
Mathematics
Material Type:
Assessment
Formative Assessment
07/10/2019
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
This resource is part of Tools4NCTeachers.
Students are asked to solve a real world problem, show their work using pictures, numbers or words, and record a matching equation. The task may be used for instructional or assessment purposes.
Subject:
Mathematics
Material Type:
Assessment
Formative Assessment | 1,925 | 8,160 | {"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} | 2.828125 | 3 | CC-MAIN-2024-30 | latest | en | 0.879461 |
http://spmath81608.blogspot.com/2009/02/eunices-pythagoras-post.html | 1,532,035,546,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591296.46/warc/CC-MAIN-20180719203515-20180719223515-00041.warc.gz | 344,131,884 | 16,173 | ### Eunice's pythagoras post
Monday, February 23, 2009
Eunice Anne
February 26, 2009
8-16 (:
Pythagoras
is a Greek mathematician and scientist, born between 580 and 572, then he died between 500 and 490 BC. he founded the religion movement called "pythagoreanism". he's best known for the Pythagorean theorem. Pythagoras and his students believed that everything is related to mathematics and that numbers were ultimate reality and everything could be predicted and measured in rhythmic patterns or cycle. with very strict rules of conduct Pythagoras undertook a reform of the cultural life of croton, urging the citizens to follow virtue and form an elite circle of followers around himself called Pythagoreans. he and his students lived at the school he opened, the student is required to be vegetarians. Pythagoras is also the first man to use music as medicine. his disciples would sing hymns to Apollo together regularly. they used lyre to cure illness of the soul or body; they recited poems after sleeping to aid the memory. he died around 90 years old for unknown reason. it was sad that the influence of Pythagoras on Plato and other was so great that he could be considered the most influential of all western philosophers.
According to Pythagoras...
To understand Pythagoras, you must know the meaning of the sides of he right triangle and the right triangle itself.
• Right Triangle- a triangle in which one of the sides measure 90 degrees
• Legs- the two sides of the triangle that make up the right angle of the right triangle
• Hypotenuse- is the longest side of the right triangle
• R.A.T- is the right angle triangle
• Greek- is the native language of Greece
• Theorem- is a law or principle that is tested and argued
I care about grade 8 math because it is a preparation or introduction for the future lessons especially in math and in higher sciences. Also, I just want to have a good mark.
Square
- a square has all equal sides
- all sides of the square is 90 degrees
- a whole square has 360 degrees because if you add all of the 90s you get 360
- internal angles
R.A.T: Right Angle Triangle
- all right angles are 90 degrees
- whole triangle is equal to 180 degrees.
- right angle triangle
- right angle = 90 degrees
- triangle=90 degrees When two angles equal up to 90 degrees they are called COMPLIMENTARY angles that make a straight line equal 180 degrees and the two angles are called SUPPLEMENTARY.
RECTANGLE:
- all sides equal
- internal angles
-sides are 90 degrees
- rectangle= 360 degrees
TRIANGLE:
- a and b make up the legs of the right triangle
- c= hypotenuse
= it is the longest legs of the right angle
- is always across the right angle
Pythagorean Theorem
- the Pythagorean theorem was started by Pythagoras
- it helps you find the c or the hypotenuse or the longest side of the triangle
- 4x4+3x3=25
this square used to be 2 squares. I added them both up to make an equal square. 4x4+3x3=25 then after I have to find a number that ...... (ask mom for help)
Pythagoras Theorem
Proof:
Another one proof:
Finding the sides of the square:
Pythagoras Homework:
A checkerboard is made of 64 small squares that each have a dimension of 3 cmx3cm. The 64 small squares are arranged in eight rows of eight.
My videos:
Those are all the things that I have learnt in this whole Pythagoras unit. I hoped you liked the pictures and enjoyed my presentation. I apologize for future drawing mistakes if I have. I hoped I helped you understand Pythagoras. I sure did enjoy explaining and making pictures. x) Until next time... BYE! :) :D <3 | 861 | 3,585 | {"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-2018-30 | latest | en | 0.97059 |
https://economics.stackexchange.com/questions/41587/is-it-worth-betting-on-this-case/41588 | 1,627,857,185,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154277.15/warc/CC-MAIN-20210801221329-20210802011329-00465.warc.gz | 232,902,465 | 37,561 | # Is it worth betting on this case?
Let's imagine a coin-flip game, which uses an unbiased coin.
Starting with X dollars, your total increases 50% every time you flip heads. But if the coin lands on tails, you lose 40% of your total. You can play this game N times. On each turn, you must bet the total amount you had on the last turn.
Is it worth betting on this case? How can we formalize this in order to have this answer in the general case?
Is it easier to formalize this for a specific value of X and N? Let us assume, for example X = 100 dollars and N = 100 turns.
PS: This scenario appears in an article discussing some ideas of Ole Peters, a theoretical physicist who is claiming that Everything we know about modern economics is wrong.
• It's more that everything Ole Peters claims to know about modern economics is wrong. – VARulle Dec 16 '20 at 1:55
The game has positive expected value, so you should play (assuming constant utility of each dollar, which tends to break down somewhat for very large sums of money). If you start with \$X, after 1 round you have a 50% chance of having \$1.5X, and a 50% chance of having \$0.6X, for a total expected value of \$1.05X. That is, for every dollar you bet, on average, you'll win \$1.05. The amount of money and number of trials doesn't matter at all for this analysis - every round is effectively identical, so you should always bet. In addition, there is absolutely no risk of "gambler's ruin" here, since you can never go bankrupt. In games where you can lose your entire wager, playing even a positive expected value game can lead to bankruptcy over time, since you'll eventually run into a losing streak that outstrips your bankroll. But here, since you will always have some amount of money on hand, you can keep playing forever no matter how bad of an unlucky streak you get. • So, I think that you will find this interesting. bloomberg.com/news/articles/2020-12-11/… – Zaratruta Dec 15 '20 at 19:20 • @Zaratruta Yep, that's the disconnect with utility in large sums of money - having \$500 is 5 times better than having \$100, but having \$500M isn't really 5 times better than having \\$100M, since you already have enough money to do anything you want anyway. In practice, people downweight high-reward, low probability events, since the utility does not scale linearly for large sums. – Nuclear Hoagie Dec 15 '20 at 19:34 | 582 | 2,396 | {"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} | 3.171875 | 3 | CC-MAIN-2021-31 | latest | en | 0.953827 |
http://users.stat.ufl.edu/~mripol/3024/Spring2019_Syllabus.htm | 1,586,389,853,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371826355.84/warc/CC-MAIN-20200408233313-20200409023813-00123.warc.gz | 184,094,791 | 14,085 | STA 3024 Introduction to Statistics II Spring 2019
Instructor: Maria Ripol
office: Griffin Floyd 117 C
email: mripol@stat.ufl.edu
phone: (352) 273-2976
web: http://www.stat.ufl.edu/~mripol
office hours: MWF 3rd and 4th periods (9:35 – 11:30 am) or by appointment
Lectures: MWF 2nd pd, TUR L011, course number 20428
MWF 5th pd, CSE A101, course number 20427
Teaching Assistants: TBA in Canvas
Course Description and Objectives
In this course, students learn how to summarize data, analyze it, and make appropriate decisions based on it.
The sequence of courses STA 2023-3024 provides students with a firm foundation in the basics of applied statistical methods. The prerequisite for this course is STA 2023, which covered chapters 1-9 in the textbook (data collection, graphical and numerical summaries, probability and an introduction to statistical inference). Concepts from STA 2023 will be reviewed as needed.
The course focuses on the following four topics:
1. Analysis of Variance to compare three or more population means.
2. Simple Linear Regression and Multiple Regression to predict a quantitative response.
3. Analysis of Two-Way Tables to study the relationship between two categorical variables.
4. Nonparametric Statistics that do not require a Normal distribution of the response variable.
Materials:
1. Recommended Textbook: Statistics,The Art and Science of Learning from Data, by Agresti, Franklin and Klingenberg, 4th edition, Prentice Hall.
2. Required Scientific Calculator (around \$10 to \$15) that has some basic statistical functions like
mean and standard deviation Graphing calculators are not allowed during the exams.
3. A shell of the lecture notes can be purchased at Target Copy (1412 W University Ave across from Library West). These will have the computer output for examples done in class and an outline of the lecture notes that we will then complete in class, so it will be your class notebook. If you prefer, an electronic version is posted in Canvas.
Course Website in CANVAS:
This is the portal for UF’s E-learning website. You must log on using your gatorlink username and password, and access the course webpages from there. Important information about the course will be posted here including this syllabus, announcements, your grades throughout the semester and computer output to supplement the examples done in class. Online quizzes will be administered there.
Online Quizzes - There will be approximately four online quizzes, administered through E-Learning. You will have three tries for each quiz (with questions randomly generated) over a period of several days. Each quiz will be worth 10 points, for a total of 40 points. Hopefully these quizzes will serve the purpose of improving your grade in the class, as well as be an important tool in learning the material for the course. Quiz details will be announced in class and the course website, see calendar on last page of syllabus for dates.
Suggested Homework Problems will appear on the website. They will help you master the material but will not be collected.
Projects - There will be several small data analysis projects to be completed during the semester. All projects together will be worth 60 points. Project details will be given in class and the course website, see calendar on last page of syllabus for dates.
Exams - There will be three exams given during the semester, each worth 100 points. Regular exams are in multiple choice format. Students are required to take the exam in the section they are registered for. All students must bring to the exam: their student ID number, picture ID, a non-graphing calculator, and pencils. In case of conflict or illness, if a student is unable to take an exam at the scheduled time, they must get in touch with the instructor prior to the exam time for any arrangements to be made for a makeup. Each case will be reviewed individually. Valid and detailed documentation is a prerequisite under such extenuating circumstances. Makeup exams may not be multiple choice. A grade of zero is the minimum punishment of any type of dishonesty on an exam.
Exam 1 - Fri February 15 in class Ch 10 and 14 Comparing Groups
Exam 2 - Fri March 29 in class Ch 12 and 13 Regression
Exam 3 - Wed April 24 in class Ch11 and 15 Chi Squared and Nonparametrics
Exam 1 100 points
Exam 2 100 points
Exam 3 100 points
Projects 60 points
Quizzes 40 points
TOTAL 400 points
A 90% to 100% A- 87% to 89% B+ 84% to 86% B 80% to 83% B- 77% to 79% C+ 74% to 76% C 70% to 73% C- 64% to 69% D 60% to 63% (No D+ or D- given) E 59% and below
Course Policies:
Email to Instructor – will be answered within one working day in most cases. Please be aware that statistical questions should be answered in person (in class or during office hours) since they often require pictures and formulas that make it very hard to communicate through email.
Attendance – although not required, is very highly recommended. This class is NOT offered online. If you miss a class for any reason, it is your responsibility to get a copy of the notes and all information given in class from another student. Please be aware that if there are other sections of this course with a different instructor, they may cover the material in different order. Requirements for class attendance and make-up exams, assignments and other work in this course are consistent with university policies that can be found at https://catalog.ufl.edu/ugrad/1617/regulations/info/attendance.aspx.
Classroom Behavior - during class students should turn off their cellular phones and refrain from eating, drinking, reading newspapers, doing homework, listening to music, excessive talking and all other behaviors that are distracting and disrespectful to the instructor and their fellow students.
University’s Honesty Policy: UF students are bound by The Honor Pledge which states, “We, the members of the University of Florida community, pledge to hold ourselves and our peers to the highest standards of honor and integrity by abiding by the Honor Code. On all work submitted for credit by students at the University of Florida, the following pledge is either required or implied: “On my honor, I have neither given nor received unauthorized aid in doing this assignment.” The Honor Code (http://www.dso.ufl.edu/sccr/process/student-conduct-honor-code/) specifies a number of behaviors that are in violation of this code and the possible sanctions. Furthermore, you are obligated to report any condition that facilitates academic misconduct to appropriate personnel.
Grading – grades will be changed only when an error has been made. Negotiation is not appropriate.
Incompletes
are only assigned when extraordinary circumstances, arising after the date for dropping the course, prevent the student from completing the course requirements. Having a failing grade in the course is not a valid reason for requesting an Incomplete.
Students with Disabilities
- Students with disabilities requesting accommodations should first register with the Disability Resource Center (352-392-8565, www.dso.ufl.edu/dre/ ) by providing appropriate documentation. Once registered, students will receive an accommodation letter which must be presented to the instructor when requesting accommodations. Please see the instructor in person during office hours, as early as possible in the semester, to discuss your accommodation letter confidentially.
Instructor / Course Evaluations: Students are expected to provide feedback on the quality of instruction in this course based on 10 criteria. These evaluations are conduced online at https://evaluations.ufl.edu/. Evaluations are typically open during the last two or three weeks of the semester, but students will be given specific times when they are open. Summary results of these assessments are available to students at https://evaluations.ufl.edu/results.
Other University Services:
U Matter, We Care: Your well-being is important to the University of Florida. The U Matter, We Care
initiative is committed to creating a culture of care on our campus by encouraging members of our
community to look out for one another and to reach out for help if a member of our community is in
need. If you or a friend is in distress, please contact umatter@ufl.edu so that the U Matter, We Care
Team can reach out to the student in distress.
A nighttime and weekend crisis counselor is available by phone at 352-392-1575. The U Matter, We
Care Team can help connect students to the many other helping resources available including, but not limited to, Victim Advocates, Housing staff, and the Counseling and Wellness Center. Please remember that asking for help is a sign of strength. In case of emergency, call 9-1-1.
Sexual Assault Recovery Services(SARS): Student Health Center, 392-1161
University Police Department, 392-1111 ( or 9-1-1 for emergencies) http://www.police.ufl.edu
Weekly Schedule – subject to change if needed.
Monday Wednesday Friday 1/07 Intro / Start Review Stats 1 1/09 Review Ch 7-10 CI and SigTests 1/11 Continue Review Stats 1 1/14 Continue Review Stats 1 1/16 Continue Review Stats 1 1/18 More + Ch 4 Designing Experiments Q1 starts 1/21 No Class – MLK Day 1/23 Ch 10 – ANOVA Theory 1/25 One-Way ANOVA examples Q1 ends 1/28 Multiple Comparisons 1/30 Bonferroni 2/01 More One-Way ANOVA examples 2/04 One/ Two-Way ANOVA 2/06 Two-Way ANOVA Q2 starts Q2 starts 2/08 Two-Way ANOVA 2/11 Review 2/13 Review Q2 ends 2/15 EXAM 1 2/18 Review Ch 3 – Simple Linear Regression 2/20 Ch 12- Regression Analysis 2/22 Continue Inference Reg 2/25 Continue Inference Reg 2/27 Continue Inference Reg 3/01 Flex Day 3/04 SPRING BREAK 3/06 SPRING BREAK 3/08 SPRING BREAK 3/11 Ch 13 Multiple Regression Q3 starts 3/13 Regression with Dummy Variables 3/15 More Reg. with Dummy Vars Project 1 INFO 3/18 Quadratic Regression Q3 ends 3/20 More Regression Examples 3/22 More Regression Examples 3/25 Review 3/27 Review Project 1 DUE 3/29 EXAM 2 4/01 Ch 11 Contingency Tables 4/03 Contingency Tables 4/05 Contingency Tables Project 2 INFO 4/08 Sec 13.6 Logistic Regression Q4 starts 4/10 Ch 14 Nonparametric Methods 4/12 Nonparametric Methods 4/15 Nonparametric Methods Q4 ends ECQ starts 4/17 Nonparametric Methods Project 2 DUE 4/19 Review 4/22 Review ECQ ends 4/24 EXAM 3 | 2,483 | 10,525 | {"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-2020-16 | latest | en | 0.827328 |
http://www.cfd-online.com/Forums/cfx/136607-energy-equation-modeling-turbine-rotor.html | 1,455,046,132,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701157443.43/warc/CC-MAIN-20160205193917-00030-ip-10-236-182-209.ec2.internal.warc.gz | 333,964,571 | 16,384 | # Energy equation modeling in turbine rotor?
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May 31, 2014, 20:48 Energy equation modeling in turbine rotor? #1 New Member mosi Join Date: May 2014 Location: College Station, Texas Posts: 9 Rep Power: 3 Hi everyone, I have faced a very fundamental question in modeling energy equation in turbine rotor which I can not find the answer to. In a real turbine, the power used to rotate the rotor blades, is extracted from the hot fluid. So the fluid loses temperature and enthalpy in the rotor. But in CFX when we are modeling the rotating stage, we just enter the rotation speed and it seems the solver does not ask you where the power to rotate the blades is coming from. so the results show that the temperature and enthalpy is not decreased in the rotor stage. so the power to rotate the rotor is not extracted from the fluid which should be wrong. Anybody has any explanation for this problem? Thanks, Mosi
June 1, 2014, 00:29 #2 New Member mosi Join Date: May 2014 Location: College Station, Texas Posts: 9 Rep Power: 3 I think I might have figured out the problem. It is likely that it is caused by my boundary conditions. Because when I checked the results of CFX Tutorial #14 it showed a reasonable decrease in temperature and enthalpy in rotor. I will change my boundary conditions and update the results here. Thanks
June 1, 2014, 02:53 #3 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 11,573 Rep Power: 90 If your simulation is accurate and there is no enthalpy drop then I bet there is no torque on the rotor either. Then you have no energy drop in the fluid and no energy extracted. Hey presto energy is conserved.
June 1, 2014, 11:55 #4 New Member mosi Join Date: May 2014 Location: College Station, Texas Posts: 9 Rep Power: 3 OK, as I guessed the problem was fixed by changing the boundary conditions. Before, I was using mass flow inlet=0.11 kg/s and static pressure outlet=0 atm, and I saw no enthalpy drop in rotor. Now I've changed it to P_tot inlet=0 atm and mass flow outlet=0.11 kg/s and I see a 22000 J/kg total enthalpy drop in rotor which is reasonable. @Glenn, thank you for your reply. You are right, the torque was zero and it's non-zero now. But I still don't know why the former boundary conditions created no power. Do you have any idea?
June 1, 2014, 20:50 #5 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 11,573 Rep Power: 90 At the operating point the rotor will generate torque. If the rotor rotational speed is too fast it will generate negative torque (ie the rotor will want to slow down). At some point between it will generate zero torque.
June 1, 2014, 22:37 #6 New Member mosi Join Date: May 2014 Location: College Station, Texas Posts: 9 Rep Power: 3 That is absolutely true. Thank you.
Tags cfx, energy, rotor, temperature, turbine
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2019-06-23, 11:44 #56
kar_bon
Mar 2006
Germany
23·32·41 Posts
Quote:
Originally Posted by sweety439 If a prime in the list is also other classes of primes, I think it should be in the notes, like the article Riesel 3, ...
Sorry for not including hundreds of years of prime searching in a 6 months running Wiki and having a real life, too but only (here on earth) 24 hours a day to work on such pages.
2019-06-23, 14:02 #57
sweety439
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3,389 Posts
Quote:
Originally Posted by kar_bon Sorry for not including hundreds of years of prime searching in a 6 months running Wiki and having a real life, too but only (here on earth) 24 hours a day to work on such pages.
However, ALL Williams MP 5 are also Generalized Fermat primes, but only "20462" and "70130" has this note.
2019-06-23, 14:25 #58
kar_bon
Mar 2006
Germany
23×32×41 Posts
Quote:
Originally Posted by sweety439 However, ALL Williams MP 5 are also Generalized Fermat primes, but only "20462" and "70130" has this note.
Posting over and over again the same isn't of any help. I only used the comments from the Top5000 entries there. I also have to determine the best way how to put those many different prime type you mentioned in the comments of any n-value of any base and prime type.
2019-06-24, 20:54 #59 sweety439 "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 338910 Posts In the page Williams prime, the "Available Online Sequences" section only lists 2<=b<=10, however, currently Williams MM has available OEIS sequences for all 2<=b<=14, and Williams MP has available OEIS sequences for all 3<=b<=12 (with the exception of b=4). Also, currently Cullen has available OEIS sequences for all 2<=b<=18 (with the exception of b=5, b=11, b=13 and b=17), and Woodall has available OEIS sequences for all 2<=b<=20. Last fiddled with by sweety439 on 2019-06-24 at 20:55
2019-06-24, 20:57 #60 sweety439 "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 3,389 Posts There are also OEIS sequences: Smallest Williams MP: A305531 Smallest Cullen: A240234 Smallest Woodall: A240235
2019-06-26, 18:28 #61 sweety439 "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 3,389 Posts @kar_bon, you missed a sequence: Williams PP 171, it also has no primes with exponent n<=1024. Done. Last fiddled with by kar_bon on 2019-06-26 at 20:16
2019-07-02, 15:46 #62 Dylan14 "Dylan" Mar 2017 25016 Posts I've generated a template for showing the history of sieve files. This template could be useful for the CRUS sequences as well as projects like the Carol/Kynea search. Calling is fairly simple: there are 5 parameters (date, upper range, contributors, lower range and post), of which the last 2 are optional. If the 4th parameter is not given, it is assumed that the file starts from 1.
2019-07-03, 07:48 #63 kar_bon Mar 2006 Germany 23·32·41 Posts Thanks, I've changed the template a little bit. No need to give the range start as separate parameter, range given at all, so this template is more than the others in handling parameters: date, param(mostly), contributor, post ID(optional). I also added this entry to CK50.
2019-07-04, 13:52 #64 kar_bon Mar 2006 Germany 23×32×41 Posts I've created an alternate table for Carol/Kynea: - no sieve files listed - no contributors listed - n-values listed in one line (mostly sufficient) Done so far: - all (hope so) available historical data from -- 3 threads in this forum -- old page from S.Harvey -- some old Yahoo posts (A short table for copying the data into an editor is available here.) Links see here, this page has to be updated after the following is done: - some bases without Carol or Kynea to insert Done. For all bases <= 3000 there are currently -- 96 without a Carol prime -- 78 without a Kynea prime - remaining data for bases<=3000 from my page/search - two tables with least Carol or Kynea n-value If more bases will be filled in it's better to change the template for categorizing any base into different parts (like 1-999, 1000-1999, 2000-2999,...). Nice to have: - some statistics - some graphs History entries and reserving any base (see base 2 for example) are also available as well as inserting a sieve file (see base 50). The same procedure can be done for other prime types. Last fiddled with by kar_bon on 2019-07-05 at 13:25 Reason: links to "without"'s
2019-07-23, 18:11 #65 sweety439 "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 3,389 Posts For Leyland primes, why the pages are "Leyland_prime P x y" and "Leyland_prime M x y" (a page for a single (probable) prime)? Not "Leyland_prime P x" and "Leyland_prime M x" (list all y's such that x^y+y^x or |x^y-y^x| are (probable) primes)? Like the pages for Williams primes and the Carol/Kynea primes and the Cullen/Woodall primes? e.g. for the Carol/Kynea primes, the pages list all n such that (b^n-1)^2-2 or (b^n+1)^2-2 are primes for a fixed b, why not (for Leyland primes) list all y such that x^y+y^x or |x^y-y^x| are (probable) primes for a fixed x?
2019-07-24, 11:35 #66
kar_bon
Mar 2006
Germany
23×32×41 Posts
Quote:
Originally Posted by sweety439 For Leyland primes, why the pages are "Leyland_prime P x y" and "Leyland_prime M x y" (a page for a single (probable) prime)?
Those numbers are more individual than Carol/Kynea. Finder/prover and dates are more different.
On the other hand most of those numbers (not yet all of more than 1400) are with different x-/y-values so most categories as "Leyland_prime P x" would contain only one member. This is not very clear and a table is also harder to create.
If you want such list, sorting the table by x- or y-value you get this.
Similar Threads Thread Thread Starter Forum Replies Last Post Uncwilly mersennewiki 16 2018-10-23 23:16 kriesel mersennewiki 0 2018-08-13 13:02 Xyzzy mersennewiki 3 2011-02-18 03:31 ixfd64 mersennewiki 6 2006-05-22 12:05 delta_t PSearch 2 2006-05-21 07:05
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Wed May 25 16:59:55 UTC 2022 up 41 days, 15:01, 0 users, load averages: 1.14, 1.25, 1.23 | 1,874 | 6,150 | {"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-2022-21 | latest | en | 0.90466 |
http://tywkiwdbi.blogspot.com/2017/08/babylonian-trigonometry.html | 1,575,559,477,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540481076.11/warc/CC-MAIN-20191205141605-20191205165605-00247.warc.gz | 148,055,333 | 23,975 | ## 25 August 2017
### Babylonian trigonometry
At least two articles this week about new insights in the history of mathematics, derived from a cuneiform tablet:
At least 1,000 years before the Greek mathematician Pythagoras looked at a right angled triangle and worked out that the square of the longest side is always equal to the sum of the squares of the other two, an unknown Babylonian genius took a clay tablet and a reed pen and marked out not just the same theorem, but a series of trigonometry tables which scientists claim are more accurate than any available today.
The 3,700-year-old broken clay tablet survives in the collections of Columbia University, and scientists now believe they have cracked its secrets.
The team from the University of New South Wales in Sydney believe that the four columns and 15 rows of cuneiform – wedge shaped indentations made in the wet clay – represent the world’s oldest and most accurate working trigonometric table, a working tool which could have been used in surveying, and in calculating how to construct temples, palaces and pyramids...
As far back as 1945 the Austrian mathematician Otto Neugebauer and his associate Abraham Sachs were the first to note that Plimpton 322 has 15 pairs of numbers forming parts of Pythagorean triples: three whole numbers a, b and c such that a squared plus b squared equal c squared. The integers 3, 4 and 5 are a well-known example of a Pythagorean triple, but the values on Plimpton 322 are often considerably larger with, for example, the first row referencing the triple 119, 120 and 169.
Of added importance is that the Babylonians used a base 60, which gave greater mathematical precision for fractions than our current base 10. | 377 | 1,726 | {"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-2019-51 | latest | en | 0.9431 |
https://bair.berkeley.edu/blog/2021/10/25/coms_mbo/?ref=assemblyai.com | 1,716,735,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058956.26/warc/CC-MAIN-20240526135546-20240526165546-00103.warc.gz | 97,403,145 | 10,540 | Figure 1: Offline Model-Based Optimization (MBO): The goal of offline MBO is to optimize an unknown objective function $f(x)$ with respect to $x$, provided access to only as static, previously-collected dataset of designs.
Machine learning methods have shown tremendous promise on prediction problems: predicting the efficacy of a drug, predicting how a protein will fold, or predicting the strength of a composite material. But can we use machine learning for design? Conventionally, such problems have been tackled with black-box optimization procedures that repeatedly query an objective function. For instance, if designing a drug, the algorithm will iteratively modify the drug, test it, then modify it again. But when evaluating the efficacy of a candidate design involves conducting a real-world experiment, this can quickly become prohibitive. An appealing alternative is to create designs from data. Instead of requiring active synthesis and querying, can we devise a method that simply examines a large dataset of previously tested designs (e.g., drugs that have been evaluated before), and comes up with a new design that is better? We call this offline model-based optimization (offline MBO), and in this post, we discuss offline MBO methods and some recent advances.
# Offline Model-Based Optimization (Offline MBO)
Formally, the goal in offline model-based optimization is to maximize a black-box objective function $f(x)$ with respect to its input $x$, where the access to the true objective function is not available. Instead, the algorithm is provided access to a static dataset $\mathcal{D} = {(x_i, y_i)}$ of designs $x_i$ and corresponding objective values $y_i$. The algorithm consumes this dataset and produces an optimized candidate design, which is evaluated against the true objective function. Abstractly, the objective for offline MBO can be written as $\arg\max_{x = \mathcal{A}(D)} f(x)$, where $x = \mathcal{A}(D)$ indicates the design $x$ is a function of our dataset $\mathcal{D}$.
## What makes offline MBO challenging?
The offline nature of the problem prevents the algorithm from querying the ground truth objective, which makes the offline MBO problem much more difficult than the online counterpart. One obvious approach to tackle an offline MBO problem is to learn a model $\hat{f}(x)$ of the objective function using the dataset, and then applying methods from the more standard online optimization problem by treating the learned objective model as the true objective.
Figure 2: Overestimation at unseen inputs in the naive objective model fools the optimizer. Our conservative model prevents overestimation, and mitigates the optimizer from finding bad designs with erroneously high values.
However, this generally does not work: optimizing the design against the learned proxy model will produce out-of-distribution designs that “fool” the learned objective model into outputting a high value, similar to adversarial examples (see Figure 2 for an illustration). This is because that the learned model is trained on the dataset and therefore is only accurate for in-distribution designs. A naive strategy to address this out-of-distribution issue is to constrain the design to stay close to the data, but this is also problematic, since in order to produce a design that is better than the best training point, it is usually necessary to deviate from the training data, at least somewhat. Therefore, the conflict between the need to remain close to the data to avoid out-of-distribution inputs and the need to deviate from the data to produce better designs is one of the core challenges of offline MBO. This challenge is often exacerbated in real-world settings by the high dimensionality of the design space and the sparsity of the available data. A good offline MBO method needs to carefully balance these two sides, producing optimized designs that are good, but not too far from the data distribution.
## What prevents offline MBO from simply copying over the best design in the dataset?
One of the fundamental requirements for any effective offline MBO method is that it must improve over the best design observed in the training dataset. If this requirement is not met, one could simply return the best design from the dataset, without needing to run any kind of learning algorithm. When is such an improvement achievable in offline MBO problems? Offline MBO methods can improve over the best design in the dataset when the underlying design space exhibits “compositional structure”. For gaining intuition, consider an example, where the objective function can be represented as a sum of functions of independent partitions of the design variables, i.e., $f(x) = f_1(x[1]) + f_2(x[2]) + \cdots + f_N(x[N]))$, where $x[1], \cdots, x[N]$ denotes disjoint subsets of design variables $x$. The dataset of the offline MBO problem contains optimal design variable for each partition, but not the combination. If an algorithm can identify the compositional structure of the problem, it would be able to combine the optimal design variable for each partition together to obtain overall optimal design and therefore improving the performance over the best design in the dataset. To better demonstrate this idea, we created a toy problem in 2 dimensions and applied a naive MBO method that learns a model of the objective function via supervised regression, and then optimizes the learned estimate, as shown in the figure below. We can clearly see that the algorithm obtains the combined optimal $x$ and $y$, outperforming the best design in the dataset.
Figure 3: Offline MBO finds designs better than the best in the observed dataset by exploiting compositional structure of the objective function $f(x, y) = -x^2 - y^2$ . Left: datapoints in a toy quadratic function MBO task over 2D space with optimum at $(0,0)$ in blue, MBO found design in red. Right: Objective value for optimal design is much higher than that observed in the dataset.
# Prior Algorithms for Offline MBO
Given an offline dataset, the obvious starting point is to learn a model $\hat{f}_\theta(x)$ of the objective function from the dataset. Most offline MBO methods would indeed employ some form of learned model $\hat{f}_\theta(x)$ trained on the dataset to predict the objective value and guide the optimization process. As discussed previously, a very simple and naive baseline for offline MBO is to treat $\hat{f}_\theta(x)$ as the proxy to the true objective model and use gradient ascent to optimize $\hat{f}_\theta(x)$ with respect to $x$. However, this method often fails in practice, as gradient ascent can easily find designs that “fool” the model to predict a high objective value, similar to how adversarial examples are generated. Therefore, a successful approach using the learned model must prevent out-of-distribution designs that cause the model to overestimate the objective values, and the prior works have adopted different strategies to accomplish this.
A straightforward idea for preventing out-of-distribution data is to explicitly model the data distribution and constraint our designs to be within the distribution. Often the data distribution modeling is done via a generative model. CbAS and Autofocusing CbAS use a variational auto-encoder to model the distribution of designs, and MINs use a conditional generative adversarial network to model the distribution of designs conditioned on the objective value. However, generative modeling is a difficult problem. Furthermore, in order to be effective, generative models must be accurate near the tail ends of the data distribution as offline MBO must deviate from being close to the dataset to find improved designs. This imposes a strong feasibility requirement on such generative models.
# Conservative Objective Models
Can we devise an offline MBO method that does not utilize generative models, but also avoids the problems with the naive gradient-ascent based MBO method? To prevent this simple gradient ascent optimizer from getting “fooled” by the erroneously high values $\hat{f}_\theta(x)$ at out-of-distribution inputs, our approach, conservative objective models (COMs) performs a simple modification to the naive approach of training a model of the objective function. Instead of training a model $\hat{f}_\theta(x)$ via standard supervised regression, COMs applies an additional regularizer that minimizes the value of the learned model $\hat{f}_\theta(x^-)$ on adversarial designs $x^-$ that are likely to attain erroneously overestimated values. Such adversarial designs are the ones that likely appear falsely optimistic under the learned model, and by minimizing their values $\hat{f}_\theta(x^-)$, COMs prevents the optimizer from finding poor designs. This procedure superficially resembles a form of adversarial training.
How can we obtain such adversarial designs $x^-$? A straightforward approach for finding such adversarial designs is by running the optimizerwhich will be used to finally obtain optimized designs after training on a partially trained function $\hat{f}_\theta$. For example, in our experiments on continuous-dimensional design spaces, we utilize a gradient-ascent optimizer, and hence, run a few iterations of gradient ascent on a given snapshot of the learned function to obtain $x^-$. Given these designs, the regularizer in COMs pushes down the learned value $\hat{f}_\theta(x^-)$. To counter balance this push towards minimizing function values, COMs also additionally maximizes the learned $\hat{f}_\theta(x)$ on the designs observed in the dataset, $x \sim \mathcal{D}$, for which the ground truth value of $f(x)$ is known. This idea is illustratively depicted below.
Figure 4: A schematic procedure depicting training in COMs: COM performs supervised regression on the training data, pushes down the value of adversarially generated designs and counterbalances the effect by pushing up the value of the learned objective model on the observed datapoints
Denoting the samples found by running gradient-ascent in the inner loop as coming from a distribution $\mu(x)$, the training objective for COMs is given by:
$\theta^* \leftarrow \arg \min_\theta {\alpha \left(\mathbb{E}_{x^- \sim \mu(x)}[\hat{f}_\theta(x^-)] - \mathbb{E}_{x \sim \mathcal{D}}[\hat{f}_\theta(x)] \right)} + \frac{1}{2} \mathbb{E}_{(x, y) \sim \mathcal{D}} [(\hat{f}_\theta(x) - y)^2].$
This objective can be implemented as shown in the following (python) code snippet:
def mine_adversarial(x_0, current_model):
x_i = x_0
for i in range(T):
# gradient of current_model w.r.t. x_i
x_i = x_i + grad(current_model, x_i)
return x_i
def coms_training_loss(x, y):
mse_loss = (model(x) - y)**2
regularizer = model(mine_adversarial(x, model)) - model(x)
return mse_loss * 0.5 + alpha * regularizer
Non-generative offline MBO methods can also be designed in other ways. For example, instead of training a conservative model as in COMs, we can instead train model to capture uncertainty in the predictions of a standard model. One example of this is NEMO, which uses a normalized maximum likelihood (NML) formulation to provide uncertainty estimates.
# How do COMs Perform in Practice?
We evaluated COMs on a number of design problems in biology (designing a GFP protein to maximize fluorescence, designing DNA sequences to maximize binding affinity to various transcription factors), materials design (designing a superconducting material with the highest critical temperature), robot morphology design (designing the morphology of DâKitty and Ant robots to maximize performance) and robot controller design (optimizing the parameters of a neural network controller for the Hopper domain in OpenAI Gym). These tasks consist of domains with both discrete and continuous design spaces and span both low and high-dimensional tasks. We found that COMs outperform several prior approaches on these tasks, a subset of which is shown below. Observe that COMs consistently find a better design than the best in the dataset, and outperforms other generative modeling based prior MBO approaches (MINs, CbAS, Autofocusing CbAS) that pay a price for modeling the manifold of the design space, especially in problems such as Hopper Controller ($\geq 5000$ dimensions).
Table 1: Comparing the performance of COMs with prior offline MBO methods. Note that COMs generally outperform prior approaches, including those based on generative models, which especially struggle in high-dimensional problems such as Hopper Controller.
Empirical results on other domains can be found in our paper. To conclude our discussion of empirical results, we note that a recent paper devises an offline MBO approach to optimize hardware accelerators in a real hardware-design workflow, building on COMs. As shown in Kumar et al. 2021 (Tables 3, 4), this COMs-inspired approach finds better designs than various prior state-of-the-art online MBO methods that access the simulator via time-consuming simulation. While, in principle, one can always design an online method that should perform better than any offline MBO method (for example, by wrapping an offline MBO method within an active data collection strategy), good performance of offline MBO methods inspired by COMs indicates the efficacy and the potential of offline MBO approaches in solving design problems.
# Discussion, Open Problems and Future Work
While COMs present a simple and effective approach for tackling offline MBO problems, there are several important open questions that need to be tackled. Perhaps the most straightforward open question is to devise better algorithms that combine the benefits of both generative approaches and COMs-style conservative approaches. Beyond algorithm design, perhaps one of the most important open problems is designing effective cross-validation strategies: in supervised prediction problems, a practitioner can adjust model capacity, add regularization, tune hyperparameters and make design decisions by simply looking at validation performance. Improving the validation performance will likely also improve the test performance because validation and test samples are distributed identically and generalization guarantees for ERM theoretically quantify this. However, such a workflow cannot be applied directly to offline MBO, because cross-validation in offline MBO requires assessing the accuracy of counterfactual predictions under distributional shift. Some recent work utilizes practical heuristics such as validation performance computed on a held-out dataset consisting of only âspecialâ designs (e.g., only the top-k best designs) for cross-validation of COMs-inspired methods, which seems to perform reasonably well in practice. However, it is not clear that this is the optimal strategy one can use for cross-validation. We expect that much more effective strategies can be developed by understanding the effects of various factors (such as the capacity of the neural network representing $\hat{f}_\theta(x)$, the hyperparameter $\alpha$ in COMs, etc.) on the dynamics of optimization of COMs and other MBO methods.
Another important open question is characterizing properties of datasets and data distributions that are amenable to effective offline MBO methods. The success of deep learning indicates that not just better methods and algorithms are required for good performance, but that the performance of deep learning methods heavily depends on the data distribution used for training. Analogously, we expect that the performance of offline MBO methods also depends on the quality of data used. For instance, in the didactic example in Figure 3, no improvement could have been possible via offline MBO if the data were localized along a thin line parallel to the x-axis. This means that understanding the relationship between offline MBO solutions and the data-distribution, and effective dataset design based on such principles is likely to have a large impact. We hope that research in these directions, combined with advances in offline MBO methods, would enable us to solve challenging design problems in various domains.
We thank Sergey Levine for valuable feedback on this post. We thank Brandon Trabucco for making Figures 1 and 2 of this post. This blog post is based on the following paper:
Conservative Objective Models for Effective Offline Model-Based Optimization
Brandon Trabucco*, Aviral Kumar*, Xinyang Geng, Sergey Levine.
In International Conference on Machine Learning (ICML), 2021. arXiv code website
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# Fixed number of split?
840
3
05-18-2017 01:51 PM
by
New Contributor III
Hi,
Is there a way to specify the exact number of times a shape is split?
For example (from the manual):
A--> split(x){ 2 : X(2) | 1 : Y(1) }*
will give as many of the "2 1 2 1 ..." pattern as can fit.
But what if I want to have exactly 3, as in "2 1 2 1 2 1" and leave the rest of the shape unused?
I can always do:
A--> split(x){ 2 : X(2) | 1 : Y(1) | 2 : X(2) | 1 : Y(1) | 2 : X(2) | 1 : Y(1) }
But I want the number of times to repeat to be an attribute. Something like:
A--> split(x){ 2 : X(2) | 1 : Y(1) } * repeat
Thanks!
3 Replies
Esri Regular Contributor
No, unfortunately, there is no easy way to specify how many times you want the pattern to repeat.
You would have to calculate the length of the last part and put that in the split.
OR
You would have to write it recursively to keep splitting until the desired count is reached.
by
New Contributor III
Thanks. How does one go about making a feature request?
Esri Regular Contributor
GeoNet users can make CityEngine feature requests on ArcGIS Ideas - CityEngine | 350 | 1,171 | {"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} | 2.703125 | 3 | CC-MAIN-2024-30 | latest | en | 0.883844 |
https://math.stackexchange.com/questions/4336940/find-n-given-average-distance-between-0-le-a-le-n-1-and-1-le-b-le-n-1/4337141 | 1,701,241,384,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100056.38/warc/CC-MAIN-20231129041834-20231129071834-00696.warc.gz | 449,710,232 | 41,015 | # Find $n$ given average distance between $0 \le a \le n-1$ and $1 \le b \le n-1$
Given a natural number $$n>1$$,
natural numbers $$a$$ are chosen uniform randomly between $$0$$ and $$n-1$$ i.e.: $$0 \le a \le n - 1$$
and
natural numbers $$b$$ are chosen uniform randomly between $$1$$ and $$n-1$$ i.e.: $$1 \le b \le n - 1$$ The average $$\delta$$ of the distance $$|a-b|$$ between numbers $$a$$ and $$b$$ equals $$6.5$$.
What is the given value of $$n$$?
First, my true apologies for, initially, not having been conform forum guidelines.
Have a look at answer provided, also upon request (by @lulu), where formula is derived.
Thanks all for getting me on rails.
I personally believe the formula is strikingly simple. Some people would perhaps expect numerical methods to be required for general reasoning, but not so.
Here is the motivation: average distance in this particular case (note $$0 \le a$$ and $$1 \le b$$) turns out to have a very simple formula in terms of $$n$$, allowing to, conversely, easily find $$n$$ in function of it. As it so happens, linear Gauss formula, and its quadratic generalization, in this particular case, both contain a common factor that perfectly cancels out for probability computation.
I have used (in disguise) the formulas for two puzzles (112210 112821) on puzzling exchange where also, conversely, some amount is asked in function of some average. So now I thought to present the math.
Have a nice day. I hope to contribute in simplicity.
To be honest: I doubted myself about the validity of such answer and therefore presented the question to the kind exchange community.
• What have you tried? Dec 18, 2021 at 16:19
• I suggest: pick a number like $n=10$ or whatever and compute the expected value of $|a-b|$. That ought to give you a pretty good idea.
– lulu
Dec 18, 2021 at 17:22
• I would ideally want to see equation(s) that express $n$ in function of the average distance, and/or vice versa, so one can find $n$ by solving such equation(s). Dec 18, 2021 at 19:07
• Please edit your post to include your efforts. I would not expect a simple closed formula, numerical methods will almost certainly be required.
– lulu
Dec 18, 2021 at 21:55
• I think this question and its answer have some interest, but its format is very unusual and confusing. If it was written in a more standard way, I would vote to re-open. Dec 21, 2021 at 0:26
I will start with a few points aimed at the original poster, which they may find useful for the future; the actual problem at hand is going to be dealt with at the very end.
The original problem we are looking at can be summarized as follows: Two integers, $$a$$ and $$b$$, such that $$0\leq a\leq (n-1)$$ and $$1\leq b\leq (n-1)$$, are chosen uniformly randomly and we compute their distance $$\delta=|a-b|$$. For which values of $$n$$ is the expected value of $$\delta$$ equal to $$6.5$$?
Now, if one has such a problem, there are multiple questions that can be asked:
• What should I do? I have no idea at all.
If this is the case, don't forget to mention that last bit! The comments / answers you get are likely to be small hints or ideas how to start rather than fully-fledged complete solutions. If you manage to solve the problem, feel free to add your own answer.
• I think I am on the right track but I got stuck. How to deal with this ugly sum?
Provide the steps you took as part of the question statement. The answers / comments you get might provide a further hint (or a full solution). Again, if you happen to solve the problem yourself in the meantime, feel free to answer it too.
• I think I have solved this problem but I am not sure if I did it correctly. Can you check my solution?
Remember, your proposed solution should be part of the question here; with the request for verification of the solution (or certain parts of it) being included explicitly. There is even a nice tag one can use: [solution-verification]. The answer should either point out a flaw in the reasoning or confirm the validity of the approach, possibly filling some gaps if necessary.
• I have a solution but it feels way too complicated. Can you find a better one?
This is not that much different from the previous one: Show how you did it and ask for possible improvements. Of course, if you discover a simpler approach to solving the problem yourself, it can be a valid answer here too.
Note that in the last two cases, your original solution to the problem is not an answer; it is part of the question. In general, stating a problem and providing one's own answer without explicitly requesting further feedback from the audience is unlikely to elicit response, especially if the existing answer seems to be answering the question satisfactorily (people won't be adding their own answers unless they (substantially) improve upon the existing one; this is where is might be different from the puzzling SE).
In this particular case, the solution provided by the original poster is correct and is also reasonably simple; hard to be improved upon.
However, if one prefers a pull-out-of-a-hat magic trick, one can observe that for any $$1\leq k\leq (n-1)$$ there are exactly $$(n-k)^2$$ pairs of numbers $$a$$, $$b$$ for which $$|a-b|\geq k$$. In the matrix shown in OP's solution, such pairs form a bigger triangle in the upper-left corner and a tad bit smaller one in the lower-right corner; if we put them together, they add up to a square. Thus, the total sum of all numbers in the matrix is the same as sum of squares from $$1$$ to $$(n-1)^2$$. But there is a well-known formula for this sum: $$\frac{1}{6}n(n-1)(2n-1)$$. Dividing by $$n(n-1)$$ gives us the desired formula $$\frac{1}{6}(2n-1)$$ which yields $$n=20$$.
• Thanks for detailed alternatives about how to ask questions one knows the answer for. I was relatively new to the math exchange back then and originally asked and answered the question rather as a puzzle where knowing the answer is common and hiding details in answer is trying to not spoil others ones efforts in advance. I apologize again for all that. Aug 7, 2022 at 11:04
$$n = 20$$
To calculate average distance between $$a$$ and $$b$$ there are a total amount of $$n(n-1)$$ pairs $$(a,b)$$ with: $$0<=a<=n-1$$ and $$1<=b<=n-1$$ so consider the $$n*(n-1)$$ matrix:
$$\begin{bmatrix}|0-(n-1)| & .. & |(n-1)-(n-1)|\\: & |a-b| & :\\|0-1| & ... & |(n-1)-1|\end{bmatrix}$$
The total of all distances between $$a$$ and $$b$$ is the total of the distances between all pairs in the square $$n*n$$ matrix:
$$\begin{bmatrix}|0-(n-1)| & .. & |(n-1)-(n-1)|\\: & |a-b| & :\\|0-0| & ... & |(n-1)-0|\end{bmatrix}$$
minus the total of the distances between all pairs from the excluded $$n*1$$ bottom row:
$$\begin{bmatrix}|0-0| & .. & |a-b| & .. & |(n-1)-0|\end{bmatrix}$$
giving (using standard formulas) : $$(n-1)n(n+1)/3 - (n-1)n/2$$ where the total amount of pairs $$n(n-1)$$ nicely cancels out in both terms and the average of the distance between $$a$$ and $$b$$ therefore is $$(n+1)/3 - 1/2 = (2n-1)/6$$ such gives $$n=20$$ for $$\delta=6.5$$
• Do we need all these spoiler tags? This is a maths Q&A website. People know they would be "spoiled" when reading an answer. Dec 21, 2021 at 0:27
• @Taladris ... I am very sorry and apologise ... I come from puzzling exchange where spoilers are 'part of the game' etiquette and the challenge is to find an answer rather than getting one. I will remove spoilers if you think that would help draw attention conform to math exchange policy. Again, sorry for clumsyness, I really wanted to show and have verified an IMO cool mathematical fact. Dec 21, 2021 at 0:36
• All right :-) spoilers removed. Thanks for the constructive comments. $\delta = (2n-1)/6$ is (hopefully) the correct formula I humbly wanted to present... Dec 21, 2021 at 0:48
• still sad :-( the simplicity of the result got negative votes, but, oh well :-) I enjoy this exchange Jul 12, 2022 at 1:29 | 2,113 | 7,937 | {"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": 60, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-50 | latest | en | 0.920325 |
https://www.hrwhisper.me/2015/12/ | 1,601,303,970,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401601278.97/warc/CC-MAIN-20200928135709-20200928165709-00059.warc.gz | 809,910,310 | 13,519 | ## leetcode Wiggle Sort II
### leetcode Wiggle Sort II
Given an unsorted array `nums`, reorder it such that `nums[0] < nums[1] > nums[2] < nums[3]...`.
Example:
(1) Given `nums = [1, 5, 1, 1, 6, 4]`, one possible answer is `[1, 4, 1, 5, 1, 6]`.
(2) Given `nums = [1, 3, 2, 2, 3, 1]`, one possible answer is `[2, 3, 1, 3, 1, 2]`.
Note:
You may assume all input has valid answer.
Can you do it in O(n) time and/or in-place with O(1) extra space?
## leetcode Coin Change
### leetcode Coin Change
You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return `-1`.
Example 1:
coins = `[1, 2, 5]`, amount = `11`
return `3` (11 = 5 + 5 + 1)
Example 2:
coins = `[2]`, amount = `3`
return `-1`.
Note:
You may assume that you have an infinite number of each kind of coin.
## Github student pack 申请
• digitalocean vps 50\$ (之前貌似是100 ,博主一年VPS不用钱了~)
• 5个github 私有仓库
• 等等
## leetcode Create Maximum Number
### leetcode Create Maximum Number
Given two arrays of length `m` and `n` with digits `0-9` representing two numbers. Create the maximum number of length `k <= m + n` from digits of the two. The relative order of the digits from the same array must be preserved. Return an array of the `k` digits. You should try to optimize your time and space complexity.
Example 1:
nums1 = `[3, 4, 6, 5]`
nums2 = `[9, 1, 2, 5, 8, 3]`
k = `5`
return `[9, 8, 6, 5, 3]`
Example 2:
nums1 = `[6, 7]`
nums2 = `[6, 0, 4]`
k = `5`
return `[6, 7, 6, 0, 4]`
Example 3:
nums1 = `[3, 9]`
nums2 = `[8, 9]`
k = `3`
return `[9, 8, 9]`
## digitalocean 下 centos 搭建 wordpress过程
• centos下安装wordpress过程
• 开启mod_rewrite模块
• 开启GZIP压缩网页
• 上传图片/文件失败解决办法
• YOAST SEO sitemap 404 解决办法
## leetcode Bulb Switcher
### leetcode Bulb Switcher
There are n bulbs that are initially off. You first turn on all the bulbs. Then, you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it’s off or turning off if it’s on). For the nth round, you only toggle the last bulb. Find how many bulbs are on after n rounds.
Example:
Given n = 3.
At first, the three bulbs are [off, off, off].
After first round, the three bulbs are [on, on, on].
After second round, the three bulbs are [on, off, on].
After third round, the three bulbs are [on, off, off].
So you should return 1, because there is only one bulb is on.
## 写在前面
• WEB开发的时候，你会更加的规范
• 知识上的扩充
• 让过多人看到你的博客站点 | 889 | 2,574 | {"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} | 3.484375 | 3 | CC-MAIN-2020-40 | latest | en | 0.659082 |
http://massivealgorithms.blogspot.com/2016/10/finding-sum-of-digits-of-number-until.html | 1,508,547,893,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824537.24/warc/CC-MAIN-20171021005202-20171021025202-00021.warc.gz | 233,383,359 | 34,672 | ## Saturday, October 8, 2016
### Finding sum of digits of a number until sum becomes single digit - GeeksforGeeks
Finding sum of digits of a number until sum becomes single digit - GeeksforGeeks
Given a number n, we need to find the sum of its digits such that:
```If n < 10
digSum(n) = n
Else
digSum(n) = Sum(digSum(n))
```
`int` `digSum(``int` `n)`
`{`
` ``return` `(n % 9 == 0) ? 9 : (n % 9);`
`}`
`int` `digSum(``int` `n)`
`{`
` ``int` `sum = 0;`
` `
` ``while``(n > 0 || sum > 9)`
` ``{`
` ``if``(n == 0)`
` ``{`
` ``n = sum;`
` ``sum = 0;`
` ``}`
` ``sum += n % 10;`
` ``n /= 10;`
` ``}`
` ``return` `sum;`
`}`
Read full article from Finding sum of digits of a number until sum becomes single digit - GeeksforGeeks | 301 | 801 | {"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-2017-43 | longest | en | 0.566447 |
https://find-error.com/questions/20449427/how-can-i-read-inputs-as-numbers | 1,670,599,020,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711417.46/warc/CC-MAIN-20221209144722-20221209174722-00769.warc.gz | 294,998,044 | 9,695 | # How can I read inputs as numbers?
Viewed 995.74 k times
Why are `x` and `y` strings instead of ints in the below code?
(Note: in Python 2.x use `raw_input()`. In Python 3.x use `input()`. `raw_input()` was renamed to `input()` in Python 3.x)
``````play = True
while play:
x = input("Enter a number: ")
y = input("Enter a number: ")
print(x + y)
print(x - y)
print(x * y)
print(x / y)
print(x % y)
if input("Play again? ") == "no":
play = False
``````
### Solution
Since Python 3, `input` returns a string which you have to explicitly convert to `int`s, with `int`, like this
``````x = int(input("Enter a number: "))
y = int(input("Enter a number: "))
``````
You can accept numbers of any base and convert them directly to base-10 with the `int` function, like this
``````>>> data = int(input("Enter a number: "), 8)
Enter a number: 777
>>> data
511
>>> data = int(input("Enter a number: "), 16)
Enter a number: FFFF
>>> data
65535
>>> data = int(input("Enter a number: "), 2)
Enter a number: 10101010101
>>> data
1365
``````
The second parameter tells what is the base of the numbers entered and then internally it understands and converts it. If the entered data is wrong it will throw a `ValueError`.
``````>>> data = int(input("Enter a number: "), 2)
Enter a number: 1234
Traceback (most recent call last):
File "<input>", line 1, in <module>
ValueError: invalid literal for int() with base 2: '1234'
``````
For values that can have a fractional component, the type would be `float` rather than `int`:
``````x = float(input("Enter a number:"))
``````
### Differences between Python 2 and 3
Summary
• Python 2's `input` function evaluated the received data, converting it to an integer implicitly (read the next section to understand the implication), but Python 3's `input` function does not do that anymore.
• Python 2's equivalent of Python 3's `input` is the `raw_input` function.
Python 2.x
There were two functions to get user input, called `input` and `raw_input`. The difference between them is, `raw_input` doesn't evaluate the data and returns as it is, in string form. But, `input` will evaluate whatever you entered and the result of evaluation will be returned. For example,
``````>>> import sys
>>> sys.version
'2.7.6 (default, Mar 22 2014, 22:59:56) \n[GCC 4.8.2]'
>>> data = input("Enter a number: ")
Enter a number: 5 + 17
>>> data, type(data)
(22, <type 'int'>)
``````
The data `5 + 17` is evaluated and the result is `22`. When it evaluates the expression `5 + 17`, it detects that you are adding two numbers and so the result will also be of the same `int` type. So, the type conversion is done for free and `22` is returned as the result of `input` and stored in `data` variable. You can think of `input` as the `raw_input` composed with an `eval` call.
``````>>> data = eval(raw_input("Enter a number: "))
Enter a number: 5 + 17
>>> data, type(data)
(22, <type 'int'>)
``````
Note: you should be careful when you are using `input` in Python 2.x. I explained why one should be careful when using it, in this answer.
But, `raw_input` doesn't evaluate the input and returns as it is, as a string.
``````>>> import sys
>>> sys.version
'2.7.6 (default, Mar 22 2014, 22:59:56) \n[GCC 4.8.2]'
>>> data = raw_input("Enter a number: ")
Enter a number: 5 + 17
>>> data, type(data)
('5 + 17', <type 'str'>)
``````
Python 3.x
Python 3.x's `input` and Python 2.x's `raw_input` are similar and `raw_input` is not available in Python 3.x.
``````>>> import sys
>>> sys.version
'3.4.0 (default, Apr 11 2014, 13:05:11) \n[GCC 4.8.2]'
>>> data = input("Enter a number: ")
Enter a number: 5 + 17
>>> data, type(data)
('5 + 17', <class 'str'>)
``````
##### joey30
In Python 3.x, `raw_input` was renamed to `input` and the Python 2.x `input` was removed.
This means that, just like `raw_input`, `input` in Python 3.x always returns a string object.
To fix the problem, you need to explicitly make those inputs into integers by putting them in `int`:
``````x = int(input("Enter a number: ")) | 1,201 | 4,026 | {"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} | 2.671875 | 3 | CC-MAIN-2022-49 | latest | en | 0.782979 |
http://www.chem.utoronto.ca/coursenotes/analsci/stats/BasicPlots.html | 1,540,290,491,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516123.97/warc/CC-MAIN-20181023090235-20181023111735-00054.warc.gz | 426,474,262 | 4,935 | ## Plotting & Charts:
A useful way to view and present your data is with a proper graph, which provides a visual summary of your experiment. Historically, Excel has been very poor at producing scientific graphs, although the situation has improved considerably. Excel provides many different types of charts, which are primarily intended for business use. The chart type appropriate for calibration curves is the X-Y scatter plot. We will use this tool to plot the data generated in the preceding exercises for the equations y = 2x + 5 and y = x2 + 2. In the process, you will learn how to:
• produce simple graphs (exercise 1)
• specify titles, legends, and axis labels (exercise 1)
• change the axis numbering scheme (exercise 2)
• specify the correct number of decimal places (exercise 2)
View a sample of a properly formatted graph
### Exercise 1:
Open the spreadsheet from the previous exercise, and select the data for the straight-line equation. You can do this by (a) click-dragging across the cells containing the data (`B1` to `C16`); or (b) by click-dragging (or clicking while holding down the shift key) on the label buttons for columns `B` and `C` at the top of the spreadsheet window. Note that the values for the x variable are in the selected column on the left, while the values for the y variable are in the selected column on the right:
Use a scatter plot with no connecting lines for a calibration curve
Do not use Excel's Line chart types for calibration curves - this type does not do what you think!
1. Select Insert→Chart. The Chart Wizard dialog box will appear.
2. There are many types of charts to choose from. Select the XY (Scatter) plot and click the Next button.
3. We can choose to plot multiple graphs on the same chart. We won’t do that here. In fact, Excel has already specified the proper data series for each axis, so we do not need to change anything. You can, however, change the legend text in the Series tab. Click on the Next button to continue.
4. You can enter chart titles, axis labels, and other display characteristics for the chart. Change what you want, then click Next.
5. Finally, you can specify whether the chart should be shown in the current worksheet, or whether it should stand alone in a separate worksheet. Click Finish and the chart should appear.
The other scatter plot types interpolate line segments between points in order; sort the data by increasing x value to avoid strange effects!
### Exercise 2:
Repeat exercise 1, only this time use the parabola data in columns `E` and `F`. Give the graph a title such as “Potential Energy Well”, and give the axes the following labels:
x-axis:
Separation r (nm)
y-axis:
Potential energy U (x1E-19 J)
Once a chart is produced, you can change all aspects of its appearance. For instance, right-clicking on either axis allows you to access the Format Axis... dialog (you can also click on the axis to select it, then use the Format→Selected Axis... menu item):
One weakness of Excel is that it still does not allow mixed fonts, subsrcipts, or superscripts in chart titles or axis labels.
Use letter u for µ, and 1E-06 for 10-6, etc.
Note that it is almost impossible to use the more correct ‘Quantity symbol / unit’ (quantity divided by unit) notation.
Using the fields in the Scale tab, set the y-axis to display values from 1.5 to 3.5, with major divisions every 0.5 units. Likewise, use the Number tab to display the y-axis numbers to 2 decimal places.
Similarly, set the x-axis to display numbers to three decimal places, and to show negative values as -n.nnn rather than in accounting style as (n.nnn)
Finally, right-click (Mac: cmd-click) on the different parts of the chart and explore the various contextual menu items that come up to remove the grid lines (Chart Options...) and background colour (Format Plot Area), and change the symbols to open, dark blue squares (Format Data Series...)
Note the Logarithmic scale and Values in reverse order items - useful for IR spectra!
The basic plots for both exercises should look like this before adjusting the scales and format: | 921 | 4,107 | {"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} | 3.28125 | 3 | CC-MAIN-2018-43 | latest | en | 0.881346 |
https://codereview.stackexchange.com/questions/247376/interpolation-across-different-dataframes-with-pandas | 1,618,466,394,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038083007.51/warc/CC-MAIN-20210415035637-20210415065637-00150.warc.gz | 281,740,720 | 40,551 | # Context of the Problem
I am running a discrete event simulation where at the end of each event I store the state of the system in a row of a dataframe. Each simulation run has a clock which determines which events are run and when, all the simulation have the same initial clock (0) and the same end clock (simulation end). but the number of rows for the dataframes may be different because the simulation has several stochastic components. The clock column is then converted to timestamp and use as index of the dataframe.
Whenever one does a simulation, it is good practice to run it many times and then average over all the replicates, doing so in this setup is a bit complicated because each simulation produced a dataframe which different indexes.
# Proposed Solution
So far this is the solution I found:
# Auxiliary functions
def get_union_index(dfs):
index_union = dfs[0].index
for df in dfs[1:]:
index_union = index_union.union(df.index)
return pd.Series(index_union).drop_duplicates().reset_index(drop=True)
def interpolate(df, index, method='time'):
aux = df.astype(float)
aux = aux.reindex(index).interpolate(method=method).fillna(method='ffill').fillna(method='bfill')
return aux
# Simulation
dfs = []
replicates = 30
seeds = list(range(40, 40 + replicates))
dfs = [simulate(seed=seed, **parameters) for seed in seeds]
# Main Code
union_index = get_union_index(dfs)
dfs_interpolates = [interpolate(df, union_index) for df in dfs]
df_concat = pd.concat(dfs_interpolates)
by_row_index = df_concat.groupby(df_concat.index)
# Averaging
df_means = by_row_index.mean()
df_std = by_row_index.std()
## Explanation
First, it is necessary to combine all the indexes, then this combined index is used to re-index all the dataframes and the nan values are filled using interpolation.
## Questions
1. Is there a native pandas function that could simplify this?
2. (If not 1) Is there an alternative way to combine the datasets directly, since the majority of the index are disjoint, union_index has a length of approximately len(df) * len(dfs) which is actually huge.
3. Which should be the best interpolation method to use in the interpolate? Since each row is an event and only a few variables are changed per event, from one row to the next only a few columns are modified, making it possible to have several consecutive rows with identical values in several columns (but not all). | 536 | 2,401 | {"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} | 2.640625 | 3 | CC-MAIN-2021-17 | longest | en | 0.767019 |
http://ls1tech.com/forums/ls1-domestic-forums/t-16020.html | 1,386,425,190,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163054576/warc/CC-MAIN-20131204131734-00014-ip-10-33-133-15.ec2.internal.warc.gz | 112,229,726 | 5,379 | # Nitrous Oxide - nitrous wiring pics
View Full Version : nitrous wiring pics
nitrousc5
07-07-2002, 06:38 AM
here's a link to my posted pics at Corvetteforum.com. easier than figuring out how to post pics here. http://forums.corvetteforum.com/zerothread?id=330002
2000 Camaro SS
08-01-2002, 05:47 PM
[QUOTE]Originally posted by nitrousc5:
<strong>here's a link to my posted pics at Corvetteforum.com. easier than figuring out how to post pics here. http://forums.corvetteforum.com/zerothread?id=330002</strong>[/QUOTE
DAMN!!!! <img border="0" title="" alt="[Eek!]" src="gr_eek2.gif" />
I guess I'm the only one that cought that you gained 323 pounds of Torque with a twin 75 shot!
<img border="0" alt="[cheers]" title="" src="graemlins/gr_cheers.gif" />
Tell me how you got it set up and run it please!
<small>[ August 01, 2002, 04:48 PM: Message edited by: 2000 Camaro SS ]</small>
nitrousc5
08-05-2002, 03:10 AM
first stage is active whenever within the RPM window, then i turn the 2nd stage on via latching pushbutton switch right after first stage kicks on (.5 second pause to allow fuel pressure to bounce back up). anyway, with all this HP coming on at 3200-3800rpm's that's where the big Tq numbers come from. here's the formula to help make sense of it:
TQ = (hp x 5252)/rpm
so, let's say your hp peaks 500rwhp at 3800 rpm's (shortly after you turn it on), then:
500 x 5252 = 2626000/3800 = 691rwtq
if you bring on the same 500 hp at, say, 3000 rpms instead of 3800rpm, the tq soars to 875rwtq!!!!
<small>[ August 05, 2002, 02:11 AM: Message edited by: nitrousc5 ]</small> | 499 | 1,592 | {"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-2013-48 | latest | en | 0.823095 |
https://www.aqua-calc.com/convert/density/slug-per-cubic-decimeter-to-stone-per-metric-cup | 1,610,745,290,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703496947.2/warc/CC-MAIN-20210115194851-20210115224851-00043.warc.gz | 680,105,255 | 10,721 | # Convert slugs per (cubic decimeter) to stones per (metric cup)
## sl/dm³ to st/metric cup (st/metric c) (sl:slug, dm:decimeter, st:stone, c:cup)
### slugs per cubic decimeter to stones per metric cup conversion cards
• 1
through
20
slugs per cubic decimeter
• 1 sl/dm³ to st/metric c = 0.574536581 st/metric c
• 2 sl/dm³ to st/metric c = 1.149073161 st/metric c
• 3 sl/dm³ to st/metric c = 1.723609742 st/metric c
• 4 sl/dm³ to st/metric c = 2.298146323 st/metric c
• 5 sl/dm³ to st/metric c = 2.872682904 st/metric c
• 6 sl/dm³ to st/metric c = 3.447219484 st/metric c
• 7 sl/dm³ to st/metric c = 4.021756065 st/metric c
• 8 sl/dm³ to st/metric c = 4.596292646 st/metric c
• 9 sl/dm³ to st/metric c = 5.170829226 st/metric c
• 10 sl/dm³ to st/metric c = 5.745365807 st/metric c
• 11 sl/dm³ to st/metric c = 6.319902388 st/metric c
• 12 sl/dm³ to st/metric c = 6.894438968 st/metric c
• 13 sl/dm³ to st/metric c = 7.468975549 st/metric c
• 14 sl/dm³ to st/metric c = 8.04351213 st/metric c
• 15 sl/dm³ to st/metric c = 8.618048711 st/metric c
• 16 sl/dm³ to st/metric c = 9.192585291 st/metric c
• 17 sl/dm³ to st/metric c = 9.767121872 st/metric c
• 18 sl/dm³ to st/metric c = 10.341658453 st/metric c
• 19 sl/dm³ to st/metric c = 10.916195033 st/metric c
• 20 sl/dm³ to st/metric c = 11.490731614 st/metric c
• 21
through
40
slugs per cubic decimeter
• 21 sl/dm³ to st/metric c = 12.065268195 st/metric c
• 22 sl/dm³ to st/metric c = 12.639804775 st/metric c
• 23 sl/dm³ to st/metric c = 13.214341356 st/metric c
• 24 sl/dm³ to st/metric c = 13.788877937 st/metric c
• 25 sl/dm³ to st/metric c = 14.363414518 st/metric c
• 26 sl/dm³ to st/metric c = 14.937951098 st/metric c
• 27 sl/dm³ to st/metric c = 15.512487679 st/metric c
• 28 sl/dm³ to st/metric c = 16.08702426 st/metric c
• 29 sl/dm³ to st/metric c = 16.66156084 st/metric c
• 30 sl/dm³ to st/metric c = 17.236097421 st/metric c
• 31 sl/dm³ to st/metric c = 17.810634002 st/metric c
• 32 sl/dm³ to st/metric c = 18.385170582 st/metric c
• 33 sl/dm³ to st/metric c = 18.959707163 st/metric c
• 34 sl/dm³ to st/metric c = 19.534243744 st/metric c
• 35 sl/dm³ to st/metric c = 20.108780325 st/metric c
• 36 sl/dm³ to st/metric c = 20.683316905 st/metric c
• 37 sl/dm³ to st/metric c = 21.257853486 st/metric c
• 38 sl/dm³ to st/metric c = 21.832390067 st/metric c
• 39 sl/dm³ to st/metric c = 22.406926647 st/metric c
• 40 sl/dm³ to st/metric c = 22.981463228 st/metric c
• 41
through
60
slugs per cubic decimeter
• 41 sl/dm³ to st/metric c = 23.555999809 st/metric c
• 42 sl/dm³ to st/metric c = 24.130536389 st/metric c
• 43 sl/dm³ to st/metric c = 24.70507297 st/metric c
• 44 sl/dm³ to st/metric c = 25.279609551 st/metric c
• 45 sl/dm³ to st/metric c = 25.854146132 st/metric c
• 46 sl/dm³ to st/metric c = 26.428682712 st/metric c
• 47 sl/dm³ to st/metric c = 27.003219293 st/metric c
• 48 sl/dm³ to st/metric c = 27.577755874 st/metric c
• 49 sl/dm³ to st/metric c = 28.152292454 st/metric c
• 50 sl/dm³ to st/metric c = 28.726829035 st/metric c
• 51 sl/dm³ to st/metric c = 29.301365616 st/metric c
• 52 sl/dm³ to st/metric c = 29.875902196 st/metric c
• 53 sl/dm³ to st/metric c = 30.450438777 st/metric c
• 54 sl/dm³ to st/metric c = 31.024975358 st/metric c
• 55 sl/dm³ to st/metric c = 31.599511939 st/metric c
• 56 sl/dm³ to st/metric c = 32.174048519 st/metric c
• 57 sl/dm³ to st/metric c = 32.7485851 st/metric c
• 58 sl/dm³ to st/metric c = 33.323121681 st/metric c
• 59 sl/dm³ to st/metric c = 33.897658261 st/metric c
• 60 sl/dm³ to st/metric c = 34.472194842 st/metric c
• 61
through
80
slugs per cubic decimeter
• 61 sl/dm³ to st/metric c = 35.046731423 st/metric c
• 62 sl/dm³ to st/metric c = 35.621268003 st/metric c
• 63 sl/dm³ to st/metric c = 36.195804584 st/metric c
• 64 sl/dm³ to st/metric c = 36.770341165 st/metric c
• 65 sl/dm³ to st/metric c = 37.344877746 st/metric c
• 66 sl/dm³ to st/metric c = 37.919414326 st/metric c
• 67 sl/dm³ to st/metric c = 38.493950907 st/metric c
• 68 sl/dm³ to st/metric c = 39.068487488 st/metric c
• 69 sl/dm³ to st/metric c = 39.643024068 st/metric c
• 70 sl/dm³ to st/metric c = 40.217560649 st/metric c
• 71 sl/dm³ to st/metric c = 40.79209723 st/metric c
• 72 sl/dm³ to st/metric c = 41.36663381 st/metric c
• 73 sl/dm³ to st/metric c = 41.941170391 st/metric c
• 74 sl/dm³ to st/metric c = 42.515706972 st/metric c
• 75 sl/dm³ to st/metric c = 43.090243553 st/metric c
• 76 sl/dm³ to st/metric c = 43.664780133 st/metric c
• 77 sl/dm³ to st/metric c = 44.239316714 st/metric c
• 78 sl/dm³ to st/metric c = 44.813853295 st/metric c
• 79 sl/dm³ to st/metric c = 45.388389875 st/metric c
• 80 sl/dm³ to st/metric c = 45.962926456 st/metric c
• 81
through
100
slugs per cubic decimeter
• 81 sl/dm³ to st/metric c = 46.537463037 st/metric c
• 82 sl/dm³ to st/metric c = 47.111999617 st/metric c
• 83 sl/dm³ to st/metric c = 47.686536198 st/metric c
• 84 sl/dm³ to st/metric c = 48.261072779 st/metric c
• 85 sl/dm³ to st/metric c = 48.83560936 st/metric c
• 86 sl/dm³ to st/metric c = 49.41014594 st/metric c
• 87 sl/dm³ to st/metric c = 49.984682521 st/metric c
• 88 sl/dm³ to st/metric c = 50.559219102 st/metric c
• 89 sl/dm³ to st/metric c = 51.133755682 st/metric c
• 90 sl/dm³ to st/metric c = 51.708292263 st/metric c
• 91 sl/dm³ to st/metric c = 52.282828844 st/metric c
• 92 sl/dm³ to st/metric c = 52.857365424 st/metric c
• 93 sl/dm³ to st/metric c = 53.431902005 st/metric c
• 94 sl/dm³ to st/metric c = 54.006438586 st/metric c
• 95 sl/dm³ to st/metric c = 54.580975167 st/metric c
• 96 sl/dm³ to st/metric c = 55.155511747 st/metric c
• 97 sl/dm³ to st/metric c = 55.730048328 st/metric c
• 98 sl/dm³ to st/metric c = 56.304584909 st/metric c
• 99 sl/dm³ to st/metric c = 56.879121489 st/metric c
• 100 sl/dm³ to st/metric c = 57.45365807 st/metric c
#### Foods, Nutrients and Calories
GOURMET SALSA VERDE, UPC: 858102004118 weigh(s) 253.61 gram per (metric cup) or 8.47 ounce per (US cup), and contain(s) 33 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
Foods high in Riboflavin, foods low in Riboflavin, and Recommended Dietary Allowances (RDAs) for Riboflavin
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Blue Ridge weighs 1 826.1 kg/m³ (113.9997 lb/ft³) with specific gravity of 1.8261 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Calcium hydrogen phosphate [CaHPO4] weighs 2 920 kg/m³ (182.28964 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Tung oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
The troy ounce per square centimeter surface density measurement unit is used to measure area in square centimeters in order to estimate weight or mass in troy ounces
Entropy is an extensive state function that describes the disorder in the system.
Tm/min² to nm/min² conversion table, Tm/min² to nm/min² unit converter or convert between all units of acceleration measurement.
#### Calculators
Volume to Weight conversions for sands, gravels and substrates | 3,138 | 7,564 | {"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-2021-04 | latest | en | 0.20082 |
https://www.airmilescalculator.com/distance/stl-to-pbg/ | 1,627,148,398,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150307.84/warc/CC-MAIN-20210724160723-20210724190723-00191.warc.gz | 626,395,454 | 43,770 | # Distance between St Louis, MO (STL) and Plattsburgh, NY (PBG)
Flight distance from St Louis to Plattsburgh (St. Louis Lambert International Airport – Plattsburgh International Airport) is 962 miles / 1548 kilometers / 836 nautical miles. Estimated flight time is 2 hours 19 minutes.
Driving distance from St Louis (STL) to Plattsburgh (PBG) is 1115 miles / 1795 kilometers and travel time by car is about 20 hours 27 minutes.
## Map of flight path and driving directions from St Louis to Plattsburgh.
Shortest flight path between St. Louis Lambert International Airport (STL) and Plattsburgh International Airport (PBG).
## How far is Plattsburgh from St Louis?
There are several ways to calculate distances between St Louis and Plattsburgh. Here are two common methods:
Vincenty's formula (applied above)
• 961.981 miles
• 1548.158 kilometers
• 835.938 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 960.130 miles
• 1545.180 kilometers
• 834.330 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A St. Louis Lambert International Airport
City: St Louis, MO
Country: United States
IATA Code: STL
ICAO Code: KSTL
Coordinates: 38°44′55″N, 90°22′12″W
B Plattsburgh International Airport
City: Plattsburgh, NY
Country: United States
IATA Code: PBG
ICAO Code: KPBG
Coordinates: 44°39′3″N, 73°28′5″W
## Time difference and current local times
The time difference between St Louis and Plattsburgh is 1 hour. Plattsburgh is 1 hour ahead of St Louis.
CDT
EDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 148 kg (327 pounds).
## Frequent Flyer Miles Calculator
St Louis (STL) → Plattsburgh (PBG).
Distance:
962
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
962
Round trip? | 529 | 2,045 | {"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-2021-31 | latest | en | 0.822429 |
https://www.studypug.com/algebra-help/divide-functions | 1,721,770,276,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00212.warc.gz | 857,444,872 | 53,438 | # Dividing functions
Get the most by viewing this topic in your current grade. Pick your course now.
##### Examples
###### Lessons
1. Determine the Quotient of Two Functions and State Its Domain
Write an expression in the simplest form for $(\frac{f}{g})(x)$
State the domain restrictions
1. $f(x)=3x-8$
$g(x)=x+2$
2. $f(x)=4x^3+5x^2$
$g(x)=x$
3. $f(x)=2x^2+4x-30$
$g(x)=2x-6$
4. $f(x)=x-3$
$g(x)=x^2+2x-15$
2. Operations of Functions – In a Nutshell
Consider the functions $f(x)=\frac{x}{x+5}$
and $g(x)=\frac{3x}{x-2}$
1. state the domain of $f(x)$
and $g(x)$
2. write an expression in the simplest form for each
of the following and state the domains
i) $(f+g)(x)$
ii) $(f-g)(x)$
iii) $(fg)(x)$
iv) $(\frac{f}{g})(x)$
3. evaluate $(\frac{f}{g})(4)$
in 2 different ways
3. Sketch the Quotient of Two Functions
Consider the functions $f(x)=2x^2+4x-30$
$g(x)=20x-60$
1. Determine the equation of the function
$h(x)=(\frac{g}{f})(x)$ and state the domain restrictions
2. Sketch the graph of
$h(x)=(\frac{g}{f})(x)$
###### Topic Notes
Dividing functions: $(\frac{f}{g})(x)=\frac{f(x)}{g(x)}$ | 410 | 1,091 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 23, "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} | 4.21875 | 4 | CC-MAIN-2024-30 | latest | en | 0.546597 |
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Subjects
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### PRODUCT DESCRIPTION
Simplifying Radical Expressions WITHOUT Variables by Multiplying Maze Activity
This resource includes 3 Leveled Mazes to make differentiating instruction simple:
LEVEL 1: Monomial x Monomial
LEVEL 2: Monomial x Binomial
LEVEL 3: Binomial x Binomial
This resource allows for student self-checking and works well as independent work, homework assignment, or even to leave with a substitute to review the concept.
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Simplifying Radical Expressions without Variables Coloring Activity
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http://www.icoachmath.com/forums/Viewdetailed.aspx?view=all&Qid=62899 | 1,685,271,610,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643663.27/warc/CC-MAIN-20230528083025-20230528113025-00028.warc.gz | 73,295,193 | 5,845 | Categoty :
elaine : Which symbol makes the number sentence true?
4 ___ 4 = 8
: Which symbol makes the number sentence true? 4 ___ 4 = 8 ÷#0#-#0#×#0#+#0# | 58 | 162 | {"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} | 2.875 | 3 | CC-MAIN-2023-23 | latest | en | 0.773213 |
https://www.coursehero.com/file/6561126/CS-1371-Review-Session-31-2/ | 1,495,590,966,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607726.24/warc/CC-MAIN-20170524001106-20170524021106-00239.warc.gz | 845,640,246 | 462,148 | CS 1371 Review Session 3[1] (2)
# CS 1371 Review Session 3[1] (2) - CS 1371 Review Session 3...
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CS 1371 Review Session 3 Electronic Test Bring your Buzzcard and show up to your usual Recitation time. If your computer does not meet the following requirements, you should e-mail your TA ASAP, since you will need to take your test in help desk: Laptop with a battery life > 1.0 hours CD Drive Wireless Connection to LAWN Study the Test Prep coding questions from the last homework, practice test, and ABCs online quizzes. Created By: Dilan D. Manatunga [email protected] ©2009
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CS 1371 Test 3 Topics Matrices Images Numerical Methods Sound Sorting
Matrices Three Main Topics to Matrices: Matrix Math Solving Systems of Equations 2-D Rotation Matrix Matrices are really just arrays. The usual stuff done with arrays, such as creation, indexing, splicing, concatenation, etc., applies the same to Matrices. The only real difference between matrices and arrays is when doing mathematical operations.
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Matrix Math When doing Matrix Math in MATLAB, remember to not use the dot when doing multiply, divide, or exponentiation. Matrix Multiply C = A*B; NOT C = A.*B; Matrix Divide C = A/B; NOT C = A./B; Matrix Exponentiation C = A^(-1); NOT C = A.^(-1);
Matrix Multiplication Rules (C = A*B) The order of the multiplication matters. A*B ≠ B*A The number of columns of A must equal the number of rows of B. So, if A is a MxN matrix, then B should be a NxP matrix. The dimensions of the resultant matrix will be the number of rows of A by the number of columns of B. So, if A is a MxN matrix and B is a NxP matrix, then C will be a matrix of dimensions MxP.
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Solving Systems of Equations MATLAB can solve Systems of Equations similar to the methods used in Linear Algebra. Essentially, we are taking the systems of equations and describing it in terms of matrix multiplication. n m nm n n m m b b b x x x a a a a a a a a a 2 1 2 1 2 1 2 22 21 1 12 11 * b x A n m nm n n m m m m b x a x a x a b x a x a x a b x a x a x a 2 2 1 1 2 2 2 22 1 21 1 1 2 12 1 11
Finding the Matrices The first step to solving the systems of equations is to find the coefficient matrix and the vector of constants. Example: Coefficient Matrix Each row contains the coefficients of a specific equation, while each column contains the coefficients of a specific variable, such as x, y, or z. A = [2, 3; 1 -2]; Vector of Constants Each row contains the constant portion of the equation. b = [9; -13];
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## This note was uploaded on 11/21/2011 for the course CS 1371 taught by Professor Stallworth during the Fall '08 term at Georgia Tech.
### Page1 / 98
CS 1371 Review Session 3[1] (2) - CS 1371 Review Session 3...
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Ask a homework question - tutors are online | 875 | 3,375 | {"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} | 3.9375 | 4 | CC-MAIN-2017-22 | longest | en | 0.837174 |
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# Accounting homework help
Accounting homework help.
discussion format
q1
According to Lasher (2013), the internal rate of return also known as IRR focuses on the rates of return. A projects IRR is just the return it earns on investment funds; this is what we have been discussing over the last few weeks (Lasher, 2013). Utilizing the IRR is most helpful when attempting to evaluate the rate of return on a project or investment. Doing so, helps investors decide or project whether they will break even or make a profit. Understanding the outcome of the IRR is important for a project because it can give you information about whether to accept or reject. On the other hand with interests’ rates and banks you are looking at collecting interest at the end of a period which cannot create additional profit only set interest. Both IRR and bank interest are similar because they both have cash inflows; however they differ in the sense that one is gradually overtime and the other is one lump sum. In addition, another difference is the risk! With bank deposits you know exactly what is transpiring as far as interest, whereas with IRR it is not constant where you can track what the exact inflows will be. This adds a risk factor which is also in turn why it is very much possible to face a failing IRR which can result in rejection. In my opinion the best option for someone who is looking to be in “control” is to go with “bank interest” that way you know what your looking at with your investment and the outcome will be a win, whereas with IRR there is more risk for a flop.
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Q2
IRR, Internal Rate of Return focuses solely on rates of return. In terms of a business project, the initial project is an investment of the company’s money and is seen as a financial asset (Lasher, 2017). The cash that is being spent on the project in the early stages of development is similar to an investor paying cash to purchase a stock or a bond. The cash flow that is coming into the project is the interest, the dividend payments recieved by the investor (Lasher, 2017). Bank’s interest is the amount the investor or lender charges for the ues of the assets expressed as a percentage that is often known as APR, or annual percentage rate. The assets borrowed are usually cash, consumer goods, vehicles or real estate which is different from IRR. Internal Rate of Return reflects what is being spent and what flows back into the project, whereas annual percentage rate charges borrowers for the consumer goods, cash, vehicles or real estate borrowed – this APR can go up dependent on the item borrowed by lender and their credit history. IRR is like bank’s interest because regardless of where the cashflow begins or ends there is still interest being charged and/or paid over a certain period of time no matter the project or bank in which the lender uses.
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Q3
The internal rate of return (IRR) is a metric used in capital budgeting to estimate the profitability of potential investments. The internal rate of return is a discount rate that makes the net present value (NPV) of all cash flows from a particular project equal to zero (Fernando, 2020). It can be thought of as the break even point. The interest rate is the external rate at which money can be borrowed from lenders. Both IRR and interest from a bank deposit will provide cash flows. For instance, one provides a cash flow in the form of interest income over the deposit period and the other provides a large one-time inflow.
The IRR is the interest on capital expenditures of investments ignoring external factors. Determined as a percentage, it is the interest rate that makes the cash flow outflows spent on an investment equal the cash inflows that come into the company as a result of the investment (Wright, 2017). Alternatively, the IRR is the interest rate that occurs when the net cash related to the investment equals zero.
At the end, it can be said that both the IRR and a bank’s interest rate are very similar. The main different is that one will provide a precise return, while the other can fluctuate.
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Q4
How would the incentive plans offered change with the age or marital status of the employee…or would it?
*** half page 2 references
Q5
In this post, I mainly analyze the compensation model and incentive plan of Bank of America. Bank of America (BAC) is one of the world’s largest investment banks and financial institutions, providing services to a wide range of customers.
First, the incentive plan is the establishment of a competitive incentive plan to achieve these goals by attracting and retaining productive employees (Snell & Bohlander, 2016). Bank of America’s incentive program aims to help customers, customers, colleagues, and communities succeed by providing the power of the company (BAC, n.d.). Just as Snell & Bohlander (2016) mentioned that a good incentive plan needs to determine important organizational indicators that encourage employee behavior and engage employees to provide fair and clear incentive relationships. Incentive plans are divided into individual incentive plans, group incentive plans, and corporate incentive plans. Bank of America’s performance-based governance framework provides employees with competitive compensation based on job requirements and personal performance. It also regularly compares salaries with other companies to improve competitiveness in the industry and market (BAC, n.d.).
From the perspective of individual incentive plans, Bank of America’s minimum wage strategy demonstrates its commitment to paying competitive wages to all employees and monitoring hourly wages to help keep their wages in line with industry trends Consistent. As BAC (n.d.) pointed out that the minimum hourly wage for US employees in 2019 has been raised to \$17, and will continue to increase until it reaches \$20 in 2021.
From the perspective of corporate incentive plans, especially high-level salaries. Bank of America’s companies uses a part of their total compensation as basic salary and the rest as variable salaries, most of which are provided as incentives based on deferred equity. As mentioned by BAC (n.d.), this helps to motivate executives to achieve sustained shareholder value and responsible growth.
The incentive plan strategy establishes performance thresholds for employees to qualify for rewards and emphasizes a common focus on organizational goals (Fotsch & Case, 2018). In other words, by expanding opportunities to motivate employees, create an operating environment that supports mutual commitments, and corporate culture and values recognized by employees. Bank of America’s performance compensation model helps them focus on specific goals in the company’s work and also focuses on supporting the company’s cultural goals. The core of Bank of America’s growth strategy is its commitment to “act responsibly,” which includes our commitment to ethical behavior, acting with integrity, and complying with the laws, rules, regulations, and policies that strengthen this behavior. In other words, Bank of America has formed a responsible corporate culture while earnestly implementing its reward strategy.
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Q6
A clear compensation system is designed to convey to employees what is most valuable in the organization, thus enabling employees to focus their efforts and behaviors in a direction that helps the organization compete and survive in the market. The salary system can not only satisfy the staff’s material conditions but also be an important symbol of the staff’s social status and self-worth recognition. Next, I will use Microsoft’s incentive plan to analyze the relationship between incentive plan and corporate culture, organizational structure, and compensation system.
The establishment of corporate culture is consistent with the development of corporate strategic goals. It can strengthen the values in the organization and thus shape the behaviors of team members. Compensation can support and help managers to promote the development of values and corporate culture. Therefore, the design and establishment of a compensation system should punish and reward employees’ specific knowledge and skills according to the goals and expectations of the company’s values. (Wagewatch, 2013) Microsoft advocates the value of “Achieving our Mission all depends on building trust with people and organizations around the Globe. Microsoft wants to be a magnet for the best people by paying smarter, also, they want to attract and retain employees by offering real ownership and great long-term financial incentives. (Microsoft, 2003) Therefore, Microsoft offers long-term stock awards to employees to promote trust and harmonious working relationship with employees.
In 2011, Microsoft used an integrated approach to implement a more streamlined, flexible reward system across the company. The system provides employees with salary increases, annual bonuses, and Russ awards, which allow companies to target their compensation investments to key positions in disciplines that are most important to the company’s success, such as engineering research and development. (Miller, 2012) For employees, a clear and targeted salary system can provide a clear direction for employees’ career development goals, and employees’ satisfaction with a salary can also be improved. For organizations, the adjustment of the salary system can provide rewards for employees with excellent performance so as to attract and retain talents. Those with high ability can always remain competitive, which can directly promote the efficiency of the implementation of organizational plans.
The reason for the successful implementation of incentive plans lies in the fact that enterprises can accurately cater to and meet the real needs and wants of employees. More and more, we observe that organizations are starting to pay employees based on the relevant skills they bring to their roles rather than traditional factors such as tenure, education (pedigree), or years of experience in the industry. (Hoof, 2017) Therefore, when making compensation strategies, managers need to understand the skills and needs of talents in detail and have an open and transparent dialogue.
*** half page 2 references
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Author Message
Courk
Daedalian Member
Posted: Wed Jun 13, 2012 6:05 am Post subject: 1 Link to the aforementioned puzzle. Happy solving!
Oscar
Daedalian Member
Posted: Thu Jun 14, 2012 10:47 am Post subject: 2 Hmmm... I've assembled the pieces and worked out what they mean, but they don't give a cast-iron final word. You could choose any of matador or toreador or bullfighter by my reckoning. I won't post a spoiler of the method yet since I'm sure some people will want to have the fun of getting the little cut-out pieces to stay in one place while trying not to sneeze. On the other hand, if there are sufficient folk already fed up with the diabolical jigsaw then I'll go ahead... After further research it appears that matador is the technically correct option[/edit]
novice
No harm. Pun intended!
Posted: Thu Jun 14, 2012 1:30 pm Post subject: 3 I've assembled some pieces of the puzzle but I don't think I've managed to fit everything together correctly yet. I haven't figured out how to interpret the result yet. Is it easy to recognize the solved jigsaw when you find it?
Oscar
Daedalian Member
Posted: Thu Jun 14, 2012 1:47 pm Post subject: 4 I wouldn't say it was *easy*, because my first attempt wasn't completely correct. But there was sufficient repetition for me to notice complete elements, which, when I'd found their origin via Google, meant I could rejig to get the desired result.
novice
No harm. Pun intended!
Posted: Fri Jun 15, 2012 5:53 pm Post subject: 5 Yeah I'm not sure what those elements are. Flags, perhaps? Btw I found this, which was intriguing but probably irrelevant: http://www.colourlovers.com/blog/2010/01/18/color-identifying-system-for-the-color-blind/
Oscar
Daedalian Member
Posted: Sat Jun 16, 2012 8:30 am Post subject: 6 Interpreting that as two questions: correct and correct(intriguing and irrelevant). Hint: the jigsaw makes 30 flags in total
novice
No harm. Pun intended!
Posted: Sat Jun 16, 2012 10:39 am Post subject: 7 A quick arts & crafts session with my kids revealed this: Now if someone can tell me what the flags mean...
Suspence
Daedalian Member
Posted: Sat Jun 16, 2012 10:57 am Post subject: 8 They look to be maritime alphabet flags: http://en.wikipedia.org/wiki/International_maritime_signal_flags#Letter_flags_.28with_ICS_meaning.29_________________I hate people who try to write interesting things in their signature.
novice
No harm. Pun intended!
Posted: Sat Jun 16, 2012 10:59 am Post subject: 9 Yup, found it too... Rotated my picture 180 degrees to make it fit. P E R S O N W A V I N G A R E D F L A G I N A C O R R I D A PERSON WAVING A RED FLAG IN A CORRIDA
Dej Mar*
Guest
Posted: Thu Aug 09, 2012 4:50 am Post subject: 10 Absent of color with the need to rotate many of the "cut-out" pieces, the "quilt" of images is revealed to be a 5x7 image of 30 maritime signal flags surrounded by a dark border. The maritime signal flags spell out: PERSO NWAVI NGARE DFLAG INACO RRIDA With spacing, the message reads "PERSON WAVING A RED FLAG IN A CORRIDA". Such a person is called an ESPADA, the swordsman or chief bull-fighter. An espada is one of the several TOREROs (bull-fighters) that participate in a bull-fight, and is often the matador, i.e., the person who kills the bull.
Beartalon
'Party line' kind of guy
Posted: Mon Nov 26, 2012 5:02 pm Post subject: 11 No-one has said anything since August? "Corrida" means "race" as in "running of the bulls" (corrida del toros). That's an important festival in Pamplona, Spain, which I assume might be the solution. The puzzle is called "Colorblind" and no-one has mentioned how it's related to the puzzle. Cattle are red/green colourblind. Bulls bred for these events attack the moving cape, not the colour red supporting the answer above.
Suspence
Daedalian Member
Posted: Mon Nov 26, 2012 5:30 pm Post subject: 12 I thought Oscar solved this in Post 2 - MATADOR Given the puzzle flavor - The message is very short: "Someone in this café knows where you should go next." - I assume there's a fellow matching that description in the cafe._________________I hate people who try to write interesting things in their signature.
Beartalon
'Party line' kind of guy
Posted: Mon Nov 26, 2012 5:55 pm Post subject: 13
Suspence wrote: I thought Oscar solved this in Post 2 - MATADOR Given the puzzle flavor - The message is very short: "Someone in this café knows where you should go next." - I assume there's a fellow matching that description in the cafe.
I was reacting to the "where you should go next." I would have hoped that if the puzzle had been solved that it would have been confirmed by now.
referee
June 21st, 2004 Member
Posted: Mon Nov 26, 2012 6:40 pm Post subject: 14 de toros. Source: is Spanish._________________Jan 21st, 2008: The pillaging continues. Mar 4th, 2008: Rest in Peace, Gary Gygax. May your dice always roll a natural 20 wherever you are. Be the Ultimate Ninja! Play Billy Vs. SNAKEMAN today!
groza528
No Place Like Home
Posted: Mon Nov 26, 2012 11:11 pm Post subject: 15 The intended answer is indeed MATADOR The reason that I haven't put the next puzzle up is because I thought one of the Top GLer puzzles was slotted for the front page, so I was expecting MNO would get something up there himself or had someone who was going to help him do so.
Beartalon
'Party line' kind of guy
Posted: Tue Nov 27, 2012 6:07 am Post subject: 16 Thanks for the confirmation, groza!
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m. malik January 31, 2006 11:24
Normal - Helical Surface
Want to know the equations for the unit normal on a helical surface (e.g., surface of a screw teeth). Thanks.
Ananda Himansu February 3, 2006 03:53
Re: Normal - Helical Surface
First, take a look at eqn.(11), the last equation in this posting. It might be the only thing you need. Think of u as the cylindrical radial coordinate, and v as the polar angular coordinate, and p as the pitch of the screw.
We use capital letters to denote vectors. Assume rectangular Cartesian coordinates x, y, z. Also, assume cylindrical polar coordinates r, t, z, such that the positive x axis (y = 0, x > 0) coincides with the r axis when the angular coordinate t is zero (t = 0, r > 0). Select the z axis as the axis of screw motion of the helix or helical surface. Define a helix as the curve traced by the screw motion of a base point (a generator point). Define a helical surface as the surface traced by the screw motion of a rigid base curve segment (or generator curve). This seems likely to be what you had in mind by the term "helical surface". A different type of "helical surface" might be one in which each point on the base curve "climbs" in the z direction at the same angle to the (x,y) plane. The analysis for such a surface would be different from but similar to the analysis that follows.
Let p be the pitch of the helix. This means that any point on the generator advances by a distance p in the z direction for every complete turn (2*pi radian rotation, where pi = 4*arctan(1)) around the screw axis. A positive value of p will correspond to a right-handed screw motion if (x,y,z) is a right-handed coordinate system.
We represent the helical surface parametrically. The base curve segment is parametrized by a parameter u belonging to the interval [u1, u2]. The screw motion is parametrized by a parameter v belonging to an infinite interval.
The base curve is specified by you in the representation:
x = x0(u); y = y0(u); z = z0(u) ------ (1)
The representation is transformed to cylindrical polar coordinates r0(u), t0(u), z0(u) (or alternatively, you may find it more convenient to directly specify the base curve in cylindrical polar coordinates, skipping eqn.(1)):
r = r0(u) = sqrt(x0^2 + y0^2); r0*cos(t0) = x0(u), r0*sin(t0) = y0(u); z = z0(u) ------ (2)
Next, the rigid base curve is subject to a screw motion to generate the helical surface. The screw motion is most conveniently described in the cylindrical polar coordinates:
r = rh(u,v) = r0(u); z = zh(u,v) = z0(u) + p*v/(2*pi); t = th(u,v) = t0(u) + v ------ (3)
In order to avoid ambiguity about the quadrant of v, we use cos(th) = cos(t0)*cos(v)-sin(t0)*sin(v) and sin(th) = sin(t0)*cos(v)+cos(t0)*sin(v) ------ (4)
Next, the cylindrical polar representation of the helical surface is transformed to rectangular Cartesians:
x = xh(u,v) = rh*cos(th); y = yh(u,v) = rh*sin(th); z = zh(u,v) ------ (5)
Now, the position vector of any point (u,v) on the helical surface can be written as
S(u,v) = xh(u,v)*I + yh(u,v)*J + zh(u,v)*K ------ (6)
where I, J, K are the unit vectors along the x, y, z directions.
Next, at any point (u,v) on the surface, the tangent plane can be characterized by two vectors:
U = dS/du; V = dS/dv ------ (7)
where the derivatives are partial derivatives. U is tangent to the surface in the direction of the helically transported image of the base curve, and V is tangent to the surface in the direction of the helix generated by the transported image of a point on the base curve.
Next, a vector normal to the tangent plane and thus to the helical surface is obtained as the cross product of the previous two vectors:
N = U x V ------ (8)
This can be normalized to yield a unit normal vector: N1 = N / sqrt(dot(N,N)) ------ (9)
where dot() is the dot product or euclidean inner product of vectors. If the parametrization of the base curve is such that increasing u implies increasing r, then dot(N,K) > 0. Equations (1) through (9) can be programmed quite easily on a computer.
Finally, as a special case, we mention the simplest helical surface in this coordinate representation, which is obtained by a base curve segment that coincides with a segment of the x axis. This case typifies the load-bearing surface of a screw thread, and it might be case that you are most interested in. For this case,
x0(u) = u; y0(u) = 0; z0(u) = 0 ------ (10)
The expressions simplify in this case to give
N = p/(2*pi)*(sin(v)*I - cos(v)*J) + u*K ------ (11)
Note that eqn.(9) should be used to calculate a unit normal N1. For u > 0, eqn.(11) gives the outward pointing normal to the load-bearing upper surface, but for u < 0 it gives the inward pointing normal of the upper surface. This is a consequence of r decreasing for u increasing when u < 0, according to the parametrization in eqn.(10). To get the outward pointing normal for u < 0, multiply the normal N from eqn.(11) by -1.
m. malik February 3, 2006 10:09
Re: Normal - Helical Surface
Ananda: Thank you very much for your posting. I am impressed by your detailed and methodical description. I guess this serves my purpose.
It may be of interest to you: The cfd problems that I am working on have polymer flow between rotating screw(s) and cylindrical barrel(s). I need surface normals to account for slip at the solid, moving and stationary, interfaces. Slip at solid interfaces is a common problem with many types of polymers.
Best regards.
Ananda Himansu February 3, 2006 13:56
Re: Normal - Helical Surface
Malik, glad the posting is helpful. Sounds like an interesting problem that you are working on. The special case in eqn.(11) of my posting should apply to your geometry. For the vertical part of the screw or the barrel surface, the general procedure can be used with the base curve:
x0(u) = rc; y0(u) = 0; z0(u) = -u
where rc is the inner radius of the barrel. Then follow the previous procedure to find N. But of course, the answer simplifies to what you would expect for the cylindrical surface, namely:
N = rc*(cos(v)*I+sin(v)*J)
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I have the following:
int x = 2;
int *ip;
ip = &x;
now as I understand *ip equals 2.
Is it possible to say that *ip equals a value and & is the adress for that
value?
JS
Jul 23 '05 #1
9 1743
JS wrote:
I have the following:
int x = 2;
int *ip;
ip = &x;
now as I understand *ip equals 2.
Is it possible to say that *ip equals a value and & is the adress for that
value?
ip is a variable that stores the memory address of variable x.
When you dereference ip like this: *ip, you access x.
--
Ioannis Vranos
http://www23.brinkster.com/noicys
Jul 23 '05 #2
"Ioannis Vranos" <iv*@remove.thi s.grad.com> skrev i en meddelelse
news:1110923535 .5379@athnrd02. ..
JS wrote:
I have the following:
int x = 2;
int *ip;
ip = &x;
now as I understand *ip equals 2.
Is it possible to say that *ip equals a value and & is the adress for that value?
ip is a variable that stores the memory address of variable x.
When you dereference ip like this: *ip, you access x.
ok so the value of x and *ip is both 2 if int x = 2 right?
Jul 23 '05 #3
JS wrote:
I have the following:
int x = 2;
int *ip;
ip = &x;
now as I understand *ip equals 2.
Is it possible to say that *ip equals a value and & is the adress for that
value?
'&' is an operator. It is not an address of anything. Was it a typo?
Anyway... *ip designates an object. The original name of that object is
'x'. The value of 'x' and, consequently, of *ip, is 2, since they both
designate the same object.
V
Jul 23 '05 #4
JS wrote:
ok so the value of x and *ip is both 2 if int x = 2 right?
Right.
--
Ioannis Vranos
http://www23.brinkster.com/noicys
Jul 23 '05 #5
> ok so the value of x and *ip is both 2 if int x = 2 right?
A clear distinction needs to be made here. The two variables are one and the
same entity. It matters not which is modified or initialized. The correct
statement would be that *ip is x, regardless of what value it happens to
hold.
Jul 23 '05 #6
codigo wrote:
A clear distinction needs to be made here. The two variables are one and the
same entity. It matters not which is modified or initialized. The correct
statement would be that *ip is x, regardless of what value it happens to
hold.
ip and x are *two* different objects with a separate memory address each
one. ip is an int pointer variable, that is, it stores memory addresses
of int variables, while x is an int variable.
--
Ioannis Vranos
http://www23.brinkster.com/noicys
Jul 23 '05 #7
Yes, but *ip -- the location ip points to -- is indistinguishab le from
x, at least as far as I can tell.
Jul 23 '05 #8
ev****@gmail.co m wrote:
Yes, but *ip -- the location ip points to -- is indistinguishab le from
x, at least as far as I can tell.
ip dereferenced, accesses x. We use very accurate terminology in clc++. :-)
--
Ioannis Vranos
http://www23.brinkster.com/noicys
Jul 23 '05 #9
<ev****@gmail.c om> wrote in message
news:11******** **************@ g14g2000cwa.goo glegroups.com.. .
Yes, but *ip -- the location ip points to -- is indistinguishab le from
x, at least as far as I can tell.
That is, so long as ip isn't changed to point elsewhere!
-Howard
Jul 23 '05 #10
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# N TP MN TON LP 10 (NNG CAO) HC K I NM HC 2009 - 2010 (Thi gian lm bi: 90 pht)
NI DUNG V d minh ha A. I S (6,5 ) I. Hm s th v phng trnh Cho (P): y = x - 4x + m bc hai (3) 1) V th (P) ng vi m = 3 1. Kho st v th hm s bc hai 2) Tm m (P) ct trc honh ti 2 2. S tng giao ca Parabol v im phn bit A, B vi OA = 3OB ng thng (quy v phng trnh bc hai v nh l Vi-et). II. Gii v bin lun (1,5) Gii v bin lun theo tham s m ca Phng trnh bc nht (c phn phng trnh sau: + =2 thc) hoc h phng trnh bc nht 2 n. III. Phng trnh quy v bc hai 1) Gii phng trnh: h phng trnh bc hai (2) = 2x - 5x - 1 Phng trnh cha cn, phng 2) Gii h phng trnh: trnh hu t (c th phi t n ph) H phng trnh i xng loi I, loi II. B. HNH HC (3,5) - Phn tch vect theo 2 vect khng Bi 1: Cho ABC bit AB = 3, AC = cng phng 4, = 60. Gi D, E l 2 im sao cho - Tnh di, tnh gc, chng minh = , = vung gc, chng minh ng thc a) Phn tch , theo 2 vect vect, ng thc di (da vo tch ,. v hng hoc cc cng thc trong b) Tnh di DE h thc lng tam gic) c) Tm tp hp cc im M tha - Tm tp hp im (da vo cc bi MB + MC = BC ton qu tch c bn Sch gio Bi 2: Gi R, r l bn knh ng trn khoa). ngoi tip v ni tip ca ABC. Gi S l din tch ABC. Chng minh: S = Rr (SinA + SinB + SinC) | 541 | 1,200 | {"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} | 2.515625 | 3 | CC-MAIN-2019-22 | latest | en | 0.235512 |
https://www.acousticfields.com/blog/page/184/ | 1,539,709,561,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510853.25/warc/CC-MAIN-20181016155643-20181016181143-00254.warc.gz | 838,862,943 | 42,722 | ## Resonant Frequencies
Wikipedia defines resonance as ” the tendency of a system to oscillate at a greater amplitude at some frequencies than at others. These are known as the system’s resonant frequencies or resonance frequencies. At these frequencies, even small, periodic driving forces can produce large amplitude oscillation, because the system stores vibrational energy.” It goes further when it talks about resonance in specific. It states,” Resonance occurs when a system is able to store and easily transfer energy between two or more different storage modes (such as kinetic energy and potential energy in the case of a pendulum). However, there are some losses from cycle to cycle, called damping. When damping is small, the resonant frequency is approximately equal to the natural frequency of the system, which is a frequency of unforced vibrations. Some systems have multiple, distinct resonant systems.”
Our listening, home theater, and mixing room are all systems that have a resonant frequency. Our room due mainly to its dimensions and parallel surfaces is definitely a system that has resonances. Can our room store sound energy and easily transfer that energy between certain storage modes. A room can “load” with excess low frequency energy and then store for brief time intervals and then transfer that energy into the next room passing through the existing room physical structures. There is some type of an energy transfer cycle that the room exhibits and the frequency of resonance is room dimension dependent. Our room can also have multiple resonance systems within itself.
How do we damp our system or room to minimize these resonances. We want to cause energy losses within our room’s energy transfer cycle. We must use the physical structure of the room. Can we make it larger , so we can spread out the multiple resonance systems within our room and provided more time and distance between the peaks and valleys in our room’s energy transfer cycle. This is usually not an option. We now must go into the room and work within the physical shell of the room. We must use damping from the inside out.
Our damping methodology must focus on the energy created from within the room. Can we reduce sound pressure at major room boundary reflections by “venting” or providing pressure release valves such as physical holes at surface pressure points? If we do that, we must prevent that energy from causing other acoustical issues in adjoining rooms and then we must also keep unwanted external energy from getting through our “vents” Can we absorb excess sound pressure energy close to the vibrational producing source such as our speakers to reduce the bandwidth that our system or room operates at? Can we reduce the mechanical resonant frequencies within the room’s wall structure? I think the answer is yes to all of these if we view the room as a pressurized box with a resonant system.
## Response of Our Ears To Sound
We have all heard really loud sounds. A jet taking off, a dragster or funny car blasting away from a dead stop, or an explosion all generate large amounts of sound energy. The perceived loudness of this energy by our ears depends on the particular frequency we are addressing and the intensity at which we perceive or hear that frequency. To interpret this frequency/intensity ratio and put it in a form that more closely resembles actual human hearing, we go to what is termed the Fletcher-Munson loudness curves.
The Fletcher-Munson curves take frequency and intensity and apply this data to a predetermined domain of measurement. The Fletcher-Munson frequency start point is 1,000 cycles. The F/M loudness scale determines that the ear is the most reactive in the frequency range that starts with 3,000 cycles and goes through 4,000 cycles. This loudness scale also shows that the threshold of hearing is more reactive at lower frequencies. For example, the threshold of hearing at 60 cycles is 48 db higher than at 1,000 cycles.
Loudness of any sound is a ratio of the perceived magnitude of that sound energy by the live organism it encounters. The units or intervals the loudness scale uses must reflect real human reaction points. The units on the scale must match common human hearing experience and also match the sensation magnitude. The scale also must be constructed so that when the units on the scale increase by a certain factor the sensation magnitude of human hearing increases proportionately. If the units are quadrupled then the corresponding human hearing sensation must also quadruple.
Pitch is defined as frequency that reacts with the medium in which it is transmitted in. When sound travels through the air in a room, we have the frequency produced by the sound source and we also have that frequency reacting with the air and producing another sound. It is source sound and air sound combined. Pitch is not an objective quality but rather a subjective one. Pitch is the subjective quality humans assign to a sound in order to place it in its appropriate position on the music scale. There is a measurable difference between frequency and pitch.
## Orchestra and Stage “Shells”
Harry Olson wrote a book entitled “Acoustical Engineering”. It was published in 1991 but with only a new introduction from a 1957 original publication. The book “Acoustical Engineering” was based on an earlier work entitled, “Elements of Acoustical Engineering” copyrighted 1940 and 1947. I like to look at older acoustic books and read what the current thinking was at the time. It is amazing how some things have really changed and some things have not at all. I like one section entitled, “Orchestra and Stage Shell”. It talks about acoustic treatment in outdoor theaters.
The article focus states, “When orchestra and stage productions are conducted in outdoor theaters it is desirable to provide a shell to augment and direct the sound to the audience, to surround the orchestra with reflecting surfaces and to protect the performers and instruments against wind, dew, and other undesirable atmospherics” I like the use of, “undesirable atmospherics”. It is a nice way to say bad.
In this necessary “shell”, we must focus on the acoustical treatment that will line our shell with, in order to maximize the sound quality for all parties concerned. It appears from this section, that most of the shells in those days were concave in shape and design. This concave shape produced what the book calls, “intense and sharp concentrations of reflected sound in both the shell and audience area”. It also goes on to say that “these acoustic effects are particularly undesirable when the sound is picked up by microphones on the stage for sound reinforcement and broadcasting”. This reflected energy produced by this concave shell can not achieve a balance sound for the conductor’s position. Therefore, without the orchestra leader hearing what the audience hears, we have a definite acoustical issue when it comes to the treatment used inside our shell.
Poly-cylindrical shell or a concave shell lined with poly-cylindrical structures will produce the best sound for all parties concerned concludes this section of thought. It will be good at the microphone positions, the orchestra leader, and finally the audience. A poly-cylindrical structure is shaped by an 180 degree arch and then a flat surface for mounting. The arch is not a diffusor. It is a sound re-director. It uses the angle of incident equals angel of refraction physical law. Numerous poly-cylindrical devices installed in a wall, would redirect the energy that strikes them in different directions opposite to their original striking direction, thus creating a sound redirected sound field.
I have never heard a shell lined with poly-cylindrical devices, but I wish I could get that chance. Most concert shells I have heard are lined with quadratic diffusors which are usually positioned in both vertical and horizontal planes thus, providing two dimensions to our sound field which would be directed at the conductor and audience. It would be interesting to compare quadratic diffusion sound with poly-cylindrical sound just to hear the difference. It could be different in many ways.
With two dimensions of sound created by quadratic diffusors positioned both “vertically” and “horizontally” I would know that sound because I have created it on numerous occasions. It is characterized by a smooth and equal frequency spread across the room plane it is focused on. The diffused sound field would be equal parts “air” and equal parts sound. If it were a solid, it would look like a very loosely woven tapestry spread across a room boundary surface. I hope sound redirection through a poly-cylindrical lined shell retains this feature but adds something of its own. I do not know what that would be, but I would sure like to hear it.
## Acoustical Treatment in Our Studios
In a professional recording studio, we have the control or monitoring room. We also have the “live” room. A live room is a room about which we seek to acoustically create a larger sound through higher than normal reverberation times. In a live room, it is necessary to acoustically treat some reflections from room boundary surfaces and let the other reflections run free. We can accomplish this variable type of acoustic presentation with acoustical treatments that allow the engineer to vary the surface of the treatment from absorption to diffusion and even sound redirection. Our acoustic goal for the room and for our microphones is to compliment the individual vocals and instruments. A live room needs to provide at the microphone, an additional amount of analog energy through room sound into our electronic, digital signal path.
In our “live room”, we must have a complete understanding of the size and locations of any room modes. Room modes occur between the parallel surfaces of our room. Because sound waves have certain lengths and travel at a given speed, they become influenced by the room dimensions. Certain parallel room dimensions are more problematic than others, but they exist in almost every room. A room mode will cause an energy field drop out or an energy field exaggeration at certain frequencies or certain frequency groupings. A microphone placed within a room mode that is a null will not let the microphone “hear” all the frequencies produced by the vocals or instruments. If the microphone is placed in a room mode that exaggerates certain frequencies, then one will have distortion at the microphone surface.
Reverberation is another acoustic variable that we need to address. Reverberation in our control room is almost non existent. We don’t want a lot of reverb in the in our control room because we don’t want it to smother or destroy any of the ambiance in our recordings created by our “live room”. In a “live room”, we want to have reverberation, not in our control room. To achieve the proper amount of reverberation is both an art and a science. All one has to do to verify this statement is tour different facilities and listen to their rooms. One will have a big sound that one could record drums in and add a large amount of ambiance to the recording. Another could be good for piano because of its acoustic surface treatment and rate and level of absorption material used.
## Surround Sound Monitoring Room Acoustics
Tomlinson Holman, in his book entitled “5.1 Surround Sound Up and Running”, discusses the room acoustic requirements for monitoring a 5.1 mix. He states,
” Multichannel affects the desired room acoustics of control rooms only in some areas. In particular, control over first reflections from each of the channels means that half-live, half-dead room acoustics are not useful. Acoustical designers concentrate on balancing diffusion and absorption in the various planes to produce a good result.” If live end/dead end is not useful as it is in two channel monitoring, what are our options with multiple channel monitoring in terms of acoustical treatment?
It appears that primary reflections off of our multichannel monitors are as much of a concern as they are in two channel. Primary reflections distort the more pure (without the room), direct sound from the monitors. The engineer wants to hear all the vocals and instruments from the speaker without the room reflections. Room reflections add the sound of the room into the mix and that is not acceptable. Near field monitoring is a process to minimize room sound by sitting closer to the speakers to hear the wanted direct sound. It is important in two channel audio and it is even more important with multiple channels that are all producing energy at the same time. No need to add room sound to 2 channels, let alone 5.
He states that acoustical designers, “concentrate on balancing diffusion and absorption in the various planes to produce a good result”. If we look at each plane or room boundary surface for two channel sound, we can get an idea on how to balance absorption and diffusion in a multiple channel presentation. The ceiling must be a balance of diffusion and absorption in two channel sound as well as multiple channel monitoring and playback. We definitely do not want ceiling reflections in our mix, just as we don’t want them in our two channel presentations.
The same will apply to all the room surfaces. A balance of diffusion and absorption will work for all the room boundary surfaces. Each surface should have the same amount and type of diffusion and absorption to provide the engineer with a uniform and consistent room sound. In two channel sound, we tend to absorb the primary reflections and use diffusion in other areas. With multiple channel monitoring consisting of multiple, mono sources, we must have a balanced acoustical field for all channels.
Low frequency issues must also be addressed, so that the low frequency effects channel can be monitored correctly and produces the necessary energy level to fill a room with explosions and other special effects. Muddy and bloated bass energy will only confuse the listener and smother the middle and high frequencies. Vocals will be lost in a low frequency muck that may prevent localization of the vocals with the image on the screen. Video movements on the screen must have a corresponding audio movement from channel to channel. Low frequency absorbers whether freestanding or built in are a must.
Thanks
Mike
## How To Choose The Correct Speaker
Working out how to choose the correct speaker for the room you are going to be using it in involves many factors. All of these factors must be considered when selecting a loudspeaker for your listening room. We are just referring to two channel sound sound with a left and right channel speaker. We need […]
## What is Good Sound?
What is Good Sound?
I walk through trade show rooms and listen to the exhibitors tell me what they consider good sound. They tell me their room is kind of their idea of good sound. Some tell me that their room sounds pretty good but does not fall into “good sound”. A hotel room is a compromise and one must work around many variables to achieve a sound that demonstrates what their product could sound like in a good room some say.
Some rooms have too much bass energy. Some rooms are bright and have so much specular reflections that it is difficult to hear all the vocals in a three part harmony or hear two bass instruments each producing their own sound.These rooms are characterized by low definition and a small image. Some rooms have the listening chair up against the back wall. Some say,”Our room will have the best sound of show”. Really?
I guess everyone has a different idea of what “good sound” is. Is good sound the type of sound where their is an emotional attachment immediately to the music? Is good sound the type of sonic presentation where one can hear every instrument and vocal in a balanced presentation? Is good sound the type of sound where the speakers and amplifiers disappear and one can only hear and “see” only the music? Is good sound a combination of some of these variables and not others?
For us, it is removing the room from the sound and having the ability to hear all the instruments and vocals in a balanced presentation.To achieve this objective, all low frequencies are heard without any bass bloat. There are layers to the bass and the bass attack and decay is as tight and clean as the attack and decay of middle and high frequencies. Middle and high frequencies are layered like our bass presentation and their is a distinct separation between the instruments and vocals.Comb filtering of middle and high frequencies is under control. Their is air present and instruments and vocals float in the room all across our sound stage. No speakers are seen or heard.
How does one achieve good sound? I am sure their are many approaches as there are opinions on what constitutes “good sound”. We choose to reduce low frequency pressure in the room from all low frequency producing devices. One must first deal with low frequency pressure in order for the middle and high frequencies to come through without being smothered by excessive low frequency energy. Excessive low frequency energy can be controlled at the source or at room boundary surfaces. Middle and high frequency reflections off of room walls can be controlled through the use and application of absorption or diffusion technologies.All of this control must be applied in a way that produces a balanced sound stage with a height, width, and depth.
## Headphone Sound or Room Sound
Head phone sound is different than sound in our rooms. Both have their good and bad points.
Head phones seal our ears with a small oval shaped room. Inside that “room” is a small speaker that produces sound energy and fires it directly into our our outside ear. From there, the sound travels into our ear canal and then finally into our inner ear. We hear bass energy, middle, and high frequency energy all evenly portrayed in the two channels that are wrapped around our head.
Were our ears designed to have sound that close to it and at those pressure levels produced by headphones. If they were wouldn’t they be smaller. No need for three or four inch ears if the sound we hear is generated from such a short distance away as with headphones. No, our ears were designed to hear sounds from all directions and at some distances. Men have larger ears than women. This is probably due to the way men used to hunt for food. Hearing everything from all directions when hunting for food, enabled man to bring home the dinner rather than be the dinner himself. A larger sound receiving instrument was needed to allow for selective hearing and localization of predators.
Do headphones produce a sound stage in front of our head. I have never experienced that with headphones. I have experienced both left and right channel separation, but no sound stage. Headphones do not have a sound stage and therefore do not have any height, width, or depth to their sonic presentation.
They do have definition. Headphones portray every sound in detail. One can hear all the instruments and vocals with distinct separation between them. If we can get our rooms to have that instrument and vocal separation coupled with headphone like clarity, we will have achieved our room acoustic objectives.
A room has space and the sound source is at a much farther distance from our ears than headphones allow for. We can have a small as is the case with near field listening or monitoring. We can also have a large sound stage depending on room dimensions and the correct balance in the room of acoustical treatments. A room and the sound produced in it is easier for the design of our ears to interpret. The music or sound reaches our ears in a way and manner that allows for localization and spaciousness to occur. With this spaciousness, we now can have information going into our brains that has much more data in it than just headphone sound. We have room sound and room sound is closer to free space sound which is probably how our ears evolved into the size and shape they have today.
## Db Meter
A popular search term in audio is “db meter”. Sometimes people search for terms that they think go together or they have seen or heard it used somewhere. Sometimes there is a combination of words into a phrase that tries to illustrate an audio point or concept. Lets examine each word within the search term “db meter” and see if it the right word or group of words to use in this situation.
Db stands for decibels and is a unit of measure that those in the audio world use to describe intervals or amounts of energy within in a given environment. It is only that: a unit of measure. It does not have any value of its own other than to say it is a unit of measure that is calculated and formed to correspond to the human ear hearing range. A large db number can cause inner ear damage. A smaller db number may be too low to hear the difference in gain jumps. A db is part of a scientifically calculated scale or ratio for human hearing comparisons and even regulations.
What does the db unit mean when it is attached to a number? A db meter measures sound pressure levels.The sound pressure level can be assigned many different units of measurement. Therefore, using a ratio is better for human sense of hearing comparisons. A db unit is a ratio of acoustic power levels expressed in decibels that “comply” within our human hearing range. These are decibels that express a power ratio made for human hearing measurements. For example, a Saturn rocket has a sound pressure (Pa) of around 100,000 Its sound pressure level measured in db is 194. Normal conversational speech has a sound pressure of .02 while the sound level is 60.
Searching for a db meter may be confusing for the clerk at the store, sine we are really measuring sound pressure levels expressed in decibels in our room. Although, if you go to Radio Shack and ask for a sound pressure meter, they will search their product data base and will find no entries in it for “sound pressure meter” If you change your request for a db meter, they will have one for you quickly. It is funny how things work.
## Building a Sound Room with a Living Roof
We have a client in Arizona who wanted us to build him a sound room with a living roof. A sound room you are all familiar with. A living roof may be an other issue. It was also a pleasant surprise for us.
A living roof is a roof designed to support 18″ of top soil and the watering and drainage system necessary to maintain this miniature ecosystem. Supporting 18″ of earth and water is no easy task, especially when the roof size is 25′ x 50′. That is 1875 cubic feet of earth at approximately 20 lbs. / cu. ft. is 37,500 pounds of earth, not to mention the piping for water and drainage. The roof must support 16 tons of earth and pipes.
Earth is an excellent barrier to external noise. Go into your basement and sit quietly. There you are surrounded on 4 sides by earth and concrete. In this project, the roof and 6′ up the 12′ side walls will also be covered with earth. So, in this project we have 1 1/2′ earth on the roof and six more feet of earth on each wall side. Now, we need concrete walls at the correct thickness to match the acoustical properties of 1 1/2′ of earth on the roof.
We determined that an 8″ poured concrete wall all around will meet all our structural issues for ceiling support and acoustical issues for sound transmission class ratings and all external noise measured calculations. The 8″ concrete shell will build a room that is 25’wide and 50′ long. The ceiling height is 12′. One could not ask for a better room size when it comes to acoustical issues that must be dealt with.
At 50′ in the length dimension, even a 20 Hz.wave, which is the lowest wave we usually work with in rooms has some room to run. No low frequency issues or any others for that matter when it comes to the 50″ length dimension.The 25′ width is also good for low frequency, but will give us a few issues. Those issues will be resolved through the use of our activated carbon technology which will be added to the inside walls. A ceiling height of 12′ only increases our room volume and is welcome for all forms of sound playback and recording. | 4,873 | 24,341 | {"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-2018-43 | longest | en | 0.956586 |
https://brilliant.org/problems/coefficient-4/ | 1,524,504,869,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946120.31/warc/CC-MAIN-20180423164921-20180423184921-00412.warc.gz | 574,052,663 | 10,809 | # Coefficient
Find the coefficient of $$x^{15}$$ in the expansion of $$(1-x)(1-2x)(1-4x)(1-8x)\cdots(1-32768x)$$
× | 51 | 116 | {"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} | 2.875 | 3 | CC-MAIN-2018-17 | latest | en | 0.538684 |
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