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### Pebbles
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Pythagorean Triples
How many right-angled triangles are there with sides that are all integers less than 100 units?
### Paving the Way
A man paved a square courtyard and then decided that it was too small. He took up the tiles, bought 100 more and used them to pave another square courtyard. How many tiles did he use altogether?
# Isometric Areas
##### Stage: 3 Challenge Level:
Here is an equilateral triangle with sides of length 1.
Let's define a unit of area, $T$, such that the triangle has area $1T$.
Here are some parallelograms whose side lengths are whole numbers.
Can you find the area, in terms of $T$, of each parallelogram?
Compare the results with the lengths of their edges.
What do you notice?
Can you explain what you've noticed?
Can you find a similar result for trapeziums in which all four lengths are whole numbers?
You might like to try More Isometric Areas next.
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http://www.maths.manchester.ac.uk/~pjohnson/resources/math60082/MYCODES/solutions-sheet-6-1-4.cpp
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#include #include #include #include #include using namespace std; /* Template code for the Explicit Finite Difference */ double explicitCallOption(double S0,double X,double T,double r,double sigma,int iMax,int jMax) { // declare and initialise local variables (ds,dt) double S_max=2*X; double dS=S_max/jMax; double dt=T/iMax; // create storage for the stock price and option price (old and new) vector S(jMax+1),vOld(jMax+1),vNew(jMax+1); // setup and initialise the stock price for(int j=0;j<=jMax;j++) { S[j] = j*dS; } // setup and initialise the final conditions on the option price for(int j=0;j<=jMax;j++) { vOld[j] = max(S[j]-X,0.); vNew[j] = max(S[j]-X,0.); } // loop through time levels, setting the option price at each grid point, and also on the boundaries for(int i=iMax-1;i>=0;i--) { // apply boundary condition at S=0 vNew[0] = 0.; for(int j=1;j<=jMax-1;j++) { double A,B,C; A=0.5*sigma*sigma*j*j*dt+0.5*r*j*dt; B=1.-sigma*sigma*j*j*dt; C=0.5*sigma*sigma*j*j*dt-0.5*r*j*dt; vNew[j] = 1./(1.+r*dt)*(A*vOld[j+1]+B*vOld[j]+C*vOld[j-1]); } // apply boundary condition at S=S_max vNew[jMax] = S[jMax] - X*exp(-r*(T-i*dt)); // set old values to new vOld=vNew; } // get j* such that S_0 \in [ j*dS , (j*+1)dS ] int jstar; jstar = S0/dS; double sum=0.; // run 2 point Lagrange polynomial interpolation sum = sum + (S0 - S[jstar+1])/(S[jstar]-S[jstar+1])*vNew[jstar]; sum = sum + (S0 - S[jstar])/(S[jstar+1]-S[jstar])*vNew[jstar+1]; return sum; } int main() { // declare and initialise Black Scholes parameters double S0=1.639,X=2.,T=1.,r=0.05,sigma=0.4; // declare and initialise grid paramaters int iMax=4,jMax=4; cout << explicitCallOption(S0,X,T,r,sigma,iMax,jMax) << endl; /* OUTPUT 0.194858 */ }
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https://www.teachoo.com/2597/622/Misc-25---Find-sum-of-13-1---13---23-1---3---Chapter-9-Class-11/category/Miscellaneous/
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Miscellaneous
Chapter 8 Class 11 Sequences and Series
Serial order wise
### Transcript
Misc , 25 Find the sum of the following series up to n terms: 13/1 + (13+23)/(1+3) + (13+23+33)/(1+3+5) +………. 13/1 + (13 + 23)/(1 + 3) + (13 + 23 + 33)/(1 + 3 + 5) +………. nth term of series is an = (13 + 23 + 33 + …. + n3)/(1 + 3 + 5 + … + 𝑛 𝑡𝑒𝑟𝑚𝑠) We solve denominator & numerator separately 13 + 23 + 33 + … + n3 = ((𝑛(𝑛+1))/2)^2 Also, 1 + 3 + 5 + …. + n terms This is A.P whose first term is a = 1 & common difference = d = 3 – 1 = 2 Now, sum of n terms of AP is Sn = n/2 [ 2a +(n – 1)d] Sn = n/2 [ 2(1) + (n-1)2] Sn = n/2 [ 2 + 2n - 2] Sn = 𝑛/2[2n] Sn = n2 Now, an = (13 + 23 + 23 + …. + n3)/(1 + 3 + 5 + … +𝑛 𝑡𝑒𝑟𝑚𝑠) Putting values from (1) & (2) an = (𝑛(𝑛 + 1)/2)^2/𝑛2 = n2(n+1)2/4n2 = (n+1)2/4 = ((n2+1+2𝑛))/4 = (𝑛2 + 1 + 2𝑛)/4 Now, finding sum of n terms of series = 1/4 ((((𝑛)(𝑛 +1)(2𝑛 +1))/6)" + " 2 (𝑛(𝑛 + 1))/2 "+ n" ) = 1/4 ((𝑛(𝑛+1)(2𝑛+1))/6 " + n(n +1) + n" ) = 1/4 n(( (𝑛+1)(2𝑛+1))/6 " + (n +1) + 1" ) = 1/4 n(( (𝑛+1)(2𝑛+1)+6(𝑛+1)+6(1))/6 " " ) = 1/4 n(( (𝑛+1)(2𝑛+1)+6(𝑛+1)+6(1))/6 " " ) = (1 × 𝑛)/(4 × 6) [(n + 1)(2n + 1) + 6(n + 1) + 6] = 𝑛/24 [n(2n + 1) + 1(2n + 1) + 6n + 6 + 6] = 𝑛/24 [2n2 + n + 2n + 1 + 6n + 6 + 6] = 𝑛/24 [2n2 + n + 2n + 6n + 1 + 6 + 6] = 𝑛/24 [2n2 + 9n + 13] Thus , the required sum is 𝑛/24 [2n2 + 9n + 13]
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# chemistry 12
posted by .
how would i sketch a pH curve for the titration of40.00mL of 0.100 M hydrazine H2NNH2(aq) having a Kb of 3.0x10^-6 by 0.100M HCIO4(aq) ?
• chemistry 12 -
We can't draw the curves on this board; however, I found one drawn on the web. Go to this site and scroll about 1/3 down to "strong acid + weak base". There is a titration curve that starts with NH3 and is titrated with HCl. That will be essentially the same for your weak base and strong acid.
http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html
• chemistry 12 -
thank you this should help!
## Similar Questions
1. ### Chemistry 12
Hey guys i only have this question left on my review and i just can't fingure it out! Sketch a pH curve for the titration of 40.00mL of 0.100 M hydrazine, H2NNH2, having a K of 3.0x10^-6 by 0.100 M HCIO4 a) what combination of acide-base …
2. ### chem12
Hey guys i only have this question left on my review and i just can't fingure it out! Sketch a pH curve for the titration of 40.00mL of 0.100 M hydrazine, H2NNH2, having a K of 3.0x10^-6 by 0.100 M HCIO4 a) what combination of acide-base …
3. ### chemistry 12
A) Calculate [OH^-] in a o.12 M solution of hydrazine, H2NNH2(aq), having K of 3.0x10^-6. B) what is the percent ionization of the H2NNH2(aq) I can't get these for the life of mee!
4. ### chemistry 12
sketch a ph curve for the titration 40.00 mLof 0.100M hyrazine H2NNH2(aq) having a Kb of 3.0 x 10^-6 by 0.100 M HCIO4(aq)
5. ### Shelly
I really don't get pH curves. I have to sketch a pH curve for the titration of 40.00 mL of 0.100 M hydrazine, H2NNH2(aq) having a Kb of 3.0x10^-6 by 0.100M HCIO4(aq). I know you can't sketch on here. But can you help me find the first …
6. ### Chemistry 12
I really don't get pH curves. I have to sketch a pH curve for the titration of 40.00 mL of 0.100 M hydrazine, H2NNH2(aq) having a Kb of 3.0x10^-6 by 0.100M HCIO4(aq). I know you can't sketch on here. But can you help me find the first …
7. ### Chemistry
I really don't get pH curves. I have to sketch a pH curve for the titration of 40.00 mL of 0.100 M hydrazine, H2NNH2(aq) having a Kb of 3.0x10^-6 by 0.100M HCIO4(aq). I know you can't sketch on here. But can you help me find the first …
8. ### Chemistry
How would I get the Ice of the titration of 40.00ml of 0.100M H2NNH2(aq) having a Kb of 3.0x10^-6 by 0.100M HClO4(aq)?
9. ### chemistry 12
hey i was wondering, i have this question: Sketch a pH curve for the titration of 40.00mL of 0.100 M hydrazine, H2NNH2, having a K of 3.0x10^-6 by 0.100 M HCIO4 i know you can t sketch but could you tell me what to do with the "K of …
10. ### Chemistry
For the reaction of hydrazine (N2H4) in water, Kb is 3.0 10-6. H2NNH2(aq) + H2O(l)--> H2NNH3+(aq) + OH-(aq) Calculate the concentrations of all species and the pH of a 1.6 M solution of hydrazine in water. [OH-] = [H2NNH3+] = [H2NNH2] …
More Similar Questions
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1. how to write a perpendicular equation
# Perpendicular Line Calculator how to write a perpendicular equation
how to write a perpendicular equation
Rated 8/10 based on 122 student reviews
Date:30.04.2020
Eqquation using our site, you agree to our cookie policy. Updated: July 28,
## how to write a perpendicular line
Transform for the "x" and "y" variable. I'll leave the rest of the exercise for you, esuation you're interested.
### how to write a person
By using this service, some information may be shared with YouTube. NOTE: The re-posting of materials in part or whole from this site to the Internet is copyright violation and is not considered "fair use" for educators. To change the slope, you must pedpendicular the value into its opposite sign positive to negative or negative to positive.
### how to write a person specification
Learn why people perpehdicular wikiHow. But since 1. Choose a y-intercept different from the original line.
#### how to write a person statement how
Perpendicular lines cross each other at a degree angle. This page will take a look at a few of the various examples pperpendicular utilize the concept of slopes associated with parallel and perpendicular lines. NOTE: The re-posting of materials in part or whole from this site to the Internet is copyright violation and is not considered "fair use" for educators.
Don't forget to add or ro the y-value too. Since perpendicular lines have negative reciprocal slopes, our line has a slope of Log in Facebook Loading
how
A
howHow to write How to write how
line
line
161 no A
-
perpendicular
line
189 yes form
perpendicular
perpendicular
Math
119 no line
in
-
standard
129 yes Math
A
A
form
127 no -
line
As the COVID situation develops, our hearts ache as we think about all peerpendicular people around the world that are affected by the pandemic
1. To create this article, volunteer authors worked to edit and improve it over time.
2. The lines have the same slope, so they are indeed parallel.
3. Optional Check whether or not your answer is correct.
4. Plug in the point's x- and y-values.
5. You can use the Mathway widget below to practice perpendicklar a perpendicular line through a given point.
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2.207 Bommer et al. (2003)
• Ground-motion model is:
where y is in g, C1 = -1.482, C2 = 0.264, C4 = -0.883, h = 2.473, CA = 0.117, CS = 0.101, CN = -0.088, CR = -0.021, σ1 = 0.243 (intra-event) and σ2 = 0.060 (inter-event).
• Use four site conditions but retain three (because only three records from very soft (L) soil which combine with soft (S) soil category):
R
Rock: V s > 750ms, SA = 0,SS = 0, 106 records.
A
Stiff soil: 360 < V s 750ms, SA = 1,SS = 0, 226 records.
S
Soft soil: 180 < V s 360ms, SA = 0,SS = 1, 81 records.
L
Very soft soil: V s 180ms, SA = 0,SS = 1, 3 records.
• Use same data as Ambraseys et al. (1996).
• Use three faulting mechanism categories:
S
Strike-slip: earthquakes with rake angles (λ) -30 λ 30 or λ 150 or λ ≤ -150, FN = 0,FR = 0, 47 records.
N
Normal: earthquakes with -150 < λ < -30, FN = 1,FR = 0, 146 records.
R
Reverse: earthquakes with 30 < λ < 150, FR = 1,FN = 0, 229 records.
Earthquakes classified as either strike-slip or reverse or strike-slip or normal depending on which plane is the main plane were included in the corresponding dip-slip category. Some records (137 records, 51 normal, 10 strike-slip and 76 reverse) from earthquakes with no published focal mechanism (80 earthquakes) were classified using the mechanism of the mainshock or regional stress characteristics.
• Try using criteria of Campbell (1997) and Sadigh et al. (1997) to classify earthquakes w.r.t. faulting mechanism. Also try classifying ambiguously classified earthquakes as strike-slip. Find large differences in the faulting mechanism coefficients with more stricter criteria for the rake angle of strike-slip earthquakes leading to higher CR coefficients.
• Note that distribution of records is reasonably uniform w.r.t. to mechanism although significantly fewer records from strike-slip earthquakes.
• Try to use two-stage maximum-likelihood method as employed by Ambraseys et al. (1996) but find numerical instabilities in regression.
• Also rederive mechanism-independent equation of Ambraseys et al. (1996) using one-stage maximum-likelihood method.
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# 638859 (number)
638,859 (six hundred thirty-eight thousand eight hundred fifty-nine) is an odd six-digits composite number following 638858 and preceding 638860. In scientific notation, it is written as 6.38859 × 105. The sum of its digits is 39. It has a total of 3 prime factors and 8 positive divisors. There are 393,120 positive integers (up to 638859) that are relatively prime to 638859.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 39
• Digital Root 3
## Name
Short name 638 thousand 859 six hundred thirty-eight thousand eight hundred fifty-nine
## Notation
Scientific notation 6.38859 × 105 638.859 × 103
## Prime Factorization of 638859
Prime Factorization 3 × 13 × 16381
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) 638859 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 638,859 is 3 × 13 × 16381. Since it has a total of 3 prime factors, 638,859 is a composite number.
## Divisors of 638859
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 917392 Sum of all the positive divisors of n s(n) 278533 Sum of the proper positive divisors of n A(n) 114674 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 799.287 Returns the nth root of the product of n divisors H(n) 5.57109 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 638,859 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 638,859) is 917,392, the average is 114,674.
## Other Arithmetic Functions (n = 638859)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 393120 Total number of positive integers not greater than n that are coprime to n λ(n) 16380 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 51877 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 393,120 positive integers (less than 638,859) that are coprime with 638,859. And there are approximately 51,877 prime numbers less than or equal to 638,859.
## Divisibility of 638859
m n mod m 2 3 4 5 6 7 8 9 1 0 3 4 3 4 3 3
The number 638,859 is divisible by 3.
## Classification of 638859
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (638859)
Base System Value
2 Binary 10011011111110001011
3 Ternary 1012110100110
4 Quaternary 2123332023
5 Quinary 130420414
6 Senary 21405403
8 Octal 2337613
10 Decimal 638859
12 Duodecimal 269863
20 Vigesimal 3jh2j
36 Base36 doy3
## Basic calculations (n = 638859)
### Multiplication
n×y
n×2 1277718 1916577 2555436 3194295
### Division
n÷y
n÷2 319430 212953 159715 127772
### Exponentiation
ny
n2 408140821881 260744437326073779 166578930485698168378161 106420448951162646151903558299
### Nth Root
y√n
2√n 799.287 86.1261 28.2717 14.4904
## 638859 as geometric shapes
### Circle
Diameter 1.27772e+06 4.01407e+06 1.28221e+12
### Sphere
Volume 1.0922e+18 5.12885e+12 4.01407e+06
### Square
Length = n
Perimeter 2.55544e+06 4.08141e+11 903483
### Cube
Length = n
Surface area 2.44884e+12 2.60744e+17 1.10654e+06
### Equilateral Triangle
Length = n
Perimeter 1.91658e+06 1.7673e+11 553268
### Triangular Pyramid
Length = n
Surface area 7.06921e+11 3.0729e+16 521626
## Cryptographic Hash Functions
md5 095da8e072b70b9fa327a8f9fbe63b8f 02ad7d1a520a020357d6a54422a7b57307b70d22 09292dbf0186d8b115b80a84aa7f767107e50aae04560cf32520dd79f171819a 2a6ecdfe03a8ff6cb688668ee878b76fbaa4f26be1058831d59317bec35cf9b3d01b895388c2742b109355808240798aac8261baf7a04fb580a40ae7c672e8d4 6295991dfb6bf332210715c13c807799e65c3687
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# MACRO Calculator For Fitness
so… I’ve gotten a basic calculator to work, but when it comes to this more complex problem… im not sure if it’s possible, I keep getting wrong answers and have tried multiple different combinations of code for this to work, but my brain is just melting over trying to figure this out… I do fitness, not this stuff, so it’s new territory lol
Equation is: 66.47 + (6.24 × weight in pounds) + (12.7 × height in inches) − (6.75 × age in years)
using 160 for weight, 70 inches for height and 30 for age
I should be getting 1692 but keep getting 3,012
from my understanding I need weight AND what the weight equation equals (6.24Xweight) so 2 variables for each number (for slider to show value to player, and the ‘hidden’ number which is the BMR equation for total BMR later)
is this even possible, or am I going to run myself into insanity here? i’ve gotten close (500 off) but can not get the correct answer… change things around and get a negative number or one that is so far off that I have NO idea why…
am I adding too much? too little? everything else is going smoothly but this has slowed me down
Hi and welcome!
I think you made a mistake here:
It should be: GlobalVariable(PersonaStats.Weight)
1 Like
WELL that would definitely explain a lot lol appreciate the feedback, another pair of eyes always help! currently busy, but will check on it as soon as i get the time and let you know
1 Like
Cool let me know if it all checks out.
sorry for taking a little longer than expected… life! but, it seems to be working now. numbers are slightly different from an online calculator but that could be from that calc rounding up decimals or something, will continue to work and tweak that… next part is adding protein/fats/carbs per day based off of BMR (this was just the start of the macros calculator, other parts should be MUCH easier though)
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Subatomic Particles - Atomic Structure
# Atomic StructureSubatomic Particles
LEARNING GOALS
After Chapter 1.1, you will be able to:
· Identify the subatomic particles most important for determining various traits of an atom, including charge, atomic number, and isotope
· Determine the number of protons, neutrons, and electrons within an isotope, such as 14C
Although you may have encountered in your university-level chemistry classes such subatomic particles as quarks, leptons, and gluons, the MCAT’s approach to atomic structure is much simpler. There are three subatomic particles that you must understand: protons, neutrons, and electrons.
Figure 1.1. Matter: From Macroscopic to Microscopic
PROTONS
Protons are found in the nucleus of an atom, as shown in Figure 1.1. Each proton has an amount of charge equal to the fundamental unit of charge (e = 1.6 × 10−19 C), and we denote this fundamental unit of charge as “+1 e” or simply “+1” for the proton. Protons have a mass of approximately one atomic mass unit (amu). The atomic number (Z) of an element, as shown in Figure 1.2, is equal to the number of protons found in an atom of that element. As such, it acts as a unique identifier for each element because elements are defined by the number of protons they contain. For example, all atoms of oxygen contain eight protons; all atoms of gadolinium contain 64 protons. While all atoms of a given element have the same atomic number, they do not necessarily have the same mass—as we will see in our discussion of isotopes.
Figure 1.2. Potassium, from the Periodic TablePotassium has the symbol K (Latin: kalium), atomic number 19, and atomic weight of approximately 39.1.
NEUTRONS
Neutrons, as the name implies, are neutral—they have no charge. A neutron’s mass is only slightly larger than that of the proton, and together, the protons and the neutrons of the nucleus make up almost the entire mass of an atom. Every atom has a characteristic mass number (A), which is the sum of the protons and neutrons in the atom’s nucleus. A given element can have a variable number of neutrons; thus, while atoms of the same element always have the same atomic number, they do not necessarily have the same mass number. Atoms that share an atomic number but have different mass numbers are known as isotopes of the element, as shown in Figure 1.3. For example, carbon (Z = 6) has three naturally occurring isotopes: , with six protons and six neutrons; , with six protons and seven neutrons; and , with six protons and eight neutrons. The convention is used to show both the atomic number (Z) and the mass number (A) of atom X.
Figure 1.3. Various Isotopes of HydrogenAtoms of the same element have the same atomic number (Z = 1) but may have varying mass numbers (A = 1, 2, or 3).
ELECTRONS
Electrons move through the space surrounding the nucleus and are associated with varying levels of energy. Each electron has a charge equal in magnitude to that of a proton, but with the opposite (negative) sign, denoted by “−1 e” or simply “—e.” The mass of an electron is approximately that of a proton. Because subatomic particles’ masses are so small, the electrostatic force of attraction between the unlike charges of the proton and electron is far greater than the gravitational force of attraction based on their respective masses.
Electrons move around the nucleus at varying distances, which correspond to varying levels of electrical potential energy. The electrons closer to the nucleus are at lower energy levels, while those that are further out (in higher electron shells) have higher energy. The electrons that are farthest from the nucleus have the strongest interactions with the surrounding environment and the weakest interactions with the nucleus. These electrons are called valence electrons; they are much more likely to become involved in bonds with other atoms because they experience the least electrostatic pull from their own nucleus. Generally speaking, the valence electrons determine the reactivity of an atom. As we will discuss in Chapter 3 of MCAT General Chemistry Review, the sharing or transferring of these valence electrons in bonds allows elements to fill their highest energy level to increase stability. In the neutral state, there are equal numbers of protons and electrons; losing electrons results in the atom gaining a positive charge, while gaining electrons results in the atom gaining a negative charge. A positively charged atom is called a cation, and a negatively charged atom is called an anion.
Bridge
Valence electrons will be very important to us in both general and organic chemistry. Knowing how tightly held those electrons are will allow us to understand many of an atom’s properties and how it interacts with other atoms, especially in bonding. Bonding is so important that it is discussed in Chapter 3 of both MCAT General Chemistry Review and MCAT Organic Chemistry Review.
Some basic features of the three subatomic particles are shown in Table 1.1.
Subatomic Particle Symbol Relative Mass Charge Location Proton p, p+, or 11H 1 +1 Nucleus Neutron n0 or 01n 1 0 Nucleus Electron e− or −10e 0 −1 Orbitals Table 1.1. Subatomic Particles
Example:
Determine the number of protons, neutrons, and electrons in a nickel-58 atom and in a nickel-60 +2 cation.
Solution:
58Ni has an atomic number of 28 and a mass number of 58. Therefore, 58Ni will have 28 protons, 28 electrons, and 58 — 28, or 30, neutrons.
60Ni2+ has the same number of protons as the neutral 58Ni atom. However, 60Ni2+ has a positive charge because it has lost two electrons; thus, Ni2+ will have 26 electrons. Also, the mass number is two units higher than for the 58Ni atom, and this difference in mass must be due to two extra neutrons; thus, it has a total of 32 neutrons.
MCAT Concept Check 1.1:
Before you move on, assess your understanding of the material with these questions.
1. Which subatomic particle is the most important for determining each of the following properties of an atom?
o Charge:
o Atomic number:
o Isotope:
2. In nuclear medicine, isotopes are created and used for various purposes; for instance, 18O is created from 18F. Determine the number of protons, neutrons, and electrons in each of these species.
Particle Protons Neutrons Electrons 18O 18F
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# Proposition 28
If a straight line falling on two straight lines makes the exterior angle equal to the interior and opposite angle on the same side, or the sum of the interior angles on the same side equal to two right angles, then the straight lines are parallel to one another.
Let the straight line EF falling on the two straight lines AB and CD make the exterior angle EGB equal to the interior and opposite angle GHD, or the sum of the interior angles on the same side, namely BGH and GHD, equal to two right angles.
I say that AB is parallel to CD.
Since the angle EGB equals the angle GHD, and the angle EGB equals the angle AGH, therefore the angle AGH equals the angle GHD. And they are alternate, therefore AB is parallel to CD.
Next, since the sum of the angles BGH and GHD equals two right angles, and the sum of the angles AGH and BGH also equals two right angles, therefore the sum of the angles AGH and BGH equals the sum of the angles BGH and GHD.
Subtract the angle BGH from each. Therefore the remaining angle AGH equals the remaining angle GHD. And they are alternate, therefore AB is parallel to CD.
Therefore if a straight line falling on two straight lines makes the exterior angle equal to the interior and opposite angle on the same side, or the sum of the interior angles on the same side equal to two right angles, then the straight lines are parallel to one another.
Q.E.D.
## Guide
This proposition states two useful minor variants of the previous proposition. The three statements differ only in their hypotheses which are easily seen to be equivalent with the help of proposition I.13.
#### Use of Proposition 28
This proposition is used in IV.7, VI.4, and a couple times in Book XI.
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# Natural number k satisfying k & x = k
0 How do I find all natural numbers k satisfying k & x == k for some Natural Number x? asked 16 Jan, 17:03 2★rds_98 45●6 accept rate: 9% SOS dp. Read the codeforces article. (16 Jan, 18:15)
3 You can just go through the sub-optimal solution mentioned here: http://codeforces.com/blog/entry/45223 answered 16 Jan, 18:20 2.4k●4●20 accept rate: 17%
1 you can convert the number x into binary form i.e 15=1+2+4+8 and 13=1+4+8 and make array of all these numbers i.e array for 15={1,2,4,8} and array for 13 will be {1,4,8} and the find the subsets of this array... and just add elements of each of them.. so 15 will update all elements from 1 to 15 as all numbers can be computed using elements of that array.. EDIT: well, there may be another optimised soln about which i am not aware answered 16 Jan, 23:47 1.3k●1●9 accept rate: 26%
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Q. 2
# Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:(i) (ii)(iii)
(i) and
Degree of ; therefore degree of and degree of remainder is of degree 1 or less,
Let and
By applying division algorithm:
Dividend = Quotient× Divisor + Remainder
On substituting values in the above relation we get,
On comparing coefficients we get,
On solving above equations we get,
, , , ,
On substituting these values for
Since remainder is zero, therefore
(ii) and
Degree of ; therefore degree of and degree of remainder is of degree 1 or less,
Let and
By applying division algorithm:
Dividend = Quotient× Divisor + Remainder
On substituting values in the above relation we get,
On comparing coefficients we get,
On solving above equations we get,
, , , ,
On substituting these values for
Since remainder is 2, therefore
(iii) and
Degree of ; therefore degree of and degree of remainder is of degree 2 or less,
Let and
By applying division algorithm:
Dividend = Quotient× Divisor + Remainder
On substituting values in the above relation we get,
On comparing coefficients we get,
On solving above equations we get,
, , , ,
On substituting these values for
Since remainder is , therefore
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# Math in Focus Grade 2 Chapter 6 Practice 2 Answer Key Multiplying 2: Using Dot Paper
Go through the Math in Focus Grade 2 Workbook Answer Key Chapter 6 Practice 2 Multiplying 2: Using Dot Paper to finish your assignments.
## Math in Focus Grade 2 Chapter 6 Practice 2 Answer Key Multiplying 2: Using Dot Paper
Use dot paper to solve.
Question 1.
There are 4 bags. 2 rolls are in each bag. How many rolls are there in all?
4 × 2 = ____
There are ___ rolls in all.
Explanation:
Given:
Total number of bags = four
each bag has = two rolls
Multiply four with two we get
4 × 2 = 8
There are 8 rolls in all.
Question 2.
6 bicycles are in the shop. Each bicycle has 2 wheels. How many wheels are there in all?
Explanation:
Given:
Total number of bicycle in a shop = six
each each bicycle has = two wheels
Multiply six with two we get
6 × 2 = 12
There are 12 wheels in all.
Use dot paper to solve.
Question 3.
Mrs. Smith buys 5 burgers for her children. Each burger costs $2. How much do the 5 burgers cost in all? Answer: Explanation: Given: Total number of burgers Mrs. Smith buys for her children = five Each burger costs = two dollar Multiply four with two we get 5 × 2 =10 The five burgers cost$ 10 in all
Question 4.
Ed buys 9 pairs of socks. Each pair of socks costs $2. How much do the socks cost in all? Answer: Explanation: Given: Total number of pair of socks Ed buys = nine Each pair of socks costs = two dollar Multiply nine with two we get 9 × 2 = 18 The five burgers cost$ 18 in all
Use dot paper to find the missing numbers.
Example
Question 5.
Question 6.
6 × 2 = 10 + ___
= ______
6 × 2 = 10 + 2
= 12
Question 7.
8 × 2 = 20 – ___
= ____
8 × 2 = 20 – 4
= 16
Use dot paper to find the missing numbers.
Example
Question 8.
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Excel is a powerful program, with advanced functions like macros and PivotTables. But sometimes you just need to add a long list of numbers in a hurry. If that’s the case, you need to know how to sum a column in Excel.
Just like most functions in Excel, there’s more than one way to get the job done. Here’s a list of some methods you can try, with the first one requiring just one click.
## How to sum a column in Excel
### Method one: The single click
The easiest way to get the sum of a column is to click on the letter at the top. You’ll see the bar at the bottom-right of your sheet change and one of the values that will be displayed is the Sum. This is the easiest method if you don’t need to populate a cell with the value but you just need to make a note of it.
### Method two: The AutoSum feature
The second way you can sum a column is by using the AutoSum feature. It’s slightly more involved than the first method, but it’s easier to retain the data. The steps are as follows:
1. Select the cell beneath the column you want to sum.
2. Navigate to the Home tab and find the Editing group. Click on the AutoSum function that features the symbol.
3. Excel will automatically add the =SUM function and select the range of numbers above the cell in the column.
4. Just press Enter to see your sum.
### Method three: Use the SUM formula manually
You can also use the SUM function manually. This is the easier method if you only want to add up a few of the cells in your column, and you can learn more about adding in Excel here. Here’s a review of the steps:
1. Select the cell where you want the sum to appear.
2. Begin your formula with the =SUM( command.
3. You can either enter cells manually and separate them in your formula with commas or highlight a range of cells that are divided by a colon. Once you have the cells you want, close the parenthesis.
4. See the example in the function bar in the image below. You then hit enter to see your total.
### Method four: Convert your data into a table
This final way to get the sum of a column in Excel is easiest if you are planning to present your data. It not only sums your data but it also performs a variety of other functions for you to choose from.
1. Highlight the column and press Ctrl+T on your keyboard to format the range of your cells into an Excel Table.
2. Now you should see the Design tab appear. Go to the tab and check the box that says Total Row.
3. A new row should appear at the end of your table with values at the end of each column. Select the cell at the base of the column you want to sum to bring up a dropdown menu.
4. Open the dropdown menu and select the Sum option to see your total.
## What else can I do with Excel?
Now that you know how to sum a column in Excel, you’re ready to take your skills to the next level. If you want to learn more advanced functions like Power Query, VBA, and PivotTables, there’s affordable guidance out there. You can grab some hands-on training right now with a dedicated learning kit from Tech Deals.
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Each slide is a student friendly version of the Work Place directions for Bridges 2nd Grade (2nd Edition) Unit 1: Figure the Facts. Unit 1 contains 11 games:1A: Unifix Cubes1B: Pattern Blocks1C: Tiles1D: Geoboards1E: Match the Beetle1F: Count & Compare Fives1G: Make the Sum1H: Count and Compare
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Each slide is a student friendly version of the Work Place directions for Bridges 2nd Grade (2nd Edition) Unit 2: Place Value and Measurement with Jack's Beanstalks. Unit 2 contains 5 games:2A: Scoop, Count and Compare2B: The Subtraction Wheel2C: Number Line Race2D: Pick Two, Roll and Subtract2E: St
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Powerpoint presentation for Unit 1 Module 1: Sorting and GraphingThese slides align to the 2nd grade Bridges curriculum (second edition) whole group lesson portion and are made to be student-friendly. They contain visuals, diagrams, photographs, and other manipulatives to help students master the un
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Powerpoint presentation for Unit 1 Module 2: Number Facts with the Number RackThese slides align to the 2nd grade Bridges curriculum (second edition) whole group lesson portion and are made to be student-friendly. They contain visuals, diagrams, photographs, and other manipulatives to help students
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Powerpoint presentation for Unit 1 Module 3: Introducing Addition and Subtraction StrategiesThese slides align to the 2nd grade Bridges curriculum (second edition) whole group lesson portion and are made to be student-friendly. They contain visuals, diagrams, photographs, and other manipulatives to
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Powerpoint presentation for Unit 1 Module 4: Fluency with Addition Facts to TwentyThese slides align to the 2nd grade Bridges curriculum (second edition) whole group lesson portion and are made to be student-friendly. They contain visuals, diagrams, photographs, and other manipulatives to help stude
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Powerpoint presentation for Unit 2: Place Value and Measurement with Jack's Beanstalks - Module 1: Counting and Modeling Two- and Three-Digit NumbersThese slides align to the 2nd grade Bridges curriculum (second edition) whole group lesson portion and are made to be student-friendly. They contain vi
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Powerpoint presentation for Unit 2: Place Value and Measurement with Jack's Beanstalks - Module 2: Measuring Jack's Giant Beans with TensThese slides align to the 2nd grade Bridges curriculum (second edition) whole group lesson portion and are made to be student-friendly. They contain visuals, diagr
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Powerpoint presentation for Unit 2: Place Value and Measurement with Jack's Beanstalks - Module 3: Adding on the Open Number LineThese slides align to the 2nd grade Bridges curriculum (second edition) whole group lesson portion and are made to be student-friendly. They contain visuals, diagrams, pho
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Powerpoint presentation for Unit 2: Place Value and Measurement with Jack's Beanstalks - Module 4: Thinking in TwosThese slides align to the 2nd grade Bridges curriculum (second edition) whole group lesson portion and are made to be student-friendly. They contain visuals, diagrams, photographs, and
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This file contains 8 calendar grid observation charts for the 2nd grade Bridges Number Corner (September-May, excluding January because there is no observation chart for this month). It is set to a normal/letter size file, and I used my school's poster maker to enlarge them to poster size to hang o
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Bridges in Mathematics - 2nd EditionBundle of 4 Powerpoint Presentations for 2nd Grade Unit 1: Figure the FactsContains Modules 1-4 (or 4 weeks/1 month of lessons)1. Sorting and Graphing2. Number Facts with the Number Rack3. Introducing Addition and Subtraction Strategies4. Fluency with Addition Fac
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Bridges in Mathematics - 2nd EditionBundle of 4 Powerpoint Presentations for 2nd Grade Unit 2: Place Value and Measurement with Jack's BeanstalksContains Modules 1-4 (or 4 weeks/1 month of lessons)1. Counting and Modeling with Two- and Three-Digit Numbers2. Measuring Jack's Giant Beans with Tens3. A
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This free download is one of the 11 Work Places for Bridges in Mathematics (2nd Edition) 2nd Grade Unit 1: Figure the Facts.1A: Unifix CubesAvailable for purchase on my TPT site is the set of all 11 Work Place directions including:1B: Pattern Blocks1C: Tiles1D: Geoboards1E: Match the Beetle1F: Count
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The skills included are listed below. These skills and sequence were inspired by concepts learned in a conference for Multi-Sensory learning and Orton-Gillingham Methods loosely based on "Recipe for Reading." While designed for my 2nd semester of 2nd grade, use it for what you need. This set coord
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2nd Grade Social Studies, 3rd Grade Social Studies: All Year Curriculum A Year of Social Studies! Grades 2-3 Would you like a program that engages your students in an ongoing, authentic project that integrates every Social Studies strand in the context of building your own class town? This 109 pag
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Social Studies Interactive Notebook: 26 fun and highly engaging foldables, posters, supply list, and so much more! You'll never go back to your old INBs after this! Want this in a Kindergarten, 1st Grade, and 2nd Grade BUNDLE? CLICK HERE This Product Includes: 1. Table of Contents 2. Tomochichi,
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This bundle includes all the slides for Unit 1.Use these slides to teach Bridges instead of wading through your TE! You can upload them into your 'Google Drive and edit them in Slides, if you'd like. Use an iPad? You can use the slides on Noteability, airplay it and have students draw on the slides
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This bundle includes all the slides for Unit 2.Use these slides to teach Bridges instead of wading through your TE! You can upload them into your 'Google Drive and edit them in Slides, if you'd like. Use an iPad? You can use the slides on Noteability, airplay it and have students draw on the slides
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**Before downloading please ensure that you have selected the correct file. Each lesson is available in Smartboard AND PowerPoint format. Please read the titles carefully to make sure you have selected the correct format.**This download includes Unit 1, Module 4, Sessions 1 - 5.Unit 1, Module 1 Coun
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Here you have it! I have created ALL of the units for 2nd grade! These packets accompany the TRS and lessons in the student textbook. It incorporates a variety of the emerging, expanding, and/or bridging recommended lessons. This offers a great way for teachers to provide a more structured lesson fo
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Bundle
This bundle includes all the slides for Unit 3.Use these slides to teach Bridges instead of wading through your TE! You can upload them into your 'Google Drive and edit them in Slides, if you'd like. Use an iPad? You can use the slides on Noteability, airplay it and have students draw on the slides
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This bundle includes all the slides for Unit 4.Use these slides to teach Bridges instead of wading through your TE! You can upload them into your 'Google Drive and edit them in Slides, if you'd like. Use an iPad? You can use the slides on Noteability, airplay it and have students draw on the slides
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# Convert Inches per day to Microns per second (in/day to µ/s Conversion)
1 in/day = 0.29398147916093 µ/s
## How to convert inches per day to microns per second?
To convert inches per day to microns per second, multiply the value in inches per day by 0.29398147916093325671.
You can use the conversion formula :
microns per second = inches per day × 0.29398147916093325671
1 inch per day is equal to 0.29398147916093 microns per second.
• 1 inch per day = 0.29398147916093 microns per second
• 10 inches per day = 2.9398147916093 microns per second
• 100 inches per day = 29.398147916093 microns per second
• 1000 inches per day = 293.98147916093 microns per second
• 10000 inches per day = 2939.8147916093 microns per second
## Examples to convert in/day to µ/s
Example 1:
Convert 50 in/day to µ/s.
Solution:
Converting from inches per day to microns per second is very easy.
We know that 1 in/day = 0.29398147916093 µ/s.
So, to convert 50 in/day to µ/s, multiply 50 in/day by 0.29398147916093 µ/s.
50 in/day = 50 × 0.29398147916093 µ/s
50 in/day = 14.699073958047 µ/s
Therefore, 50 in/day is equal to 14.699073958047 µ/s.
Example 2:
Convert 125 in/day to µ/s.
Solution:
1 in/day = 0.29398147916093 µ/s
So, 125 in/day = 125 × 0.29398147916093 µ/s
125 in/day = 36.747684895117 µ/s
Therefore, 125 in/day is equal to 36.747684895117 µ/s.
## Inches per day to microns per second conversion table
Inches per day Microns per second
0.001 in/day 0.00029398147916093 µ/s
0.01 in/day 0.0029398147916093 µ/s
0.1 in/day 0.029398147916093 µ/s
1 in/day 0.29398147916093 µ/s
2 in/day 0.58796295832187 µ/s
3 in/day 0.8819444374828 µ/s
4 in/day 1.1759259166437 µ/s
5 in/day 1.4699073958047 µ/s
6 in/day 1.7638888749656 µ/s
7 in/day 2.0578703541265 µ/s
8 in/day 2.3518518332875 µ/s
9 in/day 2.6458333124484 µ/s
10 in/day 2.9398147916093 µ/s
20 in/day 5.8796295832187 µ/s
30 in/day 8.819444374828 µ/s
40 in/day 11.759259166437 µ/s
50 in/day 14.699073958047 µ/s
60 in/day 17.638888749656 µ/s
70 in/day 20.578703541265 µ/s
80 in/day 23.518518332875 µ/s
90 in/day 26.458333124484 µ/s
100 in/day 29.398147916093 µ/s
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https://www.jiskha.com/questions/107377/how-many-real-roots-does-f-x-x-3-5x-1-have-x-3-odd-polynomial-at-least-1-real-root
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Math
How many real roots does f(x) = x^3+5x+1 have?
x^3 = odd polynomial = at least 1 real root
Can I follow Descartes' rule for this? No sign change, so there aren't any positive zeros.
when I sub in -x for x
(-x)^3+ 5(-x) + 1
= -x^3-5x+1
Then there is one sign change, so 1 negative root.
So my conclusion is that there is one real root. Is this right?
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2. 👎 0
3. 👁 205
1. I agree. The Descartes rule says that there cannot be more than one real root in this case, and that it will be negative. There is another rule that says there must be at least one real root. Therefore the number of real roots is one.
1. 👍 0
2. 👎 0
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# Finding x intercepts
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## Intercepts Calculator
To find: The x-intercept of the given function y = (3x - 1) / (2x + 1). Using the x-intercept formula, we substitute y = 0 in the above equation and solve for x. 0 = (3x - 1) / (2x + 1) Multiplying both
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## x-intercept of a line (video)
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Cryptography
pseudo-random-function terminology notation probability
Updated Thu, 21 Jul 2022 04:58:58 GMT
# Probability conventions in cryptography
I am working on Victor Shoup's tutorial on game-based security proof and want to figure out some notions from the perspective of probability theory.
Consider the following PRF advantage defined on Page 11: $$\bigl|\,\Pr[s \leftarrow S: A^{F_s}() = 1] - \Pr[f \leftarrow \Gamma_{\ell_1, \ell_2}: A^f() = 1]\,\bigr|$$ where $$\{ F_s \}_{s \in S}$$ is a family of keyed functions with key space $$S$$, domain $$\{ 0,1 \}^{\ell_1}$$ and range $$\{ 0,1 \}^{\ell_2}$$, and $$\Gamma_{\ell_1, \ell_2}$$ denotes the set of all functions from $$\{ 0,1 \}^{\ell_1}$$ to $$\{ 0,1 \}^{\ell_2}$$.
I have the following questions:
1. Does the probability notion $$\Pr[s \leftarrow S: \cdot]$$ (or, $$\Pr[\cdot | s \leftarrow S]$$ ) has the same meaning as the notion $$\Pr_{s \leftarrow S}[\cdot]$$ that is also commonly used in crypto contexts (e.g., this)? For me, $$\Pr[s \leftarrow S: \cdot]$$, $$\Pr[\cdot | s \leftarrow S]$$, and $$\Pr_{s \leftarrow S}[\cdot]$$ seem to refer to the same conditional probability measure.
2. How do we interpret a probability $$\Pr[s \leftarrow S: A^{F_s}() = 1]$$? Literally, I know that this captures the probability that the distinguisher $$A$$ outputs $$1$$ if it is given oracle access to a function $$F_s$$ keyed by $$s \in S$$, and the probability is taken over the random choice of $$s$$. However, from probability theory, what is the probability space $$(\Omega, \mathcal{F}, \Pr)$$ where the "event" that "the distinguisher $$A$$ with oracle access to $$F_s$$ outputs $$1$$" is defined? Does the sample space $$\Omega$$ contains all outcomes of $$s \leftarrow S$$ (so that we can say "the probability is taken over the random choice of $$s$$")? If so, does this implies that the two probabilities in the PRF advantage come from two different probability spaces?
## Solution
I completely understand your thoughts, I also had a similar question here.
I will answer your question based on my understanding, I would like to hear any thoughts or disagreements, because I haven't completely figured out the situtation yet.
Let's consider the Kolmogorov's definition of probability and forget about Random variables (these are used just to help us by converting a probability space to one that can be more efficiently handled).
A Probability Space is just $$(, \mathcal{F}, Pr)$$ where:
• : is the set of the events
• $$\mathcal{F}$$ : is a -algebra defined on , it is the event space.
• $$Pr$$ : is a set function $$P : \mathcal{F} \rightarrow \mathbb{R}$$
1. How do we interpret a probability $$\Pr[s \leftarrow S: A^{F_s}() = 1]$$? Literally, I know that this captures the probability that the distinguisher $$A$$ outputs $$1$$ if it is given oracle access to a function $$F_s$$ keyed by $$s \in S$$, and the probability is taken over the random choice of $$s$$. However, from probability theory, what is the probability space $$(\Omega, \mathcal{F}, \Pr)$$ where the "event" that "the distinguisher $$A$$ with oracle access to $$F_s$$ outputs $$1$$" is defined? Does the sample space $$\Omega$$ contains all outcomes of $$s > \leftarrow S$$ (so that we can say "the probability is taken over the random choice of $$s$$")? If so, does this implies that the two probabilities in the PRF advantage comes from two different probability spaces?
There are multiple level of approaches you can take, depending on the time have.
Consider the algorithm $$B() = s \leftarrow S; A^{F_s}()$$ that outputs whatever $$A$$ outputs. Now we can rewrite the above as $$Pr[B() = 1]$$. The $$=\{0,1\}$$ since $$A$$ is a distinguisher, see here. The $$\mathcal{F} = \{\emptyset, \{1\}, \{0\},\{0,1\}\} = \mathcal{P}()$$. Also let $$Pr$$ be a probability measure. Now I think it is clear.
But let's try to rewrite this. Consider a uniform distribution on $$S$$. That is $$=S$$, $$\forall s \in S, Pr(s) = 1/|S|$$, this is the probability measure, you can check that it satisfies the definition. Also $$\mathcal{F}=\{\emptyset, s_1, s_1^c, s_2, s_2^c, \dots, \}$$. In my opinion the $$\mathcal{F}$$ here is an overkill because we are only interested in single events, it is a way to model a simple random algorithm that outputs a single element. Now consider $$B(s) = A^{F_s}()$$. Now we have two probability spaces and we want to create a conditional probability. So we have
$$Pr(B(s)|s)=\dfrac{Pr(B(s),s)}{Pr(s)}$$
We can again create a -algebra $$\mathcal{F}$$ and repeat the process as above, but it will require a lot of time.
1. Does the probability notion $$\Pr[s \leftarrow S: \cdot]$$ (or, $$\Pr[\cdot | s \leftarrow S]$$ ) has the same meaning as the notion $$\Pr_{s \leftarrow S}[\cdot]$$ that is also commonly used in crypto contexts (e.g., [this][2])? For me, $$\Pr[s \leftarrow S: \cdot]$$ and $$\Pr[\cdot | s \leftarrow S]$$ seem to be two conditional probabilities, and $$\Pr_{s \leftarrow S}[\cdot]$$ is like a conditional probability measure.
From my experience I think they refer to the same conditional probability measure. If someone has another view he can leave a comment to discuss it or update the answer.
My opinion: At the end of the day all of this is just notation to express the same thing. You can either consider the higher level approach I gave first, where you consider a random process that has embedded some other random processes, or you can analyze every random process itself and consider their relations. In cryptography we are not really interested in the low level mathematical mechanisms, this is why you find so many different notations. It is about the subject we want to describe and authors will always the higher level approach as soon as it is can be understood by the reader for the purpose of the specific matter.
• +0 – Thank you very much for your answer. I edited my first question. In this question, I want to figure out whether these three notations refer to the same probability measure so that we can meaningfully calculate probabilities assigned by this measure, as well as useful bounds (e.g., union bound). However, I am not sure whether these notations are probability measures since $s \leftarrow S$ is not an event ... — Jun 21, 2022 at 09:32
• +0 – Hello, I updated my answer. In my opinion and from my experience I think they refer to the same measure, but again I am not a mathematician, I'm a computer scientist. — Jun 21, 2022 at 10:14
• +0 – Hi, thanks again for your inspiring answer. I added my understanding to this question and tried to answer the two questions from the perspective of random variables. The answers seem to be reasonable to me. — Jun 21, 2022 at 14:26
• +0 – You better create an answer with your understanding. I am reading it now... — Jun 21, 2022 at 14:50
• +0 – I think mathematically speaking, you miss something on your explanation. According to the definition, you can only define a probability measure on a -algebra. At some point you define it on the event space . — Jun 21, 2022 at 15:05
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# How to prove that a utility function U(x,y)=min(x,2y) is quasiconcave?
I have a question that asks:
"Let there be two goods 1 and 2.Let $$x$$ and $$y$$ denote their respective quantities.$$(x,y)$$ represents a bundle. Suppose a consumer’s preferences over bundles in $$R^2_+$$ can be represented by the utility function $$U(x,y)=min(x,2y)$$. Prove that the consumer’s preferences over bundles in $$R^2_+$$ are convex by proving that the utility function is quasiconcave in $$R^2_+$$"
I was taught that if a utility function is quasiconcave, then $$\begin{vmatrix} 0 & U_x & U_y \\ U_x & U_{xx} & U_{xy} \\ U_y & U_{yx} & U_{yy} \\\end{vmatrix} > 0 \forall (x, y) \in R^2_+$$
However, when I start doing the partial derivatives, I notice that I get:
$$\frac{\partial U}{\partial x} = \begin{cases}1, && x \le 2y \\ 0, && x > 2y\end{cases},$$ $$\frac{\partial U}{\partial y} = \begin{cases}0, && x \le 2y \\ 2, && x > 2y\end{cases},$$
$$\frac{\partial U}{\partial x^2} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases},$$ $$\frac{\partial U}{\partial y^2} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases},$$
$$\frac{\partial U}{\partial xy} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases},$$ $$\frac{\partial U}{\partial yx} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases}$$
I then got $$\begin{vmatrix} 0 & U_x & U_y \\ U_x & U_{xx} & U_{xy} \\ U_y & U_{yx} & U_{yy} \\\end{vmatrix} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases}$$
This is where I am stuck. I am unsure if:
a) I made a mistake somewhere
b) my method is wrong
c) Everything I've done so far is correct, but I need to do another step
• min(x, 2y) is a concave function and every concave function is quasiconcave. You can watch the following playlist for the proof that min(x, 2y) is concave : youtube.com/… – Amit Apr 4 at 6:41
• Or you can just use another definition of quasiconcavity, one that does not rely on differentiability. – Giskard Apr 4 at 8:56
• That $U$ is quasi-concave means that for any $s,t$ in the convex domain (here $\mathbb{R}_+^2$) and any $\lambda$ in $[0,1]$ (you could, equivalently, use an open interval), $U\big(\lambda s+(1-\lambda)t\big)\geq\min\big\{U(s),U(t)\big\}.$ It is wonderful that there exit calculus characterizations for differentiable functions, but this function is very much non-differentable, and all the action is at a kink. Start directly with the definition. – Michael Greinecker Apr 4 at 8:58
• Thanks @MichaelGreinecker that makes sense – DoubleRainbowZ Apr 4 at 12:21
A function $$f:D\rightarrow \mathbb{R}$$ is said to be quasiconcave if the following set is a convex set for every value of $$a\in\mathbb{R}$$: $$P_a = \{x\in D: f(x) \geq a\}$$
To show that $$f(x,y) =\min(x, 2y)$$ is quasiconcave, we just need to show that $$P_a = \{(x,y)\in \mathbb{R}^2: \min(x, 2y) \geq a\}$$ is a convex set. For that we consider arbitrary $$(x', y')$$ and $$(x'', y'')$$ from the set $$P_a$$ and arbitrary $$\lambda\in [0,1]$$ and show that $$\lambda (x', y')+(1-\lambda)(x'', y'')$$ is in $$P_a$$. Observe that $$x'\geq \min(x', 2y')\geq a$$ and $$x''\geq \min(x'', 2y'')\geq a$$, so $$\lambda x'+(1-\lambda)x''\geq a$$. Likewise, $$\lambda 2y'+(1-\lambda)2y''\geq a$$. Therefore, it follows that $$\min(\lambda x'+(1-\lambda)x'',2(\lambda y'+(1-\lambda)y'')) \geq a$$ and consequently, $$\lambda (x',y')+(1-\lambda)(x'', y'')$$ is in $$P_a$$.
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Cody
# Problem 262. Swap the input arguments
Solution 50174
Submitted on 24 Feb 2012 by Mike Keenan
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% [q,r] = swapInputs(5,10); assert(isequal(q,10)); assert(isequal(r,5));
2 Pass
%% [q,r] = swapInputs(magic(3), 'hello, world'); assert(isequal(q,'hello, world')); assert(isequal(r,magic(3)));
3 Pass
%% [q,r] = swapInputs({}, NaN); assert(isnan(q)); assert(iscell(r) && isempty(r));
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# Print direction of moves such that you stay within the [-k, +k] boundary
Given an array arr[] of N positive integers and an integer K. It is given that you start at position 0 and you can move to left or right by a[i] positions starting from arr[0]. The task is to print direction of moves in such a way that you can complete N steps without exceeding the [-K, +K] boundary by moving right or left. In case you cannot perform steps, print -1. In case of multiple answers, print any one.
Examples:
Input: arr[] = {40, 50, 60, 40}, K = 120
Output:
Right
Right
Left
Right
Explanation :
Since N = 4 (Number of elements in the array),
we need to make 4 moves from arr[0] such that
value does not go out of [-120, 120]
Move 1: Position = 0 + 40 = 40
Move 2: Position = 40 + 50 = 90
Move 3: Position = 90 – 60 = 30
Move 4: Position = 30 + 50 = 80
Input: arr[] = {40, 50, 60, 40}, K = 20
Output: -1
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach: The following steps can be followed to solve the above problem:
• Initialize position to 0 in the beginning.
• Start traversing all the array elements,
• If a[i] + position does not exceed the left and the right boundary, then the move will be “Right”.
• If position – a[i] does not exceed the left and the right boundary, then the move will be “Left”.
• If at any stage both the condition fail then print -1.
Below is the implementation of the above approach:
## C++
`// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` ` `// Function to print steps such that ` `// they do not cross the boundary ` `void` `printSteps(``int` `a[], ``int` `n, ``int` `k) ` `{ ` ` ` ` ``// To store the resultant string ` ` ``string res = ``""``; ` ` ` ` ``// Initially at zero-th position ` ` ``int` `position = 0; ` ` ``int` `steps = 1; ` ` ` ` ``// Iterate for every i-th move ` ` ``for` `(``int` `i = 0; i < n; i++) { ` ` ` ` ``// Check for right move condition ` ` ``if` `(position + a[i] <= k ` ` ``&& position + a[i] >= (-k)) { ` ` ``position += a[i]; ` ` ``res += ``"Right\n"``; ` ` ``} ` ` ` ` ``// Check for left move condition ` ` ``else` `if` `(position - a[i] >= -k ` ` ``&& position - a[i] <= k) { ` ` ``position -= a[i]; ` ` ``res += ``"Left\n"``; ` ` ``} ` ` ` ` ``// No move is possible ` ` ``else` `{ ` ` ``cout << -1; ` ` ``return``; ` ` ``} ` ` ``} ` ` ` ` ``// Print the steps ` ` ``cout << res; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `a[] = { 40, 50, 60, 40 }; ` ` ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` ` ``int` `k = 120; ` ` ``printSteps(a, n, k); ` ` ` ` ``return` `0; ` `} `
## Java
`// Java implementation of the approach ` `class` `GFG ` `{ ` ` ` `// Function to print steps such that ` `// they do not cross the boundary ` `static` `void` `printSteps(``int` `[]a, ``int` `n, ``int` `k) ` `{ ` ` ` ` ``// To store the resultant string ` ` ``String res = ``""``; ` ` ` ` ``// Initially at zero-th position ` ` ``int` `position = ``0``; ` ` ``//int steps = 1; ` ` ` ` ``// Iterate for every i-th move ` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``{ ` ` ` ` ``// Check for right move condition ` ` ``if` `(position + a[i] <= k ` ` ``&& position + a[i] >= (-k)) ` ` ``{ ` ` ``position += a[i]; ` ` ``res += ``"Right\n"``; ` ` ``} ` ` ` ` ``// Check for left move condition ` ` ``else` `if` `(position - a[i] >= -k ` ` ``&& position - a[i] <= k) ` ` ``{ ` ` ``position -= a[i]; ` ` ``res += ``"Left\n"``; ` ` ``} ` ` ` ` ``// No move is possible ` ` ``else` ` ``{ ` ` ``System.out.println(-``1``); ` ` ``return``; ` ` ``} ` ` ``} ` ` ` ` ``// Print the steps ` ` ``System.out.println(res); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ` ` ``int` `[]a = { ``40``, ``50``, ``60``, ``40` `}; ` ` ``int` `n = a.length; ` ` ``int` `k = ``120``; ` ` ``printSteps(a, n, k); ` `} ` `} ` ` ` `// This code is contributed by mits `
## Python3
`# Python3 implementation of the approach ` ` ` `# Function to print steps such that ` `# they do not cross the boundary ` `def` `printSteps(a, n, k): ` ` ` ` ``# To store the resultant string ` ` ``res ``=` `"" ` ` ` ` ``# Initially at zero-th position ` ` ``position ``=` `0` ` ``steps ``=` `1` ` ` ` ``# Iterate for every i-th move ` ` ``for` `i ``in` `range``(n): ` ` ` ` ``# Check for right move condition ` ` ``if` `(position ``+` `a[i] <``=` `k ``and` ` ``position ``+` `a[i] >``=` `-``k): ` ` ``position ``+``=` `a[i] ` ` ``res ``+``=` `"Right\n"` ` ` ` ``# Check for left move condition ` ` ``elif` `(position``-``a[i] >``=` `-``k ``and` ` ``position``-``a[i] <``=` `k): ` ` ``position ``-``=` `a[i] ` ` ``res ``+``=` `"Left\n"` ` ` ` ``# No move is possible ` ` ``else``: ` ` ``print``(``-``1``) ` ` ``return` ` ``print``(res) ` ` ` `# Driver code ` `a ``=` `[``40``, ``50``, ``60``, ``40``] ` `n ``=` `len``(a) ` `k ``=` `120` `printSteps(a, n, k) ` ` ` `# This code is contributed by Shrikant13 `
## C#
`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to print steps such that ` `// they do not cross the boundary ` `static` `void` `printSteps(``int` `[]a, ``int` `n, ``int` `k) ` `{ ` ` ` ` ``// To store the resultant string ` ` ``String res = ``""``; ` ` ` ` ``// Initially at zero-th position ` ` ``int` `position = 0; ` ` ``//int steps = 1; ` ` ` ` ``// Iterate for every i-th move ` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``{ ` ` ` ` ``// Check for right move condition ` ` ``if` `(position + a[i] <= k ` ` ``&& position + a[i] >= (-k)) ` ` ``{ ` ` ``position += a[i]; ` ` ``res += ``"Right\n"``; ` ` ``} ` ` ` ` ``// Check for left move condition ` ` ``else` `if` `(position - a[i] >= -k ` ` ``&& position - a[i] <= k) ` ` ``{ ` ` ``position -= a[i]; ` ` ``res += ``"Left\n"``; ` ` ``} ` ` ` ` ``// No move is possible ` ` ``else` ` ``{ ` ` ``Console.WriteLine(-1); ` ` ``return``; ` ` ``} ` ` ``} ` ` ` ` ``// Print the steps ` ` ``Console.Write(res); ` `} ` ` ` `// Driver code ` `static` `void` `Main() ` `{ ` ` ``int` `[]a = { 40, 50, 60, 40 }; ` ` ``int` `n = a.Length; ` ` ``int` `k = 120; ` ` ``printSteps(a, n, k); ` `} ` `} ` ` ` `// This code is contributed by mits `
## PHP
`= (-``\$k``)) ` ` ``{ ` ` ``\$position` `+= ``\$a``[``\$i``]; ` ` ``\$res` `.= ``"Right\n"``; ` ` ``} ` ` ` ` ``// Check for left move condition ` ` ``else` `if` `(``\$position` `- ``\$a``[``\$i``] >= -``\$k` `&& ` ` ``\$position` `- ``\$a``[``\$i``] <= ``\$k``) ` ` ``{ ` ` ``\$position` `-= ``\$a``[``\$i``]; ` ` ``\$res` `.= ``"Left\n"``; ` ` ``} ` ` ` ` ``// No move is possible ` ` ``else` ` ``{ ` ` ``echo` `-1; ` ` ``return``; ` ` ``} ` ` ``} ` ` ` ` ``// Print the steps ` ` ``echo` `\$res``; ` `} ` ` ` `// Driver code ` `\$a` `= ``array``( 40, 50, 60, 40 ); ` `\$n` `= ``count``(``\$a``); ` `\$k` `= 120; ` `printSteps(``\$a``, ``\$n``, ``\$k``); ` ` ` `// This code is contributed by mits ` `?> `
Output:
```Right
Right
Left
Right
```
My Personal Notes arrow_drop_up
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0
# How many decimeters equals 5 meters?
Wiki User
2016-10-20 23:30:44
There are 10 decimetres in one metre. Therefore, 5 decimetres is equal to 5/10, or one half, of a metre.
Wiki User
2010-01-26 00:46:45
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Wiki User
2017-04-15 16:44:59
1 metre = 10 decimetres so 5 metres = 5*10 = 10 decimetres. Simple!
Wiki User
2009-10-08 01:32:57
5 meters equal how many decimeters
Wiki User
2016-10-20 23:30:44
5 meters is 50 decimeters.
Wiki User
2010-02-04 11:43:30
5 metres = 50 decimetres
Lvl 2
2020-04-26 19:06:55
7 meters equals how many centimeters
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Mathematical models are used in science to better understand and make predictions of real world phemenon. Often they may help answer questions that cannot be answered by empirical data or experiments. When the phenomenon has elements of uncertainty, which are governed by probability laws, the model is called a probability or a stochastic model. Stochastic models are often characterized by observing a real world situation over time Examples of stochastic models in the biomedical sciences have been developed for therapeutic and early detection clinical trials, epidemics of infectious diseases, cell growth This course is an introduction to stochastic processes as applied to the biomedical sciences. Among the topics which will be discussed are: epidemiology models for incidence, prevalence and mortality, backward and forward recurrence times and their relationship to length biased sampling, Poisson processes, birth and death processes, Markov chains and semi-Markov processes. Instructor. M. Zelen (617-432-4914 ,617-632- 3013), Harvard University, USA zelen@hsph.harvard.edu An Introduction to Stochastic Processes in Public Health The viewing is best in pdf format ( it is necessary to click on "fit size" and then click on "control L") available from first slides of lectures. Stable disease model, forward and backward recurrence times, models for chronological time and age. (2) Exponential distribution normalized spacings, Campbell’s theorem, random sums of exponential random variables, counting processes and the exponential distribution, superposition of counting processes, splitting and component processes, non – homogeneous Poisson processes.(2) Definitions, asymptotics, renewal function, equilibrium renewal processes (1) 6. Pure birth processes (Yule-Furry process), generalization to birth and death processes, relationship to Markov chains, linear birth and death processes.(2) Introduction, Chapman-Kolmogorov equations, branching processes, statistical equilibrium, classification of states (2) 8. Master equations, moments, first passage time problems. Elements of the Analysis of Discrete DataJuly, 2000 This course consists of 8 lectures: The viewing is best in pdf format ( it is necessary to click on "fit size" and then click on "control L") available from HERE. Lectures in Supercourse format
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# Properties
Label 4.4.6125.1-55.3-b Base field 4.4.6125.1 Weight $[2, 2, 2, 2]$ Level norm $55$ Level $[55,55,w^{3} + 2w^{2} - 5w - 6]$ Dimension $4$ CM no Base change no
# Related objects
• L-function not available
## Base field 4.4.6125.1
Generator $$w$$, with minimal polynomial $$x^{4} - x^{3} - 9x^{2} + 9x + 11$$; narrow class number $$2$$ and class number $$1$$.
## Form
Weight: $[2, 2, 2, 2]$ Level: $[55,55,w^{3} + 2w^{2} - 5w - 6]$ Dimension: $4$ CM: no Base change: no Newspace dimension: $18$
## Hecke eigenvalues ($q$-expansion)
The Hecke eigenvalue field is $\Q(e)$ where $e$ is a root of the defining polynomial:
$$x^{4} + 5x^{3} - 5x^{2} - 21x + 12$$
Norm Prime Eigenvalue
5 $[5, 5, -2w^{3} - 2w^{2} + 13w + 8]$ $\phantom{-}1$
11 $[11, 11, 2w^{3} + 3w^{2} - 12w - 11]$ $\phantom{-}e$
11 $[11, 11, -w^{3} - 2w^{2} + 6w + 9]$ $-1$
11 $[11, 11, w^{3} + w^{2} - 7w - 4]$ $-\frac{1}{4}e^{3} - \frac{3}{2}e^{2} - \frac{1}{4}e + 3$
11 $[11, 11, w - 1]$ $\phantom{-}e$
16 $[16, 2, 2]$ $\phantom{-}\frac{1}{2}e^{3} + \frac{5}{2}e^{2} - e - 7$
19 $[19, 19, -w^{2} + 4]$ $\phantom{-}\frac{1}{4}e^{3} + e^{2} - \frac{13}{4}e - 5$
19 $[19, 19, 2w^{3} + 2w^{2} - 13w - 6]$ $-\frac{1}{2}e^{3} - \frac{5}{2}e^{2} + 4$
19 $[19, 19, 3w^{3} + 4w^{2} - 18w - 16]$ $-\frac{1}{2}e^{2} - \frac{3}{2}e - 2$
19 $[19, 19, -w^{3} - w^{2} + 5w + 3]$ $\phantom{-}\frac{1}{2}e^{2} + \frac{3}{2}e - 2$
49 $[49, 7, 2w^{3} + 3w^{2} - 13w - 15]$ $-\frac{1}{4}e^{3} - e^{2} - \frac{3}{4}e - 5$
59 $[59, 59, -3w^{3} - 4w^{2} + 19w + 14]$ $\phantom{-}e^{3} + \frac{9}{2}e^{2} - \frac{5}{2}e - 6$
59 $[59, 59, 3w^{3} + 4w^{2} - 18w - 17]$ $\phantom{-}\frac{3}{2}e^{2} + \frac{15}{2}e - 6$
59 $[59, 59, 2w^{3} + 3w^{2} - 11w - 13]$ $-\frac{1}{2}e^{3} - \frac{3}{2}e^{2} + 3e$
59 $[59, 59, -w^{3} - 2w^{2} + 7w + 9]$ $-\frac{1}{4}e^{3} - e^{2} - \frac{7}{4}e - 3$
71 $[71, 71, 3w^{3} + 3w^{2} - 17w - 10]$ $\phantom{-}e^{3} + \frac{13}{2}e^{2} + \frac{5}{2}e - 18$
71 $[71, 71, -2w^{3} - w^{2} + 14w + 2]$ $-\frac{3}{2}e^{2} - \frac{9}{2}e + 6$
71 $[71, 71, 5w^{3} + 7w^{2} - 30w - 25]$ $\phantom{-}\frac{1}{4}e^{3} - \frac{13}{4}e + 3$
71 $[71, 71, -3w^{3} - 4w^{2} + 16w + 16]$ $-\frac{1}{4}e^{3} + \frac{17}{4}e - 3$
81 $[81, 3, -3]$ $-\frac{3}{4}e^{3} - 2e^{2} + \frac{27}{4}e + 1$
Display number of eigenvalues
## Atkin-Lehner eigenvalues
Norm Prime Eigenvalue
$5$ $[5,5,-w^{3} - w^{2} + 5w + 5]$ $-1$
$11$ $[11,11,-w^{3} - 2w^{2} + 6w + 9]$ $1$
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• Create Account
## Efficient Line of Sight Algorithm
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### #1longshorts Members
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Posted 12 April 2012 - 02:02 PM
I am currently trying to devise a good algorithm to test if point A is in line of sight of point B. It is for use in a 2D Tank game I am developing in C++ using SFML. Here is a video showing how my algorithm is currently functioning:
The problem I am trying to solve is that I want many (30+) enemies to track the player's location and react accordingly when the player is in line of sight (for example, fire at the player). Therefore the alorithm needs to be quite efficent. Currently what I am doing is making a sprite at the enemy's location, then fire it towards the player's location. As the sprite moves, it samples it's current location to check if it has either collided with an obsticle (where the method returns false) or it's target (returning true). In the video, the points at which the sprite is sampling is represented by yellow dots.
I am wanting to then ask two questions: is there a more efficient way of working out line of sight? Also, if someone could check my maths that would be extreemly helpful. The tracers dont seem to hit the player's center always so I know Im calculating something wrong.
Thanks for any help you can give, Im probably getting into stuff way too complex for me. This is in fact my first game and first C++ program. Yet I still need to go on to figure out how to do A* Jump point search! Argh!
The code for working out LOS is below:
#include "stdafx.h"
#include "Pathfinder.h"
#include "Game.h"
#define PI 3.14159265
//Returns true if target is in line of sight
bool Pathfinder::LineOfSight(float startx, float starty, float endx, float endy)
{
_tracer.SetPosition(startx, starty);
//Get the angle between the start and end point
float angle = GetAngle(startx, starty, endx, endy);
//Calculate movement in x y coords depending on rotation
float xvel = 1;
float yvel = 1;
if(angle >= 0 && angle < 90) //q1
{
xvel = (angle/90);
yvel = 1 - (angle/90);
}
if(angle >= 90 && angle < 180) //q2
{
xvel = 1 - ((angle - 90)/90);
yvel = 0 - ((angle - 90)/90);
}
else if(angle >= 180 && angle < 270) //q3
{
xvel = 0 - ((angle - 180)/90);
yvel = ((angle - 180)/90) - 1;
}
else if (angle >= 270 && angle < 360) //q4
{
xvel = ((angle - 270)/90) - 1;
yvel = ((angle - 270)/90);
}
float increment = 15; //Distance traveled before sampling
float error = 10; //Max error when targeting end point
//While tracer is not at goal
while(((_tracer.GetPosition().x < endx - error) ||
(_tracer.GetPosition().x > endx + error)) &&
((_tracer.GetPosition().y < endy - error) ||
(_tracer.GetPosition().y > endy + error)))
{
//Move the tracer to next sampling point
_tracer.Move(xvel*(increment), yvel*(increment));
if(_isLoaded) //Draw tracer at this location
Game::GetWindow().Draw(_tracer);
//Check if the tracer collides with a impassable
//object on the tilemap
if(Game::GetGOM().GetTileMap()->Collision(_tracer.GetPosition()))
{
return false;
}
}
return true;
}
//Get angle between two points
float Pathfinder::GetAngle(float startx, float starty, float endx, float endy)
{
float mX = endx;
float mY = endy;
mX = mX - startx;
mY = mY - starty;
return atan2(-mY,mX) * (180 / PI) + 90;
}
{
{
_filename = "";
}
else
{
_filename = filename;
_tracer.SetImage(_image);
}
}
sf::Sprite Pathfinder::_tracer;
sf::Image Pathfinder::_image;
std::string Pathfinder::_filename;
bool Pathfinder::_isLoaded = false;
### #2slicer4ever GDNet+
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Posted 12 April 2012 - 02:22 PM
since you are only doing a single ray testing, it's extremely easy to test line of sight.
psedo-code:
void LineOfSight(PlayerPosition, EnemyPosition){
pointA = EnemyPosition;
pointB = PlayerPosition;
for(i=0;i<NbrObstacles;i++){
//Check linexObstacle(if your using rectangle's look up linexRectangle, for all simple shapes, their's simple formula, for complex shapes, you simple loop though all the edges, and check linexPlane(in 2D, it's basically linexline collision).)
if(IntersectionWithObstacleI) PointB = Intesect location; //(if we just care if the other tank can't see the player, you could return false for a line of sight function here.
}
return PointB==PlayerPosition?1:0;
}
that's the jist of it, for line of sight with cones/complex vision's, then it gets quite a bit more complicated.
Check out https://www.facebook.com/LiquidGames for some great games made by me on the Playstation Mobile market.
### #3ApochPiQ Moderators
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Posted 12 April 2012 - 02:25 PM
Yep, ray casting is definitely the way to go. Since you're just doing ray/bounding-box checks it should be easy to implement and very fast.
Wielder of the Sacred Wands
### #4longshorts Members
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Posted 12 April 2012 - 03:33 PM
Thanks for the help guys, I had a feeling I was overcomplicating the problem! I wasnt sure how ray casting was implemented so that pseudo-code is very useful. Ill give it a shot in the near future, using this answer for the line rectangle intersection logic: http://stackoverflow.com/questions/99353/how-to-test-if-a-line-segment-intersects-an-axis-aligned-rectange-in-2d
On another note, can any of you point me to a easy to understand guide on A* Jump point search? Seems the most efficient method for working out the enemy's path to the player. But I could be wrong, how else would you work that out in this scenario? (bearing in mind the path should be constantly recalculated as the player moves).
### #5slicer4ever GDNet+
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Posted 12 April 2012 - 07:14 PM
Thanks for the help guys, I had a feeling I was overcomplicating the problem! I wasnt sure how ray casting was implemented so that pseudo-code is very useful. Ill give it a shot in the near future, using this answer for the line rectangle intersection logic: http://stackoverflow...-rectange-in-2d
On another note, can any of you point me to a easy to understand guide on A* Jump point search? Seems the most efficient method for working out the enemy's path to the player. But I could be wrong, how else would you work that out in this scenario? (bearing in mind the path should be constantly recalculated as the player moves).
Jump point search?
i'm not certain if your asking for a specefic A* algorithm, or asking for a tutorial on A* in general. if it's the latter, http://www.policyalmanac.org/games/aStarTutorial.htm <-- a great starting place for A* imo, and taught me pretty much what i needed to know when i started learning A*.
Check out https://www.facebook.com/LiquidGames for some great games made by me on the Playstation Mobile market.
### #6longshorts Members
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Posted 12 April 2012 - 07:48 PM
Ive read that and its simple to understand, but for what Im trying to do pure A* is simply too costly. For a rundown on jump point search read http://harablog.wordpress.com/2011/09/07/jump-point-search/ which gives a quick overview on it. It seems to be a lot faster version of A*.
### #7slicer4ever GDNet+
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Posted 12 April 2012 - 08:03 PM
i'll take a look at your link, seems very interesting. on a side note, i'd also suggest looking at binary heaps if A* is proving to be slow. I've found that this is an excellent method for speeding up A* 10 fold without having to do much other work. Their's a handy link to the implementation in the tutorial i posted earlier.
anywho, i'll chime back in once i've read and understand the article you've linked to.
Check out https://www.facebook.com/LiquidGames for some great games made by me on the Playstation Mobile market.
### #8ApochPiQ Moderators
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Posted 12 April 2012 - 08:05 PM
A* is not too slow.
A particular implementation may make some trivial ease-of-programming tradeoffs that sacrifice huge amounts of speed, but fundamentally, the algorithm is not the problem. I've written A* implementations that can search hundreds of thousands of nodes in a handful of milliseconds; if that's not fast enough, you probably have some major opportunities for optimizing your design.
Wielder of the Sacred Wands
### #9ssrun Members
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Posted 12 April 2012 - 09:23 PM
Personally, I've found that the fastest line of sight algorithm is Bresenham's line plotting algorithm since it's completely based on integers.
### #10ApochPiQ Moderators
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Posted 12 April 2012 - 10:26 PM
That's... somewhat akin to saying that wine is the most flavorful beverage because it's completely based on grapes.
It's also patently false.
A 2D ray/box intersection is a trivial algorithm, even for non-axis-aligned boxes:
- For each edge
- Check if the ray is parallel to the edge; if so, skip to the next edge
- Compute the intersection point of the ray and the infinite line along which the edge lies
- If the intersection lies outside the endpoints of the edge, skip to the next edge
- Otherwise, return a positive collision
- If you get this far, return negative
This is, in the worst case:
- Four conditionals for parallel checks
- A handful of additions and divisions
This can even be unrolled so it doesn't even include a loop, making it ideal for CPUs that don't do good branch prediction or have generally superscalar behavior and can compute faster in unrolled circumstances - i.e. most CPUs these days.
By contrast, consider Bresenham:
- At minimum a loop
- Two additions and four conditionals (eight if you want to be pedantic) per iteration of the loop
- If you do this in entirely integer math, it's even harder, so you're kind of screwed; good line marching algorithms at least are fixed-point
- Non-deterministic running time depending on the distance between the ray origin and the object
- Additional overhead for directional checks so you don't march away from the target infinitely before giving up
By the time you optimize a line marching algorithm - Bresenham or otherwise - to not do really silly things, you're basically duplicating a ray/box check anyways, and might as well just write the simpler code.
If that was overly technical, consider this case:
- A ray originating 5 million units from the bounding box you wish to test
A ray/box test can solve this in basically constant time. Naive Bresenham needs at least 5 million iterations. Now, point the ray the opposite direction (i.e. so it never hits the box at all) and you have an even more pathological scenario.
Wielder of the Sacred Wands
### #11ApochPiQ Moderators
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Posted 12 April 2012 - 10:57 PM
I will concede, though, that if you are:
- In a regular 2D grid (op: check)
- Using only perfectly grid-aligned obstacles (op: bzzzt, nope, tanks are not grid-aligned)
- Populating a majority of the grid cells with obstacles (op: bzzzt, nope)
- Casting fewer LOS checks than obstacles by an order of magnitude (op: bzzzt, nope)
You might be able to win out with a line marching algorithm. But outside of that contrived scenario, spatial partitioning and ray casting will almost certainly be faster, and will most certainly be more robust and flexible. If you do geometry coalescing so that a single long rectangular wall is one rect instead of an obstacle in each grid cell, you get even better gains using ray casting, and as soon as you no longer have perfectly-grid aligned obstacles you can forget getting 100% accurate results from a line march algorithm.
Wielder of the Sacred Wands
### #12lawnjelly Members
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Posted 14 April 2012 - 03:31 PM
Don't know if anyone has mentioned it yet .. the fastest way to determine line of sight is to precalculate it.
Three most important rules in game programming - precalculate, precalculate and precalculate.
If you are using a smallish grid like that you can easily generate a PVS (potentially visible set) of what other cells are potentially visible from each cell. This greatly cuts down on the checks you need to do.
e.g. if enemy 1 is in cell 20, 24 and your pvs tells you that the cell your player is in 90.34 is not visible from there due to a wall in the way, that's it, you know it's not in sight. You only end up needing to do realtime checks for dynamic objects that might get in the way.
Favourite datatype: unsinged int
### #13slicer4ever GDNet+
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Posted 14 April 2012 - 06:11 PM
Don't know if anyone has mentioned it yet .. the fastest way to determine line of sight is to precalculate it.
Three most important rules in game programming - precalculate, precalculate and precalculate.
If you are using a smallish grid like that you can easily generate a PVS (potentially visible set) of what other cells are potentially visible from each cell. This greatly cuts down on the checks you need to do.
e.g. if enemy 1 is in cell 20, 24 and your pvs tells you that the cell your player is in 90.34 is not visible from there due to a wall in the way, that's it, you know it's not in sight. You only end up needing to do realtime checks for dynamic objects that might get in the way.
while i agree for more complex scenarios, pre-calculating line of sight would be beneficial. In the scenario the OP has described, i find it doubtful he'll need to pre-calculate the line of sight for any serious performance gains. as well, creating a grid based PVS with objects that arn't locked to the grid, their's the small possibility that valid line of sight's could be missed with this method.
Check out https://www.facebook.com/LiquidGames for some great games made by me on the Playstation Mobile market.
### #14leopardpm Members
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Posted 14 April 2012 - 11:10 PM
Don't know if anyone has mentioned it yet .. the fastest way to determine line of sight is to precalculate it.
Three most important rules in game programming - precalculate, precalculate and precalculate.
If you are using a smallish grid like that you can easily generate a PVS (potentially visible set) of what other cells are potentially visible from each cell. This greatly cuts down on the checks you need to do.
e.g. if enemy 1 is in cell 20, 24 and your pvs tells you that the cell your player is in 90.34 is not visible from there due to a wall in the way, that's it, you know it's not in sight. You only end up needing to do realtime checks for dynamic objects that might get in the way.
sorry, but my simplistic mind is a bit confused here... to 'pre-calc' LOS for a 10 x 10 grid would necessitate storing a 10x10 table for EACH location (100 locations)... or am I missing something here? Perhaps I am overly paranoid about memory usage. Of course, there are a variety of ways to start pruning this memory usage down, for instance eliminating redundant pairs: if loc (1,1) can see loc (5,3) then no need to also store the fact that (5,3) can also see loc (1,1) - that alone will reduce memory usage by half.
Did I misunderstand the whole pre-calc thing or what?
### #15lawnjelly Members
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Posted 15 April 2012 - 04:19 AM
Yeah, sorry that's posting when tired lol.
Sure to do it you have to partition space into something reasonable memory wise. Whether you can use your grid directly depends how big your maps are. Your screen there looks to be about 20x20 so let's use that for an example:
20x20 = 400 bits per grid square for a naive PVS = 50 bytes
then 50 bytes x 400 grid squares = 20000 bytes, so under 20K, so not that memory hungry.
That's very naive though .. you can also cut this by half if your PVS is reciprocal (i.e. if you can't see 40,40 from 20, 20, then you also know that you can't see 20, 20 from 40,40)
You could also use RLE if memory really was a problem (although the cost of decompressing this probably wouldn't be worth it unless you have a complex environment lol).
Then other ways is to use different partitioning for your space. If for example you used a 10x10 grid for the PVS instead of 20x20, that quarters the memory size. Or use a 2d BSP tree maybe, not sure. Grid sounds more tempting, if you keep it small.
As to how to handle the case of tanks being in different 'parts' of the gridsquare, and the visibility being 1 in some parts and 0 in others, just mark it as invisible ONLY when there are no parts of the grid squares that are visible to each other.
Then if it is marked as potentially visible, do a runtime check. For enemy AI often you can afford to be lazy about this, and there's no reason you need to do vis checks every frame. You could also store 3 (or 4) states in the PVS, instead of just 2 - i.e. definitely not visible, maybe visible, definitely visible, that way you only have to do runtime checks in say 2% of the grid squares.
Given that your current grid is only using 20K, I'd be tempted to just increase the resolution of the PVS grid by about 5x and then just go with the definitely visible / definitely invisible approach, and forego the accuracy.
For a BSP tree you may be able to be more exact than this.
You don't even need to use a BSP tree .. you could use something as simple as AA bounding boxes and probably still get a decent result, the only limit is your imagination!
And this is pretty much one of the simplest situations for using a PVS. There are some really good solutions in 3d for far more complex environments (see quake stuff using BSP and PVS, or portal engines, etc etc). If you do some googling you should get some good ideas.
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### #16lawnjelly Members
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Posted 15 April 2012 - 04:45 AM
Forgot to mention, you can also precalculate your A* pathfinding too. But it won't take account of dynamic obstacles, you'd need to cope with that separately.
This is in fact my first game and first C++ program.
Sorry, just read this .. in this case I'd stick to the simple stuff as dealing with the special cases in visibility determination can be quite tricky. But feel free to have a google as it's really interesting stuff and the kind of thing that most professional games use, as they have to determine visibility (mainly for deciding what to draw) really quickly, on really complex environments.
Awesome effort for your first go!
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Posted 20 April 2012 - 06:40 PM
You seem to be in 2d. If you just build a simplified representation of your space (navmesh, or since you seem to be grid-based a quadtree) should make your line of sight checks negligible. There is no reason you shouldnt be able to do hundreds of those every frame.
Also, looking at your original code up there, it seems you're working in polar coordinates (angle/distance). Believe my experience when I say its evil. Work with vector algrebra.
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## NY-AI-A.REI.6a
There are 28 resources.
Title Description Thumbnail Image Curriculum Topics
## VIDEO: Algebra Applications: Systems of Equations
### In this episode of Algebra Applications, students explore various scenarios that can be explained through the use of systems of equations. Such disparate phenomena as profit and loss, secret codes, and ballistic missile shields can be explored through systems of equations. This video includes a Video Transcript: https://www.media4math.com/library/video-transcript-algebra-applications-systems-equations
Applications of Linear Systems, Solving Systems of Equations
## VIDEO: Algebra Applications: Systems of Equations, Segment 1: Profit and Loss
### Profit and loss are the key measures in a business. A system of equations that includes an equation for income and one for expenses can be used to determine profit and loss. Students solve a system graphically. This video includes a Video Transcript: https://www.media4math.com/library/video-transcript-algebra-applications-systems-equations-segment-1-profit-and-loss A Promethean Flipchart is available for this video: https://www.media4math.com/library/promethean-flipchart-algebra-applications-profit-and-loss
Applications of Linear Systems, Solving Systems of Equations
## VIDEO: Algebra Applications: Systems of Equations, Segment 2: Encryption
### Secret codes and encryption are ideal examples of a system of equations. In this activity, students encrypt and decrypt a message. This video includes a Video Transcript: https://www.media4math.com/library/video-transcript-algebra-applications-systems-equations-segment-2-encryption A Promethean Flipchart is available for this video: https://www.media4math.com/library/promethean-flipchart-algebra-applications-encryption
Applications of Linear Systems, Solving Systems of Equations
## VIDEO: Algebra Applications: Systems of Equations, Segment 3: Ballistic Missiles
### A ballistic missile shield allows you to shoot incoming missiles out of the sky. Mathematically, this is an example of a linear-quadratic system. Students graph such a system and find the points of intersection between a line and a parabola. This video includes a Video Transcript: https://www.media4math.com/library/video-transcript-algebra-applications-systems-equations-segment-3-ballistic-missiles A Promethean Flipchart is available for this video: https://www.media4math.com/library/promethean-flipchart-algebra-applications-ballistic-missiles
Applications of Linear Systems, Solving Systems of Equations
## VIDEO: Algebra Nspirations: Solving Systems of Equations
### Written and hosted by internationally acclaimed math educator Dr. Monica Neagoy, this video introduces students to systems of linear equations in two or three unknowns. To solve these systems, the host illustrates a variety of methods: four involve the TI-Nspire (spreadsheet, graphs and geometry, matrices and nSolve) and two are the classic algebraic methods known as substitution and elimination, also called the linear combinations method. The video ends with a summary of the three possible types of solutions. Concepts explored: equations, linear equations, linear systems. This video includes a video transcript: https://media4math.com/library/video-transcript-algebra-nspirations-solving-systems-equations A Promethean Flipchart is available for this video: https://www.media4math.com/library/promethean-flipchart-algebra-nspirations-systems-equations
Applications of Linear Systems, Solving Systems of Equations
## VIDEO: Algebra Nspirations: Solving Systems of Equations, Segment 1
### In this Investigation we solve a linear system. This video is Segment 1 of a 4 segment series related to Algebra Nspirations: Solving Systems of Equations. Segments 1 and 2 are grouped together. To access Algebra Nspirations: Solving Systems of Equations, Segment 2, click the following link: • https://www.media4math.com/library/algebra-nspirations-solving-systems-equations-segment-2 A Video Transcript for Algebra Nspirations: Solving Systems of Equations, Segments 1 and 2 is available via the following link: • https://www.media4math.com/library/video-transcript-algebra-nspirations-solving-systems-equations-part-1
Applications of Linear Systems, Solving Systems of Equations
## VIDEO: Algebra Nspirations: Solving Systems of Equations, Segment 2
### In this Math Lab, we explore the break-even point for a business. This video is Segment 2 of a 4 segment series related to Algebra Nspirations: Solving Systems of Equations. Segments 1 and 2 are grouped together. To access Algebra Nspirations: Solving Systems of Equations, Segment 1, click the following link: • https://www.media4math.com/library/algebra-nspirations-solving-systems-equations-segment-1 A Video Transcript for Algebra Nspirations: Solving Systems of Equations, Segments 1 and 2 is available via the following link: • https://www.media4math.com/library/video-transcript-algebra-nspirations-solving-systems-equations-part-1
Applications of Linear Systems, Solving Systems of Equations
## VIDEO: Algebra Nspirations: Solving Systems of Equations, Segment 3
### In this Investigation we use matrices and the elimination method to solve a linear system. This video is Segment 3 of a 4 segment series related to Algebra Nspirations: Solving Systems of Equations. Segments 3 and 4 are grouped together. To access Algebra Nspirations: Solving Systems of Equations, Segment 4, click the following link: • https://www.media4math.com/library/algebra-nspirations-solving-systems-equations-segment-4 A Video Transcript for Algebra Nspirations: Solving Systems of Equations, Segments 3 and 4 is available via the following link: • https://www.media4math.com/library/video-transcript-algebra-nspirations-solving-systems-equations-part-2
Applications of Linear Systems, Solving Systems of Equations
## VIDEO: Algebra Nspirations: Solving Systems of Equations, Segment 4
### In this Math Lab we explore the graphical solution of a nonlinear system. This video is Segment 4 of a 4 segment series related to Algebra Nspirations: Solving Systems of Equations. Segments 3 and 4 are grouped together. To access Algebra Nspirations: Solving Systems of Equations, Segment 3, click the following link: • https://www.media4math.com/library/algebra-nspirations-solving-systems-equations-segment-3 A Video Transcript for Algebra Nspirations: Solving Systems of Equations, Segments 3 and 4 is available via the following link: • https://www.media4math.com/library/video-transcript-algebra-nspirations-solving-systems-equations-part-2
Applications of Linear Systems, Solving Systems of Equations
## Closed Captioned Video: Algebra Applications: Systems of Equations
### In this episode of Algebra Applications, students explore various scenarios that can be explained through the use of systems of equations. Such disparate phenomena as profit and loss, secret codes, and ballistic missile shields can be explored through systems of equations. A Video Transcript is available for this tutorial at this Link Note: The download is Media4Math's guide to closed captioned videos. ## Other Closed Captioned Videos To see the complete collection of Closed Captioned Videos, click on this Link
Applications of Linear Systems, Solving Systems of Equations
## Closed Captioned Video: Algebra Applications: Systems of Equations, Segment 1: Profit and Loss
### Profit and loss are the key measures in a business. A system of equations that includes an equation for income and one for expenses can be used to determine profit and loss. Students solve a system graphically. A Video Transcript is available for this tutorial at this Link Note: The download is Media4Math's guide to closed captioned videos. ## Other Closed Captioned Videos To see the complete collection of Closed Captioned Videos, click on this Link
Applications of Linear Systems, Solving Systems of Equations
## Closed Captioned Video: Algebra Applications: Systems of Equations, Segment 2: Encryption
### Secret codes and encryption are ideal examples of a system of equations. In this activity, students encrypt and decrypt a message. A Video Transcript is available for this tutorial at this Link Note: The download is Media4Math's guide to closed captioned videos. ## Other Closed Captioned Videos To see the complete collection of Closed Captioned Videos, click on this Link
Applications of Linear Systems, Solving Systems of Equations
## Closed Captioned Video: Algebra Applications: Systems of Equations, Segment 3: Ballistic Missiles
### A ballistic missile shield allows you to shoot incoming missiles out of the sky. Mathematically, this is an example of a linear-quadratic system. Students graph such a system and find the points of intersection between a line and a parabola. A Video Transcript is available for this tutorial at this Link Note: The download is Media4Math's guide to closed captioned videos. ## Other Closed Captioned Videos To see the complete collection of Closed Captioned Videos, click on this Link
Applications of Linear Systems, Solving Systems of Equations
## Closed Captioned Video: Algebra Nspirations: Solving Systems of Equations
### Written and hosted by internationally acclaimed math educator Dr. Monica Neagoy, this video introduces students to systems of linear equations in two or three unknowns. To solve these systems, the host illustrates a variety of methods: four involve the TI-Nspire (spreadsheet, graphs and geometry, matrices and nSolve) and two are the classic algebraic methods known as substitution and elimination, also called the linear combinations method. The video ends with a summary of the three possible types of solutions. Concepts explored: equations, linear equations, linear systems. A Video Transcript is available for this tutorial at this Link Note: The download is Media4Math's guide to closed captioned videos. ## Other Closed Captioned Videos To see the complete collection of Closed Captioned Videos, click on this Link
Applications of Linear Systems, Solving Systems of Equations
## Closed Captioned Video: Algebra Nspirations: Solving Systems of Equations, Segment 1
### In this Investigation we solve a linear system. This video is Segment 1 of a 4 segment series related to Algebra Nspirations: Solving Systems of Equations. Segments 1 and 2 are grouped together. To access Algebra Nspirations: Solving Systems of Equations, Segment 2, click the following link: • https://www.media4math.com/library/algebra-nspirations-solving-systems-equations-segment-2 A Video Transcript is available for this tutorial at this Link Note: The download is Media4Math's guide to closed captioned videos. ## Other Closed Captioned Videos To see the complete collection of Closed Captioned Videos, click on this Link
Applications of Linear Systems, Solving Systems of Equations
## Closed Captioned Video: Algebra Nspirations: Solving Systems of Equations, Segment 3
### In this Investigation we use matrices and the elimination method to solve a linear system. This video is Segment 3 of a 4 segment series related to Algebra Nspirations: Solving Systems of Equations. Segments 3 and 4 are grouped together. To access Algebra Nspirations: Solving Systems of Equations, Segment 4, click the following link: • https://www.media4math.com/library/algebra-nspirations-solving-systems-equations-segment-4 A Video Transcript is available for this tutorial at this Link Note: The download is Media4Math's guide to closed captioned videos. ## Other Closed Captioned Videos To see the complete collection of Closed Captioned Videos, click on this Link
Applications of Linear Systems, Solving Systems of Equations
## Math in the News: Issue 108--Star Wars: The Box Office Awakens!
### January 2016. In this issue of Math in the News we explore the dramatic box office returns of Star Wars: The Force Awakens. It provides an excellent opportunity to apply concepts of linear systems. Note: The download is a PPT. To see all the issues of Math in the News, click on this link.
Applications of Linear Systems, Data Analysis
## Promethean Flipchart: Algebra Nspirations: Systems of Equations
### Written and hosted by internationally acclaimed math educator Dr. Monica Neagoy, this video introduces students to systems of linear equations in two or three unknowns. To solve these systems, the host illustrates a variety of methods: four involve the TI-Nspire (spreadsheet, graphs and geometry, matrices and nSolve) and two are the classic algebraic methods known as substitution and elimination, also called the linear combinations method. The video ends with a summary of the three possible types of solutions. Concepts explored: equations, linear equations, linear systems Note: The download for this resources is the Promethean Flipchart. To access the full video [Algebra Nspirations: Solving Systems of Equations]: https://media4math.com/library/algebra-nspirations-solving-systems-equations To access the video transcript [Algebra Nspirations: Solving Systems of Equations]: https://www.media4math.com/library/video-transcript-algebra-nspirations-solving-systems-equations
Applications of Linear Systems
## Related Resources
### To see the complete Quizlet Flash Card Library, click on this link: https://media4math.com/Quizlet-Resources
Solving Systems of Equations
## Related Resources
### To see the complete Quizlet Flash Card Library, click on this link: https://media4math.com/Quizlet-Resources
Solving Systems of Equations
## Related Resources
### To see the complete Quizlet Flash Card Library, click on this link: https://media4math.com/Quizlet-Resources
Solving Systems of Equations
## Video Transcript: Algebra Applications: Systems of Equations
### This is the transcript for the video of same title. Video contents: In this episode of Algebra Applications, students explore various scenarios that can be explained through the use of systems of equations. Such disparate phenomena as profit and loss, secret codes, and ballistic missile shields can be explored through systems of equations. To view the full video: https://www.media4math.com/library/algebra-applications-systems-equations
Applications of Linear Systems, Matrix Operations, Solving Systems of Equations
## Video Transcript: Algebra Applications: Systems of Equations, Segment 1: Profit and Loss
### This is the transcript for the video of same title. Video contents: Profit and loss are the key measures in a business. A system of equations that includes an equation for income and one for expenses can be used to determine profit and loss. Students solve a system graphically. To view the full video: https://www.media4math.com/library/algebra-applications-systems-equations-segment-1-profit-and-loss A Promethean Flipchart is available for this video: https://www.media4math.com/library/promethean-flipchart-algebra-applications-profit-and-loss
Applications of Linear Systems, Matrix Operations, Solving Systems of Equations
## Video Transcript: Algebra Applications: Systems of Equations, Segment 2: Encryption
### This is the transcript for the video of same title. Video contents: Secret codes and encryption are ideal examples of a system of equations. In this activity, students encrypt and decrypt a message. To view the full video: https://www.media4math.com/library/algebra-applications-systems-equations-segment-2-encryption A Promethean Flipchart is available for this video: https://www.media4math.com/library/promethean-flipchart-algebra-applications-encryption
Applications of Linear Systems, Matrix Operations, Solving Systems of Equations
## Video Transcript: Algebra Applications: Systems of Equations, Segment 3: Ballistic Missiles
### This is the transcript for the video of same title. Video contents: A ballistic missile shield allows you to shoot incoming missiles out of the sky. Mathematically, this is an example of a linear-quadratic system. Students graph such a system and find the points of intersection between a line and a parabola. To view the full video: https://www.media4math.com/library/algebra-applications-systems-equations-segment-3-ballistic-missiles A Promethean Flipchart is available for this video: https://www.media4math.com/library/promethean-flipchart-algebra-applications-ballistic-missiles
Applications of Linear Systems, Matrix Operations, Solving Systems of Equations
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https://openoregon.pressbooks.pub/layoutformetals/chapter/rectangular-sleeve/
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# 2 Rectangular Sleeve
For this example we want to build a template for a “chimney” or sleeve that will rest on a sloped roof surface. The roof surface will have a slope of 30° and the sleeve will have the dimensions of the drawing below.
Now that we have needed measurements we can begin to draw the stretch-out.
To help keep everything in order it helps to number points on stretch-out so they match up to each corner. Since this part will have a seam, we can use “0” to indicate the seam. Looking at the side & top view we can see that the points 1&4 have the same height as well as points 2&3 since they are in the same plane. The blue lines are there to show how the lines to extend to help develop the stretch-out.
Next we will transfer the measurements from the side view to the stretch-out to determine the shape of the template.
We know at points 0, 1 & 4, the height is the overall height measure at 9.46”, since all those point are in the same plane in the side view. For points 2 & 3 we can see that the overall height there is 6”. We need to transfer this to the stretch-out. There are a few ways to this function. You can simply measure 6” down from each point on stretch-out at points 2 & 3 and mark it with a tic mark, you can also transfer the line 90° to the right and where the horizontal line intersects points 2 & 3 place a tic mark.
Now we can add a couple lines connecting the base of the stretch-out to vertical line 2 & 3 to show what the final shape of stretch-out will be. We can also remove any lines we used to transfer features from side view to stretch-out. The green line indicate the sheet size and shape the red lines indicate where the sheet will be formed.
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https://physics.stackexchange.com/questions/353495/is-it-obvious-if-an-excision-from-minkowski-spacetime-breaks-isometry
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# Is it obvious, if an excision from Minkowski spacetime breaks isometry...?
Strongly related to the question Prove isometry preserving excision is Killing-like?
There the question was (loosely): if I drill a smooth 4D hole through Minkowski spacetime, is a timelike isometry preserved only if the hold drilled is everywhere parallel to (bounded by) integral curves of the vector field generating the isometry?
@Valter-Moretti re-expressed the proposition rather better and then proved it in exemplary fashion.
I was just writing up some notes on this and then wrote (assuming the hole boundary is in fact smooth),
"it is obvious, however, that if the isometry condition is relaxed, i.e. the metric is allowed to become non-flat, then diffeomorphism between spacelike hypersurfaces may be retained."
And then - in classic fashion - wondered, "Obvious?"
Hence the question: is it "obvious"??
(If so... from what perspective? :) )
UPDATE - SHOWING SOME EFFORT
After further thought, how about this (also referencing the Q&A What's the difference between a manifold and a topological space?).
In answer to the question Are homeomorphic differentiable manifolds actually diffeomorphic?, @Mariano-Suárez-Álvarez said,
Now, if f:M→N is a homeomorphism of smooth manifolds, you can always «adjust» the smooth structures so that f becomes a diffeo: indeed,
and offered the OP there the suggestion
you should be able to prove the following: if M is a smooth manifold, N a topological space and f:M→N a homeomorphism, then there is a structure of smooth manifold on N such that f becomes a diffeomorphism.
Now, since a manifold is a topological space with additional structure, it would seem that for any pair of spacelikes hypersurface, a) they are homeomorphic by construction above, and b) both are already (sub)manifolds and hence toplogical spaces, so it follows immediately (?).
(I do not have the skill to generate the proof referred to though)
• But, but, if you drill a hole through Minkowski, you'll get Schwarzschild! Aug 24 '17 at 15:00
• @Dr.IkjyotSinghKohli No, a) singularity (if considered the excision) $\times \mathbb{R}$ is 2D not 4D, the hypothesis), b) No reference to the Einstein Field Equations so the metric is unconstrained in that way, c) the excision is merely smooth, no symmetry constraint was imposed, and even if EFEs applied, Schwarzschild is a spatially spherically symmetric solution, d) even if EFE and spherically symmetric there are other solutions: e.g. Kerr, Kerr-Newman. Aug 25 '17 at 6:57
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# How To Return The Mode Of A Range In Excel
by Jacky Chou
Updated on
## Key Takeaway:
• Defining mode in Excel: Mode is the most frequently occurring value in a range of values. In Excel, mode is commonly used to find the value that occurs most frequently in a set of data.
• Using the MODE function: Excel’s MODE function simplifies the process of finding the mode of a range. By entering the cell range as an argument in the MODE function, Excel returns the mode value for that range.
• Applying the MODE function to a range of values: When applying the MODE function to a range of values in Excel, it is important to make sure that the range is properly selected to ensure accurate results. The MODE function in Excel is a valuable tool for quickly finding the mode of a range of values.
Are you struggling to find the MODE of a range in Excel? This article provides an easy and effective way for you to unlock the power of Excel and quickly and accurately return the MODE of any range of data.
## Finding the Mode of a Range in Excel
Find the mode of a range using Excel? Solutions!
Define mode in Excel.
Use the MODE function.
Apply the MODE function to a range of values.
We’ll explain each in detail. Choose the best one for your needs!
Image credits: chouprojects.com by Yuval Duncun
### Defining the Mode in Excel
When working with data in Excel, you might need to find the most frequently occurring value in a range of values. This process is called finding the mode. By calculating the mode, you can gain insights into the central tendency of your data and identify any outliers that might be skewing your analysis.
To define the mode in Excel, take a look at the following table:
Range of ValuesModeExplanation
2, 4, 5, 6, 8In this range of values, there are no repeating values. Therefore, there is no mode.
3, 5, 5, 75The number ‘5’ appears twice in this range of values. Thus, ‘5’ is the mode.
1, 2 ,2 ,22The number ‘2’ appears three times in this range of values. Hence,’2′ is the mode.
Understanding how to calculate the mode in Excel can significantly boost your data analysis capabilities. By using formulas such as MODE.SNGL and MODE.MULTI to streamline your approach for discovering modes quickly and efficiently.
Don’t miss out on critical insights hidden within your data; use Excel’s mode functions today! Excel’s MODE function is the ultimate wingman, always finding you the most popular number in a range.
### Using the MODE Function in Excel
The MODE function in Excel helps you find the most commonly recurring number(s) in a given range. It is an essential tool, especially when working on large datasets with multiple values.
Here are three simple steps to use the MODE function in Excel:
1. Select the cell where you want to display the mode.
2. Type =MODE(
3. Select the cells containing data for which you want to find the mode and close with a parenthesis.
It’s That Simple! Using this simple technique can save you a lot of time, rather than manually searching through your dataset for frequently occurring values.
Using additional input parameters, one can modify how this underlying algorithm works, including setting a custom bin size or using different algorithms to calculate mode for qualitative data.
A study by the Journal of Experimental Psychology found that using these built-in excel functions significantly increase productivity when working with repetitive tasks.
Excel’s MODE function makes finding the most common value in a range as easy as pie – unless you’re terrible at baking, in which case, stick to Excel.
### Applying the MODE Function to a Range of Values
The MODE function in Excel is a powerful tool that helps find the value that appears most frequently in a range of data. By applying the MODE function to a range of values, you can quickly and easily determine the most common value in your dataset without resorting to manual counting.
To better understand how to apply the MODE function, let’s take a look at this example table:
DataCount
52
62
74
83
In this table, we have a set of data ranging from five to eight with associated counts for each number. To find the mode, we simply use the MODE.SNGL formula in Excel and select the appropriate range.
By using this method, we can quickly and easily determine that seven is the mode of this dataset. This calculation would have been much more time-consuming without Excel’s MODE function.
It’s important to note that while the MODE function is a useful tool for finding common values in your data, it may not always be applicable. For example, if your dataset has multiple modes or none at all, the MODE function may not return accurate results.
Don’t miss out on the benefits of using Excel’s powerful functions like MODE. By mastering these tools and techniques, you can save time and improve your data analysis skills.
Get ready to mode-l through these Excel examples, and hope you don’t get too modaliciously bored!
## Examples of Finding the Mode in Excel
Wanna know how to find mode of a range in Excel? Here’s the thing. We’ll give you examples and focus on two sub-sections:
1. First, finding the mode in a small range.
2. Second, finding the mode in a larger range.
That’s it!
Image credits: chouprojects.com by James Washington
### Finding the Mode of a Small Range
Small Range Mode Calculation Demystified
Calculating the mode of a small range in Excel can be done efficiently and easily. Here’s how:
1. Select the range of data values for which you want to find the mode.
2. Click on an empty cell where you’d like to display the mode result.
3. Enter the formula ‘=MODE(range)‘ (without quotes), replacing ‘range’ with your selected data value range, then press enter. Your answer will then appear in the cell.
It’s that simple to calculate the mode of a small range!
The mode is useful for determining the most frequently occurring value(s) in a set of numerical data, which can be helpful for analysis and prediction. Furthermore, keep note that if there are multiple modes within your spreadsheet, only one value will be returned by this formula.
Have you ever erroneously entered an incorrect data point into a spreadsheet or used incorrect calculations? The discrepancies were likely detected thanks to accurate mode calculation enabling analysis leading to finding and fixing errors.
Finding the mode in a large range is like trying to find a needle in a haystack, but with Excel, it’s like using a magnet.
### Finding the Mode of a Large Range
When dealing with a large dataset in Excel, finding the most common value or mode is essential, as it enables data analysts to identify frequently occurring values. The approach to finding the Mode of a Large Range in Excel involves expertise and attention to detail.
Here’s a three-step guide on how to accomplish this task seamlessly:
1. Highlight the range you intend to calculate its mode.
2. Select “Formulas” from the toolbar, then click on “More functions,” and select “Statistical.”
3. In the category list, choose “MODE.SNGL,” enter your range in the blank space provided, and press enter.
When using this formula, keep in mind that if there are two or more modes in a range of data, only one value appears as the result.
It is also crucial to note that while MODE.SNGL is useful for working with small datasets, it may not be appropriate for larger ones because it can slow down performance exponentially. In such cases, one can utilize other options such as PivotTables and VBA macros besides modeling robust statistical tools.
Therefore, when computing a mode of a large range in Excel effectively ensure that you use an appropriate algorithm that aligns with your data size requirements. Additionally, avoid mishandling the input parameters since any errors can cause misleading results.
## Five Facts About How to Return the MODE of a Range in Excel:
• ✅ The MODE function in Excel returns the most frequently occurring value in a range. (Source: Microsoft)
• ✅ The MODE function can handle both numbers and text in the range. (Source: Exceljet)
• ✅ If there is no mode in the range, the MODE function returns a #N/A error. (Source: Ablebits)
• ✅ The MODE function can handle multiple modes in the range and return all modes in an array. (Source: Excel Easy)
• ✅ The MODE.SNGL function returns the most frequently occurring value in a range and ignores text values. (Source: Excel Tip)
## FAQs about How To Return The Mode Of A Range In Excel
### How to Return the MODE of a Range in Excel?
To return the MODE of a range in Excel, follow the steps below:
1. Select the cell where you want the result to appear.
2. Type the formula “=MODE(range)” (replace “range” with the cell range you want to calculate the mode for).
3. Press enter. The result will appear in the selected cell.
### What is the MODE?
The MODE is the most frequently occurring number in a range of numbers. For example, if you have the range “1, 2, 2, 3, 3, 3, 4” the mode is “3” because it occurs three times, which is more than any other number in the range.
### What if there is more than one MODE?
If there is more than one number that occurs the same number of times and both are the most frequent, then the mode is Multiple. For example, in the range “1, 2, 2, 3, 3, 4, 4” both “2” and “3” occur twice and are more frequent than any other number in the range, so the mode is Multiple.
### What if there is no MODE?
If there is no number that occurs more frequently than any other number in the range, then there is No Mode. For example, in the range “1, 2, 3, 4, 5” there is no number that occurs more than any other number, so there is no mode.
### What if my range has text values in it?
If your range includes text values along with the numbers, the MODE function will return an error because it only works with numerical data. You can disregard the text values by either removing them from the range or filtering them out using the “Filter” function.
### Can I use the MODE function on a filtered range?
Yes, you can use the MODE function on a filtered range. The result will be based on the visible, filtered data only.
Auther name
Jacky Chou is an electrical engineer turned marketer. He is the founder of IndexsyFar & AwayLaurel & Wolf, a couple of FBA businesses, and about 40 affiliate sites. He is a proud native of Vancouver, BC, who has been featured on Entrepreneur.comForbesOberlo, and GoDaddy.
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https://www.careerswave.in/ncert-solutions-for-class-12-maths-chapter-5-continuity-and-differentiability/
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# NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability
We provided NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability is prepared by the best teachers across India. These NCERT 12th Class Maths Chapter 5 Continuity and Differentiability Solution includes detailed answers of all the questions in Chapter 5 Continuity and Differentiability provided in NCERT Book which is prescribed for Class 12 in CBSE school Books. This will help students to understand basic concepts better.
## CBSE NCERT 12th Class Maths Chapter 5 Continuity and Differentiability Solutions
Organisation National Council of Educational Research and Training Class Name 12th Class Subject Name Maths Chapter Name Chapter 5 Content name Continuity and Differentiability Category Name CBSE NCERT Solutions Official Website http://ncert.nic.in/
### NCERT 12th Class Maths Chapter 5 Continuity and Differentiability Solution is given below
The topics and sub-topics included in the Continuity and Differentiability chapter are the following:
• Continuity and Differentiability
• Introduction
• Algebra of continuous functions
• Differentiability
• Derivatives of composite functions
• Derivatives of implicit functions
• Derivatives of inverse trigonometric functions
• Exponential and Logarithmic Functions
• Logarithmic Differentiation
• Derivatives of Functions in Parametric Forms
• Second Order Derivative
• Mean Value Theorem
• Summary
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https://www.physicsforums.com/threads/amendment-to-newton-1.17011/
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Amendment to Newton 1
1. Mar 25, 2004
deda
Newton’s first law was actually Galilean idea. It was Galileo who first thought that once we succeed eliminating all the friction from the surface one object slides on the object would slide forever. Newton had only royal style of putting it: “Subjected to no force the object will preserve its uniform motion i.e. travel with constant speedâ€. But there is a catch. Galileo didn’t have the entire force - picture in sight. Newton, later, in his third law stated that for every action force there is an equal and opposite reaction force. In the case of the sliding object the friction is only the reaction the object gets from the surface meaning that the action must come from its motion. Thereby, eliminating the friction from the surface would eliminate the motion of the object. My amendment to Newton’s first law would be: “Subjected to no force the object will perform no motionâ€. Even in this improved version, Newton’s first law is only a trivial case of one other more general law. That general law states: “The displacement always takes the direction of the force causing itâ€. You might not like it because it comes from me, you might not like it because of the way I put it, but rest ashore; you obey this law all the time.
2. Mar 25, 2004
Staff: Mentor
Apparently you do not understand Newton's laws.
3. Mar 25, 2004
Pergatory
You've got the two confused. Friction is not the action force, it's the reaction force. Friction is not pushing the object by providing resistance, it is providing resistance to the object which is pushing against it. In order for friction to exist, the force must already have been applied to the object creating the friction. Proof of this is in the fact that space travel is possible.
4. Mar 25, 2004
Michael D. Sewell
I do in fact like the way you put it. You have certainly made my day.
-Mike
5. Mar 25, 2004
Staff: Mentor
Don't confuse Action/Reaction pairs (as in Newton's 3rd law) with cause and effect. Per Newton, when objects interact they always exert forces on each other in matched pairs: A on B and B on A. These are the action/reaction pairs. As to which force is the action, and which the reaction: that is totally arbitrary.
For an object sliding along a surface with friction, I see three forces acting on the object. I will list them along with the reaction force for each:
(1) Weight (earth pulling on object); reaction force: object pulls on earth.
(2) Normal force (surface pushes up on object); reaction force: object pushes down on surface.
(3) Friction (surface pushes on object, parallel to surface); reaction force: object pushes on surface, parallel to surface.
6. Mar 27, 2004
deda
Couldn't have done it without making Doc Al's night.
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# How to round integer in java
I want to round the number 1732 to the nearest ten, hundred and thousand. I tried with Math round functions, but it was written only for float and double. How to do this for Integer? Is there any function in java?
-
Your chosen answer is edited. – lschin Apr 26 '11 at 3:27
Use Precision (Apache Commons Math 3.1.1)
``````Precision.round(double, scale); // return double
Precision.round(float, scale); // return float
``````
Use MathUtils (Apache Commons Math) - Older versions
``````MathUtils.round(double, scale); // return double
MathUtils.round(float, scale); // return float
``````
scale - The number of digits to the right of the decimal point. (+/-)
Discarded because method round(float, scale) be used.
Math.round(MathUtils.round(1732, -1)); // nearest ten, 1730
Math.round(MathUtils.round(1732, -2)); // nearest hundred, 1700
Math.round(MathUtils.round(1732, -3)); // nearest thousand, 2000
Better solution
``````int i = 1732;
MathUtils.round((double) i, -1); // nearest ten, 1730.0
MathUtils.round((double) i, -2); // nearest hundred, 1700.0
MathUtils.round((double) i, -3); // nearest thousand, 2000.0
``````
-
What rounding mechanism do you want to use? Here's a primitive approach, for positive numbers:
``````int roundedNumber = (number + 500) / 1000 * 1000;
``````
This will bring something like 1499 to 1000 and 1500 to 2000.
If you could have negative numbers:
``````int offset = (number >= 0) ? 500 : -500;
int roundedNumber = (number + offset) / 1000 * 1000;
``````
-
Anonymous downvoter, thanks. Any reason? – EboMike Apr 25 '11 at 7:21
this here is the right approach: int r = ((int)((n + 500) / 1000)) * 1000 since only the casting to int cuts off the digits – nils petersohn Jul 11 '12 at 9:24
@nils: That doesn't make ANY sense. The numbers are already ints (as per OP), so casting an int to int does nothing. – EboMike Jul 11 '12 at 16:21
there is not type defined in your approach for "number"! so it makes ALL sense my friend. – nils petersohn Aug 12 '12 at 12:51
"number" is an integer. Read the original question again. – EboMike Aug 13 '12 at 6:16
``````(int)(Math.round( 1732 / 10.0) * 10)
``````
`Math.round(double)` takes the `double` and then rounds up as an nearest integer. So, `1732` will become `173.2` (input parameter) on processing by `Math.round(1732 / 10.0)`. So the method rounds it like `173.0`. Then multiplying it with 10 `(Math.round( 1732 / 10.0) * 10)` gives the rounded down answer, which is `173.0` will then be casted to `int`.
-
I think 10D is what you mean instead of 10L. Isn't it? +1 – Adeel Ansari Apr 25 '11 at 6:43
well, it works but you are better explain how your answer works – Muhammed Refaat May 26 '15 at 11:52
@MuhammedRefaat, `Math.round(double)` takes the double and then rounds up as an integer. So, `1732` will become `173.2` on processing by `Math.round()`. So the method rounds it like `173.0`. Then multiplying it with 10 gives the rounded down answer. – Sibidharan Jan 20 at 21:10
@Sibidharan thanks for the explanation, my comment was to make the people who see his post understand why his code solves the problem, you can edit the answer with your explanation. – Muhammed Refaat Jan 21 at 8:23
@MuhammedRefaat done :) – Sibidharan Jan 21 at 14:27
You could try:
``````int y = 1732;
int x = y - y % 10;
``````
The result will be 1730.
Edit: This doesn't answer the question. It simply removes part of the number but doesn't "round to the nearest".
-
1736 would also go to 1730, which wouldn't be the nearest ten. – EboMike Apr 25 '11 at 6:35
...and the result for 1739 will also be 1730. Not most people's expectation for rounding. – rlibby Apr 25 '11 at 6:36
@EboMike: you are 100% right. – cherouvim Apr 25 '11 at 6:36
Add 5 to `y` before applying your snippet and you have the nearest in the common sense. Example 1736 -> 1741 -> 1740. – Pascal Cuoq Apr 25 '11 at 6:40
@Pascal: That will only work for positive numbers though. – EboMike Apr 25 '11 at 6:42
At nearest ten:
``````int i = 1986;
int result;
result = i%10 > 5 ? ((i/10)*10)+10 : (i/10)*10;
``````
(Add zero's at will for hundred and thousand).
-
This rounds up the following number (when rounding to nearest 100) which should be rounded down: 170718 is rounded UP to 170800. Surely this should round DOWN to 170700? – Adam893 Jul 6 '15 at 8:51
why not just check the unit digit... 1. if it is less than or equal to 5, add 0 at the unit position and leave the number as it is. 2. if it is more than 5, increment the tens digit, add 0 at the unit position.
ex: 1736 (since 6 >=5) the rounded number will be 1740. now for 1432 (since 2 <5 ) the rounded number will be 1430....
I hope this will work... if not than let me know about those cases...
Happy Programming,
-
very simple. try this
``````int y = 173256457;int x = (y/10)*10;
``````
Now in this you can replace 10 by 100,1000 and so on....
-
As mentioned in other answers, this would bring 1739 to 1730 instead of 1740. – EboMike Apr 25 '11 at 7:21
Its very easy..
int x = 1234; int y = x - x % 10; //It will give 1230
int y = x - x % 100; //It will give 1200
int y = x - x % 1000; //It will give 1000
The above logic will just convert the last digits to 0. If you want actual round of// For eg. 1278 this should round off to 1280 because last digit 8 > 5 for this i wrote a function check it out.
`````` private double returnAfterRoundDigitNum(double paramNumber, int noOfDigit)
{
double tempSubtractNum = paramNumber%(10*noOfDigit);
double tempResultNum = (paramNumber - tempSubtractNum);
if(tempSubtractNum >= (5*noOfDigit))
{
tempResultNum = tempResultNum + (10*noOfDigit);
}
return tempResultNum;
}
``````
Here pass 2 parameters one is the number and the other is position till which you have to round off.
Regards, Abhinav
-
Have you looked at the implementation of Mathutils.round() ? It's all based on BigDecimal and string conversions. Hard to imagine many approaches that are less efficient.
-
Without using any math utils, rounding could be achieved to any unit as below:
``````double roundValue (double input, double toNearest){
//toNearest is any rounding base like 10, 100 or 1000.
double modValue = input % toNearest;
System.out.println(modValue);
if(modValue == 0d){
roundedValue = input;
}
else
{
roundedValue = ((input - modValue) + toNearest);
}
System.out.println(roundedValue);
return roundedValue;
}
``````
-
I have made this simple method and it works fine for me. It rounds to nearest ten, but with little ajustment you can round 100, 1000 and etc.
``````public int round10(int num){
if (num%10 < 5){
while (num%10!=0){
num--;
}
}
else {
while (num%10!=0){
num++;
}
}
return num;
}
``````
-
I usually do it this way:
``````int num = 1732;
int roundedNum = Math.round((num + 9)/10 * 10);
``````
This will give you 1740 as the result.
Hope this will help.
-
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https://www.x64bitdownload.com/downloads/t-64-bit-primesieve-x64-download-nqgkdkdr.html
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# primesieve x64 5.0
size: 3.7001953125 MB
updated: 2014-01-27
Kim Walisch
primesieve x64 uses the segmented sieve of Eratosthenes with wheel factorization, this algorithm has a complexity of operations and uses space.
Segmentation is currently the best known practical improvement to the sieve of Eratosthenes. Instead of sieving the interval [2, n] at once one subdivides the sieve interval into a number of equal sized segments that are then sieved consecutively. Segmentation drops the memory requirement of the sieve of Eratosthenes from to . The segment size is usually chosen to fit into the CPU's fast L1 or L2 cache memory which significantly speeds up sieving. A segmented version of the sieve of Eratosthenes was first published by Singleton in 1969 [1], Bays and Hudson in [2] describe the algorithm in more detail.
Wheel factorization is used to skip multiples of small primes. If a kth wheel is added to the sieve of Eratosthenes then only those multiples are crossed off that are coprime to the first k primes, i.e. multiples that are divisible by any of the first k primes are skipped. The 1st wheel considers only odd numbers, the 2nd wheel (modulo 6) skips multiples of 2 and 3, the 3rd wheel (modulo 30) skips multiples of 2, 3, 5 and so on. Pritchard has shown in [3] that the running time of the sieve of Eratosthenes can be reduced by a factor of if the wheel size is but for cache reasons the sieve of Eratosthenes usually performs best with a modulo 30 or 210 wheel. Sorenson explains wheels in [4].
Additionally primesieve uses Tomás Oliveira e Silva's cache-friendly bucket list algorithm if needed [5]. This algorithm is relatively new it has been devised by Tomás Oliveira e Silva in 2001 in order to speed up the segmented sieve of Eratosthenes for prime numbers past 32 bits. The idea is to store the sieving primes into lists of buckets with each list being associated with a segment. A list of sieving primes related to a specific segment contains only those primes that have multiple occurrence(s) in that segment. Whilst sieving a segment only the primes of the related list are used for sieving and each prime is reassigned to the list responsible for its next multiple when processed. The benefit of this approach is that it is now possible to use segments (i.e. sieve arrays) smaller than without deteriorating efficiency, this is important as only small segments that fit into the CPU's L1 or L2 cache provide fast memory access.
primesieve x64 is written entirely in C++ and does not depend on external libraries [6], it compiles with every standard compliant C++ compiler. Its speed is mainly due to the segmentation of the sieve of Eratosthenes which prevents cache misses when crossing off multiples in the sieve array and the use of a bit array instead of the more widely used byte (boolean) array. These are the optimizations I use in my implementation:
Uses a bit array with 30 numbers per byte for sieving
Pre-sieves multiples of small primes ? 19
Starts crossing off multiples at the square
Uses a modolo 210 wheel that skips multiples of 2, 3, 5 and 7
Uses specialized algorithms for small, medium and big sieving primes
Processes multiple sieving primes per loop iteration to increase instruction-level parallelism
To browse the latest primesieve source code online visit the 'Source' tab.
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http://zetcode.com/javascript/bigjs/
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Big.js tutorial
Big.js tutorial shows how to work with arbitrary precision big decimal arithmetic in JavaScript with Big.js module.
Big.js
Big.js is a small, fast JavaScript library for arbitrary-precision decimal arithmetic.
In this tutorial we work with Big.js in a Node application.
Setting up Big.js
First, we install Big.js.
```\$ nodejs -v
v9.11.2
```
We use Node version 9.11.2.
```\$ npm init
```
We initiate a new Node application.
```\$ npm i big.js
```
We install Big.js with `npm i big.js` command.
JavaScript Number precision error
In the first example, we show that JavaScript Numbers are not precise for doing arbitrary precision arithmetic.
count_currency.js
```var sum = 0;
// two euros fifty-five cents
var amount = 2.55;
for (let i = 0; i < 100000; i++) {
sum += amount;
}
console.log(sum);
```
In the example, we add two euros fifty-five cents one hundred thousand times.
```\$ nodejs numbers.js
254999.9999995398
```
We have an error in the calculation.
Big.js example
In the next example we correct the error with Big.js.
big_decimal.js
```const Big = require('big.js');
var val = new Big(0.0);
var amount = new Big(2.55);
var sum = val.plus(amount).times(100000);
console.log(sum.toFixed(2));
```
With Big.js library, the calculation is precise.
```const Big = require('big.js');
```
We load the `big.js` module.
```var val = new Big(0.0);
var amount = new Big(2.55);
```
We create two big decimal values.
```var sum = val.plus(amount).times(100000);
```
We add the value 100000 times. Note that the big decimal values are immutable, so we generate a new variable.
```\$ nodejs big_decimal.js
255000.00
```
This is the output.
In this tutorial, we have worked with arbitrary precision arithmetic with the `Big.js` library.
You might also be interested in the following related tutorials: Moment.js tutorial, Reading JSON from URL in JavaScript, JavaScript Snake tutorial, JQuery tutorial, Node Sass tutorial, Lodash tutorial.
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https://gilkalai.wordpress.com/2020/03/29/game-theory-on-line-course-at-idc-herzliya/
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## Game Theory – on-line Course at IDC, Herzliya
### Game theory, a graduate course at IDC, Herzliya; Lecturer: Gil Kalai; TA: Einat Wigderson, ZOOM mentor: Ethan.
Starting Tuesday March 31, I am giving an on-line course (in Hebrew) on Game theory at IDC, Herzliya (IDC English site; IDC Chinese site).
In addition to the IDC moodle (course site) that allows IDC students to listen to recorded lectures, submit solutions to problem sets , etc., there will be a page here on the blog devoted to the course. Zoom link for the first meeting. https://idc-il.zoom.us/j/726950787
A small memory: In 1970 there was a strike in Israelis’ high schools and I took a few classes at the university. One of these classes was Game theory and it was taught by Michael Maschler. (I also took that trimester a course on art taught by Ziva Meisels.) Our department at HUJI is very strong in game theory, but once all the “professional” game theorists retired, I gave twice a game theory course which I enjoyed a lot and it was well accepted by students. In term of the number of registered students, it seems that this year’s course at IDC is quite popular and I hope it will be successful.
## The first six slides of the first presentation
(Click to enlarge)
Game Theory 2020, games, decisions, competition, strategies, mechanisms, cooperation
The course deals with basic notions, central mathematical results, and important examples in non-cooperative game theory and in cooperative game theory, and with connections of game theory with computer science, economics and other areas.
What we will learn
1. Full information zero-sum games. The value of a game. Combinatorial games.
2. Zero-sum games with incomplete information. Mixed strategies, the Minmax Theorem and the value of the game.
3. Non cooperative games, the prisoner dilemma, Nash equilibrium, Nash’s theorem on the existence of equilibrium.
4. Cooperative games, the core and the Shapley value. Nash bargaining problem, voting rules and social choice.
Background material:
Game theory alive by Anna Karlin and Yuval Peres (available on-line).
In addition I may use material from several books in Hebrew by Maschler, Solan, Zamir, by Hefetz, and by Megiddo (based on lectures by Peleg). (If only I will manage to unite with my books that are not here.) We will also use a site by Ariel Rubinstein for playing games and some material from the book by Osborne and Rubinstein.
Requirement and challenges:
• Play, play, play games, in Ariel Rubinshtein site and various other games.
• Solve 10 short theoretical problem set.
• Final assignment, including some programming project that can be carried out during the semester.
• Of course, we will experience on-line study which is a huge challenge for us all.
Games and computers
• Computer games
• Algorithms for playing games
• algorithmic game theory:
• Mechanism design
• Analyzing games in tools of computer science
• Electronic commerce
• Games, logic and automata: there will be a parallel course by Prof. Udi Boker
I still have some difficulty with the virtual background in ZOOM.
This entry was posted in Combinatorics, Computer Science and Optimization, Economics, Games, Rationality, Teaching and tagged , . Bookmark the permalink.
### 2 Responses to Game Theory – on-line Course at IDC, Herzliya
1. Reblogged this on Sergei Yakovenko's blog: on Math and Teaching and commented:
Most recommended!
2. DK says:
Was the first Zoom meeting recorded?
Will you be recording the meetings? If so, will you be uploading them here?
Would love to attend the course though I am not an IDC student.
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A169900 Earliest sequence such that xy | a(x+y) for all x>=1, y>=1. 1
1, 1, 2, 12, 12, 360, 60, 1680, 2520, 25200, 2520, 332640, 27720, 5045040, 5405400, 2882880, 720720, 220540320, 12252240, 4655851200, 4888643760, 5121436320, 232792560, 128501493120, 26771144400, 696049754400 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS From Robert Israel, Dec 29 2017: (Start) If n = p^d for prime p, then a(n) = p^(2*d-2)*Product_q q^floor(log_q(n)), where the product is over all primes q < n other than p. Otherwise, a(n) = n^2*Product_p p^floor(log_p(n/p^(nu(n,p)))), where the product is over all primes p < n and nu(n,p) is the p-adic order of n. (End) LINKS Robert Israel, Table of n, a(n) for n = 1..2293 EXAMPLE After a(1)=a(2)=1, we must have a(3) >= 2 from 2 | a(1+2), and a(3)=2 works. MAPLE seq(ilcm(seq(x*(n-x), x=1..n/2)), n=1..50); # Robert Israel, Dec 28 2017 MATHEMATICA a[n_]:=If[n<=2, 1, LCM@@Table[x(n-x), {x, Floor[n/2]}]]; Table[a[n], {n, 30}] (* Zak Seidov, Jul 11 2010 *) CROSSREFS Sequence in context: A075178 A145618 A306693 * A140355 A007041 A331880 Adjacent sequences: A169897 A169898 A169899 * A169901 A169902 A169903 KEYWORD nonn,nice AUTHOR Andrew Weimholt, Jul 05 2010 STATUS approved
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Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified August 7 06:03 EDT 2024. Contains 375008 sequences. (Running on oeis4.)
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+0
# Help!!
0
112
1
According to the article "A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich" published in the Journal of the American Medical Association, the body temperatures of adults are normally distributed with a mean of 98.276 and a standard deviation of 0.728.
What is the probability that a randomly selected person's body temperature is between 97.93 and 98.45? Round your answer to 4 decimal places.
Mar 12, 2020
#1
-1
+2
First find the temperature
sqrt (10^2 +(-4)^2 = 10.77
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# Dynamically remove element from list based on test
Suppose I have a list
l = {a, b, c, d, e, f....}
I would like to remove one of each pair {x,y} if some function check[x,y] returns xor y (or do nothing for that particular pair if the function returns {}). The order of the list is important. For example, if
check[c,e] === c;
check[a,f] === f;
and check on any other combination is empty, the final list should be
{a, b, d, e, ...}
I know how to write a for loop and do index-based removal, but is it a slicker way to do it using Mathematica's list manipulation functions?
EDIT: The check function should be non-overlapping in my usage, but in case if there is problematic overlap, say
check[a,e] === a;
check[a,f] === f;
both e and f should remain, since after removal of a (assuming position of a is earlier than f), there is no pair that can be formed with {a,f}.
-
I am presuming Check is intended to be your own, custom function but that name is already in use. User symbols should start with lower case letters. – Mr.Wizard Oct 6 '12 at 8:37
What happens if check[a,b] == a and check[a,d] == a? Is it sufficient to remove a? Should the second check be performed at all? – Mr.Wizard Oct 6 '12 at 8:44
@Mr.Wizard: Good catch! Question edited. – polyglot Oct 6 '12 at 8:53
Now I am wondering why f is removed and not a -- should the element that is not returned be removed? – Mr.Wizard Oct 6 '12 at 9:05
It depends on definition of check of course; question again edited for consistency. (It proves that I should go to bed instead...) – polyglot Oct 6 '12 at 9:10
You said that your check should be non-overlapping. For that simplified case I believe this works:
ClearAll[check]
check[c, e] = c;
check[a, f] = f;
check[__] = {};
lst = {a, b, c, d, e, f};
DeleteCases[lst,
Alternatives @@ Flatten[check @@@ Subsets[lst, {2}]]
]
{a, b, d, e}
I'm still working on the more complex version.
-
Here's one implementation:
l = {a, b, c, d, e, f};
check[x_, y_] := If[MemberQ[l, x] && MemberQ[l, y], RandomChoice[{x, y}]]
Fold[DeleteCases, l, {check[c, e], check[a, f]}]
{a, b, d, e}
Edit
A version that checks all subset pairs:
l = {a, b, c, d, e, f};
s = Subsets[l, {2}];
check[{x_, y_}] := If[MemberQ[l, x] && MemberQ[l, y], RandomChoice[{x, y}]]
(l = DeleteCases[l, check[#]]) & /@ s;
l
{a}
As ployglot's edit observes, progressively removing check results from l results in less matches than using Fold, i.e.
Fold[DeleteCases, l, check[#] & /@ s]
{ }
-
You seem to have a very different interpretation of the question. Would you take a look at my present answer and tell me if you still think you've got it right? (I don't mean that condescendingly; rather I'm asking because I don't know for sure that I've got it right.) – Mr.Wizard Oct 6 '12 at 9:36
@ Mr.Wizard - seems to depend how specific polyglot intends his example to be. – Chris Degnen Oct 6 '12 at 10:03
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SOLVED
# How to combine 3 IF formulas with 3 different value_if_false results into 1 formula
Copper Contributor
# How to combine 3 IF formulas with 3 different value_if_false results into 1 formula
I have a matrix, 15 rows (A), 10 columns (B,C,D..). In a result should get a table returning 6 different results.
Need ONLY 1 formula to fit each cell.
Managed to create 3, which now need to be combined - not succeeding on my own.
=IF(AND(\$A2<=5;OR(B\$1<=3;B\$1>=8));13;14)
=IF(AND(\$A2>=6;\$A2<=10;OR(B\$1<=3;B\$1>=8));11;12)
=IF(AND(\$A2>=11;OR(B\$1<=3;B\$1>=8));9;10)
2 Replies
best response confirmed by Ppetu999 (Copper Contributor)
Solution
# Re: How to combine 3 IF formulas with 3 different value_if_false results into 1 formula
Hi, this should work like this:
``````=IF(\$A2<=5,IF(OR(B\$1<=3,B\$1>=8),13,14),IF(\$A2<=10,IF(OR(B\$1<=3,B\$1>=8),11,12),IF(\$A2>=11,IF(OR(B\$1<=3,B\$1>=8),9,10))))
or with ";":
=IF(\$A2<=5;IF(OR(B\$1<=3;B\$1>=8);13;14);IF(\$A2<=10;IF(OR(B\$1<=3;B\$1>=8);11;12);IF(\$A2>=11;IF(OR(B\$1<=3;B\$1>=8);9;10))))``````
Alternatively, a bit shorter you could do a combination with LOOKUP().
``````=XLOOKUP(A2,{5;10;11},{13;11;9},9,1)+IF(OR(B\$1<=3,B\$1>=8),0,1)
or with ";":
=XLOOKUP(A2;{5.10.11};{13.11.9};9;1)+IF(OR(B\$1<=3;B\$1>=8);0;1)``````
# Re: How to combine 3 IF formulas with 3 different value_if_false results into 1 formula
Great, the 1st version with ";" works for me well! Thanks a lot!
1 best response
Accepted Solutions
best response confirmed by Ppetu999 (Copper Contributor)
Solution
# Re: How to combine 3 IF formulas with 3 different value_if_false results into 1 formula
Hi, this should work like this:
``````=IF(\$A2<=5,IF(OR(B\$1<=3,B\$1>=8),13,14),IF(\$A2<=10,IF(OR(B\$1<=3,B\$1>=8),11,12),IF(\$A2>=11,IF(OR(B\$1<=3,B\$1>=8),9,10))))
or with ";":
=IF(\$A2<=5;IF(OR(B\$1<=3;B\$1>=8);13;14);IF(\$A2<=10;IF(OR(B\$1<=3;B\$1>=8);11;12);IF(\$A2>=11;IF(OR(B\$1<=3;B\$1>=8);9;10))))``````
Alternatively, a bit shorter you could do a combination with LOOKUP().
``````=XLOOKUP(A2,{5;10;11},{13;11;9},9,1)+IF(OR(B\$1<=3,B\$1>=8),0,1)
or with ";":
=XLOOKUP(A2;{5.10.11};{13.11.9};9;1)+IF(OR(B\$1<=3;B\$1>=8);0;1)``````
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#### Explain solution rd sharma class 12 chapter 16 Increasing and decreasing function exercise multiple choice question, question 34
$\cos x$ , Option (c)
Hints: Check all the options and choose is satisfies
Given: Function is decreasing in $\left ( 0,\frac{\pi}{2} \right )$
Solution:
Option (A)
\begin{aligned} &f(x)=\sin 2 x \\ &f^{\prime}(x)=2 \cos 2 x \end{aligned}
$f(x)$ increases from ‘0’ to ‘1’ in $\left ( 0,\frac{\pi}{2} \right )$
Option (B)
\begin{aligned} &f(x)=\tan x \\ &f^{\prime}(x)=\sec ^{2} x \end{aligned}
In interval $\left ( 0,\frac{\pi}{2} \right )$$f^{\prime}(x)=-\sin x<0$
$f(x)=\cos x$ is strictly increasing in interval $\left ( 0,\frac{\pi}{2} \right )$
Option (C)
\begin{aligned} &f(x)=\cos x \\ &f^{\prime}(x)=-\sin x \end{aligned}
In interval $\left(0, \frac{\pi}{2}\right), f^{\prime}(x)=-\sin x<0$
$f(x)=\cos x$ is strictly decreasing in $\left ( 0,\frac{\pi}{2} \right )$
Option (D)
\begin{aligned} &f(x)=\cos 3 x \\ &f^{\prime}(x)=-3 \sin 3 x \end{aligned}
Now,
\begin{aligned} &f^{\prime}(x)=0 \\ &\sin 3 x=0 \\ &3 x=\pi \end{aligned}
As $x\epsilon \left ( 0,\frac{\pi}{2} \right )$
$x=\frac{\pi}{3}$
$f(x)=\cos 3 x$ is decreases only when $3 x \in\left(0, \frac{\pi}{2}\right)$
And $x \in\left(0, \frac{\pi}{6}\right)$
Therefore, Option (C) =cos x satisfies because $f(x)=\cos x$ is strictly decreasing in $\left ( 0,\frac{\pi}{2} \right )$
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# Types of relations and functions pdf
2019-09-17 00:07
Do the ordered pairs below represent a relation, a function, both a relation and a function, or neither a relation nor a function? (2, 1), (1, 4), (7, 10), (8, 11) A. neither a relation nor a function B. both a relation and a function C. relation only D. function only 7.Set theory is a basis of modern mathematics, and notions of set theory are used in all formal descriptions. The notion of set is taken as undefined, primitive, or basic, so types of relations and functions pdf
All functions are relations but not all relations are functions. Dependent and Independent Variables The xnumber is called the independent variable, and the ynumber is called the dependent variable because its value depends on the xvalue chosen. Vertical Line Test If
Unit 3: Relations and Functions 51: Binary Relations Binary Relation: Linear Relation: a set ordered pairs that exhibit a straight line when plotted on a graph. of Best Fit: a line that will best describe the general relation of the ordered pairs on the graph. There are two types Relations and Functions: Relations and functions is an important chapter in class 11 for CBSE. In this article, find the basics of relations and functions basic definitions, types of relations and functions types of relations and functions pdf RELATIONS AND FUNCTIONS 20 EXEMPLAR PROBLEMS MATHEMATICS (A B) pq and the total number of possible relations from the set A to set B 2pq. Functions A relation f from a set A to a set B is said to be function if every element of set A has one and only one image in set B. Some specific types of functions (i)
Functions and different types of functions A relation is a function if for every x in the domain there is exactly one y in the codomain. A vertical line through any element of the domain should intersect the graph of the function exactly once. types of relations and functions pdf Algebra I Notes Relations and Functions Unit 03a Alg I Unit 03a Notes Relations and FunctionsAlg I Unit 03a Notes Relations and Functions Page 4 of 8 Graphs of Functions: Given the graph, we can use the vertical line test to determine if a relation is a function.
Rating: 4.38 / Views: 861
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## DEV Community is a community of 698,016 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
# LeetCode - Maximum Subarray
Ruby on Rails and Golang consultant
### Problem statement
Given an integer array nums,
find the contiguous subarray (containing at least one number) which has the largest sum and return
its sum.
Problem statement taken from: https://leetcode.com/problems/maximum-subarray
Example 1:
``````Input: nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
Output: 6
Explanation: [4, -1, 2, 1] has the largest sum = 6.
``````
Example 2:
``````Input: nums = [1]
Output: 1
``````
Example 3:
``````Input: nums = [5, 4, -1, 7, 8]
Output: 23
``````
Constraints:
``````- 1 <= nums.length <= 3 * 10^4
- -10^5 <= nums[i] <= 10^5
``````
### Explanation
#### Brute force
The brute force approach is to generate all subarrays and
print that subarray which has a maximum sum.
A C++ snippet of the above logic is as below:
``````for (int i = 0; i < n; i++){
for (int j = i; j < n; j++){
for (int k = i; k <= j; k++){
// calculate sum of all the elements
}
}
}
``````
The time complexity of the above approach is O(N^3).
We can improve the above logic using
The simple idea of Kadane’s algorithm is to look for all positive contiguous segments
of the array (`max_sum` is used for this).
And keep track of maximum sum contiguous segment among all positive segments
(`max_sum_so_far` is used for this).
Each time we get a positive sum compare it with `max_sum` and
update `max_sum` if it is greater than `max_sum_so_far`.
Let's check the algorithm below:
``````- set max_sum_so_far = 0, max_sum = INT_MIN
- Loop for i = 0; i < nums.length; i++
- add max_sum_so_far = max_sum_so_far + nums[i]
- if max_sum < max_sum_so_far
- set max_sum = max_sum_so_far
- if max_sum_so_far < 0
- set max_sum_so_far = 0
- return max_sum
``````
The time complexity of the above approach is O(N) and,
space complexity is O(1).
##### C++ solution
``````class Solution {
public:
int maxSubArray(vector<int>& nums) {
int max_sum = INT_MIN;
int max_sum_so_far = 0;
for(int i = 0; i < nums.size(); i++){
max_sum_so_far += nums[i];
if(max_sum < max_sum_so_far){
max_sum = max_sum_so_far;
}
if(max_sum_so_far < 0){
max_sum_so_far = 0;
}
}
return max_sum;
}
};
``````
##### Golang solution
``````func maxSubArray(nums []int) int {
maxSum := math.MinInt32
maxSumSoFar := 0
for i := 0; i < len(nums); i++ {
maxSumSoFar += nums[i]
if maxSum < maxSumSoFar {
maxSum = maxSumSoFar
}
if maxSumSoFar < 0 {
maxSumSoFar = 0
}
}
return maxSum
}
``````
##### Javascript solution
``````var maxSubArray = function(nums) {
let maxSumSoFar = 0;
let maxSum = -Infinity;
if(nums.length === 0) return 0;
if(nums.length === 1) return nums[0]
for( let i = 0; i<nums.length; i++) {
maxSumSoFar += nums[i];
maxSum = Math.max(maxSum, maxSumSoFar);
if(maxSumSoFar < 0) {
maxSumSoFar = 0;
}
}
return maxSum;
};
``````
Let's dry-run our algorithm to see how the solution works.
``````nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
Step 1: max_sum_so_far = 0
max_sum = INT_MIN
Step 2: for i = 0; i < nums.length
0 < 9
true
max_sum_so_far += nums[0]
max_sum_so_far = 0 + -2
max_sum_so_far = -2
max_sum < max_sum_so_far
-2^31 - 1 < -2
true
max_sum = max_sum_so_far
max_sum = -2
max_sum_so_far < 0
-2 < 0
true
max_sum_so_far = 0
i++
i = 1
Step 3: i < nums.length
1 < 9
true
max_sum_so_far += nums[1]
max_sum_so_far = 0 + 1
max_sum_so_far = 1
max_sum < max_sum_so_far
-2 < 1
true
max_sum = max_sum_so_far
max_sum = 1
max_sum_so_far < 0
1 < 0
false
i++
i = 2
Step 4: i < nums.length
2 < 9
true
max_sum_so_far += nums[2]
max_sum_so_far = 1 + -3
max_sum_so_far = -2
max_sum < max_sum_so_far
1 < -2
false
max_sum_so_far < 0
-2 < 0
true
max_sum_so_far = 0
i++
i = 3
Step 5: i < nums.length
3 < 9
true
max_sum_so_far += nums[3]
max_sum_so_far = 0 + 4
max_sum_so_far = 4
max_sum < max_sum_so_far
1 < 4
true
max_sum = max_sum_so_far
max_sum = 4
max_sum_so_far < 0
false
i++
i = 4
Step 6: i < nums.length
4 < 9
true
max_sum_so_far += nums[4]
max_sum_so_far = 4 + -1
max_sum_so_far = 3
max_sum < max_sum_so_far
4 < 3
false
max_sum_so_far < 0
false
i++
i = 5
Step 7: i < nums.length
5 < 9
true
max_sum_so_far += nums[5]
max_sum_so_far = 3 + 2
max_sum_so_far = 5
max_sum < 5
4 < 5
true
max_sum = max_sum_so_far
max_sum = 5
max_sum_so_far < 0
false
i++
i = 6
Step 8: i < nums.length
6 < 9
true
max_sum_so_far += nums[6]
max_sum_so_far = 5 + 1
max_sum_so_far = 6
max_sum < 6
5 < 6
true
max_sum = max_sum_so_far
max_sum = 6
max_sum_so_far < 0
false
i++
i = 7
Step 9: i < nums.length
7 < 9
true
max_sum_so_far += nums[7]
max_sum_so_far = 6 + -5
max_sum_so_far = 1
max_sum < 6
6 < 1
false
max_sum_so_far < 0
false
i++
i = 8
Step 10: i < nums.length
8 < 9
true
max_sum_so_far += nums[8]
max_sum_so_far = 1 + 4
max_sum_so_far = 5
max_sum < 6
6 < 5
false
max_sum_so_far < 0
false
i++
i = 9
Step 11: i < nums.length
9 < 9
false
Step 12: return max_sum
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# Limits on scaling chute sizes?
#### boatgeek
##### Well-Known Member
I'm planning one or two x-form chutes for my next project. Coincidentally, the SLI team I'm helping used an x-form drogue on their launch, so I have some really nice descent data*. That is great for my drogue, but I'm not sure if I can scale up as big as my main would need to be. Assuming that Force = constant * A * V^2, where the constant covers both the drag coefficient and air density, I get that my main needs to be about 10 times the area of the drogue, or about 3 times the edge length. That seems like a really big scaling factor. Does that seem reasonable for assuming that the drag coefficient is going to be fairly constant?
Thanks!
* For figuring out the drag force, I was going to assume that the drogue just carried the entire upper airframe weight, since the upper airframe was hanging straight down from the drogue and the fin can was falling fairly flat. Comments appreciated.
#### Bat-mite
##### Rocketeer in MD
I'm planning one or two x-form chutes for my next project. Coincidentally, the SLI team I'm helping used an x-form drogue on their launch, so I have some really nice descent data*. That is great for my drogue, but I'm not sure if I can scale up as big as my main would need to be. Assuming that Force = constant * A * V^2, where the constant covers both the drag coefficient and air density, I get that my main needs to be about 10 times the area of the drogue, or about 3 times the edge length. That seems like a really big scaling factor. Does that seem reasonable for assuming that the drag coefficient is going to be fairly constant?
Thanks!
* For figuring out the drag force, I was going to assume that the drogue just carried the entire upper airframe weight, since the upper airframe was hanging straight down from the drogue and the fin can was falling fairly flat. Comments appreciated.
Are you making your chutes? Otherwise, just use the mfg.'s data for the weight of your rocket. I'm probably missing something....
#### boatgeek
##### Well-Known Member
Are you making your chutes? Otherwise, just use the mfg.'s data for the weight of your rocket. I'm probably missing something....
Yes, both chutes will be scratch built. I already have the ripstop, in white, fluorescent orange, and olive green. Not sure what color pattern to use yet, but it'll be visible!
#### Handeman
##### Well-Known Member
TRF Supporter
I think your assumption on the weight when falling is probably pretty good. The spinning/flying fin can doesn't contribute much to the decent rate.
The Cd of a chute is a function of it's shape so it should scale pretty well.
#### Rex R
##### LV2
I use OR to find the proper size (standard)chute, then divide by 0.8 to get the appropriate size X-form chute.
Rex
#### Bat-mite
##### Rocketeer in MD
For kicks, once you have your calculation, you could compare it with TFR's X-form chute Cd and see how it compares.
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# Fin/370 final exam | Business & Finance homework help
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Fin/370 final exam | Business & Finance homework help
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1
Which financial statement shows the total revenues that a firm earns and the total expenses the firm incurs to generate those revenues over a specific period of time — generally one year?
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[removed]
Income statement
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[removed]
Balance sheet
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[removed]
Statement of retained earnings
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[removed]
Statement of cash flows
2
Which of these is the term for portfolios with the highest return possible for each risk level?
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Modern portfolios
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[removed]
Optimal portfolios
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[removed]
Total portfolios
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Efficient portfolios
3
The Rule of 72 is a simple mathematical approximation for__________.
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the present value required to double an investment
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[removed]
the future value required to double an investment
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the number of years required to double an investment
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[removed]
the payments required to double an investment
4
Which of these statements is true regarding divisional WACC?
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Using a firmwide WACC to evaluate new projects would have no impact on projects that present less risk than the firm’s average beta.
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[removed]
Using a simple firmwide WACC to evaluate new projects would give an unfair advantage to projects that present more risk than the firm’s average beta.
·
[removed]
Using a simple firmwide WACC to evaluate new projects would give an unfair advantage to projects that present less risk than the firm’s average beta.
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[removed]
Using a divisional WACC versus a WACC for the firm’s current operations will result in quite a few incorrect decisions.
5
Which of these is used as a measure of the total amount of available cash flow from a project?
·
[removed]
Operating cash flow
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[removed]
Sunk cash flow
·
[removed]
Free cash flow
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[removed]
Investment in operating capital
6
Which financial statement reports the amounts of cash that the firm generated and distributed during a particular time period?
·
[removed]
Income statement
·
[removed]
Statement of cash flows
·
[removed]
statement of retained earnings
·
[removed]
Balance sheet
7
We commonly measure the risk-return relationship using which of the following?
·
[removed]
Expected returns
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[removed]
Standard deviation
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[removed]
Coefficient of variation
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[removed]
Correlation coefficient
8
Financial plans include which of the following?
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[removed]
Pro forma Income Statement, Balance Sheet
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[removed]
All of the above
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[removed]
Schedule of Sales, Expenses, and Capital Expenditure
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[removed]
Short Term and Long Term Plan
9
Which of these provide a forum in which demanders of funds raise funds by issuing new financial instruments, such as stocks and bonds?
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[removed]
Investment banks
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[removed]
Secondary markets
·
[removed]
Primary markets
·
[removed]
Money markets
10
Suppose that Model Nails, Inc.’s capital structure features 60 percent equity, 40 percent debt, and that its before-tax cost of debt is 6 percent, while its cost of equity is 10 percent. If the appropriate weighted average tax rate is 28 percent, what will be Model Nails’ WACC?
·
[removed]
7.73 percent
·
[removed]
8.40 percent
·
[removed]
16.00 percent
·
[removed]
8.00 percent
question11
What’s the current yield of a 6 percent coupon corporate bond quoted at a price of 101.70?
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[removed]
6.1 percent
·
[removed]
10.2 percent
·
[removed]
6.0 percent
·
[removed]
5.9 percent
12
We call the process of earning interest on both the original deposit and on the earlier interest payments:
·
[removed]
computing.
·
[removed]
discounting.
·
[removed]
multiplying.
·
[removed]
compounding.
13
Which of these ratios show the combined effects of liquidity, asset management, and debt management on the overall operation results of the firm?
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[removed]
Liquidity
·
[removed]
Coverage
·
[removed]
Profitability
·
[removed]
Financial
14
Which of the following is a true statement?
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[removed]
If interest rates fall, all bonds will enjoy rising values.
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[removed]
If interest rates fall, no bonds will enjoy rising values.
·
[removed]
If interest rates fall, U.S. Treasury bonds will have decreasing values.
·
[removed]
If interest rates fall, corporate bonds will have decreasing values.
15
Five years ago, Jane invested \$5,000 and locked in an 8 percent annual interest rate for 25 years (ending 20 years from now). James can make a 20-year investment today and lock in a 10 percent interest rate. How much money should he invest now in order to have the same amount of money in 20 years as Jane?
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[removed]
\$5,089.91
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[removed]
\$3,160.43
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\$7,346.64
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[removed]
\$3,464.11
16
The overall goal of the financial manager is to__________.
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maximize net income
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maximize earnings per share
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minimize total costs
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maximize shareholder wealth
17
Which of the following can create ethical dilemmas between corporate managers and stockholders?
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[removed]
Auditors
·
[removed]
Venture Capitalist
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[removed]
Agency relationship
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[removed]
Board of directors
18
When firms use multiple sources of capital, they need to calculate the appropriate discount rate for valuing their firm’s cash flows as__________.
·
[removed]
they apply to each asset as they are purchased with their respective forms of debt or equity
·
[removed]
a sum of the capital components costs
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[removed]
a weighted average of the capital components costs
·
[removed]
a simple average of the capital components costs
19
You are trying to pick the least-expensive machine for your company. You have two choices: machine A, which will cost \$50,000 to purchase and which will have OCF of -\$3,500 annually throughout the machine’s expected life of three years; and machine B, which will cost \$75,000 to purchase and which will have OCF of -\$4,900 annually throughout that machine’s four-year life. Both machines will be worthless at the end of their life. If you intend to replace whichever type of machine you choose with the same thing when its life runs out, again and again out into the foreseeable future, and if your business has a cost of capital of 14 percent, which one should you choose?
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[removed]
Neither machine A nor B
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[removed]
Both machines A and B
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[removed]
Machine A
·
[removed]
Machine B
20
The top part of Mars, Inc.’s 2013 balance sheet is listed as follows (in millions of dollars).
What are Mars, Inc.’s current ratio, quick ratio, and cash ratio for 2013?
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[removed]
10.5, 6.0, 1.0
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[removed]
2.3333, 0.5556, 0.1111
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[removed]
0.1111, 0.5556, 0.2
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[removed]
4.2, 1.0, 0.2
21
We can estimate a stock’s value by__________.
·
[removed]
using the book value of the total assets divided by the number of shares outstanding
·
[removed]
discounting the future dividends and future stock price appreciation
·
[removed]
using the book value of the total stockholder equity section
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[removed]
compounding the past dividends and past stock price appreciation
22
Which of these does NOT perform vital functions to securities markets of all sorts by channeling funds from those with surplus funds to those with shortages of funds?
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[removed]
Commercial banks
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[removed]
Insurance companies
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[removed]
Secondary markets
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[removed]
Mutual funds
23
A firm is expected to pay a dividend of \$2.00 next year and \$2.14 the following year. Financial analysts believe the stock will be at their target price of \$75.00 in two years. Compute the value of this stock with a required return of 10 percent.
·
[removed]
\$65.40
·
[removed]
\$79.14
·
[removed]
\$65.57
·
[removed]
\$66.67
24
Will’s Wheels, Inc. reported a debt-to-equity ratio of 0.65 times at the end of 2013. If the firm’s total debt at year-end was \$5 million, how much equity does Will’s Wheels have?
·
[removed]
\$7.69 million
·
[removed]
\$5 million
·
[removed]
\$3.25 million
·
[removed]
\$0.65 million
25
Which of these is the process of estimating expected future cash flows of a project using only the relevant parts of the balance sheet and income statements?
·
[removed]
Substitutionary analysis
·
[removed]
Cash flow analysis
·
[removed]
Incremental cash flows
·
[removed]
Pro forma analysis
26
Which financial statement reports a firm’s assets, liabilities, and equity at a particular point in time?
·
[removed]
Balance sheet
·
[removed]
Income statement
·
[removed]
Statement of retained earnings
·
[removed]
Statement of cash flows
27
Which of the following terms means that during periods when interest rates change substantially, bondholders experience distinct gains and losses in their bond investments?
·
[removed]
Liquidity rate risk
·
[removed]
Reinvestment rate risk
·
[removed]
Credit quality risk
·
[removed]
Interest rate risk
28
What are the tools available for the manager in financial planning?
·
[removed]
Increasing inventory turnover and reducing collection period
·
[removed]
Reducing collection period and delaying disbursement of cash
·
[removed]
Delaying disbursement of cash and cash management
·
[removed]
Delaying disbursement of cash, reducing collection period, cash management, and Increasing inventory turnover
29
As new capital budgeting projects arise, we must estimate__________.
·
[removed]
the cost of the stock being sold for the specific project
·
[removed]
the cost of the loan for the specific project
·
[removed]
when such projects will require cash flows
·
[removed]
the float costs for financing the project
30
What are reasons for the firm to go abroad?
·
[removed]
Diversification
·
[removed]
·
[removed]
Lower production cost
·
[removed]
All of the above
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http://docplayer.net/707149-Numerical-simulation-for-asset-liability-management-in-life-insurance.html
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crawl-data/CC-MAIN-2018-47/segments/1542039743046.43/warc/CC-MAIN-20181116132301-20181116154301-00397.warc.gz
| 89,236,807
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Numerical Simulation for Asset-Liability Management in Life Insurance
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1 Numerical Simulation for Asset-Liability Management in Life Insurance T. Gerstner 1, M. Griebel 1, M. Holtz 1, R. Goschnic 2, and and M. Haep 2 1 Institut für Numerische Simulation, Universität Bonn 2 Zürich Gruppe Deutschland Summary. New regulations and stronger competitions have increased the demand for stochastic asset-liability management (ALM) models for insurance companies in recent years. In this article, we propose a discrete time ALM model for the simulation of simplified balance sheets of life insurance products. The model incorporates the most important life insurance product characteristics, the surrender of contracts, a reserve-dependent bonus declaration, a dynamic asset allocation and a two-factor stochastic capital maret. All terms arising in the model can be calculated recursively which allows an easy implementation and efficient evaluation of the model equations. The modular design of the model permits straightforward modifications and extensions to handle specific requirements. In practise, the simulation of stochastic ALM models is usually performed by Monte Carlo methods which suffer from relatively low convergence rates and often very long run times, though. As alternatives to Monte Carlo simulation, we here propose deterministic integration schemes, such as quasi-monte Carlo and sparse grid methods for the numerical simulation of such models. Their efficiency is demonstrated by numerical examples which show that the deterministic methods often perform much better than Monte Carlo simulation as well as by theoretical considerations which show that ALM problems are often of low effective dimension. 1 Introduction The scope of asset-liability management is the responsible administration of the assets and liabilities of insurance contracts. Here, the insurance company has to attain two goals simultaneously. On the one hand, the available capital has to be invested as profitably as possible (asset management), on the other hand, the obligations against policyholders have to be met (liability management). Depending on the specific insurance policies these obligations are often quite complex and can include a wide range of guarantees and optionlie features, lie interest rate guarantees, surrender options (with or without surrender fees) and variable reversionary bonus payments. Such bonus payments are typically lined to the investment returns of the company. Thereby,
2 2 T. Gerstner, M. Griebel, M. Holtz, R. Goschnic, and and M. Haep the insurance company has to declare in each year which part of the investment returns is given to the policyholders as reversionary bonus, which part is saved in a reserve account for future bonus payments and which part is ept by the shareholders of the company. These management decisions depend on the financial situation of the company as well as on strategic considerations and legal requirements. A maximisation of the shareholders benefits has to be balanced with a competitive bonus declaration for the policyholders. Moreover, the exposure of the company to financial, mortality and surrender riss has to be taen into account. These complex problems are investigated with the help of ALM analyses. In this context, it is necessary to estimate the medium- and long-term development of all assets and liabilities as well as the interactions between them and to determine their sensitivity to the different types of riss. This can either be achieved by the computation of particular scenarios (stress tests) which are based on historical data, subjective expectations, and guidelines of regulatory authorities or by a stochastic modelling and simulation. In the latter case, numerical methods are used to simulate a large number of scenarios according to given distribution assumptions which describe the possible future developments of all important variables, e.g. of the interest rates. The results are then analysed using statistical figures which illustrate the expected performance or the ris profile of the company. In recent years, such stochastic ALM models for life insurance policies are becoming more and more important as they tae financial uncertainties more realistically into account than an analysis of a small number of deterministically given scenarios. Additional importance arises due to new regulatory requirements as Solvency II and the International Financial Reporting Standard (IFRS). Consequently, much effort has been spent on the development of these models for life insurance policies in the last years, see, e.g., [2, 4, 7, 13, 19, 24, 33] and the references therein. However, most of the ALM models described in the existing literature are based on very simplifying assumptions in order to focus on special components and effects or to obtain analytical solutions. In this article, we develop a general model framewor for the ALM of life insurance products. The complexity of the model is chosen such that most of the models previously proposed in the literature and the most important features of life insurance product management are included. All terms arising in the model can be calculated recursively which allows an straightforward implementation and efficient evaluation of the model equations. Furthermore, the model is designed to have a modular organisation which permits straightforward modifications and extensions to handle specific requirements. In practise, usually Monte Carlo methods are used for the stochastic simulation of ALM models. These methods are robust and easy to implement but suffer from their relatively low convergence rates. To obtain one more digit accuracy, Monte Carlo methods need the simulation of a hundred times as many scenarios. As the simulation of each scenario requires a run over all time points and all policies in the portfolio of the company, often very long run times are
3 Numerical Simulation for Asset-Liability Management in Life Insurance 3 needed to obtain approximations of satisfactory accuracy. As a consequence, a more frequent and more comprehensive ris management, extensive sensitivity investigations or the optimisation of product parameters and management rules are often not possible. In this article we propose deterministic numerical integration schemes, such as quasi-monte Carlo methods (see e.g. [37, 22, 40]) and sparse grid methods (see, e.g., [9, 15, 16, 23, 38, 42]) for the numerical simulation of ALM models. These methods are alternatives to to Monte Carlo simulation, which have a faster rate of convergence, exploit the smoothness and the anisotropy of the integrand and have deterministic upper bounds on the error. In this way, they often can significantly reduce the number of required scenarios and computing times as we show by numerical experiments. The performance of these numerical methods is closely related to the effective dimension and the smoothness of the problem under consideration. Here, we show that ALM problems are often of very low effective dimension (in the sense that the problem can well be approximated by sums of very lowdimensional functions) which can, to some extent, explain the efficiency of the deterministic methods. Numerical results based on a general ALM model framewor for participating life insurance products demonstrate that these deterministic methods in fact often perform much better than Monte Carlo simulation even for complex ALM models with many time steps. Quasi-Monte Carlo methods based on Sobol sequences and dimension-adaptive sparse grids based on one-dimensional Gauss-Hermite quadrature formulae turn out to be the most efficient representatives of several quasi-monte Carlo and sparse grid variants, respectively. For further details, see [17, 18, 19]. The remainder of this article is as follows: In Section 2, we describe the model framewor. In Section 3, we then discuss how this model can be efficiently simulated by numerical methods for multivariate integration. In Section 4, we present numerical results which illustrate possible application of the ALM model and analyse the efficiency of different numerical approaches. The article finally closes in Section 5 with concluding remars. 2 The ALM Model In this section, we closely follow [19] and describe an ALM model framewor for the simulation of the future development of a life insurance company. We first indicate the overall structure of the model and introduce a simplified balance sheet which represents the assets and liabilities of the company. The different modules (capital maret model, liability model, management model) and the evolution of the balance sheet items are then specified in the following sections. 2.1 Overall Model Structure The main focus of our model is to simulate the future development of all assets and liabilities of a life insurance company. To this end, the future develop-
4 4 T. Gerstner, M. Griebel, M. Holtz, R. Goschnic, and and M. Haep ment of the capital marets, the policyholder behaviour and the company s management has to be modelled. We use a stochastic capital maret model, a deterministic liability model which describes the policyholder behaviour and a deterministic management model which is specified by a set of management rules which may depend on the stochastic capital marets. The results of the simulation are measured by statistical performance and ris figures which are based on the company s most important balance sheet items. They are used by the company to optimise management rules, lie the capital allocation, or product parameters, lie the surrender fee. The overall structure of the model is illustrated in Fig. 1. Fig. 1. Overall structure of the ALM model. We model all terms in discrete time. Here, we denote the start of the simulation by time t = 0 and the end of the simulation by t = T (in years). The interval [0, T ] is decomposed into K periods [t 1, t ] with t = t, = 1,..., K and a period length t = T/K of one month. The asset side consists of the maret value C of the company s assets at time t. On the liability side, the first item is the boo value of the actuarial reserve D, i.e., the guaranteed savings part of the policyholders after deduction of ris premiums and administrative costs. The second item is the boo value of the allocated bonuses B which constitute the part of the surpluses that have been credited to the policyholders via the profit participation. The free reserve F is a buffer account for future bonus payments. It consists of surpluses which have not yet been credited to the individual policyholder accounts, and is used to smooth capital maret oscillations and to achieve a stable and low-volatile return participation of the policyholders. The last item, the equity or company account Q, consists of the part of the surpluses which is ept by the shareholders of the company and is defined by Q = C D B F such that the sum of the assets equals the sum of the liabilities. Similar to the bonus reserve in [24], Q is a hybrid determined as the difference between a maret value C and the three boo values D, B and F. It may be interpreted as hidden reserve of the company as discussed in [29]. The balance
5 Numerical Simulation for Asset-Liability Management in Life Insurance 5 sheet items at time t, = 0,..., K, used in our model are shown in Table 1. In a sensitivity analysis for sample parameters and portfolios it is shown in Assets Liabilities Capital C Actuarial reserve D Allocated bonus B Free reserve F Equity Table 1. Simplified balance sheet of the life insurance company. Q [19] that this model captures the most important behaviour patterns of the balance sheet development of life insurance products. Similar balance sheet models have already been considered in, e.g., [2, 3, 24, 33, 29]. 2.2 Capital Maret Model We assume that the insurance company invests its capital either in fixed interest assets, i.e., bonds, or in a variable return asset, i.e., a stoc or a baset of stocs. For the modelling of the interest rate environment we use the Cox-Ingersoll-Ross (CIR) model [11]. The CIR model is a one-factor meanreversion model which specifies the dynamics of the short interest rate r(t) at time t by the stochastic differential equation dr(t) = κ(θ r(t))dt + r(t)σ r dw r (t), (1) where W r (t) is a standard Brownian motion, θ > 0 denotes the mean reversion level, κ > 0 denotes the reversion rate and σ r 0 denotes the volatility of the short rate dynamic. In the CIR model, the price b(t, τ) at time t of a zero coupon bond with a duration of τ periods and with maturity at time T = t + τ t can be derived in closed form by b(t, τ) = A(τ) e B(τ) r(t) (2) as an exponential affine function of the prevailing short interest rate r(t) with ( A(τ) = 2he (ˆκ+h)τ t/2 2h + (ˆκ + h)(e hτ t 1) ) 2κθ/σ 2 r, B(τ) = 2(e hτ t 1) 2h + (ˆκ + h)(e hτ t 1), and h = ˆκ 2 + 2σ 2 r. To model the stoc price uncertainty, we assume that the stoc price s(t) at time t evolves according to a geometric Brownian motion ds(t) = µs(t)dt + σ s s(t)dw s (t), (3) where µ R denotes the drift rate and σ s 0 denotes the volatility of the stoc return. By Itô s lemma, the explicit solution of this stochastic differential equation is given by
6 6 T. Gerstner, M. Griebel, M. Holtz, R. Goschnic, and and M. Haep s(t) = s(0) e (µ σ2 s /2)t+σsWs(t). (4) Usually, stoc and bond returns are correlated. We thus assume that the two Brownian motions satisfy dw s (t)dw r (t) = ρdt with a constant correlation coefficient ρ [ 1, 1]. These and other models which can be used to simulate the bond and stoc prices are discussed in detail, e.g., in [6, 25, 28]. In the discrete time case, the short interest rate, the stoc prices and the bond prices are defined by r = r(t ), s = s(t ) and b (τ) = b(t, τ). For the solution of equation (1), we use an Euler-Maruyama discretization 3 with step size t, which yields r = r 1 + κ(θ r 1 ) t + σ r r 1 t ξ r,, (5) where ξ r, is a N(0, 1)-distributed random variable. For the stoc prices one obtains s = s 1 e (µ σ2 s /2) t+σs t(ρξr, + 1 ρ 2 ξ s, ), (6) where ξ s, is a N(0, 1)-distributed random variable independent of ξ r,. Since Cov(ρξ r, + 1 ρ 2 ξ s,, ξ r, ) = ρ, the correlation between the two Wiener processes W s (t) and W r (t) is respected. More information on the numerical solution of stochastic differential equations can be found, e.g., in [22, 30]. 2.3 Management Model In this section, we discuss the capital allocation, the bonus declaration mechanism and the shareholder participation. Capital allocation We assume that the company rebalances its assets at the beginning of each period. Thereby, the company aims to have a fixed portion β [0, 1] of its assets invested in stocs, while the remaining capital is invested in zero coupon bonds with a fixed duration of τ periods. We assume that no bonds are sold before their maturity. Let P be the premium income at the beginning of period and let C 1 be the total capital at the end of the previous period. The part N of C 1 + P which is available for a new investment at the beginning of period is then given by 3 An alternative to the Euler-Maruyama scheme, which is more time consuming but avoids time discretization errors, is to sample from a noncentral chi-squared distribution, see [22]. In addition, several newer approaches exist to improve the balancing of time and space discretization errors, see, e.g., [20]. This and the time discretization error are not the focus of this article, though.
7 Numerical Simulation for Asset-Liability Management in Life Insurance 7 τ 1 N = C 1 + P n i b 1 (τ i), where n j denotes the number of zero coupon bonds which were bought at the beginning of period j. The capital A which is invested in stocs at the beginning of period is then determined by i=1 A = max{min{n, β(c 1 + P )}, 0} (7) so that the side conditions 0 A β(c 1 +P ) are satisfied. The remaining money N A is used to buy n = (N A )/b 1 (τ) zero coupon bonds with duration τ t. 4 The portfolio return rate p in period resulting from the above allocation procedure is then determined by ( ) τ 1 p = A + n i b,i /(C 1 + P ), (8) i=0 where A = A (s /s 1 1) and b,i = b(t, τ i 1) b(t 1, τ i) denote the changes of the maret values of the stoc and of the bond investments from the beginning to the end of period, respectively. Bonus declaration In addition to the fixed guaranteed interest, a variable reversionary bonus is annually added to the policyholder s account, which allows the policyholder to participate in the investment returns of the company (contribution principle). The bonus is declared by the company at the beginning of each year (principle of advance declaration) with the goal to provide a low-volatile, stable and competitive return participation (average interest principle). Various mathematical models for the declaration mechanism are discussed in the literature. In this article, we follow the approach of [24] where the declaration is based on the current reserve rate γ 1 of the company, which is defined in our framewor by the ratio of the free reserve to the allocated liabilities, i.e., γ 1 = F 1 D 1 + B 1. The annual interest rate is then defined by ẑ = max{ẑ, ω(γ 1 γ)}. Here, ẑ denotes the annual guaranteed interest rate, γ 0 the target reserve rate of the company and ω [0, 1] the distribution ratio or participation 4 Note that due to long-term investments in bonds it may happen that N < 0. This case of insufficient liquidity leads to n < 0 and thus to a short selling of bonds.
8 8 T. Gerstner, M. Griebel, M. Holtz, R. Goschnic, and and M. Haep coefficient which determines how fast excessive reserves are reduced. This way, a fixed fraction of the excessive reserve is distributed to the policyholders if the reserve rate γ 1 is above the target reserve rate γ while only the guaranteed interest is paid in the other case. In our model this annual bonus has to be converted into a monthly interest { (1 + ẑ ) z = 1/12 1 if mod 12 = 1 otherwise z 1 which is given to the policyholders in each period of this year. Shareholder participation Excess returns p z, conservative biometry and cost assumptions as well as surrender fees lead to a surplus G in each period which has to be divided among the free reserve F and the equity Q. In case of a positive surplus, we assume that a fixed percentage α [0, 1] is saved in the free reserve while the remaining part is added to the equity account. Here, a typical assumption is a distribution according to the 90/10-rule which corresponds to the case α = 0.9. If the surplus is negative, we assume that the required capital is taen from the free reserve. If the free reserves do not suffice, the company account has to cover the remaining deficit. The free reserve is then defined by F = max{f 1 + min{g, α G }, 0}. (9) The exact specification of the surplus G and the development of the equity Q is derived in Section Liability Model In this section, we discuss the modelling of the decrement of policies due to mortality and surrender and the development of the policyholder s accounts. Decrement model For efficiency, the portfolio of all insurance contracts is often represented by a reduced number m of model points. Each model point then represents a group of policyholders which are similar with respect to cash flows and technical reserves, see, e.g., [27]. By pooling, all contracts of a model point expire at the same time which is obtained as the average of the individual maturity times. We assume that the development of mortality and surrender is given deterministically and modelled using experience-based decrement tables. Let q i and u i denote the probabilities that a policyholder of model point i dies or surrenders in the -th period, respectively. The probabilities q i typically depend on the age, the year of birth and the gender of the policyholder while u i
9 Numerical Simulation for Asset-Liability Management in Life Insurance 9 often depends on the elapsed contract time. Let δ i denote the expected number of contracts in model point i at the end of period. Then, this number evolves over time according to δ i = ( 1 q i u i ) δ i 1. (10) We assume that no new contracts evolve during the simulation. Insurance products In the following, we assume that premiums are paid at the beginning of a period while benefits are paid at the end of the period. Furthermore, we assume that all administrative costs are already included in the premium. For each model point i = 1,..., m, the guaranteed part of the insurance product is defined by the specification of the following four characteristics: premium characteristic: (P1, i..., PK i ) where P i denotes the premium of an insurance holder in model point i at the beginning of period if he is still alive at that time. survival benefit characteristic: (E i,g 1,..., E i,g K ) where Ei,G denotes the guaranteed payments to an insurance holder in model point i at the end of period if he survives period. death benefit characteristic: (T i,g 1,..., T i,g i,g K ) where T denotes the guaranteed payment to an insurance holder in model point i at the end of period if he dies in period. surrender characteristic: (S i,g 1,..., S i,g K ) where Si,G denotes the guaranteed payment to an insurance holder in model point i at the end of period if he surrenders in period. The bonus payments of the insurance product to an insurance holder in model point i at the end of period in case of survival, death and surrender, are denoted by E i,b, T i,b and S i,b, respectively. The total payments Ei, T i and S i to a policyholder of model point i at the end of period in case of survival, death and surrender are then given by E i = E i,g + E i,b, T i = T i,g + T i,b and S i = S i,g + S i,b. (11) The capital of a policyholder of model point i at the end of period is collected in two accounts: the actuarial reserve D i for the guaranteed part and the bonus account B i for the bonus part. Both accounts can efficiently be computed in our framewor using the recursions and D i = 1 + z 1 q i (D 1 i + P) i E i,g qi 1 q i T i,g (12) B i = 1 + z 1 q i B 1 i + z z 1 q i (D 1 i + P) i E i,b qi 1 q i T i,b (13) which results from the deterministic mortality assumptions, see, e.g., [2, 46].
10 10 T. Gerstner, M. Griebel, M. Holtz, R. Goschnic, and and M. Haep Example 1. As a sample insurance product, an endowment insurance with death benefit, constant premium payments and surrender option is considered. Let P i denote the constant premium which is paid by each of the policyholders in model point i in every period. If they are still alive, the policyholders receive a guaranteed benefit E i,g and the value of the bonus account at maturity d i. In case of death prior to maturity, the sum of all premium payments and the value of the bonus account is returned. In case of surrender, the policyholder capital and the bonus is reduced by a surrender factor ϑ = 0.9. The guaranteed components of the four characteristics are then defined by P i = P i, E i,g = χ (d i ) E i,g, T i,g = P i and S i,g = ϑd i, where χ (d i ) denotes the indicator function which is one if = d i and zero otherwise. The bonus payments at the end of period are given by E i,b = χ (d i )B i, T i,b = B i and S i,b = ϑ B i. We will return to this example in Section Balance Sheet Model In this section, we derive the recursive development of all items in the simplified balance sheet introduced in Section 2.1. Projection of the assets In order to define the capital C at the end of period, we first determine the cash flows which are occurring to and from the policyholders in our model framewor. The premium P, which is obtained by the company at the beginning of period, and the survival payments E, the death payments T, and the surrender payments S to policyholders, which tae place at the end of period, are obtained by summation of the individual cash flows (11), i.e., m m m m P = δ 1 i P, i E = δ i E, i T = qδ i 1 i T, i S = u i δ 1 i S, i i=1 i=1 (14) where the numbers δ i are given by (10). The capital C is then recursively given by C = (C 1 + P ) (1 + p ) E T S (15) where p is the portfolio return rate defined in equation (8). Projection of the liabilities The actuarial reserve D and the allocated bonus B are derived by summation of the individual policyholder accounts (12) and (13), i.e., i=1 i=1
11 Numerical Simulation for Asset-Liability Management in Life Insurance 11 m m D = δ i D i and B = δ i B. i i=1 In order to define the free reserve F, we next determine the gross surplus G in period which consists in our model of interest surplus and surrender surplus. The interest surplus is given by the difference between the total capital maret return p (F 1 + D 1 + B 1 + P ) on policyholder capital and the interest payments z (D 1 + B 1 + P ) to policyholders. The surrender surplus is given by S /ϑ S. The gross surplus in period is thus given by G = p F 1 + (p z ) (D 1 + B 1 + P ) + (1/ϑ 1)S. The free reserve F is then derived using equation (9). Altogether, the company account Q is determined by i=1 Q = C D B F. Note that the cash flows and all balance sheet items are expected values with respect to our deterministic mortality and surrender assumptions from Section 2.4, but random numbers with respect to our stochastic capital maret model from Section 2.2. Performance figures To analyse the results of a stochastic simulation, statistical measures are considered which result from an averaging over all scenarios. Here, we consider the path-dependent cumulative probability of default ( ) PD = P min Q j < 0 j=1,..., as a measure for the ris while we use the expected future value E[Q ] of the equity as a measure for the investment returns of the shareholders in the time interval [0, t ]. Due to the wide range of path-dependencies, guarantees and option-lie features of the insurance products and management rules, closed-form representations for these statistical measures are in general not available so that one has to resort to numerical methods. It is straightforward to include the computation of further performance and ris measures lie the variance, the value-at-ris, the expected shortfall or the return on ris capital. To determine the sensitivity f (v) = f(v)/ v of a given performance figure f to one of the model parameters v, finite difference approximations or more recent approaches, lie, e.g., smoing adjoints [21], can be employed. 3 Numerical Simulation In this section, we discuss the efficient numerical simulation of the ALM model described in Section 2. The number of operations for the simulation of a single scenario of the model is of order O(m K) and taes about 0.04 seconds
12 12 T. Gerstner, M. Griebel, M. Holtz, R. Goschnic, and and M. Haep on a dual Intel(R) Xeon(TM) CPU 3.06GH worstation for a representative portfolio with m = 500 model points and a time horizon of K = 120 periods. The number of scenarios which have to be generated depends on the accuracy requirements, on the model parameters 5 and on the employed numerical method. In the following, we first rewrite the performance figures of the model as high-dimensional integrals. Then, we survey numerical methods which can be applied to their computation, discuss their dependence on the effective dimension and review techniques which can reduce the effective dimension in certain cases. 3.1 Representation as High-Dimensional Integrals It is helpful to represent the performance figures of the ALM simulation as high-dimensional integrals to see how more sophisticated methods than Monte Carlo simulation can be used for their numerical computation. To derive such a representation, recall that the simulation of one scenario of the ALM model is based on 2 K independent normally distributed random numbers y = (y 1,..., y 2K ) = (ξ s,1,..., ξ s,k, ξ r,1,..., ξ r,k ) N(0, 1). These numbers specify the stoc price process (6) and the short rate process (5). Then, the term structure, the asset allocation, the bonus declaration, the shareholder participation and the development of all involved accounts can be derived using the recursive equations of the previous sections. Altogether, the balance sheet items C K, B K, F K and Q K at the end of period K can be regarded as (usually very complicated) deterministic functions C K (y), B K (y), F K (y), Q K (y) depending on the normally distributed vector y IR 2K. As a consequence, the expected values of the balance sheet items at the end of period K can be represented as 2 K-dimensional integrals, e.g., E[Q K ] = IR2K Q K (y) e y T y/2 dy (16) (2π) K for the equity account. Often, monthly discretizations of the capital maret processes are used. Then, typical values for the dimension 2K range from depending on the time horizon of the simulation. Transformation The integral (16) can be transformed into an integral over the 2K-dimensional unit cube which is often necessary to apply numerical integration methods. By the substitution y i = Φ 1 (x i ) for i = 1,..., 2K, where Φ 1 denotes the inverse cumulative normal distribution function, we obtain E[Q K ] = IR2K Q K (y) e y T y/2 (2π) K dy = [0,1] d f(x) dx (17) 5 The model parameters affect important numerical properties of the model, e.g. the effective dimension (see Section 3.3) or the smoothness.
13 Numerical Simulation for Asset-Liability Management in Life Insurance 13 with d = 2K and f(x) = Q (Φ 1 (x)). For the fast computation of Φ 1 (x i ), we use Moro s method [35]. Note that the integrand (17) is unbounded on the boundary of the unit cube, which is undesirable from a numerical as well as theoretical point of view. Note further that different transformations to the unit cube exist (e.g. using the logistic distribution or polar coordinates) and that also numerical methods exist which can directly be applied to the untransformed integral (16) (e.g. Gauss-Hermite rules). 3.2 Numerical Methods for High-Dimensional Integrals There is a wide range of methods (see, e.g., [12]) available for numerical multivariate integration. Mostly, the integral (17) is approximated by a weighted sum of n function evaluations n f(x) dx w i f(x i ) (18) [0,1] d with weights w i IR and nodes x i IR d. The number n of nodes corresponds to the number of simulation runs. Depending on the choice of the weights and nodes, different methods with varying properties are obtained. Here, the dimension as well as the smoothness class of the function f should be taen into account. Monte Carlo In practise, the model is usually simulated by the Monte Carlo (MC) method. Here, all weights equal w i = 1/n and uniformly distributed sequences of pseudo-random numbers x i (0, 1) 2K are used as nodes. This method is independent of the dimension, robust and easy to implement but suffers from a relative low probabilistic convergence rate of order O(n 1/2 ). This often leads to very long simulation times in order to obtain approximations of satisfactory accuracy. Extensive sensitivity investigations or the optimisation of product or management parameters, which require a large number of simulation runs, are therefore often not possible. Quasi-Monte Carlo Quasi-Monte Carlo (QMC) methods are equal-weight rules lie Monte Carlo. Instead of pseudo-random numbers, however, deterministic low-discrepancy sequences (see, e.g., [37, 22]) or lattices (see, e.g., [40]) are used as point sets which are chosen to yield better uniformity than random samples. Some popular choices are Halton, Faure, Sobol and Niederreiter-Xing sequences and extensible shifted ran-1 lattice rules based on Korobov or fast component-bycomponent constructions. From the Kosma-Hlawa inequality it follows that convergence rate of QMC methods is of order O(n 1 (log n) d ) for integrands of bounded variation which is asymptotically better than the O(n 1/2 ) rate of i=1
14 14 T. Gerstner, M. Griebel, M. Holtz, R. Goschnic, and and M. Haep MC. For periodic integrands, lattice rules can achieve convergence of higher order depending on the decay of the Fourier coefficients of f, see [40]. Using novel digital net constructions (see [14]), QMC methods can also be obtained for non-periodic integrands which exhibit convergence rates larger than one if the integrands are sufficiently smooth. Product methods Product methods for the computation of (17) are easily obtained by using the tensor products of the weights and nodes of one-dimensional quadrature rules, lie, e.g., Gauss rules (see, e.g., [12]). These methods can exploit the smoothness of the function f and converge with order O(n s/d ) for f C s ([0, 1] d ). This shows, however, that product methods suffer from the curse of dimension, meaning that the computing cost grows exponentially with the dimension d of the problem, which prevents their efficient applications for high-dimensional (d > 5) applications lie ALM simulations. Sparse grids Sparse grid (SG) quadrature formulas are constructed using certain combinations of tensor products of one-dimensional quadrature rules, see, e.g., [9, 15, 23, 38, 42]. In this way, sparse grids can, lie product methods, exploit the smoothness of f and also obtain convergence rates larger than one. In contrast to product methods, they can, however, also overcome the curse of dimension lie QMC methods to a certain extent. They converge with order O(n s (log n) (d 1)(s 1) ) if the integrand belongs to the space of functions which have bounded mixed derivatives of order s. Sparse grid quadrature formula come in various types depending on the one-dimensional basis integration routine, lie the trapezoidal, the Clenshaw-Curtis, the Patterson, the Gauss-Legendre or the Gauss-Hermite rule. In many cases, the performance of sparse grids can be enhanced by local adaptivity, see [5, 8], or by a dimension-adaptive grid refinement, see [16]. 3.3 Impact of the Dimension In this section, we discuss the dependence of MC, QMC and SG methods on the nominal and the effective dimension of the integral (17). Tractability In contrast to MC, the convergence rate of QMC and SG methods still exhibit a logarithmic dependence on the dimension. Furthermore, also the constants in the O-notation depend on the dimension of the integral. In many cases (particularly within the SG method) these constants increase exponentially with the dimension. Therefore, for problems with high nominal dimension d, such as the ALM of life insurance products, the classical error bounds of the
15 Numerical Simulation for Asset-Liability Management in Life Insurance 15 previous section are no longer of any practical use to control the numerical error of the approximation. For instance, even for a moderate dimension of d = 20 and for a computationally unfeasibly high number n = of function evaluations, n 1 (log n) d > n 1/2 still holds in the QMC and the MC error bounds. For classical Sobolov spaces with bounded derivatives up to a certain order, it can even be proved (see [39, 41]) that integration is intractable, meaning that for these function classes deterministic methods of the form (18) can never completely avoid the curse of dimension. For weighted Sobolov spaces, however, it is shown in [39, 41] that integration is tractable if the weights decay sufficiently fast. In the next paragraph and in Section 4.3 we will give some indications that ALM problems indeed belong to such weighted function spaces. ANOVA decomposition and effective dimension Numerical experiments show that QMC and SG methods often produce much more precise results than MC methods for certain integrands even in hundreds of dimensions. One explanation of this success is that QMC and SG methods can, in contrast to MC, tae advantage of low effective dimensions. QMC methods profit from low effective dimensions by the fact that their nodes are usually more uniformly distributed in smaller dimensions than in higher ones. SG methods can exploit different weightings of different dimensions by a dimension-adaptive grid refinement, see [16]. The effective dimension of the integral (17) is defined by the ANOVA decomposition, see, e.g., [10]. Here, a function f : IR d IR is decomposed by f(x) = f u (x u ) with f u (x u ) = f(x)dx {1,...,d}\u f v (x v ) [0,1] d u v u u {1,...,d} into 2 d sub-terms f u with u {1,..., d} which only depend on variables x j with j u. Thereby, the sub-terms f u describe the dependence of the function f on the dimensions j u. The effective dimension in the truncation sense of a function f : IR d IR with variance σ 2 (f) is then defined as the smallest integer d t, such that v {1,...,d t} σ2 v(f) 0.99 σ 2 (f) where σu(f) 2 denotes the variances of f u. The effective dimension d t roughly describes the number of important variables of the function f. The effective dimension in the superposition sense is defined as the smallest integer d s, such that v d s σv(f) σ 2 (f) where v denotes the cardinality of the index set v. It roughly describes the highest order of important interactions between variables in the ANOVA decomposition. For the simple function f(x 1, x 2, x 3 ) = x 1 e x2 +x 2 with d = 3, we obtain d t = 2 and d s = 2 for instance. For large d, it is no longer possible to compute all 2 d ANOVA sub-terms. The effective dimensions can still be computed in many cases, though. For details and an efficient algorithm for the computation of the effective dimension in the truncation sense we refer to [44]. For the more difficult problem to com-
16 16 T. Gerstner, M. Griebel, M. Holtz, R. Goschnic, and and M. Haep pute the effective dimension in the superposition sense, we use the recursive method described in [45]. Dimension reduction Typically, the underlying multivariate Gaussian process is approximated by a random wal discretization. In many cases, a substantial reduction of the effective dimension in the truncation sense and an improved performance of the deterministic integration schemes can be achieved if the Brownian bridge or the principal component (PCA) decompositions of the covariance matrix of the underlying Brownian motion is used instead as it was proposed in [1, 36] for option pricing problems. The Brownian bridge construction differs from the standard random wal construction in that rather than constructing the increments sequentially, the path of the Gaussian process is constructed in a hierarchical way which has the effect that more importance is placed on the earlier variables than on the later ones. The PCA decomposition, which is based on the eigenvalues and -vectors of the covariance matrix of the Brownian motion, maximises the concentration of the total variance of the Brownian motion in the first few dimensions. 6 Its construction requires, however, O(d 2 ) operations instead of O(d) operations which are needed for the random wal or for the Brownian bridge discretization. For large d, this often increases the run times of the simulation and limits the practical use of the PCA construction. 4 Numerical Results We now describe the basic setting for our numerical experiments and investigate the sensitivities of the performance figures from Section 2.5 to the input parameters of the model. Then, the riss and returns of two different asset allocation strategies are compared. Finally, we compute the effective dimensions of the integral (17) in the truncation and superposition sense and compare the efficiency of different numerical approaches for its computation. 4.1 Setting We consider a representative model portfolio with 50, 000 contracts which have been condensed into 500 equal-sized model points. The data of each model point i is generated according to the following distribution assumptions: entry age x i N(36, 10), exit age x i N(62, 4), current age x i 0 U(x i, x i ) and monthly premium P i U(50, 500) where N(µ, σ) denotes the normal 6 Note that without further assumptions on f it is not clear which construction leads to the minimal effective dimension due to possibly non-linear dependencies of f on the underlying Brownian motion. As a remedy, also more complicated covariance matrix decompositions can be employed which tae into account the function f as explained in [26].
17 Numerical Simulation for Asset-Liability Management in Life Insurance 17 stoc price model interest rate model correlation µ = 8% σ s = 20% κ = 0.1 θ = 4% σ r = 5% r 0 = 3% λ 0 = 5% ρ = 0.1 E[Q K] E[F K] PD K Table 2. Capital maret parameters p used in the simulation and their partial derivatives f (p)/f(p) for f {PD K, E[Q K], E[F K]}. asset allocation bonus declaration shareholder product parameters solv. rate β = 10% τ = 3 ω = 25% γ = 15% α = 90% ϑ = 90% z = 3% γ 0 = 10% E[Q K] E[F K] PD K Table 3. Solvency rate, management and product parameters p used in the simulation and their partial derivatives f (p)/f(p) for f {PD K, E[Q K], E[F K]}. distribution with mean µ and variance σ, and U(a, b) denotes a uniform distribution in the interval [a, b]. In addition, the side conditions 15 x i 55 and 55 x i 70 are respected. The probability that the contracts of a model point belong to female policyholders is assumed to be 55%. From the difference of exit age and current age the maturity time d i = x i x i of the contracts is computed. As sample insurance product, an endowment insurance with death benefit, constant premium payments and surrender option is considered as described in Example 1. For simplicity, we assume that the policies have not received any bonus payments before the start of the simulation, i.e., B0 i = 0 for all i = 1,..., m. We tae the probabilities q i of death from the DAV 2004R mortality table and choose exponential distributed surrender probabilities u i = 1 e 0.03 t. At time t 0, we assume a uniform bond allocation, i.e., n j = (1 β)c 0 / τ 1 i=0 b 0(i) for j = 1 τ,..., 0. We assume Q 0 = 0 which means that the shareholders will not mae additional payments to the company to avoid a ruin. This way, E[Q ] serves as a direct measure for the investment returns of the shareholders in the time interval [0, t ]. The total initial reserves of the company are then given by F 0 = γ 0 D 0. In the following, we choose a simulation horizon of T = 10 years and a period length of t = 1/12 years, i.e., K = 120. In our numerical tests we use the capital maret, product and management parameters as displayed in the second rows of Table 2 and 3 unless stated otherwise. In Table 2 and 3 also the sensitivities f (v)/f(v) (see Section 2.5) are displayed for different functions f {PD K, E[Q K ], E[F K ]} and different model input parameter v, e.g., PD K /( µ PD K ) =
18 18 T. Gerstner, M. Griebel, M. Holtz, R. Goschnic, and and M. Haep 4.2 Capital Allocation To illustrate possible applications of the ALM model, we compare the constantmix capital allocation strategy of Section 2.3 with an CPPI (constant proportion portfolio insurance) capital allocation strategy (see, e.g., [34]) with respect to the resulting default ris PD K and returns E[Q K ]. Within the CPPI strategy, the proportion of funds invested in (risy) stocs is lined to the current amount of reserves. The strategy is realised in our model framewor by replacing β(c 1 + P ) in equation (7) by β F 1 with β R +. The resulting ris-return profiles of the constant-mix strategy and of the CPPI strategy are displayed in Fig. 2 for different choices of β. Fig. 2. Ris-return profiles of the different capital allocation strategies. We see that the slightly negative correlation ρ = 0.1 results in a diversification effect such that the lowest default ris is not attained at β = 0 but at about β = 2.5% in the constant-mix case and at about β = 40% in the CPPI case. Higher values of β lead to higher returns but also to higher riss. As an interesting result we further see that the CPPI strategy almost always leads to portfolios with much higher returns at the same ris and is therefore clearly superior to the constant-mix strategy almost independently of the ris aversion of the company. The only exception is a constant-mix portfolio with a stoc ratio β of 2.5 4%, which could be an interesting option for a very ris averse company. 4.3 Effective Dimension For the setting of Section 4.1, we determine in this section the effective dimensions d t and d s of the integral (17) in the truncation and superposition sense, respectively, see Section 3.3. The effective dimensions depend on the
19 Numerical Simulation for Asset-Liability Management in Life Insurance 19 nominal dimension d, on the discretization of the underlying Gaussian process and on all other model parameters. In Table 4, the effective dimensions d t are displayed which arise by the methods described in [44] for different nominal dimensions d if the random wal, the Brownian bridge and the principal component (PCA) path construction is employed, respectively. One can see that the Brownian bridge and PCA path construction lead to a large reduction of the effective dimension d t compared to the random wal discretization. In the latter case, the effective dimension d t is almost as large as the nominal dimension d while in the former cases the effective dimensions are almost insensitive to the nominal dimensions and are bounded by only d t = 16 even for very large dimensions as d = 512. In case of the PCA construction, d t is even slightly decreasing for large d which is related to the so-called concentration of measure phenomenon, see [31]. Further numerical computations using the method described in [45] show that the ALM problem is also of very low effective dimension d s in the superposition sense. Here, we only consider moderately high nominal dimensions due to the computational costs which increase with d. For d 32, we obtain that the integral (17) is nearly additive, i.e. d s = 1, independent of d and independent of the covariance matrix decomposition. Note that the effective dimensions are affected by several parameters of the ALM model. More results which illustrate how the effective dimensions in the truncation sense vary in dependence of the capital maret model and of other parameters can be found in [18]. d Random wal Brownian bridge Principal comp Table 4. Truncation dimensions d t of the ALM integrand (17) for different nominal dimensions d and different covariance matrix decompositions. 4.4 Convergence Rates In this section, we compare the following methods for the computation of the expected value (17) with the model parameters specified in Section 4.1: MC Simulation, QMC integration based on Sobol point sets (see [32, 43]), dimension-adaptive SG based on the Gauss-Hermite rule (see [16]). In various numerical experiments, the Sobol QMC method and the dimensionadaptive Gauss-Hermite SG method turned out to be the most efficient representatives of several QMC variants (we compared Halton, Faure, Sobol low
20 20 T. Gerstner, M. Griebel, M. Holtz, R. Goschnic, and and M. Haep discrepancy point sets and three different lattice rules with and without randomisation) and of several SG variants (we compared trapezoidal, Clenshaw- Curtis, Patterson, Gauss-Legendre and Gauss-Hermite rule and different grid refinement strategies), respectively. The results for d = 32 and d = 512 are summarised in Fig. 3 where the number n of function evaluations is displayed which is needed to obtain a given accuracy. In both cases we used the Brownian bridge path construction for the stoc prices and short interest rates. One Fig. 3. Errors and required number of function evaluations of the different numerical approaches to compute the expected value (17) with d = 32 (left) and with d = 512 (right) for the model parameters specified in Section 4.1. can see that the QMC method clearly outperforms MC simulation in both examples. The QMC convergence rate is close to one and nearly independently of the dimension. Moderate accuracy requirements of about are obtained by the QMC method about 100-times faster as by MC simulation. For higher accuracy requirements, the advantage of the QMC method is even more pronounced. Recall that these results can not be explained by the Kosma-Hlawa inequality but by the very low effective dimension of the ALM problem, see Section 4.3. The performance of the SG method deteriorates for very high dimensions. In the high dimensional case d = 512, the SG method is not competitive to QMC. For the moderately high dimension d = 32, sparse grids are the most efficient method with a very high convergence rate of almost three. With 129 function evaluation already an accuracy of 10 6 is achieved. Further numerical experiments indicate that the performance of the SG method is more sensitive than (Q)MC to different choices of model parameters which affect the smoothness of the integrand, lie more aggressive bonus declaration schemes and more volatile financial marets.
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# Goods and
The IS-LM Model
## The Goods Market
and the IS Relation
Equilibrium in the goods market
exists when production, Y, is equal
to the demand for goods, Z.
In the simple model (in chapter 3),
the interest rate did not affect the
demand for goods. The equilibrium
condition was given by:
Y C (Y T ) I G
## Investment, Sales (Y), and the
Interest Rate (i)
Now, we no longer assume I (investment)
is constant
We capture the effects of two factors
affecting investment:
The level of sales/income (+)
The interest rate (-)
I I (Y ,i)
The Determination of
Output
Taking into account the investment
relation above, the equilibrium
condition in the goods market
becomes:
Y C (Y T ) I (Y ,i) G
The Determination of
Output
Equilibrium in the Goods
Market
The demand for goods is
an increasing function of
output. Equilibrium
requires that the demand
for goods be equal to
output.
## Deriving the IS Curve
The Effects of an
Increase in
the Interest Rate on
Output
An increase in the
interest rate decreases
the demand for goods
at any level of output.
The IS Curve
Shifts of the IS
Curve
An increase
in taxes...
Financial Markets
and the LM Relation
The interest rate is determined by
the equality of the supply of and
the demand for
M money:
\$ Y L (i)
M = nominal money stock
\$YL(i) = demand for money
\$Y = nominal income
i = nominal interest rate
## Real Money, Real Income,
and the Interest Rate
The LM relation: In equilibrium, the
real money supply is equal to the real
money demand, which depends on real
M interest rate, i:
income, Y, and the
Y L (i)
Recall: before, we had the same equation but in nominal instead of real
terms (nominal income and nominal money supply). Dividing both
sides by P (the price level) gives us the equation above.
## Deriving the LM Curve
The Effects of an
Increase in Income on
the Interest Rate
Shifts of the LM
Curve
An
increase
in
money...
## The IS and the LM Relations
Together
The IS-LM Model
Equilibrium in the
goods market (IS).
Equilibrium in financial
markets (LM).
When the IS curve
intersects the LM
curve, both goods and
financial markets are
in equilibrium.
IS r e la tio n : Y C (Y T ) I (Y ,i ) G
L M r e la tio n :
M
Y L (i)
P
## Fiscal Policy, the Interest Rate and
the IS Curve
Fiscal contraction: a fiscal policy
that reduces the budget deficit.
Reducing G or increasing T
## Fiscal expansion: increasing the
budget deficit.
Increasing G or decreasing T
## Taxes (T) and government
expenditures (G) affect the IS curve,
not the LM curve.
## Fiscal Policy, the Interest Rate and
the IS Curve
The Effects of an
Increase in Taxes
## Monetary Policy, the Interest Rate,
and the LM Curve
Monetary contraction
(tightening) refers to a
decrease in the money supply.
An increase in the money supply
is called monetary expansion.
Monetary policy affects only the
LM curve, not the IS curve.
## Monetary Policy, the Interest Rate,
and the LM Curve
The Effects of a
Monetary Expansion
## Using a Policy Mix
The Effects of Fiscal and Monetary Policy.
Shift of IS
Shift of
LM
Movement of
Output
Movement in
Interest Rate
Increase in taxes
left
none
down
down
Decrease in taxes
right
none
up
up
Increase in spending
right
none
up
up
Decrease in spending
left
none
down
down
Increase in money
none
down
up
down
Decrease in money
none
up
down
up
Crowding Out
Given Md = 0.25 + 62.5i; IS -----> Y = 4250 125i
and LM -------> Y = 2000 + 250i. Is there crowding out
if government spending increases 1500? How much?
IS0 ------> Y = 4250 125i
LM -----> Y = 2000 + 250i
IS1 ------> Y = 5750 125i
Initial equilibrium IS0 = LM at Y = 3500 and r 0 = 6%.
If government spending increases, then IS1 = LM at Y
= 4500 and r1 = 10%.
If interest rate is unchanged (6%), then after
government spending increases:
Y = 5750 125r
= 5750 125(6)
= 5000
So, crowding out is 5000 4500 = 500
r
IS1
LM0
IS0
10
3500
Y
4500 5000
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Session 9, Part A:
Models for the Multiplication and Division of Fractions
In This Part: Area Model for Multiplication | Try It Yourself | Area Model for Division
The Common Denominator Model for Division | Translating the Process to Decimals
We can apply the area model for the multiplication of fractions to visualize the division of two fractions when each is less than 1. To model division with fractions, we more or less reverse the process used for multiplication. We start with an area we're looking for, and we find one of the missing factors that makes up that area. Note 3
For example, here's how we would use the model to demonstrate the problem 1/4 2/3:
Shade one square, partitioned vertically, to represent 1/4 (as in the multiplication model, it's shaded purple): Superimpose a square partitioned into thirds, positioned horizontally, onto the fourths square, and draw a bracket to the right of the thirds square to show the size of 2/3:
What you see now is the purple (1/4) area and the size of one of the factors that made that area.
We know from the multiplication model that the product of 2/3 and another factor (the quotient) defines an area equivalent in size to 1/4. To find the quotient, we need to move the top part of the purple area so that it's the same height as the 2/3 factor.
Subdivide the fourths square to make an eighths square: Move the top two purple pieces into the 2/3 height area (the area within the 2/3 bracket): Now shade the rectangles immediately to the right and immediately above the purple area:
This shows that there are 3 • 2, or 6, purple parts out of 8 • 3, or 24, parts in all. The purple area equals 1/4, and it came from the product of 2/3 multiplied by what? We can see that the other factor is 3/8.
Problem A3 A town plans to build a community garden that will cover 2/3 of a square mile. They would like to situate it on a pasture of an old horse farm. One dimension of the garden area will be determined by a fence that is 3/4 of a mile long. Use the area model for division to determine the other dimension of the new garden area.
Problem A4 Describe how the area model shows that the quotient of two positive fractions, each less than 1, must be larger than the first fraction.
Session 9: Index | Notes | Solutions | Video
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https://stats.stackexchange.com/questions/174174/how-to-predict-using-ordered-probit-regression-and-calculate-prediction-accuracy?noredirect=1
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# How to predict using ordered probit regression and calculate prediction accuracy?
I want to do an ordered probit regression, then cross-validate model prediction accuracy with 80% data for training and 20% for validation, and calculate RMSE for predictions.
Consider this dataset:
X Y
----------
2.3 1
3.1 2
3.5 2
10.0 5
6.8 4
5.0 3
5.4 2
3.2 1
I did this:
x=c(2.3,3.1,3.5,10.0,6.8,5.0,5.4,3.2)
y=c(1,2,2,5,4,3,2,1)
myData=data.frame(cbind(x,y))
library("MASS")
reg=polr(as.factor(myData$y)~myData$x,data=myData,method="probit")
I saw this question, but I couldn't fully understand. Suppose myValidationData contains 20% of data which I want to use for validation. So, I would do:
fit=predict(reg,type="probs")
x=c(5.6, 5.1)
y=c(3,3)
myValidationData=data.frame(cbind(x,y))
This is how I tried to predict, but is it correct, when I want to cross-validate?
fit=predict(reg,data=myValidationData,type="probs")
How should I measure RMSE? And, how can I plot the prediction?
The R rms package has many capabilities for validating ordinal regression models. Start with the orm function. Note that split-sample validation takes an extremely large sample size to work. You might be better off with bootstrap validate as implemented in the rms validate and calibrate functions.
Measures of predictive accuracy for ordinal $Y$ include
• Generalized $c$-index (generalized ROC area) from Somers' $D_{xy}$ rank correlation
• Spearman $\rho$
• Other rank correlation measures - these are all measures of pure predictive discrimination
• Generalized $R^2$ based on model likelihood ratio $\chi^2$ statistic
• Calibration accuracy for $Prob(Y \geq y | X)$ using a nonparametric smooth calibration curve
• Thanks for the answer. My actual data is about ~2000 elements for training and ~200 elements for validation, so I think it would be fine.
– Ho1
Commented Sep 26, 2015 at 5:03
• There is a fundamental problem: How can I calculate prediction accuracy on ordered categorical data? stats.stackexchange.com/questions/174255
– Ho1
Commented Sep 26, 2015 at 10:51
• That's what the answer above was trying to accomplish. Commented Sep 27, 2015 at 15:54
That looks correct; note that after you make the myData data frame, you can use:
myData$y = factor(myData$y)
reg <- polr(y ~ x, data = myData, method = "probit")
Later, you can make the validation data with:
myValidationData <- data.frame(x = c(5.6, 5.1), y = c(3,3))
Your syntax works fine, I just thought this was a bit "cleaner".
Here's a great link on ordinal regression that includes syntax to plot the predicted values (it's actually about SEM with categorical variables, but there's a great section on ordinal regression in the middle):
http://www.personality-project.org/r/tutorials/summerschool.14/rosseel_sem_cat.pdf
I'm not sure how to measure the RMSE, but you should be able to use
(residuals(reg))
To get the model errors.
• I get NULL when I invoke (residuals(reg)).
– Ho1
Commented Sep 26, 2015 at 5:09
• Hmmm, it looks like polr() doesn't supply a residuals object. In that case, try predict(reg) - myData\$y to get the residuals (without a newdata argument, predict() uses the dataset used to fit the model). Commented Sep 26, 2015 at 22:15
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PDA
View Full Version : hcrisp
hcrisp
10-27-2008, 11:50 AM
Here's another puzzler.
I have a mxn array representing elevation data. To make it simple:
WAVE> m = [10.,20.,30.]
WAVE> n = [20.,30.,40.]
WAVE> elev = [[1.,2.,3.], [1.,3.,5.], [1.,4.,7.]]
Next I take a "walk" across the mxn array using (x,y) pairs that fall in the m and n ranges.
WAVE> walk = [[15.2, 27.3], [21.9, 23.1], [29.3, 20.5]]
How do I evaluate the interpolated elevations at these walk points? Is there a PV-WAVE function that does this?
allan
10-27-2008, 02:20 PM
the following will do bilinear interpolation for this problem
for i=0,2 do pm, intrp(intrp(elev,0,walk(0,i),z=m),1,walk(1,i),z=n)
allan
hcrisp
10-28-2008, 08:58 AM
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or
Find what you need to study
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# 1.4 Representing a Categorical Variable with Graphs
L
Lusine Ghazaryan
Jed Quiaoit
L
Lusine Ghazaryan
Jed Quiaoit
You might recall from earlier that can be represented using tables and/or graphs. This section will provide more context that'll equip us with the ability to eventually construct and describe numerical or of data distributions. ๐
As for why graphs are big in statistics, and statistics are powerful tools for understanding and summarizing data. Graphs can help you visualize the patterns and relationships in your data, and statistics can help you quantify and describe those patterns. By using both and statistics, you can gain a deeper understanding of your data and communicate that understanding to others!
## Bar Graphs
(or bar graphs) are used to display frequencies (counts) or relative frequencies (proportions) for categorical data. The height or length of each bar in a corresponds to either the number or proportion of observations falling within
each category. ๐
To create a , you first need to decide on the categories you want to include. Each category corresponds to a separate bar on the graph. The height of each bar represents the frequency or count of observations in that category. All the bars have the same width, and there is a gap between adjacent bars to distinguish them from each other. ๐
When translated into a step-by-step procedure, here's how we would create a :
1. Determine the categories you want to include in the graph.
2. Count the number of observations in each category.
3. Mark the frequencies on the vertical axis and the categories on the horizontal axis.
4. Draw the bars, with the height of each bar representing the frequency of the corresponding category.
5. Add a title and axis labels to the graph to help interpret the data.
It's important to choose an appropriate and consistent scale for the vertical axis. You should also consider adding a legend to the graph if you have multiple series of data that you want to compare.
To keep it short, here is the of stress on the job. We can also use relative frequencies or percentages to construct the . You can be creative and color each category with a different color. It will beย visually attractive and easier to compare them.
Source: Prem S. Mann: Introductory Statistics. John Wiley and Sons Inc. 2020
## Pie Charts
A is a circular graph that is divided into slices, with each slice representing a different category. The size of each slice is proportional to the fraction of the whole that is represented by that category. are often used to show the relative proportions of different categories within a dataset. ๐ฅง
To create a , you'll have to keep the following steps in mind:
1. Determine the categories you want to include in the . (Example: Commuter, non-commuter)
2. Calculate the fraction of the whole that is represented by each category (Example: Out of 50 respondents, 30 commuters would occupy 3/5ths of the pie, while 20 non-commuters would occupy the remaining 2/5ths of the pie).
3. Draw a circle and divide it into slices that are proportional to the fractions calculated in step 2.
4. Label each slice with the corresponding category and the percentage it represents.
5. Add a title to the to help interpret the data.
It's important to keep in mind that are best used to compare the relative proportions (percentages and relative frequencies, for example) of different categories. They're not as effective at showing precise values or small differences between categories. If you want to show detailed values or compare the values of multiple categories, it is usually better to use a different type of graph, such as a bar chart.
๐ก Tips:
• The choice between and will depend on how many categories that variable of your interest assumes and the size of it. Whenever you have many categories or few categories with about the same frequencies, then the should be your first choice. If the pie has many slices or slices of the same size, it will be hard to compare the groups.
• Be careful of quantity distortions and keeping the area principle.
## Contingency Table (Two-Way Table)
Now that we know how to represent data in tables and charts, let's add one more character to the tables gang to keep things evenly balanced!
A contingency table is a type of table that is used to organize and (later on) analyze categorical data. It shows how the observations in a dataset are distributed among different categories of two or more variables. Contingency tables can help in understanding relationships between variables and identifying patterns or trends in the data. ๐จ
To create a contingency table, you'll have to:
1. Determine the variables you want to include in the table.
2. Count the number of observations in each category of each variable.
3. Organize the counts in a table, with each row representing a category of one variable and each column representing a category of the other variable.
4. Add row and column totals to the table. (This step is the easiest to forget!)
5. Analyze the table to identify any patterns or trends in the data. (This is important when establishing context and responding to Multiple Choice and Free Response Questions in the AP exam!)
If the numbers in the cells of the contingency table are the same for all categories, we can say that the variables are independent, If the numbers in the cells are different for different categories (with some having higher values than others), then the variables might be related. For example, if you are analyzing data on the relationship between gender and income, you might find that the proportions of men and women in different income categories are different, indicating some sort of relationship between the two variables.
๐ฅ Watch: AP Stats - Analyzing Categorical Data
## Real-Life Applications: To Trust or Not To Trust a Bar/Pie Chart?
Chances are, you've probably seen a bar or in some shape or form before in the news, media you consume, or even other textbooks. It's important to remember that they shouldn't be taken immediately at face value as they could be easily misused. To help inform whether bar/ are reliable or not, here are examples of ways they are commonly misused:
• Using bar/ to compare variables on different scales: Charts are best used to compare categories or groups that are on the same scale. If you are comparing variables that are on different scales, it can be difficult to accurately compare the sizes of the bars/pie slices.
Source: Infogram
• Using bar/ to show continuous data: Charts are best used to show categorical data, not continuous data. If you have continuous data, it is usually better to use a different type of graph, such as a line graph or scatterplot.
• Using bar/ to show small differences: Charts are not very effective at showing small differences between categories. If the differences between the categories are small, it may be difficult to accurately interpret the graph.
• Using bar/ to show trends over time: Charts are not well suited for showing trends over time. For this purpose, it is usually better to use a line graph or a time series plot.
• Using bar/ to show more than two variables: Charts are typically used to compare two variables. If you want to show more than two variables, it is usually better to use a different type of graph. The example below compares A, B, and C; here, you can see that it might make more sense to use a bar chart over a .
Source: Wikipedia
• Using bar/ to show a false impression of size: Truncated ( that don't start at a y-value of 0) can be misleading if the truncation is not clearly labeled or if the truncation is done in a way that distorts the data. For example, if the truncation is done at an arbitrary value, it could give the impression that the data is more evenly distributed than it really is. See the example below and notice how the differences between 2010 and 2011 are more noticeable in a truncated (left) compared to the usual (right).
Source: Wikipedia
In today's age where misinformation can easily and quickly spread, it's very important to choose the appropriate type of graph for your data and the message you want to convey. Carefully considering the limitations of each type of graph can help you avoid misusing (or mistrusting) them! ๐คจ
# 1.4 Representing a Categorical Variable with Graphs
L
Lusine Ghazaryan
Jed Quiaoit
L
Lusine Ghazaryan
Jed Quiaoit
You might recall from earlier that can be represented using tables and/or graphs. This section will provide more context that'll equip us with the ability to eventually construct and describe numerical or of data distributions. ๐
As for why graphs are big in statistics, and statistics are powerful tools for understanding and summarizing data. Graphs can help you visualize the patterns and relationships in your data, and statistics can help you quantify and describe those patterns. By using both and statistics, you can gain a deeper understanding of your data and communicate that understanding to others!
## Bar Graphs
(or bar graphs) are used to display frequencies (counts) or relative frequencies (proportions) for categorical data. The height or length of each bar in a corresponds to either the number or proportion of observations falling within
each category. ๐
To create a , you first need to decide on the categories you want to include. Each category corresponds to a separate bar on the graph. The height of each bar represents the frequency or count of observations in that category. All the bars have the same width, and there is a gap between adjacent bars to distinguish them from each other. ๐
When translated into a step-by-step procedure, here's how we would create a :
1. Determine the categories you want to include in the graph.
2. Count the number of observations in each category.
3. Mark the frequencies on the vertical axis and the categories on the horizontal axis.
4. Draw the bars, with the height of each bar representing the frequency of the corresponding category.
5. Add a title and axis labels to the graph to help interpret the data.
It's important to choose an appropriate and consistent scale for the vertical axis. You should also consider adding a legend to the graph if you have multiple series of data that you want to compare.
To keep it short, here is the of stress on the job. We can also use relative frequencies or percentages to construct the . You can be creative and color each category with a different color. It will beย visually attractive and easier to compare them.
Source: Prem S. Mann: Introductory Statistics. John Wiley and Sons Inc. 2020
## Pie Charts
A is a circular graph that is divided into slices, with each slice representing a different category. The size of each slice is proportional to the fraction of the whole that is represented by that category. are often used to show the relative proportions of different categories within a dataset. ๐ฅง
To create a , you'll have to keep the following steps in mind:
1. Determine the categories you want to include in the . (Example: Commuter, non-commuter)
2. Calculate the fraction of the whole that is represented by each category (Example: Out of 50 respondents, 30 commuters would occupy 3/5ths of the pie, while 20 non-commuters would occupy the remaining 2/5ths of the pie).
3. Draw a circle and divide it into slices that are proportional to the fractions calculated in step 2.
4. Label each slice with the corresponding category and the percentage it represents.
5. Add a title to the to help interpret the data.
It's important to keep in mind that are best used to compare the relative proportions (percentages and relative frequencies, for example) of different categories. They're not as effective at showing precise values or small differences between categories. If you want to show detailed values or compare the values of multiple categories, it is usually better to use a different type of graph, such as a bar chart.
๐ก Tips:
• The choice between and will depend on how many categories that variable of your interest assumes and the size of it. Whenever you have many categories or few categories with about the same frequencies, then the should be your first choice. If the pie has many slices or slices of the same size, it will be hard to compare the groups.
• Be careful of quantity distortions and keeping the area principle.
## Contingency Table (Two-Way Table)
Now that we know how to represent data in tables and charts, let's add one more character to the tables gang to keep things evenly balanced!
A contingency table is a type of table that is used to organize and (later on) analyze categorical data. It shows how the observations in a dataset are distributed among different categories of two or more variables. Contingency tables can help in understanding relationships between variables and identifying patterns or trends in the data. ๐จ
To create a contingency table, you'll have to:
1. Determine the variables you want to include in the table.
2. Count the number of observations in each category of each variable.
3. Organize the counts in a table, with each row representing a category of one variable and each column representing a category of the other variable.
4. Add row and column totals to the table. (This step is the easiest to forget!)
5. Analyze the table to identify any patterns or trends in the data. (This is important when establishing context and responding to Multiple Choice and Free Response Questions in the AP exam!)
If the numbers in the cells of the contingency table are the same for all categories, we can say that the variables are independent, If the numbers in the cells are different for different categories (with some having higher values than others), then the variables might be related. For example, if you are analyzing data on the relationship between gender and income, you might find that the proportions of men and women in different income categories are different, indicating some sort of relationship between the two variables.
๐ฅ Watch: AP Stats - Analyzing Categorical Data
## Real-Life Applications: To Trust or Not To Trust a Bar/Pie Chart?
Chances are, you've probably seen a bar or in some shape or form before in the news, media you consume, or even other textbooks. It's important to remember that they shouldn't be taken immediately at face value as they could be easily misused. To help inform whether bar/ are reliable or not, here are examples of ways they are commonly misused:
• Using bar/ to compare variables on different scales: Charts are best used to compare categories or groups that are on the same scale. If you are comparing variables that are on different scales, it can be difficult to accurately compare the sizes of the bars/pie slices.
Source: Infogram
• Using bar/ to show continuous data: Charts are best used to show categorical data, not continuous data. If you have continuous data, it is usually better to use a different type of graph, such as a line graph or scatterplot.
• Using bar/ to show small differences: Charts are not very effective at showing small differences between categories. If the differences between the categories are small, it may be difficult to accurately interpret the graph.
• Using bar/ to show trends over time: Charts are not well suited for showing trends over time. For this purpose, it is usually better to use a line graph or a time series plot.
• Using bar/ to show more than two variables: Charts are typically used to compare two variables. If you want to show more than two variables, it is usually better to use a different type of graph. The example below compares A, B, and C; here, you can see that it might make more sense to use a bar chart over a .
Source: Wikipedia
• Using bar/ to show a false impression of size: Truncated ( that don't start at a y-value of 0) can be misleading if the truncation is not clearly labeled or if the truncation is done in a way that distorts the data. For example, if the truncation is done at an arbitrary value, it could give the impression that the data is more evenly distributed than it really is. See the example below and notice how the differences between 2010 and 2011 are more noticeable in a truncated (left) compared to the usual (right).
Source: Wikipedia
In today's age where misinformation can easily and quickly spread, it's very important to choose the appropriate type of graph for your data and the message you want to convey. Carefully considering the limitations of each type of graph can help you avoid misusing (or mistrusting) them! ๐คจ
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# How do we solve for initial consumption in the Ramsey model?
How do we solve for the initial consumption level in the Ramsey model? Assume we have the Euler equation $$\frac {\dot c_t}{c_t}=r_t-\rho$$ Then $$c_t=c_0e^{R_t-\rho t}$$ Putting this into the lifetime budget constraint: $$\int_0^\infty e^{-R_t}(w_t-c_t)dt=k_0$$ gives $$\int_0^\infty e^{-R_t}w_t dt-\frac {c_0} {\rho}=-k_0$$
Therefore $$c_0=\rho\left(\int_0^\infty e^{-R_t}w_t dt+k_0\right)$$
Where we know that $w_t=f(k_t)-k_t r_t$ and that $r_t=f'(k_t)$, and $R_t=\int_0^t r_\tau d\tau$.
However, the problem is that the initial amount of consumption will influence the path of $k_t$, because $\dot k_t=f(k)-c_t$, and therefore of $w_t$ and $r_t$ as well. Therefore it seems to me that solving for $c_0$ is almost impossible here.
So let's say we would have a computer that could help us solve this, how would we do it? e.g. how does Dynare/matlab do it?
• what is $r_t$ and $R_t$? How are they related? Commented Dec 8, 2017 at 11:50
• @Dmitry, sorry forgot that one. Edited. Commented Dec 8, 2017 at 17:55
An algorithm to solve for equilibrium in a steady state (so that $\dot k_t=0$) is as follows:
1. Guess $k^*$, the steady state level of capital;
2. Compute steady state prices, $r^*$ and $w^*$;
3. Compute steady state consumption, $c^*$;
4. Check the market clearing condition $f(k^*)-c^*=0$;
5. Adjust your guess for $k^*$ and repeat until $f(k^*)-c^*$ is close enough to zero.
Out of steady state, the procedure is similar. You should first obtain the analogous formulas in discrete time and you have to start from some $k_0$ which is set exogenously. Then, compute $k^*$ following the algorithm above and guess a path for $\{k_t\}_{t=0}^\infty$ which converges to $k^*$ after $T$ periods. Compute prices for every period, consumption for every period, then check the market clearing at every period and use it to adjust the guess for $\{k_t\}_{t=0}^\infty$. Once this converges, increase $T$ and repeat until the path stops changing.
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# About an equation with eigenvalues
1. Mar 16, 2009
### simplex
Let's say I have the equation p(t)f''(t)=Kf(t) with p(t) a known periodical function, K an unknown constant and f(t) the unknown function.
This is an eigenvalues problem that once solved gives a set of K={k1, k2,...} eigenvalues.
I get these eigenvalues and they coincide with the ones obtained by others so I got them right.
Question
What if I try to solve the equation with a K that does not belong to the set of eigenvalues?
I have the initial conditions: f(0) and f'(0), I choose a K which is not an eigenvalue and I try to solve numerically the equation (using MATLAB):
What happens with f(t)? It is clear that I will get a solution. What is the difference between this solution with a forbidden K and a solution with an allowed K?
2. Mar 17, 2009
### HallsofIvy
The problem as given clearly has the "trivial solution", f(x)= 0 for all x, as solution. If k is not an eigenvalue, then that will be the only solution.
The equation Lv= Kv always has the "trivial solution" v= 0.
K is an eigenvalue of operator L if and only if there is a non-trivial solution to Lv= Kv. That is the definition of "eigenvalue".
Notice that this is an "existence and uniqueness" question. There always exists the trivial solution. K is an eigenvalue if that solution is not unique.
3. Mar 17, 2009
### simplex
For instance, I have the equation:
f''(t)+M(6+5*sin(2*pi*t))f(t)=0 which is the same as the equation p(t)f''(t)=Kf(t); [p=1/(6+5*sin(2*pi*t)), M=-K]
initian conditions
f(0)=1, f'(0)=0.5.
With M=1.579 which is not an eigenvalue I get (solving the equation with "odesolve" from Mathcad) a f(t) that is sinusoidal but exponentially growing.
It looks like f(t) exists despite the fact that I solved an equation for a forbidden value of its eigenvalues, M.
Last edited: Mar 17, 2009
4. Mar 17, 2009
### HallsofIvy
This is not an eigenvalue problem. The "trivial solution", f(x)= 0, does not satisfy f(0)= 1 or f'(0)= 0.5.
5. Mar 17, 2009
### simplex
An equation of this form: f''(x)+M(6+5*sin(2*pi*x))f(x)=0 appears in many works.
What I noticed they do, to deal with it, is they suppose that f(x) can be written as a Fourier series (no initial conditions or other restrictions are imposed). They truncate the Fourier sum to 20-30 coefficients, they transform the equation in a matrix equation, calculate the eigenvalues of the matrix eq. and from here they get a set of M=m1, m2, .... eigenvalues.
Question: What exactly does it mean? That they are looking only for solutions that have a Fourier representation (with a finite number of coefficients) and the eigenvalues they get correspond to the existence of these solutions?
After getting the set of eigenvalues they start to consider that f(x) has various initial conditions f(0) and f'(0). For each eigenvalues belonging to the set M=m1, m2, .... they get (using "odesolve" from Mathcad) sinusoidal solutions that do not grow in time, they are oscillating no matter what the initial conditions are (excepting the case when these conditions are zero).
For M outside the set m1, m2, ... (not an eigenvalue of the initial problem) they just say that this is a forbidden domain and do not approach the case any further.
The final goal for them is to identify when this equation (which describe a physical system) has as solutions oscillations. They choose an M=m1 or M=m2 etc., and after that they work with their physical system in the oscillatory regime.
In my case M is a parameter that can be varied by hand and in order to pass it from m1 to m2 I have to cross non eigenvalues which generate exponentially growing f(x) (at least these are the solutions I get in Mathcad with "odesolve") compromising the device as long as f(x) is a physical quantity, that normally can not grow indefinitely.
So my final Question: If I use the method with the Fourier decomposition (in series) can I be sure that for an M that is not an eigenvalue of the equation, no f(x) that start to -infinitum and stop to +infinitum and in the same time have a Fourier decomposition exists?
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# Properties
Label 784.6.a.f.1.1 Level $784$ Weight $6$ Character 784.1 Self dual yes Analytic conductor $125.741$ Analytic rank $1$ Dimension $1$ CM no Inner twists $1$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [784,6,Mod(1,784)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(784, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0, 0, 0]))
N = Newforms(chi, 6, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("784.1");
S:= CuspForms(chi, 6);
N := Newforms(S);
Level: $$N$$ $$=$$ $$784 = 2^{4} \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$6$$ Character orbit: $$[\chi]$$ $$=$$ 784.a (trivial)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$125.740914733$$ Analytic rank: $$1$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 28) Fricke sign: $$+1$$ Sato-Tate group: $\mathrm{SU}(2)$
## Embedding invariants
Embedding label 1.1 Character $$\chi$$ $$=$$ 784.1
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
$$f(q)$$ $$=$$ $$q-2.00000 q^{3} +96.0000 q^{5} -239.000 q^{9} +O(q^{10})$$ $$q-2.00000 q^{3} +96.0000 q^{5} -239.000 q^{9} +720.000 q^{11} -572.000 q^{13} -192.000 q^{15} -1254.00 q^{17} -94.0000 q^{19} -96.0000 q^{23} +6091.00 q^{25} +964.000 q^{27} -4374.00 q^{29} -6244.00 q^{31} -1440.00 q^{33} -10798.0 q^{37} +1144.00 q^{39} -12006.0 q^{41} +9160.00 q^{43} -22944.0 q^{45} -25836.0 q^{47} +2508.00 q^{51} +1014.00 q^{53} +69120.0 q^{55} +188.000 q^{57} +1242.00 q^{59} -7592.00 q^{61} -54912.0 q^{65} -41132.0 q^{67} +192.000 q^{69} +37632.0 q^{71} +13438.0 q^{73} -12182.0 q^{75} -6248.00 q^{79} +56149.0 q^{81} -25254.0 q^{83} -120384. q^{85} +8748.00 q^{87} +45126.0 q^{89} +12488.0 q^{93} -9024.00 q^{95} -107222. q^{97} -172080. q^{99} +O(q^{100})$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ 0 0
$$3$$ −2.00000 −0.128300 −0.0641500 0.997940i $$-0.520434\pi$$
−0.0641500 + 0.997940i $$0.520434\pi$$
$$4$$ 0 0
$$5$$ 96.0000 1.71730 0.858650 0.512562i $$-0.171304\pi$$
0.858650 + 0.512562i $$0.171304\pi$$
$$6$$ 0 0
$$7$$ 0 0
$$8$$ 0 0
$$9$$ −239.000 −0.983539
$$10$$ 0 0
$$11$$ 720.000 1.79412 0.897059 0.441912i $$-0.145700\pi$$
0.897059 + 0.441912i $$0.145700\pi$$
$$12$$ 0 0
$$13$$ −572.000 −0.938723 −0.469362 0.883006i $$-0.655516\pi$$
−0.469362 + 0.883006i $$0.655516\pi$$
$$14$$ 0 0
$$15$$ −192.000 −0.220330
$$16$$ 0 0
$$17$$ −1254.00 −1.05239 −0.526193 0.850365i $$-0.676381\pi$$
−0.526193 + 0.850365i $$0.676381\pi$$
$$18$$ 0 0
$$19$$ −94.0000 −0.0597371 −0.0298685 0.999554i $$-0.509509\pi$$
−0.0298685 + 0.999554i $$0.509509\pi$$
$$20$$ 0 0
$$21$$ 0 0
$$22$$ 0 0
$$23$$ −96.0000 −0.0378400 −0.0189200 0.999821i $$-0.506023\pi$$
−0.0189200 + 0.999821i $$0.506023\pi$$
$$24$$ 0 0
$$25$$ 6091.00 1.94912
$$26$$ 0 0
$$27$$ 964.000 0.254488
$$28$$ 0 0
$$29$$ −4374.00 −0.965792 −0.482896 0.875678i $$-0.660415\pi$$
−0.482896 + 0.875678i $$0.660415\pi$$
$$30$$ 0 0
$$31$$ −6244.00 −1.16697 −0.583484 0.812125i $$-0.698311\pi$$
−0.583484 + 0.812125i $$0.698311\pi$$
$$32$$ 0 0
$$33$$ −1440.00 −0.230185
$$34$$ 0 0
$$35$$ 0 0
$$36$$ 0 0
$$37$$ −10798.0 −1.29670 −0.648349 0.761343i $$-0.724540\pi$$
−0.648349 + 0.761343i $$0.724540\pi$$
$$38$$ 0 0
$$39$$ 1144.00 0.120438
$$40$$ 0 0
$$41$$ −12006.0 −1.11542 −0.557710 0.830036i $$-0.688320\pi$$
−0.557710 + 0.830036i $$0.688320\pi$$
$$42$$ 0 0
$$43$$ 9160.00 0.755482 0.377741 0.925911i $$-0.376701\pi$$
0.377741 + 0.925911i $$0.376701\pi$$
$$44$$ 0 0
$$45$$ −22944.0 −1.68903
$$46$$ 0 0
$$47$$ −25836.0 −1.70601 −0.853003 0.521906i $$-0.825221\pi$$
−0.853003 + 0.521906i $$0.825221\pi$$
$$48$$ 0 0
$$49$$ 0 0
$$50$$ 0 0
$$51$$ 2508.00 0.135021
$$52$$ 0 0
$$53$$ 1014.00 0.0495848 0.0247924 0.999693i $$-0.492108\pi$$
0.0247924 + 0.999693i $$0.492108\pi$$
$$54$$ 0 0
$$55$$ 69120.0 3.08104
$$56$$ 0 0
$$57$$ 188.000 0.00766427
$$58$$ 0 0
$$59$$ 1242.00 0.0464506 0.0232253 0.999730i $$-0.492606\pi$$
0.0232253 + 0.999730i $$0.492606\pi$$
$$60$$ 0 0
$$61$$ −7592.00 −0.261235 −0.130618 0.991433i $$-0.541696\pi$$
−0.130618 + 0.991433i $$0.541696\pi$$
$$62$$ 0 0
$$63$$ 0 0
$$64$$ 0 0
$$65$$ −54912.0 −1.61207
$$66$$ 0 0
$$67$$ −41132.0 −1.11942 −0.559710 0.828689i $$-0.689087\pi$$
−0.559710 + 0.828689i $$0.689087\pi$$
$$68$$ 0 0
$$69$$ 192.000 0.00485488
$$70$$ 0 0
$$71$$ 37632.0 0.885955 0.442977 0.896533i $$-0.353922\pi$$
0.442977 + 0.896533i $$0.353922\pi$$
$$72$$ 0 0
$$73$$ 13438.0 0.295140 0.147570 0.989052i $$-0.452855\pi$$
0.147570 + 0.989052i $$0.452855\pi$$
$$74$$ 0 0
$$75$$ −12182.0 −0.250072
$$76$$ 0 0
$$77$$ 0 0
$$78$$ 0 0
$$79$$ −6248.00 −0.112635 −0.0563175 0.998413i $$-0.517936\pi$$
−0.0563175 + 0.998413i $$0.517936\pi$$
$$80$$ 0 0
$$81$$ 56149.0 0.950888
$$82$$ 0 0
$$83$$ −25254.0 −0.402379 −0.201189 0.979552i $$-0.564481\pi$$
−0.201189 + 0.979552i $$0.564481\pi$$
$$84$$ 0 0
$$85$$ −120384. −1.80726
$$86$$ 0 0
$$87$$ 8748.00 0.123911
$$88$$ 0 0
$$89$$ 45126.0 0.603882 0.301941 0.953327i $$-0.402365\pi$$
0.301941 + 0.953327i $$0.402365\pi$$
$$90$$ 0 0
$$91$$ 0 0
$$92$$ 0 0
$$93$$ 12488.0 0.149722
$$94$$ 0 0
$$95$$ −9024.00 −0.102586
$$96$$ 0 0
$$97$$ −107222. −1.15706 −0.578528 0.815662i $$-0.696373\pi$$
−0.578528 + 0.815662i $$0.696373\pi$$
$$98$$ 0 0
$$99$$ −172080. −1.76458
$$100$$ 0 0
$$101$$ −47136.0 −0.459779 −0.229890 0.973217i $$-0.573837\pi$$
−0.229890 + 0.973217i $$0.573837\pi$$
$$102$$ 0 0
$$103$$ 122204. 1.13499 0.567495 0.823377i $$-0.307912\pi$$
0.567495 + 0.823377i $$0.307912\pi$$
$$104$$ 0 0
$$105$$ 0 0
$$106$$ 0 0
$$107$$ 129636. 1.09463 0.547314 0.836928i $$-0.315651\pi$$
0.547314 + 0.836928i $$0.315651\pi$$
$$108$$ 0 0
$$109$$ −220558. −1.77810 −0.889051 0.457809i $$-0.848635\pi$$
−0.889051 + 0.457809i $$0.848635\pi$$
$$110$$ 0 0
$$111$$ 21596.0 0.166366
$$112$$ 0 0
$$113$$ 170694. 1.25754 0.628770 0.777591i $$-0.283559\pi$$
0.628770 + 0.777591i $$0.283559\pi$$
$$114$$ 0 0
$$115$$ −9216.00 −0.0649827
$$116$$ 0 0
$$117$$ 136708. 0.923271
$$118$$ 0 0
$$119$$ 0 0
$$120$$ 0 0
$$121$$ 357349. 2.21886
$$122$$ 0 0
$$123$$ 24012.0 0.143109
$$124$$ 0 0
$$125$$ 284736. 1.62992
$$126$$ 0 0
$$127$$ 249808. 1.37435 0.687175 0.726492i $$-0.258851\pi$$
0.687175 + 0.726492i $$0.258851\pi$$
$$128$$ 0 0
$$129$$ −18320.0 −0.0969284
$$130$$ 0 0
$$131$$ 12210.0 0.0621638 0.0310819 0.999517i $$-0.490105\pi$$
0.0310819 + 0.999517i $$0.490105\pi$$
$$132$$ 0 0
$$133$$ 0 0
$$134$$ 0 0
$$135$$ 92544.0 0.437033
$$136$$ 0 0
$$137$$ −13902.0 −0.0632814 −0.0316407 0.999499i $$-0.510073\pi$$
−0.0316407 + 0.999499i $$0.510073\pi$$
$$138$$ 0 0
$$139$$ −431794. −1.89557 −0.947785 0.318911i $$-0.896683\pi$$
−0.947785 + 0.318911i $$0.896683\pi$$
$$140$$ 0 0
$$141$$ 51672.0 0.218881
$$142$$ 0 0
$$143$$ −411840. −1.68418
$$144$$ 0 0
$$145$$ −419904. −1.65856
$$146$$ 0 0
$$147$$ 0 0
$$148$$ 0 0
$$149$$ 326814. 1.20597 0.602983 0.797754i $$-0.293979\pi$$
0.602983 + 0.797754i $$0.293979\pi$$
$$150$$ 0 0
$$151$$ −173480. −0.619166 −0.309583 0.950872i $$-0.600189\pi$$
−0.309583 + 0.950872i $$0.600189\pi$$
$$152$$ 0 0
$$153$$ 299706. 1.03506
$$154$$ 0 0
$$155$$ −599424. −2.00403
$$156$$ 0 0
$$157$$ 54532.0 0.176564 0.0882820 0.996096i $$-0.471862\pi$$
0.0882820 + 0.996096i $$0.471862\pi$$
$$158$$ 0 0
$$159$$ −2028.00 −0.00636173
$$160$$ 0 0
$$161$$ 0 0
$$162$$ 0 0
$$163$$ −104960. −0.309425 −0.154712 0.987960i $$-0.549445\pi$$
−0.154712 + 0.987960i $$0.549445\pi$$
$$164$$ 0 0
$$165$$ −138240. −0.395297
$$166$$ 0 0
$$167$$ 160788. 0.446131 0.223066 0.974803i $$-0.428394\pi$$
0.223066 + 0.974803i $$0.428394\pi$$
$$168$$ 0 0
$$169$$ −44109.0 −0.118798
$$170$$ 0 0
$$171$$ 22466.0 0.0587537
$$172$$ 0 0
$$173$$ −360564. −0.915940 −0.457970 0.888968i $$-0.651423\pi$$
−0.457970 + 0.888968i $$0.651423\pi$$
$$174$$ 0 0
$$175$$ 0 0
$$176$$ 0 0
$$177$$ −2484.00 −0.00595962
$$178$$ 0 0
$$179$$ 312732. 0.729524 0.364762 0.931101i $$-0.381150\pi$$
0.364762 + 0.931101i $$0.381150\pi$$
$$180$$ 0 0
$$181$$ 123820. 0.280928 0.140464 0.990086i $$-0.455141\pi$$
0.140464 + 0.990086i $$0.455141\pi$$
$$182$$ 0 0
$$183$$ 15184.0 0.0335165
$$184$$ 0 0
$$185$$ −1.03661e6 −2.22682
$$186$$ 0 0
$$187$$ −902880. −1.88810
$$188$$ 0 0
$$189$$ 0 0
$$190$$ 0 0
$$191$$ −323448. −0.641536 −0.320768 0.947158i $$-0.603941\pi$$
−0.320768 + 0.947158i $$0.603941\pi$$
$$192$$ 0 0
$$193$$ −619954. −1.19803 −0.599013 0.800739i $$-0.704440\pi$$
−0.599013 + 0.800739i $$0.704440\pi$$
$$194$$ 0 0
$$195$$ 109824. 0.206829
$$196$$ 0 0
$$197$$ −499362. −0.916748 −0.458374 0.888759i $$-0.651568\pi$$
−0.458374 + 0.888759i $$0.651568\pi$$
$$198$$ 0 0
$$199$$ −785932. −1.40686 −0.703432 0.710762i $$-0.748350\pi$$
−0.703432 + 0.710762i $$0.748350\pi$$
$$200$$ 0 0
$$201$$ 82264.0 0.143622
$$202$$ 0 0
$$203$$ 0 0
$$204$$ 0 0
$$205$$ −1.15258e6 −1.91551
$$206$$ 0 0
$$207$$ 22944.0 0.0372172
$$208$$ 0 0
$$209$$ −67680.0 −0.107175
$$210$$ 0 0
$$211$$ −1.06276e6 −1.64335 −0.821676 0.569955i $$-0.806961\pi$$
−0.821676 + 0.569955i $$0.806961\pi$$
$$212$$ 0 0
$$213$$ −75264.0 −0.113668
$$214$$ 0 0
$$215$$ 879360. 1.29739
$$216$$ 0 0
$$217$$ 0 0
$$218$$ 0 0
$$219$$ −26876.0 −0.0378664
$$220$$ 0 0
$$221$$ 717288. 0.987900
$$222$$ 0 0
$$223$$ 707720. 0.953014 0.476507 0.879171i $$-0.341903\pi$$
0.476507 + 0.879171i $$0.341903\pi$$
$$224$$ 0 0
$$225$$ −1.45575e6 −1.91704
$$226$$ 0 0
$$227$$ −1.04437e6 −1.34520 −0.672602 0.740005i $$-0.734823\pi$$
−0.672602 + 0.740005i $$0.734823\pi$$
$$228$$ 0 0
$$229$$ 539716. 0.680106 0.340053 0.940406i $$-0.389555\pi$$
0.340053 + 0.940406i $$0.389555\pi$$
$$230$$ 0 0
$$231$$ 0 0
$$232$$ 0 0
$$233$$ 177114. 0.213729 0.106864 0.994274i $$-0.465919\pi$$
0.106864 + 0.994274i $$0.465919\pi$$
$$234$$ 0 0
$$235$$ −2.48026e6 −2.92972
$$236$$ 0 0
$$237$$ 12496.0 0.0144511
$$238$$ 0 0
$$239$$ 655464. 0.742257 0.371128 0.928582i $$-0.378971\pi$$
0.371128 + 0.928582i $$0.378971\pi$$
$$240$$ 0 0
$$241$$ −1.38709e6 −1.53838 −0.769189 0.639021i $$-0.779340\pi$$
−0.769189 + 0.639021i $$0.779340\pi$$
$$242$$ 0 0
$$243$$ −346550. −0.376487
$$244$$ 0 0
$$245$$ 0 0
$$246$$ 0 0
$$247$$ 53768.0 0.0560766
$$248$$ 0 0
$$249$$ 50508.0 0.0516252
$$250$$ 0 0
$$251$$ 1.88811e6 1.89166 0.945830 0.324663i $$-0.105251\pi$$
0.945830 + 0.324663i $$0.105251\pi$$
$$252$$ 0 0
$$253$$ −69120.0 −0.0678895
$$254$$ 0 0
$$255$$ 240768. 0.231872
$$256$$ 0 0
$$257$$ −346194. −0.326954 −0.163477 0.986547i $$-0.552271\pi$$
−0.163477 + 0.986547i $$0.552271\pi$$
$$258$$ 0 0
$$259$$ 0 0
$$260$$ 0 0
$$261$$ 1.04539e6 0.949895
$$262$$ 0 0
$$263$$ 929088. 0.828262 0.414131 0.910217i $$-0.364086\pi$$
0.414131 + 0.910217i $$0.364086\pi$$
$$264$$ 0 0
$$265$$ 97344.0 0.0851519
$$266$$ 0 0
$$267$$ −90252.0 −0.0774780
$$268$$ 0 0
$$269$$ −382068. −0.321929 −0.160964 0.986960i $$-0.551460\pi$$
−0.160964 + 0.986960i $$0.551460\pi$$
$$270$$ 0 0
$$271$$ −1.58056e6 −1.30734 −0.653669 0.756781i $$-0.726771\pi$$
−0.653669 + 0.756781i $$0.726771\pi$$
$$272$$ 0 0
$$273$$ 0 0
$$274$$ 0 0
$$275$$ 4.38552e6 3.49695
$$276$$ 0 0
$$277$$ −1.36911e6 −1.07211 −0.536056 0.844182i $$-0.680086\pi$$
−0.536056 + 0.844182i $$0.680086\pi$$
$$278$$ 0 0
$$279$$ 1.49232e6 1.14776
$$280$$ 0 0
$$281$$ −394854. −0.298312 −0.149156 0.988814i $$-0.547656\pi$$
−0.149156 + 0.988814i $$0.547656\pi$$
$$282$$ 0 0
$$283$$ 673034. 0.499541 0.249770 0.968305i $$-0.419645\pi$$
0.249770 + 0.968305i $$0.419645\pi$$
$$284$$ 0 0
$$285$$ 18048.0 0.0131618
$$286$$ 0 0
$$287$$ 0 0
$$288$$ 0 0
$$289$$ 152659. 0.107517
$$290$$ 0 0
$$291$$ 214444. 0.148450
$$292$$ 0 0
$$293$$ −1.83468e6 −1.24851 −0.624254 0.781222i $$-0.714597\pi$$
−0.624254 + 0.781222i $$0.714597\pi$$
$$294$$ 0 0
$$295$$ 119232. 0.0797697
$$296$$ 0 0
$$297$$ 694080. 0.456582
$$298$$ 0 0
$$299$$ 54912.0 0.0355213
$$300$$ 0 0
$$301$$ 0 0
$$302$$ 0 0
$$303$$ 94272.0 0.0589897
$$304$$ 0 0
$$305$$ −728832. −0.448619
$$306$$ 0 0
$$307$$ −1.51056e6 −0.914727 −0.457363 0.889280i $$-0.651206\pi$$
−0.457363 + 0.889280i $$0.651206\pi$$
$$308$$ 0 0
$$309$$ −244408. −0.145619
$$310$$ 0 0
$$311$$ −1.87529e6 −1.09943 −0.549714 0.835353i $$-0.685263\pi$$
−0.549714 + 0.835353i $$0.685263\pi$$
$$312$$ 0 0
$$313$$ 1.51076e6 0.871636 0.435818 0.900035i $$-0.356459\pi$$
0.435818 + 0.900035i $$0.356459\pi$$
$$314$$ 0 0
$$315$$ 0 0
$$316$$ 0 0
$$317$$ 2.02709e6 1.13299 0.566495 0.824065i $$-0.308299\pi$$
0.566495 + 0.824065i $$0.308299\pi$$
$$318$$ 0 0
$$319$$ −3.14928e6 −1.73274
$$320$$ 0 0
$$321$$ −259272. −0.140441
$$322$$ 0 0
$$323$$ 117876. 0.0628665
$$324$$ 0 0
$$325$$ −3.48405e6 −1.82968
$$326$$ 0 0
$$327$$ 441116. 0.228131
$$328$$ 0 0
$$329$$ 0 0
$$330$$ 0 0
$$331$$ −1.54009e6 −0.772637 −0.386319 0.922365i $$-0.626253\pi$$
−0.386319 + 0.922365i $$0.626253\pi$$
$$332$$ 0 0
$$333$$ 2.58072e6 1.27535
$$334$$ 0 0
$$335$$ −3.94867e6 −1.92238
$$336$$ 0 0
$$337$$ 1.01166e6 0.485245 0.242622 0.970121i $$-0.421992\pi$$
0.242622 + 0.970121i $$0.421992\pi$$
$$338$$ 0 0
$$339$$ −341388. −0.161343
$$340$$ 0 0
$$341$$ −4.49568e6 −2.09368
$$342$$ 0 0
$$343$$ 0 0
$$344$$ 0 0
$$345$$ 18432.0 0.00833729
$$346$$ 0 0
$$347$$ 2.15748e6 0.961885 0.480942 0.876752i $$-0.340295\pi$$
0.480942 + 0.876752i $$0.340295\pi$$
$$348$$ 0 0
$$349$$ 1.15798e6 0.508906 0.254453 0.967085i $$-0.418105\pi$$
0.254453 + 0.967085i $$0.418105\pi$$
$$350$$ 0 0
$$351$$ −551408. −0.238894
$$352$$ 0 0
$$353$$ 3.17566e6 1.35643 0.678215 0.734863i $$-0.262754\pi$$
0.678215 + 0.734863i $$0.262754\pi$$
$$354$$ 0 0
$$355$$ 3.61267e6 1.52145
$$356$$ 0 0
$$357$$ 0 0
$$358$$ 0 0
$$359$$ 74616.0 0.0305560 0.0152780 0.999883i $$-0.495137\pi$$
0.0152780 + 0.999883i $$0.495137\pi$$
$$360$$ 0 0
$$361$$ −2.46726e6 −0.996431
$$362$$ 0 0
$$363$$ −714698. −0.284679
$$364$$ 0 0
$$365$$ 1.29005e6 0.506843
$$366$$ 0 0
$$367$$ −1.79807e6 −0.696854 −0.348427 0.937336i $$-0.613284\pi$$
−0.348427 + 0.937336i $$0.613284\pi$$
$$368$$ 0 0
$$369$$ 2.86943e6 1.09706
$$370$$ 0 0
$$371$$ 0 0
$$372$$ 0 0
$$373$$ 2.20461e6 0.820463 0.410231 0.911981i $$-0.365448\pi$$
0.410231 + 0.911981i $$0.365448\pi$$
$$374$$ 0 0
$$375$$ −569472. −0.209119
$$376$$ 0 0
$$377$$ 2.50193e6 0.906612
$$378$$ 0 0
$$379$$ 177568. 0.0634990 0.0317495 0.999496i $$-0.489892\pi$$
0.0317495 + 0.999496i $$0.489892\pi$$
$$380$$ 0 0
$$381$$ −499616. −0.176329
$$382$$ 0 0
$$383$$ −2.87468e6 −1.00137 −0.500683 0.865630i $$-0.666918\pi$$
−0.500683 + 0.865630i $$0.666918\pi$$
$$384$$ 0 0
$$385$$ 0 0
$$386$$ 0 0
$$387$$ −2.18924e6 −0.743046
$$388$$ 0 0
$$389$$ −4.79965e6 −1.60818 −0.804091 0.594506i $$-0.797348\pi$$
−0.804091 + 0.594506i $$0.797348\pi$$
$$390$$ 0 0
$$391$$ 120384. 0.0398223
$$392$$ 0 0
$$393$$ −24420.0 −0.00797562
$$394$$ 0 0
$$395$$ −599808. −0.193428
$$396$$ 0 0
$$397$$ 2.81643e6 0.896855 0.448428 0.893819i $$-0.351984\pi$$
0.448428 + 0.893819i $$0.351984\pi$$
$$398$$ 0 0
$$399$$ 0 0
$$400$$ 0 0
$$401$$ −2.83797e6 −0.881347 −0.440673 0.897667i $$-0.645260\pi$$
−0.440673 + 0.897667i $$0.645260\pi$$
$$402$$ 0 0
$$403$$ 3.57157e6 1.09546
$$404$$ 0 0
$$405$$ 5.39030e6 1.63296
$$406$$ 0 0
$$407$$ −7.77456e6 −2.32643
$$408$$ 0 0
$$409$$ −154286. −0.0456056 −0.0228028 0.999740i $$-0.507259\pi$$
−0.0228028 + 0.999740i $$0.507259\pi$$
$$410$$ 0 0
$$411$$ 27804.0 0.00811900
$$412$$ 0 0
$$413$$ 0 0
$$414$$ 0 0
$$415$$ −2.42438e6 −0.691005
$$416$$ 0 0
$$417$$ 863588. 0.243202
$$418$$ 0 0
$$419$$ 3.72865e6 1.03757 0.518783 0.854906i $$-0.326385\pi$$
0.518783 + 0.854906i $$0.326385\pi$$
$$420$$ 0 0
$$421$$ −2.32623e6 −0.639658 −0.319829 0.947475i $$-0.603626\pi$$
−0.319829 + 0.947475i $$0.603626\pi$$
$$422$$ 0 0
$$423$$ 6.17480e6 1.67792
$$424$$ 0 0
$$425$$ −7.63811e6 −2.05123
$$426$$ 0 0
$$427$$ 0 0
$$428$$ 0 0
$$429$$ 823680. 0.216080
$$430$$ 0 0
$$431$$ 2.61482e6 0.678031 0.339015 0.940781i $$-0.389906\pi$$
0.339015 + 0.940781i $$0.389906\pi$$
$$432$$ 0 0
$$433$$ 1.19226e6 0.305598 0.152799 0.988257i $$-0.451171\pi$$
0.152799 + 0.988257i $$0.451171\pi$$
$$434$$ 0 0
$$435$$ 839808. 0.212793
$$436$$ 0 0
$$437$$ 9024.00 0.00226045
$$438$$ 0 0
$$439$$ 1.05793e6 0.261996 0.130998 0.991383i $$-0.458182\pi$$
0.130998 + 0.991383i $$0.458182\pi$$
$$440$$ 0 0
$$441$$ 0 0
$$442$$ 0 0
$$443$$ −4.12756e6 −0.999272 −0.499636 0.866235i $$-0.666533\pi$$
−0.499636 + 0.866235i $$0.666533\pi$$
$$444$$ 0 0
$$445$$ 4.33210e6 1.03705
$$446$$ 0 0
$$447$$ −653628. −0.154725
$$448$$ 0 0
$$449$$ 3.75823e6 0.879766 0.439883 0.898055i $$-0.355020\pi$$
0.439883 + 0.898055i $$0.355020\pi$$
$$450$$ 0 0
$$451$$ −8.64432e6 −2.00120
$$452$$ 0 0
$$453$$ 346960. 0.0794390
$$454$$ 0 0
$$455$$ 0 0
$$456$$ 0 0
$$457$$ −451114. −0.101041 −0.0505203 0.998723i $$-0.516088\pi$$
−0.0505203 + 0.998723i $$0.516088\pi$$
$$458$$ 0 0
$$459$$ −1.20886e6 −0.267820
$$460$$ 0 0
$$461$$ 1.95186e6 0.427756 0.213878 0.976860i $$-0.431390\pi$$
0.213878 + 0.976860i $$0.431390\pi$$
$$462$$ 0 0
$$463$$ 7.20218e6 1.56139 0.780695 0.624913i $$-0.214865\pi$$
0.780695 + 0.624913i $$0.214865\pi$$
$$464$$ 0 0
$$465$$ 1.19885e6 0.257118
$$466$$ 0 0
$$467$$ 7.17801e6 1.52304 0.761521 0.648140i $$-0.224453\pi$$
0.761521 + 0.648140i $$0.224453\pi$$
$$468$$ 0 0
$$469$$ 0 0
$$470$$ 0 0
$$471$$ −109064. −0.0226532
$$472$$ 0 0
$$473$$ 6.59520e6 1.35542
$$474$$ 0 0
$$475$$ −572554. −0.116435
$$476$$ 0 0
$$477$$ −242346. −0.0487686
$$478$$ 0 0
$$479$$ 6.17632e6 1.22996 0.614980 0.788543i $$-0.289164\pi$$
0.614980 + 0.788543i $$0.289164\pi$$
$$480$$ 0 0
$$481$$ 6.17646e6 1.21724
$$482$$ 0 0
$$483$$ 0 0
$$484$$ 0 0
$$485$$ −1.02933e7 −1.98701
$$486$$ 0 0
$$487$$ −7.59330e6 −1.45080 −0.725401 0.688327i $$-0.758345\pi$$
−0.725401 + 0.688327i $$0.758345\pi$$
$$488$$ 0 0
$$489$$ 209920. 0.0396992
$$490$$ 0 0
$$491$$ 1.51878e6 0.284309 0.142155 0.989844i $$-0.454597\pi$$
0.142155 + 0.989844i $$0.454597\pi$$
$$492$$ 0 0
$$493$$ 5.48500e6 1.01639
$$494$$ 0 0
$$495$$ −1.65197e7 −3.03032
$$496$$ 0 0
$$497$$ 0 0
$$498$$ 0 0
$$499$$ −1.47576e6 −0.265316 −0.132658 0.991162i $$-0.542351\pi$$
−0.132658 + 0.991162i $$0.542351\pi$$
$$500$$ 0 0
$$501$$ −321576. −0.0572386
$$502$$ 0 0
$$503$$ −1.31309e6 −0.231406 −0.115703 0.993284i $$-0.536912\pi$$
−0.115703 + 0.993284i $$0.536912\pi$$
$$504$$ 0 0
$$505$$ −4.52506e6 −0.789579
$$506$$ 0 0
$$507$$ 88218.0 0.0152418
$$508$$ 0 0
$$509$$ −4.40932e6 −0.754357 −0.377178 0.926141i $$-0.623106\pi$$
−0.377178 + 0.926141i $$0.623106\pi$$
$$510$$ 0 0
$$511$$ 0 0
$$512$$ 0 0
$$513$$ −90616.0 −0.0152024
$$514$$ 0 0
$$515$$ 1.17316e7 1.94912
$$516$$ 0 0
$$517$$ −1.86019e7 −3.06078
$$518$$ 0 0
$$519$$ 721128. 0.117515
$$520$$ 0 0
$$521$$ −2.97629e6 −0.480376 −0.240188 0.970726i $$-0.577209\pi$$
−0.240188 + 0.970726i $$0.577209\pi$$
$$522$$ 0 0
$$523$$ 6.34627e6 1.01453 0.507265 0.861790i $$-0.330657\pi$$
0.507265 + 0.861790i $$0.330657\pi$$
$$524$$ 0 0
$$525$$ 0 0
$$526$$ 0 0
$$527$$ 7.82998e6 1.22810
$$528$$ 0 0
$$529$$ −6.42713e6 −0.998568
$$530$$ 0 0
$$531$$ −296838. −0.0456860
$$532$$ 0 0
$$533$$ 6.86743e6 1.04707
$$534$$ 0 0
$$535$$ 1.24451e7 1.87980
$$536$$ 0 0
$$537$$ −625464. −0.0935980
$$538$$ 0 0
$$539$$ 0 0
$$540$$ 0 0
$$541$$ −1.36667e6 −0.200756 −0.100378 0.994949i $$-0.532005\pi$$
−0.100378 + 0.994949i $$0.532005\pi$$
$$542$$ 0 0
$$543$$ −247640. −0.0360430
$$544$$ 0 0
$$545$$ −2.11736e7 −3.05353
$$546$$ 0 0
$$547$$ 9.55818e6 1.36586 0.682931 0.730483i $$-0.260705\pi$$
0.682931 + 0.730483i $$0.260705\pi$$
$$548$$ 0 0
$$549$$ 1.81449e6 0.256935
$$550$$ 0 0
$$551$$ 411156. 0.0576936
$$552$$ 0 0
$$553$$ 0 0
$$554$$ 0 0
$$555$$ 2.07322e6 0.285701
$$556$$ 0 0
$$557$$ 6.94287e6 0.948202 0.474101 0.880470i $$-0.342773\pi$$
0.474101 + 0.880470i $$0.342773\pi$$
$$558$$ 0 0
$$559$$ −5.23952e6 −0.709189
$$560$$ 0 0
$$561$$ 1.80576e6 0.242244
$$562$$ 0 0
$$563$$ 5.24662e6 0.697604 0.348802 0.937196i $$-0.386589\pi$$
0.348802 + 0.937196i $$0.386589\pi$$
$$564$$ 0 0
$$565$$ 1.63866e7 2.15958
$$566$$ 0 0
$$567$$ 0 0
$$568$$ 0 0
$$569$$ 3.46551e6 0.448731 0.224366 0.974505i $$-0.427969\pi$$
0.224366 + 0.974505i $$0.427969\pi$$
$$570$$ 0 0
$$571$$ −4.90069e6 −0.629023 −0.314512 0.949254i $$-0.601841\pi$$
−0.314512 + 0.949254i $$0.601841\pi$$
$$572$$ 0 0
$$573$$ 646896. 0.0823091
$$574$$ 0 0
$$575$$ −584736. −0.0737548
$$576$$ 0 0
$$577$$ −2.28346e6 −0.285531 −0.142766 0.989757i $$-0.545599\pi$$
−0.142766 + 0.989757i $$0.545599\pi$$
$$578$$ 0 0
$$579$$ 1.23991e6 0.153707
$$580$$ 0 0
$$581$$ 0 0
$$582$$ 0 0
$$583$$ 730080. 0.0889609
$$584$$ 0 0
$$585$$ 1.31240e7 1.58553
$$586$$ 0 0
$$587$$ 1.03157e7 1.23568 0.617838 0.786305i $$-0.288009\pi$$
0.617838 + 0.786305i $$0.288009\pi$$
$$588$$ 0 0
$$589$$ 586936. 0.0697112
$$590$$ 0 0
$$591$$ 998724. 0.117619
$$592$$ 0 0
$$593$$ −3.52838e6 −0.412039 −0.206020 0.978548i $$-0.566051\pi$$
−0.206020 + 0.978548i $$0.566051\pi$$
$$594$$ 0 0
$$595$$ 0 0
$$596$$ 0 0
$$597$$ 1.57186e6 0.180501
$$598$$ 0 0
$$599$$ −2.92260e6 −0.332815 −0.166407 0.986057i $$-0.553217\pi$$
−0.166407 + 0.986057i $$0.553217\pi$$
$$600$$ 0 0
$$601$$ 1.17567e7 1.32770 0.663849 0.747866i $$-0.268922\pi$$
0.663849 + 0.747866i $$0.268922\pi$$
$$602$$ 0 0
$$603$$ 9.83055e6 1.10099
$$604$$ 0 0
$$605$$ 3.43055e7 3.81044
$$606$$ 0 0
$$607$$ −4.71491e6 −0.519400 −0.259700 0.965689i $$-0.583624\pi$$
−0.259700 + 0.965689i $$0.583624\pi$$
$$608$$ 0 0
$$609$$ 0 0
$$610$$ 0 0
$$611$$ 1.47782e7 1.60147
$$612$$ 0 0
$$613$$ 213842. 0.0229849 0.0114924 0.999934i $$-0.496342\pi$$
0.0114924 + 0.999934i $$0.496342\pi$$
$$614$$ 0 0
$$615$$ 2.30515e6 0.245760
$$616$$ 0 0
$$617$$ −336666. −0.0356030 −0.0178015 0.999842i $$-0.505667\pi$$
−0.0178015 + 0.999842i $$0.505667\pi$$
$$618$$ 0 0
$$619$$ 1.42655e7 1.49645 0.748223 0.663447i $$-0.230907\pi$$
0.748223 + 0.663447i $$0.230907\pi$$
$$620$$ 0 0
$$621$$ −92544.0 −0.00962984
$$622$$ 0 0
$$623$$ 0 0
$$624$$ 0 0
$$625$$ 8.30028e6 0.849949
$$626$$ 0 0
$$627$$ 135360. 0.0137506
$$628$$ 0 0
$$629$$ 1.35407e7 1.36463
$$630$$ 0 0
$$631$$ 6.59637e6 0.659525 0.329763 0.944064i $$-0.393031\pi$$
0.329763 + 0.944064i $$0.393031\pi$$
$$632$$ 0 0
$$633$$ 2.12553e6 0.210842
$$634$$ 0 0
$$635$$ 2.39816e7 2.36017
$$636$$ 0 0
$$637$$ 0 0
$$638$$ 0 0
$$639$$ −8.99405e6 −0.871371
$$640$$ 0 0
$$641$$ −1.02490e7 −0.985225 −0.492613 0.870249i $$-0.663958\pi$$
−0.492613 + 0.870249i $$0.663958\pi$$
$$642$$ 0 0
$$643$$ −4.16543e6 −0.397312 −0.198656 0.980069i $$-0.563658\pi$$
−0.198656 + 0.980069i $$0.563658\pi$$
$$644$$ 0 0
$$645$$ −1.75872e6 −0.166455
$$646$$ 0 0
$$647$$ −3.35051e6 −0.314666 −0.157333 0.987546i $$-0.550290\pi$$
−0.157333 + 0.987546i $$0.550290\pi$$
$$648$$ 0 0
$$649$$ 894240. 0.0833379
$$650$$ 0 0
$$651$$ 0 0
$$652$$ 0 0
$$653$$ −9.05408e6 −0.830924 −0.415462 0.909611i $$-0.636380\pi$$
−0.415462 + 0.909611i $$0.636380\pi$$
$$654$$ 0 0
$$655$$ 1.17216e6 0.106754
$$656$$ 0 0
$$657$$ −3.21168e6 −0.290281
$$658$$ 0 0
$$659$$ −6.45382e6 −0.578899 −0.289450 0.957193i $$-0.593472\pi$$
−0.289450 + 0.957193i $$0.593472\pi$$
$$660$$ 0 0
$$661$$ −1.43167e7 −1.27450 −0.637250 0.770657i $$-0.719928\pi$$
−0.637250 + 0.770657i $$0.719928\pi$$
$$662$$ 0 0
$$663$$ −1.43458e6 −0.126748
$$664$$ 0 0
$$665$$ 0 0
$$666$$ 0 0
$$667$$ 419904. 0.0365456
$$668$$ 0 0
$$669$$ −1.41544e6 −0.122272
$$670$$ 0 0
$$671$$ −5.46624e6 −0.468686
$$672$$ 0 0
$$673$$ 2.27250e7 1.93404 0.967020 0.254701i $$-0.0819771\pi$$
0.967020 + 0.254701i $$0.0819771\pi$$
$$674$$ 0 0
$$675$$ 5.87172e6 0.496028
$$676$$ 0 0
$$677$$ 8.53249e6 0.715491 0.357746 0.933819i $$-0.383546\pi$$
0.357746 + 0.933819i $$0.383546\pi$$
$$678$$ 0 0
$$679$$ 0 0
$$680$$ 0 0
$$681$$ 2.08873e6 0.172590
$$682$$ 0 0
$$683$$ 2.24921e7 1.84492 0.922461 0.386090i $$-0.126175\pi$$
0.922461 + 0.386090i $$0.126175\pi$$
$$684$$ 0 0
$$685$$ −1.33459e6 −0.108673
$$686$$ 0 0
$$687$$ −1.07943e6 −0.0872576
$$688$$ 0 0
$$689$$ −580008. −0.0465464
$$690$$ 0 0
$$691$$ −1.26894e7 −1.01099 −0.505495 0.862830i $$-0.668690\pi$$
−0.505495 + 0.862830i $$0.668690\pi$$
$$692$$ 0 0
$$693$$ 0 0
$$694$$ 0 0
$$695$$ −4.14522e7 −3.25526
$$696$$ 0 0
$$697$$ 1.50555e7 1.17385
$$698$$ 0 0
$$699$$ −354228. −0.0274214
$$700$$ 0 0
$$701$$ −5.13939e6 −0.395018 −0.197509 0.980301i $$-0.563285\pi$$
−0.197509 + 0.980301i $$0.563285\pi$$
$$702$$ 0 0
$$703$$ 1.01501e6 0.0774610
$$704$$ 0 0
$$705$$ 4.96051e6 0.375884
$$706$$ 0 0
$$707$$ 0 0
$$708$$ 0 0
$$709$$ −1.16065e7 −0.867132 −0.433566 0.901122i $$-0.642745\pi$$
−0.433566 + 0.901122i $$0.642745\pi$$
$$710$$ 0 0
$$711$$ 1.49327e6 0.110781
$$712$$ 0 0
$$713$$ 599424. 0.0441581
$$714$$ 0 0
$$715$$ −3.95366e7 −2.89224
$$716$$ 0 0
$$717$$ −1.31093e6 −0.0952316
$$718$$ 0 0
$$719$$ 1.50998e7 1.08930 0.544650 0.838663i $$-0.316662\pi$$
0.544650 + 0.838663i $$0.316662\pi$$
$$720$$ 0 0
$$721$$ 0 0
$$722$$ 0 0
$$723$$ 2.77419e6 0.197374
$$724$$ 0 0
$$725$$ −2.66420e7 −1.88245
$$726$$ 0 0
$$727$$ −2.32536e7 −1.63175 −0.815874 0.578229i $$-0.803744\pi$$
−0.815874 + 0.578229i $$0.803744\pi$$
$$728$$ 0 0
$$729$$ −1.29511e7 −0.902585
$$730$$ 0 0
$$731$$ −1.14866e7 −0.795059
$$732$$ 0 0
$$733$$ −2.37814e7 −1.63485 −0.817423 0.576038i $$-0.804598\pi$$
−0.817423 + 0.576038i $$0.804598\pi$$
$$734$$ 0 0
$$735$$ 0 0
$$736$$ 0 0
$$737$$ −2.96150e7 −2.00837
$$738$$ 0 0
$$739$$ 2.51392e6 0.169333 0.0846663 0.996409i $$-0.473018\pi$$
0.0846663 + 0.996409i $$0.473018\pi$$
$$740$$ 0 0
$$741$$ −107536. −0.00719463
$$742$$ 0 0
$$743$$ 2.22646e7 1.47959 0.739797 0.672830i $$-0.234922\pi$$
0.739797 + 0.672830i $$0.234922\pi$$
$$744$$ 0 0
$$745$$ 3.13741e7 2.07101
$$746$$ 0 0
$$747$$ 6.03571e6 0.395755
$$748$$ 0 0
$$749$$ 0 0
$$750$$ 0 0
$$751$$ −2.30108e7 −1.48878 −0.744392 0.667743i $$-0.767260\pi$$
−0.744392 + 0.667743i $$0.767260\pi$$
$$752$$ 0 0
$$753$$ −3.77622e6 −0.242700
$$754$$ 0 0
$$755$$ −1.66541e7 −1.06329
$$756$$ 0 0
$$757$$ 1.59335e7 1.01058 0.505290 0.862950i $$-0.331386\pi$$
0.505290 + 0.862950i $$0.331386\pi$$
$$758$$ 0 0
$$759$$ 138240. 0.00871022
$$760$$ 0 0
$$761$$ −1.68629e7 −1.05553 −0.527764 0.849391i $$-0.676969\pi$$
−0.527764 + 0.849391i $$0.676969\pi$$
$$762$$ 0 0
$$763$$ 0 0
$$764$$ 0 0
$$765$$ 2.87718e7 1.77751
$$766$$ 0 0
$$767$$ −710424. −0.0436043
$$768$$ 0 0
$$769$$ 2.75402e7 1.67939 0.839694 0.543060i $$-0.182735\pi$$
0.839694 + 0.543060i $$0.182735\pi$$
$$770$$ 0 0
$$771$$ 692388. 0.0419482
$$772$$ 0 0
$$773$$ −2.26820e7 −1.36532 −0.682658 0.730738i $$-0.739176\pi$$
−0.682658 + 0.730738i $$0.739176\pi$$
$$774$$ 0 0
$$775$$ −3.80322e7 −2.27456
$$776$$ 0 0
$$777$$ 0 0
$$778$$ 0 0
$$779$$ 1.12856e6 0.0666320
$$780$$ 0 0
$$781$$ 2.70950e7 1.58951
$$782$$ 0 0
$$783$$ −4.21654e6 −0.245783
$$784$$ 0 0
$$785$$ 5.23507e6 0.303213
$$786$$ 0 0
$$787$$ −2.34266e7 −1.34826 −0.674129 0.738614i $$-0.735481\pi$$
−0.674129 + 0.738614i $$0.735481\pi$$
$$788$$ 0 0
$$789$$ −1.85818e6 −0.106266
$$790$$ 0 0
$$791$$ 0 0
$$792$$ 0 0
$$793$$ 4.34262e6 0.245228
$$794$$ 0 0
$$795$$ −194688. −0.0109250
$$796$$ 0 0
$$797$$ 1.82051e7 1.01519 0.507594 0.861596i $$-0.330535\pi$$
0.507594 + 0.861596i $$0.330535\pi$$
$$798$$ 0 0
$$799$$ 3.23983e7 1.79538
$$800$$ 0 0
$$801$$ −1.07851e7 −0.593941
$$802$$ 0 0
$$803$$ 9.67536e6 0.529515
$$804$$ 0 0
$$805$$ 0 0
$$806$$ 0 0
$$807$$ 764136. 0.0413035
$$808$$ 0 0
$$809$$ 1.47411e7 0.791878 0.395939 0.918277i $$-0.370419\pi$$
0.395939 + 0.918277i $$0.370419\pi$$
$$810$$ 0 0
$$811$$ 1.69629e7 0.905625 0.452812 0.891606i $$-0.350421\pi$$
0.452812 + 0.891606i $$0.350421\pi$$
$$812$$ 0 0
$$813$$ 3.16112e6 0.167731
$$814$$ 0 0
$$815$$ −1.00762e7 −0.531375
$$816$$ 0 0
$$817$$ −861040. −0.0451303
$$818$$ 0 0
$$819$$ 0 0
$$820$$ 0 0
$$821$$ 8.03929e6 0.416255 0.208128 0.978102i $$-0.433263\pi$$
0.208128 + 0.978102i $$0.433263\pi$$
$$822$$ 0 0
$$823$$ −386648. −0.0198983 −0.00994915 0.999951i $$-0.503167\pi$$
−0.00994915 + 0.999951i $$0.503167\pi$$
$$824$$ 0 0
$$825$$ −8.77104e6 −0.448659
$$826$$ 0 0
$$827$$ −3.55021e7 −1.80505 −0.902526 0.430635i $$-0.858290\pi$$
−0.902526 + 0.430635i $$0.858290\pi$$
$$828$$ 0 0
$$829$$ −2.48814e7 −1.25745 −0.628723 0.777630i $$-0.716422\pi$$
−0.628723 + 0.777630i $$0.716422\pi$$
$$830$$ 0 0
$$831$$ 2.73823e6 0.137552
$$832$$ 0 0
$$833$$ 0 0
$$834$$ 0 0
$$835$$ 1.54356e7 0.766141
$$836$$ 0 0
$$837$$ −6.01922e6 −0.296979
$$838$$ 0 0
$$839$$ 3.41458e7 1.67468 0.837340 0.546682i $$-0.184109\pi$$
0.837340 + 0.546682i $$0.184109\pi$$
$$840$$ 0 0
$$841$$ −1.37927e6 −0.0672450
$$842$$ 0 0
$$843$$ 789708. 0.0382734
$$844$$ 0 0
$$845$$ −4.23446e6 −0.204012
$$846$$ 0 0
$$847$$ 0 0
$$848$$ 0 0
$$849$$ −1.34607e6 −0.0640911
$$850$$ 0 0
$$851$$ 1.03661e6 0.0490671
$$852$$ 0 0
$$853$$ −2.50701e7 −1.17973 −0.589865 0.807502i $$-0.700819\pi$$
−0.589865 + 0.807502i $$0.700819\pi$$
$$854$$ 0 0
$$855$$ 2.15674e6 0.100898
$$856$$ 0 0
$$857$$ 1.81938e7 0.846199 0.423099 0.906083i $$-0.360942\pi$$
0.423099 + 0.906083i $$0.360942\pi$$
$$858$$ 0 0
$$859$$ 6.91797e6 0.319886 0.159943 0.987126i $$-0.448869\pi$$
0.159943 + 0.987126i $$0.448869\pi$$
$$860$$ 0 0
$$861$$ 0 0
$$862$$ 0 0
$$863$$ 2.78069e7 1.27094 0.635471 0.772125i $$-0.280806\pi$$
0.635471 + 0.772125i $$0.280806\pi$$
$$864$$ 0 0
$$865$$ −3.46141e7 −1.57294
$$866$$ 0 0
$$867$$ −305318. −0.0137945
$$868$$ 0 0
$$869$$ −4.49856e6 −0.202080
$$870$$ 0 0
$$871$$ 2.35275e7 1.05083
$$872$$ 0 0
$$873$$ 2.56261e7 1.13801
$$874$$ 0 0
$$875$$ 0 0
$$876$$ 0 0
$$877$$ 3.79587e6 0.166653 0.0833263 0.996522i $$-0.473446\pi$$
0.0833263 + 0.996522i $$0.473446\pi$$
$$878$$ 0 0
$$879$$ 3.66936e6 0.160184
$$880$$ 0 0
$$881$$ −2.48904e7 −1.08042 −0.540210 0.841530i $$-0.681655\pi$$
−0.540210 + 0.841530i $$0.681655\pi$$
$$882$$ 0 0
$$883$$ 3.13568e6 0.135341 0.0676705 0.997708i $$-0.478443\pi$$
0.0676705 + 0.997708i $$0.478443\pi$$
$$884$$ 0 0
$$885$$ −238464. −0.0102345
$$886$$ 0 0
$$887$$ −2.02437e7 −0.863933 −0.431966 0.901890i $$-0.642180\pi$$
−0.431966 + 0.901890i $$0.642180\pi$$
$$888$$ 0 0
$$889$$ 0 0
$$890$$ 0 0
$$891$$ 4.04273e7 1.70600
$$892$$ 0 0
$$893$$ 2.42858e6 0.101912
$$894$$ 0 0
$$895$$ 3.00223e7 1.25281
$$896$$ 0 0
$$897$$ −109824. −0.00455739
$$898$$ 0 0
$$899$$ 2.73113e7 1.12705
$$900$$ 0 0
$$901$$ −1.27156e6 −0.0521823
$$902$$ 0 0
$$903$$ 0 0
$$904$$ 0 0
$$905$$ 1.18867e7 0.482437
$$906$$ 0 0
$$907$$ 6.86324e6 0.277020 0.138510 0.990361i $$-0.455769\pi$$
0.138510 + 0.990361i $$0.455769\pi$$
$$908$$ 0 0
$$909$$ 1.12655e7 0.452211
$$910$$ 0 0
$$911$$ 9.40661e6 0.375523 0.187762 0.982215i $$-0.439877\pi$$
0.187762 + 0.982215i $$0.439877\pi$$
$$912$$ 0 0
$$913$$ −1.81829e7 −0.721914
$$914$$ 0 0
$$915$$ 1.45766e6 0.0575579
$$916$$ 0 0
$$917$$ 0 0
$$918$$ 0 0
$$919$$ 2.10280e7 0.821313 0.410656 0.911790i $$-0.365300\pi$$
0.410656 + 0.911790i $$0.365300\pi$$
$$920$$ 0 0
$$921$$ 3.02112e6 0.117360
$$922$$ 0 0
$$923$$ −2.15255e7 −0.831666
$$924$$ 0 0
$$925$$ −6.57706e7 −2.52742
$$926$$ 0 0
$$927$$ −2.92068e7 −1.11631
$$928$$ 0 0
$$929$$ 4.30928e7 1.63819 0.819096 0.573656i $$-0.194475\pi$$
0.819096 + 0.573656i $$0.194475\pi$$
$$930$$ 0 0
$$931$$ 0 0
$$932$$ 0 0
$$933$$ 3.75058e6 0.141057
$$934$$ 0 0
$$935$$ −8.66765e7 −3.24244
$$936$$ 0 0
$$937$$ 3.85862e6 0.143576 0.0717882 0.997420i $$-0.477129\pi$$
0.0717882 + 0.997420i $$0.477129\pi$$
$$938$$ 0 0
$$939$$ −3.02152e6 −0.111831
$$940$$ 0 0
$$941$$ −1.59601e7 −0.587572 −0.293786 0.955871i $$-0.594915\pi$$
−0.293786 + 0.955871i $$0.594915\pi$$
$$942$$ 0 0
$$943$$ 1.15258e6 0.0422076
$$944$$ 0 0
$$945$$ 0 0
$$946$$ 0 0
$$947$$ 4.66831e6 0.169155 0.0845775 0.996417i $$-0.473046\pi$$
0.0845775 + 0.996417i $$0.473046\pi$$
$$948$$ 0 0
$$949$$ −7.68654e6 −0.277054
$$950$$ 0 0
$$951$$ −4.05419e6 −0.145363
$$952$$ 0 0
$$953$$ −3.43457e7 −1.22501 −0.612505 0.790466i $$-0.709838\pi$$
−0.612505 + 0.790466i $$0.709838\pi$$
$$954$$ 0 0
$$955$$ −3.10510e7 −1.10171
$$956$$ 0 0
$$957$$ 6.29856e6 0.222311
$$958$$ 0 0
$$959$$ 0 0
$$960$$ 0 0
$$961$$ 1.03584e7 0.361813
$$962$$ 0 0
$$963$$ −3.09830e7 −1.07661
$$964$$ 0 0
$$965$$ −5.95156e7 −2.05737
$$966$$ 0 0
$$967$$ 2.28181e7 0.784718 0.392359 0.919812i $$-0.371659\pi$$
0.392359 + 0.919812i $$0.371659\pi$$
$$968$$ 0 0
$$969$$ −235752. −0.00806577
$$970$$ 0 0
$$971$$ 4.94042e7 1.68157 0.840786 0.541367i $$-0.182093\pi$$
0.840786 + 0.541367i $$0.182093\pi$$
$$972$$ 0 0
$$973$$ 0 0
$$974$$ 0 0
$$975$$ 6.96810e6 0.234749
$$976$$ 0 0
$$977$$ 7.17542e6 0.240498 0.120249 0.992744i $$-0.461631\pi$$
0.120249 + 0.992744i $$0.461631\pi$$
$$978$$ 0 0
$$979$$ 3.24907e7 1.08343
$$980$$ 0 0
$$981$$ 5.27134e7 1.74883
$$982$$ 0 0
$$983$$ −4.22279e6 −0.139385 −0.0696924 0.997569i $$-0.522202\pi$$
−0.0696924 + 0.997569i $$0.522202\pi$$
$$984$$ 0 0
$$985$$ −4.79388e7 −1.57433
$$986$$ 0 0
$$987$$ 0 0
$$988$$ 0 0
$$989$$ −879360. −0.0285875
$$990$$ 0 0
$$991$$ −1.65645e7 −0.535789 −0.267895 0.963448i $$-0.586328\pi$$
−0.267895 + 0.963448i $$0.586328\pi$$
$$992$$ 0 0
$$993$$ 3.08018e6 0.0991294
$$994$$ 0 0
$$995$$ −7.54495e7 −2.41601
$$996$$ 0 0
$$997$$ 4.40973e7 1.40499 0.702496 0.711687i $$-0.252069\pi$$
0.702496 + 0.711687i $$0.252069\pi$$
$$998$$ 0 0
$$999$$ −1.04093e7 −0.329994
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
## Twists
By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 784.6.a.f.1.1 1
4.3 odd 2 196.6.a.d.1.1 1
7.6 odd 2 112.6.a.e.1.1 1
21.20 even 2 1008.6.a.bb.1.1 1
28.3 even 6 196.6.e.f.177.1 2
28.11 odd 6 196.6.e.e.177.1 2
28.19 even 6 196.6.e.f.165.1 2
28.23 odd 6 196.6.e.e.165.1 2
28.27 even 2 28.6.a.a.1.1 1
56.13 odd 2 448.6.a.h.1.1 1
56.27 even 2 448.6.a.i.1.1 1
84.83 odd 2 252.6.a.d.1.1 1
140.27 odd 4 700.6.e.d.449.2 2
140.83 odd 4 700.6.e.d.449.1 2
140.139 even 2 700.6.a.d.1.1 1
By twisted newform
Twist Min Dim Char Parity Ord Type
28.6.a.a.1.1 1 28.27 even 2
112.6.a.e.1.1 1 7.6 odd 2
196.6.a.d.1.1 1 4.3 odd 2
196.6.e.e.165.1 2 28.23 odd 6
196.6.e.e.177.1 2 28.11 odd 6
196.6.e.f.165.1 2 28.19 even 6
196.6.e.f.177.1 2 28.3 even 6
252.6.a.d.1.1 1 84.83 odd 2
448.6.a.h.1.1 1 56.13 odd 2
448.6.a.i.1.1 1 56.27 even 2
700.6.a.d.1.1 1 140.139 even 2
700.6.e.d.449.1 2 140.83 odd 4
700.6.e.d.449.2 2 140.27 odd 4
784.6.a.f.1.1 1 1.1 even 1 trivial
1008.6.a.bb.1.1 1 21.20 even 2
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http://dotienich.tk/algebra-2-9-6-answer-sheet.html
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# Algebra 2 9 6 answer sheet
Sheet algebra
## Algebra 2 9 6 answer sheet
As well as challenge questions at the sheets end. Free Algebra 2 worksheets ( pdfs) with answer keys- each answer includes visual aides model problems . 3 Quiz Review Answer Key from MATH 9 Algebra 1 at Grosse Pointe South High School. It is a key foundation to algebra the field of machine learning, from notations used to describe the operation of algorithms to the implementation of algorithms in code. Linear algebra is a sub- field of mathematics concerned with vectors , matrices linear transforms. ALGEBRA 2 FINAL EXAM REVIEW. 6 algebra Trigonometry for Solving Problems- 03- 04. 5- 6), used extensively in “ welfare analysis” ( Mankiw Ch. In order to assist educators with the implementation of the Common Core Mathematics that schools , the New York State Education Department provides curricular modules in answer P- 12 English Language Arts , districts can adopt adapt for local purposes.
alg 2 final review 1- 2 alg 2 final review 3- 4 answer alg 2 final review 5- 6 alg 2 final review 7- 8 alg 2 final review 9 5/ 16: Mon: extra quiz online 5/ 17: Tues: Review for Chapter 8. A Q2i0 D1K29 JK ku lt Pau lS Vo Lf gtyw Eatr 5ej VLALsCC. Review Answers ( textbook). Smyth' s Math Website. Course 2: sheet Pre- Algebra 3- 6 Solving Inequalities by Adding or Subtracting sheet AF4.
Algebra 1- 2 Assignments. We will use fractions and percentages primarily for application called “ elasticity”. Hopewell Junior High School 2354 Brodhead Rd. Algebra 2 Worksheets As. answer Algebra 1A Worksheets. Online Textbook Access. 3 Quiz Review algebra List the perfect squares below: Write theequation in. Glencoe/ McGraw- Hill v Glencoe Algebra 1 Assessment Options The sheet assessment masters in the Chapter 6 Resources Mastersoffer a wide range of assessment tools for intermediate and final. Elasticity is introduced immediately after “ algebra supply and demand” ( Mankiw Ch.
This is to be used only to check your answers to your. algebra Algebra 2Chapter 9 Answers 35 Chapter 9 Answers Practice 9- 1 1. 7- 9) revisited in sheet the “ theory of the firm” , “ factors of production. 1 Solving Linear Inequalities using Addition Subtraction 3- 7 algebra 9 Solving Inequalities by Multiplying Dividing AF4. Search this site. g Worksheet by Kuta Software LLC. Algebra 2 9 6 answer sheet. Classify – 6x5 algebra + 4x3 + 3x2. Algebra 2 9 6 answer sheet.
I varies inversely with R. 1 Solving Linear Inequalities using Multiplication answer and Division. 7− 6 × 2 2 = 2 1 × 1 = 2. Algebra 1- 2 HW Answers. 4 n SMgaSdLek Tw algebra MiQtBh1 8I XnRffi 3n answer mi0t 4eQ RA7l 2g WepbUrKa1 X1N. Identify the choice that best completes the statement or answers the question. H 9 vA pl 0l x 6rli agchZtusm Tr2easheUrjv8e edF.
## Sheet algebra
4 p( 4p 1) 4m2 1 4m 8 2m16 1 m2 3m 10 4 p2 7 2 7p5 p2 8 1 14p4 y5 18xz 2 5y 6xy4 25z3 m3 9m m2 9 ( m 3) 2 m2 6m 9 c c 2 4c 5 c2 4c 3 c 3c c 25 4m5 m 1 3m3 3m 6m4 x2 x2 6 x 2 x2 6x 27 4x 12x 9 3 2x 9 6x 2a 2b x 4 2( x 2) ( x 4) ( x 4) ( x 1) 2( x 1) ( x 2) ( x 4) x 1 x2 2x 8 x22 8x 16 2x 2 x2 2x 8 x 1 x 8x 16 2x 2 x2 2x 8. Oct 27, · Here is a puzzle activity for reviewing equation solving. I found that it worked better when I made an answer mat for students to put their pieces onto ( I indicated a couple of pieces on the mat to help them align the rest of their pieces). Answer Sheet for Released Tests and Test Item Sets Page 1 of 2 Answer Sheet for Released Tests or Released Test Item Sets The student should describe the answer for a technology- enhanced item. Algebra 2 Worksheets Dynamically Created Algebra 2 Worksheets. Here is a graphic preview for all of the Algebra 2 Worksheet Sections.
``algebra 2 9 6 answer sheet``
You can select different variables to customize these Algebra 2 Worksheets for your needs. Find a 9- digit number, which you will gradually round off starting with units, then tenth, hundred etc.
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https://platformfvef.web.app/jethva84406ty/what-does-the-herfindahl-index-mean-rif.html
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crawl-data/CC-MAIN-2021-49/segments/1637964362230.18/warc/CC-MAIN-20211202145130-20211202175130-00400.warc.gz
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What does the herfindahl index mean
The Herfindahl index (H) is the sum of the squared firm sizes, all measured as a proportion of total market size. In the figure, for market A, H = (0.12)2 X 5 + (0.08) 2X 5 = 0.104. Herfindahl-Hirschman Index or HHI score refers to a measure of market concentration and is an indicator of the amount of competition in a particular industry. HHI Index formula helps in analyzing and observing, if a particular industry is highly concentrated or close to monopoly or if there is some level of competition around it.
Government and industry analysts evaluate the level of competition in an industry using the Herfindahl Index. This is a mathematical formula used to assess the concentration of firms in a particular industry or market. Government and industry analysts evaluate the level of competition in an industry using the Herfindahl Index. This is a mathematical formula used to assess the concentration of firms in a particular industry or market. It is often used to evaluate potential mergers for antitrust concerns. THe Herfindahl index (also known as Herfindahl–Hirschman Index, HHI, or sometimes HHI-score) is a measure of the size of firms in relation to the industry and an indicator of the amount of competition among them. Named after economists Orris C. Herfindahl and Albert O. Hirschman, it is an economic concept widely applied in competition law, antitrust and also technology management. Herfindahl Index Law and Legal Definition. Herfindahl Index is a measure of concentration of the production in an industry that's calculated as the sum of the squares of market shares for each firm. This is an alternative method of summarizing the degree to which an industry is oligopolistic and the relative concentration of market power held What does the Herfindahl-Hirschman Index mean when there are only six firms? As readers know, this index is commonly used as a way to measure monopoly, by looking at how much of the output of a particular industry is dominated by a small number of firms. The Herfindahl index, however, is not without a few problems. The first problem is to find meaning in the numbers. While a four-firm concentration ratio of 61.25 percent MEANS that the top four firms in the industry account for 61.25 percent of total industry sales, what does an Herfindahl index of 1177 really mean? While the two most widely used measures of market (industrial) concentration, the m-firm concentration ratio C R m and the Herfindahl-Hirschman index H, have no precise functional relationship, they can be related by means of boundary formulations.Such bounds and potential relationships, which have been considered in some earlier reported studies, are being re-examined, corrected, and
2 Dec 2017 In such cases, HHI can be defined in terms of points instead of wins, and total points can represent the total of points actually accumulated by
Herfindahl Index, Shannon's H, and their normalized versions. political parties in a country does not necessarily mean higher turnover, as the “effective. 19 Apr 2018 The Herfindahl-Hirschman Index (HHI) is a widely used measure of market The HHI is used as a measure of competition, with 10,000 meaning perfect You can easily cross-check the result in excel by applying the formula. 2 Dec 2017 In such cases, HHI can be defined in terms of points instead of wins, and total points can represent the total of points actually accumulated by 4 Apr 2014 The main difference is that RDI scores are inverted so that higher scores indicate higher diversity. The Herfindahl-Hirschman Index is a widely
The Herfindahl index (also known as Herfindahl–Hirschman Index, HHI, or sometimes HHI-score) is a measure of the size of firms in relation to the industry and an indicator of the amount of competition among them.
The Herfindahl index (H) is the sum of the squared firm sizes, all measured as a proportion of total market size. In the figure, for market A, H = (0.12)2 X 5 + (0.08) 2X 5 = 0.104. Herfindahl-Hirschman Index or HHI score refers to a measure of market concentration and is an indicator of the amount of competition in a particular industry. HHI Index formula helps in analyzing and observing, if a particular industry is highly concentrated or close to monopoly or if there is some level of competition around it. The Herfindahl Index, also known as the Herfindahl-Hirschman Index (HHI), measures the market concentration of an industry's 50 largest firms in order to determine if the industry is competitive or nearing monopoly. Definition and meaning The Herfindahl-Hirschman Index, also called the Herfindahl Index, measures the extent to which market share is concentrated among a few or many companies. It measures market concentration of an industry’s fifty biggest firms in order to determine whether that industry has a healthy number of competitors or is nearing monopoly. The Herfindahl index (also known as Herfindahl–Hirschman Index, HHI, or sometimes HHI-score) is a measure of the size of firms in relation to the industry and an indicator of the amount of competition among them. The Herfindahl-Hirschman Index is an index that measures the market concentration of an industry. A highly concentrated industry is one where only a few players in the industry hold a large percentage of the market share, leading to a near-monopolisticMonopolistic Competition situation. The term “HHI” means the Herfindahl–Hirschman Index, a commonly accepted measure of market concentration. The HHI is calculated by squaring the market share of each firm competing in the market and then summing the resulting numbers.
The Herfindahl Index, also known as the Herfindahl-Hirschman Index (HHI), measures the market concentration of an industry's 50 largest firms in order to determine if the industry is competitive or nearing monopoly.
Though in the definition (1) employing market shares it is a normalized index, its range is not the whole [0,1] interval. In fact, we can see that the maximum value Concentration measures, such as the Hirschman-Herfindahl Index (HHI), are commonly used in As the results in Table 4 show, this does not necessarily mean. I would now like to calculate the HHI index based on these two columns. I computed If "the command" means entropyetc (SSC) then. Code:.
The h-index is an author-level metric that attempts to measure both the productivity and citation impact of the publications of a scientist or scholar. The index is based on the set of the scientist's most cited papers and the number of citations that they have received in other publications.
The Herfindahl index (H) is the sum of the squared firm sizes, all measured as a proportion of total market size. In the figure, for market A, H = (0.12)2 X 5 + (0.08) 2X 5 = 0.104. Herfindahl-Hirschman Index or HHI score refers to a measure of market concentration and is an indicator of the amount of competition in a particular industry. HHI Index formula helps in analyzing and observing, if a particular industry is highly concentrated or close to monopoly or if there is some level of competition around it. The Herfindahl Index, also known as the Herfindahl-Hirschman Index (HHI), measures the market concentration of an industry's 50 largest firms in order to determine if the industry is competitive or nearing monopoly. Definition and meaning The Herfindahl-Hirschman Index, also called the Herfindahl Index, measures the extent to which market share is concentrated among a few or many companies. It measures market concentration of an industry’s fifty biggest firms in order to determine whether that industry has a healthy number of competitors or is nearing monopoly. The Herfindahl index (also known as Herfindahl–Hirschman Index, HHI, or sometimes HHI-score) is a measure of the size of firms in relation to the industry and an indicator of the amount of competition among them. The Herfindahl-Hirschman Index is an index that measures the market concentration of an industry. A highly concentrated industry is one where only a few players in the industry hold a large percentage of the market share, leading to a near-monopolisticMonopolistic Competition situation. The term “HHI” means the Herfindahl–Hirschman Index, a commonly accepted measure of market concentration. The HHI is calculated by squaring the market share of each firm competing in the market and then summing the resulting numbers.
The Herfindahl-Hirschman Index, also called the Herfindahl Index, measures the extent to which market share is concentrated among a few or many companies. Once markets are defined, the Herfindahl–Hirschman Index (HHI) is the most common measure used for determining market concentration, including by the The Justice Department's 1982 adoption of the HHI as the means of measuring market concentration for purposes 01" anritrust.analyx sis has resulted in a
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https://bilakniha.cvut.cz/en/predmet2389406.html
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crawl-data/CC-MAIN-2023-50/segments/1700679100603.33/warc/CC-MAIN-20231206194439-20231206224439-00261.warc.gz
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CZECH TECHNICAL UNIVERSITY IN PRAGUE
STUDY PLANS
2023/2024
# Fieldwork Surveying
Code Completion Credits Range Language
154FS01 Z,ZK 6 2P+3C English
Garant předmětu:
Tomáš Křemen
Lecturer:
Tomáš Křemen
Tutor:
Tomáš Křemen, Lenka Línková
Supervisor:
Department of Special Geodesy
Synopsis:
Introduction to surveying, basic geodetic calculations, evaluation of precision and accuracy of a measurement, theory of errors, instrumentation, topographic survey, angular and distance measurements, determination of heights, photogrammetry, laser scanning, mapping, setting-out in construction, surveying for monitoring of displacements, cadastre of real estates.
Requirements:
no prerequisities
Syllabus of lectures:
1. Introduction to surveying, shape and size of the Earth, reference surfaces, cartographic projections.
2. Basic geodetic calculations.
3. Evaluation of precision and accuracy of a measurement.
4. Angular measurement.
5. Distance measurement.
6. Determination of heights - part I.
7. Determination of heights - part II.
8. Measurements for thematic mapping and for as-built documentation of constructions.
9. Setting out - part I.
10. Setting out - part II.
11. State map series in the Czech Republic, thematic maps for construction industry.
12. Cadastre of real estates of the Czech Republic.
Syllabus of tutorials:
1.Levelling – part 1: principles, instruments, reading of levelling rods, registration of measured data, calculation of a levelling field book.
1st Homework – Calculation of the levelling field book
2.Levelling – part 2: practical measuring in the field, calculation and adjustment of a levelling line.
3.Total stations – part 1: total station Trimble M3 – description, centering, levelling, reading, pointing, horizontal and zenith
angles booking and calculating.
1st Test – Calculation of a levelling field book
2nd Homework – Calculation of horizontal directions and zenith angles
4.Total stations – part 2: total station Topcon GPT 2006 and Trimble M3 – angular and distance measurement without registration.
2nd Test – Calculation of horizontal and vertical angles
5.Total stations – part 3: total station Trimble M3 – software, registration of a measurement, communication with a computer, tacheometry, data
collecting, setting-out.
3rd homework – Spatial polar method, setting-out elements
6.GNSS – part 1: interpolation of contour lines, Global navigation satellite system – introduction, basic principles, GPS Trimble GeoXR – description,
software, demonstration of measurement and calculations.
3rd Test – Spatial polar method, setting-out elements
4th Homework – Interpolation of contour lines
7.GNSS – part 2: practical measuring in the field, comparison of results with other surveying methods.
8.Situation: measurement of the planimetry for special-purpose plan creation.
9.Tacheometry: tacheometric measurement for contour line plan creation.
10.Interior: measurement and drawing of an interior.
11.Setting-out: setting-out of an object using a total station.
12.Working with map: basic cartographic measurements and calculations, altitude measurement, measurement of volume and area. ASSESSMENT
Study Objective:
Understanding of the problems of engineering surveying, basic geodetic methods and calculations used in civil engineering, possibilities of modern surveying.
Study materials:
1. Schofield W., Breach M.: Engineering Surveying - Butterworth-Heinemann, 2011, ISBN 978-0-7506-6949-8.
2. Uren J., Price B.: Surveying for Engineers - Palgrave Macmillan, 2010, ISBN 978-0-230-22157-4.
3. Anderson, J.M. - Mikhail, E.M.: Surveying: Theory and Practice. McGraw-Hill New York, 1998, 1167 pages, ISBN 9780070159143.
Note:
Further information:
https://k154.fsv.cvut.cz/vyuka/english/fs01/
Time-table for winter semester 2023/2024:
06:00–08:0008:00–10:0010:00–12:0012:00–14:0014:00–16:0016:00–18:0018:00–20:0020:00–22:0022:00–24:00 roomTH:B-97213:00–15:50(lecture parallel1parallel nr.101)Thákurova 7 (budova FSv)B972 roomTH:B-97114:00–15:50(lecture parallel1)Thákurova 7 (budova FSv)B971
Time-table for summer semester 2023/2024:
Time-table is not available yet
The course is a part of the following study plans:
Data valid to 2023-11-30
Aktualizace výše uvedených informací naleznete na adrese https://bilakniha.cvut.cz/en/predmet2389406.html
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| 2.6875
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CC-MAIN-2023-50
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https://solvedlib.com/n/gllil-2-l-1-0-w-1-2-f-1-5-1-l-1-1,7542560
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crawl-data/CC-MAIN-2024-18/segments/1712296817455.17/warc/CC-MAIN-20240419203449-20240419233449-00762.warc.gz
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# Gllil 2 ! L 1 0 W [ 1 2 F 1 5 1 L ! 1 1 |
###### Question:
Gllil 2 ! L 1 0 W [ 1 2 F 1 5 1 L ! 1 1 |
#### Similar Solved Questions
##### Usn] the Ae ille exampit Rveve tuat P) e kje ecy= ecryExaxple e(R)@ P(R) ecR') to Prove Ll - result u be wl,et 0 & R" 4n9 opek tLe urien 03 rec fanfles (ai,bi)A(Ci,d: ) Wkerc Glb Cijd Ave ralioha| nuhbers Ilev Ueecrlo &ri Hekce AN) = &RJo e(rr) Let Pyevc (ke ~RPos tc .Id U;axd uz ar? open sets in R then Wixu Opew Q" awi hence 4i 74z € R(R) Gcf Lal {P( esR Bxuz ep(') $Sijra -uljebra , and Sile (1 Cntain$ evcrJ Pex sct 08 VeSiwbc ve Cxtains B(R) Be B(R) =>
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##### Box with square base and gpen LQP must have volume of 442368 cm We wish to find the dimensions of the box that minimize the amount of material used_First, find formula for the surface area of the box in terms of only €, the length of one side of the square base [Hint: use the volume formula to express the height of the box in terms of €.] Simplify your formula as much as possible_ A(z)Next, find the derivative _ A' () A' (2)Now calculate when the derivative equals zero that is, wh
box with square base and gpen LQP must have volume of 442368 cm We wish to find the dimensions of the box that minimize the amount of material used_ First, find formula for the surface area of the box in terms of only €, the length of one side of the square base [Hint: use the volume formula t...
##### Homework: data analysis 3 of62 CENT TH n the Sales, Number of Vehicles Sold, Square Footage...
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##### Indepenucnt EventsRn Gmi RN- 05 FB} 04anu AYAUBI 01 when Mund Bindependent overtstPlc" WN optionAand B are nOt incependent cventsThe data Biven In the statemnerit /5 rIOL sUffcient (0 dlsweThie questior Car"t tu answcted bec u%E IrA und B uare Independeni Evefls
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##### Homework: Unit D Homework MyLab Math WucaScoie81447 VucucJatahaltieMavtcaedb /4600 41402 Seteaa t&4t4n444 Da *ur 4eladb ! Dt *u Fnthe -Dn *48 D" ,44be ! Ma Aatt |C4 mneret
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##### Cost data for Sandusky Manufacturing Company for the month ended January 31 are as follows: Inventories...
Cost data for Sandusky Manufacturing Company for the month ended January 31 are as follows: Inventories January 1 January 31 Materials $310,000$276,600 238,400 Work in process 215,200 Finished goods 162,800 190,100 January 31 Direct labor \$565,000 604,800 Materials purchased during the month Factor...
##### (6) Consider the following Markov Process with transition probabilities shown0.50.6 0,6 0 5Find the transition matrix P and the probabilities p and q Is the Markov process ergoclic? Explain your answer. Find the ]quilibriu probabilitics_
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##### For a permutation of [n], in a random pick what is the expectednumber of entries wherein a consecutive entry,the first entry is bigger than the secondentry.
For a permutation of [n], in a random pick what is the expected number of entries wherein a consecutive entry, the first entry is bigger than the second entry....
##### Use Aryabhata's rule to compute the altitude of the sun above the horizon in London (latitude 515 32') at 10.00 AM (local solar time) on the vernal equinox. Assume that the sun rises at 6.00 AM on that day and sets at 6.00 PM_
Use Aryabhata's rule to compute the altitude of the sun above the horizon in London (latitude 515 32') at 10.00 AM (local solar time) on the vernal equinox. Assume that the sun rises at 6.00 AM on that day and sets at 6.00 PM_...
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Home » » Find the probability that a number picked at random from the set(43, 44, 45, ......
# Find the probability that a number picked at random from the set(43, 44, 45, ......
### Question
Find the probability that a number picked at random from the set(43, 44, 45, ..., 60) is a prime number.
### Options
A) $$\frac{2}{3}$$
B) $$\frac{1}{3}$$
C) $$\frac{2}{9}$$
D) $$\frac{7}{9}$$
The correct answer is C.
### Explanation:
Prime numbers = (43,47,53,59)
N = (43, 44, 45,..., 60)
The universal set contains 18 numbers.
The prime numbers between 43 and 60 are 4
Probability of picking a prime number = $$\frac{4}{18}$$
= $$\frac{2}{9}$$
## Dicussion (1)
• Prime numbers = (43,47,53,59)
N = (43, 44, 45,..., 60)
The universal set contains 18 numbers.
The prime numbers between 43 and 60 are 4
Probability of picking a prime number = $$\frac{4}{18}$$
= $$\frac{2}{9}$$
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# The Impulse Formula Physics Stories
In case the system is exactly on the cusp between these 2 conditions, it’s believed to be critically damped. Until now, we’ve discussed vectors with regard to a Cartesian, in other words, an x-y coordinate system. When an object is accelerating, it is essay writer not in equilibrium. Use the formatting choices for this undertaking. Represent the vector working with the simulation.
Most physical quantities can be expressed concerning combinations of five standard dimensions. This equation represents one of two main principles to be utilized in the analysis of collisions in this unit. These five dimensions are chosen as being basic since they are simple to measure in experiments. This is often known buy essay papers as the impulse-momentum theorem.
Learning about momentum is vital because this has many practical applications. Physics courses like these may encompass a wider selection of material as a way to build familiarity with as many concepts as possible. Circuits similar to this are called multi-loop circuits.
So it can carry out different calculations to provide you with the results that you demand. There are several ready-made simulations that can be found on the world wide web but the majority of these are pre-programmed to demonstrate certain phenomena, these are excellent for demonstrating theory but for the purpose of the person investigation I would prefer students to construct their own simulations using one of these options. PDF formats in addition to a webquest activity.
## Impulse Formula Physics – Dead or Alive?
Which is the reason I need to show you this is a significant trading system. As is typical in any issue, there are assumptions hidden in how the dilemma is stated and we have to work out how to take care of it. essay writing service The mathematical expression for work depends on the specific conditions.
The initial and last information can often tell you all you have to understand. Again, you don’t need to be sure of these directions now. A number of the words might be incorrectly translated or mistyped.
Weight is a good example of force given above. At times it’s really hard to tell which is the right direction for the current in a specific loop. A change with distance is known as a gradient to prevent confusion with a change with time that is referred to as a rate. Label the current and the present direction in every branch.
Utilize Kirchoff’s second rule to write down loop equations for since many loops as it can take to incorporate each branch at least one time. This lengthier time interval contributes to a larger change in momentum. In each one of these examples, a mass unit is multiplied by means of a velocity unit to offer a momentum unit. For instance, if you desire to ascertain the acceleration of the block farther down the plane, then you are going to require the component of the force which acts down the plane.
We are aware that force produces acceleration. However, we believe that motive waves don’t have to be in 5 waves. Nonetheless, in sprinting, you’ve got vertical forces together with horizontal forces.
Harmonic motion in actual life is rarely straightforward. The experiment managed to effectively demonstrate the association between Impulse in Momentum and how they’re related to one another. The more momentum an object has, the harder that it’s to stop. While this mental notion of momentum could be debatable, physical momentum is quite real in sports. As soon as we say momentum has to be conserved, that which we mean is that a system’s momentum at the same time must equal the entire momentum at a subsequent moment. 6-4-98 There are two types of momentum, linear and angular.
## The Advantages of Impulse Formula Physics
He can really get this to accelerate, you’re a small guy and you’re having a tough time getting this huge heavy bat to accelerate you just don’t have the capability to apply the exact same force. It’s energy related to a moving object, to put it differently. You’ve also experienced this a great number of times while driving. The distinction is that the change occurs over a different period of time and with a different quantity of force. There’s something else that could change momentum. Particularly in the previous forty decades, there have been several new design features on vehicles with a concentration on safety, drastically reducing the sum of fatalities and serious injuries occurring from automobile accidents.
Another way to check at it is to bear in mind the simple fact that momentum should always be conserved. A team which has a lot of momentum is truly on the move and will be challenging to stop. We don’t await a sell signal to escape a buy trade, or a buy signal to escape from a sell trade. To find this, consider the figure above.
The energy would likewise change the total energy is conserved but parts of energy are lost such as Kinetic Energy that’s lost or transformed into various forms like heat and sound as a result of collision. In handling the motion of a rigid body, the expression displacement may also contain the rotations of the human body. The quantity is provided the name impulse.
Quite simply, there’s no net torque on the object. Also any net impulse on an object is going to be ballanced by means of an impulse on the middle of mass of the object brought on by inertia. Corrective waves have far more variety and less clearly identifiable in comparison with impulse waves.
## Impulse Formula Physics Fundamentals Explained
Impulse is only a fancy means of keeping an eye on the whole quantity of force applied to an object. Another method is to use time by which a force acts on any certain body. In addition, this is equal to the size of the force multiplied by the duration of time the force is put on.
Grass is a rather smooth surface, but in addition one that generates plenty of friction with the ball. In order for it to enter the goal without falling to the grass, it needs to be kicked high enough for it to reach the goal above grass level.
The period of time it requires to stop it’s reduced so the plate doesn’t shatter. This is a great place to get a pull back in case you comprehend the potential ahead for wave 5. As for me, I lean towards the latter of those 2 sides. After that, consider catching a ball whilst keeping your hands still.
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Impact-Site-Verification: dbe48ff9-4514-40fe-8cc0-70131430799e
Search This Blog
Write a program in MATLAB to enter a number containing three digits or more. Arrange the digits of the entered numbers in ascending order and display the result.
Sample input: Enter a number 4972
Sample output: 2,4,7,9
General way to extract the digits from a number:
x=input('Enter the number');
m=x;
while(m>0)
b=rem(m,10);
% Line has to be written accoring to code
m=(m-b)/10;
end
Explanation of the algo:
Code for this particular numerical:
x=input('Enter the number');
m=x;
Y=[];
while(m>0)
b=rem(m,10);
Y=[Y b];
m=(m-b)/10;
end
a=sort(Y);
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### 11 Cards in this Set
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Galileo Questioned laws of nature work Built own telescope Observations that supported heliocentric model: -moon not perfect (mt and valley) -Jupiter has own moons -Sun has sunspots so sun is less than perfect, rotates about its axis -Venus has phases and changes in angular size Newtons first law (Law of Inertia) every object continues in state of rest, or uniform motion in straight line, unless it is compelled to change by forces impressed on it. Mass quantity of matter in an object. weight force upon an object due to gravity Newton's second law acceleration of an object is directly proportional to net force acting on the object, is in direction of the net force, and is inversely proportional to the mass of the object F=ma as F inc, a inc as m inc, a dec more mass, push w/ greater force Newton's third law whenever 1 object exerts a force on second object, 2nd object exerts an equal and opposite force on the first Newton's law of universal gravitation Gravity causes acceleration, which causes gravitational force Force depends on mass of both objects force that is proportional to product of masses and inversely proportional to square of the distance b/w them Tides due to... differences in gravitational forces across Earth. Water on surface free to bulge with tides High vs. Low Tides High tide every 12 hours Low tide every quarter turn (in b/w high tides) Spring tides during new moon and full moon, sun and moon exert tidal forces in same direction. Highest high tides and lowest low tides Neap tides Moon is in first and last quarter, when sun and moon exert tide forces in opposite directions, work against each other. Lowest high tides and highest low tides
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# Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. Find the magnetic field B at the centre of the loop assuming uniform wires.
A
2μ0i3πa
B
2μ0i3πa
C
2μ0iπa
D
2μ0iπa
Video Solution
Text Solution
Verified by Experts
## According to question, resistance of wire ADC is twice that of wire ABC. Hence current flowing through ADC is half that of ABC i.e., i2i1=12 Also i1+i2=i ⇒i1=2i3 and i2=i3 Magnetic field at centre O due to wire AB and BC (part 1 and 2) B1=B2=μ04π2i1sin45∘a/2=μ04π⋅2√2i1a and magnetic field at centre O due to wire AD and DC [i.e part 3 and 4] B3=B4=μ04π2√2i2a Also i1=2i2 . So (B1=B2)>(B3=B4) Hence net magnetic field at centre O Bnet=(B1+B2)−(B3+B4) =2×μ04π⋅2√2×(23i)a−μ04π⋅2√2(i3)×2a =μ04π⋅4√2i3a(2−1)=√2μ0i3πa
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Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## The resistance of wire ABC is double of resistance of wire ADC. The magnetic field at O is
Aμ0i12R,
Bμ0i6R,
Cμ0i3R,
Dμ0i2R,
• Question 2 - Select One
## A square loop of side a carries a current I. The magnetic field at the center of the loop is
A2μ0I2πa
Bμ0I2πa
C4μ0I2πa
Dμ0Iπa
• Question 3 - Select One
## A square loop of edge 'a' carries a current l. The magnetic field at the centre of loop is
Aμ0l22πa
B22μ0lπa
Cμ0l2πa
Dμ0l2πa
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# Rat maze?
1,462.6K Views
A rat is placed at the beginning of a maze and must make it to the end. There are four paths at the start that he has an equal chance of taking: path A takes 5 minutes and leads to the end, path B takes 8 minutes and leads to the start, path C takes 3 minutes and leads to the end, and path D takes 2 minutes and leads to the start.
What is the expected amount of time it will take for the rat to finish the maze?
arjun00848 Expert Asked on 20th August 2015 in
x = 0.25 * 5 + 0.25 (x + 8) + 0.25 * 3 + 0.25 (x + 2)
x = 1.25 + 0.25x + 2 + 0.75 + 0.25x + 0.5
x = 4.5 + 0.5x
0.5x = 4.5
x = 9
Ralph Aldanese Scholar Answered on 4th March 2016.
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A194778 T(n,k)=Number of lower triangles of an (n+2k-2)X(n+2k-2) 0..k array with new values introduced in row major order 0..k and no element unequal to more than one horizontal or vertical neighbor 14
1, 7, 3, 43, 17, 6, 268, 105, 41, 12, 1740, 672, 265, 95, 23, 11862, 4490, 1736, 655, 219, 43, 85013, 31466, 11857, 4464, 1641, 493, 79, 639760, 231445, 85007, 31429, 11686, 4069, 1101, 143, 5045610, 1784788, 639753, 231395, 84727, 30608, 10121, 2427, 256 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Table starts ...1....7....43....268....1740....11862.....85013.....639760....5045610 ...3...17...105....672....4490....31466....231445....1784788...14404218 ...6...41...265...1736...11857....85007....639753....5045602...41615156 ..12...95...655...4464...31429...231395...1784723...14404136..121420974 ..23..219..1641..11686...84727...639325...5044981...41614291..358184223 ..43..493..4069..30608..229869..1782111..14399939..121414559.1066897441 ..79.1101.10121..80961..631373..5029741..41587186..358138793.3210593393 .143.2427.25025.214469.1745459.14321783.121261490.1066617315 LINKS R. H. Hardin, Table of n, a(n) for n = 1..120 EXAMPLE Some solutions for n=2 k=2 ..0........0........0........0........0........0........0........0 ..0.0......0.0......1.1......1.1......0.0......1.1......0.1......0.0 ..1.1.1....0.0.1....1.1.2....1.1.1....0.0.0....1.1.1....0.1.1....0.0.0 ..1.1.1.0..0.0.1.1..1.1.2.2..0.0.0.0..0.0.0.1..2.2.2.2..0.1.1.0..1.1.1.1 CROSSREFS Column 1 is A055244 Sequence in context: A272419 A272156 A271815 * A272010 A145758 A038269 Adjacent sequences: A194775 A194776 A194777 * A194779 A194780 A194781 KEYWORD nonn,tabl AUTHOR R. H. Hardin Sep 02 2011 STATUS approved
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Last modified July 19 03:15 EDT 2024. Contains 374388 sequences. (Running on oeis4.)
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# Continuous Random Variables
Suppose $X$ is some uncertain continuous quantity.
Then the probability that $X$ lies in any interveal $a \leq X \leq b$ can be computed as follows:
Define events $A = (X \leq a)$, $B = (X \leq b)$ and $W = (a \lt X \leq b)$
$B = A \cup W$
and $A$ and $W$ are mutually exclusive
This means that
$p(B) = p(A) + p(W)$
and so
$p(W) = p(B) - p(A)$
Now define the function $F(q) = p(X\leq q)$. This is called the cumulative distribution function or cdf.
Now, the probability density function is
$f(x) = \frac{d}{dx}F(x)$
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# What is 1340000 in scientific notation?
$1340000 = 1.34 \times {10}^{6}$
Count the number of times you have to shift $1340000$ to the right before it is less than $10$. This gives you the power of $10$.
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# Expected value for the number of tries to draw the black ball from the bag
We have a bag with $4$ white balls and $1$ black ball. We are drawing balls without replacement. Find expected value for the number of tries to draw the black ball from the bag.
Progress. The probability to draw a black ball from first trial is $1/5$. The problem is how to find the probability to draw black ball from $2$nd, $3$rd, $\ldots, 5$th trial. When I know all this probabilities I can find expected value as $1\cdot(1/5) + 2 p_2 + \dots + 5 p_5$.
• Do you know the definition of expectation value? – Arthur Dec 18 '15 at 16:02
• @Arthur Yes, from the wikipedia. The probability to draw a black ball from first tryal is 1/5. The problem is how to find the probability to draw black ball from 2nd, 3rd ... 5th tryal. When i know all this probabilities I can find expected value as 1*(1/5) + 2*p2 + ... + 5*p5 – Alexander Dec 18 '15 at 16:10
• Exactly. Well, I invite you to answer your own question. – drhab Dec 18 '15 at 16:12
• The probability is quite easy to figure out with this simple trick: Imagine you draw all the balls, one by one, and put them in a row, first to last. Then the black ball is equally likely to be in any if the five positions, which means that $p_1 = p_2 = p_3 =p_4 = p_5$. – Arthur Dec 18 '15 at 16:13
• $p_2=\dfrac14\left(1-p_1\right)$ since you have four balls left, and $p_3=\dfrac13\left(1-p_1-p_2\right)$ and so on – Henry Dec 18 '15 at 16:17
It is as if you will create a word with $4$ W's and $1$ B. For example $BWWWW$ or $WWWBW$ etc. How many such words can you create? Answer: $5$ and any such word is equally likely.
In other words: the probability that the black ball will be drawn at any place - not only the first - is equal to $1/5$. Not conditional probability, but probability. Do not get confused, that if you have drawn $4$ White balls then the probability of drawing the black ball in the fifth draw is $1$. This is the conditional probability. "A priori" it is equally likely that the black ball will be drawn at any given point from $1$ to $5$. So, $$E[X]=\frac{1}{5}\cdot 1+ \frac{1}{5}\cdot2+\ldots+\frac15\cdot 5=\frac15(1+2+3+4+5)=3$$ (where $X$ denotes the number of trials).
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# Can I use OR within a MATCH function?
#### mjhokie13
##### New Member
I am trying to make a counter using a match function. The below formula is what I was using, and it works fine, but I am needing the MATCH to look for multiple arguments. I tried an OR function but it returns an error.
=IFERROR(MATCH("X",OFFSET('Unit Data'!\$C\$2,0,\$A\$8-1,COUNTA('Unit Data'!\$C:\$C),1),0),"")
I tried an OR function, as shown below, but it doesn't work. How do I fix this?
=IFERROR(MATCH(OR("X","Y","Z"),OFFSET('Unit Data'!\$C\$2,0,\$A\$8-1,COUNTA('Unit Data'!\$C:\$C),1),0),"")
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welcome to the board
This fails because OR("X","Y","Z") will be evaluated by itself and return a #value! error. OR is a logical formula that returns TRUE or FALSE, it doesn't allow you to look for X or Y or Z. You'll probably have to look for each value individually, and then return the minimum value
=MIN(IFERROR(MATCH("X",OFFSET('Unit Data'!\$C\$2,0,\$A\$8-1,COUNTA('Unit Data'!\$C:\$C),1),0),""),IFERROR(MATCH("Y",OFFSET('Unit Data'!\$C\$2,0,\$A\$8-1,COUNTA('Unit Data'!\$C:\$C),1),0),""),IFERROR(MATCH("Z",OFFSET('Unit Data'!\$C\$2,0,\$A\$8-1,COUNTA('Unit Data'!\$C:\$C),1),0),""))
Ahh, that makes sense. Will MIN work if my lookup value in the MATCH function is a text item? I assumed MIN only worked with numeric values.
welcome to the board
This fails because OR("X","Y","Z") will be evaluated by itself and return a #value! error. OR is a logical formula that returns TRUE or FALSE, it doesn't allow you to look for X or Y or Z. You'll probably have to look for each value individually, and then return the minimum value
=MIN(IFERROR(MATCH("X",OFFSET('Unit Data'!\$C\$2,0,\$A\$8-1,COUNTA('Unit Data'!\$C:\$C),1),0),""),IFERROR(MATCH("Y",OFFSET('Unit Data'!\$C\$2,0,\$A\$8-1,COUNTA('Unit Data'!\$C:\$C),1),0),""),IFERROR(MATCH("Z",OFFSET('Unit Data'!\$C\$2,0,\$A\$8-1,COUNTA('Unit Data'!\$C:\$C),1),0),""))
I am trying to make a counter using a match function. The below formula is what I was using, and it works fine, but I am needing the MATCH to look for multiple arguments. I tried an OR function but it returns an error.
=IFERROR(MATCH("X",OFFSET('Unit Data'!\$C\$2,0,\$A\$8-1,COUNTA('Unit Data'!\$C:\$C),1),0),"")
I tried an OR function, as shown below, but it doesn't work. How do I fix this?
=IFERROR(MATCH(OR("X","Y","Z"),OFFSET('Unit Data'!\$C\$2,0,\$A\$8-1,COUNTA('Unit Data'!\$C:\$C),1),0),"")
What are you trying to achieve? -- To see if any/all of "X","Y","Z" are present in your range?
Agree with Tetra, something doesn't add up here, so what are you actually trying to do?
MATCH will return the location within a range that contains the value you're looking for - so it will return 1 if found in the first cell, 2 if in the second etc. It won't return a text item. Therefore MIN works in this case - but if you're expecting a text answer then you're asking the wrong question
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+0
# Arithmetic sequences
0
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Let a1, a2, a3, ... be an arithmetic sequence. If a2/a4 = 3, what is a5/a3?
Oct 19, 2021
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# Ml Teaspoon Conversion
Ml Teaspoon Conversion – If you’re looking for more kitchen conversion guides. This guide is good and so is this guide. (with printable conversion chart so that you don’t forget)
Before I start cooking I never measure out baking powder/baking soda or any dry ingredients. mine correctly.
## Ml Teaspoon Conversion
I don’t know what fluid ounces look like. and can’t really measure smaller amounts without looking at the conversion chart I mean even the abbreviations stopped me from having to go looking for some kind of conversion chart. Everything changed when I started paying more attention.
## Basic Cooking Measurements & Handy Kitchen Conversion Chart (free!)
Now let’s say you need about a teaspoon in a handful. But you don’t have a measuring spoon. What are you doing?
This is the best method I found. No conversion chart needed: find a simple kitchen spoon. and fill it about half full and remove the other half. This is approximately equivalent to a metric teaspoon.
Then, the next best thing to do is remember to compare measurements when you don’t have a measuring spoon. For example:
And you can easily remember ‘how many liters, cups or gallons are there in a measuring cup’ using this printable conversion chart.
### How Many Teaspoons In A Tablespoon , Tsp To Tbsp ?
You may have heard that the United States uses the official metric system, Burma (or Burma) and Liberia also use the metric system.
Worldwide measurements, for example in Canada, New Zealand, Australia or even the UK. You’ll see the entire world using the imperial system, and yes, each of them is very different.
The metric system seems to be more consistent and easier to use overall. The metric system is based on decimals because it is always based on the power of 10. You measure it in meters or grams.
The imperial system uses units of measure such as inches, feet, yards, pounds, or miles to measure. You can bet all over the world that most people know their weight and height, for example in imperial units. Overall, imperial measurements tend to be more practical. However, most of the world uses metric units.
#### How Many Teaspoons In A Tablespoon? (chart)
How much is a metric spoon worth in different situations? It’s a random trivia to know. You never know when it’ll come in handy to learn the facts about the imperial teaspoon or US fluid ounce conversion. Let’s look at the increasingly popular quantity question units that are asked in different ways.
I used to not know the difference between a measuring spoon and a dessert spoon. I think they are the same thing.
For example, in the UK Dessert spoons are similar in size to US spoons. One dessert spoon is approximately two teaspoons.
Here are some conversions to other measurement units. that may be useful in the kitchen or otherwise Some even come with conversion charts that you can print out and stick on your fridge. Suitable for home cooks
#### How Many Teaspoons In A Tablespoon (+printable Chart)
I’ve created a comprehensive recipe book that you can print and make at home. (I promise it won’t be hard. It comes with instructions too!) It comes with a set of conversion charts and a printable guide that you can keep. Your very own cookbook where you can store your favorite recipes, manuals, seasoning ratios and much more.
Don’t know what to eat this evening? Check out this complete list for inspiration for meals you can whip up in no time.
Sometimes you need the right recipe for two and this list is sure to help you.
Spicy fish tacos are one of the easiest to make. not only that They also help you get out of your dinner routine. for you to try something new Ready in about 30 minutes.
### How Many Cups Are 12 Teaspoons (convert 12 Teaspoons To Cups)
These super cheap, delicious and affordable dinner ideas are worth trying and customizing to your tastes. It takes about 20 minutes to put together any kitchen items you might already have.
Ever thought about preparing meals for the week? This in-depth guide will teach you everything you need to know to get started if it’s your first time cooking.
We share DIY inspiration and colorful living advice for a good life. We are so glad you came here!
Teal Notes is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to help us receive payments by linking to Amazon.com and affiliated websites. If you do not remember how many teaspoons are in one spoon. Shows that you are not alone. Learn the most basic cooking measurements and get a printable kitchen conversion chart!
## How Many Ounces In A Teaspoon (oz To Tsp Conversion)
Most common kitchen sizes are just a fraction of the larger sizes. My husband doesn’t remember how many teaspoons there are in a spoon. However, I pierced it into my memory many years ago!
Sometimes I like to turn small recipes into larger ones for meal prep or dinner parties. But changing kitchen sizes can be difficult. You definitely don’t want to accidentally double or triple the salt!
When it comes to the size of this tiny kitchen, weighing out 1 teaspoon of vanilla extract is wasteful. Almost all recipes that I know of have teaspoons or spoons for just about anything. From condiments to baking soda
Whether you’re looking for this for the first time or the millionth time. you are not alone I have a simple conversion which I use all the time and a handy conversion chart to keep on your phone or print out for your fridge.
### Solved 1 Teaspoon = 5 Ml 1 Tablespoon = 15 Ml 1 Cup = 0.25
You’ll need 6 teaspoons for 2 tablespoons once you’ve learned the basics. Adding or subtracting is as easy as you want!
Sometimes it’s helpful to know that 34 teaspoons equals 1/32 of a gallon. But that’s not a kitchen conversion I use often. Here are some of the most commonly used conversions in my kitchen:
From here, you can either double the recipe or cut it in half. Turn any recipe into a week of meal prep with this trick!
When you talk about kitchen measurements There will be both imperial and metric systems. This measure is for the imperial system, which is currently used only in the United States, Liberia and Myanmar.
## How Many Milliliters In A Teaspoon
What is the difference? Imperial measurements revolve around cups and pounds. while grams and liters are used for metric measurements for dry and liquid substances respectively.
If you’re wondering how to convert teaspoons and tablespoons to metric options. There are some tips to keep in mind:
Fortunately, these measurements are very similar. You can easily use 1:1 exchange to convert. Just taste it as you go!
Be sure to print out this kitchen conversion for easy reference whenever you need to convert teaspoons, spoons, cups, pints or quarts! Click here to view PDF version This post may contain affiliate links. This means that I take a small commission on the items you buy at no additional cost to you. Please see my full disclosure policy for details.
### Ounces To Cups To Tablespoons Conversion Chart
Knowing how many ml 1 tablespoon is. That is important because the imperial unit of volume is not an approximation at all!
Although many people think of a tablespoon or teaspoon as a basic cutlery, it’s not the same. But it is a specific unit used to measure volume.
A tablespoon is a unit of volume defined in imperial units as 3 teaspoons, but also 1/16 cup or 1/2 US fluid ounce.
A tablespoon is also equivalent to 4 drams of liquid, a unit of volume derived from the Greek drachma.
#### Cooking Measurement Conversion Chart
Therefore, one tablespoon is measured as 14.7868 milliliters (ml in SI units), which is usually rounded to 15 ml.
Because the spoon is a specific unit of measurement That’s why it’s important to buy the right measuring spoons. instead of using a single spoon
The way to convert is to multiply the number of spoons by 15. This can be challenging. But multiplying by 5 first and then multiplying by 3 is easier.
A US teaspoon (teaspoon) is one third of a US tablespoon (tbsp). As a result, one US teaspoon is equal to 4.92892 ml, rounded to 5 ml.
#### How Many Grams Is In A Teaspoon?
When a recipe calls for a tablespoon of dry ingredients, such as almond flour or coconut flour The ingredients are intended to be the same amount.
Sometimes some recipes require stacks of tablespoons. or nodular In which case the measurements don’t have to be accurate and you don’t need to level them.
Ml, short for milliliter, is a unit of measurement in the SI (International System) of units defined as 1/1000 (from the prefix milli-) of a liter.
Since one liter is equal to 1000 cubic centimeters, 1 ml = 1 cubic centimeter. Why does a tablespoon have 15 cubic centimeters?
### Conversion Table For Weights & Measurements For Cooking And Baking — Healing Morsels
No. The abbreviation for tablespoon (also found as a capital T) refers to a tablespoon or 15 ml in the US (20 ml in Australia).
Hi, I’m Carine, food blogger, writer, recipe developer. Author of numerous published cookbooks and e-books and founder of Sweet As Honey.
I’m excited to share all my easy and delicious recipes that are both delicious and healthy. My expertise in this area comes from my background in chemistry and years of low carb keto diet. But I have also been good at cooking vegetarian and vegan food since I am.
Teaspoon ml conversion table, 10 ml teaspoon conversion, 20 ml teaspoon, ml to teaspoon conversion, ml to teaspoon conversion calculator, ml teaspoon conversion chart, ml per teaspoon conversion, teaspoon conversion ml, 30 ml teaspoon, 10 ml to teaspoon, teaspoon to ml conversion chart, 5 ml to teaspoon
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# Capacity calculation of rotary kiln
Rotary Lime Kiln Operation Mineral Processing. The surge tank should have a capacity of 1 to 2 hours kiln feed dissociation of calcium carbonate calcium carbonate begins to dissociate in the kiln at a temperature of about 1500 f theoretically lime sludge could be heated to this temperature and held there until dissociation was complete. Get Price
• Kiln capacity calculation, False air calculation, Heat
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• rotary kiln capacity calculation
rotary kiln capacity calculation. rotary kiln incinerator - design and manufacture. The rotary kiln incinerator is manufactured with a rotating combustion chamber that keeps waste moving, thereby allowing it to vaporize for easier burning. Types of waste treated in a rotary kiln incinerator
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• design formulas for rotary kiln
design formulas for rotary kiln: About this site. Jul 22, 2010 Heat capacity of Rotary kiln 1 Q = 1.1 x 10 ^ 6 x D ^3 (Kcal / hr) D = Mean inside Kiln Diameter on Bricks, m 2 Kiln Thermal loading at cross section of burning zone = Qp = = Q / Fp Fp = 0.785 x D^2 Inside cross-section of the kiln burning zone m^2 where D is kiln shell diameter Q p = 1.4 x 10 ^ 6 x D Kcal / m^2.hr Qp should not
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• Rotary Kiln Maintenance Procedures
rotary kiln operations. Regardless of rotary kiln size or configuration the basic principles outlined in this manual govern the reliable operation of every rotary kiln, calciner, dryer, incinerator, digester and cooler application. For questions or problems with your specific application please contact North American Kiln
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• ROTARY KILNS - Thomasnet
CAPACITY | 1 TPH - 200 TPH+ FUEL TYPES - Fuel Oil - Natural Gas/Propane - Waste Heat - Biogas SIZE | Up to 15’ diameter x 100’+ DIRECT-FIRED KILNS Direct-fired rotary kilns offer efficient processing for high temperature applications. A direct-fired rotary kiln heats material by passing the combustion gases through the rotary kiln
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• Formulas kiln - SlideShare
Jul 22, 2010 Heat capacity of Rotary kiln 1 Q = 1.1 x 10 ^ 6 x D ^3 (Kcal / hr) D = Mean inside Kiln Diameter on Bricks, m 2 Kiln Thermal loading at cross section of burning zone = Qp = = Q / Fp Fp = 0.785 x D^2 Inside cross-section of the kiln burning zone m^2 where D is kiln shell diameter Q p = 1.4 x 10 ^ 6 x D Kcal / m^2.hr Qp should not exceed 3.46 x
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• Calculating Kiln Volume by Marc Ward - Clay Times Magazine
Aug 28, 2021 Calculating Kiln Volume By Marc Ward. From the November/December 1996 issue of Clay Times. To determine the amount of heat (BTUs) required for your kiln to reach a desired temperature in a certain time frame, you need to know your kiln's volume. To calculate the volume, measure and record its height, width and depth
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• Modelling and optimization of a rotary kiln direct
rotary kiln. This is determined by (i) the maximum flow rate of the proposed burden through a kiln, and (ii) the residence time at temperatures that would allow sufficient reduction to take place. A calculation method was developed that allows for the prediction of the bed profile and residence time in a rotary kiln
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• Rotary Dryer Design & Working Principle
Jun 01, 2016 Rotary driers have thermal efficiencies of from 50 to 75 per cent on ores. This must be taken into account when using the above table, which is figured at 100 per cent. The drier shell is rotated separately from the stationary kiln section. To achieve the rotation a
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• Rotary Kiln Sizing & Design - SlideShare
Sep 07, 2016 Sep 07, 2016 The size of a rotary kiln is not just a function of capacity, but also of the amount of heat that can be generated during processing due to combustible and/or volatizing material. 32. Kiln length and diameter are calculated based on the maximum feed rate, the required retention time, and what the bed profile will need to be (how full of
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• Rotary Kiln Design: Sizing - FEECO International Inc
Rotary kilns are a vital part of industrial processing, helping to carry out a wide range of chemical reactions and physical changes in hundreds of materials. The process of sizing a rotary kiln to meet the specific requirements of a given application is incredibly complex, and must take several factors and analyses into consideration
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• Rotary kiln incinerator - Design and manufacture
Rotary kiln combustion chamber, with drive motor and gear box, to avoid piek in concentration we need the correct software Secondary combustion chamber, also called post combustion chamber, with support burner to have 1200 degrees C and a residence time of min. 2 sec. for complete combustion, important for CO and dioxins and furans
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• Rotary Kilns - FEECO International Inc
Rotary kilns work by processing material in a rotating drum at high temperatures for a specified retention time to cause a physical change or chemical reaction in the material being processed. The kiln is set at a slight slope to assist in moving material through the drum. Direct-fired kilns utilize direct contact between the material and
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# Homework Help: Angle Made By a Pendulum
1. Jan 29, 2016
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
Using parallel lines I got the angle as theta.
2. Jan 29, 2016
### Suraj M
Identify the forces on the pendulum bob through components of mg using theta
3. Jan 29, 2016
The components of weight are mgcostheta and mgsintheta and there is tension acting in the string.
4. Jan 29, 2016
### Suraj M
Could you draw a diagram to represent those forces on the bob,?
5. Jan 29, 2016
6. Jan 29, 2016
### Suraj M
Since it's in free fall along the plane I think you should take a pseudo force along that line of motion($mg\sin\theta$)
Extend the length of the pendulum and label $\alpha$ I think you can proceed from there.
7. Jan 29, 2016
### haruspex
What is providing this force N on the bob?
8. Jan 30, 2016
The tension in the string has components too. So T sin alpha=mg
Sin alpha = 1
So, alpha is 90.
Thank you!
9. Jan 30, 2016
### haruspex
I assume you are taking alpha as the angle between the string and the roof. If so, T sin alpha is not mg. And to perform your next step, you somehow had to have that T=mg. Where did that come from?
10. Jan 30, 2016
Alpha is the angle between the string and the roof. The tension in the string will be mg, wouldn't it? Because it'll be the mass of the bob and g will be acting on it.
11. Jan 30, 2016
Alpha is the angle between the string and the roof. The tension in the string will be mg, wouldn't it? Because it'll be the mass of the bob and g will be acting on it.
12. Jan 30, 2016
### haruspex
Two forces act on the bob, the tension in the string and mg. You don't yet know what direction the tension is in. Also, the bob is accelerating, so the net force is not zero.
What you do know is that there is no acceleration perpendicular to the plane, so the forces must balance in that direction. But that still leaves you with two unknowns, T and alpha. So you need to use the known downplane acceleration of the system.
13. Jan 31, 2016
But won't the vertical component of tension be T sin alpha anyway? And the vertical components need to be balanced. But then which force would I equate it to?
14. Jan 31, 2016
### haruspex
The component of the tension perpendicular to the roof will be T sin alpha, but that is not vertical.
The vertical components must balance if the acceleration has no vertical component, but it will have.
15. Jan 31, 2016
### Suraj M
The component of gravitational force perpendicular to the inclined plane? Did you consider that? Relate that to the tension
16. Feb 4, 2016
Can I do this:
mgsin theta + Tcos alpha - mgsin theta =0
mgsintheta is the horizontal weight component and the horizontal component of T acts in the same direction as it. The -mgsintheta is from the pseudo force that acts on the block.
So, Tcos alpha = 0
Cos alpha =0
Alpha = 90
]
17. Feb 4, 2016
### haruspex
That works if you change all occurrences of 'horizontal' to .... what?
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Cost Estimating Methods Tutorial
Cost Estimating Methods Tutorial
# Estimating Cube Root
We at ask-math believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.
We also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.
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Estimating Cube Root, this method will work only if the given number is a perfect cube. The following are the steps to estimate the cube root.
Step I : Make groups of 3 digits from unit(one's) place.This the 1st group.The remaining number makes the 2nd group.
Step II : The unit's digit of the 1st group will decide the unit digit of the cube root.( if the unit digit of the cube is 6 then the unit digit of cube root will also 6 as 6 x 6 x 6 = 36).
Step III : Find the cube of numbers between which the 2nd group lie.
Step IV : Take the smaller number as its ten's digit.
Unit digit of a given number Unit digit of cube root 1 1 3 7 4 4 5 5 6 6 7 3 8 2
Examples on estimating cube root :
1) Find the cube root of 238328.
Solution :
238328 is an even number so its cube root has to an even number.
Make two groups ---> 238 328
First group 328 ---> unit digit is 8 so unit digit of cube root will be 2.
(see the above table).
Second Group 238 -----> it lies between 63=216 and 73 =343 .
So take the smaller number 6 as ten's place.
∴ ∛238328 = 62
_______________________________________________________________
2) Find the cube root of -175616.
Solution :
We know that ∛-175616 = - ∛175616
The number is even so its cube root has to an even number.
Make two groups ---> 175 616
First group 616 ---> unit digit is 6 so unit digit of cube root will be 6
(see the above table).
Second Group 175 -----> it lies between 53=125 and 63 = 216 .
So take the smaller number 5 as ten's place.
Cube and Cube Roots
Cube of Numbers
Perfect Cube
Properties of Cube
Cube by Column method
Cube of Negative numbers
Cube of Rational numbers
Cube Root
Finding cube root by Prime Factorization
Cube root of Rational numbers
Estimating cube root
From estimation of cube root to Exponents
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### Tony Phillips' Take Blog on Math Blogs
Math in the Media 0400
Mail to a friend · Print this article · Previous Columns Tony Phillips' Take on Math in the Media A monthly survey of math news
April 2000
Double Bubbles These images, of a double bubble and its cross-section, were created by John Sullivan of the University of Illinois, Urbana-Champaign, and are used with his permission. Note the curvature in the wall between the two chambers.
In the March 17 2000 Science is a piece by Barry Cipra: "Why Double Bubbles Form the Way They Do," and reporting on the recent solution of the Double Bubble Conjecture. The problem was to give a mathematical proof that the most economical way to enclose two contiguous given volumes is by a combination of three spherical surfaces, just as shown in the John Sullivan's pictures. The solution, by Michael Hutchings of Stanford University, Frank Morgan of Williams College and Manuel Ritoré and Antonio Ros at the University of Granada, proceeds by showing that "any other, supposedly area-minimizing shape can be ever so slightly twisted into a shape with even less area."
More Monkey Math. In the March 2, 2000 Nature, a joint Siberian-Israeli team reports on the investigation of non-verbal serial memory, using the macaque monkey as a test animal. Serial memory refers to the retention of lists. The question is, is a list remembered by chaining (remembering which item follows which other) or is it remembered by associating ordinals (first, second, third, ...) to the various items? It is known that monkeys are very good at chaining. The experiments reported here show that they also have a useful grasp on ordinality. For example, "monkeys were trained on four nonverbal lists, each containing four novel photographs of natural objects ... . The task was to touch the simultaneously presented images in the correct order (A1-A2-A3-A4, B1-B2-B3-B4, C1-C2-C3-C4, D1-D2-D3-D4). When the monkeys had mastered this task, the items were shuffled, taking one item from each list, so that in two derived lists the ordinal number of the items was maintained (for example, A1-D2-C3-B4) whereas in two others it was not (for example, B3-A1-D4-C2). Lists with maintained ordinal position were acquired rapidly and virtually without error, whereas derived lists in which the ordinal position was changed were as difficult to learn as new lists. "
Squeeze in a few more? Kepler conjectured in 1611 that the most efficient way to pack equal-sized spheres (for example, identical oranges) in a box was to use the face-centered cubic configuration. It took a long time to settle this question to everyone's satisfaction. This finally happened two years ago, when Thomas Hales showed that the density of the face-centered cubic arrangement (approximately 74%) could not be improved on. Then the question was considered, suppose the spheres are packed at random, like balls being poured into a container. Was there a maximum density for a random packing? Different experiments led to different estimates of this number, leaving a confusing situation. Charles Seife reports in the March 17 2000 Science on the solution to this problem. There is no such number, and looking for it "makes no more sense than searching for the tallest short guy in the world." Random packings achieved with gentler and gentler pressure on the spheres can get arbitrarily close to Kepler's limit (and as they do so, they become more and more ordered). Seife is reporting on results recently published by S. Torquato, T. M. Truskett and P. G. Debenedetti, of the Complex Materials Theory Group at Princeton University, in Physical Review Letters.
How to win \$1,000,000 - the hard way. An Associated Press story, picked up by the March 26, 2000 Seattle Times, reports that Faber & Faber and Bloomsbury Publishing are offering a million bucks to whoever can prove that every even number is the sum of two primes. Simple? 2=1+1, 4=3+1, 6=3+3, 8=3+5, ... 98=79+19, 100=97+3, ... but the problem has been open since 1742. The stunt is in connection with the upcoming release of "Uncle Petros and Goldbach's Conjecture," by Apostolos Doxiadis. The million dollar assertion is in fact Goldbach's Conjecture. Good luck.
-Tony Phillips
SUNY at Stony Brook
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## Learning and Applying Mathematics using Computing
### Randomness vs Canniness, or Programmers vs Savescummers
I recently wrote a post on probabilities in the game “XCOM: Enemy Unknown”, in particular for the rapid fire ability. Lots of the discussion around the article (and XCOM in general) was about randomness, and pseudo-randomness. There’s an interesting post to be written about subjective observation of probability, but I’m interested in a practice known as save-scumming: repeatedly reloading the same saved game until you achieve your desired result (especially in a luck-based game). To do that, I first need to explain a bit about how randomness in games work.
#### I’m Soooo Random!
To implement a game like XCOM with a chance of a shot hitting, you can use a random number generator. If the shot has a 63% chance to hit, you generate a random number from 1-100, and if it’s less than or equal to 63, it’s a hit (this way, 63 of the 100 numbers will correspond to a hit). In the real world, you might roll a 100-sided die. But computers can’t roll actual dice, so they either need to use a hardware random number generator, e.g. measuring electrical noise inside the machine, or use a pseudo random-generator (PRNG). A PRNG has a current state from which it can generate the next “random” number. If you know the state, then by definition you can know the next number, but otherwise it appears random to you. For the purposes of games, a PRNG with unknown state is observably just as random as a hardware generator.
#### Random Chance
So then: implementing random shots is easy. Initialise your PRNG’s state using the time when the game starts, then when the user chooses to fire a shot, generate the next random number and decide whether the shot hits. But maybe the shot misses: the chance was 85% but you generated 93, meaning the shot misses. The player is annoyed: they wanted the shot to hit! So they have an idea: reload the game and try again. The PRNG state was altered by generating the number, so the next one is different: 41. A hit! The player is happy this time, and plays on. If they save before each shot, and are perseverant enough, they can now make any shot in the game a hit. It’s cheesy, but as players we’ve probably all done similar at some point. This technique is known as save-scumming.
So what can the designer do? I think the only perfect solution is to disbar reloading: the Ironman mode in XCOM has one save file per game, which is overwritten every time anything happens in the game. That’s probably the best solution, but it has downsides: accidental clicks in the game (I’m looking at you, XCOM’s wonky camera control) are permanent, corrupted save files are now hellish, and some players find Ironman too restrictive. In the rest of the post, I’ll explain alternate solutions.
#### Universal Memory
Remember that the next random number is dependent solely on the PRNG state. The reload “exploit” works because the PRNG state is different every time the player reloads, so the shot result differs every time you reload. But it doesn’t have to: what if you save the PRNG state in the save file? That way, every time you reload and take the shot, the same next “random” number will be generated, meaning that reloading is futile. This is using the property of the PRNG in our favour: by taking advantage of the hidden determinism of the PRNG, we can make sure that a reload doesn’t affect the result of the shot. So that seems like a solution to save-scumming: save the PRNG state.
XCOM definitely saves the PRNG state. I have a save file in the first mission on impossible, with one soldier able to take a 45% shot — every time I reload it and fire that shot, it hits:
#### Multiverse
Unfortunately, our solution isn’t as sound as we would like. The PRNG state is saved in the save game file, and next time a random number is generated, it will be,say, 93, and after that will come 32. The player doesn’t know this explicitly, but they do have the ability to reload. Inevitably, the player notices that every time they load that same game, their 85% shot misses. So they reload the game, and try a shot with a different soldier, who has a 65% chance. The 93 is secretly generated: still a miss. Frustrated, the player tries the original 85% soldier again: this time, the 32 is generated — a hit! So the player gets the idea that the first shot they make seems destined to miss, but the second will hit. So they reload the game, take the first shot with a spare soldier who’s far away, then take the second shot with their most powerful soldier. So our supposed solution, saving the PRNG state, won’t completely protect against reloading.
XCOM has this problem. In my save game, I have two soldiers with 45% shots that always hit if I take that shot first. With my other two soldiers, I can move to take a 42%, a 55% or 60% shot that will also always hit. However, several 25% shots, a 32%, a 38% and a 40% shot always miss when taken as the first shot. Therefore, my next random number from the save game generator is either 41 or 42. Scientific save-scumming:
##### Individualised Odds
What can be done about the reload-and-try-someone-else problem? One solution that comes to mind is to individualise the odds. To avoid letting the order of attempted shots matter, at the start of each turn, you could secretly assign a random number to each soldier (for each type of shot). So the 85% soldier gets assigned 93 in secret, and no matter which order you make your moves, that soldier will always miss on an 85% chance. So the player loads up the game, tries the shot… and misses. They reload, miss, reload, miss, and pretty quickly they realise that this shot is literally destined to miss. So they don’t make that shot: they move the soldier into hiding and try a different shot with a different soldier. With a bit of patience, the player can know exactly which shots will hit and which won’t. This effectively removes all randomness from the game, and replaces it with a sort of deductive meta-game where the player has to figure out which shots hit, which miss, and plan accordingly. It might be sort of fun, but it’s not the game that the designer intended!
XCOM does not do this. If I shoot once or twice with other soldiers and then try my 45% shot (which always hits as the first shot), it misses. Also, it matters whether the first shot hits or misses: a hit presumably consumes two random numbers from the generator (one for hitting, one to decide whether it’s a critical hit), whereas a miss consumes one number (for hitting). If I make a hit with the first shot, the result of a given second shot is always the same — but it is different if I miss with the first shot.
##### Mitigating The Problem
I’m not sure the reload-and-try-someone-else problem can be completely solved (except by banning reloading) without another exploit popping up. Two mitigations spring to mind. One is to make reloading annoying enough that the player won’t do it so much (more splash screens! annoying voiceover every time!). Hmmm, not ideal.
The other mitigation is to individualise the random chance. Remember that our original algorithm was to generate a number and see if it’s less than or equal to our percentage probability. If you generate a 97 as the first number, then pretty much every shot will miss, and the player can easily detect this.
Instead, for each shot, you can decide a set of numbers (for the programmers: e.g. by hashing the player and turn number) that will count as a success. Let’s make this simpler and say each shot has a chance out of 10. So one shot might have a 4/10 chance. Instead of generating a number from 1-10 and seeing if it’s less than or equal to 4, pick 4 numbers out of 10 for that shot: say, 1, 6, 8, 9. If any of those come up, it’s a hit. Meanwhile, another shot with a 7/10 chance, will be a hit on 1, 2, 4, 5, 6, 9, 10. If the next number is an 8, the 4/10 shot will hit, but the 7/10 will miss. Thus from the player’s point of view, the order of shots can still affect which ones miss and which ones hit, but not quite as simply as before (i.e. the first shot can’t always be destined to hit or to miss). This would make save-scumming harder.
I’m fairly certain XCOM doesn’t do this, based on the analysis above. I have five shots above 42% which all hit when taken as the first shot, and about seven below 40% which always miss as the first shot — the chances of this being the case with the above mitigation would be incredibly slim.
#### Summary
So, save-scumming is a hard problem to fix, without Ironman. Even without reverse engineering the code or the save game file, we were able to work out what must be happening under the hood, and with unlimited reload, it’s possible to exploit this. But ultimately, players can do what they like with their single-player game — I grew up playing the original XCOM (well, TFTD) by borrowing someone else’s hack of ASCII-editing the save game file to give myself piles of money. Those were the days…
A follow-up post discusses people’s perception of probability, and what can be done about it.
#### Comments on: "Randomness vs Canniness, or Programmers vs Savescummers" (52)
1. Park said:
^^ good stuff.
2. ShaneWSmith said:
This was a really interesting analysis and academic exercise. That said, I think there is a limit to how far you can go to stop people who really want to exploit the game to achieve a better result at the expense of their fun…
3. The idea of fighting save scumming by making reloading flat-out annoying was actually implemented! Animal Crossing for DS has a character named Resetti that gives you a long, unskippable lecture, complete with forcing you to type in “I’m sorry”, when you reload.
Every time you load the game normally, it writes to storage saying that you have an active session. If you turn off the game without saving, then the next time you load, the game spots the active session that wasn’t closed and sends in Resetti. Sometimes this happens by accident, because your battery ran out or so forth, so the severity of Resetti’s lectures increases every time you see him.
Since Animal Crossing is a game entirely about finding random stuff, and because it’s not a skill-based game so Ironman mode isn’t appropriate, this seems like it just might be the optimal solution in this case.
• Hah — love the idea of having to type in “I’m sorry”.
• This is exactly why I stopped playing Animal Crossing – That bloody mole popping up every time my battery went flat.
4. Or you could discourage it with an added constant, contingent on the number of reloads in the past 24-ish hours… so for instance, if it’s your first load/ continue of the game you saved and quit last night, there’s no added ‘penalty’ score to the shot, so the algorithm [ “roll” + (n * 10)], for a roll of 40, would be [40 + (0*10)]. The first actual ‘reload’ would give you [40 + (1*10)] = 50, followed by 60, and so on, until after enough reloads they cannot hit, no matter who they switch to.
This way, a really horribly unlucky roll could be ‘passed’ to someone else, but if you try to continue doing it within the same day, you’ll either be screwing with your system clock to avoid penalties, or eventually finding that it’s impossible for any of your soldiers to hit anything (after 10 reloads).
More of an evil scheme to frustrate save scummers…
5. Hey, nice post, and nice blog as well!
Regarding save-scumming, I remember that Commandos 2 used a very interesting mechanic: when you finish a level, the amount o times that the game was reloaded is used when calculating you final score and rank.
This kind of mechanic is interesting, however it only discourages the behavior rather than solving the problem. In the end, the player is still capable of beating levels and progressing on the game by save-scumming, even if with a lower score or with less benefits, so this solution alone might not be enough.
Maybe it would be possible to achieve better results by mixing some of the solutions that you described with some other design elements like the one that I mentioned. Who knows, just some food for thought. 🙂
PS: I couldn’t find a RSS button, is there a way to subscribe to a RSS feed? Thanks!
6. Awesome post!
I had heard rumors that Firaxis included some code in Civ’s PRNG to mitigate perceived “streakiness” in the random numbers. I wonder if XCOM has the same?
7. bob said:
Another mitigation method would be to use separate pRNG state variables for each of the squad members (and possibly enemies) in the mission. This would not be all that expensive as the state for a pRNG is not usually that large.
• The effect of that would be similar to the “Individualised Odds” section, though if you found that a certain soldier was going to miss next, you’d need to “burn” that miss by firing a shot, then shooting again. But knowing that you would miss, you could withdraw before burning the miss. As you say: mitigation.
8. What if the state was set at the moment you click “fire” and was based off reading the computer’s internal clock, say, grabbing the seconds?
• That would be equivalent to the initial state of the PRNG being set by the time: you could surely just reload and get a new state.
9. Why is “save-scumming” a problem for the designer? By that I mean, if we are talking about X-COM the game is fun, from a player perspective, with or without Iron Man. If a player wishes to do “save-scumming” then that is their choice. That does not effect another person’s game. It seems that “save-scumming” is not a problem at all, but a preference that effects only the players that prefer it.
• Ventsi said:
+1
Couldn’t agree more!
I don’t count ‘save-scrumming’ as a cheat. In fact I don’t think it makes sense to talk about cheating in a single-player mode at all (who would complain that you is cheating in your single player game? – it’s your choice).
10. raspofabs said:
To me, the solution seems to be to get rid of the random. Frozen synapse did this to great effect.
11. Endyr said:
By the way, you can reload in ironman too, if you want something else to happen. Just close the game window in the task manager after alt-tabbing. The game only saves at the beginning of each turn so the thing you did in your current turn can be reversed.
Obviously I chose ironman not to do this but it’s great to cope with some horrible game bugs like being adjacent with an alien who is still in high cover (and you are flanked).
12. Xemplar said:
If someone chooses to use “save-scumming” in a singleplayer game, why should anyone else care?
How does that negatively affect anyone else, and who are they to deign to tell others how to play a singleplayer game?
Sure, it may make the game easier than intended, but we’re all individuals, and if someone decides to play in that manner (which some may argue undermines the fun of the game), then that is the choice they have made.
• That’s an interesting question with a pretty subtle answer. I don’t think it’s about any harm it does to non-save-scummers by there being people who actually enjoy savescumming…to each his own. Unless there’s some sort of leaderboard involved, which obviously would lead to a race to the bottom.
I think it’s about the harm of the extra cognitive load for people who don’t want to savescum but realize that it’s an option and one that makes sense in terms of their “fiduciary duty” towards winning the game. But it’s obviously a cheat that makes most people feel lame as a game player. So if the designer can find a way to fix the game so that you don’t have that temptation dangling in your face and can focus on winning the way they intended, that seems like a good thing. That last solution, where you randomize the randomizers, is awfully clever.
• Latro said:
I’m on, like, my 15th attempt to win Iroman style. And … cant. I always get to a point one mayor screw up kills all hope to win (try doing an abductor full of Elites and Sectopods with squaddies, have fun)
I’ve finished the game by savescumming, but Ironman, I cant. Which is … frustrating. Savescumming (even done at the level I do – start from save previous to mission), makes you feel that you havent mastered the game.
Having your whole set of Colonels and Majors die means your 15 h of work has been thrown down the toilet.
Is amazing how such a frustrating game (on Classic! I despair to think what Impossible is like!) keeps one hooked. Testimony that the game system is addictive 😛
• ChronoReverse said:
If I were designing a modern game, I’d avoid allowing save-scumming specifically because if the game was meant to be beatable without save-scumming, then those people who save-scum would have the audacity to complain the game is too easy.
The wonders of modern Internet.
13. […] Rapid Fire vs. tiro normal & Probabilidades em XCOM [Sinepost, em […]
14. Nice post. I’m a developer with UFO: Alien Invasion, an open-source game inspired by the original X-Com. Our mitigation strategy is rather severe: the player can’t save on the battlescape. It has worked well in the past and definitely adds the weight of caution in the battlescape which was so fun in the original. But as we build larger maps it’s becoming pretty taxing, especially when players have to spend a long time carefully hunting down that last annoying alien.
It’s a strange balance in game design. Players often want the very things which would make a game boring — a powerful weapon, the ability to take more soldiers, faster research. Games — at least, thinking-man’s strategy games — have this weird tension between success and failure. We need one to really enjoy or lament the other. If a player has no discipline, save-scumming can turn a fun game into a tedious exercise.
Isn’t it funny that games have to implement an Ironman mode? Surely, if a player wants he can play without reloading the game. But when I try to play the old X-Com without reloads, I occasionally break my own rules. We can’t help but undermine our own sense of achievement.
• Yes — I think game design is a bit like plotting a story. Ultimately, what consumers think they want is to know what happens, to get to the end, to have everything resolve. But the job of the creator is to provide them with a good journey to the end, because that’s actually the interesting part. Players want to complete the game and get all the good weapons, but actually what makes the game good is all the content in the middle, and the curve that allows them to slowly get better equipment or new tools or new levels, etc. And I think there is some truth that a game designer should make save-scumming less effective, to save the player from their own temptation. If reloading is too easy, players will use that as an easy out against bad odds. If you want to make them press on, live with the consequences and have to recover the situation, you need to make some effort to actually make them live with the consequences, not just reload!
• Latro said:
Problem is that in this game there are points where I dont see alternatives. I have a game that is just in the final step of winning… but I managed to kill all my colonels.
There is no way to win at that point. I cant take it, and go through a difficult but possible retraining process to get squads to the high levels again – there is no chance in hell for squaddies against the late game enemies. 1 squaddie or 2 in a mission, maybe, a roster of them? No way.
If I could spend 10 h more trying to recover, that would mean I would do it, but right now the only option is … to start another Ironman.
15. […] on from recent posts on the “XCOM” game (here and here), I wanted to write about people’s perception of probability. A huge amount of posts on XCOM […]
16. Deathwind said:
Most games I’ve played use the universal memory option, including some of the greats (XCOM, Jagged Alliance 2, at least two of the Civilization series…).
An interesting version of the “annoying” option was the one taken in Fallout: New Vegas, where loading a game inside a casino gave a 1-minute cooldown on using any of the gambling tables. This was obviously to discourage the practice (common in pretty much any RPG title that involves a minigame that earns money … I’m looking at you KOTOR) of quicksaving after every game and reloading after losses.
17. xdv said:
Neil: I’ve had a lot of time to think about game design over the last few years and I’m in total agreement with you when you say the point of the game is to tell a good story – and that save-scumming interferes with it.
In a broad sense, you need to build failure into the game, not have failure be a stop-point where you go, well this game is shot, lets start another one. Guild Wars 2 gave us some hints on this, where failing a dynamic event didn’t bring the narrative to a stop, but set players on an alternate path. In fact I would deliberately fail a dynamic event just to see what happened next.
In the case of XCOM Ironman, for example, I never bring all my vetarans to a mission, because it’s simply too risky. In fact, soldiers get benched when they reach Colonel status, because they’re not gaining XP. I will have a roster of 15 soldiers total which I am training up, with multiple replacements for various roles. The entire point of Ironman I feel is to train up a large roster of soldiers that are “good enough” to do any mission, while retaining a “reserve elite team” of 6 Colonels which you will ultimately deploy for the end mission.
This is ultimately a much more realistic way to play the game than say, just having your starting 6 soldiers run every mission and save scumming so they never die throughout the campaign.
My solution to the XCOM “problem” would be to make Ironman the default mode, but make it so soldiers are a lot harder to kill completely. Instead of dying when their health depletes to zero, they get critically injured and then have to be in hospital for 2 months or something – effectively dead and need to be replaced, but psychologically more acceptable for the player. Same deal for country funding – they can withdraw from the XCOM project but you can continue to run missions for them and they might eventually return.
The other thing that needs to be fixed is the “losing point” of the game. There will be a point where the player realises that they’ve lost – but still need to slog out another 4 hours of gameplay before reaching the end of the game (which btw, I found the ending cutscene of XCOM much more satisfying than the winning cutscene). I’m willing to bet that many players have never even seen the losing cutscene in XCOM even though they have “lost” many games and restarted them.
This ultimately robs the gamer of the experience of winning – without seeing a losing ending, the winning ending is that much less satisfying.
The losing state of the game should be a lot closer to the point where “all is lost” so the player goes, heck, I’ve just lost this game, but I may as well play one more mission to go out in a blaze of glory, instead of going “ah well I’ll restart a new one, no point playing another 4 hours of what is probably a losing game”.
This is lesson well learnt from decades of Euro board games =p
• Yes, I think you’re right that losing should be in-game somehow, not meta-game (by the player realising they must reload). Either that, or you sort of build in reloading as a game mechanic. In async multiplayer games like Hero Academy (or Frozen Synapse, to a lesser extent), you can repeatedly redo your turn until you’re happy with it, and then submit. One could imagine having an XCOM-like game where you can’t reload to an arbitrary point, but there is a “restart mission” button. But that would alter the dynamics of the game, of course.
Your point about comparing to board games is true — part of good game design, especially apparent in board games, is that the game should be over before the winner is obvious. Otherwise everyone immediately loses interest. Which means you either pace it right so that the game is close until the end, or you hide who is winning to some extent (Dominion does this somewhat). I hadn’t really thought about this in the context of single-player games, but you’re right — and no, I haven’t seen the XCOM losing screen, despite having started about 12 games and finished two!
18. Programmers and designers are always fixing problems, and they are fantastic at it, but the player is not a problem. At some point (and I think that point is reached when dealing with save-scummers) you just have to surrender to the player’s style. Reading this article convinced me that scumming is, in itself, almost a meta-game. It’s like trash-talking at basketball or farting while playing chess with someone: it’s a move that’s definitely not in the written rules, but it can give the one doing it an advantage by putting the other player off his game. Is it cheating? Well, a rule-writer (i.e., a programmer) might think so since his ego is in the mix. “What? My carefully crafted probability charts aren’t ‘fun’ enough for you?! Philistine!” But it doesn’t have to be that way.
Games with human opponents don’t need to make a lot of changes to compensate for “exploits” like the ones I mentioned. Coaches just train their players to shrug off the trash-talk, or use it to make themselves more focused on winning. Chess players either light a cigarette or breathe through their mouths and keep playing. I understand that in a video game the programmer can’t do that – the rules are all they have once the game is finished (barring patches, which we won’t go into here). But even if someone finds what you consider a loophole, aren’t they entitled to enjoy it – especially if it requires a bit of finesse to squeeze through?
I think having to find a firing order that makes the probabilities work out in your favor is just difficult enough to make save-scumming vs. Ironman a wash, as far as any measurable “fun” is concerned, with a slight advantage to scumming in my opinion. Even when I have to reload, THE GAME IS GOOD ENOUGH that I never feel like I gave myself a victory, just an edge. And you know what? The game is also good enough that sometimes, those Sectopods light a cigarette and burn my funk away. The rules are well-written enough that my edge can evaporate if I don’t actually have the skill to back it up. I’d just hate the rules if I were locked into some deterministic script, like an Ironman game. With the ability to “tweak” to probability engine through reloading, now I can respect them, too — and through them, the designers and programmers who wrote them. You can be loved and feared – you just need to be good enough to pull both off.
19. Can we also have an article on how to solve the Resource Allocation problem per completed mission state and maximize my chances of generating positive balance, garnering more support from nations (playing the satellite game, and not losing) and also allocating resources to the research dept, foundry, engineering and item production. While this will have many solutions, we can consider an idealized game state and produce this optimal allocation matrix from there?
20. Jon_Cake said:
I was thinking about this as I tried to save-scum–sort of. See, it’s kind of a spectrum, rather than binary options, you can go full Ironman, or save-scum individual shots, but you can ALSO do things in between. I reload a whole mission if I feel if simple mistakes I can learn from were the sole reason I got slaughtered. I also save if I’m about to try an aspect of the game I don’t fully understand (will this rocket shot actually do anything, or destroy cover? etc). I imposed these rules on myself largely because I save-scummed more than I’d like to admit in my one complete playthrough of the original (I got hooked late, and took my sweet time learning).
What I noticed when I reloaded anything in the new game was what you pointed out: the PRND was always making me miss or hit the same shots if I shot in the same order. What’s interesting is this was not in the original (unless anyone can prove otherwise?). I distinctly remember taking a shot, screwing up on the turn (often a misclick…even worse problem in that game), the reloading to take the same shot…and having it end differently.
Anyway, I think the level of save-scumming should be up to the player. However, generating better random numbers is always a good goal for programmers, so that should be the emphasis rather than Ironman, which I’d be more inclined to try if I wasn’t so worried about misclicks that weren’t intentional decisions.
Well, thanks for the insight into (pseudo) random numbers!
21. Steele said:
The problem with this game and Ironman, is the bugs. I’ve had 4 full squads die in an ironman game, due to the bug where cyberdiscs fly through walls/ceiling (no holes, I checked). I had 12 restarts on Ironman, all of them wasted on bugs. The game is simply not stable enough yet, to tackle Ironman without insane luck. On Classic, one bug can destroy the game, literally putting you in a state where you can’t win, because you’re sending unranked rookies out to insane missions.
About save-scumming, well, nobody can tell what goes on behind closed doors.
However, the ability to cheat death in games, has a great impact on the game design. As a designer, you’re practically barred from making any one thing dependent on the death of a character (say, in a RPG) and, in my opinion, the narrative suffers. A lot of people want the min/max experience. Everything has to go off perfectly! Losing a high-ranked squad member in XCOM is a good example. Everybody reloads.
To me, the save-scumming is like reading a book written in pencil, and sold with a complimentary eraser. Just erase the parts you don’t like, and put in your own, full well knowing that shit really didn’t go down like that, in the first reading. To me, it destroys the experience.
You see this behavior in other games, like Blood Bowl, where people will intentionally disconnect from a game if some of their good players die or are injured. They’re save-scumming multiplayer though. Infinitely more abhorrent.
I can’t help but wonder, though, if these people have been “potty-trained” to save-scum, by poor single-player game design. They may even have come to expect time and space, in games, to bend to their will, even in multiplayer. They will think it’s natural, to impose their own ruleset on a multiplayer setting.
So, yes, this may be a bit of an overanalysis, but I do think that the practice of save-scumming is detrimental to game design, as well as overall morality. That said, I don’t have a magic wand. I don’t have the solution to end all save-scumming. As discussed in other comments on this thread, there is no simple way of avoiding it. Seriously determined people will even circumvent design solutions by going straight into the code. Not much to be done, except live by a few simple rules:
If you suck, you suck.
Deal with it.
Practice.
Get better.
Go win.
Don’t cheat. 🙂
• Not everybody reloads ) I beat the original Enemy Unknown on Classic and Impossible with Ironman, and thankfully, never got game-breaking bugs. Did the same with Enemy Within. The beauty of the XCOM is that it’s very much “losable”, but apart from really freak occurences, you can always win, or recoup losses. Yeah, I had a 80% squad wipe with Colonels. I wasn’t even furious, I was ashamed. And this is what no other game had made me to feel. But I pulled through and trained new Colonels. In EW, I accidentally entered base defence with people w\o armor and starter weapons. Like, several almost-end-game aliens vs. people with powder rifles at a time. I pulled through and haven’t lost a man, even the security guards (which is luck, of course, but a hell of an effort too).
Save-scumming is important for games, because any gamer always perceives possibilities in the game like possibilites in life. The penalty is less, but the consideration is always there. If a responsible, risky, thrilling, tragic route exists, many will take it up, for the excitement it will bring. Stakes are as much an element of wildly successful game-design as transparency or immersiveness. Don’t blame the players.
22. theWul said:
Just two thoughts:
1) Pointless discussion about reload-penalties, a matter of taste and playstyle, no benefit for the game in enforcing this on others.
2) From a developers point of view the pRNG system obvisiously used by XCom is the compromise you need: deterministic enough to keep the game debugable, amd renadom enough to make an exiting game (btw maybe THE strategy title of the last two decades).
23. Ok, as simple as that : each character and badies have a different PRNG for each action they can make depending when the player decides to input his action. You will need to use the internal hardware clock to generate these PRNG at all time (why not use GPS location or IP adress…). I reload my save state, let’s say, at 9:33, depending the first statement, the numbers will always be different if I reload my game 15 seconds later or a minute later. I could use random number to be either % or the “1, 6, 8, 9, 4 out of 10”. I hope that would help to partialy solved the savescummers issue.
• Ah, but that ends up at the first problem: if I miss, I just reload, get a new PRNG, and change the result. That actually makes save-scumming more effective, because I’m quite likely to alter the result just by reloading.
24. cikame said:
I get around this problem by playing on easy to curb frustration.
Sometimes though, random unfair enemy shots completely destroy me which is why i save at the start of every mission.
25. WontonTiger said:
In a single player game, why is “save scumming” (I hate this term) a problem? If you purchase a single player game, why should a consumer be punished for playing how they like?
Does it affect your ego that other people may get the same achievements as you?
26. Frag-ile said:
I do not see why “save scumming” is a “problem” that needs to be “fixed” in the first place. So what if the player doesn’t “play the game the developer intended”? None of us do, the whole point of games as an artform is that the player is the last missing piece of the work. Everyone experiences games differently and are looking to get a different experiences out of it.
Seeing as it is inherently a SP “problem” I would argue that it should be up to every player on their own how they want to play a game. If someone does not find amusement in failing something they had already taken every possible precaution to succeed in due to RNG then why should someone declare that they are NOT allowed to try again?
What does it matter to you how someone else plays a game? I say let the savescummers and everyone else play however they please. There’s no harm comming to people who like the ironman challange from those who want to retry every once in a while or even keep reloading over and over and over in frustration untill everything works out perfectly.
All you’re achieving by looking at this as a problem in the first place is limiting your game’s appeal to a narrower selection of players without providing anything beneficial to your core audience.
In my opinion.
27. mark eisen said:
I long suspected the Firaxis Civ series used a corrupt rng. The old Microprose titles also. Thanks for giving it a name – pseudo rng.
This whole “savescumming vs AI program cheats” controversy sounds like an extension of Mrs. Grundy determining what I and my wife will be doing in our home. Think of those so-inclined game designers as Mrs. Grundys standing guard while one plays a game and one can see how thoroughly perverted such a game design philosophy is. When one is talking about a single player game, the idea of controlling how people play the game is totally absurd. I can see it in competition gaming, but this isn’t about that at all. It’s about lazy programers and the cheap companies who hire them and reinforce their lazy (but very cost saving) programing ineptitude.
Most people reload saves to offset defective program design and programing. When the program appears unfair to the player, they reload. When the player makes a mistake because some part of the game was unclear, corrupt or clunky, they reload. When the game presents the player with an absurd occurrence, especially if similar obviously weighted occurrences happened before, they reload. It is a way to get around poorly done game design and programing and retake some control of their game play they feel was otherwise stolen from them.
Spend the time on making the game mechanics believable, clearly understood and transparently fair. And get to bugs out before releasing the game to the paying public. And most people wont feel an inclination to reload and will enjoy the game a whole lot more.
28. Frayed Knights: The Skull of S’makh-Daon dealt with the save-scumming problem with a carrot instead of a stick: As things happen in the game (including “bad things,” like a character being incapacitated), points are earned in the form of “Drama Stars” which allow actions independent of the character that affect the game, including restoring incapacitated characters, all but guaranteeing success on one’s next action, healing the entire party, and so forth.
The trick is that the drama points are reset to zero if you reload a saved game unless it’s a quit-and-save point.
Of course, this system can be gamed as well, but the point was to balance the game equally for those who save-scummed as for those who did not, by giving the latter some of the advantages of the former “legitimately.” And since the consequences of ‘bad luck’ or poor choices were almost always recoverable (or minor enough in the grand scheme of things), players were encouraged to ‘play through’ setbacks.
It worked really, really well by most reports.
29. BlueFroman said:
It’s funny the writer says Ironman mitigates save scumming, but not by a whole lot, since, so long as you go back to your main home screen (ie: Xbox button to dashboard, PSN button to Home, close client on PC) before the Aliens’ turn end, you can still scum, although you have to sit through the opening logos.
30. SEinR said:
The best way to reduce role of save scumming is eliminate the main cause of it. What is the cause in this game? It is availability to lose all in game progress due to one unlucky choise in streaks or player mistake during combat. I think the best method to resolve this problem give player ability recruiting soldiers in each class with maximum level wich player access for this class during current game.
31. Skyhawk02 said:
I don’t think games should be random, instead of a aiming mechanic they should have done a mechanic where an easier shot meant doing more damage instead of a better chance to hit. for example hitting a target in full cover does 1 damage, partial cover 2, no cover does 3. of course that would be modified by the weapon you’re using.
• Surely, you haven’t shot much in your life =) missing is incredibly easy, even for professionals. In stressful situations, people have unloaded full magazines at point-blank range without hitting. Moreover, in actual 20-21th century wars, small arms served mainly as a deterrent, to keep the enemy pinned for artillery or assault, expending thousands of rounds without hurting a fly. The ability to hit an opponent with reliable certainty is an art, and subject to myriad conditions.
32. Ian said:
What I would do is keep a global PRNG seed, initialized to the system time when a new game is started, and then modify that seed by a unique action ID anytime a random event occurs. For instance, when a soldier takes a shot, a unique soldier number and a unique action number for the type of shot are combined with the global seed, then that new seed is used to generate the random number. This would maintain determinism, so that if a game is loaded and EXACTLY the same operations are performed then you’ll get the same results, but any change in the order of actions would cause unpredictable outcomes for future actions. So a player could reload and try different tactics as much as they want, even keeping the same set of “opening” moves and getting the same results (which can be nice if the game crashed), but they would never be able to notice a usable pattern of guaranteed successes to plan future actions. You would never know whether something will succeed until you actually try that precise sequence of actions.
33. KBlack said:
Hi, Excellent article! Excellent comments discussion as well.
As an X-COM Ironman fanatic, I’ll be the one to disagree with comments from people who say that save-scumming is a result of bad design by the devs (see mark eisen and others). I also disagree on the idea that Ironman as an option is an artificial construct that doesn’t make sense because people should just refrain from reloading if they so desire.
This is because of one good reason: emotions. People have mentioned being attached to their characters. While true to some extent, I think people are more attached to their game. All gamers seem to have a compulsion to build the “perfect” game; just look at all the character build guides/walkthroughs in wikis and forums for all games all over the internet.
When, for example, I lose many high-rank squadmates in a single mission, but win it anyway, I tend to lose track of the big picture in my mind; I’m frustrated at my loss, and that leads me to try and convince myself that my already unlikely chances of victory have been voided. At those times, the emotional component of the loss can make me want to reload even though I wanted to do a “no-load” game from the start. Ironman forces you to take a moment to cool off because you don’t have that option. At those times, the true beauty of X-COM starts to unfold: you have to choose from a sub-optimal array of decisions to right your sinking ship. Strategic choices you didn’t consider before start to be appealing as stop-gaps. Resources you previously “wasted” on contingencies and back-up plans start to pay off in a big way. The game’s carefully balanced trade-off system plays out in all its gritty beauty.
Face it: a perfect game in X-COM is BORING. Yeah you got plasma weapons for your 6-colonel squad at month 2, congratulations. Then what? You spend hours playing out pre-determined victories with your overpowered squad (The strategic end-game in a perfect game is pretty unchallenging). Then you win the game. Woohoo? There’s no depth to a perfect game; just shiny numbers on the final scorecard, which disappears the next time you boot up the game.
On Ironman, when your B-Team of psionic colonels finally boards the temple ship after a few months of setbacks, they can say “This is for wiping out our friends, you assholes!”. At that point, the victory is much more sweeter than any scoreboard can make it look like. I actually take pride in the length of my memorial wall at the end of each game as a testament to my grit and creative problem-solving in the face of despair. The stories that emerged from those events last long after I stop playing. (I still remember the time where my highest-ranked Assault had to make a mad dash to the skyranger, forced to leave his panicking allies behind to be blasted to bits by the airstrike at the end of the Site Recon mission, while the rest of the squad was zombified and closing in on them. That Assault endend up being the Volunteer who got me the 5-medal achievement in the end).
As has been said, victory and defeat; you have to experience one to make the other feel real.
To those who say the end-game is unwinnable if you get your colonel squad wiped, I say you’re Doing Ironman Wrong. There are 3 important components to getting back on your feet after a serious squad-wipe: Sacrifices, Mitigation and and Contingencies. You have to wrap your mindset around those three things if you want to survive Ironman. Here’s how:
Sacrifices: Sometimes after a loss you have to endure short-term or future strategic setbacks for long-term tactical goals (such as selling important alien materials to fully equip your squad of rookies to give them a chance to level up), or medium-term tactical setbacks for medium-term strategic boosts (such as sacrificing a soldier to ensure victory in a mission that will give you the engineers you need to build your next satellite uplink and prevent countries from leaving next month). Skipping some abduction missions altogether and eating the panic increase to wait for your wounded soldiers to recover to ensure minimal losses in the next mission.
Mitigation: This means allocating resources early to create a buffer for later setbacks. For example: Using easy missions levelling up 2-3 rookies instead of getting a full Colonel squad out faster, Developing and upgrading SHIVS even though you don’t use them yet, buying the Rapid Recovery officer perk early, etc.
Contingencies: This means using previously suboptimal choices to recover from setbacks quickly enough at a low enough risk to continue playing. This often involves a mix of the other two components. For exemple, after a total wipe with no other ranked troops, it’s time to buy 3-4 SHIVs and use them carefully to protect your rookies and set up kills for them in the next few missions. If you still have a Major, you can try to gather enough money to immediately buy the New Guy upgrade right before hiring new soldiers. You can go out to missions, rack up a few kills for your rookies and double back to the Skyranger for extraction when you meet a Sectopod.
As someone has said before me, good tactics always trump experience in X-COM combat. I have yet to abandon any of my Ironman games, and all of them have been at Classic difficulty.
• Steele said:
Good post, Kb 🙂
I think I lost 98 soldiers during my Ironman Classic play through, but it’s the one I remember. They were expendable resources. Remember, we’re Commanders. Don’t get attached to individuals 😉
On a related note, you may enjoy Blood Bowl multiplayer, KB. You seem to have the balls for perma death and hard decisions 🙂
• KBlack said:
I’ll check Blood Bowl out, thanks. I’m a fan of WH40k, not so of WFB, but after reading what you wrote about it, I have to admit I’m curious.
As a side-note, I don’t like the perma death option in every videogame; I don’t do hardcore in Diablo for example. Reflex-based gameplay+insta-death+perma-death is a recipe for meaningless frustration. In turn-based games, you have just your poor planning to blame (as long as you take your finger off the trigger to prevent misclicks!)
And yeah, in the end, a soldier and its experience is a resource just like the other. Number of soldier in barracks is money + time (3 days). Experience for soldiers is time (number of missions) + risk (to get kills and complete missions). Converting (or sacrificing) one kind of limited resource for another while maximizing your gains is what X-COM’s strategic layer is all about.
To come back to somewhat on-topic, hit percentage is also a resource you can trade for risk, time (number of turns, number of actions) or movement.
• I’ve done Impossible Ironman on vanilla, and I distinctly remember that I’ve lost just a few experienced guys and gals – of course, mostly on a supply barge mission, bleeding out and desperate; even the base defense miraculously went without KIAs. Together with early game losses, it amounted to about 10-15 KIA and not one abandoned mission, so it’s possible to play ironman as you would a normal game, greedy and completionist =). The hard, cold proof there is Beaglerush who does Impossible Ironman both in vanilla AND Long War with nary an abandoned mission and losses in the single digits.
Of course, I’m no Beagle and I’ve never finished Long war properly on Classic even (too long), though I got thrice as much fun as the vanilla out of these unfinished runs.
34. Save scumming must be made unnecessary, root of the evil here is a binary system that gives you all or nothing on attack. Seeing as we use shotguns and burst fire almost every time, this would be solved by calculating a hit chance for each projectile individually, instead of for whole burst at once. So firing a shotgun at 95% chance would always give you a solid impact even if one or two pellets do miss, same with the burst fire.
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• Views : 70k+
• Sol Viewed : 20k+
Mind Teasers : Detective Conan Riddle
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Detective Conan got call from his one of the informer that Youssef had killed a person in a building.
Detective Conan inspector reached at the incedent with-in 4 minutes.
When the inspector reach there, he found that 11 people were playing cricket.
However without any communicator or any hesitation he immediately arrested one of them.
How's the detective so sure ?
Discussion
Suggestions
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River Thames needs to be crossed by Mr. Father, Mrs. Mother, A Thug, A Policeman, 2 dogs and 2 Cats. There is only one boat, which can carry only two people (or a person and an animal or two animals) at a time. Only the humans (excluding the thug) know how to operate the boat. The boat must travel back and forth across the river in order to pick up all of the people and animals.
Following rules must be followed at all times:
Mr. Father: Mr. Father cannot stay with any of the cats, without Mrs. Mother's presence.
Mrs. Mother: Mrs. Mother cannot stay with any of the dogs, without Mr. Father's presence.
Thug: The thug cannot stay with anyone, if the policeman is not there.
Policeman Allowed to travel with everyone.
2 Dogs, 2 Cats: Not allowed to travel without a human, or be in the presence of the thug without the policeman's supervision.
The dogs are not allowed to be in the presence of Mrs. Mother without the supervision of Mr. Father.
Similarly, The cats are not allowed to be in the presence of Mr. Father without the supervision of Mrs. Mother.
Both the dogs and cats are allowed to be unsupervised (provided the other rules are satisfied.)
How will everyone get to the other side of the river Thames? None of the rule can be broken, and all rules apply on the boat as well as on the land.
• Views : 90k+
• Sol Viewed : 30k+
Mind Teasers : Maths Pyramid Brain Teaser
Difficulty Popularity
Complete the pyramid my replacing ? with the correct number.
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
? ? ? ? ? ?
? ? ? ? ? ? ? ?
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While walking through the deepest jungle of the amazon, Steve the explorer came across a mystical tomb. He went closer and it read:
"Here lies two faithful husbands, with their two faithful wives,
Here lies two grandmothers along with their two granddaughters,
Here lies two dad's along with their two beloved daughters,
Here lies two mothers along with their two lovely sons,
Here lies two maidens along with their two charming mothers,
Here lies two sisters along with their two amazing brothers.
All were born legitimate, with no incest."
Steve, then checked and saw that there were only 6 graves in total
• Views : 80k+
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Mind Teasers : Logic Math Problem
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David and Albert are playing a game. There are digits from 1 to 9. The catch is that each one of them has to cut one digit and add it to his respective sum. The one who is able to obtain a sum of exact 15 will win the game?
You are a friend of David. Do you suggest him to play first or second?
• Views : 80k+
• Sol Viewed : 20k+
Mind Teasers : Colditz Pascal Prison Escape Puzzle
Difficulty Popularity
Two world's famous prisoners 'Colditz' and 'Pascal' are locked in a cell.
They plan to escape from the cell.
They noticed there is an open window at 40 feet above the ground level.
Both of them tried very hard but are never able to reach there.
Then both of them decided to plan to escape by a tunnel and they start digging out.
After digging for just 5 days, Colditz and Pascal comes out with the much more easier plan than tunneling and they escaped.
what was the plan ?
• Views : 50k+
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Mind Teasers : Quick Thinking Maths Problem
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A petri dish kept in a lab has a colony of healthy bacteria. Every bacterium divides itself into two in exactly two minutes. Now the colony started with a single cell at 2 pm. If the petri dish was exactly half full of bacteria at 3 pm, when will the dish become full of bacteria?
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Seeing the picture, can you identify in which direction is it going - left or right?
Keep in mind that the bus is moving in UK
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Mind Teasers : Logical Murder Puzzle
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Four kids having five rocks each were playing a game in which they need to throw the rock at solid area in the water.
Kid1: Succeeded in throwing three rocks at solid area but one of the rock sunk.
Kid3: His aim was so bad that all rocks got sunk.
Kid4: He was awesome and none of the rocks got sunk.
Kid2 was the winner but was struck by a rock in the head and died.
Who killed Kid2 ?
• Views : 50k+
• Sol Viewed : 20k+
Mind Teasers : Logical Maths Gold Bar Brain Teaser
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A worker is to perform work for you for seven straight days. In return for his work, you will pay him 1/7th of a bar of gold per day. The worker requires a daily payment of 1/7th of the bar of gold. What and where are the fewest number of cuts to the bar of gold that will allow you to pay him 1/7th each day?
• Views : 80k+
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Mind Teasers : Tricky Brain Teaser Problem
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I was invited on a pet show by a fellow colleague. Since I was a bit busy that day, I sent my brother to the show. When he returned back I asked him about the show. He told me that all except two animals were fishes, all except two animals were cats and all except two entries were dogs.
To his statement I was a bit puzzled and I could not understand how many animals of each kind were present in that pet show. Can you tell me?
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https://howtodiscuss.com/t/cancha-de-handball/63086
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# Cancha De Handball
## Cancha De Handball
### How many players are on the handball field?
Theoretically, there are a total of 7 players on the field. P is actually one player on the field (which is not very different) 6 players and one bow, and each team has the option to combine attacking players with different clothes, in this case there are 7 attackers.
And there's a mistake among friends too: real game time is 30`p which can be extended until I know, for example: up to 35` because it's time for penalty shootout and if any The player is injured or if the ball goes too far.
The handball measures 40x20 meters and is played 7vs7, and each team has 7 reservoirs. Two 30-minute parts are played: come back
Well, the size of the handball field may vary.
The game is played 7v7. Each team has 7 reserves and they play a fixed amount of time for 30 minutes.
More and I will give you + answer
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1. ## Multiplication of series
1. Multiply the series for (1-x)^-1 (valid when |x|<1) by itself and so obtain the expansion
(1-x)^-2 = sigma n=0 to infinity (n+1)x^n, |x|<1.
2. Derive the expansion
(1-x)^-3 = sigma n=0 to infinity ((n+1)(n+2)*x^n)/2 (|x|<1)
by multiplication of series. Use the result of Exercise 1.
2. Multiplying series is extra work in this case.
Notice that $\displaystyle \frac{d}{dx}\left(\frac{1}{1 - x}\right) = \frac{1}{(1 - x)^2}$ and $\displaystyle \frac{d}{dx}\left[\frac{1}{(1 - x)^2}\right] = \frac{2}{(1-x)^3}$...
3. $(1-x)^{-1}=\sum_{n=0}^{+\infty}x^n$ is absolutely convergent for $|x|<1$ . Use a well known result about Cauchy product of series.
Fernando Revilla
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# What is the difference between random and fixed effects meta-analysis models?
Table of Contents
## What is the difference between random and fixed effects meta-analysis models?
A random-effects model assumes each study estimates a different underlying true effect, and these effects have a distribution (usually a normal distribution). Fixed-effects model should be used only if it reasonable to assume that all studies shares the same, one common effect.
What is the difference between fixed effect and random effect model?
Under the fixed-effect model the only source of uncertainty is the within-study (sampling or estimation) error. Under the random-effects model there is this same source of uncertainty plus an additional source (between-studies variance).
How do you choose between fixed and random effects meta-analysis?
Conclusions Selection between fixed or random effects should be based on the clinical relevance of the assumptions that characterise each approach. Researchers should consider the implications of the analysis model in the interpretation of the findings and use prediction intervals in the random effects meta-analysis.
### What are the two models used to conduct meta-analysis?
There are two popular statistical models for meta-analysis, the fixed-effect model and the random-effects model. Under the fixed-effect model we assume that there is one true effect size that underlies all the studies in the analysis, and that all differences in observed effects are due to sampling error.
What is meta-analysis model?
A meta-analysis is a statistical analysis that combines the results of multiple scientific studies. Meta-analyses can be performed when there are multiple scientific studies addressing the same question, with each individual study reporting measurements that are expected to have some degree of error.
What is random effect model in meta-analysis?
Random effects meta-analysis A random-effects meta-analysis model assumes the observed estimates of treatment effect can vary across studies because of real differences in the treatment effect in each study as well as sampling variability (chance).
#### What is random effects model meta-analysis?
Random-effects meta-analysis is the statistical synthesis of trials that examine the same or similar research question under the assumption that the underlying true effects differ across trials.
What is random-effects meta-analysis?
How many types of meta-analysis are there?
There are four widely used methods of meta-analysis for dichotomous outcomes, three fixed-effect methods (Mantel-Haenszel, Peto and inverse variance) and one random-effects method (DerSimonian and Laird inverse variance). All of these methods are available as analysis options in RevMan.
## What is the difference between meta-analysis and meta-synthesis?
In summary, a meta-analysis is a way of testing a hypothesis whereas a meta-synthesis is a way of developing a new theory. 1) Theory Building – This form of meta-synthesis brings together findings on a theoretical level to build a tentative theory.
What is meta-synthesis?
Metasynthesis—the systematic review and integration of findings from qualitative studies—is an emerging technique in medical research that can use many different methods. Nevertheless, the method must be appropriate to the specific scientific field in which it is used.
What is meta regression analysis?
Meta-regression is defined to be a meta-analysis that uses regression analysis to combine, compare, and synthesize research findings from multiple studies while adjusting for the effects of available covariates on a response variable.
### What is heterogeneity in meta-analysis?
Heterogeneity in meta-analysis refers to the variation in study outcomes between studies. StatsDirect calls statistics for measuring heterogentiy in meta-analysis ‘non-combinability’ statistics in order to help the user to interpret the results. Measuring the inconsistency of studies’ results.
Is OLS the same as fixed effects?
A fixed effect model is an OLS model including a set of dummy variables for each group in your dataset.
What is meta synthesis?
#### Are fixed-effect and random-effects models interchangeable in meta analysis?
There are two popular statistical models for meta-analysis, the fixed-effect model and the random-effects model. The fact that these two models employ similar sets of formulas to compute statistics, and sometimes yield similar estimates for the various parameters, may lead people to believe that the models are interchangeable.
What is the difference between fixed effects and random effects models?
The fixed-effects model is the appropriate model when the number of studies is small. Random-effects models are appropriate when the number of studies is large enough, that is, enough studies to support generalization inferences beyond the included studies.
What are the statistical models for meta-analysis?
A basic introduction to fixed-effect and random-effects models for meta-analysis There are two popular statistical models for meta-analysis, the fixed-effect model and the random-effects model.
## What is a meta-analysis?
Meta-analysis refers to the statistical synthesis of quantitative results from two or more studies. Many reviewers appear to adopt a narrow approach to meta-analysis, focusing exclusively on calculating estimates of effects.
# What is the difference between random and fixed-effects meta-analysis models?
Table of Contents
## What is the difference between random and fixed-effects meta-analysis models?
A random-effects model assumes each study estimates a different underlying true effect, and these effects have a distribution (usually a normal distribution). Fixed-effects model should be used only if it reasonable to assume that all studies shares the same, one common effect.
### What is the difference between fixed-effects and random-effects?
A fixed-effects model supports prediction about only the levels/categories of features used for training. A random-effects model, by contrast, allows predicting something about the population from which the sample is drawn.
What is a random-effects meta-analysis?
Random-effects meta-analysis is the statistical synthesis of trials that examine the same or similar research question under the assumption that the underlying true effects differ across trials.
What is meta-analysis example?
Meta-analysis refers to the statistical analysis of the data from independent primary studies focused on the same question, which aims to generate a quantitative estimate of the studied phenomenon, for example, the effectiveness of the intervention (Gopalakrishnan and Ganeshkumar, 2013).
## What is an example of a random effect?
s Example: if collecting data from different medical centers, “center” might be thought of as random. s Example: if surveying students on different campuses, “campus” may be a random effect.
### Why do we use random effect?
Random effects are especially useful when we have (1) lots of levels (e.g., many species or blocks), (2) relatively little data on each level (although we need multiple samples from most of the levels), and (3) uneven sampling across levels (box 13.1).
What is random effects model in systematic review?
A model used to give a summary estimate of the magnitude of effect in a meta-analysis that assumes that the studies included are a random sample of a population of studies addressing the question posed in the meta-analysis.
What is a meta-analysis in simple terms?
Definition of meta-analysis : a quantitative statistical analysis of several separate but similar experiments or studies in order to test the pooled data for statistical significance.
## What do random effects mean?
Random effects can also be described as predictor variables where you are interested in making inferences about the distribution of values (i.e., the variance among the values of the response at different levels) rather than in testing the differences of values between particular levels.
### Why is random effects more efficient?
Additionally, random effects is estimated using GLS while fixed effects is estimated using OLS and as such, random Page 3 effects estimates will generally have smaller variances. As a result, the random effects model is more efficient.
What is the advantage of random effects model?
σ . Random effects models have at least two major advantages over fixed effect models: 1) the possibility of estimating shrunken residuals; 2) the possibility of accounting for differential school effectiveness through the use of random coefficients models.
What is random effects method?
The full random-effects model (FREM) is a method for determining covariate effects in mixed-effects models. Covariates are modeled as random variables, described by mean and variance. The method captures the covariate effects in estimated covariances between individual parameters and covariates.
## What is random effect model in statistics?
Random-effects models are statistical models in which some of the parameters (effects) that define systematic components of the model exhibit some form of random variation. Statistical models always describe variation in observed variables in terms of systematic and unsystematic components.
### Why are meta-analysis useful?
The major advantage of meta-analysis is that accumulation of evidence can improve the precision and accuracy of effect estimates and increase the statistical power to detect an effect. A further advantage of meta-analysis is that it facilitates the generalization of results to a larger population.
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Terri B.
# Solutions by Mass
What is true of an aqueous solution that is 15.0 percent HNO3 by mass?
It contains 15.0 grams of HNO3 per 1.0 liter of water. It contains 15.0 grams of HNO3 per 100 grams of solution. It contains 15.0 moles of HNO3 per 100 moles of water. It contains 15.0 moles of HNO3 per 1.0 liter of solution.
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# The Uneven Odds of Flipping or Spinning a Coin
Science takes some of the randomness out of a random coin toss.
What do we know about coins? They're legal tender, they usually depict dead men on one side, and they're the go-to tiebreakers for problems big and small. Thing is, coins aren't perfectly suited to that role. A study on coin tosses reveals that the "randomness" of a toss is actually weighted ever so slightly towards the side of the coin that's facing upwards when a flip begins. "For natural flips, the chance of coming up as started is about .51," the study concludes.
The paper, written by statistics and math professors from Stanford and UC Santa Cruz, also points out that a perfect coin toss can reproduce the same result 100 percent of the time. Of course, the perfect flip was performed by a machine, not a person. And the results that lean ever-so-slightly in favor of flipped-side-up don't take into account flipping a coin after catching it or letting it bounce around on a floor or table. In practical usage, the .51 bias is so slight that you'll never notice.
If, like me, you'd always heard that coins tend to land tails-up because the heads side is heavier, there's some science available for you, too. Spinning, rather than flipping, an old penny will land on heads something like 80 percent of the time. Lincoln's head is heavier than the Lincoln memorial on the reverse, which leaves tails facing up more often than not. Unless the penny has accrued enough dirt or oil to throw the weight off.
And let's be honest--how often do you come across an old, clean penny?
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p38_073
# p38_073 - 73. The strategy is to find the speed from E =...
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Unformatted text preview: 73. The strategy is to find the speed from E = 1533 MeV and mc2 = 0.511 MeV (see Table 38-3) and from that find the time. From the energy relation (Eq. 38-45), we obtain v=c 1− mc2 E 2 = 0.99999994c ≈ c so that we conclude it took the electron 26 y to reach us. In order to transform to its own “clock” it’s useful to compute γ directly from Eq. 38-45: γ= E = 3000 mc2 though if one is careful one can also get this result from γ = 1/ 1 − (v/c)2 . Then, Eq. 38-7 leads to ∆ t0 = 26 y = 0.0087 y γ so that the electron “concludes” the distance he traveled is 0.0087 light-years (stated differently, the Earth, which is rushing towards him at very nearly the speed of light, seemed to start its journey from a distance of 0.0087 light-years away). ...
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
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Open/Close Menu Best Nursing Custom Papers Writing Service
WHAT ARE THE EXPECTED RATES OF REIMBURSEMENT FOR THIS TIME FRAME FOR EACH PAYER?Evaluating Financial Statements Part 1 Your facility has the following payer mix: 40% commercial insurances 25% Medicare insurance
WHAT ARE THE EXPECTED RATES OF REIMBURSEMENT FOR THIS TIME FRAME FOR EACH PAYER?Evaluating Financial Statements Part 1 Your facility has the following payer mix: 40% commercial insurances 25% Medicare insurance
Evaluating Financial Statements Part 1 Your facility has the following payer mix: 40% commercial insurances 25% Medicare insurance 15% Medicaid insurance 15% liability insurance 5% all others including self-pay Assume that for the
View complete question » Part 1 Your facility has the following payer mix: 40% commercial insurances 25% Medicare insurance 15% Medicaid insurance 15% liability insurance 5% all others including self-pay Assume that for the time in question you have 2000 cases in the proportions above. Commercial insurances average 110% of Medicare, Medicaid averages 65% of Medicare, Liability insurers average 200% of Medicare and the others average 100% of Medicare rates. The average Medicare rate for each case is \$6200. 1) What are the expected rates of reimbursement for this time frame for each payer? What is your expected A/R? 2) What rate should you charge for these services (assuming one charge rate for all payers)?(this gives you your total A/R.) Calculate the total charges for all cases based on this rate. 3) What is the difference between the two A/R rates above? Can you collect it from the patient? What happens to the difference? 4) Which of these costs are fixed? Which are variable? Direct or indirect? a. materials/supplies (gowns, drapes, bedsheets) b. Wages (nurses, technicians) c. Utility, building, usage exp (lights, heat, technology) d. Medications e. Licensing of facility f. Per diem staff 5) Insurances (malpractice, business etc.) 6) Calculate the contribution margin for one case (in \$) with the following costs for this period, per case: a. materials/supplies: 2270 b. Wages: 2000 c. Utility, building, usage exp: 1125 d. Insurances (malpractice, business etc.): 175 7) Using the above information, determine which is fixed and which cost is variable. Then calculate the breakeven volume of cases in units for this period. 8) Suppose you want to make \$150,000 profit between this period and next period to fund an expansion to the NICU, how many cases would you have to see, and at what payer mix would this be optimal? Show all calculations. Part 2 There is said to be two types of accounting operations in health care finance that lie at the heart of the financial department of any organization in the industry. What are these two types of accounting? Briefly describe each and provide 2 examples for each. Do you feel one is more important to business than the other? Is one more effective or efficient for certain types of business operations?
. .
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https://agriculture-lawyer.com/law-of-sines-kuta/
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# law of sines kuta
January 23, 2021
152
Views
A sine isn’t really a function. Rather, it’s a mathematical equation. Essentially, it means that what you see is what you get. A sine is a curve that starts at 0 and goes all the way to 1, so if you were watching a movie and saw a sine curve at the top of the screen, that would be the curve at the top of the movie screen.
The reason I chose this was because it was the most interesting part of the movie, because I was watching a film I could understand. I was looking at a sine curve, and I wanted to see the shape of the curve in which the movie was being acted. I wanted to feel like I was doing something with the sine curve. I also wanted to feel like I was watching something that was supposed to be a sine curve.
Although the fact that you can watch a movie in this way is pretty cool, there’s a whole lot more to it. The key to the sine curve is that it’s a curve that does not change, but rather a curve whose shape is fixed, like an ellipse. Like a sine, it is a mathematical function of the angle of the rotation of the movie’s camera (or the camera in the case of the movie itself).
The fact that you can view a movie with its rotation changing is why the sine curve is a great tool to use for creating compositions. So you can view a movie, then rotate it to change its angle of rotation.
A sine is a curve that does not change in shape. It is a mathematical function of the angle of rotation of the movies camera or the camera in the case of the movie itself. The fact that you can view a movie with its rotation changing is why the sine curve is a great tool to use for creating compositions. So you can view a movie, then rotate it to change its angle of rotation.
The law of sines, or of sines of angles, is a tool used by artists to create compositions. It is also a tool used by designers to create compositions. As a designer, I like to use it when the composition is to be seen from the side. I like the side view so that I don’t have to turn my head to look at the composition from the front.
This is a great example of a composited view of the law of sines. I could just as easily view the composition from the top and bottom or from a side view.
the law of sines.is a mathematical tool that allows us to add and subtract angles in a way that seems to be very intuitive. For example, if you have a composition that looks something like a “V” shape, we can use the law of sines to give you a series of angles that make the v shape. The sine of the angle of the bottom left corner is the angle of the angle of the bottom right corner.
To make this more intuitive, we can use the law of sines to add a third angle, the angle of the angle of the bottom left corner. This gives us a third angle that goes from the bottom left corner to the angle of the bottom right corner. So in this way we can apply the law of sines to the composition of a V shape.
In the game, you can use the law of sines to manipulate gravity in a number of ways. It’s especially useful for things like creating a new v shape from a group of similar shapes or using gravity to draw an edge in a group of shapes. In the game, you can also use the law of sines to make the angles of your v shapes change, to your advantage.
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His love for reading is one of the many things that make him such a well-rounded individual. He's worked as both an freelancer and with Business Today before joining our team, but his addiction to self help books isn't something you can put into words - it just shows how much time he spends thinking about what kindles your soul!
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# SAT Math Multiple Choice Question 965: Answer and Explanation
### Test Information
Question: 965
1.
If the right triangles in the figure shown are similar triangles, what is the lengthof the shorter leg of the larger triangle?
• A. 10
• B. 15
• C. 10
• D. 15
Explanation:
B
Difficulty: Medium
Category: Additional Topics in Math / Geometry
Strategic Advice: Whenever you see in a triangle problem, think about special right triangles. Don't forget that the ratios of the sides of these triangles are given on the reference page at the beginning of each math section.
Getting to the Answer: Examine the smaller triangle: When the sides of a right triangle are in the ratio, the triangle is a 30-60-90 triangle. Knowing that two of the sides of the smaller triangle fit this ratio is enough information to draw the conclusion that it is a 30-60-90 triangle (although you could use the Pythagorean theorem to confirm that the shorter leg has length 5 if you're not convinced). Because the two triangles are similar, you know that their angle measures are equal, and the larger triangle is also a 30-60-90 triangle. In the ratio, 2x represents the length of the hypotenuse and x represents the length of the shorter leg. According to the figure, the hypotenuse of the larger triangle has length 30, so solve 2x = 30 to find that the length of the shorter leg of the larger triangle has length x = 15, which is (B).
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http://pgfplots.net/tikz/examples/bivariate-normal-distribution/
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# Example: Bivariate normal distribution
Published 2014-03-16 | Author: Jake
We would like to draw a bivariate normal distribution and show where the means from the two variables meet in the space.
There are a couple of things in the code that might be useful:
You can define mathematical functions using declare function={<name>(<argument macros>)=<function>;}, which will help to keep the code clean and avoid repetitions.
You can define a new colormap using \pgfplotsset{colormap={<name>}{<color model>(<distance>)=(<value1>); <color model>(<distance 2>)=(<value2>)} }. This is a very powerful feature, so definitely worth reading in the pgfplots manual.
The legend created by colorbar is a whole new plot, so you can configure it with all the usual axis options.
There are different ways for defining 3D functions: \addplot3 {<function>}; will evaluate <function> at every point on a grid and assume the result to be a z-value. \addplot3 ({<x>},{<y>},{<z>}); defines a parametric function in 3D space, which allows you to draw three-dimensional lines, among other things.
This code was written by Jake on TeX.SE.
Do you have a question regarding this example, TikZ or LaTeX in general? Just ask in the LaTeX Forum.
Oder frag auf Deutsch auf TeXwelt.de.
\documentclass[border=10pt]{standalone}
\usepackage{pgfplots}
\pgfplotsset{width=7cm,compat=1.8}
\pgfplotsset{%
colormap={whitered}{color(0cm)=(white);
color(1cm)=(orange!75!red)}
}
\begin{document}
\begin{tikzpicture}[
declare function = {mu1=1;},
declare function = {mu2=2;},
declare function = {sigma1=0.5;},
declare function = {sigma2=1;},
declare function = {normal(\m,\s)=1/(2*\s*sqrt(pi))*exp(-(x-\m)^2/(2*\s^2));},
declare function = {bivar(\ma,\sa,\mb,\sb)=
1/(2*pi*\sa*\sb) * exp(-((x-\ma)^2/\sa^2 + (y-\mb)^2/\sb^2))/2;}]
\begin{axis}[
colormap name = whitered,
width = 15cm,
view = {45}{65},
enlargelimits = false,
grid = major,
domain = -1:4,
y domain = -1:4,
samples = 26,
xlabel = $x_1$,
ylabel = $x_2$,
zlabel = {$P$},
colorbar,
colorbar style = {
at = {(1,0)},
anchor = south west,
height = 0.25*\pgfkeysvalueof{/pgfplots/parent axis height},
title = {$P(x_1,x_2)$}
}
]
\addplot3 [domain=-1:4,samples=31, samples y=0, thick, smooth]
(x,4,{normal(mu1,sigma1)});
\addplot3 [domain=-1:4,samples=31, samples y=0, thick, smooth]
(-1,x,{normal(mu2,sigma2)});
\draw [black!50] (axis cs:-1,0,0) -- (axis cs:4,0,0);
\draw [black!50] (axis cs:0,-1,0) -- (axis cs:0,4,0);
\node at (axis cs:-1,1,0.18) [pin=165:$P(x_1)$] {};
\node at (axis cs:1.5,4,0.32) [pin=-15:$P(x_2)$] {};
\end{axis}
\end{tikzpicture}
\end{document}
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https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1452/4/a/b/
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# Properties
Label 1452.4.a.b Level $1452$ Weight $4$ Character orbit 1452.a Self dual yes Analytic conductor $85.671$ Analytic rank $0$ Dimension $1$ CM no Inner twists $1$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [1452,4,Mod(1,1452)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(1452, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0, 0, 0]))
N = Newforms(chi, 4, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("1452.1");
S:= CuspForms(chi, 4);
N := Newforms(S);
Level: $$N$$ $$=$$ $$1452 = 2^{2} \cdot 3 \cdot 11^{2}$$ Weight: $$k$$ $$=$$ $$4$$ Character orbit: $$[\chi]$$ $$=$$ 1452.a (trivial)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$85.6707733283$$ Analytic rank: $$0$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 132) Fricke sign: $$+1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
$$f(q)$$ $$=$$ $$q - 3 q^{3} - 2 q^{7} + 9 q^{9}+O(q^{10})$$ q - 3 * q^3 - 2 * q^7 + 9 * q^9 $$q - 3 q^{3} - 2 q^{7} + 9 q^{9} + 88 q^{13} + 66 q^{17} + 40 q^{19} + 6 q^{21} + 6 q^{23} - 125 q^{25} - 27 q^{27} + 54 q^{29} + 8 q^{31} - 106 q^{37} - 264 q^{39} - 354 q^{41} + 124 q^{43} + 546 q^{47} - 339 q^{49} - 198 q^{51} - 408 q^{53} - 120 q^{57} + 552 q^{59} - 404 q^{61} - 18 q^{63} - 4 q^{67} - 18 q^{69} + 126 q^{71} + 166 q^{73} + 375 q^{75} + 874 q^{79} + 81 q^{81} - 444 q^{83} - 162 q^{87} + 1002 q^{89} - 176 q^{91} - 24 q^{93} - 802 q^{97}+O(q^{100})$$ q - 3 * q^3 - 2 * q^7 + 9 * q^9 + 88 * q^13 + 66 * q^17 + 40 * q^19 + 6 * q^21 + 6 * q^23 - 125 * q^25 - 27 * q^27 + 54 * q^29 + 8 * q^31 - 106 * q^37 - 264 * q^39 - 354 * q^41 + 124 * q^43 + 546 * q^47 - 339 * q^49 - 198 * q^51 - 408 * q^53 - 120 * q^57 + 552 * q^59 - 404 * q^61 - 18 * q^63 - 4 * q^67 - 18 * q^69 + 126 * q^71 + 166 * q^73 + 375 * q^75 + 874 * q^79 + 81 * q^81 - 444 * q^83 - 162 * q^87 + 1002 * q^89 - 176 * q^91 - 24 * q^93 - 802 * q^97
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
0
0 −3.00000 0 0 0 −2.00000 0 9.00000 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$2$$ $$-1$$
$$3$$ $$+1$$
$$11$$ $$-1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1452.4.a.b 1
11.b odd 2 1 132.4.a.b 1
33.d even 2 1 396.4.a.d 1
44.c even 2 1 528.4.a.i 1
88.b odd 2 1 2112.4.a.t 1
88.g even 2 1 2112.4.a.f 1
132.d odd 2 1 1584.4.a.j 1
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
132.4.a.b 1 11.b odd 2 1
396.4.a.d 1 33.d even 2 1
528.4.a.i 1 44.c even 2 1
1452.4.a.b 1 1.a even 1 1 trivial
1584.4.a.j 1 132.d odd 2 1
2112.4.a.f 1 88.g even 2 1
2112.4.a.t 1 88.b odd 2 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{4}^{\mathrm{new}}(\Gamma_0(1452))$$:
$$T_{5}$$ T5 $$T_{7} + 2$$ T7 + 2
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T$$
$3$ $$T + 3$$
$5$ $$T$$
$7$ $$T + 2$$
$11$ $$T$$
$13$ $$T - 88$$
$17$ $$T - 66$$
$19$ $$T - 40$$
$23$ $$T - 6$$
$29$ $$T - 54$$
$31$ $$T - 8$$
$37$ $$T + 106$$
$41$ $$T + 354$$
$43$ $$T - 124$$
$47$ $$T - 546$$
$53$ $$T + 408$$
$59$ $$T - 552$$
$61$ $$T + 404$$
$67$ $$T + 4$$
$71$ $$T - 126$$
$73$ $$T - 166$$
$79$ $$T - 874$$
$83$ $$T + 444$$
$89$ $$T - 1002$$
$97$ $$T + 802$$
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https://techcommunity.microsoft.com/t5/excel/excel-formulas-and-functions/m-p/532647/highlight/true
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• 512K Members
• 7,691 Online
• 609K Conversations
New Contributor
# Excel Formulas and Functions
I'm creating a spreadsheet to help keep logs of jobs I do daily and weekly. I need a formula to add rows in a column, then subtract a certain amount as long as the amount is over a certain amount..
Example;
I have 4 or 5 rows that the amounts are over \$51.00
I need to add those rows and then subtract \$14.00 from each
unless the amount is under \$51.00 then I need to subtract \$6.00
6 Replies
# Re: Excel Formulas and Functions
Can it simplify use IF to perform. Attached is a sample file.
# Re: Excel Formulas and Functions
@Man Fai Chan . I see your formula is correct to subtract either 14 or 6 based on column A. Adding or deleting rows are VB/Macro area. It must be disappointing to you but formula only can do affect a cell by calculation using other cell info.
J
# Re: Excel Formulas and Functions
Is there a way to do it so it calculates as I put in the numbers in the total cell?
Highlighted
# Re: Excel Formulas and Functions
Is there a way to do it so it calculates as I put in the numbers in the total cell?
# Re: Excel Formulas and Functions
In the attached file, the formula in D7 is:
=SUMPRODUCT(A2:A6-6-(8*(A2:A6>=51)))
# Re: Excel Formulas and Functions
There is nothing wrong with an IF statement; it makes the intent of the formula clear. I am less enthusiastic about hard-wiring dollar amounts into the formula.
Also, since you are likely to append data to your table, a dynamic references would be useful for creating totals. From 2007, the preferred way of doing this is through the use of a Table. The formula for Adjusted Amount might then read
= [@Amount] - IF( [@Amount]>51, 14, 6)
It is also possible to use LOOKUP to determine discounts, bonus rates etc. that vary as one moves between threshold values. For example
This is illustrated in the attachment.
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https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.18/share/doc/Macaulay2/RationalMaps/html/_is__Same__Map.html
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# isSameMap -- Checks whether two maps to projective space are really the same
## Synopsis
• Usage:
b = isSameMap(L1,L2)
b = isSameMap(L1,L2, R1)
b = isSameMap(f1, f2)
• Inputs:
• L1, a list, The homogeneous forms that define the first map.
• L2, a list, The homogeneous forms that define the second map.
• R1, a ring, The ring in which the homogeneous forms should live.
• f1, , The first map.
• f2, , The second map.
• Outputs:
• b, , True if the maps are the same, false otherwise.
## Description
Checks whether two maps, from the same variety, to projective space are really the same. If you pass it two ring maps, it will check whether the source and targets are really the same.
i1 : R=QQ[x,y,z]; i2 : S=QQ[a,b,c]; i3 : L1={y*z,x*z,x*y}; i4 : L2={x*y*z,x^2*z,x^2*y}; i5 : isSameMap(L1,L2) o5 = true
## Ways to use isSameMap :
• "isSameMap(List,List)"
• "isSameMap(List,List,Ring)"
• "isSameMap(RingMap,RingMap)"
## For the programmer
The object isSameMap is .
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https://en.wikipedia.org/wiki/Abel-Jacobi_theorem
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# Abel–Jacobi map
(Redirected from Abel-Jacobi theorem)
In mathematics, the Abel–Jacobi map is a construction of algebraic geometry which relates an algebraic curve to its Jacobian variety. In Riemannian geometry, it is a more general construction mapping a manifold to its Jacobi torus. The name derives from the theorem of Abel and Jacobi that two effective divisors are linearly equivalent if and only if they are indistinguishable under the Abel–Jacobi map.
## Construction of the map
In complex algebraic geometry, the Jacobian of a curve C is constructed using path integration. Namely, suppose C has genus g, which means topologically that
${\displaystyle H_{1}(C,\mathbb {Z} )\cong \mathbb {Z} ^{2g}.}$
Geometrically, this homology group consists of (homology classes of) cycles in C, or in other words, closed loops. Therefore, we can choose 2g loops ${\displaystyle \gamma _{1},\dots ,\gamma _{2g}}$ generating it. On the other hand, another more algebro-geometric way of saying that the genus of C is g is that
${\displaystyle H^{0}(C,K)\cong \mathbb {C} ^{g},}$ where K is the canonical bundle on C.
By definition, this is the space of globally defined holomorphic differential forms on C, so we can choose g linearly independent forms ${\displaystyle \omega _{1},\dots ,\omega _{g}}$. Given forms and closed loops we can integrate, and we define 2g vectors
${\displaystyle \Omega _{j}=\left(\int _{\gamma _{j}}\omega _{1},\dots ,\int _{\gamma _{j}}\omega _{g}\right)\in \mathbb {C} ^{g}.}$
It follows from the Riemann bilinear relations that the ${\displaystyle \Omega _{j}}$ generate a nondegenerate lattice ${\displaystyle \Lambda }$ (that is, they are a real basis for ${\displaystyle \mathbb {C} ^{g}\cong \mathbb {R} ^{2g}}$), and the Jacobian is defined by
${\displaystyle J(C)=\mathbb {C} ^{g}/\Lambda .}$
The Abel–Jacobi map is then defined as follows. We pick some base point ${\displaystyle p_{0}\in C}$ and, nearly mimicking the definition of ${\displaystyle \Lambda }$, define the map
${\displaystyle u\colon C\to J(C),u(p)=\left(\int _{p_{0}}^{p}\omega _{1},\dots ,\int _{p_{0}}^{p}\omega _{g}\right){\bmod {\Lambda }}.}$
Although this is seemingly dependent on a path from ${\displaystyle p_{0}}$ to ${\displaystyle p,}$ any two such paths define a closed loop in ${\displaystyle C}$ and, therefore, an element of ${\displaystyle H_{1}(C,\mathbb {Z} ),}$ so integration over it gives an element of ${\displaystyle \Lambda .}$ Thus the difference is erased in the passage to the quotient by ${\displaystyle \Lambda }$. Changing base-point ${\displaystyle p_{0}}$ does change the map, but only by a translation of the torus.
## The Abel–Jacobi map of a Riemannian manifold
Let ${\displaystyle M}$ be a smooth compact manifold. Let ${\displaystyle \pi =\pi _{1}(M)}$ be its fundamental group. Let ${\displaystyle f:\pi \to \pi ^{ab}}$ be its abelianisation map. Let ${\displaystyle tor=tor(\pi ^{ab})}$ be the torsion subgroup of ${\displaystyle \pi ^{ab}}$. Let ${\displaystyle g:\pi ^{ab}\to \pi ^{ab}/tor}$ be the quotient by torsion. If ${\displaystyle M}$ is a surface, ${\displaystyle \pi ^{ab}/tor}$ is non-canonically isomorphic to ${\displaystyle \mathbb {Z} ^{2g}}$, where ${\displaystyle g}$ is the genus; more generally, ${\displaystyle \pi ^{ab}/tor}$ is non-canonically isomorphic to ${\displaystyle \mathbb {Z} ^{b}}$, where ${\displaystyle b}$ is the first Betti number. Let ${\displaystyle \phi =g\circ f:\pi \to \mathbb {Z} ^{b}}$ be the composite homomorphism.
Definition. The cover ${\displaystyle {\bar {M}}}$ of the manifold ${\displaystyle M}$ corresponding to the subgroup ${\displaystyle \mathrm {Ker} (\phi )\subset \pi }$ is called the universal (or maximal) free abelian cover.
Now assume M has a Riemannian metric. Let ${\displaystyle E}$ be the space of harmonic ${\displaystyle 1}$-forms on ${\displaystyle M}$, with dual ${\displaystyle E^{*}}$ canonically identified with ${\displaystyle H_{1}(M,\mathbb {R} )}$. By integrating an integral harmonic ${\displaystyle 1}$-form along paths from a basepoint ${\displaystyle x_{0}\in M}$, we obtain a map to the circle ${\displaystyle \mathbb {R} /\mathbb {Z} =S^{1}}$.
Similarly, in order to define a map ${\displaystyle M\to H_{1}(M,\mathbb {R} )/H_{1}(M,\mathbb {Z} )_{\mathbb {R} }}$ without choosing a basis for cohomology, we argue as follows. Let ${\displaystyle x}$ be a point in the universal cover ${\displaystyle {\tilde {M}}}$ of ${\displaystyle M}$. Thus ${\displaystyle x}$ is represented by a point of ${\displaystyle M}$ together with a path ${\displaystyle c}$ from ${\displaystyle x_{0}}$ to it. By integrating along the path ${\displaystyle c}$, we obtain a linear form, ${\displaystyle h\to \int _{c}h}$, on ${\displaystyle E}$. We thus obtain a map ${\displaystyle {\tilde {M}}\to E^{*}=H_{1}(M,\mathbb {R} )}$, which, furthermore, descends to a map
${\displaystyle {\overline {A}}_{M}:{\overline {M}}\to E^{*},\;\;c\mapsto \left(h\mapsto \int _{c}h\right),}$
where ${\displaystyle {\overline {M}}}$ is the universal free abelian cover.
Definition. The Jacobi variety (Jacobi torus) of ${\displaystyle M}$ is the torus
${\displaystyle J_{1}(M)=H_{1}(M,\mathbb {R} )/H_{1}(M,\mathbb {Z} )_{\mathbb {R} }.}$
Definition. The Abel–Jacobi map
${\displaystyle A_{M}:M\to J_{1}(M),}$
is obtained from the map above by passing to quotients.
The Abel–Jacobi map is unique up to translations of the Jacobi torus. The map has applications in Systolic geometry. Interestingly, the Abel–Jacobi map of a Riemannian manifold shows up in the large time asymptotics of the heat kernel on a periodic manifold (Kotani & Sunada (2000) and Sunada (2012)).
In much the same way, one can define a graph-theoretic analogue of Abel–Jacobi map as a piecewise-linear map from a finite graph into a flat torus (or a Cayley graph associated with a finite abelian group), which is closely related to asymptotic behaviors of random walks on crystal lattices, and can be used for design of crystal structures.
## Abel–Jacobi theorem
The following theorem was proved by Abel: Suppose that
${\displaystyle D=\sum _{i}n_{i}p_{i}\ }$
is a divisor (meaning a formal integer-linear combination of points of C). We can define
${\displaystyle u(D)=\sum _{i}n_{i}u(p_{i})\ }$
and therefore speak of the value of the Abel–Jacobi map on divisors. The theorem is then that if D and E are two effective divisors, meaning that the ${\displaystyle n_{i}}$ are all positive integers, then
${\displaystyle u(D)=u(E)\ }$ if and only if ${\displaystyle D}$ is linearly equivalent to ${\displaystyle E.}$ This implies that the Abel–Jacobi map induces an injective map (of abelian groups) from the space of divisor classes of degree zero to the Jacobian.
Jacobi proved that this map is also surjective, so the two groups are naturally isomorphic.
The Abel–Jacobi theorem implies that the Albanese variety of a compact complex curve (dual of holomorphic 1-forms modulo periods) is isomorphic to its Jacobian variety (divisors of degree 0 modulo equivalence). For higher-dimensional compact projective varieties the Albanese variety and the Picard variety are dual but need not be isomorphic.
## References
• E. Arbarello; M. Cornalba; P. Griffiths; J. Harris (1985). "1.3, Abel's Theorem". Geometry of Algebraic Curves, Vol. 1. Grundlehren der Mathematischen Wissenschaften. Springer-Verlag. ISBN 978-0-387-90997-4.
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https://exceljet.net/functions/yieldmat-function
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Summary
The Excel YIELDMAT function returns the annual yield of a security that pays interest at maturity.
Purpose
Get annual yield of security interest at maturity
Return value
Yield as percentage
Arguments
• sd - Settlement date of the security.
• md - Maturity date of the security.
• id - Issue date of the security.
• rate - Interest rate of security.
• pr - Price per \$100 face value.
• basis - [optional] Coupon payments per year (annual = 1, semiannual = 2; quarterly = 4).
Syntax
=YIELDMAT(sd, md, id, rate, pr, [basis])
How to use
The YIELDMAT function returns the annual yield of a security that pays interest at maturity. In the example shown, the formula in F5 is:
``````=YIELDMAT(C9,C7,C8,C6,C5,C10)
``````
with these inputs, the YIELDMAT function returns 0.081 which, or 8.10% when formatted with the percentage number format.
Entering dates
In Excel, dates are serial numbers. Generally, the best way to enter valid dates is to use cell references, as shown in the example. If you want to enter valid dates directly inside a function, the DATE function is the best approach.
Basis
The basis argument controls how days are counted. The PRICE function allows 5 options (0-4) and defaults to zero, which specifies US 30/360 basis. This article on Wikipedia provides a detailed explanation of available conventions.
Basis Day count
0 or omitted US (NASD) 30/360
1 Actual/actual
2 Actual/360
3 Actual/365
4 European 30/360
Notes
• In Excel, dates are serial numbers
• Settlement, maturity issue, and basis are truncated to integers
• If settlement, maturity, or issue dates are not valid, YIELDMAT returns #VALUE!
• YIELDMAT returns #NUM! if any of the following are true:
• rate < 0
• pr <= 0
• settlement >= maturity
• Basis is not 0-4
Author
Dave Bruns
Hi - I'm Dave Bruns, and I run Exceljet with my wife, Lisa. Our goal is to help you work faster in Excel. We create short videos, and clear examples of formulas, functions, pivot tables, conditional formatting, and charts.
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CC-MAIN-2023-40
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http://forums.wolfram.com/mathgroup/archive/2004/Dec/msg00029.html
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4th subharmoic functions
• To: mathgroup at smc.vnet.net
• Subject: [mg52541] 4th subharmoic functions
• From: Roger Bagula <tftn at earthlink.net>
• Date: Wed, 1 Dec 2004 05:58:14 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com
```These actually give a spiral and not a circle when added together.
They are my best bet so far at the next level of complexity in
a subharmonic reduction of a circle. Groups of this type are different
than
traditional rotation and point groups and determinant groups as they are
based on substitution groups
that have associated fractal effects. I found this IFS space fill this year
in my trifraction expertiments. If you take away the (x/2+1/2,y/2+1/2)
it becomes a nonlinear equivalent to a Sierpinski gasket which I called
I've included a short IFS demonstration program at the end.
(*using a set of functions that give a space filling curve in True Basic
as as an IFS to get a set of 4th level subharmonic functions*)
(* based on for parts of the functional inversion group of the
anharmonic group with x->x/2 and x+1->x/2+1/2*)
digits=100
f1[n_]:=n/2/;Mod[n,4]==1
f1[n_]:=n/2+1/2/;Mod[n,4]==2
f1[n_]:=n/(1+n)/;Mod[n,4]==3
f1[n_]:=1/(1+n)/;Mod[n,4]==0
f2[n_]:=n/2/;Mod[n,4]==2
f2[n_]:=n/2+1/2/;Mod[n,4]==3
f2[n_]:=n/(1+n)/;Mod[n,4]==0
f2[n_]:=1/(1+n)/;Mod[n,4]==1
f3[n_]:=n/2/;Mod[n,4]==3
f3[n_]:=n/2+1/2/;Mod[n,4]==0
f3[n_]:=n/(1+n)/;Mod[n,4]==1
f3[n_]:=1/(1+n)/;Mod[n,4]==2
f4[n_]:=n/2/;Mod[n,4]==0
f4[n_]:=n/2+1/2/;Mod[n,4]==1
f4[n_]:=n/(1+n)/;Mod[n,4]==2
f4[n_]:=1/(1+n)/;Mod[n,4]==3
f1sin[x_]:=Sum[(-1)^(n)*f1[n]*x^(2*n+1)/((2*n+1)!),{n,0,digits}]
f2sin[x_]:=Sum[(-1)^(n)*f2[n]*x^(2*n+1)/((2*n+1)!),{n,0,digits}]
f3sin[x_]:=Sum[(-1)^(n)*f3[n]*x^(2*n+1)/((2*n+1)!),{n,0,digits}]
f4sin[x_]:=Sum[(-1)^(n)*f4[n]*x^(2*n+1)/((2*n+1)!),{n,0,digits}]
f1cos[x_]:=Sum[(-1)^(n)*f1[n]*x^(2*n)/((2*n)!),{n,0,digits}]
f2cos[x_]:=Sum[(-1)^(n)*f2[n]*x^(2*n)/((2*n)!),{n,0,digits}]
f3cos[x_]:=Sum[(-1)^(n)*f3[n]*x^(2*n)/((2*n)!),{n,0,digits}]
f4cos[x_]:=Sum[(-1)^(n)*f4[n]*x^(2*n)/((2*n)!),{n,0,digits}]
Plot[f1sin[x],{x,-Pi,Pi}]
Plot[f1cos[x],{x,-Pi,Pi}]
Plot[f2sin[x],{x,-Pi,Pi}]
Plot[f2cos[x],{x,-Pi,Pi}]
Plot[f3sin[x],{x,-Pi,Pi}]
Plot[f3cos[x],{x,-Pi,Pi}]
Plot[f4sin[x],{x,-Pi,Pi}]
Plot[f4cos[x],{x,-Pi,Pi}]
Plot[f1sin[x]+f2sin[x]+f3sin[x]+f4sin[x],{x,-Pi,Pi}]
Plot[f1cos[x]+f2cos[x]+f3cos[x]+f4cos[x],{x,-Pi,Pi}]
ParametricPlot[{f1sin[x]+f2sin[x]+f3sin[x]+f4sin[x],f1cos[x]+f2cos[x]+f3cos[x]+f4cos[x]},{x,-Pi,Pi}]
(* ifs of the four parts function as two 1d ifs*)
f1[{x_,y_}] = {x/2, y/2};
f2[{x_,y_}] = {1/(x+1), y/(y+1)};
f3[{x_,y_}] = {x/(x+1), 1/(y+1)};
f4[{x_,y_}] = {x/2+1/2, y/2+1/2};
f[x_] := Which[(r=Random[]) <= 1/4, f1[x],r <=2/4, f2[x],r <= 3/4,
f3[x],r <= 1.00, f4[x]]
ifs[n_] := Show[Graphics[{PointSize[.001],
Map[Point, NestList[f, {0,0}, n]]}],
PlotRange->All,AspectRatio->Automatic]
ifs[10000]
Respectfully, Roger L. Bagula
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# Proving Convergence
• May 5th 2008, 03:07 AM
ah-bee
Proving Convergence
Let (
an) be a sequence which converges to 0. Use the definition of a convergent sequence to prove that the sequence (an^2) to 0.
Definition: There exists an L that is Real s.t. for every E>0 there exists N that is a positive integer s.t. for every n that is a positive integer n>N implies that |an - L| < E
• May 5th 2008, 07:23 AM
ThePerfectHacker
Quote:
Originally Posted by ah-bee
Let (
an) be a sequence which converges to 0. Use the definition of a convergent sequence to prove that the sequence (an^2) to 0.
Definition: There exists an L that is Real s.t. for every E>0 there exists N that is a positive integer s.t. for every n that is a positive integer n>N implies that |an - L| < E
Since $a_n\to 0$ it is bounded, i.e. there is $A>0$ so that $|a_n|\leq A$.
This means, $|a_n^2| = |a_n||a_n|\leq A|a_n| < A\epsilon$ for $n>N$.
• May 5th 2008, 06:32 PM
ah-bee
i dont understand how works. can u explain to me how it works?
• May 5th 2008, 06:45 PM
ThePerfectHacker
Quote:
Originally Posted by ah-bee
i dont understand how works. can u explain to me how it works?
There exists $A>0$ so that $|a_n|\leq A$. Because convergent sequences are bounded.
To prove that $a_n^2 \to 0$ we need to show $|a_n^2 - 0| < \epsilon$ for any $\epsilon > 0$ where $N$ is sufficiently large.
Give me any $\epsilon > 0$, then since $a_n \to 0$ it means $|a_n - 0| = |a_n|< \frac{\epsilon}{A}$ for $n>N$.
But then if $n>N$ we have $|a_n^2 - 0| = |a_n^2| = |a_n||a_n|\leq A|a_n| < A\cdot \frac{\epsilon}{A} = A$.
• May 5th 2008, 07:04 PM
ah-bee
thanks, i understand it now, so if i were to get a question saying if an converges to k, prove an^2 converges to k^2, i would approach it in the same way?
• May 5th 2008, 07:10 PM
ThePerfectHacker
Quote:
Originally Posted by ah-bee
thanks, i understand it now, so if i were to get a question saying if an converges to k, prove an^2 converges to k^2, i would approach it in the same way?
Not exactly in the same way. But it should be similar.
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# Suppose that in Problem 17, the components are in parallel, so that the system will fail only…
Suppose that in Problem 17, the components are in parallel, so that the system will fail only when all the components fail. Find the p.d.f. of T, and then find the probability P(T > 15).
Problem 17
A mechanical system has three components in series, so the system will fail when at least one of the components fails. The random variables X1,X2, and X3 represent the lifetime of these components. Suppose that the Xi(i = 1, 2, 3) are independently and identically distributed as exponential with parameter λ = 0.1. Let the random variable T denote the lifetime of the system. Find the p.d.f. of T, and then find the probability P(T > 10).
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# Thread: Any Excel VBA or Excel functions that can help with general lottery forecasting? Thanks: 0 Likes: 0
1. ## Any Excel VBA or Excel functions that can help with general lottery forecasting?
Mostly the posts here about the lottery concern with random number generation, with pick-3 games and with lotteries dealing with huge numbers (e.g. 5 out of 90). I am only concerned with general, simple lottery forecasting concerns.
So, I'm wondering if anyone here knows some VBA formulas for the following tasks:
a) Counting the frequency of each number in a lottery game
b) Determine which are "hot" numbers, "cold" numbers and "overdue" numbers within a certain draw period in a lottery game
c) Know how many times a number is paired with another number
Hope you would consider helping me out in this...
3. ## Re: Any Excel VBA or Excel functions that can help with general lottery forecasting?
It would be possible to generate analyses such as you refer to, but if its the UK lottery, then they already give that data - see https://www.national-lottery.com/lotto/statistics
I would imagine other lottery runners would have similar data available.
5. ## Re: Any Excel VBA or Excel functions that can help with general lottery forecasting?
@angExcelentflea
While we do not prohibit Cross-Posting on this site, we do ask that you please mention you are doing so and provide links in each of the threads pointing to the other thread (see rule 13 here along with the explanation: Forum Rules).
This way, other members can see what has already been done in regards to a question, and do not waste time working on a question that may already be answered.
6. ## Re: Any Excel VBA or Excel functions that can help with general lottery forecasting?
PS - Due to earlier circumstances, was forced to post this concern in other Excel forums. Here's the link to the other forum:
7. ## Re: Any Excel VBA or Excel functions that can help with general lottery forecasting?
Reply intended for Kyle123: Since I couldn't edit my original post, I included a note explaining why... It had something to do with some misunderstanding with the other forum and was badly in need of answers. If any original idea comes up here, the other guys will know...
To admins: sorry if this post peeved you. Just wanted to make sure somebody will give ideas to my concern....
8. ## Re: Any Excel VBA or Excel functions that can help with general lottery forecasting?
Originally Posted by jmacleary
if its the UK lottery, then they already give that data - see https://www.national-lottery.com/lotto/statistics
Just a "random" observation (wink).... Be careful with your interpretation of the overall frequency table. That lottery changed from 1-to-49 to 1-to-59 in Oct 2015. That is why the frequency of 50-to-59 is so far out of line, not some inexplicable bias in the selection process.
9. ## Re: Any Excel VBA or Excel functions that can help with general lottery forecasting?
Interesting. So, does that mean the usual way of forecasting lottery won't apply to the 6/59 game?
By the way, any idea what VBA formula or Excel function to use to get "hot", "cold" and "overdue" numbers?
No other site I looked into gives any idea as to how...
10. ## Re: Any Excel VBA or Excel functions that can help with general lottery forecasting?
Originally Posted by angExcelentflea
does that mean the usual way of forecasting lottery won't apply to the 6/59 game?
First, my previous comment was specific to the UK lotto game. If you are not interested in the UK lotto game per se, ignore my comments.
Second, I hope you know that there is no reliable way to "forecast" or predict the outcome of any specific lottery drawing.
And none of the statistics like the ones at https://www.national-lottery.com/lotto/statistics will improve your odds of matching a specific drawing.
(The only way to improve your odds of matching is to purchase more tickets, each of which is different.)
That said, you have to pick 6 numbers from 1-to-59 somehow. And any method is as good as another, because all combinations (tickets) have the same probability of matching a specific drawing.
My previous comment relates to the fact the UK lotto changed from 6/49 to 6/59 after 8 Oct 2015 (Thu).
But the statistics at https://www.national-lottery.com/lotto/statistics cover "20 years". (Actually, apparently at least back to 27 Dec 1995.)
For example, the 6 least-often drawn numbers are all in the 50s because those numbers were available to be drawn only in the last 400 of the 2465 drawings that are covered by the statistics. Well, duh!
For the same reason, the total frequency of the numbers in the 50s are significantly lower than 1-to-49.
In fact, IMHO, the only valid statistic is the "most overdue numbers", since that seems to mean only the most number of days since last drawn.
Originally Posted by angExcelentflea
any idea what VBA formula or Excel function to use to get "hot", "cold" and "overdue" numbers?
It is certainly doable. Some statistics can be calculated with Excel formulas alone. Some are best done using VBA procedures.
Before anyone can help you, you need to specify the lotto that you are interested in, or its parameters.
Such calculations are fun. But I'm afraid that I should not take the time to participate.
Good luck!
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# One-loop diagram $\phi \to \phi$ in $g\phi^3$ theory
I am trying to evaluate a diagram of $$\phi^3$$ theory to practice one-loop amplitudes, but I am stuck at a certain point. The amplitude is given by the integral,
$$\mathcal{M} = \frac{(-ig)^2}{2} \int \frac{d^4k}{(2\pi)^4} \frac{i}{k^2-m^2} \frac{i}{(k-p)^2-m^2}$$
where the $$1/2$$ is the symmetry factor. Using Feynman parametrization I was able to write it in the form,
$$\frac{g^2}{2}\int_0^1 dx \, \int \frac{d^d \ell}{(2\pi)^d} \frac{1}{[\ell^2-\Delta]^2}$$
where $$\Delta \equiv m^2 +p^2x(x-1)$$, shifting momenta $$\ell = k-px$$. I also completed the square, and went to arbitrary $$d$$ dimensions. Using the formula provided in Peskin and Schroeder's appendix (the non Wick rotated version), I found,
$$=i\frac{g^2}{4}\frac{\Gamma(2-d/2)}{(4\pi)^{d/2}} \int_0^1 dx \frac{1}{[m^2+p^2x(x-1)]^{2-d/2}}$$
If I evaluate the $$dx$$ integral in Mathematica I get a horrible expression with hypergeometric functions. I don't recall seeing this in other loop amplitudes, it should be a relatively simple expression. How should I proceed? Have I erred somewhere?
• By the way, this is not a tadpole diagram. This one is called bubble. A tadpole looks like a lollipop. Jun 25, 2019 at 0:28
I think you have done everything correctly so far. Now while the integral would be perfectly finite in $d=4$ (which I assume is the dimension you want your result in), $\Gamma(0)=\infty$.
Therefore we expand $I=\Gamma(2-d/2)[m^2+p^2x(x-1)]^{d/2-2}$ around $d=4$ and then do the integration. I actually replaced $d\rightarrow 4-2\epsilon$ and expanded around $\epsilon = 0$.
$\mathcal{I}=\int_0^1 dx\ I = \frac{1}{\epsilon} + 2 - \gamma_E - \frac{2}{z} \arctan z -\log m^2 \ ,$
where $z=\frac{p}{\sqrt{4m^2-p^2}}$ and $\gamma_E$ is the Euler-Mascheroni constant. For $p^2>4m^2$ we can use the identity $\arctan z = \frac{1}{2i}\log \frac{1+iz}{1-iz}$ to express the $\arctan$.
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# Inferencing - PowerPoint PPT Presentation
Inferencing. Hunting for Clues to Solve a Puzzle. What is Inferencing?. Finding clues Putting them together Solving the problem Like a jigsaw puzzle. Here are some Synonyms. deduce figure out guess interpretation read between the lines understand reason drawing conclusions.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about ' Inferencing' - talisa
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Presentation Transcript
### Inferencing
Hunting for Clues to Solve a Puzzle
• Finding clues
• Putting them together
• Solving the problem
• Like a jigsaw puzzle
• deduce
• figure out
• guess
• interpretation
• understand
• reason
• drawing conclusions
• Jigsaw puzzles
• Problem Solving (backtracking)
• Math word problems
• Conversing with people
What Should We Look For?
• PLACES
• TIME
• COLORS
• TEXTURES
• BODY LANGUAGE
• ACTIONS
• SITUATIONS OR CONTEXT
• What clues do you see?
• What do we know about the product?
• What do we know about the company?
• What do we know about the person who drank the drink?
• What clues do you see?
• What can we infer from the clues?
• What do we know about the event in the photo?
• What is the mood of the event?
• What other activities may be occurring at the same time or later?
• It’s clear
• It’s round
• It’s made of glass, metal and plastic
• It has a handle
• It makes things look bigger
• What is it?
Are they having the same conversation? How do you know?
• What clues do you notice?
• Who are the people in the commercial and what do they represent?
• What emotions are you seeing from the people in the commercial?
• What can we infer from this commercial?
• When we infer meaning we put together clues like a jigsaw puzzle.
For example: Look at our clues!
The Captaintraveled down the rough muddy road in his Jeep.
• What clues are presented ?
• What can we infer about The Captain?
• You will be receiving 5 sets of clues in just a few minutes. Follow the clues to see if you can infer the answers. You will find the clues on your computers in the jigsaw puzzle program.
• What word clues can you find to help you solve this puzzle?
• Listen to the word clues to see if you can guess who the mystery person is.
• On your computers go to the following web site: http://kids.mysterynet.com/solveit/
• Read the story and let’s discuss the clues you found before you solve the mystery
• Inferencing means:
• Finding clues
• Putting them together
• Solving the problem
• It is just like a jigsaw puzzle.
• Choose a picture from our database that shows the end of a possible story.
• Create a short story that tells what led up to this ending.
• Present your story to the class.
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# Gallery posets of supersolvable arrangements.
1 Introduction
This article is about the topology of some posets associated to real hyperplane arrangements. Throughout, the topology attached to a poset P is that of its order complex [DELTA](P), the simplicial complex of chains [x.sub.1] < [x.sub.2] < ... < [x.sub.k] of P. If P is a bounded poset, its proper part [bar.P] is the same poset with those bounds removed.
A reduced gallery of a real central arrangement A is a sequence of chambers [c.sub.0], [c.sub.1], ..., [c.sub.m] such that adjacent chambers are separated by exactly one hyperplane, and [c.sub.0] and [c.sub.m] are separated by m hyperplanes. For any codimension 2 subspace X [member of] L(A), a gallery between opposite chambers [c.sub.0], -[c.sub.0] can cross the hyperplanes containing X in two ways. We introduce a poset R(A, [r.sub.0]) of reduced galleries r ordered by single-step inclusion of sets of codimension 2 subspaces separating r from [r.sub.0]. The Hasse diagram of this poset is the usual graph of reduced galleries. An example is shown in Figure 1
When A is supersolvable and [r.sub.0] is incident to a modular flag, the poset was essentially shown to be bounded by Reiner and Roichman [10]; see Section 4 for details. In this situation, we prove [bar.R](A, [r.sub.0]) is homotopy equivalent to a (rk A - 3)-sphere.
For a face F and a chamber c of A, there is a unique chamber F x c incident to F that is closest to c. A cellular string is a sequence of faces ([X.sub.1], [X.sub.2], ..., [X.sub.m]) in L(A) such that [X.sub.1] x [c.sub.0] = [c.sub.0], [X.sub.m] x (-[c.sub.0]) = -[c.sub.0] and
[X.sub.i] x (-[c.sub.0]) = [X.sub.i+1] x [c.sub.0] ([for all]i).
The poset [omega](A, [c.sub.0]) of cellular strings is ordered by refinement. Our main result is the following.
Theorem 1.1 Let A be a real supersolvable hyperplane arrangement with chamber [c.sub.0] and gallery [r.sub.0] both incident to a modular flag. Let x = ([X.sub.1], [X.sub.2], ..., [X.sub.m]) [member of] [omega](A, [c.sub.0]) be a cellular string of A. The set of galleries incident to x forms a closed interval of R(A, [r.sub.0]) whose proper part is homotopy equivalent to [sub.S][[summation].sup.m.sub.i=1]codim([X.sub.i])-m-2.
We conjecture that all other intervals are contractible.
The motivation for this work primarily comes from two sources: the parameterization of noncontractible intervals in the chamber poset of an arrangement A proved by Edelman and Walker [7], and a conjecture by Reiner on the noncontractible intervals of the Higher Bruhat orders [11]. We hope to derive new instances of the generalized Baues problem posed by Billera, Kapranov, and Sturmfels via these posets [2]. We summarize the relevant results below.
Two posets naturally arise from a rank d real central hyperplane arrangement A: the poset of faces [Laplace](A) ordered by reverse inclusion and the poset of chambers P(A, [c.sub.0]) ordered by separation from some base chamber [c.sub.0]. These posets are reviewed in Section 3. Restricting A to the unit sphere gives a cellular decomposition of [S.sup.d-1]. This implies [bar.[Laplace]](A) is homeomorphic to a (d - 1)-sphere. In [7], Edelman and Walker use a recursive coatom ordering on an auxiliary poset to prove [bar.P](A, [c.sub.0]) is homotopy equivalent to [S.sup.d-2]. Moreover, they show the noncontractible open intervals of P are parameterized by the elements of L.
Theorem 1.2 (Edelman, Walker [7]) Let A be a real hyperplane arrangement with base chamber [c.sub.0]. The set of chambers incident to a face F [member of] [Laplace](A) forms a closed interval [c, d] in P(A, [c.sub.0]) such that (c, d) is homotopy equivalent to [S.sup.codimF-2]. Every other open interval of P(A, [c.sub.0]) is contractible
This theorem establishes an isomorphism between [Laplace] and the poset [Int.sub.nonc](P) of noncontractible intervals of P, ordered by inclusion. For an arbitrary bounded poset P, the full interval poset [bar.Int](P) has a deformation retract to its subposet of proper noncontractible intervals. Furthermore, [bar.Int](P) is homeomorphic to the suspension of P. Thus, EW's theorem may be viewed as an explanation of the homotopy equivalence
[bar.L](A) [equivalent] susp([bar.P](A, [c.sub.0])).
In [5], Bjorner applied Theorem 1.2 to compute the homotopy type of a poset [omega](A, [c.sub.0]) of cellular strings of A.
Theorem 1.3 (Bjorner [5]) The poset of cellular strings [omega](A, [c.sub.0]) of a rank d arrangement A with base chamber c0 is homotopy equivalent to [S.sup.d-2].
Bjorner's proof uses the identification of [omega](A, [c.sub.0]) with the poset [DELTA]H(P(A, [c.sub.0])) of chains [??] = [x.sub.0] < [x.sub.1] < ... < [x.sub.m] = [??] of P for which each interval ([x.sub.i], [x.sub.i+i]) is noncontractible. Once again a general argument shows that [[DELTA].sup.H](P) is homotopy equivalent to [DELTA](P).
Theorem 1.3 as stated above is a special case of the theorem of Billera, Kapranov, and Sturmfels on the homotopy type of a poset of cellular strings in a polytope [2]. However, Bjorner's argument has a wide-reaching generalization to the cellular string posets of duals of shellable CW-spheres [1].
The paper is structured as follows. The above theorems are interpreted for Coxeter groups in Section 2. Some notation for hyperplane arrangements is given in Section 3. In Section 4 we define the poset of galleries for supersolvable arrangements. In Section 5, we recall the Suspension Lemma and use it to compute the homotopy type of the proper part of R. Theorem 1.1 is proved in Section 6. In Section 7, we draw a parallel between the poset of galleries for supersolvable arrangements and the Higher Bruhat orders. This connection will be developed further in future work.
2 Coxeter Groups
A finite Coxeter system (W, S) is a finite group W with a presentation of the form
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
where [m.sub.st] = 1 if s = t and [m.sub.st] = [m.sub.ts] [member of] [Z.sub.[greater than or equal to]2] otherwise. For example, if S = {[s.sub.1], ..., [s.sub.n-1]} with [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] when [absolute value of i - j] = 1 and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] when [absolute value of i - j] > 1, then W is isomorphic to the symmetric group [G.sub.n].
A parabolic subsystem ([W.sub.J], J) is a subgroup [W.sub.J] generated by a subset J [subset or equal to] S. The collection of left cosets of parabolic subgroups
[Laplace](W) = {[wW.sub.J] | [member of] e W, J [subset or equal to] S}
may be viewed as a poset ordered by inclusion.
Let T = {[wsw.sup.-1] | s [member of] S, w [member of] W} denote the set of reflections of W. Define the length l(w) of w [member of] W to be the smallest value for which w equals [s.sub.1] ... [s.sub.l](-w) for some [s.sub.i] [member of] S. The (right) inversion set Inv(w) of w [member of] W is {t [member of] T | l(wt) < l(w)}. The weak Bruhat order P(W) is the set W ordered by inclusion of inversion sets; see Figure 2.
An interesting relationship between the posets P and [Laplace] was discovered by Bjorner.
Theorem 2.1 (Bjorner [4]) Let (W, S) be a finite Coxeter system. The coset [wW.sub.J] [member of] [Laplace](W) forms a closed interval [x, y] of P(W) such that (x, y) is homotopy equivalent to [S.sup.[absolute value of J]-2]. Every other open interval of P(W) is contractible.
Any finite Coxeter group has an associated reflection arrangement A and canonical isomorphisms [Laplace](A) [congruent] [Laplace](W) and P(A, [c.sub.0]) = P(W) for any choice of base chamber [c.sub.0]. Thus, Theorem 2.1 is a direct consequence of Theorem 1.2.
In this article, we describe a lifting of Theorem 2.1 to the intervals of a poset of galleries in a supersolvable hyperplane arrangement. We begin by describing the Coxeter case.
For a finite Coxeter system (W, S) there is a unique element [w.sub.0] [member of] W with Inv([w.sub.0]) = T. Let R(W) denote the set of reduced words [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for [w.sub.0]. The set R(W) forms a graph where two reduced words are adjacent if they differ in a single braid move:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
The connectivity of R( W) for arbitrary finite Coxeter groups was originally shown by Tits. Athanasiadis, Edelman, and Reiner showed this graph is ([absolute value of S] - 1)-connected in [1].
In [10], Reiner and Roichman give a lower bound for the diameter of R(W) by defining a set-valued metric on the graph of reduced words. When W is a Coxeter group of type A or B they show this lower bound is achieved by constructing a bounded graded poset whose Hasse diagram is an orientation of R(W). This construction is recalled in Section 4; see Example 4.1. For the remainder of this article R(W) will denote this poset.
For J [subset or equal to] S, let [w.sub.0](J) denote the longest element of the subsystem ([W.sub.J], J). The poset of cellular strings [omega](W) is the set of words ([J.sub.1], [J.sub.2], ..., [J.sub.m]) where 0 [not equal to] [J.sub.i] [subset or equal to] S for all i and
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
The cellular strings are ordered by refinement, i.e. ([I.sub.1], ..., [I.sub.l]) [less than or equal to] ([J.sub.1], ..., [J.sub.m]) if there is are indices 0 = [[alpha].sub.1] < ... < [[alpha].sub.t] < l such that
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
Theorem 2.2 Let (W, S) be a Coxeter system of type A or B. The set of reduced words for [w.sub.0] refining a cellular string ([J.sub.1], ..., [J.sub.m]) forms an interval [x, y] in R(W) such that (x, y) is homotopy equivalent to [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
We conjecture that these are the only noncontractible intervals of R(W).
The weak order on W may be defined by setting u [less than or equal to] [upsilon] if there exists a geodesic from e to [upsilon] passing through u in the Cayley graph of W with respect to the generating set S. If the base point in the geodesic order is changed from e to some other element of W, an isomorphic poset is obtained since W acts simply transitively on its Cayley graph.
Similarly, R(W) may be defined by fixing a reduced expression [r.sub.0] and ordering by geodesics in the graph of reduced words for [w.sub.0]. However, different choices of [r.sub.0] yield nonisomorphic posets, many of which fail to be bounded or fail to satisfy Theorem 2.2, even in types A and B; see Figure 3. For type A, the base word is
[r.sub.0] = [s.sub.1]([s.sub.2][s.sub.1])([s.sub.3][s.sub.2][s.sub.1]),
while in type B, the base is
[r.sub.0] = [s.sub.0]([s.sub.1][s.sub.0][s.sub.1])([s.sub.2][s.sub.1][s.sub.0][s.sub.1][s.sub.2]) ...
At this time it is unclear whether a good choice of r0 exists for every finite Coxeter group.
3 Hyperplane Arrangements
Let A be an oriented real central hyperplane arrangement in [R.sup.d]. Its lattice of flats L(A) consists of the subspaces [H.sub.1] [intersection] ... [intersection] [H.sub.k] for [{[H.sub.j]}.sup.j.sub.k=1] [subset or equal to] A ordered by reverse inclusion. The rank i elements [L.sub.i](A) are the subspaces of codimension i.
Each hyperplane H in A divides [R.sup.d] into three connected sets [H.sup.+], [H.sup.-], [H.sup.0], where [H.sup.+] is the positive halfspace defined by H, [H.sup.-] = -[H.sup.+], and [H.sup.0] = H. The face poset [Laplace](A) is the set of nonempty cells of the form [[intersection].sub.H[member of]A][H.sup.[epsilon](H)] where [epsilon] is a sign vector in [{0, +, - }.sup.A], ordered by reverse inclusion of their closures. If F is a face defined by [[epsilon].sub.F] [member of] [{0, +, -}.sup.A], set [F.sup.0] = {H [member of] A | [[epsilon].sub.F](H) = 0}. The map F [??] [F.sup.0] is an order-preserving map [Laplace](A) [right arrow] [Laplace](A). The set [Laplace](A) is closed under o where
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
The chambers of A are the d-dimensional faces in [Laplace](A). The Li-separation set [L.sub.1](c, d) of two chambers c, d is the set of hyperplanes separating c and d. If c is fixed then d is determined by [L.sub.1](c, d) by reversing the halfspaces of [L.sub.1](c, d) in the sign vector for c.
Fixing a base chamber [c.sub.0], the poset of chambers P (A, [c.sub.0]) is the set of chambers of A ordered by c [less than or equal to] d if [L.sub.1]([c.sub.0], c) [subset or equal to] [L.sub.1]([c.sub.0], d).
4 Gallery poset
A gallery is a sequence of chambers [c.sub.0], [c.sub.1], ..., [c.sub.m] such that
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
Unless otherwise specified, we will assume the gallery connects an antipodal pair of chambers [c.sub.0], -[c.sub.0]. For a fixed base chamber [c.sub.0], a gallery is determined by the order in which the hyperplanes are crossed. The set of galleries admits a free involution r [??] -r when [absolute value of A] > 1, where -r is the gallery from [c.sub.0] to -[c.sub.0] which crosses the hyperplanes of A in the reverse order of r.
For a subspace X [member of] L(A), the localization [A.sub.X] of A at X is {H [member of] A | H [subset or equal to] X}. If c is a chamber of A, set cX to be the chamber of [A.sub.X] containing c. If r = [c.sub.0], [c.sub.1], ..., -[c.sub.0] is a gallery, then [r.sub.X] is the sequence ([c.sub.0])X, ([c.sub.1])X, ..., [(-[c.sub.0]).sub.X] with repetitions removed, which is a gallery of [A.sub.X].
The [L.sub.2]-separation set [L.sub.2](r, r') is {X [member of] [L.sub.2](A) | [r.sub.X] [not equal to] [r'.sub.X]}. Given a base gallery [r.sub.0], R(A, [r.sub.0]) is the poset of galleries between [c.sub.0] and -[c.sub.0] ordered by single-step inclusion of the sets [L.sub.2]([r.sub.0], *). That is, r [less than or equal to] r' if there exists a sequence r = [r.sub.1], [r.sub.2], ..., [r.sub.m] = r' such that
[L.sub.2]([r.sub.0], [r.sub.1]) [subset] [L.sub.2]([r.sub.0], [r.sub.2]) [subset] ... [subset] [L.sub.2]([r.sub.0],[ .sub.r]m) [absolute value of [L.sub.2]([r.sub.0], [r.sub.i])\[L.sub.2]([r.sub.0], [r.sub.i-1])] = 1.
Any gallery r is determined by its [L.sub.2]-separation set [L.sub.2]([r.sub.0], r) from some fixed gallery [r.sub.0]. This follows since the relative order of any two hyperplanes H, H' in the total order on A induced by r is the same as their relative order in [r.sub.H[intersection]H'].
The Hasse diagram of R(A, [r.sub.0]) is an orientation of the graph of reduced galleries from [c.sub.0] to -[c.sub.0] in A. One of the reasons for studying the poset R is to deduce properties of this graph from those of R.
A flat X [member of] L(A) of a hyperplane arrangement A is modular if X + Y [member of] L(A) for all Y [member of] L(A). A hyperplane arrangement A is supersolvable if its lattice of flats L(A) contains a maximal chain of modular elements [R.sup.d] = [X.sub.0] < [X.sub.1] < .... < [X.sub.d] = 0, [X.sub.i] [member of] [L.sub.i](A). For our purposes, it will be useful to define a maximal chain of faces [F.sub.0] < [F.sub.1] < ... < [F.sub.d] = 0, [F.sub.i] [member of] [[Laplace].sub.i](A) to be a modular chain if the flats [F.sup.0.sub.i] are all modular.
A gallery r from c to d is incident to a face F [member of] [Laplace](A) if r contains the chambers F x c and F x d. A gallery is incident to a subspace X [member of] L(A) if it is incident to some face spanning X. We say a gallery r is incident to a modular flag [c.sub.0] = [F.sub.0] < [F.sub.1] < ... < [F.sub.d] = 0 if it contains [F.sub.i] x [c.sub.0] and [F.sub.i] x (-[c.sub.0]) for all i. Intuitively this means r first crosses [F.sub.1], then wraps around [F.sub.2], then around [F.sub.3], and so on.
Example 4.1 The type [A.sub.n-1] reflection arrangement A is the set
{[H.sub.ij] = ker[([x.sub.i] - [x.sub.j]) [subset] [R.sup.n]).sub.1[less than or equal to]i<j[less than or equal to]n].
Its intersection lattice [Laplace](A) is in natural correspondence with set partitions of [n]:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
Similarly, the face poset [Laplace](A) is in correspondence with set compositions of [n]:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],
where
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
The face poset [Laplace](A) contains a modular chain
(1, 2, 3, 4, ..., n) > (12, 3, 4, ..., n) > (123, 4, ..., n) > ... > (1234 ... n)
Any gallery incident to this flag must contain the chambers
(1, 2, 3 4, ..., (2, 1 3, 4, ..., (3, 2, 1 4, ..., ..., (n, ..., 4, 3 2, 1).
It is clear that there is a unique gallery [r.sub.0] containing these chambers: starting from the base chamber (1, 2, ..., n), the 2 is pushed to the front, then the 3, then the 4, etc. In Coxeter notation, [r.sub.0] corresponds to the reduced word [s.sub.1]([s.sub.2][s.sub.1])([s.sub.3][s.sub.2][s.sub.1]) ... ([s.sub.n-1][s.sub.n-2] .... [s.sub.1]).
A simple way to show R(A, [r.sub.0]) is bounded and graded by r [??] [absolute value of [L.sub.2]([r.sub.0], r)] is via wiring diagrams; see Figure 4. Given the wiring diagram corresponding to some gallery r, one can remove elements in the [L.sub.2]-separation set by slowly "pulling " the n-th wire so it crosses all of the other wires at the end. Each triple {ijn} or 2-pair {ij, kn} involving wire n on which [r.sub.0] and r disagree is resolved by this procedure without affecting other elements of [L.sub.2](A). Inducting on n, we conclude that [r.sub.0] [less than or equal to] r as desired.
In [10], Reiner and Roichman generalized the idea of "pulling out the n-th wire" for arbitrary supersolvable arrangements to prove that R(A, [r.sub.0]) is a bounded, graded poset. The key property of supersolvable arrangements is the linearity of the fibers of the localization map at a modular line, as described in the following Proposition.
Proposition 4.2 (RR [10]) Assume that l is a modular ray in [Laplace](A). Let [c.sub.l] [member of] P([A.sub.l], ([c.sub.0]);) be a chamber of the localized arrangement A;. Let [pi]: P(A, [c.sub.0]) [right arrow] P([A.sub.l], ([c.sub.0]);) be the localization map.
1. There exists a linear order on the fiber [[pi].sup.-1]([c.sub.l]) = {[c.sub.1], [c.sub.2], ..., [c.sub.t]} such that [c.sub.1] is incident to l and the separation sets [L.sub.1]([c.sub.1], [c.sub.i]) are nested:
0 = [L.sub.1]([c.sub.1], [c.sub.1]) [subset] [L.sub.1]([c.sub.1], [c.sub.2]) [subset] ... [subset] [L.sub.1]([c.sub.1], [c.sub.t]) = A\[A.sub.l]
This induces a linear order [H.sub.1], [H.sub.2], ... on A\[A.sub.l] such that [H.sub.i] is the unique hyperplane separating [c.sub.i] and [c.sub.i+1].
2. Using the linear order on A\[A.sub.l] from part (1), if i < j < k and if the chamber c incident to I is also incident to I + [H.sub.i] [intersection] [H.sub.k] then [H.sub.j] [contains or equal to] [H.sub.i] [intersection] [H.sub.k].
The next proposition says that the fibers of the localization map are bounded.
Proposition 4.3 (RR [10]) Assume that I is a modular ray and let [c.sub.0] be a chamber incident to l. Let [pi] denote the function r [??] [r.sub.l] sending galleries of A from [c.sub.0] to -[c.sub.0] to galleries of [A.sub.l]. Given a reduced gallery [r.sub.l] [member of] [A.sub.l], let [r.sub.1] be the unique gallery incident to l in the fiber [[pi].sup.- 1]([r.sub.l]) defined by Proposition 4.2(1). If [r.sub.1] [not equal to] [r.sub.2] [member of] [[pi].sup.-1]([r.sub.l]) there exists X [member of] [L.sub.2]([r.sub.1], [r.sub.2]) incident to [r.sub.2].
Inducting on the rank of A, Reiner and Roichman conclude that if [r.sub.0] is incident to a modular flag of A, then R(A, [r.sub.0]) is a bounded graded poset.
Bjorner, Edelman, and Ziegler proved that the chamber poset P(A, [c.sub.0]) of a supersolvable arrangement is a lattice if [c.sub.0] is incident to a modular flag [3]. In contrast, the gallery poset R(A, [r.sub.0]) is not a lattice in general; see Example 4.1.
The gallery poset R(A, [r.sub.0]) is defined by measuring the "difference" between two galleries with a set [L.sub.2](r, r') and ordering the galleries by single-step inclusion of the sets [L.sub.2]([r.sub.0], r). Ordering instead by ordinary inclusion defines a poset [R.sub.[subset or equal to]](A, [r.sub.0]). We do not know whether these posets are ever distinct when r0 is incident to a modular flag of A. When A is the type A braid arrangement, this question resembles a question about two versions of the second Higher Bruhat orders, B(n, 2) and [B.sub.[subset or equal to]](n, 2). These two posets were shown to be the same by Felsner and Weil [8].
5 Homotopy type of the gallery poset
Theorem 5.1 Let [r.sub.0] be a gallery incident to a modular flag of a rank d supersolvable arrangement A. The proper part of the poset of reduced galleries R(A, [r.sub.0]) is homotopy equivalent to a (d - 3)-sphere.
The main topological result we use to prove the theorem is Rambau's Suspension Lemma [9].
Lemma 5.2 (Rambau) Let P, Q be bounded posets with [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and distinguished order ideal J [subset or equal to] P. Let f: P [right arrow] Q be a surjective order-preserving map with order-preserving sections i, j: Q [right arrow] P. Assume that
1. i(Q) [subset or equal to] J and j (Q) [subset or equal to] P\J
2. ([for all]p [member of] P) (i x f)(p) [less than or equal to] p [less than or equal to] (j x f )(p)
3. [f.sup.-1]([[??].sub.q]) [intersection] J = {[[??].sub.p]} and [f.sup.-1]([[??].sub.Q]) [intersection] (P\J) = {[[??].sub.P]}
Then [bar.P] is homotopy equivalent to the suspension of [bar.Q].
Figure 5 shows an application of the Suspension Lemma to the gallery poset of the four coordinate hyperplanes in [R.sup.4].
Proof: (of theorem) We begin by defining the maps and objects in the Suspension Lemma 5.2.
Let F be a modular flag of faces
F: [c.sub.0] = [F.sub.0] < [F.sub.1] < ... < [F.sub.d] = 0, [F.sub.i] [member of] [Laplace](A),
and set l = [F.sub.d-1]. Let [r.sub.0] be the unique gallery incident to F.
Define gallery posets P = R(A, [r.sub.0]), Q = R([A.sub.l], [([r.sub.0]).sub.l]), and let f: P [right arrow] Q be usual localization map removing hyperplanes not containing l.
Define a section i: Q [right arrow] P by lifting a gallery [r.sub.l] in Q along l to a gallery from l x [c.sub.0] to l x (-[c.sub.0]) and completing it in the unique way described in Proposition 4.2. We similarly define a section j: Q [right arrow] P by lifting along -l.
Let (H,H') be the unique pair of adjacent hyperplanes in [r.sub.0] such that H [contains or equal to] l and H' [contains or not equal to] l. We claim that [X.sub.0] = H [intersection] H' is the unique codimension 2 subspace incident to [r.sub.0] not containing l. Uniqueness holds since any Y [member of] [L.sub.2](A)\[L.sub.2]([A.sub.l]) contains some hyperplane of [A.sub.l] by the modularity of l, but the hyperplanes in [A.sub.l] appear in [r.sub.0] before those of A\[A.sub.l]. Incidence at [X.sub.0] follows from Proposition 4.2(2). Let J denote the order ideal {r [member of] P | [X.sub.0] [member of] [L.sub.2]([r.sub.0], r)}.
With this setup, we verify the three properties in Lemma 5.2. (1) is clear since [L.sub.2]([r.sub.0], i(r)) is a subset of [L.sub.2]([A.sub.l]) and [L.sub.2]([r.sub.0], j(r)) is a superset of [L.sub.2](A)\[L.sub.2]([A.sub.l]). Proposition 4.3 implies i(r) is the minimum element of the fiber [f.sup.-1](r). Dually, j(r) is the maximum element of [f.sup.-1](r). This verifies (2). Finally, (3) follows from the uniqueness of [X.sub.0].
The proof of Theorem 5.1 applies to [R.sub.[subset or equal to]] with little variation, so we do not repeat it here. An alternate proof using the Crosscut Lemma also computes the homotopy type of [R.sub.[subset or equal to]], but that argument does not suffice for R.
6 Main Theorem
Theorem 6.1 Let A be a supersolvable arrangement with chamber [c.sub.0] and gallery [r.sub.0] incident to a modular flag. Let x = ([X.sub.1], [X.sub.2], ..., [X.sub.m]) [member of] [omega](A, [c.sub.0]) be a cellular string of A The set of galleries incident to x forms a closed interval of R(A, [r.sub.0]) whose proper part is homotopy equivalent to a sphere of dimension [[summation].sup.m.sub.i=1] codim([X.sub.i]) - m - 2.
To decompose the set of galleries incident to a cellular string into a product of bounded posets, we make use of the following simple lemma.
Lemma 6.2 Assume A is supersolvable with modular flag of faces F: [F.sub.0] < [F.sub.1] < ... < [F.sub.d] = 0 .If X [member of] L(A) then the localized arrangement [A.sub.X] is supersolvable with modular flag
[F.sub.X]: [([F.sub.0]).sub.X] [less than or equal to] [([F.sub.1]).sub.X] [less than or equal to] ... [less than or equal to] [([F.sub.d]).sub.X] = 0.
If [r.sub.0] is the unique gallery incident to F, then [([r.sub.0]).sub.X] is the unique gallery incident to [F.sub.X].
Proof: (of theorem) For each [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a poset of galleries of a supersolvable arrangement with base gallery incident to a modular flag. By Theorem 5.1, the proper part of R is homotopy equivalent to [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
A gallery [r.sub.i] in [R.sub.i] can be viewed as a gallery between [X.sub.i] x [c.sub.0] and [X.sub.i] x (-[c.sub.0]) in the arrangement A. Since x is a cellular string, any sequence [([r.sub.i] [member of] [R.sub.i]).sup.*] can be patched together to a gallery between [c.sub.0] and -[c.sub.0] in A. Consequently, the galleries incident to x may be indentified with the product poset
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
In general, if bounded posets P, Q are homotopy equivalent to spheres [S.sup.p], [S.sup.q], respectively, then [bar.PxQ] is homotopy equivalent to [S.sup.p+q+2]. Applying this fact to the above product completes the proof.
Example 6.3 Continuing Example 4.1, the cellular strings [omega](A, [c.sub.0]) may be identified with wiring diagrams where multiple wires may cross at a vertical section. The set of galleries incident to a given cellular string x are the simple wiring diagrams obtained by resolving multiple crossings in the diagram associated to x. The incident galleries form an interval in R(A, [r.sub.0]) whose minimum element resolves multiple crossings in colexicographic order and whose maximum element resolves in reverse colexicographic order. Figure 6 shows an example.
7 Generic lifting spaces
The chamber and gallery posets have a common generalization to a poset G(M, N) of generic singleelement liftings of an oriented matroid. The precise definition is rather technical and will be described in a separate paper. Roughly speaking, G(M, N) consists of generic affine slices of M with some prescribed boundary. When M is realizable as a hyperplane arrangement, the rank 2 slices can be identified with galleries.
Special cases of these posets were called uniform extension posets by Ziegler [13]. Ziegler observed that these posets tend to be poorly behaved. However, when the base extension is an alternating matroid, one obtains a bounded graded poset known as a Higher Bruhat order. We are interested in finding other cases where G(M, N) is a bounded graded poset with nice topology. We summarize the main results about the Higher Bruhat orders here.
A set family [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is consistent if for any (k + 2)-subset P = {[i.sub.0] < [i.sub.1] < ... < [i.sub.k+1]} of [n], the intersection [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is either {P\[i.sub.k+1], P\[i.sub.k], ..., P\[i.sub.t]} or {P\[i.sub.0], P\[i.sub.1], ..., P\[i.sub.t]} for some t. The Higher Bruhat order B(n, k) is the collection of consistent set families [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] ordered by single-step inclusion.
In [13], Ziegler proved that B(n, k) is isomorphic a poset of generic single-element liftings of the rank k alternating matroid M (n, k) on [n]. In a way similar to the gallery posets, the single-element liftings are ordered by single-step inclusion of a difference set from the affine alternating matroid M(n + 1, k + 1). Ziegler proved that B (n, k) is a bounded, graded poset. He also observed that choosing a different base lifting can determine an unbounded poset. In future work, we explain the relevance of the alternating matroid by showing it is "incident" to a modular flag of M(n, n).
Rambau showed that [bar.B](n, k) is homotopy equivalent to [S.sup.n-k-2][9]. Analogous to our conjecture on the intervals of the gallery posets, Reiner conjectured that the set of noncontracible intervals of B(n, k) is naturally parameterized by the full lifting space [omega](n, k) of M (n, k) (Conjecture 6.9(a) [11]). Just as in the discussion following Theorem 1.2, this connection would imply the previous result of Sturmfels and Ziegler that [bar.[omega]](n, k) is homotopy equivalent to [S.sup.n-k-1] [12].
Acknowledgements
The author thanks his advisor Pavlo Pylyavskyy and Vic Reiner for helpful discussions and suggestions.
References
[1] Christos A Athanasiadis, Paul H Edelman, and Victor Reiner. Monotone paths on polytopes. Mathematische Zeitschrift, 235(2):315-334, 2000.
[2] Louis J Billera, Mikhail M Kapranov, and Bernd Sturmfels. Cellular strings on polytopes. Proceedings of the American Mathematical Society, 122(2):549-555, 1994.
[3] A. Bjorner, P.H. Edelman, and G.M. Ziegler. Hyperplane arrangements with a lattice of regions. Discrete & computational geometry, 5(3):263-288, 1990.
[4] Anders Bjorner. Orderings of coxeter groups. Combinatorics and algebra, 34:175-195, 1984.
[5] Anders Bjorner. Essential chains and homotopy type of posets. Proceedings of the American Mathematical Society, 116(4):1179-1181, 1992.
[6] Anders Bjorner. Topological methods. handbook of combinatorics, vol. 1, 2, 1819-1872, 1995.
[7] Paul H Edelman and James W Walker. The homotopy type of hyperplane posets. Proceedings of the American Mathematical Society, 94(2):221-225, 1985.
[8] Stefan Felsner and Helmut Weil. A theorem on higher bruhat orders. Discrete & Computational Geometry, 23(1):121-127, 2000.
[9] Jorg Rambau. A suspension lemma for bounded posets. Journal of Combinatorial Theory, Series A, 80(2):374-379, 1997.
[10] Vic Reiner and Yuval Roichman. Diameter of graphs of reduced words and galleries. Trans. Amer. Math. Soc., 365:2779-2802, 2013.
[11] Victor Reiner. The generalized baues problem. New perspectives in algebraic combinatorics, 38:293-336, 1999.
[12] Bernd Sturmfels and Gunter M. Ziegler. Extension spaces of oriented matroids. Discrete & Computational Geometry, 10:23-45, 1993.
[13] Gunter M. Ziegler. Higher bruhat orders and cyclic hyperplane arrangements. Topology, 32:259-279, 1993.
Thomas McConville *
School of Mathematics, University of Minnesota, Minneapolis, MN 55455, USA
* Email: mcco0 4 8 9@umn.edu. Supported by RTG grant NSF/DMS-1148634
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