url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://www.numbersaplenty.com/2886817 | 1,606,333,710,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141184123.9/warc/CC-MAIN-20201125183823-20201125213823-00070.warc.gz | 788,700,637 | 3,400 | Search a number
2886817 = 9729761
BaseRepresentation
bin1011000000110010100001
312102122222011
423000302201
51214334232
6141512521
733352243
oct13006241
95378864
102886817
1116a19aa
12b72741
137a0c9b
14552093
153c0547
hex2c0ca1
2886817 has 4 divisors (see below), whose sum is σ = 2916676. Its totient is φ = 2856960.
The previous prime is 2886809. The next prime is 2886833. The reversal of 2886817 is 7186882.
It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 7186882 = 23593441.
It can be written as a sum of positive squares in 2 ways, for example, as 839056 + 2047761 = 916^2 + 1431^2 .
It is a cyclic number.
It is not a de Polignac number, because 2886817 - 23 = 2886809 is a prime.
It is an Ulam number.
It is a Duffinian number.
It is a self number, because there is not a number n which added to its sum of digits gives 2886817.
It is not an unprimeable number, because it can be changed into a prime (2886217) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 14784 + ... + 14977.
It is an arithmetic number, because the mean of its divisors is an integer number (729169).
22886817 is an apocalyptic number.
It is an amenable number.
2886817 is a deficient number, since it is larger than the sum of its proper divisors (29859).
2886817 is an equidigital number, since it uses as much as digits as its factorization.
2886817 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 29858.
The product of its digits is 43008, while the sum is 40.
The square root of 2886817 is about 1699.0635656149. The cubic root of 2886817 is about 142.3878999366.
The spelling of 2886817 in words is "two million, eight hundred eighty-six thousand, eight hundred seventeen".
Divisors: 1 97 29761 2886817 | 589 | 1,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-50 | latest | en | 0.895074 |
https://www.physicsforums.com/threads/ellipse-and-a-sine-curve.193258/ | 1,531,713,222,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589172.41/warc/CC-MAIN-20180716021858-20180716041858-00327.warc.gz | 975,511,075 | 13,533 | # Homework Help: Ellipse and a sine curve
1. Oct 23, 2007
### ehrenfest
1. The problem statement, all variables and given/known data
an ellipse whose semi axes have lengths a and b rolls without slipping on the curve y =c sin (x/a), find the relationship between a, b, and c. Assume that the ellipse completes one revolution per period of the sine curve.
The answer is b^2 = a^2 + c^2 and you find it by requiring that the arclengths be the same for one period.
Why is it wrong to just require that a = c and b = pi a /2 ? That would seem natural to me because then one half of the ellipse would fit perfectly into one "hump" of the sine curve?
2. Relevant equations
3. The attempt at a solution
Last edited: Oct 23, 2007
2. Oct 23, 2007
### ehrenfest
do people understand the problem?
3. Oct 23, 2007
### ehrenfest
should I draw a picture?
4. Oct 23, 2007
### Dick
An ellipse does not fit perfectly into a sine curve. I don't know what you are talking about.
5. Oct 23, 2007
### ehrenfest
My approach was to make the ellipse have minor axis equal to half the period of the sine curve and a semi-major axis equal to the amplitude of sine curve. All I want to know is why that approach produces ellipses that are different from the ones in the answer.
6. Oct 23, 2007
### Dick
Because they don't fit. The profile of an ellipse only resembles a sine curve. It's not an exact match.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 391 | 1,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-30 | latest | en | 0.915196 |
https://www.onlinemath4all.com/conditions-of-collinearity.html | 1,542,606,303,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745281.79/warc/CC-MAIN-20181119043725-20181119065725-00450.warc.gz | 941,348,079 | 14,433 | # CONDITIONS OF COLLINEARITY OF THREE POINTS
## About "Conditions of collinearity of three points"
Conditions of collinearity of three points :
The following are the conditions of collinearity of three points.
## Concept of slope
Let A, B and C be the three points.
If we want A, B and C be collinear, the following conditions have to be met.
(i) Slope of AB = Slope of BC
(ii) There must be a common point between AB and BC.
(In AB and BC, the common point is B)
If the above two conditions are met, then the three points A, B and C are collinear.
## Concept of distance between the two points
Let A, B and C be the three points.
We have to find the three lengths AB, BC and AC among the given three points A, B and C.
The three points A, B and C are collinear, if the sum of the lengths of any two line segments among AB, BC and AC is equal to the length of the remaining line segment.
That is,
AB + BC = AC
(or)
AB + AC = BC
(or)
AC + BC = AB
## Concept of area of triangle
Let A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) be the three points.
If the three points, A, B and C are collinear, they will lie on the same and they cannot form a triangle.
Hence, the area of triangle ABC = 0
1/2 x { (x₁y₂ + xy + xy) - (xy + xy + xy) } = 0
(or)
x₁y₂ + x₂y₃ + x₃y = xy + x₃y₂ + xy₃
## Concept of equation of line
Let A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) be the three points.
Let us find the equation through any two of the given three points. If the third point satisfies the equation, then the three points A, B and C will be collinear.
## Conditions of collinearity of three points - Practice problems
Problem 1 :
Using the concept of distance between two points, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.
Solution :
We know the distance between the two points (x₁, y₁) and (x₂, y₂) is
d = √ (x₂ - x₁) ² + (y₂ - y₁) ²
Let us find the lengths AB, BC and AC using the above distance formula.
AB = √ [(4 - 5)² + (-1 + 2)²]
AB = √ [(-1)² + (1)²]
AB = √ [1 + 1]
AB = √2
BC = √ [(1 - 4)² + (2 + 1)²]
BC = √ [(-3)² + (3)²]
BC = √ [9 + 9]
BC = √[2X9]
BC = 3√2
AC = √ [(1 - 5)² + (2 + 2)²]
AC = √ [(-4)² + (4)²]
AC = √ [16 + 16]
AC = √[2x16]
AC = 4√2
Therefore, AB + BC = √2 + 3√2 = 4√2 = AC
Thus, AB + BC = AC
Hence, the given three points A, B and C are collinear.
Let us look at the next problem on "Conditions of collinearity of three points"
Problem 2 :
Using the concept of slope, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.
Solution :
Slope of the line joining (x₁, y₁) and (x₂, y₂) is,
m = (y₂ - y₁) / (x₂ - x₁)
Using the above formula,
Slope of the line AB joining the points A (5, - 2) and B (4- 1) is
= (-1 + 2) / (4 - 5)
= - 1
Slope of the line BC joining the points B (4- 1) and C (1, 2) is
= (2 + 1) / (1 - 4)
= - 1
Thus, slope of AB = slope of BC.
And also, B is the common point.
Hence, the points A , B and C are collinear.
Let us look at the next problem on "Conditions of collinearity of three points"
Problem 3 :
Using the concept of area of triangle, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.
Solution :
Using the concept of area of triangle, if the three points A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) are collinear, then
x₁y₂ + x₂y₃ + x₃y = xy + x₃y₂ + xy₃
Comparing the given points to A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃), we get
(x₁, y₁) = (5, -2)
(x₂, y₂) = (4, -1)
(x₃, y₃) = (1, 2)
x₁y₂ + x₂y₃ + x₃y₁ = 5x(-1) + 4x2 + 1x(-2)
x₁y₂ + x₂y₃ + x₃y₁ = -5 + 8 -2
x₁y₂ + x₂y₃ + x₃y₁ = 1 --------(1)
xy + x₃y₂ + xy₃ = 4x(-2) + 1x(-1) + 5x(2)
xy + x₃y₂ + xy₃ = -8 - 1 + 10
xy + x₃y₂ + xy₃ = 1 --------(2)
From (1) and (2), we get
x₁y₂ + x₂y₃ + x₃y = xy + x₃y₂ + xy₃
Hence, the three points A, B and C are collinear.
Let us look at the next problem on "Condition of collinearity of three points"
Problem 4 :
Using the concept of equation of line, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.
Solution :
Equation of the straight line in two-points form is
(y - y₁) / (y₂ - y₁) = (x - x₁) / (x₂ - x₁)
Using the above formula, let us get equation of the line through the points A and B.
Plugging (x₁ , y₁) = (5, -2) and (x₂, y) = (4, -1), we get
(y +2) / (-1 + 2) = (x - 5) / (4 - 5)
(y + 2) / 1 = (x - 5) / (-1) ----------> y +2 = -x + 5
y +2 = -x + 5 ----------> x + y - 3 = 0
Now, we can plug the third point C(1, 2) in the above equation.
That is, plug x = 1 and y = 2
x + y - 3 = 0 ----------> 1 + 2 - 3 = 0 -------> 0 = 0
Therefore, the third point C(1, 2) satisfies the equation.
Hence, the given points three points A, B and C are collinear.
After having gone through the stuff given above, we hope that the students would have understood "Conditions of collinearity of three points".
Apart from the stuff, "Conditions of collinearity of three points", if you need any other stuff in math, please use our google custom search here.
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Sum of all three four digit numbers formed using 1, 2, 5, 6 | 2,453 | 7,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2018-47 | longest | en | 0.877542 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/64/8/e/a/ | 1,708,740,562,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474482.98/warc/CC-MAIN-20240224012912-20240224042912-00835.warc.gz | 910,618,424 | 60,970 | # Properties
Label 64.8.e.a Level $64$ Weight $8$ Character orbit 64.e Analytic conductor $19.993$ Analytic rank $0$ Dimension $26$ CM no Inner twists $2$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [64,8,Mod(17,64)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(64, base_ring=CyclotomicField(4))
chi = DirichletCharacter(H, H._module([0, 3]))
N = Newforms(chi, 8, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("64.17");
S:= CuspForms(chi, 8);
N := Newforms(S);
Level: $$N$$ $$=$$ $$64 = 2^{6}$$ Weight: $$k$$ $$=$$ $$8$$ Character orbit: $$[\chi]$$ $$=$$ 64.e (of order $$4$$, degree $$2$$, not minimal)
## 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: no Analytic conductor: $$19.9926416310$$ Analytic rank: $$0$$ Dimension: $$26$$ Relative dimension: $$13$$ over $$\Q(i)$$ Twist minimal: no (minimal twist has level 16) Sato-Tate group: $\mathrm{SU}(2)[C_{4}]$
## $q$-expansion
The dimension is sufficiently large that we do not compute an algebraic $$q$$-expansion, but we have computed the trace expansion.
$$\operatorname{Tr}(f)(q) =$$ $$26 q + 2 q^{3} - 2 q^{5}+O(q^{10})$$ 26 * q + 2 * q^3 - 2 * q^5 $$\operatorname{Tr}(f)(q) =$$ $$26 q + 2 q^{3} - 2 q^{5} - 1202 q^{11} - 2 q^{13} + 27004 q^{15} - 4 q^{17} - 60582 q^{19} + 4372 q^{21} + 233672 q^{27} - 51690 q^{29} - 357488 q^{31} - 4 q^{33} + 252004 q^{35} + 415574 q^{37} - 569754 q^{43} + 151874 q^{45} + 2076464 q^{47} - 1647090 q^{49} - 2609508 q^{51} + 907814 q^{53} + 4865142 q^{59} + 2279886 q^{61} - 8295108 q^{63} - 1426892 q^{65} + 5564458 q^{67} - 4786076 q^{69} - 6212566 q^{75} + 7604308 q^{77} + 9598912 q^{79} - 5314414 q^{81} - 4531198 q^{83} + 7377748 q^{85} - 2587652 q^{91} - 14504144 q^{93} + 4900620 q^{95} - 4 q^{97} - 18815006 q^{99}+O(q^{100})$$ 26 * q + 2 * q^3 - 2 * q^5 - 1202 * q^11 - 2 * q^13 + 27004 * q^15 - 4 * q^17 - 60582 * q^19 + 4372 * q^21 + 233672 * q^27 - 51690 * q^29 - 357488 * q^31 - 4 * q^33 + 252004 * q^35 + 415574 * q^37 - 569754 * q^43 + 151874 * q^45 + 2076464 * q^47 - 1647090 * q^49 - 2609508 * q^51 + 907814 * q^53 + 4865142 * q^59 + 2279886 * q^61 - 8295108 * q^63 - 1426892 * q^65 + 5564458 * q^67 - 4786076 * q^69 - 6212566 * q^75 + 7604308 * q^77 + 9598912 * q^79 - 5314414 * q^81 - 4531198 * q^83 + 7377748 * q^85 - 2587652 * q^91 - 14504144 * q^93 + 4900620 * q^95 - 4 * q^97 - 18815006 * q^99
## 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 $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
17.1 0 −61.7547 61.7547i 0 160.824 160.824i 0 1202.46i 0 5440.30i 0
17.2 0 −50.5872 50.5872i 0 −364.370 + 364.370i 0 182.346i 0 2931.14i 0
17.3 0 −42.2145 42.2145i 0 46.9348 46.9348i 0 1384.27i 0 1377.13i 0
17.4 0 −26.1364 26.1364i 0 −31.6175 + 31.6175i 0 444.381i 0 820.775i 0
17.5 0 −20.1662 20.1662i 0 269.345 269.345i 0 147.771i 0 1373.65i 0
17.6 0 −8.25573 8.25573i 0 63.0500 63.0500i 0 847.676i 0 2050.69i 0
17.7 0 −2.27414 2.27414i 0 −242.456 + 242.456i 0 1642.72i 0 2176.66i 0
17.8 0 12.2643 + 12.2643i 0 −210.432 + 210.432i 0 920.183i 0 1886.17i 0
17.9 0 21.1050 + 21.1050i 0 328.807 328.807i 0 874.718i 0 1296.15i 0
17.10 0 35.6625 + 35.6625i 0 50.2015 50.2015i 0 699.226i 0 356.628i 0
17.11 0 37.3394 + 37.3394i 0 −233.214 + 233.214i 0 241.506i 0 601.466i 0
17.12 0 49.4488 + 49.4488i 0 252.094 252.094i 0 1482.88i 0 2703.37i 0
17.13 0 56.5688 + 56.5688i 0 −90.1684 + 90.1684i 0 373.897i 0 4213.06i 0
49.1 0 −61.7547 + 61.7547i 0 160.824 + 160.824i 0 1202.46i 0 5440.30i 0
49.2 0 −50.5872 + 50.5872i 0 −364.370 364.370i 0 182.346i 0 2931.14i 0
49.3 0 −42.2145 + 42.2145i 0 46.9348 + 46.9348i 0 1384.27i 0 1377.13i 0
49.4 0 −26.1364 + 26.1364i 0 −31.6175 31.6175i 0 444.381i 0 820.775i 0
49.5 0 −20.1662 + 20.1662i 0 269.345 + 269.345i 0 147.771i 0 1373.65i 0
49.6 0 −8.25573 + 8.25573i 0 63.0500 + 63.0500i 0 847.676i 0 2050.69i 0
49.7 0 −2.27414 + 2.27414i 0 −242.456 242.456i 0 1642.72i 0 2176.66i 0
See all 26 embeddings
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 17.13 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
16.e even 4 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 64.8.e.a 26
4.b odd 2 1 16.8.e.a 26
8.b even 2 1 128.8.e.a 26
8.d odd 2 1 128.8.e.b 26
16.e even 4 1 inner 64.8.e.a 26
16.e even 4 1 128.8.e.a 26
16.f odd 4 1 16.8.e.a 26
16.f odd 4 1 128.8.e.b 26
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
16.8.e.a 26 4.b odd 2 1
16.8.e.a 26 16.f odd 4 1
64.8.e.a 26 1.a even 1 1 trivial
64.8.e.a 26 16.e even 4 1 inner
128.8.e.a 26 8.b even 2 1
128.8.e.a 26 16.e even 4 1
128.8.e.b 26 8.d odd 2 1
128.8.e.b 26 16.f odd 4 1
## Hecke kernels
This newform subspace is the entire newspace $$S_{8}^{\mathrm{new}}(64, [\chi])$$. | 2,631 | 5,424 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-10 | latest | en | 0.442306 |
https://www.eeeguide.com/basic-components-of-electric-circuit/ | 1,721,415,314,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514917.3/warc/CC-MAIN-20240719170235-20240719200235-00654.warc.gz | 647,415,739 | 49,294 | ## Basic Components of Electric Circuit:
The Basic Components of Electric Circuit consists of three parts: (1) energy source, such as battery or generator, (2) the load or sink, such as lamp or motor, and (3) connecting wires as shown in Fig. 1.2. This arrangement represents a simple Circuit Basics, A battery is connected to a lamp with two wires. The purpose of the circuit is to transfer energy from source (battery) to the load (lamp). And this is accomplished by the passage of electrons through wires around the circuit.
The current flows through the filament of the lamp, causing it to emit visible light. The current flows through the battery by chemical action. A closed circuit is defined as a circuit in which the current has a complete path to flow. When the current path is broken so that current cannot flow, the circuit is called an open circuit.
More specifically, interconnection of two or more simple circuit elements and Basic Components of Electric Circuit are namely. voltage sources, resistors, inductors.
If a network contains at least one closed path, it is called an electric circuit. By definition, a simple circuit element is the mathematical model of two terminal electrical devices, and it can be completely characterised by its voltage and current. Evidently then, a physical circuit must provide means for the transfer of energy.
Broadly, Classification of Network Elements divided into four groups, viz.
• Active and Passive
• Unilateral and Bilateral
• Linear and Nonlinear Elements
• Linear and Nonlinear Elements
### Active and Passive:
Energy sources (voltage or current sources) are active elements, capable of delivering power to some external device. Passive elements are those which are capable only of receiving power. Some passive elements like inductors and capacitors are capable of storing a finite amount of energy, and return it later to an external element. More specifically, an active element is capable of delivering an average power greater than zero to some external device over an infinite time interval. For example, ideal sources are active elements. A passive element is defined as one that cannot supply average power that is greater than zero over an infinite time interval. Resistors, capacitors and inductors fall into this category.
### Unilateral and Bilateral:
In the bilateral element, the voltage-current relation is the same for current flowing in either direction. In contrast, a unilateral element has different relations between voltage and current for the two possible directions of current. Examples of bilateral elements are elements made of high conductivity materials in general. Vacuum diodes, silicon diodes, and metal rectifiers are examples of unilateral elements.
### Linear and Nonlinear Elements:
An element is said to be linear, if its voltage-current characteristic is at all times a straight line through the origin. For example, the current passing through a resistor is proportional to the voltage applied through it, and the relation is expressed as V α I or V = IR. A linear element or network is one which satisfies the principle of superposition, i.e. the principle of homogeneity and additivity. An element which does not satisfy the above principle is called a nonlinear element.
### Lumped and Distributed Elements:
Lumped elements are those elements which are very small in size and in which the Typical lumped elements are capacitors, resistors, inductors and transformers. Generally the elements are considered as lumped when their size is very small compared to the wave length of the applied signal. Distributed elements, on the other hand, are those which are not electrically separable for analytical purposes. For example, a transmission line which has distributed resistance, inductance and capacitance along its length may extend for hundreds of miles. | 755 | 3,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-30 | latest | en | 0.932421 |
https://avertigoland.com/2021/03/how-train-chess-as-you-train-to-play-soccer-1/ | 1,725,849,156,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00508.warc.gz | 97,858,858 | 45,015 | March 22, 2021
# How train Chess as you train to play Soccer (1)
In Spain the king sport is football. Thousands of children play this sport every day in schools, on weekends, are federated, play in clubs, etc. The only sport that has a king, a true king, however is chess. This article (and successive) explains how to train chess to children as if it were a football training.
I’ll explain how the pieces move and a task that can be done in class for students to learn the movement of these pieces. These tasks will be equivalent to touching the ball performed by football players. In training a professional footballer at least 10-15 people spend their time touching the ball. Chess is the equivalent of moving the piece, making a move.
Here are some silver homework so you can practice with students in class. Explaining a concept explains how it is done by a task.
## Kicking a ball in front of moving the pieces
The rules of chess are taught on the first day of school. In football they teach you the ball that day, tell you what the dimensions of the field are, what are the rules, what are the positions etc..
All the pieces in chess have rules of movement that cannot be exchanged. The bishops move diagonally, the towers in rows and columns, the L-shaped horses, the queen in all directions, as well as the king, but the latter only 1 in 1 square except when performing the redness. Pawns move forward and capture to the side if there is a rival piece.
I was thinking about how I could get the children to machine the movement of the pieces. Seeing a football team train, I came up with a way to bring that to the chess I present in this article and following. Training chess is very important in the initial learning of a future chess player.
Any player devotes much of that training to touching the ball. They’re 10-15 minutes playing the ball, passing the ball to a teammate, receiving, passing the ball again, they’re continually kicking a ball.
How can we take that to Chess? We’re going to imagine that touching the ball is equivalent to touching a piece, and kicking a ball is making a chess move. Training chess is mechanizing plays.
## First move or move, starting to train chess
In this image you can see the possible start plays of a chess game. Pawns move one or two squares, and horses can also move from their exit box because they jump.
### How a pawn moves
The pawn is the soul of chess. It’s the infantry, the soldiers trying to get into enemy terrain.
The pawn always moves forward and captures an enemy piece diagonally into the enemy field. You can’t jump on the pieces on the same side. He can’t turn around either, he can’t move backwards. From your starting position, the second row, you can advance 1 or 2 squares. If a pawn is able to reach the 8 row it is crowned, that is, it can change shape and can take the form of a horse, a bishop, a tower or a lady. He can never become king, because there is only one king.
On an empty board with only one pawn, ask your students to move it in column “a” to the coronation box. With the same pawn you put it in column “b” and do the same. And so with all the columns. Repeat that exercise with all columns a total of 5 times.
Automatism performed:
• 6 pawn movements in a column
• 8 columns.
• Repeated 5 times.
Total moves made: 7 x 8 x 5 x 280 moves.
#### Flash game
Ask students to do so as quickly as possible and measure the time it takes each to complete all 280 moves. Give an award to those who are the top 3.
#### Lazy game
Now ask students to do the same exercise but in slow motion.
### How a bishop moves
The bishop would be the equivalent of archers. They attack long distances with arrows or other sharp objects. Their power lies in that they cut the rival territory like knives.
The way to move a bishop is diagonally and over long distances. Capture a piece of the opponent taking his place. You can’t skip the pieces on the same side. You can move in all allowed directions.
On an empty board only with a bishop ask your students to move it diagonally “a1-h8” round trip(14 moves). Now place the bishop in a2 and perform the same operation. In this case you will be able to move in 2 diagonals, a2-g8 and a2-b1 (14 moves). Perform successively by changing the starting point in column a, from a1 to a8. Repeat this exercise with all other columns a total of 3 times.
#### Set of 100 meters
With a stopwatch, ask students to do the exercise and in a 1-minute time measure which student is the fastest. They have to start at a1, go up column a and then switch to column b. The one in the column and the largest number wins.
#### Game the tennis bishop
On the same board, put a student in front of another student. One that starts with a bishop in a1 and the other a8 with a black bishop. The student with bishop a1 can only use the black boxes. The student with the bishop in a8 can only use the white boxes. Before
#### Game the great wall and small wall
This time we ask students to do the same as in task 2, move a bishop from a1 but this time we put pieces in the middle. If it is a piece of the same color the bishop cannot skip that piece (the great wall). But if it is a piece of another color the capture and continues forward (the murallite).
Example will be this video
## Summary of train chess
Over the next few days I am going to upload articles on how to train chess as if it were soccer. In this first chapter we talk about the pawn and bishops and how you can train with them.
If you are a chess monitor and you carry out what I just put, please I would love for you to leave a comment to know how you have done and thus enrich this article. You can also contribute your own ideas. Thanks.
### Avelino Dominguez
??? Biologist ??? Teacher ??? Technologist ? Statistician ? #SEO #SocialNetwork #Web #Data ♟Chess ? Galician
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### The match that everyone expects CLASH OF CLAIMS: Vladimir Kramnik vs José Martínez Alcántara
The world of chess has lived…
### Cheating in online chess threatens to destroy the integrity of the game
In the complicated world of chess,… | 1,443 | 6,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.971889 |
http://www.maths.usyd.edu.au/ub/sums/puzzlehunt/2014/solutions/A1S3_Cancelled | 1,566,360,487,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315750.62/warc/CC-MAIN-20190821022901-20190821044901-00149.warc.gz | 284,846,451 | 2,502 | Home Register Teams Puzzles Solve Rules FAQ Credits Archives
### Solutions for Act I Scene 3 - Cancelled
The answer is: firefly This puzzle provides a long list of fractions and not much else. The title is probably the best place to begin. "Cancelled" heavily implies we want to cancel these fractions somehow, and they certainly all seem to be not yet in their most simplified forms. But converting them to simplest forms doesn't look particularly promising, and isn't really emphasising cancellation. At this point it might seem prudent to turn to the smaller/simpler-looking fractions. The 16/64 and 19/95 examples simplify to 1/4 and 1/5 respectively, which is interesting as these simpler fractions could also have been achieved by deleting the common digits. In fact these pairs might be recognisable to some as the source of an age-old maths joke to the same effect, claiming you can cancel digits to simplify fractions. So it seems possible this trick can be applied to all the provided fractions. Note for example that 154/451 = 14/41, and 712/4717 = 72/477 (= 8/53). In some cases we have to cancel two digits, e.g. 3514/26104 = 35/260. In other rare cases there is actually no way to cancel digits and keep the same fraction - the very first example doesn't even have common digits between the numerator and denominator (933/7464). At any rate we can note that in the cases where two digits are removed, they always appear in the same order on the top and bottom, and the first digit is always a 1 or 2. In fact when concatenated, we can see there is no fraction whose removed digits make a number exceeding 26. So it seems likely we want to map the cancelled digits to their positions in the alphabet. Doing so, we can treat "uncancellable" fractions as spaces. For example, the first four fractions are 3514/26104 = 35/260, 1575/1953 = 75/93, 2346/24378 = 46/478, and 933/7464 cannot be cancelled. The cancelled digits (concatenated) in each case are 14, 15, 23, [none], which would map to the letters "NOW ". Altogether this gives the message NOW ADD A HALF TO EACH UNUSED FRACTION. So let's follow this instruction. We must make the assumption here that when adding a half, the denominator remains the same even if cancellation could occur. Thus our first unused (space) fraction 933/7464 becomes 933/7464 + 1/2 = 4665/7464. This fraction now can be digit-cancelled, since 4665/7464 = 465/744. Doing the same trick to the other unused fractions, we get all up that 4665/7464 = 465/744, 7936/8928 = 736/828, 4186/6188 = 46/68, 3757/6358 = 377/638, 8632/9628 = 832/928, 7172/9128 = 77/98, 12586/23548 = 186/348. This time the cancelled digits correspond to the letters FIREFLY, an infamously unfairly-cancelled television show. | 710 | 2,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-35 | latest | en | 0.933993 |
http://www.cinematography.com/index.php?showtopic=33270 | 1,477,404,009,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720153.61/warc/CC-MAIN-20161020183840-00316-ip-10-171-6-4.ec2.internal.warc.gz | 372,170,574 | 14,920 | # Logarithmic vs Linear???
5 replies to this topic
### #1 Stephen Price
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Posted 02 September 2008 - 11:03 AM
I have heard Panavision's Genesis uses logarithmic processing, other cameras, such as the Red One use a linear system. From what I think I understand, they are two different algorithm codes used to process and store the dynamic range of the sensor output.
Is there any truth in this? What do these terms referrer to? What is the difference between these? And what are their advantages?
Thanks
SP
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### #2 Phil Rhodes
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Posted 02 September 2008 - 06:58 PM
This is actually quite a complex subject because things which are described as linear rarely are and things that are described as logarithmic even more rarely are.
So to simplify, in essence, and technical nitpickers don't jump on me for this, it works as follows.
A linear representation of something - the loudness of a sound, the brightness of a light - will double in value as the amount of the thing it's measuring doubles. As the pressure waves creating a sound double in intensity, the number used to represent them doubles. As the light gets twice as bright, the number used to represent it doubles. A light of 10 candela is (overlooking some minutiae) twice as bright as one of 5. Straightforward.
In terms of imaging, this means that if that an object in an image has an apparent luminosity due to reflected light from the environment of, say, 10 candela, it might be represented as the digital number 100. An object under twice as much light, or with half the surface absorbtion (being a lighter colour), would appear to have a luminosity of 20 candela, and might be represented as the digital number 200.
So far so simple.
(this is complicated by the fact that humanity doesn't have any image sensors that are anywhere near linear in terms of electrical or chemical output per number of incident photons, but ignore that for now)
The problem with this is that human visual system is very much better at discerning the differences between comparatively bright objects than it is at discerning the differences between comparatively dim objects. See here for a graph of perceived against actual brightness - at the dim end, the curve is almost vertical, indicating that we don't see the difference between "really dim" and "not quite so really dim" very well; whereas, we see the difference between "bright" and "really bright" much better.
The problem with linear storage for cinematography is that it uses the same number of digital bits to store the difference between 1 and 2 candela as it uses to store the difference between 99 and 100 candela, so we end up wasting bits encoding brightness changes we can't see. In fact, as that graph I linked shows, for every time we double how bright something actually is, it only looks about 1.5 times brighter, and even more so at the "comparatively dim" end.
Clearly the apparent fidelity of a reproduced image - "image quality" - can be improved for no storage penalty by allocating bits more wisely. The logarithmic function is one way of doing this and it is quite easy to understand; Wikipedia explains it well but briefly, the log(n) of a given number is what it must be raised to in order to reproduce the number, so the base-ten logarithm of 1000 is 3 (10*10*10 or 10^3 is 1000). This is obviously handy for representing big numbers in small space and it allocates bits a lot more as we'd like them to be allocated.
So back to our example; we now have our 10 candela object being represented as log(10) 100, which is 2; our 20 candela object being represented as log(10) 200, which is about 2.3, and a hypothetical 30 candela object being represented as almost 2.78. As you can see the numbers get much closer together the brighter the object is; a log(10) 1000 is only 3, which in our example would be a 100 candela object through a perfectly linear imaging system. A tenfold increase in linear brightness is represented in a 1.0 deviation in the logarithm, so we end up concentrating data in the brighter regions of the image just as we wanted.
This is greatly complicated in practice by the tendency of manufacturers to use custom "log" functions in their camera and processing products which often approximate the shape of a log graph (look at the wiki article again), but include proprietary deviations of the manufacturer's own divising which are intended to enhance the performance of the device. Add to this the fact that no imaging device is perfectly linear in response to light in any case, and the processing you choose to apply to log (or indeed linear) images tends to be based partially on mathematics and partially on what makes a subjectively nice picture.
P
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### #3 Satsuki Murashige
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Posted 03 September 2008 - 01:13 AM
To add to Phil's excellent post, here's a link to several articles that discuss log vs. linear gamma: http://prolost.blogs...l/Image Nerdery
I just wish I understood it all!
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### #4 Hal Smith
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Posted 03 September 2008 - 03:36 AM
Sound perception works exactly the same. Decibels are the logarithmic ratio of the intensity of two sounds. For voltage: 20 X Log a/b, for power: 10 X Log a/b. (The technically knowledgeable will know why the 10 versus 20, power is proportional to the square of voltage so Log Voltage a/b^2 becomes 2 X Log a/b and therefore Decibel (voltage) = 2 X Decibel (power). The Great Designer knew to build us logarithmically to be able to handle the huge variation of sensory input intensity in the natural environment (Yah, I know about evolution and don't disagree but being an engineer I like the idea of a Great Designer).
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### #5 Paul Bruening
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Posted 03 September 2008 - 12:12 PM
The Great Designer knew to build us logarithmically to be able to handle the huge variation of sensory input intensity in the natural environment (Yah, I know about evolution and don't disagree but being an engineer I like the idea of a Great Designer).
Me, too, Hal. Me, too.
Linear is computers. But, log is people. It's peeeeeopllllllllllllllllle!!
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### #6 Stephen Price
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Posted 03 September 2008 - 04:15 PM | 1,553 | 6,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-44 | longest | en | 0.955217 |
https://madhavamathcompetition.com/category/indian-statistical-institute-entrance-exam/page/3/ | 1,643,416,684,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299894.32/warc/CC-MAIN-20220129002459-20220129032459-00155.warc.gz | 419,172,988 | 34,328 | Uses of mathematical induction to prove inequalities
Some classic examples are presented below to illustrate the use of mathematical induction to prove inequalities:
Example 1:
Prove the inequality $n<2^{n}$ for all positive integers n.
Proof 1:
Let $P(n)$ be the proposition that $n<2^{n}$.
Basis step:
$P(1)$ is true, because $1<2^{1}=2$. This completes the basis step.
Inductive step:
We first assume the inductive hypothesis that $P(k)$ is true for the positive integer k. That is, the inductive hypothesis $P(k)$ is the statement that $k<2^{k}$. To complete the inductive step, we need to show that if $P(k)$ is true, then $P(k+1)$, which is the statement that $k+1<2^{k+1}$ is also true. That is, we need to show that if $k<2^{k}$, then $k+1<2^{k+1}$. To show that this conditional statement is true for the positive integer k, we first add 1 to both sides of $k<2^{k}$ and then note that $1\leq 2^{k}$. This tells us that
$k+1<2^{k}+1 \leq 2^{k} + 2^{k} = 2.2^{k}=2^{k+1}$.
This shows that $P(k+1)$ is true; namely, that $k+1<2^{k+1}$, based on the assumption that $P(k)$ is true. The induction step is complete.
Therefore, because we have completed both the basis step and the inductive step, by the principle of mathematical induction we have shown that $n<2^{n}$ is true for all positive integers n. QED.
Example 2:
Prove that $2^{n} for every positive integer n with $n \geq 4$. (Note that this inequality is false for $n=1, 2, 3$).
Proof 2:
Let $P(n)$ be the proposition that $2^{n}
Basis Step:
To prove the inequality for $n \geq 4$ requires that the basis step be $P(4)$. Note that $P(4)$ is true because $2^{4} =16<24=4!$.
Inductive Step:
For the inductive step, we assume that $P(k)$ is true for the positive integer k with $k \geq 4$. That is, we assume that $2^{k} with $k \geq 4$. We must show that under this hypothesis, $P(k+1)$ is also also true. That is, we must show that if $2^{k} for the positive integer $k \geq 4$, then $2^{k+1}<(k+1)!$. We have
$2^{k+1}=2.2^{k}$ (by definition of exposition)
which $<2.k!$ (by the inductive hypothesis)
which $<(k+1).k!$ because $2
which equals $(k+1)!$ (by definition of factorial function).
This shows that $P(k+1)$ is true when $P(k)$ is true. This completes the inductive step of the proof.
We have completed the basis step and the inductive step. Hence, by mathematical induction $P(k)$ is true for all integers n with $n \geq 4$. That is, we have proved that $2^{n} is true for all integers n with $n \geq 4$. QED.
An important inequality for the sum of the reciprocals of a set of positive integers will be proved in the example below:
Example 3:
An inequality for Harmonic Numbers
The harmonic numbers $H_{j}, j=1,2,3, \ldots$ are defined by $H_{j}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{j}$.
For instance, $H_{4}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{25}{12}$. Use mathematical induction to show that $H_{2*{n}} \geq 1 + \frac{n}{2}$, whenever n is a nonnegative integer.
Proof 3:
To carry out the proof, let $P(n)$ be the proposition that $H_{2^{n}} \geq 1 + \frac{n}{2}$.
Basis Step:
$P(0)$ is true because $H_{2^{0}}=H_{1}=1 \geq 1 + \frac{0}{2}$.
Inductive Step:
The inductive hypothesis is the statement that $P(1)$ is true, that is, $H_{2^{k}} \geq 1+\frac{k}{2}$, where k is a nonnegative integer. We must show that if $P(k)$ is true, then $P(k+1)$, which states that $H_{2^{k+1}} \geq 1 + \frac{k+1}{2}$, is also true. So, assuming the inductive hypothesis, it follows that
$H_{2^{k+1}}=1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{2^{k}}+\frac{1}{2^{k}+1}+\ldots + \frac{1}{2^{k+1}}$ (by the definition of harmonic number)
which equals $H_{2^{n}}+\frac{1}{2^{k}+1} + \ldots + \frac{1}{2^{k+1}}$ (by the definition of the $2^{k}th$ harmonic number)
$\geq (1+\frac{k}{2})+\frac{1}{2^{k}+1}+\ldots + \frac{1}{2^{k+1}}$ (by the inductive hypothesis)
$\geq (1+\frac{k}{2}) + 2^{k}\frac{1}{2^{k+1}}$ (because there are $2^{k}$ terms each greater than or equal to $\frac{1}{2^{k+1}}$)
$\geq (1+\frac{k}{2})+\frac{1}{2}$ (canceling a common factor of $2^{k}$ in second term), which in turn, equals $1+ \frac{k+1}{2}$.
This establishes the inductive step of the proof.
We have completed the basis step and the inductive step. Thus, by mathematical induction $P(n)$ is true for all nonnegative integers. That is, the inequality $H_{2^{n}} \geq 1 + \frac{n}{2}$ for the harmonic number is valid for all non-negative integers n. QED.
Remark:
The inequality established here shows that the harmonic series $1+\frac{1}{2}+\frac{1}{3}+\ldots \frac{1}{n}+\ldots$ is a divergent infinite series. This is an important example in the study of infinite series.
More later,
Nalin Pithwa
Reference: Discrete Mathematics and its Applications by Kenneth H. Rosen, Seventh Edition.
Heights and Distances — II — problems for RMO and IITJEE maths
This is one of the prime utilities of trigonometry —- to calculate heights and distances, called as surveying in civil engineering.
1. From the extremities of a horizontal base line AB, whose length is 1 km, the bearings of the foot C of a tower are observed and it is found that $\angle {CAB}$ is 56 degrees and 23 minutes and $\angle{CBA}$ is 47 degrees and 15 minutes, and the elevation of the tower from A is 9 degrees and 25 minutes, find the height of the tower.
2. A man in a balloon observes that the angle of depression of an object on the ground bearing due north is 33 degrees; the balloon drifts 3 km due west and the angle of depression is now found to be 21 degrees. Find the height of the balloon.
3. A tower PN stands on level ground. A base AB is measured at right angles to AN, the points A, B, and N being in the same horizontal plane, and the angles PAN and PBN are found to be $\alpha$ and $\beta$ respectively. Prove that the height of the tower is
AB$\frac{\sin {\alpha}\sin{\beta}}{\sqrt{\sin{(\alpha-\beta)}\sin{(\alpha+\beta)}}}$
If AB is 100m, $\alpha = 70$ degrees and $\beta = 50$ degrees, calculate the height.
4. At each end of a horizontal base of length 2a, it is found that the angular height of a certain peak is $\theta$ and that at the middle point it is $\phi$. Prove that the vertical height of the peak is
$\frac{a \sin {\theta}\sin{\phi}}{\sqrt{\sin{(\phi+\theta)}\sin{(\phi-\theta)}}}$
5. To find the distance from A to P, a distance AB of 1 km. is measured for a convenient direction. At A the angle PAB is found to be 41 degrees 18 min and at B, the angle PBA is found to be 114 degrees and 38 min. What is the required distance to the nearest metre?
More later,
Nalin Pithwa
Heights and Distances — I — problems for RMO and IITJEE Maths
1. A man observes that at a point due south of a certain tower its angle of elevation is 60 degrees. He then walks 100 meters due west on a horizontal plane and finds that the angle of elevation is 30 degrees. Find the height of the tower and his original distance from it.
2. At the foot of a mountain the elevation of its summit is found to be 45 degrees; after ascending 1000 m towards the mountain up a slope of 30 degree inclination, the elevation is found to be 60 degrees. Find the height of the mountain.
3. A square tower stands upon a horizontal plane, from which three of its upper corners are visible, their angular elevations are respectively 45 degrees, 60 degrees and 45 degrees. Show that the height of the tower is to the breadth of one of its sides as $\sqrt{6}(\sqrt{5}+1)$ to 4.
4. A lighthouse, facing north, sends out a fan-shaped beam of light extending from north-east to north-west. An observer on a steamer, sailing due west, first sees the light when is 5 km, away from the lighthouse and continues to see it for $30\sqrt{2}$ minutes. What is the speed of the steamer?
5. A man stands at a point X on the bank XY of a river with straight and parallel banks, and observes that the line joining X to a point Z on the opposite bank makes an angle of 30 degrees with XY. He then goes along the bank a distance of 200 meters to Y and finds that the angle ZYX is 60 degrees. Find the breadth of the river.
6. A man, walking due north, observes that the elevation of a balloon, which is due east of him and is sailing toward the north-west, is then 60 degrees; after he has walked 400 meters the balloon is vertically over his head; find its height supposing it to have always remained the same.
The above are some tricky trig problems ! What is the main trick ? I will suggest a very simple hint: draw as good diagrams as possible!
Nalin Pithwa
Combinatorics for RMO and IITJEE maths
Problem:
How many arrangements of 5 $\alpha$‘s, 5 $\beta$‘s and $\gamma$‘s are there with at least one $\beta$ and at least one $\gamma$ between each successive pair of $\alpha$‘s?
Solution:
There are three cases:
1. Exactly one $\beta$ and one $\gamma$ between each pair of $\alpha$‘s: Between each of the four pairs of $\alpha$‘s, the $\beta$ or the $\gamma$ can be first — $2^{4}$ ways. The fifth $\beta$ and fifth $\gamma$ along with the sequence of the rest of the letters can be considered as 3 objects to be arranged — $3!$ ways. Altogether, $2^{4} \times 3!=96$ ways.
2. Exactly, one $\beta$ between each pair of $\alpha$‘s and two $\gamma$‘s between some pair of $\alpha$‘s (or two $\beta$‘s between some pair of $\alpha$‘s and exactly one $\gamma$ between each pair of $\alpha$‘s): there are four choices for between which pair of $\alpha$‘s the two $\gamma$‘s go and 3 ways to arrange the two $\gamma$‘s and one $\beta$ there. There are two $2^{3}$ choices for whether the $\beta$ or the $\gamma$ goes first between the other 3 pairs of $\alpha$‘s and 2 choices for at which end of the arrangement the fifth $\beta$ goes. Multiplying by 2 for the case of two $\beta$‘s between some pair of $\alpha$‘s, we obtain $2 \times (4 \times 3 \times 2^{3} \times 2)=384$ ways.
3. Two $\beta$‘s between some pair of $\alpha$‘s and two $\gamma$‘s between some pair of $\alpha$‘s. There are two subcases. If the two $\beta$‘s and two $\gamma$‘s are between the same pair of $\alpha$‘s, there are 4 choices for which pair of $\alpha$‘s, $C(4,2)$ ways to arrange them between this pair of $\alpha$‘s, and $2^{3}$ choices for whether the $\beta$ or the $\gamma$ goes first between the other 3 pairs of $\alpha$‘s. If two $\beta$‘s and two $\gamma$‘s are between the different pairs of $\alpha$‘s, there are $4 \times 3$ ways to pick between which $\alpha$‘s the two $\beta$‘s and then between which $\alpha$‘s the two $\gamma$‘s go, $3^{2}$ ways to arrange the two $\gamma$‘s and one $\beta$ and to arrange the one $\gamma$ and two $\beta$‘s, and $2^{2}$ choices for whether the $\beta$ or the $\gamma$ goes first between the other 2 pair of $\alpha$‘s. Together, $4 \times C(4,2) \times 2^{3} + 4 \times 3 \times 3^{2} \times 2^{2}=1056$ ways.
All together, the three cases give us a total of $96+384+1056=1536$ arrangements.
More later,
Nalin Pithwa
The Pigeon Hole Principle : Practice Problems
Problem 1:
A bag contains beads of two colours: black and white. What is the smallest number of beads which must be drawn from the bag, without looking, so that among these beads, there are two of the same colour?
Problem 2:
One million pine trees grow in a forest. It is known that no pine tree has more than 600000 pine needles on it. Show that two pine trees in the forest must have the same number of pine needles.
Problem 3:
Given twelve integers, show that two of them can be chosen whose difference is divisible by 11.
Nalin Pithwa
Basic Mathematical Logic I for IITJEE and Math and Physics Olympiads
It is necessary to think clearly and to communicate precisely. Whereas in the arts and other disciplines, there is no room for ambiguity in the mathematical sciences. Here’s a primer on logic for aspirants of IITJEE, and Mathematics and Physics Olympiads.
Reference:
1) Introduction to General Topology by K.D. Joshi
2) Topology by Munkres
Mathematics is a language. For example, language of physics. If Math is regarded as a language, then logic is its grammar. In other words, logical precision has the same importance in Math as grammatical correctness in a language.
I. Statements and their Truth Values:
A statement is a declarative sentence, conveying a definite meaning, which may be either true or false but not both simultaneously. Incomplete sentences, questions and exclamations are not statements.
Some examples of statements are:
i) John is intelligent.
ii) If there is life on Mars, then the postman delivers a letter.
iii) Either grandmother chews gum, or missiles are costly.
iv) Every man is mortal.
v) All men are mortal.
vi) There is a man, who is eight feet tall.
vii) Every even integer greater than 2 can be expressed as a sum of 2 prime numbers.
viii) Every man with six legs is intelligent.
Some remarks on the above:
Statements 4 and 5 are statements about all the members of a class. So, in order to produce a counter-example, we say “there exists a man who is not mortal” or in polished English, “there exists a man who is immortal”. By truth value of a statement, we mean, unambiguously, whether it is true or false (no grey areas here; it’s all black and white!!), Truth value is known to electrical engineers or computer scientists/engineers as digital logic or binary logic. (Actually, math students will recall here boolean laws of thought or set theory). A conjecture is a statement whose truth value is not known at present. Thus, statement (vii) is the famous Goldbach conjecture. Statement six talks about a six-legged man, and of course, there is no such man (at least on earth!) and hence, we cannot produce a counter example, and hence, the statement is said to be vacuously true. Note also says that statement (ii) sounds strange. This statement is only mathematical “if then” statement; it is not a statement of cause and effect in the physical universe subject to laws of physics/chemistry/biology ! So, also the next statement.
Can the following be valid statements?
• Will it rain
• Oh! those heavy rains.
• I am telling you a lie.
Out of the above, the first two are not statements, obviously. The third statement refers to itself, and hence, is circular, and is called a paradox.
II) Negation, Conjunction, Disjunction and their values:
IIa) Negation: To negate a statement, technically speaking, simply put a NOT in front of the whole statement. The negation of a statement is its counter-example.
Here’s an interesting example: consider the statement, “John is very intelligent”. The negation is not “John is very dull”. The original statement refers to degrees of intelligence. So, the correct negation is “John is not very intelligent”.
Consider the following statement: For every $x \in A$,, statement P holds. The negation of this statement is as follows: For at least one $x \in A$, statement P does not hold. Equivalently, there exists some $x \in A$ such that statement P does not hold.
IIb) Disjunction or the meaning of “OR”:
In ordinary, everyday, colloquial English, the word “OR” is ambiguous. Sometimes, the statement “P or Q” means “P or Q or both” and sometimes, it means “P or Q but not both”. (EEs and CS engineers know this as inclusive OR and Exclusive OR). Usually, one decides from the context, which meaning is intended. For example, suppose I spoke two students as follows:
“Mr. Smith, every student registered for this course has taken either a course in linear algebra or a course in analysis.”
“Mr. Jones, either you get a grade of at least 70 on the final exam, or you will flunk this course.”
In the context, Mr. Smith knows perfectly well that I mean ” everyone has had linear algebra or analysis or both”, and Mr. Jones knows I mean “either he gets at least 70 or he flunks, but not both.” Indeed, Mr. Jones would be exceedingly unhappy if both statements turned out to be true!!
In math, one cannot tolerate such ambiguity. In math, “or” always means “P or Q or both”. If we mean “P or Q or both, but not both”, we have to state it explicitly.
Meaning of “if …then”:
In everyday English, a statement of the form “if…then” is ambiguous. It always means if P is true, then Q is true. Sometimes, that is all it means, other times it means something more; that, if P is false, Q must be false. Usually, one decides from the context which interpretation is correct.
Examples:
“Mr. Smith, if any student registered for this course has not taken a course in linear algebra, then he has taken a course in analysis.”
“Mr. Jones, if you get a grade below 70 on the final, you are going to flunk this course.”
In the context, Mr. Smith understands that if a student in this course has not had linear algebra, then he has taken analysis, but if he has had linear algebra, he may or may not have taken analysis as well. And, Mr. Jones knows that if he gets a grade below 70, he will flunk the course, but if he gets a grade of at least 70, he will pass.
In math, if p, then q means the following: if p is true, q is true. But, if p is false, q may be either true or false.
Contrapositive of a statement:
A statement of the form “if p, then q” is same as “if NOT q, then NOT p”. The latter is called its contrapositive. Many a times, it is easier to prove the contrapositive of a statement rather than the original statement!!
Example: If $x < 0$, then $x^{3} \neq 0$.
Contrapositive: if $x ^{3}=0$, then it is not true that $x > 0$.
Example: If $x^{2}<0$, then $x=23$.
Contrapositive: if $x \neq 23$, then it is not true that $x^{2}<0$
Converse of a statement:
The converse of a statement of the form “if p, then q” is “if q, then p”. Note that a statement and its converse are not the same.
Example: If a function is differentiable, then it is continuous. But, the converse is true. (that, if a function is continuous, it is differentiable).
Note: A definition is always if and only if; that is, if p, then q AND if q, then p.
Real Numbers, Sequences and Series: part 9
Definition.
We call a sequence $(a_{n})_{n=1}^{\infty}$ a Cauchy sequence if for all $\varepsilon >0$ there exists an $n_{0}$ such that $|a_{m}-a_{n}|<\varepsilon$ for all m, $n > n_{0}$.
Theorem:
Every Cauchy sequence is a bounded sequence and is convergent.
Proof.
By definition, for all $\varepsilon >0$ there is an $n_{0}$ such that
$|a_{m}-a_{n}|<\varepsilon$ for all m, $n>n_{0}$.
So, in particular, $|a_{n_{0}}-a_{n}|<\varepsilon$ for all $n > n_{0}$, that is,
$a_{n_{0}+1}-\varepsilon for all $n>n_{0}$.
Let $M=\max \{ a_{1}, \ldots, a_{n_{0}}, a_{n_{0}+1}+\varepsilon\}$ and $m=\min \{ a_{1}, \ldots, a_{n_{0}+1}-\varepsilon\}$.
It is clear that $m \leq a_{n} \leq M$, for all $n \geq 1$.
We now prove that such a sequence is convergent. Let $\overline {\lim} a_{n}=L$ and $\underline{\lim}a_{n}=l$. Since any Cauchy sequence is bounded,
$-\infty < l \leq L < \infty$.
But since $(a_{n})_{n=1}^{\infty}$ is Cauchy, for every $\varepsilon >0$ there is an $n_{0}=n_{0}(\varepsilon)$ such that
$a_{n_{0}+1}-\varepsilon for all $n>n_{0}$.
which implies that $a_{n_{0}+1}-\varepsilon \leq \underline{\lim}a_{n} =l \leq \overline{\lim}a_{n}=L \leq a_{n_{0}+1}+\varepsilon$. Thus, $L-l \leq 2\varepsilon$ for all $\varepsilon>0$. This is possible only if $L=l$.
QED.
Thus, we have established that the Cauchy criterion is both a necessary and sufficient criterion of convergence of a sequence. We state a few more results without proofs (exercises).
Theorem:
For sequences $(a_{n})_{n=1}^{\infty}$ and $(b_{n})_{n=1}^{\infty}$.
(i) If $l \leq a_{n} \leq b_{n}$ and $\lim_{n \rightarrow \infty}b_{n}=l$, then $(a_{n})_{n=1}^{\infty}$ too is convergent and $\lim_{n \rightarrow \infty}a_{n}=l$.
(ii) If $a_{n} \leq b_{n}$, then $\overline{\lim}a_{n} \leq \overline{\lim}b_{n}$, $\underline{\lim}a_{n} \leq \underline{\lim}b_{n}$.
(iii) $\underline{\lim}(a_{n}+b_{n}) \geq \underline{\lim}a_{n}+\underline{\lim}b_{n}$
(iv) $\overline{\lim}(a_{n}+b_{n}) \leq \overline{\lim}{a_{n}}+ \overline{\lim}{b_{n}}$
(v) If $(a_{n})_{n=1}^{\infty}$ and $(b_{n})_{n=1}^{\infty}$ are both convergent, then $(a_{n}+b_{n})_{n=1}^{\infty}$, $(a_{n}-b_{n})_{n=1}^{\infty}$, and $(a_{n}b_{n})_{n=1}^{\infty}$ are convergent and we have $\lim(a_{n} \pm b_{n})=\lim{(a_{n} \pm b_{n})}=\lim{a_{n}} \pm \lim{b_{n}}$, and $\lim{a_{n}b_{n}}=\lim {a_{n}}\leq \lim {b_{n}}$.
(vi) If $(a_{n})_{n=1}^{\infty}$, $(b_{n})_{n=1}^{\infty}$ are convergent and $\lim_{n \rightarrow \infty}b_{n}=l \neq 0$, then $(\frac{a_{n}}{b_{n}})_{n=1}^{\infty}$ is convergent and $\lim_{n \rightarrow \frac{a_{n}}{b_{n}}}= \frac{\lim {a_{n}}}{\lim{b_{n}}}$.
Reference: Understanding Mathematics by Sinha, Karandikar et al. I have used this reference for all the previous articles on series and sequences.
More later,
Nalin Pithwa
High School Geometry: Chapter I: Basic Theorem
If two chords of a circle intersect anywhere. at any angle, what can be said about the segments of each cut off by the other? The data seem to be too for any conclusion, yet an important and far reaching theorem can be formulated from only this meager amount of “given”:
Theorem 2.
If two chords intersect, the product of the segments of the one equals the product of the segments of the other.
This theorem says that in Fig. 3A (attached as a jpeg picture), $PA.PB=PC.PD$ (where the dot is the symbol for algebraic multiplication). What do we mean when we talk about “the product of two line segments”? (There is a way to multiply two lines with a ruler and a compass, but we won’t discuss it here.) What the theorem means is that the product of the respective lengths of the segments are equal. Whenever we put a line segment like PA into an equation, we shall mean the length of PA. (Although we shall soon have to come to grips with the idea of a “negative length” for the present we make no distinction between the length of PA and the length of AP; they are the same positive number).
The Greek geometers took great care to enumerate the different cases of the above theorem. Today, we prefer, when possible, to treat all variants together in one compact theorem. In Fig 3A, the chords intersect inside the circle, in 3B outside the circle, and in 3C one of the chords has become a tangent. Theorem 2 holds in all three cases, and the proofs are so much alike that one proof virtually goes for all.
You may object, and say that in Fig. 3B the chords don’t intersect. But, they do when extended, and in these blogs, we shall say that one line intersects a second when it in fact only intersects the extension of the second line. This is just part of a terminology, in quite general use today, that may be slightly more free-wheeling than what you have been accustomed to. In the same vein, P divided AB internally in Fig 3A, whereas in Fig 3B, P is said to divide the chord AB externally, and we still talk about the two segments PA and PB.
To prove theorem 2, we need to construct two lines, indicated in Fig 3A. Perhaps, you should draw them also in figures 3B and 3C, and, following the proof letter for letter, see for yourself how few changes are required to complete the proofs of these diagrams. In Fig 3A, $\angle {1}= \angle {2}$ because they are inscribed in the same circular arc. And, $\angle {5} = \angle {4}$. Therefore, triangles PCA and PBD are similar, and hence have proportional sides:
$\frac {PA}{PD} = \frac {PC}{PB}$ or $PA. PB = PC.PD$
In the next blog, we will discuss means and another basic theorem.
Till then, aufwiedersehen,
Nalin Pithwa
Real Numbers, Sequences and Series: part 8
Definition. A sequence is a function $f:N \rightarrow \Re$. It is usual to represent the sequence f as $(a_{n})_{n=1}^{\infty}$ where $f(n)=a_{n}$.
Definition. A sequence $(a_{n})_{n=1}^{\infty}$ is said to converge to a if for each $\varepsilon>0$ there is an $n_{0}$ such that
$|a_{n}-a|<\varepsilon$ for all $n > n_{0}$.
It can be shown that this number is unique and we write $\lim_{n \rightarrow \infty}a_{n}=a$.
Examples.
(a) The sequence $\{ 1, 1/2, 1/3, \ldots, 1/n \}$ converges to 0. For $\varepsilon>0$, let $n_{0}=[\frac{1}{\varepsilon}]+1$. This gives
$|\frac{1}{n}-0|=\frac{1}{n}<\varepsilon$ for all $n>n_{0}$.
(b) The sequence $\{ 1, 3/2, 7/4, 15/8, 31/16, \ldots\}$ converges to 2. The nth term of this sequence is $\frac{2^{n}-1}{2^{n-1}}=2-\frac{1}{2^{n-1}}$. So $|2-(2-\frac{1}{2^{n-1}})|=\frac{1}{2^{n-1}}$. But, $2^{n-1} \geq n$ for all $n \geq 1$. Thus, for a given $\varepsilon >0$, the choice of $n_{0}$ given in (a) above will do.
(c) The sequence $(a_{n})_{n=1}^{\infty}$ defined by $a_{n}=n^{\frac{1}{n}}$ converges to 1.
Let $n^{\frac{1}{n}}=1+\delta_{n}$ so that for $n >1$, $\delta_{n}>0$. Now, $n=(1+\delta_{n})^{n}=1+n\delta_{n}+\frac{n(n-1)}{2}(\delta_{n})^{2}+\ldots \geq 1+\frac{n(n-1)}{2}(\delta_{n})^{2}$, thus, for $n-1>0$, we have $\delta_{n} \leq \sqrt{\frac{2}{n}}$. For any $\varepsilon > 0$, we can find $n_{0}$ in N such that $n_{0}\frac{(\varepsilon)^{2}}{2}>1$. Thus, for any $n > n_{0}$, we have $0 \leq \delta_{n} \leq \sqrt{\frac{2}{n}} \leq \sqrt{\frac{2}{n_{0}}}<\varepsilon$. This is the same as writing
$|n^{\frac{1}{n}}-1|<\varepsilon$ for $n > n_{0}$ or equivalently, $\lim_{n \rightarrow \infty}n^{\frac{1}{n}}=1$
(d) Let $a_{n}=\frac{2^{n}}{n!}$. The sequence $(a_{n})_{n=1}^{\infty}$ converges to o. Note that
$a_{n}=\frac{2}{1}\frac{2}{2}\frac{2}{3}\ldots\frac{2}{n}<\frac{4}{n}$ for $n>3$.
For $\varepsilon>0$, choose $n_{1}$ such that $n_{1}\varepsilon>4$. Now, let $n_{0}=max \{ 3, n\}$ so that $|a_{n}|<\varepsilon$ for all $n > n_{0}$.
(e) For $a_{n}=\frac{n!}{n^{n}}$, the sequence $(a_{n})_{n=1}^{\infty}$ converges to 0. (Exercise!)
In all the above examples, we somehow guessed in advance what a sequence converges to. But, suppose we are not able to do that and one asks whether it is possible to decide if the sequence converges to some real number. This can be helped by the following analogy: Suppose there are many people coming to Delhi to attend a conference. They might be taking different routes. But, as soon as they come closer and closer to Delhi the distance between the participants is getting smaller.
This can be paraphrased in mathematical language as:
Theorem. If $(a_{n})_{n=1}^{\infty}$ is a convergent sequence, then for every $\varepsilon>0$, we can find an $n_{0}$ such that
$|a_{n}-a_{m}|<\varepsilon$ for all $m, n > n_{0}$.
Proof. Suppose $(a_{n})_{n=1}^{\infty}$ converges to a. Then, for every $\varepsilon$ we can find an $n_{0}$ such that
$|a_{n}-a|<\varepsilon/2$ for all $n > n_{0}$.
So, for $m, n > n_{0}$, we have
$|a_{m}-a_{n}|=|a_{n}-a+a-a_{m}|\leq |a_{n}-a|+|a-a_{m}|<\varepsilon$. QED.
The proof is rather simple. This very useful idea was conceived by Cauchy and the above theorem is called Cauchy criterion for convergence. What we have proved above tells us that the criterion is a necessary condition for convergence. Is it also sufficient? That is, given a sequence $(a_{n})_{n=1}^{\infty}$ which satisfies the Cauchy criterion, can we assert that there is a real number to which it converges? The answer is yes!
We call a sequence $(a_{n})_{n=1}^{\infty}$ monotonically non-decreasing if $a_{n+1} \geq a_{n}$ for all n.
Theorem. A monotonically non-decreasing sequence which is bounded above converges.
Proof. Suppose $(a_{n})_{n=1}^{\infty}$ is monotonic and non-decreasing. We have $a_{1} \leq a_{2} \leq \ldots \leq a_{n} \leq a_{n+1} \leq \ldots$. Since the sequence is bounded above, $\{ a_{k}: k=1, \ldots\}$ has a least upper bound. Let it be a. By definition, $a_{n} \leq a$ for all n, but for $\varepsilon>0$ there is at least one $n_{0}$ such that $a_{n_{0}}+\varepsilon>a$. Therefore, for $n > n_{0}$, $a_{n}+\varepsilon \geq a_{n_{0}}+\varepsilon>a\geq a_{n}$. This gives us $|a_{n}-a|<\varepsilon$ for all $n \geq n_{0}$. QED.
We can similarly prove that: A monotonically non-increasing sequence which is bounded below is convergent.
Suppose we did not have the condition of boundedness below or above for a monotonically non-increasing or non-decreasing sequence respectively, then what would happen? If a sequence is monotonically non-decreasing and is not bounded above, then given any real number $M>0$ there exists at least one $n_{0}$ such that $a_{n_{0}}>M$, and hence $a_{n}>M$ for all $n > n_{0}$. In such a case, we say that $(a_{n})_{n=1}^{\infty}$ diverges to $\infty$. We write $\lim_{n \rightarrow \infty}a_{n}=\infty$. More generally, (that is, even when the sequence is not monotone), the same criterion above allows us to say that $\{ a_{n}\}_{n=1}^{\infty}$ diverges to $\infty$ and we write $\lim_{n \rightarrow \infty}a_{n}=\infty$. We can similarly define divergence to $-\infty$.
We make a digression here: In case a set is bounded above, then we have the concept of least upper bound. For any set A or real numbers, we define supremum of A as
$sup \{ x: x \in A\}=sup A = least \hspace{0.1in} upper \hspace{0.1in}bound \hspace{0.1in} of \hspace{0.1in} A$, if A is bounded above and is equal to $\infty$ if A is not bounded above.
Similarly, we define infimum of a set A as
$inf \{ x: x \in A\}= inf A= greatest \hspace{0.1in} lower \hspace{0.1in} bound \hspace{0.1in} of \hspace{0.1in}A$, if A is bounded below, and is equal to $-\infty$, if A is not bounded below.
This would help us to look for other criteria for convergence. For any bounded sequence (a_{n})_{n=1}^{\infty} of real numbers, let
$b_{n} = \sup \{ a_{n}, a_{n+1}, a_{n+2}, \ldots\}=\sup \{ a_{k}: k \geq n\}$.
It is clear that $(b_{n})_{n=1}^{\infty}$ is now a non-increasing sequence. So, it converges. We set
$\lim_{n \rightarrow \infty}b_{n}=limit \hspace{0.1in} superior \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} sequence \hspace{0.1in} (a_{n})_{n=1}^{\infty}= \lim \sup a_{n}= \overline{\lim}_{n}a_{n}$,
Similarly, if we write
$a_{n}=\inf \{ a_{n}, a_{n+1}, \ldots\}=\inf \{ a_{k}: k \geq n\}$,
then $(a_{n})_{n=1}^{\infty}$ is a monotonically non-decreasing sequence. So we write
$\lim_{n \rightarrow \infty}c_{n}=limit \hspace{0.1in} inferior \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} sequence \hspace{0.1in} (a_{n})_{n=1}^{\infty}= \lim \inf a_{n}= \underline{\lim}_{n}a_{n}$,
We may not know a sequence to be convergent or divergent, yet we can find its limit superior and limit inferior.
We have in fact the following result:
Theorem.
For a sequence $(a_{n})_{n=1}^{\infty}$ of real numbers, $\underline{\lim}a_{n} \leq \overline{\lim}a_{n}$.
Further, if $l=\underline{\lim}a_{n}$, and $L=\overline{\lim}a_{n}$ are finite, then $l=L$ if and only if the sequence is convergent.
Proof.
It is easy to see that $l \leq L$. Now, suppose l, L are finite.If $l=L$ are finite. If $l=L$, then for all $\varepsilon >0$ there exists $n_{1}, n_{2}$ such that
$l - \varepsilon< \sup_{k \geq n} a_{k}< l + \varepsilon$ for all $n \geq n_{1}$ and $l - \varepsilon < \inf_{k \geq n} < l + \varepsilon$ for all $n > n_{2}$.
Thus, with $n_{0}=max{n_{1},n_{2}}$ we have
$l - \varepsilon for all $n > n_{0}$.
This proves that equality holds only when it is convergent. Conversely, suppose $(a_{n})_{n=1}^{\infty}$ converges to a. For every $\varepsilon > 0$, we have $n_{0}$ such that $a-\varepsilon for all $n > n_{0}$. Therefore, $\sup_{k \geq n}a_{k} \leq a+\varepsilon$ and
$a-\varepsilon \leq \inf a_{n}$ for all $n \geq n_{0}$. Hence, $a-\varepsilon \leq < L \leq a+\varepsilon$. Since this is true for every $\varepsilon > 0$, we have $L=l=a$.
QED.
Real Numbers, Sequences and Series: Part 7
Exercise.
Discover (and justify) an essential difference between the decimal expansions of rational and irrational numbers.
Giving a decimal expansion of a real number means that given $n \in N$, we can find $a_{0} \in Z$ and $0 \leq a_{1}, \ldots, a_{n} \leq 9$ such that
$|x-\sum_{k=0}^{n}\frac{a_{k}}{10^{k}}|< \frac{1}{10^{n}}$
In other words if we write
$x_{n}=a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots +\frac{a_{n}}{10^{n}}$
then $x_{1}, x_{2}, x_{3}, \ldots, x_{n}, \ldots$ are approximate values of x correct up to the first, second, third, …, nth place of decimal respectively. So when we write a real number by a non-terminating decimal expansion, we mean that we have a scheme of approximation of the real numbers by terminating decimals in such a way that if we stop after the nth place of decimal expansion, then the maximum error committed by us is $10^{-n}$.
This brings us to the question of successive approximations of a number. It is obvious that when we have some approximation we ought to have some notion of the error committed. Often we try to reach a number through its approximate values, and the context determines the maximum error admissible. Now, if the error admissible is $\varepsilon >0$, and $x_{1}, x_{2}, x_{3}, \ldots$ is a scheme of successive is approximation of a number x, then we should be able to tell at which stage the desired accuracy is achieved. In fact, we should find an n such that $|x-x_{n}|<\varepsilon$. But this could be a chance event. If the error exceeds $\varepsilon$ at a later stage, then the scheme cannot be a good approximation as it is not “stable”. Instead, it would be desirable that accuracy is achieved at a certain stage and it should not get worse after that stage. This can be realized by demanding that there is a natural number $n_{0}$ such that $|x-x_{n}|<\varepsilon$ for all $n > n_{0}$. It is clear that $n_{0}$ will depend on $varepsilon$. This leads to the notion of convergence, which is the subject of a later blog.
More later,
Nalin Pithwa | 10,398 | 33,585 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 365, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2022-05 | latest | en | 0.852622 |
https://pythonpedia.com/en/knowledge-base/736043/checking-if-a-string-can-be-converted-to-float-in-python | 1,591,203,038,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435238.60/warc/CC-MAIN-20200603144014-20200603174014-00558.warc.gz | 498,502,776 | 5,223 | # Checking if a string can be converted to float in Python
### Question
I've got some Python code that runs through a list of strings and converts them to integers or floating point numbers if possible. Doing this for integers is pretty easy
``````if element.isdigit():
newelement = int(element)
``````
Floating point numbers are more difficult. Right now I'm using `partition('.')` to split the string and checking to make sure that one or both sides are digits.
``````partition = element.partition('.')
if (partition[0].isdigit() and partition[1] == '.' and partition[2].isdigit())
or (partition[0] == '' and partition[1] == '.' and partition[2].isdigit())
or (partition[0].isdigit() and partition[1] == '.' and partition[2] == ''):
newelement = float(element)
``````
This works, but obviously the if statement for that is a bit of a bear. The other solution I considered is to just wrap the conversion in a try/catch block and see if it succeeds, as described in this question.
Anyone have any other ideas? Opinions on the relative merits of the partition and try/catch approaches?
1
150
9/15/2017 11:18:51 AM
I would just use..
``````try:
float(element)
except ValueError:
print "Not a float"
``````
..it's simple, and it works
Another option would be a regular expression:
``````import re
if re.match("^\d+?\.\d+?\$", element) is None:
print "Not float"
``````
247
4/9/2009 9:55:39 PM
## Python method to check for float:
``````def isfloat(value):
try:
float(value)
return True
except ValueError:
return False
``````
## Don't get bit by the goblins hiding in the float boat! DO UNIT TESTING!
What is, and is not a float may surprise you:
``````Command to parse Is it a float? Comment
-------------------------------------- --------------- ------------
print(isfloat("")) False
print(isfloat("1234567")) True
print(isfloat("NaN")) True nan is also float
print(isfloat("NaNananana BATMAN")) False
print(isfloat("123.456")) True
print(isfloat("123.E4")) True
print(isfloat(".1")) True
print(isfloat("1,234")) False
print(isfloat("NULL")) False case insensitive
print(isfloat(",1")) False
print(isfloat("123.EE4")) False
print(isfloat("6.523537535629999e-07")) True
print(isfloat("6e777777")) True This is same as Inf
print(isfloat("-iNF")) True
print(isfloat("1.797693e+308")) True
print(isfloat("infinity")) True
print(isfloat("infinity and BEYOND")) False
print(isfloat("12.34.56")) False Two dots not allowed.
print(isfloat("#56")) False
print(isfloat("56%")) False
print(isfloat("0E0")) True
print(isfloat("x86E0")) False
print(isfloat("86-5")) False
print(isfloat("True")) False Boolean is not a float.
print(isfloat(True)) True Boolean is a float
print(isfloat("+1e1^5")) False
print(isfloat("+1e1")) True
print(isfloat("+1e1.3")) False
print(isfloat("+1.3P1")) False
print(isfloat("-+1")) False
print(isfloat("(1)")) False brackets not interpreted
`````` | 826 | 3,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-24 | longest | en | 0.778036 |
https://www.easytrans.org/en/?q=total | 1,695,366,815,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506339.10/warc/CC-MAIN-20230922070214-20230922100214-00037.warc.gz | 795,270,013 | 15,425 | # Dictionary
## 4 result(s) found for total:
[Show/Hide Search option]
Search option
Results mode: advanced simple yes no
Results found in English Dictionary
total change [a]: fullstendig, hel, totaldegree [a]: absolutt, total, fullkommelig, gjennom-, fullstendigquantity [a]: sammenlagt, totalentire [a]: hel, fullstendig, total, fullkommelig, komplettmathematics [n]: sum [u], sluttsum [u]money [n]: sum [u], sluttsum [u], beløp [n]mathematics [v]: regne sammen, addere, summerequantity [v]: regne, omfatte change [a]: total, completo, generaldegree [a]: completo, total, cabalquantity [a]: total, agregadoentire [a]: entero, total, integral, completomathematics [n]: suma [f], total [m]money [n]: importe [m]mathematics [v]: sumar, totalizarquantity [v]: ascender a, sumar change [a]: fullständig, hel, totaldegree [a]: absolut, total, fullkomlig, genom-, fullständigquantity [a]: sammanlagd, totalentire [a]: hel, fullständig, total, fullkomlig, komplettmathematics [n]: summa [u], slutsumma [u]money [n]: summa [u], slutsumma [u], belopp [n]mathematics [v]: räkna ihop, addera, summeraquantity [v]: räkna, omfatta
Synonyms: [adj]: entire, full, whole [st][adj]: overall, whole, gross [st][adj]: absolute, unconditioned, unconditional [st][adj]: full, complete [st][n]: sum, totality, aggregate, whole [gt], unit [gt][n]: sum, amount, quantity [gt][v]: number, add up, come, amount, be [gt][v]: tot, tot up, sum, sum up, summate, tote up, add, add together, tally, add up, count [gt], number [gt], enumerate [gt], numerate [gt][v]: damage [gt]
Derived terms: be totally ignored by somebody, ignore someone totally, subtotal, teetotal, teetotaler, teetotaller, total amount, total loss, totalitarian, totalitarianism, totality, totalizator, totally, totally lost, totally unfounded, grand total, sum total, teetotaling, teetotalism, teetotalist, total aphasia, total darkness, total depravity, total eclipse, total heat, total hysterectomy, total parenteral nutrition, totaled, totalisator, totalise, totaliser, totalism, totalistic, totalize, totalizer
Example: How much money did you spend in total?Our total debts amount to ten thousand dollars.If its going to require a total restructuring, Im sure they will be satisfied with the old system.If there was just 1,000 yen more, he would have taken 10,000 yen in total.The total at the bottom of the page is carried forward.Whats the total population of France?How did a total stranger know his name?The man was a total stranger.The total came to ten dollars.The total expense for the project amounts to one hundred million yen.The task was total agony.The desk is in a state of total disorder.GNP is measured as the total market value of all the goods and service produced by a nation during a specified period.Americans who are over sixty-five make up 12.5% of the total population.What is the total amount?The total is one hundred.The total is approximately ten thousand dollars.Gross national product is a nations total output of goods and services as measured in monetary value.Gross national product is a nations total output of goods and services during a given period of time as measured in monetary value.Casualties are said to total up to 1,000.
Results found in Norwegian dictionary
total endring [a]: complete, total, generalgrad [a]: thorough, total, complete, arrant, thoroughgoing, out-and-out, all-outquantity [a]: aggregate, totalhel [a]: entire, whole, complete, total, integral endring [a]: total, completo, generalgrad [a]: completo, total, cabalquantity [a]: total, agregadohel [a]: entero, total, integral, completo endring [a]: total
Synonyms:
Derived terms: total-, totalødelegge, totalisator, totalitær, totalitarisme, totalitet, totalskadd, totalskadd bil, totalsum, totalt, totalt oppslukt, totalfrede, totalskade
Wiki: Total NYSE: TOTAL er et fransk oljeselskap, etablert i 1924. Det er et av de fire største oljeselskapene i verden, sammen med Royal Dutch Shell, BP og Exxon Mobil. Selskapet ble slått sammen med belgiske Petrofina i 1999 og endret navn til Total Fina.
Example: Jeg var innelåst i en total ensomhet.Vil du se en total solformørkelse som vil legge Svalbard i mørke 20. mars 2015.?
Results found in Spanish dictionary
total general [a]: uttertonterías [a]: stark, utter, unadulterated, complete, unmitigated, sheer, plain, outright, blatant, downright, purementira [a]: outright, blatant, downright, out-and-outreinforcing word [a]: pure, sheer, absolute, uttercambio [a]: complete, total, generalgrado [a]: thorough, total, complete, arrant, thoroughgoing, out-and-out, all-out, crashing [informal]quantity [a]: aggregate, totalabsoluto [a]: absolute, perfect, completeentero [a]: entire, whole, complete, total, integralgeneral [n]: wholemathematics [n]: sum, totaldinero [n]: price, amount, sum, figure general [a]: fullstendig, fullkommeligtonterías [a]: rein, fullstendig, pur, fullkommeligmentira [a]: rein, fullkommelig, åpenbarreinforcing word [a]: rein, absolutt, fullstendigcambio [a]: fullstendig, hel, totalgrado [a]: absolutt, total, fullkommelig, gjennom-, fullstendig, komplettquantity [a]: sammenlagt, totalabsoluto [a]: absolutt, perfekt, fullkommeligentero [a]: hel, fullstendig, total, fullkommelig, komplettgeneral [n]: helhet [u]mathematics [n]: sum [u], sluttsum [u]dinero [n]: sifre [u], pris [n], sum [u], beløp [n] general [a]: fullständig, fullkomligtonterías [a]: ren, fullständig, pur, fullkomligmentira [a]: ren, fullkomlig, uppenbarreinforcing word [a]: ren, absolut, fullständigcambio [a]: fullständig, hel, totalgrado [a]: absolut, total, fullkomlig, genom-, fullständig, komplettquantity [a]: sammanlagd, totalabsoluto [a]: absolut, perfekt, fullkomligentero [a]: hel, fullständig, total, fullkomlig, komplettgeneral [n]: helhet [u]mathematics [n]: summa [u], slutsumma [u]dinero [n]: siffra [u], pris [n], summa [u], belopp [n]
Synonyms:
Derived terms: diversión total, en su totalidad, en total, estaba totalmente silencioso, estar totalmente aburrido, estar totalmente asustado, estar totalmente enamorado, estar totalmente equivocado, felicidad total, ignorar totalmente a alguien, matanza total, pérdida total, pobreza total, ser ignorado totalmente por alguien, ser totalmente igual a, ser totalmente olvidado, ser totalmente parecido a, suma total, total parcial, totalidad, totalitario, totalitarismo, totalizador, totalizar, totalmente, totalmente desaparecido, totalmente desnudo, totalmente despierto, totalmente desproporcionado, totalmente quieto, totalmente silencioso, totalmente sin fundamentos, longitud total, rendimiento total, subtotal
Example: El problema preguntaba si ocurriría una reflexión total de la luz.Son 3.000 yenes en total.Es una total pérdida de tiempo.Yo soy Windowsero total.¿Sabe usted la población total del Japón?El problema decía si ocurriría reflexión total de la luz.El eclipse de sol total de mañana será visible desde el hemisferio sur.El comportamiento de Viernes muestra su total sumisión y devoción hacia Robinson porque el inglés le salvó de los caníbales.¿Puedes calcular el costo total del viaje?Él es en verdad un total idiota.¿Cuál es la población total de Francia?Había diez huevos en total.El negocio de John resultó ser un fracaso total.Somos ocho en total.Costa Rica es considerado uno de los 20 países con mayor biodiversidad, posee un 4% del total de las especies estimadas a nivel mundial.Tengo pocos alumnos, no más que cinco en total.Hay unos mil estudiantes en total.Es un lío total y me pone de los nervios.¿Cuánto es en total?La cuenta total por las bebidas fue cerca de 7.000 dólares.
Results found in Swedish dictionary
total ändring [a]: complete, total, generalgrad [a]: thorough, total, complete, arrant, thoroughgoing, out-and-out, all-outquantity [a]: aggregate, totalhel [a]: entire, whole, complete, total, integral ändring [a]: total ändring [a]: total, completo, generalgrad [a]: completo, total, cabalquantity [a]: total, agregadohel [a]: entero, total, integral, completo
Synonyms:
Anagrams: lotta
Wiki: Total är världens fjärde största oljebolag, efter ExxonMobil, Shell och BP. Företaget tog över Petrofina 1999 och bytte då namn till Total Fina. 2000 köpte man Elf Aquitaine och ändrade namnet till TotalFinaElf.
Example: En min av total tillfredsställelse.|Total katastrof.|"Total säkerhet|It's a total mess.|Det medicinska ordet för en sådan en är "total psykopat".|Det är en total katastrof!|Drottningens besök har medfört total utförsäljning!|"att lägga dem under total despotism"-|Total blockering.
Similar words:
tonal tonal total- totalt tonal tonal total-
1. total
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www.easytrans.org © 2006-2023 - Page generated in 0.2409 - seconds. | 2,983 | 10,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-40 | latest | en | 0.458692 |
https://simpy.readthedocs.io/en/latest/examples/bank_renege.html | 1,719,040,087,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862252.86/warc/CC-MAIN-20240622045932-20240622075932-00695.warc.gz | 457,896,543 | 5,700 | # Bank Renege
Covers:
• Resources: Resource
• Condition events
A counter with a random service time and customers who renege. Based on the program bank08.py from TheBank tutorial of SimPy 2. (KGM)
This example models a bank counter and customers arriving at random times. Each customer has a certain patience. She waits to get to the counter until she’s at the end of her tether. If she gets to the counter, she uses it for a while before releasing it.
New customers are created by the source process every few time steps.
"""
Bank renege example
Covers:
- Resources: Resource
- Condition events
Scenario:
A counter with a random service time and customers who renege. Based on the
program bank08.py from TheBank tutorial of SimPy 2. (KGM)
"""
import random
import simpy
RANDOM_SEED = 42
NEW_CUSTOMERS = 5 # Total number of customers
INTERVAL_CUSTOMERS = 10.0 # Generate new customers roughly every x seconds
MIN_PATIENCE = 1 # Min. customer patience
MAX_PATIENCE = 3 # Max. customer patience
def source(env, number, interval, counter):
"""Source generates customers randomly"""
for i in range(number):
c = customer(env, f'Customer{i:02d}', counter, time_in_bank=12.0)
env.process(c)
t = random.expovariate(1.0 / interval)
yield env.timeout(t)
def customer(env, name, counter, time_in_bank):
"""Customer arrives, is served and leaves."""
arrive = env.now
print(f'{arrive:7.4f} {name}: Here I am')
with counter.request() as req:
patience = random.uniform(MIN_PATIENCE, MAX_PATIENCE)
# Wait for the counter or abort at the end of our tether
results = yield req | env.timeout(patience)
wait = env.now - arrive
if req in results:
# We got to the counter
print(f'{env.now:7.4f} {name}: Waited {wait:6.3f}')
tib = random.expovariate(1.0 / time_in_bank)
yield env.timeout(tib)
print(f'{env.now:7.4f} {name}: Finished')
else:
# We reneged
print(f'{env.now:7.4f} {name}: RENEGED after {wait:6.3f}')
# Setup and start the simulation
print('Bank renege')
random.seed(RANDOM_SEED)
env = simpy.Environment()
# Start processes and run
counter = simpy.Resource(env, capacity=1)
env.process(source(env, NEW_CUSTOMERS, INTERVAL_CUSTOMERS, counter))
env.run()
The simulation’s output:
Bank renege
0.0000 Customer00: Here I am
0.0000 Customer00: Waited 0.000
3.8595 Customer00: Finished
10.2006 Customer01: Here I am
10.2006 Customer01: Waited 0.000
12.7265 Customer02: Here I am
13.9003 Customer02: RENEGED after 1.174
23.7507 Customer01: Finished
34.9993 Customer03: Here I am
34.9993 Customer03: Waited 0.000
37.9599 Customer03: Finished
40.4798 Customer04: Here I am
40.4798 Customer04: Waited 0.000
43.1401 Customer04: Finished | 769 | 2,647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-26 | latest | en | 0.764567 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-4-quadratic-functions-and-factoring-4-7-complete-the-square-4-7-exercises-skill-practice-page-289/52 | 1,722,690,289,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00422.warc.gz | 626,130,256 | 16,904 | ## Algebra 2 (1st Edition)
The solutions are $-4$ and $-5$.
$x^{2}+9x+20=0\qquad$ ...Write left side in the form $x^{2}+bx.$(add $-20$ to each side) $x^{2}+9x=-20\qquad$ ...square half the coefficient of $x$. $(\displaystyle \frac{9}{2})^{2}=\frac{81}{4}\qquad$ ...complete the square by adding$\displaystyle \frac{81}{4}$ to each side of the expression $x^{2}+9x+\displaystyle \frac{81}{4}=-20+\frac{81}{4}\qquad$ ...Write left side as a binomial squared. $(x+\displaystyle \frac{9}{2})^{2}=\frac{1}{4}\qquad$ ...take square roots of each side. $x+\displaystyle \frac{9}{2}=\pm\sqrt{\frac{1}{4}}\qquad$ ...evaluate$\sqrt{\frac{1}{4}}$ $x+\displaystyle \frac{9}{2}=\pm\frac{1}{2}\qquad$ ...add $-\displaystyle \frac{9}{2}$ to each side. $x=-\displaystyle \frac{9}{2}\pm\frac{1}{2}$ | 306 | 782 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-33 | latest | en | 0.4757 |
https://www.learningoutofthebox.org/shop/problem-sums-guide/problem-sums-guide-for-parents-volume-2/ | 1,726,612,736,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00296.warc.gz | 777,037,255 | 49,193 | Select Page
# P4-6. Problem Sums Guide for Parents (Volume 2 out of 2)
\$29.00
#### \$36 SGD
Most problem sums types come with at least 2 levels. Some have 4 levels. Does your child know these? (Most have no idea. Let it not be your child.)
Meet the Problem Sums Guide (Vol 2), your personal handy smart assistant that tells you exactly what problem sums concepts/ types your child is being tested in his tests/ exams. This clever easy-to-keep personal assistant can be next to you during home revision, in school.
Just bring it anywhere you want! With a clear breakdown of the tough problem sums into simpler groups (using our smart techniques and tricks), it serves as your reliable home math coach who is always by your side.
With this compact Little Purple Book with all our done-for-you working steps with clear and concise explanation, there is no need to do it yourself (which can possibly take you weeks or even months.)
Going to help your child reduce his careless mistakes, remember math concepts more easily and eventually higher marks!
Planning to get one? Would you want to know the difference between Problem Sums Guide Vol 1 and 2?
• Volume 1 covers the 11 must-know level one problem sums types that all P5s must know by Term 1. (Students started to learn some of these concepts since P3 and all these concepts will be tested till your child is in P6.)
• Volume 2 covers the level 2 problem sums types that P5s will eventually see in their exam papers as they get tougher.
119 in stock
SKU: PSGV2
## Description
Here is our 2nd volume for – “Problem Sums Guide for Parents“.
Why do we create this?
There are so many math books out there.
So what makes this different?
As you know, most assessment books just provide your child the solutions (without explaining much)
While children can make out these solution, 8 to 9 out of 10 children won’t understand the type of problem sum that is being tested.
As a result, they just solve the questions as individual questions. Then, they move on to the next one, treating it as a new question again.
This old method of solving is too slow.
The good news is there is a faster and more effective way.
While the way the questions always changes, the types of problem sums is fixed.
So we specially designed this book just for our children and parents.
This proven study method has worked for hundreds of our students. Even our parents, who used to find teaching math hard, told us that they finally understood how to solve the problem sums the easy way.
Here are the 11 Types of Problem Sums with more Advanced levels (your child should know all these by P6)
1. Make a whole unit (2 and 3 identities)
2. Repeated Identity (Type 1 and 2)
3. Equal Concept (Type 1, 2 and 3)
4. Equal Concept (Type 4 – 3 Identities)
5. Equal Fraction Concept (Type 1 and 2)
6. Remainder Concept (Type 2)
7. Same Difference (Type 3)
8. Units and Parts (Type 2 and 3)
9. Guess and Check (Type 2 – Grouping)
10. 2 Ifs (Type 2 and 3)
11. Working Backwards
How can you get hold of this book?
If you know your child can start benefiting by learning math in a smarter way, you can order this math resource book from us.
All you need to do now is click on the “Add to Cart” button below.
119 in stock | 795 | 3,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.947372 |
https://leanprover-community.github.io/mathlib4_docs/Std/Data/Option/Lemmas.html | 1,679,663,506,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945282.33/warc/CC-MAIN-20230324113500-20230324143500-00605.warc.gz | 422,408,782 | 7,846 | # Documentation
Std.Data.Option.Lemmas
theorem Option.mem_iff {α : Type u_1} {a : α} {b : } :
a b b = some a
theorem Option.some_ne_none {α : Type u_1} (x : α) :
some x none
theorem Option.forall {α : Type u_1} {p : Prop} :
((x : ) → p x) p none ((x : α) → p (some x))
theorem Option.exists {α : Type u_1} {p : Prop} :
(x, p x) p none x, p (some x)
theorem Option.get_mem {α : Type u_1} {o : } (h : ) :
o
theorem Option.get_of_mem {α : Type u_1} {a : α} {o : } (h : ) :
a o = a
theorem Option.not_mem_none {α : Type u_1} (a : α) :
¬a none
@[simp]
theorem Option.some_get {α : Type u_1} {x : } (h : ) :
some () = x
@[simp]
theorem Option.get_some {α : Type u_1} (x : α) (h : ) :
Option.get (some x) h = x
theorem Option.getD_of_ne_none {α : Type u_1} {x : } (hx : x none) (y : α) :
some () = x
theorem Option.mem_unique {α : Type u_1} {o : } {a : α} {b : α} (ha : a o) (hb : b o) :
a = b
theorem Option.ext {α : Type u_1} {o₁ : } {o₂ : } :
(∀ (a : α), a o₁ a o₂) → o₁ = o₂
theorem Option.eq_none_iff_forall_not_mem :
∀ {α : Type u_1} {o : }, o = none ∀ (a : α), ¬a o
@[simp]
theorem Option.isSome_none {α : Type u_1} :
@[simp]
theorem Option.isSome_some :
∀ {α : Type u_1} {a : α},
theorem Option.isSome_iff_exists :
∀ {α : Type u_1} {x : }, a, x = some a
@[simp]
theorem Option.isNone_none {α : Type u_1} :
@[simp]
theorem Option.isNone_some :
∀ {α : Type u_1} {a : α},
@[simp]
theorem Option.not_isSome :
∀ {α : Type u_1} {a : },
theorem Option.eq_some_iff_get_eq :
∀ {α : Type u_1} {o : } {a : α}, o = some a h, = a
theorem Option.eq_some_of_isSome {α : Type u_1} {o : } (h : ) :
o = some ()
theorem Option.not_isSome_iff_eq_none :
∀ {α : Type u_1} {o : }, o = none
theorem Option.ne_none_iff_isSome :
∀ {α : Type u_1} {o : }, o none
theorem Option.ne_none_iff_exists :
∀ {α : Type u_1} {o : }, o none x, some x = o
theorem Option.ne_none_iff_exists' :
∀ {α : Type u_1} {o : }, o none x, o = some x
theorem Option.bex_ne_none {α : Type u_1} {p : Prop} :
(x x_1, p x) x, p (some x)
theorem Option.ball_ne_none {α : Type u_1} {p : Prop} :
((x : ) → x nonep x) (x : α) → p (some x)
@[simp]
theorem Option.bind_some {α : Type u_1} (x : ) :
Option.bind x some = x
@[simp]
theorem Option.bind_eq_some :
∀ {α : Type u_1} {b : α} {α_1 : Type u_2} {x : Option α_1} {f : α_1}, = some b a, x = some a f a = some b
@[simp]
theorem Option.bind_eq_none {α : Type u_1} {β : Type u_2} {o : } {f : α} :
= none ∀ (b : β) (a : α), a o¬b f a
theorem Option.bind_comm {α : Type u_1} {β : Type u_2} {γ : Type u_3} {f : αβ} (a : ) (b : ) :
(Option.bind a fun x => Option.bind b (f x)) = Option.bind b fun y => Option.bind a fun x => f x y
theorem Option.bind_assoc {α : Type u_1} {β : Type u_2} {γ : Type u_3} (x : ) (f : α) (g : β) :
Option.bind () g = Option.bind x fun y => Option.bind (f y) g
theorem Option.join_eq_some :
∀ {α : Type u_1} {a : α} {x : Option ()}, = some a x = some (some a)
theorem Option.join_ne_none :
∀ {α : Type u_1} {x : Option ()}, none z, x = some (some z)
theorem Option.join_ne_none' :
∀ {α : Type u_1} {x : Option ()}, ¬ = none z, x = some (some z)
theorem Option.join_eq_none :
∀ {α : Type u_1} {o : Option ()}, = none o = none o = some none
theorem Option.bind_id_eq_join {α : Type u_1} {x : Option ()} :
@[simp]
theorem Option.map_eq_map :
∀ {α α_1 : Type u_1} {f : αα_1},
theorem Option.map_none :
∀ {α α_1 : Type u_1} {f : αα_1}, f <$> none = none theorem Option.map_some : ∀ {α α_1 : Type u_1} {f : αα_1} {a : α}, f <$> some a = some (f a)
@[simp]
theorem Option.map_eq_some' :
∀ {α : Type u_1} {b : α} {α_1 : Type u_2} {x : Option α_1} {f : α_1α}, = some b a, x = some a f a = b
theorem Option.map_eq_some :
∀ {α α_1 : Type u_1} {f : αα_1} {x : } {b : α_1}, f <$> x = some b a, x = some a f a = b @[simp] theorem Option.map_eq_none' : ∀ {α : Type u_1} {x : } {α_1 : Type u_2} {f : αα_1}, = none x = none theorem Option.map_eq_none : ∀ {α α_1 : Type u_1} {f : αα_1} {x : }, f <$> x = none x = none
theorem Option.map_eq_bind {α : Type u_1} :
∀ {α : Type u_2} {f : αα} {x : }, = Option.bind x (some f)
theorem Option.map_congr {α : Type u_1} :
∀ {α : Type u_2} {f g : αα} {x : }, (∀ (a : α), a xf a = g a) → =
@[simp]
theorem Option.map_id' {α : Type u_1} :
= id
@[simp]
theorem Option.map_map {β : Type u_1} {γ : Type u_2} {α : Type u_3} (h : βγ) (g : αβ) (x : ) :
Option.map h () = Option.map (h g) x
theorem Option.comp_map {β : Type u_1} {γ : Type u_2} {α : Type u_3} (h : βγ) (g : αβ) (x : ) :
Option.map (h g) x = Option.map h ()
@[simp]
theorem Option.map_comp_map {α : Type u_1} {β : Type u_2} {γ : Type u_3} (f : αβ) (g : βγ) :
theorem Option.mem_map_of_mem {α : Type u_1} {β : Type u_2} {a : α} {x : } (g : αβ) (h : a x) :
g a
theorem Option.bind_map_comm {α : Type u_1} {β : Type u_2} {x : Option ()} {f : αβ} :
theorem Option.join_map_eq_map_join {α : Type u_1} {β : Type u_2} {f : αβ} {x : Option ()} :
theorem Option.join_join {α : Type u_1} {x : Option (Option ())} :
= Option.join (Option.map Option.join x)
theorem Option.mem_of_mem_join {α : Type u_1} {a : α} {x : Option ()} (h : a ) :
some a x
@[simp]
theorem Option.some_orElse {α : Type u_1} (a : α) (x : ) :
(HOrElse.hOrElse (some a) fun x => x) = some a
@[simp]
theorem Option.none_orElse {α : Type u_1} (x : ) :
(HOrElse.hOrElse none fun x => x) = x
@[simp]
theorem Option.orElse_none {α : Type u_1} (x : ) :
(HOrElse.hOrElse x fun x => none) = x
@[simp]
theorem Option.guard_eq_some :
∀ {α : Type u_1} {p : αProp} {a b : α} [inst : ], = some b a = b p a
theorem Option.liftOrGet_eq_or_eq {α : Type u_1} {f : ααα} (h : ∀ (a b : α), f a b = a f a b = b) (o₁ : ) (o₂ : ) :
Option.liftOrGet f o₁ o₂ = o₁ Option.liftOrGet f o₁ o₂ = o₂
@[simp]
theorem Option.liftOrGet_none_left {α : Type u_1} {f : ααα} {b : } :
Option.liftOrGet f none b = b
@[simp]
theorem Option.liftOrGet_none_right {α : Type u_1} {f : ααα} {a : } :
Option.liftOrGet f a none = a
@[simp]
theorem Option.liftOrGet_some_some {α : Type u_1} {f : ααα} {a : α} {b : α} :
Option.liftOrGet f (some a) (some b) = some (f a b)
theorem Option.elim_none {β : Sort u_1} {α : Type u_2} (x : β) (f : αβ) :
Option.elim none x f = x
theorem Option.elim_some {β : Sort u_1} {α : Type u_2} (x : β) (f : αβ) (a : α) :
Option.elim (some a) x f = f a
@[simp]
theorem Option.getD_map {α : Type u_1} {β : Type u_2} (f : αβ) (x : α) (o : ) :
Option.getD () (f x) = f ()
noncomputable def Option.choice (α : Type u_1) :
An arbitrary some a with a : α if α is nonempty, and otherwise none.
Equations
• = if h : then else none
theorem Option.choice_eq {α : Type u_1} [inst : ] (a : α) :
@[simp]
theorem Option.to_list_some {α : Type u_1} (a : α) :
= [a]
@[simp]
theorem Option.to_list_none (α : Type u_1) : | 2,644 | 6,640 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-14 | longest | en | 0.411277 |
https://ww2.mathworks.cn/matlabcentral/answers/367638-error-using-integral2-and-arrayfun?s_tid=srchtitle | 1,675,334,075,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00397.warc.gz | 651,968,775 | 29,656 | # Error using integral2 and arrayfun
3 次查看（过去 30 天）
Hello, I'm having some problems with integral2 and the arrayfun function. In my problem, I have a function that takes as inputs vectors of size 5000x1 (lets say it uses only 3 vectors), and returns a vector of size 5000x1. This function has multiple inputs and cannot be separated, and I have to integrate over two dimensions (vectors), and the other vectors have information relative to each individual in every element. I'm trying to do write my function with arrayfun, and then integrate with integrate2, but I get the following error:
"Error using arrayfun All of the input arguments must be of the same size and shape. Previous inputs had size 5000 in dimension 1. Input #3 has size 1"
I have "tested" the function using vectors and I have no problem. Here is a minimal working example:
Y=mvnrnd(0,1,5000);
normal=@(y,f1,f2) exp(y+f1+f2);
g= @(f1,f2) arrayfun(normal,Y,f1,f2);
test=g(zeros(5000,1),zeros(5000,1));
T = integral2(@(f1,f2)g(f1,f2),-inf,inf,-inf,inf);
Does somebody knows how to solve it? I am also considering doing this using the integral function with the 'ArrayValued' option, in order to avoid using arrayfun, but I haven't found a way to do this. I also have tried using the int function, but I have problems integrating over symbolic vectors.
### 采纳的回答
Walter Roberson 2017-11-18
"Integrand, specified as a function handle, defines the function to be integrated over the planar region xmin ≤ x ≤ xmax and ymin(x) ≤ y ≤ ymax(x). The function fun must accept two arrays of the same size and return an array of corresponding values. It must perform element-wise operations."
That gets you the same size (but not necessarily square) arrays for f1 and f2, to be passed to g(). The size of those arrays will change between calls: you can just count on the two being the same size as each other for any one call.
Inside g you have arrayfun(normal,Y,f1,f2) . arrayfun() requires that all of the arguments after the function be either scalars or arrays that are the same size as each other. As indicated, f1 and f2 are going to be the same size as each other, but the exact size is going to change. But your Y is fixed size 1 x 5000 and your f1 and f2 are very unlikely to be exactly that size.
It is not possible to use integral2() to do an array integral directly. You cannot get integral2() to return a 5000 x 1 result: integral2() always returns a scalar result.
Perhaps you can use
T = arrayfun(@(y) integral2(@(f1,f2) normal(y,f1,f2),-inf,inf,-inf,inf), Y);
##### 2 个评论显示隐藏 1更早的评论
Correction, it worked!! The problem was with the example of the "normal" function. Now I defined a more real example, using a normal pdf, and now the code works. Before the integral exploded because of how it was defined. Here is a working code:
Y=mvnrnd(0,1,50);
normal=@(y,f1,f2)(2*pi)^(-0.5)* exp( -(y - f1- f2).^2/(2));
T = arrayfun(@(y) integral2(@(f1,f2) normal(y,f1,f2),-inf,inf,-inf,inf), Y);
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Translated by | 856 | 3,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-06 | latest | en | 0.820577 |
https://community.deeplearning.ai/t/custom-loss-function/282659 | 1,708,937,386,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474653.81/warc/CC-MAIN-20240226062606-20240226092606-00524.warc.gz | 186,321,841 | 10,151 | # Custom Loss function
Hello learners. I want to define a costume loss function rather than using the predefined loss function that Keras provides and want the loss function to use my input training examples X_tain besides the default inputs y_pred and y_true. To do so, I wrote a loss function that itself has an inner loss function. However, the model does not work; I receive an error in the first iteration of the fitting stage. I suspect maybe this is because of the shape or type of the X_train (it is a NumPy matrix), since the code works when I delete terms associated with X_train. I tried to change the shape and format of X_train to a tensor because I use TensorFlow as the backend but still does not work. I was wondering if you had a similar experience or have suggestions.
Here is my code:
``````import tensorflow as tf
from tensorflow.keras.models import Sequential
from tensorflow.keras.layers import Dense
def costum_loss(X_loss):
def loss(y_true, y_pred):
q1 = y_pred[:, 1] + y_pred[:, 4] - X_loss [:,0]
q2 = y_pred[:, 2] + y_pred[:, 3] + y_pred[:, 7] + y_pred[:, 17] + y_pred[:, 22] - X_loss [:,1]
q3 = y_pred[:, 4] + y_pred[:, 5] + y_pred[:, 7] + y_pred[:, 8] + y_pred[:, 9] + y_pred[:, 19] - X_loss [:,2]
q4 = y_pred[:, 3] + y_pred[:, 10] + y_pred[:, 11] + y_pred[:, 13]
q5 = y_pred[:, 14]
q6 = y_pred[:, 5] + y_pred[:, 7] + y_pred[:, 10] + y_pred[:, 15] + y_pred[:, 17] + y_pred[:, 20] + y_pred[:, 22]
q7 = y_pred[:, 11] + y_pred[:, 16] + y_pred[:, 19] + y_pred[:, 21]
q8 = y_pred[:, 5] + y_pred[:, 7] + y_pred[:, 8] + y_pred[:, 20] + y_pred[:, 21] + y_pred[:, 22] + y_pred[:, 23]
loss_t = tf.reduce_mean(tf.square(y_true - y_pred)) + 1000 * tf.reduce_mean(tf.square(q1)) + \
1000 * tf.reduce_mean(tf.square(q2)) + 1000 * tf.reduce_mean(tf.square(q3)) + \
1000 * tf.reduce_mean(tf.square(q4)) + 1000 * tf.reduce_mean(tf.square(q5)) + \
1000 * tf.reduce_mean(tf.square(q6)) + 1000 * tf.reduce_mean(tf.square(q7)) + \
1000 * tf.reduce_mean(tf.square(q8))
return loss_t
return loss
model = Sequential([
tf.keras.Input(shape=(n2,)),
Dense(16, activation='relu'),
Dense(64, activation='relu'),
Dense(128, activation='relu'),
Dense(64, activation='relu'),
Dense(25, activation='relu')
], name="Model")
# Compile the model with the custom loss function
# Train the model
model.fit(X_train, Y_train, epochs=5000)
``````
Hello @Nasim_Deljouyi
Is your `X_loss` a constant? Or does it vary from mini-batch to mini-batch?
Raymond
It is actually constant. Later, in the compile section, I will use X_train as the input argument of custom loss, which is a NumPy array with constant values (defined before the neural network modeling) and its dimension is 7000*8, in which 8 is the number of input features and 7000 is the number of training samples.
Please share a copy of the full error traceback here. A screenshot is better, but text copy should also do.
`X_loss[:,0]` has a shape of `(7000, )` right? And `y_pred[:, 4]` has a shape of `(batch_size, )`, right?
The batch size here is the number of all samples, and it is actually 6999. Yes, X_loss[:, 0] (X_train[:,0]) has a shape of (6999, ) (I round it earlier when I said 7000), and I donât know how to determine the shape of y_pred.
The first number in the shape of `y_pred` is the batch size. It is defaulted to 32 (see this doc).
So you are subtracting with incompatible shapes.
1 Like
Oh, thanks for your help. So, do you have any suggestion to modify that? Do I need to put batch_size = 6999, in the model.fit?
You can do it that way, but do you want to learn mini-batch wise?
Yes, I really want to learn about batch-size; I donât have much information about it. Also, need to go more into detail about TensorFlow , I always have problems when it comes to changing the shape and type of NumPy arrays to tensors. I would be happy if you are able to recommend me some easy-to-understand references.
By the way, thank you so much for your help.
A few things:
1. If you want to do it mini-batch wise, then this mean `X_loss` is no longer a constant because it should change from batch to batch. Therefore, we also canât pass `X_loss` into the loss function in the way you are doing it now. Here is what you can change:
a. Learn to use Tensorflow Functional API to build a model instead of using `Sequential`. See this for an introduction, or search for more if needed. For example, we can actually build a model without `Sequential` like in below. See how we use `x` from place to place to make all the âconnectionsâ.
b. Take `X_loss` as a second Input. (Yes, you can have more than one Input to a model).
c. Build a `Lambda` layer which take `X_loss` and the output from your last Dense layer as inputs. A `Lambda` layer lets you define a custom function to process the inputs. You can do these things in the Lambda layer to produce a `lambda_layer_output `. (see the end of this reply).
d. Concatenate the `lambda_layer_output` with the output from your last Dense layer to form a new, final output.
e. In the loss function, it will still take y_true and y_pred as inputs, but your y_pred contains two things: (1) `lambda_layer_output` and (2) the output from your last Dense layer. Unpack them back, and do the last step of computing the loss. (see the end of this reply)
2. Regarding shapes, this is how I generally do this:
a. implement a custom loss function
b. add `tf.print` to each print each intermediate variableâs shape (e.g.
`tf.print(tf.shape(variable))`
c. find some small samples of `y_true` and `y_pred`, pass it through the loss function, and check if it produces expected returns.
d. Donât use it for model training until you have done step c to verify that it works as expected
Finally, for the steps I suggest in point number 1, you might not understand it until you try it out, and you might not be able to finish them unless you have studied enough examples from the internet. You might not make it work in your first try. However, these are all learning processes that I recommend you to go through.
Cheers,
Raymond
## What your lambda function can do:
`````` q1 = y_pred[:, 1] + y_pred[:, 4] - X_loss [:,0]
q2 = y_pred[:, 2] + y_pred[:, 3] + y_pred[:, 7] + y_pred[:, 17] + y_pred[:, 22] - X_loss [:,1]
q3 = y_pred[:, 4] + y_pred[:, 5] + y_pred[:, 7] + y_pred[:, 8] + y_pred[:, 9] + y_pred[:, 19] - X_loss [:,2]
q4 = y_pred[:, 3] + y_pred[:, 10] + y_pred[:, 11] + y_pred[:, 13]
q5 = y_pred[:, 14]
q6 = y_pred[:, 5] + y_pred[:, 7] + y_pred[:, 10] + y_pred[:, 15] + y_pred[:, 17] + y_pred[:, 20] + y_pred[:, 22]
q7 = y_pred[:, 11] + y_pred[:, 16] + y_pred[:, 19] + y_pred[:, 21]
q8 = y_pred[:, 5] + y_pred[:, 7] + y_pred[:, 8] + y_pred[:, 20] + y_pred[:, 21] + y_pred[:, 22] + y_pred[:, 23]
lambda_layer_output = 1000 * tf.reduce_mean(tf.square(q1)) + \
1000 * tf.reduce_mean(tf.square(q2)) + 1000 * tf.reduce_mean(tf.square(q3)) + \
1000 * tf.reduce_mean(tf.square(q4)) + 1000 * tf.reduce_mean(tf.square(q5)) + \
1000 * tf.reduce_mean(tf.square(q6)) + 1000 * tf.reduce_mean(tf.square(q7)) + \
1000 * tf.reduce_mean(tf.square(q8))
``````
## What your new custom loss function can do:
``````# unpack y_pred to extract lambda_layer_output and the output from your last Dense layer
loss_t = tf.reduce_mean(tf.square(y_true - output_from_your_last_dense_layer)) + lambda_layer_output
``````
1 Like
I appreciate your help. Thanks so much.
You are welcome @Nasim_Deljouyi!
Hi @rmwkwok. I have rebuilt my model using functional API as follows:
input_1 = keras.Input(shape=(n2,))
input_2 = keras.Input(shape=(n2,))
dense_1 = layers.Dense(16, activation = âreluâ)(input_1)
dense_2 = layers.Dense(64, activation = âreluâ)(dense_1)
dense_3 = layers.Dense(128, activation = âreluâ)(dense_2)
dense_4 = layers.Dense(64, activation = âreluâ)(dense_3)
output = layers.Dense(25, activation = âreluâ)(dense_4)
However, I have some questions. Can you please elaborate steps 1.c to 1.e a little more? I got confused. Shouldnât a lambda layer be in the format of " tf.keras.layers.Lambda(custom function)(inputs) "? The custom function is what you wrote at the end of the reply; does it return lambda_layer_output? Therefore, I donât need to define a custom loss function anymore?
And I donât need to pass input 2 to dense_1 to dense_4, right?
Hello @Nasim_Deljouyi
If you read carefully, there are two functions and none of them alone can totally replace your loss function.
I believe I have written down pretty clearly that there will be a lambda function, and there will be a loss function.
it makes sense to return something, and do you need to return `lambda_layer_output`?
I donât understand how you come to that conclusion.
I think you will need to figure it out yourself. Think why you need input 2, and where you want to use it.
@Nasim_Deljouyi, the approach I share with you will let you supply a small batch of `X_loss` each time, since you want to do it mini-batch wise. However, you will still need to figure out how to connect all the dots up. Good luck!
Cheers,
Raymond
Thanks so much for the clarification.
Best,
Nasim
Hi Raymond. I hope you are doing well.
If you recall, you suggested this approach for training models that incorporate the input of training examples into the loss function in a mini-batch-wise manner. I have implemented this approach and another approach to do so, and I get the same results from both when using a fixed seed for the random number generators. However, when I set the batch size equal to the number of training examples, I got the same results from both approaches but different from the approach I previously used and provided here, in which X_loss was assumed to be constant and was only able to work for the batch size equal to the training examples. I was wondering if you or anyone here has dealt with this issue before and can help me with it. | 2,749 | 9,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-10 | latest | en | 0.766367 |
https://www.middleschoolmathmoments.com/2013/12/prime-factorization-and-gcf.html | 1,726,718,424,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00107.warc.gz | 820,809,769 | 25,325 | ### Prime Factorization and GCF
What a week! One snow day and 2 late starts so far!
I just realized that I haven't written in almost 10 days.....I guess thinking about writing doesn't get the words on the page, huh?
Anyway, we worked on prime factorization at the end of last week, using both the factor tree method and the ladder method. Most of the students prefer the factor tree (I think that's because they've seen it before and feel comfortable with it), but a few like the ladder method. We've been using Prime Factorization Footloose to practice this skill; those who are finished have moved on to using the Factors and GCF Footloose cards.
We have discussed GCF briefly, several times, and it is really a review for them (based on their pretest results); the biggest problem here is that they often skip factors when they list them, and miss the GCF because of missing factors. Today, we're going to discuss using prime factorizations to find the GCF. I never learned this way when I was growing up, but I really like it because it eliminates the need to find every factor of a number. We'll see how they like it...
Have a great day!
### Differentiation and the Brain - Introduction
It's summer-time and time to get some reading done! Myself and my Tools for Teaching Teens collaborators are going to read and review Differentiation and the Brain, How Neuroscience Supports the Learner-Friendly Classroom , by David A. Sousa and Carol Ann Tomlinson.We will each be reviewing different chapters, and those blog posts will be linked together as we go. If you're interested in learning more about this book, check back and follow the links to the different chapters:) I'm going to give a quick review of the book introduction here, and then later today I'll be reviewing Chapter 1. According to the authors, differentiation is brain-friendly and brain-compatible! They describe the rise, fall, and rise of differentiation, starting with the one-room schoolhouses, where teachers taught all subjects to all students, of all ages, and HAD to differentiate - there was no other way! As the country's population grew, public schools grew, and students were separat
### Love to Doodle (and a freedbie)
Exponents Color by Number For most of my school life as a student (and even as an adult, during PD), I have really liked doodling! During lectures, discussions...it would help me focus, but also give me something to make me look busy, so I wouldn't get called on in class! I always hated being called on and almost never participated voluntarily:) I liked to draw cubes, rectangles, squiggly lines, etc, and color in different parts of the doodles. Download this freebie:-) I really wanted to make some color by number activities. Since I am not good at creating actual pictures, I decided to make my color by numbers similar to my random drawing/doodling. My Exponent Color by Number is most similar to my past doodles, but I thought it was a little too random, so I started using actual shapes. The Integer Operations Color by Number (freebie), as well as most of my other color by numbers are more structured, but so much fun for me to make! Computerized doodling! Anyone else | 691 | 3,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-38 | latest | en | 0.962535 |
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# Fractions NF Bundle - 4th Grade Common Core Math -Spiral Bound HARD COPY
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Fourth Grade Common Core Numbers and Operations Fractions (NF) Bundle Spiral Bound HARD COPY. This product listing is for a hard copy of this product.
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Click here to see the HUGE YEAR-LONG BUNDLE of ALL FOURTH GRADE Common Core Math Standard Packets.
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Fourth Grade Common Core Math Test Prep and Skills Review Bundle.
Don't let test prep or skills review stress you out! This bundle includes review and practice sheets to help you prepare your students and to track their areas of mastery. ALL 12 packets cover the 4th Grade Common Core standards of:
Numbers and Operations - Fractions (NF)
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The bundle includes these 12 Fractions (NF) 4th Grade Common Core Math Packets:
Each of the 12 packets includes:
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Student Mastery Checklist
The open-ended worksheets have the same problems as the multiple choice sheets. The 2 types of worksheets are offered for differentiated levels of learning.
You can purchase the packets separately here:
Equivalent Fractions (4th Grade Common Core Math: 4.NF.1)
Simplifying Fractions (4th Grade Common Core Math: 4.NF.1)
Common Denominators (4th Grade Common Core Math: 4.NF.1)
Comparing Fractions (4th Grade Common Core Math: 4.NF.2)
Order and Compare Fractions (4th Grade Common Core Math: 4.NF.2)
Fraction Subtraction (4th Grade Common Core Math: 4.NF.3)
Fractions and Mixed Numbers (4th Grade Common Core Math: 4.NF.3)
Fraction Multiplication (4th Grade Common Core Math: 4.NF.4)
Fractions and Decimals (4th Grade Common Core Math: 4.NF.6)
Fractions and Decimal Equivalency (4th Grade Common Core Math: 4.NF.5)
Decimal Comparison (4th Grade Common Core Math: 4.NF.7)
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Fractions Games, Posters, and Activity Packet
Fraction Multiplication Task Card and Poster Set
Fraction Division Task Card and Poster Set
Equivalent Fraction Fortune - Color the Path Activity Packet
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lecture_11
# lecture_11 - MIT OpenCourseWare http/ocw.mit.edu 2.161...
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MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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F m m Massachusetts Institute of Technology Department of Mechanical Engineering 2.161 Signal Processing - Continuous and Discrete Fall Term 2008 Lecture 11 1 Reading: Class Handout: Sampling and the Discrete Fourier Transform Class Handout: The Fast Fourier Transform Proakis and Manolakis (4th Ed.) Ch. 7 Oppenheim, Schafer & Buck (2nd Ed.) Chs. 8 & 9 1 The Discrete Fourier Transform continued In Lecture 10 the DFT pair associated with a sample set { f n } of length N was defined as N 1 F m = f n e j 2 πmn/N n =0 N 1 1 f n = F m e j 2 πmn/N N m =0 The value F m was interpreted as F (j Ω) evaluated at Ω = 2 π/N Δ T . 1.1 Organization of the DFT The N components in a DFT represent one period of a periodic spectrum. The first N/ 2 lines in the spectrum represent physical frequencies 0 . . . ( π/ Δ T ) radians/second. The components in the upper half of the sequence, F N/ 2+1 . . . F N 1 , may be considered to be the negative frequency components F N/ 2+1 . . . F 1 in the spectrum. It is common to translate the upper half of the record to the the left side of a plot to enhance the physical meaning. F N e q u i - s p m i n o n e p e N - 1 a c e d s a m p l e s r i o d o f F m i n t e r p r e t t o p N / 2 v a l u e s a s r e p r e s e n t i n g n e g a t i v e f r e q u e n c i e s i n F * ( j 9 ) . 0 N / 2 N - 1 N / 2 - 1 0 1 copyright c D.Rowell 2008 11–1
1.2 Spectral Resolution of the DFT The DFT pair provide a transform relationship between a pair of (complex) data sets { f n } and { F m } , each of length N . If the sampling interval associated with { f n } is Δ T units, the record duration is T = N Δ T. The frequency resolution, or line spacing, ΔΩ, in the DFT is 2 π 2 π 1 ΔΩ = = rad/s, or Δ F = Hz. (1) N Δ T T T and the frequency range spanned by the N lines in the DFT is 2 π 1 N ΔΩ = rad/s, or N Δ F = Hz. (2) Δ T Δ T The sequence { F m } represents both the positive and negative frequencies in a two-sided spectrum. The highest (positive) frequency component in the spectrum is half of this range (the Nyquist frequency), that is π 1 Ω max = rad/s, or F max = Hz. (3) Δ T T We conclude therefore, that the resolution within the DFT depends on the duration T of the data record, and the maximum frequency depends on the sampling interval Δ T . 1.3 Properties of the Discrete Fourier Transform Because the DFT is derived directly as a sampled continuous Fourier transform, it inherits most of the properties of the Fourier transform. We repeat some of the important properties here. In addition other properties are based on the assumed periodicity of { f n } and { F m } : 1. Linearity: If { f n } and { g n } are both length N , and { f n } DFT ⇐⇒ { F m } and { g n } DFT ⇐⇒ { G m } then DFT a { f n } + b { g n } ⇐⇒ a { F m } + b { G m } 2. Symmetry Properties of the DFT: If { f n } is a real-valued sequence then { F m } = F m from which it follows that {{ F m }} is an even function of m and {{ F m }} is an odd function of m .
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Ask a homework question - tutors are online | 1,125 | 3,880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-13 | latest | en | 0.773898 |
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Introduction
In the previous post I looked at some simple methods for calling C code from R using a simple Gibbs sampler as the motivating example. In this post we will look again at the same Gibbs sampler, but now implemented in Java, and look at a couple of options for calling that code from an R session.
Stand-alone Java code
Below is some Java code for implementing the bivariate Gibbs sampler discussed previously. It relies on Parallel COLT, which must be installed and in the Java CLASSPATH in order to follow the examples.
```import java.util.*;
import cern.jet.random.tdouble.*;
import cern.jet.random.tdouble.engine.*;
class Gibbs
{
public static void main(String[] arg)
{
if (arg.length != 3) {
System.err.println("Usage: java Gibbs <Iters> <Thin> <Seed>");
System.exit(1);
}
int N=Integer.parseInt(arg[0]);
int thin=Integer.parseInt(arg[1]);
int seed=Integer.parseInt(arg[2]);
DoubleRandomEngine rngEngine=new DoubleMersenneTwister(seed);
Normal rngN=new Normal(0.0,1.0,rngEngine);
Gamma rngG=new Gamma(1.0,1.0,rngEngine);
double x=0,y=0;
System.out.println("Iter x y");
for (int i=0;i<N;i++) {
for (int j=0;j<thin;j++) {
x=rngG.nextDouble(3.0,y*y+4);
y=rngN.nextDouble(1.0/(x+1),1.0/Math.sqrt(x+1));
}
System.out.println(i+" "+x+" "+y);
}
}
}
```
It can be compiled and run stand-alone from an OS shell with the following commands:
```javac Gibbs.java
java Gibbs 10 1000 1
```
As discussed in the previous post, it is possible to call any command-line program from inside an R session using the system() command. A small wrapper function for conveniently running this code from within R can be written as follows.
```gibbs<-function(N=10000,thin=500,
seed=trunc(runif(1)*1e6),
exec="Gibbs",
tmpfile=tempfile())
{
command=paste("java",exec,N,thin,seed,">",tmpfile)
system(command)
}
```
This can then be run from within an R session with a simple call to gibbs(). Note that a random seed is being generated within R to be passed to the Java code to be used to seed the COLT random number generator used within the Java code. As previously discussed, for many long running codes, this approach can be quite effective, and is clearly very simple. However, there is an overhead associated with the system() call, and also with writing output to disk and then reading it back again.
Using rJava
It is possible to avoid the overheads associated with the above approach by directly calling the Java code from R and having the return values returned directly into the R session from memory. There isn’t really direct support for this within the core R language, but there are couple of different solutions provided by R packages. The simplest and most popular approach seems to be the rJava package. This package can be installed with a simple
```install.packages("rJava")
```
This should “just work” on some OSs (eg. Windows), but may fail on other OSs if R is not aware of the local Java environment. If the installation fails, check the error message carefully for advice/instructions. On most Linux systems, the problem can be fixed by quitting R, then running the following command from the shell
```sudo R CMD javareconf
```
before re-starting R and re-attempting the installation. rJava provides a mechanism for starting a JVM within the running R session, creating objects, calling methods and having method return values returned to R. It is actually much more flexible than the .C() function for C code discussed in the previous post.
In order to use this package for our example, we must first re-factor the code slightly in the following way.
```import java.util.*;
import cern.jet.random.tdouble.*;
import cern.jet.random.tdouble.engine.*;
class GibbsR
{
public static void main(String[] arg)
{
if (arg.length != 3) {
System.err.println("Usage: java GibbsR <Iters> <Thin> <Seed>");
System.exit(1);
}
int N=Integer.parseInt(arg[0]);
int thin=Integer.parseInt(arg[1]);
int seed=Integer.parseInt(arg[2]);
double[][] mat=gibbs(N,thin,seed);
System.out.println("Iter x y");
for (int i=0;i<N;i++) {
System.out.println(""+i+" "+mat[0][i]+" "+mat[1][i]);
}
}
public static double[][] gibbs(int N,int thin,int seed)
{
DoubleRandomEngine rngEngine=new DoubleMersenneTwister(seed);
Normal rngN=new Normal(0.0,1.0,rngEngine);
Gamma rngG=new Gamma(1.0,1.0,rngEngine);
double x=0,y=0;
double[][] mat=new double[2][N];
for (int i=0;i<N;i++) {
for (int j=0;j<thin;j++) {
x=rngG.nextDouble(3.0,y*y+4);
y=rngN.nextDouble(1.0/(x+1),1.0/Math.sqrt(x+1));
}
mat[0][i]=x; mat[1][i]=y;
}
return mat;
}
}
```
This code can be compiled and run from the command-line just as the previous code could.
```javac GibbsR.java
java GibbsR 10 1000 1
```
However, we have now separated out the code we want to be able to call from R into a static method called gibbs, which runs the Gibbs sampler and stores the result in a 2-dimensional array which is its return value. We can now see how to call this code from within a running R session. We first need to set up the R environment ready to call the code.
```library(rJava)
.jinit()
obj=.jnew("GibbsR")
```
Line 1 loads the package, line 2 starts up the JVM, and line 3 creates a link to the the GibbsR class (in general this is used to create a new Java object of the given type, but here we are using static methods). Java methods are called on Java objects using .jcall(). We can write a simple R function to conveniently call the method as follows.
```jgibbs<-function(N=10000,thin=500,seed=trunc(runif(1)*1e6))
{
result=.jcall(obj,"[[D","gibbs",as.integer(N),as.integer(thin),as.integer(seed))
mat=sapply(result,.jevalArray)
mat=cbind(1:N,mat)
colnames(mat)=c("Iter","x","y")
mat
}
```
This can now be called with a simple jgibbs(). The first line of the function body carries out the actual method call. The return type of the method must be explicitly declared – “[[D” means a 2-dimensional array of doubles, using JNI notation. Care must also be taken to coerce the method parameters into the correct type that the Java method expects to receive. .jcall() is generally quite good at unpacking basic Java types into corresponding R types. However, the two dimensional array is here returned as an R list consisting of one-dimensional Java array objects. The unpacking is completed using the subsequent call to jevalArray() using sapply(), before the resulting matrix is tidied up and returned to the R session.
We have looked at a couple of very simple methods for calling Java code from an R session. The rJava package is a very flexible mechanism for integrating Java code into R.
I haven’t found a lot of tutorial-level material on the web for the rJava package. However, the package itself has very good documentation associated with it. Start with the information on the rJava home page. From an R session with the rJava package loaded, help(package="rJava") lists the available functions, all of which have associated documentation. ?.jinit, ?.jnew, ?.jcall and ?.jevalArray provide further background and information on the example covered here.
After that, the source code of R packages which use rJava are a useful source of further inspiration – look at the reverse-depends list for rJava in CRAN. In particular, the helloJavaWorld package is a tutorial for how to include Java code in an R package (read the associated vignette).
23 Responses to “Calling Java code from R”
1. application performance monitoring Says:
This is my first time i visit here. I wanted to thank you for this great read!! I definitely enjoying every little bit of it and love learning more on this topic. This is my first time i visit here. I found so many interesting stuff in your blog, it’s really helpful for me.
2. Ben Says:
Hi, Darren! I found your blog accidentally. You really made excellent posts on many topics. I’m also very interested about using Java in R. I’ve question about your code here. You said we can use the following code:
library(rJava)
.jinit()
obj=.jnew(“GibbsR”)
My question is that where should we put the file “GibbsR.class”? In the current R working directory? I tried this, but it seems that it doesn’t work. R says no definition for class “MyClass” is found. Could you explain a little bit more on this for me?
3. darrenjw Says:
I guess just anywhere in your Java CLASSPATH. For me it certainly works in the current directory, but then I have “.” in my \$CLASSPATH.
4. Gibbs sampler in various languages (revisited) « Darren Wilkinson's research blog Says:
[…] It is easy to extend R using C, C++ and Java. I have shown in previous posts how to do this using Java and using C, and the recent post by Dirk shows how to extend using C++. Although interesting, this […]
5. Faster Gibbs sampling MCMC from within R « Darren Wilkinson's research blog Says:
[…] options turn out to be. The post draws heavily on my previous posts on calling C from R and calling Java from R, as well as Dirk Eddelbuettel’s post on calling C++ from R, and it may be helpful to consult […]
6. Joe Hightower Says:
I am having similar difficulty to one of the previous posters. I have specified a ,jinit(classpath… that seems to return a code indicating success. When I run .jnew however i get the error
java.lang.ClassNotFOundException
My code:
library(rJava)
.jinit(classpath=”C:/Users/jch1748/Documents/Projects/W2009028 – Pattern Value Simulation Study/Rsimple.jar”)
s <- .jnew(“Rsimple”)
Trying to figure out why I get the error. The Java programmer did not use the same statements inside the Java code to define class and methods. Are those critical or can they be anything that Java allows?
7. Catalogue of my first 25 blog posts « Darren Wilkinson's research blog Says:
[…] Calling Java code from R: how to call a Gibbs sampler written in Java from […]
8. Sankha Narayan Guria Says:
sometimes it may be required to call R code from Java… then we can use Java2R. Its a relatively new project at https://github.com/sankha93/java2r
9. Jess Lim Says:
i cannot get ur point, can u show other simple example?
my task is : get an input value from java and pass the value to R . Then in R, add input value with one then pass the (input+1) value back to java.
this is what i wan to do…..
• darrenjw Says:
The post is about calling Java from R. It seems that you want to call R from Java. This is also possible, but not something I’ve done. I think you can do it using RSJava, and the previous commenter mentioned another possibility.
10. Jess Lim Says:
i set an example in http://www.mail-archive.com/r-help@stat.math.ethz.ch/msg76159.html
but an error occur
Error in .jcall(“my_convolve”, “[D”, “convolve”, x, y) :
RcallMethod: cannot determine object class
because the directory of classpath?
11. R Meets Java: An Absolute Beginners’ Introduction | Gage Theory Says:
[…] fire up R and load the rJava package. The helloJavaWorld vignette and Darren Wilkinson offer good examples on how to get started for people with some experience with Java. We’ll follow […]
Thank you for the great article. It has been a huge help. For those of you having a few problems, here are some things that helped me. I have been working with R a lot, and am pretty new to java. Some of these may be obvious.
1) to pass a 2d matrix, try using .jarray(MyMatrix,dispatch=T). The “dispatch=T” seems to make sure it does go in as 2d.
2) as.double(MyMatrix) does not seem to always do the trick. Each element must be of the correct format.
3) .jclassPath() shows you all the current class paths. Another can easily be added using .jaddClassPath(path). You can easily add reference to a “.jar” this way. Don’t forget to reference the class with “package.class”. The class name alone will not do it.
i am not able to call my java code in R..i hv installed RJava package..
library(rJava)
> .jinit(classpath=”C:\Code-sequentialpattern Mining\SPMF.jar”)
Error: unexpected input in “.jinit(classpath=””
> s <- .jnew(“SPMF.jar”)
Hi..i am getting above..when i am calling java from R..i have installed RJava package
I have given correct path
Kindly help me..
15. Yanchang Zhao Says:
Thanks a lot. Successfully called Java code from R, after a lot of trials and errors on calling a Java function which returning a vector of complex objects. However, still haven’t figured out how to call from R a Java function which returns a list of objects.
Thank you so much, I have been looking for an example on how to translate a Java matrix to a R matrix. You sapply example did it.
Reblogged this on Thad is NOT food and commented:
This blog helped me solve an issue that I was having with rJava.
18. Saisai Says:
Thank you for the great article. It has been a huge help. But I encountered some problem, when calling .jar file from R.
First, I wrote a very simple example in java, and generated a .jar file named Add_add.jar.
public static void main(String[] args) {
… // initial variable a and b
int result = calc(a,b);
}
public static int calc(int a, int b){
int c = a + b;
return c;
}
}
Second I set up the R environment ready to call the code.
>library(rJava)
> jobject .jcall(jobject ,”I”,”calc”,as.integer(2),as.integer(3))
Then there is an error like this:
Error in .jcall(jobject, “I”, “calc”, as.integer(2), as.integer(3)) : method calc with signature (II)I not found
I am so confused that why it cannot be found. Could you please give me a suggestion how to fix it?
Thank you very much!
19. Moi Says:
You really made excellent posts on many topics.
Thank you
20. Amina Says:
classes I need are in a package called “spaces” which is in a subfolder of my “app.jar”. And I can not access it:
> library (rjava)
>. jinit (classpath = “app.jar”)
>. jpackage (spaces, “app / microbes / app2″)
> obj =. jnew (“Class1″)
Here “app / microbes / app2″ is the way to access the package spaces
but R found neither the package nor spaces Class1
Could someone help me?
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Inverted Siphons (Depressed Sewers)
Stormwater and wastewater sewers often encounter obstructions such as rivers, other pipes, subways, tunnels, or valleys. To pass these obstructions, a common method is for the sewer pipe to drop sharply, then run horizontal under the obstruction, and finally rise to the desired elevation. The piping going under the obstruction is traditionally called an "inverted siphon", but since the pipe is not actually acting as a siphon, a better term is "depressed sewer" (Metcalf and Eddy, 1981).
Unlike the main sewer pipe, the depressed sewer pipe(s) flow under pressure. Special care must be taken in inverted siphon design since losses are greater for pressurized flow, and the velocity in the depressed sewer must be at least 4 ft/s (1.2 m/s) for storm water or 3 ft/s (0.9 m/s) for sewage (Metcalf and Eddy, 1981). Therefore, even if there is only one main sewer pipe, several depressed pipes may be required.
We are currently completing a calculation to determine the diameters of depressed sewer pipes (inverted siphons) based on the discharge. The calculation will compute pipe diameters and velocities, as well as pipe inlet invert elevations and wall heights in the inlet chamber. If you are unfamiliar with inverted siphons, we will have diagrams on our calculation page.
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
http://www.LMNOeng.com
Reference:
Metcalf and Eddy, Inc. George Tchobanoglous, editor. Wastewater Engineering: Collection and Pumping of Wastewater. McGraw-Hill, Inc. 1981.
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com.
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, Ohio 45701 USA +1(740) 592-1890
LMNO@LMNOeng.com
Static Pressure - New calculation (free)
http://www.LMNOeng.com/Statics/pressure.htm
Often people ask us, "How do you convert 40 psi to feet of water?" or "How many meters are in 1 atmosphere?" We decided to write a calculation that performs these conversions.
Conversions between pressure and elevation are known as static pressure equivalences, and the equation is P=dgh; where P=pressure, d=mass density, g=gravitational acceleration, and h=elevation. Solving the equation for h allows us to determine that 92.4 ft. of water is equivalent to 40 psi. Since the density of water varies somewhat with temperature, it is important to state the temperature used for the density. I used 20C. You might have noticed that the unit conversions can be cumbersome. Our calculation takes care of the unit conversions for you.
If you are curious how many meters are in one atmosphere, compute it yourself. The calculation does not require registration. Just remember that you must decide what fluid you want to determine the meters of. One atmosphere is equivalent to more meters of oil than meters of water (because oil has a lower density).
Please see http://www.LMNOeng.com/Statics/pressure.htm to run the calculation.
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
http://www.LMNOeng.com
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com.
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, Ohio 45701 USA (740) 592-1890
LMNO@LMNOeng.com
New Culvert Design Calculation using Inlet and Outlet Control
http://www.LMNOeng.com/Pipes/hds.htm
Culverts have been utilized for thousands of years as a means to transmit water under walkways and roads. Too often, culverts are selected without sufficient thought of how much water needs to be convey under extreme conditions. If a culvert cannot convey all of the incoming water, then the water will flow over or around the pipe - or simply back up behind the culvert creating a pond or reservoir. If any of these conditions are unacceptable, then the proper culvert diameter and number of culverts must be selected prior to installation in order to convey all of the anticipated water through the pipe(s).
Our new culvert design calculation aids the designer in selecting the number of culverts and culvert diameter. It also plots headwater depth vs. discharge so that the designer can view culvert performanace over a wide range of flows. Our calculation is primarily based on the methodology presented in Hydraulic Design of Highway Culverts by Normann (1985) and published by the USA Department of Transportation's Federal Highway Administration.
Please see http://www.LMNOeng.com/Pipes/hds.htm to run the calculation and to see equations, diagrams, and additional description.
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
http://www.LMNOeng.com
Reference:
Normann, J. M. 1985. Hydraulic design of highway culverts. HDS-5 (Hydraulic Design Series 5). FHWA-IP-85-15. NTIS publication PB86196961. Obtainable at http://www.ntis.gov.
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com .
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, Ohio 45701 USA (740) 592-1890
LMNO@LMNOeng.com
Culvert Design
We will soon be loading a new culvert design calculation. It will aid in design and analysis of circular culverts that flow under a road or dam. The calculation uses equations for inlet control and outlet control. Inlet control means that flow through the culvert is limited by culvert entrance characteristics. Outlet control means that flow through the culvert is limited by friction between the flowing water and the culvert barrel.
Inlet control most often occurs for short, smooth, or greatly downward sloping culverts. Outlet control governs for long, rough, or slightly sloping culverts. The type of control also depends on the flowrate. For a given culvert installation, inlet control may govern for a certain range of flows while outlet control may govern for other flowrates.
The calculation also will compute headwater depth for high flowrates that exceed the capacity of the culvert(s), resulting in flow over a road or dam. It will also have a graphing feature where the user enters minimum and maximum flowrates. Then, a graph of headwater depth vs. flowrate is shown.
Since 1998, we have had a calculation for circular culvert flow using Manning's equation ( http://www.LMNOeng.com/CircularCulvert.htm ). Manning's equation is the equation most commonly used for simulating outlet control. Our new calculation will implement Manning's equation for outlet control and weir/orifice equations for inlet control.
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
http://www.LMNOeng.com
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com.
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, Ohio 45701 USA (740) 592-1890
LMNO@LMNOeng.com
Before founding LMNO Engineering in 1998, I taught fluid mechanics to third year civil and mechanical engineering students at Ohio University (USA). The concept of "head loss" always seemed to be difficult for students to understand.
Head loss is part of the energy equation. One of our first (and free) calculation pages, http://www.LMNOeng.com/energy.htm, shows the energy equation in one of its simpler forms. The equation is valid for flow in pressure pipes as well as open channels. Head loss is the sum of energy in the fluid at an upstream location minus the sum of energy at a downstream location. Energy consists of elevation, pressure, and velocity. Head loss has a positive value.
Introductory courses in fluid mechanics often begin with a discussion of inviscid fluids. An inviscid fluid has no viscosity. No such fluid exists in reality, but the concept is useful as a first step in explaining the Bernoulli and energy equations. An inviscid fluid has no head losses; upstream energy minus downstream energy equals zero. Head losses arise from fluid viscosity and the friction it causes between the moving fluid and the stationary pipe (or channel) walls.
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
http://www.LMNOeng.com
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com.
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, Ohio 45701 USA (740) 592-1890
LMNO@LMNOeng.com
Parshall Flume - Submerged and Free Flow
http://www.LMNOeng.com/Flumes/parshall.htm
LMNO Engineering's newest flume calculation computes discharge and rating curves for free flowing or submerged Parshall flumes. A free flowing flume can be identified by the drop in water depth at the flume throat. In submerged flow, the downstream water backs up into the throat swallowing the drop - making the drop difficult or impossible to identify. Analysis of submerged flow requires two head measurements - one in the approach channel and one in the throat; whereas, free flow requires only the upstream head measurement. Our Parshall flume calculation is based on the ISO 9826 (1992) standard.
Graphs of discharge versus head and discharge versus submergence ratio can be prepared on the web page. You can see that increasing the submergence ratio causes the discharge to decrease for a constant approach head. (Submergence ratio is defined as throat head divided by approach head.) The Parshall flume equations and methodology are described on the web page.
Our other flume calculation ( http://www.LMNOeng.com/Flumes/flumes.htm ) analyzes free flowing trapezoidal, rectangular, U-shape, and Parshall flumes. Both parshall.htm and flumes.htm use identical equations for free flowing Parshall flumes.
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
http://www.LMNOeng.com
Reference:
International Organization of Standards (ISO 9826). 1992. Measurement of liquid flow in open channels - Parshall and SANIIRI flumes. Reference number: ISO 9826:1992(E).
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com.
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, Ohio 45701 USA (740) 592-1890
LMNO@LMNOeng.com
Price increase
We anticipate raising the rates for our password-protected calculations, and wanted to give advance notice to our newsletter readers - in case you wish to register at the current rates. Currently, we charge \$20 (US Dollars) for 7 days' access to the password-protected calculations. We anticipate raising this to \$30. For a 1 year registration, our current fee is \$100, and we anticipate raising this to \$300 [actually changed to \$200 instead of \$300- KE, Aug 2001]. We will also be changing our 6 month rate, which is currently \$70.
If you wish to register at the current rates, please register within the next few days since the rates will be increasing in the next few weeks. Our registration page is http://www.LMNOeng.com/register.htm .
We are continually developing new calculations for the website, and appreciate your past and future suggestions.
Regards,
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
http://www.LMNOeng.com
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com.
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, Ohio 45701 USA (740) 592-1890
LMNO@LMNOeng.com
Open channel flow measurement
We offer calculations for three commonly used methods for open channel flow measurement - weirs, flumes, and end depth. The end depth method is the simplest because a structure does not need to be built - water drops freely from the downstream end of a culvert or channel. All that is needed are the dimensions of the culvert or channel and the water depth. Freely discharging culverts are widely used as discharge structures, and a picture of one is shown at http://www.LMNOeng.com/Waterfall/CulvertDischarge.htm.
Weirs are used for flow measurement when large head losses are acceptable and free discharge can be accommodated. Weirs are relatively inexpensive to construct, install, and operate. However, weirs will back up the flow since they are obstructions across the channel width and cause low velocities upstream of the weir. Sediment will build up behind the weir. A simple triangular (or V-notch) weir is shown at http://www.LMNOeng.com/Weirs/vweir.htm.
Flumes have been the topic of our last two newsletters. They are more expensive than weirs but have the advantage of much less head loss. They are flow-through devices that do not cause the water to back up like weirs do. There are various types of flumes which are designed to allow varying ranges of discharge through them while minimizing sediment build-up and head loss. A flume photograph can be found at http://www.LMNOeng.com/Flumes/flumes.htm.
Thank you for your interest in the LMNO Engineering website,
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
http://www.LMNOeng.com
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com.
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, Ohio 45701 USA (740) 592-1890
LMNO@LMNOeng.com
Flume calculations and rating curves
http://www.LMNOeng.com/Flumes/flumes.htm
Our first graphing calculator is now on-line. It has our usual calculation format plus a graph below the calculation for plots of flowrate versus water depth (head).
In the calculation, flowrate can be computed for four different types of flumes - Parshall, trapezoidal, rectangular, and U shaped. Flumes, like weirs, require a water depth measurement. Then, equations or tables are used to obtain discharge. Flumes are mathematically more complicated to analyze than weirs and have more complicated construction; however, they offer less energy loss and can pass sediment much more readily than weirs.
Flumes have been studied for many decades. The International Organization of Standardization (ISO) has published complex methodologies relating flume discharge to water depth. The methodology is suited to a computer since it is not a simple algebraic equation. Discharge and rating curves can be obtained with a click of your mouse button on our flumes page, but we also present the ISO methodology so that the calculation is not simply a "black box."
Thank you for your interest in the LMNO Engineering website,
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
http://www.LMNOeng.com
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com.
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, Ohio 45701 USA (740) 592-1890
LMNO@LMNOeng.com
Flumes for open channel flow measurement
Flumes, as used in this newsletter, refer to open channel flow measuring devices. Within the next few weeks, we will have a new calculation on our website for four different flume types - Parshall, Rectangular, Trapezoidal, and U-shaped.
Flumes are available in various widths. Usually, the maximum expected depth is fixed by the channel characteristics where the flume is installed, and in no case should exceed 2 m.
Flumes are designed to force a transition from sub-critical to super-critical flow. Such a transition causes flow to pass through critical depth at the flume throat. At the critical depth, energy is minimized and there is a direct relationship between water depth and velocity (and flowrate). However, it is physically very difficult to measure critical depth in a flume because its exact location is difficult to determine and may vary with flowrate. Through mass conservation, the upstream depth is related to the critical depth. Therefore, flowrate can be determined by measuring the upstream depth, which is a highly reliable measurement.
Our flume calculators will be based on ASTM and ISO standards for flumes. These standards were developed from theoretical relationships and modified by experimental observations.
Thank you for your interest in the LMNO Engineering website,
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
http://www.LMNOeng.com
ASTM is American Society for Testing and Materials.
ISO is International Standards Organization.
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com.
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, OH 45701 USA (740) 592-1890
LMNO@LMNOeng.com
New Calculation - Packed Bed
http://www.LMNOeng.com/Groundwater/PackedBed.htm
Chemical engineers know this as a packed or porous bed. Groundwater hydrologists call this a permeameter or flow through porous media. It is a column containing particles that any fluid can be flowing through. The column can be of any orientation (upflow, downflow, sideways flow). The calculation can compute flowrate and velocity through the column, or pressure loss, or column length.
We provide two computation methods - Idelchik method and Darcy's law. Hydrogeologists and civil engineers are usually more familiar with Darcy's law while chemical and mechanical engineers may be more familiar with the Idelchik approach. The Idelchik method is valid for laminar or turbulent flow while Darcy's law is valid only for laminar flow. Darcy's law requires entering the permeability while Idelchik relies only on particle size and porosity. If the Idelchik method is selected, permeability will be back-calculated (based on Darcy's law) in case one wishes to compare a bed material with a soil type.
The calculation also computes a minor loss coefficient for flow through the bed in case the bed is part of a longer pipeline that you are modeling. Any liquid or gas can be used with either method so long as the fluid's density and viscosity are known. Several fluids (e.g. water, gasoline, oil, air, nitrogen, mercury, etc.) have properties built into the program. The calculation works in demonstration mode for Mercury as the fluid.
Thank you for your interest in the LMNO Engineering website,
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
http://www.LMNOeng.com
References
Darcy's law: Freeze, R. A. and J. A. Cherry. 1979. Groundwater. Prentice Hall, Inc.
Idelchik method: Fried, E. and I. E. Idelchik. 1989. Flow Resistance: A Design Guide for Engineers. Hemisphere Pub. Corp.
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com.
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, OH 45701 USA (740) 592-1890
LMNO@LMNOeng.com
Pipe Network Example
http://www.LMNOeng.com/Pipes/PipeNetwork.htm
I will present an example of how to determine the height of a city water tower using our Pipe Network calculation. Fluid: water at 20C. Pipe material: ductile iron (cast iron). Flow in gpd (gallons (US)/day), Elevations in ft, Diameters in inch, Pressure in psi, Head loss in psi, Lengths in ft, Z+P/S in ft. The water tower will be at node D.
In this example, the water tower is placed on a hill. The hill elevation is 50 ft. above all of the other nodes. All other nodes have the same elevation. All pipes are 10 inch inside diameter and 1000 ft. long. Each outfow node represents a collection of businesses or houses. For simplicity, let's say the required flows out of nodes A-C and E-I is 500,000 gpd each and the pressure requirement is 100 psi at each node. Therefore, set the pressure at the furthest node from the water tower to 100 psi to guarantee that all nodes will have at least 100 psi pressure. The node furthest from the water tower will be C or I (both are the same distance from node D since all pipes have the same length). I'll use node I.
Summary of inputs: Select "P known at node I". Enter Q node for nodes A-C and E-I as -5e5 gpd (be sure to use the negative sign since these are withdrawals). Enter Q node for node D as 4e6 gpd (this number is positive since it is the inflow to the system. I got 4e6 from 5e5 x 8 nodes). Enter the elevation of node D as 50 ft and all other node elevations as 0.0 ft. Enter the pressure at node I as 100 psi. Enter the diameter of each pipe as 10 inches and the length of each pipe as 1000 ft.
After making the proper data entries, click the "Calculate" button and look at the results. All node pressures A-C and E-I are at least 100 psi as required. Look at the node D results: the water tower required height is 238.71964 ft - 50 ft = 189 ft. (rounding to the nearest ft). The pressure of 81.7 psi is at the base of the tank (at an elevation of 50 ft). The pipe "H,V,Re pipe" fields are scrollable with your arrow keys, so you can see the head loss, velocity, and Reynolds number for each pipe. You can see the flowrate in each pipe and the direction of flow from the arrows. You might try reducing the diameters for pipes 5 and 10 to save money since they don't carry much flow.
A copy of this example is viewable at http://www.LMNOeng.com/Pipes/example3(4).htm. Note that the "H,V,Re pipe" field is not scrollable in the gif file.
I hope this example has been helpful,
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
http://www.LMNOeng.com
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com .
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, OH 45701 USA (740) 592-1890
LMNO@LMNOeng.com
Pipe Network Calculation
http://www.LMNOeng.com/Pipes/PipeNetwork.htm
Many of you asked for a pipe network calculation - it is now completed and on-line. It allows up to 12 pipes and 9 nodes. You can simulate a system that has many series and parallel pipes or you can model a single pipeline (or manifold) that has up to 12 pipes of different diameters and lengths in series with inflows or outflows at the nodes (junctions) between the pipes. You do not have to use all the pipes or nodes.
The program can simulate flow of any liquid or gas through the pipes. The Darcy-Weisbach (Moody diagram method) or Hazen-Williams method may be selected for losses. After entering pipe diameters, lengths, node elevations, node inflows, node outflows, and pressure at a single node, the program computes flowrate, loss, velocity, and Reynolds number for each pipe and the pressure and hydraulic head at all existing nodes.
The calculation has a demonstration mode, so it will run even if you are not a registered user. The demonstration mode allows all pipes and nodes but only works for mercury flowing through wood pipes with SI (metric) units.
Experiment with PipeNetwork. If you would like to purchase it for stand-alone use to run from your hard disk without an internet connection, the cost is \$150 (US Dollars). To use all features on-line, the cost is just \$20/week or \$100/year; the on-line fee enables all calculations (not just PipeNetwork).
Enjoy the site! Fluid flow is an exciting and challenging field!
Sincerely,
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
http://www.LMNOeng.com
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com.
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, OH 45701 USA (740) 592-1890
LMNO@LMNOeng.com
What calculation should I use? Have you ever visited our site for a specialized calculation and didn't see it listed?
Need the flowrate through a pinhole in a leaky pipe? Try our Bernoulli calculator: http://www.LMNOeng.com/Flow/bernoulli.htm.
Need to know if a pump is required to carry water through your pipe? Try our Darcy-Weisbach or Hazen-Williams calculations without pump curve. Solve for pump head. If a pump is required, pump head will be positive. http://www.LMNOeng.com/DarcyWeisbach.htm or http://www.LMNOeng.com/HazenWilliamsDesign.htm.
Need to know if a blower is required to carry air through your pipe? Try our Darcy-Weisbach calculation without pump curve: http://www.LMNOeng.com/DarcyWeisbach.htm. Solve for pump head like above. Note that you can't use Hazen-Williams for air since Hazen-Williams is only valid for water.
Need to know the flowrate through a pipe with a pump already installed? Try Darcy-Weisbach with pump curve (any liquid or gas) or Hazen-Williams with pump curve (water only): http://www.LMNOeng.com/Pipes/DWpump.htm or http://www.LMNOeng.com/Pipes/HWpump.htm.
Need to know pressure change in a pipe due to an expansion or contraction? Try our Bernoulli calculator: http://www.LMNOeng.com/Flow/bernoulli.htm.
Need a Moody friction factor? Try our Moody friction factor calculation: http://www.LMNOeng.com/moody.htm.
Need to determine velocity using a pitot tube? Try our Bernoulli calculation: http://www.LMNOeng.com/Flow/bernoulli.htm.
Need to determine discharge over a dam but our rectangular weir calculation gives "parameter out of range" messages? Try our Bernoulli calculation - not as accurate but no limits on the variables: http://www.LMNOeng.com/Flow/bernoulli.htm.
Need the flowrate through an orifice plate, but our orifice calculation gives you a "parameter out of range" message? Try our Bernoulli calculator - not as accurate but no limits on the variables: http://www.LMNOeng.com/Flow/bernoulli.htm.
Need to know pond storage volume required to attenuate a flood? Try our Detention basin storage calculation: http://www.LMNOeng.com/Hydrology/storage.htm.
Need to analyze a network of pipes? By the time our next newsletter is published, we expect to have a pipe network calculation on the website. It will be able to handle up to 12 pipes with 9 inflow (or outflow) nodes. You will have a choice of Darcy-Weisbach or Hazen-Williams for the friction losses.
Have another question? Send me an e-mail ( LMNO@LMNOeng.com) or give me a call (USA: 740/592-1890).
Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
LMNO Engineering, Research, and Software, Ltd.
You received this free newsletter because you requested it at our website. If you no longer wish to receive it, send a message stating "Discontinue Newsletter" to LMNO@LMNOeng.com.
(c) 2001 LMNO Engineering, Research, and Software, Ltd.
LMNO Engineering, Research, and Software, Ltd.
The fluid flow calculations website: http://www.LMNOeng.com
7860 Angel Ridge Rd. Athens, Ohio 45701 USA (740) 592-1890
LMNO@LMNOeng.com
Liquid or Gas Flow in Pipes with Pump Curve
http://www.LMNOeng.com/Pipes/DWpump.htm
In our newsletter of October 31, 2000, we introduced a calculation for water flow in pipes using the Hazen-Williams equation with a pump curve. We have now completed a similar program except it can handle any liquid or gas.
The new program uses the Darcy-Weisbach equation for friction losses (major losses) and also allows minor losses (valves, pipe bends, etc.). The calculation automatically intersects a system curve with a pump curve to tell you the operating point (flowrate and total dynamic head). Alternatively, if you know the flowrate or velocity, you can solve for diameter, pipe length, pressure difference, elevation difference, or the sum of the minor loss coefficients.
To keep the input data relatively simple, we only require you to enter two points on the pump curve - flow at zero head and head at zero flow. A parabolic curve is then formed between the two points. All equations and methodology are described on the web page http://www.LMNOeng.com/Pipes/DWpump.htm.
Thank you for your interest in the LMNO Engineering website, http://www.LMNOeng.com
Newsletter written by Ken Edwards, Ph.D., P.E. (Owner/Engineer/Programmer)
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(c) 2001 LMNO Engineering, Research, and Software, Ltd. | 7,577 | 31,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-34 | longest | en | 0.901573 |
https://socialresearchmethods.net/kb/randomized-block-designs/ | 1,582,357,132,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145654.0/warc/CC-MAIN-20200222054424-20200222084424-00530.warc.gz | 564,802,691 | 4,852 | Randomized Block Designs
# Randomized Block Designs
The Randomized Block Design is research design’s equivalent to stratified random sampling. Like stratified sampling, randomized block designs are constructed to reduce noise or variance in the data (see Classifying the Experimental Designs). How do they do it? They require that the researcher divide the sample into relatively homogeneous subgroups or blocks (analogous to “strata” in stratified sampling). Then, the experimental design you want to implement is implemented within each block or homogeneous subgroup. The key idea is that the variability within each block is less than the variability of the entire sample. Thus each estimate of the treatment effect within a block is more efficient than estimates across the entire sample. And, when we pool these more efficient estimates across blocks, we should get an overall more efficient estimate than we would without blocking.
Here, we can see a simple example. Let’s assume that we originally intended to conduct a simple posttest-only randomized experimental design. But, we recognize that our sample has several intact or homogeneous subgroups. For instance, in a study of college students, we might expect that students are relatively homogeneous with respect to class or year. So, we decide to block the sample into four groups: freshman, sophomore, junior, and senior. If our hunch is correct, that the variability within class is less than the variability for the entire sample, we will probably get more powerful estimates of the treatment effect within each block (see the discussion on Statistical Power). Within each of our four blocks, we would implement the simple post-only randomized experiment.
Notice a couple of things about this strategy. First, to an external observer, it may not be apparent that you are blocking. You would be implementing the same design in each block. And, there is no reason that the people in different blocks need to be segregated or separated from each other. In other words, blocking doesn’t necessarily affect anything that you do with the research participants. Instead, blocking is a strategy for grouping people in your data analysis in order to reduce noise – it is an analysis strategy. Second, you will only benefit from a blocking design if you are correct in your hunch that the blocks are more homogeneous than the entire sample is. If you are wrong – if different college-level classes aren’t relatively homogeneous with respect to your measures – you will actually be hurt by blocking (you’ll get a less powerful estimate of the treatment effect). How do you know if blocking is a good idea? You need to consider carefully whether the groups are relatively homogeneous. If you are measuring political attitudes, for instance, is it reasonable to believe that freshmen are more like each other than they are like sophomores or juniors? Would they be more homogeneous with respect to measures related to drug abuse? Ultimately the decision to block involves judgment on the part of the researcher.
## How Blocking Reduces Noise
So how does blocking work to reduce noise in the data? To see how it works, you have to begin by thinking about the non-blocked study. The figure shows the pretest-posttest distribution for a hypothetical pre-post randomized experimental design. We use the ‘X’ symbol to indicate a program group case and the ‘O’ symbol for a comparison group member. You can see that for any specific pretest value, the program group tends to outscore the comparison group by about 10 points on the posttest. That is, there is about a 10-point posttest mean difference.
Now, let’s consider an example where we divide the sample into three relatively homogeneous blocks. To see what happens graphically, we’ll use the pretest measure to block. This will assure that the groups are very homogeneous. Let’s look at what is happening within the third block. Notice that the mean difference is still the same as it was for the entire sample – about 10 points within each block. But also notice that the variability of the posttest is much less than it was for the entire sample.
Remember that the treatment effect estimate is a signal-to-noise ratio. The signal in this case is the mean difference. The noise is the variability. The two figures show that we haven’t changed the signal in moving to blocking โ โ there is still about a 10-point posttest difference. But, we have changed the noise โ โ the variability on the posttest is much smaller within each block that it is for the entire sample. So, the treatment effect will have less noise for the same signal.
It should be clear from the graphs that the blocking design in this case will yield the stronger treatment effect. But this is true only because we did a good job assuring that the blocks were homogeneous. If the blocks weren’t homogeneous โ โ their variability was as large as the entire sample’s โ โ we would actually get worse estimates than in the simple randomized experimental case. We’ll see how to analyze data from a randomized block design in the Statistical Analysis of the Randomized Block Design. | 1,023 | 5,170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-10 | latest | en | 0.931241 |
https://plainmath.org/secondary/calculus-and-analysis/precalculus/trigonometry | 1,718,932,195,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00457.warc.gz | 402,178,136 | 26,055 | # Practice Exercises for Trigonometry
Recent questions in Trigonometry
Monverdeitum 2023-03-27
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## How to find the value of $\mathrm{csc}\left(\frac{5\pi }{4}\right)$?
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## $\left(\mathrm{sin}4\theta \right)$ can be written as:A) $4\mathrm{sin}\theta \left(1-2{\mathrm{sin}}^{2}\theta \right)\left(\sqrt{1-{\mathrm{sin}}^{2}\theta }\right)$B) $2\mathrm{sin}\theta \mathrm{cos}\theta {\mathrm{sin}}^{2}\theta$ C) $4\mathrm{sin}\theta –6{\mathrm{sin}}^{3}\theta$ D) None of these
Sawyer Burch 2023-03-22 | 750 | 2,042 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-26 | latest | en | 0.452472 |
https://studysoup.com/tsg/16208/conceptual-physics-12-edition-chapter-35-problem-1rq | 1,631,873,733,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055632.65/warc/CC-MAIN-20210917090202-20210917120202-00420.warc.gz | 613,358,208 | 12,246 | ×
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# If you walk at 1 km/h down the aisle of a train that moves
ISBN: 9780321909107 29
## Solution for problem 1RQ Chapter 35
Conceptual Physics | 12th Edition
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Problem 1RQ
If you walk at 1 km/h down the aisle of a train that moves at 60 km/h, what is your speed relative to the ground?
Step-by-Step Solution:
Solution 1RQ Step 1: This is the case of a person walking with 1 km/hr speed inside a 60 km/hr speed train. So, for an external observer at ground, the speed of the train will be 60 km/hr. The speed of the man walking inside the train is given as, v ‘ = 1 km /hr Therefore, the exact speed of the person inside the train relative to the observer at ground will be, V = v + v ‘ Where, v - velocity of the train v ‘ - velocity of the person walking inside the train Therefore, V = 60 km/hr + 1 km / hr = 61 km /hr Conclusion: So, the speed of the person relative to the ground will be 61 km/hr.
Step 2 of 1
##### ISBN: 9780321909107
Conceptual Physics was written by and is associated to the ISBN: 9780321909107. This full solution covers the following key subjects: aisle, down, Ground, moves, relative. This expansive textbook survival guide covers 45 chapters, and 4650 solutions. This textbook survival guide was created for the textbook: Conceptual Physics, edition: 12. The full step-by-step solution to problem: 1RQ from chapter: 35 was answered by , our top Physics solution expert on 04/03/17, 08:01AM. Since the solution to 1RQ from 35 chapter was answered, more than 409 students have viewed the full step-by-step answer. The answer to “If you walk at 1 km/h down the aisle of a train that moves at 60 km/h, what is your speed relative to the ground?” is broken down into a number of easy to follow steps, and 25 words.
#### Related chapters
Unlock Textbook Solution | 575 | 2,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2021-39 | latest | en | 0.87352 |
https://stat.ethz.ch/pipermail/r-help/2007-June/134361.html | 1,436,023,324,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375096773.65/warc/CC-MAIN-20150627031816-00172-ip-10-179-60-89.ec2.internal.warc.gz | 965,572,633 | 2,657 | # [R] [Not R question]: Better fit for order probit model
Robert A LaBudde ral at lcfltd.com
Sat Jun 16 04:51:15 CEST 2007
```At 09:31 PM 6/15/2007, adschai wrote:
>I have a model which tries to fit a set of data with 10-level
>ordered responses. Somehow, in my data, the majority of the
>observations are from level 6-10 and leave only about 1-5% of total
>observations contributed to level 1-10. As a result, my model tends
>to perform badly on points that have lower level than 6.
>
>I would like to ask if there's any way to circumvent this problem or
>not. I was thinking of the followings ideas. But I am opened to any
>suggestions if you could please.
>
>1. Bootstrapping with small size of samples each time. Howevever, in
>each sample basket, I intentionally sample in such a way that there
>is a good mix between observations from each level. Then I have to
>do this many times. But I don't know how to obtain the true standard
>error of estimated parameters after all bootstrapping has been done.
>Is it going to be simply the average of all standard errors
>estimated each time?
>
>2. Weighting points with level 1-6 more. But it's unclear to me how
>to put this weight back to maximum likelihood when estimating
>parameters. It's unlike OLS where your objective is to minimize
>error or, if you'd like, a penalty function. But MLE is obviously
>not a penalty function.
>
>3. Do step-wise regression. I will segment the data into two
>regions, first points with response less than 6 and the rest with
>those above 6. The first step is a binary regression to determine if
>the point belongs to which of the two groups. Then in the second
>step, estimate ordered probit model for each group separately. The
>question here is then, why I am choosing 6 as a cutting point
>
>Any suggestions would be really appreciated. Thank you.
You could do the obvious, and lump categories such as 1-6 or 1-7
together to make a composite category.
You don't mention the size of your dataset. If there are 10,000 data,
you might live with a 1% category. If you only have 100 data, you
have too many categories.
Also, next time plan your study and training better so that next time
your categories are fully utilized. And don't use so many categories.
People have trouble even selecting responses on a 5-level scale.
================================================================
Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: ral at lcfltd.com
Least Cost Formulations, Ltd. URL: http://lcfltd.com/
824 Timberlake Drive Tel: 757-467-0954
Virginia Beach, VA 23464-3239 Fax: 757-467-2947
"Vere scire est per causas scire"
``` | 696 | 2,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2015-27 | longest | en | 0.936258 |
https://translate.academic.ru/constant%20factor/en/ru/ | 1,600,440,712,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187899.11/warc/CC-MAIN-20200918124116-20200918154116-00380.warc.gz | 705,692,243 | 16,787 | ## Перевод: с английского на русский
с русского на английский
# constant factor
• 1 constant factor
постоянный множитель
[Л.Г.Суменко. Англо-русский словарь по информационным технологиям. М.: ГП ЦНИИС, 2003.]
### EN
• 2 constant factor
• 3 constant factor
• 4 constant factor
• 5 constant factor
• 6 constant factor
• 7 constant factor
1) Компьютерная техника: постоянный коэффициент
• 8 constant factor
постоянный коэффициент; постоянный множитель
• 9 constant factor
• 10 constant factor
постоянный коэффициент; постоянный множитель
• 11 constant factor
• 12 constant factor
• 13 constant factor
постоянный множитель
• 14 constant factor
• 15 constant factor
постоянно действующий фактор; постоянный коэффициент
• 16 constant factor
• 17 constant factor
постоянный коэффициент; постоянный множитель
см. тж. gain
• 18 constant factor
• 19 constant factor
• 20 constant factor
1. постоянно действующий фактор
2. постоянный коэффициент
### См. также в других словарях:
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• Constant voltage method — is a type of MPPT algorithm.[1] This method makes use of the fact that the ratio of maximum power point voltage and the open circuit voltage is 0.76.[2] It is the simplest MPPT control method.[3] Algorithm The operating point of the PV array is… … Wikipedia
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• constant of proportionality — noun the constant value of the ratio of two proportional quantities x and y; usually written y = kx, where k is the factor of proportionality • Syn: ↑factor of proportionality • Hypernyms: ↑factor, ↑constant • Hyponyms: ↑Planck s constant, ↑ … Useful english dictionary | 964 | 3,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-40 | longest | en | 0.375613 |
https://leanprover-community.github.io/archive/stream/113489-new-members/topic/Difference.20between.20set.20and.20subtype.html | 1,719,179,523,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864850.31/warc/CC-MAIN-20240623194302-20240623224302-00109.warc.gz | 320,965,314 | 5,092 | Stream: new members
Topic: Difference between set and subtype
bottine (Jan 27 2022 at 08:28):
Hey! I'm struggling to understand the difference between sets and subtypes, and when should one be used instead of the other. I understand the type theoretic notion of considering a sigma type, and that if the dependent part is a proposition, then there is essentially no further information from an element lying in the sigma type, but I fail to understand the subtleties going on here. (Also, not sure what's the threshold for simple questions being considered "spam", so tell me before I reach it)
Yaël Dillies (Jan 27 2022 at 08:43):
One easy answer is you can take unions and intersections of sets, but you can't do that with subtypes. THe other way around, a set can always be coerced to the corresponding subtype.
Riccardo Brasca (Jan 27 2022 at 08:47):
I would say that the one of the main differences is that is S : set α and x ∈ S, then x is of type α. This is not true for subtypes. For example, if S = {n : ℝ | 0 ≤ n}, then (3 : ℝ) ∈ S as expected, but if you use S = {n : ℝ // 0 ≤ n} and take x : S, then x is not a real number, so you cannot use directly all the result you have for real numbers (you can of course coerc it to a real number, but this is sometimes a pain).
bottine (Jan 27 2022 at 08:47):
ah, mmh, yes, sets are all of a common type, hence the ability to have operations on them (that was RE Yaël's response)
Riccardo Brasca (Jan 27 2022 at 08:47):
Do you have a specific example where you don't know which one to use?
bottine (Jan 27 2022 at 08:49):
I have an example where I was wondering, but it's not really an important one (just trying to figure things out:)
variables {G : Type} [group G] {S : set G}
def signed (s : S) : bool → G
| tt := s
| ff := s⁻¹
def words (T : set G) := list (T × bool)
def val {T : set G} : words T → G
| [] := 1
| (⟨s , sgn⟩ :: tail) := (signed s sgn) * (val tail)
def words_for (T : set G) (g : G) : set (words T)
:= { w : (words T) | val w = g }
def geods_for (T : set G) (g : G) : set (words T)
:= { w ∈ (words_for T g) | ∀ u ∈ words_for T g, (length u) ≥ (length w) }
where I'm trying to define the word metric on a group.
Riccardo Brasca (Jan 27 2022 at 08:56):
I would say that s : S in the definition of signed is not a very good choice. s⁻¹ doesn't make sense, since the type of s (i.e. S) is not a group. Here probably Lean is smart enough to automatically insert the coercion for you, and s⁻¹ means "send s to G and then take the inverse", but if you start with (s : G) (hs : s ∈ S) this wouldn't be necessary.
Riccardo Brasca (Jan 27 2022 at 08:57):
But the best way is to try and see what happens.
Yaël Dillies (Jan 27 2022 at 08:58):
I mean, you can define signed on all of G, so why even bother mentioning S?
bottine (Jan 27 2022 at 08:59):
That's right, yes (though this snippet was just to show a place where I'm not sure of the best choice between subtype and set).
bottine (Jan 27 2022 at 08:59):
So, if I'm understanding the source right, subtype is really just a sigma type.
Riccardo Brasca (Jan 27 2022 at 09:00):
Here you are fixing a group G and working with its elements, so I suggest to take everything of type G, so using sets.
bottine (Jan 27 2022 at 09:01):
Now, when I do \forall x : subtype, I presumably have that x is a pair consisting of an element of the supertype, plus a proof of its satisfying the subtype-defining-predicate. What about \forall x \in set ?
bottine (Jan 27 2022 at 09:01):
Am I right that in the end, coercion and syntax sugar makes a lot of things look simpler than they are?
bottine (Jan 27 2022 at 09:05):
Also, what about the {{_}} notation?
Yaël Dillies (Jan 27 2022 at 09:08):
This is a semi implicit argument. It behaves like an explicit argument until you pass a later explicit argument, in which case it acts as an implicit one.
Yaël Dillies (Jan 27 2022 at 09:08):
Maybe looking at docs#monotone or docs#convex will make it make sense
bottine (Jan 27 2022 at 09:09):
OK, so nothing to do with sets/subtypes/membership… good to know
Last updated: Dec 20 2023 at 11:08 UTC | 1,238 | 4,136 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.928254 |
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Online calculators are a convenient and versatile tool for performing complex mathematical calculations without the need for physical calculators or specialized software. With just a few clicks, users can access a wide range of online calculators that can perform calculations in a variety of fields, including finance, physics, chemistry, and engineering. These calculators are often designed with user-friendly interfaces that are easy to use and provide clear and concise results. | 772 | 3,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-50 | latest | en | 0.769738 |
https://answerofmath.com/solved-feature-interaction-and-its-applications/ | 1,679,959,539,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00422.warc.gz | 128,549,700 | 19,533 | # Solved – Feature Interaction and its applications
What is a Feature Interaction?
Are Feature Interactions used for Feature Selection or Feature Generation?
Contents
5.4 Feature Interaction When features interact with each other in a prediction model, the prediction cannot be expressed as the sum of the feature effects, because the effect of one feature depends on the value of the other feature. Aristotle's predicate "The whole is greater than the sum of its parts" applies in the presence of interactions.
5.4.1 Feature Interaction? If a machine learning model makes a prediction based on two features, we can decompose the prediction into four terms: a constant term, a term for the first feature, a term for the second feature and a term for the interaction between the two features. The interaction between two features is the change in the prediction that occurs by varying the features after considering the individual feature effects.
I would like to be a little pedantic and alter the last sentence above:
The interaction between two features is the change in the prediction that occurs by varying the features while considering the individual feature effects.
Another way to think about an interaction is that it occurs when the effect of one feature depends on the value of another feature. Note that interaction are a natural result of considering the general model:
$$Y = f(x,Z)$$
where $$x$$ is a matrix of continuous explanatory variables (features) and $$Z$$ is a random variable which we can think of as normally distributed around zero, but it does not have to be. If we expand this around $$x_0$$, with a 2nd order taylor series we obtain:
$$Y approx beta_0 + sum_{i = 1}^{p} beta_{i}(x_i-x_{i0}) + sum_{i = 1}^{p}sum_{j = 1}^{p} beta_{ij}(x_i-x_{i0})(x_j-x_{j0}) + left( sigma + sum_{i = 1}^{p} gamma_i(x_i-x_{i0}) right) Z + sigma Z^2$$
where the 3rd term contains cross products of two linear terms – which are the interactions.
Are Feature Interactions used for Feature Selection or Feature Generation?
I would consider interactions as Feature Generation, and they are a very useful way of modelling a natural form of nonlinearity.
Edit: Of course, once you have generated the interaction feature, there is then the question of whether to include it in your model, and that is where feature selection comes into play.
Rate this post | 528 | 2,375 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-14 | latest | en | 0.932852 |
https://www.teachoo.com/2190/584/Ex-3.3--18---Prove-sin-x---sin--y--cos-x---cos-y--tan-(x---y)-2/category/Ex-3.3/ | 1,726,723,915,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00538.warc.gz | 920,927,909 | 22,168 | Ex 3.3
Chapter 3 Class 11 Trigonometric Functions
Serial order wise
### Transcript
Ex 3.3, 18 Prove that 〖sin x − 〗siny /(𝑐𝑜𝑠x + 𝑐𝑜𝑠y ) = tan (x − y)/2 Solving L.H.S 〖sin x − 〗siny /(𝑐𝑜𝑠x + 𝑐𝑜𝑠y ) = 〖𝟐 𝒄𝒐𝒔((𝒙 + 𝒚 )/𝟐)〗〖 𝒔𝒊𝒏((𝒙 − 𝒚 )/𝟐) 〗/(𝟐 𝒄𝒐𝒔((𝒙 + 𝒚 )/𝟐) 𝒄𝒐𝒔((𝒙 − 𝒚 )/𝟐) ) = 𝑠𝑖𝑛" " ((𝑥 − 𝑦 )/2)/(𝑐𝑜𝑠((𝑥 − 𝑦 )/2) ) = tan (𝒙 − 𝒚 )/𝟐 = R.H.S Hence L.H.S. = R.H.S. Hence proved | 317 | 392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-38 | latest | en | 0.254821 |
https://math.stackexchange.com/questions/2874974/juggling-three-non-archimedean-fields | 1,680,127,010,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949035.66/warc/CC-MAIN-20230329213541-20230330003541-00291.warc.gz | 458,526,772 | 37,607 | # Juggling three non-Archimedean fields
I'm comparing the field of hyperreals, the Levi-Civita field and the Dehn's field for the first time. I'm not very familiar with their properties, so I'm looking for ways to understand and distinguish them.
I think Dehn's field is just a Pythagorean extension of the ordered field $\mathbb R(x)$.
The construction of the Levi-Civita field makes it appear to be an extension of $\mathbb R((x))$, where the coefficients of $x$ can be well-founded sets of rational numbers, rather than just well-founded sets of integers. So this would also extend $\mathbb R(x)$.
Does the Levi-Civita field extend Dehn's field?
I'm not comfortable with the construction of the hyperreals $^\ast\mathbb R$ yet, so I am treading lightly on how it compares to the above. Being constructed with ultrafilters, I feel like it should be huge, and I half expect that it might contain both Dehn's field and the Levi-Civita field.
Do the hyperreals extend the Levi-Civita and/or Dehn fields?
I've found unclear indications that the hyperreals extend the Levi-Civita numbers, but I wasn't able to find information on the comparison with the Dehn field.
• The Dehn field is an algebraic ordered field extension of $\mathbb{R}(x)$, as such it is contained in the Levi-Civita field which contains its real closure (the Levi-Civita field is the Cauchy-completion of the real closure of the Cauchy-completion of $\mathbb{R}(x)$). The other intermediate steps are from smallest to biggest Laurent series and Puiseux series. (I used several key-words which I suggest you look up: and I would add Hahn series) Aug 7, 2018 at 16:55
• (continued) As for hyperreal numbers defined using the ultrapower construction, they contain any ordered field of germs at infinity of real valued functions, hence the Dehn field. The Levi-Civita field is not such a field as far as I know, but since the field of hyperreal numbers is countably saturated and real closed, it also contains Cauchy-completions of its subfields with countable cofinality, so it contains the Levi-Civita field, but not in a canonical way (as far as I know). I can explain this in more detail if you tell me which of those notions you know. Aug 7, 2018 at 17:00
• @nombre Oh! I had another question I was about to ask about the Levi-Civita construction asking what to call that completion process, but it seems you've anticipated it! I didn't have a real term for "making the passage from k[x] to k[[x]] to k((x))" that works for (semi)group rings instead of just the semigroup ring k[x] :) Aug 7, 2018 at 17:31
• @nombre So is this what you mean: in the Levi-Civita style construction, you could use well-founded sequences of integers (Laurent series) and then well-founded Puiseux sequences, and then these well-founded sequences of rationals? Aug 7, 2018 at 17:38
• @nombre and also a field of Hahn series between Levi-Civita and the hyperreals? You really must write a solution... please! I can't believe nobody ever talked with me about this circle of ideas before. It's very interesting., Aug 7, 2018 at 17:39
$\DeclareMathOperator{\cof}{cof}$The Dehn field embeds naturally into the Levi-Civita field and fields $^*\mathbb{R}$ of hyperreal numbers defined by ultrafilters on $\mathbb{N}$, and the Levi-Civita field embeds into $^*\mathbb{R}$.
I'll try to explain this in some detail. I assume you are familiar with the notion of ordered Hahn series fields and I first introduce what I called Cauchy-completion. You'll find a discussion about it here.
Any linear order $X$ is equipped with a topology, called the order topology, whose open sets are the unions of open intervals. In ZFC, $X$ has a cofinality $\cof(X)$ which is the least order type of a cofinal subset of $X$. If $F$ is an ordered field, then there is also a natural uniform structure on $F$ and in particular a notion of Cauchy sequence (same as regular Cauchy sequences but indexed by $\cof(F)$). The topology is sequential, in that closed sets are sets which contains the limits of their convergent sequences (indexed by $\cof(F)$), and thus in general every topological notion can be stated in terms of sequences indexed by $\cof(F)$, with the usual properties: continuity is sequential continuity, and so on....
An ordered field $F$ is said Cauchy-complete if its Cauchy sequences are convergent in $F$, or equivalently if it has no proper dense (ordered field) extension. Every ordered field $F$ has a dense Cauchy-complete extension $(\widetilde{F},\varphi)$, with the following initial and terminal properties:
(IP:) If $(I,\mu)$ is a Cauchy-complete continuous (for the order topology, equivalently, the embedding is cofinal) ordered field extension of $F$, then there is a unique morphism $\sigma: \widetilde{F} \rightarrow I$ with $\sigma \circ \varphi=\mu$.
(TP:) If $(T,\mu)$ is a dense extension of $F$, then there is a unique morphism $\sigma: T \rightarrow \widetilde{F}$ with $\varphi=\sigma \circ \mu$.
The extension $(\widetilde{F},\varphi)$ is called the Cauchy-completion of $F$.
For instance $\mathbb{R}$ is the Cauchy completion of $\mathbb{Q}$, and the field $\mathcal{L}=\mathbb{R}((\varepsilon^{\mathbb{Z}}))$ of Laurent series is the Cauchy-completion of $\mathbb{R}(\varepsilon)$. To prove this, one needs only check that this is a dense ordered field extension of this field, and that each Cauchy sequence in $\mathbb{R}(\varepsilon)$ converges in $\mathcal{L}$.
Cauchy-completion is similar real closure in that they correspond to a specific version of reflective subcategories linked to properties of extensions: algebraic extensions and dense extensions. In this answer, I give some (quite poorly written) explanation. What matters here is that the Cauchy-completion is a functorial construction.
Using the Newton polygon method, one can prove that the real closure of $\mathcal{L}$ is the ordered field $\mathcal{P}=\bigcup \limits_{n \in \mathbb{N}^{>0}} \mathbb{R}((\varepsilon^{\frac{1}{n}.\mathbb{Z}}))$ of Puiseux series. Its Cauchy-completion, which is the Levi-Civita field $\mathcal{C}$ (sorry Levi...) can be construed as the field of Hahn series $s=\sum \limits_{n \in \mathbb{N}} s_n \varepsilon^{q_n}$ where $(s_n)_{n \in \mathbb{N}}$ is a sequence of real numbers and $(q_n)_{q \in \mathbb{N}}$ is a strictly increasing and cofinal sequence of rationnal numbers. As the Cauchy-completion of a real-closed field, $\mathcal{C}$ is automatically real-closed.
Since the Dehn field $\mathcal{D}$ is an algebraic extension of $\mathbb{R}(\varepsilon)$, by the terminal property of real closure, there is a unique embedding of $\mathcal{D}$ into the real closure of $\mathbb{R}(\varepsilon)$ extending the given real closure morphism. By the initial property of real closure, this real closure enjoys a unique embedding in $\mathcal{C}$ extending the embedding $\mathbb{R}(\varepsilon) \rightarrow \mathcal{C}$.
Now let's turn to germs of real valued functions. The ring $\mathcal{G}$ of germs of real valued functions is the quotient of the set of real valued functions defined on intervals $[a,+\infty)$ for some real number $a$ by the equivalence relation $f \sim g$ iff $f(x)=g(x)$ for sufficiently big $x$. It is a partially ordered ring under poitwise sum and product, and eventual comparison. Any (linearly) ordered subfield $F$ of $\mathcal{G}$ embeds naturally in the ultrapower $^*\mathbb{R}$ (given a free ultrafilter $U$ on $\mathbb{N}$) by sending the germ of a function $f$ defined on $[n,+\infty)$ to the class of $(0,0,...,0,f(n),f(n+1),...)$ modulo $U$ (where there are $n$ zeroes for instance). The Dehn field is such an ordered field, or more precisely it is a field or representatives of germs of real valued functions. Thus it also embeds naturally in $^*\mathbb{R}$.
The field of Laurent series, and for that matter the Levi-Civita field, are not fields of germs of real valued functions, at least not that I know of (but maybe every Laurent series is Borel summable?). Thus I don't see how to naturally embed them into $^*\mathbb{R}$. It is possible that the saying "$^*\mathbb{R}$ extends the Levi-Civita field" is better understood as a thematic remark: the Levi-Civita can be used to do analysis although it is not archimedean, and thus "not standard" analysis, with nice properties (see this here found on Wikipedia), and fields of hyperreal numbers can be seen as the completion of this goal.
In ZFC, there are embeddings of $\mathcal{C}$ into $^*\mathbb{R}$. To see this, we can use the fact that $^*\mathbb{R}$ is real-closed and countably saturated: if $L,R$ are countable sets of hyperreal numbers with $L<R$, then there is a hyperreal number $a$ with $L<a<R$. This implies that $(i)$: $^*\mathbb{R}$ contains a canonical copy of the real closure of each of its subfields, and
$(ii)$: $^*\mathbb{R}$ contains a copy of the Cauchy-completion of each of its subfields with countable cofinality.
The first statement follows from the initial property of real closure.
To prove the second one, consider a subfield $K$ of $^*\mathbb{R}$ with countable cofinality. The set of cofinal extensions of $K$ in $^*\mathbb{R}$ is inductive for the relation of inclusion, and has a maximal element $F$ by Zorn's lemma. Since any algebraic extension is cofinal and by the first statement, $F$ must be real-closed. Assume for contradiction that there is a non-convergent Cauchy sequence $u$ in $F$. We may also assume that we have $u_{2m}<u_{2n+1}$ for all $m,n \in \mathbb{N}$ (I let you figure this out). Let $L$ (resp. $R$) be the set of elements of the sequence which lie below (resp. above) infinitely many elements of the sequence. We have $L<R$ so there is a hyperreal number $a$ with $L<a<R$. Note that $a$ leis outside of $F$, otherwise $u$ would converge to it. I claim that the subfield $F(a)$ of $^*\mathbb{R}$ is a dense extension of $F$ where $u$ converges to $a$.
In fact, we only require that it is a cofinal extension of $F$, but density follows. Indeed, it will follow that $u$ converges to $a$. Then since $F$ is real closed, no polynomial in $F[X]$ annihilates $a$, and thus by continuity of fractions in $F(X)$ on $F(a)$ outside of their poles (true for any ordered field), for such fraction $f(X)$, the sequence $(f(u_n))_{n \in \mathbb{N}}$ converges to $f(a)$.
So let's prove that $F$ is cofinal in $F(a)$. It is easy to see that each number $P(a)$ for $P \in F[X]$ is bounded by elements of $F$. We must only check that for any non zero polynomial $Q \in F[X]$, the fraction $\frac{1}{Q(a)}$ is also bounded in $F$, that is, we must check that $Q(a)$ may not be infinitesimal with respect to $F$, denoted $Q(a)\prec_F 1$. We do so by valuation-theoretic arguments. Since $^*\mathbb{R}$ is countably saturated, in particular $F$ is bounded in $F$, and its convex hull in $^*\mathbb{R}$ is a proper convex valuation ring on $^*\mathbb{R}$ which contains $a$. By real closure of $^*\mathbb{R}$, the corresponding valued field is henselian (see for instance Theorem 3.5.16 in ADH). Assume towards a contradiction that there is $Q \in F[X]$ which is non zero with $Q(a) \prec_F 1$, and choose such polynomial $Q$ with minimal degree, hence $Q'(a)$ is not infinitesimal with respect to $F$. By henselianity, this means that there is $b \in ^*\mathbb{R}$ with $Q(b)=0$ and $a-b\prec_F 1$. The first relation yields $b \in F$ (by real closure of $F$), and the second implies that $u$ converges to $b$ in $F$: a contradiction.
Thus $F(a)$ is a cofinal extension of $F$, which contradicts the maximality of $F$. So $F$ must be Cauchy-complete. The initial property of Cauchy completion then implies that the Cauchy-completion of $K$ embeds in $F$ and thus in $^*\mathbb{R}$.
Applying those two results and starting with the fields-of-germs-style embedding $\mathbb{R}(\varepsilon) \rightarrow ^*\mathbb{R}$ with $f(\varepsilon)\sim (0,1,\frac{1}{2},\frac{1}{3},...)$, we get an embedding of Laurent series, then Puiseux series, then "Levi-Civita series" into $^*\mathbb{R}$.
(Using similar arguments as above, one can prove that the maximal subfields $F$ in $^*\mathbb{R}$ are in fact almost countably saturated (countably saturated but at $+\infty$ and $-\infty$), hence in particular spherically complete. This implies that $^*\mathbb{R}$ also contains a copy of the Hahn series field $\mathbb{R}((x^{\mathbb{R}}))$.)
• Better answer than I ever dared hope for :) Do you happen to know of any good texts/articles which also cover this family of fields? Aug 9, 2018 at 2:21
• @rschwieb: Not really. I don't know any book dedicated to ordered fields, and although it is certain that there are many texts written on fields of hyperreal numbers and probably a few about the Levi-Civita field, I don't know any of them. (to continue) Aug 9, 2018 at 15:17
• @rschwieb: (continued) Maybe you could find the book Super-real fields by Dales and Woodin where they introduce and describe various big ordered field extending the reals, including hyperreal fields. Regarding topological and uniform notions, there is the first Bourbaki of topology. For more focus on ordered field, there are articles by Sikorski that are for instance cited in [Generalized Archimedean fields][1] which you may find interesting. [1]: projecteuclid.org/euclid.ndjfl/1093870226 Aug 9, 2018 at 15:18
• The Levi-Civita field is just the field of fractions of the monoid ring $\mathbb R[M]$ where $M$ is the additive monoid of nonnegative rationals, right? I suppose it is equally true that it's the field of fractions of the group ring $\mathbb R[\mathbb Q]$ for that matter. Dec 30, 2020 at 3:37
• @rschwieb It is the Cauchy-completion of that fraction field, any left-finite set of rational numbers can appear as the support of a Levi-Civita series. Dec 31, 2020 at 8:59 | 3,769 | 13,773 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-14 | latest | en | 0.940297 |
https://www.oyohyee.com/post/AOJ/862/ | 1,555,654,564,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527148.46/warc/CC-MAIN-20190419061412-20190419083412-00336.warc.gz | 756,702,444 | 8,619 | 406
3
1 1
1 2
2 2
1.0000
# 题解
dfs(l,r) 来表示 [l,r) 的最近的两个点的距离
1. 最近的两个点可能全部在 mid 左侧的范围里,也即 dfs(l,mid)
2. 最近的两个点可能全部在 mid 右侧的范围里,也即 dfs(mid,r)
3. 最近的两个点一个在 mid 左侧,一个在 mid 右侧
(存在极端情况导致答案错误)
(需要注意,如果这里比较距离的平方,上面遍历需要比较 (x1-x2)^2)
# 代码
//*/
#define debug
#include <ctime>
//*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
using namespace std;
const int maxn = 60005;
const double INF = 9e9;
const int K = 5;
struct Point {
int x, y;
Point(int _x = 0, int _y = 0) : x(_x), y(_y) {}
bool operator<(const Point &rhs) const { return x < rhs.x; }
};
Point point[maxn];
inline double pow(double a, double n) {
if (n == 0)
return 1.0;
if (n == 1)
return a;
return pow(a, n / 2) * pow(a, n - n / 2);
}
inline double dis(Point a, Point b) {
return pow(a.x - b.x, 2) + pow(a.y - b.y, 2);
}
//[l,r)
double dfs(int l, int r) {
int mid = (l + r) / 2;
if (r - l < 2) {
return INF;
}
if (r - l == 2) {
return dis(point[l], point[l + 1]);
}
double mm = min(dfs(l, mid), dfs(mid, r));
for (int i = mid - 1; i >= l && pow(point[mid].x - point[i].x,2) < mm; i--)
for (int j = mid; j < r && pow(point[r].x - point[mid-1].x,2) < mm; j++)
mm = min(mm, dis(point[i], point[j]));
return mm;
}
int main() {
#ifdef debug
freopen("in.txt", "r", stdin);
int START = clock();
#endif
cin.tie(0);
cin.sync_with_stdio(false);
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%d%d", &point[i].x, &point[i].y);
sort(point, point + n);
printf("%.4f\n", sqrt(dfs(0, n)));
#ifdef debug
printf("Time:%.3fs.\n", double(clock() - START) / CLOCKS_PER_SEC);
#endif
return 0;
}
• 点击查看/关闭被识别为广告的评论 | 652 | 1,631 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2019-18 | longest | en | 0.154231 |
http://www.mathworksheets4kids.com/multiplication-properties.php | 1,484,876,190,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280763.38/warc/CC-MAIN-20170116095120-00178-ip-10-171-10-70.ec2.internal.warc.gz | 578,227,109 | 8,698 | Home > Basics > Multiplication > Properties
# Multiplication Properties Worksheets
The properties worksheets in this page contain commutative and associative property of multiplication; distributive property; identifying equivalent statement; multiplicative inverse and identity; and more.
## Basic Multiplication Properties
Commutative Property of Multiplication
Identify the correct choice from the list of multiplication properties.
Commutative Property 1
Commutative Property 2
Associative Property of Multiplication
Associative Property: a * (b * c) = (a * b) * c
Find out which of the following choices describes associative property of multiplication.
Associative Property 1
Associative Property 2
## Distributive Property Worksheets
Identify the Equivalent Statement
You have 4 choices with 3 distractors in each question. Identify the correct equivalent statement.
Equivalent Statement 1
Equivalent Statement 2
Distributing Numbers
Apply distributive principle of multiplication to rewrite the multiplication sentence.
Distributive Property Worksheet 1
Distributive Property Worksheet 2
Solve using Distributive Property
Use distributive property to solve the multiplication problems.
Solving Worksheet 1
Solving Worksheet 2
## Multiplication Properties: Special Focus
Missing Number
Use commutative and associative property of multiplication to find the missing number.
Missing Number 1
Missing Number 2
Use multiplicative identity and multiplicative inverse to find the missing number.
Identity or Inverse: Easy
Identity or Inverse: Moderate
Multiplicative Inverse
Multiplicative inverse of a is 1/a; a/b is b/a and vice versa. Identify the multiplication inverse for each problem.
Multiplicative Inverse: Easy
Multiplicative Inverse: Moderate
Multiplicative Inverse: Difficult
Identifying the Property
Multiplication properties worksheets include multiplication equation in each question. Identify what property is used in each question.
Multiplication Property Worksheet 1
Multiplication Property Worksheet 2 | 401 | 2,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-04 | latest | en | 0.759535 |
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# Module 4 slides
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### Module 4 slides
1. 1. MODULE 4 Development ofAssessment Tools
2. 2. LESSON 1: MULTIPLE CHOICE TEST(MCT) The multiple choice test is a form of assessment in which the students are asked to select the correct or best answer out of the choices from the list. Is composed of a STEM, which sets up the problem, followed by the number of ALTERNATIVE RESPONSES. The other alternatives are DISTRACTORS or FOILS. Can be presented in 3 ways: direct question, incomplete statement and mathematical equation.
3. 3. ADVANTAGES AND DISADVANTAGES OF MCT Advantages Disadvantages Has a great Multiple choices are versatility in ambiguous measuring objectives Can just select Requires less time random Writing is minimized answers/guessing Scoring is objective More amenable to item analysis
4. 4. REFERENCES: Reganit, Arnulfo, Elicay, Ronaldo, & Laguerta, Cresencial (2010). Assessment of Student Learning 1 (Cognitive Learning). C & E Publishing, Inc. Garcia, Carlito D. (2008). Measuring & Evaluating Educational Outcomes. Books atbp. Publishing Corp., Mandaluyong
5. 5. LESSON 2: TRUE OR FALSE TEST This type of test requires the examinee to recognize and mark an item as true or false. Other possible options are agree or disagree, yes or no, valid or invalid, fact or opinion, and cause or effect. Are utilized to assess a student’s ability to recognize the accuracy of a declarative statement. (http://www.park.edu.cetl2/quicktips/truefalse. html)
6. 6. ADVANTAGES AND DISADVANTAGES OFTFT Advantages Disadvantages Answers tend to be Emphasis on Rote short Memorization Ease of Dependence on Construction Absolute Judgment Ease of Scoring Likelihood of Guessing
7. 7. POINTERS ON WRITING TRUE OR FALSEITEMS Construct items that measure important objectives. Avoid using determiners. Avoid using trick questions. Limit each item to the point that is being tested. Avoid excess use of negative words and phrases. Approximately half of the statements are false. Avoid qualitative terms like best, some, many, and several.
8. 8. LESSON 3: MATCHING-TYPE TEST
9. 9. ADVANTAGES AND DISADVANTAGES OF MTT Advantages Disadvantages The matching-type test is It tends to ask students to simple to construct score. It is associate trivial information. well suited in measuring Unfortunately, most associations. Like a multiple- matching-type test choice test, it presents the emphasizes memorization, student with questions and although it is impossible to alternatives. construct items that measure It reduces the effects of more complex cognitive skills. guessing, although the chance of guessing increases as the student progress in answering In case of commercial answer items. This, however, is easily sheets, matching items can remedied by adding more accommodate no more than options. five options.
10. 10. POINTERS ON WRITING MATCHING-TYPE TESTS If possible, the response list should consist of short phrase, single words, or numbers. Use homogenous options and items. Have more options than the given items. Initially, a matching- item test decrease the students tendencies to guess but as the students progress in answering the test, the guessing tendencies increase. This can be avoided by increasing the options. Arrange the options and items alphabetically, numerically, or magnitudinally. This is one way to help the examinees since they can maximize their time by not searching for the correct answers, especially if there are many options. Limit the number of items within each set. Ideally, the minimum is five and the maximum is ten per set. Place the shorter responses on column B. This time-saving practice allows the students to read the longer items first in column A and then search quickly through the shorter options to locate the correct alternative.
11. 11. 7.PROVIDE COMPLETE DIRECTIONS.DIRECTIONS SHOULDSTIPULATE WHETHER OPTIONS CAN BE USED ONLY ONCE ORMORE THAN ONCE. THEY SHOULD ALSO INSTRUCT THESTUDENTS ON HOW TO RESPOND. THE INSTRUCTIONS SHOULDALSO CLARIFY WHAT COLUMN A AND B ARE ABOUT.8. PLACE THE LIST OF OPTIONS ON THE SAME PAGE AS THELIST OF ITEMS. TIME IS WASTED IF THE STUDENTS HAVE TOFLIP PAGES TO SEARCH THROUGH ALL OPTIONS TO LOCATE THECORRECT ONES. ADDITIONALLY, SOME STUDENTS MAYOVELOOK THAT THERE ARE STILL SOME OPTIONS ON THE NEXTPAGE.9. AVOID SPECIFIC DETERMINERS AND TRIVIAL INFORMATIONTHAT CAN HELP THE STUDENTS FIND THE CORRECT RESPONSEWITHOUT ANY EFFORT ON THEIR PART. THE USE OT "NONE OFTHE ABOVE" A AN OPTION IS RECOMMENDED IF IT IS ONLYCORRECT ANSWER.10. CLEARLY EXPLAIN THE BASIS ON WHICH IS TO BE MADE.
12. 12. SUGGESTIONS FOR MEASURING COMPLEXOBJECTIVES WITH MATCHING-TYPE TESTS Match examples with terminologies. Perhaps this is the most direct and simplest method of increasing the thought content of matching tests provide that the example has not yet been taught before. Use novel pictorial materials.
13. 13. LESSON 4: COMPLETION TESTAcompletion test is a format oftesting that requires thestudents to complete with thecorrect word or phrase.
14. 14. ADVANTAGES AND DISADVANTAGES OFCT Advantages Disadvantages Construction of Completion test are relatively test is difficult to score. relative easy. They typically Guessing is measure rote eliminated. memory. Item sampling is improved.
15. 15. POINTERS OF COMPLETION TEST Write items that clearly imply of response desired. Use only one blank per item. Put the blank at the end of the item if possible. Avoid specific determiners. Structure an item so that they require response should be concise. Blanks provided for answer should be be equal in length. Provide sufficient space for the answer.
16. 16. REFERENCES:Gracia, Carlito D. (2008). Measuring andEvalauting Outcomes. Books Atbp. PublishingCorp.,MandaluyongReganit, Artulfo, Elicay, Ronaldo & Laguerta,Cresencia (2010)http://www.businessdictionary.com/definition/completiontesthttp://ph.adsfn.com/money-wordsterm6490completion test
17. 17. LESSON 5: CLOZE TEST Cloze or cloze deletion test is an exercise. Test. Or assessment consisting of a portion of a text with certain words removed (cloze text) and the students are asked to replace the missing words. The cloze test requires the ability to understand context and vocabulary to be able to identify the correct words or type of words that belong in the deleted passages of the text. Words may be deleted from the text in question either mechanically or selectively, depending on what aspect the test intends to give emphasis to.
18. 18. EXAMPLE: TODAY I WENT TO THE _________________AND BOUGHT SOME MILK AND EGGS. I KNEW IT WASGOING TO RAIN, BUT I FORGOT TO TAKE MY_________________, AND ENDED UP GETTING WETON MY WAY ____________.ADVANTAGES OF CLOZE TEST. TESTS MADE BY THECLOSE PROCEDURE HAVE DONE MUCH TO SOLVE THESEPROBLEMS. A CLOZE TEST CAN BE MADE OVER ANYPASSAGE BY REPLACING EVERY FIFTH WORD WITH ANUNDERLINED BLANK SPACE OF A STANDARD OF A LENGTH.
19. 19. REFERENCES:*ASSESSMENT OF STUDENT LEARNING 1*JOURNAL OF EDUCATIONALMEASUREMENT
20. 20. LESSON 6: ESSAY TEST (ET) An essay test permits direct assessment of the attainment of numerous goals and objectives. In contrast to with the objective test item types, an essay test demands less construction time per fixed unit of student time but a significant increase in labor and time for scoring.
21. 21. ADVANTAGES OF ESSAY TESTS Essays give students freedom to respond within broad limits. Essay examinations allow students to express their ideas with relatively few restraints. Guessing is eliminated. Essays involve recall and there are no options to select from. The student is expected to supply rather that select the proper response. Essay items are practical for testing a small number of students. However, as the number of students increases, the advantage of essay tests decrease. Essay test reduce assembling time. Less time is required for typing, mimeographing, and assembling. If only a few questions are asked, the teacher can just write them on the board. They can measure divergent thinking. Divergent thinking is indicated by unconventional, creative, relatively free responses. Because they allow great freedom in answering, the opportunity to obtain unusual responses is increased.
22. 22. DISADVANTAGES AND LIMITATIONS OF ESSAYTESTS Essays are difficult to score objectively because students have greater freedom of expression. Extended essays measure only the limited aspects of student knowledge. Essay questions are time-consuming for teachers and students. Essays eliminate guessing but not bluffing. Most essays require a little more than rote memory. Essay tests place a premium on writing.
23. 23. THE USE OF ESSAY TESTS TO FACILITATELEARNING Favorable Disadvantages Raises the quality of Essay tests do not writing allow students to Teaches students to revise and rewrite organize, outline and their work since time summarize is limited. assignments rather The teachers’ over- than simply look for facts, dates and attention to details details expected in T- can destroy the F or multiple choice themes of essays. tests.
24. 24. SITUATIONS THAT SUGGEST THE USE OFESSAY QUESTIONS If the test objectives specify that students have to write, recall or supply information, an essay type of test may be necessary. When the class size is small, a teacher can afford to spend more time to read and check the essay responses. If a test can be used only once, an essay examination may be more convenient than a multiple-choice one.
25. 25. POINTERS ON WRITING ESSAY QUESTIONS Specify limitations. The teachers should tell the students the length of the desired response and the weight each question will be given when determining scores. With these, the students will be able to know the time to be spent on each item, the maximum points per item, the approximate number of words and space for each item. Structure the task. The instructions in the essay test should clearly specify the task. Most of the questions in the essay tests are so vague that the real intentions of the instructors are lost. Make each item relatively short and increase the number of items. The more items there are, the greater chance there is of the sampling of knowledge. Give all the students the same essay questions if content is relevant. Sometimes, teachers give the students the opportunity to deal with one or two items from a set of essay questions. Ask questions in a direct manner. Avoid deviousness and pedanticism when framing questions. | 2,666 | 11,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-43 | latest | en | 0.817356 |
https://bseodisha.guru/chse-odisha-class-11-math-solutions-chapter-15-ex-15/ | 1,725,756,001,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00079.warc.gz | 128,324,176 | 47,638 | # CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15
Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 Textbook Exercise Questions and Answers.
APOLLOTYRE Pivot Point Calculator
## CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Exercise 15
Question 1.
If the values observed are 1, 2, …..,n each with frequency 1, find
(i) the mean value
Solution:
Mean of 1, 2, 3, ….. n
= $$\frac{1+2+3 \ldots . . .+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2}$$
(ii) the mean deviation from the mean separately for two cases when n is odd and when n is even.
Solution:
If n is even, let n = 2m.
Question 2.
For the same set of values as in (1) above, find the variance and standard deviation.
Solution:
x: 1, 2, 3, ….., n
Question 3.
From the table below, find the mean value and the variance.
(a) Values: 1 2 3 ….. n
Frequency: 1 2 3 …. n
Solution:
x: 1 2 3 ….. n
y: 1 2 3 …. n
Question 4.
From the table below, find the mean and the variance.
Solution:
(a) Values: 1 2 5 ….. (2n – 1)
Frequency: 1 1 1 1
(b) Values: 2 4 6 …..2n
Frequency: 1 1 1 1
Question 5.
From the table below, calculate the mean and the variance
$$\text { Values } \quad \mathbf{0} \quad 1 \quad 2 \ldots \quad r \ldots n$$
$$\text { Frequency: } \quad{ }^n \mathbf{C}_0{ }^n \mathbf{C}_1{ }^n \mathbf{C}_2{ }^n \mathbf{C}_r \ldots . .{ }^n \mathbf{C}_n$$
Solution:
Question 6.
From the following table calculate the mean, mean deviation from the mean, and variance.
Marks Number of students 30-35 5 35-40 7 40-45 8 45-50 20 50-55 16 55-60 12 60-65 7 65-70 5
Solution:
C. I f Mid value (x) d = x – A 30-35 5 32.5 -15 35-40 7 37.5 -10 40-45 8 42.5 -5 45-50 20 47.5 0 50-55 16 52.5 5 55-60 12 57.5 10 60-65 7 62.5 15 65-70 5 67.5 20 ∑f = 80
Let A (working mean) = 47.5, i = 5
u = d/i fu u2 fu2 -3 -15 9 45 -2 -14 4 28 -1 -8 1 8 0 0 0 0 1 16 1 16 2 24 4 48 3 21 9 63 4 20 16 80 ∑fu = 44 ∑fu2 = 44
Question 7.
In a soccer league, two teams A and B have the following records
A: Goals scored: 0 1 2 3 4
Number of matches: 11 18 8 6 2
B: Goals scored: 0 1 2 3 4 5
Number of matches: 5 20 10 6 3 1
Which team is more consistent? Which is a better team.
Solution:
∴ The mean of B is more than that of A, so B is the better team. A is more consistent as its variance is less than that of B.
Question 8.
The coefficient of variation is defined as $$\sigma / \bar{x}$$, that is the standard deviation divided by the mean value. Find the coefficient of variation c.v. for each of the following sets of observations.
(i) 2, 3, 4, 2, 5, 7, 8, 9
Solution:
(ii) 5, 7, 9, 10, 7, 5, 8, 9, 3
Solution:
(iii) 3, 3, 3, 4, 4, 4, 5, 5, 5
Solution:
Question 9.
Suppose the values x1, x2, …. xn having frequency f1, f2, …. fn respectively having mean value x̄ and variance σ2. Let a be a fixed real number
x1 + a, x2 + a, ….. , xn + a with frequency f1, f2, ….. fn respectively will have mean value x̄ + a and variance σ.
Solution:
Question 10.
Find the mean and deviation from the mean and the standard deviation of a, a + d, a + 2d, …. , a + 2nd assume that d > 0.
Solution:
Question 11.
Let x1, x2, …. xn be a set of observations with mean value 0 and variance σ2x and y1, y2, …. ym be another set of observations with mean value 0 and variance σ2y. Find the mean value and variance of the set of observations x1, x2, …. xn , y1, y2, …. ym combined.
Solution:
x1, x2, …. xn be a set of observations with mean value 0 and variance σ2x and y1, y2, …. ym be another set of observations with mean value 0 and variance σ2y
Question 12.
Find which group of the following data is more dispersed :
Range 10-20 20-30 30-40 40-50 50-60 (Group A) Frequency 5 1 3 2 1 (Group A) Frequency 1 3 2 3 1
Solution:
Let us find the mean and standard deviation for the given two distributions.
M. D = $$\frac{1}{N} \sum_{i=1}^n f_i\left|x_i-\bar{x}\right|$$
M. D = $$\frac{1}{N} \sum_{i=1}^n f_i\left|x_i-M\right|$$
(iii) variance
Variance is the mean of squared deviations from the mean.
(iv) Standard deviation
Standard deviation is the square root of the mean of squared deviations from the mean.
∴ Standard deviation
Question 13.
The price of land per square meter and that of gold per ten grams over five consecutive years is given below. Decide, which price maintains better stability. [Hint: Stability ⇔ Consistency]
Price of land/Sq.meter(₹) 1500 2500 2600 3000 4000 Price of gold/10 gms(₹) 2500 2600 2750 2900 2850
Solution: | 1,668 | 4,473 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-38 | latest | en | 0.555312 |
https://forums.itpro.tv/topic/3555/ipv6-addresses | 1,675,545,179,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00308.warc.gz | 278,695,306 | 13,337 | Hi Guys,
Have finished the online video's in preparation for my A+ exams and am doing extra study as well as reading my notes and doing lots of tests as want to smash the exams when I sit them; however, am having a hard time trying to get my head around IPv6 when it comes to using the "/" after an address.
I understand that in IPv4 if you use something like 192.168.0.1/16 it means that 192.168 is the network portion and 0.1 is the host (it's the equivalent of a subnet mask of 255.255.0.0) That I'm clear on. I'm also ok with the fact you can use non-standard subnet masks (adjacent 1's in binary) so you could see 255.128.0.0 or /9 after the IP. So I can see a clear split between the network part and the host part. However, from what I'm reading here IPv6 doesn't use a subnet mask but the same convention for stating the network length holds true. A few examples of what I'm looking at are:
2001:db8:3c4d:12::56ab/48 - ok with (see below)
2000::/3 Global unicast address range for use on the Internet?
FC00::/7 Unique local unicast address range?
FF00::/8 Multicast range?
I understand that :: means add in "0000" between colons until the address is 8 fields in length (128 bits) and you add back in any leading zero's but what I can't get my head around is if each field is 16 bits and theirs four values which make up a field (0-F) then each value is 4 bits. The first address seems to work if I'm doing it right? The address 'rebuilt' is;
2001:0db8:3c4d:0012:0000:0000:1234:56ab/48
So the interface ID/host portion is 000056ab (last three fields - 48bits). However, I don't seem to be able to work out the other examples using my logic.
In second example:
2000:0000:0000:0000:0000:0000:0000:0000/3
What's the 3?? I know /16 would be "0000" (last field or 16bits)
Do you have to convert them to binary and then do the /x bit? Maybe this is a bit advanced and may be covered on CompTIA network plus? If not, can you try and explain it so I'm a little less muddled please ? Thanks guys and hope I've explained this well enough
• @Andrew-Gaskell Great questions, and I do feel that these topics go well beyond A+ and even Network+, however that being said, what is important for the exam is to be able to identify the types of addresses there are:
Global Unicast
Unique Local | 624 | 2,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-06 | latest | en | 0.953867 |
https://www.tensorflow.org/versions/r2.1/api_docs/python/tf/eig?hl=TR | 1,632,217,827,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057199.49/warc/CC-MAIN-20210921070944-20210921100944-00661.warc.gz | 1,005,191,039 | 44,477 | # tf.eig
Computes the eigen decomposition of a batch of matrices.
The eigenvalues and eigenvectors for a non-Hermitian matrix in general are complex. The eigenvectors are not guaranteed to be linearly independent.
Computes the eigenvalues and right eigenvectors of the innermost N-by-N matrices in `tensor` such that `tensor[...,:,:] * v[..., :,i] = e[..., i] * v[...,:,i]`, for i=0...N-1.
`tensor` `Tensor` of shape `[..., N, N]`. Only the lower triangular part of each inner inner matrix is referenced.
`name` string, optional name of the operation.
`e` Eigenvalues. Shape is `[..., N]`. Sorted in non-decreasing order.
`v` Eigenvectors. Shape is `[..., N, N]`. The columns of the inner most matrices contain eigenvectors of the corresponding matrices in `tensor`
[{ "type": "thumb-down", "id": "missingTheInformationINeed", "label":"Missing the information I need" },{ "type": "thumb-down", "id": "tooComplicatedTooManySteps", "label":"Too complicated / too many steps" },{ "type": "thumb-down", "id": "outOfDate", "label":"Out of date" },{ "type": "thumb-down", "id": "samplesCodeIssue", "label":"Samples / code issue" },{ "type": "thumb-down", "id": "otherDown", "label":"Other" }]
[{ "type": "thumb-up", "id": "easyToUnderstand", "label":"Easy to understand" },{ "type": "thumb-up", "id": "solvedMyProblem", "label":"Solved my problem" },{ "type": "thumb-up", "id": "otherUp", "label":"Other" }] | 396 | 1,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-39 | latest | en | 0.686023 |
https://realpython.com/lessons/math-and-statistics-functions/ | 1,723,581,979,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641085898.84/warc/CC-MAIN-20240813204036-20240813234036-00013.warc.gz | 385,396,598 | 15,269 | # Math and Statistics Functions
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Happy Pythoning!
In this lesson, you’ll learn about new and improved `math` and `statistics` functions in Python 3.8. Python 3.8 brings many improvements to existing standard library packages and modules. `math` in the standard library has a few new `functions`. `math.prod()` works similarly to the built-in `sum()`, but for multiplicative products:
Python
``````>>> import math
>>> math.prod((2, 8, 7, 7))
784
>>> 2 * 8 * 7 * 7
784
``````
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The two statements are equivalent. `prod()` will be easier to use when you already have the factors stored in an iterable.
Another new function is `math.isqrt()`. You can use `isqrt()` to find the integer part of square roots:
Python
``````>>> import math
>>> math.isqrt(9)
3
>>> math.sqrt(9)
3.0
>>> math.isqrt(15)
3
>>> math.sqrt(15)
3.872983346207417
``````
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The square root of 9 is 3. You can see that `isqrt()` returns an `integer` result, while `math.sqrt()` always returns a `float`. The square root of 15 is almost 3.9. Note that `isqrt()` truncates the answer down to the next integer, in this case `3`.
Finally, you can now more easily work with n-dimensional points and vectors in the standard library. You can find the distance between two points with `math.dist()`, and the length of a vector with `math.hypot()`:
Python
``````>>> import math
>>> point_1 = (16, 25, 20)
>>> point_2 = (8, 15, 14)
>>> math.dist(point_1, point_2)
14.142135623730951
>>> math.hypot(*point_1)
35.79106033634656
>>> math.hypot(*point_2)
22.02271554554524
``````
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This makes it easier to work with points and vectors using the standard library. However, if you will be doing many calculations on points or vectors, you should check out NumPy.
The `statistics` module also has several new functions:
The following example shows the functions in use:
Python
``````>>> import statistics
>>> data = [9, 3, 2, 1, 1, 2, 7, 9]
>>> statistics.fmean(data)
4.25
>>> statistics.geometric_mean(data)
3.013668912157617
>>> statistics.multimode(data)
[9, 2, 1]
>>> statistics.quantiles(data, n=4)
[1.25, 2.5, 8.5]
``````
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In Python 3.8, there is a new `statistics.NormalDist` class that makes it more convenient to work with the Gaussian normal distribution. To see an example of using `NormalDist`, you can try to compare the speed of the new `statistics.fmean()` and the traditional `statistics.mean()`:
Python
``````>>> import random
>>> import statistics
>>> from timeit import timeit
>>> # Create 10,000 random numbers
>>> data = [random.random() for _ in range(10_000)]
>>> # Measure the time it takes to run mean() and fmean()
>>> t_mean = [timeit("statistics.mean(data)", number=100, globals=globals())
... for _ in range(30)]
>>> t_fmean = [timeit("statistics.fmean(data)", number=100, globals=globals())
... for _ in range(30)]
>>> # Create NormalDist objects based on the sampled timings
>>> n_mean = statistics.NormalDist.from_samples(t_mean)
>>> n_fmean = statistics.NormalDist.from_samples(t_fmean)
>>> # Look at sample mean and standard deviation
>>> n_mean.mean, n_mean.stdev
(0.825690647733245, 0.07788573997674526)
>>> n_fmean.mean, n_fmean.stdev
(0.010488564966666065, 0.0008572332785645231)
>>> # Calculate the lower 1 percentile of mean
>>> n_mean.quantiles(n=100)[0]
0.6445013221202459
``````
Copied!
In this example, you use `timeit` to measure the execution time of `mean()` and `fmean()`. To get reliable results, you let `timeit` execute each function 100 times, and collect 30 such time samples for each function. Based on these samples, you create two `NormalDist` objects. Note that if you run the code yourself, it might take up to a minute to collect the different time samples.
`NormalDist` has many convenient attributes and methods. See the documentation for a complete list. Inspecting `.mean` and `.stdev`, you see that the old `statistics.mean()` runs in 0.826 ± 0.078 seconds, while the new `statistics.fmean()` spends 0.0105 ± 0.0009 seconds. In other words, `fmean()` is about 80 times faster for these data.
If you need more advanced statistics in Python than the standard library offers, check out `statsmodels` and `scipy.stats`.
to join the conversation. | 1,170 | 4,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-33 | latest | en | 0.761441 |
https://gorelik.net/2019/10/28/sometimes-you-dont-really-need-a-legend/ | 1,685,370,362,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644867.89/warc/CC-MAIN-20230529141542-20230529171542-00531.warc.gz | 328,714,765 | 26,974 | # Sometimes, you don’t really need a legend
This is another “because you can” rant, where I claim that the fact that you can do something doesn’t mean that you necessarily need to.
This time, I will claim that sometimes, you don’t really need a legend in your graph. Let’s take a look at an example. We will plot the GDP per capita for three countries: Israel, France, and Italy. Plotting three lines isn’t a tricky task. Here’s how we do this in Python
```plt.plot(gdp.Year, gdp.Israel, '-', label='Israel')
plt.plot(gdp.Year, gdp.France, '-', label='France')
plt.plot(gdp.Year, gdp.Italy, '-', label='Italy')
plt.legend()
```
The last line in the code above does a small magic and adds a nice legend
In Excel, we don’t even need to do anything, the legend is added for us automatically.
### So, what is the problem?
What happens when a person wants to know which line represents which country? That person needs to compare the line color to the colors in the legend. Since our working memory has a limited capacity, we do one of the following. We either jump from the graph to the legends dozens of times, or we try to find a heuristic (a shortcut). Human brains don’t like working hard and always search for shortcuts (I recommend reading Daniel Kahneman’s “Think Fast and Slow” to learn more about how our brain works).
What would be the shortcut here? Well, note how the line for Israel lies mostly below the line for Italy which lies mostly below the line for France. The lines in the legend also lie one below the other. However, the line order in these two pieces of information isn’t conserved. This results in a cognitive mess; the viewer needs to work hard to decipher the graph and misses the point that you want to convey.
And if we have more lines in the graph, the situation is even worse.
### Can we improve the graph?
Yes we can. The simplest way to improve the graph is to keep the right order. In Python, we do that by reordering the plotting commands.
```plt.plot(gdp.Year, gdp.Australia, '-', label='Australia')
plt.plot(gdp.Year, gdp.Belgium, '-', label='Belgium')
plt.plot(gdp.Year, gdp.France, '-', label='France')
plt.plot(gdp.Year, gdp.Italy, '-', label='Italy')
plt.plot(gdp.Year, gdp.Israel, '-', label='Israel')
plt.legend()
```
We still have to work hard but at least we can trust our brain’s shortcut.
### If we have more time
If we have some more time, we may get rid of the (classical) legend altogether.
```countries = [c for c in gdp.columns if c != 'Year']
fig, ax = plt.subplots()
for i, c in enumerate(countries):
ax.plot(gdp.Year, gdp[c], '-', color=f'C{i}')
x = gdp.Year.max()
y = gdp[c].iloc[-1]
ax.text(x, y, c, color=f'C{i}', va='center')
seaborn.despine(ax=ax)
```
(if you don’t understand the Python in this code, I feel your pain but I won’t explain it here)
Isn’t it better? Now, the viewer doesn’t need to zap from the lines to the legend; we show them all the information at the same place. And since we already invested three minutes in making the graph prettier, why not add one more minute and make it even more awesome.
This graph is much easier to digest, compared to the first one and it also provides more useful information.
.
I agree that this is a mess. The life is tough. But if you have time, you can fix this mess too. I don’t, so I won’t bother, but Randy Olson had time. Look what he did in a similar situation.
I also recommend reading my older post where I compared graph legends to muttonchops.
### In conclusion
Sometimes, no legend is better than legend.
This post, in Hebrew: [link]
## By Boris Gorelik
Machine learning, data science and visualization http://gorelik.net. | 892 | 3,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-23 | latest | en | 0.899599 |
https://learnsoc.org/math-worksheets-go/ | 1,550,804,167,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247512461.73/warc/CC-MAIN-20190222013546-20190222035546-00414.warc.gz | 608,652,061 | 6,910 | Browse All Plans
Just Plans
# Math Worksheets Go
By Felix Glockner at December 12 2018 16:30:10
These students have weak immediacy. This means that compared to other students, they are slow to come up with the right answers to basic math problems like 6 x 8, or 35 divided by 7. While other kids have the answer stored in their memory, these kids do not. This causes big problems when they try to work through the more complex problems they encounter in junior high math.
Times are different these days. Kids are growing up in a world of microwaves, fast food chains, Nintendo, Wifi, iPads, along with a ton of other technical marvels.
How much of an apple pie has been eaten? The answer to this question can be expressed in percentages, 50%; or in decimals, 0.5; or in fraction, ½. In other words, half of mom's delicious apple pie is gone.
So what kinds of worksheets should you get? Anything where you feel that your child needs further drill. We often have this notion that worksheets are just for math. This, of course, is not true. While they are excellent tools for reviewing math facts such as the multiplication tables and division facts, they are just as useful for reviewing parts of speech or the states in the union.
When I was growing up we didn't have home computers let alone PlayStation to entertain ourselves. Handheld camcorders were barely coming to the retail market by the time I was in 8th grade, but still a long ways away from the YouTube and Facebook arena we now see today. Times were extremely different back then and so was school.
Therefore it is our job as the teacher to make sure that when we need to utilize a worksheet, we provide the students with one that is as inspiring as can be. | 384 | 1,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-09 | latest | en | 0.97537 |
http://www.ux1.eiu.edu/~cfadd/1150-05/Hmwk/Ch01/Ch01.html | 1,516,517,877,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890314.60/warc/CC-MAIN-20180121060717-20180121080717-00279.warc.gz | 585,438,602 | 3,132 | PHY 1151
Doug Davis
## Ch 1, Introduction
D1.5. A tabletop measures 4 ft by 5 ft. What is the area of the tabletop in square meters?
A = l w
A = (4 ft) (5 ft) = 20 ft2
A = 20 ft2 [(12 in / 1 ft) (2.54 cm / 1 in) (1 m / 100 cm)]2
A = 1.86 m2
Or, as an alternative method, we could change the lengths first and then multiply,
l = 4 ft (12 in / 1 ft) (2.54 cm / 1 in) (1 m / 100 cm) = 1.22 m
w = 5 ft (12 in / 1 ft) (2.54 cm / 1 in) (1 m / 100 cm) = 1.52 m
A = l w
A = (1.22 m) (1.52 m)
A = 1.86 m2
D1.12. Express 65 mi/h in kilometers per hour.
v = 65 (mi/h) (1.61 km/mi)
v = 105 km/h
D1.19. Being careful to give the answer to the appropriate number of significant figures, add the following numbers: 29.34, 35.452, 123.92, and 8.934 .
If we just plug these numbers into a calculator, like
we will get an answer of 197.646. However, we do not know all the numbers to the same accuracy. It may help to re-write this addition with the unknown parts shown explicityl, like
Now we can see that the answer is really just 197.64 .
This answer, for addition, is only good -- or significant -- for the number of decimal places beyond the decimal point as the smallest number of decimal places beyond the decimal point in the numbers that are being added.
D1.23. The length of a card is measured as 42.3 cm; its width, as 12.764 cm. To the correct number of significant figures, what is the area of the card?
A = l w
Again, if we just enter the numbers into a calculator, we get
A = (42.3 cm) (12.764 cm) = 539.9172 cm2
However, for multiplication (and division), the number of significant figures in the answer is equal to the smallest (or smaller) number of significant figures in the factors that are being multiplied. The length, 42.3 cm, is known only to three significant figures so that determines the number of significant figures in the answer. Since we have 539.9172 cm2 when we round that to three significant figures we get 540 cm2 .
A = 540 cm2
D1.33. Light from a helium-neon laser has a wavelength of 632.8 nm. Express this in meters, using scientific notation.
wavelength = 632.8 nm [ 1 x 10 - 9 m / nm ] = 632.8 x 10 - 9 m
wavelength = 6.328 x 10 - 7 m
D1.34. Add the following numbers, being careful to keep only the appropriate significnt figures in your answer: 9.135 x 102, 3.375 x 104, 1.934 x 103.
` `
But, as before, it may help to explicitly show what digits are unknown,
```
365.9 x 102 ```
D1.41. A mechanical watch is advertised as not gaining or losing more than 5 s a month. What is the percent accuracy of this watch? That is, if it gains (or loses) 5 s in a 30-cay month, how accurate -- expressed as a percentage -- is the time it reads?
Inaccuracy = uncertainty / number
Inaccuracy = (5 s) / (1 mo)
1 mo = 1 mo (30 da/1 mo) (24 h/1 da) (60 min/1 h) (60 s/1 min)
1 mo = 2 592 000 s = 2.592 x 106 s
Inaccuracy = (5 s) / (2.592 x 106 s)
Inaccuracy = 1.9 x 10 - 6
Inaccuracy = 1.9 x 10 - 6 (100%)
Inaccuracy = 1.9 x10 - 4 %
Inaccuracy = 0.000 19 %
Accuracy = 100% - Inaccuracy
Accuracy = 100 % = 0.000 19 %
Accuracy = 99.999 81 % | 1,009 | 3,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-05 | latest | en | 0.903773 |
https://nl.mathworks.com/matlabcentral/cody/problems/44144-duration-of-a-trip-in-minutes/solutions/1190290 | 1,603,393,286,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880014.26/warc/CC-MAIN-20201022170349-20201022200349-00290.warc.gz | 470,580,896 | 17,390 | Cody
# Problem 44144. Duration of a trip in minutes
Solution 1190290
Submitted on 16 May 2017 by Said BOUREZG
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
x = '1h20'; y_correct = 80; assert(isequal(tripinMinutes(x),y_correct))
ans = 1×2 cell array '1' '20' ans = 80
2 Pass
x = '19h20'; y_correct = 1160; assert(isequal(tripinMinutes(x),y_correct))
ans = 1×2 cell array '19' '20' ans = 1160
3 Pass
x = '5h45'; y_correct = 345; assert(isequal(tripinMinutes(x),y_correct))
ans = 1×2 cell array '5' '45' ans = 345
4 Pass
x = '1h'; y_correct = 60; assert(isequal(tripinMinutes(x),y_correct))
5 Pass
x = '1h05'; y_correct = 65; assert(isequal(tripinMinutes(x),y_correct))
ans = 1×2 cell array '1' '05' ans = 65
6 Pass
x = '1h5'; y_correct = 65; assert(isequal(tripinMinutes(x),y_correct))
ans = 1×2 cell array '1' '5' ans = 65
7 Pass
x = '6h35'; y_correct = 395; assert(isequal(tripinMinutes(x),y_correct))
ans = 1×2 cell array '6' '35' ans = 395
8 Pass
x = '2h60'; y_correct = 180; assert(isequal(tripinMinutes(x),y_correct))
ans = 1×2 cell array '2' '60' ans = 180
9 Pass
x = '29'; y_correct = 29; assert(isequal(tripinMinutes(x),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 497 | 1,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2020-45 | latest | en | 0.44213 |
https://www.askiitians.com/forums/Differential-Calculus/25/6111/limits.htm | 1,722,852,989,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640436802.18/warc/CC-MAIN-20240805083255-20240805113255-00278.warc.gz | 525,776,710 | 42,842 | # Lt (x^2-a^2)/(x-a)
148 Points
14 years ago
Dear ramakrishna
I think you want to ask
Ltx→a (x^2-a^2)/(x-a)
factorize
Ltx→a (x-a)(x+a)/(x-a)
or Ltx→a (x+a)
apply limit
Ltx→a (x+a) =2a
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. | 132 | 400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-33 | latest | en | 0.914136 |
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# Reading constant battery voltage and using the obtained value in sensor formula
#### anon7548
##### Newbie
Hi, i am using 3.7~4.2V lithium battery. I am using internal voltage reference to read constant battery voltage as we know that the battery voltage level depletes overtime. The problem is that my sensor (mini solar panel) reads max value under little light and does not go beyond that level no matter how much light falls onto it in the later stage. I need my logic to be <<< if read sensor voltage less than 3V detect night and do something whereas if voltage level is above 3V detect day and go to sleep. The entire project is ready i just need to figure out this.
Code:
void loop() {
printVolts();
//REFS1 AND REFS0 to 1 1 -> internal 1.1V refference
analogReference( INTERNAL);
DIDR0 = 0;
// Detect end-of-conversion
val = val * 5.7; //Multiply by the inverse of the divider
Serial.println("val: ");
Serial.println(val);
}
Last edited by a moderator:
#### betwixt
##### Super Moderator
Staff member
The unloaded PV voltage tends to flatten out at around 0.7V per cell (there may be many cells in series in a single panel) even under low light conditions. As you have noticed, as the light level increases, the voltage rises, hits a limit then goes no higher. It's an electrical phenomenon and nothing in software can avoid it.
The solution is to load the PV so the cells have to 'push' harder and show a better light to voltage relationship. You can do it by wiring a resistor across the panel or if you are trying to conserve the power, use another MCU pin to periodically switch a load on, measure the voltage then switch the load off again.
The light level to output power relationship of PV panels is complicated and the reason why maximum power point tracking (MPPT) is used on larger systems.
Brian.
#### KlausST
##### Super Moderator
Staff member
Hi,
I don't understand "... read constant battery voltage...". What does "constant" mean?
You talk about "battery voltage" and "sensor". But in your code I see you only measure "A1" ... whatever A1 is...
I don't see how you can measure two things here (battery and sensor).
How is your variable "val" declared. First you use it as integer, but then you use it a float.
Please give complete informations and draw a flow chart about the problem.
Btw: if you go to sleep, how do you decide to wake up?
Some informatiin about microcontroller type and compiler could be useful, also schematic and some clear comments in your code.
You could put a comment what "val" means or you could use a more descriptive variable name, like "SolarVolt" or "BattVolt".
Klaus
Edit: Corrected "A0" to "A1"
Last edited:
#### anon7548
##### Newbie
Hi,
I don't understand "... read constant battery voltage...". What does "constant" mean?
You talk about "battery voltage" and "sensor". But in your code I see you only measure "A0" ... whatever A0 is...
I don't see how you can measure two things here (battery and sensor).
How is your variable "val" declared. First you use it as integer, but then you use it a float.
Please give complete informations and draw a flow chart about the problem.
Btw: if you go to sleep, how do you decide to wake up?
Some informatiin about microcontroller type and compiler could be useful, also schematic and some clear comments in your code.
You could put a comment what "val" means or you could use a more descriptive variable name, like "SolarVolt" or "BattVolt".
Klaus
Okay, by reading constant battery voltage, i mean, the voltage level should remain constant (same/unchanged)irrespective of what the voltage level is at the battery and yes i am able to read same voltage level.
Code:
float val;
float voltage;
int led = 8;
void setup(){
Serial.begin(9600);
pinMode (A0, INPUT);
pinMode (A1, INPUT);
pinMode (led, INPUT);
}
void loop() {
printVolts();
//REFS1 AND REFS0 to 1 1 -> internal 1.1V refference
analogReference( INTERNAL);
DIDR0 = 0;
// Detect end-of-conversion
val = val * 5.7; //Multiply by the inverse of the divider
Serial.println("val: ");
Serial.println(val);
delay(1000);
}
void printVolts()
{
voltage = (sensorValue/ val) * 1024.;
delay(1000);
Serial.println( "voltage: ");
Serial.print(voltage);
}
check this, val is the value obtained from battery or battery voltage (A0) and (A1) is reading solar pin
Last edited by a moderator:
#### anon7548
##### Newbie
The unloaded PV voltage tends to flatten out at around 0.7V per cell (there may be many cells in series in a single panel) even under low light conditions. As you have noticed, as the light level increases, the voltage rises, hits a limit then goes no higher. It's an electrical phenomenon and nothing in software can avoid it.
The solution is to load the PV so the cells have to 'push' harder and show a better light to voltage relationship. You can do it by wiring a resistor across the panel or if you are trying to conserve the power, use another MCU pin to periodically switch a load on, measure the voltage then switch the load off again.
The light level to output power relationship of PV panels is complicated and the reason why maximum power point tracking (MPPT) is used on larger systems.
Brian.
I tried using resistor also but unfortunately, i could not find much differences, below i have attached schematic
#### Attachments
• solar charge.png
162.2 KB · Views: 23
#### betwixt
##### Super Moderator
Staff member
Try a load of about half the PV rating, 45/4 = 22.5mA so the load resistor should be 5/.0225 = 222 Ohms. Use a standard 220 Ohm value and see what the PV voltage on A1 measures under different light conditions.
A word of caution: The PV might produce more than 5V unloaded and in bright light, without a full specification it's hard to tell. It may be rated to produce 5V at 45mA load, in which case it might be more than 5V unloaded and therefore could damage the MCU. For safety, wire a schottky diode from the positive side of the PV to the 5V line on the Arduino board. It will have no effect on the reading but will ensure the PV voltage cant go much higher than 5V and overload the A1 input.
Brian.
#### anon7548
##### Newbie
Try a load of about half the PV rating, 45/4 = 22.5mA so the load resistor should be 5/.0225 = 222 Ohms. Use a standard 220 Ohm value and see what the PV voltage on A1 measures under different light conditions.
A word of caution: The PV might produce more than 5V unloaded and in bright light, without a full specification it's hard to tell. It may be rated to produce 5V at 45mA load, in which case it might be more than 5V unloaded and therefore could damage the MCU. For safety, wire a schottky diode from the positive side of the PV to the 5V line on the Arduino board. It will have no effect on the reading but will ensure the PV voltage cant go much higher than 5V and overload the A1 input.
Brian.
Yes, i have used schottky diode for protection | 1,700 | 6,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-14 | longest | en | 0.890779 |
http://microsoft.newsgroups.archived.at/public.excel.misc/200605/060507423637.html | 1,571,336,292,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986675598.53/warc/CC-MAIN-20191017172920-20191017200420-00169.warc.gz | 127,295,198 | 2,529 | # Which formula is correct for calculating times?
Subject: Which formula is correct for calculating times? Posted by: djm (djm.27gm1m_1147042502.264@excelforum-nospam.com) Date: Sun, 7 May 2006
I built a spreadsheet to calculate hours worked; I did it over a couple
of weeks, reading and learning as I went. During the time I was doing
this, somehow I ended up with 2 different formula for calculating hours
worked...these are:
=IF(N31>O31,MIN(("24:00"-N31+O31)*24,7.25),MIN((O31-N31)*24,7.25))
and
=IF(N32>O32,CEILING(("24:00"-N32+O32),7.25),MIN((O32-N32)*24,7.25))
where N and O are the start & finsh times respectively.
Both seem to calculate properly, but does anyone know which is the
better formula to use...of does it matter ? Thanks.
--
djm
------------------------------------------------------------------------
djm's Profile:http://www.excelforum.com/member.php?action=getinfo&userid=4793 | 256 | 908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-43 | latest | en | 0.870885 |
http://algebralab.org/practice/practice.aspx?file=Algebra2_8-4.xml | 1,642,971,760,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00145.warc.gz | 3,246,562 | 5,196 | Algebra II Exercises: Solving Exponential Equations With Real Exponents
Question Group #1
Directions and/or Common Information: Solve each equation.
4x = 64
1
9x = 27
2
8x = 16x-1
3
7x = 343
4
3x = 9x/2
5
252x-9 = 5x
6
512(2x/3) = 64x
7
4x = 1/256
8
2x = 1/32
9
9x = 3x+4
10
B Horn
Show Related AlgebraLab Documents | 163 | 347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-05 | longest | en | 0.699343 |
https://writersanswers.com/questions/ab-and-cd-are-two-chords-of-circle-such-that-ab-6/ | 1,679,726,974,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00229.warc.gz | 693,411,649 | 20,660 | # Ab And Cd Are Two Chords Of Circle Such That Ab 6 Cm Cd 12 Cm And Ab В€Ґ Cd
AB and CD are two chords of a circle such that AB = 6 cm, CD = 12 cm and AB ∥ CD. If the … AB and CD is 3 cm, find the radius of…
Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm CD = 12 cm. Let the radius of the circle be r cm. Drwa OP ⊥ AB and OQ ⊥ CD. Since, AB ∥ CD and OP ⊥ AB, OQ ⊥ CD. Therefore points O, Q and P are collinear. Clearly, PQ = 3 cm.
Let OQ = x cm. Then, OP = (x + 3) cm
In right triangles OAP and OCQ, we have
OA2 = OP2 + AP2 and OC2 = OQ2 + CQ2
⇒ r2 = (x + 3)2 + 32 and r2 = x2 + 62
[∵ AP = ½ AB = 3 cm and CQ = ½ CD = 6 cm
⇒ (x + 3)2 + 32 = x2 + 62 (on equating the value of r2)
⇒ x2 + 6x + 9 + 9 = x2 + 36
⇒ 6x = 18 ⇒ x = 3 cm
Putting the values of x in r2 = x2 + 62, we get
r2 = 32 + 62 = 45
⇒ r = √45 cm = 6.7 cm
Hence, the radius of the circle is 6.7 cm. | 401 | 928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-14 | latest | en | 0.724483 |
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https://www.adventuresinmachinelearning.com/upgrade-your-loop-incrementing-for-loops-by-2-in-python/ | 1,723,665,670,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00270.warc.gz | 488,748,688 | 16,799 | Programming can seem daunting at first with all the complex computations and syntax to be memorized. However, for loops are a fundamental part of programming that can make code more efficient and allow for looping through data structures in a concise manner.
In this article, we will explore various ways to increment a for loop by 2. Incrementing a for loop by 2 can be useful when there is a need to skip every other item in a data set or sequence.
## Incrementing For Loops by 2 in Python
The following are ways to achieve this in Python.
### 1. Using the Range Function
The range function is a built-in function in Python that generates a sequence of numbers, starting from zero by default, and increments by 1 for every subsequent number. To increment by 2, the third parameter of the range function is used as the step parameter.
### The following code snippet demonstrates this:
``````for i in range(0, 10, 2):
# code block
print(i)
``````
The output of this code snippet will be 0, 2, 4, 6, 8. The first parameter specifies the starting point (inclusive), and the second parameter specifies the ending point (exclusive).
### 2. Using Slicing Syntax
Slicing syntax can also be used to increment a for loop by 2. Slicing is essentially the process of creating a new object that contains a subset of an existing object.
In this case, we create a new sequence by skipping every other element in the original sequence. The following code demonstrates this:
``````data = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
for i in data[::2]:
# code block
print(i)
``````
The `::2` in the index refers to a step of 2.
This produces the same results as the previous code snippet.
### 3. With a List, Use the Range() and Len() Functions
If there is a need to iterate through a list and increment by 2, the `range()` and `len()` functions can be utilized.
The `len()` function is used to obtain the length of the list, which is then utilized as the second parameter of the `range()` function. The code snippet below demonstrates this:
``````data = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
for i in range(0, len(data), 2):
# code block
print(data[i])
``````
Output: 0, 2, 4, 6, 8
### 4. Using the Step Parameter of the Range Function
In addition to incrementing by 1, the `range()` function can also increment the loop by any other given interval using the `step` parameter. The `step` parameter must be an integer and must be specified as the third parameter of the `range()` function.
### The code snippet below demonstrates this:
``````for i in range(0, 10, 2):
# code block
print(i)
``````
Output: 0, 2, 4, 6, 8
## Conclusion
In summary, for-loops are important in programming and can be made more efficient by incrementing them by 2. There are various approaches to achieving this such as using the range() function with a step parameter, the slicing syntax, and using a list with range() and len() functions.
By utilizing these methodologies, efficiency in data processing can be improved, as only the required data is obtained. For-loops are a vital aspect of programming, as they allow us to iterate through data structures and perform operations on individual elements.
Incrementing by 2 is a useful technique, especially when dealing with data sets whose items are grouped in pairs or when we only need to access every other element. In this article, we will explore two more ways to increment for-loops by 2: using slicing syntax and using the range() and len() functions with lists containing only numbers.
### Using the Slicing Syntax
Python’s slicing syntax allows us to extract a portion of a list or string by specifying the starting and stopping indices. We can also add a third parameter, which determines how many elements to skip, thereby partially incrementing the loop.
### Here is an example:
``````numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for number in numbers[::2]:
print(number)
``````
In this code snippet, we slice the list by appending `::2` to the end, which means we start from index 0 and move in increments of 2, stopping when the end of the list is reached. The output produced upon executing this code will be:
``````1
3
5
7
9
``````
This method is useful when you need to partially increment the loop, or if you don’t want to access all the elements of the list.
### With a List, Use the Range() and Len() Functions
Using the range() and len() functions, we can loop over a list explicitly and increment by two.
The len() function is used to compute the length of the list, which is then fed as an argument to the range() function’s second argument. Here is an example:
``````numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for i in range(0, len(numbers), 2):
print(numbers[i])
``````
In this code snippet, we use a for-loop with a range.
The first parameter of the range function specifies the starting index, which, in this case, is 0. The second parameter is the length of the list, which we obtain by invoking the len() function on the numbers list.
The third parameter specifies the increment value, which is set to 2. The output upon executing this code will be:
``````1
3
5
7
9
``````
### Working Only with Lists Containing Numbers
Another way to increment for-loops by 2 when working with lists containing only numbers is to use the NumPy library. NumPy is a library in Python designed for scientific computation, and it has highly optimized functions that can greatly enhance the performance of iterable operations.
To install it, you can run the following command in your console:
``````pip install numpy
``````
After installing NumPy, we can use its array object to create and operate on arrays. Here’s how we can use NumPy to increment by 2.
``````import numpy as np
numbers = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
for number in numbers[::2]:
print(number)
``````
In this code snippet, we use NumPy to create an array, which we then slice to produce the required output. The `::2` indexing ensures we increment the loop by 2.
The output upon executing this code will be:
``````1
3
5
7
9
``````
## Conclusion
For-loops are a fundamental part of programming, and incrementing by 2 can improve code efficiency, especially when looped items are grouped in pairs or when we only need to access every other element. The methods demonstrated in this article, using the slicing syntax and the range() and len() functions with lists of numbers and NumPy arrays, provide an effective way of partially incrementing loops and accessing only the required data.
In conclusion, for-loops are an essential part of programming, allowing us to iterate through data structures and perform operations on individual elements. Incrementing a loop by 2 can improve code efficiency, especially when looped items are grouped in pairs or when we only need to access every other element.
The article discussed three methods for incrementing for-loops by 2: using the range() function with a step parameter, slicing syntax, and using the range() and len() functions with lists containing only numbers or NumPy arrays. These methods provide an effective way of accessing only the required data.
By employing these techniques, programmers can achieve increased efficiency and accuracy in data processing. | 1,766 | 7,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-33 | latest | en | 0.849426 |
https://philoid.com/question/105579-write-the-domain-and-the-range-of-the-function-fx-x- | 1,686,057,758,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652569.73/warc/CC-MAIN-20230606114156-20230606144156-00076.warc.gz | 504,574,214 | 11,240 | ## Book: RS Aggarwal - Mathematics
### Chapter: 3. Functions
#### Subject: Maths - Class 11th
##### Q. No. 21 of Exercise 3F
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
21
##### Write the domain and the range of the function, f(x) = –|x|.
(i) Domain
|x | is defined for all real values.
Hence -|x| is also defined for all real values.
The domain is R.
(ii) Range
Range for |x| is [0, ∞)
Therefore, range for - |x| is ( -∞,0].
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21 | 186 | 527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2023-23 | longest | en | 0.747015 |
https://math.stackexchange.com/questions/375283/a-principal-open-set-of-an-affine-algebraic-set-is-an-affine-variety | 1,566,507,094,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317359.75/warc/CC-MAIN-20190822194105-20190822220105-00360.warc.gz | 555,391,364 | 31,291 | # A principal open set of an affine algebraic set is an affine variety
Notations
• $k$ is an algebraic closed field and $\mathbb A^n(k)$ is the topological space $k^n$ with the Zariski topology
• If $X\subseteq\mathbb A^n(k)$ is an affine algebraic set and $f\in\Gamma(X)$, then $D(f)=\{x\in X\,:\, f(x)\neq0\}$
• An affine variety for me is a ringed space $(X,\mathcal O_X)$, where $\mathcal O_X$ is a sheaf of $k$-valued function, that is isomorphic (as ringed space) to $(V,\mathcal O_V)$, where $V$ is an IRREDUCIBLE affine algebraic set and $\mathcal O_V$ is the sheaf of regular functions on $V$.
I want to find an affine variety that is not an irreducible affine algebraic set. If $X\subseteq\mathbb A^n(k)$ is an irreducible affine algebraic set, and $f\in\Gamma(X)$, I would prove that $(D(f),\mathcal O_{X|D(f)})$ is an affine variety:
Let's consider $\mathfrak a=I(X)\subseteq k[T_1,\ldots, T_n]$ as a subring of $k[T_1,\ldots, T_n, T_{n+1}]$ and let $F\in k[T_1,\ldots, T_n]$ be a representative of $f$. If $Y=V(\big<\mathfrak a, FT_{n+1}-1\big>)$, the function $$j: Y\longrightarrow D(f)$$ that is the restriction of the projection of $\mathbb A^{n+1}(k)$ on $\mathbb A^n(k)$ is clearly bijective and continuous with inverse $j^{-1}:(x_1,\ldots, x_n)\longmapsto (x_1,\ldots, x_n,\frac{1}{f(x_1,\ldots,x_n)})$. Now
1. How can I prove that $Y$ is irreducible?
2. Why is $j^{-1}$ continuous? (so $j$ is a homeomorphism)
3. Why is $j$ an isomorphism of ringed spaces? If $U\subset D(f)$ is open, should be proved that for all $g\in \mathcal O_X(U)$, then $g\circ f_{|f^{-1}(U)}\in\mathcal O_X(f^{-1}(U))$.
1) $X$ is irreducible and $D(f)$ is open in $X$ implies that $D(f)$ is also irreducble. Now, since $j: Y \rightarrow D(f)$ is continuous and bijective, it is a homeomorphism. This implies that $Y$ is also irreducible (as it is topologically homeomorphic to $D(f)$).
3) To show that $j$ is an isomorphism of ringed spaces, you need to construct $j^{\#}: \mathcal{O}_{D(f)} \rightarrow j_{\ast} \mathcal{O}_Y$ and verify that it gives you an isomorphism of sheaves for each open subset $U \subset D(f)$. This can be done as follows. For any section $g \in \mathcal{O}_{D(f)}(U)$ that is represented by $G \in k[T_1, ...., T_n]$, we define its image to be the section represented by $H(T_1,...,T_n, T_{n+1}) = G(T_1,...,T_n)$. There is a little bit detailed and routine checking required here that $H$ does represent validly a unique section in $j_{\ast \mathcal{O}_Y(U)} = \mathcal{O}_Y(j^{-1}(U))$ and vice versa, i.e. a section in the $\mathcal{O}_Y(j^{-1}(U))$ has a preimage in $\mathcal{O}_{D(f)}(U)$. It probably uses some fact in step 1 but all in all, it is not too bad.
• I don't understand why if $j$ Is bijective and continuous, then it is a homeomorphism. – Dubious Apr 28 '13 at 18:10
• You're right, I was missing a detail, say $j^{-1}$ is also continuous. – mr.bigproblem Apr 28 '13 at 18:32
• To do this, take a distinguished open subset $D(h) \subset k^{n+1}$ that gives a distinguished open subset $D(h) \cap Y$ in $Y$, and compute the preimage via $j^{-1}$, which is $j(D(h) \cap Y) = j(D(h)) \cap D(f)$ and verify that this is open in $X$. – mr.bigproblem Apr 28 '13 at 18:37 | 1,102 | 3,214 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-35 | latest | en | 0.829777 |
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# Polar Coordinates help watch
1. Hi guys,
Can someone explain why r=(1/1+2cosϴ) for -2/3Pi < ϴ , 2/3 Pidoes not have a pole? I was able to differentiate it so I'm quite confused. I got ϴ = 0
Thanks
2. (Original post by ChemBoy1)
Hi guys,
Can someone explain why r=(1/1+2cosϴ) for -2/3Pi < ϴ , 2/3 Pidoes not have a pole? I was able to differentiate it so I'm quite confused. I got ϴ = 0
Thanks
The pole is where and clearly the function is never 0
3. Thank you very much, I seem to having difficulties with this part of the topic. It said in the book that if F(Thetha) = 0 then the theta is tangent to the graph
(Original post by RDKGames)
The pole is where and clearly the function is never 0
4. Also if there was a situation with more than 1 trig function e..g. 2+ cos theta - cos^2 theta. How are you meant to find out the tangents to the pole?
5. (Original post by ChemBoy1)
Also if there was a situation with more than 1 trig function e..g. 2+ cos theta - cos^2 theta. How are you meant to find out the tangents to the pole?
Try to use the correct terminology. The pole is a point, so it wouldn't make sense to have something 'tangent' to it.
Here you are interested in straight lines going through the pole, which are tangent to your polar eq. . These lines have equation for (or between 0 and 2 pi, if you want to use that).
The first step is to obvious set and solve it for the values. But make sure that .
For the example above, we would simply need to solve but note that this is just a quadratic in which also factorises.
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aliplayer1
This is the description of the problem I am working on:
public static void hittingIntegerPowers(int a, int b, int t, int[] out)
Define two positive integers to be “close enough for government work” if their absolute difference multiplied by the given tolerance level t is at most equal to the smaller of those two numbers. For example, the integers 2000 and 2007 are “close enough” when t = 100, since the difference 7 multiplied by 100 gives 700, which is less than 2000. On the other hand, the integers 2000 and 2050 would not be close enough for t = 100, although they would be close enough for a wider tolerance of t = 10. Tolerance of t = 100 corresponds to the notion of being within one percent, but without using any division or floating point numbers to compute this. The more narrow tolerance of t = 100000 would require these powers to be within one thousandth of a percent of each other, which ought to satisfy even the most ardent six sigma advocate auditing this course from the business school. Given two positive integers a and b and the desired tolerance t, this method should find and return find the smallest integer powers pa and pb so that when a is raised to the power of pa, and b is raised to the power of pb, the resulting two numbers are close enough for government work. Since these powers can get pretty big, as you can see in the table below, you need to perform these calculations using the BigInteger type. However, this is only for the actual powers that might end up having tens of thousands of digits; the exponents pa and pb are guaranteed to fit inside the int type for all the test cases given to your method.
This is the solution I have come up with so far:
Java:
``````public static void hittingIntegerPowers(int a, int b, int t, int[] out) {
BigInteger bigA = BigInteger.valueOf(a);
BigInteger bigB = BigInteger.valueOf(b);
BigInteger currentPowerA = bigA;
BigInteger currentPowerB = bigB;
BigInteger tolerance = BigInteger.valueOf(t);
int pa = 1;
int pb = 1;
while (pa < 1000) {
while (pb <= pa){
BigInteger diff = currentPowerA.subtract(currentPowerB).abs();
if (diff.multiply(tolerance).compareTo(bigA.min(bigB)) <= 0) {
out[0] = pa;
out[1] = pb;
return;
}
pb++;
currentPowerB = currentPowerB.multiply(bigB);
}
pa++;
currentPowerA = currentPowerA.multiply(bigA);
pb = 1;
currentPowerB = bigB;
}
}``````
There has to be some sort of logical problem in the while loop because using inputs a = 2, b = 3 and t = 100, the array out stays at [0,0]. The code seems to work fine with examples where the final pa and pb are different in value by one (a = 2, b = 4, t = 100).
I appreciate any help with the logic of the program or other approaches to solve the problem.
WhiteCube
Code:
``if (diff.multiply(tolerance).compareTo(bigA.min(bigB)) <= 0)``
The comparison should use currentPowerA and currentPowerB, not bigA and bigB.
WhiteCube
I think you can remove your inner loop.
Code:
``````pa=1
pb=1
LOOP
IF currentPowerB too small
pb++
ELSE
IF currentPowerB too large
pa++
ELSE
solution found
exit loop
too small means:
currentPowerB*(t+1) < currentPowerA*t
too large means:
currentPowerA*(t+1) < currentPowerB*t``````
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# Texas 1/31 New Evening extra numbers
Topic closed. 6 replies. Last post 12 years ago by Texasman.
Page 1 of 1
Glergenflergen, Texas
United States
Member #7809
October 13, 2004
962 Posts
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Posted: January 31, 2005, 5:05 pm - IP Logged
Well, 345 box hit at Midday Monday. That kind of diminishes my 245 for tonight. I did some revamping and adjusting on part of my workout and got some other numbers for Monday night 1/31/05.
Looks like "7" is the strongest with 0,1, 3, 4. So, It's good to play 017, 037, 047, 137, 147, 347 and (007, just because), along with 245 and 246.
The reasoning for the 7 is because it's in a column that shows me the 2, 7 and 8. History tells me that when three or more number are in this particular column that at least one of them is drawn. So, at least 2, 7 or 8 will come tonight. My other workout shows the 7 a lot. Therefore 7 is strongest.
Amarillo/Austin
United States
Member #1424
April 25, 2003
696 Posts
Offline
Posted: January 31, 2005, 5:10 pm - IP Logged
Texasman:
Seven shows up in Mike K's Candy Number system, the WheelWorld.net Key Number system, Steve Player's Pyramid system and Lotto Buster 2010 in the Hot, Pattern and Super categories. Eight is also strong, but not as strong as 7. Use 7 and 8 as key numbers and eliminate number 6. Play it as a nine-number wheel and only play numbers containing 8 or 7.
Orangeman
Glergenflergen, Texas
United States
Member #7809
October 13, 2004
962 Posts
Offline
Posted: January 31, 2005, 5:17 pm - IP Logged
Thanks, Orangeman. I'll watch for the 8. Tonight, I will just play the 7 and see what happens. Keep in touch about Tuesday night as well, in case 8 shows up tonight and I need your advice again on your sources for best numbers. Thanks.
Amarillo/Austin
United States
Member #1424
April 25, 2003
696 Posts
Offline
Posted: January 31, 2005, 5:25 pm - IP Logged
Texasman:
I listed my sources. I use multiple programs to pick numbers. By the way, when you eliminate one number and use two key numbers tonight, you will have 42 combinations to play. I changed my mind about 6 and plan to eliminat 9 instead. I have found that a high percentage profit is better than none.
Forty-two played six-way is a \$252 bet to win \$500. A lot of people are unwilling to play that much, but it means a \$248 profit when you win. My advice is to start slow with 50 cent bets and work your way up.
I test a lot and don't put down that much money. I would rather win when I play and not lose money.
Orangeman
Glergenflergen, Texas
United States
Member #7809
October 13, 2004
962 Posts
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Posted: January 31, 2005, 7:44 pm - IP Logged
bumpeme
Houston, Texas
United States
Member #3507
January 26, 2004
60 Posts
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Posted: January 31, 2005, 7:46 pm - IP Logged
Don't forget to watch out for doubles.
Glergenflergen, Texas
United States
Member #7809
October 13, 2004
962 Posts
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Posted: January 31, 2005, 10:31 pm - IP Logged
Yep,...always watch for the doubles. Tonight? 022, 002, 077, 007 and 111?
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Topic: Making a square matrix from two vector
Replies: 22 Last Post: Mar 19, 2013 12:37 AM
Messages: [ Previous | Next ]
Bruno Luong Posts: 9,822 Registered: 7/26/08
Re: Making a square matrix from two vector
Posted: Feb 14, 2013 7:18 AM
dpb <none@non.net> wrote in message <kfi6r9\$lc6\$1@speranza.aioe.org>...
>
> But even 'doc colon' is essentially narrative examples of what it does
> w/ certain values--it doesn't say a thing about complex--nor does it
> intimate there's a problem with it. So, it isn't completely defined or
> at least user-documented;
But I never say the contrary dpb.
IIRC, the behavior of a:b:c when a, b, c are (pure real) arrays are documented recently, (after the thread I provide the like earlier).
But the behavior for complex ins't (yet?) documenetd. Though the same behavior is reproductible in all MATLAB versions: it operates only on real part first element of the inputs. That's logical and meaningful IMO.
There are many things that are not documented. Just take a look at Yair's blog. Are things discussed there are meaningless to you? I thing Yair's blog is the most useful blog concerning MATLAB.
In many languages - even a strict one like C - the behavior of many function is not completely defined (such as malloc(0), or something as simple as the bit shift (a >> b)). The specific compiler will implement specific behavior, yet all of them are meaningful. Just like MATLAB colon.
Bruno
Date Subject Author
2/8/13 Jerry
2/8/13 Barry Shaw
2/8/13 dpb
2/8/13 james bejon
2/9/13 Bruno Luong
2/9/13 james bejon
2/9/13 Bruno Luong
2/12/13 james bejon
2/12/13 Steven Lord
2/12/13 Bruno Luong
2/12/13 dpb
2/12/13 Bruno Luong
2/12/13 dpb
2/13/13 Bruno Luong
2/13/13 dpb
2/14/13 Bruno Luong
2/14/13 dpb
2/14/13 Bruno Luong
2/14/13 dpb
2/14/13 dpb
2/13/13 james bejon
3/18/13 James
3/19/13 Bruno Luong | 612 | 2,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-17 | latest | en | 0.870319 |
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Vaughan Pratt pratt at cs.stanford.edu
Mon Aug 31 16:41:45 EDT 2009
```Bill Taylor wrote:
> The main idea is that one has a countable collection of sets which
> are each coverable by (necessarily countable) collections of intervals
> of arbitrarily small total finite length. Then when one comes to try
> to cover the union set with a collection of intervals, one must pick
> a countable (and small-lengthed) bunch of intervals from the countable
> union of countable sets. This requires making a countable number of
> choices to do so, hence AC.
Thanks, Bill. The reasoning that lured me to my incorrect conclusion (that a countable union of null sets must be null even without Choice) was that the property of having zero measure for a countable set was independent of the particular enumeration selected, whence one should be allowed to infer that the sum of a constantly zero series 0+0+0+... is its limit, namely zero.
The Feferman-Levy result shows that this argument cannot be made to work, by exhibiting a sequence of sets each of zero measure whose total measure is nonzero (say 1 in the case of the unit interval (0,1)). We then obtain 0+0+0+... = 1.
I seem to have committed a variant of the error made by Cauchy in 1821 when he proved that the pointwise limit of a sequence of continuous functions is continuous, by neglecting the possibility that the convergence might not be uniform in the sense appropriate to this variant.
In the variant at hand the source of the difficulty would seem to be the sensitivity of Lebesgue measure to the structure of the set being measured. One would naturally expect the Lebesgue measure of an increasing sequence of sets whose measures converge to a limit to be that limit; Feferman-Levy shows that this does not follow in the absence of Choice.
The Wikipedia article on Lebesgue measure mentions (more than once) that the existence of non-measurable sets depends on Choice. However in its list of 14 properties of Lebesgue measure on R^n it does not indicate which of those do not hold in the absence of Choice. It would appear from the above that property 2 (concerning the disjoint union of countably many measurable sets) does not. Which others fail?
(As an aside, is the Solovay result that every set could be Lebesgue-measurable consistent with the Feferman-Levy result that the reals could be a countable union of countable sets? I don't right away see how either possibility implies the negation of the other.)
For those who find these counterexamples unreasonably weird yet do not want to commit to the possibility of well-ordering the continuum, what does locale theory have to offer? Recently I argued for the naturalness of the axiom that the complete ultrafilters (aka complete homomorphisms to 2) of a complete atomic Boolean algebra (CABA) are its atoms, but I can see some being bothered by the choices implicit in the complete distributivity of CABAs, namely of infinite unions over infinite meets and vice versa. The advantage of frames (dual to locales in the same way Boolean algebras are dual to Stone spaces) over CABAs is that they allow infinite joins (of any cardinality) while avoiding the sorts of choices entailed by complete distributivity. Are there any locale theorists on this list who'd like to expand on the relationship between locales and choice? In particular what is the connection between Brouwer's proposal to drop excluded middle and Isbell's proposal half a century later to substitute frame homomorphisms for CABA homomorphisms (which is what the inverse-image operation used in defining continuity for topological spaces amounts to).
Vaughan Pratt
``` | 836 | 3,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-10 | latest | en | 0.951144 |
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Use these 32 multiple choice ordering decimals to help your students practice this important skill. Each of these cards includes series' of four decimals for students to order. The decimals range from ones to thousands and are mixed on the cards so that students can develop fluency with this skill.
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Includes 20 task cards that support 4.9B: Solve one- and two-step problems using data in whole numbers, decimal, and fraction form in a frequency table, dot plot (line plot), or stem and leaf plot. Includes gameboard, answer key, and answer document.
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Capture the Decimals is an exciting math game for comparing and ordering decimals. There are two variations of the game, a whole class version and a partner version, and there are three levels of game cards. Students take turns placing decimal cards in sequential order. If any player is able to plac
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Comparing decimals has never been easier! This 20 page file contains 1 activity where students shade in decimal models to develop an understanding of decimal number sense and how to compare decimals. There are also 2 sets of 35+ printable, self-checking cards that are ideal for math centers, flashc
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A great tool to reinforce fluency between fraction, decimal, and percent conversions. I used these in all middle school grades and found students to be successful when they were able to memorize their basic conversions. Aligned to CCSS and TEKS. Quizzes can be implemented quickly as a warm up, qui
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This classroom tested game helps your students practice converting fractions to decimals to percents, find equivalents between the three, and have fun doing it! Your students will play a traditional game of Go Fish!, however instead of making matching suits, the students will match the equivalent
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This resource can be found in my FIFTH GRADE MATH HUGE BUNDLE OF RESOURCES *This game is part of my Let's Make a Deal math and science review games bundle.**If you are looking for the fourth grade version, click here* ************************************************************************** Le
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Comparing Decimal Numbers - Order Up! Welcome to a great resource that will allow your students to practice KEY skills in a self-checking, hands-on, and self-paced way. Check out the FREE PREVIEW to see what is included and for more information on how it works! This set of ORDER UP! focuses on
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***math word problems following Common Core*** Teach your students to use math language/vocabulary in their written, extended responses. This math journal includes 18, open-ended (extended response) questions to use with your math instruction, to collect data or to assess. All questions follow 5
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RATIONAL NUMBERS Word Problems - Task Cards Use these 40 task cards with your students to help them practice solving real-word RATIONAL NUMBERS word problems. What is Included: This resource includes 40 task cards, a student answer sheet, and an answer key. Your students will love working with
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Rounding Decimals 'Clip and Flip' Cards contains 70 self-correcting cards to help students practice rounding decimals to the hundreds, tens, ones, tenths, hundredths, and thousandths place. These cards were created to support Common Core standard 5.NBT.A.4: Use place value understanding to round de
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Decimal Place Value Scoot Activity/Task Cards Scoot, scoot, scoot! Don't your kids just love to scoot!! This Decimal Place Value Scoot Activity gives 3rd, 4th, and 5th graders a fantastic opportunity to move around while learning. This will really help your students who NEED to move. In this scoot
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This 5.NBT.2 CCSS aligned place value packet includes 17 practice sheets to teach/reinforce multiplying and dividing whole numbers and decimals by powers of 10. These sheets can be used to introduce a concept, review a concept, or send home as reinforcement for homework. An answer key is provided.--
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Also included in: 5.NBT.2 BUNDLE: Multiply & Divide by Powers of 10
\$4.25
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This CCSS aligned place value packet includes 12 practice sheets to teach/reinforce identifying place value of whole numbers and decimals as well as comparing place values (10 times the value, 100 times the value, 1/10 the value, and 1/100 the value). These sheets can be used to introduce a concept,
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Also included in: 5.NBT.1 BUNDLE: Understanding Place Value
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Decimal Puzzles: Use these easy-to-cut decimal place value puzzles as a fun way for students to practice base-ten place value, powers of ten, decimal forms, comparing and ordering decimals, and rounding decimals. These decimal puzzles make great math centers! Aligned to fifth grade common core stand
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Also included in: Decimal Place Value Math Centers Bundle
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Fractions and Decimals PowerPoint: Are you looking for a systematic PowerPoint that will introduce your students to the fraction skills identified in the 4th Grade Common Core State Standards? If so, this highly interactive 72-slide PowerPoint may be exactly what you are looking for! It is fully a
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Also included in: Fractions Bundle 2 (based on 4th Grade CCSS)
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The following are task cards focus on interpreting decimal points on a number line. The following set includes... -36 task cards -Number Lines with Decimals -Cards with rigorous questions -Answer Response sheet -Answer Key The questions on the task cards are mirrored to questions in the STAAR Math T
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THIS BUNDLE NOW INCLUDES 5.3H Included in this bundle are 8 sets of task cards which cover 9 different TEKS. Save money when you purchase the bundle set! (Please note this bundle does not include 5.3A, or 5.3K) All of the task cards available for individual purchase. Please see the links below.
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Need no prep math test prep resources? This resource is a no prep way to review decimal word problems with your students.Types of Problems:Constructed response problems and multiple choice problems to help prepare the students for multiple testing situationsSkill Covered:This resource has word probl
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Ordering rational numbers can be a VERY hard concept for students to understand, so what's better than a fun color by answer activity to have students practice and forget they're even doing "work". :) You may also like my Ordering Rational Numbers with QR Codes Task Cards as well! This activity is
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,334 | 9,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-30 | latest | en | 0.868296 |
https://www.physicsforums.com/threads/optics-convolution.795963/ | 1,619,031,262,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039546945.85/warc/CC-MAIN-20210421161025-20210421191025-00511.warc.gz | 1,032,074,874 | 18,833 | Optics convolution
Homework Statement
According to my notes, if we have a sinusoidal aperture/transmission function of the form a(x)=1+sin(wx) and a 'top-hat' aperture function given by b(x)=1, -0.5d≤x≤0.5d, b(x)=0 otherwise, then their convolution should give a finite sinusoidal aperture function, i.e sinusoidal but only over a certain range - outside of this it is zero.
Homework Equations
Convolution of a(x) and b(x) is ∫-∞a(y)b(x-y)dy
The Attempt at a Solution
We need ∫-∞[1+sin(wy)]b(x-y)dy, and my thinking is that the integrand is zero everywhere apart from between x-0.5d and x+0.5d, where it is simply [1+sin(wy)]. Then the convolution integral reduces to ∫x-0.5dx+0.5d[1+sin(wy)]dy giving a-{cos[w(x+d/2)]+cos[w(x-d/2)]}/w which is not the finite sinusoidal aperture function I've been told I should get. Is this right?
Last edited:
mfb
Mentor
The convolution of a signal with a "box" is not the signal restricted to a specific range. If you want that, you have to multiply the two functions.
"a" seems to have two different meanings which is confusing.
The convolution of a signal with a "box" is not the signal restricted to a specific range. If you want that, you have to multiply the two functions.
"a" seems to have two different meanings which is confusing.
Changed a to d, didn't realise that.
I'm not clear - the function b(x-y) is a box centred on y=x with width d, right? Then the integrand [1+sin(wy)]b(x-y) surely vanishes everywhere but for where the box is non-zero, and so basically we can write b(x-y)=1 as long as we integrate over the box, from x-0.5d to x+0.5d. Then I just do the integral. I'm not sure what you mean...
Ray Vickson
Homework Helper
Dearly Missed
Homework Statement
According to my notes, if we have a sinusoidal aperture/transmission function of the form a(x)=1+sin(wx) and a 'top-hat' aperture function given by b(x)=1, -0.5d≤x≤0.5d, b(x)=0 otherwise, then their convolution should give a finite sinusoidal aperture function, i.e sinusoidal but only over a certain range - outside of this it is zero.
Homework Equations
Convolution of a(x) and b(x) is ∫-∞a(y)b(x-y)dy
The Attempt at a Solution
We need ∫-∞[1+sin(wy)]b(x-y)dy, and my thinking is that the integrand is zero everywhere apart from between x-0.5d and x+0.5d, where it is simply [1+sin(wy)]. Then the convolution integral reduces to ∫x-0.5dx+0.5d[1+sin(wy)]dy giving a-{cos[w(x+d/2)]+cos[w(x-d/2)]}/w which is not the finite sinusoidal aperture function I've been told I should get. Is this right?
Yes, except for a sign. If ##C(x)## is the convolution we have
$$C(x) = a+\frac{\cos(wx - wa/2) -\cos(wx+wa/2)}{w}$$
However, using standard trigonometric additional formulas, this can be manipulated to give ##C(x) = a + K \sin(w x)##, where ##K## is some constant that I will leave you to compute.
Note, however: the formula for ##C(x)## applies on the whole real line ## -\infty < x < +\infty##; it does not vanish outside of some finite interval. This follows from the fact that
$$C(x) = \int_{-\infty}^{\infty} a(y) b(x-y) \, dy = \int_{-\infty}^{\infty} a(x-y) b(y) \, dy$$
(by a simple change of variables). The second form gives
$$C(x) = \int_{-a/2}^{a/2} [1 + \sin(w(x-y)) \, dy$$
Yes, except for a sign. If ##C(x)## is the convolution we have
$$C(x) = a+\frac{\cos(wx - wa/2) -\cos(wx+wa/2)}{w}$$
However, using standard trigonometric additional formulas, this can be manipulated to give ##C(x) = a + K \sin(w x)##, where ##K## is some constant that I will leave you to compute.
Note, however: the formula for ##C(x)## applies on the whole real line ## -\infty < x < +\infty##; it does not vanish outside of some finite interval. This follows from the fact that
$$C(x) = \int_{-\infty}^{\infty} a(y) b(x-y) \, dy = \int_{-\infty}^{\infty} a(x-y) b(y) \, dy$$
(by a simple change of variables). The second form gives
$$C(x) = \int_{-a/2}^{a/2} [1 + \sin(w(x-y)) \, dy$$
Thanks for your reply! So my notes say that it should give me a sinusoidal function, BUT, only over a finite interval - so we agree that they must be wrong then? | 1,262 | 4,071 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-17 | longest | en | 0.907645 |
https://www.mymoneyblog.com/make-money-from-credit-cards-0-balance-transfer-profit-calculator-tool.html | 1,708,464,593,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00589.warc.gz | 937,277,970 | 38,858 | Make Money From Credit Cards: 0% Balance Transfer Profit Calculator Tool
My Money Blog has partnered with CardRatings and may receive a commission from card issuers. Some or all of the card offers that appear on this site are from advertisers and may impact how and where card products appear on the site. MyMoneyBlog.com does not include all card companies or all available card offers. All opinions expressed are the author’s alone.
My series of articles on How To Make “Free” Money From 0% APR Balance Transfers has been very popular and many readers have also jumped in. Despite the risks, I’m still happily earning some money from the credit card companies for a change, and haven’t missed any payments. From the beginning, people have asked me to make a spreadsheet or calculator in order to estimate the potential profit from such endeavors. I initially decided against doing so because there are lots of different variables at stake that make an exact prediction close to impossible. However, I think it may still be useful to obtain some more realistic numbers.
Without further ado, I present to you the…
0% Balance Transfer Profit Calculator
Enter savings account APY: % Enter starting balance: \$ Enter the monthly minimum payment percentage (2%) % Your interest earned: \$ (See assumptions and definitions below)
Inputs and Definitions
1. Arbitraged Interest Rate (APY) – Where are you putting the money you’re borrowing for free? This is the interest rate of the investment vehicle (savings account, CD, Treasury bond) you are using, or perhaps the interest rate of the existing loan (car, home equity, student) that you are paying down.
2. Starting Balance (dollars) – How much money are you transferring?
3. Monthly Minimum Payment (%) – Usually you must still make a monthly minimum payment on the outstanding balance during the 0% period, which will decrease your profit potential slightly. This is usually around 2%, but may vary between 1.5% and 4%.
Assumptions
1. The balance transfer is for 12 months at 0% APR, with no balance transfer fee. You can find my list of the best 0% APR offers here with low or no balance transfer fees here.
2. The interest is assumed to compound monthly, which allows me to convert from APY to APR, and then to a periodic rate. Compounding frequency is a variable here, but doesn’t change the numbers too much.
3. I am ignoring the time required to actually convert the balance transfer into cash earning interest. Sometimes this can take up to a few weeks, sometimes it is much faster. Instead of guessing, I just leave it be.
4. I am also ignoring things like grace periods and the timing of statement cycles and due dates, which can actually increase the time that your borrowed money is earning interest, and thus your profit.
5. If you are earning interest in a taxable bank account, you will likely owe income tax on that interest at your marginal rate. This is not accounted for in the calculator, but is a simple calculation.
Example Profit Calculation
Let’s say you obtain \$15,000 and place it in a bank account paying 5.25% APY, with a 2% monthly payment. Using our assumptions, the 5.25% APY is equivalent to 5.13% APR, or earning 0.4273% of the balance each month.
Beginning of Month #1: You have \$15,000 in the bank. Total balance left on credit card: \$15,000. Nothing is due yet.
End of Month #1: You earn \$64.10 in interest, but also need to pay back \$300 (2% of \$15,000) out of your bank balance for the minimum payment.
Beginning of Month #2: Total in bank:\$14,764.10. Total balance left on credit card: \$14,700.
End of Month #2: You earn \$63.09 in interest, but also need to pay back \$294 (2% of 14,700).
This continues for 12 months, as shown below:
At the end of the 12th month, your bank balance is \$12,477.87, and you still owe \$11,770.75 on the card. You pay it off completely, leaving you with the resulting estimated profit of \$707.12.
Play around with the calculator. Some people actually have over \$100,000 out at once, earning them thousands of dollars a year. My credit limits aren’t quite that high…. yet!
My Money Blog has partnered with CardRatings and may receive a commission from card issuers. Some or all of the card offers that appear on this site are from advertisers and may impact how and where card products appear on the site. MyMoneyBlog.com does not include all card companies or all available card offers. All opinions expressed are the author’s alone, and has not been provided nor approved by any of the companies mentioned.
MyMoneyBlog.com is also a member of the Amazon Associate Program, and if you click through to Amazon and make a purchase, I may earn a small commission. Thank you for your support.
User Generated Content Disclosure: Comments and/or responses are not provided or commissioned by any advertiser. Comments and/or responses have not been reviewed, approved or otherwise endorsed by any advertiser. It is not any advertiser's responsibility to ensure all posts and/or questions are answered.
1. Eric says
I can’t tell you how great this post is. I’m send a link to my wife. I’ve been trying to get her to consider letting us do this. We already have about \$15,000 in the bank and with this on top of that, the 5.05 of our HSBC account will bring in EVEN MORE money. Thanks so much!
2. MM says
This is a great post. One thing you might consider to add is the tax impact: one will pay income tax for interest income, which will eat into the arbitrage profit.
3. McWilliams says
Man, I am seriously thinking of doing this. I’ll have to pour over your tutorials. Thanks!
4. Esther says
When are you going to pay off the credit cards? Shouldn’t you do it well in advance of purchasing a home, which might be right around the corner? Carrying such high balances can’t be good for your credit score.
5. Susana says
After reading your post and tutorials on making money on credit cards, i have decided to use this credit card process to pay off a student loan. Can you tell me which of the credit cards you suggested actually have the 2% minimum payment, since we will be actually paying it with our own funds, the minimum payment is important. Thanks for your help!
6. heather says
Are you considering doing a margin loan down the road, by sort of the same rational?
7. alf says
There is also a “forced savings” from the monthly payments if you put the balance transfer into a separate savings account and make the minimum payments from your checking account. If you don’t have to use the balance transfer principal to make the monthly payments, that’s a good \$3600 you’ve forced yourself to save.
8. zeb says
So based on your balances of \$28,875, you are making \$1,200 a year on this….pretty nice.
But it seems too risky for me.
And the wife would NEVER go for it…
9. yd says
I slowly converted my credit card debt into 0% balances and as I did that I put the amount I would have been paying on the cards into savings. It took me a while, but I’m up to almost the full amount I owe in savings and earning almost \$100/month just from interest. You’re right – it’s great to be making money off of the CC companies for a change! I’ll never break even on what I paid them over the years, but at least I’m ending on my terms.
Once I pay all of this off I’ll probably get transfers just for the purpose of earning a little extra money. The amount isn’t that great, but the satisfaction is certainly there. Properly set up with automatic payments, and you have no worries about missing a payment.
10. Rick says
I just got an AmEx Blue card offering 0% for 15 months.
11. Sam says
I have taken some 0% balance transfers following tips from your blog and it worked out well. BTW does any know how long the 6000 bonus points for ATT Universal Credit Card take to appear in your account. The official site says 3 months..
12. Rand says
Thank you! I’d always just been using online interest calculators but it’s nice to have a calculator more geared to this specific scenario.
13. MillionDollarJourney.com says
I don’t think this type of arbitrage is for everyone. Question to ask yourself: Is the hit to your credit score worth the couple thousand bucks a year?
14. Dan says
I just pulled out about 80K from Balance transfer for 6-15 months of interest. After fees and taxes (low this year) I’m estimating taking about 3 grand for the year. My credit score dipped for a couple of weeks, but since I kept all my personal balances under 50% of the credit line, and business balances 80%, overall I’ve only lowered my score 10-20 pts. My Experian score hasn’t changed at all. In fact, 3 months after starting, I applied for and was approved for a car loan at the lowest rate available.
15. guapo says
It does damage the credit score, all depending on what percentage of your revolving credit balances you carry.
So, I had 2 credit cards for spending which I paid off every month. So, the limit ratio was very low, under 10% of my limits. I added a 3rd card with the balance transfer offer. It pushed my limit ratio into the 30-40% range, which lowered my score a bit.
Bankrate has a nifty FICO score calculator that you can plug your information into and get a credit score range. So, you can see how much your credit limit %’s affect your score. Just Google “credit score calculator”, it’ll be the first link.
So, if you are going to purchase a home (especially if your credit is not excellent to begin with), definitely pay off those balance transfers before you apply.
16. Dennis says
Jonathan, have you seen your credit score hurt from doing this? I’ve been considering this for a very long time, but at my age, I don’t want a hit to my credit score just yet.
17. Jake says
I often take advantage of 0% offers myself, so I know the benefits of doing so. I think that your analysis is missing one major factor however, and that is TAXES. Your “real” profits are actually ~30% less than you have shown depending on your tax situation. Still worthwhile in my opinion though.
18. Ken says
Eric is “trying to get his wife to consider letting us do this.”
19. cubes says
From reading other posts, I think there is general consensus that there is a little, but not major harm to FICO scores if this strategy is executed correctly.
My only concern is credit-based insurance, where we don’t know the formulas the insurance companies use. Has any noticed a change in insurance rates?
20. hokie96 says
I’m curious… have you thought of taking advantage of the float with respect to paying off a mortgage. (i know, i know.. tons of ‘software’ out there that does this but i’m not looking to pay 3500 to someone for it) I’m just merely saying.. get a 1st mortgage and then a 2nd equity line of credit and then put your total paycheck towards the second and then utilize the credit card float for the month to pay off your monthly expenses. you’ll get about 30 of ‘free’ money using the credit card that is working towards keeping down the interest calculation on your 2nd. i’m thinking of doing this but have been scouring the ‘net trying to find some type of calculator that could provice me with the interest saved utilizing this float method. was wondering if your balance transfer calculator could be retooled to do this? you seem real interested in capturing profit with credit card companies utilizing the float idea….
hokie96
21. MillionDollarJourney.com says
Another issue is how is the interest income taxed in the US? In Canada, interest is taxed at 100% of your marginal rate, which for some of us means that half of the gains would vanish. For me, high taxation + the high probability of reducing your credit score doesn’t make this strategy viable.
But again, I’m in Canada, so that will probably make all the difference.
22. Brian says
The credit score hit isn’t really that bad, even if you get the cards all the way up to the limit (rather than at
23. Brian says
Weird, it cut off my post? Or is it just not showing up?
24. Rahul says
If you are not doing a balance transfer, some credit card companies charge a transaction fee for cash transfers…about 3%. So that may lower your earnings to barely anything.
25. jack says
ok this is great but how could you get the money frome the CC company for no fee?
26. Ellie says
Don’t use the less than or greater than symbols as the blog thinks its html stuff.
27. Kevin says
I got slammed by my insurance company AMICA after I did a 0% balance transfer. When my auto policy was renewed, they told me that my credit score was low and that I was not getting the cheapest rate possible for my Auto. After checking Equifax and waiting for my credit score to raise (782), I called AMICA back up and told them I had corrected “some things” on my credit report. They checked again and still my rate came out the same. Be careful.
28. Jonathan says
Yes, carrying a balance will lower your credit score. This is part of the many warnings listed at the very beginning of the tutorial, and is also addressed in the “Question and Answer” section (which I admit, is a bit long and does need to be cleaned up.) One way to dampen the hit is to keep the amount borrowed to less than 50% of the available credit limit. For example, if you have a limit of \$15,000, only borrow \$7,000. If you have all your credit cards maxed out, the credit issuers will get nervous.
The actual effect depends greatly on your own unique credit history and how “resilient” it is. I haven’t seen any ill effects yet myself, besides getting less promo offers in the mail at times, but I also started out with a good credit score and am using only a portion of my overall available credit. My auto insurance rates actually got lowered last time around. I wouldn’t do this if I was going to shop around for mortgage rates soon.
The cool thing is that in our situation we are going to have just my wife apply for the mortgage (part of our “live on one salary” approach). I do this, my wife doesn’t. She’s just not into it, and I don’t want to do it with her cards since I do call in the credit cards to change statement dates and confirm offer info, and they always want to speak to her. So she just applies up for the occasional signup bonus since it’s easier.
Yep, sorry the comments got cut off. If the blog sees what looks like an HTML tag, it tries to read it as such.
The “less than” symbol, <, should be typed in HTML format, “<”.
Similarly, “greater than” is “”>”
29. Jonathan says
Is the AmEx Blue 15 months at 0% APR for balance transfer or just purchases? I think it’s just purchases (according to tthis application) unless you got a special targeted offer.
30. Rick says
It was a mailed offer for balance transfers, so maybe not generally available.
31. NP says
I have been using 0% BT offers to pay off student loan debt, and also put some in the bank, and it worked out really well for the past couple of years. However, I think good offers, targeted and otherwise, are drying up, so I don’t see that it will be possible to do this for a long time. Many issuers introduced 3% fee with no cap for balance transfers , which kind of makes those offers worthless. And, once you have 2-3 cards from every single major bank, they are not that interested in luring you in anymore :). Jonathan, all of the no fee offers you cite now come from Citi, and they are not too keen on you having 5-6 cards with them, and also gives low limits for subsequent ones. Bank of America, Chase, Discover, and many other no longer do no fee transfers, and few offers are capped at \$75 (alhtough occasionally you can try to waive the fee over the phone, this is not part of standard T&C anymore as it used to be). So I see profits from this going down in the future… gotta find some other way to make some money on the side.
32. Paul says
Just got a mailed notice from Discover that they are increasing the minimum payment from 2%-4% for customers who have balance transfers that are at least 50% of their credit line. Guess who is also reading these posts 🙂
33. Jonathan says
Discover may also be under pressure from people who think the minimum payment % is too low, not because of this type of thing, but because it would take the average person with a high interest rate something like 7 years to pay back a loan if they only paid the minimum each month.
34. Alex says
Not sure if anyone saw Suze Orman’s show last weekend but she totally bashed this “App-o-rama” way of making money and totally advised against doing this (mainly for all of the negative reasons that Jonathan has already posted). First, you’re playing russian roulette with your credit. The high balance to available credit ratio will surely lower your FICO scores. Second, miss one payment and you just lost any profit you might have made from the scam. Third, when it’s all said and done, after you pay income taxes on the interest you’ve earned, you’re only looking at 1 or 2% profit. Fourth, you gotta pay attention to the fine print of these 0% balance transfers. Some include transfer fees of up to 3% of the money borrowed and the credit card companies, by law, can end the 0% promotion at any time. So even if you sign up for 12 months, if XYZ credit card decides they are losing money, they can yank the 0% away and if you’re like me and don’t always pay attention to every little piece of junk mail you get each day, you could easily miss this and end up having a high percentage APR on your once 0% card. Personally….I think there are easier (and more honest) ways of making money…but that’s just my opinion.
Just a side note – regarding Jonathan’s mortgage situation – having a mortgage on your credit report will greatly increase your FICO score and it looks good history-wise (especially if you ever end up getting divorced – you’ll never have that if you only go with your wife on the mortgage). She’ll get all the credit. Might want to think about doing a joint mortgage instead.
35. Jonathan says
From what I can tell from the few times I watched her show, I think Suze Orman’s advice is probably right on for her target audience. Most people that call it have problems dealing with credit already. This activity is not for those without (1) discipline, (2) good credit, and (3) an interest for doing such things. If you don’t have all three, it’ll be a drag. If you do, then it can end up being a small thing that take 15 minutes a month to maintain, and helps fund a big chunk of your Roth IRA for the year. Know thyself…
I don’t see how this isn’t honest. If they ask me what I’m doing, I tell them it’s going into a bank account. Not only am I not hiding, I posting my activities in free, public blog. Would they rather me spend it on a Rolex? Put another way, how is it different from taking out a mortgage – somebody loans me money, I hope to invest it in something that will appreciate and make me come out ahead in the long run.
Although it may help slightly, people can have awesome credit scores without any mortgage at all. I like the idea of being able qualify for another mortgage (maybe a rental?) all on my own later on.
36. joelkton says
How do you all find your credit score? It sounds as if some of you check it quite often, but that costs money, doesn’t it?
37. Jonathan says
I agree that the offers are getting less easily available.
Are 0% Balance Transfer Offers Coming To An End?
It also includes some ways that one can “save” an offer for later, buying you an extra 12 months at 0% in the future.
38. Brian says
Yeah, I think it was a less-than sign in my last post that screwed it up. I will watch out for that HTML!
What I was trying to say before was that my wife maxed out a few cards for the 0% offers, since we will not need her credit in the near-term, and since I would still have a top-tier score. We checked her credit score to see what the impact was after maxing out \$55k on the cards, and the credit report correctly indicated the new cards and balances, with her newly raised debt-to-credit ratio of 77%, with a still respectable FICO score of 707. Her score was previously a little over 760 when we first applied for the cards. This was much less of a hit than we thought it would be!
We also used the score estimator to see what would happen if the cards were immediately paid off, and sure enough, her score would bounce right back over 760. So it isn’t really too bad of a credit hit as long as you don’t plan on buying a house or taking out an equity line in the next couple weeks. 🙂
39. Kevin Spring says
I am having trouble with the 0% At&T Citibank card you put on your site. I had \$15,000 from my Amex Blue with 0% balance transfer moved to the Citibank card. First they tell me it will take 2 weeks and then they tell me they are conducting an investigation of my account. I’m going to call them tomorrow and demand my money.
40. Damian says
It’s been mentioned here before, but if you’re into this sort of thing, having a WAMU Mastercard provides free access to one’s FICO score with monthly updates. Very handy.
41. Damian says
Oops! The WAMU card that offers your FICO score is a Visa, not Mastercard. But, is there actually any difference?
42. Jonathan says
I think all WaMu cards (formerly Providian) offer card issuers a free FICO score. It is only the Transunion FICO score, however, not Equifax or Experian.
43. TW says
BUSTED! I tried to transfer much of my CITI credit limit from a high limit card that I had already done a 12 month 0% on to a lower limit CITI card that I have not yet done a 12 month 0% on and they denied the limit exchange (first time they’ve done this denial) because they know that I can get a no fee transfer on the lower card!
The card companies are definitely getting more cautious since more people are doing the 0% money making deals. Any ideas how I can transfer the limit? I talked to a guy today and he only offered to try to get an increase on the lower card, but I didn’t want to do that.
44. SanDance says
Good Calculator Jonathan, just one change to make it more accurate (and I know I am splitting hairs here): the minimum payment is always rounded up to the nearest whole dollar amount. In my Excel spreadsheets for my balances I use the function =ROUNDUP(2%*balance) It also helps to know that I have to pay \$55.00 instead of \$54.11 (don’t want any late payment fees or that will eat up the whole profit!)
45. SanDance says
TW – correct me if I am wrong, but most credit card companies won’t let you transfer money from on of their cards to another one of their cards. You must transfer it to a card backed or issued by a bank other than CITI. Jonathan describes this caveat in his series mentioned at the beginning of this post.
I also love my Capitol One pruchase checks! It is free money for two months at a time, but I can keep using long after the BT offer expired. They keep sending me purchase checks with every statement. (Note: just keep an eye on the due date! My CC statements just moved the normal date due earlier by a couple days than it had been.)
46. Teri says
IT depends how you do it though (on the credit score). I took a card for BT and dh took one this year. We will earn a good \$1k or so on \$20k this year. It has not affected my score at all. If it was my only card it would fare worse. But I already had a score in the 800s and it didn’t really change much at all with the balance transfer. Now with the monthly payments it is going up (helps since I usually have little credit to up my score). Having an excellent credit score I have always found taking a loan and paying it off has upped my score considerably (I took a car loan for a month last year and shot up to the 800s all of a sudden). It will be different for everyone, but I wouldn’t assume one Balance Transfer would kill your score, particularly if it is already excellent and the percentage your borrow is less than 30% of your entire credit available to you. Which is what we are doing. (& if you have a good score you will be easily be approved for a higher BT from one card – makes the whole process easier).
That being said, I have another warning I haven’t seen mentioned here. We are not very spendy and I we have never paid credit card fees in our life so though I was initially a little nervous by the idea (which I did get from this blog) I decided credit card arbitrage was really perfect for us. I mean it seems easy enough to work the system (we have been for years with rewards and stuff) and to pay your bill on time, etc. BUT I did have my identity stolen recently – a month or 2 after my balance transfer. I found out right away and so far nothing looks like it will hit my credit report. But I just wanted to add a warning that you should check your credit report religiously when you do something like a balance transfer. I am writing a letter to my credit card company to let them know my identity was stolen and to please not apply universal default if something bad shows up on my credit report. This is just something I never foresaw and really sucks. So far I may luck out, but the whole point of doing this for me was because it was “easy” and no hassle. I would probably rather pay it back now than plead my case if stuff starts showing up on my credit report. I immediately thought of universal default when this happened – yeouch! Not much you can do when someone mucks up your credit – I have no idea how my identity was stolen, etc. Really not much I could have done. Also, when the first fraudulent card hit my account my credit score dropped 20 points because it pushed my outstanding credit ratio above 30%. The card was removed though and my credit score shot back up immediately. So yeah I was trying to control the ratio but wasn’t thinking about outside factors…
Just a general warning. Something I never thought of when running through all the risks…
47. goldsink says
There are many ways to play the 0% game… I’ve posted this before but the ultimate way (IMO) is to use cash back or reward cards to INITIALLY fund new bank accounts or new credit union accounts. I did this with a CU in the Albany, NY area to the tune of \$24,000 with a 0% APY on purchases for 12 months… once the funds were in the CU they got moved to FNBO at 6%. I make the min mothly payment and the cash sits and collects interest. Best part was that I got 24,000 reward points which I immeadiatly cashed out for a \$200 check. I suppose even better is that the CU allows you to open more than one account so my plan is to do the same with about \$26,000 from a Citi Pro Cash card that pays me 1% cash back (this card has no Intro/promo deals). So I PURCHASE my initial funding into the CU then move the money out, pay off the balance to Citi in full during that statement cycle… Citi cuts me a check for \$260.
My advice in all of this is be very aware of the terms of all offers and new account openings. You will need a good credit score to build Limits within one family of card issuers. Always watch for fees associated with BT’s… somebody above mentioned BT’s with AmEx… watch the 3%. I have yet to do a BT offer but I will… just want to use purchases (like w/ the CU) before I tie into haveing to move money through a bank or make a request to have it deposited to my account.
And to comment on Alex… where is any of this dishonest? In all actuallity it shows attention to detail, above average intelligence and savvy financial manipulation. These cunning moves can fund your Roth IRA (and then some each year)… you’ve heard of compound interest right? I wouldn’t “use” an individual person like this but if Big Business wants to offer me a free ride I’ll ride that bus everytime. If you can elaborate on means to make 3-5K a year in free cash/gift card/merchandise I think everyone on this board is ready to listen.
My last comment would be to document EVERYTHING that pertains to your individual finacial manuvers. Since I began to play the game of card aquisition and strategic offer manipulation I’ve got over \$1600 in bonus’s, gift cards and merchandise alone. And that does not take into account the interest I’m making on the \$24,000 Chase deal or the move of my large amount of idle cash to higher yeild accounts such as FNBO, my previously mentioned CU (5.5%), or Amtrust Direct.
48. TW says
I must correct you SanDance. I was talking about credit limit moving, not transferring balances. This is the first time they have denied me moving limit amounts. I’ve done limit moves once or twice before.
49. Shak says
I guess I am going to sound like the bad guy here but this whole scheme sounds like it is extremely prone to error. If you make one mistake, you will pay a heavy price. I fear at least one of Jonathan’s “fans” is unknowingly going to jump in this pit and cause him or herself severe problems.
Also, I have to say this–Jonathan, I hope you are aware that you are not fooling all the financial institutions that you partake in this scheme.
I know for a fact that they all read this blog along with several others on the www and really do enjoy your ploy. With that being said, they are also not loosing money as it appears from some of your readers comments.
From reading your blogs, most people who do not live in the U.S. can tell a lot about the American person’s (both male and female’s) psyche.
50. Shak JR says
Shak – What’s the scheme again? Is it accepting the credit cards offer of credit? And how is Johnathan fooling the credit card companies? Again is it by accepting their offer of credit? Just curious
51. Kevin Spring says
I don’t see why people are calling this a scam or scheme. This is a very legitimate way to make money. Financial institutions do this all the time with currency fluctuations. Arbitrage is a way in which the market corrects itself. If someone wants to give out a loan at 0% interest then I have a right to take that money and do whatever I want with it. The credit card business is very competetive and this 0% interest will increase even more as the competition gets even more fierce. I’m sure for every app-o-rama person out there, there are 100 people that use the 0% interest cards and charge up things they could never have afforded. I on the other hand keep the money in a bank and make interest off of it. Credit cards are not losing any money as this is only a marketing gimmick to get me to use the card.
52. Phillip says
I also find it hilarious that anyone would think this is a scam. Credit card, loan, and payday loan companies are scamming the ignorant masses by the millions. This form of arbitrage is completely legal and logical to any sane observer.
53. Tex says
Suze Orman on app-o-rama was just about the last straw for me. She acted like this was the worst idea ever because she just knew you weren’t going to have the money to pay it back, and she just knew you were going to be late on a payment. She seems to have no problem with car loans or student loans. But 0% credit card debt–with the matching savings account–revolts her.
I wonder if Suze’s position is also the official Fair Issac brand party line. I mean credit card companies want you to take out huge balances and accounts–that’s why they offer the 0% in the first place! If they are getting scammed, they would stop. Clearly they are making money somehow on these offers, just not on Jonathan. 🙂
54. nisha says
Great blog , I have also created a lens in a same niche hope u like it ?.. ..Are you fed up with work? Maybe your boss is a grade-A pain in the behind. Well, don’t fret about your current situation; it’s time to look for an alternative route to income. That company office cubicle isn’t the only way to earn a living. These days there are opportunities opening up all the time. With the World-Wide-Web in full swing, many individuals are turning to the Internet in search of a money making business. Have you ever considered this new-age road of opportunity?
55. random says
TW Says:
August 10th, 2007 at 7:40 am
“I must correct you SanDance. I was talking about credit limit moving, not transferring balances. This is the first time they have denied me moving limit amounts. I?ve done limit moves once or twice before”
I work at Citigroup, it’s a new policy that started about a month or so ago, Cards can’t consolidate limits at all if the account that is having activity moving has an offer for a 0-3% BT offer on it. Nor can limits be moved, There is no way around it as a system will automatically stop you, i’d suggest asking for retention or a supervisor next time you call in to speak with someone about an important move.
56. Mark says
Regarding Credit Score and Insurance Companies.
I just recieved a letter yesterday saying my insurance company (American Family). pulled my credit score resulting in my “premium being higher”….
I had borrowed about 24k from CITI (at 0%) for the last year ending in July having paid all of this back to CITI…. however AMFAM pulled my report or score? in June. I have viewed my credit report every year since the annualcreditreport.com site went up and as far as I know AMFAM has never done this before.
btw, my credit should be perfect; it was around 750 when I did my last mortgage three years ago as I recall.
I am calling AMFAM today..!
word to the wise; consider calling your insurance company before you do the arbitrage.
57. SanDance says
Thanks for the correction TW. I see noe that you meant moving credit limit, not balance. Sounds like CitiGroup is getting wiser to this play. I need to do a Credit Limit move at Chase soon and wonder if I will be able.
Note to all the skeptics, like Shak, who say: “If you make one mistake, you will pay a heavy price.” My wife forgot to make the first payment on a new BT card recently. She got hit with late payment fee and all that. She just made the payment as fast as she could, called Bank of America, got the fees refunded, and even got the 0% back for the full term of the offer. (Most of ) These companies would much rather keep you as a customer than lose you for a \$25 late payment fee. Most banks will give you at least one strike before you are out, all you have to do is pick up the phone or go to a local branch and ask nicely.
58. Dave says
I would strongly advise against this strategy for those who will soon be seeking credit (auto loan/mortgage). Depending on your credit limits and situation of course, this could drive up the utilization of your revolving credit… dropping your credit scores. Not really that big of a deal if you’re not shopping around. At would say at least try to keep your utilization below 50%. Thoughts?
59. savingeverything says
SanDance… not all the time will “most banks give you at least one strike before you are out.” In my situation (when i was was exactly 71hrs late, ) I called, and tried to get Chase to waive both the late fee and finance charges on my balance and even said that they will lose my business if they dont. However, the only thing the lady said was they will waive the \$39 late fee; and they apologized that they wont do anything else and hope i still stay with them. Unfortunately, I was not doing a BT thing, and was charged almost 17.99%.
60. Kevin Spring says
I don’t have to worry about my credit score. I don’t borrow money to buy things. I don’t understand why anyone would finance a depreciating asset such as a car, boat, or electronic device. I can understand buying a house but when most people finance things they buy more than they can really afford. I am going to buy a house and car the same way my grandfather did, with cash.
61. christy says
Is the 50 bonus still good for the AT&T Card? I did not see that on their site.
Thanks!
62. Joe says
I just learned that Citi doesn’t report business cards to the credit agencies. About 9 months ago they switched my citi personal card into a business card when it came up for renewel. I really didn’t think much about it until I found out that my 5 years of credit history with this card was wiped clean from my credit report. I obviously took a hit on my credit score but it allowed me to take out a balance transfer which doesn’t affect my credit score. Some food for thought.
63. Angela says
I have two credit cards that are maxed out and on one I’m paying 24.9% interest and the other 20%. It’s disgusting. I make payments of \$500 to each of them but it only goes towards interest so I never pay my balance down. I have tried getting approved for balance transfers to lower interest rate cards but I get denied because I’m maxed out on them both. It’s a vicious cycle that I feel like I’ll never get of. Does ANYONE have any suggestions? Are there credit card companies out there who will cut me a break? Making monthly payments is not the problem as I can easily do that…
64. Emil says
I’ve been doing this for 5-6 years. My cousin did not pay rent for a year while I was in college, and I ended up with \$4k in debt. Since I had stellar credit, I kept using 0% offers (albeit usually w/2-3% balance transfer fees). It’s still the equivalent of the best loan you can get — it matches inflation. Now I’m out of school and working — but I still use this strategy and it’s worked amazingly well — saving thousands.. NOTE that THIS strategy only works in the long run if you have incredible discipline and don’t overspend what your expected salary will be. For example if you accumulate \$20k in debt and make \$20k per year, you are going to be in deep trouble over the long run. I’ve invested all my savings, and never turn down 0% financing or credit. 😉
65. Emil says
It’s absolutely true that this strategy can hurt your credit score, which goes down if you have maxxed out your cards. But not necessarily by that much — I am an extreme example, but I had up to \$20-30k of maxxed credit cards when I got my mortgage last year, but otherwise had stellar credit — and my FICO score was still 729 even with the credit debt. I do not recommend this if you are trying to get a loan (ie. buy a car or house) soon. If your credit score is not as high — the drop in points could dramatically affect you. But you should be able “fix” this immediately by simplying paying off the credit cards and allowing the FICO credit score to increase — about 3-6 months in advance of trying to get a big loan, etc. | 8,804 | 38,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-10 | latest | en | 0.9556 |
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In the last weeks there have been some discussions regarding various ways of determining Bernoulli numbers on the 41-series and other calculators. The usual formulas included powers with exponents greater than 100, leading to reduced accuracy since an exact result would require at least twelve or thirteen digits for the base as opposed to the ten we have. Another problem is the available working range, so that the used algorithm has to make sure no intermediate result exceeds the limit at 9,999...E99.
So I wondered if there might be a way of evaluating all possible Bernoulli numbers within the working range sufficiently fast and, more important, as accurately as possible. Which leads to the question: how close can you get in the face of accumulating roundoff errors? Even a simple multiplication can be surprisingly inaccurate, the result may be off by up to 5 units in the last place. Try a simple $$\pi·\pi$$ or $$e·e$$, and the result on a correctly working 10-digit calculator is 3 ULP high or low. So far, so bad.
Here is the approach used in the following program. As usual, the lower Bernoulli numbers B0 to B8 are given directly. For n = 2...8 a simple quadratic equation can do the trick. For n = 10 to 116 (largest value within the 41's working range) the following formula was used:
$$\large B_n = 4 \pi · \zeta(n) · e^{(0,5 + \frac{1}{12n} - \frac{1}{360n^3} + \frac{1}{1260n^5} + ...)} · (\frac{n}{2 \pi e})^{n+0,5}$$
For n ≥ 10 and 10-digit accuracy three terms of the series in the exponent of the e-function are sufficient.
The expression $$\frac{1}{12n} - \frac{1}{360n^3} + \frac{1}{1260n^5}$$ can be evaluated as $$\frac{210n^4 - 7n^2 + 2}{2520n^5}$$.
A literal implementation of the complete formula would yield results with substantial errors. At least the last two digits would be off. So a different way to handle this formula had to be found.
Within the relevant domain, the factors at the left ($$4\pi=10·0,4\pi, \zeta$$ and the exponential function) all start with 1. They do not vary much for n = 10...116:
$$B_n = 10 · 1,256... · 1,000... · 1,65... · (\frac{n}{2 \pi e})^{n+0,5}$$
The basic idea now is to evaluate all three factors minus one so that one additional digit is gained. Obtaining $$\zeta - 1$$ is trivial, and for the exponential function there is a dedicated $$e^x-1$$ command. The multiplication of three values close to 1 can be done in a way that preserves one additional digit of working precision. Since the product of the three factors is something between 2,07 and 2,09, the program even tries to calculate half of this minus 1 (and finally multiplies this +1 with twice the power), so that again a precious digit is saved. The program uses a 9-digit approximation of $$0,4\pi - 1 = rad(72°)-1$$ which is slightly low, so a correction term is applied. Its exact value should be near 7,2E-10, but tests showed that in this case even better accuracy is obtained with a slighty lower value close to 6E-10 (cf. line 89).
Now let's look at the power at the right. For a correct 10-digit result, the base would have to carry at least 12 or 13 digits. Here is how this is accomplished in the program:
$$(\frac{n}{2 \pi e})^{n+0,5}$$
$$= (n · 0,05854983152432)^{n+0,5}$$
$$\approx (n · 0,05854983)^{n+0,5} + 1,52432·10^{-9}·n·(n+0,5)·(n · 0,05854983)^{n-0,5}$$
For n = 10...116 the base of the first power carries at most 9 digits, so both the base and the exponent are exact. However, the 41's power function sometimes truncates its result instead of rounding it, so the constant 1,52432E-9 is better rounded up to 1,5244E-9.
Here is the 41C code:
Code:
01 LBL"BN" 02 ABS 03 INT 04 STO 00 05 SIGN 06 RCL 00 07 X>Y? 08 GTO 01 09 1,5 10 * 11 - 12 GTO 99 13 LBL 01 14 2 15 MOD 16 - 17 X=0? 18 GTO 99 19 9 20 RCL 00 21 X>Y? 22 GTO 02 23 6 24 - 25 X^2 26 3 27 * 28 42 29 - 30 ABS 31 1/X 32 GTO 98 33 LBL 02 34 5 E-10 35 RCL 00 36 CHS 37 1/X 38 Y^X 39 INT 40 STO 01 41 0 42 ISG Y 43 LBL 03 44 RCL Y 45 RCL 00 46 CHS 47 Y^X 48 + 49 DSE Y 50 DSE 01 51 GTO 03 52 RCL 00 53 X^2 54 ENTER 55 ENTER 56 210 57 * 58 7 59 - 60 * 61 2 62 + 63 2520 64 / 65 RCL 00 66 ENTER 67 X^2 68 X^2 69 * 70 / 71 ,5 72 + 73 E^X-1 74 ST* Z 75 + 76 + 77 ENTER 78 ENTER 79 1 80 - 81 72 82 D-R 83 FRC 84 + 85 X<>Y 86 LASTX 87 * 88 + 89 5,8 E-10 90 + 91 2 92 / 93 STO 01 94 RCL 00 95 ,05854983 96 * 97 ENTER 98 ENTER 99 RCL 00 100 ,5 101 + 102 Y^X 103 ST+ X 104 ENTER 105 ENTER 106 R^ 107 / 108 1,5244 E-9 109 * 110 RCL 00 111 2 112 / 113 RCL 00 114 X^2 115 + 116 * 117 + 118 RCL 01 119 X<>Y 120 * 121 LASTX 122 + 123 10 124 * 125 LBL 98 126 RCL 00 127 -4 128 MOD 129 SIGN 130 * 131 CHS 132 LBL 99 133 END
One may now ask if the result is worth all the effort. I think it is. In total there are 60 possible non-zero results within the 41's working range (n = 0, 1, 2, 4, 6, 8, ..., 114, 116). The program returns 45 of these correctly rounded or truncated after 10 digits. The rest is 1 ULP high or low. I did not find any larger errors. In other words: the results are close to machine accuracy.
BTW, while the largest possible result is B116, the program can also provide B118. The expected OUT OF RANGE error appears in the last calculation step when the program tries to multiply X by 10. At this point, pressing [X<>Y] reveals B118 as 6,116052000E+100. ;-)
Of course suggestions for improvements are always welcome.
Dieter
Did you ever consider to write the program in MCODE? That provides the possibility to use 13 digits for your calculations and only round the result to 10 digits.
Cheers
Thomas
(03-10-2014 12:20 AM)Thomas Klemm Wrote: [ -> ]Did you ever consider to write the program in MCODE? That provides the possibility to use 13 digits for your calculations and only round the result to 10 digits.
Getting 10 valid digits out of 13 digits working precision is not quite a challenge. ;-)
The basic question here was: how close can you get without any additional guard digits. Otherwise things are trivial: I tried the basic formula on a 12-digit 35s. Even with sloppy programming 10 valid digits ±1 ULP are not very hard to achieve. So with 13 digit precision things become trivial if the goal is merely a correct 10-digit result.
But nevertheless I would love to try MCODE programming. Years ago I always enjoyed combining efficient x86 assembler code with the beauty of Pascal. But honestly, I do not know the slightest thing about MCODE. I do not even how what is required to build a working programming environment. Maybe there is some basic info somewhere to start with?!
BTW, in the meantime there is a shorter and even slightly faster Bernoulli program with comparable accuracy: ±1 ULP over the whole working range. Yes, with a standard 10-digit 41C. ;-)
Dieter
(03-10-2014 08:23 PM)Dieter Wrote: [ -> ]Getting 10 valid digits out of 13 digits working precision is not quite a challenge. ;-)
(03-12-2014 01:45 AM)Paul Dale Wrote: [ -> ]Having extra guard digits during a computation is a godsend.
But nevertheless I would love to try MCODE programming.
You'll find all you need on Warren Furlow's HP-41 web-site.
Have fun
Thomas
It's slightly more involved than just applying a cookie-cutter recipy, even if it's all published somewhere (like TOS). The 13-digit routines allow for intermediate calculations done with more precision, so the final result is more accurate (or less inaccurate, in a more strict speak). There are many 13-digit versions of the standard math routines in the 41 OS, a good starting point could be the mini-paper I prepared a while ago; available at TOS (can't paste the link here as you probably know).
I've attached it to this post as well (hope it works)
Cheers,
ÁM
(03-12-2014 04:34 PM)Thomas Klemm Wrote: [ -> ]You'll find all you need on Warren Furlow's HP-41 web-site.
(03-14-2014 10:05 AM)Ángel Martin Wrote: [ -> ]It's slightly more involved than just applying a cookie-cutter recipy, even if it's all published somewhere (like TOS).
Yes, I am absolutely sure it is all "somewhere". ;-) But what I am looking for is something like a structured introduction to MCODE programming: what are the hardware options, what else is required, what is covered by the instruction set and what are basic programming techniques. I am quite familiar with FOCAL programming, and I also appreciate the the benefits of doing things in machine code, so MCODE sounds like an interesting new project to me. However, I need some basic info first.
Quote:The 13-digit routines allow for intermediate calculations done with more precision, so the final result is more accurate (or less inaccurate, in a more strict speak). There are many 13-digit versions of the standard math routines in the 41 OS, a good starting point could be the mini-paper I prepared a while ago; available at TOS (can't paste the link here as you probably know).
I've attached it to this post as well (hope it works)
Thank you very much, and yes, it worked. That's exactly what I am looking for. In a way I am a kind of "accuracy junkie" - that's why I love the 34s project and the option of 34-digit precision for certain things. Defining special functions like the Normal distribution quantile, Lambert's W or just the Bernoulli numbers will full 10-digit accuracy is virtually impossible in FOCAL. With 13 digits all this becomes easy.
Now - where and how to start?
Dieter
By the way, do you know of a kind of overview with the actual accuracy of the 41's internal 13-digit routines?
(03-14-2014 09:37 PM)Dieter Wrote: [ -> ]Now - where and how to start?
Software Development Kit
For the HP-41 Release 6
PROGRAMMER'S MANUAL
cf. APPENDIX I – Step by Step Example
HTH
Thomas
(03-14-2014 10:05 AM)Ángel Martin Wrote: [ -> ]available at TOS (can't paste the link here as you probably know).
Not a problem: HP41 OS 13-digit Math Routines
Thanks again for the compilation!
Cheers
Thomas
(03-14-2014 09:37 PM)Dieter Wrote: [ -> ]By the way, do you know of a kind of overview with the actual accuracy of the 41's internal 13-digit routines?
Probably not what you want:
HEWLETT-PACKARD JOURNAL
cf. Algorithms and Accuracy in the HP-35
But I just remembered that picture of the saw-tooth diagram.
Cheers
Thomas
There's an interesting article (written by Dennis W. Harms) that broadly describes the accuracy of the math algorithms on the HP-67 - which in my understanding were pretty much the same ones also used by the 41C. You can read it in the November 1976 issue of the HP Journal, pages 16 and 17.
Cheers,
'AM
(03-15-2014 09:32 AM)Ángel Martin Wrote: [ -> ]You can read it in the November 1976 issue of the HP Journal, pages 16 and 17.
Original article in Advanced Pocket Calculators
Dennis W. Harms, pictured 30 years later:
Thomas, thanks for the links, I didn't have the original article.
Your historian archivist skills largely exceed mine
(03-15-2014 12:26 PM)Thomas Klemm Wrote: [ -> ]
(03-15-2014 09:32 AM)Ángel Martin Wrote: [ -> ]You can read it in the November 1976 issue of the HP Journal, pages 16 and 17.
Original article in Advanced Pocket Calculators
Thank you very much, especially for the "Advanced Pocket Calculators" version. I read the HP journal version of this article many years ago, but it does not answer the question regarding the accuracy of the 13-digit internal functions of the HP-41, -67 or any other calculator. What the article essentially says is that 13 digits (HP-91 and newer) allow (nearly) correct 10-digit results, which had errors before when no guard digits were used. Or simply "13 digits give better results than 10 digits". Which nobody will deny. ;-)
Dieter
(03-15-2014 06:12 PM)Dieter Wrote: [ -> ]but it does not answer the question regarding the accuracy of the 13-digit internal functions of the HP-41
There's most probably no information available concerning this question. Thus there is much to do. I suggest to use the trace which is available in Nonpareil to figure out the 13-digit internal results of the operations you are interested in. You probably have some corner cases already in mind such as cos(x) close to $$\frac{\pi}{2}$$ or ln(x) close to 1. Or maybe the SDK41 provides a decent debugger? Let me know if you need detailed instructions. IIRC Nonpareil doesn't support tracing out of the box. You have to compile it setting the option has_debugger_gui.
Cheers
Thomas
V41 does have a tracer and breakpoint capability. I do lots of troubleshooting on it, and it can also be used to see the intermediate results - provided you have the source code of course. So it's a bit of a catch-22, but once you're in the game it's very rewarding.
Thomas and Ángel: Thank you very much for some first information on how to start with MCODE. It seems to be a long way to go. On the other hand I just read through the Sandmath manual, and all this looks very promising.
BTW, Ángel: Sandmath seems to lack a Normal Quantile function. I do not think it's as complicated as Appendix 9 suggests: on the 34s it can be done with 30+ digit accuracy in merely two iterations. At the moment I am playing around a bit with V41. If there is a way to have Sandmath's ERF in V41 I could see what can be done in user code. ;-) How fast does ERF execute on a real 41?
Dieter
Hi Dieter, glad to see you find the SandMath worth looking at. Which revision are you using? Are there two appendices 9a and 9b, or only one appendix 9?
http://hp41.claughan.com/file/SANDMATH_4...Manual.pdf
I say this because in the latest versions (Revision "M", see above link) there are a couple of functions directly related to the quantile, one (ICPF) using the inverse error function on a general Normal distribution (s,m), and another (QNTL) using Halley's iterative method on a standard Normal (0,1). The functions are a bit buried in the -FACTORIAL group, in the sub-functions FAT
The manual has room for improvement, but isn't the quantile function QNTL what you're asking about? It is briefly (and poorly) documented in page 48. Below you can see the FOCAL code for this function, which uses Halley's method to solve for a CPF (Cumulative Probability Function) equation equal to the quantile's value. The convergence criteria is a poor E-7, probably not what you're looking for I'm afraid.
Code:
01 LBL "QNTL" 02 STO 03 03 CLX 04 STO 04 05 LBL 01 06 0 07 ENTER^ 08 1 09 RCL 04 10 CPF 11 RCL 03 12 - 13 STO 05 14 0 15 ENTER^ 16 1 17 RCL 04 18 PDF 19 RCL 05 20 X<>Y 21 ST* Y 22 X^2 23 LASTX 24 RCL 04 25 * 26 RCL 05 27 * 28 2 29 / 30 + 31 / 32 ST- 04 33 ABS 34 E-7 35 X<Y? 36 GTO 01 37 END
As per ICPF, it's a much simpler code since IERF does all the work in there:
Code:
01 LBL "ICPF" 02 ST+ X 03 E 04 - 05 IERF 06 2 07 SQRT 08 * 09 * 10 + 11 END
I use the CUDA library algorithms to calculate the inverse error function - it's a very fast implementation (yes, even on a normal HP-41) that takes a huge MCODE listing code.
I may be confusing subjects, statistics is not my forte (I'm really a jack of all trades and master of none, as I'm sure you have noticed ;-) I think this function is equivalent to the 38E's, but haven't looked at the 34S at all (don't have one myself).
l hope this clarifies, but let me know if you see any issues or would like additions/changes made.
Best,
'AM
(03-19-2014 06:47 AM)Ángel Martin Wrote: [ -> ]Hi Dieter, glad to see you find the SandMath worth looking at. Which revision are you using? Are there two appendices 9a and 9b, or only one appendix 9?
The version I had was 4.44.44, now updated to 5.55.55 that indeed has additional information on the Normal quantile function.
Actually I do not use Sandmath. I heard about it quite often here, and when I tried a first look at MCODE I came across the manual. At the moment this is all what I got. ;-)
(03-19-2014 06:47 AM)Ángel Martin Wrote: [ -> ]I say this because in the latest versions (Revision "M", see above link) there are a couple of functions directly related to the quantile, one (ICPF) using the inverse error function on a general Normal distribution (s,m), and another (QNTL) using Halley's iterative method on a standard Normal (0,1).
Halley's method is nice, but it can be done better and faster. ;-)
I now remember that some time ago I set up an algorithm specially taylored for the HP41 and other 10-digit machines. It first calculates an initial estimate with a very simple (1, 2) rational function, requiring just four constants with few digits, and then one single correction step is sufficient for a result with error less than 0,055 ULP, i.e. 5,5 units in the 12th significant digit. All it needs is a sufficiently accurate CDF (cumulative distribution function).
The following method assumes p < 0,5. The rest is handled by CDF(-x) = 1 - CDF(x).
$$u := \sqrt{-2 ln p}$$
$$x := max( 0 , u - \large \frac{2,374 + 0,39 u}{1 + 1,11 u + 0,07157 u^2} )$$
$$t := \large \frac{CDF(x) - p}{PDF(x)}$$
Now apply a third order (!) correction, cf. Abramowitz & Stegun, 10th ed., p. 954:
$$x := x + t + \frac{x}{2}t^2 + \frac{2x^2+1}{6}t^3$$
If evaluated exactly, the result has an error within approx. +0 and -5,5 units in the 12th significant digit throughout the 41's working range (even down to 1E-121).
Here is an implementation in Visual Basic for Excel:
Code:
Function icdf10(p) signflag = (p > 0.5) If signflag Then q = 1 - p Else q = p u = Sqr(-2 * Log(q)) x = u - (2.374 + 0.39 * u) / ((0.07157 * u + 1.11) * u + 1) If x < 0 Then x = 0 t = (normcdf(x) - q) / normpdf(x) x = t * t * t * (2 * x * x + 1) / 6 + t * t * x / 2 + t + x If signflag Then x = -x icdf10 = x End Function
Here normcdf() and normpdf() refer to the Normal CDF resp. PDF. Please note that the PDF refers to the upper integral, i.e. from x to infinity. Otherwise two small adjustments are required.
The essential magic is in the combination of a dedicated initial guess and a very effective correction afterwards. If the term at t³ is omitted, the result quite exactly matches that of Halley's method and the error is less than 2 units in the 9th digit.
All this should work for p = 0,2 as well as p = 1E-12 or 1E-99 (that's why $$erf^{-1}(2p-1)$$ is not a good idea). Cases close to the distribution center and and the lower working limit (1E-99) require some care at the calculation of CDF(x) - p since this difference may become numerically zero due to underflow or digit cancellation. But there are ways to handle this.
Dieter
Edit: corrected an error and tweaked one of the constants for even better accuracy.
Will look at your method as soon as I get a chance, thanks for sharing it.
I only included the Halley's method QNTL for comparison reasons; the preferred approach is no doubt to use IPFC (based on the inverse error function) - much faster (really several orders of magnitude) and generally more accurate.
BTW you can use the SandMath on V41 when you want, the MOD file is available at TOS.
Cheers,
'AM
(03-19-2014 07:32 PM)Ángel Martin Wrote: [ -> ]...the preferred approach is no doubt to use IPFC (based on the inverse error function) - much faster (really several orders of magnitude) and generally more accurate.
"Fast and accurate" always sounds very promising. ;-) Do you have any figures here? How fast is, for instance, qf(0,01) on an actual HP41? You say you used the inverse error function as implemented in the CUDA library. Do you have a link to this algorithm? I found some information on the library, but not on the actual algorithms.
Just for comparison: I currently use a method similar to the one posted above in combination with an optimized method for the CDF. In a regular FOCAL program, the CDF runs in about 3 - 12 seconds, the quantile function requires roughly 3 seconds more. Maybe you can estimate how fast this would run in MCODE.
Dieter
Pages: 1 2
Reference URL's
• HP Forums: https://www.hpmuseum.org/forum/index.php
• : | 6,192 | 20,294 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-33 | latest | en | 0.893284 |
http://www.chegg.com/homework-help/calculus-early-transcendentals-single-variable-10th-edition-chapter-8.3-problem-25e-solution-9780470647684 | 1,474,813,132,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660214.50/warc/CC-MAIN-20160924173740-00289-ip-10-143-35-109.ec2.internal.warc.gz | 378,629,083 | 16,505 | Solutions
Calculus Early Transcendentals Single Variable
# Calculus Early Transcendentals Single Variable (10th Edition) View more editions Solutions for Chapter 8.3 Problem 25E
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Chapter: Problem:
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 3
(a)
Consider the initial value problem,
If Euler’s method is used to evaluate using n steps, then will be approximately equal to .
As the value of steps n keep on increasing will more closely to the value of. So if thenwill become equal to.
Therefore the exact value of will be which can be calculated by solving equation.
Therefore,
Take integration both sides
Use the initial condition
Therefore,
And
• Chapter , Problem is solved.
Corresponding Textbook
Calculus Early Transcendentals Single Variable | 10th Edition
9780470647684ISBN-13: 047064768XISBN: Authors: | 251 | 1,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-40 | latest | en | 0.774885 |
http://oeis.org/A040273 | 1,571,509,585,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986697439.41/warc/CC-MAIN-20191019164943-20191019192443-00331.warc.gz | 142,062,590 | 3,549 | This site is supported by donations to The OEIS Foundation.
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A040273 Continued fraction for sqrt(291). 1
17, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34, 17, 34 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 LINKS Index entries for linear recurrences with constant coefficients, signature (0, 1). FORMULA a(n)=(1/2)*[51+17*(-1)^n]17*[C(2*n,n) mod 2], with n>=0 [From Paolo P. Lava, Apr 24 2009] MAPLE with(numtheory): Digits := 300: convert(evalf(sqrt(291)), confrac); MATHEMATICA Block[{\$MaxExtraPrecision=1000}, ContinuedFraction[Sqrt[291], 100]] (* or *) PadRight[{17}, 100, {34, 17}] (* Harvey P. Dale, Aug 25 2016 *) CROSSREFS Sequence in context: A060360 A081702 A238235 * A022351 A146873 A146793 Adjacent sequences: A040270 A040271 A040272 * A040274 A040275 A040276 KEYWORD nonn,cofr,easy AUTHOR STATUS approved
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Last modified October 19 13:01 EDT 2019. Contains 328222 sequences. (Running on oeis4.) | 546 | 1,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-43 | latest | en | 0.684129 |
https://lookformedical.com/faq.php?lang=1&q=Chi-Square+Distribution | 1,548,206,622,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583879117.74/warc/CC-MAIN-20190123003356-20190123025356-00445.warc.gz | 562,167,100 | 17,615 | • ###### cumulative
• They find the critical values using a chart and then confirm the area between the critical values and in each tail using the cumulative Chi-Square command. (ti.com)
• Johnson and Kotz give a comprehensive review on the approximations to the cumulative distribution function (cdf) of the F distribution. (scirp.org)
• Computer code for evaluating the cumulative distribution function of the generalized chi-squared distribution has been published, but some preliminary manipulation of the parameters of the distribution is usually necessary. (wikipedia.org)
• ###### goodness
• The chi-squared distribution is used in the common chi-squared tests for goodness of fit of an observed distribution to a theoretical one, the independence of two criteria of classification of qualitative data, and in confidence interval estimation for a population standard deviation of a normal distribution from a sample standard deviation. (wikipedia.org)
• ###### gamma
• There are several other such variants for which the same term is sometimes used, or which clearly are generalizations of the chi-squared distribution, and which are treated elsewhere: some are special cases of the family discussed here, for example the noncentral chi-squared distribution and the gamma distribution, while the generalized gamma distribution is outside this family. (wikipedia.org)
• ###### differences
• Whereas group differences indicate differing score distributions on Y, DIF explicitly involves conditioning on θ. (wikipedia.org)
• ###### contingency
• So, we construct a contingency table that shows the distribution of one variable at each level of the other variable. (symynet.com)
• ###### value
• The Chi-Square distribution is based on a sum of squares, therefore the value of X^2 will always be larger than (or equal to) zero. (uva.nl)
• If we wish to reject H o at the .05 level, we will determine if our value of chi square is greater than the critical value of chi square that cuts off the upper 5% of the distribution at our particular degrees of freedom value. (symynet.com)
• If our value of chi square from the formula is greater than the critical value of chi square, we reject H o and conclude that the obtained frequencies differ from the expected frequencies more than would be predicted by chance. (symynet.com)
• Define a new random variable Q. To generate a random sample from Q, take a sample from Z and square the value. (wikipedia.org)
• The expected value of a chi-square random variable with 8 degrees of freedom is 8. (wikipedia.org)
• ###### population
• The oldest of these tests are used to detect whether two or more population distributions differ from one another. (biology-online.org)
• ###### standard
• Some important special cases relating to this particular form either omit the additional standard normal term and/or have central rather than non-central chi-squared distributions for the components of the summation. (wikipedia.org)
• ###### sample
• If the number of deaths at each hospital is large (say, five or more), then the usual chi-square distribution or other large-sample approach may be used, avoiding the need for simulations that would require a large amount of computer time. (thefreedictionary.com)
• ###### mean
• In practice it is rare - if not impossible - for an increase of X in a group mean to occur via an increase of each member's score by X.) This will shift the distribution X units in the positive direction, but will not have any impact on the variability within the group. (wikipedia.org)
• The most general form of generalized chi-squared distribution is obtained by extending the above consideration in two ways: firstly, to allow z to have a non-zero mean and, secondly, to include an additional linear combination of z in the definition of X. Note that, in the above formulation, A and B need not be positive definite. (wikipedia.org)
• ###### parameters
• distribution, where r is the degrees of freedom, which is the difference in the number of unconstrained parameters being estimated and the number of constrained parameters being estimated. (scirp.org) | 848 | 4,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-04 | latest | en | 0.834186 |
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### Pulleys and gears chapter 5
TermDefinition
Lever A rigid bar that is supported at one point
Fulcrum The point of the lever
First class lever Always has the fulcrum between the input and the output forces
Second class lever In this situation the fulcrum the very end of the opener and that remains in contact with the bottle cap
Third class lever If you hold the top-of-the-range stationary with your left hand ( the falcon) and then move the rake with the right your hand and your right hand is the input force
Pulley Consists of a grooved wheel with a rope or cable looped around it
Wheel and axle Consists of a shorter axle that is attached to a larger disk called the wheel
Inclined planes A ramp is another name for an inclined plane
Screw Is simply an inclined plane wrapped around a rod
Wedge Is an inclined plane that travels through the object or material
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https://aakashsrv1.meritnation.com/ask-answer/question/1-state-newton-s-law-of-gravitation-hence-define-universal-g/gravitation/1825451 | 1,669,769,109,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710712.51/warc/CC-MAIN-20221129232448-20221130022448-00438.warc.gz | 113,328,290 | 11,428 | # 1. State Newton's law of gravitation. Hence define universal gravitational constant . Give the value and dimensions of G. 2. Define acceleration due to gravity. Show that the value of 'g' decreases with altitude or height . 3. Discuss the variation of ‘g' with depth. What happens to 'g' at the centre of earth? 4. Write down the formula of gravitational potential energy and obtain from it an expression for gravitational potential . 5. What do you mean by gravitational potential energy of a body? Obtain an expression for it for a body of mass m lying at distance r from the centre of the earth . 6. Define the term orbital speed. Establish a relation for orbital speed of a satellite orbiting very close to the surface of the earth. Find the ratio of this orbital speed and escape speed. 7. What are geostationary satellites? Calculate the height of the orbit above the surface of the earth in which a satellite, if placed, will appear stationary. 8. State Kepler's law of planetary motion. 9. What is a polar satellite? Explain how does it scan the entire earth in its each revolution? Give two important uses of a polar. Satellite. 10. What do you mean by the term weightlessness? Explain the state of weightlessness of (i) a freely falling body (ii) an astronaut in a satellite orbiting the earth. 11. Obtain an expression for the acceleration due to gravity on the surface of the earth in terms of mass of the earth and its radius. Discuss the variation of acceleration due to gravity with altitude and depth. 12. State the conditions necessary for a satellite to appear stationary.
Due to paucity of time it would not be possible for us to provide the answers to all your queries. We are providing solutions to some your good queries. Try solving rest of the questions yourself and if you face any difficulty then do get back to us.
1.Newton's law of gravitation states that every body in this universe attracts every other body with a force which is directly proportional to the product of their masses and is inversely proportional to the squares of the distance between them.
This force depends upon the masses of the bodies which attracts each other and the distance between them.
According to Newton' law of gravitation :
F =
Thus, universal gravitational constant is equal to the force of attraction acting between two bodies each of unit mass, whose centre are placed unit distance apart.
3. Variation of g with depth (d) is given by :
g' = g
​where R is the radius of earth.
At centre of earth, d = R
Hence, at centre of earth, the value of acceleration due to gravity (g) is zero.
• 1
• -1
1. (i) Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
(ii) The gravitational constant denoted by letter G, is an empirical physical constant involved in the calculation(s) of gravitational force between two bodies.
2. gravitational acceleration is the acceleration on an object caused by gravity. Neglecting friction such as air resistance, all small bodies accelerate in a gravitational field at the same rate relative to the center of mass.
Friend you are very curious man. I can't give you the all answers. Everything is simply given on our NCERT books. But I want to suggest you that "Pradeep Books" are best for XI physics.
• 3
help us palz...we need ur help..
• -3
derive the necessary relation for variation of g with depth?Answer dis pleasee>>>
• -2
Newton's law of gravitaion states that every body in the universe attracts evry other body by an attractive force which is directly proportional to the product of the massess of the bodies and is inversely proportional to the square of the distance between them.
• 1
i think u should read out ur book
• 0
hey plz search their ans on google .or use pra deep.these are the best options to get ans....
thank u..
• 1
8. (a) Law of orbits- All planets revolve around the sun in an elliptical orbit with sun at one of its focii.
(b) Law of area- It states that the line joining the centre of sun and centre of planet sweeps out equal area at equal intervals of time i.e. aerial velocity of planet remains constant.
(c) Law of timeperiods- It states that the square of timeperiod of a planet around the sun is directly proportional to the cube of average distance between the sun and planet.
• -2
1. (i) Newton 's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
(ii) The gravitational constant denoted by letter G, is an empirical physical constant involved in the calculation(s) of gravitational force between two bodies.
2. gravitational acceleration is the acceleration on an object caused by gravity. Neglecting friction such as air resistance, all small bodies accelerate in a gravitational field at the same rate relative to the center of mass.
F=Gm'.m"/r^2
• -1
refer pradeeps , dinesh ,etcits clearly written and described in them
• 0
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# Structural Equation Modelling (SEM) Part 1
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This presentation is an introduction to the concept and theory of Structural Equation Modelling.
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### Structural Equation Modelling (SEM) Part 1
1. 1. Structural Equation Modelling (SEM) An Introduction (Part 1)
2. 2. What is Structural Equation Modelling? • SEM is a general statistical modelling technique used to establish relationship among variables. • SEM is a confirmatory data analysis technique, i.e. it tests models that are conceptually derived, beforehand it tests if the theory fits the data • SEM can be thought of as a combination of factor analysis and multiple regression it can simultaneously test measurement and structural relationships • SEM is a family of related procedures. It is alternately defined by the following terms Path Analysis, Path Modelling, Causal Modelling, Analysis of Covariance Structures, Latent Variable Analysis, Linear Structural Relations
3. 3. Covariance: At the Heart of SEM • Covariance is a measure of how much two random variables change together. Alternately, it can be defined as the strength of association between the two variables and their variabilities. 𝑛 𝑖=1(𝑋 𝑖 − 𝑋)(𝑌 𝑖 − 𝑌) 𝑁−1 OR 𝑐𝑜𝑣 𝑥𝑦 = 𝑟 𝑥𝑦 𝑆𝐷 𝑥 𝑆𝐷 𝑦 • The basic statistic of SEM Understanding patterns of correlations among a set of variables Explain as much of their variance as possible with the model specified
4. 4. Logic of SEM • Every theory (model) implies a set of correlations And why variables are correlated • Necessary (but insufficient) condition for the validity of the theory is that it should be able to reproduce the correlations that are actually observed i.e., the implied covariance matrix should = the actual covariance matrix
5. 5. Why SEM over Regression? • Regression allows for only a single dependent variable, whereas SEM allows for multiple dependent variables. • SEM allows for variables to correlate, whereas regression adjusts for other variables in the model. • Regression assumes perfect measurement, whereas SEM accounts for measurement error.
6. 6. USES OF SEM • Theory testing Strength of prediction/association in models with multiple DVs Model fit • Mediation/tests of indirect effects • Group differences Multiple-sample analysis • Longitudinal models • Multilevel nested models
7. 7. Looking for Online SEM Training? Contact us: info@costarch.com Visit: http://tinyurl.com/costarch-sem www.costarch.com | 672 | 2,738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-13 | latest | en | 0.869362 |
https://codereview.stackexchange.com/questions/149757/shortest-path-bfs-through-a-maze | 1,713,655,743,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817688.24/warc/CC-MAIN-20240420214757-20240421004757-00274.warc.gz | 161,053,270 | 42,355 | # Shortest Path (BFS) through a maze
## Background
I have decided to use this year's Advent Of Code to learn Haskell. I feel like I vaguely understand the language and can solve most of the problems with relative ease. However, the code I produce is not the most readable and possibly has inefficiencies. Any suggestions on how to improve readability and performance would be much appreciated.
## Problem
The problem is Day 13 of AoC. It consists of a maze where each cell (x,y) is either a wall or an empty space, dictated by the population count of the following equation:
x*x + 3*x + 2*x*y + y + y*y + key (where key is some arbitrary integer)
The cell is a wall if the population count is odd. The problem then comes in two parts:
1. Find shortest distance between (1,1) and (31,39).
2. Find number of cells that can be visited in 50 or less steps (from (1,1)).
## Code
The code uses BFS for both parts. I am aware that A* may quicker for Part1.
import Data.Bits (popCount)
import qualified Data.Map as Map
import Debug.Trace
-- Define all types
type Index = (Int, Int) -- (x,y)
type Edges = [Index] -- list of connecting edges
type Prob = (Int, Int, Int) -- (key, width, height)
type State = (Index, Int, Map.Map Index Bool) -- (current, steps, visited)
-- Determine if a specific cell index represents a wall
isWall :: Int -> Index -> Bool
isWall key (x,y) = odd $popCount num where num = x*x+3*x+2*x*y+y+y*y+key -- Generate all edges of a specific cell mkEdges :: Int -> Index -> Edges mkEdges key (x,y) = filter (not.isWall key) adjs where adjs = wB [(x, y-1), (x+1, y), (x,y+1), (x-1,y)] wB = filter (\(a,b) -> not (a<0 || b<0)) -- Find the shortest path length bfsPathLength :: Int -> Index -> [State] -> Int bfsPathLength key goal t@((curr, steps, visited):rest) | goal==curr = steps | otherwise = bfsPathLength key goal newStates where newStates = (filter (not.isQueued)$ filter (not.isVisited) $map mkStates$ mkEdges key curr) ++ rest
mkStates s = (s, steps+1, Map.insert curr True visited)
isVisited (s,_,_) = Map.member s visited
isQueued (s,_,_) = elem s $map (\(x,_,_) -> x) t -- Calculate the number of reachable nodes from a starting position bfsReachableLocations :: Int -> [(Int, Index)] -> [Index] -> Int bfsReachableLocations key a@((lim,curr):rest) visited | lim < 0 = 0 | otherwise = 1 + bfsReachableLocations key newNodes (visited++[curr]) where neighbours = filter (not.isQueued)$ filter (not.(\x -> elem x visited)) $mkEdges key curr isQueued n = elem n$ map (\(_,x) -> x) a
newNodes = rest ++ map (\x -> (lim-1, x)) neighbours
main = do
print $bfsPathLength 1362 (31,39) [((1,1), 0, Map.empty)] print$ bfsReachableLocations 1362 [(50, (1,1))] []
The code is not as readable as I'd like. There are probably easier ways of performing some of the steps that I have no yet encountered on my short Haskell journey. Any recommendations would be appreciated.
mkEdges
I would probably just call this function edges. mk (which I assume is short for make) has an imperative ring to it.
bfsPathLength :: Int -> Index -> [State] -> Int
You are forcing users of bfsPathLength to pass an initial state. This forces them to be aware of the internal details of the algorithm, which is not a good design. You could get rid of the [State] parameter and have bfsPathLength delegate to a helper function that takes the initial state. Public interfaces should never expose details that aren't necessary.
bfsReachableLocations :: Int -> [(Int, Index)] -> [Index] -> Int
• What's the purpose of this function? It doesn't appear to be used for computation of the actual answer to the riddle. If it's for debugging purposes, you should add a comment that says so (be sure to include what it actually does, and why it's useful for debugging as well).
• Also, once again you force users to incorporate knowledge about the internal state of the algorithm.
# Performance
You perform a linear search for nodes in the queue and the visited set. You should use a more appropriate datastructure, such as a set, that can do the search in $O(\log{n})$, or even $O(1)$, time.
# Abstraction
• Consider separating the graph generation from the search algorithm. This will allow you to use BFS search to solve other problems. You have some options here, including:
1. Turn mkEdges into a parameter to the search function.
2. Use a typeclass.
For example, if you turn mkEdges into a parameter (and remove the internal state parameter), bfsPathLength might have the following type signature:
bfsPathLength :: (a -> [a]) -> a -> Int
bfsPathLength edgeGenerator goalLocation = ...
• You should parameterize the start location:
bfsPathLength :: (a -> [a]) -> a -> a -> Int
bfsPathLength edgeGnerator startLocation goalLocation = ...
• It seems like it would be useful to be able to find the actual path in most cases, rather than just the length:
bfsPath :: (a -> [a]) -> a -> [a]
Then you can write bfsPathLength as follows:
bfsPathLength :: (a -> [a]) -> a -> a-> Int
bfsPathLength edgeGenerator startLocation = length . (bfsPath edgeGenerator startLocation) | 1,339 | 5,168 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-18 | latest | en | 0.846082 |
https://www.leovegas.com/en-ca/blog/live-casino/blackjack/blackjack-strategies/blackjack-card-counting | 1,721,756,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00103.warc.gz | 752,791,888 | 124,583 | Back
# Blackjack Card Counting: Everything You Need To Know
Summary:
Card counting in blackjack involves assigning values to each card in the deck to predict when players have an advantageous position. Methods like the popular Hi-Lo system use high, low, or neutral values for cards. The True Count system adjusts for the number of decks in play. Advanced strategies like Omega II, Wong Halves, Red 7, KO, and Zen Count offer more accuracy but require greater focus.
Table of Content:
## What Is Card Counting In Blackjack?
Did the movie get you? We’re talking about 21, the 2008 drama based on the true story of MIT students who took casinos by storm with their card counting antics and movie worthy profit haul. Or, are we talking Rain Man and Dustin Hoffman’s sharp memory? Either way, learning how to master blackjack card counting isn’t only for MITers, nor does it have to be so dramatic.
Blackjack card counting strategies begin with giving each card in the deck a high, low or neutral value. Players keep a running count of cards dealt at the table in order to predict when they hold an advantageous position over the house. Bets are adjusted based on this knowledge; lowered when favour falls toward the house and increased if the player holds the advantage.
It should come as no surprise that the grandfather of card counting was a mathematician. Edward Thorp originally outlined a card counting strategy back in the early 1960’s in his book, Beat the Dealer. Were there others before him? Absolutely. But they either failed to write a book on the subject or their strategies neglected to tie in betting recommendations. Since then, numerous card counting systems have come about for players of all levels.
There’s no need to get discouraged about the level of tracking and quick math required when implementing these strategies as card counting can be learned, and with practice and patience, brought to life at the blackjack tables. And keep in mind that blackjack basic strategy is always a great sidekick to any blackjack card counting system to help you gain an even greater advantage.
PLAY BLACKJACK ONLINE
## How to count cards in blackjack?
Before we get into detailed card counting strategies, here’s the just of it: every card in the deck is given a value. When a new shoe begins, it’s time to start a running count in the hopes of determining when you, the player, hold the advantage. When it’s determined you do, react by throwing stronger bets on the table and doing the opposite, lightning wagers, when the advantage swings to the house.
The actual value assigned to each card, and whether or not the running count starts at zero, depends on the blackjack card counting strategy – and there are many to choose from. Stay with us as we dig into all the systems and possibilities of card counting.
## How does blackjack card counting work?
### Hi-Lo System
The Hi Lo system is the most popular of the card counting bunch and one of the easiest for beginners. It breaks down the deck into three main value camps, -1, 0 and +1. This makes it a balanced system, as the three values equate to zero. Let’s have a look:
• High Cards, 10, J, Q, K, A are valued at -1.
• Neutral Cards, 7, 8, 9 are worth 0.
• Low Cards, 2, 3, 4, 5, 6 are valued at +1.It begins when a new shoe is introduced – a clean slate – and starts with a running count of zero. As the cards are dealt, keep a running tally using the value chart above. As the count increases, so should your wager. When it dips into the negative realm, ease back on bets to minimize losses.
### True Count System
Hi Lo and other blackjack card counting systems work well with a single shoe deck, but as you’ll soon find out perusing blackjack games, many tables use 4, 6 or even 8 shoes. This is where the True Count system comes into play. It’s a simple equation that aims to increase the accuracy of your count by taking into account the number of decks in play.
• True count = running count / decks remainingOverall, even with the True Count equation, the premise is the same. When your tally increases, it’s due time to increase your bet and work the advantage. When it does the opposite and falls lower, ride through the house advantage by minimizing risk with smaller bets.
Feeling like a MIT wizard with the Hi Lo system under your belt? Good stuff. Now you’re ready to kick it up a notch with more advanced ones that improve accuracy but also require much more focus.
### Omega II
Remember our Hi Lo system? A three value camp ranging from -1 to +1. Well, Omega II takes it to five values ranging from -2 to +2. Still a balanced system, but as you’ll see by the value chart below, it’s a little tricker with more to remember, learn and track. Here’s the breakdown:
• Cards 4, 5, 6 hold a value of 2.
• Cards 2, 3, 7 hold a value of 1
• Cards 8 and Ace are 0.
• 9 is worth -1
• Cards 10, J, Q, K hold a value of -2.Everything else, including the starting point of zero and the use of True Count when additional shoes are in play, is similar to the Hi Lo system except for one variation. Omega II also asks you to keep a side count for Aces to help determine when a blackjack is possible.
### Wong Halves
Do you want MIT level expertise? You’ve come to the right blackjack card counting place. Wong Halves blackjack card counting starts at zero, keeps a running tally and pulls in True Count when multiple decks are in play. However, Wong Halves’ three value camps feature a twist – fractions. And yes, this is definitely an ‘f’ word for card counters.
• 5 holds a value of +1.5
• 3, 4, 6 count as +1.
• 2, 7 count as +.5
• 8, 10, J, Q, K & Ace are 0.
• 9 holds a value of -.5Just remember, with great difficulty comes great accuracy! And that’s the goal – accurate measures of when to hammer your bets and when to pull back. The overall tells are the same – lower when the count lowers, and raise your bets when they’re higher.
### Red 7 System
The Red 7 blackjack card counting system works similar to Hi Lo and is also suitable for beginners. To get started, take the number of decks in play and multiple it by x -2. And that my friends is your starting number. Let’s say you’re playing blackjack with a four deck shoe, that’s 4 x (-2) for a starting count of -8. Work from there using the following card values:
• Cards 2, 3, 4, 5, 6 are valued at +1
• Red 7 is +1
• Black 7 is 0
• Cards 8, 9 are valued at 0
• Cards 10 - Ace are valued at -1 When the count gets into a positive realm, +1, +2 etc, that’s when you can assume advantage is falling your way and increase betting. This intertwines with a blackjack strategy of hitting and standing in certain situations, as follows:
• When the count is 0 or higher, stand with 16 if dealer’s got 10.
• When the count is 0 or higher, stand with 12 if the dealer has 3.
• When the count is +2 or higher, stand with 15 if the dealer has 10.
• When the count is +2 or higher, stand with 12 if the dealer has 2.
• When count is +2 or higher, double down versus Ace. Known to be incredibly accurate with starting count adjustments and incorporation of the above blackjack strategy, Red 7 could be the one for you!
### KO System
The blackjack card counting system known as KO, or Knock Out, uses a balanced three value camp system that’s actually quite easy to follow. It also knocks out the need for side tallies or true count, but we’ll get to that later.
• High Cards, 10, J, Q, K, Ace are valued at -1
• Neutral Cards 8, 9 are worth 0.
• Low Cards, 2, 3, 4, 5, 6, 7 are valued at +1Why no true count? Your starting count isn’t necessarily zero with the KO system because of this formula {4 - (4 x the number of decks in play)} that determines your starting count using the number of decks in play. So, for a single deck blackjack game, you’d start at zero but for a two shoe game, -4, a six deck game, -20 etc. Sidebar tip – if you’re practicing this card counting system, final count should rest at -4.
### Zen Count
It’s only fitting that zen would be balanced, right? The Zen Count system of blackjack card counting is a balanced count, meaning it equals zero, and as such, requires true count when multiple decks are in play to help with accuracy. Perfect for beginners to intermediates, it provides a solid level of accuracy to adjust your bets to. So without further ado, let’s move on to its card values:
• Cards 4, 5, 6 are valued at +2
• Cards 2, 3, 7 are valued at +1
• Cards 8, 9 are 0
• Ace is -1
• 10, J, Q, K are valued at -2From here, it follows the system of increasing your bet to take advantage of the table when the count gets high and decreasing bets when the count gets low to minimize your risk. Choose the Zen Count and see if it brings you profitable peace.
## How to count cards in single deck blackjack
If you’re feeling overwhelmed with the notion of card counting or are under the impression it's a tactic only for Elon Musks, fear not. Blackjack card counting systems are best put to use, especially for beginners, at single deck/shoe blackjack tables. The count itself will depend on the card counting system you choose and the values associated with each card.
Even without card counting tactics, single deck blackjack games offer the lowest house odds, at just .15%. And, card counting itself is easiest with only one deck in play, negating any need for true count or multiple deck adjustments. You have your choice of any of the systems, just remember, it’s accuracy you’re going for. Taking on a difficult card counting system that may impair your accuracy will negate the entire purpose! Start with simplicity and work up from there.
## How to count cards in multiple deck blackjack?
Single deck blackjack games make card counting easier, that’s a fact. And it’s also the reason they may be difficult to find on a casino floor, virtual or otherwise. So, let’s talk card counting when multiple decks are in play. Every single one of the systems we cover can be applied to a multiple deck setup, they’ll just require more focus and attention to keep an accurate count. Even right out of the starting get, playing multiple decks raises the house edge from a single deck’s .04% to an 8 deck’s .70%.
The key is also to choose a counting system that works best for you, given the multiple deck factor. If it's a balanced system like Hi Lo, Wong Halves, Omega II or Zen, you’ll need to adjust your starting count with the help of True Count (True count = running count / decks remaining). Then, set a point to recognize when favour falls your way.
The moral of the multiple deck story is to seek out players with the lowest amount of decks.
## Common Mistakes in Card Counting
Here are some common mistakes to avoid while card counting in blackjack game:
Screwing up the running count: Card counting requires accuracy, and mistakes in the running count can lead to incorrect decisions about how much to bet or how to play a hand
Counting games with poor playing conditions: Some games have rules that make it difficult to count cards accurately, such as not allowing doubling after splitting pairs or only allowing doubling on certain totals. It's important to know which games to avoid.
Lack of practice: It is best to have enough practice in free to play games before trying your card counting skills at a blackjack table.
## Tips for Successful Blackjack Card Counting
Learn a counting system: There are several counting systems to choose from, such as the Hi-Lo Count or the Omega II. Choose a system that works for you and practice it until you can count cards accurately.
Practice, practice, practice: The more you practice, the better you will become at counting cards. Start with a single deck and work your way up to multiple decks.
Manage your bankroll: Don't overbet your bankroll. Know how many sessions and what long-term bankroll you should have for your betting level to minimize the risk of going broke.
## Blackjack Counting FAQ
### Is it possible to count cards in online blackjack?
To circumvent card counting, many online casino blackjack tables use continuous shuffle machines, making card counting essentially ineffective. Even with live dealers that don’t offer the machine version, dealer’s are known to shuffle mid shoe to throw off your tally!
### Can you win at blackjack without counting cards?
Absolutely. Improving your blackjack knowledge, doable using our very own blackjack basic strategy guide, increases your chances of winning. At live casinos, you’ll often hear ‘what does the book say to do’ - they’re referring to the basic playing strategy and chart that tells you in general terms when to hit, stand and fold. Or you can employ a variety of strategies for wagering and play that may improve your game and chances. In fact, we’ve compiled a list of the best blackjack strategies to help you along in this regard too!
### How does counting cards help in blackjack?
The goal of card counting is to be aware and predict when the advantage shifts in your favour. When that happens, it’s time to adjust your betting strategy and increase wagers to make the most of it. It’s a system of prediction and betting strategies all mixed into one.
### Do blackjack dealers count cards?
Many can, and many can recognize players doing so too. But for the most part, you’re likely to encounter a dealer who isn’t counting cards and is simply focusing on their job.
### What is the best blackjack card counting strategy?
The key is to find a strategy that you can manage at the table. Accuracy is the name of the game with card counting. So, if you take on a system that is ‘better’ but you can’t maintain a solid, accurate count, it’s pointless. Give simple strategies a go first, like the Hi Lo strategy, and advance from there.
### How hard is it to count cards in blackjack?
Well, it doesn't seem that hard if you’ve watched 21 and the boys from MIT, but for the rest of us who may not be so mathematically talented, it’s a tough skill to learn. Beyond picking up the skill and knowledge behind card counting, you’ve also got to have a bankroll that can sustain swings and a casino table that allows, or won’t catch you making excessive profits doing it. If you’re keen on putting a card counting strategy to use, seek out a single deck shoe game as your starting point!
### How do you practice blackjack card counting?
A great starting point is to get your hours in with one deck practise.Train yourself by removing one card from the deck. Then, flip cards over while keeping a running count. By the time you reach the end, you should know what card (value) is missing. Once you get the hang of the count, ramp up your speed and time yourself to get faster in an effort to keep up to a dealer’s pace and table game.
### How long does it take to learn how to count cards?
Well, not quite the 10,000 hours Gladwell states it takes to become an expert, but it will take hours and hours of dedication and practice to master card counting. If you’re got the focus and the drive though, maybe quicker!
### Who invented blackjack card counting?
That title goes to a man by the name of Edward Thorp. And not surprisingly, he’s a mathematician. If you fancy yourself a history buff, have a read of his 1962 book ‘Beat the Dealer’ for an overview of his play and bet strategies in detail. Fine, there’s likely an audible book to simply listen to if you wish.
### Is counting cards in blackjack the same as poker?
You can definitely count cards in both card games, but they’re quite a different approach. In Poker, you’d be counting cards to gain an advantage over your fellow table players while in blackjack, you’re attempting to gain an advantage over the house. Plus, remember it’s a completely different system given you’re not privy to seeing other player’s hole cards in most poker games.
### What is the best method for counting cards in blackjack?
KISS. Keep it simple silly. The best method is to maintain focus, attention and be disciplined in the process. Blackjack card counting takes much focus and never underestimate the distractions of the casino floor. For this reason, seek out single deck tables and start with a simple system that you can build up from. Accuracy is your goal.
### How effective is card counting in improving the odds?
Card counting can be effective in improving the odds in blackjack, but it requires practice, concentration, and analytical skills. It is not a guaranteed way to win and may not be worth the effort for most players.
Learn about other strategies in our guides:
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Gambling can be addictive. Play responsibly. | 3,809 | 16,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-30 | latest | en | 0.949023 |
https://docs.scipy.org/doc/scipy-1.2.1/reference/generated/scipy.special.sph_harm.html | 1,620,505,512,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00601.warc.gz | 237,999,523 | 3,509 | # scipy.special.sph_harm¶
scipy.special.sph_harm(m, n, theta, phi) = <ufunc 'sph_harm'>
Compute spherical harmonics.
The spherical harmonics are defined as
$Y^m_n(\theta,\phi) = \sqrt{\frac{2n+1}{4\pi} \frac{(n-m)!}{(n+m)!}} e^{i m \theta} P^m_n(\cos(\phi))$
where $$P_n^m$$ are the associated Legendre functions; see lpmv.
Parameters: m : array_like Order of the harmonic (int); must have |m| <= n. n : array_like Degree of the harmonic (int); must have n >= 0. This is often denoted by l (lower case L) in descriptions of spherical harmonics. theta : array_like Azimuthal (longitudinal) coordinate; must be in [0, 2*pi]. phi : array_like Polar (colatitudinal) coordinate; must be in [0, pi]. y_mn : complex float The harmonic $$Y^m_n$$ sampled at theta and phi.
Notes
There are different conventions for the meanings of the input arguments theta and phi. In SciPy theta is the azimuthal angle and phi is the polar angle. It is common to see the opposite convention, that is, theta as the polar angle and phi as the azimuthal angle.
Note that SciPy’s spherical harmonics include the Condon-Shortley phase [2] because it is part of lpmv.
With SciPy’s conventions, the first several spherical harmonics are
$\begin{split}Y_0^0(\theta, \phi) &= \frac{1}{2} \sqrt{\frac{1}{\pi}} \\ Y_1^{-1}(\theta, \phi) &= \frac{1}{2} \sqrt{\frac{3}{2\pi}} e^{-i\theta} \sin(\phi) \\ Y_1^0(\theta, \phi) &= \frac{1}{2} \sqrt{\frac{3}{\pi}} \cos(\phi) \\ Y_1^1(\theta, \phi) &= -\frac{1}{2} \sqrt{\frac{3}{2\pi}} e^{i\theta} \sin(\phi).\end{split}$
References
[1] Digital Library of Mathematical Functions, 14.30. https://dlmf.nist.gov/14.30
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scipy.special.lpmv
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scipy.special.clpmn | 553 | 1,715 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-21 | latest | en | 0.723228 |
http://gri.sourceforge.net/gridoc/html/tst_suite/tst_rpn.html | 1,516,260,391,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887077.23/warc/CC-MAIN-20180118071706-20180118091706-00280.warc.gz | 137,855,649 | 2,053 | # This html document was prepared by gri2html based on the Gri script named
# tst_rpn.gri
```show "doc/tst_suite/tst_rpn.gri ..." ...
rpnfunction same - abs 1e-5 > # Are numbers virtually same?
# Simple arithmetic
.a. = 0
assert {rpn .a. !} " failed test 1.1"
.a. += 1
assert .a. " failed test 1.2"
.a. += 1
assert {rpn .a. 2 same} " failed test 1.3"
.a. *= 2
assert {rpn .a. 4 same} " failed test 1.4"
.a. /= 4
assert {rpn .a. 1 same} " failed test 1.5"
assert {rpn 2 1 - 1 same} " failed test 1.6"
assert {rpn 2 1 + 3 same} " failed test 1.7"
assert {rpn 3 2 * 6 same} " failed test 1.8"
assert {rpn 4 2 / 2 same} " failed test 1.9"
# Conversions (lower-case ok on input, but output is upper-case)
assert {rpn "aa" hex2dec 170 ==} " failed test 2.1"
assert {rpn "AB" hex2dec 171 ==} " failed test 2.2"
assert {rpn 63 dec2hex "3F" ==} " failed test 2.3"
assert {rpn 193 dec2hex "C1" ==} " failed test 2.4"
# Logic
assert {rpn 1 0 or} " failed test 3.1"
assert {rpn 0 1 or} " failed test 3.2"
assert {rpn 1 0 |} " failed test 3.3"
assert {rpn 0 1 |} " failed test 3.4"
assert {rpn 1 0 and not} " failed test 3.5"
assert {rpn 1 0 & !} " failed test 3.6"
# Logs, powers
.a. _= 10
assert {rpn .a. 0 same} " failed test 4.1"
.a. = 2
.a. ^= 8
assert {rpn .a. 256 same} " failed test 4.2"
assert {rpn -2 4 power 16 ==} " failed test 4.3"
assert {rpn -2 3 power -8 ==} " failed test 4.4"
assert {rpn -2 2 power 4 ==} " failed test 4.5"
assert {rpn 2 3 power 8 ==} " failed test 4.6"
# String operations
\a = {rpn "file" ".dat" strcat}
assert {rpn "\a" "file.dat" ==} " failed test 5.1"
\sentence = "This sentence has five words"
\w1 = word 0 of "\sentence "
assert {rpn "\w1" "This" ==} " failed test 5.2"
\w2 = word 1 of "\sentence "
assert {rpn "\w2" "sentence" ==} " failed test 5.3"
assert {rpn 0 4 "hello" substr "hell" ==} " failed test 5.4"
# Q: will the below work on all OS?????
# NB. better to switch with something more universal
\six = system "date | wc | awk '{print \$2}'"
assert {rpn \six 6 ==} " failed test 6.1"
# Statistical operations
1
3
2
9
3
assert {rpn x mean 3.6 same} " failed test 7.1"
assert {rpn x stddev 3.1305 same} " failed test 7.2"
assert {rpn x skewness 0.882432 same} " failed test 7.3"
assert {rpn x kurtosis 1.88008 same} " failed test 7.4"
# ----------- FILL IN LATER -----------------
# Math functions (e.g. sin, ...)
# Stack operations push, pop, and exch.
assert {rpn 45 cos 0.7071 same} " failed test 8.1"
assert {rpn 45 sin 0.7071 same} " failed test 8.2"
assert {rpn 45 tan 1 same} " failed test 8.3"
# Missing-values
set missing value -99
1 -99
assert {rpn .y. ismissing} " failed test 9.1"
assert {rpn .y. -99 == } " failed test 9.2"
assert {rpn .y. -99 != !} " failed test 9.3"
assert {rpn .y. 0 != } " failed test 9.4"
# some boolean tests need not check that both values exist
assert {rpn .y. 0 & !} " failed test 9.5"
assert {rpn 0 .y. & !} " failed test 9.6"
assert {rpn .y. 1 & } " failed test 9.7"
assert {rpn .y. 1 | } " failed test 9.8"
assert {rpn 1 .y. | } " failed test 9.9"
if {rpn .y. -99 !=}
show " failed test 9.10"
end if
show " passed"
``` | 1,324 | 3,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-05 | latest | en | 0.702851 |
https://programmer.ink/think/area-covered-by-hdu-1255-scan-line.html | 1,718,529,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00468.warc.gz | 442,729,089 | 5,581 | Area covered by (HDU - 1255) (scan line)
Posted by wherertheskips on Thu, 03 Feb 2022 02:01:39 +0100
Given several rectangles on the plane, find the area of the area covered by these rectangles at least twice
Input
Lower left and upper right coordinates
The first line of input data is a positive integer t (1 < = T < = 100), representing the number of test data The first row of each test data is a positive integer N (1 < = N < = 1000), representing the number of rectangles, and then N rows of data. Each row contains four floating-point numbers, representing the coordinates of the lower left corner and the upper right corner of a rectangle on the plane. The upper and lower edges of the rectangle are parallel to the X axis, and the left and right sides are parallel to the Y axis The coordinates range from 0 to 100000
Note: there are many input data in this question. It is recommended to use scanf to read the data
Output
For each set of test data, please calculate the area covered by these rectangles at least twice Keep the result to two decimal places
Sample Input
```2
5
1 1 4 2
1 3 3 7
2 1.5 5 4.5
3.5 1.25 7.5 4
6 3 10 7
3
0 0 1 1
1 0 2 1
2 0 3 1```
Sample Output
```7.63
0.00```
This problem is still solved by scanline. If you don't understand the principle of scanline, you can read my previous blog and attach the address of scanline blog below: Paint area (scan line)_ AC__dream blog - CSDN blog
The topic of paint area is to find the area of the union of rectangles, and this topic is to find the area of the intersection of rectangles. The principle is basically the same. In the topic of paint area, we maintain the maximum length of covering a section at least once, and this problem obviously can't only maintain this one, If you think about it carefully, it's easy to think that this problem requires us to maintain the maximum length of covering a section at least twice, because the area itself is the integral of the line segment. When the line segment is covered only once, the integrated area will be covered only once, and when the line segment is covered multiple times, the integrated area will be covered multiple times, Speaking of this, the idea of this topic is relatively clear. Let me talk about how to calculate the maximum length covered twice in the interval.
First, let's review how to calculate the maximum length covered at least once: (it needs to be reminded again that the point interval [l,r] represents the distance between the L-th point and the r+1 point)
First, if the current interval is covered once as a whole, it is obvious that the maximum length of the interval covered at least once is the length of the interval itself. If the current interval is not covered once, there are two clear situations. One is that the current interval is a point, and the actual length is 1, Then it is certain that the maximum length of the interval with length one covered at least once is 0. When the interval is an interval with length and length not 1, the maximum length of the interval covered at least once is the sum of the maximum length of the interval covered at least once of the left and right sub intervals of the interval. Here is the code of this part:
void pushup(int id)
{
if(cnt[id]) len[id]=alls[r[id]]-alls[l[id]-1];// Point interval [l[id],r[id]] actually represents segment interval [all [l [ID] - 1], all [R [ID]]]
else if(l[id]==r[id]) len[id]=0;
else len[id]=len[id<<1]+len[id<<1|1];
}
Here's how to calculate the maximum length of being covered at least twice:
Firstly, if the current interval is covered twice as a whole, it is obvious that the maximum length of the interval covered at least twice is the length of the interval itself. Moreover, if the current interval is a point, the actual length is 1. It is certain that the maximum length of the interval covered at least twice with a length of one is 0, which is relatively easy to get, Next, there are two cases. The interval length represented by these two cases is not 1. One is that the number of times the current length is covered is 1. Since we did not update the child node through the parent node when updating the segment tree, the maximum length covered by the child interval of the current interval is independent of the current interval, Then, the maximum length of the current interval covered at least twice is the sum of the interval lengths of the left and right sub intervals covered at least once. In addition, the current interval has been completely covered once, so the public part is covered at least twice. This place is prone to errors. We must understand it well. The following is the last case, that is, the number of times the current length is covered is 0, There's nothing to consider. It's the sum of the length of the interval where the left and right intervals are covered at least twice.
Corresponding code (this code is a combined code that updates the longest length of the interval covered at least once and the longest length of the interval covered at least twice):
void pushup(int id)
{
/ / find the length of the longest interval whose coverage times are greater than or equal to 1
if(cnt[id]) len[id]=alls[r[id]]-alls[l[id]-1];
else if(l[id]==r[id]) len[id]=0;
else len[id]=len[id<<1]+len[id<<1|1];
/ / find the length of the longest interval whose coverage times are greater than or equal to 2
if(cnt[id]>=2) len2[id]=alls[r[id]]-alls[l[id]-1];
else if(l[id]==r[id]) len2[id]=0;
else if(cnt[id]==1) len2[id]=len[id<<1]+len[id<<1|1];// If the current interval is covered once, the maximum length of the current interval covered twice is the sum of the maximum lengths of the left and right intervals covered once or more
else len2[id]=len2[id<<1]+len2[id<<1|1];
}
Another thing that needs special attention is rounding. Pay attention to the accuracy
The following code is:
```#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
const int N=1e4+10;
int l[N],r[N],cnt[N];//cnt[i] record the number of times the i-th interval is covered
double len[N],len2[N];//len[i] records the maximum length in which the i-th interval is covered more than or equal to 1, and len2[i] records the maximum length in which the i-th interval is covered more than or equal to 2
vector<double>alls;
struct node{
double x,yn,yx;
int k;
}p[N];
bool cmp(node a,node b)
{
return a.x<b.x;
}
double find(double x)
{
return lower_bound(alls.begin(),alls.end(),x)-alls.begin()+1;
}
void pushup(int id)
{
//Find the longest interval length with coverage times greater than or equal to 1
if(cnt[id]) len[id]=alls[r[id]]-alls[l[id]-1];
else if(l[id]==r[id]) len[id]=0;
else len[id]=len[id<<1]+len[id<<1|1];
//Find the longest interval length with coverage times greater than or equal to 2
if(cnt[id]>=2) len2[id]=alls[r[id]]-alls[l[id]-1];
else if(l[id]==r[id]) len2[id]=0;
else if(cnt[id]==1) len2[id]=len[id<<1]+len[id<<1|1];//If the current interval is covered once, the maximum length of the current interval covered twice is the sum of the maximum lengths of the left and right intervals covered once or more
else len2[id]=len2[id<<1]+len2[id<<1|1];
}
void build(int id,int L,int R)
{
l[id]=L;r[id]=R;cnt[id]=len[id]=0;
if(L==R) return ;
int mid=L+R>>1;
build(id<<1,L,mid);
build(id<<1|1,mid+1,R);
}
void update_interval(int id,int L,int R,int k)
{
if(l[id]>=L&&r[id]<=R)//The current interval is completely in the target interval
{
cnt[id]+=k;
pushup(id);
return ;
}
int mid=l[id]+r[id]>>1;
if(mid>=L) update_interval(id<<1,L,R,k);
if(mid+1<=R) update_interval(id<<1|1,L,R,k);
pushup(id);
}
int main()
{
int T,n;
cin>>T;
while(T--)
{
scanf("%d",&n);
double x1,y1,x2,y2;
int cnt=0;
alls.clear();
for(int i=1;i<=n;i++)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
p[++cnt]={x1,y1,y2,1};
p[++cnt]={x2,y1,y2,-1};
alls.push_back(y1);
alls.push_back(y2);
}
sort(p+1,p+cnt+1,cmp);
sort(alls.begin(),alls.end());
alls.erase(unique(alls.begin(),alls.end()),alls.end());
for(int i=1;i<=cnt;i++)
{
p[i].yn=find(p[i].yn);
p[i].yx=find(p[i].yx);
}
build(1,1,alls.size());
double ans=0;
for(int i=1;i<=cnt;i++)
{
if(i>1) ans+=len2[1]*(p[i].x-p[i-1].x);
update_interval(1,(int)p[i].yn,(int)p[i].yx-1,p[i].k);
}
printf("%.2lf\n",ans+0.0001);
}
return 0;
}```
Topics: data structure | 2,232 | 8,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-26 | latest | en | 0.907048 |
https://www.theinternet.io/articles/ask-ai/how-far-is-32410-the-the-power-of-14/ | 1,686,010,411,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652184.68/warc/CC-MAIN-20230605221713-20230606011713-00600.warc.gz | 1,125,677,539 | 8,193 | The Internet
## Ask AI on The Internet
Ask Another Question Browse Other Questions
Question: How far is 3.24×10 the the power of 14
1 Point
3.24 x 10^14 is an expression in scientific notation. To describe it in standard notation: 3.24 x 10^14 = 324,000,000,000,000 So, it is 324 trillion units in magnitude. Please note that you need to define the units you are using (e.g., meters, kilometers) to provide a specific distance or comparison.
April 18, 2023, 8:36 p.m.
0 Points | 137 | 483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-23 | latest | en | 0.831644 |
http://mathoverflow.net/questions/81012/a-simple-stopping-time-problem/93841 | 1,462,132,513,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860116886.38/warc/CC-MAIN-20160428161516-00039-ip-10-239-7-51.ec2.internal.warc.gz | 182,873,335 | 16,708 | # A simple stopping time problem.
This should be rather standard so I hope somebody with a good background in probability theory would give me a quick solution or a reference.
We are given a threshold positive integer $T>0$. Let $a_1=1$ and for all $k$ with probability one half set $a_k=3a_{k-1}$ or else $a_k=2a_{k-1}$. We will stop the process at smallest time $\tau$ when $a_{\tau} \geq T$. We would like to compute the constant $c$ defined to be,
$E[ \sum_{i=1}^{\tau} a_i ] = c T + o(T)$
Could you estimate $c$ ?
-
From the way you ask, I conclude that you can prove that the limit exists (which by itself is by no means trivial), so I'll just show how to compute it under this assumption.
Let $v(t)$ be $\frac 1t$ times the expectation in question if we stop after we exceed $t>0$ (not necessarily an integer). Then $v(t)=\frac 1t$ for $0<t<1$ and $v(t)=\frac 1t+\frac 12[v(t/2)+v(t/3)]$ for $t\ge 1$. Now let $F(s)=\int_1^\infty t^{-s}v(t)\frac{dt}{t}$. Using the recurrence, we get that for every $s>0$, $$F(s)=\frac 1{s+1}+\frac 12\left(\int_{1/2}^1 \frac 1t t^{-s}\frac{dt}t+\int_{1/3}^1 \frac 1t t^{-s}\frac{dt}t\right)+\frac 12(2^{-s}+3^{-s})F(s)$$ The limit we are interested in is the same as $\lim_{s\to 0+}sF(s)$. Putting all terms with $F(s)$ to one side, dividing, and passing to the limit, we get $\frac{5}{\log 6}$, which differs from Will's heuristic answer a bit. I cannot say that I really understood his post but it is quite fascinating that he was somehow right with $\log 6$ in the denominator :).
I apologize for computational mistakes in the original post.
-
Could you comment on how you would establish the existence of the limit ? – Nick B. Nov 16 '11 at 0:57
I think some factors of $t$, etc., may be missing from the working shown here. When I apply the method given above I find $c=5/\log 6=2.79055$, which is in good agreement with an experimental value of $2.79 \pm 0.01$. – David Moews Nov 16 '11 at 1:30
yes I think that is not a $1$ ; rather it is a $\frac{1}{s+1}$ but beside that I don't any possible trivial calculation problem. do you ? – Nick B. Nov 16 '11 at 1:31
Fedja: I got the same answer using a purely probabilistic arguments. I'll try to write it up later... – Ori Gurel-Gurevich Nov 16 '11 at 5:21
please do, Ori ! – Nick B. Nov 16 '11 at 5:52
I think it is possible to find the result using renewal theory. Indeed, the process $(\ln(a_i))$ is a random walk with i.i.d. increments ($\ln(2)$ or $\ln(3)$ with probability $1/2$). The renewal theorem will tell you the structure of the walk when it jumps over a large time (here $\ln(T)$). More precisely when $T \to \infty$ the jump that goes over $\ln(T)$ is a size biaised version of the orignal jump measure, i.e. $\ln(3)$ with probability $\ln(3)/\ln(6)$ and $\ln(2)$ with probability $\ln(2)/\ln(6)$. Furthermore, knowing this jump, the actual position of $\ln(T)$ is uniform in the jump. Easy calculations (if correct) then yield $E[a_\tau] = 3T/\ln(6)$ and $E[a_{\tau-1}]=7T/(6 \ln(6))$. But going down from $a_{\tau-1}$ is easy (the walk is asymptotically the reversed version) and we can compute $E[a_{\tau-1}+ a_{\tau-2}+...]= 12/7 E[a_{\tau-1}]$.
-
I don't believe it. You have $a_k = X_k a_{k-1}$, with$X = 3$or $2$ etc since $log(X)$ actually has positive expectation we'll have $a_k \approx e^{k\mu}$ and $\tau$ no worse than about log(T).
- | 1,098 | 3,369 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2016-18 | longest | en | 0.867651 |
https://holifestivalz.com/weather-worksheets-cut-and-paste/ | 1,603,886,921,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107898499.49/warc/CC-MAIN-20201028103215-20201028133215-00414.warc.gz | 355,097,076 | 8,625 | # Weather Worksheets: Cut-and-Paste
Published at Thursday, June 25th 2020, 13:07:03 PM. English Worksheets. By Baylie Deschamps.
If you must use worksheets, then be sure you do the following things: Know what you are buying. If you can not see it (there is no sample shown), then do not buy it. There are many people out there trying to make a buck off the current popularity of worksheets. Many, if not most, of these people know nothing about mathematics, teaching, or how the brain learns. Anyone can type columns of addition, subtraction, multiplication, etc. problems; but these worksheets will be bad for your child. Do not trust what you can not see.
You can find several types of sheets online and offline. You can choose among multiplication, Addition, Subtraction, Division, Geometry, Decimal, Shapes and Space worksheets. These sheets help the users to practice mathematical problems. Solving these problems become much easier with the help of mathematical worksheets. Parents can easily help their kids with the help of printable sheets. Printable work sheets add fun to the kid`s learning process. Nowadays parents and teachers are using colorful sheets for teaching their children. These sheets help in learning new skills. Colorful sheets are easy to read and understand.
### Jumbled Sentences Worksheets For Grade 6
##### Year 2 Grammar Worksheets
###### Minimal Pairs Vowels Worksheet
Once you have a scope and sequence book, make a list of each area in math that he needs to work on for the school year. For example for grades three and four, by the end of the year in subtraction, your child should be able to: Solve vertical and horizontal computation problems, Review subtraction of 2 numbers whose sums would be 18 or less, subtract 1- or 2-digit number from a 2-digit number with/without renaming, subtract 1-, 2-, or 3-digit numbers from 3- and 4-digit number with/without renaming, Subtract 1-, 2-, 3-, 4-, or 5-digit number from a 5-digit number. When you have this list, begin searching online for free math worksheets that fit your child has scope and sequence for the year and the goals you have set for your child.
Once you have a scope and sequence book, make a list of each area in math that he needs to work on for the school year. For example for grades three and four, by the end of the year in subtraction, your child should be able to: Solve vertical and horizontal computation problems, Review subtraction of 2 numbers whose sums would be 18 or less, subtract 1- or 2-digit number from a 2-digit number with/without renaming, subtract 1-, 2-, or 3-digit numbers from 3- and 4-digit number with/without renaming, Subtract 1-, 2-, 3-, 4-, or 5-digit number from a 5-digit number. When you have this list, begin searching online for free math worksheets that fit your child has scope and sequence for the year and the goals you have set for your child.
### Trending Today
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###### Possessive Nouns Sentences Exercises
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Popular Galleries | 730 | 3,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-45 | latest | en | 0.905564 |
http://www.mathopenref.com/trigfunctions.html | 1,545,039,087,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828501.85/warc/CC-MAIN-20181217091227-20181217113227-00509.warc.gz | 412,871,572 | 6,544 | # Trigonometry functions - introduction
There are six functions that are the core of trigonometry. There are three primary ones that you need to understand completely:
• Sine (sin)
• Cosine (cos)
• Tangent (tan)
The other three are not used as often and can be derived from the three primary functions. Because they can easily be derived, calculators and spreadsheets do not usually have them.
• Secant (sec)
• Cosecant (csc)
• Cotangent (cot)
All six functions have three-letter abbreviations (shown in parentheses above).
## Definitions of the six functions
Consider the right triangle on the left. For each angle P or Q, there are six functions, each function is the ratio of two sides of the triangle. The only difference between the six functions is which pair of sides we use.
In the following table
• a is the length of the side adjacent to the angle (x) in question.
• o is the length of the side opposite the angle.
• h is the length of the hypotenuse.
"x" represents the measure of ther angle in either degrees or radians.
Sine The three primary functions Cosine Tangent
Cosecant Notice how each is the reciprocal of sin, cos or tan. Secant Cotangent
For example, in the figure above, the cosine of x is the side adjacent to x (labeled a), over the hypotenuse (labeled h): If a=12cm, and h=24cm, then cos x = 0.5 (12 over 24).
## Soh Cah Toa
These 9 letters are a memory aid to remember the ratios for the three primary functions - sin, cos and tan. Pronounced a bit like "soaka towa". See Sohcahtoa.
## The ratios are constant
Because the functions are a ratio of two side lengths, they always produce the same result for a given angle, regardless of the size of the triangle.
In the figure above, drag the point C. The triangle will adjust to keep the angle C at 30°. Note how the ratio of the opposite side to the hypotenuse does not change, even though their lengths do. Because of that, the sine of 30° does not vary either. It is always 0.5.
Remember: When you apply a trig function to a given angle, it always produces the same result. For example tan 60° is always 1.732.
## Using a calculator
Most calculators have buttons to find the sin, cos and tan of an angle. Be sure to set the calculator to degrees or radians mode depending on what units you are using.
## Inverse functions
For each of the six functions there is an inverse function that works in reverse. The inverse function has the letters 'ARC' in front of it.
For example the inverse function of COS is ARCCOS. While COS tells you the cosine of an angle, ARCCOS tells you what angle has a given cosine. See Inverse trigonometric functions.
On calculators and spreadsheets, the inverse functions are sometimes written acos(x) or cos-1(x).
## Trigonometry functions of large and/or negative angles
The six functions can also be defined in a rectangular coordinate system. This allows them to go beyond right triangles, to where the angles can have any measure, even beyond 360°, and can be both positive and negative. For more on this see Trigonometry functions of large and negative angles.
## Identities - replacing a function with others
Trigonometric identities are simply ways of writing one function using others. For example, from the table above we see that This equivalence is called an identity. If we had an equation with sec x in it, we could replace sec x with
one over cos x if that helps us reach our goals. There are many such identities. For more see Trigonometric identities.
## Not just right triangles
These functions are defined using a right triangle, but they have uses in other triangles too. For example the Law of Sines and the Law of Cosines can be used to solve any triangle - not just right triangles.
## Graphing the functions
The functions can be graphed, and some, notably the SIN function, produce shapes that frequently occur in nature. For example see the graph of the SIN function, often called a sine wave, on the right. For more see
Pure audio tones and radio waves are sine waves in their respective medium.
## Derivatives of the trig functions
Each of the functions can be differentiated in calculus. The result is another function that indicates its rate of change (slope) at a particular values of x. These derivative functions are stated in terms of other trig functions. For more on this see Derivatives of trigonometric functions. See also the Calculus Table of Contents.
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Become a patron of the site at patreon.com/mathopenref | 1,132 | 4,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2018-51 | latest | en | 0.868304 |
https://www.onlinemath4all.com/least-common-multiple-by-prime-factorization-worksheet.html | 1,670,045,204,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710924.83/warc/CC-MAIN-20221203043643-20221203073643-00523.warc.gz | 986,262,497 | 8,182 | # LEAST COMMON MULTIPLE BY PRIME FACTORIZATION WORKSHEET
In each case, find the least common multiple of the numbers by prime factorization.
Problem 1 :
4 and 6
Problem 2 :
2 and 3
Problem 3 :
4 and 5.
Problem 4 :
3 and 6
Problem 5 :
12 and 18
Problem 6 :
24 and 60
Problem 7 :
4, 5 and 8
Problem 8 :
48, 72 and 108
Problem 9 :
4, 6, 8 and 12
Problem 10 :
2, 3, 4 and 5
Problem 11 :
2, 4, 5 and 8
Problem 12 :
3, 6, 7 and 14
4 and 6
Resolve the given numbers into their prime factors.
From the above division,
4 = 2 x 2
6 = 2 x 3
The different prime factors are 2 and 3.
The prime factor 2 appears a maximum of 2 times in the prime factorization of 4.
The prime factor 3 appears a maximum of 1 time in the prime factorization of 6.
Therefore, the least common multiple of 4 and 6 is
= 2 x 2 x 3
= 12
2 and 3
There is no common divisor for 2 and 3 other than 1. So, 2 and 3 are relatively prime.
To get least common multiple of relatively prime numbers, we have to multiply them.
Therefore, the least common multiple of 2 and 3 is
= 2 x 3
= 6
4 and 5
There is no common divisor for 4 and 5 other than 1. So, 4 and 5 are relatively prime.
To get least common multiple of relatively prime numbers, we have to multiply them.
Therefore, the least common multiple of 4 and 5 is
= 4 x 5
= 20
3 and 6
Since 3 is a prime number, we don't have to resolve it into prime factors anymore.
Resolve 6 into its prime factors.
From the above division,
6 = 2 x 3
3 is already a prime number.
3 = 3
The different prime factors are 2 and 3.
The prime factor 2 appears a maximum of 1 time in the prime factorization of 6.
The prime factor 3 appears a maximum of 1 time in the prime factorization of 6 and 3.
Therefore, the least common multiple of 3 and 6 is
= 2 x 3
= 6
12 and 18
Resolve 12 and 18 into their prime factors.
From the above division,
12 = 2 x 2 x 3
18 = 2 x 3 x 3
The different prime factors are 2 and 3.
The prime factor 2 appears a maximum of 2 times in the prime factorization of 12.
The prime factor 3 appears a maximum of 2 times in the prime factorization of 18.
Therefore, the least common multiple of 12 and 18 is
= 2 x 2 x 3 x 3
= 36
24 and 60
Resolve 24 and 60 into their prime factors.
From the above division,
24 = 2 x 2 x 2 x 3
60 = 2 x 2 x 3 x 5
The different prime factors are 2, 3 and 5.
The prime factor 2 appears a maximum of 3 times in the prime factorization of 24.
The prime factor 3 appears a maximum of 1 time in the prime factorization of 24 and 60.
The prime factor 5 appears a maximum of 1 time in the prime factorization of 60.
Therefore, the least common multiple of 24 and 60 is
= 2 x 2 x 2 x 3 x 5
= 120
4, 5 and 8
Since 5 is a prime number, we don't have to resolve it into prime factors anymore.
Resolve 4 and 8 into their prime factors.
From the above division,
4 = 2 x 2
8 = 2 x 2 x 2
5 is already a prime number.
5 = 5
The different prime factors are 2 and 5.
The prime factor 2 appears a maximum of 3 times in the prime factorization of 8.
The prime factor 5 appears a maximum of 1 time in the prime factorization of 5.
Therefore, the least common multiple of 4, 5 and 8 is
= 2 x 2 x 2 x 5
= 40
48, 72 and 108
Resolve 48, 72 and 108 into their prime factors.
From the above division,
48 = 2 x 2 x 2 x 2 x 3
72 = 2 x 2 x 2 x 3 x 3
108 = 2 x 2 x 3 x 3 x 3
The different prime factors are 2 and 3.
The prime factor 2 appears a maximum of 4 times in the prime factorization of 48.
The prime factor 3 appears a maximum of 3 times in the prime factorization of 108.
Therefore, the least common multiple of 48, 72 and 108
= 2 x 2 x 2 x 2 x 3 x 3 x 3
= 432
4, 6, 8 and 12
Resolve 4, 6, 8 and 12 into their prime factors.
From the above division,
4 = 2 x 2
6 = 2 x 3
8 = 2 x 2 x 2
12 = 2 x 2 x 3
The different prime factors are 2 and 3.
The prime factor 2 appears a maximum of 3 times in the prime factorization of 8.
The prime factor 3 appears a maximum of 1 time in the prime factorization of 6 and 12.
Therefore, the least common multiple of 4, 6, 8 and 12 is
= 2 x 2 x 2 x 3
= 24
2, 3, 4 and 5
There is no common divisor for all the four numbers 2, 3, 4 and 5. So, the given numbers are relatively prime.
To get least common multiple of relatively prime numbers, we have to multiply them.
Therefore, the least common multiple of 2, 3, 4 and 5 is
= 2 x 3 x 4 x 5
= 120
2, 4, 5 and 8
Since 2 and 5 are prime numbers, we don't have to resolve them into prime factors anymore.
Resolve 4 and 8 into their prime factors.
From the above division,
4 = 2 x 2
8 = 2 x 2 x 2
2 and 5 are already prime numbers.
2 = 2
5 = 5
The different prime factors are 2 and 5.
The prime factor 2 appears a maximum of 3 times in the prime factorization of 8.
The prime factor 5 appears a maximum of 1 time in the prime factorization of 5.
Therefore, the least common multiple of 2, 4, 5 and 8 is
= 2 x 2 x 2 x 5
= 40
3, 6, 7 and 14
Since 3 and 7 are prime numbers, we don't have to resolve them into prime factors anymore.
Resolve 6 and 14 into their prime factors.
From the above division,
6 = 2 x 3
14 = 2 x 7
3 and 7 are already prime numbers.
3 = 3
7 = 7
The different prime factors are 2, 3 and 7.
The prime factor 2 appears a maximum of 1 time in the prime factorization of 6 and 14.
The prime factor 3 appears a maximum of 1 time in the prime factorization of 6 and 3.
The prime factor 7 appears a maximum of 1 time in the prime factorization of 14 and 7.
Therefore, the least common multiple of 3, 6, 7 and 14 is
= 2 x 3 x 7
= 42
Kindly mail your feedback to v4formath@gmail.com
## Recent Articles
1. ### Unit Rates
Dec 02, 22 04:18 PM
Unit Rates - Concept - Examples with step by step explanation
2. ### Adding and Subtracting Fractions Worksheet
Dec 02, 22 07:27 AM | 1,907 | 5,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2022-49 | latest | en | 0.939649 |
https://www.education.com/activity/article/Fact_Family_third/ | 1,675,686,455,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500339.37/warc/CC-MAIN-20230206113934-20230206143934-00590.warc.gz | 758,392,523 | 40,553 | # Find the Math Fact Family
### What You Need:
• Pen
• 20-30 Strips of Paper
• Clock or timer
### What You Do:
Step 1: To set up the game, you will write 4 numbers onto each strip of paper. 3 will be part of a fact family, and one will not be part of the fact family. A fact family is 3 numbers that are connected through multiplication and division. For instance, 2, 4 and 8 are a fact family because 2 x 4 = 8, 4 x 2 = 8, 8 ÷ 4 = 2 and 8 ÷ 2 = 4.
Step 2: Explain the concept of the fact family to your child and give several examples, such as 3, 9 and 27. Ask your child to go through the multiplication and division relationships between the example you gave.
Fact Family Examples:
2, 3, 6
6, 8, 48
4, 7, 28
5, 8, 40
3, 7, 21
8, 9, 72
Step 3: Explain to your child that they will select a strip of paper with 4 numbers on it. 3 numbers will be part of a fact family and one will not. They will need to tell you which one does not belong and then create a multiplication or division equation using the 3 numbers that are in the fact family.
Step 4: Ask your child to start by choosing a strip of paper and trying to identify the number that does not belong.
Step 5: Once they have done one for practice, set a timer for 2 minutes and tell them to go through as many as they can in 2 minutes.
Step 6: After 2 minutes have passed, count the number of strips they completed. Now you can challenge your learner to try to beat their record and complete even more strips in 2 minutes this time. Play several rounds and see whether they are able to improve.
It may be beneficial to return to the game several days later to see whether your child can get an even better time. Another version of this game is to compete against someone else. Show the strip of paper to two people at the same time, and the first person to call out the number that isn’t part of the fact family wins that strip and earns one point. The person with the most points at the end of the game wins. | 524 | 1,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2023-06 | latest | en | 0.957964 |
http://www.baigacademy.in/2017/03/basic-mathematicsem.html | 1,529,844,194,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866937.79/warc/CC-MAIN-20180624121927-20180624141927-00638.warc.gz | 374,498,545 | 28,871 | ### BASIC MATHEMATICS (English Medium)
BASIC CONCEPT OF MATHEMATICS
--» Mathematics: The counting of of numbers is called Mathematics .
*Digit : To represent counting 0,1,2,3,4,5,6,7,8 & 9 symbols are used .it is called digits.
Note : digit is only unit.
*Number : Combining of two or more digits then we get numbers.
Ex: 23,567,8932.
Here
number "23" is form by digits "2" & "3".
Numeral : Symbolic representation of numbers is called "Numeral".
Ex:
I V X L C D M 1 5 10 50 100 500 1000
Place Value: Value of digit is contains according to its place.it is called place Value.
Ex: 1) In 210 the place Value of "2"is 200.because "2" is in the place of hundred.
2) In 120 the place Value of "2" is 20. because "2" is in the place of tens.
3) In 102 the place Value of "2" is only 2. because "2" is in the place of units.
Order of numbers:
1) Ascending order: Increasing order of numbers is called "ascending order".
Ex: 1<2<3<4<5<6.....
14<24<45<67<90
2) Descending order: Decreasing order of numbers is called descending order.
Ex: 7>6>5>4>3>2>1...
98>76>54>32
For more Click Below Link
click here | 363 | 1,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-26 | longest | en | 0.809687 |
https://dfpaintingredding.com/m%C3%A4n/n%C6%A1i-b%C3%A1n-gi%C3%A0y-th%E1%BB%83-thao-nike-air-max-2016-gi%C3%A1-r%E1%BA%BB-nh%E1%BA%A5t-th%C3%A1ng-112018-websosanh-vn | 1,669,802,552,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710734.75/warc/CC-MAIN-20221130092453-20221130122453-00513.warc.gz | 244,409,401 | 5,616 | # Math proof solver
In this blog post, we will show you how to work with Math proof solver. Let's try the best math solver.
## The Best Math proof solver
There is Math proof solver that can make the process much easier. Math home work can be a tricky thing for some students. Math is a difficult subject for some, so doing homework on it can be frustrating. Some tips to help with math homework are to get a tutor, practice at home, and try to understand the concepts. A tutor can help go over the material and help with any confusion. Also, practicing math problems at home can be helpful. Doing a few problems each night can help solidify the material. Lastly, trying to understand the concepts can be very helpful. If a student understands why they are doing a certain math problem, it can make the problem much easier. Math homework can be tough, but these tips can make it a little bit easier.
A 3 equation solver is a tool that can be used to solve three equations simultaneously. This can be helpful in a variety of situations, such as when trying to solve a system of linear equations or when attempting to find the roots of a polynomial equation. There are many different 3 equation solvers available online, and each one has its own advantages and disadvantages. However, all 3 equation solvers work by using a set of algorithms to find a solution that satisfies all three equations. In most cases, the 3 equation solver will return more than one possible solution, so it is important to examine each solution carefully before choosing the one that best meets your needs.
Solving a system of equations by graphing is a process of finding the points of intersection of the lines represented by the equations. This can be done by graphing both equations on the same coordinate plane and then finding the x and y coordinates of the points where the lines intersect. Solve system of equations by graphing can be used to solve problems in a variety of fields, including mathematics, physics, and engineering. In physics, for example, solving system of equations by graphing can be used to calculate the trajectory of a projectile. In engineering, it can be used to determine the load-bearing capacity of a structure. And in mathematics, it can be used to find the solutions to problems that cannot be solved using algebraic methods. Solve system of equations by graphing is a versatile tool that can be used to solve a wide variety of problems.
A complex number solver is a mathematical tool that can be used to solve equations that involve complex numbers. Complex numbers are numbers that have both a real and imaginary component, and they can be represented in the form a+bi, where a is the real component and b is the imaginary component. Many equations that involve variables raised to a power or roots cannot be solved using real numbers alone, but can be solved by adding or subtracting complex numbers. A complex number solver can be used to find the value of an unknown variable in such an equation. In addition, a complex number solver can also be used to graph complex numbers on a coordinate plane. This can be helpful in visualizing the solutions to equations or in understanding the behavior of complex numbers.
Math can be a difficult subject for many students, but it doesn't have to be. There are now many online math solvers that can help you work through problems and even show you the work that was done to get the answer. This can be a great way to check your work or to see how to do a problem if you are stuck. Some math solvers will even give you step-by-step instructions on how to solve a problem. All you need is a internet connection and you can get help with math anytime, anywhere. So if you are struggling with math, don't hesitate to use one of these online solvers. You might just find it makes math a lot easier.
## Math checker you can trust
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Yvonne Hayes
This is truly the best math app. It's so easy, instead of typing all the numbers and all, just scan and that's it! And, the solutions are 100% correct and very well explained. The steps are clear and it helped me to be better at math!! Install it everybody it will make your day a lot better.
Esther Mitchell
Math problem scanner Basic algebra questions How to solve for arc length App that does your math Math question help free | 931 | 4,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2022-49 | latest | en | 0.956301 |
https://mxnet.apache.org/versions/master/api/python/docs/api/np/random/generated/mxnet.np.random.chisquare.html | 1,716,436,878,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058588.75/warc/CC-MAIN-20240523015422-20240523045422-00065.warc.gz | 371,784,447 | 19,038 | # mxnet.np.random.chisquare¶
chisquare(df, size=None, dtype=None, device=None)
Draw samples from a chi-square distribution.
When df independent random variables, each with standard normal distributions (mean 0, variance 1), are squared and summed, the resulting distribution is chi-square (see Notes). This distribution is often used in hypothesis testing.
Parameters
• df (float or ndarray of floats) – Number of degrees of freedom, must be > 0.
• size (int or tuple of ints, optional) – Output shape. If the given shape is, e.g., (m, n, k), then m * n * k samples are drawn. If size is None (default), a single value is returned if df is a scalar. Otherwise, np.array(df).size samples are drawn.
• dtype ({'float16', 'float32', 'float64'}, optional) – Data type of output samples. Default is ‘float32’.
• device (Device, optional) – Device context of output. Default is current device.
Returns
out – Drawn samples from the parameterized chi-square distribution 1.
Return type
ndarray or scalar
Raises
ValueError – When df <= 0 or when an inappropriate size is given.
Notes
The variable obtained by summing the squares of df independent, standard normally distributed random variables:
$Q = \sum_{i=0}^{\mathtt{df}} X^2_i$
is chi-square distributed, denoted
$Q \sim \chi^2_k.$
The probability density function of the chi-squared distribution is
$p(x) = \frac{(1/2)^{k/2}}{\Gamma(k/2)} x^{k/2 - 1} e^{-x/2},$
where $$\Gamma$$ is the gamma function,
$\Gamma(x) = \int_0^{-\infty} t^{x - 1} e^{-t} dt.$
References
1
NIST “Engineering Statistics Handbook” https://www.itl.nist.gov/div898/handbook/eda/section3/eda3666.htm
Examples
>>> np.random.chisquare(2,4)
array([ 1.89920014, 9.00867716, 3.13710533, 5.62318272]) # random | 495 | 1,752 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-22 | latest | en | 0.69152 |
myrodow.currclickblog.com | 1,571,283,553,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672548.33/warc/CC-MAIN-20191017022259-20191017045759-00546.warc.gz | 619,831,728 | 5,366 | # Write a decimal that is greater than 100
This isbecause when turning a decimal into a percent, you must multiply itbywhich is 2 decimal places. Sometimes the decimal will repeat only after a certain number of places.
If required, the number is padded with zeros to its left to produce the number of digits given by the precision specifier. The short cut for repeated addition is multiplication.
Net Worth Calculator Balance sheet software: We have pages to copy at 11 cents per page. A set of predefined steps applicable to a class of problems that gives the correct result in every case when the steps are carried out correctly.
Multiplication and division within You can edit them any way you want to fit your practice. Calculate current snapshots of assets and liabilities held by everyone; net worth, balance sheet, asset allocations, and after-capital gains tax investment values, using up to assets and 20 liabilities.
If so, then it's easiest to use the site map. Dual RWR is the professional version, and has the most extensive draw-down analysis function available.
Care must be taken when identifying values across distinct primitive datatypes. The numerator represents a number of equal parts, and the denominator indicates how many of those parts make up a unit or a whole. How do you write Those settings are used to initialize the NumberFormatInfo object associated with the current thread culture, which provides values used to govern formatting.
Consider the number 0.
A tape diagram, number line diagram, or area model. It contains a script that suggests what to say for some slides.
For example, expressed as a percentage of the whole, 0. You just move the decimal point back two places and slap on the percentage point, and vice versa for the opposite. If a negative value was simply represented as the inverse of the positive number, 0 would have two representations: This is a basic marketing must if you practice asset allocation, use portfolio models, or want to start.
Size[ edit ] A typical book can be printed with zeros around pages with 50 lines per page and 50 zeros per line. There may be in the future.
This works vice versa for decimal to percent. The value spaces and the values therein are abstractions. Comprehensive Asset Allocation Software: You can make any changes you want to it. Other uses for fractions are to represent ratios and division.
How to write percent into a decimal? It treats all of one's investments in one pie, like they should be in the Real World; so you can control, evaluate, discuss, and implement the whole scenario without leaving anything out.
There are detailed instructions for all financial planning software, so anyone with basic Microsoft Excel experience can use them. Investment Portfolio Benchmarking Software: In pure mathematics[ edit ] In pure mathematicsthere are several notational methods for representing large numbers by which the magnitude of a googolplex could be represented, such as tetrationhyperoperationKnuth's up-arrow notationSteinhaus—Moser notationor Conway chained arrow notation.
A multi-digit number is expressed in expanded form when it is written as a sum of single-digit multiples of powers of ten. A list of length one containing a value V1 and an atomic value V2 are equal if and only if V1 is equal to V2.
Divide 60 by Familiarize yourself with the situation. It works great using the "bucket approach" to set up asset accounts according to how they pay out, so you can deplete one investment bucket before tapping another e. It considers all cash flows, income taxes, depreciation, basis, purchases, sales, improvements, commissions, will handle one refinancing, it has an underwater rental calculator, and more.
When configuring power tables, a command must be sent for each modulation type. There are only 8 power levels used in the CCx. Write the decimal as a percent?When a denominator has no prime factors other than 2's and 5's, we can find decimal notation by multiplying by 1. We multiply to get a denominator that is a power of ten, like 10,or Example 1: Find decimal notation for 3/5.
In decimal numbers greater than 1 (such as ), the fractional part of the number is expressed by the digits to the right of the decimal (with a value of in this case). can be written either as an improper fraction, /, or as a mixed number.
kcc1 Count to by ones and by tens. kcc2 Count forward beginning from a given number within the known sequence (instead of having to begin at 1). kcc3 Write numbers from 0 to Represent a number of objects with a written numeral (with 0 representing a count of no objects).
kcc4a When counting objects, say the number names in the standard order, pairing each object with one and only. Earlier this week I got tangled up doing a Roman Numeral conversion in my head.
So of course my second thought, right after “Doh!”, was “I bet I can write a SQL statement to do this for me next time.”. Standard numeric format strings are supported by: Some overloads of the ToString method of all numeric types.
For example, you can supply a numeric format string to the IntToString(String) and IntToString(String, IFormatProvider) methods. currclickblog.com composite formatting feature, which is used by some Write and WriteLine methods of the Console and StreamWriter classes, the currclickblog.com Decimal Greater Than And Less Than.
Showing top 8 worksheets in the category - Decimal Greater Than And Less Than. Some of the worksheets displayed are Comparing decimals, 14 3 percents greater than and less than 1, Decimal inequalities 1 directions greater than less than, 45 41 63 63 33 42, Comparing numbers up to 2 digit s1, Chapter 1 place value, Percent competency packet, Fourth grade.
Write a decimal that is greater than 100
Rated 4/5 based on 75 review | 1,224 | 5,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-43 | longest | en | 0.895173 |
https://in.mathworks.com/matlabcentral/cody/problems/44096-matrix-game-winner-takes-all/solutions/2117476 | 1,600,748,479,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400203096.42/warc/CC-MAIN-20200922031902-20200922061902-00474.warc.gz | 457,726,524 | 16,236 | Cody
# Problem 44096. Matrix game: Winner takes all
Solution 2117476
Submitted on 4 Feb 2020 by Asif Newaz
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
A = [5 4 3;1 5 6;3 0 9]; a_correct = [1 0 0;0 1 0;0 0 1]; assert(isequal(Winner_takes_all(A),a_correct))
a = 3×3 logical array 1 0 0 0 1 0 0 0 1
2 Pass
A = [0.9572 0.1419 0.7922 0.0357; 0.4854 0.4218 0.9595 0.8491; 0.8003 0.9157 0.6557 0.9340]; a_correct = [1 0 0 0; 0 0 1 0; 0 1 0 1]; assert(isequal(Winner_takes_all(A),a_correct))
a = 3×4 logical array 1 0 0 0 0 0 1 0 0 1 0 1 | 304 | 661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-40 | latest | en | 0.632758 |
http://ludumdare.com/compo/author/joecarrot/ | 1,571,443,352,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986685915.43/warc/CC-MAIN-20191018231153-20191019014653-00375.warc.gz | 125,503,770 | 13,452 | ## Entries
1,500,000Ludum Dare 25
## How a ladder could work.
Posted by
Saturday, April 26th, 2014 11:42 am
Okay, writing this here to get down some thoughts for later (after a shower), and also in case it might help someone else down the line.
A ladder is a bit more complex a scenario than I originally anticipated. It’s not enough to just say so and so object is ‘nearLadder’ and allow them to float around on the Y axis. There are a few different scenarios that happen when a player gets near a ladder, which all need their own behaviors.
## 1: Player walks next to bottom of ladder
In this most basic scenario, the player could have some kind of variable set that it is over a ladder and can climb. Being able to climb should allow the player to modify their Y location via input. The basic climbing stuff.
For this scenario, some basic climbing animation could play. In my case, I am setting the gravity scale for the player to 0, but that has it’s own issues.
## 3. Player walks next to top of ladder
And here is the issue. If you set the gravity scale to 0 when near a ladder, coming from the top makes the character float! Instead, I think a variable isClimbing could get set if the player presses up or down while over a ladder, otherwise, the player falls as normal.
## 4. Player climbs partially up ladder then jumps
This should set isClimbing to false, and since I’m using grounded checks to allow climbing, ladders should ground the player the whole way.
So a state machine would be something like:
Well that’s a start anyway.
## Day One progress.
Posted by
Saturday, April 26th, 2014 2:24 am
Gonna have to go to bed now. I think things are coming together well. Still need to get that ladder situation settled though :/
## I’m In, plus some art
Posted by
Saturday, April 26th, 2014 12:37 am
Using Unity 4.3, 2D Toolkit, Pixen, and Ableton Live
Here’s a jump blend tree!
And here’s a half-done tileset:
## With 4 hours left, I finally overcome my biggest hurdle. Unity’s Character Motor and double jumping
Posted by
Sunday, August 25th, 2013 2:15 pm
WHOOO!
WHOOOOOOO!
Here’s the new character motor if anyone ever wants it.
Full Disclosure: I had to bork the part that changes the angle of the jump if you jump off an angle ground, but I don’t personally use that much anyways.
http://pastebin.com/t3bvNvNd
## Probably going to have to quit
Posted by
Saturday, August 24th, 2013 8:49 pm
My greatest nemesis has proven impossible to beat again: making a character double jump in Unity while using character controller (not a rigidbody).
I don’t think it makes sense for me to completely rework the game just because I can’t get a feature to work, I guess not having a double jump upgrade is not going to make the game NO FUN, but when you are talking about metroidvanias, it’s kind of a dick move to not have a double jump.
Oh well, maybe I’ll try again in the morning. I’m too aggravated to do any good work.
Posted by
Saturday, August 24th, 2013 1:06 am
## The last change I made was a bit of text at the intro, at 8 on the dot.
Posted by
Sunday, December 16th, 2012 7:10 pm
Now when I export my textures are all messed up :(. Gonna fix that and up.
## For the first time, I feel the pressure mounting
Posted by
Sunday, December 16th, 2012 5:44 pm
I’ve only got 4 things left:
2. Better boss fight song
3. Start and end levels
4. Goat!
woof.
## Finally implemented the game mechanic. All that’s left is polish, bug fixes, and an end boss
Posted by
Sunday, December 16th, 2012 1:19 pm
## Food just arrived. Time to take an hour break
Posted by
Saturday, December 15th, 2012 10:17 pm
My jaw aches. I’m gonna watch some tv and try to think about anything other than games for a bit.
## Some music and sound fx
Posted by
Saturday, December 15th, 2012 7:37 pm
## Lose condition, textures
Posted by
Saturday, December 15th, 2012 5:31 pm
## Points added, guns now properly look at the mouse, and PARTICLES
Posted by
Saturday, December 15th, 2012 3:11 pm
http://dl.dropbox.com/u/96086/ld25/ld25_v_0.0.html
I can never get particles to not look junky
Via this handy little script I sort of figured out:
#pragma strict
function Update () {
var gunPlane : Plane = new Plane(Vector3.up, transform.position);
var hitdist : float = 0.0;
var ray = Camera.main.ScreenPointToRay(Input.mousePosition);
if(gunPlane.Raycast(ray, hitdist))
{
var targetPosition : Vector3 = ray.GetPoint(hitdist);
transform.LookAt(targetPosition);
}
}
## How not to do a screen shake
Posted by
Saturday, December 15th, 2012 1:53 pm
[cache: storing page] | 1,220 | 4,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-43 | latest | en | 0.948425 |
https://www.flightpedia.org/convert/21000-acre-foot-to-bushels.html | 1,722,807,743,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00751.warc.gz | 617,828,217 | 5,781 | # Convert 21,000 Acre Foot to Bushels
21,000 Acre Foot (ac⋅ft)
1 ac⋅ft = 33,916.0 bsh
=
712,235,896.15 Bushels (bsh)
1 bsh = 2.9e-05 ac⋅ft | 73 | 139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-33 | latest | en | 0.443121 |
https://betterlesson.com/lesson/556451/formative-assessment-matrices?from=mtp_lesson | 1,607,037,284,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141732835.81/warc/CC-MAIN-20201203220448-20201204010448-00537.warc.gz | 182,944,434 | 19,338 | # Formative Assessment: Matrices
1 teachers like this lesson
Print Lesson
## Objective
SWBAT use matrices and matrix operations to solve a variety of problems.
#### Big Idea
Assess your students' understanding of matrices with a quiz.
## Quiz Review
15 minutes
Gaining understanding of a Markov chain during yesterday's class may have been a difficult task for some students. I will give students about 5 minutes to go through the homework assignment with their table groups and I will circulate around the class to see if there are any pressing concerns.
After that, I will usually open up the floor for any questions about matrices before our quiz. Usually students will want some clarification about matrix multiplication and how you find the value of each entry. My students also want some more information about communication matrices and how interpret their results.
If no one in the class has any questions, here are some that I may pose to the class before they take their quiz.
1. Can you add any two matrices together? Multiply?
2. What is a communication matrix? What can it tell you?
3. What happens when you raise a communication matrix to a power? What can it tell you?
4. What would be a context where we would subtract two matrices? Add two matrices? Multiply two matrices?
## Quiz
35 minutes
Today's quiz is going to give you an idea of how well students understand the concepts of matrices before we move on to systems of equations and how those relate to matrices. This is a good stopping point to assess your students and to intervene with some if it is necessary. Some sample questions are given in this document.
I allow students to use graphing calculators on this quiz - but you can see from some of the questions that the focus is on interpreting and explication. I discuss more in the video below. | 372 | 1,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-50 | longest | en | 0.94911 |
http://www.enotes.com/homework-help/how-solve-this-chemistry-stoichiometry-sum-369804 | 1,448,929,444,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398464386.98/warc/CC-MAIN-20151124205424-00199-ip-10-71-132-137.ec2.internal.warc.gz | 413,393,659 | 10,240 | # How to solve this chemistry stoichiometry sum?A compund has O=61.32%,S=11.15%,H=4.88% and Zn=22.65%.The relative molecular mass of the compound is 287 a.m.u.Find the molecular formula of the...
Topic:
## Science
How to solve this chemistry stoichiometry sum?
A compund has O=61.32%,S=11.15%,H=4.88% and Zn=22.65%.The relative molecular mass of the compound is 287 a.m.u.Find the molecular formula of the compound,assuming that all the hydrogen is present as water of crystallisation.
Posted on (Answer #1)
Lets assume the compound to be `Zn_xS_yO_z*nH_2O` ` `
Let us say we have 100g of the compound.
Then;
Mass of S = 11.15g
Mass of O = 61.32g
Mass of H = 4.88g
Mass of Zn = 22.65g
Molic weights;
S = 32g/mol
O = 16g/mol
H = 1g/mol
Zn = 65g/mol
It is given that all the hydrogen in the compound is retain as water.
H moles present `= 4.88/1 = 4.88`
one water crystal is formed by two H moles.
Amount of water crystals moles present `= 4.88/2 = 2.44`
Mole ratio of O:H in water `= 1:2`
Amount of O moles used for water crystals `= 2.44`
Total O moles present in 100g of compound `= 61.32/16 = 3.83`
O moles in compound not as water `= 3.83-2.44 = 1.39`
Number of Zn moles in compound `= 22.65/65 = 0.348 mol`
Number of S moles in compound `=11.15/32 = 0.348 mol`
Mole ratio:
`x:y:z = 0.348:0.348:1.39`
Usually we keep the ratios in whole numbers.
`x:y:z = 1:1:3.99`
Approximately we can say;
`x:y:Z = 1:1:4`
So the compound will be;
`ZnSO_4*nH_2O`
It is given that the molar mass of the compound is 287 a.m.u.
`65+32+16*4+n(2+16) = 287`
` n = 7`
So the formula of the compound is;
`ZnSO_4*7H_2O`
Sources:
We’ve answered 288,409 questions. We can answer yours, too. | 609 | 1,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2015-48 | longest | en | 0.900951 |
http://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-5-section-5-2-negative-exponents-and-scienti-c-notation-practice-page-349/9 | 1,524,237,227,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125938462.12/warc/CC-MAIN-20180420135859-20180420155859-00165.warc.gz | 413,806,042 | 13,685 | ## Algebra: A Combined Approach (4th Edition)
$27x^{12}$
$\frac{(3x^{5})^{3}x}{x^{4}}=\frac{3^{3}\times(x^{5})^{3}\times x}{x^{4}}=\frac{27\times x^{5\times3}\times x}{x^{4}}=\frac{27\times x^{15}\times x}{x^{4}}=\frac{27x^{15+1}}{x^{4}}=\frac{27x^{16}}{x^{4}}=27x^{16-4}=27x^{12}$ | 136 | 282 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-17 | latest | en | 0.336694 |
realworldfsharp.org | 1,575,564,407,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540481281.1/warc/CC-MAIN-20191205164243-20191205192243-00401.warc.gz | 529,568,797 | 34,161 | # Solving Sudoku with F#
One of my more recent additions to the life of {Redacted} is a thing we call the Thursday Kata. Every Thursday, I pop a message into our slack channel, with a challenge for the week. Dev team members noodle over the problem, slack a gist with their solution, and then on Wednesday, during our weekly team meeting, we vote on a submitted solution and have the author present what they did, how they did it, and what they learned.
It’s a lot of fun, and is gaining quite a bit of traction within the development org. Developers are walking around discussing the kata, and strategies for solving it, and it interrupts what can be some of the more mundane day-to-day tasks.
The tasks aren’t meant to take more than an hour or two. A more recent example was a Sudoku solver. Devs were challenged to come up with a solver, in whatever language they wanted to look at.
Unfortunately, as the emcee of this event, my own solutions are off limits for voting… but that’s when I remembered I have readers of this blog (even if it is only myself and my mom. Hi Mom!) Here’s the gist.
Basically, I had two ideas. First, be able to parse and render a puzzle grid, and second, solve the grid using a simple recursive algorithm.
I started with a few types that would help me with the domain.
1. Sector. A type that represented the 9 blocks (tic-tac-toe blocks) of a Sudoku grid.
2. Position. A type that represented an individual location of a Sudoku grid. The position was made of a Rank and File, which indicated the X and Y coordinate (in similar terms to chess, a game I’m more familiar with.)
3. Grid. A type simply mapping a Position (Rank and File) with integers.
The first thing I did was create a simple set of all available positions, and second I created a Map of positions with a unique set of positions to “check”. For example: the position at Rank.One and File.A (the upper left square), would be mapped to a distinct set of positions that included:
• All positions in Rank.One (A1, B1, C1… I1)
• All positions in File.A (A1, A2, A3… A9)
• All positions in the same TopLeft sector (A1, B1, C1, A2, B2, C2, A3, B2, C3)
The next three functions were some simple rendering / parsing functions, enabling me to quickly test my solver, by entering grids as simple strings.
``` let easy = parseGrid "5,-,-,-,1,-,-,-,4,2,7,4,-,-,-,6,-,-,-,8,-,9,-,4,-,-,-,8,1,-,4,6,-,3,-,2,-,-,2,-,3,-,1,-,-,7,-,6,-,9,1,-,5,8,-,-,-,5,-,3,-,1,-,-,-,5,-,-,-,9,2,7,1,-,-,-,2,-,-,-,3";; val easy : Grid option = Some (map [({Rank = One; File = A;}, 5; .... ```
```let rendered = renderGrid e;; val rendered : string = " 5 - - | - 1 - | - - 4 2 7 4 | - - - | 6 - - - 8 - | 9 - 4 | - - - ********************* 8 1 - | 4 6 - | 3 - 2 - - 2 | - 3 - | 1 - - 7 - 6 | - 9 1 | - 5 8 ********************* - - - | 5 - 3 | - 1 - - - 5 | - - - | 9 2 7 1 - - | - 2 - | - - 3 "```
Ah UI code… my favorite 🙂
I created a few helper functions to make some logical tests work as well.
• getAvailableValues : Grid -> position -> (Position * int[])
This function took a grid and a position, and returned the array of possible number values that were available for the position in question.
• isNotSolvable : (‘a * ‘b []) list -> bool
This was a simple test to look at a list of things, and if there were any items that came in with an empty array (as the b value).
• isSolved : (‘a * ‘b []) list -> bool
This simple test ensures that the incoming list, the max length of the b array is 1 value.
All together, this made the solve method easy to write.
Solve : Grid -> Map<Position, int> option.
If solve returned a None value, no solution was found. Otherwise, it would return Some solution (see what I did there.)
In that function, it defined a recursive function called ‘createSolution’, taking a grid g.
That function would first iterate through allGrid positions, and then create a list of Position * int [], mapping over the getAvailableValues function and the grid passed in. That gave us a list of positions, and all positions potential answers. It caled the isNotSolvable method, to test if ANY of the positions couldn’t be solved (because they had no avialable options in the array.) If so, the function returned None. Otherwise, it checked if the puzzle was solved, using the isSolved function above, and if so, it created a Grid map, and returned Some with that grid.
Finally, the meat of the function assumed that at least one of the Grid squares had more than one possible answer. It then did a simple query for the square with the smallest number of possible answers, and then simply called ‘createSolution’ recursively with one of those possible answers set for the square in question.
What ends up here is a depth first solution, but one that always returns (eventually.)
Here are some simple results I got. Fire up FSI and see what you get! Have a good puzzle everyone!
``` // easy (solve time 0.167 seconds) let easy = Option.get (parseGrid "5,-,-,-,1,-,-,-,4,2,7,4,-,-,-,6,-,-,-,8,-,9,-,4,-,-,-,8,1,-,4,6,-,3,-,2,-,-,2,-,3,-,1,-,-,7,-,6,-,9,1,-,5,8,-,-,-,5,-,3,-,1,-,-,-,5,-,-,-,9,2,7,1,-,-,-,2,-,-,-,3");;```
``` // hard (6.401 seconds) (38x easy) let hard = Option.get (parseGrid "-,-,5,-,-,-,9,-,-,-,-,4,6,9,-,1,-,-,7,9,-,-,-,-,-,-,-,-,1,-,2,-,-,-,-,3,-,7,-,-,-,6,-,8,-,-,-,-,-,1,4,6,-,2,2,3,-,-,-,8,-,-,-,-,-,-,-,5,-,-,-,7,-,-,-,4,-,3,-,1,-");;```
``` // extreme (20 minutes, 50.464 seconds) (7,487x easy) let extreme = Option.get (parseGrid "8,-,-,-,-,-,-,-,-,-,-,3,6,-,-,-,-,-,-,7,-,-,9,-,2,-,-,-,5,-,-,-,7,-,-,-,-,-,-,-,4,5,7,-,-,-,-,-,1,-,-,-,3,-,-,-,1,-,-,-,-,6,8,-,-,8,5,-,-,-,1,-,-,9,-,-,-,-,4,-,-");;```
# Sisters
“I believe that children are our future. Teach them well, and let them lead the way…” – Whitney Houston
After an evening of library time, Daddy’s Crossfit, and a Lacrosse pickup, my two girls, Zoe (the oldest) and Lydia were tired and hungry. During a quick evening meal of leftover lasagna, I popped open the laptop to work on a quick kata, when my oldest asked “Whatcha doin’?”
Zoe’s interest in my work is rare, so I happily showed her some of my F# test code. She looked around. Even asked a few questions.
So I popped open the FSI and showed her how stuff works.
```Microsoft (R) F# Interactive version 10.2.3 for F# 4.5
For help type #help;;
> let zoe = "awesome";;
val zoe : string = "awesome"
> zoe = "not awesome";;
val it : bool = false
>```
She was pretty happy that Zoe != “not awesome”.
I gave her her own FSI to try out, and I happily present to you, the very first code in what I’m sure is a very long lucrative functional programming career.
```Microsoft (R) F# Interactive version 10.2.3 for F# 4.5
For help type #help;;
> let lydia = "annoying";;
val lydia : string = "annoying"
>```
I guess sisters will be sisters.
# What do you mean it has to be done in a half hour?
Sometimes, you can see problems coming.
A vendor of ours provides security information monthly in a series of text files. Typically, those text files are in the 10s of megabytes large. Those files get viewed cleaned up, and imported into our system by portfolio managers monthly.
The system to do the loading is homegrown, not ETL based (someone thought users should ‘upload’ the files) into an ASP.NET MVC app, and then render UIs with the data. :eyeroll:
Naturally, as our data requirements got larger the files got bigger. And bigger.
This last month, the system eventually exploded. A 1 GB text file proved to be too much for our cute little “not-quite-an-ETL” tool. The portfolio managers needed a solution, and in order to quickly get ’em one, I present (with relevant bits redacted) a 30 minute “lets play with F#” script to solve the problem.
Nothing too fancy. The function simply takes a file path, creates a couple of functions to help it along, and then loads up the file and splits it up into distinct new files. Whole thing took 30 minutes to do. Yes, the complexity is O(n-squared), but when you’ve got panicked users, and all of a half hour to hit it, getting it ‘working’ first is the best way to go.
# Greensleeves
Yesterday afternoon, I had a bit of time on my hands, and a report that someone desperately needed redone. The original author was a support engineer who was trying to learn Ruby and didn’t realize that Ruby hash-maps are case sensitive. In the interest of time, I corrected the issue by downcasing all the keys, and reran the script (which ran successfully, but inelegantly), but I looked at the problem, and once again, found a perfectly lovely use case for the language I love so much, F#.
First off, some of the basics.
```open System
// these may not be strictly necessary, but they do help describe
// precisely what data I'm dealing with
module UpperCaseString =
type UpperCaseString = UpperCaseString of string
let private upper (a:string) =
a.ToUpper()
let create (a:string) =
if String.IsNullOrEmpty(a) then None
else Some (UpperCaseString (upper a))
let value (UpperCaseString s) =
s
open UpperCaseString
module FloatingDecimalBetweenZeroAndOne =
type FloatingDecimalBetweenZeroAndOne = FloatingDecimalBetweenZeroAndOne of decimal
let create (a:decimal) =
if a > Decimal.One || a < Decimal.Zero then None else Some (FloatingDecimalBetweenZeroAndOne a) let value (FloatingDecimalBetweenZeroAndOne f) = f let make a = (create a |> Option.get)
open FloatingDecimalBetweenZeroAndOne
```
Generally, I’ll do this for most business application coding nowadays. Creating a simple type that encapsulates correctness from the get-go (including basic validation) makes it so I don’t have to try to figure out weird results. I know the sorts of things I’m dealing with, and I deal with them from the beginning. Also, given the fundamental problem of the original script I had (the case sensitive hash-maps), ensuring I had a type that enforced upper-casing seemed worthwhile.
Next, the domain.
```type Model = Model of UpperCaseString
with static member Value (Model (UpperCaseString s)) = s
type Ticker = Ticker of UpperCaseString
with static member Value (Ticker (UpperCaseString s)) = s
type Holding = { Ticker : Ticker; Allocation : FloatingDecimalBetweenZeroAndOne }
type ModelAllocations = { Model : Model; Holdings : Holding list }
type ModelWeight = { Model : Model; Weight : FloatingDecimalBetweenZeroAndOne }
type SleevedHolding = { Ticker : Ticker; Model : Model; Allocation : FloatingDecimalBetweenZeroAndOne }
with override x.ToString () = sprintf "%s,%s,%5M" (Ticker.Value x.Ticker) (Model.Value x.Model) (FloatingDecimalBetweenZeroAndOne.value x.Allocation)
// This is the simplest description of what we're doing.
// taking a list of holdings, a list of models and their holdings
// a list of weights to apply to the model, and returning a set of 'sleeved' holdings.
type SleeveProcess = Holding list -> ModelAllocations list -> ModelWeight list -> SleevedHolding list
```
The problem the original script was attempting to solve was to assign an appropriate ‘Model’ to a holding. A Model is, for sake of brevity, simply an identity (representing the “name” of a standard financial benchmark (like the S&P 500), and a Ticker an identity representing a stock ticker.
A Holding represents an individual stock in a of a portfolio, with the Allocation representing the percentage of the portfolio. ModelAllocations represent a model, and the list of holdings it has (what stocks are in the S&P 500, etc.) ModelWeight represents the approximate weight a portfolio is associated to a given model or benchmark.
E.G. S&P 500 -> 50% , and Russell 3000 -> 50%.
```let makeModel str =
UpperCaseString.create str |> Option.get |> Model
let makeTicker str =
UpperCaseString.create str |> Option.get |> Ticker
let holding str all =
{ Ticker = makeTicker str; Allocation = make all }
let makeSleeve t m a = { Ticker = t; Model = m; Allocation = a }
let modelWeight str a = { Model = makeModel str; Weight = make a }
let modelAllocation str h = { Model = makeModel str; Holdings = h }
```
One of the problems of heavy use of custom types is that you’ll typically want ‘shorthand’ functions for creating the types as necessary. It can be a pain, but difficult to explain results after the fact is worse.
Finally, the meat of it.
```let sleeveHoldings holdings models weights =
let findHolding t =
models |>
List.collect (fun n -> n.Holdings
|> List.filter (fun n -> n.Ticker = t)
|> List.map (fun x -> (n.Model, x.Allocation)))
let sleeve h =
let t = h.Ticker
let m = findHolding t // a list of models that hold the security
match m with
| [] ->
weights |>
List.map (fun w ->
let weightTimesAllocation = (value w.Weight * value h.Allocation)
makeSleeve t w.Model (make weightTimesAllocation))
| [(x,_)] -> [makeSleeve t x h.Allocation]
| xs ->
let weightsMap = weights |> List.map (fun n -> (n.Model, value n.Weight)) |> Map.ofList
let total = xs |> List.sumBy (fun (m, w) -> weightsMap.[m] * value (w))
xs |> List.map (fun (m, w) ->
let weightTimesAllocationOverTotal = (value w * weightsMap.[m]) / total
makeSleeve t m (make (weightTimesAllocationOverTotal * value h.Allocation)))
holdings |> List.collect sleeve
```
First, we define a function to find the models with a given holding, and return that model and the allocation. Then we define a function to sleeve an individual holding. Finally, we call that function over the list of holdings passed in.
A simple set of tests:
```let testHoldings = [
holding "A" 0.20m;
holding "B" 0.20m;
holding "C" 0.20m;
holding "D" 0.20m;
holding "E" 0.20m; ]
let testWeights = [
modelWeight "MODEL1" 0.5m;
modelWeight "MODEL2" 0.3m;
modelWeight "MODEL3" 0.2m;
]
let testModelAllocations = [
modelAllocation "MODEL1" [
holding "A" 1.0m;
];
modelAllocation "MODEL2" [
holding "A" 0.5m;
holding "B" 0.3m;
holding "D" 0.2m;
];
modelAllocation "MODEL3" [
holding "B" 0.5m;
holding "C" 0.3m;
holding "D" 0.2m;
]
]
sleeveHoldings testHoldings testModelAllocations testWeights |> List.map (fun m -> m.ToString ())
val it : string list =
["A,MODEL1,0.1538461538461538461538461538";
"A,MODEL2,0.0461538461538461538461538462";
"B,MODEL2,0.0947368421052631578947368421";
"B,MODEL3,0.1052631578947368421052631579"; "C,MODEL3, 0.20";
"D,MODEL2,0.120"; "D,MODEL3,0.080"; "E,MODEL1,0.100"; "E,MODEL2,0.060";
"E,MODEL3,0.040"]
```
Explaining the image: At the financial services firm I work for, the concept of attributing holdings is called ‘sleeving.’ The photo is from an original 19th century (1872) painting by Dante Rossetti, of “Lady Greensleeves.”
# Poker Hands and Discriminated Unions
Discriminated union types are pretty damned powerful, especially when modeling business domains. A business domain I enjoy immensely is Poker. The below Fun Friday post examples modeling poker hands.
First thing, I started with the below enums to start modeling out the cards themselves.
```type Suit = | Clubs = 0 | Diamonds = 1 | Hearts = 2 | Spades = 3
type Rank = | Two = 2 | Three = 3 | Four = 4 | Five = 5
| Six = 6 | Seven = 7 | Eight = 8 | Nine = 9 | Ten = 10
| Jack = 11 | Queen = 12 | King = 13 | Ace = 14
type Card = { Rank : Rank; Suit: Suit }```
This makes the problem space fairly easy to put together, but preferred to have a ‘deck’ of cards to deal with. Note, because I used standard enumerations for the Suit and Ranks, I was able to write a quick ‘toList’ function to grab enumeration values and stick them in a standard F# list.
```let toList = [for i in System.Enum.GetValues(typedefof)
do yield i] |>
List.map (fun n -> downcast n : 'a )
let deck = List.allPairs toList toList |>
List.map (fun (s,r) -> { Suit = s; Rank = r; })```
OK. So we have a deck to deal with. Now, to deal with Poker Hands.
I chose to model the Poker Hand with a discriminated union type mainly to example a bounded set scenario. The fact is, for my game here (or my ‘business scenario’), I need to treat ‘poker hands’ as singular types, but the individual types differ quite a lot. In practice, they became distinctly organized sets of ‘Rank’ objects.
```type PokerHand =
| StraightFlush of HighestCard : Rank
| FullHouse of TripsRank : Rank * PairRank : Rank
| Flush of Card1 : Rank * Card2 : Rank * Card3 : Rank * Card4 : Rank * Card5 : Rank
| Straight of HighestCard : Rank
| Trips of TripsRank : Rank * Kickers : (Rank * Rank)
| TwoPair of HighPairRank : Rank * SecondPairRank : Rank * Kicker: Rank
| Pair of PairRank : Rank * Kickers : (Rank * Rank * Rank)
| HighCard of Kickers : (Rank * Rank * Rank * Rank * Rank)```
A lot of the reasons I chose to model out the Poker Hands individually like this was to make comparing poker hands relatively easy to reason about. I’ll assume you know how poker hands compare, but if not, click here.
Comparing two poker hands is a fairly simple operation. First, you start with the ranks of the hand types themselves, and if one hand is ‘bigger’ than the other, that’s your winner. A function-style expression works with that nicely, as we can bind directly to the value. This style expression takes an implied parameter of type PokerHand (you’ll see usage below.)
```let handRank = function
| StraightFlush _ -> 8 | Quads _ -> 7 | FullHouse _ -> 6
| Flush _ -> 5 | Straight _ -> 4 | Trips _ -> 3
| TwoPair _ -> 2 | Pair _ -> 1 | HighCard _ -> 0
```
Second, if the hands are the same, you need to be able to compare the ranks of the cards, in order of relevance to the hand. It is this relevance that I decided to model in the PokerHand union type above. Still, comparing ranks was easy.
```let compareRanks x y =
if x > y then 1
else if y > x then -1
else 0```
So we have the baseline functions all set here. Putting it all together, we end up with the following:
```let compare hand1 hand2 =
let handRank = function
| StraightFlush _ -> 8 | Quads _ -> 7 | FullHouse _ -> 6 | Flush _ -> 5 | Straight _ -> 4
| Trips _ -> 3 | TwoPair _ -> 2 | Pair _ -> 1 | HighCard _ -> 0
let compareRanks x y =
if x > y then 1
else if y > x then -1
else 0
match hand1, hand2 with
| (x, y)
when (handRank x) - (handRank y) 0 -> Some (sign ((handRank x) - (handRank y)))
| (FullHouse (c, _), FullHouse (c2, _))
| (Flush (c,_, _ ,_ ,_), Flush (c2,_, _ ,_ ,_)) | (Flush (_,c, _ ,_ ,_), Flush (_,c2, _ ,_ ,_))
| (Flush (_,_,c,_,_), Flush (_,_,c2,_,_)) | (Flush (_,_, _ ,c ,_), Flush (_,_, _ ,c2,_))
| (Trips (c,_), Trips (c2, _)) | (Trips (_,(c,_)), Trips(_,(c2,_)))
| (TwoPair (c, _, _), TwoPair (c2, _ , _)) | (TwoPair (_, c, _), TwoPair (_, c2 , _))
| (Pair (c, _), Pair (c2, _)) | (Pair (_, (c, _, _)), Pair (_, (c2, _, _)))
| (Pair (_, (_, c, _)), Pair (_, (_, c2, _)))
| (HighCard (c, _, _, _, _), HighCard (c2, _, _, _, _)) | (HighCard (_, c, _, _, _), HighCard (_, c2, _, _, _))
| (HighCard (_, _, c, _, _), HighCard (_, _, c2, _, _)) | (HighCard (_, _, _, c, _), HighCard (_, _, _, c2, _))
when compareRanks c c2 0
-> Some (compareRanks c c2)
| (StraightFlush c, StraightFlush c2)
| (Straight c, Straight c2)
| (FullHouse (_, c), FullHouse (_, c2))
| (Flush (_,_, _ ,_ ,c), Flush (_,_, _ ,_ ,c2))
| (Trips (_,(_,c)), Trips(_,(_,c2)))
| (TwoPair (_, _, c), TwoPair (_, _, c2))
| (Pair (_, (_, _, c)), Pair (_, (_, _, c2)))
| (HighCard (_, _, _, _, c), HighCard (_, _, _, _, c2))
-> Some (compareRanks c c2)
| _ -> None```
I will admit, it’s got a weighty match expression, but match expressions can be broken down by their guard clauses, so altogether, it can be reasoned about as:
match hand1, hand2 with
| when the handRanks are different, return the bigger one.
| when the card ranks are different in the relevant order for the hand type in question, return the bigger one.
| when one relevant card rank remains, return the comparison between that last card rank.
| when passed anything else, return None.
Take comparing two Two Pair hands.
```TwoPair (Rank.King, Rank.Four, Rank.Eight)
TwoPair (Rank.King, Rank.Six, Rank.Seven)```
The first relevant rank here compares as 0 (King vs King), but the second relevant rank is non-zero (4 vs 6) so the second hand wins.
Finally, creating PokerHand instances requires 5 distinct cards.
```let getHand card1 card2 card3 card4 card5 =
let ranks = [card1.Rank
card2.Rank
card3.Rank
card4.Rank
card5.Rank] |>
List.groupBy id |>
List.sortByDescending (fun (m,n) -> (List.length n) * 100 + (int) m);
match List.length ranks with
| 4 -> // pair
let card = Pair (fst ranks.Head, (fst ranks.[1], fst ranks.[2], fst ranks.[3]))
Some card
| 2 -> // quads or fullhouse
if(List.length (snd ranks.Head) = 4) then
Some card
else
let card = FullHouse (fst ranks.Head, fst ranks.[1])
Some card
| 3 -> // trips or twopair
if(List.length (snd ranks.Head) = 3) then
let card = Trips (fst ranks.Head, (fst ranks.[1], fst ranks.[2]))
Some card
else
let card = TwoPair (fst ranks.Head, fst ranks.[1], fst ranks.[2])
Some card
| 5 -> // a flush or straight flush
let r = [card1.Rank; card2.Rank; card3.Rank; card4.Rank; card5.Rank]
|> List.sortByDescending id
let suits = [card1.Suit
card2.Suit
card3.Suit
card4.Suit
card5.Suit] |>
List.distinct
match (List.length suits, r) with
| (1, Rank.Ace::Rank.Five::_) -> Some (StraightFlush Rank.Five)
| (1, x::xs) when x - (List.last xs) = Rank.Four -> Some (StraightFlush x)
| (1, _) -> Some (Flush (r.[0], r.[1], r.[2], r.[3], r.[4]))
| (_, Rank.Ace::Rank.Five::_) -> Some (Straight Rank.Five)
| (_, x::xs) when x - (List.last xs) = Rank.Four -> Some (Straight x)
| _ -> Some (HighCard (r.[0], r.[1], r.[2], r.[3], r.[4]))
| _ ->
None
```
This method works fairly simply. We take the ranks of the incoming cards, group the like ranks together, then sorting by the rank count (times 100) plus the rank value. (Hence, 2 Kings, and 2 sixes and a 4 will end up as 213, 206, and 4, respectively.)
Then, by matching against the number of a distinct ranks, we can create simple logic to determine the hand type. When 5 cards have 4 distinct ranks, the hand must be a pair, so we simply set the values for the PokerHand instance accordingly. When there are 2 distinct ranks, the hand must be either a FullHouse (3 of one rank, 2 of the other), or Quads (4 of one rank, 1 of the other.) Trips and TwoPair will both have 3 distinct ranks. If there are 5 distinct ranks, we need to check for a bit more. In the case of 5 distinct ranks, we need to sort them in order, and get a count of distinct suits. If there is 1 suit, then we’re dealing with a Flush or a StraightFlush. Straights are 5 cards in a row (rank order), OR Ace, Two, Three, Four, Five. If we have more than one suit, we are dealing with a HighCard hand (the most common hand), or a Straight. If we get any other number of distinct ranks, we have an invalid set of arguments, so we return the option value None.
The last function will give us a random hand to play with.
```let getRandomHand() =
let r = System.Random();
let rec makeHand length list =
let m = r.Next(0, 52)
if (List.contains m list) then
makeHand length list
else
let newList = m::list
if (List.length newList = length) then
newList
else
makeHand length newList
let idx = makeHand 5 []
getHand deck.[idx.[0]] deck.[idx.[1]] deck.[idx.[2]] deck.[idx.[3]] deck.[idx.[4]]```
It’s getting late. Time to head to the card room, but I encourage you to try this out in FSI, and think about how discriminated unions could make your OWN business modeling a little easier. See ya!
# I’m still alive
December came and went. January came and went. February came and went…
Yes, I’m alive. I’ve been going through a whirlwind this past few months. Here’s a few things that have gone on.
#### A promotion!
Yep, the wonderful folks I work for at {redacted} have given me the grand title of “Principal Software Developer.”
Having worked here for 5 years now, I am officially to blame for most all the code here. I’ve been here long enough to own it. It’s my bad, folks, but it’ll get better, I promise.
#### Softball Season
I help run a local girls’ slowpitch league, and have been coaching and umpiring for 5 years now. Softball season starts around the beginning of March, but the work up to the season starting is substantial.
Coaching and helping out my league is meaningful for me. I don’t have better words for it. It’s just an amazing thing that I love to do.
#### Conferences
I was at Agile Open Northwest and a local programmer introduced a concept he was excited about called “tiny objects”, wherein he used C# to make objects that had:
1. No state.
2. Content that was immutable.
3. Just methods.
Finally, he had constructed a few rules as well, to make mocking and testing easier. Specifically, there was to be no static methods on an object.
```var example = "some string";
var stringUpper = new StringUppercaser(example).Result;
Assert.That(stringUpper, Is.EqualTo("SOME STRING"));```
It. Was. Surreal.
This developer was exactly where I had been. Frustrated by state logic. Completely tired of dealing with new ways to worry about async and threading issues. He just had not heard of or even considered that the language itself was part of the problem.
That is simply why this blog exists.
Immediately, I setup a quick session in the conference to go over how functional programming works, and where to learn about it. Hopefully, I helped some folks see the light, but I doubt it, as I didn’t have a great talk already ‘set up.’ My note to self after that session was that putting together a half-hour talk on What and Why F# is just something I should have in my back pocket.
So there it is. With a new promotion and getting my team ready to go, we are knee deep in a hundred different maintenance projects. Too many projects, not enough F#!
Thanks to the folks at F# Weekly finding my Fishful of Dollars post!
# A Fishful of Dollars
First of all, allow me to apologize for a solid month of failed updates. The world of coaching my son’s football team and bugbashing caused me to want to spend my time at home, asleep, rather than writing about my various misadventures in F#.
To update on the application rewrite, upcoming tax legislation has my firm in quite a tizzy, and is requiring an all-hands on deck approach for the next few weeks. Assuming the bill passes, we may be changing a lot of software, quite quickly, and absolutely none of that has anything to do with an old poorly-written report. The upcoming business need will be drastic, and preparing for that change is important to do.
That said, let’s play around a bit in F#!
I was watching an old episode of Futurama, when a question popped into my mind. If Fry left a bank account with \$0.93 in it, at 2.25% interest, would he actually be sitting on a cool \$4.3 billion after being frozen for 1000 years? Time to find out!
`val fry : float = 4283508450.0`
Looks like the math works… but man oh man does it take a while to get there. Using FSharp Charting and iterating over each year…
```let fryOverTime = [ 1.0 .. 1000.0 ]
|> List.map (fun t -> compound { Compounding = TimesPerYear 1.0
Rate = 0.0225
Principal = 0.93
TermInYears = t } )
Chart.Line(fryOverTime
, Name="Balance over Years"
, XTitle = "Years"
, YTitle = "Balance in Dollars")
.WithXAxis(Min=0.0, Max=1000.0)
// because it's almost impossible to see any change until year 650 or so...
Chart.Line(fryOverTime
, Name="Balance over Years"
, XTitle = "Years"
, YTitle = "Balance in Dollars")
.WithXAxis(Min=0.0, Max=1000.0)
.WithYAxis(Log = true)
```
It looks like Fry’s bank account doesn’t get interesting until about 300 years or so. Still, compound interest is a wonder, despite whether or not Albert Einstein said so.
We have a reporting application at {Redacted}, one I’ve spent more than a few hours maintaining. A colleague of mine and I (the junior developer I spoke of in a prior post) met up with the business owner of the reporting application, to talk about rebuilding the application in more functional way.
This application is not a terribly complicated one. In short, it takes data from multiple data sources, and re-aggregates it into a series of reports. It has however, had multiple developers come and “peek in”, drop code into it, and leave. In the current C# implementation, it’s 5 distinct assemblies. There are 8 test assemblies to go with it. It has a lot of business value riding on it being robust and correct, so the tests do seem warranted.
A word to the wise though… simple things that are more complicated than they should be AND that have a ton of tests are REFACTORING GOLD. Nothing feels better than taking something complex and unwinding it to it’s core essentials, and nothing feels better than doing it in F#!
We took a page out of For Fun and Profit’s DDD page and focused for nearly an hour on the objects and processes involved in this report. Naturally, the process seemed extremely solvable in a functional way.
Unfortunately, due to time constraints on the rest of the day we were only able to start with some very high level type definitions, but those type definitions described our problem in such a way that the business owner was able to see and understand.
We did the whole thing with an instance of VS Code and Ionide, and had types describing the objects and functions involved, all with just a simple “domain” setup in the process. Did we implement anything particularly? Well, kinda yeah, this is perfectly compile-able F# code, which as opposed to Gherkin or some other spec-based “code”, does not need a secondary interpreter.
That’s the crux of this application and function. I think we may be able to get this down to a few pages of code… as opposed to the novella you’d call it now. #FeelingHopeful
# Fractions Cont’d, Factoring a Number
As we last spoke about Fractions, we had a simple task after getting a Fraction object. We wanted to reduce them. I was taught to rit educe a fraction by factorizing the numerator and denominator, and then crossing out the common factors. Then, multiplying the numbers together to get the reduced value. E.G.
``` 60 2 * 2 * 3 * 5 2 * 2 * 3 * 5 2
---- = ----------------- = --------------- = ---
90 2 * 3 * 3 * 5 2 * 3 * 3 * 5 3
```
The example above is a simple one.
So the question is, how do we best factor a number?
First, we define the mechanisms we require. We need a way to define that a number is a factor of another.
```let isFactor i pf =
i % pf = 0I```
What does this method buy us? First, it gives us a simple True/False flag defining whether or not a number is a factor of another. Namely, if the module of i and pf is 0I (specifically, zero, as a System.Numerics.BigInteger), then we know that pf is a factor of i.
Next, we define a recursive function that takes two parameters, start and i. The idea of this function is to return a sequence of numbers from start to i, counting by twos. The start value is an option type, so if not present, it assumes 2 is a valid option, and starts the sequence there, then goes with every odd number. Note: I’m accepting as a given that this will do some checks against numbers it’s unnecessary to do that against.
Finally, I define my internal recursive loop to accumulate the resulting found factors. An interesting thing we’ve got here is some issues of partial application for function calls. The “findFunc” function is a partially applied isFactor call, with the i value already passed in. The reason I did this was to make the Seq.tryFind call easy to use, because the Seq.tryFind call is shaped like this:
`('a -> bool) -> seq 'a -> Option 'a`
But to get to my “bool”, I needed more than just a single ‘a parameter. That’s why partial application was so valuable here! Instead of changing the function signature to meet my needs, I simply made a quick, easy to reference function by supplying some arguments ahead of time.
The rest of the function is fairly apparent. The loop recurses over the sequence, finding factors and then appending them to as the head of the accumulator (f::acc), until the Seq.tryFind call eventually returns None, and the accumulated list of factors is returned.
The last bits of the function just wrap up return values. In this case, we check for negative numbers, or a 0I being passed in. Negative numbers are fairly easy to factor as they only require normal factorization of the inverse with -1I appended, and 0I simply returns an empty list.
With that, we get our results (mapped to ints, for ease of readability.)
```getFactors 64000I |> List.map int;;
val it : int list = [5; 5; 5; 2; 2; 2; 2; 2; 2; 2; 2; 2]
getFactors 6423453000I |> List.map int;;
val it : int list = [8599; 83; 5; 5; 5; 3; 3; 2; 2; 2]
getFactors 9536923853063I |> List.map int64;;
val it : int64 list = [354041L; 178393L; 151L]
getFactors 953692385306332I |> List.map int64;;
// this one's still chugging.```
The last thing to do here is to figure out how to avoid unnecessary iterations but that’s a topic for another blog.
# Fun Friday (Monday version) – Fractions
This will be a slightly longer series of posts. I thought about treating fractions as a distinct problem, and wanted to find a way to code reducing fractions in a logical and meaningful way.
The first problem was describing the type. Naturally, you can do it a as simple pair:
```type Fraction = int * int
let a = (-1, -2 : Fraction) // negatives?```
But that design describes some issues I’m not super keen on handling, namely negative denominators. Mathematically, negative denominators aren’t all that complicated to deal with, but in my head, it seemed unnecessary to solve them, so that led to the following:
```type Fraction = int * uint32
let a = (-1, 2u : Fraction) // ugh, not as clean as I'd like```
Now we’ve dealt with the negative denominator problem, but that leaves us describing fractions in a sort of “planar” way. Great for talking to a professor, but I’ve always been something of a “make it clear” personality. Record types work well for this.
```type Fraction = { Numerator : int; Denominator : uint32 }
{ Numerator = 1 ; Denominator = 2u; }```
Nice and pretty. Except 0 is a valid uint32 value. Shit.
```type BiggerThanZero = private BiggerThanZero of uint32
module BiggerThanZero =
let create uintValue =
if (uintValue > 0u ) then BiggerThanZero uintValue
else failwith "No zeros"
let value (BiggerThanZero u) =
u
type Fraction = { Numerator: int; Denominator: BiggerThanZero }```
Alright. Something is breaking my “F# is less verbose” than other languages spidey-sense here, in C#.
```using System;
public class Fraction
{
public struct BiggerThanZero
{
public BiggerThanZero(UInt32 u)
{
if(u == 0u) then
throw new ArgumentException("No zeros");
Value = u;
}
public UInt32 Value
{
get;
}
}
public Fraction(int numerator, BiggerThanZero denominator)
{
Numerator = numerator;
Denominator = denominator;
}
public int Numerator { get; }
public BiggerThanZero Denominator { get; }
}
```
Nope, spidey-sense was off. C# is still more code. Phew… thought I was going to have to go back to OO land. 🙂
OK… so now we have a domain object. We cannot represent an object in the domain that is “invalid” in any way, so fundamentally, our functions should be easy to reason about.
So first thing… decimals to my new “Fraction” type.
```let getFraction decimalValue =
let rec fract dm =
let a = (d * m)
let r = a % 1.0m
match r with
| 0.0m -> { Numerator = (int a) ; Denominator = BiggerThanZero.create (uint32 m)}
| _ -> fract d (m * 10.0m)
fract decimalValue 10.0m```
This function goes back to our old friend, the recursive inner function. We take our input decimal value, and multiplying it by 10, and calling the result ‘a’ (shorthand for ‘amount’). Then we take ‘a’, and get the decimal part of that value by applying the modulo function to it and 1.0, and naming that value ‘r’ (shorthand for remainder). Assuming that ‘r’ is non-zero, we recurse into the loop, updating the multiplier to another factor of 10 greater than what we had before. Otherwise, we simply return a Fraction object, with the Numerator set to ‘a’, and the Denominator set to the multiple. E.G.
```getFraction 0.4m;;
val it : Fraction = { Numerator = 4; Denominator = BiggerThanZero 10u }
getFraction 0.542m;;
val it : Fraction = { Numerator = 542; Denominator = BiggerThanZero 1000u }
getFraction 0.8675421m;;
val it : Fraction = { Numerator = 8675421; Denominator = BiggerThanZero 10000000u }```
This is the start of our Fractions work, and although it’s correct, it’s certainly got some potential error cases. Our int and uint32 bases for values could be overflowed. That’s fixed in the following.
Next time, we’ll deal with reducing our fractions. | 9,997 | 36,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-51 | longest | en | 0.952054 |
http://mathematicssolution.com/every-group-is-isomorphic-to-a-group-of-permutations/ | 1,503,350,937,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109670.98/warc/CC-MAIN-20170821211752-20170821231752-00429.warc.gz | 281,013,312 | 10,566 | BIGtheme.net http://bigtheme.net/ecommerce/opencart OpenCart Templates
Tuesday , August 22 2017
Home / Modern Abstract Algebra / Every group is isomorphic to a group of permutations.
# Every group is isomorphic to a group of permutations.
PROOF: To prove this, let G be any group. We must find a group of permutations that we believe is isomorphic to G. Since G is all we have to work with, we will have to use it to construct . For any g in G, define a function Tg from G to G by Tg (x) = gx for all x in G.
Tg is just multiplication by g on the left. We have to prove that Tg is a permutation on the set of elements of G. Now, let G ={Tg | g ∈G}. Then, is a group under the operation of function composition. To verify this, we first observe that for any g and h in G we have TgTh (x) = Th(Th(x)) = Tg(hx) = g(hx) =(gh)x = Tgh(x), so that Tg Th = Tgh. From this it follows that Te is the identity and (Tg)-1 = Tg-1 is associative, we have verified all the conditions for to be a group.
The isomorphism ∅ between G and is now ready-made. For every g in G, define ∅(g) = Tg. If Tg = Th, then Tg(e) = Th(e) or ge = he.
Thus, g = h and ∅ is one-to-one. By the way was constructed, we see that f is onto. The only condition that remains to be checked is that ∅ is operation-preserving. To this end, let a and b belong to G. Then
∅ (ab) = Tab = TaTb = ∅ (a) ∅ (b).
## Lecture – 4| Abstract Algebra
Problem – 1 : Show that if every element of the group G except the ... | 440 | 1,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-34 | latest | en | 0.924688 |
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# mathematical formulation造句
### 例句与造句
1. the mathematical formulation of this hypothesis is a persuasive one .
这一假设的数学形式是诱人的。
2. the mathematical formulation of the quantum theory has required the use of imaginary numbers .
量子理论的数学公式需要运用虚数。
3. these concepts were firmly entrenched in the mathematical formulation of classical mechanics .
这些概念在经典力学的数学表述中根深蒂固。
4. the mathematical formulations are more concise and easier to handle if they are properly applied .
数学公式比较简练，而且如果运用得当，也比较容易处理。
5. the mathematical formulation becomes lengthy and does not appear to offer much advantage over the scheme incorporated in equation(2) .
但是这种数学方程式很长，而且它比并入方程式（2）的图式显不出更多的优点。
6. It's difficult to find mathematical formulation in a sentence. 用mathematical formulation造句挺难的
7. the first mathematical formulation was established by von neumann in 1928
最初的数学形式由冯?诺伊曼于1928年建立。
8. the mathematical formulation becomes lengthy and does not appear to offer much advantage over the scheme incorporated in equation ( 2 )
但是这种数学方程式很长，而且它比并入方程式（2）的图式显不出更多的优点。
9. topics include : mathematical formulations; finite difference and finite volume discretizations; finite element discretizations; boundary element discretizations; direct and iterative solution methods
课程主题有：数学列式、有限差分与有限体积离散、有限元素离散、边界元素离散、直接与叠代解法。
10. at the same time, as more and more radio sources have been set up, electromagnetic pollution in urban area becomes more serious, which is harmful to people's healthy and communication system so it is necessary to simulate the electromagnetic propagation in urban area by computer being an approximate hf method in evaluating em scattering ， the complex ray method （ crm ） is valuable in actual applications ， mainly benefited from its simple physical model ， convenient mathematical formulation and computational efficiency, especially with the scenes of scattering from complex objects based on the condition mentioned above, a hybrid ray model of the urban area electromagnetic wave propagation prediction was established with the foundation of ray tracing theory and complex ray theory
复射线技术作为一种求解波场问题的高频近似方法，由于其具有物理模型简单、数学处理方便、计算效率高等优点，在复杂的目标散射特性分析等应用领域中有着重要的应用价值。基于以上的情况，本文在射线追踪理论和复射线理论的基础上，为城区环境建立了混合射线预测模型。本文采用椭圆模型对已有的射线追踪方法进行加速，并将复射线理论应用到城区环境电波传播预测中，提出新的预测方法混合射线方法。
11. e there is one-to-one map between the mathematical formulation and circuit realization, and that the method is systematic and suited for designing the network function with zeros and high-order filter . at last, the limitation and applied fields of the methods for realizing one-dimension cwt based on state-space log-domain filter are pointed out, it is also proved by the theoretical analysis and calculation results that the methods are suited for synthesizing the wavelet functions with very small scale or the frequence-transloction version of the mother wavelet ( with much translocation in frequence-domain )
论文还给出了用状态空间对数域滤波器实现一维连续小波函数的限制条件，并指出了这种实现方法的应用范围，理论分析与计算结果表明该方法适合用于实现尺度很小的小波函数或由原基小波经频谱搬移(搬移程度较大)后的小波函数。
12. many works on 2d non-linear transient magnetic field coupling with mechanical movement, including theoretically mathematical formulation, algorithm, program and experiment, are mainly researched in this paper with the fem theory, and some valuable results to theory analysis and engineering designing have been gained
本文以有限元理论为基础，具有机械运动耦合的二维非线性瞬态磁场为研究对象，进行了理论上数学模型的建立，算法与程序的实现，计算与实验验证的研究工作，并取得一些具有理论意义和工程实际价值的结果。
## 相关阅读
Copyright © 2020 WordTech Co. | 966 | 3,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-33 | latest | en | 0.785111 |
https://www.physicsforums.com/threads/unitary-ball-exploring-metrics-possibilities.583225/ | 1,726,087,444,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651400.96/warc/CC-MAIN-20240911183926-20240911213926-00421.warc.gz | 859,811,455 | 18,962 | # Unitary Ball: Exploring Metrics & Possibilities
• zendani
In summary, a unit ball is a ball of radius 1 in a given metric space. It is represented as B(x;1) and is defined as the set of all points y in the metric space that are within a distance of 1 from a given point x. The shape of a unit ball depends on the chosen distance function, with the most common one being a circular shape in 2-D and a spherical shape in 3-D. However, using alternative distance functions can result in different shapes such as squares or cubes.
zendani
I don't know anything about Unitary Ball
what is a Unitary ball? how make a unitary ball consider to a metric
(example: d(x,y) = max |xi-yi|)
Probably you mean "unit ball". This only means "a ball of radius 1". So, if (X,d) is a metric space, and x is a point in X, then the unit ball in (X,d) entered around x is the set
B(x;1) = {y in X | d(x,y)<1}
Thank you qusar987
so unit ball for d(x,y) = max |xi-yi| => B(x,1)= {y in x | maxi |xi-yi|<1}
maxi |xi-yi|<1 => |xi-yi|<1 => -1 <xi-yi<1 => 0<= xi-yi <1 => maxi (xi-yi)< 1
Why don't you try some simple examples, e.g., with xi=0, for xi real, then for xi in
R^2?
thank you Balce2
i want it for a paper about friction
If you use the max function as your distance function, then I think the "unit ball" is actually a square/cube/whatever you call one in higher dimensions.
Thank you Matterwave
my answer for find unit ball is wrong?
so i exactly can't recognize that unit ball will get which shape?
The usual distance function on R^n is d^2=sqrt(x^2+y^2+...), but that's not the only one you can use. You can certainly use your max function distance function.
The terminology "unit balls" comes from the usual distance function in which case, in 3-D you would get "balls". If you use other distance functions, you can get different shapes for your "unit balls".
## 1. What is a unitary ball?
A unitary ball is a mathematical concept that refers to a set of points in a multi-dimensional space that are all the same distance from the origin. It can also be thought of as a sphere with a radius of 1.
## 2. How is a unitary ball used in mathematics?
Unitary balls are commonly used in mathematics to explore different metrics and possibilities within a specific space. They can be used to define distances and measure properties of objects within the space.
## 3. What are some examples of metrics that can be explored using a unitary ball?
Some examples of metrics that can be explored using a unitary ball include Euclidean distance, Manhattan distance, and Minkowski distance. These metrics can be used to measure the distance between points in a space and can provide insights into the properties of the space.
## 4. How does the concept of a unitary ball relate to other mathematical concepts?
The concept of a unitary ball is closely related to other mathematical concepts such as normed vector spaces, metric spaces, and topology. These concepts all involve exploring and measuring properties within a space.
## 5. What are the real-world applications of studying unitary balls?
The study of unitary balls has many real-world applications in fields such as data analysis, computer science, and physics. It can be used to analyze and visualize complex data sets, optimize algorithms, and model physical systems.
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# Parallel Algorithms - PowerPoint PPT Presentation
Parallel Algorithms. Sung Yong Shin TC Lab CS Dept. KAIST. Contents. 1. Background 2. Parallel Computers 3. PRAM 4. Parallel Algorithms. 1. Background. Von Neumann Machines sequential executing one instruction at a time
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### Parallel Algorithms
Sung Yong Shin
TC Lab
CS Dept. KAIST
1. Background
2. Parallel Computers
3. PRAM
4. Parallel Algorithms
1.Background
• Von Neumann Machines
• sequential
• executing one instruction at a time
• Inherent limitation “ not faster than electrical signals ” 1 ft / 1 nanosecond ( 10-9 sec )
• Parallelism or Concurrency Carrying out many operations simultaneously
• partition a complex problem in such a way that various parts of the work can be carried out independently and in parallel, and combine the results when all subcomputation are complete.
• need parallel computers to support this approach.
• Hardware-oriented
• A parallel architecture of a specific kind is built.
• The parallel algorithms for solving different problems are developed to make use of these hardware features to the best advantage.
• Problem-oriented
• Whether the parallel algorithms can truly enhance the speed of obtaining a solution to a given problem, or not.
• If so, how much ?
(i) The usefulness of parallel computers depends greatly on :
• suitable parallel algorithms
• parallel computer languages
“ A major rethinking needed”
(ii) Practical limitations by parallel computers “ too many factors to be considered” How to abstract ingredient from complex reality !!!
Which problems can be solved substantially faster using many processors rather than one processor ?
• Nicholas Pippenger (1976)
“ NC-class problems” ( Nick’s Class )
“ ultra-fast on a parallel computer with feasible amount of hardware”
( independent of the particular parallel model chosen )
• Inherent Parallelism
probably not possible now but for the future !!!
“ fascinating research topics”
P
P(n) processors
NC
(log n)m
P-complete
P = NC ?
Applications (needs ) processors rather than one processor ?
• Computer vision / Image processing
• Computer Graphics
• Searching huge databases
• Artificial Intelligence
· · · · · · · ·
2. processors rather than one processor ?Parallel Computers
SIMD ( Single Instruction Multiple Data Stream )
MIMD ( Multiple Instruction Multiple Data Stream )
What does SISD stand for ?
Program processors rather than one processor ?
Result x+y
Data
Source
x
Function
Unit
y
SISD
Program processors rather than one processor ?
Result x+y
x
Function
Unit
y
Data
Source
Result s+q
s
Function
Unit
q
Result v+w
v
Function
Unit
w
• SIMD
• array processors
• vector processors (pipelining)
Process3 processors rather than one processor ?
Process4
Process1
Process2
Data
Source
Branch
Function
Unit
NO
Result x · y
YES
x
Multiply
Function
Unit
Data
Source
y
Result w+v
w
Function
Unit
Data
Source
v
Result s/q
s
Divide
Function
Unit
Data
Source
q
MIMD
Array Processors processors rather than one processor ?
instructions (for multiple data)
master
slave
slave
slave
Control
Processor
Arithmetic
Processor
Arithmetic
Processor
Arithmetic
Processor
PE
Memory
Memory
Memory
Memory
Communication network
Identical processors rather than one processor ?
Processors
· · ·
P
P
P
Interconnection network
· · ·
M
M
M
tightly coupled multiprocessors
P processors rather than one processor ?
P
P
Identical
Processing
Elements
( PEs )
· · ·
M
M
M
Interconnection network
loosely coupled multiprocessors
Vector ( pipe-line ) processors processors rather than one processor ?
functional unit
Operand one
exponents
and multiply
mantissas
Compare
components
Align
operands
accordingly
Determine
normalization
factor
Result
Operand two
Normalize
results
Stage 1
Stage 2
Stage 3
Stage 4
Stage 5
A simplified pipeline for floating-point multiplication
3. processors rather than one processor ?PRAM (Parallel Random Access Machine)
Processors
(i) p general-purpose processors
(ii) Each processor is connected to a large shared, random access memory M.
(iii) Each processor has a private (or local) memory for its own computation.
(iv) All communications among processors take place via the shared memory.
(v) The input for an algorithm is assumed to be the 1stn memory cells,
and output is to be placed in the 1st cell.
(vi) All memory cells are initialized to be “0”.
P1
P2
P3
Pp
· · ·
Interconnection
· · ·
M
· · ·
1
m
Memory
[A PRAM]
( processors rather than one processor ?vii) All processors run the same program.
(viii) Each processor knows its own index.
(ix) A PRAM program may instruct processors to do different things
depending on their indices.
write
computation
three phases
Major Assumption processors rather than one processor ?
(i) PRAM processors are synchronized !!!
(1) processors begin each step at the same time.
(2) All the processors that write at any step write at the same time.
(ii) Any number of processors may read the same memory cell
concurrently !!!
Variants of PRAM’s processors rather than one processor ?
CREW ( Concurrent Read Exclusive Write )
CRCW ( Concurrent Read Concurrent Write )
– Common-write
– Priority-write
Why not EREW ?
yes, if you want !!!
Other Models processors rather than one processor ?
[Other parallel architectures]
(a) A hypercube (dimension = 3)
(b) A bounded degree network (degree = 4)
···
···
· · · · · · · · · · · · · · · ·
(c) Octree model
4. processors rather than one processor ?Parallel Algorithms
• Binary Fan-in Technique
• Matrix multiplication
• Handling write conflicts
• Merging & Sorting
Binary Fan-in Technique processors rather than one processor ?
P7
P5
P3
P1
x1
Compute
Write
Compute
Write
Compute
Write
x7
x3
x5
x8
x2
x4
x6
comparison
write
save
M[1] = max
[A parallel tournament]
( finding Max )
Processors: processors rather than one processor ?
Step 0
Step 1
big := max (big, temp)
write big
Step 2
big := max (big, temp)
write big
Step 3
big := max (big, temp)
write big
P1 P2 P3 P4 P5 P6 P7 P8
M
16 12 1 17 23 19 4 8
16 12 1 17 23 19 4 8
M
16 12 1 17 23 19 4 8
12 1 17 23 19 4 8 –
16 12 17 23 23 19 8 8
M
16 12 17 23 23 19 8 8
17 23 23 19 8 8 – –
17 23 23 23 23 19 8 8
M
17 23 23 23 23 19 8 8
23 19 8 8 – – – –
23 23 23 23 23 19 8 8
M
23 23 23 23 23 19 8 8
max
[A tournament example showing the activity of all the processors.]
read processors rather than one processor ?M[i] into big ;
incr := 1 ;
write – { some very small value } into M[n+i] ;
for step := 1 to lg n do
big := max (big, temp) ;
incr := 2 * incr ;
write big into M[i]
end { for }
O( log n ) using n/2 processors
no write conflicts
Matrix Multiplication processors rather than one processor ?
O(n) using n2 processors
What if using n3 processors ?
O( log n )
Why ?
Handling write conflict processors rather than one processor ?
Algorithm:Computing the or of n Bits
Input : Bits x1, · · · · ,xn in M[1],· · · ·, M[n].
Output : x1 · · · · xn in M[1].
If xi=1, then Pi writes 1 in M[1].
O(1) using n processors
write conflict !!!
CRCW
– Common-write
– Priority-write
Fast algorithm for finding Max processors rather than one processor ?
Initial memory contents (n = 4).
Input
loser
2 7 3 6 0 0 0 0
8
1
P14
After Step 2
P13
P12
P23
P24
P34
1
1
1
1
1
1
2 7 3 6 1 0 1 1
P23
After Step 3
7
7
[Example for the fast max-finding algorithm]
O(1) using processors
common-write
Algorithm : Finding the Largest of processors rather than one processor ?n Keys
Input : n keys x1, x2,···, xn, initially in memory cells M[1], M[2],···, M[n] (n>2).
Output : The largest key will be left in M[1].
Comment : For clarity, the processors will be numbered Pi.j for 1 i j n.
Step 1
Step 2
Pi.j compares xi and xj.
Let k be the index of the smaller key.
(If the keys are equal, let k be the smaller index.)
Pi.j writes 1 in loser[k].
{At this point, every key other than the largest has lost a comparison. }
Step 3
Any processor that read a 0 writes xi in M[1]. (P1.n would write xn.)
{ Pi.i+1 already has xi in its local memory ; P1.n has xn. }
Merging and Sorting processors rather than one processor ?
merging
P1
Pn
Pn/2+1
Pn/2
x1
xn/2
yn
y1
· · ·
· · ·
M[1]
M[n]
M[n/2]
(a) Assignment of processors to keys.
Pi
yj
xi
>yj
>xi
<xi
<yj
(b) Binary search steps; Pi finds j such that yj-1<xi<yj.
binary search
Pi
x1,…, xi-1 and y1,…, yj-1 (merged)
xi
M[i+j-1]
(c) Output step.
[Parallel merging]
O(log n) using n processors
no write conflict
Algorithm processors rather than one processor ?:Parallel Merging
Input : Two sorted lists of n/2 keys each, in the first n cells of memory.
Output : The merged list, in the first n cells of memory.
Comment : Each processor Pihas a local variable x (if in/2) or y (if i>n/2) and other local variables for conducting its binary search. Each processor has a local variable position that will indicate where to write its key.
Initialization :
Pi reads M[i] into x (if in/2) or into y (if i>n/2).
Pi does initialization for its binary search.
Binary search steps :
Processors Pi, for 1in/2, do binary search in M[n/2+1],…, M[n]
to find the smallest j such that x<M[n/2+j], and assign i+j–1 to
position. If there is no such j, Piassigns n/2+i to position.
Processors Pn/2+i, for 1in/2, do binary search in M[1],…, M[n/2]
to find the smallest j such that y<M[j], and assign i+j–1 to position.
If there is no such j, Pi assigns n/2+i to position.
Output step :
Each Pi(for 1in) writes its key (x or y) in M[position].
Break the list into two halves. processors rather than one processor ?
Sort the two halves (recursively).
Merge the two sorted halves.
Algorithm : Sorting by Merging
Input : A list of n keys in M[1],…,M[n].
Output : The n key sorted in nondecreasing order in M[1],…,M[n].
Comment : The indexing in the algorithm is easier if the number of keys is a power of 2, so the first step will “pad” the input with large keys at the end. We still use only n processors.
Piwrites (some large key) in M[n+i] ;
for t := 1 to lg n do
k := 2t-1 ; { the size of the lists being merged }
Pi,…, Pi+2k-1 merge the two sorted lists of size k beginning at M[i];
end { for }
O((log n)2) using n processors | 3,095 | 11,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-47 | latest | en | 0.837192 |
http://www.chegg.com/homework-help/introductory-algebra-10th-edition-chapter-2.6-problem-24e-solution-9780321269478 | 1,469,545,648,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824994.73/warc/CC-MAIN-20160723071024-00319-ip-10-185-27-174.ec2.internal.warc.gz | 356,633,871 | 16,792 | View more editions
# TEXTBOOK SOLUTIONS FOR Introductory Algebra 10th Edition
• 7236 step-by-step solutions
• Solved by publishers, professors & experts
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Chapter: Problem:
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 3
Let be the first angle
Since the second angle is four times as the first angle.
So the second angle is
Since the third angle is less than the sum of the other two angles.
So the third angle is
.
• Chapter , Problem is solved.
Corresponding Textbook
Introductory Algebra | 10th Edition
9780321269478ISBN-13: 0321269470ISBN: Marvin L. BittingerAuthors:
Alternate ISBN: 9780321305985, 9780321305992, 9780321306005, 9780321690166 | 217 | 783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-30 | longest | en | 0.766844 |
http://conceptmap.cfapps.io/wikipage?lang=en&name=Rolling_checksum | 1,606,787,303,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141542358.71/warc/CC-MAIN-20201201013119-20201201043119-00618.warc.gz | 25,033,136 | 18,221 | # Rolling hash
(Redirected from Rolling checksum)
A rolling hash (also known as recursive hashing or rolling checksum) is a hash function where the input is hashed in a window that moves through the input.
A few hash functions allow a rolling hash to be computed very quickly—the new hash value is rapidly calculated given only the old hash value, the old value removed from the window, and the new value added to the window—similar to the way a moving average function can be computed much more quickly than other low-pass filters.
One of the main applications is the Rabin–Karp string search algorithm, which uses the rolling hash described below. Another popular application is the rsync program, which uses a checksum based on Mark Adler's adler-32 as its rolling hash. Low Bandwidth Network Filesystem (LBFS) uses a Rabin fingerprint as its rolling hash. FastCDC (Fast Content-Defined Chunking) uses a compute-efficient Gear fingerprint as its rolling hash.
At best, rolling hash values are pairwise independent[1] or strongly universal. They cannot be 3-wise independent, for example.
## Polynomial rolling hash
The Rabin–Karp string search algorithm is often explained using a rolling hash function that only uses multiplications and additions:
${\displaystyle H=c_{1}a^{k-1}+c_{2}a^{k-2}+c_{3}a^{k-3}+...+c_{k}a^{0},}$
where ${\displaystyle a}$ is a constant, and ${\displaystyle c_{1},...,c_{k}}$ are the input characters (but this function is not a Rabin fingerprint, see below).
In order to avoid manipulating huge ${\displaystyle H}$ values, all math is done modulo ${\displaystyle n}$ . The choice of ${\displaystyle a}$ and ${\displaystyle n}$ is critical to get good hashing; see linear congruential generator for more discussion.
Removing and adding characters simply involves adding or subtracting the first or last term. Shifting all characters by one position to the left requires multiplying the entire sum ${\displaystyle H}$ by ${\displaystyle a}$ . Shifting all characters by one position to the right requires dividing the entire sum ${\displaystyle H}$ by ${\displaystyle a}$ . Note that in modulo arithmetic, ${\displaystyle a}$ can be chosen to have a multiplicative inverse ${\displaystyle a^{-1}}$ by which ${\displaystyle H}$ can be multiplied to get the result of the division without actually performing a division.
## Rabin fingerprint
The Rabin fingerprint is another hash, which also interprets the input as a polynomial, but over the Galois field GF(2). Instead of seeing the input as a polynomial of bytes, it is seen as a polynomial of bits, and all arithmetic is done in GF(2) (similarly to CRC32). The hash is the result of the division of that polynomial by an irreducible polynomial over GF(2). It is possible to update a Rabin fingerprint using only the entering and the leaving byte, making it effectively a rolling hash.
Because it shares the same author as the Rabin–Karp string search algorithm, which is often explained with another, simpler rolling hash, and because this simpler rolling hash is also a polynomial, both rolling hashes are often mistaken for each other. The backup software restic uses a Rabin fingerprint for splitting files, with blob size varying between 512 bytes and 8MiB.[2]
## Cyclic polynomial
Hashing by cyclic polynomial[3]—sometimes called Buzhash—is also simple, but it has the benefit of avoiding multiplications, using barrel shifts instead. It is a form of tabulation hashing: it presumes that there is some hash function ${\displaystyle h}$ from characters to integers in the interval ${\displaystyle [0,2^{L})}$ . This hash function might be simply an array or a hash table mapping characters to random integers. Let the function ${\displaystyle s}$ be a cyclic binary rotation (or circular shift): it rotates the bits by 1 to the left, pushing the latest bit in the first position. E.g., ${\displaystyle s(10011)=00111}$ . Let ${\displaystyle \oplus }$ be the bitwise exclusive or. The hash values are defined as
${\displaystyle H=s^{k-1}(h(c_{1}))\oplus s^{k-2}(h(c_{2}))\oplus \ldots \oplus s(h(c_{k-1}))\oplus h(c_{k}),}$
where the multiplications by powers of two can be implemented by binary shifts. The result is a number in ${\displaystyle [0,2^{L})}$ .
Computing the hash values in a rolling fashion is done as follows. Let ${\displaystyle H}$ be the previous hash value. Rotate ${\displaystyle H}$ once: ${\displaystyle H\leftarrow s(H)}$ . If ${\displaystyle c_{1}}$ is the character to be removed, rotate it ${\displaystyle k}$ times: ${\displaystyle s^{k}(h(c_{1}))}$ . Then simply set
${\displaystyle H\leftarrow s(H)\oplus s^{k}(h(c_{1}))\oplus h(c_{k+1}),}$
where ${\displaystyle c_{k+1}}$ is the new character.
Hashing by cyclic polynomials is strongly universal or pairwise independent: simply keep the first ${\displaystyle L-k+1}$ bits. That is, take the result ${\displaystyle H}$ and dismiss any ${\displaystyle k-1}$ consecutive bits.[1] In practice, this can be achieved by an integer division ${\displaystyle H\rightarrow H\div 2^{k-1}}$ .
## Content-based slicing using a rolling hash
One of the interesting use cases of the rolling hash function is that it can create dynamic, content-based chunks of a stream or file. This is especially useful when it is required to send only the changed chunks of a large file over a network and a simple byte addition at the front of the file would cause all the fixed size windows to become updated, while in reality, only the first "chunk" has been modified.
The simplest approach to calculate the dynamic chunks is to calculate the rolling hash and if it matches a pattern (like the lower N bits are all zeroes) then it’s a chunk boundary. This approach will ensure that any change in the file will only affect its current and possibly the next chunk, but nothing else.
When the boundaries are known, the chunks need to be compared by their hash values to detect which one was modified and needs transfer across the network.[4] The backup software Attic uses a Buzhash algorithm with a customizable chunk size range for splitting file streams.[5]
## Content-based slicing using moving sum
Several programs, including gzip (with the --rsyncable option) and rsyncrypto, do content-based slicing based on this specific (unweighted) moving sum:[6]
${\displaystyle S(n)=\sum _{i=n-8195}^{n}c_{i},}$
${\displaystyle H(n)=S(n)\mod 4096,}$
where
• ${\displaystyle S(n)}$ is the sum of 8196 consecutive bytes ending with byte ${\displaystyle n}$ (requires 21 bits of storage),
• ${\displaystyle c_{i}}$ is byte ${\displaystyle i}$ of the file,
• ${\displaystyle H(n)}$ is a "hash value" consisting of the bottom 12 bits of ${\displaystyle S(n)}$ .
Shifting the window by one byte simply involves adding the new character to the sum and subtracting the oldest character (no longer in the window) from the sum.
For every ${\displaystyle n}$ where ${\displaystyle H(n)==0}$ , these programs cut the file between ${\displaystyle n}$ and ${\displaystyle n+1}$ . This approach will ensure that any change in the file will only affect its current and possibly the next chunk, but no other chunk.
## Gear fingerprint and content-based chunking algorithm FastCDC
The Content-Defined Chunking (CDC) algorithm needs to compute the hash value of a data stream byte by byte and split the data stream into chunks when the hash value meets a predefined value. However, comparing a string byte-by-byte will introduce the heavy computation overhead. FastCDC [7] proposes a new and efficient Content-Defined Chunking approach. It uses a fast rolling Gear hash algorithm [8], skipping the minimum length, normalizing the chunk-size distribution, and last but not the least, rolling two bytes each time to speed up the CDC algorithm, which can achieve about 10X higher throughput than Rabin-based CDC approach. [9]
The basic version pseudocode is provided as follows:
algorithm FastCDC
input: data buffer src,
data length n,
output: cut point i
MinSize ← 2KB // split minimum chunk size is 2 KB
MaxSize ← 64KB // split maximum chunk size is 64 KB
fp ← 0
i ← MinSize
// buffer size is less than minimum chunk size
if n ≤ MinSize then
return n
if n ≥ MaxSize then
n ← MaxSize
while i < n do
fp ← (fp << 1 ) + Gear[src[i]]
return i
return i
Where Gear array is a pre-calculated hashing array. Here FastCDC uses Gear hashing algorithm which can calculate the rolling hashing results quickly and keep the uniform distribution of the hashing results as Rabin. Compared with the traditional Rabin hashing algorithm, it achieves a much faster speed. Experiments suggest that it can generate nearly the same chunk size distribution in the much shorter time (about 1/10 of rabin-based chunking [9]) when segmenting the data stream.
## Computational complexity
All rolling hash functions are linear in the number of characters, but their complexity with respect to the length of the window (${\displaystyle k}$ ) varies. Rabin–Karp rolling hash requires the multiplications of two ${\displaystyle k}$ -bit numbers, integer multiplication is in ${\displaystyle O(k\log k2^{O(\log ^{*}k)})}$ .[10] Hashing ngrams by cyclic polynomials can be done in linear time.[1] | 2,235 | 9,252 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 48, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-50 | latest | en | 0.852988 |
https://www.queryhome.com/puzzle/34859/these-clues-find-where-they-dont-waste-second-start-working?show=34889 | 1,585,659,813,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500482.27/warc/CC-MAIN-20200331115844-20200331145844-00044.warc.gz | 1,121,501,130 | 32,107 | # Use these clues to find out where they are. Now, don't waste a second, and start working!
23 views
You wake up on your birthday, only to find that all your presents are gone!
The only thing there is a card.
You open it, and it says: I have hidden your presents.
Use these clues to find out where they are. Now, don't waste a second, and start working!
Lap At Star Fit Act
posted Mar 18
Second letters give the word "Attic". Everyone is there with presents
Use the second letters of each word, and they form the word ATTIC. You walk up to your attic, and there your family waits, along with all your friends and all your presents.
Similar Puzzles
Three friends all live in houses with painted doors. Can you use these clues to find out what colour door they have and what road they live on?
Friends - Carly, Harry, Alan
Colours - Red, Blue, Pink
Streets - Oak St, South Ave, Drury Ave
1) Carly just loves the colour pink and she was ecstatic when her mom painted the door for her.
2) Alan's always around at Harry's house. It's easy to get there because Oak St is only just around the corner.
3) South Ave has mostly white doors, so the red one, belonging to one of the friends, stands out easily to the passing cars.
4) Alan thinks that his friend's blue door clashes badly with their house but he doesn't want to tell them in case he hurts their feelings.
+1 vote
There are four different integers. Here are eight clues to help you find out what they are.
The sum of the three lowest numbers is 0.
The two lowest numbers are negative, and consecutive.
The highest number is the second-lowest number squared.
The difference of all the numbers (in order from greatest to least) is 9.
The difference of all the numbers (in order from least to greatest) is -17.
The difference of the two highest numbers is 2.
The sum of the two lowest numbers is -7.
The second-lowest number plus 10 is the second-highest.
What are the four integers?
A shipment of butterflies was mixed up by the dock workers, and they could not find who bought which species, where it was from, and what was the price. All the workers know is that Alejandro, Faye, Yvette, Sophie, and Zachary could have each bought butterflies that cost \$60, \$75, \$90, \$105, or \$120. Each could have bought the Clearwing, the Emperor, the Grayling, the Swallowtail, or the Torturix butterflies. Each butterfly could have lived in Australia, Jordan, Luxembourg, Panama, or Qatar. It is up to you to find out who bought which butterfly, what was the price, and where did it come from with the provided clues:
1. Neither the butterfly from Luxembourg nor the one from Australia sold for \$90.
2. The Emperor butterfly cost \$30 more than the Torturix butterfly.
3. Zachary's purchase was \$75.
4. The butterfly from Australia cost less than the one from Luxembourg.
5. Alejandro's purchase was from Luxembourg.
6. Of Yvette's purchase and the purchase for \$60, one was from Qatar and the other was the Torturix.
7. The butterfly that sold for \$120 was not from Panama.
8. The insect from Australia was not the Torturix.
9. Faye bought the Torturix.
10. Sophie did not buy the Grayling.
11. Of the Emperor and the insect worth \$105, one was won by Yvette and the other was from Luxembourg.
12. The insect that sold for \$105 was the Swallowtail.
+1 vote
These are 15 things which we use daily correct the spelling:
Ex :1) Dogo :Good
2) elviesnoit
3) Miet
4) Boilme
5) Ishtr
6) Agdrne
7) Solhoc
8) sjdmai
9) Tsere
10) Yclecs
11) Neplci
12) Odosr
13) Thgli
14) Sglas
15) Owilpl | 893 | 3,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-16 | latest | en | 0.965253 |
https://forum.powerscore.com/viewtopic.php?f=452&t=16965&sid=e33c68c65f6514d7beeeec9af032bdd8 | 1,709,115,875,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474700.89/warc/CC-MAIN-20240228080245-20240228110245-00501.warc.gz | 273,518,992 | 8,791 | # LSAT and Law School Admissions Forum
Get expert LSAT preparation and law school admissions advice from PowerScore Test Preparation.
## #14 - Local, Must Be True, Maximum
Dave Killoran
• PowerScore Staff
• Posts: 5839
• Joined: Mar 25, 2011
#45685
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=13387)
The correct answer choice is (D)
This question dictates the Cities-to-Countries numerical distribution in this case to be 2-2-2 (two cities in each country), so Hannah will be visiting exactly two cities in country X. To maximize her number of days in those two cities, we should consider the Days-to-Cities numerical distributions:
• With a 4-2-2-2-2-2 distribution, she could spend 4 days in one X city, and 2 days in another X city, for a total of six days.
With a 3-3-2-2-2-2 distribution, she could spend 3 days in each of two cities in X.
Either way, the most time Hannah could possibly spend in two cities would be six days, so the correct answer choice is (D).
Analyze and track your performance with our Testing and Analytics Package. | 280 | 1,111 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-10 | latest | en | 0.894436 |
https://www.oreilly.com/library/view/the-complete-idiots/9781615641444/ | 1,544,404,802,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823228.36/warc/CC-MAIN-20181209232026-20181210013526-00487.warc.gz | 1,008,371,970 | 11,494 | ## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.
No credit card required
## Book Description
- Follows a standard course curriculum.
- Includes both polar coordinates and complex numbers, unlike the competition.
## Table of Contents
1. Cover
2. Title Page
3. Dedication
4. Copyright
5. Contents
6. Introduction
7. Part 1: The Basic Tools
1. 1 What Is Trigonometry?
2. 2 Geometry Tools Needed to Study Trigonometry
1. Right Triangles
2. More on Triangles
3. Circles, Arcs, and Chords
3. 3 Algebra Tools Needed to Study Trigonometry
1. Cartesian Coordinates
2. What Are Functions?
3. Complex Numbers
8. Part 2: Triangle Trigonometry
1. 4 Trigonometric Functions and Right Triangles
1. The Tangent Ratio
2. Introducing the Sine and Cosine Ratios
3. Practical Applications of Trigonometric Ratios
4. Practice Problems
2. 5 Relations Among Trigonometric Ratios
1. Six Trigonometric Relatives
2. Determining Values of Trigonometric Functions
3. Solving Right Triangles
4. Practice Problems
3. 6 The Law of Sines
1. Oblique Triangles
2. The Sine of an Obtuse Angle
3. The Ambiguous Case
4. Solving Oblique Triangles Using the Law of Sines
5. Practice Problems
4. 7 The Law of Cosines
1. The Law of Cosines
2. Two Cases That the Law of Cosines Helps to Solve
3. Application of Trigonometry to Navigation and Surveying
4. Hero’s Formula for the Area of a Triangle
5. Practice Problems
9. Part 3: Trigonometric Functions and the Unit Circle
1. 8 Angles and Rotations
1. Measuring Angles
2. Sectors of Circles
3. Practice Problems
2. 9 The Unit Circle Approach
1. What Is the Unit Circle?
2. Placing Points on the Unit Circle
3. Making Sense of the Unit Circle
4. Evaluating Trigonometric Functions of Real Numbers
5. Practice Problems
3. 10 Trigonometric Functions of Any Angle
1. Calculating Values of Trigonometric Functions
2. Reference Angles
3. Evaluating Trigonometric Functions
4. Looking Closely at Trigonometric Functions
5. Practice Problems
10. Part 4: Graphs of Trigonometric Functions
1. 11 Graphs of Sine and Cosine Functions
1. Basic Sine Curve
2. Other Sinusoidal Curves
3. Translations of the Sine Curve
4. The Cosine Curve
5. Modeling Periodic Behavior
6. Practice Problems
2. 12 Graphs of Other Trigonometric Functions
1. Graphs of Tangent Function
2. Graphs of Cotangent Function
3. Graphs of the Reciprocal Functions
4. Graphing Combinations of Functions
5. Practice Problems
3. 13 Graphs of Inverse Trigonometric Functions
1. Inverse Sine Function
2. Other Inverse Trigonometric Functions
1. Inverse Cosine and Tangent Functions
2. Inverses of Reciprocal Trigonometric Functions
3. Sample Problem 2
4. Sample Problem 3
5. Compositions of Trig and Inverse Trig Functions
6. Practice Problems
11. Part 5: Trigonometric Identities and Equations
1. 14 Kaleidoscope of Identities
1. Basic Identities
2. Many Uses of the Fundamental Identities
3. Practice Problems
2. 15 Verifying Trigonometric Identities
1. Important Guidelines
2. Different Tactics for Verifying Identities
3. Practice Problems
3. 16 Solving Trigonometric Equations
1. What It Means to Solve a Trigonometric Equation
2. Solving Techniques
3. Equations of Quadratic Type
4. When Multiple Angles Are Involved
5. Practice Problems
4. 17 Sum and Difference Formulas
1. Sum and Difference Formulas for Sine and Cosine
2. Applying Formulas to Various Trig Problems
3. Practice Problems
5. 18 Double-Angle and Power-Reducing Formulas
1. Double-Angle Formulas
2. Power-Reducing Formulas
3. Practice Problems
6. 19 Half-Angle and Product-to-Sum Formulas
1. Half-Angle Formulas
2. Product–to-Sum Formulas
3. Sum-to-Product Formulas
4. How to Avoid Frustration While Working with Formulas
5. Practice Problems
12. Part 6: Polar Coordinates and Complex Numbers
1. 20 Polar Coordinates
1. Introducing Polar Coordinates
2. Coordinate Conversions
3. Converting Polar Equations to Rectangular Form
4. Graphs of Polar Equations
5. Practice Problems
2. 21 Complex Numbers and Operations with Them
1. Geometric Representation of Complex Numbers
2. Converting from One Form to Another
3. Product of Two Complex Numbers in Polar Form
4. De Moivre’s Theorem and Powers of Complex Numbers
5. Roots of Complex Numbers
6. Practice Problems
3. 22 Trigonometry and Calculators?
1. Calculators That Can Help You with Trigonometry
2. Evaluating Trigonometric Functions
3. Evaluating Inverse Trigonometric Functions
4. Sketching Graphs of Trigonometric Functions
5. Solving Trig Equations with a Calculator
13. Appendixes
14. Index | 1,247 | 4,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-51 | latest | en | 0.669112 |
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# Wiki Contributions
This reminds me of a bit from Feynman's Lectures on Physics:
"What is this law of gravitation? It is that every object in the universe attracts every other object with a force which for any two bodies is proportional to the mass of each and varies inversely as the square of the distance between them. This statement can be expressed mathematically by the equation F=Gmm'/r^2. If to this we add the fact that an object responds to a force by accelerating in the direction of the force by an amount that is inversely proportional to the mass of the object, we shall have said everything required, for a sufficiently talented mathematician could then deduce all the consequences of these two principles."
Like Feynman, however, I think his next sentence is important:
"However, since you are not assumed to be sufficiently talented yet, we shall discuss the consequences in more detail, and not just leave you with these two bare principles."
"The average shareholder definitely does not care about the value of R&D to the firm long after their deaths, or I suspect any time at all after they sell the stock."
This was addressed in the post: the price of the stock today (when its being sold) is a prediction of its future value. Even if you only care about the price that you can sell it at today, that means that you care about at least the things that can lead to predictably greater value in the future, including R&D, because the person you're selling to cares about those things.
Also worth noting: the reason that the 2% value is meaningful is that if firms captured 100% of the value, they would be incentivized to increase the amount produced such that the amount they create would be maximally efficient. When they only capture 2% of the value, they are no longer incentivized to create the maximally efficient amount (stop producing it when cost to produce = value produced). This is basically why externalities lead to market inefficiencies. The issue isn't that they won't produce it at all, it's that they will underproduce it.
Spandrels certainly exist. But note the context of what X is in the quoted text:
"a chunk of complex purposeful functional circuitry X (e.g. an emotion)"
a chunk of complex purposeful functional circuitry cannot be a spandrel. There are edge cases that are perhaps hard to distinguish, but the complexity of a feature is a sign of its adaptiveness. Eyes can't be spandrels. The immune system isn't a spandrel. Even if we didn't understand what they do, the very complexity and fragility of these systems necessitates that they are adaptive and were selected for (rather than just being byproducts of something else that was selected for).
Complex emotions (not specific emotional responses) fall under this category.
The wealthy may benefit from the existence of low-skilled labour, but compared to what? Do they benefit more than they would from the existence of high-skilled labour?
Yes, they benefit from low skilled labour as compared to no labour at all, but high skilled labour, being more productive, is an even greater benefit. If it weren't, it couldn't demand a higher wage.
If "the wavefunction is real, but it is a function over potential configurations, only one of which is real." then you have the real configuration interacting with potential configurations. I don't see how you can say something isn't real (if only one of them is real then the others aren't) is interacting with something that is. If that "potential" part of the wave function can interact with the other parts of the wave function, then it's clearly real in every sense that the word "real" means anything at all.
I know they're just cartoons and I get the gist, but the graphs labelled "naive scenario" and "actual performance" are a little confusing.
The X axis seems to be measuring performance, with benchmarks like "high schooler" and "college student", but in that case, what's the Y axis? Is it the number of tasks that the model performs at that particular level? Something like that?
I think it would be helpful if you labeled the Y axis, even with just a vague label.
Re: the dark matter analogy. I think the analogy works well, but would just like to point out that even in theories where dark matter doesn't interact even with the weak force, and there is some other force that it does interact with that's analogous to electromagnetism, so it could bind together to form an earth-like planet, it still interacts with gravity, and if this earth-sized dark matter planet really did overlap with ours, we'd feel it's gravity and the earth would seem to be twice as massive as it is. Or, to state it slightly differently, the actual earth would be half as massive as we measure it to be. But that would be inconsistent with what we know of its composition and density. We know the mass of rocks, and the measurement of the mass of a rock of a particular size wouldn't be subject to this error, so we can rule out there being a dark matter Earth coincident with ours.
This isn't in any way a criticism of what I found to be a brilliant piece. And I'm not even sure that it's reason enough not to use that particular analogy, which otherwise works great.
Related to this topic, with a similar outlook but also more discussion of specific approaches going forward, is Vitalik's recent post on techno-optimism:
https://vitalik.eth.limo/general/2023/11/27/techno_optimism.html
There is a lot at the link, but just to give a sense of the message here's a quote:
"To me, the moral of the story is this. Often, it really is the case that version N of our civilization's technology causes a problem, and version N+1 fixes it. However, this does not happen automatically, and requires intentional human effort. The ozone layer is recovering because, through international agreements like the Montreal Protocol, we made it recover. Air pollution is improving because we made it improve. And similarly, solar panels have not gotten massively better because it was a preordained part of the energy tech tree; solar panels have gotten massively better because decades of awareness of the importance of solving climate change have motivated both engineers to work on the problem, and companies and governments to fund their research. It is intentional action, coordinated through public discourse and culture shaping the perspectives of governments, scientists, philanthropists and businesses, and not an inexorable "techno-capital machine", that had solved these problems."
I've no real insight to add, but would just like to comment that this generally lines up with the picture Steven Pinker paints in books like "Better Angels of Our Nature" and "Enlightenment Now".
Thanks for a good comment. My oversimplified thought process was that a 10x increase in energy usage for the brain would equate to a ~2x increase in total energy usage. Since we're able to maintain that kind of energy use during exercise, and elite athletes can maintain that for many hours/day, it seems reasonable that the heart and other organs could maintain this kind of output.
However, the issue you bring up, of actually getting that much blood to the brain, evacuating waste products, doing the necessary metabolism there, and dealing with so much heat localized in the small area of the brain, are all valid. While it seems like the rest of the body wouldn't be constrained by this level of energy use, a 10x power output in the brain probably might be a problem.
It's worth a more detailed analysis of exactly where the max. power output constraint on the brain, without any major changes, lie. | 1,653 | 7,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-18 | latest | en | 0.966754 |
https://www.symbolab.com/study-guides/collegealgebra1/key-concepts-glossary-14.html | 1,721,066,284,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00884.warc.gz | 883,906,414 | 46,104 | We've updated our
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# Key Concepts & Glossary
## Key Equations
probability of an event with equally likely outcomes $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$ probability of the union of two events $P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)-P\left(E\cap F\right)$ probability of the union of mutually exclusive events $P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)$ probability of the complement of an event $P\left(E\text{'}\right)=1-P\left(E\right)$
## Key Concepts
• Probability is always a number between 0 and 1, where 0 means an event is impossible and 1 means an event is certain.
• The probabilities in a probability model must sum to 1.
• When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in the sample space for the experiment.
• To find the probability of the union of two events, we add the probabilities of the two events and subtract the probability that both events occur simultaneously.
• To find the probability of the union of two mutually exclusive events, we add the probabilities of each of the events.
• The probability of the complement of an event is the difference between 1 and the probability that the event occurs.
• In some probability problems, we need to use permutations and combinations to find the number of elements in events and sample spaces.
## Glossary
complement of an event
the set of outcomes in the sample space that are not in the event $E$
event
any subset of a sample space
experiment
an activity with an observable result
mutually exclusive events
events that have no outcomes in common
outcomes
the possible results of an experiment
probability
a number from 0 to 1 indicating the likelihood of an event
probability model
a mathematical description of an experiment listing all possible outcomes and their associated probabilities
sample space
the set of all possible outcomes of an experiment
union of two events
the event that occurs if either or both events occur | 480 | 2,090 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-30 | latest | en | 0.910494 |
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